Question

The results of Theorem \(4.1\) do not hold for time-varying systems as shown by the solution of $$ \frac{d}{d t}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{cc} 4 a & -3 a e^{8 a t} \\ a e^{-8 a t} & 0 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) $$ The eigenvalues of the system matrix are \(\lambda_{1}=a, \lambda_{2}=3 a\) (they happen to be constants, i.e. they do not depend on time) and hence for \(a<0\) both eigenvalues have real parts less than zero. However, the exact solution is (with initial condition \(\left.x_{1}(0)=x_{10}, x_{2}(0)=x_{20}\right)\) : $$ \begin{aligned} &x_{1}(t)=\frac{3}{2}\left(x_{10}+x_{20}\right) e^{5 a t}-\frac{1}{2}\left(x_{10}+3 x_{20}\right) e^{7 a t} \\ &x_{2}(t)=\frac{1}{2}\left(x_{10}+3 x_{20}\right) e^{-a t}-\frac{1}{2}\left(x_{10}+x_{20}\right) e^{-3 a t} \end{aligned} $$ which is unstable for any nonzero real \(a\).

Short answer

The system is unstable for any non-zero \(a\) due to exponentially growing terms in the solution.

Step-by-step solution

Understand the Problem

We're given a time-varying linear system expressed as a differential equation with a time-dependent matrix. The problem states that, despite the system's eigenvalues indicating stability for \(a < 0\), the actual solution is unstable for any non-zero \(a\). Our goal is to understand why this discrepancy occurs.

Analyze the Matrix Eigenvalues

The system matrix is: \[A(t) = \begin{pmatrix} 4a & -3ae^{8at} \ ae^{-8at} & 0 \end{pmatrix}\]The eigenvalues given are \(\lambda_1 = a\) and \(\lambda_2 = 3a\), both of which do not depend on time. For \(a < 0\), the eigenvalues have negative real parts, which typically suggests stability if these were constant coefficients.

Examine the Exact Solution

The exact solution to the system is provided:\[\begin{align*}x_1(t) &= \frac{3}{2}(x_{10} + x_{20})e^{5at} - \frac{1}{2}(x_{10} + 3x_{20})e^{7at} \x_2(t) &= \frac{1}{2}(x_{10} + 3x_{20})e^{-at} - \frac{1}{2}(x_{10} + x_{20})e^{-3at}\end{align*}\]The presence of terms like \(e^{5at}\) and \(e^{7at}\) suggests potential growth (instability) when \(a > 0\) due to positive exponents. Note here how both positive and negative exponential terms coexist, showing complexity in behavior for various \(a\).

Understand the Instability

For any nonzero \(a\), examining the coefficients in the solution, at least one term in each \(x_1(t)\) and \(x_2(t)\) contains an exponential growth term. Specifically, if \(a < 0\), terms like \(e^{5at}\) and \(e^{-at}\) with positive coefficients will dominate, leading to solutions that grow without bound, thus indicating instability.

Conclusion

Even though eigenvalues would traditionally suggest stability (for \(a < 0\)), the presence of exponentially growing terms in the exact solution shows that the system actually behaves unstably for any non-zero value of \(a\). The time-dependence of the matrix elements complicates the stability analysis, going against simple eigenvalue predictions.

Key concepts

Eigenvalues

In linear algebra, eigenvalues provide insight into a system's behavior. For a system described by a matrix, eigenvalues represent the rates of expansion or contraction. When dealing with time-varying systems, eigenvalues can initially mislead us about stability since they don't account for matrix variation over time.

- **Definition**: Eigenvalues are solutions to the characteristic equation, allowing us to understand directional behaviors of dynamical systems.
- **Role in Stability**: Typically, for constant matrices, negative real parts of eigenvalues suggest stability. However, in time-varying systems, the fixed eigenvalues might not reflect true system behavior.

Even though in this exercise, the eigenvalues (\(\lambda_1 = a\) and \(\lambda_2 = 3a\)) appear constant and negative for \(a < 0\), suggesting stability, they don't capture the complexity added by time-dependent matrix variations.

Stability Analysis

Stability analysis is the process of determining whether a system will converge to an equilibrium or diverge over time. For systems with time-invariant matrices, eigenvalues provide a clear picture.

In this exercise, despite negative eigenvalues indicating potential stability, the presence of terms like \(e^{5at}\) in the solution shows instability for any non-zero \(a\). This is because:

• Stability traditionally considers constant matrices.
• Time-varying systems result in coefficients affecting the system differently over time.
• Exponential terms with positive growth indicate instability, regardless of eigenvalue analysis.

Hence, for time-varying matrices, the eigenvalue method's reliability decreases, requiring more complex, often numerical methods, to predict stability correctly.

Differential Equations

Differential equations describe the relationship between a function and its derivatives, indicating how a system evolves over time. In this context, they serve as the foundation explaining the dynamics of the system.

- **Time-Varying Differential Equations**: Unlike time-invariant systems, variables or parameters change over time, complicating solutions. This exercise demonstrates a system where the coefficients of the differential equation matrix explicitly involve time terms \(e^{8at}\) and \(e^{-8at}\).
- **Solution Dynamics**: Solutions to these equations involve complex expressions, evidenced by calculating \(x_1(t)\) and \(x_2(t)\) through exponential dynamics, indicating potentially different behavior over different time intervals.

The art of solving these equations involves understanding how parameters interact over time, often requiring numerical solutions or advanced analytical techniques to uncover system behavior and ensure accurate predictions.