Simple version of Gronwall lemma is as follows. Suppose $x(t) \in \mathbb{R}$ is differentiable and satisfies $\dot{x}(t)\leq kx(t)$, then $x(t) \leq x_0e^{kt}$ for all $t \in \mathbb{R}$ where $x_0=x(0)$. For the case $t \geq 0$ it is straight forward as follows but there is a subtlety for $t<0$.

To show this we $\dot{x}(t)\leq kx(t)$, then multiply by $e^{-kt}$ to get $$ e^{-kt}\dot{x}(t)\leq x(t)e^{-kt} \Rightarrow e^{-kt}\dot{x}(t) - x(t)e^{-kt} \leq 0 $$

$$ \frac{d}{dt}(x(t)e^{-kt}) \leq 0 $$

To remove $\frac{d}{dt}$ we need to take integral but we need an assumption on $t$ to proceed. Let $t \geq 0$.

$\int_a^b f(t)dt \leq \int_a^b g(t)dt \tag{1}$ where $b \geq a$.

Because $t\geq 0$, using $(1)$, we can remove the integral $x(t)e^{-kt}|_0^t=x(t)e^{-kt} -x(0)\leq 0$, hence the result.

However, assuming $t<0$, again using $(1)$ we have $x(t)e^{-kt}|_t^0=x(0) - x(t)e^{-kt}\leq 0$ which contradicts with the Gronwall lemma.

Which part am I missing? What is the problem for the case $t<0$? I think I cannot use $(1)$ as I did but why?


Gronwall's inequality states that $x(t)\le x(a)e^{kt}$ for any $t$ in the interval $[a,\infty)$. Gronwall's inequality is only applicable on left-bounded intervals.

One counterexample is

$x(t)=\begin{cases} t^2+2t+2 \ \text{ for } t\le0 \\ 2e^t \ \text{ for } t\ge 0 \end{cases}$

$x(t)$ is differentiable and satisfies $x'(t)\le x(t)$. For any $x_0$, $x_0e^t\rightarrow 0$ as $t\rightarrow -\infty$, whereas $x(t)\rightarrow \infty$ as $t\rightarrow -\infty$.

  • $\begingroup$ Assume it is left bounded and we want to show Gronwall inequality for $[-a,a]$ where $a>0$. For $[0,a]$ we're done. How about $[-a,0)$? $\endgroup$ – Sepide Mar 10 at 7:10

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