Rectilinear Drawing 71
e =(cos φ- г sin φ)e ,
and then proceeding essentially as in the case m = 1 treated
above.
We therefore content ourselves with stating the final
result and giving in an appended Note the series of re-
ductions involved. The reader who wishes a more detailed
deduction will find it in my article referred to at the outset.
In the case of a given harmonic distribution function of the
mth order,
Fm(r, θ) =Fm(r) cos τnθ+Gm(r) sin mθ,
where Hm(r) =Fm(r) +iGm(f) is supposed to vanish to the
{m-l)st order at r =Q so that Hm(r)∕rm~' remains finite, there
exists a corresponding continuous distribution function.
fm(s, <p) =fm(s) COS mφ +gm(ʃ) sin m<f>,
likewise harmonic of the mth order, if and only if the integral
in the equation
<v, ⅛ im d fsH (^+Vu^-∖y + {u-^u^-F)m
(5) hm(s) Ts jo 77m(∕)-------------------------dt,
J-
is continuous and admits a continuous derivative as to s, in
which case the unique solution is derived by setting hm(s') in
(5) equal to fn(fi) +igm(s).
10. EXPLICIT SOLUTION IN THE N0N-HARM0NIC CASE
In virtue of the known relationship between continuous
functions and their Fourier series we are now in a position
to formulate the general solution of our problem:
If the given continuous density function F(r, 0) be expanded
in a Fourier series