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+---
+title: "Sturm-Liouville theory"
+date: 2021-02-23
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**Sturm-Liouville theory** defines the analogue of Hermitian matrix
+eigenvalue problems for linear second-order ODEs.
+
+It states that, given suitable boundary conditions, any linear
+second-order ODE 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 $x \in [a,b]$ which are non-zero for all $x \in ]a, b[$:
+
+$$\begin{aligned}
+    \hat{L} \{u(x)\} = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x)
+\end{aligned}$$
+
+We now define the **adjoint** or **Hermitian** operator
+$\hat{L}^\dagger$ analogously to matrices:
+
+$$\begin{aligned}
+    \inprod{f}{\hat{L} g}
+    = \inprod{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+What is $\hat{L}^\dagger$, given the above definition of $\hat{L}$?
+We start from the inner product $\inprod{f}{\hat{L} g}$:
+
+$$\begin{aligned}
+    \inprod{f}{\hat{L} g}
+    &= \int_a^b f^*(x) \hat{L}\{g(x)\} \dd{x}
+    = \int_a^b (f^* p_0) g'' + (f^* p_1) g' + (f^* p_2) g \dd{x}
+    \\
+    &= \big[ (f^* p_0) g' + (f^* p_1) g \big]_a^b - \int_a^b (f^* p_0)' g' + (f^* p_1)' g - (f^* p_2) g \dd{x}
+    \\
+    &= \big[ f^* \big( p_0 g' \!+\! p_1 g \big) \!-\! (f^* p_0)' g \big]_a^b + \int_a^b \! \big( (f p_0)'' - (f p_1)' + (f p_2) \big)^* g \dd{x}
+    \\
+    &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x}
+\end{aligned}$$
+
+We now have an expression for $\hat{L}^\dagger$, but are left with an
+annoying boundary term:
+
+$$\begin{aligned}
+    \inprod{f}{\hat{L} g}
+    &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \inprod{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+To fix this,
+let us demand that $p_1(x) = p_0'(x)$ and that
+$[p_0(f^* g' - (f^*)' g)]_a^b = 0$, leaving:
+
+$$\begin{aligned}
+    \inprod{f}{\hat{L} g}
+    &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \Inprod{\hat{L}^\dagger f}{g}
+    = \inprod{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+Using the aforementioned restriction $p_1(x) = p_0'(x)$,
+we then take a look at the definition of $\hat{L}^\dagger$:
+
+$$\begin{aligned}
+    \hat{L}^\dagger \{f\}
+    &= (p_0 f)'' - (p_1 f)' + (p_2 f)
+    \\
+    &= p_0 f'' + (2 p_0' - p_1) f' + (p_0'' - p_1' + p_2) f
+    \\
+    &= p_0 f'' + p_0' f' + p_2 f
+    \\
+    &= (p_0 f')' + p_2 f
+\end{aligned}$$
+
+The original operator $\hat{L}$ reduces to the same form,
+so it is **self-adjoint**:
+
+$$\begin{aligned}
+    \hat{L} \{f\}
+    &= p_0 f'' + p_0' f' + p_2 f
+    = (p_0 f')' + p_2 f
+    = \hat{L}^\dagger \{f\}
+\end{aligned}$$
+
+Consequently, every such second-order linear operator $\hat{L}$ is self-adjoint,
+as long as it satisfies the constraints $p_1(x) = p_0'(x)$ and $[p_0 (f^* g' - (f^*)' g)]_a^b = 0$.
+
+Let us ignore the latter constraint for now (it will return later),
+and focus on the former: what if $\hat{L}$ does not satisfy $p_0' \neq p_1$?
