Some simple mathematical models - sacema.org Maths...dt = g kv : Some simple mathematical models....
-
Upload
truongduong -
Category
Documents
-
view
221 -
download
3
Transcript of Some simple mathematical models - sacema.org Maths...dt = g kv : Some simple mathematical models....
Some simple mathematical models
Some simple mathematical models
July 2, 2012
Some simple mathematical models
Some simple mathematical models
The birth of modern science
�Philosophy is written in this grand book the universe, which standscontinually open to our gaze. But the book cannot be understoodunless one �rst learns to comprehend the language and to read thealphabet in which it is composed. It is written in the language ofmathematics, and its characters are triangles, circles and othergeometric �gures, without which it is humanly impossible tounderstand a single word of it; without these, one wanders about ina dark labyrinth.�
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Galileo's assumption/hypothesis
�When ... I observe a stone initially at rest falling from an elevatedposition and continually acquiring new increments of speed, whyshould I not believe that such increases take place in a mannerwhich is exceedingly simple? ...A motion is said to be uniformly accelerated when starting fromrest, it acquires, during equal increments of time, equal incrementsof speed. That is the concept of accelerated motion that is mostsimple and easy, and it is con�rmed to be that actually occurring inNature, by exact correspondence with experimental results basedupon it.�
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Galileo's deduction/prediction:
Distances are to each other as the squares of the times.
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Galileo's experiment:
The ball and the inclined plane ...
time t 1 2 3 4 5
distance y 1 4 9 16 25
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Galileo's other assumption:
�In order to handle this matter in a scienti�c way, it is necessary tocut loose from such irksome di�culties [as friction or airresistance], otherwise it is not possible to give any exactdescription. Once we have discovered our theorems about thestone, assuming there is no resistance, we can correct them orapply them with limitations as experience will teach. I can show byexperiments that these external and incidental resistances arescarcely observable for bodies travelling over short distances.�
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Our modern version of Galileo's simple mathematical model:
Acceleration a(t) =d2y
dt2= g ,
so velocity v(t) =dy
dt=
∫g dt = gt,
hence distance y(t) =
∫gt dt =
1
2gt2.
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Ball rolling on an inclined plane:
θ
����
mg cos θ
mg sin θ
F
N
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Falling stone with air resistance:
Taking into account the air resistance −kmv should improve ourreal-life modelling.
mass× accel. = force,
so ma = md2y
d2t= mg − km
dy
dt,
hence ma = mdv
dt= mg − kmv ,
and therefore a(t) =dv
dt= g − kv .
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
Separating variables and integrating:∫dv
g − kv=
∫dt,
hence − 1
kln(g − kv) = t + c1,
so that ln(g − kv) = −kt + c2.
Taking exponentials, g − kv = c3e−kt ,
so v(t) =g
k− c4e
−kt ,
→ g
kas t →∞.
This equation tells us that in the long run the velocity will approacha constant limiting value g/k , called the terminal velocity. But wecould have found the terminal velocity much more easily. How?
Some simple mathematical models
Some simple mathematical models
Galileo's falling stone
How can we test our model?
What were our assumptions?
When might it be necessary to improve our model?
How could we do this?
Some simple mathematical models
Some simple mathematical models
Galileo's and Huygens' simple pendulum
Galileo's �rst observation
�One must observe that each pendulum has its own time ofvibration so de�nite and determinate that it is not possible to makeit move with any other period than that which nature has given it.For let any one take in his hand the cord to which the weight isattached and try, as much as he pleases, to increase or decrease thefrequency of its vibrations; it will be time wasted.�
Some simple mathematical models
Some simple mathematical models
Galileo's and Huygens' simple pendulum
Galileo's assumptions:
That the angle of swing is small, and that friction/resistance maybe ignored.
Galileo's experimental conclusion:
�As to the times of vibration of bodies suspended by threads ofdi�erent lengths, they bear to each other the same proportion asthe square roots of the lengths of the thread; or one might say thatthe lengths are to each other as the squares of the times....�
Some simple mathematical models
Some simple mathematical models
Galileo's and Huygens' simple pendulum
Our modern modelling of the pendulum:
θ
F
mg
M Sy
L
s
From the geometry, y ≈ arc MS = Lθ.
