1
Mathematical Modelling
Worksheet 6
1. Exponential Growth.
Assume that a population contains N t individuals at time t, where t is measured in
years. Suppose that the number of births in a year is a fraction  and the number
of deaths in a year is a fraction  of the total population. We can write down an
expression for N t 1 in terms of N t ,  and  and, without using calculus, obtain an
expression for N t given that N t = N 0 at t = 0.
Suppose N  Population at time t.
In a small time interval t, the population size will change because of
(1) births  
(2) deaths  
Assume the number of births and deaths depend on
(1) size of t
(2) size of population
 number of births    N t  t
number of deaths    N t  t
The increase in

so population in
time t


  Number of   Number of 



 Deaths

 Births

N t t  t   N t t     N t  t    N t  t
putting t  1 i.e 1 year  we form a difference equation
N t t  1  N t t 1     
2
Using suffix notation we have
N t 1  N t 1     
Now using given condition N t  N 0 at t  0 we get
N 1  N 0 1     
N 2  N 1 1     
 N 0 1     1     
.
.
.
.
.
.
.
.
N t  N 0 1     
t
This model for population growth would be appropriate for modelling the effect of
natural disasters or the lack of resources.
2. The Logistic Curve.
In an environment that will support a limited population it is assumed that the rate
of growth of population decreases as the limiting population is approached. An
appropriate model (the Verhulst-Pearl model) is given by

N 

 rN 1 


dt
N
m 

dN
N m is the maximum population which can be sustained and r is the intrinsic
growth rate.
We can solve this equation for N, assuming N  N 0 at t = 0.
3

dN
N 

 rN 1 
dt
 Nm 
N
Let x 
ie. N  N m x
Nm
Nm
dx
 rN m x1  x 
dt
dx
 rx 1  x 
dt
dx
 rdt
x1  x 
1 
1
that is  
dx  rdt
 x 1  x 
Integrating both sides we get
ln
x
 rt  c
1- x
Applying boundary conditions we have
t  0,

ln
x  x0
 N0 
 

 Nm 
x 1  x0

 rt
1  x x0
taking natural logarithms of both sides
x 1  x0

 e rt
1  x x0
 1 
  1  x e rt
x
 x0  1 
x
 x  1  x e rt
x0
0
x
 1  x e rt  x
x0
 1

0  x  e rt  1  e rt
 x0

4
e rt
x 
 1

  e rt  1
 x0

to tidy up expression multiply both sides by e  rt
1
e  rt x 
 1

  e rt  1
 x0

x
Now putting x 
1
 1

1    1e  rt
 x0

N
N
and x0  0 back into the above expression , we get
Nm
Nm
N

Nm
N 
1
N

1   m  1e  rt
 N0

Nm
N

1   m  1e  rt
 N0

A sketch can illustrate this result.
N
Nm
Below a critical
value of the
population, the
population cannot
be sustained and
diminishes.
N0
The
population
grows until it
reaches a
maximum
limit.
t
5
3. Finite difference solution.
In finite difference terms, the differential equation of question (2) becomes
rN t2
. This equation is not easily solved but a slight
N t 1  1  r N t 
Nm
modification (replacing N t2 by N t N t 1 ) produces an equation, which is tractable
1
by making the substitution Vt 
.
Nt
rN t2
Nm
rN N
 1  r N t  t t 1
Nm
N t 1  1  r N t 
Replacing N t2 by N t N t 1 ,
Substituting Vt 
1
, Vt 1
Nt
N t 1

rN 
N t 1  N t 1  r   t 1 
Nm 

1
1

and Vm 
we get
N t 1
Nm
rV 
1
1 
 1  r   m 
Vt 1 Vt 
Vt 1 
Vt  1  r Vt 1  rVm
This equation can now be solved as a recurrence relation.
Firstly find the homogeneous solution to the equation:
Vt  1  r Vt 1  0
If Vt  q t
q  0
q t  1  r q t 1  0
1  1  r q  0
q 
1
1 r
 1 
Vt  A

1 r 
t
6
Secondly for the particular solution try Vt  B
B  1  r B  rVm
 B  Vm
Therefore the general solution of the equation is:
t
 1 
Vt  A
  Vm
1 r 
Applying the boundary condition Vt  V0 when t  0
V0  A  Vm
 A  V0  Vm
Therefore the solution of the equation is:
t
 1 
Vt  V0  Vm 
  Vm
1 r 
Now that the equation is solved, we can rewrite it in terms of N.
t

