31.5 Systems of Differential Equations 1031 Where dx dt = 0but dy dt = 0, only y is changing with t so the trajectory’s tangent line is vertical. The sign of dy dt indicates how y changes with t; we display this information by orienting the vertical tangents up or down. Where dy dt = 0but dx dt = 0, only x is changing with t so the trajectory’s tangent line is horizontal. The sign of dx dt indicates how x changes with t; we display this information by orienting the horizontal tangents right or left. Where dx dt = 0 and dy dt = 0 simultaneously, the system is at equilibrium. Equilibrium points are the points of intersection of nullclines for which dx dt = 0 and those for which dy dt = 0. The nullclines partition the phase plane into regions in which neither dx dt nor dy dt changes sign. In each region we’ll have one of the following cases: dx dt > 0, dy dt > 0 as t increases, both x and y increase dx dt > 0, dy dt < 0 as t increases, x increases and y decreases dx dt < 0, dy dt > 0 as t increases, x decreases and y increases dx dt < 0, dy dt < 0 as t increases, both x and y decrease Sketch possible trajectories using the information gathered. Fact The limit point of a trajectory must be an equilibrium point. By this we mean the following. Suppose lim t→∞ x(t) and lim t→∞ y(t) both exist and are finite. Denote the limits by A and B, respectively. Then (A, B) must be an equilibrium point. (Verify that this is indeed the case in Examples 31.24 and 31.25.) Note the analogy to the case of solutions to autonomous differential equations. If y 1 (t) is a solution to an autonomous equation and lim t→∞ y 1 (t) = L, where L is finite, then y(t) = L is an equilibrium solution. The slope of a trajectory at any point is dy dx = dy/dt dx/dt = g(x,y) f(x,y) evaluated at that point (provided f(x,y) = 0). Sometimes explicitly solving for a relationship between x and y is enlightening. ◆ EXAMPLE 31.26 Analyze the following system of differential equations, sketching solution trajectories in the xy-plane. dx dt = y dy dt =−x SOLUTION We begin by finding the nullclines. The trajectories are horizontal where dy dt =−x=0,i.e., along the y-axis. The trajectories are vertical where dx dt = y = 0, i.e., along the x-axis. We can figure out the direction in which the trajectories are traveled along the nullclines by looking back at the original equations. For example, dx dt = y, so on the part of the y-axis with y>0weknow dx dt > 0; the trajectories are traveled from left to right, x increasing with t. On the section of the y-axis with y<0, dx dt < 0 so the trajectories are traveled from 1032 CHAPTER 31 Differential Equations right to left, x decreasing with t. Vertical and horizontal tangents are oriented as shown in Figure 31.23. The only equilibrium point is the origin. y x II I III IV Figure 31.23 The nullclines partition the plane into the four regions labeled. In each region we look at the signs of dy dt and dx dt to determine the basic direction of the trajectory. Region I: dx dt = y>0 x increases , dy dt =−x<0 y decreases Region III: dx dt = y<0 x decreases , dy dt =−x>0 y increases Region II: dx dt = y>0 x increases , dy dt =−x>0 y increases Region IV: dx dt = y<0 x decreases , dy dt =−x<0 y decreases This information is collected in Figure 31.24. y x II I III IV Figure 31.24 At this point we have some idea of what the trajectories look like. However, we are not sure, for instance, if they form closed circles, or ellipses, or if they spiral in or spiral out. We can look at dy dx and try to solve the resulting differential equation to uncover the shapes of the trajectories. dy dx = dy dt dx dt = −x y , y = 0 so 31.5 Systems of Differential Equations 1033 dy dx = −x y . Separate variables and solve. ydy=−xdx ydy= −xdx y 2 2 =− x 2 2 + C 1 ⇒ x 2 + y 2 = C Therefore, the trajectories in the xy-plane are circles centered at the origin; the picture in the phase plane is given in Figure 31.25. y x Figure 31.25 Through every point in the xy-plane there is one trajectory passing through that point. Each trajectory is a circle, except for the origin, which can be thought of as a degenerate circle with radius 0. As t