Chapter 22: Problem 12
Consider GF for the Helmholtz operator \(\nabla^{2}+\mu^{2}\) in two dimensions. (a) Show that $$ G\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=-\frac{i}{4} H_{0}^{(1)}\left(\mu\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)+H\left(\mathbf{r}, \mathbf{r}^{\prime}\right) $$ where \(H\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) satisfies the homogeneous Helmholtz equation. (b) Separate the variables and use the fact that \(H\) is regular at \(\mathbf{r}=\mathbf{r}^{\prime}\) to show that \(H\) can be written as $$ H\left(\mathbf{r}, \mathbf{r}^{\prime}\right)=\sum_{n=0}^{\infty} J_{n}(\mu r)\left[a_{n}\left(\mathbf{r}^{\prime}\right) \cos n \theta+b_{n}\left(\mathbf{r}^{\prime}\right) \sin n \theta\right] . $$ (c) Now assume a circular boundary of radius \(a\) and the \(\mathrm{BC} G\left(\mathbf{a}, \mathbf{r}^{\prime}\right)=0\), in which a is a vector from the origin to the circular boundary. Using this \(\mathrm{BC}\), show that $$ \begin{aligned} a_{0}\left(\mathbf{r}^{\prime}\right)=& \frac{i}{8 \pi J_{0}(\mu a)} \int_{0}^{2 \pi} H_{0}^{(1)}\left(\mu \sqrt{a^{2}+r^{\prime 2}-2 a r^{\prime} \cos \left(\theta-\theta^{\prime}\right)}\right) d \theta, \\ a_{n}\left(\mathbf{r}^{\prime}\right)=& \frac{i}{4 \pi J_{n}(\mu a)} \\ & \times \int_{0}^{2 \pi} H_{0}^{(1)}\left(\mu \sqrt{a^{2}+r^{\prime 2}-2 a r^{\prime} \cos \left(\theta-\theta^{\prime}\right)}\right) \cos n \theta d \theta \\ b_{n}\left(\mathbf{r}^{\prime}\right)=& \frac{i}{4 \pi J_{n}(\mu a)} \\ & \times \int_{0}^{2 \pi} H_{0}^{(1)}\left(\mu \sqrt{a^{2}+r^{\prime 2}-2 a r^{\prime} \cos \left(\theta-\theta^{\prime}\right)}\right) \sin n \theta d \theta \end{aligned} $$ These equations completely determine \(H\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\) and therefore \(\bar{G}\left(\mathbf{r}, \mathbf{r}^{\prime}\right)\).
Short Answer
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