Advanced Mathematics for Economics, course 2013-2014
Juan Pablo Rincón–Zapatero
Contents
1. Review of Matrices and Determinants
1.1. Square matrices
1.2. Determinants
2. Diagonalization of matrices
1. Introduction
2. Systems of first order difference equations
3. First order linear difference equations
4. Second order linear difference equations
4.1. The nonhomogeneous case
5. Linear systems of difference equations
5.1. Homogeneous systems
5.2. Nonhomogeneous systems
5.3. Stability of linear systems
6. The nonlinear first order equation
6.1. Phase diagrams
1. Introduction. Definitions and classifications of ODEs
2. Elementary integration methods of first order ODEs
2.1. Separable equations
2.2. Exact equations. Integrating factors
2.3. Linear equations
2.4. Phase diagrams
3. Applications
4. Second order linear ODEs
4.1. Stability of second order ODEs with constant coefficients
5. Systems of first order ODEs
5.1. Linear systems
5.2. Nonlinear systems
1
2
2
3
5
12
13
16
19
20
23
23
24
25
29
31
33
35
35
36
39
41
42
48
51
52
52
56
2
Topic 1: Matrix diagonalization
1. Review of Matrices and Determinants
Definition 1.1. A matrix is a rectangular array of real numbers


a11 a12 · · · a1m
 a21 a22 · · · a2m 
A=
.
..
.. 
..
 ...
.
.
. 
an1 an2 · · · anm
The matrix is said to be of order n×m if it has n rows and m columns. The set of matrices
of order n × m will be denoted Mn×m .
The element aij belongs to the ith row and to the jth column. Most often we will write
in abbreviated form A = (aij )i=1,...,n
j=1,...,m or even A = (aij ).
The main or principal, diagonal of a matrix is the diagonal from the upper left– to the
lower right–hand corner.
Definition 1.2. The transpose of a matrix
changing the rows and columns of A

a11 a21

a
a22
12
AT = 
.
..
 ..
.
A, denoted AT , is the matrix formed by inter···
···
..
.
a1m a2m · · ·

an1
an2 
∈ Mm×n .
.. 
. 
anm
We can define two operations with matrices, sum and multiplication. The main properties
of these operations as well as transposition are the following. It is assumed that the
matrices in each of the following laws are such that the indicated operation can be performed
and that α, β ∈ R.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(AT )T = A.
(A + B)T = AT + B T .
A + B = B + A (commutative law).
A + (B + C) = (A + B) + C (associative law).
α(A + B) = αA + αB.
(α + β)A = αA + βA.
Matrix multiplication is not always commutative, i.e., AB 6= BA.
A(BC) = (AB)C (associative law).
A(B + C) = AB + AC (distributive with respect to addition).
1.1. Square matrices. We are mainly interested in square matrices. A matrix is square
if
Pnn = m. The trace of a square matrix A is the sum of its diagonal elements, trace (A) =
i=1 aii .
3
Definition 1.3. The identity matrix of order n is

1 0 ...
 0 1 ...
In = 
 ... ... . . .

0
0 
.
.. 
. 
0 0 ... 1
The square matrix of order n with all its entries null is the null matrix, and will be denoted
On . It holds that In A = AIn = A and On A = AOn = On .
Definition 1.4. A square matrix A is called regular or invertible if there exists a matrix B
such that AB = BA = In . The matrix B is called the inverse of A and it is denoted A−1 .
Theorem 1.5. The inverse matrix is unique.
Uniqueness of A−1 can be easily proved. For, suppose that B is another inverse of matrix
A. Then BA = In and
B = BIn = B(AA−1 ) = (BA)A−1 = In A−1 = A−1 ,
showing that B = A−1 .
Some properties of the inverse matrix are the following. It is assumed that the
matrices in each of the following laws are regular.
(1) (A−1 )−1 = A.
(2) (AT )−1 = (A−1 )T .
(3) (AB)−1 = B −1 A−1 .
1.2. Determinants. To a square matrix A we associate a real number called the determinant, |A| or det (A), in the following way.
For a matrix of order 1, A = (a), det (A)
= a.
a b
, det (A) = ad − bc.
For a matrix of order 2, A =
c d
For a matrix of order 3
a11 a12 a13 a12 a13 a12 a13 a22 a23 − a21 det (A) = a21 a22 a23 = a11 a32 a33 + a31 a22 a23 .
a
a
32
33
a31 a32 a33 This is known as the expansion of the determinant by the first column, but it can be done
for any other row or column, giving the same result. Notice the sign (−1)i+j in front of the
element aij .
Before continuing with the inductive definition, let us see an example.
Example 1.6. Compute the following determinant expanding by the second column.
1 2 1 4 3 5 = (−1)1+2 2 4 5 + (−1)2+2 3 1 1 + (−1)2+3 1 1 1 3 3 3 3 4 5 3 1 3 = −2 · (−3) + 3 · (0) − (1) · 1 = 5
4
For general n the method is the same that for matrices of order 3, expanding the determinant by a row or a column and reducing in this way the order of the determinants that must
be computed. For a determinant of order 4 one has to compute 4 determinants of order 3.
Definition 1.7. Given a matrix A of order n, the complementary minor of element aij is
the determinant of order n − 1 which results from the deletion of the row i and the column
j containing that element. The adjoint Aij of the element aij is the minor multiplied by
(−1)i+j .
According to this definition, the determinant of matrix A can be defined as
|A| = ai1 Ai1 + ai2 Ai2 + · · · + ain Ain
(by row i)
or, equivalently
|A| = a1j A1j + a2j A2j + · · · + anj Anj
Example 1.8. Find the value of the determinant
1 2 0 3
4 7 2 1
1 3 3 1
0 2 0 7
Answer: Expanding the determinant by the
1 2 0 3 1 2
4 7 2 1 3+2
= (−1) 2 1 3
1 3 3 1 0 2
0 2 0 7 (by column j).
.
third column, one gets
1 2 3 3 1 + (−1)3+3 3 4 7 1 .
0 2 7 7 The main properties of the determinants are the following. It is assumed that the
matrices A and B in each of the following laws are square of order n and λ ∈ R.
(1) |A| = |AT |.
(2) |λA| = λn |A|.
(3) |AB| = |A||B|.
1
(4) A matrix A is regular if and only if |A| =
6 0; in this case |A−1 | = |A|
.
(5) If in a determinant two rows (or columns) are interchanged, the value of the determinant is changed in sign.
(6) If two rows (columns) in a determinant are identical, the value of the determinant is
zero.
(7) If all the entries in a row (column) of a determinant are multiplied by a constant λ,
then the value of the determinant is also multiplied by this constant.
(8) In a given determinant, a constant multiple of the elements in one row (column) may
be added to the elements of another row (column) without changing the value of the
determinant.
The next result is very useful to check if a given matrix is regular or not.
Theorem 1.9. A square matrix A has an inverse if and only |A| =
6 0.
5
2. Diagonalization of matrices
Definition 2.1. Two matrices A and B of order n are similar if there exists a matrix P
such that
B = P −1 AP.
Definition 2.2. A matrix A is diagonalizable if it is similar to a diagonal matrix D, that
is, there exists D diagonal and P invertible such that D = P −1 AP .
Of course, D diagonal means that every element out of the diagonal is null


λ1 0 . . . 0
 0 λ2 . . . 0 
D=
.. . .
.  , λ1 , . . . , λn ∈ R.
 ...
. .. 
.
0 0 . . . λn
Proposition 2.3. If A is diagonalizable, then for all m ≥ 1
Am = P Dm P −1 ,
(2.1)
where
λm
0
1
m

0
λ
2
m

D =  ..
..
.
.
0
0

...
...
..
.

