Improve knowledge base

9 files changed, 339 insertions, 301 deletions

diff --git a/source/know/concept/capillary-action/index.md b/source/know/concept/capillary-action/index.md
deleted file mode 100644
index fea6ef8..0000000
--- a/source/know/concept/capillary-action/index.md
+++ /dev/null
@@ -1,126 +0,0 @@
----
-title: "Capillary action"
-sort_title: "Capillary action"
-date: 2021-03-29
-categories:
-- Physics
-- Fluid mechanics
-- Fluid statics
-- Surface tension
-layout: "concept"
----
-
-**Capillary action** refers to the movement of liquid
-through narrow spaces due to surface tension, often against gravity.
-It occurs when the [Laplace pressure](/know/concept/young-laplace-law/)
-from surface tension is much larger in magnitude than the
-[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity.
-
-Consider a spherical droplet of liquid with radius $$R$$.
-The hydrostatic pressure difference
-between the top and bottom of the drop
-is much smaller than the Laplace pressure:
-
-$$\begin{aligned}
-    2 R \rho g \ll 2 \frac{\alpha}{R}
-\end{aligned}$$
-
-Where $$\rho$$ is the density of the liquid,
-$$g$$ is the acceleration due to gravity,
-and $$\alpha$$ is the energy cost per unit surface area.
-Rearranging the inequality yields:
-
-$$\begin{aligned}
-    R^2 \ll \frac{\alpha}{\rho g}
-\end{aligned}$$
-
-From the right-hand side we define the **capillary length** $$L_c$$,
-so gravity is negligible if $$R \ll L_c$$:
-
-$$\begin{aligned}
-    \boxed{
-        L_c
-        \equiv \sqrt{\frac{\alpha}{\rho g}}
-    }
-\end{aligned}$$
-
-In general, for a system with characteristic length $$L$$,
-the relative strength of gravity compared to surface tension
-is described by the **Bond number** $$\mathrm{Bo}$$
-or **Eötvös number** $$\mathrm{Eo}$$:
-
-$$\begin{aligned}
-    \boxed{
-        \mathrm{Bo}
-        \equiv \mathrm{Eo}
-        \equiv \frac{L^2}{L_c^2}
-        = \frac{m g}{\alpha L}
-    }
-\end{aligned}$$
-
-The rightmost side gives an alternative way of understanding $$\mathrm{Bo}$$:
-$$m$$ is the mass of a cube with side $$L$$, such that the numerator is the weight force,
-and the denominator is the tension force of the surface.
-In any case, capillary action can be observed when $$\mathrm{Bo \ll 1}$$.
-
-The most famous example of capillary action is **capillary rise**,
-where a liquid "climbs" upwards in a narrow vertical tube with radius $$R$$,
-apparently defying gravity.
-Assuming the liquid-air interface is a spherical cap
-with constant [curvature](/know/concept/curvature/) radius $$R_c$$,
-then we know that the liquid is at rest
-when the hydrostatic pressure equals the Laplace pressure:
-
-$$\begin{aligned}
-    \rho g h
-    \approx \alpha \frac{2}{R_c}
-    = 2 \alpha \frac{\cos\theta}{R}
-\end{aligned}$$
-
-Where $$\theta$$ is the liquid-tube contact angle,
-and we are neglecting variations of the height $$h$$ due to the curvature
-(i.e. the [meniscus](/know/concept/meniscus/)).
-By isolating the above equation for $$h$$,
-we arrive at **Jurin's law**,
-which predicts the height climbed by a liquid in a tube with radius $$R$$:
-
-$$\begin{aligned}
-    \boxed{
-        h
-        = 2 \frac{L_c^2}{R} \cos\theta
-    }
-\end{aligned}$$
-
-Depending on $$\theta$$, $$h$$ can be negative,
-i.e. the liquid might descend below the ambient level.
-
-
-An alternative derivation of Jurin's law balances the forces instead of the pressures.
-On the right, we have the gravitational force
-(i.e. the energy-per-distance to lift the liquid),
-and on the left, the surface tension force
-(i.e. the energy-per-distance of the liquid-tube interface):
-
-$$\begin{aligned}
