Suppose $\Omega$ be a bounded open set in $\mathbb{R}^{n}$. We assume that $u$ is sufficiently smooth so that the calculations make sense. Then is it true that
$$\int_{\Omega}|\nabla u|^{2}dx = -\int_{\Omega}u\Delta udx,$$
or is there something missing?
This is a question that arises from an equation I've seen here
Thanks in advance.
Recall the divergence theorem \begin{align} \iiint_R \operatorname{div}(\mathbf{F})\ dV = \iint_{\partial R} \mathbf{F}\cdot\mathbf{n}\ dS. \end{align} Let $\mathbf{F}= u\nabla u$, then we see \begin{align} \iiint_U \{|\nabla u|^2+u\Delta u\}\ dV=\iiint_U \operatorname{div}(u\nabla u)\ dV = \iint_{\partial U} u\frac{\partial u}{\partial \mathbf{n}} dS = 0 \end{align} since $u\equiv 0$ on the boundary.