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Chapter 6: Problem 2

For the following, find all the equilibrium solutions. (a) \(\frac{d X}{d t}=3 X-2 X Y, \quad \frac{d Y}{d t}=X Y-Y\), (b) \(\frac{d X}{d t}=2 X-X Y, \quad \frac{d Y}{d t}=Y-X Y\), (c) \(\frac{d X}{d t}=Y-2 X Y\), \(\frac{d Y}{d t}=X Y-Y^{2}\)

Short Answer

Expert verified
Equilibrium solutions: (a) (0,0), (1,0), (1,3/2); (b) (0,0), (1,0), (1,2); (c) (0,0), (1/2,0), (1/2,1/2).

Step by step solution

01

Understanding Equilibrium Solutions

An equilibrium solution occurs when the system of differential equations is set to zero, meaning that the derivatives are zero. For the given systems, this implies finding values of \(X\) and \(Y\) at which the right-hand sides of all the equations equal zero simultaneously.
02

Solve Part (a) Equations

For (a), set \( \frac{dX}{dt} = 3X - 2XY = 0 \) and \( \frac{dY}{dt} = XY - Y = 0 \). Solving the first equation, \(X(3 - 2Y) = 0\), gives solutions \(X = 0\) or \(Y = \frac{3}{2}\). From the second equation, \(Y(X - 1) = 0\), implies \(Y = 0\) or \(X = 1\). Combining these gives equilibrium points: \((X, Y)\) is \((0, 0)\), \((1, 0)\), and \((1, \frac{3}{2})\).
03

Solve Part (b) Equations

For (b), set \( \frac{dX}{dt} = 2X - XY = 0 \) and \( \frac{dY}{dt} = Y - XY = 0 \). Solving the first equation, \(X(2 - Y) = 0\), gives solutions \(X = 0\) or \(Y = 2\). From the second equation, \(Y(1 - X) = 0\), implies \(Y = 0\) or \(X = 1\). Combining these gives equilibrium points: \((X, Y)\) is \((0, 0)\), \((1, 0)\), and \((1, 2)\).
04

Solve Part (c) Equations

For (c), set \( \frac{dX}{dt} = Y - 2XY = 0 \) and \( \frac{dY}{dt} = XY - Y^2 = 0 \). Solving the first equation, \(Y(1 - 2X) = 0\), gives solutions \(Y = 0\) or \(X = \frac{1}{2}\). From the second equation, \(Y(X - Y) = 0\), implies \(Y = 0\) or \(X = Y\). Combining these gives equilibrium points: \((X, Y)\) is \((0, 0)\), \((\frac{1}{2}, 0)\), and \((\frac{1}{2}, \frac{1}{2})\).

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Key Concepts

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

Differential Equations
Differential equations are crucial in mathematical modeling and describe how a system changes over time. They consist of equations that involve rates at which quantities change with respect to others. In this exercise, we have systems of two differential equations, each with respect to time, which describe interactions between variables such as \(X\) and \(Y\).

To understand equilibrium solutions, imagine a scenario where the system has reached a point where it no longer changes. Mathematically, this means setting the derivatives to zero, as the change rates must be zero for equilibrium. By doing this, we identify specific points where these conditions hold true.

Working with differential equations involves finding solutions that satisfy these equations under given conditions. It is common to encounter systems that describe real-world phenomena such as population dynamics, chemical reactions, or even economics.
Phase Plane Analysis
Phase plane analysis is a visual tool used in studying systems of differential equations. It helps us understand how the variables like \(X\) and \(Y\) evolve over time by plotting them against each other in a graph, known as the phase plane. This analysis makes it easier to visualize the trajectory of the system and how it moves from one state to another.

Equilibrium points play a crucial role in phase plane analysis. These are points where the system stays still, as described earlier. In our exercise, such points were calculated for different sets of equations, like \((0,0)\), \((1,0)\), etc. These points can be plotted on the phase plane.

By examining the phase plane around these equilibrium points, one can determine the nature of the equilibrium. This could mean studying whether trajectories in the plane converge to, diverge from, or loop around these points.
Stability Analysis
Stability analysis is an extension of phase plane analysis that tells us more about the nature of equilibrium points. After finding the equilibrium solutions, it's essential to understand their behavior. Do small changes near these points grow, fade, or remain constant over time?

There are several types of stability in such systems:
  • Stable Equilibrium: Small disturbances diminish over time, returning the system to equilibrium.
  • Unstable Equilibrium: Small disturbances grow, pushing the system further away from equilibrium.
  • Saddle Points: Disturbances behave differently in various directions, showing stability in one and instability in another.

