Chapter 1 Mathematical Modeling - Chulalongkorn...
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Chapter 1 Mathematical Modeling Paisan Nakmahachalasint
Transcript of Chapter 1 Mathematical Modeling - Chulalongkorn...
Chapter 1 Mathematical Modeling
Paisan Nakmahachalasint
Modeling Process
RealReal--world world behaviorbehavior ModelModel
Mathematical Mathematical conclusionsconclusions
RealReal--world world conclusionsconclusions
Observation
Simplification
Analysis
Interpretation
Trials
Discrete Dynamical Systems
We are often interested in building models to explain behavior or make predictions.In this chapter we direct our attention to modeling change.
change = future value –
present valueData will be collected over a period of time and plotted to capture the trend of the change.Discrete time leads to difference equations.Continuous time leads to differential equations.
Growth of a Yeast CultureTime in hours
n
Observed yeast
biomassPn
Change in
biomasspn+1
–
pn
0 9.6 8.7
1 18.3 10.7
2 29.0 18.2
3 47.2 23.9
4 71.1 48.0
5 119.1 55.5
6 174.6 82.7
7 257.3
y = 0.4495x + 5.273
0102030405060708090
0 50 100 150 200
BiomassCh
ange
in b
iom
ass
1 0.45n n n np p p p+Δ = − ≈
Growth of a Yeast Culturen pn pn+1
–
pn
0 9.6 8.7
1 18.3 10.7
2 29.0 18.2
3 47.2 23.9
4 71.1 48.0
5 119.1 55.5
6 174.6 82.7
7 257.3 93.4
8 350.7 90.3
9 441.0 72.3
10 513.3 46.4
11 559.7 35.1
12 594.8 34.6
13 629.4 11.4
14 640.8 10.3
15 651.1 4.8
16 655.9 3.7
17 659.6 2.2
18 661.8
( )1 665n n n n np p p k p p+Δ = − = −
0
100
200
300
400
500
600
700
0 5 10 15 20
Time in hours
Yeas
t bio
mas
s
Growth of a Yeast Culturepn+1
–
pn pn
(665 -
pn
)
8.7 6291.84
10.7 11834.61
18.2 18444.00
23.9 29160.16
48.0 42226.29
55.5 65016.69
82.7 85623.84
93.4 104901.21
90.3 110225.01
72.3 98784.00
46.4 77867.61
35.1 58936.41
34.6 41754.96
11.4 22406.64
10.3 15507.36
4.8 9050.29
3.7 5968.69
2.2 3561.84
y = 0.00082x - 0.3079
0102030405060708090
100
0 50000 100000 150000
pn(665-pn)
p n+1
- p n
( )1 0.00082 665n n n np p p p+ − = −
Growth of a Yeast Culturen Observations Predictions
0 9.6 9.60
1 18.3 14.72
2 29.0 22.52
3 47.2 34.31
4 71.1 51.93
5 119.1 77.87
6 174.6 115.10
7 257.3 166.65
8 350.7 234.30
9 441.0 316.49
10 513.3 406.32
11 559.7 491.93
12 594.8 561.27
13 629.4 608.69
14 640.8 636.61
15 651.1 651.33
16 655.9 658.58
17 659.6 662.02
0
100
200
300
400
500
600
700
0 5 10 15 20Time in hours
Yeas
t bio
mas
s
ObservationsPredictions
Spotted Owl Population
Unconstrained growthAssume the population changes by only births and deaths.
k
may be either positive or negative constant.
1n n n
n n
n
p p p
bp dp
kp
+Δ = −
= −
=
Spotted Owl Population
Constrained growthSuppose the habitat can only support an owl population of size M.
k
is a positive constant.
( )1n n n
n n
p p p
k M p p
+Δ = −
= −
Spotted Owl Population
Competing speciesSuppose a second species lives in the habitat. Denote the population of the competing species by cn.
k1
, k2
, α1
, α2
are positive constants.
1 1 1
1 2 2
n n n n n n
n n n n n n
p p p k p c p
c c c k c c p
α
α+
+
Δ = − = −
Δ = − = −
Spotted Owl Population
Predator-Prey speciesAssume that the spotted owl’s primary food source is a single prey, say mice. Denote the size of the mouse population after n periods by mn.
k1
, k2
, α1
, α2
are positive constants.