 1 
1
1 


N t  





 N 0  N m  1  r  N m 
1
If we compare this equation with that of question (2) we find that the solution of this
equation behaves differently. Initially the population N rises rapidly, until the
population can no longer be sustained. Then the population level falls until it begins
to level out at the value of N m the maximum population, which can be sustained.
A graph can illustrate this result.
N
Nm
7
t
4. Fish-hatchery problem.
A fish hatchery is stocked with a population of N 0 juvenile fish, which is subject
to a depletion rate r. After a time T (to allow the fish to grow) fishing commences
and causes an additional depletion rate q. We can obtain an expression for the
population at time t (t>T) and determine the proportion of the original N 0 fish
which will be recovered by fishing.
t T
dN
 rN
dt
N  Ae  rt
when t  0 N  N 0
N  N 0 e  rt
for t  T
 A  N0
Now when t  T
N  N 0 e  rT
t T
dN
 r  q N
dt
N  Be r  q t
for t  T
when t  T
N  N 0 e  rT  Be r  q t
 B  N 0 e  rT e rT e qT
 N 0 e qT
N  N 0 e qT e r  q t
for t  T
The proportion of the original N 0 fish, which will be recovered by fishing.
8
P
qN t
N0

qN t
dt
N0

qN 0 e qT e r  q t
dt
N0
 qe qT  e r  q t dt
qe qT e r  q t
A
 r  q 
t  T and P  0

A

qe  rt
rq
P

qe qT e  rT e  qT
rq
qe  rt e  r  q T qe  rT

 r  q 
rq
qe  rT
1  e r  q t
rq


5. Growth of fish.
Experiment and observation indicate that the surface area of a fish is proportional
the square of its length and that its weight is proportional to the cube of its length.
The energy used in seeking food is proportional to its weight, while the energy
provided by the food is proportional to its surface area. The rate of increase in
weight is proportional to the difference between these two energies. We can
obtain an equation showing how the weight of the fish varies with time.
9
dw
 E 2  E1
dt
a  kl 2
E 1  Dw
w  Cl 3
E 2  Ga
dw
 M Ga  Dw 
dt
since w  l 3
w
2
3
 l2
a  kl 2
2
dw
 M  Nw 3  Dw 


dt
2
dw
 w 3   w
dt
1 2
1
call   MN and   MD
vw 3 w 3
2
dv 1  2 3 dw 1  2 3 
 w
 w  w 3  w 


dt 3
dt 3
dv   13
  w
dt 3 3
dv 

 v
dt 3
3

Integrating factor  e
3
dt

 e3


dv  3 t   3 t
ve   e
dt 
 3

ve
3
t

 3 t
  e3 T
3 
t
10
w
1
3

 v   Te


t
3
3

t 


wt     Te 3 


 The total weight of the fish recovered in question (4)
p
qe  rT
1  e r  q t
rq



t 


Total weight of fish after t  p    Te 3 


3
11
Mathematical Modelling
Worksheet 7
1.
Insects in a Limited Environment.
The study of an insect population in a limited environment revealed that as the
population increased the available food per insect declined. In consequence
the number of eggs laid N e decreased in proportion to the population size
according to the law N e  A  N t where N t is the size of the population and
A and  are constants. None of the adult insects survive the winter but a
fraction  of the eggs hatch out the following year.
Equilibrium is when N t  1  N t for following years
N e   A  N t 
N t 1   A  N t 
N t 1  N t
 N t    A  N t 
Determining the equilibrium value N of N t
N t  N t  A
N t 1     A
Nt 
A
N
1  
Writing N t  N  nt , we can obtain and solve a difference equation for nt and hence
obtain an expression for N t assuming N t  N 0 at t = 0. Also N t 1  N  nt 1


N  nt 1   A   N  nt

nt 1  A   N  nt  N
 A  nt  1   N
nt 1  nt
n1  n0
n2  n1    n0
2
n3  n2    n0
3
12
 nt   1   n0
t
t
N t  N  nt ie nt  N t  N
n0  N 0  N

N t  N   1   N 0  N
t
t



N t   1   N 0  N  N
t
t
A
t
t
N t   1    N 0 
1  


A
 
 1  
We can draw a graph showing the behaviour of N t
(i)
(ii)
when <1
when >1
(i)
Nt
N
From the relationship
above
tends
Nt
towards
<1
N
when
t
(ii)
Nt
N
Over time N t tends
towards zero when
>1
t
2.
Predator-Prey Situation.
A predator-prey situation occurs with two populations, when encounters
between members of the populations benefit the members of one population
(predators) but adversely affect the members of the other population (prey).
For a simple model assume that the prey has an adequate food supply and that
13
in the absence of interaction there will be a positive rate of increase
proportional to the population size. Likewise, assume that in the absence of the
prey, the predator’s food supply will be inadequate and that there will be a
negative rate of increase proportional to the population size. The incidence of
interaction will depend on the size of both populations and may be taken to be
proportional to their product, and the effect will increase the rate of growth of
the predator population and decrease the rate of growth of the prey population.
Letting X, Y denote the size of the prey and predator populations respectively,
we can write down two simultaneous differential equations to represent the
situation.
dX
 kX
dt
dY
 lY
dt
dX
 kX  XY
dt
 kX 1  Y   0
dY
 lY  XY
dt
 lY 1  X   0
In terms of the constants in these equations, we can obtain the equilibrium values
X and Y .
X 0
1  Y  0
1
Y  Y

