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You may copy it, give it away or % +% re-use it under the terms of the Project Gutenberg License included % +% with this eBook or online at www.gutenberg.org % +% % +% % +% Title: An Elementary Treatise on Fourier's Series and Spherical, Cylindrical, and Ellipsoidal Harmonics +% With Applications to Problems in Mathematical Physics % +% % +% Author: William Elwood Byerly % +% % +% Release Date: August 19, 2009 [EBook #29779] % +% % +% Language: English % +% % +% Character set encoding: ISO-8859-1 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK TREATISE ON FOURIER'S SERIES *** +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{29779} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% An Elementary Treatise on Fourier's Series, by William Elwood Byerly %% +%% %% +%% Packages and substitutions: %% +%% %% +%% book : Document class. %% +%% amsmath: Basic AMS math package. %% +%% amssymb: Basic AMS symbols %% +%% tabularx Tabulation. %% +%% graphicx Basic graphics for images. %% +%% inputenc: Encoding %% +%% needspace Conditional new page %% +%% wrapfig textflow round images %% +%% verbatim Preformated text %% +%% wasysym WASY2 fonts (upright integrals) %% +%% %% +%% PDF Pages: 309 %% +%% %% +%% 144 underfull hboxes %% +%% %% +%% 14 includegraphics calls included as nnn.png %% +%% %% +%% Compile sequence: %% +%% pdflatex x 2 %% +%% %% +%% Compile History: %% +%% %% +%% Jul 09: Laverock. 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You may copy it, give it away or +re-use it under the terms of the Project Gutenberg License included +with this eBook or online at www.gutenberg.org + + +Title: An Elementary Treatise on Fourier's Series and Spherical, Cylindrical, and Ellipsoidal Harmonics + With Applications to Problems in Mathematical Physics + +Author: William Elwood Byerly + +Release Date: August 19, 2009 [EBook #29779] + +Language: English + +Character set encoding: ISO-8859-1 + +*** START OF THIS PROJECT GUTENBERG EBOOK TREATISE ON FOURIER'S SERIES *** + + +Produced by Laura Wisewell, Carl Hudkins, Keith Edkins and +the Online Distributed Proofreading Team at +http://www.pgdp.net (The original copy of this book was +generously made available for scanning by the Department +of Mathematics at the University of Glasgow.) +\end{verbatim} + +% -----File: 001.png +\newpage +% title page +\begin{center} + +{\Large AN ELEMENTARY TREATISE}\\[3ex] + +{\scriptsize ON}\\[3ex] + +{\Huge FOURIER'S SERIES}\\[3ex] + +{\scriptsize AND}\\[3ex] + +{\Large SPHERICAL, CYLINDRICAL, AND ELLIPSOIDAL\\ + +HARMONICS,}\\[3ex] + +{\scriptsize WITH}\\[3ex] + +APPLICATIONS TO PROBLEMS IN MATHEMATICAL PHYSICS.\\[5ex] + +{\scriptsize BY}\\ + +WILLIAM ELWOOD BYERLY, \textsc{Ph.D.},\\ + +{\scriptsize PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY.}\\[4ex] + +\hspace{1.5in} \hrulefill \hspace*{1.7in} + +\vspace{5ex} + +{\Large GINN \& COMPANY}\\ + +BOSTON · NEW YORK · CHICAGO · LONDON\\ + +\end{center} + +% -----File: 002.png + +\bigskip\bigskip + +\begin{center} +{\footnotesize Copyright, 1893,\\ +By WILLIAM ELWOOD BYERLY.} + +\par\rule{5em}{0.5pt}\par + +{\scriptsize ALL RIGHTS RESERVED.} +\end{center} + +\bigskip\bigskip +\scriptsize \noindent \textsc{Transcriber's Note:} \emph{A few typographical errors have been corrected - +these are noted at the end of the text.} \normalsize + +% -----File: 003.png + +\frontmatter +\pagestyle{myheadings} +\newpage +\setcounter{footnote}{0} +\newpage +\medskip\begin{center}{\Large PREFACE.}\par\rule{5em}{0.5pt}\par\end{center} +\markright{} + +About ten years ago I gave a course of lectures on Trigonometric Series, +following closely the treatment of that subject in Riemann's ``Partielle +Differentialgleichungen,'' to accompany a short course on The Potential +Function, given by Professor B.~O. Peirce. + +My course has been gradually modified and extended until it has become an +introduction to Spherical Harmonics and Bessel's and Lamé's Functions. + +Two years ago my lecture notes were lithographed by my class for their +own use and were found so convenient that I have prepared them for +publication, hoping that they may prove useful to others as well as to my +own students. Meanwhile, Professor Peirce has published his lectures on +``The Newtonian Potential Function'' (Boston, Ginn \& Co.), and the two +sets of lectures form a course (Math.~10) given regularly at Harvard, and +intended as a partial introduction to modern Mathematical Physics. + +Students taking this course are supposed to be familiar with so much of the +infinitesimal calculus as is contained in my ``Differential Calculus'' (Boston, +Ginn \& Co.)\ and my ``Integral Calculus'' (second edition, same publishers), +to which I refer in the present book as ``Dif.\ Cal.''\ and ``Int.\ Cal.'' Here, +as in the ``Calculus,'' I speak of a ``derivative'' rather than a ``differential +coefficient,'' and use the notation $D_x$ instead of $\frac{\delta}{\delta x}$ for ``partial derivative with +respect to $x$.'' + +The course was at first, as I have said, an exposition of Riemann's ``Partielle +Differentialgleichungen.'' In extending it, I drew largely from Ferrer's +``Spherical Harmonics'' and Heine's ``Kugelfunctionen,'' and was somewhat +indebted to Todhunter (``Functions of Laplace, Bessel, and Lamé''), Lord +Rayleigh (``Theory of Sound''), and Forsyth (``Differential Equations''). + +In preparing the notes for publication, I have been greatly aided by the +criticisms and suggestions of my colleagues, Professor B.~O. Peirce and Dr.\ +Maxime Bôcher, and the latter has kindly contributed the brief historical +sketch contained in Chapter~IX. + +\rightline{W.~E. BYERLY.\qquad} + +{\small \textsc{Cambridge, Mass.}, Sept.\ 1893.} +% -----File: 004.png +%[Blank Page] +% -----File: 005.png + +\setcounter{footnote}{0} +\newpage +\medskip\begin{center}{\Large ANALYTICAL TABLE OF CONTENTS.}\par\rule{5em}{0.5pt}\par\end{center} +\markright{} + +\begin{center}CHAPTER I.\end{center} + +\hfill \textsc{pages} + +\textsc{Introduction} \hfill \pageref{ch1start}--\pageref{ch1end}\\ + +\textsc{Art.}\ 1. List of some important homogeneous linear partial differential equations +of Physics.---\textsc{Arts.}\ 2--4. Distinction between the general solution and a particular +solution of a differential equation. Need of additional data to make the solution +of a differential equation determinate. Definition of linear and of linear and +homogeneous.---\textsc{Arts.}\ 5--6. Particular solutions of homogeneous linear differential +equations may be combined into a more general solution. Need of development +in terms of normal forms.---\textsc{Art.}\ 7. Problem: Permanent state of temperatures +in a thin rectangular plate. Need of a development in sine series. Example.---\textsc{Art.}\ +8. Problem: Transverse vibrations of a stretched elastic string. A development +in sine series suggested.---\textsc{Art.}\ 9. Problem: Potential function due to the +attraction of a circular ring of small cross-section. Surface Zonal Harmonics +(Legendre's Coefficients). Example.---\textsc{Art.}\ 10. Problem: Permanent state of +temperatures in a solid sphere. Development in terms of Surface Zonal Harmonics +suggested.---\textsc{Arts.}\ 11--12. Problem: Vibrations of a circular drumhead. +Cylindrical Harmonics (Bessel's Functions). Recapitulation.---\textsc{Art.}\ 13. Method +of making the solution of a linear partial differential equation depend upon solving +a set of ordinary differential equations by assuming the dependent variable equal +to a product of factors each of which involves but one of the independent variables. +\textsc{Arts.}\ 14--15 Method of solving ordinary homogeneous linear differential equations +by development in power series. Applications.---\textsc{Art.}\ 16. Application to +Legendre's Equation. Several forms of general solution obtained. Zonal +Harmonics of the second kind.---\textsc{Art.}\ 17. Application to Bessel's Equation. +General solution obtained for the case where $m$ is not an integer, and for the case +where $m$ is zero. Bessel's Function of the second kind and zeroth order.---\textsc{Art.}\ +18. Method of obtaining the general solution of an ordinary linear differential +equation of the second order from a given particular solution. Application to +the equations considered in Arts.~14--17.\\[1ex] + +\begin{center}CHAPTER II.\end{center} + +\textsc{Development in Trigonometric Series} \hfill \pageref{ch2start}--\pageref{ch2end}\\ + +\textsc{Arts.}\ 19--22. Determination of the coefficients of $n$ terms of a sine series so that +the sum of the terms shall be equal to a given function of $x$ for $n$ given values +of $x$. Numerical example.---\textsc{Art.}\ 23. Problem of development in sine series +treated as a limiting case of the problem just solved.---\textsc{Arts.}\ 24--25. Shorter +method of solving the problem of development in series involving sines of whole +multiples of the variable. Working rule deduced. Recapitulation.---\textsc{Art.}\ 26. A +% -----File: 006.png +few important sine developments obtained. Examples.---\textsc{Arts.}\ 27--28. Development +in cosine series. Examples.---\textsc{Art.}\ 29. Sine series an odd function of the +variable, cosine series an even function, and both series periodic functions.---\textsc{Art.}\ +30. Development in series involving both sines and cosines of whole +multiples of the variable. Fourier's series. Examples.---\textsc{Art.}\ 31. Extension of +the range within which the function and the series are equal. Examples.---\textsc{Art.}\ +32. Fourier's Integral obtained.\\[1ex] + +\markright{TABLE OF CONTENTS} +\begin{center}CHAPTER III.\end{center} + +\textsc{Convergence of Fourier's Series} \hfill \pageref{ch3start}--\pageref{ch3end}\\ + +\textsc{Arts.}\ 33--36. The question of the convergence of the sine series for unity considered +at length.---\textsc{Arts.}\ 37--38. Statement of the conditions which are sufficient +to warrant the development of a function into a Fourier's series. Historical note. +\textsc{Art.}\ 39. Graphical representation of successive approximations to a sine series. +Properties of a Fourier's series inferred from the constructions.---\textsc{Arts.}\ 40--42. +Investigation of the conditions under which a Fourier's series can be differentiated +term by term.---\textsc{Art.}\ 43. Conditions under which a function can be expressed as +a Fourier's Integral.\\[1ex] + +\begin{center}CHAPTER IV.\end{center} + +\textsc{Solution of Problems in Physics by the Aid of Fourier's Integrals and +Fourier's Series } \hfill \pageref{ch4start}--\pageref{ch4end}\\ + +\textsc{Arts.}\ 44--48. Logarithmic Potential. Flow of electricity in an infinite plane, +where the value of the Potential Function is given along an infinite straight line; +along two mutually perpendicular straight lines; along two parallel straight lines. +Examples. Use of Conjugate Functions. Sources and Sinks. Equipotential +lines and lines of Flow. Examples.---\textsc{Arts.}\ 49--52. One-dimensional flow of heat. +Flow of heat in an infinite solid; in a solid with one plane face at the temperature +zero; in a solid with one plane face whose temperature is a function of the time +(Riemann's solution); in a bar of small cross section from whose surface heat +escapes into air at temperature zero. Limiting state approached when the temperature +of the origin is a periodic function of the time. Examples.---\textsc{Arts.}\ 53--54. +Temperatures due to instantaneous and to permanent heat sources and sinks, +and to heat doublets. Examples. Application to the case where there is +leakage.---\textsc{Arts.}\ 55--56. Transmission of a disturbance along an infinite stretched +elastic string. Examples.---\textsc{Arts.}\ 57--58. Stationary temperatures in a long +rectangular plate. Temperature of the base unity. Summation of a Trigonometric +series. Isothermal lines and lines of flow. Examples.---\textsc{Art.}\ 59. Potential +Function given along the perimeter of a rectangle. Examples.---\textsc{Arts.}\ 60--63. +One-dimensional flow of heat in a slab with parallel plane faces. Both faces at +temperature zero. Both faces adiathermanous. Temperature of one face a +function of the time. Examples.---\textsc{Art.}\ 64. Motion of a stretched elastic string +fastened at the ends. Steady vibration. Nodes. Examples.---\textsc{Art.}\ 65. Motion +of a string in a resisting medium.---\textsc{Art.}\ 66. Flow of heat in a sphere whose +surface is kept at a constant temperature.---\textsc{Arts.}\ 67--68. Cooling of a sphere in +air. Surface condition given by a differential equation. Development in a Trigonometric +series of which Fourier's Sine Series is a special case. Examples.---\textsc{Arts.}\ +% -----File: 007.png +69--70. Flow of heat in an infinite solid with one plane face which is +exposed to air whose temperature is a function of the time. Solution for an +instantaneous heat source when the temperature of the air is zero. Examples.---\textsc{Arts.}\ +71--73. Vibration of a rectangular drumhead. Development of a function +of two variables in a double Fourier's Series. Examples. Nodal lines in a +rectangular drumhead. Nodal lines in a square drumhead.\\[1ex] + +\textsc{Miscellaneous Problems} \hfill \pageref{probstart}--\pageref{probend}\\ + +I. Logarithmic Potential. Polar Coördinates.---II. Potential Function in Space. +III. Conduction of heat in a plane.---IV. Conduction of heat in Space.\\[1ex] + +\begin{center}CHAPTER V.\end{center} + +\textsc{Zonal Harmonics} \hfill \pageref{ch5start}--\pageref{ch5end}\\ + +\textsc{Art.}\ 74. Recapitulation. Surface Zonal Harmonics (Legendrians). Zonal Harmonics +of the second kind.---\textsc{Arts.}\ 75--76. Legendrians as coefficients in a Power +Series. Special values.---\textsc{Art.}\ 77. Summary of the properties of a Legendrian. +List of the first eight Legendrians. Relation connecting any three successive +Legendrians.---\textsc{Arts.}\ 78--81. Problems in Potential. Potential Function due to +the attraction of a material circular ring of small cross section. Potential Function +due to a charge of electricity placed on a thin circular disc. Examples: Spheroidal +conductors. Potential Function due to the attraction of a material homogeneous +circular disc. Examples: Homogeneous hemisphere; Heterogeneous sphere; +Homogeneous spheroids. Generalisation.---\textsc{Art.}\ 82. Legendrian as a sum of +cosines.---\textsc{Arts.}\ 83--84. Legendrian as the $m$th derivative of the $m$th power of +$x^2--1$.---\textsc{Art.}\ 85. Equations derivable from Legendre's Equation.---\textsc{Art.}\ 86. +Legendrian as a Partial Derivative.---\textsc{Art.}\ 87. Legendrian as a Definite Integral. +\textsc{Arts.}\ 88--90. Development in Zonal Harmonic Series. Integral of the product of +two Legendrians of different degrees. Integral of the square of a Legendrian. +Formulas for the coefficients of the series.---\textsc{Arts.}\ 91--92. Integral of the product +of two Legendrians obtained by the aid of Legendre's Equation; by the aid of +Green's Theorem. Additional formulas for integration. Examples.---\textsc{Arts.}\ 93--94. +Problems in Potential where the value of the Potential Function is given on a +spherical surface and has circular symmetry about a diameter. Examples.---\textsc{Art.}\ +95. Development of a power of $x$ in Zonal Harmonic Series.---\textsc{Art.}\ 96. +Useful formulas.---\textsc{Art.}\ 97. Development of $\sin n\theta$ and $\cos n\theta$ in Zonal Harmonic +Series. Examples. Graphical representation of the first seven Surface Zonal +Harmonics. Construction of successive approximations to Zonal Harmonic Series. +\textsc{Arts.}\ 98--99. Method of dealing with problems in Potential when the density is +given. Examples.---\textsc{Art.}\ 100. Surface Zonal Harmonics of the second kind. +Examples: Conal Harmonics.\\[1ex] + +\begin{center}CHAPTER VI.\end{center} + +\textsc{Spherical Harmonics} \hfill \pageref{ch6start}--\pageref{ch6end}\\ + +\textsc{Arts.}\ 101--102. Particular Solutions of Laplace's Equation obtained. Associated +Functions. Tesseral Harmonics. Surface Spherical Harmonics. Solid Spherical +Harmonics. Table of Associated Functions. Examples.---\textsc{Arts.}\ 103--108. Development +in Spherical Harmonic Series. The integral of the product of two +% -----File: 008.png +Surface Spherical Harmonics of different degrees taken over the surface of the +unit sphere is zero. Examples. The integral of the product of two Associated +Functions of the same order. Formulas for the coefficients of the series. Illustrative +example. Examples.---\textsc{Arts.}\ 109--110. Any homogeneous rational integral +Algebraic function of $x$, $y$, and $z$ which satisfies Laplace's Equation is a Solid +Spherical Harmonic. Examples.---\textsc{Art.}\ 111. A transformation of axes to a new +set having the same origin will change a Surface Spherical Harmonic into another +of the same degree.---\textsc{Arts.}\ 112--114. Laplacians. Integral of the product of a +Surface Spherical Harmonic by a Laplacian of the same degree. Development in +Spherical Harmonic Series by the aid of Laplacians. Table of Laplacians. Example.---\textsc{Art.}\ +115. Solution of problems in Potential by direct integration. +Examples.---\textsc{Arts.}\ 116--118. Differentiation along an axis. Axes of a Spherical +Harmonic.---\textsc{Art.}\ 119. Roots of a Zonal Harmonic. Roots of a Tesseral Harmonic. +Nomenclature justified.\\[1ex] + +\Needspace*{4\baselineskip} + +\begin{center}CHAPTER VII.\end{center} + +\textsc{Cylindrical Harmonics (Bessel's Functions)} \hfill \pageref{ch7start}--\pageref{ch7end}\\ + +\textsc{Art.}\ 120. Recapitulation. Cylindrical Harmonics (Bessel's Functions) of the +zeroth order; of the $n$th order; of the second kind. General solution of Bessel's +Equation.---\textsc{Art.}\ 121. Bessel's Functions as definite integrals. Examples.---\textsc{Art.}\ +122. Properties of Bessel's Functions. Semi-convergent series for a Bessel's +Function. Examples.---\textsc{Art.}\ 123. Problem: Stationary temperatures in a +cylinder (\emph{a}) when the temperature of the convex surface is zero; (\emph{b}) when the +convex surface is adiathermanous; (\emph{c}) when the convex surface is exposed to air +at the temperature zero.---\textsc{Art.}\ 124. Roots of Bessel's functions.---\textsc{Art.}\ 125. +The integral of $r$ times the product of two Cylindrical Harmonics of the zeroth +order. Example.---\textsc{Art.}\ 126. Development in Cylindrical Harmonic Series. +Formulas for the coefficients. Examples.---\textsc{Art.}\ 127. Problem: Stationary +temperatures in a cylindrical shell. Bessel's Functions of the second kind +employed. Example: Vibration of a ring membrane.---\textsc{Art.}\ 128. Problem: +Stationary temperatures in a cylinder when the temperature of the convex surface +varies with the distance from the base. Bessel's Functions of a complex variable. +Examples.---\textsc{Art.}\ 129. Problem: Stationary temperatures in a cylinder when +the temperatures of the base are unsymmetrical. Bessel's Functions of the $n$th +order employed. Miscellaneous examples. Bessel's Functions of fractional +order.\\[1ex] + +\begin{center}CHAPTER VIII.\end{center} + +\textsc{\spreadout{Laplace's Equation in Curvilinear Coördinates. Ellipsoidal} +}\\ +\textsc{Harmonics} \hfill \pageref{ch8start}--\pageref{ch8end}\\ + +\textsc{Arts.}\ 130--131. Orthogonal Curvilinear Coördinates in general. Laplace's Equation +expressed in terms of orthogonal curvilinear coördinates by the aid of Green's +theorem.---\textsc{Arts.}\ 132--135. Spheroidal Coördinates. Laplace's Equation in +spheroidal coördinates, in normal spheroidal coördinates. Examples. Condition +that a set of curvilinear coördinates should be normal. Thermometric Parameters. +Particular solutions of Laplace's Equation in spheroidal coördinates. Spheroidal +Harmonics. Examples. The Potential Function due to the attraction of an +oblate spheroid. Solution for an external point. Examples.---\textsc{Arts.}\ 136--141. +% -----File: 009.png +\emph{Ellipsoidal Coördinates.} Laplace's Equation in ellipsoidal coördinates. Normal +ellipsoidal coördinates expressed as Elliptic Integrals. Particular solutions of +Laplace's Equation. Lamé's Equation. Ellipsoidal Harmonics (Lamé's Functions). +Tables of Ellipsoidal Harmonics of the degrees 1, 2, and 3. Lamé's +Functions of the second kind. Examples. Development in Ellipsoidal Harmonic +series. Value of the Potential Function at any point in space when its value is +given at all points on the surface of an ellipsoid.---\textsc{Art.}\ 142. \emph{Conical Coördinates.} +The product of two Ellipsoidal Harmonics a Spherical Harmonic.---\textsc{Art.}\ 143. +\emph{Toroidal Coördinates.} Laplace's Equation in toroidal coördinates. Particular +solutions. Toroidal Harmonics. Potential Function for an anchor ring.\\[1ex] + +\begin{center} +CHAPTER IX. +\end{center} + +\textsc{Historical Summary} \hfill \pageref{ch9start}--\pageref{ch9end}\\ + +\begin{center}APPENDIX.\end{center} + +\textsc{Tables} \hfill\pageref{tablestart}--\pageref{tableend}\\ +Table I\@. Surface Zonal Harmonics. Argument $\theta$ \hfill\pageref{tableI}\\ +Table II\@. Surface Zonal Harmonics. Argument $x$ \hfill\pageref{tableII}\\ +Table III\@. Hyperbolic Functions \hfill\pageref{tableIII}\\ +Table IV\@. Roots of Bessel's Functions \hfill\pageref{tableIV}\\ +Table V\@. Roots of Bessel's Functions \hfill\pageref{tableV}\\ +Table VI\@. Bessel's Functions \hfill\pageref{tableVI} + +% -----File: 010.png +%[Blank Page] +% -----File: 011.png +\mainmatter +%\pagestyle{myheadings} +\mychap{CHAPTER I.}{INTRODUCTION.} +\label{ch1start} + +\mypara{1.} In many important problems in mathematical physics we are obliged +to deal with \emph{partial differential equations} of a comparatively simple form. + +For example, in the Analytical Theory of Heat we have for the change of +temperature of any solid due to the flow of heat within the solid, the equation +\[ +\tag*{\smallrom{I}} +D_t u = a^2(D_x^2 u + D_y^2 u + D_z^2 u),\footnote + {For the sake of brevity we shall often use the symbol $\nabla^2$ for the operation $D_x^2 + D_y^2 + D_z^2$; + and with this notation equation~\smallrom{I} would be written $D_t u = a^2\nabla^2 u$.} +\] +where $u$ represents the temperature at any point of the solid and $t$ the time. + +In the simplest case, that of a slab of infinite extent with parallel plane +faces, where the temperature can be regarded as a function of one coördinate, +\smallrom{I} reduces to +\[ +\tag*{\smallrom{II}} +D_t u = a^2 D_x^2 u, +\] +a form of considerable importance in the consideration of the problem of the +cooling of the earth's crust. + +In the problem of the permanent state of temperatures in a thin rectangular +plate, the equation~\smallrom{I} becomes +\[ +\tag*{\smallrom{III}} +D_x^2u + D_y^2u = 0. +\] + +In \emph{polar} or \emph{spherical coördinates} \smallrom{I} is less simple, it is +\[ +\tag*{\smallrom{IV}} +D_t u = \dfrac{a^2}{r^2} + \left[ D_r(r^2D_r u) + + \dfrac{1}{\sin\theta} D_\theta (\sin\theta D_\theta u) + + \dfrac{1}{\sin^2\theta} D_\phi^2 u \right]. +\] + +In the case where the solid in question is a sphere and the temperature +at any point depends merely on the distance of the point from the centre +\smallrom{IV} reduces to +\[ +\tag*{\smallrom{V}} +D_t(ru) = a^2D_r^2(ru). +\] + +In \emph{cylindrical coördinates} \smallrom{I} becomes +\[ +\tag*{\smallrom{VI}} +D_t u = a^2[D_r^2 u + \dfrac{1}{r} D_r u + \dfrac{1}{r^2} D_\phi^2 u + D_z^2 u]. +\] + +In considering the flow of heat in a cylinder when the temperature at +any point depends merely on the distance $r$ of the point from the axis +\smallrom{VI} becomes +\[ +\tag*{\smallrom{VII}} +D_t u = a^2(D_r^2 u + \dfrac{1}{r} D_r u). +\] +% -----File: 012.png + +In Acoustics in several problems we have the equation +\[ +\tag*{\smallrom{VIII}} + D_t^{2}y = a^{2}D_{x}^{2}y; +\] +for instance, in considering the transverse or the longitudinal vibrations of a +stretched elastic string, or the transmission of plane sound waves through +the air. + +\markright{INTRODUCTION.} +If in considering the transverse vibrations of a stretched string we take +account of the resistance of the air \smallrom{VIII} is replaced by +\[\tag*{\smallrom{IX}} +D_t^2 y + 2kD_t y = a^2 D_x^2 y. +\] + +In dealing with the vibrations of a stretched elastic membrane, we have the +equation +\[\tag*{\smallrom{X}} +D_t^{2}z = c^2(D_x^{2}z + D_y^{2}z), +\] +or in \emph{cylindrical coördinates} +\[\tag*{\smallrom{XI}} +D_{t}^2 z = c^2 (D_{r}^2 z + \dfrac{1}{r} D_{r}z + \dfrac{1}{r^2}D_{\phi}^2 z). +\] + +In the theory of \emph{Potential} we constantly meet Laplace's Equation + +\begin{flalign*}\tag*{\smallrom{XII}} +&\hfill & D_x^{2}V + &D_y^{2}V + D_z^{2}V=0&& \\ +&\text{or}\hfill & &\nabla^2 V = 0&&\hfill\phantom{or} +\end{flalign*} +which in \emph{spherical coördinates} becomes +\[\tag*{\smallrom{XIII}} + \dfrac{1}{r^2} \left[ + rD_r^{2}(rV) + + \dfrac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta}V) + + \dfrac{1}{\sin^{2}\theta}D_{\phi}^{2}V \right] = 0, +\] +and in \emph{cylindrical coördinates} +\[\tag*{\smallrom{XIV}} +D_r^{2}V + \dfrac{1}{r} D_r V + \frac{1}{r^2} D_{\phi}^{2}V + D_z^{2} V = 0. +\] + +In \emph{curvilinear coördinates} it is +\[\tag*{\smallrom{XV}} + h_{1}h_{2}h_{3} \left[ + D_{\rho_1} \left( \frac{h_1}{h_2 h_3} D_{\rho_1} V \right) + + D_{\rho_2} \left( \frac{h_2}{h_3 h_1} D_{\rho_2} V \right) + + D_{\rho_3} \left( \frac{h_3}{h_1 h_2} D_{\rho_3} V \right) \right] = 0;\\ +\] +where \hfill +$f_1 (x,y,z) = \rho_{1}$, +$f_2 (x,y,z) = \rho_{2}$, +$f_3 (x,y,z) = \rho_{3}$ \hfill\phantom{where}\\[1ex] +represent a set of surfaces which cut one another at right angles, no matter +what values are given to $\rho_1$, $\rho_2$, and $\rho_3$; and where +\begin{align*} + h_1^2 &= (D_{x}\rho_1)^2 + (D_{y}\rho_{1})^2 + (D_{z}\rho_1)^2 \\ + h_2^2 &= (D_{x}\rho_2)^2 + (D_{y}\rho_{2})^2 + (D_{z}\rho_2)^2 \\ + h_3^2 &= (D_{x}\rho_3)^2 + (D_{y}\rho_{3})^2 + (D_{z}\rho_3)^2, +\end{align*} +and, of course, must be expressed in terms of $\rho_1$, $\rho_2$, and $\rho_3$. + +If it happens that $\nabla^{2}\rho_1 = 0$, $\nabla^{2}\rho_{2}= 0$, and $\nabla^{2}\rho_3=0$, then Laplace's +Equation~\smallrom{XV} assumes the very simple form +\[\tag*{\smallrom{XVI}} +h_1^2 D_{\rho_1}^2 V + h_2^2 D_{\rho_2}^2 V + h_3^2 D_{\rho_3}^2 V= 0. +\] +% -----File: 013.png + +\mypara{2.} A \emph{differential equation} is an equation containing derivatives or differentials +with or without the primitive variables from which they are derived. + +The \emph{general solution} of a differential equation is the equation expressing the +most general relation between the primitive variables which is consistent with +the given differential equation and which does not involve differentials or +derivatives. A general solution will always contain arbitrary (\emph{i.\,e.}, undetermined) +\emph{constants} or \emph{arbitrary functions}. + +A \emph{particular solution} of a differential equation is a relation between the +primitive variables which is consistent with the given differential equation, +but which is less general than the general solution, although included in it. + +Theoretically, every particular solution can be obtained from the general +solution by substituting in the general solution particular values for the arbitrary +constants or particular functions for the arbitrary functions; but in +practice it is often easy to obtain particular solutions directly from the differential +equation when it would be difficult or impossible to obtain the general +solution. + +\mypara{3.} If a problem requiring for its solution the solving of a differential equation +is \emph{determinate}, there must always be given in addition to the differential +equation enough outside conditions for the determination of all the arbitrary +constants or arbitrary functions that enter into the general solution of the +equation; and in dealing with such a problem, if the differential equation can +be readily solved the natural method of procedure is to obtain its general +solution, and then to determine the constants or functions by the aid of the +given conditions. + +It often happens, however, that the general solution of the differential equation +in question cannot be obtained, and then, since the problem \emph{if determinate} +will be solved if by any means a solution of the equation can be found which +will also satisfy the given outside conditions, it is worth while to try to get +\emph{particular solutions} and so to combine them as to form a result which shall +satisfy the given conditions without ceasing to satisfy the differential equation. + +\mypara{4.} A differential equation is \emph{linear} when it would be of the first degree if +the dependent variable and all its derivatives were regarded as algebraic +unknown quantities. If it is linear and contains no term which does not +involve the dependent variable or one of its derivatives, it is said to be linear +and \emph{homogeneous}. + +All the differential equations collected in Art.~1 are linear and homogeneous. + +\mypara{5.} \emph{If a value of the dependent variable has been found which satisfies a +given homogeneous, linear, differential equation, the product formed by multiplying +this value by any constant will also be a value of the dependent variable +which will satisfy the equation.} +% -----File: 014.png + +For if all the terms of the given equation are transposed to the first member, +the substitution of the first-named value must reduce that member to +zero; substituting the second value is equivalent to multiplying each term of +the result of the first substitution by the same constant factor, which therefore +may be taken out as a factor of the whole first member. The remaining +factor being zero, the product is zero and the equation is satisfied. + +\emph{If several values of the dependent variable have been found each of which +satisfies the given differential equation, their sum will satisfy the equation;} for +if the sum of the values in question is substituted in the equation each term +of the sum will give rise to a set of terms which must be equal to zero, and +therefore the sum of these sets must be zero. + +\mypara{6.} It is generally possible to get by some simple device \emph{particular solutions} +of such differential equations as those we have collected in Art.~1. The +object of the branch of mathematics with which we are about to deal is to +find methods of so combining these particular solutions as to satisfy any given +conditions which are consistent with the nature of the problem in question. + +This often requires us to be able to develop any given function of the variables +which enter into the expression of these conditions in terms of \emph{normal +forms} suited to the problem with which we happen to be dealing, and suggested +by the form of particular solution that we are able to obtain for the +differential equation. + +These normal forms are frequently sines and cosines, but they are often +much more complicated functions known as \emph{Legendre's Coefficients,} or \emph{Zonal +Harmonics; Laplace's Coefficients,} or \emph{Spherical Harmonics: Bessel's Functions,} +or \emph{Cylindrical Harmonics; Lamé's Functions,} or \emph{Ellipsoidal Harmonics,} \&c. + +\mypara{7.} As an illustration, let us take Fourier's problem of the permanent state +of temperatures in a thin rectangular plate of breadth $\pi$ and of infinite length +whose faces are impervious to heat. We shall suppose that the two long +edges of the plate are kept at the constant temperature zero, that one of the +short edges, which we shall call the base of the plate, is kept at the temperature +unity, and that the temperatures of points in the plate decrease indefinitely +as we recede from the base; we shall attempt to find the temperature +at any point of the plate. + +Let us take the base as the axis of $X$ and one end of the base as the origin. +Then to solve the problem we are to find the temperature $u$ of any point from +the equation +\[ + D_x^{2}u +D_y^{2}u= 0 \tag*{[III] Art.~1} +\] +subject to the conditions +\begin{alignat*}{3} + u &= 0\quad &&\text{when}\quad & x &= 0 \tag{1} \\ + u &= 0 &&\quad\text{``} & x &= \pi \tag{2} \\ + u &= 0 &&\quad\text{``} & y &= \infty \tag{3} \\ + u &= 1 &&\quad\text{``} & y &= 0. \tag{4} +\end{alignat*} +% -----File: 015.png + +We shall begin by getting a particular solution of \smallrom{III}, and we shall use +a device which always succeeds when the equation is \textit{linear} and \textit{homogeneous} +and has \textit{constant coefficients}. + +Assume\label{notep5}\footnote + {This assumption must be regarded as purely tentative. It must be tested by substituting + in the equation, and is justified if it leads to a solution.} +$u = e^{\alpha y + \beta x}$, where $\alpha$ and $\beta$ are constants, substitute in \smallrom{III} and +divide by $e^{\alpha y + \beta x}$, and we have $\alpha^2 + \beta^2 = 0$. If, then, this condition is satisfied +$u = e^{\alpha y + \beta x}$ is a solution. + +Hence $u = e^{\alpha y \pm \alpha xi}$ \footnote + {We shall regularly use the symbol $i$ for $\sqrt{-1}$.} +is a solution of \smallrom{III}, no matter what value may be +given to $\alpha$. + +This form is objectionable, since it involves an imaginary. We can, however, +readily improve it. + +Take $u=e^{\alpha y} e^{\alpha xi}$, a solution of \smallrom{III}, and $u = e^{\alpha y} e^{-\alpha xi}$, another solution +of \smallrom{III}; add these values of $u$ and divide the sum by 2 and we have +$e^{\alpha y} \cos\alpha x$. (v.~Int.\ Cal.\ Art.~35,\ [1].) Therefore by Art.~5 +\[ +u = e^{\alpha y} \cos \alpha x \tag{5} +\] +is a solution of \smallrom{III}. Take $u=e^{\alpha y} e^{\alpha xi}$ and $u = e^{\alpha y} e^{-\alpha xi}$, subtract the +second value of $u$ from the first and divide by $2i$ and we have $e^{\alpha y} \sin\alpha x$. +(v.~Int.\ Cal.\ Art.~35,\ [2]). Therefore by Art.~5 +\[ +u = e^{\alpha y} \sin\alpha x \tag{6} +\] +is a solution of \smallrom{III}. + +Let us now see if out of these particular solutions we can build up a solution +which will satisfy the conditions (1), (2), (3), and (4). +\begin{flalign*} +&\text{\indent Consider }& +& u = e^{\alpha y} \sin\alpha x. &&\phantom{\indent Consider } \tag{6} +\end{flalign*} +It is zero when $x = 0$ for all values of $\alpha$. It is zero when $x = \pi$ if $\alpha$ is a +whole number. It is zero when $y = \infty$ if $\alpha$ is negative. If, then, we write +$u$ equal to a sum of terms of the form $Ae^{-my} \sin mx$, where $m$ is a positive +integer, we shall have a solution of \smallrom{III} which satisfies conditions (1), (2) +and (3). Let this solution be +\[ +u = A_1e^{-y} \sin x + A_2e^{-2y} \sin 2x + A_3e^{-3y} \sin 3x + A_4e^{-4y} \sin 4x + \cdots \tag{7} +\] +$A_1$, $A_2$, $A_3$, $A_4$, \&c., being undetermined constants. + +When $y = 0$ (7) reduces to +\[ +u = A_1 \sin x + A_2 \sin 2x + A_3 \sin 3x + A_4 \sin 4x + \cdots . \tag{8} +\] + +If now it is possible to develop unity into a series of the form (8), our +problem is solved; we have only to substitute the coefficients of that series for +$A_1$, $A_2$, $A_3$, \&c.\ in (7). +% -----File: 016.png + +It will be proved later that +\[ +1 = \frac{4}{\pi} \left( \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + + \frac{1}{7} \sin 7x + \cdots \right) +\] +for all values of $x$ between 0 and $\pi$; hence our required solution is +\[ + u = \frac{4}{\pi} \left[ e^{-y}\sin x + \frac{1}{3} e^{-3y} \sin 3x + + \frac{1}{5} e^{-5y} \sin 5x + \frac{1}{7} e^{-7y} \sin 7x + \cdots \right] +\tag{9} +\] +for this satisfies the differential equation and all the given conditions. + +If the given temperature of the base of the plate instead of being unity +is a function of $x$, we can solve the problem as before if we can express the +given function of $x$ as a sum of terms of the form $A \sin m x$, where $m$ is a +whole number. + +The problem of finding the value of the \textit{potential function} at any point of +a long, thin, rectangular conducting sheet, of breadth $\pi$, through which an +electric current is flowing, when the two long edges are kept at potential zero, +and one short edge at potential unity, is mathematically identical with the +problem we have just solved. + +\Example{} + +Taking the temperature of the base of the plate described above as 100° +centigrade, and that of the sides of the plate as 0°, compute the temperatures +of the points +\begin{center} +(\textit{a}) $\displaystyle \left( \frac{\pi}{6}, 1 \right)$; +(\textit{b}) $\displaystyle \left( \frac{\pi}{3}, 2 \right)$; +(\textit{c}) $\displaystyle \left( \frac{\pi}{2}, 3 \right)$, +\end{center} +correct to the nearest degree. \hfill \emph{Ans.}\ (\textit{a}) 26°; (\textit{b}) 15°; (\textit{c}) 6°. + +\mypara{8.} As another illustration, we shall take the problem of the transverse +vibrations of a stretched string fastened at the ends, initially distorted into +some given curve and then allowed to swing. + +Let the length of the string be $l$. Take the position of equilibrium of the +string as the axis of $\text{X}$, and one of the ends as the origin, and suppose the +string initially distorted into a curve whose equation $y = f(x)$ is given. + +We have then to find an expression for $y$ which will be a solution of the +equation +\[ +D_t^2 y = a^2 D_x^2 y \tag*{\smallrom{VIII}\ Art.~1,} +\] +while satisfying the conditions +\begin{alignat}{4} + y &= 0 && \text{when}\quad& x &= 0 \tag{1} \\ + y &= 0 && \quad\text{``} & x &= l \tag{2} \\ + y &= f(x) && \quad\text{``} & t &= 0 \tag{3} \\ + D_t y &= 0 && \quad\text{``} & t &= 0, \tag{4} +\end{alignat} +the last condition meaning merely that the string starts from rest. +% -----File: 017.png + +As in the last problem let\footnote{See note on page \pageref{notep5}.} $y = e^{\alpha x+\beta t}$ and substitute in \smallrom{VIII}. Divide +by $e^{\alpha x +\beta t}$ and we have $\beta^2=a^2\alpha^2$ as the condition that our assumed value of +$y$ shall satisfy the equation. +\[ +y= e^{\alpha x \pm a\alpha t} \tag{5} +\] +is, then, a solution of \smallromr{VIII} whatever the value of $\alpha$. + +It is more convenient to have a trigonometric than an exponential form to +deal with, and we can readily obtain one by using an imaginary value for $\alpha$ in (5). +Replace $\alpha$ by $\alpha i$ and (5) becomes $y=e^{(x\pm at)\alpha i}$, a solution of \smallrom{VIII}. Replace +$\alpha$ by $-\alpha i$ and (5) becomes $y=e^{-(x\pm at)\alpha i}$, another solution of \smallrom{VIII}. Add +these values of $y$ and divide by 2 and we have $\cos\alpha(x\pm at)$. Subtract the +second value of $y$ from the first and divide by $2i$ and we have $\sin\alpha(x\pm at)$. +\begin{align*} +& y = \cos \alpha(x + at)\\ +& y = \cos \alpha(x - at)\\ +& y = \sin \alpha(x + at)\\ +& y = \sin \alpha(x - at) +\end{align*} +are, then, solutions of \smallrom{VIII}. Writing $y$ successively equal to half the sum +of the first pair of values, half their difference, half the sum of the last +pair of values, and half their difference, we get the very convenient particular +solutions of \smallrom{VIII}. +\begin{align*} +&y = \cos \alpha x \cos \alpha at\\ +&y = \sin \alpha x \sin \alpha at\\ +&y = \sin \alpha x \cos \alpha at\\ +&y = \cos \alpha x \sin \alpha at. +\end{align*} + +If we take the third form +\begin{align*} +&y = \sin \alpha x \cos \alpha at +\end{align*} +it will satisfy conditions (1) and (4), no matter what value may be given to +$\alpha$, and it will satisfy (2) if $\alpha = \dfrac{m\pi}{l}$ +where $m$ is an integer. + +If then we take +\[ +y=A_1\sin\frac{\pi x}{l}\cos\frac{\pi at}{l} + A_2\sin\frac{2\pi x}{l}\cos\frac{2\pi at}{l} + A_3\sin\frac{3\pi x}{l}\cos\frac{3\pi at}{l} + \cdots \tag{6} +\] +where $A_{1}$, $A_{2}$, $A_{3}$ $\cdots$ are undetermined constants, we shall have a solution of +\smallrom{VIII} which satisfies (1), (2), and (4). When $t=0$ it reduces to +\[ +y=A_1\sin\frac{\pi x}{l} + A_2\sin\frac{2\pi x}{l} + A_3\sin\frac{3\pi x}{l} + \cdots \tag{7} +\] + +If now it is possible to develop $f(x)$ into a series of the form (7), we can +solve our problem completely. We have only to take the coefficients of this +series as values of $A_1$, $A_2$, $A_3$ $\cdots$ in (6), and we shall have a solution of +\smallrom{VIII} which satisfies all our given conditions. +% -----File: 018.png + +In each of the preceding problems the \emph{normal function}, in terms of which a +given function has to be expressed, is the sine of a simple multiple of the +variable. It would be easy to modify the problem so that the \emph{normal form} +should be a cosine. + +We shall now take a couple of problems which are much more complicated +and where the normal function is an unfamiliar one. + +\mypara{9.} Let it be required to find the potential function due to a circular wire +ring of small cross section and of given radius $c$, supposing the matter of the +ring to attract according to the law of nature. + +We can readily find, by direct integration, the value of the potential function +at any point of the axis of the ring. We get for it +\[ +V=\frac{M}{\sqrt{c^2 + x^2}} \tag{1} +\] +where $M$ is the mass of the ring, and $x$ the distance of the point from the +centre of the ring. + +Let us use spherical coördinates, taking the centre of the ring as origin and +the axis of the ring as the polar axis. + +To obtain the value of the potential function at any point in space, we must +satisfy the equation +\[ +rD_r^2(rV) + \frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V) + \frac{1}{\sin^2\theta}D_\phi^2V=0, \tag*{\smallrom{XIII} Art.~1,} +\] +subject to the condition +\[ +V=\frac{M}{(c^2+r^2)^{\frac{1}{2}}} \quad \text{when} \quad \theta=0. \tag{1} +\] + +From the symmetry of the ring, it is clear that the value of the potential +function must be independent of $\phi$, so that \smallrom{XIII} will reduce to +\[ +rD_r^2(rV) + \frac{1}{\sin\theta} D_\theta(\sin\theta D_\theta V)=0. \tag{2} +\] + +We must now try to get particular solutions of (2), and as the coefficients +are not constant, we are driven to a new device. + +Let\footnote{See note on page \pageref{notep5}.} $V=r^m P$, where $P$ is a function of $\theta$ only, and $m$ is a positive integer, +and substitute in (2), which becomes +\[ +m(m+1)r^mP + \frac{r^m}{\sin\theta}D_\theta(\sin\theta D_\theta P) =0. +\] +% -----File: 019.png +Divide by $r^m$ and use the notation of ordinary derivatives since $P$ depends +upon $\theta$ only, and we have the equation +\[ +m(m+1)P + \frac{1}{\sin\theta}\frac{d\Big(\sin\theta\dfrac{dP}{d\theta}\Big)}{d\theta}=0, \tag{3} +\] +from which to obtain $P$. + +Equation (3) can be simplified by changing the independent variable. Let +$x=\cos\theta$ and (3) becomes +\[ +\frac{d}{dx}\left [(1-x^2)\frac{dP}{dx}\right ] + m(m+1)P = 0. \tag{4} +\] + +Assume\footnote{See note on page \pageref{notep5}.} now that $P$ can be expressed as a sum or as a series of terms +involving whole powers of $x$ multiplied by constant coefficients. + +Let $P=\sum a_nx^n$ and substitute this value of $P$ in (4). We get +\[ +\textstyle\sum[n(n-1)a_n x^{n-2} - n(n+1) a_n x^n + m(m+1) a_nx^n] = 0, \tag{5} +\] +where the symbol $\sum$ indicates that we are to form all the terms we can by +taking successive whole numbers for $n$. + +As (5) must be true no matter what the value of $x$, the coefficient of any +given power of $x$, as for instance $x^k$, must vanish. Hence +\begin{flalign*} +&&(k+2)(k+1)a_{k+2}-k(k+1)a_k+m(m+ {}& 1)a_k = 0 & \tag{6}\\ +&\text{and} & a_{k+2} = -\frac{m(m+1)-k(k+1)}{(k+1)(k+2)}a_k.& & \tag{7} +\end{flalign*} +If now any set of coefficients satisfying the relation (7) be taken, $P = \sum a_k x^k$ +will be a solution of (4). +\begin{flalign*} +&\indent\text{If }& & k=m,\quad a_{k+2}=0,\quad a_{k+4}=0,\quad \text{\&c.} &&\phantom{\indent If } +\end{flalign*} +Since it will answer our purpose if we pick out the simplest set of coefficients +that will obey the condition (7), we can take a set including $a_m$. + +Let us rewrite (7) in the form +\[ +a_k=-\frac{(k+2)(k+1)}{(m-k)(m+k+1)}a_{k+2}. \tag{8} +\] +We get from (8), beginning with $k=m-2$, +\begin{align*} +a_{m-2}& = -\frac{m(m-1)}{2.(2m-1)}a_m\\ +a_{m-4}& = \frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}a_m\\ +a_{m-6}& = -\frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{2.4.6.(2m-1)(2m-3)(2m-5)}a_m, \text{\quad\&c.} +\end{align*} +% -----File: 020.png +If $m$ is even we see that the set will end with $a_0$, if $m$ is odd, with $a_1$. +\[ +P = a_m \left[ x^m + - \frac{m(m-1)}{2.(2m-1)} x^{m-2} + + \frac{m(m-1)(m-2)(m-3)} + {2.4.(2m-1)(2m-3)} x^{m-4} + - \cdots\,\right] +\] +where $a_m$ is entirely arbitrary, is, then, a solution of (4). It is found convenient +to take $a_m$ equal to +\[ +\frac{(2m - 1)(2m - 3)\cdots 1}{m!} +\] +and it can be shown that with this value of $a_m$ $P = 1$ when $x = 1$. + +$P$ is a function of $x$ and contains no higher powers of $x$ than $x^m$. It is +usual to write it as $P_m(x)$. + +We proceed to compute a few values of $P_m(x)$ from the formula +\begin{align*} +P_m(x) &= \frac{(2m-1)(2m-3) \cdots 1}{m!} \left[x^m - \frac{m(m-1)}{2.(2m-1)}x^{m-2} \right. \\[1ex] +&\left. + \frac{m(m-1)(m-2)(m-3)} + {2.4.(2m-1)(2m-3)} x^{m-4} - \cdots \,\right]. \tag{9} +\end{align*} + +We have: +{ +\[ +\left.\begin{alignedat}{4} +P_0(x) &= 1 && \text{or}\qquad & P_0(\cos\theta) &= 1 \\ +P_1(x) &= x && \:\text{``} & P_1(\cos\theta) &= \cos\theta \\ +P_2(x) &= \tfrac{1}{2}(3x^2 - 1) && \:\text{``} & P_2(\cos\theta) &= \tfrac{1}{2}(3 \cos^2\theta - 1) \\ +P_3(x) &= \tfrac{1}{2}(5x^3 - 3x) \quad && \:\text{``} & P_3(\cos\theta) &= \tfrac{1}{2}(5 \cos^3\theta - 3 \cos\theta) \\ +P_4(x) &= \rlap{$\tfrac{1}{8}(35x^4 - 30x^2 + 3)$ or} \\ + &&&& \llap{$P_4(\cos\theta) = \tfrac{1}{8}(35$} &\cos^4\theta - 30 \cos^2\theta + 3) \\ +P_5(x) &= \rlap{$\tfrac{1}{8}(63x^5 - 70x^3 + 15x)$ or} \\ + &&&& \llap{$P_5(\cos\theta) = \tfrac{1}{8}(63 \cos^5\theta $} &-70 \cos^3\theta + 15 \cos\theta). +\end{alignedat} +\tag{10} +\quad\right\} +\]} + +We have obtained $P = P_m(x)$ as a particular solution of (4) and +$P = P_m(\cos\theta)$ as a particular solution of (3). $P_m(x)$ or $P_m(\cos\theta)$ is a +new function, known as a \textit{Legendre's Coefficient}, or as a \textit{Surface Zonal Harmonic}, +and occurs as a normal form in many important problems. + +$V = r^mP_m(\cos\theta)$ is a particular solution of (2) and $r^mP_m(\cos\theta)$ is sometimes +called a \textit{Solid Zonal Harmonic}. + +We can now proceed to the solution of our original problem. +\[ +V=A_0r^0P_0(\cos\theta) + A_1rP_1(\cos\theta) + A_2r^2P_2(\cos\theta) + A_3r^3P_3(\cos\theta) + \cdots \tag{11} +\] +where $A_0$, $A_1$, $A_2$, \&c., are entirely arbitrary, is a solution of (2) (v.\ Art.~5). +When $\theta = 0$ (11) reduces to +\[ +V=A_0 + A_1 r + A_2 r^2 + A_3 r^3 + \cdots\,, +\] +since, as we have said, $P_m(x) = 1$ when $x = 1$, or $P_m(\cos\theta) = 1$ when $\theta = 0$. + +By our condition (1) +\[ +V=\frac{M}{(c^2 + r^2)^{\frac{1}{2}}} +\] +when $\theta = 0$. +% -----File: 021.png + +By the Binomial Theorem +\[ +\frac{M}{(c^2+r^2)^{\frac{1}{2}}} += \frac{M}{c} \left[ 1 - \frac{1}{2} \frac{r^2}{c^2} + + \frac{1.3}{2.4} \frac{r^4}{c^4} + - \frac{1.3.5}{2.4.6} \frac{r^6}{c^6} + + \cdots \,\right] +\] +provided $r < c$. Hence +% recast to fit line +\begin{align*} +V = \frac{M}{c} \left[ P_0(\cos\theta) + - \frac{1}{2} \frac{r^2}{c^2} P_2(\cos\theta)\right. +& + \frac{1.3}{2.4} \frac{r^4}{c^4} P_4(\cos\theta) \\ +& \left. - \frac{1.3.5}{2.4.6} \frac{r^6}{c^6} P_6(\cos\theta) + + \cdots\, \right] \tag{12} +\end{align*} +is our required solution if $r < c$; for it is a solution of equation~(2) and satisfies +condition~(1). + +\Example{} + +Taking the mass of the ring as one pound and the radius of the ring as one +foot, compute to two decimal places the value of the potential function due to +the ring at the points +{\def\bit#1#2#3{\text{(\textit{#1})\ }\left(r = #2, \, \theta = #3 \right);} +\begin{align*} +& \bit{a}{.2}{0} & +& \bit{d}{.6}{0} & +& \bit{f}{.6}{\frac{\pi}{3}} \\ +% +& \bit{b}{.2}{\frac{\pi}{4}} & +& \bit{e}{.6}{\frac{\pi}{6}} & +& \bit{g}{.6}{\frac{\pi}{2}} \\ +% +& \bit{c}{.2}{\frac{\pi}{2}} +\end{align*} +\hfill\smash{\raisebox{3ex}{\begin{tabular} +{l@{\hfill}} +\llap{\emph{Ans.} }(\emph{a}) .98; (\emph{b}) .99; (\emph{c}) 1.01; (\emph{d}) .86;\\ + (\emph{e}) .90; (\emph{f}) 1.00; (\emph{g}) 1.10. +\end{tabular}}}}\break +The unit used is the potential due to a pound of mass concentrated at a point +and attracting a second pound of mass concentrated at a point, the two points +being a foot apart. + +\mypara{10.} A slightly different problem calling for development in terms of Zonal +Harmonics is the following: + +Required the permanent temperatures within a solid sphere of radius~1, +one half of the surface being kept at the constant temperature zero, and the +other half at the constant temperature unity. + +Let us take the diameter perpendicular to the plane separating the unequally +heated surfaces as our axis and let us use spherical coördinates. As in the +last problem, we must solve the equation +\[ +rD_r^2(ru) + \frac{1}{\sin\theta} D_\theta (\sin \theta D_\theta u) + \frac{1}{\sin^2\theta} D_\phi^2 u = 0 \tag*{\smallrom{XIII} Art.~1} +\] +which as before reduces to +\[ +r D_r^2 (ru) + \frac{1}{\sin\theta} D_\theta (\sin\theta D_\theta u) = 0 \tag{1} +\] +from the consideration that the temperatures must be independent of $\phi$. + +Our equation of condition is +\[ +u = 1 \text{ from } \theta = 0 \text{ to } \theta = \frac{\pi}{2} \text{ and } u = 0 \text{ from } \theta = \frac{\pi}{2} \text{ to } \theta = \pi, \tag{2} +\] +when $r = 1$. +% -----File: 022.png + +As we have seen $u = r^mP_m(\cos \theta)$ is a particular solution of (1), $m$ being +any positive whole number, and +\[ + u = A_0r^0P_0 (\cos \theta) + + A_1r P_1 (\cos \theta) + + A_2r^2P_2 (\cos \theta) + + A_3r^3P_3 (\cos \theta) + \cdots +\tag{3} +\] +where $A_0$, $A_1$, $A_2$, $A_3 \cdots$ are undetermined constants, is a solution of (1). + +When $r = 1$ (3) reduces to +\[ +\tag{4} +u = A_0P_0 (\cos \theta) + A_1P_1 (\cos \theta) + A_2P_2(\cos \theta) + A_3P_3 (\cos \theta) + \cdots +\] +If then we can develop our function of $\theta$ which enters into equation~(2) in +a series of the form (4), we have only to take the coefficients of that series +as the values of $A_0$, $A_1$, $A_2$, \&c., in (3) and we shall have our required solution. + +\mypara{11.} As a last example we shall take the problem of the vibration of a stretched +circular membrane fastened at the circumference, that is, of an ordinary drumhead. +We shall suppose the membrane initially distorted into any given form +which has circular symmetry\label{notep12}\footnote + {A function of the coördinates of a point has \textit{circular symmetry} about an axis when its + value is not affected by rotating the point through any angle about the axis. A surface has + circular symmetry about an axis when it is a surface of revolution about the axis.} +about an axis through the centre perpendicular +to the plane of the boundary, and then allowed to vibrate. + +Here we have to solve +\[ +\tag*{\smallrom{XI} Art.~1} +D_t^2z=c^2\left(D_r^2z + \frac{1}{r}D_rz + \frac{1}{r^2} D_\phi^2z \right) +\] +subject to the conditions +\begin{alignat*}{3} +\tag{1} + z &= f(r) \quad && \text{when}\quad & t &= 0 \\ +\tag{2} + D_t z &= 0 && \quad\text{``} & t &= 0 \\ +\tag{3} + z &= 0 && \quad\text{``} & r &= a. +\end{alignat*} + +From the symmetry of the supposed initial distortion $z$ must be independent +of $\phi$, therefore \smallrom{XI} reduces to +\[ +\tag{4} +D_t^2 z = c^2\left(D_r^2 z + \frac{1}{r} D_r z\right) +\] +and this is the equation for which we wish to find a particular solution. + +We shall employ a device not unlike that used in Art.~9. + +Assume\footnote + {See note on page \pageref{notep5}.} +$z = R.T$ where $R$ is a function of $r$ alone and $T$ is a function of +$t$ alone. Substitute this value of $z$ in (4) and we get +\[ +R D_t^2 T = c^2 T\left(D_r^2 R + \frac{1}{r} D_r R\right) +\] +\begin{flalign*} +\tag{5} +&\text{or }&& +\frac{1}{c^2T} \frac{d^2T}{dt^2} = \frac{1}{R} \left(\frac{d^2R}{dr^2} + \frac{1}{r} \frac{dR}{dr}\right). && +\end{flalign*} +The second member of (5) does not involve $t$, therefore its equal the first +member must be independent of $t$. The first member of (5) does not involve +% -----File: 023.png +$r$, and consequently since it contains neither $t$ nor $r$, it must be constant. Let +it equal $-\mu^2$, where $\mu$ of course is an undetermined constant. + +Then (5) breaks up into the two differential equations +\begin{gather} +\frac{d^2T}{dt^2} + \mu^2c^2T=0 \tag{6} \\ +\frac{d^2R}{dr^2} + \frac{1}{r}\frac{dR}{dr} + \mu^2R = 0. \tag{7} +\end{gather} +(6) can be solved by familiar methods, and we get $T = \cos \mu ct$ and $T = \sin \mu ct$ +as simple particular solutions (v.\ Int.\ Cal.\ p.~319, §~21). + +To solve (7) is not so easy. We shall first simplify it by a change of independent +variable. Let $r = \dfrac{x}{\mu}$. (7) becomes +\[ +\frac{d^2R}{dx^2} + \frac{1}{x}\frac{dR}{dx}+R=0. \tag{8} +\] + +Assume, as in Art.~9, that $R$ can be expressed in terms of whole powers of +$x$. Let $R = \sum a_n x^n$ and substitute in (8). We get +\[ + \textstyle\sum [n(n - 1)a_nx^{n-2} + na_nx^{n-2} + a_nx^n] = 0, \notag +\] +an equation which must be true no matter what the value of $x$. The coefficient +of any given power of $x$, as $x^{k-2}$, must, then, vanish, and +\begin{flalign*} +&& k(k - 1)a_k + k&a_k + a_{k-2} = 0 &&\phantom{whence we obtain } \\ +&\text{or }& k^2a_k + a&_{k-2} = 0 && \\ +&\text{whence we obtain }& + a_{k-2} ={}& - k^2a_k && \tag{9} +\end{flalign*} +as the only relation that need be satisfied by the coefficients in order that +$R = \sum a_k x^k$ shall be a solution of (8). +\[ +\hbox to \displaywidth {\rlap{\indent If}\hfil $k = 0$,\quad $a_{k-2}= 0$,\quad $a_{k-4}=0$,\quad \&c.\hfil} +\] +We can then begin with $k = 0$ as our lowest subscript. +\begin{flalign*} +&\text{\indent From (9) }& a_k &=-\frac{a_{k-2}}{k^2}. &&\phantom{\indent From\ (9) }\\ +&\text{Then }& a_{2} &= -\frac{a_0}{2^{2}} \\ +&& a_{4} &= \frac{a_0}{2^2.4^2} \\ +&& a_{6} &= -\frac{a_0}{2^2.4^2.6^2}, \text{ \&c.} \\ +&\text{Hence }& + R = a_0\left[1-\frac{x^2}{2^2} \right.+{}& \frac{x^4}{2^2.4^2}- \left. \frac{x^6}{2^2.4^2.6^2} + \cdots\,\right] +\end{flalign*} +where $a_0$ may be taken at pleasure, is a solution of (8), provided the series is +convergent. +% -----File: 024.png + +Take $a_0=1$, and then $R=J_0(x)$ where +\[ +J_0(x)=1-\frac{x^2}{2^2} + \frac{x^4}{2^2.4^2} - \frac{x^6}{2^2.4^2.6^2} + \frac{x^8}{2^2.4^2.6^2.8^2} - \cdots \tag{10} +\] +is a solution of (8). + +$J_0(x)$ is easily shown to be convergent for all values real or imaginary of $x$, +since the series made up of the moduli of the terms of $J_0(x)$ (v.\ Int.\ Cal.\ +Art.~30) +\[ +1 + \frac{r^2}{2^2} + \frac{r^4}{2^2.4^2} + \frac{r^6}{2^2.4^2.6^2} + \cdots, +\] +where $r$ is the modulus of $x$, is convergent for all values of $r$. For the ratio +of the $n + 1$st term of this series to the $n$th term is $\dfrac{r^2}{4n^2}$ and approaches +zero as its limit as $n$ is indefinitely increased, no matter what the value of $r$. +Therefore $J_0(x)$ is \textit{absolutely convergent}. + +$J_0(x)$ is a new and important form. It is called a \textit{Bessel's Function} of the +zeroth order, or a \textit{Cylindrical Harmonic}. + +Equation (8) was obtained from (7) by the substitution of $x=\mu r$, therefore +\[ +R = J_0(\mu r) = 1 - \frac{(\mu r)^2}{2^2} + \frac{(\mu r)^4}{2^2.4^2} - \frac{(\mu r)^6}{2^2.4^2.6^2} + \cdots +\] +is a solution of (7), no matter what the value of $\mu$, and $z=J_0(\mu r)\cos\mu ct$ +or $z=J_0(\mu r)\sin\mu ct$ is a solution of (4). + +$z=J_0(\mu r)\cos\mu ct$ satisfies condition~(2) whatever the value of $\mu$. In +order that it should also satisfy condition~(3) $\mu$ must be so taken that +\[ +J_0(\mu a)=0; \tag{11} +\] +that is, $\mu$ must be a root of (11) regarded as an equation in $\mu$. + +It can be shown that $J_0(x)=0$ has an infinite number of real positive +roots, any one of which can be obtained to any required degree of approximation +without serious difficulty. Let $x_1$, $x_2$, $x_3$, $\cdots$ be these roots. Then if +\begin{gather*} +\frac{x_1}{a}=\mu_1, \quad \frac{x_2}{a}=\mu_2, \quad \frac{x_3}{a}=\mu_3, \quad \text{\&c.\,}\\ +z=A_1J_0(\mu_1r)\cos\mu_1ct + A_2J_0(\mu_2r)\cos\mu_2ct + A_3J_0(\mu_3r)\cos\mu_3ct + \cdots, \tag{12} +\end{gather*} +where $A_1$, $A_2$, $A_3$, \&c., are any constants, is a solution of (4) which satisfies +conditions (2) and (3). + +When $t=0$ (12) reduces to +\[ +z=A_1J_0(\mu_1 r) + A_2J_0(\mu_2 r) + A_3J_0(\mu_3 r) + \cdots. \tag{13} +\] +If then $f(r)$ can be expressed as a series of the form just given, the solution +of our problem can be obtained by substituting the coefficients of that series +for $A_1$, $A_2$, $A_3$, \&c., in (12). +% -----File: 025.png + +\Example{} + +The temperature of a long cylinder is at first unity throughout. The +convex surface is then kept at the constant temperature zero. Show +that the temperature of any point in the cylinder at the expiration +of the time $t$ is +\[ +u=A_{1}e^{-a^2 \mu_1^2t}J_{0}(\mu_1r) + + A_{2}e^{-a^2 \mu_2^2t}J_{0}(\mu_2r) + + A_{3}e^{-a^2 \mu_3^2t}J_{0}(\mu_3r) + \cdots +\] +where $\mu_{1}$, $\mu_{2}$, \&c., are the roots of +$J_{0}(\mu c) = 0$, and where +\[ +1=A_{1}J_{0}(\mu_{1}r) + A_{2}J_{0}(\mu_{2}r) +A_{3}J_{0}(\mu_{3}r) ++ \cdots, +\] +$c$ being the radius of the cylinder. + +\mypara{12.} Each of the five problems which we have taken up forces upon us +the consideration of the development of a given function in terms of +some \emph{normal form}, and in two of them the normal form suggested is an +unfamiliar function. It is clear, then, that a complete treatment of +our subject will require the investigation of the properties and +relations of certain new and important functions, as well as the +consideration of methods of developing in terms of them. + +\mypara{13.} In each of the problems just taken up we have to deal with a +homogeneous linear partial differential equation involving two +independent variables, and we are content if we can obtain +particular solutions. In each case the assumption made in the last +problem, that there exists a solution of the equation in which the +dependent variable is the product of two factors each of which +involves but one of the independent variables, will reduce the +question to solving two ordinary differential equations which can be +treated separately. + +If these equations are familiar ones their solutions can be written +down at once; if unfamiliar, the device used in problems 3 and 5 is +often serviceable, namely, that of assuming that the dependent +variable can be expressed as a sum or series of terms involving +whole powers of the independent variable, and then determining the +coefficients. + +Let us consider again the equations used in the first, second and +third problems. +\begin{flalign} +&\text{\indent(\textit{a})}&& D_x^2 u + D_y^2 u = 0 && \quad \tag{1} +\end{flalign} +Assume $u = X.Y$ where $X$ involves $x$ but not $y$, and $Y$ involves $y$ but +not $x$. +\begin{flalign*} +&\text{Substitute in (1), }&& YD_x^2 X + XD_y^2 Y = 0, &&\phantom{Substitute in (1), } +\end{flalign*} +or, since +we are now dealing with functions of a single variable, +\begin{flalign} +&& \frac{1}{X}\frac{d^2X}{dx^2} &+ \frac{1}{Y}\frac{d^2Y}{dy^2}=0, &&\notag\\ +&\text{or}& \frac{1}{Y}\frac{d^2Y}{dy^2} &= -\frac{1}{X}\frac{d^2X}{dx^2}. \tag{2} +\end{flalign} +% -----File: 026.png + +Since the first member of (2) does not contain $x$, and the second member +does not contain $y$, and the two members must be identically equal, neither of +them can contain either $x$ or $y$, and each must be equal to a constant, say $\alpha^2$. +\begin{flalign*} +&\text{Then}& \frac{d^2Y}{dy^2} - \alpha^2 Y& = 0 &\tag{3}\\ +&\text{and}& \frac{d^2X}{dx^2} + \alpha^2 X &= 0; &\tag{4} +\end{flalign*} +and if (3) and (4) can be solved, we can solve (1). They have for their complete +solutions +\begin{flalign*} +& & Y &= Ae^{\alpha y} + Be^{-\alpha y}\\ +&\text{and}& X &= C \sin {\alpha x} + D \cos {\alpha x}. \quad\text{ (v.\ Int.\ Cal.\ p.~319, \S~21.)} +\end{flalign*} +Hence $Y = e^{\alpha y}$ and $Y=e^{-\alpha y}$ are particular solutions of (3), $X=\sin {\alpha x}$ +and $X = \cos {\alpha x}$ are particular solutions of (1), and consequently +\[ +u =e^{\alpha y} \sin{\alpha x},\quad u = e^{\alpha y} \cos {\alpha x}, \quad u = e^{-\alpha y} \sin {\alpha x}, \text{ and }u =e^{-\alpha y} \cos {\alpha x} +\] +are particular solutions of (1). These agree with the results of Art.~7. +\begin{flalign*} +&\text{\indent(\textit{b})} & D_t^2 y &= a^2 D_x^2 y &\quad\tag{1} +\end{flalign*} + +Assume $y = T.X$ where $T$ is a function of $t$ only and $X$ a function of $x$ +only; substitute in (1) and divide by $a^2 TX$. We get +\begin{flalign*} +&\phantom{(2)}&& \frac{1}{a^2 T}\frac{d^2T}{dt^2} = \frac{1}{X}\frac{d^2X}{dx^2}; \tag{2}\\ +\intertext{hence as in the last case $\dfrac{1}{X}\dfrac{d^2X}{dx^2}$ is a constant; call it $- \alpha^2$, and (2) breaks up into} +&& &\frac{d^2X}{dx^2} + \alpha^2 X = 0 &\tag{3}\\ +&\phantom{(4)}& &\frac{d^2T}{dt^2} + \alpha^2a^2T = 0.&\tag{4} +\end{flalign*} +The complete solutions of (3) and (4) are +\begin{flalign*} +&& &X = A \sin {\alpha x} + B \cos {\alpha x} \\ +&\text{and }& &T = C \sin {\alpha at} + D\cos {\alpha at}, \quad\text{ (v.\ Int.\ Cal.\ p.~319, \S~21).} +\end{flalign*} +\[ + y = \sin {\alpha x}\cos {\alpha at}, \ + y = \sin {\alpha x}\sin {\alpha at}, \ + y = \cos {\alpha x}\cos {\alpha at}, \ + y = \cos {\alpha x}\sin {\alpha at} +\] +are particular solutions of (1), and agree with the results of Art.~8. +\begin{flalign*} +&\text{\indent (\textit{c})}& rD_r^2(rV) + \dfrac{1}{\sin{\theta}} D_\theta (\sin{\theta} D_{\theta} V)& = 0. &\quad\tag{1} +\end{flalign*} + +Assume $V= R.\Theta$ where $R$ involves $r$ alone, and $\Theta$ involves $\theta$ alone; substitute +in (1), divide by $R.\Theta$, and transpose; we get +\[ +\frac{r}{R} \frac{d^2(rR)}{dr^2} = - \frac{1}{\Theta \sin \theta}\frac{d\Big(\sin \theta \dfrac{d\Theta}{ d\theta}\Big)}{d\theta}. \tag{2} +\] +% -----File: 027.png +Since by the reasoning used in (\textit{a}) and (\textit{b}) each member of (2) must be a constant, +say $\alpha^2$, we have +\begin{flalign*} +&& r\frac{d^2(rR)}{dr^2} ={}& \alpha^2 R \tag{3} \\ +&\text{and} &\frac{1}{\sin\theta}\frac{d\Big(\sin\theta \dfrac{d\Theta}{d\theta}\Big)}{d\theta} &+ \alpha^2\Theta = 0. &&\phantom{and}\tag{4} +\end{flalign*} +(3) can be expanded into +\[ + r^2 \frac{d^2 R}{dr^2} + 2r \frac{dR}{dr} - \alpha^2 R = 0. \tag{5} +\] +(5) can be solved (v.\ Int.\ Cal.\ p.~321, \S~23), and has for its complete solution +\[ + R = Ar^m + Br^n, +\] +where \hfill $m = -\frac{1}{2} + \sqrt{\alpha^2 + \frac{1}{4}}$ \quad and \quad $n = -\frac{1}{2} - \sqrt{\alpha^2 + \frac{1}{4}}$. \hfill\phantom{where }\\[1ex] +Hence $n = - m - 1$, and $\alpha^2$ may be written $m(m + 1)$, $m$ being wholly +arbitrary; and +\begin{gather*} + R = Ar^m + Br^{-m-1}. +\\ + R = r^m,\quad \text{ and }\quad R = \frac{1}{r^{m + 1}} +\end{gather*} +are, then, particular solutions of +\[ + r^2 \frac{d^2R}{dr^2} + 2r\frac{dR}{dr} - m(m + 1) R = 0. \tag{6} +\] +With the new value of $\alpha^2$ (4) becomes +\[ + \frac{1}{\sin\theta}\frac{d\Big(\sin\theta \dfrac{d\Theta}{d\theta}\Big)}{d\theta} + m(m + 1) \Theta = 0. \tag{7} +\] +which has been treated in Art.~9 for the case where $m$ is a positive integer, +and the particular solution $\Theta = P_m (\cos\theta)$ has been obtained. +\begin{flalign*} +&\indent\text{Hence }& + V &= r^m P_m (\cos \theta) &&\phantom{\indent Hence }\\ +&\text{and }& + V &= \frac{1}{r^{m + 1}} P_m (\cos \theta), && +\end{flalign*} +$m$ being a positive integer, are particular solutions of (1). The first of these +was obtained in Art.~9, but the second is new and exceedingly important. + +\mypara{14.} The method of obtaining a particular solution of an ordinary linear +differential equation, which we have used in Articles 9 and 11, is of very +extensive application, and often leads to the general solution of the equation +in question. +% -----File: 028.png + +As a very simple example, let us take the equation Art.~13 (\emph{a}) (4), which +we shall write +\[ +\frac{d^2z}{dx^2}+\alpha^2z = 0. \tag{1} +\] + +Assume that there is a solution which can be expressed in terms of powers +of $x$; that is, let $z = \sum{a_nx^n}$, where the coefficients are to be determined. +Substitute this value for $z$ in (1) and we get +\[ +\textstyle \sum{[n(n - 1)a_nx^{n-2} + \alpha^2a_nx^n]} = 0. +\] +Since this equation must be true from its form, without reference to the value +of $x$, that is, since it must be an identical equation, the coefficient of each +power of $x$ must equal zero, and we have +\begin{flalign*} +&& (n + 1)(n + 2)a_{n+2} + \alpha^2a_n = 0; & &&\phantom{whence }\\ +&\text{whence }& a_n = -\frac{(n+1)(n+2)}{\alpha^2}a_{n + 2}& && +\end{flalign*} +is the only relation that need hold between the coefficients in order that +$z = \sum{a_nx^n}$ should be a solution of (1). + +If $n + 2 = 0$ or $n + 1 = 0$, $a_n$ will be zero and $a_{n-2}$, $a_{n-4}$, \&c., will be +zero. In the first case the series will begin with $a_0$, in the second with $a_1$. +\[ +a_{n+2} = - \frac{\alpha^2}{(n+1)(n+2)}a_n. +\] +If we begin with $a_0$ we have +\begin{align*} + a_2 &= -\frac{\alpha^2}{2!}a_0, & + a_4 &= \frac{\alpha^4}{4!}a_0, & + a_6 &= -\frac{\alpha^6}{6!}a_0, \text{ \&c}., \:\cdots\\[-5ex] +\end{align*} +\begin{flalign*} +&\text{and }& + z = a_0 \left( 1 + - \frac{\alpha^2x^2}{2!} \right.& + + \frac{\alpha^4x^4 }{4!} - \left. + \frac{\alpha^6x^6}{6!} + \cdots \right) &&\tag{2} +\\ +&\text{or }& z = {}& a_0 \cos\alpha x &&\tag{3} +\end{flalign*} +is a particular solution of (1). + +If we begin with $a_1$ we have +\begin{align*} + a_3 &= -\frac{\alpha^2}{3!}a_1, & + a_5 &= \frac{\alpha^4}{5!}a_1, & + a_7 &= -\frac{\alpha^6}{7!}a_1, \text{ \&c}., \:\cdots\\[-5ex] +\end{align*} +\begin{flalign*} +&\text{and }& + z &= a_1 \left( x + - \frac{\alpha^2x^3}{3!} + + \frac{\alpha^4x^5}{5!} + - \frac{\alpha^6x^7}{7!} + \cdots \right) &&\tag{4} +\end{flalign*} +% -----File: 029.png +is a solution of (1); $a_1$ can be taken at pleasure. Let $a_1 = \alpha$, (4) becomes +\begin{flalign*} +&&z &= \alpha x - \frac{\alpha^3x^3}{3!} + \frac{\alpha^5x^5}{5!} - \frac{\alpha^7x^7}{7!} + \cdots &&\phantom{or}\\ +&\text{or} & z &= \sin\alpha x& +\end{flalign*} +which, then, is a particular solution of (1). +\[ +z = A\sin\alpha x + B\cos\alpha x \tag{5} +\] +is, then, a solution of (1), and since it contains two arbitrary constants it is +the general solution. + +\mypara{15.} As another example we will take the equation +\[ +x^2\frac{d^2z}{dx^2} + 2x\frac{dz}{dx} - m(m+1)z = 0, \tag{1} +\] +which is in effect equation (6), Art.~13~(\emph{c}), and let $m$ be a positive integer. + +Assume $z = \sum a_nx^n$ and substitute in (1). We get +\[ + \textstyle\sum [n(n+1) - m(m+1)]a_nx^n = 0. +\] +This is an identical equation, therefore +\[ +[n(n+1) - m(m+1)]a_n = 0. +\] +Hence $a_n=0$ for all values of $n$ except those which make +\[ +n(n+1) - m(m+1) = 0, +\] +that is, for all values of $n$ except $n=m$ and $n=-m-1$. Then +\[ +z = Ax^m + Bx^{-m-1} \tag{2} +\] +is the general solution of (1) and +\[ +z=x^m \quad \text{and} \quad z=\frac{1}{x^{m+1}} +\] +are particular solutions. If $m$ is not a positive integer this method will not +lead to a result, and we are driven back to that employed in Art.~13~(\emph{c}). + +\mypara{16.} Let us now take the equation +\[ +\frac{d}{dx}\left [(1-x^2)\frac{dz}{dx}\right ] + m(m+1)z = 0 \tag{1} +\] +which is in effect equation (4), Art.~9, and is known as \emph{Legendre's Equation}. +(1) may be written +\[ +(1-x^2) \frac{d^2z}{dx^2}-2x\frac{dz}{dx} + m(m+1)z = 0. \tag{2} +\] +% -----File: 030.png + +Assume $z=\sum a_nx^n$ and substitute in (2). We get +\[ + \textstyle\sum \{n(n-1)a_nx^{n-2} + [m(m+1) - n(n+1)]a_nx^n\} = 0. +\] +\begin{flalign*} +&\text{Hence } & (&n+1)(n+2)a_{n+2} + [m(m+1)-n(n+1)]a_n = 0, &&\phantom{Hence } +\end{flalign*} +\begin{flalign*} +&\text{or} & a_n &= -\frac{(n+1)(n+2)}{m(m+1)-n(n+1)}a_{n+2}. && \tag{3} +\intertext{If $a_n=0$, then $a_{n-2}=0$, $a_{n-4}=0$, \&c.; but $a_n=0$ if $n=-2$ or $n=-1$. +For the first case we have the sequence of coefficients} +&&a_2&=-\frac{m(m+1)}{2!}a_0\\ +&&a_4&= \frac{m(m-2)(m+1)(m+3)}{4!}a_0\\ +&&a_6&=-\frac{m(m-2)(m-4)(m+1)(m+3)(m+5)}{6!}a_0, \quad \text{\&c.} +\end{flalign*} +Let us take $a_0$, which is arbitrary, as 1. Then $z=p_m(x)$ where +\[ + p_m(x) = \left[ 1 - \frac{m(m+1)}{2!}x^2 + \frac{m(m-2)(m+1)(m+3)}{4!}x^4 - \cdots \right] \tag{4} +\] +is a solution of Legendre's Equation if $p_m(x)$ is a finite sum or a convergent +series. + +For the second case we have the sequence of coefficients +\begin{align*} +&a_3=-\frac{(m-1)(m+2)}{3!}a_1\\ +&a_5= \frac{(m-1)(m-3)(m+2)(m+4)}{5!}a_1\\ +&a_7=-\frac{(m-1)(m-3)(m-5)(m+2)(m+4)(m+6)}{7!}a_1, && \text{\&c.} +\end{align*} +Let us take $a_1$, which is arbitrary, as 1. Then $z=q_m(x)$ where +\[ +q_m(x) = \left[ x - \frac{(m-1)(m+2)}{3!}x^3 + \frac{(m-1)(m-3)(m+2)(m+4)}{5!}x^5 - \cdots \!\right]\!\text{(5)} +\] +is a solution of Legendre's Equation if $q_m(x)$ is a finite sum or a convergent +series. +% -----File: 031.png + +If $m$ is a positive even whole number, $p_m(x)$ will terminate with the term +containing $x^m$, and is easily seen to be identical with +\[ + (-1)^\frac{m}{2} + \frac{ 2^m\Big [\Gamma \Big(\dfrac{m}{2}+1\Big)\Big]^2 } + { \Gamma(m+1) } P_m(x). \tag*{[v.\ Art.~9~(9)]} +\] +For all other values of $m$, $p_m(x)$ is a series. + +The ratio of the $(n+1)$st term of $p_m(x)$ to the $n$th, when $m$ is not a positive +even integer, is +\[ +\frac{(2n-2-m)(2n-1+m)}{(2n-1)(2n)}x^2. +\] +Its limiting value, as $n$ is increased, is $x^2$, and the series is therefore convergent +if $-1 < x < 1$. It is divergent for all other values of $x$. + +If $m$ is a positive odd whole number $q_m(x)$ will terminate with the term +containing $x^m$, and is easily seen to be identical with +\[ + (-1)^\frac{m-1}{2} + \frac{ 2^{m-1} \Big[\Gamma\Big(\dfrac{m+1}{2} \Big) \Big]^2 } + { \Gamma(m+1)} P_m(x). +\] +For all other values of $m$, $q_m(x)$ is a series, and can be shown to be convergent +if $-1 < x < 1$, and divergent for all other values of $x$. +\[ +z=Ap_m(x)+Bq_m(x) \tag{6} +\] +is the general solution of Legendre's Equation if $-1 < x < 1$, no matter +what the value of $m$. From Art.~13~(\emph{c}) it follows that +\[ +\left. +\begin{aligned} + V &= r^mp_m(\cos\theta)\\ + V &= \frac{1}{r^{m+1}}p_m(\cos\theta)\quad\\ + V &= r^mq_m(\cos\theta)\\ + V &= \frac{1}{r^{m+1}}q_m(\cos\theta)\\ +\end{aligned} +\right\} \tag{7} +\] +are particular solutions of +\[ +rD_r^2(rV) + \frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V)=0, +\] +no matter what the value of $m$, provided $\cos\theta$ is neither one nor minus one. + +In the work we shall have to do with Laplace's and Legendre's Equations, +it is generally possible to restrict $m$ to being a positive integer, and hereafter +we shall usually confine our attention to that case. +% -----File: 032.png + +With this understanding let us return to (3), which may be rewritten +\[ +a_{n+2}=-\frac{(m-n)(m+n+1)}{(n+1)(n+2)}a_n.\\[-3ex] +\] +\begin{flalign*} +&\text{If} && \text{$a_{n+2}=0$, then $a_{n+4}=0$,\; $a_{n+6} = 0$, \&c.;} &&\qquad \\[1ex] +&\text{but} && \text{$a_{n+2}=0$\phantom{, }if\; $n=m$,\; or\; $n=-m-1$.} && +\end{flalign*} +If in (3) we begin with $n=m-2$, we get the sequence of coefficients already +obtained in Art.~9, and we have $z=P_m(x)$, where +\begin{align*} + P_m(x) &= \frac{(2m-1)(2m-3)\cdots 1}{m!} + \left[ x^m - \frac{m(m-1)}{2(2m-1)} x^{m-2} \right. +\\[1ex] +&+ \frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)} + x^{m-4} +\\[1ex] +&- \left.\frac{m(m-1)(m-2)(m-3)(m-4)(m-5)} + {2.4.6.(2m-1)(2m-3)(2m-5)} + x^{m-6} + \cdots \right], \tag{8} +\end{align*} +as a particular solution of Legendre's Equation. + +If, however, we begin with $n=-m-3$, we have +\begin{align*} +a_{-m-3} &= \frac{(m+1)(m+2)}{2(2m+3)}a_{-m-1} \\[1ex] +a_{-m-5} &= \frac{(m+1)(m+2)(m+3)(m+4)} + {2.4.(2m+3)(2m+5)}a_{-m-1} +\\[1ex] +a_{-m-7} +&= \frac{(m+1)(m+2)(m+3)(m+4)(m+5)(m+6)} + {2.4.6.(2m+3)(2m+5)(2m+7)} a_{-m-1}, \quad\text{\&c.} +\end{align*} +$a_{-m-1}$ may be taken at pleasure, and is usually taken as $\dfrac{m!}{1.3.5.\cdots(2m+1)}$, +and $z=Q_m(x)$ where +\begin{align*} +Q_m(x) &= \frac{m!}{(2m+1)(2m-1)\cdots1} + \left[ \frac{1}{x^{m+1}} + + \frac{(m+1)(m+2)}{2.(2m+3)} \frac{1}{x^{m+3}} \right. +\\[1ex] +&+ \left. \frac{(m+1)(m+2)(m+3)(m+4)} + {2.4.(2m+3)(2m+5)} + \frac{1}{x^{m+5}} + \cdots \right] \tag{9} +\end{align*} +is a second particular solution of Legendre's Equation, provided the series is +convergent. $Q_m(x)$ is called a \emph{Surface Zonal Harmonic} of the \emph{second kind}. +% -----File: 033.png +It is easily seen to be convergent if $x < - 1$ or $x > 1$, and divergent if +$- 1 < x < 1$. + +Hence if $m$ is a positive integer, +\[ +z = A P_m(x) + B Q_m(x) \tag{10} +\] +is the general solution of Legendre's Equation if $x < - 1$ or $x > 1$. + +We have seen that for $- 1 < x < 1$ +\begin{align*} + P_m(x) +&= (-1)^\frac{m}{2} + \frac{ \Gamma (m + 1) } + { 2^m\left[\Gamma\left( \dfrac{m}{2} + 1 \right) \right]^2 } + p_m(x) \tag{11} +\intertext{if $m$ is an even integer, and } + P_m(x) +&= (- 1)^\frac{m-1}{2} + \frac{ \Gamma (m + 1) } + { 2^{m-1}\Big[\Gamma\Big( \dfrac{m+1}{2} \Big) \Big]^2 } + q_m(x) \tag{12} +\end{align*} +if $m$ is an odd integer. + +If now we define $Q_m(x)$ as follows when $- 1 < x < 1$ +\begin{align*} + Q_m(x) +&= (-1)^\frac{m+1}{2} + \frac{ 2^{m-1} \Big[\Gamma\Big( \dfrac{m+1}{ 2} \Big)\Big]^2 } + { \Gamma(m+1) } p_m (x) \tag{13} +\intertext{if $m$ is an odd integer, and } + Q_m(x) +&= (-1)^\frac{m}{2} + \frac{ 2^m \Big[\Gamma\Big( \dfrac{m}{2}+1 \Big) \Big]^2 } + { \Gamma(m+1) } q_m(x) \tag{14} +\end{align*} +if $m$ is an even integer, then (10) will be the general solution of Legendre's +Equation if $m$ is a positive integer when $-1 < x < 1$, as well as when $x < -1$ +or $x > 1$. + +\mypara{17.} Let us last consider the equation +\[ +\frac{d^2z}{dx^2} + \frac{1}{x} \frac{dz}{dx} + \left (1 - \frac{m^2}{ x^2} \right ) z = 0 \tag{1} +\] +which is known as Bessel's Equation, and which reduces to (8) Art.~11, +that is, to +\[ +\frac{d^2 z}{ dx^2} + \frac{1}{x} \frac{dz}{dx} + z = 0 +\] +when $m$ = 0;\footnote + {This equation was first studied by Fourier in considering the cooling of a cylinder. We + shall designate it as ``Fourier's Equation.''} +(1) can be simplified by a change of the dependent variable. +% -----File: 034.png +Let $z = x^m v$ and we get +\[ +\frac{d^2 v}{ dx^2} + \frac{2m +1 }{ x}\frac{ dv}{dx} + v = 0 \tag{2} +\] +to determine $v$. + +Assume $v = \sum a_n x^n$, and substitute in (2). We get +\[ + \textstyle\sum [n(2m + n)a_n x^{n-2} + a_n x^n] = 0; +\] +\begin{flalign*} +&\text{whence }& &a_{n-2} = -n(2m + n)a_n. &&\phantom{whence } +\end{flalign*} +If we begin with $n = 0$, then $a_{n-2} = 0$, $a_{n-4} = 0$, \&c., and we have the +set of values +\begin{align*} +a_2 &= - \frac{a_0}{ 2(2m+2)} = - \frac{a_0}{ 2^2 (m+1)}\\[1ex] +a_4 &= \frac{a_0}{ 2.4(2m + 2)(2m + 4)} + = \frac{a_0}{ 2^4.2!(m + 1)(m + 2)} +\end{align*}\\[-3ex] +\[ +a_6 = - \frac{a_0} {2.4.6(2m + 2)(2m + 4)(2m + 6)} + = - \frac{a_0}{ 2^6.3!(m + 1)(m + 2)(m + 3)}; +\]\\[-2ex] +\begin{flalign*} +&\text{whence }& z = a_0 x^m \left[ 1 \vphantom{\dfrac{x^2}{(0)}}\right. +&- \frac{x^2}{ 2^2 (m+1)} + \frac{x^4}{ 2^4.2!(m + 1)(m + 2)} &&\\[1ex] +&& &- \left. \frac{x^6}{ 2^6.3!(m + 1)(m + 2)(m + 3)} + \cdots \right ] && \tag{3} +\end{flalign*} +is a solution of Bessel's Equation. $a_0$ is usually taken as $\dfrac{1}{2^m m!}$ if $m$ is a positive +integer, or as $\dfrac{1}{ 2^m \Gamma (m + 1)}$ if $m$ is unrestricted in value, and the second +member of (3) is represented by $J_m(x)$ and is called a \textit{Bessel's Function} of the +$m$th order, or a \textit{Cylindrical Harmonic} of the $m$th order. + +If $m = 0$, $J_m(x)$ becomes $J_0(x)$ and is the value of $z$ obtained in Art.~11 +as the solution of equation~(8) of that article. + +If in equation (1) we substitute $x^{-m }v$ in place of $x^m v$ for $z$, we get in place +of (2) the equation +\[ +\frac{d^2 v}{ dx^2} + \frac{1-2m}{x} \frac{dv}{dx} + v = 0 +\] +and in place of (3) +\begin{align*} +z = a_0 x^{-m} \left[ 1 \vphantom{\dfrac{x^2}{(0)}}\right. +&- \frac{x^2}{ 2^2 (1-m)} + \frac{x^4}{ 2^4.2!(1-m)(2-m)} \\[1ex] +&- \left. \frac{x^6}{ 2^6.3!(1 - m)(2 - m)(3 - m)} + \cdots \right ] \tag{4} +\end{align*} +% -----File: 035.png + +If $a_0$ is taken equal to $\dfrac{1}{2^{-m} \Gamma (1 - m)}$ the second member of (4) is the same +function of $-m$ and $x$ that $J_m (x)$ is of $+m$ and $x$ and may be written +$J_{-m}(x)$. +\begin{flalign*} +&\text{\indent Therefore }& + z &= A J_m(x) + B J_{-m}(x) &&\phantom{Therefore } \tag{5} +\end{flalign*} +is the general solution of (1) unless $J_m(x)$ and $J_{-m}(x)$ should prove not to be +independent. + +It is easily seen that when $m = 0$, $J_{-m}(x)$ and $J_m(x)$ become identical +and (5) reduces to +\[ +z = (A + B) J_0(x) +\] +and contains but a single arbitrary constant and is not the general solution of +Fourier's Equation~(8) Art.~(11). + +It can be shown that $J_{-m} (x) = (-1)^m J_m(x)$ whenever $m$ is an integer, +and consequently that the solution~(5) is general only when $m$ if real is fractional +or incommensurable. + +The general solution for the important case where $m = 0$ is, however, easily +obtained. Let $F(m,x)$ be the value which the second member of (3) assumes +when $a_0 = 1$; then the value which the second member of (4) assumes +when $a_0 = 1$ will be $F(-m,x)$, and it has been shown that $z = F(m,x)$ and +$z = F(-m,x)$ are solutions of Bessel's Equation; $z = F(m,x) - F(-m,x)$ +is, then, a solution, as is also +\[ + z = \frac{F(m,x) - F(-m,x)}{ 2m}, \tag{6} +\] +but the limiting value which $\dfrac{F(m,x) - F(-m,x)}{ 2m}$ approaches as $m$ approaches +zero is $[D_m F(m,x)]_{m=0}$ and consequently +\[ +z=[D_m F(m,x)]_{m=0} \tag{7} +\] +is a solution of the equation +\[ + \frac{d^2 z}{dx^2} + \frac{1}{x} \frac{dz}{dx} + z = 0, \tag{8} +\] +and the general solution of (8) is +\begin{align*} + z = {}& A J_0(x) + B[D_m F(m,x)]_{m=0}. +\\[1ex] + F(m,x) = x^m \bigg[ 1 +&- \frac{x^2}{ 2^2 (m + 1)} + + \frac{x^4}{ 2^4.2! (m + 1)(m + 2)} +\\[1ex] +&- \frac{x^6}{2^6.3!(m+1)(m+2)(m+3)} + \cdots \bigg]\\ +% -----File: 036.png +D_mF(m,x) = x^m\log x \bigg[ 1&-\frac{x^2}{2^2(m+1)} + \frac{x^4}{2^4.2!(m+1)(m+2)} - \cdots \bigg] & \\ + +x^mD_m \bigg[ 1 &- \frac{x^2}{2^2(m+1)} + \frac{x^4}{2^4.2!(m+1)(m+2)} + \cdots \bigg].& +\end{align*} +The general term of the last parenthesis can be written +\[ +(-1)^k\frac{x^{2k}}{2^{2k}.k!(m+1)(m+2)\cdots(m+k)}, +\] +and its partial derivative with respect to $m$ is +\begin{gather*} +(-1)^k\frac{x^{2k}}{2^{2k}.k!} D_m\frac{1}{(m+1)(m+2)\cdots(m+k)}.\\ +\log\frac{1}{(m+1)(m+2)\cdots(m+k)} = +\begin{aligned}[t] + -[ &\log(m+1)+\log(m+2)+\cdots\\ + &+ \log(m+k)]. +\end{aligned} +\end{gather*} + +Take the $D_m$ of both members and we have +\begin{align*} +D_m & \frac{1}{(m+1)(m+2)\cdots(m+k)}\\ + &=-\frac{1}{(m+1)(m+2)\cdots(m+k)} + \left[ \frac{1}{m+1} + \frac{1}{m+2} + \cdots\frac{1}{m+k} \right]. +\\ +D_m\left[1\vphantom{\frac{x^2}{(0)}}\right. & -\frac{x^2}{2^2(m+1)} + \frac{x^4}{2^4.2!(m+1)(m+2)} - \frac{x^6}{2^6.3!(m+1)(m+2)(m+3)} +\\ +& +\cdots\left.\vphantom{\frac{x^2}{(0)}}\right] = \frac{x^2}{2^2}\frac{1}{(m+1)^2} - \frac{x^4}{2^4.2!}\frac{1}{(m+1)(m+2)} \left[ \frac{1}{m+1} + \frac{1}{m+2} \right] +\\ +& +\frac{x^6}{2^6.3!}\frac{1}{(m+1)(m+2)(m+3)} \left[ \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{m+3}\right] + \cdots +\end{align*} +and we have +% recast to fit line +\begin{align*} +[D_mF(m,x)]_{m=0} &= J_0(x)\log x + \frac{x^2}{2^2(1!)^2}\frac{1}{1} - \frac{x^4}{2^4(2!)^2} \left(\frac{1}{1} + \frac{1}{2}\right) +\\ &+ \frac{x^6}{2^6(3!)^2} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \right) +-\frac{x^8}{2^8(4!)^2} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) + \cdots; +\end{align*} +\begin{flalign*} +&\text{and} && \hspace{6em} z = AJ_0(x) + BK_0(x), \tag{9} +\\ +% recast to fit line +&\text{where} & K_0&(x) = J_0(x)\log x + \frac{x^2}{2^2} - \frac{x^4}{2^4(2!)^2} \left( \frac{1}{1}+\frac{1}{2}\right) +\\ +&&& + \frac{x^6}{2^6(3!)^2}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\right) +-\frac{x^8}{2^8(4!)^2} \left( \frac{1}{1} + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} \right) + \cdots \tag{10} +\end{flalign*} +is the general solution of Fourier's Equation (8). + +$K_0(x)$ is known as a \emph{Bessel's Function} of the \emph{Second Kind}. +% -----File: 037.png + +\mypara{18.} It is worth while to confirm the results of the last few articles by +getting the general solutions of the equations in question by a different and +familiar method. + +The general solution of any ordinary linear differential equation of the +second order can be obtained when a particular solution of the equation has +been found [v.\ Int.\ Cal.\ p.~321, §~24~(\emph{a})]. + +The most general form of a homogeneous ordinary linear differential equation +of the second order is +\[ +\frac{d^2y}{dx^2} + P \frac{dy}{dx} + Qy = 0 \tag{1} +\] +where $P$ and $Q$ are functions of $x$. Suppose that +\[ +y = v \tag{2} +\] +is a particular solution of (1). Substitute $y = vz$ in (1) and we get +\[ +v\frac{d^2z}{dx^2}+\left(2\frac{dv}{dx} + Pv\right)\frac{dz}{dx} = 0. \tag{3} +\] +Call $\dfrac{dz}{dx} = z'$. Then (3) becomes +\[ +v\frac{dz'}{dx}+\left(2\frac{dv}{dx}+ Pv\right)z'=0, \tag{4} +\] +a differential equation of the first order in which the variables can be separated. +Multiply by $dx$ and divide by $vz'$ and (4) reduces to +\[ +\frac{dz'}{z'} + 2\frac{dv}{v} + Pdx = 0. +\] +Integrate and we have +\begin{flalign*} +&&\log{}&z' + \log v^2 + \int Pdx = C +\\ +&\text{or } & z'v^2 &= e^{C-\int\!Pdx}=Be^{-\int\!Pdx}, &&\phantom{or }\\ +&&z'&=\frac{dz}{dx}=B\frac{e^{-\int\!Pdx}}{v^2},&\\ +&&z &=A+B\int\frac{e^{-\int\!Pdx}}{v^2}dx;&\\ +&\text{and } & y=&\ v\bigg(A+B\int \frac{e^{-\int\!Pdx}}{v^2}dx\bigg) && \tag{5} +\end{flalign*} +is the general solution of (1), the only arbitrary constants in the second member +of (5) being those explicitly written, namely, $A$ and $B$.\\ + +(\emph{a})\quad Apply this formula to (1) Art.~14, +\[ +\frac{d^2z}{dx^2} + \alpha^2z=0; \tag{1} +\] +% -----File: 038.png +given: $z=\cos\alpha x$, as a particular solution. Substituting in (5) we have +since $P=0$ +\begin{align*} +z &= \cos\alpha x\left(A+B\int\frac{dx}{\cos^2\alpha x}\right) \\ + &= \cos\alpha x\left(A+\frac{B}{\alpha}\tan\alpha x\right) \\ + &= A\cos\alpha x + B_1\sin\alpha x, \tag{2} +\end{align*} +as the general solution of (1), and this agrees perfectly with (5) Art.~14.\\ + +(\emph{b})\quad Take equation (1) Art.~15. +\[ +x^2\frac{d^2z}{dx^2}+2x\frac{dz}{dx}-m(m+1)z=0; \tag{1} +\] +given: $z=x^m$, as a particular solution. + +\noindent Here $P=\dfrac{2}{x}$, $\displaystyle\int Pdx=2\log x =\log x^2$, and $e^{-\int\!Pdx}=\dfrac{1}{x^2}$. Hence by (5) +\[ +z = x^m \left( A + B\int\frac{dx}{x^{2m+2}} \right) + = x^m \left( A + \frac{B}{-2m-1} \frac{1}{x^{2m+1}} \right),\\[-3ex] +\] +\begin{flalign*} +&\text{that is} & z= &\ Ax^m+\frac{B_1}{x^{m+1}} &&\phantom{that is} \tag{2} +\end{flalign*} +is the general solution of (1), and agrees with (2) Art.