An Elementary Treatise on Fourier's Series and Spherical, Cylindrical, and Ellipsoidal Harmonics, With Applications to Problems in Mathematical Physics

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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 ***

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AN ELEMENTARY TREATISE ON

FOURIER’S SERIES AND

SPHERICAL, CYLINDRICAL, AND ELLIPSOIDAL HARMONICS, WITH

APPLICATIONS TO PROBLEMS IN MATHEMATICAL PHYSICS.

BY

WILLIAM ELWOOD BYERLY, Ph.D., PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY.

GINN & COMPANY BOSTON · NEW YORK · CHICAGO · LONDON

Copyright, 1893, By WILLIAM ELWOOD BYERLY. ALL RIGHTS RESERVED.

Transcriber’s Note: A few typographical errors have been corrected - these are noted at the end of the text.

i

PREFACE. 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 δ for “partial derivative with respect to x.” use the notation Dx instead of δx 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 Bˆ ocher, and the latter has kindly contributed the brief historical sketch contained in Chapter IX. W. E. BYERLY. Cambridge, Mass., Sept. 1893.

ii

ANALYTICAL TABLE OF CONTENTS. CHAPTER I. pages 1–29

Introduction

Art. 1. List of some important homogeneous linear partial differential equations of Physics.—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.—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.—Art. 7. Problem: Permanent state of temperatures in a thin rectangular plate. Need of a development in sine series. Example.—Art. 8. Problem: Transverse vibrations of a stretched elastic string. A development in sine series suggested.—Art. 9. Problem: Potential function due to the attraction of a circular ring of small cross-section. Surface Zonal Harmonics (Legendre’s Coefficients). Example.—Art. 10. Problem: Permanent state of temperatures in a solid sphere. Development in terms of Surface Zonal Harmonics suggested.—Arts. 11–12. Problem: Vibrations of a circular drumhead. Cylindrical Harmonics (Bessel’s Functions). Recapitulation.—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. Arts. 14–15 Method of solving ordinary homogeneous linear differential equations by development in power series. Applications.—Art. 16. Application to Legendre’s Equation. Several forms of general solution obtained. Zonal Harmonics of the second kind.—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.—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.

CHAPTER II. Development in Trigonometric Series

30–55

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.—Art. 23. Problem of development in sine series treated as a limiting case of the problem just solved.—Arts. 24–25. Shorter

TABLE OF CONTENTS

iii

method of solving the problem of development in series involving sines of whole multiples of the variable. Working rule deduced. Recapitulation.—Art. 26. A few important sine developments obtained. Examples.—Arts. 27–28. Development in cosine series. Examples.—Art. 29. Sine series an odd function of the variable, cosine series an even function, and both series periodic functions.— Art. 30. Development in series involving both sines and cosines of whole multiples of the variable. Fourier’s series. Examples.—Art. 31. Extension of the range within which the function and the series are equal. Examples.—Art. 32. Fourier’s Integral obtained.

CHAPTER III. Convergence of Fourier’s Series

56–69

Arts. 33–36. The question of the convergence of the sine series for unity considered at length.—Arts. 37–38. Statement of the conditions which are sufficient to warrant the development of a function into a Fourier’s series. Historical note. Art. 39. Graphical representation of successive approximations to a sine series. Properties of a Fourier’s series inferred from the constructions.— Arts. 40–42. Investigation of the conditions under which a Fourier’s series can be differentiated term by term.—Art. 43. Conditions under which a function can be expressed as a Fourier’s Integral.

CHAPTER IV. Solution of Problems in Physics by the Aid of Fourier’s Integrals and Fourier’s Series 70–135 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.—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.—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.—Arts. 55–56. Transmission of a disturbance along an infinite stretched elastic string. Examples.—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.—Art. 59.

TABLE OF CONTENTS

iv

Potential Function given along the perimeter of a rectangle. Examples.—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.—Art. 64. Motion of a stretched elastic string fastened at the ends. Steady vibration. Nodes. Examples.—Art. 65. Motion of a string in a resisting medium.—Art. 66. Flow of heat in a sphere whose surface is kept at a constant temperature.—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.—Arts. 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.—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. Miscellaneous Problems

135–143

I. Logarithmic Potential. Polar Coördinates.—II. Potential Function in Space. III. Conduction of heat in a plane.—IV. Conduction of heat in Space.

CHAPTER V. Zonal Harmonics

143–195

Art. 74. Recapitulation. Surface Zonal Harmonics (Legendrians). Zonal Harmonics of the second kind.—Arts. 75–76. Legendrians as coefficients in a Power Series. Special values.—Art. 77. Summary of the properties of a Legendrian. List of the first eight Legendrians. Relation connecting any three successive Legendrians.—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.—Art. 82. Legendrian as a sum of cosines.—Arts. 83–84. Legendrian as the mth derivative of the mth power of x2 − −1.—Art. 85. Equations derivable from Legendre’s Equation.—Art. 86. Legendrian as a Partial Derivative.—Art. 87. Legendrian as a Definite Integral. 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.— 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.—Arts. 93–94. Problems in Potential where the value of the Potential Function is given on a spherical surface and has circular

TABLE OF CONTENTS

v

symmetry about a diameter. Examples.—Art. 95. Development of a power of x in Zonal Harmonic Series.—Art. 96. Useful formulas.—Art. 97. Development of sin nθ and cos nθ in Zonal Harmonic Series. Examples. Graphical representation of the first seven Surface Zonal Harmonics. Construction of successive approximations to Zonal Harmonic Series. Arts. 98–99. Method of dealing with problems in Potential when the density is given. Examples.—Art. 100. Surface Zonal Harmonics of the second kind. Examples: Conal Harmonics.

CHAPTER VI. 196–219

Spherical Harmonics

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.—Arts. 103– 108. Development in Spherical Harmonic Series. The integral of the product of two 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.—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.—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.—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.—Art. 115. Solution of problems in Potential by direct integration. Examples.—Arts. 116–118. Differentiation along an axis. Axes of a Spherical Harmonic.—Art. 119. Roots of a Zonal Harmonic. Roots of a Tesseral Harmonic. Nomenclature justified.

CHAPTER VII. Cylindrical Harmonics (Bessel’s Functions)

220–238

Art. 120. Recapitulation. Cylindrical Harmonics (Bessel’s Functions) of the zeroth order; of the nth order; of the second kind. General solution of Bessel’s Equation.—Art. 121. Bessel’s Functions as definite integrals. Examples.— Art. 122. Properties of Bessel’s Functions. Semi-convergent series for a Bessel’s Function. Examples.—Art. 123. Problem: Stationary temperatures in a cylinder (a) when the temperature of the convex surface is zero; (b) when the convex surface is adiathermanous; (c) when the convex surface is exposed to air at the temperature zero.—Art. 124. Roots of Bessel’s functions.—Art. 125. The integral of r times the product of two Cylindrical Harmonics of the zeroth order.

TABLE OF CONTENTS

vi

Example.—Art. 126. Development in Cylindrical Harmonic Series. Formulas for the coefficients. Examples.—Art. 127. Problem: Stationary temperatures in a cylindrical shell. Bessel’s Functions of the second kind employed. Example: Vibration of a ring membrane.—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.—Art. 129. Problem: Stationary temperatures in a cylinder when the temperatures of the base are unsymmetrical. Bessel’s Functions of the nth order employed. Miscellaneous examples. Bessel’s Functions of fractional order.

CHAPTER VIII. ¨ rdinates. Ellipsoidal Laplace’s Equation in Curvilinear Coo Harmonics 239–266 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.—Arts. 132–135. Spheroidal Coördinates. Laplace’s Equation in spheroidal co¨ ordinates, 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.—Arts. 136–141. Ellipsoidal Co¨ ordinates. Laplace’s Equation in ellipsoidal coördinates. Normal ellipsoidal co¨ ordinates 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.—Art. 142. Conical Co¨ ordinates. The product of two Ellipsoidal Harmonics a Spherical Harmonic.— Art. 143. Toroidal Co¨ ordinates. Laplace’s Equation in toroidal coördinates. Particular solutions. Toroidal Harmonics. Potential Function for an anchor ring.

CHAPTER IX. 267–274

Historical Summary APPENDIX. Tables Table I. Surface Zonal Harmonics. Argument θ Table II. Surface Zonal Harmonics. Argument x

274–285 276 278

TABLE OF CONTENTS Table Table Table Table

III. Hyperbolic Functions IV. Roots of Bessel’s Functions V. Roots of Bessel’s Functions VI. Bessel’s Functions

vii 280 284 284 285

1

CHAPTER I. INTRODUCTION. 1. In many important problems in mathematical physics we are obliged to deal with 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 Dt u = a2 (Dx2 u + Dy2 u + Dz2 u), 1

[I]

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, [I] reduces to Dt u = a2 Dx2 u, [II] 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 [I] becomes Dx2 u + Dy2 u = 0. In polar or spherical co¨ ordinates [I] is less simple, it is a2 1 1 2 Dθ (sin θDθ u) + D u . Dt u = 2 Dr (r2 Dr u) + φ r sin θ sin2 θ

[III]

[IV]

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 [IV] reduces to Dt (ru) = a2 Dr2 (ru). [V] In cylindrical co¨ ordinates [I] becomes 1 1 Dt u = a2 [Dr2 u + Dr u + 2 Dφ2 u + Dz2 u]. r r

[VI]

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 [VI] becomes 1 Dt u = a2 (Dr2 u + Dr u). r

[VII]

In Acoustics in several problems we have the equation Dt2 y = a2 Dx2 y;

[VIII]

1 For the sake of brevity we shall often use the symbol ∇2 for the operation D 2 + D 2 + D 2 ; x y z and with this notation equation [I] would be written Dt u = a2 ∇2 u.

INTRODUCTION.

2

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. If in considering the transverse vibrations of a stretched string we take account of the resistance of the air [VIII] is replaced by Dt2 y + 2kDt y = a2 Dx2 y.

[IX]

In dealing with the vibrations of a stretched elastic membrane, we have the equation Dt2 z = c2 (Dx2 z + Dy2 z), [X] or in cylindrical co¨ ordinates 1 1 Dt2 z = c2 (Dr2 z + Dr z + 2 Dφ2 z). r r In the theory of Potential we constantly meet Laplace’s Equation Dx2 V +Dy2 V + Dz2 V = 0

[XI]

[XII]

2

∇ V =0

or

which in spherical co¨ ordinates becomes 1 1 1 2 2 D V = 0, D (sin θD V ) + rD (rV ) + θ θ φ r r2 sin θ sin2 θ

[XIII]

and in cylindrical co¨ ordinates 1 1 Dr2 V + Dr V + 2 Dφ2 V + Dz2 V = 0. [XIV] r r In curvilinear co¨ ordinates it is h2 h3 h1 Dρ V + Dρ2 Dρ V + Dρ3 Dρ V = 0; h1 h2 h3 Dρ1 h2 h3 1 h3 h1 2 h1 h2 3 [XV] where f1 (x, y, z) = ρ1 , f2 (x, y, z) = ρ2 , f3 (x, y, z) = ρ3 represent a set of surfaces which cut one another at right angles, no matter what values are given to ρ1 , ρ2 , and ρ3 ; and where h21 = (Dx ρ1 )2 + (Dy ρ1 )2 + (Dz ρ1 )2 h22 = (Dx ρ2 )2 + (Dy ρ2 )2 + (Dz ρ2 )2 h23 = (Dx ρ3 )2 + (Dy ρ3 )2 + (Dz ρ3 )2 , and, of course, must be expressed in terms of ρ1 , ρ2 , and ρ3 . If it happens that ∇2 ρ1 = 0, ∇2 ρ2 = 0, and ∇2 ρ3 = 0, then Laplace’s Equation [XV] assumes the very simple form h21 Dρ21 V + h22 Dρ22 V + h23 Dρ23 V = 0.

[XVI]

INTRODUCTION.

3

2. A differential equation is an equation containing derivatives or differentials with or without the primitive variables from which they are derived. The 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 (i. e., undetermined) constants or arbitrary functions. A 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. 3. If a problem requiring for its solution the solving of a differential equation is 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 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 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. 4. A differential equation is 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 homogeneous. All the differential equations collected in Art. 1 are linear and homogeneous. 5. 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. 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

INTRODUCTION.

4

taken out as a factor of the whole first member. The remaining factor being zero, the product is zero and the equation is satisfied. 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. 6. It is generally possible to get by some simple device 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 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 Legendre’s Coefficients, or Zonal Harmonics; Laplace’s Coefficients, or Spherical Harmonics: Bessel’s Functions, or Cylindrical Harmonics; Lamé’s Functions, or Ellipsoidal Harmonics, &c. 7. As an illustration, let us take Fourier’s problem of the permanent state of temperatures in a thin rectangular plate of breadth π 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 Dx2 u + Dy2 u = 0 [III] Art. 1 subject to the conditions u=0

when x = 0

(1)

u=0

“

x=π

(2)

u=0

“

y=∞

(3)

u=1

“

y = 0.

(4)

We shall begin by getting a particular solution of [III], and we shall use a device which always succeeds when the equation is linear and homogeneous and has constant coefficients.

INTRODUCTION.

5

Assume2 u = eαy+βx , where α and β are constants, substitute in [III] and divide by eαy+βx , and we have α2 + β 2 = 0. If, then, this condition is satisfied u = eαy+βx is a solution. Hence u = eαy±αxi 3 is a solution of [III], no matter what value may be given to α. This form is objectionable, since it involves an imaginary. We can, however, readily improve it. Take u = eαy eαxi , a solution of [III], and u = eαy e−αxi , another solution of [III]; add these values of u and divide the sum by 2 and we have eαy cos αx. (v. Int. Cal. Art. 35, [1].) Therefore by Art. 5 u = eαy cos αx

(5)

is a solution of [III]. Take u = eαy eαxi and u = eαy e−αxi , subtract the second value of u from the first and divide by 2i and we have eαy sin αx. (v. Int. Cal. Art. 35, [2]). Therefore by Art. 5 u = eαy sin αx

(6)

is a solution of [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). Consider

u = eαy sin αx.

(6)

It is zero when x = 0 for all values of α. It is zero when x = π if α is a whole number. It is zero when y = ∞ if α 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 [III] which satisfies conditions (1), (2) and (3). Let this solution be u = A1 e−y sin x + A2 e−2y sin 2x + A3 e−3y sin 3x + A4 e−4y sin 4x + · · ·

(7)

A1 , A2 , A3 , A4 , &c., being undetermined constants. When y = 0 (7) reduces to u = A1 sin x + A2 sin 2x + A3 sin 3x + A4 sin 4x + · · · .

(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 A1 , A2 , A3 , &c. in (7). It will be proved later that 1 1 1 4 sin x + sin 3x + sin 5x + sin 7x + · · · 1= π 3 5 7 2 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. 3 We shall regularly use the symbol i for −1.

INTRODUCTION.

6

for all values of x between 0 and π; hence our required solution is 4 −y 1 1 1 u= e sin x + e−3y sin 3x + e−5y sin 5x + e−7y sin 7x + · · · π 3 5 7

(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 mx, where m is a whole number. The problem of finding the value of the potential function at any point of a long, thin, rectangular conducting sheet, of breadth π, 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 π π π , 1 ; (b) , 2 ; (c) ,3 , (a) 6 3 2 correct to the nearest degree. Ans. (a) 26◦ ; (b) 15◦ ; (c) 6◦ . 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 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 Dt2 y = a2 Dx2 y [VIII] Art. 1, while satisfying the conditions y=0

when

y=0 “ y = f (x) “ Dt y = 0

“

x=0

(1)

x=l t=0

(2) (3)

t = 0,

(4)

the last condition meaning merely that the string starts from rest. As in the last problem let4 y = eαx+βt and substitute in [VIII]. Divide by αx+βt e and we have β 2 = a2 α2 as the condition that our assumed value of y shall satisfy the equation. y = eαx±aαt (5) 4 See

note on page 5.

INTRODUCTION.

7

is, then, a solution of (VIII) whatever the value of α. 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 α in (5). Replace α by αi and (5) becomes y = e(x±at)αi , a solution of [VIII]. Replace α by −αi and (5) becomes y = e−(x±at)αi , another solution of [VIII]. Add these values of y and divide by 2 and we have cos α(x ± at). Subtract the second value of y from the first and divide by 2i and we have sin α(x ± at). y = cos α(x + at) y = cos α(x − at) y = sin α(x + at) y = sin α(x − at) are, then, solutions of [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 [VIII]. y = cos αx cos αat y = sin αx sin αat y = sin αx cos αat y = cos αx sin αat. If we take the third form y = sin αx cos αat it will satisfy conditions (1) and (4), no matter what value may be given to α, mπ and it will satisfy (2) if α = where m is an integer. l If then we take πat 2πx 2πat 3πx 3πat πx cos + A2 sin cos + A3 sin cos + · · · (6) y = A1 sin l l l l l l where A1 , A2 , A3 · · · are undetermined constants, we shall have a solution of [VIII] which satisfies (1), (2), and (4). When t = 0 it reduces to πx 2πx 3πx + A2 sin + A3 sin + ··· (7) l l l 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 A1 , A2 , A3 · · · in (6), and we shall have a solution of [VIII] which satisfies all our given conditions. In each of the preceding problems the 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 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. y = A1 sin

INTRODUCTION.

8

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 =√

M + x2

c2

(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 co¨ ordinates, 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 rDr2 (rV ) +

1 1 Dθ (sin θDθ V ) + Dφ2 V = 0, sin θ sin2 θ

[XIII] Art. 1,

subject to the condition V =

M (c2

1

+ r2 ) 2

when θ = 0.

(1)

From the symmetry of the ring, it is clear that the value of the potential function must be independent of φ, so that [XIII] will reduce to rDr2 (rV ) +

1 Dθ (sin θDθ V ) = 0. sin θ

(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. Let5 V = rm P , where P is a function of θ only, and m is a positive integer, and substitute in (2), which becomes m(m + 1)rm P +

rm Dθ (sin θDθ P ) = 0. sin θ

Divide by rm and use the notation of ordinary derivatives since P depends upon θ only, and we have the equation dP d sin θ 1 dθ = 0, (3) m(m + 1)P + sin θ dθ from which to obtain P . Equation (3) can be simplified by changing the independent variable. Let x = cos θ and (3) becomes d dP (1 − x2 ) + m(m + 1)P = 0. (4) dx dx 5 See

note on page 5.

INTRODUCTION.

9

Assume6 now that P can be expressed as a sum or as a series of terms involving whole of x multiplied by constant coefficients. P powers Let P = an xn and substitute this value of P in (4). We get P [n(n − 1)an xn−2 − n(n + 1)an xn + m(m + 1)an xn ] = 0, (5) P where the symbol 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 xk , must vanish. Hence (k + 2)(k + 1)ak+2 − k(k + 1)ak + m(m + 1)ak = 0 ak+2 = −

and

(6)

m(m + 1) − k(k + 1) ak . (k + 1)(k + 2)

(7)

If now any set of coefficients satisfying the relation (7) be taken, P = will be a solution of (4). If

k = m,

ak+2 = 0,

ak+4 = 0,

P

ak xk

&c.

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 am . Let us rewrite (7) in the form ak = −

(k + 2)(k + 1) ak+2 . (m − k)(m + k + 1)

(8)

We get from (8), beginning with k = m − 2, m(m − 1) am 2.(2m − 1) m(m − 1)(m − 2)(m − 3) = am 2.4.(2m − 1)(2m − 3) m(m − 1)(m − 2)(m − 3)(m − 4)(m − 5) =− am , 2.4.6.(2m − 1)(2m − 3)(2m − 5)

am−2 = − am−4 am−6

&c.

If m is even we see that the set will end with a0 , if m is odd, with a1 . m(m − 1) m−2 m(m − 1)(m − 2)(m − 3) m−4 m P = am x − x + x − ··· 2.(2m − 1) 2.4.(2m − 1)(2m − 3) where am is entirely arbitrary, is, then, a solution of (4). It is found convenient to take am equal to (2m − 1)(2m − 3) · · · 1 m! and it can be shown that with this value of am P = 1 when x = 1. 6 See

note on page 5.

INTRODUCTION.

10

P is a function of x and contains no higher powers of x than xm . It is usual to write it as Pm (x). We proceed to compute a few values of Pm (x) from the formula (2m − 1)(2m − 3) · · · 1 m m(m − 1) m−2 x − x Pm (x) = m! 2.(2m − 1) m(m − 1)(m − 2)(m − 3) m−4 + x − ··· . (9) 2.4.(2m − 1)(2m − 3) We have: P0 (x) = 1

or

P0 (cos θ) = 1

P1 (x) = x

“

P1 (cos θ) = cos θ

“

P2 (cos θ) = 12 (3 cos2 θ − 1)

“

P3 (cos θ) = 12 (5 cos3 θ − 3 cos θ)

P2 (x) = P3 (x) = P4 (x) =

2 1 2 (3x − 1) 3 1 2 (5x − 3x) 1 4 2 8 (35x − 30x

+ 3) or

P4 (cos θ) = 18 (35 cos4 θ − 30 cos2 θ + 3) P5 (x) = 18 (63x5 − 70x3 + 15x) or P5 (cos θ) = 18 (63 cos5 θ − 70 cos3 θ + 15 cos θ).

               

(10)

              

We have obtained P = Pm (x) as a particular solution of (4) and P = Pm (cos θ) as a particular solution of (3). Pm (x) or Pm (cos θ) is a new function, known as a Legendre’s Coefficient, or as a Surface Zonal Harmonic, and occurs as a normal form in many important problems. V = rm Pm (cos θ) is a particular solution of (2) and rm Pm (cos θ) is sometimes called a Solid Zonal Harmonic. We can now proceed to the solution of our original problem. V = A0 r0 P0 (cos θ) + A1 rP1 (cos θ) + A2 r2 P2 (cos θ) + A3 r3 P3 (cos θ) + · · · (11) where A0 , A1 , A2 , &c., are entirely arbitrary, is a solution of (2) (v. Art. 5). When θ = 0 (11) reduces to V = A0 + A1 r + A2 r 2 + A3 r 3 + · · · , since, as we have said, Pm (x) = 1 when x = 1, or Pm (cos θ) = 1 when θ = 0. By our condition (1) M V = 1 2 (c + r2 ) 2 when θ = 0. By the Binomial Theorem M 1

(c2 + r2 ) 2

M 1 r2 1.3 r4 1.3.5 r6 = 1− + − + ··· c 2 c2 2.4 c4 2.4.6 c6

INTRODUCTION.

11

provided r < c. Hence 1.3 r4 M 1 r2 P (cos θ) + P4 (cos θ) V = P0 (cos θ) − 2 c 2 c2 2.4 c4 1.3.5 r6 − P6 (cos θ) + · · · 2.4.6 c6

(12)

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 π (a) (r = .2, θ = 0) ; (d ) (r = .6, θ = 0) ; (f ) r = .6, θ = ; 3 π π π (b) r = .2, θ = ; (e) r = .6, θ = ; (g) r = .6, θ = ; 4 6 2 π ; (c) r = .2, θ = Ans. (a) .98; (b) .99; (c) 1.01; (d ) .86; 2 (e) .90; (f ) 1.00; (g) 1.10. 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. 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 rDr2 (ru) +

1 1 Dθ (sin θDθ u) + Dφ2 u = 0 sin θ sin2 θ

[XIII] Art. 1

which as before reduces to rDr2 (ru) +

1 Dθ (sin θDθ u) = 0 sin θ

(1)

from the consideration that the temperatures must be independent of φ. Our equation of condition is u = 1 from θ = 0 to θ =

π π and u = 0 from θ = to θ = π, 2 2

(2)

INTRODUCTION.

12

when r = 1. As we have seen u = rm Pm (cos θ) is a particular solution of (1), m being any positive whole number, and u = A0 r0 P0 (cos θ) + A1 rP1 (cos θ) + A2 r2 P2 (cos θ) + A3 r3 P3 (cos θ) + · · · (3) where A0 , A1 , A2 , A3 · · · are undetermined constants, is a solution of (1). When r = 1 (3) reduces to u = A0 P0 (cos θ) + A1 P1 (cos θ) + A2 P2 (cos θ) + A3 P3 (cos θ) + · · ·

(4)

If then we can develop our function of θ 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 A0 , A1 , A2 , &c., in (3) and we shall have our required solution. 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 symmetry7 about an axis through the centre perpendicular to the plane of the boundary, and then allowed to vibrate. Here we have to solve 1 2 1 2 2 2 [XI] Art. 1 Dt z = c Dr z + Dr z + 2 Dφ z r r subject to the conditions z = f (r)

when

t=0

(1)

Dt z = 0

“

t=0

(2)

z=0

“

r = a.

(3)

From the symmetry of the supposed initial distortion z must be independent of φ, therefore [XI] reduces to 1 2 2 2 (4) Dt z = c Dr z + Dr z r 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. Assume8 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 1 RDt2 T = c2 T Dr2 R + Dr R r 7 A function of the co¨ ordinates of a point has 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. 8 See note on page 5.

INTRODUCTION.

13 1 1 d2 T = 2 2 c T dt R

or

d2 R 1 dR + dr2 r dr

.

(5)

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 r, and consequently since it contains neither t nor r, it must be constant. Let it equal −µ2 , where µ of course is an undetermined constant. Then (5) breaks up into the two differential equations d2 T + µ2 c2 T = 0 (6) dt2 d2 R 1 dR + + µ2 R = 0. (7) dr2 r dr (6) can be solved by familiar methods, and we get T = cos µct and T = sin µct 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 indepenx dent variable. Let r = . (7) becomes µ d2 R 1 dR + + R = 0. (8) dx2 x dx Assume, P as in Art. 9, that R can be expressed in terms of whole powers of x. Let R = an xn and substitute in (8). We get P [n(n − 1)an xn−2 + nan xn−2 + an xn ] = 0, an equation which must be true no matter what the value of x. The coefficient of any given power of x, as xk−2 , must, then, vanish, and k(k − 1)ak + kak + ak−2 = 0 k 2 ak + ak−2 = 0

or whence we obtain

ak−2 = − k 2 ak

(9)

as P the konly relation that need be satisfied by the coefficients in order that R = ak x shall be a solution of (8). If

k = 0, ak−2 = 0, ak−4 = 0,

&c.

We can then begin with k = 0 as our lowest subscript. ak−2 From (9) ak = − 2 . k a0 Then a2 = − 2 2 a0 a4 = 2 2 2 .4 a0 a6 = − 2 2 2 , &c. 2 .4 .6 x2 x4 x6 Hence R = a0 1 − 2 + 2 2 − 2 2 2 + · · · 2 2 .4 2 .4 .6

INTRODUCTION.

14

where a0 may be taken at pleasure, is a solution of (8), provided the series is convergent. Take a0 = 1, and then R = J0 (x) where J0 (x) = 1 −

x4 x6 x8 x2 + − + − ··· 22 22 .42 22 .42 .62 22 .42 .62 .82

(10)

is a solution of (8). J0 (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 J0 (x) (v. Int. Cal. Art. 30) 1+

r4 r6 r2 + + + ··· , 22 22 .42 22 .42 .62

where r is the modulus of x, is convergent for all values of r. For the ratio of r2 the n + 1st term of this series to the nth term is and approaches zero as 4n2 its limit as n is indefinitely increased, no matter what the value of r. Therefore J0 (x) is absolutely convergent. J0 (x) is a new and important form. It is called a Bessel’s Function of the zeroth order, or a Cylindrical Harmonic. Equation (8) was obtained from (7) by the substitution of x = µr, therefore R = J0 (µr) = 1 −

(µr)2 (µr)4 (µr)6 + 2 2 − 2 2 2 + ··· 2 2 2 .4 2 .4 .6

is a solution of (7), no matter what the value of µ, and z = J0 (µr) cos µct or z = J0 (µr) sin µct is a solution of (4). z = J0 (µr) cos µct satisfies condition (2) whatever the value of µ. In order that it should also satisfy condition (3) µ must be so taken that J0 (µa) = 0;

(11)

that is, µ must be a root of (11) regarded as an equation in µ. It can be shown that J0 (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 x1 , x2 , x3 , · · · be these roots. Then if x1 x2 x3 = µ1 , = µ2 , = µ3 , &c. a a a z = A1 J0 (µ1 r) cos µ1 ct + A2 J0 (µ2 r) cos µ2 ct + A3 J0 (µ3 r) cos µ3 ct + · · · , (12) where A1 , A2 , A3 , &c., are any constants, is a solution of (4) which satisfies conditions (2) and (3). When t = 0 (12) reduces to z = A1 J0 (µ1 r) + A2 J0 (µ2 r) + A3 J0 (µ3 r) + · · · .

(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 A1 , A2 , A3 , &c., in (12).

INTRODUCTION.

15 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 = A1 e−a

2

µ21 t

2

J0 (µ1 r) + A2 e−a

µ22 t

J0 (µ2 r) + A3 e−a

2

µ23 t

J0 (µ3 r) + · · ·

where µ1 , µ2 , &c., are the roots of J0 (µc) = 0, and where 1 = A1 J0 (µ1 r) + A2 J0 (µ2 r) + A3 J0 (µ3 r) + · · · , c being the radius of the cylinder. 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 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. 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. (a)

Dx2 u + Dy2 u = 0

(1)

Assume u = X.Y where X involves x but not y, and Y involves y but not x. Substitute in (1),

Y Dx2 X + XDy2 Y = 0,

or, since we are now dealing with functions of a single variable,

or

1 d2 X 1 d2 Y + = 0, 2 X dx Y dy 2 1 d2 Y 1 d2 X =− . 2 Y dy X dx2

(2)

INTRODUCTION.

16

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 α2 . Then and

d2 Y − α2 Y = 0 dy 2 d2 X + α2 X = 0; dx2

(3) (4)

and if (3) and (4) can be solved, we can solve (1). They have for their complete solutions Y = Aeαy + Be−αy and

X = C sin αx + D cos αx.

(v. Int. Cal. p. 319, § 21.)

Hence Y = eαy and Y = e−αy are particular solutions of (3), X = sin αx and X = cos αx are particular solutions of (1), and consequently u = eαy sin αx,

u = eαy cos αx,

u = e−αy sin αx, and u = e−αy cos αx

are particular solutions of (1). These agree with the results of Art. 7. (b)

Dt2 y = a2 Dx2 y

(1)

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 a2 T X. We get 1 d2 T 1 d2 X = ; 2 dt X dx2

(2)

a2 T

1 d2 X hence as in the last case is a constant; call it −α2 , and (2) breaks up X dx2 into d2 X + α2 X = 0 dx2 d2 T + α2 a2 T = 0. dt2

(3) (4)

The complete solutions of (3) and (4) are X = A sin αx + B cos αx and

T = C sin αat + D cos αat,

(v. Int. Cal. p. 319, § 21).

y = sin αx cos αat, y = sin αx sin αat, y = cos αx cos αat, y = cos αx sin αat are particular solutions of (1), and agree with the results of Art. 8. (c)

rDr2 (rV ) +


(1)

INTRODUCTION.

17

Assume V = R.Θ where R involves r alone, and Θ involves θ alone; substitute in (1), divide by R.Θ, and transpose; we get dΘ d sin θ 2 1 r d (rR) dθ . =− (2) R dr2 Θ sin θ dθ Since by the reasoning used in (a) and (b) each member of (2) must be a constant, say α2 , we have d2 (rR) r = α2 R dr2 dΘ d sin θ 1 dθ + α2 Θ = 0. and sin θ dθ (3) can be expanded into

(3)

(4)

dR d2 R − α2 R = 0. (5) + 2r dr2 dr (5) can be solved (v. Int. Cal. p. 321, § 23), and has for its complete solution r2

R = Arm + Brn , where

m = − 21 +

q

α2 +

1 4

and

n = − 12 −

q α2 + 41 .

Hence n = −m − 1, and α2 may be written m(m + 1), m being wholly arbitrary; and R = Arm + Br−m−1 . R = rm ,

and

R=

1 rm+1

are, then, particular solutions of r2

d2 R dR + 2r − m(m + 1)R = 0. dr2 dr

(6)

With the new value of α2 (4) becomes dΘ d sin θ 1 dθ + m(m + 1)Θ = 0. (7) sin θ dθ which has been treated in Art. 9 for the case where m is a positive integer, and the particular solution Θ = Pm (cos θ) has been obtained. V = rm Pm (cos θ) 1 and V = m+1 Pm (cos θ), r 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. Hence

INTRODUCTION.

18

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. As a very simple example, let us take the equation Art. 13 (a) (4), which we shall write d2 z + α2 z = 0. (1) dx2 Assume that there isPa solution which can be expressed in terms of powers of x; that is, let z = an xn , where the coefficients are to be determined. Substitute this value for z in (1) and we get P [n(n − 1)an xn−2 + α2 an xn ] = 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 (n + 1)(n + 2)an+2 + α2 an = 0; an = −

whence

(n + 1)(n + 2) an+2 α2

is the only relation that need hold between the coefficients in order that z = P an xn should be a solution of (1). If n + 2 = 0 or n + 1 = 0, an will be zero and an−2 , an−4 , &c., will be zero. In the first case the series will begin with a0 , in the second with a1 . an+2 = −

α2 an . (n + 1)(n + 2)

If we begin with a0 we have a2 = −

α2 a0 , 2!

a4 =

and

z = a0

α4 a0 , 4!

a6 = −

α6 a0 , &c., · · · 6!

α 4 x4 α 6 x6 α 2 x2 + − + ··· 1− 2! 4! 6!

or

z = a0 cos αx

(2) (3)

is a particular solution of (1). If we begin with a1 we have a3 = −

and

α2 a1 , 3!

a5 =

α4 a1 , 5!

a7 = −

α6 a1 , &c., · · · 7!

α2 x3 α 4 x5 α 6 x7 z = a1 x − + − + ··· 3! 5! 7!

(4)

INTRODUCTION.

19

is a solution of (1); a1 can be taken at pleasure. Let a1 = α, (4) becomes α 3 x3 α 5 x5 α 7 x7 + − + ··· 3! 5! 7! z = sin αx z = αx − or

which, then, is a particular solution of (1). z = A sin αx + B cos αx

(5)

is, then, a solution of (1), and since it contains two arbitrary constants it is the general solution. 15.

As another example we will take the equation x2

dz d2 z + 2x − m(m + 1)z = 0, dx2 dx

(1)

which is in effectPequation (6), Art. 13 (c), and let m be a positive integer. Assume z = an xn and substitute in (1). We get P [n(n + 1) − m(m + 1)]an xn = 0. This is an identical equation, therefore [n(n + 1) − m(m + 1)]an = 0. Hence an = 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 = Axm + Bx−m−1

(2)

is the general solution of (1) and z = xm

and z =

1 xm+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 (c). 16.

Let us now take the equation d 2 dz (1 − x ) + m(m + 1)z = 0 dx dx

(1)

which is in effect equation (4), Art. 9, and is known as Legendre’s Equation. (1) may be written d2 z dz (1 − x2 ) 2 − 2x + m(m + 1)z = 0. (2) dx dx

INTRODUCTION.

20

P Assume z = an xn and substitute in (2). We get P {n(n − 1)an xn−2 + [m(m + 1) − n(n + 1)]an xn } = 0. (n + 1)(n + 2)an+2 + [m(m + 1) − n(n + 1)]an = 0,

Hence

or

an = −

(n + 1)(n + 2) an+2 . m(m + 1) − n(n + 1)

(3)

If an = 0, then an−2 = 0, an−4 = 0, &c.; but an = 0 if n = −2 or n = −1. For the first case we have the sequence of coefficients m(m + 1) a0 2! m(m − 2)(m + 1)(m + 3) a4 = a0 4! m(m − 2)(m − 4)(m + 1)(m + 3)(m + 5) a0 , a6 = − 6! a2 = −

&c.

Let us take a0 , which is arbitrary, as 1. Then z = pm (x) where m(m + 1) 2 m(m − 2)(m + 1)(m + 3) 4 pm (x) = 1 − x + x − ··· 2! 4!

(4)

is a solution of Legendre’s Equation if pm (x) is a finite sum or a convergent series. For the second case we have the sequence of coefficients (m − 1)(m + 2) a1 3! (m − 1)(m − 3)(m + 2)(m + 4) a5 = a1 5! (m − 1)(m − 3)(m − 5)(m + 2)(m + 4)(m + 6) a7 = − a1 , 7!

a3 = −

&c.

Let us take a1 , which is arbitrary, as 1. Then z = qm (x) where (m − 1)(m + 2) 3 (m − 1)(m − 3)(m + 2)(m + 4) 5 qm (x) = x − x + x − · · · (5) 3! 5! is a solution of Legendre’s Equation if qm (x) is a finite sum or a convergent series. If m is a positive even whole number, pm (x) will terminate with the term containing xm , and is easily seen to be identical with h m i2 m 2 Γ + 1 m 2 Pm (x). [v. Art. 9 (9)] (−1) 2 Γ(m + 1)

INTRODUCTION.

21

For all other values of m, pm (x) is a series. The ratio of the (n + 1)st term of pm (x) to the nth, when m is not a positive even integer, is (2n − 2 − m)(2n − 1 + m) 2 x . (2n − 1)(2n) Its limiting value, as n is increased, is x2 , 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 qm (x) will terminate with the term containing xm , and is easily seen to be identical with h m + 1 i2 m−1 2 Γ m−1 2 Pm (x). (−1) 2 Γ(m + 1) For all other values of m, qm (x) is a series, and can be shown to be convergent if −1 < x < 1, and divergent for all other values of x. z = Apm (x) + Bqm (x)

(6)

is the general solution of Legendre’s Equation if −1 < x < 1, no matter what the value of m. From Art. 13 (c) it follows that V = rm pm (cos θ) 1 V = m+1 pm (cos θ) r V = rm qm (cos θ) 1 V = m+1 qm (cos θ) r

       

(7)

      

are particular solutions of rDr2 (rV ) +

1 Dθ (sin θDθ V ) = 0, sin θ

no matter what the value of m, provided cos θ 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. With this understanding let us return to (3), which may be rewritten an+2 = −

(m − n)(m + n + 1) an . (n + 1)(n + 2)

If

an+2 = 0, then an+4 = 0, an+6 = 0, &c.;

but

an+2 = 0 if n = m, or n = −m − 1.

INTRODUCTION.

22

If in (3) we begin with n = m − 2, we get the sequence of coefficients already obtained in Art. 9, and we have z = Pm (x), where (2m − 1)(2m − 3) · · · 1 m m(m − 1) m−2 Pm (x) = x − x m! 2(2m − 1) +

m(m − 1)(m − 2)(m − 3) m−4 x 2.4.(2m − 1)(2m − 3)

m(m − 1)(m − 2)(m − 3)(m − 4)(m − 5) m−6 x + ··· , − 2.4.6.(2m − 1)(2m − 3)(2m − 5)

(8)

as a particular solution of Legendre’s Equation. If, however, we begin with n = −m − 3, we have a−m−3 =

(m + 1)(m + 2) a−m−1 2(2m + 3)

a−m−5 =

(m + 1)(m + 2)(m + 3)(m + 4) a−m−1 2.4.(2m + 3)(2m + 5)

a−m−7 =

(m + 1)(m + 2)(m + 3)(m + 4)(m + 5)(m + 6) a−m−1 , 2.4.6.(2m + 3)(2m + 5)(2m + 7)

a−m−1 may be taken at pleasure, and is usually taken as

&c.

m! , 1.3.5. · · · (2m + 1)

and z = Qm (x) where m! (m + 1)(m + 2) 1 1 Qm (x) = + (2m + 1)(2m − 1) · · · 1 xm+1 2.(2m + 3) xm+3 (m + 1)(m + 2)(m + 3)(m + 4) 1 + + ··· 2.4.(2m + 3)(2m + 5) xm+5

(9)

is a second particular solution of Legendre’s Equation, provided the series is convergent. Qm (x) is called a Surface Zonal Harmonic of the second kind. 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 = APm (x) + BQm (x)

(10)

is the general solution of Legendre’s Equation if x < −1 or x > 1. We have seen that for −1 < x < 1 m

Pm (x) = (−1) 2

Γ(m + 1) h m i2 pm (x) 2m Γ +1 2

(11)

if m is an even integer, and Pm (x) = (−1)

m−1 2

Γ(m + 1) h m + 1 i2 qm (x) 2m−1 Γ 2

(12)

INTRODUCTION.

23

if m is an odd integer. If now we define Qm (x) as follows when −1 < x < 1 h m + 1 i2 m−1 2 Γ m+1 2 pm (x) Qm (x) = (−1) 2 Γ(m + 1)

(13)

if m is an odd integer, and h m i2 m 2 Γ + 1 m 2 Qm (x) = (−1) 2 qm (x) Γ(m + 1)

(14)

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. 17.

Let us last consider the equation 1 dz m2 d2 z + + 1− 2 z =0 dx2 x dx x

(1)

which is known as Bessel’s Equation, and which reduces to (8) Art. 11, that is, to d2 z 1 dz + +z =0 dx2 x dx when m = 0;9 (1) can be simplified by a change of the dependent variable. Let z = xm v and we get d2 v 2m + 1 dv +v =0 (2) + dx2 x dx to determine v. P Assume v = an xn , and substitute in (2). We get P [n(2m + n)an xn−2 + an xn ] = 0; an−2 = −n(2m + n)an .

whence

If we begin with n = 0, then an−2 = 0, an−4 = 0, &c., and we have the set of values a2 = − a4 =

a0 a0 =− 2 2(2m + 2) 2 (m + 1)

a0 a0 = 4 2.4(2m + 2)(2m + 4) 2 .2!(m + 1)(m + 2)

9 This equation was first studied by Fourier in considering the cooling of a cylinder. We shall designate it as “Fourier’s Equation.”

INTRODUCTION.

a6 = −

24

a0 a0 =− 6 ; 2.4.6(2m + 2)(2m + 4)(2m + 6) 2 .3!(m + 1)(m + 2)(m + 3)

whence

z = a0 xm 1 −

x2 x4 + 4 + 1) 2 .2!(m + 1)(m + 2)

22 (m

x6 − 6 + ··· 2 .3!(m + 1)(m + 2)(m + 3)

(3)

1 is a solution of Bessel’s Equation. a0 is usually taken as m if m is a positive 2 m! 1 integer, or as m if m is unrestricted in value, and the second member 2 Γ(m + 1) of (3) is represented by Jm (x) and is called a Bessel’s Function of the mth order, or a Cylindrical Harmonic of the mth order. If m = 0, Jm (x) becomes J0 (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 xm v for z, we get in place of (2) the equation d2 v 1 − 2m dv + +v =0 dx2 x dx and in place of (3) x2 x4 −m z = a0 x 1− 2 + 4 2 (1 − m) 2 .2!(1 − m)(2 − m) x6 − 6 + ··· (4) 2 .3!(1 − m)(2 − m)(3 − m) 1 the second member of (4) is the same 2−m Γ(1 − m) function of −m and x that Jm (x) is of +m and x and may be written J−m (x). If a0 is taken equal to

Therefore

z = AJm (x) + BJ−m (x)

(5)

is the general solution of (1) unless Jm (x) and J−m (x) should prove not to be independent. It is easily seen that when m = 0, J−m (x) and Jm (x) become identical and (5) reduces to z = (A + B)J0 (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 Jm (x) whenever m is an integer, and consequently that the solution (5) is general only when m if real is fractional or incommensurable.

INTRODUCTION.

25

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 a0 = 1; then the value which the second member of (4) assumes when a0 = 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 F (m, x) − F (−m, x) , (6) z= 2m F (m, x) − F (−m, x) but the limiting value which approaches as m approaches 2m zero is [Dm F (m, x)]m=0 and consequently z = [Dm F (m, x)]m=0

(7)

1 dz d2 z + + z = 0, 2 dx x dx

(8)

is a solution of the equation

and the general solution of (8) is z = AJ0 (x) + B[Dm F (m, x)]m=0 . x4 x2 + 4 F (m, x) = xm 1 − 2 2 (m + 1) 2 .2!(m + 1)(m + 2) x6 − 6 + ··· 2 .3!(m + 1)(m + 2)(m + 3) x2 x4 m Dm F (m, x) = x log x 1 − 2 + − ··· 2 (m + 1) 24 .2!(m + 1)(m + 2) x4 x2 m +x Dm 1 − 2 + + ··· . 2 (m + 1) 24 .2!(m + 1)(m + 2) The general term of the last parenthesis can be written (−1)k

x2k , 22k .k!(m + 1)(m + 2) · · · (m + k)

and its partial derivative with respect to m is (−1)k log

x2k 22k .k!

Dm

1 . (m + 1)(m + 2) · · · (m + k)

1 = −[ log(m + 1) + log(m + 2) + · · · (m + 1)(m + 2) · · · (m + k) + log(m + k)].

INTRODUCTION.

26

Take the Dm of both members and we have 1 (m + 1)(m + 2) · · · (m + k) 1 1 1 1 + + ··· . =− (m + 1)(m + 2) · · · (m + k) m + 1 m + 2 m+k x2 x4 x6 Dm 1 − 2 + 4 − 6 2 (m + 1) 2 .2!(m + 1)(m + 2) 2 .3!(m + 1)(m + 2)(m + 3) 1 x4 x2 1 1 1 − 4 + ··· = 2 + 2 (m + 1)2 2 .2! (m + 1)(m + 2) m + 1 m + 2 x6 1 1 1 1 + 6 + + + ··· 2 .3! (m + 1)(m + 2)(m + 3) m + 1 m + 2 m + 3 Dm

and we have x4 1 1 x2 1 − + [Dm F (m, x)]m=0 = J0 (x) log x + 2 2 (1!)2 1 24 (2!)2 1 2 x6 1 1 1 x8 1 1 1 1 + 6 + + − + + + + ··· ; 2 (3!)2 1 2 3 28 (4!)2 1 2 3 4 and where

z = AJ0 (x) + BK0 (x), (9) 2 4 x x 1 1 K0 (x) = J0 (x) log x + 2 − 4 + 2 2 2 (2!) 1 2 x6 x8 1 1 1 1 1 1 1 + 6 + + − + + + + · · · (10) 2 (3!)2 1 2 3 28 (4!)2 1 2 3 4

is the general solution of Fourier’s Equation (8). K0 (x) is known as a Bessel’s Function of the Second Kind. 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 (a)]. The most general form of a homogeneous ordinary linear differential equation of the second order is d2 y dy + Qy = 0 (1) +P 2 dx dx where P and Q are functions of x. Suppose that y=v is a particular solution of (1). Substitute y = vz in (1) and we get d2 z dv dz v 2+ 2 + Pv = 0. dx dx dx

(2)

(3)

INTRODUCTION. Call

27

dz = z 0 . Then (3) becomes dx dv dz 0 + 2 + P v z 0 = 0, v dx dx

(4)

a differential equation of the first order in which the variables can be separated. Multiply by dx and divide by vz 0 and (4) reduces to dv dz 0 +2 + P dx = 0. 0 z v Integrate and we have log z 0 + log v 2 + z0 v2 = e

or

r C− P dx

w

P dx = C

= Be−

r

P dx

,

r

e− P dx dz =B , dx vr2 w e− P dx dx; z =A+B vr2 w e− P dx y = v A+B dx v2 z0 =

and

(5)

is the general solution of (1), the only arbitrary constants in the second member of (5) being those explicitly written, namely, A and B. (a)

Apply this formula to (1) Art. 14, d2 z + α2 z = 0; dx2

(1)

given: z = cos αx, as a particular solution. Substituting in (5) we have since P =0 w dx z = cos αx A + B cos2 αx B = cos αx A + tan αx α = A cos αx + B1 sin αx,

(2)

as the general solution of (1), and this agrees perfectly with (5) Art. 14. (b)

Take equation (1) Art. 15. x2

d2 z dz − m(m + 1)z = 0; + 2x dx2 dx

given: z = xm , as a particular solution.

(1)

INTRODUCTION. Here P =

that is

28

r 2 w 1 , P dx = 2 log x = log x2 , and e− P dx = 2 . Hence by (5) x x w dx B 1 m = x A + , z = xm A + B x2m+2 −2m − 1 x2m+1 B1 z = Axm + m+1 x

(2)

is the general solution of (1), and agrees with (2) Art. 15. (c)

Take Legendre’s Equation, (2) Art. 16. (1 − x2 )

d2 z dz − 2x + m(m + 1)z = 0; 2 dx dx

given: z = Pm (x), as a particular solution. r 1 −2x w , P dx = log(1 − x2 ), and e− P dx = . Here P = 2 1−x 1 − x2 w dx Hence by (5) z = Pm (x) A + B (1 − x2 )[Pm (x)]2

(1)

(2)

is the general solution of (1) and must agree with (10) Art. 16, if m is an integer, and therefore w dx Qm (x) = CPm (x) (3) 2 (1 − x )[Pm (x)]2 where C is as yet undetermined, and no constant term is to be understood with the integral in the second member. (d )

Take Bessel’s Equation, (1) Art. 17. d2 z 1 dz m2 + + 1 − z = 0; dx2 x dx x2

given: z = Jm (x), as a particular solution. r 1 w 1 Here P = , P dx = log x, and e− P dx = . Hence by (5) x x w dx z = Jm (x) A + B x[Jm (x)]2 is the general solution of Bessel’s Equation. If m = 0 (2) becomes w z = J0 (x) A + B

dx x[J0 (x)]2

and must agree with (9) Art. 17. Therefore w K0 (x) = CJ0 (x)

dx , x[J0 (x)]2

(1)

(2)

(3)

(4)

INTRODUCTION.

29

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).

30

CHAPTER II. DEVELOPMENT IN TRIGONOMETRIC SERIES.

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 b0 + b1 cos x + b2 cos 2x + · · · + a1 sin x + a2 sin 2x + · · · 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 = π. 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 = a1 sin x + a2 sin 2x + a3 sin 3x + · · · , of a curve which shall coincide with so much of the given curve as lies between the points corresponding to x = 0 and x = π. It is clear that in the equation y = a1 sin x

(1)

a1 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 a1 is the only unknown quantity and which will therefore give us one and only one value for a1 . In like manner the curve y = a1 sin x + a2 sin 2x may be made to pass through any two arbitrarily chosen points whose abscissas lie between 0 and π provided that the abscissas are not equal; and y = a1 sin x + a2 sin 2x + a3 sin 3x + · · · + an sin nx may be made to pass through any n arbitrarily chosen points whose abscissas lie between 0 and π 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 = π it has one and only one value, and if between

DEVELOPMENT IN TRIGONOMETRIC SERIES.

