How to know whether a solution for a set of coupled non-linear differential equations are stable or not

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I have 3-coupled ordinary non-linear differential equations, one second order in time and other two first order in time, with 3-dependant variables say x(t), y(t) and z(t).

I have a particular solution, for a particular choice of initial conditions. Can I tell whether this solution is stable or not ?

Can any body help me on this, if possible by explaining the situation and how to carry out the analysis, else by suggesting reading materials that are good for a physics student who is absolutely not familiar with the mathematical notations and conventions.

Thanking in Advance

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First, reduce the order of your differential equation to one by introducing a new dynamical variable corresponding to the temporal derivative of that variable whose second temporal derivative appears in the equation. E.g., if you have an equation $\ddot{x} = f(x,y,z)$, replace this equation by the two equations $\dot{w} = f(x,y,z)$ and $\dot{x} = w$. See, e.g., this Wikipedia article for more information.

Case 1: Fixed point

If your solution is a fixed point, calculate the eigenvalues of the Jacobian of $f$, where $f: ℝ^4 → ℝ^4$ comprises the right-hand side of all differential equations. If any of those eigenvalues has a positive real part, the fixed point is unstable, otherwise it’s stable. For more information, see the Scholarpedia on equilibria.

Case 2: Periodic solution

If your solution is periodic, things become more complicated. You can calculate the Jacobian along the entire orbit and see whether there is any divergence orthogonal to the trajectory, but usually that’s infeasible. It’s likely more easy to numerically integrate the differential equations and see what happens. If you need better proof for stability, you can estimate the Lyapunov exponents numerically, test the periodicity of your outcome, and so on.

Case 3: Something else

Essentially the same as for periodic solutions, but analytic methods are even less likely to work.