1 1 1
1 2 2
n n n n n n
n n n n n n
p p p k p m p
m m m k m m p
α
α+
+
Δ = − = − +
Δ = − = −
Spread of a Contagious Disease
There are 400 students in a college dormitory.
Some students has a severe case of the flu.
some interaction between those infected and those not infected is required to pass on the disease and all are susceptible to disease.
Let in represent the number of infected students after n time periods.
( )1 400n n n n ni i i k i i+Δ = − = −
Heating of a Cold Object
A cold can of soda is taken from a refrigerator and placed in a warm classroom.
Suppose the can is initially at 5°C and the room temperature is 30°C .Let tn represent the temperature of the soda after n time periods.
( )1 30n n n nt t t k t+Δ = − = −
Long-Term Behavior
Grows large
without bound
Grows negative
without bound
Approaches a limit
Long-Term Behavior
Oscillates
Oscillates while
approaching a limit
Periodic
Long-Term Behavior
( )1 01 , 0.2n n na r a a a+ = − =
0
0.2
0.4
0.6
0.8
1
0 5 10 15 200
0.2
0.4
0.6
0.8
1
0 5 10 15 20
0
0.2
0.4
0.6
0.8
1
0 5 10 15 200
0.2
0.4
0.6
0.8
1
0 5 10 15 20
r
= 2 r
= 3
r
= 3.6 r
= 3.7
Competitive Hunter Models
Spotted owls and hawks are competing for resources.Let On and Hn denote the number of owls and hawks, respectively, at the end of day n.
( )
( )1 1 1
1 2 2
1
1
n n n n
n n n n
O k O O H
H k H O H
α
α
+
+
= + −
= + −
Competitive Hunter Models
Choose specific values for k1, k2, α1, α2.
If an equilibrium point exists, let (O,H) be an equilibrium point.
The solutions are
1
1
1.2 0.001
1.3 0.002
n n n n
n n n n
O O O H
H H O H
+
+
= −
= −
1.2 0.001
1.3 0.002
O O OH
H H OH
= −
= −
( , ) (0, 0) or (150,200).O H =
Competitive Hunter Modelsn Owls Hawks
0 150 2001 150 200
2 150 200
3 150 2004 150 200
5 150 200
6 150 2007 150 200
8 150 200
9 150 20010 150 200
11 150 200
12 150 20013 150 200
14 150 200
15 150 20016 150 200
17 150 200
18 150 20019 150 200
20 150 200
21 150 20022 150 200
23 150 200
24 150 20025 150 200
26 150 200
27 150 20028 150 200
29 150 200
30 150 200
0
50
100
150
200
250
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Haw ks
Competitive Hunter Modelsn Owls Hawks0 151 1991 151 199
2 151 198
3 152 1984 152 197
5 152 196
6 153 1957 154 194
8 155 193
9 156 19110 157 188
11 159 186
12 161 18213 164 178
14 168 173
15 172 16716 178 159
17 185 151
18 194 14019 206 127
20 221 113
21 240 9722 265 80
23 297 61
24 338 4325 391 27
26 459 14
27 544 528 650 1
29 779 0
30 935 0
0100200300400500600700800900
1000
0 5 10 15 20 25 30 35
Days
Bird
sOw ls
Haw ks
Competitive Hunter Modelsn Owls Hawks
0 149 2011 149 201
2 149 202
3 148 2024 148 203
5 148 204
6 147 2057 146 206
8 145 208
9 144 21010 143 212
11 141 215
12 139 21913 136 224
14 133 230
15 129 23716 124 247
17 119 260
18 111 27619 103 298
20 93 326
21 81 36322 68 413
23 54 480
24 39 57325 24 701
26 12 877
27 4 111928 0 1447
29 0 1880
30 0 2444
0
500
1000
1500
2000
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Haw ks
Competitive Hunter Modelsn Owls Hawks
0 10 101 12 13
2 14 16
3 17 214 20 26
5 23 33
6 27 427 31 52
8 36 64
9 41 7910 46 96
11 51 116
12 55 13913 58 165
14 60 196
15 60 23116 58 273
17 54 322
18 48 38419 39 463
20 29 566
21 18 70322 9 889
23 3 1139
24 0 147525 0 1917
26 0 2492
27 0 324028 0 4212
29 0 5474
30 1 7120
0
200
400
600
800
1000
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Haw ks
Predator-Prey Models
Assume mice are the sole source of food for spotted owls.Let On and Mn denote the number of owls and hawks, respectively, at the end of day n.
