Lecture 10: Logistic growth models #2

Transcription

1 Lecture 1: Logistic growth moels #2 Fugo Takasu Dept. Information an Computer Sciences Nara Women s University takasu@ics.nara-wu.ac.jp 6 July 29 1 Analysis of the stochastic process of logistic growth We have implemente the stochastic logistic growth process in a C program an confirme that the stochastic ynamics exhibits a feature that is similar to the eterministic logistic growth. The process is that for N iniviuals (N is now non-negative integer), 1) a new iniviual is born with probability birth(n) t, 2) the iniviual ies an is remove from the population with probability eath(n) t, an 3) the iniviual neither gives birth nor ies with probability 1 birth(n) t eath(n) t. The birth an eath rate, birth(n) an eath(n), are given as some functions of the population size N. In this lecture we explore the stochastic ynamics from analytic viewpoint. 2 Master equation We assume that the time interval t is so small that the change of the population size n uring the interval is at most ±1, i.e., transition to a state n is possible either from n 1 or n + 1 (n 1). Then the probability that the population size is n at time t + t, P n (t + t), is given as P n (t + t) =P n (t) {1 birth(n)n t eath(n)n t} + P n 1 (t)birth(n 1)(n 1) t + P n+1 (t)eath(n + 1)(n + 1) t As in the birth an eath process, transition from n = to n = 1 is now impossible. This means that empty (extinct) population cannot prouce offspring anymore. The bounary n = is an absorbing bounary that separates positive an negative region of n. By assuming that P n (t) for negative n is always zero, equation (1) is vali for all integers of n. (1) Arranging equation (1) an letting t, we obtain P n (t) =birth(n 1)(n 1)P n 1 (t) + eath(n + 1)(n + 1)P n+1 (t) {birth(n) + eath(n)} np n (t) (2) 1

2 Equation (2) is the master equation of the stochastic logistic growth. Once the functional forms of birth(n) an eath(n) are given, P n (t) can be solve with a certain initial conition, e.g., P () = 1, P n () = for n 1, but it is in general not easy. In the next section we explore some properties of the process using moment ynamics. Hereafter we assume a general case that the per-capita birth rate is a linearly ecreasing function of N an the per-capita eath rate is a linearly increasing function of N birth(n) = b 1 b 2 N eath(n) = N where b 1, b 2, 1, 2 are positive. If b 2 = an 2 = this is the birth-eath process we learne in previous lectures. In eterministic worl the birth an eath rate assume above gives the ODE N = {b 1 1 (b )N} N ( ) = (b 1 1 ) 1 N b 1 1 b N This is a logistic growth with the intrinsic rate of increase r = b 1 1 an the carrying capacity K = (b 1 1 )/(b ). 3 Moment ynamics From the master equation we now try to erive moment ynamics, especially of the first an the secon moment. The master equation is now P n (t) = {b 1 b 2 (n 1)} (n 1)P n 1 (t) + { (n + 1)} (n + 1)P n+1 (t) {b 1 b 2 n n)} np n (t) (3) We multiply equation (3) with n an taking summation for n we have n = {b 1 b 2 (n 1)} n(n 1)P n 1 (t) + { (n + 1)} n(n + 1)P n+1 (t) {b 1 b 2 n n)} n 2 P n (t) =(b 1 1 ) n (b ) n 2 (4) 2

3 Note that the secon moment n 2 appears in the ODE of the first moment. In the same way we multiply equation (3) with n 2 an taking summation for n we have n2 = {b 1 b 2 (n 1)} n 2 (n 1)P n 1 (t) + { (n + 1)} n 2 (n + 1)P n+1 (t) {b 1 b 2 n n)} n 3 P n (t) =(b ) n + (2b 1 b ) n 2 2(b ) n 3 (5) Note again that the thir moment n 3 appears in the ODE of the secon moment. We have erive the first an secon moment ynamics as follows. n = (b 1 1 ) n (b ) n 2 (6) n2 = (b ) n + (2b 1 b ) n 2 2(b ) n 3 (7) These two equations are not close with respect to n an n 2 an they cannot be solve without the knowlege of n 3. But we will fin that the ODE for the thir moment contains the fourth moment, the ODE for the fourth moment contains the fifth, an we cannot erive a set of ODE in a close form. This is a general property when birth(n) an eath(n) are function of N, i.e., transition probability becomes non-linear. Then how can we solve the ynamics? One way to resolve this problem is to erive an approximation with an assumption that higher-orer moment be given as some function of lower-oer moment. But we will not step into such etails further here. Remember that V ar[n] = n 2 n 2, then equation (6) can be arrange as n = (b 1 1 ) n (b ) n 2 = (b 1 1 ) n (b ) n 2 (b )V ar[n] = (b 1 1 ) ( 1 n b 1 2 b ) n (b )V ar[n] (8) We fin in equation (8) that the ynamics of the first moment n, or the expecte value of population size n (ensemble average of n), obeys a ynamics that is no longer logistic growth because of the aitional term in the right han sie (V ar[n] ). Although we have not yet etermine V ar[n] this is a remarkable result we have never observe in the previous moels of immigration-emigration an birth-eath where the first moment ynamics obeys the same eterministic ynamics. In the previous moels, we obtaine the ynamics of the first moment that is exactly the same as the corresponing eterministic ynamics. But in the stochastic logistic growth where birth(n) 3

