Problem understanding Chetaev theorem

Question

I am studying control theory, and I am focusing on the Lyapunov stability. In particular, I am looking the Chetaev theorem, but I have some problems understanding it well.

I know that the Cheatev theorem gives a result for instability, and in articular, it sayst that an equilibrium point $x_e=0$ is unstable if it exists a Lyapunov function in $C^1$ such that $V(x)>0$ has $x_e$ as accumulation point (not sure what it means) and $\dot{V}(x)>0$ in a neigborhood $U$.

In the notes of my professor is present the following example, which I do not understand:

consider the system

$\dot{x_1}=x_1+g_1(x)$

$\dot{x_2}=-x_2+g_2(x)$

with $x_e=\begin{pmatrix} 0\\ 0 \end{pmatrix}$.

the example starts by saying that $g(0)=0$ so the origin is an equilibrium point, and $|g_i(x)|<||x||^{2}$

so far I don't understand why he does this and don't know what it means.

It continues by chosing a Lyapunov function as follows:

$V(x) = \frac{1}{2}(x_1^{2}-x_2^{2})$

ans then takes its derivative:

$\dot{V}(x)=x_1^{2}+x_2^{2}+x_1g_1(x)-x_2g_2(x)$

but I dont' understand from where this comes from.

Moreover it coninues by saying that since:

$|x_1g_1(x)-x_2g_2(x)|\leq \sum_{i=1}^{2}|x_i||g_i(x)|< 2k||x||^2$

we have

$\dot{V}(x)\geq||x|^2 - 2k||x||^3|=||x||^2(1+2k||x||)$

and if I consider a ball $S(0,r)$ with $r=\frac{1}{2k}$ the system is unstable.

I don't understand what it has been done here. Can somebody please help me?

[EDIT] If it can be useful, I have also found that this example is in the Hassan K. Khalil book

Answer 1

You have $g_i(\cdot )$ upper bounded by $|g_i(x)| \le k\lVert x \rVert_2^2$ , clearly this implies that $g_i(0)=0$ which makes $(0,0) $ as your equilibrium point. Now define the Chetaev/Lyapunov function $V(x_1,x_2):=\frac{1}{2}(x_1^2-x_2^2)$ according to Theorem $(3.3)$. Now take the Lie derivative along the vector field you get $$\begin{align}\dot{V}=(\mathcal{L}_f)(V)&=\dot{x}_1x_1-\dot{x}_2x_2=x_1(x_1+g_1(x))-x_2(-x_2+g_2(x))\\&=x_1^2+x_2^2+(x_1g_1(x)-x_2g_2(x))\end{align}$$

Next using the defining property of $g_i(x)$ you have $$\begin{align}|x_1g_1(x)+(-x_2)g_2(x)|&<|x_1||g_1(x)|+|x_2||g_2(x)|\\&\le k\lVert x \rVert_2^3+k\lVert x \rVert_2^3=2k\lVert x \rVert_2^3 \tag{1}\end{align}$$ Now construction of the set $U$ is clearly explained in Khalil $$U=\left \{ x \in \mathcal{B}_r:V(\mathbf{x})>0\right\}=\left \{ x \in \mathcal{B}_r:x_1^2>x_2^2\right\}$$ see Fig.$(3.5)$, notice the boundary of the set is $\partial U:=\left\{x_2=\pm |x_1|\right\}$. Now from the Theorem $(3.3)$ and eq $(1)$ you have $$\mathcal{L}_fV \ge \lVert x \rVert_2^2-2k\lVert x \rVert_2^3=\lVert x\rVert_2^2 \left(1-2k\lVert x \rVert_2 \right)\tag{2}$$ Note: to satisfy the conditions of the Theorem choose $r$ such that the right hand side of the preceding inequality is positive definite which requires $2k \lVert x\rVert_2 -1<0 $ which gives you the required condition to find $r$, which is $r<\frac{1}{2k}$

Reference : Nonlinear systems, Hasan Khalil, 2nd Edition.