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I was reading a text regarding to PDE, and I encountered a step which I didn't understand.

Let $U$ be an open and bounded set in $\mathbb R^n$ with smooth boundary. Let $u\in W^{2,2}_0(U).$

Now it says that by the divergence theorem, $$\int_U|D^2u|^2dx=\int_U\sum_{i,j=1}^n u_{x_i x_j}u_{x_i x_j}dx=\int_U\sum_{i,j=1}^n u_{x_i x_i}u_{x_j x_j}dx=\int_U |\Delta u|^2dx.$$

I don't really understand why do we have $$\int_U\sum_{i,j=1}^n u_{x_i x_j}u_{x_i x_j}dx=\int_U\sum_{i,j=1}^n u_{x_i x_i}u_{x_j x_j}dx.$$

The integrands look very different from each other and they have many terms which the other one doesn't have. Can any one help me with this?

Thanks for help.

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1 Answer 1

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Integration by parts (a corollary of the divergence theorem) yields \begin{equation} \int_U u_{x_ix_j} u_{x_ix_j}\,dx = \int_{\partial U} u_{x_i} u_{x_ix_j}\nu^j \,dS(y) -\int_U u_{x_i} u_{x_ix_jx_j}\,dx, \end{equation} where $\nu(y) = (\nu^1(y),\nu^2(y),\cdots,\nu^n(y))$ is the unit outward normal of $\partial U$ at $y\in\partial U$ and $dS(y)$ is the surface element of $\partial U$. Notice that the boundary term vanishes since $u\in W^{2,2}_0(U)$. Integrating by parts again, we have \begin{align} \int_U u_{x_ix_j} u_{x_ix_j}\,dx =& - \int_U u_{x_i} u_{x_ix_jx_j}\,dx \\ =& - \int_U u_{x_i} u_{x_jx_jx_i}\,dx \\ =& - \int_{\partial U}u_{x_i}u_{x_jx_j}\nu^i \,dS(y)+\int_{U}u_{x_ix_i}u_{x_jx_j}\,dx \\ =&\int_{U}u_{x_ix_i}u_{x_jx_j}\,dx, \end{align} which implies \begin{equation} \int_U \sum\limits_{i,j=1}^n u_{x_ix_j} u_{x_ix_j}\,dx= \int_U \sum\limits_{i,j=1}^n u_{x_ix_i}u_{x_jx_j}\,dx. \end{equation}

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