-     \pi R^2 \rho g h
-     \approx 2 \pi R (\alpha_{sg} - \alpha_{sl})
-\end{aligned}$$
-
-Where $$\alpha_{sg}$$ and $$\alpha_{sl}$$ are the energy costs
-of the solid-gas and solid-liquid interfaces.
-Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/),
-we can rewrite this as follows:
-
-$$\begin{aligned}
-     R \rho g h
-     = 2 \alpha \cos\theta
-\end{aligned}$$
-
-Isolating this for $$h$$ simply yields Jurin's law again, as expected.
-
-
-
-## References
-1.  B. Lautrup,
-    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
-    CRC Press.
diff --git a/source/know/concept/capillary-length/index.md b/source/know/concept/capillary-length/index.md
new file mode 100644
index 0000000..4dbb788
--- /dev/null
+++ b/source/know/concept/capillary-length/index.md
@@ -0,0 +1,79 @@
+---
+title: "Capillary length"
+sort_title: "Capillary length"
+date: 2021-03-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+- Surface tension
+layout: "concept"
+---
+
+**Capillary action** refers to the movement of liquid
+through narrow spaces due to surface tension, often against gravity.
+It occurs when the [Laplace pressure](/know/concept/young-laplace-law/)
+from surface tension is much larger in magnitude than the
+[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity.
+
+Consider a spherical droplet of liquid with radius $$R$$.
+The hydrostatic pressure difference
+between the top and bottom of the drop
+is much smaller than the Laplace pressure:
+
+$$\begin{aligned}
+    2 R \rho g \ll 2 \frac{\alpha}{R}
+\end{aligned}$$
+
+Where $$\rho$$ is the density of the liquid,
+$$g$$ is the acceleration due to gravity,
+and $$\alpha$$ is the energy cost per unit surface area.
+Rearranging the inequality yields:
+
+$$\begin{aligned}
+    R^2 \ll \frac{\alpha}{\rho g}
+\end{aligned}$$
+
+From this, we define the **capillary length** $$L_c$$
+such that gravity is negligible if $$R \ll L_c$$:
+
+$$\begin{aligned}
+    \boxed{
+        L_c
+        \equiv \sqrt{\frac{\alpha}{\rho g}}
+    }
+\end{aligned}$$
+
+In general, for a system with characteristic length $$L$$,
+the relative strength of gravity compared to surface tension
+is described by the **Bond number** $$\mathrm{Bo}$$
+or **Eötvös number** $$\mathrm{Eo}$$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{Bo}
+        \equiv \mathrm{Eo}
+        \equiv \frac{L^2}{L_c^2}
+    }
+\end{aligned}$$
+
+Capillary action is observed when $$\mathrm{Bo \ll 1}$$,
+while for $$\mathrm{Bo} \gg 1$$ surface tension is negligible.
+
+For an alternative interpretation of $$\mathrm{Bo}$$,
+let $$m \equiv \rho L^3$$ be the mass of a cube with side $$L$$
+such that its weight is $$m g$$.
+The tension force on its face is $$\alpha L$$,
+so $$\mathrm{Bo}$$ is simply the force ratio:
+
+$$\begin{aligned}
+    \mathrm{Bo}
+    = \frac{m g}{\alpha L}
+\end{aligned}$$
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/source/know/concept/cartesian-coordinates/index.md b/source/know/concept/cartesian-coordinates/index.md
index d198e84..a0bfc39 100644
--- a/source/know/concept/cartesian-coordinates/index.md
+++ b/source/know/concept/cartesian-coordinates/index.md
@@ -11,8 +11,8 @@ layout: "concept"
 This article is a supplement to the ones on
 [orthogonal curvilinear systems](/know/concept/orthogonal-curvilinear-coordinates/),
 [spherical coordinates](/know/concept/spherical-coordinates/),
-[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/),
-and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/).
+[polar cylindrical coordinates](/know/concept/polar-cylindrical-coordinates/),
+and [parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coordinates/).
 
