Bilinear Harmonic Function

u(x) is harmonic in \Omega\subset\mathbb{R}^n, and suppose B_c(x_0)\subset\subset\Omega. If a\le b\le c, ac=b^2, then

\displaystyle \int_{|\omega|=1}u(x_0+a\omega)u(x_0+c\omega)\mathrm{d}\omega=\int_{|\omega|=1}u^2(x_0+b\omega)\mathrm{d}\omega

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I like this theorem, it is beautiful  and concise. The proof was kind of trivial if apply Poisson Kernel bravely. Let’s make it more general.

Set S=\{|\omega|=1\}, thus define \xi(\rho_1,\rho_2)=\int_{S}\phi_1(\rho_1 x)\phi_2(\rho_2 x), if we are given that \alpha_1\alpha_2=\beta_1\beta_2, then we are going to prove:

\xi(\alpha_1,\alpha_2)=\xi(\beta_1,\beta_2)

consider unit vector a,b, by Poisson Integral,

\phi(\alpha a)=\int_{S_b}K(\alpha a,\beta b)\phi(\beta b),

where

\displaystyle K(\alpha a,\beta b)=\frac{\beta^{n-2}(\beta^2-\alpha^2)}{C_n(\alpha^2+\beta^2-2\alpha\beta\cos(a,b))^{n/2}},

since \alpha_1\alpha_2=\beta_1\beta_2, thus we assume \kappa=\frac{\alpha_1}{\beta_2}=\frac{\beta_1}{\alpha_2}, we know that K is homogeneous, hence

K(\alpha_1 a,\beta_1 b)=K(\kappa\beta_2 a,\kappa\alpha_2 b)=K(\beta_2 a,\alpha_2 b).

\xi(\alpha_1,\alpha_2)=\int_{S_a}\int_{S_b}K(\alpha_1 a,\beta_1 b)\phi_1(\beta_1 b)\phi_2(\alpha_2 a)

which is

\xi(\alpha_1,\alpha_2)=\int_{S_b}\int_{S_a}K(\beta_2 a,\alpha_2 b)\phi_2(\alpha_2 a)\phi_1(\beta_1 b)

The inner integral is \phi_2(\beta_2 a).

Done.

\Box

About YiMin

This is just a nerd PhD student of Math@UT Austin.

One thought on “Bilinear Harmonic Function

  1. YiMin says:

    It seems that this conclusion cannot be generalized. because the Poisson Kernel only involves two surfaces, not N surfaces.

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