In a rectangular domain $R$ with sides $a$ and $b$, with $b^2/a^2$ irrational, we look at the differential equation $- \Delta u = \lambda u$ with Dirichlet boundary conditions. How to show that the eigenvalues are given by $\lambda_{m,n} = \frac{m^2}{a^2} + \frac{n^2}{b^2}$ and the eigenfunctions are given by $u_{m,n}(x,y) = \sin (\frac{mx}{a}) \sin(\frac{ny}{b})$ where $u_{m,n}(x, y)$ is solution of the differential equation ?
I used the separation of variables : $u_{m,n}(x, y) = X(x)Y(y)$ and then : $$\frac{X''(x)}{X(x)} - \frac{Y''(y)}{Y(y)} = \lambda$$ Then we have $$\frac{X''(x)}{X(x)} = - \lambda - \frac{Y''(y)}{Y(y)} $$ We assume that $\frac{X''(x)}{X(x)} = -k^2$. Donc $X(x) = \sin(kx)$ On the other side, we have $$\frac{Y''(y)}{Y(y)} = k^2 - \lambda.$$ It implies that $Y(y) = \sin(\sqrt{k^2-\lambda} y)$
With the Dirichlet conditions, $u_{m,n}(x, y) = X(x)Y(y) = 0$ if $x = 0$ or $a$, $y = 0$ or $b$.
Then for $X(x) = \sin(kx)$, $k = \frac{m \pi}{a}$ and for $Y(y) = \sin(\sqrt{k^2-\lambda} y)$, $\sqrt{k^2-\lambda} = n\pi$. It implies that $\lambda = k^2 - \frac{n^2 \pi^2}{b^2} = \frac{m^2 \pi^2}{a^2} - \frac{n^2 \pi^2}{b^2}. $ I don't know how i can delete $\pi^2$ from the solution...
After i have to show that $\mu (u_{m,n}) = mn$ where $\mu$ counts the number of nodal components in the rectangle. But how ?