/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Each of Problems I through 6 can... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 2

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

Expert verified
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

01

Identify the system of differential equations

The given system of differential equations is: $$ \begin{cases} \frac{dx}{dt} = x(1.5 - x - 0.5y)\\ \frac{dy}{dt} = y(2 - 0.5y - 1.5x) \end{cases} $$
02

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 \((1, 1)\).
03

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 \((1, 1)\): $$ J(1,1) = \begin{bmatrix} -1 & -0.5 \\ -1.5 & 1 \end{bmatrix} $$
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{ll}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x}\)

The motion of a certain undamped pendulum is described by the equations $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion with an angular displacement \(A\) and no initial velocity, then the initial conditions are \(x(0)=A, y(0)=0\) (a) Let \(A=0.25\) and plot \(x\) versus \(t\). From the graph estimate the amplitude \(R\) and period \(T\) of the resulting motion of the pendulum. (b) Repeat part (a) for \(A=0.5,1.0,1.5,\) and \(2.0 .\) (c) How do the amplitude and period of the pendulum's motion depend on the initial position \(A^{7}\) Draw a graph to show each of these relationships. Can you say anything about the limiting value of the period as \(A \rightarrow 0 ?\) (d) Let \(A=4\) and plot \(x\) versus \(t\) Explain why this graph differs from those in parts (a) and (b). For what value of \(A\) does the transition take place?

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=x-x^{2}-x y, \quad d y / d t=3 y-x y-2 y^{2} $$

Prove Theorem 9.7 .2 by completing the following argument. According to Green's theorem in the plane, if \(C\) is a sufficiently smooth simple closed curve, and if \(F\) and \(G\) are continuous and have continuous first partial derivatives, then $$ \int_{C}[F(x, y) d y-G(x, y) d x]=\iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A $$ where \(C\) is traversed counterclockwise and \(R\) is the region enclosed by \(C .\) Assume that \(x=\phi(t), y=\psi(t)\) is a solution of the system ( 15) that is periodic with period \(T\). Let \(C\) be the closed curve given by \(x=\phi(t), y=\psi(t)\) for \(0 \leq t \leq T\). Show that for this curve the line integral is zero. Then show that the conclusion of Theorem 9.7 .2 must follow.

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=(1+x) \sin y, \quad d y / d t=1-x-\cos y $$
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