( )
( )1 1 1
1 2 2
1
1
n n n n
n n n n
O k O O M
M k M O M
α
α
+
+
= − +
= + −
Predator-Prey Models
Choose specific values for k1, k2, α1, α2.
If an equilibrium point exists, let (O,M) be an equilibrium point.
The solutions are
1
1
0.7 0.002
1.2 0.001
n n n n
n n n n
O O O M
M M O M
+
+
= +
= −
0.7 0.002
1.2 0.001
O O OM
M M OM
= +
= −
( , ) (0, 0) or (200,150).O M =
Predator-Prey Modelsn Owls Mice
0 200 1501 200 150
2 200 150
3 200 1504 200 150
5 200 150
6 200 1507 200 150
8 200 150
9 200 15010 200 150
11 200 150
12 200 15013 200 150
14 200 150
15 200 15016 200 150
17 200 150
18 200 15019 200 150
20 200 150
21 200 15022 200 150
23 200 150
24 200 15025 200 150
26 200 150
27 200 15028 200 150
29 200 150
30 200 150
0
50
100
150
200
250
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Mice
Predator-Prey Modelsn Owls Mice
0 220 1301 211 127
2 202 126
3 192 1264 183 127
5 174 129
6 167 1327 161 137
8 157 142
9 154 14810 154 155
11 155 162
12 159 16913 165 176
14 174 183
15 185 18716 199 190
17 215 190
18 232 18719 250 181
20 265 172
21 277 16122 283 149
23 283 136
24 275 12525 261 116
26 243 109
27 223 10428 203 101
29 183 101
30 165 103
0
50
100
150
200
250
300
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Mice
Predator-Prey Modelsn Owls Mice
0 180 1701 187 173
2 196 176
3 206 1764 217 175
5 228 172
6 238 1687 246 161
8 252 154
9 254 14610 252 138
11 245 131
12 236 12513 224 120
14 211 117
15 197 11616 184 117
17 171 118
18 161 12219 152 127
20 144 133
21 139 14022 137 149
23 136 158
24 138 16825 143 178
26 152 188
27 163 19828 179 205
29 198 209
30 222 209
0
50
100
150
200
250
300
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Mice
Predator-Prey Modelsn Owls Mice
0 150 2001 165 210
2 185 217
3 210 2214 239 219
5 272 210
6 305 1957 332 174
8 348 151
9 349 12910 334 110
11 307 95
12 274 8513 238 79
14 204 76
15 174 7516 148 77
17 126 81
18 109 8719 95 95
20 85 105
21 77 11722 72 132
23 69 149
24 69 16825 72 190
26 78 214
27 87 24128 103 268
29 128 294
30 164 315
050
100150200250300350400
0 5 10 15 20 25 30 35
Days
Bird
s
Ow ls
Mice
A Car Rental Company
A car rental company has distributorships in Los Angeles and San Francisco.
Records show that only 60% of cars rented in LA are returned to LA, and only 70% of cars rented in SF are returned to SF.
30%
LA SF
40%
70%60%
A Car Rental Company
Let Ln and Sn denote the number of cars in LA and SF, respectively, at the end of day n.
If an equilibrium point exists, let (L,S) be an equilibrium point.
We obtain
1
1
0.6 0.3
0.4 0.7
n n n
n n n
L L S
S L S
+
+
= +
= +
0.6 0.3
0.4 0.7
L L S
S L S
= +
= +
3.
