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/calculus-of-variations/index.md   | 130 ++++++++++-----------
 1 file changed, 65 insertions(+), 65 deletions(-)

(limited to 'source/know/concept/calculus-of-variations')

diff --git a/source/know/concept/calculus-of-variations/index.md b/source/know/concept/calculus-of-variations/index.md
index 0b1d070..7701358 100644
--- a/source/know/concept/calculus-of-variations/index.md
+++ b/source/know/concept/calculus-of-variations/index.md
@@ -11,24 +11,24 @@ layout: "concept"
 The **calculus of variations** lays the mathematical groundwork
 for [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
 
-Consider a **functional** $J$, mapping a function $f(x)$ to a scalar value
-by integrating over the so-called **Lagrangian** $L$,
-which represents an expression involving $x$, $f$ and the derivative $f'$:
+Consider a **functional** $$J$$, mapping a function $$f(x)$$ to a scalar value
+by integrating over the so-called **Lagrangian** $$L$$,
+which represents an expression involving $$x$$, $$f$$ and the derivative $$f'$$:
 
 $$\begin{aligned}
     J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
 \end{aligned}$$
 
-If $J$ in some way measures the physical "cost" (e.g. energy) of
-the path $f(x)$ taken by a physical system,
-the **principle of least action** states that $f$ will be a minimum of $J[f]$,
+If $$J$$ in some way measures the physical "cost" (e.g. energy) of
+the path $$f(x)$$ taken by a physical system,
+the **principle of least action** states that $$f$$ will be a minimum of $$J[f]$$,
 so for example the expended energy will be minimized.
 In practice, various cost metrics may be used,
-so maxima of $J[f]$ are also interesting to us.
+so maxima of $$J[f]$$ are also interesting to us.
 
-If $f(x, \varepsilon\!=\!0)$ is the optimal route, then a slightly
+If $$f(x, \varepsilon\!=\!0)$$ is the optimal route, then a slightly
 different (and therefore worse) path between the same two points can be expressed
-using the parameter $\varepsilon$:
+using the parameter $$\varepsilon$$:
 
 $$\begin{aligned}
     f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x)
@@ -36,19 +36,19 @@ $$\begin{aligned}
     \delta f = \varepsilon \eta(x)
 \end{aligned}$$
 
-Where $\eta(x)$ is an arbitrary differentiable deviation.
-Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$,
+Where $$\eta(x)$$ is an arbitrary differentiable deviation.
+Since $$f(x, \varepsilon)$$ must start and end in the same points as $$f(x,0)$$,
 we have the boundary conditions:
 
 $$\begin{aligned}
     \eta(x_0) = \eta(x_1) = 0
 \end{aligned}$$
 
-Given $L$, the goal is to find an equation for the optimal path $f(x,0)$.
+Given $$L$$, the goal is to find an equation for the optimal path $$f(x,0)$$.
 Just like when finding the minimum of a real function,
-the minimum $f$ of a functional $J[f]$ is a stationary point
-with respect to the deviation weight $\varepsilon$,
-a condition often written as $\delta J = 0$.
+the minimum $$f$$ of a functional $$J[f]$$ is a stationary point
+with respect to the deviation weight $$\varepsilon$$,
+a condition often written as $$\delta J = 0$$.
 In the following, the integration limits have been omitted:
 
 $$\begin{aligned}
@@ -63,14 +63,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The boundary term from partial integration vanishes due to the boundary
-conditions for $\eta(x)$. We are thus left with:
+conditions for $$\eta(x)$$. We are thus left with:
 
 $$\begin{aligned}
     0
     = \int \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f'} \Big) \bigg) \dd{x}
 \end{aligned}$$
 
-This holds for all $\eta$, but $\eta$ is arbitrary, so in fact
+This holds for all $$\eta$$, but $$\eta$$ is arbitrary, so in fact
 only the parenthesized expression matters:
 
 $$\begin{aligned}
@@ -79,27 +79,27 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This is known as the **Euler-Lagrange equation** of the Lagrangian $L$,
-and its solutions represent the optimal paths $f(x, 0)$.
+This is known as the **Euler-Lagrange equation** of the Lagrangian $$L$$,
+and its solutions represent the optimal paths $$f(x, 0)$$.
 
