$u\in C_0^3(\Omega)$ satisfis $ -\Delta u=f(x), x\in\Omega. $Prove that $ \sum_{i,j=1}^{n}\int_\Omega u^2_{x_i x_j} dx\leq n\int_{\Omega}f^2 dx. $

Suppose $u\in C_0^3(\Omega)$ satisfis $$ -\Delta u=f(x), x\in\Omega. $$ Prove that $$ \sum_{i,j=1}^{n}\int_\Omega u^2_{x_i x_j} \mathrm{d}x\leq n\int_{\Omega}f^2\mathrm{d}x. $$

I have tried as follows: Let $v=\Delta u$ in the Green's formula $\int_\Omega \nabla u\cdot\nabla v \mathrm{d}x=-\int_\Omega v\Delta u \mathrm{d}x+\int_{\partial\Omega}\frac{\partial u}{\partial n}v\mathrm{d}S$, then we get $$ \int_\Omega \nabla u\cdot\nabla(\Delta u)dx=-\int_\Omega f^2+\int_{\partial\Omega}\frac{\partial u}{\partial n}(-f)\mathrm{d}S. $$ By the Integration by parts formula or Green's formula, I find the left side can be written as $$ LHS=\sum_{i,j=1}^n\left[-\int_\Omega(u_{x_ix_j})^2\mathrm{d}x+\int_{\partial\Omega}u_{x_ix_j}\cdot u_{x_i}\cdot n^j\mathrm{d}S\right]\\=-\sum_{i,j=1}^n\int_\Omega(u_{x_ix_j})^2\mathrm{d}x+\int_{\partial\Omega}\mathrm{div}u\cdot\dfrac{\partial \mathrm{div}u}{\partial n}\mathrm{d}S. $$ I do not know how to proceed. Where to find the coeffcient $n$? Also, how to estimate the term $\int_{\partial\Omega}\frac{\partial u}{\partial n}(-f)\mathrm{d}S$?

Did I get wrong somewhere? Appreciate any help!

Solution 1:

In fact a stronger statement holds and it follows directly from integration by parts. Indeed, integrating by parts twice we obtain \begin{align*}\sum_{i,j=1}^n \int_\Omega u^2_{x_ix_j} \, dx &= -\sum_{i,j=1}^n \int_\Omega u_{x_ix_ix_j}u_{x_j} \, dx \\ &=\sum_{i,j=1}^n \int_\Omega u_{x_ix_i}u_{x_jx_j} \, dx \\ &= \int_\Omega (\Delta u )^2 \, dx \\ &=\int_\Omega f^2 \, dx. \end{align*} There are no boundary terms since $u$ has compact support in $\Omega$.