0
$\begingroup$

Consider the unit square $\,\Omega = (0,1) \times (0,1) $ and the normal eigenvalue problem for Laplace's equation $$ -\Delta u = \lambda u $$ with the boundary conditions that on the vertical sides of the square and on the bottom of the square $$u = 0$$ and on the top of the square $$\frac{\partial u}{\partial n} = 0$$ where $n$ is the outward normal. I have come up with the answer $\sin{\pi k x_1} \, \sin{\frac{\pi k}{2} x_2}$ with eigenvalues $\lambda = \pi^2 k^2 + \frac{\pi^2k^2}{4}$ where $k \in \mathbb{Z}\backslash\{0\}$. However, the question asks to show that the eigenvalues are the roots of the equation $s - \tan s = 0$, but I plugged my solution into wolfram and it matches all the boundary conditions and solves the PDE. Any ideas?

$\endgroup$
0
$\begingroup$

I see where the error is, where you have $\sin(\frac{\pi}{2}x_2)$, what you should have is $(\frac{\pi}{2}+k\pi)$ for $k\in\Bbb Z$

Overall resulting in $\lambda_k=\pi^2k^2+(\frac{\pi}{2}+k\pi)^2=\pi^2k^2+\frac{\pi^2}{4}+k\pi^2+k^2\pi^2=\frac{\pi^2}{4}+k\pi^2+2k^2\pi^2$.

Check your working and you will see that for $k=0$, $\alpha=0$ does not give you a non trivial solution for $X_2(x_2)$.

$\endgroup$
  • $\begingroup$ Oh I see. Is $\frac{\pi^2}{4} + k\pi^2 + 2k^2\pi^2$ the solution to the equation $s -\tan s = 0$? $\endgroup$ – user137302 May 6 '14 at 23:01
  • $\begingroup$ @user137302 it is not quite! but I may have made mistakes, go back to your eigenvalues, taking into account the correction have put, and tell me what happens, I feel we are close to the desired result $\endgroup$ – Ellya May 6 '14 at 23:05
  • $\begingroup$ I have seen situations similar to this, but it required $du/dn=0$ on the top and bottom boundaries, are you sure that is not the case here? $\endgroup$ – Ellya May 6 '14 at 23:09
  • $\begingroup$ I have checked, and it definitely solves the PDE, but not $\tan(s)=s$ which is very strange. $\endgroup$ – Ellya May 6 '14 at 23:15
  • $\begingroup$ Just looking into it, the boundary conditions that cause this situation are exactly the same but it should be $du/dn+u=0$ on the upper boundary. $\endgroup$ – Ellya May 6 '14 at 23:21

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.