Lecture 5: Equilibrium solutions and stability. 1 Mathematical models of population dynamics

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1 Lecture 5: Equilibrium solutions and stability 1 Mathematical models of population namics The stu of populations namics is a big application of differential equations that we have been waiting to discuss until now. Let y(t) be the size of population of a species at time t. This could be any organism (people, people that have disease, people that know a rumor, bugs, bacteria, fish, other animals, plants, etc.). The (t) means rate of change of population with respect to time. For example, if y(t) is in bacteria and t is in minutes, then the units of will be bacteria/minute. So can, roughly, be thought of as the number of bacteria that will be added to the population over the next minute (I say roughly because it actually is the instantaneous rate which is not exactly the same as the average rate over the next minute, but they would be very, very close). Typically, the rate of change of a population is only dependent on the current population size in some way. In other word, we assume that the rate of change of y(t) only depends on y. Thus, we consider the autonomous equation: (t) = F(y), y > 0, t > 0. (1.1) We will consider four models of population namics, using differential equations. Model I: Unrestricted (Natural) Growth. The change of rate of y(t) is propotional to y(t) itself, i.e. (t) where a > 0 is a constant, called relative growth rate. = ay(t), (1.2) The basic idea of this model is that for small population, we have small growth rate while for large population, we have large growth rate. It is true if there is plenty of food and space. 1 Copy right reserved by Yingwei Wang

2 Given initial value y(0) = y 0, the solution to the problem (1.2) is y(t) = y 0 e at. (1.3) For the plots of solution curves, see Figures in your textbook. Model II: Logistic Growth. Insead of constant growth rate, let us assume the relative growth rate depends on y, i.e., (t) where a > 0,b > 0 and K = a b > 0. The basic idea behind this model is as follows: ( = ay 1 y ) = ay by 2, (1.4) K 1. Forsmall 0 < y 1, therateay by 2 ay which is thesame asunrestricted growth. 2. For large y 1, the rate ay by 2 by 2 < 0. It means that the population y(t) can not be too large due to wars or limitation of food, enviroment situation, etc. In other word, there is a carrying capacity K that the population size can t exceed. Given initial value y(0) = y 0, the solution to the problem (1.4) is y(t) = Ky 0 y 0 +(K y 0 )e at. (1.5) For the plots of solution curves, see Figures 2.1.2, and 2.2.3, 2.2.4, in your textbook. Model III: Growth With A Critical Threshold. The differential equation is (t) ( y ) = ay T 1, (1.6) where T > 0 is called critical threshold. Assume the initial conion is y(0) = y 0, then we have 1. If 0 < y 0 < T, then lim t y(t) = 0, which means the species will die out (extinction). 2 Copy right reserved by Yingwei Wang

3 2. If y 0 > T, then lim t y(t) =, which means unrestricted growth. Given initial value y(0) = y 0, the solution to the problem (1.6) is y(t) = Ty 0 y 0 +(T y 0 )eat. (1.7) For the plots of solution curves, see Figures and in your textbook. Model IV: Logistic Growth With A Threshold. The differential equation is (t) where 0 < T < K and a > 0. ( = ay 1 y )( y ) K T 1, (1.8) 1. Ifthepopulationsizeisbelowacertainthresholdsize, T, thenthepopulation will decrease to zero. 2. If the population is above this threshold, then the population behaves like logistic growth. For the plots of solution curves, see Figures and in your textbook. A special case with a = 1,T = 1/2,K = 1 gives 2 Stability analysis 2.1 Useful concepts y = y(1 y)(2y 1). Recall that an equilibrium solution (also called critical point or stea state ) is any constant (horizontal) function y(t) = c that is a solution to the differential equation. Notice that the derivative of a constant function is always 0, so we find equilibrium solutions by solving for y in the equation = f(t,y) = 0. For autonomous equation(1.1), using the Theorem 2.1(about the existence/uniqueness of the solution) in Lecture note 2, if F(y) and df have no discontinuities, then we are guaranteed that no other solution can intersect an equilibrium solution (if they did we wouldnt have uniqueness). Thus, we can classify the equilibrium solutions to (1.1) as follows: 3 Copy right reserved by Yingwei Wang

