Estimating Eigenvalues from linear ODE

[member 428835]

Homework Statement

Given $$u''(x)+\lambda u = 0\\
u(-1)=u(1)=0.$$
If $\lambda_0$ is the lowest eigenvalue, show that $4 \lambda_0 = \pi^2$.

Homework Equations

$$\lambda_0 = glb\frac{(L(u),u)}{(u,u)}$$ where $glb$ denotes greatest lower bound and $L$ is the Sturm-Louiville operator. I found this equation in the book though I am not sure it is needed.

The Attempt at a Solution

We have $L \equiv -d^2_x$, so $$\lambda_0 = glb\frac{(L(u),u)}{(u,u)}\\ = glb\frac{\int_{-1}^1 u''u\,dx}{\int_{-1}^1 u^2\,dx}\\=glb\frac{\int_{-1}^1 u'^2\,dx}{\int_{-1}^1 u^2\,dx}$$
but from here I'm stuck. I know that last integral should be $\pi/4$ but I'm unsure how to proceed. Perhaps I'm not on the correct track to start? Any ideas?

 

[member 428835]

The solution is to guess $u=\cos n\pi x/2$ and from there it all works out...inspection.

 

[LCKurtz]

Or you could just solve the eigenvalue problem. Look at the cases $$
\lambda = \mu^2 > 0,~\lambda = -\mu^2 < 0, \lambda = 0$$Show only the first case yields non-zero solutions and find the eigenvalues.