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I am stuck with the following problem:

The PDE
$u_{xx}+u_{yy}+\lambda u=0, 0
$u(x,0)=u(x,1)=0; 0\leq x \leq 1$
$u(0,y)=u(1,y)=0; 0\leq y \leq 1$
has

(a)a unique solution u for any $\lambda \in \mathbb R ,$
(b)infinitely many solutions for some $\lambda \in \mathbb R ,$
(c)a solution for countably many values of $\lambda \in \mathbb R ,$
(d)infinitely many solutions for all $\lambda \in \mathbb R .$

I do not know how to progress with it.Could someone point me in the right direction( e.g. a certain theorem or property i have to use?) Thanks in advance for your time.

  • 0
    Thanks a lot Jon and Paul.2012-12-16

2 Answers 2

1

Consider the basis $ B=\{\phi_{00},\ \theta_{0n},\ \xi_{n0},\ \phi_{mn},\ \psi_{mn},\ \theta_{mn},\xi_{mn}\}_{(m,n) \in \mathbb{N}^2} $ of $L^2([0,1]^2)$ given by \begin{eqnarray} \phi_{00}(x,y)&=&1,\ \theta_{0n}(x,y)=\sin(2\pi ny),\ \xi_{n0}(x)=\sin(2\pi nx),\\ \phi_{mn}(x,y)&=&\cos(2\pi mx)\cos(2\pi ny),\ \psi_{mn}(x,y)=\sin(2\pi mx)\sin(2\pi ny),\\ \theta_{mn}(x,y)&=&\cos(2\pi mx)\sin(2\pi ny),\ \xi_{mn}(x,y)=\sin(2\pi mx)\cos(2\pi ny). \end{eqnarray} Then for every $m,n \ge 0$ we have $ -\Delta\phi_{mn}=\lambda_{mn}\phi_{mn},\ -\Delta\psi_{mn}=\lambda_{mn}\psi_{mn},\ -\Delta\theta_{mn}=\lambda_{mn}\theta_{mn},\ -\Delta\xi_{mn}=\lambda_{mn}\xi_{mn}\ \forall\ m,n \ge 0, $ where $ \lambda_{mn}=4\pi^2(m^2+n^2) \quad \forall\ m, n\ge 0. $ This shows that the number $\lambda_{mn}$ is an eigenvalue of $-\Delta$ with corresponding eigenfunctions $\phi_{mn},\psi_{mn},\theta_{mn},\xi_{mn}$.

Because of the boundary conditions \begin{eqnarray} u(x,0)=u(x,1)&=&0 \quad \forall x \in [0,1],\\ u(0,y)=u(1,y)&=&0 \quad \forall y \in [0,1], \end{eqnarray} it is clear that a pair $(u,\lambda)$ satisfies the given PDE precisely when it belongs to $\{(c\psi_{mn},\lambda_{mn}): \ m,n \ge 1, \ c \in \mathbb{R}\setminus\{0\}\}$.

  • 0
    I was trying to understand your solution . But I can not. Can you please explain to me in simple language?@Mercy King2018-05-10
0

Hint: See the Dirichlet eigenvalues.

  • 0
    @user52976: You are welcome.2012-12-16