So I have these two equations:
$\ dx /dt = y-x^3+x$
$\ dy /dt = -x-y^3+y$
I'm trying to show the system has at least one periodic solution between:
$\ x^2+y^2 = 1$
$\ x^2+y^2 = 2$
I set $\ V(x,y)=x^2+y^2$ and find that $\ dV /dt > 0$ when $\ x^2+y^2 < 1$
My question is: Why does this ($\ dV /dt > 0$) tell me that trajectories starting inside the region $\ x^2+y^2 < 1$ leave the region enclosed by $\ x^2+y^2 = 1$?
Nite that on the circle, $x^2+y^2=1$
$$ \frac {dV}{dt}= 4x^2y^2 $$ which is positive except at $(\pm 1,0)$ or $(0,\pm 1)$.
We can check the slopes at these points and conclude that the trajectory leaves the circle at these points as well.
for example at $(1,0)$, we have $ \frac{dx}{dt} = 0, \frac{dy}{dt} = -1$.
Thus the trajectory is moving out the circle vertically downward.
Similarly for the other three points.