# Gronwall's Inequality with RHS absolute value

Wikipedia states Gronwall's inequality (simplest case on compact domain) as follows. Let $$u$$ and $$\beta$$ be continuous on $$I=[a,b]$$ with $$\beta$$ continuous on $$I$$. If $$u$$ is differentiable on $$I^{0}$$, the interior of $$I$$, and satifies the differential inequality \begin{align} \label{eqn:first} u'(t) \leq \beta(t) u(t), \end{align} then \begin{align} u(t) \leq u(a) exp\left(\int_{a}^{t} \beta(s) ds\right). \end{align}

What happens if I replace the right-hand side of my differential inequality with $$|u(t)|$$? That is, change none of the assumptions to the theorem except for weakening the inequality to \begin{align} u'(t) \leq \beta(t) \color{red}{|u(t)|}. \end{align} My question: if I change only the absolute value term in red above, does the conclusion \begin{align} u(t) \leq u(a) exp\left(\int_{a}^{t} \beta(s)ds\right) \end{align} still hold?

For further context: In my lecture notes for proving continuous dependence for simple ODE, we used a Lipschitz bound to derive the differential inequality \begin{align} E'(t) &\leq L |E(t)| \\ E(0) &= \epsilon \end{align} for some constant $$L$$. Then apply Gronwall's inequality. My concern with this has to do with the fact that $$|u(t)|$$ is not necessarily differentiable if I were to have $$u(t)=0$$. At that point, I am violating the assumptions of Gronwall's lemma.

• $|u|$ is not differentiable if $u(t)=0$. Commented May 5, 2022 at 19:08
• Yes, I'm looking to weaken the assumption of Gronwall's inequality. The usual Gronwall's inequality cannot be applied directly to conclude the answer to my question. Commented May 5, 2022 at 20:06
• Direct application would lead to the differentiability issue that you've mentioned. Commented May 5, 2022 at 20:45

Your formula for the conclusion of the Bellman Gronwall lemma is incorrect, it should read $$u(t) \le u(a) e^{\int_a^t \beta(s)ds}$$.
Choose $$\beta = 1, a=0, u(t) = -1-t$$.
Then $$u'(t) = -1$$, and $$u'(t) \le \beta |u(t)|$$, but the conclusion would read $$u(t) = -1-t \le (-1) e^t$$ which is clearly violated for a finite value of $$t>a$$.
Aside: It is not clear what 'continuous dependence' means in your question, but one nice approach to proving continuity is to show, under appropriate assumptions, that the fixed point solution of an operator $$P_y$$ (that is, the solution to $$x = P_y(x)$$) is continuously dependent (in a prescribed sense) on the parameter $$y$$.