0
0 
.
.. 
. 
. . . λm
n
Proof. Since A is diagonalizable
m
Am = (P DP −1 )(P DP −1 ) · · · (P DP −1 )
= P D(P −1 P )D · · · D(P −1 P )DP −1
= P DIn D · · · DIn DP −1 = P Dm P −1 .
The expression for Dm is readily obtained by induction on m.
Example 2.4. At a given date, instructor X can teach well or teach badly. After a good day,
the probability of doing well for the next class is 1/2, whilst after a bad day, the probability
of doing well is 1/9. Let gt (bt ) the probability of good (poor) teaching at day t. Suppose
that at time t = 1 the class has been right, that is, g1 = 1, b1 = 0. Which is the probability
that the 5th class go fine (bad)?
Answer: The data lead to the following equations that relate the probability of a
good/bad class with the performance showed by the teacher the day before
1
gt+1 = gt +
2
1
bt+1 = gt +
2
1
bt ,
9
8
bt .
9
6
In matrix form
gt+1
bt+1
1
9
8
9
1
2
1
2
=
!
gt
bt
g1
b1
.
Obviously
g5
b5
1
2
1
2
=
1
9
8
9
!4 .
If the matrix were diagonalizable and we could find matrices P and D, then the computation
of the 10th power of the matrix would be easy using Proposition 2.3. We will come back to
this example afterwards.
Definition 2.5. Let A be a matrix of order n. We say that λ ∈ R is an eigenvalue of A and
that u ∈ Rn , u 6= 0, is an eigenvector of A associated to λ if
Au = λu.
The set of eigenvalues of A, σ(A) = {λ1 , . . . , λk }, is called the spectrum of A. The set of all
eigenvectors of A associated to the same eigenvalue λ, including the null vector, is denoted
S(λ), and is called the eigenspace or proper subspace associated to λ.
The following result shows that an eigenvector can only be associated to a unique eigenvalue.
Proposition 2.6. Let 0 6= u ∈ S(λ) ∩ S(µ). Then λ = µ.
Proof. Suppose 0 6= u ∈ S(λ) ∩ S(µ). Then
Au = λu
Au = µu.
Subtracting both equations we obtain 0 = (λ−µ)u and, since 0 6= u, we must have λ = µ. Recall that for an arbitrary matrix A, the rank of the matrix is the number of linearly
independent columns or rows (both numbers necessarily coincide). It is also given by the
order of the largest non null minor of A.
Theorem 2.7. The real number λ is an eigenvalue of A if and only if
|A − λIn | = 0.
Moreover, S(λ) is the set of solutions (including the null vector) of the linear homogeneous
system
(A − λIn )u = 0,
and hence it is a vector subspace, which dimension is
dim S(λ) = n − rank(A − λIn ).
7
Proof. Suppose that λ ∈ R is an eigenvalue of A. Then the system (A − λIn )u = 0 admits
some non–trivial solution u. Since the system is homogeneous, this implies that the determinant of the system is zero, |A − λIn | = 0. The second part about S(λ) follows also from
the definition of eigenvector, and the fact that the set of solutions of a linear homogenous
system is a subspace (the sum of two solutions is again a solution, as well as it is the product
of a real number by a solution). Finally, the dimension of the space of solutions is given by
the Theorem of Rouche–Frobenius.
Definition 2.8. The characteristic polynomial of A is the polynomial of order n given by
pA (λ) = |A − λIn |.
Notice that the eigenvalues of A are the real roots of pA . This polynomial is of degree n.
The Fundamental Theorem of Algebra estates that a polynomial of degree n has n complex
roots (not necessarily different, some of the roots may have multiplicity grater than one). It
could be the case that some of the roots of pA were not real numbers. For us, a root of pA (λ)
which is not real is not an eigenvalue of A.
Example 2.9. Find the eigenvalues and the proper subspaces of


0 −1 0
0 0 .
A= 1
0
0 1
Answer:

−λ −1
0
0 ;
A − λI =  1 −λ
0
0 1−λ

−λ −1
p(λ) = (1 − λ) 1 −λ
= (1 − λ)(λ2 + 1).
The characteristic polynomial has only one real root, hence the spectrum of A is σ(A) =
{1}. The proper subspace S(1) is the set of solutions of the homogeneous linear system
(A − I3 )u = 0, that is, the set of solutions of
   

0
x
−1 −1 0
(A − I3 )u =  1 −1 0   y  =  0 
0
0 0
z
0
Solving the above system we obtain
S(1) = {(0, 0, z) : z ∈ R} =< (0, 0, 1) > (the subspace generated by (0, 0, 1)).
Notice that pA (λ) has other roots that are not reals. They are the complex numbers ±i,
that are not (real) eigenvalues of A. If we would admit complex numbers, then they would
be eigenvalues of A in this extended sense.
Example 2.10. Find the eigenvalues and the proper subspaces of


2 1 0
B =  0 1 −1  .
0 2 4
8
Answer: The eigenvalues are obtained solving
2−λ
1
0
0
1
−
λ
−1
0
2
4−λ
= 0.
The solutions are λ = 3 (simple root) and λ = 2 (double root). To find S(3) = {u ∈ R3 :
(B − 3I3 )u = 0} we compute the solutions to

   
−1 1
0
x
0
(B − 3I3 )u =  0 −2 −1   y  =  0  ,
0
2
1
z
0
which are x = y and z = −2y, and hence S(3) =< (1, 1, −2) >.
system
  

x
0 1
0



y =
0
−1
−1
(B − 2I3 )u =
z
0 2
2
To find S(2) we solve the

0
0 ,
0
from which x = y = 0 and hence S(2) =< (1, 0, 0) >.
Example 2.11. Find the eigenvalues and the

1
C = 0
1
proper subspaces of

2 0
2 0
1 3
Answer: To compute the eigenvalues we solve the characteristic equation
1 − λ
2
0
2−λ
0 0 = |C − λI3 | = 0
1
1
0 − λ
1 − λ
0 = 2 − λ = (2 − λ)(1 − λ)(3 − λ)
1
3 − λ
So, the eigenvalues are λ1 = 1, λ2 = 2 and λ3 = 3. We now compute the eigenvectors. The
eigenspace S(1) is the solution of the homogeneous linear system whose associated matrix is
C − λI3 with λ = 1. That is, S(1) is the solution of the following homogeneous linear system

