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Given the equation $x''(t)+x'(t)+x^3(t)=0$ with $x(0)=1$ and $x'(0)=0$ how to prove that

  1. $ \mathop {\lim }\limits_{t \to + \infty } x\left( t \right) = 0 $.
  2. For every $t>0$ it is $x(t)>0$.

the second question is what I really need. This problem is from G. Gallavotti "The elements of Mechanics"

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Consider the energy function $E=\frac12 x'^2+\frac14x^4$, then the time derivative of that is $$ E'=x'(x''+x^3)=-x'^2\le 0. $$ The only solution that has $x'=0$ constantly is the stationary solution $x=0$. All other solutions continuously lose energy and fall towards the state $E=0$.

The second question is more complicated, you have to show that the friction is over-critical, so that no oscillation occurs. The equation $x''+x^3=0$ without friction is conservative and oscillates along the level curves of $E$, for a small friction coefficient this oscillation persists, giving a spiral in phase space.

enter image description here

As this plot of $x(t)$ shows, a friction coefficient of $0.8$ still crosses the zero line, any proof method for 2. must be quite specific for $c>0.9$.

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  • $\begingroup$ For the first question I got the same conclusion of your answer. The problem is exactly the second. How to prove that we have no oscillations. $\endgroup$ – Luca Goldoni Ph.D. Feb 16 at 18:15
  • $\begingroup$ I got all these conclusions by myself but the problem is how to prove the last statement analytically. According to the author of the book it is possible but I'm stuck. $\endgroup$ – Luca Goldoni Ph.D. Feb 17 at 5:41
  • $\begingroup$ Anyway, thank you for your job. $\endgroup$ – Luca Goldoni Ph.D. Feb 17 at 5:48
  • $\begingroup$ In zooms for $c\in[0.9,1]$ it looks like it has a boundary layer. You can write it as system $x'+cx=v$, $v'=-x^3$, where in some sense the first is the "fast" and the second the "slow" equation. Then when $x$ is small enough, $v$ is slow moving while the first equation restores its equilibrium at $x=v/c$. Insert this approximation into the second to get $cx'=-x^3\implies x^{-2}(t)-x^{-2}(t_1)=2(t-t_1)/c$ as long-time behavior. I'm just not sure how to prove how exact this is, what "small enough" is, and how to connect to the initial piece. $\endgroup$ – Lutz Lehmann Feb 17 at 8:47

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