Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 9: Problem 13

a. Sketch the nullclines and describe how the critical points move as \(\alpha\) increases. b. Find the critical points. c. Let \(\alpha=2\). Classify each critical point by investigating the corresponding approximate linear system. Draw a phase portrait in a rectangle containing the critical points. d. Find the bifurcation point \(\alpha_{0}\) at which the critical points coincide. Locate this critical point, and find the eigenvalues of the approximate linear system. Draw a phase portrait. e. For \(\alpha>\alpha_{0},\) there are no critical points. Choose such a value of \(\alpha\) and draw a phase portrait. $$x^{\prime}=-4 x+y+x^{2}, \quad y^{\prime}=-\alpha-x+y$$

Short Answer

Expert verified
Short Answer: To analyze the given two-dimensional nonlinear system, first, we find the nullclines and the critical points by solving for x and y when \(x^{\prime}=0\) and \(y^{\prime}=0\). The behavior of the critical points can be determined by calculating the eigenvalues of the Jacobian matrix. We can then draw the phase portrait for the system with a chosen value of \(\alpha\), such as 2. To find the bifurcation point \(\alpha_0\) where the critical points coincide, we solve for \(\alpha\) in terms of the critical points and analyze the system's behavior at this value. Finally, we draw the phase portrait for a chosen value of \(\alpha>\alpha_0\) to analyze the system's behavior.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Find the Nullclines

To find the nullclines, set each equation in the system equal to 0 separately and solve for x and/or y. $$ x^{\prime}=-4x+y+x^{2}=0 \quad \Rightarrow \quad y=(4-x)x $$ $$ y^{\prime}=-\alpha-x+y=0 \quad \Rightarrow \quad y=\alpha+x $$
02

Sketch the Nullclines

Now sketch the nullclines on a graph with x-axis as x and y-axis as y. You should see that the nullclines intersect at the critical points of the system, and the behavior of the critical points changes as \(\alpha\) increases. #b. Critical Points#
03

Find the Critical Points

To find the critical points, set both \(x^{\prime}\) and \(y^{\prime}\) equal to 0 simultaneously and solve for x and y. Equate the nullclines which will give: $$ (4-x)x = \alpha+x $$ Now, solve this equation for x and substitute back into one of the nullclines to find y. There may be multiple critical points depending on the value of \(\alpha\). #c. Classify Critical Points and Phase Portrait for \(\alpha=2\)#
04

Find the Approximate Linear System

Let \(\alpha=2\). Before classifying the critical points, find the Jacobian matrix evaluated at the critical points (\(x_0\), \(y_0\)): $$ J(x_0, y_0) = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix} = \begin{pmatrix} 1-2x_0 & 1 \\ -1 & 1 \end{pmatrix} $$
05

Classify the Critical Points

Now, we can classify each critical point by finding the eigenvalues of the Jacobian matrix. The behavior of the critical point will depend on the eigenvalues.
06

Draw the Phase Portrait

Draw the phase portrait for the system with \(\alpha=2\), showing the classified critical points and the behavior of the corresponding approximate linear system in a rectangle containing all critical points. #d. Bifurcation Point and Phase Portrait#
07

Find the Bifurcation Point \(\alpha_0\)

To find the bifurcation point \(\alpha_0\) at which the critical points coincide, we need to find the value of \(\alpha\) for which the system has multiple solutions for the critical points. Solve the equation for x in terms of \(\alpha\) and find the bifurcation point.
08

Analyze the System at the Bifurcation Point

Now, analyze the behavior of the system at the bifurcation point by finding the eigenvalues of the Jacobian matrix evaluated at the coinciding critical points.
09

Draw the Phase Portrait

Draw the phase portrait for the system at the bifurcation point, showing the coinciding critical points and the behavior of the corresponding approximate linear system. #e. Phase Portrait for \(\alpha>\alpha_0\)#
10

Choose a Value of \(\alpha>\alpha_0\)

Now, choose a value of \(\alpha\) that is greater than the bifurcation point \(\alpha_0\). Since there are no critical points for these values of \(\alpha\), we can analyze the behavior of the system by looking at the phase portrait.
11

Draw the Phase Portrait

Draw the phase portrait for the chosen value of \(\alpha>\alpha_0\), showing the behavior of the nonlinear system.

