If $x_1 = x_2=0$ does this imply that the origin is the equilibrium?

Question

Can someone please help me solve the following? Thank you for your time and help.

Given the nonlinear oscillator $$\ddot{x} + F(x,\dot{x})\dot{x} + x = 0,$$ where $F(x,\dot{x}) < 0$ if $x^2 + (\dot{x})^2 < a$ and $F(x,\dot{x}) > 0$ if $x^2 + (\dot{x})^2 >b.$ I want to show that this system has a positive periodic orbit.

$\textbf{Solution:}$ Let $\dot x_1 = x_2$ and $\dot x_2 = - x_2 F(x_1,x_2)-x_1$ as we make this system first order. Multiplying across by $x_1$ we get $x_1\dot x_1 = x_1x_2$ and multiply by $x_2$ we arrive at $x_2\dot x_2 = - x_2^2 F(x_1,x_2)-x_1x_2$.

If $x_1 = x_2=0$ does this imply that the origin is the equilibrium? I am unsure how to proceed from here.

Answer 1

At any stationary/equilibrium point you have $\dot x=0$, $\ddot x=0$ (nothing moves) and consequently by the equations also $x=0$. As there is no other possibility, this is the only stationary/equilibrium point of the system