Control Theory – Exponential Stability and Lasalle’s Invariance Theorem

control theorylyapunov-functionsnonlinear dynamicsnonlinear systemstability-theory

It is well known that a system $\dot{x}=f(x)$ with $x \in \mathbb{R}^n$ is exponentially stable if there exists a Lyapunov function $V(x)$ which satisfies

\begin{align}
k_1\Vert x \Vert \leq V(x) &\leq k_2\Vert x \Vert \\
\dot{V} &\leq -k_3\Vert x \Vert.
\end{align}

Can we say something about exponential convergence to the set $S = \{ x \in \mathbb{R}^n \mid \dot{V}=0 \}$ if we replace the second equation by a positive semidefinite condition, i.e.

$$\dot{V} \leq -k_3\Vert x_1 \Vert, $$

where $x_1$ contains only some of the components of $x$, e.g. $x = (x_1^\top , x_2^\top)^\top$. From the invariance principle, we know in fact that the system converges to the set $S$, but not if it converges exponentially. I could not find anything in this regard in the standard nonlinear systems books by Khalil, Sepulchre, etc.

Can we say anything about the exponential convergence to $S$?

Best Answer

No. Let us rewrite the conditions of the theorem in an alternative, more convenient form (as in the Khalil's book): $$\tag{1} k_1\Vert x \Vert^a \leq V(x) \leq k_2\Vert x \Vert^a $$ $$\tag{2} \dot{V} \leq -k_3\Vert x \Vert^a, $$ $k_1,k_2,k_3,a>0$. Consider the system $$\tag{3} \dot x_1=-x_1,\quad \dot x_2=-f(x_2), $$ where $f(x)$ is some function that satisfies the existence and uniqueness conditions, $\forall x\ne0\;\;xf(x)> 0$, $f(0)=0$. Let's choose $V(x_1,x_2)=x_1^2+x_2^2$. Then $$ \dot V= -2x_1^2-2x_2f(x_2)\le -2x_1^2. $$ Noticing that condition (1) is satisfied ($k_1=1$, $k_2=1$, $a=2$) and $\dot V\le -2\Vert x_1\Vert^a$, we should obtain (if our guess is true) exponential stability for any $f(x_2)$ that satisfies the above conditions. This is obviously false. It is easy to come up with a differential equation of type $\dot x_2=-f(x_2)$, solutions of which do not tend to the equilibrium at an exponential rate.

(Update) Here the set $S$ contains only equilibrium because $$\dot V=-2x_1^2-2x_2f(x_2)=0 \;\Leftrightarrow\; x_1=x_2=0;$$ therefore, there is no exponential tendency towards $S$. As for the exponential tendency of some of the variables to zero, this is called exponential stability with respect to part of the variables. It was described in some articles and books. For example, there is an article (it is, however, in Russian): https://www.mathnet.ru/eng/dan5135

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