In the book Introduction to Partial Differential Equations by Folland, he states the claim
Let $ \Omega$ be a bounded domain in $\mathbb{R}^n$ with smooth boundary $ S$. The Green's function $G$ for $\Omega$ exists and for each $x \in \Omega , G(x, \cdot) \in C^\infty(\overline{\Omega} \backslash \{x\})$.
Then he claims
$\forall x,y \in \Omega : G(x,y) = G(y,x)$.
He provides the following formal argument:
\begin{align*}
G(x,y) – G(y,x) &= \int_{\Omega} G(x,z) \delta(y – z) – G(y,z) \delta(x – z) \mathrm{\,d} z \\
&= \int_{S} G(x,z) \partial_{\nu _z} G(y,z) – G(y,z) \partial_{\nu _z} G(x,z) \mathrm{\,d} \sigma(z) = 0
\end{align*}
as $ \forall x \in \Omega : \Delta G(x,\cdot) = \delta (x – \cdot)$ and $ G(x, S) = 0$. A strategy for formalising he suggest is that we can excise small balls of $x,z$ from $\Omega$ and then let their radii shrink to zero.
Approach
As $ G(x , \cdot)\in C^\infty(\overline{\Omega} \backslash \{x\})$ so $\forall x,y \in \Omega : $
\begin{align*}
G(x, y) – G(y,x) &= G(x, \cdot) \star \delta (y) – G(y, \cdot) \star \delta (x) \\&= \int_{\Omega} G(x,z) \delta(y – z) -G(y,z) \delta(x – z) \mathrm{\,d} z \\
&= \int_{\Omega} G(x,z) \Delta G(y,z) – G(y,z) \Delta G(x,z) \\
&= \lim_{\varepsilon \to 0} \int_{\Omega \backslash B_\varepsilon(x,z)} G(x,z) \Delta G(y,z) – G(y,z) \Delta G (x,z) \mathrm{\,d}z \\
&= \int_{\partial \Omega} – \lim_{\varepsilon \to 0} \int_{\partial B_\varepsilon(x) + \partial B_\varepsilon(z)} G(x,z) \partial_{\nu _z} G(y,z) – G(y,z) \partial_{\nu _z} G(x,z) \mathrm{\,d} \sigma(z) \\
&= 0 – \lim_{\varepsilon \to 0} \int_{\partial B_\varepsilon(x) + \partial B_\varepsilon(y)} G(x,z) \partial_{\nu _z} G(y,z) – G(y,z) \partial_{\nu _z} G(x,z) \mathrm{\,d} \sigma(z) \\
&= – \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \int_{ \partial B_\varepsilon(x)} (x – z) \cdot (G(x,z) \nabla_z G(y,z) – G(y,z) \nabla _z G(x,z)) \mathrm{\,d} \sigma(z) \\
&\phantom{sds} – \lim_{\varepsilon \to 0}\frac{1}{\varepsilon}\int_{ \partial B_\varepsilon(y)} (y – z) \cdot (G(x,z) \nabla_z G(y,z) – G(y,z) \nabla _z G(x,z)) \mathrm{\,d} \sigma(z)
\end{align*}
where we have used that $ G(x,\cdot) , G(y,\cdot) \in C^\infty(\overline{\Omega} \backslash \{x,y\})$ so Green's identities are applicable, and $ G(x,\partial \Omega) = G(y, \partial \Omega) = 0$.
Can someone tell me why would the end limit be $0$? The last expression looks like Lebesgue Differentiation, but neither the dimensions are right nor is the integrand well-behaved (could be singular at $\{x,y\}$).
Best Answer
To do this more rigorously, you should avoid all mention of the $\delta$ function, and start with the excised domain $\Omega\backslash B_\epsilon(x, y)$. Here $\Delta G(x, z)=0$ and $\Delta G(y, z)=0$, so we have \begin{align} 0 &= -\lim_{\epsilon\to 0}\int_{\Omega\backslash B_\epsilon(x, y)} G(x, z)\Delta G(y, z) - G(y, z)\Delta G(x, z)\, dz\\ &= -\int_{\partial \Omega} 0 + \lim_{\epsilon\to 0}\int_{\partial B_\epsilon(x)\cup \partial B_\epsilon(y)} G(x, z)\partial_\nu G(y, z) - G(y, z)\partial_\nu G(x, z)\, d\sigma\\ &= \lim_{\epsilon\to 0}\int_{\partial B_\epsilon(y)} G(x, z)\partial_\nu N(z-y) \,d\sigma - \lim_{\epsilon\to 0}\int_{\partial B_\epsilon(x)} G(y, z)\partial_\nu N(z-x) \,d\sigma \\ &= G(x, y) - G(y, x). \end{align} Here we are using that $G(x, z)-N(x, z)$ is smooth (see the definition of Green's function), and $N(x, z)=N(x-z)=N(z-x)$ is the Newton potential. In particular, $$ \int_{\partial B_\epsilon(x)} \partial_\nu N(z-x)\,d\sigma = 1. $$ Since $x$ is the only singularity of $N(x, z)$, we also have $$ \lim_{\epsilon\to 0}\int_{\partial B_\epsilon(y)} \partial_\nu N(z-x)\,d\sigma = 0. $$ It is actually in this sense, we say that $\Delta N(x, z)=\delta_x$. As Folland says, see his proof of (2.8) the mean value theorem, for more details.