# A dynamical system of differential equations - periodic solutions?

I am solving a physical problem with a known periodic solution. When I simulate the behaviour of the system numerically, with full blown differential equations, I get stable, but rather complicated looking oscillations about a certain point in parameter space. I constructed a simplified set of differential equations by perturbing the stable position by a small change and then obtained its equations of motion.

I have 2 coordinates $\epsilon$ and $\beta$, 1 characteristic angular velocity $\Omega$ and a small constant $q$ ($0 \lt q \ll 1$). The set looks as follows:

$\ddot{\epsilon} = 2 \Omega \dot{\beta} + 3 \Omega^2 \epsilon - \frac{9}{4} q \Omega^2 \beta$

$\ddot{\beta} = -2 \Omega \dot{\epsilon} -\frac{9}{4} q \Omega^2 \epsilon + \frac{9}{4} q \Omega^2 \beta$

I was wondering whether it can be shown that it allows for periodic solutions (or stable) and what their form could be. If it only were for the second derivatives and the actual variables, I could try to diagonalise the system, but I am confused by the first derivatives.

Thanks a lot.

SSF

• The linear system is not very helpful in answering the question. For periodic solutions you need purely imaginary roots of the characteristic system (or equation of degree 4). Numerically, this is an unstable situation; except perhaps for time.symmetric methods. -- In short, for somewhat stable periodic orbits that survive under numeric methods, you need non-linearity. – LutzL Apr 8 '15 at 12:03
• Did you try to fit the parameters $(A,B,C,D,\omega)$ such that the functions $$\epsilon(t)=A\cos(\omega t)+B\sin(\omega t),\qquad \beta(t)=C\cos(\omega t)+D\sin(\omega t),$$ solve the system? – Did Apr 8 '15 at 12:36
• It seems that periodic solutions exist exactly when $q>\frac1{18}(1+4\sqrt{3}+\sqrt{1+8\sqrt{3}})\approx0.655$, hence, not for small positive values of $q$. – Did Apr 8 '15 at 16:07

A system of the form $$\mathbf{\ddot{x}}+\mathbf{A}\mathbf{\dot{x}}+\mathbf{B}\mathbf{x}=0\,,$$ where $\mathbf{x}$ is a vector and $\mathbf{A}, \mathbf{B}$ are matrices, can be solved by substituting $\mathbf{x}=\mathbf{x_0} e^{\lambda t}$ with constant $\mathbf{x_0}$. Then you are solving $$\left(\lambda^2 \mathbf{I}+\lambda \mathbf{A}+\mathbf{B}\right)\mathbf{x_0}=0\,.$$ In two dimensions you get four linearly independent solutions. You find the $\lambda$s by solving $$\det\left(\lambda^2 \mathbf{I}+\lambda \mathbf{A}+\mathbf{B}\right)=0\,,$$ which in your case is a quartic. The system is stable if $\Re{\lambda}\leq0$ for every root.