1 competing species Math 1280 Notes 4 Last section revised, 1/31, 9:30 pm. This section and the next deal with the subject of population biology. You will already have seen examples of this. Most calculus books include the topic of exponential growth. Suppose that x (t) denotes the number of bacteria in a petri dish where there is plenty of food and plenty of space. Then the rate of growth of x is assumed to be proportional to x: This gives the equation x 0 = kx; where k is a positive constant. If the initial population is P; then x (t) = P e kt : Later on, at least by ode I, you improve the model by recognizing that there are only limited food and space in that petri dish, so some term is need to re ect this. The most common ode including this e ect is x 0 = kx (b x) ; for some positive b denoting the maximum population the environment can support. This equation can be solved, or we can look at the phase line, that is, the x -axis, considering x (t) as a point on that axis. If 0 < x < b; then x 0 > 0; which we can show graphically with an arrow to the right in that interval. If x > b; then the arrow is to the left. Thus, if the initial population is positive, whether above or below the carrying capacity b; we see that lim t!1 x (t) = b: This can also be seen from the formula for the solution. In section 9.4 it is assumed that there are two species in the environment, denoting their population at time t by x (t) and y (t). They are assumed to eat the same food. The ode for x now includes a term showing that the presence of y decreases the amount of food available for x; and so decreases the carrying capacity. The ode for x is now x 0 = kx (b x cy) ; (1) for some constant c: Similarly, the population y (t) satis es y 0 = ry (d y ex) ; (2) 1
where some other constants have been introduced. The resulting system of two equations is called a model for competing species. (Di erent letters are used in the text for the various constants.) The system (1)-(2) has a number of di erent phase planes, depending on the constants. In class we will discuss several possibilities. 2 Bifurcation 2.1 bifurcation for one di erential equation The text uses the exercises in 9.4 to explore an important idea in applied mathematics, that of bifurcation. This was rst introduced in the exercises in section 2.5. We will discuss one of these here and assign another as homework. Then we will turn to related exercises in 9.4. In exercise 25 of section 2.5 the equation considered is y 0 = a y 2 : (3) Observe that we have only one equation, not two as in chapter 9. The quantity a is a parameter, and it is easy to see that the solutions are very di erent according to whether a < 0 or a > 0: One way this can be seen is by solving the equation using separation of variables. We will do this for a = 1 and a = 1: For a = 1 we get dy dx = 1 y2 ; Z 1 1 y 2 dy = Z 1dx I won t go through the details. The result is : y + 1 ln = 2 (x + c) : y 1 We can solve for y and get y = 1 + ke2x ke 2x 1 ; assuming that the denominator is not zero. 2
On the other hand, for a = and the integral is very di erent: 1 we get Z 1 dy = x + c; 1 + y2 arctan y = x c y = tan (x + c) : Notice that this solution only exists on an interval of length, no matter what c is chosen. It is clear that the transition between an exponential solution and a trig solution occurs at a = 0: We can see this clearly if we draw a phase line, which will be done in class. But notice that if a < 0 then there are no equilibrium points, while if a > 0 there are equilibrium points at y = p a and y = p a. This appearance of two new equilibrium points as a crosses from below 0 to above 0 is called bifurcation, and a = 0 is called a bifurcation point. 2.2 Two di erential equations Bifurcation can be more complicated for a system of two rst order ode s. We will discuss an example di erent from those in the text. In this chapter, the parameter is denoted by instead of by a: Consider the system x 0 = y y 0 = x x 3 : (4) Again we look for equilibrium points. First suppose = 1: We want y = 0 x 1 x 2 = 0: Clearly there is only one equilibrium point, (x 0 ; y 0 ) = (0; 0) : If we linearize around this point (with = 1); we get a familiar system: u 0 = v v 0 = u: (Just drop the x 3 term from (4) : ) We should know by now that (0; 0) is a center for the linearized system, and either a center or a spiral for (4) : I ll leave it to the homework for you to gure out which. 3
