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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.

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  • $\begingroup$ $|u|$ is not differentiable if $u(t)=0$. $\endgroup$
    – copper.hat
    Commented May 5, 2022 at 19:08
  • $\begingroup$ 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. $\endgroup$ Commented May 5, 2022 at 20:06
  • $\begingroup$ Direct application would lead to the differentiability issue that you've mentioned. $\endgroup$ Commented May 5, 2022 at 20:45

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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 $y$.

Such an approach can be found in Functional Analysis, 2nd Ed. by Kantorovich & Alikov, ch. XVI, Section 1, Theorem 3 (continuity of fixed point) and ch. XVI, Section 4, Theorem 1 (continuity of solution of ODE on initial condition).

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  • $\begingroup$ Great thanks. This counterexample is simple. And the sense of continuous dependence I was looking at is the sense of your second source, continuity w.r.t. initial conditions. $\endgroup$ Commented May 6, 2022 at 18:07

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