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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 are at (0, 0) and \((\frac{3}{4}, \frac{5}{8})\). Sketch nullclines and indicate direction of trajectories.

Step by step solution

01

Find Nullclines

Nullclines are determined by setting the right-hand side of each differential equation to zero.For \( \frac{dx}{dt} = x(1 - \frac{x}{2} - y) = 0 \), the nullclines are:1. \( x = 0 \)2. \( 1 - \frac{x}{2} - y = 0 \Rightarrow y = 1 - \frac{x}{2} \).For \( \frac{dy}{dt} = y(1 - \frac{y}{3} - x) = 0 \), the nullclines are:1. \( y = 0 \)2. \( 1 - \frac{y}{3} - x = 0 \Rightarrow x = 1 - \frac{y}{3} \).
02

Find Equilibrium Points

Equilibrium points occur where the nullclines intersect. Solve the equations\( x = 0 \) and \( y = 0 \) to find one equilibrium point at \((0, 0)\).Next, solve \( y = 1 - \frac{x}{2} \) and \( x = 1 - \frac{y}{3} \).By substitution: set \( y = 1 - \frac{x}{2} \) into \( x = 1 - \frac{y}{3} \).So, \( x = 1 - \frac{1 - \frac{x}{2}}{3} = 1 - \frac{3 - x}{6} = \frac{3 + x}{6} \).Solving gives \( x = \frac{3}{4} \).Substitute back to find \( y = 1 - \frac{3/4}{2} = \frac{5}{8} \).Thus, another equilibrium point is \((\frac{3}{4}, \frac{5}{8})\).
03

Sketch the Nullclines

Plot the nullclines on the phase plane:1. The line \( x = 0 \) is the y-axis.2. The line \( y = 1 - \frac{x}{2} \) is a line with a negative slope intersecting the y-axis at 1.3. The line \( y = 0 \) is the x-axis.4. The line \( x = 1 - \frac{y}{3} \) is a line with a negative slope intersecting the x-axis at 1.
04

Indicate Direction of Trajectories

Analyze each region created by the nullclines for the direction of the vector field by checking signs of derivatives.- If \( x \) nullcline, check \( x = 0.5 \) in \( y = 1 - \frac{x}{2} \), then \(0 < y < 1 - \frac{0.5}{2}\), \(\frac{dx}{dt} > 0\).- If \( y \) nullcline, \( x < 1 - \frac{y}{3} \), then \(\frac{dy}{dt} > 0\).Mark vectors indicating \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) directions around equilibrium points.

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Key Concepts

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

Understanding Nullclines
In phase plane analysis, nullclines are crucial for understanding the behavior of a dynamical system. They are curves where the rate of change of a variable is zero. For our equations, nullclines are found by setting \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to zero and solving for the variables.
  • For \( \frac{dx}{dt} = 0\), nullclines are \( x = 0 \) and \( y = 1 - \frac{x}{2} \). This means along these lines, the rate of change \( x(t) \) is zero.
  • For \( \frac{dy}{dt} = 0\) nullclines are \( y = 0 \) and \( x = 1 - \frac{y}{3} \). Here, the rate of change \( y(t) \) is zero.
These lines divide the phase plane into regions that help predict how solutions will behave. They are significant because they act as boundaries, indicating potential changes in trajectory directions.
Equilibrium Points
Equilibrium points are where the system is at rest, meaning both variables do not change over time. To find these points, see where the nullclines intersect. For the system at hand, two intersections are present:
  • At \((0, 0)\), where both \(x\) and \(y\) equal zero.
  • At \(\left(\frac{3}{4}, \frac{5}{8}\right)\), found by intersecting the lines \(y = 1-\frac{x}{2}\) and \(x = 1-\frac{y}{3}\).
The equilibrium points are crucial because they represent steady states of the system, where there is no net change in the variables. These points act as attractors or repellers, influencing nearby trajectories in the phase plane.
Direction Fields and Trajectories
Direction fields, also known as vector fields, help illustrate the behavior of trajectories in the phase plane. By analyzing small arrows or vectors on a phase plane, we can see how the system evolves over time. Examine the signs of derivatives to predict these directions:
  • Check a point in each region determined by nullclines. For instance, when \(x < 1-\frac{y}{3}\), \(\frac{dy}{dt} > 0\), indicating an upward movement.
  • Similarly, if \(y < 1-\frac{x}{2}\), \(\frac{dx}{dt} > 0\), showing movement to the right.
By combining these elements, direction fields sketch potential paths that solutions of the system can trace over time, helping us to understand how solutions move in relation to the equilibrium points.

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Most popular questions from this chapter

The amount of radioactive carbon- 14 in a sample is measured using a Geiger counter, which records each disintegration of an atom. Living tissue disintegrates at a rate of about 13.5 atoms per minute per gram of carbon. In 1977 a charcoal fragment found at Stonehenge, England, recorded 8.2 disintegrations per minute per gram of carbon. Assuming that the half-life of carbon- 14 is 5730 years and that the charcoal was formed during the building of the site, estimate the date at which Stonehenge was built.

Give \(k, L, A,\) a formula for \(P\) as a function of time \(t,\) and the time to the peak value of \(d P / d t\) $$\frac{d P}{d t}=0.02 P-0.0025 P^{2}, \quad P_{0}=1$$

When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of \(0.002 \mathrm{m}^{3} / \mathrm{min}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?

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 $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

The systems of differential equations model the interaction of two populations \(x\) and \(y .\) In each case, answer the following two questions: (a) What kinds of interaction (symbiosis, \(^{30}\) competition, predator-prey) do the equations describe? (b) What happens in the long run? (For one of the systems, your answer will depend on the initial populations.) Use a calculator or computer to draw slope fields. $$\begin{aligned} &\frac{1}{x} \frac{d x}{d t}=1-\frac{x}{2}-\frac{y}{2}\\\ &\frac{1}{y} \frac{d y}{d t}=1-x-y \end{aligned}$$
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