2
$\begingroup$

Let $\Omega$ be a bounded domain with sufficiently smooth(say piece-wise smooth) boundary and consider the Laplacian eigenvalue problem with the given boundary condition(Cauchy BV ) \begin{cases} -\Delta u=\lambda u \text{, $\,\,\,\, \Omega$} \\ \frac{\partial u}{\partial n}=u=0 \text{,$\,\,\,\,\, \partial\Omega$.} \end{cases} What are some examples of $\Omega$ with explicitly known spectrum? In particular, what are the spectra for $\Omega$=disk, circular annulus, regular polygon?

Any references would be highly appreciated.

$\textbf{Remark:}$ The above problem is a generalization of this question.

$\endgroup$
  • 1
    $\begingroup$ I have a feeling that because the boundary conditions force both the normal derivative AND the function values to zero on the boundary, there will be no solutions. For the classical round drum-head problem for example, you can have zero crossings of the bessel functions at the boundary, but then the derivative is not zero there. Generalizing this to other smooth boundaries is just based my on intuition, however, and I may be wrong. $\endgroup$ – rajb245 Sep 21 '16 at 17:40
  • 1
    $\begingroup$ Forcing assumption on the boundary value and its normal derivative makes the problem overdetermined. There are plenty of overdetermined problems with non-zero solutions, but again this does not refute your answer nor gives anything fruitful. $\endgroup$ – BigM Sep 21 '16 at 17:49
  • $\begingroup$ @rajb245 You are right.I managed to show that the only solution is the zero function. At the moment I am not able to prove this cheaply. $\endgroup$ – BigM Sep 22 '16 at 17:13
3
$\begingroup$

The given BVP has no non-trivial solution. This follows immediately from Rellich's formula which states a non-zero eigenfunction $u$ corresponding to an eigenvalue $\lambda$ for the Dirichlet eigenvalue problem on $\Omega$ satisfies the following

$$\lambda=\frac{\int_{\partial\Omega}(\frac{\partial u}{\partial n})^2\frac{\partial(r^2)}{\partial n}ds}{4\int_\Omega u^2dA},$$ where $r^2=x_1^2+x_2^2,$ $dA$ is the area measure and $ds$ is arc-length measure.

$\endgroup$
  • 1
    $\begingroup$ I'm glad you were able to find a result that directly implies that $u=0$ is the only solution to your problem! $\endgroup$ – rajb245 Sep 22 '16 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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