What is a model
Some notations
– Independent variables:
• Time variable: t, n
• Space variable: x in one dimension (1D), (x,y) in 2D or (x,y,z) in 3D
– State variables or dependent variables: is a scalar or vector
valued variables, generally a function of the independent variables
• u(t): the total capital invested; w(t): total dividend to the shareholders
• Parameters: The parameters are physical quantities that characterize the
physical system to be modeled. These quantities can be defined either in a
dimensional or in a dimensionless form and can be obtained either by direct
measurements on the system itself or by comparisons between the
predictions of the model and the behavior of the real system.
• c: speed of light; g: gravitational acceleration; profit rate, ….
What is a model
A mathematical model is an equation, or a set of
equations, whose solution provides the time-space
evolution of the state variable, that is the physical
behavior, in the frame work of the mathematical model,
of the related physical system.
The equation that defines the mathematical model can
also be called the state equation
The Process of Mathematical Modeling
Problem identification: The questions to be answered must be
clarified. The underlying mechanism at work in the physical situation must be
identified as accurately as possible. Formulate the problem in words, and
document the relevant data.
Assumptions and physical laws: The must be analyzed to decide
which factors are important and which factors are to be ignored so that
realistic assumptions can be made.
Model construction or selection: This is the translation of the
problem into mathematical language which normally results in a collection of
questions and/or inequalities after the variables has been identified. The
``word’’ problem is transformed into an abstract mathematical problem.
The Process of Mathematical Modeling
Model analysis, solution & reduction: The mathematical
problem is solved so that the unknown variables are expressed in terms of
known quantities, and/or it is analyzed to obtain information about parameters.
In some cases, the model may need simplified.
Model interpretation: The solution to the abstract mathematical
problem must be compared to the original ``word’’ problem to see if it makes in
the real-world situation. If not, go back to formulate more realistic
assumptions, and construct a new model.
Model validation, revision & modification: Check whether the
solution agrees with the data of the real-world problem. If the correlation is
unsatisfactory, return to the ``word’’ problem for a re-appraisal of the data and
the assumptions. Modify or add assumptions and construct a new model.
The Process of Mathematical Modeling
Model Implementation: If the solution agrees with the data, then the
model can be used to predict what will happen in the future, or conclusions
can be drawn to help in future planning, etc. In the case of predictions care
should be taken to determine the time interval in which the predictions are
valid.
Flow chart of the modelling process
identification  assumptions and physical laws
construction or selection analysis, solution & reduction
 interpretation  validation, revision & modification
 Implementation
Model classification
No unique classification
Classification by the state variables
– Discrete models vs continuous models – state variables depend
space variables?
– Dynamic model vs static models – state variables depend on time t
???
Classification by the type of state equation
f (u; x, r , t )  0
– Algebra model –
– ODE models -- y  f (t , y )
– PDE models ---  t u ( x , t )  Lu
Model classification
Classification by structure of the state equation
– Linear models vs nonlinear models
Classification by parameters and stochasticity
– Optimization and/or static models; dynamic models;
probability models; ….
Justify a model
– Usually use good vs bad (or valid vs invalid)
– No absolutely right vs wrong!!!
– A model maybe good in some situations & bad otherwise!!
Main focus
Main focus:
– How to convert real-world into mathematical models
– Basic mathematical analysis techniques
• Dimensionless, analytical solutions, steady states, stability,
equilibrium, bifurcation, ….
– Some basic numerical methods
Not focus
– Theory of ODE & PDE, general numerical methods
Some basic knowledge
– Calculus, linear algebra, basic physical laws
Another example: World population
World population
time
1650 1800 1918-27 1960 1974 1987 1999 2010 2022
population 0.5
1
2
3
4
5
6
7
8
Questions:
– What is the population in 2020??
– What is the population in 2050???
Malthus’s model
Consider the time interval
[t , t  t ]
N (t  t )  N (t )  # of birth  # of death+miration=b N t  d N t  H t
 /t
N (t  t )  N (t )
bN d N H
t
 t  0
dN(t)
 (b  d ) N (t )  H
dt
births
 number of births per unit time per unit population
N
deaths
d
 number of deaths per unit time per unit population
N
b
Malthus’s model
Malthus’s assumption:
– Unlimited resource & no migration
– Birth rate and death rate are both constants
The equation
d N (t )
 (b0  d 0 ) N (t ) : r0 N (t ),
dt
N (0)  N 0  0

