• $B_r(x_0)$ and $\overline{B}_r(x_0)$ be the open and closed ball in $\mathbb{R}^n$, respectively
  • $u\in C^2(B_r(x_0))$ and $\rho\in (0,r)$
  • $\lambda_n$ be the Lebesgue-measure on the Borel-$\sigma$-algebra on $\mathbb{R}^n$
  • $S_M$ denote the "surface measure" on a submanifold $M$

I've read in a book that we've got $$\int_{B_\rho(x_0)}\Delta u\;d\lambda_n=\int_{\partial B_\rho(x_0)}\frac{\partial u}{\partial\nu}dS_{\partial B_\rho(x_0)}\color{red}{\stackrel{(1)}{=}\int_{\partial B_1(0)}\frac{\partial u}{\partial\rho}(x_0+\rho\omega)\rho^{n-1}\;d\omega}$$ where $\nu$ is the "outer normal field" of $B_\rho$. How can we prove the right equation?

I've tried the following: Let $$\phi:\partial B_\rho(x_0)\to\partial B_1(0)\;,\;\;\;x\mapsto\frac{x-x_0}\rho .$$ Moreover, let $$F:\partial B_\rho(x_0)\to\mathbb{R}\;,\;\;\;x\mapsto\frac{\partial u}{\partial \nu}(x)$$ and $$f:\partial B_1(0)\to\mathbb{R}\;,\;\;\;\omega\mapsto F(x_0+\rho\omega) .$$ Obviously, we've got $F=f\circ\phi$ and thereby $$\int_{\partial B_\rho(x_0)}F\;dS_{\partial B_\rho(x_0)}=\int_{\partial B_1(0)}f\;d\left(S_{\partial B_\rho(x_0)}\circ\phi^{-1}\right)\tag{2}$$ where $S_{\partial B_\rho(x_0)}\circ\phi^{-1}$ is the pushforward measure. Now, that doesn't look too bad. We may need to notice that $$\nu(x)=\frac{1}{\rho}(x-x_0)\;\;\;\text{for all }x\in\partial B_\rho(x_0) .$$ So, $$\nu\left(\phi^{-1}(\omega)\right)=\omega\;\;\;\text{for all }\omega\in\partial B_1(0) .$$ Hence, $$\frac{\partial u}{\partial\nu}\left(\phi^{-1}(\omega)\right)=\langle\nabla u\left(\phi^{-1}(\omega)\right),\nu\left(\phi^{-1}(\omega)\right)\rangle=\langle\nabla u(x_0+\rho\omega),\omega\rangle\;\;\;\text{for all }\omega\in\partial B_1(0)$$

I've got stuck at this point. How can we use the last equation in $(2)$ to achieve $(1)$? Please note, that I'm not familiar with the notation "$d\omega$". I assume that it is exactly the pushforward measure $S_{\partial B_\rho(x_0)}\circ\phi^{-1}$, but I may be wrong.


The notation $d\omega$ is being used for the standard (hyper)area measure on the unit sphere $\partial B_1$. The pushforward measure is in fact $d\left(S_{\partial B_\rho (x_0)} \circ \phi^{-1}\right) = \rho^{n-1} d\omega$, since the scaling $x \to \rho x$ scales hyperareas by $\rho^{n-1}$ and the translation leaves them invariant.

Thus the only missing piece is $$f(\omega) = \frac{\partial u}{\partial \nu} (x_0 + \rho \omega) = \frac{\partial}{\partial \rho} u(x_0 + \rho \omega),$$ which follows from the fact that the outwards derivative $$\frac{\partial u}{\partial \nu} (x_0 + \rho \omega) = \langle \nabla u(x_0 + \rho \omega), \omega\rangle$$ is simply $\partial/\partial \rho$ in spherical coordinates.

  • $\begingroup$ So, $d\omega=S_{\partial B_1(0)}$, right? Could you derive in more detail why $$\frac{\partial u}{\partial\nu}(x_0+\rho\omega)=\frac{\partial u}{\partial \rho}\;?$$ $\endgroup$ – 0xbadf00d Apr 20 '15 at 21:05
  • 1
    $\begingroup$ @0xbadf00d: Right. The multivariable chain rule gives $\frac{\partial}{\partial \rho} u(x_0 + \rho \omega)= \langle\nabla u(x_0 + \rho \omega),\frac{\partial}{\partial \rho}(x_0 + \rho \omega)\rangle = \langle\nabla u(x_0 + \rho \omega),\omega\rangle $ which is exactly the expression you found for $\frac{ \partial u }{\partial \nu}$. $\endgroup$ – Anthony Carapetis Apr 21 '15 at 6:24

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