I have two questions:
Problem 1: Let $V (x)$ be the Lyapunov function candidate with $x \in \mathbf{R}$, and the time derivative of $V(x)$ is given by
$\dot{V} (x) \le - x ( x - \alpha (t) )$
where $| \alpha (t) | \le \alpha^{*}$, for all $t \ge t_{0}$, $\alpha^{*} > 0$. Then, for $|x| > \alpha^{*}$, we have $\dot{V} (x) < 0$.
Consequently, can we say that, for any $x \in \mathbf{R}$, $x$ converges to the interval $[ - \alpha^{*} , \alpha^{*}]$ in finite time ?
Problem 2: Let $|x (t) | \le z (t)$, for all $t \ge t_{0}$, $x(t) \in \mathbf{R}^{n}$, $z(t) \in \mathbf{R}_{\ge 0}^{n}$, and $\dot{z} (t) = A z(t)$, where $A$ is a Hurwitz and Metzler matrix.
Then, can we say that $x$ converges to the origin in finite time ?
For the problem 1:
Example 1: Non-convergence in finite time
Consider the following system:
$$\dot{x} = -x(x - \alpha(t))$$
where $x \in \mathbb{R}$ and $|\alpha(t)| \leq \alpha^*$ for all $t \geq t_0$, where $\alpha^* > 0$.
Let's assume $\alpha(t) = \alpha^*$ for simplicity. Then the system becomes:
$$ \dot{x} = -x(x - \alpha^*) $$
To analyze the convergence behavior, we can examine the Lyapunov function candidate:
$$ V(x) = \frac{1}{2}x^2 $$
Taking the derivative of V(x) along the trajectories of the system:
\begin{aligned} \dot{V}(x) &= \frac{d}{dt}\left(\frac{1}{2}x^2\right) \\ &= x\dot{x} \\ &= -x^2(x - \alpha^*) \end{aligned}
Since $\alpha^*$ is a constant, we can rewrite the equation as:
$$ \dot{V}(x) = -x^2(x - \alpha^*) = -x^3 + \alpha^*x^2 $$
From this expression, we can see that $\dot{V}(x) < 0$ for $|x| > \alpha^*$, which implies that the Lyapunov function $V(x)$ is strictly decreasing for $|x| > \alpha^*$.
However, we cannot conclude that $x$ converges to the interval $- \alpha^*, \alpha^*$ in finite time. The negative derivative $\dot{V}(x) < 0$ for $|x| > \alpha^*$ only tells us that $V(x)$ decreases over time for values of $x$ outside the interval $- \alpha^*, \alpha^*$. It does not provide information about the convergence or the finiteness of time for reaching the interval.
For problem 2: Example 2: Stability but no finite-time convergence
Consider the following system:
$$ \dot{x} = Ax $$
where $x \in \mathbb{R}^n$, $A$ is a Hurwitz and Metzler matrix, and $x(t) \geq 0$ for all $t \geq t_0$.
Let's assume $n = 2$ for simplicity, and consider the following matrix:
A = \begin{bmatrix} -1 & 2 \\ -2 & -3 \\ \end{bmatrix}
The eigenvalues of matrix $A$ are $\lambda_1 = -0.7913$ and $\lambda_2 = -3.2087$, both of which have negative real parts. Therefore, A is Hurwitz and guarantees stability.
Now, let's assume $x(t)$ is a vector with non-negative components:
x(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \\ \end{bmatrix}
We can define a Lyapunov function candidate as:
$$ V(x) = \frac{1}{2}\left(x_1^2 + x_2^2\right) $$
Taking the derivative of V(x) along the trajectories of the system
:
\begin{aligned} \dot{V}(x) &= \frac{d}{dt}\left(\frac{1}{2}\left(x_1^2 + x_2^2\right)\right) \\ &= x_1\dot{x}_1 + x_2\dot{x}_2 \\ &= x_1(A_{11}x_1 + A_{12}x_2) + x_2(A_{21}x_1 + A_{22}x_2) \\ &= x_1^2A_{11} + x_1x_2(A_{12} + A_{21}) + x_2^2A_{22} \end{aligned}
Since $A$ is a Hurwitz matrix, we can conclude that $\dot{V}(x) < 0$ for all $x \neq 0$. This implies that $V(x)$ is strictly decreasing along the trajectories of the system, indicating stability.
However, we cannot guarantee finite-time convergence to the origin. The stability property ensures that $(x(t)$ approaches the origin as $t$ tends to infinity, but it does not provide information about the finiteness of time for reaching the origin.