Prove that $u = |\nabla v|^2$ reaches its maximum in $\partial \Omega$, $v$ being a solution to the given Poisson pde

Question

Let $\Omega \subset \mathbb{R}^2$ and $v: \mathbb{R}^2 \to \mathbb{R}$ be a solution to:

$$\begin{cases} v_{xx}+v_{yy} = -2 &\text{in $\Omega$}\\ v = 0 &\text{in $\partial \Omega$} \end{cases}$$ And let $u = |\nabla v|^2$. I need to show that $u$ reaches its maximum in $\partial \Omega$. My first line of thought would be to show that $u$ is harmonic in $\Omega$, which by the maximum principle would show what I need. I proceeded as follows:

$u = v_x^2+v_y^2$, so $u_x = 2(v_xv_{xx}+v_yv_{yx})$, and $u_y = 2(v_xv_{xy}+v_yv_{yy})$.

So, $u_{xx} = 2(v_{xx}^2+v_xv_{xxx} + v_{yx}^2+v_yv_{yxx})$, and $u_{yy} = 2(v_{xy}^2+v_xv_{xyy}+v_{yy}^2+v_yv_{yyy})$.

Now, $u_{xx}+u_{yy} = 2(v_{xx}^2+2v_{yx}^2+v_{yy}^2+v_y\Delta v_y + v_x\Delta v_x)$.

Finally, $\Delta v_y = \Delta v_x = 0$ (This follows easily since $\Delta v$ is constant), we have $u_{xx} + u_{yy} = 2(v_{xx}^2 + 2v_{yx}^2 + v_{yy}^2)$, which i'd like to show to be $0$. I know that $v_{xx}^2 + 2v_{yx}^2 + v_{yy}^2 = (\Delta v)^2 - 2v_{xx}v_{yy} + 2v_{yx}^2 = 4 - 2v_{xx}v_{yy} + 2v_{yx}^2$.

So ideally, $-v_{xx}v_{yy} + v_{yx}^2 = 2$ and i'm done, but I dont see how this is true.

Answer 1

You've actually pretty much solved the problem. Your computation that $$ u_{xx}+u_{yy} = 2 (v_{xx}^2+v_{yy}^2+2v_{xy}^2) $$ is correct. Now, you say that you want to show that $v_{xx}^2+v_{yy}^2+2v_{xy}^2=0$, but this is not the way to go. Indeed, if this were the case, then this would imply that $v_{xx}=v_{yy}=v_{xy}=0$ which is only true if $v(x,y)=ax+by+c$ for some $a,b,c\in \mathbb R$. Instead, you should notice that $v_{xx}^2+v_{yy}^2+2v_{xy}^2\geqslant 0$ for any $v$. Hence, you have proven that $$ u_{xx}+u_{yy} \geqslant 0,$$ that is, $u$ is subharmonic. Then the result follows from the maximum principle for subharmonic functions.

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As an extra note: I mentioned in the comments about Bochner's formula. I agree that the wiki article makes it seems way more involved than is necessary for your problem, but I just wanted to mention that you have in fact, in your specific situation, proved Bochner's formula. Since we are in $\mathbb R^2$, Ricci curvature is zero (whatever this means), so Bochner's formula states that $$ \frac12 \Delta \vert \nabla v\vert^2 = \nabla \Delta v \cdot \nabla v + \vert D^2 v \vert^2 $$ where $\Delta v = v_{xx}+v_{yy}$ and $\vert D^2 v \vert^2 = v_{xx}^2+v_{xy}^2+v_{yx}^2+v_{yy}^2=v_{xx}^2+v_{yy}^2+2v_{xy}^2$. In your case, $\Delta v=-2$, so $\nabla \Delta v=0$ (precisely the observation you made in your computation), so $$\frac12 \Delta \vert \nabla v\vert^2 = \vert D^2 v \vert^2 $$ which is the identity that you obtained. :)