二維逆問題/黎曼對映定理

For every proper simply connected open subset U of the complex plane C there exists a one-to-one conformal Riemann map from U onto the open unit disk D. Since the composition of harmonic and analytic function is harmonic, the Riemann map provides a one-to-one correspondence b/w harmonic functions defined on the set U and on the disc D. Therefore, one can transfer a solution of Dirichlet boundary problem on the set D to the domain U. 

Let $f:U\rightarrow D$ be a Riemann map for the region U, the kernel of the Dirichlet-to-Neumann map for the region U can be expressed in terms of the kernel for the disc.

練習 (*)。證明 $K_{U}(\phi ,\theta )=|f'(\phi )|K_{D}(f(\phi ),f(\theta ))|f'(\theta )|$ 在對角線之外。

The Cayley transform ${\frac {1-z}{1+z}}:C^{+}\rightarrow D$ maps the complex right half-plane onto the unit disc.

練習 (**)。推匯出單位圓盤D的DN對映核的公式：$K_{D}(\phi ,\theta )={\frac {-1}{\pi (1-cos(\phi -\theta ))}}$ 使用半平面核公式和泊松核的徑向導數，求解圓盤上的狄利克雷邊界問題

$K_{P}(z,\theta )={\frac {1-|z|^{2}}{2\pi |1-ze^{-i\theta }|^{2}}}.$

In order to solve a continuous inverse problem by data discretisation, one may obtain Dirichlet-to-Neumann (DN) matrix by uniform sampling of the kernel off the diagonal, and by defining the diagonal entries so, that rows and columns of the DN matrix sum up to zero. This leads to the following definition of the matrix in the case of the unit disc:

$\Lambda _{kl}={\begin{cases}{\frac {2n(n+1)}{3}},{\mbox{ }}k=l,\\{\frac {-1}{1-\cos {\frac {2\pi (k-l)}{2n+1}}}},{\mbox{ }}k\neq l,\end{cases}}$ 其中n是自然數，k,l = 1,2, ... 2n+1。

練習 (**)。證明矩陣的特徵值是自然數 (!) $2n,4n-2,6n-6,\ldots ,n(n+1)$ 重數為二，0 重數為一。