上一节，我们得到了圆内的狄利克雷（Dirichlet）问题\[\begin{cases}u_{xx}+u_{yy}=0\\ u(R,\varphi)\end{cases}\]的解由泊松（Poison）公式给出 \[u(r,\varphi)=\frac{1}{2\pi}\int_0^{2\pi}g(\theta)\cdot\frac{R^2-r^2}{R^2+r^2-2Rr\cos(\theta-\varphi)}d\theta\]
现在我们来看一个例子。
例：求解狄利克雷问题\[\begin{cases}u_{xx}+u_{yy}=0,\qquad&\\ u(x,y)=0,& y=\sqrt{1-x^2}\\ u(x,y)=1,&y=-\sqrt{1-x^2}\end{cases}\]
解：将 \[g(\varphi)=\begin{cases}0,\qquad & 0<\varphi\leq \pi\\ 1, & \pi<\varphi\leq 2\pi\end{cases}\] 代入 Poison 公式，
\begin{array}{lll}u(r,\varphi)&=&\frac{1}{2\pi}\int_0^{2\pi}g(\theta)\cdot\frac{1-r^2}{1+r^2-2r\cos(\theta-\varphi)}d\theta\\ &=&\frac{1}{2\pi}\int_0^{\pi}0\cdot\frac{1-r^2}{1+r^2-2r\cos(\theta-\varphi)}d\theta+\frac{1}{2\pi}\int_{\pi}^{2\pi}1\cdot\frac{1-r^2}{R^2+r^2-2Rr\cos(\theta-\varphi)}d\theta\\ &=&\frac{1}{2\pi}\int_{\pi}^{2\pi}\frac{1-r^2}{1+r^2-2r\cos(\theta-\varphi)}d\theta\end{array}
我们考虑积分\[\int \frac{1-r^2}{1+r^2-2r\cos t}dt\]作半角代换 \(u=\tan\frac{t}{2}\)，相应地有\[\sin t=\frac{2u}{1+u^2},\cos t=\frac{1-u^2}{1+u^2},dt=\frac{2du}{1+u^2}\]代入积分，有
\begin{array}{lll}\int \frac{1-r^2}{1+r^2-2r\cos t}dt&=&\int\frac{1-r^2}{1+r^2-2r\cdot\frac{1-u^2}{1+u^2}}\cdot\frac{2du}{1+u^2}\\ &=&2\int\frac{1(1-r^2)du}{(1+r^2)(1+u^2)-2r(1-u^2)}\\&=&2\int\frac{1-r^2}{u^2(1+r^2)+(1-r^2)}du\\&=&\frac{2(1-r^2)}{(1-r)^2}\int\frac{du}{1+\left(\frac{1+r}{1-r\cdot u}\right)^2}\\&=&2\arctan\frac{1+r}{1-r}u+C\\&=&2\arctan\left[\frac{1+r}{1-r}\cdot\tan\frac{\theta-\varphi}{2}\right]\end{array}
所以\begin{array}{lll}u&=&\frac{1}{2\pi}\int_{\pi}^{2\pi}\frac{1-r^2}{1+r^2-2r\cos(\theta-\varphi)}d\theta\\&=&\frac{1}{\pi}\arctan\left[\frac{1+r}{1-r}\cdot\tan\frac{\theta-\varphi}{2}\right]\Bigg|_{\pi}^{2\pi}\\ &=&\frac{1}{\pi}\arctan\left[\frac{1+r}{1-r}\cdot\tan\frac{2\pi-\varphi}{2}\right]-\frac{1}{\pi}\arctan\left[\frac{1+r}{1-r}\cdot\tan\frac{\pi-\varphi}{2}\right]\end{array}