31

x = 0 and x = π it is finite and continuous, or if discontinuous has only finite discontinuities (v. Int. Cal. Art. 83, p. 78), the coefficients in y = a1 sin x + a2 sin 2x + a3 sin 3x + · · · + an sin nx

(2)

can be determined so that the curve represented by (2) will pass through any n arbitrarily chosen points of the curve y = f (x)

(3)

whose abscissas lie between 0 and π 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. π = ∆x; then the coördinates of the n points will be [∆x, f (∆x)], Call n+1 [2∆x, f (2∆x)], [3∆x, f (3∆x)], · · · [n∆x, f (n∆x)]. Substitute them in (2) and we have  f (∆x) = a1 sin ∆x + a2 sin 2∆x + a3 sin 3∆x + · · · + an sin n∆x     f (2∆x) = a1 sin 2∆x + a2 sin 4∆x + a3 sin 6∆x + · · · + an sin 2n∆x     f (3∆x) = a1 sin 3∆x + a2 sin 6∆x + a3 sin 9∆x + · · · + an sin 3n∆x (4)   .. .. .. .. ..   . . . . .     2 f (n∆x) = a1 sin n∆x + a2 sin 2n∆x + a3 sin 3n∆x + · · · + an sin n ∆x, n equations of the first degree to determine the n coefficients a1 , a2 , a3 , · · · an . 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 a1 , a2 , a3 , a4 , and a5 , so that y = a1 sin x + a2 sin 2x + a3 sin 3x + a4 sin 4x + a5 sin 5x

(5)

shall pass through the five points of the line y=x 5π π π 2π 3π 4π which have the abscissas , , , , and , here being ∆x. 6 6 6 6 6 6 We must now solve the equations π 6 2π 6 3π 6 4π 6 5π 6

π 6 2π = a1 sin 6 3π = a1 sin 6 4π = a1 sin 6 5π = a1 sin 6 = a1 sin

2π 6 4π + a2 sin 6 6π + a2 sin 6 8π + a2 sin 6 10π + a2 sin 6 + a2 sin



 5π    6    10π     + a5 sin  6    15π + a5 sin 6    20π     + a5 sin  6    25π   + a5 sin . 6 + a5 sin

(6)


32

π 2π Multiply the first equation by 2 sin , the second by 2 sin , the third by 6 6 3π 4π 5π 2 sin , the fourth by 2 sin , the fifth by 2 sin and add the equations. 6 6 6 The coefficient of a2 is 2 sin

but

π 2π 2π 4π 3π 6π 4π 8π sin + 2 sin sin + 2 sin sin + 2 sin sin 6 6 6 6 6 6 6 6 5π 10π + 2 sin sin ; 6 6 2π π 3π π = cos − cos , &c. 2 sin sin 6 6 6 6

Hence the coefficient of a2 becomes  π 2π 3π 4π 5π   + cos + cos + cos + cos 6 6 6 6 6 3π 6π 9π 12π 15π   − cos − cos − cos − cos − cos 6 6 6 6 6 cos

(7)

and this may be reduced by the aid of an important Trigonometric formula which we proceed to establish. 20.

Lemma. θ 1 1 sin(2n + 1) 2 . cos θ + cos 2θ + cos 3θ + · · · + cos nθ = − + θ 2 2 sin 2

(1)

For let S = cos θ + cos 2θ + cos 3θ + · · · + cos nθ and multiply by 2 cos θ. 2S cos θ = 2 cos2 θ + 2 cos θ cos 2θ + 2 cos θ cos 3θ + · · · + 2 cos θ cos nθ = 1 + cos θ + cos 2θ + · · · + cos(n − 1)θ + cos 2θ + cos 3θ + cos 4θ + · · · + cos(n + 1)θ = 2S + 1 + cos(n + 1)θ − cos θ − cos nθ. 1 cos nθ − cos(n + 1)θ S=− + 2 2(1 − cos θ) θ 1 1 sin(2n + 1) 2 S=− + . θ 2 2 sin 2

or

21.

Hence

Q.E.D.

Applying (1) Art. 20 to (7) Art. 19 the coefficient of a2 reduces to 11π 33π sin 12 − 12 ; π 3π 2 sin 2 sin 12 12 sin


33

11π π 33π 3π = π − , and = 3π − ; 12 12 12 12

but

3π π sin 3π − sin π − 12 − 12 = 1 − 1 = 0, π 3π 2 2 2 sin 2 sin 12 12

therefore

and a2 vanishes. In like manner it may be shown that the coefficients of a3 , a4 , and a5 vanish. The coefficient of a1 is 2 sin2

π 2π 3π 4π 5π + 2 sin2 + 2 sin2 + 2 sin2 + 2 sin2 6 6 6 6 6

=1

+ 1 + 1 + 1 + 1 2π 4π 6π 8π 10π − cos − cos − cos − cos − cos 6 6 6 6 6 11π π sin sin 2π − 1 6 = 51 − 6 = 6. =5+ − 2 2 π π 2 sin 2 sin 6 6 The first member of the final equation is 2π π 2π 2π 3π 3π 4π 4π 5π 5π sin + 2 sin +2 sin +2 sin +2 sin . Hence 6 6 6 6 6 6 6 6 6 6 k=5 √ 2 X kπ kπ π a1 = sin = (2 + 3) = 2 approximately. 6 6 6 6 k=1

2π If we multiply the first equation of (6) Art. 19 by 2 sin , the second by 6 4π 6π 8π 10π 2 sin , the third by 2 sin , the fourth by 2 sin , the fifth by 2 sin , add 6 6 6 6 and reduce as before we shall find k=5

a2 =

2 X kπ 2kπ π√ sin =− 3 = −0.9; 6 6 6 6 k=1

and in like manner we get k=5

2 X kπ 3kπ π a3 = sin = = 0.5 6 6 6 6 k=1 √ k=5 2 X kπ 4kπ π 3 a4 = sin =− = −0.3 6 6 6 18 k=1 k=5

a5 =

√ 2 X kπ 5kπ π sin = (2 − 3) = 0.1. 6 6 6 6 k=1


34

Therefore y = 2 sin x − 0.9 sin 2x + 0.5 sin 3x − 0.3 sin 4x + 0.1 sin 5x

(1)

π 2π 3π 4π cuts the curve y = x at the five points whose abscissas are , , , , and 6 6 6 6 5π . 6 22. The equations (4) Art. 19 can be solved by exactly the same device. To find any coefficient am multiply the first equation by 2 sin m∆x, the second by 2 sin 2m∆x, the third by 2 sin 3m∆x, &c. and add. The coefficient of any other a as ak in the resulting equation will be 2 sin k∆x sin m∆x + 2 sin 2k∆x sin 2m∆x + 2 sin 3k∆x sin 3m∆x + · · · + 2 sin nk∆x sin nm∆x = cos(m − k)∆x + cos 2(m − k)∆x + cos 3(m − k)∆x + · · · + cos n(m − k)∆x − cos(m + k)∆x − cos 2(m + k)∆x − cos 3(m + k)∆x − · · · − cos n(m + k)∆x 2n + 1 2n + 1 (m − k)∆x sin (m + k)∆x sin 2 2 − ; by (1) Art. 20. = (m − k)∆x (m + k)∆x 2 sin 2 sin 2 2 2n + 1 1 =n+1− and 2 2 Hence the coefficient of ak may be written

(n + 1)∆x = π.

h h (m − k)∆x i (m + k)∆x i sin (m − k)π − sin (m + k)π − 2 2 − (m − k)∆x (m + k)∆x 2 sin 2 sin 2 2 1 1 1 1 but this is equal to − or − + according as m − k is odd or even and so 2 2 2 2 is zero in either case. The coefficient of am will be 2 sin2 m∆x + 2 sin2 2m∆x + 2 sin2 3m∆x + · · · + 2 sin2 nm∆x =

1

+

1

+

1

+

···

+

1

− cos 2m∆x − cos 4m∆x − cos 6m∆x − · · · − cos 2nm∆x =n+

But therefore

1 sin(2n + 1)m∆x − , by (1) Art. 20. 2 2 sin m∆x

(2n + 1)m∆x = 2m(n + 1)∆x − m∆x = 2mπ − m∆x, sin(2mπ − m∆x) 1 sin(2n + 1)m∆x = =− , 2 sin m∆x 2 sin m∆x 2


35

and the coefficient of am is n + 1. The first member of our final equation will be 2

k=n X

f (k∆x) sin km∆x.

k=1

Hence am =

k=n 2 X f (k∆x) sin km∆x, n+1

(1)

k=1

and the curve y = a1 sin x + a2 sin 2x + · · · + an sin nx,

(2)

where the coefficients are given by (1) will pass through the n points of the π . curve y = f (x) whose abscissas are ∆x, 2∆x, 3∆x, · · · n∆x. ∆x being 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 a1 , a2 , a3 , · · · an that can satisfy these equations. Consequently the solution which we have obtained is the only solution possible. 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 = π, 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). Let us see what limiting value am approaches as n is indefinitely increased. am =

k=n 2 X f (k∆x) sin km∆x n+1

(1) Art. 22.

k=1

k=n 2∆x X = f (k∆x) sin km∆x π k=1

2 = π

"

2 = π

"

f (∆x) sin m∆x.∆x + f (2∆x) sin 2m∆x.∆x + · · ·

#

+ f (n∆x) sin nm∆x.∆x

f (∆x) sin m∆x.∆x + f (2∆x) sin 2m∆x.∆x + · · ·

since ∆x =

+f (π − ∆x) sin m(π − ∆x).∆x π . n+1

#


36

As n is increased indefinitely ∆x approaches zero as a limit. Hence the limiting value of am as n increases indefinitely is " # f (∆x) sin m∆x.∆x + f (2∆x) sin 2m∆x.∆x + · · · 2 1 limit . π ∆x=0 +f (π − ∆x) sin m(π − ∆x).∆x =

π 2w f (x) sin mx.dx. π

[v. Int. Cal. Arts. 80, 81.]

0

Hence

f (x) = a1 sin x + a2 sin 2x + a3 sin 3x + · · · ,

(2)

where any coefficient am is given by the formula am =

π 2w f (x) sin mx.dx, π

(3)

0

is a true development of f (x) for all values of x between x = 0 and x = π provided that the series (2) 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 = π 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 = π; but even when f (x) is not zero for x = 0 and x = π the development given above holds good for any value of x between zero and π no matter how near it may be taken to either of these values. 24. Instead of actually performing the elimination in equations (4) Art. 19 and getting a formula for am in terms of n, and then letting n increase indefinitely, we might have saved labor by the following method. Return to equations (4) Art. 19 and multiply the first by ∆x sin m∆x, the second by ∆x sin 2m∆x, and so on, that is multiply each equation by ∆x times the coefficient of am in that equation, and then add the equations. We get as the coefficient of ak sin k∆x sin m∆x.∆x + sin 2k∆x sin 2m∆x.∆x + · · · + sin nk∆x sin nm∆x.∆x. Let us find its limiting value as n is indefinitely increased. It may be written, since (n + 1)∆x = π, " # sin k∆x sin m∆x.∆x + sin 2k∆x sin 2m∆x.∆x + · · · limit . + sink(π − ∆x) sin m(π − ∆x).∆x ∆x=0 =

wπ

sin kx sin mx.dx;

0 1 We

. . shall use the sign = for approaches. ∆x = 0 is read ∆x approaches zero.

DEVELOPMENT IN TRIGONOMETRIC SERIES. wπ

but

sin kx sin mx.dx =

1 2

0

wπ

37

[cos(m − k)x − cos(m + k)x]dx

0

= 0 if m and k are not equal. The coefficient of am is ∆x(sin2 m∆x + sin2 2m∆x + sin2 3m∆x + · · · + sin2 nm∆x). Its limiting value 2 2 2 limit sin m∆x.∆x + sin 2m∆x.∆x + · · · + sin m(π − ∆x)∆x . ∆x=0

=

wπ

sin2 mx.dx =

0

π . 2

The first member is f (∆x) sin m∆x.∆x + f (2∆x) sin 2m∆x.∆x + · · · + f (n∆x) sin mn∆x.∆x and its limiting value is

wπ

f (x) sin mx.dx.

0

Hence the limiting form approached by the final equation as n is increased is

wπ

f (x) sin mx.dx =

0

Whence

am =

π am . 2

π 2w f (x) sin mx.dx π

as before.

0

This method is practically the same as multiplying the equation f (x) = a1 sin x + a2 sin 2x + a3 sin 3x + · · ·

(1)

by sin mx.dx and integrating both members from zero to π. It is exceedingly important to realize that the short method of determining any coefficient am of the series (1) which has just been described in the italicized paragraph, is essentially the same as that of obtaining am 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.


38

25. We see, then, that any function of x which is single-valued, finite, and continuous between x = 0 and x = π, 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) = a1 sin x + a2 sin 2x + a3 sin 3x + · · ·

where

am =

(1)

π π 2w 2w f (x) sin mx.dx = f (α) sin mα.dα; π π 0

(2)

0

and the series and the function will be identical for all values of x between x = 0 and x = π, not including the values x = 0 and x = π 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 above2 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. 26. (a) We have where

Let us obtain a few sine developments. Let

f (x) = x.

(1)

x = a1 sin x + a2 sin 2x + a3 sin 3x + · · · π 2w x sin mx.dx am = π

(2) (3)

0

w

x sin mx.dx = wπ 0

and

x=2

1 (sin mx − mx cos mx), m2

x sin mx.dx = −

(−1)m π , m

sin x sin 2x sin 3x sin 4x − + − + ··· 1 2 3 4

(4)

2 Provided the function has not an infinite number of maxima and minima in the neighborhood of a point. v. Arts. 37–38.

DEVELOPMENT IN TRIGONOMETRIC SERIES. (b)

Let

39

f (x) = 1. π 2w sin mx.dx; π

am = w wπ

(2)

0

sin mx.dx = −

sin mx.dx =

0

(1)

cos mx , m

1 1 (1 − cos mπ) = [1 − (−1)m ] m m

= 0 if m is even 2 = if m is odd. m 4 1= π

Hence

sin x sin 3x sin 5x sin 7x + + + + ··· 1 3 5 7

.

(3)

It is to be noticed that (3) gives at once a sine development for any constant c. It is, 4c sin x sin 3x sin 5x + + + ··· . (4) c= π 1 3 5 π If we substitute x = in (4) (a) or (3) (b) we get a familiar result, namely 2 1 1 1 1 π = − + − + ··· , 4 1 3 5 7

(5)

a formula usually derived by substituting x = 1 in the power series for tan−1 x. (v. Dif. Cal. Art. 135.) (4) (a) does not hold good when x = π, and (3) (b) fails when x = 0 and when x = π, for in all these cases the series reduces to zero. π Let f (x) = x from x = 0 to x = and 2 π f (x) = π − x from x = to x = π. That is, let 2 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 (c)

wπ 0

π

f (x) sin mx.dx

into

w2 0

f (x) sin mx.dx +

wπ π 2

f (x) sin mx.dx.


40

We have, then, π

am

π 2 2w 2w = x sin mx.dx + (π − x) sin mx.dx π ππ 0

(1)

2

4 π = 2 sin m . m π 2 But

sin m

π =1 2 =0

if “

m=1 m=2

or 4k + 1 “ 4k + 2

= −1 “

m=3

“

4k + 3

=0

m=4

“

4k.

“

Hence if y = f (x) represents our broken line, 4 sin x sin 3x sin 5x sin 7x f (x) = − + − + · · · . π 12 32 52 72 When x =

π π f (x) = and we have 2 2 π2 1 1 1 1 = 2 + 2 + 2 + 2 + ··· 8 1 3 5 7

(d )

(2)

(3)

As a case where the function has a finite discontinuity, let f (x) = 1

from

f (x) = 0

“

x=0 x=

π 2

to x = “

π 2

and

x = π.

y = f (x) will in this case represent the locus in the figure. As before wπ

π

f (x) sin mx.dx =

0

+

w2 0 wπ

f (x) sin mx.dx f (x) sin mx.dx.

π 2 π

am

π 2 2w 2w = sin mx.dx + 0. sin mx.dx. π ππ 0

π

am

2

2 2w 2 1 π = sin mx.dx = 1 − cos m . π πm 2

0

(1)


But

cos m

π = 0 2 = −1

if “

m=1 m=2

or 4k + 1 “ 4k + 2

=

0

“

m=3

“

4k + 3

=

1

“

m=4

“

4k.

41

Hence f (x) =

2 π

If x =

sin x 2 sin 2x sin 3x sin 5x 2 sin 6x sin 7x + + + + + + ··· 1 2 3 5 6 7

. (2)

π 1 the second member of (2) reduces to , for 2 2 1 2 1 1 1 1 − + − + ··· = by (5) (b); π 1 3 5 7 2

and we see that the series represents the function completely for all values of π x between x = 0 and x = π except for x = and there it has a value which 2 π is the mean of the values approached by the function as x approaches from 2 opposite sides. EXAMPLES. Obtain the following developments:— 2 2 π 4 π 4 π2 π2 2 (1) x2 = − 3 sin x − sin 2x + − 3 sin 3x − sin 4x π 1 1 2 3 3 4 2 π 4 + − 3 sin 5x − · · · . 5 5 3 3 3 π 6π π 6π π 6π 2 3 − 3 sin x − − 3 sin 2x + − 3 sin 3x (2) x = π 1 1 2 2 3 3 3 6π π − 3 sin 4x + · · · . − 4 4 2 sin x π sin 3x 2π sin 5x (3) f (x) = + 2 sin 2x − − 2 sin 4x + π 12 2 32 4 52 3π + 2 sin 6x − · · · , 6 if f (x) = x from x = 0 to x =

π π and f (x) = 0 from x = to x = π. 2 2


(4)

sin µx =

42

2 2 sin 2x 3 sin 3x 4 sin 4x sin x − + − + · · · sin µπ 2 π 1 − µ2 22 − µ2 32 − µ2 42 − µ2

if µ is a fraction. 2 1 2 3 (5) ex = (1 + eπ ) sin x + (1 − eπ ) sin 2x + (1 + eπ ) sin 3x π 2 5 10 4 + (1 − eπ ) sin 4x + · · · . 17 2 sinh π 1 2 3 4 (6) sinh x = sin x − sin 2x + sin 3x − sin 4x + · · · . π 2 5 10 17 2 1 2 (7) cosh x = (1 + cosh π) sin x + (1 − cosh π) sin 2x π 2 5 3 + (1 + cosh π) sin 3x + · · · . 10 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 = π, that it does not become infinite between x = 0 and x = π, and that if discontinuous it has only finite discontinuities. Assume f (x) = b0 + b1 cos x + b2 cos 2x + b3 cos 3x + · · · (1) To determine any coefficient bm multiply (1) by cos mx.dx and integrate each term from 0 to π. wπ

b0 cos mx.dx = 0.

0

wπ

bk cos kx cos mx.dx =

0

π bk w [cos(m − k)x + cos(m + k)x]dx 2 0

= 0 if m and k are not equal. w

bm cos2 mx.dx = wπ 0

Hence

bm =

bm cos2 mx.dx =

π bm , 2

if m is not zero.

π π 2w 2w f (x) cos mx.dx = f (α) cos mα.dα, π π 0

if m is not zero.

bm (mx + cos mx sin mx), 2m

0

(2)


43

To get b0 multiply (1) by dx and integrate from zero to π.

Hence

b0 =

1 π

wπ

b0 dx = b0 π,

0 wπ

bk cos kx.dx = 0.

0 wπ

f (x)dx =

0

π 1w f (α)dα, π

(3)

0

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) = 21 b0 + b1 cos x + b2 cos 2x + b3 cos 3x + · · ·

(4)

and then the formula bm =

π π 2w 2w f (x) cos mx.dx = f (α) cos mα.dα π π 0

(2)

0

will give b0 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 = 21 b0 + b1 cos x + b2 cos 2x + · · · + bn cos nx

(5)

in turn the co¨ ordinates of the n + 1 points of the curve y = f (x) whose projections on the axis of X are equidistant, determining b0 , b1 , b2 , · · · bn by elimination from the n + 1 resulting equations, and then taking the limiting values they approach as n is indefinitely increased. (v. Art. 24.) π If ∆x = the abscissas of the n + 1 points used are 0, ∆x, 2∆x, 3∆x, n+1 · · · n∆x, so that we should expect our cosine development to hold for x = 0 as well as for values of x between zero and π. 28. (a)

Let us take one or two examples: Let

f (x) = x.

b0 =

π 2w 2 π2 x dx = = π. π π 2 0

bm

π 2w 2 2 = x cos mx.dx = 2 (cos mπ − 1) = 2 [(−1)m − 1]. π m π m π 0

(1)


Hence

x=

π 4 − 2 π

cos 3x cos 5x cos 7x + + + · · · . cos x + 32 52 72

44

(2)

(2) holds good not only for values of x between zero and π but for x = 0 and x = π as well, since for these values we have 1 1 π 4 1 (3) 0= − 1 + 2 + 2 + 2 + ··· 2 π 3 5 7 π 1 1 4 1 and π= + (4) 1 + 2 + 2 + 2 + ··· 2 π 3 5 7 which are true by Art. 26 (c)(3).

(b)

Let

f (x) = x sin x.

b0 =

(1)

π 2w 2 x sin x.dx = π = 2, π π 0

π π 1w 1 2w x sin x cos x.dx = x sin 2x.dx = − , b1 = π π 2 0

bm

0

π π 2w 1w = x sin x cos mx.dx = [x sin(m + 1)x − x sin(m − 1)x]dx π π 0

0

2 = if m is odd (m − 1)(m + 1) 2 =− if m is even. (m − 1)(m + 1) Hence x sin x = 1 − If x =

π we have 2

cos x 2 cos 2x 2 cos 3x 2 cos 4x − + − + ··· 2 1.3 2.4 3.5

1 1 1 1 π = + − + − ··· . 4 2 1.3 3.5 5.7

(2)

(3)

DEVELOPMENT IN TRIGONOMETRIC SERIES. EXAMPLES. Obtain the following developments: π 2 cos 2x cos 6x cos 10x cos 14x + + + + · · · (1) f (x) = − 4 π 12 32 52 72 π π if f (x) = x from x = 0 to x = and f (x) = π − x from x = to x = π. 2 2 1 2 cos x cos 3x cos 5x cos 7x (2) f (x) = + − + − + ··· , 2 π 1 3 5 7 π π if f (x) = 1 from x = 0 to x = and f (x) = 0 from x = to x = π. 2 2 cos x cos 2x cos 3x cos 4x π2 2 −4 − + − + ··· , (3) x = 3 12 22 32 42 2 2 6 π 4 π2 π 4 π3 − − cos x − cos 2x + − cos 3x (4) x3 = 4 π 12 14 22 32 34 2 π2 π 4 − 2 cos 4x + − cos 5x − · · · , 4 52 54 π π 2 2 1 3π (5) f (x) = + − 1 cos x − 2 cos 2x − 2 + 1 cos 3x 8 π 2 2 3 2 1 5π 2 + 2 − 1 cos 5x − 2 cos 6x − · · · , 5 2 6 π π if f (x) = x from x = 0 to x = and f (x) = 0 from x = to x = π. 2 2 2 1 1 1 (6) ex = (eπ − 1) − (eπ + 1) cos x + (eπ − 1) cos 2x π 2 1 + 12 1 + 22 1 π − (e + 1) cos 3x + · · · , 1 + 32 2 sinh π 1 1 1 1 (7) cosh x = − cos x + cos 2x − cos 3x π 2 2 5 10 1 + cos 4x − · · · , 17 2 1 1 (8) sinh x = (cosh π − 1) − (cosh π + 1) cos x π 2 2 1 1 + (cosh π − 1) cos 2x − (cosh π + 1) cos 3x + · · · , 5 10 2µ sin µπ 1 cos x cos 2x cos 3x (9) cos µx = − 2 + 2 − 2 π 2µ2 µ − 12 µ − 22 µ − 32 cos 4x + 2 − ··· , µ − 42 if µ is a fraction.

45


46

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 π, there is a decided difference in the two series for other values of x. Both series are periodic functions of x having the period 2π. If then we let y equal the series in question and construct the portion of the corresponding curve which lies between the values x = −π and x = π the whole curve will consist of repetitions of this portion. Since sin mx = − sin(−mx) the ordinate corresponding to any value of x between −π 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 = a1 sin x + a2 sin 2x + a3 sin 3x + · · · (1) is symmetrical with respect to the origin. Since cos mx = cos(−mx) the ordinate corresponding to any value of x between −π 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 = 21 b0 + b1 cos x + b2 cos 2x + b3 cos 3x + · · ·

(2)

is symmetrical with respect to the axis of Y . If then f (x) = −f (−x), that is if f (x) is an odd function the sine series corresponding to it will be equal to it for all values of x between −π and π, 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 even function the cosine series corresponding to it will be equal to it for all values of x between x = −π and x = π, 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. sin 2x sin 3x sin 4x y = 2 sin x − + − + ··· (3) 2 3 4 4 cos 3x cos 5x cos 7x π y= − cos x + + + + · · · (4) 2 π 32 52 72 [v. Art. 26(a) and Art. 28(a)]. (3) represents the curve


47

and (4) the curve

Both coincide with y = x from x = 0 to x = π, (3) coincides with y = x from x = −π to x = π, and neither coincides with y = x for values of x less than −π or greater than π. Moreover (3), in addition to the continuous portions of the locus represented in the figure, gives the isolated points (−π, 0) (π, 0) (3π, 0) &c. 30. We have seen that if f (x) is an odd function its development in sine series holds for all values of x from −π to π, as does the development of f (x) in cosine series if f (x) is an even function. Thus the developments of Art. 26(a), Art. 26 Exs. (2), (4), (6); Art. 28(b), Art. 28 Exs. (3), (7), (9) are valid for all values of x between −π and π. Any function of x can be developed into a Trigonometric series to which it is equal for all values of x between −π and π. 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) =

f (x) + f (−x) f (x) − f (−x) + 2 2

(1)

f (x) + f (−x) is not changed by reversing the sign of x and is 2 f (x) − f (−x) therefore an even function of x; and when we reverse the sign of x, 2 is affected only to the extent of having its sign reversed and is consequently an odd function of x. Therefore for all values of x between −π and π identically; but

f (x) + f (−x) 1 = b0 + b1 cos x + b2 cos 2x + b3 cos 3x + · · · 2 2 where

bm =

π 2 w f (x) + f (−x) cos mx.dx; π 2 0

f (x) − f (−x) = a1 sin x + a2 sin 2x + a3 sin 3x + · · · 2 where

am =

π 2 w f (x) − f (−x) sin mx.dx. π 2 0

and


48

bm and am can be simplified a little.

bm =

π 2 w f (x) + f (−x) cos mx.dx π 2 0

=

1 π

wπ

f (x) cos mx.dx +

0

wπ

f (−x) cos mx.dx ,

0

but if we replace x by −x, we get wπ

f (−x) cos mx.dx = −

0

−π w 0

and we have

bm =

w0

f (x) cos mx.dx =

f (x) cos mx.dx,

−π

π 1 w f (x) cos mx.dx. π −π

In the same way we can reduce the value of am to π 1 w f (x) sin mx.dx. π −π

Hence    f (x) = 1 b + b cos x + b cos 2x + b cos 3x + · · · 0 1 2 3 2   + a1 sin x + a2 sin 2x + a3 sin 3x + · · ·

(2)

where

bm =

π π 1 w 1 w f (x) cos mx.dx = f (α) cos mα.dα. π −π π −π

(3)

and

am =

π π 1 w 1 w f (x) sin mx.dx = f (α) sin mα.dα. π −π π −π

(4)

and this development holds for all values of x between −π and π. The second member of (2) is known as a Fourier’s Series. EXAMPLES. 1. Obtain the following developments, all of which are valid from x = −π to x = π:— 2 sinh π 1 1 1 1 1 (1) ex = − cos x + cos 2x − cos 3x + cos 4x + · · · π 2 2 5 10 17 2 sinh π 1 2 3 4 + sin x − sin 2x + sin 3x − sin 4x + · · · . π 2 5 10 17

DEVELOPMENT IN TRIGONOMETRIC SERIES. (2)

f (x) =

49

π 2 cos 3x cos 5x cos 7x + + + · · · − cos x + 4 π 32 52 72 sin x sin 2x sin 3x sin 4x + − + − + ··· , 1 2 3 4

where f (x) = 0 from x = −π to x = 0 and f (x) = x from x = 0 to x = π. 3π 2 1 1 1 1 (3) f (x) = − cos x + 2 cos 2x + 2 cos 3x + 2 cos 5x + 2 16 π 1 2 3 5 2 + 2 cos 6x + · · · 6 1 3π 3π 3π 1 + − 1 sin x − sin 2x + + 2 sin 3x π 2 4 6 3 3π 3π 1 − sin 4x + − 2 sin 5x − · · · , 8 10 5 where

f (x) = x from x = −π to x = 0,

and

f (x) = x −

2.

where

1 c0 cos β0 + c1 cos(x − β1 ) + c2 cos(2x − β2 ) + c3 cos(3x − β3 ) + · · · 2 1 am cm = (a2m + b2m ) 2 and βm = tan−1 . bm

Show that formula (2) Art. 30 can be written

f (x) = where

π π from x = to x = π. 2 2

π , 2


f (x) =

3.

f (x) = 0 from x = 0 to x =

1 c0 sin β0 + c1 sin(x + β1 ) + c2 sin(2x + β2 ) + c3 sin(3x + β3 ) + · · · 2 1 bm cm = (a2m + b2m ) 2 and βm = tan−1 . am

31. In developing a function of x into a Trigonometric series it is often inconvenient to be held within the narrow boundaries x = −π and x = π. 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 π z = x, c which is equal to −π when x = −c and to π when x = c. c f (x) = f z can be developed in terms of z by Art. 30 (2), (3), and (4). π


50

We have f

where

 c 1 z = b0 + b1 cos z + b2 cos 2z + b3 cos 3z + · · · π 2  + a1 sin z + a2 sin 2z + a3 sin 3z + · · · π 1 w c bm = f z cos mz.dz. π −π π

and

am =

π 1 w c f z sin mz.dz. π −π π

(1) (2)

(3)

and (1) holds good from z = −π to z = π. Replace z by its value in terms of x and (1) becomes  πx 2πx 3πx 1  b0 + b1 cos + b2 cos + b3 cos + · · · 2 c c c (4) πx 2πx 3πx  + a1 sin + a2 sin + a3 sin + · · · c c c 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. π c 1 w mπx π 1 w c f z cos mz.dz = f (x) cos dx bm = π −π π π −c c c f (x) =

or

bm =

c c 1w mπx 1w mπλ f (x) cos dx = f (λ) cos dλ. c −c c c −c c

(5)

In like manner we can transform (3) into am =

c c mπx mπλ 1w 1w f (x) sin f (λ) sin dx = dλ. c −c c c −c c

(6)

By treating in like fashion formulas (1) and (2) Art. 25 and formulas (4) and (2) Art. 27 we get f (x) = a1 sin

where

and where

am =

2πx 3πx πx + a2 sin + a3 sin + ··· c c c

mπx mπλ 2w 2w f (x) sin dx = f (λ) sin dλ. c c c c c

c

0

0

1 πx 2πx 3πx b0 + b1 cos + b2 cos + b3 cos + ··· 2 c c c c c 2w mπx 2w mπλ = f (x) cos dx = f (λ) cos dλ. c c c c

f (x) = bm

0

0

and (7) and (9) hold good from x = 0 to x = c.

(7)

(8)

(9) (10)


51

EXAMPLES. 1.

Obtain the following developments: 4 πx 1 3πx 1 5πx (1) 1 = sin + sin + sin + ··· π c 3 c 5 c

(2)

from x = 0 to x = c. 2c πx 1 2πx 1 3πx 1 4πx x= sin − sin + sin − sin + ··· π c 2 c 3 c 4 c from x = −c to x = c. πx 3πx 5πx 7πx c 4c 1 1 1 x = − 2 cos + 2 cos + 2 cos + 2 cos + ··· 2 π c 3 c 5 c 7 c

from x = −c to x = c. 2 2 π π 4 2πx 4 2c2 πx π 2 3πx − 3 sin − sin + − 2 sin (3) x2 = 3 π 1 1 c 2 c 3 3 c 2 2 π 4πx π 4 5πx − sin + − 3 sin + ··· 4 c 5 5 c from x = 0 to x = c. c2 4c2 πx 1 2πx 1 3πx 1 4πx x2 = − 2 cos − 2 cos + 2 cos − 2 cos + ··· 3 π c 2 c 3 c 4 c from x = −c to x = c. 1 + ec πx 2(1 − ec ) 2πx (4) ex = 2π 2 sin + 2 sin 2 c +π c c + 4π 2 c c 3(1 + e ) 3πx 4(1 − ec ) 4πx + 2 sin + sin + · · · , c + 9π 2 c c2 + 16π 2 c c 1e −1 ec + 1 πx ec − 1 2πx ex = 2c − 2 cos + 2 cos 2 2 2 c c +π c c + 4π 2 c c e +1 3πx − 2 cos + ··· c + 9π 2 c from x = 0 to x = c.

(5) f (x) =

4c 1 1 πx 3πx 5πx − + + · · · sin sin sin π2 c 32 c 52 c from x = 0 to x = c,

where f (x) = x from x = 0 to x =

c c and f (x) = c − x from x = to x = c. 2 2

DEVELOPMENT IN TRIGONOMETRIC SERIES. 2.


1 f (x) = c0 cos β0 + c1 cos 2

πx − β1 c

1

cm = (a2m + b2m ) 2

where 3.

52

2πx + c2 cos − β2 c 3πx + c3 cos − β3 + · · · c

and βm = tan−1

am . bm


1 f (x) = c0 sin β0 + c1 sin 2

πx + β1 c

1

cm = (a2m + b2m ) 2

where

+ c2 sin

2πx + β2 c 3πx + c3 sin + β3 + · · · c

and βm = tan−1

am . bm

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 b0 , b1 , b2 , · · · a1 , a2 , · · · by their values given in Art. 31 (5) and (6). c 1 1w f (x) = f (λ) dλ c 2 −c wc πλ πx 2πλ 2πx cos dλ + f (λ) cos cos dλ + · · · c c c c −c −c wc wc πλ πx 2πλ 2πx + f (λ) sin sin dλ + f (λ) sin sin dλ + · · · c c c c −c −c c 1w 1 πλ πx πλ πx = f (λ) dλ + cos cos + sin sin c −c 2 c c c c 2πλ 2πx 2πλ 2πx + cos cos + sin sin + ··· c c c c +

wc

f (λ) cos


53

c 1 1w π 2π f (λ) dλ + cos (λ − x) + cos f (x) = (λ − x) + · · · c −c 2 c c c π 1 w 2π f (λ) dλ 1 + cos (λ − x) + cos = (λ − x) + · · · 2c −c c c π 2π + cos − (λ − x) + cos − (λ − x) + · · · c c since cos(−φ) = cos φ. f (x) =

c 2π π π π 1 w (λ − x) + cos − (λ − x) f (λ) dλ · · · + cos − 2π −c c c c c π 0π π π cos (λ − x) + cos (λ − x) c c c c π 2π + cos (λ − x) + · · · c c +

(1)

As c is indefinitely increased the limiting value approached by the parenthesis in (1) is w∞ cos α(λ − x).dα. −∞

Hence the limiting form approached by (1) is ∞ w∞ 1 w f (x) = f (λ) dλ cos α(λ − x).dα, 2π −∞ −∞

(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 Fourier’s Integral, and since it is a limiting form of 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.3 (2) is sometimes given in a slightly different form. w∞ w0 w∞ Since cos α(λ − x).dα = cos α(λ − x).dα + cos α(λ − x).dα −∞

−∞

0

and w0

cos α(λ − x).dα =

−∞

cos(−α)(λ − x).d(−α) = −

∞

w∞ −∞ 3 See

w0

note on page 38.

cos α(λ − x).dα = 2

w0

cos α(λ − x).dα

∞

w∞ 0

cos α(λ − x).dα


54

and (2) may be written ∞ w∞ 1 w f (λ) dλ cos α(λ − x).dα. f (x) = π −∞

(3)

0

If f (x) is an even function or an odd function (3) can be still further simplified. Let f (x) = −f (−x). Since the limits of integration in (3) do not contain α or λ the integrations may be performed in whichever order we choose. That is w∞

f (λ) dλ

w∞

−∞

w∞

cos α(λ − x).dα =

0

0

dα

w∞

f (λ) cos α(λ − x).dλ.

−∞

Now w∞

f (λ) cos α(λ − x).dλ =

−∞

w0

f (λ) cos α(λ − x).dλ +

−∞

w0

0

f (λ) cos α(λ − x).dλ =

−∞

w∞ f (λ) cos α(λ − x).dλ.

w0

f (−λ) cos α(−λ − x).d(−λ)

∞

=−

w∞

f (λ) cos α(λ + x).dλ

0

and (3) becomes f (x) =

∞ w∞ 1w dα f (λ)[cos α(λ − x) − cos α(λ + x).dλ π 0

0

∞ w∞ 2w = dα f (λ) sin αλ sin αx.dλ π 0

or

f (x) =

0

∞ w∞ 2w f (λ)dλ sin αλ sin αx.dα. π 0

(4)

0

If f (x) = f (−x) (3) can be reduced in like manner to ∞ w∞ 2w f (x) = f (λ) dλ cos αλ cos αx.dα. π 0

(5)

0

Although (4) holds for all values of x only in case f (x) is an odd function, and (5) only in case f (x) is an even, function, both (4) and (5) hold for all positive values of x in the case of any function.

DEVELOPMENT IN TRIGONOMETRIC SERIES. EXAMPLE. (1)

Obtain formulas (4) and (5) directly from (7) and (9) Art. 31.

55

56

CHAPTER III. CONVERGENCE OF FOURIER’S SERIES.

33. The question of the 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 4 sin 3x sin 5x sin 7x sin x + + + + ··· , [v. (3) Art. 26(b).] π 3 5 7 and prove that for all values of x between zero and π its sum is absolutely equal to unity; that is, that the limit approached by the sum of n terms of the series wπ wπ wπ 2 sin x sin α.dα + sin 2x sin 2α.dα + sin 3x sin 3α.dα + · · · , π 0

0

0

as n is indefinitely increased, is 1, provided that x lies between zero and π. Let wπ wπ wπ 2 Sn = sin x sin α.dα + sin 2x sin 2α.dα + sin 3x sin 3α.dα + · · · π 0 0 0 wπ + sin nx sin nα.dα . (1) 0

Then Sn =

π 2w [sin α sin x + sin 2α sin 2x + sin 3α sin 3x + · · · + sin nα sin nx]dα π 0

π 1w = [ cos(α − x) − cos(α + x) + cos 2(α − x) − cos 2(α + x) + · · · π 0

+ cos n(α − x) − cos n(α + x)] dα π w 1 = [cos(α − x) + cos 2(α − x) + cos 3(α − x) + · · · + cos n(α − x)]dα π 0

π 1w − [cos(α + x) + cos 2(α + x) + cos 3(α + x) + · · · + cos n(α + x)]dα. π 0

CONVERGENCE OF FOURIER’S SERIES.

57

Therefore by Art. 20 (1) α−x π 1w 1 1 sin(2n + 1) 2 Sn = − + dα α−x π 2 2 sin 0 2 α+x π 1w 1 1 sin(2n + 1) 2 − dα. − + α+x π 2 2 sin 0 2 α − x α+x π π 1 w sin(2n + 1) 2 1 w sin(2n + 1) 2 Sn = dα − dα. α−x α+x 2π 2π sin sin 0 0 2 2 In the first integral substitute β for α+x . stitute β for 2 We get π

α−x , and in the second integral sub2

π

x

x

2−2 2+2 1 w sin(2n + 1)β 1 w sin(2n + 1)β Sn = dβ − dβ. π x sin β π x sin β

−2

(2)

2

It remains to find the limit approached by Sn as n is indefinitely increased. π

34.

w2 sin(2n + 1)β π dβ = . sin β 2

(1)

0

For sin(2n + 1)β = 2 sin β

1 2

+ cos 2β + cos 4β + · · · + cos 2nβ,

by Art. 20.

π

and

w2

cos 2kβ.dβ = 0.

0

Let us construct the curve y=

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.


58

Each arch has for its altitude unity and for its base π and is symmetrical with respect to the ordinate 2n + 1 of its highest or lowest point. sin(2n + 1)x If now we form the curve y = from sin x the curve y = sin(2n + 1)x, it is clear that, since sin x π increases as x increases from 0 to , the ordinate of 2 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 = sin(2n + 1)x will be less than that of the arch before sin x it. If a0 , a1 , a2 , · · · an−1 are the areas of the successive arches and an that of the incomplete arch terminated π by the ordinate corresponding to x = 2 π

w2 sin(2n + 1)x dx = a0 − a1 + a2 − a3 + · · · . sin x 0

But π

π

w2 sin(2n + 1)x w2 sin(2n + 1)β π dx = dβ = sin x sin β 2 0

by (1).

0

Hence π = a0 − a1 + a2 − a3 + a4 − · · · + an 2

if n is even,

or π = a0 − a1 + a2 − a3 + a4 − · · · − an 2

if n is odd.

These equations can be written π = a0 + (−a1 + a2 ) + (−a3 + a4 ) 2 + (−a5 + a6 ) + · · · + (−an−1 + an ) if n is even, and π = a0 + (−a1 + a2 ) + (−a3 + a4 ) 2 + (−a5 + a6 ) + · · · + (−an−2 + an−1 ) + (−an ) if n is odd.


59

In either case each parenthesis is a negative quantity since a0 > a1 > a2 > a3 · · · > an , π and it follows that a0 is greater than . 2 Again π = a0 − a1 + (a2 − a3 ) + (a4 − a5 ) + · · · + (an−2 − an−1 ) + an 2 if n is even and π = a0 − a1 + (a2 − a3 ) + (a4 − a5 ) + · · · + (an−1 − an ) 2 if n is odd. In either case each parenthesis is positive and it follows that a0 − a1 is is less π than . 2 Since π a0 > > a0 − a1 , 2 π by less than they differ from each other, that is, a0 and a0 − a1 differ from 2 by less than a1 . π In like manner we can show that a0 − a1 and a0 − a1 + a2 differ from by 2 π less than a2 ; and in general that a0 − a1 + a2 − a3 + · · · ± ak differs from by 2 less than ak ; or even that ak a0 − a1 + a2 − a3 + · · · ± p π by less than ak no matter the value of p, provided p is greater differs from 2 than unity. 35.

From what has been proved in the last article it follows that wb sin(2n + 1)x dx, sin x 0

π π π and , differs from by less than the 2n + 1 2 2 sin(2n + 1)x area of the arch in which the ordinate of y = corresponding to sin x 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. π The area of the arch in question is less than , its base, multiplied by 2n + 1 1 π , a value greater than the length of its longest ordinate. sin b − 2n + 1 where b is some value between


60

wb sin(2n + 1)x dx sin x 0 π π 1 . differs from by less than 2 2n + 1 sin b − π 2n + 1 π 1 approaches zero If now n is indefinitely increased 2n + 1 sin b − π 2n + 1 as its limit, and we get the very important result Therefore

wb sin(2n + 1)x π limit dx = n=∞ sin x 2

(1)

0

if 0 < b
1.

1 1−x 1+x + tan−1 tan−1 . π y y

1 1 1 log[(1 − z)i] = log[(1 − x − yi)i] = log[y + (1 − x)i] π π π 1 i 1−x = log[(1 − x)2 + y 2 ] + tan−1 , 2π π y

Now

and −

1 1 1 log[(−1 − z)i] = − log[(−1 − x − yi)i] = − log[y − (1 + x)i] π π π 1 i 1+x 2 2 =− log[(1 + x) + y ] + tan−1 . 2π π y

(1)

SOLUTION OF PROBLEMS IN PHYSICS.

75

i 1 1−z 1 (1 − x)2 + y 2 −1 1 + x −1 1 − x + log = log tan + tan . π −1 − z 2π (1 + x)2 + y 2 π y y Hence 1 π

1+x 1−x tan−1 + tan−1 y y

and

1 (1 − x)2 + y 2 log 2π (1 + x)2 + y 2

are conjugate functions: 1 and 1 1+x 1−x tan−1 + tan−1 =a π y y

(2)

is any equipotential line, and (1 − x)2 + y 2 1 log =b 2π (1 + x)2 + y 2

(3)

any line of flow for the system described at the beginning of this article; and V1 =

(1 − x)2 + y 2 1 log 2π (1 + x)2 + y 2

(4)

is the solution of a new problem for which (3) represents any equipotential line and (2) any line of flow. 1 The

function conjugate to » – 1 1+x 1−x tan−1 + tan−1 π y y

might have been found as follows. If φ is the required function and ψ the given function we have by Int. Cal. Arts. 211, 212, and 213 the relations Dx φ = Dy ψ Here and

and

Dy φ = −Dx ψ.

1 π

»

Dy ψ = −

1−x 1+x + (1 + x)2 + y 2 (1 − x)2 + y 2

–

1 π

»

−Dx ψ = −

– y y − . (1 + x)2 + y 2 (1 − x)2 + y 2

If now we integrate Dy ψ with respect to x treating y as a constant and add an arbitrary function of y we shall have φ. So that  ff 1 φ=− log[(1 + x)2 + y 2 ] − log[(1 − x)2 + y 2 ] + f (y). 2π » – 1 y y df (y) Dy φ = − − + π (1 + x)2 + y 2 (1 − x)2 + y 2 dy Comparing this with its equal −Dx ψ above we find therefore

df (y) = 0 and f (y) = C a constant dy

1 (1 − x)2 + y 2 log + C, 2π (1 + x)2 + y 2

where C may be taken at pleasure, is our required conjugate function.


76

(2) reduces to x2 or

and (3) to

or

or

2y = tan aπ + y2 − 1

x2 + (y − ctn aπ)2 = csc2 aπ;

x2 + y 2 + 2

(5)

e2bπ + 1 x+1=0 e2bπ − 1

2 2 bπ ebπ + e−bπ e + e−bπ 2 + y −1 x + bπ = e − e−bπ ebπ − e−bπ (x + ctnh bπ)2 + y 2 = csch2 bπ.

(6)

(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.

EXAMPLES. 1. Show that if f (x) = a1 when x < −b, f (x) = a2 when −b < x < b, f (x) = a3 when x > b, a1 + a3 1 b+x b−x V = + (a2 − a1 ) tan−1 + (a2 − a3 ) tan−1 . 2 π y y


77

2. Show that if f (x) = 0 if x < 0, f (x) = a1 if 0 < x < b1 , f (x) = a2 if b1 < x < b2 , f (x) = a3 if b2 < x < b3 , &c., 1 x b1 − x b2 − x V = a1 tan−1 + (a1 − a2 ) tan−1 + (a2 − a3 ) tan−1 π y y y −1 b3 − x + (a3 − a4 ) tan + ··· . y 3. Show that if f (x) = −1 if x < −1, f (x) = x if −1 < x < 1, f (x) = 1 if x > 1, 1 y (1 − x)2 + y 2 −1 1 + x −1 1 − x . V = (1 + x) tan − (1 − x) tan + log π y y 2 (1 + x)2 + y 2 4. Show that if f (x) = −1 if x < −1, f (x) = 0 if −1 < x < 1, f (x) = 1 if x > 1, 1 −1 1 + x −1 1 − x V = tan − tan . π y y 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. 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, 1 π −1 a − x −1 b − x −1 x V = − tan − tan − tan . π 2 y y y The conjugate function V =

x2 + y 2 1 log 2 2π [(a − x) + 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 x2 + y 2 = ab.


78

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 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 ∞ w∞ 2w dα e−αy f (λ) sin αx sin αλ.dλ. π

(1)

∞ y y 1w f (λ)dλ 2 − . π y + (λ − x)2 y 2 + (λ + x)2

(2)

V =

0

0

v. (9) Art. 44. This reduces to V =

0

v. Ex. 3 Art. 45. EXAMPLES. 1.

If V = 0 when y = 0 and V = F (y) when x = 0 show that ∞ w∞ 2w dα e−αx F (λ) sin αy sin αλ.dλ π 0 0 ∞ w x x 1 F (λ)dλ 2 − 2 . = π x + (λ − y)2 x + (λ + y)2

V =

0

2.

If V = f (x) when y = 0 and V = F (y) when x = 0 show that V =

3.

∞ 1w y y f (λ) − π y 2 + (λ − x)2 y 2 + (λ + x)2 0 x x + F (λ) − 2 dλ. x2 + (λ − y)2 x + (λ + y)2

If F (y) = b the result of Ex. 2 reduces to V =

∞ 2b y 1w y y tan−1 + f (λ)dλ 2 − . π x π y + (λ − x)2 y 2 + (λ + x)2 0

4. 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 1−x 1+x y 1 tan−1 − tan−1 + 2 tan−1 V = π y y x 1 − y 1 + y x + tan−1 − tan−1 + 2 tan−1 . x x y


79

5. If one edge of the conducting sheet treated in Art. 47 is insulated, so that Dx V = 0 if x = 0 and V = f (x) when y = 0 ∞ w∞ 2w V = dα e−αy f (λ) cos αx cos αλ.dλ π 0 0 ∞ w y y 1 + 2 . f (λ)dλ 2 = π y + (λ + x)2 y + (λ − x)2 0

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 = ∞ V = eαy sin ax and V = eαy cos ax are available as particular solutions of the equation Dx2 V + Dy2 V = 0

(1)

as well as V = e−αy sin αx and V = e−αy cos αx. eαy + e−αy sin αx = cosh αy sin αx 2 eαy − e−αy sin αx = sinh αy sin αx 2

Consequently and and

cosh αy cos αx and

[Int. Cal. Art. 43 (2)] [Int. Cal. Art. 43 (1)]

sinh αy cos αx

are now available values of V and can be used precisely as e−αy cos αx and e−αy sin αx are used in Art. 44. Following the same course as in Art. 44 we get ∞ w∞ sinh αy 1w V = dα F (λ) cos α(λ − x).dλ π sinh αb −∞

(2)

0

as a solution of (1) which will reduce to V = F (x) when y = b and to

V =0

when y = 0, since sinh 0 =

1−1 = 0, 2

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 =

∞ w∞ sinh α(b − y) 1w dα f (λ) cos α(λ − x).dλ. π sinh αb −∞ 0

(3)


80

If V = f (x) when y = 0 and V = F (x) when y = b then V =

∞ w∞ sinh α(b − y) 1w dα f (λ) cos α(λ − x).dλ π sinh αb −∞ 0

∞ w∞ sinh αy 1w + dα F (λ) cos α(λ − x).dλ. π sinh αb −∞ 0

This can be considerably simplified by the aid of the formula pπ sin π q cos rx.dx = sinh qx 2q cos pπ + cosh rπ q q

w∞ sinh px 0

if p2 < q 2 . [Bierens de Haan, Tables of Def. Int. (7) 265] and becomes w∞ 1 π sin (b − y) f (λ) 2b b −∞

dλ π π(b − y) + cosh (λ − x) cos b b ∞ 1 dλ πy w or + F (λ) sin π πy 2b b −∞ + cosh (λ − x) cos b b ∞ 1 πy w fλ Fλ V = sin + π πy dλ. (5) 2b b −∞ cosh π (λ − x) − cos πy cosh (λ − x) + cos b b b b V =

EXAMPLES. 1.

