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---
title: "Grönwall-Bellman inequality"
firstLetter: "G"
publishDate: 2021-11-07
categories:
- Mathematics
date: 2021-11-07T09:51:57+01:00
draft: false
markup: pandoc
---
# Grönwall-Bellman inequality
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:
$$\begin{aligned}
u'(t)
\le \beta(t) \: u(t)
\end{aligned}$$
Where $\beta(t)$ is known.
Then **Grönwall's inequality** states that the solution $u(t)$ is bounded:
$$\begin{aligned}
\boxed{
u(t)
\le u(0) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-original"/>
<label for="proof-original">Proof</label>
<div class="hidden">
<label for="proof-original">Proof.</label>
We define $w(t)$ to equal the upper bounds above
on both $w'(t)$ and $w(t)$ itself:
$$\begin{aligned}
w(t)
\equiv u(0) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
\quad \implies \quad
w'(t)
= \beta(t) \: w(t)
\end{aligned}$$
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 $u(t)$
in order to satisfy the inequality.
We thus calculate:
$$\begin{aligned}
\dv{t} \bigg( \frac{u}{w} \bigg)
= \frac{u' w - u w'}{w^2}
= \frac{u' w - u \beta w}{w^2}
= \frac{u' - u \beta}{w}
\end{aligned}$$
Since $u' \le \beta u$ as a condition,
the above derivative is always negative.
</div>
</div>
Grönwall's inequality can be generalized to non-differentiable functions.
Suppose we know:
$$\begin{aligned}
u(t)
\le \alpha(t) + \int_0^t \beta(s) \: u(s) \dd{s}
\end{aligned}$$
Where $\alpha(t)$ and $\beta(t)$ are known.
Then the **Grönwall-Bellman inequality** states that:
$$\begin{aligned}
\boxed{
u(t)
\le \alpha(t) + \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-integral"/>
<label for="proof-integral">Proof</label>
<div class="hidden">
<label for="proof-integral">Proof.</label>
We start by defining $w(t)$ as follows,
which will act as shorthand:
$$\begin{aligned}
w(t)
\equiv \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg) \bigg( \int_0^t \beta(s) \: u(s) \dd{s} \bigg)
\end{aligned}$$
Its derivative $w'(t)$ is then straightforwardly calculated to be given by:
$$\begin{aligned}
w'(t)
&= \bigg( \dv{t}\! \int_0^t \beta(s) \: u(s) \dd{s} - \beta(t)\int_0^t \beta(s) \: u(s) \dd{s} \bigg)
\exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
\\
&= \beta(t) \bigg( u(t) - \int_0^t \beta(s) \: u(s) \dd{s} \bigg)
\exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
\end{aligned}$$
The parenthesized expression it bounded from above by $\alpha(t)$,
thanks to the condition that $u(t)$ is assumed to satisfy,
for the Grönwall-Bellman inequality to be true:
$$\begin{aligned}
w'(t)
\le \alpha(t) \: \beta(t) \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
\end{aligned}$$
Integrating this to find $w(t)$ yields the following result:
$$\begin{aligned}
w(t)
\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s}
\end{aligned}$$
In the initial definition of $w(t)$,
we now move the exponential to the other side,
and rewrite it using the above inequality for $w(t)$:
$$\begin{aligned}
\int_0^t \beta(s) \: u(s) \dd{s}
&= w(t) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
\\
&\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s}
\\
&\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
\end{aligned}$$
Insert this into the condition under which the Grönwall-Bellman inequality holds.
</div>
</div>
In the special case where $\alpha(t)$ is non-decreasing with $t$,
the inequality reduces to:
$$\begin{aligned}
\boxed{
u(t)
\le \alpha(t) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
}
\end{aligned}$$
<div class="accordion">
<input type="checkbox" id="proof-special"/>
<label for="proof-special">Proof</label>
<div class="hidden">
<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:
$$\begin{aligned}
u(t)
&\le \alpha(t) + \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
\\
&\le \alpha(t) + \alpha(t) \int_0^t \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
\end{aligned}$$
Now, consider the following straightfoward identity, involving the exponential:
$$\begin{aligned}
\dv{s} \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
&= - \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
\end{aligned}$$
By inserting this into Grönwall-Bellman inequality, we arrive at:
$$\begin{aligned}
u(t)
&\le \alpha(t) - \alpha(t) \int_0^t \dv{s} \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
\\
&\le \alpha(t) - \alpha(t) \bigg[ \int \dv{s} \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s} \bigg]_{s = 0}^{s = t}
\end{aligned}$$
Where we have converted the outer integral from definite to indefinite.
Continuing:
$$\begin{aligned}
u(t)
&\le \alpha(t) - \alpha(t) \bigg[ \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \bigg]_{s = 0}^{s = t}
\\
&\le \alpha(t) - \alpha(t) \exp\!\bigg( \int_t^t \beta(r) \dd{r} \bigg) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
\\
&\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
\end{aligned}$$
</div>
</div>
## References
1. U.H. Thygesen,
*Lecture notes on diffusions and stochastic differential equations*,
2021, Polyteknisk Kompendie.
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