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Set $D:=\{(x,y,z)\in \Bbb R^2:x^2+y^2+z^2 \le 1\}$. Denote $\Delta:= \frac{\partial f^2}{\partial x^2}+\frac{\partial f^2}{\partial y^2}+\frac{\partial f^2}{\partial z^2}$. Suppose that a $C^2$-function $\phi(x,y,z)$ vanishes in the neighborhood of $\partial D$. I want to prove that $$\iiint_D (x^2+y^2+z^2)^{-1/2}(\Delta \phi)(x,y,z)dxdydz=-4\pi\phi(0,0,0).$$ Using the spherical coordinates I can write $$\iiint_D (x^2+y^2+z^2)^{-1/2}(\Delta \phi)(x,y,z)dxdydz=\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^1 (\Delta \phi)(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\rho \sin \varphi d\rho d\theta d\varphi.$$ However I don't know where to use the assumption $\phi$ vanishes in the neighborhood of $\partial D$.

Any help would be much appreciated.

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  • $\begingroup$ You should try to use some Gauss/Stokes at some point. $\endgroup$
    – Fabian
    Commented Oct 5, 2016 at 8:47
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    $\begingroup$ $\nabla^{2}\left(1 \over r\right) = -4\pi\,\delta\left(\vec{r}\right)$ $\endgroup$ Commented Oct 6, 2016 at 2:23

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Use Green's second identity, $$\iiint_D (\psi\Delta\phi-\phi\Delta\psi)dV=\iint_{\partial D^+}\left(\psi\frac{\partial \phi}{\partial {\bf n}}-\phi\frac{\partial \psi}{\partial {\bf n}}\right)dS$$ with $\psi(x,y,z)=(x^2+y^2+z^2)^{-1/2}$.

Note that $\Delta\psi(x,y,z)=-4\pi\delta(x,y,z)$ and the integral on the right-hand side is zero because $\phi$ and $\frac{\partial \phi}{\partial {\bf n}}$ vanish in the neighborhood of $\partial D$.

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  • $\begingroup$ Does the fact that $\frac{\partial \phi}{\partial {\bf n}}$ vanishes in the neighborhood of $\partial D$ follow from ${\partial \phi \over \partial {\mathbf {n}}}=\nabla \phi \cdot {\mathbf {n}}=\nabla _{{\mathbf {n}}}\phi$? $\endgroup$
    – user
    Commented Oct 5, 2016 at 10:31
  • $\begingroup$ Yes, also $\nabla \phi=0$ on $\partial D$. That is the reason why you need $\phi=0$ in neighborhood of $\partial D$ and not simply $\phi=0$ on $\partial D$. $\endgroup$
    – Robert Z
    Commented Oct 5, 2016 at 11:02
  • $\begingroup$ So what we get from your computation is $\iiint_D \psi\Delta\phi dV=\iiint_D \phi\Delta\psi dV=-4\pi\iiint_D \phi\delta(x,y,z)dV$. How can you prove that $\iiint_D \phi\delta(x,y,z)dV=\phi(0,0,0)$? $\endgroup$
    – user
    Commented Oct 5, 2016 at 12:51
  • $\begingroup$ This is a property of the 3D-delta function, see here: en.wikipedia.org/wiki/… $\endgroup$
    – Robert Z
    Commented Oct 5, 2016 at 13:02
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    $\begingroup$ See the first integral en.wikipedia.org/wiki/Dirac_delta_function#Generalizations $\endgroup$
    – Robert Z
    Commented Oct 5, 2016 at 13:23

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