# Prove that the solution is defined on $[0,+\infty)$

I'm currently stuck on the following problem:

Let $n\in\mathbb N$ be a fixed natural number $$t\mapsto \left(x(t),y(t)\right)$$ be one of the solutions to the differential system $$\cases{x'=e^{-y^2}\sin(x^n+y^n),\\ y'=x^n\sin(x^n+y^n).}$$ Then prove that all such maps are defined on $[0,+\infty).$

I don't know how to conclude neatly the problem I admit, in particulare my plan is focused on the condition of strict decrease of $x'$. Then one would like to show that if the maximal interval of definiton were of the form $[0,\alpha),\alpha<+\infty$, then we would find a compact subset which contains the solution. But I cannot delineate it neatly, as i said, so any help is welcomed. Thank you and regards.

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It may help to show that $\mathcal{I}^\prime(t) = 0$, where $\mathcal{I}(t) = \frac{x(t)^{n+1}}{n+1}- \frac{\sqrt{\pi}}{2} \operatorname{erf}(y(t))$. Then, seeing that $\operatorname{erf}: \mathbb{R} \to (-1,1)$ is bounded, $x(t)$ is bounded as well. – Sasha Mar 16 '12 at 18:05

You have directly that $|x'(t)| \leq 1$ thus x is 1-lip. On the other and $|y'(t)| \leq |x(t)|^n \leq t^n +C$ where C is a constant. And $y$ is bounded on every compact because if $[a,b]$ is a compact, $|y(t)| \leq b^nt + C_2 \leq b^{n+1} +C_2$ where $C_2$ is another constant. Then $(x,y)$ is bounded on every compact. Thus your definition interval are not bounded, i,e your solution is defined on $\mathbb{R}$. In fast if you have a differential equation like $X'=f(X)$ where $f$ is $C^1$ the maximal solution couldn't be bounded in time $\textbf{and}$ in space. "If you are bounded you are defined". If you want more details I can proove it for you.

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Nono that was exactly my thoughts and as i hinted at in my post i was sure this had to be the expanation. But, i would like to know how to write it neatly so if you want to add more details feel free to do it. BTW thank you. – uforoboa Mar 17 '12 at 2:09