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 at \((0,0)\) is a stable spiral.

Step 1: Calculate the Characteristic Equation

The first step is to find the characteristic equation of matrix \( A \). The characteristic equation is given by \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. For the given matrix \( A \), this becomes: \[ \begin{vmatrix} 2 - \lambda & -3 \ 3 & -2 - \lambda \end{vmatrix} = 0 \]Calculate the determinant: \[ (2 - \lambda)(-2 - \lambda) - (-3)(3) = \lambda^2 + 2\lambda + 4 + 9 = \lambda^2 + 2\lambda + 13 = 0 \]

Step 2: Solve for Eigenvalues

In this step, we solve the quadratic characteristic equation \( \lambda^2 + 2\lambda + 13 = 0 \) using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 1, b = 2, c = 13 \) into the formula: \[ \lambda = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 13}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 52}}{2} = \frac{-2 \pm \sqrt{-48}}{2} \]This simplifies to: \[ \lambda = -1 \pm i \sqrt{12} = -1 \pm 2i\sqrt{3} \]

Step 3: Determine the Nature of the Eigenvalues

The eigenvalues are \( \lambda = -1 \pm 2i\sqrt{3} \) which are complex with both real and imaginary components. The real part of the eigenvalues is \( -1 \).

Step 4: Classify the Equilibrium

Since the real part of the complex eigenvalues is negative (\( -1 \)), the equilibrium point at \( (0,0) \) is a **Stable Spiral**. A stable spiral describes an equilibrium where trajectories spiral into the equilibrium point as time progresses.

Key Concepts

Differential Equations

Differential equations form the core of many mathematical models that describe how variables change with respect to one another over time. They are equations that involve an unknown function and its derivatives. In our exercise, the differential equation is presented in matrix form:\[\frac{d \mathbf{x}}{d t}=A \mathbf{x}(t)\]where \( A \) is a matrix and \( \mathbf{x}(t) \) is the state vector that changes over time.
This form is specifically applicable to systems of linear differential equations and helps us understand dynamic systems. In essence, the matrix \( A \) determines how the system evolves. By examining the properties of \( A \)—in particular, its eigenvalues—we can predict the behavior of the system.
The eigenvalues provide insights into whether the state approaches stability, diverges, or oscillates over time. They are critical in understanding the behavior near equilibrium points. Thus, analyzing differential equations through the lens of linear algebra allows us to quantify and visualize the dynamic behavior of complex systems.

Stability Analysis

Stability analysis is fundamental when evaluating the behavior of systems conceptualized by differential equations. The primary goal is to determine if solutions converge to an equilibrium over time. In our matrix \( A \), the eigenvalues play a pivotal role.

• If the real parts of all eigenvalues are negative, the equilibrium is considered stable, meaning perturbations diminish, leading the system back to equilibrium.
• If at least one eigenvalue has a positive real part, the system is unstable, causing perturbations to grow over time, diverging from the equilibrium.

For the matrix given in the problem:\[A=\begin{bmatrix}2 & -3 \3 & -2\end{bmatrix}\]we found the eigenvalues \( \lambda = -1 \pm 2i\sqrt{3} \). With both eigenvalues containing a negative real part, \( -1 \), the equilibrium (0,0) is deemed stable. This stability indicates that the system will naturally return to equilibrium after small disturbances occur.
Ultimately, the stability analysis provides a tool to assess how resilient a system is to change, relying heavily on the characteristics of eigenvalues within the differential equation's matrix representation.

Equilibrium Classification

Equilibrium classification helps in understanding the qualitative nature of solution trajectories near an equilibrium point. Different types of equilibria describe how solutions behave over time.
For systems with eigenvalues:

• Real and negative eigenvalues indicate a node, leading to solutions that slowly converge to equilibrium without oscillations.
• Real and positive eigenvalues suggest an unstable node, with solutions diverging away from equilibrium.
• Complex eigenvalues with negative real parts describe a stable spiral, meaning solutions spiral into equilibrium.
• Complex eigenvalues with positive real parts lead to an unstable spiral, where solutions spiral out and away.

In our exercise, the equilibrium is classified as a stable spiral since the complex eigenvalues \(-1 \pm 2i\sqrt{3}\) have a negative real part. This classification implies that any displaced state will spiral back toward the point \((0,0)\).
Such classification not only gives a clear visual interpretation of the equilibrium's nature but also highlights the predictability of the system's future states.