i could use just a bit of help with the justification, i believe i know why this is true but a small explination beyond my intuition would be great.
Consider the Helmholtz equation in the form
$$(\nabla^2 + \alpha^2)u(x,y) = f(x,y), \qquad -\infty < x,y < \infty$$
where $\alpha > 0$ and $f(x,y)$ is the forcing with compact support (ie vanishing outside some finite region D of $\mathbb{R}^2$) at infinity, u and its derivatives tend to zeroDefine an appropriate Green's function for this problem and use it in Greens 2nd identity to show that the function is symmetrical and that
$$u(\xi,\eta) = \int_{D}g(x,y;\xi,\eta)f(x,y)dxdy$$
So, lets begin with the greens function. we define the Green's Function $g(x,y;\xi,\eta)$ to be one such that it satisfies $$(Lg)(x,y;\xi,\eta) = \delta(x-\xi)\delta(y-\eta)$$
then to show it's symmetrical we note greens second identity is
$$ \int_{V}(v \nabla^2 w – w\nabla^2v)~dxdy = \int_{\partial V}(v\nabla w-w\nabla v)\cdot \mathbf{\hat{n}}~dS$$
Letting $ v = g(x,y;\xi,\eta) ~\&~ w = g(\xi,\eta;x,y)$
$$g(x,y;\xi,\eta) – g(\xi,\eta;x,y) = \int_{D}(g(x,y;\xi,\eta) \nabla^2 g(\xi,\eta;x,y) – g(\xi,\eta;x,y)\nabla^2 g(x,y;\xi,\eta))~dxdy = \int_{\partial D}(g(x,y;\xi,\eta)\nabla g(\xi,\eta;x,y)-g(\xi,\eta;x,y)\nabla g(x,y;\xi,\eta))\cdot \mathbf{\hat{n}}~dS = 0$$
Now my rationale for why this is equal to zero is due to the assumption of u vanishing under the Dirichlet and Neumann boundary conditions as x,y tend to infinity (in either direction)
Is this sufficient justification or am i missing something more subtle and thus could someone kindly expand for me. Thank you very much.
essentially i believe i want to prove that either $$g(x,y;\xi,\eta) \bigg \rvert_{(x,y) \in \partial D} = 0$$ or $$\mathbf{\hat{n}}\cdot \nabla g(x,y;\xi,\eta) \bigg \rvert_{(x,y) \in \partial D} = 0$$
Best Answer
Yes, what you wrote seems correct. The definition of the Green's function will depend on what information you know at the boundary. From the definition of the Green's function and Green's second identity, \begin{equation} \iint_{D} (g Lu - u Lg)dxdy = \iint_{D} (g \nabla^2u - u \nabla^2g)dxdy = \int_{\partial D} (g \nabla u - u \nabla g) \cdot \mathbf{n} ds, \end{equation} that is \begin{equation} \iint_{D}fgdxdy-u(\xi,\eta) = \int_{\partial D} \left( g \dfrac{\partial u}{\partial n} - u \dfrac{\partial g}{\partial n} \right) ds. \end{equation} So you have two possibilities: Dirichlet or Neumann boundary conditions. That is, you either know the value of the solution at the boundary (eg. clapped membrane), or you have some information on the flux leaving the boundary (eg. insulation). Based on what you have prescribed, you use the Green's function to "kill off" whatever you don't know. So for instance, if you're given $u(x,y)$ on $\partial D$, you want to remove dependence on the normal derivative in the integral and you set $g(x,y;\xi,\eta)=0$ on $\delta D$. Vice versa if $\partial u/ \partial n$ is instead prescribed.
Either way, the solution is a convolution of the Green's function and the forcing function as required.
I think it's best to think of this as the "first part" of the problem - you haven't fully determined the Green's function until you've specified what it does at the boundaries. Then, once you know that either $g$ or its normal derivative vanish at the boundary, you can argue symmetry.