Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Determine whether the following systems are stable. a. \(\quad \begin{aligned} x_{t+1} &=0.48 x_{t}+0.2 y_{t} \\ y_{t+1} &=0.128 x_{t}+0.72 y_{t} \end{aligned}\) b. \(\quad \begin{aligned} x_{t+1} &=0.8 x_{t}-0.5 y_{t} \\ y_{t+1} &=0.2 x_{t}+0.9 y_{t} \end{aligned}\) c. \(\quad \begin{aligned} x_{t+1} &=1.1 x_{t}-0.5 y_{t} \\ y_{t+1} &=0.2 x_{t}+0.9 y_{t} \end{aligned}\) d. \(\quad \begin{aligned} x_{t+1} &=0.8 x_{t}-0.1 y_{t} \\ y_{t+1} &=0.1 x_{t}+0.6 y_{t} \end{aligned}\) e. \(\quad \begin{aligned} x_{t+1} &=1.2 x_{t}-0.1 y_{t} \\ y_{t+1} &=0.9 x_{t}+0.6 y_{t} \end{aligned}\) f. \(\quad \begin{aligned} x_{t+1} &=0.56 x_{t}+0.4 y_{t} \\ y_{t+1} &=0.256 x_{t}+1.04 y_{t} \end{aligned}\)

Step-by-Step Solution

Answer

a and d are stable; b and f are borderline stable; c and e are unstable.

Step 1: Write the system matrix

For each system, write the mathematical system in matrix form \[ \begin{bmatrix} x_{t+1} \ y_{t+1} \end{bmatrix} = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} x_{t} \ y_{t} \end{bmatrix} \].

Step 2: Determine the eigenvalues of the matrix

To determine the stability of each system, calculate the eigenvalues of the matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) by solving the characteristic equation \( \det(\lambda I - A) = 0 \). The characteristic polynomial is \( \lambda^2 - (a+d)\lambda + (ad-bc) = 0 \).

Step 3: Analyze the eigenvalues

Calculate the solutions to the characteristic equation. If the modulus of all eigenvalues is less than one, the system is stable. If any eigenvalue has a modulus that is equal to or greater than one, the system is unstable.

Step 4: Apply the procedure to part a

Matrix: \( A = \begin{bmatrix} 0.48 & 0.2 \ 0.128 & 0.72 \end{bmatrix} \). Calculate its eigenvalues: \( \lambda^2 - 1.2\lambda + 0.3264 = 0 \). Eigenvalues \( \lambda_1 = 0.6 \, \lambda_2 = 0.54 \). Both are less than 1: Stable.

Step 5: Apply the procedure to part b

Matrix: \( A = \begin{bmatrix} 0.8 & -0.5 \ 0.2 & 0.9 \end{bmatrix} \). Eigenvalues from \( \lambda^2 - 1.7\lambda + 0.82 = 0 \). Solving gives \( \lambda_1 = 1 \, \lambda_2 = 0.7 \). Since one eigenvalue is 1: Borderline stable.

Step 6: Apply the procedure to part c

Matrix: \( A = \begin{bmatrix} 1.1 & -0.5 \ 0.2 & 0.9 \end{bmatrix} \). Eigenvalues from \( \lambda^2 - 2\lambda + 1.28 = 0 \). Roots: \( \lambda_1 = 1.3 \, \lambda_2 = 0.7 \). With \( \lambda_1 > 1 \): Unstable.

Step 7: Apply the procedure to part d

Matrix: \( A = \begin{bmatrix} 0.8 & -0.1 \ 0.1 & 0.6 \end{bmatrix} \). Eigenvalues are solutions of \( \lambda^2 - 1.4\lambda + 0.47 = 0 \). Roots: \( \lambda_1, \lambda_2 \approx 0.85, 0.55 \). Both less than 1: Stable.

Step 8: Apply the procedure to part e

Matrix: \( A = \begin{bmatrix} 1.2 & -0.1 \ 0.9 & 0.6 \end{bmatrix} \). Eigenvalues from \( \lambda^2 - 1.8\lambda + 0.78 = 0 \). Solutions: \( \lambda_1 = 1.2 \, \lambda_2 = 0.6 \). With \( \lambda_1 > 1 \): Unstable.

Step 9: Apply the procedure to part f

Matrix: \( A = \begin{bmatrix} 0.56 & 0.4 \ 0.256 & 1.04 \end{bmatrix} \). Calculate its eigenvalues using \( \lambda^2 - 1.6\lambda + 0.8064 = 0 \). Eigenvalues: \( \lambda_1, \lambda_2 \approx 1, 0.6 \). One eigenvalue is 1: Borderline stable.

Key Concepts

Eigenvalues

Eigenvalues are crucial in determining the stability of a system modeled by linear equations. An eigenvalue is a scalar that helps indicate how a matrix transforms a vector. When you multiply a matrix by its eigenvector, the result is the same eigenvector scaled by the eigenvalue. This gives us valuable insight into how systems evolve over time. To find the eigenvalues of a given matrix, you need to solve its characteristic equation. It's essential to check the modulus (absolute value) of every eigenvalue to determine system stability:

• All eigenvalues with modulus less than 1 imply the system is stable.
• At least one eigenvalue with modulus greater than 1 implies instability.
• If any eigenvalue’s modulus is exactly 1, the system is considered borderline stable.

Characteristic Equation

The characteristic equation is a polynomial generated from a matrix, and it plays a pivotal role in finding eigenvalues. By setting up the equation \( \det(\lambda I - A) = 0 \) for a square matrix \( A \), we arrive at the characteristic polynomial. In the context of 2x2 matrices, this equation is typically \( \lambda^2 - (a+d)\lambda + (ad-bc) = 0 \). Solving this polynomial provides us with the eigenvalues. It's important to remember that the roots of the characteristic equation, which are the solutions \( \lambda \), determine the nature of the system's stability. Solving it can require techniques like factoring or using the quadratic formula, particularly when the equation is quadratic as is common with 2x2 matrices.
Understanding the characteristic equation is crucial because it directly impacts our ability to evaluate the dynamics of the system represented by the matrix.

Matrix Analysis

Matrix analysis involves examining the properties and behavior of matrices, often in the context of system stability. Understanding matrix operations can help us predict how systems of linear equations will behave. In system stability analysis, we specifically look for how matrices transform vectors over iterations:

• Writing the system in matrix form allows for efficient computations and theoretical analysis.
• Finding eigenvalues through the characteristic equation reveals information about matrix stability.

This analysis can be applied to systems of linear equations found in dynamic systems models, whether it's economic, physical, or biological systems. Analyzing matrices with an emphasis on eigenvectors and eigenvalues enables us to derive critical insights into how systems might evolve, stabilize, or become unstable over time.