Prove an inequality in Poisson equation

Question

Let $u$ be a $C^2$ solution for $-\Delta u=f$ in a bounded set $\Omega$. Show that $$sup_{\bar \Omega} \vert u\vert\le sup_{\bar \Omega}\,f+sup_{\partial\Omega} \vert u\vert$$

If the maximum of $u$ is attained on the boundary $\partial\Omega$, then it's trivial. So I consider the case that the maximum of $u$ is attained in $\Omega$, denoting the maximum point by $x_0$.

What I know is that $-\Delta u(x_0)\le 0$ and so $f(x_0)\le 0$. And I don't know how to go on. I guess we need to use the comparison principle(or equvilantly, the maximum principle). But I don't know how to proceed.

Any help will be appreciated.

Answer 1

Here are two hints. First, I don't believe the estimate you're after is true as stated. You need a constant on the $f$ term on the right, i.e. your target estimate should be $$ \sup_{\bar{\Omega}} |u| \le C \sup_{\bar{\Omega}} |f| + \sup_{\partial \Omega} |u|. $$ This is needed because there are nontrivial solutions to $-\Delta u = \lambda u$ with $u=0$ on $\partial \Omega$ and $\lambda>0$ arbitrarily large (the eigenfunctions of the Laplacian). If we normalize so that $\sup_{\bar{\Omega}} |u| =1$, then it's clear that we need some constant on the $f$ term on the right in order to compensate.

The second hint is to indeed use the maximum principle, but not on $u$ directly. Define the function $$ v(x) = u(x) + \alpha |x|^2 $$ for some constant $\alpha >0$ and study $-\Delta v$.

Answer 2

The auxiliary function which @Glitch proposed should be modified as $v(x)=u(x)+\alpha \vert x\vert^2,$ i.e. we change the sign of the second term on the right hand side. We will see the reason in a moment.

Proof:

Let $M:=sup_{\bar\Omega}\vert f\vert \ge0$

Then $-\Delta u=f\le M$

Since $\Delta (\vert x\vert^2)=2n,$ we let $w(x):=\frac{M}{2n}\vert x\vert^2$ and note that $\Delta w(x)=M.$

Let $v(x):=u(x)+w(x)$

Then $-\Delta v(x)=-\Delta u\,-\Delta w\le M-M=0.$ And so we have the maximum principle. i.e. $$sup_{\bar\Omega}v\le sup_{\partial\Omega}v\,.$$ (Remark: If we don't change the sign of the second term in the auxiliary function $v$, then what we get will be the minimum principle, which can not be used to deduce the following step.)

Now, $sup_{\bar\Omega}u\le sup_{\bar\Omega}v\le sup_{\partial\Omega}v \le sup_{\partial\Omega}u \,+\, sup_{\partial\Omega}w=sup_{\partial\Omega}u\,+\frac{M}{2n}D=sup_{\partial\Omega}u\, + \frac{D}{2n}sup_{\bar\Omega}\vert f\vert,$ where $D:=sup_{\partial\Omega}\vert x\vert^2.$

Q.E.D