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This problem is asking to demonstrate that the solutions to the system 

\[
\dot{x} = Ax + bw(t)
\]

where \( x(t) \in \mathbb{R}^n \), \( b \in \mathbb{R}^n \), \( A \in \mathbb{R}^{n \times n} \) (with all eigenvalues having negative real parts), and \( w(t) \) is bounded noise, are uniformly bounded and uniformly ultimately bounded.

### Approach

1. **Homogeneous system analysis (no noise term):**  
   The system can first be analyzed without the noise term (\( w(t) = 0 \)). In this case, the equation reduces to:
   \[
   \dot{x} = Ax
   \]
   Since \( A \) is a matrix whose eigenvalues have negative real parts, the solutions of this homogeneous system will decay exponentially. Specifically, if \( \lambda_1, \lambda_2, \dots, \lambda_n \) are the eigenvalues of \( A \), then the general solution will behave like:
   \[
   x(t) = e^{At} x(0)
   \]
   where \( e^{At} \) is the matrix exponential. Since the eigenvalues have negative real parts, \( \|e^{At}\| \) decays to zero as \( t \to \infty \), which implies that the homogeneous system is stable and the solutions are bounded.

2. **Non-homogeneous system (with noise term):**  
   Now, consider the non-homogeneous equation:
   \[
   \dot{x} = Ax + bw(t)
   \]
   Using the method of variation of parameters, we can write the solution as:
   \[
   x(t) = e^{At} x(0) + \int_0^t e^{A(t - \tau)} b w(\tau) \, d\tau
   \]
   The first term, \( e^{At} x(0) \), decays to zero because of the stability of the homogeneous system, as established earlier. The second term involves the bounded noise \( w(t) \).

   Since \( w(t) \) is bounded, say \( |w(t)| \leq M \) for some constant \( M \), the integral term can be bounded by:
   \[
   \left| \int_0^t e^{A(t - \tau)} b w(\tau) \, d\tau \right| \leq M \int_0^t \|e^{A(t - \tau)} b\| \, d\tau
   \]
   Given that \( \|e^{A(t - \tau)}\| \) decays exponentially as \( t - \tau \) increases, the integral converges and remains bounded for all \( t \). This shows that the total solution \( x(t) \) is uniformly bounded.

3. **Uniform ultimate boundedness:**  
   The decay of \( e^{At} \) ensures that the contribution of the initial condition decays over time. For large \( t \), the solution is dominated by the integral term involving the noise. Since \( w(t) \) is bounded and \( e^{At} \) decays, there exists a bound \( B \) such that:
   \[
   \|x(t)\| \leq B \quad \text{for all} \ t \geq T
   \]
   for some sufficiently large \( T \). This shows that the system is uniformly ultimately bounded.

### Conclusion:
The system is both uniformly bounded and uniformly ultimately bounded due to the stability of the matrix \( A \) (which has eigenvalues with negative real parts) and the bounded nature of the noise \( w(t) \).

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Would you like more details or have any specific questions?

### Related questions:
1. What is the method of variation of parameters used in non-homogeneous systems?
2. How does the matrix exponential \( e^{At} \) behave when the eigenvalues of \( A \) have negative real parts?
3. What is the concept of uniform boundedness in dynamical systems?
4. How does bounded noise affect the stability of a linear system?
5. What conditions guarantee that a system is uniformly ultimately bounded?

**Tip:** In stability analysis, negative real eigenvalues of the system matrix ensure that perturbations decay over time, which is crucial for the system's boundedness.

Consider the system ẋ = Ax + bw, where x(t) ∈ ℝⁿ, b ∈ ℝⁿ is a constant vector, and A ∈ ℝⁿˣⁿ is a constant matrix with all eigenvalues having negative real parts, and w(t) ∈ R is a bounded noise. Show that the solutions of this system are uniformly bounded and uniformly ultimately bounded.

Explore how linear systems of the form ẋ = Ax + bw, with all eigenvalues of A having negative real parts and w(t) being bounded noise, ensure that solutions are uniformly bounded and uniformly ultimately bounded. The solution involves techniques like matrix exponentials, variation of parameters, and stability analysis.

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