~15.\\ + +(\emph{c})\quad Take Legendre's Equation, (2) Art.~16. +\[ +(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+m(m+1)z=0; \tag{1} +\] +given: $z=P_m(x)$, as a particular solution. + +Here $P=\dfrac{-2x}{1-x^2}$, $\displaystyle\int Pdx=\log(1- x^2)$, and $e^{-\int\!Pdx}=\dfrac{1}{1-x^2}$. +\begin{flalign*} +&\text{Hence by (5)} && + z = P_m(x) \left( A + B\int\frac{dx}{(1-x^2)[P_m(x)]^2} \right) &&\phantom{Hence by } \tag{2} +\end{flalign*} +is the general solution of (1) and must agree with (10) Art.~16, if $m$ is an +integer, and therefore +\[ +Q_m(x)=CP_m(x)\int\frac{dx}{(1-x^2)[P_m(x)]^2} \tag{3} +\] +where $C$ is as yet undetermined, and no constant term is to be understood with +the integral in the second member.\\ + +(\emph{d})\quad Take Bessel's Equation, (1) Art.~17. +\[ +\frac{d^2z}{dx^2} + \frac{1}{x}\frac{dz}{dx} + \left(1-\frac{m^2}{x^2}\right)z = 0; \tag{1} +\] +given: $z=J_m(x)$, as a particular solution. +% -----File: 039.png + +Here $P=\dfrac{1}{x}$, $\displaystyle\int Pdx=\log x$, and $e^{-\int\!Pdx}=\dfrac{1}{x}$. Hence by (5) +\[ +z=J_m(x)\left(A+B\int\frac{dx}{x[J_m(x)]^2}\right) \tag{2} +\] +is the general solution of Bessel's Equation. + +If $m=0$ (2) becomes +\[ +z=J_0(x)\left(A+B\int\frac{dx}{x[J_0(x)]^2}\right) \tag{3} +\] +and must agree with (9) Art.~17. Therefore +\[ +K_0(x)=CJ_0(x)\int\frac{dx}{x[J_0(x)]^2}, \tag{4} +\] +where $C$ is at present undetermined, and no constant term is to be taken +with the integral. + +The first considerable subject suggested by the problems which we have +taken up in this introductory chapter is that of development in Trigonometric +Series (v.\ Arts.~7 and 8). +\label{ch1end} +% -----File: 040.png + +\mychap{CHAPTER II.}{DEVELOPMENT IN TRIGONOMETRIC SERIES.} +\label{ch2start} + +\mypara{19.} We have seen in Chapter I. that it is sometimes important to be able +to express a given function of a variable $x$, in terms of the sines or of the +cosines of multiples of $x$. The problem in its general form was first solved +by Fourier in his ``Analytic Theory of Heat'' (1822), and its solution plays a +very important part in most branches of modern Physics. Series involving +only sines and cosines of whole multiples of $x$, that is series of the form +\[ +b_0 + b_1\cos x + b_2\cos 2x + \cdots + a_1\sin x + a_2\sin 2x + \cdots +\] +are generally known as Fourier's series. + +Let us endeavor to develop a given function of $x$ in terms of $\sin x$, $\sin 2x$, +$\sin 3x$, \&c., in such a way that the function and the series shall be equal for +all values of $x$ between $x=0$ and $x=\pi$. + +To fix our ideas let us suppose that we have a curve, +\[ +y=f(x), +\] +given, and that we wish to form the equation, +\[ +y = a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots, +\] +of a curve which shall coincide with so much of the given curve as lies between +the points corresponding to $x=0$ and $x=\pi$. +It is clear that in the equation +\[ +y = a_1 \sin x \tag{1} +\] +$a_1$ may be determined so that the curve represented shall pass through any +given point. For if we substitute in (1) the coördinates of the point in question +we shall have an equation of the first degree in which $a_1$ is the only +unknown quantity and which will therefore give us one and only one value +for $a_1$. + +In like manner the curve +\[ +y = a_1 \sin x + a_2 \sin 2x +\] +may be made to pass through any two arbitrarily chosen points whose abscissas +lie between 0 and $\pi$ provided that the abscissas are not equal; and +\[ +y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots + a_n \sin nx +\] +may be made to pass through any $n$ arbitrarily chosen points whose abscissas +lie between 0 and $\pi$ provided as before that their abscissas are all different. + +If, then, the given function $f(x)$ is of such a character that for each value of $x$ +between $x=0$ and $x=\pi$ it has one and only one value, and if between +$x=0$ and $x=\pi$ it is finite and continuous, or if discontinuous has only +\emph{finite discontinuities} (v.\ Int.\ Cal.\ Art.~83, p.~78), the coefficients in +\[ +y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots + a_n \sin nx \tag{2} +\] +% -----File: 041.png +can be determined so that the curve represented by (2) will pass through any +$n$ arbitrarily chosen points of the curve +\[ +y = f(x) \tag{3} +\] +whose abscissas lie between $0$ and $\pi$ and are all different, and these coefficients +will have but one set of values. + +For the sake of simplicity suppose that the $n$ points are so chosen that their +projections on the axis of $X$ are equidistant. + +\markright{DEVELOPMENT IN TRIGONOMETRIC SERIES.} +Call $\dfrac{\pi}{ n+1} = \Delta x$; then the coördinates of the $n$ points will be $[\Delta x,f(\Delta x)],$ +$[2\Delta x,f(2\Delta x)]$, $[3\Delta x,f(3\Delta x)]$, $\cdots\ [n\Delta x,f(n\Delta x)]$. Substitute them in (2) and +we have +\begin{equation*} +\left.\begin{alignedat}{5} +f( \Delta x) &= {}& a_1 \sin \Delta x &+{}& a_2 \sin 2 \Delta x &+{} +& a_3 \sin 3 \Delta x &+ \cdots +{}& a_n \sin n \Delta x\phantom{,} & +\\ +f(2\Delta x) &= {}& a_1 \sin 2\Delta x &+{}& a_2 \sin 4 \Delta x &+{} +& a_3 \sin 6 \Delta x &+ \cdots +{}& a_n \sin 2n \Delta x\phantom{,} & +\\ +f(3\Delta x) &= {}& a_1 \sin 3\Delta x &+{}& a_2 \sin 6 \Delta x &+{} +& a_3 \sin 9 \Delta x &+ \cdots +{}& a_n \sin 3n \Delta x\phantom{,} & +\\ + \vdots \quad && \vdots \qquad && \vdots \qquad & +& \vdots \qquad && \vdots \qquad & +\\ +f(n\Delta x) &= {}& a_1 \sin n\Delta x &+{}& a_2 \sin 2n \Delta x &+{} +& a_3 \sin 3n \Delta x &+ \cdots +{}& a_n \sin n^2 \Delta x, & +\end{alignedat} +\right\} +\tag{4} +\end{equation*} +$n$ equations of the first degree to determine the $n$ coefficients $a_1$, $a_2$, $a_3$, $\cdots$ $a_n$. + +Not only can equations (4) be solved in theory, but they can be actually +solved in any given case by a very simple and ingenious method due to +Lagrange. + +Let us take as an example the simple problem to determine the coefficients +$a_1$, $a_2$, $a_3$, $a_4$, and $a_5$, so that +\[ +y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + a_4 \sin 4x + a_5 \sin 5x \tag{5} +\] +shall pass through the five points of the line +\[ +y = x +\] +which have the abscissas $\dfrac{\pi}{6}$, $\dfrac{2\pi}{6}$, $\dfrac{3\pi}{6}$, $\dfrac{4\pi}{6}$, and $\dfrac{5\pi}{6}$, $\dfrac{\pi}{6}$ here being $\Delta x$. + +We must now solve the equations +\begin{equation*} +\left.\begin{alignedat}{6} +\frac{\pi}{6}\: &= {}& a_1 \sin \:\frac{ \pi}{6}\: &+{} +& a_2 \sin \:\frac{2\pi}{6}\: &+{}& a_3 \sin \:\frac{3\pi}{6}\: &+{} +& a_4 \sin \:\frac{4\pi}{6}\: &+{}& a_5 \sin \:\frac{5\pi}{6}\:\phantom{.} +\\ +\frac{2\pi}{6} &= & a_1 \sin \frac{2\pi}{6} &+{} +& a_2 \sin \:\frac{4\pi}{6}\: &+{}& a_3 \sin \:\frac{6\pi}{6}\: &+{} +& a_4 \sin \:\frac{8\pi}{6}\: &+{}& a_5 \sin \frac{10\pi}{6}\phantom{.} +\\ +\frac{3\pi}{6} &= & a_1 \sin \frac{3\pi}{6} &+{} +& a_2 \sin \:\frac{ 6\pi}{6}\: &+{}& a_3 \sin \:\frac{ 9\pi}{6}\: &+{} +& a_4 \sin \frac{12\pi}{6} &+{}& a_5 \sin \frac{15\pi}{6}\phantom{.} +\\ +\frac{4\pi}{6} &= {}& a_1 \sin \frac{4\pi}{6} &+{} +& a_2 \sin \:\frac{ 8\pi}{6}\: &+{}& a_3 \sin \frac{12\pi}{6} &+{} +& a_4 \sin \frac{16\pi}{6} &+{}& a_5 \sin \frac{20\pi}{6}\phantom{.} +\\ +\frac{5\pi}{6} &= {}& a_1 \sin \frac{5\pi}{6} &+{} +& a_2 \sin \frac{10\pi}{6} &+{}& a_3 \sin \frac{15\pi}{6} &+{} +& a_4 \sin \frac{20\pi}{6} &+{}& a_5 \sin \frac{25\pi}{6}. +\end{alignedat} +\right\} +\tag{6} +\end{equation*} +% -----File: 042.png + +Multiply the first equation by $2 \sin \dfrac{\pi}{6}$, the second by $2 \sin \dfrac{2 \pi }{6}$, the third +by $2\sin\dfrac{3 \pi}{6}$, the fourth by $2\sin\dfrac{4\pi}{6}$, the fifth by $2 \sin \dfrac{5 \pi }{6}$ and add the +equations. + +The coefficient of $a_2$ is +\begin{flalign*} +&&2 \sin \frac{\pi}{6} \sin \frac{2 \pi}{ 6} &+ 2 \sin \frac{2 \pi}{6} \sin \frac{4 \pi}{6} + 2 \sin \frac{3 \pi }{6} \sin \frac{6 \pi}{6} + 2 \sin \frac{4 \pi}{6} \sin \frac{8 \pi}{6} & \phantom{\text{but}}\\ +&&&+ 2 \sin \frac{5 \pi}{6} \sin \frac{10 \pi}{6}; \\ +&\text{but} && \qquad 2 \sin \dfrac{\pi }{6} \sin \dfrac{2 \pi}{6} = \cos \dfrac{\pi}{6} - \cos \dfrac{3 \pi }{6}, \text{\&c.} +\end{flalign*} + +Hence the coefficient of $a_2$ becomes +\begin{equation*} +\left.\begin{alignedat}{5} +&\cos \:\frac{\pi}{6} &+{}& \cos \frac{2 \pi}{6} &+{}& \cos \frac{3 \pi}{6} &+{}& \cos \:\frac{4 \pi}{6} &+{}& \cos \:\frac{5 \pi}{6}\\ +-{}&\cos \frac{3 \pi}{6} &-{}& \cos \frac{6 \pi}{6} &-{}& \cos \frac{9 \pi}{6} &-{}& \cos \frac{12 \pi}{6} &-{}& \cos \frac{15 \pi}{6} +\end{alignedat} +\right\} +\tag{7} +\end{equation*} +and this may be reduced by the aid of an important Trigonometric formula +which we proceed to establish. + +\mypara{20.} \textsc{Lemma.} +\begin{align*} +\cos \theta &+ \cos 2 \theta + \cos 3 \theta + \cdots + \cos n \theta = - \frac{1}{2} + \frac{1}{2} \cfrac{\sin(2n+1) \dfrac{\theta}{2}}{\sin \dfrac{\theta}{2}}. \tag{1} +\end{align*} +For let $S=\cos \theta + \cos 2 \theta + \cos 3 \theta + \cdots + \cos n \theta$ and multiply by $2 \cos \theta$. +\begin{align*} +2S \cos \theta &= 2 \cos^2 \theta + 2 \cos \theta \cos 2 \theta + 2 \cos \theta \cos 3 \theta + \cdots + 2 \cos \theta \cos n \theta&&\hspace{1em}\\ +&= 1 + \cos \theta + \cos 2 \theta + \cdots + \cos (n - 1) \theta\\ +&\hspace{1.85em}+ \cos 2\theta + \cos 3\theta + \cos 4\theta + \cdots + \cos (n + 1)\theta\\ +&= 2S + 1 + \cos (n + 1) \theta - \cos \theta - \cos n \theta. \qquad \text{Hence}\\[-5.5ex] +\end{align*} +\begin{flalign*} +&& S &= - \frac{1}{2} + \frac{\cos n \theta - \cos (n + 1) \theta }{ 2(1- \cos \theta )} && \\ +&\text{or }& S &= - \frac{1}{2} + \frac{1}{2} \cfrac{\sin (2n + 1) \dfrac{\theta}{2}}{ \sin \dfrac{\theta}{ 2 }}. && \tag*{\scriptsize Q.E.D.} +\end{flalign*} +% -----File: 043.png + +\mypara{21.} Applying (1) Art.~20 to (7) Art.~19 the coefficient of $a_2$ reduces to +\[ + \cfrac{\sin\dfrac{11\pi}{12}}{2\sin\dfrac{ \pi}{12}} +- \cfrac{\sin\dfrac{33\pi}{12}}{2\sin\dfrac{3\pi}{12}}; +\] +\begin{flalign*} +&\text{but} && \frac{11\pi}{12} = \pi - \frac{\pi}{12},\text{ and }\frac{33\pi}{12} = 3\pi - \frac{3\pi}{12}; &&\phantom{\text{but}} +\end{flalign*} +\begin{flalign*} +&\text{therefore} && \frac{\sin \Big( \pi- \dfrac{\pi}{12} \Big) }{2\sin\dfrac{\pi}{12}} +- \frac{\sin \Big(3\pi- \dfrac{3\pi}{12} \Big) }{2\sin\dfrac{3\pi}{12}} = \frac12 - \frac12 = 0, &&\phantom{\text{therefore}} +\end{flalign*} +and $a_2$ vanishes. + +In like manner it may be shown that the coefficients of $a_3$, $a_4$, and $a_5$ +vanish. + +The coefficient of $a_1$ is +\begin{align*} + &\ 2\sin^2 \frac{ \pi}{6} + 2\sin^2 \frac{2\pi}{6} + + 2\sin^2 \frac{3\pi}{6} + 2\sin^2 \frac{4\pi}{6} + + 2\sin^2 \frac{5\pi}{6} +\\[1ex] += {}& \makebox[20.5em]{1 \hfill + \hfill 1 \hfill + \hfill 1 \hfill + \hfill 1 \hfill + \hfill 1} +\\ + &\!\!- \cos \frac{2\pi}{6} - \cos \frac{4\pi}{6} + - \cos \frac{6\pi}{6} - \cos \frac{8\pi}{6} + -\cos \frac{10\pi}{6} +\\[1ex] += {}& 5 + \frac12 + - \cfrac{\sin \dfrac{11\pi}{6}}{2 \sin \cfrac{\pi}{6}} += 5\tfrac12 - \cfrac{ \sin \left( 2\pi-\dfrac{\pi}{6} \right) } + { 2\sin \cfrac{\pi}{6} } = 6. +\end{align*} + +The first member of the final equation is +\begin{gather*} + \frac{2\pi}{6} \sin \frac{ \pi}{6} ++ 2 \frac{2\pi}{6} \sin \frac{2\pi}{6} ++ 2 \frac{3\pi}{6} \sin \frac{3\pi}{6} ++ 2 \frac{4\pi}{6} \sin \frac{4\pi}{6} ++ 2 \frac{5\pi}{6} \sin \frac{5\pi}{6}. \quad\text{Hence}\\ + a_1 = \frac26 \sum_{k=1}^{k=5} \frac{k\pi}{6} \sin \frac{k\pi}{6} += \frac{\pi}{6} (2 + \surd3) = 2\text{\quad approximately.} +\end{gather*} + +If we multiply the first equation of (6) Art.~19 by $2\sin\dfrac{2\pi}{6}$, the second by +$2\sin\dfrac{4\pi}{6}$, the third by $2\sin\dfrac{6\pi}{6}$, the fourth by $2\sin\dfrac{8\pi}{6}$, the fifth +by $2\sin\dfrac{10\pi}{6}$, add and reduce as before we shall find +\begin{align*} + a_2 &= \frac26 \sum_{k=1}^{k=5} \frac{k\pi}{6} \sin \frac{2k\pi}{6} += - \frac{\pi}{6} \surd3 = -0.9; +% -----File: 044.png +\intertext{and in like manner we get} +a_3 &= \frac{2}{6}\sum_{k=1}^{k=5} \frac{k\pi}{6}\sin\frac{3k\pi}{6} += \frac{\pi}{6} = 0.5\\ +a_4 &= \frac{2}{6}\sum_{k=1}^{k=5} \frac{k\pi}{6}\sin\frac{4k\pi}{6} +=-\frac{\pi\surd3}{18} = -0.3\\ +a_5 &= \frac{2}{6}\sum_{k=1}^{k=5} \frac{k\pi}{6}\sin\frac{5k\pi}{6} += \frac{\pi}{6}(2-\surd3) = 0.1. +\end{align*} + +Therefore +\[ +y=2\sin x - 0.9\sin 2x + 0.5\sin 3x - 0.3\sin 4x + 0.1\sin 5x \tag{1} +\] +cuts the curve $y=x$ at the five points whose abscissas are $\dfrac{\pi}{6}$, $\dfrac{2\pi}{6}$, $\dfrac{3\pi}{6}$, +$\dfrac{4\pi}{6}$, and $\dfrac{5\pi}{6}$. + +\mypara{22.} The equations (4) Art.~19 can be solved by exactly the same device. +To find any coefficient $a_m$ multiply the first equation by $2\sin m\Delta x$, the +second by $2\sin 2m\Delta x$, the third by $2\sin 3m\Delta x$, \&c.\ and add. + +The coefficient of any other $a$ as $a_k$ in the resulting equation will be +\begin{align*} + 2&\sin k\Delta x\sin m\Delta x ++ 2\sin 2k\Delta x\sin 2m\Delta x ++ 2\sin 3k \Delta x\sin 3m\Delta x + \cdots\\ +&\qquad+ 2\sin nk\Delta x\sin nm\Delta x\\ +={}& \cos(m-k)\Delta x + \cos 2(m-k)\Delta x + \cos 3(m-k)\Delta x + \cdots + \cos n(m-k)\Delta x\\ +-{}&\cos(m+k)\Delta x - \cos 2(m+k)\Delta x - \cos 3(m+k)\Delta x - \cdots - \cos n(m+k)\Delta x\\ +={}& \cfrac{\sin\dfrac{2n+1}{2}(m-k)\Delta x}{2\sin\dfrac{(m-k)\Delta x}{2}} +-\cfrac{\sin\dfrac{2n+1}{2}(m+k)\Delta x}{2\sin\dfrac{(m+k)\Delta x}{2}}; +\text{\qquad by (1) Art.~20.} +\end{align*} +\[ +\frac{2n+1}{2}=n+1-\frac{1}{2} \quad \text{and} \quad (n + 1)\Delta x = \pi . +\] +Hence the coefficient of $a_k$ may be written +\[ + \cfrac{ \sin \Big[ (m-k)\pi - \dfrac{(m-k)\Delta x}{2} \Big] } + {2\sin\dfrac{(m-k)\Delta x}{2} } +-\cfrac{ \sin \Big[ (m+k)\pi - \dfrac{(m+k)\Delta x}{2} \Big] } + {2\sin\dfrac{(m+k)\Delta x}{2} } +\] +but this is equal to $\dfrac{1}{2}-\dfrac{1}{2}$ or $-\dfrac{1}{2}+\dfrac{1}{2}$ according as $m-k$ is odd or even +and so is zero in either case. +% -----File: 045.png + +The coefficient of $a_m$ will be +\begin{align*} +&\;2\sin^2m\Delta x + 2\sin^22m\Delta x + +2\sin^23m\Delta x + \cdots + 2\sin^2nm\Delta x\\ +&= \qquad 1 \qquad + \qquad 1 \qquad + \qquad 1 \qquad + \quad \cdots\quad + \qquad 1\\ +&\quad-\cos2m\Delta x - \cos4m\Delta x - \cos6m\Delta x - \;\cdots\; - \cos2nm\Delta x\\ +&=n+\frac{1}{2}-\frac{\sin(2n+1)m\Delta x}{2\sin m\Delta x}, \text{ by (1) Art.~20.} +\end{align*} +\begin{flalign*} +&\text{\indent But } & + (2n+1)m\Delta x = 2m(n & +1)\Delta x-m\Delta x = 2m\pi-m\Delta x, &&\phantom{therefore } +\\ +&\text{therefore } & + \frac{\sin(2n+1)m\Delta x}{2\sin m\Delta x} & += \frac{\sin(2m\pi-m\Delta x)}{2\sin m\Delta x} = -\frac{1}{2}, && +\end{flalign*} +and the coefficient of $a_m$ is $n+1$. + +The first member of our final equation will be +\[ +2\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x. +\] + +Hence +\[ +a_m=\frac{2}{n+1}\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x, \tag{1} +\] +and the curve +\[ +y = a_1\sin x + a_2\sin 2x + \cdots + a_n\sin nx, \tag{2} +\] +where the coefficients are given by (1) will pass through the $n$ points of the +curve $y =f(x)$ whose abscissas are $\Delta x$, $2\Delta x$, $3\Delta x$, $\cdots$ $n\Delta x$. $\Delta x$ being $\dfrac{\pi}{n+1}$. + +It should be noted that since the $n$ equations (4) Art.~19 are all of the first +degree there will exist only one set of values for the $n$ quantities $a_1$, $a_2$, $a_3$, +$\cdots$ $a_n$ that can satisfy these equations. Consequently the solution which we +have obtained is the only solution possible. + +\mypara{23.} The result just obtained obviously holds good no matter how great a +value of $n$ may be taken. + +If now we suppose $n$ indefinitely increased the two curves (2) Art.~22 and +$y=f(x)$ will come nearer and nearer to coinciding throughout the whole of +their portions between $x=0$ and $x=\pi$, and consequently the limiting +form that equation (2) Art.~22 approaches as $n$ is indefinitely increased will +represent a curve absolutely coinciding between the values of $x$ in question +with $y=f(x)$. +% -----File: 046.png + +Let us see what limiting value $a_m$ approaches as $n$ is indefinitely increased. +\begin{align*} +\phantom{(1) Art.~22.}a_m &=\frac{2}{n+1}\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x \tag*{(1) Art.~22.}\\ + &=\frac{2\Delta x}{\pi}\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x +\end{align*} +% recast to fit line +\begin{flalign*} += \frac{2}{\pi} \left[ +\begin{alignedat}{1} + f( \Delta x)\sin m\Delta x.\Delta x + + f(2\Delta x)\sin 2m\Delta x.\Delta x & + \cdots \\ + & + f(n\Delta x)\sin nm\Delta x.\Delta x +\end{alignedat} +\right] +\end{flalign*} +\begin{flalign*} += \frac{2}{\pi} \left[ +\begin{alignedat}{1} +f( \Delta x)\sin m\Delta x.\Delta x ++ f( 2\Delta x)\sin2m\Delta x.\Delta &x + \cdots\\ ++ f(&\pi-\Delta x)\sin m(\pi-\Delta x).\Delta x +\end{alignedat} +\right] +\end{flalign*} +since $\Delta x=\dfrac{\pi}{n+1}$. + +As $n$ is increased indefinitely $\Delta x$ approaches zero as a limit. Hence the +limiting value of $a_m$ as $n$ increases indefinitely is +\begin{flalign*} +& \frac{2}{\pi}\limit_{\Delta x\doteq 0}\left[ +\begin{alignedat}{1} +f( \Delta x)\sin m\Delta x.\Delta x ++ f( 2\Delta x)\sin 2m\Delta x.\Delta x &+ \cdots\\ ++ f(\pi-\Delta x)&\sin m(\pi-\Delta x).\Delta x +\end{alignedat} +\right]\footnotemark &&\\[-8ex] +\end{flalign*} +\footnotetext{We shall use the sign $\doteq$ for \emph{approaches}. $\Delta x\doteq 0$ is read $\Delta x$ approaches zero.} +\begin{flalign*} +&&= \frac{2}{\pi}\int\limits_0^\pi f(x)\sin mx.dx. \quad\text{[v.\ Int.\ Cal.\ Arts.~80, 81.]}\\[-5ex] +\end{flalign*} +\begin{flalign*} +&\text{\indent Hence} & f(x)=a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots, & \tag{2}& +\end{flalign*} +where any coefficient $a_m$ is given by the formula +\[ +a_m=\frac{2}{\pi}\int\limits_0^\pi f(x)\sin mx.dx, \tag{3} +\] +is a true development of $f(x)$ for all values of $x$ between $x=0$ and $x=\pi$ +\emph{provided that the series} (2) \emph{is convergent}, for it is in that case only that we can +assume that the limiting value of the second member of (2) Art.~22 can be obtained +by adding the limiting values of the several terms. + +When $x=0$ and when $x=\pi$ every term in the second member of (2) +is zero, and the second member is zero and will not be equal to $f(x)$ unless $f(x)$ +is itself zero when $x=0$ and $x=\pi$; but even when $f(x)$ is not zero for +$x=0$ and $x=\pi$ the development given above holds good for any value +of $x$ between zero and $\pi$ no matter how near it may be taken to either of these +values. + +\mypara{24.} Instead of actually performing the elimination in equations (4) Art.~19 +and getting a formula for $a_m$ in terms of $n$, and then letting $n$ increase +indefinitely, we might have saved labor by the following method. + +% -----File: 047.png + +Return to equations (4) Art.~19 and multiply the first by $\Delta x\sin m\Delta x$, +the second by $\Delta x\sin 2m\Delta x$, and so on, that is multiply each equation by $\Delta x$ +times the coefficient of $a_m$ in that equation, and then add the equations. + +We get as the coefficient of $a_k$ +\[ + \sin k\Delta x\sin m\Delta x.\Delta x ++\sin 2k\Delta x\sin 2m\Delta x.\Delta x + \cdots ++\sin nk\Delta x\sin nm\Delta x.\Delta x. +\] + +Let us find its limiting value as $n$ is indefinitely increased. It may be +written, since $(n+1)\Delta x=\pi$, +\begin{flalign*} +\limit_{\Delta x\doteq 0} \left[ +\begin{alignedat}{1} +\sin k\Delta x\sin m\Delta x.\Delta x ++ \sin2k\Delta x\sin2m&\Delta x.\Delta x +\cdots\\ ++ \sin &k(\pi-\Delta x)\sin m(\pi-\Delta x).\Delta x +\end{alignedat} +\right] +\end{flalign*} +\[ + = \int\limits_0^\pi\sin kx\sin mx.dx; +\] +\begin{flalign*} +&\text{but } & \int\limits_0^\pi\sin kx\sin mx.dx &=\tfrac{1}{2}\int\limits_0^\pi[\cos(m-k)x-\cos(m+k)x]dx &&\qquad{}\\ +&&& = 0 \text{ if $m$ and $k$ are not equal.} && +\end{flalign*} + +The coefficient of $a_m$ is +\[ +\Delta x(\sin^2 m\Delta x + \sin^2 2m\Delta x + \sin^2 3m\Delta x + \cdots + \sin^2 nm\Delta x). +\] + +Its limiting value +\begin{gather*} +\limit_{\Delta x\doteq 0} \bigg[ + \sin^2 m\Delta x.\Delta x ++ \sin^2 2m\Delta x.\Delta x + \cdots ++ \sin^2 m(\pi - \Delta x)\Delta x \bigg] +\\ += \int\limits_0^\pi\sin^2 mx.dx = \frac{\pi}{2}. +\end{gather*} + +The first member is +\[ + f( \Delta x)\sin m\Delta x.\Delta x ++ f(2\Delta x)\sin 2m\Delta x.\Delta x + \cdots ++ f(n\Delta x)\sin mn \Delta x.\Delta x +\] +and its limiting value is +\[ +\int\limits_0^\pi f(x)\sin mx.dx. +\] + +Hence the limiting form approached by the final equation as $n$ is increased is +\[ +\int\limits_0^\pi f(x)\sin mx.dx = \frac{\pi}{2}a_m. +\] +\begin{flalign*} +&\text{\indent Whence } & a_m = \frac{2}{\pi}\int\limits_0^\pi f(x)\sin mx.dx && \text{ as before.}& +\end{flalign*} + +This method is practically the same as \emph{multiplying the equation} +\[ +f(x) = a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots \tag{1} +\] +\emph{by $\mathop\textit{sin} mx.dx$ and integrating both members from zero to $\pi$}. +% -----File: 048.png + +It is exceedingly important to realize that the short method of determining +any coefficient $a_m$ of the series (1) which has just been described in the italicized +paragraph, is essentially the same as that of obtaining $a_m$ by actual +elimination from the equations (4) Art.~19, and then supposing $n$ to increase +indefinitely, thus making the curves (3) Art.~19 and (2) Art.~19 absolutely +coincide between the values of $x$ which are taken as the limits of the +definite integration. + +\mypara{25.} We see, then, that any function of $x$ which is single-valued, finite, and +continuous between $x = 0$ and $x = \pi$, or if discontinuous has only finite +discontinuities each of which is preceded and succeeded by continuous portions, +can probably be developed into a series of the form +\[ +f(x) = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots \tag{1} +\] +\begin{flalign*} +&\text{where }& a_m += \frac{2}{\pi}\int\limits_{0}^{\pi} f(x) \sin mx.dx += \frac{2}{\pi}\int\limits_{0}^{\pi} f(\alpha) \sin m\alpha.d\alpha; &&\phantom{where } \tag{2} +\end{flalign*} +and the series and the function will be identical for all values of $x$ between +$x = 0$ and $x = \pi$, not including the values $x = 0$ and $x = \pi$ unless +the given function is equal to zero for those values. + +An elaborate investigation of the question of the convergence of the series +(1), for which we have not space, entirely confirms the result formulated +above\label{notep38}\footnote + {Provided the function has not an infinite number of maxima and minima in the neighborhood + of a point. v.\ Arts.~37--38.} +and shows in addition that at a point of finite discontinuity the series +has a value equal to half the sum of the two values which the function +approaches as we approach the point in question from opposite sides. + +The investigation which we have made in the preceding sections establishes +the fact that the curve represented by $y = f(x)$ need not follow the same +mathematical law throughout its length, but may be made up of portions of +entirely different curves. For example, a broken line or a locus consisting of +finite parts of several different and disconnected straight lines can be +represented perfectly well by $y =$ a sine series. + +\mypara{26.} Let us obtain a few sine developments. +\begin{flalign*} +&\text{\indent (\emph{a})\quad Let }&f(x)=x.& &&\phantom{\indent (a)\quad Let\ } \tag{1} +\end{flalign*} +\begin{flalign*} +&\text{We have }&x = a_1 \sin x + {}& a_2 \sin 2x + a_3\sin 3x + \cdots &&\phantom{We\ have } \tag{2}\\ +&\text{where }&a_m = \frac{2}{ \pi} & \int\limits_{0}^{\pi} x \sin mx.dx&& \tag{3} +\end{flalign*} +% -----File: 049.png +\[ +\int x\sin mx.dx = \frac{1}{m^2}(\sin mx - mx\cos mx), +\] +\[ +\int\limits_0^\pi x\sin mx.dx = -\frac{(-1)^m \pi}{m}, +\] +\begin{flalign*} +&\text{and } & x = 2 \left( \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \frac{\sin 4x}{4} + \cdots \right) & \tag{4}&& +\end{flalign*} +\begin{flalign*} +&\indent \text{(\emph{b})\quad Let } & f(x) = 1. & \tag{1} &&\phantom{\indent(b)\quad Let\ } +\end{flalign*} +\[ +a_m = \frac{2}{\pi}\int\limits_0^\pi\sin mx.dx; \tag{2} +\] +\[ +\int\sin mx.dx = -\frac{\cos mx}{m}, +\] +\begin{align*} +\int\limits_0^\pi\sin mx.dx &= \frac{1}{m}(1-\cos m\pi) = \frac{1}{m}[1-(-1)^m]\\ +&= 0 \text{ if $m$ is even}\\ +&= \frac{2}{m} \text{ if $m$ is odd.} +\end{align*} +\begin{flalign*} +&\text{Hence } & 1 = \frac{4}{\pi} \left( \frac{\sin x}{1} + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \frac{\sin 7x}{7} + \cdots \right). & \tag{3} &&\phantom{Hence} +\end{flalign*} + +It is to be noticed that (3) gives at once a sine development for any constant +$c$. It is, +\[ +c = \frac{4c}{\pi} \left( \frac{\sin x}{1} + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots \right). \tag{4} +\] + +If we substitute $x=\dfrac{\pi}{2}$ in (4)\,(\emph{a}) or (3)\,(\emph{b}) we get a familiar result, namely +\[ +\frac{\pi}{4}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots, \tag{5} +\] +a formula usually derived by substituting $x=1$ in the power series for +$\tan^{-1}x$. (v.\ Dif.\ Cal.\ Art.~135.) + +(4)\,(\emph{a}) does not hold good when $x=\pi$, and (3)\,(\emph{b}) fails when $x=0$ and +when $x=\pi$, for in all these cases the series reduces to zero.\\ + +%[Illustration] +\pngright{001.png}{626}{390}{-24} + +(\emph{c})\quad Let $f(x)=x$ from $x=0$ to $x=\dfrac{\pi}{2}$ +and $f(x)=\pi-x$ from $x=\dfrac{\pi}{2}$ to $x=\pi$. +That is, let $y=f(x)$ represent the broken +line in the figure. + +As the mathematical expression for +$f(x)$ is different in the two halves of the +curve we must break up +% -----File: 050.png +\[ +\int\limits_0^\pi f(x)\sin mx.dx \quad \text{into}\quad \int\limits_0^\frac{\pi}{2}f(x)\sin mx.dx + \int\limits_\frac{\pi}{2}^\pi f(x)\sin mx.dx. +\] +We have, then, +\begin{align*} +a_m &= \frac{2}{\pi} \int\limits_0^\frac{\pi}{2}x\sin mx.dx + \frac{2}{\pi} \int\limits_\frac{\pi}{2}^\pi(\pi-x)\sin mx.dx \tag{1}\\ +&= \frac{4}{m^2\pi}\sin m \frac{\pi}{2}.\\[-6ex] +\end{align*} + +\begin{flalign*} +&\text{\indent But}& +\begin{alignedat}[t]{5} +\sin m \dfrac{\pi}{2} &= 1 && \text{if}\quad && m=1\quad && \text{or}\quad && 4k+1 \\[-1ex] + &= 0 && `` && m=2 && \,`` && 4k+2 \\ + &= -1\;\; && `` && m=3 && \,`` && 4k+3 \\ + &= 0 && `` && m=4 && \,`` && 4k. \\ +\end{alignedat} +&&\phantom{\indent But} +\end{flalign*} +Hence if $y=f(x)$ represents our broken line, +\[ +f(x)=\frac{4}{\pi}\left(\frac{\sin x}{1^2}-\frac{\sin 3x}{3^2}+\frac{\sin 5x}{5^2}-\frac{\sin 7x}{7^2}+\cdots\right). \tag{2} +\] + +When $x=\dfrac{\pi}{2}$ $f(x)=\dfrac{\pi}{2}$ and we have +\[ +\frac{\pi^2}{8} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \tag{3} +\] + +(\emph{d})\quad As a case where the function has a finite discontinuity, let +\begin{alignat*}{4} +f(x)=1 \quad& \text{from} \quad&& x=0\quad && \text{to}\quad && x=\dfrac{\pi}{2} \quad \text{and}\\ +f(x)=0 \quad& \;\;\text{``} && x=\dfrac{\pi}{2} && \;\text{``} && x=\pi. +\end{alignat*} +$y=f(x)$ will in this case represent the locus in the figure. + +%[Illustration] +\pngrightl{002.png}{675}{3}{-24} + +As before +\begin{align*} +\int\limits_0^\pi f(x)\sin mx.dx & = \int\limits_0^\frac{\pi}{2} f(x)\sin mx.dx \\ +&+ \int\limits_\frac{\pi}{2}^\pi f(x)\sin mx.dx. +\end{align*} +\[ +a_m = \frac{2}{\pi} \int\limits_0^\frac{\pi}{2}\sin mx.dx + \frac{2}{\pi}\int\limits_\frac{\pi}{2}^\pi 0 +.\sin mx.dx. \tag{1} +\] +% -----File: 051.png +\[ +a_m=\frac{2}{\pi} \int\limits_0^\frac{\pi}{2}\sin mx.dx +=\frac{2}{\pi}\frac{1}{m}\left(1-\cos m\frac{\pi}{2}\right). +\] + +\begin{flalign*} +&\text{\indent But}& +\begin{alignedat}[t]{5} +\cos m \dfrac{\pi}{2} &= \phantom{-}0\quad && \text{if}\quad && m=1\quad && \text{or}\quad && 4k+1 \\[-1ex] + &=-1 && `` && m=2 && \,`` && 4k+2 \\ + &= \phantom{-}0 && `` && m=3 && \,`` && 4k+3 \\ + &= \phantom{-}1 && `` && m=4 && \,`` && 4k. \\ +\end{alignedat} +&&\phantom{\indent But} +\end{flalign*} + +Hence +\[ +f(x) = \frac{2}{\pi} + \left( + \frac{ \sin x}{1} + \frac{2\sin 2x}{2} + \frac{ \sin 3x}{3} + + \frac{ \sin 5x}{5} + \frac{2\sin 6x}{6} + \frac{ \sin 7x}{7} + + \cdots \right). \tag{2} +\] + +If $x=\dfrac{\pi}{2}$ the second member of (2) reduces to $\dfrac{1}{2}$, for +\[ + \frac{2}{\pi} + \left( \frac{1}{1} - \frac{1}{3} + + \frac{1}{5} - \frac{1}{7} + \cdots \right) += \frac{1}{2}\qquad \text{by (5) ($b$)}; +\] +and we see that the series represents the function completely for all values of +$x$ between $x=0$ and $x=\pi$ except for $x=\dfrac{\pi}{2}$ and there it has a +value which is the mean of the values approached by the function as $x$ +approaches $\dfrac{\pi}{2}$ from opposite sides. + +\EXAMPLE{S} + +Obtain the following developments:--- +\begin{flalign*} +&\indent\text{(1)}\quad x^2 = \dfrac{2}{\pi} \bigg[ + \bigg( \dfrac{\pi^2}{1} - \dfrac{4}{1^3} \bigg) \sin x + - \dfrac{\pi^2}{2}\sin 2x ++ \bigg( \dfrac{\pi^2}{3} - \dfrac{4}{3^3} \bigg) \sin 3x + - \dfrac{\pi^2}{4}\sin 4x \\ +& \hspace{8em} + \bigg( \dfrac{\pi^2}{5} - \dfrac{4}{5^3} \bigg) \sin 5x - \cdots \bigg].\\ +% +&\indent\text{(2)}\quad x^3 = \dfrac{2}{\pi} \bigg[ + \bigg( \dfrac{\pi^3}{1} - \dfrac{6\pi}{1^3} \bigg) \sin x +- \bigg( \dfrac{\pi^3}{2} - \dfrac{6\pi}{2^3} \bigg) \sin 2x ++ \bigg( \dfrac{\pi^3}{3} - \dfrac{6\pi}{3^3} \bigg) \sin 3x \\ +&\hspace{8em} - \bigg( \dfrac{\pi^3}{4} - \dfrac{6\pi}{4^3} \bigg) \sin 4x + \cdots \bigg]. \\ +% +&\indent\text{(3)}\quad f(x) = \dfrac{2}{\pi} \bigg[ + \dfrac{\sin x}{1^2} + \dfrac{ \pi}{2^2}\sin 2x +- \dfrac{\sin 3x}{3^2} - \dfrac{2\pi}{4^2}\sin 4x ++ \dfrac{\sin 5x}{5^2} \\ +&\hspace{8em} + \dfrac{3\pi}{6^2}\sin 6x - \cdots \bigg], +\end{flalign*} +% -----File: 052.png +if $f(x)=x$ from $x=0$ to $x=\dfrac{\pi}{2}$ and $f(x)=0$ from $x=\dfrac{\pi}{2}$ to $x=\pi$. + +\begin{flalign*} +&\indent\text{(4)}\quad \sin\mu x=\dfrac{2}{\pi}\sin\mu\pi \bigg[ + \dfrac{ \sin x}{1^2-\mu^2} - \dfrac{2\sin2x}{2^2-\mu^2} ++ \dfrac{3\sin3x}{3^2-\mu^2} - \dfrac{4\sin4x}{4^2-\mu^2} + \cdots \bigg]\\ +&\text{if $\mu$ is a fraction.}\\ +% +&\indent\text{(5)}\quad e^x = \dfrac{2}{\pi} \bigg[ + \dfrac{1}{ 2}(1+e^\pi)\sin x + \dfrac{2}{5}(1-e^\pi)\sin 2x ++ \dfrac{3}{10}(1+e^\pi)\sin 3x \\ +&\hspace{8em} + \dfrac{4}{17}(1-e^\pi)\sin 4x + \cdots \bigg].\\ +% +&\indent\text{(6)}\quad \sinh x = \dfrac{2\sinh\pi}{\pi} \bigg[ + \dfrac{1}{ 2}\sin x - \dfrac{2}{5}\sin 2x ++ \dfrac{3}{10}\sin3x - \dfrac{4}{17}\sin4x + \cdots \bigg].\\ +% +&\indent\text{(7)}\quad \cosh x = \dfrac{2}{\pi} \bigg[ + \dfrac{1}{2}(1+\cosh\pi)\sin x ++ \dfrac{2}{5}(1-\cosh\pi)\sin 2x \\ +&\hspace{8em} + \dfrac{3}{10}(1+\cosh\pi)\sin 3x + \cdots \bigg]. +\end{flalign*} + +\mypara{27.} Let us now try to develop a given function of $x$ in a series of cosines. + +As before suppose that $f(x)$ has a single value for each value of $x$ between +$x=0$ and $x=\pi$, that it does not become infinite between $x=0$ and + $x=\pi$, and that if discontinuous it has only finite discontinuities. + +Assume +\[ +f(x) = b_0 + b_1\cos x + b_2\cos 2x + b_3\cos 3x + \cdots \tag{1} +\] + +To determine any coefficient $b_m$ +multiply (1) by $\cos mx.dx$ and integrate +each term from $0$ to $\pi$. +\begin{align*} +\int\limits_0^\pi &b_0\cos mx.dx = 0. \\ + \int\limits_0^\pi b_k\cos kx\cos mx.dx +&= \frac{b_k}{2}\int\limits_0^\pi[\cos(m-k)x+\cos(m+k)x]dx\\ +&= 0 \text{ if $m$ and $k$ are not equal.} +\end{align*} +\[ +\int b_m\cos^2 mx.dx = \frac{b_m}{2m}(mx+\cos mx\sin mx), +\] +\[ +\phantom{if\ m\ is\ not\ zero.} +\int\limits_0^\pi b_m\cos^2 mx.dx = \frac{\pi}{2}b_m, \tag*{if $m$ is not zero.} +\] +\begin{flalign*} +&\text{Hence} & b_m += \frac{2}{\pi} \int\limits_0^\pi f(x)\cos mx.dx += \frac{2}{\pi} \int\limits_0^\pi f(\alpha)\cos m\alpha.d\alpha, & \tag{2}&& +\end{flalign*} +if $m$ is not zero. +% -----File: 053.png + +To get $b_0$ multiply (1) by $dx$ and integrate from zero to $\pi$. +\begin{flalign*} +&&&\int\limits_0^\pi b_0 dx = b_0 \pi,&&\\ +&&&\int\limits_0^\pi b_k\cos kx.dx=0.&&\\ +\label{err053} +&\text{\indent Hence} & b_0=\frac{1}{\pi} &\int\limits_0^\pi f(x)dx=\frac{1}{\pi}\int\limits_0^\pi f(\alpha)d\alpha, & \tag{3} &\phantom{Hence} +\end{flalign*} +which is just half the value that would be given by formula~(2) if zero were +substituted for $m$. + +To save a separate formula (1) is usually written +\[ +f(x) = \tfrac{1}{2} b_0 + b_1\cos x + b_2\cos 2x + b_3\cos 3x + \cdots \tag{4} +\] +and then the formula +\[ +b_m = \frac{2}{\pi} \int\limits_0^\pi f(x)\cos mx.dx += \frac{2}{\pi}\int\limits_0^\pi f(\alpha)\cos m\alpha.d\alpha \tag{2} +\] +will give $b_0$ as well as the other coefficients. + +It is important to see clearly that what we have just done in determining +the coefficients of (1) is equivalent to taking $n+1$ terms of (4), +substituting in +\[ +y = \tfrac{1}{2} b_0 + b_1\cos x + b_2\cos 2x + \cdots + b_n\cos nx \tag{5} +\] +in turn the coördinates of the $n+1$ points of the curve +\[ +y=f(x) +\] +whose projections on the axis of $X$ are equidistant, determining $b_0$, $b_1$, $b_2$, $\cdots$ $b_n$ +by elimination from the $n+1$ resulting equations, and then taking the limiting +values they approach as $n$ is indefinitely increased.\quad (v.\ Art.~24.) + +If $\Delta x = \dfrac{\pi}{n+1}$ the abscissas of the $n+1$ points used are 0, $\Delta x$, $2\Delta x$, +$3\Delta x$, $\cdots$ $n\Delta x$, so that we should expect our cosine development to hold for + $x=0$ as well as for values of $x$ between zero and $\pi$. + +\mypara{28.} Let us take one or two examples: +\begin{flalign*} +&\text{\indent($a$)\quad Let} & f(x)=x. & \tag{1} &&\phantom{\indent\quad Let} +\end{flalign*} +\begin{gather*} +b_0 = \frac{2}{\pi} \int\limits_0^\pi x\,dx + = \frac{2}{\pi} \frac{\pi^2}{2} = \pi. +\\ +b_m = \frac{2}{\pi}\int\limits_0^\pi x\cos mx.dx + = \frac{2}{m^2\pi} (\cos m\pi-1) + = \frac{2}{m^2\pi} [(-1)^m-1]. +\end{gather*} +% -----File: 054.png + +\begin{flalign*} +&\text{\indent Hence} & x &= \frac{\pi}{2} - \frac{4}{\pi} \left( + \cos x + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2} + + \frac{\cos 7x}{7^2} + \cdots \right). &&\phantom{Hence} \tag{2} +\end{flalign*} + +(2) holds good not only for values of $x$ between zero and $\pi$ but for $x = 0$ +and $x = \pi$ as well, since for these values we have +\begin{flalign*} +&& 0 &= \frac{\pi}{ 2} - \frac{4}{\pi} \left( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{ 1}{7^2} + \cdots \right) &&\tag{3}\\ +&\text{and }& + \pi &= \frac{\pi}{2} + \frac{4}{\pi} \left(1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{ 1}{7^2} + \cdots \right) && \tag{4} +\end{flalign*} +which are true by Art.~26 (\emph{c})(3). + +\begin{flalign*} +&\text{\indent($b$)\quad Let}& f(x) &= x\sin x. &&\phantom{\indent Let} \tag{1} +\end{flalign*} +\begin{gather*} +b_0 += \frac{2}{\pi}\int\limits_{0}^{\pi} x\sin x.dx += \frac{2}{\pi}\pi = 2,\\ +b_1 += \frac{2}{\pi}\int\limits_{0}^{\pi} x\sin x \cos x.dx += \frac{1}{\pi}\int\limits_{0}^{\pi} x\sin 2x.dx = -\frac{1}{2},\\ +b_m += \frac{2}{\pi}\int\limits_{0}^{\pi} x\sin x \cos mx.dx += \frac{1}{\pi}\int\limits_{0}^{\pi} [x\sin (m+1)x - x \sin (m-1)x]dx\\ +\begin{split} +&= \frac{2}{(m-1)(m+1)} \quad \text{if $m$ is odd}\\ +&=-\frac{2}{(m-1)(m+1)} \quad \text{if $m$ is even.}\\ +\end{split} +\end{gather*} + +Hence +\[ +x \sin x = 1 - \frac{\cos x}{2} + - \frac{2\cos 2x}{1.3} + + \frac{2\cos 3x}{2.4} + - \frac{2\cos 4x}{3.5} + \cdots \tag{2} +\] +If $x = \dfrac{\pi}{2}$ we have +\[ +\frac{\pi}{4} = \frac{1}{ 2} + + \frac{1}{1.3} + - \frac{1}{3.5} + + \frac{1}{5.7} - \cdots. \tag{3} +\] + +\newpage +\EXAMPLE{S} + +Obtain the following developments: +\begin{flalign*} +&\text{\indent(1)\quad} f(x) = \frac{\pi}{ 4} - \frac{2}{\pi} \bigg[ + \frac{\cos 2x}{1^2} + \frac{\cos 6x}{3^2} + + \frac{\cos 10x}{5^2} + \frac{\cos 14x}{7^2} + \cdots \bigg] &\\ +&\text{if \quad $f(x) = x$ from $x = 0$ to $x = \dfrac{\pi}{ 2 }$ and $f(x) = \pi - x$ from $x = \dfrac{\pi}{ 2}$ to $x = \pi$.}\\ +% -----File: 055.png +% +&\text{\indent(2)\quad} f(x) = \dfrac{1}{2} + \dfrac{2}{\pi}\bigg[\dfrac{\cos x}{1} - \dfrac{\cos 3x}{3} + \dfrac{\cos 5x}{5} - \dfrac{\cos 7x}{7} +\cdots\bigg], \\ +&\text{if \quad $f(x) = 1$ from $x = 0$ to $x = \dfrac{\pi}{ 2}$ and $f(x) = 0$ from $x = \dfrac{\pi }{2}$ to $x = \pi$.}\\ +% +&\text{\indent(3)\quad} x^2 = \dfrac{\pi^2}{ 3} - 4 \bigg[\dfrac{\cos x}{1^2} - \dfrac{\cos 2x}{2^2} + \dfrac{\cos 3x}{3^2} - \dfrac{\cos 4x}{4^2} +\cdots\bigg], \\ +% +&\text{\indent(4)\quad} x^3 = \dfrac{\pi^3 }{4} - \dfrac{6}{\pi}\bigg[\bigg(\dfrac{\pi^2}{1^2} - \dfrac{4}{1^4}\bigg)\cos x - \dfrac{\pi^2}{2^2}\cos 2x + \bigg(\dfrac{\pi^2}{3^2} - \dfrac{4}{3^4}\bigg)\cos 3x \\ +&\hspace{10em} - \dfrac{\pi^2}{4^2} \cos 4x + \bigg(\dfrac{\pi^2}{5^2} - \dfrac{4}{5^4}\bigg)\cos 5x - \cdots\bigg], \\ +% +&\text{\indent(5)\quad} f(x) = \dfrac{\pi }{ 8} + \dfrac{2}{\pi}\bigg[\bigg( \dfrac{\pi}{2} - 1\bigg)\cos x - \dfrac{2}{2^2} \cos 2x - \dfrac{1}{3^2} \bigg( \dfrac{3\pi}{2}+1\bigg) \cos 3x \\ +&\hspace{11em} +\dfrac{1}{5^2}\bigg(\dfrac{5\pi}{2} -1\bigg)\cos 5x -\dfrac{2 }{6^2} \cos 6x - \cdots\bigg], \\ +&\text{if \quad $f(x) = x$ from $x = 0$ to $x = \dfrac{\pi }{2}$ and $f(x) = 0$ from $x = \dfrac{\pi}{ 2}$ to $x = \pi$.}\\ +% +&\text{\indent(6)\quad} e^x = \dfrac{2}{\pi}\bigg[\dfrac{1}{2}(e^\pi-1)- \dfrac{1}{1+1^2} (e^\pi+1)\cos x + \dfrac{1}{1+2^2} (e^\pi-1)\cos 2x \\ +&\hspace{7em} - \dfrac{1}{1+3^2} (e^\pi+1)\cos 3x + \cdots\bigg], \\ +% +&\text{\indent(7)\quad} \cosh x = \dfrac{2 \sinh \pi}{\pi} \bigg[\dfrac{1}{2} - \dfrac{1}{2}\cos x + \dfrac{1}{5}\cos 2x - \dfrac{1}{10}\cos 3x \\ +&\hspace{13em} + \dfrac{1}{17}\cos 4x - \cdots\bigg], \\ +% +&\text{\indent(8)\quad} \sinh x =\dfrac{ 2}{\pi}\bigg[\dfrac{1}{2} (\cosh\pi-1) - \dfrac{1}{2}(\cosh\pi +1)\cos x \\ +&\hspace{8em} + \dfrac{1}{5} (\cosh \pi-1)\cos 2x - \dfrac{1}{10}(\cosh\pi +1) \cos 3x + \cdots \bigg], \\ +% +&\text{\indent(9)\quad} \cos \mu x = \dfrac{2\mu \sin \mu \pi}{\pi} \bigg[ \dfrac{1}{2\mu^2} - \dfrac{\cos x}{ \mu^2-1^2} + \dfrac{\cos 2x}{\mu^2-2^2} - \dfrac{\cos 3x}{\mu^2-3^2} \\ +% +&\hspace{12.5em} + \dfrac{\cos 4x}{\mu^2-4^2} - \cdots \bigg], \\ +&\indent\phantom{(9)} \quad \text{if $\mu$ is a fraction.} +\end{flalign*} + +\mypara{29.} Although any function can be expressed both as a sine series and as a +cosine series, and the function and either series will be equal for all values of +$x$ between zero and $\pi$, there is a decided difference in the two series for other +values of $x$. + +Both series are periodic functions of $x$ having the period $2\pi$. If then we +let $y$ equal the series in question and construct the portion of the corresponding +% -----File: 056.png +curve which lies between the values $x = -\pi$ and $x = \pi$ the whole +curve will consist of repetitions of this portion. + +Since $\sin mx = -\sin(- mx)$ the ordinate corresponding to any value of +$x$ between $-\pi$ and zero in the sine curve; will be the negative of the ordinate +corresponding to the same value of $x$ with the positive sign. In other words +the curve +\[ + y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots \tag{1} +\] +is symmetrical with respect to the origin. + +Since $\cos mx = \cos (-mx)$ the ordinate corresponding to any value of $x$ +between $-\pi$ and zero in the cosine curve will be the same as the ordinate +belonging to the corresponding positive value of $x$. In other words the curve +\[ + y = \tfrac{1}{2}b_0 +b_1\cos x + b_2\cos 2x + b_3\cos 3x + \cdots \tag{2} +\] +is symmetrical with respect to the axis of $Y$. + +If then $f(x) = -f(-x)$, that is if $f(x)$ is an \textit{odd} function the sine series +corresponding to it will be equal to it for all values of $x$ between $-\pi$ and $\pi$, +except perhaps for the value $x = 0$ for which the series will necessarily be +zero. + +If $f(x) =f(-x)$, that is if $f(x)$ is an \textit{even} function the cosine series corresponding +to it will be equal to it for all values of $x$ between $x = -\pi$ and +$x = \pi$, not excepting the value $x = 0$. + +As an example of the difference between the sine and cosine developments +of the same function let us take the series for $x$. +\begin{align*} + y &= 2\left[\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + - \frac{\sin 4x}{4} + \cdots \right] \tag{3} +\\ + y &= \frac{\pi}{2} - \frac{4}{\pi} \left[ + \cos x + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2} + + \frac{\cos 7x}{7^2} + \cdots \right] \tag{4} +\end{align*} +[v.\ Art.~26(\emph{a}) and Art.~28(\emph{a})]. (3) represents the curve +%[Illustration] +\pngcent{003.png}{1266}%{346} + +\newpage\noindent and (4) the curve +%[Illustration] +\pngcent{004.png}{1266}%{236} +% -----File: 057.png + +Both coincide with $y = x$ from $x = 0$ to $x = \pi$, (3) coincides with + $y = x$ from $x = -\pi$ to $x = \pi$, and neither coincides with $y = x$ for +values of $x$ less than $-\pi$ or greater than $\pi$. Moreover (3), in addition to +the continuous portions of the locus represented in the figure, gives the isolated +points $(-\pi,0)$ $(\pi,0)$ $(3\pi,0)$ \&c. + +\mypara{30.} We have seen that if $f(x)$ is an \textit{odd} function its development in sine +series holds for all values of $x$ from $-\pi$ to $\pi$, as does the development of +$f(x)$ in cosine series if $f(x)$ is an \textit{even} function. + +Thus the developments of Art.~26(\emph{a}), Art.~26 Exs.~(2), (4), (6); Art.~28(\emph{b}), +Art.~28 Exs.~(3), (7), (9) are valid for all values of $x$ between $-\pi$ and $\pi$. + +Any function of $x$ can be developed into a Trigonometric series to which it +is equal for all values of $x$ between $-\pi$ and $\pi$. + +Let $f(x)$ be the given function of $x$. It can be expressed as the sum of an +even function of $x$ and an odd function of $x$ by the following device. +\[ +f(x)=\frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} \tag{1} +\] +identically; but $\dfrac{f(x)+f(-x)}{2}$ is not changed by reversing the sign of $x$ and +is therefore an \emph{even} function of $x$; and when we reverse the sign of $x$, +$\dfrac{f(x)-f(-x)}{2}$ is affected only to the extent of having its sign reversed and +is consequently an \textit{odd} function of $x$. + +Therefore for all values of $x$ between $-\pi$ and $\pi$\\[-1ex] +\[ + \frac{f(x)+f(-x)}{2} = \frac{1}{2}b_0 + + b_1\cos x + b_2\cos 2x + b_3\cos3x + \cdots \\[1ex] +\] +where \hfill$\displaystyle + b_m = \frac{2}{\pi} \int\limits_{0}^{\pi} \frac{f(x) + f(-x)}{2} + \cos mx.dx; +$ \hfill\quad and \\[1ex] +\[ + \frac{f(x)-f(-x)}{2} = a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots \\[1ex] +\] +where \hfill$\displaystyle + a_m = \frac{2}{\pi} \int\limits_{0}^{\pi} \frac{f(x) - f(-x)}{2} + \sin mx.dx. +$ \hfill\phantom{where}\\[2ex] +\newpage\noindent $b_m$ and $a_m$ can be simplified a little. \\[-1ex] +\begin{align*} + b_m &= \frac{2}{\pi} \int\limits_{0}^{\pi} \frac{f(x)+f(-x)}{2} + \cos mx.dx +\\[1ex] + &= \frac{1}{\pi} + \bigg[ \int\limits_{0}^{\pi} f( x)\cos mx.dx + + \int\limits_{0}^{\pi} f(-x)\cos mx.dx \bigg], +\end{align*} +% -----File: 058.png +but if we replace $x$ by $-x$, we get +\[ + \int\limits_{0}^{ \pi} f(-x) \cos mx.dx += -\int\limits_{0}^{-\pi} f( x) \cos mx.dx += \int\limits_{-\pi}^{0} f( x) \cos mx.dx,\\[-4ex] +\] +\begin{flalign*} +&\text{and we have }&& + b_m = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \cos mx.dx. &&\phantom{and\ we\ have } +\end{flalign*} + +In the same way we can reduce the value of $a_m$ to +\[ + \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \sin mx.dx. +\] + +Hence +\[ + \left\{ \begin{aligned} + f(x) = \frac{1}{2} b_0 &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots\\ + &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots +\end{aligned} + \right \}\tag{2}\\[-3ex] +\] +\begin{flalign*} +&\text{where }& + b_m &= \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \cos mx.dx + = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \cos m\alpha.d\alpha. &&\phantom{where} \tag{3} +\\ +&\text{and }& + a_m &= \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \sin mx.dx + = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \sin m\alpha.d\alpha. && \tag{4} +\end{flalign*} +and this development holds for all values of $x$ between $-\pi$ and $ \pi$. + +The second member of (2) is known as a Fourier's Series. + +\EXAMPLE{S} + +1.\quad Obtain the following developments, all of which are valid from $x = -\pi$ +to $x=\pi$:--- +\begin{flalign*} +&\indent\text{(1)}& +e^x +={}& \frac{2 \sinh \pi}{\pi} + \left[ \frac{1}{2} - \frac{1}{2} \cos x + \frac{1}{5} \cos 2x + - \frac{1}{10} \cos 3x + \frac{1}{17} \cos 4x + \cdots \right]&& +\\ +&&&+ \frac{2 \sinh \pi}{ \pi} + \left[ \frac{1}{2}\sin x - \frac{2}{5} \sin 2x + + \frac{3}{10} \sin 3x - \frac{4}{17} \sin 4x + \cdots \right]. +\end{flalign*} +\begin{flalign*} +&\indent\text{(2)}& +f(x) +={}& \frac{\pi}{ 4} - \frac{2}{ \pi} + \left[ \cos x + \frac{\cos 3x }{ 3^2} + \frac{\cos 5x}{ 5^2} + + \frac{\cos 7x }{ 7^2} + \cdots \right]&&\hspace{5.5em} +\\ +&&&+ \frac{\sin x}{ 1} - \frac{\sin 2x}{ 2} + \frac{\sin 3x}{ 3} + - \frac{\sin 4x}{ 4} + \cdots, +\end{flalign*} +where $f(x) = 0$ from $x = -\pi$ to $x=0$ and $f(x) = x$ from $x=0$ to $x=\pi$. +% -----File: 059.png +\begin{flalign*} +&\indent\text{(3)} &f(x)={}&- \frac{3\pi}{16} + \frac{1}{\pi}\bigg[\frac{1}{1^2}\cos x + \frac{2}{2^2}\cos2x + \frac{1}{3^2}\cos3x + \frac{1}{5^2}\cos5x\\ +&&&\hspace{5em} + \frac{2}{6^2}\cos 6x + \cdots\bigg]&&\hspace{2em}\\ +&&&+ \frac{1}{\pi}\bigg[\biggl(\frac{3\pi}{2} - 1\biggr)\sin x - \frac{3\pi}{4}\sin2x + \biggl(\frac{3\pi}{6} + \frac{1}{3^2}\biggr)\sin3x\\ +&&&\hspace{3em} - \frac{3\pi}{8}\sin4x + \biggl(\frac{3\pi}{10} - \frac{1}{5^2}\biggr)\sin5x - \cdots\bigg],\\[-6ex] +\end{flalign*} +\begin{flalign*} +&\text{where } &f(x)&=x \text{ from } x=-\pi \text{ to } x=0, \quad f(x)=0 \text{ from } x=0 \text{ to } x=\frac\pi2, &&\\ +&\text{and } &f(x)&=x - \frac{\pi}{2} \text{ from } x=\frac{\pi}{2} \text{ to } x=\pi. +\end{flalign*} + +2.\quad Show that formula~(2) Art.~30 can be written +\[ +f(x)=\frac{1}{2} c_0\cos \beta_0 + c_1\cos(x - \beta_1) + c_2\cos(2x - \beta_2) + c_3\cos(3x - \beta_3) + \cdots +\] +\vspace{-4ex} +\begin{flalign*} +&\text{where} +&c_m=(a_m^2 + b_m^2)^\frac{1}{2} \text{ and } \beta_m=\tan^{-1} \frac{a_m}{b_m}.&&\phantom{where} +\end{flalign*} + +3.\quad Show that formula~(2) Art.~30 can be written +\[ +f(x)=\frac{1}{2} c_0\sin \beta_0 + c_1\sin(x + \beta_1) + c_2\sin (2x + \beta_2) + c_3\sin (3x + \beta_3) + \cdots\\[-4ex] +\] +\begin{flalign*} +&\text{where} +&c_m=(a_m^2 + b_m^2)^\frac{1}{2} \text{ and } \beta_m=\tan^{-1} \frac{b_m}{a_m}.&&\phantom{where} +\end{flalign*} + +\mypara{31.} In developing a function of $x$ into a Trigonometric series it is often +inconvenient to be held within the narrow boundaries $x=-\pi$ and $x=\pi$. +Let us see if we cannot widen them. + +Let it be required to develop a function of $x$ into a Trigonometric series +which shall be equal to $f(x)$ for all values of $x$ between $x=-c$ and $x=c$. + +Introduce a new variable +\[ +z=\frac{\pi}{c} x, +\] +which is equal to $-\pi$ when $x=-c$ and to $\pi$ when $x=c$. + +$f(x)=f\left(\dfrac{c}{\pi}z\right)$ can be developed in terms of $z$ by Art.~30 (2), (3), and (4). + +We have +\begin{equation} +\left. +\begin{aligned} +f\left(\frac{c}{\pi}z\right)=\frac{1}{2} b_0 &+ b_1\cos z + b_2\cos 2z + b_3\cos 3z + \cdots\\ +&+ a_1\sin z + a_2\sin 2z + a_3\sin 3z + \cdots +\end{aligned}\right\}\tag{1} +\end{equation} +\vspace{-3ex} +\begin{flalign*} +&\text{where} +&b_m&=\frac{1}{\pi} \int\limits_{-\pi}^{\pi} f\left(\frac{c}{\pi} z\right)\cos mz.dz.&\phantom{where} \tag{2}\\ +% -----File: 060.png +&\text{and}& + a_m &= \frac{1}{\pi} + \int\limits_{-\pi}^{\pi} f\left(\frac{c}{\pi} z\right) + \sin mz.dz. & \tag{3} +\end{flalign*} +and (1) holds good from $z = -\pi$ to $z = \pi$. + +Replace $z$ by its value in terms of $x$ and (1) becomes +\[ \left . \begin{aligned} +f(x)=\frac{1}{2}b_0 +&+ b_1 \cos\frac{\pi x}{c} + b_2 \cos \frac{2\pi x}{c} + + b_3 \cos \frac{3\pi x}{c} + \cdots +\\ +&+a _1 \sin \frac{\pi x}{c} + a_2 \sin \frac{2\pi x}{c} + a_3 \sin \frac{3\pi x}{c} + \cdots +\end{aligned} +\right \} +\tag{4} +\] + +The coefficients in (4) are the same as in (1), and (4) holds good from + $x = -c$ to $x = c$. + +Formulas (2) and (3) can be put into more convenient shape. +\begin{flalign*} +&& b_m +&= \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f\left(\frac{c}{\pi} z\right) \cos mz.dz + = \frac{1}{\pi} \int\limits_{-c}^{c} f(x) \cos \frac{m\pi x}{c} \frac{\pi}{c} dx&& +\\[1ex] +&\text{or}& b_m +&= \frac{1}{c} \int\limits_{-c}^{c} f(x) \cos \frac{m\pi x}{c} dx + = \frac{1}{c} \int\limits_{-c}^{c} f(\lambda) \cos\frac{m\pi\lambda}{c} d\lambda. && \tag{5} +\\ +\intertext{\indent In like manner we can transform (3) into} +&& a_m +&= \frac{1}{c} \int\limits_{-c}^{c} f(x) \sin \frac{m\pi x}{c} dx + = \frac{1}{c} \int\limits_{-c}^{c} f(\lambda) \sin \frac{m\pi \lambda}{c} d\lambda. && \tag{6} +\end{flalign*} + +By treating in like fashion formulas (1) and (2) Art.~25 and formulas (4) +and (2) Art.~27 we get +\[ + f(x) = a_1 \sin \frac{\pi x}{c} + a_2 \sin \frac{2\pi x}{c} + + a_3 \sin \frac{3\pi x}{c} + \cdots \tag{7} +\] +\begin{flalign*} +&\text{where }& a_m +&= \frac{2}{c} \int\limits_{0}^{c} f(x) \sin \frac{m\pi x}{c} dx + = \frac{2}{c} \int\limits_{0}^{c} f(\lambda) \sin \frac{m\pi \lambda}{c} d\lambda. && \tag{8} +\\[1ex] +&\text{and}& f(x) +&= \frac{1}{2 } b_0 + b_1 \cos \frac{\pi x}{c} + + b_2 \cos \frac{2\pi x}{c} + b_3 \cos \frac{3\pi x}{c} + \cdots && \tag{9} +\\[1ex] +&\text{where}& b_m +&= \frac{2}{c} \int\limits_{0}^{c} f(x) \cos \frac{m\pi x}{c} dx + = \frac{2}{c} \int\limits_{0}^{c} f(\lambda) \cos \frac{m\pi \lambda}{c} d\lambda. && \tag{10} +\end{flalign*} +and (7) and (9) hold good from $x = 0$ to $x = c$. +% -----File: 061.png + +\EXAMPLE{S} + +1.\quad Obtain the following developments: +\begin{flalign*} +&\rlap{\indent(1)\quad} & 1={}&\dfrac{4}{\pi}\bigg[\sin\dfrac{\pi x}{c}+\dfrac{1}{3}\sin\dfrac{3\pi x}{c}+\dfrac{1}{5}\sin\dfrac{5\pi x}{c}+\cdots\bigg]\\ +&& &\text{from } x=0 \text{ to } x=c.\\ +&\rlap{\indent(2)\quad} & x={}&\dfrac{2c}{\pi}\bigg[\sin\dfrac{\pi x}{c}-\dfrac{1}{2}\sin\dfrac{2\pi x}{c}+\dfrac{1}{3}\sin\dfrac{3\pi x}{c}-\dfrac{1}{4}\sin\dfrac{4\pi x}{c}+\cdots\bigg]\\ +&& &\text{from } x=-c \text{ to } x=c.\\ +&& x={}&\dfrac{c}{2}-\dfrac{4c}{\pi^2}\bigg[\cos\dfrac{\pi x}{c}+\dfrac{1}{3^2}\cos\dfrac{3\pi x}{c}+\dfrac{1}{5^2}\cos\dfrac{5\pi x}{c}+\dfrac{1}{7^2}\cos\dfrac{7\pi x}{c}+\cdots\bigg]\\ +&& &\text{from } x=-c \text{ to } x=c.\\ +&\rlap{\indent(3)\quad} & x^2={}&\dfrac{2c^2}{\pi^3}\bigg[\bigg(\dfrac{\pi^2}{1}-\dfrac{4}{1^3}\bigg)\sin\dfrac{\pi x}{c}-\dfrac{\pi^2}{2}\sin\dfrac{2\pi x}{c}+\bigg(\dfrac{\pi^2}{3}-\dfrac{4}{3^2}\bigg)\sin\dfrac{3\pi x}{c}\\ +&& &\hspace{3em}-\dfrac{\pi^2}{4}\sin\dfrac{4\pi x}{c}+\bigg(\dfrac{\pi^2}{5}-\dfrac{4}{5^3}\bigg)\sin\dfrac{5\pi x}{c}+\cdots\bigg]\\ +&& &\text{from } x=0 \text{ to } x=c.\\ +&& x^2={}&\dfrac{c^2}{3}-\dfrac{4c^2}{\pi^2}\bigg[\!\cos\dfrac{\pi x}{c}-\dfrac{1}{2^2}\cos\dfrac{2\pi x}{c}+\dfrac{1}{3^2}\cos\dfrac{3\pi x}{c}-\dfrac{1}{4^2}\cos\dfrac{4\pi x}{c}+\cdots\!\bigg]\:\\ +&& &\text{from } x=-c \text{ to } x=c.\\ +&\rlap{\indent(4)\quad} & e^x={}&2\pi\bigg[\dfrac{1+e^c}{c^2+\pi^2}\sin\dfrac{\pi x}{c}+\dfrac{2(1-e^c)}{c^2+4\pi^2}\sin\dfrac{2\pi x}{c}\\ +&& &\hspace{3em}+\dfrac{3(1+e^c)}{c^2+9\pi^2}\sin\dfrac{3\pi x}{c}+\dfrac{4(1-e^c)}{c^2+16\pi^2}\sin\dfrac{4\pi x}{c}+\cdots\bigg],\\ +&& e^x={}&2c\bigg[\dfrac{1}{2}\dfrac{e^c-1}{c^2}-\dfrac{e^c+1}{c^2+\pi^2}\cos\dfrac{\pi x}{c}+\dfrac{e^c-1}{c^2+4\pi^2}\cos\dfrac{2\pi x}{c}\\ +&& &\hspace{3em}-\dfrac{e^c+1}{c^2+9\pi^2}\cos\dfrac{3\pi x}{c}+\cdots\bigg]\\ +&& &\text{from } x=0 \text{ to } x=c. +\end{flalign*} +\begin{flalign*} +&\rlap{\indent(5)\quad} & f(x)={}&\dfrac{4c}{\pi^2}\bigg[\sin\dfrac{\pi x}{c}-\dfrac{1}{3^2}\sin\dfrac{3\pi x}{c}+\dfrac{1}{5^2}\sin\dfrac{5\pi x}{c}+\cdots\bigg]\hspace{8em}\\ +&& &\text{from } x=0 \text{ to } x=c, +\end{flalign*} +where $f(x)=x$ from $x=0$ to $x=\dfrac{c}{2}$ and $f(x)=c-x$ from $x=\dfrac{c}{2}$ to +$x=c$.\\ +% -----File: 062.png + +2.\quad Show that formula (4) Art.~31 can be written +\begin{multline*} + f(x) = \frac{1}{2} c_0 \cos\beta_0 + + c_1 \cos \bigg( \frac{ \pi x}{c} - \beta_1 \bigg) + + c_2 \cos \bigg( \frac{2\pi x}{c} - \beta_2 \bigg) +\\ + + c_3 \cos \bigg( \frac{3\pi x}{c} - \beta_3 \bigg) + \cdots +\end{multline*} +\vspace{-3ex} +\begin{flalign*} +&\text{where}&& + c_m = (a_m^2 + b_m^2)^{\frac{1}{2}} \quad \text{and} \quad + \beta_m = \tan^{-1}\frac{a_m}{b_m}.&&\phantom{where} +\end{flalign*} + +3.\quad Show that formula (4) Art.~31 can be written +\begin{multline*} + f(x) = \frac{1}{2} c_0 \sin\beta_0 + + c_1 \sin \bigg( \frac{ \pi x}{c} + \beta_1 \bigg) + + c_2 \sin \bigg( \frac{2\pi x}{c} + \beta_2 \bigg) +\\ + + c_3 \sin \bigg( \frac{3\pi x}{c} + \beta_3 \bigg) + \cdots +\end{multline*} +\vspace{-3ex} +\begin{flalign*} +&\text{where}&& + c_m = (a_m^2 + b_m^2)^{\frac{1}{2}} \quad \text{and} \quad + \beta_m = \tan^{-1}\frac{a_m}{b_m}.&&\phantom{where} +\end{flalign*} + +\mypara{32.} In the formulas of Art.~31 $c$ may have as great a value as we please, +so that we can obtain a Trigonometric Series for $f(x)$ that will represent the +given function through as great an interval as we may choose to take. If, +then, we can obtain the limiting form approached by the series (4) Art.~31 as +$c$ is indefinitely increased the expression in question ought to be equal to the +given function of $x$ for all values of $x$. Equation (4) Art.~31 can be written +as follows if we replace $b_0$, $b_1$, $b_2$, $\cdots$ $a_1$, $a_2$, $\cdots$ by their values given in +Art.~31 (5) and (6). +\begin{align*} +f(x) &= \frac{1}{c} \biggl[ +\begin{aligned}[t] +& \frac{1}{2} \int\limits_{-c}^{c} f(\lambda) \,d\lambda +\\ +&+ \int\limits_{-c}^{c} f(\lambda) + \cos\frac{ \pi\lambda}{c} \cos\frac{ \pi x}{c} d\lambda + + \int\limits_{-c}^{c} f(\lambda) + \cos\frac{2\pi\lambda}{c} \cos\frac{2\pi x}{c} d\lambda + \cdots +\\ +&+ \int\limits_{-c}^{c} f(\lambda) + \sin\frac{ \pi\lambda}{c} \sin\frac{ \pi x}{c} d\lambda + + \int\limits_{-c}^{c} f(\lambda) + \sin\frac{2\pi\lambda}{c} \sin\frac{2\pi x}{c} d\lambda + \cdots \biggr] +\end{aligned} +\\ +&= \frac{1}{c} \int\limits_{-c}^{c} f(\lambda) \,d\lambda \biggl[ +\begin{aligned}[t] +& \frac{1}{2} + + \cos\frac{ \pi\lambda}{c} \cos\frac{\pi x}{c} + + \sin\frac{ \pi\lambda}{c} \sin\frac{\pi x}{c} +\\[1ex] +&+ \cos\frac{2\pi\lambda}{c} \cos\frac{2\pi x}{c} + + \sin\frac{2\pi\lambda}{c} \sin\frac{2\pi x}{c} + \cdots \biggr ] +\end{aligned} +\end{align*} +% -----File: 063.png +\begin{flalign*} + f(x) &= \frac{1}{c} \int\limits_{-c}^c f(\lambda) \,d\lambda \biggl[ + \frac{1}{2} + \cos\frac{ \pi}{c} (\lambda-x) + + \cos\frac{2\pi}{c} (\lambda-x) + \cdots \biggr] +\\ +&= \frac{1}{2c} \int\limits_{-c}^c f(\lambda) \,d\lambda \biggl[ +\begin{aligned}[t] +& 1 + \cos\frac{ \pi}{c} (\lambda-x) + + \cos\frac{2\pi}{c} (\lambda-x) + \cdots +\\ +&+ \cos \Bigl( -\frac{ \pi}{c} \Bigr) (\lambda-x) + + \cos \Bigl( -\frac{2\pi}{c} \Bigr) (\lambda-x) + \cdots \biggr] +\end{aligned} +\end{flalign*} +since $\cos(-\phi)=\cos\phi$. +\begin{flalign*} + f(x) = \frac{1}{2\pi} \int\limits_{-c}^c f(\lambda) \,d\lambda \biggl[ \cdots +&+ \frac{\pi}{c} \cos \Bigl( -\frac{2\pi}{c} \Bigr) (\lambda-x) + + \frac{\pi}{c} \cos \Bigl( -\frac{ \pi}{c} \Bigr) (\lambda-x) +\\ +&+ \frac{\pi}{c} \cos \frac{0\pi}{c} (\lambda-x) + + \frac{\pi}{c} \cos \frac{ \pi}{c} (\lambda-x) +\\ +&+ \frac{\pi}{c} \cos \frac{2\pi}{c} (\lambda-x) + \cdots \biggr] \tag{1} +\end{flalign*} + +As $c$ is indefinitely increased the limiting value approached by the +parenthesis in (1) is +\[ +\int\limits_{-\infty}^{\infty}\cos\alpha(\lambda-x).d\alpha. +\] + +Hence the limiting form approached by (1) is +\[ + f(x) += \frac{1}{2\pi} + \int\limits_{-\infty}^{\infty} f(\lambda) \,d\lambda + \int\limits_{-\infty}^{\infty} \cos\alpha(\lambda-x).d\alpha, \tag{2} +\] +and the second member of (2) must be equal to $f(x)$ for all values of $x$. + +The double integral in (2) is known as \emph{Fourier's Integral}, and since it is a +limiting form of \emph{Fourier's Series} it is subject to the same limitations as the +series. + +That is, in order that (2) should be true $f(x)$ must be finite, continuous, and +single valued for all values of $x$, or if discontinuous, must have only finite +discontinuities.\footnote{See note on page \pageref{notep38}.} + +(2) is sometimes given in a slightly different form. + +Since \quad $\displaystyle + \int\limits_{-\infty}^{\infty} \cos\alpha(\lambda-x).d\alpha += \int\limits_{-\infty}^0 \cos\alpha(\lambda-x).d\alpha ++ \int\limits_0^{\infty} \cos\alpha(\lambda-x).d\alpha$\\ +and +\begin{gather*} + \int\limits_{-\infty}^0 \cos\alpha(\lambda-x).d\alpha += \int\limits_{ \infty}^0 \cos(-\alpha)(\lambda-x).d(-\alpha) +=-\int\limits_{ \infty}^0 \cos\alpha(\lambda-x).d\alpha +\\ + \int\limits_{-\infty}^{\infty} \cos\alpha(\lambda-x).d\alpha +=2\int\limits_0^{\infty} \cos\alpha(\lambda-x).d\alpha +\end{gather*} +% -----File: 064.png +and (2) may be written +\[ +f(x) = \frac{1}{\pi} + \int\limits_{-\infty}^{\infty} f(\lambda) \,d\lambda + \int\limits_{0}^{\infty} \cos \alpha(\lambda-x).d\alpha. \tag{3} +\] + +If $f(x)$ is an \textit{even} function or an \textit{odd} function (3) can be still further simplified. + +Let \hfill $ f(x) = -f(-x)$. \hfill\phantom{\indent Let} + +Since the limits of integration in (3) do not contain $\alpha$ or $\lambda$ the integrations +may be performed in whichever order we choose. That is +\[ + \int\limits_{-\infty}^{\infty} f(\lambda) \,d\lambda + \int\limits_{0}^{\infty} \cos \alpha(\lambda-x).d\alpha += \int\limits_{0}^{\infty} d\alpha + \int\limits_{-\infty}^{\infty} f(\lambda) \cos\alpha(\lambda-x).d\lambda. +\] + +Now +\begin{gather*} + \int\limits_{-\infty}^{\infty} \!f(\lambda) \cos\alpha(\lambda-x).d\lambda += \int\limits_{-\infty}^{0} \!f(\lambda) \cos\alpha(\lambda-x).d\lambda ++ \int\limits_{0}^{\infty} \!f(\lambda) \cos\alpha(\lambda-x).d\lambda. +\\ +\begin{aligned} + \int\limits_{-\infty}^{0} f(\lambda) \cos\alpha(\lambda-x).d\lambda +&= \int\limits_{\infty}^{0} f(-\lambda) \cos\alpha(-\lambda-x).d(-\lambda) +\\ +&=-\int\limits_{0}^{\infty} f(\lambda) \cos\alpha(\lambda+x).d\lambda +\end{aligned} +\end{gather*} +and (3) becomes +\begin{align*} +f(x) +&= \frac{1}{\pi} \int\limits_{0}^{\infty} d\alpha \int\limits_{0}^{\infty} + f(\lambda) [\cos\alpha(\lambda-x) - \cos\alpha(\lambda+x).d\lambda +\\ +&= \frac{2}{\pi} \int\limits_{0}^{\infty} d\alpha \int\limits_{0}^{\infty} + f(\lambda) \sin\alpha\lambda \sin\alpha x.d\lambda +\end{align*} +\begin{flalign*} +&\text{or}&& + f(x) = \frac{2}{\pi} \int\limits_{0}^{\infty} f(\lambda)d\lambda \int\limits_{0}^{\infty} \sin\alpha\lambda \sin\alpha x.d\alpha. && \tag{4} +\end{flalign*} + +If $f(x) = f(-x)$ (3) can be reduced in like manner to +\[ +f(x) = \frac{2}{\pi} \int\limits_{0}^{\infty} f(\lambda) \,d\lambda \int\limits_{0}^{\infty} \cos\alpha\lambda \cos\alpha x.d\alpha. \tag{5} +\] + +Although (4) holds for all values of $x$ only in case $f(x)$ is an \textit{odd} function, +and (5) only in case $f(x)$ is an \textit{even}, function, both (4) and (5) hold for all +\textit{positive} values of $x$ in the case of any function. + +\EXAMPLE{} + +(1)\quad Obtain formulas (4) and (5) directly from (7) and (9) Art.~31. +\label{ch2end} + +% -----File: 065.png + +\mychap{CHAPTER III.}{CONVERGENCE OF FOURIER'S SERIES.} +\label{ch3start} + +\mypara{33.} The question of the \textit{convergence} of a Fourier's Series is altogether too +large to be completely handled in an elementary treatise. We will, however, +consider at some length one of the most important of the series we have +obtained, namely +\[ + \frac{4}{\pi} \biggl[ \sin x + \frac{\sin 3x}{3} + + \frac{\sin 5x}{5 } + \frac{\sin 7x}{7} + \cdots\biggr], +\tag*{[v.~(3) Art.~26(\emph{b}).]} +\] +and prove that for all values of $x$ between zero and $\pi$ its sum is absolutely +equal to unity; that is, that the limit approached by the sum of $n$ terms of the +series +\[ + \frac{2}{\pi} \biggl[ + \sin x \int\limits_{0}^{\pi} \sin \alpha.d\alpha + + \sin 2x \int\limits_{0}^{\pi} \sin 2\alpha.d\alpha + + \sin 3x \int\limits_{0}^{\pi} \sin 3\alpha.d\alpha + \cdots \biggr ], +\] +as $n$ is indefinitely increased, is $1$, provided that $x$ lies between zero and $\pi$. + +Let +\begin{align*} +S_n = \frac{2}{\pi}\biggl[ + & \sin x \int\limits_0^{\pi} \sin \alpha.d\alpha + + \sin 2x \int\limits_0^{\pi} \sin 2\alpha.d\alpha + + \sin 3x \int\limits_0^{\pi} \sin 3\alpha.d\alpha + \cdots +\\ + &+ \sin nx \int\limits_0^{\pi} \sin n\alpha.d\alpha \biggr ]. \tag{1} +\end{align*} + +Then +\begin{align*} + S_{n} &= \frac{2}{\pi} \int\limits_0^{\pi} [ + \sin\alpha\sin x + \sin 2\alpha\sin 2x + \sin 3\alpha\sin 3x + + \cdots + \sin n\alpha\sin nx ] d\alpha +\\ +&= \frac{1}{\pi} \int\limits_0^{\pi} +\begin{aligned}[t] +[ & \cos (\alpha-x) - \cos (\alpha+x) + + \cos 2(\alpha-x) - \cos 2(\alpha+x) + \cdots +\\[3ex] + &+\cos n(\alpha-x) - \cos n(\alpha+x) ] \,d\alpha +\end{aligned} +\\ +&= \frac{1}{\pi} \int\limits_0^{\pi} + [ \cos (\alpha-x) + \cos 2(\alpha-x) + \cos 3(\alpha-x) + \cdots + + \cos n(\alpha-x) ] d\alpha +\\ +&-\frac{1}{\pi} \int\limits_0^{\pi} + [ \cos (\alpha+x) + \cos 2(\alpha+x) + \cos 3(\alpha+x) + \cdots + + \cos n(\alpha+x) ] d\alpha. +\end{align*} +% -----File: 066.png + +Therefore by Art.~20 (1) +\begin{gather*} +\begin{aligned} +S_n ={}& \frac{1}{\pi} \int\limits_{0}^{\pi} \biggl[ -\frac{1}{2} + + \frac{1}{2} \frac{ \sin (2n+1)\dfrac{\alpha-x}{2} } + { \sin \dfrac{\alpha-x}{2} } \biggr] d\alpha +\\ +&- \frac{1}{\pi} \int\limits_{0}^{\pi} \biggl[ -\frac{1}{2} + + \frac{1}{2} \frac{ \sin (2n+1)\dfrac{\alpha+x}{2} } + { \sin \dfrac{\alpha+x}{2} } \biggr] d\alpha. +\end{aligned} +\\ +S_n = \frac{1}{2\pi} \int\limits_{0}^{\pi} + \frac{ \sin (2n+1)\dfrac{\alpha-x}{2} } + { \sin \dfrac{\alpha-x}{2} } d\alpha + - \frac{1}{2\pi} \int\limits_{0}^{\pi} + \frac{ \sin (2n+1)\dfrac{\alpha+x}{2} } + { \sin \dfrac{\alpha+x}{2} } d\alpha. +\end{gather*} + +In the first integral substitute $\beta$ for $\dfrac{\alpha-x}{2}$, and in the second integral substitute +$\beta$ for $\dfrac{\alpha+x}{2}$. + +We get +\[ +S_n = \frac{1}{\pi} \int\limits_{-\frac{x}{2}}^{\frac{\pi}{2}-\frac{x}{2}} + \frac{\sin(2n+1)\beta }{ \sin\beta } \,d\beta + - \frac{1}{\pi} \int\limits_{ \frac{x}{2}}^{\frac{\pi}{2}+\frac{x}{2}} + \frac{\sin(2n+1)\beta }{ \sin\beta } \,d\beta . \tag{2} +\] + +It remains to find the limit approached by $S_{n}$ as $n$ is indefinitely increased. + +\markright{CONVERGENCE OF FOURIER'S SERIES.} +\mypara{34.}\vspace{-2ex} +\[ +\int\limits_0^{\frac{\pi}{2}} \frac{ \sin(2n+1)\beta}{ \sin\beta} d\beta = \frac{\pi}{2}. \tag{1} +\] + +For +\begin{flalign*} +&\phantom{fr}& \frac{\sin(2n+1)\beta }{ 2 \sin\beta} = \tfrac{1}{2 }+ \cos 2\beta + \cos 4\beta &+ \cdots + \cos 2n\beta, && \text{by Art.~20.} +\\ +&\text{and }& \int\limits_0^{\frac{\pi}{2}} \cos 2k\beta.d\beta &= 0.&& +\end{flalign*} + +Let us construct the curve +\[ +y = \frac{\sin(2n+1)x}{\sin x}. +\] + +We have only to draw the curve $y = \sin (2n + 1)x$ and then to divide +the length of each ordinate by the value of the sine of the corresponding +abscissa. + +In $y = \sin (2n + 1)x$ the successive arches into which the curve is +divided by the axis of $X$ are equal, and consequently their areas are equal. +% -----File: 067.png + +%[Illustration] +\pngrightl{005.png}{476}{28}{-12} + +Each arch has for its altitude unity and for its base + $\dfrac{\pi}{2n+1}$ and is symmetrical with respect to the ordinate +of its highest or lowest point. + +If now we form the curve $y = \dfrac{\sin(2n+1)x }{\sin x}$ from +the curve $y= \sin(2n + 1)x$, it is clear that, since +$\sin x$ increases as $x$ increases from $0$ to $\dfrac{\pi}{2}$, the ordinate +of any point of the new curve will be shorter +than the ordinate of the corresponding point in the +preceding arch, and that consequently the area of +each arch $y=\dfrac{\sin(2n + 1)x}{\sin x}$ will be less than +that of the arch before it. + +If $a_{0}$, $a_{1}$, $a_{2}$, $\cdots a_{n-1}$ are the areas of the successive +arches and $a_n$ that of the incomplete arch terminated +by the ordinate corresponding to $x =\dfrac{\pi}{2}$ +\[ +\int\limits_0^{\frac{\pi}{2}} \frac{\sin(2n+1)x}{\sin x} \,dx = a_0-a_1+a_2-a_3+\cdots . +\] +But +\[ + \int\limits_0^{\frac{\pi}{2}} \frac{\sin(2n+1)x }{\sin x } \,dx += \int\limits_0^{\frac{\pi}{2}} \frac{\sin(2n+1)\beta}{\sin \beta} \,d\beta += \frac{\pi}{2} \tag*{by (1).} +\] + +Hence +\[ +\frac{\pi}{2}= a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - \cdots + a_{n} \quad \text{if $n$ is even,} +\] +or +\[ +\frac{\pi}{2}= a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - \cdots - a_{n} \quad \text{if $n$ is odd.} +\] + +These equations can be written +\begin{align*} +\frac{\pi}{2} ={}& a_{0} + (-a_{1} + a_{2}) + (-a_{3} + a_{4})\\ + &+ (-a_{5} + a_{6}) + \cdots +(-a_{n-1} + a_{n}) +\end{align*} +if $n$ is even, and +\begin{align*} +\frac{\pi}{2} ={}& a_{0} + (-a_{1} + a_{2}) + (-a_{3} + a_{4})\\ + &+ (-a_{5} + a{_6}) + \cdots + (-a_{n-2}+a_{n-1})+(-a_{n}) +\end{align*} +if $n$ is odd. +% -----File: 068.png + +In either case each parenthesis is a negative quantity since +\[ + a_0 > a_1 > a_2 > a_3 \cdots > a_n, +\] +and it follows that $a_0$ is greater than $\dfrac{\pi}{2}$. + +Again +\begin{align*} +&\frac{\pi}{2} = a_{0} - a_{1} + (a_2-a_3) + (a_4-a_5) + \cdots + (a_{n-2}-a_{n-1}) + a_n +\intertext{if $n$ is even and} +&\frac{\pi}{2} = a_{0} - a_{1} + (a_2-a_3) + (a_4-a_5) + \cdots + (a_{n-1}-a_{n}) +\end{align*} +if $n$ is odd. + +In either case each parenthesis is positive and it follows that $a_0-a_1$ is +is less than $\dfrac{\pi}{2}$. + +Since +\[ + a_0 > \frac{\pi}{2} > a_0 - a_1, +\] +$a_{0}$ and $a_{0} - a_{1}$ differ from $\dfrac{\pi}{2}$ by less than they differ from each other, that +is, by less than $a_1$. + +In like manner we can show that $a_0-a_1$ and $a_0-a_1+a_2$ differ from +$\dfrac{\pi}{2}$ by less than $a_2$; and in general that $a_0-a_1+a_2-a_3+\cdots \pm a_k$ differs +from $\dfrac{\pi}{2}$\vphantom{\fbox{$\dfrac00$}} +%[**F2: Spacer added to keep the dfrac from colliding with the line above.] +by less than $a_k$; or even that +\[ + a_0 - a_1 + a_2 - a_3 + \cdots \pm \frac{a_k}{p} +\] +differs from $\dfrac{\pi}{2}$ by less than $a_k$ no matter the value of $p$, provided $p$ is +greater than unity. + +\mypara{35.} From what has been proved in the last article it follows that +\[ + \int\limits_0^b \frac{\sin(2n+1)x}{\sin x} \,dx, +\] +where $b$ is some value between $\dfrac{\pi}{2n+1}$ and $\dfrac{\pi}{2}$, differs from $\dfrac{\pi}{2}$ by less than +the area of the arch in which the ordinate of $y= \dfrac{\sin(2n+1)x}{\sin x}$ corresponding +to $x = b$ falls if this ordinate divides an arch, or by less than the area +of the arch next beyond the point $(b, 0)$ if the curve crosses the axis of $X$ at +that point. +% -----File: 069.png + +The area of the arch in question is less than $\dfrac{\pi}{2n+1}$, its base, multiplied by +$\dfrac{1}{ \sin\Big( b-\dfrac{\pi}{ 2n+1} \Big)}$, a value greater than the length of its longest ordinate. + +Therefore \hfill $\displaystyle + \int\limits_0^b \frac{\sin(2n+1)x }{ \sin x } \,dx $\hfill\hfill\quad\\ +differs from $\dfrac{\pi}{2}$ by less than $\dfrac{\pi}{2n+1} \dfrac{ 1}{\sin\Big( b-\dfrac{\pi}{ 2n+1} \Big)}$ . + +If now $n$ is indefinitely increased $\dfrac{\pi}{2n+1} \dfrac{ 1}{ \sin\Big( b-\dfrac{\pi}{2n+1} \Big)}$ approaches +zero as its limit, and we get the very important result +\begin{flalign*} +&& \limit_{n=\infty} \biggl[\int\limits_0^b \frac{\sin(2n+1)x }{ \sin x } \,dx \biggr] += \frac{\pi}{2} && \phantom{\text{if } 0<b< \dfrac{\pi}{2}}\tag{1} +\\ +&\text{if } 0<b< \dfrac{\pi}{2}.\\ +\end{flalign*} + +\mypara{36.}\vspace{-7.9ex} +\begin{flalign*} +&&S_n ={} +& \frac{1}{\pi} \int\limits_{-\frac{x}{2}}^{\frac{\pi}{2}-\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta +- \frac{1}{\pi} \int\limits_{ \frac{x}{2}}^{\frac{\pi}{2}+\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta. +\hspace{6pt}\text{[Art.~33. (2)]}\\[1ex] +&&={}& \frac{1}{\pi} \int\limits_{-\frac{x}{2}}^0 + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta + + \frac{1}{\pi} \int\limits_0^{\frac{\pi}{2}-\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta +\\[1ex] +&&&- \frac{1}{\pi} \int\limits_0^{\frac{\pi}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta + + \frac{1}{\pi} \int\limits_0^{\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta +\\[1ex] +&&&- \frac{1}{\pi} \int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}+\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta. +\end{flalign*} + +This last value for $S_n$ can be somewhat simplified. + +Substituting $\gamma = -\beta$ we get +\[ + \int\limits_{-\frac{x}{2}}^0 \frac{ \sin(2n+1)\beta }{\sin\beta } \,d\beta +=-\int\limits_{ \frac{x}{2}}^0 \frac{ \sin(2n+1)\gamma}{\sin\gamma} \,d\gamma += \int\limits_0^{\frac{x}{2}} \frac{\sin(2n+1)\beta }{\sin\beta } \,d\beta. +\] +% -----File: 070.png + +Substituting $\gamma = \pi - \beta$ in +\[ +\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}+\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta \tag*{we have} +\] +\begin{flalign*} && + \int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}+\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta +&= -\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}-\frac{x}{2}} + \frac{\sin(2n+1)\gamma}{ \sin\gamma} \,d\gamma + = \int\limits_{\frac{\pi}{2}-\frac{x}{2}}^{\frac{\pi}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta && +\\[1ex] +&&&= \int\limits_{0}^{\frac{\pi}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta + - \int\limits_{0}^{\frac{\pi}{2}-\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta. && +\end{flalign*} + +Hence +\[ +S_n = \frac{2}{\pi} \int\limits_{0}^{\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta + + \frac{2}{\pi} \int\limits_{0}^{\frac{\pi}{2}-\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta + - \frac{2}{\pi} \int\limits_{0}^{\frac{\pi}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta. +\] +\[ + \int\limits_{0}^{\frac{\pi}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} d\beta = \frac{\pi}{2 } +\tag*{by (1) Art.~34.} +\] +\begin{flalign*} && + \limit_{n=\infty} \biggl[ \int\limits_{0}^{\frac{x}{2}} + \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta \biggr] +&= \frac{\pi}{2 } \quad \text{if} \quad 0<x<\pi &&\text{by (1) Art.~35} +\\ +\intertext{and} +&& \limit_{n=\infty} \biggl[ \int\limits_0^{\frac{\pi}{2}-\frac{x}{2}} + \frac{ \sin(2n+1)\beta}{ \sin\beta} \,d\beta \biggr] +&= \frac{\pi}{2} \quad \text{if} \quad 0 < x < \pi &&\text{by (1) Art.~35.} +\end{flalign*} +Therefore \hfill $\displaystyle\limit_{n=\infty} [S_n] = 1 + 1 - 1 = 1$ \quad if \quad $0 < x < \pi$ \hfill and +\[ + \frac{4}{\pi} \left[ \sin x + \frac{\sin 3x}{ 3} ++ \frac{\sin 5x}{ 5} + \frac{\sin 7x}{ 7} + \cdots \right] = 1 +\] +for all values of $x$ between zero and $ \pi$. + +\mypara{37.} By a somewhat long but not especially difficult extension of the reasoning +just given it can be shown that if $f(x)$ is \textit{single-valued} and \textit{finite} +between $x = -\pi$ and $x = \pi$, and has only a \textit{finite number} of \textit{discontinuities} +and of \textit{maxima} and \textit{minima} between $x = -\pi$ and $x = \pi$ the +\textit{Fourier's Series} +\begin{flalign*} +&& \frac{1}{2} b_0 &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots\\ +&& &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots\\ +% -----File: 071.png +&\text{where}& &\quad a_m = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \sin m\alpha.d\alpha &&\phantom{where}\\ +&\text{and}& &\quad b_m = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \cos m\alpha.d\alpha, && +\end{flalign*} +\textit{and that Fourier's Series only} is equal to $f(x)$ for all values of $x$ between + $x = -\pi$ and $x = \pi$, \textit{excepting the values of $x$ corresponding to the +discontinuities} of $f(x)$, and the values $\pi$ and $-\pi$ if $f(\pi)$ is not equal to $f(-\pi)$; +and that if $c$ is a value of $x$ corresponding to a discontinuity of $f(x)$, the value +of the series when $x = c$ is +\[ +\frac{1}{2} \limit_{\epsilon \doteq 0}[f(c-\epsilon) + f(c+\epsilon)]; +\] +and that if $f(\pi)$ is not equal to $f(-\pi)$ the value of the series when $x = -\pi$ +and when $x = \pi$ is +\[ + \frac{1}{2} [f(-\pi) + f(\pi)]. +\] + +If $f(x)$ while satisfying the conditions named in the preceding paragraph +except for a finite number of values of $x$, becomes infinite for those values, the +series is equal to the function except for the values of $x$ in question provided +that $\displaystyle\int\limits_{-\pi}^{\pi} f(x)\,dx$ is finite and determinate. \quad (v.\ Int.\ Cal.\ Arts.~83 and 84.) + +\mypara{38.} The question of the convergency of a Fourier's Series and the conditions +under which a function may be developed in such a series was first +attacked successfully by Dirichlet in 1829, and his conclusions have been +criticised and extended by later mathematicians, notably by Riemann, Heine, +Lipschitz, and du Bois Reymond. It may be noted that the criticisms relate +not to the sufficiency but to the necessity of Dirichlet's conditions. + +An excellent résumé of the literature of the subject is given by Arnold +Sachse in a short dissertation published by Gauthier--Villars, Paris, 1880, +entitled ``Essai Historique sur la Représentation d'une Fonction Arbitraire +d'une seule variable par une Série Trigonométrique." + +\mypara{39.} A good deal of light is thrown on the peculiarities of trigonometric +series by the attempt to construct approximately the curves corresponding to them. + +If we construct $y = a_1 \sin x$ and $y = a_2 \sin 2x$ and add the ordinates +of the points having the same abscissas we shall obtain points on the curve +% -----File: 072.png +\[ +y = a_1 \sin x + a_2 \sin 2x. +\] + +If now we construct $y = a_3 \sin 3x$ and add the ordinates to those of + $y = a_1 \sin x + a_2 \sin 2x$ we shall get the curve +\[ +y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x. +\] +By continuing this process we get successive approximations to +\[ +y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + a_4 \sin 4x + \cdots +\] +Let us apply this method to a few of the series which we have obtained in +Chapter~II. + +Take +\begin{align*} + y &= \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots \tag{1} +\\ + &= 0 \text{ when } x=0,\ + \frac{\pi}{4} \text{ from } x=0 \text{ to } x=\pi, \text{ and } + 0 \text{ when } x=\pi, +\\ +\intertext{v.\ Art.~26 [\emph{b}](3).} + y &= 2\left(\sin x - \frac{1}{2} \sin 2x + \frac{1}{3} \sin 3x + - \frac{1}{4} \sin 4x + \cdots \right) \tag{2} +\\ + &= x \text{ from } x=0 \text{ to } x=\pi, \text{ and } + 0 \text{ when } x=\pi, +\\ +\intertext{Art.~26 [\emph{a}](4).} + y &= \frac{4}{\pi}\left[\frac{1}{1^2} \sin x - \frac{1}{3^2}\sin 3x + + \frac{1}{5^2}\sin 5x - \frac{1}{7^2} \sin 7x + \cdots \right] \tag{3} +\\ + &= x \text{ from } x=0 \text{ to } x=\frac{\pi}{2}, \text{ and } + \pi-x \text{ from } x=\frac{\pi}{2} \text{ to } x=\pi, +\intertext{Art.~26 [\emph{c}](2).} + y &= \frac{1}{1} \sin x + \frac{2}{2} \sin 2x + \frac{1}{3}\sin 3x + + \frac{1}{5} \sin 5x - \frac{2}{6} \sin 6x + \frac{1}{7}\sin 7x + \cdots \tag{4} +\\ + &= 0 \text{ when } x=0, \ + \frac{\pi}{2} \text{ from } x=0 \text{ to } x=\frac{\pi}{2}, \text{ and } + 0 \text{ from } x=\frac{\pi}{2} \text{ to } x=\pi, +\end{align*} +v.\ Art.~26 [\emph{d}](2). + +It must be borne in mind that each of these curves is periodic having the +period $2\pi$, and is symmetrical with respect to the origin. + +The following figures I, II, III, and IV represent the first four approximations +to each of these curves. + +In each figure the curve $y =$ the series, and the approximation in question +are drawn in continuous lines, and the preceding approximation and the +curve corresponding to the term to be added are drawn in dotted lines. + +%[Illustration +\pngcent{006.png}{1466}%{2170} + +\label{was63} + +Figs.~I, II, III, and IV immediately suggest the following facts: + +(\emph{a}) The curve representing each approximation is continuous even when +the curve representing the series is discontinuous. + +(\emph{b}) When the curve representing the series is discontinuous the portion of +each successive approximate curve in the neighborhood of the point whose +abscissa is a value of $x$ for which the series curve is discontinuous approaches +more and more nearly a straight line perpendicular to the axis of $X$ and connecting +the separate portions of the series curve. + +%[Illustration] +\pngcent{007.png}{1466} + +\label{was64} + +(\emph{c}) The curves representing successive approximations do not necessarily +tend to lose their wavy character, since each is obtained from the preceding +one by superposing upon it a wave line whose waves are shorter each time but +do not necessarily lose their sharpness of pitch. This is the case in Figures +I, II, and IV\@. In Fig.~III the waves of the superposed curves grow rapidly +flatter. + +It follows from this that in such cases as those represented in Figures I, II, +and IV the direction of the approximate curve at a point having a given +abscissa does not in general approach the direction of the series curve at the +corresponding point, or indeed, approach any limiting value, as the approximation +is made closer and closer; and that the length of any portion of the +approximate curve will not in general approach the length of the corresponding +portion of the series curve. + +% -----File: 073.png + +Analytically this amounts to saying that the derivative of a function of $x$ +cannot in general be obtained by differentiating term by term the Fourier's +Series which represents the function. + +(\emph{d}) The area bounded by a given ordinate, the approximate curve, the axis of +$X$, and any second ordinate will approach as its limit the corresponding area of +the series curve if the series curve is continuous between the ordinates in +question; and will approach the area bounded by the given ordinate, the series +curve, the axis of $X$, any second ordinate, and a line perpendicular to the axis +of $X$, and joining the separate portions of the series curve if the latter has a +discontinuity between the ordinates in question. + +% -----File: 074.png + +% -----File: 075.png + +Analytically this amounts to saying that the Fourier's Series corresponding +to any given function can be integrated term by term and the resulting series +will represent the integral of the function even when the function is +discontinuous (v.\ Int.\ Cal.\ Art.~83). + +We may note in passing that if the function curve is continuous a curve +representing the integral of the function will be continuous and will not +change its direction abruptly at any point; while if the function curve is discontinuous +the curve representing the integral will still be continuous but will +change its direction abruptly at points corresponding to the discontinuities of +the given function. +% -----File: 076.png + +\mypara{40.} The facts that the derivative of a Fourier's Series cannot in general be +obtained by differentiating the series term by term and that its integral can be +obtained by integrating the series term by term are so important that it is +worth while to look at the matter a little more closely. Let us consider the +differentiation of the series represented in Art.~39 Figure~I. + +Let +\[ +S_n = \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots + \frac{1}{2n+1} \sin (2n + 1)x. +\] +\begin{flalign*} +&\text{\indent Then}&\frac{dS_n}{dx} = \cos x + \cos 3x + \cos 5x + \cdots + \cos (2n + 1)x.&&\phantom{\indent Then} +\end{flalign*} + +If $x = \dfrac{\pi}{2}$ +\[ +\frac{dS_n}{dx} = 0 +\] +and the curve is parallel to the axis of $X$ for $x = \dfrac{\pi}{2}$ no matter what the +value of $n$. + +If $x = 0$ or $x = \pi$ +\[ +\frac{dS_n}{dx} = 1 + 1 + 1 + 1 + \cdots + 1 = n + 1 +\] +and the curve $y = S_n$ becomes more nearly perpendicular to the axis of $X$ +at the origin and for $x = \pi$ as we increase $n$. + +If $x = \dfrac{\pi}{3}$ +\begin{flalign*} +&& \frac{dS_n}{dx} +&= \frac{1}{2} - 1 + \frac{1}{2} + \frac{1}{2} - 1 + \frac{1}{2} + \cdots \\ +&\text{That is }& \frac{dS_n}{dx}& \! +\begin{alignedat}[t]{6} +&= &&\frac{1}{2} &\ \ &\text{ if } &\ \ n&=0 &\ \ &\text{ or } &\ n&=3k \\ +&= -&&\frac{1}{2} & &\text{ `` } & n&=1 & &\text{ `` } & n&=3k+1 \\ +&= &&\,0 & &\text{ `` } & n&=2 & &\text{ `` } & n&=3k+2. +\end{alignedat} &&\phantom{That\ is} +\end{flalign*} +Consequently when $x = \dfrac{\pi}{3}$ +$\dfrac{dS_n}{dx}$ does not approach any limiting value as $n$ is +indefinitely increased. Indeed, in the successive approximations the point +whose abscissa is $\dfrac{\pi}{3}$ is successively on the rear, on the front, and on the crest +or in the trough of a wave, and although the waves are getting smaller they +do +not lose their sharpness of pitch. + +If $x$ has any other value between $0$ and $\pi$ \ $\dfrac{dS_n}{dx}$ will change abruptly as $n$ is +changed and will not approach any limiting value as $n$ is increased. +% -----File: 077.png + +\mypara{41.} In general if we differentiate a Fourier's Series +\begin{flalign*} + S = \frac{1}{2} b_0 + &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots \\ + &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots\\[-2ex] +\intertext{we get}\\[-6ex] +&\! +\begin{alignedat}{4} + &- b_1 \sin x &&- 2b_2 \sin 2x &&- 3b_3 \sin 3x &&- \cdots \\[1ex] + &+ a_1 \cos x &&+ 2a_2 \cos 2x &&+ 3a_3 \cos 3x &&+ \cdots. +\end{alignedat} +\intertext{Differentiate again and we get} + &- b_1 \cos x - 2^2b_2 \cos 2x - 3^2b_3 \cos 3x - \cdots \\ + &- a_1 \sin x - 2^2a_2 \sin 2x - 3^2b_3 \sin 3x \,- \cdots . +\end{flalign*} + +We see that each time we differentiate we multiply the coefficient of $\sin kx$ +and of $\cos kx$ by $k$ while the term still involves $\cos kx$ or $\sin kx$. + +Since the series +\begin{alignat*}{3} + \cos x &+ \cos 2x &&+ \cos 3x &&+ \cdots \\ ++ \sin x &+ \sin 2x &&+ \sin 3x &&+ \cdots +\end{alignat*} +is not convergent, and a Fourier's Series converges only because its coefficients +decrease as we advance in the series, the differentiation of a Fourier's Series +must make its convergence less rapid if it does not actually destroy it, and +repetitions of the process will usually eventually make the derived series +diverge. + +It is to be observed that the derived series are Fourier's Series, but of somewhat +special form, that is they lack the constant term. (v.\ Art.~30.) + +If now we integrate a Fourier's Series +\begin{align*} +\frac{1}{2} b_0 &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots \\ + &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots \\[-6ex] +\end{align*} +\begin{flalign*} +&\text{we get } && +\begin{alignedat}[t]{4} + C + \frac{1}{2} b_0x + &+ b_1 \sin x &&+ \frac{1}{2} b_2 \sin 2x &&+ \frac{1}{3} b_3 \sin 3x &&+ \cdots\\ + &- a_1 \cos x &&- \frac{1}{2} a_2 \cos 2x &&- \frac{1}{3} a_3 \cos 3x &&- \cdots, +\end{alignedat} +&&\phantom{we\ get} +\end{flalign*} +a Trigonometric Series which converges more rapidly than the given series. + +It is to be observed that the series obtained by integrating a Fourier's +Series is not in general a Fourier's Series owing to the presence of the term +$\frac{1}{2} b_0x$. (v.