Showing inexistance of periodic orbits

Question

I need help with the last bit of the following problem

Consider the system $$ \begin{aligned} & \dot{x}=y+a x-b x^3, \\ & \dot{y}=x^3-x, \end{aligned} $$ where $a$ and $b$ are constants and $a \neq 0$. (a) Find and classify the fixed points, and state their Poincaré indices. (b) By considering a suitable function $H(x, y)$, show that any periodic orbit $\Gamma$ satisfies $$ \oint_{\Gamma}\left(x-x^3\right)\left(a x-b x^3\right) d t=0 $$ where $x(t)$ is evaluated along the orbit. (c) Deduce that if $b / a<1$ then the second-order differential equation $$ \ddot{x}-\left(a-3 b x^2\right) \dot{x}+x-x^3=0 $$ has no periodic solutions.

I wish to answer (c). I did the previous parts successfully finding that the fixed points are $(0,0)$ (node); $(-1, a-b)$ and $(1,b-a)$ (saddles). Therefore I know that if such $\Gamma$ were to exist it must only contain the origin out of those. Now I need to seek a contradiction with $(b)$. I tried showing that the integrand is never zero, however, I do not think this is true since $x$ can be unbounded. Note that also the ode system is due to the dynamical systemsince \begin{align*} \ddot{x} =\dot{y} + (a-3bx^{2}) \dot{x} \end{align*} by def of $x$.

Question: How do I show there doesn't exist a periodic solution?

Answer 1

Let's assume that $$ \ddot{x}-\left(a-3 b x^2\right) \dot{x}+x-x^3=0 \tag{1} $$ has a (nonconstant) periodic solution. Such a solution oscillates between the points $x_{\min}$ and $x_{\max}$, where its first derivative vanishes. In addition, $\ddot{x}<0$ at $x_{\max}$ --- otherwise, $x_{\max}$ would be a fixed point (if $\ddot{x}=0$ at $x_{\max}$) or $x(t)$ would grow beyond $x_{\max}$ (if $\ddot{x}>0$ at $x_{\max}$). Therefore, according to $(1)$, $$ \ddot{x}|_{x=x_{\max}}=x_{\max}^3-x_{\max}<0 \implies 0<x_{\max}<1. \tag{2} $$ By an analogous argument, $\ddot{x}>0$ at $x_{\min}$, hence $$ \ddot{x}|_{x=x_{\min}}=x_{\min}^3-x_{\min}>0 \implies -1<x_{\min}<0. \tag{3} $$ From $(2)$ and $(3)$ it follows that, if $x(t)$ is a periodic solution to $(1)$, then $|x(t)|<1$. Therefore $$ \oint_{\Gamma}(x-x^3)(ax-bx^3)\,dt =a\oint_{\Gamma}x^2(1-x^2)\left(1-\frac{b}{a}x^2\right)dt\neq 0 \tag{4} $$ if $\frac{b}{a}<1$, since the integrand in the last integral vanishes only at $x=0$ and is positive in $\Gamma\setminus\{0\}$. It then follows from the result (b) that $x(t)$ is not a periodic solution to $(1)$, in contradiction to our initial hypothesis. We conclude, therefore, that $(1)$ does not have a (nonconstant) periodic solution if $\frac{b}{a}<1$.