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We have the following equation:

$$mx'' + cx' +k(x^3-x)=0$$

By choosing m,c,k properly, we may simplify it to have the following system:

$$x' = y,\qquad y' = x - x^3 - ay$$

a) Find fixed points of the system

b) Characterize the fixed points' stability by linearization.

I have no idea how to find fixed points (i.e, equilibrium points) of this problem. Any help?

Legendary
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  • If we take $y=0$ then we find $x=0, 1$ and $-1$ . Hence the equilibrium points are $(0,0), (1,0)$ and $(-1,0)$. – Albert Dec 16 '15 at 17:09
  • Thank you! then how can I characterize the fixed points' stability by linearization? – Legendary Dec 16 '15 at 17:12
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    Ok, first consider the Linearized System at any fixed points I mean $Z'=Df(x^{\ast},y^{\ast})Z$ where $(x^{\ast},y^{\ast})$ is a fixed point. Then find the eigenvalues of $Df(x^{\ast},y^{\ast})$. If all eigenvalueshas have positive real part then the fixed point is unstable and if the real parts are negative then it is stable and if the eigenvalues have zero real part then it is a center (and hence stable) and if one eigenvalue is positive and the other one is negative we have a saddle (which is unstable). – Albert Dec 16 '15 at 17:24
  • @Albert Equilibrium which has linearization with zero real part eigenvalues is not necessarily a center :) to some extent, it's almost never a center (all Lyapunov quantities should be zero for return map, which is very restrictive). Check this for examples: http://math.stackexchange.com/questions/1380801/stability-for-higher-dimensional-dynamical-systems/1380930#1380930 – Evgeny Dec 17 '15 at 00:34
  • Yes, you are right and I refer you to "Differential Equations and Dynamical Systems" by L. Perko: http://www.amazon.com/Differential-Equations-Dynamical-Systems-Mathematics/dp/0387951164. – Albert Dec 17 '15 at 06:25

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