Grönwall-Bellman inequality

Proof. We define $w(t)$ as equal to 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 $w(0) = u(0)$. Then 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.

Proof. 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 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:

$$\begin{aligned} w'(t) \le \alpha(t) \: \beta(t) \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg) \end{aligned}$$

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 $w(t)$:

$$\begin{aligned} \int_0^t \beta(s) \: u(s) \dd{s} &= w(t) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg) \\ &\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s} \\ &\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \end{aligned}$$

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

Proof. 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 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 normal 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:

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