4L S=
A Car Rental Company
n LA SF
0 7000 0
1 4200 2800
2 3360 3640
3 3108 3892
4 3032 3968
5 3010 3990
6 3003 3997
7 3001 3999
8 3000 4000
9 3000 4000
10 3000 4000
0
1000
2000
3000
4000
5000
6000
7000
0 2 4 6 8
Days
Car
s
LA
SF
A Car Rental Company
n LA SF
0 5000 2000
1 3600 3400
2 3180 3820
3 3054 3946
4 3016 3984
5 3005 3995
6 3001 3999
7 3000 4000
8 3000 4000
9 3000 4000
10 3000 4000
0
1000
2000
3000
4000
5000
6000
7000
0 2 4 6 8
Days
Car
s
LA
SF
A Car Rental Company
n LA SF
0 2000 5000
1 2700 4300
2 2910 4090
3 2973 4027
4 2992 4008
5 2998 4002
6 2999 4001
7 3000 4000
8 3000 4000
9 3000 4000
10 3000 4000
0
1000
2000
3000
4000
5000
6000
7000
0 2 4 6 8
Days
Car
s
LA
SF
A Car Rental Company
n LA SF
0 0 7000
1 2100 4900
2 2730 4270
3 2919 4081
4 2976 4024
5 2993 4007
6 2998 4002
7 2999 4001
8 3000 4000
9 3000 4000
10 3000 4000
0
1000
2000
3000
4000
5000
6000
7000
0 2 4 6 8
Days
Car
s
LA
SF
Voting Tendency
There are three political parties: Republicans, Democrats, and Independents.
In the next election, the voters change according to the diagram below with nobody entering or leaving the system.
Republicans Democrats Independents
20%
20%
60%
20%
75%
20%20%
5%
60%
Voting Tendency
Let Rn, Dn, and In denote the number of Republican, Democrats, and Independents voters, respectively, in the nth election.
If there exists an equilibrium points (R,D,I), then R:D:I = 2.2221:0.77777:1 approximately.
1
1
1
0.75 0.20 0.40
0.05 0.60 0.20
0.20 0.20 0.40
n n n n
n n n n
n n n n
R R D I
D R D I
I R D I
+
+
+
= + +
= + +
= + +
Voting Tendency
0
50,000
100,000
150,000
200,000
250,000
0 2 4 6 8 10
n th election
Vote
rs
Republicans Democrats Independents
n Republicans Democrats Independents
0 222,221 77,777 100,000
1 222,221 77,777 100,000
2 222,221 77,777 100,000
3 222,221 77,777 100,000
4 222,221 77,777 100,000
5 222,221 77,777 100,000
6 222,221 77,777 100,000
7 222,221 77,777 100,000
8 222,221 77,777 100,000
9 222,221 77,777 100,000
10 222,221 77,777 100,000
0
50,000
100,000
150,000
200,000
250,000
0 2 4 6 8 10
n th election
Vote
rs
Republicans Democrats Independents
Voting Tendency
n Republicans Democrats Independents
0 227,221 82,777 90,000
1 222,971 79,027 98,000
2 222,234 78,165 99,600
3 222,148 77,930 99,920
4 222,165 77,850 99,984
5 222,187 77,815 99,996
6 222,202 77,797 99,999
7 222,210 77,788 99,999
8 222,215 77,783 99,999
9 222,218 77,781 99,999
10 222,219 77,779 99,999
Voting Tendency
n Republicans Democrats Independents
0 100,000 100,000 199,998
1 174,999 105,000 119,999
2 200,249 95,750 103,999
3 210,936 88,262 100,799
4 216,175 83,664 100,159
5 218,927 81,039 100,031
6 220,416 79,576 100,006
7 221,230 78,768 100,001
8 221,676 78,322 100,000
9 221,921 78,077 100,000
10 222,056 77,942 100,000
0
50,000
100,000
150,000
200,000
250,000
0 2 4 6 8 10
n th election
Vote
rs
Republicans Democrats Independents
Voting Tendency
n Republicans Democrats Independents
0 0 0 399,998
1 159,999 80,000 159,999
2 199,999 88,000 111,999
3 212,399 85,200 102,399
4 217,299 82,220 100,479
5 219,610 80,293 100,095
6 220,804 79,175 100,019
7 221,446 78,549 100,003
8 221,795 78,202 100,000
9 221,987 78,011 100,000
10 222,092 77,906 100,000
050,000
100,000150,000200,000250,000300,000350,000400,000
0 2 4 6 8 10
n th election
Vote
rs
Republicans Democrats Independents
Modeling with Differential Equations
represents the instantaneous rate of change.
The derivative is used in two distinct roles:To represent the instantaneous rate of change in continuous problems.
To approximate an average rate of change in discrete problems, where calculus can be used to help discovering functional relations.
