/*! 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 18 Analyze the phase plane of the d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 18

Analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-\frac{x}{2}-y\right)\\\&\frac{d y}{d t}=y\left(1-\frac{y}{3}-x\right)\end{aligned}$$

Short Answer

Expert verified
Equilibrium points: (0, 0), (2, 0), (0, 3), (1, 1). Phase plane shows trajectories towards (1, 1).

Step by step solution

01

Identify the Nullclines

The nullclines are found by setting the derivatives equal to zero. \( \frac{dx}{dt} = 0 \) gives the x-nullcline; therefore, \( x(1 - \frac{x}{2} - y) = 0 \). This implies \( x = 0 \) or \( 1 - \frac{x}{2} - y = 0 \), leading to the equations: \( x = 0 \) or \( y = 1 - \frac{x}{2} \). For \( \frac{dy}{dt} = 0 \), the y-nullcline is \( y(1 - \frac{y}{3} - x) = 0 \), providing \( y = 0 \) or \( x + \frac{y}{3} = 1 \).
02

Find the Equilibrium Points

The equilibrium points occur at intersections of the nullclines. For the x-nullcline equations \( x = 0 \) and \( y = 1 - \frac{x}{2} \), and y-nullcline equations \( y = 0 \) and \( x + \frac{y}{3} = 1 \): the points are \( (0, 0), (2, 0), (0, 3), (1, 1) \).
03

Sketch the Nullclines

Graph the nullclines on the phase plane. \( x = 0 \) is the vertical axis, \( y = 0 \) is the horizontal axis, \( y = 1 - \frac{x}{2} \) is a line with slope \(-0.5\) through the y-intercept at 1, and \( x + \frac{y}{3} = 1 \) is a line with slope \(-3\) through the x-intercept at 1.
04

Determine the Direction of Trajectories

Pick test points in each region divided by the nullclines and substitute into the differential equations to determine trajectories' direction. For instance, in the region where \( (x, y) \) is \( (0.5, 0.5) \), evaluate \( \frac{dx}{dt} = + \) and \( \frac{dy}{dt} = + \), indicating motion towards positive x and y. Repeat this for major regions divided by the nullclines.
05

Sketch the Phase Plane

Using the information from previous steps, sketch the phase plane. Mark the equilibrium points and draw arrows showing the direction of trajectories in different regions. These should converge to equilibria such as \( (1, 1) \) and showcase behavior towards axes and boundary.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Nullclines
Nullclines are essential in understanding the dynamics of a system described by differential equations. They represent the curves where the derivative of a variable is zero. For the given system:
  • The x-nullcline: Obtained by setting \( \frac{dx}{dt} = 0 \), leading to either \( x = 0 \) or \( y = 1 - \frac{x}{2} \). These are the lines where the rate of change for \( x \) is zero and help partition the plane into regions.
  • The y-nullcline: Found by setting \( \frac{dy}{dt} = 0 \), resulting in \( y = 0 \) or \( x + \frac{y}{3} = 1 \). These lines highlight where the rate of change for \( y \) is zero.
By sketching these nullclines, you can visualize crucial boundaries in the phase-plane that delineate different behavior zones.
Equilibrium Points
Equilibrium points are the intersections of nullclines, where both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are zero simultaneously. This means there is no net change in \( x \) or \( y \) at these points. For our equations:
  • Solving \( x = 0 \) and \( y = 0 \) yields \( (0, 0) \).
  • Combining \( x = 0 \) and \( x + \frac{y}{3} = 1 \) gives \( (0, 3) \).
  • From \( y = 0 \) and \( y = 1 - \frac{x}{2} \), we obtain \( (2, 0) \).
  • Intersecting \( y = 1 - \frac{x}{2} \) and \( x + \frac{y}{3} = 1 \) finds the point \( (1, 1) \).
These equilibrium points are crucial as they often represent steady-states or repeated patterns in the system's behavior.
Differential Equations
Differential equations describe how variables change over time. In this system, each equation represents the dynamic rate of change for one variable in relation to the other. Here's a breakdown:
  • \( \frac{dx}{dt} = x(1 - \frac{x}{2} - y) \) describes how \( x \) evolves. It combines logistic growth and inhibitory effects depending on \( y \).
  • \( \frac{dy}{dt} = y(1 - \frac{y}{3} - x) \) does the same for \( y \), where growth is modulated by \( x \).
Understanding how these equations work helps when predicting the system's behavior. The terms within the equations control growth, indicate self-restriction due to population limits, and show interactions between \( x \) and \( y \).
Trajectories
Trajectories depict the path that a system's state follows over time in the phase plane. They are the real-world representations of the dynamics described by differential equations. To assess them, it's helpful to:
  • Choose test points within regions defined by nullclines, and determine the direction of movement using the sign of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
  • In regions where \( \frac{dx}{dt} > 0 \) and \( \frac{dy}{dt} > 0 \), the system moves towards higher \( x \) and \( y \).
  • Opposite signs indicate motion towards lower values as one variable decreases.
By systematically checking major regions, you can map out where the trajectories are headed and whether they converge, diverge, or loop around particular points.
Sketching Phase Planes
Drawing the phase plane provides a visual summary of a system's dynamics, enhancing understanding of how nullclines, equilibrium points, and trajectories interact. Here's how to do it effectively:
  • First, sketch the nullclines, labeling axes and major lines like \( x = 0 \) and \( y = 0 \), along with lines like \( y = 1 - \frac{x}{2} \) and \( x + \frac{y}{3} = 1 \).
  • Mark the equilibrium points you found earlier, such as \( (0, 0) \), \( (0, 3) \), \( (2, 0) \), and \( (1, 1) \).
  • Use arrows to illustrate trajectory directions in regions divided by nullclines, ensuring clarity in how the system鈥檚 state progresses or stabilizes over time.
By integrating these elements seamlessly, the phase plane becomes an informative map of the differential equation system's potential behavior.

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

Find the general solution to the given differential equation. $$9 z^{\prime \prime}-z=0$$

The motion of a mass on the end of a spring satisfies the differential equation $$ \frac{d^{2} s}{d t^{2}}+7 \frac{d s}{d t}+10 s=0 $$ (a) If the mass \(m=10,\) what is the spring coefficient \(k ?\) What is the damping coefficient a? (b) Solve the differential equation if the initial conditions are \(s(0)=-1\) and \(s^{\prime}(0)=-7\) (c) How low does the mass at the end of the spring go? How high does it go? (d) How long does it take until the spring stays within 0.1 unit of equilibrium?

Find the general solution to the given differential equation. $$y^{\prime \prime}-3 y^{\prime}+2 y=0$$

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester's analysis to this battle, with \(x\) representing the number of US troops and \(y\) the number of Japanese troops, it has been estimated \(^{35}\) that \(a=0.05\) and \(b=0.01\) (a) Using these values for \(a\) and \(b\) and ignoring reinforcements, write a differential equation involving \(d y / d x\) and sketch its slope field. (b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500 draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted?

Find the general solution to the given differential equation. $$z^{\prime \prime}+2 z^{\prime}=0$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

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.