Rectilinear Drawing 67
which we write in the form
fo'⅛(r sin χ)(cos χ cos χ - sin χ sin χ')dχ=0.
On integrating by parts the first term evidently takes the
form
1 [aτ
- ∕2w (r sin χ) sin χdχ,
rJo
while the second is seen to be equal to
d faτ
-jf I fz'^'' (r sɪn χ) sin χdχ.
Hence if we write
W = X*7?0 (r sin χ) sin χdχ,
we conclude that
1jγ-jγ≡o,
r
i.e., W = cτ. But since /Fz(0) =0 we must have c =0, and so
F≡0.
Employing now the same argument as in the case m = 1,
for the equation W = Owe find that f∙ι'∖s) =0 and so ∕a(j,)≡0.
Likewise we readily infer that gs(∕)≡O.
More generally, we may obtain the stated result for
m = 1, 2 ■ ∙ ∙ in succession, by writing at the mth stage
cos mχ =cos χ cos (tn — l)χ —sin χ sin (m-l')χ
sin mχ =sin χ cos (m — l)χ+ cos χ sin (m-l)χ
in the equations involving ∕m(j) and gm(j∙) respectively, and
thus reducing the question to one of the type considered at
the (tn —l)st stage.
In this manner the stated result is readily established.
We are now in a position to formulate preliminary neces-
sary and sufficient conditions for a continuous solution of the
general harmonic case of the mth order.