In Arnold's Classical Mechanics of Classical Mechanics, he refers to Lyapunov stability in many of the problems in the second chapter.

E.g. on page 20: "Problem: Consider a periodic motion along the closed phase curve corresponding to the energy level $E$. Is it stable in the sense of Lyapunov? Answer: No."

On page 36: "Problem: For which values of $\alpha$ is motion along a circular orbit in the field with potential energy $U = r^\alpha$, $-2\leq\alpha\leq\infty$, Lyapunov stable?"

In the footnote he refers to his own book on Ordninary Differential Equations for a definition of Lyapunov stability. In that book he defines Lyapunov stability ONLY for equilibrium points of the equation, i.e. if

$$ \dot{\mathbf{x}} = \mathbf{v} \left( \mathbf{x} \right) $$ where $\mathbf{v}$ is an at least twice differentiable vector field. Then we can only talk about Lyapunov stability if we are looking at a solution, i.e. a solution where the vector field is zero.

(I.e. if $\mathbf{v}(\mathbf x_0) = \mathbf 0$ we obviously will get a solution $\mathbf x(t) = \mathbf x_0$ if the initial condition is $\mathbf x(0) = \mathbf x_0$.)

My confusion is that none of the two systems I in the problems above seem to be in equilibrium positions! For example, the motion in a central potential is described by

$$ \left( \begin{array}{c} \dot{\mathbf r} \\ \dot{\mathbf v} \end{array} \right) = \left( \begin{array}{c} \mathbf v \\ - \Phi (r) \mathbf{e}_r \end{array} \right) $$ So wouldn't an equilibrium point be one for which the right hand side of these equations are zero?

Using polar coordinates I cannot even manage to write the equations for $r$ and $\phi$ in the form above...

As you can tell from my rambling I am clearly confused about all this and I am most likely missing something essential here.


2 Answers 2


let say you have this vector differential equations:

$$\vec{\ddot{x}}=-\vec{f}(\vec{x},\vec{\dot{x})}\tag 1$$

and you want to check the stability at a stable point $\vec{x}_0$

according to Lyapunov theory, you have to linearized equation (1) and calculate the eigenvalues of the linearized system.

we first transformed equation (1) to first order differential equations :


$\vec{\dot{x}}=\vec{{y}}_1$ and $\vec{{x}}=\vec{{y}}_2$ you get $$\underbrace{\begin{bmatrix} \vec{\dot{y}}_1\\ \vec{\dot{y}}_2\\ \end{bmatrix}}_{\vec{\dot{y}}}=\underbrace{\begin{bmatrix} -\vec{f}(\vec{y}_2,\vec{y}_1)\\ \vec{y}_1\\ \end{bmatrix}}_{\vec{g}(\vec{y})}\tag 2$$

if you linearized equation (2) at $\vec{y}_2=\vec{x}_0$ you get:

$$\vec{\dot{y}}_L= \underbrace{\left[ \begin {array}{cc} -{\frac {\partial }{\partial {\it \vec{y}_1}}}f \left( {\it \vec{y}_2},{\it \vec{y}_1} \right) &-{\frac {\partial }{\partial {\it \vec{y}_2}}}f \left( {\it \vec{y}_2},{\it \vec{y}_1} \right) \\1&0 \end {array} \right]}_{A(\vec{x}_0)}\vec{y}_L+\vec{b}$$

The eigenvalues of the matrix $A$ can be zero, real or conjugate complex

-if all real eigenvalues are lees then zero and the real part of the complex eigenvalues also less then zero,you get stable system.

-If one of your eigenvalues has a positive real part your system is unstable.

-If one of your eigenvalues is zero your system is semi stable

  • 1
    $\begingroup$ I think the key point I missed was that I needed to linearise around the orbit in question. $\endgroup$ Sep 12, 2019 at 20:46
  • $\begingroup$ I'm still confused, immediately my calculation gives $$A=\left[\begin{smallmatrix}0& -\alpha r^{\alpha-1} + M^2 r^{-3} \\ 1 & 0 \end{smallmatrix}\right]$$ with eigenvalues $\pm \sqrt{-\alpha r^{\alpha-1} + M^2 r^{-3}}$ which has nonnegative eigenvalues whenever $\alpha r^{\alpha+2}\geq M^2$ which is true along some range of $\alpha$ seemingly dependent on how big the orbit is. $\endgroup$ Jan 17, 2020 at 14:51

Partial answer

I know Lyapunov stability better from chaotic systems in one dimension like the logistic map and I can confirm that one can calculate the Lyapunov constant for non-stable points.

The key thing we test with Lyapunov is whether two points close together move together or apart in a dynamical system. If they move together it is stable, if it they move apart it is unstable.

For example if we move in discrete time steps for a one dimensional system

$$ x_{i+1} = f(x_i) $$

now if we have some fixed point $x^*$

$$ x^* = f(x^*)$$

Now we define a point close to a stable point $x_i=x^*+\epsilon_i$

$$ x_{i+1} = f(x_i) = f(x^* +\epsilon_i) \\x^*+\epsilon_{i+1}=f(x^*)+\epsilon_i f'(x^*)n+ {\epsilon_i^2 \over 2} f''(x^*) ~~~~~etc. $$

provided $\epsilon_i$ is small, and remembering $f(x^)=x^ we can simplify to

$$ x^*+\epsilon_{i+1}=x^*+\epsilon_i f'(x^*)$$


$$ \epsilon_{i+1}=\epsilon_i f'(x^*)$$


$$ \epsilon_{i}=\epsilon_0 f'(x^*)^i$$

thus if $|f'(x^*)|<1$ then $\epsilon_i$ will get smaller and smaller, but if it is greater than one it will get bigger... and so we have a measure for whether or not two points get closer together or further apart.

The final thing is to take logs - the Lyapunov constant, $\lambda$, is defined with

$$\lambda={1 \over n} \Sigma_{i=0}^n ln |f'(x^*)|$$

a negative value is stable - a positive value is unstable.

Note that we can use this formula even for points which are not stable.

$$\lambda={1 \over n} \Sigma_{i=0}^n ln |f'(x_i)|$$

and, I imagine, along a trajectory in two dimensions one would want to monitor two close points to see if they move together or move apart.

hope this is helpful,


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