1

Let $f:\mathbb R\times \mathbb R\longrightarrow\mathbb R$ be a $C^1$ function, periodic in the first variable, and such that $f(t,1)\leq > 0\leq f(t,0)$ for all $t$.

Consider the differential equation $$u'(t)=f(t,u(t))$$ and prove that there exists a periodic solution $u$ such that $0\leq u\leq 1$.

It seems that we need the condition $f(t,1)\leq 0\leq f(t,0)$ for periodicity too, since, for example, $f(t,x)=\sin t+2$ provides only non-periodic solutions (correct?).

Moreover, I can't understand how the $C^1$ condition is involved: it assures us for example that $f$ is globally Lipschitz, and that we have $C^1$-dependence on the initial values, but I can't see how this can help us.

W. Rether
  • 3,110
  • 1
    You can refer to this answer of mine to prove the existence of solution http://math.stackexchange.com/questions/1515353/stability-of-periodic-solution/1519112#1519112 . The key picture is the first drawing which shows how vector field is pointing at the borders of some rectangle on phase plane. Feel free to ask any questions. – Evgeny Jul 04 '16 at 06:08
  • Thank you very much. The idea involving Brouwer's theorem is very nice. I have just one problem in transferring your proof to my situation: your function $f=x(1-p(t))(1-x)$ takes strictly negative values for $x$ near to $1$ and strictly positive values for $x$ near to $0$; hence the graph of the vector field. But in my case the integral curves seem to be allowed the rectangle before (i.e. meeting the horizontal sides), since the condition $f(t,1)\leq 0\leq f(t,0)$ doesn't seem to exclude, for example, that $x(t)$ increases near the higher side (with an inflection point on the side). – W. Rether Aug 07 '16 at 20:02
  • Yeah, it seems that they could go out of rectangle, but Bony-Brezis theorem doesn't allow them to do so. This theorem covers the case when tangencies are allowed and very-very simple naive explanation needs some correction :) Even in the previous answer I've implicitly used this theorem. – Evgeny Aug 08 '16 at 06:10
  • Thank you! I'd like to know whether the use of this theorem (which, by the way, I didn't know and turns out to be quite useful) can be avoided: the problem comes from a 2015 test for students who begin the fourth year of Mathematics, and I'm pretty sure that the knowledge of such results is not requested. – W. Rether Aug 08 '16 at 09:48
  • To me in your particular problem Bony-Brezis theorem is pretty much intuitive. If you try to draw a trajectory that starts in the outside of the rectangle, touches its border and continues in the outside of the rectangle, how nearby trajectories should behave? Well, (intuitively) there should be at least one trajectory that goes inside of the rectangle and goes to the outside of it. And this will lead to contradiction with $f(t, 1) \leqslant 0 \leqslant f(t, 0)$ condition. – Evgeny Aug 08 '16 at 14:02

0 Answers0