N(t)=N 0 e r0 t ,
t 0
t  0,
Malthus’s model
Phenomena
– Population `explosion’: ``story of Prof. Yanchu Ma’’
b0  d0  r0  0
– Population distinction
b0  d0  r0  0
 N (t )   when t  
 N (t )  0 when t  
– No change b0  d0  r0  0  N (t )  N0 for any t
World population
time
1650 1800 1918-27 1960 1974 1987 1999 2010 2022
population 0.5
1
2
3
4
5
6
7
8
Malthus’s model
Prediction for World population by Malthus’s model
r0 t
N(t)=N 0 e ,
t 0
For prediction, we need determine the parameters
N0 & r0 with N (t  0)  N0
Choose two data in the population & set year 1650 as t=0
– Time
– Population
1650 (t=0) 1800 (t=150)
0.5
1
Determine the parameters
N 0  0.5 &1  N (t  150)  N 0 e r0t  0.5 e150 r0  r0 
1
ln 2  0.0046
150
World population
Prediction:
N(t)=0.5 er0 t  0.5 e0.0046t ,
t 0
time
1650 1800 1918-27 1960 1974 1987 1999 2010 2022
population 0.5
1
2
3
4
5
6
7
8
prediction 0.5
1
1.73-1.80 2.09 2.23 2.37 2.51 2.64 2.79
Observations:
– Under predictions & possible reasons:
• Ways to determine parameters!!
• Model errors, ……..
Population of Singapore
1960
1961
1962
1963
1964
1965
1966
1967
1968
1969
1970
1,646.40
1,702.40
1,750.20
1,795.00
1,841.60
1,886.90
1,934.40
1,977.60
2,012.00
2,042.50
2,074.50
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
2,112.90
2,152.40
2,193.00
2,229.80
2,262.60
2,293.30
2,325.30
2,353.60
2,383.50
2,413.90
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
2,532.80
2,646.50
2,681.10
2,732.20
2,736.00
2,733.40
2,774.80
2,846.10
2,930.90
3,047.10
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
3,135.10
3,230.70
3,313.50
3,419.00
3,524.50
3,670.70
3,796.00
3,927.20
3,958.70
4,027.90
2001
2002
2003
2004
2005
2006
4,138.00
4,176.00
4,186.10
4,238.30
4,341.80
4,483.90
Malthus’s model prediction: case 1
Choose two data in the population
– Time
– Population
1960 (t0=0) 1970 (t1=10)
1646.40
2074.50
Determine the parameters
N 0  1646.4 & 2074.5  N (t1  10)  N 0 e r0t1  1646.4 e10 r0 
r0 
1 2074.5
ln
 0.02311
10 1646.4
Prediction
N(t)=1646.4 er0 t  1646.4 e0.02311t ,
t0
Singapore population
Population vs Predicted Population
6,000.00
Population
4,000.00
3,000.00
Predicted
Population
2,000.00
1,000.00
t,year
45
40
35
30
25
20
15
10
5
0.00
0
N(t) '000
5,000.00
Singapore population
Percentage Error
-10.00
-15.00
t,year
45
40
35
30
25
20
15
10
-5.00
5
0.00
0
N(t)'000
5.00
Percentage Error
Prediction
Predicted population from year 2007-2050
10000.00
Predicted
population from
year 2007-2050
5000.00
t,year
89
83
77
71
65
59
53
0.00
47
N(t)
15000.00
Malthus’s model prediction: case 2
Choose two data in the population
– Time
– Population
1960 (t0=0) 1980 (t1=20)
1646.40
2413.90
Determine the parameters
N 0  1646.4 & 2413.9  N (t1  20)  N 0 e r0t1  1646.4 e 20 r0 
r0 
1 2413.9
ln
 0.019133
20 1646.4
Prediction
N(t)=1646.4 er0 t  1646.4 e0.019133t ,
t 0
Singapore population
Population vs Predicted Population
Population
4,000.00
2,000.00
Predicted
Population
t,year
45
40
35
30
25
20
15
10
5
0.00
0
N(t)'000
6,000.00
Singapore population
Percentage error
10.00
5.00
Percentage error
t,year
42
36
30
24
18
12
-5.00
6
0.00
0
Error %
15.00
Prediction
Predicted population in year 2007-2050
Predicted
population in year
2007-2050
5000.00
t,year
89
83
77
71
65
59
53
0.00
47
N(t)
10000.00
Another example: mixing problem
The physical mixing of two species
– Chemical system
– Biological system
– Environmental engineering
A solid material is dissolved in a fluid
– Solid is referred as solute; Fluid is referred as solvent
– mixture is referred as solution; e.g. salt is dissolved in water
Consider a reservoir or container with inflow and outflow
– Inflow:
• solution flows into the container at rate qi (liter of solution /sec: L/s)
• Concentrations of the solute (units of mass of solute per unit volume of
solution): ci (kg of solute/ liter of solution: kg/L)
Another example: mixing problem
– Outflow:
• solution flows out of the container at rate qe (liter of solution /sec: L/s)
• Concentrations of the solute (units of mass of solute per unit volume of
solution): ce  c(t ) (kg of solute/ liter of solution: kg/L)
– Solution in the container is kept thoroughly mixed by stirring: uniform mixing
Question:
– What is the amount of solute in the container?
– What is the concentration of the solute in the container?
Mathematical Model:
– Variables:
• t : time; x (t ) : amount of solute in the container
• V(t): volume of solution in the container
x(t )
• c(t )  V (t ) : concentration of the solute
Another example: mixing problem
– Physical law or first principle: mass conservation of solute
 mass change of solute  solute flow into  solute flow out of 
in the container
   the container    the container