4 an eath(n) epen on N, we no longer have such coincience. This is typical to cases when transition probabilities n n+1, n, n 1 are non-linear with respect to n. Transition probabilities in immigration-emigration an birth-eath moels are linear so that the eterministic ynamics an the first moment ynamics exactly match with each other (Table). Prob[n n + 1] Prob[n n 1] Prob[ n n] Immigration-emigration α t β t 1 α t β t Birth-eath βn t δn t 1 βn t δn t Logistic (b 1 b 2 n)n t ( n)n t 1 (b 1 b 2 n)n t ( n)n t In the simulation of the logistic process we foun that the ensemble average of the population size E[n] an the variance of population size V ar[n] are eventually stabilize at certain levels (this is actually a quasi-stationary state, not a true stationary state as we will see in the next lecture). If the ensemble average an variance of population size converge to constant, n e, an σe, 2 respectively, they shoul satisfy equation (8) with the time erivative be zero ( ) = (b 1 1 ) 1 n e b 1 2 b n e (b )σ 2 e This is a quaratic equation of n e an we can solve n e as follows if the variance is small enough where we use an approximation 1 + x 1 + x/2 when x 1. n e = b 1 1 b b b 1 1 σ 2 e (9) In the eterministic logistic growth the population size converges to the carrying capacity K = b 1 1 b but equation (9) shows that the equilibrium ensemble average E[n] in the stochastic logistic growth n e is lowere by the amount proportional to the variance V ar[n] at equilibrium σ 2 e. n e = K b b 1 1 σ 2 e (1) We have not yet etermine the ynamics of the variance V ar[n] because it contains the thir moment an the ynamics of the thir moment contains the fourth moment, an the fourth moment contains the fifth,. In the next section we see how much the variance will be. 4 Linear approximation In the logistic process, the transition probability Prob[n n+1] an Prob[n n 1] is (b 1 b 2 n)n an( n)n, respectively. The former is concave an the latter is convex an these are non-linear 4

5 function of n. Due to the non-linearity we coul not erive moment ynamics in a close form. We now linearlize these functions aroun a state at which both the probabilities, (b 1 b 2 n)n an ( n)n, equal, i.e., at n = n = (b 1 1 )/(b ) = K. Note that K is the carrying capacity of the eterministic logistic moel. Prob[n to n 1] Prob[n to n+1] n* Pop. size n We focus on an approximate stochastic process where the transition probabilities, Prob[n n+1] an Prob[n n 1], are linear function of n. Let the slope of (b 1 b 2 n)n at n = n be A an that of ( n)n be B. Then the linearize (approximate) transient probabilities are given as where (b 1 b 2 n)n A(n K) + C (11) ( n)n B(n K) + C (12) A = b 1b 2 + b b 2 1 b B = 2b b b C = (b 1 1 )(b b 1 2 ) (b ) 2 We expect that this linearization will work successfully if eviation from K is not large (variance, or stanar erivation, is small relative to K). Base on the linearize transient probabilities we construct master equation of the approximate linear system. P t (t) = (A(n 1 K) + C)P n 1 + (B(n + 1 K) + C)P n+1 (A(n K) + C + B(n K) + C)P n (13) From this master equation the first moment ynamics is erive as n = (A B) n (A B)K = (b 1 1 )(K n ) (14) 5

6 This is easily solve an we see n K if b 1 > 1. In the same way the secon moment ynamics is obtaine after some calculus. n 2 = 2(A B) n 2 + (A + B 2KA 2KB) n (A + B)K + 2C (15) Using the relationship V ar[n] = n 2 n 2, we erive the ynamics of the variance V ar[n] = 2(A B)V ar[n] + (A + B) n (A + B)K + 2C (16) ODE (16) can be reaily solve, but if the variance converges to a constant σ 2 e, it must satisfy = 2(A B)σ 2 e + (A + B)K (A + B)K + 2C = 2(A B)σ 2 e + 2C because n K in this linear process. The variance at quasi-equilibrium is then σ 2 e = C B A = b b 2 1 (b ) 2 (17) Remember that this is erive from the approximate linear process an is not exact evaluation of the variance. But we will see this gives goo estimate of the variance of the logistic process. 5 Problem 1. Carry out simulation with appropriate parameter values of b 1, b 2, 1, 2 to see if the simulation is in goo agreement with the analytical results in terms of the variance (equation 1 an 17). 6