path: root/source/know/concept/sturm-liouville-theory/index.md

---
title: "Sturm-Liouville theory"
sort_title: "Sturm-Liouville theory"
date: 2021-02-23
categories:
- Mathematics
- Physics
layout: "concept"
---

**Sturm-Liouville theory** extends
the concept of Hermitian matrix eigenvalue problems
to linear second-order ordinary differential equations.

It states that, given suitable boundary conditions,
any such equation can be rewritten using the **Sturm-Liouville operator**,
and that the corresponding eigenvalue problem,
known as a **Sturm-Liouville problem**,
will give real eigenvalues and a complete set of eigenfunctions.



## General operator

Consider the most general form of a second-order linear
differential operator $$\hat{L}$$, where $$p_0(x)$$, $$p_1(x)$$, and $$p_2(x)$$
are real functions of $$x \in [a,b]$$ and are nonzero for all $$x \in \,\,]a, b[$$:

$$\begin{aligned}
    \hat{L} \{u(x)\}
    \equiv p_2(x) \: u''(x) + p_1(x) \: u'(x) + p_0(x) \: u(x)
\end{aligned}$$

Analogously to matrices,
we now define its **adjoint** operator $$\hat{L}^\dagger$$ as follows:

$$\begin{aligned}
    \inprod{\hat{L}^\dagger f}{g}
    \equiv \inprod{f}{\hat{L} g}
\end{aligned}$$

What is $$\hat{L}^\dagger$$, given the above definition of $$\hat{L}$$?
We start from the inner product $$\inprod{f}{\hat{L} g}$$:

$$\begin{aligned}
    \inprod{f}{\hat{L} g}
    &= \int_a^b f^*(x) \hat{L}\{g(x)\} \dd{x}
    = \int_a^b (f^* p_2) g'' + (f^* p_1) g' + (f^* p_0) g \dd{x}
    \\
    &= \Big[ (f^* p_2) g' + (f^* p_1) g \Big]_a^b - \int_a^b (f^* p_2)' g' + (f^* p_1)' g - (f^* p_0) g \dd{x}
    \\
    &= \Big[ f^* (p_2 g' + p_1 g) - (f^* p_2)' g \Big]_a^b + \int_a^b \! \Big( (f p_2)'' - (f p_1)' + (f p_0) \Big)^* g \dd{x}
    \\
    &= \Big[ f^* \big( p_2 g' + (p_1 - p_2') g \big) - (f^*)' p_2 g \Big]_a^b + \int_a^b \Big( \hat{L}^\dagger\{f\} \Big)^* g \dd{x}
\end{aligned}$$

The newly-formed operator on $$f$$ must be $$\hat{L}^\dagger$$,
but there is an additional boundary term.
To fix this, we demand that $$p_1(x) = p_2'(x)$$
and that $$\big[ p_2 (f^* g' - (f^*)' g) \big]_a^b = 0$$, leaving:

$$\begin{aligned}
    \inprod{f}{\hat{L} g}
    &= \Big[ f^* \big( p_2 g' + (p_1 - p_2') g \big) - (f^*)' p_2 g \Big]_a^b + \inprod{\hat{L}^\dagger f}{g}
    \\
    &= \Big[ p_2 \big( f^* g' - (f^*)' g \big) \Big]_a^b + \inprod{\hat{L}^\dagger f}{g}
    \\
    &= \inprod{\hat{L}^\dagger f}{g}
\end{aligned}$$

Let us look at the expression for $$\hat{L}^\dagger$$ we just found,
with the restriction $$p_1 = p_2'$$ in mind:

$$\begin{aligned}
    \hat{L}^\dagger \{f\}
    &= (p_2 f)'' - (p_1 f)' + (p_0 f)
    \\
    &= (p_2'' f + 2 p_2' f' + p_2 f'') - (p_1' f + p_1 f') + (p_0 f)
    \\
    &= p_2 f'' + (2 p_2' - p_1) f' + (p_2'' - p_1' + p_0) f
    \\
    &= p_2 f'' + p_1 f' + p_0 f
    \\
    &= \hat{L}\{f\}
\end{aligned}$$

So $$\hat{L}$$ is **self-adjoint**, i.e. $$\hat{L}^\dagger$$ is the same as $$\hat{L}$$!
Indeed, every such second-order linear operator is self-adjoint
if it satisfies the constraints $$p_1 = p_2'$$ and $$\big[ p_2 (f^* g' - (f^*)' g) \big]_a^b = 0$$.

