Does the plane $x_3 = 1$ contain any periodic solutions? 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!
 A: 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
streamplot[{x^2+y^2-1,y(1-2x)}, {x,-3,5}, {y,-4,4}]

and zoom-in 
streamplot[{x^2+y^2-1,y(1-2x)}, {x,0.4,0.6}, {y,0.75,0.95}]

gives the phase portraits


which looks negative for any periodic solution.
A: To add on to LutzL's answer, I think we can also use some integral theorems argument.
Let $V = (x_1^2 + x_2^2 - 1, x_2(1-2x_1))$ be your vector field on the $x_3 = 1$ plane.
Then $div(V) = 2x_1 + 1 - 2x_1 = 1$.
Now suppose $\gamma$ is an integral curve of $V$ encircling some region $D$.
Then (since $\dot\gamma =V$), we get:
$0 = \int_\gamma (-V_2, V_1) \cdot\dot \gamma  = \int_D div(V) dA = area(D)$
So that there are no periodic orbits!
