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I have a question to the following problem.

Let $f\in C(K(0,1))\subset \mathbb{R}^2.$ Prove that $\sup_{K(0,1)}|u|\leq \frac{1}{4}\sup_{K(0,1)}|f|,$ if $u(x,y)$ is a solution to the problem:

$\Delta u=f, \ \mbox{on} \ K(0,1),$ $u(x,y)=0 \ \mbox{the boundary of} \ K(0,1).$

Let $G(x,y,x',y')$ be the appropriate Green function for $K(0,1).$

I tried to solve the problem, but came to a problem. My solution goes like this: $|u(x,y)|=|\int_{K(0,1)}G(x,y,x',y')f(x',y')~dx'~dy'|\leq \sup_{K(0,1)}|f|\int_{K(0,1)}|G(x,y,x',y')|~dx'~dy'.$

But now I have a problem. How to get rid of the absolute value inside the integral. Because, if it were not there, then $\int_{K(0,1)}G(x,y,x',y')~dx'~dy'=\frac{x^2+y^2-1}{4},$ which is the unique solution to the problem: $\Delta u=1, \ \mbox{on} \ K(0,1),$ $u(x,y)=0 \text{ the boundary of } K(0,1).$ Then it is easy to establish, that $|\frac{x^2+y^2-1}{4}|\leq \frac{1}{4}$ on $K(0,1).$

Can someone please tell me how to write a fully correct solution to this problem.

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    +1 for good title. What do you mean by $K(0,1)$?2012-10-06

1 Answers 1

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It is enough to prove than $G$ is non-positive. This can be done using the maximum principle and the fact that $\lim_{(x,y)\to(x',y')}G(x,y,x',y')=-\infty,\quad (x',y')\in K.$

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    @user35058 it is true for Green functions of the Dirichlet problem for the second order elliptic equations. I don't know about all, some conditions perhaps are needed, like a possibility to apply the maximum principle.2012-07-04