Given the formula r w 2 b−a x dx −1 √ = tanh tan a + b cosh x b+a 2 b2 − a2

show that if V = 1 when y = 0 and V = 0 when y = b V =

if b > a, 1 (b − y). b

2. 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 πx # " tanh 2 2b V = tan−1 πy π tan 2b The solution for the conjugate system, that is, for a strip having a source at (0, 0) and an infinitely distant sink is 1 πx πy V = − log cosh2 − cos2 . π 2b 2b


81

3. 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, 2 π πx 2 π πx −1 −1 V = tan tan (b − y) tanh + tan tan y tanh π 2b 2b π 2b 2b πx # " sinh 2 b = tan−1 πy . π sin b 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 πy # πx " + cos cosh 1 b b V = log πy . πx π − cos cosh b b 4. If V = 0 when x = 0, V = f (x) when y = 0 and x > 0, and V = 0 when y = b and x > 0. ∞ w∞ sinh α(b − y) 1w dα [cos α(λ − x) − cos α(λ + x)]f (λ)dλ π sinh αb 0 0 ∞ 1 1 πy w 1 = sin π π πy − πy f (λ)dλ 2b b cosh (λ + x) − cos 0 cosh (λ − x) − cos b b b b

V =

for positive values of x and for values of y between 0 and b. 5. If V1 = 0 when x = 0, V1 = F (x) when y = b and x > 0, and V1 = 0 when y = 0 and x > 0 ∞ 1 1 πy w 1 V1 = sin π πy − π πy F (λ)dλ 2b b cosh (λ + x) − cos 0 cosh (λ − x) + cos b b b b for positive values of x and values of y between 0 and b. 6. If V2 = 0 when x = 0, V2 = f (x) when y = 0 and x > 0, and V2 = F (x) when y = b and x > 0 V2 = V + V1

for x > 0

and

0 < y < b. (v. Exs. 4 and 5)

7. If one edge of the strip described in Art. 48 is insulated so that we have V = f (x) when y = 0 and Dy V = 0 when y = b show that V =

∞ w∞ cosh α(b − y) 1w dα f (λ) cos α(λ − x).dλ. π cosh αb −∞ 0


82

By the aid of the formula rπ pπ cosh cos π 2q 2q cos rx.dx = pπ rπ cosh qx q cos + cosh q q

w∞ cosh px 0

if p < q,

[Bierens de Haan, Def. Int. Tables (6) 265], reduce this to π ∞ f (λ) cosh (λ − x) πy w 1 2b V = sin dλ. b 2b −∞ cosh π (λ − x) − cos πy b b 8. If V = 0 when y = 0 or b and x < −a, V = 1 when y = 0 or b and −a < x < a, and V = 0 when y = 0 or b and x > a π(a − x) π(a + x) # " sinh sinh 1 −1 −1 b b tan + tan . V = πy πy π sin sin b b 9. If V = 0 when y = 0 or b and x < −a, V = 1 when y = 0 and −a < x < a, V = 0 when y = 0 or b and x > a, and V = −1 when y = b and −a < x < a π(a + x) # π(a − x) " tanh tanh 1 −1 −1 b b V = + tan . tan πy πy π tan tan b b 10. A system conjugate to that of Ex. 9 is V = +∞ when y = 0 or b and x = −a, V = −∞ when y = 0 or b and x = a. In this case πy π(a − x) sin2 + sinh2 1 b b V = log . 2π 2 π(a + x) 2 πy + sinh sin b b 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 Dt u = a2 Dx2 u

(1)

[v. Art. 1 (II)] subject to the condition u = f (x)

when

t = 0.

(2)


83

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 = eβt+αx and substitute in (1). We get β = a2 α2 as the only relation which need hold between β and α. u = eαx+a

Hence

2

α2 t

2

= ea

α2 t αx

e

(3)

is a solution of (1) no matter what value is given to α. To get a trigonometric form replace α by αi. u = e−a

Then

2

α2 t αxi

e

.

If in (3) we replace α by −αi we get 2

u = e−a

α2 t −αxi

e

.

As in Arts. 7 and 8 we get from these values 2

u = e−a

α2 t

2

sin αx and u = e−a

α2 t

cos αx

as particular solutions of (1), α 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). We have

∞ w∞ 1w dα f (λ) cos α(λ − x).dλ π −∞

f (x) =

(4)

0

v. Art. 32 (3), and by proceeding as in Art. 44 we get ∞ w∞ 2 2 1w e−a α t f (λ) cos α(λ − x).dλ dα u= π −∞

(5)

0

as our required value of u. This can be considerably simplified. Changing the order of integration ∞ w∞ 2 2 1 w f (λ)dλ e−a α t cos α(λ − x).dα. π −∞ 0 r ∞ w 2 1 π − (λ−x) −a2 α2 t e cos α(λ − x).dα = .e 4a2 t 2a t

u=

(6)

(7)

0

by the formula w∞

√

0

Hence

2

e−a

u=

x2

cos bx.dx =

π − b22 e 4a 2a

w∞ (λ−x)2 1 √ f (λ)e− 4a2 t dλ. 2a πt −∞

[Int. Cal. Art. 94 (2)] (8)


84

λ−x √ , 2a t √ λ = x + 2a t.β ∞ √ 2 1 w f (x + 2a t.β)e−β dβ. u= √ π −∞

Let now

β=

then and

(9)

EXAMPLES. 1. Let the solid be of infinite extent and let the temperature be equal to a constant c at the time t = 0. ∞ ∞ 2c w −β 2 c w −β 2 e dβ = √ e dβ = c. u= √ π −∞ π

Then

0

v. Int. Cal. Art. 92 (2). 2.

Let u = x when t = 0. ∞ √ 2 1 w (x + 2a t.β)e−β dβ = x. u= √ π −∞

Then 3.

Let u = x2 when t = 0. u = x2 + 2a2 t.

Then 4.

Let u = 0 if x < −b, u = 1 if −b < x < b, and u = 0 if x > b, when t = 0.

Then b−x √ 2a t 2 b3 + 3bx2 b5 + 10b3 x2 + 5bx4 1 w −β 2 b √ − √ √ e dβ = √ + − · · · . u= √ π b+x π 2a t 3(2a t)3 5.2!(2a t)5

− 2a√t

5. Let u = 0 if x < 0 and u = 1 if x > 0 when t = 0. Then x√

x√

2aw t ∞ 2a t w∞ 2 2 1 w −β 2 1 1 w −β 2 1 u= √ e dβ = √ e−β dβ + e−β dβ = √ e dβ + 2 π x√ π π − 2a

0

t

0

0

x x3 x5 x7 1 1 √ − √ √ √ + − + · · · . +√ 2 π 2a t 3.(2a t)3 5.2!(2a t)5 7.3!(2a t)7

=

6. 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


85

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 a2 = 0.185 in C.G.S. units. Ans., 21◦ .6. 7. 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. Ans., 70◦ .8, 29◦ .2. 8. Find a particular solution of Dt u = a2 Dx2 u 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. 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 co¨ ordinates in the plane face. Then we have as before the equation Dt u = a2 Dx2 u, (1) but our conditions are u=0

when

u = f (x)

“

x=0

(2)

t=0

(3)

and we are concerned only with positive values of x. We may then use the form (4) Art. 32 f (x) =

∞ w∞ 2w dα f (λ) sin αx sin αλ.dλ, π 0

(4)

0

and proceeding as in the last section we get u=

∞ w∞ 2 2 2w dα e−a α t f (λ) sin αx sin αλ.dλ π 0

(5)

0

as our required solution. This may be reduced considerably. u=

∞ w∞ 2 2 1w f (λ)dλ e−a α t [cos α(λ − x) − cos α(λ + x)]dα, π 0

0

(λ−x)2 (λ+x)2 1 w √ f (λ)(e− 4a2 t − e− 4a2 t )dλ 2a πt 0

∞

or u =

by (7) Art. 49, and this may be reduced to the form w∞ w∞ √ √ 1 −β 2 −β 2 u= √ e f (x + 2a t.β)dβ − e f (−x + 2a t.β)dβ . π x√ x√ − 2a

t

2a

t

(6)

(7)


86

EXAMPLES. 1. Let the initial temperature be constant and equal to c. Then w∞ w∞ 2 2 c u= √ e−β dβ − e−β dβ π x√ x√ − 2a

t

2a

t

x√ 2a t

2c w −β 2 e dβ =√ π 0 2c x x3 x5 x7 √ − √ √ √ =√ + − + · · · . π 2a t 3.(2a t)3 5.2!(2a t)5 7.3!.(2a t)7 2. 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 a2 (400) the foot, the Fahrenheit degree, and the year being taken as units. (Thomson and Tait’s Nat. Phil. Vol. II. Appendix.) 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.

Let the initial temperature be constant and equal to −b, then by Ex. 1 x√

2a t 2b w −β 2 u = −√ e dβ. π

0

4. 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 x√ 2a t 2 w −β 2 u=b 1− √ e dβ . π

0

5. Let u = b when x = 0 and u = f (x) when t = 0. Then x√

∞ 2a t (λ−x)2 (λ+x)2 2 w −β 2 1 w u=b 1− √ e dβ + √ f (λ)[e− 4a2 t − e− 4a2 t ]dλ π 2a πt 0 0


87

by (6) Art. 50. 6.

Let u = b when x = 0 and u = c when t = 0. x√

Then

2a t 2 w −β 2 u = b + (c − b) √ e dβ. π

0

7. 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 a2 = 400. 8. Show that if the plane face of the solid considered in Art. 50 instead of being kept at temperature zero is impervious to heat ∞ (λ+x)2 (λ−x)2 1 w v. (6) Art. 50. u= √ f (λ)(e− 4a2 t + e− 4a2 t )dλ. 2a πt 0 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 Dt u = a2 Dx2 u

(1)

subject to the conditions u = F (t)

when

x=0

“

t = 0.

u=0 We know that

) (2)

x√

2a t 2 w −β 2 u= √ e dβ π

0

is a solution of (1), v. Ex. 1 Art. 50. It is easily shown that x √

2a t−c 2 w −β 2 √ e dβ, u= π

0

where c is any constant, is a solution of (1). For 2 x2 3 − 2 x 1 x − 4a2x(t−c) Dt u = − √ = − √ (t − c)− 2 e 4a2 (t−c) 3 e π 2a 2(t − c) 2 2a π 2 x 2 1 − √ Dx u = √ e 4a2 (t−c) π 2a t − c x2 x2 3 − 2 1 2x x − √ Dx2 u = − √ e 4a2 (t−c) = − 3 √ (t − c)− 2 e 4a2 (t−c) 2 π 2a t − c 4a (t − c) 2a π

(3)


88

Dt u = a2 Dx2 u.

and

Let φ(x, t) be a function of x and t which shall be equal to zero if t is negative and shall be equal to x√ 2a t 2 w −β 2 1− √ e dβ π 0

if t is equal to or greater than zero; so that if x = 0 φ(x, t) = 1 and if t = 0 φ(x, t) = 0. We shall now attack the following problem, to solve equation (1) subject to the conditions u=0

if t = 0

u = F (0) “

x=0

and

u = F (kτ ) “

x=0

“

0 t. If, then, nτ is the greatest whole multiple of τ not exceeding t, u=

k=n X

F (kτ )[φ(x, t − kτ ) − φ(x, t − (k + 1)τ )].

(6)

k=0

If now we decrease τ 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=

k=n X k=0

F (kτ )

φ(x, t − kτ ) − φ(x, t − (k + 1)τ ) τ τ

(7)


89

and if τ is indefinitely decreased the limiting form of (7) is u=−

wt

F (λ)Dλ φ(x, t − λ)dλ.

(8)

0

Since t − λ is positive between the limits of integration √x

2a t−λ 2 2 w e−β dβ, φ(x, t − λ) = 1 − √ π

0

x2 3 x − Dλ φ(x, t − λ) = − √ e 4a2 (t−λ) (t − λ)− 2 ; 2a π

and

and (8) may be written u=

t x2 3 x w − √ F (λ)e 4a2 (t−λ) (t − λ)− 2 dλ, 2a π

(9)

0

or if we let

x √ 2a t − λ ∞ w −β 2 e F t−

β= 2 u= √ π

x√ 2a t

x2 4a2 β 2

dβ.

(10)

EXAMPLES. 1.

If u = nt when x = 0 and u = 0 when t = 0 x2 u=n t+ 2 2a

x√ √ 2a t x2 nx t 2 w −β 2 e dβ − √ e− 4a2 t . 1− √ π a π

0

2. 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; a2 = .185. Ans., 20◦ .1. 3.

If u = F (t) when x = 0 and u = f (x) when t = 0, then

∞ ∞ 2 (λ+x)2 2 w −β 2 x2 1 w − (λ−x) u= √ e F t − 2 2 dβ + √ (e 4a2 t − e− 4a2 t )f (λ)d(λ). 4a β π x√ 2a πt 0 2a

t

v. (6) Art. 50.


90

4. 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

or

F (t) =

m=∞ X 1 b0 + [am sin mαt + bm cos mαt], 2 m=1

F (t) =

m=∞ X 1 + ρm sin(mαt + λm ), 2b0 m=1

where

ρm cos λm = am

where

and ρm sin λm = bm .

α=

2π , T

v. Art. 31 Ex. 3.

Show that with this value of F (t) (10) Art. 51. becomes m=∞ w∞ w∞ 2 2 X mαx2 1 −β 2 e e−β cos 2 2 dβ dβ + √ ρm sin(mαt + λm ) u = √ b0 4a β π π m=1 x√ x√ 2a

t

2a

t

w∞

− cos(mαt + λm )

2

e−β sin

x√ 2a t

mαx2 dβ 4a2 β 2

and that as t increases u approaches the value u=

m=∞ √ mα X x x 1 b0 + ρm e− a 2 sin(mαt − 2 a m=1

r

mα + λm ). 2

Given that w∞ 0

−x2

e

√ √ π −b√2 b2 e sin b 2; sin 2 dx = x 2

w∞ 0

e

−x2

√ √ π −b√2 b2 e cos b 2. cos 2 dx = x 2

v. Riemann, Lin. par. dif. gl. § 54. 5. 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 Dt u = a2 Dx2 u − b2 u.

v. Fourier, Heat § 105.

Show that for this case 2

u = e−(b

+a2 α2 )t

sin αx

and

2

u = e−(b

+a2 α2 )t

cos αx

are particular solutions, and that if u = f (x) when t = 0 u=

2 2 ∞ ∞ 2 √ e−b t w −β 2 e−b t w − (λ−x) √ e f (x + 2a t.β)dβ. e 4a2 t f (λ)dλ = √ π −∞ 2a πt −∞

cf. (8) and (9) Art. 49.


91

If u = 0 when x = 0 and u = f (x) when t = 0 2

e−b t u= √ π

w∞

−β 2

e

√

f (x + 2a t.β)dβ −

− 2ax√t

w∞

e

−β 2

√ f (−x + 2a t.β)dβ .

x√ 2a t

cf. (7) Art. 50. bx If u = −e− a when t = 0 and u = 0 when x = 0 w∞ w∞ √ √ 2 2 bx bx 1 u= √ ea e−(b t+β) dβ − e− a e−(b t+β) dβ , π x√ x√ 2a

− 2a

t

t

bx

and if u = 1 when x = 0 and u = 0 when t = 0 we have only to add e− a to the bx second member of the last equation, since u = e− a satisfies the equation Dt u = a2 Dx2 u − b2 u. If u = F (t) when x = 0 and u = 0 when t = 0 we can employ the method of Art. 51. w∞ w∞ √ √ bx 1 − bx −(b t−λ+β)2 −(b t−λ+β)2 − bx a a a φ(x, t − λ) = e e e +√ e dβ − e dβ , π √x √x 2a

− 2a

t−λ

− 23

x(t − λ) √ −Dλ φ(x, t − λ) = 2a π

and

u=

e

−b

2

t−λ

2 (t−λ)− 4a2x(t−λ)

;

t x2 3 −b2 (t−λ)− x w 4a2 (t−λ) F (λ)dλ, √ (t − λ)− 2 e 2a π 0

cf. (9) Art. 51, ∞ b2 x2 2 w −β 2 − 4a x2 2 β2 √ u= e F t − 2 2 dβ, 4a β π x√

or

2a

t

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 √ m=∞ √ X x 2 1 − bx − x2a2 p a + sin mαt − q + λm , u = b0 e ρm e 2 2a m=1 where

p = (b2 +

p 1 b4 + m2 α2 ) 2

and q = (−b2 +

p 1 b4 + m2 α2 ) 2 .

SOLUTION OF PROBLEMS IN PHYSICS. w∞

92

√

π −2a e 2 0 √ w∞ 2 a2 b2 π −2c e−x − x2 sin 2 dx = e sin 2d x 2 0 √ w∞ a2 π −2c b2 −x2 − x 2 e cos 2 dx = e cos 2d, x 2

Given

and

2

e

a −x2 − x 2

dx =

0

where √

√

1 2 2 p 4 c= (a + a + b4 ) 2 2

and d =

p 1 2 (−a2 + a4 + b4 ) 2 . 2

¨ Angstrom’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. 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 = eβt+αx is a solution of Dt u = a2 Dx2 u

(1)

provided only that β = a2 α2 we have x

u = eβt± a

√ β

as a solution. Replacing β by ±βi this becomes x

u = e±βti± a

√ √ β ±i

√β x u = e±βti± a 2 (1±i) √ 1√ i=± 2(1 + i) 2 √ 1√ −i = ± 2(1 − i). 2

or since and Hence √β

r x β u=e sin βt − , a 2 r √β x x β u = e a 2 sin βt + , a 2 −x a

2

√β

r x β u=e cos βt − , a 2 r √β x x β u = e a 2 cos βt + , a 2 −x a

2

(2) (3)


93

are particular solutions of (1). From these we get readily −x a

u = ρm e

√ mα 2

r x mα + λm sin mαt − a 2

(4)

as a solution. (4) reduces to

and to

u = ρm sin(mαt + λm ) when x = 0 r √ mα x mα −x a 2 u = ρm e sin λm − when t = 0. a 2

If we add a term which and which is equal to zero when x = 0 r (1) satisfies √ mα x mα −x when t = 0 (v. Art. 50) we shall have and to −ρm e a 2 sin λm − a 2 a solution of (1) which is zero when t = 0 and which is ρm sin(mαt + λm )

when 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 = eβt+αx is a solution of Dt u = a2 Dx2 u − b2 u if β = a2 α2 − b2 , and hence that √2 √2 x x x (p±qi) ±βti± a√ 2 u = eβt± a b +β , u = e±βti± a b ±βi , u = e , qx qx √ √ ± px ± px u = e a 2 sin βt ± √ , and u = e a 2 cos βt ± √ , a 2 a 2 where

p 1 p = [ β 2 + b4 + b2 ] 2

p 1 and q = [ β 2 + b4 − b2 ] 2 ,

are solutions. Hence u = ρm e

√ − apx 2

qx sin βt − √ + λm a 2

is a solution. If β = mα this last result reduces to u = ρm sin(mαt + λm ) 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.


94

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 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 quantity Q of heat is suddenly generated the point is called an 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 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 = λ 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 = λ to x = λ + ∆λ, where ∆λ is any arbitrary length, the temperature Q , and we shall have by (8) Art. 49 of that portion will be suddenly raised to ∆λ λ+∆λ (λ−x)2 Q 1 w u= √ e− 4a2 t dλ (1) 2a πt ∆λ λ

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 ∆λ is made to approach zero we get u=

(λ−x)2 Q √ e− 4a2 t 2a πt

(2)

as the solution for the case where we have an instantaneous source at the point x = λ. Q √ It is to be observed that in (2) u = 0 when t = 0 and u = when 2a πt x = λ 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 (λ)dλ 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 (λ)dλ in every element of the bar whose distance to the right of the origin is λ, and an instantaneous sink of strength f (λ)dλ in every element of the bar whose distance to the left of the origin is λ. If we have an instantaneous source at the origin (2) reduces to u=

x2 Q √ e− 4a2 t 2a πt

(3)


95

For a permanent source of constant strength Q at the origin (3) gives t 2 1 Q w − 4a2x(t−τ ) (t − τ )− 2 dτ √ u= e 2a π

(4)

0

and for a permanent source of variable strength f (t) t 2 1 1 w − 4a2x(t−τ ) (t − τ )− 2 f (τ )dτ. u= √ e 2a π

(5)

0

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 doublet. 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 doublet of strength P whose axis is tangent to the line of approach and points from sink to source. A 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 = η and an equal sink at x = −η we have u=

(η+x)2 (η−x)2 Q √ (e− 4a2 t − e− 4a2 t ), 2a πt

or if 2ηQ = P , (η 2 +x2 ) ηx ηx P √ e− 4a2 t (e 2a2 t − e− 2a2 t ) 4aη πt (η 2 +x2 ) P ηx √ e− 4a2 t sinh 2 . = 2a t 2aη πt

u=

If η is made to approach zero ηx x 1 sinh 2 = 2 , limit η 2a t 2a t

and

u=

x2 Px √ e− 4a2 t 3 πt

4a3

(1)

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 = λ we have 2 P (x − λ) − (x−λ) √ u= e 4a2 t . (2) 4a3 πt3


96

For a permanent doublet of constant strength P placed at the origin we have u=

t 2 3 P x w − 4a2x(t−τ ) (t − τ )− 2 dτ ; √ e 4a3 π

(3)

0

and for a permanent doublet of variable strength f (t) t 2 3 x w − 4a2x(t−τ ) (t − τ )− 2 f (τ )dτ, √ e 3 4a π 0 ∞ w 2 1 x2 u = 2√ e−β f t − 2 2 dβ 4a β a π x√

u= or

2a

(4) (5)

t

if x > 0, and

u=

−∞ w 1 x2 −β 2 √ e dβ f t − 4a2 β 2 a2 π x√ 2a

if x < 0, if we let β =

(6)

t

x √ . 2a t − τ

From (5) and (6) we see readily that u = 0 when t = 0 and that u =

f (t) 2a2

f (t) when x = 0 if we approach the origin from the right and that u = − 2 when 2a 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 2a2 b placed at the origin. To solve the problem treated in Art. 51 we have only to suppose a permanent doublet of strength 2a2 F (t) placed at x = 0 and from (5) we get at once (10) Art. 51. EXAMPLE. Show that if Dt u = a2 Dx2 u − b2 u and an instantaneous source of strength Q is placed at x = λ u=

(λ−x)2 2 Q √ e−b t− 4a2 t 2a πt

v. Art. 51, Ex. 5.

Show that if an instantaneous doublet of strength P is placed at the point x=0 2 x2 Px √ u= e−b t− 4a2 t . 4a3 πt3


97

If a permanent doublet of strength f (t) is placed at x = 0 t 2 3 x w −b2 (t−τ )− 4a2x(t−τ ) (t − τ )− 2 f (τ )dτ √ e 3 4a π 0 ±∞ w b2 x2 x2 1 −β 2 − 4a 2 β2 f t − 2 2 dβ, = 2√ e 4a β a π x√

u=

2a

t

f (t) when x = 0. 2a2 Hence if we place at x = 0 a permanent doublet of strength 2a2 F (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

whence u = 0 when t = 0 and x > 0 or x < 0 and u = ±

∞ b2 x2 2c w −β 2 − 4a 2 β2 u= √ dβ. e π x√ 2a

t

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 Dt2 y = a2 Dx2 y

(1)

[v. (VIII) Art. 1] subject to the conditions y = f (x) Dt y = 0

when t = 0 “

t = 0.

(2) (3)

As in Art. 8 we find y = cos α(x ± at)

and y = sin α(x ± at)

as particular solutions of (1). From these we must build up a value that will reduce to f (x) =

∞ w∞ 1w dα f (λ) cos α(λ − x).dλ π −∞ 0

(4)


98

when t = 0 and will at the same time satisfy (3). y = cos αλ cos α(x + at) + sin αλ sin α(x + at) y = cos α(λ − x − at)

or

is a solution of (1). ∞ w∞ 1w y= dα f (λ) cos α(λ − x − at).dλ π −∞

Hence

(5)

0

is also a solution of (1). (5) reduces to y = f (x) when t = 0 but it gives Dt y =

∞ w∞ aw αdα f (λ) sin α(λ − x).dλ π −∞ 0

when t = 0 and consequently does not satisfy equation (3). If in forming (5) we use cos α(x−at) and sin α(x−at) instead of cos α(x+at) and sin α(x + at) we get ∞ w∞ 1w dα f (λ) cos α(λ − x + at).dλ y= π −∞

(6)

0

which is a solution of (1), and reduces to y = f (x) when t = 0, but it gives Dt y = −

∞ w∞ aw αdα f (λ) sin α(λ − x).dλ π −∞ 0

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 " ∞ w∞ 1 1w f (λ) cos α(λ − x − at).dλ y= dα 2 π −∞ 0 # ∞ w∞ 1w + dα f (λ) cos α(λ − x + at).dλ , (7) π −∞ 0

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 ∞ w∞ 1w dα f (λ) cos α(λ − x − at).dλ π −∞ 0

∞ w∞ 1w = dα f (λ) cos α(λ − z).dλ = f (z) = f (x + at); π −∞ 0


99

and in like manner we can show that ∞ w∞ 1w dα f (λ) cos α(λ − x + at).dλ = f (x − at). π −∞ 0

Hence our solution becomes 1 (8) y = [f (x + at) + f (x − at)]. 2 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. 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 Dt2 y = a2 Dx2 y (1) subject to the conditions y=0

when t = 0

Dt y = F (x) “

t= 0.

(2) (3)

We get by the process used in Art. 55 ∞ w∞ sin α(λ − x + at) sin α(λ − x − at) 1 w dα F (λ) − dλ y= 2πa α α −∞ 0 ∞ w∞ sin α(λ − x + at) sin α(λ − x − at) 1 w F (λ)dλ − = dα; 2πa −∞ α α 0

but

w∞ sin α(λ − x + at)

dα −

w∞ sin α(λ − x − at)

dα = π α α 0 if x − at < λ < x + at, and is equal to zero for all other values of λ; since 0

w∞ sin mx 0

x

π 2 π =− 2 = 0

dx =

if m > 0 if m < 0 if

m = 0.

v. Int. Cal. Art. 92 (3). Hence is our required solution.

y=

x+at 1 w F (λ)dλ 2a x−at

(4)


100

EXAMPLES. 1. If the string is initially distorted and starts with initial velocity so that y = f (x) and Dt y = F (x) when t = 0 y=

x+at 1 w 1 F (λ)dλ. [f (x + at) + f (x − at)] + 2 2a x−at

2. 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 b c p 2 and a length equal to 4a + b2 . Of course a wave having a front of the 2a 2a b will be seen to run to the right along same length with a slope equal to − 2a the string, and the effect of the two waves will be to lift the string bodily and bc permanently to a distance above its original position. 2a 57. We shall now take up a few examples of the use of 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. Since

where we have

f (x) =

am =

2 π

m=∞ X

(am sin mx)

m=1 wπ

f (α) sin mα.dα

0

m=∞ wπ 2 X −my u= e sin mx f (α) sin mα.dα . π m=1

(1)

0

If the breadth of the plate is a instead of π a m=∞ 2 X − mπy mπλ mπx w a u= e sin f (λ) sin dλ . a m=1 a a

(2)

0

58. If the temperature of the base is unity and the breadth of the plate is π the solution is, as we have seen in Art. 7, 4 −y 1 −3y 1 −5y u= e sin x + e sin 3x + e sin 5x + · · · . (1) π 3 5 This series can be summed without difficulty. We have the development log(1 + z) =

z z2 z3 z4 − + − + ··· 1 2 3 4


101

if the modulus of z is less than 1. Int. Cal. Art. 221 (4). z z2 z3 z4 log(1 − z) = − − − − − ··· 1 2 3 4

Hence if mod. z < 1.

z z3 z5 1 [log(1 + z) − log(1 − z)] = + + + ··· 2 1 3 5

and

(2)

if mod. z < 1. But log(1 + z) = log[1 + r(cos φ + i sin φ)] 1 r sin φ = log[(1 + r cos φ)2 + (r sin φ)2 ] + i tan−1 2 1 + r cos φ 1 r sin φ 2 −1 = log (1 + 2r cos φ + r ) + i tan , 2 1 + r cos φ and log(1 − z) =

1 r sin φ log(1 − 2r cos φ + r2 ) − i tan−1 , 2 1 − r cos φ

[Int. Cal. Art. 33 (2)], and (2) becomes 1 + 2r cos φ + r2 1 1 −1 2r sin φ log + i tan 2 2 1 − 2r cos φ + r2 1 − r2 r(cos φ + i sin φ) r3 (cos 3φ + i sin 3φ) = + + ··· 1 3

(3)

From (3) we get two equations 1 + 2r cos φ + r2 r cos φ r3 cos 3φ r5 cos 5φ 1 log = + + + ··· 4 1 − 2r cos φ + r2 1 3 5 1 2r sin φ r sin φ r3 sin 3φ r5 sin 5φ tan−1 = + + + ··· 2 1 − r2 1 3 5

(4) (5)

both valid for all values of φ provided r < 1. e−y is less than 1 if y is positive. Hence from (5) e−y sin x e−3y sin 3x e−5y sin 5x 1 2e−y sin x + + + · · · = tan−1 1 3 5 2 1 − e−2y 1 2 sin x 1 sin x = tan−1 y = tan−1 , 2 e − e−y 2 sinh y


102

and (1) may be written u=

2 sin x tan−1 . π sinh y

(6)

If we replace r by e−y and φ by x in log[1 + r( cos φ + i sin φ)] it becomes

log[1 + e−y cos x + ie−y sin x]

or

log[1+ cos z + i sin z]

v. Int. Cal. Art. 35 (3) and (4) a function of z as a whole; and log[1 − r( cos φ + i sin φ)] log(1− cos z − i sin z);

becomes

hence by Int. Cal. Arts. 212 and 213, 1 + 2e−y cos x + e−2y 1 2e−y sin x 1 log and tan−1 −y −2y 4 1 − 2e cos x + e 2 1 − e−2y 1 cosh y + cos x 1 sin x or log and tan−1 4 cosh y − cos x 2 sinh y are conjugate functions, and u1 =

cosh y + cos x 1 log π cosh y − cos x

(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

or

2 sin x tan−1 =a π sinh y sin x aπ = tan sinh y 2

(8)

and the lines of flow by

or

1 cosh y + cos x log = b, π cosh y − cos x cosh y + cos x = eπb . cosh y − cos x

(9)


103

EXAMPLES. Dx2 u

1. If when x = a,

2. x=a

+

Dy2 u

= 0, and u = 1 when y = 0, and u = 0 when x = 0 and

πx 1 − 3πy 3πx 1 − 5πy 5πx 4 − πy a a a e sin + e sin + e sin + ··· u= π a 3 a 5 a πx sin 2 a = tan−1 πy . π sinh a If u = φ(x) when y = 0, u = f (y) when x = 0, and u = F (y) when

a m=∞ mπx w mπλ 2 X − mπy φ(λ) sin dλ e a sin a m=1 a a 0 ∞ 1 1 πx w 1 − + sin π πx π πx f (λ) dλ 2a a cosh (λ − y) − cos cosh (λ + y) − cos 0 a a a a ∞ 1 πx w 1 1 + sin π πx − π πx F (λ) dλ 2a a cosh (λ − y) + cos cosh (λ + y) + cos 0 a a a a v. Art. 48, Exs. 4, 5, and 6.

u=

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:— Dx2 V + Dy2 V = 0.

(1)

V =0

when

x = 0.

(2)

V =0

“

x = a.

(3)

V =0 “ V = f (x) “

y = b. y = 0.

(4) (5)

Working as in Art. 48 we get mπ (b − y) sinh mπx a sin mπb a sinh a as a value of V which satisfies equations (1), (2), (3), and (4) if m is an integer. Therefore mπ a m=∞ sinh (b − y) 2 X mπx w mπλ a V = sin f (λ) sin dλ (6) mπb a m=1 a a 0 sinh a


104

is our required solution. EXAMPLES. 1.

If f (x) = 1 Eq. (6) Art. 59 reduces to π 3π " 4 sinh a (b − y) πx 1 sinh a (b − y) 3πx V = sin + sin πb 3πb π a 3 a sinh sinh a a 5π # 5πx 1 sinh a (b − y) sin + + ··· 5πb 5 a sinh a

2. If V = 0 when x = 0, V = 0 when x = a, V = 0 when y = 0, and V = F (x) when y = b, then mπy " # a m=∞ mπx w mπλ 2 X sinh a sin F (λ) sin dλ . V = mπb a m=1 a a 0 sinh a 3.

If F (x) = 1 the answer of Ex. 2 reduces to

πy 3πy 5πy " # 4 sinh a πx 1 sinh a 3πx 1 sinh a 5πx V = sin + sin + sin + ··· . πb 3πb 5πb π a 3 a 5 a sinh sinh sinh a a a 4. 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 " m=∞ mπx 2 X sin V = a m=1 a

5.

mπ (b − y) wa mπλ a dλ f (λ) sin mπb a 0 sinh a mπy a !# w sinh mπλ a + F (λ) sin dλ . mπb a 0 sinh a sinh

If f (x) = F (x) the answer of Ex. 4 reduces to

V =

2 a

"

m=∞ X m=1

mπ b # wa −y mπλ mπx a 2 f (λ) sin sin dλ . mπb a a 0 cosh 2a

cosh

SOLUTION OF PROBLEMS IN PHYSICS. 6.

105

If f (x) = F (x) = 1 the answer of Ex. 5 reduces to πb 3π b " cosh − y cosh − y πx 1 3πx 4 a 2 a 2 sin sin V = + πb 3πb π a 3 a cosh cosh 2a 2a 5π b # 1 cosh a 2 − y 5πx + sin + ··· . 5πb 5 a cosh 2a

7. If V = f (x) when y = 0, V = F (x) when y = b, V = φ(y) when x = 0, and V = χ(y) when x = a, then mπ " a m=∞ mπx sinh a (b − y) w mπλ 2 X sin f (λ) sin dλ V = mπb a m=1 a a 0 sinh a mπy a !# w sinh mπλ a dλ + F (λ) sin mπb a 0 sinh a mπ " b m=∞ 2 X mπy sinh b (a − x) w mπλ + sin φ(λ) sin dλ mπa b m=1 b b sinh 0 b mπx b !# w sinh mπλ b + mπa χ(λ) sin b dλ . sinh b 0 8. If f (x) = φ(y) = 0 and F (x) = χ(y) = 1 the answer of Ex. 7 may be reduced to π a 2π a " − x − x sinh cosh 2πy 2 πy πy 1 b 2 b 2 sin V = − sin + πa 2πa π 2b b 2 b sinh cosh 2b 2b 3π a 4π a # sinh − x cosh − x 3πy 1 4πy 1 b 2 b 2 sin + sin − ··· . − 3πa 4πa 3 b 4 b sinh cosh 2b 2b 9. 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. Ans., (1) 25◦ ; (2) 50◦ ; (3) 50◦ .


106

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 Dt u = a2 Dx2 u

(1)

subject to the conditions u=0

when

x=0

(2)

u=0

“

x=c

(3)

u = f (x)

“

t = 0.

(4)

In Art. 49 we have found 2

u = e−a and

u=e

α2 t

−a2 α2 t

sin αx cos αx

as particular solutions of (1). 2 2 u = e−a α t sin αx satisfies (2) whatever value is given to α. It satisfies (3) if mπ provided m is an integer. Let us try to build a value of u out of terms α= c a2 m2 π 2 t mπx of the form Ae− c2 sin which shall satisfy (4). c We have f (x) =

c m=∞ 2 Xh mπx w mπλ i sin f (λ) sin dλ . c m=1 c c

(5)

0

c m=∞ 2 X h − m2 a22 π2 t mπλ i mπx w c e u= sin f (λ) sin dλ , c m=1 c c

(6)

0

reduces to (5) when t = 0 and is our required solution. EXAMPLES. 1.

If f (λ) = b, a constant, (6) Art. 60 reduces to i πx 1 − 9a22π2 t 3πx 1 − 25π22a2 t 5πx 4b h − a2 π22 t c e c sin + e c sin + e sin + ··· . u= π c 3 c 5 c 2. 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


107

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 a2 = 0.185 in C.G.S. units. Ans., 10◦ .3. 3. 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. Ans., 22◦ .8; 15◦ .1; 17◦ .2. 4. 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. Ans., 22◦ .8; 15◦ .6; 16◦ .7. 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 Dt u = a2 Dx2 u subject to the conditions Dx u = 0

when

x=0

Dx u = 0

“

x=c

“

t = 0.

u = f (x)

We have only to use the particular solution u = e−a as we used

u=e

2

α2 t

−a2 α2 t

cos αx sin αx

in Art. 60. We get u=

c c m=∞ X − m 2 a2 π 2 t mπx w mπλ 2 1w c2 f (λ)dλ + e cos f (λ) cos dλ . c 2 c c m=1 0

0

(1)


108

EXAMPLES. 1. 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. Ans., 33◦ .3; 33◦ .3. 2. Solve example 3 Art. 60 on the hypothesis just stated, getting in addition the temperatures of points on the outer surfaces. Ans., 50◦ ; 33◦ .9; 66◦ .1; 27◦ .2; 72◦ .8. 3. 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. Ans., 50◦ ; 39◦ .7; 60◦ .3; 35◦ .5; 64◦ .5. 4. Show that if u = 0 when x = 0, Dx u = 0 when x = c, and u = f (x) when t = 0, c m=∞ (2m+1)2 a2 π 2 t (2m + 1)πx w (2m + 1)πλ 2 X − 2 4c e sin f (λ) sin dλ . u= c m=0 2c 2c 0

Suggestion: Assume u = 0 when x = 2c and f (2c − x) = f (x), and see (6) Art. 60. 62. If the temperature of the right-hand face of the slab considered in Art. 60 is a constant γ instead of zero we have only to add to the second member of (6) Art. 60 a term u1 which shall satisfy the conditions Dt u1 = a2 Dx2 u1 u1 = 0

(1)

when x = 0

(2)

u1 = 0

“

t=0

(3)

u1 = γ

“

x = c.

(4)

γx obviously satisfies (1), (2), and (4); to make it satisfy (3) as well u1 = c we must add a term u2 which shall be equal to zero when x = 0 and when x = c γx and to − when t = 0, while always satisfying (1). It is given immediately by c (6) Art. 60 and is u2 = −

c m=∞ 2 2 2 mπx w mπλ 2γ X − m ac2 π t sin λ sin dλ . e c2 m=1 c c 0

wc 0

λ sin

mπλ c2 c2 dλ = − cos mπ = (−1)m+1 , c mπ mπ

(5)


and

109

m=∞ 2γ X (−1)m − m2 a22 π2 t mπx c u2 = sin e . π m=1 m c

(6)

m=∞ mπx x 2 X (−1)m − m2 a22 π2 t c sin u1 = γ + e . c π m=1 m c

(7)

Hence

If the left-hand face of the slab considered in Art. 60 is to be kept at a constant temperature β and the right-hand face at the temperature zero we can get the term u3 which must be added to the second member of (6) Art. 60 by replacing γ by β and x by c − x in (7). We then have m=∞ 2 X 1 − m2 a22 π2 t mπx c−x c − e . sin u3 = β c π m=1 m c

(8)

EXAMPLES. 1. t=0

Show that if u = β when x = 0, u = γ when x = c, and u = f (x) when m=∞ x 2 X (−1)m − m2 π22 a2 t mπx c u = β + (γ − β) + e sin c π m=1 m c c m=∞ 2 2 2 mπx w 2 X mπλ − m ac2 π t sin + e [f (λ) − β] sin dλ . c m=1 c c 0

2. Show that if u = β when x = 0, u = 0 when t = 0, and Dx u = 0 when x=c m=∞ (2m+1)2 a2 π 2 t 4 X 1 (2m + 1)πx 4c2 u=β 1− e− sin π m=0 2m + 1 2c 2 2 4 πx 1 − 9a2 π22 t 3πx 1 − 25a22π2 t 5πx − a 4cπ2 t sin =β 1− e + e 4c sin + e 4c sin + ··· . π 2c 3 2c 5 2c 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. Let φ(x, t) be a function of x and t which shall be zero if t is less than zero and shall be equal to m=∞ 2 X (−1)m − m2 a22 π2 t mπx x c + e sin c π m=1 m c


110

[v. (7) Art. 62] if t is equal to or greater than zero. So that φ(x, t) = 0

if

t a,

π 1 w f (φ) sin mφ.dφ π −π

Show that if V satisfies (1) Ex. 2 and V = f (r) when φ = 0 and r > 0 ∞ w∞ cosh α(π − φ) 1 w λ V = f (e )dλ cos α(λ − log r).dα π −∞ cosh απ 0

1 ∞ cosh (λ − log r) 1 φ w λ 2 = sin f (e ) dλ. π 2 −∞ cosh(λ − log r) − cos φ

MISCELLANEOUS PROBLEMS. 5.

136

If V = 1 when φ = 0 and 0 < r < 1, and V = 0 when φ = 0 and r > 1 1 V = π

6.

(

log r #) " " sinh 1 π π −1 2 = − tan − tan−1 φ 2 π 2 sin 2

r−1 √ φ 2 r. sin 2

!# .

If V = f (r) when φ = 0 and V = 0 when φ = β V =

∞ w∞ sinh(β − φ)α 1 w f (eλ )dλ cos α(λ − log r).dα π −∞ sinh βα 0

∞ f (eλ )dλ 1 πφ w , = sin 2β β −∞ cosh π (λ − log r) − cos π φ β β

if 0 < φ < β. 7.

If V = 0 when φ = 0 and V = F (r) when φ = β ∞ w∞ sinh φα 1 w λ V = F (e )dλ cos α(λ − log r).dα π −∞ sinh βα 0

∞ 1 πφ w F (eλ )dλ = sin . π 2β β −∞ cosh (λ − log r) + cos π φ β β

8. If V = χ(r) when φ = 0 and r < a, V = 0 when φ = β, and V = 0 when r=a " 0 πφ w 1 dλ sin χ(aeλ ) V = π r πφ 2β β −∞ cosh λ − log − cos β a β # dλ − . π r πφ cosh λ + log − cos β a β 9.

If V = 0 when r = 1, V = 1 when φ = 0, V = 0 when φ = V =

10.

π 2

1 − r2 2 tan−1 ctn φ . π 1 + r2

If V = 0 when r = 1, V = 1 when φ = 0, V = 1 when φ = V =

2 1 − r4 tan−1 . π 2r2 sin 2φ

π 2

MISCELLANEOUS PROBLEMS. 11.

137

If V = f (φ) when r = a, V = 0 when φ = 0, and V = 0 when φ = β V = V =

m=∞ X m=1 m=∞ X

am am

r mπ β a a mπ β

m=1

where

am

r

sin

mπφ β

if r < a

sin

mπφ β

if r > a

β 2w mπφ = f (φ) sin dφ β β

and

0 < φ < β.

0

12. If V = f (φ) when r = a, V = 0 when r = b, V = 0 when φ = 0, and V = 0 when φ = β, then if a < r < b and 0 < φ < β mπ mπ mπ m=∞ X a mπ β b β r β b β mπφ − am sin V = 2mπ 2mπ b r β a β −b β m=1

where

am =

β 2w mπφ dφ. f (φ) sin β β 0

13. If V = F (φ) when r = b, V = 0 when r = a, V = 0 when φ = 0, and V = 0 when φ = β, then if a < r < b and 0 < φ < β mπ mπ mπ m=∞ X a mπ β b β r β a β mπφ V = − a sin m 2mπ 2mπ a r β b β −a β m=1

where

am

β 2w mπφ = dφ. F (φ) sin β β 0

14. If V = χ(r) when φ = 0, V = 0 when φ = β, V = 0 when r = a, and V = 0 when r = b, then if a < r < b and 0 < φ < β

V =

mπ(β − φ) ) mπ(log r − log a) log b − log a sin mπβ log b − log a sinh log b − log a sinh

(

m=∞ X

am

m=1

b

where

am

log a w mπx 2 χ(aex ) sin dx. = log b − log a log b − log a 0

15. If V = ψ(r) when φ = β, V = 0 when φ = 0, V = 0 when r = a, and V = 0 when r = b, then if a < r < b and 0 < φ < β (

V =

m=∞ X m=1

mπφ ) mπ(log r − log a) log b − log a am sin mπβ log b − log a sinh log b − log a sinh

MISCELLANEOUS PROBLEMS.

138 b

where

am

log a w 2 mπx ψ(aex ) sin = dx. log b − log a log b − log a 0

II. Potential Function in Space. 1.

Show that f (x, y) =

∞ w∞ w∞ w∞ 1 w dα dβ dλ f (λ, µ) cos α(λ − x) cos β(µ − y).dµ, π2 0

0

0

0

for all values of x and y. Find particular solutions of Dx2 V + Dy2 V + Dz2 V = 0 in the forms √ 2 2 V = e±z α +β cos(αx ± βy) √ 2 2 V = e±z α +β sin(αx ± βy) p V = sinh z α2 + β 2 . sin(αx ± βy) p V = cosh z α2 + β 2 . sin(αx ± βy)

2.

&c. 3. Given Dx2 V + Dy2 V + Dz2 V = 0, and V = f (x, y) when z = 0, solve for positive values of z. ∞ w∞ 1 w zf (λ, µ)dµ Result: V = . dλ 2 2π −∞ −∞ [z + (λ − x)2 + (µ − y)2 ] 23 4. Confirm the result of the last example by showing that if f (x, y) is independent of y ∞ 1 w zf (λ, µ)dλ V = (v. Ex. 3 Art. 45). π −∞ z 2 + (λ − x)2 5. If Dx2 V + Dy2 V + Dz2 V = 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 2πV = b−y p

(b − y)2

π 1 (a − x)2 (b − y)2 − z 2 [(a − x)2 + (b − y)2 + z 2 ] + sin−1 2 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 [(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 [(a − x)2 + (b + y)2 + z 2 ] (a − x)2 (b + y)2 + z 2 [(a − x)2 + (b + y)2 + z 2 ]

1 (a + x)2 (b + y)2 − z 2 [(a + x)2 + (b + y)2 + z 2 ] sin−1 2 (a + x)2 (b + y)2 + z 2 [(a + x)2 + (b + y)2 + z 2 ]

1 sin−1 2 b+y π 1 p + + sin−1 2 2 (b + y) 2 +

+


139

if −a < x < a, and 4πV = p

b−y (b − y)2

sin−1 − sin−1

+p

b+y

(b + y)2

sin−1 −1

− sin

(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 ] (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 ] (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 ]

(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 ]

if x < −a or x > a. 6. 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 q m=∞ n=∞ mπx nπy 4 X X −πz m22 + n22 a b sin e sin V = ab m=1 n=1 a b

wa

dλ

0

wb 0

f (λ, µ) sin

nπµ mπλ sin dµ. a b

If V = 1 on the base of the prism this reduces to q m=∞ n=∞ sin 2 (2n+1)2 16 X X −πz (2m+1) + b2 a2 V = 2 e π m=0 n=0

(2n + 1)πy (2m + 1)πx sin a b . (2m + 1)(2n + 1)

7. 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 if the value of V is given for every point of the sixth face, then for any point within the parallelopiped r m2 n2 m=∞ sinh π(c − z) + 2 X n=∞ X 2 a b sin mπx sin nπy r V = Am,n 2 2 a b m n m=1 n=1 sinh πc + 2 2 a b w 4 w mπλ nπµ dλ f (λ, µ) sin sin dµ. ab a b a

where

Am,n =

0

b

0

8. 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


140

the parallelopiped r

n2 m2 + sinh π(c − z) a2 b2 sin mπx sin nπy r V = Am,n 2 2 a b m n m=1 n=1 + sinh πc a2 b2 r 2 m n2 m=∞ + sinh πz X n=∞ X 2 b2 sin mπx sin nπy ra + Bm,n 2 a b m n2 m=1 n=1 sinh πc + 2 2 a b m=∞ X n=∞ X

w 4 w nπµ mπλ sin dµ dλ f (λ, µ) sin ab a b a

where

Am,n =

b

0

and

Bm,n

0

a wb 4 w mπλ nπµ = dλ F (λ, µ) sin sin dµ. ab a b 0

0

9. 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? III. Conduction of Heat in a Plane. 1.

Find particular solutions of Dt u = a2 (Dx2 u + Dy2 u) of the forms 2

u = e−a u=e

(α2 +β 2 )t

−a2 (α2 +β 2 )t

sin(αx ± βy) cos(αx ± βy).

2. Given the initial temperature of every point in a thin plane plate, find the temperature of any point at any time, u=

= 3.

w∞ w∞ (λ−x)2 +(µ−y)2 1 4a2 t dλ e− f (λ, µ)dµ 2 4a πt −∞ −∞ ∞ w∞ √ √ 2 1 w −β 2 e dβ e−γ f (x + 2a t.β, y + 2a t.γ)dγ. π −∞ −∞

For an instantaneous source of strength Q at (λ, µ) u=

2 Q − (λ−x)2 +(µ−y) 4a2 t e 2 4πa t

v. Art. 53.

For an instantaneous doublet of strength P at (0, µ) with its axis perpendicular to the axis of Y u=

2 P x − x2 +(µ−y) 4a2 t e 8πa4 t2

v. Art. 54.


141

For a permanent doublet of strength P at (0, µ) with its axis perpendicular to the axis of Y x2 +(µ−y)2 P x − 4a2 t . u= e 2πa2 x2 + (µ − y)2 If the strength of the doublet were P dµ and the heat were uniformly generated and absorbed along the element dµ of the axis of Y beginning at (0, µ) we should have u=

2 2 P − x2 +(µ−y) xdµ P − x2 +(µ−y) µ−y 4a2 t 4a2 t e = e d tan−1 , 2πa2 x2 + (µ − y)2 2πa2 x

µ−y is the angle ARA0 , where A and A0 are the points (0, µ) x and (0, µ+dµ) and R is the point (x, y), u = 0 when x = 0 unless µ < y < µ+dµ, P if x approaches zero from the positive side; and u = 0 in which case u = 2a2 when t = 0 except in the element dµ. If then u = 0 when t = 0 and u = f (y) when x = 0 we have only to suppose a doublet of strength 2a2 f (x)dx placed in each element of the axis of Y and then to integrate; we get and since d tan−1

u=

∞ 2 1 w − x2 +(µ−y) xf (µ) 4a2 t e dµ. 2 π −∞ x + (µ − y)2

For a permanent doublet of strength F (t) at (0, µ) we have t 2 +(µ−y)2 x w − x4a 2 (t−τ ) (t − τ )−2 F (τ )dτ. e 8πa4 0 wt x2 +(µ−y)2 x2 +(µ−y)2 1 xF 0 (τ ) xF (0) − 4a2 (t−τ ) − 2t 4a = + e e dτ . 2πa2 x2 + (µ − y)2 x2 + (µ − y)2

u=

0

From the reasoning above this must be zero when t = 0 except at the point (0, µ), must be 2a2 F (t) at the point (0, µ), 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=

∞ ∞ wt xD F (µ, τ ) x2 +(µ−y)2 x2 +(µ−y)2 1 w xF (µ, 0) 1 w − 4a2 (t−τ ) τ − 4a2 t e dµ + dµ e dτ. π −∞ x2 + (µ − y)2 π −∞ x2 + (µ − y)2 0

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. 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


142

for any point of the plate u=

” “ m=∞ n=∞ 4 X X −a2 π2 mb22 + nc22 t nπy mπx sin e sin bc m=1 n=1 b c

wb

dλ

0

wc

f (λ, µ) sin

0

nπµ mπλ sin dµ. b c

if b is the length and c the breadth of the plate. 5. 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 a2 = .185 in C.G.S. units. Ans., 52◦ .9. 6. 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. Ans., 20◦ .8. IV. Conduction of Heat in Space. 1.

Show that

∞ w∞ w∞ w∞ w∞ w∞ 1 w dα dβ dγ dλ dµ f (λ, µ, ν) π3 −∞ −∞ −∞ 0

0

0

cos α(λ − x) cos β(µ − y) cos γ(ν − z).dν = f (x, y, z) for all values of x, y, and z. 2.

Show that f (x, y, z) =

m=∞ X p=∞ X X n=∞

Am,n,p sin

m=1 n=1 p=1

where

Am,n,p =

8 abc

wa

dλ

0

wb

dµ

0

wc

f (λ, µ, ν) sin

0

mπx nπy pπz sin sin a b c mπλ nπµ pπν sin sin dν, a b c

for 0 < x < a, 0 < y < b, 0 < z < c. 3.

Obtain particular solutions of Dt u = a2 (Dx2 u + Dy2 u + Dz2 u) of the forms u = e−a

2

u = e−a

2

(α2 +β 2 +γ 2 )t 2

2

2

(α +β +γ )t

sin(αx ± βy ± γz). cos(αx ± βy ± γz).


143

4. Given the initial temperature of every point in an infinite homogeneous solid find the temperature of any point at any time.