increases, the point (x(t), y(t)) traces out a circle in the clockwise direction. ◆ REMARK The system of equations in Example 31.26 can arise when modeling the vibrations of a frictionless spring. Consider a spring and attached block positioned as shown in Figure 31.26. Let x give the position of the block (of negligible weight) attached to the end of the spring. Suppose we stretch the spring by pulling the block from its original position (x = 0) out to x 0 and release the block. x 0 x = 0 Figure 31.26 The block will oscillate back and forth about x = 0 and, in the absence of friction, will repeat its motion ad infinitum. We expect the graph of x versus t to look like Figure 31.27 on the following page. 1034 CHAPTER 31 Differential Equations x 0 x t Figure 31.27 Let y = the velocity of the block. Then y = dx dt ; velocity is the derivative of position with respect to time. By Newton’s second law we know that force = (mass) · (acceleration). We can denote the mass of the block by m and its acceleration by d 2 x dt 2 or dy dt . Hooke’slaw,an experimentally derived law, tells us that the force exerted by the spring is proportional to the displacement from its equilibrium length. Putting the two laws together gives −∝x=m dy dt or dy dt =−kx for some k>0.Acceleration, dy dt , is proportional to x but opposite in sign. The proportionality constant is determined by the mass of the block and the nature of the spring. (Think through some special cases to assure yourself that the equation dy dt =−kx makes sense in terms of the spring.) The frictionless block and spring system can therefore be modeled by the system of differential equations dx dt = y dy dt =−kx. For convenience let’s assume that k = 1. Then the equations describing the motion become dx dt = y and dy dt =−x. Look back at the phase-plane diagram (Figure 31.25) to see how the motion of the spring is reflected in this picture. If we pull the block out a distance x 0 and simply let it go, giving it no initial velocity, we should look at the point (x 0 ,0)in the phase plane. Follow the trajectory around in the direction of the arrow and note how the position of the block (the x-coordinate) and the velocity of the block (the y-coordinate) change with time. EXERCISE 31.8 It is unrealistic to ignore friction. Let x(t) be the position of the block. x(0) = x 0 x (0) = 0 (a) Sketch a possible graph of x versus t for the spring example, considering the force of friction and assuming the block vibrates back and forth several times. (b) Now, letting y = dx dt = velocity, sketch the trajectory in the xy-plane. (In the next section we will deal analytically with the friction issue.) The answer to Exercise 31.8 appears at the end of the section. Returning to Epidemic Model B: A Contagious, Fatal Disease ◆ EXAMPLE 31.27 Consider the epidemic model that was used to motivate this discussion. Residing in a large susceptible population is a small group of people with a fatal infectious disease. An infected 31.5 Systems of Differential Equations 1035 individual can immediately infect others, and any person not infected is susceptible. We make the assumption that the population remains fixed during the time interval in question except for deaths due to the disease. I (t ) = the number of infected people at time t, and S(t) = the number of susceptible people at time t. dS dt =−rSI, where r>0 As the susceptible class loses members, the infected class gains members but it also loses people due to death from the disease. dI dt = rSI − kI, where r, k>0 Analyze the system of equations dS dt =−rSI dI dt = I (rS − k) r, k>0, sketching solution trajectories in the SI-plane. SOLUTION Look for the nullclines and equilibrium points. Let’s plot S on the horizontal axis and I on the vertical axis. Then the slopes of the trajectories are given by dI dS = dI/dt dS/dt . The trajectories are horizontal where dI dt = I (rS − k) = 0, that is, for I = 0orS= k r . The trajectories are vertical where dS dt =−rSI = 0, that is, for S = 0orI=0. The system is at equilibrium if and only if dI dt = 0 and dS dt = 0 simultaneously. S cannot simultaneously be 0 and k r , so the equilibria for the system are at I = 0. Every point on the S-axis is an equilibrium point. The nullclines are S = 0, I = 0, and S = k r . These partition the first quadrant into two regions, I and II, as indicated in Figure 31.28. I S II I k r Figure 31.28 Region I Region II sign of dS dt : −− sign of dI dt : +− Weknow that the solution curves go in the direction of the arrows drawn in Figure 31.28. Further information about the shape of the trajectories in the SI-plane can be obtained by looking at dI dS . dI dS = dI/dt dS/dt = I (rS − k) −rSI = rS − k −rS =−1+ k rS . 