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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
$ \langle f, g\rangle = \int_{a}^{b} \overline{f(x)} g(x)w(x)\,dx \ = \ 0$, where f(x) and g(x) are solutions to the Sturm-Liouville equation corresponding to distinct eigenvalues and w(x) is the "weight" or "density" function.

Proof
Let f(x) and g(x) be solutions of the Sturm-Liouville equation (1)  corresponding to eigenvalues $\lambda $ and $ \mu $ respectively. Multiply the equation for g(x) by f (x) (the complex conjugate of f(x)) to get:

$-\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(x), g(x), $\lambda $, and $\mu $ may be complex; all other quantities are real.) Complex conjugate this equation, exchange f(x) and g(x), and subtract the new equation from the original:

$-\bar{f} \left( x\right) \frac{d\left( p\left( x\right) \frac{dg}{dx} \left( x\right) \right) }{dx} +g\left( x\right) \frac{d\left( p\left( x\right) \frac{d\bar{f} }{dx} \left( x\right) \right) }{dx} =\frac{d\left( p\left( x\right) \left[ g\left( x\right) \frac{d\bar{f} }{dx} \left( x\right) -\bar{f} \left( x\right) \frac{dg}{dx} \left( x\right) \right] \right) }{dx} =\left( \mu -\bar{\lambda} \right) \bar{f} \left( x\right) g\left( x\right) w\left( x\right) $ Integrate this between the limits $x=a$ and $x=b$

$\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:


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


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


 * $\bullet $ the condition cited in equation (2)  or  (3)  holds or:
 * $\bullet $ $p(x)=0$.

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 &#955; =&#955; 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. QED.