\ Art.~30.) + +\mypara{42.} We are now ready to consider the conditions under which a function of +$x$ can be developed into a Fourier's Series whose term by term derivative shall +be equal to the derivative of the function. +% -----File: 078.png + +Let the function $f(x)$ satisfy the conditions stated in Art.~37. Then there +is one Fourier's Series and but one which is equal to it. Call this series $S$. + +Let the derivative $f'(x)$\footnote + {We shall regularly use the notation $f'(x)$ for $\dfrac{df(x)}{dx}$. v.\ Dif.\ Cal.\ Art.~124.} + of the given function also satisfy the conditions +stated in Art.~37. Then $f'(x)$ can be expressed as a Fourier's Series. By Art.~39 +($d$) the integral of this latter series will be equal to the integral of $f'(x)$, +that is to $f(x)$ plus a constant, and one integral will be equal to $f(x)$. + +If this integral which is necessarily a Trigonometric Series is a Fourier's +Series it must be identical with $S$. It will be a Fourier's Series only in case +the Fourier's Series for $f'(x)$ lacks the constant term $\frac{1}{2}b_0$. +\begin{flalign*} +&\indent\text{But}& b_0 = \frac{1}{\pi} \int\limits_{-\pi}^\pi & f'(x)dx &\text{by (3) Art.~30.}&\\ +&\text{Therefore }& b_0 = \frac{1}{\pi}[f(\pi) & - f(-\pi)]; && +\end{flalign*} +and will be zero if $f(\pi)=f(-\pi)$. + +In order that $f'(x)$ shall satisfy the conditions stated in Art.~37 $f(x)$ while +satisfying the same conditions must in addition be finite and continuous +between $x=-\pi$ and $x=\pi$. + +If, then, $f(x)$ is \emph{single-valued, finite, and continuous, and has only a finite +number of maxima and minima}, between $x=-\pi$ and $x=\pi$, (the values + $x=-\pi$ and $x=\pi$ being included), \emph{and if} $f(\pi)=f(-\pi)$ $f(x)$ can be +developed into a Fourier's Series whose term by term derivative will be equal +to the derivative of the function. + +It will be observed that in this case the periodic curve $y=S$ is continuous +throughout its whole extent. + +\mypara{43.} Since a Fourier's Integral is a limiting case of a Fourier's Series the +conclusions stated in this chapter hold, \emph{mutatis mutandis} for a Fourier's +Integral. + +For example if a function of $x$ is finite and single-valued for all values of $x$ +and has not an infinite number of discontinuities or of maxima and minima in +the neighborhood of any value of $x$ it will be equal to the Fourier's Integral +\[ +\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x).d\lambda +\] +and to that Fourier's Integral only, and the integral with respect to $x$ of this +Fourier's Integral will be equal to $\displaystyle\int f(x)dx$. + +If in addition $f(x)$ is finite and continuous for all values of $x$ the derivative +of the Fourier's Integral with respect to $x$ will be equal to $\dfrac{df(x)}{dx}$. +\label{ch3end} + +% -----File: 079.png + +\mychap{CHAPTER IV.}{SOLUTION OF PROBLEMS IN PHYSICS BY THE AID OF FOURIER'S INTEGRALS AND FOURIER'S SERIES.} +\label{ch4start} + +\mypara{44.} In Art.~7 we have already considered at some length a problem in +Heat Conduction which required the use of a Fourier's Series. We shall begin +the present chapter with a problem closely analogous in its treatment to that +of Art.~7, but calling for the use of a Fourier's Integral. + +Suppose that electricity is flowing in a thin plane sheet of infinite extent +and that the value of the potential function is given for every point in some +straight line in the sheet, required the value of the potential function at any +point of the sheet. + +Let us take the line as the axis of $X$ and consider at first only those points +for which $y$ is positive: + +We have, then, to satisfy the equation +\[ + D_x^2 V + D_y^2 V = 0 \tag{1} +\] +subject to the conditions +\begin{flalign*} +&\quad& +\begin{aligned} + V &= 0 &&\text{when } & y &= \infty \\ + V &= f(x) &&\text{\quad ``} & y &= 0 +\end{aligned} +&& +\begin{aligned} (2)& \\ (3)& \end{aligned} +\end{flalign*} +where $f(x)$ is a given function, and we are not concerned with negative +values of $y$. + +As in Art.~7 we have $e^{-\alpha y} \sin \alpha x$ and $e^{-\alpha y} \cos \alpha x$ as particular values of $V$ +which satisfy (1) and (2). We must multiply them by constant coefficients +and so combine them as to satisfy condition (3). + +By (3) Art.~32 +\[ +f(x) = \frac{1}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} + f(\lambda) \cos \alpha(\lambda - x).d\lambda. \tag{4} +\] +We wish to build up a value of $V$ which will reduce to (4) when $y = 0$. +This requires a little care but not much ingenuity. +% -----File: 080.png + +Take $e^{-\alpha y} \cos \alpha x$ and $e^{-\alpha y} \sin \alpha x$ and multiply the first by $\cos {\alpha\lambda}$, and +the second by $\sin {\alpha\lambda}$; they are still values of $V$ which satisfy (1). Add +these and we get +\[ +e^{-\alpha y}\cos \alpha(\lambda- x), +\] +still a value of $V$ which satisfies (1), no matter what the values of $\alpha$ and $\lambda$. +Multiply by $f(\lambda)d\lambda$ and we have +\[ +e^{-\alpha y}f(\lambda)\cos \alpha(\lambda - x).d\lambda \tag{5} +\] +as a value of $V$ which satisfies (1). +\[ +V= \int\limits_{-\infty}^{\infty} e^{-\alpha y}f(\lambda) \cos \alpha(\lambda - x).d\lambda \tag{6} +\] +is still a solution of (1) since it is the limit of the sum of terms covered by +the form (5); and finally +\[ + V = \frac{1}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} e^{-\alpha y}f(\lambda) \cos \alpha(\lambda - x).d\lambda \tag{7} +\] +is a solution of (1) as it is $\dfrac{1}{\pi}$ multiplied by the limit of the sum of terms +formed by multiplying the second member of (6) by $d\alpha$ and giving different +values to $\alpha$. + +\markright{SOLUTION OF PROBLEMS IN PHYSICS.} +But (7) must be our required solution since while it satisfies (1) and (2), it +reduces to (4) when $y = 0$ and therefore satisfies condition (3). + +If $f(x)$ is an \textit{even} function we can reduce (7) to the form +\[ + V = \frac{2}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{0}^{\infty} e^{-\alpha y} f(\lambda) \cos \alpha x \cos \alpha\lambda.d\lambda \tag{8} +\] +and if $f(x)$ is an \textit{odd} function to the form +\[ + V = \frac{2}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{0}^{\infty} e^{-\alpha y} f(\lambda) \sin \alpha x \sin \alpha\lambda.d\lambda. \tag{9} +\] + +(7), (8), and (9) are valid only for positive values of $y$, but as the problem is +obviously symmetrical with respect to the axis of $X$, (7), (8), and (9) enable +us to get the value of the potential function at any point of the plane. + +\EXAMPLE{S} + +1.\quad Obtain forms (8) and (9) directly by the aid of (5) and (4) Art.~32.\\ + +2.\quad State a problem in statical electricity of which the solution given in +Art.~44 is the solution. +% -----File: 081.png + +\mypara{45.} As a special case under Art.~44 let us consider the problem:---To find +the value of the potential function at any point of a thin plane sheet of infinite +extent where all points of a given line which lie to the left of the origin are +kept at potential zero, and all points which lie to the right of the origin are +kept at potential unity. + +Here $f(x)=0$ if $x<0$ and $f(x)=1$ if $x>0$. + +(7) Art.~44 gives us the required solution. It is +\[ +V=\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_0^\infty e^{-\alpha y}\cos\alpha(\lambda-x).d\lambda; \tag{1} +\] +but this can be much simplified. + +We have +\begin{flalign*} +&& V= {}& \frac{1}{\pi}\int\limits_0^\infty d\lambda\int\limits_0^\infty e^{-\alpha y}\cos\alpha(\lambda-x).d\alpha. && +\\ +&\indent\text{Now}& &\int\limits_0^\infty e^{-ax}\cos mx.dx = \frac{a}{a^2+m^2} &&\phantom{\indent Now} +\end{flalign*} +if $a>0$. (Int.\ Cal.\ Art.~82, Ex.~8.) +\begin{flalign*} +&\indent\text{Hence} & \int\limits_0^\infty e^{-\alpha y} \cos\alpha(\lambda-x).d\alpha = \frac{y}{y^2+(\lambda-x)^2}, &&\phantom{\indent Hence} \\ +&\text{and} +& V = \frac{1}{\pi} \int\limits_0^\infty \frac{y\,d\lambda}{y^2+(\lambda-x)^2} += \frac{1}{\pi} \left( \frac{\pi}{2} + \tan^{-1}\frac{x}{y} \right). && +\end{flalign*} +\[ + \tan \left( \frac{\pi}{2} - \tan^{-1}\frac{x}{y} \right) += \ctn \left( \tan^{-1}\frac{x}{y} \right) = \frac{y}{x}; +\] +and consequently +\[ +V=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan^{-1}\frac{x}{y}\right) +=1-\frac{1}{\pi}\tan^{-1}\frac{y}{x}. \tag{2} +\] + +Since \hfill $\log z=\log(x+yi)=\dfrac{1}{2}\log(x^2+y^2)+i\tan^{-1}\dfrac{y}{x}$, \hfill +{\,} \mbox{[Int.\ Cal.\ Art.~33 (2)],} +\[ + i - \frac{1}{\pi}\log z += i - \frac{1}{\pi}\log(x+yi) += -\frac{1}{2\pi}\log(x^2+y^2) ++ i\left( 1 - \frac{1}{\pi}\tan^{-1}\frac{y}{x} \right) +\] +and $1-\dfrac{1}{\pi}\tan^{-1}\dfrac{y}{x}$ and $-\dfrac{1}{2\pi}\log(x^2+y^2)$ are \emph{conjugate functions}. (v.\ Int.\ +Cal.\ Arts.~209 and 210.) Hence +\[ +V_1=-\frac{1}{2\pi}\log(x^2+y^2) \tag{3} +\] +is a solution of the equation +\[ +D_x^2V_1 + D_y^2V_1=0; \tag{4} +\] +% -----File: 082.png +and the curves +\begin{flalign*} +&&&\frac{1}{\pi}\left(\frac{\pi}{2}+\tan^{-1}\frac{x}{y}\right)=a \tag{5}\\ +&\text{and}& &-\frac{1}{2\pi}\log(x^2+y^2)=b &&\phantom{and} \tag{6} +\end{flalign*} +cut each other at right angles. + +If we construct the curves obtained by giving different values to $a$ in (5) we +get a set of \emph{equipotential lines} for the conducting sheet described at the beginning +of this article, and the curves obtained by giving different values to $(b)$ in +(6) will be the \emph{lines of flow}. + +Moreover since +\[ +V_1=-\frac{1}{2\pi}\log(x^2+y^2) \tag{3} +\] +is a solution of Laplace's Equation (4), the lines of flow just mentioned will be +equipotential lines for a certain distribution of potential, for which the equipotential +lines above mentioned will be lines of flow. + +$V=a$, that is +\begin{flalign*} +&&\frac{1}{\pi}&\left(\frac{\pi}{2}+\tan^{-1}\frac{x}{y}\right)=a, \tag{5}\\ +&\text{reduces to} & y&=-x\tan a\pi. &&\phantom{reduces\ to} \tag{7} +\end{flalign*} + +If now we give to $a$ values differing by a constant amount we get a set of +straight lines radiating from the origin and at equal angular intervals. + +$V_1=b$, that is +\begin{flalign*} +&&-\frac{1}{2\pi}&\log(x^2+y^2)=b, &&\tag{6}\\[-1ex] +\intertext{reduces to}\\[-6ex] +&&x^2&+y^2=e^{-2\pi b}. \tag{8} +\end{flalign*} + +%[Illustration] +\pngright{008.png}{701}{390}{-24} + +If we give to $b$ a set of values differing by a constant amount we get a set +of circles whose centres are at the origin and whose radii form a geometrical +progression. They are the equipotential lines for a thin plane sheet of infinite +extent where the potential function is kept equal to given different constant +values on the circumferences of two given concentric circles or where we have +a \emph{source} at the origin; and for this +system the lines (7) are lines of flow, +and (3) is the complete solution. + +The figure gives the equipotential +lines and lines of flow for either system, +but only for positive values of $y$. +The complete figure has the axis of $X$ +as an axis of symmetry. + +% -----File: 083.png +\EXAMPLE{S} + +1.\quad Solve the problem of Art.~44 for the case where +\[ + f(x)=-1 \quad\text{if}\quad x<0 \quad\text{and}\quad + f(x)= 1 \quad\text{if}\quad x>0. +\] +\begin{flushright} +\textit{Ans.}, \quad $V = \dfrac{2}{\pi} \tan^{-1}\dfrac{x}{y}$. +\end{flushright} + +2.\quad Solve the problem of Art.~44 for the case where +\[ + f(x)=a \quad\text{if}\quad x<0 \quad\text{and}\quad + f(x)=b \quad\text{if}\quad x>0. +\] +\begin{flushright} +\textit{Ans.}, \quad $V = \dfrac{1}{2}(a+b) + \dfrac{1}{\pi}(b-a) \tan^{-1}\dfrac{x}{y}$. +\end{flushright} + +3.\quad Reduce (7), (8), and (9) Art.~44 to the forms +\begin{align*} + V &= \dfrac{1}{\pi} \int\limits_{-\infty}^\infty \dfrac{yf(\lambda)d\lambda}{y^2+(\lambda-x)^2}, \\ + V &= \dfrac{1}{\pi} \int\limits_0^\infty yf(\lambda)d\lambda \left[ \dfrac{1}{y^2+(\lambda-x)^2} + \dfrac{1}{y^2+(\lambda+x)^2} \right], \\ + V &= \dfrac{1}{\pi} \int\limits_0^\infty yf(\lambda)d\lambda \left[ \dfrac{1}{y^2+(\lambda-x)^2} - \dfrac{1}{y^2+(\lambda+x)^2} \right], +\end{align*} +respectively. + +\mypara{46.} An especially interesting case of Art.~44 is the following where +\[ + f(x)=0 \;\;\text{if}\;\; x<-1, \quad + f(x)=1 \;\;\text{if}\;\; -1<x< 1, \quad \text{and} \quad + f(x)=0 \;\;\text{if}\;\; x> 1. +\] +\begin{flalign*} +&\indent\text{Here} & V = \frac{1}{\pi} \left[ \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right]. &&\phantom{\indent Here} \tag{1} +\end{flalign*} +\begin{flalign*} +&\text{Now} & \frac{1}{\pi} \log[(1-z)i] +&= \frac{1}{\pi} \log[(1-x-yi)i] + = \frac{1}{\pi} \log[y+(1-x)i] &&\phantom{Now} +\\ +&&&= \frac{1}{2\pi} \log[(1-x)^2+y^2] + \frac{i}{\pi} \tan^{-1}\frac{1-x}{y}, && +\end{flalign*} +and +\begin{align*} + -\frac{1}{\pi} \log[(-1-z)i] +&= -\frac{1}{\pi} \log[(-1-x-yi)i] + = -\frac{1}{\pi} \log[y-(1+x)i] +\\ +&= -\frac{1}{2\pi} \log[(1+x)^2+y^2] + + \frac{i}{ \pi} \tan^{-1}\frac{1+x}{y}. +\end{align*} +\[ + \frac{1}{ \pi} \log\frac{1-z}{-1-z} += \frac{1}{2\pi} \log\frac{(1-x)^2+y^2}{(1+x)^2+y^2} ++ \frac{i}{\pi} \left[ \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right]. +\] +% -----File: 084.png +Hence +\[ + \frac{1}{\pi} \left( \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right) \quad\text{and}\quad \frac{1}{2\pi} \log\frac{(1-x)^2+y^2}{(1+x)^2+y^2} +\] +are \emph{conjugate functions:}\footnote + {The function conjugate to +\[ + \frac{1}{\pi} \left[ \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right] +\] +might have been found as follows. If $\phi$ is the required function and $\psi$ the given function we +have by Int.\ Cal.\ Arts.~211, 212, and 213 the relations +\[ +D_x\phi = D_y\psi \quad \text{and} \quad D_y\phi = -D_x\psi. +\] +\begin{flalign*} +&\text{Here}& D_y\psi +&= -\dfrac{1}{\pi} \left[ \dfrac{1+x}{(1+x)^2+y^2} + +\dfrac{1-x}{(1-x)^2+y^2} \right] &&\phantom{Here} +\\ +&\text{and} & -D_x\psi +&= -\dfrac{1}{\pi} \left[ \dfrac{y}{(1+x)^2+y^2} + - \dfrac{y}{(1-x)^2+y^2} \right]. &&\phantom{and} +\end{flalign*} + +If now we integrate $D_y\psi$ with respect to $x$ treating $y$ as a constant and add an arbitrary +function of $y$ we shall have $\phi$. So that +\begin{gather*} + \phi = -\dfrac{1}{2\pi} \biggl\{ \log[(1+x)^2 + y^2\,] - \log[(1-x)^2 + y^2\,]\,\biggr\} + f(y). +\\ + D_y\phi = -\dfrac{1}{\pi} \biggl[ \dfrac{y}{(1+x)^2+y^2 } - \dfrac{y}{(1-x)^2+y^2} \biggr] + \dfrac{df(y)}{dy} +\end{gather*} +Comparing this with its equal $-D_x\psi$ above we find $\dfrac{df(y)}{dy}=0$ and $f(y) = C$ a constant +\begin{flalign*} +&\text{therefore} && \dfrac{1}{2\pi} \log{ \dfrac{(1-x)^2+y^2}{(1+x)^2+y^2} } + C, +&&\phantom{therefore} +\end{flalign*} +where $C$ may be taken at pleasure, is our required conjugate function.}\ %[endfootnotetext] +and +\[ +\dfrac{1}{\pi} \biggl( \tan^{-1}\dfrac{1+x}{y} + \tan^{-1}\dfrac{1-x}{y} \biggl) = a \tag{2} +\] +is any equipotential line, and +\[ +\dfrac{1}{2\pi} \log \dfrac{(1-x)^2+y^2 }{(1+x)^2+y^2} = b \tag{3} +\] +any line of flow for the system described at the beginning of this article; and +\[ +V_1 = \dfrac{1}{2\pi} \log \dfrac{(1-x)^2+y^2}{(1+x)^2+y^2} \tag{4} +\] +is the solution of a new problem for which (3) represents any equipotential +line and (2) any line of flow. +% -----File: 085.png + +(2) reduces to +\[ +\frac{2y}{x^2+y^2-1}=\tan a\pi +\] +\begin{flalign*} +&\text{or} & x^2+(y-\ctn a\pi)^2 = \csc^2 a\pi;& \tag{5}& +\end{flalign*} +\begin{flalign*} +&\text{and (3) to} & x^2 + y^2 + 2\frac{e^{2b\pi}+1}{e^{2b\pi}-1}x + 1 = 0 &&\phantom{and\ (3)\ to} +\end{flalign*} +\begin{flalign*} +&\text{or} +& \left( x + \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}} \right)^2 + y^2 += \left( \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}} \right)^2 - 1 &&\phantom{or} +\end{flalign*} +\begin{flalign*} +&\text{or} & (x+\ctnh b\pi)^2+y^2=\csch^2 b\pi. &&\phantom{or} \tag{6} +\end{flalign*} +(5) and (6) are circles. The circles (5) have their centres in the axis of $Y$, +and pass through the points $(-1,0)$ and $(1,0)$; and the circles (6) have their +centres in the axis of $X$. + +(4) is the complete solution, (6) is any equipotential line and (5) any line of +flow for a plane sheet in which the points in the circumferences of two given +circles whose centres are further apart than the sum of their radii are kept at +different constant potentials, or where a source and a sink of equal intensity +are placed at the points $(-1,0)$ and $(1,0)$. An important practical example +is where two wires connected with the poles of a battery are placed +with their free ends in contact with a thin plane sheet of conducting material. +The figure shows the equipotential lines and lines of flow of either system. + +The complete figure would have the axis of $X$ for an axis of symmetry. + +%[Illustration] +\pngcent{009.png}{1074} + +{}\EXAMPLE{S} + +1.\quad Show that if $f(x)=a_1$ when $x<-b$, $f(x)=a_2$ when $-b<x<b$, +$f(x)=a_3$ when $x > b$, +\[ +V = \frac{a_1+a_3}{2} + \frac{1}{\pi} \left[ (a_2-a_1)\tan^{-1}\frac{b+x}{y} + (a_2-a_3)\tan^{-1}\frac{b-x}{y} \right]. +\] +% -----File: 086.png + +2.\quad Show that if $f(x) = 0$ if $x < 0$, + $f(x) = a_1$ if $0 < x < b_1$, + $f(x) = a_2$ if $b_1 < x < b_2$, + $f(x) = a_3$ if $b_2 < x < b_3$, \&c., +\begin{multline*} +V = \frac{1}{\pi}\left [a_1 \tan^{-1} \frac{x}{y }+ (a_1 - a_2) \tan^{-1} \frac{b_1 - x}{y }+ (a_2 - a_3) \tan^{-1} \frac{b_2 - x}{y} \right. \\ +\left.{} + (a_3 - a_4) \tan^{-1} \frac{b_3 - x}{y }+ \cdots \right]. +\end{multline*} + +3.\quad Show that if $f(x) = -1$ if $x < -1$, $f(x) = x$ if $-1 < x < 1$, +$f(x) = 1$ if $x > 1$, +\[ +V = \frac{1}{\pi}\left [(1 + x)\tan^{-1} \frac{1 + x}{y } - (1 - x) \tan^{-1}\frac {1 - x}{y }+ \frac{y}{2} \log \frac{(1 - x)^2 + y^2 }{ (1 + x)^2 + y^2}\right]. +\] + +4.\quad Show that if $f(x) = -1$ if $x < -1$, $f(x) = 0$ if $-1 < x < 1$, +$f(x) = 1$ if $x > 1$, +\[ +V = \frac{1}{\pi}\left [\tan^{-1} \frac{1 + x }{ y }- \tan^{-1} \frac{1 - x}{y }\right ]. +\] + +Show that the equipotential lines are equilateral hyperbolas passing through +the points $(-1, 0)$ and $(1, 0)$, and that the lines of flow are Cassinian ovals +having $(-1, 0)$ and $(1, 0)$ as foci. The lines of flow are equipotential lines +and the equipotential lines are lines of flow for the case where the points +$(-1, 0)$ and $(1, 0)$ are kept at the same infinite potential, or where very small +ovals surrounding these points are kept at the same finite potential. The case +is approximately that of a pair of wires connected with the same pole of a +battery whose other pole is grounded, and then placed with their ends in contact +with a thin plane conducting sheet.\\ + +5.\quad Show that if $f(x) = 0$ if $x < 0$, $f(x) = -1$ if $0 < x < a$, $f(x) = 0$ + if $a < x < b$, and $f(x) = 1$ if $x > b$, +\[ +V = \frac{1}{\pi}\left[\frac{\pi}{2} - \tan^{-1} \frac{a - x }{y }- \tan^{-1} \frac{b - x }{y} - \tan^{-1} \frac{x}{y}\right]. +\] + +The conjugate function +\[ +V = \frac{1}{2\pi} \log \frac{x^2 + y^2 }{[(a-x)^2 + y^2] [(b-x)^2 + y^2]} +\] +is the solution for the case where a sink and two sources of equal intensity lie +on the axis of $X$, the sink at the origin and the sources at the distances $a$ and +$b$ to the right of the origin. One of the lines of flow is easily seen to be the +circle $ x^2 + y^2 = ab$. + +\mypara{47.} If the plane conducting sheet has two straight edges at right angles +with each other and one is kept at potential zero while the value of the potential +% -----File: 087.png +function is given at each point of the second, that is if $V=0$ when + $x=0$ and $V=f(x)$ when $y=0$, the solution is readily obtained. It is +\[ +V = \frac{2}{\pi} \int\limits_0^\infty d\alpha +\int\limits_0^\infty e^{-\alpha y} +f(\lambda) \sin \alpha x \sin \alpha\lambda.d\lambda. \tag{1} +\] +v.~(9) Art.~44. + +\noindent This reduces to +\[ +V = \frac{1}{\pi} \int\limits_0^\infty f(\lambda)d\lambda \left[ \frac{y}{y^2+(\lambda-x)^2} -\frac{y}{y^2+(\lambda+x)^2} \right]. \tag{2} +\] +v.\ Ex.~3 Art.~45. + +\EXAMPLE{S} + +1.\quad If $V=0$ when $y=0$ and $V=F(y)$ when $x=0$ show that +\begin{align*} +V &= \frac{2}{\pi} \int\limits_0^\infty d\alpha \int\limits_0^\infty e^{-\alpha x} F(\lambda) \sin\alpha y \sin\alpha\lambda.d\lambda\\ + &= \frac{1}{\pi} \int\limits_0^\infty F(\lambda) d\lambda \left[ \frac{x}{x^2+(\lambda-y)^2} - \frac{x}{x^2+(\lambda+y)^2} \right]. +\end{align*} + +2.\quad If $V=f(x)$ when $y=0$ and $V=F(y)$ when $x=0$ show that +\begin{align*} +V = \frac{1}{\pi} \int\limits_0^\infty & \left[ f(\lambda) \left( \frac{y}{y^2+(\lambda-x)^2} - \frac{y}{y^2+(\lambda+x)^2} \right) \right. \\ +& \left. {}+ F(\lambda) \left( \frac{x}{x^2+(\lambda-y)^2} - \frac{x}{x^2+(\lambda+y)^2} \right) \right] d\lambda. +\end{align*} + +3.\quad If $F(y)=b$ the result of Ex.~2 reduces to +\[ +V = \frac{2b}{\pi} \tan^{-1}\frac{y}{x} + \frac{1}{\pi} \int\limits_0^\infty f(\lambda) d\lambda \left[ \frac{y}{y^2+(\lambda-x)^2} - \frac{y}{y^2+(\lambda+x)^2} \right]. +\] + +4.\quad If $F(y)=1$ for $0<y<1$ and $F(y)=0$ for $y>1$ while $f(x)=1$ + for $0<x<1$ and $f(x)=0$ for $x>1$ +\begin{align*} +V = \frac{1}{\pi} & \left[ \tan^{-1}\frac{1-x}{y} - \tan^{-1}\frac{1+x}{y} + 2\tan^{-1}\frac{y}{x} \right.\\ +& \left. {}+ \tan^{-1}\frac{1-y}{x} - \tan^{-1}\frac{1+y}{x} + 2\tan^{-1}\frac{x}{y} \right]. +\end{align*} +% -----File: 088.png + +5.\quad If one edge of the conducting sheet treated in Art.~47 is insulated, so that + $D_{x}V=0$ if $x = 0$ and $V=f(x)$ when $y = 0$ +\begin{align*} + V &= \frac{2 }{ \pi } \int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} e^{-\alpha y} f(\lambda) \cos\alpha x \cos\alpha\lambda.d\lambda\\ + &= \frac{1 }{ \pi } \int\limits_0^{\infty} f(\lambda) d\lambda \left [ \frac{y }{ y^2+(\lambda+x)^2 } + \frac{y }{ y^2+(\lambda-x)^2} \right]. +\end{align*} + +\mypara{48.} If the conducting sheet is a long strip with parallel edges one of which +is at potential zero while the value of the potential function is given at all +points of the other, that is if $V=0$ when $y = 0$ and $V=F(x)$ when + $y = b$ the problem is not a very difficult one. + +Since we are no longer concerned with the value of $V$ when $y = \infty$ \quad +$V= e^{\alpha y}\sin ax$ and $V=e^{\alpha y} \cos ax$ are available as particular solutions of the +equation +\[ +D_x^2V+D_y^2V=0 \tag{1} +\] +as well as $V=e^{-\alpha y}\sin\alpha x$ and $V=e^{-\alpha y}\cos\alpha x$. +\begin{flalign*} +&\indent\text{Consequently }&& + \frac{e^{\alpha y} + e^{-\alpha y}}{2} \sin \alpha x = \cosh\alpha y \sin\alpha x & \text{[Int.\ Cal.\ Art.~43 (2)]} +\\ +&\text{and}&& + \frac{e^{\alpha y} - e^{-\alpha y} }{2} \sin \alpha x = \sinh\alpha y \sin\alpha x & \text{[Int.\ Cal.\ Art.~43 (1)]}\\[-5ex] +\end{flalign*} +\begin{flalign*} +&\text{and}&& + \cosh \alpha y \cos \alpha x \quad \text{and}\quad \sinh\alpha y \cos\alpha x &&\phantom{and} +\end{flalign*} +are now available values of $V$ and can be used precisely as $e^{-\alpha y}\cos\alpha x$ and + $e^{-\alpha y} \sin\alpha x$ are used in Art.~44. + +Following the same course as in Art.~44 we get +\[ + V = \frac{1}{ \pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} \frac{\sinh\alpha y}{\sinh\alpha b} F(\lambda) \cos\alpha(\lambda-x).d\lambda \tag{2} + \] +as a solution of (1) which will reduce to $V= F(x)$ when $y= b$ +\begin{flalign*} +&\text{and to }&& V = 0 \quad\text{when}\quad y = 0\text{, since } \sinh 0 = \frac{1-1 }{2} = 0, &&\phantom{and\ to} +\end{flalign*} +and (2) is therefore our required solution. + +If $V$ is to be equal to zero when $y = b$ and to $f(x)$ when $y = 0$ we have +only to replace $y$ by $b-y $ and $F(x)$ by $f(x)$ in (2). We get +\[ + V = \frac{1}{ \pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty}\frac{\sinh \alpha(b-y) }{\sinh \alpha b} f(\lambda) \cos \alpha(\lambda-x).d\lambda. \tag{3} +\] +% -----File: 089.png + +If $V = f(x)$ when $y = 0$ and $V = F(x)$ when $y = b$ then +\begin{equation*} \begin{split} +V = {}& \frac{1}{\pi} \int\limits^\infty_{0} d\alpha \int\limits^\infty_{-\infty} +\dfrac{\sinh\alpha(b - y)}{\sinh\alpha b} f(\lambda)\cos {\alpha(\lambda - x)}.d\lambda \\ +&+ \frac{1}{\pi} \int\limits^\infty_{0} d\alpha \int\limits^\infty_{-\infty} \frac{\sinh\alpha y} +{\sinh\alpha b} F(\lambda)\cos {\alpha(\lambda - x)}.d\lambda. +\end{split} \end{equation*} + +This can be considerably simplified by the aid of the formula +\[ +\int \limits^{\infty}_{0} \dfrac{\sinh px}{\sinh qx}\cos{rx.dx} += \dfrac{\pi}{2q} \dfrac{\sin \dfrac{p \pi}{q}} {\cos\dfrac{p \pi}{q} + \cosh \dfrac{r \pi}{q}} +\] +if $p^2 < q^2$. [Bierens de Haan, Tables of Def.\ Int.\ (7) 265] and becomes +\begin{flalign*} && +V ={} & \dfrac{1}{2b} \sin \dfrac{\pi}{b} (b - y) \int \limits^\infty_{-\infty}f(\lambda) +\dfrac{d\lambda}{\cos \dfrac{\pi(b - y)}{b} + \cosh \dfrac{\pi}{b} (\lambda - x)} && +\\ +&&\phantom{or} +& {}+ \dfrac{1}{2b} \sin \dfrac{\pi y}{b} \int \limits^\infty_{-\infty} F(\lambda) \dfrac{d\lambda} +{\cos \dfrac{\pi y}{b} + \cosh \dfrac{\pi}{b} (\lambda - x)} &\text{or}& +\end{flalign*} +\[\tag{5} +V = \dfrac{1}{2b}\sin \dfrac{\pi y}{b} \int\limits^\infty_{-\infty} \biggl +[\dfrac{f\lambda}{\cosh \dfrac{\pi}{b} (\lambda - x) - \cos\dfrac{\pi y}{b}} ++ \dfrac{F\lambda}{\cosh\dfrac{\pi}{b} (\lambda - x) + \cos\dfrac{\pi y}{b}}\biggr] +d\lambda. +\] + +\EXAMPLE{S} + +1.\quad Given the formula +\[ +\int \dfrac{dx}{a+b \cosh x} = \dfrac{2}{\sqrt{b^2 - a^2}} \tan^{-1}\biggl(\sqrt{\dfrac{b-a}{b+a}} +\tanh \dfrac{x}{2}\biggr) \quad\text{ if } b > a, +\] +show that if $V = 1$ when $y = 0$ and $V = 0$ when $y = b$\;\;$V = \dfrac{1}{b}(b - y).$\\ + +2.\quad Show that if $V = 0$ when $y = b$, $V = -1$ when $y = 0$ and + $x < 0$, and $V = 1$ when $y = 0$ and $x > 0$ +\[ +V=\dfrac{2}{\pi}\tan^{-1}\Biggl[\dfrac{\tanh \dfrac{\pi x}{2b}} {\tan \dfrac{\pi y}{2b}}\Biggr] +\] +The solution for the conjugate system, that is, for a strip having a source at +$(0, 0)$ and an infinitely distant sink is +\[ +V = -\dfrac{1}{\pi}\log \biggl[\cosh^{2}\dfrac{\pi x}{2b} - \cos^{2}\dfrac{\pi y}{2b}\biggr]. +\] +% -----File: 090.png + +3.\quad Show that if $V= -1$ when $y = 0$ and $x<0$, $V=1$ when $y = 0$ +and $x>0$, $V= -1$ when $y = b$ and $x < 0$, and $V=1$ when + $y = b$ and $x > 0$, +\[ +\begin{split} +V & += \frac{2}{\pi}\tan^{-1}\bigg ( \tan \dfrac{\pi}{2b} (b -y) \tanh \dfrac{\pi x}{2b}\bigg ) ++ \frac{2}{\pi}\tan^{-1} \bigg ( \tan \dfrac{\pi}{2b} y \tanh \dfrac{\pi x}{2b}\bigg ) +\\ + &=\frac{2}{\pi}\tan^{-1} \Bigg[ \frac{\sinh \dfrac{\pi x}{b }} {\sin\dfrac{ \pi y}{b}} \Bigg]. + \end{split} + \] +The solution for the conjugate system, that is, for a strip having a source and a +sink at the points $(0, 0)$ and $(0, b)$ is + \[ +V = \frac{1}{\pi}\log \Bigg[ \dfrac{\cosh\dfrac{ \pi x}{b} + \cos \dfrac{ \pi y}{b}}{\cosh\dfrac{ \pi x}{b} - \cos \dfrac{ \pi y}{b} } \Bigg]. +\] + +4.\quad If $V=0$ when $x = 0$, $V = f(x)$ when $y = 0$ and $x>0$, and $V=0$ +when $y = b$ and $x > 0$. +\[ +\begin{split} +V &= \frac{1}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} \frac{\sinh\alpha(b-y)}{\sinh\alpha b} + [ \cos\alpha(\lambda-x) - \cos\alpha(\lambda+x) ] f(\lambda) d\lambda\\ +&= \frac{1}{2b} \sin \dfrac{\pi y }{b} \int\limits_0^{\infty} \biggl[ + \frac{1} {\cosh \dfrac{\pi }{b} (\lambda-x) - \cos \dfrac{\pi y}{ b}} + - \frac{1} {\cosh \dfrac{\pi }{b} (\lambda+x) - \cos \dfrac{\pi y}{ b}} \biggr] f(\lambda) d\lambda +\end{split} +\] +for positive values of $x$ and for values of $y$ between $0$ and $b$.\\ + +5.\quad If $V_1 = 0$ when $x = 0$, $V_1 = F(x)$ when $y = b$ and $x > 0$, and + $V_1 = 0$ when $y = 0$ and $x>0$ +\[ +V_1 = \frac{1}{2b} \sin \frac{\pi y}{ b} \int\limits_0^{\infty} \biggl[ + \frac{1} {\cosh \dfrac{\pi }{b} (\lambda-x) + \cos \dfrac{\pi y}{ b}} + - \frac{1} {\cosh \dfrac{\pi }{b} (\lambda+x) - \cos \dfrac{\pi y}{ b}} \biggr] F(\lambda) d\lambda + \] +for positive values of $x$ and values of $y$ between $0$ and $b$.\\ + +6.\quad If $V_2=0$ when $x = 0$, $V_2 =f(x)$ when $y = 0$ and $x > 0$, and + $V_2 = F(x)$ when $y = b$ and $x>0$ +\[ \phantom{(v.\ Exs.\ 4 and 5)} +V_2 = V+ V_1 \quad \text{for} \quad x > 0\quad \text{and} \quad 0 < y < b. \tag*{(v.\ Exs.\ 4 and 5)} +\] + +7.\quad If one edge of the strip described in Art.~48 is insulated so that we have + $V=f(x)$ when $y = 0$ and $D_y V=0$ when $y = b$ show that +\[ + V = \frac{1}{\pi} \int\limits_0^{\infty}d\alpha \int\limits_{-\infty}^{\infty} \frac{\cosh\alpha(b-y)}{\cosh\alpha b} f(\lambda) \cos\alpha(\lambda-x).d\lambda. +\] +% -----File: 091.png + +By the aid of the formula +\begin{flalign*} +&\phantom{if\ p<q}&& + \int\limits_0^\infty \frac{\cosh px}{\cosh qx} \cos rx.dx = +\frac{\pi}{q} \frac{\cosh\dfrac{r\pi}{2q}\cos\dfrac{p\pi}{2q}} + {\cos\dfrac{p\pi}{q}+\cosh\dfrac{r\pi}{q}} && \text{if } p<q,& +\end{flalign*} +[Bierens de Haan, Def.\ Int.\ Tables (6) 265],\\ +reduce this to +\[ +V=\frac{1}{b}\sin\dfrac{\pi y}{2b}\int\limits_{-\infty}^\infty +\frac{f(\lambda)\cosh\dfrac{\pi}{2b}(\lambda-x)} + {\cosh\dfrac{\pi}{b}(\lambda-x)-\cos\dfrac{\pi y}{b}}d\lambda. +\] + +8.\quad If $V=0$ when $y=0 \text{\; or \;} b$ and $x<-a$, $V=1$ when $y=0 \text{\; or \;} b$ +and $-a<x<a$, and $V=0$ when $y=0 \text{\; or \;} b$ and $x>a$ +\[ +V = \frac{1}{\pi} \Bigg[ + \tan^{-1}\frac{\sinh\dfrac{\pi(a-x)}{b}}{\sin\dfrac{\pi y}{b}} ++\tan^{-1}\frac{\sinh\dfrac{\pi(a+x)}{b}}{\sin\dfrac{\pi y}{b}} \Bigg]. +\] + +9.\quad If $V=0$ when $y=0 \text{\; or \;} b$ and $x<-a$, $V=1$ when $y=0$ and + $-a<x<a$, $V=0$ when $y=0 \text{\; or \;} b$ and $x>a$, and $V=-1$ when + $y=b$ and $-a<x<a$ +\[ +V=\frac{1}{\pi} \Bigg[ + \tan^{-1} \frac{\tanh\dfrac{\pi(a-x)}{b}}{\tan\dfrac{\pi y}{b}} + + \tan^{-1} \frac{\tanh\dfrac{\pi(a+x)}{b}}{\tan\dfrac{\pi y}{b}} \Bigg]. +\] + +10.\quad A system conjugate to that of Ex.~9 is $V=+\infty$ when $y=0 \text{\; or \;} b$ \ +and $x=-a$, $V=-\infty$ when $y=0 \text{\; or \;} b$ and $x=a$. In this case +\[ +V=\frac{1}{2\pi}\log +\frac{\sin^2\dfrac{\pi y}{b}+\sinh^2\dfrac{\pi(a-x)}{b}} + {\sin^2\dfrac{\pi y}{b}+\sinh^2\dfrac{\pi(a+x)}{b}}. +\] + +\mypara{49.} Let us take now a problem in the flow of heat. Suppose we have an +infinite solid in which heat flows only in one direction, and that at the start the +temperature of each point of the solid is given. Let it be required to find the +temperature of any point of the solid at the end of the time $t$. + +Here we have to solve the equation +\[ +D_t u = a^2D_x^2 u \tag{1} +\] +[v.\ Art.~1 \smallromr{II}] subject to the condition +\[ +u=f(x)\quad \text{when}\quad t = 0. \tag{2} +\] +% -----File: 092.png + +As the equation (1) is linear with constant coefficients we can get a particular +solution by the device used in Arts.~7 and 8. + +Let $u = e^{\beta t + \alpha x}$ and substitute in (1). We get +\[ + \beta = a^2 \alpha^2 +\] +as the only relation which need hold between $\beta$ and $\alpha$. +\begin{flalign*} +&\indent\text{Hence}&& + u = e^{\alpha x + a^2 \alpha^2 t} = e^{a^2 \alpha^2 t} e^{\alpha x} &&\phantom{Hence} \tag{3} +\end{flalign*} +is a solution of (1) no matter what value is given to $\alpha$. + +To get a trigonometric form replace $\alpha$ by $\alpha i$. +\begin{flalign*} +&\indent\text{Then}&& u = e^{-a^2 \alpha^2 t} e^{\alpha x i}. &&\phantom{\indent Then} +\end{flalign*} + +If in (3) we replace $\alpha$ by $-\alpha i$ we get +\[ + u = e^{-a^2 \alpha^2 t} e^{-\alpha x i}. +\] +As in Arts.~7 and 8 we get from these values +\[ + u = e^{-a^2 \alpha^2 t} \sin {\alpha x} \quad \text{and} \quad + u = e^{-a^2 \alpha^2 t} \cos {\alpha x} +\] +as particular solutions of (1), $\alpha$ being wholly unrestricted. + +From these values we wish to build up a value of $u$ which shall reduce to +$f(x)$ when $t = 0$ and shall still be a solution of (1). +\begin{flalign*} +&\indent\text{We have }&& + f(x) = \frac{1}{\pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} f(\lambda) \cos \alpha (\lambda - x).d\lambda +&&\phantom{We\ have} \tag{4} +\end{flalign*} +v.\ Art.~32 (3), and by proceeding as in Art.~44 we get +\[ + u = \frac{1}{\pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} e^{-a^2\alpha^2t} f(\lambda) \cos \alpha (\lambda - x).d\lambda \tag{5} +\] +as our required value of $u$. + +This can be considerably simplified. + +Changing the order of integration +\begin{gather*} + u = \frac{1}{\pi} \int\limits_{-\infty}^{\infty} f(\lambda) d\lambda \int\limits_0^{\infty} e^{-a^2\alpha^2t} \cos \alpha(\lambda - x).d\alpha. \tag{6} +\\ + \int\limits_0^{\infty} e^{-a^2 \alpha^2 t} \cos \alpha (\lambda - x).d\alpha += \frac{1}{2a } \sqrt{\frac{ \pi}{t }}.e^{-\frac{(\lambda - x)^2 }{4a^2 t}} \tag{7} +\end{gather*} +by the formula +\begin{flalign*} +&&& \int\limits_0^{\infty} e^{-a^2 x^2} \cos bx.dx += \frac{\sqrt{ \pi}}{ 2a} e^{-\frac{b^2}{4a^2}} \tag*{[Int.\ Cal.\ Art.~94 (2)]}\\ +&\text{Hence}&& + u = \frac{1}{2a \sqrt{ \pi t} } \int\limits_{-\infty}^{\infty} f(\lambda) e^{-\frac{(\lambda - x)^2 }{4a^2 t}} d\lambda. +&&\phantom{Hence} \tag{8}\\[-5ex] +\end{flalign*} +% -----File: 093.png +\begin{flalign*} +&\indent\text{Let now} && \beta = \frac{\lambda-x}{2a \sqrt{t}}, &&\phantom{\indent Let\ now} \\[1ex] +&\text{then} & \lambda &= x + 2a \sqrt{t}.\beta &&\\ +&\text{and } & u = \frac{1 }{\sqrt{\pi}}&\int\limits_{-\infty}^{\infty} f(x + 2a \sqrt{t}.\beta) e^{-\beta^2} d\beta. & \tag{9}& +\end{flalign*} + +\EXAMPLE{S} + +1.\quad Let the solid be of infinite extent and let the temperature be equal to a +constant $c$ at the time $t = 0$. +\begin{flalign*} +&\indent\text{Then }&& + u = \frac{c }{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} e^{-\beta^2} d\beta += \frac{2c }{\sqrt{\pi}} \int\limits_0^{\infty} e^{-\beta^2} d\beta = c. &&\phantom{\indent Then} +\end{flalign*} +v.\ Int.\ Cal.\ Art.~92 (2).\\ + +2.\quad Let $u = x$ when $t = 0$. +\begin{flalign*} +&\indent\text{Then}&& + u = \frac{1 }{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} (x + 2a \sqrt{t}.\beta) e^{-\beta^2} d\beta = x. +&&\phantom{\indent Then} +\end{flalign*} + +3.\quad Let $u = x^2$ when $t = 0$. +\begin{flalign*} +&\indent\text{Then}&& u = x^2 + 2a^2t. &&\phantom{\indent Then} +\end{flalign*} + +4.\quad Let $u = 0$ if $x< -b$, $u = 1$ if $-b < x < b$, and $u = 0$ if $x > b$, +when $t = 0$.\\ + +Then +\[ +u = \frac{1 }{\sqrt{\pi}} \!\!\int\limits_{-\frac{b+x}{2a\sqrt{t}}}^{\frac{b-x}{2a\sqrt{t}}}\!\! e^{-\beta^2} d\beta += \frac{2 }{\sqrt{\pi}} \left [ \frac{b}{ 2a \sqrt{ t }} - \frac{b^3+3bx^2 }{ 3(2a \sqrt{ t })^3} + + \frac{b^5 + 10b^3x^2 + 5bx^4 }{5.2!( 2a \sqrt{ t })^5} - \cdots \right]. +\] + +5.\quad Let $u = 0$ if $x < 0$ and $u = 1$ if $x > 0$ when $t = 0$. + +Then +\begin{align*} +u &= \frac{1 }{\sqrt{\pi}} \!\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} d\beta += \frac{1 }{\sqrt{\pi}} \biggl[ + \int\limits_0^{\frac{x}{2a\sqrt{t}}}\!\! e^{-\beta^2} d\beta + + \int\limits_0^{\infty} e^{-\beta^2 } d\beta \biggr] += \frac{1 }{\sqrt{\pi}} \!\!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\!\! e^{-\beta^2 } d\beta + \frac{1 }{2} +\\ + &= \frac{1}{ 2} + \frac{1 }{\sqrt{\pi}} \left [ + \frac{ x }{ 2a \sqrt{ t } } + - \frac{ x^3}{ 3.(2a \sqrt{ t })^3} + + \frac{ x^5}{ 5.2!(2a \sqrt{ t })^5} + - \frac{ x^7}{ 7.3!(2a \sqrt{ t })^7} + \cdots \right]. +\end{align*} + +6.\quad An iron slab 10~c.\,m.\ thick is placed between and in contact with two +very thick iron slabs. The initial temperature of the middle slab is 100°, and +of each of the outer slabs 0°. Required the temperature of a point in the +middle of the inner slab fifteen minutes after the slabs have been put together. +Given \ $a^2 = 0.185$ \ in C.G.S. units. \hfill \emph{Ans.}, 21°.6.\\ +% -----File: 094.png + +7.\quad Two very thick iron slabs one of which is at the temperature 0° and the +other at the temperature 100° throughout are placed together face to face. +Find the temperature of each slab 10~c.\,m.\ from their common face fifteen +minutes after they have been placed together. \hfill \emph{Ans.}, 70°.8, 29°.2.\\ + +8.\quad Find a particular solution of $D_tu = a^2D_x^2u$ on the assumption that it +is of the form $u = T.X$ where $T$ is a function of $t$ alone and $X$ is a function +of $x$ alone. + +\mypara{50.} If our solid has one plane face which is kept at the constant temperature +zero, and we start with any given distribution of heat, the problem is +somewhat modified. + +Take the origin of coördinates in the plane face. Then we have as before +the equation +\[ + D_t u = a^2D_x^2 u, \tag{1} +\] +but our conditions are +\begin{alignat*}{3} + u &= 0 && \text{when} \quad &x = 0 \tag{2}\\ + u &= f(x) && \text {\quad`` } \quad & t = 0 \tag{3} +\end{alignat*} +and we are concerned only with positive values of $x$. + +We may then use the form (4) Art.~32 +\[ +f(x) = \frac{2}{\pi}\int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} f(\lambda) \sin\alpha x \sin\alpha \lambda.d\lambda, \tag{4} +\] +and proceeding as in the last section we get +\[ +u = \frac{2}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} e^{-a^2 \alpha^2 t} f(\lambda) \sin\alpha x \sin\alpha\lambda.d\lambda \tag{5} +\] +as our required solution. This may be reduced considerably. +\[ + u = \frac{1}{\pi} \int\limits_0^{\infty} f(\lambda) d\lambda \int\limits_0^{\infty} e^{-a^2 \alpha^2 t}[\cos \alpha (\lambda - x) - \cos \alpha (\lambda + x)]d\alpha, +\] +\[ +\text{or} \quad u = \frac{1}{2a\sqrt{\pi t}} \int\limits_0^{\infty} f(\lambda)(e^{-\frac{(\lambda-x)^2 }{ 4a^2 t}} - e^{-\frac{(\lambda+x)^2 }{ 4a^2 t}} ) d\lambda \tag{6} +\] +by (7) Art.~49, and this may be reduced to the form +\[ +u = \frac{1}{\sqrt{ \pi} }\biggl [\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(x + 2a \sqrt{ t}.\beta) d\beta + - \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(-x + 2a \sqrt{ t }.\beta) d\beta \biggr ] . \tag{7} +\] + +\EXAMPLE{S} + +1.\quad Let the initial temperature be constant and equal to $c$. +% -----File: 095.png + +Then +\[ +\begin{split} + u &= \frac{ c}{\sqrt{ \pi}} \biggl [ \!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} d\beta - \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} d\beta \biggr ]\\ + &= \frac{2c}{\sqrt{ \pi}} \!\int\limits_{0}^{\frac{x}{2a\sqrt{t}}}\!e^{-\beta^2} d\beta\\ + &= \frac{2c}{\sqrt{ \pi}} \left [ \frac{x }{ 2a \sqrt{ t}} + - \frac{x^3}{3.(2a \sqrt{t})^3} + + \frac{x^5}{5.2!(2a \sqrt{t})^5} + - \frac{x^7}{7.3!.(2a \sqrt{t})^7} + \cdots \right ]. +\end{split} +\] + +2.\quad Assuming that the earth was originally at the temperature 7000° Fahrenheit +throughout, and that the surface was kept at the constant temperature 0°, +find (1) the temperature 10 miles below the surface 10,000,000 years after the +cooling began; (2) the temperature 1 mile below the surface at the same +epoch; (3) the temperature 10 miles below the surface 100,000,000 years after +the cooling began; (4) the temperature 1 mile below the surface at the same +epoch; (5) the rate at which the temperature was increasing with the distance +from the surface at each point at each epoch. + +Neglect the convexity of the earth's surface and take Sir Wm.\ Thomson's +value of $a^2 (400)$ the foot, the Fahrenheit degree, and the year being taken as +units. (Thomson and Tait's Nat.\ Phil.\ Vol.\ II. Appendix.) + +\emph{Ans.}, (1) 3114°; (2) 329°.5; (3) 1036°; (4) 103°; (5) 1° for every 20 feet, 3° +for every 50 feet, 1° for every 50 feet, 1° for every 50 feet.\\ + +3.\quad Let the initial temperature be constant and equal to $ -b$, then by Ex.~1 +\[ + u = - \frac{2b}{\sqrt{\pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2 }d\beta . +\] + +4.\quad Let the temperature of the plane face be $b$ instead of zero, and let the +initial temperature be zero. + +Then we have only to add $b$ to the second member of the solution in Ex.~3, +as we may since $u = b$ is a solution of (1) Art.~49, and we get +\[ + u = b \biggl ( 1 - \frac{2}{\sqrt{ \pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta \biggr ) . +\] + +5.\quad Let $u = b$ when $x = 0$ and $u = f(x)$ when $t = 0$. + +Then +\[ + u = b \biggl ( 1 - \frac{2}{\sqrt{ \pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta \biggr ) + + \frac{1}{ 2a \sqrt{\pi t }}\int\limits_0^{\infty} f(\lambda) + [ e^{-{\frac{(\lambda-x)^2 }{4a^2t}}} + - e^{-{\frac{(\lambda+x)^2 }{4a^2t }}}] d\lambda +\] +by (6) Art.~50.\\ +% -----File: 096.png + +6.\quad Let $u=b$ when $x=0$ and $u=c$ when $t=0$. +\begin{flalign*} +&\indent \text{Then} && u = b + (c - b) \frac{2}{\sqrt{\pi}}\!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta.&& \phantom{\indent Then} +\end{flalign*} + +7.\quad If the earth has been cooling for 200,000,000 years from a uniform temperature, +prove that the rate of cooling is greatest at a depth of about 76 +miles, and that at a depth of about 130 miles the rate of cooling has reached +its maximum value for all time. Let $a^2 = 400$.\\ + +8.\quad Show that if the plane face of the solid considered in Art.~50 instead of +being kept at temperature zero is impervious to heat +\[\phantom{v.\ (6) Art.~50.} +u = \frac{1}{2a \sqrt{\pi t}} \int\limits_0^{\infty} f(\lambda)(e^{-\frac{(\lambda - x)^2 }{ 4a^2t}} + e^{-\frac{(\lambda + x)^2}{ 4a^2t}}) d\lambda. \tag*{v.\ (6) Art.~50.} +\] + +\mypara{51.} If the temperature of the plane face of the solid described in Art.~50 +is a given function of the time and the initial temperature is zero, the solution +of the problem can be obtained by a very ingenious method due to Riemann. + +Here we have to solve the equation +\[ +D_t u = a^2 D_x^2 u \tag{1} +\] +subject to the conditions +\[ +\left . \begin{aligned} +u& = F(t) &&\text{when} &x = 0\phantom{.}\\ +u& = 0 &&\text{\quad``} &t = 0. +\end{aligned}\right \} +\tag{2} +\] +We know that +\[ +u= \frac{2}{\sqrt{ \pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta +\] +is a solution of (1), v.\ Ex.~1 Art.~50. It is easily shown that +\[ +u= \frac{2}{\sqrt{ \pi}} \!\!\int\limits_0^{\frac{x}{2a\sqrt{t-c}}}\!\! e^{-\beta^2} d\beta, \tag{3} +\] +where $c$ is any constant, is a solution of (1). + +For +\begin{align*} +D_t u &= -\frac{2}{\sqrt{ \pi}} \frac{ x}{2a}\frac{ 1}{2(t-c)^\frac{3}{2}} e^{-\frac{x^2}{4a^2(t-c)}} = -\frac{x}{2a \sqrt{ \pi} }(t - c)^{-\frac{3}{2}} e^{- \frac{x^2}{4a^2(t-c)}}\\ +D_x u &= \frac{2}{\sqrt{ \pi}} \frac{ 1}{2a\sqrt{t-c}} e^{-\frac{x^2}{4a^2(t-c)}}\\ +D_x^2u&= - \frac{2}{\sqrt{ \pi}} \frac{ 1}{2a\sqrt{t-c}}\frac{2x}{4a^2(t-c)}e^{-\frac{x^2}{ 4a^2(t-c)}} + = -\frac{x}{2a^3 \sqrt{ \pi}}(t-c)^{-\frac{3}{2}} e^{-\frac{x^2}{4a^2(t-c)}} +\end{align*} +\begin{flalign*} +\text{and} && D_t u &= a^2 D_x^2 u.&& +\end{flalign*} +% -----File: 097.png + +Let $\phi(x,t)$ be a function of $x$ and $t$ which shall be equal to zero if $t$ is +negative and shall be equal to +\[ +1-\frac{2}{\sqrt\pi}\!\int\limits_0^\frac{x}{2a\sqrt t}\! e^{-\beta^2}d\beta +\] +if $t$ is equal to or greater than zero; so that if $x=0$\;\;$\phi(x,t)=1$ and if + $t=0$\;\;$\phi(x,t)=0$. + +We shall now attack the following problem, to solve equation (1) subject to +the conditions +\begin{alignat*}{5} +&u=0 && \text{if}\quad && t=0 && &&\\ +&u=F(0) && \text{``} && x=0\quad && \text{and}\quad && 0<t<\tau\\ +&u=F(k\tau)\; && \text{``} && x=0 && \phantom{a}\text{``} && k\tau<t<(k+1)\tau, +\end{alignat*} +where $k$ is any whole number and $\tau$ is any arbitrarily chosen interval of time. + +If we form the value +\[ +u=F(k\tau)[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)] \tag{4} +\] +$u$ will satisfy equation (1) since zero, unity and +\[ +\frac{2}{\sqrt\pi}\!\!\!\int\limits_0^\frac{x}{2a\sqrt{t-k\tau}}\!\!\! e^{-\beta^2}d\beta +\] +are values of $u$ which satisfy (1). $u$ will be zero if $t<k\tau$ by the definition +of the function $\phi(x,t)$; if $x=0$ $u=0$ if $t>(k+1)\tau$ and $u=F(k\tau)$ if + $k\tau<t<(k+1)\tau$. + +Therefore +\[ +u=\sum_{k=0}^{k=\infty}F(k\tau)[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)] \tag{5} +\] +is the solution of the problem stated above. + +(5) can be simplified somewhat from the consideration that for a given value +of $t$\;\;$\phi(x,t-k\tau)=0$ if $k\tau>t$. If, then, $n\tau$ is the greatest whole multiple +of $\tau$ not exceeding $t$, +\[ +u=\sum_{k=0}^{k=n}F(k\tau)[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)]. \tag{6} +\] + +If now we decrease $\tau$ indefinitely the limiting form of (6) will be the solution +of the problem stated at the beginning of this article. + +(6) may be written +\[ +u=\sum_{k=0}^{k=n}F(k\tau)\left[\frac{\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)}{\tau}\right]\tau \tag{7} +\] +% -----File: 098.png +and if $\tau$ is indefinitely decreased the limiting form of (7) is +\[ +u=-\int\limits_0^t F(\lambda)D_\lambda\phi(x,t-\lambda)d\lambda. \tag{8} +\] + +Since $t-\lambda$ is positive between the limits of integration +\begin{flalign*} +&& \phi(x,t-\lambda)&=1-\frac{2}{\sqrt\pi}\!\!\int\limits_0^\frac{x}{2a\sqrt{t-\lambda}}\!\! e^{-\beta^2}d\beta, &&\\ +&\text{and} & D_\lambda\phi(x,t-\lambda)=&-\dfrac{x}{2a\sqrt\pi}e^{-\frac{x^2}{4a^2(t-\lambda)}}(t-\lambda)^{-\frac{3}{2}}; && \phantom{and} +\end{flalign*} +and (8) may be written +\[ +u=\frac{x}{2a\sqrt\pi}\int\limits_0^t F(\lambda)e^{-\frac{x^2}{4a^2(t-\lambda)}}(t-\lambda)^{-\frac{3}{2}}d\lambda, \tag{9} +\] +or if we let \qquad\qquad\qquad\qquad $\beta=\dfrac{x}{2a\sqrt{t-\lambda}}$ +\[ +u=\frac{2}{\sqrt\pi}\!\int\limits_{\frac{x}{2a\sqrt t}}^\infty\! e^{-\beta^2}F\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta. \tag{10} +\] + +\EXAMPLE{S} + +1.\quad If $u=nt$ when $x=0$ and $u=0$ when $t=0$ +\[ +u=n\bigg(t+\frac{x^2}{2a^2}\bigg)\bigg[1-\frac{2}{\sqrt\pi}\!\int\limits_0^\frac{x}{2a\sqrt t}\!e^{-\beta^2}d\beta\bigg]-\frac{nx\sqrt t}{a\sqrt\pi}e^{-\frac{x^2}{4a^2t}}. +\] + +2.\quad A thick iron slab is at the temperature zero throughout, one of its plane +faces is then kept at the temperature 100° Centigrade for 5 minutes, then at +the temperature zero for the next 5 minutes, then at the temperature 100° for +the next 5 minutes, and then at the temperature zero. Required the temperature +of a point in the slab 5 c.m.\ from the face at the expiration of 18 +minutes. Given; $a^2=.185$. \hfill \textit{Ans.}, 20°.1.\\ + +3.\quad If $u=F(t)$ when $x=0$ and $u=f(x)$ when $t=0$, then +\[ +u=\frac{2}{\sqrt\pi}\int\limits_\frac{x}{2a\sqrt t}^\infty e^{-\beta^2}F\bigg(t-\frac{x^2}{4a^2\beta^2}\bigg)d\beta ++\frac{1}{2a\sqrt{\pi t}}\int\limits_0^\infty(e^{-\frac{(\lambda-x)^2}{4a^2t}}-e^{-\frac{(\lambda+x)^2}{4a^2t}})f(\lambda)d(\lambda). +\] +v.~(6) Art.~50.\\ + +% -----File: 099.png +4.\quad If in Art.~(51) $F(t)$ is a periodic function of the time of period $T$ it can +be expressed by a Fourier's series of the form +\iffalse +\[ +F(t) =\frac{1}{2}b_0 + \sum\limits_{m=1}^{m=\infty}[a_m\sin m\alpha t + b_m\cos m\alpha t],\qquad \text{where}\qquad \alpha=\frac{2\pi}{T}, +\] +or \qquad\qquad $F(t)=\dfrac{1}{2b_0} + \sum\limits_{m=1}^{m=\infty}\rho_m\sin(m\alpha t + \lambda_m)$, +\begin{flalign*} +&\text{where} & \rho_m\cos\lambda_m=a_m \quad \text{and}\quad \rho_m\sin\lambda_m=b_m. && \text{v.\ Art.~31 Ex.~3.}& +\end{flalign*} +\fi +\begin{flalign*} +&&F(t) =\frac{1}{2}b_0 &+ \sum\limits_{m=1}^{m=\infty}[a_m\sin m\alpha t + b_m\cos m\alpha t],\quad \text{where} &&\alpha=\frac{2\pi}{T},\\ +&\text{or} & F(t)=\dfrac{1}{2b_0} &+ \sum\limits_{m=1}^{m=\infty}\rho_m\sin(m\alpha t + \lambda_m),\\ +&\rlap{where\hfill} && \rho_m\cos\lambda_m=a_m \quad \text{and}\quad \rho_m\sin\lambda_m=b_m. &&\text{v.\ Art.~31 Ex.~3.} +\end{flalign*} + +Show that with this value of $F(t)$ (10) Art.~51.\ +becomes +\begin{gather*} +\begin{split} +u=\frac{1}{\sqrt\pi}b_0\!\int\limits_\frac{x}{2a\sqrt t}^{\infty}\! e^{-\beta^2}d\beta + \frac{2}{\sqrt\pi}\sum\limits_{m=1}^{m=\infty}\rho_m\biggl[\sin(m \alpha t + \lambda_m)\int\limits_{\frac{x}{2a\sqrt t}}^{\infty}e^{-\beta^2}\cos\frac{m\alpha x^2}{4a^2\beta^2}d\beta\biggr.\\ +\biggl.-\cos(m\alpha t + \lambda_m)\!\int\limits_\frac{x}{2a\sqrt t}^{\infty}\! e^{-\beta^2}\sin\frac{m\alpha x^2}{4a^2\beta^2}d\beta\biggr] +\end{split} +\end{gather*} +and that as $t$ increases $u$ approaches the value +\[ +u= \frac{1}{2}b_0 + \sum\limits_{m=1}^{m=\infty}\rho_m e^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin(m\alpha t - \frac{x}{a}\sqrt\frac{m\alpha}{2}+\lambda_m). +\] + +Given that +\[ +\int\limits_{0}^{\infty}e^{-x^2}\sin\frac{b^2}{x^2}dx=\frac{\sqrt\pi}{2}e^{-b\sqrt2}\sin b\sqrt2;\quad \int\limits_{0}^{\infty}e^{-x^2}\cos\frac{b^2}{x^2}dx=\frac{\sqrt\pi}{2}e^{-b\sqrt2}\cos b\sqrt2. +\] +v.\ \textit{Riemann, Lin.\ par.\ dif.\ gl.}\ §~54.\\ + +5.\quad If we are dealing with a bar of small cross-section where the heat not +only flows along the bar but at the same time escapes at the surface of the +bar into air at the temperature zero we have to solve the differential +equation +\[ +\phantom{v.\ Fourier, Heat §~105.} +D_t u = a^2D_x^2 u - b^2u. \tag*{v.\ Fourier, Heat §~105.} +\] +Show that for this case +\[ +u = e^{-(b^2+a^2\alpha^2)t}\sin\alpha x \qquad \text{and}\qquad u=e^{-(b^2+a^2\alpha^2)t}\cos\alpha x +\] +are particular solutions, and that if $u=f(x)$ when $t=0$ +\[ +u=\frac{e^{-b^2t}}{2a\sqrt{\pi t}}\int\limits_{-\infty}^{\infty}e^{-\frac{(\lambda-x)^2}{4a^2t}}f(\lambda)d\lambda + =\frac{e^{-b^2t}}{\sqrt\pi}\int\limits_{-\infty}^{\infty} e^{-\beta^2}f(x+2a\sqrt t.\beta)d\beta. +\] +cf.~(8) and (9) Art.~49. +% -----File: 100.png + +If $u = 0$ when $x = 0$ and $u = f(x)$ when $t = 0$ + +\[ +u = \frac{e^{-b^2t}}{\sqrt{ \pi}}\biggl [\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(x+2a \sqrt{ t }.\beta) d\beta - \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(-x + 2a \sqrt{ t}.\beta) d\beta \biggr ]. +\] +cf.~(7) Art.~50. + +If $u = -e^{-\frac{bx}{a}}$ when $t = 0$ and $u = 0$ when $x = 0$ +\[ +u = \frac{1}{\sqrt{\pi}} \biggl [e^{\frac{bx}{a}} \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-(b \sqrt{ t} + \beta)^2} d\beta - e^{-\frac{bx}{a}} \!\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-(b \sqrt{ t} + \beta)^2} d\beta \biggr ], +\] +and if $u = 1$ when $x = 0$ and $u = 0$ when $t = 0$ we have only to add + $e^{-\frac{bx}{a}}$ to the second member of the last equation, since $u = e^{-\frac{bx}{a}}$ satisfies the +equation +\[ +D_t u = a^2 D_x^2 u - b^2u. +\] + +If $u = F(t)$ when $x = 0$ and $u = 0$ when $t = 0$ we can employ the +method of Art.~51. +\begin{gather*} +\phi(x,t- \lambda) = e^{-\frac{bx}{a}} + \frac{1}{\sqrt{ \pi}} \biggl [e^{\frac{bx}{a}} \hspace{-10pt}\int\limits_{\frac{x}{2a\sqrt{t-\lambda}}}^{\infty}\hspace{-10pt} e^{-(b \sqrt{ t - \lambda} + \beta)^2} d\beta - e^{-\frac{bx}{a}} \hspace{-10pt}\int\limits_{-\frac{x}{2a\sqrt{t-\lambda}}}^{\infty}\hspace{-10pt} e^{-(b \sqrt{t-\lambda} + \beta)^2} d\beta \biggr ],\\ +-D_{\lambda} \phi(x,t-\lambda) = \frac{x(t-\lambda)^{-\frac{3}{2}}} { 2a \sqrt{ \pi}} e^{-b^2(t-\lambda)- \frac{x^2}{4a^2 (t-\lambda)}}; +\end{gather*} +\begin{flalign*} +&\text{and} &&u = \frac{x}{2a\sqrt{\pi} }\int\limits_0^t (t-\lambda)^{-\frac{3}{2}} e^{-b^2(t-\lambda) -\frac{x^2}{4a^2(t-\lambda)}} F(\lambda)d\lambda,&&\phantom{and} +\end{flalign*} +cf.\ (9) Art.~51, +\begin{flalign*} +&\text{or} &&u = \frac{2}{\sqrt{ \pi}} \!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\! e^{-\beta^2 -\frac{ b^2 x^2}{4a^2\beta^2} }F\biggl (t-\frac{x^2}{4a^2\beta^2}\biggr )d\beta,&&\phantom{or} +\end{flalign*} +cf.\ (10) Art.~51. + +If $F(t)$ is periodic and has the value taken in Ex.~4, show that the value +approached by $u$ as $ t$ increases is +\[ +u = \frac{1}{2}b_0 e^{-\frac{bx}{a}}+ \sum_{m=1}^{m=\infty} \rho_m e^{-\frac{x \sqrt{ 2}}{ 2a} p} \sin \biggl (m\alpha t - \frac{x \sqrt{2}}{2a }q + \lambda_m \biggr), +\] +\begin{flalign*} +&\text{where} && p = (b^2 + \sqrt{b^4 + m^2\alpha^2})^{\frac{1}{2}} \quad \text{and} \quad q = (-b^2 + \sqrt{b^4 + m^2\alpha^2})^{\frac{1}{2}}. &&\phantom{where} +\end{flalign*} +% -----File: 101.png +\begin{flalign*} +&\indent\text{Given} & \int\limits_0^\infty e^{-x^2-\frac{a^2}{x^2}}dx&=\frac{\sqrt\pi}{2}e^{-2a}&\phantom{\indent Given}\\ +& & \int\limits_0^\infty e^{-x^2-\frac{a^2}{x^2}}\sin\frac{b^2}{x^2}dx &=\frac{\sqrt\pi}{2}e^{-2c}\sin 2d&\\ +&\text{and} & \int\limits_0^\infty e^{-x^2-\frac{a^2}{x^2}}\cos\frac{b^2}{x^2}dx &=\frac{\sqrt\pi}{2}e^{-2c}\cos 2d,& +\end{flalign*} +where +\[ +c=\frac{\sqrt2}{2}(a^2+\sqrt{a^4+b^4})^\frac{1}{2}\quad \text{and}\quad d=\frac{\sqrt2}{2}(-a^2+\sqrt{a^4+b^4})^\frac{1}{2}. +\] + +Ängstrom's method of determining the conductivity of a metal is based on +the result just given (v.\ Phil.\ Mag.\ Feb.\ 1863), and is described by Sir Wm.~Thomson +(Encyc.\ Brit.\ Article ``Heat'') as by far the best that has yet been +devised. + +\mypara{52.} If $u$ is a periodic function of the time when $x=0$ as in Art.~51 Ex.~4 +and we are concerned with the limiting value approached by $u$ as $t$ increases +we can avoid evaluating a complicated definite integral if we take the following +course. + +Since as we have seen in Art.~49 $u=e^{\beta t+\alpha x}$ is a solution of +\[ +D_tu = a^2D_x^2u \tag{1} +\] +provided only that $\beta=a^2\alpha^2$ +we have +\[ +u=e^{\beta t±\frac{x}{a}\sqrt\beta} +\] +as a solution. + +Replacing $\beta$ by $±\beta i$ this becomes +\begin{flalign*} +& & u &=e^{±\beta ti±\frac{x}{a}\sqrt{\beta}\sqrt{±i}}&\\ +&\text{or} & u &=e^{±\beta ti±\frac{x}{a}\sqrt{\frac{\beta}{2}}(1±i)}&\\ +&\text{since} & \sqrt i &=±\frac{1}{2}\sqrt2(1+i)&\\ +&\text{and} & \sqrt{-i} &=±\frac{1}{2}\sqrt2(1-i).& +\end{flalign*} + +Hence +\begin{align*} +&& u&=e^{-\frac{x}{a}\sqrt\frac{\beta}{2}}\sin\biggl(\beta t-\frac{x}{a}\sqrt\frac{\beta}{2}\biggr),& + u&=e^{-\frac{x}{a}\sqrt\frac{\beta}{2}}\cos\biggl(\beta t-\frac{x}{a}\sqrt\frac{\beta}{2}\biggr), \tag{2}&\\ +&& u&=e^{ \frac{x}{a}\sqrt\frac{\beta}{2}}\sin\biggl(\beta t+\frac{x}{a}\sqrt\frac{\beta}{2}\biggr),& + u&=e^{ \frac{x}{a}\sqrt\frac{\beta}{2}}\cos\biggl(\beta t+\frac{x}{a}\sqrt\frac{\beta}{2}\biggr), \tag{3}& +\end{align*} +are particular solutions of (1). +% -----File: 102.png + +From these we get readily +\begin{flalign*} +&&u&=\rho_me^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin\biggl( +m\alpha t-\frac{x}{a}\sqrt\frac{m\alpha}{2}+\lambda_m\biggr) &\tag{4}& +\intertext{as a solution. (4) reduces to} +&&u&=\rho_m\sin(m\alpha t+\lambda_m)\quad \text{when}\quad x=0&&\\ +&\text{and to} &u&=\rho_me^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin\biggl +(\lambda_m-\frac{x}{a}\sqrt\frac{m\alpha}{2}\biggr) \quad \text{when}\quad t=0.&& +\end{flalign*} + +If we add a term which satisfies (1) and which is equal to zero when $x=0$ +and to +$-\rho_me^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin\biggl(\lambda_m-\dfrac{x}{a}\sqrt{\dfrac{m\alpha}{2}}\biggr)$ + when $t=0$ (v.\ Art.~50) we shall +have a solution of (1) which is zero when $t=0$ and which is +\[ +\rho_m\sin(m\alpha t+\lambda_m)\quad \text{when}\quad x=0. +\] +The term in question approaches zero as $t$ increases [v.~(7) Art.~50] and we +have at once the solution given in Art.~51 Ex.~4, as our required result. + +\EXAMPLE{} + +Show that $u=e^{\beta t+\alpha x}$ is a solution of $D_tu=a^2D_x^2u-b^2u$ if $\beta=a^2\alpha^2-b^2$, +and hence that +\begin{align*} +u&=e^{\beta t \pm\frac{x}{a}\sqrt{b^2+\beta}},\quad +u =e^{ \pm\beta ti \pm\frac{x}{a}\sqrt{b^2 \pm\beta i}},\quad +u =e^{ \pm\beta ti \pm\frac{x}{a\sqrt2}(p \pm qi)},\\ +u&=e^{ \pm\frac{px}{a\sqrt2}}\sin\biggl(\beta t \pm\frac{qx}{a\sqrt2}\biggr),\quad \text{and}\quad +u =e^{ \pm\frac{px}{a\sqrt2}}\cos\biggl(\beta t \pm\frac{qx}{a\sqrt2}\biggr), +\end{align*} +where +\[ +p=[\sqrt{\beta^2+b^4}+b^2]^\frac{1}{2}\quad \text{and}\quad +q=[\sqrt{\beta^2+b^4}-b^2]^\frac{1}{2}, +\] +are solutions. Hence +\[ +u=\rho_me^{-\frac{px}{a\sqrt2}}\sin\biggl(\beta t-\frac{qx}{a\sqrt2}+\lambda_m\biggr) +\] +is a solution. + +If $\beta=m\alpha$ this last result reduces to $u=\rho_m\sin(m\alpha t+\lambda_m)$ when $x=0$ +and by the reasoning of Art.~52 it must be the value $u$ approaches as $t$ increases +if we have the same conditions as in the last part of Art.~51 Ex.~5. + +\mypara{53.} The whole problem of the flow of heat is treated by Sir William Thomson +(v.\ Math.\ and Phys.\ Papers, Vol.~II), and other recent writers from a different +and decidedly interesting point of view, which we shall briefly sketch +in connection with the problem of \textit{Linear Flow}. + +Suppose we are dealing with a bar having a small cross-section and an adiathermanous +surface, and take as our unit of heat the amount required to raise by +a unit the temperature of a unit of length of the bar. If at a point of the bar a +% -----File: 103.png +quantity $Q$ of heat is suddenly generated the point is called an \emph{instantaneous +heat source} of strength $Q$. + +If the heat instead of being suddenly generated is generated gradually and +at a rate that would give $Q$ units of heat per unit of time the point is called a +\emph{permanent heat source} of strength $Q$. + +The temperature at any point of the bar at any time due to an instantaneous +source of strength $Q$ at the point $x=\lambda$ is easily found by the aid of formula +(8) Art.~49 as follows:--- + +If a quantity of heat $Q$ is suddenly generated along the portion of the bar +from $x=\lambda$ to $x=\lambda+\Delta\lambda$, where $\Delta\lambda$ is any arbitrary length, the temperature +of that portion will be suddenly raised to $\dfrac{Q}{\Delta\lambda}$, and we shall have by +(8) Art.~49 +\[ +u=\frac{Q}{2a\sqrt{\pi t}}\frac{1}{\Delta\lambda}\int\limits_\lambda^{\lambda+\Delta\lambda}e^{-\frac{(\lambda-x)^2}{4a^2t}}d\lambda \tag{1} +\] +as the temperature of any point of the bar at any time $t$ thereafter. + +If now we write $u$ equal to the limiting value approached by the second +member of (1) as $\Delta\lambda$ is made to approach zero we get +\[ +u=\frac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda-x)^2}{4a^2t}} \tag{2} +\] +as the solution for the case where we have an instantaneous source at the +point $x=\lambda$. + +It is to be observed that in (2) $u=0$ when $t=0$ and $u=\dfrac{Q}{2a\sqrt{\pi t}}$ +when $x=\lambda$ and $t>0$. + +If we have several sources we have only to add the temperatures due to the +separate sources. + +Formula (8) Art.~49 may now be regarded as the solution for the case where +we start with an instantaneous heat source of strength $f(\lambda)d\lambda$ in every +element of length of the bar. + +A source of strength $-Q$ is called a sink of strength $Q$; and (6) Art.~50 +may be regarded as the solution for the case where we have at the start an +instantaneous source of strength $f(\lambda)d\lambda$ in every element of the bar whose distance +to the right of the origin is $\lambda$, and an instantaneous sink of strength +$f(\lambda)d\lambda$ in every element of the bar whose distance to the left of the origin is $\lambda$. + +If we have an instantaneous source at the origin (2) reduces to +\[ +u=\frac{Q}{2a\sqrt{\pi t}}e^{-\frac{x^2}{4a^2t}} \tag{3} +\] +% -----File: 104.png + +For a permanent source of constant strength $Q$ at the origin (3) gives +\[ +u=\frac{Q}{2a\sqrt\pi}\int\limits_0^t e^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{1}{2}}d\tau \tag{4} +\] +and for a permanent source of variable strength $f(t)$ +\[ +u=\frac{1}{2a\sqrt\pi}\int\limits_0^t e^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{1}{2}}f(\tau)d\tau. \tag{5} +\] + +In (4) and (5) $u$ obviously reduces to zero when $t=0$ and $x>0$, but its +value when $x=0$ is not easily determined. We can avoid the difficulty by +introducing the conception of a \emph{doublet}. + +\mypara{54.} If a source and a sink of equal strength $Q$ are made to approach each +other while $Q$ multiplied by their distance apart is kept equal to a constant $P$ +the limiting state of things is said to be due to a \emph{doublet} of strength $P$ whose +axis is tangent to the line of approach and points from sink to source. A +\emph{doublet} of strength $-P$ differs from a doublet of strength $P$ only in that its +axis has the opposite direction. + +Let us find the temperature due to an instantaneous doublet of strength $P$ +placed at the origin. For a source of strength $Q$ at $x=\eta$ and an equal sink +at $x=-\eta$ we have +\begin{align*} +u&=\frac{Q}{2a\sqrt{\pi t}}(e^{-\frac{(\eta-x)^2}{4a^2t}}-e^{-\frac{(\eta+x)^2}{4a^2t}}),\\ +\intertext{or if $2\eta Q=P$,} +u&=\frac{P}{4a\eta\sqrt{\pi t}}e^{-\frac{(\eta^2+x^2)}{4a^2t}} +(e^{\frac{\eta x}{2a^2t}}-e^{-\frac{\eta x}{2a^2t}})\\ +&=\frac{P}{2a\eta\sqrt{\pi t}}e^{-\frac{(\eta^2+x^2)}{4a^2t}}\sinh\frac{\eta x}{2a^2t}. +\end{align*} + +If $\eta$ is made to approach zero +\[ +\limit\left[\frac{1}{\eta}\sinh\frac{\eta x}{2a^2t}\right]=\frac{x}{2a^2t}, +\] +\begin{flalign*} +&\text{and} & u=\frac{Px}{4a^3\sqrt{\pi t^3}}e^{-\frac{x^2}{4a^2t}} &\tag{1}& +\end{flalign*} +is the solution for the temperature at any time and place due to an instantaneous +doublet of strength $P$ placed at the origin. For a doublet at any other +point $x=\lambda$ we have +\[ +u=\frac{P(x-\lambda)}{4a^3\sqrt{\pi t^3}}e^{-\frac{(x-\lambda)^2}{4a^2t}}. \tag{2} +\] +% -----File: 105.png + +For a permanent doublet of constant strength $P$ placed at the origin we +have +\begin{flalign*} +&&u&=\frac{Px}{4a^3\sqrt\pi}\int\limits_0^t e^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{3}{2}}d\tau;& \tag{3}&\\ +\intertext{and for a permanent doublet of variable strength $f(t)$} +&&u&=\frac{x}{4a^3\sqrt\pi}\int\limits_0^te^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{3}{2}}f(\tau)d\tau, & \tag{4}&\\ +&\text{or} & u&=\frac{1}{a^2\sqrt\pi}\int\limits_\frac{x}{2a\sqrt t}^\infty e^{-\beta^2}f\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta & \tag{5}&\\ +\intertext{if $x>0$, and} +&&u&=\frac{1}{a^2\sqrt\pi}\int\limits_\frac{x}{2a\sqrt t}^{-\infty}e^{-\beta^2}f\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta & \tag{6}& +\end{flalign*} +if $x<0$, if we let $\beta=\dfrac{x}{2a\sqrt{t-\tau}}$. + +From (5) and (6) we see readily that $u=0$ when $t=0$ and that + $u=\dfrac{f(t)}{2a^2}$ when $x=0$ if we approach the origin from the right and that + $u=-\dfrac{f(t)}{2a^2}$ when $x=0$ if we approach the origin from the left. + +If the point $x=0$ is kept at the constant temperature $b$ and we are concerned +only with positive values of $x$ we can get from (5) the solution given in +Art.~50 Ex.~4 by supposing a permanent doublet of strength $2a^2b$ placed at +the origin. + +To solve the problem treated in Art.~51 we have only to suppose a permanent +doublet of strength $2a^2F(t)$ placed at $x=0$ and from (5) we get at once +(10) Art.~51. + +\EXAMPLE{} + +Show that if $D_tu=a^2D_x^2u-b^2u$ and an instantaneous source of strength +$Q$ is placed at $x=\lambda$ +\[\phantom{v.\ Art.~51, Ex.~5.} +u=\frac{Q}{2a\sqrt{\pi t}}e^{-b^2t-\frac{(\lambda-x)^2}{4a^2t}} \tag*{v.\ Art.~51, Ex.~5.} +\] + +Show that if an instantaneous doublet of strength $P$ is placed at the point + $x=0$ +\[ +u=\frac{Px}{4a^3\sqrt{\pi t^3}}e^{-b^2t-\frac{x^2}{4a^2t}}. +\] +% -----File: 106.png + +If a permanent doublet of strength $f(t)$ is placed at $x=0$ +\begin{align*} +u&=\frac{x}{4a^3\sqrt\pi}\int\limits_0^t e^{-b^2(t-\tau)-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{3}{2}}f(\tau)d\tau\\ + &=\frac{1}{a^2\sqrt\pi}\int\limits_{\frac{x}{2a\sqrt t}}^{±\infty} e^{-\beta^2-\frac{b^2x^2}{4a^2\beta^2}}f\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta, +\end{align*} +whence $u=0$ when $t=0$ and $x>0$ or $x<0$ and $u=±\dfrac{f(t)}{2a^2}$ when + $x=0$. + +Hence if we place at $x=0$ a permanent doublet of strength $2a^2F(t)$ we +get the solution given in Art.~51 Ex.~5 for the case where $u=F(t)$ when + $x=0$ and $u=0$ when $t=0$ provided we are concerned only with positive +values of $x$. + +If $F(t)=c$ this reduces to +\[ +u=\frac{2c}{\sqrt\pi}\int\limits_{\frac{x}{2a\sqrt t}}^{\infty} e^{-\beta^2-\frac{b^2x^2}{4a^2\beta^2}}d\beta. +\] +\mypara{55.} As another example of the use of Fourier's Integral we shall consider +the transmission of a disturbance along a stretched elastic string. + +Suppose we have a stretched elastic string so long that we need not consider +what happens at its ends, that is so long that we may treat its length as +infinite. Let the string be initially distorted into some given form and then +released; to investigate its subsequent motion. + +Let us take the position of equilibrium of the string as the axis of $X$ and +any given point as origin. + +We have, then, to solve the differential equation +\[ +D_t^2y=a^2D_x^2 y \tag{1} +\] +[v.\ \smallromr{VIII} Art.~1] subject to the conditions +\begin{alignat*}{2} + y&=f(x) & \quad\text{when}\quad & t=0 \tag{2}\\ +D_ty&=0 & \quad\text{``}\quad\quad & t=0. \tag{3} +\end{alignat*} + +As in Art.~8 we find +\[ +y=\cos\alpha(x±at)\quad \text{and}\quad y=\sin\alpha(x±at) +\] +as particular solutions of (1). + +From these we must build up a value that will reduce to +\[ +f(x)=\frac{1}{\pi} \int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x).d\lambda \tag{4} +\] +% -----File: 107.png +when $t=0$ and will at the same time satisfy (3). +\begin{flalign*} +&&y&=\cos\alpha\lambda\cos\alpha(x+at)+\sin\alpha\lambda\sin\alpha(x+at)&\phantom{or}\\ +&\text{or}& y&=\cos\alpha(\lambda-x-at)& +\end{flalign*} +is a solution of (1). +\begin{flalign*} +&\indent\text{Hence}& y=\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x-at).d\lambda & \tag{5}&\hspace{2.5em} +\end{flalign*} +is also a solution of (1). + +(5) reduces to $y=f(x)$ when $t=0$ but it gives +\[ +D_ty=\frac{a}{\pi}\int\limits_0^\infty\alpha d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\sin\alpha(\lambda - x).d\lambda +\] +when $t=0$ and consequently does not satisfy equation (3). + +If in forming (5) we use $\cos\alpha(x-at)$ and $\sin\alpha(x-at)$ instead of + $\cos\alpha(x+at)$ and $\sin\alpha(x+at)$ we get +\[ +y=\frac{1}{\pi}\int\limits_0^\infty d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x+at).d\lambda \tag{6} +\] +which is a solution of (1), and reduces to $y=f(x)$ when $t=0$, but it gives +\[ +D_ty=-\frac{a}{\pi}\int\limits_0^\infty\alpha d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\sin\alpha(\lambda-x).d\lambda +\] +when $t=0$ and does not satisfy (3). + +If, however, we take one-half the sum of the values of $y$ in (5) and (6) we +get +\begin{align*} +y=\frac{1}{2}\Bigg[\frac{1}{\pi}&\int\limits_0^\infty d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x-at).d\lambda\\ ++&\frac{1}{\pi}\int\limits_0^\infty d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x+at).d\lambda\Bigg], \tag{7} +\end{align*} +a solution of (1) which satisfies both (2) and (3), and is, therefore, our required +solution. + +This result can be very much simplified. + +If we substitute $z=x+at$ +\begin{align*} + &\frac{1}{\pi}\int\limits_0^\infty d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x-at).d\lambda\\ +=&\frac{1}{\pi}\int\limits_0^\infty d\alpha +\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-z).d\lambda +=f(z)=f(x+at); +\end{align*} +% -----File: 108.png +and in like manner we can show that +\[ +\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x+at).d\lambda=f(x-at). +\] + +Hence our solution becomes +\[ +y=\frac{1}{2}[f(x+at)+f(x-at)]. \tag{8} +\] + +This result is of great importance in the theory of elastic strings and it +shows that the initial disturbance splits into two equal waves which run along +the string, one to the right and the other to the left, with a uniform velocity $a$, +and that there is nothing like a periodic motion or vibration of any sort unless +the ends of the string produce some effect. + +\mypara{56.} If the string is not initially distorted but starts from its position of +equilibrium with a given initial velocity impressed upon each point we have to +solve the equation +\[ +D_t^2y=a^2D_x^2y \tag{1} +\] +subject to the conditions +\begin{alignat}{3} + y& = 0 & \text{when\quad} & t& = 0 &\tag{2}\\ +D_ty& = F(x) & \text{``\qquad} & t& = 0. &\tag{3} +\end{alignat} + +We get by the process used in Art.~55 +\begin{align*} +y&=\frac{1}{2\pi a}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty F(\lambda) +\left[\frac{\sin\alpha(\lambda-x+at)}{\alpha} + -\frac{\sin\alpha(\lambda-x-at)}{\alpha}\right]d\lambda\\ + &=\frac{1}{2\pi a}\int\limits_{-\infty}^\infty F(\lambda)d\lambda\int\limits_0^\infty +\left[\frac{\sin\alpha(\lambda-x+at)}{\alpha} + -\frac{\sin\alpha(\lambda-x-at)}{\alpha}\right]d\alpha; +\end{align*} +but \qquad\qquad $\displaystyle\int\limits_0^\infty\frac{\sin\alpha(\lambda-x+at)}{\alpha}d\alpha +-\int\limits_0^\infty\frac{\sin\alpha(\lambda-x-at)}{\alpha}d\alpha=\pi$\\ +if $x-at < \lambda < x+at$, and is equal to zero for all other values of $\lambda$; since +\begin{alignat*}{2} +\int\limits_0^\infty\frac{\sin mx}{x}dx +&= \phantom{-}\frac{\pi}{2} &\quad\text{if}\quad & m>0\\ +&= - \frac{\pi}{2} &\quad\text{if}\quad & m<0\\ +&= \phantom{-}0 &\quad\text{if}\quad & m=0. +\end{alignat*} +v.\ Int.\ Cal.\ Art.~92~(3). +\begin{flalign*} +&\indent\text{Hence} & y=\frac{1}{2a}\int\limits_{x-at}^{x+at}F(\lambda)d\lambda &\tag{4}&\hspace{2.5em} +\end{flalign*} +is our required solution. +% -----File: 109.png + +\EXAMPLE{S} + +1.\quad If the string is initially distorted and starts with initial velocity so that + $y=f(x)$ and $D_{t}y=F(x)$ when $t=0$ +\[ +y=\frac{1}{2}[f(x+at)+f(x-at)]+\frac{1}{2a}\int\limits_{x-at}^{x+at}F(\lambda)d\lambda. +\] + +2.\quad If the initial disturbance is caused by a blow, as from the hammer in a +piano, which impresses upon all the points in a portion of the string of length +$c$ an equal transverse velocity $b$ show that the front of the wave which will be +seen to run to the left along the string will be a straight line having a slope + equal to $\dfrac{b}{2a}$ and a length equal to $\displaystyle\frac{c}{2a}\sqrt{4a^2+b^2}$. Of course a wave having +a front of the same length with a slope equal to $-\dfrac{b}{2a}$ will be seen to run to +the right along the string, and the effect of the two waves will be to lift the +string bodily and permanently to a distance $\dfrac{bc}{2a}$ above its original position. + +\mypara{57.} We shall now take up a few examples of the use of \textit{Fourier's Series}. + +In the problem of Art.~7 let the temperature of the base of the plate be a +given function of $x$, the other conditions remaining unchanged. +\begin{flalign*} +&\indent\text{Since} & f(x)& =\sum_{m=1}^{m=\infty}(a_m\sin mx)&&\phantom{\indent Since}\\[-7ex] +\end{flalign*} +\begin{flalign*} +&\text{where} & a_m =& \frac{2}{\pi} \int\limits_0^\pi f(\alpha)\sin m\alpha.d\alpha&&\phantom{where}\\[-8ex] +\end{flalign*} +\begin{flalign*} +&\text{we have} &u=\frac{2}{\pi}\sum_{m=1}^{m=\infty}\bigg[e^{-my}&\sin mx\int\limits_0^\pi f(\alpha)\sin m\alpha.d\alpha\bigg]. & \tag{1}&\hspace{1.5em} +\end{flalign*} + +If the breadth of the plate is $a$ instead of $\pi$ +\[ +u=\frac{2}{a}\sum_{m=1}^{m=\infty}\bigg[e^{-\frac{m\pi y}{a}}\sin\frac{m\pi x}{a}\int\limits_0^a f(\lambda)\sin\frac{m\pi\lambda}{a}d\lambda\bigg]. \tag{2} +\] + +\mypara{58.} If the temperature of the base is unity and the breadth of the plate is +$\pi$ the solution is, as we have seen in Art.~7, +\[ +u=\frac{4}{\pi}\bigg[e^{-y}\sin x+\frac{1}{3}e^{-3y}\sin 3x+\frac{1}{5}e^{-5y}\sin 5x+\cdots\bigg]. \tag{1} +\] + +This series can be summed without difficulty. We have the development +\[ +\log(1+z)=\frac{z}{1}-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\cdots +\] +if the modulus of $z$ is less than 1. Int.\ Cal.\ Art.~221~(4). +% -----File: 110.png +\begin{flalign*} +& \text{Hence} & \log(1 - z) = -\frac{z}{1} -\frac{ z^2}{2} -\frac{ z^3}{3} -\frac{ z^4}{4} - \cdots && \phantom{Hence} +\end{flalign*} +if $\operatorname{mod.} z<1$. +\begin{flalign*} +& \text{and} & \frac{1}{2}[\log(1+z)-\log(1-z)]= \frac{ z}{1} +\frac{ z^3}{3} + \frac{ z^5}{5} + \cdots && \,\tag{2} +\end{flalign*} +if $\operatorname{mod.} z <1$. \\ + +But +\begin{flalign*} +\log(1+z) &=\log[1+r(\cos \phi + i \sin \phi)] \\ +&=\frac{1}{2} \log[(1+r \cos \phi)^2 + (r \sin \phi)^2] + i \tan^{-1} \frac{r \sin \phi }{ 1+r \cos \phi} \\ +&=\frac{1}{2} \log{(1 + 2r \cos \phi + r^2)} + i \tan^{-1}\frac{ r \sin \phi }{1+r \cos \phi}, +\intertext{and} +\log(1-z) &= \frac{1}{2} \log(1-2r \cos\phi + r^2) - i \tan^{-1} \frac{r \sin \phi}{ 1-r \cos\phi}, +\end{flalign*} +[Int.\ Cal.\ Art.~33~(2)],\\ +and (2) becomes +\begin{multline*} +\frac{1}{2}\left[\frac{1}{2} \log\frac{ 1+2r \cos\phi+r^2 }{ 1-2r \cos\phi+r^2} + i \tan^{-1} \frac{2r \sin \phi }{1-r^2} \right]\\ += \frac{r(\cos\phi+i \sin\phi)}{1} +\frac{ r^3(\cos 3\phi + i\sin 3\phi) }{3} + \cdots \tag{3} +\end{multline*} + +From (3) we get two equations +\begin{gather*} +\frac{1}{4}\log {\frac{1+2r \cos \phi+r^2}{ 1-2r \cos \phi+r^2}} = \frac{r\cos \phi}{1} + \frac{r^3 \cos 3 \phi}{3} + \frac{r^5 \cos 5 \phi}{5} + \cdots \tag{4}\\ +\frac{1}{2} \tan^{-1}\frac{ 2r \sin\phi}{ 1-r^2} = \frac{r \sin\phi}{1} + \frac{r^3 \sin3\phi}{3} + \frac{r^5 \sin5\phi}{5}+ \cdots \tag{5} +\end{gather*} +both valid for all values of $\phi$ provided $r < 1$. + +$e^{-y}$ is less than $1$ if $y$ is positive. + +Hence from (5) +\begin{align*} +\frac{e^{-y }\sin x}{1} + \frac{e^{-3y }\sin 3x}{3}+ \frac{e^{-5y }\sin 5x}{5} &+ \cdots = \frac{1}{2} \tan^{-1}\frac{ 2e^{-y }\sin x }{ 1-e^{-2y}}\\ + &= \frac{1}{2} \tan^{-1}\frac{ 2\sin x }{e^y-e^{-y} }= \frac{1}{2}\tan^{-1}\frac{ \sin x }{ \sinh y}, +\end{align*} +and (1) may be written\\ +\[ +u = \frac{2}{\pi}\tan^{-1}\frac{ \sin x }{ \sinh y}. \tag{6} +\] + +% -----File: 111.png +If we replace $r$ by $e^{-y}$ and $\phi$ by $x$ in\\[-4ex] +\begin{flalign*} +& & \log[1 + r(&\cos \phi + i \sin \phi)]&\\ +&\text{it becomes} & \log[1 + e^{-y}&\cos x + i e^{-y} \sin x]\hfil\phantom{it becomes}&\\ +&\text{or} & \log[1 + &\cos z + i \sin z]&\\[-5ex] +\end{flalign*} +v.\ Int.\ Cal.\ Art.~35 (3) and (4)\\ +a function of $z$ as a whole; and\\[-4ex] +\begin{flalign*} +& & \log [1 - r(&\cos \phi + i \sin \phi)]&\phantom{becomes}\\ +&\text{becomes} & \log (1 - &\cos z - i\sin z);&\\[-5ex] +\end{flalign*} +hence by Int.\ Cal.\ Arts.~212 and 213,\\[-4ex] +\begin{flalign*} +&&\frac{1}{4} \log &\frac{1 + 2e^{-y}\cos x + e^{-2y}}{1 - 2e^{-y}\cos x + e^{-2y}}\quad \text{and}\quad + \frac{1}{2}\tan^{-1}\frac{2e^{-y}\sin x}{1 - e^{-2y}}&\\ +&\text{or}&& \frac{1}{4}\log\dfrac{\cosh y + \cos x}{\cosh y -\cos x}\quad \text{and}\quad + \frac{1}{2}\tan^{-1}\frac{\sin x}{\sinh y}& +\end{flalign*} +are conjugate functions, and +\[ +u_1= \frac{1}{\pi}\log\dfrac{\cosh y + \cos x}{\cosh y - \cos x} \tag{7} +\] +is the solution for the problem where the isothermal lines are the lines of flow +of the present problem and the lines of flow are the isothermal lines of the +present problem. + +For our problem, then, the isothermal lines are given by the equation +\begin{flalign*} +& & \frac{2}{\pi}\tan^{-1} &\frac{\sin x}{\sinh y} = a&&\\ +&\text{or} & \dfrac{\sin x}{\sinh y}& = \tan\dfrac{a \pi}{2}& \tag{8}& +\end{flalign*} +and the lines of flow by +\begin{flalign*} +&& \frac{1}{\pi}&\log\dfrac{\cosh y + \cos x}{\cosh y - \cos x} = b,&&\\ +&\text{or}&& \dfrac{\cosh y + \cos x}{\cosh y - \cos x} = e^{\pi b}.&\tag{9}& +\end{flalign*} + +\EXAMPLE{S} + +1.\quad If $D_x^2u+D_y^2u=0$, and $u=1$ when $y=0$, and $u=0$ when +$x=0$ and when $x=a$, +\begin{align*} +u &= \frac{4}{\pi}\bigg[e^{-\frac{\pi y}{a}} \sin\frac{\pi x}{a} + \frac{1}{3}e^{- +\frac{3 \pi y}{a}} \sin\frac{3\pi x}{a} + \frac{1}{5}e^{-\frac{5 \pi y}{a}} \sin \frac{5\pi x}{a} + + \cdots \bigg] \\ +\label{err111} +&= \frac{2}{\pi}\tan^{-1} \frac{\sin \dfrac{\pi x}{a}}{\sinh \dfrac{\pi y}{a}}. +\end{align*} + +% -----File: 112.png + +2.\quad If $u = \phi (x)$ when $y = 0$, $u =f(y)$ when $x = 0$, and $u = F(y)$ + when $x = a$ +\begin{align*} + &u= \frac{2}{a} + \sum_{m=1}^{m=\infty} e^{-\frac{m\pi y}{a}} \sin \frac{m\pi x}{a} + \int \limits_0^a \phi(\lambda) \sin \frac{m\pi\lambda}{a} d\lambda +\\ +& + \frac{1}{2a} \sin \frac{\pi x}{a} \int \limits_0^\infty \ \biggl[ \ + \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda-y) + - \cos \dfrac{\pi x}{a} } + - \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda+y) + - \cos \dfrac{\pi x}{a} } + \ \biggr] \ f(\lambda)\, d\lambda +\\ +& + \frac{1}{2a} \sin \frac{\pi x}{a} \int \limits_0^\infty \ \biggl[ \ + \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda-y) + + \cos \dfrac{\pi x}{a} } + - \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda+y) + + \cos \dfrac{\pi x}{a} } + \ \biggr] \ F(\lambda)\, d\lambda +\end{align*} +v.\ Art.~48, Exs.~4, 5, and 6. + +\mypara{59.} If three sides of a plane rectangular sheet of conducting material be +kept at potential zero and the value of the potential function at every point of +the fourth side be given; to find the value of this potential function at any +point of the sheet. + +To formulate:--- +\begin{alignat*}{3} + &&\llap{$D_x^2V$} &\rlap{$+\;D_y^2V = 0.$} \tag{1}\\ + &V = 0 && \text{when}\quad && x = 0. \tag{2}\\ + &V = 0 && \quad\text{``} && x = a. \tag{3}\\ + &V = 0 && \quad\text{``} && y = b. \tag{4}\\ + &V = f(x) && \quad\text{``} && y = 0. \tag{5} +\end{alignat*} + +Working as in Art.~48 we get +\[ + \frac{ \sinh \dfrac{m\pi }{a} (b-y) } + { \sinh \dfrac{m\pi b}{a} } + \sin \frac{m\pi x}{a} +\] +as a value of $V$ which satisfies equations (1), (2), (3), and (4) if $m$ is an integer. +Therefore +\[ + V = \frac{2}{a} \sum_{m=1}^{m=\infty} \ \biggl[ \ + \frac{ \sinh \dfrac{m\pi }{a} (b-y) } + { \sinh \dfrac{m\pi b}{a} } + \sin \frac{m\pi x}{a} + \int \limits_0^a f(\lambda) \sin \frac{m\pi\lambda}{a} d\lambda + \ \biggr] +\tag{6} +\] +is our required solution. + +% -----File: 113.png +\EXAMPLE{S} + +1.\quad If $f(x) = 1$ Eq.~(6) Art.~59 reduces to +\[ +\begin{split} +V=\frac{4}{\pi} \Biggl [ &\frac{\sinh \dfrac{\pi}{a} (b-y) }{ \sinh \dfrac{\pi b}{a }}\sin \frac{\pi x}{a} + \frac{1}{3}\frac{\sinh \dfrac{3\pi}{a} (b-y) }{ \sinh \dfrac{3\pi b}{a} }\sin \frac{3\pi x}{a} \Biggr .\\ +&+\Biggl . \frac{1}{5}\frac{\sinh \dfrac{5\pi}{a} (b-y) } {\sinh \dfrac{5\pi b}{a}} \sin \frac{5\pi x}{a}+ \cdots \Biggr ] +\end{split} +\] + +2.\quad If $V=0$ when $x = 0$, $V=0$ when $x = a$, $V= 0$ when $y = 0$, + and $V=F(x)$ when $y = b$, then +\[ +V = \frac{2}{a} \sum_{m=1}^{m=\infty}\Biggl [\frac{\sinh \dfrac{m\pi y }{ a}} {\sinh \dfrac{m\pi b}{a}} \sin \frac{m\pi x}{a }\int \limits_0^a F(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda \Biggr ]. +\] + +3.\quad If $F(x) = 1$ the answer of Ex.~2 reduces to +\[ +V = \frac{4}{\pi}\Biggl [\frac{\sinh \dfrac{\pi y}{a }}{\sinh \dfrac{\pi b}{a}} \sin \frac{\pi x}{a }+ \frac{1}{3} \frac{\sinh \dfrac{3\pi y}{a}}{ \sinh \dfrac{3\pi b}{a}} \sin \frac{3\pi x}{a} + \frac{1}{5} \frac{\sinh \dfrac{5\pi y}{a}}{ \sinh \dfrac{5\pi b}{a}} \sin \frac{5\pi x}{a} + \cdots \Biggr ]. +\] + +4.\quad If $V=0$ when $x = 0$, $V=0$ when $x = a$, $V=f(x)$ when $y = 0$, + and $V= F(x)$ when $y = b$, then +\[ +\begin{split} +V = \frac{2}{a} \sum_{m=1}^{m=\infty} \Biggl [\sin \frac{m\pi x }{ a} \Biggl ( \frac{\sinh \dfrac{m\pi}{a} (b-y)}{ \sinh \dfrac{m\pi b}{a}} \int \limits_0^a f(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda \Biggr . \Biggr .\\ ++\Biggl . \Biggl . \frac{\sinh\dfrac{ m\pi y }{ a} }{ \sinh \dfrac{m\pi b}{a }}\int \limits_0^a F(\lambda)\sin\frac{ m\pi\lambda}{a} d\lambda \Biggr ) \Biggr ]. +\end{split} +\] + +5.\quad If $f(x) = F(x)$ the answer of Ex.~4 reduces to +\[ +V = \frac{2}{a }\sum_{m=1}^{m=\infty} \Biggl [ \frac{\cosh \dfrac{m\pi}{a } \Bigl (\dfrac{b}{2}-y \Bigr) }{\cosh \dfrac{m\pi b}{2a}} \sin \frac{m\pi x}{a} \int \limits_0^a f(\lambda)\sin \frac{m\pi \lambda}{a} d\lambda \Biggr ]. +\] +% -----File: 114.png + +6.\quad If $f(x) = F(x) = 1$ the answer of Ex.~5 reduces to +\[\begin{split} +V = \frac{4}{\pi}\Biggl [ &\frac{\cosh\dfrac{ \pi}{a} \Bigl ( \dfrac{b}{2}-y \Bigr ) }{\cosh \dfrac{\pi b}{2a}} \sin\frac{ \pi x}{a} + \frac{1}{3}\dfrac{\cosh\dfrac{3 \pi}{a} \Bigl ( \dfrac{b}{2}-y \Bigr ) }{\cosh \dfrac{3\pi b}{2a}} \sin\frac{ 3\pi x}{a} \Biggr .\\ +&+\Biggl . \frac{1}{5}\dfrac{\cosh\dfrac{5 \pi}{a} \Bigl ( \dfrac{b}{2}-y \Bigr ) }{\cosh \dfrac{5\pi b}{2a}} \sin\frac{ 5\pi x}{a} + \cdots \Biggr ]. +\end{split} +\] + +7.\quad If $V=f(x)$ when $y = 0$, $V=F(x)$ when $y = b$, $V=\phi (y)$ when + $x = 0$, and $V=\chi (y)$ when $x=a$, then +\[ +\begin{alignedat}{2} +V &= \frac{2}{a} \sum_{m=1}^{m=\infty}\Biggl [&&\sin \frac{m\pi x}{a} \Biggl (\dfrac{\sinh \dfrac{m\pi}{a}(b-y) }{ \sinh \dfrac{m\pi b}{a}} \int\limits_0^a f(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda \Biggr . \Biggr .\\ +&&&+ \Biggl . \Biggl. \dfrac{ \sinh \dfrac{m\pi y}{a }} {\sinh \dfrac{m\pi b}{a}} \int\limits_0^a + F(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda\Biggr) \Biggr ]\\ + &+ \frac{2}{b}\sum_{m=1}^{m=\infty}\Biggl [&&\sin \frac{m\pi y }{b} \Biggl (\frac{\sinh \dfrac{m\pi}{b}(a-x) }{ \sinh \dfrac{m\pi a}{b}} \int\limits_0^b \phi(\lambda)\sin \frac{m\pi\lambda}{b} d\lambda \Biggr . \Biggr .\\ +&&&+ \Biggl . \Biggl . \frac{\sinh \dfrac{m\pi x}{b} }{ \sinh \dfrac{m\pi a}{b}} \int\limits_0^b + \chi(\lambda)\sin \frac{m\pi\lambda}{b} d\lambda \Biggr ) \Biggr ]. + \end{alignedat} +\] + +8.\quad If $f(x)=\phi(y)=0$ and $F(x)=\chi(y)=1$ the answer of Ex.~7 may be +reduced to +\begin{align*} +V = \frac{2}{\pi} \Biggl [ &\frac{ \pi y}{2b} - \frac{\sinh \dfrac{\pi}{b} \left (\dfrac{a}{2}-x \right ) }{ \sinh \dfrac{\pi a}{2b }}\sin \frac{\pi y}{b} + \frac{1}{2}\frac{ \cosh \dfrac{2\pi}{b} \left (\dfrac{a}{2}-x \right ) }{\cosh \dfrac{2\pi a}{2b}} \sin \frac{2\pi y}{b} \Biggr .\\ + &- \Biggl . \frac{1}{3} \frac{\sinh \dfrac{3\pi}{b} \left (\dfrac{a}{2}-x \right ) }{ \sinh \dfrac{3\pi a}{2b}} \sin \frac{3\pi y}{b} + \frac{1}{4} \frac{\cosh \dfrac{4\pi}{b} \left (\dfrac{a}{2}-x \right ) }{ \cosh \dfrac{4\pi a}{2b}} \sin \frac{4\pi y}{b} - \cdots \Biggr]. +\end{align*} +% -----File: 115.png + +9.\quad Find the temperature of the middle point of a thin square plate whose +faces are impervious to heat; 1st, when three edges are kept at the temperature +0° and the fourth edge at the temperature 100°; 2d, when two +opposite edges are kept at the temperature 0° and the other two at the temperature +100°; 3d, when two adjacent edges are kept at the temperature 0° +and the other edges at the temperature 100°. See examples 3, 6, and 8.\\ +\phantom{kindly go to the right hand margin}\hfill +\textit{Ans.}, (1) 25°; (2) 50°; (3) 50°. + +\mypara{60.} Let us pass on to the consideration of the flow of heat in one dimension. + +Suppose that we have an infinite solid with two parallel plane faces whose +distance apart is $c$. + +Take the origin in one face and the axis of $X$ perpendicular to the faces. +Let the initial temperature be any given function of $x$ and let the two faces be +kept at the constant temperature zero; to find the temperature at any point of +the slab at any time. + +We have to solve the equation +\[ +D_tu = a^2D_x^2u \tag{1} +\] +subject to the conditions +\begin{alignat*}{3} +u& = 0 && \text{when}\quad & x & = 0 \tag{2}\\ +u& = 0 && \quad\text{``} & x & = c \tag{3}\\ +u& =f(x)&& \quad\text{``} & t & = 0. \tag{4} +\end{alignat*} + +In Art.