1036 CHAPTER 31 Differential Equations The slope is a function of S; the trajectories are vertical translates. (As an exercise, solve for I in terms of S.) The second derivative of I with respect to S is given by d 2 I dS 2 =− k rS 2 , where r and k are positive constants. d 2 I dS 2 < 0, so the trajectories are concave down. See Figure 31.29. (S 0 , I 0 ) I S k r Figure 31.29 REMARKS Observe that if S 0 < k r , then I immediately decreases toward zero; the disease leaves the population without becoming an epidemic. On the other hand, if S 0 > k r , then the disease will spread and the number of infected people will increase until S drops to k r . Only when S falls below the threshold value of k r does I begin to decrease. The disease does not die out completely for lack of a susceptible population but rather for lack of infected people. Generally, some individuals will remain who have not caught the disease. ◆ The same system of differential equations that were set up in the previous example can also be used to model an epidemic such as measles or chicken pox, where instead of the disease being fatal, sickness confers immunity upon those who recover. Once an individual is infected he cannot become reinfected, so after leaving the “infected” class he does not rejoin the “susceptible” class but belongs to a “recovered” class. ◆ EXAMPLE 31.28 In an isolated community of 800 susceptible children, one child is diagnosed with chicken pox. Suppose the spread of the disease can be modeled by the system of differential equations dS dt =−0.001SI dI dt = 0.001SI − 0.3I . (a) What is the maximum number of children sick at any one moment? (b) According to our model, how many susceptible children will avoid getting chicken pox while the epidemic runs its course? SOLUTION (a) From the qualitative analysis done in Example 31.27 we know that I is maximum when S = k r , that is, when dI dS = 0. So I is maximum at S = 0.3 .001 = 300. To find I when S = 300 we need an explicit relationship between I and S. I = 800 − S because some infected children have already recovered. dI dS = dI/dt dS/dt = 0.001SI − 0.3I −0.001SI =−1+ 0.3 0.001S =−1+ 300 S 31.5 Systems of Differential Equations 1037 Separate variables and integrate to obtain I as a function of S. dI = −1 + 300 S dS dI = (−1)dS + 300 dS S I =−S+300 ln S + C, S>0, so |S|=S. Use the initial conditions, I = 1 and S = 800 when t = 0, to solve for C. 1 =−800 + 300 ln(800) + C,soC=801 − 300 ln(800) ≈−1204.38 Then I (S) ≈−S+300 ln S − 1204, and I(300) ≈−300 + 300 ln 300 − 1204 ≈ 207. There were at most about 207 children sick at one time. This is just over a quarter of the susceptible population. (b) As pointed out in Example 31.27, the epidemic ends not for lack of susceptible children but for lack of infected children. Therefore, we set I = 0 and solve for S. Because the equation involves both S and ln S we can only approximate the solution. Using a calculator, computer, or numerical methods, we find that S ≈ 70. At the end of the epidemic about 70 children will still be susceptible to chicken pox; 730 children will have caught the disease. ◆ In some sense, in the previous two examples it is the ratio k r that governs the course of the epidemic. The likelihood of the disease being passed from one individual to the next is reflected in r. For a disease such as chicken pox, communities sometimes attempt to raise the value of r (to confer immunity before adulthood) whereas for a fatal infectious disease, like AIDs, communities sometimes educate to lower the value of r. When dangerous epidemics break out