Mathematical modeling

1. Chapter 3
Modeling Discrete Dynamical Systems with
Difference Equations
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2. Modeling Discrete Dynamical
Systems with Difference Equations
Outlines
 Introduction
3.1 Discrete Dynamic Systems and Difference
Equations
3.2 Modeling with Linear First order Difference
Equations
3.3 Modeling with Systems of Difference Equations
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3.  One of the main concerns of modeling is to
predict the future development of a system.
 A system that changes over time is called
dynamical system.
 A powerful paradigm to model change is
future value = present value + change
change = future value – present value
Introduction
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4.  If a variable of our interest changes in
discrete time intervals, the above formula
leads to a difference equation and a
dynamical system we work with difference
equations is called discrete dynamical system.
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5. • The mathematical assumption is that the time
variable n is incremented discretely and
corresponds to the integers {0, 1, 2, 3, 4, . . . }. The
value of a variable x of interest is then a sequence
{x0, x1, x2, x3, x4, . . . }.
• Discrete models can be used in population growth,
interests in accounts, drug dosages, genetics and
others.
3.1 Discrete Dynamic Systems and Difference
Equations
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6. For discrete models, the difference equation
(discrete dynamical system) has the form,
Future value = Function of {Present value, Previous
values and possibly time}.
xn+1 = f (xn , xn-1, …, n), where
xn+1 = Future value,
xn = Present value,
xn-1 = Previous value,
n = time.
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7. Definitions
Definition 1: A difference equation is of first order if the
value xn+1 depends only on xn, n and constants. It is of
second order if xn+1depends on xn, xn-1, n and constants.
And so on.
Example 1: xn+1= 3xn-n2+7n+2 is first order difference
equation.
Example 2: nxn+1- (n3-0.5n+1)xn -6xn-1+8n=10
is second order difference equation.
Definition 2: A difference equation is said to be
autonomous if its calculation does not depend explicitly
on n.
Example 3: xn+1=7xn-12 is autonomous difference
equation.
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8. …Continued
Definition 3: If a difference equation involves no
products of sequence variables, no powers of
sequence variables, nor functions of sequence
variables such as exponential, logarithmic or
trigonometric functions, then we call the difference
equation linear.
For otherwise, the difference equation is nonlinear.
Example 4: xn+1= 3xn+n2 is linear difference equation.
Example 5: xn+1= (xn)2+xnxn-1 + 3n+5 is nonlinear
difference equation. 8
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9. …Continued
Definition 4: If each term of a difference
equation contains sequence of variables, then it is called
homogeneous difference equation.
For otherwise, it is called non-homogeneous difference
equation.
Example 6: (xn+1)2 +3xn-10xn-1 = 0 is homogeneous and
xn+1 +5xn- xn-2=n2+6n-10 is non-homogeneous.
Definition 5: A solution of a difference equation is a
sequence xn given by the formula in terms of n,
xn = f(n), n=0, 1, 2, …
 The solution can be given analytically, graphically or
numerically.
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10. 3.2 Modeling with Linear First Order Difference
Equations
The simplest possible difference (Linear, Autonomous) equation
is of the form
xn+1= rxn, n = 0,1,2,3,… (*)
with some initial condition x0, where r is a constant.
The solution can be found by implementing the iteration
x1=rx0,
x2=rx1=r(rx0)=r2x0,
x3=rx2=r(r2x0)=r3x0,
.
.
.
xn=rnx0.
Therefore, the solution of the Difference equation given in
(*) is xn=rnx0.
3.2.1 Difference Equation of the form xn+1= rxn
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11. Discussions on Long Term Behavior!
Long- term behavior (i.e. n→ ∞)
Suppose x0 > 0, then
0
0
0
0
0
0
0
0 0
If 1, .
If 1, .
If 0< 1, 0 .
If -1< 0, 0 .
even
If 1, ( 1)
, odd
and thus no convergence as .
If 1, , = as n
and thus n
n
n
n
n
n
n
n
n
n
n
n
n n
r x r x as n
r x x as n
r x r x as n
r x r x as n
x n
r x x
x n
n
r x r x x r x
     
   
    
    

     


