From c705ac1d7dc74709835a8c48fae4a7dd70dc5c49 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 25 Feb 2021 16:14:20 +0100
Subject: Expand knowledge base

---
 .../know/concept/calculus-of-variations/index.pdc  | 40 +++++++++++-----------
 1 file changed, 20 insertions(+), 20 deletions(-)

(limited to 'content/know/concept/calculus-of-variations/index.pdc')

diff --git a/content/know/concept/calculus-of-variations/index.pdc b/content/know/concept/calculus-of-variations/index.pdc
index fb043e0..c5280e5 100644
--- a/content/know/concept/calculus-of-variations/index.pdc
+++ b/content/know/concept/calculus-of-variations/index.pdc
@@ -29,18 +29,18 @@ 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.
 
-If $f(x, \alpha\!=\!0)$ 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 $\alpha$:
+using the parameter $\varepsilon$:
 
 $$\begin{aligned}
-    f(x, \alpha) = f(x, 0) + \alpha \eta(x)
+    f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x)
     \qquad \mathrm{or} \qquad
-    \delta f = \alpha \eta(x)
+    \delta f = \varepsilon \eta(x)
 \end{aligned}$$
 
 Where $\eta(x)$ is an arbitrary differentiable deviation.
-Since $f(x, \alpha)$ must start and end in the same points as $f(x,0)$,
+Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$,
 we have the boundary conditions:
 
 $$\begin{aligned}
@@ -50,16 +50,16 @@ $$\begin{aligned}
 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 $\alpha$,
+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}
     0
     &= \delta J
-    = \pdv{J}{\alpha} \Big|_{\alpha = 0}
-    = \int \pdv{L}{\alpha} \dd{x}
-    = \int \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f'} \pdv{f'}{\alpha} \dd{x}
+    = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \int \pdv{L}{\varepsilon} \dd{x}
+    = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x}
     \\
     &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x}
     = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \frac{d}{dx} \Big( \pdv{L}{f'} \Big) \eta \dd{x}
@@ -99,16 +99,16 @@ In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$,
 satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$:
 
 $$\begin{aligned}
-    f_n(x, \alpha) = f_n(x, 0) + \alpha \eta_n(x)
+    f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x)
 \end{aligned}$$
 
 The derivation procedure is identical to the case $N = 1$ from earlier:
 
 $$\begin{aligned}
     0
-    &= \pdv{J}{\alpha} \Big|_{\alpha = 0}
-    = \int \pdv{L}{\alpha} \dd{x}
-    = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\alpha} + \pdv{L}{f_n'} \pdv{f_n'}{\alpha} \Big) \dd{x}
+    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \int \pdv{L}{\varepsilon} \dd{x}
+    = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\varepsilon} + \pdv{L}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x}
     \\
     &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x}
     \\
@@ -140,9 +140,9 @@ Once again, the derivation procedure is the same as before:
 
 $$\begin{aligned}
     0
-    &= \pdv{J}{\alpha} \Big|_{\alpha = 0}
-    = \int \pdv{L}{\alpha} \dd{x}
-    = \int \pdv{L}{f} \pdv{f}{\alpha} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\alpha} \dd{x}
+    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \int \pdv{L}{\varepsilon} \dd{x}
+    = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\varepsilon} \dd{x}
     \\
     &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x}
 \end{aligned}$$
@@ -187,17 +187,17 @@ $$\begin{aligned}
 The arbitrary deviation $\eta$ is then also a function of multiple variables:
 
 $$\begin{aligned}
-    f(x, y; \alpha) = f(x, y; 0) + \alpha \eta(x, y)
+    f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y)
 \end{aligned}$$
 
 The derivation procedure starts in the exact same way as before:
 
 $$\begin{aligned}
     0
-    &= \pdv{J}{\alpha} \Big|_{\alpha = 0}
-    = \iint \pdv{L}{\alpha} \dd{x} \dd{y}
+    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \iint \pdv{L}{\varepsilon} \dd{x} \dd{y}
     \\
-    &= \iint \pdv{L}{f} \pdv{f}{\alpha} + \pdv{L}{f_x} \pdv{f_x}{\alpha} + \pdv{L}{f_y} \pdv{f_y}{\alpha} \dd{x} \dd{y}
+    &= \iint \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f_x} \pdv{f_x}{\varepsilon} + \pdv{L}{f_y} \pdv{f_y}{\varepsilon} \dd{x} \dd{y}
     \\
     &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y}
 \end{aligned}$$
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