# Initial value problem and grönwall's inequality

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

• Use $\|φ(t)\|-\|φ(t_0)\|\le\|φ(t)-φ(t_0)\|\le\int_{t_0}^t\|φ'(s)\|ds$. – LutzL Oct 17 '18 at 7:48
• 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 $\forall t\in \mathbb{R}$ ? – Relativo Oct 17 '18 at 21:06
• This seems wrong, as with $t$ large enough the bound becomes negative. Also, the bound should be symmetric to $t_0$. – LutzL Oct 17 '18 at 21:25

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