Non-Linear Dynamical System

$\bf {Lyapunov \ Theorem \ \& \ Stability}$

Lyapunov Functions and stability

The intuitive definition of a stable system, is one where the system tends towards a stable equilibrium point. For a mechanical system this is best represented by a system which dissipates energy and comes to rest at an equilibrium point.

This is the basis of Lyapunov direct method, where the stability of the equilibria of a system are analysed using a Lyapunov function.

for a dynamical system:

$\dot x = f(x)$

we assume an equilibrium point at x = 0;

In order to determine the stability of the equilibrium point in terms of Lyapunov, we identify a differentiable function denoted V, as a candidate for the Lyapunov function. Fundamentally, we will determine from this function whether the system will remain at the stability point if initialised a it and whether the system tends towards said equilibrium point globally or locally.

for a subset $S \rightarrow \R$ defined on some region $S \subset \R ^n$ containing the region such that:

V(0) = 0;
V(x) > 0 for all $x \in S$ with $x \neq 0$ ;
$\dot V(x) \leq 0$ for all $$x \in S$

If the above statements are true that then x = 0 is a stable equilibrium point.

it is important to note that the Lyapunov function is not unique for a system and that is can be difficult or in some case in possible to identify a suitable Lyapunov candidate.

Where we have an autonomous system, such that that the dynamics of the system are not explicitly dependent on time, the Lyapunov function will not depend on tiem either. Therefore, the derivative of sai function $\dot V(x)$ can be defined as follows:

$\frac {\delta V(x)}{\delta t} = \frac {\delta V(x)}{\delta x} f(x)$

$\bf {General \ Definitions \ of \ Stability \ According }$ $\bf {\ to \ Lyapunov \ Theorem}$

Local Stability:

function Again taking a function $\dot x = f(x)$ for which f(0) = 0 without loss of generality:

The system is locally stable about the equilibrium point where:

V(x) is positive definite;
$\dot V(x)$ is negative semi-definite.

if more precisely $\. V(x)$ is negative definite, the equilibrium point is asymptotically stable.

Lyapunov theorem of global stability:

By definition a globally asymptotically stable equilibrium point is one that is stable, as defined above, everywhere i.e. for all $\R ^n$ . Thus:

V(x) is positive definite everywhere in the state space;
$\dot V(x)$ is negative semi-definite.

If the equilibrium point is globally asymptoticaly stable, it implies that it is the only stable equilibrium point in the system.

A third condition, allows us to identify if nto only the equilibrium point but the system itsels is globally asymptotically stable:

$V(x) \rightarrow +\infty$ as $||x|| \rightarrow +\infty$ i.e. $x \rightarrow \pm \infty$

This means that V(x) tends to infinity as x tends towards positive infinity, in either the negative or positive direction. This has the effect of essentially bounding the bounding V(x).

$\bf {Invarient \ Set \ Theorem \ (LaSalle's \ Theorem) }$

Invarient Set:- A subset of the space $R^n$ $ such that if the system is initialised within the subset, it stays there and cannot leave. Once it is initialized in the subset it cannot leave the subset which comprises a region $\R^n$ .

Example of an invariant set:

The equilibrium point;
The state space within the invariant set;
any autonomous system’s path;
limit cycle: a subset within the state space in which if the system enters it will follow a path in the state space.

Reasons for the conception of invariant set theorem:

To provide a looser definition of asymptotic stability that is not constrained to a only systems with a negative definite Lyapunov function;
Allowing the study of system convergence to more general behaviours, not just the equilibrium point.

conclusions are less precise and the hypothesis of the invariant set theorem are more general.

$\underline {Global \ Invariant \ Set \ Theorem}$

the Principle of the Theorem Let S, a subset in $\R ^n$ , be a bounded invariant set. Assuming there exists a differentiable function $V : S \rightarrow \R$ such that:

$\dot V(x) \leq 0 \ \forall x \in S$

If we take B to be the largest invariant set contained within ( ${x \in S | \dot V(x) =0}$ ) the set of x values within the subset S for which the derivative of V(x) is equal to zero, then all trajectories starting in S approach the invariant set B as time tends to infinity. This means that the

According to LaSalle’s Theorem, for a dynamical system:

$\dot x = f(x)$

we no longer have to assume that Lyapunov function V is positive definite.

For a scalar function V we assume:

$\dot V \leq 0$ everywhere in state space
$V(x) \rightarrow +\infty \ as \ ||x|| \rightarrow + \infty$

all trajectories tend toward the subset $R{x, \dot V (x) = 0}$

More precisely, the system does not only tends toward R, but the largest invariant set within R.

Example:

$\ddot x + b(\dot x ) + c(x) = 0$

where

0 \ for \ x \neq 0 “>

0 \ for \ \dot x \neq 0 “>

Therefore, the trajectory of both functions b(x) and c(x) have to pass through the point zero, as they both have the same sign as their argument.

In addition, the point $x = 0$ and $\dot x=0$ is a globally asymptotically stable equilibrium point.

virtual physics:

Assuming we are working on a mechanical engineering problem we can take the total energy of the system as the obvious Lyapunov function to consider.

$V(x) = \frac {1}{2} \dot x^2 + \int_0 ^x c(y) dy$

We can see that the Lyapunov candidate above is scalar, however, the last term : $\int_0 ^x c(y) dy$

does not guarantee the following assumption

$V(x) \rightarrow +\infty \ as \ ||x|| \rightarrow + \infty$

therefore we need to prove that we still need to show that $\int_0 ^{\infty} c(y) dy = + \infty$

by taking the differential of the Lyapunov candidate (Potential energy equation)

$\dot V = \dot x \ddot x + \dot x c(x) = -\dot x b(\dot x)$

This shows that $\dot x \rightarrow 0$ , which corresponds to the set R.

For it to be globally asymptotically stable we said the system should tend toward he largest invariant set within R.

To verify that the system stays in R{ $\dot x = 0$ } if it starts there, we can do this by computing the derivative $\ddot x$ which would need to be 0 for $\dot x = 0$ .

We can see from the initial dynamics of the equation that for $\dot x = 0$ then the function $b(\dot x)$ is also equal to zero, hence:

$\ddot x = - c(x)$

The only way the system tends towards an invariant set within R, is if c(x) is equal to zero, which is at x = 0.

Hence, the system is globally asymptotically convergent to the equilibrium point:

$\left ( \begin{matrix} x = 0 \\ \dot x =0 \end{matrix} \right )$

Important Note: This is a very simplified summary of my own notes which I personally find helpful as a shortlist of fundamental definitions. However, these concepts can seem rather abstract and complicated, and they can be at times, therefore to understand them better and really get a better grasp of Nonlinear dynamical systems I would advise reading through notes and books which cover these topics in more depth and provide practical examples which are extremely helpful in building a better understanding.

Non-Linear Dynamical System

Published by Majed Yassin

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