Problem 54 - 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. $$ A=\left[\begin{array}{rr} 0 & -3 \\ 2 & 2 \end{array}\right] $$

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is an unstable spiral.

Step 1: Identify the Eigenvalues

For a matrix \( A \), eigenvalues are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \). For the matrix given: \[ A = \begin{bmatrix} 0 & -3 \ 2 & 2 \end{bmatrix} \] The characteristic equation becomes: \[ \det\begin{bmatrix} -\lambda & -3 \ 2 & 2 - \lambda \end{bmatrix} = \lambda^2 - 2\lambda + 6 = 0 \] Solve for \( \lambda \).

Step 2: Find the Complex Eigenvalues

Solve the quadratic equation \( \lambda^2 - 2\lambda + 6 = 0 \) using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -2, c = 6 \). This gives: \[ \lambda = \frac{2 \pm \sqrt{4 - 24}}{2} = \frac{2 \pm \sqrt{-20}}{2} \] \[ \lambda = 1 \pm i\sqrt{5} \] Thus, the eigenvalues are complex conjugates: \( 1 + i\sqrt{5} \) and \( 1 - i\sqrt{5} \).

Step 3: Analyze Stability

The eigenvalues \( 1 + i\sqrt{5} \) and \( 1 - i\sqrt{5} \) indicate that the real part is \( 1 \). Since the real part is positive, the equilibrium point \((0,0)\) is unstable. The presence of imaginary parts indicates a spiral.

Step 4: Classify the Equilibrium

With a positive real part and non-zero imaginary parts, the system at equilibrium \((0,0)\) is classified as an unstable spiral. This is because any perturbation will lead the system to spiral away from the origin.

Key Concepts

Eigenvalues

Eigenvalues are fundamental in understanding the properties of differential equations systems. In the context of matrix equations, they can influence how solutions to the system behave over time. To find these values, we utilize the characteristic equation, which is derived from the matrix: - Consider the matrix: \[ A = \begin{bmatrix} 0 & -3 \ 2 & 2 \end{bmatrix} \] - The characteristic equation is formed as follows: \[ \det(A - \lambda I) = 0 \]- Where \( I \) is the identity matrix, and \( \lambda \) represents the eigenvalues. In this exercise, by using the matrix values, we arrive at a characteristic equation of \[ \lambda^2 - 2\lambda + 6 = 0 \] Solving this equation using the quadratic formula, we find the eigenvalues of the matrix to be complex conjugates: \[ \lambda = 1 \pm i\sqrt{5} \] These complex eigenvalues imply certain behaviors in the system, particularly relating to oscillations around an equilibrium point, which we will explore further.

Stability Analysis

Stability analysis for a system of differential equations involves examining the eigenvalues’ real parts. This helps us determine whether an equilibrium is stable or unstable. The key is understanding how the real part of eigenvalues dictates system behavior: - If the real part is negative: The system is stable, moving towards equilibrium.- If the real part is zero: The system might be stable, as it can sustain a constant state.- If the real part is positive: The system is unstable and will diverge from equilibrium.For the differential equation \[ \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \] with the matrix \[ A = \begin{bmatrix} 0 & -3 \ 2 & 2 \end{bmatrix} \] The complex eigenvalues \( 1 + i\sqrt{5} \) and \( 1 - i\sqrt{5} \) indicate an unstable system because the real part is \( 1 \), which is positive. This analysis shows that any small disturbance or variation from the equilibrium will cause the system to spiral away from its stable state, indicating instability.

Equilibrium Classification

To classify an equilibrium, especially with non-real eigenvalues, we look at both the real and imaginary components:- The real part of the eigenvalues dictates stability. - The imaginary part indicates oscillatory behavior, causing a spiral motion. Given the eigenvalues\[ 1 + i\sqrt{5} \] and \[ 1 - i\sqrt{5} \] since the real part is positive, the equilibrium at the origin \((0, 0)\) is classified as an unstable spiral.This means any small movement away from the equilibrium will lead to the system spiraling outward over time. Thus, the equilibrium is neither a stable center nor a stable spiral, but rather an unstable spiral, suggesting a dynamic and expanding trajectory away from the equilibrium point with oscillations due to the imaginary component.