Problem 43 - Calculus for Biology and Medicine

Question

In Problems 43-56, 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. \(A=\left[\begin{array}{rr}2 & -1 \\ 3 & 0\end{array}\right]\)

Step-by-Step Solution

Answer

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

Step 1: Calculate the characteristic polynomial of A

To find the eigenvalues, we need the characteristic polynomial of matrix \(A\). The characteristic polynomial is given by \( \text{det}( \lambda I - A) \) where \(I\) is the identity matrix and \(\lambda\) is a scalar. Substituting \(A\) into the formula, we have: \[\text{det}\begin{pmatrix}\lambda - 2 & 1 \-3 & \lambda \\end{pmatrix} = (\lambda-2)(\lambda) - (1)(-3)\]This simplifies to: \[ \lambda^2 - 2\lambda + 3 \]

Step 2: Find the eigenvalues

Using the characteristic polynomial \(\lambda^2 - 2\lambda + 3\), solve for \(\lambda\) by finding the roots of the equation:\[\lambda^2 - 2\lambda + 3 = 0 \]Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = -2\), and \(c = 3\), we have:\[\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4\cdot1\cdot3}}{2\cdot1} = \frac{2 \pm \sqrt{4 - 12}}{2}= \frac{2 \pm \sqrt{-8}}{2}\]This results in complex conjugate eigenvalues:\[ \lambda = 1 \pm i\sqrt{2} \]

Step 3: Determine the nature of stability

For complex eigenvalues \( \lambda = a \pm bi \), the stability of the equilibrium point \((0,0)\) depends on the real part \(a\). - If \(a < 0\), the equilibrium is a stable spiral.- If \(a > 0\), the equilibrium is an unstable spiral.- If \(a = 0\), the equilibrium is a center.In our case, the real part is \(a = 1\). Because \(a > 0\), the equilibrium \((0,0)\) is an unstable spiral.

Key Concepts

Eigenvalues

Eigenvalues are crucial for understanding the behavior of differential equations. They are the special numbers associated with a square matrix like matrix \(A\). Formally, eigenvalues are found by solving the characteristic equation obtained from \(\text{det}(\lambda I - A)\), where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues. In this exercise, the matrix \(A\) yielded the characteristic polynomial \( \lambda^2 - 2\lambda + 3 \). Solving this polynomial equation using the quadratic formula provides us with the eigenvalues of the matrix.

• These eigenvalues determine how solutions to the differential equation behave over time.
• For our specific case, the eigenvalues were complex: \( \lambda = 1 \pm i\sqrt{2} \). Understanding the nature of these values is key to analyzing stability.

Complex eigenvalues indicate oscillatory behavior in the system, leading us to consider their real and imaginary parts during stability analysis.

Stability Analysis

Stability analysis helps us understand the long-term behavior of a system described by a differential equation. Specifically, we focus on whether solutions remain close to an equilibrium point, such as \((0,0)\), over time. The eigenvalues of the matrix \(A\) play a pivotal role here.When eigenvalues are complex, we consider their real part to determine stability:

• If the real part \(a < 0\), trajectories spiral towards the equilibrium, indicating stability.
• If \(a > 0\), trajectories spiral away from the equilibrium, showing instability.
• If \(a = 0\), the trajectories neither consistently spiral in nor out, often indicating a center.

In our example, the real part of the eigenvalues \( \lambda = 1 \pm i\sqrt{2} \) is \(a = 1\), which means the system is unstable as trajectories spiral away from equilibrium.

Complex Conjugates

Complex conjugate eigenvalues often appear in two-dimensional systems, indicating a type of rotational behavior, usually spiraling due to their oscillatory nature. These eigenvalues have the form \( a \pm bi \), where each corresponds to a rotation and scaling in the plane.

• Complex conjugates will always come in pairs due to the nature of polynomials having real coefficients.
• These pairs produce oscillations that can be stable, unstable, or neutral, depending on the real part.

Our exercise provided the complex conjugate eigenvalues \( \lambda = 1 \pm i\sqrt{2} \), supporting the fact that any rotations induced in the system will compound over time, specifically due to the positive real component.

Equilibrium Classification

Classifying the equilibrium at \((0,0)\) involves understanding how the system behaves around this point. The classification relies heavily on the nature of the eigenvalues:

• A **stable spiral** occurs when eigenvalues have a negative real part, causing trajectories to spiral towards the equilibrium.
• An **unstable spiral** occurs with positive real parts, leading trajectories away from equilibrium.
• A **center** has purely imaginary eigenvalues where oscillations are neither dampened nor amplified, and trajectories tend to encircle the equilibrium indefinitely.

In our scenario, the presence of complex eigenvalues with positive real parts \( (1 \pm i\sqrt{2}) \) classifies the equilibrium as an unstable spiral. It is crucial because it predicts behavior—knowing such classifications allows us to understand and predict the dynamics around equilibrium points better.