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We know that $\Phi(x):=\frac{1}{(n-2)\omega_n|x|^{n-2}}$ for $n\ge3$ is the fundamental solution for the Laplace's equation. Given that $v\in C^{\infty}(\mathbb{R^n})$ has compact support, how is it that we can verify the integral?

$$\int_{\mathbb{R^{n}}} \Phi(x) \Delta v(x) dx = v(0) $$

For reference, I am reading/studying Partial Differential Equations (2nd Ed) by Lawrence Evans.

MrStormy83
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By the definition of distributional derivatives and since $-\Delta\Phi = \delta_0$ in the sense of distributions, $$ \int \Phi(x)\,\Delta v(x)\,\mathrm d x = \langle \Delta\Phi, v\rangle = -\langle \delta_0, v\rangle = -v(0). $$ Does it answer your question or was your question about why $\Phi$ is the fundamental solution of Laplace equation?

LL 3.14
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  • From where does the relation $\int \Phi(x) \Delta v(x) dx = \left< \Delta\Phi(x), v(x) \right>$ come? I don't think I've ever seen that. – MrStormy83 Oct 09 '22 at 21:06
  • If $\Phi$ was a nice function, this is just integration by parts ... in this case since it is not, it follows from the definition of distributional derivative: $\langle \partial_{k}f,\varphi\rangle = -\langle f,\partial_{k}\varphi\rangle$ $$ \int \Phi(x),\Delta v(x),\mathrm d x = \sum_{k=1}^n \langle \Phi,\partial_k^2v\rangle = \sum_{k=1}^n \langle - \partial_k \Phi,\partial_kv\rangle = \sum_{k=1}^n \langle \partial_k^2 \Phi,v\rangle = \langle \Delta \Phi,v\rangle $$ – LL 3.14 Oct 09 '22 at 22:12