I am studying the following system of equations:
\begin{align}
\dot{x_1} &= x_1 – x_1x_2 – x_2^3 + x_3(x_1^2 + x_2^2 – 1 – x_1 + x_1x_2 + x_2^3)\\
\dot{x_2} &= x_1 – x_3(x_1 – x_2 + 2x_1x_2)\\
\dot{x_3} &= (x_3 -1)(x_3 + 2x_3x_2^2 + x_3^3)
\end{align}
The set/plane $x_3 = 1$ is an invariant set. I am curious whether this invariant set contains any periodic solutions, and how I can find out. I have the following definitions to work with:
Definition (periodic solution): Suppose that $x = \phi(t)$ is a solution of the equation $\dot{x} = f(x)$, $x\in D\subset\mathbb{R}^n$ and suppose there exists a positive number $T$ such that $\phi(t + T) = \phi(t)$ for all $t\in\mathbb{R}$. Then $\phi(t)$ is called a periodic solution of the equation with period $T$.
Definition (invariant set): Consider the equation $\dot{x} = f(x)$, $x\in D\subset \mathbb{R}^n$. The set $M\subset D$ is invariant if the solution $x(t)$ with $x(0)\in M$ is contained in $M$ for $-\infty <t < \infty$. If this property is valid only for $t\geq 0 (t\leq 0)$ then $M$ is called a positive (negative) invariant set.
Question: How should I find out whether the plane $x_3 =1$ contains any periodic solutions? I understand that $x_3 = 1$ is an invariant set since $\dot{x_3} = 0$ in $x_3 = 1$, but I don't really know how I should use this fact to find out whether there are any periodic solutions in $x_3 = 1$.
Thanks!
Best Answer
You can also write your system as \begin{align} \dot{x_1} &= (1-x_3)(x_1 - x_1x_2 - x_2^3) + x_3(x_1^2 + x_2^2 - 1)\\ \dot{x_2} &= (1-x_3)x_1 + x_3x_2( 1 - 2x_1)\\ \dot{x_3} &= (x_3 -1)(x_3 + 2x_3x_2^2 + x_3^3) \end{align} so that on the plane $x_3=1$ the system reduces to \begin{align} \dot{x_1} &= x_1^2 + x_2^2 - 1 \\ \dot{x_2} &= x_2( 1 - 2x_1)\\ \dot{x_3} &= 0 \end{align} Wolfram Alpha with the command for an overview
and zoom-in
gives the phase portraits
which looks negative for any periodic solution.