Linear models of structured populations Matthew Macauley Department - PowerPoint PPT Presentation

Linear models of structured populations Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4500, Spring 2015 M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 1 / 8

Motivation: Population dynamics Consider a population divided into several groups, such as children and adults egg, larva, pupa, adult For example, consider a population of insects Egg Larva Adult Dead E t = # eggs at time t L t = # larve at time t A t = # adults at time t M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 2 / 8

An example 73 Suppose we have the following data: 4% of eggs survive to become larvae .04 .39 E L A 39% of larvae make it to adulthood .61 The average adult produces 73 eggs each .96 1 Dead Each adult dies after 1 day We can write this as a system of difference equations:  E t +1 = 73 A t       0 0 73 E t E t +1    =  . L t +1 = . 04 E t . 04 0 0 L t L t +1     0 . 39 0 A t A t +1  A t +1 = . 39 L t  By back-substitution, or inspection, we can deduce the following: A t +3 = ( . 39)( . 04)(73) A t = 1 . 1388 A t Thus, this is just exponential growth. But what if instead of dying, 65% of adults survive another day? M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 3 / 8

A slighly more complicated example 73 Suppose we have the following data: .65 4% of eggs survive to become larvae .04 .39 E L A 39% of larvae make it to adulthood .61 The average adult produces 73 eggs each .96 .35 Each day, 35% of adults die. Dead This yields a more complicated system of difference equations:  E t +1 = 73 A t       0 0 73 E t E t +1    =  . L t +1 = . 04 E t . 04 0 0 L t L t +1     . 65 . 39 0 A t A t +1  A t +1 = . 39 L t + . 65 A t  Questions Best way to solve this? What is the growth rate? What is the long-term behavior? How much effect does changing the initial conditions have? M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 4 / 8

Another example Consider a forest that has 2 species of trees, A and B . Let A t and B t denote the population of each, in year t . When a tree dies, a new tree grows in its place (either species). Each year: (.75)(.01) .99 .95 1% of the A -trees die 5% of the B -trees die A B 25% of the vacant spots go to species A 75% of the vacant spots go to species B (.25)(.01) (.25)(.05) (.75)(.05) This can be written as a 2 × 2 system: � A t +1 = . 99 A t + ( . 25)( . 01) A t + ( . 25)( . 05) B t � . 9925 . 0125 � � A t � � A t +1 � = . 0075 . 9875 B t B t +1 B t +1 = . 95 A t + ( . 75)( . 01) A t + ( . 75)( . 05) B t M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 5 / 8

Solving systems of difference equations One way to solve x t +1 = P x t : x 1 = P x 0 x 2 = P x 1 = P ( P x 0 ) = P 2 x 0 x 3 = P x 2 = P 3 x 0 . . . A better method Find the eigenvalues and eigenvectors of P . Then write the initial vector x 0 using a basis of eigenvectors . Suppose x 0 = c 1 v 1 + c 2 v 2 . Then x 1 = P x 0 = P ( c 1 v 1 + c 2 v 2 ) = c 1 λ 1 v 1 + c 2 λ 2 v 2 x 2 = P x 1 = P 2 x 0 = P ( c 1 λ 1 v 1 + c 2 λ 2 v 2 ) = c 1 λ 2 1 v 1 + c 2 λ 2 2 v 2 . . . . x t = P t x 0 = c 1 λ t 1 v 1 + c 2 λ t 2 v 2 . M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 6 / 8

An example, revisted � . 9925 . 0125 � Let us revisit our “tree example”, where P = . . 0075 . 9875 The eigenvalues and eigenvectors of P are � 1 � 5 � � λ 1 = 1 , v 1 = λ 2 = . 98 , v 2 = , . 3 − 1 � 10 � A 0 � � Consider the initial condition x 0 = = . B 0 990 First step � 1 � 5 � � Write x 0 = c 1 + c 2 , i.e., solve P c = x 0 : 3 − 1 � 10 � 5 � � c 1 � � 1 = . 3 − 1 990 c 2 � � 10 � 125 c = P − 1 x 0 = − 1 � − 1 � � − 1 = 8 − 3 5 990 − 615 � 10 � 1 � � 5 � � Thus, our initial vector is x 0 = = 125 − 615 . 990 3 − 1 M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 7 / 8

An example (cont.) Solving for x t Once we have written x 0 = c 1 v 1 + c 2 v 2 , the solution x t is simply x t = P t x 0 = c 1 λ t 1 v 1 + c 2 λ t 2 v 2 . � 1 � 5 � � In our example, x 0 = 125 − 615 , and so 3 − 1 � 5 � � 1 � � 625 − (615)( . 98) t � x t = 125(1) t − 615( . 98) t = . 375 − (615)( . 98) t 3 1 The long-term behavior of this system is � 5 � � 625 � t →∞ x t = 125 lim = . 3 375 Notice that this does not depend on x 0 ! M. Macauley (Clemson) Linear models of structured populations Math 4500, Spring 2015 8 / 8