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I'm wondering if the Pohozaev identity: $$n\int_\Omega\int_0^{u(x)}f(t)\operatorname{d}t\operatorname{d}x-\frac{n-2}{2}\int_\Omega u(x)f(u(x))\operatorname{d}x=\frac{1}{2}\int_{\partial\Omega}\left|\frac{\partial u}{\partial\nu}(x)\right|^2x\cdot\nu(x)\operatorname{d}S(x)$$ where $\Omega$ is a non-empty open bounded subset of $\mathbb{R}^n$ with smooth boundary, $\nu:\partial\Omega\to\mathbb{R}^n$ is the outer normal, $S$ is the surface measure, $f\in C^1(\mathbb{R})$ and $u\in C^2(\Omega)\cap C^1_0(\bar\Omega)$ is such that $-\Delta u=f(u)$, is just a "black magic formula" or if there some kind of intuition behind it... I looked into its proof, but actually it seems just a lot of dirty tricks about rewriting terms in order to use divergence theorem several times, leaving me a bit confused about the meaning of this formula... Can anyone give some kind of (geometric?) insight about this formula and/or can show an intuitive way to prove it?

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  • $\begingroup$ Have you checked Kavian's book? He makes a lot of remarks that you may find useful. As I understand, the Pohozave identity is specially useful when proving the non-existence of nontrivial solutions. Of course this does not answer your question, but maybe this perspective could help. $\endgroup$ – Danilo Gregorin Jul 30 at 14:45

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