Key Concepts

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

Nullclines
Nullclines are special curves in the phase space of dynamical systems where either the derivative of the variable x or y equals zero. In other words, they help us find the steady states of the system where the change in one variable temporarily halts.
To find nullclines, we need to solve the given system of equations by setting each derivative to zero separately. For example, with the equations:
  • \(x^{\prime}=-4x+y+x^{2}=0\) leads to \(y=(4-x)x\)
  • \(y^{\prime}=-\alpha-x+y=0\) results in \(y=\alpha+x\)
The intersection points of these nullclines indicate potential critical points. As \(\alpha\) changes, the appearance and location of nullclines—and hence the behavior of critical points—might shift, giving insights into the system's dynamics.
Critical Points
Critical points, sometimes referred to as equilibrium points, are locations within the phase space where both derivatives of the system are zero. This means the system is in a state of balance, with no net change over time.
To determine these points, we set both equations for the nullclines equal to each other:
  • \( (4-x)x = \alpha +x \)
Solving this equation yields the critical points for various values of \(\alpha\). These points play a crucial role in understanding how the dynamical system behaves at specific parameters. Different \(\alpha\) values may result in multiple, one, or no critical points.
Uncovering these critical points allows us to deeply analyze how the system transitions between different states of equilibrium as parameters like \(\alpha\) vary.
Phase Portrait
The phase portrait is a visual representation of a dynamical system's trajectories in the phase space. It provides a map of all possible behaviors and states the system can exhibit over time.
When \(\alpha=2\), critical points can first be classified through linear approximation by examining the Jacobian matrix. This involves:
  • Finding eigenvalues of the matrix \(J(x_0, y_0) = \begin{pmatrix} 1-2x_0 & 1 \ -1 & 1 \end{pmatrix}\)
By evaluating these eigenvalues, we get insights into the stability of the critical points—whether a point is stable, a saddle, or otherwise. Drawing a phase portrait involves plotting trajectories that align with these calculations and classification.
This step-by-step analysis allows us to visualize how the system evolves with time, giving clarity on how variables like \(\alpha\) affect system dynamics.
Bifurcation Analysis
Bifurcation analysis helps us understand how small changes in system parameters can lead to significant changes in the behavior of the system. It identifies crucial points, known as bifurcation points, where the qualitative nature of the system's equilibrium changes.
Finding the bifurcation point \(\alpha_0\) involves recognizing when critical points coincide. We solve the critical point condition \((4-x)x = \alpha +x\) to find specific values of \(\alpha\). At these points, the behavior of the system dramatically shifts, often leading to the appearance or disappearance of critical points.
Once the bifurcation point is identified, the system can be further analyzed by finding the eigenvalues at these critical points. This determines the stability of coinciding critical points and can be represented in a phase portrait.
Exploring beyond \(\alpha_0\), for \(\alpha>\alpha_0\), reveals configurations where no critical points exist, helping to visualize the system's behavior in new parameter realms. Understanding bifurcation enables predicting and explaining shifts in dynamical system structures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In this problem we show how small changes in the coefficients of a system of linear equations can affect a critical point that is a center. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-1} & {0}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \pmi so that \((0,0)\) is a center. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\epsilon} & {1} \\ {-1} & {\epsilon}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrarily small. Show that the eigenvalues are \(\epsilon \pm i .\) Thus no matter how small \(|\epsilon| \neq 0\) is, the center becomes a spiral point. If \(\epsilon<0,\) the spiral point is asymptotically stable; if \(\epsilon>0,\) the spiral point is unstable.

In this problem we show how small changes in the coefficients of a system of linear equations can affect the nature of a critical point when the eigenvalues are equal. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-1} & {1} \\ {0} & {-1}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \(r_{1}=-1, r_{2}=-1\) so that the critical point \((0,0)\) is an asymptotically stable node. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{cc}{-1} & {1} \\ {-\epsilon} & {-1}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrararily small. Show that if \(\epsilon>0,\) then the eigenvalues are \(-1 \pm i \sqrt{\epsilon}\), so that the asymptotically stable node becomes an asymptotically stable spiral point. If \(\epsilon<0,\) then the roots are \(-1 \pm \sqrt{|\epsilon|},\) and the critical point remains an asymptotically stable node.

(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. $$ d x / d t=x-x^{2}-x y, \quad d y / d t=3 y-x y-2 y^{2} $$

Can be interpreted as describing the interaction of two species with population densities \(x\) and \(y .\) In each of these problems carry out the following steps. (a) Draw a direction field and describe how solutions seem to behave. (b) Find the critical points. (c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, or unstable. (d) Sketch the trajectories in the neighborhood of each critical point. (e) Draw a phase portrait for the system. (f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results in terms of the populations of the two species. $$ \begin{array}{l}{d x / d t=x(1-0.5 x-0.5 y)} \\ {d y / d t=y(-0.25+0.5 x)}\end{array} $$

Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (0.5,0.5) , corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of \(\delta a\) and \(\delta b\) for the species \(x\) and \(y,\) respectively. In this case equations ( 3 ) are replaced by $$\frac{d x}{d t}=x(1-x-y)+\delta a, \quad \frac{d y}{d t}=\frac{y}{4}(3-4 y-2 x)+\delta b$$ The question is what effect this has on the location of the stable equilibrium point. a. To find the new critical point, we must solve the equations $$\begin{aligned} x(1-x-y)+\delta a &=0 \\ \frac{y}{4}(3-4 y-2 x)+\delta b &=0 \end{aligned}$$ One way to proceed is to assume that \(x\) and \(y\) are given by power series in the parameter \(\delta ;\) thus $$x=x_{0}+x_{1} \delta+\cdots, \quad y=y_{0}+y_{1} \delta+\cdots$$ Substitute equations (44) into equations (43) and collect terms according to powers of \(\delta\). b. From the constant terms (the terms not involving \(\delta\) ), show that \(x_{0}=0.5\) and \(y_{0}=0.5,\) thus confirming that in the absence of immigration or emigration, the critical point is (0.5,0.5) . c. From the terms that are linear in \(\delta,\) show that \\[ x_{1}=4 a-4 b, \quad y_{1}=-2 a+4 b \\] d. Suppose that \(a>0\) and \(b>0\) so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free