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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.

Analyze the stability properties of the system ẋ = -ζ(t)ζᵀ(t)x, where ζ(t) ∈ R^n is a uniformly bounded and continuous vector signal (use a Lyapunov candidate V(x,t) = xᵀx).

Analyze the stability of the system ẋ = -ζ(t)ζᵀ(t)x using Lyapunov functions. Explore key concepts in stability analysis, including bounded continuous vector signals and the Lyapunov Stability Theorem. Suitable for advanced university or graduate-level studies.

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