Problem 19 - Calculus for Biology and Medicine

Question

Show that \(\left.\right|_{0} ^{0} \mid\) is an equilibrium of $$ \left[\begin{array}{l} x_{1}(t+1) \\ x_{2}(t+1) \end{array}\right]=\left[\begin{array}{ll} -1.6 & 0 \\ -0.5 & 0.1 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ and determine its stability.

Step-by-Step Solution

Answer

The equilibrium point (0,0) is unstable.

Step 1: Define Equilibrium

An equilibrium point occurs when the system does not change over time. Mathematically, this means that the state vector at the next time step is equal to the current state vector, so \( \begin{bmatrix} x_1(t+1) \ x_2(t+1) \end{bmatrix} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \). This implies that\[\begin{bmatrix} -1.6 & 0 \ -0.5 & 0.1 \end{bmatrix} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix}.\]

Step 2: Analyze the Matrix Equation

Substitute \( x_1(t) = 0 \) and \( x_2(t) = 0 \) into the matrix equation.\[\begin{bmatrix} -1.6 & 0 \ -0.5 & 0.1 \end{bmatrix} \begin{bmatrix} 0 \ 0 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]Both sides of the equation are zero, confirming that \( (0,0) \) is indeed an equilibrium point.

Step 3: Determine Stability

To determine stability, calculate the eigenvalues of the system matrix\[A = \begin{bmatrix} -1.6 & 0 \ -0.5 & 0.1 \end{bmatrix}.\]The characteristic equation is \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. This gives\[\det\begin{bmatrix} -1.6 - \lambda & 0 \ -0.5 & 0.1 - \lambda \end{bmatrix} = (-1.6 - \lambda)(0.1 - \lambda) = 0.\]Solving for \( \lambda \), we find eigenvalues \( \lambda_1 = -1.6 \) and \( \lambda_2 = 0.1 \).

Step 4: Conclude Stability

An equilibrium is stable if all eigenvalues have magnitudes less than 1. Here, \(|λ_1| = 1.6| > 1|\) and \(|λ_2| = 0.1| < 1|\). Because one eigenvalue has magnitude greater than 1, the equilibrium point \((0,0)\) is unstable.

Key Concepts

Stability Analysis

Stability analysis in dynamical systems focuses on the behavior of equilibrium points over time. An equilibrium point is said to be stable if the system returns to this point after a small disturbance. For a linear system, stability can be determined by analyzing the eigenvalues of the system's matrix.

• If all eigenvalues have a magnitude less than 1, the equilibrium point is stable.
• If any eigenvalue has a magnitude greater than or equal to 1, the equilibrium is unstable.

In our exercise, the equilibrium point \( (0,0) \) had differing eigenvalues; one with magnitude greater than 1 and the other less than 1. Therefore, \( (0,0) \) is considered unstable. This simple rule provides a clear way to evaluate stability, making stability analysis a crucial tool in understanding dynamical systems.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors have special significance in the analysis of linear transformations. For a matrix system, these components reveal intrinsic properties of the system's behavior. An eigenvalue tells us if and how an input vector (or state) will be scaled as it is transformed by the matrix.

• An eigenvalue greater than 1 indicates amplification.
• An eigenvalue less than 1 signifies reduction.
• A negative eigenvalue can indicate an oscillation or change in the direction of the system.

In our problem, the eigenvalues \( -1.6 \) and \( 0.1 \) were derived from the characteristic equation. This equation is a determinant calculation set to zero. These specific eigenvalues suggest that one component of the system's state will decrease, while the other has a potential to oscillate or grow depending on the sign. This understanding is vital for predicting how the system evolves from different initial conditions.

Matrix Equations

Matrix equations are central in expressing linear relationships in dynamical systems. They describe how each component of the system influences others over time. The matrix in our problem acts as the system's description, encapsulating dynamics compactly.

• Each column of the matrix affects a particular coordinate, illustrating how changes in one variable affect others.
• The entries in the rows show how different states contribute to state changes over time.

For the given matrix: \[A = \begin{bmatrix} -1.6 & 0 \ -0.5 & 0.1 \end{bmatrix},\]\ the diagonal elements indicate rates of scaling for each variable independently, while the off-diagonal elements indicate coupling effects across variables.
The equilibrium point is shown by setting the system equal to this matrix equation, which results in zero change, confirming when the system remains unaltered. Understanding matrix equations enables us to forecast potential behaviors and interactions within multi-dimensional spaces.