Orthogonality of Solutions to the Sturm-Liouville Equation with Distinct Eigenvalues

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Theorem

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

$(1): \quad -\map {\dfrac \d {\d x} } {\map p x \dfrac {\d y} {\d x} } + \map q x y = \lambda \map w x y$

where $y$ is a function of the free variable $x$.

The functions $\map p x$, $\map q x$ and $\map w x$ are specified.

In the simplest cases they are continuous on the closed interval $\closedint a b$.

In addition:

$(1a): \quad \map p x > 0$ has a continuous derivative

$(1b): \quad \map w x > 0$

$(1c): \quad y$ is typically required to satisfy some boundary conditions at $a$ and $b$.

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

$(2): \quad \map y a \cos \alpha - \map p a \map {y'} a \sin \alpha = 0$

$(3): \quad \map y b \cos \beta - \map p b \map {y'} b \sin \beta = 0$

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

Then:

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

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:

\(\ds -\overline {\map f x} \frac {\map \d {\map p x \map {\frac {\d g} {\d x} } x} } {\d x} + \map g x \frac {\map \d {\map p x \map {\frac {\d \bar f} {\d x} } x} } {\d x}\)

\(=\)

\(\ds \frac {\map \d {\map p x \paren {\map g x \map {\frac {\d \bar f} {\d x} } x - \overline {\map f x} \map {\frac {\d g} {\d x} } x} } } {\d x}\)

\(\ds \)

\(=\)

\(\ds \paren {\mu - \bar \lambda} \overline {\map f x} \map g x \map w x\)

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

$\ds \paren {\mu - \bar \lambda} \int_a^b \overline {\map f x} \map g x \map w x \rd x = \map p b \paren {\map g b \map {\frac {\d \bar f} {\d x} } b - \overline {\map f b} \map {\frac {\d g} {\d x} } b} - \map p a \paren {\map g a \map {\frac {\d \bar f} {\d x} } a - \overline {\map f a} \map {\frac {\d g} {\d x} } a}$

The right side 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 $(2)$ or $(3)$ holds

or:

$\map p x = 0$.

So:

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

If we set $f = g$, so that the integral surely is non-zero, then it follows that $\bar \lambda = \lambda$.

That is, the eigenvalues are real, making the differential operator in the Sturm-Liouville equation self-adjoint (hermitian).

So:

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

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

$\blacksquare$