   
0 2 0
x
0
0 2 0 y  = 0
1 1 2
z
0
Solving the above system we find that
S(1) = {(−2z, 0, z) : z ∈ R} =< (−2, 0, 1) >
9
On the other hand, S(2) is the set of solutions of the homogeneous linear system whose
associated matrix is C − λI3 with λ = 2. That is, S(2) is the solution of the following

   
−1 2 0
x
0
 0 0 0 y  = 0
1 1 1
z
0
So,
S(2) = {(2y, y, −3y) : y ∈ R} =< (2, 1, −3) >
Finally, S(3) is the set of solutions of the homogeneous linear system whose associated matrix
is A − λI3 with λ = 3. That is, S(3) is the solution of the following

   
−2 2 0
x
0
 0 −1 0 y  = 0
1
1 0
z
0
and we obtain
S(3) = {(0, 0, z) : z ∈ R} =< (0, 0, 1) >
We now start describing the procedure to diagonalize a matrix. Fix a square matrix A.
Let
λ1 , λ2 , . . . , λk
be distinct real roots of the characteristic polynomial pA (λ) an let mk be the multiplicity
of each λk (Hence mk = 1 if λk is a simple root, mk = 2 if it is double, etc.). Note that
m1 + m2 + · · · + mk ≤ n.
The following result estates that the number of independent vectors in the subspace S(λ)
can never be bigger than the multiplicity of λ.
Proposition 2.12. For each j = 1, . . . , k
1 ≤ dim S(λj ) ≤ nj .
The following theorem gives necessary and sufficient conditions for a matrix A to be
diagonalizable.
Theorem 2.13. A matrix A is diagonalizable if and only if the two following conditions
hold.
(1) Every root, λ1 , λ2 , . . . , λk of the charateristic polynomial pA (λ) is real.
(2) For each j = 1, . . . , k
dim S(λj ) = nj .
Corollary 2.14. If the matrix A has n distinct real eigenvalues, then it is diagonalizable.
Theorem 2.15. If A is diagonalizable, then the diagonal matrix D is formed by the eigenvalues of A in its main diagonal, with each λj repeated nj times. Moreover, a matrix P
such that D = P −1 AP has as columns independent eigenvectors selected from each proper
subspace S(λj ), j = 1, . . . , k.
10
Comments on the examples above.
• Matrix A of Example 1.3 is not diagonalizable, since pA has complex roots.
• Although all roots of pB are real, B of Example 1.4 is not diagonalizable, because
dim S(2) = 1, which is smaller than the multiplicity of λ = 2.
• Matrix C of Example 2.11 is diagonalizable, since pC has 3 different real roots. In
this case




