Page images
[merged small][ocr errors][merged small][merged small]

For a direct investigation of the complete solution in finite terms for the case is a positive integer, see below § (n), Example 2; and for the case i an integer, and s either not an integer or not <i, see § (o') (111).

The rational integral solution alone can enter, and it alone suffices, when the problem deals with the complete spherical surface. When there are boundaries, whether by two planes meeting in a diameter at an angle equal to a submultiple of four right angles, or by coaxal cones corresponding to certain particular values of 0, or by planes and cones, both the rational integral solution and the other are required. But when there are coaxal cones for boundaries, the values of i required by the boundary conditions [§ (1)] are not generally integral, and it is only when i-s is integral that either solution is a rational and integral function of sin and cos 0. Hence, in general, for the class of problems referred to, two solutions are required and neither is a rational integral function of sin and cos 0.

[ocr errors]

(ƒ) The ordinary process for the solution of linear differential equations in series of powers of the independent variable when the multipliers of the differential coefficients are rational algebraic functions of the independent variable leads easily from the equation (75) to any of the forms of rational integral solutions referred to above, as well as to the second solution in a form corresponding to each of them, when i and s are integers; and, quite generally, to the two particular solutions in every case, whether and s be integral or fractional, real or imaginary. Thus, putting as above, § (k),


of "La

cos =


sin 0: μg ..(77), make μ the independent variable in the first place, in order to Differential find expansions in powers of μ: thus (75) becomes


with v independent variable omitted here for brevity.


This is the form in which "Laplace's equation" has been most commonly presented. To avoid the appearance of supposing



place's functions."

form of


Obvious solution in ascending powers of μ;

why dismissed.

i and s to be integers or even real, put

"w, i(i+1)= a, s2= b,...............(79).

Using this notation, and multiplying both members by (1-), we have, instead of (78),




(1 − μ3) 1/4 [ ( 1 − μ3) — ] + [a (1 − μ3) − b] w = 0 ......... (80).


To integrate this equation, assume

w = ΣK12μ",

and in the series so found for its first member equate to zero the coefficient of μ". Thus we find (n + 1) (n + 2) K„+2 = [2n3 − a + b] K„ — [(n − 1) (n − 2) – a] K...................(81).



The first member of this vanishes for n=- - 1, and for n=-
-2, if
K and K be finite. Hence, we may put K = 0 for all negative
values of n, give arbitrary values to K and K1, and then find
K1, K, K ̧, &c., by applications of (81) with n = 0, n = 1, n = 2,...
successively. Thus if we first put K=1, and K1 = 0; then
again K=0, K1 = 1; we find two series of the forms


1 + K ̧μ2 + K ̧μ* + &c.


and therefore


μ+ K ̧μ3 + K ̧μ3 + &c.,


each of which satisfies (80); and therefore the complete solution is




w = C (1 + K ̧μ2 + K ̧μ1 + &c.) + C'' {μ + K ̧μ3 + K ̧μ3 + &c.)... (82). From the form of (81) we see that for very great values of ʼn we have

K+2=2K-K-, approximately,

[blocks in formation]

Hence each of the series in (82) converges for every value of μ less than unity.

(g) But this is a very unsatisfactory form of solution. It gives in the form of an infinite series 1+ K ̧μ2 + K ̧μ*+ &c. or μ + K ̧μ3 + K ̧μ3 + &c., the finite solution which we know exists


in the form


(1 − μ2)2 (4 ̧ + Â ̧μ2 + ... A¡-‚μ3¬3)

(1 − μ3)2(4 ̧μ +  ̧μ3 + ... ‚¬‚μ2),

when b is the square of an odd integer (s), and when a = i (i + 1),
i being an odd integer or an even integer; and, a minor defect,
but still a serious one, it does not show without elaborate veri-
fication that one or other of its constituents 1+ K ̧μ2+ &c. or
μ+K ̧μ3+ &c. consists of a finite number, i or (i + 1), of terms
when b is the square of an even integer and a = i (i + 1), i being
an even integer or an odd integer.