 
 

– Let t be infinite small time step and consider time interval [t , t  t ]
– Change of solute in the time interval is:
x : x(t  t )  x(t )  solute flow into  solute flow out of 
 qi ci t  qe ce (t ) t  [qi ci  qe c(t )] t
– Change rate:
x x(t  t )  x(t )
x(t )
x(t )
:
 qi ci  qe c(t )  qi ci  qe
 qi ci  qe
t
t
V (t )
V0  (qi  qe )t
Another example: mining problem
– Initial data:
V0 : volume of solution in the container at t  0
x0  x(0) : amount of solute in the container at t  0
– Rate of change: t  0
x(t ) :
dx
x(t  t )  x(t )
x(t )
 x(t )  lim
 qi ci  qe
, t 0

t

0
dt
t
V0  (qi  qe ) t
– First order ODE for time evolution of solute:
x(t )  qi ci  qe
x(0)  x0
x(t )
, t 0
V0  (qi  qe ) t
Another example: mixing problem
Solution:
– A special case:
ci  0, V0  0, x0  0, qi  qe  0
• Simplified first order ODE
x(t ) 
dx
x(t )
 qe
,
dt
V0
• Divide both sides by x(t )
dx
1
  qe
x dt
V0
• Integrate both sides over [0,t]
x(0)  x0
Another example: mixing problem
t
t
t
qe t
qe t
dx
dt
ln x(t )  ln x(0)  ln x 0  
dt   qe  

x dt
V0
V0 0
V0
0
0
t
– Solution
Amount of solute: x(t )  x(0) e  qe t / V0  x0 e  qe t /V0 ,
Concentration:
c(t ) 
– Interpretation
•
•
•
•
t 0
x(t ) x(t ) x0  qe t /V0

