Problem 56 - 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. \(A=\left[\begin{array}{ll}2 & -3 \\ 3 & -2\end{array}\right]\)

Step-by-Step Solution

Answer

The equilibrium \((0,0)\) is an unstable spiral because the eigenvalues have positive real parts.

Step 1: Determine Characteristic Equation

The characteristic equation of matrix \(A\) is given by \( \det(A - \lambda I) = 0 \), where \(I\) is the identity matrix and \(\lambda\) is an eigenvalue. For matrix \(A\), this becomes \[ \det \begin{pmatrix} 2 - \lambda & -3 \ 3 & -2 - \lambda \end{pmatrix} = 0 \].

Step 2: Calculate Determinant

Find the determinant of the matrix \(\begin{pmatrix} 2 - \lambda & -3 \ 3 & -2 - \lambda \end{pmatrix}\). This is computed as \((2-\lambda)(-2-\lambda) - (-3)(3)\). Simplifying, \(\lambda^2 - 4\lambda + 13 = 0\).

Step 3: Solve for Eigenvalues

Using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = 13\), we calculate the eigenvalues: \(\lambda = \frac{4 \pm i\sqrt{36}}{2}\). So, \(\lambda = 2 \pm 3i\).

Step 4: Analyze Stability

For complex eigenvalues \(\lambda = 2 \pm 3i\), the real part is \(2\), which is positive. For stability in such cases, the real part of eigenvalues determines stability. A positive real part indicates that the equilibrium is unstable.

Step 5: Classify the Equilibrium

Since the eigenvalues have positive real parts and complex conjugate imaginary parts, the equilibrium point \((0,0)\) is classified as an unstable spiral.

Key Concepts

Eigenvalues: Understanding Their Importance

Eigenvalues are a fundamental concept when studying systems of differential equations, especially linear systems. They are numbers associated with a matrix, in this case, matrix \( A \) of the system \( \frac{d \mathbf{x}}{d t}=A \mathbf{x}(t) \), that provide essential information about the system's behavior. Calculating eigenvalues involves finding the roots of the characteristic equation formed by \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix, and \( \lambda \) represents the eigenvalues.
In the given problem, after computations, we found the eigenvalues to be \( 2 \pm 3i \). These are complex numbers, indicating that the system involves oscillatory motion (as implied by the imaginary component), while the real component impacts the dynamic stability of the equilibrium point. Understanding the nature and value of these eigenvalues enables us to conduct further analyses, such as stability analysis and equilibrium classification.

Stability Analysis: Determining System Behavior

Stability in differential equations is a key property that highlights whether solutions to equations tend towards or away from equilibrium points as time progresses. For systems with complex eigenvalues, like \( 2 \pm 3i \), the stability is primarily dictated by the real part of these eigenvalues.

• If the real part is positive, solutions will diverge from the equilibrium, indicating an unstable system.
• If the real part is negative, solutions will move towards the equilibrium, suggesting a stable system.
• If the real part is zero, it's more nuanced and requires further investigation, potentially indicating "centers."

In this exercise, since the real part of the eigenvalue is \( 2 \), which is positive, the system is unstable. This means that any small perturbation or deviation from the equilibrium point \((0,0)\) will grow over time, leading the system away from this point.

Equilibrium Classification: Recognizing System Patterns

Classifying equilibrium points is crucial as it allows us to understand the qualitative behavior of a system near those points. With complex eigenvalues \( 2 \pm 3i \), the equilibrium points can often be classified into different types based on stability and oscillatory behavior.

• If the real part is positive, as in this problem, the equilibrium point \((0,0)\) is classified as an "unstable spiral." This indicates that trajectories spiral outwards, away from the equilibrium, reflecting both divergence and rotational movement.
• If the real part had been negative, you would classify it as a "stable spiral."
• Additionally, if the real part had been zero, with non-zero imaginary parts, it could be classified as a "center," implying no net growth or decay, just oscillation around the equilibrium.

Understanding these classifications helps predict and visualize how a system might evolve over time, which is vital in fields as diverse as physics, engineering, and even economics.