Categories: Mathematics.

Grönwall-Bellman inequality

Suppose we have a first-order ordinary differential equation for some function u(t)u(t), and assume that we can prove from this equation that the derivative u(t)u'(t) is bounded as follows:

u(t)β(t)u(t)\begin{aligned} u'(t) \le \beta(t) \: u(t) \end{aligned}

Where β(t)\beta(t) is known. Then Grönwall’s inequality states that the solution u(t)u(t) is bounded:

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

We define w(t)w(t) as equal to the upper bounds above on both w(t)w'(t) and w(t)w(t) itself:

w(t)u(0)exp ⁣(0tβ(s)ds)    w(t)=β(t)w(t)\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)w(0) = u(0). Then the goal is to show the following for all tt:

u(t)w(t)1\begin{aligned} \frac{u(t)}{w(t)} \le 1 \end{aligned}

For t=0t = 0, this is trivial, since w(0)=u(0)w(0) = u(0) by definition. For t>0t > 0, we want w(t)w(t) to grow at least as fast as u(t)u(t) in order to satisfy the inequality. We thus calculate:

ddt(uw)=uwuww2=uwuβww2=uuβw\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 uβuu' \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:

u(t)α(t)+0tβ(s)u(s)ds\begin{aligned} u(t) \le \alpha(t) + \int_0^t \beta(s) \: u(s) \dd{s} \end{aligned}

Where α(t)\alpha(t) and β(t)\beta(t) are known. Then the Grönwall-Bellman inequality states that:

u(t)α(t)+0tα(s)β(s)exp ⁣(stβ(r)dr)ds\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)w(t) as follows, which will act as shorthand:

w(t)exp ⁣( ⁣ ⁣ ⁣0tβ(s)ds)(0tβ(s)u(s)ds)\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)w'(t) is then straightforwardly calculated to be given by:

w(t)=(ddt0tβ(s)u(s)dsβ(t)0tβ(s)u(s)ds)exp ⁣( ⁣ ⁣ ⁣0tβ(s)ds)=β(t)(u(t)0tβ(s)u(s)ds)exp ⁣( ⁣ ⁣ ⁣0tβ(s)ds)\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 is bounded from above by α(t)\alpha(t), thanks to the condition that u(t)u(t) is assumed to satisfy, for the Grönwall-Bellman inequality to be true:

w(t)α(t)β(t)exp ⁣( ⁣ ⁣ ⁣0tβ(s)ds)\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)w(t) yields the following result:

w(t)0tα(s)β(s)exp ⁣( ⁣ ⁣ ⁣0sβ(r)dr)ds\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)w(t), we now move the exponential to the other side, and rewrite it using the above inequality for w(t)w(t):

0tβ(s)u(s)ds=w(t)exp ⁣(0tβ(s)ds)0tα(s)β(s)exp ⁣(0tβ(r)dr)exp ⁣( ⁣ ⁣ ⁣0sβ(r)dr)ds0tα(s)β(s)exp ⁣(stβ(r)dr)\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}

This yields the desired result after inserting it into the condition under which the Grönwall-Bellman inequality holds.

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

u(t)α(t)exp ⁣(0tβ(s)ds)\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 α(t)\alpha(t) is non-decreasing tells us that α(s)α(t)\alpha(s) \le \alpha(t) for all sts \le t, so:

u(t)α(t)+0tα(s)β(s)exp ⁣(stβ(r)dr)dsα(t)+α(t)0tβ(s)exp ⁣(stβ(r)dr)ds\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 straightforward identity, involving the exponential:

ddsexp ⁣(stβ(r)dr)=β(s)exp ⁣(stβ(r)dr)\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 normal Grönwall-Bellman inequality, we arrive at:

u(t)α(t)α(t)0tddsexp ⁣(stβ(r)dr)dsα(t)α(t)[ddsexp ⁣(stβ(r)dr)ds]s=0s=t\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:

u(t)α(t)α(t)[exp ⁣(stβ(r)dr)]s=0s=tα(t)α(t)exp ⁣(ttβ(r)dr)+α(t)exp ⁣(0tβ(r)dr)α(t)α(t)+α(t)exp ⁣(0tβ(r)dr)\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}


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