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I'm studying a proof from a unique solution theorem of an ODE where I encountered an inequality. I don't know if this is applied Gronwall, unfortunately I can't see it.

So $y'(s)=f(t,y(s))$ and $f$ is a continuous and Lipschitz $$\|f(t,y)-f(t,z)\|\leq L\|y-z\| $$ And we definie this norm $\|y\|_{C^0}=\max (e^{-2(s-t_0)}\|y(s)\|)$

Then this inequality occurs

$$\int_{t_0}^t L\|y(s)-z(s)\|\mathrm{d}s\leq \left(\int_{t_0}^t Le^{2L(s-t_0)}\mathrm{d}s \right)\|y-z\|_{C^0}$$

Is this Gronwall? It confuses me that there is an integral on the LHS. I computed RHS of the integral but I'm not seeing the inequality:

$$\left(\int_{t_0}^t Le^{2L(s-t_0)}\mathrm{d}s \right)\|y-z\|_{C^0}=\frac{1}{2} \left(e^{2 L (t - t_0)} - 1\right)\max (e^{-2(s-t_0)}\|y(s)-z(s)\|)$$

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    $\begingroup$ Are you sure the norm does not involve and $L$ in the exponent, or that the inequality does not have an $L$ in the exponent? $\endgroup$ – Aloizio Macedo Mar 27 '17 at 16:02
  • $\begingroup$ @AloizioMacedo atleast in my proof there is no $L$ in the norm. $\endgroup$ – WaldoRozir Mar 27 '17 at 16:06
  • $\begingroup$ If this were not a typo in the script, then there should also be no $L$ in the exponentials inside the integrals, since they stem exclusively from the application of the norm definition. $\endgroup$ – LutzL Mar 27 '17 at 18:05
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As of now, this is a non-answer. But other answers suggest that OP has an error, so I wrote it down.

If the norm is given by $\|y\|_{C^0}=\max (e^{-2L(s-t_0)}\|y(s)\|)$, then

$$\int_{t_0}^t L\|y(s)-z(s)\|ds=\int_{t_0}^t Le^{2L(s-t_0)}e^{-2L(s-t_0)}\Vert y(s)-z(s) \Vert ds$$ $$\leq \int_{t_0}^t L e^{2L(s-t_0)}\max\{e^{-2L(s-t_0)}\Vert y(s)-z(s)\Vert\}ds $$ $$ =\Vert y-z\Vert_{C^0}\int_{t_0}^t Le^{2L(s-t_0)}ds. $$

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  • $\begingroup$ Thank you very much. Yes it was probably a typo. Now everything is much clearer, especially your way is very nice. $\endgroup$ – WaldoRozir Mar 27 '17 at 19:43
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In the inequality, $y$ and $z$ are solutions of the ODE. Then \begin{align} y(t)&=y(t_0)+\int_{t_0}^tf(s,y(s))\,ds,\\ z(t)&=z(t_0)+\int_{t_0}^tf(s,z(s))\,ds. \end{align} If $y(t_0)=z(t_0)$ then $$ \|y(t)-z(t)\|\le\int_{t_0}^t\|f(s,y(s))-f(s,z(s))\|\,ds\le L\int_{t_0}^t\|y(s)-z(s)\|\,ds. $$ Can you finish from here?

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  • $\begingroup$ Unfortunately not. The last expression is the point where I'm stucked. I tried applying the integral version of Gronwall, but I failed at seeing which function is which. Then I tried to do it without Gronwall but when I replace the values from your expression I get: $$L\int_{t_0}^t \left|\left|\left(\int_{t_0}^s f(q,y(q)) -f(q,z(q)) \mathrm{d}q \right)\right|\right| \mathrm{d}s$$ Now I don't think that the $f(q,y(q)) -f(q,z(q))$ cancel out? $\endgroup$ – WaldoRozir Mar 27 '17 at 16:37
  • $\begingroup$ As Aloizio Macedo suggest in a comment, the exponentials do not look right (in one we have $2$ and in others $L$.) Also, the fact that $L$ appears on both sides of the inequality doesn't seem right. $\endgroup$ – Julián Aguirre Mar 27 '17 at 17:10
  • $\begingroup$ I see, then my script is wrong. Thank you for the input. $\endgroup$ – WaldoRozir Mar 27 '17 at 17:44
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This modified norm is part of an "alternative" proof of the Picard-Lindelöf theorem that provides global convergence if there is a global $y$-Lipschitz constant for $f$ on a domain $[a,b]×\Bbb R^n$. In the inequalities in the question, there is nothing more done than defining and applying this norm. Grönwall is only interesting in that it shows that the bound on the divergence of solutions is $e^{L(s-t_0)}$ so that the modified norm (with factor $e^{-2L(s-t_0)}$) will suppress "events" resp. the behavior of the solution at large values of $s$ in favor for values close to the initial point.

The last inequality should thus, using $\|x(s)-y(s)\|\le e^{2L(s-t_0)}\|x-y\|_{C^0}$, read as \begin{align} e^{-2(t-t_0)}\|x_+(t)-y_+(t)\|&\le e^{-2(t-t_0)}\int_{t_0}^tL\|x(s)-y(s)\|\,ds\\ &\le \|x-y\|_{C^0}\int_{t_0}^tLe^{2L(s-t)}\,ds\\ &=\frac{1-e^{-2L(t-t_0))}}2\|x-y\|_{C^0} \end{align} so that $$ \|x_+-y_+\|_{C^0}\le\frac12\|x-y\|_{C^0} $$ where $x_+(t)=x_0+\int_{t_0}^t f(s,x(s))\,ds$ etc.

The more common proof (for example reproduced here: Existence of solution to first order ODE. What is there to be proved?) only assumes the Lipschitz condition on a bounded cylindrical neighborhood of the initial point but then only gives you the existence and uniqueness of a solution on a small neighborhood $(t_0-\epsilon,t_0+ϵ)$ where $ϵ$ depends on the scale of $f$ and on its Lipschitz constant.

Both variants of the theorem have their pro's and con's, and in the end they both can serve to provide the background for maximal solutions and flows of vector fields.

For an alternative proof of the first variant without modified norm see for instance https://math.stackexchange.com/a/1587871/115115

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  • $\begingroup$ As I pointed out, the modified norm does not have a factor of $e^{-2L(s-t_0)}$, only of $e^{-2(s-t_0)}$. It may be an error, but OP says it is not. $\endgroup$ – Aloizio Macedo Mar 27 '17 at 17:55
  • $\begingroup$ @AloizioMacedo : In a comment of the first answer he conceded that it may be a printing/typing error in the script. $\endgroup$ – LutzL Mar 27 '17 at 18:01
  • $\begingroup$ Thank you very much for this detailed answer! $\endgroup$ – WaldoRozir Mar 27 '17 at 18:36

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