 
      
o convergence as .
n   11
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12. MatLab Plot For x0 > 0
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13. Definitions
Definition 1: A number x* is called an equilibrium
point or fixed point or steady state of a discrete
dynamic system, xn+1 = f(xn) if f(x*)= x*.
For xn+1=rxn, an equilibrium point satisfy the
equation x*=rx*. Thus,
if r ≠ 1, then x* = 0 the only equilibrium point.
if r = 1, every number x* is an equilibrium point.
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14. Example 1
1. For each of the following models,
a) xn+1=1.2 xn, x0=1000
b) xn+1=-0.5 xn, x0=4000
c) xn+1= xn, x0=20
(i) find x1, x2, x3, x4, x5
(ii) find the solution
(iii) determine the equilibrium point
(iv) plot xn versus n.
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15. Solutions:
a) Xn+1=1.2xn, x0 = 1000
(i) x1=1.2x0 = 1.2(1000) = 1200
x2 =1.2x1=1.2(1200) = 1440
x3 = 1.2x2 =1.2(1440) = 1728
x4 =1.2x3=1.2(1728) = 2073.6
x5 = 1.2x4 =1.2(2073.6) = 2488.32
(ii) xn=rnx0 = (1.2)n(1000)
Using this, x5=(1.2)5(1000) =2488.32
(iii) Since r ≠1, the only equilibrium point is x* = 0.
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16. 0 1 2 3 4 5 6 7 8 9 10
0
1000
2000
3000
4000
5000
6000
7000
n
xn
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17. Example 2
Suppose a certain population of owls is growing
at the rate of 2% per year. If initially we have a
population of 100 owls,
(i) Develop a model to predict the owls population.
(ii) What will the population be after 10 years?
(iii) Plot the population versus years.
(iv) After what year will the population be doubled?
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18. Solutions:
(i) Let pn denote the population of owls after n
years. Then,
pn+1= pn+0.02pn = 1.02pn
Hence the model is
pn+1 = 1.02pn , p0 = 100.
The solution is
pn =(1.02)n 100.
(ii) The population after 10 years is
p10 = =(1.02)10 100 = 122
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19. (iii)
0 10 20 30 40 50 60 70 80 90
0
100
200
300
400
500
600
700
n
xn
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20. …continued
Therefore the doubling time of the population
is about 35 years.
2
1.02
( ) 200 (1.02) 100
200
(1.02) 2
100
log 2
log 35 years
log1.02
n
n
iv
n

 
  
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21. Exercises:
1) If 5000 birr is invested at the rate of 8%
compounded annually,
(i) Develop a model to describe the sum of the
money after n years.
(ii) What is the sum of the money at the end of 10
years?
(iii) How long does it take the sum of the money to
double itself?
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22. …Continued
2) Radium is a radioactive element which decays at a
rate of 1% every 25 years. If the initial amount of
radium is 500 grams, then
(i) develop a model to describe the amount of
radium left.
(ii) find the amount left after 100 years.
(iii) plot of the amount of radium left versus years.
(iv) what is the half-life time of the radium?
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23. 3) Suppose that a bacterial colony starts with 100
bacteria and the bacteria divide every 20
minutes.
a) How will the population size change over time?
b) Plot the bacteria population versus time.
c) What will be the bacteria population after 120
minutes?
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24. 4) CIPRO is a drug for combating many infections
including anthrax. Let us assume that during one-
hour period our kidneys purify ¼ of this drug. If a
patient takes 16 mg of this drug, then
a) write a model to predict the amount of the drug
in the patient’s blood.
b) how long will the drug it take to be 6.75 mg in
the patient’s blood?
c) plot the amount of drug in the blood versus time.
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25. Consider the dynamic system
xn+1= rxn +b,
with initial condition x0, where r and b are constants.
3.2.2 Difference Equation of the form xn+1= rxn + b
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26. The solution of this system is obtained as follows:
1 0
2
2 1 0 0
2 3 2
3 2 0 0
2 1
0
0
0
( ) (1 )
( (1 )) (1 )
(1 . . . )
If 1,
( 1)
1
If 1,
n n
n
n
n
n
n
x rx b
x rx b r rx b b r x b r
x rx b r r x b r b r x b r r
x r x b r r r
r
r
x r x b
r
r
x x nb

 
       
         
     


 


 
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27. Equilibrium Points of the System:
1
Given a dynamical sysem .
If 1, * is the equilibriumpoint.
1
If 1 and 0, everynumberis the equilibriumpoint.
If 1 and 0, no the equilibrium point exists.
n n
x rx b
b
r x
r
r b
r b
  
 

 
 
