7
$\begingroup$

Given a nonlinear system \begin{equation} \dot{x}=f(x),~x(0)=x_0 \tag{1} \end{equation} where $f\in{\mathcal{C}^{1}}:D\to\mathbb{R}^{n}$. The Lasalle invariance theorem statement goes as follows:

Let $\Omega\subset{D}$ be a compact set that is positively invariant with respect to (1). Let $V:D\to{\mathbb{R}}$ be a continuously differentiable function such that $\dot{V}(x)\leq{0}$ in $\Omega$. Let $E$ be the set of all points in $\Omega$ where $\dot{V}(x)=0$. Let $M$ be the largest invariant set in $E$. Then every solution starting in $\Omega$ approaches $M$ as $t\to\infty$.

Here $M$ is the largest invariant limit set (every invariant set is a subset of M) where $x(t)$ converges to at infinite time. My question is how is $M$ and $E$ different. Since $\dot{V}$ is zero at whole of $E$, then isn't $E$ itself the largest invariant set?

$\endgroup$

3 Answers 3

3
$\begingroup$

A set is invariant with respect to its dynamics if $x(t_0)\in M \implies x(t)\in M$, this is not the case for the set $E$. The set $E = \left\lbrace x\in \Omega \mid \dot{V}(x) = 0 \right\rbrace$ does not need to be an invariant set, since it does not consider the solution of $x(t)$.

I'll show this statement using the well known pendulum example. The dynamics of a pendulum with friction are given by $$ m\ell^2 \ddot{\theta} +d\dot{\theta} + mg\ell \sin(\theta) = 0, $$ and in state space form by $$\dot{x} = f(x) = \begin{bmatrix} x_2\\ -\frac{g}{\ell} \sin(x_1) - \frac{d}{m\ell^2} x_2\end{bmatrix},$$ where $x_1 = \theta$, $x_2 = \dot{\theta}$. The equilibria are $\bar{x}_1 = k\pi$, $k\in\mathbb{Z}=\left\lbrace\ldots,-1,0,1,2,\ldots\right\rbrace$ and $\bar{x}_2 = 0$.

Now, take the energy as the Lyapunov function candidate $$V(x) =mg\ell\big(1-\cos(x_1)\big) + \frac{1}{2}m\ell^2 (x_2)^2$$ with its derivative $$\dot{V}(x) = \frac{\partial V}{\partial x}f(x) = -d(x_2)^2 \leq 0.$$

The sets in LaSalle's invariance principle are $$\begin{align} \Omega &= \Omega_c := \left\lbrace x\in \mathbb{R}^2 \mid V(x) < c\right\rbrace\\ E &= \left\lbrace x\in \Omega \mid \dot{V}(x) = 0 \right\rbrace = \left\lbrace x \in \Omega \mid x_2 = 0\right\rbrace\\ M &= \left\lbrace 0 \right\rbrace, \end{align} $$ where $M$ is the largest invariant set in $E$.

Now, consider the point $x = [1, 0]^\top \in E$, i.e. $\theta = 1$, $\dot{\theta} = 0$. This point is in the set $E$, however it will not stay in $E$ because the gravity will cause the pendulum to move, i.e. $x_2 \neq 0$ leaving $E$. Hence, for a point in the set $E$ it does not imply that $x(t_0) \in E \implies x(t)\in E$. This is visualized in the figure below by the vector field (black arrows) and the different sets. The vector field in $E$ is pointing away from $E$.

The vector field and sets below are drawn for $m = 1$, $\ell = 1$, $g = 9.81$, $d = 1$ and $c = 8$.

Sets in LaSalle's invariance principle

$\endgroup$
2
  • 1
    $\begingroup$ Thank you so very much for your time you dedicated in explaining this fact. $\endgroup$
    – jbgujgu
    Jul 1, 2020 at 17:08
  • $\begingroup$ It would be nice if you can accept one of the two answers given, such that it is clear that this question is answered to your satisfaction. $\endgroup$ Jul 1, 2020 at 20:18
2
$\begingroup$

The sets $M$ and $E$ can be different. The set $E$ only considers $\dot{V}=0$ while $M$ also considers $f(x)$. Namely, invariant set means that for all $x(0) \in M$ the solution $x(t)\in M$ for all $t>0$, which can also be written as $x(0)+\delta\,f(x(0)) \in M$ for infinitesimally small $\delta$.

For example consider the system

\begin{align} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= g(x_1,x_2), \end{align}

and there exists a $V(x_1,x_2)$ such that $\dot{V}=-x_1^2$. From this it would follow that $E$ only requires $x_1=0$ and $x_2$ can be anything. However, when considering the dynamics in order to keep $x_1=0$ it is also required that $x_2=0$, thus $M$ only consists of the point $x_1=0$ and $x_2=0$.

$\endgroup$
0
0
$\begingroup$

The above answers are great and helped me understand better but I thought I might add my intuitive/visual understanding of how this works. So using the pendulum example above by useless-machine, the Lyapunov function represents the energy of the pendulum system. Therefore, the height of the function $V(\mathbf{x})$ represents the energy at a given state. Therefore, each contour at a particular height represents all the states for a given amount of energy. We know that because of damping in the pendulum the energy should decrease over time. So if we start at some point, $x_1(t_0), x_2(t_0)$, and plot out a trajectory for it, over time we should see that $V(\mathbf{x}(t)) \leq V(\mathbf{x}(t_0))$. You can imagine the trajectories "flowing" down the surface of $V$.

Importantly, the slope of the surface is not $\dot{V}$ since we are taking a derivative with respect to time. Instead $\dot{V}$ is the slope of these trajectories on the surface. With that in mind now let's consider the point mentioned $\mathbf{x} = [1, \; 0]^T$ which corresponds to $x_1 = 1$ and $ x_2 = 0$ remembering that $x_1 = \theta$ and $x_2 = \dot{\theta}$. So in this case, since the instantaneous angular velocity is 0, the instantaneous loss in energy due to damping is also 0. As a result, the height of $V$ does not change with respect to time indicating that the slope of the trajectory is briefly along one of the contours of $V$. Despite this, we know that due to gravity both $\theta$ and $\dot{\theta}$ will be changing and in this case that is along the contour. Therefore, at an infinitesimally small step away, $\dot{\theta}$ is no longer 0 and so $\dot{V}$ is no longer 0 and the trajectory is pointing down the surface again. So this point we found is more like a point of inflection along a trajectory rather than an actual equilibrium point and clearly this means that this point does not form an invariant set with the other points that follow it on the trajectory since $\dot{V}$ is negative semi-definite (it will never climb back up again).

I am very new to this so if I have a mistake in my understanding here, please correct me!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.