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

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 $f \left({x}\right)$ and $g \left({x}\right)$ be solutions of the Sturm-Liouville equation:


 * $ (1): \quad \displaystyle - \frac {\mathrm d} {\mathrm d x} \left({p \left({x}\right) \frac {\mathrm d y} {\mathrm d x}}\right) + q \left({x}\right) y = \lambda w \left({x}\right) y$

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

The functions $p \left({x}\right)$, $q \left({x}\right)$ and $w \left({x}\right)$ are specified.

In the simplest cases they are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

In addition:


 * $(1a): \quad p \left({x}\right) > 0$ has a continuous derivative


 * $(1b): \quad w \left({x}\right) > 0$


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

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


 * $ (2): \quad \displaystyle y\left({a}\right)\cos \alpha - p\left({a}\right)y^\prime \left({a}\right)\sin \alpha = 0 $


 * $ (3): \quad \displaystyle y\left({b}\right)\cos \beta - p\left({b}\right)y^\prime \left({b}\right)\sin \beta = 0 $

where $\alpha, \beta \in \left[{0 \,.\,.\, \pi}\right)$.

Then:
 * $\displaystyle \langle f, g \rangle = \int_a^b \overline {f \left({x}\right)} q \left({x}\right) w \left({x}\right) \, \mathrm d x = 0$

where $f \left({x}\right)$ and $g \left({x}\right)$ are solutions to the Sturm-Liouville equation corresponding to distinct eigenvalues and $w \left({x}\right)$ is the "weight" or "density" function.

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


 * $ \displaystyle -\bar{f} \left({x}\right) \frac {\mathrm d \left({p\left({x}\right) \frac{\mathrm d g} {\mathrm d x}

\left({x}\right)}\right) }{\mathrm d x} +\bar{f} \left({x}\right) q \left({x}\right) g \left({x}\right) = \mu \bar{f} \left({x}\right) w \left({x}\right) g \left({x}\right)$

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

Complex conjugate this equation, exchange $f \left({x}\right)$ and $g \left({x}\right)$, and subtract the new equation from the original:

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


 * $\displaystyle \left({ \mu -\bar{\lambda} }\right) \int \nolimits_a^b \bar{f} \left({x}\right) g\left({x}\right) w \left({x}\right) \ \mathrm d x = p\left({b}\right) \left({g \left({ b}\right) \frac{\mathrm d \bar{f} }{\mathrm d x} \left({b}\right) - \bar{f} \left({b}\right) \frac{\mathrm d g}{\mathrm d x} \left({b}\right)} \right) -p \left({a}\right) \left({g \left({a}\right) \frac {\mathrm d \bar{f} }{\mathrm d x} \left({a}\right) - \bar{f} \left({a}\right) \frac{\mathrm d g}{\mathrm d x} \left({a}\right)}\right)$

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


 * periodic boundary conditions, i.e., that $f \left({x}\right)$, $g \left({x}\right)$, and their first derivatives (as well as $p \left({x}\right)$) 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:
 * $p \left({x}\right)=0$.

So:
 * $\displaystyle \left({ \mu -\bar{\lambda} }\right) \int\nolimits_a^b \bar{f} \left({x}\right) g \left({x}\right) w \left({x}\right) \ \mathrm d x = 0$

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:


 * $\displaystyle \left({ \mu -\lambda }\right) \int\nolimits_a^b \bar{f} \left({x}\right) g \left({x}\right) w \left({x}\right) \ \mathrm d x = 0$

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