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??


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??

  • $\begingroup$ If using Poisson kernel, it is trivial. Max principle is hard to apply if $f$ is negative here. $\endgroup$ – Yimin May 20 '15 at 20:28
  • $\begingroup$ The exercise requires to use the max principle. How could we apply this?? @Yimin $\endgroup$ – 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, $$ |u|\le\frac18\max_{S}|f|,x\in S. $$

  • $\begingroup$ this question: math.stackexchange.com/questions/2094252/… is similar. Would you mind giving me a hand with it? $\endgroup$ – ALannister Jan 12 '17 at 6:48
  • $\begingroup$ @JessyunBourne, I will look at it when I have time. $\endgroup$ – xpaul Jan 12 '17 at 14:26
  • $\begingroup$ thanks man I appreciate it. $\endgroup$ – ALannister Jan 12 '17 at 14:43

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