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<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