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Discrete Dynamical SystemsSuppose that A is an n n matrix and suppose that x0 is a vector in n. Then
x1 Ax0is a vector in n. Likewise,
x2 Ax1is a vector in n, and we can in fact generate an infinite sequence of vectors xkk0 inn defined recursively by
xk1 Axk.When viewed in this context, we say that the matrix A defines a discrete
dynamical system (which we will also refer to more briefly as a dynamical system)on n. For a given initial vector x0 n, the infinite sequence xkk0 is called thepositive orbit (or, for our purposes, just the orbit) of the dynamical system A throughthe initial vector x0. A fundamental problem is to determine the long term behavior ofthe orbit through each vector x0 n. Some particular questions of interest are1. Does the orbit xkk0 approach some “limiting vector” as k ?2. If the orbit xkk0 does not approach some limiting vector as k , then is
the behavior of this orbit at least in some way predictable?
Example Suppose that A is the matrix
A 12 140 1
and consider the dynamical systemxk1 Axk.
Describe the long term behaviors of the orbits of this dynamical system throughthe vectors
x0 10
, x0 12
, and x0 02
.
Solution First we study the orbit through
x0 10
by computing the first few iterates:
1
x1 Ax0 12 140 1
10
120
x2 Ax1 12 140 1
120
14
0
x3 Ax2 12 140 1
14
0
180
x4 Ax3 12 140 1
180
116
0.
There is a clear pattern here. It can be seen thatlimkxk 0.
In fact, it can be see that for each k 0, we have
xk 1k 1
2k
0.
Next, we study the orbit through
x0 12
by computing the first few iterates:
x1 Ax0 12 140 1
12
12
x2 Ax1 12 140 1
12
12
x3 Ax2 12 140 1
12
12
x4 Ax3 12 140 1
12
12
.
The pattern of this orbit is clear. It does not approach a limiting vector ask but, instead, it alternates back and forth between the two vectors
2
12
and12
.
We call this a periodic orbit with period 2. An explicit formula for this orbit is
xk 1k 11k 2
.
Finally, we study the orbit through
x0 02
by computing the first few iterates:
x1 Ax0 12 140 1
02
122
x2 Ax1 12 140 1
122
34
2
x3 Ax2 12 140 1
34
2
782
x4 Ax3 12 140 1
782
1516
2.
For this orbit, it can be seen that the second component alternates between thevalues 2 and 2. The first component alternates between positive and negativevalues that are approaching the limiting values of 1 and 1, respectively. It is thusfair to say that the limiting behavior of the orbit through the vector
x0 02
is essentially the same as the limiting behavior of the orbit through the vector
x0 12
.
With a little thought, we can write an explicit formula for the orbit through
x0 02
.
It is
3
xk 1k 2k1
2k
1k 2.
In the preceding example, we computed the orbits of a dynamical system, A,through three specific initial vectors. This computations showed that there are at leasttwo basic “behaviors” that can be expected from orbits of this dynamical system. Inparticular, we saw that some orbits (at least one!) satisfies
limkxk 0
and that some orbits approach the periodic orbit
12
12
as k .Are these the only two possible behaviors of orbits of this dynamical system, or
are there other possible behaviors. In order to answer this question, we need to usean arbitrary initial vector
x0 ab
n
and see if we can tell what happens to it. By performing a few computations, we seethat
x1 Ax0 12 140 1
ab
12 a
14 b
b
x2 Ax1 12 140 1
12 a 14 b
b
14 a
38 b
b
x3 Ax2 12 140 1
14 a
38 b
b
18 a 716 b
b
x4 Ax3 12 140 1
18 a 716 b
b
116 a
1532 b
b.
These computations seem to show that
xk 1k 1
2k a 1k 1
2 2k12k
b
1k b
1k 12k
1k 12
2k12k
0 1kab
for all k 0.Since
4
limk
1k 12k
0
and the quantity
1k 12 2k 12k
alternates between positive and negative values that approach 12 and
12 ,
respectively, we conclude that if if k is very large and even, then
xk 0 1
2
0 1ab
12 b
b 12 b
12
and if k is very large and odd, then
xk 0 120 1
ab
12 b
b 12 b
12
.
Hence, if b 0, then xk alternates between a scalar multiple (whose magnitudedepends on b) of the vectors
12
and12
as k .However, if b 0, then xk 0 for all large values of k and
limkxk 0.
For example, the analysis that we just performed shows that the orbit through theinitial vector
x0 36
approaches the periodic orbit
36
36
as k ; whereas the orbit through the initial vector
x0 180
satisfieslimkxk 0.