 The well-known Cartesian coordinate system $$(x, y, z)$$
 has trivial **scale factors**:
diff --git a/source/know/concept/drude-model/index.md b/source/know/concept/drude-model/index.md
index c4faf81..0026d90 100644
--- a/source/know/concept/drude-model/index.md
+++ b/source/know/concept/drude-model/index.md
@@ -11,7 +11,7 @@ layout: "concept"
 
 The **Drude model**, also known as
 the **Drude-Lorentz model** due to its analogy
-to the *Lorentz oscillator model*
+to the [Lorentz oscillator model](/know/concept/lorentz-oscillator-model/)
 classically predicts the [dielectric function](/know/concept/dielectric-function/)
 and electric conductivity of a gas of free charges,
 as found in metals and doped semiconductors.
@@ -59,7 +59,7 @@ $$\begin{aligned}
     = - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} \vb{E}(t)
 \end{aligned}$$
 
-The electric displacement field $$\vb{D}$$ is then as follows,
+The electric displacement field $$\vb{D}(t)$$ is then as follows,
 where the parenthesized expression is the dielectric function
 $$\varepsilon_r$$ of the material:
 
@@ -180,7 +180,7 @@ We must replace the carriers' true mass $$m$$ with their *effective mass* $$m^*$
 found from the material's electronic band structure.
 Furthermore, semiconductors already have
 a high intrinsic dielectric function $$\varepsilon_{\mathrm{int}}$$
-before being doped, so the displacement field $$\vb{D}$$ becomes:
+before being doped, so the displacement field $$\vb{D}(t)$$ becomes:
 
 $$\begin{aligned}
     \vb{D}
diff --git a/source/know/concept/jurins-law/index.md b/source/know/concept/jurins-law/index.md
new file mode 100644
index 0000000..6214477
--- /dev/null
+++ b/source/know/concept/jurins-law/index.md
@@ -0,0 +1,78 @@
+---
+title: "Jurin's law"
+sort_title: "Jurin's law"
+date: 2023-06-15
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+- Surface tension
+layout: "concept"
+---
+
+A well-known example of *capillary action* is
+when a liquid climbs up a narrow vertical tube with radius $$R$$,
+apparently defying gravity.
+Indeed, this occurs when the liquid's surface tension can overpower gravity;
+specifically, when the [capillary length](/know/concept/capillary-length/) $$L_c > R$$.
+
+Let us assume that the liquid-air interface has a spherical shape,
+which may point up or down depending on the liquid.
+This interface then has a constant [curvature radius](/know/concept/curvature/) $$r$$
+determined by the contact angle $$\theta$$ of the liquid to the tube:
+$$r = R / \cos{\theta}$$. We know that the liquid is at rest
+when the [hydrostatic pressure](/know/concept/hydrostatic-pressure/)
+equals the resulting [Laplace pressure](/know/concept/young-laplace-law/):
+
+$$\begin{aligned}
+    \rho g h
+    = \alpha \frac{2}{r}
+    = 2 \alpha \frac{\cos{\theta}}{R}
+\end{aligned}$$
+
+Note that $$h$$ is the height of interface's highest/lowest point;
+we neglect the [meniscus](/know/concept/meniscus/).
+By isolating the above equation for $$h$$, we arrive at **Jurin's law**:
+
+$$\begin{aligned}
+    \boxed{
+        h
+        = \frac{2 \alpha \cos{\theta}}{\rho g R}
+        = 2 \frac{L_c^2}{R} \cos{\theta}
+    }
+\end{aligned}$$
+
+Where $$L_c \equiv \sqrt{\alpha / \rho g}$$.
+This predicts the height climbed by a liquid in a narrow tube.
+If $$\theta > 90\degree$$, then $$h$$ is negative,
+i.e. the liquid descends below the ambient level.
+
+An alternative derivation of Jurin's law balances the forces instead of the pressures.
+On the right, we have the gravitational force
+(i.e. the energy-per-distance to lift the liquid),
+and on the left, the surface tension force
+(i.e. the energy-per-distance of the liquid-tube interface):
+
+$$\begin{aligned}
+     \pi R^2 \rho g h
+     \approx 2 \pi R (\alpha_{sg} - \alpha_{sl})
+\end{aligned}$$
+
+Where $$\alpha_{sg}$$ and $$\alpha_{sl}$$ are the energy costs
+of the solid-gas and solid-liquid interfaces.
+Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/),
+we can rewrite this as follows:
+
+$$\begin{aligned}
+     R \rho g h
+     = 2 \alpha \cos\theta
+\end{aligned}$$
+
+Isolating this for $$h$$ simply yields Jurin's law again, as expected.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/source/know/concept/orthogonal-curvilinear-coordinates/index.md b/source/know/concept/orthogonal-curvilinear-coordinates/index.md
index 669358c..4714a95 100644
--- a/source/know/concept/orthogonal-curvilinear-coordinates/index.md
+++ b/source/know/concept/orthogonal-curvilinear-coordinates/index.md
@@ -18,10 +18,10 @@ where at least one of the coordinate surfaces is curved:
 e.g. in cylindrical coordinates, the coordinate line of $$r$$ and $$z$$ is a circle.
 Here we limit ourselves to **orthogonal** systems,
 where the coordinate surfaces are always perpendicular.
-Examples of such orthogonal curvilinear systems include
+Examples of such orthogonal curvilinear systems are
 [spherical coordinates](/know/concept/spherical-coordinates/),
-[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/),
-[cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/),
+[polar cylindrical coordinates](/know/concept/polar-cylindrical-coordinates/),
+[parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coordinates/),
 and (trivially) [Cartesian coordinates](/know/concept/cartesian-coordinates/).
 
