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I have the following problem:

Consider the Initial Value Problem $x'=f(t,x), \; x(t_0)=x_0 $, where $f:I\times \mathbb{R}^n \to \mathbb{R}^n $ is locally lipschitz with respect to $x$. Suppose that exists $c_1,c_2 \geq 0$ such that $$\| f(t,x) \| \leq c_1 \|x\| + c_2 , \; \forall (t,x)\in I \times \mathbb{R}^n $$ Prove that the maximal solution $\varphi(t)$ of the IVP is defined $\forall t\in I$. Hint: Use the Grönwall's inequality.

I have done this: If $\| f(t,x) \| \leq c_1 \|x\| + c_2 , \; \forall (t,x)\in I \times \mathbb{R}^n $, in particular: $$\| f(t,\varphi(t)) \| =\|\varphi'(t) \|\leq c_1 \|\varphi(t)\| + c_2= c_1\left\|x_0+\int_{t_0}^tf(t,\varphi(s))ds \right\|+c_2 =c_2+c_1\|x_0\|+c_1 \int_{t_0}^t \left\| f(t,\varphi(s)) \right\|ds$$

Then, by the Grönwall's inequality,

$$\|\varphi'(t) \| \leq e^{c_1 |t-t_0|}\left( c_2+c_1 \|x_0 \| \right)$$

The professor said that we should prove firstly that $\varphi(t)$ is bounded, $\forall t \in I$. I have proved that $\varphi '(t) $ is bounded. It implies that $\varphi(t) $ is bounded? And how can I finish the problem?

Thank you very much

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    $\begingroup$ Use $\|φ(t)\|-\|φ(t_0)\|\le\|φ(t)-φ(t_0)\|\le\int_{t_0}^t\|φ'(s)\|ds$. $\endgroup$ – LutzL Oct 17 '18 at 7:48
  • $\begingroup$ Thanks, I have got $$ \| \varphi(t) \| \leq \| \varphi(t_0) \| + \left( \frac{c_2}{c_1} + \| x_0 \| \right) \frac{t-t_0}{|t-t_0|} \left( 1 - e ^{c_1 |t-t_0 |} \right) $$ so $\| \varphi(t) \| $ is bounded. But why for that reason I can deduce that $\varphi(t)$ is defined $\forall t\in \mathbb{R} $ ? $\endgroup$ – Relativo Oct 17 '18 at 21:06
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    $\begingroup$ This seems wrong, as with $t$ large enough the bound becomes negative. Also, the bound should be symmetric to $t_0$. $\endgroup$ – LutzL Oct 17 '18 at 21:25
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Ok, I have got this:

Using the LutzL's comment, $\| \varphi(t) - x_0 \| \leq \left( \frac{c_2}{c_1} + \| x_0 \| \right) \left( e^{c_1 (t-t_0) }-1 \right) $, assuming $t\geq t_0$

So, I have solved this exercise by contradiction. If $I=(a,b)$, suppose that the maximal interval of definition of $\varphi(t)$ is $(t_-,t_+)$, with $t_+< b $. In the interval $[t_0, t_+) $ I have proved that $$\| \varphi(t) - x_0 \| \leq \left( \frac{c_2}{c_1} + \| x_0 \| \right) \left( e^{c_1 (t_+-t_0) }-1 \right) $$ Then, $\varphi(t)\in \bar B(x_0, R)$, $R=\left( \frac{c_2}{c_1} + \| x_0 \| \right) \left( e^{c_1 (t_+-t_0) }-1 \right)$. Let $K$ be the compact $K=[t_0,t_+]\times \bar B(x_0,R)$. By the 'boundary approximation theorem' (I'm not sure what is the name of this theorem in english), $\exists t_K\in (t_-,t_+)$ such that $\forall t\geq t_K, \; (t,\varphi(t)) \notin K$. But $t\in [t_0,t_+]$, so $\varphi(t)\notin \bar B(x_0,R)$, which is a contradiction. Then $t_+=b$, and using the same argument, $t_-=a$. Is it correct?

Thanks!

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    $\begingroup$ That looks good. Note that you get from the negative direction that you can again put the absolute value into the exponential, $...(e^{c_1|t-t_0|}-1)$. $\endgroup$ – LutzL Oct 22 '18 at 11:57

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