66 Lectures on Scientific Subjects
tion. Furthermore it will follow that the function /(ʃ, φ),
if it exists, is unique.
Thus we need only prove that if a harmonic distribution
function of order m,fn(s, φ), is not identically zero it cannot
yield a harmonic density function Fm(r, θ) which is identically
zero. This has already been proved for m=0. But for
m>0 we see that Fm(r, ¢)=0 would imply
fg2τ∕m(r sin χ) cos mxdχ=f02rgm(r sin χ) sin mχdχ =0
for∕m(j) and gm(ʃ) not both identically zero.
We will only prove this to be impossible for the cases
m = 1,2 since the extension can then be made at once to the
cases m = 3,4 ∙ ∙ ∙ . Suppose that we have for m = l
f"fι(r sin χ) cos χdχ =4fo2fi(r sin χ) cos χdχ =0.
Multiply through by r and integrate; we obtain at once
∕ι'° W =θ> where we write
/ɪ'ɔ (w) =∫07ι(w)^∙
Hence we infer ∕i(r)≡0. Likewise if we have
J∕gι (r sin χ) sin χdχ=Q,
then by multiplying through by dr and integrating from
0 to r,
fa2gι'^∖r sin χ)dχ=Q.
This is a homogeneous equation of the Abel type in gɪw(ʃ),
and we find similarly
gιc'0(j)≡θ, and so gι(∙τ)≡O.
We pass now to the case m=2. Suppose that we have
f02rf2(r sin χ) cos 2χdχ = 0,