 
diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/parabolic-cylindrical-coordinates/index.md
index 58358dd..6ba19f5 100644
--- a/source/know/concept/cylindrical-parabolic-coordinates/index.md
+++ b/source/know/concept/parabolic-cylindrical-coordinates/index.md
@@ -1,6 +1,6 @@
 ---
-title: "Cylindrical parabolic coordinates"
-sort_title: "Cylindrical parabolic coordinates"
+title: "Parabolic cylindrical coordinates"
+sort_title: "Parabolic cylindrical coordinates"
 date: 2021-03-04
 categories:
 - Mathematics
@@ -8,7 +8,7 @@ categories:
 layout: "concept"
 ---
 
-**Cylindrical parabolic coordinates** extend parabolic coordinates $$(\sigma, \tau)$$ to 3D,
+**Parabolic cylindrical coordinates** extend parabolic coordinates $$(\sigma, \tau)$$ to 3D,
 by describing a point in space using the variables $$(\sigma, \tau, z)$$.
 The $$z$$-axis is the same as in the Cartesian system, (hence the name *cylindrical*),
 while the coordinate lines of $$\sigma$$ and $$\tau$$ are confocal parabolas.
@@ -38,10 +38,10 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             \sigma
-            &= \sgn(x) \sqrt{\sqrt{x^2 + y^2} - x}
+            &= \sqrt{\sqrt{x^2 + y^2} - x}
             \\
             \tau
-            &= \sqrt{\sqrt{x^2 + y^2} + x}
+            &= \sgn(y) \sqrt{\sqrt{x^2 + y^2} + x}
             \\
             z
             &= z
@@ -49,7 +49,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Cylindrical parabolic coordinates form
+Parabolic cylindrical coordinates form
 an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/),
 whose **scale factors** $$h_\sigma$$, $$h_\tau$$ and $$h_z$$ we need.
 To get those, we calculate the unnormalized local basis:
diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/polar-cylindrical-coordinates/index.md
index cf227a6..2223996 100644
--- a/source/know/concept/cylindrical-polar-coordinates/index.md
+++ b/source/know/concept/polar-cylindrical-coordinates/index.md
@@ -1,6 +1,6 @@
 ---
-title: "Cylindrical polar coordinates"
-sort_title: "Cylindrical polar coordinates"
+title: "Polar cylindrical coordinates"
+sort_title: "Polar cylindrical coordinates"
 date: 2021-07-26
 categories:
 - Mathematics
@@ -8,7 +8,7 @@ categories:
 layout: "concept"
 ---
 