=

w∞

1

u=

8a3 (πt) ∞ 1 w

π

3 2

3 2

−∞ 2

e−β dβ

−∞

w∞

dλ

−∞

w∞

w∞

dµ 2

−∞ w∞

e−γ dγ

−∞

e−

(λ−x)2 +(µ−y)2 +(ν−z) 4a2 t

f (λ, µ, ν)dν

√ √ √ 2 e−δ f (x + 2a t.β, y + 2a t.γ, z + 2a t.δ)dδ.

−∞

5. 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=

m=∞ X p=∞ X X n=∞

Am,n,p e

−a2 π 2

m=1 n=1 p=1

where

Am,n,p =

“

m2 b2

” 2 p2 +n +d 2 t c2

sin

mπx nπy pπz sin sin b c d

w w mπλ nπµ pπν 8 w dλ dµ f (λ, µ, ν) sin sin sin dν. bcd b c d b

c

0

0

d

0

6. 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. Ans., 38◦ .4. 7. 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. Ans., 9◦ .5.

144

CHAPTER V.1 ZONAL HARMONICS.

74.

In Art. 16 we obtained z = Apm (x) + Bqm (x)

(1)

[v. (6) Art. 16] as the general solution of Legendre’s Equation (1 − x2 )

dz d2 z − 2x + m(m + 1)z = 0, dx2 dx

(2)

m being wholly unrestricted in value and x lying between −1 and 1; where pm (x) = 1 −

m(m + 1) 2 m(m − 2)(m + 1)(m + 3) 4 x + x 2! 4! m(m − 2)(m − 4)(m + 1)(m + 3)(m + 5) 6 − x + ··· 6!

(3)

and qm (x) = x −

and we found

(m − 1)(m + 2) 3 (m − 1)(m − 3)(m + 2)(m + 4) 5 x + x 3! 5! (m − 1)(m − 3)(m − 5)(m + 2)(m + 4)(m + 6) 7 x + ··· ; − 7!  V = rm pm (cos θ)     1   V = m+1 pm (cos θ)  r  V = rm qm (cos θ)      1 V = m+1 qm (cos θ), r

(4)

(5)

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 φ; that is, of the equation rDr2 (rV ) +


(6)

For the important case where m is a positive integer we found z = APm (x) + BQm (x)

(7)

1 Before reading this chapter the student is advised to re-read carefully articles 9, 10, 13(c), 15, 16, and 18(c).

ZONAL HARMONICS.

145

[v. (10) Art. 16] as the general solution of Legendre’s Equation (2), whence  V = rm Pm (cos θ)     1   V = m+1 Pm (cos θ)  r (8) V = rm Qm (cos θ)       1 V = m+1 Qm (cos θ) r are particular solutions of (6) if m is a positive integer. (2m − 1)(2m − 3) · · · 1 m m(m − 1) m−2 x − x Pm (x) = m! 2(2m − 1) m(m − 1)(m − 2)(m − 3) m−4 x − ··· (9) + 2.4.(2m − 1)(2m − 3) [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 x0 if m is even. It is called a Surface Zonal Harmonic, or a Legendre’s Coefficient, or more briefly a Legendrian. 1 (m + 1)(m + 2) 1 m! + Qm (x) = m+1 (2m + 1)(2m − 1) · · · 1 x 2.(2m + 3) xm+3 (m + 1)(m + 2)(m + 3)(m + 4) 1 + ··· (10) + 2.4.(2m + 3)(2m + 5) xm+5 if x < −1 or x > 1. [v. (9) Art. 16.] It is called a Surface Zonal Harmonic of the second kind. h m + 1 i2 m−1 2 Γ m+1 2 Qm (x) = (−1) 2 pm (x) Γ(m + 1) m+1 2.4.6. · · · (m − 1) = (−1) 2 pm (x) 3.5.7. · · · m [v. (13) Art. 16] if m is odd and −1 < x < 1. h m + 1 i2 m 2 Γ m 2 Qm (x) = (−1) 2 qm (x) Γ(m + 1) m 2.4.6. · · · m = (−1) 2 qm (x) 1.3.5. · · · (m − 1)

(11)

(12)

[v. (14) Art. 16] if m is even and −1 < x < 1. In most of the work that immediately follows we shall regard x in Pm (x)as equal to cos θ and therefore as lying between −1 and 1.2 2 English writers on Spherical Harmonics generally use µ in place of x for cos θ. We shall follow them, however, only when we should thereby avoid confusion.

ZONAL HARMONICS.

146

In Article 9 the undetermined coefficient am of xm in Pm (x) was (2m − 1)(2m − 3) · · · 1 arbitrarily written in the form for reasons which shall m! now be given. In Articles 9 and 16 z = Pm (x) was obtained as a particular solution of Legendre’s Equation 75.

(1 − x2 )

dz d2 z − 2x + m(m + 1)z = 0 2 dx dx

(1)

by the device of assuming that z could be expressed as a sum or a series of terms of the form an xn 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 (x1 , y1 , z1 ) is 1 (2) V =p (x − x1 )2 + (y − y1 )2 + (z − z1 )2 and this must be a particular solution of Laplace’s Equation Dx2 V + Dy2 V + Dz2 V = 0,

(3)

as is easily verified by direct substitution. If we transform (2) to spherical coördinates using the formulas of transformation x = r cos θ y = r sin θ cos φ z = r sin θ sin φ

we get

1 V =p r2 − 2rr1 [cos θ cos θ1 + sin θ sin θ1 cos(φ − φ1 )] + r12

(4)

as a solution of Laplace’s Equation in Spherical Coördinates rDr2 (rV ) +

1 1 Dθ (sin θDθ V ) + Dφ 2 V = 0 sin θ sin2 θ

[XIII] Art. 1.

If the given point (x1 , y1 , z1 ) is taken on the axis of X, as it must be that (4) may be independent of φ, θ1 = 0, and 1 V =p r2 − 2rr1 cos θ + r12 is a solution of rDr2 (rV ) +


(5)

(6)

ZONAL HARMONICS.

147

Equation (5) may be written

or

V =

1 r r

V =

1 s r1

1

(7)

r1 r2 1 − 2 cos θ + 12 r r 1

.

(8)

r2 r 1 − 2 cos θ + 2 r1 r1

√

1 − 2z cos θ + 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 θ + z 2 = 0 namely z = cos θ + i sin θ and z = cos θ − i sin θ, both of which have the modulus unity and which are critical values. 1 is finite and continuous except for the values of z = 1 − 2z cos θ + z 2 cos θ − i sin θ and z = cos θ + i sin θ for which it becomes infinite; it is doublevalued but has as critical values only these values of z. It is then 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.) 1 If then r > r1 r can be developed into a convergent series 2r1 r12 1− cos θ + 2 r r r1 involving whole powers of . r X rm Let pm 1m be this series, pm , of course, being a function of cos θ. Then r rm 1X pm 1m V = r r √

[v. (7)] is a solution of (6). Substitute this value of V in (6) and we get X rm dpm r1m 1 d 1 m(m + 1)pm + m+1 sin θ = 0. rm+1 r sin θ dθ dθ As this must hold whatever the value of r provided r > r1 the coefficient of each power of r must be zero, and hence the equation 1 d dpm sin θ + m(m + 1)pm = 0 (9) sin θ dθ dθ must be true. But as we have seen in Art. 9 the substitution of x = cos θ in (9) reduces it to d2 pm dpm (1 − x2 ) − 2x + m(m + 1)pm = 0, 2 dx dx

ZONAL HARMONICS.

148

and therefore

z = pm

is a solution of Legendre’s Equation (1). 1 If r < r1 s can be developed into a convergent series 2r r2 1− cos θ + 2 r1 r1 2 r involving whole powers of 2 . r1 X rm Let pm m be this series. Then r1 V =

1 X rm pm m r1 r1

(v. 8) is a solution of (6); substituting in (6) we get X rm rm 1 d dpm m(m + 1)pm + m+1 sin θ = 0, sin θ dθ dθ r1m+1 r1 whence it follows as before that z = pm is a solution of Legendre’s Equation. r in the development of But pm is the coefficient of the mth power of r1 1 −2 r r r2 r1 1 − 2 cos θ + 2 according to powers of , or of the mth power of r1 r1 r1 r 1 2 −2 r1 r r1 according to powers of , or in the development of 1 − 2 cos θ + 12 r r r more briefly it is the coefficient of the mth power of z in the development of 1 (1 − 2xz + z 2 )− 2 according to powers of z, x standing for cos θ. 1

1

(1 − 2xz + z 2 )− 2 = [1 − z(2x − z)]− 2

and can be developed by the Binomial Theorem; the coefficient of z m is easily picked out and is (2m − 1)(2m − 3) · · · 1 m m(m − 1) m−2 x − x m! 2(2m − 1) m(m − 1)(m − 2)(m − 3) m−4 + x − ··· . 2.4.(2m − 1)(2m − 3) But this is precisely Pm (x). [v. Art. 74 (9)] Hence Pm (x) is equal to the coefficient of the mth power of z in the devel1 opment of [1 − 2xz + z 2 ]− 2 into a power series, the modulus of z being less than unity.

ZONAL HARMONICS.

149 1

76. If x = 1 Pm (x) = 1. For if x = 1 (1 − 2xz + z 2 )− 2 reduces to 1 (1 − 2z + z 2 )− 2 that is to (1 − z)−1 , which develops into 1 + z + z2 + z3 + z4 + · · · , and the coefficient of each power of z is unity. Therefore Pm (1) = 1.

(1)

We have seen that if m is even Pm (x) contains only even powers of x and terminates with the term involving x0 , that is with the constant term. The value of this constant term can be picked out from the formula for Pm (x) m 1.3.5. · · · (m − 1) ; or it can be found as follows:—It [v. Art. 74 (9)]. It is (−1) 2 2.4.6. · · · m is clearly the value Pm (x) assumes when x = 0; it is, then, the coefficient of z m 1 in the development of (1 + z 2 )− 2 ; but 1 1 1.3 4 1.3.5 6 1.3.5.7 8 (1 + z 2 )− 2 = 1 − z 2 + z − z + z − ··· 2 2.4 2.4.6 2.4.6.8

1.3.5 · · · (m − 1) . 2.4.6 · · · m If m is odd Pm (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 3.5.7. · · · m m−1 out from (9) Art. 74 and is (−1) 2 ; or it can be found as 2.4.6. · · · (m − 1) dPm (x) follows:—It is clearly the value assumed by when x = 0. dx z It is, then, the coefficient of z m in the development of 3 . (1 + z 2 ) 2 m

and the coefficient of z m , m being an even number, is (−1) 2

z (1 +

3 z2) 2

3 3.5 5 3.5.7 7 = z − z3 + z − z + ··· 2 2.4 2.4.6

and the coefficient of z m in this development is (−1)

m−1 2

3.5.7 · · · m , m 2.4.6 · · · (m − 1)

being an odd number. 77.

To recapitulate:

Pm (x) =

1.3.5 · · · (2m − 1) m m(m − 1) m−2 x − x m! 2(2m − 1) m(m − 1)(m − 2)(m − 3) m−4 + x 2.4.(2m − 1)(2m − 3) m(m − 1)(m − 2)(m − 3)(m − 4)(m − 5) m−6 − x + ··· , 2.4.6.(2m − 1)(2m − 3)(2m − 5)

(1)

m being a positive integer, is a Surface Zonal Harmonic or Legendrian of the mth 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.

ZONAL HARMONICS.

150

Pm (x) is the coefficient of the mth power of z in the development of (1 − 1 2xz + z 2 )− 2 into a power series. Hence if z < 1 1

(1 − 2xz + z 2 )− 2 = P0 (x) + P1 (x).z + P2 (x).z 2 + P3 (x).z 3 + P4 (x).z 4 + P5 (x).z 5 + · · · + Pm (x).z m + · · · . (2) Whence  1 r1 r12  p = P0 (cos θ) + P1 (cos θ) + 2 P2 (cos θ) + · · ·    r r r  r2 − 2rr1 cos θ + r12    m   r1  + m Pm (cos θ) + · · · if r > r1   r (3)  1 r r2   = P0 (cos θ) + P1 (cos θ) + 2 P2 (cos θ) + · · ·    r1 r1 r1    m   r  + m Pm (cos θ) + · · · if r < r1 . r1 1

z = Pm (x) is a solution of Legendre’s Equation (1 − x2 )

dz d2 z − 2x + m(m + 1)z = 0 dx2 dx

when m is a positive integer. V = rm Pm (cos θ) 1 V = m+1 Pm (cos θ) r

and

are solutions of the form of Laplace’s Equation in Spherical Coördinates which is independent of φ, namely rDr2 (rV ) +

1 Dθ (sin θDθ V ) = 0. sin(θ) Pm (1) = 1.

(5)

P2m (−x) = P2m (x).

(6)

P2m+1 (−x) = −P2m+1 (x).

(7)

P2m+1 (0) = 0.

(8)

P2m (0) = (−1)m

(4)

dP2m+1 (x) dx

x=0

1.3.5. · · · (2m − 1) . 2.4.6. · · · 2m

= (−1)m

3.5.7. · · · (2m + 1) . 2.4.6. · · · 2m

(9) (10)

ZONAL HARMONICS.

151

For convenience of reference we write out a few Zonal Harmonics. They are obtained by substituting successive integers for m in formula (1).  P0 (x) = 1       P1 (x) = x      1  2  P2 (x) = (3x − 1)    2    1  3  P3 (x) = (5x − 3x)    2    1  4 2 P4 (x) = (35x − 30x + 3) (11) 8   1  5 3  P5 (x) = (63x − 70x + 15x)    8    1  6 4 2  (231x − 315x + 105x − 5) P6 (x) =    16    1  7 5 3  P7 (x) = (429x − 693x + 315x − 35x)    16    1  8 6 4 2 P8 (x) = (6435x − 12012x + 6930x − 1260x + 35). 128 Any Surface Zonal Harmonic may be obtained from the two of next lower orders by the aid of the formula (n + 1)Pn+1 (x) − (2n + 1)xPn (x) + nPn−1 (x) = 0

(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 −(z − x) 3

(1 − 2xz + z 2 ) 2

= P1 (x) + 2P2 (x).z + 3P3 (x).z 2 + · · ·

whence −(z − x) 1

(1 − 2xz + z 2 ) 2

= (1 − 2xz + z 2 )(P1 (x) + 2P2 (x).z + 3P3 (x).z 2 + · · · ).

Hence by (2) (1 − 2xz + z 2 )(P1 (x) + 2P2 (x).z + 3P3 (x).z 2 + · · · ) + (z − x)(P0 (x) + P1 (x).z + P2 (x).z 2 + · · · ) = 0

(13)

(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.3 3 For

tables of Surface Zonal Harmonics v. Appendix Tables I and II.

ZONAL HARMONICS.

152

78. We are now able to solve completely the problem considered in Art. 9. We were to find a solution of the differential equation rDr2 (rV ) +

1 Dθ (sin θDθ V ) = 0 sin θ

(1)

subject to the condition V =

M (c2

1

+ r2 ) 2

when θ = 0.

(2)

We know (v. Art. 77) that V = rm Pm (cos θ) 1 V = m+1 Pm (cos θ) r

and are solutions of (1). For values of r < c M (c2 + r2 )

1 2

=

i Mh 1 r2 1.3 r4 1.3.5 r6 1− + − + · · · . c 2 c2 2.4 c4 2.4.6 c6

Therefore for values of r < c 1 r2 Mh P0 (cos θ) − P2 (cos θ) V = c 2 c2 i 1.3 r4 1.3.5 r6 + P (cos θ) − P (cos θ) + · · · 4 6 2.4 c4 2.4.6 c6

(3)

(4)

is our required solution; because each term satisfies equation (1), and therefore the whole value satisfies (1), and when θ = 0 Pm (cos θ) = Pm (1) = 1 [v. (5) Art. 77], and hence (4) reduces to (3) and (2) is satisfied. For values of r > c i Mh 1 c2 1.3 c4 1.3.5 c6 M 1− + − + ··· 1 = 2 4 6 r 2r 2.4 r 2.4.6 r (c2 + r2 ) 2 h 1 1 c2 i 4 1.3 c 1.3.5 c6 =M − + − + · · · . r 2 r3 2.4 r5 2.4.6 r7 Therefore for values of r > c M hc 1 c3 V = P0 (cos θ) − P2 (cos θ) c r 2 r3 i 1.3 c5 1.3.5 c7 + P (cos θ) − P (cos θ) + · · · 4 6 2.4 r5 2.4.6 r7 is our required solution. For it satisfies (1) and reduces to (2) when θ = 0.

(5)

(6)

ZONAL HARMONICS.

153

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 σ=

M √ 4aπ a2 − r2

and all points of the conductor are at the potential

πM . (v. Peirce’s Newtonian 2a

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 V = =

a rdr Mw p 2 − r 2 )(x2 + r 2 ) a (a 0

x2 − a 2 M cos−1 2 . 2a x + a2

d M x2 − a2 M cos−1 2 =− 2 2 dx 2a x +a a + x2 Mh =− 2 1− a if x < a, Mh =− 2 1− x if x > a.

i x2 x4 x6 + − + · · · a2 a4 a6 i a4 a6 a2 + − + · · · x2 x4 x6

Integrating and then determining the arbitrary constant we have i M x3 x2 − a2 M hπ x x5 x7 cos−1 2 − + = − + − · · · 2a x + a2 a 2 a 3a3 5a5 7a7 if x < a, =

i M ha a3 a5 a7 − 3 + 5 − 7 + ··· a x 3x 5x 7x

if x > a. We have, then, to solve the equation rDr2 (rV ) +


ZONAL HARMONICS.

154

subject to the conditions r5 r7 M π r r3 V = − + 3 − 5 + 7 − ··· a 2 a 3a 5a 7a when

θ=0

and r < a

M a a5 a7 a V = − 3 + 5 − 7 + ··· a r 3r 5r 7r 3

and

when

θ=0

and r > a.

The required solution is easily seen to be M π r 1 r3 1 r5 V = − P1 (cos θ) + P3 (cos θ) − P5 (cos θ) + · · · a 2 a 3 a3 5 a5 if r < a and θ
a.

EXAMPLES. 1.

Given that if a charge M of electricity is placed on an ellipsoidal conMp ductor the surface density at any point P of the conductor is equal to , 4πabc 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 x2 y2 the rotation of the ellipse 2 + 2 = 1 about its minor axis. a b Ans. (1) If the point is on the axis of revolution M bx + a2 − b2 bx − a2 + b2 −1 −1 √ √ V = √ sin − sin 2 a2 − b2 a x2 + a2 − b2 a x2 + a2 − b2 x being the distance from the centre. (2) If the point is on the surface of the spheroid 2 2b − a2 b M π M π −1 −1 √ − sin − tan V = √ =√ . a2 2 a2 − b2 2 a2 − b2 2 a2 − b2

ZONAL HARMONICS.

155

√ (3) If the distance r of the point from the centre is less than a2 − b2 and π θ< 2 M π r V =√ − 1 P1 (cos θ) 2 2 2 2 a −b (a − b2 ) 2 r5 r3 . + 3 P3 (cos θ) − 5 P5 (cos θ) + · · · 3(a2 − b2 ) 2 5(a2 − b2 ) 2 √ (4) If the distance r of the point from the centre is greater than a2 − b2 2 1 3 (a − b2 ) 2 M (a2 − b2 ) 2 P2 (cos θ) V =√ − r 3r3 a2 − b2 5 7 (a2 − b2 ) 2 (a2 − b2 ) 2 + P (cos θ) − P (cos θ) + · · · . 4 6 5r5 7r7 2.

If the conductor is the prolate spheroid generated by the rotation of the x2 y2 ellipse 2 + 2 = 1 about its major axis, show that if the point is an external a b point and is on the axis at a distance x from the centre, √ x + a2 − b2 M √ log . V = √ 2 a2 − b2 x − a2 − b2 √ If the point is not on the axis and r > a2 − b2 2 3 1 (a2 − b2 ) 2 M (a − b2 ) 2 + P2 (cos θ) V =√ r 3r3 a2 − b2 5 7 (a2 − b2 ) 2 (a2 − b2 ) 2 + P (cos θ) + P (cos θ) + · · · . 4 6 5r5 7r7 80. As a third example we will find the value of the potential function due to a thin homogeneous circular disc, of density ρ, 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 p 2M p V0 = 2πρk( x2 + a2 − x) = 2 [ x2 + a2 − x]. a If x > a 1 p a2 2 2 2 x +a =x 1+ 2 x 1 a2 1.1 a4 1.1.3 a6 1.1.3.5 a8 =x 1+ − + − + · · · 2 x2 2.4 x4 2.4.6 x6 2.4.6.8 x8 3 5 2M 1 a 1.1 a 1.1.3 a 1.1.3.5 a7 and V0 = − + − + · · · . a 2 x 2.4 x3 2.4.6 x5 2.4.6.8 x7

ZONAL HARMONICS.

156

If x < a 1 x2 2 x2 + a2 = a 1 + 2 a 1 x2 1.1 x4 1.1.3 x6 =a 1+ − + + · · · 2 a2 2.4 a4 2.4.6 a6 2M 1.1 x4 1.1.3 x6 1.1.3.5 x8 x 1 x2 V0 = − + − + ··· . 1− + a a 2 a2 2.4 a4 2.4.6 a6 2.4.6.8 a8

p

and

Hence the solution for any external point is V =

2M a

1 a 1.1 a3 − P2 (cos θ) 2r 2.4 r3 1.1.3 a5 1.1.3.5 a7 + P (cos θ) − P (cos θ) + · · · 4 6 2.4.6 r5 2.4.6.8 r7

if r > a, and V =

2M h r 1 − P1 (cos θ) a a 1.1 r4 1.1.3 r6 1 r2 P2 (cos θ) − P4 (cos θ) + P6 (cos θ) − · · · + 2 a2 2.4 a4 2.4.6 a6

if r < a and θ
a, and is M V = a

3 3r r2 + P1 (cos θ) + 2 P2 (cos θ) 2 2a a 3.1 r3 3.1.1 r5 + P3 (cos θ) − P5 (cos θ) + · · · 2.4 a3 2.4.6 a5

if r < a and θ >

π . 2

ZONAL HARMONICS.

157

2. The potential function due to a solid sphere whose density is proportional to the distance from a diametral plane is, at an external point, 8 M 5.3 a 5.3.1 a3 V = P2 (cos θ) + 15 a 2.4 r 2.4.6 r3 5.3.1.1 a5 5.3.1.1.3 a7 − P4 (cos θ) + P6 (cos θ) − · · · . 2.4.6.8 r5 2.4.6.8.10 r7 3.

The potential function due to the homogeneous oblate spheroid generx2 y2 ated by the rotation of 2 + 2 = 1 about its minor axis is, at an external a b point, 2 M x + a2 − b2 (a2 − b2 + bx) 3 sin−1 √ V = 1 2 2 2 (a − b ) 2(a2 − b2 ) 2 a x2 + a2 − b2 2 2 −1 (a − b − bx) √ + sin −x a x2 + a2 − b2 if the point is on the axis of the spheroid at a distance x from its centre. " 1 3 3M 1 (a2 − b2 ) 2 1 (a2 − b2 ) 2 V = − P2 (cos θ) 1 r 3.5 r3 (a2 − b2 ) 2 1.3 5

1 (a2 − b2 ) 2 + P4 (cos θ) − · · · 5.7 r5

#

1

if r > (a2 − b2 ) 2 , and 3M π r π r2 V = − P (cos θ) + P2 (cos θ) 1 1 1 4 (a2 − b2 ) (a2 − b2 ) 2 4 (a2 − b2 ) 2 1 r3 1 r5 − P (cos θ) + P (cos θ) − · · · 3 5 1.3 (a2 − b2 ) 32 3.5 (a2 − b2 ) 52 1

if r < (a2 − b2 ) 2 and θ < 4.

π . 2

The potential function due to the homogeneous prolate spheroid generx2 y2 ated by the rotation of 2 + 2 = 1 about its major axis is, at an external a b point, " 1 3 3M 1 (a2 − b2 ) 2 1 (a2 − b2 ) 2 V = + P2 (cos θ) 1 r 3.5 r3 (a2 − b2 ) 2 1.3 # 5 1 (a2 − b2 ) 2 + P4 (cos θ) + · · · 5.7 r5 1

if r > (a2 − b2 ) 2 .

ZONAL HARMONICS.

158

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 rDr2 (rV ) +

1 Dθ (sin θDθ V ) = 0, sin θ

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 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 Pm (cos θ) 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 absolutely convergent. 82. In addition to the form given in (1) Art. 77 for Pm (x) other forms are often useful. It ought to be possible to develop Pm (cos θ), which may be regarded as a function of θ, 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 θ may be obtained much more easily by the following device. We have seen in Art. 75 that Pm (cos θ) is the coefficient of the mth power 1 of z in the development of (1 − 2z cos θ + z 2 )− 2 in a power series, and that if 1 mod z < 1 (1 − 2z cos θ + z 2 )− 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. 1

1

(1 − 2z cos θ + z 2 )− 2 = (1 − z(eθi + e−θi ) + z 2 )− 2 1

1

= (1 − zeθi )− 2 (1 − ze−θi )− 2 . 1

(1−zeθi )− 2 may be developed into an absolutely convergent series if mod z < 1, by the Binomial Theorem. We have 1 1 1.3 2 2θi 1.3.5 3 3θi 1.3.5.7 4 4θi (1 − zeθi )− 2 = 1 + zeθi + z e + z e + z e + ··· 2 2.4 2.4.6 2.4.6.8 1 1 1.3 2 −2θi (1 − ze−θi )− 2 = 1 + ze−θi + z e 2 2.4 1.3.5 3 −3θi 1.3.5.7 4 −4θi + z e + z e + ··· 2.4.6 2.4.6.8 1

The product of these series will give a development for (1 − 2z cos θ + z 2 )− 2 in power series. The coefficient of z m is easily picked out, and must be equal to

ZONAL HARMONICS.

159

Pm (cos θ). We thus get 1.3.5. · · · (2m − 1) mθi e + e−mθi 2.4.6. · · · 2m 1 2m + . (e(m−2)θi + e−(m−2)θi ) 2 2m − 1 2m(2m − 2) 1.3 . (e(m−4)θi + e−(m−4)θi ) + · · · + 2.4 (2m − 1)(2m − 3) 1.3.5 · · · (2m − 1) 1.m Pm (cos θ) = 2 cos mθ + 2 cos(m − 2)θ 2.4.6. · · · 2m 1.(2m − 1) 1.3 m(m − 1) +2 cos(m − 4)θ 1.2(2m − 1)(2m − 3) m(m − 1)(m − 2) 1.3.5 cos(m − 6)θ + · · · . +2 1.2.3 (2m − 1)(2m − 3)(2m − 5)

Pm (cos θ) =

(1)

If m is odd the development runs down to cos θ; 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 2 1.3.5 · · · (m − 1) . 2.4.6. · · · m We write out the values of Pm (cos θ) for a few values of m  P0 (cos θ) = 1        P1 (cos θ) = cos θ     1    P2 (cos θ) = (3 cos 2θ + 1)   4    1    P3 (cos θ) = (5 cos 3θ + 3 cos θ)   8    1    P4 (cos θ) = (35 cos 4θ + 20 cos 2θ + 9)  64 (2) 1   P5 (cos θ) = [63 cos 5θ + 35 cos 3θ + 30 cos θ]   128     1   P6 (cos θ) = [231 cos 6θ + 126 cos 4θ + 105 cos 2θ + 50]   512     1  P7 (cos θ) = [429 cos 7θ + 231 cos 5θ + 189 cos 3θ + 175 cos θ]   1024     1   P8 (cos θ) = [6435 cos 8θ + 3432 cos 6θ + 2772 cos 4θ   16384    + 2520 cos 2θ + 1225]. 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 Pm (cos θ) has its maximum value when θ = 0; but we have shown in Art. 76 that Pm (1) = 1. Therefore Pm (cos θ) is never greater than unity if θ is real. It is also easily seen from (1) that Pm (cos θ) can never be less than −1.

ZONAL HARMONICS. 83.

160

Pm (x) can be very simply expressed as a derivative. We have

(2m − 1)(2m − 3) · · · 1 m m(m − 1) m−2 x − x m! 2.(2m − 1) m(m − 1)(m − 2)(m − 3) m−4 + x − ··· 2.4.(2m − 1)(2m − 3) wx (m + 1)m m−1 (2m − 1)(2m − 3) · · · 1 m+1 x − x Pm (x)dx = (m + 1)! 2.(2m − 1) Pm (x) =

0

(m + 1)m(m − 1)(m − 2) m−3 + x − ··· 2.4.(2m − 1)(2m − 3) wx

2

Pm (x)dx2 =

wx

0

0

dx

wx

Pm (x)dx

0

(2m − 1)(2m − 3) · · · 1 m+2 (m + 2)(m + 1) m x − x (m + 2)! 2.(2m − 1) (m + 2)(m + 1)m(m − 1) m−2 + x − ··· 2.4.(2m − 1)(2m − 3) (2m − 1)(2m − 3) · · · 1 2m 2m(2m − 1) 2m−2 = x x − (2m)! 2(2m − 1) 2m(2m − 1)(2m − 2)(2m − 3) 2m−4 + x − ··· 2.4.(2m − 1)(2m − 3) (2m − 1)(2m − 3) · · · 1 2m m(m − 1) 2m−4 = x x − mx2m−2 + (2m)! 2! m(m − 1)(m − 2) 2m−6 − x + ··· . 3!

=

wx

m

Pm (x)dxm

0

The quantity in brackets obviously differs from (x2 − 1)m by terms involving lower powers of x than the mth. Hence or

1.3.5 · · · (2m − 1) dm 2 (x − 1)m , (2m)! dxm 1 dm 2 Pm (x) = m (x − 1)m . 2 m! dxm

Pm (x) =

(1)

This important formula is entirely general and holds not merely when x = cos θ, but for all values of x. 84. The last result is so important that it is worth while to confirm it by obtaining it directly from Legendre’s Equation (1 − x2 ) v. (1) Art. 75.

d2 z dz − 2x + m(m + 1)z = 0 dx2 dx

(1)

ZONAL HARMONICS.

161

Let us differentiate (1) with respect to x a few times representing 3 d2 z 00 d z by z , by z 000 , &c. We get dx2 dz 3

dz by z 0 , dx

d2 z 0 dz 0 − 2.2x + [m(m + 1) − 2]z 0 = 0, 2 dx dx dz 00 d2 z 00 + [m(m + 1) − 2(1 + 2)]z 00 = 0, (1 − x2 ) 2 − 2.3x dx dx dz 000 d2 z 000 − 2.4x + [m(m + 1) − 2(1 + 2 + 3)]z 000 = 0, (1 − x2 ) dx2 dx (1 − x2 )

and in general d2 z (n) dz (n) − 2(n + 1)x + [m(m + 1) − 2(1 + 2 + 3 + · · · + n)]z (n) = 0 2 dx dx d2 z (n) dz (n) (1 − x2 ) − 2(n + 1)x + m(m + 1) − n(n + 1)]z (n) = 0. (2) dx2 dx

(1 − x2 ) or

Following the analogy of these steps it is easy to write equations that will differentiate into (1). d2 z2 dz1 d3 z3 = z, Let = z, = z, &c. Then dx dx2 dx3 (1 − x2 )

d2 z1 + m(m + 1)z1 = 0 dx2

will differentiate into (1), (1 − x2 )

d2 z2 dz2 + 2.1x + [m(m + 1) − 2.1]z2 = 0 2 dx dx

if differentiated twice will give (1), (1 − x2 )

dz3 d2 z3 + 2.2x + [m(m + 1) − 2(1 + 2)]z3 = 0 dx2 dx

if differentiated three times will give (1), and in general (1 − x2 )

d2 zn dzn + 2(n − 1)x + [m(m + 1) − n(n − 1)]zn = 0 dx2 dx

(3)

if differentiated n times with respect to x will give (1). If n = m + 1 (3) reduces to (1 − x2 )

d2 zm+1 dzm+1 + 2mx = 0, 2 dx dx

(4)

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 − x2 )

dzm + 2mxzm = 0 dx

ZONAL HARMONICS.

162

and can be readily solved by separating the variables and integrating. v. Int. Cal. (1) page 314. It gives zm = C(x2 − 1)m . Hence

z=

dm (x2 − 1)m dm zm = C dxm dxm

(5)

is a solution of Legendre’s Equation (1) and agrees with the value of Pm (x) obtained in Art. 83. 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 dz d2 z + 2mx =0 dx2 dx

(1)

d2 z dz + 2mz = 0; + 2(m − 1)x 2 dx dx

(2)

(1 − x2 ) differentiates into (1 − x2 )

that if we differentiate (2) m times we get Legendre’s Equation (1 − x2 )

d2 z dz − 2x + m(m + 1)z = 0; 2 dx dx

(3)

that if we differentiate (2) 2m times we get (1 − x2 )

d2 z dz − 2(m + 1)x = 0; dx2 dx

(4)

that if we differentiate (2) m − n times we have (1 − x2 )

dz d2 z + 2(n − 1)x + [m(m + 1) − n(n − 1)]z = 0; dx2 dx

(5)

and that if we differentiate (2) m + n times we have (1 − x2 )

d2 z dz − 2(n + 1)x + [m(m + 1) − n(n + 1)]z = 0. dx2 dx

(6)

By the aid of (1) we found in the last article a particular solution of (2), namely z = (x2 − 1)m . If we substitute in (2) z = u(x2 − 1)m following the method illustrated fully in Art. 18, we get as the general solution of (2) z = A(x2 − 1)m + B(x2 − 1)m

w

dx , (x2 − 1)m+1

(7)

ZONAL HARMONICS.

163

A and B being arbitrary constants. w dx is easily written out [v. formula (42) page 6. Table of Inte2 (x − 1)m+1 grals. Int. Cal. Appendix]. If x < 1 it vanishes when x = 0. If x > 1 it vanishes when x = ∞. If then x < 1 (7) can be written z = A(x2 − 1)m + B(x2 − 1)m

wx 0

and if x > 1 z = A(x2 − 1)m + B(x2 − 1)m

w∞ x

dx − 1)m+1

(8)

dx (x2 − 1)m+1

(9)

(x2

and in these forms unnecessary arbitrary constants are avoided. From (7) we can get the general solutions of (3), (4), (5), and (6). w dm (x2 − 1)m dm dx 2 m z=A + B (x − 1) dxm dxm (x2 − 1)m+1

(10)

is the general solution of (3). z=A

w d2m d2m (x2 − 1)m dx 2 m + B (x − 1) dx2m dx2m (x2 − 1)m+1

(11)


w dm−n (x2 − 1)m dm−n dx 2 m + B (x − 1) dxm−n dxm−n (x2 − 1)m+1

(12)


w dm+n dm+n (x2 − 1)m dx 2 m + B (x − 1) dxm+n dxm+n (x2 − 1)m+1

(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 ∞ 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 w dx (14) z = A1 + B 1 2 (x − 1)m+1 which must be equivalent to (11). 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 (x2 − 1)m ,

ZONAL HARMONICS.

164

or in other words that if we write (2) d2 z dz + 2(m − 1)x + 2mz = 0, dx2 dx dz1 d2 z1 =0 (1 − x2 ) 2 − 2(m + 1)x dx dx

(1 − x2 ) and (4) as

(2) (4)

z = z1 (x2 − 1)m ; and the substitution of this value in (2) will give (4), and z the substitution of z1 = 2 in (4) will give (2). (x − 1)m 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 z1 (x2 − 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. m

If

z = y(x2 − 1) 2

m

and

y = z1 (x2 − 1) 2

z = z1 (x2 − 1)m . m

So that if in (2) we replace z by y(x2 − 1) 2 and then repeat the operation on the resulting equation we shall get (4). Making the first substitution we find, d2 y dy m2 (1 − x2 ) 2 − 2x + m(m + 1) − y = 0, (15) dx dx 1 − x2 not Legendre’s Equation but a somewhat more general form. Of course its solution is m m w dx y = A(x2 − 1) 2 + B(x2 − 1) 2 . (16) 2 (x − 1)m+1 (2) and (4) are special forms of (5) and (6). Let us try the experiment of n y substituting in (5) z = y(1 − x2 ) 2 and in (6) z = n . We find that both (1 − x2 ) 2 substitutions give the same equation 2 dy n2 2 d y (1 − x ) 2 − 2x + m(m + 1) − y = 0. (17) dx dx 1 − x2 The solution of (17) can be obtained from either (12) or (13) and is m−n 2 w 1 d (x − 1)m dm−n dx 2 m y= + B m−n (x − 1) n A dxm−n dx (x2 − 1)m+1 (1 − x2 ) 2 (18) or w n dm+n dx dm+n (x2 − 1)m 2 2 2 m y = (1 − x ) A1 + B1 m+n (x − 1) dxm+n dx (x2 − 1)m+1 (19)

ZONAL HARMONICS.

165

which of course must be equivalent. 86. In addition to the value of Pm (x) given in (1) Art. 83 there is another important derivative form which we shall proceed to obtain. It is (−1)m m+1 m 1 Pm (cos θ) = r . (1) Dx m! r 1 1 r can be developed into r r1 r12 1 − 2 cos θ + 2 r r a convergent series if r1 < r and that the (m + 1)st term of that series is Pm (cos θ)r1m . Let us obtain this term by Taylor’s Theorem. rm+1 We have seen in Art. 75 that

1 1 1 1 r =p =p 2 2 2 2 2 r r − 2r1 r cos θ + r1 x + y + z 2 − 2xr1 + r12 r1 r 1 − 2 cos θ + 12 r r 1 =p (x − r1 )2 + y 2 + z 2 Regarding this as a function of (x − r1 ) and developing according to powers of r1 by Taylor’s Theorem we get as the (m + 1)st term (−1)m m m 1 r1 Dx p m! x2 + y 2 + z 2 Hence

(−1)m m m 1 r1 Dx . m! r (−1)m m 1 Pm (cos θ) Dx = . rm+1 m! r or

87. We have now obtained four different forms for our zonal harmonic, a polynomial in x, an expression involving cosines of multiples of θ, a form involving an ordinary mth derivative with respect to x, and a form involving a partial mth derivative with respect to x. We shall now get a form due to Laplace, involving a definite integral. wπ 0

π dφ = 1 2 a − b cos φ (a − b2 ) 2

(1)

if a2 > b2 [v. Int. Cal. page 68]. 1 1 can be expressed in the form by taking a = 1 1 (1 − 2xz + z 2√ )2 (a2 − b2 ) 2 1 − zx and b = z x2 − 1 and no matter what value x may have z can be taken

ZONAL HARMONICS.

166

so small that a2 will be greater than b2 . Then by (1) 1 1

(1 − 2xz + z 2 ) 2

π 1w dφ √ = π 1 − zx − z x2 − 1. cos φ 0 π 1w dφ √ = π 1 − z(x + x2 − 1. cos φ) 0 π w p p 1 = [1 + (x + x2 − 1. cos φ)z + (x + x2 − 1. cos φ)2 z 2 π 0 p + (x + x2 − 1. cos φ)3 z 3 + · · · ]dφ

√ if z is taken so small that the modulus of z(x + x2 − 1. cos φ) is less than 1. But by Art. 77 (2) Pm (x) is the coefficient of z m in the development of 1 1 , (1 − 2xz + z 2 ) 2 hence

Pm (x) =

π p 1w [x + x2 − 1. cos φ]m dφ. π

(2)

0

By replacing φ by π − φ in (2) we get Pm (x) =

π p 1w [x − x2 − 1. cos φ]m dφ. π

(3)

0

1 1 12 and if mod z < 1 or in other words 1 1 (1 − 2xz + 1 − 2x + 2 z z 1 if mod z > 1 can be developed into a convergent series 1 21 1 1 − 2x + 2 z z 1 m 1 involving powers of , and the coefficient of will be Pm (x); but this will z z 1 be the coefficient of z −m−1 in the development of 1 according to (1 − 2xz + z 2 ) 2 descending powers of z, mod z being greater than 1. √ If now we let a = zx − 1 and b = z x2 − 1, a2 − b2 = 1 − 2xz + z 2 and z may be taken so great that a2 − b2 > 0. Then by (1) 1

1 z2) 2

=

1 z

ZONAL HARMONICS.

1 (1 − 2xz + z 2 )

1 2

167

π 1w dφ √ π zx − 1 − z x2 − 1. cos φ 0 wπ dφ h i √ 1 √ 0 z(x − x2 − 1. cos φ) 1 − z(x − x2 − 1. cos φ) wπ 1 1 √ √ z −1 + z −2 2 2 (x − x − 1. cos φ) (x − x − 1. cos φ) 0 1 −3 √ + z + · · · dφ (x − x2 − 1. cos φ)2

=

=

1 π

=

1 π

and the coefficient of z −m−1 is

Hence

Pm (x) =

π 1w dφ √ . π [x − x2 − 1. cos φ]m+1 0 π dφ 1w √ . π [x − x2 − 1. cos φ]m+1 0

(4)

Replace φ by π − φ and we get Pm (x) =

π 1w dφ √ . π [x + x2 − 1. cos φ]m+1 0

(5)

88. In the problems in which we have already used 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 θ for some given value of r, that is on the surface of some sphere, we must develop the function of θ in terms of zonal harmonics of cos θ (v. Art. 10), and our problem becomes the following:—To develop a given function of cos θ in terms of zonal harmonics of cos θ, or to develop a given function of x in terms of the functions Pm (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. Let

f (x) = A0 P0 (x) + A1 P1 (x) + A2 P2 (x) + A3 P3 (x) + · · ·

(1)

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 ∆x between x = −1 and x = 1 so that

ZONAL HARMONICS.

168

(n + 2)∆x = 2; f (−1 + ∆x), f (−1 + 2∆x), f (−1 + 3∆x), · · · f [−1 + (n + 1)∆x] will be the corresponding values of f (x). Substitute these values in (1) and we have  f (−1 + ∆x) = A0 P0 (−1 + ∆x) + A1 P1 (−1 + ∆x)     + A2 P2 (−1 + ∆x) + · · · + An Pn (−1 + ∆x)       f (−1 + 2∆x) = A0 P0 (−1 + 2∆x) + A1 P1 (−1 + 2∆x)    + A2 P2 (−1 + 2∆x) + · · · + An Pn (−1 + 2∆x) (2)   .. .. .. .. .. .. .. ..    . . . . . . . .     f (1 − ∆x) = A0 P0 (1 − ∆x) + A1 P1 (1 − ∆x) + A2 P2 (1 − ∆x) + · · ·     + An Pn (1 − ∆x), that is, n + 1 equations from which in theory the n + 1 coefficients A0 , A1 , · · · An can be determined. Following the analogy of Art. 24 let us multiply the first equation by Pm (−1+ ∆x).∆x, the second by Pm (−1+2∆x).∆x, the third by Pm (−1+3∆x).∆x, &c., and add the equations. The first member of the resulting equation is k=n+1 X

f (−1 + k∆x)Pm (−1 + k∆x).∆x,

(3)

k=1

and the coefficient of any A as Al in the second member is k=n+1 X

Pm (−1 + k∆x)Pl (−1 + k∆x).∆x.

(4)

k=1

If now n is indefinitely increased (3) approaches as its limiting value w1

f (x)Pm (x)dx

(5)

Pm (x)Pl (x)dx.

(6)

−1

and (4) approaches

w1 −1

We have now to find the value of the integral (6) or as we shall write it for the sake of greater convenience w1 −1

Pm (x)Pn (x)dx.

ZONAL HARMONICS. w1

89.

169

Pm (x)Pn (x)dx =

w1 dm (x2 − 1)m dn (x2 − 1)n 1 . dx by 2m+n m!n! dxm dxn −1

−1

(1) Art. 83. m 2 1 w1 dm (x2 − 1)m dn (x2 − 1)n d (x − 1)m dn−1 (x2 − 1)n . dx = . dxm dxn dxm dxn−1

−1

−1

w1 dm+1 (x2 − 1)m dn−1 (x2 − 1)n − . dx dxm+1 dxn−1

(1)

−1

by integration by parts. Now if z = X(x2 − 1)n dz dX = 2nxX(x2 − 1)n−1 + (x2 − 1)n dx dx dX 2 n−1 2 = (x − 1) 2nxX + (x − 1) . dx

(2)

Hence the pth derivative with respect to x of any function of x containing (x2 − 1)n as a factor will contain (x2 − 1)n−p as a factor if p < n. dn−1 (x2 − 1)n , then, contains (x2 − 1) as a factor and is zero when x = 1 dxn−1 and when x = −1, so that (1) reduces to w1 dm+1 (x2 − 1)m dn−1 (x2 − 1)n w1 dm (x2 − 1)m dn (x2 − 1)n . dx = − . dx. dxm dxn dxm+1 dxn−1 −1

−1

It follows that w1 dm (x2 − 1)m dn (x2 − 1)n . dx dxm dxn

−1

= (−1)p

w1 dm+p (x2 − 1)m dn−p (x2 − 1)n . dx dxm+p dxn−p

−1

= (−1)p

w1 dm−p (x2 − 1)m dn+p (x2 − 1)n . dx. (3) dxm−p dxn+p

−1

If m < n we get from (3) w1 dm (x2 − 1)m dn (x2 − 1)n w1 d2m (x2 − 1)m dn−m (x2 − 1)n m . dx = (−1) . dx dxm dxn dx2m dxn−m

−1

−1

1 n−m−1 2 n d (x − 1) = 0, = (−1)m (2m)! dxn−m−1 −1

ZONAL HARMONICS.

170 d2m (x2 − 1)m = (2m)!. dx2m

since If m > n

w1 dm−n (x2 − 1)m d2n (x2 − 1)n w1 dm (x2 − 1)m dn (x2 − 1)n n . dx = (−1) . dx dxm dxn dxm−n dx2n −1

−1

m−n−1 2 1 d (x − 1)m = (−1) (2n)! = 0. dxm−n−1 n

−1

If, then, m is not equal to n w1

Pm (x)Pn (x)dx = 0.

(4)

−1

If m = n we have to find

w1

[Pm (x)]2 dx.

−1

w1

w1 dm (x2 − 1)m dm (x2 − 1)m 1 [Pm (x)] dx = 2m . dx. 2 (m!)2 dxm dxm 2

−1

−1

w1 dm (x2 − 1)m dm (x2 − 1)m w1 d2m (x2 − 1)m m . dx = (−1) .(x2 − 1)m dx dxm dxm dx2m

−1

−1

= (−1)m (2m)!

by (3),

w1

(x2 − 1)m dx.

−1

w1

(x2 − 1)m dx =

−1

w1

(x − 1)m (x + 1)m dx = −

−1

−1

= (−1)m

1 m w (x − 1)m−1 (x + 1)m+1 dx m+1

m! (m + 1)(m + 2) · · · 2m

w1

(x + 1)2m dx

−1

22m+1 m! . = (−1)m (m + 1)(m + 2) · · · (2m + 1)

Hence

w1

[Pm (x)]2 dx =

−1

or

w1 −1

1 22m (m!)2

(−1)m (2m)!(−1)m m!22m+1 (m + 1)(m + 2) · · · (2m + 1)

[Pm (x)]2 dx =

2 . 2m + 1

(5)

ZONAL HARMONICS. 90. we have

171

The solution of the problem in Art. 88 is now readily obtained, and f (x) =A0 P0 (x) + A1 P1 (x) + A2 P2 (x) + · · ·

where

Am

1 2m + 1 w f (x)Pm (x)dx. = 2

(1) (2)

−1

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 f (cos θ) = A0 P0 (cos θ) + A1 P1 (cos θ) + A2 P2 (cos θ) + · · · where

Am =

1 2m + 1 w f (cos θ)Pm (cos θ)d(cos θ) 2 −1

or if f (cos θ) = F (θ)

where

F (θ) =A0 P0 (cos θ) + A1 P1 (cos θ) + A2 P2 (cos θ) + · · · π 2m + 1 w Am = F (θ)Pm (cos θ) sin θ.dθ 2

(3) (4)

0

and the development holds good from θ = 0 to θ = π. 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 w1

f (x)P2k (x)dx = 2

−1

f (x)P2k (x)dx,

0

w1

and that

w1

f (x)P2k+1 (x)dx = 0;

−1

so that if f (−x) = f (x) f (x) = A0 P0 (x) + A2 P2 (x) + A4 P4 (x) + A6 P6 (x) + · · · where

A2k = (4k + 1)

w1

f (x)P2k (x)dx.

(5) (6)

0

If f (x) is an odd function, that is, if f (−x) = −f (x) it can be shown in like manner that f (x) = A1 P1 (x) + A3 P3 (x) + A5 P5 (x) + A7 P7 (x) + · · · where

A2k+1 = (4k + 3)

w1 0

f (x)P2k+1 (x)dx.

(7) (8)

ZONAL HARMONICS.

172

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. 91.

We can establish the fact that

w1

Pm (x)Pn (x)dx = 0 by a more general

−1

method than that used in Art. 89. Let Xm be any solution of Legendre’s Equation dz d (1 − x2 ) + m(m + 1)z = 0 dx dx

[v. (1) Art. 16].

which with its first derivative with respect to x is finite, continuous, and singlevalued for values of x between −1 and 1, −1 and 1 being included. dXm d (1 − x2 ) + m(m + 1)Xm = 0 dx dx d dXn (1 − x2 ) + n(n + 1)Xn = 0 dx dx

Then and

(1) (2)

Multiply (1) by Xn and (2) by Xm and subtract and integrate and we get [m(m + 1) − n(n + 1)]

w1

Xm Xn dx =

−1

w1

Xm

−1

dXn d (1 − x2 ) dx dx dx

−

w1

Xn

−1

d dXm (1 − x2 ) dx. dx dx

Integrate by parts,

[m(m + 1) − n(n + 1)]

w1 −1

−

w1

(1 − x2 )

−1

x=1 2 dXn 2 dXm Xm Xn dx = Xm (1 − x ) − Xn (1 − x ) dx dx x=−1

dXn dXm dx + dx dx w1

Whence

w1

(1 − x2 )

−1

dXm dXn dx. dx dx

Xm Xn dx = 0

(3)

(4)

−1

unless m = n. (3) gives at once the important formula w1 x

h dX dXn i m − Xm (1 − x2 ) Xn dx dx Xm Xn dx = m(m + 1) − n(n + 1)

(5)

ZONAL HARMONICS.

173

from which come as special cases w1 x

h dPm (x) dPn (x) i (1 − x2 ) Pn (x) − Pm (x) dx dx Pm (x)Pn (x)dx = m(m + 1) − n(n + 1)

(6)

and since P0 (x) = 1 w1

dPm (x) dx , m(m + 1)

(1 − x2 ) Pm (x)dx =

x

(7)

unless m = 0. EXAMPLES. 1.

Show that w1

Pm (x)dx = 0

if m is even and is not zero.

0

= (−1)

m−1 2

1 3.5.7. · · · m m(m + 1) 2.4.6. · · · (m − 1)

if m is odd.

v. Art. 91 (7) and Art. 77 (10). 2. w1

Show that

Pm (x)Pn (x)dx = 0

if m and n are both even or both odd.

0

= (−1)

m! n!

m+n+1 2

2m+n−1 (m − n)(m + n + 1)

m 2 n − 1 2 ! ! 2 2

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.

Show that

w1 0

[Pm (x)]2 dx =

1 2m + 1

v. Art. 89 (5).

