Symmetry of Green's Functions

Question

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 $\{x,y\}$).

Answer 1

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.

Answer 2

With discussion with @Three aggies, here is the complete solution:

Fix $x \neq y \in \Omega$ and let $ G^x := G(x, \cdot)$. Since $ \Delta G^x = \delta _x$ so $G^x$ is harmonic away from $x.$ When $ n > 2 $, we have $ \partial_{\nu _z} G^x$ on $\partial B_\varepsilon(x)$ is given by $$ \nu _x (z) \cdot \nabla _z G^x = \varepsilon ^{-1} (z - x) \cdot \nabla_z \frac{ \lvert z -x \rvert^{2 - n}}{\omega _n (2 - n)} = (\varepsilon \omega _n ) ^{-1} \lvert z - x \rvert^2 \lvert z - x \rvert^{ -n} = \varepsilon ^{1 - n} \omega _n ^{-1} = \lvert \partial B_\varepsilon \rvert^{-1} $$ Moreover, for $ \partial_{\nu _z}G^y$ on $\partial B_\varepsilon(x)$, we have $$ \lvert \nu _x(z) \cdot \nabla_z G^y \rvert = (\varepsilon \omega _n)^{-1} \lvert y - z \rvert^{-n} \lvert (z -y) \cdot ( z- x) \rvert \leq (\varepsilon \omega _n)^{-1} \lvert y - z \rvert^{1 -n} \lvert z- x \rvert. $$ As such, via Green's identities applicable due to smoothness of $G^x$ (respectively $G^y$), away from $x$ and $ G^x(\partial \Omega) = 0$, \begin{align*} 0 &= G^x(y)\mathbf{1}_{y \not \in B_\varepsilon(x,y)} - G^y(x) \mathbf{1}_{x \not \in B_\varepsilon(x,y)} = \int_{ \Omega \backslash B_\varepsilon(x,y)} G^x \Delta G^y - G^y \Delta G^x \\ &= \int_{ \partial \Omega} - \int_{\partial B_\varepsilon(x) \cup \partial B_\varepsilon(y)} G^x \partial_z G^y - G^y \partial _z G^x \\ &= - \int_{ \partial B_\varepsilon(x)} G^x \partial_{\nu _z} G^y - \lvert \partial B_\varepsilon \rvert^{-1} G^y - \int_{ \partial B_\varepsilon(y)} \lvert \partial B_\varepsilon \rvert^{-1} G^x - G^y \partial _{\nu _z} G^x \\ &= - \int_{ \partial B_\varepsilon(x)} G^x \partial_{\nu _z} G^y + \frac{1}{ \lvert \partial B_\varepsilon \rvert} \int_{ \partial B_\varepsilon(x)} G^y + \int_{ \partial B_\varepsilon(y)} G^y \partial _{\nu _z} G^x - \frac{1}{ \lvert \partial B_\varepsilon \rvert} \int_{ \partial B_\varepsilon(y)} G^x. \end{align*} Finally, $$ \left\lvert \int_{ \partial B_\varepsilon(x)}G^x \partial_{\nu _z} G^y \right\rvert \leq C \varepsilon ^{-1} \int_{ \partial B_\varepsilon(x)} \lvert x -z \rvert^{2 - n} \lvert y - z \rvert^{1 - n} \lvert z - x \rvert \leq C \varepsilon ^{-1} \int_{ \partial B_\varepsilon(x)} \varepsilon ^{3 - n} \lvert y - \frac{x}{2} \rvert^{1 - n} \leq C \varepsilon ^{2 -n + n-1} = C \varepsilon, $$
and $ \frac{1}{ \lvert B_\varepsilon \rvert} \int_{ \partial B_\varepsilon(x)} G^y = G^y(x) $ via mean-value formula as $G^y $ is harmonic away from $x$, gives us $$ 0 = - C \varepsilon + G^y(x) + C' \varepsilon - G^x(y). $$ Letting $\varepsilon \to 0$ gives our result.