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I'm studying Classical mechanics on Arnold's "Mathematical Methods of Classical Mechanics". In a problem i'm 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 angolar 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?

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?