92. Formula (4) Art. 91 can be obtained directly from Laplace’s Equation by the aid of Green’s Theorem (v. Peirce’s Newt. Pot. Func. § 48). Take the special form of Green’s Theorem, [(148) § 48 Peirce’s Newt. Pot. Func.] y w (U ∇2 V − V ∇2 U )dxdydz = (U Dn V − V Dn U )ds (1) where ∇2 stands for (Dx2 + Dy2 + D22 ), Dn is the partial derivative along the external normal, and the left-hand member is the space-integral through the

ZONAL HARMONICS.

174

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 ∇2 V = ∇2 U = 0 and (1) reduces to w (U Dn V − V Dn U )ds = 0. (2) Now rm Xm and rn Xn are solutions of Laplace’s Equation if x = cos θ (v. Art. 16). If the unit sphere is taken as the bounding surface and U = rm Xm and V = rn Xn (1) and (2) will hold good. Dn U = Dr (rm Xm ) = mrm−1 Xm , Dn V = nrn−1 Xn , ds = sin θ.dθdφ, and (2) becomes

2π w 0

wπ dφ (nXm Xn − mXm Xn ) sin θ.dθ = 0 0

2π(n − m)

or

wπ

Xm Xn sin θ.dθ = 0.

(3)

0

Since x = cos θ, sin θ.dθ = −dx and (3) reduces to w1

Xm Xn dx = 04

(4)

−1

unless m = n. 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 θ in the form A0 P0 (cos θ) + A1 P1 (cos θ) + A2 P2 (cos θ) + · · · given that and

f (θ) = 1 f (θ) = 0

from from

θ=0 θ=

π 2

to θ =

π 2

to θ = π.

This amounts to the same thing as developing F (x) into the series F (x) = A0 P0 (x) + A1 P1 (x) + A2 P2 (x) + A3 P3 (x) + · · · 4 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 rm Xm 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. § 48.)

ZONAL HARMONICS.

175

where

F (x) = 0

from

x = −1

to x = 0

and

F (x) = 1

from

x=0

to x = 1.

By Art. 90 (1) and (2) 1w 1w 1 P0 (x)dx = dx = , 2 2 2 1

A0 =

1

0

0

(2m + 1) w Pm (x)dx. = 2 1

and any coefficient

Am

0

By Art. 91, Ex. 1 w1

Pm (x)dx = 0

0

if m is even 1 3.5.7. · · · m m(m + 1) 2.4.6. · · · (m − 1) if m is even

= (−1) Hence

Am = 0

if m is odd.

2m + 1 1.3.5. · · · (m − 2) . if m is odd. 2m + 2 2.4.6. · · · (m − 1) 1 3 7 1 11 1.3 P5 (x) − · · · F (x) = + P1 (x) − . P3 (x) + . 2 4 8 2 12 2.4 = (−1)

Then

m−1 2

m−1 2

1 3 7 1 11 1.3 5 + rP1 (cos θ) − . r3 P3 (cos θ) + . r P5 (cos θ) + · · · 2 4 8 2 12 2.4 for any point within the sphere. and

u=

(1) (2)

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 symmetry5 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 rDr2 (rV ) +


(1)

subject to the conditions V = f (θ) V =0 We have where

“

r = ∞.

f (θ) = A0 P0 (cos θ) + A1 P1 (cos θ) + A2 P2 (cos θ) + · · · π (2m + 1) w Am = f (θ)Pm (cos θ) sin θ.dθ. v. Art. 90 (4). 2 0

5 See

when r = a

note on page 12.

ZONAL HARMONICS.

176

Hence V = A0 + A1

r a

P1 (cos θ) + A2

r 2 a

P2 (cos θ) + A3

r 3 a

P3 (cos θ) + · · · (2)

is the required solution for a point within the sphere, and V = A0

a r

+ A1

a 2 r

P1 (cos θ) + A2

a 3 r

P2 (cos θ) + A3

a 4 r

P3 (cos θ) + · · · (3)

is the required solution for an external point. EXAMPLES. 1. If on the surface of a sphere of radius c V is constant and equal to a ac for any external show that V = a for any point within the sphere and V = r point. 2. 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. V =

a+b 3r 7 1 r3 + (b − a) P1 (cos θ) − . 3 P3 (cos θ) 2 4c 8 2c 11 1.3 r5 P (cos θ) − · · · + . 5 12 2.4 c5

for an internal point V =

2 a+b c 3c 7 1 c4 . + (b − a) P (cos θ) − . P3 (cos θ) 1 2 r 4 r2 8 2 r4 11 1.3 c6 + . P (cos θ) − · · · 5 12 2.4 r6

for an external point. 3. If V1 = f (cos θ) when r = a and V1 = 0 when r = b show that for a n or if m + n is odd.

Therefore n! (2n + 1) (2n + 1)Pn (x) + (2n − 3) x = Pn−2 (x) 1.3.5 · · · (2n + 1) 2 (2n + 1)(2n − 1) + (2n − 7) Pn−4 (x) 2.4 (2n + 1)(2n − 1)(2n − 3) Pn−6 (x) + · · · + (2n − 11) 2.4.6 n

the second member ending with the term term

3 P1 (x) if n is odd. n+2

(4)

1 P0 (x) if n is even and with the n+1

ZONAL HARMONICS.

180

For convenience of reference we write out a few powers of x. x0 = 1 = P0 (x)

                           

x = P1 (x) 2 1 x2 = P2 (x) + P0 (x) 3 3 3 2 3 x = P3 (x) + P1 (x) 5 5 8 4 1 4 x = P4 (x) + P2 (x) + P0 (x) 35 7 5   8 4 3 5   x = P5 (x) + P3 (x) + P1 (x)   63 9 7     24 10 1 16 6   P6 (x) + P4 (x) + P2 (x) + P0 (x) x =   231 77 21 7     16 8 14 1 7   x = P7 (x) + P5 (x) + P3 (x) + P1 (x)   429 39 33 3     128 64 48 40 1 8 x = P8 (x) + P6 (x) + P4 (x) + P2 (x) + P0 (x). 6435 495 143 99 9

(5)

If a given function of x can be expressed as a terminating power series it can be developed into a Zonal Harmonic Series by the aid of (4). Given that f (x) = a0 + a1 x + a2 x2 + a3 x3 + · · · , f (x) = B0 + B1 P1 (x) + B2 P2 (x) + B3 P3 (x) + · · · ;

let

then picking out carefully the coefficient of Pm (x) we have Bm

(m + 1)(m + 2) m! am+2 am + = 1.3.5. · · · (2m − 1) 2.(2m + 3) (m + 1)(m + 2)(m + 3)(m + 4) + am+4 + · · · . (6) 2.4.(2m + 3)(2m + 5)

96.

The development of

dPn (x) is useful and is easily obtained. dx

dPn (x) =A0 P0 (x) + A1 P1 (x) + A2 P2 (x) + · · · dx 1 2m + 1 w dPn (x) Am = Pm (x) dx 2 dx

Let Then

(1)

−1

by Art. 90 (2); w1 −1

Pm (x)

h ix=1 w1 dPn (x) dPm (x) dx = Pm (x)Pn (x) − Pn (x) dx. dx dx x=−1 −1

(2)

ZONAL HARMONICS. h

181

ix=1 =0 Pm (x)Pn (x)

if m + n is even

=2

if m + n is odd.

x=−1

dPn (x) is an dx algebraic polynomial of the n − 1st degree in x. Therefore in (1) m is less than dPm (x) n; consequently is an algebraic polynomial in x of lower degree than n dx and

Since Pn (x) is an algebraic polynomial of the nth degree in x,

w1

Pn (x)

−1

We get then

Am = 2m + 1 =0

dPm (x) dx = 0 dx

by Art. 95 (3).

if m + n is odd and m < n,

if m + n is even or m > n − 1;

and

dPn (x) = (2n − 1)Pn−1 (x) + (2n − 5)Pn−3 (x) + (2n − 9)Pn−5 (x) + · · · (3) dx the second member ending with the term 3P1 (x) if n is even and with the term P0 (x) if n is odd. From (3) a number of simple formulas are readily obtained. For example dPn+1 (x) dPn−1 (x) − = (2n + 1)Pn (x) dx dx w1 1 [Pn−1 (x) − Pn+1 (x)]. Pn (x)dx = 2n +1 x (2n + 1)x

dPn (x) dPn+1 (x) dPn−1 (x) =n + (n + 1) dx dx dx

(4) (5) (6)

[v. (4) and Article 77 (12)]. (x2 − 1)

dPn (x) = nxPn (x) − nPn−1 (x) dx

(7)

[v. (5) and Article 91 (7).] 97. By the aid of the formulas of Art. 96 a number of valuable developments can be obtained. Let us get cos nθ and sin nθ, n being any positive real. z = cos nθ and z = sin nθ are solutions of the equation d2 z + n2 z = 0 dθ2

ZONAL HARMONICS.

182

or if we let x = cos θ, of the equation (1 − x2 )

dz d2 z −x + n2 z = 0. dx2 dx

(1)

a0 P0 (x) + a1 P1 (x) + a2 P2 (x) + · · ·

Let

be the required development of cos nθ or of sin nθ. m=∞ X dPm (x) d2 Pm (x) 2 − x + n Then am (1 − x2 ) P (x) =0 m dx2 dx m=0

by (1).

z = Pm (x) is a solution of Legendre’s Equation (v. Art. 77). Hence (1 − x2 )

dPm (x) dPm (x) d2 Pm (x) −x =x − m(m + 1)Pm (x), 2 dx dx dx

and (1) becomes m=∞ X m=0

dPm (x) 2 + [n − m(m + 1)]Pm (x) = 0. am x dx

Formulas (4) and (6) of Art. 96 enable us to throw (2) into the form 2 m=∞ X n − m2 dPm+1 (x) n2 − (m + 1)2 dPm−1 (x) am − = 0. 2m + 1 dx 2m + 1 dx m=0 (3) must be identically true. Therefore the coefficient of equal zero, and we have am+2 =

n2 − m2 2m + 5 . 2 am . 2m + 1 n − (m + 3)2

(2)

(3)

dPm+1 (x) must dx (4)

If we are developing cos nθ 1w cos nθ sin θ.dθ 2 π

a0 = =

and

1 4

0 wπ

by Art. 90 (4),

[sin(n + 1)θ − sin(n − 1)θ]dθ,

0

1 1 + cos nπ a0 = − . ; 2 n2 − 1 π 3w cos nθ cos θ sin θ.dθ a1 = 2

(5) by Art. 90 (4),

0

3 1 − cos nπ a1 = − . . 2 n2 − 4

(6)

ZONAL HARMONICS.

183

(4), (5), and (6) give us 1 + cos nπ h n2 P (cos θ) + 5 P2 (cos θ) 0 2(n2 − 1) n2 − 3 2 i n2 (n2 − 22 ) +9 2 P (cos θ) + · · · 4 (n − 32 )(n2 − 52 ) h n2 − 12 1 − cos nπ 3P1 (cos θ) + 7 2 P3 (cos θ) − 2 2 2(n − 2 ) n − 42 i (n2 − 12 )(n2 − 32 ) + 11 2 P (cos θ) + · · · . 5 (n − 42 )(n2 − 62 )

cos nθ = −

(7)

If n is a whole number 1 + cos nπ or 1 − cos nπ will vanish and the series will end with the term involving Pn (cos θ). For this case (7) may be rewritten cos nθ =

1 2.4.6. · · · 2n h . (2n + 1)Pn (cos θ) 2 3.5.7. · · · (2n + 1) n2 − (n + 1)2 Pn−2 (cos θ) + (2n − 3) 2 n − (n − 2)2 i [n2 − (n + 1)2 ][n2 − (n − 1)2 ] P (cos θ) + · · · . + (2n − 7) 2 n−4 [n − (n − 2)2 ][n2 − (n − 4)2 ]

(8)

If we are developing sin nθ 1w 1 sin nπ sin nθ sin θ.dθ = − . 2 , 2 2 n −1 π

a0 =

0

a1 =

π 3w

2

0

sin nθ cos θ sin θ.dθ =

3 sin nπ . 2 n2 − 22

n2 1 sin nπ h sin nθ = − . 2 P0 (cos θ) + 5 2 P2 (cos θ) 2 n −1 n − 32 i n2 (n2 − 22 ) +9 2 P (cos θ) + · · · 4 (n − 32 )(n2 − 52 ) h 1 sin nπ n2 − 12 + . 2 3P (cos θ) + 7 P3 (cos θ) 1 2 n − 22 n2 − 42 i (n2 − 12 )(n2 − 32 ) P5 (cos θ) + · · · . + 11 2 2 2 2 (n − 4 )(n − 6 )

and

(9)

If n is a whole number sin nπ = 0, and all the terms of (9) vanish except those involving Pn−1 (cos θ), Pn+1 (cos θ), Pn+3 (cos θ) &c., which become indeterminate. For this case it is necessary to compute an−1 independently.

ZONAL HARMONICS.

184

We have 2n − 1 w sin nθPn−1 (cos θ) sin θ.dθ 2 π

an−1 =

0

=

Hence

π 2n − 1 w

4

an−1 =

[cos(n − 1)θ − cos(n + 1)θ]Pn−1 (cos θ)dθ.

0

2n − 1 1.3.5. · · · (2n − 3) . π 4 2.4.6. · · · (2n − 2)

[v. Art. 82 (1)],

and π 1.3. · · · (2n − 3) h (2n − 1)Pn−1 (cos θ) sin nθ = . 4 2.4. · · · (2n − 2) n2 − (n − 1)2 + (2n + 3) 2 Pn+1 (cos θ) n − (n + 2)2 i [n2 − (n − 1)2 ][n2 − (n + 1)2 ] + (2n + 7) 2 P (cos θ) + · · · . (10) n+3 [n − (n + 2)2 ][n2 − (n + 4)2 ]

EXAMPLES. 1.

Show that 1.3 2 1.3.5 2 i 1 2 πh csc θ = P2 (cos θ) + 9 P4 (cos θ) + 13 P6 (cos θ) + · · · 1+5 2 2 2.4 2.4.6 whence √

1 2 1.3 2 1.3.5 2 i 1 πh = 1+5 P2 (x) + 9 P4 (x) + 13 P6 (x) + · · · 2 2 2.4 2.4.6 1 − x2

[v. Art. 90 (4) and Art. 82]. 2.

Show that 3 1 2 5 1.3 2 i π h 1 3 P1 (cos θ) + 7 P3 (cos θ) + 11 P5 (cos θ) + · · · ctn θ = 2 2 4 2 6 2.4 whence √

3 1 2 5 1.3 2 i 1 πh 1 = 3 P1 (x) + 7 P3 (x) + 11 P5 (x) + · · · 2 2 4 2 6 2.4 1 − x2

[v. Art. 90 (4) and Art. 82]. 3. By integrating the result of Ex. 1 and simplifying by the aid of Art. 96 (5), obtain the development

ZONAL HARMONICS.

185

1 2 π h 1 2 3 P1 (x) + 7 P3 (x) 2 2 2.4 1.3.5 2 i 1.3 2 P5 (x) + 15 P7 (x) + · · · + 11 2.4.6 2.4.6.8 h 1 2 π 1 2 θ= P0 (cos θ) − 3 P1 (cos θ) − 7 P3 (cos θ) 2 2 2.4 1.3 2 i −11 P5 (cos θ) − · · · . 2.4.6

sin−1 x =

whence

4. By integrating the result of Ex. 2 and simplifying by the aid of Art. 96 (5) obtain p

1 − x2 =

3 1 2 1 1 2 πh1 P2 (x) − 9 P4 (x) −5 2 2 4 2 6 2.4 5 1.3 2 i −13 P6 (x) + · · · 8 2.4.6

whence sin θ =

3 1 2 i 1 1 2 πh1 P2 (cos θ) − 9 P4 (cos θ) − · · · . P0 (cos θ) − 5 2 2 4 2 6 2.4

To make clearer the analogy of development in Zonal Harmonic Series with development in Fourier’s Series we give on page 186 a cut representing the first seven Surface Zonal Harmonics P1 (cos θ), P2 (cos θ), · · · P7 (cos θ), which are of course somewhat complicated Trigonometric curves resembling roughly cos θ, cos 2θ, · · · cos 7θ; and on page 187, the first four successive approximations to the Zonal Harmonic Series 1 3 7 1 11 1.3 + P1 (cos θ) − . P3 (cos θ) + . P5 (cos θ) − · · · 2 4 8 2 12 2.4

[I]

[v. (1) Art. 93], and 1 2 1 2 πh P0 (cos θ) − 3 P1 (cos θ) − 7 P3 (cos θ) 2 2 2.4 1.3 2 i − 11 P5 (cos θ) − · · · 2.4.6

[II]

(v. Ex. 3 Art. 97). π π [I] is equal to 1 from θ = 0 to θ = , and to 0 from θ = to θ = π; and [II] 2 2 is equal to θ from θ = 0 to θ = π. The figures on page 187 are constructed on precisely the same principle as those on pages 64 and 65, with which they should be carefully compared.

ZONAL HARMONICS.

186

The curves y = P0 (cos θ), y = P1 (cos θ), . . . y = P7 (cos θ).

(v. page 185.)

98. By applying Gauss’s Theorem (B. O. Peirce, Newt. Pot. Func. § 31) or the special Form of Green’s Theorem, y w y ∇2 V dxdydz = Dn V ds = −4π ρdxdydz, [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 4πρκ = Dn V1 − Dn V2 .

(1)

where ρ is the density and κ the thickness of the shell, V1 the value of the potential function due to the shell at an internal point and V2 its value at an external point, and where Dn 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 ρκ in (1) by σ where σ is the surface density, whence 4πσ = Dn V1 − Dn V2 (v. Peirce, N. P. F. §§ 45, 46, and 47).

(2)

ZONAL HARMONICS.

187

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 ρ and radius a. Since V must be a solution of Laplace’s Equation and must be finite both

ZONAL HARMONICS.

188

when r = 0 and r = ∞ we have X V1 = Am rm Pm (cos θ) X 1 V2 = Bm m+1 Pm (cos θ). r V1 and V2 must approach the same limiting values as r approaches a. Hence Bm = Am am am+1 Bm = Am a2m+1 . X Dn V1 = Dr V1 = mrm−1 Am Pm (cos θ), X Am a2m+1 Pm (cos θ). Dn V2 = Dr V2 = − (m + 1) rm+2

or

Therefore by (1) 4πρκ =

X (2m + 1)Am am−1 Pm (cos θ)

if κ is the thickness of the shell. Let

ρ = f (cos θ) =

where

Cm =

X

Cm Pm (cos θ)

1 2m + 1 w f (x)Pm (x)dx 2

by Art. 90 (2).

−1

Then Am and and

4πκCm = (2m + 1)Am am−1 , 4πκCm 4πκ = , and Bm = Cm am+2 , m−1 (2m + 1)a 2m + 1 X Cm rm V1 = 4πaκ Pm (cos θ), 2m + 1 am X Cm am+1 V2 = 4πaκ Pm (cos θ). 2m + 1 rm+1

and

(3) (4)

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 Crk Pm (cos θ). Let a be the radius of the outer surface and b be the radius of the inner surface of the shell. 1st.—Let ρ = Crk Pm (cos θ). Then for the shell of radius s and thickness ds Csk rm Pm (cos θ) 2m + 1 sm Csk sm+1 V2 = 4πsds Pm (cos θ) 2m + 1 rm+1 V1 = 4πsds

and

by (3) Art. 98, by (4) Art. 98.

ZONAL HARMONICS.

189

Then if r < b V =

wa

V1 =

b

if r > a V =

wa

4πC (ak−m+2 − bk−m+2 ) m r Pm (cos θ), (2m + 1) (k − m + 2)

(1)

4πC (ak+m+3 − bk+m+3 ) Pm (cos θ) , (2m + 1) (k + m + 3) rm+1

(2)

V2 =

b

and if b < r < a V =

wr b

V2 +

wa r

V1 =

4πC rk+m+3 − bk+m+3 2m + 1 (k + m + 3)rm+1 +

ak−m+2 − rk−m+2 m r Pm (cos θ). (3) (k − m + 2)

X 2d.—If ρ = Cm rk Pm (cos θ) the solutions will consist of sums of terms of the forms given in (1), (2), and (3). EXAMPLES. 1.

If the shell is homogeneous V = 2πρ(a2 − b2 ) if r < b, 1 M 4 if r > a, V = πρ(a3 − b3 ) = 3 r r 2b3 r2 V = 2πρ a2 − − if b < r < a. 3r 3

2. If the density is any given function of the distance from the centre M V = if r > a, and V = a constant if r < b. r 3. If the density at any point of a solid sphere is proportional to the square of the distance from a diametral plane M a 2 a3 + P2 (cos θ) if r > a. V = a r 7 r3 4. If the density at any point of a solid sphere is proportional to its distance from a diametral plane M a 1 a3 1.1 a5 1.1.3 a7 + V = P (cos θ) − P (cos θ) + P (cos θ) − · · · 2 4 6 a r 6 r3 6.8 r5 6.8.10 r7 if r > a. Compare Ex. 2 Art. 80.

ZONAL HARMONICS.

190

100.

We have seen in Art. 18 (c) (3) that w dx , (1) Qm (x) = CPm (x) (1 − x2 )[Pm (x)]2 w dx no constant term being understood with . 2 (1 − x )[Pm (x)]2 1 is a rational fraction and becomes infinite only for x = 1, (1 − x2 )[Pm (x)]2 x = −1, and for the roots of Pm (x) = 0, all of which are real and lie between −1 1 dm (x2 − 1)m and 1, as can be proved by the aid of the relation Pm (x) = m . 2 m! dxm w∞ dx is finite and determinate and contains no If x2 > 1 2 (1 − x )[Pm (x)]2 x constant term. Hence if x2 > 1 w∞ w∞ dx dx Qm (x) = −Pm (x) = P (x) m 2 2 2 (1 − x )[Pm (x)] (x − 1)[Pm (x)]2 x x

(2)

for the constant factor of Qm (x) has been chosen so that C = −1. If x2 < 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 1 1+x log 2 1−x 1+x x Q1 (x) = −1 + log 2 1−x Q0 (x) =

if x2 < 1. (2) gives us

(3) (4)

x+1 1 log 2 x−1 x x+1 Q1 (x) = −1 + log 2 x−1

(5)

Q0 (x) =

(6)

if x2 > 1. From Art. 85 (10) it follows that wx dm dx if x2 < 1, Qm (x) = C m (x2 − 1)m dx (x2 − 1)m+1 0 ∞ w m d dx 2 m = C m (x − 1) if x2 > 1. 2 − 1)m+1 dx (x x C can be determined and is equal to

(−1)m+1 2m m! if x2 < 1, and is equal to (2m)!

(−1)m 2m m! if x2 > 1. (2m)! Hence

Qm (x) =

wx (−1)m+1 2m m! dm dx 2 m (x − 1) (2m)! dxm (x2 − 1)m+1 0

(7)

ZONAL HARMONICS.

191

if x2 < 1, and

w∞ (−1)m 2m m! dm dx 2 m Qm (x) = (x − 1) (2m)! dxm (x2 − 1)m+1 x

(8)

if x2 > 1. (7) and (8) give us for Q0 (x) and Q1 (x) the values already written in (3), (4), (5), and (6). By the repeated application of the formula (m + 1)Qm+1 (x) − (2m + 1)xQm (x) + mQm−1 (x) = 0,

(9)

which may be obtained for the case where x2 < 1 from Art. 16 (13) and (14), and for the case where x2 > 1 from Art. 16 (9), any Surface Zonal Harmonic of the Second Kind can be obtained from Q0 (x) and Q1 (x) as given in (3), (4), (5), and (6). Analogous formulas for pm (x) and qm (x) can be obtained without difficulty from Art. 16 (4) and (5). They are

and

(m + 1)2 qm+1 (x) − (2m + 1)xpm (x) − m2 qm−1 (x) = 0

(10)

pm+1 (x)+(2m + 1)xqm (x) − pm−1 (x) = 0

(11)

and they hold good for any value of m. EXAMPLES. 1. Confirm the values of Q0 (x) and Q1 (x) given in Art. 100 (3), (4), (5), and (6) by expanding them and comparing them with Art. 16 (13), (14), and (9). 2. 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. X Bm If V = Am rm + m+1 when θ = α then r X Bm Pm (cos θ) m V = Am r + m+1 . r Pm (cos α) X Bm m 3. If V = Am r + m+1 when θ = α, and V = 0 when θ = β, r X Bm Qm (cos β)Pm (cos θ) − Pm (cos β)Qm (cos θ) V = Am rm + m+1 . r Pm (cos α)Qm (cos β) − Pm (cos β)Qm (cos α) 4. Find V for points corresponding to values of θ between α and β when V can be given in terms of whole powers of r for θ = α and for θ = β.

ZONAL HARMONICS. 5. form

192

Find by the method of Art. 16 solutions of Legendre’s Equation of the

m(m + 1) (m − 1)m(m + 1)(m + 2) (x − 1)2 (x − 1) + 2 22 (2!)2 (m − 2)(m − 1)m(m + 1)(m + 2)(m + 3) + (x − 1)3 + · · · , 23 (3!)2 m(m + 1) (m − 1)m(m + 1)(m + 2) (x + 1)2 z = −1 Pm (x) = 1 − (x + 1) + 2 22 (2!)2 (m − 2)(m − 1)m(m + 1)(m + 2)(m + 3) + (x + 1)3 + · · · . 23 (3!)2 z = 1 Pm (x) = 1 +

If m is a whole number, 1 Pm (x) = Pm (x) and −1 Pm (x) = (−1)m Pm (x). No matter what the value of m, 1 Pm (x) is absolutely convergent for −1 < x < 3, and −1 Pm (x) is absolutely convergent for −3 < x < 1. 6.

By the aid of (7) Art. 16 show that 1 V = √ sin(n log r)kn (cos θ), r

1 V = √ sin(n log r)ln (cos θ), r

1 V = √ cos(n log r)kn (cos θ), r

1 V = √ cos(n log r)ln (cos θ), r

are solutions of Laplace’s Equation 1 Dθ (sin θDθ V ) = 0, if sin θ 1 2 h 1 2 ih 5 2 i n2 + n2 + n2 + 2 2 2 2 kn (x) = p− 12 +ni (x) = 1 + x + x4 2! 4! h 1 2 ih 5 2 ih 9 2 i n2 + n2 + n2 + 2 2 2 x6 + · · · + 6! 3 2 h 3 2 ih 7 2 i and n2 + n2 + n2 + 2 x3 + 2 2 ln (x) = −q− 21 +ni (x) = x + x5 3! 5! 7 2 ih 11 2 i h 3 2 ih n2 + n2 + n2 + 2 2 2 x7 + · · · + 7! kn (x) and ln (x) are convergent if x2 < 1, but are divergent if x2 = 1. rDr2 (rV ) +

7.

Show by the aid of Example 5 that 1 V = √ sin(n log r)Kn (cos θ), r 1 V = √ cos(n log r)Kn (cos θ), r

1 V = √ sin(n log r)Kn (− cos θ), r 1 V = √ cos(n log r)Kn (− cos θ), r

ZONAL HARMONICS.

193

rDr2 (rV ) +

are solutions of

1 Dθ (sin θDθ V ) = 0 sin θ n2 +

Kn (x) =1 P− 12 +ni (x) = 1 −

if

h

n2 +

1 2 ih

2 2

n2 +

(x − 1)

3 2 i

2 (x − 1)2 22 (2!)2 h 1 2 ih 3 2 ih 5 2 i n2 + n2 + n2 + 2 2 2 (x − 1)3 + · · · − 22 (3!)2

+

2

1 2

and n2 +

1 2

2 (x + 1) Kn (−x) =−1 P− 12 +ni (x) = 1 + 2 h 1 2 ih 3 2 i n2 + n2 + 2 2 + (x + 1)2 22 (2!)2 h 1 2 ih 3 2 ih 5 2 i n2 + n2 + n2 + 2 2 2 (x + 1)3 + · · · . + 23 (3!)2 Kn (cos θ) is convergent except for θ = π, and Kn (− cos θ) is convergent except for θ = 0. kn (x), ln (x), Kn (x), and Kn (−x) are sometimes called Conal Harmonics. They are particular values of z which satisfy Legendre’s Equation written in the form d2 z dz 2 1 (1 − x2 ) 2 − 2x − n + z = 0. dx dx 4 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.

If V = f (r) when θ = β,

V = 9.

1 √

w∞

π r −∞

dλ

w∞ 0

λ

e 2 f (eλ )

Kα (cos θ) cos[α(λ − log r)]dα; Kα (cos β)

if θ < β.

If V = f (r) when θ = β and r < a, and V = 0 when r = a,

2 V = π

r

0 w∞ λ a w Kα (cos θ) r dλ e 2 f (aeλ ) sin αλ sin α log dα; r −∞ Kα (cos β) a 0

if θ < β.

ZONAL HARMONICS.

194

10. If V = f (r) when θ = β and a < r < b, and V = 0 when r = a and when r = b, V =

m=∞ X

Am

m=1

where

Km0 (cos θ) mπ(log r − log a) sin Km0 (cos β) log b − log a mπ m0 = log b − log a

and

b

Am

2 = log b − log a

r

log a x a w mπx e 2 f (aex )sin dx; r log b − log a

if θ < β.

0

11.

If θ > β cos θ must be replaced by (− cos θ) in examples 8, 9, and 10.

12.

If V = f (r) when θ = β, and V = 0 when θ = γ,

V =

w∞

1 √

π r −∞

dλ

w∞

λ

e 2 f (eλ )

0

kα (cos θ)lα (cos γ) − kα (cos γ)lα (cos θ) cos[α(λ − log r)]dα; kα (cos β)lα (cos γ) − kα (cos γ)lα (cos β) if β < θ < γ. 13. If V = f (r) when θ = β and a < r < b, V = 0 when θ = γ and a < r < b, and V = 0 when r = a and when r = b, V =

m=∞ X m=1

where

Am

mπ(log r − log a) km0 (cos θ)lm0 (cos γ) − km0 (cos γ)lm0 (cos θ) sin , 0 0 0 0 km (cos β)lm (cos γ) − km (cos γ)lm (cos β) log b − log a mπ m0 = and log b − log a

Am =

2 log b − log a

r

a r

b log a w 0

x

e 2 f (aex ) sin

mπx dx; log b − log a

if β < θ < γ and a < r < b. 14. If V = f (r) when θ = β and a < r < b, and V = 0 when r = a and Dr V + hV = 0 when r = b, V =

m=∞ X m=1

Am

Kαm (cos θ) r sin αm log , Kαm (cos β) a b

Am

where

log a w 2 x 2(αm + h2 b2 ) = 2 e 2 f (aex ) sin αm x.dx αm (log b − log a) + hb[hb(log b − log a) + 1] 0

ZONAL HARMONICS. and αm is a root of the equation b b α cos α log + hb sin α log =0 a a

195

v. Art. 68 Ex. 5.

196

CHAPTER VI. SPHERICAL HARMONICS. 101. When we are dealing with problems in finding the potential function due to forces which have not circular symmetry1 about an axis and are using Spherical Co¨ ordinates, we have to solve Laplace’s Equation in the form rDr2 (rV ) +

1 1 Dφ2 V = 0 Dθ (sin θDθ V ) + sin θ sin2 θ

(1)

[v. (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.Θ.Φ; where R involves r only, Θ involves θ only, and Φ φ only. Substitute in (1) and we get dΘ d sin θ 2 1 1 d2 Φ r d (rR) dθ + + =0 (2) 2 2 R dr Θ sin θ dθ Φ sin θ dφ2 dΘ 2 d sin θ 2 2 r sin θ d (rR) sin θ dθ = − 1 d Φ . or + R dr2 Θ dθ Φ dφ2 As the first member does not contain φ the second member cannot contain φ, and as it contains no other variable it must be constant; call it n2 . Equation (2) is then equivalent to the two equations

and

d2 Φ + n2 Φ = 0 dφ2 h dΘ i d sin θ 2 2 r d (rR) 1 dθ − n = 0 + 2 R dr2 Θ sin θ dθ sin θ

(3)

(4)

(3) has been solved before and gives us Φ = A cos nφ + B sin nφ

(5)

[v. Art. 13(a)]. The first term of (4) does not involve θ and the second and third terms do not involve r. r d2 (rR) must, then, be a constant; we shall call it m(m+1) as in Art. 13(c). R dr2 Then (4) breaks up into r 1 See

note, page 12.

d2 (rR) = m(m + 1)R dr2

(6)

SPHERICAL HARMONICS.

and

197

h dΘ i d sin θ 2 1 dθ + m(m + 1) − n Θ = 0. sin θ dθ sin2 θ

(7)

(6) was solved in Art. 13(c) and gives R = A1 rm + B1 r−m−1 . If in (7) we replace cos θ by µ we get d n2 2 dΘ Θ = 0, (1 − µ ) + m(m + 1) − dµ dµ 1 − µ2

(8)

(9)

the equivalent of (1 − x2 )

d2 z dz n2 − 2x + m(m + 1) − z = 0, dx2 dx 1 − x2

(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) n

Θ = (1 − µ2 ) 2

dn Pm (µ) dn Pm (µ) = sinn θ , n dµ dµn

(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 n

Θ = (1 − µ2 ) 2

dn Qm (µ) . dµn

Combining our results we have as important particular solutions of (1) dn Pm (µ) , dµn 1 dn Pm (µ) V = m+1 (A cos nφ + B sin nφ) sinn θ , r dµn V = rm (A cos nφ + B sin nφ) sinn θ

and

(12) (13)

where m and n are positive integers and n < m + 1. n dn Pm (µ) dn Pm (µ) or (1 − µ2 ) 2 is a new function of µ, that n dµ dµn n is of cos θ, and we shall represent it by Pm (µ)2 and shall call it an associated function of the nth order and mth degree. It is a value of Θ satisfying equation (9) Art. 101.

102.

2 Most

sinn θ

n (µ). of the English writers represent this function by Tm


198

By differentiating the value of Pm (x) given in (9) Art. 74 we get the formula n Pm (µ)

(2m)! sinn θ (m − n)(m − n − 1) m−n−2 = m µm−n − µ 2 m!(m − n)! 2.(2m − 1) (m − n)(m − n − 1)(m − n − 2)(m − n − 3) m−n−4 + µ − ··· (1) 2.4.(2m − 1)(2m − 3)

the expression in the parenthesis ending with the term involving µ0 if m − n is even and with the term involving µ if m − n is odd. For convenience of reference we give on the next page a table from which n Pm (µ) can be readily obtained for values of m and n from 1 to 8. n n cos nφPm (µ) and sin nφPm (µ), that is, cos nφ sinn θ

dn Pm (µ) dµn

and

sin nφ sinn θ

dn Pm (µ) dµn

are called Tesseral Harmonics of the mth degree and nth order, and are values of V which satisfy the equation m(m + 1)V +

1 1 Dθ (sin θDθ V ) + Dφ2 V = 0 sin θ sin2 θ

(2)

1 D2 V = 0. 1 − µ2 φ

(3)

or its equivalent m(m + 1)V + Dµ [(1 − µ2 )Dµ V ] +

There are obviously 2m + 1 Tesseral Harmonics of the mth degree, namely dPm (µ) dµ d2 Pm (µ) sin 2φ sin2 θ dµ2 3 d Pm (µ) sin 3φ sin3 θ dµ3

dPm (µ) , dµ d2 Pm (µ) cos 2φ sin2 θ , dµ2 d3 Pm (µ) cos 3φ sin3 θ , dµ3

Pm (µ),

sin φ sin θ

cos φ sin θ

.

.

.

cos mφ sinm θ

.

.

.

dm Pm (µ) , dµm

.

.

.

.

.

sin mφ sinm θ

.

.

dm Pm (µ) dµm

If each of these is multiplied by a constant and their sum taken, this sum is called a Surface Spherical Harmonic of the mth degree, and is a solution of equations (2) and (3). We shall represent it by Ym (µ, φ) or by Ym (θ, φ).


199

n Table for cscn θPm (µ) =

dn Pm (µ) . dµn

m

n = 1.

1

1

2

3µ

3

3

3 (5µ2 − 1) 2

15µ

4

5 (7µ3 − 3µ) 2

15 (7µ2 − 1) 2

5

15 (21µ4 − 14µ2 + 1) 8

105 (3µ3 − µ) 2

6

21 (33µ5 − 30µ3 + 5µ) 8

105 (33µ4 − 18µ2 + 1 8

7

7 (429µ6 − 495µ4 + 135µ2 − 5) 16

63 (143µ5 − 110µ3 + 15µ) 8

8

9 (715µ7 − 1001µ5 + 385µ3 − 35µ) 16

315 (143µ6 − 143µ4 + 33µ2 − 1) 16

m

n = 3.

n = 2.

n = 4.

1 2 3

15

4

105µ

105

5

105 (9µ2 − 1) 2

945µ

6

315 (11µ3 − 3µ) 2

945 (11µ2 − 1) 2

7

315 (143µ4 − 66µ2 + 3) 8

3465 (13µ3 − 3µ) 2

8

3465 (39µ5 − 26µ3 + 3µ) 8

10395 (65µ4 − 26µ2 + 1) 8


m

200

n = 5.

n = 6.

n = 7.

n = 8.

1 2 3 4 5

945

6

10395µ

10395

7

10395 (13µ2 − 1) 2

135135µ

135135

8

135135 (5µ3 − µ) 2

135135 (15µ2 − 1) 2

2027025µ

2027025

1 Ym (µ, φ) are called Solid Spherical Harmonics of the rm+1 mth degree, and are solutions of Laplace’s Equation (1) Art. 101. To formulate:— n=m i n Xh dn Pm (µ) n d Pm (µ) + B sin nφ sin θ Ym (µ, φ) = An cos nφ sinn θ (4) n dµn dµn n=0 n=m X n n or Ym (µ, φ) = A0 Pm (µ) + [An cos nφPm (µ) + Bn sin nφPm (µ)] (5) rm Ym (µ, φ) and

n=1

is a Surface Spherical Harmonic of the mth degree. A Tesseral Harmonic is a special case of a Surface Spherical Harmonic, and a Zonal Harmonic a special case of a Tesseral Harmonic; Pm (µ) being the Tesseral 0 Harmonic of the zeroth order and the mth degree; it might be written Pm (µ). EXAMPLES. 1.

Show that dz n2 d2 z + m(m + 1) − z=0 (1 − x ) 2 − 2x dx dx 1 − x2 2

reduces to (1 − x2 )

d2 y dy − 2(n + 1)x + [m(m + 1) − n(n + 1)]y = 0 dx2 dx n

if we substitute (1 − x2 ) 2 y for z, even when m and n are unrestricted.

SPHERICAL HARMONICS. 2.

Show that if in the second equation of Ex. 1 we let y = ak+2 = −

(m − n − k)(m + n + 1 + k) ak (k + 1)(k + 2)

201 P

ak xk we get (v. Art. 16)

n whence z = pnm (x) and z = qm (x) are solutions of the first equation of Ex. 1, no matter what the values of m and n, if

n (m − n)(m + n + 1) 2 x pnm (x) = (1 − x2 ) 2 1 − 2! (m − n)(m − n − 2)(m + n + 1)(m + n + 2) 4 + x − ··· 4! and n (m − n − 1)(m + n + 2) 3 n qm (x) = (1 − x2 ) 2 x − x 3! (m − n − 1)(m − n − 3)(m + n + 2)(m + n + 4) 5 x − ··· . + 5! n (x) will terminate with the term If m − n is a positive integer, pnm (x) or qm m−n involving x , and in that case

n (m − n)(m − n − 1) m−n−2 z = (1 − x2 ) 2 xm−n − x 2.(2m − 1) (m − n)(m − n − 1)(m − n − 2)(m − n − 3) m−n−4 + x − ··· . 2.4.(2m − 1)(2m − 3) the parenthesis ending with a term involving x0 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 2m m!(m − n)! n Pm (x). value of z is (2m)! 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 θ, that is of µ, in terms of Zonal Harmonics of µ. So it is often desirable to express a given function of µ and φ in terms of Tesseral Harmonics of µ and φ. If, for example, we are trying to find the 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 θ and φ and if we can express it in terms of Spherical Harmonics of θ and φ 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 θ and φ on the surface of the given sphere.


202

104. Suppose that we have a function of µ and φ given for all points on the unit sphere, that is, for all values of µ from −1 to 1 and for all values of φ from 0 to 2π, µ and φ being independent variables, and that we wish to express it in terms of Surface Spherical Harmonics. Assume that m=∞ n=m i Xh X n n f (µ, φ) = A0,m Pm (µ)+ An,m cos nφPm (µ) + Bn,m sin nφPm (µ) . (1) m=0

n=1

Let us consider first a finite case, and attempt to determine the coefficients so that m=p n=m i Xh X n n f (µ, φ) = A0,m Pm (µ)+ An,m cos nφPm (µ) + Bn,m sin nφPm (µ) (2) m=0

n=1

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 mth 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 µ = −1 to µ = 1 be divided into p + 2 parts each of which is ∆µ so that (p + 2)∆µ = 2, and let the interval from φ = 0 to φ = 2π be divided into p + 2 parts each of which is ∆φ so that (p + 2)∆φ = 2π. Then if we substitute in equation (2) in turn the values (−1 + ∆µ, ∆φ), (−1 + 2∆µ, ∆φ), · · · [−1 + (p + 1)∆µ, ∆φ]; (−1 + ∆µ, 2∆φ), (−1 + 2∆µ, 2∆φ), · · · [−1 + (p + 1)∆µ, 2∆φ]; · · · [−1 + ∆µ, (p + 1)∆φ], [−1 + 2∆µ, (p + 1)∆φ], · · · [−1+(p+1)∆µ, (p+1)∆φ]; 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 ∆µ∆φ 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 dµdφ = − sin θ.dθdφ is the element of surface of the unit sphere, and as the limits −1 and 1 of µ correspond to π and 0 of θ the integration is a surface integration over the surface of the unit sphere. In determining any coefficient as An,m in (1) the first member of the limiting form of our resulting equation will be 2π w 0

dφ

w1 −1

n f (µ, φ) cos nφPm (µ)dµ.


203

In the second member we shall come across terms of the forms 2π w 0

dφ

w1

n sin lφ cos nφPml (µ)Pm (µ)dµ,

−1 2π w

2π w

dφ

dφ

n sin nφ cos nφ[Pm (µ)]2 dµ,

−1

0

n cos lφ cos nφPml (µ)Pm (µ)dµ,

−1

0

w1

w1

2π w

dφ

w1

n cos2 nφ[Pm (µ)]2 dµ,

−1

0

and other terms all of which come under the form 2π w

w1

dφ

Yl (µ, φ)Ym (µ, φ)dµ,

−1

0

where Ym (µ, φ) and Yl (µ, φ) are Surface Spherical Harmonics of different degrees. If we are determining a coefficient Bn,m the only difference is that sin nφ and cos nφ will be interchanged in the forms just specified. 105. The integral over the surface of the unit sphere of the product of two Surface Spherical Harmonics of different degrees is zero. 2π w

That is

dφ

w1

Yl (µ, φ)Ym (µ, φ)dµ = 0.

(1)

−1

0

For as we have seen U = rl Yl (µ, φ) and V = rm Ym (µ, φ) are solutions of Laplace’s Equation. Hence by Green’s Theorem w (U Dn V − V Dn U )ds = 0 v. Art. 92. Dn V =Dr V = mrm−1 Ym (µ, φ), Dn U = Dr U = lrl−1 Yl (µ, φ); U Dn V − V Dn U = (m − l)rl+m−1 Yl (µ, φ)Ym (µ, φ), = (m − l)Yl (µ, φ)Ym (µ, φ) on the surface of the unit sphere; and (m − l)

w

Yl (µ, φ)Ym (µ, φ)ds = (m − l)

2π w 0

dφ

w1


1

Hence unless l = m 2π w 0

dφ

w1 −1



204 EXAMPLES.

1.

Obtain (1) Art. 105 directly from the equation

m(m + 1)Ym (µ, φ) + Dµ [(1 − µ2 )Dµ Ym (µ, φ)] +

1 D2 Ym (µ, φ) = 0 1 − µ2 φ

v. (3) Art. 102, and Art. 91. 2. 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. Suggestion: 2π w

sin kφ cos lφ.dφ =

0

w1

106.

2π w

sin kφ sin lφ.dφ =

0

2π w

cos kφ cos lφ.dφ = 0.

0

n Pln (µ)Pm (µ)dµ = 0

unless l = m

−1

=

(m + n)! 2 2m + 1 (m − n)!

if l = m.

For w1

n Pln (µ)Pm (µ)dµ =

−1

w1

(1 − µ2 )n

−1

= (1 − µ2 )n

dn Pl (µ) dn Pm (µ) . dµ dµn dµn

1 dn Pm (µ) dn−1 Pl (µ) i . dµn dµn−1

−1

w1

−

−1

=−

n dn−1 Pl (µ) d 2 n d Pm (µ) . (1 − µ ) dµ, dµn−1 dµ dµn

w1 dn−1 P (µ) d n l 2 n d Pm (µ) . (1 − µ ) dµ; dµn−1 dµ dµn

−1

by integration by parts. Replacing n by n − 1 in equation (2) Art. 84 and remembering that dn−1 Pm (x) is a possible value of z (n−1) we get dxn−1 dn+1 Pm (µ) dn Pm (µ) dn−1 Pm (µ) − 2nµ + [m(m + 1) − n(n − 1)] = 0, (1 − µ2 ) dµn+1 dµn dµn−1 or if we multiply by (1 − µ2 )n−1 (1 − µ2 )n

dn+1 Pm (µ) dn Pm (µ) − 2nµ(1 − µ2 )n−1 n+1 dµ dµn + (m + n)(m − n + 1)(1 − µ2 )n−1

dn−1 Pm (µ) = 0, dµn−1


205

or n−1 Pm (µ) d dn Pm (µ) 2 n−1 d = −(m + n)(m − n + 1)(1 − µ ) . (1 − µ2 )n n dµ dµ dµn−1 Hence follows the reduction formula w1

(1 − µ2 )n

−1

dn Pl (µ) dn Pm (µ) . dµ dµn dµn w1

= (m + n)(m − n + 1)

(1 − µ2 )n−1

−1

dn−1 Pl (µ) dn−1 Pm (µ) . dµ. dµn−1 dµn−1

Using this formula n times we get w1

n Pln (µ)Pm (µ)dµ =

−1

1 (m + n)! w Pl (µ)Pm (µ)dµ (m − n)! −1

=0 =

unless l = m

2 (m + n)! 2m + 1 (m − n)!

if l = m

v. Art. 89 (4) and (5). 107.

We are now able to complete the solution of the problem in Art. 104 2π 2π 2π w w w 2 2 and since cos nφ.dφ = sin nφ.dφ = π and dφ = 2π we get as the 0

0

0

coefficients in (1) Art. 104 2π w1 2m + 1 w = dφ f (µ, φ)Pm (µ)dµ, 4π

(1)

2π w1 2m + 1 (m − n)! w n = . dφ f (µ, φ) cos nφPm (µ)dµ, 2π (m + n)!

(2)

2π w1 2m + 1 (m − n)! w n = . dφ f (µ, φ) sin nφPm (µ)dµ, 2π (m + n)!

(3)

A0,m

−1

0

An,m

0

Bn,m

0

−1

−1

whence f (µ, φ) =

m=∞ X

n=m X

m=0

n=1

A0,m Pm (µ) +

n (An,m cos nφ + Bn,m sin nφ)Pm (µ)

(4)

and the development holds good for all values of µ and φ 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 Fourier’s Series.


206

If we use µ1 and φ1 in place of µ and φ in (1), (2), and (3), we can write (4) in the form w 2π m=∞ w1 1 X 1 f (µ, φ) = (2m + 1) dφ1 f (µ1 , φ1 )Pm (µ)Pm (µ1 )dµ1 2π m=0 2 −1

0

+

n=m X n=1

2π w1 (m − n)! w n n dφ1 f (µ1 , φ1 )Pm (µ)Pm (µ1 ) cos n(φ − φ1 )dµ1 . (5) (m + n)! −1

0

Formulas (1), (2), (3), and (4) are convenient for actual work; (5) is rather more compactly written. 108. As an example let us express sin2 θ cos2 θ sin φ cos φ in terms of Surface Spherical Harmonics. Here

f (µ, φ) =

A0,m

1 2 µ (1 − µ2 ) sin 2φ. 2

1 2π w 2m + 1 w 2 2 = µ (1 − µ )Pm (µ)dµ sin 2φ.dφ = 0, 8π −1

An,m

0

1 2π w 2m + 1 (m − n)! w 2 2 n = . µ (1 − µ )Pm (µ)dµ sin 2φ cos nφ.dφ = 0, 4π (m + n)! −1

0

w 2m + 1 (m − n)! w 2 n . µ (1 − µ2 )Pm (µ)dµ sin 2φ sin nφ.dφ, 4π (m + n)! 1

Bn,m =

2π

0

=0

2π w

If n = 2

0

unless n = 2.

sin 2φ sin nφ.dφ =

0

B2,m =

2π w

sin2 2φ.dφ = π,

0

1 d2 Pm (µ) 2m + 1 (m − 2)! w 2 . µ (1 − µ2 )2 dµ 4 (m + 2)! dµ2 −1

=

1 dm+2 (µ2 − 1)m 1 2m + 1 (m − 2)! w 2 µ (1 − µ2 )2 dµ. m 2 m! 4 (m + 2)! dµm+2 −1

w1 −1

w dm−4 (µ2 − 1)m dm+2 (µ2 − 1)m dµ = 720 dµ m+2 dµ dµm−4 1

µ2 (1 − µ2 )2

−1

and


207

by repeated integration by parts, =0

if m > 4,

= 720

w1

(µ2 − 1)4 dµ =

−1

and

4096 7

if

m = 4,

1 9 2! 4096 1 . . . = . 24 4! 4 6! 7 105

B2,4 =

By a like process we find B2,3 = 0 sin2 θ cos2 θ sin φ cos φ = =

and B2,2 =

1 . 42

Hence

1 2 1 2 P (µ) sin 2φ + P (µ) sin 2φ, 42 2 105 4

(1)

2 1 d2 P2 (µ) 1 2 d P4 (µ) sin 2φ sin2 θ + sin 2φ sin θ , 42 dµ2 105 dµ2

(2)

1 1 sin2 θ sin 2φ + sin2 θ(7µ2 − 1) sin 2φ. (3) 14 14 The required expression might have been obtained without using the formulas of Art. 107, by a very simple device, as follows: =

1 2 2 µ sin θ sin 2φ. (4) 2 X d2 Pm (µ) If now we can express µ2 in the form the work will be done. dµ2 sin2 θ cos2 θ sin φ cos φ =

µ2 =

1 d2 (µ4 ) , 4.3 dµ2

µ4 =

4 1 8 P4 (µ) + P2 (µ) + P0 (µ), 35 7 5

(5) Art. 95.

d2 (µ4 ) 8 d2 P4 (µ) 4 d2 P2 (µ) = + ; 2 dµ 35 dµ2 7 dµ2 µ2 =

whence

2 d2 P4 (µ) 1 d2 P2 (µ) + , 2 105 dµ 21 dµ2

and substituting this value in (4) we get (2). EXAMPLES. 1.

Show that

cos3 θ sin3 θ sin φ cos2 φ =

1 1 P63 (µ) + P43 (µ) sin 3φ 6930 1540 2 1 1 1 1 − P6 (µ) − P4 (µ) − P21 (µ) sin φ. 693 770 63

SPHERICAL HARMONICS. 2.

208

Show that cos 2φ = 2 cos 2φ

5 2 9.2! 2 13.4! 2 P2 (µ) + P4 (µ) + P6 (µ) + · · · . 4! 6! 8!

