A "paradox" in classical dynamics

Question

Consider the following trajectory

$$\dot{x}(t) = 1-\frac{x(t)^2}{x^2_0}.$$

The solution to the above is as follows

$$x(t) = x_0\coth\left[x_0t+\coth^{-1}\left(\frac{x_i}{x_0}\right)\right] \quad x > x_0$$ $$x(t) = x_0\tanh\left[x_0t+\tanh^{-1}\left(\frac{x_i}{x_0}\right)\right] \quad x < x_0.$$

From the above solution, one can see that $x = x_0$ is a stable point. However, if you take another derivative of the trajectory, we get $$\ddot{x}(t) = -\frac{2}{x^2_0}x(t)+\frac{2}{x^4_0}x(t)^3.$$ If you take the RHS in the above as as $-V'(x)$ then the potential we get is $$V(x) = -\frac{1}{2}\left(1-\frac{x^2}{x^2_0}\right)^2.$$ As you can see, in the above potential $x = x_0$ are unstable point as $V''(x_0) < 0$. How does one resolve this contradiction? Is taking a second derivative of the trajectory like that incorrect?

Answer 1

The catch is that the extended phase space that we obtain after the differentiation, contains the trajectories of the initial system, but also many other trajectories. This is why the intuition based on analyzing potential fails.

Indeed, but squaring the first equation we see that it corresponds to $$ \frac{\dot{x}^2}{2}-\frac{1}{2}\left(1-\frac{x^2}{x_0^2}\right)^2= \frac{\dot{x}^2}{2}+V(x)=0 $$ If we consider the solutions of the extended equation for total mechanical energy $0$, we will get that the point indeed stops at $x=x_0$.

In more formal terms, the trajectory described by the first differential equation is a heteroclinic orbit for the second equation: whenever the initial conditions $\dot{x}(0), x(0)$ are related via $\dot{x}(t) = 1-\frac{x(t)^2}{x^2_0}$, the system would end in its equilibrium state. However, for the continuum of all the other initial conditions, the system would evolve away from the equilibrium, regardless of how close we started. In other words, yes, the state at $x=x_0$ is unstable for the second order differential equation, but it is stable for the first order one, because it is a special trajectory.

Note also that this trajectory is also a separatrix, as it separates the areas with different topological behavior in teh phase diagram (particle oscillating between two peaks, and a high energy particle passing above them.)

Answer 2

This problem has two equilibria, namely

$$\overline{x}_{1,2} = \mp x_0 \ .$$

Linear stability (either modal, or response to variation of initial conditions) can be evaluated using linearized system around the equilibria,

$$\delta \dot{x} = \partial_x f(\overline{x}_{1,2}) \delta x(t) \ ,$$

with

$$\partial_x f(\overline{x}_{1,2}) = -\dfrac{2 \overline{x}_{1,2}}{ x_0^2} = \pm \dfrac{2}{x_0} \ .$$

Thus, equilibrium $x_{1} = - x_0$ is unstable as $\partial_x f(\overline{x}_1) > 0$, while equilibrium $x_{2} = x_0$ is stable as $\partial_x f(\overline{x}_2) < 0$.

I'd like to point out two mistakes in your answer:

1. You can't interpret a first order system as a second order systems coming from some mechanical systems in general, that can be analysed with potential energy. Here you have a relation between the first order time derivative of what you'd like to interpret as the position of a system equal to a force you're interpreting as the space derivative of a pontential energy. But you can't do that as you should have the time derivative of velocity equal to space derivative of potential energy,

$$\dot{v} = f(x) \qquad , \qquad \dot{x} = v$$

2. I guess your integration is wrong (if it's not some strange equivalent expression, I won't check) or at least can be written in a much simpler way, as integrand function can be rewritten as sum of fractions

   $$\dfrac{d x}{1 - \frac{x^2}{x_0^2}} = \dfrac{1}{2} \left( \dfrac{1}{1 - \frac{x}{x_0}} + \dfrac{1}{1 + \frac{x}{x_0}} \right) \, dx \ ,$$

   whose integral gives

   $$\dfrac{x_0}{2} \ln \dfrac{\left| \frac{x}{x_0} + 1\right|}{\left| \frac{x}{x_0} - 1\right|} + C \ ,$$

   and thus the expression $x(t)$ can be written in a form

   $$x(t) = - \dfrac{1 + k e^{a t}}{1 - k e^{at}}$$

Answer 3

$\def \c {\boldsymbol} \def \p {\dot} \def \pp {\ddot} \def \b {\mathbf}$

First order ODE

\begin{align*} \dot x&=\underbrace{1-\frac{x^2}{x_0^2}}_{f(x)} \end{align*}

The fixed points are $~\dot x=0\quad\Rightarrow x=\pm x_0~$ thus \begin{align*} f'(x)&=-\frac{2\,x}{x_0^2}\quad\Rightarrow\\ f'(x=x_0)&=-\frac{2}{x_0}\quad\text{stable}\\ f'(x=-x_0)&=\frac{2}{x_0}\quad\text{unstable}\\ \end{align*}

Second order ODE

\begin{align*} \ddot x&=\underbrace{2\,{\frac {x}{{x_{{0}}}^{2}}}-2\,{\frac {{x}^{3}}{{x_{{0}}}^{4}}}}_{f(x)}\\ f'(x)&=-V''(x) \end{align*} The fixed points are $~\ddot x=0\quad\Rightarrow x=0~,\pm x_0~$ thus

\begin{align*} f'(x)&=\frac{2}{x_0^2}-\,{\frac {6\,{x}^{2}}{{x_{{0}}}^{4}}}\quad\Rightarrow\\ f'(x=x_0)&=-4\,{x_{{0}}}^{-2}\quad\text{stable}\\ f'(x=-x_0)&=-4\,{x_{{0}}}^{-2}\quad\text{stable}\\ f'(x=0)&=2\,{x_{{0}}}^{-2}\quad\text{unstable}\\ \end{align*}

with $~f'(x) > 0~$ unstable and $~f'(x) < 0~$ stable

You can’t just compare the 1D system to the 2D one

The second-order system: $~\ddot x=f(x)~$

• Is a dynamical system in 2D phase space $~x~,\dot x~$
• The same function $f(x)$ now behaves like a force in a mechanical system.
• The fixed point $~(x_0,0)~$ corresponds to zero velocity and equilibrium position.