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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. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-x-\frac{y}{3}\right)\\\&\frac{d y}{d t}=y\left(1-y-\frac{x}{2}\right)\end{aligned}$$

Short Answer

Expert verified
Equilibrium points are (0,0), (0,1), (1,0), and \((\frac{2}{5}, \frac{3}{5})\). Nullclines divide regions where trajectories' directions vary.

Step by step solution

01

Identify Nullclines

To find the nullclines, set \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \). For \( x \)-nullclines: \( x(1-x-\frac{y}{3}) = 0 \). So, \( x = 0 \) or \( 1-x-\frac{y}{3} = 0 \), which gives two lines: \( x = 0 \) and \( y = 3(1-x) \). For \( y \)-nullclines: \( y(1-y-\frac{x}{2}) = 0 \). So, \( y = 0 \) or \( 1-y-\frac{x}{2} = 0 \), which gives: \( y = 0 \) and \( x = 2(1-y) \).
02

Find Equilibrium Points

Equilibrium points occur where both \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \). Solving the system of equations \( x(1-x-\frac{y}{3}) = 0 \) and \( y(1-y-\frac{x}{2}) = 0 \), we find the equilibrium points: \((0,0), (0,1), (1,0),\) and \( (\frac{2}{5}, \frac{3}{5}) \).
03

Determine Directions of Trajectories

To assess the direction of trajectories, analyze the sign of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) in regions divided by nullclines. Above the \( x \)-nullcline \( y = 3(1-x) \), \( \frac{dx}{dt} > 0 \); below, \( \frac{dx}{dt} < 0 \). Likewise, to the left of the \( y \)-nullcline \( x = 2(1-y) \), \( \frac{dy}{dt} > 0 \); to the right, \( \frac{dy}{dt} < 0 \). This indicates the directions of arrows in different regions of the phase plane.
04

Sketch the Phase Portrait

Combine the nullclines, equilibrium points, and directions of trajectories. Plot the lines corresponding to each nullcline: \( x = 0 \), \( y = 0 \), \( y = 3(1-x) \), and \( x = 2(1-y) \). Place equilibrium points: \((0,0), (0,1), (1,0), (\frac{2}{5}, \frac{3}{5})\). Sketch arrows indicating flow directed according to the trajectory directions determined in Step 3.

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Key Concepts

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

Differential Equations
Differential equations represent mathematical expressions where derivatives of functions are involved. These equations describe how functions change and are fundamental in modeling real-world phenomena. In this exercise, we have two differential equations, each representing a rate of change for two variables, \(x\) and \(y\). The equations are:
  • \( \frac{dx}{dt} = x(1-x-\frac{y}{3}) \)
  • \( \frac{dy}{dt} = y(1-y-\frac{x}{2}) \)
These expressions define a system with interactions between \(x\) and \(y\). Here, \(\frac{dx}{dt}\) refers to how \(x\) changes over time, and \(\frac{dy}{dt}\) refers to \(y\). We aim to understand how these variables influence each other as time progresses. Differential equations in this context help us analyze and predict such dynamics, integral to studying phase planes.
Equilibrium Points
Equilibrium points are crucial in phase plane analysis and occur where the system doesn't change over time. In other words, these are the points where both \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). To find them, we solve the system of equations:
  • \( x(1-x-\frac{y}{3}) = 0 \)
  • \( y(1-y-\frac{x}{2}) = 0 \)
From our solution, these equilibrium points are \((0,0), (0,1), (1,0),\) and \((\frac{2}{5}, \frac{3}{5})\). At these locations, neither \(x\) nor \(y\) is changing, meaning they are constant over time. These points are often analyzed to determine their stability and how they influence surrounding trajectories in the phase plane.
Nullclines
Nullclines are essential for understanding the behavior of differential equations in a phase plane. A nullcline is a curve where one of the derivatives, either \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\), is zero. For our system, they are found by setting each derivative to zero:For the \(x\)-nullclines, \( \frac{dx}{dt} = 0 \), we obtain:
  • \( x = 0 \)
  • \( y = 3(1-x) \)
For the \(y\)-nullclines, \( \frac{dy}{dt} = 0 \), we have:
  • \( y = 0 \)
  • \( x = 2(1-y) \)
Nullclines partition the phase plane into regions with different qualitative behaviors. They help indicate where the direction of change for \(x\) or \(y\) switches sign, crucial for determining the flow of trajectories.
Trajectories
Trajectories in the context of differential equations are paths traced by solutions on the phase plane. They show how the system state evolves over time from a given initial condition. Determining trajectory directions involves assessing the sign of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) in regions defined by nullclines.For example:
  • Above the \(x\)-nullcline \( y = 3(1-x) \), \( \frac{dx}{dt} > 0 \).
  • Below this nullcline, \( \frac{dx}{dt} < 0 \).
  • Similarly, to the left of the \(y\)-nullcline \( x = 2(1-y) \), \( \frac{dy}{dt} > 0 \).
  • To the right, \( \frac{dy}{dt} < 0 \).
Trajectories aid in visualizing how the system transitions between different states over time. By sketching these paths, we can better understand the dynamics at play, including movement toward or away from equilibrium points and how variables interact within the system.

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