# Symmetry of Green's function on the general case

Let's consider the differential equation

$$\nabla\cdot[p(\mathbf{r})\nabla u(\mathbf{r})]-s(\mathbf{r})u(\mathbf{r})=-f(\mathbf{r}).$$

I want to show that the Green's function is symmetric, so that $G(\mathbf{r}_1,\mathbf{r}_2)=G(\mathbf{r}_2,\mathbf{r}_1)$.

I tried one argument similar to that used with the Helmholtz equation. In that case the equation is:

$$\nabla^2\psi+k^2\psi=f,$$

In that case the Green's function satisfies

$$\nabla^2G(\mathbf{r},\mathbf{r}_0)+k^2G(\mathbf{r},\mathbf{r}_0)=\delta(\mathbf{r}-\mathbf{r}_0)$$

If we then write this equation for $G(\mathbf{r},\mathbf{r}_i)$ with $i=1,2$ and then we multiply the first by $G(\mathbf{r},\mathbf{r}_2)$, multiply the second by $G(\mathbf{r},\mathbf{r}_1)$ and subtract, we get emplyoing Green's second identity that:

$$G(\mathbf{r}_1,\mathbf{r}_2)-G(\mathbf{r}_2,\mathbf{r}_1)=\int_{S}[G(\mathbf{r},\mathbf{r}_2)\nabla G(\mathbf{r},\mathbf{r}_1)-G(\mathbf{r},\mathbf{r}_1)\nabla G(\mathbf{r},\mathbf{r}_2)]\cdot \operatorname{d\mathbf{S}}$$

Thus, considering Dirichlet or Neumann boundary conditions on $S$ we see that the integral vanishes so that $G(\mathbf{r}_1,\mathbf{r}_2)=G(\mathbf{r}_2,\mathbf{r}_1)$.

Now, I tried to mimic this to the more general case, but the $p$ function stops us from applying Green's second identity. I've been trying this for some time, but I haven't got any far.

So, how can I show the symmetry of the Green function in this case? Any help is appreciated!

• maybe there is a minus missing in your second sentence ? Or this "antisymmetric" is antisymmetric to my usual meaning. Commented May 3, 2016 at 3:32
• Sorry, it was my mistake here. The correct property is that the function is symmetric, not antisymmetric. I thought the right thing and wrote the wrong one. Thanks for pointing this out. Anyway, do you have any idea on this matter? I believe there should be one very simple way to do it that I'm not figuring it out.
– Gold
Commented May 3, 2016 at 3:35
• I don't know, but, that doesn't mean too much on this topic :) Commented May 3, 2016 at 3:42

\begin{align} G(\vec r_1,\vec r_2)-G(\vec r_2,\vec r_1)&=\int_V \left(G(\vec r,\vec r_2)\nabla \cdot(p(\vec r)\nabla G(\vec r,\vec r_1))-G(\vec r,\vec r_1)\nabla \cdot(p(\vec r)\nabla G(\vec r,\vec r_2))\right)\,dV\\\\ &=\oint_S p(\vec r)\left(G(\vec r,\vec r_2)\nabla G(\vec r,\vec r_1)-G(\vec r,\vec r_1)\nabla G(\vec r,\vec r_2)\right)\cdot \hat n\,dS\\\\ &=0 \end{align}
if either $\left. G(\vec r,\vec r')\right|_{\vec r\in S}=0$ or $\left. \hat n\cdot \nabla G(\vec r,\vec r')\right|_{\vec r\in S}=0$.