From 7ec42764de400df4db629780f3c758f553ac5a93 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 9 Jun 2023 23:33:22 +0200
Subject: Improve knowledge base

---
 .../know/concept/sturm-liouville-theory/index.md   | 58 ++++++++++++----------
 1 file changed, 31 insertions(+), 27 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 0ac7476..bff57af 100644
--- a/source/know/concept/sturm-liouville-theory/index.md
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -18,6 +18,7 @@ and that the corresponding eigenvalue problem, known as a
 of eigenfunctions.
 
 
+
 ## General operator
 
 Consider the most general form of a second-order linear
@@ -65,7 +66,7 @@ $$[p_0(f^* g' - (f^*)' g)]_a^b = 0$$, leaving:
 
 $$\begin{aligned}
     \inprod{f}{\hat{L} g}
-    &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \Inprod{\hat{L}^\dagger f}{g}
+    &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \inprod{\hat{L}^\dagger f}{g}
     = \inprod{\hat{L}^\dagger f}{g}
 \end{aligned}$$
 
@@ -115,18 +116,19 @@ $$\begin{aligned}
 
 The latter is a differential equation for $$p(x)$$, which we solve by integration:
 
-$$\begin{gathered}
-    \frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x}
-    \quad \implies \quad
-    \frac{p_1(x)}{p_0(x)} \dd{x} = \frac{1}{p(x)} \dd{p}
+$$\begin{aligned}
+    \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)
+    \int \frac{p_1(x)}{p_0(x)} \dd{x}
+    &= \int \frac{1}{p} \dd{p}
+    = \ln\!\big( p(x) \big)
     \\
-    \implies \quad
-    p(x) = p(a) \exp\!\Big( \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} \Big)
-\end{gathered}$$
+    \implies \qquad\qquad
+    p(x)
+    &= \exp\!\bigg( \int \frac{p_1(x)}{p_0(x)} \dd{x} \bigg)
+\end{aligned}$$
 
 Now that we have $$p(x)$$ and $$q(x)$$, we can define a new operator $$\hat{L}_p$$ as follows:
 
@@ -153,6 +155,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Eigenvalue problem
 
 A **Sturm-Liouville problem** (SLP) is analogous to a matrix eigenvalue problem,
@@ -166,7 +169,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 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.
+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 $$\hat{L}_{SL}$$,
@@ -212,7 +215,7 @@ $$\begin{aligned}
     &= (\lambda_m^* - \lambda_n) \int_a^b u_n u_m^* w \:dx
     \\
     \inprod{u_m}{\hat{L}_{SL} u_n} - \inprod{\hat{L}_{SL} u_m}{u_n}
-    &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
+    &= (\lambda_m^* - \lambda_n) \inprod{u_m}{w u_n}
 \end{aligned}$$
 
 The operator $$\hat{L}_{SL}$$ is self-adjoint by definition,
@@ -220,17 +223,17 @@ so the left-hand side vanishes, leaving us with:
 
 $$\begin{aligned}
     0
-    &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
+    &= (\lambda_m^* - \lambda_n) \inprod{u_m}{w u_n}
 \end{aligned}$$
 
-When $$m = n$$, the inner product $$\Inprod{u_n}{w u_n}$$ is real and positive
+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.
+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/).
@@ -240,8 +243,8 @@ and all the corresponding eigenfunctions $$u(x)$$ are mutually orthogonal**:
 
 $$\begin{aligned}
     \boxed{
-        \Inprod{u_m(x)}{w(x) u_n(x)}
-        = \Inprod{u_n}{w u_n} \delta_{nm}
+        \inprod{u_m(x)}{w(x) u_n(x)}
+        = \inprod{u_n}{w u_n} \delta_{nm}
         = A_n \delta_{nm}
     }
 \end{aligned}$$
@@ -257,6 +260,7 @@ 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
@@ -287,18 +291,18 @@ By integrating we get inner products on both the left and the right:
 
 $$\begin{aligned}
     \int_a^b f(x) w(x) u_m^*(x) \dd{x}
-    &= \int_a^b \Big(\sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)\Big) \dd{x}
+    &= \int_a^b \bigg( \sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x) \bigg) \dd{x}
     \\
-    \Inprod{u_m}{w f}
-    &= \sum_{n = 0}^\infty a_n \Inprod{u_m}{w u_n}
+    \inprod{u_m}{w f}
+    &= \sum_{n = 0}^\infty a_n \inprod{u_m}{w u_n}
 \end{aligned}$$
 
 Because the eigenfunctions of an SLP are mutually orthogonal,
 the summation disappears:
 
 $$\begin{aligned}
-    \Inprod{u_m}{w f}
-    &= \sum_{n = 0}^\infty a_n \Inprod{u_m}{w u_n}
+    \inprod{u_m}{w f}
+    &= \sum_{n = 0}^\infty a_n \inprod{u_m}{w u_n}
     = \sum_{n = 0}^\infty a_n A_n \delta_{nm}
     = a_m A_m
 \end{aligned}$$
@@ -310,8 +314,8 @@ function $$f(x)$$ onto the normalized eigenfunctions $$u_n(x) / A_n$$:
 $$\begin{aligned}
     \boxed{
         a_n
-        = \frac{\Inprod{u_n}{w f}}{A_n}
-        = \frac{\Inprod{u_n}{w f}}{\Inprod{u_n}{w u_n}}
+        = \frac{\inprod{u_n}{w f}}{A_n}
+        = \frac{\inprod{u_n}{w f}}{\inprod{u_n}{w u_n}}
     }
 \end{aligned}$$
 
@@ -321,10 +325,10 @@ after inserting the expression for $$a_n$$:
 
 $$\begin{aligned}
     f(x)
-    &= \sum_{n = 0}^\infty \frac{1}{A_n} \Inprod{u_n}{w f} u_n(x)
-    = \int_a^b \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) f(\xi) u_n(x) \Big) \dd{\xi}
+    &= \sum_{n = 0}^\infty \frac{1}{A_n} \inprod{u_n}{w f} u_n(x)
+    = \int_a^b \bigg(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) f(\xi) u_n(x) \bigg) \dd{\xi}
     \\
-    &= \int_a^b f(\xi) \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) \Big) \dd{\xi}
+    &= \int_a^b f(\xi) \bigg(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) \bigg) \dd{\xi}
 \end{aligned}$$
 
 Upon closer inspection, the parenthesized summation
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