Classifying Equilibrium Points (Stability of the
equilibrium point):
 If all solutions of 𝒙𝒏+𝟏=r𝒙𝒏 +b approach to the
equilibrium point 𝒙∗ =
𝒃
𝟏−𝒓
as 𝒏 ⟶ ∞, then 𝒙∗ is called
stable equilibrium point. For otherwise, it is called
unstable. 27
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28. Thus, depending on the value of r, we
obtain the following long-term behavior
for the given system:
Value of r Long–term behavior observed
𝑟 < 1 Stable equilibrium point
𝑟 > 1 Unstable equilibrium point
𝑟 = 1 No equilibrium point
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29. Example 1
Consider the following dynamical systems:
a) xn+1 = 2xn-1, x0 =3
b) xn+1 =-0.5xn + 6, x0 =2
(i) Find the solutions.
(ii) Find the equilibrium points and check their
stability.
(iii) Plot the solution to observe long-term behavior
and stability, determine the limit of solutions.
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30. Solutions:
0
1
.( ) The solution of thedifference equation is
( 1)
1
1(2 1)
2 (3) 3(2 ) 2 1
2 1
2 1
( ) The equilibriumpoint
1
* 1
1 1 2
Since 2 2 1, * 1 is unstable equilibriumpoint.
n
n
n
n
n n n
n
n
n
a i
b r
x r x
r
x
x
ii
b
x
r
r x


 

 
    

 

  
 
   
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31. (iii) MatLab Plot of the System
0 1 2 3 4 5 6 7
0
20
40
60
80
100
120
140
n
xn
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32. 0
.( ) The solution of thedifference equation is
( 1)
1
6(( 0.5) 1)
( 0.5) (2)
0.5 1
2( 0.5) 4
( ) The equilibriumpoint
6
* 4
1 1 ( 0.5)
Since 0.5 0.5 1, * 4 is stable equilibriumpoint.
n
n
n
n
n
n
n
n
b i
b r
x r x
r
x
x
ii
b
x
r
r x

 

 
  
 
   
  
  
    
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33. (iii) MatLab Plot of the System
0 2 4 6 8 10 12 14 16 18 20
2
2.5
3
3.5
4
4.5
5
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34. Exercises
1) You currently have $5,000 in a saving
account that pays 0.5% interest each
month. If you add $400 each month, then
a) develop a model to calculate the amount in
the account.
b) how much is in the account after 3 years?
c) determine when the amount in the account
reaches $10,000.
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35. … Continued
2) Every day a person consumes 5 micrograms of
toxin which leaves the body at a rate of 2% per
day.
a) Develop a model to describe the
accumulation of toxin in the body.
b) How much toxin is accumulated in 30 days?
c) How much toxin is accumulated in the body
in the long run?
d) Plot toxin accumulation versus time.
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36. Assignment I Question 1
1) Suppose that there are currently 25,000
unemployed workers in Bahir Dar city. Each
month 8% of all those unemployed find jobs
but another 1500 become unemployed.
a) How many will be unemployed 6 months
from now?
b) At what level will the number of unemployed
workers stabilize over time?
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37. Assignment I Question 2
2) A certain drug is effective in treating a disease if the
concentration remains above 100 𝑚𝑔
𝑙 . The initial
concentration is 640 𝑚𝑔
𝑙. It is known from
laboratory experiments that the drug decays at the
rate of 20% of the amount present each hour.
a) Formulate a discrete model that describes the
concentration after each hour.
b) At what hour does the concentration reach
100 𝑚𝑔
𝑙?
c) Determine the maintenance doses that will keep
the concentration above the minimum effective
level of 100 𝑚𝑔
𝑙 and below the maximum safe
level of 800 𝑚𝑔
𝑙.
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38. 3.3 Modeling with Systems of Difference Equations
 Motivational problems (Systems)
 Population interaction (Prey-Predator interaction)
 Commodity distribution
 Disease spread
Consider system of first order linear homogeneous difference
equation
xn+1 = a11xn+a12yn +a13zn
yn+1 = a21xn+a22yn +a23zn
zn+1 = a31xn+a32yn +a33zn
with initial conditions x0, y0, and z0.
3.3.1 Systems of Linear homogeneous Difference Equations
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39. In matrix form, we can write the above
system as
39
1 11 12 13
1 21 22 23
1 31 32 33
0
0
0
with initial condition
n n
n n
n n
x a a a x
y a a a y
z a a a z
x
y
z



     
     

     
     
     
 
 
 
 
 
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40. This can be written more concisely as
1
1
1 1
1
11 12 13
21 22 23
31 32 33
0
0 0
0
,
,
with initial condition
n n
n
n n
n
n
n n
n
X RX
x
where X y
z
a a a
R a a a
a a a
x
X y
z
x
X y
z


