## A bilinear integral identity for harmonic functions

$\mathbf{Thm:}$ Suppose $\phi_1$ and $\phi_2$ are harmonic function in $\Omega_1$ and $\Omega_2$ respectively. For any $x_1\in\Omega_1,x_2\in\Omega_2$ and $0\leq a,b,c,d\leq \min\{dist(x_1,\Omega_1), dist(x_2,\Omega_2)\}$ with $ab=cd$, we have

$\displaystyle \int_{|w|=1}\phi_1(x_1+aw)\phi_2(x_2+bw)dw=\int_{|w|=1}\phi_{1}(x_1+cw)\phi_2(x_2+dw)dw$

$\mathbf{Proof:}$ We give two different proofs. The first is using possion integral formula

$\displaystyle u(x)=\int_{\partial B} \frac{R^2-|x|^2}{n\omega_nR|x-y|^n}\phi(y)ds_y=\int_{|y|=1} \frac{R^{n-2}(R^2-|x|^2)}{n\omega_n|x-y|^n}\phi(y)dy$

Here $\displaystyle R$ is the radius of $B$

WLOG, assume $x_1,x_2$ are the origin. And $a, $a=\lambda d$, $c=\lambda b$. Define

$\displaystyle L(a,b)=\int_{|w|=1}\psi_1(aw)\psi_2(bw)dw$

We have to prove $L(a,b)=L(c,d)$. Apply the possion integral formula

$\displaystyle \psi_1(aw)=\int_{|y|=1}P(aw,cy)\psi_1(y)dy$

where $\displaystyle P(aw,cy)=\frac{c^{n-2}(c^2-a^2)}{n\omega_n|cy-aw|^n}$. Noticing that

$\displaystyle P(aw,cy)=P(\lambda dw,\lambda b y)=P(dw,by)=P(dy,bw)$

So $\displaystyle L(a,b)=\int_{|w|=1}\psi_1(aw)\psi_2(bw)dw=\int_{|w|=1}\int_{|y|=1}P(aw,by)\psi_1(cy)dy\psi_2(bw)dw$

$\displaystyle =\int_{|y|=1}\int_{|w|=1}P(dy,bw)\psi_2(bw)dw\psi_1(cy)dy=\int_{|y|=1}\psi_2(dy)\psi_1(cy)dy=L(c,d)$

The second proof relies on Green formula. Let $\rho=\sqrt{ab}$, we will prove $L(\lambda \rho,\lambda^{-1} \rho)$ is independent of $\lambda$. Here we require $\lambda \rho,\lambda^{-1}\rho$ are still inside the domain of $\Omega_1$ and $\Omega_2$

$\displaystyle \frac{dL(\lambda \rho,\lambda^{-1} \rho)}{d\lambda}=\int_{|w|=1}\left[\frac{\partial\psi_1(\lambda\rho w)}{\partial\lambda}\psi_2(\lambda^{-1}\rho w)+\psi_1(\lambda\rho w)\frac{\partial\psi_2(\lambda^{-1}\rho w)}{\partial\lambda}\right]dw$

Since $\displaystyle \frac{\partial \psi_1(\lambda\rho w)}{\partial \lambda}=\frac{\rho}{\lambda}\frac{\partial \psi_1(\lambda\rho w)}{\partial\rho}$, $\displaystyle \frac{\partial \psi_2(\lambda^{-1}\rho w)}{\partial \lambda}=-\frac{\rho}{\lambda}\frac{\partial \psi_1 (\lambda^{-1}\rho w)}{\partial\rho}$, plug in this fact

$\displaystyle \frac{dL(\lambda \rho,\lambda^{-1} \rho)}{d\lambda}=\frac{\rho}{\lambda}\int_{|w|=1}\left[\frac{\partial\psi_1(\lambda\rho w)}{\partial\rho} \psi_2(\lambda^{-1}\rho w)-\psi_1(\lambda\rho w)\frac{\partial\psi_2(\lambda^{-1}\rho w)}{\partial\rho} \right]dw$

Define $u_1(r\theta)=\psi_1(\lambda r\theta)$,  $u_2(r\theta)=\psi_2(\lambda^{-1} r\theta)$ for any $0\leq r\leq \rho$, $|\theta|=1$, then $u_1,u_2$ are harmonic in the disc $B_\rho(0)$

$\displaystyle \frac{\partial u_1}{\partial \nu}=\frac{\partial u_1}{\partial r}=\frac{\partial \psi_1}{\partial \rho}$, $\displaystyle \frac{\partial u_2}{\partial \nu}=\frac{\partial u_2}{\partial r}=\frac{\partial \psi_2}{\partial \rho}$ on $\partial B_\rho(0)$.

$\displaystyle \frac{dL(\lambda \rho,\lambda^{-1} \rho)}{d\lambda}=\frac{\rho}{\lambda}\int_{|\theta|=1}\left[\frac{\partial u_1}{\partial\nu} u_2-u_1\frac{\partial u_2}{\partial\nu} \right]d\theta=\frac{1}{\lambda\rho^{n-2}}\int_{\partial B_\rho(0)}\left[\frac{\partial u_1}{\partial\nu} u_2-u_1\frac{\partial u_2}{\partial\nu} \right]ds$

By the Green second identity,  we know that the end of last equation is 0.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ See Gustin’s paper: A bilinear integral identity for harmonic functions. And Gilbarg Trudinger’s book, chapter 2,2.18