This problem asks to analyze the stability of the system described by the differential equation:

$\dot{x} = -\zeta(t) \zeta^T(t) x,$ where $\zeta(t) \in \mathbb{R}^n$ is a uniformly bounded and continuous vector signal, and a Lyapunov function $V(x,t) = x^T x$ is suggested.

Solution Outline:

1. Define the Lyapunov Candidate: The suggested Lyapunov function is: $V(x,t) = x^T x,$ which is a positive definite function. This is a common choice because it represents the squared norm of the state vector $x$, meaning $V(x,t) > 0$ for all $x \neq 0$ and $V(x,t) = 0$ when $x = 0$.

2. Compute the Time Derivative of $V(x,t)$: To analyze stability, we need to compute the derivative of $V(x,t)$ with respect to time, $\dot{V}(x,t)$.

   Using the chain rule: $\dot{V}(x,t) = \frac{d}{dt} (x^T x) = 2x^T \dot{x}.$

   From the system equation $\dot{x} = -\zeta(t) \zeta^T(t) x$, substitute for $\dot{x}$: $\dot{V}(x,t) = 2x^T (-\zeta(t) \zeta^T(t) x).$

   Simplifying: $\dot{V}(x,t) = -2 (x^T \zeta(t)) (\zeta^T(t) x).$

   Since $(x^T \zeta(t)) (\zeta^T(t) x)$ is a scalar and can be written as: $\dot{V}(x,t) = -2 (x^T \zeta(t))^2.$

   This shows that $\dot{V}(x,t) \leq 0$, meaning that the Lyapunov function is non-increasing over time.

3. Conclusion on Stability: Since $V(x,t)$ is positive definite and $\dot{V}(x,t) \leq 0$, we can conclude that the system is Lyapunov stable. However, to establish asymptotic stability, we need to check whether $\dot{V}(x,t) = 0$ implies that $x = 0$.

   From the expression $\dot{V}(x,t) = -2 (x^T \zeta(t))^2$, if $\dot{V}(x,t) = 0$, then $(x^T \zeta(t))^2 = 0$, which implies $x^T \zeta(t) = 0$. This shows that $x$ is orthogonal to $\zeta(t)$.

   However, since $\zeta(t)$ is only known to be uniformly bounded and continuous, there is no guarantee that $x$ must be zero. Therefore, asymptotic stability cannot be directly concluded without additional conditions on $\zeta(t)$.

Summary:

• The system is Lyapunov stable.
• Asymptotic stability cannot be guaranteed without further information about the properties of $\zeta(t)$.

Would you like further clarification or more details on specific parts of the solution? Here are some related questions:

1. How does the choice of Lyapunov function impact the stability analysis?
2. What additional conditions on $\zeta(t)$ could ensure asymptotic stability?
3. Can we use other types of Lyapunov functions for more general systems?
4. What role does uniform boundedness of $\zeta(t)$ play in stability?
5. How does the continuity of $\zeta(t)$ affect the system's behavior?

Tip:

When analyzing stability using Lyapunov functions, always check whether the conditions for both Lyapunov stability and asymptotic stability are satisfied, as they have different implications for system behavior.