Given that $u$ is harmonic. Prove $\Delta v \geq 0$ where $v = |\nabla u |^2$

Question

Let $\Omega \subset \Re^2$ be open, and let $B_r(x) \subset \Omega$ be any open disc in $\Omega$.

Assume that $u \in C^2(\Omega)$ is harmonic in $\Omega$

Let $v = |\nabla u |^2$. Prove $\Delta v \geq 0$ in $\Omega$

My Attempt:

So what I have tried is that

By integrating $v = |\nabla u |^2$ over $\Omega$ I get that

$$\int_\Omega v \, dx = \int_\Omega |\nabla u|^2 \, dx \geq \frac{1}{C}\int_\Omega |u|^2 \, dx$$

By using the Poincare Inequality.

So What I need to do is that I need to show $\int_\Omega |u|^2 \, dx$ is equal to $0$, but I don't really know how to do about this. Can I just say that this must be greater than or equal to zero?

Edit I misread the question so now I am going to attempt this again

$$\int_\Omega \Delta v \, dx = \int_\Omega div(\nabla v) \, dx = \int_{\delta\Omega} \nabla v \cdot n \, ds $$

Now what I am saying is that $\nabla v = \nabla(|\nabla u|^2) = 2\nabla u \Delta u $ but I don't think this is correct?

Assuming this is correct I get that

$$ = \int_{\delta\Omega} 2\nabla u \Delta u \, ds$$

But once again I am stuck here

Answer 1

Useful general formula: If $u,v\in C^2,$ then

$$\Delta (uv) = u\Delta v + v\Delta u + 2 \langle \nabla u, \nabla v\rangle.$$

Thus if $u$ is harmonic, then $\Delta (u^2) = 2 |\nabla u|^2.$ In your problem, apply this to each $(D_k u)^2,$ and recall that if $u$ is harmonic, then so is $D_ku.$

Answer 2

Hint: $$ v(x_1,x_2) = \left| \frac{\partial u}{\partial x_1} \right|^2 + \left| \frac{\partial u}{\partial x_2} \right|^2. $$ Now you have to compute $$ \Delta v = \frac{\partial^2 v}{\partial x_1^2} + \frac{\partial^2 v}{\partial x_2^2}. $$ You can proceed formally, although you should somehow justify the fact that $v$ is twice differentiable.