 
 ## Multiple functions
 
-Suppose that the Lagrangian $L$ depends on multiple independent functions
-$f_1, f_2, ..., f_N$:
+Suppose that the Lagrangian $$L$$ depends on multiple independent functions
+$$f_1, f_2, ..., f_N$$:
 
 $$\begin{aligned}
     J[f_1, ..., f_N] = \int_{x_0}^{x_1} L(f_1, ..., f_N, f_1', ..., f_N', x) \dd{x}
 \end{aligned}$$
 
-In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$,
-satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$:
+In this case, every $$f_n(x)$$ has its own deviation $$\eta_n(x)$$,
+satisfying $$\eta_n(x_0) = \eta_n(x_1) = 0$$:
 
 $$\begin{aligned}
     f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x)
 \end{aligned}$$
 
-The derivation procedure is identical to the case $N = 1$ from earlier:
+The derivation procedure is identical to the case $$N = 1$$ from earlier:
 
 $$\begin{aligned}
     0
@@ -113,8 +113,8 @@ $$\begin{aligned}
     + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \dv{}{x}\Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x}
 \end{aligned}$$
 
-Once again, $\eta_n(x)$ is arbitrary and disappears at the boundaries,
-so we end up with $N$ equations of the same form as for a single function:
+Once again, $$\eta_n(x)$$ is arbitrary and disappears at the boundaries,
+so we end up with $$N$$ equations of the same form as for a single function:
 
 $$\begin{aligned}
     \boxed{
@@ -127,7 +127,7 @@ $$\begin{aligned}
 
 ## Higher-order derivatives
 
-Suppose that the Lagrangian $L$ depends on multiple derivatives of $f(x)$:
+Suppose that the Lagrangian $$L$$ depends on multiple derivatives of $$f(x)$$:
 
 $$\begin{aligned}
     J[f] = \int_{x_0}^{x_1} L(f, f', f'', ..., f^{(N)}, x) \dd{x}
@@ -144,9 +144,9 @@ $$\begin{aligned}
     &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x}
 \end{aligned}$$
 
-The goal is to turn each $\eta^{(n)}(x)$ into $\eta(x)$, so we need to
-partially integrate the $n$th term of the sum $n$ times. In this case,
-we will need some additional boundary conditions for $\eta(x)$:
+The goal is to turn each $$\eta^{(n)}(x)$$ into $$\eta(x)$$, so we need to
+partially integrate the $$n$$th term of the sum $$n$$ times. In this case,
+we will need some additional boundary conditions for $$\eta(x)$$:
 
 $$\begin{aligned}
     \eta'(x_0) = \eta'(x_1) = 0
@@ -161,7 +161,7 @@ $$\begin{aligned}
     &= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dvn{n}{}{x}\Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x}
 \end{aligned}$$
 
-Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes:
+Once again, because $$\eta(x)$$ is arbitrary, the Euler-Lagrange equation becomes:
 
 $$\begin{aligned}
     \boxed{
@@ -172,16 +172,16 @@ $$\begin{aligned}
 
 ## Multiple coordinates
 
-Suppose now that $f$ is a function of multiple variables.
-For brevity, we only consider two variables $x$ and $y$,
+Suppose now that $$f$$ is a function of multiple variables.
+For brevity, we only consider two variables $$x$$ and $$y$$,
 but the results generalize effortlessly to larger amounts.
-The Lagrangian now depends on all the partial derivatives of $f(x, y)$:
+The Lagrangian now depends on all the partial derivatives of $$f(x, y)$$:
 
 $$\begin{aligned}
     J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
 \end{aligned}$$
 
-The arbitrary deviation $\eta$ is then also a function of multiple variables:
+The arbitrary deviation $$\eta$$ is then also a function of multiple variables:
 
 $$\begin{aligned}
     f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y)
@@ -199,7 +199,7 @@ $$\begin{aligned}
     &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y}
 \end{aligned}$$
 
-We partially integrate for both $\eta_x$ and $\eta_y$, yielding:
+We partially integrate for both $$\eta_x$$ and $$\eta_y$$, yielding:
 
 $$\begin{aligned}
     0
@@ -208,7 +208,7 @@ $$\begin{aligned}
     &\quad + \iint \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big) \bigg) \dd{x} \dd{y}
 \end{aligned}$$
 
-But now, to eliminate these boundary terms, we need extra conditions for $\eta$:
+But now, to eliminate these boundary terms, we need extra conditions for $$\eta$$:
 
 $$\begin{aligned}
     \forall y: \eta(x_0, y) = \eta(x_1, y) = 0
@@ -216,15 +216,15 @@ $$\begin{aligned}
     \forall x: \eta(x, y_0) = \eta(x, y_1) = 0
 \end{aligned}$$
 
-In other words, the deviation $\eta$ must be zero on the whole "box".
-Again relying on the fact that $\eta$ is arbitrary, the Euler-Lagrange
+In other words, the deviation $$\eta$$ must be zero on the whole "box".
+Again relying on the fact that $$\eta$$ is arbitrary, the Euler-Lagrange
 equation is:
 
 $$\begin{aligned}
     0 = \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big)
 \end{aligned}$$
 
-This generalizes nicely to functions of even more variables $x_1, x_2, ..., x_N$:
+This generalizes nicely to functions of even more variables $$x_1, x_2, ..., x_N$$:
 
 $$\begin{aligned}
     \boxed{
@@ -235,14 +235,14 @@ $$\begin{aligned}
 
 ## Constraints
 
-So far, for multiple functions $f_1, ..., f_N$,
-we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$.
-Suppose that we now have $M < N$ constraints $\phi_m$
-that all $f_n$ need to obey, introducing implicit dependencies between them.
+So far, for multiple functions $$f_1, ..., f_N$$,
+we have been assuming that all $$f_n$$ are independent, and by extension all $$\eta_n$$.
+Suppose that we now have $$M < N$$ constraints $$\phi_m$$
+that all $$f_n$$ need to obey, introducing implicit dependencies between them.
 