in livestock populations, farmers and ranchers often remove sick animals, effectively raising the value of k, in order to end up with more uninfected animals. Recurrent Epidemics Many diseases recur in various populations with some regularity. For instance, in London in the early 1900s, measles epidemics recurred approximately every two years. In 1929, when the mathematical biologist H. E. Soper was attempting to model this cyclic measles epidemic, he dropped the assumption that the population remains fixed for the time interval being observed. Instead he assumed that the population of susceptibles grows at a constant rate µ and arrived at this system of differential equations: dS dt =−rSI + µ dI dt = rSI − kI for r, k, and µ positive constants In fact this system of equations does not predict recurrent outbreaks of the epidemic. Instead it predicts that the disease will reach a steady state level since the cycles are heavily damped. The problem with this model is the assumption that the susceptible population grows at a constant rate. Try to modify this system of equations. If you assume the population grows at 1038 CHAPTER 31 Differential Equations a rate proportional to itself, then the solutions to the equations are cyclic and undamped, as desired. Think about the repeating cycle of measles epidemics again after working through Example 31.29. Modeling Population Interactions The more closely we look at the world the more clearly we see its interconnected nature. We can use the tools we’ve developed in this section to model interactions between distinct populations. We can model symbiotic interactions, such as the interaction between sea anemones and clown fish, or competitive interactions, such as the interaction between lion and hyena populations competing for small prey. We can model predator-prey interactions, such as that between hyenas and the Thomson’s gazelle, lions and water buffalo, or cats and mice. In the examples in this section we’ll simplify our models to focus on just two populations. ◆ EXAMPLE 31.29 Consider the predator-prey relationship between hyenas and the Thomson’s gazelle, a small gazelle native to Africa. Let’s make the following simplifying assumptions. i. Assume hyenas are the gazelles’ major predator and that in the hyenas’ absence the gazelle population would grow exponentially. ii. Assume gazelles are the major food source for hyenas; with no gazelles the hyenas would die off. Model this interaction with a system of differential equations. SOLUTION Let h(t) = the number of hundreds of hyenas at time t. Let g(t) = the number of hundreds of gazelles at time t. We can model the interaction by a system of differential equations of the form dh dt =−k 1 h+k 2 hg dg dt = k 3 g − k 4 hg where k 1 , k 2 , k 3 , and k 4 are positive constants. Just as the rate of transmission of disease is proportional to interactions between the infected and the susceptible, which is in turn proportional to the product of their numbers, so too is the rate of nourishing/fatal interaction between hyena and gazelle proportional to the product of their population sizes. Observe that if h = 0 then dg dt = k 3 g and if g = 0 then dh dt =−k 1 h.For the sake of concreteness, we’ll work with the values of k 1 , k 2 , k 3 , and k 4 given below and analyze solutions in the gh-plane. dh dt =−0.3h + 0.1gh = 0.1h(−3 + g) dg dt =+0.4g − 0.4gh = 0.4g(1 − h) Nullclines: dh dt = 0 when h = 0org=3. In the gh-plane this is where trajectories have horizontal tangent lines. dg dt = 0 when g = 0orh=1. In the gh-plane this is where trajectories have vertical tangent lines. Equilibrium points: dh dt = 0ath=0org=3, so at an equilibrium point either h = 0or g=3. 