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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:


 * $ \displaystyle -\frac{d}{dx}\left[{p(x)\frac{dy}{ dx} }\right]+q\left({x}\right)y=\lambda w\left({x}\right)y $

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

Assume that the Sturm-Liouville equation is regular, that is, p(x)&minus;1 > 0, q(x), and w(x) > 0 are real-valued integrable functions over the finite interval [a, b], with separated boundary conditions of the form:


 * $ \displaystyle y\left({a}\right)\cos \alpha - p\left({a}\right)y^\prime \left({a}\right)\sin \alpha = 0 $ ... (1)


 * $ \displaystyle y\left({b}\right)\cos \beta - p\left({b}\right)y^\prime \left({b}\right)\sin \beta = 0 $ ... (2)

where $\alpha, \beta \in [0, \pi) $

Then,


 * $ \displaystyle \langle f, g\rangle = \int_{a}^{b} \overline{f(x)} g(x)w(x)\,dx \ = \ 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{d\left({p\left({x}\right) \frac{dg}{dx}

\left({x}\right)}\right) }{dx} +\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) \ dx =p\left({ b}\right) \left[ g\left({ b}\right) \frac{d\bar{f} }{dx} \left({ b}\right) -\bar{f} \left({ b}\right) \frac{dg}{dx} \left({ b}\right) \right] -p\left({ a}\right) \left[ g\left({ a}\right) \frac{d\bar{f} }{dx} \left({ a}\right) -\bar{f} \left({ a}\right) \frac{dg}{dx} \left({ a}\right) \right] $.

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


 * $ \displaystyle \bullet $ 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


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


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


 * $ \displaystyle \bullet $ $p \left({x}\right)=0$.

So:
 * $\left({ \mu -\bar{\lambda} }\right) \int\nolimits_{a}^{b}\bar{f} \left({x}\right) g\left({x}\right) w\left({x}\right) \ dx =0$

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:


 * $\left({ \mu -\lambda }\right) \int\nolimits_{a}^{b}\bar{f} \left({x}\right)

g\left({x}\right) w\left({x}\right) \ dx =0$

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