3. If in a problem on the Potential Function V = f (µ, φ) when r = a, we shall obviously have V =

m=∞ X m=0

n=m X rm n A0,m Pm (µ) + (An,m cos nφ + Bn,m sin nφ)Pm (µ) am n=1

at an internal point and V =

m=∞ X m=0

n=m X am+1 n A0,m Pm (µ) + (An,m cos nφ + Bn,m sin nφ)Pm (µ) rm+1 n=1

at an external point, where A0,m , An,m , and Bn,m have the values given in (1), (2), and (3) Art. 107. 4. Solve problems (3), (4), and (5) of Art. 94 for the case where V is not symmetrical with respect to an axis. 109. Any Solid Spherical Harmonic rm Ym (µ, φ) being a value of V that satisfies Laplace’s Equation in Spherical Coördinates will transform into a function of x, y, and z satisfying ∇2 V = 0 if we change to a set of rectangular 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 mth degree. n (µ) is of the form For each term of rm cos nφPm Crm cosn−2k φ sin2k φ sinn θ cosm−2l−n θ where

2k < n + 1

and

2l < m − n + 1.

This may be written Cr2l .rm−2l−n cosm−2l−n θ.rn−2k sinn−2k θ cosn−2k φ.r2k sin2k θ sin2k φ which becomes

C(x2 + y 2 + z 2 )l xm−2l−n y n−2k z 2k ,

and is a homogeneous rational integral Algebraic function of x, y, and z of the n mth degree. The same thing may be shown of each term of rm sin nφPm (µ). m Consequently r Ym (µ, φ) is a homogeneous rational integral Algebraic function of the mth degree in x, y, and z.


209

110. Any homogeneous rational integral Algebraic function Sm (x, y, z) of the mth degree in x, y, and z, which is a value of V satisfying ∇2 V = 0 contains 2m + 1 arbitrary constant coefficients. (m + 1)(m + 2) terms and will thereFor Sm (x, y, z) will in general consist of 2 (m + 1)(m + 2) fore contain coefficients. 2 2 ∇ Sm (x, y, z) will be homogeneous of the (m − 2)d degree and will contain m(m − 1) coefficients, which, of course, will be functions of the coefficients 2 in Sm (x, y, z). Since ∇2 Sm (x, y, z) = 0 independently of the numerical values m(m − 1) of x, y, and z the coefficients in ∇2 Sm (x, y, z) must be separately 2 m(m − 1) equations of condition between the zero, and that fact will give us 2 (m + 1)(m + 2) m(m − 1) (m + 1)(m + 2) original coefficients and will leave − 2 2 2 or 2m + 1 of them undetermined. Sm (x, y, z) contains, then, the same number of arbitrary coefficients as rm Ym (µ, φ). We can then choose the coefficients in rm Ym (µ, φ) so that it will transform into any given Sm (x, y, z). Consequently a Solid Spherical Harmonic of the mth degree might be defined as a homogeneous rational integral Algebraic function of x, y, and z, Sm (x, y, z), of the mth degree satisfying the equation ∇2 Sm (x, y, z) = 0; and a Surface Spherical Harmonic of the mth degree as such a function divided by (x2 + y 2 + m z 2 ) 2 , that is by rm . EXAMPLES. 1.

Show that if Sm (x, y, z) is a Solid Spherical Harmonic of the mth degree ∇2 [rn Sm (x, y, z)] = n(2m + n + 1)rn−2 Sm (x, y, z).

Suggestion: ∇2 Sm = 0. ∇2 r =

2 . r

Dr Sm =

mSm . (Dx r)2 + (Dy r)2 + (Dz r)2 = 1. r

2. Show that if fn (x, y, z) is a rational integral homogeneous function of x, y, and z of the nth degree it can be expressed in the form fn (x, y, z) = Sn (x, y, z) + r2 Sn−2 (x, y, z) + r4 Sn−4 (x, y, z) + · · · ,

(1)

terminating with rn−1 S1 (x, y, z) if n is odd, and with rn S0 (x, y, z) if n is even. Suggestion: If a term rSn−1 were present in the second member of (1), and we were to operate with ∇2 on both members we should by Ex. 1 have a term 2n Sn−1 which would be irrational when all the other terms of the resulting r


210

equation were rational. No such term, then, could occur. In the same way it may be shown by operating twice on (1) with ∇2 that there can be no term r3 Sn−3 in (1); and thus step by step we can reach the result formulated in (1). 3.

Express x2 yz in the form S4 + r2 S2 + r4 S0 . x2 yz = S4 + r2 S2 + r4 S0

Suggestion: let

and take ∇2 of both members we get 2yz = 14S2 + 20r2 S0 . Operate again with ∇2 .

Whence

1 1 yz, and S4 = (6x2 − y 2 − z 2 )yz. 7 7 Express sin2 θ cos2 θ sin φ cos φ in terms of Surface Spherical Harmonics. S0 = 0,

4.

0 = 120S0 .

Suggestion:

S2 =

sin2 θ cos2 θ sin φ cos φ =

x2 yz . r4

For result v. Art. 108 (3). 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 ∇2 V = 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 α1 + y cos α2 + z cos α3 , x cos β1 + y cos β2 + z cos β3 and x cos γ1 + y cos γ2 + z cos γ3 respectively where the cosines are the direction cosines of the new axes, and it will leave the function a homogeneous function of the mth degree in the new variables, and on dividing this by the mth power of the unchanged radius vector we shall have a Surface Spherical Harmonic of the mth degree. 112. We have seen in Art. 75 that if (x1 , y1 , z1 ) are the coördinates of a given point 1 V =p (1) 2 (x − x1 ) + (y − y1 )2 + (z − z1 )2 is a solution of Laplace’s Equation ∇2 V = 0, and transforming to spherical co¨ ordinates that 1 V =p r2 − 2rr1 [cos θ cos θ1 + sin θ sin θ1 cos(φ − φ1 )] + r12

(2)

is a solution of rDr2 (rV ) +

1 1 Dθ (sin θDθ V ) + Dφ2 V = 0. sin θ sin2 θ

(3)


211

If γ is the angle between the radii vectores r and r1 of the points (x, y, z) and (x1 , y1 , z1 ) (1) can be written 1 V =p 2 r − 2rr1 cos γ + r12

(4)

which must be equivalent to (2), and hence cos γ = cos θ cos θ1 + sin θ sin θ1 cos(φ − φ1 ). (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 = Pm (cos γ) is a solution of the equation 1 1 Dθ (sin θDθ V ) + Dφ2 V = 0 sin θ sin2 θ 1 V = rm Pm (cos γ) and V = m+1 Pm (cos γ) r

m(m + 1)V + and that

(5)

are solutions of (3). If we transform our co¨ ordinates keeping the origin unchanged and taking as our new polar axis the radius vector of (x1 , y1 , z1 ) γ becomes our new θ and Pm (cos γ) reduces to Pm (cos θ), a Surface Zonal Harmonic, or a Legendrian,3 of the mth degree. It is then a Legendrian having for its axis not the original polar axis but the radius vector of (x1 , y1 , z1 ). Since a Legendrian is a Surface Spherical Harmonic, Pm (cos γ) = Pm [cos θ cos θ1 + sin θ sin θ1 cos(φ − φ1 )] is a Surface Spherical Harmonic of the mth degree. It is, however, of very special form, since being a determinate function of µ, φ, µ1 , and φ1 it contains but two arbitrary constants if we regard it as a function of µ and φ, instead of containing 2m + 1. It is known as a Laplace’s Coefficient, or briefly as a Laplacian, of the mth degree. We shall soon express it in the regulation form of a Surface Spherical Harmonic. The radius vector of (x1 , y1 , z1 ) is called the axis of the Laplacian and the point where the axis cuts the surface of the unit sphere is the pole of the Laplacian. We shall represent the Laplacian Pm (cos γ) by Lm (µ, φ, µ1 , φ1 ). Of course Lm (µ, φ, 1, φ1 ) = Pm (µ) = Pm (cos θ), and is really independent of φ. 3 v.

Art. 74.


212

113. If the product of a Surface Spherical Harmonic of the mth degree by a Laplacian of the same degree is integrated over the surface of the unit sphere, 4π the result is equal to multiplied by the value of the Spherical Harmonic 2m + 1 at the pole of the Laplacian. That is, 2π w 0

dφ

w1

Ym (µ, φ)Lm (µ, φ, µ1 , φ1 )dµ =

−1

4π Ym (µ1 , φ1 ). 2m + 1

(1)

Transform to the axis of the Laplacian as a new polar axis, and let Zm (µ, φ) be the transformed Spherical Harmonic. Lm (µ, φ, µ1 , φ1 ) will become Pm (µ), and (1) will be proved if we can show that 2π w

w1

dφ

Zm (µ, φ)Pm (µ)dµ =

−1

0

Zm (µ, φ)Pm (µ) = A0 [Pm (µ)]2 +

n=m X

4π Zm (1, 0). 2m + 1

(2)

n (An cos nφ + Bn sin nφ)Pm (µ)Pm (µ)

n=1

(v. (5) Art. 102). 2π w

Zm (µ, φ)Pm (µ)dφ = 2πA0 [Pm (µ)]2

and

0

w1

dµ

−1

2π w

Zm (µ, φ)Pm (µ)dφ =

0

4π A0 2m + 1

(v. (5) Art. 89). n

n (1) contains (1 − 1) 2 as a factor But Zm (1, 0) = A0 , since Pm (1) = 1 and Pm and is equal to zero. Hence (2) is proved.

114. We can now express a Laplacian in the regulation form as a Spherical Harmonic, by the formulas of Art. 107. Lm (µ, φ, µ1 , φ1 ) =Pm (cos γ) = Pm [cos θ cos θ1 + sin θ sin θ1 cos(φ − φ1 )] k=∞ n=k X X n = A0,k Pk (µ) + (An,k cos nφ + Bn,k sin nφ)Pk (µ) n=1

k=0

where

A0,m =

2π w1 2m + 1 w dφ Lm (µ, φ, µ1 , φ1 )Pm (µ)dµ. 4π 0

−1

2m + 1 4π = Pm (µ1 ) = Pm (µ1 ) 4π 2m + 1

by (1) Art. 113,


An,m =

213

2π w1 2m + 1 (m − n)! w n dφ Lm (µ, φ, µ1 , φ1 ) cos nφPm (µ)dµ 2π (m + n)! −1

0

Bn,m

2(m − n)! n cos nφ1 Pm (µ1 ) by (1) Art. 113, and = (m + n)! 2π w1 2m + 1 (m − n)! w n = dφ Lm (µ, φ, µ1 , φ1 ) sin nφPm (µ)dµ 2π (m + n)! −1

0

2(m − n)! n = sin nφ1 Pm (µ1 ) (m + n)!

by (1) Art. 113,

and A0,k = An,k = Bn,k = 0 by Art. 105 unless k = m. Hence Lm (µ, φ, µ1 , φ1 ) = Pm (µ)Pm (µ1 ) + 2

n=m X n=1

(m − n)! n n P (µ)Pm (µ1 ) cos n(φ − φ1 ) . (1) (m + n)! m

Each term of a Laplacian involves a numerical coefficient, a factor which is a function of µ, a second factor which is the same function of µ1 , and a third factor which is of the form cos k(φ − φ1 ). 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 µ1 . By the aid of (1) we can write (5) Art. 107 more compactly. It becomes f (µ, φ) =

2π m=∞ w w1 1 X (2m + 1) dφ1 f (µ1 , φ1 )Lm (µ, φ, µ1 , φ1 )dµ1 4π m=0

(2)

2π m=∞ w wπ 1 X (2m + 1) dφ1 F (θ1 , φ1 )Pm (cos γ) sin θ1 dθ1 . 4π m=0

(3)

0

or

F (θ, φ) =

0

−1

0

LAPLACIANS. coef. of cos 0(φ − φ1 ) L0

1

L1

µ

L2 L3 L4

1 (3µ2 − 1) 4 1 (5µ3 − 3µ) 4

coef. of cos(φ − φ1 )

coef. of cos 2(φ − φ1 )

1

(1 − µ2 ) 2 1

3µ(1 − µ2 ) 2

3 (1 − µ2 ) 4 15 µ(1 − µ2 ) 4

3 1 (1 − µ2 ) 2 (5µ2 − 1) 8 1 5 5 1 (35µ4 − 30µ2 + 3) (1 − µ2 ) 2 (7µ3 − 3µ) (1 − µ2 )(7µ2 − 1) 64 8 16


214

coef. of cos 3(φ − φ1 ) coef. of cos 4(φ − φ1 ) L0 L1 L2 5 3 (1 − µ2 ) 2 8 35 3 µ(1 − µ2 ) 2 8

L3 L4

35 (1 − µ2 )2 64

EXAMPLE. Work the problems of Art. 108 and Art. 108 Exs. 1 and 2 by the aid of (3) Art. 114. 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. ρ = Cr1k Ym (µ1 , φ1 ).

Let

Then using our ordinary notation we have V =

wa

dr1

2π w

0

dφ1

0

w1 Crk Y (µ , φ )r2 dµ p 1 m 1 1 1 1 . r2 − 2rr1 cos γ + r12 −1

But by (3) Art. 77 1 r1 = P0 (cos γ) + P1 (cos γ) 2 2 r r r − 2rr1 cos γ + r1 1

p

+ if

r12 r1m P (cos γ) + · · · + Pm (cos γ) + · · · 2 r2 rm

r > r1 .

Consequently since 2π w 0

dφ1

w1 −1

Ym (µ1 , φ1 )Yn (µ1 , φ1 )dµ1 = 0,


215

V reduces to the single term V =

=

C rm+1 C rm+1

wa 0 wa

r1m+k+2 dr1

2π w 0

r1m+k+2

0

∴ V =

dφ1

w1

Ym (µ1 , φ1 )Pm (cos γ)dµ1

−1

4π Ym (µ, φ) dr1 2m + 1

by Art. 113.

4πC am+k+3 Ym (µ, φ) . . . 2m + 1 m + k + 3 rm+1 EXAMPLES.

1. Solve by direct integration the problems worked in Arts. 98 and 99 and Examples 1, 2, 3, and 4 of Art. 99. 2. 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. ρ = kr14 cos2 θ1 sin2 θ1 cos2 φ1 1 1 4 1 = kr1 P0 (µ1 ) + P2 (µ1 ) + cos 2φ1 P22 (µ1 ) 15 21 42 1 4 2 − P4 (µ1 ) + cos 2φ1 P4 (µ1 ) . 35 105 M a a3 1 1 V = + 3 P2 (µ) + cos 2φP22 (µ) a r r 9 18 a5 4 1 − 5 P4 (µ) − cos 2φP42 (µ) . r 33 99 Ans.

3.

Solve Example 2 by an extension of the method of Arts. 98 and 99.

4. 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. Suggestion: Let V1 be the potential function due to the point, and V2 that due to the induced charge, and let b be the distance of the point from the centre


216

of the sphere. Then V1 = √

m b2

− 2br cos θ + r2 m r r2 = if r P0 (cos θ) + P1 (cos θ) + 2 P2 (cos θ) + · · · b b b m b b2 = if r P0 (cos θ) + P1 (cos θ) + 2 P2 (cos θ) + · · · r r r r r2 V2 = A0 P0 (cos θ) + A1 P1 (cos θ) + A2 2 P2 (cos θ) + · · · if a a a3 a a2 = A0 P0 (cos θ) + A1 2 P1 (cos θ) + A2 3 P2 (cos θ) + · · · r r r

< b. > b. r < a. if r > a.

When r = a V1 + V2 = 0. Hence A0 = −

m , b

A1 = −

ma , b2

A2 = −

ma2 ,··· b3

and r r2 m P0 (cos θ) + P1 (cos θ) + 2 P2 (cos θ) + · · · if r < a b b b ma a2 a4 =− P0 (cos θ) + P1 (cos θ) + 2 2 P2 (cos θ) + · · · if r > a. br br b r

V2 = −

Hence the effect of the induced charge is precisely the same at an external ma units of negative electricity concenpoint as if the sphere were replaced by b a2 trated at the point r = , θ = 0. v. Peirce, Newt. Pot. Func., § 66. b 116. If the two points P and P 0 are taken on the line OH whose direction 0 cosines are λ, µ, and ν, and if u and u0 are the valuesat P and P of any 0 u −u continuous function of the space coördinates, then limit is called the . PP0 P P 0 =0 partial derivative of u along the line OH and will be represented by Dh u. Let x, y, z be the co¨ ordinates of P and x+∆x, y+∆y, z+∆z the coördinates of P 0 ; then u0 − u = Dx u.∆x + Dy u.∆y + Dz u.∆z + where is an infinitesimal of higher order than the first if ∆x, ∆y, and ∆z are infinitesimal (v. Dif. Cal. Art. 198). Hence But Therefore

∆x ∆y ∆z u0 − u = Dx u. + Dy u. + Dz u. + . PP0 PP0 PP0 PP0 PP0 ∆x ∆y ∆z = λ, = µ, and = ν. PP0 PP0 PP0 Dh u = λDx u + µDy u + νDz u.

(1)


217

If ∇2 u = 0, Dxp Dyq Dzr u is a solution of Laplace’s Equation. ∇2 (Dxp Dyq Dzr u) = Dxp Dyq Dzr (∇2 u) = 0.

For

Hence if ∇2 u = 0 Dh u is a solution of Laplace’s Equation, and if OH1 , OH2 , OH3 , · · · are a set of lines through the origin Dh1 Dh2 Dh3 · · · u is a solution of Laplace’s Equation. 117. If Hk is a rational integral homogeneous Algebraic function of x, y, and z of the kth degree Hk 1 Hk = D Dx r + l Dx (Hk ) Dx r l l r r r lxHk Hk−1 lxHk r2 Hk−1 = − l+2 + = − l+2 + , l r r r rl+2 Hk+1 . rl+2 H H k k The same thing can be proved of Dy and Dz and therefore holds l l r r H k . good of Dh rl If u is a homogeneous function of x, y, and z of the degree −m − 1 and ∇2 u = 0 then ∇2 (r2m+1 u) = 0. and is of the form

∇2 (r2m+1 u) = (2m + 1)(2m + 2)r2m−1 u + 2(2m + 1)r2m−1 (xDx u + yDy u + zDz u) + r2m+1 ∇2 u = 0, since

xDx u + yDy u + zDz u = −(m + 1)u

by Euler’s Theorem (v. Dif. Cal. Art. 220). M M =p is a solution of Laplace’s Equation (v. Art. 75) 2 r x + y2 + z2 H0 . and is of the form r M Dh1 Dh2 Dh3 · · · Dhm is then a solution of Laplace’s Equation by Art. r Hm 116; it is of the form 2m+1 by Art. 117 and is a homogeneous function of the r degree −m − 1. M Therefore r2m+1 Dh1 Dh2 Dh3 · · · Dhm is a solution of Laplace’s Equar tion, and is a rational integral homogeneous Algebraic function of x, y, and z of the mth degree, and is consequently a Solid Spherical Harmonic of the mth M degree (v. Art. 110); and rm+1 Dh1 Dh2 Dh3 · · · Dhm is a Surface Spherical r Harmonic of the mth degree. 118.


218

Moreover since the direction of each of the lines OH1 , OH2 , · · · OHm 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 rm+1 Dh1 Dh2 Dh3 · · · M may be any given Surface Spherical Harmonic. Dhm r Consequently any given Surface Spherical Harmonic may be regarded as M formed by differentiating successively along m determinate lines OH1 , OH2 r · · · OHm , and is given except for the undetermined factor M when these lines are given. The lines OH1 , OH2 , OH3 , · · · OHm are called the axes of the Harmonic, and the points where they meet the surface of the unit sphere the 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). 119. Any Surface Zonal Harmonic Pm (µ) is equal to zero for m real and distinct values of µ which lie between −1 and 1; and any Associated Function n (µ) is equal to zero for m − n real and distinct values of µ, which lie between Pm −1 and 1. Pm (µ) =

1 2m m!

.

dm (µ2 − 1)m . dµm

v. Art. 83 (1).

dk (µ2 − 1)m contains (µ2 − 1)m−k as a factor. v. Art. 89. dµk From Rolle’s Theorem, “If f (x) is continuous and single-valued and is equal df (x) to zero for the real values a and b of x, is equal to zero for at least one dx real value of x between a and b,” (v. Dif. Cal. Art. 126) it follows that since d(µ2 − 1)m = 0 for at least one (µ2 − 1)m = 0 when µ = −1 and when µ = 1 dµ d(µ2 − 1)m cannot be equal to zero for more than value of µ between −1 and 1. dµ one value of µ between −1 and 1, for it contains (µ2 − 1)m−1 as a factor and is a rational Algebraic polynomial of the 2m − 1st degree. d2 (µ2 − 1)m In like manner we can show that = 0 has m − 2 roots equal to dµ2 −1, m − 2 roots equal to 1 and two real roots between −1 and 1 which separate d(µ2 − 1)m the three distinct roots of = 0; and in general if k < m + 1 that dµ k 2 m d (µ − 1) = 0 has m − k roots equal to −1, m − k roots equal to 1, and k dµk dk−1 (µ2 − 1)m = 0. real roots separating the k + 1 distinct roots of dµk−1 1 dm (µ2 − 1)m Hence Pm (µ) = 0 or m . = 0 has m real and distinct roots 2 m! dµm between −1 and 1, and it has no more since it is of the mth degree.


219

dm+n (µ2 − 1)m = 0 has m − n distinct real dµm+n n roots between −1 and 1, and therefore that Pm (µ) is equal to zero for m − n n distinct real values of µ between −1 and 1. Since Pm (µ) contains sinn θ as a factor it is also equal to zero when µ = −1 and when µ = 1. cos nφ is equal to zero for 2n equidistant values of φ, and sin nφ is equal n to zero for 2n values of φ. Hence any Tesseral Harmonic sin nφPm (µ) or n cos nφPm (µ) is equal to zero for 2n equidistant values of φ, for µ = 1, for µ = −1, and for m − n real and different values of µ between −1 and 1. It follows that the value of any Surface Zonal Harmonic Pm (µ) 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 zones into which the surface of the sphere is divided by the m circles of latitude corresponding to the m roots of Pm (µ) = 0, and will change sign whenever the point passes from one of these zones into an adjoining n one; and that the value of any Tesseral Harmonic sin nφPm (µ) 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 tesserae into which the surface of the sphere is divided by n (µ) = 0 and the the m − n circles of latitude corresponding to the roots of Pm 2n meridians corresponding to the roots of sin nφ = 0, and will change sign whenever the point passes from one of these tesserae into an adjoining one. The same reasoning shows that

220

CHAPTER VII.1 CYLINDRICAL HARMONICS (BESSEL’S FUNCTIONS).

120.

In Arts. 11 and 17 we obtained z = AJ0 (x) + BK0 (x)

(1)

as the general solution of Fourier’s Equation

where

d2 z 1 dz + + z = 0, dx2 x dx x4 x6 x2 J0 (x) = 1 − 2 + 2 2 − 2 2 2 + · · · 2 2 .4 2 .4 .6

(2) (3)

and is called a Cylindrical Harmonic or Bessel’s Function of the zeroth order; and where x2 x4 1 1 x6 1 1 1 K0 (x) = J0 (x) log x + 2 − 2 2 + + + + 2 2 2 − · · · (4) 2 2 .4 1 2 2 .4 .6 1 2 3 and is called a Cylindrical Harmonic or Bessel’s Function of the Second Kind, and of the zeroth order. In Art. 17 we found that z = Jn (x) is a particular solution of Bessel’s Equation 1 dz n2 d2 z + + 1 − z = 0, (5) dx2 x dx x2 where if n is unrestricted in value xn x2 x4 Jn (x) = n 1− 2 + 4 2 Γ(n + 1) 2 (n + 1) 2 .2!(n + 1)(n + 2) x6 − 6 + ··· (6) 2 .3!(n + 1)(n + 2)(n + 3) and is called a Cylindrical Harmonic or Bessel’s Function of the nth order; and that unless n is an integer z = AJn (x) + BJ−n (x) is the general solution of Bessel’s Equation. If n is an integer it can be shown that Jn (x) = (−1)n J−n (x), 1 The

student should re-read carefully Arts. 11, 17, and 18(d) before beginning this chapter.

CYLINDRICAL HARMONICS.

221

(v. Forsyth’s Diff. Eq. Art. 102), and then z = AJn (x) + B{Kn (x)} is the general solution of Bessel’s Equation and k=n−1 1 x −n X (n − k − 1)! x 2k 2 2 k! 2 k=0 k=∞ 1 x n X (−1)k 1 1 1 − 1 + + + ··· + 2 2 (n + k)!k! 2 3 k

{Kn (x)} = Jn (x) log x −

k=0

1 1 1 x 2k +1 + + + · · · + 2 3 n+k 2

(7)

v. M. Bˆ ocher, Ann. Math. Vol. VI, No. 4. 121. A useful expression for Jn (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 = xn v is a solution of Bessel’s Equation if v satisfies the equation 2n + 1 dv d2 v + + v = 0. 2 dx x dx wb v = T cos(xt)dt

Assume

(1) (2)

a

where x and t are independent, T is an unknown function of t, and a and b are at present undetermined. wb dv = − tT sin(xt)dt dx a

Then

wb d2 v = − t2 T cos(xt)dt. dx2 a

and

Substituting in (1) after multiplying through by x, we have wb

wb (1 − t2 )T x cos(xt)dt − (2n + 1)tT sin(xt)dt = 0.

a

(3)

a

By integration by parts we find that wb

h ib (1 − t2 )T x cos(xt)dt = (1 − t2 )T sin(xt)

a

a

−

i wb h dT (1 − t2 ) − 2tT sin(xt)dt, dt a


222

and (3) reduces to h ib wb h i dT (1 − t2 )T sin(xt) − + (2n − 1)tT sin(xt)dt = 0. (1 − t2 ) dt a

(4)

a

If we determine T so that dT + (2n − 1)tT = 0, dt h ib (1 − t2 )T sin(xt) = 0

(1 − t2 ) and a and b so that

(5) (6)

a

(4) will be satisfied and our problem will be solved. (5) gives 1

T = C(1 − t2 )n− 2 ,

(7)

and (6) will obviously be satisfied if a = −1 and b = 1. Hence

v=C

w1 (1 − t2 )n cos(xt)dt √ 1 − t2 −1

z = Cxn

and

is a solution of (1),

w1 (1 − t2 )n cos(xt)dt √ 1 − t2 −1

(8)

is a solution of Bessel’s Equation. If we let t = cos φ in (8) we get z = Cxn

wπ

sin2n φ cos(x cos φ)dφ.

0

Expand cos(x cos φ) into a series involving powers of x cos φ, integrate term by term by the aid of the formulas π √ Γ n+1 w2 π . 2 sinn x.dx = [Int. Cal. (1) Art. 99], 2 Γ n +1 0 2 π m + 1 n + 1 w2 Γ Γ sinn x cosm x.dx = 2m + n 2 2Γ +1 0 2 (Int. Cal. Art. 99 Ex. 2), and compare with (6) Art. 120, and we get w xn sin2n φ cos(x cos φ)dφ. √ 1 n 0 2 π Γ n+ 2 π

Jn (x) =

(9)


223

If n is a positive integer (9) reduces to w 1 xn sin2n φ cos(x cos φ)dφ. Jn (x) = . π 1.3.5. · · · (2n − 1) π

(10)

0

Let n = 0 in (9) or (10) and we get J0 (x) =

π 1w cos(x cos φ)dφ. π

(11)

0

EXAMPLES. 1.

Obtain Formula (11) directly from Fourier’s Equation, (2) Art. 120.

2.

Prove by integration by parts that if n > − wπ

x w sin2n+2 φ cos(x cos φ)dφ. 2n + 1 π

sin2n φ cos φ sin(x cos φ)dφ =

0

3.

0

Prove by integration by parts that if n > wπ

1 2

1 2

sin2n φ cos φ sin(x cos φ)dφ

0

1w [2n sin2n φ − (2n − 1) sin2n−2 φ] cos(x cos φ)dφ. x π

=

0

122. We can now readily obtain a number of useful formulas. Differentiate (11) Art. 121 with respect to x and we get π 1w dJ0 (x) =− cos φ sin(x cos φ)dφ dx π 0

π xw =− sin2 φ cos(x cos φ)dφ π

by Ex. 2 Art. 121.

0

Hence by (10) Art. 121

dJ0 (x) = −J1 (x). dx

(1)

In like manner by the aid of Exs. 3 and 2, Art. 121, we can obtain the relations d[xn Jn (x)] = xn Jn−1 (x) (2) dx

CYLINDRICAL HARMONICS. if n >

1 , 2

d[x−n Jn (x)] = −x−n Jn+1 (x) dx

224

(3)

1 if n > − . 2 (2) can be written wx

1 if n > . 2

xn Jn−1 (x)dx = xn Jn (x)

(4)

0

(2) and (3) can be written xn and

x−n

dJn (x) + nxn−1 Jn (x) = xn Jn−1 (x) dx

dJn (x) − nx−n−1 Jn (x) = −x−n Jn+1 (x), dx dJn (x) n = Jn−1 (x) − Jn (x) dx x

(5)

n dJn (x) = − Jn+1 (x) + Jn (x); dx x

(6)

or and whence

2

dJn (x) = Jn−1 (x) − Jn+1 (x) dx

(7)

and

2n Jn (x) = Jn−1 (x) + Jn+1 (x). x

(8)

The repeated use of formula (8) will enable us to get from J0 (x) and J1 (x) any of Bessel’s Functions whose order is a positive integer. For example, we have 2 J2 (x) = J1 (x) − J0 (x) x 8 4 J3 (x) = − 1 J1 (x) − J0 (x). 2 x x From a table giving the values of J0 (x) and J1 (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 Jn (x) can be expressed in terms of Jn (x) and Jn+1 (x). For example d2 Jn (x) n(n − 1) 1 = − 1 Jn (x) + Jn+1 (x). dx2 x2 x


225

If we write J0 (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 integration by parts, we get dJ0 (x) w + xJ0 (x)dx = 0; dx x

x

0

wx

whence by (1)

xJ0 (x)dx = xJ1 (x).

(9)

0

If we write J0 (x) for z in Fourier’s Equation, then multiply through by 2 dJ0 (x)

x dx we get

dx and integrate from zero to x, simplifying by integration by parts

x2 2 whence by (1)

wx 0

dJ0 (x) dx

2 + (J0 (x))

2

−

wx

x(J0 (x))2 dx = 0;

0 2

x(J0 (x))2 dx =

x [(J0 (x))2 + (J1 (x))2 ]. 2

(10)

In like manner we can get from Bessel’s Equation [(5) Art. 120] the formula wx 0

2 1 2 dJn (x) 2 2 2 x(Jn (x)) dx = x + (x − n )(Jn (x)) 2 dx 2

(11)

which (6) enables us to reduce to the form wx 0

x(Jn (x))2 dx =

x2 [(Jn (x))2 + (Jn+1 (x))2 ] − nxJn (x)Jn+1 (x). 2

(12)

Formulas (9), (10), (11), and (12) will prove useful when we attempt to develop in terms of Cylindrical Harmonics. Values of Jn (x) for larger values of x than those given in Table VI., Appendix, may be computed very easily from the formula r 2 (12 − 4n2 )(32 − 4n2 ) Jn (x) = 1− πx 2!(8x)2 (12 − 4n2 )(32 − 4n2 )(52 − 4n2 )(72 − 4n2 ) π π + − · · · cos x − − n 4 4!(8x) 4 2 r 2 2 2 1 − 4n + πx 1!8x (12 − 4n2 )(32 − 4n2 )(52 − 4n2 ) π π − + · · · sin x − − n . (13) 3!(8x)3 4 2


226

v. Lommel, Studien u ¨ber 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 Jn (x) by less than the last term included, and consequently the formula can safely be used for numerical computation. EXAMPLES. 1. Confirm (1), (2), and (3), Art. 122, by obtaining them from (3) and (6), Art. 120. 2. 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.

Show that (9), Art. 122, is a special case of (4), Art. 122.

4. Show that the limit approached by Jn (x) as n increases indefinitely is zero, and by the aid of this fact and of (8), Art. 122, prove that Jn−1 (x) = 5.

2 [nJn (x) − (n + 2)Jn+2 (x) + (n + 4)Jn+4 (x) + · · · ]. x

Prove that dJn (x) 2 = [ 12 nJn (x) − (n + 2)Jn+2 (x) + (n + 4)Jn+4 (x) − · · · ]. dx x

6.

Show that the substitution of

y2 1− 2 n

12 for x in Legendre’s Equation

will reduce it to the form 1 2y dz 1 y 2 d2 z + − + 1+ z = 0, 1− 2 n dy 2 y n2 dy n and that the limiting form approached by this equation as n is indefinitely increased is Fourier’s Equation, and hence that J0 (x) can be regarded as some 1 x2 2 constant factor multiplied by the limiting value approached by Pn 1 − 2 n as n is indefinitely increased. 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) = A1 J0 (µ1 r) + A2 J0 (µ2 r) + A3 J0 (µ3 r) + · · · where µ1 , µ2 , µ3 , &c., are the roots of the transcendental equation J0 (µa) = 0; and in Art. 11, Ex. the development of unity in a series of precisely the same


227

form was needed. (a) 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 ([XIV] Art. 1). 1 1 (1) Dr2 u + Dr u + 2 Dφ2 u + Dz2 u = 0 r r subject to the conditions u=0

when z = 0

u=0

“

r=a

u = f (r)

“

z = b,

and from the symmetry of the problem we know that Dφ2 u = 0. Assuming as usual u = R.Z we break (1) up into the equations d2 Z − µ2 Z = 0 dz 2 d2 R 1 dR + + µ2 R = 0, dr2 r dr whence

u = sinh(µz)J0 (µr)

(2)

and

u = cosh(µz)J0 (µr)

(3)

are particular solutions of (1).

If µk is a root of

J0 (µa) = 0

(4)

u = sinh(µk z)J0 (µk r) satisfies (1) and two of the three equations of condition. If then

f (r) = A1 J0 (µ1 r) + A2 J0 (µ2 r) + A3 J0 (µ3 r) + · · ·

(5)

µ1 , µ2 , µ3 , &c., being roots of (4), u = A1

sinh(µ2 z) sinh(µ3 z) sinh(µ1 z) J0 (µ1 r)+A2 J0 (µ2 r)+A3 J0 (µ3 r)+· · · (6) sinh(µ1 b) sinh(µ2 b) sinh(µ3 b)

satisfies (1) and all of the equations of condition, and is the required solution.


228

(b) 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 Dr u = 0 when r = a, or if u = sinh(µz)J0 (µr), by that is by by

dJ0 (µr) =0 dr 0 µJ0 (µa) = 02

or (v. (1) Art. 122)

J1 (µa) = 0.

(7)

when r = a,

If now in (5) and (6) µ1 , µ2 , µ3 , &c., are roots of (7), (6) will be the solution of our new problem. (c) 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 Dr u + hu = 0 when r = a, or if u = sinh(µz)J0 (µr) µJ00 (µr) + hJ0 (µr) = 0

by

when

r=a

that is by

µaJ00 (µa) + ahJ0 (µa) = 0

or (v. (1) Art. 122)

by

µaJ1 (µa) − ahJ0 (µa) = 0.

(8)

If now in (5) and (6) µ1 , µ2 , µ3 , &c., are roots of (8), (6) will be the solution of our present problem. 124.

It can be shown that J0 (x) = 0

(1)

J1 (x) = 0

(2)

xJ00 (x) + λJ0 (x) = 0

(3)

and

have each an infinite number of real positive roots (v. Riemann, Par. Dif. Gl., § 97). The earlier roots of these equations can be computed without serious difficulty from the table for the values of J0 (x) (Table VI., Appendix). The first twelve roots of J0 (x) = 0 and J1 (x) = 0 are given in Table IV., Appendix, a table due to Stokes. Large roots of J0 (x) = 0 and of J1 (x) = 0 may be very easily computed from the formulas x

(s) 0

π x

(s) 1

π 2 We

119).

= s − .25 +

.050661 .053041 .262051 − + − ··· 4s − 1 (4s − 1)3 (4s − 1)5

(4)

= s + .25 −

.151982 .015399 .245270 + − + ··· 3 4s + 1 (4s + 1) (4s + 1)5

(5)

shall find it convenient to use the familiar notation of f 0 (x) =

df (x) (v. Dif. Cal., p. dx


229

given by Stokes in Camb. Phil. Trans., Vol. IX., x of J0 (x) = 0, and

(s) x1

(s) 0

representing the sth root

the sth root of J1 (x) = 0.

125. We have seen in Art. 123 that U = sinh(µk z)J0 (µk r) and V = sinh(µl z)J0 (µl r) are solutions of ∇2 U = 0 and ∇2 V = 0 if we express Laplace’s Equation in terms of Cylindrical Coördinates (v. (1) Art. 123). r Hence, if dS represents the surface integral over any closed surface, we have w (U Dn V − V Dn U )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 wa

rJ0 (µk r)J0 (µl r)dr =

0

=

−1 [µk aJ0 (µl a)J00 (µk a) − µl aJ0 (µk a)J00 (µl a)] µk 2 − µl 2

−1 [µk aJ0 (µl a)J1 (µk a) − µl aJ0 (µk a)J1 (µl a)]. (1) µl − µk 2 2

Hence if µk and µl are different roots of J0 (µa) = 0, or of or of then

J1 (µa) = 0, µaJ1 (µa) − λJ0 (µa) = 0, wa

rJ0 (µk r)J0 (µl r)dr = 0.

(2)

0

EXAMPLE. Obtain (1) Art. 125 directly from Fourier’s Equation d2 J0 (µr) 1 dJ0 (µr) + + µ2 J0 (µr) = 0. dr2 r dr 126. Let

We are now able to obtain the developments called for in Art. 123. f (r) = A1 J0 (µ1 r) + A2 J0 (µ2 r) + A3 J0 (µ3 r) + · · ·

(1)

µ1 , µ2 , µ3 , &c., being roots of J0 (µa) = 0, or of J1 (µa) = 0, or of µaJ1 (µa) − λJ0 (µa) = 0. To determine any coefficient Ak multiply (1) by rJ0 (µk r)dr and integrate from zero to a. The first member will become wa rf (r)J0 (µk r)dr. 0


230

Every term of the second member will vanish by (2) Art. 125 except the term Ak

wa

r(J0 (µk r))2 dr.

0

wa

r(J0 (µk r))2 dr =

0

µwk a

1 µ2k

x(J0 (x))2 dx =

0

a2 [(J0 (µk a))2 + (J1 (µk a))2 ] 2

by (10) Art. 122. w 2 Ak = 2 rf (r)J0 (µk r)dr. a [(J0 (µk a))2 + (J1 (µk a))2 ] a

Hence

(2)

0

The development (1) holds good from r = 0 to r = a (v. Arts. 24, 25, and 88). If µ1 , µ2 , µ3 , &c., are roots of J0 (µa) = 0, (2) reduces to Ak =

wa 2 rf (r)J0 (µk r)dr. a2 (J1 (µk a))2

(3)

0

If µ1 , µ2 , µ3 , &c., are roots of J1 (µa) = 0, (2) reduces to Ak =

wa 2 rf (r)J0 (µk r)dr. a2 (J0 (µk a))2

(4)

0

If µ1 , µ2 , µ3 , &c., are roots of µaJ1 (µa) − λJ0 (µa) = 0, (2) reduces to Ak =

wa 2µ2k rf (r)J0 (µk r)dr. (λ2 + µ2k a2 )(J0 (µk a))2

(5)

0

For the important case where f (r) = 1 wa

rf (r)J0 (µk r)dr =

0

wa

rJ0 (µk r)dr =

0

µk a a 1 w xJ0 (x)dx = J1 (µk a) 2 µk µk

(6)

0

by (9) Art. 122, and (3) reduces to

Ak =

2 , µk aJ1 (µk a)

(7)

(4) reduces to Ak = 0 except for k = 1 when µk = 0 and we have A1 = 1,

(5) reduces to

Ak =

2λ . (λ2 + µ2k a2 )J0 (µk a)

(8)


231 EXAMPLES.

1. Show that in (12) Art. 11 any coefficient Ak has the value given in (3) Art. 126; and in the answer to Art. 11, Ex. the value given in (7) Art. 126. 2. 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 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 J0 (x) = 0; that the possible nodes for such vibrations are concentric circles whose radii are proportional to the roots of J0 (x) = 0. 3. 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 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. Ans., 29◦ .6; 47◦ .6; 71◦ .2; 25◦ .8. 4. 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. Ans., 15◦ .4; 40◦ .85; 72◦ .8; 15◦ .3; 40◦ .7; 72◦ .5; 0◦ .0; 0◦ .0; 1◦ .3. 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 u=0

when z = 0,

u = f (r)

“

z = b,

u=0 u=0

when r = a, “

r = c.

Then (v. Art. 123) J0 (µk c) K0 (µk r) u = sinh(µk z) J0 (µk r) − K0 (µk c)


232

where µk is a root of the equation J0 (µa) −

J0 (µc) K0 (µa) = 0 K0 (µc)

(1)

will satisfy Laplace’s Equation [(1) Art. 123] and all of the equations of condition except the second. k=∞ X J0 (µk c) sinh(µk z) Hence u = J0 (µk r) − K0 (µk r) (2) Ak sinh(µk b) K0 (µk c) k=1

is the required solution if f (r) =

k=∞ X k=1

J0 (µk c) K0 (µk r) . Ak J0 (µk r) − K0 (µk c)

(3)

The development (3) is easily obtained. Call the parenthesis for the sake of brevity B0 (µk r). Then by the method of Art. 125 we get if we integrate over our cylindrical shell wc

rB0 (µk r)B0 (µl r)dr = 0

(4)

a

if µk and µl are roots of (1); and by an easy extension of (10) Art. 122 wc

r[B0 (µk r]2 dr = 21 {c2 [B00 (µk c)]2 − a2 [B00 (µk a)]2 }.

(5)

a

Determining the coefficients in (3) as in Art. 124 and simplifying by the aid of (4) we have i h wc J0 (µk c) K0 (µk r) dr 2 rf (r) J0 (µk r) − K0 (µk c) a Ak = h i i2 . h 2 J0 (µk c) 0 J0 (µk c) 0 c2 J00 (µk c) − K0 (µk c) − a2 J0 0 (µk a) − K0 (µk a) K0 (µk c) K0 (µk c) (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 k=∞ X J0 (µk b) K0 (µk r) y= Ak cos(µk ct) J0 (µk r) − K0 (µk b) k=1

where Ak is obtained from (6) Art. 127 by replacing c by b.


233

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 µ is replaced by µi we can readily obtain z = sin(µz)J0 (µri) and

z = cos(µz)J0 (µri)

as particular solutions of Laplace’s Equation [(1) Art. 123]; and J0 (xi) = 1 +

x2 x4 x6 + + + ··· 22 22 .42 22 .42 .62

(1)

and is real. f (z) =

k=∞ X

Ak sin

k=1

kπz b

2w kπz f (z) sin dz ba b b

where

Ak =

(2)

by Art. 31 (7) and (8).

Hence

u=

k=∞ X k=1

kπri J kπz 0 b Ak sin kπai b J0 b

(3)

is our required solution.

EXAMPLES. 1. 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

u=

k=∞ X k=1

 Ak sin

kπz   b

J0 J0

kπri b kπci − b

K0 K0

kπri   b kπci  b   ÷

J0 J0

kπai b kπci − b

K0 K0

kπai   b kπci  b


234

where Ak has the value given in (2) Art. 128, and K0 (xi) = K0 (xi) − J0 (xi) log i = J0 (xi) log x −

x4 x2 x6 − 2 2 ( 11 + 21 ) − 2 2 2 ( 11 + 2 2 2 .4 2 .4 .6

1 2

+ 31 ) − · · ·

[v. (4) Art. 120], and is real. 2. 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 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. Ans., 49◦ .6; 20◦ .2. 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, θ) when z = b, we have to get particular solutions of Laplace’s Equation [(1) Art. 123] for the case where Dφ2 u is not equal to zero. We readily find that u = sinh (µz)[A cos nφ + B sin nφ]Jn (µr) and

u = cosh (µz)[A cos nφ + B sin nφ]Jn (µr)

are such solutions, and that u=

n=∞ X k=∞ X n=0 k=1

sinh µk z [An,k cos nφ + Bn,k sin nφ]Jn (µk r) sinh µk b

(1)

is the solution of the given problem if f (r, φ) =

n=∞ X X k=∞

(An,k cos nφ + Bn,k sin nφ)Jn (µk r)

(2)

n=0 k=1

where µk is a root of the equation Jn (µa) = 0. µn an

(3)

EXAMPLES. 1. wa 0

Show that

rJn (µk r)Jn (µl r)dr a [µl Jn (µk a)Jn0 (µl a) − µk Jn (µl a)Jn0 (µk a)] µk 2 − µl 2 a = 2 [µk Jn (µl a)Jn+1 (µk a) − µl Jn (µk a)Jn+1 (µl a)]. µk − µl 2

=

CYLINDRICAL HARMONICS. 2.

Show that

wa

r[Jn (µk r)]2 dr

0

235

i n2 1h 2 0 a (Jn (µk a))2 + a2 − 2 (Jn (µk a))2 2 µk 2 a na = [(Jn (µk a))2 + (Jn+1 (µk a))2 ] − Jn (µk a)Jn+1 (µk a). 2 µk =

3.

Show that in Art. 129

A0,k

1 = π

2π w

dφ

0

wa

rf (r, φ)J0 (µk r)dr

0

a2 [J1 (µk a)]2

,

B0,k = 0,

An,k

Bn,k

2 = π

2 = π

2π w

dφ

0

wa

rf (r, φ) cos nφJn (µk r)dr

0

a2 [Jn+1 (µk a)]2 2π w 0

dφ

wa

,

rf (r, φ) sin nφJn (µk r)dr

0

a2 [Jn+1 (µk a)]2

.

4. Obtain the coefficients for the case where the convex surface of the cylinder is impervious to heat. 5. Obtain the coefficients for the case where the convex surface of the cylinder is exposed to air at the temperature zero. 6. Show that if in a drumhead problem of Art. 11 the initial distortion is unsymmetrical, so that we have to solve the equation [XI] Art. 1 subject to the conditions z = f (r, φ) when t = 0, Dt z = 0 when t = 0, z = 0 when r = a, the solution is z=

n=∞ X k=∞ X

cos(µk ct)(An,k cos nφ + Bn,k sin nφ)Jn (µk r)

n=0 k=1

where A0,k , B0,k , An,k , and Bn,k have the values given in Ex. 3. 7. 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).

CYLINDRICAL HARMONICS. 8. case.

236

Solve the problem of Art. 127 and of Art. 127. Ex. for the unsymmetrical Suggestion: AJn (x) + BKn (x) is a solution of Bessel’s Equation.

9. Solve the problem of Art. 128 and of Art. 128. Ex. 1. for the case where u = f (z, φ) when r = a. Suggestion: u = sin µz(A cos nφ + B sin nφ)Jn (µri) is a solution of Laplace’s Equation, and f (z, φ) can be developed into a double Fourier’s Series [v. (15) Art. 71]. 10. 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. 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 Dt2 y = c2 (xDx2 y +Dx y), subject to the conditions Dt y = 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.

Show that

y=

k=∞ X

Ak cos µk ct B0 (µ2k x),

k=1

where

x2 x3 x + − + ··· 12 12 .22 12 .22 .32 √ = J0 (2 x)

B0 (x) = 1 −

and µk is a root of the equation √ B0 (µ2 a) = J0 (2µ a) = 0, wa and

Ak =

wa

f (x)B0 (µ2k x)dx

0

µ2 a2 [B00 (µ2k a)]2

=

0

√ f (x)J0 (2µk x)dx √ a[J1 (2µk a)]2

.

. 12. 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 nφ and by taking m n = . The required solution is 2 z=

m=∞ X k=∞ X m=1 k=1

Bm,k cos µk ct sin

mφ J m (µk r) 2 2


237 J m2 (µa)

where µk is a root of

and

Bm,k =

m

m

µ2a2

2 π

2π w 0

dφ

wa

=0

rf (r, φ) sin

0

mφ J m (µk r)dr 2 2

a2 [J m2 0 (µk a)]2

.

For the terms in which m is odd, J m2 (x) can be readily obtained from (13) Art. 122, which will become a finite sum. For example, (13) Art. 122 gives the values r r 2 2 1 sin x; J 32 (x) = sin x − cos x ; J 12 (x) = πx πx x r 2 3 3 J 25 (x) = − 1 + 2 sin x + cos x ; &c. πx x x 13. 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 a2 1 1 2 2 Dt u = 2 Dr (r Dr u) + Dφ u (1) Dθ (sin θDθ u) + r sin θ sin2 θ ([IV] Art. 1) subject to the conditions u=0

when r = c,

u = f (r, θ, φ)

when t = 0.

If we assume u = T.R.V where T is a function of t only, R of r only, and V of θ and φ only, (1) can be broken up into dT + a2 α2 T = 0 dt 1 1 m(m + 1)V + Dθ (sin θDθ V ) + Dφ2 V = 0 sin θ sin2 θ d2 R 2 dR m(m + 1) 2 + α − R = 0. + dr2 r dr r2

and 2

2

(2) (3)

(4)

Hence T = e−a α t , V = Ym (µ, φ)√[v. Art. 102 (2)], and R is still to be found. If in (4) we let x = αr and z = R αr it becomes (m + 21 )2 d2 z 1 dz + + 1 − z=0 dx2 x dx x2


238

which is satisfied by z = Jm+ 12 (x). (v. Art. 17.) 1 R = √ Jm+ 12 (αr). αr

Therefore

f (r, θ, φ) =

2π m=∞ w wπ 1 X (2m + 1) dφ1 f (r, θ1 , φ1 )Pm (cos γ) sin θ1 dθ1 4π m=0 0

0

by (3) Art. 114, =

m=∞ X n=m X

n [Am,n fm,n (r) cos nφ + Bm,n Fm,n (r) sin nφ]Pm (µ).

m=0 n=0

√

rfm,n (r) =

k=∞ X

Cm,n,k Jm+ 12 (αk r)

k=0

where αk is a root of the equation Jm+ 12 (αc) 1

(αc)m+ 2

2 and

wc

=0

3

r 2 fm,n (r)Jm+ 12 (αk r)dr

0

Cm,n,k =

0 2 c2 [Jm+ 1 (αk c)]

.

2

√

rFm,n (r) =

m=∞ X

Dm,n,k Jm+ 12 (αk r)

m=0

2 where

wc

Dm,n,k =

3

r 2 Fm,n (r)Jm+ 12 (αk r)dr

0 0 2 c2 [Jm+ 1 (αk c)]

.

2

The final solution is m=∞ n=m k=∞ X 1 X X n √ Pm (µ) (Am,n Cm,n,k cos nφ u= r m=0 n=0 k=1

+ Bm,n Dm,n,k sin nφ)e cf. Riemann, Par. Dif. Gl., §§ 72 and 73.

−a2 α2k t

Jm+ 12 (αk r)

239

CHAPTER VIII. ¨ LAPLACE’S EQUATION IN CURVILINEAR COORDINATES. ELLIPSOIDAL HARMONICS. 130.

