Level sets of a conserved quantity are trajectories of differential equation

Question

If we have a differential equation $\mathbf{\dot{x}}=\mathbf{F}(\mathbf{x})$ and we have conserved quantity $E(\mathbf{x})$, which means $\dot{E}=0$, then I don't understand why level sets of $E$ are trajectories of the differential equation. So for some constant $C$ look at $E^{-1}(C)$. In the book of Robinson (introduction to dynamical system, continuous and discrete) it says this is true in an example, but explains not why it is true. I really want to understand this. Thanks.

Answer 1

The correct statement is: every trajectory is contained in some level set of $E$. Indeed, suppose $x=f(t)$ is a trajectory passing through $x_0=f(t_0)$. Since $E$ is conserved, the quantity $ E(f(t)) $ is independent of $t$. Thus, this trajectory is contained in the level set $\{x: E(x)=E(x_0)\}$.

As Hans Lundmark said, in two dimensions the level set may actually coincide with a trajectory, since both are one-dimensional in this case. However, level sets may be disconnected and may have branching at critical points of $E$; so in general a level set may be larger than a trajectory, even in two dimensions.