User:Dan Nessett/Sandboxes/Sandbox 3

This article proves that solutions to the Sturm-Liouville equation corresponding to distinct eigenvalues are orthogonal. Note that when the Sturm-Liouville problem is regular, distinct eigenvalues are guaranteed. For background see Sturm-Liouville Theory.

Orthogonality Theorem
Let $\map f x$ and $\map g x$ be solutions of the Sturm-Liouville equation:


 * $-\map {\dfrac \d {\d x} } {\map p x \dfrac {\d y} {\d x} } + \map q x y = \lambda \map w x y$

where $y$ is a function of the free variable $x$ and the functions $\map p x > 0$ has a continuous derivative, $\map q x$, and $\map w x > 0$ are specified at the outset, and in the simplest of cases are continuous on the finite closed interval $\closedint a b$.

Assume that the Sturm-Liouville equation is regular, that is, $\map p x^{-1} > 0$, $\map q x$, and $\map w x > 0$ are real-valued integrable functions over the finite interval $\closedint a b$, with separated boundary conditions of the form:


 * $(1)$ $: \quad \map y a \cos \alpha - \map p a \map {y'} a \sin \alpha = 0$


 * $(2)$ $: \quad \map y b \cos \beta - \map p b \map {y'} b \sin \beta = 0$

where $\alpha, \beta \in \hointr 0 \pi$

Then:


 * $\ds \innerprod f g = \int_a^b \overline {\map f x} \map g x \map w x \rd x = 0$

where $\map f x$ and $\map g x$ are solutions to the Sturm-Liouville equation corresponding to distinct eigenvalues and $\map w x$ is the "weight" or "density" function.

Proof
Multiply the equation for $\map g x$ by $\overline {\map f x}$ (the complex conjugate of $\map f x$) to get:


 * $-\overline {\map f x} \dfrac {\map \d {\map p x \map {\dfrac {\d g} {\d x} } x} } {\d x} + \overline {\map f x} \map q x \map g x =\mu \overline {\map f x} \map w x \map g x$

(Only $\map f x$, $\map g x$, $\lambda$, and $\mu$ may be complex; all other quantities are real.)

Complex conjugate this equation, exchange $\map f x$ and $\map g x$, and subtract the new equation from the original:

Integrate this between the limits $x = a$ and $x = b$:


 * $\ds \paren {\mu - \overline \lambda} \int \nolimits_a^b \overline {\map f x} \map g x \map w x \rd x = \map p b \paren {\map g b \map {\frac {\d \overline f} {\d x} } b - \overline {\map f b}

\map {\frac {\d g} {\d x} } b} - \map p a \paren {\map g a \map {\frac {\d \overline f} {\d x} } a - \overline {\map f a} \map {\frac {\d g} {\d x} } a}$

The of this equation vanishes because of the boundary conditions, which are either:


 * periodic boundary conditions, that is, that $\map f x$, $\map g x$, and their first derivatives (as well as $\map p x$) have the same values at $x=b$ as at $x=a$, or


 * that independently at $x=a$ and at $x=b$ either:


 * the condition cited in equation $(1)$ or $(2)$ holds or:


 * $\map p x = 0$.

So:
 * $\ds \paren {\mu - \overline \lambda} \int \nolimits_a^b \overline {\map f x} \map g x \map w x \rd x = 0$

If we set $f = g$, so that the integral surely is non-zero, then it follows that $\overline \lambda = \lambda$; that is, the eigenvalues are real, making the differential operator in the Sturm-Liouville equation self-adjoint (hermitian); so:


 * $\ds \paren {\mu - \lambda} \int \nolimits_a^b \overline {\map f x} \map g x \map w x \rd x = 0$

It follows that, if $f$ and $g$ have distinct eigenvalues, then they are orthogonal.