Problem involving divergence theorem and laplacian squared

Question

Let $B=\{x\in\mathbb{R}^m:|x|<1\}$ and $u\in C^3(\bar{B})$. Suppose that $u=0$ on the boundary $\partial B$ of $B$. Show that

$$ \int_B(|\Delta u|^2-\sum_{i,j=1}^m|D_{ij}u|^2)dx=(m-1)\int_{\partial B}|\nabla u\cdot n|^2 dS, $$ where $n$ is the unit outer normal vector to $\partial B$ (i.e. $n(x)=x$) and $dS$ is the element of the surface measure.

It seems like we need to use the divergence theorem at some point, but I'm struggling to get it in a form where we can actually use divergence theorem. So far I've tried evaluating the terms on the left hand side. Here's what I got:

$$ |\Delta u|^2=\left(\frac{\partial^2 u}{\partial x_1 ^2}+\cdots+\frac{\partial^2 u}{\partial x_m ^2}\right)^2\\ =\sum_{i=1}^m \left( \frac{\partial^2 u}{\partial x_i^2} \right)^2+\sum_{i,j=1, i\neq j}^m \frac{\partial^2 u}{\partial x_i^2}\frac{\partial^2 u}{\partial x_j^2} $$ and $$ \sum_{i,j=1}^m|D_{ij}u|^2=\sum_{i=1}^m\left( \frac{\partial^2 u}{\partial x_i^2} \right)^2+\sum_{i,j=1, i\neq j}^m \left(\frac{\partial^2 u}{\partial x_i \partial x_j}\right)^2 $$ which gives $$ LHS=\int_B \sum_{i,j=1, i\neq j}^m \frac{\partial^2 u}{\partial x_i^2}\frac{\partial^2 u}{\partial x_j^2}-\left(\frac{\partial^2 u}{\partial x_i \partial x_j}\right)^2 $$

And now I'm stuck. Any ideas?

Thanks,

Tom

Update: I've tried using integration by parts and Green's identities with no luck yet.

Answer 1

So now $$LHS = \int_\Omega\sum_{i,j=1,i\ne j}^m\frac{\partial^2 u}{\partial x_i^2}\frac{\partial^2u}{\partial x_j^2}-\left(\frac{\partial^2u}{\partial x_ix_j}\right)^2\,dx$$ $$=\int_\Omega\sum_{i,j=1,i\ne j}^m-\frac{\partial^3 u}{\partial x_i^2dx_j}\frac{\partial u}{\partial x_j}-\left(\frac{\partial^2u}{\partial x_i\partial x_j}\right)^2\,dx+boundary\,\,terms$$ $$=\int_\Omega\sum_{i,j=1,i\ne j}^m\left(\frac{\partial^2 u}{\partial x_i\partial x_j}\right)^2-\left(\frac{\partial^2u}{\partial x_i\partial x_j}\right)^2\,dx+boundary\,\,terms$$ $$= 0 +boundary\,\,terms.$$ Can you take it from here? (note that the above steps where all done by integrating by parts)

Answer 2

Here are a couple of hints:

(i) Let $\eta$ be a smooth cutoff function such that on the ball of radius 1/4 it is equal to one, outside the ball of radius 1/2 it is equal to zero, and radially monotone decreasing. For any $C^3$ function $u$, consider the product $v(x) = u(x)\eta(x)$.

(ii) Consider $u(x) = 1 - |x|^2$.

(iii) Consider $u(x) = e^{\frac{-1}{1-x^2}}$.

It might be helpful to see what the desired equality says in each case.