4 Stable: The equilibrium solution y(t) = c is stable if all solutions with initial conditions y 0 near y = c approach c as t. Unstable: The equilibrium solution y(t) = c is unstable if all solutions with initial conditions y 0 near y = c do NOT approach c as t. Semistable: The equilibrium solution y(t) = c is semistable if initial conditions y 0 on one side of c lead to solutions y(t) that approach c as t, while initial conditions y 0 on the other side of c do NOT approach c. 2.2 Classifying equilibrium solutions Consider the autonomous equation (1.1), in which F(y) and df 1. Solve F(y) = 0 to get the equilibrium solutions (critical points). are continuous. 2. Stu F(y) around the equilibrium values as follows (drawing the phase diagram: F(y) vs y might help) (a) Draw a vertical line (the phase line) and make tick marks at equilibrium values. (b) Between tick marks determine if F(y) is positive or negative. (c) If F(y) is positive, then will be positive so any solution in this region will be increasing. (d) If F(y) is negative, then will be negative so any solution in this region will be decreasing. 3. Classify: (a) Increasing below and decreasing above stable. (b) Decreasing below and increasing above unstable. (c) Decreasing below and above OR increasing below and above semistable. Now let us classify the equilibrium solutions (critical points) for the Models I-IV in previous section. Model I: Unrestricted Growth. It is easy to know that f(y) > 0 for all y > 0. There is only one equilibrium solution (critical point) to the equation (1.2), y = 0, which is unstable. 4 Copy right reserved by Yingwei Wang

5 Model II: Logistic Growth. Since if 0 < y < K, f(y) > 0, if y > K, f(y) < 0, there are two equilibrium solutions (critical points) to the equation (1.4): y = 0 is unstable; y = K is stable. Model III: Growth With A Critical Threshold. Since if 0 < y < T, f(y) < 0, if y > T, f(y) > 0, there are two equilibrium solutions (critical points) to the equation (1.6): y = 0 is stable; y = T is unstable. Model IV: Logistic Growth With A Threshold. Since if 0 < y < T, f(y) < 0, if T < y < K, f(y) > 0, if y > K, f(y) < 0, there are three equilibrium solutions (critical points) to the equation (1.6): y = 0 is stable; y = T is unstable; y = K is stable. 2.3 Adional examples Find and classify the equilibrium points of the following equations. Example 1. (t) = (1 y)(3 y). (2.1) 5 Copy right reserved by Yingwei Wang

6 1. y(t) = 1 and y(t) = 3 are the equilibrium solutions. 2. The sign of : if y > 3, if 1 < y < 3, if y < 1, > 0, < 0, > 0, 3. y(t) = 1 is stable and y(t) = 3 is unstable. Example 2. (t) 1. y(t) = 4 and y(t) = 10 are the equilibrium solutions. 2. The sign of : if y > 10, if 4 < y < 10, if y < 4, = (y 4)(y 10) 2. (2.2) < 0, < 0, > 0, 3. y(t) = 4 is stable and y(t) = 10 is semistable. Example 3. (t) 1. y(t) = 0 and y(t) = 2 are the equilibrium solutions. 2. The sign of : if y > 2, if 0 < y < 2, if y < 0, = (y 3 8)(e y 1). (2.3) > 0, < 0, > 0, 6 Copy right reserved by Yingwei Wang

7 3. y(t) = 0 is stable and y(t) = 2 is unstable. Example 4 (Harvesting equation). (t) = y(4 y) h, (2.4) where h > 0 is a positive parameter and h is called harvesting rate, which will reduce the growth rate. The right hand side is The roots of the quadratic polynomial (2.5) are f(y) = y(4 y) h = y 2 +4y h. (2.5) y 1,2 = 2± 4 h. Then we have three cases for different values of h. 1. Under harvesting. If h < 4, then the differential equation (2.4) has two critical points. For example, if h = 3, then (2.4) becomes (2.1). 2. Critival harvesting. If h = 4, then the differential equation (2.4) has only one critical point. 3. Over harvesting. If h > 4, then the differential equation (2.4) has no critical point. For more details, see the examples 4 and 6 in Chapter 2.2 in your textbook. 7 Copy right reserved by Yingwei Wang