There's something I'm slightly confused about regarding drawing the trajectory of a particle in state space $(x,v)$, where $v:=x'$. Here, I'm only working with $x\in\mathbb{R}$. Suppose a particle satisfies Newton’s law in the form $mx'' = F(x)$. Let $V(x)=-\int F(x)dx$ and $E(x,x')=\frac{1}{2}m(x'(t))^2+V(x(t))$. Then the energy $E$ is conserved:
Proof. We verify this by differentiation, using the chain rule: $\frac{d}{dt}E(x(t), x'(t)) = \frac{d}{dt}\frac{1}{2}m(x'(t))^2 + V(x(t))= mx'(t)x''(t) + \frac{dV}{dx}x'(t) = x'(t)[mx''(t) − F(x(t))]$
This is $=0$ by Newton's law, so $E(x,x')$ is conserved (this of course only applies to forces that are independent of time). Now drawing a trajectory in $(x,v)$ space, I know two trajectories cannot cross due to uniqueness of solutions to IVPs. My book claims that, if we start at $(x_0,v_0)$ with initial condition $(x_0',v_0')$, then the trajectory is a the level curve $S:=\{(x,v):E(x,v)=E(x_0,v_0)\}$.
Here's my confusion: it seems to me that the trajectory is a subset of $S$, and not necessarily equal to it. If that is the case, then looking at a trajectory in state space doesn't tell you whether or not $E$ is conserved. For example, looking at the picture below, there is a stable limit cycle. I would look at it and think the that a trajectory that starts outside radius $=1$ has a monotonically decreasing energy $E(x,v)$, and this trajectory converges to a periodic trajectory on which the energy is constant $E_0$. This would only apply to forces $F=(x,t)$ that are functions of both position and time. Alternatively, if $E$ is not known and it isn't known if $F$ depends on time, it could be, just looking at the picture, that the periodic trajectory at radius $=1$ and the trajectories outside it are part of the same level curve of energy $E_0$: while the particle is spiraling, it has constant energy $E_0$, and it converges to a periodic trajectory is the same constant energy $E_0$.
Is this correct, or am I not understanding something?
