Sturm-Liouville Variational Problem

Question

I'm entirely clueless with this problem. No formal training in variational methods.

Show that for function $\phi\left ( x \right )$ with

$$\phi\left ( a \right )=\phi\left ( b \right )=0$$

and with the further constraint

$$I = \int_{a}^{b}r\left ( x \right )\left ( \phi'\left ( x \right ) \right )^{2} dx=1 $$

that the functional

$$J\left ( \phi\left ( x \right ) \right )=\int_{a}^{b}\left [ p\left ( x \right )\left ( \phi'\left ( x \right ) \right )^{2}-q\left ( x \right )\left ( \phi^{2}\left ( x \right ) \right ) \right ]dx$$

is rendered stationary by the solution $y\left ( x, \lambda \right )$ of the Sturm-Liouville system

$$\left ( p\left ( x \right )y'\left ( x \right ) \right )'+\left ( q\left ( x \right ) +\lambda r\left ( x \right )y\left ( x \right ) \right )=0$$

in $a< x< b$ with $y\left ( a \right )=y\left ( b \right )=0$.

and that the minimum value of J is given by $\lambda_{1}$, the smallest eigen value of the Sturm-Liouville system

This popped up in a past year paper so I'm expecting something like this to spring up on the exam paper.

Help is appreciated

Answer 1

I am afraid your exam is long past, but maybe somebody else is interested. Your functional $I$ should have $\phi$ not $\phi'$ in it. The functional is stationary at $y$ (more or less by definition) if there is a Lagrange multiplier $\lambda$ such that the derivative of $$ F(s)=J(y+s\theta)-\lambda I(y+s\theta) $$ at $s=0$ vanishes for all smooth functions $\theta$ vanishing at $a$ and $b$. Since $F$ is a quadratic polynomial (for each $\theta$) it is easy to differentiate, and one obtains the condition $$ \int_a^b(py'\theta'-(q+\lambda r)y\theta)=0. $$ Now integrate by parts to move the derivative on $\theta$ to the other factor and you are done!