# A clarifying question on Lasalle invariance principle

Given a nonlinear system $$$$\dot{x}=f(x),~x(0)=x_0 \tag{1}$$$$ where $$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?

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 $$c = 8$$.

• Thank you so very much for your time you dedicated in explaining this fact. Jul 1 '20 at 17:08
• 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. Jul 1 '20 at 20:18

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 $$x_2=0$$.