The approach that we have taken in studying the dynamical system defined by
5
A 12 140 1
was, in some sense, completely successful, because we were able to determine thelong term behaviors of all orbits of this dynamical system. In particular, we determinedthat there are essentially only two possible behaviors and we were able to identifywhich orbits exhibit which of the behaviors. Clearly, however, the approach that weused here is not efficient in general. This approach generally does not yield such apleasing and complete set of results if the matrix A is of size larger than 2 2 and/orthe matrix A is not triangular. Clearly, we need to make more efficient use of thepowerful theory of linear algebra that we have been developing in this course. Themain ideas that we need are those of eigenvalues, eigenvectors, similarity, anddiagonalization.
Dynamical Systems Defined by DiagonalizableMatrices
If A is an n n matrix, then the orbit of a vector x0 n for the dynamical systemdefined by A is
x1 Ax0x2 Ax1 AAx0 A2x0x3 Ax2 AA2x0 A3x0
and in general
xk Akx0.Thus, understanding the long term behavior of an orbit depends on understanding
the nature of Ak for large values of k.If A is diagonalizable, then A PDP1 where D is an n n diagonal matrix whose
entries along the main diagonal are the eigenvalues of A and P is a matrix whosecolumns are an eigenvector basis for n (consisting of n linearly independenteigenvectors of A that correspond to the eigenvalues of A). As we have seen in ourprevious work,
Ak PDkP1
for any positive integer k. Thus, for the orbit through the vector x0 of the dynamicalsystem defined by A, we have
xk PDkP1x0for all positive integers k.
Example Use the fact that the matrix
6
A 12 140 1
is diagonalizable to compute Ak (for any positive integer k). Then obtain an explicitformula for the orbit of the dynamical system defined by A through an arbitraryinitial vector
x0 ab
.
Solution Since A is a triangular matrix, it can be seen that the eigenvalues of A are 12 and 1. Since these eigenvalues are distinct (and there are two ofthem), then A is diagonalizable. Furthermore, it can be seen that an eigenvectorcorresponding to 12 is
v1 10
and that an eigenvector corresponding to 1 is
v2 12
.
This means that A PDP1 where
D 12 0
0 1
and
P 1 10 2
P1 1 120 1
2
.
Therefore
7
Ak PDkP1
1 10 2
12k 0
0 1k1 120 1
2
12
k1k
0 21k1 120 1
2
12
k 12 12
k 12 1
k
0 1k.
For the initial vector
x0 ab
,
we havexk Akx0
12
k 12 12
k 12 1
k
0 1kab
12
ka 12 12
k 12 1
k b
1kb
a 12
k
0 b
12 12
k 12 1
k
1k
a 12
k
0 12 b
12k1 1k
1k 2.
Example In a previous example (in the Section 5.3 notes), we showed that the matrix
A
1 3 33 5 33 3 1
is diagonalizable. In particular, we showed that A PDP1 where
8
D
1 0 00 2 00 0 2
P
1 1 11 1 01 0 1
P1 1 1 11 2 11 1 0
.
Use this information to determine the long term behavior of the orbit throughthe initial vector
x0 104
for the dynamical systemxk1 Axk.
Solution We know thatxk Akx0
for all positive integers k. Also,Ak PDkP1
1 1 11 1 01 0 1
1 0 00 2k 00 0 2k
1 1 11 2 11 1 0
1 2k 2k
1 2k 01 0 2k
1 1 11 2 11 1 0
1 1 2k 1 2k
1 2k 1 22k 1 2k
1 2k 1 2k 1
.
This gives us
9
xk Akx0
1 1 2k 1 2k
1 2k 1 22k 1 2k
1 2k 1 2k 1
104
3 42k
3 32k
3 2k.
It can now be seen that the components of xk grow beyond all bounds andalternate in sign as k . For example
x10 3 4210
3 3210
3 210
4,0933,0691,027
x11 3 4211
3 3211
3 211
8,1956,1472,045
x12 3 4212
3 3212
3 212
16,38112,2854,099
.
An Introductory Application of Discrete DynamicalSystems in the Study of Age–Structured Population
ModelsSuppose that we have a population of some type of animal that has an average
life span of two years. Suppose also that these animals are capable of reproduction,on average, when they get to be one year old. Finally, suppose that each of theseanimals produces two offspring during its lifetime but that, due to harsh conditions,only about half of the offspring survive long enough to reach maturity (meaningreproductive age). What do you think will happen to this animal population over thelong run?
We can write a simple mathematical model of this situation as follows:Let xk be the juvenile population at time k, and let yk be the adult population at time
k. Then the population changes (approximately) according to the equations
10
xk1 2ykyk1 1
2 xk.
