Green's function method for Helmholtz equation with an obstacle

Question

Suppose I wish to solve the 2D inhomogeneous Helmholtz equation $$ (-\nabla^2 +k^2)\psi(\vec{r})=-V(\vec{r})\psi(\vec{r})$$ for a potential $V(\vec{r})$ which corresponds to an impenetrable obstacle, i.e. $V(\vec{r})=\infty$ inside some domain in the plane and $V(\vec{r})=0$ outside it (this corresponds to stationary scattering of waves by the obstacle).

I think the Green function method leads to the identity $$ \psi(\vec{r})=\psi_0(\vec{r})-\int_{\mathbb{R}^2} G(\vec{r},\vec{q}) V(\vec{q})\psi(\vec{q})d^2q,$$ where $$ (-\nabla^2 +k^2)\psi_0(\vec r)=0,\quad (-\nabla^2 +k^2)G(\vec{r},\vec{q})=\delta(\vec{r}-\vec{q}).$$

This seems simple, but: The product $V(\vec{q})\psi(\vec{q})$ must be zero inside the obstacle because $\psi$ vanishes there, and it must also be zero outside the obstacle, because $V$ vanishes there.

So how am I supposed to understand this integral? I suppose it somehow reduces to the boundary of the obstacle, but I am not able to see this exactly.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The "correct" equation involves de Gauss-Green Theorem. Namely $\ds{\pars{~\mbox{by the way,}\ \psi\pars{\vec{r}} = 0\quad \forall\ \vec{r} \in \mathbb{R}^{3}\quad\mbox{is the}\ solution~}}$,

\begin{align} & \mbox{With}\quad\vec{r}\ \mbox{and}\ \vec{q}\quad \mbox{"inside the domain"}\ \color{red}{\mc{D}}\,,\quad \left\{\begin{array}{rcl} \ds{\pars{-\nabla^{2} + k^{2}}\psi\pars{\vec{r}}} & \ds{=} & \ds{0} \\[2mm] \ds{\pars{-\nabla^{2} + k^{2}}\mrm{G}\pars{\vec{r},\vec{q}}} & \ds{=} & \ds{\delta\pars{\vec{r} - \vec{q}}} \end{array}\right. \\[1cm] & \mbox{Then,}\ \mrm{G}\pars{\vec{r},\vec{q}}\pars{-\nabla^{2} + k^{2}}\psi\pars{\vec{r}} - \psi\pars{\vec{r}}\pars{-\nabla^{2} + k^{2}}\mrm{G}\pars{\vec{r},\vec{q}} = -\psi\pars{\vec{r}}\delta\pars{\vec{r} - \vec{q}} \\[5mm] & -\nabla\cdot\bracks{\mrm{G}\pars{\vec{r},\vec{q}}\nabla\psi\pars{\vec{r}} - \psi\pars{\vec{r}}\nabla\mrm{G}\pars{\vec{r},\vec{q}}} = -\psi\pars{\vec{r}}\delta\pars{\vec{r} - \vec{q}} \\[5mm] & \mbox{Integrating over}\ \vec{r} \in \color{red}{\mc{D}}: \\ & \psi\pars{\vec{q}} = \int_{\partial\color{red}{\mc{D}}}\bracks{\mrm{G}\pars{\vec{r},\vec{q}}\nabla\psi\pars{\vec{r}} - \psi\pars{\vec{r}}\nabla\mrm{G}\pars{\vec{r},\vec{q}}}\cdot \dd\vec{S_{\,\vec{r}}} \end{align}

Now, you set $\ds{\mrm{G}\pars{\vec{r},\vec{q}} \equiv \mrm{G}_{D}\pars{\vec{r},\vec{q}}}$ where $\ds{\left.\rule{0pt}{5mm}\mrm{G}_{D}\pars{\vec{r},\vec{q}} \right\vert_{\ \vec{r}\ \in\ \partial\color{red}{\mc{D}}} = 0}$:

$$ \bbx{\psi\pars{\vec{r}} = -\int_{\partial\color{red}{\mc{D}}} \psi\pars{\vec{q}}\nabla\mrm{G}_{D}\pars{\vec{q},\vec{r}}\cdot \dd\vec{S_{\,\vec{q}}}\,,\qquad\vec{r} \in \color{red}{\mc{D}}\,,\quad \left.\rule{0pt}{5mm}\mrm{G}_{D}\pars{\vec{q},\vec{r}} \right\vert_{\ \vec{q}\ \in\ \partial\color{red}{\mc{D}}} = 0} $$