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 $||\nabla u_{L}||\leqslant ||\nabla u_{W}||$(where $u_{L}$ is the solution of the first equation, and $u_{W}$ is another)?
I know almost nothing about partial differential equations, and I encountered this problem while working on other problems.
This is an interesting question, but, unfortunately, it’s false. Let $T\subset \mathbb R^2$ be the equilateral triangle with vertices $(-2,0)$, $(1,\sqrt 3)$, and $(1,-\sqrt 3)$. It is known that $$ u_T(x,y)=-\frac 1 {12} (x-1)(x-\sqrt 3 y +2) (x+\sqrt 3 y +2) $$ satisfies $-\Delta u_T = 1$ in $T$ and $u_T=0$ on $\partial T$. Moreover, if $B_R:= \{ (x,y) \in \mathbb R^2 \text {s.t. } x^2+y^2 <R^2\}$ then $$ u_{B_R}(x,y) = \frac14 (R^2-x^2-y^2)$$ solves $-\Delta u_{B_R}=1$ in $B_R$ and $u_{B_R}=0$ on $\partial B_R$. Suppose that $R$ is chosen to sufficiently large so that $T \subset B_R$.
Now, by a direct computation, $$\vert \nabla u_T(x,y) \vert^2= \frac 1{ 16} (x^2+2x-y^2)^2+\frac14 y^2 (x-1)^2 $$ and $$ \vert \nabla u_{B_R}(x,y)\vert^2=\frac14 (x^2+y^2). $$ Then $\vert \nabla u_T(0,1) \vert^2= \frac 5{16}$ and $\vert \nabla u_{B_R}(0,1)\vert^2=\frac14 $, so $$\vert \nabla u_T(x,y) \vert \not \leq \vert \nabla u_{B_R}(x,y)\vert \qquad \text{in } T. $$