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authorPrefetch2022-12-17 18:19:26 +0100
committerPrefetch2022-12-17 18:20:50 +0100
commita39bb3b8aab1aeb4fceaedc54c756703819776c3 (patch)
treeb21ecb4677745fb8c275e54f2ad9d4c2e775a3d8 /source/know/concept/gronwall-bellman-inequality
parent49cc36648b489f7d1c75e1fde79f0990e08dd514 (diff)
Rewrite "Lagrange multiplier", various improvements
Diffstat (limited to 'source/know/concept/gronwall-bellman-inequality')
-rw-r--r--source/know/concept/gronwall-bellman-inequality/index.md17
1 files changed, 9 insertions, 8 deletions
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)