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<c, 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

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