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778 CHAPTER 23 Conformal Mappings and Applications

If w = T (z), then z = T (w), and we define
−1
−1
˜ g(w) = g(T (w)) = g(z).
In the w-plane, we now have a Dirichlet problem for the unit disk:
2
2
∂ ˜u ∂ ˜u
˜
x
+ = 0for (˜, ˜y) in D
2
∂x 2 ∂ y
and
x
˜
˜ u(˜, ˜y) =˜g(˜, ˜y) for (˜, ˜y) in C.
x
x
By equation (23.6), the solution is

1 ζ + w 1
˜ u(w) = Re( f (w)) = Re ˜ g(ζ) dζ .
˜
2πi ˜ C ζ − w ζ
Because T maps C onto C and ζ =T (ξ) for ξ on C, the solution of the Dirichlet problem for D is
˜
1 T (ξ) + T (z) 1

u(x, y) = Re( f (z)) = Re ˜ g(T (ξ)) T (ξ)dξ .
2πi ˜ C T (ξ) − T (z) T (ξ)
−1
Since ˜g(T (ξ)) = g(T (T (ξ))) = g(ξ), we can write the solution
1 T (ξ) + T (z) T (ξ)

u(x, y) = Re( f (z)) = Re g(ξ) dξ . (23.7)
2πi C T (ξ) − T (z) T (ξ)
In this integral, C need not be a closed curve. If D is (for example) the right quarter-plane
x > 0, y > 0, then C will consist of two segments of the real and imaginary axes.
EXAMPLE 23.19
We will demonstrate this technique for a Dirichlet problem for the right half-plane:
2
2
∂ u ∂ u
+ = 0for x > 0,−∞ < y < ∞
2
∂x 2 ∂ y
and
u(0, y) = g(y) for −∞ < y < ∞.
We need a conformal mapping from the right half-plane to the unit disk. There are many such
mappings, but we found one in Example 23.13:
z − 1

w = T (z) =−i .
z + 1
Compute
−2i

T (z) = .
(z + 1) 2
From equation (23.7), the solution is the real part of
f (z) =
1 −i(ξ − 1)/(ξ + 1) − i(z − 1)/(z + 1) 1 −2i
u(ξ) dξ
2πi C −i(ξ − 1)/(ξ + 1) + i(z − 1)/(z + 1) −i(ξ − 1)/(ξ + 1) (ξ + 1) 2
1 ξz − 1 1
= u(ξ) dξ.
2
πi C ξ − z ξ − 1

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October 14, 2010 15:39 THM/NEIL Page-778 27410_23_ch23_p751-788