Problem 14 - Calculus for Biology and Medicine

Question

Assume that \(a>0 .\) Find all point equilibria of $$ \begin{array}{l} \frac{d x_{1}}{d t}=1-a x_{1} x_{2} \\ \frac{d x_{2}}{d t}=a x_{1} x_{2}-x_{2} \end{array} $$ and characterize their stability.

Step-by-Step Solution

Answer

The system has one equilibrium point at \( \left( \frac{1}{a}, 1 \right) \), which is a stable spiral.

Step 1: Understand Equilibrium Points

An equilibrium point of a system of differential equations occurs where the time derivatives are zero. For our system, this means setting \( \frac{dx_1}{dt} = 0 \) and \( \frac{dx_2}{dt} = 0 \). This leads to the equations: \( 1 - ax_1x_2 = 0 \) and \( ax_1x_2 - x_2 = 0 \).

Step 2: Solve the First Equation

The equation \( 1 - ax_1x_2 = 0 \) simplifies to \( ax_1x_2 = 1 \), which suggests that \( x_1x_2 = \frac{1}{a} \).

Step 3: Solve the Second Equation

The equation \( ax_1x_2 - x_2 = 0 \) can be factored as \( x_2(ax_1 - 1) = 0 \). This gives two possibilities: \( x_2 = 0 \) or \( ax_1 = 1 \).

Step 4: Find the Equilibrium Points

From Step 3, consider \( x_2 = 0 \), which makes the first equation \( 1 = 0 \), leading to no solution for this case. \(ax_1 = 1 \) leads to \( x_1 = \frac{1}{a} \). Substitute \( x_1 = \frac{1}{a} \) in the equation from Step 2: \( \frac{1}{a} x_2 = \frac{1}{a} \), implying \( x_2 = 1 \). Thus, the only equilibrium point is \( (x_1, x_2) = \left( \frac{1}{a}, 1 \right) \).

Step 5: Analyze Stability

To analyze stability, compute the Jacobian matrix of the system at the equilibrium point \( \left( \frac{1}{a}, 1 \right) \). The Jacobian \( J \) is calculated from partial derivatives: \ \[ J = \begin{pmatrix} \frac{\partial}{\partial x_1}(1 - ax_1x_2) & \frac{\partial}{\partial x_2}(1 - ax_1x_2) \ \frac{\partial}{\partial x_1}(ax_1x_2 - x_2) & \frac{\partial}{\partial x_2}(ax_1x_2 - x_2) \end{pmatrix} \] \ Evaluating these, \ \[ J = \begin{pmatrix} -ax_2 & -ax_1 \ ax_2 & ax_1 - 1 \end{pmatrix} \] \ at \( \left( \frac{1}{a}, 1 \right) \), we get \ \[ J = \begin{pmatrix} -a & -\frac{1}{a} \ a & 0 \end{pmatrix} \] \ Calculate the eigenvalues from \( \det(J - \lambda I) = 0 \): \ \[ 0 = \begin{vmatrix} -a - \lambda & -\frac{1}{a} \ a & -\lambda \end{vmatrix} \] \ This simplifies to \( \lambda^2 + a\lambda + 1 = 0 \). The discriminant \( a^2 - 4 \) is negative for all \( a > 0 \), indicating complex eigenvalues with a positive real part. Therefore, the equilibrium is a stable spiral.

Key Concepts

Stability Analysis

In a dynamics context, stability analysis helps us understand whether a system's equilibrium will maintain its state when subjected to small disturbances. It tells whether small deviations from the equilibrium grow or diminish over time.

For instance, if an equilibrium is stable, a minor disturbance will cause the variables to return to their original state. On the contrary, if it is unstable, even the smallest disturbance will lead the system away from equilibrium.

The method frequently employed involves examining eigenvalues of the Jacobian matrix, derived from the system's differential equations. Here's what can happen when analyzing an equilibrium point:

• Stable node: All eigenvalues have negative real parts. Perturbations diminish over time, returning to equilibrium.
• Unstable node: Any eigenvalue has a positive real part. Perturbations grow over time, moving away from equilibrium.
• Saddle point: Some eigenvalues have negative real parts, while others have positive real parts. Perturbations can either grow or diminish.
• Spiral or circle: Complex eigenvalues with negative real parts result in a spiral into equilibrium, indicating stability; complex eigenvalues with positive real parts mean a spiral out and instability.

Systems of Differential Equations

Systems of differential equations are mathematical models that describe how variables evolve over time. These systems consist of multiple interrelated functions, each depending on independent variables like time. In our case, we have two variables, denoted as \(x_1\) and \(x_2\), showing how they change over time.

Generally, a system of differential equations given in the form:

• \( \frac{d x_1}{d t} \) describes how \(x_1\) changes concerning time.
• \( \frac{d x_2}{d t} \) shows the rate at which \(x_2\) alters.

In many practical scenarios, these equations represent interactions between components or populations, like predator-prey models in biology or economic indicators interacting over time. Solving these systems typically requires finding equilibrium points and determining their nature—whether the system leans towards stability or instability under small changes in variables.

Jacobian Matrix

The Jacobian matrix is a pivotal tool in analyzing systems of differential equations, representing the first partial derivatives of a vector-valued function. It gives insight into the behavior of the system around an equilibrium point by linearizing the system.

For a two-equation system like ours, the Jacobian matrix \( J \) comprises the partial derivatives of all functions concerning all variables. For example:

• \( J_{11} = \frac{\partial}{\partial x_1}(1 - ax_1x_2) \)
• \( J_{12} = \frac{\partial}{\partial x_2}(1 - ax_1x_2) \)
• \( J_{21} = \frac{\partial}{\partial x_1}(ax_1x_2 - x_2) \)
• \( J_{22} = \frac{\partial}{\partial x_2}(ax_1x_2 - x_2) \)

By evaluating this matrix at equilibrium points, one obtains a linear approximation of the system near those points. Analyzing the eigenvalues of the Jacobian helps determine the nature of stability at the equilibria. Specifically, if all obtained eigenvalues have negative real parts, the point is stable. Positive real parts indicate instability, as noted in system behavior.