But what if $$p_1 \neq p_2'$$?
Let us multiply $$\hat{L}$$ by an unknown $$p(x) \neq 0$$
and divide by $$p_2(x) \neq 0$$:

$$\begin{aligned}
    \frac{p}{p_2} \hat{L} \{u\}
    = p u'' + p \frac{p_1}{p_2} u' + p \frac{p_0}{p_2} u
\end{aligned}$$

We now demand that the derivative $$p'(x)$$ of the unknown $$p(x)$$ satisfies:

$$\begin{aligned}
    p'(x)
    = p(x) \frac{p_1(x)}{p_2(x)}
    \quad \implies \quad
    \frac{p_1(x)}{p_2(x)} \dd{x}
    = \frac{1}{p(x)} \dd{p}
\end{aligned}$$

Taking the indefinite integral of this differential equation
yields an expression for $$p(x)$$:

$$\begin{aligned}
    \int \frac{p_1(x)}{p_2(x)} \dd{x}
    = \int \frac{1}{p} \dd{p}
    = \ln\!\big( p(x) \big)
    \quad \implies \quad
    \boxed{
        p(x)
        = \exp\!\bigg( \int \frac{p_1(x)}{p_2(x)} \dd{x} \bigg)
    }
\end{aligned}$$

We define an additional function $$q(x)$$
based on the last term of $$(p / p_2) \hat{L}$$ shown above:

$$\begin{aligned}
    \boxed{
        q(x)
        \equiv p(x) \frac{p_0(x)}{p_2(x)}
    }
    = \frac{p_0(x)}{p_2(x)} \exp\!\bigg( \int \frac{p_1(x)}{p_2(x)} \dd{x} \bigg)
\end{aligned}$$

When rewritten using $$p$$ and $$q$$,
the modified operator $$(p / p_2) \hat{L}$$ looks like this:

$$\begin{aligned}
    \frac{p}{p_2} \hat{L} \{u\}
    = p u'' + p' u' + q u
    = (p u')' + q u
\end{aligned}$$

This is the self-adjoint form from earlier!
So even if $$p_1 \neq p_2'$$, any second-order linear operator
with $$p_2(x) \neq 0$$ can easily be made self-adjoint.
The resulting general form is called the **Sturm-Liouville operator** $$\hat{L}_\mathrm{SL}$$,
for nonzero $$p(x)$$:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \hat{L}_\mathrm{SL} \{u(x)\}
            &= \Big( p(x) \: u'(x) \Big)' + q(x) \: u(x)
            \\
            &= \hat{L}_\mathrm{SL}^\dagger \{u(x)\}
        \end{aligned}
    }
\end{aligned}$$

Still subject to the constraint $$\big[ p (f^* g' - (f^*)' g) \big]_a^b = 0$$
such that $$\inprod{f}{\hat{L}_\mathrm{SL} g} = \inprod{\hat{L}_\mathrm{SL}^\dagger f}{g}$$.



## Eigenvalue problem

An eigenvalue problem of $$\hat{L}_\mathrm{SL}$$
is called a **Sturm-Liouville problem** (SLP).
The goal is to find the **eigenvalues** $$\lambda$$
and corresponding **eigenfunctions** $$u(x)$$ that fulfill:

$$\begin{aligned}
    \boxed{
        \hat{L}_\mathrm{SL}\{u(x)\} = - \lambda \: w(x) \: u(x)
    }
\end{aligned}$$

Where $$w(x)$$ is a real weight function satisfying $$w(x) > 0$$ for $$x \in \,\,]a, b[$$.
By convention, the trivial solution $$u = 0$$ is not valid.
Some authors have the opposite sign for $$\lambda$$ and/or $$w$$.