~49 we have found +\begin{flalign*} +&&u &= e^{-a^2\alpha^2t}\sin\alpha x&\\ +&\text{and} & u&=e^{-a^2\alpha^2t}\cos\alpha x&\\ +\end{flalign*} +as particular solutions of (1). + +$u=e^{-a^2\alpha^2t}\sin\alpha x$ satisfies (2) whatever value is given to $\alpha$. It satisfies (3) +if $\alpha=\dfrac{m\pi}{c}$ provided $m$ is an integer. Let us try to build a value of $u$ out of +terms of the form $Ae^{-\frac{a^2m^2\pi^2t}{c^2}}\sin\dfrac{m\pi x}{c}$ which shall satisfy (4). + +We have +\begin{gather*} +f(x)=\frac{2}{c}\sum_{m=1}^{m=\infty}\Bigl[\sin\frac{m\pi x}{c} +\int\limits_0^c f(\lambda)\sin\frac{m\pi\lambda}{c}d\lambda\Bigr]. \tag{5}\\ +u=\frac{2}{c}\sum_{m=1}^{m=\infty}\Bigl[e^{-\frac{m^2a^2\pi^2t}{c^2}}\sin\frac{m\pi x}{c} +\int\limits_0^c f(\lambda)\sin\frac{m\pi\lambda}{c}d\lambda\Bigr], \tag{6} +\end{gather*} +reduces to (5) when $t=0$ and is our required solution. +% -----File: 116.png + +\EXAMPLE{S} + +1.\quad If $f(\lambda)=b$, a constant, (6) Art.~60 reduces to +\[ +u=\frac{4b}{\pi}\Bigl[e^{-\frac{ a^2\pi^2t}{c^2}}\sin\frac{ \pi x}{c} + +\frac{1}{3}e^{-\frac{ 9a^2\pi^2t}{c^2}}\sin\frac{3\pi x}{c} + +\frac{1}{5}e^{-\frac{25\pi^2a^2t}{c^2}}\sin\frac{5\pi x}{c} ++\cdots\Bigr]. +\] + +2.\quad An iron slab 10~cm.\ thick is placed between and in contact with two +other iron slabs each 10~cm.\ thick. The temperature of the middle slab is at +first 100° throughout, and of the outside slabs 0° throughout. The outer faces +of the outside slabs are kept at the temperature 0°. Required the temperature +of a point in the middle of the middle slab fifteen minutes after the slabs have +been placed in contact. Given $a^2=0.185$ in C.G.S. units. \hfill \textit{Ans.}, 10°.3.\\ + +3.\quad Two iron slabs each 20~cm.\ thick one of which is at the temperature 0° +and the other at the temperature 100° throughout, are placed together face to +face, and their outer faces are kept at the temperature 0°. Find the temperature +of a point in their common face and of points 10~cm.\ from the common +face fifteen minutes after the slabs have been put together.\\ +\phantom{kindly go to the right hand margin}\hfill +\textit{Ans.}, 22°.8; 15°.1; 17°.2.\\ + +4.\quad One face of an iron slab 40~cm.\ thick is kept at the temperature 0° and +the other face at the temperature 100° until the permanent state of temperatures +is set up. Each face is then kept at the temperature 0°. Required +the temperature of a point in the middle of the slab, and of points 10~cm.\ from +the faces fifteen minutes after the cooling has begun. \hfill \textit{Ans.}, 22°.8; 15°.6; 16°.7. + +\mypara{61.} If the faces of the slab treated in Art.~60 instead of being kept at the +temperature zero are rendered impervious to heat, the solution of the problem +is easy. + +In this case we have to solve the equation +\[ +D_tu = a^2D_x^2u +\] +subject to the conditions +\begin{alignat*}{3} +D_xu& = 0 &&\text{when}\quad & x&=0\\ +D_xu& = 0 &&\text{\quad``} & x&=c\\ + u& = f(x)&&\text{\quad``} & t&=0. +\end{alignat*} + +We have only to use the particular solution +\begin{flalign*} +& & u& = e^{-a^2\alpha^2t}\cos\alpha x&\\ +&\text{as we used}& u& = e^{-a^2\alpha^2t}\sin\alpha x&\phantom{as we used} +\end{flalign*} +in Art.~60. We get +\[ +u=\frac{2}{c}\biggl[\frac{1}{2}\int\limits_0^c f(\lambda)d\lambda ++\sum_{m=1}^{m=\infty} +\Bigl(e^{-\frac{m^2a^2\pi^2t}{c^2}}\cos\frac{m\pi x}{c} +\int\limits_0^c f(\lambda)\cos\frac{m\pi\lambda}{c}d\lambda\Bigr)\biggr]. \tag{1} +\] +% -----File: 117.png + +\EXAMPLE{S} + +1.\quad Solve example 2 Art.~60 supposing that the outer surfaces are blanketed +after the slabs are placed together so that heat can neither enter nor escape. +Find in addition the temperature of the outer surfaces fifteen minutes after +the slabs are placed in contact. \hfill \textit{Ans.}, 33°.3; 33°.3.\\ + +2.\quad Solve example 3 Art.~60 on the hypothesis just stated, getting in addition +the temperatures of points on the outer surfaces.\\ +\phantom{kindly go to the right hand margin}\hfill +\textit{Ans.}, 50°; 33°.9; 66°.1; 27°.2; 72°.8.\\ + +3.\quad Solve example 4 Art.~60 supposing that heat neither enters nor escapes +at the outer surfaces after the permanent state of temperatures has been set +up. Find also the temperatures of points in the outer surfaces.\\ +\phantom{kindly go to the right hand margin}\hfill +\textit{Ans.}, 50°; 39°.7; 60°.3; 35°.5; 64°.5.\\ + +4.\quad Show that if $u = 0$ when $x = 0$, $D_xu = 0$ when $x = c$, and $u=f(x)$ + when $t = 0$, +\[ +u = \frac{2}{c}\sum_{m=0}^{m=\infty} \biggl ( e^{-\frac{(2m+1)^2a^2\pi^2t}{4c^2}} \sin\frac{(2m+1)\pi x}{2c} \int\limits_0^c f(\lambda) \sin \frac{(2m+1)\pi\lambda}{2c} d\lambda \biggr ). +\] + +\textit{Suggestion}: Assume $u = 0$ when $x = 2c$ and $f(2c - x) =f(x)$, and see +(6) Art.~60. + +\mypara{62.} If the temperature of the right-hand face of the slab considered in Art.~60 +is a constant $\gamma$ instead of zero we have only to add to the second member +of (6) Art.~60 a term $u_1$ which shall satisfy the conditions +\begin{alignat*}{3} +&&\llap{$D_tu_1\:$} &\rlap{$= a^2D_x^2u_1$}\tag{1}\\ +u_1 &= 0\quad && \text{when} \quad & x &= 0 \tag{2}\\ +u_1 &= 0 && \quad \text{``} & t &= 0 \tag{3}\\ +u_1 &= \gamma && \quad \text{``} & x &= c. \tag{4} +\end{alignat*} + +$u_1 = \dfrac{\gamma x}{c}$ obviously satisfies (1), (2), and (4); to make it satisfy (3) as well +we must add a term $u_2$ which shall be equal to zero when $x = 0$ and when + $x = c$ and to $-\dfrac{\gamma x}{c}$ when $t = 0$, while always satisfying (1). It is given +immediately by (6) Art.~60 and is +\begin{gather*} +u_2=-\frac{2\gamma}{c^2} \sum_{m=1}^{m=\infty} \biggl (e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \int\limits_0^c \lambda \sin\frac{m\pi\lambda}{c} d\lambda \biggr ). \tag{5}\\ +\int\limits_0^c \lambda \sin\frac{m\pi\lambda}{c} d\lambda =-\frac{c^2}{m\pi}\cos m\pi = (- 1)^{m+1} \frac{c^2}{m\pi}, +\end{gather*} +% -----File: 118.png +\begin{flalign*} +&\text{and} &&u_2=\frac{2\gamma}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{(-1)^m}{m }e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \right ). && \tag{6}\\[-5ex] +\end{flalign*} +\begin{flalign*} +&\indent \text{Hence} && u_1 = \gamma \biggl [ \frac{x}{c} + \frac{2}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{(-1)^m}{m }e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \right ) \biggr ]. &&\hspace{2em} \tag{7} +\end{flalign*} + +If the left-hand face of the slab considered in Art.~60 is to be kept at a +constant temperature $\beta$ and the right-hand face at the temperature zero we +can get the term $u_3$ which must be added to the second member of (6) Art.~60 +by replacing $\gamma$ by $\beta$ and $x$ by $c-x$ in (7). We then have +\[ +u_3= \beta \biggl [\frac{c-x}{c} - \frac{2}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{1}{m}e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \right ) \biggr ]. \tag{8} +\] + +\EXAMPLE{S} + +1.\quad Show that if $u = \beta$ when $x = 0$, $u = \gamma$ when $x = c$, and $u=f(x)$ +when $t = 0$ +\[ +\begin{split} +u=\beta + (\gamma - \beta) \biggl [ \frac{x}{c} + \frac{2}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{(-1)^m}{m }e^{-\frac{m^2\pi^2a^2t}{c^2}} \sin \frac{m\pi x}{c} \right ) \biggr ]\\ ++\frac{2}{c} \sum_{m=1}^{m=\infty} \biggl ( e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \int\limits_0^c [f(\lambda)-\beta] \sin \frac{m\pi\lambda}{c} d\lambda \biggr ). +\end{split} +\] + +2.\quad Show that if $u = \beta$ when $x = 0$, $u = 0$ when $t = 0$, and $D_xu = 0$ + when $x = c$ +\begin{align*} +&u =\beta \biggl [1 - \frac{4}{\pi}\sum_{m=0}^{m=\infty} \left (\frac{1}{2m+1} e^{-\frac{(2m+1)^2a^2\pi^2t}{4c^2}}\sin \frac{(2m+1)\pi x}{2c} \right ) \biggr ]\\ +&=\beta \left [1 -\frac{4}{\pi} \left (e^{-\frac{ a^2\pi^2t}{4c^2}} \!\sin \frac{ \pi x}{2c} + + \frac{1}{3}e^{-\frac{ 9a^2\pi^2t}{4c^2}} \!\sin \frac{3\pi x}{2c} + + \frac{1}{5}e^{-\frac{25a^2\pi^2t}{4c^2}} \!\sin \frac{5\pi x}{2c} + \cdots \!\right ) \right ]. +\end{align*} + +\mypara{63.} If the temperature of the right-hand face of the slab just considered is +a function of the time instead of a constant and the temperature of the left-hand +face is zero the problem can be solved by a method nearly identical with +that of Art.~51. +% -----File: 119.png + +Let $\phi(x,t)$ be a function of $x$ and $t$ which shall be zero if $t$ is less than zero +and shall be equal to +\[ +\frac{x}{c}+\frac{2}{\pi}\sum_{m=1}^{m=\infty} +\left(\frac{(-1)^m}{m}e^{-\frac{m^2a^2\pi^2t}{c^2}}\sin\frac{m\pi x}{c}\right) +\] +[v.\ (7) Art.~62] if $t$ is equal to or greater than zero. So that +\begin{alignat*}{4} +\phi(x,t)& =0 \quad&&\text{if}\quad & t &< 0 &&\\ +\phi(x,t)& =0 &&\text{``} & t &= 0 & \quad\text{unless}\quad x &= c\\ +\phi(x,t)& =1 &&\text{``} & t &= 0 & \text{and}\quad x &= c\\ +\phi(x,t)& =1 &&\text{``} & x &= c &&\\ +\phi(x,t)& =0 &&\text{``} & x &= 0.&& +\end{alignat*} + +Precisely as in Art.~51 we get +\[ +u=\limit_{\tau\doteq 0} \sum_{k=0}^{k=n}\left[F(k\tau)\frac{[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)]\tau}{\tau}\right] \tag{1} +\] +as the required solution of our problem, $n$ being as in Art.~51 the largest +integer in $\dfrac{t}{\tau}$ where $t$ is any given value of the time. + +On our hypothesis the last term of (1), that is, $-F(n\tau)\phi[x,t-(n+1)\tau]=0$; +the next to the last term $F(n\tau)\phi(x,t-n\tau)$ has for its limiting value +\[ +F(t)\phi(x,0)=F(t)\left[\frac{x}{c}+\frac{2}{\pi}\sum_{m=1}^{m=\infty}\left(\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}\right)\right], +\] +while as in Art.~51 the limiting value of the rest of the sum is +\begin{gather*} +-\int\limits_0^t F(\lambda)D_\lambda\phi(x,t-\lambda)d\lambda.\\ +D_\lambda\phi(x,t-\lambda)=\frac{2a^2\pi}{c^2}\sum_{m=1}^{m=\infty}\bigg[(-1)^m me^{-\frac{m^2a^2\pi^2}{c^2}(t-\lambda)}\sin\frac{m\pi x}{c}\bigg]. +\end{gather*} + +Hence +\begin{align*} +u=F(t)\biggl[\frac{x}{c}&+\frac{2}{\pi}\sum_{m=1}^{m=\infty}\biggl(\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}\biggr)\biggr]\\ +&-\frac{2a^2\pi}{c^2}\sum_{m=1}^{m=\infty}\biggl((-1)^m m\sin\frac{m\pi x}{c}\int\limits_0^t F(\lambda)e^{-\frac{m^2a^2\pi^2}{c^2}(t-\lambda)}d\lambda\biggr), +\end{align*} +% -----File: 120.png +\begin{multline*} +u=\frac{x}{c}F(t)+\frac{2}{\pi}\sum_{m=1}^{m=\infty}\biggl[ +\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}\biggl(F(t)\\ +-\frac{m^2a^2\pi^2}{c^2}\int\limits_0^t F(\lambda)e^{-\frac{m^2a^2\pi^2}{c^2}(t-\lambda)}d\lambda\biggr)\biggr]. +\tag{2} +\end{multline*} + +If we substitute $\beta=\dfrac{m^2a^2\pi^2}{c^2}(t-\lambda)$ we get +\[ +u=\frac{x}{c}F(t)+\frac{2}{\pi}\sum_{m=1}^{m=\infty} +\biggl[\frac{(-1)^m}{m}\sin\frac{m\pi x}{c} +\biggl(F(t)-\hspace{-10pt}\int\limits_0^{\frac{m^2a^2\pi^2t}{c^2}}\hspace{-10pt}e^{-\beta} +F\biggl(t-\frac{\beta c^2}{m^2a^2\pi^2}\biggr)d\beta\biggr)\biggr]. \tag{3} +\] + +\EXAMPLE{S} + +1.\quad If the temperature of the left-hand face is a function of $t$ and the temperature +of the right-hand face is zero and the initial temperature is zero +\[ +u=\biggl(1-\frac{x}{c}\biggr)F(t)-\frac{2}{\pi}\sum_{m=1}^{m=\infty} +\biggl[\frac{1}{m}\sin\frac{m\pi x}{c} +\biggl(F(t)-\hspace{-10pt}\int\limits_0^{\frac{m^2a^2\pi^2t}{c^2}}\hspace{-10pt}e^{-\beta} +F\biggl(t-\frac{\beta c^2}{m^2a^2\pi^2}\biggr)d\beta\biggr)\biggr]. +\] + +2.\quad If the temperature of the left-hand face is a function of $t$, the initial +temperature is zero, and the right-hand face is impervious to heat +\begin{multline*} +u=F(t)-\frac{4}{\pi}\sum_{m=0}^{m=\infty} +\biggl[\frac{1}{2m+1} \sin\frac{(2m+1)\pi x}{2c} +\biggl(F(t)\\ +-\frac{(2m+1)^2a^2\pi^2}{4c^2}\int\limits_0^t +F(\lambda)e^{-\frac{(2m+1)^2a^2\pi^2}{4c^2}(t-\lambda)}d\lambda\biggr)\biggr]. +\end{multline*} + +3.\quad If in Arts.~60--63 we are dealing with a bar of small cross-section and of +length $c$ and heat is radiating from the surface of the bar into air at the temperature +zero so that $D_tu=a^2D_x^2u-b^2u$, show that: (\textit{a}) the second members +of (6) Art.~60 and (1) Art.~61 must be multiplied by $e^{-b^2t}$; (\textit{b}) equation +(7) Art.~62 becomes +\[ +u_1=\gamma\Biggl\{\frac{\sinh\dfrac{bx}{a}}{\sinh\dfrac{bc}{a}}+2a^2\pi e^{-b^2t} +\sum_{m=1}^{m=\infty}\left[(-1)^m\frac{m}{b^2c^2+m^2a^2\pi^2} +e^{-\frac{m^2a^2\pi^2t}{c^2}}\sin\frac{m\pi x}{c}\right]\Biggr\}; +\] +% -----File: 121.png +(\textit{c}) equation (2) Art.~63 becomes +\begin{multline*} +u=\frac{\sinh \dfrac{bx}{a}} {\sinh \dfrac{bc}{a}} F(t)+2a^2\pi \sum_{m=1}^{m=\infty} \bigg \{ \frac{(-1)^m m}{b^2c^2+m^2a^2\pi^2} \sin\frac{ m\pi x}{c} \bigg [ F(t) \\ +- \frac{b^2c^2+m^2a^2\pi^2}{c^2} \int\limits_0^t e^{-\frac{b^2c^2+m^2a^2\pi^2}{c^2}(t-\lambda)}F(\lambda)d\lambda \bigg ] \bigg \}. +\end{multline*} + +\mypara{64.} The problem of the motion of a finite stretched elastic string of length +$l$ fastened at the ends and distorted at first into some given curve $y=f(x)$, + and then allowed to swing, has been treated and partially solved in Art.~8. + +The complete solution is easily seen to be +\[ +y=\frac{2}{l} \sum_{m=1}^{m=\infty} \sin \frac{m\pi x}{l} \cos \frac{m\pi at}{l }\int\limits_0^l f(\lambda) \sin \frac{m\pi\lambda}{l }d\lambda. \tag{1} +\] + +The second member of (1) is a periodic function of $t$ having the period +$\dfrac{2l}{a}$. The motion, then, unlike that in the case of an infinite string (Art.~55) is +a true vibration, a periodic motion. The period $\dfrac{2l}{a}$ is the time it takes a disturbance +to travel twice the length of the string (v.\ Art.~55). + +A careful examination of (1) will show that the actual motion is a good deal +like that in the case considered in Art.~55. The original disturbance breaks +up into two waves one of which runs to the right until it reaches the end of +the string and is then reflected, and runs back to the left or the under side of +the string, while the other wave runs to the left and is reflected at the left-hand +end of the string and runs back to the right under the string and is +again reflected, runs back to the left over the string and so on indefinitely. + +If the curve into which the string is distorted at the start is of the form + $y=b \sin \dfrac{m\pi x}{l}$ the solution is +\[ +y=b \sin \frac{m\pi x}{l} \cos\frac{ m\pi at}{l}. \tag{2} +\] + +No matter what value $t$ may have the curve is always of the form +\[ +y=A \sin \frac{m\pi x}{l}; +\] +that is, for different values of $t$ we have a set of sine curves differing only in +the amplitude and not at all in the period of the curve. In this case either +the whole string if $m=1$, or each $m$th +of the string if $m$ is not equal to +one, rises and falls, and there is no apparent onward motion. When this is +the case we are said to have a \textit{steady} vibration. +% -----File: 122.png + +If $m = 1$ we get steady motion of the string as a whole and if the vibration +is rapid enough to give a musical note the note is said to be the pure fundamental +note of the string. If $m = 2$ the vibration is twice as rapid as when + $m = 1$, the middle point of the string does not move and is called a node, the +two halves of the string are in opposite phases of vibration at any instant, and +the note given is an octave higher than the fundamental note and is called its +pure \textit{first harmonic}. + +If $m = 3$ the vibration is three times as rapid as in the first case, there are +two nodes $x=\dfrac{l}{3}$ and $x=\dfrac{2l}{3}$, and the note is the pure \textit{second harmonic} of +the fundamental note. + +For any value of $m$ the vibration is $m$ times as rapid as when $m = 1$, there +are ${m -1}$ nodes at the points $x=\dfrac{l}{m}, x=\dfrac{2l}{m}, \cdots x=\dfrac{m-1}{m}l$, and we get the +${m - 1}$st harmonic of the fundamental note. + +It is clear from (1) that no matter what the original form of the string the +resulting vibration can be regarded as a combination of steady vibrations each +of which alone would give the fundamental note of the string or one of its +harmonics, and that the complex note resulting is really a concord of the fundamental +note and some of its harmonics. + +A finely trained ear can often recognize in a complex note the fundamental +note of the string and some of its harmonics and is capable of analyzing a +complex note into its component pure notes precisely as Fourier's Theorem +enables us to analyze the complex function representing the initial form of the +string into the simpler sine-functions which must be combined to form it. + +\EXAMPLE{S} + +1.\quad Show that if a point whose distance from the end of a harp string is +$\dfrac{1}{n}$th the length of the string is drawn aside by the player's finger to a distance +$b$ from its position of equilibrium and then released, the form of the vibrating +string at any instant is given by the equation +\[ +y=\frac{2bn^2}{(n-1)\pi^2} \sum_{m=1}^{m=\infty} \left ( \dfrac{1}{m^2} \sin \frac{m\pi}{n} \sin \frac{ m\pi x}{l} \cos \frac{m\pi at}{l} \right ). +\] +Show from this that all the harmonics of the fundamental note of the +string which correspond to forms of vibration having nodes at the point +drawn aside by the finger will be wanting in the complex note actually +sounded.\\ +% -----File: 123.png + +2.\quad If a stretched string starts from its position of equilibrium, each of its +points having a given initial velocity, so that we have +\begin{alignat*}{3} + y&=0 & \text{when}\quad && t&=0\\ +D_ty&=F(x) & \text{``}\qquad && t&=0\\ + y&=0 & \text{``}\qquad && x&=0\\ + y&=0 & \text{``}\qquad && x&=l, +\end{alignat*} +the solution of the problem of its vibration is easy and gives +\[ +y=\frac{2}{a\pi}\sum_{m=1}^{m=\infty}\bigg(\frac{1}{m}\sin\frac{m\pi x}{l}\sin\frac{m\pi at}{l} +\int\limits_0^l F(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\bigg). +\] + +3.\quad Write down the solution for the case where the string is initially distorted +and each point has a given initial velocity. + +\mypara{65.} If we do not neglect the resistance of the air in the problem of the +vibration of a stretched string the differential equation is rather more complicated +and the solution is not so easily obtained. The equation is given as \smallromr{IX} +Art.~1. + +Let us solve the problem for the case where there is no initial velocity. +\begin{flalign*} +&\indent\text{Here we have} & D_t^2y+2kD_ty=a^2D_x^2y.\qquad\qquad & \tag{1}& +\end{flalign*} +\vspace{-6.5ex} +\begin{alignat*}{3} + y&=0 && \text{when}\quad & x&=0 \tag{2}\\ + y&=0 && \quad\text{``} & x&=l \tag{3}\\ + y&=f(x) && \quad\text{``} & t&=0 \tag{4}\\ +D_ty&=0 && \quad\text{``} & t&=0. \tag{5} +\end{alignat*} + +We get particular solutions of (1) in the usual way. Assume $y=e^{\alpha x+\beta t}$ + and substitute in (1). We have +\[ +\beta^2+2k\beta=a^2\alpha^2 +\] +as the only necessary relation between $\beta$ and $\alpha$. This gives +\begin{flalign*} +&&\beta=-k±\sqrt{a^2\alpha^2+k^2}.&&\\ +&\indent\text{Hence}& y=e^{\alpha x-kt±t\sqrt{a^2\alpha^2+k^2}}& \tag{6}& +\end{flalign*} +is a solution of (1) no matter what the value of $\alpha$. + +To throw it into Trigonometric form replace $\alpha$ by $\alpha i$, and since in actual +problems $k$, which is proportional to the resistance, is very small, take $-1$ +out as a factor of the radical. We have +\[ +y=e^{-kt}e^{(\alpha x±t\sqrt{a^2\alpha^2-k^2)}i}. +\] +% -----File: 124.png + +Since $\alpha$ +may be positive or negative we can get +\begin{flalign*} +& &y = e^{-kt}\sin(\alpha x ± t\sqrt{a^2\alpha^2-k^2)}&&\\ +&\text{and} &y = e^{-kt}\cos(\alpha x ± t\sqrt{a^2\alpha^2-k^2)}&& +\end{flalign*} +as solutions of (1), or by combining these +\begin{align*} +y& = e^{-kt}\sin \alpha x \cos t\sqrt{a^2\alpha^2-k^2} \tag{7}\\ +y& = e^{-kt}\sin \alpha x \sin t\sqrt{a^2\alpha^2-k^2} \tag{8}\\ +y& = e^{-kt}\cos \alpha x \cos t\sqrt{a^2\alpha^2-k^2} \tag{9}\\ +y& = e^{-kt}\cos \alpha x \sin t\sqrt{a^2\alpha^2-k^2} \tag{10} +\end{align*} + +(7) and (8) satisfy (1) and (2) for all values of $\alpha$. They satisfy (3) if +$\alpha=\dfrac{m\pi}{l}$. Let us see if out of them we cannot build up a value that will satisfy +(4) and (5) as well. +\begin{gather*} +f(x)=\frac{2}{l}\sum_{m=1}^{m=\infty}\biggl(\sin\frac{m\pi x}{l}\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr). \tag{11}\\ +y=\frac{2}{l}e^{-kt}\sum_{m=1}^{m=\infty}\biggl(\sin\frac{m\pi x}{l}\cos t +\sqrt{\frac{m^2\pi^2 a^2}{l^2}-k^2}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr) \tag{12} +\end{gather*} +reduces to (11) when $t=0$ and therefore satisfies (4). +\begin{align*} +% recast to fit line +D_t y=&-\frac{2}{le^{kt}}\sum_{m=1}^{m=\infty}\biggl(\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}. +\sin\frac{m\pi x}{l}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\\ +&\hspace{18.5em}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr)\\ +&-\frac{2k}{le^{kt}}\sum_{m=1}^{m=\infty}\biggl( +\sin\frac{m\pi x}{l}\cos t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr). \tag{13} +\end{align*} + +When $t=0$ the first line of the second member of (13) vanishes but the +second line reduces to +\[ +-\frac{2k}{l}\sum_{m=1}^{m=\infty}\biggl(\sin\frac{m\pi x}{l}\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr). +\] +We must, then, introduce into (12) an additional term which shall equal zero +when $t=0$ and whose derivative with respect to $t$ shall cancel the term above +when $t=0$. +% -----File: 125.png + +This is easily seen to be +\[ +\frac{2k}{l}e^{-kt}\sum_{m=1}^{m=\infty}\frac{1}{\sqrt{\dfrac{m^2\pi^2a^2}{l^2}-k^2}} +\sin\frac{m\pi x}{l}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda. +\] + +Hence our complete solution is +\begin{multline*} +y=\frac{2}{l}e^{-kt}\sum_{m=1}^{m=\infty}\bigg[\bigg(\cos t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\\ ++\frac{k}{\sqrt{\dfrac{m^2\pi^2a^2}{l^2}-k^2}}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\bigg) + \sin\frac{m\pi x}{l}\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\bigg]. \tag{14} +\end{multline*} + +Here the fact that $e^{-kt}$, which decreases rapidly as $t$ increases, is a factor of +the whole second member shows that the amplitude of the vibration rapidly +decreases. + +Comparing this solution with that given in Art.~64 for the case where there +is no resistance we see that the period of any given term +\[ +A\sin\frac{m\pi x}{l}\cos t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}, +\] +is greater than that of the corresponding term $A_1\sin\dfrac{m\pi x}{l}\cos\dfrac{m\pi at}{l}$ in Art.~64. +In other words the effect of the resistance of the air is to flatten somewhat +each component part of the note given by the string. More than this +since the periods of the different terms of (14) are no longer exact submultiples +of the period of the first term, the component notes are no longer in perfect +harmony with the fundamental note of the string, and the ideal perfect harmony +between the fundamental note and its harmonics is not quite realized in +any actual case. + +When $k$ is very small, as in the case of a fine string, the departure from +perfect harmony is very slight; but in the case of a coarse string or worse still +of an elastic ribbon, where the resistance of the air is considerable, the +unmusical character of the sound is very noticeable. + +\EXAMPLE{S} + +1.\quad Solve Ex.~1 Art.~64 allowing for the resistance of the air.\\ + +2.\quad Solve Ex.~2 Art.~64 allowing for the resistance of the air; +\begin{multline*} +% recast to fit line +y=\frac{2}{l}e^{-kt}\sum_{m=1}^{m=\infty}\biggl(\frac{1}{\sqrt{\dfrac{m^2\pi^2a^2}{l^2}-k^2}} +\sin\frac{m\pi x}{l}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\\[-5ex] +. +\int\limits_0^l F(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr). +\end{multline*} +% -----File: 126.png + +3.\quad Find a particular solution of (1) Art.~65 on the assumption that it is of +the form $y=T.X$, where $T$ is a function of $t$ alone and $X$ a function of $x$ +alone. + +\mypara{66.} We pass on now to a couple of problems that require the modification +and extension of Fourier's Theorem, \emph{the cooling of a sphere in air}, and the +\emph{vibration of a stretched rectangular membrane}, but as an introduction to the +former we shall first consider the following very simple problem; to find the +temperature of any point of a sphere whose initial temperature is any given +function of $r$ the distance of the point from the centre, and whose surface is +kept at the constant temperature $b$. + +Here we are to solve +\[ +D_t(ru) = a^2D_r^2(ru), \tag{1} +\] +see \smallrom{V} Art.~1, subject to the conditions +\begin{alignat*}{3} +u &=f(r)\ && \text{when}\quad &t&=0 \tag{2}\\ +u &=b && \quad\text{``} &r&=c \tag{3} +\end{alignat*} +if $c$ is the radius. + +Let $v=ru$, then our equations become +\begin{alignat*}{3} +&&\llap{$D_tv =\:$} &\rlap{$a^2D_r^2v$} \tag{4}\\ +v &=rf(r)\ && \text{when}\quad &t&=0 \tag{5}\\ +v &=bc && \quad\text{``} &r&=c \tag{6}\\ +v &=0 && \quad\text{``} &r&=0. \tag{7} +\end{alignat*} +Our problem is now precisely that of Art.~62 and we have as our solution +\begin{align*} +ru=&\frac{2}{c}\sum_{m=1}^{m=\infty}\biggl(e^{-\frac{m^2a^2\pi^2}{c^2}t}\sin\frac{m\pi r}{c} +\int\limits_0^c\lambda f(\lambda)\sin\frac{m\pi\lambda}{c}d\lambda\biggr)\\ +&+b\biggl[r+\frac{2c}{\pi}\sum_{m=1}^{m=\infty}\biggl(\frac{(-1)^m}{m}e^{-\frac{m^2a^2\pi^2}{c^2}t}\sin\frac{m\pi r}{c}\biggr)\biggr]. \tag{8} +\end{align*} + +\EXAMPLE{S} + +1.\quad If $f(r)=b$ (8) Art.~66 reduces to $u=b$ and there is no change of +temperature. + +2.\quad If the initial temperature is constant and equal to $\beta$ +\begin{multline*} +u= b + \frac{2c}{\pi r}(\beta-b) + \bigg[e^{-\frac{ a^2\pi^2}{c^2}t}\sin\frac{ \pi r}{c} +-\frac{1}{2}e^{-\frac{4a^2\pi^2}{c^2}t}\sin\frac{2\pi r}{c}\\ ++\frac{1}{3}e^{-\frac{9a^2\pi^2}{c^2}t}\sin\frac{3\pi r}{c}-\cdots\biggr]. +\end{multline*} +% -----File: 127.png + +3.\quad An iron sphere 40~cm.\ in diameter is heated to the temperature 100° +centigrade throughout; its surface is then kept at the constant temperature 0°. +Find the temperature of a point 10~cm.\ from the centre, and find the temperature +of the centre, 15 minutes after cooling has begun. Given $a^2 = 0.185$ + in C.G.S.\ units. \hfill \textit{Ans.}, 2°.1; 3°.3. + +\mypara{67.} If instead of having the temperature of the surface of the sphere +constant, the sphere is placed in air which is kept at the constant temperature +zero, the problem is much more complicated. For in this case the +surface temperature can no longer be simply expressed but is given by a new +differential equation +\[ +D_{r}u + hu = 0 \quad \text{when} \quad r = c, \tag{1} +\] +where $h$ is an experimental constant depending upon what is called the surface +conductivity of the sphere. + +Our equations, then, are +\begin{gather*} +D_{t}(ru) = a^{2}D_{r}^{2}(ru) \tag{2}\\ +u =f(r) \qquad \text{when}\qquad t = 0 \tag{3}\\ +D_{r}u + hu = 0 \qquad \text{when} \qquad r = c. \tag{4} +\end{gather*} +As in Art.~66 let $v = ru$; then we have +\begin{alignat*}{3} +&&\llap{$D_{t}v =\:$} & \rlap{$a^{2}D_{r}^{2}v$} \tag{5}\\ +v &= rf(r) \quad && \text{when} \quad & t &= 0 \tag{6}\\ +v &= 0 &&\quad \text{``} & r &= 0 \tag{7}\\[-5ex] +\end{alignat*} +\[ +D_{r}v + \left ( h - \frac{1}{c} \right )v = 0 \quad \text{when} \quad r = c. \tag{8} +\] + +$v = e^{-a^2\alpha^{2}t} \cos \alpha r$ and $v = e^{-a^2\alpha^{2}t} \sin \alpha r$ have already been found as particular +solutions of (5) (see Art.~60). +\[ +v = e^{-a^2\alpha^{2}t} \sin \alpha r \tag{9} +\] +satisfies (7) for all values of $\alpha$. + +Substitute this value of $v$ in (8) and we have +\[ +\alpha c \cos \alpha c + (hc - 1) \sin \alpha c = 0. \tag{10} +\] + +If $\alpha_{k}$ is a value of $\alpha$ which is a root of the transcendental equation (10) +\[ +v = e^{-a^{2}\alpha_{k}^{2}t} \sin \alpha_{k}r \tag{11} +\] +will satisfy (5), (7), and (8). + +It remains to see whether out of terms of the form given in (11) we can +build up a value of $v$ which will satisfy (6). +% -----File: 128.png + +When $t = 0$ the second member of (11) reduces to $\sin \alpha_kr$. If then we +can express $rf(r)$ as a sum of terms of the form $b_k \sin \alpha_k r$ where $\alpha_k$ is a root +of (10) +\[ +v = \sum b_k e^{-a^2\alpha_k^2t} \sin \alpha_k r \tag{12} +\] +will satisfy all of the equations (5), (6), (7), and (8), and will be the required +solution. + +Here, then, we have a new problem analogous to that of developing in a +Fourier's Series, but rather more complicated, namely, to develop any function +of $x$ in a series of the form $\displaystyle\sum a_m \sin \alpha_m x$ where $\alpha_m$ is a root of the equation +(10); or if we call $ac = \phi$ and $hc-1=p$, where $a_m = \dfrac{\phi_m}{c}$, $\phi_m$ being a root +of the equation +\begin{gather*} +\phi \cos \phi + p \sin \phi = 0 \tag{13} +\intertext{or more simply of} +\phi + p \tan \phi = 0; \tag{14} +\end{gather*} +remembering that the series and the function must be equal for all values of $x$ +between zero and $c$. + +If $\phi_m$ is a root of (14) $- \phi_m$ is also a root. + +Since $\sin \dfrac{\phi_m}{c} x = - \sin \left (- \dfrac{\phi_m}{c} x \right )$ the terms of the required development +which correspond to negative roots may be combined with those corresponding +to positive roots, and therefore we need consider only positive roots. + +$\phi = 0$ is a root of (14) but as $\sin 0 = 0$ there will be no corresponding +term in the development. +If we construct the curve +\begin{align*} +y &= - \frac{1}{p} x \tag{15} +\intertext{and the curve} +y &= \tan x \tag{16} +\end{align*} +the abscissas of their points of intersection are values of $x$ which satisfy + $\dfrac{x}{p} + \tan x = 0$, that is, are roots of equation (14). It is easy to see that +there will always be an infinite number of real positive roots, one for each of +the branches of the periodic curve $y = \tan x$ which lie to the right of the +origin. The numerical values of these roots can be obtained by an easy computation. +The construction suggested above shows that as $m$ increases $\phi_m$ +will rapidly approach the value $(2m - 1) \dfrac{\pi}{2}$ if $p$ is positive or if $p$ is negative +and numerically less than unity, and $(2m + 1) \dfrac{\pi}{2}\tablestrut$ if $p$ is negative and numerically +greater than unity. +% -----File: 129.png + +There exist, then, an infinite number of positive real roots of $\phi + p \tan \phi = 0$ + and consequently of +\[ +\alpha c \cos \alpha c + (hc - 1) \sin \alpha c = 0. +\] + +\mypara{68.} The development called for in the last article can be obtained very +easily from a simpler one which we shall now consider, namely, to develop $f(x)$ +into a series of the form +\[ +f(x) = a_1 \sin\phi_1 x + a_2 \sin\phi_2 x + a_3 \sin\phi_3 x + \cdots \tag{1} +\] +where $\phi_1$, $\phi_2$, $\phi_3 \cdots$ are roots of the equation +\[ +\phi \cos\phi +p \sin\phi = 0, \tag{2} +\] +the development to hold good for all values of $x$ between $x = 0$ and $x = 1$. + +Let us proceed as in Arts.~24 and 27. Call $\dfrac{1}{n+1} = \Delta x$ and form $n$ equations +by substituting for $x$ in turn in the equation +\[ +f(x) = a_1 \sin\phi_1 x + a_2 \sin\phi_2 x + a_3 \sin\phi_3 x + \cdots + a_n \sin\phi_n x \tag{3} +\] +the values $\Delta x$, $2\Delta x$, $3\Delta x$, $\cdots n\Delta x$; this being equivalent to making the values +of the sum and the function coincide for the n values of $x$ substituted. + +To determine any coefficient $a_m$ multiply the first equation by $\Delta x.\sin(\phi_m$ $\Delta x)$, + the second by $\Delta x.\sin(2\phi_m\Delta x)$, the third by $\Delta x.\sin(3\phi_m \Delta x)$, and so on, the +$n$th equation by $\Delta x.\sin(n\phi_m\Delta x)$; add the equations and compute the limiting +values of the terms of the resulting equation as $n$ is indefinitely increased. +This as in Art.~24 is seen to be equivalent to multiplying (3) by $\sin\phi_m x.dx$ +and integrating between the limits $x = 0$ and $x = 1$. + +The first member of the resulting equation is +\[ +\int\limits_0^1 f(x) \sin \phi_m x.dx; +\] + +The coefficient of $a_k$ is +\[ +\int\limits_0^1 \sin\phi_k x \sin\phi_m x.dx, +\] +and of $a_m$ is +\[ +\int\limits_0^1 \sin^2\phi_m x.dx. +\] +% -----File: 130.png +\begin{align*} +\int\limits_0^1 \sin\phi_k x \sin\phi_m x.dx &= \frac{1}{2}\int\limits_0^1 [\cos(\phi_k-\phi_m)x - \cos(\phi_k+\phi_m)x]dx\\ +&=\frac{1}{2} \left [ \frac{\sin(\phi_k-\phi_m) }{ \phi_k-\phi_m }- \frac{\sin(\phi_k+\phi_m) }{ \phi_k+\phi_m} \right ]\\ +&= - \frac{\phi_k \cos\phi_k \sin\phi_m - \phi_m \sin\phi_k \cos\phi_m }{ \phi_k^2 - \phi_m^2 } \tag{4} +\end{align*} +\begin{flalign*} +& \indent \text{But} & \phi_k \cos\phi_k &+ p \sin\phi_k = 0&&\\ +& \text{and} & \phi_m \cos\phi_m &+ p \sin\phi_m = 0\qquad\text{by (2)}. +\end{flalign*} + +Hence the numerator of the second member of (4) is zero, and the coefficient +of $a_k$ vanishes if $k$ is not equal to $m$. + +\[ +\int\limits_0^1 \sin^2\phi_m x.dx = \frac{1}{2\phi_m }[\phi_m - \sin\phi_m \cos\phi_m] = \frac{1}{2} \left [1 - \frac{\sin 2\phi_m }{ 2\phi_m} \right ]. \tag{5} +\] +\begin{flalign*} +&\text{Therefore}& a_m &= \frac{2 }{1 - \dfrac{\sin 2\phi_m }{ 2\phi_m}} \int\limits_0^1 f(x) \sin\phi_m x.dx. &&\hspace{2em}\tag{6} +\end{flalign*} + +The coefficient of the integral in (6) can be transformed as follows so as not +to involve trigonometric functions. +\begin{gather*} +\phantom{\qquad by (2)}\phi_m \cos\phi_m + p \sin\phi_m = 0, \qquad \text{by (2)}\\ +\phi_m \cos^2\phi_m + \frac{p}{2} \sin 2\phi_m = 0,\\ +\frac{\sin 2\phi_m }{2\phi_m} = -\frac{\cos^2\phi_m }{p}. \tag{7}\\ +\phi_m^2 \cos^2\phi_m = p^2 \sin^2\phi_m,\\[1ex] +(\phi_m^2 + p^2) \cos^2\phi_m = p^2,\\[1ex] +\frac{\cos^2\phi_m }{p} = \frac{p }{ \phi_m^2 + p^2}. \tag{8} +\end{gather*} + +Hence by (7) and (8) +\begin{flalign*} +& &&1-\frac{\sin 2\phi_m }{ 2\phi_m} = \frac{\phi_m^2+p(p+1) }{ \phi_m^2 + p^2},&&\\ +&\text{and} &a_m = &\frac{2(\phi_m^2+p^2) }{ \phi_m^2+p(p+1)} \int\limits_0^1 f(\alpha)\sin \phi_m \alpha.d\alpha.&& \tag{9} +\end{flalign*} + +Therefore our required development is +\[ +f(x)=\sum_{m=1}^{m=\infty} \biggl ( \frac{2(\phi_m^2+p^2) }{ \phi_m^2+p(p+1)} \sin\phi_m x \int\limits_0^1 f(\alpha) \sin\phi_m \alpha.d\alpha \biggr ). \tag{10} +\] +% -----File: 131.png + +From (10) it easily follows that for values of $x$ between $0$ and $c$ +\begin{flalign*} +&&f(x) = &a_1 \sin a_1 x + a_2\sin a_2 x + a_3 \sin a_3 x + \cdots \tag{11}\\ +&\text{where} && a_m = \frac{2}{c}.\frac{\alpha_m^2 c^2 + p^2}{\alpha_m^2 c^2 + p(p+1)} \int\limits_0^c f(\lambda) \sin \alpha_m \lambda.d\lambda, \hspace{2.5em}&&\tag{12} +\end{flalign*} +and $\alpha_m$ is a root of the equation +\[ +\alpha c \cos \alpha c + p \sin \alpha c =0. \tag{13} +\] + +It is to be observed that if $p$ is infinite (13) reduces to $\sin \alpha c=0$, $\alpha_m$ +becomes $\dfrac{m \pi}{c}$ and (11) and (12) give our regulation Fourier sine series (v.\ Art.\ +31), and therefore the ordinary Fourier development in sine series is merely a +special case of the problem just solved. + +Moreover since the Fourier method of determining the coefficients of such a +series requires that\\[-4ex] +\begin{flalign*} +&&\int\limits_0^c \sin \alpha_m x &\sin \alpha_n x.dx = 0,&&\\ +&\text{that is that} &\frac{\sin(\alpha_m-\alpha_n)c}{\alpha_m-\alpha_n} - &\frac{\sin(\alpha_m+\alpha_n)c}{\alpha_m+\alpha_n} = 0 &&\\[1ex] +&\text{or reducing, that} &\frac{\alpha_m c \cos \alpha_mc }{ \sin \alpha_mc} &= \frac{\alpha_n c \cos \alpha_n c }{ \sin \alpha_n c}, &&\phantom{or reducing, that} +\end{flalign*} +or that $\alpha_m$ and $\alpha_n$ should be roots of the equation +\[ + \frac{\alpha c \cos \alpha c }{ \sin \alpha c} = p +\] +where $p$ is some constant, it follows that we have obtained in (11) the most +general sine development that can be obtained by Fourier's method. + +\EXAMPLE{S} + +1.\quad Show that the solution of the problem of Art.~67 is +\begin{flalign*} +& && ru = \sum_{m=1}^{m=\infty} b_m e^{-a^2\alpha_m^2 t}\sin \alpha_m r, &&\\ +&\text{where} &b_m = \frac{2}{c }.&\frac{\alpha_m^2c^2+(hc-1)^2 }{ \alpha_m^2c^2+hc(hc-1)} \int\limits_0^c \lambda f(\lambda) \sin \alpha_m \lambda.d\lambda &&\phantom{where} +\end{flalign*} +and $\alpha_m$ is a root of +\[ +\alpha c \cos \alpha c + (hc - 1) \sin \alpha c = 0. +\] + +% -----File: 132.png +2.\quad If the initial temperature of the sphere is constant and equal to $\beta$ +\begin{flalign*} +&&ru &= \sum_{m=1}^{m=\infty} b_m e^{-a^2\alpha_m^2t} \sin \alpha_m r &&\\ +&\text{where} & b_m &= 2\beta h.\frac{ \alpha_m^2c^2+(hc-1)^2}{\alpha_m^2c^2+hc(hc-1)}.\frac{\sin \alpha_mc}{\alpha_m^2} &&\phantom{where}\\ +&&& = \frac{2\beta hc}{\alpha_m}.\frac{ [\alpha_m^2c^2+(hc-1)^2]^{\frac{1}{2}}}{\alpha_m^2c^2+hc(hc-1)}. && +\end{flalign*} + +3.\quad If the temperature of the air is a constant $\gamma$ instead of zero the surface +equation of condition is +\[ +D_r u + h (u - \gamma) = 0 \quad \text{when} \quad r = c. +\] + +The substitution of $u_1 = u - \gamma$, however, brings the problem under Ex.~1 +and we get +\begin{flalign*} +&& r(u - \gamma) = \sum_{m=1}^{m=\infty} b_m e^{-a^2\alpha_m^2t} \sin \alpha_m &r && \\ +&\text{where} & b_m = \frac{2}{c}.\frac{ \alpha_m^2c^2+(hc-1)^2}{\alpha_m^2c^2+hc(hc-1)} \int\limits_0^c \lambda[f(\lambda) - \gamma] &\sin \alpha_m \lambda.d\lambda. &&\phantom{where} +\end{flalign*} + +4.\quad An iron sphere 40 cm.\ in diameter is heated to the temperature 100° +centigrade throughout; it is then allowed to cool in air which is kept at the +constant temperature 0°. Find the temperature at the centre; at a point 10 +cm.\ from the centre; and at the surface; 15 minutes after cooling has begun. +Given $a^2 = 0.185$ and $h = \dfrac{1}{800}$ in C.G.S. units. (v.\ Ex.~3, Art.~66.)\\ +\phantom{kindly go to the right hand margin}\hfill +\emph{Ans}., 97°.67; 97°.36; 96°.46.\\ + +5.\quad Show that if in the slab considered in Art.~60 one face is exposed to air +at the temperature zero, so that we have $D_t u = a^2D_x^2u$, $u = 0$ when $x = 0$, + $u=f(x)$ when $t = 0$, and $D_x u + hu = 0$ when $x = c$, then +\begin{flalign*} +&&& u = \sum_{m=1}^{m=\infty} a_m e^{-a^2\alpha_m^2t} \sin \alpha_m x &&\\ +&\text{where} & a_m & = 2 \frac{\alpha_m^2+h^2}{\alpha_m^2c+h(hc+1)} \int\limits_0^c f(\lambda) \sin \alpha_m \lambda.d\lambda, && \phantom{where} +\end{flalign*} +$\alpha_m$ being a root of $\alpha c \cos \alpha c + hc \sin \alpha c = 0$.\\ + +% -----File: 133.png +6.\quad If in the problem of Art.~57 heat escapes from one side of the plate into +air at the temperature zero so that we have $D_x^2u+D_y^2u = 0$, $u = 0$ when +$x = 0$, $u=f(x)$ when $y = 0$, and $D_{x}u + hu = 0$ when $x = a$, then +\begin{flalign*} +&&u = \sum_{m=1}^{m=\infty} a_m e^{-\alpha_m y} &\sin \alpha_m x &&\\ +&\text{where} & a_m = 2 \frac{\alpha_m^2 + h^2}{\alpha_m^2 a + h(ha + 1)} &\int\limits_0^a f(\lambda) \sin \alpha_m\lambda.d\lambda,&& \phantom{where} +\end{flalign*} +$\alpha_m$ being a root of $\alpha a \cos \alpha a + ha \sin \alpha a = 0$.\\ + +7.\quad If in the problem of Art.~59 there is leakage at one side of the sheet so +that we have $D_x^2 V + D_y^2 V = 0$, $V = 0$ when $x = 0$, $V = 0$ when $y = b$, + $V=f(x)$ when $y = 0$, and $D_xV+hV=0$ when $x = a$, then +\[ +V=\sum_{m=1}^{m=\infty} a_m \frac{\sinh \alpha_m(b-y)}{\sinh \alpha_m b }\sin \alpha_m x, +\] +where $a_m$ has the value given in Ex. 6. + +\mypara{69.} If we have an infinite solid with one plane face which is exposed to air +at the temperatures $U = F(t)$ and heat can flow only at right angles to this +face, we can solve the problem readily for the case where the initial temperatures +are zero. We have +\[ +D_tu=a^2D_x^2u\\ +\] +subject to the conditions +\begin{flalign*} +&&\multispan{2}{\hfill $u = 0 \quad \text{when} \quad t = 0$ \hfill}\\ +&\text{and} &D_xu+ h(U - &u) = 0 \quad \text{when} \quad x = 0. && \phantom{and}\\ +&\text{Let} & v ={}&u - \frac{1}{h}D_xu. && \tag{1} +\end{flalign*} +Then $v$ will satisfy the equation +\[ +D_tv = a^2D_x^2v, +\] +and we shall also have $v = U$ when $x = 0$. +\begin{flalign*} +&\text{\indent Since } U=F(t) \qquad v = \frac{2}{\sqrt{\pi}}\!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\!e^{-\beta^2} F \left (t-\frac{x^2}{4a^2\beta^2} \right )d\beta && \,\tag{2} +\end{flalign*} +by Art.~51 (10). +\begin{flalign*} +& &&D_xu - hu = - hv \qquad\text{by (1)}.&&\\ +&\indent \text{Hence} \quad &&ue^{-hx} = - h \int e^{-hx} vdx+C; &&\phantom{\indent \text{Hence}} +\end{flalign*} +v.\ Int.\ Cal.\ \S~4, page 314. +% -----File: 134.png + +Determining $C$ by the fact that $ue^{-hx} = 0$ when $x = \infty$ we have +\[ +u=he^{hx} \int\limits_x^{\infty} e^{-hx} vdx. \tag{3} +\] + +Substituting the value of $v$ from (2) we have +\[ +u= \frac{2he^{hx}}{\sqrt{\pi}}\int\limits_x^{\infty} e^{-hx} dx \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2}F\left ( t- \frac{x^2}{4a^2\beta^2} \right ) d\beta, \tag{4} +\] +as our required solution. + +For an extension of this method to the flow of heat in two and three dimensions +and for the interpretation of the results by the aid of the theory of +\textit{Images}, see E.~W. Hobson, Proc.\ Lond.\ Math.\ Soc., Vol.\ XIX. + +\EXAMPLE{S} + +1.\quad If the temperature of the air is a periodic function of the time, say + $\rho_{m}\sin (m\alpha t + \lambda_{m})$ and we care only for the limiting value of $u$ as $t$ increases, +show that this value is +\begin{align*} +\frac{h\rho_m e^{-\frac{x}{a}\frac{\sqrt{m\alpha}}{2}}}{\Big( h +\dfrac{1}{a} \sqrt{\dfrac{m\alpha}{2}}\Big)^2+\dfrac{m\alpha}{2a^2}} &\left [ \left ( h+\dfrac{1}{a} \sqrt{\frac{m\alpha}{2}}\right ) \sin \left (m\alpha t - \frac{x}{a} \sqrt{\frac{m\alpha}{2}}+\lambda_m \right ) \right .\\ +& \quad-\left . \frac{1}{a} \sqrt{\frac{m\alpha}{2}}\cos \left (m\alpha t - \frac{x}{a} \sqrt{\frac{m\alpha}{2}}+\lambda_m \right ) \right ]. +\end{align*} +v.\ Art.~52 and Art.~51 Ex.~4. +\begin{flalign*} +&\text{\indent Note that } & \int e^{ax}\sin bx.dx &= \frac{e^{ax}(a \sin bx - b \cos bx)}{a^2+b^2} && \phantom{\indent Note that}\\ +&\text{and } & \int e^{ax}\cos bx.dx &=\frac{e^{ax}(a \cos bx + b \sin bx)}{a^2+b^2} && +\end{flalign*} +v.\ Int.\ Cal.\ Table of Int.\ (235) and (236).\\ + +2.\quad If $D_{x}^{2}V+D_{y}^{2}V=0$, $V=0$ when $y = 0$ and $D_{x}V+ h[F(y) - V] = 0$ + when $x = 0$ show that +\[ +V=\frac{he^{hx}}{\pi} \int\limits_x^{\infty}e^{-hx}dx \int\limits_0^{\infty}F(\lambda)d\lambda \left [ \frac{x}{x^2+(\lambda-y)^2}-\frac{x}{x^2+(\lambda+y)^2} \right ];\\ +\] +v.\ Art.~47 Ex.~1. + +\mypara{70.} The solution for an instantaneous heat source of strength $Q$ at the +point $x = \lambda$ if heat escapes at the origin into air at the temperature zero, so +that $D_{x}u - hu = 0$ when $x = 0$, can be obtained by the aid of Art.~53. +% -----File: 135.png + +Let $u = u_1 + u_2$ where $u_1$ is the temperature that would be due to the given +source if we had no boundary at the origin, so that +\[ +u_1= \frac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda -x)^2}{4a^2t} }. \tag*{ [Art.~53 (2)]} +\] +\vspace{-5ex} +\begin{flalign*} +&&D_{x}u - hu = D_{x}u_{1} - hu_{1} &+ D_{x}u_{2} - hu_{2} = 0 \quad \text{when }\quad x=0. && \\ +&\text{Therefore} & D_{x}u_{2} - hu_{2} &= -(D_{x}u_{1} - hu_{1}) && \tag{1}\\ +\multispan{3}{when $x = 0$. \hfill}\\[-6ex] +\end{flalign*} +\begin{flalign*} +&\indent \text{But} & -(D_x u_1-hu_1)&=- \frac{Q}{2a\sqrt{\pi t}} \left ( \frac{\lambda -x}{2a^2t} -h \right )e^{-\frac{(\lambda -x)^2}{4a^2t} } && \phantom{But}\\ +&&& = -\frac{Q}{2a\sqrt{\pi t}} \left( \frac{\lambda}{2a^2t} -h \right ) e^{-\frac{\lambda^2}{4a^2t}} && \\ +\multispan{3}{when $x = 0$.\hfill} +\end{flalign*} + +This is easily seen to be the value to which +\[ +-\frac{Q}{2a\sqrt{\pi t}} \left( \frac{\lambda+x}{2a^2t} -h \right ) e^{-\frac{(\lambda+x)^2}{4a^2t}} +\] +reduces when $x = 0$, and this last expression is +\[ +(D_x+h)\frac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda+x)^2}{4a^2t}} +\] +and therefore satisfies the equation +\[ + D_{t}u=a^{2}D_{x}^{2}u; \tag{2} +\] +since $\dfrac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda+x)^2}{4a^2t}}$ is the temperature due to a source at $x = - \lambda$.\\ + +If, then, we determine $u_2$ from the condition that +\[ +D_x u_2-hu_2= -\frac{Q}{2a\sqrt{\pi t}} \left ( \frac{\lambda+x}{2a^2t}-h \right ) e^{-\frac{(\lambda+x)^2}{4a^2t}} \tag{3} +\] +taking care not to introduce any arbitrary constant or arbitrary function of $t$ +in our integration, $u_2$ will satisfy equation (2) and condition (1). + +Integrating (3) [v.\ Int.\ Cal.\ \S~4, page 314] and determining the constants of +integration suitably we get +\[ +u_2= \frac{Q}{2a\sqrt{\pi t}} \biggl [ e^{-\frac{(\lambda + x)^2}{4a^2t} } - 2he^{hx} \int\limits_x^{\infty} e^{-hx -\frac{(\lambda + x)^2}{4a^2t} } dx \biggr ]. \tag{4} +\] +Therefore the solution of our problem is +\[ +u= \frac{Q}{2a\sqrt{\pi t}} \biggl [ e^{-\frac{(\lambda -x)^2}{4a^2t} } +e^{-\frac{(\lambda+x)^2}{4a^2t}}- 2he^{hx} \int\limits_x^{\infty} e^{-hx -\frac{(\lambda + x)^2}{4a^2t} } dx \biggr ]. \tag{5} +\] +% -----File: 136.png + +If we replace $Q$ by $f(\lambda)d\lambda$ and integrate from $0$ to $\infty$ we get as the solution +for the case where $u=f(x)$ when $t = 0$ and $x > 0$, and $D_{x}u - hu = 0$ + when $x = 0$ +\[ +u=\frac{1}{2a\sqrt{\pi t}}\int\limits_0^{\infty}f(\lambda)d\lambda \biggl [ e^{-\frac{(\lambda-x)^2}{4a^2t}}+ e^{-\frac{(\lambda+x)^2}{4a^2t}} -2he^{hx} \int\limits_x^{\infty} e^{- hx -\frac{(\lambda+x)^2}{4a^2t} } dx \biggr ]. \tag{6} +\] + +For an interpretation of this result by the theory of Images and the +extension of the method to the conduction of heat in $n$ dimensions see G.~H. +Bryan, Proc.\ Lond.\ Math.\ Soc., Vol.\ XXII. + +\EXAMPLE{} + +Show that if $u=f(x)$ when $t = 0$ and $D_{x}u + h[F(t) - u] = 0$ when + $x = 0$ we must take $u$ equal to the sum of the second members of (6) Art.~70 +and of (4) Art.~69. + +\mypara{71.} As another problem requiring a slight extension of Fourier's Theorem +let us consider the vibration of a rectangular stretched elastic membrane +fastened at the edges, that is of a rectangular drumhead. + +If two of the sides are taken as axes and the plane of equilibrium of the +membrane as the plane of $XY$ the equation for the motion of the membrane is +\[ + D_{t}^{2}z= c^{2}(D_{x}^{2}z+D_{y}^{2}z) \tag{1} +\] +see \smallrom{X} Art.~1. + +Let the membrane be distorted at the start into some given form $z =f(x,y)$ + and then allowed to swing. Our equations of conditions are then +\begin{alignat*}{3} + z &= 0 && \text{when}\qquad & x &= 0 \tag{2}\\ + z &= 0 &&\quad \text{``} & x &= a \tag{3}\\ + z &= 0 &&\quad \text{``} & y &= 0 \tag{4}\\ + z &= 0 &&\quad \text{``} & y &= b \tag{5}\\ + z &=f(x,y) &&\quad \text{``} & t &= 0 \tag{6}\\ + D_{t}z&=0 &&\quad \text{``} & t &= 0. \tag{7} +\end{alignat*} + +We can get a particular solution of (1) by our usual device. Assume +\[ +z=e^{\alpha x + \beta y + \gamma t} +\] +and substitute in (1). We get $\gamma^{2} = c^{2}(\alpha^{2} + \beta^{2})$ as the only +relation that +need hold between $\alpha$, $\beta$, and $\gamma$, in order that $z=e^{\alpha x + \beta y + \gamma t}$ may be a +solution. This gives +\begin{flalign*} +&& \gamma = &\pm c \sqrt{ \alpha^2 + \beta^2}. &&\\ +&\text{\indent Therefore} &z = e&^{\alpha x + \beta y \pm ct \sqrt{ \alpha^2 + \beta^2}}&&\phantom{Therefore} +\end{flalign*} +is a solution of (1) no matter what values are given to $\alpha$ and $\beta$. +% -----File: 137.png + +Replace $\alpha$ and $\beta$ by $ \alpha i$ and $\beta i$ and we have +\[ +z=e^{(\alpha x + \beta y \pm ct \sqrt{\alpha^2+\beta^2 })i} +\] +as a solution, and from this we get +\begin{flalign*} +& &z = \sin (\alpha x + \beta y \pm ct \sqrt{ \alpha^{2}+\beta^{2}}) && \tag{8}\\ +&\text{and}\qquad &z = \cos (\alpha x + \beta y \pm ct \sqrt{ \alpha^{2}+\beta^{2}}) && \tag{9} +\end{flalign*} +as particular solutions of (1), $\alpha$ and $\beta$ being unrestricted. + +From (8) and (9) we can get solutions of the following forms +\[ +\left . \begin{alignedat}{3} + z &= \sin \alpha x &&\sin \beta y &&\sin ct \sqrt{\alpha^2 + \beta^2}\\ + z &= \sin \alpha x &&\sin \beta y &&\cos ct \sqrt{\alpha^2 + \beta^2}\\ + z &= \sin \alpha x &&\cos \beta y &&\sin ct \sqrt{\alpha^{2} + \beta^{2}}\\ + z &= \sin \alpha x &&\cos \beta y &&\cos ct \sqrt{\alpha^{2} + \beta^{2}}\\ + z &= \cos \alpha x &&\sin \beta y &&\sin ct \sqrt{\alpha^{2} + \beta^{2}}\\ + z &= \cos \alpha x &&\sin \beta y &&\cos ct \sqrt{\alpha^{2} + \beta^{2}}\\ + z &= \cos \alpha x &&\cos \beta y &&\sin ct \sqrt{\alpha^{2} + \beta^{2}}\\ + z &= \cos \alpha x &&\cos \beta y &&\cos ct \sqrt{\alpha^{2} + \beta^{2}}, +\end{alignedat} +\right \} + \qquad \text{(10)} +\] +each of which will satisfy equation (1). The second of these will satisfy also +(2), (4) and (7) whatever values be taken for $\alpha$ and $\beta$. It will satisfy (3) and +(5) if $ \alpha$ and $\beta$ are equal $\dfrac{m\pi}{a}$ and $\dfrac{n\pi}{b}$ respectively. + +If, then, we can so combine terms of the form +\[ +\sin \frac{m\pi x}{a}\sin\frac{n\pi y}{b}\cos c\pi t \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}} +\] +as to satisfy (6) our problem will be completely solved. + +This can be done if we can express $f(x,y)$ as a sum of terms of the form + $A \sin \dfrac{m\pi x}{a}\sin \dfrac{n\pi y}{b}$, the sum and the function being equal when $x$ lies +between $0$ and $a$ and $y$ between $0$ and $b$. + +$f(x,y)$ can be expressed in terms of $\sin \dfrac{m\pi x}{a}$ by Fourier's Theorem if we +regard $y$ as constant. We have +\begin{flalign*} +&&f(x,y)&= \sum_{m=1}^{m=\infty} a_m \sin \frac{m\pi x}{a} && \tag{11}\\ +% -----File: 138.png +&\text{where} &a_m=\frac{2}{a} &\int\limits_0^a f(\lambda,y)\sin \frac{m\pi\lambda}{a} d\lambda. \tag{12}&&\; +\end{flalign*} +$f(\lambda,y)$ in (12) is a function of $y$ and may be developed by Fourier's Theorem. +\begin{flalign*} +&\text{We have} & \quad f(\lambda,y)&= \sum_{n=1}^{n=\infty} b_n \sin \frac{n\pi y}{b } \tag{13} &&\\ +&\text{where} & b_n=\frac{2}{b} &\int\limits_0^b f(\lambda,\mu)\sin \frac{n\pi\mu}{b} d\mu. \hspace{3em}\tag{14} +\end{flalign*} + +Substituting for $f(\lambda,y)$ in (12) the value just obtained we have +\[ +a_m =\frac{2}{a}\frac{2}{b}\sum_{n=1}^{n=\infty}\biggl (\int\limits_0^a d\lambda \int\limits_0^b f(\lambda,\mu) \sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b} d\mu \biggr ) \sin\frac{n\pi y}{b} +\] +and +\[ +f(x,y) =\frac{4}{ab} \sum_{m=1}^{m=\infty}\sum_{n=1}^{n=\infty}\biggl ( \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} \int\limits_0^a d\lambda \int\limits_0^b f(\lambda,\mu) \sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b}d\mu\biggr ) .\tag{15} +\] +\begin{flalign*} +&\text{Hence} & z=&\sum_{m=1}^{m=\infty}\sum_{n=1}^{n=\infty}\biggl ( A_{m,n} \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} \cos c\pi t \sqrt{ \frac{m^2}{a^2}+\frac{n^2}{b^2} }\biggr ), && \tag{16}\\ +&\text{where} &&\quad A_{m,n}=\frac{4}{ab} \int\limits_0^ad\lambda \int\limits_0^bf(\lambda,\mu)\sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b}d\mu. &&\tag{17} +\end{flalign*} +is our required solution. + +\EXAMPLE{S} + +1.\quad Show that if the membrane starts from its position of equilibrium but +with a given initial velocity impressed upon each point so that $z = 0$ when + $t = 0$ and $D_{t}z= F(x,y)$ when $t = 0$ the solution is +\[ +z=\frac{1}{c\pi}\sum_{m=1}^{m=\infty}\sum_{n=1}^{n=\infty}\biggl ( A_{m,n} \dfrac{1}{\sqrt{ \dfrac{m^2}{a^2}+\dfrac{n^2}{b^2} }} \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} \sin c\pi t \sqrt{ \frac{m^2}{a^2}+\frac{n^2}{b^2} }\biggr )\\[-2ex] +\] +\begin{flalign*} +&\text{where} &A_{m,n}&=\frac{4}{ab} \int\limits_0^ad\lambda \int\limits_0^bF(\lambda,\mu)\sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b}d\mu. && \phantom{where} +\end{flalign*} + +% -----File: 139.png +2.\quad If there is both initial distortion and initial velocity +\begin{align*} +% recast to fit line +z = \dfrac{4}{ab}\sum_{m=1}^{m=\infty}\,\sum_{n=1}^{n=\infty}\sin\dfrac{m\pi x}{a} +\sin\dfrac{n\pi y}{b}\biggl[&A_{m,n}\cos{c\pi t}\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}\\ + + &B_{m,n}\sin{c\pi t}\sqrt{\dfrac{m^2}{a^2} + \dfrac{n^2}{b^2}}\biggr]\\[-4ex] +\end{align*} +\begin{flalign*} +&\text{where}\hfil &&A_{m,n} = \int\limits_0^a d\lambda\int\limits_0^bf +(\lambda,\mu)\sin\dfrac{m\pi\lambda}{a}\sin\dfrac{n\pi\mu}{b}d\mu,\hfil\phantom{where}&&\\[-1ex] +&\text{and}\hfil &B_{m,n} =& \dfrac{1}{c\pi\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}} +\int\limits_0^a d\lambda \int\limits_0^b F(\lambda,\mu)\sin\dfrac{m\pi\lambda}{a} +\sin\dfrac{n\pi\mu}{b}d\mu.\hfil\phantom{and} +\end{flalign*} + +3.\quad Obtain a particular solution of (1) Art.~71 by assuming $z = T.X.Y$ + where $T$ is a function of $t$ alone, $X$ of $x$ alone, and $Y$ of $y$ alone. + +\mypara{72.} A number of interesting conclusions can be drawn from the results of +Art.~71 and Exs.~1 and 2. + +(\emph{a}) No one of the three values of $z$ is in general a periodic function of $t$, +and consequently a vibrating rectangular membrane will not in general give a +musical note. + +(\emph{b}) A stretched rectangular membrane can be made to give a musical note +by starting the vibration properly. For if the initial circumstances are such +that the solution reduces to a single term, as will be the case if the initial distortion +in the problem of Art.~71 be such that $f(x,y) = A_{m,n} \sin\dfrac{m\pi x}{a} \sin\dfrac{n\pi y}{b}$, + or the initial velocity in Ex.~1 be such that $F(x,y) = B_{m,n} \sin\dfrac{m\pi x}{a}\tablestrut\sin\dfrac{n\pi y}{b}$, + or the initial distortion and initial velocity in Ex.~2 be the values just given, +then the vibration will be periodic and will have the period + +\[\tag{1} +T = \dfrac{2}{c\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}}. +\] + +Since $T$ is a function of $m$ and $n$ and $m$ and $n$ are any whole numbers, the +same membrane is capable of giving a great variety of musical notes of different +pitches. If $m$ and $n$ are both unity we get the lowest note the membrane +can give, which is called its fundamental note. Its period +\[\tag{2} +T_1 = \dfrac{2}{c\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}}}=\dfrac{2ab}{c\sqrt{a^2+b^2}} +\] +If $m$ and $n$ are both equal to $k$ we get + +\[\tag{3} +T_k = \dfrac{2ab}{kc\sqrt{a^2+b^2}}; +\] +% -----File: 140.png +therefore the membrane can be made to give any harmonic of its fundamental +note. + +More than this, since as we have seen +\[ +T_{m,n}=\frac{2}{c\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}} +\] +is the period of any note the membrane can give, and since if $m$ and $n$ are +replaced by $mk$ and $nk$ we get +\[ +T_{mk,nk}= \frac{2}{ck\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}} +\] +the membrane can sound all the harmonics of any note which it can give. + +(\emph{c}) In the case considered above, where the solution reduces to the single +term +\[ +z=\sin\frac {m\pi x}{a}\sin \frac {n\pi y}{b} \left [A_{m,n}\cos c\pi t \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}+B_{m,n}\sin c\pi t \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}} \right ], +\] +if $x = \dfrac{a}{m}$, or $\dfrac{2a}{m}$, or $\dfrac{3a}{m} \cdots$ or $ \dfrac{(m-1)a}{m}$, $z = 0$ for all values of $t$, and +the lines $x = \dfrac{a}{m}$, $x = \dfrac{2a}{m}\tablestrut$, $\cdots$ $x = \dfrac{(m-1)a}{m}$ remain at rest during the whole +vibration and are nodes. The same thing is true of the lines +\[ +y=\frac{b}{n},\; y=\frac{2b}{n},\; y=\frac{3b}{n},\; \cdots\; y=\frac{(n-1)b}{n}. +\] + +\mypara{73.} If the membrane is square it may have much more complicated nodes +than if the length and breadth are unequal, as in this case the period of any +term of the general solution reduces to +\[ +T=\frac{2a}{c\sqrt{m^2+n^2}} \tag{1} +\] +and there will in general be two terms having the same period, and a musical +note of the pitch corresponding to that period may be produced by initial circumstances +that bring in both terms. Thus +\begin{align*} +z=&\sin\frac {m\pi x}{a}\sin \frac{ n\pi y}{a} \left [A_{m,n}\cos\frac{ c\pi t}{a} \sqrt{m^2+n^2}+B_{m,n}\sin \frac{c\pi t}{a} \sqrt{m^2+n^2} \right ]\\ +&+ \sin\frac {n\pi x}{a}\sin \frac{ m\pi y}{a} \left [A_{n,m} \cos\frac{c\pi t}{a} \sqrt{m^2+n^2}+B_{n,m} \sin \frac{c\pi t}{a} \sqrt{m^2+n^2}\right ] +\end{align*} +% -----File: 141.png +is a form of vibration that will give a musical note. Let us write this +\begin{align*} +z = &\cos \frac{c\pi t}{a} \sqrt{m^2+n^2}\bigg [ A \sin \frac{m\pi x}{a} \sin \frac{n\pi y}{a} + B \sin\frac{n\pi x}{a} \sin \frac{m\pi y}{a} \bigg ]\\ +&+ \sin\frac{c\pi t}{a} \sqrt{m^2+n^2}\bigg [ C \sin \frac{m\pi x}{a} \sin \frac{n\pi y}{a} + D \sin\frac{n\pi x}{a} \sin \frac{m\pi y}{a} \bigg ] \tag{2} +\end{align*} +and in studying the forms of musical vibration of which the membrane is +capable we may take $A$, $B$, $C$, and $D$ at pleasure. Consider the simple case +where $A = C$ and $B = D$; then (2) reduces to +\begin{multline*} +z = \bigg ( A \sin\frac{m\pi x}{a} \sin \frac{n\pi y}{a} + B \sin\frac{n\pi x}{a} \sin\frac{m\pi y}{a} \bigg ) \biggl (\cos \frac{c\pi t}{a} \sqrt{m^2+n^2} \biggr .\\ +\biggl .+ \sin \frac{c\pi t}{a} \sqrt{m^2+n^2} \biggr ). \tag{3} +\end{multline*} +Values of $x$ and $y$ that will reduce the first parenthesis in (3) to zero will correspond +to points of the membrane remaining motionless during the vibration. + +Let us consider a few cases at length. + +(\emph{a})\quad If $m= 1$ and $n = 1$, the first parenthesis in (3) becomes +\[ +(A + B) \sin\frac{\pi x}{a} \sin \frac{\pi y}{a}, +\] +which is equal to zero only when $x = 0$ or $y = 0$, or $x = a$ or $y = a$, + that is, for the four edges of the membrane. If, then, the membrane is sounding +its fundamental note it has no nodes. + +(\emph{b})\quad If $m = 1$ and $n = 2$, we have +\[ +A \sin\frac{\pi x}{a} \sin \frac{2\pi y}{a} + B \sin\frac{2\pi x}{a} \sin \frac{\pi y}{a} = 0 +\] +to give the nodes. + +Let $B = 0$, then $\sin\dfrac{\pi x}{a} \sin \dfrac{2\pi y}{a} = 0$, which is satisfied by $y = \dfrac{a}{2}$; and +in addition to the edges the line $y = \dfrac{a}{2}$ is at rest and is a node. + +If $A = 0$ \quad $x = \dfrac{a}{2}$ is a node. + +If $A = B$ +\begin{gather*} +\sin\frac{\pi x}{a} \sin \frac{2\pi y}{a} + \sin\frac{2\pi x}{a} \sin \frac{\pi y}{a} = 0\\[1ex] +2 \sin\frac{\pi x}{a} \sin \frac{\pi y}{a} \cos \frac{\pi y}{a} + 2 \sin\frac{\pi x}{a} \cos\frac{\pi x}{a} \sin \frac{\pi y}{a} = 0\\[1ex] +\sin\frac{\pi x}{a} \sin \frac{\pi y}{a} \left(\cos \frac{\pi y}{a} + \cos \frac{\pi x}{a} \right) = 0. +\end{gather*} +% -----File: 142.png +The first factor gives the four edges of the membrane. The second written +equal to zero gives +\begin{gather*} +\cos \frac{\pi y}{a} = -\cos \frac{\pi x}{a} = \cos\left (\pi - \frac{\pi x}{a} \right )\\ +\frac{\pi y}{a}=\pi-\frac{\pi x}{a}\\ +x + y= a, +\end{gather*} +which is a diagonal of the square. + +If $B = - A$ +\begin{gather*} +\sin \frac{\pi x}{a}\sin \frac{2\pi y}{a} - \sin \frac{2\pi x}{a} \sin\frac{\pi y}{a}=0\\ + \cos\frac{\pi y}{a} = \cos\frac{\pi x}{a}\\ +x-y=0, +\end{gather*} +which is the other diagonal of the square. + +Other relations between $A $ and $B$ will give Trigonometric curves of the form +\[ +\cos \frac{\pi y}{a} =-\frac{B}{A} \cos \frac{\pi x}{a} +\] +which are easily constructed and which obviously all agree in passing through +the middle point of the square. + +We give the figures for a few of the cases +%[Illustration] +\pngcent{010.png}{1288} + +% -----File: 143.png + +{}(\emph{c})\quad If $m = n = 2$ we have\\[-4ex] +\pngrightl{011.png}{344}{10}{-24} + +\[ +(A + B)\sin \frac{2\pi x}{a}\sin \frac{2\pi y}{a} = 0 +\] +to give the nodes, which are merely the lines +\[ +x = \frac{a}{2}, \quad \text{and} \quad y = \frac{a}{2}. +\] +This form gives the octave of the fundamental note. + +(\emph{d})\quad If $m = 1$ and $n = 3$ we have +\[ +A \sin \frac{\pi x}{a} \sin \frac{3\pi y}{a} + B \sin \frac{3\pi x}{a} \sin \frac{\pi y}{a} = 0 +\] +to give the nodes. +\begin{flalign*} +&\text{\indent If} \quad A = 0 \quad \text{we get} &x &= \frac{a}{3}\quad \text{and} \quad x = \frac{2a}{3} && \phantom{\indent If A = 0 we get} \tag{1} \\ +&\text{\indent If} \quad B = 0 \quad \text{we get} &y &= \frac{a}{3} \quad \text{and} \quad y = \frac{2a}{3} && \tag{2} +\end{flalign*} + +If \quad $A = -B$ \quad we get +\begin{gather*} + \sin \frac{\pi x}{a} \sin \frac{3\pi y}{a} - \sin \frac{3\pi x}{a} \sin \frac{\pi y}{a} = 0 \\[1ex] + \sin \frac{\pi x}{a} \sin \frac{\pi y}{a} \left [4 \cos^2 \frac{\pi y}{a} - 1 - 4 \cos^2 \frac{\pi x}{a} + 1 \right ] = 0 \\[1ex] + \cos^2 \frac{\pi y}{a} - \cos^2 \frac{\pi x}{a} = 0 \\[1ex] + \left ( \cos \frac{\pi y}{a} - \cos \frac{\pi x}{a} \right ) \left ( \cos \frac{\pi y}{a} +\cos \frac{\pi x}{a} \right ) = 0 \\[-5ex] +\end{gather*} +\begin{flalign*} +&\text{or}\hspace{1em} & x - y = 0 \quad &\text{and} \quad x + y = a. \tag{3} && \\[-5ex] +\end{flalign*} +\begin{flalign*} +&\text{\indent If} \quad A = B \quad \text{we get} & \cos^2 \frac{\pi y}{a} &+\cos^2 \frac{\pi x}{a} = \frac{1}{2} && \phantom{\indent If \quad A = B \quad we get}\\[-6ex] +\end{flalign*} +\begin{flalign*} +&\text{or}\hspace{1em} & \cos \frac{2\pi y}{a} +{}& \cos \frac{2\pi x}{a} = -1, \tag{4} && +\end{flalign*} +a Trigonometric curve easily constructed. + +For other relations between $A$ and $B$ we get more complicated Trigonometric +curves coming under the general form +\[ +A \cos \frac{2\pi y}{a} + B \cos \frac{2\pi x}{a} = - \frac{A + B}{ 2} \tag{5} +\] +% -----File: 144.png +which all agree in containing the points +\[ +\Bigl(\dfrac{a}{3}, \dfrac{a}{3}\Bigr), +\Bigl(\dfrac{a}{3}, \dfrac{2a}{3}\Bigr), +\Bigl(\dfrac{2a}{3}, \dfrac{a}{3}\Bigr), +\text{ and } \Bigl(\dfrac{2a}{3}, \dfrac{2a}{3}\Bigr). +\] +%[Illustration] +\pngcent{012.png}{1275} + +\newpage +% -----File: 145.png + +\label{ch4end} + +\newpage +\begin{center}{MISCELLANEOUS PROBLEMS.}\end{center} +\markright{} +\label{probstart} + +\begin{center}{I. \emph{Logarithmic Potential. Polar Coördinates.}}\end{center} + +\vspace{\baselineskip} + +1.\quad Show that $D_x^2V+D_y^2V=0$ becomes +\[ +D_r^2V + \dfrac{1}{r}D_rV+ \dfrac{1}{r^2}D_\phi^2V = 0 +\] +if we transform to Polar Coördinates. + +\begin{flalign*} +&\text{\indent 2.\quad If in} && D_r^2V + \dfrac{1}{r}D_rV + \dfrac{1}{r^2}D_\phi^2V = 0 && \phantom{\indent 2.\quad If in}(1) +\end{flalign*} +we let $V = R.\Phi$ we get +\begin{align*} +\begin{aligned} +\Phi &= A \cos \alpha\phi + B \sin \alpha\phi\ \\ + R&= A_1r^\alpha+B_1r^{-\alpha} +\end{aligned} +\bigg\} +\quad\text{or}\quad +\begin{aligned} +\Phi &= A e^{\alpha\phi} + B e^{-\alpha\phi}\\ + R& = A_1 \cos(\alpha \log r) + B_1 \sin(\alpha \log r);\ +\end{aligned} +\bigg\} +\end{align*} +whence\rule{0em}{1.3em} +\[ + \begin{array}{l | l | l} +V = r^\alpha \cos \alpha\phi & V = e^{\alpha\phi} \cos(\alpha \log r) & V = \cosh \alpha\phi \cos(\alpha \log r)\\[0.5em] +V = r^\alpha \sin \alpha\phi & V= e^{\alpha\phi} \sin(\alpha \log r) & V = \cosh \alpha\phi \sin(\alpha \log r)\\[0.5em] +V = \dfrac{1}{r^\alpha} \cos \alpha\phi & V = e^{-\alpha\phi} \cos(\alpha \log r) & V = \sinh \alpha\phi \cos(\alpha \log r)\\[0.5em] +V = \dfrac{1}{r^\alpha} \sin \alpha\phi & V = e^{-\alpha\phi}\sin(\alpha \log r) & V = \sinh \alpha\phi \sin(\alpha \log r) +\end{array} +\] +are particular solutions of (1).\\ + +3.\quad Show that if $V$ satisfies (1) Ex. 2 and $V = f(\phi)$ when $r = a$ +\begin{flalign*} +& &V &= \dfrac{1}{2} b_0 + \sum^{m=\infty}_{m=1} \left(\dfrac{r}{a}\right)^m (b_m \cos m\phi + a_m \sin m\phi)\quad\text{for}\quad r < a\rule{3em}{0em}\\[0.5em] +&\text{and} & V &= \dfrac{1}{2} b_0 + \sum^{m=\infty}_{m=1} \left(\dfrac{a}{r}\right)^m (b_m \cos m\phi + a_m \sin m\phi)\quad\text{for}\quad r > a,\\[0.5em] +&\text{where} &b_m& = \dfrac{1}{\pi} \int \limits ^\pi_{-\pi} f(\phi)\cos m\phi.d\phi\quad\text{and}\quad a_m = \dfrac{1}{\pi} \int \limits ^\pi_{-\pi} f(\phi) \sin m\phi.d\phi && +\end{flalign*} +% -----File: 146.png + +4.\quad Show that if $V$ satisfies (1) Ex.~2 and $V=f(r)$ when $\phi=0$ and $r>0$ +\begin{align*} +V&=\frac{1}{\pi}\int\limits_{-\infty}^\infty f(e^\lambda)d\lambda + \int\limits_0^\infty\frac{\cosh\alpha(\pi-\phi)}{\cosh\alpha\pi}\cos\alpha(\lambda-\log r).d\alpha\\ + &=\frac{1}{\pi}\sin\frac{\phi}{2}\int\limits_{-\infty}^\infty f(e^\lambda) +\frac{\cosh\dfrac{1}{2}(\lambda-\log r)} + {\cosh(\lambda-\log r)-\cos\phi}d\lambda. +\end{align*} + +\markright{MISCELLANEOUS PROBLEMS.} +5.\quad If $V=1$ when $\phi=0$ and $0<r<1$, and $V=0$ when $\phi=0$ and + $r>1$ +\[ +V=\frac{1}{\pi}\Bigg\{\frac{\pi}{2}-\tan^{-1}\Bigg[\frac{\sinh\dfrac{\log r}{2}}{\sin\dfrac{\phi}{2}}\Bigg]\Bigg\} + =\frac{1}{\pi}\Bigg[\frac{\pi}{2}-\tan^{-1}\Bigg(\frac{r-1}{2\sqrt{r}.\sin\dfrac{\phi}{2}}\Bigg)\Bigg]. +\] + +6.\quad If $V=f(r)$ when $\phi=0$ and $V=0$ when $\phi=\beta$ +\begin{align*} +V &=\frac{1}{\pi}\int\limits_{-\infty}^\infty f(e^\lambda) d\lambda\int\limits_0^\infty\frac{\sinh(\beta-\phi)\alpha}{\sinh\beta\alpha}\cos\alpha(\lambda-\log r).d\alpha\\ + &=\frac{1}{2\beta}\sin\frac{\pi\phi}{\beta}\int\limits_{-\infty}^\infty +\frac{f(e^\lambda)d\lambda}{\cosh\dfrac{\pi}{\beta}(\lambda-\log r)-\cos\dfrac{\pi}{\beta}\phi}, +\end{align*} +if $0<\phi<\beta$.\\ + +7.\quad If $V=0$ when $\phi=0$ and $V=F(r)$ when $\phi=\beta$ +\begin{align*} +V &=\frac{1}{\pi}\int\limits_{-\infty}^\infty F(e^\lambda)d\lambda\int\limits_0^\infty\frac{\sinh\phi\alpha}{\sinh\beta\alpha}\cos\alpha(\lambda-\log r).d\alpha\\ + &=\frac{1}{2\beta}\sin\frac{\pi\phi}{\beta}\int\limits_{-\infty}^\infty +\frac{F(e^\lambda)d\lambda}{\cosh\dfrac{\pi}{\beta}(\lambda-\log r)+\cos\dfrac{\pi}{\beta}\phi}. +\end{align*} + +8.\quad If $V=\chi(r)$ when $\phi=0$ and $r<a$, $V=0$ when $\phi=\beta$, and + $V=0$ when $r=a$ +\begin{align*} +V=\frac{1}{2\beta}\sin\frac{\pi\phi}{\beta}\int\limits_{-\infty}^0\chi(ae^\lambda) +\Bigg[&\frac{d\lambda}{\cosh\dfrac{\pi}{\beta}\Big(\lambda-\log\dfrac{r}{a}\Big)-\cos\dfrac{\pi\phi}{\beta}}\\ + &-\frac{d\lambda}{\cosh\dfrac{\pi}{\beta}\Big(\lambda+\log\dfrac{r}{a}\Big)-\cos\dfrac{\pi\phi}{\beta}}\Bigg]. +\end{align*} + +% -----File: 147.png +9.\quad If $V=0$ when $r=1$, $V=1$ when $\phi=0$, $V=0$ when $\phi=\dfrac{\pi}{2}$ +\[ +V = \frac{2}{\pi}\tan^{-1}\biggl[\frac{1-r^2}{1+r^2}\ctn\phi\biggr]. +\] + +10.\quad If $V=0$ when $r=1$, $V=1$ when $\phi=0$, $V=1$ when $\phi=\dfrac{\pi}{2}$ +\[ +V = \frac{2}{\pi}\tan^{-1}\biggl[\frac{1-r^4}{2r^2\sin 2\phi}\biggr]. +\] + +11.\quad If $V=f(\phi)$ when $r=a$, $V=0$ when $\phi=0$, and $V=0$ when $\phi=\beta$ +\begin{flalign*} +&&V&= \sum_{m=1}^{m=\infty} a_m\Bigl(\frac{r}{a}\Bigr)^\frac{m\pi}{\beta} \sin\frac{m\pi\phi}{\beta}\quad \text{if} \quad r<a &\\ +&&V&= \sum_{m=1}^{m=\infty} a_m\Bigl(\frac{a}{r}\Bigr)^\frac{m\pi}{\beta} \sin\frac{m\pi\phi}{\beta}\quad \text{if} \quad r>a &\\ +&\text{where} & a_m &= \frac{2}{\beta}\int\limits_0^\beta f(\phi)\sin\frac{m\pi\phi}{\beta}d\phi \quad \text{and}\quad 0<\phi<\beta.& +\end{flalign*} + +12.\quad If $V=f(\phi)$ when $r=a$, $V=0$ when $r=b$, $V=0$ when $\phi=0$, + and $V=0$ when $\phi=\beta$, then if $a<r<b$ and $0<\phi<\beta$ +\[ +V = \sum_{m=1}^{m=\infty}\biggl\{\frac{a^{\frac{ m\pi}{\beta}} b^{\frac{ m\pi}{\beta}}} + {a^{\frac{2m\pi}{\beta}}-b^{\frac{2m\pi}{\beta}}} + \biggl[\Bigl(\frac{r}{b}\Bigr)^{\frac{m\pi}{\beta}} + -\Bigl(\frac{b}{r}\Bigr)^{\frac{m\pi}{\beta}}\biggr] +a_m\sin\frac{m\pi\phi}{\beta}\biggr\} +\] +\begin{flalign*} +&\text{where} & a_m = \frac{2}{\beta}\int\limits_0^\beta f(\phi) +\sin\frac{m\pi\phi}{\beta}d\phi.&& +\end{flalign*} + +13.\quad If $V=F(\phi)$ when $r=b$, $V=0$ when $r=a$, $V=0$ when $\phi=0$, + and $V=0$ when $\phi=\beta$, then if $a<r<b$ and $0<\phi<\beta$ +\[ +V = \sum_{m=1}^{m=\infty}\biggl\{\frac{a^{\frac{ m\pi}{\beta}} b^{\frac{ m\pi}{\beta}}} + {b^{\frac{2m\pi}{\beta}}-a^{\frac{2m\pi}{\beta}}} + \biggl[\Bigl(\frac{r}{a}\Bigr)^\frac{m\pi}{\beta} + -\Bigl(\frac{a}{r}\Bigr)^\frac{m\pi}{\beta}\biggr] +a_m\sin\frac{m\pi\phi}{\beta}\biggr\} +\] +\begin{flalign*} +&\text{ where }& a_m = \frac{2}{\beta}\int\limits_0^\beta F(\phi) +\sin\frac{m\pi\phi}{\beta} d\phi.&& +\end{flalign*} + +14.\quad If $V=\chi(r)$ when $\phi=0$, $V=0$ when $\phi=\beta$, $V=0$ when $r=a$, + and $V=0$ when $r=b$, then if $a<r<b$ and $0<\phi<\beta$ +\[ +V = \sum_{m=1}^{m=\infty}\Biggl\{ a_m\ +\frac{\sinh\dfrac{m\pi(\beta-\phi)}{\log b - \log a}} + {\sinh\dfrac{m\pi\beta}{\log b - \log a}} +\sin +\frac{m\pi(\log r - \log a)} + {\log b - \log a}\Biggl\} +\] +% -----File: 148.png +\begin{flalign*} +&\text{where} & a_m = \frac{2}{\log b - \log a} \int\limits_0^{\log\frac{b}{a}} \chi(ae^x) +\sin\frac{m\pi x}{\log b - \log a} dx.&& +\end{flalign*} + +15.\quad If $V=\psi(r)$ when $\phi=\beta$, $V=0$ when $\phi=0$, $V=0$ when $r=a$, + and $V=0$ when $r=b$, then if $a<r<b$ and $0<\phi<\beta$ +\[ +V = \sum_{m = 1}^{m = \infty}\Biggl\{ a_m +\frac{\sinh\dfrac{m\pi\phi} {\log b - \log a}} + {\sinh\dfrac{m\pi\beta}{\log b - \log a}} +\sin\frac{m\pi(\log r - \log a)}{\log b - \log a}\Biggr\} +\] +\begin{flalign*} +&\text{where} & a_m =\frac{2}{\log b - \log a} \int\limits_0^{\log\frac{b}{a}} \psi(ae^x) +\sin\frac{m\pi x}{\log b - \log a} dx.&& +\end{flalign*} + +\vspace{1ex} + +\begin{center}{II. \textit{Potential Function in Space.}}\end{center} + +1.\quad Show that +\[ +f(x,y) = \frac{1}{\pi^2}\int\limits_0^\infty d\alpha + \int\limits_0^\infty d\beta + \int\limits_0^\infty d\lambda + \int\limits_0^\infty f(\lambda,\mu)\cos\alpha(\lambda-x)\cos\beta(\mu-y).d\mu, +\] +for all values of $x$ and $y$.\\ + +2.\quad Find particular solutions of $D_x^2V+D_y^2V+D_z^2V=0$ in the forms +\begin{align*} +V& = e^{±z\sqrt{\alpha^2+\beta^2}}\cos(\alpha x ± \beta y)&\\ +V& = e^{±z\sqrt{\alpha^2+\beta^2}}\sin(\alpha x ± \beta y)&\\ +V& =\sinh z\sqrt{\alpha^2+\beta^2}.\sin(\alpha x ± \beta y)&\\ +V& =\cosh z\sqrt{\alpha^2+\beta^2}.\sin(\alpha x ± \beta y)&\\ +& \hspace{15ex}\text{\&c.} +\end{align*} + +3.\quad Given $D_x^2V+D_y^2V+D_z^2V=0$, and $V=f(x,y)$ when $z=0$, solve for +positive values of $z$. +\begin{flalign*} +&\text{\emph{Result:}} & V=\frac{1}{2\pi} +\int\limits_{-\infty}^\infty d\lambda +\int\limits_{-\infty}^\infty \frac{zf(\lambda,\mu)d\mu}{[z^2+(\lambda-x)^2+(\mu-y)^2]^\frac{3}{2}}.&& \phantom{Result:} +\end{flalign*} + +4.\quad Confirm the result of the last example by showing that if $f(x,y)$ is independent +of $y$ +\[ +\phantom{(v.\ Ex.~3 Art.~45)}V = \frac{1}{\pi}\int\limits_{-\infty}^\infty \frac{zf(\lambda,\mu)d\lambda}{z^2+(\lambda-x)^2} \tag*{\text{(v.\ Ex.~3 Art.~45)}.} +\] + +% -----File: 149.png +5.\quad If $D_x^2V + D_y^2V + D_z^2V = 0$, and $V=1$ when $z=0$ for all points +within the rectangle bounded by the lines $x=a$, $x=-a$, $y=b$, and + $y=-b$; and $V=0$ when $z=0$ for all points outside of this rectangle, +then +\begin{flalign*} +&\multispan{3}{$2\pi V = $\hfill} \\ + && \frac{b-y}{\sqrt{(b-y)^2}} \bigg\{\dfrac{\pi}{2} & + \dfrac{1}{2} \sin^{-1}\dfrac{(a-x)^2(b-y)^2 - z^2[(a-x)^2 + (b-y)^2 + z^2]}{(a-x)^2(b-y)^2 + z^2[(a-x)^2 + (b-y)^2 + z^2]}\\ + && & + \dfrac{1}{2} \sin^{-1}\dfrac{(a+x)^2(b-y)^2 - z^2[(a+x)^2 + (b-y)^2 + z^2]}{(a+x)^2(b-y)^2 + z^2[(a+x)^2 + (b-y)^2 + z^2]}\bigg\}\\ + && + \frac{b+y}{\sqrt{(b+y)^2}} \bigg\{\dfrac{\pi}{2} & + \dfrac{1}{2} \sin^{-1}\dfrac{(a-x)^2(b+y)^2 - z^2[(a-x)^2+(b+y)^2+z^2]}{(a-x)^2(b+y)^2 + z^2[(a-x)^2 + (b+y)^2 + z^2]}\\ + && & + \dfrac{1}{2} \sin^{-1}\dfrac{(a+x)^2(b+y)^2 - z^2[(a+x)^2+(b+y)^2+z^2]}{(a+x)^2(b+y)^2 + z^2[(a+x)^2 + (b+y)^2 + z^2]}\bigg\} +\end{flalign*} +if $-a<x<a$, and +\begin{alignat*}{2} +4\pi V &= \frac{b-y}{\sqrt{(b-y)^2}} \bigg\{&\phantom{- }&\sin^{-1}\dfrac{(a-x)^2(b-y)^2 - z^2[(a-x)^2 + (b-y)^2 + z^2]}{(a-x)^2(b-y)^2 + z^2[(a-x)^2 + (b-y)^2 + z^2]} \\ + & & -{}&\sin^{-1}\dfrac{(a+x)^2(b-y)^2 - z^2[(a+x)^2 + (b-y)^2 + z^2]}{(a+x)^2(b-y)^2 + z^2[(a+x)^2 + (b-y)^2 + z^2]}\bigg\}\\ + &+ \frac{b+y}{\sqrt{(b+y)^2}} \bigg\{& &\sin^{-1}\dfrac{(a-x)^2(b+y)^2 - z^2[(a-x)^2 + (b+y)^2 + z^2]}{(a-x)^2(b+y)^2 + z^2[(a-x)^2 + (b+y)^2 + z^2]} \\ + & & -{}&\sin^{-1}\dfrac{(a+x)^2(b+y)^2 - z^2[(a+x)^2 + (b+y)^2 + z^2]}{(a+x)^2(b+y)^2 + z^2[(a+x)^2 + (b+y)^2 + z^2]}\bigg\}; +\end{alignat*} +if $x<-a$ or $x>a$.\\ + +6.\quad If the value of the potential function $V$ is given at every point of the base +of an infinite rectangular prism and if the sides of the prism are at potential +zero the value of $V$ at any point within the prism is\label{err149} +\begin{multline*} +% recast to fit line +V = \dfrac{4}{ab} \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} e^{-\pi z \sqrt{ \frac{m^2}{a^2} + \frac{n^2}{b^2}}} \sin \dfrac{m\pi x}{a}\sin \dfrac{n\pi y}{b}\\ +\int\limits^a_0 d\lambda \int\limits^b_0 f(\lambda,\,\mu) \sin \dfrac{m\pi\lambda}{a}\sin \dfrac{n\pi\mu}{b} d\mu. +\end{multline*} +If $V=1$ on the base of the prism this reduces to +\[ +V = \dfrac{16}{ \pi^2}\sum\limits^{m=\infty}_{m=0} \sum\limits^{n=\infty}_{n=0} e^{-\pi z \sqrt{ \frac{(2m+1)^2 }{a^2} + \frac{(2n+1)^2}{b^2}}}\:\dfrac{\sin\dfrac{ (2m+1)\pi x}{a} \sin \dfrac{(2n+1)\pi y}{ b}}{ (2m+1)(2n+1)}. +\] + +7.\quad If the value of the potential function on five faces of a rectangular +parallelopiped, whose length, breadth, and height are $a$, $b$, and $c$, is zero, and +% -----File: 150.png +if the value of $V$ is given for every point of the sixth face, then for any +point within the parallelopiped +\[ +V =\sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} A_{m,n} \dfrac{\sinh \pi (c - z) \sqrt{\dfrac{ m^2}{a^2} + \dfrac{n^2}{ b^2}}}{\sinh \pi c \sqrt{\dfrac{ m^2 }{a^2} + \dfrac{n^2 }{ b^2}}} \sin \dfrac{m \pi x}{a} \sin \dfrac{n \pi y}{ b} +\] +\begin{flalign*} +&\text{where} &A_{m,n}&= \dfrac{4}{ab} \int\limits^a_0 d\lambda \int\limits^b_0 f(\lambda,\;\mu) \sin \dfrac{m\pi\lambda}{a} \sin \dfrac{n\pi\mu}{b} d\mu.& +\end{flalign*} + +8.\quad If the value of the potential function is given on two opposite faces of a +rectangular parallelopiped and is zero on the four remaining faces, then within +the parallelopiped +\begin{align*} +V =& \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} A_{m,n} \dfrac{\sinh \pi (c - z) \sqrt{\dfrac{ m^2}{a^2} + \dfrac{n^2}{ b^2}}}{\sinh \pi c \sqrt{\dfrac{ m^2 }{a^2} + \dfrac{n^2 }{ b^2}}} \sin \dfrac{m \pi x}{a} \sin \dfrac{n \pi y}{ b}\\ + &+ \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} B_{m,n} \dfrac{\sinh \pi z \sqrt{\dfrac{ m^2}{a^2} + \dfrac{n^2}{ b^2}}}{\sinh \pi c \sqrt{\dfrac{ m^2 }{a^2} + \dfrac{n^2 }{ b^2}}} \sin \dfrac{m \pi x}{a} \sin \dfrac{n \pi y}{ b} +\end{align*} +\begin{flalign*} +&\text{where} &A_{m,n}&=\dfrac{4}{ab}\int\limits^a_0 d\lambda \int\limits^b_0 f (\lambda,\,\mu) \sin \dfrac{m\pi\lambda}{a}\sin \dfrac{n\pi\mu}{b} d\mu&\\ +&\text{and} &B_{m,n}&=\dfrac{4}{ab}\int\limits^a_0 d\lambda \int\limits^b_0 F(\lambda,\,\mu) \sin \dfrac{m\pi\lambda}{a}\sin \dfrac{n\pi\mu }{b} d\mu.& +\end{flalign*} + +9.\quad If the value of the potential function is given at every point on the surface +of a rectangular parallelopiped, what is its value at any point within the +parallelopiped? + +\vspace{1ex} +\begin{center}{III. \textit{Conduction of Heat in a Plane.}}\end{center} + +1.\quad Find particular solutions of $D_tu = a^2(D_x^2 u + D_y^2 u)$ of the forms +\begin{align*} +&u = e^{-a^2(\alpha^2 + \beta^2)t} \sin(\alpha x\pm \beta y)\\ +&u = e^{-a^2(\alpha^2 + \beta^2)t} \cos(\alpha x\pm \beta y). +\end{align*} + +2.\quad Given the initial temperature of every point in a thin plane plate, find +the temperature of any point at any time, +% -----File: 151.png +\begin{align*} +u &= \dfrac{1}{4a^2 \pi t}\int\limits^{\infty}_{-\infty} d\lambda \int\limits^{\infty}_{-\infty} e^{-\frac{(\lambda-x)^2+(\mu -y)^2}{4a^2t}} f(\lambda,\, \mu)d\mu\\ + &= \dfrac{1}{\pi}\int\limits^{\infty}_{-\infty}e^{-\beta^2}d\beta\int\limits^{\infty}_{-\infty}e^{-\gamma^2}f(x+2a\sqrt{t}.\beta,\,y+2a\sqrt{t}.\gamma)d\gamma. +\end{align*} + +3.\quad For an instantaneous \emph{source} of strength $Q$ at $(\lambda, \mu)$ +\begin{flalign*} +&\phantom{v.\ Art.~53.}&u &= \dfrac{Q}{ 4\pi a^2t}e^{-\frac{(\lambda-x)^2+(\mu -y)^2}{4a^2t}} & \text{v.\ Art.~53}. +\end{flalign*} + +For an instantaneous \emph{doublet} of strength $P$ at $(0, \mu)$ with its axis perpendicular +to the axis of $Y$ +\begin{flalign*} +&\phantom{v.\ Art.~54.}&u &= \dfrac{Px}{ 8\pi a^4t^2}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} & \text{v.\ Art.~54}. +\end{flalign*} + +For a permanent doublet of strength $P$ at $(0, \mu)$ with its axis perpendicular +to the axis of $Y$ +\[ +u = \dfrac{P}{2\pi a^2} \frac{x}{x^2+(\mu-y)^2}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} . +\] + +If the strength of the doublet were $Pd\mu$ and the heat were uniformly +generated and absorbed along the element $d\mu$ of the axis of $Y$ beginning at +(0, $\mu$) we should have +\[ +u = \dfrac{P }{2\pi a^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} \dfrac{xd\mu}{ x^2+(\mu-y)^2} = \dfrac{P }{ 2\pi a^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} d \tan^{-1}\dfrac{ \mu-y}{x}, +\] +and since $d \tan^{-1}\dfrac{ \mu-y}{x}$ is the angle $ARA'$, where $A$ and $A'$ are the points (0, $\mu$) + and (0, $\mu + d\mu$) and $R$ is the point $(x, y)$, $u = 0$ when $x = 0$ unless + $\mu < y < \mu + d\mu$, in which case $u = \dfrac{P}{2a^2}$ if $x$ approaches zero from the +positive side; and $u = 0$ when $t = 0$ except in the element $d\mu$. If then + $u = 0$ when $t = 0$ and $u=f(y)$ when $x=0$ we have only to suppose a +doublet of strength $2a^2f(x)dx$ placed in each element of the axis of $Y$ and +then to integrate; we get +\[ +u = \dfrac{1}{ \pi} \int\limits^{\infty}_{-\infty}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} \dfrac{xf(\mu)}{x^2+(\mu -y)^2}{}d\mu. +\] + +For a permanent doublet of strength $F(t)$ at (0, $\mu$) we have +\begin{align*} +u &= \dfrac{x}{ 8\pi a^4} \int\limits^t_0 e^{-\frac{x^2+(\mu-y)^2}{4a^2(t-\tau)}} (t-\tau)^{-2}F(\tau)d\tau.\\ + &= \dfrac{1}{2\pi a^2}\bigg[\dfrac{xF(0)}{x^2+(\mu-y)^2}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}}+\int\limits^t_0\dfrac{xF'(\tau)}{x^2+(\mu-y)^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2(t-\tau)}} d\tau\bigg]. +\end{align*} + +% -----File: 152.png +From the reasoning above this must be zero when $t = 0$ except at the point +(0, $\mu$), must be $2a^2F(t)$ at the point (0, $\mu$), and 0 at every other point of the +axis of $Y$ when $t$ is not zero. + +Hence if $u = 0$ when $t = 0$ and $u = F(y,t)$ when $x = 0$ +\[ +u = \dfrac{1}{\pi} \!\int\limits^{\infty}_{-\infty}\! \dfrac{xF(\mu,0)}{x^2 + (\mu - y)^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} d\mu + \dfrac{1}{\pi} \!\int\limits^{\infty}_{-\infty}\! d\mu \int\limits^t_0 \dfrac{xD_\tau F(\mu,\tau)}{x^2 + (\mu - y)^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2(t-\tau)}}d\tau. +\] + +For an extension of this solution by the method of images to the case where +there are other rectilinear boundaries and for its application to the corresponding +problems in the flow of heat in three dimensions see E.~W. Hobson in Vol.\ +XIX. Proc.\ Lond.\ Math.\ Soc.\\ + +4.\quad If the perimeter of a thin plane rectangular plate is kept at the temperature +zero and the initial temperatures of all points of the plate are given, +then for any point of the plate +\begin{multline*} +% recast to fit line +u = \dfrac{4 }{ bc} \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1}e^{-a +^2\pi^2\left(\frac{m^2}{b^2} +\frac{ n^2}{c^2}\right)t} \sin \dfrac{m\pi x}{b} \sin \dfrac{n\pi y}{c}\\ +\int\limits^b_0d\lambda \int\limits^c_0 f(\lambda,\mu) \sin \dfrac{m\pi\lambda}{b} \sin \dfrac{n\pi\mu}{c} d\mu. +\end{multline*} +if $b$ is the length and $c$ the breadth of the plate.\\ + +5.\quad A large mass of iron at the temperature 0° contains an iron core in the +shape of a long prism 40 cm.\ square. The core is removed and heated to the +temperature of 100° throughout and then replaced. Find the temperature of a +point in the axis of the core fifteen minutes afterward. Given $a^2 = .185$ in +C.G.S. +units. \hfill \emph{Ans}., 52°.9. + +6.\quad If the prism described in Ex.~5 after being heated to 100° has its lateral +faces kept for 15 minutes at the temperature 0° find the temperature of a point +in its axis. \hfill \emph{Ans}., 20°.8. + +\vspace{1ex} +\begin{center}IV.~ \textit{Conduction of Heat in Space.}\end{center} + +1.\quad Show that +\begin{multline*} +% recast to fit line +\dfrac{1}{\pi^3}\int\limits^\infty_0 d\alpha \int\limits^\infty_0 d\beta \int\limits^\infty_0 d\gamma \int\limits^\infty_{-\infty} d\lambda \int \limits^\infty_{-\infty} d\mu \int\limits^\infty_{-\infty} f(\lambda,\mu,\nu) \\ +\cos \alpha(\lambda - x) \cos \beta(\mu - y) \cos \gamma(\nu - z).d\nu += f(x,y,z) +\end{multline*} +for all values of $x$, $y$, and $z$.\\ + +2.\quad Show that +\begin{flalign*} +&&f(x,y,z) = \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1}\sum\limits^{p=\infty}_{p=1}A_{m,n,p} &\sin \dfrac{m\pi x}{a} \sin \dfrac{n\pi y}{b} \sin \dfrac{p\pi z}{c}&&\\ +&\text{where}\quad & A_{m,n,p} = \dfrac{8}{abc}\displaystyle \int \limits^a_0 d\lambda \int \limits^b_0 d\mu \int \limits^c_0 f(\lambda,\mu,\nu) &\sin \dfrac{m\pi \lambda}{a} \sin \dfrac{n\pi \mu}{b} \sin \dfrac{p\pi \nu}{c} d\nu, +\end{flalign*} +for $0 < x < a$, $0 < y < b$, $0 < z < c$.\\ + +% -----File: 153.png +3.\quad Obtain particular solutions of $D_tu = a^2(D_x^2u + D_y^2u + D_z^2u)$ of the +forms +\begin{align*} +u &= e^{-a^2(\alpha^2 + \beta^2 + \gamma^2)t} \sin(\alpha x \pm\beta y \pm \gamma z).\\ +u &= e^{-a^2(\alpha^2 + \beta^2 + \gamma^2)t} \cos(\alpha x \pm \beta y \pm \gamma z). +\end{align*} + +4.\quad Given the initial temperature of every point in an infinite homogeneous +solid find the temperature of any point at any time. +\begin{align*} +&u=\dfrac{1}{8a^3(\pi t)^{\frac{3}{2}}} \int\limits^{\infty}_{-\infty} d\lambda \int\limits^{\infty}_{-\infty} d\mu \int\limits^{\infty}_{-\infty} e^{-\frac{(\lambda-x)^2+(\mu-y)^2+(\nu-z)}{4a^2t}} f(\lambda,\mu,\nu)d\nu\\ + &=\dfrac{1}{\pi^{\frac{3}{2}}} \!\int\limits^{\infty}_{-\infty}\! e^{-\beta^2} d\beta \!\int\limits^{\infty}_{-\infty}\! e^{-\gamma^2} d\gamma \!\int\limits^{\infty}_{-\infty}\! e^{-\delta^2} f(x+2a \sqrt{t}.\beta,\,y+2a \sqrt{t}.\gamma,\,z+2a \sqrt{t}.\delta)d\delta. +\end{align*} + +5.\quad If the surface of a rectangular parallelopiped is kept at the temperature +zero and the initial temperatures of all points of the parallelopiped are given, +then for any point of the parallelopiped +\[ +u=\sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} \sum\limits^{p=\infty}_{p=1} A_{m,n,p}e^{-a^2\pi^2\left(\frac{m^2}{b^2}+\frac{n^2}{c^2}+\frac{p^2}{d^2}\right)t} \sin \dfrac{m\pi x}{b} \sin \dfrac{n\pi y}{c} \sin \dfrac{p\pi z}{d}\\[-5ex] +\] +\begin{flalign*} +&\text{where}&A_{m,n,p}&=\dfrac{8}{bcd}\int \limits^b_0 d \lambda \int \limits^c_0 d\mu \int \limits^d_0 f(\lambda,\,\mu,\,\nu)\sin \dfrac{m\pi\lambda}{b} \sin \dfrac{n\pi\mu}{c} \sin \dfrac{p\pi\nu}{d} d\nu.& +\end{flalign*} + +6.\quad An iron cube 40 cm.\ on an edge is heated to the uniform temperature of +100° Centigrade and then tightly enclosed in a large iron mass which is at the +uniform temperature of 0°. Find the temperature of the centre of the cube +fifteen minutes afterwards. \hfill \emph{Ans}., 38°.4.\\ + +7.\quad An iron cube 40 cm.\ on an edge is heated to the uniform temperature of +100° and then its surface is kept for fifteen minutes at the temperature 0°. +Required the temperature of its centre. \hfill \emph{Ans.}, 9°.5. + +\label{probend} + +% -----File: 154.png +\label{ch5start} +\mychap{CHAPTER V.\footnotemark}{ZONAL HARMONICS.} + +\footnotetext{ Before reading this chapter the student is advised to re-read carefully articles 9, 10, 13(\emph{c}), +15, 16, and 18(\emph{c}).} + +\mypara{74.} In Art.