How to show that the infinity norm obtained in the problem given below is correct?

Question

I found the following example problem in page 178 of the book "Nonlinear System" by Hassan K. Khalil, Third Edition. The objective of this problem is to show whether a system is input-to-state stable (ISS). There is one specific part of the solution given to the example problem, which I am not getting.

Show the following system \begin{equation}\tag{1} \dot{x}_{1} =-x_{1}+x^{2}_{2},\\ \dot{x}_{2} =-x_{2}+u \end{equation} is ISS. They consider a Lyapunov function $V(x)=\dfrac{1}{2}x^{2}_{1}+\dfrac{1}{4}x^{4}_{2}$ which is a positive definite function, ofcourse. Next, they evaluate $\dot{V}(x)$ along the trajectories of the system in Eq. (1) and they attempt to find conditions for which $\dot{V}(x)<0$. Then, it will be straightforward to determine the bounds on $\|x(t)\|$ and the system (1) is ISS.

$\dot{V}(x)$ along the system trajectories in Eq. (1) become $\dot{V}(x)\leq -\dfrac{1}{2}(1-\theta)(x^{2}_{1}+x^{4}_{2})-\dfrac{1}{2}\theta(x^{2}_{1}+x^{4}_{2})+|x_{2}|^3u,\ 0\leq{\theta}\leq{1}$.

$\dot{V}(x)\leq{0}$ if $|x_{2}|\geq\dfrac{2|u|}{\theta}$ or $|x_{2}|\leq\dfrac{2|u|}{\theta}$ and $|x_{1}|\geq\left(\dfrac{2|u|}{\theta}\right)^2$.

This condition is implied by \begin{equation}\tag{2} \max\{|x_{1}|,|x_2|\}\geq\max\left\{\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right\} \end{equation}

My question is : how does author arrive at Eq. (2) from the conditions satisfying $\dot{V}(x)\leq{0}$? In other words, how both of the conditions obtained from $\dot{V}(x)\leq{0}$ are captured in Eq. (2)?

Any suggestions in this regard are greatly appreciated.

Answer 1

Suppose $(2)$ holds. Separate cases according to whether $|x_{2}|\geq\dfrac{2|u|}{\theta}$ or not.

• If $|x_{2}|\ge \dfrac{2|u|}{\theta}$,then we know $\dot{V}(x)\leq{0}$.

• If $|x_{2}|< \dfrac{2|u|}{\theta}$, then (2) implies that $$ |x_{1}| \geq\max\left\{\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right\}$$ which again yields that $\dot{V}(x)\leq{0}$.

Answer 2

What I tried is as follows. Based on the second condition of $\dot{V}(x)\leq{0}$, we obtain

\begin{align*} \max\left\{|x_1|,|x|_2\right\}=|x_1|,\ \text{when}\ 0\leq|x_2|\leq\dfrac{2|u|}{\theta}\leq\left(\dfrac{2|u|}{\theta}\right)^2\leq|x_1|,\tag{1.1}\\ \max\left\{|x_1|,|x|_2\right\}=|x_2|,\ \text{when}\ 0\leq\left(\dfrac{2|u|}{\theta}\right)^2\leq|x_1|\leq|x_2|\leq\dfrac{2|u|}{\theta}.\tag{1.2}\\ \end{align*}

From Eqs. (1.1) and (1.2), we further obtain \begin{align*} \max\left\{|x_1|,|x|_2\right\} &=|x_1|\geq \left(\dfrac{2|u|}{\theta}\right)^2\geq\dfrac{2|u|}{\theta},\tag{1.3}\\ \max\left\{|x_1|,|x|_2\right\} &=|x_2|\geq\ \left(\dfrac{2|u|}{\theta}\right)^2.\tag{1.4} \end{align*}

From Eqs. (1.3) and (1.4), we again obtain \begin{align*} \max\left\{|x_1|,|x|_2\right\} &\geq\dfrac{2|u|}{\theta}=\min\left\{\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right\},\tag{1.5}\\ \max\left\{|x_1|,|x|_2\right\} &\geq\ \left(\dfrac{2|u|}{\theta}\right)^2=\min\left\{\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right\}.\tag{1.6} \end{align*}

So I guess in Eq. (2) of the question $\max()$ on the right hand side should be replaced with $\min()$. ISin't it? Furthermore, Eqs. (1.3) and (1.4) respectively yield \begin{align*} \max\left\{|x_1|,|x|_2\right\} &\geq \max\left(\left(\dfrac{2|u|}{\theta}\right)^2,\dfrac{2|u|}{\theta}\right),\\ \max\left\{|x_1|,|x|_2\right\} &\ngeq \max\left(\left(\dfrac{2|u|}{\theta}\right)^2,\dfrac{2|u|}{\theta}\right). \end{align*} That's the reason why I think Eq. (2) in the question is not quite correct.

Now on the other hand, based on the first condition of $\dot{V}(x)\leq 0$, we obtain \begin{align*} \max\left\{|x_1|,|x|_2\right\}=|x_1|,\ \text{when}\dfrac{2|u|}{\theta}\leq|x_2|\leq|x_1|,\tag{1.7}\\ \max\left\{|x_1|,|x|_2\right\}=|x_2|,\ \text{when}\ |x_1|\leq\dfrac{2|u|}{\theta}\leq|x_2|,\tag{1.8}\\ \max\left\{|x_1|,|x|_2\right\}=|x_2|,\ \text{when}\ \dfrac{2|u|}{\theta}\leq|x_1|\leq|x_2|.\tag{1.9} \end{align*}

From Eqs. (1.7)-(1.9), we further obtain \begin{align*} \max\left\{|x_1|,|x|_2\right\}=|x_1|\geq\dfrac{2|u|}{\theta}\geq\min\left(\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right),\tag{1.10}\\ \max\left\{|x_1|,|x|_2\right\}=|x_2|\geq\dfrac{2|u|}{\theta}\geq\min\left(\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right),\tag{1.11}\\ \max\left\{|x_1|,|x|_2\right\}=|x_2|\geq\dfrac{2|u|}{\theta}\geq\min\left(\dfrac{2|u|}{\theta},\left(\dfrac{2|u|}{\theta}\right)^2\right).\tag{1.12} \end{align*}