- #1
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?