I was looking for a proof of this identity: \begin{equation} \int_{\mathbb{R}^n}\phi \Delta\phi = - \int_{\mathbb{R}^n}|\nabla\phi|^2 \end{equation} given $\phi \in C_0^{\infty}(\mathbb{R}^n)$.
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I'm assuming $\phi: \Bbb{R}^n\to\Bbb{R}$? In that case write $\Delta \phi=\nabla\boldsymbol{\cdotp}\nabla\phi$ and use the divergence theorem. – K.defaoite Dec 18 '20 at 15:52
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The difference is a surface term,$$\int_{\Bbb R^n}(\phi\nabla^2\phi+\nabla\phi\cdot\nabla\phi)d^nx=\int_{\Bbb R^n}\nabla\cdot(\phi\nabla\phi)d^nx.$$
J.G.
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