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

Short answer

Question: Determine and classify the critical points of the given system of differential equations: $$ \begin{cases} \frac{dx}{dt} = x(1.5 - x - 0.5y)\\ \frac{dy}{dt} = y(2 - 0.5y - 1.5x) \end{cases} $$ Answer: The critical points of the given system are (0, 0), (0, 4), and (1, 1). The classifications of the critical points are as follows: the point (0, 0) is unstable, the point (0, 4) is asymptotically stable, and the point (1, 1) is a saddle point and also unstable.

Step-by-step solution

Identify the system of differential equations

The given system of differential equations is: $$ \begin{cases} \frac{dx}{dt} = x(1.5 - x - 0.5y)\\ \frac{dy}{dt} = y(2 - 0.5y - 1.5x) \end{cases} $$

Find the critical points

To find the critical points, we need to find the values of \(x\) and \(y\) for which both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are zero. We get the following equations to solve for critical points: $$ \begin{cases} x(1.5 - x - 0.5y) = 0 \\ y(2 - 0.5y - 1.5x) = 0 \end{cases} $$ By solving the above equations, we find the critical points \((0, 0)\), \((0, 4)\), and \((1, 1)\).

Linearize the system at each critical point

To find the linear system associated with each critical point, we compute the Jacobian matrix of the system at each critical point: $$ J(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} 1.5-2x-0.5y & -0.5x \\ -1.5y & 2-y \end{bmatrix} $$ Now, we will evaluate the Jacobian matrix at each critical point: At \((0, 0)\): $$ J(0,0) = \begin{bmatrix} 1.5 & 0 \\ 0 & 2 \end{bmatrix} $$ At \((0, 4)\): $$ J(0,4) = \begin{bmatrix} -1 & 0 \\ -6 & -2 \end{bmatrix} $$ At \((1, 1)\): $$ J(1,1) = \begin{bmatrix} -1 & -0.5 \\ -1.5 & 1 \end{bmatrix} $$

Classify the critical points

To classify the critical points, we will find the eigenvalues and eigenvectors of the Jacobian matrices evaluated at the critical points. For the critical point \((0, 0)\), the Jacobian matrix is diagonal, so the eigenvalues are the diagonal elements, \(\lambda_1 = 1.5\) and \(\lambda_2 = 2\). Both the eigenvalues are positive, so the critical point \((0, 0)\) is unstable. For the critical point \((0, 4)\), the eigenvalues are \(\lambda_1 = -2\) and \(\lambda_2 = -1\). Both eigenvalues are negative, so the critical point \((0, 4)\) is asymptotically stable. For the critical point \((1, 1)\), the eigenvalues are \(\lambda_1 = 1.5\) and \(\lambda_2 = -2.5\). Since one eigenvalue is positive and the other is negative, the critical point \((1, 1)\) is a saddle point and hence unstable. Now that we have classified all the critical points, we can proceed with the other tasks in the exercise, such as sketching trajectories and computing and plotting solutions, to further analyze the behavior of the given system of differential equations.

Key concepts

Critical Points

Critical points in a system of differential equations are where the system's derivatives equal zero. More simply, these are the points at which the behavior of the system changes or pivots. For a system to have a critical point, both rate of change functions must be zero simultaneously.

If we consider the system of differential equations provided:

• \( \frac{dx}{dt} = x(1.5 - x - 0.5y) \)
• \( \frac{dy}{dt} = y(2 - 0.5y - 1.5x) \)

To find the critical points, we solve:

• \( x(1.5 - x - 0.5y) = 0 \)
• \( y(2 - 0.5y - 1.5x) = 0 \)

For this system, solving these equations gives us critical points at coordinates \((0, 0)\), \((0, 4)\), and \((1, 1)\). Each critical point represents a potential point of equilibrium for the system, where the population levels of species do not change.

Jacobian Matrix

The Jacobian matrix is a powerful tool in the analysis of differential equations. It offers a way to approximate the behavior of a nonlinear system near a critical point by providing a linear approximation.

The Jacobian matrix \( J(x, y) \) is formed by taking the first partial derivatives of each function in the system, and arranging them in a matrix:\[J(x, y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix}\]For our example, the Jacobian matrix is:\[J(x, y) = \begin{bmatrix} 1.5-2x-0.5y & -0.5x \-1.5y & 2-y \end{bmatrix}\]Evaluating the Jacobian matrix at each critical point provides us with a linear system to study their local behavior. This step is crucial for understanding how the system behaves in the vicinity of these points.

Eigenvalues

Eigenvalues are essential in determining the nature of a critical point. They are derived from the Jacobian matrix. When analyzing a linearized system, the eigenvalues inform the stability and type of critical points.

To find the eigenvalues, calculate the determinant of the Jacobian matrix subtracted by a scalar \( \lambda \) times the identity matrix, set it to zero, and solve:\[det(J(x, y) - \lambda I) = 0\]The roots \( \lambda_1 \) and \( \lambda_2 \) of this polynomial give us the eigenvalues. For example, at the critical point \( (0, 0) \), the eigenvalues are \( \lambda_1 = 1.5 \) and \( \lambda_2 = 2 \).

Analyzing the signs of the eigenvalues helps classify the type of critical point:

• Both positive eigenvalues indicate an unstable node.
• Both negative eigenvalues mean the point is asymptotically stable.
• A mix of negative and positive suggests a saddle point, indicating instability.

Stability Analysis

Stability analysis uses eigenvalues from the Jacobian matrix to determine the nature of critical points in dynamical systems.

A critical point can be:

• **Asymptotically stable**: All solutions nearby converge towards it when all eigenvalues are negative.
• **Unstable**: Solutions diverge; this occurs when any eigenvalues are positive.
• **Saddle point**: At least one positive and one negative eigenvalue suggest a mix of converging and diverging behavior around the point.

For instance, the critical point \((0, 0)\) is unstable, as indicated by positive eigenvalues (1.5 and 2). In contrast, \((0, 4)\) is asymptotically stable with negative eigenvalues (-1 and -2). Meanwhile, \((1, 1)\) is a saddle point due to having both positive and negative eigenvalues.

Stability analysis helps predict the system's long-term behavior, thus understanding how populations might evolve over time.