-**Cylindrical polar coordinates** extend polar coordinates $$(r, \varphi)$$ to 3D,
+**Polar cylindrical coordinates** extend polar coordinates $$(r, \varphi)$$ to 3D,
 by describing the location of a point in space
 using the variables $$(r, \varphi, z)$$.
 The $$z$$-axis is unchanged from the Cartesian system,
@@ -52,7 +52,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Cylindrical polar coordinates form
+Polar cylindrical coordinates form
 an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/),
 whose **scale factors** $$h_r$$, $$h_\varphi$$ and $$h_z$$ we need.
 To get those, we calculate the unnormalized local basis:
diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md
index bff57af..d7984b5 100644
--- a/source/know/concept/sturm-liouville-theory/index.md
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -8,14 +8,15 @@ categories:
 layout: "concept"
 ---
 
-**Sturm-Liouville theory** defines the analogue of Hermitian matrix
-eigenvalue problems for linear second-order ODEs.
+**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 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.
+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.
 
 
 
@@ -23,18 +24,19 @@ of eigenfunctions.
 
 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 nonzero for all $$x \in ]a, b[$$:
+are real functions of $$x \in [a,b]$$ and are nonzero for all $$x \in \,\,]a, b[$$:
 
 $$\begin{aligned}
-    \hat{L} \{u(x)\} = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x)
+    \hat{L} \{u(x)\}
+    \equiv p_2(x) \: u''(x) + p_1(x) \: u'(x) + p_0(x) \: u(x)
 \end{aligned}$$
 
-We now define the **adjoint** or **Hermitian** operator
-$$\hat{L}^\dagger$$ analogously to matrices:
+Analogously to matrices,
+we now define its **adjoint** operator $$\hat{L}^\dagger$$ as follows:
 
 $$\begin{aligned}
-    \inprod{f}{\hat{L} g}
-    = \inprod{\hat{L}^\dagger f}{g}
+    \inprod{\hat{L}^\dagger f}{g}
+    \equiv \inprod{f}{\hat{L} g}
 \end{aligned}$$
 
 What is $$\hat{L}^\dagger$$, given the above definition of $$\hat{L}$$?
@@ -43,146 +45,155 @@ We start from the inner product $$\inprod{f}{\hat{L} g}$$:
 $$\begin{aligned}
     \inprod{f}{\hat{L} g}
     &= \int_a^b f^*(x) \hat{L}\{g(x)\} \dd{x}
-    = \int_a^b (f^* p_0) g'' + (f^* p_1) g' + (f^* p_2) g \dd{x}
+    = \int_a^b (f^* p_2) g'' + (f^* p_1) g' + (f^* p_0) g \dd{x}
     \\
-    &= \big[ (f^* p_0) g' + (f^* p_1) g \big]_a^b - \int_a^b (f^* p_0)' g' + (f^* p_1)' g - (f^* p_2) g \dd{x}
+    &= \Big[ (f^* p_2) g' + (f^* p_1) g \Big]_a^b - \int_a^b (f^* p_2)' g' + (f^* p_1)' g - (f^* p_0) g \dd{x}
     \\
-    &= \big[ f^* \big( p_0 g' \!+\! p_1 g \big) \!-\! (f^* p_0)' g \big]_a^b + \int_a^b \! \big( (f p_0)'' - (f p_1)' + (f p_2) \big)^* g \dd{x}
+    &= \Big[ f^* (p_2 g' + p_1 g) - (f^* p_2)' g \Big]_a^b + \int_a^b \! \Big( (f p_2)'' - (f p_1)' + (f p_0) \Big)^* g \dd{x}
     \\
-    &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x}
-\end{aligned}$$
-
-We now have an expression for $$\hat{L}^\dagger$$, but are left with an
-annoying boundary term:
-
-$$\begin{aligned}
-    \inprod{f}{\hat{L} g}
-    &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \inprod{\hat{L}^\dagger f}{g}
+    &= \Big[ f^* \big( p_2 g' + (p_1 - p_2') g \big) - (f^*)' p_2 g \Big]_a^b + \int_a^b \Big( \hat{L}^\dagger\{f\} \Big)^* g \dd{x}
 \end{aligned}$$
 
