Let $Q = \{ (x, y): 0 < x < a , 0 < y <b \}$. Solve the problem: $$ - \Delta u (x, y) = \lambda u(x, y), ~ u \Bigr|_{ \delta Q} = 0. $$
P.S. I know the theorem about eigenvalue of Dirichlet problem: any eigenvalue of the Laplace operator $\lambda$ is real, positive and each $\lambda$ has finite multiplicity. But I don't know how to find a solution of this task.
consider the Fourier series on the segment $(0, 2a)$: $ \{ \sin \left({ \frac{\pi n }{a}x} \right) \}_{n=1}^{\infty} $ and the Fourier series on the segment $(0, 2b)$: $ \{ \sin \left({ \frac{\pi m }{b}y} \right) \}_{m=1}^{\infty}$. Any $f(x , y) \in C^2( \overline{Q}) , f \bigr|_{\delta Q} = 0 $ is provided as $$ f(x, y) = \sum_{1 \leq n, m < + \infty } a(n, m) \sin \left({ \frac{\pi n }{a}x} \right) \sin \left({ \frac{\pi m }{b}y} \right) $$ find $- \Delta f(x, y)$ : $$ -\Delta f(x, y) = \sum_{1 \leq n, m < + \infty } \lambda(n, m) a(n, m) \sin \left({ \frac{\pi n }{a}x} \right) \sin \left({ \frac{\pi m }{b}y} \right), ~ \lambda(n, m) = \pi^2 \left( \frac{n^2}{a^2} + \frac{m^2}{b^2} \right) $$ Finally, we get if $ -\Delta f(x, y) = \lambda f(x, y)$, then $\lambda(n, m) = \lambda$ , therefore only $\phi_{nm}(x , y)$ are eigenfunctions and $\lambda(n , m)$ are eigenvalues in this task.