1 0 0
−2 2 0
1 0 .
D = 0 2 0 ,
P = 0
0 0 3
1 −3 1
Example 2.16. Returning to Example 2.4, we compute
1
1
−λ
9
2 1
= 0,
8
−λ 2
9
7
. Now, S(1) is the solution set of
or 18λ − 25λ + 7 = 0. We get λ1 = 1 and λ2 = 18
! − 12 91
x
0
=
.
1
1
y
0
−
2
2
We find y =
9
x,
2
9
7
so that S(1) =< (2, 9) >. In the same way, S( 18
) is the solution set of
! 1
1
x
0
9
9
=
.
1
1
y
0
2
We find y = −x, so that
and
7
S( 18
)
2
=< (1, −1) >. Hence the diagonal matrix is
!
1 0
D=
7
0 18
2 1
,
P =
9 −1
P
−1
1
=
11
1 1
.
9 −2
Thus,
!
1
0
2 1
1 1
.
7 n
9 −1
9 −2
0 ( 18
)
1
A =
11
n
In particular, for n = 4 we obtain
A4 =
Hence
g4
b4
= A4
g1
b1
=
!
0.1891 0.1802
0.8111 0.8198
.
! 0.1891 0.1802
1
0.1891
=
.
0
0.8111
0.8111 0.8198
This means that probability that the 5th class goes right, conditioned to the event that the
first class was also right is of 0.1891.
11
We can wonder what happens in the long run, that is, supposing that the course lasts
forever (oh no!). In this case
!
!
2
2
1
0
11
lim An = P ( lim Dn )P −1 = P
P −1 = 11
,
9
9
n→∞
n→∞
0 0
11
11
to find that the stationary distribution of probabilities is
g∞
0.1818
=
.
b∞
0.8182
12
Topic 2: Difference Equations
1. Introduction
In this chapter we shall consider systems of equations where each variable has a time
index t = 0, 1, 2, . . . and variables of different time–periods are connected in a non–trivial
way. Such systems are called systems of difference equations and are useful to describe
dynamical systems with discrete time. The study of dynamics in economics is important
because it allows to drop out the (static) assumption that the process of economic adjustment
inevitable leads to an equilibrium. In a dynamic context, this stability property has to be
checked, rather than assumed away.
Let time be a discrete denoted t = 0, 1, . . .. A function X : N −→ Rn that depends on
this variable is simply a sequence of vectors of n dimensions
X0 , X1 , X2 , . . .
If each vector is connected with the previous vector by means of a mapping f : Rn −→ Rn
as
Xt+1 = f (Xt ),
t = 0, 1, . . . ,
then we have a system of first–order difference equations. In the following definition, we
generalize the concept to systems with longer time lags and that can include t explicitly.
Definition 1.1. A kth order discrete system of difference equations is an expression of the
form
(1.1)
Xt+k = f (Xt+k−1 , . . . , Xt , t),
t = 0, 1, . . . ,
where every Xt ∈ Rn and f : Rnk × [0, ∞) −→ Rn . The system is
• autonomous, if f does not depend on t;
• linear, if the mapping f is linear in the variables (Xt+k−1 , . . . , Xt );
• of first order, if k = 1.
Definition 1.2. A sequence {X0 , X1 , X2 , . . .} obtained from the recursion (1.1) with initial
value X0 is called a trajectory, orbit or path of the dynamical system from X0 .
In what follows we will write xt instead of Xt if the variable Xt is a scalar.
Example 1.3. [Geometrical sequence] Let {xt } be a scalar sequence, xt+1 = qxt , t = 0, 1, . . .,
with q ∈ R. This a first–order, autonomous and linear difference equation. Obviously
xt = q t x0 . Similarly, for arithmetic sequence, xt+1 = xt + d, with d ∈ R, xt = x0 + td.
Example 1.4.
• xt+1 = xt + t is linear, non–autonomous and of first order;
• xt+2 = −xt is linear, autonomous and of second order;
• xt+1 = x2t + 1 is non–linear, autonomous and of first order;
13
Example 1.5. [Fibonacci numbers (1202)] “How many pairs of rabbits will be produced
in a year, beginning with a single pair, if every month each pair bears a new pair which
becomes productive from the second month on?”. With xt denoting the pairs of rabbits in
month t, the problem leads to the following recursion
xt+2 = xt+1 + xt ,
t = 0, 1, 2, . . . , with x0 = 1 and x1 = 1.
This is an autonomous and linear second–order difference equation.
2. Systems of first order difference equations
Systems of order k > 1 can be reduced to first order systems by augmenting the number
of variables. This is the reason we study mainly first order systems. Instead of giving a
general formula for the reduction, we present a simple example.
Example 2.1. Consider the second–order difference equation yt+2 = g(yt+1 , yt ). Let x1,t =
yt+1 , x2,t = yt , then x2,t+1 = yt+1 = x1,t and the resulting first order system is
x1,t+1
g(x1,t , x2,t )
=
.
x2,t+1
x1,t
x1,t
g(Xt )
If we denote Xt =
, f (Xt ) =
, then the system can be written Xt+1 =
x2,t
x1,t
f (Xt ).
For example, yt+2 = 4yt+1 + yt2 + 1 can be reduced to the first order system
4x1,t + x22,t + 1
x1,t+1
=
,
x2,t+1
x1,t
and the Fibonacci equation of Example 1.5 is reduced to
x1,t+1
x1,t + x2,t
=
,
x2,t+1
x1,t
For a function f : Rn −→ Rn , we shall use the following notation: f t denotes the t–fold
composition of f , i.e. f 1 = f , f 2 = f ◦ f and, in general, f t = f ◦ f t−1 for t = 1, 2, . . .. We
also define f 0 as the identity function, f 0 (X) = X.
Theorem 2.2. Consider the autonomous first order system Xt+1 = f (Xt ) and suppose that
there exists some subset D such that for any X ∈ D, f (X) ∈ D. Then, given any initial
condition X0 ∈ D, the sequence {Xt } is given by
Xt = f t (X0 ).
14
Proof. Notice that
X1 = f (X0 ),
X2 = f (X1 ) = f (f (X0 )) = f 2 (X0 ),
..
.
Xt = f (f · · · f (X0 ) · · · )) = f t (X0 ).
The theorem provides the current value of X, Xt , in terms of the initial value, X0 . We
are interested what is the behavior of Xt in the future, that is, in the limit
lim f t (X0 ).
t→∞
Generally, we are more interested in this limit that in the analytical expression of Xt . Nevertheless, there are some cases where the solution can be found explicitly, so we can study
the above limit behavior quite well. Observe that if the limit exists, limt→∞ f t (X0 ) = X 0 ,
say, and f is continuous
f (X 0 ) = f ( lim f t (X0 )) = lim f t+1 (X0 ) = X 0 ,
t→∞
t→∞
0
hence the limit X is a fixed point of map f . This is the reason fixed points play a distinguished role in dynamical systems.
Definition 2.3. A point X 0 ∈ D is called a fixed point of the autonomous system f if,
starting the system from X 0 , it stays there:
If X0 = X 0 , then Xt = X 0 ,
t = 1, 2, . . . .
Obviously, X 0 is also a fixed point of map f . A fixed point is also called equilibrium,
stationary point, or steady state.
Example 2.4. In Example 1.3 (xt+1 = qxt ), if q = 1, then every point is a fixed point; if
q 6= 1, then there exists a unique fixed point: x0 = 0. Notice that the solution xt = q t x0 has
the following limit (x0 6= 0) depending the value of q.
−1 < q < 1 ⇒ lim q t x0 = 0,
t→∞
q = 1 ⇒ lim q t x0 = x0 ,
t→∞
q ≤ −1 ⇒ the sequence oscillates between + and − and the limit does not exist
In Example 1.5, x0 = 0 is the unique fixed point. Consider now the difference equation
xt+1 = x2t − 6. Then, the fixed points are the solutions of x = x2 − 6, that is, x0 = −2 and
x0 = 3.
In the following definitions, kX − Y k stands for the Euclidean distance between X and
Y . For example, if X = (1, 2, 3) and Y = (3, 6, 7), then