-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:
+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:
 
 $$\begin{aligned}
     \inprod{f}{\hat{L} g}
-    &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \inprod{\hat{L}^\dagger f}{g}
-    = \inprod{\hat{L}^\dagger f}{g}
+    &= \Big[ f^* \big( p_2 g' + (p_1 - p_2') g \big) - (f^*)' p_2 g \Big]_a^b + \inprod{\hat{L}^\dagger f}{g}
+    \\
+    &= \Big[ p_2 \big( f^* g' - (f^*)' g \big) \Big]_a^b + \inprod{\hat{L}^\dagger f}{g}
+    \\
+    &= \inprod{\hat{L}^\dagger f}{g}
 \end{aligned}$$
 
-Using the aforementioned restriction $$p_1(x) = p_0'(x)$$,
-we then take a look at the definition of $$\hat{L}^\dagger$$:
+Let us look at the expression for $$\hat{L}^\dagger$$ we just found,
+with the restriction $$p_1 = p_2'$$ in mind:
 
 $$\begin{aligned}
     \hat{L}^\dagger \{f\}
-    &= (p_0 f)'' - (p_1 f)' + (p_2 f)
+    &= (p_2 f)'' - (p_1 f)' + (p_0 f)
     \\
-    &= p_0 f'' + (2 p_0' - p_1) f' + (p_0'' - p_1' + p_2) f
+    &= (p_2'' f + 2 p_2' f' + p_2 f'') - (p_1' f + p_1 f') + (p_0 f)
     \\
-    &= p_0 f'' + p_0' f' + p_2 f
+    &= p_2 f'' + (2 p_2' - p_1) f' + (p_2'' - p_1' + p_0) f
     \\
-    &= (p_0 f')' + p_2 f
+    &= p_2 f'' + p_1 f' + p_0 f
+    \\
+    &= \hat{L}\{f\}
 \end{aligned}$$
 
-The original operator $$\hat{L}$$ reduces to the same form,
-so it is **self-adjoint**:
+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 $$p_2(x) \neq 0$$:
 
 $$\begin{aligned}
-    \hat{L} \{f\}
-    &= p_0 f'' + p_0' f' + p_2 f
-    = (p_0 f')' + p_2 f
-    = \hat{L}^\dagger \{f\}
+    \frac{p}{p_2} \hat{L} \{u\}
+    = p u'' + p \frac{p_1}{p_2} u' + p \frac{p_0}{p_2} u
 \end{aligned}$$
 
-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 $$p_0(x) \neq 0$$:
+We now demand that the derivative $$p'(x)$$ of the unknown $$p(x)$$ satisfies:
 
 $$\begin{aligned}
-    \frac{p(x)}{p_0(x)} \hat{L} \{u\} = p(x) u'' + p(x) \frac{p_1(x)}{p_0(x)} u' + p(x) \frac{p_2(x)}{p_0(x)} u
+    p'(x)
+    = p(x) \frac{p_1(x)}{p_2(x)}
+    \quad \implies \quad
+    \frac{p_1(x)}{p_2(x)} \dd{x}
+    = \frac{1}{p(x)} \dd{p}
 \end{aligned}$$
 