+We multiply it by an unknown $p(x) \neq 0$, and divide by $p_0(x) \neq 0$:
+
+$$\begin{aligned}
+    \frac{p(x)}{p_0(x)} \hat{L} \{u\} = p(x) u'' + p(x) \frac{p_1(x)}{p_0(x)} u' + p(x) \frac{p_2(x)}{p_0(x)} u
+\end{aligned}$$
+
+We now define $q(x)$,
+and demand that the derivative $p'(x)$ of the unknown $p(x)$ satisfies:
+
+$$\begin{aligned}
+    q(x) = p(x) \frac{p_2(x)}{p_0(x)}
+    \qquad
+    p'(x) = p(x) \frac{p_1(x)}{p_0(x)}
+\end{aligned}$$
+
+The latter is a differential equation for $p(x)$, which we solve by integration:
+
+$$\begin{gathered}
+    \frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x}
+    \quad \implies \quad
+    \frac{p_1(x)}{p_0(x)} \dd{x} = \frac{1}{p(x)} \dd{p}
+    \\
+    \implies \quad
+    \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} = \int_{p(a)}^{p(x)} \frac{1}{f} \dd{f}
+    = \ln\!\Big( \frac{p(x)}{p(a)} \Big)
+    \\
+    \implies \quad
+    p(x) = p(a) \exp\!\Big( \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} \Big)
+\end{gathered}$$
+
+Now that we have $p(x)$ and $q(x)$, we can define a new operator $\hat{L}_p$ as follows:
+
+$$\begin{aligned}
+    \hat{L}_p \{u\}
+    = \frac{p}{p_0} \hat{L} \{u\}
+    = p u'' + p' u' + q u
+    = (p u')' + q u
+\end{aligned}$$
+
+This is the self-adjoint form from earlier!
+So even if $p_0' \neq p_1$, any second-order linear operator with $p_0(x) \neq 0$
+can easily be put in self-adjoint form.
+
+This general form is known as the **Sturm-Liouville operator** $\hat{L}_{SL}$,
+where $p(x)$ and $q(x)$ are non-zero real functions of the variable $x \in [a,b]$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{L}_{SL} \{u(x)\}
+        = \frac{d}{dx}\Big( p(x) \frac{du}{dx} \Big) + q(x) u(x)
+        = \hat{L}_{SL}^\dagger \{u(x)\}
+    }
+\end{aligned}$$
+
+
+## Eigenvalue problem
+
+A **Sturm-Liouville problem** (SLP) is analogous to a matrix eigenvalue problem,
+where $w(x)$ is a real weight function, $\lambda$ is the **eigenvalue**,
+and $u(x)$ is the corresponding **eigenfunction**:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{L}_{SL}\{u(x)\} = - \lambda w(x) u(x)
+    }
+\end{aligned}$$
+
+Necessarily, $w(x) > 0$ except in isolated points, where $w(x) = 0$ is allowed;
+the point is that any inner product $\Inprod{f}{w g}$ may never be zero due to $w$'s fault.
+Furthermore, the convention is that $u(x)$ cannot be trivially zero.
+
+In our derivation of $\hat{L}_{SL}$,
+we removed a boundary term to get self-adjointness.
+Consequently, to have a valid SLP, the boundary conditions for
+$u(x)$ must be as follows, otherwise the operator cannot be self-adjoint:
+
+$$\begin{aligned}
+    \Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0
+\end{aligned}$$
+
+There are many boundary conditions (BCs) which satisfy this requirement.
+Some notable ones are listed here non-exhaustively:
+
++ **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"**: $p(a) = p(b) = 0$
+
+Once this requirement is satisfied, Sturm-Liouville theory gives us
+some very useful information about $\lambda$ and $u(x)$.
