# Sturm-Liouville theory/Proofs

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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

${\displaystyle \langle f,g\rangle =\int _{a}^{b}{\overline {f(x)}}g(x)w(x)\,dx\ =\ 0}$ , 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 ${\displaystyle \lambda }$ and ${\displaystyle \mu }$ respectively. Multiply the equation for g(x) by f(x) (the complex conjugate of f(x)) 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(x), g(x), ${\displaystyle \lambda }$, and ${\displaystyle \mu }$ 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:

${\displaystyle -{\bar {f}}\left(x\right){\frac {d\left(p\left(x\right){\frac {dg}{dx}}\left(x\right)\right)}{dx}}+g\left(x\right){\frac {d\left(p\left(x\right){\frac {d{\bar {f}}}{dx}}\left(x\right)\right)}{dx}}={\frac {d\left(p\left(x\right)\left[g\left(x\right){\frac {d{\bar {f}}}{dx}}\left(x\right)-{\bar {f}}\left(x\right){\frac {dg}{dx}}\left(x\right)\right]\right)}{dx}}=\left(\mu -{\bar {\lambda }}\right){\bar {f}}\left(x\right)g\left(x\right)w\left(x\right)}$

Integrate this between the limits ${\displaystyle x=a}$ and ${\displaystyle x=b}$

${\displaystyle \left(\mu -{\bar {\lambda }}\right)\int \nolimits _{a}^{b}{\bar {f}}\left(x\right)g\left(x\right)w\left(x\right)dx=p\left(b\right)\left[g\left(b\right){\frac {d{\bar {f}}}{dx}}\left(b\right)-{\bar {f}}\left(b\right){\frac {dg}{dx}}\left(b\right)\right]-p\left(a\right)\left[g\left(a\right){\frac {d{\bar {f}}}{dx}}\left(a\right)-{\bar {f}}\left(a\right){\frac {dg}{dx}}\left(a\right)\right]}$ .

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

${\displaystyle \bullet }$ periodic boundary conditions, i.e., that f(x), g(x), and their first derivatives (as well as p(x)) have the same values at ${\displaystyle x=b}$ as at ${\displaystyle x=a}$, or
${\displaystyle \bullet }$ that independently at ${\displaystyle x=a}$ and at ${\displaystyle x=b}$ either:
${\displaystyle \bullet }$ the condition cited in equation (2) or (3) holds or:
${\displaystyle \bullet }$ ${\displaystyle p(x)=0}$.

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

If we set ${\displaystyle f=g}$ , 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:

${\displaystyle \left(\mu -\lambda \right)\int \nolimits _{a}^{b}{\bar {f}}\left(x\right)g\left(x\right)w\left(x\right)dx=0}$

It follows that, if ${\displaystyle f}$ and ${\displaystyle g}$ have distinct eigenvalues, then they are orthogonal. QED.