Problem 16 - 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. $$\begin{aligned} &\frac{d x}{d t}=x\left(1-y-\frac{x}{3}\right)\\\ &\frac{d y}{d t}=y\left(1-\frac{y}{2}-x\right) \end{aligned}$$

Step-by-Step Solution

Answer

The equilibrium point is \( (\frac{3}{5}, \frac{4}{5}) \).

Step 1: Finding the Nullclines for dx/dt

To find the nullclines for the equation \( \frac{dx}{dt} = x(1 - y - \frac{x}{3}) \), we set \( \frac{dx}{dt} = 0 \). This gives two conditions: either \( x = 0 \) or \( 1 - y - \frac{x}{3} = 0 \). The first condition \( x = 0 \) likely represents a vertical line at \( x = 0 \). The second condition can be rewritten as \( y = 1 - \frac{x}{3} \). This nullcline is a line with a negative slope.

Step 2: Finding the Nullclines for dy/dt

For the equation \( \frac{dy}{dt} = y(1 - \frac{y}{2} - x) \), set \( \frac{dy}{dt} = 0 \). This gives two conditions: \( y = 0 \) or \( 1 - \frac{y}{2} - x = 0 \). The condition \( y = 0 \) represents a horizontal line at \( y = 0 \). The second condition can be expressed as \( x = 1 - \frac{y}{2} \), which is another line with a negative slope.

Step 3: Identifying Equilibrium Points

Equilibrium points occur where both \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \). From the nullclines, solve \( y = 1 - \frac{x}{3} \) and \( x = 1 - \frac{y}{2} \) simultaneously. Substituting \( y = 1 - \frac{x}{3} \) into \( x = 1 - \frac{y}{2} \), we find that \( x = \frac{3}{5} \) and \( y = \frac{4}{5} \). Therefore, the equilibrium point is \( (\frac{3}{5}, \frac{4}{5}) \).

Step 4: Sketching Nullclines and Equilibria

Draw the nullclines on the phase plane: a vertical line at \( x = 0 \), a horizontal line at \( y = 0 \), a diagonal line for \( y = 1 - \frac{x}{3} \), and another for \( x = 1 - \frac{y}{2} \). Mark the equilibrium point \( (\frac{3}{5}, \frac{4}{5}) \) on this graph.

Step 5: Determining Direction of Trajectories

Determine the sign of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) in each region divided by the nullclines. Above the nullcline \( y = 1 - \frac{x}{3} \), \( \frac{dx}{dt} < 0 \) and below it, \( \frac{dx}{dt} > 0 \). Similarly, to the right of \( x = 1 - \frac{y}{2} \), \( \frac{dy}{dt} < 0 \) and to the left, \( \frac{dy}{dt} > 0 \). Use these signs to sketch arrows indicating trajectory directions.

Key Concepts

Nullclines

In phase plane analysis, nullclines are curves where the rate of change of the system variables is zero. For \( \frac{dx}{dt} = 0 \), we have two cases:

• \( x = 0 \)
• \( y = 1 - \frac{x}{3} \)

The first is a vertical line, indicating that the change in \(x\) is zero regardless of \(y\)'s value. The second is a diagonal line with a negative slope. Here, \( dx/dt \) changes sign as \( (x,y) \) crosses this line.
Similarly, nullclines for \( \frac{dy}{dt} = 0 \) emerge from:

• \( y = 0 \)
• \( x = 1 - \frac{y}{2} \)

The horizontal line \( y = 0 \) means no change in \(y\) regardless of \(x\), while the other line slopes negatively and divides the plane into regions where \(dy/dt\) changes sign. Nullclines provide pathways along which trajectories touch but do not cross, forming distinct regions in the phase plane.

Equilibrium Points

Equilibrium points are the core of phase plane analysis. They are points where both system equations have zero rates of change, \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \). Solving the nullcline equations together, \( y = 1 - \frac{x}{3} \) and \( x = 1 - \frac{y}{2} \), helps identify these points.
Here, the solution reveals \( x = \frac{3}{5} \) and \( y = \frac{4}{5} \), giving a single equilibrium point at \( (\frac{3}{5}, \frac{4}{5}) \). At this point, there is no dynamic movement in the system, meaning trajectories converge here if conditions align.
Equilibrium points help us predict stable or unstable dynamics, acting as destinations or points of origin for different trajectories. Understanding their location and stability is crucial for studying system behavior over time.

Differential Equations

Differential equations describe how a dynamic system evolves over time. In our given system: \(\frac{dx}{dt}=x\left(1-y-\frac{x}{3}\right)\) and \(\frac{dy}{dt}=y\left(1-\frac{y}{2}-x\right)\),
these equations indicate how populations \(x\) and \(y\) interact. \( \frac{dx}{dt} \) refers to the change in population \(x\) dependent on its current size and \(y\). Conversely, \( \frac{dy}{dt} \) describes the same for \(y\) with an added competitive effect from \(x\).
By setting derivatives to zero, we find the nullclines and, subsequently, equilibrium points. Differential equations in this form, with variables multiplying each other, often signal predator-prey dynamics, competition, or cooperative interactions. Breaking these down into simpler components helps to predict long-term behavior and system interactions.

Direction of Trajectories

The direction of trajectories on the phase plane reveals dynamic pathways that solutions might follow. After identifying nullclines, understanding trajectory directions involves determining signs of \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) in each region they create.
For example, consider the line \( y = 1 - \frac{x}{3} \). Above this line, \( \frac{dx}{dt} \) is negative, meaning \(x\) is decreasing, while below, it is positive, signifying growth. Similarly, for \( x = 1 - \frac{y}{2} \), \( \frac{dy}{dt} \) is negative to the right and positive to the left.
By sketching arrows based on these signs, we can visualize how trajectories approach or move away from equilibrium points. These directions give critical insights into system behavior, stability of equilibrium points, and potential oscillations or cycles.