11.3: Finding Lyapunov Functions
Lyapunov’s method and the LaSalle invariance principle are very powerful techniques, but the obvious question always arises, "how do I find the Lyapunov function? The unfortunate answer is that given an arbitrary ODE there is no general method to find a Lyapunov function appropriate for a given ODE for the application of these methods.
In general, to determine a Lyapunov function appropriate for a given ODE the ODE must have a structure that lends itself to the construction of the Lyapunov function. Therefore the next question is "what is this structure?" If the ODE arises from physical modeling there may be an "energy function" that is "almost conserved". What this means is that when certain terms of the ODE are neglected the resulting ODE has a conserved quantity, i.e. a scalar valued function whose time derivative along trajectories is zero, and this conserved quantity may be a candidate to for a Lyapunov function. If that sounds vague it is because the construction of Lyapunov functions often requires a bit of "mathematical artistry". We will consider this procedure with some examples. Energy methods are important techniques for understanding stability issues in science and engineering; see, for example see the book by Langhaar and the article by Maschke.
To begin, we consider Newton’s equations for the motion of a particle of mass m under a conservative force in one dimension:
\[m\ddot{x} = -\frac{d\Phi}{dx} (x), x \in \mathbb{R}, \label{C.1}\]
Writing this as a first order system gives:
\[\begin{align} \dot{x} &= y \\[4pt] \dot{y} &= -\frac{1}{m}\frac{d\Phi}{dx} (x). \label{C.2} \end{align}\]
It is easy to see that the time derivative of the following function is zero
\[E = \frac{my^2}{2} + \Phi (x), \label{C.3}\]
since
\[ \begin{align} \dot{E} &= my\dot{y}+\frac{d \Phi}{dx}(x)\dot{x} \\[4pt] &= -y\frac{d \Phi}{dx}(x) + y\frac{d \Phi}{dx}(x) = 0. \label{C.4} \end{align}\]
In terms of dynamics, the function in Equation \ref{C.3} has the interpretation as the conserved kinetic energy associated with (C.1).
Now we will consider several examples. In all cases we will simplify matters by taking \(m = 1\).
Example \(\PageIndex{38}\)
Consider the following autonomous vector field on \(\mathbb{R}^2\):
\(\dot{x} = y\),
\[\dot{y} = -x-\delta y, \delta \ge 0, (x, y) \in \mathbb{R}^2. \label{C.5}\]
For \(\delta = 0\) Equtation \ref{C.5} has the form of (C.1):
\(\dot{x} = y\),
\[\dot{y} = x, (x, y) \in \mathbb{R}^2. \label{C.6}\]
with
\[E = \frac{y^2}{2}+\frac{x^2}{2}. \label{C.7}\]
It is easy to verify that \(\frac{dE}{dt} = 0\) along trajectories of (C.6).
Now we differentiate E along trajectories of (C.5) and obtain:
\[\frac{dE}{dt} = -\delta y^2. \label{C.8}\]
(C.6) has only one equilibrium point located at the origin. E is clearly positive everywhere, except for the origin, where it is zero. Using E as a Lyapunov function we can conclude that the origin is Lyapunov stable. If we use E to apply the LaSalle invariance principle, we can conclude that the origin is asymptotically stable. Of course, in this case we can linearize and conclude that the origin is a hyperbolic sink for \(\delta > 0\).
Example \(\PageIndex{39}\)
Consider the following autonomous vector field on \(\mathbb{R}^2\):
\(\dot{x} = y\),
\[dot{y} = x-x^3-\delta y, \delta \ge 0, (x, y) \in \mathbb{R}^2. \label{C.9}\]
For \(\delta = 0\), Equation \ref{C.9} has the form of (C.1):
\(\dot{x} = y\),
\[dot{y} = x-x^3, (x, y) \in \mathbb{R}^2. \label{C.10}\]
with
\[E = \frac{y^2}{2}-\frac{x^2}{2}+\frac{x^4}{4}. \label{C.11}\]
It is easy to verify that \(\frac{dE}{dt} = 0\) along trajectories of (C.10).
The question now is how we will use E to apply Lyapunov’s method or the LaSalle invariance principle? (C.9) has three equilibrium points, a hyperbolic saddle at the origin for \(\delta \ge 0\) and hyperbolic sinks at (x, y) = \((\pm1, 0)\) for \(\delta > 0\) and centers for \(\delta = 0\). So linearization gives us complete information for \(\delta > 0\). For \(\delta = 0\) linearization is sufficient to allow is to conclude that the origin is a saddle. The equilibria (x, y) = \((\pm 1, 0)\) are Lyapunov stable for \(\delta = 0\), but an argument involving the function E would be necessary in order to conclude this. Linearization allows us to conclude that the equilibria (x, y) = \((\pm 1, 0)\) are asymptotically stable for \(\delta > 0\).
The function E can be used to apply the LaSalle invariance principle to conclude that for \(\delta > 0\) all trajectories approach one of the three equilibria as \(t \rightarrow \infty\).