p
√
kX − Y k = (3 − 1)2 + (6 − 2)2 + (7 − 3)2 = 36 = 6.
15
Definition 2.5.
• A fixed point X 0 is called stable if for any close enough initial state X0 , the resulting
trajectory {Xt } exists and stays close forever to X 0 , that is, for any positive real ε,
there exists a positive real δ(ε) such that if kX0 − X 0 k < δ(), then kXt − X 0 k < ε
for every t.
• A stable fixed point X 0 is called locally asymptotically stable (l.a.s.) if the trajectory
{Xt } starting from any initial point X0 close to enough to X 0 , converges to the fixed
point.
• A stable fixed point is called globally asymptotically stable (g.a.s.) if any trajectory
generated by any initial point X0 converges to it.
• A fixed point is unstable if it is not stable or asymptotically stable.
Remark 2.6.
• If X 0 is stable, but not l.a.s., {Xt } need not approach X 0 .
• A g.a.s. fixed point is necessarily unique.
• If X 0 is l.a.s., then small perturbations around X 0 decay and the trajectory generated
by the system returns to the fixed point as the time grows.
Definition 2.7. Let P be an integer larger than 1. A series of vectors X0 , X1 , . . . , XP −1 is
called a P –period cycle of system f if a trajectory starting from X0 goes through X1 , . . . , XP −1
and returns to X0 , that is
Xt+1 = f (Xt ),
t = 0, 1, . . . , P − 1,
XP = X0 .
Observe that the series of vectors X0 , X1 , . . . , XP repeats indefinitely in the trajectory,
{Xt } = {X0 , X1 , . . . , XP −1 , X0 , X1 , . . . , XP −1 , . . .}.
For this reason, the trajectory itself is called a P –cycle.
Example 2.8. In Example 1.3 (xt+1 = qxt ) with q = −1 all the trajectories contains 2–
cycles, because a typical path is
{x0 , −x0 , x0 , −x0 , . . .}.
Example 2.9. In Example 1.4 where yt+2 = −yt , to find the possible cycles of the equation,
first we write it as first order system using Example 2.1, to obtain
x1,t+1
−x2,t
Xt+1 =
=
≡ f (Xt ).
x2,t+1
x1,t
Let X0 = (2, 4). Then
X1 = f (X0 ) = (−4, 2),
X2 = f (X1 ) = (−2, −4),
X3 = f (X2 ) = (4, −2),
X4 = f (X3 ) = (2, 4) = X0 .
Thus, a 4–cycle appears starting at X0 . In fact, any trajectory is a 4–cycle.
16
3. First order linear difference equations
The linear equation is of the form
(3.1)
xt ∈ R,
xt+1 = axt + b,
a, b ∈ R.
Consider first the case b = 0 (homogeneous case). Then, by Theorem 2.2 the solution is
xt = at x0 , t = 0, 1, . . .. Consider now the non–homogeneous case, b 6= 0. Let us find the
fixed points of the equation. They solve (see Definition 2.3)
x0 = ax0 + b,
hence there is no fixed point if a = 1. However, if a 6= 1, the unique fixed point is
b
.
1−a
Define now yt = xt − x0 and replace xt = yt + x0 into (3.1) to get
x0 =
yt+1 = ayt ,
hence yt = at y0 . Returning to the variable xt we find that the solution of the linear equation
is
xt = x0 + at (x0 − x0 )
b
b
t
=
+ a x0 −
.
1−a
1−a
Theorem 3.1. In (3.1), the fixed point x0 =
b
1−a
is g.a.s. if and only if |a| < 1.
Proof. Notice that limt→∞ at = 0 iff |a| < 1 and hence limt→∞ xt = limt→∞ x0 + at (x0 − x0 ) =
x0 iff |a| < 1, independently of the initial x0 .
The convergence is monotonous if 0 < a < 1 and oscillating if −1 < a < 0.
Example 3.2 (A Multiplier–Accelerator Model of Growth). Let Yt denote national income,
It total investment, and St total saving—all in period t. Suppose that savings are proportional to national income, and that investment is proportional to the change in income from
period t to t + 1. Then, for t = 0, 1, 2, . . .,
St = αYt ,
It+1 = β(Yt+1 − Yt ),
St = It .
The last equation is the equilibrium condition that saving equals investment in each period.
Here β > α > 0. We can deduce a difference equation for Yt and solve it as follows. From
the first and third equation, It = αYt , and so It+1 = αYt+1 . Inserting these into the second
equation yields αYt+1 = β(Yt+1 − Yt ), or (α − β)Yt+1 = −βYt . Thus,
α
β
Yt+1 =
Yt = 1 +
Yt ,
t = 0, 1, 2, . . . .
β−α
β−α
17
The solution is
t
α
Yt = 1 +
Y0 ,
t = 0, 1, 2, . . . .
β−α
Thus, Y grows at the constant proportional rate g = α/(β − α) each period. Note that
g = (Yt+1 − Yt )/Yt .
Example 3.3 (A Cobweb Model). Consider a market model with a single commodity where
producer’s output decision must be made one period in advance of the actual sale—such as
in agricultural production, where planting must precede by an appreciable length of time
the harvesting and sale of the output. Let us assume that the output decision in period t is
based in the prevailing price Pt , but since this output will no be available until period t + 1,
the supply function is lagged one period,
Qs,t+1 = S(Pt ).
Suppose that demand at time t is determined by a function that depends on Pt ,
Qd,t+1 = D(Pt ).
Supposing that functions S and D are linear and that in each time period the market clears,
we have the following three equations
Qd,t = Qs,t ,
Qd,t+1 = α − βPt+1 ,
Qs,t+1 = −γ + δPt ,
α, β > 0,
γ, δ > 0.
By substituting the last two equations into the first the model is reduced to the difference
equation for prices
δ
α+γ
Pt+1 = − Pt +
.
β
β
The fixed point is P 0 = (α + γ)/(β + δ), which is also the equilibrium price of the market,
that is, S(P 0 ) = D(P 0 ). The solution is
t
δ
0
Pt+1 = P + −
(P0 − P 0 ).
β
Since −δ/β is negative, the solution path is oscillating. It is this fact which gives rise to the
cobweb phenomenon. There are three oscillations patterns: it is explosive if δ > β (S steeper
than D), uniform if δ = β, and damped if δ < β (S flatter than D). The three possibilities
are illustrated in the graphics below. The demand is the downward–slopping line, with slope
−β. The supply is the upward–slopping line, with slope δ. When δ > β, as in Figure 3, the
interaction of demand and supply will produce an explosive oscillation as follows: Given an
initial price P0 , the quantity supplied in the next period will be Q1 = S(P0 ). In order to
clear the market, the quantity demanded in period 1 must be also Q1 , which is possible if
and only if price is set at the level of P1 given by the equation Q1 = D(P1 ). Now, via the S
curve, the price P1 will lead to Q2 = S(P1 ) as the quantity supplied in period 2, and to clear
the market, price must be set at the level of P2 according to the demand curve. Repeating
this reasoning, we can trace out a “cobweb” around the demand and supply curves.
18
Figure 1. Cobweb diagram with damped oscillations
Figure 2. Cobweb diagram with uniform oscillations
Figure 3. Cobweb diagram with explosive oscillations
19
4. Second order linear difference equations
The second–order linear difference equation is
xt+2 + a1 xt+1 + a0 xt = bt ,
where a0 and a1 are constants and bt is a given function of t. The associated homogeneous
equation is
xt+2 + a1 xt+1 + a0 xt = 0,
and the associated characteristic equation is
r2 + a1 r + a0 = 0.
This quadratic equation has solutions
q
q
1
1
1
1
2
a1 − 4a0 ,
r2 = − a1 −
a21 − 4a0 .
r1 = − a1 +
2
2
2
2