 


 
 
  
 
 
 
 
  
 
 
 
 
  
 
 
 
 
  
 
 
0
The solution for the system is
n
n
X R X
 40
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41. Remarks:
1) Although the above solution is correct, it presents
a daunting practical problem.
It is difficult (if not impossible ) to calculate Xn for
large values of n.
(As such calculation involves excessive computation
of matrix multiplications to evaluate Rn.)
2) In the next section we use a systematic way of
obtaining the general solution, using the eigenvalues
and eigenvectors of the matrix R.
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42.  Let λ1, λ2 , . . . ,λn be n distinct eigenvalues of
R, and let v1, v2,. . ., vn be the corresponding
eigenvectors.
 An eigenvalue λ of multiplicity k has k linearly
independent eigenvectors, then the contribution
to the general solution will be of the form:
42
Then the general solution is:
𝑿𝒏=𝑐1𝜆1
𝑛
𝒗𝟏+𝑐2𝜆2
𝑛
𝒗𝟐 + ⋯ + 𝑐𝑛𝜆𝑛
𝑛
𝒗𝒏
where 𝒄𝟏, 𝒄𝟐, … . 𝒄𝒏 are real constants such that
𝑿𝟎 = 𝑐1𝒗𝟏+𝑐2𝒗𝟐 + ⋯ + 𝑐𝑛𝒗𝒏.
𝒄𝟏𝝀𝒏𝒗𝟏+𝒄𝟐𝒏𝝀𝒏𝒗𝟐 + ⋯ + 𝒄𝒌𝒏𝒌−𝟏𝝀𝒏𝒗𝒌.
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43. Equilibria and stability analysis of System of
Linear Homogeneous Difference Equations
The equilibrium vector of the system is X* such that
X*=RX*
⇒ (I-R)X*= 0, where 0 is the zero vector.
NB:
1) The zero vector X* = 0 is always an equilibrium vector.
2) If I−R ≠0, then the system has no nonzero vector as
equilibrium vector.(i.e. X* = 0 is the only equilibrium
vector.)
3) If I−R = 0, then the system has nonzero vector as
equilibrium vector. 43
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44. Definition 3.3.1
An equilibrium vector X* is stable if the general
solution Xn tends to X* regardless of the initial
conditions. Otherwise, it is called unstable.
Criteria of stability in terms of nature of eigenvalues:
Definition 3.3.2
If λ1, λ2 , . . . ,λn are eigenvalues of the matrix R, then
𝜌=max { 𝜆1 , 𝜆2 , . . . , 𝜆𝑛 } is called the spectral radius of R.
 If 𝜌 >1, then the solution will grow without bounds. Hence, the equilibrium
vector is unstable.
 If 𝜌 <1, then the solution tends to the zero vector. Thus, the equilibrium vector
is stable.
 If 𝜌 =1, then the solution converges to a multiple eigenvector associated to λ
where 𝜌 = 𝜆. Hence, an equilibrium vector is unstable.
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45. Example 1
A battle is to start between 10,000 troops of army
x and 5,000 troops of army y. Given that the
destroying rate of x is 0.1 and that of y is 0.15.
Develop a model to predict the outcome of the
battle.
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46. Solution:
• Let 𝑥𝑛 and 𝑦𝑛 be the number of troops of army x
and y after n time interval respectively.
Thus,
46
1
1
1
1
0 0
0.15
0.1 0.1
1 0.15
0.1 1
10,000, 5,000
n n n
n n n n n n
n n
n n
x x y
y y x x y
x x
y y
x y




 
    

   
 

   
 

 
   
 
1
1
1
0
1 0.15
, ,
0.1 1
10000
,
5000
n
n
n
n
n
n
x
Let X R
y
x
X X
y




   
 
   

 
 
   
 
   
 
 
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47. 1 0
Then, the model can be written as
, with intial condition
n n
X RX X
 
Let us use MATLAB to compute eigenvalues and eigenvectors of R.
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48. 1 2
1 2
Eigenvalues of R are
0.8775, 1.1225
The correspondingeigenvectorsare
1.2247 1.2247
,
1 1
v v
 
 

   
 
   
   
1 1 1 2 2 2
1 1 2 2 0
Thesolutionis
where
n n
n
X c v c v
c v c v X
 
 
 
