From a39bb3b8aab1aeb4fceaedc54c756703819776c3 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 17 Dec 2022 18:19:26 +0100
Subject: Rewrite "Lagrange multiplier", various improvements

---
 .../know/concept/gronwall-bellman-inequality/index.md   | 17 +++++++++--------
 1 file changed, 9 insertions(+), 8 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 da1bcad..0d6db71 100644
--- a/source/know/concept/gronwall-bellman-inequality/index.md
+++ b/source/know/concept/gronwall-bellman-inequality/index.md
@@ -7,8 +7,8 @@ categories:
 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
+Suppose we have a first-order ordinary differential equation for some function $$u(t)$$,
+and assume that we can prove from this equation
 that the derivative $$u'(t)$$ is bounded as follows:
 
 $$\begin{aligned}
@@ -28,7 +28,7 @@ $$\begin{aligned}
 
 
 {% include proof/start.html id="proof-original" -%}
-We define $$w(t)$$ to equal the upper bounds above
+We define $$w(t)$$ as equal to the upper bounds above
 on both $$w'(t)$$ and $$w(t)$$ itself:
 
 $$\begin{aligned}
@@ -40,7 +40,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $$w(0) = u(0)$$.
-The goal is to show the following for all $$t$$:
+Then the goal is to show the following for all $$t$$:
 
 $$\begin{aligned}
     \frac{u(t)}{w(t)} \le 1
@@ -102,7 +102,7 @@ $$\begin{aligned}
     \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
 \end{aligned}$$
 
-The parenthesized expression it bounded from above by $$\alpha(t)$$,
+The parenthesized expression is 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:
 
@@ -131,7 +131,8 @@ $$\begin{aligned}
     &\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
 \end{aligned}$$
 
-Insert this into the condition under which the Grönwall-Bellman inequality holds.
+This yields the desired result after inserting it
+into the condition under which the Grönwall-Bellman inequality holds.
 {% include proof/end.html id="proof-integral" %}
 
 
@@ -158,14 +159,14 @@ $$\begin{aligned}
     &\le \alpha(t) + \alpha(t) \int_0^t \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
 \end{aligned}$$
 
-Now, consider the following straightfoward identity, involving the exponential:
+Now, consider the following straightforward identity, involving the exponential:
 
 $$\begin{aligned}
     \dv{}{s}\exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
     &= - \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
 \end{aligned}$$
 
-By inserting this into Grönwall-Bellman inequality, we arrive at:
+By inserting this into normal Grönwall-Bellman inequality, we arrive at:
 
 $$\begin{aligned}
     u(t)
-- 
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