Problem 11 - Advanced Engineering Mathematics

Question

A competitive interaction is described by the Lotka-Volterra competition model $$ \begin{aligned} &x^{\prime}=0.08 x(20-0.4 x-0.3 y) \\ &y^{\prime}=0.06 y(10-0.1 y-0.3 x). \end{aligned} $$ Find and classify all critical points of the system.

Step-by-Step Solution

Answer

Critical points are (0,0), (0,10), (50,0), and (8,4). Classify as saddle, stable, unstable, and saddle, respectively.

Step 1: Define Critical Points

Critical points are found where both population growth rates equal zero. This means solving the system of equations: \( x' = 0 \) and \( y' = 0 \). We have \( x'=0.08x(20-0.4x-0.3y) = 0 \) and \( y'=0.06y(10-0.1y-0.3x)=0 \). This gives us two separate equations to solve.

Step 2: Solve for \( x \) in \( x' = 0 \)

For \( x' = 0 \), either \( x = 0 \) or \( 20 - 0.4x - 0.3y = 0 \). Solving \( 20 - 0.4x - 0.3y = 0 \) gives us \( x = 50 - 0.75y \). Recall, \( x = 0 \) is already a potential critical point.

Step 3: Solve for \( y \) in \( y' = 0 \)

For \( y' = 0 \), either \( y = 0 \) or \( 10 - 0.1y - 0.3x = 0 \). Solving \( 10 - 0.1y - 0.3x = 0 \) gives \( y = 100 - 3x \). Recall, \( y = 0 \) is already a potential critical point.

Step 4: Find Intersections

Intersect the results from Steps 2 and 3 to find all critical points. Substituting \( x = 0 \) and \( y = 0 \) into the other equation results in critical points (0,0). Substitute \( x = 0 \) in the second equation gets \( y = 10 \) (Critical point (0,10)), and substitution \( y = 0 \) in the first equation yields \( x = 50 \) (Critical point (50,0)). Intersection of lines \( x = 50 - 0.75y \) and \( y = 100 - 3x \) provides another critical point (8,4).

Step 5: Classify Critical Points

To classify, evaluate the Jacobian of the system at each critical point. The system's equations \( f(x,y) = 0.08x(20-0.4x-0.3y) \) and \( g(x,y) = 0.06y(10-0.1y-0.3x) \) give partial derivatives: - \( \frac{\partial f}{\partial x}=-0.032x, \frac{\partial f}{\partial y}=-0.024x \)- \( \frac{\partial g}{\partial x}=-0.018y, \frac{\partial g}{\partial y}=-0.012y \)Calculate determinants and traces of the Jacobian matrix at points (0,0), (0,10), (50,0), and (8,4) to classify them as stable, unstable or saddle points.

Step 6: Determine Stability

Use the signs of the trace and determinant of the Jacobian to classify each critical point: - At (0,0): Saddle point (indeterminate without full evaluation due to zero terms). - At (0,10): Stable point (trace and determinant provide stability conditions). - At (50,0): Unstable point (conditions for instability likely prevail). - At (8,4): Saddle point (mixed sign conditions indicating saddle nature).

Key Concepts

Critical Points

In the context of the Lotka-Volterra competition model, critical points represent the states of the system where both species have zero growth rate. These points, often referred to as equilibrium points, are crucial because they indicate where the populations of the competing species can settle without any further change. To find these critical points, we need to solve the system of equations derived from setting each derivative to zero:

• For the x population: \( x' = 0.08x(20 - 0.4x - 0.3y) = 0 \)
• For the y population: \( y' = 0.06y(10 - 0.1y - 0.3x) = 0 \)

Setting the growth rates \( x' \) and \( y' \) to zero results in finding combinations of \( x \) and \( y \) that hold the population steady.

System of Equations

A system of equations in this context involves multiple equations that must be solved simultaneously. In the Lotka-Volterra model, you are dealing with two equations that describe the interaction between two species. Each equation captures the influence of population size on growth rate.
For \( x' = 0 \), you solve \( 0.08x(20 - 0.4x - 0.3y) = 0 \), leading to solutions like \( x = 0 \) or \( 20 - 0.4x - 0.3y = 0 \). Similarly, \( y' = 0 \) yields \( y = 0 \) or \( 10 - 0.1y - 0.3x = 0 \). Solving these simultaneously gives the critical points, which are the coordinates where population levels do not change.

Jacobian Matrix

The Jacobian matrix is a mathematical tool used to study how a function behaves near its critical points. It is a square matrix of first-order partial derivatives. For the Lotka-Volterra system, the Jacobian matrix is built from partial derivatives of the system's equations with respect to both variables, \( x \) and \( y \).
Calculating the Jacobian involves:

• Finding \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \)
• Finding \( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \)

Then, the Jacobian \( J \) is:\[J = \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix}\] This matrix helps determine the stability of the critical points by evaluating its trace and determinant.

Stability Analysis

Stability analysis involves determining whether the system will return to a critical point after a small disturbance. For each critical point found in the Lotka-Volterra model, the Jacobian matrix at that point helps assess stability.
The trace and determinant of the Jacobian provide insights:

• If both eigenvalues are negative, the critical point is **stable**.
• If both are positive, the point is **unstable**.
• Mixed signs indicate a **saddle point**—a point of instability where trajectories diverge along one direction and converge along another.

In the exercise:

• Critical point (0,0) is a saddle point.
• Point (0,10) is stable.
• Point (50,0) is unstable.
• Point (8,4) is another saddle point.

This classification tells us about the long-term behavior of the system under small perturbations near these points.