[Math] rayleigh quotient of eigenvalue problem (sturm liouville theory and partial differential equations)

Question

I am reading "A First Course in Partial Differential Equations with Complex Variables and Transform Methods" (Weinberger, p. 168).

if we have the eigenvalue problem
$$ (pu')'- qu + \lambda \rho u = 0 $$
$$ u(0) = 0 $$
$$ p(1)u'(1) + au(1) = 0, a \ge 0 $$

we find that the eigenvalues are defined by the minimum principles

$$ \lambda_k = \min_{\phi \in S_k} \frac{\int_{0}^1 (p\phi'^2 + q\phi^2) dx + a\phi(1)^2} {\int_{0}^1 \rho \phi^2 dx} $$

where $ S_k $ is the set of all continuous and piecewise continuously differentiable functions satisfying

$$ \phi(0) = 0 $$
$$ \int_{0}^1 \rho \phi u_j dx = 0 $$
for $ j = 1, …, (k-1) $

Note that the eigenvalues are arranged in increasing order.

Could you please explain how the eigenvalues $ \lambda_k $ of the problem are given by minima of the aforementioned Rayleigh quotient.

Answer 1

Theorem: Suppose $\mathcal{H}$ is a Hilbert space, and suppose $A : \mathcal{D}(A)\subseteq \mathcal{H}\rightarrow \mathcal{H}$ is a densely-defined symmetric linear operator such that $$ \inf_{x \in \mathcal{D}(A)\setminus\{0\}}\frac{(Ax,x)}{(x,x)} = \mu > -\infty. $$ If $(Ax,x)=\mu(x,x)$ for some $x\in \mathcal{D}(A)\setminus\{0\}$, then $Ax=\mu x$.

Note: densely-defined means $\mathcal{D}(A)$ is a dense subspace of $\mathcal{H}$. A proof of this theorem can be reduced to $\mu=0$ by considering $B=A-\mu I$ instead; then $$ (Bx,x) \ge 0,\;\;\; x \in \mathcal{D}(A), $$ and you can show that $(Bx,x)=0$ for some $x\ne 0$ iff $Bx=0$. This can be shown by observing that $[x,y]=(Bx,y)$ is a pseudo inner product (pseudo means that $[x,x]=0$ may occur even if $x \ne 0$.) Because of this, the Cauchy-Schwarz inequality holds: $$ |[x,y]|^2 \le [x,x][y,y] \\ |(Bx,y)|^2 \le (Bx,x)(By,y) $$ Therefore, if $(Bx,x)=0$ for some $x$, then $(Bx,y)=0$ for all $y\in\mathcal{D}(B)$, which forces $Bx=0$ because $\mathcal{D}(B)=\mathcal{D}(A)$ is dense in $\mathcal{H}$.