Question

In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=2 x-x y, y^{\prime}(t)=-y+x y$$

Short answer

The prey equation is \(x^{\prime}(t)=2 x-x y\), and the predator equation is \(y^{\prime}(t)=-y+x y\). b. What are the lines along which \(x^{\prime}(t)=0\) and \(y^{\prime}(t)=0\)? The lines are \(x = 0, y = 2\) for \(x^{\prime}(t)=0\) and \(y = 0, x = 1\) for \(y^{\prime}(t)=0\). c. What are the equilibrium points for the system? The equilibrium points are \((0, 0), (1, 0), (0, 2),\) and \((1, 2)\). d. In the first quadrant of the xy-plane, identify the four regions where \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. 1. Region I: \(x<1\) and \(y<2\), \(x^{\prime} > 0\) and \(y^{\prime} > 0\). 2. Region II: \(x>1\) and \(y<2\), \(x^{\prime} < 0\) and \(y^{\prime} < 0\). 3. Region III: \(x<1\) and \(y>2\), \(x^{\prime} < 0\) and \(y^{\prime} < 0\). 4. Region IV: \(x>1\) and \(y>2\), \(x^{\prime} > 0\) and \(y^{\prime} > 0\). e. How does the representative solution curve in the xy-plane look, and what is the direction in which the solution evolves? The representative solution curve is a closed loop around the equilibrium point \((1, 2)\), moving in a counterclockwise direction.

Step-by-step solution

Identifying predator and prey equations

We have two equations given: $$x^{\prime}(t)=2 x-x y \text{ and } y^{\prime}(t)=-y+x y$$ Since the prey typically grows faster in the absence of a predator, we can say that \(x^{\prime}(t)=2 x-x y\) corresponds to the prey equation (where x is the prey), and \(y^{\prime}(t)=-y+x y\) corresponds to the predator equation (where y is the predator).

Finding when \(x^{\prime}(t) = 0\) and \(y^{\prime}(t) = 0\)

Now, we will find the lines along which \(x'(t) = 0\) and \(y'(t) = 0.\) For \(x^{\prime}(t)=0,\) we have: $$2x - xy = 0 \Rightarrow x(2 - y) = 0$$ There are two lines: \(x = 0\) and \(y = 2.\) For \(y^{\prime}(t)=0,\) we have: $$-y + xy = 0 \Rightarrow y(1 - x) = 0$$ There are two lines: \(y = 0\) and \(x = 1.\)

Finding equilibrium points

Equilibrium points occur when both \(x^{\prime}(t) = 0\) and \(y^{\prime}(t) = 0.\) We can find these points by solving these equations simultaneously: $$x(2 - y) = 0 \text{ and } y(1 - x) = 0$$ The equilibrium points are \((0, 0),\) \((1, 0),\) \((0, 2),\) and \((1, 2).\)

Determining regions where \(x^{\prime}\) and \(y^{\prime}\) are positive or negative

We will examine the signs of \(x^{\prime}\) and \(y^{\prime}\) in the four regions formed by the lines \(x = 1\) and \(y = 2\) in the first quadrant of the xy-plane: 1. Region I: \(x<1\) and \(y<2.\) In this region, \(x^{\prime} = 2x - xy > 0\) and \(y^{\prime} = (1-x)y > 0.\) 2. Region II: \(x>1\) and \(y<2.\) In this region, \(x^{\prime} = 2x - xy < 0\) and \(y^{\prime} = (1-x)y < 0.\) 3. Region III: \(x<1\) and \(y>2.\) In this region, \(x^{\prime} = 2x - xy < 0\) and \(y^{\prime} = (1-x)y < 0.\) 4. Region IV: \(x>1\) and \(y>2.\) In this region, \(x^{\prime} = 2x - xy > 0\) and \(y^{\prime} = (1-x)y > 0.\)

Sketching the solution curve and indicating direction

To get an idea of the solution curve in the xy-plane and the direction, we can set up an xy-plane with the equilibrium points and divide the plane into four regions using the lines \(x=1\) and \(y=2\). Next, we can indicate the direction of the prey and predator populations growth in each region using arrows or vector fields. A representative solution curve would typically follow a closed loop around the interior equilibrium point \((1, 2).\) The solution starts from an initial population and moves in a counterclockwise direction throughout the xy-plane.

Key concepts

Equilibrium Points in Predator-Prey Models

In the study of predator-prey models, equilibrium points serve as critical references for understanding population dynamics. These points represent the states where the populations of both predators and prey remain constant over time, meaning neither the prey’s nor the predator’s population is changing. Formally, equilibrium points are those solutions to the system where the derivatives, or the rates of change, equal zero.

To find the equilibrium points in our given differential equations, we set both \(x'(t)\) and \(y'(t)\) to zero and solve for \(x\) and \(y\). This results in certain combinations of \(x\) and \(y\) that are particularly important as they may represent stable states (where the system tends to return to) or unstable states (where the system tends to move away from).

In the exercise, we ascertain the equilibrium points by solving the equations simultaneously, obtaining points such as \(0, 0)\), \(1, 0)\), \(0, 2)\), and \(1, 2)\). Each of these points can inform us about the conditions necessary for balance or imbalance between the predator and prey populations. For instance, the point \(0, 0)\) suggests the absence of both species, whereas \(1, 2)\) implies a certain stability with sustained populations of both.

Differential Equations and Population Dynamics

Differential equations are a powerful tool in mathematically modeling the interactions between predator and prey populations. These equations take into account the rate of change in populations over time. In ecological models like the one in our exercise, \(x'(t) = 2x - xy\) reflects the growth rate of the prey, and \(y'(t) = -y + xy\) reflects the growth rate of the predator.

Understanding these equations requires recognizing that the prey population, represented by \(x\), would grow exponentially without predators (due to the \(2x\) term). The \(xy\) term indicates the negative impact the predator population (represented by \(y\)) has on the prey. Similarly, the predator population decreases in the absence of prey (due to the \( -y\)) but increases with the availability of prey (the \(xy\) term).

Solving these equations doesn't yield a simple formula but rather leads to the construction of phase plane portraits that visually represent interactions, stability, and cyclical nature of the predator-prey relationship.

Phase Plane Analysis in Predator-Prey Models

Phase plane analysis is the visual component of studying predator-prey systems. By graphing both populations on an xy-plane, we can depict the trajectories of the populations over time. These trajectories, called solution curves, show how the predator and prey populations evolve.

In our example, after finding equilibrium points and determining where the rates of change are positive or negative in different regions, we can sketch the trajectories to gain insights into the system's behavior. These sketches are qualitative but reveal a lot about the system, such as stability and possible oscillatory patterns.

The direction field, or arrows showing the direction of population changes, illustrates how populations grow or shrink in each region of the plane. For instance, near the equilibrium point \(1, 2)\), we might observe a closed loop, indicating a cyclical dynamic where the populations of predators and prey fluctuate but return to this point, suggesting a stable coexistence under ideal conditions.