In our derivation of $$\hat{L}_\mathrm{SL}$$ above,
we imposed the constraint $$\big[ p (f^* g' - (f')^* g) \big]_a^b = 0$$ to ensure that
$$\inprod{\hat{L}_\mathrm{SL}^\dagger f}{g} = \inprod{f}{\hat{L}_\mathrm{SL} g}$$.
Consequently, to have a valid SLP,
the boundary conditions (BCs) on $$u$$ must be such that,
for any two (possibly identical) eigenfunctions $$u_m$$ and $$u_n$$, we have:

$$\begin{aligned}
    \Big[ p(x) \big( u_m^*(x) \: u_n'(x) - \big(u_m'(x)\big)^* u_n(x) \big) \Big]_a^b = 0
\end{aligned}$$

There are many boundary conditions that satisfy this requirement.
Some notable ones are listed non-exhaustively below.
Verify for yourself that these work:

+ **Dirichlet BCs**: $$u(a) = u(b) = 0$$
+ **Neumann BCs**: $$u'(a) = u'(b) = 0$$
+ **Robin BCs**: $$\alpha_1 u(a) + \beta_1 u'(a) = \alpha_2 u(b) + \beta_2 u'(b) = 0$$ with $$\alpha_{1,2}, \beta_{1,2} \in \mathbb{R}$$
+ **Periodic BCs**: $$p(a) = p(b)$$, $$u(a) = u(b)$$, and $$u'(a) = u'(b)$$
+ **Legendre "BCs"**: $$p(a) = p(b) = 0$$

If this is fulfilled, Sturm-Liouville theory gives us
useful information about $$\lambda$$ and $$u$$.
By definition, the following must be satisfied
for two arbitrary eigenfunctions $$u_m$$ and $$u_n$$:

$$\begin{aligned}
    0
    &= \hat{L}_\mathrm{SL}\{u_m^*\} + \lambda_m^* w u_m^*
    \\
    &= \hat{L}_\mathrm{SL}\{u_n\} + \lambda_n w u_n
\end{aligned}$$

We multiply each by the other eigenfunction,
subtract the results, and integrate:

$$\begin{aligned}
    0
    &= \int_a^b u_m^* \big(\hat{L}_\mathrm{SL}\{u_n\} + \lambda_n w u_n\big)
    - u_n \big(\hat{L}_\mathrm{SL}\{u_m^*\} + \lambda_m^* w u_m^*\big) \dd{x}
    \\
    &= \int_a^b u_m^* \hat{L}_\mathrm{SL}\{u_n\} - u_n \hat{L}_\mathrm{SL}\{u_m^*\}
    + (\lambda_n - \lambda_m^*) u_m^* w u_n \dd{x}
    \\
    &= \inprod{u_m}{\hat{L}_\mathrm{SL} u_n} - \inprod{\hat{L}_\mathrm{SL} u_m}{u_n}
    + (\lambda_n - \lambda_m^*) \inprod{u_m}{w u_n}
\end{aligned}$$

The operator $$\hat{L}_\mathrm{SL}$$ is self-adjoint of course,
so the first two terms vanish, leaving us with:

$$\begin{aligned}
    0
    &= (\lambda_n - \lambda_m^*) \inprod{u_m}{w u_n}
\end{aligned}$$

When $$m = n$$, we get $$\inprod{u_n}{w u_n} > 0$$,
so the equation is only satisfied if $$\lambda_n^* = \lambda_n$$,
meaning the eigenvalue $$\lambda_n$$ is real for any $$n$$.
When $$m \neq n$$, then $$\lambda_n - \lambda_m^*$$
may or may not be zero depending on the degeneracy.
If there is no degeneracy, then $$\lambda_n - \lambda_m^* \neq 0$$,
meaning $$\inprod{u_m}{w u_n} = 0$$, i.e. the eigenfunctions are orthogonal.
In case of degeneracy, manual orthogonalization is needed,
which is guaranteed to be doable using the [Gram-Schmidt method](/know/concept/gram-schmidt-method/).