Force towards centre F = mg sin θ ≈ mgθ ≈ mg
Ly
Some simple mathematical models
Some simple mathematical models
Galileo's and Huygens' simple pendulum
Force = mass× acceleration,
so that − mg
Ly = F = m
d2y
dt2,
givingd2y
dt2= −g
Ly .
General solution: y(t) = A sin(√
gLt + α
),
where A is the �amplitude�, α is the �phase�,√
gLis the �frequency�.
The �period� of this function is 2π√
Lg, in harmony with Galileo's
experimental conclusion.
Some simple mathematical models
Some simple mathematical models
Galileo's and Huygens' simple pendulum
How can we test our model?
What assumptions have we made in this model?
When might it fail to model vibrations accurately?
How could we improve it?
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
Liber Abbaci (Book of Counting), 1202, by Leonardo of Pisa, alsocalled Leonardo Fibonacci � son of Bonacci.
Problem: A certain person places one pair of rabbits in a certainplace surrounded on all sides by a wall. We want to know howmany pairs can be bred from that pair in one year, assuming that itis in their nature that each month they give birth to another pair,and in the second month after birth each new pair can also breed.
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, · · ·
Fibonacci's answer is 377 pairs of rabbits, and his sequence ofnumbers is now known as the Fibonacci sequence.
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
Leonardo's solution:
�After the �rst month there will be two pairs, after the second,three. In the third month, two pairs will produce, so at the end ofthat month there will be �ve pairs. In the fourth month, three pairswill produce, so there will be eight pairs. Continuing thus, in thesixth month there will be �ve plus eight equals thirteen, in theseventh month, eight plus thirteen equals twenty-one, etc. Therewill be 377 pairs at the end of the twelfth month. For the sequenceof numbers is as follows, where each is the sum of the twopredecessors, and thus you can do it in order for an in�nite numberof months.�
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
A (2, 1, 2)
(3, 2, 3)
5
8
B
B
B
Y
Y
A B
A (1, 1, 1) 3
A (1, 0, 1) 2
Y (0, 1, 0) 1
B (1, 0, 0)
(B,Y ,A)
1
Total
??
??
??
??
-
-
-
R
R R -
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
After n months, let there be (Bn,Yn,An) pairs of babies, youngrabbits and adults, respectively. Then
An+1 = An + Yn
Bn+1 = An + Yn
Yn+1 = Bn
so that An+1 = Bn+1 = Yn+2.
Let the total number of (pairs of) rabbits at n months be given by
Fn = An + Bn + Yn.
Then we deduce that
Fn+2 = Fn+1 + Fn, for n ≥ 0, F0 = F1 = 1.
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, · · ·
The number of pairs of rabbits after Fibonacci's 12th month is F13,as he starts with F1, not F0, with a pair of young rabbits, notbabies. We have F13 = 377. In the next month after that,F11 = 233 pairs will produce;and in the 100th month, F98 pairs will produce.
Leonardo got there without much fuss. But our notation andtechnique is very powerful and can be applied to a host of otherproblems.
Is there a quick way of arriving at this number F13, or higherFibonacci numbers like F100?
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
(Fn+2
Fn+1
)=
(1 11 0
)(Fn+1
Fn
), for n ≥ 0.
Hence (F2F1
)=
(1 11 0
)(F1F0
)=
(1 11 0
)(11
),(
F3F2
)=
(1 11 0
)2(11
),(
F13F12
)=
(1 11 0
)12(11
)=
(377233
).
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
This same model for population growth can be applied to othersituations. Assuming that each branch of a tree gives rise to a newbranch, but only after skipping a season's maturation period, weobtain a visually very plausible model:
1
2358
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
In certain species of bee, the forebears follow a Fibonacci law.Female bees are born from the mating of male and female, malebees are born asexually to single females.
F
F
M
F
M
F
M
3 great grandparents
2 grandparents
1 parent
1 male bee
Exercise: Prove that the number Bn of bee ancestors in the nthgeneration backward, satis�es the Fibonacci generating equation.(Let Bn = Mn + Fn, the sum of the number of males and females,and write down the equations connecting Mn, Fn, Mn+1, Fn+1.)
