Using the Rayleigh Quotient

Boundary conditions are:

$\displaystyle \alpha_1\phi(a) + \alpha_2\phi'(a) = 0$

$\displaystyle \beta_1\phi(b) = \beta_2\phi'(b) = 0$

Prove that, if u(x) satisfies the BC (but not necessarily the DE), then

$\displaystyle \lambda_1^2 = \min_i{\lambda_i^2} \leq \min_u{RQ(u)}$

Where RQ(u) is the Rayleigh Quotient where $\displaystyle \phi = u$

by expanding u(x) in eigenfunctions: $\displaystyle u \approx \sum_{n=1}^{\infty} c_n\phi_n$

and using the operator: $\displaystyle L[\phi] \equiv \frac{d}{dx}\left(p(x)\frac{d\phi}{dx}\right) + q(x)\phi$

My professor told us that the first eigenvalue is the smallest eigenvalue, but did not show us how to begin to prove it with the stuff provided. Any help would be appreciated.