Question

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-0.5 x-y)} \\ {d y / d t=y(2-y-1.125 x)}\end{array} $$

Short answer

Answer: The critical point \((\frac{8}{5},\frac{4}{5})\) is a saddle point. This implies that the point is unstable, as trajectories will approach and move away from this point.

Step-by-step solution

(a) Draw a direction field and describe the solutions' behavior.

We are given the following system of equations: $$\frac{dx}{dt} = x(1.5 - 0.5x - y)$$ $$\frac{dy}{dt} = y(2 - y - 1.125x)$$ To draw a direction field, create a grid of points and indicate the direction of the solution at each point. It's best to use software like MATLAB, Mathematica, or an online tool for this. Observe the direction field and describe how the solutions behave based on the arrows.

(b) Find the critical points.

To find the critical points, set both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) equal to zero and solve for \(x\) and \(y\): $$x(1.5 - 0.5x - y) = 0$$ $$y(2 - y - 1.125x) = 0$$ We get three critical points: \((0,0), (3,0),\) and \((\frac{8}{5},\frac{4}{5})\).

(c) Linearize, find eigenvalues and eigenvectors, classify the critical points, and determine their stability.

To linearize the system at each critical point, we will find the Jacobian matrix: $$ J(x, y) = \begin{bmatrix} \frac{\partial}{\partial x}(1.5 - 0.5x - y) & \frac{\partial}{\partial y}(1.5 - 0.5x - y)\\ \frac{\partial}{\partial x}(2 - y - 1.125x) & \frac{\partial}{\partial y}(2 - y - 1.125x) \end{bmatrix} = \begin{bmatrix} -0.5x - y + 1.5 & -x \\ -1.125y & -x - y + 2 \end{bmatrix} $$ Now, evaluate the Jacobian at each critical point, find eigenvalues and eigenvectors, classify the critical points, and determine their stability. For the critical point \((0,0)\): $$ J(0,0) = \begin{bmatrix} 1.5 & 0 \\ 0 & 2 \end{bmatrix} $$ Eigenvalues: \(\lambda_1 = 1.5\) and \(\lambda_2 = 2\). Since both eigenvalues are positive, the critical point is an unstable node. For the critical point \((3,0)\): $$ J(3,0) = \begin{bmatrix} -1.5 & -3 \\ 0 & -1 \end{bmatrix} $$ Eigenvalues: \(\lambda_1 = -1.5\) and \(\lambda_2 = -1\). Since both eigenvalues are negative, the critical point is a stable node. For the critical point \((\frac{8}{5},\frac{4}{5})\): $$ J\left(\frac{8}{5},\frac{4}{5}\right) = \begin{bmatrix} -0.2 & -\frac{8}{5} \\ -\frac{9}{5} & 0.8 \end{bmatrix} $$ Eigenvalues: \(\lambda_1 = -0.5\) and \(\lambda_2 = 1.1\). Since one eigenvalue is positive and the other is negative, the critical point is a saddle point and is unstable.

(d) Sketch the trajectories near the critical points.

Based on the classifications and stabilities of the critical points, sketch the trajectories near each critical point: \((0,0)\) will have trajectories moving away; \((3,0)\) will have trajectories converging and \((\frac{8}{5},\frac{4}{5})\) will have trajectories that approach as well as move away from the point.

(e) Compute and plot enough trajectories of the system to show the behavior of the solutions.

You can use software like MATLAB, Python, Mathematica, or an online tool to solve the system of equations numerically and plot several trajectories with different initial conditions. This will provide a clear demonstration of how solutions behave in different regions of the \(xy\)-plane.

(f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\).

By analyzing the eigenvalues of the Jacobian matrix at each critical point, we can determine the long-term behavior of populations \(x\) and \(y\) as \(t \rightarrow \infty\). Since the critical point \((3,0)\) is a stable node, most solutions will converge to this point as \(t \rightarrow \infty\). In this case, population \(x\) will tend to 3, and population \(y\) will tend to 0. This means that as time goes on, the population of species \(x\) will stabilize at a constant non-zero value, while the population of species \(y\) will ultimately go extinct.

Key concepts

Critical Points

In a system of differential equations, critical points are where the system's derivatives equal zero. For our exercise, we determine critical points by setting \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \). This tells us where the system is in equilibrium, or in other words, where the populations cease to change. For the given equations, the critical points are \( (0,0), (3,0),\) and \( (\frac{8}{5}, \frac{4}{5})\). Each critical point represents a scenario with specific population values of the two species involved.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors help us understand how a system behaves around critical points. To find them, we linearize the system using the Jacobian matrix evaluated at each critical point. Eigenvalues determine stability, with:

• Positive eigenvalues indicating instability.
• Negative eigenvalues indicating stability.
• Mixed signs indicating saddle points.

For example, at \( (0,0)\), the eigenvalues are 1.5 and 2, leading to an unstable node. These calculations help predict the behavior near equilibrium points.

Stability Analysis

Stability analysis involves determining whether a critical point is stable, asymptotically stable, or unstable. This is critical in predicting the long-term behavior of the species in our model. Here's how we categorize them:

• An unstable point (like \( (0,0)\)) has all positive eigenvalues.
• A stable point (like \( (3,0)\)) has all negative eigenvalues.
• A saddle point (like \( (\frac{8}{5}, \frac{4}{5})\)) has both positive and negative eigenvalues, making it inherently unstable.

This analysis helps us understand the possible long-term outcomes of the species interaction.

Linearization

Linearization simplifies a nonlinear system near a critical point. This is achieved by using the Jacobian matrix, which describes how small changes affect the system. For our model, the matrix is evaluated at each critical point to understand local behavior. By knowing how the system behaves when slightly perturbed, we can derive stability insights without solving the whole system. It allows for more manageable approximations in a complex model, especially around equilibrium points.

Direction Field

A direction field offers a visual representation of a differential equation's solution trends. By plotting arrows on a grid, it shows the direction solutions tend to move over the plane. It gives an intuitive grasp of the system dynamics. In our exercise, drawing a direction field for the given system will illustrate:

• How solutions behave near each critical point.
• Which directions the populations may grow or decline.
• Overall interaction patterns between the two species.

This field is fundamental in visualizing and predicting system behavior without solving the equations fully.