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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

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  • $\begingroup$ The locally Lipschitz functions $g_i(x), i=1,2$ are defined the way they are right ? and by $\lVert \cdot \rVert$ which norm you are specifying? Also as far as i recall $V(x_1,x_2):=x_1^2-x_2^2$ is called a Chetaev function ? Can you mention the the page in Khalil where it's given ? $\endgroup$ – Siddhartha May 5 at 9:12
  • $\begingroup$ Thanks for answering. The page of the Khalil for this example is pag. 126. The norm it is specified is the L2 norm, sorry I forgot to write it. $\endgroup$ – J.D. May 5 at 9:36
  • $\begingroup$ You should probably specify what exactly is not clear. I.e., start from the very beginning, then study it step by step and let us known what is the first step that you do not understand. It will be simpler to answer. Note also that it should be $|x_1g_1(x) - x_2g_2(x)|\le 2k|x|^3$ and not $2k|x|^2$, where I use $|x|$ instead of $\|x\|_2$. $\endgroup$ – Arastas May 5 at 10:44
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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.

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