Orthogonal Curvilinear Co¨ ordinates.  F1 (x, y, z) = ρ1   F2 (x, y, z) = ρ2   F3 (x, y, z) = ρ3

If

(1)

are the equations in rectangular coördinates of three surfaces that are mutually perpendicular no matter what the values of ρ1 , ρ2 , and ρ3 , the parameters ρ1 , ρ2 , and ρ3 , may be regarded as a set of coördinates for a point of intersection of the three surfaces, in the sense that when ρ1 , ρ2 , ρ3 are given the point in question is determined, and when the point is given the corresponding values of ρ1 , ρ2 , ρ3 , can be found. From equations (1) x, y, and z can be expressed in terms of ρ1 , ρ2 , and ρ3 . Suppose this done. If now x, y, z are the rectangular coördinates of the point ρ1 = a, ρ2 = b, ρ3 = c, the rectangular coördinates of the points ρ1 = a + dρ1 , ρ2 = b, ρ3 = c, are obviously x+Dρ1 x.dρ1 +1 , y+Dρ1 y.dρ1 +2 , z+Dρ1 z.dρ1 +3 , where 1 , 2 , and 3 are infinitesimals of higher order than dρ1 . Hence the square of the distance between the points will differ by an infinitesimal of higher order than that of dρ21 from dn21 where dn21 = [(Dρ1 x)2 + (Dρ1 y)2 + (Dρ1 z)2 ]dρ21 . Let

 1 = (Dρ1 x)2 + (Dρ1 y)2 + (Dρ1 z)2   2  h1     1 2 2 2 = (D x) + (D y) + (D z) ρ2 ρ2 ρ2 h22     1  2 2 2   = (D x) + (D y) + (D z) . ρ ρ ρ 3 3 3 h23

(2)

Then if dn1 is the element of length normal to the surface ρ1 = a, dn2 normal to ρ2 = b, and dn3 normal to ρ3 = c dn1 =

dρ1 , h1

dn2 =

dρ2 , h2

dn3 =

dρ3 . h3

(3)

The element of surface dS1 on the surface ρ1 = a is easily seen to be dρ2 dρ3 ; h2 h3

(4)

dρ1 dρ2 dρ3 . h1 h2 h3

(5)

dS1 = and the element of volume dv is dv =

ELLIPSOIDAL HARMONICS.

240 EXAMPLE.

Show that

h21 = (Dx ρ1 )2 + (Dy ρ1 )2 + (Dz ρ1 )2 h22 = (Dx ρ2 )2 + (Dy ρ2 )2 + (Dz ρ2 )2 h23 = (Dx ρ3 )2 + (Dy ρ3 )2 + (Dz ρ3 )2 .

Dx ρ1 Dy ρ1 Dz ρ1 , , are the direch1 h1 h1 tion cosines of the normal at any given point of ρ1 = a. (v. Int. Cal. page 161.) Then Dx ρ1 D y ρ1 D z ρ1 1 dn1 = dx + dy + dz = dρ1 . h1 h1 h1 h1 Suggestion: If h1 has the value just given

131. Laplace’s Equation in orthogonal curvilinear co¨ ordinates. If we apply the special form of Green’s Theorem y w ∇2 V dxdydz = Dn V dS (v. Art. 98) to the space bounded by the surfaces ρ1 = a, ρ2 = b, ρ3 = c, ρ1 = a + dρ1 , ρ2 = b + dρ2 , ρ3 = c + dρ3 , we have ∇2 V dρ1 dρ2 dρ3 = h1 h2 h3 dρ2 dρ3 −h1 Dρ1 V + h1 Dρ1 V h2 h3 dρ3 dρ1 + h2 Dρ2 V −h2 Dρ2 V h3 h1 dρ1 dρ2 −h3 Dρ3 V + h3 Dρ3 V h1 h2

dρ2 dρ3 h1 + Dρ1 Dρ V dρ1 dρ2 dρ3 h2 h3 h2 h3 1 h2 dρ3 dρ1 + Dρ2 Dρ V dρ1 dρ2 dρ3 h3 h1 h3 h1 2 dρ1 dρ2 h3 + Dρ3 Dρ V dρ1 dρ2 dρ3 ; h1 h2 h1 h2 3

whence h2 h3 h1 ∇2 V = h1 h2 h3 Dρ1 Dρ1 V + Dρ2 Dρ2 V + Dρ3 Dρ3 V , h2 h3 h3 h1 h1 h2 (6) and Laplace’s Equation in our curvilinear system is h1 h2 h3 h1 h2 h3 Dρ1 Dρ V + Dρ2 Dρ V + Dρ3 Dρ V = 0. (7) h2 h3 1 h3 h1 2 h1 h2 3 If it happens that ∇2 ρ1 = 0, V = ρ1 will satisfy (7) and we shall have h1 h2 h1 h2 h3 Dρ1 = 0. In like manner if ∇2 ρ2 = 0 we have Dρ2 = 0, h2 h3 h 3 h1 h3 = 0; and therefore (7) reduces to and if ∇2 ρ3 = 0 we have Dρ3 h1 h2 h21 Dρ21 V + h22 Dρ22 V + h23 Dρ23 V = 0

(8)


241

when ∇2 ρ1 = 0, ∇2 ρ2 = 0, and ∇2 ρ3 = 0. 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 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 co¨ ordinates. 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 spheroid and the hyperboloid orthogonally. The equations of the three surfaces can be written:— y2 z2 x2 + + −1=0 λ2 λ2 − b2 λ2 y2 z2 x2 + 2 −1=0 F2 (x, y, z, µ) = 2 + 2 2 µ µ −b µ F3 (x, y, z, ν) = z − νx = 0, F1 (x, y, z, λ) =

(1) (2) (3)

where λ2 > b2 > µ2 , 2b being the distance between the foci. For all values of λ, µ, and ν consistent with the inequality above written the surfaces (1), (2), (3) intersect in real points and cut orthogonally. λ, µ, and ν 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 [XV] Art. 1. From (1), (2), and (3) we readily get  λ2 µ2  2  x = 2    b (1 + ν 2 )   2 2 2 2  (λ − b )(b − µ ) 2 (4) y =  b2     λ2 µ2 ν 2   z2 = 2 ;  b (1 + ν 2 ) r µ λ b2 − µ2 µν whence Dλ x = √ , Dλ y = , Dλ z = √ ; 2 − b2 2 b λ b 1+ν b 1 + ν2 and

1 λ2 − µ2 = 2 2 h1 λ − b2

(5)

[v. 130 (2)]. In like manner we get 1 λ2 − µ2 = 2 2 h2 b − µ2

(6)


242

λ2 µ2 1 = 2 , 2 h3 b (1 + ν 2 )2

and

(7)

and [XV] Art. 1 becomes p µ p Dλ [λ λ2 − b2 .Dλ V ] b(1 + ν 2 ) b2 − µ2 p λ √ Dµ [µ b2 − µ2 .Dµ V ] + b(1 + ν 2 ) λ2 − b2 b(λ2 − µ2 ) + p Dν [(1 + ν 2 )Dν V ] = 0, λµ (λ2 − b2 )(b2 − µ2 )

(8)

which is Laplace’s Equation in terms of our Spheroidal Co¨ ordinates λ, µ, and ν. If now in place of λ, µ, and ν we can introduce some function of λ, some function of µ, and some function of ν which, therefore, will represent the same set of orthogonal surfaces, and if we can choose these functions α, β, and γ, which of course are functions of x, y, and z, so that ∇2 α = 0, ∇2 β = 0, and ∇2 γ = 0, equation (8) must reduce to the simple and symmetrical form given in [XVI] Art. 1. These functions α, β, and γ are easily found. Equation (8) is ∇2 V = 0 expressed in terms of λ, µ, and ν. Assume that V is a function of λ only; then Dµ V = 0, and Dν V = 0, and (8) reduces to p Dλ [λ λ2 − b2 .Dλ V ] = 0 p dV whence λ λ2 − b2 = c1 , dλ c1 dλ , dV = √ λ λ2 − b2 c1 λ and V = sec−1 , b b and is a function of λ which satisfies Laplace’s Equation. Take this as α leaving c1 at present undetermined, so that c1 dλ dα = √ λ λ2 − b2

and α =

c1 λ sec−1 . b b

and β =

c2 µ sech−1 , b b

In the same way we get dβ =

c dµ p2 µ b2 − µ2

(v. Int. Cal. Art. 46, Ex.) dγ =

c3 dν , 1 + ν2

and γ = c3 tan−1 ν.


243

Substituting these values in (8) and taking c1 = −c2 = b, and c3 = 1, (8) reduces at once to Dβ2 V λ2 − µ2 2 Dα2 V + + Dγ V = 0, λ2 µ2 λ2 µ2 λ = b sec α, µ = b sech β, and ν = tan γ,

or since

cos2 αDα2 V + cosh2 βDβ2 V + (cosh2 β − cos2 α)Dγ2 V = 0

to

(9) (10) (11)

which is Laplace’s Equation in terms of what we may call Normal Oblate Spheroidal Co¨ ordinates. In using (11) it is to be noted that the point whose coördinates are (α, β, γ) is the point of intersection of an oblate spheroid whose semi-axes are b sec α and b tan α, an unparted hyperboloid of revolution whose semi-axes are b sech β and b tanh β, and a plane containing the axis of the system and making the angle γ 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 (α, β, γ) are x = b sec α sech β cos γ,

y = b tan α tanh β,

z = b sec α sech β sin γ

(12)

[v. (4)]. π If now we let α range from 0 to , β from −∞ to ∞, and γ from 0 to 2π, we 2 shall be able to represent all points in space; and if we agree that negative values of β shall belong to points below a plane through the origin and perpendicular to the axis of revolution and positive values of β 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. EXAMPLES. 1. If the spheroid is a 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, x2 y2 z2 + + −1=0 λ2 λ2 − b2 λ2 − b2 y2 z2 x2 + 2 + 2 −1=0 2 2 µ µ −b µ − b2 z − νy = 0 λ2 > b2 > µ2 .

where h21 =

λ2 − b2 , λ2 − µ2

h22 =

b2 − µ2 , λ2 − µ2

h23 =

(λ2

b2 (1 + ν 2 )2 . − b2 )(b2 − µ2 )


244

Laplace’s Equation becomes 1 1 Dλ [(λ2 − b2 )Dλ V ] + 2 Dµ [(b2 − µ2 )Dµ V ] b2 (1 + ν 2 ) b (1 + ν 2 ) λ2 − µ2 Dν [(1 + ν 2 )Dν V ] = 0. (1) + 2 (λ − b2 )(b2 − µ2 ) (1) reduces to where

Dβ2 V Dα2 V λ2 − µ2 + 2 + 2 D2 V = 0, 2 2 2 λ −b b −µ (λ − b2 )(b2 − µ2 ) γ dα = −

bdλ , λ 2 − b2

α = ctnh−1 Since

λ , b

λ = b ctnh α,

dβ =

bdµ , b2 − µ2

β = tanh−1

µ , b

µ = b tanh β,

dγ =

(2)

dν , 1 + ν2

and γ = tan−1 ν. and ν = tan γ

(2) can be reduced to sinh2 αDα2 V + cosh2 βDβ2 V + (sinh2 α + cosh2 β)Dγ2 V = 0.

(3)

In using (3) it is to be noted that the point (α, β, γ) is the point of intersection of a prolate spheroid whose semi-axes are b ctnh α and b csch α, a biparted hyperboloid of revolution whose semi-axes are b tanh β and b sech β, and a plane containing the axis of revolution and making the angle γ with a fixed plane. If the fixed plane is that of (XY ) the rectangular coördinates of any point (α, β, γ) are x = b ctnh α tanh β,

y = b csch α sech β cos γ,

z = b csch α sech β sin γ,

and α may range from ∞ to 0, β from −∞ to ∞, and γ from 0 to 2π. Negative values of β are to be taken for points lying to the left of a plane through the origin perpendicular to the axis of revolution. 2. Transform Laplace’s Equation in Spherical Coördinates [XIII] Art. 1 to the symmetrical form α2 Dα2 V + cosh2 βDβ2 V + cosh2 βDγ2 V = 0 θ 1 , β = log tan , and γ = φ. r 2 3. Transform Laplace’s Equation in Cylindrical Coördinates [XIV] Art. 1 to the symmetrical form

where

α=

Dα2 V + Dβ2 V + e2α Dγ2 V = 0 where

α = log r,

β = φ,

and γ = z.


245

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 [XVI] Art. 1; and it is evident that our new co¨ ordinate α is a value of V corresponding to such a distribution that the surfaces obtained by giving particular values to ρ1 are equipotential surfaces; that β is a value of V corresponding to such a distribution that the surfaces obtained by giving particular values to ρ2 are equipotential surfaces; and that γ is a value of V corresponding to such a distribution that the surfaces obtained by giving particular values to ρ3 are equipotential surfaces. α, β, and γ are called by Lamé “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 α. It is merely the condition that V in Laplace’s Equation may be a function of ρ1 alone. If V is a function of ρ1 alone Dx V =

dV Dx ρ1 , dρ1

Dy V =

dV D y ρ1 , dρ1

Dz V =

dV Dz ρ1 , dρ1

d2 V dV 2 2 2 (Dx ρ1 ) + dρ Dx ρ1 dρ1 1 2 dV 2 d V (Dy ρ1 )2 + Dy2 V = D ρ1 dρ1 y dρ1 2 dV 2 d2 V (Dz ρ1 )2 + D ρ1 . Dz2 V = dρ1 z dρ1 2

Dx2 V =

Therefore [(Dx ρ1 )2 + (Dy ρ1 )2 + (Dz ρ1 )2 ] whence

or

d2 V dV + [Dx2 ρ1 + Dy2 ρ1 + Dz2 ρ1 ] =0 2 dρ1 dρ1

Dx2 ρ1 + Dy2 ρ1 + Dz2 ρ1 d2 V dV =− ÷ . 2 2 2 (Dx ρ1 ) + (Dy ρ1 ) + (Dz ρ1 ) dρ1 2 dρ1 ∇2 ρ 1 = F1 (ρ1 ) h21

where F1 (ρ1 ) may be any function of ρ1 alone. Our required conditions are then  ∇2 ρ1   = F (ρ ) 1 1   h21     2 ∇ ρ2 (1) = F (ρ ) 2 2  h22      ∇2 ρ3   = F (ρ ) 3 3 2 h3 and when they are fulfilled the original curvilinear coördinates ρ1 , ρ2 , ρ3 , correspond to possible equipotential or isothermal surfaces, thermometric parameters


246

α, β, and γ exist, and the reduction of Laplace’s Equation to the symmetrical form [XVI] Art. 1 is possible. 134. Returning to our Oblate Spheroid problem of Art. 132 we can proceed as usual to break up our equation (11) Art. 132. Assume that V = L.M.N , where L is a function of α only, M of β only, and N of γ only. (11) Art. 132 becomes [cosh2 β − cos2 α] d2 N cos2 α d2 L cosh2 β d2 M + + =0 L dα2 M dβ 2 N dγ 2 1 d2 L 1 d2 M 1 d2 N cos2 α cosh2 β + = − . 2 2 L cosh β − cos2 α dα2 M cosh β − cos2 α dβ 2 N dγ 2

or

The first member is independent of γ, and the second member is independent of α and β, and the two members are identically equal. The second member is then independent of α, β, and γ and must be constant; call it n2 . We have, then, d2 N + n2 N = 0 dγ 2 cos2 α d2 L cosh2 β d2 M and + − n2 (cosh2 β − cos2 α) = 0. L dα2 M dβ 2 (1) gives us N = A cos nγ + B sin nγ.

(1) (2) (3)

(2) can be written

whence and

cosh2 β d2 M cos2 α d2 L 2 2 2 2 + n cos α = n cosh β − = m(m + 1), L dα2 M dβ 2 d2 L cos2 α 2 + [n2 cos2 α − m(m + 1)]L = 0 (4) dα d2 M cosh2 β + [m(m + 1) − n2 cosh2 β]M = 0. (5) dβ 2

If we introduce x = tanh β in (5) it becomes 2 n2 dM 2 d M (1 − x ) 2 − 2x + m(m + 1) − M =0 dx dx 1 − x2

(6)

where since x = tanh β and β may have any value from −∞ to ∞, x may have any value between −1 and 1. (6) is a familiar equation having for a particular solution n n d Pm (x) n n M = (1 − x2 ) 2 = Pm (x) = Pm (tanh β). (7) dxn (v. Arts. 101 and 102). If we introduce in (4) x = tan α it reduces to d2 L dL n2 (1 + x2 ) 2 + 2x + − m(m + 1) L = 0. (8) dx dx 1 + x2


247

(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. y This labor can be saved, however, by noting that if we let x = (8) becomes i dL n2 d2 L L=0 + m(m + 1) − (1 − y 2 ) 2 − 2y dy dy 1 − y2 and is identical in form with (6). Hence n L = Pm (y)

n

and L = (1 − y 2 ) 2

dn Qm (y) dy n

(v. Art. 101),

where y = i tan α, are particular solutions of (4). We can avoid imaginaries if we use the values n L = (−i)m−n Pm (y)

n

and L = im+n+1 (1 − y 2 ) 2

dn Qm (y) . dy n

(9)

Since we assumed V = L.M.N we have n n V = (A cos nγ + B sin nγ)Pm (tanh β)(−i)m−n Pm (i tan α)

and

n V = (A cos nγ + B sin nγ)Pm (tanh β)im+n+1 secn α

 

dn Qm (i tan α) (10)  (d(i tan α))n

as particular solutions of (11) Art. 132. If the problem is symmetrical with respect to the axis of the spheroid Dγ2 V = 0, n2 = 0 and our particular solutions (10) reduce to ) V = (−i)m Pm (i tan α)Pm (tanh β) (11) V = im+1 Qm (i tan α)Pm (tanh β). and If, then, V is given on the surface of a spheroid as a function of β and γ, we must express it as a function of tanh β and γ, and shall be obliged to develop it in terms of Spherical Harmonics of tanh β and γ 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 β by the formulas of Chapter VI. A convenient form for Qm (i tan α) is obtained from (2) Art. 100; it is Qm (i tan α) = −iPm (i tan α)

w∞ tan α

Hence

Q0 (i tan α) = −i

w∞ tan α

dx . (1 + x2 )[Pm (xi)]2

π dx = −i − α . 1 + x2 2

(12) (13)


248 EXAMPLES.

1. A conductor in the form of an oblate spheroid whose semi-axes are b sec α0 and b tan α0 is charged with electricity and is found to be at potential V0 ; find the value of the potential function at any internal or external point. Here V0 = V0 P0 (tanh β). Hence at an internal point V = V0

P0 (i tan α) P0 (tanh β) = V0 , P0 (i tan α0 )

(1)

and at an external point π −α Q0 (i tan α) . V = V0 P0 (tanh β) = V0 π2 Q0 (i tan α0 ) − α0 2

(2)

Since V in (2) involves α only, the equipotential surfaces are all spheroids confocal with the conductor. 2. The upper half of an oblate spheroid whose semi-axes are b sec α0 and b tan α0 is kept at the temperature unity, and the lower half at the temperature zero. Find the permanent temperature at any internal point. Ans.

u=

1 3 P1 (i tan α) 7 1 P3 (i tan α) + P1 (tanh β) − . P3 (tanh β) + · · · 2 4 P1 (i tan α0 ) 8 2 P3 (i tan α0 )

(v. Art. 93). u may be expressed in terms of x, y, and z without serious difficulty [v. (12) Art. 132]. u=

1 3 y 7 1 1 [25y 3 − 15y(x2 + y 2 + z 2 − b2 ) − 9b2 y] + − . . + ··· 2 4c 8 2 2 5c3 + 3b2 c

if 2c = 2b tan α0 = minor axis of spheroid. 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 α = φ and α = φ + dφ. By (1) Art. 98 we have 4πρκ = [Dn V1 − Dn V2 ]α=φ ,

(1)

where ρ is the density and κ the thickness of the shell, V1 the value of the potential function at an internal point, and V2 the value of the potential function at an external point. X Let V1 = Am (−i)m Pm (i tan α)Pm (tanh β) X and V2 = Bm im+1 Qm (i tan α)Pm (tanh β) [v. (11) Art. 134].


249

Since V1 and V2 must have the same value when α = φ Am = Bm i2m+1

w∞ Qm (i tan φ) dx = (−1)m Bm 2 Pm (i tan φ) (1 + x )[Pm (xi)]2

(2)

tan φ

[v. (12) Art. 134]. Hence

and

V1 =

X

V2 =

X

im Bm Pm (tanh β)Pm (i tan α) m

i Bm Pm (tanh β)Pm (i tan α)

w∞ tan φ w∞ tan α

Dn V1 = Dα V1 .Dn α.

      

dx (1 + x2 )[Pm (xi)]2

(3)

  dx  .  2 2 (1 + x )[Pm (xi)] 

Dn V2 = Dα V2 .Dn α

[Dn V1 − Dn V2 ]α=φ = [Dα V1 − Dα V2 ]α=φ (Dn α)α=φ = [Dα (V1 − V2 )]α=φ [Dn α]α=φ . V1 − V2 =

X

im Bm Pm (tanh β)Pm (i tan α)

tan wα tan φ

Dα (V1 − V2 ) =

X

dx (1 +

x2 )[P

2 m (xi)]

.

sec2 α (1 + tan α)[Pm (i tan α)]2 tan α dPm (i tan α) w dx + . dα (1 + x2 )[Pm (xi)]2

i Bm Pm (tanh β) Pm (i tan α) m

2

tan φ

Dα [V1 − V2 ]α=φ =

X

Dn α =

im Bm

Pm (tanh β) . Pm (i tan φ)

dα dn

p q dρ1 dλ λ2 − µ2 = = √ dλ = b sec α tan2 α + tanh2 β.dα h1 h1 λ2 − b2 v. Art. 130 (3), and Art. 132 (5) and (10). dn =

[Dn α]α=φ =

1 p . b sec φ tan2 φ + tanh2 β

Hence X 1 Pm (tanh β) p . im Bm 2 2 Pm (i tan φ) b sec φ tan φ + tanh β q = b sec φ tan2 φ + tanh2 β.dφ

[Dn V1 − Dn V2 ]α=φ = κ = [dn]α=φ

(4)


250

by (4), and (1) may be written 4πρb2 sec2 φ(tan2 φ + tanh2 β)dφ =

X

im Bm

Pm (tanh β) . Pm (i tan φ)

(5)

tanh2 β = 31 P0 (tanh β) + 23 P2 (tanh β)

Since

by (5) Art. 95, to satisfy (5) we must give m the values 0 and 2 and B0 = 43 πρb2 sec2 φ(3 tan2 φ + 1)dφ B2 = 43 πρb2 sec2 φ(3 tan2 φ + 1)dφ.

and So that by (3) V1 =

2 4 3 πρb

2

2

sec φ(3 tan φ + 1)dφ

w∞ tan φ

dx 1 + x2

− P2 (tanh β)P2 (i tan α)

w∞ tan φ

and

dx (1 + x2 )[P2 (xi)]2

(6)

V2 = 34 πρb2 sec2 φ(3 tan2 φ + 1)dφ[iQ0 (i tan α) + i3 P2 (tanh β)Q2 (i tan α)].

(7)

The potential function at an external point due to the solid spheroid for which α = α0 is V =

φ=α w 0

V2 = 34 πρb2 sec2 α0 tan α0 [iQ0 (i tan α) + i3 P2 (tanh β)Q2 (i tan α)]. (8)

φ=0

If 2a is the major axis and 2c the minor axis of the spheroid 2 4 3 πρb

sec2 α0 tan α0 =

4 3

πρa2 c M = b b

where M is the mass of the spheroid. Therefore V =

M [iQ0 (i tan α) + i3 P2 (tanh β)Q2 (i tan α)] b

(9)

is the required value. (9) can be reduced to i M π 1 h π 2 2 V = −α+ − α (3 tan α + 1) − 3 tan α [3 tanh β − 1] . (10) b 2 4 2


251 EXAMPLES.

1. Break up the equation (3) Ex. 1, Art. 132, for the prolate spheroid, and obtain particular solutions of the term n n V = (A cos nγ + B sin nγ)Pm (tanh β)Pm (ctnh α), n

n V = (A cos nγ + B sin nγ)Pm (tanh β)(−1) 2 cschn α

dn Qm (ctnh α) . (d ctnh α)n

2. Break up and solve the equations of Exs. 2 and 3, Art. 132, and show that they lead to familiar forms. 3. If in Ex. 1, Art. 132, the conductor is a prolate spheroid whose semi-axes are b ctnh α0 and b csch α0 show that V = V0 at an internal point.

V = V0

α at an external point. α0

4. Show that the potential function at an external point due to the attraction of a homogeneous solid prolate spheroid is V =

M [Q0 (ctnh α) − P2 (tanh β)Q2 (ctnh α)]. b Ellipsoidal Harmonics.

136. If we are dealing with an ellipsoid instead of a spheroid, we can take as our orthogonal system of surfaces a set of confocal quadrics;  y2 z2 x2   + + − 1 = 0   λ2 λ2 − b2 λ2 − c2    2 2 2 x y z (1) + + − 1 = 0  µ2 µ2 − b2 µ2 − c2     x2 y2 z2   + + − 1 = 0 ν2 ν 2 − b2 ν 2 − c2 where λ2 > c2 > µ2 > b2 > ν 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. λ, µ, and ν will be our curvilinear co¨ ordinates, and are known as Ellipsoidal Coördinates. We find without difficulty that x2 =

λ2 µ2 ν 2 , b2 c 2

y2 =

(λ2 − b2 )(µ2 − b2 )(b2 − ν 2 ) , b2 (c2 − b2 ) (λ2 − c2 )(c2 − µ2 )(c2 − ν 2 ) z2 = , (2) c2 (c2 − b2 )

ELLIPSOIDAL HARMONICS. h21 =

(λ2 − b2 )(λ2 − c2 ) , (λ2 − µ2 )(λ2 − ν 2 )

252

h22 =

(µ2 − b2 )(c2 − µ2 ) , (µ2 − ν 2 )(λ2 − µ2 ) (b2 − ν 2 )(c2 − ν 2 ) . (3) h23 = 2 (λ − ν 2 )(µ2 − ν 2 )

p To avoid ambiguity, we shall suppose that of the nine semi-axes in (1) c2 − µ2 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; b2 − ν 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; ν 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 Co¨ ordinates have the disadvantage that to fully fix a point we need p to know not merely the values of its co¨ ordinates λ, µ, and ν, but the signs of c2 − µ2 , and √ b2 − ν 2 as well. We shall see later, Art. 139, when we come to introduce what we may call the Normal Ellipsoidal Co¨ ordinates α, β, and γ that they are free from this disadvantage. It is to be observed that λ may range from c to ∞, µ from b to c, and ν from −b to b. The element of length perpendicular to the Ellipsoid is s dλ (λ2 − µ2 )(λ2 − ν 2 ) dn = .dλ. (4) = h1 (λ2 − b2 )(λ2 − c2 ) The element of Ellipsoidal surface is s (λ2 − µ2 )(λ2 − ν 2 ) dµdν .dµdν, dS = = (µ2 − ν 2 ) h2 h3 (µ2 − b2 )(c2 − µ2 )(b2 − ν 2 )(c2 − ν 2 ) and the element of volume is dλdµdν dv = h1 h2 h3 (λ2 − µ2 )(λ2 − ν 2 )(µ2 − ν 2 ) =p dλdµdν. (λ2 − b2 )(λ2 − c2 )(µ2 − b2 )(c2 − µ2 )(b2 − ν 2 )(c2 − ν 2 )

(5)

(6)

The surface integral of any given function of µ and ν taken over the ellipsoid is w

f (µ, ν)dS =

wb −b

dν

wc

[f1 (µ, ν) + f2 (µ, ν) + f3 (µ, ν)

b

s + f4 (µ, ν)](µ2 − ν 2 )

(λ2 − µ2 )(λ2 − ν 2 ) .dµ, (7) (µ2 − b2 )(c2 − µ2 )(b2 − ν 2 )(c2 − ν 2 )


253

where f1 (µ, ν), f2 (µ, ν), f3 (µ, ν) and f4 (µ, ν) 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 (µ2 − ν 2 )Dα2 V + (λ2 − ν 2 )Dβ2 V + (λ2 − µ2 )Dγ2 V = 0

where

α=c

wλ c

dλ p

(λ2 − b2 )(λ2 − c2 )

,

β=c

γ=c

wµ

p

b wν

(c2 − µ2 )(µ2 − b2 )

p

(b2

0

dµ

dν − ν 2 )(c2 − ν 2 )

(8)

,

.

(9)

α, β, and γ can be expressed as Elliptic Integrals of the first class and are

α=F

whence

b π b c , − F , sin−1 , c 2 c λ

λ=

β=F

r

v u b2 u u1 − 2 2 u b µ 1 − 2 , sin−1 u , t c b2 1− 2 c b −1 ν ; γ = F , sin c b

c b dn α b mod =c mod , sn(K − α) c cn α c b b b2 12 µ= , ν = b sn γ mod mod 1 − 2 dn β c c

(10)

(11)

(v. Int. Cal. Arts. 179, 192, and 196). 137. If in (8) Art. 136 we assume V = L.M.N where L involves α only, M involves β only, and N involves γ only, (8) can be written µ2 − ν 2 d2 L λ2 − ν 2 d2 M λ2 − µ2 d2 N + + = 0. L dα2 M dβ 2 N dγ 2

(1)

(1) is too complicated to be broken up by our usual method. If, however, we let 1 d2 L P = ak λk , L dα2

P 1 d2 M = bk µk , M dβ 2

P 1 d2 N = ck ν k , N dγ 2

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 a0 = −b0 = c0 , and a2 = −b2 = c2 .


254

Therefore (1) can be broken up into the three equations d2 L = (a0 + a2 λ2 )L dα2 d2 M = −(a0 + a2 µ2 )M dβ 2 d2 N = (a0 + a2 ν 2 )N. dγ 2 We shall find it convenient to take a2 as m(m + 1) and a0 as −(b2 + c2 )p; whence  d2 L  2 2 2  − [m(m + 1)λ − (b + c )p]L = 0   dα2      2 d M 2 2 2 + [m(m + 1)µ − (b + c )p]M = 0 (2) dβ 2        d2 N 2 2 2   − [m(m + 1)ν − (b + c )p]N = 0. dγ 2 If now in (2) we replace α, β, and γ by their values in terms of λ, µ, and ν, we get  dL d2 L   (λ2 − b2 )(λ2 − c2 ) 2 + λ(λ2 − b2 + λ2 − c2 )   dλ dλ    2 2 2 −[m(m + 1)λ − (b + c )p]L = 0      2   d M dM 2 2 2 2 2 2 2 2  (µ − b )(µ − c ) 2 + µ(µ − b + µ − c ) dµ dµ (3)   −[m(m + 1)µ2 − (b2 + c2 )p]M = 0      2  d N dN  2 2 2 2 2 2 2 2  (ν − b )(ν − c ) 2 + ν(ν − b + ν − c )    dν dν   2 2 2 −[m(m + 1)ν − (b + c )p]N = 0. p p p (µ) and N = Em (ν), and that (λ), it follows that M = Em Whence if L = Em p p p V = Em (λ)Em (µ)Em (ν)

(4)

is a solution of Laplace’s Equation, (8) Art. 136. The equation (x2 − b2 )(x2 − c2 )

dz d2 z + x(x2 − b2 + x2 − c2 ) 2 dx dx − [m(m + 1)x2 − (b2 + c2 )p]z = 0

(5)

p is known as Lamé’s Equation, and Em (x) as a Lamé’s Function or an Ellipsoidal Harmonic. We shall suppose m a positive integer.


255

To get a particular solution of (5) let z = reduce and we get

P

ak xk . Substitute in (5) and

[k(k + 1) − m(m + 1)]ak − (b2 + c2 )[(k + 2)2 − p]ak+2 + b2 c2 (k + 3)(k + 4)ak+4 = 0. (6) We have now only to choose a sequence of coefficients satisfying (6), and we may take any two consecutive coefficients arbitrarily. (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 am+2 = 0, am+4 , am+6 , &c., will vanish. Let am = 1. If m is even the coefficient of a0 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 am−2 , am−4 , &c., by the aid of (6) we see that am−2 is of the first degree in p, am−4 of the second degree in p, &c., and a−2 of the m m + 1 in p. There are then + 1 values of p which we shall call p1 , degree 2 2 p2 , p3 , &c., for which a−2 will vanish, and for which our solutions will be of the form p Em (x) = xm + am−2 xm−2 + am−4 xm−4 + · · · + a0 if m is even. If m is odd, the coefficient of a1 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 m+1 m+1 is of the degree in p, and there will be values p1 , p2 , p3 , &c., of p 2 2 for which p Em (x) = xm + am−2 xm−2 + am−4 xm−4 + · · · + a1 x. p (x) so that Following Heine we shall call the solution just obtained Km p Km (x) = xm + am−2 xm−2 + am−4 xm−4 + · · ·

(7)

terminating with a0 if m is even, and with a1 x if m is odd. If m is even, there m+1 m p1 p2 (x), &c., and there are are + 1 of these functions Km (x), Km of them 2 2 if m is odd. The coefficients can be computed √ by the aid of (6). If in Lamé’s Equation (5) we let z = v x2 − b2 we get the equation (x2 − b2 )(x2 − c2 )

Letting v =

P

d2 v dv + x[x2 − b2 + 3(x2 − c2 )] dx2 dx − [(m + 2)(m − 1)x2 + c2 − (b2 + c2 )p]v = 0. (8)

ak xk we obtain the relation

[k(k + 3) − (m + 2)(m − 1)]ak − {(b2 + c2 )[(k + 2)2 − p] + c2 (2k + 5)}ak+2 + b2 c2 (k + 3)(k + 4)ak+4 = 0. (9)


256

m values q1 , q2 , q3 , &c., 2 m+1 values of p for which v = xm−1 + am−3 xm−3 + · · · + a1 x if m is even, and 2 m−1 for which v = + am−3 xm−3 + · · · + a0 if m is odd. √x Calling v x2 − b2 Lpm (x) so that p Lpm (x) = x2 − b2 [xm−1 + am−3 xm−3 + am−5 xm−5 + · · · ], (10) Proceeding exactly as before, we find that there are

m values 2 m+1 p of Em (x), namely Lqm1 (x), Lqm2 (x), &c., of the form (10) if m is even and 2 values if m is odd. By interchanging b and c in (8), (9), and (10) we may show that if p p Mm (x) = x2 − c2 [xm−1 + am−3 xm−3 + am−5 xm−5 + · · · ] (11)

terminating with a1 x if m is even and with a0 if m is odd, we have

m r3 r2 r1 p (x), &c., of the form (x), Mm (x), Mm (x), namely Mm values of Em 2 m+1 (11) if m is even and values if m is odd. 2 p Finally if in Lamé’s Equation (5) we let z = v (x2 − b2 )(x2 − c2 ) we get

there are

(x2 − b2 )(x2 − c2 )

If now we let v =

P

d2 v dv + 3x(x2 − b2 + x2 − c2 ) dx2 dx − [(m + 3)(m − 2)x2 − (b2 + c2 )(p − 1)]v = 0. (12) ak xk we obtain the relation

[k(k + 5) − (m − 2)(m + 3)]ak − (b2 + c2 )[(k + 2)(k + 4) + 1 − p]ak+2 + b2 c2 (k + 3)(k + 4)ak+4 = 0. (13) m values s1 , s2 , s3 , &c., of p for 2 m+1 which v = xm−2 + am−4 xm−4 + am−6 xm−6 + · · · + a0 if m is even, and 2 m−2 m−4 values for which v = x + a x + · · · + a x if m is odd. m−4 1 p p Calling v (x2 − b2 )(x2 − c2 ) Nm (x) so that p p Nm (x) = (x2 − b2 )(x2 − c2 )[xm−2 + am−4 xm−4 + am−6 xm−6 + · · · ] (14) Proceeding as before we find that there are

m values of 2 p s1 s2 s3 Em (x), namely Nm (x), Nm (x), Nm (x), &c., of the form (14) if m is even and m−1 values if m is odd. 2 Summing up our results we see that there are 2m + 1 Ellipsoidal Harmonics √ p 2 2 Em (x) each √ of which is a finite sum√ of the mth degree √ in x, or in x and x − b , or in x and x2 − c2 , or in x and x2 − b2 and x2 − c2 . terminating with a0 if m is even and with a1 x if m is odd, we have


257

It was proved by Lamé that the 2m + 1 values of p, namely p1 , p2 , p3 , &c., q1 , q2 , q3 , &c., r1 , r2 , r3 , &c., s1 , s2 , s3 , &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). E0 (x)

E1 (x)

K0 (x) = 1 M0 (x)= 0

K1 (x) = x √ L1 (x) = x2 − b2 √ M1 (x)= x2 − c2

N0 (x) = 0

N1 (x) = 0

L0 (x) = 0

E2 (x) K2p1 (x)= x2 − 31 [b2 + c2 −

p

(b2 + c2 )2 − 3b2 c2 ]

K2p2 (x)= x2 − 31 [b2 + c2 + (b2 + c2 )2 − 3b2 c2 ] √ L2 (x) = x x2 − b2 √ M2 (x) = x x2 − c2 p N2 (x) = (x2 − b2 )(x2 − c2 ) p

E3 (x) p x K3p1 (x) = x3 − [2(b2 + c2 ) − 4(b2 + c2 )2 − 15b2 c2 ] 5 p x p2 K3 (x) = x3 − [2(b2 + c2 ) + 4(b2 + c2 )2 − 15b2 c2 ] 5 √ p q1 2 L3 (x) = x − b2 [x2 − 15 (b2 + 2c2 − (b2 + 2c2 )2 − 5b2 c2 )] √ p Lq32 (x) = x2 − b2 [x2 − 15 (b2 + 2c2 + (b2 + 2c2 )2 − 5b2 c2 )] √ p M3r1 (x)= x2 − c2 [x2 − 15 (2b2 + c2 − (2b2 + c2 )2 − 5b2 c2 )] √ p M3r2 (x)= x2 − c2 [x2 − 15 (2b2 + c2 + (2b2 + c2 )2 − 5b2 c2 )] p N3 (x) = x (x2 − b2 )(x2 − c2 )

It is to be noted that since in the solution (4) of Laplace’s Equation, p p p V = Em (λ)Em (µ)Em (ν),

we have the same m and p in each of the three factors, we shall have to deal


258

merely with products made up of factors of the same form, for example, pk pk pk (λ)Km (µ)Km (ν), Km

Lqmk (λ)Lqmk (µ)Lqmk (ν),

&c.;

and that in a solution of the form X p p p V = Am,p Em (λ)Em (µ)Em (ν) we shall have for a given m just 2m + 1 terms. 138. From the particular solution of Lamé’s Equation [(5) Art. 137] z = p Em (x), we can get by formula (5), Art. 18, the general solution. p p z = AEm (x) + BEm (x)

It is

w∞ x

dx p . p 2 2 (x)]2 (x − b )(x2 − c2 )[Em

(1)

Making A = 0 and B = 2m + 1 we get a second form of particular solution of p (x) where Lamé’s Equation, z = Fm p p Fm (x) = (2m + 1)Em (x)

w∞ x

dx p . p 2 2 (x)]2 (x − b )(x2 − c2 )[Em

(2)

p (x) a Lamé’s Function of the second kind. We shall call Fm It is easily seen to approach the value zero as x is indefinitely increased.

EXAMPLES. 1. If an ellipsoidal conductor is charged with electricity, and is found to be at potential V0 , show that since V0 = V0 K0 (λ), V = V0 K0 (λ)K0 (µ)K0 (ν) = V0 at an internal point, and w∞ V = V0 K0 (µ)K0 (ν) K0 (λ) p λ

dx (x2

−

b2 )(x2

÷ K0 (λ0 )

w∞ λ0

= V0

w∞ λ

whence

dx p

(x2 − b2 )(x2 − c2 )

÷

w∞ λ0

− c2 )[K0 (x)]2

dx p

(x2 − b2 )(x2 − c2 )[K0 (x)]2

b c F , sin−1 λ , p = V0 c b (x2 − b2 )(x2 − c2 ) −1 c F , sin c λ0 dx

b π F , −α V = V0 c 2 b π F , − α0 c 2

v. (10) Art. 136.


259

2. 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 (l2 − µ2 )(l2 − ν 2 ) = 13 [3l4 − 2(b2 + c2 )l2 + b2 c2 ]K0 (µ)K0 (ν) b2 + c2 − 3l2 K2p1 (µ)K2p1 (ν) + 21 1 + p (b2 + c2 )2 − 3b2 c2 b2 + c2 − 3l2 1 p + 2 1− K2p2 (µ)K2p2 (ν); 2 2 2 2 2 (b + c ) − 3b c and that wλ0

3l 4 3 πρ

0

4

− 2(b2 + c2 )l2 + b2 c2 p dl = 43 πρλ0 (l2 − b2 )(l2 − c2 )

q (λ20 − b2 )(λ20 − c2 ) = M

where M is the mass of the ellipsoid. Ans.

V =M

w∞ λ

dx p

(x2 − b2 )(x2 − c2 )

3 − p 2 (b2 + c2 )2 − 3b2 c2

w∞ K2p1 (µ)K2p1 (ν)K2p1 (λ) p λ

− K2p2 (µ)K2p2 (ν)K2p2 (λ)

w∞ λ

139.

dx (x2

−

in the formulas (11) Art. 136 we have dn α (modk), cn α

µ=

− c2 ).(K2p1 (x))2

dx p (x2 − b2 )(x2 − c2 ).(K2p2 (x))2

If for the sake of brevity we represent

λ=c

b2 )(x2

.

b2 21 b by k, and 1 − 2 by k 0 c c

b , dn β(modk 0 )

ν = b sn γ(modk)

(1)

and from these we get without difficulty (v. Int. Cal. Art. 192) p λ2 − b2 =

ck 0 , cn α(modk)

p b2 − ν 2 = b cn γ(modk), p ck 0 cn β (modk 0 ), c2 − µ2 = dn β

 bk 0 sn β  (modk 0 ),    dn β    0 p ck sn α 2 2 λ −c = (modk),  cn α    p   2 2  c − ν = c dn γ(modk).

p

µ2 − b2 =

(2)

If we let α range from 0 to K, and β from 0 to 2K 0 ,and γ from 0to 4K, π π and F k 0 , where K and K 0 are the complete Elliptic Integrals F k, 2 2


260

respectively, (α, β, γ) may represent any point in space, and there will be no ambiguity in sign (v. Art. 136). We may note that if 0 < β < K 0 , z is positive; if K 0 < β < 2K 0 , z is negative; if 0 < γ < K, x and y are both positive; if K < γ < 2K, x is positive and y negative; if 2K < γ < 3K, x and y are both negative; and if 3K < γ < 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 α, β, and γ. We get 1p 2 (λ − µ2 )(λ2 − ν 2 )dα, c p 1 dS = 2 (µ2 − ν 2 ) (λ2 − µ2 )(λ2 − ν 2 )dβdγ, c 1 dv = 3 (λ2 − µ2 )(λ2 − ν 2 )(µ2 − ν 2 )dαdβdγ. c dn =

(3) (4) (5)

For the integral of any function of α, β, and γ over the ellipsoid α = α0 , we shall have 2K 0 4K w w p 1 w F (α, β, γ)dS = 2 dβ F (α0 , β, γ)(µ2 − ν 2 ) (λ2 − µ2 )(λ2 − ν 2 )dγ. (6) c 0

140.

0

If we make use of the formula (2) Art. 92 w (U Dn V − V Dn U )dS = 0

(1)

and take as our closed surface any given ellipsoid, we can get a very important result. If then

p p p U = Em (λ)Em (µ)Em (ν)

and V = Enq (λ)Enq (µ)Enq (ν)

∇2 U = ∇2 V = 0. p p Dn U = Dα U Dn α = Em (µ)Em (ν)

and

Dn V = Dα V Dn α = Enq (µ)Enq (ν)

p (λ) dEm c p , 2 2 dα (λ − µ )(λ2 − ν 2 )

dEnq (λ) c p , 2 2 dα (λ − µ )(λ2 − ν 2 )

p p U Dn V − V D n U = Em (µ)Em (ν)Enq (µ)Enq (ν) p dEnq (λ) dEm (λ) c p q p Em (λ) − En (λ) . 2 2 dα dα (λ − µ )(λ2 − ν 2 )

Integrating U Dn V − V Dn U over the whole ellipsoid, and writing the result


261

equal to zero, we have 0

2K 4K w 1 w p p dβ Em (µ)Em (ν)Enq (µ)Enq (ν) c 0 0 p dEm (λ) dEnq (λ) q p − En (λ) (µ2 − ν 2 )dγ = 0. Em (λ) dα dα

Hence

0 2K w

dβ

0

4K w

p p Em (µ)Em (ν)Enq (µ)Enq (ν)(µ2 − ν 2 )dγ = 0

(2)

dE p (λ) dEnq (λ) − Enq (λ) m = 0. dα dα

(3)

0 p Em (λ)

unless

But as our ellipsoid may be taken at pleasure, λ and α are unrestricted, and if (3) is true it must be true identically. p (λ)]2 it becomes If we divide (3) by [Em d Enq (λ) Enq (λ) = 0 and = a constant; p p (λ) dα Em (λ) Em and this obviously cannot be true unless n = m and q = p. EXAMPLES. 1.

Show that it follows from (2) Art. 140 that K w

0

−K 0

dβ

wK

p p Em (µ)Em (ν)Enq (µ)Enq (ν)(µ2 − ν 2 )dγ = 0.

−K

Suggestion: 0 2K w

p Em (µ)Enq (µ)(µ2

2

− ν )dβ =

0

K w

0

p Em (µ)Enq (µ)(µ2 − ν 2 )dβ

0

+

0 2K w

p Em (µ)Enq (µ)(µ2 − ν 2 )dβ.

K0

If in the last integral we replace β by β + 2K 0 it becomes ±

w0

p Em (µ)Enq (µ)(µ2 − ν 2 )dβ

−K 0

v. Arts. 136 and 139 and Int. Cal. Art. 196.

ELLIPSOIDAL HARMONICS. 2. 0 2K w

dβ

0

262

Show that 4K w

p p [Em (µ)Em (ν)]2 (µ2

2

− ν )dγ = 8

0

K w 0

0

wK p p dβ [Em (µ)Em (ν)]2 (µ2 − ν 2 )dγ. 0

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 α = α0 . We have first to develop in Ellipsoidal Harmonics a function of µ and ν or rather of α and β 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 m=∞ X k=2m+1 X pk pk f (α0 , β, γ) = Am,pk Em (µ)Em (ν), (1) m=0 0 2K w

where

Am,pk =

4K w

dβ

0

k=1

pk pk f (α0 , β, γ)Em (µ)Em (ν)(µ2 − ν 2 )dγ

0

8

K0 w

wK pk pk dβ [Em (µ)Em (ν)]2 (µ2 − ν 2 )dγ

0

.

(2)

0

Our final solution is V =

m=∞ X k=2m+1 X

Am,pk

m=0

k=1

pk Em (λ) pk pk E (µ)Em (ν) pk Em (λ0 ) m

(3)

pk Fm (λ) pk pk E (µ)Em (ν) pk Fm (λ0 ) m

(4)

at an internal point; V =

m=∞ X X k=2m+1

Am,pk

m=0

k=1

at an external point. Lamé has proved rather ingeniously that K w 0

0

wK pk pk dβ [Em (µ)Em (ν)]2 (µ2 − ν 2 )dγ 0

π multiplied by a rational integral 2 b 2 pk function of the coefficients of Em (x) and of c2 and . c Of course the labor of obtaining even a few terms of the development of a function that is in the least complicated is enormous.

can always be found and that it is equal to


263

p If in Laplace’s Equation (8) Art. 136 we let V = Em (λ)U supposing p 1 d2 Em (λ) by U to be a function of β and γ only, we get after replacing p Em (λ) dα2 2 2 2 its value m(m + 1)λ − (b + c )p [v. (2) Art. 137]

142.

(λ2 − ν 2 )Dβ2 U + (λ2 − µ2 )Dγ2 U + (µ2 − ν 2 )[m(m + 1)λ2 − (b2 + c2 )p]U = 0; (1) and since by hypothesis U is independent of λ, the coefficient of λ2 in (1) must vanish. Hence Dβ2 U + Dγ2 U + (µ2 − ν 2 )m(m + 1)U = 0. (2) p p Of course U = Em (µ)Em (ν) will satisfy (2).

EXAMPLES. p p 1. Substitute U = Em (µ)Em (ν) in (2) Art. 142 and by the aid of (2) Art. 137 show that the equation (2) Art. 142 is satisfied.

2.

Obtain (2) Art. 140 directly from (2) Art. 142.

3. Conical Co¨ ordinates. Consider the system of coördinates defined by the equations  x2 + y 2 + z 2 = r 2     2 2 2   y z x + + = 0 (1) µ2 µ2 − b2 µ2 − c2    2 2 2  x y z  + 2 + 2 =0 ν2 ν − b2 ν − c2 where c2 > µ2 > b2 > ν 2 . Show that x2 =

r2 µ2 ν 2 , b2 c 2 h21 =

y2 =

r2 (µ2 − b2 )(ν 2 − b2 ) , b2 (b2 − c2 )

(µ2 − b2 )(c2 − µ2 ) , r2 (µ2 − ν 2 )

h22 =

z2 =

r2 (µ2 − c2 )(ν 2 − c2 ) ; c2 (c2 − b2 )

(b2 − ν 2 )(c2 − ν 2 ) , r2 (µ2 − ν 2 )

h23 = 1.

Laplace’s Equation is Dα2 V + Dβ2 V + (µ2 − ν 2 )Dr (r2 Dr V ) = 0 where

α=

wµ b

dµ p

(µ2

−

b2 )(c2

−

µ2 )

and β =

wν 0

dν p , 2 (b − ν 2 )(c2 − ν 2 )

(2)


264

If V = U.R (2) breaks up into d 2 dR r = m(m + 1)R, dr dr Dα2 U + Dβ2 U + m(m + 1)(µ2 − ν 2 )U = 0. (3) gives

R = Arm + Br−m−1 .

(4) gives

p p U = Em (µ)Em (ν)

(3) (4)

(v. Art. 142).

So that a solution of (2) is p p V = Arm Em (µ)Em (ν).

But since (2) is Laplace’s Equation, V = Arm Ym (µ, φ), if expressed in Conp p ical Co¨ ordinates, must satisfy it, consequently Em (µ)Em (ν) must be simply a Spherical Harmonic of the mth degree. Toroidal Co¨ ordinates. 143. Any pair of circles belonging to the orthogonal system obtained and figured in Art. 46 can be represented by the equations  x2 + y 2 + a2  2ax  =  sinh α cosh α (1) x2 + y 2 − a2  2ay   = sin β cos β 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 Toroidal Co¨ ordinates. If we take the axis of the system as the axis of Z, the equations of a set of the surfaces may be written  4a2 (x2 + y 2 ) [x2 + y 2 + z 2 + a2 ]2   =    sinh2 α cosh2 α  2 2 2 2 2az x +y +z −a (2) =    sin β cos β    y = x tan γ α, β, and γ being regarded as the coördinates of a point of intersection of the three surfaces.


