Using the function: $V(x) = x_1^2 + \frac{1}{2}x_2^2$ prove that the following system satisfies the Input-to-State property with respect to the input $u$:
\begin{equation} \dot{x}_1 = -(1+2x_1^2)\cdot x_1 + x_2 \\ \dot{x}_2 = -2x_1 - (1+3x_1^2)\cdot x_2+3u \end{equation}
I am trying to solve this problem and this is the procedure I followed:
In order for the system to satisfy the ISS property, it is enough to show that $V(x)$ is an ISS Lyapunov function. In view of this:
- $V(x)$ is continuous and differentiable
- $V(x)$ is positive definite
- $V(x)$ is radially unbounded
And last thing is to show that there exist a function $\mu\in K$ (of class K) such that:
\begin{equation} \mu(|u|)\leq V(x) \\ x \neq 0 \Rightarrow \dot{V}(x)<0 \end{equation}
\begin{equation} \dot{V}(x) = -2x_1^2\cdot (1+2x_2^2) - x_2^2\cdot (1+3x_1^2) + x_2\cdot 3u = \\ = -2x_1^2 - 7x_1^2x_2^2-x_2^2 + x_2\cdot 3u \Rightarrow \\ \Rightarrow \text{Using Young's Inequality } (ab \leq \frac{1}{p}a^p+\frac{1}{q}b^q, \ \frac{1}{p}+\frac{1}{q} = 1) \Rightarrow \\ \Rightarrow x_2\cdot 3u \leq \frac{1}{2}x_2^2 + \frac{9}{2}u^2 \leq \frac{1}{2}x_2^2 + \frac{9}{2}|u|^2 \Rightarrow \\ \Rightarrow \dot{V} \leq -2x_1^2 - 7x_1^2x_2^2-\frac{1}{2}x_2^2 + \frac{9}{2}|u|^2 \end{equation}
Now, in order to include $V(x)$ in the inequality: $$ 2x_1^2 + 7x_1^2x_2^2 \geq x_1^2 \Rightarrow -2x_1^2-7x_1^2x_2^2\leq -x_1^2 $$ and the former inequality becomes: \begin{equation} \dot{V}(x) \leq -x_1^2 - \frac{1}{2}x_2^2 +\frac{9}{2}|u|^2 \Rightarrow \dot{V}(x) \leq -V(X) + \frac{9}{2}|u|^2 \end{equation}
Define $\rho(s) = \lambda s$ and set: $$ -V(x) + \frac{9}{2}|u|^2 \leq -\rho (V) = -\lambda \cdot V(x) \Rightarrow V(x) \geq \frac{9}{2(1-\lambda)}\cdot |u|^2, \ \lambda\in (0 \ \ 1) $$
And let $\mu(|u|) = \frac{9}{2(1-\lambda)}\cdot |u|^2 \Rightarrow \mu(s) = \frac{9}{2(1-\lambda)}\cdot s^2$
Now check whether $\mu\in K$:
- $\mu(s)$ is continuous
- $\mu(0) = 0$
- $\mu(s)$ is increasing
So, $\mu(s)$ is of class $K$ (I think also of class $K_{\infty}$).
Now, check if $\dot{V} < 0, \ \forall x\neq 0$. Going back to the inequality:
\begin{equation} \dot{V}(x) \leq -V(x) + \frac{9}{2}|u|^2 \leq -V(x) + V(x)\cdot (1-\lambda) \Rightarrow \\ \Rightarrow \dot{V}(x) \leq -\lambda\cdot V(x) \end{equation}
which $\forall x\neq 0$ is negative since it equals to $0$ only when $V(x) = 0$, which happens only for $x=0$
I would like some feedback regarding the procedure. Is it correct ? Is there an alternate solution ?