I am having difficulty understanding how the following identity is suppose to come from the divergence theorem:


$\int_{B(y, r)} \Delta u(x) dx = \int_{\partial B(y, r)} {\partial u(x) \over \partial \nu} do(x) $

Where $do(x)$ is a volume element of the boundary ${\partial B(y, r)}$, and $\nu$ is exterior normal of the boundary.

I don't know where the partial derivative with respect to $\nu$ is coming from, and how the divergence theorem applies.

Divergence Theorem:

$\int_{\Omega}div V(x)dx=\int_{\partial \Omega} V(z)\cdot v(z)do(z)$

Other notation:

$\Delta=\nabla^2 $

div$V(x):= \sum_{i=1}^{d}\frac{\partial V^i}{\partial x^i}(x)$

Thanks for any help that can be provided.


Write $\Delta u = div(\nabla u)$, then

$$\int_B \Delta u dx= \int_B div (\nabla u) dx = \int_{\partial B} \nabla u \cdot \nu do(x) = \int_{\partial B} \frac{\partial u}{\partial \nu} do(x)$$

  • $\begingroup$ Thanks very much, it is a lot more clear. However, I do not understand equality of the last two terms. Is this a special identity? I tried to write u in 3 variables, and take the gradient of u (which produces a vector), then I take the dot product with the normal vector \nu (which produces a scalar). But I don't understand how this will change into the partial derivative w.r.t \nu. What happens to the spatial derivatives that came from grad(u)? Thanks again, $\endgroup$ – BBaire Dec 2 '13 at 6:07
  • $\begingroup$ @BBaire: $\frac{\partial u}{\partial \nu}$ is just the directional derivative of $u$ (in the $\nu$ direction). It is just Chain rule there: $\nabla_v u = \langle \nabla u, v\rangle$. $\endgroup$ – user99914 Dec 2 '13 at 6:37

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