# Arnold's Mathematical Methods of Classical Mechanics and Lyapunov stability

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.

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 :

with:

$$\vec{\dot{x}}=\vec{{y}}_1$$ 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

• I think the key point I missed was that I needed to linearise around the orbit in question. – JezuzStardust Sep 12 '19 at 20:46
• 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. – Christian Chapman Jan 17 '20 at 14:51

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

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

hence

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

and

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

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.