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.

  • $\begingroup$ maybe do something like $\int u \Delta \phi = \int \Delta u \cdot \phi + \int_{\partial \Omega} u = \int f\cdot \phi+\int_{\partial \Omega} u$ for $\phi \in C^\infty(\Omega)$. idk $\endgroup$ – mathworker21 Oct 11 '18 at 8:47
  • $\begingroup$ It turns out to have nothing to do with the Green's identity, but there are still a lot of thanks to you. :) $\endgroup$ – Sam Wong Oct 12 '18 at 3:56

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 $-\Delta v$.

  • $\begingroup$ Thanks Glitch. But for your first hint, I think the problem is that $\lambda$ can be arbitrarily small, not due to that it can be arbitrarily large, right? $\endgroup$ – Sam Wong Oct 11 '18 at 21:24
  • $\begingroup$ @SamWong In fact the eigenvalues form a non-decreasing sequence $0 < \lambda_1 \le \lambda_2 \le \cdots$, so the only value we really need to worry about is $\lambda_1$. $\endgroup$ – Glitch Oct 12 '18 at 3:29
  • $\begingroup$ Yeah, what we need to do is to find a eigenvalue-problem which has a very small eigenvalue $\lambda_1$. I have successfully figured out a counter example which is in $\mathbb R$. And also, I think the auxiliary function should be $v=u+\alpha \vert x\vert^2$, as I argued in my post. However, still a lot of thanks to you Glitch. $\endgroup$ – Sam Wong Oct 12 '18 at 3:39
  • $\begingroup$ Oops, you're right about my sign error. However, you said that your domain is bounded. It is a theorem in this case that the smallest eigenvalue is positive. Of course, when you switch to unbounded sets like $\mathbb{R}$ then things are going to change, but even the maximum / comparison principles fall apart there. $\endgroup$ – Glitch Oct 12 '18 at 4:15
  • $\begingroup$ Emmm, I think we still have the maximum principle even when we move to unbounded sets, since what the proof of the maximum principle relies are merely the connectivity and openness of the underlying set. And so we will also have the comparison principle, which is a corollary of the maximum principle. Both principles are referred to the case of the Laplace equation. $\endgroup$ – Sam Wong Oct 12 '18 at 4:33

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.


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 $D:=sup_{\partial\Omega}\vert x\vert^2.$



Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.