I'm studying Classical mechanics on Arnold's "Mathematical Methods of Classical Mechanics". In a problem I am asked to find for which $\alpha$ the circular orbits in the central field problem are Lyapunov stable with the potential in the form $$U(r)=r^{\alpha}, -2 \leq \alpha < \infty$$ I know that the answer is $\alpha =2$ but i cannot find out why. I know the definition of Lyapunov stability (the solution $\varphi(t)$ defined on her, for instance, right maximal interval of existence, $J_{\varphi}^{+}=[\tau,+\infty)$, with initial condition $(\tau,\xi)$ for an autonomous system is Lyapunov stable if $\forall \varepsilon, \exists \delta (\varepsilon)$ such that if $\|\xi -\eta \|<\delta$ the solution $\psi(t)$ with initial condition $(\tau,\eta)$ is defined on $J_{\psi}^{+}=J_{\varphi}^{+}$ and $\| \psi(t)-\varphi(t) \|<\epsilon, \forall t \in [\tau,+\infty)$ ) and I know also how to use the Lyapunov function to prove this property. The problem is that I cannot neither find a proper Lyapunov function nor prove Lyapunov stability using the eigenvalues of the solution of the system. I cannot also understand the difference between this condition, and the stability condition that I can find with small departures from the condition $\ddot r=0$. In fact (supposing that $M$ is the angular momentum for the system and the mass $m$=1) we have that: $$\ddot r -\frac{M^2}{r^3}=-\frac{\partial U(r)}{\partial r}=f(r)$$ and for $\ddot r=0$ we have the circular orbits for the system where the radius $r_c$ is given by the relation $$-\frac{M^2}{r{_c}^{3}}=-\frac{\partial U(r)}{\partial r}\bigg|_{r=r_c}=f(r_c)$$ For small departures $r_c+\varepsilon$ substituting in the previous equation and using Taylor expansion we have $$f(r_c+\varepsilon)=\ddot \varepsilon -\frac{M^2}{(r_c+\varepsilon)^3}\Rightarrow \ddot \varepsilon= \bigg(f'(r_c)+\frac{3f(r_c)}{r_C} \bigg)\varepsilon $$ Now if the the term between brackets is negative I have the equation of an harmonic oscillator and the circular orbits are thus stable in this sense. With the potential in the form $U(r)=r^\alpha$ this condition is $\alpha>-3$. Summing up, my questions are:

  • What is the difference between these two types of stability?
  • Is Lyapunov one stronger, and in which sense it is?
  • How i can prove that if $\alpha=2$ circular orbits are Lyapunov stable?

1 Answer 1


Let me call radial stability the stability of $r$ around $r_0$, where $r_0$ is the radius of the circular orbit. The difference between this one and the Lyapunov stability is that the latter looks not only to $r$ but also to the polar angle $\theta$ (for a central force) and their conjugated momenta. So in this sense I would say Lyapunov is stronger.

Basically an orbit $\lambda(t)$ on the phase space is Lyapunov stable if it remains arbitrarily close to the orbit $\lambda_0(t)$ (the circular orbit) as long as we make the initials conditions of both cases arbitrarily close. Now if we restrict this Lyapunov criteria only for the radial coordinate then it coincides with the radial stability criteria.

I will not prove the result claimed by Arnold but I can give you some intuition. Consider for example the attractive inverse square force whose potential is $U=-r^{-1}$. A perturbed (around the circular) orbit is an ellipse. By Kepler's third law we have that the period of this perturbed orbit is $$T_p^2\propto a^3.$$ Since $a>r_0$, this period is greater than the period $T_0$ of the circular orbit. This implies that the difference between the angles of the two orbits, $|\theta_p(t)-\theta_0(t)|$ grows indefinitely with time. They do not remains close to each other, so this orbit is not Lyapunov stable. It is radially stable though as you have just shown in the question.

Now consider an isotropic harmonic oscillator, $U=r^2$. The orbit can be obtained by a superposition of two harmonic motions, with same frequency, orthogonally placed on the plane. Since the period of each harmonic motion is independent of the amplitude, the period of the isotropic oscillator also is independent of amplitude. The difference $|\theta_p(t)-\theta_0(t)|$ is therefore constant and can be done arbitrarily small by choosing arbitrarily close initial conditions. The isotropic oscillator is Lyapunov stable.

I think that a proof of Arnold's claim would be to propose attractive and radially stable power-law potentials, $U\propto r^n$, $n>-3$, and then show that only for $n=2$ the period of the perturbed orbit equals the period of the circular orbit.

  • 3
    $\begingroup$ This is the best answer I've read on SE in a long time. Thank you for both answering it and not just giving it away. $\endgroup$ Jul 8, 2016 at 19:28

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