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23.3 Conformal Mapping Solutions of Dirichlet Problems 777

Substitute these expressions for the coefficients into the Maclaurin series for f (z) to get
∞ ∞
1 1 1
n −n−1 n
f (z) = a n z = u(ζ) dζ + u(ζ)ζ dζ z
2πi ζ πi
n=0 γ n=1 γ

∞ n

1 z u(ζ)
= 1 + 2 dζ.
2πi γ n=1 ζ ζ
Since |z| < 1 and |ζ|= 1, then |z/ζ| < 1, and the geometric series in this equation converges:
∞ n
z z/ζ z

= = .
ζ 1 − z/ζ ζ − z
n=1
Insert this into the last equation to obtain
1
2z 1 1 ζ + z 1
f (z) = 1 + dζ = dζ.
2πi γ ζ − z ζ 2πi γ ζ − z ζ
Since ζ is on γ , u(ζ) = g(ζ). Therefore, for |z| < 1,

1 ζ + z 1
u(x, y) = Re( f (z)) = Re g(ζ) dζ . (23.6)
2πi γ ζ − z ζ
iθ
This is an integral solution of the Dirichlet problem for the unit disk. If z = re and ζ = e iϕ
are inserted into this solution, the Poisson integral formula results.
Equation (23.6) is well suited to solving certain Dirichlet problems using conformal map-
˜
˜
pings. Suppose we know a conformal mapping T : D → D where D is the unit disk |w| < 1
(Figure 23.31). Assume that T maps C, which is the boundary of D, onto the unit circle C ˜
˜
bounding D, and that T −1 is also a conformal mapping.
x
To clarify the discussion, we will use ζ for a point of C, ξ for a point on C, and (˜, ˜y) for a
˜
point in the w-plane.
Now consider a Dirichlet problem for D:
2
2
∂ u ∂ u
+ = 0for (x, y) in D
2
∂x 2 ∂ y
and
u(x, y) = g(x, y) for (x, y) in C = ∂ D.
The idea is to map this problem to one for the unit disk D for which we have the solution of
˜
equation (23.6), then use the inverse map to convert this integral into the solution for D.

z w
y v
T ~
i D : w < 1
D
1
x u
~
C
C T −1
FIGURE 23.31 Solving a Dirichlet problem by a con-
formal mapping.

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October 14, 2010 15:39 THM/NEIL Page-777 27410_23_ch23_p751-788