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
Fibonacci numbers also arise from the following problem. Why?
Problem: Taking either one or two steps at a time, how manydi�erent ways are there to climb n steps?
Some simple mathematical models
Some simple mathematical models
Fibonacci's rabbits
How accurate is the Fibonacci model for the biologicalsituations we are modelling?
What were the assumptions we made?
What could be done to improve the model?
Some simple mathematical models
Some simple mathematical models
Red blood cells
These give your blood its red colour.Their function is to carry oxygen around the body.When there are too few in your body you su�er from anaemia.They are manufactured in your bone marrow, and each cell lives forabout 4 months before dying, probably in your kidney or spleen.
You should normally have about 2× 1013 red blood cells in yourbody at any moment, unless you live in Lesotho. Why?
Your bone marrow produces about 2× 106 red blood cells persecond, or 1728× 108 per day.
Some simple mathematical models
Some simple mathematical models
Red blood cells
Assume that a �xed number b of cells are produced every day, andthat a �xed proportion m of existing cells die each day.
Problem 1: What percentage of cells normally survive each day inthe equilibrium situation?
Problem 2: Is this equilibrium stable? On our assumptions above,what will happen if the red blood cell count is suddenly di�erentfrom normal?
Some simple mathematical models
Some simple mathematical models
Red blood cells
Let Cn be the number of red blood cells present at day n.Let m be the proportion of red blood cells that die each day.Let b be the number of red blood cells produced each day in thebone marrow.Then
Cn+1 = (1−m)Cn + b.
For equilibrium, Cn = C , say, for all n. Substituting in the equationwe have
C = (1−m)C + b, hence mC = b,
m =b
C=
1728× 108
2× 1013= 0.00864.
Thus, the percentage of red blood cells that die each day is0.864%, so that 99.136% of them survive each day.
Some simple mathematical models
Some simple mathematical models
Red blood cells
Next, suppose Cn 6= C , and let Dn = Cn − C , so Dn is thedeviation from the normal. Then
Dn+1 = Cn+1 − C
= (1−m)Cn + b − C
= (1−m)Cn +mC − C
= (1−m)(Cn − C )
= (1−m)Dn.
Hence Dn = (1−m)nD0 → 0, so Cn → C .
The equilibrium situation is a stable one: after a small deviationthe system will return to the equilibrium.
Some simple mathematical models
Some simple mathematical models
Red blood cells
How can we test our model?
What assumptions have we made in this model?
When might it fail to model blood cell count accurately?
How could we improve it?
Could we make a continuous model?
Some simple mathematical models
Some simple mathematical models
Red blood cells
Cn Cn+1(1−m)Cn
b
mCn
DISCRETE
C (t)b
mC
CONTINUOUS
Consider a continuous function C (t) of time t,constant birth rate b,and �per capita mortality rate� m
Some simple mathematical models
Some simple mathematical models
Red blood cells
dC = b dt −mC dt,
hencedC
dt= b −mC .
Integrating, C (t) =b
m− ke−mt .
But C (0) =b
m− k ,
hence C (t) =b
m−(b
m− C (0)
)e−mt .
Some simple mathematical models
Some simple mathematical models
Red blood cells
C (t) → b/m
dC
dt→ 0.
If C (0) = b/m then C (t) is constant.
-
6
b/m
tC (0)
C C (t) = bm−(bm− C (0)
)e−mt
C (0) = bm
will give stable equilibrium
Some simple mathematical models
Some simple mathematical models
Red blood cells
What assumptions have we made in this model?
Does the continuous model have any advantages over thediscrete model?
Some simple mathematical models
Some simple mathematical models
Leonhard Euler & human populations
In 1761, in the 2nd ed. of Johann Süssmilch's The Divine Order onthe Changes of the Human Generation, Through the Births, Deathsand the Procreation of the Same Established, appears Euler'sapplication to human population of Fibonacci's idea.
Euler's assumptions are:Start with one (male-and-female) couple aged 20 years;assume people marry at age 20 and die at age 40;assume each couple has 6 children:
1 boy and 1 girl at age 22,1 boy and 1 girl at age 24,1 boy and 1 girl at age 26.