Many differential equations cannot be solved analytically.
dPdt
Population Growth: Malthusian Model
Thomas Malthus (1766-1834) used an exponential growth to model human population and is referred to as the Malthusian model.
where k
is a constant. (k
= birth rate –
death rate)
0 0, ( )dP
kP P t Pdt
= =
Population Growth: Malthusian Model
Solving the model, we get
The number of population is doubled
every period of
The population will grow exponentially, which is unreasonable over the long term.
0( )0( ) .k t tP t Pe −=
ln 2.
k
Population Growth: Logistic Model
Refining the model to reflect limited growth due to limited resources:
where r
is a constant.The model was introduced by Pierre-Francois Verhulst (1804-1849) and is referred to as logistic growth.
( ) 0 0, ( )dP
r M P P P t Pdt
= − =
Population Growth: Logistic Model
1
2
ln
rMt
dP dPrM dt
P M PP
rMt CM P
PC e
M P
+ =−
= +−
=− 0
2
2
12
( )3
1
1
1
rMt
rMt
rMt
rM t t
MC eP
C e
MC e
MC e
− −
− −
=+
=+
=+
0
0( )
0 0
( )( ) rM t t
MPP t
P M P e− −=+ −
Population Growth: Logistic Model
t
M
2M
0P
0t *t
P
( )( )
2 2 ( 2 )
d dP dP r M P P
dt dt dt
rMP rPP P M P
⎛ ⎞⎟⎜′′ = = −⎟⎜ ⎟⎜⎝ ⎠′ ′ ′= − = −
*0
0
1ln 1
Mt t
rM P
⎛ ⎞⎟⎜ ⎟= + −⎜ ⎟⎜ ⎟⎜⎝ ⎠
*( )( )
1 rM t t
MP t
e− −=
+
Prescribing Drug Dosage
We want to determine how much of a drug dosage to prescribe and how often the dosage should be administered.
For most drugs there is a concentration below which the drug is ineffective and a concentration above which the drug is dangerous.
Prescribing Drug Dosage
Let H denote the highest safe level of the drug, and L its lowest effective level.It would be desirable to prescribe a dosage C0 with time T between doses so that the concentration remains between L and H over each dose period.
0
0( ) ( ) ( ) kt
C H L
C t kC t C t C e−
= −
′ = − ⇒ =
Prescribing Drug Dosage0 0
20 0 0 0
2 2 30 0 0 0 0 0
( 1)0 0 0 0
kT
kT kT kT
kT kT kT kT kT
n kT kT nkT
C C e
C C e C e C e
C C e C e C e C e C e
C C e C e C e
−
− − −
− − − − −
− − − −
+ +
+ + + +
+ + + +nR
1R
0C
1C
2C3C
2R3R 4R
L
H
t
( )
( )
( 1)0
0
11
11
n kTn kT
nkTn kT
CC e
eC
R ee
− +−
−
= −−
= −−
Prescribing Drug Dosage
( )0 0lim lim 11 1
nkTn kT kTn n
C CR R e
e e−
→∞ →∞= = − =
− −
0,1kT
RH L
LL C H Le
= = − ⇒−
=−
1ln
HT
k L=
Solutions of Differential Equations
-3 -2 -1 0 1 2 3
-1.5
-1
-0.5
0
0.5
1
1.5
2
2y y x′ = −
2
21
xyy
x′ = −
+
(1 )2x
y x y′ = − +
Slope Fields
-3 -2 -1 0 1 2 3
-1.5
-1
-0.5
0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
-1.5
-1
-0.5
0
0.5
1
1.5
2
Autonomous Differential Equations
A differential equation for which
is called an autonomous
differential equation.The values of y for which g(y) = 0 are called equilibrium values or rest points.
( )dy
g ydx
=
Phase Lines
( 1)( 2)dy
y ydx
= + −
y1− 2
0y ′ >0y ′ <0y ′ >
-1 -0.5 0.5 1
-3
-2
-1
1
2
3
unstable
equilibrium
stable
equilibrium
Autonomous Systems
The system
is called an autonomous
system of differential equations. Note that t
is
absent from the right side.
( , )
( , )
dxf x y
dtdy
g x ydt
=
=
Autonomous Systems
A solution of an autonomous system ia a pair of parametric equations (x(t), y(t)).As t varies over time, the solution curve is called a trajectory path or orbit of the system. The xy plane is referred to as the phase plane.The points for which f(x,y) = 0 and g(x,y) = 0 are called rest points or equilibrium points.