-We now define $$q(x)$$,
-and demand that the derivative $$p'(x)$$ of the unknown $$p(x)$$ satisfies:
+Taking the indefinite integral of this differential equation
+yields an expression for $$p(x)$$:
 
 $$\begin{aligned}
-    q(x) = p(x) \frac{p_2(x)}{p_0(x)}
-    \qquad
-    p'(x) = p(x) \frac{p_1(x)}{p_0(x)}
+    \int \frac{p_1(x)}{p_2(x)} \dd{x}
+    = \int \frac{1}{p} \dd{p}
+    = \ln\!\big( p(x) \big)
+    \quad \implies \quad
+    \boxed{
+        p(x)
+        = \exp\!\bigg( \int \frac{p_1(x)}{p_2(x)} \dd{x} \bigg)
+    }
 \end{aligned}$$
 
-The latter is a differential equation for $$p(x)$$, which we solve by integration:
+We define an additional function $$q(x)$$
+based on the last term of $$(p / p_2) \hat{L}$$ shown above:
 
 $$\begin{aligned}
-    \frac{p_1(x)}{p_0(x)} \dd{x}
-    &= \frac{1}{p(x)} \dd{p}
-    \\
-    \implies \quad
-    \int \frac{p_1(x)}{p_0(x)} \dd{x}
-    &= \int \frac{1}{p} \dd{p}
-    = \ln\!\big( p(x) \big)
-    \\
-    \implies \qquad\qquad
-    p(x)
-    &= \exp\!\bigg( \int \frac{p_1(x)}{p_0(x)} \dd{x} \bigg)
+    \boxed{
+        q(x)
+        \equiv p(x) \frac{p_0(x)}{p_2(x)}
+    }
+    = \frac{p_0(x)}{p_2(x)} \exp\!\bigg( \int \frac{p_1(x)}{p_2(x)} \dd{x} \bigg)
 \end{aligned}$$
 
-Now that we have $$p(x)$$ and $$q(x)$$, we can define a new operator $$\hat{L}_p$$ as follows:
+When rewritten using $$p$$ and $$q$$,
+the modified operator $$(p / p_2) \hat{L}$$ looks like this:
 
 $$\begin{aligned}
-    \hat{L}_p \{u\}
-    = \frac{p}{p_0} \hat{L} \{u\}
+    \frac{p}{p_2} \hat{L} \{u\}
     = p u'' + p' u' + q u
     = (p u')' + q u
 \end{aligned}$$
 
 This is the self-adjoint form from earlier!
-So even if $$p_0' \neq p_1$$, 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 nonzero real functions of the variable $$x \in [a,b]$$:
+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 $$p(x)$$:
 
 $$\begin{aligned}
     \boxed{
-        \hat{L}_{SL} \{u(x)\}
-        = \frac{d}{dx}\Big( p(x) \frac{du}{dx} \Big) + q(x) u(x)
-        = \hat{L}_{SL}^\dagger \{u(x)\}
+        \begin{aligned}
+            \hat{L}_\mathrm{SL} \{u(x)\}
+            &= \hat{L}_\mathrm{SL}^\dagger \{u(x)\}
+            \\
+            &= \Big( p(x) \: u'(x) \Big)' + q(x) \: u(x)
+        \end{aligned}
     }
 \end{aligned}$$
 
+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
 
-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**:
+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:
 
 $$\begin{aligned}
     \boxed{
-        \hat{L}_{SL}\{u(x)\} = - \lambda w(x) u(x)
+        \hat{L}_\mathrm{SL}\{u(x)\} = - \lambda \: w(x) \: u(x)
     }
 \end{aligned}$$
 
-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.
+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 $$\hat{L}_{SL}$$,
-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:
+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:
 
 $$\begin{aligned}
-    \Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0
+    \Big[ p(x) \big( u_m^*(x) \: u_n'(x) - \big(u_m'(x)\big)^* u_n(x) \big) \Big]_a^b = 0
 \end{aligned}$$
 