(h) A form of solution which turns out to be much simpler in every case is suggested by our primary knowledge [§ (j) above] of integral solutions. Put

dv άμε

Assume now

(1 − μ3)



in (80) and divide the first member by (1-μ3) 2. Thus we find

[ocr errors]

2 (√b +.1) μ




dv άμ

+ [a − √b (√b + 1)]v = 0................(84).

v = ΣA ̧μ” .


equating to zero the coefficient of μ" in the first member of (84) gives (n+1)(n+2) A„+,− [(n−1) n + 2 (√b+1)n−a + √b (√b+1)] 4 ̧=0... (86), its solution or (n+1)(n+2) 4+ (n + + 8 + a) (n + 1 + 8 − a) 4..... (87), 1 if we put a = √(a + }), 8 = √b ..(88),

in powers of μ:


and with this notation (84) becomes


The second member of (87) shows that if the series (85) is in

descending powers of ሥ its first term must have either

2 (8 + 1)μ



[ocr errors]
[ocr errors]

Geometrical antecedents suggest modified

form of solution;

+ [a3 − (8 + 1)2] v = 0 .............. (84').

correspondingly modified differential equation:

or n=


order dis


n = − 1 − s + a, the expansion thus obtained would, if not finite, be convergent descending when μ> 1 and divergent when <1, and they are there- missed. fore not suited for the physical applications. On the other chosen. hand, the first member of (87) shows that if the series (85) is in ascending powers of μ, its first term must have either n =

0 or

Complete solution for tesseral harmonics.

Alternative solution by changing sign of s:

its import


n=1: the expansions thus obtained are necessarily convergent when μ < 1, and it is therefore these that are suited for our purposes. Taking then 41 and 4, 0, and denoting by p the series so found, and again 40 and 4, 1, and q the series; so that we have










q = μ + A ̧μ3 + Â ̧μ3 + etc.


A,, 4, etc. and 4, 4,, etc. being found by two sets of successive applications of (87); then the complete solution of (84) is v = Cp + C'q....

(90). This solution is identical with (38) of § (1) above, as we see by (88) and (79), which give

a = i + 1/2


(i) The sign of either a or s may be changed, in virtue of (88). No variation however is made in the solution by changing the sign of a [which corresponds to changing i into i-1, and verifies (13) (g) above]: but a very remarkable variation is made by changing the sign of 8, from which, looking to (88), (83), (87), we infer that if p and q denote what p and q become when is substituted for s in (89), we have


1 + Â ̧μ3 + Â ̧μ1 + etc. )





and the prescribed modification of (89) gives

p=1+ A ̧μ2 + A ̧μa + etc.


q=μ+ Ã ̧μ3 + A ̧μ3 + etc.



[ocr errors]
[ocr errors]


A,, A, etc., and A, A,, etc. being found by successive applica

tions of

(n + − 8 + a) (n + 1 - 8 - a)
(n + 1)(n+2)




(j) In the case of "complete harmonics" s is zero or an integer, and the p or q solution expressing the result of multiplying the already finite and integral p or q solution by the integral polynomial (1 μ3), is only interesting on account of the way of obtaining it from (87), etc. in virtue of (88). But when either a-or s is not an integer, the possession of the alternative solutions, p or p, q or q may come to be of great intrinsic importance, in respect to obtaining results in finite form. For, supposing a and s to be both positive, it is impossible that both p and q can be finite polynomials, but one or both of p and q may be so; or

[ocr errors]


of finite

one of the p, q forms and the other of the p, q forms may be Cases finite. This we see from (87) and (94), which show as follows:-algebraic


1. If +8-a is positive, p and q must each be an infinite series; but por q will be finite if either + 8-a or + 8 + a is a positive integer*; and both p and q will be finite if +8-a and +8+a are positive integers differing by unity or any odd number.

is zero or a positive

2. If a 8+, one of the two series p, q must be infinite; and if a - 8 - is zero or a positive integer, one of the two series p, q is finite. If, lastly, a +8integer, one of the two p, q is finite. a-s-is zero or even, q if it is odd: a+sis zero or even, q if it is odd.


is p that is finite if and p that is finite if Hence it is p and P, or


and q that are finite if 2s be zero or even; but it is Р and q, or 9 and P that are finite if 2s be odd. Hence in this latter case the complete solution is a finite algebraic function of μ.

[ocr errors]

(k) Remembering that by a and s we denote the positive values of the square roots indicated in (88), we collect from (j') 1 and 2, that, if F denote a rational integral function of μ and (1-2), the character of the solution of (80) is as follows in the several cases indicated:-


(A; a <s+; if s and a - are integers.

I. B; as+; if s+ and a are integers.

The complete solution is F.

·A; a < s + 1; if s±(a) is an integer, but a – not an integer.

II. B; a8+ 1; if a-s is an integer, but 8+ 1⁄2 not an integer.

A particular solution is F; but the complete solution is not F.

() "Complete Spherical Harmonics," or "Laplace's Coefficients," are included in the particular solution F of Case II. B.

(m) Differentiate (84) and put

dv du

= u


Unity being understood as included in the class of "positive integers."

[ocr errors]
« PreviousContinue »