phase portrait for a conservative system - understanding the interpretation of the potential energy graph

Question

I specifically want to understand how to interpret the potential graph $V(x)$ to sketch the phase portrait in the $xp$-plane. Consider the potential energy function $V(x) = -\frac{x^2}{2} + \frac{x^3}{3}$ and kinetic energy $\frac{p^2}{2}$.

We may sketch the graph of $V(x)$ and observe a potential energy "well" between $x=0$ and $x = \frac{3}{2}$. Also, $V(x) \to \infty$ to the right of $x = \frac{3}{2}$ and likewise $V(x) \to -\infty$ to the left of $x = 0$. At $x = 0$ corresponds at a saddle node and at $x = 1$ corresponds a center.

As I want to understand the interpretation of the graph of $V(x)$ for sketching the phase portrait, I consider a cyclist starting at some point on $V(x)$.

Consider that the cyclist starts at $x = 0$ and is moving right. We begin with $ p = 0$ (no kinetic energy) and high potential energy. The cyclist moves to the right down the well towards the minimum at $x = 1$. We have positive kinetic energy and small potential energy. At the minimum of the well, we have the maximum kinetic energy and zero potential energy. Now the cyclist continues upward towards the point $x = \frac{3}{2}$. The kinetic energy is decreasing and the potential energy is increasing. At the point $x = \frac{3}{2}$ there is zero kinetic energy and maximum potential energy. Hence, a homoclinic trajectory.

Now, what happens if the cyclist moves to the left of $x = 0$? Likewise, to the right of $x = \frac{3}{2}$?

If someone could help me understand this interpretation via this cyclist I would be quite appreciative.

Here is my attempt:

Consider the cyclist starting at $x = 0$ and moving left. As the cyclist descends the potential energy mountain, their kinetic energy increases quadratically. However, the cyclist is doomed to continue down the mountain forever and so therefore the corresponding trajectory curves downwards like a parabola. It is not closed.

Consider the cyclist starting at $x = \frac{3}{2}$ and moving right. The same scenario, but the trajectory curves upward always. However, this doesn't seem to make sense when we consider the center. I expected the trajectories to curve around it, downwards, move left, then go downwards to the abyss.

Answer 1

The easy way to understand is to plot the total energy $E=V(x)+K$. Further, assume that the cyclist does not continue to put energy into the system. That means that the total energy is constant (a horizontal line on the graph). From conservation of energy, the trajectory of the cyclist is always below this line (it can touch it when $K=0$). So you have two distinct cases: the line intersects the curve at two points between $0$ and $3/2$ or it does not. Let's look at the first case. The cyclist starts in the valley, and does not have enough energy to get out of it. It will always oscillate around $x=1$. The second case is more interesting: it does not matter where the cyclists starts, or in which way is his initial direction of motion, it will end up moving down the slope moving left of $0$, towards $-\infty$. Let's say it moves initially in the positive direction. The kinetic energy will transform into potential energy. The cyclist continues moving to the right until $K=0$, or $V(x)=E$. At that point it has a lot of potential energy that transform into kinetic energy. But it can't go right any more ($V(x)$ would increase), so it goes left. It might slow down between $1$ and $0$, but it has enough kinetic energy to continue moving left. Once it passes $x=0$ and moves left, there is nothing that will force the cyclist to change direction. There are two special cases. 1. $E=0$. If the cyclist moves to the right at some point it will get to $x=0$. At that point the kinetic energy is $0$, and the derivative of $V(x)$ is zero, so there is no force acting on the cyclist. But it's an unstable equilibrium. Give it just a little kinetic energy, and he ends up down the hill towards negative values of $x$. 2. The other special case is at $x=1$ if $E=-1/6$. Then the cyclist is once again in equilibrium. If you add just a small amount of kinetic energy, it will oscillate around $x=1$. But if you give more than $-1/6$ it will end up down the hill again. So this is a metastable equilibrium point.