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| 1. | $U=\{z:\operatorname{Im}z≥0\},\quad u(x, 0)=0$ when $x>0, \quad u(x, 0)=1$ when $x<0$.
| | 2. | $U=\{z:\operatorname{Im}z≥0\},\quad u(x, 0)=0$ when $\abs{x}>1, \quad u(x, 0)=1$ when $\abs{x}< 1$. |
To find a conformal map $T$, transform a solution of your second Dirichlet problem to a solution of your first problem.
提问者原文:
I presume there is some way to use a conformal map sending the positive real axis to the real numbers with $|x| > 1$ and the negative to those with $|x|<1$.
这里的回答:
考虑Möbius transformation
$$T(z) = \frac{1+z}{1-z}$$
把$1$映射到$∞$, 把$-1$映射到$0$,
把$-1<z<1$映射到$z>0$, 把$z<-1\vee z>1$映射到$z<0$.
对于$U$的余下部分, 对$x,y∈\Bbb R$,
$$\operatorname{Im}T(x+yi)=\frac{2y}{(1-x)^2+y^2}$$
当$y>0$时, $\operatorname{Im}T(x+yi)>0$, 所以$T(U)=U$. [最后这步也可以用$ad-bc > 0$判断,省得展开计算.] |
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