To determine stability, we often evaluate the Jacobian matrix at an equilibrium point. The eigenvalues of this matrix give clues on the stability nature. For each equilibrium point in our exercise, conducting such an analysis will reveal whether we have a stable condition, or if the system tends towards instability.

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Most popular questions from this chapter

In a long-range battle, neither army can see the other soldiers, but fires into a known area. A simple mathematical model describing this battle is given by the coupled differential equations $$ \frac{d R}{d t}=-c_{1} R B, \quad \frac{d B}{d t}=-c_{2} R B, $$ where \(c_{1}\) and \(c_{2}\) are positive constants. (a) Use the chain rule to find a relationship between \(R\) and \(B\), given the initial numbers of soldiers for the two armies are \(r_{0}\) and \(b_{0}\), respectively. (b) Draw a sketch of typical phase-plane trajectories. (c) Explain how to estimate the parameter \(c_{1}\) given that the blue army fires into a region of area \(A\)

A population of sterile rabbits \(X(t)\) is preyed upon by a population of foxes \(Y(t) .\) A model for this population interaction is the pair of differential equations $$ \frac{d X}{d t}=-a X Y, \quad \frac{d Y}{d t}=b X Y-c Y, $$ where \(a, b\) and \(c\) are positive constants. (a) Use the chain rule to obtain a relationship between the density of foxes and the density of rabbits. (b) Sketch typical phase-plane trajectories, indicating the direction of movement along the trajectories. (c) According to the model, is it possible for the foxes to completely wipe out the rabbit population? Give reasons.

Adapted from Brauer and CastilloCh脿vez (2001). For the standard SIR epidemic model, $$ S^{\prime}=-\beta S I, \quad I^{\prime}=\beta S I-\gamma I $$ From the equation for the phase-plane solution, $$ I=-S+\frac{\gamma}{\beta} \ln (S)+K $$ (a) If \(s_{f}\) denotes the remaining number of susceptibles when there are no remaining infectives, show that $$ \frac{\beta}{\gamma}=\frac{\ell \mathrm{n}\left(s_{0} / S_{f}\right)}{S_{0}+i_{0}-s_{f}} $$ where \(s_{0}\) and \(i_{0}\) are the initial numbers of susceptibles and infectives, respectively. (b) A study at Yale university in 1982 described an influenza epidemic with initial proportions of susceptibles of the student population as \(91.1 \%\) and final proportion of susceptibles as \(51.3 \%\). (Assume, initially, that no one had recovered). Given that the mean infectious period for influenza \(\gamma^{-1}\) is approximately 3 days, estimate the combination \(\beta N\), where \(N\) is the total population size, and hence estimate \(R_{0}\), the basic reproduction number.

Consider the aimed fire battle model developed in the text: $$ \frac{d R}{d t}=-a_{1} B, \quad \frac{d B}{d t}=-a_{2} R . $$ The exact solution can be found using theoretical techniques as follows: (a) Take the derivative of the first equation to get a second-order differential equation, and then eliminate \(d B / d t\) from this equation by substituting the second equation (given above) into this second-order equation. (b) Now assume the solution to be an exponential of the form \(e^{\lambda t} .\) Substitute it into the secondorder equation and solve for the two possible values of \(\lambda\). The general solution for \(R\) will be of the form $$ R(t)=c_{1} e^{\lambda_{1}}+c_{2} e^{\lambda_{2}} $$ where \(c_{1}\) and \(c_{2}\) are the arbitrary constants of integration. The solution for \(B\) is then found using the equation \(d R / d t=-a_{1} B\). Write the solutions in terms of hyperbolic functions cosh and sinh (this makes it more convenient to solve for the arbitrary constants). (c) Now find the arbitrary constants by applying the initial conditions \(R(0)=r_{0}\) and \(B(0)=b_{0}\), when \(t=0\). (d) Using Maple or MATLAB (with symbolic toolbox), check the solution above. Note: The software may give the solution in terms of exponential functions, in which case you will need to convert to check. (Note: Further details about methods for solving second-order differential equations, in particular for differential equations with constant coefficients, as used here, can be found in Appendix A.5.)

Consider the competition population model with density-dependent growth $$ \frac{d X}{d t}=X\left(\beta_{1}-c_{1} Y-d_{1} X\right), \quad \frac{d Y}{d t}=Y\left(\beta_{2}-c_{2} X-d_{2} Y\right) $$ Find all four equilibrium points for the system (either using Maple or MATLAB, or, if you like algebra, by hand).
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