## Tag Archives: green formula

### Cut-off function and interior holder estimate

Suppose we know the following theorem

$\bf{Thm 1:}$ Let $v\in C^2_0(\mathbb{R}^n)$, $g\in C^\alpha_0(\mathbb{R}^n)$ satisfy the Possion equation $\Delta v=g$ in $\mathbb{R}^n$. Then $v\in C^{2,\alpha}_0(\mathbb{R}^n)$ and if $\text{supp }g\subset B=B_{R}(x_0)$, we have

1. $\displaystyle |D^2v|'_{0,\alpha;B}\leq C|g|'_{0,\alpha;B}$, $C=C(n,\alpha)$
2. $|v|'_{1;B}\leq CR^2|g|_{0;B}$, $C=C(n)$

Use the above theorem and cut-off function

$\eta(|x|)\in C^2(\mathbb{R}^n)$,

$\eta(|x|)=1$ when $\displaystyle |x|<\frac{3}{2}$ and $\eta(|x|)=0$ when $|x|>2$

to prove the interior holder estimate of possion equation

$\bf{Thm 2:}$ Let $\Omega$ be a domain in $\mathbb{R}^n$, $\Delta u=f$ in $\Omega$. If $u\in C^2(\Omega)$ and $f\in C^\alpha(\Omega)$, then $u\in C^{2,\alpha}(\Omega)$ and for any two concentric balls $B_{R}\subset B_{2R}(x_0)\subset \Omega$, we have

$\displaystyle |u|'_{2,\alpha;B_{R}}\leq C(|u|_{0;B_{2R}}+R^2|f|'_{0,\alpha;B_{2R}})\quad (1)$

$\bf{Proof:}$ WLOG assume $x_0=0$, let $v(x)=u(x)\eta(|x|/R)$, $\eta$ is the cut off fuction.

then $\displaystyle v_i(x)=u_i(x)\eta\left(\frac{|x|}{R}\right)+u(x)\eta'\left(\frac{|x|}{R}\right)\frac{x_i}{|x|}\frac 1R$.

$\displaystyle v_{ii}(x)=u_{ii}(x)\eta\left(\frac{|x|}{R}\right)+2u_i(x)\eta'\left(\frac{|x|}{R}\right)\frac{x_i}{|x|}\frac 1R+u(x)\eta''\left(\frac{|x|}{R}\right)\frac{x^2_i}{|x|^2}\frac {1}{R^2}+u(x)\eta'\left(\frac{|x|}{R}\right)\frac 1R\frac{|x|^2-x^2_i}{|x|^3}$

$\displaystyle \Delta v(x)=\Delta u\cdot\eta\left(\frac{|x|}{R}\right)+2\nabla u\cdot\nabla |x|\eta'\left(\frac{|x|}{R}\right)\frac 1R+u(x)\eta''\left(\frac{|x|}{R}\right)\frac{1}{R^2}+u(x)\eta'\left(\frac{|x|}{R}\right)\frac{1}{R}\frac{n-1}{|x|}:=g$

Since $v$ has compact support, we have representation $\displaystyle v=\int_{B_{2R}}\Gamma(x-y)g(y)dy$

Let $\displaystyle w_1=\int_{B_{2R}}\Gamma(x-y)\Delta u\cdot\eta\left(\frac{|y|}{R}\right)dy$

$\displaystyle w_2=\int_{B_{2R}}\Gamma(x-y)\nabla u\cdot\nabla |y|\eta'\left(\frac{|y|}{R}\right)\frac 1Rdy$

$\displaystyle w_3=\int_{B_{2R}}\Gamma(x-y)u(y)\left[\eta''\left(\frac{|y|}{R}\right)\frac{1}{R^2}+\eta'\left(\frac{|y|}{R}\right)\frac{1}{R}\frac{n-1}{|y|}\right]dy$

Then $v=w_1+2w_2+w_3$ from the representation. We will bound them respectively.

For $w_1$, note that $\displaystyle |w_1|_{0;B_{R}}\leq |u|_{0;B_{2R}}$, and theorem 4.5 implies $\displaystyle R|Dw_1|_{0;B_{R}}+R^2|D^2w_1|_{0,\alpha;B_{R}}\leq CR^2|f\eta|'_{0,\alpha;B_{2R}}\leq CR^2|f|'_{0,\alpha;B_{2R}}$

So $w_1$ satisfies (1)

For $w_3$, since $\eta'(x)=0$ when $\displaystyle |x|<\frac 32$, we get that in fact

$\displaystyle w_3=\int_{B_{2R}\backslash B_{\frac 32 }R}\Gamma(x-y)u(y)\left[\eta''\left(\frac{|y|}{R}\right)\frac{1}{R^2}+\eta'\left(\frac{|y|}{R}\right)\frac{1}{R}\frac{n-1}{|y|}\right]dy$

Since we have $\displaystyle |x-y|\geq \frac 12 R$ when $x\in B_R$ and $\displaystyle y\in B_{2R}\backslash B_{\frac 32 }R$, this means

$\displaystyle |w_3|\leq C\frac{|u|_{0;B_{2R}}}{R^2}\int_{B_{2R}\backslash B_{\frac 32 }R}\Gamma(x-y)dy\leq C|u|_{0;B_{2R}}$

$\displaystyle |Dw_3|'_{1,\alpha;B_{2R}}\leq C\frac{|u|_{0;B_{2R}}}{R^2}\int_{B_{2R}\backslash B_{\frac 32 }R}|D\Gamma(x-y)|'_{1,\alpha;B_{2R}}dy\leq C|u|_{0;B_{2R}}$

So $w_3$ satisfies (1)

For $w_2$, by the same reason,

$\displaystyle w_2=\int_{B_{2R}\backslash B_{\frac 32 R}}\Gamma(x-y)\nabla u\cdot\nabla |y|\eta'\left(\frac{|y|}{R}\right)\frac 1Rdy$

Apply the Green identity on the domin $B_{2R}\backslash B_{\frac 32 R}$

$\displaystyle w_2+\int_{B_{2R}\backslash B_{\frac 32 R}}\nabla \Gamma(x-y)\cdot \nabla |y|u(y)\eta'\left(\frac{|y|}{R}\right)\frac 1Rdy+\int_{B_{2R}\backslash B_{\frac 32 R}}\Gamma(x-y)u(y)\eta''\left(\frac{|y|}{R}\right)\frac {1}{R^2}dy$

$\displaystyle +\int_{B_{2R}\backslash B_{\frac 32 R}}\Gamma(x-y)u(y)\eta'\left(\frac{|y|}{R}\right)\frac{n-1}{|y|}\frac 1Rdy=\left(\int_{\partial B_{2R}}-\int_{\partial B_{\frac 32R}}\right)\Gamma(x-y)u(y)\eta'\left(\frac{|y|}{R}\right)\frac 1Rds_y=0$

So we can estimate $w_2$ as $w_3$, which means $w_2$ also satisfies (1).

$\bf{Remark:}$ Gilbarg Trudinger’s book. chapter 4, exercise 4.3.

### 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