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Question:

Consider the following two special Poisson equations: $$\Delta u=-2 \quad \text{in} \quad L$$ $$u=0 \quad \text{on} \quad \partial L$$

and

$$\Delta u=-2 \quad \text{in} \quad W$$ $$u=0 \quad \text{on} \quad \partial W.$$

If $L\subseteq W$, can we get $||\nabla u_{L}(x)||\leqslant ||\nabla u_{W}(y)||$, where $x \in \partial L$, $y \in \partial W$ and $x,y $ are in the same direction?

I know almost nothing about partial differential equations, and I encountered this problem while working on other problems.

zeyu hao
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  • I believe that if this question is made a bit more precise, this is true and follows from the fact that a $u$ is convex iff its gradient is monotone, c.f. https://math.stackexchange.com/questions/1717542/a-function-is-convex-if-and-only-if-its-gradient-is-monotone#:~:text=Let%20a%20convex%20U%E2%8A%82,f%3AU%E2%86%92R%20differentiable. – whpowell96 Feb 07 '24 at 14:29
  • Related: https://math.stackexchange.com/questions/4855598/the-relationship-between-the-norm-of-the-gradient-solution-of-poissons-equation/4856418?noredirect=1#comment10351990_4856418 – JackT Feb 07 '24 at 21:40
  • I assume be $x,y$ having the same direction you mean $x/\vert x\vert =y /\vert y\vert$. This assumption is actually redundant since two points are always co-linear and your problem is invariant under translations, so you can always translate the origin onto this line – JackT Feb 07 '24 at 21:45

1 Answers1

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No.

Given $\Omega \subset \mathbb R^n$ open, bounded with $C^1$ boundary, define $\psi_\Omega :\overline \Omega\to\mathbb R$ to be the unique solution in $C^2(\Omega)\cap C^0(\overline \Omega)$ satisfying $-\Delta \psi_\Omega = 1 $ in $\Omega$ and $\psi_\Omega=0$ on $\partial \Omega$.

Now, consider $L=B_{1/2}$ (the ball of radius $1/2$ centred at $0$) and $W= B_1 \cup B_R((2+R)e_1)$ where $e_1=(1,0,\dots,0)\in \mathbb R^n$ and $R>0$. Then, since $B_1$ and $B_R((2+R)e_1)$ are disjoint, \begin{align*}\psi_W (x) &= \begin{cases} \psi_{B_1}(x), &\text{in } B_1 \\ \psi_{B_R((2+R)e_1)}(x), &\text{in } B_R((2+R)e_1) \end{cases} \\ &=\frac1 {2n}\begin{cases} 1-\vert x \vert^2, &\text{in } B_1 \\ R^2-\vert x - (2+R)e_1\vert^2, &\text{in } B_R((2+R)e_1) . \end{cases} \end{align*} Then, for $x\in \overline{B_R((2+R)e_1)} $, $$\nabla \psi_W(x) =-\frac1n (x-(2+R)e_1) ,$$ so $$\vert \nabla \psi_W((2+2R)e_1) \vert^2 = \frac {R^2} {n^2}. $$ Similarly, we have that $$\vert \nabla \psi_L (1/2 e_1) \vert^2 =\frac 1 {4n^2}.$$ Thus, $L \subset W$, $1/2 e_1$ and $2(1+R)e_1$ point in the same direction, but for $0<R<1/2$, we have that $$\vert \nabla \psi_L (1/2 e_1) \vert^2 =\frac 1 {4n^2} > \frac{R^2}{n^2} = \vert \nabla \psi_W((2+2R)e_1) \vert^2. $$

You might complain that this isn’t really a satisfactory counter-example because $W$ isn’t connected. However, from the previous example you should be able to construct a counter-example with $W$ connected. Indeed, let $\varepsilon>0$ and $W_\varepsilon$ be the union of $B_1$ and $B_R((2+R)e_1)$ (as above) with a ‘neck’ connecting the right hand side of$B_1$ to the left hand side of $B_R((2+R)e_1)$ in a smooth way and such that the neck is contained in a cylinder with cross-sectional radius $\varepsilon$. Then consider $\psi_{W_\varepsilon}$. Now, as $\varepsilon \to 0^+$, $\psi_{W_\varepsilon} \to \psi_W$ in $C^1$ provided that you are away from the neck i.e. outside some small ball that contains the join of the neck and the ball. (If you have to contrust this counter-example rigorously then you have to prove my previous statement, but there exists literature somewhere on this kind of thing). But now, by continuity, for $R$ as above and $\varepsilon$ sufficiently small, $$\vert \nabla \psi_L (1/2 e_1) \vert^2 > \vert \nabla \psi_W((2+2R)e_1) \vert^2+\varepsilon \geq \vert \nabla \psi_{W_\varepsilon}((2+2R)e_1)\vert^2. $$

JackT
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