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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 λ =λ 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.