Let Bn, Dn be the number of births, and of deaths, during year 2n.
Some simple mathematical models
Some simple mathematical models
Leonhard Euler & human populations
Then
Bn = Bn−11 + Bn−12 + Bn−13, n ≥ 1,
and Dn = Bn−20, n ≥ 0,
The number of people alive in year n is given by
Pn = Pn−1 + Bn − Dn.
Initial conditions are
Bn = 0, − 9 ≤ n ≤ 0, B−12 = B−11 = 0, and B−10 = 2.
Dn = 0, n ≤ 9.
Some simple mathematical models
Some simple mathematical models
Leonhard Euler & human populations
Thus Euler made a table with columns (Bn, Dn, Pn), forn = 1, 2, . . . , 119
Said Süssmilch:
The great disorder that seems to prevail in Euler's table does notprevent the number of births from following a kind of progressionthat one calls recurrent series [...] Whatever the initial disorder ofthese progressions, they turn into a geometric progression if theyare not interrupted and the disorders of the beginning fade little bylittle and vanish almost completely.
Some simple mathematical models
Some simple mathematical models
Leonhard Euler & human populations
How could we test such a human demographic model?
How realistic are Euler's assumptions?
How could we improve this model?
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
An attempt at a new analysis of the mortality caused by smallpoxand of the advantages of inoculation to prevent it � paper of 1760read to the Paris Academy of Sciences.
The problem: How to compare the long-term bene�t ofinoculation with the immediate risk of dying as a result of it.
Bernoulli's assumptions were:
1 an infected person (of any age) dies with probability p, andsurvives with probability 1− p;
2 every person has probability q of being infected in any year;
3 people surviving from smallpox are immunised;
4 the background mortality rate at age t is m(t).
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Consider a cohort of P0 people born the same year.Let S(t) be the (expected) number alive and susceptible at age t;let R(t) be the (expected) number who have survived smallpox, atage t;then P(t) = S(t) + R(t) is the total number alive at age t.
Initial conditions are: P(0) = S(0) = P0, R(0) = 0.
In a small interval of time (t, t + dt),the probability of an individual being infected is qdt;the probability of an individual dying is m(t)dt. Thus
in small interval of time dt, dS = −qSdt −mSdt
hencedS
dt= −qS −mS .
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
S Rq(1− p)S
?? ?mS qpS mR
-
dS
dt= −qS −m(t)S , (1)
dR
dt= q(1− p)S −m(t)R, (2)
hence, adding,dP
dt= −pqS −m(t)P. (3)
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Homework: Follow Daniel Bernoulli, in �rst eliminating m fromequations (1) and (3), then setting f (t) = S(t)
P(t) , to get:
df
dt= −qf + pqf 2, �Jakob Bernoulli's equation�.
Next, divide by f 2, and set g(t) = 1/f (t), to get:
dg
dt= qg(t)− pq, with g(0) = 1,
hence g(t) = p + (1− p)eqt ,
so thatS(t)
P(t)= f (t) =
1
p + (1− p)eqt. (4)
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Then Bernoulli took Edmund Halley's Life Table for t = 1, 2, . . . , 24and added four columns, calculated from Halley's values of P(t):
Age Alive Susceptible Immune Smallpox No smallpoxt P(t) S(t) R(t) deaths Q(t)
0 1300 1300 0 17.2 13001 1000 896 104 12.3 10122 855 685 170 9.8 8793 798 571 227 8.2 8304 760 485 275 7.0 7995 732 416 316 6.1 7776 710 359 351 5.2 760...
......
......
...
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Bernoulli's methods:He estimated p = 1/8, from observations of his day;he chose q = 1/8, probably after many trials, in order to end upwith the observed ratio, smallpox deaths : all deaths = 1/13.S(t) is calculated directly from P(t) using the derived formula;R(t) is just the di�erence P(t)− S(t).He calculated smallpox deaths as follows:
6
-������t t + 1
Spq∫ t+1
tS(t) dt ≈ pq
(S(t)+S(t+1)
2
)
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Finally, the numbers alive after time t in the absence of smallpoxwould be given by:
dQ
dt= −m(t)Q(t) (5)
from which Bernoulli deduced the connection between P and Q:
P(t)/Q(t) = 1− p + pe−qt (6)
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Homework: Follow Bernoulli in eliminating m(t) betweenequations (3) and (5), then setting h(t) = P(t)/Q(t) and usingequation (4) to get
1
h
dh
dt= −pq e−qt
1− p + pe−qt,
that is,d
dtln(t) =
d
dtln(1− p + pe−qt),
hence h(t) = 1− p + pe−qt , since h(0) = 1.
Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
Some more homework: You might also enjoy proving hisequations (4), (6) by a simpler method he did not �nd:
Solve equations (1) and (5), resp., to get
S(t) = P0e−qt exp
(−∫ t
0
m(x) dx
),
and Q(t) = P0 exp
(−∫ t
0
m(x) dx
).
Substitute from the �rst into equation (2) to get
R(t) = P0(1− p)(1− e−qt) exp
(−∫ t
0
m(x) dx
).
Now you can easily educe equations (4) and (6).Some simple mathematical models
Some simple mathematical models
Daniel Bernoulli & smallpox inoculation
So, what was Bernoulli's conclusion about smallpox inoculation?
He took the columns for P and Q in his table, extended to 84years, and found:
life-expectancy with smallpox is 26 years and 7 months;life-expectancy with no smallpox is 29 years and 8 months.
He guessed that the probability of dying from the inoculation wasless than 1 percent; therefore, he concluded, everyone should beinoculated.
Some simple mathematical models
Some simple mathematical models
Crop yields and bacteria cultures
In the Chinese classic Jiu zhang suan shu (The Nine Chapters ofthe Mathematical Art), written over 2000 years ago, the followingproblem is solved on the Chinese counting board:
Problem: Now there are 3 classes of paddy: top, medium andlow grade. Given 3 bundles of top grade paddy, 2 bundles ofmedium grade paddy and 1 bundle of low grade paddy, the yield is39 dou. For 2 bundles of top grade, 3 bundles of medium grade and1 bundle of low grade, the yield is 34 dou. And for 1 bundle of topgrade, 2 bundles of medium grade and 3 bundles of low grade, theyield is 26 dou. How much does one bundle of each grade yield?Answer: Top grade paddy yields nine and a quarter dou perbundle; medium grade paddy four and a quarter dou; and low gradepaddy two and three quarters dou.
Some simple mathematical models
Some simple mathematical models
Crop yields and bacteria cultures
high grade 1 2 3medium grade 2 3 2low grade 3 1 1yield 26 34 39
3x+2y+z = 39 (1)2x+3y+z = 34 (2)x+2y+3z = 26 (3)
Exercise: Solve the Chinese problem by doing row operations onthe matrix of coe�cients.
Some simple mathematical models
Some simple mathematical models
Crop yields and bacteria cultures
In the 21st century, three species of bacteria B1, B2, B3 are mixedtogether in the same culture in a laboratory. It is known that eachrequires three types of food, F1, F2, F3, and:
B1 consumes F1, F2, F3 at the rates (1, 2, 2);B2 consumes F1, F2, F3 at the rates (3, 2, 1);B3 consumes F1, F2, F3 at the rates (8, 4, 1).
It is found experimentally that when 2750 units of F1, 1500 units ofF1, and 500 units of F3 are supplied per day, a constant situation isreached in which all food is consumed.
Problem 1: Find the amounts of bacteria present.Problem 2: Which set of constant states is possible, and which isnot, regardless of the food supply?
Some simple mathematical models
Some simple mathematical models
Crop yields and bacteria cultures
Solution: Suppose that there are amounts X1, X2, X3,respectively, of the three bacteria present. Then 1 3 8
2 2 42 1 1
X1
X2
X3
=
27501500500
, or AX = F .
Now det(A) = 0, so there are many possible solutions:
(X1,X2,X3) = (t − 250, 1000− 3t, t), t ∈ R.
Notice that for positive values of the Xi we must have250 ≤ t ≤ 1000
3.
Some simple mathematical models
Some simple mathematical models
Crop yields and bacteria cultures
For the second question, consider the map
f : R3 → R3, de�ned by f (X ) = AX .
