Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question might sound naive but on p.34 of his book, he is considering the Poisson-dirichlet problem with $\Delta u=-f$ on $U$ with $u=g$ on $\partial U$. He then derives a formula for the general solution but says that it is not much useful since $\frac{\partial u}{\partial \nu}$ on $\partial U$ is unknown.

But this seems strange to me because isn't it that $\frac{\partial u}{\partial \nu} = \frac{\partial g}{\partial \nu}$ on $\partial U$ since $\frac{\partial u}{\partial \nu}$ on $\partial U$? Here, $\frac{\partial u}{\partial \nu} = Du\cdot \nu$ where $\nu$ is the unit normal to $U$.

If indeed I misunderstood the above, then if I had two boundary conditions $g_1$ and $g_2$ then could it be possible that $\frac{\partial u}{\partial \nu}$ on $\partial U$ is different for each boundary condition?


share|cite|improve this question
In general $g$ is not defined off the boundary so taking the directional derivative in the normal direction is not well defined. – Evan Sep 23 '13 at 4:28

Evan is rigth. I will try to clarify his comment with an example.

Let's assume $g=0$, $f=-1$ and $U=\mathbb{D}$. Then $u(x,y)=-0.25(x^2+y^2)+0.25$, but, easily, the normal derivative is not zero.

In general, the operator $$ T_U:g\rightarrow \partial u/\partial \nu, $$ i.e., such that, given $g$, it solves for $u$, and then return the normal derivative is called the Dirichlet-to-Neumann map for the laplacian operator and the domain $U$. These operator have a lot of importance in inverse problems and fluid dynamics.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.