31.5 Systems of Differential Equations 1039 Suppose h = 0. Then in order for dg dt to be zero we must have g = 0. (0, 0) is an equilibrium point. Suppose g = 3. Then in order for dg dt to be zero we must have h = 1. The point (3, 1) is an equilibrium point. Use the differential equations to orient the vertical and horizontal tangents. For instance, when g = 3 we know that dg dt > 0 for 0 <h<1and dg dt < 0 for h>1. h g 1 3 Figure 31.30 The h and g axes are oriented as shown in Figure 31.30. The nullclines divide the relevant first quadrant region into four subregions. The direction of trajectories in each region is indicated in Figure 31.30. From Figure 31.30 we see that for g(0)>0and h(0)>0the trajectories either spiral in toward (3, 1) or spiral outward, or are closed curves. From the slope field it appears that the curves are closed. Looking at dh dg = 0.1h(−3+0.1g) 0.4g(1−h) enables us to distinguish between these options. Separating variables, we find that ln(h) − h =−3ln(g) + g + C. Suppose we start at the point (5, 1). We can solve for C and show that the trajectory through (5, 1) intersects the line g = 3 exactly twice, once for h>1and once for h<1. Similarly we can show that it intersects the line h = 1 exactly twice, once for g<3and once for g>3. The trajectory through (5, 1) is a closed curve. In fact, all the trajectories in the first quadrant are closed. g h 3 1 Figure 31.31 Our model predicts that the hyena and gazelle populations will oscillate cyclically. When there are few gazelle the hyena population decreases due to lack of food. The decrease 1040 CHAPTER 31 Differential Equations in the number of hyenas allows the gazelle population to thrive, but as the gazelle population increases the hyenas’ food source is replenished, allowing the hyena population to flourish. This flourishing takes its toll on the gazelles, and the cycle repeats. ◆ To the extent that a model reflects observed population dynamics, it is a good model. Models that don’treflect observed behavior must be modified. There are various ways this predator-prey model can be modified. For instance, there may be competition among gazelle for limited grazing land so that in the absence of hyena the gazelle population exhibits logistic growth. This can be reflected in the system of differential equations by inserting a −k 5 g 2 term as shown. dh dt =−k 1 h+k 2 hg dg dt = k 3 g − k 4 hg − k 5 g 2 , where the −k 5 g 2 term (k 5 > 0) reflects competition among gazelles. The predator-prey system of differential equations given in Example 31.29 is sometimes referred to as Volterra’s model after the Italian mathematician Vito Volterra (1860-1940) who was encouraged to analyze the predator-prey relationship between sharks and the fish they prey upon by his son-in-law, the biologist Humberto D’Ancona. 10 By inserting a term in each equation to account for fishing, Volterra was able to explain why fishing conducted in the Adriatic Sea was raising the average number of prey and lowering the average number of predators over any cycle. Similar analysis has been successfully used to analyze unexpected results of introducing DDT into a predator-prey system, leaving predator populations lowered and prey populations elevated. Competition between species can be modeled by differential equations of the form dx dt = k 1 x − k 2 xy dy dt = k 4 y − k 5 xy or dx dt = k 1 x − k 2 xy − k 3 x 2 dy dt = k 4 y − k 5 xy − k 6 y 2 where k 1 , k 2 , ,k 6 are positive constants. The latter set of differential equations takes into account competition between members of the same species in addition to competition between species. Answers to Selected Exercises Answers to Exercise 31.8 x x y(b)(a) x 0 x 0 t 10 For more information on a very interesting story see Martin Braun, Differential Equations and their Applications, 2nd ed. New York. Springer Verlag, 1978 or G. F. Gause, The Struggle for Existence, New York. Hafner, 1964, or Umberto D’Ancona, The Struggle for Existence, Leiden, Brill, 1954. . the analogy to the case of solutions to autonomous differential equations. If y 1 (t) is a solution to an autonomous equation and lim t→∞ y 1 (t) = L, where L is finite, then y(t) = L is an equilibrium. infected and the susceptible, which is in turn proportional to the product of their numbers, so too is the rate of nourishing/fatal interaction between hyena and gazelle proportional to the product of. trajectories are vertical where dS dt =−rSI = 0, that is, for S = 0orI=0. The system is at equilibrium if and only if dI dt = 0 and dS dt = 0 simultaneously. S cannot simultaneously be 0 and k r ,