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'

---
 .../concept/gronwall-bellman-inequality/index.md   | 45 +++++++++++-----------
 1 file changed, 23 insertions(+), 22 deletions(-)

(limited to 'source/know/concept/gronwall-bellman-inequality')

diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md
index 417b033..8096aaf 100644
--- a/source/know/concept/gronwall-bellman-inequality/index.md
+++ b/source/know/concept/gronwall-bellman-inequality/index.md
@@ -8,16 +8,16 @@ layout: "concept"
 ---
 
 Suppose we have a first-order ordinary differential equation
-for some function $u(t)$, and that it can be shown from this equation
-that the derivative $u'(t)$ is bounded as follows:
+for some function $$u(t)$$, and that it can be shown from this equation
+that the derivative $$u'(t)$$ is bounded as follows:
 
 $$\begin{aligned}
     u'(t)
     \le \beta(t) \: u(t)
 \end{aligned}$$
 
-Where $\beta(t)$ is known.
-Then **Grönwall's inequality** states that the solution $u(t)$ is bounded:
+Where $$\beta(t)$$ is known.
+Then **Grönwall's inequality** states that the solution $$u(t)$$ is bounded:
 
 $$\begin{aligned}
     \boxed{
@@ -31,8 +31,8 @@ $$\begin{aligned}
 <label for="proof-original">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-original">Proof.</label>
-We define $w(t)$ to equal the upper bounds above
-on both $w'(t)$ and $w(t)$ itself:
+We define $$w(t)$$ to equal the upper bounds above
+on both $$w'(t)$$ and $$w(t)$$ itself:
 
 $$\begin{aligned}
     w(t)
@@ -42,15 +42,15 @@ $$\begin{aligned}
     = \beta(t) \: w(t)
 \end{aligned}$$
 
-Where $w(0) = u(0)$.
-The goal is to show the following for all $t$:
+Where $$w(0) = u(0)$$.
+The goal is to show the following for all $$t$$:
 
 $$\begin{aligned}
     \frac{u(t)}{w(t)} \le 1
 \end{aligned}$$
 
-For $t = 0$, this is trivial, since $w(0) = u(0)$ by definition.
-For $t > 0$, we want $w(t)$ to grow at least as fast as $u(t)$
+For $$t = 0$$, this is trivial, since $$w(0) = u(0)$$ by definition.
+For $$t > 0$$, we want $$w(t)$$ to grow at least as fast as $$u(t)$$
 in order to satisfy the inequality.
 We thus calculate:
 
@@ -61,7 +61,7 @@ $$\begin{aligned}
     = \frac{u' - u \beta}{w}
 \end{aligned}$$
 
-Since $u' \le \beta u$ as a condition,
+Since $$u' \le \beta u$$ as a condition,
 the above derivative is always negative.
 </div>
 </div>
@@ -74,7 +74,7 @@ $$\begin{aligned}
     \le \alpha(t) + \int_0^t \beta(s) \: u(s) \dd{s}
 \end{aligned}$$
 
-Where $\alpha(t)$ and $\beta(t)$ are known.
+Where $$\alpha(t)$$ and $$\beta(t)$$ are known.
 Then the **Grönwall-Bellman inequality** states that:
 
 $$\begin{aligned}
@@ -89,7 +89,7 @@ $$\begin{aligned}
 <label for="proof-integral">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-integral">Proof.</label>
-We start by defining $w(t)$ as follows,
+We start by defining $$w(t)$$ as follows,
 which will act as shorthand:
 
 $$\begin{aligned}
@@ -97,7 +97,7 @@ $$\begin{aligned}
     \equiv \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg) \bigg( \int_0^t \beta(s) \: u(s) \dd{s} \bigg)
 \end{aligned}$$
 
-Its derivative $w'(t)$ is then straightforwardly calculated to be given by:
+Its derivative $$w'(t)$$ is then straightforwardly calculated to be given by:
 
 $$\begin{aligned}
     w'(t)
@@ -108,8 +108,8 @@ $$\begin{aligned}
     \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
 \end{aligned}$$
 
-The parenthesized expression it bounded from above by $\alpha(t)$,
-thanks to the condition that $u(t)$ is assumed to satisfy,
+The parenthesized expression it bounded from above by $$\alpha(t)$$,
+thanks to the condition that $$u(t)$$ is assumed to satisfy,
 for the Grönwall-Bellman inequality to be true:
 
 $$\begin{aligned}
@@ -117,16 +117,16 @@ $$\begin{aligned}
     \le \alpha(t) \: \beta(t) \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
 \end{aligned}$$
 
-Integrating this to find $w(t)$ yields the following result:
+Integrating this to find $$w(t)$$ yields the following result:
 
 $$\begin{aligned}
     w(t)
     \le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s}
 \end{aligned}$$
 
-In the initial definition of $w(t)$,
+In the initial definition of $$w(t)$$,
 we now move the exponential to the other side,
-and rewrite it using the above inequality for $w(t)$:
+and rewrite it using the above inequality for $$w(t)$$:
 
 $$\begin{aligned}
     \int_0^t \beta(s) \: u(s) \dd{s}
@@ -141,7 +141,7 @@ Insert this into the condition under which the Grönwall-Bellman inequality hold
 </div>
 </div>
 
-In the special case where $\alpha(t)$ is non-decreasing with $t$,
+In the special case where $$\alpha(t)$$ is non-decreasing with $$t$$,
 the inequality reduces to:
 
 $$\begin{aligned}
@@ -157,8 +157,8 @@ $$\begin{aligned}
 <div class="hidden" markdown="1">
 <label for="proof-special">Proof.</label>
 Starting from the "ordinary" Grönwall-Bellman inequality,
-the fact that $\alpha(t)$ is non-decreasing tells us that
-$\alpha(s) \le \alpha(t)$ for all $s \le t$, so:
+the fact that $$\alpha(t)$$ is non-decreasing tells us that
+$$\alpha(s) \le \alpha(t)$$ for all $$s \le t$$, so:
 
 $$\begin{aligned}
     u(t)
@@ -194,6 +194,7 @@ $$\begin{aligned}
     \\
     &\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
 \end{aligned}$$
+
 </div>
 </div>
 
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