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author | Prefetch | 2022-10-20 18:25:31 +0200 |
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committer | Prefetch | 2022-10-20 18:25:31 +0200 |
commit | 16555851b6514a736c5c9d8e73de7da7fc9b6288 (patch) | |
tree | 76b8bfd30f8941d0d85365990bcdbc5d0643cabc /source/know/concept/gronwall-bellman-inequality | |
parent | e5b9bce79b68a68ddd2e51daa16d2fea73b84fdb (diff) |
Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'
Diffstat (limited to 'source/know/concept/gronwall-bellman-inequality')
-rw-r--r-- | source/know/concept/gronwall-bellman-inequality/index.md | 45 |
1 files changed, 23 insertions, 22 deletions
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> |