+From the definition of an SLP, we know that, given two arbitrary (and possibly identical)
+eigenfunctions $u_n$ and $u_m$, the following must be satisfied:
+
+$$\begin{aligned}
+    0 = \hat{L}_{SL}\{u_n\} + \lambda_n w u_n = \hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*
+\end{aligned}$$
+
+We subtract these expressions, multiply by the eigenfunctions, and integrate:
+
+$$\begin{aligned}
+    0
+    &= \int_a^b u_m^* \big(\hat{L}_{SL}\{u_n\} + \lambda_n w u_n\big) - u_n \big(\hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*\big) \:dx
+    \\
+    &= \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} + u_n u_m^* w (\lambda_n - \lambda_m^*) \:dx
+\end{aligned}$$
+
+Rearranging this a bit reveals that these are in fact three inner products:
+
+$$\begin{aligned}
+    \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} \:dx
+    &= (\lambda_m^* - \lambda_n) \int_a^b u_n u_m^* w \:dx
+    \\
+    \inprod{u_m}{\hat{L}_{SL} u_n} - \inprod{\hat{L}_{SL} u_m}{u_n}
+    &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
+\end{aligned}$$
+
+The operator $\hat{L}_{SL}$ is self-adjoint by definition,
+so the left-hand side vanishes, leaving us with:
+
+$$\begin{aligned}
+    0
+    &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
+\end{aligned}$$
+
+When $m = n$, the inner product $\Inprod{u_n}{w u_n}$ is real and positive
+(assuming $u_n$ is not trivially zero, in which case it would be disqualified anyway).
+In this case we thus know that $\lambda_n^* = \lambda_n$,
+i.e. the eigenvalue $\lambda_n$ is real for any $n$.
+
+When $m \neq n$, then $\lambda_m^* - \lambda_n$ may or may not be zero,
+depending on the degeneracy. If there is no degeneracy, we
+see that $\Inprod{u_m}{w u_n} = 0$, i.e. the eigenfunctions are orthogonal.
+
+In case of degeneracy, manual orthogonalization is needed, but as it turns out,
+this is guaranteed to be doable, using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method/).
+
+In conclusion, **a Sturm-Liouville problem has real eigenvalues $\lambda$,
+and all the corresponding eigenfunctions $u(x)$ are mutually orthogonal**:
+
+$$\begin{aligned}
+    \boxed{
+        \Inprod{u_m(x)}{w(x) u_n(x)}
+        = \Inprod{u_n}{w u_n} \delta_{nm}
+        = A_n \delta_{nm}
+    }
+\end{aligned}$$
+
+When you're 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 of SLPs is that they always
+have an infinite number of discrete eigenvalues.
+Furthermore, the eigenvalues always ascend to $+\infty$;
+in other words, there always exists a *lowest* eigenvalue $\lambda_0 > -\infty$,
+known as the **ground state**.
+
+
+## Completeness
+
+Not only are the eigenfunctions $u_n(x)$ of an SLP orthogonal, they
+also form a **complete basis**, meaning that any well-behaved function $f(x)$ 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 will converge significantly faster if $f(x)$
+satisfies the same BCs as $u_n(x)$. In that case the
+expansion will even be 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 $w(x) u_m^*(x)$ for an arbitrary $m$:
+
+$$\begin{aligned}
+    f(x) w(x) u_m^*(x)
+    &= \sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)
+\end{aligned}$$
+
+By integrating we get inner products on both the left and the right:
+
+$$\begin{aligned}
+    \int_a^b f(x) w(x) u_m^*(x) \dd{x}
+    &= \int_a^b \Big(\sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)\Big) \dd{x}
+    \\
+    \Inprod{u_m}{w f}
+    &= \sum_{n = 0}^\infty a_n \Inprod{u_m}{w u_n}
+\end{aligned}$$
+
+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 $a_n$, we see that
+the coefficients are given by the projection of the target
+function $f(x)$ onto the normalized eigenfunctions $u_n(x) / A_n$:
+
+$$\begin{aligned}
+    \boxed{
+        a_n
+        = \frac{\Inprod{u_n}{w f}}{A_n}
+        = \frac{\Inprod{u_n}{w f}}{\Inprod{u_n}{w u_n}}
+    }
+\end{aligned}$$
+
+As a final remark, we can see something interesting
+by rearranging the generalized Fourier series
+after inserting the expression for $a_n$:
+
+$$\begin{aligned}
+    f(x)
+    &= \sum_{n = 0}^\infty \frac{1}{A_n} \Inprod{u_n}{w f} u_n(x)
+    = \int_a^b \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) f(\xi) u_n(x) \Big) \dd{\xi}
+    \\
+    &= \int_a^b f(\xi) \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) \Big) \dd{\xi}
+\end{aligned}$$
+
+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.
+This is in fact 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.