~16 we obtained +\[ +z = Ap_m(x) + Bq_m(x) \tag{1} +\] +[v.\ (6) Art.~16] as the general solution of Legendre's Equation +\[ +(1-x^2)\dfrac{d^2z}{dx^2}-2x\dfrac{dz}{dx}+m(m+1)z = 0, \tag{2} +\] +$m$ being wholly unrestricted in value and $x$ lying between $-1$ and 1; where +\begin{multline*} +p_m(x)=1-\dfrac{m(m +1)}{2!} x^2 + \dfrac{m(m -2)(m + 1)(m + 3)}{4!} x^4\\ +-\dfrac{m(m - 2)(m - 4)(m + 1)(m + 3)(m + 5)}{6!} x^6+ \cdots \tag{3} +\end{multline*} +and +\begin{multline*} +q_m(x) = x-\dfrac{(m-1)(m + 2)}{3!} x^3 + \dfrac{(m-1)(m-3)(m + 2)(m + 4)}{5!} x^5\\ +-\dfrac{(m - 1)(m - 3)(m - 5)(m + 2)(m + 4)(m + 6)}{7!} x^7 +\cdots; \tag{4} +\end{multline*} +\begin{flalign*} +\left.\!\! +\begin{aligned} +&\text{and we found}\hspace{5em}& V&=r^mp_m(\cos\theta)\\ +&& V&=\dfrac{1}{r^{m+1}} p_m(\cos\theta)\\ +&& V&=r^mq_m(\cos\theta)\\ +&& V&=\dfrac{1}{r^{m+1}} q_m(\cos\theta), +\end{aligned} +\right\}&&\hfill\tag{5} +\end{flalign*} +$m$ being unrestricted in value, as particular solutions of the special form +assumed by Laplace's Equation in spherical coördinates when $V$ is independent +of $\phi$; that is, of the equation +\[ +rD_r^2(rV) + \dfrac{1}{\sin\theta} D_\theta(\sin\theta D_\theta V) = 0. \tag{6} +\] + +% -----File: 155.png +For the important case where $m$ is a positive integer we found +\[ +z = AP_m(x) + BQ_m(x) \tag{7} +\] +[v.\ (10) Art.~16] as the general solution of Legendre's Equation (2), whence +\[\left . +\begin{aligned} +V&= r^mP_m(\cos \theta)\\ +V&=\frac{1}{r^{m+1}} P_m(\cos\theta)\\ +V&=r^mQ_m(\cos \theta)\\ +V&=\frac{1}{r^{m+1}} Q_m(\cos\theta) +\end{aligned} +\right \} +\tag{8} +\] +are particular solutions of (6) if m is a positive integer. +\begin{multline*} +P_m(x)=\frac{(2m-1)(2m-3) \cdots 1}{m!} \left [x^m-\frac{m(m-1)}{2(2m-1)}x^{m-2} \right . \\ ++\left . \frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3) }x^{m-4} - \cdots \right ] \tag{9} +\end{multline*} +[v.\ (8) Art.~16] and is a finite sum terminating with the term which involves +$x$ if $m$ is odd and with the term involving $x^0$ if $m$ is even. + +\markright{ZONAL HARMONICS.} + +It is called a \textit{Surface Zonal Harmonic}, or a \textit{Legendre's Coefficient}, or more +briefly a \textit{Legendrian}. +\begin{multline*} +Q_m(x)=\frac{m!}{(2m+1)(2m-1)\cdots 1} \biggl [\frac{1}{x^{m+1}} + \frac{(m+1)(m+2)}{2.(2m+3)} \frac{1}{x^{m+3}} \biggr . \\ ++ \biggl . \frac{(m+1)(m+2)(m+3)(m+4)}{2.4.(2m+3)(2m+5)} \frac{1}{x^{m+5}} + \cdots \biggr ] \tag{10} +\end{multline*} +if $x < -1$ or $x > 1.$ [v.\ (9) Art.~16.] + +It is called a \textit{Surface Zonal Harmonic} of the \textit{second kind}. + \begin{align*} +Q_m(x)&=(-1)^\frac{m+1}{2} \frac{2^{m-1}\Big[\Gamma\Big(\dfrac{m+1}{2}\Big)\Big]^2}{\Gamma(m+1)} p_m(x) \\ + &=(-1)^\frac{m+1}{2} \frac{2.4.6.\cdots (m-1)}{3.5.7.\cdots m} p_m(x) \tag{11} +\end{align*} +[v.\ (13) Art.~16] if $m$ is odd and $-1 < x < 1$. + \begin{align*} +Q_m(x)&=(-1)^\frac{m}{2} \frac{2^m\Big[\Gamma\Big(\dfrac{m+1}{2}\Big)\Big]^2}{\Gamma(m+1)} q_m(x) \\ + &=(-1)^\frac{m}{2} \frac{2.4.6.\cdots m}{1.3.5.\cdots (m-1)} q_m(x) \tag{12} +\end{align*} +[v.\ (14) Art.~16] if $m$ is even and $-1 < x < 1$. + +% -----File: 156.png +In most of the work that immediately follows we shall regard $x$ in $P_m(x) $as +equal to $\cos\theta$ and therefore as lying between $-1$ and 1.\footnote{English writers on Spherical Harmonics generally use $\mu$ in place of $x$ for $\cos \theta$. We +shall follow them, however, only when we should thereby avoid confusion.} + +\mypara{75.} In Article 9 the undetermined coefficient $a_m$ of $x^m$ in $P_m(x)$ was +arbitrarily written in the form $\dfrac{(2m-1)(2m-3)\cdots 1}{m!}$ for reasons which shall +now be given. + +In Articles 9 and 16 $z = P_m(x)$ was obtained as a particular solution of +Legendre's Equation +\[ +(1-x^2)\dfrac{ d^2z}{dx^2} - 2x\dfrac{ dz}{dx} + m(m+1)z = 0 \tag{1} +\] +by the device of assuming that z could be expressed as a sum or a series of +terms of the form $a_nx^n$ and then determining the coefficients. We can, however, +obtain a particular solution of Legendre's Equation by an entirely different +method. + +The potential function due to a unit of mass concentrated at a given point +$(x_1, y_1, z_1)$ is +\[ +V= \dfrac{1}{\sqrt{ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2}}\tag{2} +\] +and this must be a particular solution of Laplace's Equation +\[ +D_x^2 V+D_y^2V+D_z^2 V= 0, \tag{3} +\] +as is easily verified by direct substitution. + +If we transform (2) to spherical coördinates using the formulas of +transformation +\begin{flalign*} +&&x&=r\cos\theta & \\ +&&y&=r\sin\theta\cos\phi&\\ +&&z&=r\sin\theta\sin\phi& \text{we get} +\end{flalign*} +\[ +V = \dfrac{1 }{\sqrt{ r^2 - 2rr_1[\cos\theta \cos\theta_1 + \sin\theta \sin\theta_1 \cos(\phi-\phi_1)] + r_1^2}} \tag{4} +\] +as a solution of Laplace's Equation in Spherical Coördinates +\begin{flalign*} +&&&&rD_r^2(rV)+\dfrac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta} V)+\dfrac{1}{\sin^2\theta} {D_\phi}^2V=0 \quad \text{ [{\scriptsize XIII}] Art.~1}. +\end{flalign*} + +If the given point $(x_1, y_1, z_1)$ is taken on the axis of $X$, as it must be that +(4) may be independent of $\phi$, $\theta_1 = 0$, and +\[ +V = \dfrac{1}{ \sqrt{ r^2-2rr_1 \cos\theta + r_1^2}}\tag{5} +\] +% -----File: 157.png +is a solution of +\[ +rD_{r}^{2}(rV) + \dfrac{1}{\sin\theta} D_\theta(\sin\theta D_{\theta}V) = 0. \tag{6} +\] +Equation (5) may be written +\begin{flalign*} +& &V&=\frac{1}{r}\frac{1}{\sqrt {1-2 \dfrac{r_1}{r} \cos\theta + \dfrac{r_1^2}{r^2}} } \tag{7} && \\[1ex] +&\text{or} &V&=\frac{1}{r_1}\frac{ 1}{\sqrt{ 1-2 \dfrac{r}{r_1} \cos\theta + \dfrac{r^2}{r_1^2}}}. \tag{8} && +\end{flalign*} + +$\sqrt{ 1- 2z \cos \theta + z^{2}}$ is finite and continuous for all values real or complex of +$z$. It is double-valued but the two branches of the function are distinct except +for the values of $z$ which make $1 - 2z \cos \theta + z^{2} = 0$ namely $z = \cos \theta + i \sin \theta$ + and $z = \cos \theta - i\sin \theta$, both of which have the modulus unity and which are +\emph{critical} values.\\ + +$\dfrac{1}{\sqrt{ 1- 2z\cos\theta + z^2}}$ is finite and continuous except for the values of + $z = \cos \theta -i\sin \theta$ and $z=\cos\theta + i\sin\theta$ for which it becomes infinite; it is +double-valued but has as critical values only these values of $z$. It is then +\textit{holomorphic} within a circle described with the origin as centre and the radius +unity, and can be developed into a power series which will be convergent for +all values of $ z$ having moduli less than one. (Int.\ Cal.\ Arts.~207, 212, 214, +220.) + +If then $r>r_1$ $\displaystyle\frac{1}{\sqrt{1-\dfrac{2r_1}{r}\cos \theta + \dfrac{r_1^2}{r^2}}}$ can be developed into a convergent series +involving whole powers of $\dfrac{r_1}{r}$. + +Let $\displaystyle\sum p_m \dfrac{r_1^m}{r^m}$ be this series, $p_{m}$, of course, being a function of $\cos\theta$. +Then +\[ +V = \frac{1}{r }\sum p_m \frac{r_1^m}{r^m} +\] +[v.\ (7)] is a solution of (6). Substitute this value of $V$ in (6) and we get +\[ +\sum \left [\frac{r_1^m}{r^{m+1}} m(m+1)p_m + \frac{r_1^m}{r^{m+1}} \frac{1}{\sin\theta} \frac{d}{d\theta} \left ( \sin\theta \frac{dp_m}{d\theta} \right ) \right ] = 0. +\] +As this must hold whatever the value of r provided $ r > r_1$ the coefficient of +each power of $r$ must be zero, and hence the equation +\[ +\frac{1}{\sin \theta} \frac{d}{d\theta} \left (\sin \theta \frac{dp_{m}}{d\theta} \right )+m(m+1)p_{m}=0 \tag{9} +\] +must be true. + +% -----File: 158.png +But as we have seen in Art.~9 the substitution of $x = \cos \theta$ in (9) reduces +it to +\[ +(1 - x^2) \dfrac{d^2p_m}{ dx^2} - 2x \dfrac{dp_m }{ dx} + m(m + 1)p_m = 0, +\] +\begin{flalign*} +&\text{and therefore} & z &= p_m & \phantom{and therefore} +\end{flalign*} +is a solution of Legendre's Equation (1). + +If $r < r_1$ $\dfrac{1 }{\sqrt{ 1 - \dfrac{2r}{r_1}\cos \theta + \dfrac{r^2}{r_1^2}} }$ can be developed into a convergent series +involving whole powers of $\dfrac{r^2}{r_1^2}$. + +Let $\displaystyle \sum p_m \dfrac{r^m}{r_1^m}$ be this series. Then +\[ +V = \dfrac{1}{r_1}\sum p_m\dfrac{r^m}{r_1^m} +\] +(v.\ 8) is a solution of (6); substituting in (6) we get +\[ + \sum\left[\dfrac{r^m}{r_1^{m+1}} m(m + 1) p_m + \dfrac{r^m}{r_1^{m+1}}\dfrac{ 1}{\sin \theta} \dfrac{d}{d \theta} \left(\sin \theta \dfrac{dp_m}{d \theta}\right)\right] = 0, +\] +whence it follows as before that +\[ +z = p_m +\] +is a solution of Legendre's Equation. + +But $p_m$ is the coefficient of the $m$th power of $\dfrac{r}{r_1}$ in the development of +$\left(1 - 2 \dfrac{r}{r_1} \cos \theta + \dfrac{r^2}{r_1^2}\right)^{-\frac{1}{2}}$ according to powers of $\dfrac{r}{r_1}$, or of the $m$th power of $\dfrac{r_1}{r }$ in +the development of $\left(1 - 2 \dfrac{r_1}{r} \cos \theta + \dfrac{r_1^2}{r^2}\right)^{-\frac{1}{2}}$ according to powers of $\dfrac{r_1}{r }$, or +more briefly it is the coefficient of the $m$th power of z in the development of +$(1 - 2xz + z^2)^{-\frac{1}{2}}$ according to powers of $z$, $x$ standing for $\cos \theta$. +\[ +(1 - 2xz + z^2)^{-\frac{1}{2}} = [1 - z(2x - z)]^{-\frac{1}{2}} +\] +and can be developed by the Binomial Theorem; the coefficient of $z^m$ is easily +picked out and is +\begin{multline*} +\dfrac{(2m - 1)(2m - 3)\cdots 1 }{ m!}\left[x^m - \dfrac{m(m - 1)}{ 2(2m - 1)} x^{m-2}\right.\\ +\left.+\dfrac{m(m - 1)(m - 2)(m - 3)}{2.4.(2m - 1)(2m - 3)} x^{m-4} - \cdots\right]. +\end{multline*} +But this is precisely $P_m(x)$. [v.\ Art.~74 (9)] + +Hence $P_m(x)$ is equal to the coefficient of the $m$th power of $z$ in +the development of $[1 - 2xz + z^2]^{-\frac{1}{2}}$ into a power series, the modulus of $z$ +being less than unity. +% -----File: 159.png + +\mypara{76.} If $x=1$\;\;$P_m(x)=1$. For if $x=1$\;\;$(1-2xz+z^2)^{-\frac{1}{2}}$ reduces to +$(1-2z+z^2)^{-\frac{1}{2}}$ that is to $(1-z)^{-1}$, which develops into +\[ +1 + z + z^2 + z^3 + z^4 + \cdots, +\] +and the coefficient of each power of $z$ is unity. Therefore +\[ +P_m(1) = 1. \tag{1} +\] + +We have seen that if $m$ is even $P_m(x)$ contains only even powers of $x$ and +terminates with the term involving $x^0$, that is with the constant term. + +The value of this constant term can be picked out from the formula for +$P_m(x)$ [v.\ Art.~74 (9)]. It is +$(-1)^{\frac{m}{2}}\dfrac{1.3.5.\cdots(m-1)} + {2.4.6.\cdots m}$; or it can be found as +follows:---It is clearly the value $P_m(x)$ assumes when $x=0$; it is, then, the +coefficient of $z^m$ in the development of $(1+z^2)^{-\frac{1}{2}}$; but +\[ +(1+z^2)^{-\frac{1}{2}}=1 + -\frac{1}{2}z^2 + +\frac{1.3}{2.4}z^4 + -\frac{1.3.5}{2.4.6}z^6 + +\frac{1.3.5.7}{2.4.6.8}z^8 - \cdots +\] +and the coefficient of $z^m$, $m$ being an even number, is +$(-1)^{\frac{m}{2}}\dfrac{1.3.5\cdots(m-1)} + {2.4.6\cdots m}$. + +If $m$ is odd $P_m(x)$ contains only odd powers of $x$ and terminates with the +term involving $x$ to the first power. The coefficient of this term can be +picked out from (9) Art.~74 and is +$(-1)^{\frac{m-1}{2}}\dfrac{3.5.7.\cdots m} + {2.4.6.\cdots(m-1)}$; or it can be +found as follows:---It is clearly the value assumed by $\dfrac{dP_m(x)}{dx}$ when $x=0$. + +It is, then, the coefficient of $z^m$ in the development of $\dfrac{z}{(1+z^2)^{\frac{3}{2}}}$. +\[ +\frac{z}{(1+z^2)^{\frac{3}{2}}} = z + -\frac{3}{2}z^3 + +\frac{3.5}{2.4}z^5 + -\frac{3.5.7}{2.4.6}z^7 + +\cdots +\] +and the coefficient of $z^m$ in this development is +$(-1)^{\frac{m-1}{2}}\dfrac{3.5.7\cdots m} + {2.4.6\cdots(m-1)}$, + $m$ being an odd number. + +\mypara{77.} To recapitulate: +\begin{align*} +P_m(x) ={}&\frac{1.3.5\cdots(2m-1)}{m!}\bigg[ +x^m-\frac{m(m-1)}{2(2m-1)}x^{m-2}\\ + &+\frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}x^{m-4}\\ + &-\frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{2.4.6.(2m-1)(2m-3)(2m-5)}x^{m-6} + +\cdots\bigg], \tag{1} +\end{align*} +% -----File: 160.png +$m$ being a positive integer, is a \textit{Surface Zonal Harmonic} or \textit{Legendrian} of the +$m$th order. It is a finite sum terminating with the first power of $x$ if $m$ is +odd, and with the zeroth power of $x$ if $m$ is even. + +$P_m(x)$ is the coefficient of the $m$th power of $z$ in the development of +$(1 - 2xz + z^2)^{-\frac{1}{2}}$ into a power series. Hence if $z < 1$ +\begin{multline*} +(1 - 2xz + z^2)^{-\frac{1}{2}} = P_0(x) + P_1(x).z + P_2(x).z^2 + P_3(x).z^3\\ ++ P_4(x).z^4 + P_5(x).z^5 + \cdots + P_m(x).z^m + \cdots. \tag{2} +\end{multline*} +Whence +\begin{flalign*} + \left . \begin{aligned} +\frac{1}{\sqrt{ r^2 - 2rr_1 \cos\theta + r_1^2}} = \frac{1}{ r } \left [ P_0 (\cos \theta) + \frac{r_1 }{ r }P_1 (\cos \theta) + \frac{r_1^2 }{r^2 }P_2 (\cos \theta) + \cdots \right . \\ ++ \left . \frac{r_1^m }{ r^m} P_m (\cos \theta) + \cdots \right ] \quad \text{if} \quad r > r_1 \\ + = \frac{1}{ r_1} \left [P_0(\cos \theta) +\frac{ r }{ r_1} P_1(\cos \theta) +\frac{ r^2 }{ r_1^2} P_2(\cos \theta) + \cdots \right . \\ ++\left . \frac{ r^m }{r_1^m} P_m(\cos \theta) + \cdots \right ]\;\text{ if } \;r < r_1. +\end{aligned} \right \} (3) +\end{flalign*} +\[ +z = P_m(x) +\] +is a solution of Legendre's Equation +\[ +(1 - x^2) \frac{d^2z }{ dx^2} - 2x \frac{dz}{ dx} + m(m + 1)z = 0 +\] +when $m$ is a positive integer. +\begin{flalign*} + & & V&= r^mP_m(\cos \theta) && \\ +&\text{and} & V&= \frac{1}{ r^{m+1}} P_m (\cos \theta) && \phantom{and} +\end{flalign*} +are solutions of the form of Laplace's Equation in Spherical Coördinates +which is independent of $\phi$, namely +\begin{gather*} +rD^2_r (rV) + \frac{1}{ \sin(\theta) }D_\theta (\sin \theta D_\theta V) = 0. \tag{4}\\ +P_m(1) = 1. \tag{5}\\[1ex] +P_{2m}(-x) = P_{2m}(x). \tag{6}\\[1ex] +P_{2m+1} (-x) =-P _{2m+1}(x). \tag{7}\\[1ex] +P_{2m+1}(0) = 0. \tag{8}\\ +P_{2m}(0) = (-1)^m \frac{1.3.5. \cdots (2m - 1) }{2.4.6. \cdots 2m}. \tag{9}\\[1ex] +% -----File: 161.png +\left[\frac{dP_{2m+1}(x)}{dx}\right]_{x=0} + =(-1)^m\frac{3.5.7.\cdots(2m+1)} + {2.4.6.\cdots 2m}. \tag{10} +\end{gather*} + +For convenience of reference we write out a few Zonal Harmonics. They +are obtained by substituting successive integers for $m$ in formula (1). +\[\left. +\begin{aligned} +P_0(x) &= 1\\[1ex] +P_1(x) &= x\\ +P_2(x) &= \frac{1}{2}(3x^2 - 1)\\ +P_3(x) &= \frac{1}{2}(5x^3 - 3x)\\ +P_4(x) &= \frac{1}{8}(35x^4 - 30x^2 + 3)\\ +P_5(x) &= \frac{1}{8}(63x^5 - 70x^3 + 15x)\\ +P_6(x) &= \frac{1}{16}(231x^6 - 315x^4 + 105x^2 - 5)\\ +P_7(x) &= \frac{1}{16}(429x^7 - 693x^5 + 315x^3 - 35x)\\ +P_8(x) &= \frac{1}{128}(6435x^8 - 12012x^6 + 6930x^4 - 1260x^2 + 35). +\end{aligned}\right\} \tag{11} +\] + +Any Surface Zonal Harmonic may be obtained from the two of next lower +orders by the aid of the formula +\[ +(n+1)P_{n+1}(x)-(2n+1)xP_n(x)+nP_{n-1}(x)=0 \tag{12} +\] +which is easily obtained and is convenient when the numerical value of $x$ is +given. + +Differentiate (2) with respect to $z$ and we get +\[ +\frac{-(z-x)}{(1-2xz+z^2)^{\frac{3}{2}}} + =P_1(x) + 2P_2(x).z + 3P_3(x).z^2 + \cdots +\] +whence +\[ +\frac{-(z-x)}{(1- 2xz + z^2)^{\frac{1}{2}}} + =(1-2xz+z^2)(P_1(x) + 2P_2(x).z + 3P_3(x).z^2 + \cdots). +\] +Hence by (2) +\begin{align*} +(1-2xz+z^2)(P_1(x) & ++2P_2(x).z + 3P_3(x).z^2 + \cdots)\\ +&+(z-x)(P_0(x)+P_1(x).z + + P_2(x).z^2 + \cdots) = 0 \tag{13} +\end{align*} +% -----File: 162.png + +(13) is identically true, hence the coefficient of each power of $z$ must vanish. +Picking out the coefficient of $z^n$ and writing it equal to zero we have formula +(12) above.\footnote +{For tables of Surface Zonal Harmonics v.\ Appendix Tables I and II.} + +\mypara{78.} We are now able to solve completely the problem considered in Art.~9. + +We were to find a solution of the differential equation +\[ +rD_r^2(rV) + \frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V) = 0 \tag{1} +\] +subject to the condition +\[ +V = \frac{M}{(c^2 + r^2)^\frac{1}{2}}\quad \text{when}\quad \theta = 0. \tag{2} +\] + +We know (v.\ Art.~77) that +\begin{flalign*} +&&V = r^mP_m(\cos\theta)&&\\ +&\text{and} & V=\frac{1}{r^{m+1}}P_m(\cos\theta)&&&\phantom{and} +\end{flalign*} +are solutions of (1). + +For values of $r<c$ +\[ +\frac{M}{(c^2+r^2)^\frac{1}{2}} = \frac{M}{c} +\Bigl[1-\frac{1}{2}\frac{r^2}{c^2} + +\frac{1.3}{2.4}\frac{r^4}{c^4} + -\frac{1.3.5}{2.4.6}\frac{r^6}{c^6} + +\cdots\Bigr]. \tag{3} +\] + +Therefore for values of $r<c$ +\begin{align*} +V = \frac{M}{c}\Big[P_0(\cos\theta)-\frac{1}{2}\frac{r^2}{c^2}&P_2(\cos\theta)\\ +&+\frac{1.3}{2.4}\frac{r^4}{c^4}P_4(\cos\theta) +-\frac{1.3.5}{2.4.6}\frac{r^6}{c^6}P_6(\cos\theta)+\cdots\Big]\tag{4} +\end{align*} +is our required solution; because each term satisfies equation (1), and therefore +the whole value satisfies (1), and when $\theta=0$ +\[ +P_m(\cos\theta) = P_m(1) = 1 +\] +[v.\ (5) Art.~77], and hence (4) reduces to (3) and (2) is satisfied. + +For values of $r>c$ +\begin{align*} +\frac{M}{(c^2+r^2)^\frac{1}{2}} &= \frac{M}{r} + \Bigl[1-\frac{1}{2}\frac{c^2}{r^2} + +\frac{1.3}{2.4}\frac{c^4}{r^4} + -\frac{1.3.5}{2.4.6}\frac{c^6}{r^6} + +\cdots\Bigr] \tag{5}\\ +&=M\Bigl[\frac{1}{r}-\frac{1}{2}\frac{c^2}{r^3} + +\frac{1.3}{2.4}\frac{c^4}{r^5} + -\frac{1.3.5}{2.4.6}\frac{c^6}{r^7} + +\cdots\Bigr]. +\end{align*} + +% -----File: 163.png +Therefore for values of $r>c$ +\begin{align*} +V=\frac{M}{c}\Big[\frac{c}{r}P_0(\cos\theta)-\frac{1}{2}&\frac{c^3}{r^3}P_2(\cos\theta)\\ +&+\frac{1.3}{2.4}\frac{c^5}{r^5}P_4(\cos\theta) + -\frac{1.3.5}{2.4.6}\frac{c^7}{r^7}P_6(\cos\theta) + +\cdots\Big] \tag{6} +\end{align*} +is our required solution. For it satisfies (1) and reduces to (2) when $\theta=0$. + +\mypara{79.} As another example let us suppose a conductor in the form of a thin +circular disc charged with electricity, and let it be required to find the value +of the potential function at any point in space. + +If the magnitude of the charge is $M$ and the radius of the plate is $a$ the +surface density at a point of the plate at a distance $r$ from the centre is +\[ +\sigma = \frac{M}{4a\pi\sqrt{a^2-r^2}} +\] +and all points of the conductor are at the potential $\dfrac{\pi M}{2a}$. (v.\ Peirce's Newtonian +Potential Function, §~61.) + +The value of the potential function at a point in the axis of the plate at the +distance $x$ from the plate is easily seen to be +\begin{align*} +V &= \frac{M}{a}\int\limits_0^a\frac{r dr}{\sqrt{(a^2-r^2)(x^2+r^2)}}\\ + &= \frac{M}{2a}\cos^{-1}\frac{x^2-a^2}{x^2+a^2}. +\end{align*} +\begin{alignat*}{2} +&\frac{d}{dx}\Bigl(\frac{M}{2a}\cos^{-1}\frac{x^2-a^2}{x^2+a^2}\Bigr) &&= -\frac{M}{a^2+x^2}\\ +&&&= -\frac{M}{a^2}\Bigl[1- \frac{x^2}{a^2}+\frac{x^4}{a^4}-\frac{x^6}{a^6}+\cdots\Bigr]\\ +&\text{if} \quad x<a,\\ +&&&= -\frac{M}{x^2}\Bigl[1- \frac{a^2}{x^2}+\frac{a^4}{x^4}-\frac{a^6}{x^6}+\cdots\Bigr]\\ +&\text{if} \quad x>a. +\end{alignat*} + +Integrating and then determining the arbitrary constant we have +\begin{alignat*}{2} +&\frac{M}{2a}\cos^{-1}\frac{x^2-a^2}{x^2+a^2} +&&= \frac{M}{a}\Bigl[\frac{\pi}{2}-\frac{x}{a} + +\frac{x^3}{3a^3}-\frac{x^5}{5a^5}+\frac{x^7}{7a^7}-\cdots\Big]\\[1ex] +&\quad\text{if} \quad x<a,\\ +&&&= \frac{M}{a}\Big[\frac{a}{x}-\frac{a^3}{3x^3} + +\frac{a^5}{5x^5}-\frac{a^7}{7x^7}+\cdots\Bigr]\\ +&\quad\text{if} \quad x>a. +\end{alignat*} +% -----File: 164.png + +We have, then, to solve the equation +\[ +rD^2_r(rV)+\frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V)=0 +\] +subject to the conditions +\begin{flalign*} +&&V&=\frac{M}{a}\bigg[\frac{\pi}{2}-\frac{r}{a} + +\frac{r^3}{3a^3}-\frac{r^5}{5a^5} + +\frac{r^7}{7a^7}-\cdots\bigg]&\\ +&&&\text{when}\quad \theta=0\quad \text{and}\quad r<a&\\ +&\text{and} & V &=\frac{M}{a}\bigg[\frac{a}{r}-\frac{a^3}{3r^3} + +\frac{a^5}{5r^5}-\frac{a^7}{7r^7}+\cdots\bigg]&\phantom{and}\\ +&&&\text{when}\quad \theta=0\quad \text{and}\quad r>a.& +\end{flalign*} + +The required solution is easily seen to be +\begin{flalign*} +&&V&=\frac{M}{a}\bigg[ +\frac{\pi}{2}-\frac{r}{a}P_1(\cos\theta) + +\frac{1}{3}\frac{r^3}{a^3}P_3(\cos\theta) + -\frac{1}{5}\frac{r^5}{a^5}P_5(\cos\theta)+\cdots\bigg] +\intertext{if $r<a$ and $\theta<\dfrac{\pi}{2}$,} +&\text{and}& V&=\frac{M}{a}\bigg[\frac{a}{r} +-\frac{1}{3}\frac{a^3}{r^3}P_2(\cos\theta) ++\frac{1}{5}\frac{a^5}{r^5}P_4(\cos\theta) +-\frac{1}{7}\frac{a^7}{r^7}P_6(\cos\theta)+\cdots\bigg] +\intertext{if $r>a$.} +\end{flalign*} + +\EXAMPLE{S} + +1.\quad Given that if a charge $M$ of electricity is placed on an ellipsoidal conductor +the surface density at any point $P$ of the conductor is equal to $\dfrac{Mp}{4\pi abc}$, + where $p$ is the distance from the centre of the conductor to the tangent plane at +$P$ (v.\ Peirce, New.\ Pot.\ Func.\ §~61); find the value of the potential function at +any external point when the conductor is the oblate spheroid generated by the +rotation of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ about its minor axis. + +\textit{Ans.} (1) If the point is on the axis of revolution +\[ +V=\frac{M}{2\sqrt{a^2-b^2}}\bigg[\sin^{-1}\bigg(\frac{bx+a^2-b^2}{a\sqrt{x^2+a^2-b^2}}\bigg) + -\sin^{-1}\bigg(\frac{bx-a^2+b^2}{a\sqrt{x^2+a^2-b^2}}\bigg)\bigg] +\] +$x$ being the distance from the centre. + +(2) If the point is on the surface of the spheroid +\[ +V=\frac{M}{2\sqrt{a^2-b^2}}\bigg[\frac{\pi}{2}-\sin^{-1}\bigg(\frac{2b^2-a^2}{a^2}\bigg)\bigg] + =\frac{M}{ \sqrt{a^2-b^2}}\bigg[\frac{\pi}{2}-\tan^{-1}\bigg(\frac{b}{\sqrt{a^2-b^2}}\bigg)\bigg]. +\] + +% -----File: 165.png +(3) If the distance $r$ of the point from the centre is less than $\sqrt{ a^2 - b^2}$ and +$\theta < \dfrac{\pi}{2}\tablestrut$ +\begin{multline*} +V = \frac{M }{\sqrt{ a^2 - b^2}} \left [ \frac{\pi }{2} - \frac{r }{(a^2-b^2)^{\frac{1}{2}}} P_1(\cos \theta) \right . \\ +\left . + \frac{r^3}{ 3(a^2-b^2)^{\frac{3}{2}} }P_3(\cos \theta) - \frac{r^5 }{5(a^2-b^2)^{\frac{5}{2}}} P_5(\cos \theta)+ \cdots \right ]. +\end{multline*} + +(4) If the distance $ r$ of the point from the centre is greater than $\sqrt{ a^2 - b^2}$ +\begin{multline*} +V = \frac{M }{\sqrt{ a^2 - b^2}} \biggl [ \frac{(a^2-b^2)^{\frac{1}{2}}}{r} - \frac{(a^2-b^2)^{\frac{3}{2}}}{3r^3} P_2(\cos \theta) \biggr .\\ +\biggl . + \frac{(a^2-b^2)^{\frac{5}{2}}}{5r^5} P_4(\cos \theta) - \frac{(a^2-b^2)^{\frac{7}{2}}}{7r^7}P_6(\cos \theta) + \cdots \biggr ]. +\end{multline*} + +2.\quad If the conductor is the prolate spheroid generated by the rotation of the +ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ about its major axis, show that if the point is an external +point and is on the axis at a distance $x$ from the centre, +\[ +V = \frac{M }{2\sqrt{ a^2 - b^2}} \log\frac{ x + \sqrt{ a^2 - b^2}} {x - \sqrt {a^2 - b^2}}. +\] + +If the point is not on the axis and $r > \sqrt{a^2 - b^2}$ +\begin{multline*} +V = \frac{M }{\sqrt{ a^2 - b^2}} \biggl [ \frac{(a^2-b^2)^{\frac{1}{2}}}{r} + \frac{(a^2-b^2)^{\frac{3}{2}}}{3r^3} P_2(\cos \theta) \biggr .\\ +\biggl . + \frac{(a^2-b^2)^{\frac{5}{2}}}{5r^5} P_4(\cos \theta) + \frac{(a^2-b^2)^{\frac{7}{2}}}{7r^7}P_6(\cos \theta) + \cdots \biggr ]. +\end{multline*} + +\mypara{80.} As a third example we will find the value of the potential function due +to a thin homogeneous circular disc, of density $\rho$, thickness $k$, and radius $a$. + +The value of $V$ at a point in the axis of the disc at a distance $x$ from its +centre is readily found and proves to be +\[ +V_0 = 2 \pi \rho k(\sqrt{x^2 + a^2} - x) = \frac{2M}{ a^2} [\sqrt{x^2 + a^2} - x]. +\] + +If $x>a$ +\begin{flalign*} +% recast to fit line +&&\sqrt{x^2+a^2} &= x \left (1+\frac{a^2}{x^2} \right )^{\frac{1}{2}} \\ +&& &= x \left [1+\frac{1}{2} \frac{a^2}{x^2} - \frac{1.1}{2.4} \frac{a^4}{x^4}+ \frac{1.1.3}{2.4.6} \frac{a^6}{x^6} - \frac{1.1.3.5}{2.4.6.8} \frac{a^8}{x^8} + \cdots \right ] &&\\ +&\text{and} &V_0 &=\frac{ 2M}{a} \left [ \frac{1}{2} \frac{a}{x} - \frac{1.1}{2.4} \frac{a^3}{x^3} + \frac{1.1.3}{2.4.6} \frac{a^5}{x^5} - \frac{1.1.3.5}{2.4.6.8} \frac{a^7}{x^7} + \cdots \right ]. && +\end{flalign*} + +% -----File: 166.png +If $x < a$ +\begin{flalign*} +% recast to fit line +&&\sqrt{x^2+a^2} &= a\left(1+\dfrac{x^2}{a^2}\right)^{\frac{1}{2}}\\ +&&& = a\left[1+\dfrac{1}{2} \dfrac{x^2}{a^2} - \dfrac{1.1}{2.4}\dfrac{ x^4}{a^4} + \dfrac{1.1.3}{2.4.6} \dfrac{x^6}{a^6} +\cdots\right]&\\ +&\text{and}& V_0 = \dfrac{2M}{a}{}&\left[1 -\dfrac{ x}{a} + \dfrac{1}{2} \dfrac{x^2}{a^2} - \dfrac{1.1}{2.4}\dfrac{ x^4}{a^4} + \dfrac{1.1.3}{2.4.6} \dfrac{x^6}{a^6} - \dfrac{1.1.3.5}{2.4.6.8}\dfrac{ x^8}{a^8} + \cdots \right].& +\end{flalign*} + +Hence the solution for any external point is +\begin{multline*} +V = \dfrac{2M}{a} \left[\dfrac{1}{2} \dfrac{a}{r} - \dfrac{1.1}{2.4} \dfrac{a^3}{r^3} P_2(\cos\theta)\right.\\ + +\left. \dfrac{1.1.3}{2.4.6} \dfrac{a^5}{r^5}P_4(\cos\theta) - \dfrac{1.1.3.5}{2.4.6.8}\dfrac{ a^7}{r^7}P_6(\cos\theta) + \cdots \right] +\end{multline*} +if $r > a$, and +\begin{multline*} +V = \dfrac{2M}{a}\left[1-\dfrac{r}{a}P_1(\cos\theta)\right.\\ ++\left.\dfrac{1}{2} \dfrac{r^2}{a^2}P_2(\cos\theta) - \dfrac{1.1}{2.4}\dfrac{ r^4}{a^4} P_4(\cos\theta) + \dfrac{1.1.3}{2.4.6} \dfrac{r^6}{a^6}P_6(\cos\theta) -\cdots\right] +\end{multline*} +if $r < a$ and $\theta < \dfrac{\pi}{2}$. + +\EXAMPLE{S} + +1.\quad The potential function due to a homogeneous hemisphere whose axis is +taken as the polar axis, is +\begin{multline*} +V = \dfrac{M}{a} \left[\dfrac{a}{r} + \dfrac{3.1}{2.4} \dfrac{a^2}{r^2}P_1(\cos\theta) \right.\\ + -\left. \dfrac{3.1.1}{2.4.6} \dfrac{a^4}{r^4} P_3(\cos\theta) + \dfrac{3.1.1.3}{2.4.6.8} \dfrac{a^6}{r^6}P_5(\cos\theta) -\cdots \right] +\end{multline*} +if $r > a$, and is +\begin{multline*} +V = \dfrac{M}{a} \left[\dfrac{3}{2} + \dfrac{3}{2} \dfrac{r}{a}P_1(\cos\theta) + \dfrac{r^2}{a^2}P_2(\cos\theta)\right.\\ + +\left. \dfrac{3.1}{2.4}\dfrac{ r^3}{a^3} P_3(\cos\theta) - \dfrac{3.1.1}{2.4.6} \dfrac{r^5}{a^5}P_5(\cos\theta)+ \cdots \right] +\end{multline*} +if $r < a$ and $\theta > \dfrac{\pi}{2}$.\\ + +2.\quad The potential function due to a solid sphere whose density is proportional +to the distance from a diametral plane is, at an external point, +\begin{multline*} +V = \dfrac{8}{15}\dfrac{ M}{a} \left[\dfrac{5.3}{2.4}\dfrac{ a}{r} + \dfrac{5.3.1}{2.4.6} \dfrac{a^3}{r^3}P_2(\cos\theta)\right.\\ + -\left. \dfrac{5.3.1.1}{2.4.6.8} \dfrac{a^5}{r^5} P_4(\cos\theta) + \dfrac{5.3.1.1.3}{2.4.6.8.10}\dfrac{ a^7}{r^7}P_6(\cos\theta) - \cdots \right]. +\end{multline*} + +% -----File: 167.png +3.\quad The potential function due to the homogeneous oblate spheroid generated +by the rotation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ about its minor axis is, at an external point, +\begin{multline*} +V = \frac{3}{2}\frac{ M}{(a^2-b^2)} \left [ \frac{x^2+a^2-b^2 }{ 2(a^2-b^2)^{\frac{1}{2}}} \left ( \sin^{-1} \frac{(a^2-b^2+bx)}{a \sqrt{ x^2+a^2-b^2}} \right . \right . \\ +\left . \left . + \sin^{-1} \frac{(a^2-b^2-bx)}{a \sqrt{ x^2+a^2-b^2}} \right ) - x \right ] +\end{multline*} +if the point is on the axis of the spheroid at a distance $x$ from its centre. +\begin{multline*} +V = \frac{ 3M}{(a^2-b^2)^{\frac{1}{2} }} \left [\frac{1}{1.3}\frac{ (a^2-b^2)^{\frac{1}{2}}}{ r} - \frac{1}{3.5}\frac{ (a^2-b^2)^{\frac{3}{2}}} {r^3 }P_2(\cos \theta) \right . \\ +\left . + \frac{1}{5.7}\frac{(a^2-b^2)^{\frac{5}{2}}} {r^5} P_4(\cos \theta) - \cdots \right ] +\end{multline*} +if $r > (a^2 - b^2)^{\frac{1}{2}}$, and +\begin{multline*} +V = \frac{3M}{(a^2-b^2)^{\frac{1}{2}}} \left [ \frac{\pi}{4} -\frac{ r}{(a^2-b^2)^{\frac{1}{2}}} P_1(\cos \theta) + \frac{\pi}{4} \frac{r^2}{(a^2-b^2)} P_2(\cos \theta) \right .\\ +\left . - \frac{1}{1.3} \frac{r^3}{(a^2-b^2)^{\frac{3}{2}}} P_3(\cos \theta) + \frac{1}{3.5}\frac{ r^5}{(a^2-b^2)^{\frac{5}{2}}} P_5(\cos \theta) - \cdots \right ] +\end{multline*} +if $r<(a^2-b^2)^{\frac{1}{2}}$ and $\theta < \dfrac{\pi }{2}$.\\ + +4.\quad The potential function due to the homogeneous prolate spheroid +generated by the rotation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ about its major axis is, at an +external point, +\begin{gather*} +V = \frac{3M}{(a^2-b^2)^{\frac{1}{2}}} \left [\frac{1}{1.3}\frac{ (a^2-b^2)^{\frac{1}{2}}}{r } + \frac{1}{3.5}\frac{ (a^2-b^2)^{\frac{3}{2}}}{r^3} P_2(\cos \theta) \right . \\ ++\left . \frac{1}{5.7}\frac{ (a^2-b^2)^{\frac{5}{2}}}{ r^5} P_4(\cos \theta) + \cdots \right ] +\end{gather*} +if $r>(a^2 -b^2)^{\frac{1}{2}}$. + +\mypara{81.} The method employed in the last three articles may be stated in +general as follows:---Whenever in a problem involving the solving of the +special form of Laplace's Equation +\[ +rD^2_r (rV) + \frac{1}{\sin\theta }D_\theta (\sin \theta D_\theta V) = 0, +\] +the value of $V$ is given or can be found for all points on the axis of $X$ and +this value can be expressed as a sum or a series involving only whole powers +positive or negative of the radius vector of the point, the solution for a point +% -----File: 168.png +not on the axis can be obtained by multiplying each term by the appropriate +Zonal Harmonic, subject only to the condition that the result if a series must +be convergent. + +It will be shown in the next article that $P_m(\cos\theta)$ is never greater than +one nor less than minus one. Hence the series in question will be convergent +for all values of $r$ for which the original series was \textit{absolutely convergent}.\ + +\mypara{82.} In addition to the form given in (1) Art.~77 for $P_m(x)$ other forms +are often useful. + +It ought to be possible to develop $P_m(\cos\theta)$, which may be regarded as a +function of $\theta$, into a Fourier's Series, and such a development may be obtained, +though with much labor, by the methods of Chapter II. + +The development in terms of cosines of multiples of $\theta$ may be obtained +much more easily by the following device. + +We have seen in Art.~75 that $P_m(\cos\theta)$ is the coefficient of the $m$th power +of $z$ in the development of $(1 - 2z \cos\theta + z^2)^{-\frac{1}{2}}$ in a power series, and that +if mod $z < 1$\;\;$(1 - 2z~\cos\theta + z^2)^{-\frac{1}{2}}$ can be developed into such a series. We +know by the Theory of Functions that only one such series exists, so that the +method by which we may choose to obtain the development will not affect the +result. +\begin{align*} +(1 - 2z\cos\theta + z^2)^{-\frac{1}{2}} &= (1-z(e^{\theta i} + e^{-\theta i}) + z^2)^{-\frac{1}{2}}\\ + &= (1-ze^{\theta i})^{-\frac{1}{2}}(1-ze^{-\theta i})^{-\frac{1}{2}}. +\end{align*} +$(1-ze^{\theta i})^{-\frac{1}{2}}$ may be developed into an absolutely convergent series if + mod $z <1$, by the Binomial Theorem. We have +\[ +(1-ze^{\theta i})^{-\frac{1}{2} }= 1 + \frac{1}{2} ze^{\theta i} + \frac{1.3}{2.4} z^2e^{2\theta i } ++ \frac{1.3.5}{2.4.6} z^3e^{3\theta i} + \frac{1.3.5.7}{2.4.6.8 }z^4e^{4\theta i }+ \cdots\\[-4ex] +\] +\begin{multline*} +(1-ze^{-\theta i})^{-\frac{1}{2} }= 1 + \frac{1}{2} ze^{-\theta i} + \frac{1.3}{2.4} z^2e^{-2\theta i } \\ ++ \frac{1.3.5}{2.4.6} z^3e^{-3\theta i} + \frac{1.3.5.7}{2.4.6.8 }z^4e^{-4\theta i }+ \cdots +\end{multline*} + +The product of these series will give a development for $(1-2z\cos\theta+z^2)^{-\frac{1}{2}}$ + in power series. The coefficient of $z^m$ is easily picked out, and must be equal +to $P_m(\cos\theta)$. We thus get +\begin{align*} +% recast to fit line +P_m(\cos\theta)={}&\frac{ 1.3.5. \cdots (2m-1) }{ 2.4.6.\cdots 2m} \bigg[e^{m\theta i} + e^{-m\theta i} \\ +&\quad + \frac{1}{2}.\frac{2m}{2m-1}(e^{(m-2)\theta i} + e^{-(m-2)\theta i}) \\ +&\quad + \frac{1.3}{2.4}.\frac{ 2m(2m-2) }{ (2m-1)(2m-3)} (e^{(m-4)\theta i} + e^{-(m-4)\theta i}) + \cdots \bigg]\\[1ex] +% -----File: 169.png +P_m(\cos\theta)={}&\frac{1.3.5\cdots(2m-1)}{2.4.6.\cdots 2m} +\bigg[2\cos m\theta + 2\frac{1.m}{1.(2m-1)}\cos(m-2)\theta\\ +&\quad +2\frac{1.3\,m(m-1)}{1.2(2m-1)(2m-3)}\cos(m-4)\theta\\ +&\quad +2\frac{1.3.5}{1.2.3}\frac{m(m-1)(m-2)}{(2m-1)(2m-3)(2m-5)}\cos(m-6)\theta +\cdots\bigg]. \tag{1} +\end{align*} + +If $m$ is odd the development runs down to $\cos\theta$; if $m$ is even to $\cos(0)$, but +in that case the coefficient of $\cos(0)$, that is, the constant term, will not contain +the factor 2 which is common to all the other terms, but will be simply +$\bigg[\dfrac{1.3.5\cdots(m-1)}{2.4.6.\cdots m}\bigg]^2$. + +We write out the values of $P_m(\cos\theta)$ for a few values of $m$ +\[\left. +\begin{aligned} +P_0(\cos\theta) &=1\\[1ex] +P_1(\cos\theta) &=\cos\theta\\ +P_2(\cos\theta) &=\frac{1}{4}(3\cos 2\theta+1)\\ +P_3(\cos\theta) &=\frac{1}{8}(5\cos 3\theta+3\cos\theta)\\ +P_4(\cos\theta) &=\frac{1}{64}(35\cos 4\theta+20\cos 2\theta+9)\\ +P_5(\cos\theta) &=\frac{1}{128}[63\cos 5\theta+35\cos 3\theta+30\cos\theta]\\ +P_6(\cos\theta) &=\frac{1}{512}[231\cos 6\theta+126\cos 4\theta+105\cos 2\theta+50]\\ +P_7(\cos\theta) &=\frac{1}{1024}[429\cos 7\theta+231\cos 5\theta+189\cos 3\theta+175\cos\theta]\\ +P_8(\cos\theta) &=\frac{1}{16384}[6435\cos 8\theta+3432\cos 6\theta+2772\cos 4\theta\\ + &\qquad\qquad +2520\cos 2\theta+1225]. +\end{aligned}\right\} \tag{2} +\] + +Since all the coefficients in the second member of (1) are positive, and since +each cosine has unity for its maximum value it is clear that $P_m(\cos\theta)$ has +its maximum value when $\theta=0$; but we have shown in Art.~76 that $P_m(1)=1$. + Therefore $P_m(\cos\theta)$ is never greater than unity if $\theta$ is real. It is also easily +seen from (1) that $P_m(\cos\theta)$ can never be less than $-1$. + +% -----File: 170.png +\mypara{83.} $P_m(x)$ can be very simply expressed as a derivative. We have +\begin{multline*} +\hspace{12pt}P_m(x)=\frac{(2m-1)(2m-3)\cdots 1}{m!}\bigg[x^m-\frac{m(m-1)}{2.(2m-1)}x^{m-2}\hspace{8em}\\ +\shoveright{+\frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}x^{m-4}-\cdots\bigg]\phantom{.}\quad}\\ +% +\shoveleft{\int\limits_0^x P_m(x)dx=\frac{(2m-1)(2m-3)\cdots 1}{(m+1)!}\bigg[x^{m+1}-\frac{(m+1)m}{2.(2m-1)}x^{m-1}}\\ +\shoveright{+\frac{(m+1)m(m-1)(m-2)}{2.4.(2m-1)(2m-3)}x^{m-3}-\cdots\bigg]}\\ +% +\shoveleft{\int\limits_0^x{}^2 P_m(x)dx^2= \int\limits_0^x dx \int\limits_0^x P_m(x)dx}\\ +\shoveleft{\hspace{62pt}=\frac{(2m-1)(2m-3)\cdots 1}{(m+2)!}\bigg[x^{m+2}-\frac{(m+2)(m+1)}{2.(2m-1)}x^m}\\ +\shoveright{+\frac{(m+2)(m+1)m(m-1)}{2.4.(2m-1)(2m-3)}x^{m-2}-\cdots\bigg]}\\ +% +\shoveleft{\int\limits_0^x{}^m P_m(x)dx^m =\frac{(2m-1)(2m-3)\cdots 1}{(2m)!}\bigg[x^{2m}-\frac{2m(2m-1)}{2(2m-1)}x^{2m-2}}\\ +\shoveright{+\frac{2m(2m-1)(2m-2)(2m-3)}{2.4.(2m-1)(2m-3)}x^{2m-4}-\cdots\bigg]}\\ +\shoveleft{\hspace{68pt}=\frac{(2m-1)(2m-3)\cdots 1}{(2m)!}\bigg[x^{2m}-mx^{2m-2}+\frac{m(m-1)}{2!}x^{2m-4}}\\ +\shoveright{-\frac{m(m-1)(m-2)}{3!}x^{2m-6}+\cdots\bigg].\quad}\\[-4ex] +\end{multline*} +\indent The quantity in brackets obviously differs from $(x^2-1)^m$ by terms involving +lower powers of $x$ than the $m$th. +\begin{flalign*} +&\text{Hence} +& P_m(x)&=\frac{1.3.5\cdots(2m-1)}{(2m)!}\frac{d^m}{dx^m}(x^2-1)^m,&&\phantom{Hence}\\ +&\text{or} +&& P_m(x)=\frac{1}{2^mm!}\frac{d^m}{dx^m}(x^2-1)^m. & \tag{1}& +\end{flalign*} + +This important formula is entirely general and holds not merely when + $x=\cos\theta$, but for all values of $x$. +% -----File: 171.png + +\mypara{84.} The last result is so important that it is worth while to confirm it by +obtaining it directly from Legendre's Equation +\[ +(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+m(m+1)z=0 \tag{1} +\] +v.\ (1) Art.~75. + +Let us differentiate (1) with respect to $x$ a few times representing +$\dfrac{dz}{dx}$ by $z'$, $\dfrac{d^2z}{dx^2}$ by $z''$, $\dfrac{d^3z}{dz^3}$ by $z'''$, \&c. We get +\begin{gather*} +(1-x^2)\frac{d^2z'}{dx^2}-2.2x\frac{dz'}{dx}+[m(m+1)-2]z'=0,\\ +(1-x^2)\frac{d^2z''}{dx^2}-2.3x\frac{dz''}{dx}+[m(m+1)-2(1+2)]z''=0,\\ +(1-x^2)\frac{d^2z'''}{dx^2}-2.4x\frac{dz'''}{dx}+[m(m+1)-2(1+2+3)]z'''=0, +\end{gather*} +and in general +\[ +(1-x^2)\frac{d^2z^{(n)}}{dx^2}-2(n+1)x\frac{dz^{(n)}}{dx}+[m(m+1)-2(1+2+3+\cdots+n)]z^{(n)}=0\\[-3ex] +\] +\begin{flalign*} +&\text{or} & (1-x^2)\frac{d^2z^{(n)}}{dx^2}-2(n+1)x\frac{dz^{(n)}}{dx}+m(m+1)-n(n+1)]z^{(n)}=0. \hspace{15pt}(2) +\end{flalign*} + +Following the analogy of these steps it is easy to write equations that will +differentiate into (1). + +Let $\dfrac{dz_1}{dx}=z$, $\dfrac{d^2z_2}{dx^2}=z$, $\dfrac{d^3z_3}{dx^3}=z$, \&c. Then +\[ +(1-x^2)\frac{d^2z_1}{dx^2}+m(m+1)z_1=0 +\] +will differentiate into (1), +\[ +(1-x^2)\frac{d^2z_2}{dx^2}+2.1x\frac{dz_2}{dx}+[m(m+1)-2.1]z_2=0 +\] +if differentiated twice will give (1), +\[ +(1-x^2)\frac{d^2z_3}{dx^2}+2.2x\frac{dz_3}{dx}+[m(m+1)-2(1+2)]z_3=0 +\] +if differentiated three times will give (1), and in general +\[ +(1-x^2)\frac{d^2z_n}{dx^2}+2(n-1)x\frac{dz_n}{dx}+[m(m+1)-n(n-1)]z_n=0 \tag{3} +\] +if differentiated $n$ times with respect to $x$ will give (1). + +If $n=m+1$ (3) reduces to +\[ +(1-x^2)\frac{d^2z_{m+1}}{dx^2}+2mx\frac{dz_{m+1}}{dx}=0, \tag{4} +\] +% -----File: 172.png +and the $(m+1)$st derivative with respect to $x$ of any function of $x$ which +satisfies (4) will be a solution of (1). (4) can be written +\[ +(1-x^2)\frac{dz_m}{dx}+2mxz_m=0 +\] +and can be readily solved by separating the variables and integrating. v.\ Int.\ +Cal.~(1) page 314. It gives +\begin{flalign*} +&&&\quad z_m = C(x^2-1)^m.&\phantom{Hence}\\ +&\text{Hence} & z=&\frac{d^mz_m}{dx^m}=C\frac{d^m(x^2-1)^m}{dx^m} &\tag{5} +\end{flalign*} +is a solution of Legendre's Equation (1) and agrees with the value of $P_m(x)$ +obtained in Art.~83. + +\mypara{85.} The equations obtained in Art.~84 are so curious and so simply related +that it is worth while to consider them a little more fully. + +We have seen that +\[ +(1-x^2)\frac{d^2z}{dx^2}+2mx\frac{dz}{dx}=0 \tag{1} +\] +differentiates into +\[ +(1-x^2)\frac{d^2z}{dx^2}+2(m-1)x\frac{dz}{dx}+2mz=0; \tag{2} +\] +that if we differentiate (2) $m$ times we get Legendre's Equation +\[ +(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+m(m+1)z=0; \tag{3} +\] +that if we differentiate (2) $2m$ times we get +\[ +(1-x^2)\frac{d^2z}{dx^2}-2(m+1)x\frac{dz}{dx}=0; \tag{4} +\] +that if we differentiate (2) $m-n$ times we have +\[ +(1-x^2)\frac{d^2z}{dx^2}+2(n-1)x\frac{dz}{dx}+[m(m+1)-n(n-1)]z=0; \tag{5} +\] +and that if we differentiate (2) $m+n$ times we have +\[ +(1-x^2)\frac{d^2z}{dx^2}-2(n+1)x\frac{dz}{dx}+[m(m+1)-n(n+1)]z=0. \tag{6} +\] +By the aid of (1) we found in the last article a particular solution of (2), +namely +\[ +z=(x^2-1)^m. +\] +% -----File: 173.png + +If we substitute in (2) $z=u(x^2-1)^m$ following the method illustrated +fully in Art.~18, we get as the general solution of (2) +\[ +z=A(x^2-1)^m+B(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}, \tag{7} +\] +$A$ and $B$ being arbitrary constants. + +$\displaystyle\int\frac{dx}{(x^2-1)^{m+1}}$ is easily written out [v.\ formula (42) page 6. Table of Integrals. +Int.\ Cal.\ Appendix]. If $x<1$ it vanishes when $x=0$. If $x>1$ it +vanishes when $x=\infty$. If then $x<1$ (7) can be written +\[ +z=A(x^2-1)^m+B(x^2-1)^m\int\limits_0^x\frac{dx}{(x^2-1)^{m+1}} \tag{8} +\] +and if $x>1$ +\[ +z=A(x^2-1)^m+B(x^2-1)^m\int\limits_x^\infty\frac{dx}{(x^2-1)^{m+1}} \tag{9} +\] +and in these forms unnecessary arbitrary constants are avoided. + +From (7) we can get the general solutions of (3), (4), (5), and (6). +\[ +z=A\frac{d^m(x^2-1)^m}{dx^m}+B\frac{d^m}{dx^m}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{10} +\] +is the general solution of (3). +\[ +z=A\frac{d^{2m}(x^2-1)^m}{dx^{2m}}+B\frac{d^{2m}}{dx^{2m}}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{11} +\] +is the general solution of (4). +\[ +z=A\frac{d^{m-n}(x^2-1)^m}{dx^{m-n}}+B\frac{d^{m-n}}{dx^{m-n}}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{12} +\] +is the general solution of (5). +\[ +z=A\frac{d^{m+n}(x^2-1)^m}{dx^{m+n}}+B\frac{d^{m+n}}{dx^{m+n}}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{13} +\] +is the general solution of (6). + +In each of these forms $A$ and $B$ are arbitrary constants and the integral is +to be taken from $0$ to $x$ if $x<1$ and from $x$ to $\infty$ if $x>1$. + +Of course (10) must be identical with the forms already obtained in Arts.~16 +and 18 as general solutions of Legendre's Equation. + +Equation (4) is so simple that it can be solved directly, and we get its +solution in the form +\[ +z=A_1+B_1\int\frac{dx}{(x^2-1)^{m+1}} \tag{14} +\] +which must be equivalent to (11). +% -----File: 174.png + +Comparing (14) with (7), the solution of (2), we see that every solution of (4) +can be obtained from a solution of (2) by dividing the latter by $(x^2 - 1)^m$, or +in other words that if we write (2) +\begin{flalign*} +&&(1 - x^{2})&\dfrac{d^2 z}{dx^2} + 2(m-1)x\dfrac{dz}{dx} + 2mz = 0,\tag{2}\\ +&\text{and (4) as} & (1 -{}&x^2)\dfrac{d^2 z_1}{dx^2} - 2(m + 1) +x\dfrac{dz_1}{dx} = 0 +\qquad\qquad\hfill\phantom{and (4) as}\tag{4} +\end{flalign*} + +$z = z_1 (x^2 - 1)^m$; and the substitution of this value in (2) will give (4), and +the substitution of $z_1 = \dfrac{z}{(x^2 - 1)^m}$ in (4) will give (2). + +We have, then, two ways of obtaining (4) from (2); we may differentiate (2) +$2m$ times with respect to $x$, or we may replace $z$ in (2) by $z_1 (x^2 - 1)^m$. + +If we use the first method we have seen that Legendre's Equation (3) is +midway between (2) and (4). That is if we differentiate (2) $m$ times we get +(3) and if we then differentiate (3) $m$ times we get (4). Let us see if the +half-way equation in our second process is Legendre's Equation. +\begin{flalign*} +&\indent\text{If} &&z = y (x^2 - 1)^{\tfrac{m}{2}}&&\\ +&\text{and} &&y = z_1 (x^2 - 1)^{\tfrac{m}{2}}\hfill\phantom{and}&&\\ +& &&z = z_1 (x^2 - 1)^{m}.&& +\end{flalign*} + +So that if in (2) we replace $z$ by $y(x^2 - 1)^{\tfrac{m}{2}}$ and then repeat the operation +on the resulting equation we shall get (4). Making the first substitution we +find, +\[\tag{15} +(1 - x^2)\dfrac{d^2 y}{dx^2} - 2x\dfrac{dy}{dx} + \biggl[m(m + 1) - +\dfrac{m^2}{1 - x^2}\biggr] y = 0, +\] +not Legendre's Equation but a somewhat more general form. Of course its solution is +\[\tag{16} +y = A(x^2 - 1)^{\tfrac{m}{2}} + B(x^2 - 1)^{\tfrac{m}{2}}\int\dfrac{dx}{(x^2 - 1)^{m+1}}. +\] + +(2) and (4) are special forms of (5) and (6). Let us try the experiment of +substituting in (5) $z = y(1 - x^2)^{\tfrac{n}{2}}$ and in (6) + $z = \dfrac{y}{(1 - x^2)^{\tfrac{n}{2}}}$. +We find that both substitutions give the same equation +\[\tag{17} +(1 - x^{2})\dfrac{d^2 y}{dx^{2}} - 2x \dfrac{dy}{dx} + \biggl[m(m + 1) - +\dfrac{n^{2}}{1 - x^{2}}\biggr] y = 0. +\] +% -----File: 175.png + +The solution of (17) can be obtained from either (12) or (13) and is +\[\tag{18} +y = \dfrac{1}{(1 - x^2)^{\tfrac{n}{2}}} \biggl\lbrace A \dfrac{d^{m - n}(x^2 - 1)^m}{dx^{m-n}} + +B \dfrac{d^{m - n}} {dx^{m - n}} \biggl[(x^2 - 1)^{m} \int\dfrac{dx}{(x^2 - 1)^{m + 1}}\biggr]\biggr\rbrace +\] +or +\[\tag{19} +y = (1 - x^2)^{\tfrac{n}{2}} \biggl\lbrace A_1 \dfrac{d^{m+n} (x^2 - 1)^m}{dx^{m + n}} + B_1 +\dfrac{d^{m+n}}{dx^{m + n}}\biggl[(x^2 -1)^{m} \int\dfrac{dx}{(x^2 -1)^{m+1}}\biggr]\biggr\rbrace +\] +which of course must be equivalent. + +\mypara{86.} In addition to the value of $P_{m}(x)$ given in (1) Art.~83 there is another +important derivative form which we shall proceed to obtain. It is +\[\tag{1} +P_m (\cos \theta) = \dfrac{(-1)^m}{m!} r^{m+1} D_x^m \biggl(\dfrac{1}{r}\biggr). +\] + +We have seen in Art.~75 that $\dfrac{1}{r}\dfrac{1}{\sqrt{1 - 2\dfrac{r_1}{r}\cos\theta + +\dfrac{r_1^2}{r^2}}}$ can be developed into a convergent series if $r_1 < r$ and that the +$(m + 1)$st term of that series is $\dfrac{P_m (\cos\theta) r_1^m} {r^{m+1}}$. +Let us obtain this term by Taylor's Theorem. + +\begin{align*} +\dfrac{1}{r}\dfrac{1}{\sqrt{1-2\dfrac{r_1}{r} \cos\theta + \dfrac{r_1^2}{r^2}}} +&= \dfrac{1}{\sqrt{r^2-2r_1 r\cos\theta +r_1^2}} = \dfrac{1}{\sqrt{x^2+y^2+z^2-2xr_1+r_1^2}}\\ +&=\dfrac{1}{\sqrt{(x-r_1)^2+y^2+z^2}} +\end{align*} + +Regarding this as a function of $(x - r_1)$ and developing according to powers +of $r_1$ by Taylor's Theorem we get as the $(m + 1)$st term \\ +\begin{flalign*} +&&\dfrac{(-1)^m}{m!} r_1^m D_x^m&\biggl[\dfrac{1}{\sqrt{x^2 + y^2 + z^2}} +\biggr]\quad\text{or}\quad \dfrac{(-1)^m}{m!}{ r_1^{m}} D_x^{m} \biggl(\dfrac{1}{r}\biggr).&& \\[1ex] +&\text{Hence}&&\dfrac{P_m (\cos \theta)}{r^{m + 1}} = \dfrac{(-1)^m}{m!} +D_x^m \biggl(\dfrac{1}{r}\biggr).&&\hfil\phantom{Hence} +\end{flalign*} + +\mypara{87.} We have now obtained four different forms for our \emph{zonal harmonic}, +a polynomial in $x$, an expression involving cosines of multiples of $\theta$, a form +involving an ordinary $m$th derivative with respect to $x$, and a form involving +a partial $m$th derivative with respect to $x$. We shall now get a form due +to Laplace, involving a definite integral. + +\[\tag{1} +\int\limits_0^\pi +\dfrac{d\phi}{a-b\cos\phi} = \dfrac{\pi}{(a^2-b^2)^{\frac{1}{2}}} +\] +if $a^2 > b^2$ [v.\ Int.\ Cal.\ page 68]. +% -----File: 176.png + +$\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}$ can be expressed in the form $\dfrac{1}{(a^2-b^2)^{\frac{1}{2}}}$ by taking $a = 1 - zx$ +and $b = z \sqrt{ x^2 - 1}$ and no matter what value $x$ may have $z$ can be taken so small +that $a^2$ will be greater than $b^2$. Then by (1) +\begin{align*} +% recast to fit line +\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}&= \frac{1}{\pi} \int\limits_0^{\pi} \frac{d\phi}{1-zx-z\sqrt{ x^2 -1}.\cos \phi} \\ +&= \frac{1}{\pi} \int\limits_0^{\pi} \frac{d\phi}{1-z(x+\sqrt{ x^2 -1}.\cos \phi)}\\ +&= \frac{1}{\pi} \int\limits_0^{\pi} [1+(x+ \sqrt{ x^2 -1}.\cos \phi)z+(x+ \sqrt{ x^2 -1}.\cos \phi)^2 z^2\\ +&\qquad\qquad+ (x+ \sqrt{ x^2 -1}.\cos \phi)^3 z^3 + \cdots ]d\phi +\end{align*} +if $z$ is taken so small that the modulus of $z(x + \sqrt{ x^2 -1}.\cos \phi)$ is less than 1. But +by Art.~77 (2) $P_m(x)$ is the coefficient of $z^m$ in the development of $\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}$, +\begin{flalign*} +&\text{hence} & & P_m(x)=\frac{1}{\pi} \int\limits_0^{\pi} [x+ \sqrt{ x^2 -1}.\cos\phi]^m d\phi. \tag{2}&&\hspace{1em} +\end{flalign*} +By replacing $\phi$ by $\pi - \phi$ in (2) we get +\[ +P_m(x)=\frac{1}{\pi} \int\limits_0^{\pi} [x- \sqrt{ x^2 -1}.\cos \phi]^m d\phi. \tag{3} +\] + +$\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}} = \dfrac{1}{z}\dfrac{1}{\Big(1-2x \dfrac{1}{z }+ \dfrac{1}{z^2}\Big)^{\frac{1}{2} }}$ and if mod $\dfrac{1}{z}<1$ or in other words if +mod $z>1$\;\;$\dfrac{1}{\Big(1-2x \dfrac{1}{z} + \dfrac{1}{z^2} \Big)^{\frac{1}{2}}}$ can be developed into a convergent series involving +powers of $\dfrac{1}{z}$, and the coefficient of $\Big(\dfrac{1}{z} \Big)^m$ will be $P_m(x)$; but this will be +the coefficient of $z^{-m-1}$ in the development of $\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}$ according to +descending powers of $z$, mod $z$ being greater than 1. + +If now we let $a = zx - 1$ and $b=z\sqrt{x^2 -1}$, $a^2 - b^2 = 1 - 2xz +z^2$ and +$z$ may be taken so great that $a^2 - b^2 > 0$. Then by (1) + +\begin{align*} +\frac{1}{(1-2xz+z^2)^{\frac{1}{2}}} &= \frac{1}{\pi} \int\limits_0^{\pi}\frac{ d\phi}{zx-1-z \sqrt{ x^2 -1}.\cos \phi}&\\ + ={}&\frac{1}{\pi}\int\limits_0^{\pi}\frac{ d\phi}{z(x-\sqrt{x^2-1}.\cos \phi) \Big[1- \dfrac{1}{ z(x-\sqrt{x^2-1}.\cos \phi)} \Big]}&\\ + ={}&\frac{1}{\pi}\int\limits_0^{\pi}\frac{ 1}{(x- \sqrt{ x^2 -1}.\cos \phi) }\left [z^{-1}+ \frac{1}{(x- \sqrt{ x^2 -1}.\cos \phi) } z^{-2} \right .&\\ + & \left . \qquad \qquad \qquad \qquad + \frac{1}{(x- \sqrt{ x^2 -1}.\cos \phi)^2} z^{-3} + \cdots \right ] d\phi & +\end{align*} +% -----File: 177.png +and the coefficient of $z^{-m-1}$ is +$\displaystyle\frac{1}{\pi}\int\limits_0^\pi\frac{d\phi}{[x-\sqrt{x^2-1}.\cos\phi]^{m+1}}$. +\begin{flalign*} +&\text{Hence} & P_m(x)=\frac{1}{\pi}\int\limits_0^\pi\frac{d\phi}{[x-\sqrt{x^2-1}.\cos\phi]^{m+1}}. & \tag{4}& +\end{flalign*} + +Replace $\phi$ by $\pi-\phi$ and we get +\[ +P_m(x) =\frac{1}{\pi}\int\limits_0^\pi\frac{d\phi}{[x+\sqrt{x^2-1}.\cos\phi]^{m+1}}. \tag{5} +\] + +\mypara{88.} In the problems in which we have already used \emph{Zonal Harmonics} +(v.\ Arts.~78--81) we have been able to start with the value of the Potential +Function at any point on the axis of $X$, and it has been necessary to develop +the expression for $V$ on that axis in terms of ascending or descending powers +of $x$. If, however, we start with the value of $V$ in terms of $\theta$ for some given +value of $r$, that is on the surface of some sphere, we must develop the function +of $\theta$ in terms of \emph{zonal harmonics} of $\cos\theta$ (v.\ Art.~10), and our problem becomes +the following:---To develop a given function of $\cos\theta$ in terms of zonal harmonics +of $\cos\theta$, or to develop a given function of $x$ in terms of the functions +$P_m(x)$, $x$ lying between 1 and $-1$. + +The problem resembles closely that of developing in a Fourier's series, +which we have already considered at such length. +\begin{flalign*} +&\indent\text{Let} & f(x) = A_0P_0(x) + A_1P_1(x) + A_2P_2(x) + A_3P_3(x) + \cdots & \tag{1}& +\end{flalign*} +for all values of $x$ from $-1$ to 1 and let it be required to determine the +coefficients. + +If $f(x)$ is single-valued and has only finite discontinuities between $x=-1$ + and $x=1$ we may proceed as in Art.~19. + +Let us take $n+1$ terms of (1) and attempt to determine the coefficients. +Take $n+1$ values of $x$ at equal intervals $\Delta x$ between $x=-1$ and $x=1$ + so that $(n+2)\Delta x=2$; $f(-1+\Delta x)$, $f(-1+2\Delta x)$, $f(-1+3\Delta x)$, $\cdots$ +$f[-1+(n+1)\Delta x]$ will be the corresponding values of $f(x)$. Substitute +these values in (1) and we have +\begin{equation*}\left. +\begin{aligned} +f&(-1+\Delta x) = A_0 P_0(-1+\Delta x)+A_1 P_1(-1+\Delta x)\\ +&\hspace{22ex}+ A_2 P_2(-1+\Delta x)+ \cdots + A_n P_n(-1+\Delta x)\\ +f&(-1+2\Delta x) = A_0 P_0(-1+2\Delta x)+A_1 P_1(-1+2\Delta x)\\ +&\hspace{20ex}+ A_2 P_2(-1+2\Delta x)+ \cdots + A_n P_n(-1+2\Delta x)\\ +&\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots +\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\\ +f&(1-\Delta x) = A_0 P_0(1-\Delta x) + A_1 P_1(1-\Delta x) + A_2 P_2(1-\Delta x) + \cdots\\ +&\hspace{48ex}+ A_n P_n(1-\Delta x), +\end{aligned}\right\}\tag{2} +\end{equation*} +that is, $n+1$ equations from which in theory the $n+1$ coefficients +$A_0$, $A_1$, $\cdots$ $A_n$ can be determined. + +% -----File: 178.png +Following the analogy of Art.~24 let us multiply the first equation by +$P_{m}(- 1 + \Delta x).\Delta x$, the second by $P_{m}(-1 + 2\Delta x).\Delta x$, the third by +$P_{m}(- 1 +3\Delta x).\Delta x$, \&c., and add the equations. The first member of the +resulting equation is +\[ +\sum_{k=1}^{k=n+1} f(-1+k\Delta x)P_m(-1+k\Delta x).\Delta x, \tag{3} +\] +and the coefficient of any $A$ as $A_l$ in the second member is +\[ + \sum_{k=1}^{k=n+1} P_m(-1+k\Delta x)P_l(-1+k\Delta x).\Delta x. \tag{4} +\] +If now $n$ is indefinitely increased (3) approaches as its limiting value +\begin{flalign*} +&&&\int\limits_{-1}^1 f(x) P_m(x) dx \tag{5}\\ +&\text{and (4) approaches}&& \int\limits_{-1}^1 P_m(x) P_l(x)dx. && \phantom{and (4) approaches}\tag{6} +\end{flalign*}\\ +\indent We have now to find the value of the integral (6) or as we shall write +it for the sake of greater convenience +\[ + \int\limits_{-1}^1 P_m(x) P_n(x) dx. + \] + +\mypara{89.} $\displaystyle\int\limits_{-1}^1 P_m(x) P_n(x) dx = \frac{1}{2^{m+n} m!n!} \int\limits_{-1}^1 \frac{d^m(x^2-1)^m}{dx^m}.\frac{d^n(x^2-1)^n}{dx^n}dx$ +by (1) Art.~83. +\begin{flalign*} +\int\limits_{-1}^1 \frac{d^m(x^2-1)^m}{dx^m}.\frac{d^n(x^2-1)^n}{dx^n}dx &= \bigg[\frac{d^m(x^2-1)^m}{dx^m}.\frac{d^{n-1}(x^2-1)^n}{dx^{n-1}} \mathop{\bigg]}_{-1}^1 \\ + &-\int\limits_{-1}^1 \frac{d^{m+1}(x^2-1)^m}{dx^{m+1}}.\frac{d^{n-1}(x^2-1)^n}{dx^{n-1}}dx \tag{1} +\end{flalign*} +by \textit{integration by parts}. + +Now if $z = X(x^2 - 1)^n$ +\begin{align*} +% recast to fit line +\frac{dz}{dx} &= 2nxX(x^2-1)^{n-1} + (x^2-1)^n \frac{dX}{dx} \\ + &= (x^2-1)^{n-1} \left [2nxX+(x^2-1)\frac{dX}{dx} \right ]. \tag{2} +\end{align*} + +Hence the $p$th derivative with respect to $x$ of any function of $x$ containing + $(x^2- 1)^n$ as a factor will contain $(x^2- 1)^{n-p}$ as a factor if $p < n$. + +% -----File: 179.png +$\dfrac{d^{n-1} (x^2 -1)^n }{ dx^{n-1}}$, then, contains $(x^2 - 1)$ as a factor and is zero when $x = 1$ + and when $x = -1$, so that (1) reduces to +\[ +\int\limits^1_{-1}\dfrac{d^m (x^2 -1)^m }{ dx^m}.\dfrac{d^n (x^2 -1)^n}{dx^n} dx = - \int\limits^1_{-1} \dfrac{d^{m+1} (x^2 -1)^m }{dx^{m+1}}.\dfrac{ d^{n-1} (x^2 -1)^n }{ dx^{n-1}} dx. +\] +It follows that +\begin{align*} +% recast to fit line +\int\limits^1_{-1} \dfrac{d^m (x^2 -1)^m}{dx^m}.\dfrac{d^n (x^2 -1)^n }{ dx^n} &dx \\ +={}&(-1)^p \int \limits^1_{-1} \dfrac{d^{m+p} (x^2 -1)^m }{ dx^{m+p}}.\dfrac{ d^{n-p} (x^2 -1)^n }{ dx^{n-p} }dx\\ +={}&(-1)^p \int \limits^1_{-1} \dfrac{d^{m-p} (x^2 -1)^m }{dx^{m-p}}.\dfrac{d^{n+p}(x^2 -1)^n}{dx^{n+p}} dx. \tag{3} +\end{align*} + +If $m < n$ we get from (3) +\begin{align*} +\int\limits^1_{-1} \dfrac{d^m (x^2 -1)^m }{ dx^m}.\dfrac{d^n (x^2 -1)^n }{dx^n} dx &= (-1)^m \int\limits^1_{-1} \dfrac{d^{2m} (x^2 -1)^m }{ dx^{2m} }.\dfrac{d^{n-m} (x^2 -1)^n }{ dx^{n-m}} dx\\ + &= (-1)^m (2m)! \bigg[ \dfrac{d^{n-m-1} (x^2 -1)^n}{ dx^{n-m-1} }\mathop{\bigg]}^1_{-1} = 0,\\[-6ex] +\end{align*} +\begin{flalign*} +&\text{since}&& \dfrac{d^{2m} (x^2 -1)^m }{ dx^{2m}} = (2m)!.&\phantom{since} +\end{flalign*} + +If $m > n$ +\begin{align*} +\int \limits^1_{-1} \dfrac{d^m (x^2 -1)^m }{dx^m }.\dfrac{d^n (x^2 -1)^n }{ dx^n}dx &= (-1)^n \int \limits^1_{-1} \dfrac{d^{m-n} (x^2 -1)^m }{ dx^{m-n}}.\dfrac{d^{2n} (x^2 -1)^n }{ dx^{2n}} dx\\ + &= (-1)^n (2n)! \bigg[ \dfrac{d^{m-n-1} (x^2 -1)^m }{dx^{m-n-1}} \mathop{\bigg]}^1_{-1} = 0. +\end{align*} + +If, then, $m$ is not equal to $n$ +\[ +\int\limits^1_{-1} P_m (x) P_n (x) dx = 0. \tag{4} +\] +If $m = n$ we have to find $\displaystyle \int\limits^1_{-1} [P_m (x)]^2 dx$. +\[ +\int\limits^1_{-1} [P_m (x)]^2 dx = \dfrac{1}{2^{2m} (m!)^2} \int\limits^1_{-1} \dfrac{ d^m (x^2 -1)^m }{ dx^m}.\dfrac{ d^m (x^2 -1)^m }{ dx^m} dx. +\] +\begin{flalign*} +&&\int\limits^1_{-1} \dfrac{d^m (x^2 -1)^m }{dx^m }.\dfrac{d^m (x^2 -1)^m }{ dx^m} dx &= (-1)^m \int\limits^1_{-1} \dfrac{ d^{2m} (x^2 -1)^m }{ dx^{2m}}.(x^2 - 1)^m dx&&\\ +&\multispan{2}{by (3),\hfill}&= (-1)^m (2m)! \int\limits^1_{-1} (x^2 - 1)^m dx.&& +\end{flalign*} + +% -----File: 180.png +\begin{align*} +\int\limits_{-1}^1(x^2 -1)^m dx &= \int\limits_{-1}^1 (x-1)^m (x+1)^m dx = -\frac{m}{m+1} \int\limits_{-1}^1 (x-1)^{m-1} (x+1)^{m+1} dx\\ +&= (-1)^m \frac{m!}{(m+1)(m+2) \cdots 2m} \int\limits_{-1}^1 (x+1)^{2m} dx\\ +&= (-1)^m \frac{2^{2m+1} m!}{(m+1)(m+2) \cdots (2m+1)}. +\end{align*} +\begin{flalign*} +&\text{Hence} &\int\limits_{-1}^1 [P_m (x)]^2 dx &= \frac{1}{2^{2m} (m!)^2}\frac{ (-1)^m (2m)!(-1)^m m!2^{2m+1}}{(m+1)(m+2) \cdots (2m+1)} &&\phantom{Hence}\\ +&\text{or} && \int\limits_{-1}^1 [P_m(x)]^2 dx = \frac{2}{2m+1}. \tag{5}&& +\end{flalign*} + +\mypara{90.} The solution of the problem in Art.~88 is now readily obtained, and +we have +\begin{flalign*} +&&f(x) = &A_0P_0(x) + A_1P_1(x) + A_2P_2(x) + \cdots \tag{1} &&\\ +&\text{where}&&A_m = \frac{2m+1}{2 }\int\limits_{-1}^1 f(x)P_m(x)dx. \tag{2} &&\hspace{1em} +\end{flalign*} + +The function and the series are equal for all values of $x$ from $x = -1$ to + $x = 1$, and $f(x)$ is subject to no conditions save those which would enable us +to develop it in a Fourier's Series. [v.\ Chapter III.] + +Of course (1) can be written +\begin{flalign*} +&&f(\cos \theta) &= A_0P_0(\cos \theta) +A_1P_1(\cos \theta) + A_2P_2(\cos \theta) + \cdots &&\\ +&\text{where} && A_m = \frac{2m+1}{2} \int\limits_{-1}^1 f(\cos \theta)P_m(\cos \theta)d(\cos \theta) && \phantom{where} +\intertext{or if $f(\cos \theta) = F(\theta)$} +&&F(\theta) = &A_0P_0(\cos \theta) + A_1P_1(\cos \theta) + A_2P_2(\cos \theta) + \cdots \tag{3} \\ +&\text{where} && A_m = \frac{2m+1}{2} \int\limits_{0}^{\pi} F(\theta)P_m(\cos \theta)\sin\theta.d\theta && \tag{4} +\end{flalign*} +and the development holds good from $ \theta = 0$ to $\theta = \pi$. + +If $f(x)$ is an even function, that is, if $f(-x) =f(x)$ (1) and (2) can be +somewhat simplified. For in that case it can be easily shown (v.\ Art.~77) that +\begin{flalign*} +&&\int\limits_{-1}^1 f(x)&P_{2k}(x)dx = 2 \int\limits_0^1 f(x)P_{2k}(x)dx,\\ +% -----File: 181.png +&\text{and that}& &\int\limits_{-1}^1 f(x)P_{2k+1} (x)dx = 0; && \phantom{and that} +\intertext{so that if $f(-x) =f(x)$} +&&f(x) = A_0P_0(x) &+A_2P_2(x) + A_4P_4(x) + A_6P_6(x) + \cdots \tag{5} \\ +&\text{where}& A_{2k} &= (4k+1)\int\limits_0^1 f(x)P_{2k}(x)dx. && \tag{6} +\end{flalign*} + +If $f(x)$ is an odd function, that is, if $f(-x) = -f(x)$ it can be shown in +like manner that +\begin{flalign*} +&&f(x) = A_1P_1&(x) +A_3P_3(x) + A_5P_5(x) + A_7P_7(x) + \cdots &&\tag{7}\\ +&\text{where}&& A_{2k+1} = (4k+3) \int\limits_0^1 f(x)P_{2k+1}(x)dx. && \phantom{where} \tag{8} +\end{flalign*} + +If it is only necessary that the development should hold for $0 < x < 1$ any +function may be expressed in form (5) or (7) at pleasure. + +\mypara{91.} We can establish the fact that $\displaystyle\int\limits_{-1}^1 P_m(x)P_n(x)dx = 0$ by a more general +method than that used in Art.~89. + +Let $X_m$ be any solution of Legendre's Equation +\begin{flalign*} +&\hspace{6em}&&\frac{d}{dx} \left [ (1-x^2)\frac{ dz}{dx} \right ] + m(m+1)z = 0 && \tag*{[v.\ (1) Art.~16].} +\intertext{which with its first derivative with respect to $x$ is finite, continuous, and +single-valued for values of $x$ between $-1$ and $1$, $-1$ and $1$ being included.} +&\text{\indent Then}&&\frac{d}{dx} \left [ (1-x^2) \frac{dX_m }{ dx} \right ] + m(m+1)X_m = 0 \tag{1}&&\\ +&\text{and}&&\frac{d}{dx} \left [ (1-x^2) \frac{dX_n }{ dx} \right ] + n(n+1)X_n = 0 \tag{2}&&\phantom{and} +\end{flalign*} + +Multiply (1) by $X_n$ and (2) by $X_m$ and subtract and integrate and we get +\begin{multline*} +[m(m+1)-n(n+1)] \int\limits_{-1}^1 X_mX_n dx = \int\limits_{-1}^1 X_m \frac{d}{dx} \left [ (1-x^2) \frac{dX_n }{ dx} \right ] dx\\ +- \int\limits_{-1}^1 X_n \frac{d}{dx} \left [ (1-x^2)\frac{ dX_m }{ dx} \right ] dx. +\end{multline*} +% -----File: 182.png + +Integrate by parts, +\begin{multline*} +[m(m + 1) - n(n + 1)] \int\limits_{-1}^1 X_m X_n dx = \bigg[X_m(1-x^2)\frac{dX_n}{dx} - X_n(1-x^2)\frac{dX_m}{dx} \!\!\!\!\!\mathop{\bigg]}_{x=-1}^{x=1} \\ + -\int\limits_{-1}^1 (1-x^2)\frac{dX_n}{dx}\frac{ dX_m}{dx} dx + \int\limits_{-1}^1 (1-x^2)\frac{dX_m}{dx} \frac{dX_n}{dx} dx. \tag{3}\\[-5ex] +\end{multline*} +\begin{flalign*} +&\text{Whence}&&\int\limits_{-1}^1 X_m X_n dx = 0 && \phantom{Whence} \tag{4}\\[-5ex] +\end{flalign*} +unless $m = n$. + +(3) gives at once the important formula +\[ +\int\limits_x^1X_mX_ndx =\frac{ (1-x^2) \Big [X_n \dfrac{dX_m}{dx} - X_m\dfrac{dX_n}{dx} \Big ]}{m(m + 1) - n(n + 1) } \tag{5} \\ +\] +from which come as special cases +\[ +\int\limits_x^1P_m(x)P_n(x)dx = \frac{(1-x^2) \Big [P_n(x)\dfrac{dP_m(x)}{dx} - P_m(x)\dfrac{dP_n(x)}{dx} \Big ] }{ m(m + 1) - n(n + 1)} \tag{6} +\] +and since $P_0(x) = 1$ +\[ +\int\limits_x^1P_m(x)dx = \frac{(1-x^2)\dfrac{dP_m(x)}{dx}}{ m(m + 1)}, \tag{7}\\ +\] +unless $m = 0$. + +\EXAMPLE{S} + +1.\quad Show that +\begin{align*} +\int\limits_0^1 P_m(x)dx &= 0 \quad\text{ if $m$ is even and is not zero.}\\ +&=(-1)^{\frac{m-1}{2}}\frac{ 1}{m(m+1)}\frac{ 3.5.7.\cdots m }{ 2.4.6.\cdots (m-1) } \quad\text{if $m$ is odd. } +\end{align*} +v.\ Art.~91 (7) and Art.~77 (10).\\ + +2.\quad Show that +\begin{multline*} +\int\limits_0^1P_m(x)P_n(x)dx = 0 \quad \text{if $m$ and $n$ are both even or both odd.}\\ + = (-1)^{\frac{m+n+1}{2}} \frac{m!\;n!}{ 2^{m+n-1} (m-n)(m+n+1) \left (\dfrac{m}{2} ! \right )^2 \Big (\dfrac{n-1}{2} ! \Big )^2} + \end{multline*} + if $m$ is even and $n$ odd. v.\ Art.~91 (6) and Art.~77 (8), (9), and (10). cf.\ J.~W. Strutt +(Lord Rayleigh) Lond.\ Phil.\ Trans.\ 1870, page 579.\\ + +3.\quad Show that \quad $\displaystyle\int\limits_0^1[P_m(x)]^2dx = \dfrac{1}{2m+1}$ \quad v.\ Art.~89 (5). + +% -----File: 183.png +\mypara{92.} Formula (4) Art.~91 can be obtained directly from Laplace's Equation +by the aid of \emph{Green's Theorem} (v.\ Peirce's Newt.\ Pot.\ Func.\ \S~48). + +Take the special form of Green's Theorem, [(148) \S~48 Peirce's Newt.\ Pot.\ +Func.] +\[ +\iiint(U\nabla^2V - V\nabla^2U)dxdydz = \int (UD_n V - VD_n U)ds \tag{1} +\] +where $\nabla^2$ stands for $(D_x^2 + D_y^2 + D_2^2)$, $D_n$ is the partial derivative along the +external normal, and the left-hand member is the space-integral through the +space bounded by any closed surface, and the right-hand member is the surface +integral taken over the same surface. (v.\ Int.\ Cal.\ Chapter XIV.) + +If $U$ and $V$ are solutions of Laplace's Equation $\nabla^2 V = \nabla^2 U = 0$ and (1) +reduces to +\[ +\int(UD_n V - V D_n U) ds = 0. \tag{2} +\] + +Now $r^mX_m$ and $r^nX_n$ are solutions of Laplace's Equation if $x = \cos \theta$ + (v.~Art.~16). + +If the unit sphere is taken as the bounding surface and $U = r^mX_m$ and + $V = r^nX_n$ (1) and (2) will hold good. +\begin{gather*} +D_nU = D_r(r^mX_m) = mr^{m-1}X_m,\\ +D_nV = nr^{n-1}X_n,\\ +ds = \sin \theta.d\theta d\phi, +\end{gather*} +and (2) becomes $\displaystyle \int\limits^{2\pi}_0 d\phi \int\limits^{\pi}_0 (n X_m X_n - m X_m X_n) \sin \theta.d \theta = 0$ +\begin{flalign*} +&\text{or}& 2\pi(n - m) \int\limits^{\pi}_0 X_m X_n \sin \theta.d \theta &= 0. & \tag{3} +\end{flalign*} +Since $x = \cos\theta$, $\sin\theta.d\theta = - dx$ and (3) reduces to +\[ +\int\limits^1_{-1} X_m X_n dx = 0\tag{4}\footnotemark +\] +unless $m = n$. + +\footnotetext{It should be noted that this proof is no more general than that of the last article, for, in +order that Green's Theorem should apply to $r^mX_m$ this function and its first derivatives must +be finite continuous and single-valued within and on the surface of the unit sphere. (v.\ Peirce, +Newt.\ Pot.\ Func.\ \S~48.) } + +\mypara{93.} We can now solve completely the problem of Art.~10 which was in +that article carried to the point where it was only necessary to develop a +certain function of $\theta$ in the form +\[ +A_0P_0(\cos\theta) + A_1P_1(\cos\theta) + A_2P_2(\cos\theta) + \cdots\\[-2ex] +\] +% -----File: 184.png +\begin{flalign*} +&\text{given that}& f(\theta) = 1 \quad\text{from}\quad\theta = 0\quad\text{to}\quad\theta = \dfrac{\pi}{2} &&\phantom{given that}\\ +&\text{and} & f(\theta) = 0 \quad\text{from}\quad\theta = \dfrac{\pi}{2}\quad\text{to}\quad\theta = \pi. && +\end{flalign*} + +This amounts to the same thing as developing $F(x)$ into the series +\[ +F(x) = A_0 P_0(x) + A_1 P_1(x) + A_2 P_2(x) + A_3 P_3(x) + \cdots\\[-2ex] +\] +\begin{flalign*} +&\text{where} &&F(x) = 0\quad\text{from}\quad x = -1\quad\text{to}\quad x = 0&&\phantom{where}\\ +&\text{and} &&F(x) = 1\quad\text{from}\quad x = 0\phantom{-}\quad\text{to}\quad x = 1.&& +\end{flalign*} +By Art.~90 (1) and (2) +\[ +A_0 = \dfrac{1}{2} \int\limits^1_0 P_0(x)dx = \dfrac{1}{2} \int\limits^1_0 dx = \dfrac{1}{2} , +\] +\begin{flalign*} +&\text{and any coefficient}&&A_m = \dfrac{(2m + 1)}{2} \int\limits^1_0 P_m(x)dx. &&\phantom{and any coefficient} +\end{flalign*} +By Art.~91, Ex.~1 +\[ +\int\limits^1_0 P_m(x)dx = 0\quad \text{ if $m$ is even}\rule{14.7em}{0em}\\[-5ex] +\] +\begin{flalign*} +&& &= (-1)^{\tfrac{m-1}{2}} \dfrac{1}{m(m+1)}\dfrac{ 3.5.7.\cdots m}{2.4.6.\cdots (m-1)} \quad \text{if $m$ is odd.}&\\ +&\text{\indent Hence}& A_m &= 0\quad \text{ if $m$ is even}&\\ +&& &= (-1)^{\tfrac{m-1}{2}} \dfrac{2m+1}{2m+2}.\dfrac{ 1.3.5.\cdots(m-2)}{2.4.6.\cdots (m-1)}\quad \text{ if $m$ is odd}.&\\ +&\text{\indent Then}& F(x) &= \dfrac{1}{2} + \dfrac{3}{4} P_1 (x) - \dfrac{7}{8}.\dfrac{1}{2} P_3(x) + \dfrac{11}{12}.\dfrac{1.3}{2.4} P_5(x) -\cdots &\tag{1} +\end{flalign*} +and \quad $u = \dfrac{1}{2} + \dfrac{3}{4} r P_1(\cos\theta) - \dfrac{7}{8}.\dfrac{1}{2} r^3 P_3 (\cos\theta) + \dfrac{11}{12}.\dfrac{1.3}{2.4} r^5 P_5(\cos \theta) + \cdots$\hfill(2)\\[1ex] +for any point within the sphere. + +\mypara{94.} If in a problem on the Potential Function the value of $V$ is given at +every point of a spherical surface and has circular symmetry\footnote{See note on page \pageref{notep12}.} about a diameter +of that surface the value of $V$ at any point in space can be obtained. + +We have to solve Laplace's Equation in the form +\[ +r D_r^2(rV) + \dfrac{1}{\sin\theta} D_\theta (\sin\theta D_\theta V) = 0 \tag{1} +\] +% -----File: 185.png +subject to the conditions +\begin{alignat*}{3} +&V=f(\theta)\quad && \text{when}\quad &&r=a\\ +&V=0 &&\quad \text{``} &&r=\infty. +\end{alignat*} +\begin{flalign*} +&\text{We have}& f(\theta) &= A_0 P_0(\cos\theta) + A_1 P_1(\cos\theta) + A_2 P_2(\cos\theta) +\cdots&&\\ +&\text{where}&& A_m = \frac{(2m+1)}{2} \int\limits_0^{\pi} f(\theta) P_m(\cos\theta) \sin\theta.d\theta. && \tag*{v.\ Art.~90 (4).} +\end{flalign*} +Hence +\[ +V=A_0 + A_1 \left (\frac{r}{a} \right )P_1(\cos\theta) + A_2 \left (\frac{r}{a} \right )^2 P_2(\cos\theta) + A_3 \left (\frac{r}{a} \right )^3 P_3(\cos\theta) + \cdots (2) +\] +is the required solution for a point within the sphere, and +\[ +V=A_0\left (\frac{a}{r} \right ) + A_1 \left (\frac{a}{r} \right ) ^2P_1(\cos\theta) + A_2 \left (\frac{a}{r} \right ) ^3 P_2(\cos\theta) + A_3 \left (\frac{a}{r} \right ) ^4 P_3(\cos\theta) + \cdots \tag{3} +\] +is the required solution for an external point. + +\EXAMPLE{S} + +1.\quad If on the surface of a sphere of radius $c$ $V$ is constant and equal to a +show that $V=a$ for any point within the sphere and $V=\dfrac{ac}{r}$ for any +external point.\\ + +2.\quad Two equal thin hemispherical shells of radius $c$ placed together to form +a spherical surface are separated by a thin non-conducting layer. Charges of +statical electricity are placed on the two hemispheres one of which is then +found to be at potential $a$ and the other at potential $b$. Find the value of the +potential function at any point. +\begin{multline*} +V= \frac{a+b}{2} + (b-a) \left [\frac{3}{4} \frac{r}{c} P_1(\cos\theta)- \frac{7}{8}.\frac{ 1}{2}\frac{ r^3}{c^3} P_3(\cos\theta) \right.\\ + \left . + \frac{11}{12}.\frac{1.3}{2.4}\frac{ r^5}{c^5} P_5(\cos\theta)- \cdots \right ] +\end{multline*} +for an internal point +\begin{multline*} +V= \frac{a+b}{2}.\frac{c}{r} + (b-a) \left [\frac{3}{4} \frac{c^2}{r^2} P_1(\cos\theta)- \frac{7}{8}.\frac{ 1}{2}\frac{ c^4}{r^4} P_3(\cos\theta) \right.\\ + \left . + \frac{11}{12}.\frac{1.3}{2.4}\frac{ c^6}{r^6} P_5(\cos\theta)- \cdots \right ] +\end{multline*} +for an external point.\\ + +% -----File: 186.png +3.\quad If $V_1 =f(\cos\theta)$ when $r = a$ and $V_1 = 0$ when $r = b$ show that for + $a < r < b$ +\begin{flalign*} +&&V_1 = \sum_{m=0}^{m=\infty} &A_m \left ( \frac{b^{m+1}}{r^{m+1}} - \frac{r^m}{b^m} \right ) \left (\frac{b^{m+1}}{a^{m+1}} -\frac{ a^m}{b^m} \right )^{-1} P_m(\cos\theta)\\ +&\text{where} &&A_m = \frac{2m+1}{2} \int\limits_{-1}^1 f(x)P_m(x)dx.&& \phantom{where} +\end{flalign*} + +4.\quad If $V_2 =F(\cos\theta)$ when $r = b$ and $V_2 = 0$ when $r = a$ show that for + $a < r < b$ +\begin{flalign*} +&&V_2 = \sum_{m=0}^{m=\infty} &B_m \left ( \frac{r^{m}}{a^{m}} - \frac{a^{m+1}}{r^{m+1}} \right ) \left (\frac{b^{m}}{a^{m}} -\frac{ a^{m+1}}{b^{m+1}} \right )^{-1} P_m(\cos\theta)\\ +&\text{where} &&B_m = \frac{2m+1}{2} \int\limits_{-1}^1 F(x)P_m(x)dx.&& \phantom{where} +\end{flalign*} + +5.\quad If the value of the potential function is given arbitrarily on the surfaces +of a spherical shell but has circular symmetry\footnote{See note on page 12} about a diameter $V= V_1 + V_2$ + (v.\ Exs.~3 and 4).\\ + +6.\quad Two concentric hollow spherical conductors are insulated and charged. +The inner one of radius $a$ is at potential $p$, and the outer one of radius $b$ is at +potential $q$. Find $V$ for any point in space. +\begin{align*} +V&=p \quad\text{if}\quad r < a,\\ +V&= \frac{pa}{b-a}\bigg ( \frac{b}{r}-1 \bigg ) +\frac{ qb}{b-a} \bigg (1- \frac{a}{r} \bigg ) \quad\text{if} \quad a<r<b,\\ +V&=\frac{qb}{r} \quad\text{if} \quad r>b. +\end{align*} + +7.\quad If $V=0$ on the base of a hemisphere and $V=f(\cos\theta)$ on the convex +surface, show that for a point within the hemisphere +\begin{flalign*} +&&V={}&\sum_{k=0}^{k=\infty} A_{2k+1}\bigg (\frac{r}{a} \bigg )^{2k+1} P_{2k+1} (\cos\theta)\\ +&\text{where}& A_{2k+1} &= (4k + 3) \int\limits_0^1 f(x)P_{2k+1}(x)dx \tag*{ [v.\ Art.~90 (8)].}&& +\end{flalign*} + +8.\quad If the convex surface of a solid hemisphere of radius $a$ is kept at the +constant temperature unity and the base at the constant temperature zero +show that after the permanent state of temperatures is set up the temperature +of any internal point is +\[ +u = \frac{3}{2} \frac{ r}{a} P_1(\cos\theta) - \frac{ 7}{4}.\frac{1}{2} \frac{ r^3}{a^3} P_3(\cos\theta) + \frac{11}{6}.\frac{ 1.3}{2.4 } \frac{r^5}{a^5} P_5(\cos\theta) - \cdots +\] + +% -----File: 187.png +9.\quad A sphere of radius $a$ and with blackened surface is exposed to the direct +rays of the sun in air at the temperature zero. Find the \textit{stationary temperature} +of any internal point.\\ + +\textit{Suggestion:}\quad $D_ru + hu - Mf(\theta) = 0$ \quad when \quad $r = a$. +\begin{flalign*} +&\text{Let} &u= \sum A_m\frac{ r^m }{a^m} P_m(\cos\theta), \quad \text{and}\quad f(\theta) = \sum B_m P_m(\cos\theta).&& +\end{flalign*} +Then we have +\begin{flalign*} +&&\sum m \frac{A_m}{a} P_m(\cos\theta) + h \sum &A_m P_m (\cos\theta) - M \sum B_m P_m (\cos\theta) = 0,\\ +&\multispan{2}{whence\hfill}&A_m = \frac{MB_m }{ h+ \dfrac{m}{a}}. &&\phantom{whence} +\end{flalign*} + +Here $f(\theta) = \cos\theta$ if $0 < \theta <\dfrac{\pi}{2}$ and $f(\theta) = 0$ if $\dfrac{\pi}{2} < \theta < \pi$. +\begin{multline*} +f(\theta) = \frac{1}{4} + \frac{1}{2} P_1(\cos\theta) + \frac{5}{16} P_2(\cos\theta) - \frac{3}{32} P_4(\cos\theta) + \cdots\\ ++ (-1)^{k+1}\frac{ (4k+1)(2k)!}{(4k+4)(2k-1)2^{2k} (k!)^2} P_{2k} (\cos \theta) + \cdots +\end{multline*} +v.\ Art.~91 Exs.~(2) and (3). cf. J.~W. Strutt (Lord Rayleigh), Lond.\ Phil.\ +Trans.\ vol.\ 160, page 587. + +\mypara{95.} The formulas of Art.~90 enable us to develop a given function of $x$ in +terms of \textit{Zonal Surface Harmonics}, the development holding true for values of +$x$ between $ -1$ and $+1$. If, however, we can show by outside considerations +that a given function of $x$ can be expressed in Zonal Surface Harmonics, the +development holding true for all values of $ x$, the formulas of Art.~90 will give +us the development in question. + +For example if $n$ is a positive integer $x^n$ can be expressed in terms of Zonal +Surface Harmonics no matter what the value of $x$, and no Harmonic of higher +order than $n$ will enter. For the formulas giving the values of $P_1(x)$, $P_2(x), \cdots +P_n(x)$ (v.\ Art.~77) may be regarded as $n$ algebraic equations of the first degree +in terms of $x$, $x^2$, $x^3, \cdots x^n$ and $P_1(x)$, $P_2(x), \cdots P_n(x)$. + +From these equations the $n -1$ quantities $x$, $x^2$, $x^3, .\cdots x^{n-1}$, can be eliminated, +and there will result an equation of the first degree in $x^n$ and $P_1(x)$, +$P_2(x), \cdots P_n(x)$, which will enable us to express $x^n$ in the form +\[ +A_0 +A_1 P_1(x) + A_2 P_2(x) + \cdots + A_n P_n(x), +\] +no matter what the value of $x$, and we shall have the same formula when + $-1 < x < 1$ as when $x > 1$ or $x < -1$. + +% -----File: 188.png +Let us obtain this development. By Art.~90 (1) and (2) +\begin{flalign*} +&&x^n &= A_0 P_0(x) + A_1 P_1(x) + A_2 P_2(x) + \cdots\tag{1}\\ +&\text{where}& &\qquad A_m = \dfrac{2m + 1}{2}\int\limits^1_{-1} x^n P_m(x)dx. &\phantom{where}\tag{2}\\ +&\text{Then}\hspace{2em}& A_m &= \dfrac{2m + 1}{2}\dfrac{ 1}{2^m m!} \int\limits^1_{-1} x^n \dfrac{d^m(x^2-1)^m }{dx^m} dx & \tag*{by (1) Art.~83.} +\end{flalign*} + +By \emph{integration by parts} we get +\begin{alignat*}{2} +\int\limits^1_{-1} x^n \dfrac{d^m(x^2-1)^m}{dx^m} dx &\multispan{3}{$\displaystyle{}= n(n-1)(n-2) \cdots(n-m+1) \int\limits^1_{-1} x^{n-m} (1-x^2)^m dx,$} \\[-2ex] + &&& \text{if}\quad m < n + 1,\tag{3}\\ +&=0 \quad &&\text{if}\quad m > n. +\end{alignat*} + +By \emph{integration by parts} we readily obtain the reduction formula +\begin{flalign*} +&&\int\limits^1_{-1}x^p(1 - x^2)^q dx &= \dfrac{2q}{p+1} \int\limits^1_{-1} x^{p+2} (1 -x^2)^{q-1} dx & \text{whence}\\[-5ex] +\end{flalign*} +\[ +\int\limits^1_{-1} x^{n-m} (1- x^2)^m dx = \dfrac{2^m m!}{(n - m + 1)(n - m + 3)\cdots (n + m - 1)} \int\limits^1_{-1} x^{n+m} dx.\\[-5ex] +\] +\begin{flalign*} +\int\limits^1_{-1} x^{n+m} dx &= \dfrac{2}{(n + m + 1)}\quad \text{if}\quad n + m \quad\text{is even,}\\ + &= 0\quad\text{if}\quad n + m \quad\text{is odd.} +\end{flalign*} +\begin{flalign*} +&\text{Hence} &A_m &= \dfrac{(2m + 1)n(n - 1)(n - 2)\cdots (n - m + 1) }{(n - m + 1)(n - m + 3) (n - m + 5) \cdots (n + m + 1)}&\phantom{hence}\\[1ex] +&&& \hspace{6em}\text{if } m < n + 1 \text{ and } m + n \text{ is even,}\\[1ex] +&&&= 0 \quad \text{if } m > n \text{ or if } m + n \text{ is odd.} +\end{flalign*} + +Therefore +\begin{align*} +&\rlap{$\displaystyle x^n = \dfrac{n! }{1.3.5 \cdots (2n+1)}\left [(2n+1) P_n(x) + (2n-3) \dfrac{(2n+1)}{2} P_{n-2} (x)\right.$}\\ +&&&+ (2n-7) \dfrac{(2n+1)(2n-1)}{2.4 } P_{n-4}(x)\\ +&\hspace{7em}&&+ \left.(2n-11) \dfrac{(2n+1)(2n-1)(2n-3)}{2.4.6 } P_{n-6}(x) + \cdots \right] && \tag{4} +\end{align*} +the second member ending with the term $\dfrac{1}{n+1} P_0(x)$ if $n$ is even and with +the term $\dfrac{3}{n+2} P_1(x)$ if $n$ is odd. +% -----File: 189.png + +For convenience of reference we write out a few powers of $x$. +\[ \left . \begin{aligned} +x^0 &= 1 = P_0(x)\\ +x &= P_1(x)\\ +x^2&=\frac{2}{3} P_2(x)+ \frac{1}{3}P_0(x)\\ +x^3&=\frac{2}{5} P_3(x) + \frac{3}{5} P_1(x)\\ +x^4 &= \frac{8}{35} P_4(x) + \frac{4}{7 }P_2(x)+ \frac{1}{5} P_0(x)\\ +x^5 &= \frac{8}{63} P_5(x) + \frac{4}{9} P_3(x) + \frac{3}{7}P_1(x)\\ +x^6 &= \frac{16}{231} P_6(x) +\frac{24}{77} P_4(x) +\frac{10}{21 }P_2(x) +\frac{1}{7} P_0(x)\\ +x^7 &= \frac{16}{429} P_7(x) +\frac{8}{39} P_5(x) +\frac{14}{33} P_3(x) +\frac{1}{3 }P_1 (x)\\ +x^8 &= \frac{128}{6435} P_8(x) + \frac{64}{495} P_6(x) +\frac{48}{143} P_4(x) + \frac{40}{99 }P_2 (x) + \frac{1}{9} P_0(x).\\ +\end{aligned} +\right \} +\tag{ 5} +\] + +If a given function of $x$ can be expressed as a \textit{terminating power series} it can +be developed into a Zonal Harmonic Series by the aid of (4). Given that +\begin{flalign*} +&&f(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots,\\[1ex] +&\text{let} &f(x) = B_0 &+ B_1P_1(x) + B_2P_2(x) + B_3P_3(x) + \cdots ;&&\phantom{let} +\end{flalign*} +then picking out carefully the coefficient of $P_m(x)$ we have +\begin{multline*} +B_m = \frac{m! }{ 1.3.5.\cdots (2m - 1)} \left [ a_m + \frac{(m+1)(m+2)}{ 2.(2m+3)} a_{m+2} \right .\\ +\left . + \frac{(m + 1)(m + 2)(m + 3)(m + 4) }{2.4.(2m + 3)(2m + 5)} a_{m+4} + \cdots \right ]. \tag{6} +\end{multline*} + +\mypara{96.} The development of $\dfrac{dP_n(x) }{dx}$ is useful and is easily obtained. +\begin{flalign*} +&\text{\indent Let}&\frac{dP_n(x) }{dx} = &A_0P_0(x) + A_1P_1(x) + A_2P_2(x) + \cdots &&\phantom{\indent Let}\\ +&\text{Then}& A_m &= \frac{2m+1 }{2} \int\limits_{-1}^1 P_m(x)\frac{ dP_n(x) }{dx} dx && \tag{1} +\end{flalign*} +by Art.~90 (2); +\[ +\int\limits_{-1}^1 P_m(x) \frac{dP_n(x)}{dx } dx = \Big[P_m(x)P_n(x)\Big]_{x=-1}^{x=1}-\int\limits_{-1}^1 P_n(x) \frac{dP_m(x) }{dx} dx. \tag{2} +\] +% -----File: 190.png +\begin{align*} +\Big[P_m(x)P_n(x)\Big]_{x=-1}^{x=1}& =0 \quad \text{if $m + n$ is even}\\ + & =2 \quad \text{if $m + n$ is odd.} +\end{align*} +Since $P_n(x)$ is an algebraic polynomial of the $n$th degree in $x$, $\dfrac{dP_n(x)}{dx}$ is an +algebraic polynomial of the $n-1$st degree in $x$. Therefore in (1) $m$ is less +than $n$; consequently $\dfrac{dP_m(x)}{dx}$ is an algebraic polynomial in $x$ of lower degree +than $n$ and +\begin{flalign*} +&\phantom{by Art. 95 (3).} &&\int\limits_{-1}^1 P_n(x) \frac{dP_m(x)}{dx} dx = 0 && \tag*{by Art.~95 (3).} +\end{flalign*} +\begin{flalign*} +&\text{We get then}&A_m & = 2m+1 \quad \text{if $m+n$ is odd and $m<n$,}&&\phantom{We get then}\\[1ex] + &&& = 0 \quad \text{if $m+n$ is even or $m > n-1$;} &&\tag*{and} +\end{flalign*} +\[ +\frac{dP_n(x)}{dx }= (2n-1)P_{n-1}(x) + (2n-5)P_{n-3}(x) + (2n-9)P_{n-5}(x) + \cdots \tag{3} +\] +the second member ending with the term $3P_1(x)$ if $n$ is even and with the +term $P_0(x)$ if $n$ is odd. + +From (3) a number of simple formulas are readily obtained. For example +\begin{gather*} +\frac{dP_{n+1}(x)}{dx} - \frac{dP_{n-1}(x)}{dx} =(2n+1)P_n(x) \tag{4}\\ +\int\limits_x^1P_n(x) dx = \frac{1}{2n+1} [P_{n-1}(x)-P_{n+1}(x)]. \tag{5}\\ +(2n+1)x \frac{dP_n(x)}{dx} = n \frac{dP_{n+1}(x)}{dx} + (n+1) \frac{dP_{n-1}(x)}{dx } \tag{6} +\intertext{[v.\ (4) and Article 77 (12)].} +(x^2-1) \frac{dP_n(x)}{dx} = nxP_n(x)-nP_{n-1}(x) \tag{7} +\end{gather*} +[v.\ (5) and Article 91 (7).] + +\mypara{97.} By the aid of the formulas of Art.~96 a number of valuable developments +can be obtained. + +Let us get $\cos n\theta$ and $\sin n\theta$, $n$ being any positive real. + +$z=\cos n\theta$ and $z=\sin n\theta$ are solutions of the equation +\[ +\frac{d^2z}{d\theta^2} +n^2z =0 +\] +% -----File: 191.png +or if we let $x = \cos \theta$, of the equation +\[ +(1-x^2)\dfrac{d^2z}{dx^2} -x\dfrac{dz}{dx} +n^2z=0. \tag{1} +\] +\begin{flalign*} +&\text{\indent Let}&& a_0P_0(x) + a_1P_1(x)+a_2P_2(x) + \cdots&&\phantom{\indent Let} +\end{flalign*} +be the required development of $\cos n\theta$ or of $\sin n\theta$. +\begin{flalign*} +&\text{\indent Then} &&\sum\limits^{m=\infty}_{m=0} a_m\left[(1-x^2)\dfrac{d^2P_m(x)}{dx^2} - x\dfrac{ dP_m(x)}{dx} + n^2P_m(x)\right] = 0 \;&&\text{ by (1).} +\end{flalign*} +$z = P_m(x)$ is a solution of Legendre's Equation (v.\ Art.~77). Hence +\[ +(1-x^2)\dfrac{d^2P_m(x)}{dx^2}-x\dfrac{dP_m(x)}{dx}=x\dfrac{dP_m(x)}{dx}-m(m+1)P_m(x), +\] +and (1) becomes +\[ +\sum\limits^{m=\infty}_{m=0} a_m\left[x\dfrac{dP_m(x)}{dx}+[n^2-m(m+1)]P_m(x)\right]=0. \tag{2} +\] + +Formulas (4) and (6) of Art.~96 enable us to throw (2) into the form +\[ +\sum\limits^{m=\infty}_{m=0}a_m\left[\dfrac{n^2-m^2}{2m+1}\dfrac{dP_{m+1}(x)}{dx} -\dfrac{ n^2-(m+1)^2}{2m+1} +\dfrac{dP_{m-1}(x)}{dx}\right]=0.\tag{3} +\] + +(3) must be identically true. Therefore the coefficient of $\dfrac{dP_{m+1}(x)}{dx}$ must +equal zero, and we have +\[ +a_{m+2}=\dfrac{2m+5}{2m+1}.\dfrac{n^2-m^2}{n^2-(m+3)^2} a_m. \tag{4} +\] + +If we are developing $\cos n\theta$ +\begin{flalign*} +&&a_0&=\dfrac{1}{2}\int\limits^{\pi}_0 \cos n\theta \sin\theta.d\theta &&\tag*{ by Art.~90 (4),}\\ +&& &=\dfrac{1}{4}\int\limits^{\pi}_0 [\sin(n+1)\theta-\sin(n-1)\theta]d\theta,& +\end{flalign*} +\begin{flalign*} +&&a_0&=-\dfrac{1}{2}.\dfrac{1+\cos n\pi}{n^2-1}; && \tag{5}\\ +&\text{and}\hspace{6em}& a_1={}&\dfrac{3}{2}\int\limits^{\pi}_0 \cos n\theta\cos\theta\sin\theta.d\theta &&\tag*{by Art.~90 (4),}\\ +&&a_1&=-\dfrac{3}{2}.\dfrac{1-\cos n\pi }{n^2-4}. && \tag{6} +\end{flalign*} +% -----File: 192.png + +(4), (5), and (6) give us +\begin{align*} +\cos n\theta=-\frac{1+\cos n\pi}{2(n^2-1)} +\Big[ & P_0(\cos\theta)+5\frac{n^2}{n^2-3^2}P_2(\cos\theta)\\ + & +9\frac{n^2(n^2-2^2)}{(n^2-3^2)(n^2-5^2)}P_4(\cos\theta)+\cdots\Big]\\ +-\frac{1-\cos n\pi}{2(n^2-2^2)} +\Big[ & 3P_1(\cos\theta)+7\frac{n^2-1^2}{n^2-4^2}P_3(\cos\theta)\\ + & +11\frac{(n^2-1^2)(n^2-3^2)}{(n^2-4^2)(n^2-6^2)}P_5(\cos\theta)+\cdots\Big]. \tag{7} +\end{align*} +If $n$ is a whole number $1+\cos n\pi$ or $1-\cos n\pi$ will vanish and the series +will end with the term involving $P_n(\cos\theta)$. For this case (7) may be rewritten +\begin{align*} +\cos n\theta={}&\frac{1}{2}.\frac{2.4.6.\cdots 2n}{3.5.7.\cdots(2n+1)} +\Big[(2n+1)P_n(\cos\theta)\\ +& +(2n-3)\frac{n^2-(n+1)^2}{n^2-(n-2)^2}P_{n-2}(\cos\theta)\\ +& +(2n-7)\frac{[n^2-(n+1)^2][n^2-(n-1)^2]}{[n^2-(n-2)^2][n^2-(n-4)^2]}P_{n-4}(\cos\theta)+\cdots\Big]. &&\tag{8} +\end{align*} + +If we are developing $\sin n\theta$ +\begin{align*} +a_0=\frac{1}{2}\int\limits_0^\pi\sin n\theta\sin\theta.d\theta +=-\frac{1}{2}.\frac{\sin n\pi}{n^2-1},\\ +a_1=\frac{3}{2}\int\limits_0^\pi\sin n\theta\cos\theta\sin\theta.d\theta +=\frac{3}{2}.\frac{\sin n\pi}{n^2-2^2} \tag*{and}\\[-5ex] +\end{align*} +\begin{align*} +\sin n\theta=-\frac{1}{2}.\frac{\sin n\pi}{n^2-1} +\Big[ & P_0(\cos\theta)+5\frac{n^2}{n^2-3^2}P_2(\cos\theta)\\ + & + 9\frac{n^2(n^2-2^2)}{(n^2-3^2)(n^2-5^2)}P_4(\cos\theta)+\cdots\Big]\\ + +\frac{1}{2}.\frac{\sin n\pi}{n^2-2^2} +\Big[ &3P_1(\cos\theta)+7\frac{n^2-1^2}{n^2-4^2}P_3(\cos\theta)\\ + & + 11\frac{(n^2-1^2)(n^2-3^2)}{(n^2-4^2)(n^2-6^2)}P_5(\cos\theta)+\cdots\Big]. \tag{9} +\end{align*} + +If $n$ is a whole number $\sin n\pi=0$, and all the terms of (9) vanish except +those involving $P_{n-1}(\cos\theta)$, $P_{n+1}(\cos\theta)$, $P_{n+3}(\cos\theta)$ \&c., which become indeterminate. +For this case it is necessary to compute $a_{n-1}$ +independently. + +% -----File: 193.png +We have +\begin{align*} +a_{n-1} & = \frac{2n-1}{2}\int\limits_0^\pi\sin n\theta P_{n-1}(\cos\theta)\sin\theta.d\theta\\ + & = \frac{2n-1}{4}\int\limits_0^\pi[\cos(n-1)\theta-\cos(n+1)\theta]P_{n-1}(\cos\theta)d\theta. +\end{align*} +\begin{flalign*} +&\text{Hence} & a_{n-1}=\frac{2n-1}{4}.\frac{1.3.5.\cdots(2n-3)}{2.4.6.\cdots(2n-2)}\pi& \tag*{\text{[v.\ Art.~82 (1)],}}& +\end{flalign*} +and +\begin{align*} +\sin n\theta= & \frac{\pi}{4}.\frac{1.3.\cdots(2n-3)}{2.4.\cdots(2n-2)}\Big[(2n-1)P_{n-1}(\cos\theta)\\ + &\quad\quad +(2n+3)\frac{n^2-(n-1)^2}{n^2-(n+2)^2}P_{n+1}(\cos\theta)\\ + &\quad\quad +(2n+7)\frac{[n^2-(n-1)^2][n^2-(n+1)^2]}{[n^2-(n+2)^2][n^2-(n+4)^2]}P_{n+3}(\cos\theta)+\cdots\Big]. \tag{10} +\end{align*} + +\EXAMPLE{S} + +1.\quad Show that +\[ +\csc\theta=\frac{\pi}{2}\Big[1 ++5 \Big(\frac{1}{2}\Big)^2P_2(\cos\theta) ++9 \Big(\frac{1.3}{2.4}\Big)^2P_4(\cos\theta) ++13\Big(\frac{1.3.5}{2.4.6}\Big)^2P_6(\cos\theta)+\cdots\Big] +\] +whence +\[ +\frac{1}{\sqrt{1-x^2}}=\frac{\pi}{2}\Big[1 ++5 \Big(\frac{1}{2}\Big)^2P_2(x) ++9 \Big(\frac{1.3}{2.4}\Big)^2P_4(x) ++13\Big(\frac{1.3.5}{2.4.6}\Big)^2P_6(x)+\cdots\Big] +\] +[v.\ Art.~90 (4) and Art.~82].\\ + +2.\quad Show that +\[ +\ctn\theta=\frac{\pi}{2}\Big[3\Big(\frac{1}{2}\Big)P_1(\cos\theta) + +7 \Big(\frac{3}{4}\Big)\Big(\frac{1}{2}\Big)^2P_3(\cos\theta) + +11\Big(\frac{5}{6}\Big)\Big(\frac{1.3}{2.4}\Big)^2P_5(\cos\theta)+\cdots\Big] +\] +whence +\[ +\frac{1}{\sqrt{1-x^2}}=\frac{\pi}{2}\Big[3\Big(\frac{1}{2}\Big)P_1(x) + +7 \Big(\frac{3}{4}\Big)\Big(\frac{1}{2}\Big)^2P_3(x) + +11\Big(\frac{5}{6}\Big)\Big(\frac{1.3}{2.4}\Big)^2P_5(x)+\cdots\Big] +\] +[v.\ Art.~90 (4) and Art.~82].\\ + +3.\quad By integrating the result of Ex.~1 and simplifying by the aid of Art.~96 +(5), obtain the development + +\begin{align*} +\sin^{-1}x=\frac{\pi}{2}\Big[3 & + \Big(\frac{1}{2}\Big)^2P_1(x) ++7 \Big(\frac{1}{2.4}\Big)^2P_3(x)\\ +&+11\Big(\frac{1.3}{2.4.6}\Big)^2P_5(x) ++15\Big(\frac{1.3.5}{2.4.6.8}\Big)^2P_7(x)+\cdots\Big] +\end{align*} +% -----File: 194.png +whence \qquad +$\displaystyle\theta=\frac{\pi}{2}\Big[P_0(\cos\theta) + -3\Big(\frac{1}{2}\Big)^2P_1(\cos\theta) + -7\Big(\frac{1}{2.4}\Big)^2P_3(\cos\theta)$ +\[ +-11\Big(\frac{1.3}{2.4.6}\Big)^2P_5(\cos\theta)-\cdots\Big]. +\] + +4.\quad By integrating the result of Ex.~2 and simplifying by the aid of Art.