This mathematical model can be written in matrix form:
xk1yk1
0 212 0
xkyk
or more simply asxk1 Axk
where
xk xkyk
is called the age–structured population vector.Now, to ask a specific question, suppose that we have an initial population
consisting of 200 juveniles and 1000 adults. What will happen to this population overtime?
To answer this question, we observe (via the usual diagonalization process) thatthe matrix
A 0 212 0
is diagonalizable. In fact, after a little work, we can see that
A 0 212 0
PDP1 2 21 1
1 00 1
1412
14
12
.
The population vector, starting with initial population vector
x0 2001000
thus satisfies
11
xk PDkP1x0
2 21 1
1k 00 1
1412
14
12
2001000
21k 21k 1
1412
14
12
2001000
12 1
k 12 1k 1
14 1k 1
412 1
k 12
2001000
9001k 11004501k 550
.
Thus, the juvenile population in any year k 0 will bexk 900 1k 1100
and the adult population will beyk 4501k 550.
The total (juvenile and adult) population will bexk yk 450 1k 1650.
The total population and the juvenile and adult populations fluctuate from year toyear. This is summarized in the table below
Year Juveniles Adults Total0 200 1000 12001 2000 100 21002 200 1000 12003 2000 100 21004 200 1000 1200
.
Now let us change our model slightly by adding more age structure. In particular, letus assume that juveniles take two years (instead of one year) to mature. This give ustwo juvenile age classes, which we will call x and y, and one adult age class, which wewill call z. Let us further assume that only half of the juveniles in the younger juvenileage class survive and reach the older juvenile age class, but that all juveniles whoreach the older juvenile age class mature into adults. Our model then becomes
12
xk1 2zkyk1 1
2 xk
zk1 ykwhich can be written in matrix form as
xk1yk1zk1
0 0 212 0 0
0 1 0
xkykzk
or simply asxk1 Axk
where
xk xkykzk
and
A
0 0 212 0 0
0 1 0
.
We would like to see whether or not the matrix A is diagonalizable. (If it is, then wewill be happy because we can proceed as before in trying to understand the long termpopulation dynamics.)
The characteristic polynomial of A is
0 0 212 0 0
0 1 0
1 3,
which means that the characteristic equation of A is1 3 0
and that the only (real) eigenvalue of A is 1. (The other two eigenvalues of A areimaginary numbers.)
To find the eigenvectors of A that are associate with the eigenvalue 1, weconsider
1 0 212 1 0
0 1 1
x1x2x3
000
.
13
Since
1 0 212 1 0
0 1 1
~1 0 20 1 10 0 0
,
we see that the general solution of the above matrix equation is
x1x2x3
2ttt
t211
.
Therefore, eig1 is only one–dimensional and we do not have an eigenvector basis.This means that the matrix A is not diagonalizable.
Fortunately, even though A is not diagonalizable, this model is sufficiently simpleto allow us to determine the long term behavior or populations by direct computation.For example, suppose that we start with 100 juveniles in each of the younger andolder juvenile age classes and with 1000 adults: That is,
x0 1001001000
.
Then
x1 Ax0 0 0 212 0 0
0 1 0
1001001000
200050100
x2 Ax1 0 0 212 0 0
0 1 0
200050100
200100050
x3 Ax2 0 0 212 0 0
0 1 0
200100050
1001001000
x4 Ax3 0 0 212 0 0
0 1 0
1001001000
200050100
.
We see that the age distribution in this population repeats itself over and overagain in four year cycles.
14
The basic formulation for an age-structured population model is as follows:Suppose that juveniles are divided into r age classes, x1,x2, ,xr, with correspondingsurvival rates s1, s2, , sr, and suppose that adults are divided into p age classes,y1,y2, ,yp, with corresponding fertility rates f1, f2, , fp, and survival rates t1, t2, tp(where tp 0). Then the model for the population growth is
x1k1 f1y1k f2y2k fpypkx2k1 s1x1k
xrk1 sr1xr1ky1k1 srxrky2k1 t1y1k
ypk1 tp1yp1k.In matrix form, this model becomes
x1k1x2k1
xrk1y1k1y2k1
ypk1
0 0 0 f1 f2 fps1 0 0 0 0 0 0
0 0 0 0 0 0 00 0 sr 0 0 0 00 0 0 0 t1 0 0 0
0 0 0 0 0 0 tp1 0
x1kx2k
xrky1ky2k
ypk
.
The r p r p matrix used in this model is called a Leslie matrix.
15
