# 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, 2020 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, 2020 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 $$x_2=0$$.