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.
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, $$\vert \nabla \psi_L (1/2 e_1) \vert^2 > \vert \nabla \psi_W((2+2R)e_1) \vert^2+\varepsilon \geq \vert \nabla \psi_{W_\varepsilon}((2+2R)e_1)\vert^2. $$