There are three different cases depending of the sign of the discriminant a21 − 4a0 of the
equation. When it is negative, the solutions are (conjugate) complex numbers.
Recall that
√
a complex number is z = a + ib, where a and b are real numbers and i = −1 is called the
imaginary unit, so that i2 = −1. The real part of z is a, and the imaginary part of z is b. The
conjugate of z = a+ib is z = a−ib. Complex numbers can be added, z+z 0 = (a+a0 )+i(b+b0 ),
and multiplied,
zz 0 = (a + ib)(a0 + ib0 ) = aa0 + iab0 + ia0 b + i2 bb0 = (aa0 − bb0 ) + i(ab0 + a0 b).
√
For the following theorem we need the modulus of z, ρ = |z| = a2 + b2 , and the argument
of z, which is the angle θ ∈ (−π/2, π/2] such that tan θ = b/a. It is useful to recall the
following table of trigonometric values
θ sin θ cos θ tan θ
0
π
6
π
3
π
2
0
1
2
√
3
2
1
1
0
√
3
2
√
3
3
1
2
√
3
0
∞
For the negative values of the argument θ, observe that sin (−θ) = − sin θ and cos (−θ) =
cos θ, so that tan (−θ) = − tan θ.
√
For example, the modulus and argument of 1 − i is ρ = 2 and θ = −π/2, respectively,
since tan θ = −1/1 = −1.
Theorem 4.1. The general solution of
(4.1)
is as follows:
xt+2 + a1 xt+1 + a0 xt = 0
(a0 6= 0)
20
(1) If a21 − 4a0 > 0 (the characteristic equation has two distinct real roots),
q
1
1
t
t
r1,2 = − a1 ±
xt = Ar1 + Br2 ,
a21 − 4a0 .
2
2
(2) If a21 − 4a0 = 0 (the characteristic equation has one real double roots),
xt = (A + Bt)rt ,
1
r = − a1 .
2
(3) If a21 − 4a0 < 0 (the characteristic equation has no real roots),
p
√
4a0 − a21
xt = ρt (A cos θt + B sin θt), ρ = a0 , tan θ = −
,
a1
θ ∈ [0, π].
Remark 4.2. When the characteristic equation has complex roots, the solution of (4.1)
involves oscillations. Note that when ρ < 1, ρt tends to 0 as t → ∞ and the oscillations are
damped. If ρ > 1, the oscillations are explosive, and in the case ρ = 1, we have undamped
oscillations.
Example 4.3. Find the general solutions of
(a) xt+2 − 7xt+1 + 6xt = 0,
(b) xt+2 − 6xt+1 + 9xt = 0,
(c) xt+2 − 2xt+1 + 4xt = 0.
Solution: (a) The characteristic equation is r2 − 7r + 6 = 0, whose roots are r1 = 6 and
r2 = 1, so the general solution is
xt = A6t + B,
A, B ∈ R.
(b) The characteristic equation is r2 − 6r + 9 = 0, which has a double root r = 3. The
general solution is
xt = 3t (A + Bt).
√
(c) The characteristic equation is r2 −2r +4 = 0, with complex
solutions r1 = 21 (2+ −12) =
√
√
√
√
12
= 3. this means that θ = π/3.
(1 + i 3), r2 = (1 − i 3). Here ρ = 2 and tan θ = − −2
The general solution is
π
π xt = 2t A cos t + B sin t .
3
3
4.1. The nonhomogeneous case. Now consider the nonhomogeneous equation
(4.2)
xt+2 + a1 xt+1 + a0 xt = bt ,
and let x∗t be a particular solution. It turns out that solutions of the equation have an
interesting structure, due to the linearity of the equation.
Theorem 4.4. The general solution of the nonhomogeneous equation (4.2) is the sum of
the general solution of the homogeneous equation (4.1) and a particular solution x∗t of the
nonhomogeneous equation.
21
Example 4.5. Find the general solution of xt+2 − 4xt = 3.
Solution: Note that x∗t = −1 is a particular solution. To find the general solution of
the homogeneous equation, consider the solutions of the characteristic equation, m2 − 4 = 0,
m1,2 = ±2. Hence, the general solution of the nonhomogeneous equation is
xt = A(−2)t + B2t − 1.
Example 4.6. Find the general solution of xt+2 − 4xt = t.
Solution: Now it is not obvious how to find a particular solution. We can try with the
method of undetermined coefficients and try with some expression of the form x∗t = Ct + D.
Then, we look for constants a, b such that x∗t is a solution. This requires
C(t + 2) + D − 4(Ct + D) = t,
∀t = 0, 1, 2, . . .
One must have C − 4C = 1 and 2C + D − 4D = 0. It follows that C = −1/3 and D = −2/9.
Thus, the general solution is
xt = A(−2)t + B2t − t/3 − 2/9.
Example 4.7. Find the solution of xt+2 − 4xt = t satisfying x0 = 0 and x1 = 1/3.
Solution: Using the general solution found above, we have two equations for the two
unknown parameters A and B:
A + B + 29
=0
.
−2A + 2B − 31 + 29 = 13
The solution is A = −2/9 and B = 0. Thus, the solution of the nonhomogeneous equation
is
2
t 2
xt = − (−2)t − + .
9
2 9
The method of undetermined coefficients for solving equation (4.2) suppose that a particular solution has the form of the nonhomogeneous term, bt . The method works quite well
when this term is of the form
at ,
tm ,
cos at,
sin at
or linear combinations of them.
Example 4.8. Solve the equation xt+2 − 5xt+1 + 6xt = 4t + t2 + 3.
Solution: The homogeneous equation has characteristic equation r2 − 5r + 6 = 0, with
two different real roots r1,2 = 2, 3. Its general solution is, therefore, A2t + B3t . To find a
particular solution we look for constants C, D, E and F such that a particular solution is
x∗t = C4t + Dt2 + Et + F.
22
Plugging this into the equation we find
C4t+2 + D(t + 2)2 + E(t + 2) + F − 5(C4t+1 + D(t + 1)2 + E(t + 1) + F )
+ 6(C4t + Dt2 + Et + F ) = 4t + t2 + 3.
Expanding and rearranging yields
2C4t + 2Dt2 + (−6D + 2E)t + (−D − 3E + 2F ) = 4t + t2 + 3.
This must hold for every t = 0, 1, 2, . . . thus,
2C
2D
−6D + 2E
−D − 3E + 2F
= 4,
= 1,
= 0,
= 3.
It follows that C = 1/2, D = 1/2, E = 3/3 and F = 4. The general solution is
1
3
1
xt = A2t + B3t + 4t + t2 + t + 4.
2
2
2
Example 4.9 (A Multiplier–Accelerator Growth Model). Let Yt denote national income,
Ct total consumption, and It total investment in a country at time t. Assume that for
t = 0, 1, . . . ,
(i) Yt = Ct + It
(ii) Ct+1 = aYt + b
(income is divided between consumption and investment)
(consumption is a linear function of previous income)
(iii) It+1 = c(Ct+1 − Ct ) (investment is proportional to to the change in consumption),
where a, b, c > 0. Find a second order difference equation describing this national economy.
Solution: We eliminate two of the unknown functions as follows. From (i), we get (iv)
Yt+2 = Ct+2 + It+2 . Replace now t by t + 1 in (ii) and (iii) to get (v) Ct+2 = aYt+1 + b and
(vi) It+2 = c(Ct+2 − Ct+1 ), respectively. Then, inserting (iii) and (v) into (vi) gives It+2 =
ac(Yt+1 − Yt ). Inserting this result and (v) into (iv) we get Yt+2 = aYt+1 + b + ac(Yt+1 − Yt )
and rearranging we arrive to
Yt+2 − a(1 + c)Yt+1 + acYt = b,
t = 0, 1, . . .
The form of the solution depends on the coefficients a, b, c.
23
5. Linear systems of difference equations
Now we suppose that the dynamic variables are vectors, Xt ∈ Rn . A first order system of
linear difference equations with constant coefficients is given by
x1,t+1 = a11 x1,t + · · · + a1n xn,t + b1,t
..
.
xn,t+1 = an1 x1,t + · · · + ann xn,t + bn,t
An example is
x1,t+1 = 2x1,t − x2,t + 1
x2,t+1 = x1,t + x2,t + e−t .
Most often we will rewrite systems omitting subscripts using different letters for different
variables, as in
xt+1 = 2xt − yt + 1
yt+1 = xt + yt + e−t .
A linear system is equivalent to the matrix equation
Xt+1 = AXt + Bt ,
where