Y 0
1  X  0
1
X  X

Let x  X  X , y  Y  Y , denote the deviations of the population sizes
from their equilibrium values. Assuming these deviations are small compared
to the size of the populations, we can rewrite the differential equations in
terms of x and y and solve the resulting equations, neglecting the second order
terms. Hence we can obtain approximate solutions for X and Y.
14
X  x X  x
Y  y Y  y 
1

1


dx
1 
1 

 k  x  1    y   
dt
 
 



1
 k  x   y 


k
 kxy 
y



k
y

dy l

x
dt 
Renaming
dx
dy
  Ay
 Bx
dt
dt
d 2x
dy
 A
  ABx
2
dt
dt
d 2x
d 2x
2


w
x

 w2 x  0
2
2
dt
dt
x  p cos wt  q sin wt
let AB  w 2 
1 dx
A dt
1
y    wp sin wt  wq cos wt
A
w
y   p sin wt  q cos wt 
A
y
15
Taking axes to represent X and Y we can sketch a graph to show the variation in
populations.
X, Y
t
3.
Finite Difference Equations.
The differential equations from question (2) may be represented as finite difference
equations in the form:
X t 1  a11 X t  a12 X t Yt
Yt 1  a 21Yt  a 22 X t Yt
dX t X t 1  X t

dt
t 1 t
dYt Yt 1  Yt

dt
t 1 t
dX t
 a11 X t  a 12 X t Yt  X t
dt
dX t
 a11  1X t  a12 X t Yt
dt
dX t
 aX t  bX t Yt
dt
And similarly
dYt
 a 21  1Yt  a 22 X t Yt
dt
dYt
 cX t Yt  dY t
dt
16
Taking a11 =1.5, a12  0.025 , a21  0.75 , a22  0.0025 , and with initial values
X=100, Y=20, we can show that equilibrium is maintained.
Population maintained in equilibrium
120
100
Population size
80
Prey (X)
Predator (Y)
60
40
20
0
1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79
Time
By using different initial values, we can show that the system is not stable. Firstly by
considering the values X = 90, Y = 20, we can plot the population of the predators and
preys.
Unstable System
400
350
Population Size
300
250
Prey (X)
Predator (Y)
200
150
100
50
0
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51
Time
17
Secondly we can try another set of initial values for the population of the predators
and preys. Therefore taking X = 100, Y = 25, we can plot the resulting population over
time.
Unstable System
600
500
Population Size
400
Prey (X)
Predator (Y)
300
200
100
0
1
3
5
7
9
11 13 15 17 19 21 23 25 27 29 31 33 35 37 39
Time
4.
Competing Species.
The differential equation representing growth of population with a limited
food supply requires modification if a second species also competes for the
same food resources. In this case the equation becomes:
dN1
 r1 N1 M 1  N1  c2 N 2 
dt
where N1 is the number of species 1, r1 is the intrinsic growth rate, and M 1 is
the saturation population for species 1, N 2 is the number of species 2 and c2
dN 2
is the inhibitory factor of species 2 on species 1. The growth rate
is
dt
given by the corresponding equation:
dN 2
 r2 N 2 M 2  N 2  c1 N1 
dt
By taking axes to represent N1 and N 2 , we can draw lines representing
equilibrium conditions for species 1 and 2. Any point ( N1 , N 2 ) represents a
possible population configuration. By considering the directions in which this
point would tend to move, we can show under what initial conditions:
18
(a)
Species 1 would be eliminated.
N2
dN 2
dt
2
1+2
(b)
N1
dN1
0
dt
Species 2 would be eliminated.
N2
dN1
dt
1
1+2
N1
dN 2
0
dt
(c)
Both species would co-exist in stable equilibrium.
N2
M
2
dN 1
0
dt
1
1+2
2
dN 2
0
dt
M1
N1
19
(d)
Equilibrium would be unstable.
N2
dN 2
0
dt
2
1+2
dN 1
0
dt
1
N1