Say for the following problem, suppose boundary of $\Omega$ is $C^{1,1}$: $$ \left\{ \begin{aligned} -\Delta \phi &= \mathrm{div} \,\vec{u}\quad \text{ in } \Omega \\ \phi&=0 \quad \text{ on }\partial \Omega \end{aligned} \right. $$ Could we deduce $\nabla \phi\cdot \vec{n}$ on $\partial \Omega$ by the following reasoning? Multiply a test function $v\in H^1(\Omega)$ and by doing integration by parts we have: $$ \int_{\Omega} \nabla \phi\cdot \nabla v -\int_{\partial \Omega} (\nabla \phi\cdot\vec{n})\,v = -\int_{\Omega} \vec{u}\cdot \nabla v +\int_{\partial \Omega} (\vec{u}\cdot\vec{n})\,v $$ Could we say that by the arbitrariness of $v$ that $\nabla \phi\cdot \vec{n} = -\vec{u}\cdot \vec{n}$?


1 Answer 1


I think you got the signs wrong. (It wouldn't have hurt to spell out the integration by parts).

No, you can't deduce this from the arbitrariness of $v$, because $\nabla v$ isn't independent of $v$. Also, you didn't, as the title suggests, deduce the Neumann boundary data from the Dirichlet boundary data; you tried to deduce them from the differential equation (at least I don't see where you used $\phi=0$). That can't work, since the boundary data are not determined by the differential equation.

Another way to see that this can't be right is that the solenoidal (divergence-free) part of $\vec u$ doesn't enter into the equation at all, whereas it does enter into the boundary condition you deduced.

In another sense, however, the Neumann boundary data are indeed determined by the Dirichlet boundary data, since both determine and are determined by the solution (up to an additive constant).

  • $\begingroup$ Thanks, and sign corrected. $\endgroup$
    – Shuhao Cao
    Oct 2, 2011 at 16:31
  • $\begingroup$ which, btw, tells about us that we cannot impose Neumann and Dirichlet boundary conditions simultaniously. $\endgroup$
    – shuhalo
    Nov 27, 2011 at 12:59

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