Autonomous Systems
There is at most one trajectory through any point in the phase plane.A trajectory that starts at a point other than a rest point cannot reach a rest point in a finite amount of time.
No trajectory can cross itself unless it is a closed curve. If it is a closed curve, it is a periodic solution.
Autonomous Systems
From a starting point that is not a rest point, the resulting motion
will move along the same trajectory regardless of the starting time;
cannot return to the starting point unlessthe motion is periodic;
can never cross another trajectory; and
can only approach (never reach) a rest point.
Autonomous Systems
The rest point (x0
, y0
) isstable if any trajectory that starts close to the point stays close to it for all future time.asymptotically stable if it is stable and if any trajectory that starts close to it approaches the point as t → ∞.unstable if it is not stable.
A Linear Autonomous System
dxx y
dtdy
x ydt
= − +
= − −
0
0
sin( )
cos( )
t
t
x Ae t t
y Ae t t
−
−
= −
= −
(0,0) is an asymptotically stable rest point. -1 -0.5 0.5 1
-1
-0.5
0.5
1
All points (0,a) are rest points and such a point is
unstable if a ≥ 0.stable, but not asymptotically stable, if a < 0.
A Nonlinear Autonomous System
2
dxxy
dtdy
xdt
=
=-2 -1 1 2
-2
-1
1
2
A Competitive-Hunter Model
There are two species, X and Y, competing for common resources.
Rest points are (0,0) and
( )
( )
dxax bxy a by x
dtdy
my nxy m nx ydt
= − = −
= − = −
( , ).m an b
A Competitive-Hunter Model
mn
ab
x
y
mn
ab
x
y
Limitation of a Graphical Analysis
periodicmotion
asymptoticallystable rest point
unstablerest point
It is possible that a graphical analysis is insufficient to determine the nature of the motion near a rest point.
Limitation of a Graphical Analysis
2 2
2 2
( )
( )
dxy x x x y
dtdy
x y y x ydt
= + − +
= − + − +x
y
The point (0,0) is the only rest point and it is unstable.
Any trajectory starting on the circle x 2
+ y2
= 1 will
traverse the circle. Any other trajectory starting from any point not on the circle and not on the origin will spiral toward the circle. It is thus called a limit cycle.
A Predator-Prey Model
There are two species, X and Y, in which X is the primary food for Y.
Rest points are (0,0) and
( )
( )
dxax bxy a by x
dtdy
my nxy m nx ydt
= − = −
= − + = − +
( , ).m an b
Lotka-Volterramodel
A Predator-Prey Model
mn
ab
x
y
A Predator-Prey Model
Find an analytic solution of the model.
/ ( )/ ( )
dy dt m nx ydydx dx dt a by x
− += =
−
a mb dy n dx
y x
⎛ ⎞ ⎛ ⎞⎟⎜ ⎟⎜− = −⎟⎜ ⎟⎜⎟ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠a m
by nx
y xK
e e
⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜ =⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎠1ln lna y by nx m x k− = − +
( )f y ( )g x
A Predator-Prey Model
( )g x
xmn
xMm
nx
xe
( )f y
yab
yMa
by
ye
If K
> Mx
My
, there is no solution.If K
= Mx
My
, there is exactly one solution , .m a
x yn b
= =
A Predator-Prey Model
x
y
mn
ab
x
t
t
y
A Predator-Prey Model
0 0an
1( ) (d
1)
T T
x x t dt y y t dtT T
= =∫ ∫
( )0 0
0
1
ln ( ) ln (0)
T T
T
dxdt a by dt
x dt
x T x aT b y dt
⎛ ⎞⎟⎜ = −⎟⎜ ⎟⎜⎝ ⎠
− = −
∫ ∫
∫
anda m
y xb n
= =
A Predator-Prey Model
If the prey, X, is killed at a rate rx(t), and it is assumed that the predator, Y, decreases at a rate ry(t), then
[ ]
[ ]
( ) ( )
( ) ( )
dxa by x by x
dtdy
m nx y nx
rx a r
ry m yd
rt
− −
− +
= − = −
= − + = − +
andm r a r
x yn b+ −
= =Volterra’s
Principle