-There are many boundary conditions (BCs) which satisfy this requirement.
-Some notable ones are listed here non-exhaustively:
+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**: $$u'(a) = u'(b) = 0$$
@@ -190,108 +201,103 @@ Some notable ones are listed here non-exhaustively:
 + **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:
+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 $$u_n$$:
 
 $$\begin{aligned}
     0
-    &= \int_a^b u_m^* \big(\hat{L}_{SL}\{u_n\} + \lambda_n w u_n\big) - u_n \big(\hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*\big) \:dx
+    &= \hat{L}_\mathrm{SL}\{u_m^*\} + \lambda_m^* w u_m^*
     \\
-    &= \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} + u_n u_m^* w (\lambda_n - \lambda_m^*) \:dx
+    &= \hat{L}_\mathrm{SL}\{u_n\} + \lambda_n w u_n
 \end{aligned}$$
 
-Rearranging this a bit reveals that these are in fact three inner products:
+We multiply each by the other eigenfunction,
+subtract the results, and integrate:
 
 $$\begin{aligned}
-    \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} \:dx
-    &= (\lambda_m^* - \lambda_n) \int_a^b u_n u_m^* w \:dx
+    0
+    &= \int_a^b u_m^* \big(\hat{L}_\mathrm{SL}\{u_n\} + \lambda_n w u_n\big)
+    - u_n \big(\hat{L}_\mathrm{SL}\{u_m^*\} + \lambda_m^* w u_m^*\big) \dd{x}
     \\
-    \inprod{u_m}{\hat{L}_{SL} u_n} - \inprod{\hat{L}_{SL} u_m}{u_n}
-    &= (\lambda_m^* - \lambda_n) \inprod{u_m}{w u_n}
+    &= \int_a^b u_m^* \hat{L}_\mathrm{SL}\{u_n\} - u_n \hat{L}_\mathrm{SL}\{u_m^*\}
+    + (\lambda_n - \lambda_m^*) u_m^* w u_n \dd{x}
+    \\
+    &= \inprod{u_m}{\hat{L}_\mathrm{SL} u_n} - \inprod{\hat{L}_\mathrm{SL} u_m}{u_n}
+    + (\lambda_n - \lambda_m^*) \inprod{u_m}{w u_n}
 \end{aligned}$$
 
-The operator $$\hat{L}_{SL}$$ is self-adjoint by definition,
-so the left-hand side vanishes, leaving us with:
+The operator $$\hat{L}_\mathrm{SL}$$ is self-adjoint of course,
+so the first two terms vanish, leaving us with:
 
 $$\begin{aligned}
     0
-    &= (\lambda_m^* - \lambda_n) \inprod{u_m}{w u_n}
+    &= (\lambda_n - \lambda_m^*) \inprod{u_m}{w u_n}
 \end{aligned}$$
 
-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.
+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 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, an SLP has **real eigenvalues**
+and **orthogonal eigenfunctions**: for all $$m$$, $$n$$:
 
 $$\begin{aligned}
     \boxed{
-        \inprod{u_m(x)}{w(x) u_n(x)}
-        = \inprod{u_n}{w u_n} \delta_{nm}
+        \lambda_n \in \mathbb{R}
+    }
+    \qquad\qquad
+    \boxed{
+        \inprod{u_m}{w u_n}
         = A_n \delta_{nm}
     }
 \end{aligned}$$
 
-When you're solving a differential eigenvalue problem,
-knowing that all eigenvalues are real is a *huge* simplification,
+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 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**.
+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**.
 
 
 
-## Completeness
+## Complete basis
 
-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 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 $$a_n$$:
 
 $$\begin{aligned}
     \boxed{
         f(x)
         = \sum_{n = 0}^\infty a_n u_n(x)
-        \quad \mathrm{for}\: x \in ]a, b[
+        \quad \mathrm{for} \: x \in \,\,]a, b[
     }
 \end{aligned}$$
 
-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 $$x \in [a, b]$$.
+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_