From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 .../know/concept/sturm-liouville-theory/index.md   | 118 ++++++++++-----------
 1 file changed, 59 insertions(+), 59 deletions(-)

(limited to 'source/know/concept/sturm-liouville-theory/index.md')

diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md
index 23922d5..75daae3 100644
--- a/source/know/concept/sturm-liouville-theory/index.md
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -21,23 +21,23 @@ 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 $x \in ]a, b[$:
+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:
+$$\hat{L}^\dagger$$ 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 $\inprod{f}{\hat{L} g}$:
+What is $$\hat{L}^\dagger$$, given the above definition of $$\hat{L}$$?
+We start from the inner product $$\inprod{f}{\hat{L} g}$$:
 
 $$\begin{aligned}
     \inprod{f}{\hat{L} g}
@@ -51,7 +51,7 @@ $$\begin{aligned}
     &= \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
+We now have an expression for $$\hat{L}^\dagger$$, but are left with an
 annoying boundary term:
 
 $$\begin{aligned}
@@ -60,8 +60,8 @@ $$\begin{aligned}
 \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:
+let us demand that $$p_1(x) = p_0'(x)$$ and that
+$$[p_0(f^* g' - (f^*)' g)]_a^b = 0$$, leaving:
 
 $$\begin{aligned}
     \inprod{f}{\hat{L} g}
@@ -69,8 +69,8 @@ $$\begin{aligned}
     = \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 $\hat{L}^\dagger$:
+Using the aforementioned restriction $$p_1(x) = p_0'(x)$$,
+we then take a look at the definition of $$\hat{L}^\dagger$$:
 
 $$\begin{aligned}
     \hat{L}^\dagger \{f\}
@@ -83,7 +83,7 @@ $$\begin{aligned}
     &= (p_0 f')' + p_2 f
 \end{aligned}$$
 
-The original operator $\hat{L}$ reduces to the same form,
+The original operator $$\hat{L}$$ reduces to the same form,
 so it is **self-adjoint**:
 
 $$\begin{aligned}
@@ -93,19 +93,19 @@ $$\begin{aligned}
     = \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 $[p_0 (f^* g' - (f^*)' g)]_a^b = 0$.
+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 $p_0(x) \neq 0$:
+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:
+We now define $$q(x)$$,
+and demand that the derivative $$p'(x)$$ of the unknown $$p(x)$$ satisfies:
 
 $$\begin{aligned}
     q(x) = p(x) \frac{p_2(x)}{p_0(x)}
@@ -113,7 +113,7 @@ $$\begin{aligned}
     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:
+The latter is a differential equation for $$p(x)$$, which we solve by integration:
 
 $$\begin{gathered}
     \frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x}
@@ -128,7 +128,7 @@ $$\begin{gathered}
     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:
+Now that we have $$p(x)$$ and $$q(x)$$, we can define a new operator $$\hat{L}_p$$ as follows:
 
 $$\begin{aligned}
     \hat{L}_p \{u\}
@@ -138,11 +138,11 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This is the self-adjoint form from earlier!
-So even if $p_0' \neq p_1$, any second-order linear operator with $p_0(x) \neq 0$
+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 $x \in [a,b]$:
+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 $$x \in [a,b]$$:
 
 $$\begin{aligned}
     \boxed{
@@ -156,8 +156,8 @@ $$\begin{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**:
+where $$w(x)$$ is a real weight function, $$\lambda$$ is the **eigenvalue**,
+and $$u(x)$$ is the corresponding **eigenfunction**:
 
 $$\begin{aligned}
     \boxed{
@@ -165,14 +165,14 @@ $$\begin{aligned}
     }
 \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.
+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 $\hat{L}_{SL}$,
+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:
+$$u(x)$$ must be as follows, otherwise the operator cannot be self-adjoint:
 
 $$\begin{aligned}
     \Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0
@@ -181,16 +181,16 @@ $$\begin{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"**: $p(a) = p(b) = 0$
++ **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 $u(x)$.
+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:
+eigenfunctions $$u_n$$ and $$u_m$$, the following must be satisfied:
 
 $$\begin{aligned}
     0 = \hat{L}_{SL}\{u_n\} + \lambda_n w u_n = \hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*
@@ -215,7 +215,7 @@ $$\begin{aligned}
     &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
 \end{aligned}$$
 
-The operator $\hat{L}_{SL}$ is self-adjoint by definition,
+The operator $$\hat{L}_{SL}$$ is self-adjoint by definition,
 so the left-hand side vanishes, leaving us with:
 
 $$\begin{aligned}
@@ -223,20 +223,20 @@ $$\begin{aligned}
     &= (\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 $$m = n$$, 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,
+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.
+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**:
+In conclusion, **a Sturm-Liouville problem has real eigenvalues $$\lambda$$,
+and all the corresponding eigenfunctions $$u(x)$$ are mutually orthogonal**:
 
 $$\begin{aligned}
     \boxed{
@@ -252,16 +252,16 @@ 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 $\lambda_0 > -\infty$,
+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 $a_n$:
+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 $$a_n$$:
 
 $$\begin{aligned}
     \boxed{
@@ -271,12 +271,12 @@ $$\begin{aligned}
     }
 \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 $x \in [a, b]$.
+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 $m$:
+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 $$m$$:
 
 $$\begin{aligned}
     f(x) w(x) u_m^*(x)
@@ -303,9 +303,9 @@ $$\begin{aligned}
     = a_m A_m
 \end{aligned}$$
 
-After isolating this for $a_n$, we see that
+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 $u_n(x) / A_n$:
+function $$f(x)$$ onto the normalized eigenfunctions $$u_n(x) / A_n$$:
 
 $$\begin{aligned}
     \boxed{
@@ -317,7 +317,7 @@ $$\begin{aligned}
 
 As a final remark, we can see something interesting
 by rearranging the generalized Fourier series
-after inserting the expression for $a_n$:
+after inserting the expression for $$a_n$$:
 
 $$\begin{aligned}
     f(x)
@@ -328,7 +328,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Upon closer inspection, the parenthesized summation
-must be the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$
+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:
 
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