I am trying to prove this:

$S$ is a bounded surface in $\mathbb{R}^2$ and $a$ a given scalar field

$u,v$ are such that $\nabla^2 u=0$ on $S$ and $u=v=a$ on $\partial S$. Then:

$$\int_S |\nabla v|^2\;\mathrm{d}S\geqslant \int_S |\nabla u|^2\;\mathrm{d}S$$

I would know how to do this is if $S$ were a volume, because I can use the divergence theorem. So I am wondering: is it true that for a bounded surface in $\mathbb{R}^2,$

$$\int_{\partial S}\mathbf{F}\cdot\mathrm{d}\mathbf{r}=\int_S \text{div}\,\mathbf{F}\;\mathrm{d}S?$$

  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/130081/… $\endgroup$ – David H Jun 1 '14 at 14:16
  • $\begingroup$ Oh man, you really seem to have problems with the English language. Even a flat surface is still not restricted to $\mathbb R^2$. The non-mathematical meaning the word "surface" is the outer face of an object. What you probably mean is called an "area" or a (two dimensional) "domain". $\endgroup$ – Thomas Klimpel Jun 1 '14 at 19:21

No, the divergence theorem cannot be restricted to surfaces in the way you hope. The Stokes theorem for surfaces would involve vector fields and the rotation.

As a simple counterexample to your proposed statement, consider a simple bounded surface $S$ like the unit disk in the x-y plane, and two simple functions $u$ and $v$. Let $a(x,y,z)=v(x,y,z)=0$ and $u(x,y,z)=z$. It's easy to see that $\nabla^2u=0$ and $u=v=a=0$ on $\partial S$. Because $|\nabla v|^2=0$ and $|\nabla u|^2=1$, we have

$$\int_S |\nabla v|^2\;\mathrm{d}S < \int_S |\nabla u|^2\;\mathrm{d}S$$

  • $\begingroup$ This is why I asked about flat surfaces (in which case I think the initial statement holds) $\endgroup$ – custodia Jun 1 '14 at 14:25
  • $\begingroup$ For flat surfaces, yes, your initial statement does hold, because it reduces to just Green's theorem in this case. $\endgroup$ – David H Jun 1 '14 at 14:31

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