# Oscillations about equilibrium for coupled differentail equations

I have the following system of equations:

\begin{align} \frac{dX}{dt} &= 2Y-2\\ \frac{dY}{dt} &= 9X-X^3 \end{align}

I would like to study the property of solutions to this function about the point $(3,1)$. Namely, I'd like to find a linear approximation for small oscillations about the equilibrium point, and I'd like to estimate the period of oscillation.

I don't exactly know what is meant by "find a linear approximation" for a coupled system, though. I've taken the Jacobian of the system, yielding eigenvalues $6i,-6i$ and corresponding eigenvectors $[1,3i]$ and $[1,-3i]$.

But from here, I don't exactly know where to go.

-

First write: $$X=x+3,\ \ \ Y=y+1$$ so that the equilibrium point moves to $(0,0)$. The equations become: $$\begin{array}{l}\dot{x}=2y\\\dot{y}=9x+27-(x+3)^3=-x^3-9x^2-18x\end{array}$$ Now linearize in both $x$ and $y$ (i.e.: take the Taylor expansion up to first order). You get: $$\begin{array}{l}\dot{x}=2y\\\dot{y}=-18x\end{array}$$ This equation determinates the dominant behavior of your system near the equilibrium point. I think you can take it from here. Good luck.
Also, notice that you could directly take the Taylor expansion in $X$ and $Y$, but you should do it around $X=3$ and $Y=1$. – Daniel Robert-Nicoud Jul 21 '14 at 8:05
AH! Since $x'=2y$ we have $x''=2y'=-36x$ which has solution $x = A \cos (-6t - \phi)$. Similarly we can find $y = B \cos (-6t-\phi )$. So the period is $\frac{\pi}{3}$. Am I correct? – user2899162 Jul 21 '14 at 8:08
@user2899162 Also, you should be able to find some relations between the phase shifts and amplitudes: your system has only two degrees of freedom (corresponding to the choice of initial values for $x$ and $y$) and, as it is right now, your solution admits three (namely $\phi$, $A$ and $B$). – Daniel Robert-Nicoud Jul 21 '14 at 8:13
Kind-of. I'm trying to do the taylor expansion directly as a sanity and comprehension check. I get $\left. \frac{dX}{dt} \right|_{(3,1)} \approx 2(X-3)$ and $\left. \frac{dY}{dt} \right|_{(3,1)} \approx -18(Y-1)$. Which seems to be what you got after correcting for the coordinate shift! So I think I didn't mess up and can waddle my way through. – user2899162 Jul 21 '14 at 8:18