The rank is easily found to be 2, so the kernel of f has dimension 1(it's a line) and the range is the 2-dimensional plane through theorigin, spanned by the two image vectors:
f
100
=
122
, f
010
=
321
.
Taking the vector product of these two, we get a vector in direction(−2, 5,−4) perpendicular to the required plane, hence the equationof the plane is
2x − 5y + 4z = 0.
Some simple mathematical models
Some simple mathematical models
Crop yields and bacteria cultures
The range of f is the plane
2x − 5y + 4z = 0.
One point on this plane is that experimentally discoveredequilibrium state (2750, 1500, 500).
But all possible equilibrium states must lie on this plane!
Some simple mathematical models
Some simple mathematical models
Age-strati�ed populations & a butter�y's life cycle
Consider a population of individuals each of which gives birth, livesfor a while, then dies. Assume there is a maximum life span, anddivide the population into k age-groups, numberinga1(n), a2(n), . . . , ak(n) individuals at time n, measured in years orbreeding seasons. Suppose that
1 everyone in stage k dies at the end of the year/season;
2 the probability that an individual in class j survives anotheryear/season is pj ;
3 the probability that an individual in class j gives birth toexactly one individual is fj (the fertility rate).
Some simple mathematical models
Some simple mathematical models
Age-strati�ed populations & a butter�y's life cycle
For example, take k = 4. The numbers in the next season are givenby the Leslie matrix:
a1(n + 1)a2(n + 1)a3(n + 1)a4(n + 1)
=
f1 f2 f3 f4p1 0 0 00 p2 0 00 0 p3 0
a1(n)a2(n)a3(n)a4(n)
.
After N years we will havea1(N)a2(N)a3(N)a4(N)
=
f1 f2 f3 f4p1 0 0 00 p2 0 00 0 p3 0
N
a1(0)a2(0)a3(0)a4(0)
.
Some simple mathematical models
Some simple mathematical models
A butter�y's life cycle
Consider the four stages of a butter�y's life cycle: egg, caterpillar,pupa (or chrysalis) and adult butter�y.
EGG
PUPA
CATERPILLARBUTTERFLY
~
/
o
> p
qr
(CHRYSALIS)
f
Clearly f1 = f2 = f3 = 0. Assume that a butter�y survives to layeggs with probability f , and then always lays N eggs, so f4 must bereplaced by fN. Let the respective probabilities of egg, caterpillarand pupa surviving to move into next class be p, q, r .
Some simple mathematical models
Some simple mathematical models
A butter�y's life cycle
Let En, Cn, Pn, Bn be the numbers of individual eggs, caterpillars,pupa, and butter�ies, in a population after the nth season. Thenthe numbers for the next season will be
En+1
Cn+1
Pn+1
Bn+1
=
fNBn
pEnqCn
rPn
=
0 0 0 fNp 0 0 00 q 0 00 0 r 0
EnCn
Pn
Bn
.
Some simple mathematical models
Some simple mathematical models
A butter�y's life cycle
The parameters f , p, q, r could be estimated from experimentaldata. The long term fate of the population will depend on theirvalues, and that of N. To see how, we look at powers of the matrix:
0 0 0 fNp 0 0 00 q 0 00 0 r 0
4
= fNpqr
1 0 0 00 1 0 00 0 1 00 0 0 1
En+4
Cn+4
Pn+4
Bn+4
= fNpqr
EnCn
Pn
Bn
Some simple mathematical models
Some simple mathematical models
A butter�y's life cycle
It's easy to check that matrix equation, but how do we arrive at itin the �rst place? When powers of a matrix are in view, it is naturalto �nd the eigenvalues.
The Cayley-Hamilton theorem says that the matrix will satisfy itscharacteristic equation, and the eigenvalues of this Leslie matrix areeasily seen to be the solutions of the quartic equation
λ4 = fNpqr
Some simple mathematical models
Some simple mathematical models
A butter�y's life cycle
Thus we conclude:
if fNpqr < 1, ultimate extinction occurs;
if fNpqr = 1, there is unstable equilibrium;
if fNpqr > 1, there is unsustainable expansion.
How realistic is this model?
What were our assumptions?
How can we test the model?
How can we improve it?
What can we learn from it anyway?Some simple mathematical models