Math 56

September 13, 2007 [corrected Sept. 18]

Problem Set 1 (due September 20)

These problems relate to the interplay between discrete and continuous population models.

1.  (“Hard” exponential model)

a.  Consider the discrete model

xk+1 = xk (1+R)     for k = 0, 1, 2, …                           (1)

where  x0  and  R  are parameters of the model (i.e., treat them

as if they are known constants).

Show that this model has the analytic solution

xk = x0 (1+R)k      for every integer k ≥ 0.                   (2)

b.  Consider the continuous model

x’(t) = r x(t)       for t ≥ t0                                            (3)

x(t0) = x0

where  t0  is given and  x0  and  r  are parameters of the model.

Show that this model has the analytic solution

x(t) = x0  exp ( r(t-t0) ) for all real  t ≥ t0.                     (4)

[corrected 9/18 to insert “-t0”]

[ Note:  “exp(z)” means ez  but is easier to type.  ]

c.  Reconcile these two models.  Show that these models agree exactly if

t0, t1, t2, …  are evenly spaced times with

increment   Dt = tk+1 - tk

xk means x(tk)

and if  r  and  R  are related by

exp ( r Dt ) = 1 + R.                                         (5)

2.  (“Soft” exponential model)

Note that the discrete model

xk+1 = xk (1 + Rk)                                                         (6)

(where x0 is a parameter but Rk can vary with k)

has the (useless) analytic solution

xk = x0 ( 1+R0 ) ( 1+R1 ) … ( 1 + Rk-1 ).                      (7)

a.  Consider the continuous model

x’(t) = r(t) x(t)                                                             (8)

x(t0) = x0

(where r(t) can vary with time).

Show that this model has the analytic solution

x(t) = x0 exp (  )                                         (9)

for all real  t ≥ t0.

b.  Show that these models (equations 6-7 and equations 8-9) agree if

1 + Rk = exp (  )     for each k.                   (10)

[corrected 9/18; l.h.s. was just “Rk”]

3.  (Rate changing linearly)

a.  Consider the continuous model

x’(t) = r(t) x(t)

x(t0) = x0

(as before)

r(t) = r0 – at                                                                 (11)

where  r0  and  a  (and x0)  are parameters of the model.  That is,

the growth rate decreases linearly.

Show that this model has the analytic solution

x(t) = x0 exp ( r0 t – a (t2 – t02) /2 ).                             (12)

b.  Suppose that  x0,  r0, and  a  are all positive.  Show that x(t) is an increasing

function of  t  when  t is slightly larger than t0, but that then it decreases.

What is the limit (as t -> infinity) of  x(t) ?

c.  Show that if we turn this model into a discrete model by observing x(t)

only when t = t0, t1, …, and if we define  Rk as in the previous problem

then  Rk  is NOT LINEAR as a function of  k.  (Computing three values of

Rk in an example might be enough to show this.)

4.  (Discrete logistic model)

a.  Consider the discrete logistic model

xk+1 = xk (1 + Rk)                                             (13)

with

Rk = R* ( (M – xk ) / M ).                                (14)

Here  R*  and  M  are parameters of the model.

Show that this implies that

xk+1 = (1+R*) xk (1 – Z xk )                             (15)

where  Z  is some combination of  R* and M.

b.  Suppose that   0 < x0 < M  and that  1 + R* = 1.02.  Show that xk increases

as a function of k, from x0, and that it approaches a limit of M as

t approaches infinity.

c.  Suppose that  x0 = 1 million,  M = 2 million,  and  1 + R* = 3.5.  Show

that xk DOES NOT increase as a function of k (at least, not for all k)

and that xk DOES NOT approach a limit as t -> infinity.

(A computational illustration is enough, if it is convincing.)

5.  (Continuous logistic model)

a.  Consider the continuous model

x’(t) = r(t) x(t)

x(t0) = x0

with

r(t) = r* ( (M – x(t) ) / M ).                              (16)

(Now r* and M are parameters of the model.)

Show that this model is equivalent to the equation

x’(t) = r* x(t) – (r*/M) x(t)2 .                           (17)

b.  Show that this model has the analytic solution

x(t) =                                      (18)

for all real  t ≥ t0.

c.  Show that  if  0 < x(t) < M  and  r* is positive, then x(t) increases

as a function of  t  and has limit M as t approaches infinity,

regardless of the exact value of  r*.