-Let us consider constraints $\phi_m$ of the two forms below.
+Let us consider constraints $$\phi_m$$ of the two forms below.
 It is important that they are **holonomic**,
-meaning they do not depend on any derivatives of any $f_n(x)$:
+meaning they do not depend on any derivatives of any $$f_n(x)$$:
 
 $$\begin{aligned}
     \phi_m(f_1, ..., f_N, x) = 0
@@ -250,14 +250,14 @@ $$\begin{aligned}
     \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m
 \end{aligned}$$
 
-Where $C_m$ is a constant.
-Note that the first form can also be used for $\phi_m = C_m \neq 0$,
-by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.
+Where $$C_m$$ is a constant.
+Note that the first form can also be used for $$\phi_m = C_m \neq 0$$,
+by simply redefining the constraint as $$\phi_m^0 = \phi_m - C_m = 0$$.
 
-To solve this constrained optimization problem for $f_n(x)$,
-we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$.
-In the former case $\lambda_m(x)$ is a function of $x$, while in the
-latter case $\lambda_m$ is constant:
+To solve this constrained optimization problem for $$f_n(x)$$,
+we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $$\lambda_m$$.
+In the former case $$\lambda_m(x)$$ is a function of $$x$$, while in the
+latter case $$\lambda_m$$ is constant:
 
 $$\begin{aligned}
     \int \lambda_m(x) \: \phi_m(\{f_n\}, x) \dd{x} = 0
@@ -265,15 +265,15 @@ $$\begin{aligned}
     \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m
 \end{aligned}$$
 
-The reason for this distinction in $\lambda_m$
-is that we need to find the stationary points with respect to $\varepsilon$
+The reason for this distinction in $$\lambda_m$$
+is that we need to find the stationary points with respect to $$\varepsilon$$
 of both constraint types. Written in the variational form, this is:
 
 $$\begin{aligned}
     \delta \int \lambda_m \: \phi_m \dd{x} = 0
 \end{aligned}$$
 
-From this, we define a new Lagrangian $\Lambda$ for the functional $J$,
+From this, we define a new Lagrangian $$\Lambda$$ for the functional $$J$$,
 with the contraints built in:
 
 $$\begin{aligned}
@@ -283,7 +283,7 @@ $$\begin{aligned}
     &= \int L + \sum_{m} \lambda_m \phi_m \dd{x}
 \end{aligned}$$
 
-Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$:
+Then we derive the Euler-Lagrange equation as usual for $$\Lambda$$ instead of $$L$$:
 
 $$\begin{aligned}
     0
@@ -297,15 +297,15 @@ $$\begin{aligned}
     + \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x}
 \end{aligned}$$
 
-Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$:
+Using the same logic as before, we end up with a set of Euler-Lagrange equations with $$\Lambda$$:
 
 $$\begin{aligned}
     0
     = \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big)
 \end{aligned}$$
 
-By inserting the definition of $\Lambda$, we then get the following.
-Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$:
+By inserting the definition of $$\Lambda$$, we then get the following.
+Recall that $$\phi_m$$ is holonomic, and thus independent of all derivatives $$f_n'$$:
 
 $$\begin{aligned}
     \boxed{
@@ -316,18 +316,18 @@ $$\begin{aligned}
 
 These are **Lagrange's equations of the first kind**,
 with their second-kind counterparts being the earlier Euler-Lagrange equations.
-Note that there are $N$ separate equations, one for each $f_n$.
+Note that there are $$N$$ separate equations, one for each $$f_n$$.
 
-Due to the constraints $\phi_m$, the functions $f_n$ are not independent.
-This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold,
-i.e. solving a system of $M$ equations for $\lambda_m$:
+Due to the constraints $$\phi_m$$, the functions $$f_n$$ are not independent.
+This is solved by choosing $$\lambda_m$$ such that $$M$$ of the $$N$$ equations hold,
+i.e. solving a system of $$M$$ equations for $$\lambda_m$$:
 
 $$\begin{aligned}
     \dv{}{x}\Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n}
     = \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
 \end{aligned}$$
 
-And then the remaining $N - M$ equations can be solved in the normal unconstrained way.
+And then the remaining $$N - M$$ equations can be solved in the normal unconstrained way.
 
 
 
-- 
cgit v1.2.3