d We have, as will be proved presently, d2u" (1-μ2) -2(8+1)μ + [(a±1)3 − (s + 1 )3]u”= 0... (100). αμε d du" αμ The operation performed on a solid harmonic of degree dz -a-, and type H{≈, √(x2 + y2)} sin COS 8p, and transformed to polar co-ordinates r, μ, ø, with attention to (83), gives the transition from v to u", as expressed in (99), and thus (100) is proved by (g) (15). Similarly the operation d dx dy transformed to co-ordinates r, p, p, gives (97), and thus (98) is proved by (g) (15). Thus it was that (97) (98), and (99) (100) were found. But, assuming (97) and (99) arbitrarily as it were, we prove (98) and (100) most easily as follows. Let Lastly, applying (87), we find that the corresponding equation is satisfied by B', B', with a 1 and 8+1 instead of a and 8; and by B" ÷B", with a 1 instead of n+8 but with s unchanged. a, derivation. As to (95) and (96), they merely express for the generalized Examples of surface harmonics the transition from s to s+ 1 without change of i shown for complete harmonics by Murphy's formula, § 782 (6) below. (n) Examples of (95) (96), and (99) (100). Example 1. Let a=8+1. This is the particular finite solution indicated in § (k′) II. A. The liberty we now have to let a be negative as well as positive allows us now to include in our formula for u the cases represented by the double sign in II. A of (k). Example 2. By m successive applications of (99) (100), with the upper sign, to v of (103), we find for the complete integral of where ƒ(μ), F(μ), F(μ) denote rational integral algebraic functions of με Of this solution the part C'F(u) is the particular finite solution indicated in § (k') II. B. We now see that the complete solution involves no other transcendent than integer, this is reducible to the form αμ When s is an crease of i, with 8 unchanged, Examples of derivation continued. Algebraic case of last exam ple. a being a constant and f(u) a rational integral algebraic function The process of Example 2, § (n'), gives the complete integral of this equation when i-8 is a positive integer. When also 8, and therefore also i, is an integer, the transcendent involved becomes log1+: in this case the algebraic part of the solution 1 μ [or C'F(μ) (1—μ3) according to the notation of (105) and (78′')] is the ordinary "Laplace's Function" of order and type (i, s); the ",", &c. of our previous notations of $$ (j), (y). It is interesting to know that the other particular solution which we now have, completing the solution of the differential equation for these functions, involves nothing of transcendent but (o') Examples of (99) (100), and (95) (96) continued. Example 3. Returning to (n'), Example 2, let s + be an is algebraic. Thus we have the integer: the integral (1 αμ case of (k') I. B, in which the complete solution is algebraic. (p') Returning to (n), Example 1: let a = and s= 0; (103) becomes Zonal of order zero: of which the complete integral is 1 + μ v = 1C log +C" 1 μ growing into one sectorial by augmenta tion of s, with order still zero: As before, apply (95) (96) n times successively: we find (103'). ..(106) To find the other: treat (106) by (99) (100) with the lower sign; the effect is to diminish a from to, and therefore to make no change in the differential equation, but to derive from (106) another particular solution, which is as follows: the other solution for Giving any different values to C in (106) and (106'), and, using Complete K, K' to denote two arbitrary constants, adding we have the com- tesserals of plete solution of (96′), which we may write as follows: (2) That (107) is the solution of (96') we verify in a moment by trial, and in so doing we see farther that it is the complete solution, whether n be integral or not. order zero: from it of both tes serals of (r) Example 4. Apply (99) (100) with upper sign i times to derivation (107) and successive results. We get thus the complete solution of (84') for a-i any integer, if n is not an integer. But if n every inteis an integer we get the complete solution only provided i<n: this is case I. A of § (k). If we take in-1, the result, algebraic as it is, may be proved to be expressible in the form. gral order; (108): except case of s an inte ger and i8, when place's being the case of (84') for which a=s-, and s=n an integer: only Laapplying to this (99) (100) with upper sign, the constant C dis- function is appears, and we find u'= C' as a solution of found. Hence, for in one solution is lost. The other, found by Examples of derivation continued. continued applications of (99) (100) with upper sign, is the regular "Laplace's function" growing from C' sin" sin COS no, which is the case represented by u'- C' in (109). But in this continuation we are only doing for the case of n an integer, part of what was done in § (n'), Example 2, where the other part, from the other part of the solution of (109) now lost, gives the other part of the complete solution of Laplace's equation subject to the limitation i-n (or i-s) a positive integer, but not to the limitation of an integer or n an integer. (s) Returning to the commencement of § (r), with s put for n, we find a complete solution growing in the form f denoting an integral algebraic function of the th degree, readily found by the proper successive applications of (99) (100). Hence, by (83) (79), we have Finite algebraic expression of complete solution for tesseral harmonics of integral order. for the case of i an integer without any restriction as to the value of 8, which may be integral or fractional, real or imaginary, with no failure except the case of s an integer and i>s, of which the complete treatment is included in § (m'), Example 2, above. |