265

Finding Laplace’s Equation in the usual manner we get x=

a sinh α cos γ , cosh α ∓ cos β

r=

p

h1 =

x2 + y 2 =

y=

a sinh α sin γ , cosh α ∓ cos β

a sinh α , cosh α ∓ cos β

cosh α ∓ cos β , a

h2 =

z=

a sin β , cosh α ∓ cos β

a + z ctn β =

a cosh α ; cosh α ∓ cos β

cosh α ∓ cos β , a

h3 =

cosh α ∓ cos β ; a sinh α

and Laplace’s Equation becomes a sinh α a sinh α Dα Dα V + Dβ Dβ V cosh α ∓ cos β cosh α ∓ cos β a Dγ V = 0, + Dγ sinh α(cosh α ∓ cos β)

or

Dα (rDα V ) + Dβ (rDβ V ) +

1 rDγ2 V = 0. sinh2 α

(1)

(2)

We cannot proceed further by our usual method, for the assumption that V is a function of α alone, or that V is a function of β alone, proves to be inadmissible. Indeed, not only are α, β, and γ not 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 √

√ √ √ 1 Dγ2 (V r) − V (Dα2 r + Dβ2 r) = 0. (3) sinh2 α √ √ √ √ r 2 2 ; hence if U = V r (3) becomes Dα r + Dβ r proves equal to − 4 sinh2 α Dα2 (V

r) + Dβ2 (V

√

r) +

sinh2 α(Dα2 U + Dβ2 U ) + Dγ2 U + 41 U = 0,

(4)

for which particular solutions can readily be found by our usual process. (4) can be broken up into the three equations d2 N + (m + 12 )2 N = 0 dγ 2 d2 M + n2 M = 0 dβ 2 d2 L sinh2 α 2 − [m(m + 1) + n2 sinh2 α]L = 0. dα N = A cos(m + 21 )γ + B sin(m + 12 )γ M = A1 cos nβ + B1 sin nβ.

(5) (6) (7)


266

If we introduce into (7) x = ctnh α it becomes d2 L dL n2 (1 − x2 ) 2 − 2x L = 0, + m(m + 1) − dx dx 1 − x2 a solution of which is n

n L = Pm (x) = (1 − x2 ) 2

dn Pm (x) dxn

(v. Art. 102).

It is to be noted that since ctnh α is greater than 1 n

n Pm (ctnh α) = i 2 cschn α

dn Pm (ctnh α) . (d ctnh α)n

n

The constant coefficient i 2 can be rejected and we get U = [A cos(m+ 12 )γ+B sin(m+ 12 )γ](A1 cos nβ+B1 sin nβ) cschn α

dn Pm (ctnh α) (d ctnh α)n

as a particular solution of (4). n 1 n n d Pm (ctnh α) n Pm (ctnh α) = csch α (d ctnh α)n i2

has been called a Toroidal Harmonic. EXAMPLES. 1. 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. Suggestion: If V = f (β, γ) when α = α0 , the function to be developed is √

r.f (β, γ) i.e.

a sinh α0 cosh α0 ∓ cos β

12 f (β, γ)

and the development will be in a double Fourier’s Series (v. Art. 71). 2. Show that if we let α range from 0 to ∞, β from −π to π, and γ from 0 to 2π, each of the double signs on page 264 may be replaced by the minus sign without loss of generality.

267

CHAPTER IX.1 HISTORICAL SUMMARY. 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 17532 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.3 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 Euler4 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 Etrangers for 1785, first introduced the zonal harmonics Pm 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 written5 determined the potential of a solid differing but little from a sphere by means of the development according to the spherical harmonics Ym . Very closely related to this problem is Gauss’s celebrated treatment of the theory of terrestrial magnetism,6 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: Théorie du mouvement 1 See

preface. two articles by Bernoulli and one by Euler in the Memoirs of the Academy of Berlin for this year. 3 See the Transactions of the Academy of St. Petersburg for 1732-33, 1734 and 1781. 4 Transactions of the Academy of St. Petersburg. 5 “Th´ eorie des attractions des sph´ ero¨ıdes et de la figure des Plan` etes” M´ emoires de l’acad´ emie 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 co¨ ordinates. 6 Resultate aus den Beobachtungen des magnetischen Vereins im Jahre 1838. Leipzig, 1839. Reprinted in Gauss’s collected works, Vol. V., p. 121. 2 See

HISTORICAL SUMMARY.

268

de la chaleur dans les corps solides was crowned by the French Academy. Although not printed until the years 1824-26,7 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 Théorie analytique de la chaleur (1822), contains a treatment of the following problems and of practically all of their special cases: (a) The one dimensional flow of heat. (b) The two dimensional flow of heat in a rectangle. (c) The three dimensional flow of heat in a rectangular parallelopiped. (d ) The flow of heat in a sphere when the temperature depends only on the distance from the centre. (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.” 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,8 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 18209 to the general (unsymmetrical) case of the flow of heat in a sphere, and by Poisson10 (1821) to the unsymmetrical flow of heat in a cylinder. In 1835 Green published a paper11 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 these12 the equation ∇2 V = 0 is transformed to ellipsoidal co¨ ordinates and is then broken up into three ordinary differential 7 M´ emoires

de l’acad´ emie des sciences for 1819-20 and 1821-22. had practically determined these coefficients long before but failed to notice what he had got. 9 Connaissance des Temps pour l’an 1823. 10 Journal de l’Ecole ´ Polytechnique, 19e Cahier. Although the final forms to which Poisson reduces his results are similar to Fourier’s, his methods are very different. 11 “On the determination of the exterior and interior attraction of ellipsoids of variable densities.” Transactions of the Cambridge Philosophical Society. 12 M´ ´ emoires des Savants Etrangers, 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. 8 Lagrange

HISTORICAL SUMMARY.

269

equations. The rest of the solution, however, is hardly touched upon. Lamé’s most important work on this subject13 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 θ and φ, the parameters of two families of confocal cones of the second degree are used as co¨ ordinates. 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 dissertation14 (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 afterwards15 Heine and Liouville 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 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 memoirs16 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. 13 “Sur l’´ equilibre des Temp´ eratures dans un ellipso¨ıde ` a trois axes in´ egaux.” An article by the same author on the two dimensional potential will be found in Vol. I. of this Journal. 14 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. 15 Heine: Crelle’s Journal, Vol. 29, 1845. 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 1 of a development of in terms of Lam´ e’s functions, see a paper by Heine in Crelle’s Journal, r Vol. 42, 1851. 16 See especially the one in the M´ emoires de l’acad´ emie des sciences, Vol. VIII., 1829.

HISTORICAL SUMMARY.

270

He was, nevertheless, able to solve the problem of the symmetrical vibration of a free circular plate. The complete theory of the vibration of a free circular plate was first given by Kirchhoff.17 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)18 and Lamb (1882).19 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,20 to which we have already referred, deserves to be regarded as one of the most important contributions to the general theory. The way in which spherical harmonics are introduced (as homogeneous functions of the rectangular coördinates) was then new21 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 co¨ ordinates 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.22 Returning now to the Continent, we find as the next important question 17 Crelle’s

Journal, Vol. 40, 1850. Journal, Vol. 88. 19 Proc. Lond. Math. Soc. 20 First edition, 1867. This appendix was evidently written as early as 1862, as Thomson refers to it in the memoir quoted above. 21 The same method was used at about the same time by Clebsch. 22 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. 18 Crelle’s

HISTORICAL SUMMARY.

271

taken up the problem of the potential of an anchor ring. The first publication on this subject is a monograph by C. Neumann23 (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 (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 papers24 (1868) deals with the solid bounded by two intersecting spheres, while in the second25 (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 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 Mathieu26 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,27 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 23 “Theorie

der Elektricit¨ ats- und W¨ arme-Vertheilung in einem Ringe.” Halle. Journal, Vol. 68, 1868. 25 Jahresbericht des Gymnasiums zu Elbing. 26 Liouville’s Journal, Vol. XIII. 27 “Ueber die Integration der partiellen Differentialgleichung δ 2 u + δ 2 u + k 2 u = 0.” Math. δx2 δy 2 Ann., Vol. I. No physical problem is mentioned in this paper. 24 Crelle’s

HISTORICAL SUMMARY.

272

of problems here considered by treating in succession the potential problem for the paraboloid of revolution,28 the parabolic cylinder29 and the general paraboloid.30 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 degree. Wangerin31 (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, cyclidic coördinates.32 He showed, however, merely how to break up Laplace’s equation into three ordinary differential equations.33 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.34 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 K¨ orper welche von confocalen Flächen zweiten Grades begrenzt sind”35 (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.36 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 development of the method; it is rather an application of it to 28 “Ueber das Gleichgewicht und die Bewegung der W¨ arme in einem Rotationsparaboloid.” Dissertation, Halle, 1881. 29 “Die Funktion des parabolischen Cylinders,” Gymnasialprogramm C¨ ustrin, 1883. 30 “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 Lam´ e in Liouville’s Journal for 1874, Vol. XIX. 31 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. 32 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. 33 See also a paper by this author in Gr¨ unert’s Archiv for 1873, where the problem of the equilibrium of elastic solids of revolution is treated. 34 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. 35 Math. Ann., 18. 36 For an exposition of this theory see the treatise: Ueber die Reihenentwickelungen der Potentialtheorie, Leipsic, Teubner, 1894, by the writer of the present chapter.

HISTORICAL SUMMARY.

273

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. Niven: On the Conduction of Heat in Ellipsoids of Revolution. Phil. Trans., 1880. Niven: On the Induction of Electric Currents in Infinite Plates and Spherical Shells. Phil. Trans., 1881. Hicks: On Toroidal Functions. Phil. Trans., 1881. Hicks: On the Steady Motion and Small Vibrations of a Hollow Vortex. Phil. Trans., 1884, 1885. Lamb: On Ellipsoidal Current Sheets. Phil. Trans., 1887. Chree: The Equations of an Isotropic Elastic Solid in Polar and Cylindrical Co¨ ordinates, their Solution and Application. Camb. Phil. Soc. Trans., XIV., 1889. Hobson: On a Class of Spherical Harmonics of Complex Degree with Applications to Physical Problems. Camb. Phil. Soc. Trans., XIV., 1889. Chree: On some Compound Vibrating Systems. Camb. Phil. Soc. Trans., XV., 1891. 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: Fourier: Théorie Analytique de la Chaleur, 1822. Lamé: Le¸cons sur les Functions inverses des Transcendantes et les Surfaces isothermes, 1857. Lamé: Le¸cons sur les Coordonnées Curvilignes et leurs diverses Applications, 1859. Mathieu: Cours de Physique Mathématique, 1873. Riemann: Partielle Differentialgleichungen, und deren Anwendung auf physikalische Fragen (edited by Hattendorf), third edition. 1882. F. Neumann: Theorie des Potentials und der Kugelfunktionen (edited by C. Neumann), 1887. Thomson and Tait: Natural Philosophy, second edition, 1879. Rayleigh: Theory of Sound, 1877.

HISTORICAL SUMMARY.

274

Basset: Hydrodynamics, 1888. Love: Theory of Elasticity, 1892. Heine: Handbuch der Kugelfunktionen (second edition), 1878-81. Ferrers: Spherical Harmonics, 1881. 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: Todhunter: The Functions of Laplace, Lamé and Bessel, 1875. Lommel: Studien u ¨ber die Bessel’schen Funktionen, 1868. F. Neumann: Beitr¨ age zur Theorie der Kugelfunktionen, 1878. And finally concerning the question of convergence: ¨ C. Neumann: Uber die nach Kreis-, Kugel- und Cylinder-Functionen fortschreitenden Entwickelungen, 1881.

275

APPENDIX. TABLES. Table I., a table of Surface Zonal Harmonics (Legendrians), gives the values of the first seven Harmonics P1 (cos θ), P2 (cos θ), · · · P7 (cos θ) for the argument θ 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 P1 (x), P2 (x), · · · P7 (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 ex , 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)

log cosh x = log sec(gd x).

Table IV. gives the first twelve roots of J0 (x) = 0 and J1 (x) = 0 each divided by π. 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 J0 (x) = 0, J1 (x) = 0, · · · J5 (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 J0 (x) and J1 (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.

APPENDIX

276 TABLE I. — Surface Zonal Harmonics

θ

P1 (cos θ)

P2 (cos θ)

P3 (cos θ)

P4 (cos θ)

P5 (cos θ)

P6 (cos θ)

P7 (cos θ)

0◦ 1 2 3 4

1.0000 .9998 .9994 .9986 .9976

1.0000 .9995 .9982 .9959 .9927

1.0000 .9991 .9963 .9918 .9854

1.0000 .9985 .9939 .9863 .9758

1.0000 .9977 .9909 .9795 .9638

1.0000 .9967 .9872 .9713 .9495

1.0000 .9955 .9829 .9617 .9329

5 6 7 8 9

.9962 .9945 .9925 .9903 .9877

.9886 .9836 .9777 .9709 .9633

.9773 .9674 .9557 .9423 .9273

.9623 .9459 .9267 .9048 .8803

.9437 .9194 .8911 .8589 .8232

.9216 .8881 .8476 .8053 .7571

.8961 .8522 .7986 .7448 .6831

10 11 12 13 14

.9848 .9816 .9781 .9744 .9703

.9548 .9454 .9352 .9241 .9122

.9106 .8923 .8724 .8511 .8283

.8532 .8238 .7920 .7582 .7224

.7840 .7417 .6966 .6489 .5990

.7045 .6483 .5892 .5273 .4635

.6164 .5461 .4732 .3940 .3219

15 16 17 18 19

.9659 .9613 .9563 .9511 .9455

.8995 .8860 .8718 .8568 .8410

.8042 .7787 .7519 .7240 .6950

.6847 .6454 .6046 .5624 .5192

.5471 .4937 .4391 .3836 .3276

.3982 .3322 .2660 .2002 .1347

.2454 .1699 .0961 .0289 −.0443

20 21 22 23 24

.9397 .9336 .9272 .9205 .9135

.8245 .8074 .7895 .7710 .7518

.6649 .6338 .6019 .5692 .5357

.4750 .4300 .3845 .3386 .2926

.2715 .2156 .1602 .1057 .0525

.0719 .0107 −.0481 −.1038 −.1559

−.1072 −.1662 −.2201 −.2681 −.3095

25 26 27 28 29

.9063 .8988 .8910 .8829 .8746

.7321 .7117 .6908 .6694 .6474

.5016 .4670 .4319 .3964 .3607

.2465 .2007 .1553 .1105 .0665

.0009 −.0489 −.0964 −.1415 −.1839

−.2053 −.2478 −.2869 −.3211 −.3503

−.3463 −.3717 −.3921 −.4052 −.4114

30 31 32 33 34

.8660 .8572 .8480 .8387 .8290

.6250 .6021 .5788 .5551 .5310

.3248 .2887 .2527 .2167 .1809

.0234 −.0185 −.0591 −.0982 −.1357

−.2233 −.2595 −.2923 −.3216 −.3473

−.3740 −.3924 −.4052 −.4126 −.4148

−.4101 −.4022 −.3876 −.3670 −.3409

35 36 37 38 39

.8192 .8090 .7986 .7880 .7771

.5065 .4818 .4567 .4314 .4059

.1454 .1102 .0755 .0413 .0077

−.1714 −.2052 −.2370 −.2666 −.2940

−.3691 −.3871 −.4011 −.4112 −.4174

−.4115 −.4031 −.3898 −.3719 −.3497

−.3096 −.2738 −.2343 −.1918 −.1469

40 41 42 43 44

.7660 .7547 .7431 .7314 .7193

.3802 .3544 .3284 .3023 .2762

−.0252 −.0574 −.0887 −.1191 −.1485

−.3190 −.3416 −.3616 −.3791 −.3940

−.4197 −.4181 −.4128 −.4038 −.3914

−.3234 −.2938 −.2611 −.2255 −.1878

−.1003 −.0534 −.0065 .0398 .0846

45◦

.7071

.2500

−.1768

−.4062

−.3757

−.1485

.1270

APPENDIX

277 TABLE I. — Surface Zonal Harmonics

θ

P1 (cos θ)

P2 (cos θ)

P3 (cos θ)

P4 (cos θ)

P5 (cos θ)

P6 (cos θ)

P7 (cos θ)

45◦ 46 47 48 49

.7071 .6947 .6820 .6691 .6561

.2500 .2238 .1977 .1716 .1456

−.1768 −.2040 −.2300 −.2547 −.2781

−.4062 −.4158 −.4252 −.4270 −.4286

−.3757 −.3568 −.3350 −.3105 −.2836

−.1485 −.1079 −.0645 −.0251 .0161

.1270 .1666 .2054 .2349 .2627

50 51 52 53 54

.6428 .6293 .6157 .6018 .5878

.1198 .0941 .0686 .0433 .0182

−.3002 −.3209 −.3401 −.3578 −.3740

−.4275 −.4239 −.4178 −.4093 −.3984

−.2545 −.2235 −.1910 −.1571 −.1223

.0563 .0954 .1326 .1677 .2002

.2854 .3031 .3153 .3221 .3234

55 56 57 58 59

.5736 .5592 .5446 .5299 .5150

−.0065 −.0310 −.0551 −.0788 −.1021

−.3886 −.4016 −.4131 −.4229 −.4310

−.3852 −.3698 −.3524 −.3331 −.3119

−.0868 −.0510 −.0150 .0206 .0557

.2297 .2559 .2787 .2976 .3125

.3191 .3095 .2949 .2752 .2511

60 61 62 63 64

.5000 .4848 .4695 .4540 .4384

−.1250 −.1474 −.1694 −.1908 −.2117

−.4375 −.4423 −.4455 −.4471 −.4470

−.2891 −.2647 −.2390 −.2121 −.1841

.0898 .1229 .1545 .1844 .2123

.3232 .3298 .3321 .3302 .3240

.2231 .1916 .1571 .1203 .0818

65 66 67 68 69

.4226 .4067 .3907 .3746 .3584

−.2321 −.2518 −.2710 −.2896 −.3074

−.4452 −.4419 −.4370 −.4305 −.4225

−.1552 −.1256 −.0955 −.0650 −.0344

.2381 .2615 .2824 .3005 .3158

.3138 .2996 .2819 .2605 .2361

.0422 .0021 −.0375 −.0763 −.1135

70 71 72 73 74

.3420 .3256 .3090 .2924 .2756

−.3245 −.3410 −.3568 −.3718 −.3860

−.4130 −.4021 −.3898 −.3761 −.3611

−.0038 .0267 .0568 .0864 .1153

.3281 .3373 .3434 .3463 .3461

.2089 .1786 .1472 .1144 .0795

−.1485 −.1811 −.2099 −.2347 −.2559

75 76 77 78 79

.2588 .2419 .2250 .2079 .1908

−.3995 −.4112 −.4241 −.4352 −.4454

−.3449 −.3275 −.3090 −.2894 −.2688

.1434 .1705 .1964 .2211 .2443

.3427 .3362 .3267 .3143 .2990

.0431 .0076 −.0284 −.0644 −.0989

−.2730 −.2848 −.2919 −.2943 −.2913

80 81 82 83 84

.1736 .1564 .1392 .1219 .1045

−.4548 −.4633 −.4709 −.4777 −.4836

−.2474 −.2251 −.2020 −.1783 −.1539

.2659 .2859 .3040 .3203 .3345

.2810 .2606 .2378 .2129 .1861

−.1321 −.1635 −.1926 −.2193 −.2431

−.2835 −.2709 −.2536 −.2321 −.2067

85 86 87 88 89

.0872 .0698 .0523 .0349 .0175

−.4886 −.4927 −.4959 −.4982 −.4995

−.1291 −.1038 −.0781 −.0522 −.0262

.3468 .3569 .3648 .3704 .3739

.1577 .1278 .0969 .0651 .0327

−.2638 −.2811 −.2947 −.3045 −.3105

−.1779 −.1460 −.1117 −.0735 −.0381

90◦

.0000

−.5000

.0000

.3750

.0000

−.3125

.0000

APPENDIX

278 TABLE II.—Surface Zonal Harmonics.

x

P1 (x)

P2 (x)

P3 (x)

P4 (x)

P5 (x)

P6 (x)

P7 (x)

0.00 .01 .02 .03 .04

0.0000 .0100 .0200 .0300 .0400

−.5000 −.4998 −.4994 −.4986 −.4976

0.0000 −.0150 −.0300 −.0449 −.0598

0.3750 .3746 .3735 .3716 .3690

0.0000 .0187 .0374 .0560 .0744

−.3125 −.3118 −.3099 −.3066 −.3021

0.0000 −.0219 −.0436 −.0651 −.0862

.05 .06 .07 .08 .09

.0500 .0600 .0700 .0800 .0900

−.4962 −.4946 −.4926 −.4904 −.4878

−.0747 −.0895 −.1041 −.1187 −.1332

.3657 .3616 .3567 .3512 .3449

.0927 .1106 .1283 .1455 .1624

−.2962 −.2891 −.2808 −.2713 −.2606

−.1069 −.1270 −.1464 −.1651 −.1828

.10 .11 .12 .13 .14

.1000 .1100 .1200 .1300 .1400

−.4850 −.4818 −.4784 −.4746 −.4706

−.1475 −.1617 −.1757 −.1895 −.2031

.3379 .3303 .3219 .3129 .3032

.1788 .1947 .2101 .2248 .2389

−.2488 −.2360 −.2220 −.2071 −.1913

−.1995 −.2151 −.2295 −.2427 −.2545

.15 .16 .17 .18 .19

.1500 .1600 .1700 .1800 .1900

−.4662 −.4616 −.4566 −.4514 −.4458

−.2166 −.2298 −.2427 −.2554 −.2679

.2928 .2819 .2703 .2581 .2453

.2523 .2650 .2769 .2880 .2982

−.1746 −.1572 −.1389 −.1201 −.1006

−.2649 −.2738 −.2812 −.2870 −.2911

.20 .21 .22 .23 .24

.2000 .2100 .2200 .2300 .2400

−.4400 −.4338 −.4274 −.4206 −.4136

−.2800 −.2918 −.3034 −.3146 −.3254

.2320 .2181 .2037 .1889 .1735

.3075 .3159 .3234 .3299 .3353

−.0806 −.0601 −.0394 −.0183 .0029

−.2935 −.2943 −.2933 −.2906 −.2861

.25 .26 .27 .28 .29

.2500 .2600 .2700 .2800 .2900

−.4062 −.3986 −.3906 −.3824 −.3738

−.3359 −.3461 −.3558 −.3651 −.3740

.1577 .1415 .1249 .1079 .0906

.3397 .3431 .3453 .3465 .3465

.0243 .0456 .0669 .0879 .1087

−.2799 −.2720 −.2625 −.2512 −.2384

.30 .31 .32 .33 .34

.3000 .3100 .3200 .3300 .3400

−.3650 −.3558 −.3464 −.3366 −.3266

−.3825 −.3905 −.3981 −.4052 −.4117

.0729 .0550 .0369 .0185 −.0000

.3454 .3431 .3397 .3351 .3294

.1292 .1492 .1686 .1873 .2053

−.2241 −.2082 −.1910 −.1724 −.1527

.35 .36 .37 .38 .39

.3500 .3600 .3700 .3800 .3900

−.3162 −.3056 −.2946 −.2834 −.2718

−.4178 −.4234 −.4284 −.4328 −.4367

−.0187 −.0375 −.0564 −.0753 −.0942

.3225 .3144 .3051 .2948 .2833

.2225 .2388 .2540 .2681 .2810

−.1318 −.1098 −.0870 −.0635 −.0393

.40 .41 .42 .43 .44

.4000 .4100 .4200 .4300 .4400

−.2600 −.2478 −.2354 −.2226 −.2096

−.4400 −.4427 −.4448 −.4462 −.4470

−.1130 −.1317 −.1504 −.1688 −.1870

.2706 .2569 .2421 .2263 .2095

.2926 .3029 .3118 .3191 .3249

−.0146 .0104 .0356 .0608 .0859

.45 .46 .47 .48 .49

.4500 .4600 .4700 .4800 .4900

−.1962 −.1826 −.1686 −.1544 −.1398

−.4472 −.4467 −.4454 −.4435 −.4409

−.2050 −.2226 −.2399 −.2568 −.2732

.1917 .1730 .1534 .1330 .1118

.3290 .3314 .3321 .3310 .3280

.1106 .1348 .1584 .1811 .2027

.50

.5000

−.1250

−.4375

−.2891

.0898

.3232

.2231

APPENDIX

279 TABLE II.—Surface Zonal Harmonics.

x

P1 (x)

P2 (x)

P3 (x)

P4 (x)

P5 (x)

P6 (x)

P7 (x)

.50 .51 .52 .53 .54

.5000 .5100 .5200 .5300 .5400

−.1250 −.1098 −.0944 −.0786 −.0626

−.4375 −.4334 −.4285 −.4228 −.4163

−.2891 −.3044 −.3191 −.3332 −.3465

.0898 .0673 .0441 .0204 −.0037

.3232 .3166 .3080 .2975 .2851

.2231 .2422 .2596 .2753 .2891

.55 .56 .57 .58 .59

.5500 .5600 .5700 .5800 .5900

−.0462 −.0296 −.0126 .0046 .0222

−.4091 −.4010 −.3920 −.3822 −.3716

−.3590 −.3707 −.3815 −.3914 −.4002

−.0282 −.0529 −.0779 −.1028 −.1278

.2708 .2546 .2366 .2168 .1953

.3007 .3102 .3172 .3217 .3235

.60 .61 .62 .63 .64

.6000 .6100 .6200 .6300 .6400

.0400 .0582 .0766 .0954 .1144

−.3600 −.3475 −.3342 −.3199 −.3046

−.4080 −.4146 −.4200 −.4242 −.4270

−.1526 −.1772 −.2014 −.2251 −.2482

.1721 .1473 .1211 .0935 .0646

.3226 .3188 .3121 .3023 .2895

.65 .66 .67 .68 .69

.6500 .6600 .6700 .6800 .6900

.1338 .1534 .1734 .1936 .2142

−.2884 −.2713 −.2531 −.2339 −.2137

−.4284 −.4284 −.4268 −.4236 −.4187

−.2705 −.2919 −.3122 −.3313 −.3490

.0347 .0038 −.0278 −.0601 −.0926

.2737 .2548 .2329 .2081 .1805

.70 .71 .72 .73 .74

.7000 .7100 .7200 .7300 .7400

.2350 .2562 .2776 .2994 .3214

−.1925 −.1702 −.1469 −.1225 −.0969

−.4121 −.4036 −.3933 −.3810 −.3666

−.3652 −.3796 −.3922 −.4026 −.4107

−.1253 −.1578 −.1899 −.2214 −.2518

.1502 .1173 .0822 .0450 .0061

.75 .76 .77 .78 .79

.7500 .7600 .7700 .7800 .7900

.3438 .3664 .3894 .4126 .4362

−.0703 −.0426 −.0137 .0164 .0476

−.3501 −.3314 −.3104 −.2871 −.2613

−.4164 −.4193 −.4193 −.4162 −.4097

−.2808 −.3081 −.3333 −.3559 −.3756

−.0342 −.0754 −.1171 −.1588 −.1999

.80 .81 .82 .83 .84

.8000 .8100 .8200 .8300 .8400

.4600 .4842 .5086 .5334 .5584

.0800 .1136 .1484 .1845 .2218

−.2330 −.2021 −.1685 −.1321 −.0928

−.3995 −.3855 −.3674 −.3449 −.3177

−.3918 −.4041 −.4119 −.4147 −.4120

−.2397 −.2774 −.3124 −.3437 −.3703

.85 .86 .87 .88 .89

.8500 .8600 .8700 .8800 .8900

.5838 .6094 .6354 .6616 .6882

.2603 .3001 .3413 .3837 .4274

−.0506 −.0053 .0431 .0947 .1496

−.2857 −.2484 −.2056 −.1570 −.1023

−.4030 −.3872 −.3638 −.3322 −.2916

−.3913 −.4055 −.4116 −.4083 −.3942

.90 .91 .92 .93 .94

.9000 .9100 .9200 .9300 .9400

.7150 .7422 .7696 .7974 .8254

.4725 .5189 .5667 .6159 .6665

.2079 .2698 .3352 .4044 .4773

−.0411 .0268 .1017 .1842 .2744

−.2412 −.1802 −.1077 −.0229 .0751

−.3678 −.3274 −.2713 −.1975 −.1040

.95 .96 .97 .98 .99

.9500 .9600 .9700 .9800 .9900

.8538 .8824 .9114 .9406 .9702

.7184 .7718 .8267 .8830 .9407

.5541 .6349 .7198 .8089 .9022

.3727 .4796 .5954 .7204 .8552

.1875 .3151 .4590 .6204 .8003

.0112 .1506 .3165 .5115 .7384

1.00

1.0000

1.0000

1.0000

1.0000

1.0000

1.0000

1.0000

APPENDIX

280 TABLE III.—Hyperbolic Functions. x

ex

e−x

sinh x

cosh x

gd x ◦

0.00 .01 .02 .03 .04

1.0000 1.0100 1.0202 1.0305 1.0408

1.0000 0.9900 .9802 .9704 .9608

0.0000 .0100 .0200 .0300 .0400

1.0000 1.0000 1.0002 1.0004 1.0008

0.0000 0.5729 1.1458 1.7186 2.2912

.05 .06 .07 .08 .09

1.0513 1.0618 1.0725 1.0833 1.0942

.9512 .9418 .9324 .9231 .9139

.0500 .0600 .0701 .0801 .0901

1.0012 1.0018 1.0025 1.0032 1.0040

2.8636 3.4357 4.0074 4.5788 5.1497

.10 .11 .12 .13 .14

1.1052 1.1163 1.1275 1.1388 1.1503

.9048 .8958 .8869 .8781 .8694

.1002 .1102 .1203 .1304 .1405

1.0050 1.0061 1.0072 1.0085 1.0098

5.720 6.290 6.859 7.428 7.995

.15 .16 .17 .18 .19

1.1618 1.1735 1.1853 1.1972 1.2092

.8607 .8521 .8437 .8353 .8270

.1506 .1607 .1708 .1810 .1911

1.0113 1.0128 1.0145 1.0162 1.0181

8.562 9.128 9.694 10.258 10.821

.20 .21 .22 .23 .24

1.2214 1.2337 1.2461 1.2586 1.2712

.8187 .8106 .8025 .7945 .7866

.2013 .2115 .2218 .2320 .2423

1.0201 1.0221 1.0243 1.0266 1.0289

11.384 11.945 12.505 13.063 13.621

.25 .26 .27 .28 .29

1.2840 1.2969 1.3100 1.3231 1.3364

.7788 .7711 .7634 .7558 .7483

.2526 .2629 .2733 .2837 .2941

1.0314 1.0340 1.0367 1.0395 1.0423

14.177 14.732 15.285 15.837 16.388

.30 .31 .32 .33 .34

1.3499 1.3634 1.3771 1.3910 1.4049

.7408 .7334 .7261 .7189 .7118

.3045 .3150 .3255 .3360 .3466

1.0453 1.0484 1.0516 1.0549 1.0584

16.937 17.484 18.030 18.573 19.116

.35 .36 .37 .38 .39

1.4191 1.4333 1.4477 1.4623 1.4770

.7047 .6977 .6907 .6839 .6771

.3572 .3678 .3785 .3892 .4000

1.0619 1.0655 1.0692 1.0731 1.0770

19.656 20.195 20.732 21.267 21.800

.40 .41 .42 .43 .44

1.4918 1.5068 1.5220 1.5373 1.5527

.6703 .6636 .6570 .6505 .6440

.4108 .4216 .4325 .4434 .4543

1.0811 1.0852 1.0895 1.0939 1.0984

22.331 22.859 23.386 23.911 24.434

.45 .46 .47 .48 .49

1.5683 1.5841 1.6000 1.6161 1.6323

.6376 .6313 .6250 .6188 .6126

.4653 .4764 .4875 .4986 .5098

1.1030 1.1077 1.1125 1.1174 1.1225

24.955 25.473 25.989 26.503 27.015

0.50

1.6487

0.6065

0.5211

1.1276

27.524

◦

APPENDIX

281 TABLE III.—Hyperbolic Functions. x

ex

e−x

sinh x

cosh x

gd x

0.50 .51 .52 .53 .54

1.6487 1.6653 1.6820 1.6989 1.7160

0.6065 .6005 .5945 .5886 .5827

0.5211 .5324 .5438 .5552 .5666

1.1276 1.1329 1.1383 1.1438 1.1494

27.524 28.031 28.535 29.037 29.537

.55 .56 .57 .58 .59

1.7333 1.7507 1.7683 1.7860 1.8040

.5770 .5712 .5655 .5599 .5543

.5782 .5897 .6014 .6131 .6248

1.1551 1.1609 1.1669 1.1730 1.1792

30.034 30.529 31.021 31.511 31.998

.60 .61 .62 .63 .64

1.8221 1.8404 1.8589 1.8776 1.8965

.5488 .5433 .5379 .5326 .5273

.6367 .6485 .6605 .6725 .6846

1.1855 1.1919 1.1984 1.2051 1.2119

32.483 32.965 33.444 33.921 34.395

.65 .66 .67 .68 .69

1.9155 1.9348 1.9542 1.9739 1.9937

.5220 .5169 .5117 .5066 .5016

.6967 .7090 .7213 .7336 .7461

1.2188 1.2258 1.2330 1.2402 1.2476

34.867 35.336 35.802 36.265 36.726

.70 .71 .72 .73 .74

2.0138 2.0340 2.0544 2.0751 2.0959

.4966 .4916 .4867 .4819 .4771

.7586 .7712 .7838 .7966 .8094

1.2552 1.2628 1.2706 1.2785 1.2865

37.183 37.638 38.091 38.540 38.987

.75 .76 .77 .78 .79

2.1170 2.1383 2.1598 2.1815 2.2034

.4724 .4677 .4630 .4584 .4538

.8223 .8353 .8484 .8615 .8748

1.2947 1.3030 1.3114 1.3199 1.3286

39.431 39.872 40.310 40.746 41.179

.80 .81 .82 .83 .84

2.2255 2.2479 2.2705 2.2933 2.3164

.4493 .4449 .4404 .4360 .4317

.8881 .9015 .9150 .9286 .9423

1.3374 1.3464 1.3555 1.3647 1.3740

41.608 42.035 42.460 42.881 43.299

.85 .86 .87 .88 .89

2.3396 2.3632 2.3869 2.4109 2.4351

.4274 .4232 .4190 .4148 .4107

.9561 .9700 .9840 .9981 1.0122

1.3835 1.3932 1.4029 1.4128 1.4229

43.715 44.128 44.537 44.944 45.348

.90 .91 .92 .93 .94

2.4596 2.4843 2.5093 2.5345 2.5600

.4066 .4025 .3985 .3946 .3906

1.0265 1.0409 1.0554 1.0700 1.0847

1.4331 1.4434 1.4539 1.4645 1.4753

45.750 46.148 46.544 46.936 47.326

.95 .96 .97 .98 .99

2.5857 2.6117 2.6379 2.6645 2.6912

.3867 .3829 .3791 .3753 .3716

1.0995 1.1144 1.1294 1.1446 1.1598

1.4862 1.4973 1.5085 1.5199 1.5314

47.713 48.097 48.478 48.857 49.232

1.00

2.7183

0.3679

1.1752

1.5431

49.605

◦

◦

APPENDIX

282 TABLE III.—Hyperbolic Functions.

x

l sinh x

l cosh x

x

l sinh x

l cosh x

x

l sinh x

l cosh x

1.00 1.01 1.02 1.03 1.04

0.0701 .0758 .0815 .0871 .0927

0.1884 .1917 .1950 .1984 .2018

1.50 1.51 1.52 1.53 1.54

0.3282 .3330 .3378 .3426 .3474

0.3715 .3754 .3794 .3833 .3873

2.00 2.01 2.02 2.03 2.04

0.5595 .5640 .5685 .5730 .5775

0.5754 .5796 .5838 .5880 .5922

1.05 1.06 1.07 1.08 1.09

.0982 .1038 .1093 .1148 .1203

.2051 .2086 .2120 .2154 .2189

1.55 1.56 1.57 1.58 1.59

.3521 .3569 .3616 .3663 .3711

.3913 .3952 .3992 .4032 .4072

2.05 2.06 2.07 2.08 2.09

.5820 .5865 .5910 .5955 .6000

.5964 .6006 .6048 .6090 .6132

1.10 1.11 1.12 1.13 1.14

.1257 .1311 .1365 .1419 .1472

.2223 .2258 .2293 .2328 .2364

1.60 1.61 1.62 1.63 1.64

.3758 .3805 .3852 .3899 .3946

.4112 .4152 .4192 .4232 .4273

2.10 2.11 2.12 2.13 2.14

.6044 .6089 .6134 .6178 .6223

.6175 .6217 .6259 .6301 .6343

1.15 1.16 1.17 1.18 1.19

.1525 .1578 .1631 .1684 .1736

.2399 .2435 .2470 .2506 .2542

1.65 1.66 1.67 1.68 1.69

.3992 .4039 .4086 .4132 .4179

.4313 .4353 .4394 .4434 .4475

2.15 2.16 2.17 2.18 2.19

.6268 .6312 .6357 .6401 .6446

.6386 .6428 .6470 .6512 .6555

1.20 1.21 1.22 1.23 1.24

.1788 .1840 .1892 .1944 .1995

.2578 .2615 .2651 .2688 .2724

1.70 1.71 1.72 1.73 1.74

.4225 .4272 .4318 .4364 .4411

.4515 .4556 .4597 .4637 .4678

2.20 2.21 2.22 2.23 2.24

.6491 .6535 .6580 .6624 .6668

.6597 .6640 .6682 .6724 .6767

1.25 1.26 1.27 1.28 1.29

.2046 .2098 .2148 .2199 .2250

.2761 .2798 .2835 .2872 .2909

1.75 1.76 1.77 1.78 1.79

.4457 .4503 .4549 .4595 .4641

.4719 .4760 .4801 .4842 .4883

2.25 2.26 2.27 2.28 2.29

.6713 .6757 .6802 .6846 .6890

.6809 .6852 .6894 .6937 .6979

1.30 1.31 1.32 1.33 1.34

.2300 .2351 .2401 .2451 .2501

.2947 .2984 .3022 .3059 .3097

1.80 1.81 1.82 1.83 1.84

.4687 .4733 .4778 .4824 .4870

.4924 .4965 .5006 .5048 .5089

2.30 2.31 2.32 2.33 2.34

.6935 .6979 .7023 .7067 .7112

.7022 .7064 .7107 .7150 .7192

1.35 1.36 1.37 1.38 1.39

.2551 .2600 .2650 .2699 .2748

.3135 .3173 .3211 .3249 .3288

1.85 1.86 1.87 1.88 1.89

.4915 .4961 .5007 .5052 .5098

.5130 .5172 .5213 .5254 .5296

2.35 2.36 2.37 2.38 2.39

.7156 .7200 .7244 .7289 .7333

.7235 .7278 .7320 .7363 .7406

1.40 1.41 1.42 1.43 1.44

.2797 .2846 .2895 .2944 .2993

.3326 .3365 .3403 .3442 .3481

1.90 1.91 1.92 1.93 1.94

.5143 .5188 .5234 .5279 .5324

.5337 .5379 .5421 .5462 .5504

2.40 2.41 2.42 2.43 2.44

.7377 .7421 .7465 .7509 .7553

.7448 .7491 .7534 .7577 .7619

1.45 1.46 1.47 1.48 1.49

.3041 .3090 .3138 .3186 .3234

.3520 .3559 .3598 .3637 .3676

1.95 1.96 1.97 1.98 1.99

.5370 .5415 .5460 .5505 .5550

.5545 .5587 .5629 .5671 .5713

2.45 2.46 2.47 2.48 2.49

.7597 .7642 .7686 .7730 .7774

.7662 .7705 .7748 .7791 .7833

1.50

0.3282

0.3715

2.00

0.5595

0.5754

2.50

0.7818

0.7876

APPENDIX

283 TABLE III.—Hyperbolic Functions.

x

l sinh x

l cosh x

x

l sinh x

l cosh x

x

l sinh x

l cosh x

2.50 2.51 2.52 2.53 2.54

0.7818 .7862 .7906 .7950 .7994

0.7876 .7919 .7962 .8005 .8048

2.75 2.76 2.77 2.78 2.79

0.8915 .8959 .9003 .9046 .9090

0.8951 .8994 .9037 .9080 .9123

3.0 3.1 3.2 3.3 3.4

1.0008 1.0444 1.0880 1.1316 1.1751

1.0029 1.0462 1.0894 1.1327 1.1761

2.55 2.56 2.57 2.58 2.59

.8038 .8082 .8126 .8169 .8213

.8091 .8134 .8176 .8219 .8262

2.80 2.81 2.82 2.83 2.84

.9134 .9178 .9221 .9265 .9309

.9166 .9209 .9252 .9295 .9338

3.5 3.6 3.7 3.8 3.9

1.2186 1.2621 1.3056 1.3491 1.3925

1.2194 1.2628 1.3061 1.3495 1.3929

2.60 2.61 2.62 2.63 2.64

.8257 .8301 .8345 .8389 .8433

.8305 .8348 .8391 .8434 .8477

2.85 2.86 2.87 2.88 2.89

.9353 .9396 .9440 .9484 .9527

.9382 .9425 .9468 .9511 .9554

4.0 4.1 4.2 4.3 4.4

1.4360 1.4795 1.5229 1.5664 1.6098

1.4363 1.4797 1.5231 1.5665 1.6099

2.65 2.66 2.67 2.68 2.69

.8477 .8521 .8564 .8608 .8652

.8520 .8563 .8606 .8649 .8692

2.90 2.91 2.92 2.93 2.94

.9571 .9615 .9658 .9702 .9746

.9597 .9641 .9684 .9727 .9770

4.5 4.6 4.7 4.8 4.9

1.6532 1.6967 1.7401 1.7836 1.8270

1.6533 1.6968 1.7402 1.7836 1.8270

2.70 2.71 2.72 2.73 2.74

.8696 .8740 .8784 .8827 .8871

.8735 .8778 .8821 .8864 .8907

2.95 2.96 2.97 2.98 2.99

.9789 .9833 .9877 .9920 .9964

.9813 .9856 .9900 .9943 .9986

5.0 6.0 7.0 8.0 9.0

1.8704 2.3047 2.7390 3.1733 3.6076

1.8705 2.3047 2.7390 3.1733 3.6076

2.75

0.8915

0.8951

3.00

1.0008

1.0029

10.0

4.0419

4.0419

APPENDIX

284 TABLE IV.—Roots of Bessel’s Functions.

1 2 3 4 5 6

x for J0 (x) = 0 π

x for J1 (x) = 0 π

0.7655 1.7571 2.7546 3.7534 4.7527 5.7522

1.2197 2.2330 3.2383 4.2411 5.2428 6.2439

x for J0 (x) = 0 π

x for J1 (x) = 0 π

6.7519 7.7516 8.7514 9.7513 10.7512 11.7511

7.2448 8.2454 9.2459 10.2463 11.2466 12.2469

7 8 9 10 11 12

TABLE V.—Roots of Jn (x) = 0.

1 2 3 4 5 6 7 8 9

n=0

n=1

n=2

n=3

n=4

n=5

2.405 5.520 8.654 11.792 14.931 18.071 21.212 24.353 27.494

3.832 7.016 10.173 13.323 16.470 19.616 22.760 25.903 29.047

5.135 8.417 11.620 14.796 17.960 21.117 24.270 27.421 30.571

6.379 9.760 13.017 16.224 19.410 22.583 25.749 28.909 32.050

7.586 11.064 14.373 17.616 20.827 24.018 27.200 30.371 33.512

8.780 12.339 15.700 18.982 22.220 25.431 28.628 31.813 34.983

APPENDIX

285 TABLE VI.— Bessel’s Functions.

x

J0 (x)

J1 (x)

x

J0 (x)

J1 (x)

x

J0 (x)

J1 (x)

0.0 0.1 0.2 0.3 0.4

1.0000 .9975 .9900 .9776 .9604

0.0000 .0499 .0995 .1483 .1960

5.0 5.1 5.2 5.3 5.4

−.1776 −.1443 −.1103 −.0758 −.0412

−.3276 −.3371 −.3432 −.3460 −.3453

10.0 10.1 10.2 10.3 10.4

−.2459 −.2490 −.2496 −.2477 −.2434

.0435 .0184 −.0066 −.0313 −.0555

0.5 0.6 0.7 0.8 0.9

.9385 .9120 .8812 .8463 .8075

.2423 .2867 .3290 .3688 .4060

5.5 5.6 5.7 5.8 5.9

−.0068 .0270 .0599 .0917 .1220

−.3414 −.3343 −.3241 −.3110 −.2951

10.5 10.6 10.7 10.8 10.9

−.2366 −.2276 −.2164 −.2032 −.1881

−.0789 −.1012 −.1224 −.1422 −.1604

1.0 1.1 1.2 1.3 1.4

.7652 .7196 .6711 .6201 .5669

.4401 .4709 .4983 .5220 .5419

6.0 6.1 6.2 6.3 6.4

.1506 .1773 .2017 .2238 .2433

−.2767 −.2559 −.2329 −.2081 −.1816

11.0 11.1 11.2 11.3 11.4

−.1712 −.1528 −.1330 −.1121 −.0902

−.1768 −.1913 −.2039 −.2143 −.2225

1.5 1.6 1.7 1.8 1.9

.5118 .4554 .3980 .3400 .2818

.5579 .5699 .5778 .5815 .5812

6.5 6.6 6.7 6.8 6.9

.2601 .2740 .2851 .2931 .2981

−.1538 −.1250 −.0953 −.0652 −.0349

11.5 11.6 11.7 11.8 11.9

−.0677 −.0446 −.0213 .0020 .0250

−.2284 −.2320 −.2333 −.2323 −.2290

2.0 2.1 2.2 2.3 2.4

.2239 .1666 .1104 .0555 .0025

.5767 .5683 .5560 .5399 .5202

7.0 7.1 7.2 7.3 7.4

.3001 .2991 .2951 .2882 .2786

−.0047 .0252 .0543 .0826 .1096

12.0 12.1 12.2 12.3 12.4

.0477 .0697 .0908 .1108 .1296

−.2234 −.2157 −.2060 −.1943 −.1807

2.5 2.6 2.7 2.8 2.9

−.0484 −.0968 −.1424 −.1850 −.2243

.4971 .4708 .4416 .4097 .3754

7.5 7.6 7.7 7.8 7.9

.2663 .2516 .2346 .2154 .1944

.1352 .1592 .1813 .2014 .2192

12.5 12.6 12.7 12.8 12.9

.1469 .1626 .1766 .1887 .1988

−.1655 −.1487 −.1307 −.1114 −.0912

3.0 3.1 3.2 3.3 3.4

−.2601 −.2921 −.3202 −.3443 −.3643

.3391 .3009 .2613 .2207 .1792

8.0 8.1 8.2 8.3 8.4

.1717 .1475 .1222 .0960 .0692

.2346 .2476 .2580 .2657 .2708

13.0 13.1 13.2 13.3 13.4

.2069 .2129 .2167 .2183 .2177

−.0703 −.0489 −.0271 −.0052 .0166

3.5 3.6 3.7 3.8 3.9

−.3801 −.3918 −.3992 −.4026 −.4018

.1374 .0955 .0538 .0128 −.0272

8.5 8.6 8.7 8.8 8.9

.0419 .0146 −.0125 −.0392 −.0653

.2731 .2728 .2697 .2641 .2559

13.5 13.6 13.7 13.8 13.9

.2150 .2101 .2032 .1943 .1836

.0380 .0590 .0791 .0984 .1166

4.0 4.1 4.2 4.3 4.4

−.3972 −.3887 −.3766 −.3610 −.3423

−.0660 −.1033 −.1386 −.1719 −.2028

9.0 9.1 9.2 9.3 9.4

−.0903 −.1142 −.1367 −.1577 −.1768

.2453 .2324 .2174 .2004 .1816

14.0 14.1 14.2 14.3 14.4

.1711 .1570 .1414 .1245 .1065

.1334 .1488 .1626 .1747 .1850

4.5 4.6 4.7 4.8 4.9

−.3205 −.2961 −.2693 −.2404 −.2097

−.2311 −.2566 −.2791 −.2985 −.3147

9.5 9.6 9.7 9.8 9.9

−.1939 −.2090 −.2218 −.2323 −.2403

.1613 .1395 .1166 .0928 .0684

14.5 14.6 14.7 14.8 14.9

.0875 .0679 .0476 .0271 .0064

.1934 .1999 .2043 .2066 .2069

5.0

−.1776

−.3276

10.0

−.2459

.0435

15.0

−.0142

.2051

ANNOUNCEMENTS

APPLICATIONS OF THE CALCULUS TO MECHANICS By E. R. Hedrick, Professor of Mathematics in the University of Missouri, and O. D. Kellogg, Assistant Professor of Mathematics in the University of Missouri 8vo, cloth, 116 pages, with diagrams, $1.25

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.

GINN AND COMPANY Publishers

ELECTRICAL PROBLEMS By William L. Hooper, Professor of Electrical Engineering, Tufts College, Mass., and Roy T. Wells, in Charge of Electrical Engineering, State University of Iowa 8vo, cloth, 170 pages, with diagrams, $1.25

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: 1. 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. 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. Five sets of problems on calculating and making winding tables and drawings for direct and alternating current armatures, armature reactions, field windings, etc. 4. Problems on winding and operation of transformers, rotary converters, and induction motors, and on testing of dynamos and transmission of power.

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.

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THEORETICAL MECHANICS By Percey F. Smith, Professor of Mathematics in the Sheffield Scientific School, Yale University, and William R. Longley, Assistant Professor of Mathematics in the Sheffield Scientific School, Yale University 8vo, semiflexible cloth, 288 pages, with diagrams, $2.50

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.


BOOKS ON HIGHER MATHEMATICS BYERLY: ELEMENTARY TREATISE ON FOURIER’S SERIES, AND SPHERICAL, CYLINDRICAL, AND ELLIPSOIDAL HARMONICS, WITH APPLICATIONS TO PROBLEMS IN MATHEMATICAL PHYSICS. $3.00. An introduction to the treatment of some of the important linear partial differential equations which lie at the foundation of modern theories in physics. HEDRICK: GOURSAT’S COURSE IN MATHEMATICAL ANALYSIS, VOLUME I. $4.00. 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. HEDRICK AND KELLOGG: APPLICATIONS OF THE CALCULUS TO MECHANICS. $1.25. A completed summary of those parts of mechanics which occur as applications of the calculus. PEIRCE, B. O.: ELEMENTS OF THE THEORY OF THE NEWTONIAN POTENTIAL FUNCTION (Third, Revised, and Enlarged Edition). $2.50. 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. PEIRCE, J. M.: ELEMENTS OF LOGARITHMS. 50 cents. Designed to give the student a complete and accurate knowledge of the nature and use of logarithms. TAYLOR, J. M.: ELEMENTS OF THE DIFFERENTIAL AND INTEGRAL CALCULUS (Revised Edition, Enlarged and Entirely Rewritten). With Examples and Applications. $2.00. Presents clearly, scientifically, and in their true relations the three common methods in the calculus.

MATHEMATICAL TEXTS. Edited by Percey F. Smith, of the Sheffield Scientific School of Yale University. Eisenhart: Differential Geometry of Curves and Surfaces. $4.50. Granville: Plane Trigonometry and Tables. $1.00. Granville: Plane and Spherical Trigonometry, without Tables. $1.00. Granville: Plane and Spherical Trigonometry, and Tables. $1.25. Granville: Elements of Differential and Integral Calculus. $2.50. Granville: Logarithmic Tables. 50 cents. Hawkes: Advanced Algebra. $1.40. Smith and Gale: Elements of Analytic Geometry. $2.00. Smith and Gale: Introduction to Analytic Geometry. $1.25.


Typographical Errors corrected in Project Gutenberg edition p. 43 Equation (3), the second

wπ

was not printed.

0

2 p. 103 Example 1, the = sign on the second line (beginning ) was not π printed. p. 139 Example 6, second sin term sin

mπy nπy in original, amended to sin . b b

p. 240 Equation (8), Dρ23 V was printed as Dρ3 V .

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