Now consider = 1 : Equilibrium points: x 0 = y y 0 = x x 3 : y = 0 x 1 x 2 = 0; so we get (0; 0) ; (1; 0), and ( 1; 0). I will not take time to discuss the linearized systems here. But I do want to discuss what is called a bifurcation diagram. This is a graph in which the horizontal axis is ; the parameter, and the vertical axis is x 0 ; the x coordinate of any equilibrium points. This is just the graph of the equation x x 3 = 0 in the (; x) plane. Writing this as x ( x 2 ) = 0; we see that either x = 0; which is the axis, or = x 2 ; which is a parabola on its side. We plot them both on the same axes: Perhaps you can see why this is called a pitchfork bifurcation. 3 Predator-Prey models The next section is about a related but di erent population model, the so-called predator-prey equations. In this model there are also two species, but x eats grass and y eats x. There is assumed to be an unlimited amount of grass, so if there are no y 0 s around, x grows according to x 0 = ax: 4
Meanwhile, y will starve if there are no x s around, and so the y population will decrease, according to an equation y 0 = cy: But if both predator and prey are present, then each is a ected in an obvious way, leading to equations x 0 = x (a y 0 = y(dx by) c): Again the text uses di erent letters. This time, there is essentially only one picture, so long as a; b; c; d are all positive. First, there are only two equilibrium points, (0; 0) and c d ; a b. To go further, let s deal with a speci c case: x 0 = x (1 y) y 0 = y (2x 4) : (5) The equilibrium points are (0; 0) and (2; 1). around (0; 0) is u 0 = u v 0 = 4v: It is easily seen that the linearization So (0; 0) is a saddle point. Around (2; 1) it turns out the and the relevant matrix, A = u 0 = 2v v 0 = 2u; 0 2 2 0 has eigenvalues 2i: So (0; 0) is a center for the linearized system. We ask if (2; 1) is a center for the nonlinear system. As before, we can answer this if we can nd a function E (x; y) which is constant along trajectories, or on the other hand, such a function which is monotone along trajectories. To look for such an E we write the equations as dy dx = y (2x 4) x (1 y) : 5
This gives Z 1 y dy = y Integrating, we get Z 2x 4 dx x ln y y = 2x 4 ln x + c: (6) This is quite di erent from anything we got before. We know that (2; 1) is either a center or a saddle, and so solutions go around it, either coming back to where they started, or spiraling in some way. This can also be seen by drawing nullclines for (5). In that way, we see that if a solution starts at, say 2; 2 1 ; then it moves to the right, crosses the line y = 1, moves to the left and crosses the lines x = 2 and y = 1; and returns to the line x = 2 with 0 < y 1: We want to know more about where it intersects that line. Does it intersect where it started? To see this, we look at d (ln y y) = 1 1: This is less than zero if y > 1 and dy y greater than zero if y < 1. This tells us that there are only two possible values of y on the trajectory of a solution, one with 0 < y 1 and one with y > 1: This means the solution (x (t) ; y (t)) must come back to the place it started, and this shows that (2; 1) is a center for (5) : We will discuss this further in class. 4 Homework pg. section 9.4, # 2, 9 (Explain your result in part (b) by describing what changes in the phase plane.) section 2.5, # 26. (pg. 93 in both editions) Problem 4: This is a modi cation of problem 17 of section 9.4 in the 9th edition. There is no equivalent problem in the 8th edition. Consider the system x 0 = x (1 x y) y 0 1 = y y 2 x : (a) Find the equilibrium points as functions of. Assume that > 0: (b) Determine the bifurcation points. These are any positive values of where either the number of equilibrium points changes, or the type of some equilibrium point, e.g. from saddle to node. Sometimes both of these may change at the same value of. You will have to nd the eigenvalues at each equilibrium point, as 6
functions of ; to do this part, but don t bother with this if the equilibrium point is not in the region x 0 and y 0. (c) Use Pplane to plot phase planes for enough values of to include an example of each kind of phase plane except those which occur exactly at a bifurcation point. Include the nullclines in your graph. This are found from a choice under the "solutions" tab. Be sure to indicate the relevant equilibrium points. You can simply mark these on your Pplane plots if you wish. 7