Asymptotic location of the turning points of $\ddot x = -x-{\dot x}^3$ Fix a solution $x(t)$ of the following ODE: $$\ddot x=-x-\dot {x}^3.$$

*

*Does the energy $E(t) := x^2 + \dot x^2 $ decrease to zero as $t \to \infty$?


*Denote by $x_1,x_2,\ldots $ the positive values of $x(t)$ for which $\dot x(t)=0$, ordered by time. Prove that the following limit exists: $$\lim_{n \to \infty} \frac{x_n-x_{n+1}}{x_n^3}.$$
Some Remarks

*

*To prove 1 it is enough to prove that $\dot x \to 0$, since then $x\to c$ and the ODE forces $c=0$.


*If we had $\dot x \not \to 0$ then $\dot x$ would have been bounded away from $0$ "a lot", so $\dot E=-\dot x ^4$ would decrease at a rate bounded from below, hence tend to zero. This is not yet a proof because $\dot x$ might only have spikes away from zero. Perhaps this will make its second derivative large and lead to a contradiction with the ODE?


*In the linearized equation the sequence $(x_n)$ is constant, and for $t\gg 0$ we expect the system to be very close to its linearization. The limit quantifies this.


*I cross-posted from the physics site.
 A: The answer to question 1 is yes.
The ODE can be written as a system $\dot x = y$, $\dot y = -x - y^3$, and with $E=\frac12(x^2+y^2)$ we have $\dot E = - y^4 \le 0$, so $E$ is a weak Lyapunov function. The set where $\dot E=0$ is the $x$-axis, and it contains no complete trajectories except the origin. (Indeed, the vector field $(\dot x,\dot y)$ equals $(0,-x)$ when $y=0$, so the other trajectories intersect the $x$-axis orthogonally). Therefore LaSalle's theorem says that the origin is an asymptotically stable equilibrium; in fact, it's even globally asymptotically stable, since $E \to \infty$ as $\sqrt{x^2+y^2} \to \infty$. That is, all trajectories converge to the origin as $t \to +\infty$.
A: As found in https://physics.stackexchange.com/questions/622604/asymptotic-frequency-of-nonlinear-oscillator, the main trend of the solution for smaller amplitudes is, starting from $x(t_n)=x_n>0$, $y(t_n)=\dot x(t_n)=0$, $y=\dot x$,
$$
x(t)\approx \frac{x_n\cos(t-t_n)}{\sqrt{1+\frac34x_n^2(t-t_n)}},
\\
y(t)\approx \frac{x_n\sin(t-t_n)}{\sqrt{1+\frac34x_n^2(t-t_n)}}.
$$
Thus the energy is falling, the points $x_n$ are $2\pi$ apart in time and the limit does indeed exist with value $\frac34\pi$.
