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Let's assume $f: \mathbb{R}^d \to \mathbb{R}^d$ is continuous and locally lipschitz. Consider the autonomous ODE $x' = f(x)$. Assume $\exists R > 0. \forall n \in \mathbb{N}. \exists \varphi_n: \mathbb{R} \to \mathbb{R}^d$ which is a $\frac 1 n$-periodic solution and such that $\forall t \in \mathbb{R}. \varphi_n(t) \in \stackrel{-}{B}(0,R)$. Proof that $\exists p \in \stackrel{-}{B}(0,R). f(p) = 0$.

I have the intuition that maybe I could work in $(C[0,1],\|\cdot\|_{\infty})$ extract a converging partial $\varphi_n \to \varphi$ and prove that the limit has to be $0$-periodic. In the language of dynamical systems, we have a sequence of cycles with radius converging to $0$ and have to show that they have a equilibrium as a limit point. What is a nice way of proving this?

Edit

I think that using a Poincaré map can be of use here.

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  • $\begingroup$ Idea: Take the centers of gravity or some other centers of these cycles and use compactness to find an accumulation point. Use continuity and thus boundedness of $f$ to show that this point satisfies the properties for $p$. $\endgroup$ – LutzL Jun 8 at 13:48
  • $\begingroup$ @LutzL so I take an instant of time and observe the solution there. Since it is confined in a compact set there is a partial that converges to some $p$. But how would continuity of $f$ give me that $f(p) = 0$? $\endgroup$ – Javier Jun 9 at 8:08
  • $\begingroup$ I'm afraid that I'm myself not clear on. If $\|f\|<M$ in $\bar B(0,R)$, then the length of the cycle with period $\frac1n$ can not be more than $\frac{M}n$. As a cycle, it must has opposing directions, as they sum to zero. Now apply some intermediate value argument... The problem is that standard examples like $\ddot x+x=0$ have constant period around the center, perhaps $\ddot x+x^3=0$ is an illustration for the situation in the claim. $\endgroup$ – LutzL Jun 9 at 8:25
  • $\begingroup$ @LutzL I actually found the solution in Verhulst's "Nonlinear Differential Equations and Dynamical Systems", theorem 4.8, it is quite elegant. Thank you, I will check your example $\endgroup$ – Javier Jun 9 at 8:57

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