# How to show Gronwall Lemma for $t<0$?

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.
$$x(t)=\begin{cases} t^2+2t+2 \ \text{ for } t\le0 \\ 2e^t \ \text{ for } t\ge 0 \end{cases}$$
$$x(t)$$ 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 $$t\rightarrow -\infty$$.
• 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 $[-a,0)$? – Sepide Mar 10 at 7:10