x1,t
Xt =  ...  ,
xn,t


a11 . . . a1n
A =  ... . . . ...  ,
an1 . . . ann


b1,t
B =  ... 
bn,t

We will center on the case where the independent term Bt ≡ B is a constant vector.
5.1. Homogeneous systems. Consider the homogeneous system Xt+1 = AXt .
Note that X1 = AX0 , X2 = AX1 = AAX0 = A2 X0 . Thus, given the initial vector X0 , the
solution is
Xt = At X0 .
In the case that A be diagonalizable, P −1 AP = D with D diagonal, the expression above
simplifies to
Xt = P Dt P −1 X0 ,
that is easy to compute since D is diagonal.
Example 5.1. Find the general solution of the system
xt+1
4 −1
xt
=
2 1
yt+1
yt
24
4 −1
Solution: The matrix A =
has characteristic polynomial pA (λ) = λ2 −5λ+6,
2 1
with roots λ1 = 3 and λ2 = 2. Thus, the matrix is diagonalizable. It is easy to find the
eigenspaces
S(3) =< (1, 1) >,
S(2) =< (1, 2) > .
Hence, the matrix P is
1 1
P =
,
1 2
P
−1
=
2 −1
−1 1
,
D=
3 0
0 2
and the solution
t
x0
2 −1
2 3t − 2t
−3t + 2t
1 1
3 0
X0 =
Xt =
y0
−1 1
2 3t − 2t+1 − 1 −3t + 2t+1
1 2
0 2t
Supposing that the initial condition is (x0 , y0 ) = (1, 2), the solution is given by
xt = 2 3t − 2t + 2(−3t + 2t ),
yt = 2 3t − 2t+1 − 1 + 2(−3t + 2t+1 ).
5.2. Nonhomogeneous systems. Consider the system Xt+1 = AXt + B, where B is a
non–null, constant vector.
To obtain a closed-form solution of the system, we begin by noting that
X1 = AX0 + B,
X2 = AX1 + B = A(AX0 + B) + B = A2 X0 + (A + In )B,
..
.
Xt = AXt−1 + B = · · · = At X0 + (At−1 + At−2 + · · · + In )B.
Observe that
(At−1 + At−2 + · · · + In )(A − In ) = At + At−1 + · · · + A − At−1 − · · · − A − In = At − In .
Thus, assuming that (A − In ) is invertible, we find
At−1 + At−2 + · · · + In = (At − In )(A − In )−1 .
Plugging this equality into the expression for Xt above one gets
Xt = At X0 + (At − In )(A − In )−1 B.
On the other hand, note that the constant solutions of the nonhomogeneous system (or fixed
points of the system) satisfy
X 0 = AX 0 + B.
Assuming again that the matrix A − In has inverse, we can solve for X 0
(In − A)X 0 = B ⇒ X 0 = (In − A)−1 B.
25
Then, collecting all the above observations, we can write the solution of the nonhomogeneous system in a nice form as
(5.1)
Xt = At X0 − (At − In )X 0 = X 0 + At (X0 − X 0 ).
Theorem 5.2. Suppose that |A−In | =
6 0. Then, the general solution of the nonhomogeneous
system is given in Eqn. (5.1). Moreover, when A is diagonalizable, the above expression may
be written as
(5.2)
Xt = X 0 + P Dt P −1 (X0 − X 0 ),
where P −1 AP = D with D diagonal.
Proof. Eqn. (5.2) easily follows from Eqn. (5.1), taking into account that At = P Dt P −1 . Example 5.3. Find the general solution of the system
xt+1
4 −1
xt
1
=
+
yt+1
2 1
yt
−1
Solution: The fixed point X ∗ is given by
−1 1 0 −1
−3 1
1
1
1/2
−1
(I3 − A) B =
=
=
−2 0
−1
−1
5/2
2 2 −3
By the example above we already know the general solution of the homogeneous system.
The general solution of the nonhomogeneous system is then
1/2
x0 − 1/2
xt
2 3t − 2t
−3t + 2t
.
+
=
5/2
y0 − 5/2
yt
2 3t − 2t+1 − 1 −3t + 2t+1
5.3. Stability of linear systems. We study here the stability properties of a first order
system Xt+1 = AXt + B where |In − A| =
6 0.
For
the
following
theorem,
recall
that
for a complex number z = α + βi, the modulus is
p
2
2
ρ = α + β . For a real number α, the modulus is |α|.
Theorem 5.4. A necessary and sufficient condition for system Xt+1 = AXt + B to be g.a.s.
is that all roots of the characteristic polynomial pA (λ) (real or complex) have moduli less
than 1. In this case, any trajectory converges to X ∗ = (In − A)−1 B as t → ∞.
We can give an idea of the proof of the above theorem in the case where the matrix A is
diagonalizable. As we have shown above, the solution of the nonhomogeneous system in this
case is
Xt = X 0 + P Dt P −1 (X0 − X 0 ),
where
 t

