Problem 18 - 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. $$\begin{aligned} &\frac{d x}{d t}=x\left(1-\frac{x}{2}-y\right)\\\ &\frac{d y}{d t}=y\left(1-\frac{y}{3}-x\right) \end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: 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. \( 1 - \frac{y}{3} - x = 0 \Rightarrow x = 1 - \frac{y}{3} \).

Step 2: 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 \((\frac{3}{4}, \frac{5}{8})\).

Step 3: 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.

Step 4: 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.

Key Concepts

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.