Properties of Green function

As we all know, Green’s function G=\Gamma+h, where \Gamma is the fundamental solution of Laplacian operator

\displaystyle \Gamma(x,y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}, n>2\\\frac{1}{2\pi}\log|x-y|, n=2.\end{cases}

and h(x,y) is solved for each x\in \Omega by

\displaystyle \begin{cases}\Delta h(x,\cdot)=0\text{ in }\Omega\\h(x,\cdot)=-\Gamma(x,\cdot) \text{ on }\partial \Omega \end{cases}

Then for any u\in C^1(\overline{\Omega})\cap C^2(\Omega), G satisfies

\displaystyle u(x)=\int_{\partial \Omega}u\frac{\partial G}{\partial \nu}ds+\int_{\partial G}G(x,y)\Delta u(y)dy

\mathbf{Problem:} \Omega\subset \mathbb{R}^n is bounded. G(x,y) is the first kind of Green function of \Omega.

  1. G(x,y)<0 for all x,y\in \Omega, x\neq y.
  2. Let f be a bounded and locally integrable function in \Omega. Then

\displaystyle \int_\Omega G(x,y)f(y)dy\to 0 as x\to \partial \Omega.

\mathbf{Lemma:} Let u be a non-negative harmonic function in \Omega, y\in \Omega such that B_{4R}(y)\subset \Omega, then

\displaystyle \sup\limits_{B_R(y)}u\leq 3^n\inf\limits_{B_R(y)}u

We omit the proof of this lemma. The importance of this lemma is sup is bounded by inf with a constant independent of radius.

\mathbf{Proof:} Fix x\in \Omega. Denote u_x(\cdot)=-G(x,\cdot). Recall that u_x(\cdot) is a harmonic function in \Omega\backslash{x},  and u_x(y)\to \infty as y\to x, there exists  B_r(x)\subset \Omega with 0<r\leq d(x)=dist(x,\partial \Omega) such that \exists \,y_0\in\partial B_r(x) and u_x(y_0)=\epsilon.

we can cover \partial B_r(x) by finitely many balls centered at \partial B_r(x) with radius \frac{r}{8}. By using the lemma iteratively, we know that u_x(y)\leq M(n)\epsilon on the \partial B_r(x).

Since u_x is a harmonic function in \Omega\backslash B_r(x) and u_x(y)=0 on \partial \Omega, by the maximum principle

0<u_x\leq M\epsilon in \Omega\backslash B_r(x).

This implies \displaystyle \int_\Omega -G(x,y)dy=\int_{\Omega\backslash B_r(x)}u_xdy+\int_{B_r(x)}-G(x,y)dy\leq M\epsilon |\Omega|+\int_{B_r(x)}-G(x,y)dy

When x\to \partial \Omega , we must have r\to 0, then \displaystyle \int_{B_r(x)}G(x,y)dy\to 0, because G(x,y)=O(|x-y|^{2-n}).

So \displaystyle 0\leq \overline{\lim\limits_{x\to \partial \Omega}}\int_\Omega -G(x,y)dy\leq M\epsilon |\Omega|. Let \epsilon \to 0. We have

\displaystyle \int_\Omega G(x,y)dy\to 0 as x\to \partial \Omega.

This implies when x\to \partial \Omega.

\displaystyle \left|\int_\Omega G(x,y)f(y)dy\right|\leq ||f||_\infty \int_\Omega |G(x,y)|dy=-||f||_\infty \int_\Omega G(x,y)dy\to 0

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

\mathbf{Remark:} GT’s chapter 2.3.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: