Problem 53 - Calculus for Biology and Medicine

Question

We consider differential equations of the form $$ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) $$ where $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ The eigenvalues of \(A\) will be complex conjugates. Analyze the stability of the equilibrium \((0,0)\), and classify the equilibrium according to whether it is a stable spiral, an unstable spiral, or a center. $$ A=\left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is a center, indicating neutral stability with closed trajectories.

Step 1: Identify the Matrix A

Given the matrix \( A \) as \( \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \), we will analyze its stability.

Step 2: Find the Eigenvalues

The eigenvalues \( \lambda \) of matrix \( A \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). Set up the determinant: \[ \det \begin{bmatrix} -\lambda & -1 \ 1 & -\lambda \end{bmatrix} = (-\lambda)(-\lambda) - (-1)(1) = \lambda^2 + 1. \] So the characteristic equation is \( \lambda^2 + 1 = 0 \).

Step 3: Solve for the Eigenvalues

Solve \( \lambda^2 + 1 = 0 \), which gives \( \lambda = \pm i \).The eigenvalues \( i \) and \( -i \) are complex conjugates with zero real parts.

Step 4: Analyze the Stability

The eigenvalues \( \lambda = \pm i \) imply a center, since they are purely imaginary with no real part.A center describes a system that is neutrally stable, meaning it neither spirals toward nor away from equilibrium, but rather exhibits closed circular or elliptical trajectories.

Step 5: Classify the Equilibrium

Based on the fact that the eigenvalues are purely imaginary, the equilibrium point at \((0,0)\) in this system is classified as a center. This means that trajectories will form closed orbits around the equilibrium point.

Key Concepts

Eigenvalues

When dealing with a system of differential equations, eigenvalues play a crucial role in understanding the system's behavior. Simply put, eigenvalues are special numbers associated with a matrix, derived from the characteristic equation. In our problem, we have a matrix \[ A=\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} \] To find its eigenvalues, we compute the determinant of \( (A - \lambda I) \) and set it to zero: \[ \det \begin{bmatrix} -\lambda & -1 \ 1 & -\lambda \end{bmatrix} = \lambda^2 + 1 = 0 \] Solving this, we obtain the eigenvalues \( \lambda = \pm i \), which are complex numbers.

• Characteristic Equation: Derives from substituting \( A \) and \( \lambda I \) into the determinant.
• Calculation: Solving the characteristic equation gives us real or complex solutions.

Here, the imaginary unit \( i \) influences how the system behaves dynamically. In essence, these values indicate the nature of the solutions to the differential equation. This is a critical first step to predict stability.

Stability Analysis

Stability analysis helps us determine how the solutions to the differential equations behave over time, particularly near equilibrium points. For our matrix, we have the eigenvalues \( \lambda = \pm i \), which are purely imaginary. This is significant because it implies certain stability characteristics.

• If eigenvalues have positive real parts, solutions grow exponentially, indicating instability.
• If they have negative real parts, solutions decay, indicating stability.
• Purely imaginary eigenvalues, like ours, imply a neutrally stable system.

Neutrally stable systems with purely imaginary eigenvalues suggest closed orbits or cycles without growing or decaying. In practical terms, such systems don't move away from equilibrium but rotate around it. Comprehending this analysis is essential for predicting long-term behavior of dynamic systems. Stability tells us whether solutions will persistently behave in a predictable manner or diverge uncontrollably. This knowledge is indispensable for many engineering and scientific applications.

Equilibrium Classification

Equilibrium classification helps us categorize the behavior of the system at equilibrium points based on the eigenvalues derived from our differential equation matrix. After calculating the eigenvalues \( \lambda = \pm i \), we classify the equilibrium point as a center. This classification depends heavily on the nature of the eigenvalues:

• Stable Node/Sink: Real, negative eigenvalues.
• Unstable Node/Source: Real, positive eigenvalues.
• Saddle Point: Mixed real eigenvalues, some positive, some negative.
• Center: Purely imaginary eigenvalues with zero real parts.

For our system, the classification as a center means the system will exhibit a perpetually cyclical motion around the equilibrium point \((0,0)\), forming closed curves. Center classification is unique because it does not imply stable convergence to an equilibrium or divergence away from it, but rather a steady, repeating pattern. Understanding this classification aids in visualizing the trajectory behavior of the solutions over time, critical for effectively modeling and simulating real-world phenomena.