and

+ ( ± a + s + 1 ) v.

du'

n+2

Lastly, applying (87),
tion is satisfied by B',
a and s; and by B",
unchanged.

n+2

+ [(a ± 1)2 − (8 + 2)2]u' = 0...(98).

(99).

(± a + s + 1 ) μv......

d
The operation
dz
-a-1, and type H{z, √(x2 + y2)}

co-ordinates r, p, p, 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

..(97).

du"

dμ + [(a±1)3 — (s + 1 )*] u" = 0.....(100).

performed on a solid harmonic of degree
sin
COS

84, and transformed to polar

u' = ΣB'", and u" = ΣB",
"M"
Then, by (97) and (99), with (85), we find
B'n+3 = (n+2a + 8 + 1) A +2
s
n + 1 + 1 + s ± a
n+ 2

B" (= a + 8 − 1)

A

#+2

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

a+1

(102).

we find that the corresponding equa÷B', with a±1 and 8+1 instead of ÷B", with a 1 instead of a, but with s

..(101).

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 shown for complete harmonics by Murphy's formula, § 782 (6) below.

(1 − μ3)

d'u'
αμε

du

- 2 (n + s + 1) μ Jμ

dv

-2 (s+1)μ Jμ

(1

2(8 + 1)μ

as a solution of

d'u

(1-μ2)

αμε

This is the particular finite solution indicated in § (') 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 ́).

a log

=

n(n + 2s + 1)v = 0

du'

άμ

0,

Example 2. By m successive applications of (99) (100), with the upper sign, to v of (103), we find for the complete integral of

·+m(m + 28 +1)u'=0

1 μ
1-μ

.....

+f(μ),

.(103).

(104).

where ƒ(μ), F(u), F(μ) denote rational integral algebraic functions of μ.

Of this solution the part C'F(u) is the particular finite solution indicated in § (') II. B. We now see that the complete solution . When s is an

αμ

involves no other transcendent than

integer, this is reducible to the form

Tesserals from sec

(105), torial by in

crease of i, with s unchanged,

Tesserals from sectorial by in

crease of s, with order i unchanged.

Examples of derivation continued.

Algebraic case of

last example.

Zonal of order zero:

growing into one sectorial by

augmenta tion of 8, with order still zero:

a being a constant and f(μ) a rational integral algebraic function of μ. In this case, remembering that (105) is what (84) becomes when m+s+ is put for a, we may recur to our notation of $$ (g) (j), by putting i for m+s, which is now an integer and going back, by (83) to (80) or (78), put

thus (105) is equivalent to

d

dw

¿/ [α-~~] + [− + (+1)].

(1-μ2) dμ
αμ

2

αμ

μ

(78').

The process of Example 2, § (n'), gives the complete integral of this equation when i-s is a positive integer. When also 8, and therefore also i, is an integer, the transcendent involved be1+ μ. in this case the algebraic part of the solution 1-μ

comes log

[or C'F(μ) (1— μ2); according to the notation of (105) and (78')] is the ordinary "Laplace's Function" of order and type (i, 8); 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

log

(p) Returning to (n), (103) becomes

dv

(1 − μ3) dμ3

(o') Examples of (99) (100), and (95) (96) continued.

Example 3. Returning to (n'), Example 2, let s + be an αμ

integer: the integral (μ3)+1 is algebraic. Thus we have the

case of (k') I. B, in which the complete solution is algebraic.

Example 1: let a = and s=0;

of which the complete integral is

v = C log

..(83');

....

0,

1+ μ + C' 1-μ

As before, apply (95) (96) n times successively: we find

...(106)

(103′).

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:

u1 = 1.1.2... (n − 1). n. C © [ ( 1 ) + ( ~ )"]

x

u =

(106').

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:

order zero:

K
(1 - μ)"

the other
derived
from this by

n (n + 1) u = 0. ... ... ...... (96'). order to -1,

equivalent to zero.

u =

(107).

(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.

+

K'
(1 + μ)"'

d'u

(1 - p) d
αμ

from it of

serals of

() Example 4. Apply (99) (100) with upper sign i times to derivation (107) and successive results. We get thus the complete solution both tesof (84′) for a − } = i any integer, if n is not an integer. But if n is 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

every integral order;

which is therefore for n an integer the complete integral of

....

0

ger and
is when

being the case of (84) for which a = s−1, 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.

d2u'

du'

(1 − μ2) dμ3
αμε

(109).

· 2 (n + 1) μ Jp

Hence, for in one solution is lost. The other, found by

(108):

except case of s an inte

Examples of derivation continued.

Finite algebraic expression of complete solution for tesseral har

monics of

integral order.

continued applications of (99) (100) with upper sign, is the

sin

regular "Laplace's function" growing from C' 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 in (or i-8) 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

Kƒ. (μ)(1 + μ)' + (−)' K'ƒ., (— μ) (1 − μ)°
(1 μ2)

w=

(110');

f denoting an integral algebraic function of the ith degree, readily found by the proper successive applications of (99) (100). Hence, by (83) (79), we have

Kƒ.(μ) (1 + μ)' + (−)'K'ƒ. (−μ) (1 − μ)'

(1 − μ3) 1⁄2

as the complete solution of Laplace's equation

% [0-9] + [1/2 + 6 + 1)]·

d
άμ

dw
dp

1

i (i w =
20 0

(110);

(111),

(112),

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.