If $n,k$ are odd numbers, proof that all solutions of the ODE $x^{''}+x^k+x^n=0$ are periodic.

Question

I'm trying to prove this statement:

If $n,k$ are odd numbers, proof that all solutions of the differential equation $$ x^{''}+x^k+x^n=0 \ \ \ \ \ \ \ \ \ \ \ \ (\star) $$ are periodic.

First of all, calling $y=x'$, I rewrite $(\star)$ in system of ODE's notation: $$ \begin{pmatrix} x'\\ y' \end{pmatrix}= \begin{pmatrix}y\\ -f(x) \end{pmatrix}, \ \ \ \ \ \ \ \ \ \ \ \ (\blacksquare) $$ where $f(x)=x^k+x^n$.

I see that the only equilibrium point of $(\blacksquare)$ is the $(0,0)$. Furthermore, I prove that the differentiable function $$ H(x,y):=\frac{x^{k+1}}{k+1}+\frac{x^{n+1}}{n+1}+\frac{y^2}{2}, $$ is an first integral of $(\blacksquare)$, because $$ H_xy-f(x)H_y=0. $$

It's not hard to see that the level curves of the first integral $H$ are periodic. This point is where I get stuck, and I don't know how to continue. Maybe Is there any way to apply the Poncairé-Bendixon's theorem in this point?

I appreciate any help :)

Answer 1

The claim does not hold for all positive integers $k,n$. I conjecture that:

• all orbits are periodic if and only if $\max\{k,n\}$ is an odd number;
• there is no periodic obit if both $k$ and $n$ are even;
• only the orbits associated with $x(0)$ and $x'(0)$ close to $0$ are periodic when \max{k,n}$ is even and \min\{k,n\}$ is odd.

Consider $k=n=2$, then the ODE becomes $x''=-2x^2$. For illustration, is the sampling of trajectories provided by Mathematica:

from where it is apparent that the solution is not bounded from below. Notice $x''<0$ for any $x\neq 0$. Thus if $x'(t)$ is negative at some $t_1$ then it stays negative for all $t>t_1$. So there are only two cases and in non of them the solutions is periodic:

• $x'(t)$ is never negative, and so $x(t)$ grows indefinitely;
• $x'(t)$ is negative starting from some $t_1$, and so $x(t)$ falls indefinitely.