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I am studiyng the stability of the critical points of the system: \begin{cases} x'=y \\ y'=-x+x^3 .\end{cases} This system has 3 critical points, two of the critical points are easy to study, but I am stuck with the point $(0,0)$. If I linearize the system near this point I obtain the system \begin{cases} x'=y \\ y'=-x \end{cases} The eigenvalues of the corresponding matrix are $\pm i$, so I can deduce that the linearized system is stable but not asynstotically stable. The problem is that i cannot say anything about the behaviour of the original system, because the real part of the eigenvalues is zero.

Doing some numerically simulation of the system in the phase space, it seems that the point $(0,0)$ is not only stable, but asymptotically stable. How can I prove that?

Thanks!

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You can think of this system as $x''=-F'(x)$ where $F(x)=x^2/2-x^4/4$. By the usual line of reasoning about Hamiltonian dynamics, this means that (returning to your notation) $\frac{1}{2} y^2 + \frac{1}{2} x^2 - \frac{1}{4} x^4$ is a conserved quantity. The potential energy here has a minimum at $x=0$ and has maxima at $x=\pm 1$, with the value being $1/4$ at these maxima. Thus if $\frac{1}{2} y_0^2 + \frac{1}{2} x_0^2 - \frac{1}{4} x_0^4<1/4$ then the dynamics oscillate back and forth; if it is exactly equal to $1/4$ then the dynamics asymptotically approach $(1,0)$ or $(-1,0)$ (depending on the sign of $y_0$); if it is greater than $1/4$ then the dynamics get over the energy barrier (or were already over it) and then "fall" to $+\infty$ or $-\infty$ again depending on the sign of $y_0$.

If you are seeing decay to $(0,0)$ in your numerical simulations then you have some numerical dissipation going on; in this class of problems that needs to be avoided (at least for long term dynamics) by the use of symplectic integrators (which maintain a conservation law exactly, but a slightly different one than the one in the continuous dynamics).

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