📂Dynamics

Lyapunov Function

Definition¹

Given a space $X$ and a function $f : X \to X$, suppose the following vector field is given as a differential equation: $$ \dot{x} = f(x) $$ For a point $x_{0} \in X$ in such an autonomous system, a scalar function $V \in C^{1} \left( \mathcal{N} (x_{0}) , \mathbb{R} \right)$ defined in the neighborhood $\mathcal{N} \left( x_{0} \right)$ of $x_{0}$ is called a Liapunov Function if it satisfies the following conditions:

• (i): If $V(x_{0}) = 0$ and $x \ne x_{0}$, then $V(x) > 0$
• (ii): $x \in \mathcal{N} \left( x_{0} \right) \setminus \left\{ x_{0} \right\}$ implies $V ' (x) \le 0$

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• $C^{1}(A,B)$ refers to the set of functions that are differentiable, with a domain of $A$ and a codomain of $B$, having continuous derivatives.
• For $V$ to belong to $C^{1} \left( \mathcal{N} (x_{0}) , \mathbb{R} \right)$ means that it is a scalar function defined in the vicinity of $x_{0}$, is differentiable, and $v '$ is continuous.

Explanation

A Liapunov function is a possible function depending on the given system $\dot{x} = f(x)$, particularly considered for verifying the stability of the fixed point $x_{0} = \overline{x}$ when $X = \mathbb{R}^{n}$. The existence of a Liapunov function implies stability, and $V$ should be a function appropriately defined according to the system $f$. Here, $v '$ represents the differentiation with respect to time $t$, revealing the relationship with $f$ when differentiating $V$ and introducing a term related to $\dot{x}$.

Following these explanations, it may seem like Liapunov functions are a universal tool for understanding nonlinear systems, but finding such functions for nonlinear systems is not that straightforward. There is no general method for finding Liapunov functions, and finding one for a significant system can be challenging enough to become a research topic.

Example

Let’s look at a very simple process of finding a Liapunov function: $$ \begin{align*} \dot{x} =& -x + 4y \\ \dot{y} =& -x - y^{3} \end{align*} $$ In this system, a fixed point $(0,0)$ is given. As mentioned, there is no general method to find a Liapunov function, so intuition must be employed. With experience in finding Liapunov functions, it may become quicker to identify them. Here, we assume $V(x,y) = x^{2} + a y^{2}$ to be a Liapunov function and try to specify its value through $a \ge 0$.

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Part (i).

If $V(0,0) = 0$ and $(x,y) \ne (0,0)$, then $V(x,y) > 0$. Assuming $a$ is not negative at point $(0,0)$, then $V > 0$.

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Part (ii).

From $(x,y) \in \mathcal{N} \left( (0,0) \right) \setminus \left\{ (0,0) \right\}$, it follows that $V ' (x,y) \le 0$. Differentiating $V$ with respect to $t$ yields $$ \begin{align*} {{ d V } \over { d t }} =& 2xx’ + 2ay\dot{y} \\ =& 2x(-x+4y) + 2ay \left( -x-y^{3} \right) \\ =& -2x^{2} + (8-2a)xy - 2ay^{2} \end{align*} $$ where, if $a=4$, then $V ' <0$.

Therefore, when a fixed point $(0,0)$ is given, the existence of a Liapunov function as $V(x,y) = x^{2} + 4 y^{2}$ can be guaranteed, and it also implies that $(0,0)$ has Liapunov stability.