Elasticity_ Barber

18.6 Solution of harmonic boundary value problems 285
Example
Suppose that φ(a, θ)=φ 2 cos(2θ), where φ 2 is a real constant and we wish to
find the corresponding harmonic function φ(r, θ) inside the circle of radius a.
Using (18.50), we can write
φ 2 cos(2θ) = φ 2(e 2ıθ + e −2ıθ )
2
= φ 2
2
( ) s
2
a 2 + a2
s 2
and it follows that
∮
1 φ(s)ds
2πı (s − ζ) = φ ∮
2
4πa 2 ı
s 2 ds
(s − ζ) + φ 2a 2 ∮
4πı
ds
s 2 (s − ζ) = φ 2ζ 2
2a 2 ,
where we have used (18.44) for the first integral and the second integral is
zero in view of (18.42).
Using (18.55), we then have
and
f = φ 2ζ 2
2a 2 − C 0 ; f = φ 2 ¯ζ 2
2a 2 − C 0
φ = f + f = φ 2(ζ 2 + ¯ζ 2 )
2a 2 − C 0 − C 0 .
Evaluating the integral in (18.57), we obtain
C 0 + C 0 = φ 2
2π
∫ 2π
and hence the required harmonic function is
φ = φ 2(ζ 2 + ¯ζ 2 )
2a 2
0
cos(2θ)dθ = 0 ,
= φ 2r 2 cos(2θ)
a 2 ,
using (18.3). Of course, this result could also have been obtained by writing
a general harmonic function as the Fourier series
∞∑
∞∑
φ = A n r n cos(nθ) + B n r n sin(nθ)
n=0
and equating coefficients on the boundary r = a, but the present method,
though longer, is considerably more direct.
n=1
18.6.2 Direct method for the exterior problem for a circle
For the exterior problem, we again define φ as in (18.49) and apply the Cauchy
integral operator to the boundary data, obtaining