Maximum principle-estimation

Let $S=\{x \in \mathbb{R}^2 \mid |x| <1\}$. Using the maximum principle I have to show that the solution of the problem $$-\Delta u(x)=f(x), x \in S \\ u(x)=0, x \in \partial{S}$$ satisfies the estimation $$|u(x)| \leq \frac{1}{4}\max_{x \in \overline{S}} |f(x)|, x \in S$$

To use the maximum principle shouldn't it stand that $$\Delta u \geq 0$$ ??

Do we have to take cases for $f$, if it is positive or negative??

EDIT:

Is it as followed??

$$\max_{\overline{S}} u=\max_{\partial{S}}u =0$$

How can we use $f$ ?? How do we get an expression with $f$ at the inequality??

• If using Poisson kernel, it is trivial. Max principle is hard to apply if $f$ is negative here. – Yimin May 20 '15 at 20:28
• The exercise requires to use the max principle. How could we apply this?? @Yimin – Mary Star May 20 '15 at 20:31

Define $$w(x)=\frac{1}{2}(x_1-x_1^2) \max_{S}f^+, x=(x_1,x_2).$$ Here $s^+=s$ if $s\ge0$ and $s^+=0$ if $s<0$. Then, in $S$ $$-\Delta (w-u)=\max_{s}f^++\Delta u=\max_{s}f^+-f\ge0,$$ and on $\partial S$, $w-u\ge0$. By the Maximum Principle, $w-u\ge 0$ in $S$ or $$u\le w=\frac12(x_1-x_1^2) \max_{S}f^+.$$ Since $x_1\in(-1,1)$, we have $x_1(1-x_1)\le\frac14$. So $$u\le\frac18\max_Sf^+, x\in S.$$ Similarly we have $$u\ge-\frac18\max_{S}f^-,x\in S.$$ Hence, $$|u|\le\frac18\max_{S}|f|,x\in S.$$