# Bound the maximal solution of ODE system

Let $$(x,y)$$ be the maximal solution of $$\begin{cases}x' = cos(x) (t-y^2) \\ y' = xy\end{cases}$$ with $$x(0) = 0$$ and $$y(0) = 1$$.

I'm asked to prove that $$|x(t)| < \frac \pi 2, 0 < y(t) < e^{\frac \pi 2 t}$$ in the domain of $$(x,y)$$.

• What is your idea? Could you show the claim if the first equation were $x'=\cos(x)\phi(t)$ with some given function $\phi$? – LutzL Jun 12 at 15:10
• @LutzL I didn't elaborate on my ideas to not confuse readers. But in principle second could be easy after having solved the first for some monotony argument. Also it did remind me to the argument you used with the nullclines in a past question but the nullcline $t = y^2$ is difficult to plot since this time is a two dimensional system. there is another argument that was used several times in my lecture called "first-instant" lemma informing of what happens if $x$ ever reaches $\frac \pi 2$ (then there is a first time such that...). I'll think for a while your suggestion – Javier Jun 12 at 15:29
• The guidelines for this site demand the opposite "Provide context - Include your own work". You can use formatting, for instance the highlight markup, or section headings, to separate the task, your own ideas and the questions about your solution attempt. – LutzL Jun 12 at 15:39

## 1 Answer

Any solution that starts in one of the planes $$\{(t,x,y):\cos x=0\}$$ stays inside that plane.

No other solution can can cross these planes by the uniqueness theorem.

Initially, $$x(0)=0\in(-\frac\pi2,\frac\pi2)$$, so that $$x(t)\in(-\frac\pi2,\frac\pi2)$$ for the full solution.

The bound on $$y$$ follows, you might cite the Grönwall lemma for this.

The surface $$\{(t,x,y):y^2-t\}=0$$ has no influence on the claims, but the sign of the factor will influence the long-term behavior of the solution inside these bounds.

• if I'm not mistaken $y > 0$ also follows because otherwise the uniqueness property would be violated (if at some point the solution has $y = 0$ it should remain constant for all other instant) right? – Javier Jun 12 at 16:47
• Yes, that too. $y(t)=y_0\exp(\int_0^t x(s)ds)$ keeps its sign. – LutzL Jun 12 at 16:52