Problem 13 - Calculus

Question

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

Step-by-Step Solution

Answer

The nullclines are \( x=0 \), \( y=0 \), and \( x+y=1, 2 \). Equilibrium points are \((0,0), (0,1), (2,0)\).

Step 1: Identify the Nullclines

Nullclines are found by setting each differential equation to zero. For \( \frac{dx}{dt}=0 \), solve \( x(2-x-y)=0 \). This gives \( x=0 \) or \( 2=x+y \). For \( \frac{dy}{dt}=0 \), solve \( y(1-x-y)=0 \). This gives \( y=0 \) or \( 1=x+y \). These lines divide the plane into regions.

Step 2: Find Equilibrium Points

Equilibrium points occur where both nullclines intersect, meaning both \( \frac{dx}{dt}=0 \) and \( \frac{dy}{dt}=0 \). From \( x=0 \) and \( y=0 \), the first intersection is \((0,0)\). From the intersection of \( x=0 \) and \( 1=x+y \), we have \((0,1)\). From \( y=0 \) and \( 2=x+y \), we find \((2,0)\). Lastly, where \( x+y=2 \) and \( x+y=1 \), no valid intersection since this is a contradiction. Hence the equilibrium points are \((0,0), (0,1), (2,0)\).

Step 3: Determine Direction of Trajectories

Evaluate the sign of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) in each region determined by the nullcline. Above the line \( x+y=2 \), both \( dx/dt < 0 \) and \( dy/dt < 0 \) suggesting trajectories move downwards and to the left. Between \( x+y=1 \) and \( x+y=2 \), \( dx/dt > 0 \) while \( dy/dt < 0 \). Below \( x+y=1 \), both \( dx/dt > 0 \) and \( dy/dt > 0 \) suggesting movement upwards and rightward.

Step 4: Sketch the Phase Plane

Draw the nullclines on the phase plane. At each region, sketch arrows to indicate trajectories based on step 3. Arrows should clearly demonstrate movement towards, away, or parallel to nullclines. The sketch should show convergence of trajectories towards stable or unstable equilibrium points determined by stability analysis.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They are a key tool for modeling situations where the rate of change of a quantity is significant. In the exercise we're examining, we have two differential equations:

• Equation for the change in \( x \): \( \frac{d x}{d t} = x(2-x-y) \)
• Equation for the change in \( y \): \( \frac{d y}{d t} = y(1-x-y) \)

These equations describe how \( x \) and \( y \) evolve over time. By analyzing these, we can understand the system's behavior and predict future states given initial conditions. Often, the goal with differential equations in a phase plane is to determine how variables like \( x \) and \( y \) change in relation to each other rather than over time, providing a powerful visualization of dynamics.

Nullclines

Nullclines help us to decompose the complex motion described by differential equations into simpler parts. A nullcline is a curve in the phase plane where one of the differential equations equals zero.

• For \( \frac{d x}{d t} = 0 \), we have two possibilities: \( x = 0 \) or \( x+y = 2 \).
• For \( \frac{d y}{d t} = 0 \), the possibilities are: \( y = 0 \) or \( x+y = 1 \).

These nullclines divide our plane into several regions, which help us analyze the behavior of the system. By understanding how nullclines isolate parts of the phase plane, we can assess and visualize how different regions affect the dynamics between variables \( x \) and \( y \).

Equilibrium Points

Equilibrium points are steady states where the system does not change, meaning both \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \) are zero. These occur where the nullclines intersect. In our exercise:

• The point \((0,0)\) is a natural equilibrium from the existence of both nullclines at the origin.
• \((0,1)\) and \((2,0)\) arise where the nullclines \( x=0 \) meets \( x+y=1 \) and \( y=0 \) meets \( x+y=2 \) respectively.

Analyzing these points provides insight into the system's overall behavior. Recognizing equilibrium points is crucial because they often represent the long-term behavior of the system, revealing which states are sustainable over time.

Trajectory Direction

To understand how the system evolves, we need to determine the direction of trajectories in each region defined by the nullclines. The behavior of trajectories tells us how \( x \) and \( y \) change:

• In regions above \( x+y = 2 \), both \( \frac{d x}{d t} < 0 \) and \( \frac{d y}{d t} < 0 \), indicating movement downward and to the left.
• Between lines \( x+y = 1 \) and \( x+y = 2 \), we find \( \frac{d x}{d t} > 0 \) whereas \( \frac{d y}{d t} < 0 \), suggesting rightward and downward movement.
• Below \( x+y = 1 \), with both derivatives positive, the trajectories move upwards and to the right.

Understanding trajectory direction allows for predicting tendency of systems towards stability or instability and helps visualize the dynamic flow of the system.

Stability Analysis

Stability analysis helps determine whether an equilibrium point will maintain its state when the system is slightly perturbed. There are generally three types of equilibrium based on stability:

• Stable (attracting): Small disturbances lead the system to return to equilibrium.
• Unstable (repelling): Small disturbances cause the system to deviate further from equilibrium.
• Saddle points: A mix where the system is stable in one direction and unstable in another.

By sketching the phase plane, we can visually inspect which points are stable. Stable equilibria attract nearby trajectories, whereas unstable ones repel them. This visualization adds deeper insight into how systems respond to changes and how phase plane analysis predicts these responses.