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I was trying to do linear stability analysis of spring pendulum. I arrived at the differential equations which describe the system. But I am unable to proceed to linear stability analysis. Is it possible to do linear stability analysis on 2nd order differential equations by finding eigen-values of Jacobian matrix? The equations are as below:

$$mr'' - m(L+r)(\theta')^2-mg\cos(\theta)+kr' = 0$$ and

$$(L+r)\theta''+2r'\theta'+g\sin(\theta)=0.$$


UPDATE : The fixed points are $(mg/k,0)$ and $(-mg/k,\pi)$

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    $\begingroup$ I'd start by linearizing the problem about a stationary point. Then reorganize into a system of first order linear o.d.e.s, the eigenvalues of that matrix should be <1. $\endgroup$ Commented Sep 12, 2014 at 6:11
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    $\begingroup$ @user3823992 Don't you mean that the eigenvalues should have negative real part, since an eigenvalue $\lambda$ corresponds to an $\exp(\lambda\,t)$ term in the solution of the linearised DEs? ($|\lambda_j|<1$ is the criterion for discrete time systems, i.e. coupled recurrence relationships, since $\lambda_j$ then corresponds to a $\lambda_j^n$ term in the solution of linearised equations). $\endgroup$ Commented Sep 12, 2014 at 7:43
  • $\begingroup$ Yeah, quite right. Negative real parts. $\endgroup$ Commented Sep 13, 2014 at 6:27

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I take the core of the question to be

Is it possible to do linear stability analysis on 2nd order differential equations by finding eigen values of Jacobian matrix?

The answer is yes, but first you have to convert your second-order equations into first-order ones.

This is actually pretty easy to do: every time you see a second derivative, e.g. $x''$, you just introduce a new variable, $y=x'$. Then you can replace $x''$ with $y'$. The equation $x'=y$ should be added to the set of equations, so in your case your two second-order equations will become four first-order ones. Then after rearranging the equations to put all the derivatives on the left-hand side, you can write down the Jacobian and calculate its eigenvalues just as you would for any set of first-order differential equations.

(I apologise for not doing the work of actually applying this technique to your problem, but I hope this answer gives you enough information to do it yourself.)

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The answer by Nathaniel is clear and correct, yet let me add: it is obvious by inspection that this is motion in a potential, which can be derived from a Hamiltonian $$ \frac{\dot r^2+ r^2\dot\theta^2}{2} + V(r,\theta)\;. $$ All you need to do now is to establish whether, around the equilibrium points, $V$ has a minimum or a maximum/saddle: in the first case the motion is stable, otherwise not. This requires computing eigenvalues of a $2\times 2$ matrix, considerably easier than the $4\times 4$ matrix suggested by Nathaniel.

Also, it allows you to make some contact with physics, rather discussing it as a purely mathematical problem.

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You need to linearize your equations around any stationary point. Then you can indeed treat it like by finding eigen values.

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