Paper 3, Section I\mathbf{I}, B

Further Complex Methods
Part II, 2009

Suppose that the real function u(x,y)u(x, y) satisfies Laplace's equation in the upper half complex zz-plane, z=x+iy,xR,y>0z=x+i y, x \in \mathbb{R}, y>0, where

u(x,y)0 as x2+y2,u(x,0)=g(x),xR.u(x, y) \rightarrow 0 \quad \text { as } \quad \sqrt{x^{2}+y^{2}} \rightarrow \infty, \quad u(x, 0)=g(x), \quad x \in \mathbb{R} .

The function u(x,y)u(x, y) can then be expressed in terms of the Poisson integral

u(x,y)=1πyg(ξ)(xξ)2+y2dξ,xR,y>0u(x, y)=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{y g(\xi)}{(x-\xi)^{2}+y^{2}} d \xi, \quad x \in \mathbb{R}, y>0

By employing the formula

f(z)=2u(z+aˉ2,zaˉ2i)f(a)f(z)=2 u\left(\frac{z+\bar{a}}{2}, \frac{z-\bar{a}}{2 i}\right)-\overline{f(a)}

where aa is a complex constant with Ima>0\operatorname{Im} a>0, show that the analytic function whose real part is u(x,y)u(x, y) is given by

f(z)=1iπg(ξ)ξzdξ+ic,Imz>0f(z)=\frac{1}{i \pi} \int_{-\infty}^{\infty} \frac{g(\xi)}{\xi-z} d \xi+i c, \quad \operatorname{Im} z>0

where cc is a real constant.