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49. 1 1 2 2 0
1 2
1
2
1 1
0
2
implies
1.2247 -1.2247 10000
c c
1 1 50000
1.2247 1.2247 10000
1 1 50000
1.2247 1.2247
, then
1 1
6582.6
1582.6
c v c v X
c
c
If M
c
M X
c

 
     
 
     
     
  
   

 
   
   
 

 
  
 
   
 
   

 
 
49
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50. Hence , the solution is
1 1 1 2 2 2
1.2247 1.2247
6582.6(0.8775) ( 1582.6)(1.1225)
1 1
6582.6(0.8775) (1.2247) ( 1582.6)(1.1225) ( 1.2247)
6582.6(0.8775) (1) ( 1582.6)(1.1225) (1)
8061.71(0.
n n
n
n n n
n
n n
n
n n
n
n
X c v c v
x
y
x
y
x
 
 

     
  
     
   
 
   
  
 8775) 1938.21(1.1225)
6582.6(0.8775) 1582.6(1.1225)
n n
n n
n
y

 
Therefore, we observe that x army will win the battle.
Moreover, 𝐼 − 𝑅 ≠0, then 𝑥∗
=
0
0
, the only equilibrium
vector for the given system, which is unstable. (WHY?)
50
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51. MatLab plots of the Armies:
51
0 2 4 6 8 10 12 14 16 18 20
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
x 10
4
n
x
y
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52. Example 2
Suppose in a small town, on any given day 50% ill
people become healthy and 10% healthy people
become ill. If we start with 5000 healthy and 500 ill
people,
a) What will be the situation after 5 days?
b) What will happen in the long run?
Solution:
52
Let Hn = Number of healthy people after n days
In = Number of ill people after n days
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53. …continued
53
1 0
1 0
1 0
1 0
0.9 0.5 , 5000
0.1 0.5 , 500
0.9 0.5 5000
,
0.1 0.5 500
n n n
n n n
n n
n n
H H I H
I H I I
H H H
I I I




  
  
     
   
 
     
   
   
     
Therefore, the system will be modeled as :
How do we obtain this model? Justify !
Now, let us find the general solution of the system
using eigenvalue and eigenvector method:
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54. 54
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55. 1 1 1 2 2 2
5
5
1 5
416.6667(0.4) 916.6667(1)
1 1
416.6667(0.4) 4583.3
416.6667(0.4) 916.6667
4588
912
45883
917
n n
n
n n n
n
n
n
n
n
X c v c v
H
I
H
I
H
I
H
I
 


 

     
  
     
   
 
 
  




a.
b.
55
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56. 0 5 10 15 20 25 30
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
time(day)
population
healthy
ill
56
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57. Assignment I Question 3
1) Let 𝑈𝑛 and 𝑉
𝑛 be the total amount of pollutant in
lakes 𝐴 and 𝐵 respectively, in year 𝑛, and 38% of the
pollutant from lake 𝐴 and 13% of the pollutant from
lake 𝐵 are removed every year. Also, the pollutant
that is removed from lake 𝐴 is added to lake 𝐵 due to
the flow of water from lake 𝐴 to lake 𝐵. It is also
assumed that 3 𝑡𝑜𝑛𝑠 of pollutant are directly added
to lake 𝐴 and 9 𝑡𝑜𝑛𝑠 of pollutant are added to lake 𝐵.
a) Develop a discrete dynamical system model to
describe this system. Find the equilibrium points
and state whether they are stable or not.
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58. … Assignment I Question 3 Contd.
b) Suppose it is determined that an equilibrium level of
a total of 10 𝑡𝑜𝑛𝑠 of pollutant in lake 𝐴 and a total of
30 𝑡𝑜𝑛𝑠 in lake 𝐵 would be acceptable. What
restrictions should be placed upon the total amounts
of pollutants that are added directly, so that these
equilibria can be achieved?
c) Plot 𝑈𝑛vs 𝑛 and 𝑉
𝑛 vs 𝑛 using MatLab.(MatLab code
is part of the solution)
58
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59. Consider a linear systems of non-homogeneous
difference equations
Xn+1 = RXn + B, with initial condition X0
where B is a vector whose components are constants.
Therefore, the solution is of the form:
( ) ( )
( )
n ( )
1
where is the solution of the
associated homogeneous system:
and is the particular solution.
h p
n n n
h
n
p
n n n
X X X
X
X R X X

 

59
3.3.2 Systems of Linear non-homogeneous
Difference Equations (Reading Assignment!)
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60. 4/11/2022 Math 3111/484 60
End of Chapter 3!
Thank You!