~96 +(5) obtain +\begin{gather*} +\sqrt{1-x^2}=\frac{\pi}{2}\Big[\frac{1}{2} + -5\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)^2P_2(x) + -9\Big(\frac{3}{6}\Big)\Big(\frac{1}{2.4}\Big)^2P_4(x) +\\ + -13\Big(\frac{5}{8}\Big)\Big(\frac{1.3}{2.4.6}\Big)^2P_6(x)+\cdots\Big] +\end{gather*} +whence +\[ +\sin\theta=\frac{\pi}{2}\Big[\frac{1}{2}P_0(\cos\theta) + -5\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)^2P_2(\cos\theta) + -9\Big(\frac{3}{6}\Big)\Big(\frac{1}{2.4}\Big)^2P_4(\cos\theta) + -\cdots\Big]. +\] + +To make clearer the analogy of development in Zonal Harmonic Series with +development in Fourier's Series we give on page \label{was184}\pageref{was185} a cut representing the +first seven Surface Zonal Harmonics $P_1(\cos\theta)$, $P_2(\cos\theta)$, $\cdots$ $P_7(\cos\theta)$, which +are of course somewhat complicated Trigonometric curves resembling roughly +$\cos\theta$, $\cos2\theta$, $\cdots$ $\cos7\theta$; and on page \pageref{was186}, the first four successive approximations +to the Zonal Harmonic Series +\[ + \frac{1}{2}+\frac{3}{4}P_1(\cos\theta) +-\frac{7}{8}.\frac{1}{2}P_3(\cos\theta) ++\frac{11}{12}.\frac{1.3}{2.4}P_5(\cos\theta) +-\cdots \tag*{\smallrom{I}} +\] +[v.\ (1) Art.~93], and +\begin{multline} +\frac{\pi}{2}\Big[P_0(\cos\theta) + -3\Big(\frac{1}{2}\Big)^2P_1(\cos\theta) + -7\Big(\frac{1}{2.4}\Big)^2P_3(\cos\theta) +\\ +-11\Big(\frac{1.3}{2.4.6}\Big)^2P_5(\cos\theta) +-\cdots\Big] \tag*{\smallrom{II}} +\end{multline} +(v.\ Ex.~3 Art.~97). + +\smallrom{I} is equal to 1 from $\theta=0$ to $\theta=\dfrac{\pi}{2}$, and to 0 from $\theta=\dfrac{\pi}{2}$ to $\theta=\pi$; and +\smallrom{II} is equal to $\theta$ from $\theta=0$ to $\theta=\pi$. + +The figures on page \pageref{was186} are constructed on precisely the same principle as +those on pages \pageref{was63} and \pageref{was64}, with which they should be carefully compared. + +\pngcent{013.png}{1720} + +\label{was185} + +\begin{center} +{\scriptsize The curves $y=P_0(\cos\theta)$, $y=P_1(\cos\theta)$, \ldots $y=P_7(\cos\theta)$. \qquad (v.\ page \pageref{was184}.)} +\end{center} + +\mypara{98.} By applying \emph{Gauss's Theorem} (B.\ O.\ Peirce, Newt.\ Pot.\ Func.\ §~31) or +the special Form of \emph{Green's Theorem}, +\[ +\iiint\nabla^2Vdxdydz=\int D_nVds=-4\pi\iiint\rho dxdydz, +\] +% -----File: 195.png +% -----File: 196.png +% -----File: 197.png +\noindent[Peirce, N.~P.~F.\ §~49 (149)] to a box cut from an infinitely thin shell of +attracting matter by a tube of force whose end is an element of the surface of +the shell we readily obtain the important result +\[ +4\pi\rho\kappa = D_nV_1-D_nV_2. \tag{1} +\] +where $\rho$ is the density and $\kappa$ the thickness of the shell, $V_1$ the value of the +potential function due to the shell at an internal point and $V_2$ its value at an +external point, and where $D_n$ is the partial derivative along the external normal +to the outer surface of the shell. + +If we have to deal with a surface distribution of matter we have only to +replace $\rho\kappa$ in (1) by $\sigma$ where $\sigma$ is the surface density, whence +\[ +4\pi\sigma = D_nV_1-D_nV_2 \tag{2} +\] +(v.\ Peirce, N.~P.~F.\ §§~45, 46, and 47). + +\pngcent{014.png}{1321} + +\label{was186} +%[Illustration: I.] +%[Illustration: II.] + +Formulas (1) and (2) enable us to solve problems in attraction when we +know the density of the attracting mass, and problems in Statical Electricity +when we know the distribution of the charge, by methods analogous to that of +Art.~94. + +For example let us find the value of the potential function due to a thin +material spherical shell of density $\rho$ and radius $a$. + +Since $V$ must be a solution of Laplace's Equation and must be finite both +when $r=0$ and $r=\infty$ we have +\begin{align*} +V_1&=\sum A_mr^mP_m(\cos\theta)\\ +V_2&=\sum B_m \frac{1}{r^{m+1}}P_m(\cos\theta). +\end{align*} +$V_1$ and $V_2$ must approach the same limiting values as $r$ approaches $a$. Hence +\begin{flalign*} +&&\frac{B_m}{a^{m+1}}&=A_ma^m&\\ +&\text{or} & B_m=A&_ma^{2m+1}.&\\ +&&D_nV_1=D_rV_1=\sum &mr^{m-1}A_mP_m(\cos\theta),&\\ +&&D_nV_2=D_rV_2=-\sum(&m+1)\frac{A_ma^{2m+1}}{r^{m+2}}P_m(\cos\theta).& +\end{flalign*} +Therefore by (1) +\[ +4\pi\rho\kappa=\sum(2m+1)A_ma^{m-1}P_m(\cos\theta) +\] +if $\kappa$ is the thickness of the shell. +% -----File: 198.png +\begin{flalign*} +&\indent\text{Let} & \rho=f(\cos\theta)=&\sum C_mP_m(\cos\theta) && \phantom{\indent Let}\\ +&\text{where} & C_m=\frac{2m+1}{2}&\int\limits_{-1}^1 f(x)P_m(x)dx && \tag*{by Art.~90 (2).}\\ +&\indent\text{Then} & 4\pi\kappa C_m=(2&m+1)A_ma^{m-1}, && \tag*{and} \\ +&&A_m=\frac{4\pi\kappa C_m}{(2m+1)a^{m-1}},\quad &\text{and}\quad B_m=\frac{4\pi\kappa}{2m+1}C_ma^{m+2}, \\ +&\text{and} & V_1=4\pi a\kappa\sum&\frac{C_m}{2m+1}\frac{r_m}{a_m}P_m(\cos\theta), & \tag{3}\\ +&\text{and} & V_2=4\pi a\kappa\sum&\frac{C_m}{2m+1}\frac{a^{m+1}}{r^{m+1}}P_m(\cos\theta). & \tag{4} +\end{flalign*} + +\mypara{99.} We can now get the value of the potential function due to a spherical +shell of finite thickness, provided that its density can be expressed as a sum of +terms of the form $Cr^kP_m(\cos\theta)$. + +Let $a$ be the radius of the outer surface and $b$ be the radius of the inner +surface of the shell. + +1st.---Let $\rho=Cr^kP_m(\cos\theta)$. Then for the shell of radius $s$ and thickness $ds$ +\begin{flalign*} +&&V_1={}&4\pi sds\frac{Cs^k}{2m+1}\frac{r^m}{s^m}P_m(\cos\theta) && \tag*{by (3) Art.~98,}\\ +&\text{and}\hspace{6em} &V_2=4&\pi sds\frac{Cs^k}{2m+1}\frac{s^{m+1}}{r^{m+1}}P_m(\cos\theta) && \tag*{by (4) Art.~98.} +\end{flalign*} + +Then if $r<b$ +\[ +V=\int\limits_b^a V_1=\frac{4\pi C}{(2m+1)}\frac{(a^{k-m+2}-b^{k-m+2})}{(k-m+2)}r^mP_m(\cos\theta), \tag{1} +\] +if $r>a$ +\[ +V=\int\limits_b^a V_2=\frac{4\pi C}{(2m+1)}\frac{(a^{k+m+3}-b^{k+m+3})}{(k+m+3)}\frac{P_m(\cos\theta)}{r^{m+1}}, \tag{2} +\] +and if $b<r<a$ +\begin{multline} +V=\int\limits_b^r V_2+\int\limits_r^a V_1=\frac{4\pi C}{2m+1}\bigg[\frac{r^{k+m+3}-b^{k+m+3}}{(k+m+3)r^{m+1}}\\ ++\frac{a^{k-m+2}-r^{k-m+2}}{(k-m+2)}r^m\bigg]P_m(\cos\theta). \tag{3} +\end{multline} + +2d.---If $\displaystyle\rho=\sum C_mr^kP_m(\cos\theta)$ the solutions will consist of sums of terms of +the forms given in (1), (2), and (3). +% -----File: 199.png + +\EXAMPLE{S} + +1.\quad If the shell is homogeneous +\begin{align*} +V&=2\pi\rho(a^2-b^2)\quad \text{if}\quad r<b,\\ +V&=\frac{4}{3}\pi\rho(a^3-b^3)\frac{1}{r}=\frac{M}{r}\quad \text{if}\quad r>a,\\ +V&=2\pi\rho\bigg[a^2-\frac{2b^3}{3r}-\frac{r^2}{3}\bigg]\quad \text{if}\quad b<r<a. +\end{align*} + +2.\quad If the density is any given function of the distance from the centre + $V=\dfrac{M}{r}$ if $r>a$, and $V=$ a constant if $r<b$.\\ + +3.\quad If the density at any point of a solid sphere is proportional to the square +of the distance from a diametral plane +\[ +V=\frac{M}{a}\bigg[\frac{a}{r}+\frac{2}{7}\frac{a^3}{r^3}P_2(\cos\theta)\bigg]\quad \text{if}\quad r>a. +\] + +4.\quad If the density at any point of a solid sphere is proportional to its distance +from a diametral plane +\[ +V=\frac{M}{a}\bigg[\frac{a}{r} + +\frac{1}{6} \frac{a^3}{r^3}P_2(\cos\theta) + -\frac{1.1}{6.8} \frac{a^5}{r^5}P_4(\cos\theta) + +\frac{1.1.3}{6.8.10}\frac{a^7}{r^7}P_6(\cos\theta) + -\cdots\bigg] +\] +if $r>a$. Compare Ex.~2 Art.~80. + +\mypara{100.} We have seen in Art.~18 (\emph{c}) (3) that +\[ +Q_m(x)=CP_m(x)\int\frac{dx}{(1-x^2)[P_m(x)]^2}, \tag{1} +\] +no constant term being understood with $\displaystyle\int\frac{dx}{(1-x^2)[P_m(x)]^2}$. + +$\dfrac{1}{(1-x^2)[P_m(x)]^2}$ is a rational fraction and becomes infinite only for $x=1$, + $x=-1$, and for the roots of $P_m(x)=0$, all of which are real and lie +between $-1$ and 1, as can be proved by the aid of the relation + $P_m(x)=\dfrac{1}{2^mm!}\dfrac{d^m(x^2-1)^m}{dx^m}$. + +If $x^2>1$ $\displaystyle\int\limits_{x}^\infty\frac{dx}{(1-x^2)[P_m(x)]^2}$ is finite and determinate and contains no +constant term. Hence if $x^2>1$ +\[ +Q_m(x)=-P_m(x)\int\limits_{x}^\infty\frac{dx}{(1-x^2)[P_m(x)]^2}=P_m(x)\int\limits_{x}^\infty\frac{dx}{(x^2-1)[P_m(x)]^2} \tag{2} +\] +for the constant factor of $Q_m(x)$ has been chosen so that $C=-1$. + +% -----File: 200.png +If $x^2 < 1$ the second member of (2) is not finite and determinate, and we +are thrown back to the form (1), and $C$ proves to be unity. + +(1) gives us readily +\begin{flalign*} +&& &\quad Q_0(x) = \dfrac{1}{2}\log \dfrac{1+x}{1-x} &&\tag{3}\\ +&&&Q_1(x) = -1+ \dfrac{x}{2} \log \dfrac{1+x}{ 1-x} &&\tag{4} +\intertext{if $x^2<1$.}\\[-7ex] +&\text{\indent (2) gives us}&&\quad Q_0(x) = \dfrac{1}{2}\log \dfrac{x+1}{x-1}&&\tag{5}\hspace{4em}\\ +&&&Q_1(x) = -1 + \dfrac{x}{2}\log \dfrac{x+1}{ x-1}& &\tag{6} +\end{flalign*} +if $x^2>1$. + +From Art.~85 (10) it follows that +\begin{align*} +Q_m(x) &= C\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^x_0\dfrac{dx}{(x^2-1)^{m+1}}\bigg]\quad\text{if}\quad x^2 < 1,\\ + &= C\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^{\infty}_x\dfrac{dx}{(x^2-1)^{m+1}}\bigg]\quad\text{if}\quad x^2 > 1. +\end{align*} +$C$ can be determined and is equal to $\dfrac{(-1)^{m+1}2^mm!}{(2m)!}$ if $x^2 < 1$, and is equal +to $\dfrac{(-1)^m2^m m!}{(2m)!}$ if $x^2 > 1$. +\begin{flalign*} +&\text{Hence}& Q_m(x) &= \dfrac{(-1)^{m+1}2^mm!}{(2m)!}\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^x_0\dfrac{dx}{(x^2-1)^{m+1}}\bigg]& (7)\\[-2ex] +\intertext{if $x^2<1$,}\\[-6ex] +&\text{and} & Q_m(x) &= \dfrac{(-1)^m2^m m!}{(2m)!}\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^{\infty}_x\dfrac{dx}{(x^2-1)^{m+1}}\bigg]&(8)\\[-5ex] +\end{flalign*} +if $x^2>1$. + +(7) and (8) give us for $Q_0(x)$ and $Q_1(x)$ the values already written in (3), +(4), (5), and (6). + +By the repeated application of the formula +\[ +(m+1)Q_{m+1}(x) - (2m+1)xQ_m(x) + mQ_{m-1}(x)=0, \tag{9} +\] +which may be obtained for the case where $x^2 < 1$ from Art.~16 (13) and (14), +and for the case where $x^2 > 1$ from Art.~16 (9), any Surface Zonal Harmonic +of the Second Kind can be obtained from $Q_0(x)$ and $Q_1(x)$ as given in (3), (4), +(5), and (6). + +% -----File: 201.png +Analogous formulas for $p_m(x)$ and $q_m(x)$ can be obtained without difficulty +from Art.~16 (4) and (5). They are +\begin{flalign*} +&&(m + 1)^2q_{m+1}(x) &- (2m + 1)xp_m(x) - m^2 q_{m-1}(x) = 0 \tag{10}\\ +&\text{and}\hspace{1em} & p_{m+1}(x) + &(2m + 1)xq_m(x) - p_{m-1}(x) = 0 &\,\tag{11} +\end{flalign*} +and they hold good for any value of $m$. + +\EXAMPLE{S} + +1.\quad Confirm the values of $Q_0(x)$ and $Q_1(x)$ given in Art.~100 (3), (4), (5), and +(6) by expanding them and comparing them with Art.~16 (13), (14), and (9).\\ + +2.\quad If the value of $V$ on the surface of a cone of revolution can be expressed +in terms of whole powers positive or negative of $r$, $V$ can be found for any +point in space, cf.\ Art.~81. + +If $V=\displaystyle \sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)$ when $\theta = \alpha$ then +\[ +V=\sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)\dfrac{P_m(\cos \theta)}{P_m(\cos \alpha)}. +\] + +3.\quad If $V=\displaystyle \sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)$ when $\theta = \alpha$, and $V = 0$ when $\theta = \beta$, +\[ +V=\sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)\left[\dfrac{Q_m(\cos\beta)P_m(\cos\theta)-P_m(\cos\beta)Q_m(\cos\theta)}{P_m(\cos \alpha)Q_m(\cos\beta)-P_m(\cos\beta)Q_m(\cos \alpha)}\right]. +\] + +4.\quad Find $V$ for points corresponding to values of $\theta$ between $\alpha$ and $\beta$ when +$V$ can be given in terms of whole powers of $r$ for $\theta = \alpha$ and for $\theta = \beta$.\\ + +5.\quad Find by the method of Art.~16 solutions of Legendre's Equation of the +form +\begin{align*} +z = {}_1P_m(x) &= 1 + \dfrac{m(m+1) }{ 2} (x - 1) + \dfrac{(m-1)m(m+1)(m+2) }{ 2^2 (2!)^2} (x - 1)^2\\ + &\quad+ \dfrac{(m-2)(m-1)m(m+1)(m+2)(m+3) }{2^3 (3!)^2} (x-1)^3 + \cdots,\\ +z = {}_{-1}P_m(x) &= 1 - \dfrac{m(m+1)}{2}(x + 1) + \dfrac{(m-1)m(m+1)(m+2) }{2^2(2!)^2} (x + 1)^2\\ + &\quad+ \dfrac{(m-2)(m-1)m(m+1)(m+2)(m+3)}{2^3(3!)^2} (x + 1)^3 + \cdots. +\end{align*} +If $m$ is a whole number, ${}_1P_m(x) = P_m(x)$ and ${}_{-1}P _m(x) = (-1)^m P_m (x)$. No +matter what the value of $m$, ${}_1P_m (x)$ is absolutely convergent for $-1 < x < 3$, +and ${}_{-1}P_m(x)$ is absolutely convergent for $ -3 < x < 1$.\\ +% -----File: 202.png + +6.\quad By the aid of (7) Art.~16 show that +\[ +\arraycolsep=0.3em +\begin{array}{c@{\quad} |@{\quad} c} +V= \dfrac{1}{\sqrt{r}} \sin (n \log r)k_n(\cos \theta), & V= \dfrac{1}{\sqrt{r}} \sin (n \log r)l_n(\cos \theta),\\[1.3em] +V= \dfrac{1}{\sqrt{r}} \cos (n \log r)k_n(\cos \theta), & V= \dfrac{1}{\sqrt{r}} \cos (n \log r)l_n(\cos \theta), +\end{array} +\] +are solutions of Laplace's Equation +\begin{flalign*} +&&r&D_r^2(rV) + \dfrac{1}{\sin \theta} D_\theta (\sin \theta D_\theta V) = 0, &\text{if}\\ +&&k_n(x) = p_{-\frac{1}{2}+ni} (x) &= 1 + \dfrac{n^2 + \Big(\dfrac{1}{2}\Big)^2}{ 2!} x^2 + \dfrac{\Big[n^2 + \Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{5}{2}\Big)^2\Big]}{ 4!} x^4\\[1ex] +&& &+ \dfrac{\Big[n^2 + \Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{5}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{9}{2}\Big)^2\Big]}{ 6!} x^6 + \cdots +\end{flalign*} +and\\[-5ex] +\begin{flalign*} +&&l_n(x) = -q_{-\frac{1}{2}+ni} (x) &= x + \dfrac{n^2 + \Big(\dfrac{3}{2}\Big)^2}{ 3!} x^3 + \dfrac{\Big[n^2 + \Big(\dfrac{3}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{7}{2}\Big)^2\Big]}{ 5!} x^5&&\\[1ex] +&& &+ \dfrac{\Big[n^2 + \Big(\dfrac{3}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{7}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{11}{2}\Big)^2\Big]}{ 7!} x^7 + \cdots&& +\end{flalign*} +$k_n (x)$ and $l_n (x)$ are convergent if $x^2 < 1$, but are divergent if $x^2 = 1$.\\ + +7.\quad Show by the aid of Example 5 that +\[ +\arraycolsep=0.3em +\begin{array}{c@{\quad} |@{\quad} c} +V= \dfrac{1}{\sqrt{r}} \sin (n \log r)K_n(\cos \theta), & V= \dfrac{1}{\sqrt{r}} \sin (n \log r)K_n(-\cos \theta),\\[1em] +V= \dfrac{1}{\sqrt{r}} \cos (n \log r)K_n(\cos \theta), & V= \dfrac{1}{\sqrt{r}} \cos (n \log r)K_n(-\cos \theta), +\end{array} +\] +\begin{flalign*} +&\text{are solutions of}& &rD_r^2(rV) + \dfrac{1}{\sin \theta} D_\theta (\sin \theta D_\theta V) = 0&\phantom{are solutions of}\\[-5ex] +\end{flalign*} +\begin{flalign*} +&\text{if }& K_n(x) =& {}_1P_{-\frac{1}{2}+ni}(x)=1-\dfrac{n^2+\Big(\dfrac{1}{2}\Big)^2}{2}(x-1)&\\[0.5em] +& & &+ \dfrac{\Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]}{2^2(2!)^2}(x-1)^2 &\\[0.5em] +& & & - \dfrac{\Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]\Big[n^2+\Big(\dfrac{5}{2}\Big)^2\Big]}{2^2(3!)^2}(x-1)^3+\cdots& +\end{flalign*} +% -----File: 203.png +and\\[-5ex] +\begin{align*} +K_n(-x) = &{}_{-1}P_{-\frac{1}{2} + ni}(x) = 1 + \dfrac{n^2+\Big(\dfrac{1}{2}\Big)^2}{2} (x + 1)\\ +&\phantom{{}_-}+\dfrac{ + \Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big] + \Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]}{2^2(2!)^2}(x+1)^2\\ +&\phantom{{}_-}+\dfrac{ + \Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big] + \Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big] + \Big[n^2+\Big(\dfrac{5}{2}\Big)^2\Big]}{{2^3(3!)^2}} (x+1)^3 +\cdots. +\end{align*} +$K_n(\cos \theta)$ is convergent except for $\theta =\pi$, and $K_n(-\cos\theta)$ is convergent +except for $\theta = 0$. + +$k_n(x)$, $l_n(x)$, $K_n(x)$, and $K_n(-x)$ are sometimes called \emph{Conal Harmonics}. +They are particular values of $z$ which satisfy Legendre's Equation written in +the form +\[ +(1-x^2)\dfrac{d^2z}{dx^2}- 2x + \dfrac{dz}{dx}-\Big(n^2+ + \dfrac{1}{4}\Big)z=0. +\] +For an elaborate treatment of them see E.~W. Hobson on ``A Class of Spherical +Harmonics of Complex Degree.'' Trans.\ Camb.\ Phil.\ Soc., Vol.\ XIV.\\ + +8.\quad If $V=f(r)$ when $\theta=\beta,$ +\[ + V = \dfrac{1}{\pi\sqrt{r}} + \int\limits_{-\infty}^\infty d\lambda + \int\limits_0^\infty e^{\tfrac{\lambda}{2}} f(e^\lambda) + \dfrac{K_\alpha(\cos\theta)}{K_\alpha(\cos\beta)} + \cos[\alpha(\lambda-\log r)]d\alpha; + \quad\text{if}\quad\theta<\beta. +\] + +9.\quad If $V=f(r)$ when $\theta = \beta$ and $r<a$, and $V=0$ when $r=a$, +\[ + V = \dfrac{2}{\pi} + \sqrt{\frac{a}{r}} + \int\limits_{-\infty}^0d\lambda + \int\limits_0^{\infty}e^{\tfrac{\lambda}{2}} f(ae^\lambda) + \dfrac{K_\alpha(\cos\theta)}{K_\alpha(\cos\beta)} + \sin \alpha\lambda \sin\left(\alpha \log \dfrac{r}{a}\right)d\alpha; + \quad\text{if}\quad\theta<\beta. +\] + +10.\quad If $V = f(r)$ when $\theta =\beta$ and $a < r < b$, and $V=0$ when $r = a$ + and when $r = b$, + \iffalse +\[ + V = \sum\limits_{m=1}^{m=\infty}A_m + \dfrac{K_{m^\prime}(\cos\theta)}{K_{m^\prime}(\cos\beta)} + \sin\left[\dfrac{m\pi(\log r - \log a)}{\log b - \log a}\right]\\[-3ex] +\] +\begin{flalign*} + &\text{where} + &m^\prime = \dfrac{m\pi}{\log b - \log a} + &&\text{and} +\end{flalign*} +\[ + A_m = \dfrac{2}{\log b-\log a} + \sqrt{\dfrac{a}{r}} + \int\limits_{0}^{\log{\tfrac{b}{a}}} + e^{\tfrac{x}{2}} f(ae^x) + \sin \dfrac{m\pi x}{\log b-\log a}dx; + \quad\text{if}\quad\theta<\beta. +\] +\fi +\begin{flalign*} + &&V = \sum\limits_{m=1}^{m=\infty}A_m + \dfrac{K_{m^\prime}(\cos\theta)}{K_{m^\prime}(\cos\beta)} + &\sin\left[\dfrac{m\pi(\log r - \log a)}{\log b - \log a}\right]\\ + &\text{where} & m^\prime ={}&\dfrac{m\pi}{\log b - \log a} &&\text{and}\\ + &&\llap{\hfill $\displaystyle A_m = \dfrac{2}{\log b-\log a} + \sqrt{\dfrac{a}{r}} + \int\limits_{0}^{\log{\tfrac{b}{a}}} + e^{\tfrac{x}{2}} f(ae^x$} + &)\rlap{$\displaystyle \sin \dfrac{m\pi x}{\log b-\log a}dx; + \quad\text{if}\quad\theta<\beta.$\hfill} +\end{flalign*} +% -----File: 204.png + +11.\quad If $\theta > \beta$\;\;$\cos \theta$ must be replaced by $(- \cos \theta)$ in examples 8, 9, and 10.\\ + +12.\quad If $V=f(r)$ when $\theta=\beta$, and $V=0$ when $\theta=\gamma$, +\begin{multline*} +% recast to fit line + V = \dfrac{1}{\pi\sqrt{r}} + \int\limits^{\infty}_{-\infty}d\lambda + \int\limits^{\infty}_0 e^{\tfrac{\lambda}{2}} f(e^{\lambda})\\ + \dfrac{k_\alpha(\cos \theta)l_\alpha(\cos \gamma) - k_\alpha(\cos \gamma)l_\alpha(\cos \theta) } + {k_\alpha(\cos \beta)l_\alpha(\cos \gamma) - k_\alpha(\cos \gamma) l_\alpha(\cos \beta) } + \cos[\alpha(\lambda - \log r)] d\alpha; +\end{multline*} +if $\beta<\theta<\gamma$.\\ + +13.\quad If $V=f(r)$ when $\theta = \beta$ and $a<r<b$, $V=0$ when $\theta = \gamma$ and + $a<r<b$, and $V=0$ when $r = a$ and when $r = b$, +\iffalse +\[ +V = \sum^{m=\infty}_{m=1} A_m + \dfrac{k_{m'}(\cos \theta) l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \theta)} + {k_{m'}(\cos \beta) l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \beta)} + \sin\dfrac{m \pi (\log r - \log a)}{\log b - \log a}, +\] +\begin{flalign*} +&\text{where}& m' &= \dfrac{m \pi }{ \log b - \log a}&\text{and} +\end{flalign*} +\[ + A_m = \dfrac{2}{\log b-\log a} + \sqrt{\dfrac{a}{r}} + \int\limits^{\log\tfrac{b}{a}}_0 e^{\tfrac{x}{2}}f(ae^x) + \sin\dfrac{m\pi x}{\log b-\log a}dx; +\] +\fi +\begin{flalign*} +&&&&\llap{$\displaystyle V = \sum^{m=\infty}_{m=1} A_m + \dfrac{k_{m'}(\cos \theta) l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \theta)} + {k_{m'}(\cos \beta) l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \beta)} + \sin\dfrac{m \pi (\log r - \log a)}{\log b - \log a},\:$}\\ +&\text{where}& m' &= \dfrac{m \pi }{ \log b - \log a}&\text{and}\\ +&& A_m = \dfrac{2}{\log b-\log a} + \sqrt{\dfrac{a}{r}} + &\int\limits^{\log\tfrac{b}{a}}_0 e^{\tfrac{x}{2}}f(ae^x) + \sin\dfrac{m\pi x}{\log b-\log a}dx; +\end{flalign*} +if $\beta<\theta<\gamma$ and $a<r<b$.\\ + +14.\quad If $V=f(r)$ when $\theta=\beta$ and $a<r<b$, and $V=0$ when $r= a$ + and $D_rV+hV=0$ when $r = b$, +\begin{flalign*} + &\phantom{where}&V&=\sum\limits^{m=\infty}_{m=1}A_m + \dfrac{{K_{\alpha_m}(\cos \theta)}}{{K_{\alpha_m}(\cos \beta)}} + \sin \left(\alpha_m\log \dfrac{r}{a}\right), &\text{where} +\end{flalign*} +\[ + A_m = \dfrac{2(\alpha_m^2 + h^2b^2)}{\alpha_m^2(\log b -\log a)+hb[hb(\log b - \log a)+1]} + \int\limits^{\log\tfrac{b}{a}}_0 + e^{\tfrac{x}{2}}f(ae^x)\sin \alpha_mx.dx +\] +and $\alpha_m$ is a root of the equation +\begin{flalign*} +&\phantom{v.\ Art.~68 Ex.~5.}&& \alpha\cos\left(\alpha\log \dfrac{b}{a}\right)+ + hb\sin \left(\alpha\log\dfrac{b}{a}\right)=0&\text{v.\ Art.~68 Ex.~5.} +\end{flalign*} +\label{ch5end} + +% -----File: 205.png + +\mychap{CHAPTER VI.}{SPHERICAL HARMONICS.} +\label{ch6start} + +\mypara{101.} When we are dealing with problems in finding the \textit{potential function} +due to forces which have not circular symmetry\footnote{See note, page 12.} about an axis and are using +Spherical Coördinates, we have to solve Laplace's Equation in the form + +\[ +rD_r^2(rV) + \frac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta}V) + \frac{1}{\sin^2\theta }D_{\phi}^2V = 0 \tag{1} +\] +[v.\ \smallromr{XIII} Art.~1]. + +To get a particular solution of (1) we shall assume as usual that $V$ is a +product of functions each of which involves but a single variable. + +Let $V=R.\Theta.\Phi$; where $R$ involves $r$ only, $\Theta$ involves $\theta $ only, and $ \Phi$ $ \phi$ only. +Substitute in (1) and we get +\begin{flalign*} + &&\frac{r}{R}\frac{ d^2(rR)}{dr^2 }&+ + \frac{1}{\Theta \sin\theta } + \frac{d\Big (\sin\theta \dfrac{d\Theta}{d\theta}\Big )}{d\theta} + + \frac{1}{\Phi \sin^2\theta} \frac{d^2\Phi}{d\phi^2}=0 && \tag{2} +\\ +&\text{or}& + \frac{ r \sin^2\theta}{R} &\frac{ d^2(rR)}{dr^2} + + \frac{ \sin\theta}{\Theta } \frac{ d\Big (\sin\theta \dfrac{d\Theta}{d\theta}\Big )}{d\theta} = - + \frac{ 1}{\Phi } \frac{ d^2\Phi}{d\phi^2} .&&\phantom{or} +\end{flalign*} + +As the first member does not contain $\phi$ the second member cannot contain +$\phi$, and as it contains no other variable it must be constant; call it $n^2$. Equation +(2) is then equivalent to the two equations +\begin{flalign*} +&&\frac{d^2\Phi}{d\phi^2} &+ n^2\Phi = 0 && \tag{3} +\\ +&\text{and}& + \frac{r}{R}\frac{ d^2(rR)}{dr^2 }+ + \frac{1}{\Theta \sin\theta } + &\frac{d\Big [\sin\theta \dfrac{d\Theta}{d\theta}\Big ]}{d\theta} - + \frac{n^2}{\sin^2\theta} = 0 \tag{4} &&\phantom{\,} +\end{flalign*} + +(3) has been solved before and gives us +\[ +\Phi = A \cos n\phi + B \sin n\phi \tag{5} +\] +[v.\ Art.~13(\emph{a})]. + +The first term of (4) does not involve $\theta$ and the second and third terms do\\ +not involve $r$.\\ +% -----File: 206.png + +$\dfrac{r}{R}\dfrac{ d^2(rR)}{dr^2}$ must, then, be a constant; we shall call it $m(m + 1)$ as in Art.~13(\emph{c}). Then (4) breaks up into +\[ +r\frac{ d^2(rR)}{dr^2} = m(m+1) R \tag{6} +\] +\begin{flalign*} +&\text{and}&\frac{ 1}{\sin \theta }\frac{d \Big [\sin \theta \dfrac{d \Theta}{ d \theta} \Big ]}{d \theta} + \left [m(m+1) - \frac{n^2}{\sin^2 \theta} \right ] \Theta = 0. \tag{7} &&\phantom{\,} +\end{flalign*} + +\markright{SPHERICAL HARMONICS.} +(6) was solved in Art.~13(\emph{c}) and gives +\[ +R = A_1 r^m + B_1 r^{-m-1}. \tag{8} +\] + +If in (7) we replace $\cos \theta$ by $\mu$ we get +\[ +\frac{d}{d \mu} \left [(1- \mu^2) \frac{d \Theta}{d \mu} \right ] + \left [m(m+1) - \frac{n^2}{1- \mu^2} \right ] \Theta = 0, \tag{9}\\ +\] +the equivalent of +\[ +(1- x^2) \frac{d^2z}{dx^2} - 2x \frac{dz}{dx} + \left [m(m+1) - \frac{n^2}{1- x^2} \right ] z = 0, \tag{10} +\] +[v.\ (17) Art.~85], which was solved in Art.~85 for the case where $m$ and $n$ are +positive integers and $n < m + 1$. v.\ (18) and (19) Art.~85. + +From (19) Art.~85 we get as a particular solution of (9) +\[ +\Theta = (1- \mu^2)^{\tfrac{n}{2}} \frac{d^n P_m(\mu)}{d \mu^n} = \sin^n \theta \frac{d^n P_m(\mu)}{d \mu^n}, \tag{11} +\] +if we restrict ourselves to whole positive values of $m$ and $n$, as we shall do +hereafter unless the contrary is explicitly stated, and suppose $m$ not less +than $n$. + +A second but less useful particular solution of (9) is +\[ +\Theta = (1- \mu^2)^{\tfrac{n}{2}} \frac{d^n Q_m(\mu)}{d \mu^n}. +\] + +Combining our results we have as important particular solutions of (1) +\begin{flalign*} +&&&\quad V = r^m(A \cos n \phi + B \sin n \phi) \sin^n \theta \frac{d^n P_m (\mu)}{d \mu^n}, \tag{12} +\\ +&\text{and}&& +V = \frac{1}{r^{m+1}} (A \cos n \phi + B \sin n \phi) + \sin^n \theta \frac{d^n P_m (\mu)}{d \mu^n}, \tag{13} && +\end{flalign*} +where $m$ and $n$ are positive integers and $n< m + 1$. +% -----File: 207.png + +\mypara{102.} $\sin^n\theta \dfrac{d^nP_m(\mu)}{d\mu^n}$ or $(1-\mu^2)^{\tfrac{n}{2}} \dfrac{d^nP_m(\mu)}{d\mu^n}$ is a new function of $\mu$, that is +of $\cos \theta$, and we shall represent it by $P_m^n(\mu)$\footnote{Most of the English writers represent this function by $T_m^n(\mu)$.} and shall call it an \emph{associated function} of the $n$th order and $m$th degree. It is a value of $ \Theta$ satisfying +equation (9) Art.~101. + +By differentiating the value of $P_m(x)$ given in (9) Art.~74 we get the formula +\begin{multline*} +P_m^n(\mu) = + \frac{(2m)! \sin^n\theta}{2^m m!(m-n)!} + \left [\mu^{m-n} - \frac{(m-n)(m-n-1)}{2.(2m-1)} \mu^{m-n-2} \right .\\ + \left . + \frac{(m-n)(m-n-1)(m-n-2)(m-n-3)}{2.4.(2m-1)(2m-3)} \mu^{m-n-4} -\cdots \right ] \tag{1} +\end{multline*} +the expression in the parenthesis ending with the term involving $\mu^0$ if $m-n$ is +even and with the term involving $\mu$ if $m-n$ is odd. + +For convenience of reference we give on the next page a table from which +$P_m^n(\mu)$ can be readily obtained for values of $m$ and $n$ from 1 to 8. + +$\cos n\phi P_m^n(\mu)$ and $\sin n\phi P_m^n(\mu)$, that is, +\[ +\cos n\phi \sin^n\theta\frac{ d^nP_m(\mu)}{d\mu^n}\quad \text{and} \quad \sin n\phi \sin^n\theta\frac{ d^nP_m(\mu)}{d\mu^n} +\] +are called \textit{Tesseral Harmonics} of the $m$th degree and $n$th order, and are +values of $V$ which satisfy the equation +\begin{gather*} +m(m+1)V + \frac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta} V) + \frac{1}{\sin^2\theta} D_{\phi}^2 V = 0 \tag{2} +\intertext{or its equivalent} +m(m+1)V + D_{\mu} [(1-\mu^2) D_{\mu} V] + \frac{1}{1-\mu^2} D_{\phi}^2 V = 0. \tag{3} +\end{gather*} +There are obviously $2m+1$ Tesseral Harmonics of the $m$th degree, namely +\begin{align*} +P_m(\mu), \qquad\qquad \cos\phi \sin\theta \frac{dP_m(\mu)}{d\mu}, \qquad & \sin\phi \sin\theta \frac{dP_m(\mu)}{d\mu}\\ + \cos 2\phi \sin^2\theta \frac{d^2P_m(\mu)}{d\mu^2}, \qquad & \sin 2\phi \sin^2\theta \frac{d^2P_m(\mu)}{d\mu^2}\\ + \cos 3\phi \sin^3\theta \frac{d^3P_m(\mu)}{d\mu^3}, \qquad & \sin 3\phi \sin^3\theta \frac{d^3P_m(\mu)}{d\mu^3}\\[-1ex] +\intertext{\phantom{$P_m(\mu),$}\qquad\qquad\qquad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad} + \cos m\phi \sin^m\theta \frac{d^mP_m(\mu)}{d\mu^m}, \qquad & \sin m\phi \sin^m\theta \frac{d^mP_m(\mu)}{d\mu^m} +\end{align*} + +If each of these is multiplied by a constant and their sum taken, this sum +is called a \textit{Surface Spherical Harmonic} of the $m$th degree, and is a solution of +equations (2) and (3). We shall represent it by $Y_m(\mu,\phi)$ or by $Y_m(\theta,\phi)$. + +\newpage + +% -----File: 208.png + +\renewcommand{\arraystretch}{0.2} +\begin{center} +\textit{Table for} $\csc^n\theta P_m^n(\mu)=\dfrac{d^nP_m(\mu)}{d\mu^n}.$ + +\vspace{\baselineskip} + +% recast to fit page width +\begin{footnotesize} +\begin{tabular}{|c|c|c|} +\hline +\multicolumn{3}{|c|}{}\\ +\hline +\tablestrut{$m$} & $n=1.$ + & $n=2.$\\ +\hline +&&\\ +\hline +\tablestrut{1} & 1 &\\ +\hline +\tablestrut{2} & $3\mu$ + & 3 \\ +\hline +\tablestrut{3} & $\dfrac{3}{2}(5\mu^2-1)$ + & $15\mu$\\ +\hline +\tablestrut{4} & $\dfrac{5}{2}(7\mu^3-3\mu)$ + & $\dfrac{15}{2}(7\mu^2-1)$\\ +\hline +\tablestrut{5} & $\dfrac{15}{8}(21\mu^4-14\mu^2+1)$ + & $\dfrac{105}{2}(3\mu^3-\mu)$\\ +\hline +\tablestrut{6} & $\dfrac{21}{8}(33\mu^5-30\mu^3+5\mu)$ + & $\dfrac{105}{8}(33\mu^4-18\mu^2+1$\\ +\hline +\tablestrut{7} & $\dfrac{7}{16}(429\mu^6-495\mu^4+135\mu^2-5)$ + & $\dfrac{63}{8}(143\mu^5-110\mu^3+15\mu)$\\ +\hline +\tablestrut{8} & $\dfrac{9}{16}(715\mu^7-1001\mu^5+385\mu^3-35\mu)$ + & $\dfrac{315}{16}(143\mu^6-143\mu^4+33\mu^2-1)$\\ +\hline +\multicolumn{3}{|c|}{}\\ +\hline +\end{tabular} +\end{footnotesize} + +\vspace{\baselineskip} + +\begin{footnotesize} +\begin{tabular}{|c|c|c|} +\hline +\multicolumn{3}{|c|}{}\\ +\hline +\tablestrut{$m$} & $n=3.$ + & $n=4.$\\ +\hline +&&\\ +\hline +\tablestrut{1} & &\\ +\hline +\tablestrut{2} & &\\ +\hline +\tablestrut{3} & 15 &\\ +\hline +\tablestrut{4} & $105\mu$ +& 105\\ +\hline +\tablestrut{5} & $\dfrac{105}{2}(9\mu^2-1)$ +& $945\mu$\\ +\hline +\tablestrut{6} & $\dfrac{315}{2}(11\mu^3-3\mu)$ +& $\dfrac{945}{2}(11\mu^2-1)$\\ +\hline +\tablestrut{7} & $\dfrac{315}{8}(143\mu^4-66\mu^2+3)$ +& $\dfrac{3465}{2}(13\mu^3-3\mu)$\\ +\hline +\tablestrut{8} & $\dfrac{3465}{8}(39\mu^5-26\mu^3+3\mu)$ +& $\dfrac{10395}{8}(65\mu^4-26\mu^2+1)$\\ +\hline +\multicolumn{3}{|c|}{}\\ +\hline +\end{tabular} +\end{footnotesize} + +\vspace{\baselineskip} + +\begin{footnotesize} +\begin{tabular}{|c|c|c|c|c|} +\hline +\multicolumn{5}{|c|}{}\\ +\hline + \tablestrut{$m$} +&\tablestrut{$n=5.$} +&\tablestrut{$n=6.$} +&\tablestrut{$n=7.$} +&\tablestrut{$n=8.$}\\ +\hline +&&&&\\ +\hline +1& \tablestrut{ }&&&\\ +\hline +2& \tablestrut{ }&&&\\ +\hline +3& \tablestrut{ }&&&\\ +\hline +4& \tablestrut{ }&&&\\ +\hline +5& \tablestrut{945}&&&\\ +\hline +6& \tablestrut{$10395\mu$} +& 10395&&\\ +\hline +7& \tablestrut{$\dfrac{10395}{2}(13\mu^2-1)$} +& $135135\mu$ +& 135135&\\ +\hline +8& \tablestrut{$\dfrac{135135}{2}(5\mu^3-\mu)$} +& $\dfrac{135135}{2}(15\mu^2-1)$ +& $2027025\mu$ +& 2027025\\ +\hline +\multicolumn{5}{|c|}{}\\ +\hline +\end{tabular} +\end{footnotesize} +\end{center} +\renewcommand{\arraystretch}{1} + +\vspace{\baselineskip} + +$r^mY_m(\mu,\phi)$ and $\dfrac{1}{r^{m+1}}Y_m(\mu,\phi)$ are called \emph{Solid Spherical Harmonics} of the +$m$th degree, and are solutions of Laplace's Equation (1) Art.~101. + +To formulate:--- +\[ +Y_m(\mu,\phi) = \sum_{n=0}^{n=m}\Big[ + A_n\cos n\phi\sin^n\theta\frac{d^nP_m(\mu)}{d\mu^n} + + B_n\sin n\phi\sin^n\theta\frac{d^nP_m(\mu)}{d\mu^n}\Big] \tag{4}\\[-5ex] +\] +\begin{flalign*} +&\text{or} & Y_m(\mu,\phi) = A_0P_m(\mu) + \sum_{n=1}^{n=m}[ + A_n\cos n\phi P_m^n(\mu) ++ B_n\sin n\phi P_m^n(\mu)] & \tag{5} &\\[-5ex] +\end{flalign*} +is a Surface Spherical Harmonic of the $m$th degree. + +A Tesseral Harmonic is a special case of a Surface Spherical Harmonic, and +a Zonal Harmonic a special case of a Tesseral Harmonic; $P_m(\mu)$ being the +Tesseral Harmonic of the zeroth order and the $m$th degree; it might be +written $P_m^0(\mu)$. + +\EXAMPLE{S} + +1.\quad Show that +\[ +(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+\bigg[m(m+1)-\frac{n^2}{1-x^2}\bigg]z = 0 +\] +reduces to +\[ +(1-x^2)\frac{d^2y}{dx^2}-2(n+1)x\frac{dy}{dx}+[m(m+1)-n(n+1)]y=0 +\] +if we substitute $(1-x^2)^\frac{n}{2}y$ for $z$, even when $m$ and $n$ are unrestricted.\\ +% -----File: 209.png + +2.\quad Show that if in the second equation of Ex.~1 we let $y=\sum a_kx^k$ we get +\[ +a_{k+2}=-\frac{(m-n-k)(m+n+1+k)}{(k+1)(k+2)}a_k \tag*{\text{(v.\ Art.~16)}} +\] +whence $z=p_m^n(x)$ and $z=q_m^n(x)$ are solutions of the first equation of Ex.~1, +no matter what the values of $m$ and $n$, if +\begin{multline*} +p_m^n(x)=(1-x^2)^\frac{n}{2}\bigg[1-\frac{(m-n)(m+n+1)}{2!}x^2\\ ++\frac{(m-n)(m-n-2)(m+n+1)(m+n+2)}{4!}x^4-\cdots\bigg] +\end{multline*} +and +\begin{multline*} +q_m^n(x)=(1-x^2)^\frac{n}{2}\bigg[x-\frac{(m-n-1)(m+n+2)}{3!}x^3\\ ++\frac{(m-n-1)(m-n-3)(m+n+2)(m+n+4)}{5!}x^5-\cdots\bigg]. +\end{multline*} + +If $m-n$ is a positive integer, $p_m^n(x)$ or $q_m^n(x)$ will terminate with the term +involving $x^{m-n}$, and in that case +\begin{multline*} +z=(1-x^2)^\frac{n}{2}\bigg[x^{m-n}-\frac{(m-n)(m-n-1)}{2.(2m-1)}x^{m-n-2}\\ ++\frac{(m-n)(m-n-1)(m-n-2)(m-n-3)}{2.4.(2m-1)(2m-3)}x^{m-n-4}-\cdots\bigg]. +\end{multline*} +% -----File: 210.png +the parenthesis ending with a term involving $x^0$ if $m-n$ is even and $x$ if +$m-n$ is odd, is a solution of the first equation of Ex.~1. If $m$ and $n$ are +integers this value of $z$ is $\dfrac{2^mm!(m-n)!}{(2m)!}P_m^n(x)$. + +\mypara{103.} We have seen in the last chapter that in many problems it is important +to be able to express a given function of $\cos\theta$, that is of $\mu$, in terms of +Zonal Harmonics of $\mu$. So it is often desirable to express a given function of +$\mu$ and $\phi$ in terms of Tesseral Harmonics of $\mu$ and $\phi$. + +If, for example, we are trying to find the \emph{Potential Function} due to certain +forces and have the value of the function given for some given value of $r$, +that is, on the surface of some given sphere whose centre is at the origin of +coördinates, of course the given value will be a function of $\theta$ and $\phi$ and if we +can express it in terms of Spherical Harmonics of $\theta$ and $\phi$ we have only to +multiply each term by the proper power of $r$ to get the required solution of +the problem. For we shall then have a value of $V$ satisfying Laplace's +Equation and reducing to the given function of $\theta$ and $\phi$ on the surface of the +given sphere. + +\mypara{104.} Suppose that we have a function of $\mu$ and $\phi$ given for all points on +the unit sphere, that is, for all values of $\mu$ from $-1$ to 1 and for all values of +$\phi$ from 0 to $2\pi$, $\mu$ and $\phi$ being independent variables, and that we wish to +express it in terms of Surface Spherical Harmonics. + +Assume that\\[1ex] +$\displaystyle f(\mu,\phi)\!=\!\sum_{m=0}^{m=\infty}\!\Big[A_{0,m}P_m(\mu) +\!+\!\sum_{n=1}^{n=m}\!\Big(A_{n,m}\cos n\phi P_m^n(\mu) ++B_{n,m}\sin n\phi P_m^n(\mu)\Big)\Big].\hfill (1)$\\ + +Let us consider first a finite case, and attempt to determine the coefficients +so that\\[1ex] +$\displaystyle f(\mu,\phi)\!=\!\sum_{m=0}^{m=p}\!\Big[A_{0,m}P_m(\mu) +\!+\!\sum_{n=1}^{n=m}\!\Big(A_{n,m}\cos n\phi P_m^n(\mu) ++B_{n,m}\sin n\phi P_m^n(\mu)\Big)\Big] \hfill (2)$\\[1ex] +shall hold good at as many points of the sphere as possible. The expression +in brackets in the second member of (2) is a Surface Spherical Harmonic of +the $m$th degree and contains $2m+1$ constant coefficients. The whole number +of coefficients to be determined is then the sum of an Arithmetical Progression +of $p+1$ terms the first term of which is 1 and the last is $2p+1$, and is +therefore equal to $(p+1)^2$. + +Let the interval from $\mu=-1$ to $\mu=1$ be divided into $p+2$ parts each of +which is $\Delta\mu$ so that $(p+2)\Delta\mu=2$, and let the interval from $\phi=0$ to $\phi=2\pi$ +be divided into $p+2$ parts each of which is $\Delta\phi$ so that $(p+2)\Delta\phi=2\pi$. +% -----File: 211.png + +Then if we substitute in equation (2) in turn the values $(-1 + \Delta\mu$, $\Delta\phi)$, +$(-1 + 2\Delta\mu$, $\Delta\phi)$, $\cdots [-1 + (p + 1)\Delta\mu$, $\Delta\phi]$; $(-1 + \Delta\mu$, $2\Delta\phi)$, +$(-1 + 2\Delta\mu$, $2\Delta\phi)$, $\cdots[-1 + (p + 1)\Delta\mu$, $2\Delta\phi]$; $\cdots [-1 + \Delta\mu$, $(p + 1)\Delta\phi]$, +$[-1 + 2\Delta\mu$, $(p + 1)\Delta\phi]$, $\cdots [-1 + (p + 1)\Delta\mu, (p + 1)\Delta\phi]$; since the first +member in each case will be known we shall have $(p + 1)^2$ equations of the +first degree containing no unknown except the $(p + 1)^2$ coefficients, and from +them the coefficients can be determined. When they are substituted in equation +(2) it will hold good at the $(p + 1)^2$ points of the unit sphere where $p + 1$ +circles of latitude whose planes are equidistant intersect $p + 1$ meridians +which divide the equator into equal arcs. If now $p$ is indefinitely increased +the limiting values of the coefficients will be the coefficients in equation (1), +and (1) will hold good all over the surface of the unit sphere. + +To determine any particular constant we multiply each of our $(p + 1)^2$ +equations by $\Delta\mu \Delta\phi$ times the coefficient of the constant in question in that +equation and add the equations and then investigate the limiting form +approached by the resulting equation as $p$ is indefinitely increased. + +As $p$ is indefinitely increased the summation in question will approach an +integration; and since $d\mu d\phi = -\sin\theta.d\theta d\phi$ is the element of surface of the +unit sphere, and as the limits $-1$ and 1 of $\mu$ correspond to $\pi$ and 0 of $\theta$ the +integration is a \emph{surface integration} over the surface of the unit sphere. + +In determining any coefficient as $A_{n,m}$ in (1) the first member of the limiting +form of our resulting equation will be +\[ +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} f(\mu,\phi)\cos n\phi P_m^n(\mu)d\mu. +\] + +In the second member we shall come across terms of the forms +\begin{gather*} +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} \sin l\phi \cos n\phi P_m^{\,l}(\mu)P_m^n(\mu)d\mu,\quad +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} \cos l\phi \cos n\phi P_m^{\,l}(\mu)P_m^n(\mu)d\mu,\\ +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} \sin n\phi \cos n\phi[P_m^n(\mu)]^2d\mu,\quad +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} \cos^2 n\phi[P_m^n(\mu)]^2d\mu, +\end{gather*} +and other terms all of which come under the form +\[ +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} Y_l(\mu,\phi) Y_m(\mu,\phi)d\mu, +\] +where $Y_m(\mu,\phi)$ and $Y_l(\mu,\phi)$ are Surface Spherical Harmonics of different +degrees. + +If we are determining a coefficient $B_{n,m}$ the only difference is that $\sin n\phi$ +and $\cos n\phi$ will be interchanged in the forms just specified. +% -----File: 212.png + +\mypara{105.} \textit{The integral over the surface of the unit sphere of the product of two +Surface Spherical Harmonics of different degrees is zero}. +\begin{flalign*} +&\text{\indent That is}& + \int\limits_0^{2\pi} d\phi + \int\limits_{-1}^1 Y_l(\mu,\phi)Y_m(\mu,\phi)d\mu=0.\tag{1}&&\hspace{3em} +\end{flalign*} + +For as we have seen $U=r^lY_l(\mu,\phi)$ and $V=r^mY_m(\mu,\phi)$ are solutions of +Laplace's Equation. Hence by \textit{Green's Theorem} +\begin{flalign*} + \int (U&D_nV-VD_nU)ds=0 \tag*{v.\ Art.~92.}\\ + D_n V =& D_r V = m r^{m-1} Y_m (\mu,\phi),\\ + D_n U &= D_r U = l r^{l-1} Y_l (\mu,\phi);\\ +U D_n V - V D_n U &= (m-l) r^{l + m - 1} Y_l (\mu,\phi) Y_m (\mu,\phi),\\ + &= (m - l) Y_l (\mu,\phi) Y_m (\mu,\phi) +\end{flalign*} +on the surface of the unit sphere; and +\[ + (m-l) + \int Y_l (\mu,\phi) Y_m (\mu,\phi)ds = (m-l) + \int\limits_0^{2\pi} d\phi + \int\limits_{1}^1 Y_l (\mu,\phi) Y_m (\mu,\phi) d\mu = 0. +\] + +Hence unless $l = m$ +\[ + \int\limits_0^{2\pi} d \phi + \int\limits_{-1}^1 Y_l (\mu,\phi) Y_m (\mu, \phi) d \mu = 0. +\] + +\EXAMPLE{S} + +1.\quad Obtain (1) Art.~105 directly from the equation +\[ + m(m+1) Y_m(\mu,\phi) + D_\mu [(1-\mu^2)D_\mu Y_m (\mu,\phi)] + + \dfrac{1}{1- \mu^2} D_\phi^2 Y_m (\mu,\phi) = 0 +\] +v.\ (3) Art.~102, and Art.~91.\\ + +2.\quad Show that the integral over the surface of the unit sphere of the product +of two Tesseral Harmonics of the same degree but of different orders is zero.\\ + +\textit{Suggestion:} +\[ + \int\limits_0^{2\pi} \sin k \phi \cos l \phi.d\phi = + \int\limits_0^{2\pi} \sin k\phi \sin l \phi.d \phi = + \int\limits_0^{2\pi} \cos k \phi \cos l \phi.d \phi = 0. +\] + +\mypara{106.} +\vspace{-7.5ex} +\begin{align*} + &&\int\limits_{-1}^1 P_l^n (\mu)P_m^n (\mu) d \mu &= 0 \quad\text{unless}\quad l = m&\\[-1ex] + &&&=\frac{2}{2m+1}\frac{ (m+n)!}{(m-n)!}\quad \text{if} \quad l = m. +\end{align*} +% -----File: 213.png + +For +\begin{align*} + \int\limits^1_{-1}P_l^n(\mu)P_m^n(\mu)d\mu &= + \int\limits^1_{-1}(1-\mu^2)^n \dfrac{d^nP_l(\mu)}{d\mu^n}.\dfrac{d^nP_m(\mu)}{d\mu^n} d\mu\\ + &=(1-\mu^2)^n \dfrac{d^nP_m(\mu)}{d\mu^n}.\dfrac{ d^{n-1} P_l(\mu)}{d\mu^{n-1}}\mathop{\Big]}^1_{-1}\\ + &\phantom{=}-\int\limits^1_{-1}\dfrac{d^{n-1} P_l(\mu)}{d\mu^{n-1}}.\dfrac{ d}{d\mu}\left[(1-\mu^2)^n \dfrac{d^n P_m(\mu)}{d\mu^n}\right]d\mu,\\ + &= - \int\limits^1_{-1}\dfrac{ d^{n-1} P_l(\mu)}{d\mu^{n-1}}.\dfrac{d}{d\mu}\left[(1-\mu^2)^n\dfrac{ d^nP_m(\mu)}{d\mu^n}\right]d\mu; +\end{align*} +by \emph{integration by parts}. + +Replacing \hfill $n$ \hfill by \hfill $n - 1$ \hfill in \hfill equation \hfill (2) \hfill Art.~84 \hfill and \hfill remembering \hfill that\\ +$\dfrac{d^{n-1} P_m(x)}{dx^{n-1}}$ +is a possible value of $z^{(n - 1)}$ we get +\[ +(1-\mu^2)\dfrac{d^{n+1} P_m(\mu)}{d\mu^{n+1}}-2n\mu \dfrac{d^n P_m(\mu)}{d\mu^n}+[m(m+1)-n(n-1)]\dfrac{d^{n-1} P_m(\mu)}{d\mu^{n-1}}=0, +\] +or if we multiply by $(1-\mu^2)^{n-1}$ +\begin{multline*} +(1-\mu^2)^n \dfrac{d^{n+1} P_m(\mu)}{d\mu^{n+1}}-2n\mu(1-\mu^2)^{n-1}\dfrac{d^n P_m(\mu)}{d\mu^n}\\ + +(m+n)(m-n+1)(1-\mu^2)^{n-1}\dfrac{ d^{n-1} P_m(\mu)}{d\mu^{n-1}}=0, +\end{multline*} +or +\[ +\dfrac{d}{d\mu}\left[(1-\mu^2)^n \dfrac{d^nP_m(\mu)}{d\mu^n}\right]=-(m+n)(m-n+1)(1-\mu^2)^{n-1} \dfrac{d^{n-1} P_m(\mu)}{d\mu^{n-1}}. +\] + +Hence follows the \emph{reduction formula} +\begin{multline*} +\int\limits^1_{-1}(1-\mu^2)^n \dfrac{d^nP_l(\mu)}{d\mu^n}.\dfrac{d^n P_m(\mu)}{d\mu^n} d\mu\\ +=(m+n)(m-n+1) \int\limits^1_{-1} (1-\mu^2)^{n-1}\dfrac{d^{n-1}P_l(\mu)}{d\mu^{n-1}}.\dfrac{d^{n-1} P_m(\mu)}{d\mu^{n-1}} d\mu. +\end{multline*} +Using this formula $n$ times we get +\begin{align*} +\int\limits^1_{-1} P_l^n(\mu)P_m^n(\mu)d\mu&=\dfrac{(m+n)!}{ (m-n)!} \int\limits^1_{-1} P_l(\mu)P_m(\mu)d\mu\\ + &= 0\quad \text{unless}\quad l = m\\ + &=\dfrac{2}{2m+1}\dfrac{ (m+n)!}{(m-n)!}\quad \text{if}\quad l = m +\end{align*} +v.\ Art.~89 (4) and (5). +% -----File: 214.png + +\mypara{107.} We are now able to complete the solution of the problem in Art.~104 +and since $\displaystyle\int\limits_0^{2\pi}\cos^2 n\phi.d\phi= \displaystyle\int\limits_0^{2\pi} \sin^2 n\phi.d\phi=\pi$ and $\displaystyle\int\limits_0^{2\pi}d\phi=2\pi$ we get as the +coefficients in (1) Art.~104 +\begin{gather*} +A_{0,m}=\frac{2m+1}{4\pi} + \int\limits_0^{2\pi} d\phi + \int\limits_{-1}^1 f(\mu,\phi)P_m(\mu)d\mu, \tag{1}\\ +A_{n,m}=\frac{2m+1}{2\pi}.\frac{(m-n)!}{(m+n)!} + \int\limits_0^{2\pi} d\phi + \int\limits_{-1}^1 f(\mu,\phi)\cos n\phi P_m^n(\mu)d\mu, \tag{2}\\ +B_{n,m}=\frac{2m+1}{2\pi}.\frac{(m-n)!}{(m+n)!} + \int\limits_0^{2\pi} d\phi + \int\limits_{-1}^1 f(\mu,\phi)\sin n\phi P_m^n(\mu)d\mu, \tag{3} +\intertext{whence} +f(\mu,\phi)=\sum_{m=0}^{m=\infty}\bigg[ A_{0,m}P_m(\mu) + \sum_{n=1}^{n=m}(A_{n,m}\cos n\phi + B_{n,m} \sin n\phi)P_m^n(\mu) \bigg] \tag{4} +\end{gather*} +and the development holds good for all values of $\mu$ and $\phi$ corresponding to +points on the unit sphere, provided only that the given function satisfies the +conditions that would have to be satisfied if it were to be developed into a +\textit{Fourier's Series}. + +If we use $\mu_1$ and $\phi_1$ in place of $\mu$ and $\phi$ in (1), (2), and (3), we can write (4) +in the form +\begin{multline*} +f(\mu,\phi) = + \frac{1}{2\pi} + \sum_{m=0}^{m=\infty} (2m+1) \bigg [ \frac{1}{2} + \int\limits_0^{2\pi} d\phi_1 + \int\limits_{-1}^1f(\mu_1,\phi_1) P_m(\mu)P_m(\mu_1)d\mu_1 \\ + +\sum_{n=1}^{n=m}\frac{(m-n)!}{(m+n)!} + \int\limits_0^{2\pi}d\phi_1 + \int\limits_{-1}^1f(\mu_1,\phi_1)P_m^n(\mu)P_m^n(\mu_1)\cos n(\phi-\phi_1)d\mu_1\bigg ]. \tag{5} +\end{multline*} + +Formulas (1), (2), (3), and (4) are convenient for actual work; (5) is rather +more compactly written. + +\mypara{108.} As an example let us express $\sin^2 \theta \cos^2 \theta \sin \phi \cos \phi$ in terms of +Surface Spherical Harmonics. +\begin{flalign*} +&\text{\indent Here}& f(\mu,\phi)=\frac{1}{2}\mu^2(1-\mu^2)\sin2\phi.&&\phantom{\indent Here}\\[-5ex] +\end{flalign*} +\begin{gather*} +\quad A_{0,m}=\frac{2m+1}{8\pi} + \int\limits_{-1}^1 \mu^2(1-\mu^2)P_m(\mu)d\mu + \int\limits_0^{2\pi}\sin2\phi.d\phi=0,\\ +% -----File: 215.png +A_{n,m} ={} \dfrac{2m+1}{4\pi}. + \dfrac{ (m-n)!}{(m+n)!} + \int\limits_{-1}^1 \mu^2(1-\mu^2)P_m^n(\mu)d\mu + \int\limits^{2\pi}_0 \sin2\phi \cos n\phi.d\phi=0,\\ +\end{gather*} +\begin{align*} +B_{n,m}& = \dfrac{2m+1}{4\pi}. + \dfrac{(m-n)!}{(m+n)!} + \int\limits^1_0 \mu^2(1-\mu^2)P_m^n(\mu)d\mu + \int\limits^{2\pi}_0 \sin2\phi \sin n\phi.d\phi,\\ + & = 0\quad\text{unless}\quad n = 2. +\end{align*} +\begin{flalign*} +&\text{If $n = 2$} & + \int\limits^{2\pi}_0 \sin 2\phi \sin n\phi.d\phi& = + \int\limits^{2\pi}_0 \sin^2 2\phi.d\phi=\pi, &\text{and}\\[-7ex] +\end{flalign*} +\begin{align*} +B_{2,m}& = \dfrac{2m+1}{4}. + \dfrac{(m-2)!}{(m+2)!} + \int\limits_{-1}^1 \mu^2(1-\mu^2)^2 + \dfrac{d^2P_m(\mu)}{d\mu^2} d\mu\\ + & = \dfrac{1}{2^m m!}\dfrac{ 2m+1}{4}\dfrac{ (m-2)!}{(m+2)!} + \int\limits_{-1}^1 \mu^2(1-\mu^2)^2 + \dfrac{d^{m+2}(\mu^2-1)^m}{ d\mu^{m+2}} d\mu. +\end{align*} +\[ + \int\limits_{-1}^1 \mu^2(1-\mu^2)^2 \dfrac{d^{m+2}(\mu^2-1)^m }{ d\mu^{m+2}} d\mu=720 + \int\limits_{-1}^1 \dfrac{d^{m-4}(\mu^2-1)^m }{ d\mu^{m-4}} d\mu +\] +by repeated integration by parts, +\begin{flalign*} +&&& = 0\quad\text{if}\quad m>4,\\ +&&& = 720 \int\limits_{-1}^1 (\mu^2-1)^4 d\mu=\dfrac{4096}{7}\quad\text{if}\quad m=4,\qquad +\\ + &\text{and} &B_{2,4} = + \dfrac{1}{2^44!}&.\dfrac{9}{4}. + \dfrac{2!}{6!}.\dfrac{4096}{7}=\dfrac{1}{105}. +\end{flalign*} + +By a like process we find +\begin{flalign*} + &&&\qquad\qquad B_{2,3} = 0\quad\text{and}\quad + B_{2,2} = \dfrac{1}{42}. &&\tag*{Hence} +\\[1ex] +&& \sin^2\theta \cos^2\theta \sin\phi &\cos\phi = + \dfrac{1}{42} P_2^2(\mu)\sin 2\phi + + \dfrac{1}{105} P_4^2(\mu)\sin 2\phi, \tag{1} +\\[1ex] +&& &=\dfrac{1}{42}\sin 2\phi \sin^2\theta + \dfrac{d^2P_2(\mu)}{d\mu^2} + + \dfrac{1}{105}\sin 2\phi \sin^2\theta + \dfrac{d^2P_4(\mu)}{d\mu^2},&\tag{2}\\[1ex] +&& &=\dfrac{1}{14}\sin^2\theta \sin 2\phi + + \dfrac{1}{14} \sin^2\theta (7\mu^2-1)\sin 2\phi. \tag{3} +\end{flalign*} + +The required expression might have been obtained without using the +formulas of Art.~107, by a very simple device, as follows: +\[ + \sin^2\theta \cos^2\theta \sin\phi \cos\phi = + \dfrac{1}{2}\mu^2\sin^2\theta \sin 2\phi.\tag{4} +\] +% -----File: 216.png + +If now we can express $\mu^2$ in the form $\displaystyle\sum\dfrac{ d^2P_m(\mu)}{d\mu^2}$ the work will be done. +\begin{flalign*} +&& \mu^2&=\dfrac{1}{4.3}\dfrac{ d^2(\mu^4)}{d\mu^2}, &\\[0.5em] +&& \mu^4&=\dfrac{8}{35}P_4(\mu)+\dfrac{4}{7}P_2(\mu)+\dfrac{1}{5}P_0(\mu), &\text{(5) Art.~95.}\\[0.5em] +&& \dfrac{d^2(\mu^4)}{d\mu^2}&=\dfrac{8}{35}\dfrac{d^2P_4(\mu)}{d\mu^2}+\dfrac{4}{7}\dfrac{d^2P_2(\mu)}{d\mu^2}; &\\[0.5em] +&\text{whence}& \mu^2&=\dfrac{2}{105}\dfrac{d^2P_4(\mu)}{d\mu^2}+\dfrac{1}{21}\dfrac{d^2P_2(\mu)}{d\mu^2},& +\end{flalign*} +and substituting this value in (4) we get (2). + +\EXAMPLE{S} + +1.\quad Show that +\begin{multline*} +\cos^3\theta \sin^3\theta \sin\phi \cos^2\phi=\left[\dfrac{1}{6930}P_6^3(\mu)+\dfrac{1}{1540}P_4^3(\mu)\right]\sin 3\phi\\ + -\left[\dfrac{2}{693}P_6^1(\mu)-\dfrac{1}{770}P_4^1(\mu)-\dfrac{1}{63}P_2^1(\mu)\right]\sin\phi. +\end{multline*} + +2.\quad Show that +\[ +\cos 2\phi=2 \cos 2\phi\left[\dfrac{5}{4!}P_2^2(\mu)+\dfrac{9.2!}{6!}P_4^2(\mu)+\dfrac{13.4!}{8!}P_6^2(\mu)+\cdots\right]. +\] + +3.\quad If in a problem on the Potential Function $V=f(\mu,\phi)$ when $r = a$, we +shall obviously have +\[ +V= \sum\limits^{m=\infty}_{m=0}\dfrac{ r^m}{a^m}\bigg[A_{0,m} P_m(\mu) + \sum\limits^{n=m}_{n=1} (A_{n,m} \cos n\phi+B_{n,m} \sin n\phi)P_m^n(\mu)\bigg] +\] +at an internal point and +\[ +V= \sum\limits^{m=\infty}_{m=0}\dfrac{a^{m+1}}{r^{m+1}}\bigg[A_{0,m} P_m(\mu) + \sum\limits^{n=m}_{n=1} (A_{n,m} \cos n\phi+B_{n,m} \sin n\phi)P_m^n(\mu)\bigg] +\] +at an external point, where $A_{0,m}$, $A_{n,m}$, and $B_{n,m}$ have the values given in (1), +(2), and (3) Art.~107.\\ + +4.\quad Solve problems (3), (4), and (5) of Art.~94 for the case where $V$ is not +symmetrical with respect to an axis. + +\mypara{109.} Any Solid Spherical Harmonic $r^mY_m(\mu,\phi)$ being a value of $V$ that +satisfies Laplace's Equation in Spherical Coördinates will transform into a +function of $x$, $y$, and $z$ satisfying $\nabla^2 V = 0$ if we change to a set of rectangular +% -----File: 217.png +axes having the same origin and the same axis of $X$ as the polar system. +Moreover the new function will be a homogeneous rational integral Algebraic +function of $x$, $y$, $z$, of the $m$th degree. + +For each term of $r^m \cos n\phi P_m^n(\mu)$ is of the form +\begin{flalign*} +&&&Cr^m \cos^{n-2k} \phi \sin^{2k} \phi \sin^n \theta \cos^{m-2l-n} \theta &&\\[1ex] +&\text{where}&&2k < n + 1\quad\text{and}\quad 2l < m - n + 1.&&\phantom{where} +\end{flalign*} + +This may be written\\[-5ex] +\begin{flalign*} +\intertext{\qquad$Cr^{2l}.r^{m-2l-n}\cos^{m-2l-n} \theta.r^{n-2k}\sin^{n-2k} \theta \cos^{n-2k} \phi.r^{2k}\sin^{2k} \theta \sin^{2k} \phi$} +&\text{which becomes}&& C(x^2 + y^2 + z^2)^l x^{m-2l-n} y^{n-2k} z^{2k},&&\phantom{which becomes} +\end{flalign*} +and is a homogeneous rational integral Algebraic function of $x$, $y$, and $z$ of the +$m$th degree. The same thing may be shown of each term of $r^m \sin n\phi P_m^n(\mu)$. + Consequently $r^mY_m(\mu,\phi)$ is a homogeneous rational integral Algebraic function +of the $m$th degree in $x$, $y$, and $z$. + +\mypara{110.} Any homogeneous rational integral Algebraic function $S_m(x, y, z)$ of +the $m$th degree in $x$, $y$, and $z$, which is a value of $V$ satisfying $\nabla^2 V = 0$ contains +$2m + 1$ arbitrary constant coefficients. + +For $S_m(x, y, z)$ will in general consist of $\dfrac{(m+1)(m+2)}{2}$ terms and will +therefore contain $\dfrac{(m+1)(m+2)}{2}$ coefficients. + +$\nabla^2 S_m(x, y, z)$ will be homogeneous of the $(m - 2)$d degree and will contain +$\dfrac{m(m-1)}{2}$ coefficients, which, of course, will be functions of the coefficients in +$S_m(x, y, z)$. Since $\nabla^2 S_m(x, y, z) = 0$ independently of the numerical values +of $x$, $y$, and $z$ the $\dfrac{m(m-1)}{2}$ coefficients in $\nabla^2 S_m(x, y, z)$ must be separately +zero, and that fact will give us $\dfrac{m(m-1)}{2}$ equations of condition between the +$\dfrac{(m+1)(m+2)}{2}$ original coefficients and will leave $\dfrac{(m+1)(m+2)}{2} -\dfrac{ m(m-1)}{2}$ +or $2m + 1$ of them undetermined. $S_m(x, y, z)$ contains, then, the same number +of arbitrary coefficients as $r^m Y_m(\mu,\phi)$. + +We can then choose the coefficients in $r^m Y_m(\mu,\phi)$ so that it will transform +into any given $S_m(x, y, z)$. + +Consequently a Solid Spherical Harmonic of the $m$th degree might be +defined as \textit{a homogeneous rational integral Algebraic function of x, y, and z,} +$S_m(x, y, z)$, \textit{of the $m$th degree satisfying the equation} $\nabla^2 S_m(x, y, z) = 0$; and a +Surface Spherical Harmonic of the $m$th degree as such a function divided by +$(x^2 + y^2 + z^2)^{\frac{m}{2}}$, that is by $r^m$. + +% -----File: 218.png + +\EXAMPLE{S} + +1.\quad Show that if $S_m(x,y,z)$ is a Solid Spherical Harmonic of the $m$th degree +\[ +\nabla^2[r^nS_m(x,y,z)]=n(2m+n+1)r^{n-2}S_m(x,y,z). +\] + +\emph{Suggestion:} +\[ + \nabla^2S_m=0. \quad \nabla^2r=\dfrac{2}{r}.\quad + D_rS_m=\dfrac{mS_m}{r}.\quad + (D_xr)^2+(D_yr)^2+(D_zr)^2=1. +\] + +2.\quad Show that if $f_n(x,y,z)$ is a rational integral homogeneous function of $x$, +$y$, and $z$ of the $n$th degree it can be expressed in the form +\[ +f_n(x,y,z)=S_n(x,y,z) +r^2S_{n-2}(x,y,z) + r^4S_{n-4}(x,y,z) + \cdots,\tag{1} +\] +terminating with $r^{n-1}S_1(x,y,z)$ if $n$ is odd, and with $r^nS_0(x,y,z)$ if $n$ is even.\\ + +\emph{Suggestion}: If a term $rS_{n-1}$ were present in the second member of (1), and +we were to operate with $\nabla^2$ on both members we should by Ex.~1 have a term +$\dfrac{2n}{r }S_{n-1}$ which would be irrational when all the other terms of the resulting +equation were rational. No such term, then, could occur. In the same way +it may be shown by operating twice on (1) with $\nabla^2$ that there can be no term +$r^3S_{n-3}$ in (1); and thus step by step we can reach the result formulated in (1).\\ + +3.\quad Express $x^2yz$ in the form $S_4+r^2S_2+r^4S_0$. +\begin{flalign*} +&\text{\indent \emph{Suggestion}: let}& x^2yz&=S_4+r^2S_2+r^4S_0 \rule{6em}{0em}&\ +\intertext{and take $\nabla^2$ of both members we get}\\[-4ex] +&& 2yz&=14S_2+20r^2S_0.&\\[1ex] +&\text{Operate again with $\nabla^2$.}& 0&=120S_0. &\text{Whence} +\end{flalign*} +\[ +S_0=0, \quad S_2=\dfrac{1}{7}yz,\quad \text{and}\quad S_4=\dfrac{1}{7}(6x^2-y^2-z^2)yz. +\] + +4.\quad Express $\sin^2\theta \cos^2\theta \sin\phi \cos\phi$ in terms of Surface Spherical Harmonics. +\begin{flalign*} +&\text{\indent \emph{Suggestion:}}& &\sin^2\theta \cos^2\theta \sin\phi \cos\phi =\dfrac{x^2yz}{r^4}.\rule{6em}{0em}& +\end{flalign*} +For result v.\ Art.~108 (3). + +\mypara{111.} A transformation of coördinates to a new set of axes having the same +origin as the old set will change a given Surface Spherical Harmonic into +another of the same degree. For such a transformation does not change the +form of Laplace's Equation $\nabla^2V=0$ if both sets of axes are rectangular, +and it is effected by replacing $x$, $y$, and $z$ in the Solid Harmonic corresponding +to the given Surface Harmonic by $x\cos\alpha_1+y\cos\alpha_2+z\cos\alpha_3$, +$x\cos\beta_1+y\cos\beta_2+z\cos\beta_3$ and $x\cos\gamma_1+y\cos\gamma_2+z\cos\gamma_3$ respectively +where the cosines are the \emph{direction cosines} of the new axes, and it will leave +% -----File: 219.png +the function a homogeneous function of the $m$th degree in the new variables, +and on dividing this by the $m$th power of the unchanged radius vector we shall +have a Surface Spherical Harmonic of the $m$th degree. + +\mypara{112.} We have seen in Art.~75 that if $(x_1, y_1, z_1)$ are the coördinates of a +given point +\[ +V = \dfrac{1}{\sqrt{ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2}}\tag{1} +\] +is a solution of Laplace's Equation $\nabla^2V=0$, and transforming to spherical +coördinates that +\[ + V=\dfrac{1}{\sqrt{r^2-2rr_1[\cos\theta \cos\theta_1 + + \sin\theta \sin\theta_1 \cos(\phi-\phi_1)] + r_1^2}}\tag{2} +\] +is a solution of +\[ + rD_r^2(rV) + \dfrac{1}{\sin\theta} + D_\theta (\sin\theta D_\theta V) + \dfrac{1}{\sin^2 \theta} + D_\phi^2 V = 0. \tag{3} +\] + +If $\gamma$ is the angle between the radii vectores $r$ and $r_1$ of the points $(x, y, z)$ +and $(x_1, y_1, z_1)$ (1) can be written +\[ +V = \dfrac{1}{\sqrt{ r^2-2rr_1 \cos\gamma + r_1^2}}\tag{4} +\] +which must be equivalent to (2), and hence +\[ +\cos\gamma = \cos\theta \cos\theta_1 + \sin\theta \sin\theta_1 \cos(\phi - \phi_1). +\] +(4) which is a solution of (3) is of the same form as (5) Art.~75 and by +developing it as we developed (5) Art.~75 we find that +\[ +V=P_m(\cos\gamma) +\] +is a solution of the equation +\begin{flalign*} +&&m(&m+1)V + \dfrac{1}{\sin\theta} D_\theta(\sin\theta D_\theta V) + + \dfrac{1 }{ \sin^2\theta} D_\phi^2 V = 0 &&\tag{5} +\\ +&\text{and that}&& V= r^m P_m(\cos\gamma)\quad\text{and}\quad V=\dfrac{1}{r^{m+1} }P_m(\cos\gamma)&&\phantom{and that} +\end{flalign*} +are solutions of (3). + +If we transform our coördinates keeping the origin unchanged and taking as +our new polar axis the radius vector of $(x_1, y_1, z_1)$ $\gamma$ becomes our new $\theta$ and +$P_m(\cos\gamma)$ reduces to $P_m(\cos\theta)$, a Surface Zonal Harmonic, or a \emph{Legendrian},\footnote{v.\ Art.~74.} of +the $m$th degree. It is then a Legendrian having for its axis not the original +polar axis but the radius vector of $(x_1, y_1, z_1)$. Since a Legendrian is a Surface +Spherical Harmonic, +\[ +P_m(\cos\gamma) = P_m[\cos\theta \cos\theta_1 + \sin\theta \sin\theta_1 \cos(\phi-\phi_1)] +\] +is a Surface Spherical Harmonic of the $m$th degree. + +% -----File: 220.png +It is, however, of very special form, since being a determinate function of +$\mu$, $\phi$, $\mu_1$, and $\phi_1$ it contains but two arbitrary constants if we regard it as a +function of $\mu$ and $\phi$, instead of containing $2m + 1$. + +It is known as a \textit{Laplace's Coefficient}, or briefly as a Laplacian, of the $m$th +degree. + +We shall soon express it in the regulation form of a Surface Spherical +Harmonic. + +The radius vector of $(x_1, y_1, z_1)$ is called the axis of the Laplacian and the +point where the axis cuts the surface of the unit sphere is the \textit{pole} of the +Laplacian. + +We shall represent the Laplacian $P_m(\cos\gamma)$ by $L_m(\mu, \phi, \mu_1, \phi_1)$. Of course +$L_m(\mu, \phi, 1, \phi_1) = P_m(\mu) = P_m(\cos\theta)$, and is really independent of $\phi$. + +\mypara{113.} \emph{If the product of a Surface Spherical Harmonic of the $m$th degree by a +Laplacian of the same degree is integrated over the surface of the unit sphere, the +result is equal to $\dfrac{4\pi }{ 2m+1}$ multiplied by the value of the Spherical Harmonic at +the pole of the Laplacian.} + +That is, +\[ +\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} Y_m(\mu, \phi) L_m(\mu, \phi, \mu_1, \phi_1) d\mu = \dfrac{4\pi }{ 2m+1} Y_m(\mu_1, \phi_1).\tag{1} +\] + +Transform to the axis of the Laplacian as a new polar axis, and let $Z_m(\mu, \phi)$ +be the transformed Spherical Harmonic. $L_m(\mu, \phi, \mu_1, \phi_1)$ will become $P_m(\mu)$, +and (1) will be proved if we can show that +\begin{gather*} + \int \limits^{2\pi}_0 d\phi + \int \limits^1_{-1} Z_m(\mu,\phi) P_m(\mu)d\mu = + \dfrac{4\pi }{ 2m+1} Z_m\text{(1, 0)}.\tag{2}\\ + Z_m(\mu, \phi) P_m(\mu) = A_0[P_m(\mu)]^2 + + \sum \limits^{n=m}_{n=1} (A_n \cos n\phi + B_n \sin n\phi) P_m^n(\mu) P_m(\mu) +\end{gather*} +\indent(v.\ (5) Art.~102). +\begin{flalign*} + &\phantom{(v.\ (5) Art.~89).}&&\int\limits^{2\pi}_0 Z_m(\mu,\phi) + P_m(\mu)d\phi = 2\pi A_0[P_m(\mu)]^2 &\text{and}\\ + &&&\int\limits^1_{-1} d\mu \int\limits^{2\pi}_0 + Z_m(\mu,\phi) P_m(\mu)d\phi = \dfrac{4\pi}{2m+1} A_0& \text{(v.\ (5) Art.~89).} +\end{flalign*} + +But $Z_m(1,0) = A_0$, since $P_m(1) = 1$ and $P_m^n(1)$ contains $(1 - 1)^{\frac{n}{2}}$ as a factor +and is equal to zero. + +Hence (2) is proved. +% -----File: 221.png +\mypara{114.} We can now express a Laplacian in the regulation form as a Spherical +Harmonic, by the formulas of Art.~107. +\begin{flalign*} + L_m(\mu,\phi,\mu_1,\phi_1) + =& P_m(\cos \gamma) = + P_m[\cos \theta \cos \theta_1 + + \sin \theta \sin \theta_1 \cos(\phi - \phi_1)]&\\ + =&\sum\limits_{k = 0}^{k = \infty}\bigg[A_{0,k} P_k(\mu) + + \sum\limits_{n = 1}^{n = k}(A_{n,k} \cos n\phi + B_{n,k} \sin n\phi) P_k^n (\mu)\bigg]& +\end{flalign*} +\begin{flalign*} +&\text{where}& + A_{0,m} + &= \dfrac{2m + 1}{4\pi} + \int\limits_{0}^{2\pi}d\phi + \int\limits_{-1}^{1} L_m(\mu,\phi,\mu_1,\phi_1)P_m(\mu)d\mu.&&\phantom{where}\\ + &&&= \dfrac{2m + 1}{4\pi} + \dfrac{4\pi}{2m + 1}P_m(\mu_1)=P_m(\mu_1) &\tag*{by (1) Art.~113,} +\end{flalign*} +\begin{flalign*} +&& A_{n,m} &= \dfrac{2m + 1}{2\pi}\dfrac{(m - n)!}{(m + n)!} + \int\limits_{0}^{2\pi}d\phi + \int\limits_{-1}^{1}L_m(\mu,\phi,\mu_1,\phi_1) \cos n\phi P_m^n(\mu)d\mu\\ +&& &= \dfrac{2(m-n)!}{(m+n)!}\cos n\phi_1 P_m^n(\mu_1) &\tag*{by (1) Art.~113, and}\\ +&& B_{n,m} &= + \dfrac{2m+1}{2\pi} + \dfrac{(m-n)!}{(m+n)!} + \int\limits_{0}^{2\pi}d\phi + \int\limits_{-1}^{1}L_m(\mu,\phi,\mu_1,\phi_1) \sin n\phi P_m^n(\mu)d\mu\\ +&& &= \dfrac{2(m-n)!}{(m+n)!}\sin n\phi_1 P_m^n(\mu_1) &\tag*{by (1) Art.~113,} +\end{flalign*} +and $A_{0,k} = A_{n,k} = B_{n,k} = 0$ by Art.~105 unless $k = m$. Hence +\begin{multline*} + L_m(\mu,\phi,\mu_1,\phi_1) =\\ + P_m(\mu)P_m(\mu_1) + 2\sum\limits_{n=1}^{n=m} + \left[\dfrac{(m-n)!}{(m+n)!}P_m^n(\mu)P_m^n(\mu_1)\cos n(\phi - \phi_1)\right]. \tag{1} +\end{multline*} +Each term of a Laplacian involves a numerical coefficient, a factor which is +a function of $\mu$, a second factor which is the same function of $\mu_1$, and a third +factor which is of the form $\cos k(\phi - \phi_1)$. We give below a table +of the first few Laplacians, taken from Minchin's Statics, omitting in each term +for the sake of brevity the function of $\mu_1$. + +By the aid of (1) we can write (5) Art.~107 more compactly. It becomes +\begin{flalign*} +&&f(\mu,\phi) &= \dfrac{1}{4\pi} + \sum\limits_{m=0}^{m=\infty} (2m + 1) + \int\limits_{0}^{2\pi} d\phi_1 + \int\limits_{-1}^{1} f(\mu_1,\phi_1)L_m(\mu,\phi,\mu_1,\phi_1)d\mu_1 \tag{2}\\ +&\text{or}& + F(\theta,\phi) &= + \dfrac{1}{4\pi} + \sum\limits_{m=0}^{m=\infty}(2m + 1) + \int\limits_{0}^{2\pi}d\phi_1 + \int\limits_{0}^{\pi}F(\theta_1,\phi_1)P_m(\cos \gamma) \sin \theta_1 d \theta_1. &&\tag{3} +\end{flalign*} + +% -----File: 222.png + +\begin{center} +LAPLACIANS. +\end{center} +\[ +\arraycolsep=0.3em +\begin{array}{|c|c|c|c|} +\hline + \rule{3.5em}{0em} + &\text{coef.\ of }\cos 0(\phi-\phi_1) + &\text{coef.\ of}\cos(\phi-\phi_1) + &\text{coef.\ of}\cos 2(\phi-\phi_1)\rule{0em}{1.5em}\\[0.5em] +\hline +L_0& 1&&\rule{0em}{1.5em}\\[0.5em] +L_1& \mu&(1 - \mu^2)^{\frac{1}{2}}&\rule{0em}{1.5em}\\[0.5em] +L_2& \dfrac{1}{4}(3\mu^2 - 1) + & 3\mu(1-\mu^2)^{\frac{1}{2}} + & \dfrac{3}{4}(1 - \mu^2)\\[1em] +L_3& \dfrac{1}{4}(5\mu^3 - 3\mu) + & \dfrac{3}{8}(1 - \mu^2)^{\frac{1}{2}}(5\mu^2 - 1) + & \dfrac{15}{4} \mu(1 -\mu^2)\\[1em] +L_4& \dfrac{1}{64}(35\mu^4 - 30\mu^2 + 3) + & \dfrac{5}{8}(1- \mu^2)^{\frac{1}{2}}(7\mu^3 - 3\mu) + & \dfrac{5}{16}(1 - \mu^2)(7\mu^2 - 1)\\[1em] +\hline +\end{array} +\] + +\vspace{\baselineskip} + +\[ +\arraycolsep=0.3em +\begin{array}{|c|c|c|} +\hline + \rule{3.5em}{0em} + &\text{coef.\ of}\cos 3(\phi-\phi_1) + &\text{coef.\ of}\cos 4(\phi-\phi_1)\rule{0em}{1.5em}\\[0.5em] +\hline +L_0&&\rule{0em}{1.5em}\\[0.5em] +L_1&&\rule{0em}{1.5em}\\[0.5em] +L_2&&\rule{0em}{1.5em}\\[1em] +L_3& \dfrac{5}{8}(1 - \mu^2)^{\frac{3}{2}} + & \rule{0em}{1.5em}\\[1em] +L_4& \dfrac{35}{8} \mu(1 - \mu^2)^{\frac{3}{2}} + & \dfrac{35}{64} (1 - \mu^2)^2\rule{0em}{1.5em}\\[1em] +\hline +\end{array} +\] + +\EXAMPLE{} + +Work the problems of Art.~108 and Art.~108 Exs.~1 and 2 by the aid of (3) +Art.~114. + +% -----File: 223.png +\mypara{115.} Such problems as we have handled in Arts.~98 and 99, and also problems +differing from them in not having circular symmetry about an axis, can +now be solved by direct integration. + +For instance let it be required to find the value at an external point of the +potential function due to the attraction of a solid sphere whose density at any +point is proportional to the product of any power of the radius vector by a +Surface Spherical Harmonic. +\begin{flalign*} +&\text{\indent Let}& \rho = Cr_1^kY_m(\mu_1,\phi_1). &&\phantom{\indent Let} +\end{flalign*} +Then using our ordinary notation we have +\[ + V = \int\limits_0^a dr_1 + \int\limits_0^{2\pi} d\phi_1 + \int\limits_{-1}^1 + \dfrac{Cr^k_1 Y_m(\mu_1,\phi_1)r_1^2d\mu_1 }{\sqrt{r^2 - 2rr_1 \cos\gamma + r_1^2}}. +\] +But by (3) Art.~77 +\begin{flalign*} + \dfrac{1 }{\sqrt{ r^2 - 2rr_1 \cos \gamma + r_1^2}} = + \dfrac{1}{r} &\bigg [P_0(\cos\gamma) + + \dfrac{r_1}{r }P_1(\cos\gamma) \\ + & + \dfrac{r_1^2}{r^2}P_2(\cos\gamma) + \cdots + + \dfrac{r_1^m }{r^m }P_m(\cos\gamma) + \cdots \bigg]\\[1ex] +\text{if} \quad &r > r_1. +\end{flalign*} +Consequently since +\[ + \int\limits_0^{2\pi} d\phi_1 + \int\limits_{-1}^1 Y_m(\mu_1,\phi_1)Y_n(\mu_1,\phi_1)d\mu_1 = 0, +\] +$V$ reduces to the single term +\begin{align*} + V &= \frac{C}{r^{m+1}} + \int\limits_0^a r_1^{m+k+2} dr_1 + \int\limits_0^{2\pi} d\phi_1 + \int\limits_{-1}^1Y_m(\mu_1, \phi_1)P_m(\cos\gamma)d\mu_1\\ + &= \frac{C}{r^{m+1}} + \int\limits_0^a r_1^{m+k+2} + \left (\frac{4\pi }{ 2m+1} Y_m(\mu,\phi) \right )dr_1 \tag*{by Art.~113.} +\end{align*} +\[ + \therefore \:V = + \frac{4\pi C }{ 2m+1}. + \frac{a^{m + k + 3} }{m + k + 3}. + \frac{Y_m(\mu,\phi) }{r^{m+1}}. +\] +% -----File: 224.png + +\EXAMPLE{S} + +1.\quad Solve by direct integration the problems worked in Arts.~98 and 99 and +Examples 1, 2, 3, and 4 of Art.~99.\\ + +2.\quad The density of a solid sphere is proportional to the product of the +squares of the distances from two mutually perpendicular diametral planes; +find the value of the potential function at an external point. +\begin{alignat*}{3} +&\textit{\indent Ans.}\qquad &\rho&= kr_1^4 \cos^2\theta_1 \sin^2\theta_1 \cos^2\phi_1 && +\\[1ex] +&&&= kr_1^4 \bigg [\frac{1}{15} P_0(\mu_1) + \frac{1}{21} P_2(\mu_1) &&+ \frac{1}{42} \cos 2\phi_1 P_2^2(\mu_1) \\ +&&&&& - \frac{4}{35} P_4(\mu_1) + \frac{1}{105} \cos 2\phi_1 P_4^2(\mu_1) \bigg ].\\[-8ex] +\end{alignat*} +\begin{multline*} +V = \frac{M}{a} \left [ \frac{a}{r} + \frac{a^3}{r^3} \left ( \frac{1}{9}P_2(\mu) + \frac{1}{18} \cos 2\phi P_2^2(\mu) \right ) \right .\\ +\left . - \frac{a^5}{r^5} \left ( \frac{4}{33} P_4(\mu) - \frac{1}{99} \cos 2\phi P_4^2(\mu) \right ) \right ]. +\end{multline*} + +3.\quad Solve Example 2 by an extension of the method of Arts.~98 and 99.\\ + +4.\quad A conducting sphere of radius $a$ connected with the ground by a wire is +placed in the field of force due to an electrified point at which $m$ units of +electricity are concentrated. Find the value of the potential function due to +the induced charge.\\ + +\textit{Suggestion:} Let $V_1$ be the potential function due to the point, and $V_2$ that +due to the induced charge, and let $b$ be the distance of the point from the +centre of the sphere. Then +\begin{align*} +V_1 &= \frac{m}{\sqrt{\strxx b^2-2br \cos\theta + r^2} } &&\\ +&=\frac{m}{b} \left [ P_0(\cos\theta) +\frac{ r}{b}P_1(\cos\theta) + \frac{r^2}{b^2}P_2(\cos\theta) + \cdots \right ] \quad \text{if} \quad r<b.&&\\ +&=\frac{m}{r} \left [ P_0(\cos\theta) +\frac{ b}{r}P_1(\cos\theta) + \frac{b^2}{r^2}P_2(\cos\theta) + \cdots \right ] \quad \text{if} \quad r>b.&&\\ +V_2 &= A_0P_0(\cos\theta) + A_1 \frac{r}{a }P_1(\cos\theta) + A_2 \frac{r^2}{a^2 }P_2(\cos\theta) + \cdots \quad \text{if} \quad r<a. &&\\ +&= A_0\frac{a}{r}P_0(\cos\theta) + A_1 \frac{a^2}{r^2}P_1(\cos\theta) + A_2 \frac{a^3}{r^3 }P_2(\cos\theta) + \cdots \quad \text{if} \quad r>a.&& +\end{align*} +When $r = a$\;\;$V_1 + V_2 = 0$. Hence +\[ +A_0 = -\dfrac{m}{b}, \quad A_1 = -\dfrac{ma}{b^2}, \quad A_2 = -\dfrac{ma^2}{b^3}, \cdots +\] +% -----File: 225.png +and +\begin{align*} + V_2&=-\dfrac{m}{b}\left[P_0(\cos\theta)+ + \dfrac{r}{b} P_1(\cos\theta) + + \dfrac{ r^2}{b^2} P_2(\cos\theta)+ + \cdots\right] \quad \text{if}\quad r<a\\ + &=-\dfrac{ma}{br}\left[P_0(\cos\theta)+ + \dfrac{a^2}{br} P_1(\cos\theta) + + \dfrac{a^4}{b^2r^2} P_2(\cos\theta) + + \cdots\right] \quad \text{if}\quad r>a. +\end{align*} + +Hence the effect of the induced charge is precisely the same at an external +point as if the sphere were replaced by $\dfrac{ma}{b}$ units of negative electricity concentrated +at the point $r=\dfrac{a^2}{b}$, $\theta=0$. v.\ Peirce, Newt.\ Pot.\ Func., \S~66. + +\mypara{116.} If the two points $P$ and $P'$ are taken on the line $OH$ whose direction +cosines are $\lambda$, $\mu$, and $\nu$, and if $u$ and $u'$ are the values at $P$ and $P'$ of any continuous +function of the space coördinates, then $\displaystyle\limit_{PP'\doteq 0} \left[\dfrac{u'-u}{PP'}\right]$ is called the +\emph{partial derivative} of $u$ along the line $OH$ and will be represented by $D_hu$. + +Let $x$, $y$, $z$ be the coördinates of $P$ and $x + \Delta x$, $y + \Delta y$, $z + \Delta z$ the coördinates +of $P'$; then +\[ +u'-u=D_xu.\Delta x+D_yu.\Delta y+D_zu.\Delta z+\epsilon +\] +where $\epsilon$ is an infinitesimal of higher order than the first if $\Delta x$, $\Delta y$, and $\Delta z$ are +infinitesimal (v.\ Dif.\ Cal.\ Art.~198). +\begin{flalign*} +&\text{\indent Hence}& \dfrac{u'-u}{PP'} &=D_x u.\dfrac{\Delta x}{PP'} + D_y u.\dfrac{\Delta y}{PP'} + D_z u.\dfrac{\Delta z}{PP'} + \dfrac{\epsilon}{PP'}.&\phantom{\indent Henc}\\ +&\text{But}&& \dfrac{\Delta x}{PP'} = \lambda,\quad \dfrac{\Delta y}{PP'}=\mu,\quad\text{and}\quad\dfrac{ \Delta z}{PP'}=\nu.&\\ +&\text{Therefore}& &\qquad D_hu=\lambda D_xu+\mu D_yu+ \nu D_zu. & \tag{1} +\end{flalign*} + +If $\nabla^2u=0$, $D_x^pD_y^qD_z^r u$ is a solution of Laplace's Equation. +\begin{flalign*} + &\text{\indent For}& + \nabla^2(D_x^pD_y^qD_z^ru) &= + D_x^pD_y^qD_z^r(\nabla^2u) = 0.&\phantom{\indent For} +\end{flalign*} + +Hence if $\nabla^2u=0$ $D_hu$ is a solution of Laplace's Equation, and if $OH_1$, +$OH_2$, $OH_3$, $\cdots$ are a set of lines through the origin $D_{h_1}D_{h_2}D_{h_3}\cdots u$ is a solution +of Laplace's Equation. + +\mypara{117.} If $H_k$ is a rational integral homogeneous Algebraic function of $x$, $y$, +and $z$ of the $k$th degree +\begin{align*} +D_x\left(\dfrac{H_k}{r^l}\right) &= D_r\left(\dfrac{H_k}{r^l}\right)D_xr+\dfrac{1}{r^l}D_x(H_k)\\ + &=-\dfrac{lxH_k}{r^{l+2}} + \dfrac{H_{k-1}}{r^l} = -\dfrac{lxH_k}{r^{l+2}} + \dfrac{r^2H_{k-1}}{r^{l+2}}, +\end{align*} +and is of the form $\dfrac{H_{k+1}}{r^{l+2}}$. +% -----File: 226.png + +The same thing can be proved of $D_y\Big(\dfrac{H_k}{r^l}\Big)$ and +$D_z\Big(\dfrac{H_k}{r^l}\Big)$ and therefore holds +good of $D_h\Big(\dfrac{H_k}{r^l}\Big)$. + +If $u$ is a homogeneous function of $x$, $y$, and $z$ of the degree $-m-1$ and + $\nabla^2u=0$ then $\nabla^2(r^{2m+1}u)=0$. +\begin{flalign*} +&&\llap{$\displaystyle \nabla^2(r^{2m+1}u) + ={}$}&(2m+1)(2m+2)r^{2m-1}u\\ +&& &+2(2m+1)r^{2m-1}(xD_xu + yD_yu + zD_zu) + r^{2m+1}\nabla^2u\hspace{1.5em}\\ +&& ={}&0,\\ +&\text{since} &&\;xD_xu+yD_yu+zD_zu=-(m+1)u +\end{flalign*} +by Euler's Theorem (v.\ Dif.\ Cal.\ Art.~220). + +\mypara{118.} $\dfrac{M}{r}=\dfrac{M}{ \sqrt{x^2+y^2+z^2 }}$ is a solution of Laplace's Equation (v.\ Art.~75) +and is of the form $\dfrac{H_0}{r}$. + +$D_{h_1}D_{h_2}D_{h_3} \cdots D_{h_m} \Big (\dfrac{M}{r} \Big )$ is then a solution of Laplace's Equation by Art. 116; +it is of the form $\dfrac{H_m}{r^{2m+1}}$ by Art.~117 and is a homogeneous function of the +degree $ -m-1$. + +Therefore $r^{2m+1}D_{h_1}D_{h_2}D_{h_3} \cdots D_{h_m} \Big (\dfrac{M}{r} \Big )$ is a solution of Laplace's Equation, +and is a rational integral homogeneous Algebraic function of $x$, $y$, and $z$ of the +$m$th degree, and is consequently a Solid Spherical Harmonic of the $m$th +degree (v.\ Art.~110); and $r^{m+1}D_{h_1}D_{h_2}D_{h_3} \cdots D_{h_m} \Big (\dfrac{M}{r} \Big )$ is a Surface Spherical Harmonic of the $m$th degree. + +Moreover since the direction of each of the lines $OH_1$, $OH_2$, $\cdots$ $OH_m$ depends +upon two angles which may be taken at pleasure, these angles and $M$ are +$2m+1$ arbitrary constants and may be so chosen that $r^{m+1}D_{h_1}D_{h_2}D_{h_3} \cdots$ +$D_{h_m} \Big (\dfrac{M}{r} \Big )$ may be any given Surface Spherical Harmonic. + +Consequently any given Surface Spherical Harmonic may be regarded as +formed by differentiating $\dfrac{M}{r}$ successively along $m$ determinate lines $OH_1$, $OH_2$ $\cdots$ +$OH_m$, and is given except for the undetermined factor $M$ when these lines are given. + +The lines $OH_1$, $OH_2$, $OH_3, \cdots OH_m$ are called the \textit{axes} of the Harmonic, and +the points where they meet the surface of the unit sphere the \textit{poles} of the +Harmonic. The $m$ axes of a Zonal Harmonic coincide with the axis of coördinates +(v.\ Art.~86) and consequently the $m$ axes of a Laplacian coincide with +what we have called the axis of the Laplacian (v.\ Art.~112). + +% -----File: 227.png +\mypara{119.} Any \emph{Surface Zonal Harmonic} $P_m(\mu)$ is equal to zero for $m$ real and +distinct values of $\mu$ which lie between $-1$ and 1; and any \emph{Associated Function} +$P_m^n(\mu)$ is equal to zero for $m-n$ real and distinct values of $\mu$, which lie +between $-1$ and 1. +\begin{flalign*} +&\phantom{v.\ Art.~83 (1).}&P_m(\mu) &= \dfrac{1}{ 2^m m!}.\dfrac{ d^m (\mu^2 - 1)^m }{ d \mu^m}. & \text{v.\ Art.~83 (1).} +\end{flalign*} +$\dfrac{d^k (\mu^2 - 1)^m }{ d \mu^k}$ contains $(\mu^2 - 1)^{m-k}$ as a factor. v.\ Art.~89. + +From Rolle's Theorem, ``If $f(x)$ is continuous and single-valued and is equal +to zero for the real values $a$ and $b$ of $x$, $\dfrac{df(x)}{dx}$ is equal to zero for at least one +real value of $x$ between $a$ and $b$,'' (v.\ Dif.\ Cal.\ Art.~126) it follows that since + $(\mu^2 - 1)^m = 0$ when $\mu = -1$ and when $\mu = 1$ $\dfrac{d(\mu^2 - 1)^m }{ d\mu} = 0$ for at least +one value of $\mu$ between $-1$ and 1. $\dfrac{d(\mu^2 - 1)^m }{ d\mu}$ cannot be equal to zero for +more than one value of $\mu$ between $-1$ and 1, for it contains $(\mu^2 - 1)^{m-1}$ as a +factor and is a rational Algebraic polynomial of the $2m - 1$st degree. + +In like manner we can show that $\dfrac{d^2 (\mu^2 - 1)^m }{d\mu^2} = 0$ has $m-2$ roots equal to +$-1$, $m-2$ roots equal to 1 and two real roots between $-1$ and 1 which +separate the three distinct roots of $\dfrac{d(\mu^2 - 1)^m }{ d\mu} = 0$; and in general if $k < m+1$ +that $\dfrac{d^k(\mu^2 - 1)^m }{ d\mu^k} = 0$ has $m-k$ roots equal to $-1$, $m-k$ roots equal to 1, +and $k$ real roots separating the $k+1$ distinct roots of $\dfrac{d^{k-1}(\mu^2 - 1)^m}{d\mu^{k-1}} = 0$. + +Hence $P_m(\mu) = 0$ or $\dfrac{1}{2^m m!}.\dfrac{d^m (\mu^2 - 1)^m }{ d\mu^m} = 0$ has $m$ real and distinct roots +between $-1$ and 1, and it has no more since it is of the $m$th degree. + +The same reasoning shows that $\dfrac{d^{m+n}(\mu^2 - 1)^m }{ d\mu^{m+n}} = 0$ has $m-n$ distinct real +roots between $-1$ and 1, and therefore that $P_m^n(\mu)$ is equal to zero for $m-n$ +distinct real values of $\mu$ between $-1$ and 1. Since $P_m^n(\mu)$ contains $\sin^n \theta$ as a +factor it is also equal to zero when $\mu = -1$ and when $\mu = 1$. + +$\cos n\phi$ is equal to zero for $2n$ equidistant values of $\phi$, and $\sin n\phi$ is equal to +zero for $2n$ values of $\phi$. Hence any \emph{Tesseral Harmonic} $\sin n\phi P_m^n(\mu)$ or +$\cos n\phi P_m^n(\mu)$ is equal to zero for $2n$ equidistant values of $\phi$, for $\mu = 1$, for +$\mu = -1$, and for $m-n$ real and different values of $\mu$ between $-1$ and 1. + +It follows that the value of any Surface Zonal Harmonic $P_m(\mu)$ at a point +on the surface of the unit sphere will have the same sign so long as the point +remains on one of the \textit{zones} into which the surface of the sphere is divided by +% -----File: 228.png +the $m$ circles of latitude corresponding to the $m$ roots of $P_m(\mu) = 0$, and will +change sign whenever the point passes from one of these zones into an adjoining +one; and that the value of any Tesseral Harmonic $\sin n\phi P^n_m(\mu)$ at a point +on the surface of the unit sphere will have the same sign so long as the point +remains on any one of the \textit{tesserae} into which the surface of the sphere is +divided by the $m-n$ circles of latitude corresponding to the roots of $P_m^n(\mu)=0$ +and the $2n$ meridians corresponding to the roots of $\sin n\phi = 0$, and will change +sign whenever the point passes from one of these tesserae into an adjoining +one. +\label{ch6end} +% -----File: 229.png +\mychap{CHAPTER VII.\footnotemark}{CYLINDRICAL HARMONICS (BESSEL'S FUNCTIONS).} +\label{ch7start} + +\footnotetext{The student should re-read carefully Arts.~11, 17, +and 18(\emph{d}) before beginning this chapter.} + +\mypara{120.} In Arts.~11 and 17 we obtained +\[ + z = AJ_0(x) + BK_0(x) \tag{1} +\] +as the general solution of \textit{Fourier's Equation} +\begin{flalign*} + &&&\frac{d^2z}{dx^2} + + \frac{1}{x} \frac{dz}{dx}+ z = 0, \tag{2} +\\ +&\text{where}& J_0(x) ={}&1 - + \frac{x^2}{2^2} + + \frac{x^4}{2^2.4^2 } - + \frac{x^6}{2^2.4^2.6^2} + \cdots \tag{3} +&&\phantom{where} +\end{flalign*} +and is called a \textit{Cylindrical Harmonic} or \textit{Bessel's Function} of the zeroth order; +and where +\[ + K_0(x) = J_0(x)\log x + + \frac{x^2}{2^2} - + \frac{x^4}{2^2.4^2}\left ( + \frac{1}{1} + + \frac{1}{2} \right ) + + \frac{x^6}{2^2.4^2.6^2 } \left ( + \frac{1}{1} + + \frac{1}{2} + + \frac{1}{3 }\right ) - \cdots \tag{4} +\] +and is called a \textit{Cylindrical Harmonic} or \textit{Bessel's Function} of the Second Kind, +and of the zeroth order. + +In Art.~17 we found that\qquad\quad $z = J_n(x)$ + +\noindent is a particular solution of \textit{Bessel's Equation} +\[ + \frac{d^2z}{dx^2} + + \frac{1}{x}\frac{ dz}{dx} +\left (1 - + \frac{n^2}{x^2} \right ) z = 0, \tag{5} +\] +where if $n$ is unrestricted in value +\begin{multline*} + J_n(x) = \frac{x^n}{2^n \Gamma(n+1)} \left [1 - + \frac{x^2}{2^2(n+1)} + + \frac{x^4}{2^4.2!(n+1)(n+2)} \right . \\ \left . - + \frac{x^6}{2^6.3!(n+1)(n+2)(n+3)} + \cdots \right ]\tag{6} +\end{multline*} +and is called a \textit{Cylindrical Harmonic} or \textit{Bessel's Function} of the $n$th order; +and that unless $n$ is an integer +\[ + z = AJ_n(x)+ BJ_{-n}(x) +\] +is the general solution of Bessel's Equation. + +% -----File: 230.png +If $n$ is an integer it can be shown that +\[ +J_n (x) = (-1)^n J_{-n} (x), +\] +(v.\ Forsyth's Diff.\ Eq.\ Art.~102), and then +\[ +z = AJ_n(x) + B\{K_n(x)\} +\] +is the general solution of Bessel's Equation and +\begin{multline*} + \{K_n(x)\}= J_n(x)\log x - + \dfrac{1}{2}\left( + \dfrac{x}{2}\right)^{-n} + \sum\limits^{k=n-1}_{k=0} + \dfrac{(n-k-1)!}{k!} \left(\dfrac{x}{2}\right)^{2k}\\ + - \dfrac{1}{2}\left(\dfrac{x}{2}\right)^n + \sum\limits^{k=\infty}_{k=0} + \dfrac{(-1)^k}{(n+k)!k!}\left[1 + + \dfrac{1}{2}+\dfrac{1}{3}+\cdots + + \dfrac{1}{k}\right.\\ + \left.+1+\dfrac{1}{2} + + \dfrac{1}{3}+\cdots + + \dfrac{1}{n+k}\right]\left( + \dfrac{x}{2}\right)^{2k}\tag{7} +\end{multline*} +v.\ M.\ Bôcher, Ann.\ Math.\ Vol.\ VI, No.\ 4. + +\markright{CYLINDRICAL HARMONICS.} +\mypara{121.} A useful expression for $J_n (x)$ as a definite integral can be obtained +without difficulty from Bessel's Equation [(5) Art.~120] by a slight modification +of the method given by Forsyth (Diff.\ Eq.\ Art.~136). + +It was shown in Art.~17 that $z = x^nv$ is a solution of Bessel's Equation if +$v$ satisfies the equation +\begin{flalign*} + &&\dfrac{d^2v}{dx^2} &+\dfrac{2n+1}{x} + \dfrac{dv}{dx} + v = 0. \tag{1} +\\ + &\text{\indent Assume} &&v = + \int\limits^b_a T\cos (xt)dt &\phantom{\indent Assume}\tag{2} +\end{flalign*} +where $x$ and $t$ are independent, $T$ is an unknown function of $t$, and $a$ and $b$ +are at present undetermined. +\begin{flalign*} + &\text{\indent Then}& + \dfrac{dv}{dx} &= - + \int\limits^b_a tT\sin(xt)dt&\phantom{\indent Then}\\ + &\text{and}& + \dfrac{d^2v}{dx^2}& = - + \int\limits^b_at^2 T \cos(xt)dt. +\end{flalign*} +Substituting in (1) after multiplying through by $x$, we have +\[ + \int\limits^b_a(1 - t^2) Tx \cos(xt)dt - + \int\limits^b_a(2n + 1)tT\sin(xt)dt = 0. \tag{3} +\] +% -----File: 231.png +By \emph{integration by parts} we find that +\begin{multline*} + \int\limits^b_a(1 - t^2)Tx \cos(xt)dt = \Big[(1 - t^2) T\sin (xt)\mathop{\Big]}\limits^b_a\\ + -\int\limits^b_a\Big[(1 - t^2)\dfrac{dT}{dt} - 2tT\Big]\sin(xt)dt, +\end{multline*} +and (3) reduces to +\[ +\Big[\vphantom{\dfrac{X}{Y}}(1 - t^2)T\sin(xt)\mathop{\Big]}\limits^b_a - \int\limits^b_a\Big[(1 - t^2)\dfrac{dT}{dt} + (2n - 1)tT\Big]\sin(xt)dt = 0. \tag{4} +\] + +If we determine $T$ so that +\begin{flalign*} +&&&(1-t^2) \dfrac{dT}{dt} +(2n-1)tT=0, \tag{5} +\\ +&\text{and $a$ and $b$ so that}&& \Big[(1 - t^2)T\sin(xt)\mathop{\Big]}\limits^b_a =0 &\phantom{and a and b so that} \tag{6} +\end{flalign*} +(4) will be satisfied and our problem will be solved. (5) gives +\[ +T=C(1-t^2)^{n- \frac{1}{2}}, \tag{7} +\] +and (6) will obviously be satisfied if $a = -1$ and $b = 1$. +\begin{flalign*} +&\text{\indent Hence}\qquad\qquad\quad & v&=C \int\limits^1_{-1}\dfrac{ (1-t^2)^n \cos(xt)dt }{ \sqrt{1-t^2}}&\text{is a solution of (1),}\\ +&\text{and} & z&=Cx^n \int\limits^1_{-1}\dfrac{(1-t^2)^n \cos(xt)dt}{ \sqrt{1-t^2}} & (8) +\end{flalign*} +is a solution of Bessel's Equation. + +If we let $t = \cos\phi$ in (8) we get +\[ +z=Cx^n \int\limits^{\pi}_0 \sin^{2n} \phi \cos(x\cos\phi)d\phi. +\] + +Expand $\cos(x\cos\phi)$ into a series involving powers of $x\cos\phi$, integrate +term by term by the aid of the formulas +\begin{flalign*} +&\hspace{8.5em}&\int\limits^{\tfrac{\pi}{2}}_0 \sin^n x.dx &= \dfrac{\sqrt{\pi}}{2}.\dfrac{ \Gamma\Big(\dfrac{n+1}{2}\Big)}{\Gamma\Big(\dfrac{n}{2}+1\Big)} &\text{ [Int.\ Cal.\ (1) Art.~99],} +\end{flalign*} +% -----File: 232.png +\[ + \int\limits_0^{\tfrac{\pi}{2}} \sin^n x \cos^m x.dx = + \frac{\Gamma \Big (\dfrac{m+1}{2}\Big ) \Gamma \Big ( + \dfrac{n+1}{2} \Big ) }{2\Gamma \Big (\dfrac{m+n}{2} + 1 \Big )} +\] +(Int.\ Cal.\ Art.~99 Ex.~2), and compare with (6) Art.~120, and we get +\[ + J_n(x) = \frac{x^n }{2^n \sqrt{\pi}\ + \Gamma\Big(n + \dfrac{1}{2}\Big)} + \int\limits_0^{\pi}\sin^{2n} \phi \cos(x \cos\phi)d\phi. \tag{9} +\] + +If $n$ is a positive integer (9) reduces to +\[ + J_n(x) = \frac{1}{\pi}. + \frac{x^n }{1.3.5.\cdots(2n-1)} + \int\limits_0^{\pi} \sin^{2n} \phi \cos(x \cos\phi)d\phi. \tag{10} +\] + +Let $n = 0$ in (9) or (10) and we get +\[ + J_0(x) = \frac{1}{\pi} + \int\limits_0^{\pi} \cos(x \cos\phi)d\phi. \tag{11} +\] + +\EXAMPLE{S} + +1.\quad Obtain Formula (11) directly from Fourier's Equation, (2) Art.~120.\\ + +2.\quad Prove by \textit{integration by parts} that if $n > -\dfrac{1}{2}$ +\[ + \int\limits_0^{\pi} \sin^{2n} \phi \cos\phi \sin(x \cos\phi)d\phi = + \frac{x}{ 2n+1} + \int\limits_0^{\pi}\sin^{2n+2} \phi \cos(x \cos\phi)d\phi. +\] + +3.\quad Prove by \textit{integration by parts} that if $n > \dfrac{1}{2}$ +\begin{align*} + \int\limits_0^{\pi} \sin^{2n} \phi \cos\phi &\sin(x \cos\phi)d\phi\\ + &= \frac{1}{x} \int\limits_0^{\pi} [2n \sin^{2n} \phi - (2n-1)\sin^{2n-2} \phi] \cos(x \cos\phi)d\phi. +\end{align*} + +\mypara{122.} We can now readily obtain a number of useful formulas. + +\noindent Differentiate (11) Art.~121 with respect to $x$ and we get +\begin{flalign*} +\hspace{6em} \dfrac{dJ_0(x) }{dx}& = -\dfrac{1}{\pi} + \int\limits_0^{\pi} \cos\phi \sin(x \cos\phi)d\phi\\ + &= -\dfrac{x}{\pi} + \int\limits_0^{\pi} \sin^2 \phi \cos(x \cos\phi)d\phi \tag*{by Ex.~2 Art.~121.}\\[-6ex] +\end{flalign*} +% -----File: 233.png +\begin{flalign*} + &\text{\indent Hence by (10) Art.~121} \qquad + \dfrac{ dJ_0(x)}{dx} = -J_1(x).\tag{1}&& +\end{flalign*} + +In like manner by the aid of Exs.~3 and 2, Art.~121, we can obtain the +relations +\[ + \frac{d[x^n J_n(x)] }{ dx} = x^n J_{n-1}(x)\tag{2} +\] +if $n > \dfrac{1}{2}$, +\[ + \frac{d[x^{-n} J_n (x)] }{ dx }= -x^{-n} J_{n+1} (x)\tag{3} +\] +if $n > -\dfrac{1}{2}$.\\ + +(2) can be written +\[ + \int\limits_0^x x^nJ_{n-1}(x)dx=x^nJ_n(x)\tag{4}\\[-4ex] +\] +if $n > \dfrac{1}{2}$.\\ + +(2) and (3) can be written +\begin{flalign*} + &&x^n \frac{dJ_n(x)}{dx} +{}&nx^{n-1}J_n(x) = x^nJ_{n-1}(x) +\\[2ex] + &\text{and} &x^{-n} + \frac{dJ_n(x)}{dx} - n&x^{-n-1}J_n(x) = -x^{-n}J_{n+1}(x), && +\\[2ex] + &\text{or} & + \frac{dJ_n(x)}{dx} &= J_{n-1}(x) - \frac{n}{x}J_n(x)\tag{5}&& +\\[2ex] + &\text{and} & + \frac{dJ_n(x)}{dx} ={}&- J_{n+1}(x) + \frac{n}{x}J_n(x);\tag{6}&& +\\[2ex] + &\text{whence}& + 2\frac{dJ_n(x)}{dx} &= J_{n-1}(x) - J_{n+1}(x)\tag{7}&& +\\[2ex] + &\text{and}& + \frac{2n}{x}J_n(x) &= J_{n-1}(x) + J_{n+1}(x).&&\tag{8} +\end{flalign*} + +The repeated use of formula (8) will enable us to get from $J_0(x)$ and $J_1(x)$ +any of Bessel's Functions whose order is a positive integer. For example, we +have +\begin{gather*} + J_2(x) = \frac{2}{x}J_1(x) - J_0(x)\\ + J_3(x) = \Big ( \frac{8}{x^2}-1 \Big )J_1(x) - \frac{4}{x}J_0(x). +\end{gather*} +% -----File: 234.png + +From a table giving the values of $J_0(x)$ and $J_1(x)$, then, tables for the +functions of higher order are readily constructed. Such a table taken from +Rayleigh's Sound (Vol.~I., page 265) will be found in the Appendix (Table VI.). + +By the aid of (5) and (6) any derivative of $J_n(x)$ can be expressed in terms +of $J_n(x)$ and $J_{n+1}(x)$. For example +\[ + \dfrac{d^2J_n(x)}{dx^2} = \bigg [ + \dfrac{n(n-1)}{x^2} -1\bigg ] J_n(x) + + \dfrac{1}{x } J_{n+1}(x). +\] + +If we write $J_0(x)$ for $z$ in Fourier's Equation [(2) Art.~120], then multiply +through by $xdx$ and integrate from zero to $x$, simplifying the resulting equation +by \textit{integration by parts}, we get +\begin{flalign*} + &&x &\dfrac{dJ_0(x)}{dx} + + \int\limits_0^x x J_0(x)dx = 0; +\\ + &\text{whence by (1)}&& + \int\limits_0^x x J_0(x)dx = xJ_1(x). \tag{9} &&\phantom{whence by} +\end{flalign*} + +If we write $J_0(x)$ for $z$ in Fourier's Equation, then multiply through by +$x^2 \dfrac{dJ_0(x)}{dx}\,dx$ and integrate from zero to $x$, simplifying by \textit{integration by parts} +we get +\begin{flalign*} + &&\dfrac{x^2}{2} &\bigg [ \left ( + \dfrac{dJ_0(x)}{dx} \right )^2+(J_0(x))^2 \bigg ] - + \int\limits_0^x x(J_0(x))^2 dx = 0;&&\qquad +\\ + &\text{whence by (1)}& + &\int\limits_0^x x(J_0(x))^2 dx = + \dfrac{x^2}{2} [(J_0(x))^2 + (J_1(x))^2]. \tag{10}&& +\end{flalign*} + +In like manner we can get from Bessel's Equation [(5) Art.