In conclusion, an SLP has **real eigenvalues**
and **orthogonal eigenfunctions**: for all $$m$$, $$n$$:

$$\begin{aligned}
    \boxed{
        \lambda_n \in \mathbb{R}
    }
    \qquad\qquad
    \boxed{
        \inprod{u_m}{w u_n}
        = A_n \delta_{nm}
    }
\end{aligned}$$

When solving a differential eigenvalue problem,
knowing that all eigenvalues are real is a huge simplification,
so it is always worth checking whether you are dealing with an SLP.

Another useful fact:
it turns out that SLPs always have an infinite number of *discrete* eigenvalues.
Furthermore, there always exists a *lowest* eigenvalue $$\lambda_0 > -\infty$$,
called the **ground state**.



## Complete basis

Not only are an SLP's eigenfunctions orthogonal,
they also form a **complete basis**, meaning any well-behaved $$f(x)$$
can be expanded as a **generalized Fourier series** with coefficients $$a_n$$:

$$\begin{aligned}
    \boxed{
        f(x)
        = \sum_{n = 0}^\infty a_n u_n(x)
        \quad \mathrm{for} \: x \in \,\,]a, b[
    }
\end{aligned}$$

This series converges faster if $$f$$ satisfies the same BCs as $$u_n$$;
in that case the expansion is also valid for the inclusive interval $$x \in [a, b]$$.

To find an expression for the coefficients $$a_n$$,
we multiply the above generalized Fourier series by $$u_m^* w$$
and integrate it to get inner products on both sides:

$$\begin{aligned}
    u_m^* w f
    &= \sum_{n = 0}^\infty a_n u_m^* w u_n
    \\
    \int_a^b u_m^* w f \dd{x}
    &= \int_a^b \bigg( \sum_{n = 0}^\infty a_n u_m^* w u_n \bigg) \dd{x}
    \\
    \inprod{u_m}{w f}
    &= \sum_{n = 0}^\infty a_n \inprod{u_m}{w u_n}
\end{aligned}$$

Because the eigenfunctions of an SLP are mutually orthogonal,
the summation disappears:

$$\begin{aligned}
    \inprod{u_m}{w f}
    &= \sum_{n = 0}^\infty a_n \inprod{u_m}{w u_n}
    = \sum_{n = 0}^\infty a_n A_n \delta_{nm}
    = a_m A_m
\end{aligned}$$

After isolating this for $$a_m$$, we see that
the coefficients are given by the projection of the target
function $$f$$ onto the normalized eigenfunctions $$u_m / A_m$$:

$$\begin{aligned}
    \boxed{
        a_n
        = \frac{\inprod{u_n}{w f}}{A_n}
        = \frac{\inprod{u_n}{w f}}{\inprod{u_n}{w u_n}}
    }
\end{aligned}$$

As a final remark, we can see something interesting
by rearranging the generalized Fourier series
after inserting the expression for $$a_n$$:

$$\begin{aligned}
    f(x)
    &= \sum_{n = 0}^\infty \frac{1}{A_n} \inprod{u_n}{w f} u_n(x)
    \\
    &= \int_a^b \bigg(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) \: w(\xi) \: f(\xi) \: u_n(x) \bigg) \dd{\xi}
    \\
    &= \int_a^b f(\xi) \bigg(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) \: w(\xi) \: u_n(x) \bigg) \dd{\xi}
\end{aligned}$$

Upon closer inspection, the parenthesized summation
must be the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$
for the integral to work out.
In fact, this is the underlying requirement for completeness:

$$\begin{aligned}
    \boxed{
        \sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) \: w(\xi) \: u_n(x) = \delta(x - \xi)
    }
\end{aligned}$$



## References
1.  O. Bang,
    *Applied mathematics for physicists: lecture notes*, 2019,
    unpublished.