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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,f_n(s, φ), is not identically zero it cannot
yield a harmonic density function F_m(r, θ) which is identically
zero. This has already been proved for m=0. But for
m>0 we see that F_m(r, ¢)=0 would imply

f_g^2τ∕m(r sin χ) cos m_xdχ=f₀^2rg_m(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 =4f_o²f_i(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,

f_a²gι'^∖r sin χ)dχ=Q.

This is a homogeneous equation of the Abel type in gɪw(ʃ),
and we find similarly

gι^c'⁰(j)≡θ, and so gι(∙τ)≡O.

We pass now to the case m=2. Suppose that we have
f₀^2rf2(r sin χ) cos 2χdχ = 0,

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