# Eigenvalue problem with asymmetric boundary conditions

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?

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 $X_2(x_2)$.
• Oh I see. Is $\frac{\pi^2}{4} + k\pi^2 + 2k^2\pi^2$ the solution to the equation $s -\tan s = 0$? – user137302 May 6 '14 at 23:01
• 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? – Ellya May 6 '14 at 23:09
• I have checked, and it definitely solves the PDE, but not $\tan(s)=s$ which is very strange. – Ellya May 6 '14 at 23:15
• 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. – Ellya May 6 '14 at 23:21