~120] the formula +\[ + \int\limits_0^x x(J_n(x))^2 dx= + \dfrac{1}{2} \bigg [x^2 \left ( + \dfrac{dJ_n(x)}{dx} \right )^2 + + (x^2 - n^2)(J_n(x))^2 \bigg ] \tag{11} +\] +which (6) enables us to reduce to the form +\[ + \int\limits_0^x x(J_n(x))^2 dx= + \dfrac{x^2}{2} [(J_n(x))^2 + + (J_{n+1}(x))^2] - nxJ_n(x)J_{n+1}(x).\tag{12} +\] + +Formulas (9), (10), (11), and (12) will prove useful when we attempt to +develop in terms of \textit{Cylindrical Harmonics}. +% -----File: 235.png + +Values of $J_n(x)$ for larger values of $x$ than those given in Table VI., +Appendix, may be computed very easily from the formula +\begin{flalign*} + J_n(x)& = \sqrt{\frac{ 2}{\pi x} } + \left [1-\frac{(1^2-4n^2)(3^2-4n^2)}{2!(8x)^2} \right .\\ + & \quad \left . + + \frac{(1^2-4n^2)(3^2-4n^2)(5^2-4n^2)(7^2-4n^2)}{4!(8x)^4} - + \cdots \right ]\cos \left (x-\frac{\pi}{4}-n\frac{\pi}{2} \right ) \\ + & + \sqrt{ \frac{2}{\pi x}} \left [\frac{1^2-4n^2}{1!8x} \right .\\ + & \quad \left . - \frac{ (1^2-4n^2)(3^2-4n^2)(5^2-4n^2)}{3!(8x)^3} + + \cdots \right ]\sin \left (x-\frac{\pi}{4}-n\frac{\pi}{2} \right ).\tag{13} +\end{flalign*} +v.\ Lommel, Studien über die Bessel'schen Functionen, page 59.\\ + +The series terminates if $2n$ is an odd integer, but otherwise it is divergent. +It can be proved, however, that in any case the sum of $m$ terms differs from $J_n(x)$ +by less than the last term included, and consequently the formula can safely +be used for numerical computation. + +\EXAMPLE{S} + +1.\quad Confirm (1), (2), and (3), Art.~122, by obtaining them from (3) and (6), +Art.~120.\\ + +2.\quad Confirm (1), Art.~122, by showing that Fourier's Equation will differentiate +into the special form assumed by Bessel's Equation when $n=1$.\\ + +3.\quad Show that (9), Art.~122, is a special case of (4), Art.~122.\\ + +4.\quad Show that the limit approached by $J_n(x)$ as $ n$ increases indefinitely is +zero, and by the aid of this fact and of (8), Art.~122, prove that +\[ + J_{n-1}(x) = + \frac{2}{x} [nJ_n(x) - + (n+2)J_{n+2}(x) + + (n+4)J_{n+4}(x) + \cdots ]. +\] + +5.\quad Prove that +\[ + \frac{dJ_n(x)}{dx} = + \frac{2}{x} [ + \tfrac{1}{2}nJ_n(x) - + (n+2)J_{n+2}(x) + + (n+4)J_{n+4}(x) - \cdots ]. +\] + +6.\quad Show that the substitution of $\left (1-\dfrac{y^2}{n^2}\right )^{\frac{1}{2}}$ for $x$ in Legendre's Equation +will reduce it to the form +\[ + \left (1-\frac{y^2}{n^2}\right ) + \frac{d^2z}{dy^2} +\left ( + \frac{1}{y} -\frac{2y}{n^2} \right ) + \frac{dz}{dy }+ \left (1+ + \frac{1}{n} \right )z=0, +\] +and that the limiting form approached by this equation as $n$ is indefinitely +increased is Fourier's Equation, and hence that $J_0(x)$ can be regarded as some +constant factor multiplied by the limiting value approached by $P_n \bigg (1-\dfrac{x^2}{n^2} \bigg )^{\frac{1}{2}}$ +as $n$ is indefinitely increased. +% -----File: 236.png + +\mypara{123.} To complete the solution of the drumhead problem taken up in +Art. 11, we found that it would be necessary to develop a given function of $r$ +in the form +\[ + f(r) = A_1 J_0(\mu_1 r) + + A_2 J_0(\mu_2 r) + + A_3 J_0(\mu_3 r) + \cdots +\] +where $\mu_1$, $\mu_2$, $\mu_3$, \&c., are the roots of the transcendental equation $J_0(\mu a) = 0$; +and in Art.~11, Ex.\ the development of unity in a series of precisely the +same form was needed.\\ + +(\emph{a})\quad Let us consider another problem. + +The convex surface and one base of a cylinder of radius $a$ and length $b$ are +kept at the constant temperature zero, the temperature at each point of the +other base is a given function of the distance of the point from the centre of +the base; required the temperature of any point of the cylinder after the +permanent temperatures have been established. + +Here we have to solve Laplace's Equation in Cylindrical Coördinates +(\smallrom{XIV} Art.~1). +\[ + D_r^2 u + \frac{1}{r }D_r u + + \frac{1}{r^2}D_{\phi}^2 u + D_z^2 u = 0 \tag{1} +\] +subject to the conditions +\begin{alignat*}{3} + &u = 0 &&\text{when}\quad &&z = 0\\ + &u = 0 &&\quad\text{``} &&r = a\\ + &u =f(r) &&\quad\text{``} &&z = b, +\end{alignat*} +and from the symmetry of the problem we know that $D_{\phi}^2u = 0$. + +Assuming as usual $u = R.Z$ we break (1) up into the equations +\begin{flalign*} +&&& \frac{d^2Z}{dz^2} - + \mu^2 Z = 0\\[1ex] +&& \frac{d^2R}{dr^2} &+ + \frac{1}{r} + \frac{dR}{dr} + \mu^2 R =0, +\\[1ex] + &\text{whence}& u &= + \sinh (\mu z)J_0(\mu r)\tag{2}&&\phantom{whence}\\[1ex] + &\text{and}& u &= + \cosh (\mu z)J_0(\mu r)\tag{3}&&\\[-1ex] +\intertext{are particular solutions of (1).} +\end{flalign*} +\begin{flalign*} + &\text{\indent If $\mu_k$ is a root of} & J_0(\mu a) = 0 &&\phantom{\indent If mu is a root of}\tag{4} +\end{flalign*} +\[ + u = \sinh(\mu_k z)J_0(\mu_k r) +\] +satisfies (1) and two of the three equations of condition. +\begin{flalign*} + & \text{\indent If then}& f(r) = + A_1 J_0(\mu_1 r) + + A_2 J_0(\mu_2 r) + + A_3 J_0(\mu_3 r) + \cdots &&\tag{5} +\end{flalign*} +$\mu_1$, $\mu_2$, $\mu_3$, \&c., being roots of (4), +\[ + u = A_1 \frac{ \sinh(\mu_1z)}{\sinh(\mu_1b)} J_0(\mu_1 r) + + A_2 \frac{\sinh(\mu_2z)}{\sinh(\mu_2b)} J_0(\mu_2 r) + + A_3\frac{ \sinh(\mu_3z)}{\sinh(\mu_3b) }J_0(\mu_3 r) + \cdots \tag{6} +\] +satisfies (1) and all of the equations of condition, and is the required solution.\\ +% -----File: 237.png + +(\emph{b})\quad If instead of keeping the convex surface of the cylinder at the temperature +zero we surround it by a jacket impervious to heat, the equation of +condition $u = 0$ when $r = a$ will be replaced by $D_ru = 0$ when $r = a$, or if +\begin{flalign*} + &\qquad & u = + \sinh(\mu z&)J_0(\mu r), &&\\ + &\text{by}& + \frac{dJ_0(\mu r)}{dr} &= 0&&\tag*{when $r = a$,}\\ + &\text{that is by}& + \mu J'_0(\mu a) &= 0\footnotemark&&\tag*{or (v.\ (1) Art.~122)}\\[1ex] + &\text{by} & J_1(\mu a)&=0. && \tag{7} +\end{flalign*} + +\footnotetext{We shall find it convenient to use the familiar notation of $f'(x) = \dfrac{df(x)}{dx}$ (v.\ Dif. Cal., p.\ 119).} + +If now in (5) and (6) $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of (7), (6) will be the solution +of our new problem.\\ + +(\emph{c})\quad If instead of keeping the convex surface of the cylinder at the temperature +zero we allow it to cool in air at the temperature zero, the condition $u = 0$ +when $r = a$ will be replaced by $D_ru + hu = 0$ when $r = a$, or if +\begin{flalign*} +&&u = \sinh (\mu z)J_0(&\mu r)&&\\[1ex] +&\text{by}& \mu J'_0(\mu r) + hJ_0(\mu r) &= 0 && \tag*{when \quad $r = a$}\\[1ex] +&\text{that is by}& \mu a J'_0(\mu a) + ahJ_0(\mu a) &=0 && \tag*{or (v.\ (1) Art.~122)}\\[1ex] +&\text{by}& \mu a J_1(\mu a) - ahJ_0(\mu a) &= 0. && \tag{8} +\end{flalign*} + +If now in (5) and (6) $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of (8), (6) will be the +solution of our present problem. + +\mypara{124.} It can be shown that\\[-7.1ex] +\begin{flalign*} + && J_0(x) &= 0 &&\tag{1}\\ + && J_1(x) &= 0 && \tag{2}\\ + &\text{and} & xJ'_0(x) + \lambda &J_0(x) = 0&& \tag{3}\\[-4ex] +\end{flalign*} +have each an infinite number of real positive roots (v.\ Riemann, Par.\ Dif.\ Gl., +\S~97). The earlier roots of these equations can be computed without serious +difficulty from the table for the values of $J_0(x)$ (Table VI., Appendix). + +The first twelve roots of $J_0(x) = 0$ and $J_1(x) =0$ are given in Table IV., +Appendix, a table due to Stokes. Large roots of $J_0(x) = 0$ and of $J_1(x) = 0$ +may be very easily computed from the formulas +\begin{gather*} + \frac{x_{\phantom{(}0}^{(s)}}{\pi} = s-.25 + + \frac{.050661}{4s-1} - + \frac{.053041}{(4s-1)^3 } + + \frac{.262051}{(4s-1)^5 }- \cdots \tag{4}\\ + \frac{x_{\phantom{(}1}^{(s)}}{\pi} = s+.25 - + \frac{.151982}{4s+1} + + \frac{.015399}{(4s+1)^3} - + \frac{.245270}{(4s+1)^5} + \cdots \tag{5} +\end{gather*} +given by Stokes in Camb.\ Phil.\ Trans., Vol.\ IX., $x_{\phantom{(}0}^{(s)}$ representing the $s$th root +of $J_0(x) = 0$, and $x_{\phantom{(}1}^{(s)}$ the $s$th root of $J_1(x) = 0$. + +% -----File: 238.png +\mypara{125.} We have seen in Art.~123 that + $U=\sinh(\mu_k z)J_0(\mu_k r)$ and $V=\sinh(\mu_l z)J_0(\mu_l r)$ are solutions of $\nabla^2U=0$ + and $\nabla^2V=0$ if we express Laplace's Equation in terms of Cylindrical +Coördinates (v.\ (1) Art.~123). + +Hence, if $\int dS$ represents the surface integral over any closed surface, we +have +\[ +\int(UD_nV-VD_nU)dS=0 +\] +by Green's Theorem (v.\ Art.~92). + +If we take the cylinder of Art.~123 as our surface, and perform the +integrations and simplify the resulting equation, we find +\begin{multline*} +\int\limits^a_0rJ_0(\mu_k r)J_0(\mu_l r)dr=\dfrac{-1}{{\mu_k}^2-{\mu_l}^2} [\mu_kaJ_0(\mu_l a)J'_0(\mu_k a) +-\mu_laJ_0(\mu_k a)J'_0(\mu_l a)]\\ +=\dfrac{-1}{{\mu_l}^2-{\mu_k}^2} [\mu_kaJ_0(\mu_l a)J_1(\mu_k a) +-\mu_laJ_0(\mu_k a)J_1(\mu_l a)]. \tag{1} +\end{multline*} + +Hence if $\mu_k$ and $\mu_l$ are different roots of +\begin{flalign*} +& & J_0(\mu a)&=0, &&\\ +&\text{or of} & J_1(\mu a)&=0, &&\phantom{or of}\\ +&\text{or of}& \mu aJ_1(\mu a)-\lambda &J_0(\mu a)=0, &&\\ +&\text{then} & \int\limits^a_0 rJ_0(\mu_k r)J_0(&\mu_l r)dr=0. &&\tag{2} +\end{flalign*} + +\EXAMPLE{} + +Obtain (1) Art.~125 directly from Fourier's Equation +\[ +\dfrac{d^2 J_0(\mu r)}{dr^2} + \dfrac{1}{r}\dfrac{ dJ_0(\mu r)}{dr} + \mu^2 J_0 (\mu r)=0. +\] + +\mypara{126.} We are now able to obtain the developments called for in Art.~123. +\begin{flalign*} +&\text{\indent Let}& f(r)&=A_1J_0(\mu_1 r)+A_2J_0(\mu_2 r)+A_3J_0(\mu_3 r)+\cdots&\phantom{\indent Let} \tag{1} +\end{flalign*} +$\mu_1$, $\mu_2$, $\mu_3$, \&c., being roots of $J_0(\mu a)=0$, or of $J_1(\mu a)=0$, or of +\[ +\mu aJ_1(\mu a)-\lambda J_0(\mu a)=0. +\] + +To determine any coefficient $A_k$ multiply (1) by $rJ_0(\mu_k r)dr$ and integrate +from zero to $a$. The first member will become +\[ +\int\limits^a_0 rf(r)J_0(\mu_k r)dr. +\] +% -----File: 239.png + +Every term of the second member will vanish by (2) Art.~125 except the +term +\begin{gather*} + A_k \int\limits_0^a r(J_0(\mu_k r))^2dr.\\ + \int\limits_0^a r(J_0(\mu_k r))^2dr = + \frac{1}{\mu_k^2} + \int\limits_0^{\mu_ka} x(J_0(x))^2dx = + \frac{a^2}{2}[(J_0(\mu_k a))^2+(J_1(\mu_k a))^2] +\end{gather*} +by (10) Art.~122. +\begin{flalign*} + &\text{\indent Hence}& A_k = + \frac{2}{a^2[(J_0(\mu_k a))^2 + + (J_1(\mu_k a))^2] } + \int\limits_0^a rf(r)J_0(\mu_k r)dr. && \tag{2} +\end{flalign*} +The development (1) holds good from $r = 0$ to $r = a$ (v.\ Arts.~24, 25, and 88). + +If $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of $J_0(\mu a)=0$, (2) reduces to +\[ + A_k = + \frac{2}{a^2(J_1(\mu_k a))^2} + \int\limits_0^a rf(r)J_0(\mu_k r)dr. \tag{3} +\] + +If $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of $J_1(\mu a)=0$, (2) reduces to +\[ + A_k = \frac{2}{a^2(J_0(\mu_k a))^2} + \int\limits_0^a rf(r)J_0(\mu_k r)dr. \tag{4} +\] + +If $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of $\mu aJ_1(\mu a)-\lambda J_0(\mu a)=0$, (2) reduces to\\ +\[ + A_k = \frac{2\mu_k^2}{(\lambda^2+\mu_k^2a^2)(J_0(\mu_k a))^2} + \int\limits_0^a rf(r)J_0(\mu_k r)dr. \tag{5} +\] + +For the important case where $ f(r) = 1$ +\[ + \int\limits_0^a r f(r) J_0(\mu_k r) dr = + \int\limits_0^a r J_0(\mu_k r) dr = + \frac{1}{\mu_k^2} + \int\limits_0^{\mu_k a}xJ_0(x) dx = + \frac{a}{\mu_k} J_1(\mu_k a) \tag{6} +\] +by (9) Art.~122, and (3) reduces to\\ +\begin{flalign*} +&& A_k&=\frac{2}{\mu_k aJ_1(\mu_k a)},\hspace{1.6em} && \tag{7} +\end{flalign*} + +(4) reduces to $A_k = 0$ except for $k = 1$ when $\mu_k = 0$ and we have $A_1 = 1$, + +\begin{flalign*} +&\indent\text{(5) reduces to} &A_k &= + \frac{2\lambda }{(\lambda^2 + \mu_k^2 a^2)J_0(\mu_k a)}. && \phantom{(5) reduces to} \tag{8} +\end{flalign*} +% -----File: 240.png + +\EXAMPLE{S} + +1.\quad Show that in (12) Art.~11 any coefficient $A_k$ has the value given in (3) +Art.~126; and in the answer to Art.~11, Ex.~the value given in (7) Art.~126.\\ + +2.\quad Show that if a drumhead be initially distorted so that it has circular +symmetry, it will not in general give a musical note; that it may be initially +distorted so as to give a musical note; that in this case the vibration will be +a \textit{steady} vibration; that the frequencies of the various musical notes that can be +given when the distortion has circular symmetry are proportional to the roots +of $J_0(x) = 0$; that the possible nodes for such vibrations are concentric circles +whose radii are proportional to the roots of $J_0(x) = 0$.\\ + +3.\quad A cylinder of radius one meter and altitude one meter has its upper +surface kept at the temperature 100°, and its base and convex surface at the +temperature 15°, until the \textit{stationary temperature} is set up. Find the temperature +at points on the axis 25 cm., 50 cm., and 75 cm.\ from the base, and also +at a point 25 cm.\ from the base and 50 cm.\ from the axis.\\ +\phantom{kindly go to the right hand margin}\hfill +\emph{Ans.}, 29°.6; 47°.6; 71°.2; 25°.8.\\ + +4.\quad An iron cylinder one meter long and twenty centimeters in diameter has its +convex surface covered with a so-called non-conducting cement one centimeter +thick. One end and the convex surface of the cylinder thus coated are kept at the +temperature zero, the other end at the temperature of 100°. Find to the nearest +tenth of a degree the temperature of the middle point of the axis, and of the +points of the axis twenty centimeters from each end after the temperatures +have ceased to change. Given that the conductivity of iron is 0.185 and of +cement 0.000162 in C. G. S. units. Find also the temperature of a point on +the surface midway between the ends, and of points on the surface twenty +centimeters from each end. Find the temperatures of the three points of the +axis, supposing the coating a perfect non-conductor, and again, supposing the +coating absent. Neglect the curvature of the coating.\\ +\phantom{right margin}\hfill +\emph{Ans.}, 15°.4; 40°.85; 72°.8; 15°.3; 40°.7; 72°.5; 0°.0; 0°.0; 1°.3. + +\mypara{127.} If instead of considering the cooling of a cylinder as in Art.~123 we +have to deal with a cylindrical shell whose curved surfaces are co-axial +cylinders, we are obliged to use the Bessel's Functions of the second kind. +Let our equations of condition be +\begin{alignat*}{6} +&u = 0 &&\text{when}\quad &&z = 0, \qquad\qquad && u = 0\quad &&\text{when}\quad &&r = a,\\ +&u = f(r) &&\quad\text{`` } &&z = b, && u = 0 &&\quad\text{`` } &&r = c. +\end{alignat*} + +Then (v.\ Art.~123) +\[ + u = \sinh(\mu_k z) + \left [J_0(\mu_k r) - + \frac{J_0(\mu_kc)}{K_0(\mu_kc)}K_0(\mu_k r) + \right ] +\] +% -----File: 241.png +where $\mu_k$ is a root of the equation +\[ +J_0(\mu a)-\dfrac{J_0(\mu c)}{K_0(\mu c)} K_0(\mu a)=0 \tag{1} +\] +will satisfy Laplace's Equation [(1) Art.~123] and all of the equations of +condition except the second. +\begin{flalign*} +&\text{\indent Hence} & u&= \sum\limits^{k=\infty}_{k=1} A_k \dfrac{\sinh(\mu_k z)}{\sinh(\mu_k b)}\left[J_0(\mu_k r)-\dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K_0(\mu_k r)\right]&\phantom{Hence}\tag{2} +\intertext{is the required solution if} +&&&f(r) = \sum\limits^{k=\infty}_{k=1} A_k \bigg[J_0(\mu_k r)-\dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K_0(\mu_k r)\bigg]. \tag{3} +\end{flalign*} + +The development (3) is easily obtained. + +Call the parenthesis for the sake of brevity $B_0(\mu_kr)$. Then by the method +of Art.~125 we get if we integrate over our cylindrical shell +\[ +\int\limits^c_a rB_0(\mu_kr)B_0(\mu_lr)dr=0 \tag{4} +\] +if $\mu_k$ and $\mu_l$ are roots of (1); and by an easy extension of (10) Art.~122 +\[ + \int\limits^c_ar[B_0(\mu_kr]^2dr = + \tfrac{1}{2}\lbrace c^2[B'_0(\mu_kc)]^2-a^2[B'_0(\mu_ka)]^2\rbrace. \tag{5} +\] + +Determining the coefficients in (3) as in Art.~124 and simplifying by the +aid of (4) we have + +\noindent +\[ + A_k = \dfrac{\displaystyle 2\int\limits^c_a rf(r) + \Big[J_0(\mu_k r) - + \dfrac{J_0(\mu_kc)}{K_0(\mu_kc)}K_0(\mu_k r)\Big]dr} + {c^2\Big[J'_0(\mu_k c) - + \dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K'_0 + (\mu_k c)\Big]^2-a^2\Big[{J_0}'(\mu_k a) - + \dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K'_0(\mu_k a)\Big]^2}.\tag{6} +\] + +\EXAMPLE{} + +If a membrane bounded by concentric circles of radius $a$ and radius $b$, and +fastened at the edges, is initially distorted into a form symmetrical with respect +to the centre, and then allowed to vibrate +\[ +y= \sum\limits^{k=\infty}_{k=1} A_k \cos(\mu_kct)\left[J_0(\mu_k r)-\dfrac{J_0(\mu_kb)}{K_0(\mu_kb)} K_0(\mu_k r)\right] +\] +where $A_k$ is obtained from (6) Art.~127 by replacing $c$ by $b$. + +% -----File: 242.png +\mypara{128.} If in the cooling of a cylinder $u = 0$ when $z = 0$, $u = 0$ when $z = b$, +and $u = f(z)$ when $r = a$, the problem is easily solved. + +If in (2) and (3) Art.~123 $\mu$ is replaced by $\mu i$ we can readily obtain +\begin{flalign*} +& &z &= \sin (\mu z) J_0(\mu ri)&\\ +&\text{and} & z& = \cos (\mu z) J_0(\mu ri)&\phantom{and} +\end{flalign*} +as particular solutions of Laplace's Equation [(1) Art.~123]; and +\[ + J_0(xi)=1 + \dfrac{x^2}{2^2} + \dfrac{x^4}{2^2.4^2} + \dfrac{x^6}{2^2.4^2.6^2} +\cdots \tag{1} +\] +and is real. +\begin{flalign*} +&& &f(z)= \sum \limits^{k=\infty}_{k=1} A_k \sin \dfrac{k\pi z}{b} &&\\ +&\text{where} &&A_k = \dfrac{2}{b}\int \limits^b_a f(z) \sin \dfrac{k\pi z}{b} dz &&\tag{2} +\intertext{by Art.~31 (7) and (8).} +&\text{\indent Hence}&u&=\sum \limits^{k=\infty}_{k=1} A_k \sin \dfrac{k\pi z}{b} +\dfrac{ J_0\Big(\dfrac{k\pi ri}{b}\Big)} + {J_0\Big(\dfrac{k\pi ai}{b}\Big)} && \tag{3} +&\intertext{is our required solution.} +\end{flalign*} + +\EXAMPLE{S} + +1.\quad If the cylinder is hollow and we have $u = 0$ when $z = 0$, $u = 0$ when +$z = b$, $u = 0$ when $r = c$, and $u =f(z)$ when $r = a$; then +% recast to fit page width +\begin{multline*} +u=\sum\limits^{k=\infty}_{k=1} A_k \sin \dfrac{k\pi z}{b} +\left[ +\dfrac{ +J_0\Big(\dfrac{k\pi ri}{b}\Big)}{ +J_0\Big(\dfrac{k\pi ci}{b}\Big)} +-\dfrac{ +\overline{K_0}\Big(\dfrac{k\pi ri}{b}\Big)}{ +\overline{K_0}\Big(\dfrac{k\pi ci}{b}\Big)} +\right]\\ +\div +\left [ +\dfrac{ +J_0\Big(\dfrac{k\pi ai}{b}\Big)}{ +J_0\Big(\dfrac{k\pi ci}{b}\Big)} +-\dfrac{ +\overline{K_0}\Big(\dfrac{k\pi ai}{b}\Big)}{ +\overline{K_0}\Big(\dfrac{k\pi ci}{b}\Big)} +\right] +\end{multline*} +where $A_k$ has the value given in (2) Art.~128, and +\begin{align*} +\overline{K_0}(xi) &= K_0(xi) - J_0(xi) \log i\\ + &= J_0(xi) \log x - \dfrac{x^2}{2^2} - \dfrac{x^4}{2^2.4^2}(\tfrac{1}{1}+\tfrac{1}{2}) - \dfrac{x^6}{2^2.4^2.6^2}(\tfrac{1}{1}+\tfrac{1}{2}+\tfrac{1}{3}) -\cdots +\end{align*} +[v.\ (4) Art.~120], and is real.\\ + +2.\quad A hollow cylinder 6 feet long whose inner surface has the radius 3 inches, +and whose outer surface has the radius 1 foot, has its bases and outer surface +kept at the temperature 0°, and its inner surface at the temperature 100°, until +% -----File: 243.png +the permanent state of temperatures is established; find the temperatures of +two points in a plane parallel to the bases and half-way between them, one of +which is 6 inches and the other 9 inches from the axis. \hfill \emph{Ans.}, 49°.6; 20°.2. + +\mypara{129.} If in the problem of Art.~123 the temperatures of the points of the +upper base of the cylinder are unsymmetrical so that $u=f(r,\theta)$ when $z = b$, + we have to get particular solutions of Laplace's Equation [(1) Art.~123] for +the case where $D_\phi^2u$ is not equal to zero. We readily find that +\begin{flalign*} +&& u&=\sinh\;(\mu z)[A \cos n\phi +B \sin n\phi]J_n(\mu r)&\\[1ex] +&\text{and}&u&=\cosh\;(\mu z)[A \cos n\phi +B \sin n\phi]J_n(\mu r)&\phantom{and} +\end{flalign*} +are such solutions, and that +\[ +u= \sum\limits^{n=\infty}_{n=0} \sum\limits^{k=\infty}_{k=1} +\dfrac{\sinh \mu_{k} z}{\sinh \mu_k b} + [A_{n,k} \cos n\phi +B_{n,k} \sin n\phi]J_n(\mu_k r) \tag{1} +\] +is the solution of the given problem if +\[ +f(r,\phi) = \sum\limits^{n=\infty}_{n=0} \sum\limits^{k=\infty}_{k=1} + (A_{n,k} \cos n\phi +B_{n,k} \sin n\phi)J_n(\mu_k r) \tag{2} +\] +where $\mu_k$ is a root of the equation +\[ +\dfrac{J_n(\mu a)}{\mu^n a^n} = 0. \tag{3} +\] + +\EXAMPLE{S} + +1.\quad Show that +% recast to fit page width +\begin{align*} +\int\limits^a_0 +rJ_n(\mu_k r)J_n(\mu_l r)&dr\\[-2ex] +&=\dfrac{a}{{\mu_k}^2-{\mu_l}^2} +[\mu_l J_n(\mu_k a)J'_n(\mu_l a)-\mu_k J_n(\mu_l a)J'_n(\mu_k a)]\\ +&=\dfrac{a}{{\mu_k}^2-{\mu_l}^2} +[\mu_k J_n(\mu_l a)J_{n+1}(\mu_k a)-\mu_l J_n(\mu_k a)J_{n+1}(\mu_l a)]. +\end{align*} + +2.\quad Show that +% recast to fit page width +\begin{align*} +\int\limits^a_0 r [J_n(\mu_k r)]^2&dr\\[-2ex] +&=\dfrac{1}{2} \Big[a^2(J'_n(\mu_k a))^2+\Big(a^2 - \dfrac{n^2}{{\mu_k}^2}\Big) (J_n(\mu_k a))^2\Big]\\ +&=\dfrac{a^2}{2}[(J_n (\mu_k a))^2 + (J_{n+1} (\mu_k a))^2]- \dfrac{na}{\mu_k} J_n (\mu_k a)J_{n+1} (\mu_k a). +\end{align*} + +% -----File: 244.png +3.\quad Show that in Art.~129 +\begin{align*} +A_{0,k}&=\dfrac{1}{\pi} +\dfrac{\displaystyle \int\limits^{2\pi}_0d\phi\int\limits^a_0 +rf(r,\phi)J_0(\mu_kr)dr} +{a^2[J_1(\mu_ka)]^2},\\[0.5em] +B_{0,k}&=0,\\ +A_{n,k}&=\dfrac{2}{\pi} +\dfrac{\displaystyle\int\limits^{2\pi}_0 d\phi \int\limits^a_0 +rf(r,\phi)\cos n\phi J_n(\mu_k r)dr} +{a^2[J_{n+1}(\mu_ka)]^2},\\ +B_{n,k}&=\dfrac{2}{\pi} +\dfrac{\displaystyle\int\limits^{2\pi}_0 d\phi \int\limits^a_0 +rf(r,\phi)\sin n\phi J_n(\mu_kr)dr} +{a^2[J_{n+1}(\mu_ka)]^2}. +\end{align*} + +4.\quad Obtain the coefficients for the case where the convex surface of the +cylinder is impervious to heat.\\ + +5.\quad Obtain the coefficients for the case where the convex surface of the +cylinder is exposed to air at the temperature zero.\\ + +6.\quad Show that if in a drumhead problem of Art.~11 the initial distortion is +unsymmetrical, so that we have to solve the equation \smallrom{XI} Art.~1 subject to +the conditions $ z = f(r,\phi)$ when $t = 0$, $D_t z = 0$ when $t = 0$, $z = 0$ when $r = a$, +the solution is +\[ +z=\sum\limits^{n=\infty}_{n=0}\sum\limits^{k=\infty}_{k=1}\cos(\mu_k ct)(A_{n,k}\cos n\phi + B_{n,k}\sin n\phi) J_n(\mu_k r) +\] +where $A_{0,k}$, $B_{0,k}$, $A_{n,k}$, and $B_{n,k}$ have the values given in Ex.~3.\\ + +7.\quad What modifications do the statements made in Ex.~2, Art.~126, need to +make them apply to the unsymmetrical case treated in Ex.~6? + +Show that any possible nodal system in Ex.~6 is composed of concentric +circles and of radii whose outer extremities are equidistant. v.\ Rayleigh's +Sound, Vol.\ I., Arts.~(202-207).\\ + +8.\quad Solve the problem of Art.~127 and of Art.~127.\ Ex.~for the unsymmetrical +case. \quad \emph{Suggestion:} $AJ_n(x) + BK_n(x)$ is a solution of Bessel's +Equation.\\ + +9.\quad Solve the problem of Art.~128 and of Art.~128.\ Ex.~1.\ for the case where +$u = f(z,\phi)$ when $r = a$. \quad \emph{Suggestion:} $u = \sin \mu z (A \cos n\phi + B \sin n\phi) J_n(\mu ri)$ +is a solution of Laplace's Equation, and $f(z,\phi)$ can be developed into a double +Fourier's Series [v.\ (15) Art.~71].\\ +% -----File: 245.png + +10.\quad Show that in dealing with a wedge cut from a cylinder by planes +passed through the axis, or with a membrane in the form of a circular sector, +it may be necessary to use Bessel's Functions of fractional or incommensurable +orders.\\ + +11.\quad \textit{Bernouilli's Problem} (v.\ Chapter IX). In considering small transverse +vibrations of a uniform, heavy, flexible, inelastic string fastened at one end +and initially distorted into some given curve, we have to solve the equation + $D_t^2y=c^2(xD_x^2y+D_xy)$, subject to the conditions $D_ty=0$ when $t = 0$, + $y =f(x)$ when $t = 0$, $y = 0$ when $x = a$; the origin being taken at the +distance $a$ below the point of suspension and the axis of $X$ taken vertical. + +\begin{flalign*} +&\text{Show that} &&y=\sum_{k=1}^{k=\infty}A_k \cos\mu_k ct\,B_0(\mu_k^2x),&&\\ +&\text{where} & B_0(x)&=1-\frac{x}{1^2}+\frac{x^2}{1^2.2^2}-\frac{x^3}{1^2.2^2.3^2}+\cdots&&\phantom{where}\\[1ex] +&&&=J_0(2\sqrt{x})&& +\end{flalign*} +and $\mu_k$ is a root of the equation +\[B_0(\mu^2 a)=J_0(2\mu \sqrt{a})=0,\] +\begin{flalign*} +&\text{and}& A_k = \frac{\displaystyle\int\limits_0^af(x)B_0(\mu_k^2 x)dx}{\mu^2a^2[B'_0(\mu^2_ka)]^2}=\frac{\displaystyle\int\limits_0^af(x)J_0(2\mu_k\sqrt{x})dx}{a[J_1(2\mu_k\sqrt{a})]^2}.&&\\. +\end{flalign*} + +12.\quad As a simple case under Example 10 consider the vibrations of a circular +membrane fastened at the perimeter and also along a radius and then initially +distorted (v.\ Rayleigh's Sound, Art.~207). In this case we must modify the +formula given in Ex.~6 by dropping out the terms involving $\cos n\phi$ and by +taking $n = \dfrac{m}{2}$. The required solution is +\[ +z=\sum_{m=1}^{m=\infty}\sum_{k=1}^{k=\infty}B_{m,k}\cos\mu_kct\sin\frac{m\phi}{2}J_{\frac{m}{2}}(\mu_kr) +\] +\begin{flalign*} +&\text{where $\mu_k$ is a root of} &\frac{J_{\frac{m}{2}}(\mu a) }{\mu^{\frac{m}{2}}a^{\frac{m}{2}}}&=0&&\phantom{where mu is a root of} +\end{flalign*} +\begin{flalign*} +&\text{and} &B_{m,k}&=\frac{2}{\pi}\frac{\displaystyle\int\limits_0^{2\pi}d\phi\int\limits_0^a rf(r,\phi)\sin\frac{m\phi}{2}J_{\frac{m}{2}}(\mu_k r)dr}{a^2[{J_{\frac{m}{2}}}'(\mu_ka)]^2}.&&\phantom{and} +\end{flalign*} +% -----File: 246.png + +For the terms in which $m$ is odd, $J_{\frac{m}{2}}(x)$ can be readily obtained from (13) +Art.~122, which will become a finite sum. + +For example, (13) Art.~122 gives the values +\begin{gather*} +J_{\frac{1}{2}}(x)=\sqrt{\frac{2}{\pi x }}\sin x; \quad J_{\frac{3}{2}}(x) =\sqrt{\frac{2}{\pi x } }\bigg(\frac{1}{x} \sin x-\cos x \bigg);\\ +J_{\frac{5}{2}}(x)=-\sqrt{\frac{2}{\pi x }}\bigg[\bigg(1+\frac{3}{x^2}\bigg)\sin x+\frac{3}{x} \cos x \bigg]; \quad \text{\&c}. +\end{gather*} + +13.\quad The question of the flow of heat in three dimensions involves a problem +not unlike the last. + +Suppose the initial temperatures of all points in a sphere of radius $c$ given, +and let the surface be kept at the temperature zero. Then we have to solve +the equation +\[ +D_t u=\frac{a^2}{r^2}\biggr [D_r(r^2 D_r u) + \frac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta} u) + \frac{1}{\sin^2\theta} D_{\phi}^2 u\biggr] \tag{1} +\] +(\smallrom{IV} Art.~1) subject to the conditions +\begin{gather*} + u = 0 \quad \text{when}\quad r = c,\\ + u =f(r, \theta, \phi)\quad \text{when}\quad t = 0. +\end{gather*} +If we assume $u = T.R.V$ where $T$ is a function of $t$ only, $R$ of $ r$ only, and $V$ +of $\theta$ and $\phi$ only, (1) can be broken up into +\begin{gather*} +\frac{dT}{dt} + a^2 \alpha^2 T = 0 \tag{2}\\ + m(m+1)V + \frac{1}{\sin\theta }D_{\theta} (\sin\theta D_{\theta} V) + \frac{1}{\sin^2 \theta} D_{\phi}^2 V = 0 \tag{3} +\end{gather*} +\begin{flalign*} +&\text{and}&\frac{ d^2R}{dr^2} + \frac{2}{r }\frac{dR}{dr }\ + \biggr[\alpha^2 - \frac{m(m+1)}{r^2} \biggr ]R=0. && \tag{4} +\end{flalign*} +Hence $T = e^{-a^2 \alpha^2 t}$, $V=Y_m (\mu, \phi)$ [v.\ Art.~102 (2)], and $R$ is still to be found. +If in (4) we let $x=\alpha r$ and $z=R\sqrt{\alpha r}$ it becomes +\[ +\frac{d^2z}{dx^2} + \frac{1}{x} \frac{dz}{dx} + \biggr [ 1 - \frac{(m+\frac{1}{2})^2 }{x^2} \biggr ] z=0 +\] +which is satisfied by $z=J_{m+\frac{1}{2}} (x)$. (v.\ Art.~17.) +\begin{flalign*} +&\text{\indent Therefore}& R= \frac{1}{\sqrt{\alpha r }}J_{m+\frac{1}{2}} (\alpha r).&&\phantom{\indent Therefore} +\end{flalign*} +% -----File: 247.png +\begin{flalign*} +&&f(r,\theta,\phi)&=\dfrac{1}{4\pi}\sum\limits^{m=\infty}_{m=0} (2m+1) \int\limits^{2\pi}_0 d\phi_1 \int\limits^{\pi}_0 f(r,\theta_1,\phi_1)P_m(\cos \gamma)\sin\theta_1 d\theta_1&\tag*{by (3) Art.~114,}\\ +&&&=\sum\limits^{m=\infty}_{m=0} \sum\limits^{n=m}_{n=0} [A_{m,n}f_{m,n}(r)\cos n\phi + B_{m,n}F_{m,n}(r)\sin n\phi]P_m^n(\mu).& +\end{flalign*} +\[ +\sqrt{r}f_{m,n}(r)=\sum\limits^{k=\infty}_{k=0} C_{m,n,k}J_{m+\frac{1}{2}}(\alpha_kr) +\] +where $\alpha_k$ is a root of the equation +\[ +\dfrac{J_{m+\frac{1}{2}}(\alpha c)}{(\alpha c)^{m+\frac{1}{2}}}=0 +\] +\begin{flalign*} +&\text{and} &&C_{m,n,k}= \dfrac{2\displaystyle \int\limits^c_0 r^{\frac{3}{2}}f_{m,n}(r)J_{m+\frac{1}{2}}(\alpha_kr)dr}{c^2[J'_{m+\frac{1}{2}}(\alpha_kc)]^2}.&&\\ +& &&\sqrt{r}F_{m,n}(r)=\sum\limits^{m=\infty}_{m=0} D_{m,n,k}J_{m+\frac{1}{2}}(\alpha_kr)\\ +&\text{where }&& D_{m,n,k}= \dfrac{2\displaystyle\int\limits^c_0 r^{\frac{3}{2}}F_{m,n}(r)J_{m+\frac{1}{2}}(\alpha_kr)dr}{c^2[J'_{m+\frac{1}{2}}(\alpha_kc)]^2}.&&\phantom{where} +\end{flalign*} + +The final solution is +\begin{multline*} +u=\dfrac{1}{\sqrt{r}}\sum\limits^{m=\infty}_{m=0} \sum\limits^{n=m}_{n=0} \bigg[P_m^n(\mu)\sum\limits^{k=\infty}_{k=1} (A_{m,n}C_{m,n,k}\cos n\phi\\ + \vphantom{ \sum\limits^{n=m}_{n=0}} + +B_{m,n}D_{m,n,k} \sin n\phi)e^{-a^2\alpha_k^2 t} J_{m+\frac{1}{2}}(\alpha_kr)\bigg] +\end{multline*} +cf.\ Riemann, Par.\ Dif.\ Gl., \S\S~72 and 73. +\label{ch7end} + +% -----File: 248.png +\mychap{CHAPTER VIII.}{LAPLACE'S EQUATION IN CURVILINEAR COÖRDINATES. ELLIPSOIDAL HARMONICS.} +\label{ch8start} + +\mypara{130.} \textit{Orthogonal Curvilinear Coördinates.} + +\begin{flalign*} +&\raisebox{3.5ex}{If}& +\left. +\begin{gathered} +F_1(x,y,z) =\rho_1\\ +F_2(x,y,z) = \rho_2\\ +F_3(x,y,z) = \rho_3 +\end{gathered} +\right\}&&\tag{1} +\end{flalign*} +are the equations in rectangular coördinates of three surfaces that are mutually +perpendicular no matter what the values of $\rho_1$, $\rho_2$, and $\rho_3$, the parameters $\rho_1$, +$\rho_2$, and $\rho_3$, may be regarded as a set of coördinates for a point of intersection +of the three surfaces, in the sense that when $\rho_1$, $\rho_2$, $\rho_3$ are given the point in +question is determined, and when the point is given the corresponding values +of $\rho_1$, $\rho_2$, $\rho_3$, can be found. + +From equations (1) $x$, $y$, and $z$ can be expressed in terms of $\rho_1$, $\rho_2$, and $\rho_3$. +Suppose this done. If now $x$, $y$, $z$ are the rectangular coördinates of the +point $\rho_1=a$, $\rho_2=b$, $\rho_3=c$, the rectangular coördinates of the points +$\rho_1=a+d\rho_1$, $\rho_2=b$, $\rho_3=c$, are obviously $x+D_{\rho_1}x.d\rho_1+\epsilon_1$, $y+D_{\rho_1} y.d\rho_1+\epsilon_2$, +$z+D_{\rho_1}z.d\rho_1+\epsilon_3$, where $\epsilon_1$, $\epsilon_2$, and $\epsilon_3$ are infinitesimals of higher order than +$d\rho_1$. Hence the square of the distance between the points will differ by an +infinitesimal of higher order than that of $d\rho_1^2$ from $dn_1^2$ where +\[ +dn_1^2=[(D_{\rho_1}x)^2+(D_{\rho_1}y)^2+(D_{\rho_1}z)^2]d\rho_1 ^2. +\] +\begin{flalign*} +&\raisebox{6.5ex}{\indent Let}& +\left. +\begin{aligned} +\frac{1}{h_1^2}&=(D_{\rho_1}x)^2+(D_{\rho_1}y)^2+(D_{\rho_1}z)^2\\ +\frac{1}{h_2^2}&=(D_{\rho_2}x)^2+(D_{\rho_2}y)^2+(D_{\rho_2}z)^2\\ +\frac{1}{h_3^2}&=(D_{\rho_3}x)^2+(D_{\rho_3}y)^2+(D_{\rho_3}z)^2. +\end{aligned} +\right\}&&\tag{2} +\end{flalign*} + +Then if $dn_1$ is the element of length normal to the surface $\rho_1=a$, $dn_2$ +normal to $\rho_2=b$, and $dn_3$ normal to $\rho_3=c$ +\[ +dn_1=\frac{d\rho_1}{h_1},\quad dn_2=\frac{d\rho_2}{h_2},\quad dn_3=\frac{d\rho_3}{h_3}. \tag{3} +\] +% -----File: 249.png +The element of surface $dS_1$ on the surface $\rho_1=a$ is easily seen to be +\[ +dS_1=\frac{d\rho_2d\rho_3}{h_2h_3}; \tag{4} +\] +and the element of volume $dv$ is +\[ +dv=\frac{d\rho_1d\rho_2d\rho_3}{h_1h_2h_3}. \tag{5} +\] + +\EXAMPLE{} + +\begin{flalign*} +&\text{\indent Show that} & + h_1^2&=(D_x\rho_1)^2+(D_y\rho_1)^2+(D_z\rho_1)^2&\phantom{\indent Show that}\\ +&&h_2^2&=(D_x\rho_2)^2+(D_y\rho_2)^2+(D_z\rho_2)^2&\\ +&&h_3^2&=(D_x\rho_3)^2+(D_y\rho_3)^2+(D_z\rho_3)^2.& +\end{flalign*} + +\textit{Suggestion:} If $h_1$ has the value just given $\dfrac{D_x\rho_1}{h_1}$, $\dfrac{D_y\rho_1}{h_1}$, $\dfrac{D_z\rho_1}{h_1}$ are the +direction cosines of the normal at any given point of $\rho_1=a$. (v.\ Int.\ Cal.\ +page 161.) Then +\[ +dn_1 = \frac{D_x\rho_1}{h_1}dx + + \frac{D_y\rho_1}{h_1}dy + + \frac{D_z\rho_1}{h_1}dz += \frac{1}{h_1}d\rho_1. +\] + +\markright{ELLIPSOIDAL HARMONICS.} +\mypara{131.} \textit{Laplace's Equation in orthogonal curvilinear coördinates.} + +If we apply the special form of Green's Theorem +\[ +\iiint\nabla^2Vdxdydz=\int D_nVdS \tag{v.\ Art.~98} +\] +to the space bounded by the surfaces $\rho_1=a$, $\rho_2=b$, $\rho_3=c$, $\rho_1=a+d\rho_1$, + $\rho_2=b+d\rho_2$, $\rho_3=c+d\rho_3$, we have +% recast to fit page width +\begin{align*} +\frac{\nabla^2Vd\rho_1d\rho_2d\rho_3}{h_1h_2h_3}&=\\ +-h_1D_{\rho_1}&V\frac{d\rho_2d\rho_3}{h_2h_3}+h_1D_{\rho_1}V\frac{d\rho_2d\rho_3}{h_2h_3}+D_{\rho_1}\left(\frac{h_1}{h_2h_3}D_{\rho_1}V\right)d\rho_1d\rho_2d\rho_3\\ +-h_2D_{\rho_2}&V\frac{d\rho_3d\rho_1}{h_3h_1}+h_2D_{\rho_2}V\frac{d\rho_3d\rho_1}{h_3h_1}+D_{\rho_2}\left(\frac{h_2}{h_3h_1}D_{\rho_2}V\right)d\rho_1d\rho_2d\rho_3\\ +-h_3D_{\rho_3}&V\frac{d\rho_1d\rho_2}{h_1h_2}+h_3D_{\rho_3}V\frac{d\rho_1d\rho_2}{h_1h_2}+D_{\rho_3}\left(\frac{h_3}{h_1h_2}D_{\rho_3}V\right)d\rho_1d\rho_2d\rho_3; +\end{align*} +whence +\[ +\nabla^2V=h_1h_2h_3\bigg[ + D_{\rho_1}\bigg(\frac{h_1}{h_2h_3}D_{\rho_1}V\bigg) ++ D_{\rho_2}\bigg(\frac{h_2}{h_3h_1}D_{\rho_2}V\bigg) ++ D_{\rho_3}\bigg(\frac{h_3}{h_1h_2}D_{\rho_3}V\bigg)\bigg], \tag{6} +\] +and Laplace's Equation in our curvilinear system is +\[ +h_1h_2h_3\bigg[ + D_{\rho_1}\bigg(\frac{h_1}{h_2h_3}D_{\rho_1}V\bigg) ++ D_{\rho_2}\bigg(\frac{h_2}{h_3h_1}D_{\rho_2}V\bigg) ++ D_{\rho_3}\bigg(\frac{h_3}{h_1h_2}D_{\rho_3}V\bigg)\bigg]=0. \tag{7} +\] + +% -----File: 250.png +If it happens that $\nabla^2\rho_1=0$, $V=\rho_1$ will satisfy (7) and we shall have + $h_1h_2h_3D_{\rho_1}\bigg(\dfrac{h_1}{h_2h_3}\bigg)=0$. In like manner if $\nabla^2\rho_2=0$ + we have $D_{\rho_2}\bigg(\dfrac{h_2}{h_3h_1}\bigg)=0$, + and if $\nabla^2\rho_3=0$ we have $D_{\rho_3}\bigg(\dfrac{h_3}{h_1h_2}\bigg)=0$; and therefore (7) reduces to +\[ +h_1^2 D_{\rho_1}^2V + h_2^2 D_{\rho_2}^2V + h_3^2 D_{\rho_3}^2V=0\tag{8} +\label{err250} +\] +when $\nabla^2\rho_1=0$, $\nabla^2\rho_2=0$, and $\nabla^2\rho_3=0$. + +\mypara{132.} If instead of having the value of the Potential Function $V$ given on +the surface of a sphere as in our Spherical Harmonic problem, we have it +given at all the points on the surface of an \emph{oblate spheroid}, and are required to +find its value at any internal or external point, we can easily get a solution by +methods in no essential respect different from those already employed, if only +we rightly choose our system of coördinates. + +If we take an ellipse and an hyperbola having the same foci, and revolve +them about the minor axis of the ellipse, we shall get a pair of surfaces which +are mutually perpendicular; a plane through the axis of revolution will cut +both the \emph{spheroid} and the \emph{hyperboloid} orthogonally. + +The equations of the three surfaces can be written:--- +\begin{gather*} +F_1(x,y,z,\lambda)=\frac{x^2}{\lambda^2}+\frac{y^2}{\lambda^2-b^2}+\frac{z^2}{\lambda^2}-1=0 \tag{1}\\ +F_2(x,y,z,\mu) =\frac{x^2}{\mu^2} +\frac{y^2}{\mu^2 -b^2}+\frac{z^2}{\mu^2} -1=0 \tag{2}\\ +F_3(x,y,z,\nu) =z - \nu x = 0, \tag{3} +\end{gather*} +where $\lambda^2>b^2>\mu^2$, $2b$ being the distance between the foci. + +For all values of $\lambda$, $\mu$, and $\nu$ consistent with the inequality above written +the surfaces (1), (2), (3) intersect in real points and cut orthogonally. + +$\lambda$, $\mu$, and $\nu$ can be so chosen that the surfaces will intersect in any given +point, and therefore can be taken as a set of curvilinear coördinates, and +Laplace's Equation can be expressed in terms of them by the aid of Formula +\smallrom{XV} Art.~1. + +From (1), (2), and (3) we readily get +\begin{gather*}\left. +\begin{split} +x^2 &= \frac{\lambda^2\mu^2}{b^2(1+\nu^2)}\\ +y^2 &= \frac{(\lambda^2-b^2)(b^2-\mu^2)}{b^2}\\ +z^2 &= \frac{\lambda^2\mu^2\nu^2}{b^2(1+\nu^2)}; +\end{split}\right\}\tag{4} +\end{gather*} +% -----File: 251.png +whence \quad $D_\lambda x=\dfrac{\mu}{b\sqrt{1+\nu^2}}$, + \quad $D_\lambda y=\dfrac{\lambda}{b}\sqrt{\dfrac{b^2-\mu^2}{\lambda^2-b^2}}$, + \quad $D_\lambda z=\dfrac{\mu\nu}{b\sqrt{1+\nu^2}}$; +\begin{flalign*} + &\text{and} & \frac{1}{h_1^2} = + \frac{\lambda^2-\mu^2}{\lambda^2-b^2}& \tag{5}& +\end{flalign*} +[v.~130~(2)]. In like manner we get +\[ + \frac{1}{h_2^2} = + \frac{\lambda^2-\mu^2}{b^2-\mu^2} \tag{6} +\] +\begin{flalign*} + &\text{and} & \frac{1}{h_3^2} = + \frac{\lambda^2\mu^2}{b^2(1+\nu^2)^2}, &\tag{7}& +\end{flalign*} +and \smallrom{XV} Art.~1 becomes +% recast to fit page width +\begin{align*} + &\frac{\mu}{b(1+\nu^2)\sqrt{b^2-\mu^2}}D_\lambda[\lambda\sqrt{\lambda^2-b^2}.D_\lambda V]\\ + + &\frac{\lambda}{b(1+\nu^2)\sqrt{\lambda^2-b^2}}D_\mu[\mu\sqrt{b^2-\mu^2}.D_\mu V]\\ + + &\frac{b(\lambda^2-\mu^2)}{\lambda\mu\sqrt{(\lambda^2-b^2)(b^2-\mu^2)}}D_\nu[(1+\nu^2)D_\nu V]=0, +\tag{8} +\end{align*} +which is Laplace's Equation in terms of our \emph{Spheroidal Coördinates} $\lambda$, $\mu$, and $\nu$. + +If now in place of $\lambda$, $\mu$, and $\nu$ we can introduce some function of $\lambda$, some +function of $\mu$, and some function of $\nu$ which, therefore, will represent the +same set of orthogonal surfaces, and if we can choose these functions $\alpha$, $\beta$, +and $\gamma$, which of course are functions of $x$, $y$, and $z$, so that $\nabla^2\alpha=0$, + $\nabla^2\beta=0$, and $\nabla^2\gamma=0$, equation (8) must reduce to the simple and symmetrical +form given in \smallrom{XVI} Art.~1. + +These functions $\alpha$, $\beta$, and $\gamma$ are easily found. Equation (8) is $\nabla^2V=0$ + expressed in terms of $\lambda$, $\mu$, and $\nu$. Assume that $V$ is a function of $\lambda$ only; +then $D_\mu V=0$, and $D_\nu V=0$, and (8) reduces to +\begin{flalign*} +&&D_\lambda[&\lambda\sqrt{\lambda^2-b^2}.D_\lambda V]=0&&\\ +&\text{whence} && \lambda\sqrt{\lambda^2-b^2}\ \frac{dV}{d\lambda}=c_1,&&\phantom{whence}\\ +&&&\hspace{0.5em}dV=\frac{c_1d\lambda}{\lambda\sqrt{\lambda^2-b^2}},&&\\ +&\text{and} & &\hspace{1em}V=\frac{c_1}{b}\sec^{-1}\frac{\lambda}{b},&& +\end{flalign*} +and is a function of $\lambda$ which satisfies Laplace's Equation. + +% -----File: 252.png +Take this as $\alpha$ leaving $c_1$ at present undetermined, so that +\[ +d\alpha=\frac{c_1d\lambda}{\lambda\sqrt{\lambda^2-b^2}}\quad \text{and}\quad \alpha=\frac{c_1}{b}\sec^{-1}\frac{\lambda}{b}. +\] + +In the same way we get +\[ +d\beta=\frac{c_2d\mu}{\mu\sqrt{b^2-\mu^2}}\quad \text{and}\quad \beta=\frac{c_2}{b}\sech^{-1}\frac{\mu}{b}, +\] +(v.\ Int.\ Cal.\ Art.~46, Ex.) +\[ +d\gamma=\frac{c_3d\nu}{1+\nu^2},\quad \text{and}\quad \gamma=c_3\tan^{-1}\nu. +\] +Substituting these values in (8) and taking $c_1=-c_2=b$, and $c_3=1$, + (8) reduces at once to +\[ +\frac{D_\alpha^2V}{\lambda^2}+\frac{D_\beta^2V}{\mu^2}+\frac{\lambda^2-\mu^2}{\lambda^2\mu^2}D_\gamma^2V=0, \tag{9}\\[-4ex] +\] +\begin{flalign*} +&\text{or since} & \lambda&=b\sec\alpha,\quad \mu=b\sech\beta,\quad \text{and}\quad \nu=\tan\gamma, & \tag{10}&\\[1ex] +&\text{to} & \cos^2\alpha &D_\alpha^2V+\cosh^2\beta D_\beta^2V+(\cosh^2\beta-\cos^2\alpha)D_\gamma^2V=0 & \tag{11}& +\end{flalign*} +which is Laplace's Equation in terms of what we may call \emph{Normal Oblate +Spheroidal Coördinates}. + +In using (11) it is to be noted that the point whose coördinates are $(\alpha, \beta, \gamma)$ +is the point of intersection of an oblate spheroid whose semi-axes are $b\sec\alpha$ +and $b\tan\alpha$, an unparted hyperboloid of revolution whose semi-axes are +$b\sech\beta$ and $b\tanh\beta$, and a plane containing the axis of the system and +making the angle $\gamma$ with a fixed plane; and that if the axis of revolution is +the axis of $Y$ and the fixed plane is the plane of $XY$, the rectangular coördinates +of $(\alpha, \beta, \gamma)$ are +\[ +x=b\sec\alpha\sech\beta\cos\gamma,\quad y=b\tan\alpha\tanh\beta,\quad z=b\sec\alpha\sech\beta\sin\gamma \tag{12} +\] +[v.~(4)]. + +If now we let $\alpha$ range from 0 to $\dfrac{\pi}{2}$, $\beta$ from $-\infty$ to $\infty$, and $\gamma$ from 0 to $2\pi$, +we shall be able to represent all points in space; and if we agree that negative +values of $\beta$ shall belong to points below a plane through the origin and +perpendicular to the axis of revolution and positive values of $\beta$ to points +above that plane, not only shall we have no ambiguity, but also the rectangular +coördinates of any point as given in (12) will have their proper signs. +% -----File: 253.png + +\EXAMPLE{S} + +1.\quad If the spheroid is a \emph{prolate} spheroid, the ellipse and confocal hyperbola +must be revolved about the major axis of the ellipse, and the plane must contain +that axis. In place of equations (1), (2), and (3) of Art.~132 we have, +then, +\begin{align*} +\frac{x^2}{\lambda^2}+\frac{y^2}{\lambda^2-b^2}+\frac{z^2}{\lambda^2-b^2}-1&=0\\ +\frac{x^2}{\mu^2}+\frac{y^2}{\mu^2-b^2}+\frac{z^2}{\mu^2-b^2}-1&=0 +\end{align*} +\[ +z-\nu y=0 +\] +\begin{flalign*} +&\text{where} & \lambda^2>b^2>\mu^2. &&\phantom{where} +\end{flalign*} +\[ +h_1^2=\frac{\lambda^2-b^2}{\lambda^2-\mu^2},\quad h_2^2=\frac{b^2-\mu^2}{\lambda^2-\mu^2},\quad h_3^2=\frac{b^2(1+\nu^2)^2}{(\lambda^2-b^2)(b^2-\mu^2)}. +\] +Laplace's Equation becomes +\begin{align*} + \frac{1}{b^2(1+\nu^2)}D_\lambda[(\lambda^2-b^2)D_\lambda V] +&+\frac{1}{b^2(1+\nu^2)}D_\mu[(b^2-\mu^2)D_\mu V]\\ +&+\frac{\lambda^2-\mu^2}{(\lambda^2-b^2)(b^2-\mu^2)}D_\nu[(1+\nu^2)D_\nu V]=0. \tag{1} +\end{align*} +\begin{flalign*} +&\text{(1) reduces to} \quad \frac{D_\alpha^2V}{\lambda^2-b^2} + + \frac{D_\beta^2V}{b^2-\mu^2} + + \frac{\lambda^2-\mu^2}{(\lambda^2-b^2)(b^2-\mu^2)}D_\gamma^2V=0, &\hfill\tag{2}\\[-5ex] +\end{flalign*} +\begin{flalign*} +&\text{where} && d\alpha=-\frac{bd\lambda}{\lambda^2-b^2},\quad + d\beta=\frac{bd\mu}{b^2-\mu^2},\quad + d\gamma=\frac{d\nu}{1+\nu^2},&&\phantom{where} +\end{flalign*} +\[ + \alpha=\ctnh^{-1}\frac{\lambda}{b},\quad + \beta =\tanh^{-1}\frac{\mu}{b},\quad \text{and}\quad + \gamma=\tan^{-1}\nu. +\] +\begin{flalign*} +&\indent\text{Since} & \lambda=b\ctnh\alpha,\quad + \mu =b\tanh\beta,\quad \text{and}\quad + \nu = \tan\gamma&&\phantom{\indent Since} +\end{flalign*} +(2) can be reduced to +\[ +\sinh^2\alpha D_\alpha^2V +\cosh^2\beta D_\beta^2 V + (\sinh^2\alpha + \cosh^2\beta)D_\gamma^2 V=0. \tag{3} +\] + +In using (3) it is to be noted that the point $(\alpha, \beta, \gamma)$ is the point of intersection +of a prolate spheroid whose semi-axes are $b\ctnh\alpha$ and $b\csch\alpha$, a +biparted hyperboloid of revolution whose semi-axes are $b\tanh\beta$ and $b\sech\beta$, + and a plane containing the axis of revolution and making the angle $\gamma$ with a +fixed plane. +% -----File: 254.png + +If the fixed plane is that of ($XY$) the rectangular coördinates of any point +$(\alpha, \beta, \gamma)$ are +\[ +x = b \ctnh \alpha \tanh \beta, \quad y = b \csch \alpha \sech \beta \cos \gamma, \quad z = b \csch \alpha \sech \beta \sin\gamma, +\] +and $\alpha$ may range from $\infty$ to 0, $\beta$ from $-\infty$ to $\infty$, and $\gamma$ from 0 to $2\pi$. +Negative values of $\beta$ are to be taken for points lying to the left of a plane +through the origin perpendicular to the axis of revolution.\\ + +2.\quad Transform Laplace's Equation in Spherical Coördinates \smallrom{XIII} Art.~1 +to the symmetrical form +\[ +\alpha^2D_\alpha^2V + \cosh^2\beta D_\beta^2V + \cosh^2\beta D_\gamma^2V = 0 +\] +\begin{flalign*} +&\text{where}&& \alpha = \dfrac{1}{r},\quad \beta =\log\tan \dfrac{\theta}{2},\quad \text{and}\quad \gamma = \phi.&&\phantom{where} +\end{flalign*} + +3.\quad Transform Laplace's Equation in Cylindrical Coördinates \smallrom{XIV} Art.~1 +to the symmetrical form +\[ +D_\alpha^2V + D_\beta^2V + e^{2\alpha}D_\gamma^2V = 0 +\] +\begin{flalign*} +&\text{where}& &\alpha = \log r, \quad \beta = \phi, \quad \text{and}\quad \gamma = z.&&\phantom{where} +\end{flalign*} + +\mypara{133.} In each of the cases we have considered, it has been easy to pass +from Laplace's Equation in terms of the chosen coördinates representing an +orthogonal system of surfaces to the symmetrical form \smallrom{XVI} Art.~1; and it is +evident that our new coördinate $\alpha$ is a value of $V$ corresponding to such a +distribution that the surfaces obtained by giving particular values to $\rho_1$ are +\emph{equipotential} surfaces; that $\beta$ is a value of $V$ corresponding to such a +distribution that the surfaces obtained by giving particular values to $\rho_2$ are +equipotential surfaces; and that $\gamma$ is a value of $V$ corresponding to such a +distribution that the surfaces obtained by giving particular values to $\rho_3$ are +equipotential surfaces. $\alpha$, $\beta$, and $\gamma$ are called by Lamé ``\emph{thermometric +parameters}.'' + +The condition that these values should exist, for a given system of surfaces, +that is, that the distribution described above should be possible, is readily +obtained. We shall work it out for $\alpha$. It is merely the condition that $V$ in +Laplace's Equation may be a function of $\rho_1$ alone. + +If $V$ is a function of $\rho_1$ alone +\[ +D_xV = \dfrac{dV}{d\rho_1}D_x\rho_1, \quad D_yV = \dfrac{dV}{d\rho_1}D_y\rho_1,\quad D_zV = \dfrac{dV}{d\rho_1}D_z\rho_1,\\[-2ex] +\] +\begin{align*} +D_x^2 V &= \dfrac{d^2 V}{{d \rho_1}^2} (D_x \rho_1)^2 + \dfrac{dV}{d \rho_1} D_x^2 \rho_1\\ +% -----File: 255.png +D_y^2 V &= \dfrac{d^2 V}{{d \rho_1}^2} (D_y \rho_1)^2 + \dfrac{dV}{d \rho_1} D_y^2 \rho_1\\ +D_z^2 V &= \dfrac{d^2 V}{{d \rho_1}^2} (D_z \rho_1)^2 + \dfrac{dV}{d \rho_1} D_z^2 \rho_1. +\end{align*} +Therefore $[(D_x \rho_1)^2 + (D_y \rho_1)^2 + (D_z \rho_1)^2]\dfrac{d^2 V}{d {\rho_1}^2} + [D_x^2 \rho_1 + D_y^2 \rho_1 + D_z^2 \rho_1]\dfrac{ dV}{d \rho_1} = 0$ +\begin{flalign*} +&\text{whence}& &\dfrac{D_x^2 \rho_1 + D_y^2 \rho_1 + D_z^2 \rho_1}{(D_x \rho_1)^2 + (D_y \rho_1)^2 + (D_z \rho_1)^2} = - \dfrac{d^2V}{d{\rho_1}^2}\div \dfrac{ dV}{d \rho_1}.&&\phantom{whence} +\end{flalign*} +\begin{flalign*} +&\text{or} & &\dfrac{\nabla^2\rho_1}{h_1^2} = F_1 (\rho_1)&&\phantom{or} +\end{flalign*} +where $F_1(\rho_1)$ may be any function of $\rho_1$ alone. Our required conditions are +then +\[ +\left. +\begin{aligned} +\dfrac{\nabla^2 \rho_1}{h_1^2} &= F_1 (\rho_1)\\ +\dfrac{\nabla^2 \rho_2}{h_2^2} &= F_2 (\rho_2)\\ +\dfrac{\nabla^2 \rho_3}{h_3^2} &= F_3 (\rho_3) +\end{aligned} +\right\} +\tag{1} +\] +and when they are fulfilled the original curvilinear coördinates $\rho_1$, $\rho_2$, $\rho_3$, +correspond to possible \emph{equipotential} or \emph{isothermal} surfaces, \emph{thermometric +parameters} $\alpha$, $\beta$, and $\gamma$ exist, and the reduction of Laplace's Equation to the +symmetrical form \smallrom{XVI} Art.~1 is possible. + +\mypara{134.} Returning to our Oblate Spheroid problem of Art.~132 we can proceed +as usual to break up our equation (\textbf{11}) Art.~132. + +Assume that $V = L.M.N$, where $L$ is a function of $\alpha$ only, $M$ of $\beta$ only, +and $N$ of $\gamma$ only. (11) Art.~132 becomes +\begin{flalign*} +&&\dfrac{\cos^2 \alpha}{L}\dfrac{ d^2L}{d \alpha^2} + \dfrac{\cosh^2 \beta}{M}\dfrac{ d^2 M}{ d \beta^2} +\dfrac{[\cosh^2 \beta - \cos^2 \alpha]}{N}\dfrac{ d^2N}{d\gamma^2} &= 0\\ +&\text{or }& \dfrac{1}{L}\dfrac{\cos^2 \alpha}{ \cosh^2 \beta - \cos^2 \alpha} \dfrac{d^2L}{d \alpha^2} + \dfrac{1}{M}\dfrac{\cosh^2 \beta}{ \cosh^2 \beta - \cos^2 \alpha} \dfrac{d^2M}{d \beta^2} = -&\dfrac{ 1}{N}\dfrac{ d^2 N}{d \gamma^2}.&&\phantom{or} +\end{flalign*} + +The first member is independent of $\gamma$, and the second member is independent +of $\alpha$ and $\beta$, and the two members are identically equal. The second member +is then independent of $\alpha$, $\beta$, and $\gamma$ and must be constant; call it $n^2$. We have, +then, +\begin{flalign*} +&\rule{14em}{0em} \dfrac{d^2N}{d \gamma^2} + n^2 N = 0 \tag{1}&\\ +% -----File: 256.png +&\text{and} \rule{4em}{0em} \dfrac{ \cos^2\alpha}{L}\dfrac{ d^2L}{d\alpha^2} + \dfrac{\cosh^2\beta}{M}\dfrac{ d^2M}{d\beta^2} - n^2(\cosh^2\beta - \cos^2\alpha) = 0.&&&(2)\\ +&\text{(1) gives us} \rule{6.5em}{0em} N = A \cos n\gamma + B \sin n\gamma. && &\tag{3} +\end{flalign*} +(2) can be written +\begin{flalign*} +&&\dfrac{\cos^2\alpha}{L}\dfrac{ d^2L}{d\alpha^2} + n^2\cos^2&\alpha = n^2\cosh^2\beta - \dfrac{\cosh^2\beta}{M}\dfrac{d^2M}{d\beta^2} = m(m+1), &\quad\\ +&\text{whence}&\phantom{c}\cos^2\alpha \dfrac{d^2L}{d\alpha^2}& + [n^2\cos^2\alpha - m(m+1)]L = 0&\tag{4}\\ +&\text{and} &\cosh^2\beta \dfrac{d^2M}{d\beta^2} &+ [m(m+1) - n^2\cosh^2\beta]M = 0. &\tag{5} +\end{flalign*} + +If we introduce $x = \tanh\beta$ in (5) it becomes +\[ +(1-x^2)\dfrac{d^2M}{dx^2} - 2x \dfrac{dM}{dx} + \left[m(m+1) - \dfrac{n^2}{1-x^2}\right]M = 0 \tag{6} +\] +where since $x = \tanh\beta$ and $\beta$ may have any value from $-\infty$ to $\infty$, $x$ may +have any value between $-1$ and 1. (6) is a familiar equation having for a +particular solution +\[ +M = (1-x^2)^{\frac{n}{2}}\dfrac{ d^nP_m(x)}{dx^n} = P_m^n(x) = P_m^n(\tanh\beta).\tag{7} +\] +(v.\ Arts.~101 and 102). + +If we introduce in (4) $x = \tan\alpha$ it reduces to +\[ +(1+x^2)\dfrac{d^2L}{dx^2} + 2x \dfrac{dL}{dx} + \left[\dfrac{n^2}{1+x^2} - m(m+1)\right]L = 0.\tag{8} +\] +(8) is an unfamiliar equation, but it can be treated as (6) was treated if we +take the pains to go back to the beginning and follow the steps of the treatment +of Legendre's Equation. + +This labor can be saved, however, by noting that if we let $x = \dfrac{y}{i}$ (8) becomes +\[ +(1-y^2)\dfrac{ d^2L}{dy^2} - 2y\dfrac{ dL}{dy} + \left[m(m+1) - \dfrac{n^2}{1-y^2}\right]L = 0 +\] +and is identical in form with (6). Hence +\begin{flalign*} +&&\rule{4em}{0em}L &= P_m^n(y)\quad\text{and}\quad L = (1-y^2)^{\frac{n}{2}}\dfrac{d^nQ_m(y)}{dy^n}&\text{(v.\ Art.~101)}, +\end{flalign*} +where $y = i \tan\alpha$, are particular solutions of (4). + +We can avoid imaginaries if we use the values +\[ +L = (-i)^{m-n} P_m^n(y)\quad \text{and}\quad L = i^{m+n+1}(1-y^2)^{\frac{n}{2}}\dfrac{ d^nQ_m(y)}{dy^n}.\qquad \tag{9} +\] +% -----File: 257.png +Since we assumed $V= L.M.N$ we have +\begin{flalign*} +&\raisebox{-1.7ex}{and}& +\left. +\begin{aligned} +V &= (A \cos n \gamma + B \sin n \gamma) P_m^n(\tanh \beta)(-i)^{m-n} P_m^n (i \tan \alpha)\\ +V &= (A \cos n \gamma + B \sin n \gamma) P_m^n(\tanh \beta) i^{m+n+1} \sec^n \alpha \dfrac{d^n Q_m (i \tan \alpha)}{ (d(i \tan \alpha))^n} +\end{aligned} +\right\} (10)\\[-6ex] +\end{flalign*} +as particular solutions of (\textbf{11}) Art.~132. + +If the problem is symmetrical with respect to the axis of the spheroid + $D_\gamma^2 V = 0$, $n^2 = 0$ and our particular solutions (10) reduce to +\begin{flalign*} +&\raisebox{-2ex}{and}& +\left. +\begin{aligned} +V& = (-i)^m P_m (i \tan \alpha) P_m (\tanh \beta)&\\ +V&= i^{m+1} Q_m (i \tan \alpha) P_m (\tanh \beta). +\end{aligned} +\right\}&&\tag{11} +\end{flalign*} + +If, then, $V$ is given on the surface of a spheroid as a function of $\beta$ and $\gamma$, +we must express it as a function of $\tanh \beta$ and $\gamma$, and shall be obliged to +develop it in terms of \emph{Spherical Harmonics} of $\tanh \beta$ and $\gamma$ by the formulas of +Chapter VII, using the first equation in (10) for the value of $V$ at an internal +point, and the second for the value of $V$ at an external point. If the problem +is symmetrical, we must develop in Zonal Harmonics of $\tanh \beta$ by the formulas +of Chapter VI. + +A convenient form for $Q_m (i \tan \alpha)$ is obtained from (2) Art.~100; it is +\begin{flalign*} +&&Q_m (i \tan \alpha) = -i P_m &(i \tan \alpha) \int\limits^{\infty}_{\tan\alpha}\dfrac{ dx}{ (1 + x^2)[P_m(xi)]^2}.& \tag{12} \\ +&\text{Hence} &Q_0 (i \tan \alpha) = -&i \int\limits^{\infty}_{\tan\alpha}\dfrac{dx}{ 1 + x^2} = -i \left(\dfrac{\pi}{ 2} - \alpha\right).& \tag{13} +\end{flalign*} + +\EXAMPLE{S} + +1.\quad A conductor in the form of an oblate spheroid whose semi-axes are +$b \sec \alpha_0$ and $b \tan \alpha_0$ is charged with electricity and is found to be at potential +$V_0$; find the value of the potential function at any internal or external point. + +Here $V_0 = V_0P_0(\tanh \beta)$. Hence at an internal point +\[ +V = V_0 \dfrac{P_0 (i \tan \alpha) }{ P_0 (i \tan \alpha_0)} P_0 (\tanh \beta) = V_0, \tag{1} +\] +and at an external point +\[ +V = V_0 \dfrac{Q_0 (i \tan \alpha) }{Q_0 (i \tan \alpha_0)} P_0 (\tanh \beta) =V_0 \dfrac{ \left(\dfrac{\pi}{2} - \alpha\right) }{\left(\dfrac{ \pi }{ 2} - \alpha_0\right)}. \tag{2} +\] +Since $V$ in (2) involves $\alpha$ only, the equipotential surfaces are all spheroids +confocal with the conductor.\\ + +% -----File: 258.png +2.\quad The upper half of an oblate spheroid whose semi-axes are $b \sec \alpha_0$ and +$b \tan \alpha_0$ is kept at the temperature unity, and the lower half at the temperature +zero. Find the permanent temperature at any internal point. +\begin{flushright} +\textit{Ans.} \quad $u = \dfrac{1}{2} + \dfrac{3}{4} \dfrac{P_1(i \tan\alpha)}{P_1(i \tan\alpha_0)}P_1(\tanh\beta) - \dfrac{7}{8}.\dfrac{1}{2 }\dfrac{P_3(i \tan\alpha)}{P_3(i \tan\alpha_0)}P_3(\tanh\beta) + \cdots$ +\end{flushright} +(v.\ Art.~93). $u$ may be expressed in terms of $x$, $y$, and $z$ without serious +difficulty [v.\ (12) Art.~132]. +\[ +u = \frac{1}{2 }+ \frac{3}{4} \frac{y}{c} - \frac{7}{8}.\frac{1}{2}.\frac{1}{2}\frac{ [25y^3-15y(x^2+y^2+z^2-b^2)-9b^2y]}{5c^3+3b^2c} + \cdots +\] +if $2c = 2b \tan \alpha_0 = $ minor axis of spheroid. + +\mypara{135.} Let us now find the potential function at an external point due to +the attraction of a solid homogeneous oblate spheroid, using the method employed +in Arts.~98 and 99. + +Consider first the potential function due to a shell bounded by the spheroids +for which $\alpha = \phi$ and $\alpha = \phi + d\phi$. + +By (1) Art.~98 we have +\[ +4\pi\rho\kappa = [D_nV_1 - D_nV_2]_{\alpha=\phi}, \tag{1} +\] +where $\rho$ is the density and $\kappa$ the thickness of the shell, $V_1$ the value of the +potential function at an internal point, and $V_2$ the value of the potential +function at an external point. +\begin{flalign*} +&\indent \text{Let }&V_1 &= \sum A_m(-i)^mP_m(i \tan\alpha)P_m(\tanh\beta)&\\ +&\text{and}&V_2 &= \sum B_m i^{m+1}Q_m(i \tan\alpha)P_m(\tanh\beta) \tag*{[v.\ (11) Art.~134].}& +\end{flalign*} + +Since $V_1$ and $V_2$ must have the same value when $\alpha = \phi$ +\[ +A_m = B_m i^{2m+1}\frac{ Q_m(i \tan\phi)}{P_m(i \tan\phi)} = (-1)^m B_m \int\limits_{\tan\phi}^{\infty}\frac{dx}{(1+x^2)[P_m(xi)]^2} \tag{2} +\] +[v.\ (12) Art.~134]. +\begin{flalign*} +\left.\!\! +\begin{alignedat}{2} +&\indent \text{Hence\quad} & V_1 &= \sum i^mB_mP_m(\tanh\beta)P_m(i \tan\alpha) \int\limits_{\tan\phi}^{\infty}\frac{dx}{(1+x^2)[P_m(xi)]^2}\\ +&\text{and}& V_2 &= \sum i^mB_mP_m(\tanh\beta)P_m(i \tan\alpha) \int\limits_{\tan\alpha}^{\infty}\frac{dx}{(1+x^2)[P_m(xi)]^2} . +\end{alignedat} +\right \}\hspace{2.8pt}(3) +\end{flalign*} +\[ +D_nV_1 = D_{\alpha}V_1.D_n\alpha. \qquad\qquad D_nV_2 = D_{\alpha}V_2.D_n\alpha +\] +% -----File: 259.png +\begin{align*} +[D_nV_1-D_nV_2]_{\alpha=\phi}&=[D_\alpha V_1-D_\alpha V_2]_{\alpha=\phi}(D_n\alpha)_{\alpha=\phi}\\ + &\qquad =[D_\alpha(V_1-V_2)]_{\alpha=\phi}[D_n\alpha]_{\alpha=\phi}. +\end{align*} +\[ +V_1-V_2= \sum i^mB_mP_m(\tanh\beta)P_m(i \tan\alpha) \int\limits^{\tan\alpha}_{\tan\phi}\dfrac{ dx}{(1+x^2)[P_m(xi)]^2}. +\] +\begin{multline*} +D_\alpha(V_1-V_2)= \sum i^mB_mP_m(\tanh\beta)\bigg[P_m(i \tan\alpha) \dfrac{\sec^2\alpha}{(1+\tan^2\alpha)[P_m(i \tan\alpha)]^2}\\ + + \dfrac{dP_m(i \tan\alpha)}{d\alpha} \int\limits^{\tan\alpha}_{\tan\phi}\dfrac{ dx}{(1+x^2)[P_m(xi)]^2}\bigg]. +\end{multline*} +\begin{gather*} +D_\alpha[V_1-V_2]_{\alpha=\phi} = \sum i^mB_m \dfrac{P_m(\tanh\beta)}{P_m(i \tan\phi)}.\\ + D_n\alpha = \dfrac{d\alpha}{dn} +\end{gather*} +\[ +dn=\dfrac{d\rho_1}{h_1}=\dfrac{d\lambda}{h_1}= \dfrac{\sqrt{ \lambda^2-\mu^2}}{\sqrt{\lambda^2-b^2}}d\lambda=b \sec\alpha \sqrt{\tan^2\alpha+\tanh^2\beta}.d\alpha \tag{4} +\] +v.\ Art.~130 (3), and Art.~132 (5) and (10). +\begin{flalign*} +& &[D_n\alpha]_{\alpha=\phi} &=\dfrac{1}{b \sec\phi\sqrt{\tan^2\phi+\tanh^2\beta}}.& +% recast to fit page width +&\intertext{\indent Hence} +&& [D_nV_1-D_nV_2]_{\alpha=\phi}& = \dfrac{1}{b \sec\phi\sqrt{\tan^2\phi+\tanh^2\beta}}\sum i^mB_m \dfrac{P_m(\tanh\beta)}{P_m(i \tan\phi)}.&\\ +& & \kappa = [dn]_{\alpha=\phi} &= b\sec\phi \sqrt{\tan^2\phi+\tanh^2\beta}.d\phi& +\end{flalign*} +by (4), and (1) may be written +\[ +4\pi\rho b^2 \sec^2\phi(\tan^2\phi+\tanh^2\beta)d\phi = \sum i^mB_m \dfrac{P_m(\tanh\beta)}{P_m(i \tan\phi)}.\tag{5} +\] +\begin{flalign*} +&\indent\text{Since}& \tanh^2\beta &= \tfrac{1}{3}P_0(\tanh\beta)+\tfrac{2}{3}P_2(\tanh\beta)&\phantom{\indent Since} +\end{flalign*} +by (5) Art.~95, to satisfy (5) we must give $m$ the values 0 and 2 and +\begin{flalign*} +& &B_0&=\tfrac{4}{3}\pi\rho b^2 \sec^2\phi(3\tan^2\phi+1)d\phi&\\ +&\text{and}& B_2&=\tfrac{4}{3}\pi\rho b^2 \sec^2\phi(3\tan^2\phi+1)d\phi.& +\end{flalign*} + +% -----File: 260.png +So that by (3) +\begin{flalign*} +&&V_1&=\tfrac{4}{3} \pi \rho b^2 \sec^2 \phi (3 \tan^2 \phi + 1)d \phi \bigg[\int\limits^{\infty}_{\tan\phi}\dfrac{ dx}{1+x^2}\\ +&&&\hspace{7em}-P_2(\tanh \beta)P_2(i \tan \alpha) \int\limits^{\infty}_{\tan\phi} \dfrac{dx}{(1+x^2)[P_2(xi)]^2}\bigg] \tag{6} \\ +&\text{and}& V_2&=\tfrac{4}{3} \pi \rho b^2 \sec^2 \phi (3 \tan^2 \phi + 1)d \phi [i Q_0 (i \tan \alpha)&\\ +&&&\hspace{16em}+ i^3 P_2 (\tanh \beta) Q_2 (i \tan \alpha)]. &&\tag{7} +\end{flalign*} +The potential function at an external point due to the solid spheroid for +which $\alpha = \alpha_0$ is +\[ +V=\!\!\int\limits^{\phi=\alpha_0}_{\phi=0}\!\!V_2 = \tfrac{4}{3} \pi \rho b^2 \sec^2 \alpha_0 \tan \alpha_0 [i Q_0 (i \tan \alpha) + i^3 P_2 (\tanh \beta) Q_2 (i \tan \alpha)].\tag{8} +\] + +If $2a$ is the major axis and $2c$ the minor axis of the spheroid +\[ +\tfrac{4}{3} \pi\rho b^2 \sec^2\alpha_0 \tan\alpha_0 =\tfrac{4}{3} \dfrac{ \pi\rho a^2c}{b}=\dfrac{M}{b} +\] +where $M$ is the mass of the spheroid. Therefore +\[ +V = \dfrac{M}{b} [i Q_0 (i \tan \alpha) + i^3 P_2 (\tanh \beta)Q_2 (i \tan \alpha)] \tag{9} +\] +is the required value. (9) can be reduced to\\[2ex] +\phantom{.}$\displaystyle\;V = \dfrac{M}{b} \left\{\dfrac{\pi}{2} - \alpha + \dfrac{1}{4}\left[\left(\dfrac{\pi}{2} - \alpha\right)(3 \tan^2 \alpha+1) - 3 \tan \alpha\right][3 \tanh^2 \beta - 1]\right\}.$\hfill(10) + +\EXAMPLE{S} + +1.\quad Break up the equation (3) Ex.~1, Art.~132, for the prolate spheroid, and +obtain particular solutions of the term +\begin{gather*} +V= (A \cos n\gamma + B \sin n\gamma)P_m^n(\tanh\beta)P_m^n(\ctnh\alpha),\\ +V= (A \cos n\gamma + B \sin n\gamma)P_m^n(\tanh\beta)(-1)^{\frac{n}{2}} \csch^n \alpha \dfrac{d^n Q_m (\ctnh\alpha)}{(d \ctnh \alpha)^n}. +\end{gather*} + +2.\quad Break up and solve the equations of Exs.\ 2 and 3, Art.~132, and show +that they lead to familiar forms.\\ + +3.\quad If in Ex.~1, Art.~132, the conductor is a prolate spheroid whose semi-axes +are $b \ctnh\alpha_0$ and $b \csch\alpha_0$ show that +\[ +V= V_0\text{ at an internal point.}\qquad V = V_0\dfrac{ \alpha}{\alpha_0}\text{ at an external point}. +\] + +% -----File: 261.png +4.\quad Show that the potential function at an external point due to the attraction +of a homogeneous solid prolate spheroid is +\[ +V = \dfrac{M}{b} [Q_0(\ctnh \alpha) - P_2(\tanh \beta)Q_2(\ctnh \alpha)]. +\] + +\vspace{1ex}\begin{center} +\emph{Ellipsoidal Harmonics.} +\end{center}\vspace{-1ex} + +\mypara{136.} If we are dealing with an \emph{ellipsoid} instead of a spheroid, we can take +as our orthogonal system of surfaces a set of \emph{confocal quadrics}; +\[ +\left. +\begin{aligned} +\dfrac{x^2}{\lambda^2} + \dfrac{y^2}{\lambda^2 -b^2} + \dfrac{z^2}{\lambda^2 - c^2} - 1& = 0\;\\ +\dfrac{x^2}{\mu^2} + \dfrac{y^2}{\mu^2 - b^2} + \dfrac{z^2}{\mu^2 - c^2} - 1 &= 0\\ +\dfrac{x^2}{\nu^2} + \dfrac{y^2}{\nu^2 - b^2} + \dfrac{z^2}{\nu^2 - c^2} - 1 &= 0 +\end{aligned} +\right\} +\tag{1} +\] +where $\lambda^2 > c^2 > \mu^2 > b^2 > \nu^2$. Here the first surface is an ellipsoid, the +second an unparted hyperboloid, and the third a biparted hyperboloid. Each +of the three principal sections of the system consists of confocal conics, and it +is well known and is easily shown that the surfaces cut orthogonally. $\lambda$, $\mu$, +and $\nu$ will be our curvilinear coördinates, and are known as Ellipsoidal +Coördinates. + +We find without difficulty that +% recast to fit page width +\begin{multline*} +x^2 = \dfrac{\lambda^2 \mu^2 \nu^2}{b^2 c^2},\quad y^2=\dfrac{(\lambda^2 - b^2)(\mu^2 - b^2)(b^2 - \nu^2)}{ b^2(c^2 - b^2)}, \\ z^2 = \dfrac{(\lambda^2 - c^2)(c^2 - \mu^2)(c^2 - \nu^2)}{ c^2(c^2 - b^2)},\tag{2} +\end{multline*} +% recast to fit page width +\begin{multline*} +h_1^2 = \dfrac{(\lambda^2 - b^2)(\lambda^2 - c^2)}{(\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}, \quad h_2^2 = \dfrac{(\mu^2 - b^2)(c^2 - \mu^2)}{(\mu^2 -\nu^2)(\lambda^2 - \mu^2)},\\ h_3^2 = \dfrac{(b^2 - \nu^2)(c^2 - \nu^2)}{(\lambda^2 - \nu^2)(\mu^2 - \nu^2)}.\tag{3} +\end{multline*} + +To avoid ambiguity, we shall suppose that of the nine semi-axes in (1) +$\sqrt{\strxx c^2 - \mu^2}$ is to be taken with the positive sign for a point on the half of the +unparted hyperboloid on which $z$ is positive, and with the negative sign for a +point on the half on which $z$ is negative; $\sqrt{\strxx b^2 - \nu^2}$ is to be taken with the +positive sign for a point on the half of the biparted hyperboloid on which $y$ is +positive, and with the negative sign for a point on the half on which $y$ is +negative; $\nu$ is to be taken positive for a point on the half of the biparted +hyperboloid on which $x$ is positive, and negative for a point on the half on +which $x$ is negative, and that the remaining six are to be always positive. It +follows that our Ellipsoidal Coördinates have the disadvantage that to fully +fix a point we need to know not merely the values of its coördinates $\lambda$, $\mu$, and +$\nu$, but the signs of $\sqrt{\strxx c^2 - \mu^2}$, and $\sqrt{\strxx b^2 - \nu^2}$ as well. + +% -----File: 262.png +We shall see later, Art.~139, when we come to introduce what we may call +the \emph{Normal Ellipsoidal Coördinates} $\alpha$, $\beta$, and $\gamma$ that they are free from this +disadvantage. + +It is to be observed that $\lambda$ may range from $c$ to $\infty$, $\mu$ from $b$ to $c$, and $\nu$ from +$-b$ to $b$. + +The element of length perpendicular to the Ellipsoid is +\[ +dn = \dfrac{d \lambda}{h_1} = \sqrt{\dfrac{(\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}{(\lambda^2 - b^2)(\lambda^2 - c^2)}}.d \lambda. \tag{4} +\] + +The element of Ellipsoidal surface is +\[ +dS = \dfrac{d \mu d \nu }{ h_2 h_3} = (\mu^2 - \nu^2) \sqrt{\dfrac{ (\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}{(\mu^2 - b^2)(c^2 - \mu^2)(b^2 - \nu^2)(c^2 - \nu^2)}}.d \mu d\nu, \tag{5} +\] +and the element of volume is +% recast to fit page width +\begin{align*} +dv &=\dfrac{ d \lambda d \mu d \nu}{ h_1h_2h_3}\\ + &=\dfrac{ (\lambda^2 - \mu^2)(\lambda^2 - \nu^2)(\mu^2 - \nu^2)}{\sqrt{(\lambda^2 - b^2)(\lambda^2 - c^2)(\mu^2 - b^2)(c^2 - \mu^2)(b^2 - \nu^2)(c^2 - \nu^2)}} d \lambda d \mu d \nu. \tag{6} +\end{align*} + +The surface integral of any given function of $\mu$ and $\nu$ taken over the +ellipsoid is +\begin{multline*} +\int f(\mu, \nu)dS =\int\limits^{b}_{-b} d \nu \int\limits^{c}_{b} [f_1 (\mu, \nu) + f_2 (\mu, \nu) + f_3 (\mu, \nu)\\ + +f_4 (\mu, \nu)](\mu^2 - \nu^2) \sqrt{\dfrac{(\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}{(\mu^2 - b^2)(c^2 - \mu^2)(b^2 - \nu^2)(c^2 - \nu^2)}}.d \mu, \tag{7} +\end{multline*} +where $f_1(\mu, \nu)$, $f_2(\mu, \nu)$, $f_3(\mu, \nu)$ and $f_4(\mu, \nu)$ are the values of the given function +on the four quarters of the ellipsoid into which it is divided by the planes of +($XY$) and ($XZ$). + +Laplace's Equation proves reducible to +\[ +(\mu^2 - \nu^2) D_{\alpha}^2 V + (\lambda^2 - \nu^2) D_{\beta}^2 V + (\lambda^2 - \mu^2) D_{\gamma}^2 V = 0 \tag{8} +\] +\begin{flalign*} +&\text{where}& \alpha = c \int\limits^{\lambda}_c \dfrac{d\lambda}{\sqrt{(\lambda^2 - b^2)(\lambda^2 - c^2)}},\quad \beta &= c \int\limits^{\mu}_b \dfrac{d \mu}{\sqrt{(c^2 - \mu^2)(\mu^2 - b^2)}},&\\ +& & \gamma &= c \int \limits^{\nu}_0 \dfrac{d \nu }{\sqrt{(b^2 - \nu^2)(c^2 - \nu^2)}}. &\tag{9} +\end{flalign*} + +% -----File: 263.png +$\alpha$, $\beta$, and $\gamma$ can be expressed as Elliptic Integrals of the first class and are +\begin{multline*} +\alpha=F\Big(\dfrac{b}{c}, \dfrac{\pi}{2}\Big)-F\Big(\dfrac{b}{c}, \sin^{-1} \dfrac{c}{\lambda}\Big),\quad \beta=F\Big(\sqrt{ 1-\dfrac{ b^2}{c^2}},\, \sin^{-1} \sqrt{\dfrac{1- \dfrac{b^2}{\mu^2}}{1- \dfrac{b^2}{c^2}}}\Big),\\ + \gamma=F\Big(\dfrac{b}{c}, \sin^{-1} \dfrac{\nu}{b}\Big); \tag{10} +\end{multline*} +\begin{flalign*} +&\text{whence} + &\lambda = \frac{c}{\sn(K-\alpha)} &\Big(\bmod \frac{b}{c}\Big) += c\,\frac{ \dn \alpha}{\cn \alpha} \Big(\bmod \frac{b}{c}\Big),&& +\\ +&& \mu = \frac{b}{\dn \beta } + &\Big( \bmod\Big( 1-\dfrac{b^2}{c^2} \Big)^{\frac{1}{2}} \Big),\quad + \nu = b \sn \gamma\Big( \bmod \frac{b}{c} \Big)&& +\tag{11} +\end{flalign*} +(v.\ Int.\ Cal.\ Arts.~179, 192, and 196). + +\mypara{137.} If in (8) Art.~136 we assume $V=L.M.N$ where $L$ involves $\alpha$ only, +$M$ involves $\beta$ only, and $N$ involves $\gamma$ only, (8) can be written +\[ +\dfrac{\mu^2-\nu^2}{L}\dfrac{ d^2 L}{d\alpha^2} + \dfrac{\lambda^2 - \nu^2}{M} \dfrac{d^2 M}{d\beta^2} + \dfrac{\lambda^2 - \mu^2}{N}\dfrac{ d^2 N}{d\gamma^2} =0. \tag{1} +\] +(1) is too complicated to be broken up by our usual method. + +If, however, we let +\[ +\dfrac{1}{L}\dfrac{d^2L}{d\alpha^2} ={\textstyle\sum} a_k \lambda^k,\quad\dfrac{1}{M}\dfrac{ d^2M}{d\beta^2} ={\textstyle\sum} b_k \mu^k,\quad \dfrac{1}{N}\dfrac{ d^2N}{d\gamma^2} ={\textstyle\sum} c_k \nu^k, +\] +substitute in (1) and make use of the fact that the result must be identically +zero, we find that the coefficients are zero for all values of $k$ except $k = 0$ and +$k = 2$, and that $a_0 = -b_0 = c_0$, and $a_2 = -b_2 = c_2$. + +Therefore (1) can be broken up into the three equations +\begin{align*} +\dfrac{d^2L}{d\alpha^2}&= (a_0 + a_2 \lambda^2)L \\ +\dfrac{d^2M}{d\beta^2} &= -(a_0 + a_2 \mu^2)M \\ +\dfrac{d^2N}{d\gamma^2}& = (a_0 + a_2 \nu^2)N. +\end{align*} + +% -----File: 264.png +We shall find it convenient to take $a_2$ as $m(m + 1)$ and $a_0$ as $-(b^2 + c^2)p$; +whence +\[\left .\begin{aligned} +\frac{d^2 L }{d \alpha^2} &- [m(m + 1) \lambda^2 - (b^2 + c^2)p]L = 0\\[1ex] +\frac{d^2 M }{ d \beta^2} &+ [m(m + 1) \mu^2 - (b^2 + c^2)p]M = 0\; \\[1ex] +\frac{d^2 N }{d \gamma^2} &- [m(m + 1) \nu^2 - (b^2 + c^2)p]N = 0 . +\end{aligned} +\right \} +\tag{2}\] + +If now in (2) we replace $\alpha$, $\beta$, and $\gamma$ by their values in terms of $\lambda$, $\mu$, and +$\nu$, we get +\begin{equation}\left . +\begin{aligned} +(\lambda^2 - b^2) (\lambda^2 - c^2) \frac{d^2 L }{d\lambda^2} ++ \lambda (\lambda^2 - b^2 + \lambda^2 - c^2) \frac{dL }{d\lambda}\hspace{8em}\\ +- [m(m + 1) \lambda^2 - (b^2 + c^2)p] L = 0\phantom{.}\\ +(\mu^2 - b^2) (\mu^2 - c^2) \frac{d^2 M }{ d\mu^2} ++ \mu (\mu^2 - b^2 + \mu^2 - c^2) \frac{dM }{d\mu}\hspace{7em}\\ +- [m(m + 1) \mu^2 - (b^2 + c^2)p] M = 0\phantom{.}\\ +(\nu^2 - b^2)(\nu^2 - c^2)\frac{ d^2 N }{d\nu^2} ++ \nu (\nu^2 - b^2 + \nu^2 - c^2) \frac{dN}{ d\nu}\hspace{7.5em}\\ +- [m(m + 1) \nu^2 - (b^2 + c^2)p] N = 0 . +\end{aligned} +\right \} +\tag{3}\end{equation} +Whence if $L = E_m^p (\lambda)$, it follows that $M = E_m^p (\mu)$ and $N = E_m^p (\nu)$, and that +\[ +V = E_m^p (\lambda) E_m^p (\mu) E_m^p (\nu) \tag{4} +\] +is a solution of Laplace's Equation, (8) Art.~136. + +The equation +\begin{multline*} + (x^2 - b^2) (x^2 - c^2)\frac{ d^2 z }{ dx^2} ++ x(x^2 - b^2 + x^2 - c^2) \frac{dz }{dx}\\ +- [m(m + 1) x^2 - (b^2 + c^2)p] z = 0 \tag{5} +\end{multline*} +is known as Lamé's Equation, and $E_m^p (x)$ as a \textit{Lamé's Function} or an \textit{Ellipsoidal} +\textit{Harmonic}. We shall suppose $m$ a positive integer. + +To get a particular solution of (5) let $z = \sum a_k x^k$. Substitute in (5) and +reduce and we get +\begin{multline*} + [k(k + 1) - m(m + 1)] a_k - (b^2 + c^2) [(k + 2)^2 - p] a_{k+2}\\ ++ b^2 c^2 (k + 3)(k + 4) a_{k+4} = 0. \tag{6} +\end{multline*} +We have now only to choose a sequence of coefficients satisfying (6), and we +may take any two consecutive coefficients arbitrarily. + +% -----File: 265.png +(6) which is ordinarily a relation connecting three consecutive coefficients +reduces to a relation between two when $k = m$, when $k = -3$, and when + $k = -4$. If we take $a_{m + 2} = 0$, $a_{m+4}$, $a_{m+6}$, \&c., will vanish. Let $a_m = 1$. + If $m$ is even the coefficient of $a_0$ in (6) will be zero; if $p$ has such a value +that $a_{-2}$ is zero, $a_{-4}$, $a_{-6}$, \&c., will be zero, and there will be no terms in +the solution involving negative powers of $x$. + +If we write the values of $a_{m-2}$, $a_{m-4}$, \&c., by the aid of (6) we see that +$a_{m-2}$ is of the first degree in $p$, $a_{m-4}$ of the second degree in $p$, \&c., and $a_{-2}$ +of the degree $\dfrac{m}{2} +1$ in $p$. There are then $\dfrac{m}{2} +1$ values of $p$ which we shall +call $p_1$, $p_2$, $p_3$, \&c., for which $a_{-2}$ will vanish, and for which our solutions will +be of the form +\[ + E_m^p(x)=x^m+a_{m-2}x^{m-2}+a_{m-4}x^{m-4}+ \cdots+a_0 +\] +if $m$ is even. + +If $m$ is odd, the coefficient of $a_1$ in (6) will vanish and we can choose $p$ so +that $a_{-1}$ shall be zero, and then all coefficients of lower order will vanish. +$a_{-1}$ is of the degree $\dfrac{m+1}{2}$ in $p$, and there will be $\dfrac{m+1}{2}$ values $p_1$, $p_2$, $p_3$, +\&c., of $p$ for which +\[ + E_m^p(x)=x^m+a_{m-2}x^{m-2}+a_{m-4}x^{m-4}+ \cdots+a_1x. +\] + +Following Heine we shall call the solution just obtained $K_m^p(x)$ so that +\[ + K_m^p(x)=x^m+a_{m-2}x^{m-2}+a_{m-4}x^{m-4}+ \cdots \tag{7} +\] +terminating with $ a_0$ if $m$ is even, and with $a_1x$ if $m$ is odd. If $m$ is even, +there are $\dfrac{m}{2} +1$ of these functions $K_m^{p_1}(x)$, $K_m^{p_2}(x)$, \&c., and there are $\dfrac{m+1}{2}$ +of them if $m$ is odd. The coefficients can be computed by the aid of (6). + +If in Lamé's Equation (5) we let $z=v\sqrt{\strxx x^2-b^2}$ we get the equation +\begin{multline*} + (x^2-b^2) (x^2-c^2)\frac{ d^2v}{dx^2} ++ x[x^2-b^2+3(x^2-c^2)]\frac{ dv}{dx}\\ +-[(m+2)(m-1)x^2 + c^2 - (b^2+c^2)p] v = 0. \tag{8} +\end{multline*} + +Letting $v=\sum a_kx^k$ we obtain the relation +\begin{multline*} + [k(k+3)-(m+2) (m-1)]a_k +- \{(b^2+c^2) [(k+2)^2-p] + c^2(2k+5)\} a_{k+2}\\ ++ b^2c^2 (k+3) (k+4) a_{k+4} = 0. \tag{9} +\end{multline*} +% -----File: 266.png + +Proceeding exactly as before, we find that there are $\dfrac{m}{2}$ values $q_1$, $q_2$, $q_3$, \&c., +of $ p$ for which $v=x^{m-1}+a_{m-3}x^{m-3} + \cdots + a_{1}x$ if $m$ is even, and $\dfrac{m+1}{2}$ +values for which $v=x^{m-1}+a_{m-3}x^{m-3} + \cdots + a_{0}$ if $m$ is odd. + +Calling $v\sqrt{\strxx x^2-b^2}$ $L_{m}^{p}(x)$ so that +\[ +L_{m}^{p}(x) = \sqrt{ x^2-b^2}[x^{m-1}+a_{m-3}x^{m-3} + a_{m-5}x^{m-5} + \cdots], \tag{10} +\] +terminating with $a_1x$ if $m$ is even and with $a_0$ if $m$ is odd, we have $\dfrac{m}{2}$ +values of $E_m^p(x)$, namely $L_m^{q_1}(x)$, $L_m^{q_2}(x)$, \&c., of the form (10) if $m$ is even +and $\dfrac{m+1}{2}$ values if $m$ is odd. + +By interchanging $b$ and $c$ in (8), (9), and (10) we may show that if +\[ +M_{m}^{p}(x) = \sqrt{x^2-c^2}[x^{m-1}+a_{m-3}x^{m-3} + a_{m-5}x^{m-5} + \cdots ] \tag{11} +\] +there are $\dfrac{m}{2}$ values of $E_m^p(x)$, namely $M_m^{r_1}(x)$, $M_m^{r_2}(x)$, $M_m^{r_3}(x)$, \&c., of the form +(11) if $m$ is even and $\dfrac{m+1}{ 2}$ values if $m$ is odd. + +Finally if in Lamé's Equation (5) we let $z=v \sqrt{\strxx(x^2-b^2)(x^2-c^2)}$ we get +\begin{multline*} +(x^2-b^2)(x^2-c^2)\frac{d^2v}{dx^2}+3x(x^2-b^2+x^2-c^2)\frac{dv}{dx}\\ + -[(m+3)(m-2)x^2-(b^2+c^2)(p-1)]v=0. \tag{12} +\end{multline*} +If now we let $v= \sum a_kx^k$ we obtain the relation +\begin{multline*} +[k(k+5)-(m-2)(m+3)]a_k\\ +-(b^2+c^2)[(k+2)(k+4)+1-p]a_{k+2}+b^2c^2(k+3)(k+4)a_{k+4}=0. \tag{13} +\end{multline*} + +Proceeding as before we find that there are $\dfrac{m}{2}$ values $s_1$, $s_2$, $s_3$, \&c., of $ p$ +for which $v=x^{m-2} + a_{m-4}x^{m-4} + a_{m-6}x^{m-6} + \cdots + a_0$ if $m$ is even, and +$\dfrac{m+1}{2}$ values for which $ v=x^{m-2}+a_{m-4}x^{m-4} + \cdots + a_1x$ if $m$ is odd. + +Calling $v\sqrt{\strxx(x^2-b^2)(x^2-c^2) }$ $N_m^p(x)$ so that +\[ +N_m^p(x) = \sqrt{(x^2-b^2)(x^2-c^2) }[x^{m-2} + a_{m-4}x^{m-4} + a_{m-6}x^{m-6} + \cdots] \tag{14} +\] +terminating with $a_0$ if $m$ is even and with $a_1x$ if $m$ is odd, we have $\dfrac{m}{2 }$ values +of $ E_m^p(x) $, namely $N_m^{s_1}(x)$, $N_m^{s_2}(x)$, $N_m^{s_3}(x)$, \&c., of the form (14) if $m$ is even and +$\dfrac{m-1}{2}$ values if $m$ is odd. + +% -----File: 267.png +Summing up our results we see that there are $2m+1$ Ellipsoidal Harmonics +$E_m^p(x)$ each of which is a finite sum of the $m$th degree in $x$, or in $x$ and $\sqrt{\strxx x^2-b^2}$, +or in $x$ and $\sqrt{\strxx x^2-c^2}$, or in $x$ and $\sqrt{\strxx x^2-b^2}$ and $\sqrt{\strxx x^2-c^2}$. + +It was proved by Lamé that the $2m + 1$ values of $p$, namely $p_1$, $p_2$, $p_3$, \&c., +$q_1$, $q_2$, $q_3$, \&c., $r_1$, $r_2$, $r_3$, \&c., $s_1$, $s_2$, $s_3$, \&c., were all real, and by Liouville that +they were all different. + +We give tables of the Ellipsoidal Harmonics for $m = 0$, $m = 1$, $m = 2$, and +$m = 3$. The coefficients were obtained by the aid of formulas (6), (9), +and (13). + +\rule{3em}{0em} \begin{minipage}{0.25\linewidth}{ +\[ +\arraycolsep=0.7em +\begin{array}{|l@{}l|} +\multicolumn{2}{c}{E_0(x)}\\[0.5em] +\hline +K_0(x)&=1\rule{0em}{1.3em}\\[0.5em] +L_0(x)&=0\\[0.5em] +M_0(x)&=0\\[0.5em] +N_0(x)&=0\\[0.5em] +\hline +\end{array} +\] +} +\end{minipage} +\begin{minipage}{0.6\linewidth}{ +\[ +\begin{array}{|l@{}l|} +\multicolumn{2}{c}{E_1(x)}\\[0.5em] +\hline +K_1(x)&=x\rule{0em}{1.3em}\\[0.5em] +L_1(x)&=\sqrt{\strxx x^2-b^2}\\[0.5em] +M_1(x)&=\sqrt{\strxx x^2-c^2}\\[0.5em] +N_1(x)&=0\\[0.5em] +\hline +\end{array}\rule{3.7em}{0em} +\] +} +\end{minipage} +\[ +\begin{array}{|l@{}l|} +\multicolumn{2}{c}{E_2(x)}\\[0.5em] +\hline +K_2^{p_1}(x)&=x^2 - \frac{1}{3} [b^2 + c^2 - \sqrt{(b^2 + c^2)^2-3b^2c^2}]\rule{0em}{1.3em}\\[0.5em] +K_2^{p_2}(x)&=x^2 - \frac{1}{3} [b^2 + c^2 + \sqrt{(b^2 + c^2)^2-3b^2c^2}]\\[0.5em] +L_2(x)&=x \sqrt{\strxx x^2-b^2}\\[0.5em] +M_2(x)&=x \sqrt{\strxx x^2-c^2}\\[0.5em] +N_2(x)&=\sqrt{(x^2-b^2)(x^2-c^2)}\\[0.5em] +\hline +\end{array} +\] +\[ +\begin{array}{|l@{}l|} +\multicolumn{2}{c}{E_3(x)}\\[0.5em] +\hline +K_3^{p_1}(x)&=x^3 - \dfrac{x}{5} [2(b^2+c^2) - \sqrt{4(b^2+c^2)^2-15b^2c^2}] +\rule{0em}{1.3em}\\[0.9em] + +K_3^{p_2}(x)&=x^3 -\dfrac{x}{5} [2(b^2+c^2) + \sqrt{4(b^2+c^2)^2-15b^2c^2}]\\[0.9em] +L_3^{q_1}(x)&=\sqrt{\strxx x^2-b^2 }[x^2-\frac{1}{5} (b^2+2c^2 - \sqrt{(b^2+2c^2)^2-5b^2c^2})]\\[0.7em] +L_3^{q_2}(x)&=\sqrt{\strxx x^2-b^2} [x^2-\frac{1}{5} (b^2+2c^2 + \sqrt{(b^2+2c^2)^2-5b^2c^2})]\\[0.7em] +M_3^{r_1}(x)&=\sqrt{\strxx x^2-c^2} [x^2-\frac{1}{5} (2b^2+c^2 - \sqrt{(2b^2+c^2)^2-5b^2c^2})]\\[0.7em] +M_3^{r_2}(x)&=\sqrt{\strxx x^2-c^2} [x^2-\frac{1}{5} (2b^2+c^2 + \sqrt{(2b^2+c^2)^2-5b^2c^2})]\\[0.7em] +N_3(x)&=x \sqrt{(x^2-b^2)(x^2-c^2)}\\[0.5em] +\hline +\end{array} +\] +\vspace{\baselineskip} + +% -----File: 268.png +It is to be noted that since in the solution (4) of Laplace's Equation, +\[ +V = E_m^p(\lambda) E_m^p(\mu) E_m^p(\nu), +\] +we have the same $m$ and $p$ in each of the three factors, we shall have to deal +merely with products made up of factors of the same form, for example, +\[ +K_m^{p_k}(\lambda)K_m^{p_k}(\mu)K_m^{p_k}(\nu),\quad L_m^{q_k}(\lambda)L_m^{q_k}(\mu)L_m^{q_k}(\nu),\quad \text{\&c.;} +\] +and that in a solution of the form +\[ +V=\sum A_{m,p} E_m^p(\lambda) E_m^p(\mu) E_m^p(\nu) +\] +we shall have for a given $m$ just $2m+1$ terms. + +\mypara{138.} From the particular solution of Lamé's Equation [(5) Art.~137] +$z=E_m^p(x)$, we can get by formula (5), Art.~18, the general solution. +\begin{flalign*} +&\indent\text{It is} & z=AE_m^p(x)+BE_m^p(x)\int\limits_x^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[E_m^p(x)]^2}. & \tag{1}& +\end{flalign*} +Making $A=0$ and $B=2m+1$ we get a second form of particular solution of +Lamé's Equation, $z=F_m^p(x)$ where +\[ +F_m^p(x)=(2m+1)E_m^p(x)\int\limits_x^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[E_m^p(x)]^2}. \tag{2} +\] +We shall call $F_m^p(x)$ a \emph{Lamé's Function of the second kind}. + +It is easily seen to approach the value zero as $x$ is indefinitely increased. + +\EXAMPLE{S} + +1.\quad If an ellipsoidal conductor is charged with electricity, and is found to +be at potential $V_0$, show that since $V_0=V_0K_0(\lambda)$, +\[ +V=V_0K_0(\lambda)K_0(\mu)K_0(\nu)=V_0 +\] +at an internal point, and +\begin{multline*} +V=V_0K_0(\mu)K_0(\nu)\bigg[K_0(\lambda)\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[K_0(x)]^2}\\ +÷ K_0(\lambda_0)\int\limits_{\lambda_0}^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[K_0(x)]^2}\bigg] +\end{multline*} +\[ +=V_0\bigg[\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}} + ÷ \int\limits_{\lambda_0}^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}}\bigg] +=V_0\frac{F\Big(\dfrac{b}{c},\sin^{-1}\dfrac{c}{\lambda}\Big)}{F\Big(\dfrac{b}{c},\sin^{-1}\dfrac{c}{\lambda_0}\Big)}, +\] +% -----File: 269.png +\begin{flalign*} +&\text{whence} & V=V_0\dfrac{\Big(F\Big(\dfrac{b}{c},\dfrac{\pi}{2}\Big)-\alpha\Big)}{F\Big(\dfrac{b}{c},\dfrac{\pi}{2}\Big)-\alpha_0} & \tag*{\text{v.~(10) Art.~136.}}& +\end{flalign*} + +2.\quad Find the value of the potential function at an external point due to the +attraction of a solid homogeneous ellipsoid (v.\ Art.~135). + +Observe that +\begin{align*} +(l^2-\mu^2)(l^2&-\nu^2)=\tfrac{1}{3}[3l^4-2(b^2+c^2)l^2+b^2c^2]K_0(\mu)K_0(\nu)\\ + &+\tfrac{1}{2}\bigg[1+\frac{b^2+c^2-3l^2}{\sqrt{(b^2+c^2)^2-3b^2c^2}}\bigg]K_2^{p_1}(\mu)K_2^{p_1}(\nu)\\ + &+\tfrac{1}{2}\bigg[1-\frac{b^2+c^2-3l^2}{\sqrt{(b^2+c^2)^2-3b^2c^2}}\bigg]K_2^{p_2}(\mu)K_2^{p_2}(\nu); +\end{align*} +and that +\[ +\int\limits_0^{\lambda_0}\tfrac{4}{3}\pi\rho\frac{3l^4-2(b^2+c^2)l^2+b^2c^2}{\sqrt{(l^2-b^2)(l^2-c^2)}}dl +=\tfrac{4}{3}\pi\rho\lambda_0\sqrt{(\lambda_0^2-b^2)(\lambda_0^2-c^2)}=M +\] +where $M$ is the mass of the ellipsoid. +% recast to fit page width +\begin{flalign*} +&\textit{Ans.} & V=M\bigg\{&\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}}-\frac{3}{2\sqrt{(b^2+c^2)^2-3b^2c^2}}\\ +&&&\bigg[K_2^{p_1}(\mu)K_2^{p_1}(\nu)K_2^{p_1}(\lambda) +\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}.(K_2^{p_1}(x))^2}\\ +&&&-K_2^{p_2}(\mu)K_2^{p_2}(\nu)K_2^{p_2}(\lambda) +\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}.(K_2^{p_2}(x))^2}\bigg]\bigg\}. +\end{flalign*} + +\mypara{139.} If for the sake of brevity we represent $\dfrac{b}{c}$ by $k$, and $\Big(1-\dfrac{b^2}{c^2}\Big)^\frac{1}{2}$ by $k'$ in +the formulas (11) Art.~136 we have +\[ +\lambda=c\,\frac{\dn\alpha}{\cn\alpha}(\bmod k),\quad \mu=\frac{b}{\dn\beta(\bmod k')},\quad \nu=b\sn\gamma(\bmod k) \tag{1} +\] +and from these we get without difficulty (v.\ Int.\ Cal.\ Art.~192) +\begin{equation*}\left. +\begin{aligned} +\sqrt{\lambda^2-b^2}&=\frac{ck'}{\cn\alpha(\bmod k)}, &\sqrt{\mu^2-b^2} &=\frac{bk'\sn\beta}{\dn\beta}(\bmod k'),\;\\ +\sqrt{b^2-\nu^2} &=b\cn\gamma(\bmod k), &\sqrt{\lambda^2-c^2}&=\frac{ck'\sn\alpha}{\cn\alpha}(\bmod k),\\ +\sqrt{c^2-\mu^2} &=\frac{ck'\cn\beta}{\dn\beta}(\bmod k'), &\sqrt{c^2-\nu^2} &=c\dn\gamma(\bmod k). +\end{aligned}\right\} \tag{2} +\end{equation*} + +% -----File: 270.png +If we let $\alpha$ range from 0 to $K$, and $\beta$ from 0 to $2K'$, and $\gamma$ from 0 to $4K$, +where $K$ and $K'$ are the complete Elliptic Integrals $F\Big(k,\dfrac{\pi}{2}\Big)$ and $F\Big(k',\dfrac{\pi}{2}\Big)$ +respectively, $(\alpha, \beta, \gamma)$ may represent any point in space, and there will be no +ambiguity in sign (v.\ Art.~136). + +We may note that if $0<\beta<K'$, $z$ is positive; if $K'<\beta<2K'$, $z$ is +negative; if $0<\gamma<K$, $x$ and $y$ are both positive; if $K<\gamma<2K$, $x$ is +positive and $y$ negative; if $2K<\gamma<3K$, $x$ and $y$ are both negative; and if +$3K<\gamma<4K$, $x$ is negative and $y$ positive (v.\ Art.~136). + +We can write the values in (4), (5), (6), and (7), Art.~136, more neatly by +bringing in $\alpha$, $\beta$, and $\gamma$. We get +\begin{gather*} +dn=\frac{1}{c}\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}d\alpha, \tag{3}\\ +dS=\frac{1}{c^2}(\mu^2-\nu^2)\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}d\beta d\gamma, \tag{4}\\ +dv=\frac{1}{c^3}(\lambda^2-\mu^2)(\lambda^2-\nu^2)(\mu^2-\nu^2)d\alpha d\beta d\gamma. \tag{5} +\end{gather*} + +For the integral of any function of $\alpha$, $\beta$, and $\gamma$ over the ellipsoid $\alpha=\alpha_0$, we +shall have\\ +$\displaystyle\int F(\alpha,\beta,\gamma)dS +=\frac{1}{c^2}\int\limits_0^{2K'}d\beta\int\limits_0^{4K} +F(\alpha_0,\beta,\gamma)(\mu^2-\nu^2)\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}d\gamma.\hfill(6)$ + +\mypara{140.} If we make use of the formula (2) Art.~92 +\[ +\int(UD_n V - VD_n U)dS=0 \tag{1} +\] +and take as our closed surface any given ellipsoid, we can get a very important +result. +\begin{flalign*} +&\indent\text{If} & U &=E_m^p(\lambda)E_m^p(\mu)E_m^p(\nu)\quad \text{and}\quad + V =E_n^q(\lambda)E_n^q(\mu)E_n^q(\nu)&&\\ +\intertext{then\hspace*{\stretch{1}} $\nabla^2U=\nabla^2V=0$.\hspace*{\stretch{1}}} +&&D_nU&=D_\alpha U D_n \alpha +=E_m^p(\mu)E_m^p(\nu)\frac{dE_m^p(\lambda)}{d\alpha}\frac{c}{\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}},&&\\ +&\text{and}& D_nV &=D_\alpha V D_n \alpha +=E_n^q(\mu)E_n^q(\nu)\frac{dE_n^q(\lambda)}{d\alpha}\frac{c}{\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}},&& +\end{flalign*} +% recast to fit page width +\begin{multline*} +U D_n V - V D_n U +=E_m^p(\mu) E_m^p(\nu) E_n^q(\mu) E_n^q(\nu)\\ +\bigg(E_m^p(\lambda) \frac{dE_n^q(\lambda)}{d\alpha} + - E_n^q(\lambda)\frac{dE_m^p(\lambda)}{d\alpha}\bigg) +\frac{c}{\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}}. +\end{multline*} + +% -----File: 271.png +Integrating $UD_nV-VD_nU$ over the whole ellipsoid, and writing the result +equal to zero, we have +% recast to fit page width +\begin{multline*} +\frac{1}{c}\int\limits_0^{2K'}d\beta\int\limits_0^{4K} E_m^p(\mu)E_m^p(\nu)E_n^q(\mu)E_n^q(\nu)\\ +\bigg[E_m^p(\lambda)\frac{dE_n^q(\lambda)}{d\alpha}-E_n^q(\lambda)\frac{dE_m^p(\lambda)}{d\alpha}\bigg] +(\mu^2-\nu^2)d\gamma=0. +\end{multline*} +\begin{flalign*} +&\indent\text{Hence} & \int\limits_0^{2K'}d\beta\int\limits_0^{4K} E_m^p(\mu)E_m^p(\nu)&E_n^q(\mu)E_n^q(\nu)(\mu^2-\nu^2)d\gamma=0 && \tag{2}&\\ +&\text{unless} & E_m^p(\lambda)\frac{dE_n^q(\lambda)}{d\alpha}&-E_n^q(\lambda)\frac{dE_m^p(\lambda)}{d\alpha}=0. && \tag{3}& +\end{flalign*} + +But as our ellipsoid may be taken at pleasure, $\lambda$ and $\alpha$ are unrestricted, and +if (3) is true it must be true identically. + +If we divide (3) by $[E_m^p(\lambda)]^2$ it becomes +\[ +\frac{d}{d\alpha}\bigg[\frac{E_n^q(\lambda)}{E_m^p(\lambda)}\bigg]=0\quad \text{and}\quad \frac{E_n^q(\lambda)}{E_m^p(\lambda)}=\text{a constant;} +\] +and this obviously cannot be true unless $n=m$ and $q=p$. + +\EXAMPLE{S} + +1.