 e
 c0 e  qe t /V0 ,
V (t ) V0 V0
t 0
Amount of solute decays exponentially with time
Concentration decays exponentially with time
Decay rates are the same
Long time behavior t   : x(t )  0 & c(t )  0
Another example: mixing problem
– Another special case:
• Simplified first order ODE
x(t ) 
ci  0, V0  0, x0  0, qi  qe
dx
x(t )
 qe
,
dt
V0  (qi  qe ) t
• Divide both sides by x(t )
dx
1
 qe
x dt
V0  (qi  qe ) t
• Integrate both sides over [0,t]
x(0)  x0
Another example: mixing problem
t
ln
t
t
q
x(t )
dx
dt
 ln x(t )  ln x(0)  
dt   qe
  e ln[V0  (qi  qe )t ]
x(0)
x dt
V0  (qi  qe )t
qi  qe
0
0
0

qe
q
V  (qi  qe )t
[ln(V0  (qi  qe )t )  ln V0 ]   e ln 0
qi  qe
qi  qe
V0




qe
V0
V0

ln 
  ln 

qi  qe  V0  (qi  qe )t 
V

(
q

q
)
t
i
e 
 0
qe
qi  qe
– Solution


V0
x(t )  x0 

V

(
q

q
)
t
i
e 
 0
qe
qi  qe
,
x(t )
x(t )
c(t ) 

,
V (t ) V0  (qi  qe )t
t 0
Another example: mixing problem
– Interpretation
• When inflow is fast than outflow: qi  qe
– Amount of solute and concentration decay with time
– Decay rates are different
– Long time behavior t   : x(t )  0 & c(t )  0
• When inflow is slower than outflow: qi  qe
qe
qe  qi
 V0  (qe  qi )t 
x(t )
x(t )
x(t )  x0 
,
c(t ) 

,

V0
V (t ) V0  (qe  qi )t


– Finite time empty: t*  V0 /(qe  qi )  V (t )  0
– Amount of solute and concentration decay with time
t  t*: x(t )  0 & c(t )  0
t0
Another example: mixing problem
– Another special case:
ci  0, V0  0, x0  0, qi  qe  0
• Simplified first order ODE
dx
x(t )
x(t ) 
 qi ci  qe
,
dt
V0
x(0)  x0
• Solution
x(t )  ci V0   x0  ci V0  e

qe t
V0
x0
x(t )
c(t ) 
 ci  (  ci ) e
V0
V0

qe t
V0
 ci  (c0  ci ) e

qe t
V0
Another example: mixing problem
– Interpretation
• Long time behavior: t   : x(t )  ci V0 & c(t )  ci
• When initial concentration is dense: c0  ci
– Amount of solute and concentration decay with time
– Decay rates are the same
– Dilute process
• When initial concentration is dilute: c0  ci
– Amount of solute and concentration increase with time
– Increase rates are the same
– Pollution process
Another example: mining problem
– First order ODE for time evolution of solute:
x(t )  qi ci  qe
x(0)  x0
– A special case:
•
•
•
•
x(t )
, t 0
V0  (qi  qe ) t
ci  0, V0  0, x0  0, qi  qe  0
Amount of solute decays exponentially with time
Concentration decays exponentially with time
Decay rates are the same
Long time behavior t   : x(t )  0 & c(t )  0
Another example: mining problem
– Another special case:
ci  0, V0  0, x0  0, qi  qe
• Amount of solute and concentration decay with time
• Decay rates are different
• Long time behavior t   : x(t )  0 & c(t )  0
– Another special case:
ci  0, V0  0, x0  0, qi  qe  0
• Long time behavior: t   : x(t )  ci V0 & c(t )  ci
• When initial concentration is dense: c0  ci
– Dilute process
• When initial concentration is dilute:
– Pollution process
c0  ci
Another example: mixing problem
– General case:
x(t )  qi ci  qe
x(t )
, t 0
V0  (qi  qe ) t
x(0)  x0
– Analytical solution in explicit form & interpretation
Non constant inflow and outflow rates:
x(t )  qi (t ) ci (t )  qe (t )
x (t )
t
, t0
V0   [ qi ( s )  qe ( s )]ds
0
x (0)  x0
– Analytical solution in integral form & interpretation