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'

---
 source/know/concept/beltrami-identity/index.md | 48 +++++++++++++-------------
 1 file changed, 24 insertions(+), 24 deletions(-)

(limited to 'source/know/concept/beltrami-identity/index.md')

diff --git a/source/know/concept/beltrami-identity/index.md b/source/know/concept/beltrami-identity/index.md
index 3fa566c..be9a344 100644
--- a/source/know/concept/beltrami-identity/index.md
+++ b/source/know/concept/beltrami-identity/index.md
@@ -8,26 +8,26 @@ categories:
 layout: "concept"
 ---
 
-Consider a general functional $J[f]$ of the following form,
-with $f(x)$ an unknown function:
+Consider a general functional $$J[f]$$ of the following form,
+with $$f(x)$$ an unknown function:
 
 $$\begin{aligned}
     J[f]
     = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
 \end{aligned}$$
 
-Where $L$ is the Lagrangian.
-To find the $f$ that maximizes or minimizes $J[f]$,
+Where $$L$$ is the Lagrangian.
+To find the $$f$$ that maximizes or minimizes $$J[f]$$,
 the [calculus of variations](/know/concept/calculus-of-variations/)
-states that the Euler-Lagrange equation must be solved for $f$:
+states that the Euler-Lagrange equation must be solved for $$f$$:
 
 $$\begin{aligned}
     0
     = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f'} \Big)
 \end{aligned}$$
 
-We now want to know exactly how $L$ depends on the free variable $x$,
-since it is a function of $x$, $f(x)$ and $f'(x)$.
+We now want to know exactly how $$L$$ depends on the free variable $$x$$,
+since it is a function of $$x$$, $$f(x)$$ and $$f'(x)$$.
 Using the chain rule:
 
 $$\begin{aligned}
@@ -44,16 +44,16 @@ $$\begin{aligned}
     &= \dv{}{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x}
 \end{aligned}$$
 
-Although we started from the "hard" derivative $\idv{L}{x}$,
-we arrive at an expression for the "soft" derivative $\ipdv{L}{x}$,
-describing the *explicit* dependence of $L$ on $x$:
+Although we started from the "hard" derivative $$\idv{L}{x}$$,
+we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$,
+describing the *explicit* dependence of $$L$$ on $$x$$:
 
 $$\begin{aligned}
     - \pdv{L}{x}
     = \dv{}{x} \bigg( f' \pdv{L}{f'} - L \bigg)
 \end{aligned}$$
 
-What if $L$ does not explicitly depend on $x$, i.e. $\ipdv{L}{x} = 0$?
+What if $$L$$ does not explicitly depend on $$x$$, i.e. $$\ipdv{L}{x} = 0$$?
 In that case, the equation can be integrated to give the **Beltrami identity**:
 
 $$\begin{aligned}
@@ -63,19 +63,19 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $C$ is a constant.
-This says that the left-hand side is a conserved quantity in $x$,
+Where $$C$$ is a constant.
+This says that the left-hand side is a conserved quantity in $$x$$,
 which could be useful to know.
-If we insert a concrete expression for $L$,
-the Beltrami identity might be easier to solve for $f$ than the full Euler-Lagrange equation.
-The assumption $\ipdv{L}{x} = 0$ is justified;
-for example, if $x$ is time, it means that the potential is time-independent.
+If we insert a concrete expression for $$L$$,
+the Beltrami identity might be easier to solve for $$f$$ than the full Euler-Lagrange equation.
+The assumption $$\ipdv{L}{x} = 0$$ is justified;
+for example, if $$x$$ is time, it means that the potential is time-independent.
 
 
 ## Higher dimensions
 
-Above, a 1D problem was considered, i.e. $f$ depended only on a single variable $x$.
-Consider now a 2D problem, such that $J[f]$ is given by:
+Above, a 1D problem was considered, i.e. $$f$$ depended only on a single variable $$x$$.
+Consider now a 2D problem, such that $$J[f]$$ is given by:
 
 $$\begin{aligned}
     J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
@@ -87,7 +87,7 @@ $$\begin{aligned}
     0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{y} \Big( \pdv{L}{f_y} \Big)
 \end{aligned}$$
 
-Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous):
+Once again, we calculate the hard $$x$$-derivative of $$L$$ (the $$y$$-derivative is analogous):
 
 $$\begin{aligned}
     \dv{L}{x}
@@ -99,7 +99,7 @@ $$\begin{aligned}
     &= \dv{}{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
 \end{aligned}$$
 
-This time, we arrive at the following expression for the soft derivative $\ipdv{L}{x}$:
+This time, we arrive at the following expression for the soft derivative $$\ipdv{L}{x}$$:
 
 $$\begin{aligned}
     - \pdv{L}{x}
@@ -109,9 +109,9 @@ $$\begin{aligned}
 Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity,
 and therefore we use that name only in the 1D case.
 
-However, if $\ipdv{L}{x} = 0$, this equation is still useful.
+However, if $$\ipdv{L}{x} = 0$$, this equation is still useful.
 For an off-topic demonstration of this fact,
-let us choose $x$ as the transverse coordinate, and integrate over it to get:
+let us choose $$x$$ as the transverse coordinate, and integrate over it to get:
 
 $$\begin{aligned}
     0
@@ -123,7 +123,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 If our boundary conditions cause the boundary term to vanish (as is often the case),
-then the integral on the right is a conserved quantity with respect to $y$.
+then the integral on the right is a conserved quantity with respect to $$y$$.
 While not as elegant as the 1D Beltrami identity,
 the above 2D counterpart still fulfills the same role.
 
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