Categories: Mathematics.

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:

\[\begin{aligned} \boxed{ u(t) \le u(0) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg) } \end{aligned}\]

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.

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:

\[\begin{aligned} \boxed{ u(t) \le \alpha(t) + \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s} } \end{aligned}\]

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.

In the special case where \(\alpha(t)\) is non-decreasing with \(t\), the inequality reduces to:

\[\begin{aligned} \boxed{ u(t) \le \alpha(t) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg) } \end{aligned}\]

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:

\[\begin{aligned} u(t) &\le \alpha(t) - \alpha(t) \bigg[ \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \bigg]_{s = 0}^{s = t} \\ &\le \alpha(t) - \alpha(t) \exp\!\bigg( \int_t^t \beta(r) \dd{r} \bigg) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg) \\ &\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg) \end{aligned}\]

- U.H. Thygesen,
*Lecture notes on diffusions and stochastic differential equations*, 2021, Polyteknisk Kompendie.

© Marcus R.A. Newman, a.k.a. "Prefetch".
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