\quad Show that it follows from (2) Art.~140 that +\[ +\int\limits_{-K'}^{K'}d\beta\int\limits_{-K}^{K} +E_m^p(\mu)E_m^p(\nu)E_n^q(\mu)E_n^q(\nu)(\mu^2-\nu^2)d\gamma=0. +\] +\textit{Suggestion:} +\begin{align*} + \int\limits_0^{2K'} E_m^p(\mu)E_n^q(\mu)(\mu^2-\nu^2)d\beta +=\int\limits_0^{K'} E_m^p(\mu)&E_n^q(\mu)(\mu^2-\nu^2)d\beta\\ +&+\int\limits_{K'}^{2K'} E_m^p(\mu)E_n^q(\mu)(\mu^2-\nu^2)d\beta. +\end{align*} + +If in the last integral we replace $\beta$ by $\beta+2K'$ it becomes +\[ + ±\int\limits_{-K'}^0 E_m^p(\mu)E_n^q(\mu)(\mu^2-\nu^2)d\beta +\] +v.\ Arts.~136 and 139 and Int.\ Cal.\ Art.~196.\\ + +2.\quad Show that +\[ + \int\limits_0^{2K'} d\beta \int\limits_0^{4K} [E_m^p(\mu)E_m^p(\nu)]^2(\mu^2-\nu^2)d\gamma +=8\int\limits_0^{K'} d\beta \int\limits_0^{K} [E_m^p(\mu)E_m^p(\nu)]^2(\mu^2-\nu^2)d\gamma. +\] + +% -----File: 272.png +\mypara{141.} We can now solve the problem of finding the value of $V$ at any point +in space when it is given at all the points on the surface of the ellipsoid +$\alpha=\alpha_0$. + +We have first to develop in Ellipsoidal Harmonics a function of $\mu$ and $\nu$ or +rather of $\alpha$ and $\beta$ given at all points on the surface of the ellipsoid in question; +and this is now easily accomplished by our usual method, which leads us to +the result +\[ +f(\alpha_0,\beta,\gamma)=\sum_{m=0}^{m=\infty}\sum_{k=1}^{k=2m+1}\hspace{-0.7em}A_{m,p_k} E^{p_k}_m(\mu)E^{p_k}_m(\nu), \tag{1} +\] +\begin{flalign*} +&\text{where} & A_{m,p_k} = +\frac{\displaystyle\int\limits_0^{2K'}d\beta \int\limits_0^{4K}f(\alpha_0,\beta,\gamma) +E_m^{p_k}(\mu)E_m^{p_k}(\nu)(\mu^2-\nu^2)d\gamma} + {\displaystyle 8\int\limits_0^{K'}d\beta\int\limits_0^{K}[E_m^{p_k}(\mu)E_m^{p_k}(\nu)]^2(\mu^2-\nu^2)d\gamma}. & \tag{2}& +\end{flalign*} + +Our final solution is +\[ +V=\sum_{m=0}^{m=\infty}\sum_{k=1}^{k=2m+1}\hspace{-0.7em}A_{m,p_k} +\frac{E^{p_k}_m(\lambda)}{E^{p_k}_m(\lambda_0)}E^{p_k}_m(\mu)E^{p_k}_m(\nu) \tag{3} +\] +at an internal point; +\[ +V=\sum_{m=0}^{m=\infty}\sum_{k=1}^{k=2m+1}\hspace{-0.7em}A_{m,p_k}\frac{F^{p_k}_m(\lambda)}{F^{p_k}_m(\lambda_0)}E^{p_k}_m(\mu)E^{p_k}_m(\nu) \tag{4} +\] +at an external point. + +Lamé has proved rather ingeniously that +\[ +\int\limits_0^{K'}d\beta \int\limits_0^{K}[E^{p_k}_m(\mu)E^{p_k}_m(\nu)]^2(\mu^2-\nu^2)d\gamma +\] +can always be found and that it is equal to $\dfrac{\pi}{2}$ multiplied by a rational integral +function of the coefficients of $E^{p_k}_m(x)$ and of $c^2$ and $\Big(\dfrac{b}{c}\Big)^2$. + +Of course the labor of obtaining even a few terms of the development of a +function that is in the least complicated is enormous. + +\mypara{142.} If in Laplace's Equation (8) Art.~136 we let $V=E^p_m(\lambda)U$ supposing +$U$ to be a function of $\beta$ and $\gamma$ only, we get after replacing $\dfrac{1}{E_m^p(\lambda)}\dfrac{d^2E_m^p(\lambda)}{d\alpha^2}$ +by its value $m(m+1)\lambda^2-(b^2+c^2)p$\, [v.~(2) Art.~137]\\ +\[ + (\lambda^2-\nu^2)D^2_\beta U ++ (\lambda^2-\mu^2)D^2_\gamma U ++ (\mu^2-\nu^2)[m(m+1)\lambda^2-(b^2+c^2)p] U=0; \tag{1} +\] +% -----File: 273.png +and since by hypothesis $U$ is independent of $\lambda$, the coefficient of $\lambda^2$ in (1) +must vanish. Hence +\[ +D^2_\beta U + D^2_\gamma U + (\mu^2 - \nu^2) m (m + 1) U = 0. \tag{2} +\] +Of course $U= E^p_m(\mu)E^p_m(\nu)$ will satisfy (2). + +\EXAMPLE{S} + +1.\quad Substitute $U=E^p_m(\mu)E^p_m(\nu)$ in (2) Art.~142 and by the aid of (2) Art.~137 +show that the equation (2) Art.~142 is satisfied.\\ + +2.\quad Obtain (2) Art.~140 directly from (2) Art.~142.\\ + +3.\quad \textit{Conical Coördinates.} Consider the system of coördinates defined by the +equations +\begin{equation*}\left. +\begin{aligned} +x^2+y^2&+z^2=r^2\\ +\dfrac{x^2}{\mu^2}+\dfrac{y^2}{\mu^2-b^2}&+\dfrac{z^2}{\mu^2-c^2}=0\\ +\dfrac{x^2}{\nu^2}+\dfrac{y^2}{\nu^2-b^2}&+\dfrac{z^2}{\nu^2-c^2}=0\; +\end{aligned} \right\}\tag{1} +\end{equation*} +where $c^2>\mu^2>b^2>\nu^2$. + +Show that +\[ +x^2=\frac{r^2\mu^2\nu^2}{b^2c^2},\quad +y^2=\frac{r^2(\mu^2-b^2)(\nu^2-b^2)}{b^2(b^2-c^2)},\quad +z^2=\frac{r^2(\mu^2-c^2)(\nu^2-c^2)}{c^2(c^2-b^2)}; +\] +\[ +h_1^2=\frac{(\mu^2-b^2)(c^2-\mu^2)}{r^2(\mu^2-\nu^2)},\quad +h_2^2=\frac{(b^2-\nu^2)(c^2-\nu^2)}{r^2(\mu^2-\nu^2)},\quad +h_3^2=1. +\] + +Laplace's Equation is +\[ +D^2_\alpha V + D^2_\beta V +(\mu^2-\nu^2)D_r(r^2 D_r V) = 0 \tag{2} +\] +where \hspace{\stretch{1}}$\displaystyle\alpha=\int\limits_b^\mu\frac{d\mu}{\sqrt{(\mu^2-b^2)(c^2-\mu^2)}}\quad \text{and}\quad +\beta = \int\limits_0^\nu\frac{d\nu}{\sqrt{(b^2-\nu^2)(c^2-\nu^2)}},$\hspace{\stretch{1}} + +If $V=U.R$ (2) breaks up into +\begin{flalign*} +&&\frac{d}{dr}\bigg(r^2&\frac{dR}{dr}\bigg)=m(m+1)R, \tag{3}\\ +&&D^2_\alpha U + D^2_\beta U &+ m(m+1)(\mu^2-\nu^2)U=0. \tag{4}\\[1ex] +&\text{(3) gives} & R&=Ar^m+Br^{-m-1}. &&\\[1ex] +&\text{(4) gives} && U=E^p_m(\mu)E^p_m(\nu) & \tag*{(v.\ Art.~142).}& +\end{flalign*} + +So that a solution of (2) is +\[ +V=Ar^m E^p_m(\mu)E^p_m(\nu). +\] + +But since (2) is Laplace's Equation, $V=Ar^mY_m(\mu,\phi)$, if expressed in +Conical Coördinates, must satisfy it, consequently $E^p_m(\mu)E^p_m(\nu)$ must be simply +a Spherical Harmonic of the $m$th degree. +% -----File: 274.png + +\vspace{1ex}\Needspace*{5\baselineskip} +\begin{center} +\emph{Toroidal Coördinates.} +\end{center}\vspace{-1ex} + +\mypara{143.} Any pair of circles belonging to the orthogonal system obtained and +figured in Art.~46 can be represented by the equations +\[ +\left. +\begin{aligned} +\dfrac{2ax}{\sinh \alpha} &= \dfrac{x^2 + y^2 + a^2}{ \cosh \alpha}\;\\ +\dfrac{2ay }{\sin \beta} &= \dfrac{x^2 + y^2 - a^2 }{\cos \beta} +\end{aligned} +\right\} +\tag{1} +\] +if we take $2a$ instead of 2 as the distance between the points common to the +second set of circles. + +If we rotate the system about the axis of $y$ we get a set of spheres and a +set of anchor rings which cut orthogonally. These and a set of planes through +the axis of revolution will form an orthogonal system of surfaces, and the +parameters corresponding to them may be taken as a set of curvilinear +coördinates and may be called \emph{Toroidal Coördinates}. + +If we take the axis of the system as the axis of $Z$, the equations of a set of +the surfaces may be written +\[ +\left. +\begin{aligned} +\dfrac{4a^2(x^2+y^2) }{\sinh^2 \alpha} &= \dfrac{[x^2+y^2+z^2+a^2]^2}{\cosh^2 \alpha}\;\\ +\dfrac{2az}{\sin \beta} &= \dfrac{x^2+y^2+z^2-a^2}{\cos \beta}\\ +y &= x \tan\gamma +\end{aligned} +\right\} +\tag{2} +\] + +$\alpha$, $\beta$, and $\gamma$ being regarded as the coördinates of a point of intersection of the +three surfaces. + +Finding Laplace's Equation in the usual manner we get +\begin{gather*} +x= \dfrac{a \sinh\alpha \cos\gamma}{\cosh\alpha \mp \cos\beta}, \quad y=\dfrac{a \sinh\alpha \sin\gamma}{\cosh\alpha \mp \cos\beta},\quad z=\dfrac{a \sin\beta}{\cosh\alpha \mp \cos\beta},\\[0.5em] +r=\sqrt{x^2+y^2}=\dfrac{a \sinh\alpha}{\cosh\alpha \mp \cos\beta},\quad a+z \ctn\beta=\dfrac{a \cosh\alpha}{\cosh\alpha \mp \cos\beta};\\[0.5em] +h_1=\dfrac{\cosh\alpha \mp \cos\beta}{a},\quad h_2=\dfrac{\cosh\alpha \mp \cos\beta}{a},\quad h_3=\dfrac{\cosh\alpha \mp \cos\beta}{a \sinh\alpha}; +\end{gather*} +and Laplace's Equation becomes +\begin{align*} +D_\alpha\left[\dfrac{a \sinh\alpha}{\cosh\alpha \mp \cos\beta} D_\alpha V\right]&+D_\beta\left[\dfrac{a \sinh\alpha}{\cosh\alpha \mp \cos\beta} D_\beta V\right]\\[0.5em] +&+D_\gamma\left[\dfrac{a}{\sinh\alpha(\cosh\alpha \mp \cos\beta)} D_\gamma V\right] = 0, \tag{1} +\end{align*} +% -----File: 275.png +\begin{flalign*} +&\text{or}& D_\alpha(rD_\alpha V) + D_\beta(rD_\beta V) + \dfrac{1}{\sinh^2\alpha} rD_\gamma^2 V &= 0.& \tag{2} +\end{flalign*} + +We cannot proceed further by our usual method, for the assumption that $V$ +is a function of $\alpha$ alone, or that $V$ is a function of $\beta$ alone, proves to be +inadmissible. Indeed, not only are $\alpha$, $\beta$, and $\gamma$ not \textit{thermometric parameters} +(v.\ Art.~133), but no thermometric parameters exist, and no possible distribution +can make our anchor rings or our spheres a set of equipotential surfaces. + +We can, however, simplify (2). It can be written +\[ +D_\alpha^2(V\sqrt{r}) + D_\beta^2(V\sqrt{r})+ \dfrac{1}{\sinh^2\alpha} D_\gamma^2(V\sqrt{r})-V(D_\alpha^2 \sqrt{r}+D_\beta^2\sqrt{r})=0. \tag{3} +\] + +$D_\alpha^2\sqrt{r}+D_\beta^2\sqrt{r}$ proves equal to $-\dfrac{\sqrt{r}}{4\sinh^2\alpha}$; hence if $U = V\sqrt{r}$ (3) becomes +\[ +\sinh^2\alpha(D_\alpha^2U+D_\beta^2U) + D_\gamma^2 U + \tfrac{1}{4} U=0, \tag{4} +\] +for which particular solutions can readily be found by our usual process. + +(4) can be broken up into the three equations +\begin{gather*} +\dfrac{d^2N}{d\gamma^2} + (m+\tfrac{1}{2})^2N = 0 \tag{5}\\ +\dfrac{d^2M}{d\beta^2} + n^2M = 0 \tag{6}\\ +\sinh^2\alpha \dfrac{d^2L}{d\alpha^2} - [m(m+1)+n^2\sinh^2\alpha]L=0. \tag{7}\\ +N = A \cos(m+\tfrac{1}{2})\gamma + B \sin(m+\tfrac{1}{2})\gamma\\[1ex] +M = A_1 \cos n\beta + B_1 \sin n\beta. +\end{gather*} + +If we introduce into (7) $x = \ctnh \alpha$ it becomes +\[ +(1-x^2) \dfrac{d^2L}{dx^2} - 2x\dfrac{ dL}{dx} + \left[m(m+1) - \dfrac{n^2}{1-x^2}\right]L=0, +\] +a solution of which is +\begin{flalign*} +&&L &= P_m^n(x) = (1-x^2)^{\frac{n}{2}} \dfrac{d^nP_m(x)}{dx^n} &\text{(v.\ Art.~102).} +\end{flalign*} +It is to be noted that since $\ctnh \alpha$ is greater than 1 +\[ +P_m^n(\ctnh \alpha) = i^{\frac{n}{2}}\csch^n \alpha \dfrac{d^n P_m(\ctnh \alpha)}{(d \ctnh \alpha)^n}. +\] +% -----File: 276.png +The constant coefficient $i^{\frac{n}{2}}$ can be rejected and we get +\[ +U=[A \cos(m+\tfrac{1}{2})\gamma + B \sin(m+\tfrac{1}{2})\gamma](A_1 \cos n\beta + B_1 \sin n\beta)\csch^n \alpha \dfrac{d^nP_m(\ctnh \alpha)}{(d \ctnh \alpha)^n} +\] +as a particular solution of (4). +\[ +\dfrac{1}{i^{\frac{n}{2}}} P_m^n(\ctnh \alpha)=\csch^n \alpha \dfrac{d^nP_m(\ctnh \alpha)}{(d \ctnh\alpha)^n} +\] +has been called a Toroidal Harmonic. + +\EXAMPLE{S} + +1.\quad Given the value of the potential function at all points on the surface of +an anchor ring; find its value at any point within the ring. + +\emph{Suggestion}: If $V = f(\beta,\gamma)$ when $\alpha = \alpha_0$, the function to be developed is +\[ +\sqrt{r}.f(\beta,\gamma)\quad \text{\textit{i.e}.}\quad \left[\dfrac{ a \sinh \alpha_0}{ \cosh \alpha_0\mp \cos \beta} \right]^{\frac{1}{2}} f(\beta,\gamma) +\] +and the development will be in a double Fourier's Series (v.\ Art.~71).\\ + +2.\quad Show that if we let $\alpha$ range from 0 to $\infty$, $\beta$ from $-\pi$ to $\pi$, and $\gamma$ from +0 to $2\pi$, each of the double signs on page 264 may be replaced by the minus +sign without loss of generality. +\label{ch8end} + +% -----File: 277.png +\mychap{CHAPTER IX.\footnotemark}{HISTORICAL SUMMARY.} +\label{ch9start} + +\footnotetext{See preface.}The method of development in series which has enabled us in the preceding +chapters to solve problems in various branches of mathematical physics, had +its origin, as might have been expected, in the theory of the musical vibrations +of a stretched string. It was in the year 1753\footnote{See two articles by Bernoulli and one by Euler in the Memoirs of the Academy of +Berlin for this year.} that Daniel Bernoulli +enunciated the principle of the coexistence of small oscillations, which, in +connection with Taylor's and John Bernoulli's theory of the vibrating string, +led him to believe that the general solution of this problem could be put in +the form of a trigonometric series. This principle also led him and Euler to +treat in a similar manner the problems of the vibration of a column of air and +of an elastic rod. The problem of the vibration of a heavy string suspended +from one end was also treated in the same manner by these mathematicians +and deserves special mention here as in it Bessel's functions of the zeroth +order appear for the first time.\footnote{See the Transactions of the Academy of St.\ Petersburg for 1732-33, 1734 and 1781.} In none of these cases, however, was any +method given for determining the coefficients of the series. + +This last remark also applies to the more complicated problems of the +vibration of rectangular and circular membranes, which were discussed by +Euler\footnote{Transactions of the Academy of St.\ Petersburg.} in 1764, and in the last of which the general Bessel's functions of +integral orders occur. + +It is in problems connected with astronomy that the first completely +successful application of the method here considered occurs. Legendre in a +paper published in the Mémoires des Savants Étrangers for 1785, first +introduced the zonal harmonics $P_m$ and applied them to the determination of +the attraction of solids of revolution. He was followed by Laplace, who in +one of the most remarkable memoirs ever written\footnote{``Théorie des attractions des sphéroïdes et de la figure des Planètes'' Mémoires de +l'académie des sciences 1782. This article, although bearing an earlier date than that of +Legendre, was really inspired by it. It is here that ``Laplace's equation'' first appears, +occurring, however, only in polar coördinates.} determined the potential +of a solid differing but little from a sphere by means of the development +according to the spherical harmonics $Y_m$. + +% -----File: 278.png +Very closely related to this problem is Gauss's celebrated treatment of the +theory of terrestrial magnetism,\footnote{Resultate aus den Beobachtungen des magnetischen Vereins im Jahre 1838. Leipzig, +1839. Reprinted in Gauss's collected works, Vol.\ V., p.\ 121.} which we will for that reason mention here, +although it was not published until more than half a century later. This +paper is particularly noteworthy as it contains a numerical application of the +method on a larger scale than has ever been attempted before or since. + +After the researches of Legendre and Laplace there was a pause of a +quarter of a century until in 1812 Fourier's extensive memoir: \textit{Théorie du +mouvement de la chaleur dans les corps solides} was crowned by the French +Academy. Although not printed until the years 1824-26,\footnote{Mémoires de l'académie des sciences for 1819-20 and 1821-22.} the manuscript of +this work was in the meantime accessible to the other French mathematicians +presently to be mentioned. The first part of this memoir, which was reproduced +with but few alterations in the \textit{Théorie analytique de la chaleur} (1822), +contains a treatment of the following problems and of practically all of their +special cases: + +(\emph{a}) The one dimensional flow of heat. (\emph{b}) The two dimensional flow of +heat in a rectangle. (\emph{c}) The three dimensional flow of heat in a rectangular +parallelopiped. (\emph{d}) The flow of heat in a sphere when the temperature +depends only on the distance from the centre. (\emph{e}) The flow of heat in a +right circular cylinder when the temperature depends only on the distance +from the axis. In these problems not merely the simpler boundary conditions +are considered but also the question of radiation into an atmosphere. In +special cases of the first three problems just mentioned (when one or more +dimensions become infinite) the series degenerate into ``Fourier's integrals." + +\markright{HISTORICAL SUMMARY.} +More important even than any of these special problems is the great +advance which Fourier caused the theory of trigonometric series to make. +In a posthumous paper Euler had given the formulae for determining the +coefficients,\footnote{Lagrange had practically determined these coefficients long before but failed to notice +what he had got.} but Fourier was the first to assert and to attempt to prove that +any function, even though for different values of the argument it is expressed +by different analytical formulae, can be developed in such a series. The fact +that the real importance of trigonometric series was thus for the first time +shown justifies us in associating Fourier's name with them, although, as we +have seen, they were known long before his day. + +Fourier's results were extended by Laplace in 1820\footnote{Connaissance des Temps pour l'an 1823.} to the general (unsymmetrical) +case of the flow of heat in a sphere, and by Poisson\footnote{Journal de l'École Polytechnique, 19\textsuperscript{e} Cahier. Although the final forms to which Poisson +reduces his results are similar to Fourier's, his methods are very different.} (1821) to the +unsymmetrical flow of heat in a cylinder. + +% -----File: 279.png +In 1835 Green published a paper\footnote{``On the determination of the exterior and interior attraction of ellipsoids of variable +densities.'' Transactions of the Cambridge Philosophical Society.} in which the method we are considering +is employed to determine the potential of a heterogeneous ellipsoid. This +paper, in which the analysis is performed at once for space of $n$ dimensions, +anticipates much that was subsequently done by others, but has failed to +exert an influence proportional to its importance. + +At about this time Lamé began a series of publications which have connected +his name inseparably with the problem of the permanent state of +temperature of an ellipsoid. In the first of these\footnote{Mémoires des Savants Étrangers, Vol.\ V\@. Although the volume is dated 1838 this paper +(which was reprinted in Liouville's Journal, 1837) must have appeared at least as early as 1835.} the equation $\nabla^2V=0$ is +transformed to ellipsoidal coördinates and is then broken up into three +ordinary differential equations. The rest of the solution, however, is hardly +touched upon. Lamé's most important work on this subject\footnote{ +``Sur l'équilibre des Températures dans un ellipsoïde à trois axes inégaux.'' An article +by the same author on the two dimensional potential will be found in Vol.\ I. of this Journal.} was published +in Liouville's Journal in 1839, and in it the complete solution of the problem +is given. Lamé clearly shows in this paper how he arrived at his solution, by +considering first the simpler case of a sphere where, instead of the polar +coördinates $\theta$ and $\phi$, the parameters of two families of confocal cones of the +second degree are used as coördinates. This system of curvilinear coördinates, +which, when applied to the complete sphere, merely gives the old results of +Laplace in a new form, is barely mentioned in Lamé's later publications. In +the same volume of Liouville's Journal Lamé published a second paper in +which he applies his results to the special cases of ellipsoids of revolution. + +These two papers form the starting-point for a series of articles on the +same subject by Heine and Liouville. Heine in his doctor dissertation\footnote{Reprinted in Crelle's Journal, Vol.\ 26 (1843).\\ +In the same Journal for 1847 F.~Neumann discussed the related problem of the magnetisation +of a soft iron ellipsoid of revolution.} (1842) +determined the potential not merely for the interior of an ellipsoid of +revolution when the value of the potential is given on the surface, but also +for the exterior of such an ellipsoid and for the shell between two confocal +ellipsoids of revolution. Even in the first of these problems, which is +equivalent to that of Lamé, he simplified Lamé's solution materially by +showing that the functions used may be reduced to spherical harmonics, +while in the other two problems he introduced spherical harmonics of the +second kind, which were then new. Shortly afterwards\footnote{\textit{Heine:} Crelle's Journal, Vol.\ 29, 1845. \textit{Liouville:} Liouville's Journal, Vol.\ X., +1845, and Vol.\ XI., 1846. For a treatment of the problem of the potential of an ellipsoidal +shell by means of a development of $\dfrac{1}{r}$ in terms of Lamé's functions, see a paper by Heine +in Crelle's Journal, Vol.\ 42, 1851.} Heine and Liouville +% -----File: 280.png +published simultaneously two papers in which they arrived independently of +each other at about the same results. In each of these papers attention is +called to the fact that the product of two Lamé's functions is a spherical +harmonic, and this fact is made use of to throw Lamé's solution of the +problem of the permanent state of temperatures of an ellipsoid into a more +elementary form. Besides this the second solution of Lamé's equation is +introduced for the sake of solving the potential problem for the \emph{exterior} of +the ellipsoid. + +In thus following up the theory of heat and the related potential problems, +we have lost sight of the question of small vibrations, to which during the +early part of the century a great deal of attention had been devoted by +Poisson, who frequently made use of the method of development in series. +In his memoirs\footnote{See especially the one in the Mémoires de l'académie des sciences, Vol.\ VIII., 1829.} most of the problems left unfinished by Bernoulli and Euler +are thoroughly treated, as well as various slight modifications of them. +When, however, he attacked the problem of the vibration of an elastic plate +he was unable to make much progress, owing in part to the erroneous form of +his boundary conditions. He was, nevertheless, able to solve the problem of +the \emph{symmetrical} vibration of a free circular plate. The complete theory of the +vibration of a free circular plate was first given by Kirchhoff.\footnote{Crelle's Journal, Vol.\ 40, 1850.} + +Passing now to a new subject, the theory of the equilibrium of an elastic +spherical shell, we find a solution by Lamé in Liouville's Journal for 1854, +and by Sir William Thomson (1862) in the Philosophical Transactions for +1863. Both of these papers consist of an application of the spherical harmonic +analysis to this rather complicated problem. Thomson, however, considers +besides Lamé's problem certain related questions and the form of his analysis +is very different from Lamé's, being of the same nature as that used in the +Appendix B of his Natural Philosophy of which we shall have to speak +presently. These investigations form the starting point for a number of +recent memoirs among which those of G.~H. Darwin on cosmographical +questions deserve special mention. + +Closely related to this last mentioned problem is the theory of the small +vibrations of an elastic sphere. While the simplest case of this problem was +treated by Poisson in the memoir referred to above, the general solution has +been only recently obtained by Jaerisch (1879)\footnote{Crelle's Journal, Vol.\ 88.} and Lamb (1882).\footnote{Proc.\ Lond.\ Math.\ Soc.} The +functions involved are the same as those which occur in the problem of the +non-stationary flow of heat in a sphere as solved by Laplace. + +The Appendix B of Thomson and Tait's Natural Philosophy,\footnote{First edition, 1867. This appendix +was evidently written as early as 1862, as Thomson refers to it in the memoir quoted above.} to which we +have already referred, deserves to be regarded as one of the most important +% -----File: 281.png +contributions to the general theory. The way in which spherical harmonics +are introduced (as homogeneous functions of the rectangular coördinates) was +then new\footnote{The same method was used at about the same time by Clebsch.} and the solution of the potential problem for a variety of new +solids was indicated; viz., for solids whose boundaries consist of concentric +spheres, cones of revolution, and planes. We shall have more to say presently +concerning the method employed for the solution of these problems. + +Although connected only indirectly with the theory we are discussing, it +will be well to mention at this point the method of electrical images which is +also due to Sir William Thomson (1845). This method enables us to solve +many potential problems for the inverse of any solid when once we have +solved it for the solid itself. By means of this method most of the solutions +of potential problems obtained by our method may be applied at once with +very little modification to systems of curvilinear coördinates derived by +inversion from those we have used. It will not be necessary to mention +separately problems of this sort, as it is clearly immaterial whether they be +solved directly or by means of the method of inversion.\footnote{A case in point would be the potential problem for the shell between two non-intersecting +eccentric spheres, since these spheres can be inverted into concentric spheres. This problem +was treated directly by C. Neumann in a monograph published in Halle in 1862.} + +Returning now to the Continent, we find as the next important question +taken up the problem of the potential of an anchor ring. The first publication +on this subject is a monograph by C. Neumann\footnote{``Theorie der Elektricitäts- und Wärme-Vertheilung in einem Ringe.'' Halle.} (1864), but in Riemann's +posthumous papers which were not published until 1876, ten years after his +death, will be found a short fragment on this subject, which (\emph{cf.}\ the last page +of Hattendorf's edition of Riemann's lectures: ``Partielle Differentialgleichungen'') +would appear to date back to the winter 1860-61. This fragment +is of peculiar interest, as the opening paragraphs clearly show that Riemann +had in mind an extended article on the fundamental principles of our subject. + +We will next mention two papers by Mehler in which the functions known +as ``conal harmonics,'' which had already been introduced by Thomson in the +Appendix B above mentioned, were applied to the solution of two problems in +electrostatics. The first of these papers\footnote{Crelle's Journal, Vol.\ 68, 1868.} (1868) deals with the solid bounded +by two intersecting spheres, while in the second\footnote{Jahresbericht des Gymnasiums zu Elbing.} (1870) the infinite cone of +revolution is treated. Both of these problems are essentially different from +those discussed in the ``Appendix B,'' inasmuch as the infinite series which +we usually have degenerate in these cases into definite integrals, just as they +do in some simpler cases treated by Fourier. The later of the two papers +just quoted also contains valuable information concerning the nature of the +% -----File: 282.png +solution of similar problems for the hyperboloids and paraboloids of revolution. +The solutions of these problems are not, however, given. + +It remains, in order to close the history of this part of the subject, to mention +a number of memoirs which although treating entirely new problems are of far +less importance than most of those considered up to this point, partly because +the solution is not brought to a point where it can be of much immediate use, +and partly because most of the methods employed are such as could not fail +to present themselves to any one attacking these problems. + +Of these the first is a paper by Mathieu\footnote{Liouville's Journal, Vol.\ XIII.} on the vibration of an elliptic +membrane (1868), in which the functions of the elliptic cylinder occur for the +first time. + +This was followed in the same year by a paper on closely allied subjects by +H. Weber,\footnote{``Ueber die Integration der partiellen Differentialgleichung $\frac{\delta^2 u}{\delta x^2} + \frac{\delta^2 u}{\delta y^2} + k^2 u=0.$'' +Math.\ Ann., Vol.\ I\@. No \emph{physical} problem is mentioned in this paper.} in which not merely the case of the complete ellipse is briefly +considered, but also that in which the boundary consists of two arcs of +confocal ellipses and two arcs of hyperbolas confocal with them. The special +case in which the ellipses and hyperbolas become confocal parabolas is also +considered, whereby the functions of the parabolic cylinder are for the first +time introduced. + +In Mathieu's ``Cours de physique mathématique'' (1873) the problem of +the non-stationary flow of heat in an ellipsoid is touched upon, and an +elaborate though not very satisfactory treatment of the special cases where +we have ellipsoids of revolution is given. New functions appear in all of +these problems. + +Of late years C. Baer has supplied a number of missing links in the chain +of problems here considered by treating in succession the potential problem +for the paraboloid of revolution,\footnote{``Ueber das Gleichgewicht und die Bewegung der Wärme in einem Rotationsparaboloid.'' +Dissertation, Halle, 1881.} the parabolic cylinder\footnote{``Die Funktion des parabolischen Cylinders,'' Gymnasialprogramm Cüstrin, 1883.} and the general +paraboloid.\footnote{``Parabolische Coordinaten,'' Frankfurt, 1888. See also a paper by Greenhill in the +Proc.\ Lond.\ Math.\ Soc., Vol.\ XIX., 1889 (read Dec.\ 8, 1887). Also a posthumous paper by +Lamé in Liouville's Journal for 1874, Vol.\ XIX.} In the first of these problems Bessel's functions occur, as had +already been stated by Mehler, while in the last we find the functions of the +elliptic cylinder. For each of the three systems of coördinates employed the +same author also touches upon the more general problem of the non-stationary +flow of heat, in which new functions occur. + +Except in the case of the anchor ring we have found so far only such solids +treated by our method as are bounded by surfaces of the first or second +% -----File: 283.png +degree. Wangerin\footnote{Preisschriften der Jablanowski'schen Gesellschaft, No.\ XVIII., and Crelle's Journal, +Vol.\ 82. See also, concerning a still further extension, the Berliner Monatsberichten +for 1878.} (1875-76) considered in connection with the theory of +the potential, more general systems of curvilinear coördinates than had +previously been used in physical questions, namely, \textit{cyclidic} coördinates.\footnote{Cyclids are a kind of surface of the fourth order (see Salmon's Geom.\ of three Dimensions, +p.\ 527). In his first memoir Wangerin considers only cyclids of revolution.} He +showed, however, merely how to break up Laplace's equation into three +ordinary differential equations.\footnote{See also a paper by this author in Grünert's Archiv for 1873, where the problem of the +equilibrium of elastic solids of revolution is treated.} + +An important branch of our theory which we have not yet touched upon +dates back to the year 1836, when Sturm published a series of fundamentally +important papers in the first two volumes of Liouville's Journal. The +physical question which lies at the basis of these papers is the problem of the +flow of heat in a heterogeneous bar.\footnote{The similar problem of the vibration of a heterogeneous string under the action of an +external force was treated by Maggi (Giornale di Matematiche, 1880). Several special cases +are also considered here in detail.} The method here employed depends +upon the fact that the functions which occur are characterized by the number +of times they vanish in a certain interval. This same idea reappears in +Thomson and Tait's Appendix B already referred to, but first finds its full +expression in this more general field of the three dimensional potential in an +article by Klein: ``Ueber Körper welche von confocalen Flächen zweiten +Grades begrenzt sind''\footnote{Math.\ Ann., 18.} (1881). Still more recently (1889-90) Klein has in +his lectures extended this theory to the treatment of solids bounded by six +confocal cyclids, and has indicated how all the potential problems heretofore +treated by our method are special cases of this one.\footnote{For an exposition of this theory see the treatise: Ueber die Reihenentwickelungen der +Potentialtheorie, Leipsic, Teubner, 1894, by the writer of the present chapter.} + +Of late years, especially since the year 1880, the younger English mathematicians +have done a vast amount of work in the theory we are here +considering. Although much of this work is of great value, hardly any of it +can be regarded as being a real \emph{development} of the method; it is rather an +application of it to a great variety of problems. We must therefore content +ourselves with giving a mere list of a few of the more important of these +papers. + +\textit{Niven:} On the Conduction of Heat in Ellipsoids of Revolution. Phil.\ +Trans., 1880. + +\textit{Niven:} On the Induction of Electric Currents in Infinite Plates and +Spherical Shells.\ Phil.\ Trans., 1881. + +% -----File: 284.png +\emph{Hicks:} On Toroidal Functions. Phil.\ Trans., 1881. + +\emph{Hicks:} On the Steady Motion and Small Vibrations of a Hollow Vortex. +Phil.\ Trans., 1884, 1885. + +\emph{Lamb:} On Ellipsoidal Current Sheets. Phil.\ Trans., 1887. + +\emph{Chree:} The Equations of an Isotropic Elastic Solid in Polar and Cylindrical +Coördinates, their Solution and Application. Camb.\ Phil.\ Soc.\ Trans., +XIV., 1889. + +\emph{Hobson:} On a Class of Spherical Harmonics of Complex Degree with +Applications to Physical Problems. Camb.\ Phil.\ Soc.\ Trans., XIV., 1889. + +\emph{Chree:} On some Compound Vibrating Systems. Camb.\ Phil.\ Soc.\ Trans., +XV., 1891. + +\emph{Niven:} On Ellipsoidal Harmonics. Phil.\ Trans., 1892. + +The historical sketch we have just given would naturally require as a +supplement some account of the work that has been done on the question of +the convergence of the various series which occur. This, however, would +carry us too far, and we will content ourselves with mentioning the two +fundamental memoirs by Dirichlet in Crelle's Journal, one in 1829 on +Fourier's series, and one, which has been criticised to some extent by subsequent +mathematicians, in 1837 on Laplace's spherical harmonic development. + +Another subject which naturally presents itself here is the theory of the +various new functions we have met. Those properties of these functions, +however, which the physicist needs have usually been investigated by the +physicists themselves in the papers mentioned above; while any thorough +account of the development of the theory of these functions would lead us +into the vast region of the modern theory of linear differential equations. + +We will therefore close by merely giving a list of books which will be +found useful by those wishing to continue their study of the subject further. + +We begin with the books relating directly to physical questions: + +\emph{Fourier:} Théorie Analytique de la Chaleur, 1822. + +\emph{Lamé:} Leçons sur les Functions inverses des Transcendantes et les Surfaces +isothermes, 1857. + +\emph{Lamé:} Leçons sur les Coordonnées Curvilignes et leurs diverses Applications, +1859. + +\emph{Mathieu:} Cours de Physique Mathématique, 1873. + +\emph{Riemann:} Partielle Differentialgleichungen, und deren Anwendung auf +phys\-ikalische Fragen (edited by Hattendorf), third edition. 1882. + +\emph{F.~Neumann:} Theorie des Potentials und der Kugelfunktionen (edited by +C. Neumann), 1887. + +\emph{Thomson and Tait:} Natural Philosophy, second edition, 1879. + +\emph{Rayleigh:} Theory of Sound, 1877. + +\emph{Basset:} Hydrodynamics, 1888. + +\emph{Love:} Theory of Elasticity, 1892. + +% -----File: 285.png +\emph{Heine:} Handbuch der Kugelfunktionen (second edition), 1878-81. + +\emph{Ferrers:} Spherical Harmonics, 1881. + +\emph{Haentzschel:} Reduction der Potentialgleichung auf gewöhnliche Differentialgleichungen, +1893. + +These last three books would also belong in the following list of books +relating to the theory of the various functions we use: + +\emph{Todhunter:} The Functions of Laplace, Lamé and Bessel, 1875. + +\emph{Lommel:} Studien über die Bessel'schen Funktionen, 1868. + +\emph{F.~Neumann:} Beiträge zur Theorie der Kugelfunktionen, 1878. + +And finally concerning the question of convergence: + +\emph{C.~Neumann:} Über die nach Kreis-, Kugel- und Cylinder-Functionen +fort\-schreitenden Entwickelungen, 1881. +\label{ch9end} +% -----File: 286.png +%[Blank Page] +% -----File: 287.png +\label{tablestart} +\setcounter{footnote}{0} +\newpage +\medskip\begin{center}{\Large APPENDIX.}\par\rule{5em}{0.5pt}\par\bigskip{TABLES.}\end{center} +\markright{} + +Table I., a table of Surface Zonal Harmonics (Legendrians), gives the values +of the first seven Harmonics $P_1(\cos \theta)$, $P_2(\cos \theta)$, $\cdots P_7(\cos \theta)$ for the argument +$\theta$ in degrees. It is taken from the Philosophical Magazine for December, +1891, and was computed by Messrs.\ C.~E. Holland, P.~R. James, and C.~G. Lamb, under the direction of Professor John Perry. + +Table II., a table of Surface Zonal Harmonics (Legendrians), gives the +values of the first seven Harmonics $P_1(x)$, $P_2(x)$, $\cdots P_7(x)$ for the argument $x$. +It is reduced from the Tables of Legendrian Functions computed under the +direction of Dr.\ J.~W.~L. Glaisher, and published in the Report of the British +Association for the Advancement of Science for the year 1879. + +Table III., the table of Hyperbolic Functions, gives the values of $e^x$, $e^{-x}$, +$\sinh x$, $\cosh x$, and $\gd x$ (Gudermannian of $x$) for values of $x$ from 0.00 to 1.00; +and the values of $\log \sinh x$ and $\log \cosh x$ for values of $x$ from 1.00 to 10.0. +The values of $\gd x$, $\log \sinh x$, and $\log \cosh x$ are taken from the Mathematical +Tables prepared by Professor J.~M. Peirce (Boston: Ginn \& Co.). + +The $\log \sinh x$ and $\log \cosh x$ for values of $x$ between 0.00 and 1.00 can be +obtained from the values given for the Gudermannian of $x$ in the table by the +aid of the relations +\[ +\log \sinh x = \log \tan (\gd x)\qquad +\log \cosh x = \log \sec (\gd x). +\] + +Table IV.\ gives the first twelve roots of $J_0(x) = 0$ and $J_1(x) = 0$ each +divided by $\pi$. The table is taken from Lord Rayleigh's Sound, Vol.\ I., +page 274, and is due to Professor Stokes, Camb.\ Phil.\ Trans., Vol.\ IX., +page 186. + +Table V.\ gives the first nine roots of $J_0(x) = 0$, $J_1(x) = 0$, $\cdots J_5(x)=0$. +The table is taken from Rayleigh's Sound, Vol.\ I., page 274, and is due to +Professor J. Bourget, Ann.\ de l'Ecole Normale, T. III., 1866, page 82. + +Table VI., the table of Bessel's Functions, gives the values of the Bessel's +Functions $J_0(x)$ and $J_1(x)$ for the argument $x$ from $x = 0$ to $x = 15$. It is +taken from Rayleigh's Sound, Vol.\ I., page 265, and from Lommel's Bessel'sche +Functionen. + +% -----File: 288.png +\newpage +\markright{APPENDIX} +\label{tableI} +\begin{center} +TABLE I. --- \textsc{Surface Zonal Harmonics} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|c|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut\theta}& +\multicolumn{1}{c|}{P_1(\cos\theta)}& +\multicolumn{1}{c|}{P_2(\cos\theta)}& +\multicolumn{1}{c|}{P_3(\cos\theta)}& +\multicolumn{1}{c|}{P_4(\cos\theta)}& +\multicolumn{1}{c|}{P_5(\cos\theta)}& +\multicolumn{1}{c|}{P_6(\cos\theta)}& +\multicolumn{1}{c|}{P_7(\cos\theta)}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\phantom{\text{°}0}0\text{°} & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 \\ +\phantom{0}1 & .9998 & .9995 & .9991 & .9985 & .9977 & .9967 & .9955 \\ +\phantom{0}2 & .9994 & .9982 & .9963 & .9939 & .9909 & .9872 & .9829 \\ +\phantom{0}3 & .9986 & .9959 & .9918 & .9863 & .9795 & .9713 & .9617 \\ +\phantom{0}4 & .9976 & .9927 & .9854 & .9758 & .9638 & .9495 & .9329 \\[2ex] +\phantom{0}5 & .9962 & .9886 & .9773 & .9623 & .9437 & .9216 & .8961 \\ +\phantom{0}6 & .9945 & .9836 & .9674 & .9459 & .9194 & .8881 & .8522 \\ +\phantom{0}7 & .9925 & .9777 & .9557 & .9267 & .8911 & .8476 & .7986 \\ +\phantom{0}8 & .9903 & .9709 & .9423 & .9048 & .8589 & .8053 & .7448 \\ +\phantom{0}9 & .9877 & .9633 & .9273 & .8803 & .8232 & .7571 & .6831 \\[2ex] +10 & .9848 & .9548 & .9106 & .8532 & .7840 & .7045 & .6164 \\ +11 & .9816 & .9454 & .8923 & .8238 & .7417 & .6483 & .5461 \\ +12 & .9781 & .9352 & .8724 & .7920 & .6966 & .5892 & .4732 \\ +13 & .9744 & .9241 & .8511 & .7582 & .6489 & .5273 & .3940 \\ +14 & .9703 & .9122 & .8283 & .7224 & .5990 & .4635 & .3219 \\[2ex] +15 & .9659 & .8995 & .8042 & .6847 & .5471 & .3982 & .2454 \\ +16 & .9613 & .8860 & .7787 & .6454 & .4937 & .3322 & .1699 \\ +17 & .9563 & .8718 & .7519 & .6046 & .4391 & .2660 & .0961 \\ +18 & .9511 & .8568 & .7240 & .5624 & .3836 & .2002 & .0289 \\ +19 & .9455 & .8410 & .6950 & .5192 & .3276 & .1347 & -.0443 \\[2ex] +20 & .9397 & .8245 & .6649 & .4750 & .2715 & .0719 & -.1072 \\ +21 & .9336 & .8074 & .6338 & .4300 & .2156 & .0107 & -.1662 \\ +22 & .9272 & .7895 & .6019 & .3845 & .1602 & -.0481 & -.2201 \\ +23 & .9205 & .7710 & .5692 & .3386 & .1057 & -.1038 & -.2681 \\ +24 & .9135 & .7518 & .5357 & .2926 & .0525 & -.1559 & -.3095 \\[2ex] +25 & .9063 & .7321 & .5016 & .2465 & .0009 & -.2053 & -.3463 \\ +26 & .8988 & .7117 & .4670 & .2007 & -.0489 & -.2478 & -.3717 \\ +27 & .8910 & .6908 & .4319 & .1553 & -.0964 & -.2869 & -.3921 \\ +28 & .8829 & .6694 & .3964 & .1105 & -.1415 & -.3211 & -.4052 \\ +29 & .8746 & .6474 & .3607 & .0665 & -.1839 & -.3503 & -.4114 \\[2ex] +30 & .8660 & .6250 & .3248 & .0234 & -.2233 & -.3740 & -.4101 \\ +31 & .8572 & .6021 & .2887 & -.0185 & -.2595 & -.3924 & -.4022 \\ +32 & .8480 & .5788 & .2527 & -.0591 & -.2923 & -.4052 & -.3876 \\ +33 & .8387 & .5551 & .2167 & -.0982 & -.3216 & -.4126 & -.3670 \\ +34 & .8290 & .5310 & .1809 & -.1357 & -.3473 & -.4148 & -.3409 \\[2ex] +35 & .8192 & .5065 & .1454 & -.1714 & -.3691 & -.4115 & -.3096 \\ +36 & .8090 & .4818 & .1102 & -.2052 & -.3871 & -.4031 & -.2738 \\ +37 & .7986 & .4567 & .0755 & -.2370 & -.4011 & -.3898 & -.2343 \\ +38 & .7880 & .4314 & .0413 & -.2666 & -.4112 & -.3719 & -.1918 \\ +39 & .7771 & .4059 & .0077 & -.2940 & -.4174 & -.3497 & -.1469 \\[2ex] +40 & .7660 & .3802 & -.0252 & -.3190 & -.4197 & -.3234 & -.1003 \\ +41 & .7547 & .3544 & -.0574 & -.3416 & -.4181 & -.2938 & -.0534 \\ +42 & .7431 & .3284 & -.0887 & -.3616 & -.4128 & -.2611 & -.0065 \\ +43 & .7314 & .3023 & -.1191 & -.3791 & -.4038 & -.2255 & .0398 \\ +44 & .7193 & .2762 & -.1485 & -.3940 & -.3914 & -.1878 & .0846 \\[2ex] +\phantom{\text{°}}45\text{°} + & .7071 & .2500 & -.1768 & -.4062 & -.3757 & -.1485 & .1270 \\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 289.png +\begin{center} +TABLE I. --- \textsc{Surface Zonal Harmonics} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|c|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut\theta}& +\multicolumn{1}{c|}{P_1(\cos\theta)}& +\multicolumn{1}{c|}{P_2(\cos\theta)}& +\multicolumn{1}{c|}{P_3(\cos\theta)}& +\multicolumn{1}{c|}{P_4(\cos\theta)}& +\multicolumn{1}{c|}{P_5(\cos\theta)}& +\multicolumn{1}{c|}{P_6(\cos\theta)}& +\multicolumn{1}{c|}{P_7(\cos\theta)}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\phantom{\text{°}}45\text{°} + & .7071 & .2500 & -.1768 & -.4062 & -.3757 & -.1485 & .1270\\ +46 & .6947 & .2238 & -.2040 & -.4158 & -.3568 & -.1079 & .1666\\ +47 & .6820 & .1977 & -.2300 & -.4252 & -.3350 & -.0645 & .2054\\ +48 & .6691 & .1716 & -.2547 & -.4270 & -.3105 & -.0251 & .2349\\ +49 & .6561 & .1456 & -.2781 & -.4286 & -.2836 & .0161 & .2627\\[2ex] +50 & .6428 & .1198 & -.3002 & -.4275 & -.2545 & .0563 & .2854\\ +51 & .6293 & .0941 & -.3209 & -.4239 & -.2235 & .0954 & .3031\\ +52 & .6157 & .0686 & -.3401 & -.4178 & -.1910 & .1326 & .3153\\ +53 & .6018 & .0433 & -.3578 & -.4093 & -.1571 & .1677 & .3221\\ +54 & .5878 & .0182 & -.3740 & -.3984 & -.1223 & .2002 & .3234\\[2ex] +55 & .5736 & -.0065 & -.3886 & -.3852 & -.0868 & .2297 & .3191\\ +56 & .5592 & -.0310 & -.4016 & -.3698 & -.0510 & .2559 & .3095\\ +57 & .5446 & -.0551 & -.4131 & -.3524 & -.0150 & .2787 & .2949\\ +58 & .5299 & -.0788 & -.4229 & -.3331 & .0206 & .2976 & .2752\\ +59 & .5150 & -.1021 & -.4310 & -.3119 & .0557 & .3125 & .2511\\[2ex] +60 & .5000 & -.1250 & -.4375 & -.2891 & .0898 & .3232 & .2231\\ +61 & .4848 & -.1474 & -.4423 & -.2647 & .1229 & .3298 & .1916\\ +62 & .4695 & -.1694 & -.4455 & -.2390 & .1545 & .3321 & .1571\\ +63 & .4540 & -.1908 & -.4471 & -.2121 & .1844 & .3302 & .1203\\ +64 & .4384 & -.2117 & -.4470 & -.1841 & .2123 & .3240 & .0818\\[2ex] +65 & .4226 & -.2321 & -.4452 & -.1552 & .2381 & .3138 & .0422\\ +66 & .4067 & -.2518 & -.4419 & -.1256 & .2615 & .2996 & .0021\\ +67 & .3907 & -.2710 & -.4370 & -.0955 & .2824 & .2819 & -.0375\\ +68 & .3746 & -.2896 & -.4305 & -.0650 & .3005 & .2605 & -.0763\\ +69 & .3584 & -.3074 & -.4225 & -.0344 & .3158 & .2361 & -.1135\\[2ex] +70 & .3420 & -.3245 & -.4130 & -.0038 & .3281 & .2089 & -.1485\\ +71 & .3256 & -.3410 & -.4021 & .0267 & .3373 & .1786 & -.1811\\ +72 & .3090 & -.3568 & -.3898 & .0568 & .3434 & .1472 & -.2099\\ +73 & .2924 & -.3718 & -.3761 & .0864 & .3463 & .1144 & -.2347\\ +74 & .2756 & -.3860 & -.3611 & .1153 & .3461 & .0795 & -.2559\\[2ex] +75 & .2588 & -.3995 & -.3449 & .1434 & .3427 & .0431 & -.2730\\ +76 & .2419 & -.4112 & -.3275 & .1705 & .3362 & .0076 & -.2848\\ +77 & .2250 & -.4241 & -.3090 & .1964 & .3267 & -.0284 & -.2919\\ +78 & .2079 & -.4352 & -.2894 & .2211 & .3143 & -.0644 & -.2943\\ +79 & .1908 & -.4454 & -.2688 & .2443 & .2990 & -.0989 & -.2913\\[2ex] +80 & .1736 & -.4548 & -.2474 & .2659 & .2810 & -.1321 & -.2835\\ +81 & .1564 & -.4633 & -.2251 & .2859 & .2606 & -.1635 & -.2709\\ +82 & .1392 & -.4709 & -.2020 & .3040 & .2378 & -.1926 & -.2536\\ +83 & .1219 & -.4777 & -.1783 & .3203 & .2129 & -.2193 & -.2321\\ +84 & .1045 & -.4836 & -.1539 & .3345 & .1861 & -.2431 & -.2067\\[2ex] +85 & .0872 & -.4886 & -.1291 & .3468 & .1577 & -.2638 & -.1779\\ +86 & .0698 & -.4927 & -.1038 & .3569 & .1278 & -.2811 & -.1460\\ +87 & .0523 & -.4959 & -.0781 & .3648 & .0969 & -.2947 & -.1117\\ +88 & .0349 & -.4982 & -.0522 & .3704 & .0651 & -.3045 & -.0735\\ +89 & .0175 & -.4995 & -.0262 & .3739 & .0327 & -.3105 & -.0381\\[2ex] +\phantom{\text{°}}90\text{°} + & .0000 & -.5000 & .0000 & .3750 & .0000 & -.3125 & .0000\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 290.png +\label{tableII} +\begin{center} +TABLE II.---\textsc{Surface Zonal Harmonics.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut x}& +\multicolumn{1}{c|}{P_1(x)}& +\multicolumn{1}{c|}{P_2(x)}& +\multicolumn{1}{c|}{P_3(x)}& +\multicolumn{1}{c|}{P_4(x)}& +\multicolumn{1}{c|}{P_5(x)}& +\multicolumn{1}{c|}{P_6(x)}& +\multicolumn{1}{c|}{P_7(x)}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +0.00 & 0.0000 & -.5000 & 0.0000 & 0.3750 & 0.0000 & -.3125 & 0.0000\\ + .01 & .0100 & -.4998 & -.0150 & .3746 & .0187 & -.3118 & -.0219\\ + .02 & .0200 & -.4994 & -.0300 & .3735 & .0374 & -.3099 & -.0436\\ + .03 & .0300 & -.4986 & -.0449 & .3716 & .0560 & -.3066 & -.0651\\ + .04 & .0400 & -.4976 & -.0598 & .3690 & .0744 & -.3021 & -.0862\\[2ex] + .05 & .0500 & -.4962 & -.0747 & .3657 & .0927 & -.2962 & -.1069\\ + .06 & .0600 & -.4946 & -.0895 & .3616 & .1106 & -.2891 & -.1270\\ + .07 & .0700 & -.4926 & -.1041 & .3567 & .1283 & -.2808 & -.1464\\ + .08 & .0800 & -.4904 & -.1187 & .3512 & .1455 & -.2713 & -.1651\\ + .09 & .0900 & -.4878 & -.1332 & .3449 & .1624 & -.2606 & -.1828\\[2ex] + .10 & .1000 & -.4850 & -.1475 & .3379 & .1788 & -.2488 & -.1995\\ + .11 & .1100 & -.4818 & -.1617 & .3303 & .1947 & -.2360 & -.2151\\ + .12 & .1200 & -.4784 & -.1757 & .3219 & .2101 & -.2220 & -.2295\\ + .13 & .1300 & -.4746 & -.1895 & .3129 & .2248 & -.2071 & -.2427\\ + .14 & .1400 & -.4706 & -.2031 & .3032 & .2389 & -.1913 & -.2545\\[2ex] + .15 & .1500 & -.4662 & -.2166 & .2928 & .2523 & -.1746 & -.2649\\ + .16 & .1600 & -.4616 & -.2298 & .2819 & .2650 & -.1572 & -.2738\\ + .17 & .1700 & -.4566 & -.2427 & .2703 & .2769 & -.1389 & -.2812\\ + .18 & .1800 & -.4514 & -.2554 & .2581 & .2880 & -.1201 & -.2870\\ + .19 & .1900 & -.4458 & -.2679 & .2453 & .2982 & -.1006 & -.2911\\[2ex] + .20 & .2000 & -.4400 & -.2800 & .2320 & .3075 & -.0806 & -.2935\\ + .21 & .2100 & -.4338 & -.2918 & .2181 & .3159 & -.0601 & -.2943\\ + .22 & .2200 & -.4274 & -.3034 & .2037 & .3234 & -.0394 & -.2933\\ + .23 & .2300 & -.4206 & -.3146 & .1889 & .3299 & -.0183 & -.2906\\ + .24 & .2400 & -.4136 & -.3254 & .1735 & .3353 & .0029 & -.2861\\[2ex] + .25 & .2500 & -.4062 & -.3359 & .1577 & .3397 & .0243 & -.2799\\ + .26 & .2600 & -.3986 & -.3461 & .1415 & .3431 & .0456 & -.2720\\ + .27 & .2700 & -.3906 & -.3558 & .1249 & .3453 & .0669 & -.2625\\ + .28 & .2800 & -.3824 & -.3651 & .1079 & .3465 & .0879 & -.2512\\ + .29 & .2900 & -.3738 & -.3740 & .0906 & .3465 & .1087 & -.2384\\[2ex] + .30 & .3000 & -.3650 & -.3825 & .0729 & .3454 & .1292 & -.2241\\ + .31 & .3100 & -.3558 & -.3905 & .0550 & .3431 & .1492 & -.2082\\ + .32 & .3200 & -.3464 & -.3981 & .0369 & .3397 & .1686 & -.1910\\ + .33 & .3300 & -.3366 & -.4052 & .0185 & .3351 & .1873 & -.1724\\ + .34 & .3400 & -.3266 & -.4117 & -.0000 & .3294 & .2053 & -.1527\\[2ex] + .35 & .3500 & -.3162 & -.4178 & -.0187 & .3225 & .2225 & -.1318\\ + .36 & .3600 & -.3056 & -.4234 & -.0375 & .3144 & .2388 & -.1098\\ + .37 & .3700 & -.2946 & -.4284 & -.0564 & .3051 & .2540 & -.0870\\ + .38 & .3800 & -.2834 & -.4328 & -.0753 & .2948 & .2681 & -.0635\\ + .39 & .3900 & -.2718 & -.4367 & -.0942 & .2833 & .2810 & -.0393\\[2ex] + .40 & .4000 & -.2600 & -.4400 & -.1130 & .2706 & .2926 & -.0146\\ + .41 & .4100 & -.2478 & -.4427 & -.1317 & .2569 & .3029 & .0104\\ + .42 & .4200 & -.2354 & -.4448 & -.1504 & .2421 & .3118 & .0356\\ + .43 & .4300 & -.2226 & -.4462 & -.1688 & .2263 & .3191 & .0608\\ + .44 & .4400 & -.2096 & -.4470 & -.1870 & .2095 & .3249 & .0859\\[2ex] + .45 & .4500 & -.1962 & -.4472 & -.2050 & .1917 & .3290 & .1106\\ + .46 & .4600 & -.1826 & -.4467 & -.2226 & .1730 & .3314 & .1348\\ + .47 & .4700 & -.1686 & -.4454 & -.2399 & .1534 & .3321 & .1584\\ + .48 & .4800 & -.1544 & -.4435 & -.2568 & .1330 & .3310 & .1811\\ + .49 & .4900 & -.1398 & -.4409 & -.2732 & .1118 & .3280 & .2027\\[2ex] + .50 & .5000 & -.1250 & -.4375 & -.2891 & .0898 & .3232 & .2231\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 291.png +\begin{center} +TABLE II.---\textsc{Surface Zonal Harmonics.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut x}& +\multicolumn{1}{c|}{P_1(x)}& +\multicolumn{1}{c|}{P_2(x)}& +\multicolumn{1}{c|}{P_3(x)}& +\multicolumn{1}{c|}{P_4(x)}& +\multicolumn{1}{c|}{P_5(x)}& +\multicolumn{1}{c|}{P_6(x)}& +\multicolumn{1}{c|}{P_7(x)}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +.50 & .5000 & -.1250 & -.4375 & -.2891 & .0898 & .3232 & .2231\\ +.51 & .5100 & -.1098 & -.4334 & -.3044 & .0673 & .3166 & .2422\\ +.52 & .5200 & -.0944 & -.4285 & -.3191 & .0441 & .3080 & .2596\\ +.53 & .5300 & -.0786 & -.4228 & -.3332 & .0204 & .2975 & .2753\\ +.54 & .5400 & -.0626 & -.4163 & -.3465 & -.0037 & .2851 & .2891\\[2ex] +.55 & .5500 & -.0462 & -.4091 & -.3590 & -.0282 & .2708 & .3007\\ +.56 & .5600 & -.0296 & -.4010 & -.3707 & -.0529 & .2546 & .3102\\ +.57 & .5700 & -.0126 & -.3920 & -.3815 & -.0779 & .2366 & .3172\\ +.58 & .5800 & .0046 & -.3822 & -.3914 & -.1028 & .2168 & .3217\\ +.59 & .5900 & .0222 & -.3716 & -.4002 & -.1278 & .1953 & .3235\\[2ex] +.60 & .6000 & .0400 & -.3600 & -.4080 & -.1526 & .1721 & .3226\\ +.61 & .6100 & .0582 & -.3475 & -.4146 & -.1772 & .1473 & .3188\\ +.62 & .6200 & .0766 & -.3342 & -.4200 & -.2014 & .1211 & .3121\\ +.63 & .6300 & .0954 & -.3199 & -.4242 & -.2251 & .0935 & .3023\\ +.64 & .6400 & .1144 & -.3046 & -.4270 & -.2482 & .0646 & .2895\\[2ex] +.65 & .6500 & .1338 & -.2884 & -.4284 & -.2705 & .0347 & .2737\\ +.66 & .6600 & .1534 & -.2713 & -.4284 & -.2919 & .0038 & .2548\\ +.67 & .6700 & .1734 & -.2531 & -.4268 & -.3122 & -.0278 & .2329\\ +.68 & .6800 & .1936 & -.2339 & -.4236 & -.3313 & -.0601 & .2081\\ +.69 & .6900 & .2142 & -.2137 & -.4187 & -.3490 & -.0926 & .1805\\[2ex] +.70 & .7000 & .2350 & -.1925 & -.4121 & -.3652 & -.1253 & .1502\\ +.71 & .7100 & .2562 & -.1702 & -.4036 & -.3796 & -.1578 & .1173\\ +.72 & .7200 & .2776 & -.1469 & -.3933 & -.3922 & -.1899 & .0822\\ +.73 & .7300 & .2994 & -.1225 & -.3810 & -.4026 & -.2214 & .0450\\ +.74 & .7400 & .3214 & -.0969 & -.3666 & -.4107 & -.2518 & .0061\\[2ex] +.75 & .7500 & .3438 & -.0703 & -.3501 & -.4164 & -.2808 & -.0342\\ +.76 & .7600 & .3664 & -.0426 & -.3314 & -.4193 & -.3081 & -.0754\\ +.77 & .7700 & .3894 & -.0137 & -.3104 & -.4193 & -.3333 & -.1171\\ +.78 & .7800 & .4126 & .0164 & -.2871 & -.4162 & -.3559 & -.1588\\ +.79 & .7900 & .4362 & .0476 & -.2613 & -.4097 & -.3756 & -.1999\\[2ex] +.80 & .8000 & .4600 & .0800 & -.2330 & -.3995 & -.3918 & -.2397\\ +.81 & .8100 & .4842 & .1136 & -.2021 & -.3855 & -.4041 & -.2774\\ +.82 & .8200 & .5086 & .1484 & -.1685 & -.3674 & -.4119 & -.3124\\ +.83 & .8300 & .5334 & .1845 & -.1321 & -.3449 & -.4147 & -.3437\\ +.84 & .8400 & .5584 & .2218 & -.0928 & -.3177 & -.4120 & -.3703\\[2ex] +.85 & .8500 & .5838 & .2603 & -.0506 & -.2857 & -.4030 & -.3913\\ +.86 & .8600 & .6094 & .3001 & -.0053 & -.2484 & -.3872 & -.4055\\ +.87 & .8700 & .6354 & .3413 & .0431 & -.2056 & -.3638 & -.4116\\ +.88 & .8800 & .6616 & .3837 & .0947 & -.1570 & -.3322 & -.4083\\ +.89 & .8900 & .6882 & .4274 & .1496 & -.1023 & -.2916 & -.3942\\[2ex] +.90 & .9000 & .7150 & .4725 & .2079 & -.0411 & -.2412 & -.3678\\ +.91 & .9100 & .7422 & .5189 & .2698 & .0268 & -.1802 & -.3274\\ +.92 & .9200 & .7696 & .5667 & .3352 & .1017 & -.1077 & -.2713\\ +.93 & .9300 & .7974 & .6159 & .4044 & .1842 & -.0229 & -.1975\\ +.94 & .9400 & .8254 & .6665 & .4773 & .2744 & .0751 & -.1040\\[2ex] +.95 & .9500 & .8538 & .7184 & .5541 & .3727 & .1875 & .0112\\ +.96 & .9600 & .8824 & .7718 & .6349 & .4796 & .3151 & .1506\\ +.97 & .9700 & .9114 & .8267 & .7198 & .5954 & .4590 & .3165\\ +.98 & .9800 & .9406 & .8830 & .8089 & .7204 & .6204 & .5115\\ +.99 & .9900 & .9702 & .9407 & .9022 & .8552 & .8003 & .7384\\[2ex] +1.00 & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000 & 1.0000\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 292.png +\label{tableIII} +\begin{center} +TABLE III.---\textsc{Hyperbolic Functions.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut x}& +\multicolumn{1}{c|}{e^x}& +\multicolumn{1}{c|}{e^{-x}}& +\multicolumn{1}{c|}{\sinh x}& +\multicolumn{1}{c|}{\cosh x}& +\multicolumn{1}{c|}{\gd x}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\;0.00 & \;1.0000 & \;1.0000 & \;0.0000 & \;1.0000 & \;0\raisebox{0.8ex}{\text{°}}\!\!\!.0000\\ +.01 & 1.0100 & 0.9900 & .0100 & 1.0000 & 0.5729\\ +.02 & 1.0202 & .9802 & .0200 & 1.0002 & 1.1458\\ +.03 & 1.0305 & .9704 & .0300 & 1.0004 & 1.7186\\ +.04 & 1.0408 & .9608 & .0400 & 1.0008 & 2.2912\\[2ex] +.05 & 1.0513 & .9512 & .0500 & 1.0012 & 2.8636\\ +.06 & 1.0618 & .9418 & .0600 & 1.0018 & 3.4357\\ +.07 & 1.0725 & .9324 & .0701 & 1.0025 & 4.0074\\ +.08 & 1.0833 & .9231 & .0801 & 1.0032 & 4.5788\\ +.09 & 1.0942 & .9139 & .0901 & 1.0040 & 5.1497\\[2ex] +.10 & 1.1052 & .9048 & .1002 & 1.0050 & 5.720\phantom{0}\\ +.11 & 1.1163 & .8958 & .1102 & 1.0061 & 6.290\phantom{0}\\ +.12 & 1.1275 & .8869 & .1203 & 1.0072 & 6.859\phantom{0}\\ +.13 & 1.1388 & .8781 & .1304 & 1.0085 & 7.428\phantom{0}\\ +.14 & 1.1503 & .8694 & .1405 & 1.0098 & 7.995\phantom{0}\\[2ex] +.15 & 1.1618 & .8607 & .1506 & 1.0113 & 8.562\phantom{0}\\ +.16 & 1.1735 & .8521 & .1607 & 1.0128 & 9.128\phantom{0}\\ +.17 & 1.1853 & .8437 & .1708 & 1.0145 & 9.694\phantom{0}\\ +.18 & 1.1972 & .8353 & .1810 & 1.0162 & 10.258\phantom{0}\\ +.19 & 1.2092 & .8270 & .1911 & 1.0181 & 10.821\phantom{0}\\[2ex] +.20 & 1.2214 & .8187 & .2013 & 1.0201 & 11.384\phantom{0}\\ +.21 & 1.2337 & .8106 & .2115 & 1.0221 & 11.945\phantom{0}\\ +.22 & 1.2461 & .8025 & .2218 & 1.0243 & 12.505\phantom{0}\\ +.23 & 1.2586 & .7945 & .2320 & 1.0266 & 13.063\phantom{0}\\ +.24 & 1.2712 & .7866 & .2423 & 1.0289 & 13.621\phantom{0}\\[2ex] +.25 & 1.2840 & .7788 & .2526 & 1.0314 & 14.177\phantom{0}\\ +.26 & 1.2969 & .7711 & .2629 & 1.0340 & 14.732\phantom{0}\\ +.27 & 1.3100 & .7634 & .2733 & 1.0367 & 15.285\phantom{0}\\ +.28 & 1.3231 & .7558 & .2837 & 1.0395 & 15.837\phantom{0}\\ +.29 & 1.3364 & .7483 & .2941 & 1.0423 & 16.388\phantom{0}\\[2ex] +.30 & 1.3499 & .7408 & .3045 & 1.0453 & 16.937\phantom{0}\\ +.31 & 1.3634 & .7334 & .3150 & 1.0484 & 17.484\phantom{0}\\ +.32 & 1.3771 & .7261 & .3255 & 1.0516 & 18.030\phantom{0}\\ +.33 & 1.3910 & .7189 & .3360 & 1.0549 & 18.573\phantom{0}\\ +.34 & 1.4049 & .7118 & .3466 & 1.0584 & 19.116\phantom{0}\\[2ex] +.35 & 1.4191 & .7047 & .3572 & 1.0619 & 19.656\phantom{0}\\ +.36 & 1.4333 & .6977 & .3678 & 1.0655 & 20.195\phantom{0}\\ +.37 & 1.4477 & .6907 & .3785 & 1.0692 & 20.732\phantom{0}\\ +.38 & 1.4623 & .6839 & .3892 & 1.0731 & 21.267\phantom{0}\\ +.39 & 1.4770 & .6771 & .4000 & 1.0770 & 21.800\phantom{0}\\[2ex] +.40 & 1.4918 & .6703 & .4108 & 1.0811 & 22.331\phantom{0}\\ +.41 & 1.5068 & .6636 & .4216 & 1.0852 & 22.859\phantom{0}\\ +.42 & 1.5220 & .6570 & .4325 & 1.0895 & 23.386\phantom{0}\\ +.43 & 1.5373 & .6505 & .4434 & 1.0939 & 23.911\phantom{0}\\ +.44 & 1.5527 & .6440 & .4543 & 1.0984 & 24.434\phantom{0}\\[2ex] +.45 & 1.5683 & .6376 & .4653 & 1.1030 & 24.955\phantom{0}\\ +.46 & 1.5841 & .6313 & .4764 & 1.1077 & 25.473\phantom{0}\\ +.47 & 1.6000 & .6250 & .4875 & 1.1125 & 25.989\phantom{0}\\ +.48 & 1.6161 & .6188 & .4986 & 1.1174 & 26.503\phantom{0}\\ +.49 & 1.6323 & .6126 & .5098 & 1.1225 & 27.015\phantom{0}\\[2ex] +0.50 & 1.6487 & 0.6065 & 0.5211 & 1.1276 & 27\raisebox{0.8ex}{\text{°}}\!\!\!.524\phantom{0}\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 293.png +\begin{center} +TABLE III.---\textsc{Hyperbolic Functions.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut x}& +\multicolumn{1}{c|}{e^x}& +\multicolumn{1}{c|}{e^{-x}}& +\multicolumn{1}{c|}{\sinh x}& +\multicolumn{1}{c|}{\cosh x}& +\multicolumn{1}{c|}{\gd x}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\;0.50 & \;1.6487 & \;0.6065 & \;0.5211 & \;1.1276 & \;27\raisebox{0.8ex}{\text{°}}\!\!\!.524\\ +.51 & 1.6653 & .6005 & .5324 & 1.1329 & 28.031\\ +.52 & 1.6820 & .5945 & .5438 & 1.1383 & 28.535\\ +.53 & 1.6989 & .5886 & .5552 & 1.1438 & 29.037\\ +.54 & 1.7160 & .5827 & .5666 & 1.1494 & 29.537\\[2ex] +.55 & 1.7333 & .5770 & .5782 & 1.1551 & 30.034\\ +.56 & 1.7507 & .5712 & .5897 & 1.1609 & 30.529\\ +.57 & 1.7683 & .5655 & .6014 & 1.1669 & 31.021\\ +.58 & 1.7860 & .5599 & .6131 & 1.1730 & 31.511\\ +.59 & 1.8040 & .5543 & .6248 & 1.1792 & 31.998\\[2ex] +.60 & 1.8221 & .5488 & .6367 & 1.1855 & 32.483\\ +.61 & 1.8404 & .5433 & .6485 & 1.1919 & 32.965\\ +.62 & 1.8589 & .5379 & .6605 & 1.1984 & 33.444\\ +.63 & 1.8776 & .5326 & .6725 & 1.2051 & 33.921\\ +.64 & 1.8965 & .5273 & .6846 & 1.2119 & 34.395\\[2ex] +.65 & 1.9155 & .5220 & .6967 & 1.2188 & 34.867\\ +.66 & 1.9348 & .5169 & .7090 & 1.2258 & 35.336\\ +.67 & 1.9542 & .5117 & .7213 & 1.2330 & 35.802\\ +.68 & 1.9739 & .5066 & .7336 & 1.2402 & 36.265\\ +.69 & 1.9937 & .5016 & .7461 & 1.2476 & 36.726\\[2ex] +.70 & 2.0138 & .4966 & .7586 & 1.2552 & 37.183\\ +.71 & 2.0340 & .4916 & .7712 & 1.2628 & 37.638\\ +.72 & 2.0544 & .4867 & .7838 & 1.2706 & 38.091\\ +.73 & 2.0751 & .4819 & .7966 & 1.2785 & 38.540\\ +.74 & 2.0959 & .4771 & .8094 & 1.2865 & 38.987\\[2ex] +.75 & 2.1170 & .4724 & .8223 & 1.2947 & 39.431\\ +.76 & 2.1383 & .4677 & .8353 & 1.3030 & 39.872\\ +.77 & 2.1598 & .4630 & .8484 & 1.3114 & 40.310\\ +.78 & 2.1815 & .4584 & .8615 & 1.3199 & 40.746\\ +.79 & 2.2034 & .4538 & .8748 & 1.3286 & 41.179\\[2ex] +.80 & 2.2255 & .4493 & .8881 & 1.3374 & 41.608\\ +.81 & 2.2479 & .4449 & .9015 & 1.3464 & 42.035\\ +.82 & 2.2705 & .4404 & .9150 & 1.3555 & 42.460\\ +.83 & 2.2933 & .4360 & .9286 & 1.3647 & 42.881\\ +.84 & 2.3164 & .4317 & .9423 & 1.3740 & 43.299\\[2ex] +.85 & 2.3396 & .4274 & .9561 & 1.3835 & 43.715\\ +.86 & 2.3632 & .4232 & .9700 & 1.3932 & 44.128\\ +.87 & 2.3869 & .4190 & .9840 & 1.4029 & 44.537\\ +.88 & 2.4109 & .4148 & .9981 & 1.4128 & 44.944\\ +.89 & 2.4351 & .4107 & 1.0122 & 1.4229 & 45.348\\[2ex] +.90 & 2.4596 & .4066 & 1.0265 & 1.4331 & 45.750\\ +.91 & 2.4843 & .4025 & 1.0409 & 1.4434 & 46.148\\ +.92 & 2.5093 & .3985 & 1.0554 & 1.4539 & 46.544\\ +.93 & 2.5345 & .3946 & 1.0700 & 1.4645 & 46.936\\ +.94 & 2.5600 & .3906 & 1.0847 & 1.4753 & 47.326\\[2ex] +.95 & 2.5857 & .3867 & 1.0995 & 1.4862 & 47.713\\ +.96 & 2.6117 & .3829 & 1.1144 & 1.4973 & 48.097\\ +.97 & 2.6379 & .3791 & 1.1294 & 1.5085 & 48.478\\ +.98 & 2.6645 & .3753 & 1.1446 & 1.5199 & 48.857\\ +.99 & 2.6912 & .3716 & 1.1598 & 1.5314 & 49.232\\[2ex] +1.00 & 2.7183 & 0.3679 & 1.1752 & 1.5431 & 49\raisebox{0.8ex}{\text{°}}\!\!\!.605\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 294.png +\begin{center} +TABLE III.---\textsc{Hyperbolic Functions.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut x}& +l\sinh x& +l\cosh x& +\multicolumn{1}{c|}{x}& +l\sinh x& +l\cosh x& +\multicolumn{1}{c|}{x}& +l\sinh x& +l\cosh x\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\;1.00 & 0.0701 & 0.1884 & \;1.50 & 0.3282 & 0.3715 & \;2.00 & 0.5595 & 0.5754\\ +1.01 & .0758 & .1917 & 1.51 & .3330 & .3754 & 2.01 & .5640 & .5796\\ +1.02 & .0815 & .1950 & 1.52 & .3378 & .3794 & 2.02 & .5685 & .5838\\ +1.03 & .0871 & .1984 & 1.53 & .3426 & .3833 & 2.03 & .5730 & .5880\\ +1.04 & .0927 & .2018 & 1.54 & .3474 & .3873 & 2.04 & .5775 & .5922\\[2ex] +1.05 & .0982 & .2051 & 1.55 & .3521 & .3913 & 2.05 & .5820 & .5964\\ +1.06 & .1038 & .2086 & 1.56 & .3569 & .3952 & 2.06 & .5865 & .6006\\ +1.07 & .1093 & .2120 & 1.57 & .3616 & .3992 & 2.07 & .5910 & .6048\\ +1.08 & .1148 & .2154 & 1.58 & .3663 & .4032 & 2.08 & .5955 & .6090\\ +1.09 & .1203 & .2189 & 1.59 & .3711 & .4072 & 2.09 & .6000 & .6132\\[2ex] +1.10 & .1257 & .2223 & 1.60 & .3758 & .4112 & 2.10 & .6044 & .6175\\ +1.11 & .1311 & .2258 & 1.61 & .3805 & .4152 & 2.11 & .6089 & .6217\\ +1.12 & .1365 & .2293 & 1.62 & .3852 & .4192 & 2.12 & .6134 & .6259\\ +1.13 & .1419 & .2328 & 1.63 & .3899 & .4232 & 2.13 & .6178 & .6301\\ +1.14 & .1472 & .2364 & 1.64 & .3946 & .4273 & 2.14 & .6223 & .6343\\[2ex] +1.15 & .1525 & .2399 & 1.65 & .3992 & .4313 & 2.15 & .6268 & .6386\\ +1.16 & .1578 & .2435 & 1.66 & .4039 & .4353 & 2.16 & .6312 & .6428\\ +1.17 & .1631 & .2470 & 1.67 & .4086 & .4394 & 2.17 & .6357 & .6470\\ +1.18 & .1684 & .2506 & 1.68 & .4132 & .4434 & 2.18 & .6401 & .6512\\ +1.19 & .1736 & .2542 & 1.69 & .4179 & .4475 & 2.19 & .6446 & .6555\\[2ex] +1.20 & .1788 & .2578 & 1.70 & .4225 & .4515 & 2.20 & .6491 & .6597\\ +1.21 & .1840 & .2615 & 1.71 & .4272 & .4556 & 2.21 & .6535 & .6640\\ +1.22 & .1892 & .2651 & 1.72 & .4318 & .4597 & 2.22 & .6580 & .6682\\ +1.23 & .1944 & .2688 & 1.73 & .4364 & .4637 & 2.23 & .6624 & .6724\\ +1.24 & .1995 & .2724 & 1.74 & .4411 & .4678 & 2.24 & .6668 & .6767\\[2ex] +1.25 & .2046 & .2761 & 1.75 & .4457 & .4719 & 2.25 & .6713 & .6809\\ +1.26 & .2098 & .2798 & 1.76 & .4503 & .4760 & 2.26 & .6757 & .6852\\ +1.27 & .2148 & .2835 & 1.77 & .4549 & .4801 & 2.27 & .6802 & .6894\\ +1.28 & .2199 & .2872 & 1.78 & .4595 & .4842 & 2.28 & .6846 & .6937\\ +1.29 & .2250 & .2909 & 1.79 & .4641 & .4883 & 2.29 & .6890 & .6979\\[2ex] +1.30 & .2300 & .2947 & 1.80 & .4687 & .4924 & 2.30 & .6935 & .7022\\ +1.31 & .2351 & .2984 & 1.81 & .4733 & .4965 & 2.31 & .6979 & .7064\\ +1.32 & .2401 & .3022 & 1.82 & .4778 & .5006 & 2.32 & .7023 & .7107\\ +1.33 & .2451 & .3059 & 1.83 & .4824 & .5048 & 2.33 & .7067 & .7150\\ +1.34 & .2501 & .3097 & 1.84 & .4870 & .5089 & 2.34 & .7112 & .7192\\[2ex] +1.35 & .2551 & .3135 & 1.85 & .4915 & .5130 & 2.35 & .7156 & .7235\\ +1.36 & .2600 & .3173 & 1.86 & .4961 & .5172 & 2.36 & .7200 & .7278\\ +1.37 & .2650 & .3211 & 1.87 & .5007 & .5213 & 2.37 & .7244 & .7320\\ +1.38 & .2699 & .3249 & 1.88 & .5052 & .5254 & 2.38 & .7289 & .7363\\ +1.39 & .2748 & .3288 & 1.89 & .5098 & .5296 & 2.39 & .7333 & .7406\\[2ex] +1.40 & .2797 & .3326 & 1.90 & .5143 & .5337 & 2.40 & .7377 & .7448\\ +1.41 & .2846 & .3365 & 1.91 & .5188 & .5379 & 2.41 & .7421 & .7491\\ +1.42 & .2895 & .3403 & 1.92 & .5234 & .5421 & 2.42 & .7465 & .7534\\ +1.43 & .2944 & .3442 & 1.93 & .5279 & .5462 & 2.43 & .7509 & .7577\\ +1.44 & .2993 & .3481 & 1.94 & .5324 & .5504 & 2.44 & .7553 & .7619\\[2ex] +1.45 & .3041 & .3520 & 1.95 & .5370 & .5545 & 2.45 & .7597 & .7662\\ +1.46 & .3090 & .3559 & 1.96 & .5415 & .5587 & 2.46 & .7642 & .7705\\ +1.47 & .3138 & .3598 & 1.97 & .5460 & .5629 & 2.47 & .7686 & .7748\\ +1.48 & .3186 & .3637 & 1.98 & .5505 & .5671 & 2.48 & .7730 & .7791\\ +1.49 & .3234 & .3676 & 1.99 & .5550 & .5713 & 2.49 & .7774 & .7833\\[2ex] +1.50 & 0.3282 & 0.3715 & 2.00 & 0.5595 & 0.5754 & 2.50 & 0.7818 & 0.7876\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 295.png +\begin{center} +TABLE III.---\textsc{Hyperbolic Functions.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|}{\tablestrut x}& +l\sinh x& +l\cosh x& +\multicolumn{1}{c|}{x}& +l\sinh x& +l\cosh x& +\multicolumn{1}{c|}{x}& +l\sinh x& +l\cosh x\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\;2.50 & 0.7818 & 0.7876 & \;2.75 & 0.8915 & 0.8951 & 3.0 & 1.0008 & 1.0029\\ +2.51 & .7862 & .7919 & 2.76 & .8959 & .8994 & 3.1 & 1.0444 & 1.0462\\ +2.52 & .7906 & .7962 & 2.77 & .9003 & .9037 & 3.2 & 1.0880 & 1.0894\\ +2.53 & .7950 & .8005 & 2.78 & .9046 & .9080 & 3.3 & 1.1316 & 1.1327\\ +2.54 & .7994 & .8048 & 2.79 & .9090 & .9123 & 3.4 & 1.1751 & 1.1761\\[2ex] +2.55 & .8038 & .8091 & 2.80 & .9134 & .9166 & 3.5 & 1.2186 & 1.2194\\ +2.56 & .8082 & .8134 & 2.81 & .9178 & .9209 & 3.6 & 1.2621 & 1.2628\\ +2.57 & .8126 & .8176 & 2.82 & .9221 & .9252 & 3.7 & 1.3056 & 1.3061\\ +2.58 & .8169 & .8219 & 2.83 & .9265 & .9295 & 3.8 & 1.3491 & 1.3495\\ +2.59 & .8213 & .8262 & 2.84 & .9309 & .9338 & 3.9 & 1.3925 & 1.3929\\[2ex] +2.60 & .8257 & .8305 & 2.85 & .9353 & .9382 & 4.0 & 1.4360 & 1.4363\\ +2.61 & .8301 & .8348 & 2.86 & .9396 & .9425 & 4.1 & 1.4795 & 1.4797\\ +2.62 & .8345 & .8391 & 2.87 & .9440 & .9468 & 4.2 & 1.5229 & 1.5231\\ +2.63 & .8389 & .8434 & 2.88 & .9484 & .9511 & 4.3 & 1.5664 & 1.5665\\ +2.64 & .8433 & .8477 & 2.89 & .9527 & .9554 & 4.4 & 1.6098 & 1.6099\\[2ex] +2.65 & .8477 & .8520 & 2.90 & .9571 & .9597 & 4.5 & 1.6532 & 1.6533\\ +2.66 & .8521 & .8563 & 2.91 & .9615 & .9641 & 4.6 & 1.6967 & 1.6968\\ +2.67 & .8564 & .8606 & 2.92 & .9658 & .9684 & 4.7 & 1.7401 & 1.7402\\ +2.68 & .8608 & .8649 & 2.93 & .9702 & .9727 & 4.8 & 1.7836 & 1.7836\\ +2.69 & .8652 & .8692 & 2.94 & .9746 & .9770 & 4.9 & 1.8270 & 1.8270\\[2ex] +2.70 & .8696 & .8735 & 2.95 & .9789 & .9813 & 5.0 & 1.8704 & 1.8705\\ +2.71 & .8740 & .8778 & 2.96 & .9833 & .9856 & 6.0 & 2.3047 & 2.3047\\ +2.72 & .8784 & .8821 & 2.97 & .9877 & .9900 & 7.0 & 2.7390 & 2.7390\\ +2.73 & .8827 & .8864 & 2.98 & .9920 & .9943 & 8.0 & 3.1733 & 3.1733\\ +2.74 & .8871 & .8907 & 2.99 & .9964 & .9986 & 9.0 & 3.6076 & 3.6076\\[2ex] +2.75 & 0.8915 & 0.8951 & 3.00 & 1.0008 & 1.0029 & 10.0 & 4.0419 & 4.0419\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 296.png +\begin{center} +\label{tableIV}TABLE IV.---\textsc{Roots of Bessel's Functions.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|c|c|r@{\quad}|c|c|} +\hline +\tablestrut{}& +\dfrac{x}{\pi}$ for $J_0(x)=0& +\dfrac{x}{\pi}$ for $J_1(x)=0& +\qquad& +\dfrac{x}{\pi}$ for $J_0(x)=0& +\dfrac{x}{\pi}$ for $J_1(x)=0\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +1 & 0.7655 & 1.2197 & 7 & \phantom{0}6.7519 & \phantom{0}7.2448\\ +2 & 1.7571 & 2.2330 & 8 & \phantom{0}7.7516 & \phantom{0}8.2454\\ +3 & 2.7546 & 3.2383 & 9 & \phantom{0}8.7514 & \phantom{0}9.2459\\ +4 & 3.7534 & 4.2411 & 10 & \phantom{0}9.7513 & 10.2463\\ +5 & 4.7527 & 5.2428 & 11 & 10.7512 & 11.2466\\ +\;6 & 5.7522 & 6.2439 & \;12 & 11.7511 & 12.2469\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} + +\bigskip +\begin{center} +\label{tableV}TABLE V.---\textsc{Roots of $J_n(x)=0$.} +\end{center} +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|c|c|c|c|c|c|} +\hline +\tablestrut{}& +\;n=0\;& +\;n=1\;& +\;n=2\;& +\;n=3\;& +\;n=4\;& +\;n=5\;\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +1 & \phantom{0}2.405 & \phantom{0}3.832 & \phantom{0}5.135 & \phantom{0}6.379 & \phantom{0}7.586 & \phantom{0}8.780\\ +2 & \phantom{0}5.520 & \phantom{0}7.016 & \phantom{0}8.417 & \phantom{0}9.760 & 11.064 & 12.339\\ +3 & \phantom{0}8.654 & 10.173 & 11.620 & 13.017 & 14.373 & 15.700\\ +4 & 11.792 & 13.323 & 14.796 & 16.224 & 17.616 & 18.982\\ +5 & 14.931 & 16.470 & 17.960 & 19.410 & 20.827 & 22.220\\ +6 & 18.071 & 19.616 & 21.117 & 22.583 & 24.018 & 25.431\\ +7 & 21.212 & 22.760 & 24.270 & 25.749 & 27.200 & 28.628\\ +8 & 24.353 & 25.903 & 27.421 & 28.909 & 30.371 & 31.813\\ +\;9 & 27.494 & 29.047 & 30.571 & 32.050 & 33.512 & 34.983\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\newpage + +% -----File: 297.png +\label{tableVI} +\begin{center} +TABLE VI.--- \textsc{Bessel's Functions.} +\end{center} + +\begin{scriptsize} +\[ +\arraycolsep=0.2em +\begin{array}{|c|} +\hline\\[-2.1ex] +\arraycolsep=0.5em +\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|} +\hline +\multicolumn{1}{|c|} {\tablestrut x} & +\multicolumn{1}{c|} {J_0(x)} & +\multicolumn{1}{c|||}{J_1(x)} & +\multicolumn{1}{c|} {x } & +\multicolumn{1}{c|} {J_0(x)} & +\multicolumn{1}{c|||}{J_1(x)} & +\multicolumn{1}{c|} {x} & +\multicolumn{1}{c|} {J_0(x)} & +\multicolumn{1}{c|} {J_1(x)}\\ +\hline +\vrule width0ptdepth0ptheight3.5ex +\;0.0 & 1.0000 & 0.0000 & 5.0 & -.1776 & -.3276 & \;10.0 & -.2459 & .0435\\ +0.1 & .9975 & .0499 & 5.1 & -.1443 & -.3371 & 10.1 & -.2490 & .0184\\ +0.2 & .9900 & .0995 & 5.2 & -.1103 & -.3432 & 10.2 & -.2496 & -.0066\\ +0.3 & .9776 & .1483 & 5.3 & -.0758 & -.3460 & 10.3 & -.2477 & -.0313\\ +0.4 & .9604 & .1960 & 5.4 & -.0412 & -.3453 & 10.4 & -.2434 & -.0555\\[2ex] +0.5 & .9385 & .2423 & 5.5 & -.0068 & -.3414 & 10.5 & -.2366 & -.0789\\ +0.6 & .9120 & .2867 & 5.6 & .0270 & -.3343 & 10.6 & -.2276 & -.1012\\ +0.7 & .8812 & .3290 & 5.7 & .0599 & -.3241 & 10.7 & -.2164 & -.1224\\ +0.8 & .8463 & .3688 & 5.8 & .0917 & -.3110 & 10.8 & -.2032 & -.1422\\ +0.9 & .8075 & .4060 & 5.9 & .1220 & -.2951 & 10.9 & -.1881 & -.1604\\[2ex] +1.0 & .7652 & .4401 & 6.0 & .1506 & -.2767 & 11.0 & -.1712 & -.1768\\ +1.1 & .7196 & .4709 & 6.1 & .1773 & -.2559 & 11.1 & -.1528 & -.1913\\ +1.2 & .6711 & .4983 & 6.2 & .2017 & -.2329 & 11.2 & -.1330 & -.2039\\ +1.3 & .6201 & .5220 & 6.3 & .2238 & -.2081 & 11.3 & -.1121 & -.2143\\ +1.4 & .5669 & .5419 & 6.4 & .2433 & -.1816 & 11.4 & -.0902 & -.2225\\[2ex] +1.5 & .5118 & .5579 & 6.5 & .2601 & -.1538 & 11.5 & -.0677 & -.2284\\ +1.6 & .4554 & .5699 & 6.6 & .2740 & -.1250 & 11.6 & -.0446 & -.2320\\ +1.7 & .3980 & .5778 & 6.7 & .2851 & -.0953 & 11.7 & -.0213 & -.2333\\ +1.8 & .3400 & .5815 & 6.8 & .2931 & -.0652 & 11.8 & .0020 & -.2323\\ +1.9 & .2818 & .5812 & 6.9 & .2981 & -.0349 & 11.9 & .0250 & -.2290\\[2ex] +2.0 & .2239 & .5767 & 7.0 & .3001 & -.0047 & 12.0 & .0477 & -.2234\\ +2.1 & .1666 & .5683 & 7.1 & .2991 & .0252 & 12.1 & .0697 & -.2157\\ +2.2 & .1104 & .5560 & 7.2 & .2951 & .0543 & 12.2 & .0908 & -.2060\\ +2.3 & .0555 & .5399 & 7.3 & .2882 & .0826 & 12.3 & .1108 & -.1943\\ +2.4 & .0025 & .5202 & 7.4 & .2786 & .1096 & 12.4 & .1296 & -.1807\\[2ex] +2.5 & -.0484 & .4971 & 7.5 & .2663 & .1352 & 12.5 & .1469 & -.1655\\ +2.6 & -.0968 & .4708 & 7.6 & .2516 & .1592 & 12.6 & .1626 & -.1487\\ +2.7 & -.1424 & .4416 & 7.7 & .2346 & .1813 & 12.7 & .1766 & -.1307\\ +2.8 & -.1850 & .4097 & 7.8 & .2154 & .2014 & 12.8 & .1887 & -.1114\\ +2.9 & -.2243 & .3754 & 7.9 & .1944 & .2192 & 12.9 & .1988 & -.0912\\[2ex] +3.0 & -.2601 & .3391 & 8.0 & .1717 & .2346 & 13.0 & .2069 & -.0703\\ +3.1 & -.2921 & .3009 & 8.1 & .1475 & .2476 & 13.1 & .2129 & -.0489\\ +3.2 & -.3202 & .2613 & 8.2 & .1222 & .2580 & 13.2 & .2167 & -.0271\\ +3.3 & -.3443 & .2207 & 8.3 & .0960 & .2657 & 13.3 & .2183 & -.0052\\ +3.4 & -.3643 & .1792 & 8.4 & .0692 & .2708 & 13.4 & .2177 & .0166\\[2ex] +3.5 & -.3801 & .1374 & 8.5 & .0419 & .2731 & 13.5 & .2150 & .0380\\ +3.6 & -.3918 & .0955 & 8.6 & .0146 & .2728 & 13.6 & .2101 & .0590\\ +3.7 & -.3992 & .0538 & 8.7 & -.0125 & .2697 & 13.7 & .2032 & .0791\\ +3.8 & -.4026 & .0128 & 8.8 & -.0392 & .2641 & 13.8 & .1943 & .0984\\ +3.9 & -.4018 & -.0272 & 8.9 & -.0653 & .2559 & 13.9 & .1836 & .1166\\[2ex] +4.0 & -.3972 & -.0660 & 9.0 & -.0903 & .2453 & 14.0 & .1711 & .1334\\ +4.1 & -.3887 & -.1033 & 9.1 & -.1142 & .2324 & 14.1 & .1570 & .1488\\ +4.2 & -.3766 & -.1386 & 9.2 & -.1367 & .2174 & 14.2 & .1414 & .1626\\ +4.3 & -.3610 & -.1719 & 9.3 & -.1577 & .2004 & 14.3 & .1245 & .1747\\ +4.4 & -.3423 & -.2028 & 9.4 & -.1768 & .1816 & 14.4 & .1065 & .1850\\[2ex] +4.5 & -.3205 & -.2311 & 9.5 & -.1939 & .1613 & 14.5 & .0875 & .1934\\ +4.6 & -.2961 & -.2566 & 9.6 & -.2090 & .1395 & 14.6 & .0679 & .1999\\ +4.7 & -.2693 & -.2791 & 9.7 & -.2218 & .1166 & 14.7 & .0476 & .2043\\ +4.8 & -.2404 & -.2985 & 9.8 & -.2323 & .0928 & 14.8 & .0271 & .2066\\ +4.9 & -.2097 & -.3147 & 9.9 & -.2403 & .0684 & 14.9 & .0064 & .2069\\[2ex] +5.0 & -.1776 & -.3276 & 10.0 & -.2459 & .0435 & 15.0 & -.0142 & .2051\\[1ex] +\hline +\end{array}\\[-2.1ex] +{}\\ +\hline +\end{array} +\] +\end{scriptsize} +\label{tableend} +\newpage +\pagestyle{empty} + +% -----File: 298.png +%[Blank Page] +% -----File: 299.png +\markright{}\vspace*{\fill} +\begin{center} +ANNOUNCEMENTS +\end{center} +\vspace*{\fill} +\pagebreak +% -----File: 300.png +%[Blank Page] +% -----File: 301.png +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\begin{LARGE} +APPLICATIONS OF THE CALCULUS\\ +TO MECHANICS\par +\end{LARGE} + +\vspace{\baselineskip} + +By \textsc{E. R. Hedrick}, Professor of Mathematics in the University of Missouri,\\ +and \textsc{O. D. Kellogg}, Assistant Professor of Mathematics\\ +in the University of Missouri + +\rule{0.25\textwidth}{1pt} + +8vo, cloth, 116 pages, with diagrams, \$1.25 + +\rule{0.25\textwidth}{1pt} + +\end{center} + +\noindent{\Huge T}HIS book presents a completed summary of those parts +of mechanics which occur as applications of the calculus. + +Although intended primarily as a supplement to the usual +standard course in calculus, it may be used independently as a +text for a short course on the mathematical side of mechanics, +if the time allotted to the former study is not sufficient to include +this work. + +As a review it fastens in the student's mind the notions of +mechanics previously gained. It aims also to present these +topics in a new light, as articulated portions of one general +theory, and thus to make mechanics seem an integral subject. + +As a preparation for more extended courses in mechanics, or +indeed for courses dealing with any applications of the calculus, +the material presented is valuable in showing concretely +how the theoretical ideas of this subject are used in specific +practical applications. + +The course outlined in the book is the result of a number of +years' experience in presenting this material to classes in the +calculus, both at the University of Missouri and elsewhere. The +text itself is a modification of a similar text written by Professor +Hedrick and published in mimeograph for the use of +students at the Sheffield Scientific School. + +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\vspace{\baselineskip} + +\begin{LARGE} +GINN~~AND~~COMPANY~~~\textsc{Publishers} +\end{LARGE} +\end{center} +\newpage + +% -----File: 302.png +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\begin{LARGE} +ELECTRICAL PROBLEMS\par +\end{LARGE} + +\vspace{\baselineskip} + +By \textsc{William L. Hooper}, Professor of Electrical Engineering,\\ +Tufts College, Mass., and \textsc{Roy T. Wells}, in Charge of\\ +Electrical Engineering, State University of Iowa + +\rule{0.25\textwidth}{1pt} + +8vo, cloth, 170 pages, with diagrams, \$1.25 + +\rule{0.25\textwidth}{1pt} + +\end{center} + +\noindent{\Huge T}HE work contains sets of problems typical of those +met with in electrical laboratory and engineering +practice, with very brief treatment of the methods +of solution. It is intended to meet the need for a collection +of numerical problems in the classes of colleges and +technical schools. It is made up as follows:\medskip\par + +\begin{small} +1.\quad Twelve sets of problems and calculations on combinations of +electro-motive forces and resistances in series and multiple grouping; +distribution and fall of potential in railway and lighting circuits, inductance +of coils, capacity of condensers, thermo-electricity, electro-chemistry; +and output and efficiency of batteries, generators, motors, etc. + +2.\quad Four sets of problems on combinations of alternating electromotive +forces and currents and the impedance of circuits with constant +and with varying values of resistance, inductance, capacity, and frequency. + +3.\quad Five sets of problems on calculating and making winding tables +and drawings for direct and alternating current armatures, armature +reactions, field windings, etc. + +4.\quad Problems on winding and operation of transformers, rotary converters, +and induction motors, and on testing of dynamos and transmission +of power. +\end{small}\medskip\par + +Answers are given to all problems, many in the form of +curves showing the effect of varying the various constants +involved, such as temperature, frequency, capacity, resistance, +and inductance. The text contains about forty explanatory +diagrams. + +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\vspace{\baselineskip} + +\begin{LARGE} +GINN~~\&~~COMPANY~~~\textsc{Publishers} +\end{LARGE} +\end{center} +\newpage + +% -----File: 303.png +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\begin{LARGE} +THEORETICAL MECHANICS +\par +\end{LARGE} + +\vspace{\baselineskip} + +By \textsc{Percey F. Smith}, Professor of Mathematics in the Sheffield Scientific\\ School, +Yale University, and \textsc{William R. Longley}, Assistant Professor of\\ +Mathematics in the Sheffield Scientific School, Yale University + +\rule{0.25\textwidth}{1pt} + +8vo, semiflexible cloth, 288 pages, with diagrams, \$2.50 + +\rule{0.25\textwidth}{1pt} + +\end{center} + +\noindent{\Huge T}HIS book is intended for use in courses in mechanics, +which, as in many colleges and technical schools, are +based upon the calculus. For the convenience of the student, +formulas from analytic geometry and the calculus, and a table +of integrals, are included. + +The first chapter deals with centers of gravity and moments +of inertia. This is followed by chapters on kinematics and +kinetics of a particle (including impact), motion in various +fields of force (constant field, central field, harmonic field), +kinetics of a system of particles, potential, motion in a resisting +medium, dynamics of a rigid body, including uniplanar +motion, and equilibrium of coplanar forces. + +The fundamental problem---to determine the motion due +to a given force under given initial conditions---is thoroughly +discussed. + +The equations of motion obtained by integration of the force +equations have, however, been studied in a previous chapter, +and the student is therefore cognizant immediately of the significance +of his results. + +Emphasis is laid everywhere in the solution of problems upon +the general application of the force equations, the energy equation, +and the impulse equation. + +The problems are carefully selected, and numerous illustrative +examples are worked out in the text. +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\vspace{\baselineskip} +\begin{LARGE} +GINN~~AND~~COMPANY~~~\textsc{Publishers} +\end{LARGE} +\end{center} + +\newpage + +% -----File: 304.png + + \newenvironment{hang}{% + \begin{list}{}{% + \setlength{\leftmargin}{1cm}% + \setlength{\itemindent}{-1cm}% + }% + \item[]}{\end{list}} + +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\begin{LARGE} +BOOKS ON +HIGHER MATHEMATICS +\par +\end{LARGE} + +\rule{0.25\textwidth}{1pt} + +\end{center} + +\begin{hang} +\spreadout{BYERLY: ELEMENTARY TREATISE ON FOURIER'S SERIES, AND} +\\ +\spreadout{SPHERICAL, CYLINDRICAL, AND ELLIPSOIDAL HARMONICS,} +\\ +\spreadout{WITH APPLICATIONS TO PROBLEMS IN MATHEMATICAL} +\\ +PHYSICS\@. \$3.00. +\end{hang} + +\textsc{An} introduction to the treatment of some of the important linear partial differential +equations which lie at the foundation of modern theories in physics. + +\begin{hang} +HEDRICK: GOURSAT'S COURSE IN MATHEMATICAL ANALYSIS, +VOLUME I\@. \$4.00. +\end{hang} + +\textsc{The} French edition of this work at once attracted widespread attention on +account of the clearness of its style and the thoroughness and rigor with which +the subject matter was presented. + +\begin{hang} +HEDRICK AND KELLOGG: APPLICATIONS OF THE CALCULUS +TO MECHANICS\@. \$1.25. +\end{hang} + +\textsc{A completed} summary of those parts of mechanics which occur as applications +of the calculus. + +\begin{hang} +PEIRCE, B. O.: ELEMENTS OF THE THEORY OF THE NEWTONIAN +POTENTIAL FUNCTION (Third, Revised, and Enlarged Edition). \$2.50. +\end{hang} + +\textsc{Intended} for general students who wish to acquire a sound knowledge of the +properties of the Newtonian Potential Function. Nearly four hundred miscellaneous +problems have been added in this new edition. + +\begin{hang} +PEIRCE, J. M.: ELEMENTS OF LOGARITHMS\@. 50 cents. +\end{hang} + +\textsc{Designed} to give the student a complete and accurate knowledge of the +nature and use of logarithms. + +\begin{hang} +TAYLOR, J. M.: ELEMENTS OF THE DIFFERENTIAL AND INTEGRAL +CALCULUS (Revised Edition, Enlarged and Entirely Rewritten). +With Examples and Applications. \$2.00. +\end{hang} + +\textsc{Presents} clearly, scientifically, and in their true relations the three common +methods in the calculus. + +\newpage +\begin{hang} +MATHEMATICAL TEXTS. Edited by Percey F. Smith, of the Sheffield Scientific +School of Yale University. +\end{hang} + +\begin{footnotesize} +\textsc{Eisenhart: Differential Geometry of Curves and Surfaces.} \$4.50. + +\textsc{Granville: Plane Trigonometry and Tables.} \$1.00. + +\textsc{Granville: Plane and Spherical Trigonometry, without Tables.} \$1.00. + +\textsc{Granville: Plane and Spherical Trigonometry, and Tables.} \$1.25. + +\textsc{Granville: Elements of Differential and Integral Calculus.} \$2.50. + +\textsc{Granville: Logarithmic Tables.} 50 cents. + +\textsc{Hawkes: Advanced Algebra.} \$1.40. + +\textsc{Smith and Gale: Elements of Analytic Geometry.} \$2.00. + +\textsc{Smith and Gale: Introduction to Analytic Geometry.} \$1.25. + +\end{footnotesize} + +\begin{center} +\rule{\textwidth}{1pt}\\[-2ex] +\rule{\textwidth}{1pt} + +\vspace{\baselineskip} + +\LARGE +GINN~~AND~~COMPANY~~~\textsc{Publishers} +\end{center} + +\newpage +\begin{center}\textsc{Typographical Errors corrected in Project Gutenberg edition}\end{center} + +p.~\pageref{err053}~Equation (3), the second $\displaystyle \int\limits_0^\pi$ was not printed.\\ + +p.~\pageref{err111}~Example~1, the = sign on the second line (beginning $\dfrac{2}{\pi}$) was not printed.\\ + +p.~\pageref{err149}~Example~6, second sin term $\sin \dfrac{m\pi y}{b}$ in original, amended to $\sin \dfrac{n\pi y}{b}$.\\ + +p.~\pageref{err250}~Equation (8), $D_{\rho_3}^2V$ was printed as $D_{\rho_3}V$. + +\newpage +\small +\pagenumbering{Roman} +\begin{verbatim} +End of the Project Gutenberg EBook of An Elementary Treatise on Fourier's +Series and Spherical, Cylindrical, and Ellipsoidal Harmonics, by William Elwood Byerly + +*** END OF THIS PROJECT GUTENBERG EBOOK TREATISE ON FOURIER'S SERIES *** + +***** This file should be named 29779-pdf.pdf or 29779-pdf.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/2/9/7/7/29779/ + +Produced by Laura Wisewell, Carl Hudkins, Keith Edkins and +the Online Distributed Proofreading Team at +http://www.pgdp.net (The original copy of this book was +generously made available for scanning by the Department +of Mathematics at the University of Glasgow.) + + +Updated editions will replace the previous one--the old editions +will be renamed. + +Creating the works from public domain print editions means that no +one owns a United States copyright in these works, so the Foundation +(and you!) can copy and distribute it in the United States without +permission and without paying copyright royalties. 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