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Consider the following boundary value problem where $U=\{x \in \mathbb{R}^3 \mid |x|<1\}$ and $g$ is some nice bounded function, $$\Delta u = g ~~~ \text{on}~U\\ u=0 ~~~\text{on} ~\partial U.$$ Assume that $x_0 \in U$ with $|x_0|=r$ for some $0<r<1$ such that $x_0$ lies (very) close to the boundary of $U$. Is it possible to find an upper bound for $|u(x_0)|$?

Any idea is much appreciated!

mathematikos
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  • This is neither a boundary value problem, nor a Poisson equation. You're probably missing the Laplace operator somewhere? – joriki Dec 27 '19 at 00:09
  • @joriki True! Thanks for correcting. Now it is both ;) – mathematikos Dec 27 '19 at 00:11
  • @mathematikos why is there a $\frac{1}{2}$? you can just multiply both sides of the first centered equation by $2$ to rid of it (by just changing $g$ to $2g$). And change $-g$ to $g$ also. Please explain – mathworker21 Jan 04 '20 at 03:17
  • @mathworker21 Well, this is just to be consistent with the rest of the work I am doing but of course the PDE can be formulated in a "nicer" way as you explained. I will change it. – mathematikos Jan 04 '20 at 10:03
  • so you're asking for an analogue of Harnack's inequality? – mathworker21 Jan 05 '20 at 23:20
  • @mathworker21 Yeah, putting it in a nutshell, that's what I'm looking for :D – mathematikos Jan 06 '20 at 00:18
  • @mathematikos what goes wrong if you alter the proof of Harnack? – mathworker21 Jan 06 '20 at 00:34
  • @mathworker21 I suppose that the proof is very strongly based on the property that u is harmonic and many of the tools for harmonic functions won't be available. However, I have neither tried it this way, nor have I studied Harnack's proof in depth to be honest. So give me some hours to try it and I will come back if necessary. – mathematikos Jan 06 '20 at 00:44
  • What prevents you from applying the Poisson formula? It will give you the exact value $u(x_0)$. – timur Jan 09 '20 at 03:31
  • @timur What prevents me, is the fact that I end up with an integral which I neither can evaluate nor did I succeed in finding an upper bound (i.e. a bound which depends on $r$). Maybe it is possible, this would be fine, too. But I couldn't do it. That's why I thought that maybe there is another way. – mathematikos Jan 09 '20 at 10:47
  • Would a bound like $u\leq\frac{r^2-1}{6}\inf_{U}g$ be ok? – timur Jan 09 '20 at 13:36
  • @timur Yes! How does one get this? – mathematikos Jan 09 '20 at 16:47

2 Answers2

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Let $w(x)=a(|x|^2-1)$, where $a$ is a constant whose value will be adjusted shortly. By construction, we have $$ w|_{\partial U} = 0. $$ On the other hand, we compute $$ \Delta w = 6a, $$ and so by choosing $$ a=\frac16\inf_Ug, $$ we ensure $$ \Delta u\geq\Delta w\qquad\textrm{in}\,\, U. $$ Thus the comparison principle gives $$ u\leq w\qquad\textrm{in}\,\, U, $$ that is, $$ u(x)\leq\frac{|x|^2-1}6\inf_Ug. $$

timur
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Yes you can. In fact, there is a theorem :

Theorem (a priori bound) : Let $u\in C^2(\Omega)\cap C^0(\overline{\Omega})$ be a solution of the Dirichlet problem $Lu=f$ in $\Omega$, $u=g$ on $\partial\Omega$, with $\Omega$ bounded, and $Lu=-\sum_{i,j=1}^na_{ij}(x)\partial_{x_i,x_j}u(x)+\sum_{i=1}^nb_i(x)\partial_{x_i}u(x)+c(x)u(x)$ is uniformly elliptic with $c\geq 0$. Then $$||u||_{C^0(\overline{\Omega})}\leq||g||_{C^0(\partial\Omega)}+C\sup_{x\in\Omega}\frac{|f(x)|}{\lambda(x)}$$ with $C=e^{2(B+1)}-1$, and $B:=\sup_\Omega \frac{|b(x)|}{\lambda(x)}$.

Where uniform ellipticity of the operator is defined as :

Definition : $L$ is said to be elliptic in $\Omega$ if there exists positive functions $\lambda,\Lambda :\Omega\rightarrow \mathbb{R}_+$ such that $$0<\lambda(x)|\xi|^2\leq \xi^TA(x)\xi=\sum_{i,j}a_{i,j}(x)\xi_i\xi_j\leq\Lambda(x)|\xi^2|,$$ for all $x\in\Omega$, $\xi=(\xi_1,\cdots,\xi_n)\in\mathbb{R}^n$. $L$ is uniformly elliptic if there exists $\lambda_0,M\in\mathbb{R}$ such that $$\lambda(x)\geq \lambda_0 >0\text{ and }\frac{\Lambda(x)}{\lambda(x)}\leq M\leq \infty,$$ for all $x\in\Omega$.

So In this case, (by the other theorem we know the solution exists) $g=0$, $\Omega=S^3$, and $L=\Delta$, which is uniformly elliptic with $\lambda(x)=1$. Therefore, we have a priori bound : $$||u||_{C^0(\overline{\Omega})}\leq (e^2-1)||g||_{C^0(\Omega)}.$$ But this bound covers all $x\in\Omega$, so almost surely not a optimal bound for $x_0$. Similar to @timur did, you can have a super solution that is explicit, and try to get a better bound.

Cat
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  • Thanks a lot! Can you give a reference for the mentioned theorem? – mathematikos Jan 19 '20 at 19:41
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    @mathematikos David Gilbarg, Niel S. Trudinger, Elliptic Partial Differential Equations of Second Order, Springer-Verlag, 2nd edition, 2001., see Theorem 17.9. – Cat Jan 20 '20 at 17:18