λ1 0 . . . 0
 0 λt2 . . . 0 
D=
,
.. . .
.. 
 ...
.
.
. 
0 0 . . . λtn
26
and λ1 , . . . , λn are the real roots (possibly repeated) of pA (λ). Since |λj | < 1 for all j, the
diagonal elements of Dt tends to 0 as t goes to ∞, since λtj ≤ |λj |t → 0. Hence
lim Xt = X 0 .
t→∞
Example 5.5. Study the stability of the system
1
xt+1 = xt − yt + 1,
2
yt+1 = xt − 1.
1 −1/2
Solution: The matrix of the system is
, with characteristic equation λ2 −
1
0
p
λ+1/2
=
0.
The
(complex)
roots
are
λ
=
1/2±i/2.
Both
have
modulus
ρ
=
1/4 + 1/4 =
1,2
√
1/ 2 < 1, hence the system is g.a.s. and the limit of any trajectory is the equilibrium point,
−1 1 − 1 0 − (−1/2)
1
3
0
X =
=
.
0−1
1−0
−1
2
Example 5.6. Study the stability of the system
xt+1 = −xt + yt ,
yt+1 = −xt /2 − yt /2.
1
3
Solution: The matrix of the system is
, with characteristic equation λ2 −
1/2 1/2
(3/2)λ−1 = 0. The roots are λ1 = 2 and λ2 = −1/2. The system is not g.a.s. However, there
are initial conditions X0 such that the trajectory converges to the fixed point X 0 = (0, 0).
This can be seen once we find the solution
Xt = P Dt P −1 X0 .
The eigenspaces are S(2) =< (3, 1) > and S(−1/2) =< (2, −1) >, thus
3 2
1/5 2/5
−1
P =
,
P =
.
1 −1
1/5 −3/5
The solution is
xt+1
yt+1
=
2t 35 (x0 + 2y0 ) + 21−t 15 (x0 − 3y0 )
!
.
2t 51 (x0 + 2y0 ) − 2−t 15 (x0 − 3y0 )
If the initial conditions are linked by the relation x0 + 2y0 = 0, then the solution converges
to (0, 0). For this reason, the line x + 2y = 0 is called the stable manifold. Notice that the
stable manifold is in fact the eigenspace associated to the eigenvalue λ2 = −1/2, since
S(−1/2) =< (2, −1) >= {x + 2y = 0}.
For any other initial condition (x0 , y0 ) ∈
/ S(−1/2), the solution does not converge.
27
Example 5.7 (Dynamic Cournot adjustment). The purpose of this example is to investigate
under what conditions a given adjustment process converges to the Nash equilibrium of the
Cournot game.
Consider a Cournot duopoly in which two firms produce the same product and face constant marginal costs c1 > 0 and c2 > 0. The market price Pt is a function of the total
quantity of output produced Q = q1 + q2 in the following way
P = α − βQ,
α > ci ,
i = 1, 2,
β > 0.
In the Cournot duopoly model each firm chooses qi to maximize profits, taking as given the
production level of the other firm, qj . At time t, firm i’s profit is
πi = q i P − ci q i .
i
As it is well–known, taking ∂π /∂qi = 0 we obtain the best response of firm i, which depends
on the output of firm j as follows1
br1 = a1 − q2 /2,
br2 = a2 − q1 /2,
α − ci
, i = 1, 2. We suppose that a1 > a2 /2 and that a2 > a1 /2 in order to have
2β
positive quantities in equilibrium, as will be seen below.
The Nash equilibrium of the game, (q1N , q2N ), is a pair of outputs of the firms such that
none firm has incentives to deviate from it unilaterally, that is, it is the best response against
itself. This means that the Nash equilibrium of the static game solves
where ai =
q1N = br1 (q2N ),
q2N = br2 (q1N ).
In this case
q1N = a1 − q2N /2,
q2N = a2 − q2N /2.
Solving, we have
4
a1 a2 −
,
3
2
4
a2 q2N =
a1 −
,
3
2
which are both positive by assumption. As a specific example, suppose for a moment that
α−c
the game is symmetric, with c1 = c2 = c. Then, a1 = a2 =
and the Nash equilibrium
2β
is the output
α−c
q1N =
,
3β
α−c
q2N =
.
3β
q1N =
1Actually,
the best response map is bri = max{ai − qj /2, 0}, since negative quantities are not allowed.
28
Now we turn to the general asymmetric game and introduce a dynamic component in the
game as follows. Suppose that each firm does not choose its Nash output instantaneously, but
they adjust gradually its output qi towards its best response bri at each time t as indicated
below
(
q1,t+1 = q1,t + d1 (br1,t − q1,t ) = q1,t + d1 (a1 − 12 q2,t − q1,t ),
(5.3)
q2,t+1 = q1,t + d2 (br2,t − q2,t ) = q2,t + d2 (a2 − 12 q1,t − q2,t ),
where d1 and d2 are positive constants. The objective is to study whether this tattonement
process converges to the Nash equilibrium.
To simplify notation, let us rename x = q1 and y = q2 . Then, rearranging terms in the
system (5.3) above, it can be rewritten as
(
xt+1 = (1 − d1 )xt − d21 yt + d1 a1 ,
yt+1 = (1 − d2 )yt −
d2
x
2 t
+ d 2 a2 .
It is easy to find the equilibrium points by solving the system
(
x = (1 − d1 )x − d21 y + d1 a1 ,
y = (1 − d2 )y −
d2
x
2
+ d 2 a2 .
The only solution is precisely the Nash equilibrium,
4
a1 4 a2 N
N
a2 −
,
a1 −
(x , y ) =
.
3
2
3
2
Under what conditions this progressive adjustment of the produced output does converge to
the Nash equilibrium? According to the theory, it depends on the module of the eigenvalues
being smaller than 1. Let us find the eigenvalues of the system. The matrix of the system is
1 − d1 − d21
− d22 1 − d2
To simplify matters, let us suppose that the adjustment parameter is the same for both
players, d1 = d2 = d. The eigenvalues of the matrix are
d
3d
λ1 = 1 − ,
λ2 = 1 − ,
2
2
which only depend on d. We have
|λ1 | < 1 iff 0 < d < 4,
|λ2 | < 1 iff 0 < d < 4/3,
therefore |λ1 | < 1 and |λ2 | < 1 iff 0 < d < 4/3. Thus, 0 < d < 4/3 is a necessary and
sufficient condition for convergence to the Nash equilibrium of the one shot game from any
initial condition (g.a.s. system).
29
6. The nonlinear first order equation
We investigate here the stability of the solutions of an autonomous first order difference
equation
xt+1 = f (xt ),
t = 0, 1, . . . ,
where f : I → I is nonlinear and I is an interval of the real line. Recall that a function f is
said to be of class C 1 in an open interval, if f 0 exists and it is continuous in that interval. For
example, the functions x2 , cos x or ex are C 1 is the whole real line, but |x| is not differentiable
at 0, so is not C 1 in any open interval that contains 0.
Theorem 6.1. Let x0 ∈ I a fixed point of f , and suppose that f is C 1 in an open interval
around x0 , Iδ = (x0 − δ, x0 + δ).
(1) If |f 0 (x0 )| < 1, then x0 is locally asymptotically stable;
(2) If |f 0 (x0 )| > 1, then x0 is unstable.
Proof. Since f 0 is continuous in Iδ and f 0 (x0 ) < 1, there exists some open interval Iδ =
(x0 − δ, x0 + δ) and a positive number k < 1 such that |f 0 (x)| ≤ k for any x ∈ Iδ .
(1) By the mean value theorem (also called Theorem of Lagrange), there exists some c
between x0 and x0 such that
f (x0 ) − f (x0 ) = f 0 (c)(x0 − x0 ),
or
x1 − x0 = f 0 (c)(x0 − x0 ),
since x0 = f (x0 ) by definition of fixed point. Consider an initial condition x0 ∈ Iδ .
Then any c between x0 and x0 belongs to Iδ and thus taking absolute values in the
equality above we get
|x1 − x0 | = |f 0 (c)||x0 − x0 | ≤ k|x0 − x0 |.
(6.1)
Also note that |x1 − x0 | ≤ kδ < δ, thus x1 ∈ Iδ . Reasoning as above, one gets
|x2 − x0 | = |f (x1 ) − x0 | = |f 0 (c)||x1 − x0 | ≤ k|x1 − x0 | ≤ k 2 |x0 − x0 |.
where c is a number between x1 and x0 that belongs to Iδ (and thus |f 0 (c)| ≤ k).
Continuing in this fashion we get after t steps
|xt − x0 | ≤ k t |x0 − x0 | → 0,
as t → ∞.
0
So xt converges to the fixed point x as t → ∞, and x0 is l.a.s.
(2) Now suppose that |f 0 (x0 )| > 1. Again by continuity of f 0 , there exists δ > 0 and
K > 1 such that |f 0 (x)| > K for any x ∈ Iδ . By equation (6.1) one has
|x1 − x0 | = |f 0 (c)||x0 − x0 | > K|x0 − x0 |
and after t steps
|xt − x0 | > K t |x0 − x0 |.
Since K t tends to ∞ as t → ∞, then xt departs more and more from x0 at each step,
and the fixed point x0 is unstable.
30
Remark 6.2. If |f 0 (x)| < 1 for every point x ∈ I, then the fixed point x0 is globally
asymptotically stable.
Example 6.3 (Population growth models). In the Malthus model of population growth it
is postulated that a given population x grows at constant rate r,
xt+1 − xt
= r,
or
xt+1 = (1 + r)xt .
xt
This is a linear equation and the population grows unboundedly if the per capita growth rate
r is positive2. This is not realistic for large t. When the population is small, there are ample
environmental resources to support a high birth rate, but for later times, as the population
grows, there is a higher death rate as individuals compete for space and food. Thus, the
growth rate should be decreasing as the population increases. The simplest case is to take a
linearly decreasing per capita rate, that is
xt r 1−
,
K
where K is the carrying capacity. This modification is known as the Verhulst’ law. Then
the population evolves as
r xt+1 = xt 1 + r − xt ,
K
which is not linear. The function f is quadratic, f (x) = x(1 + r − rx/K). In Fig. 6.3 is
depicted a solution with x0 = 5, r = 0.5 and K = 20.
We observe that the solution converges to 20. In fact, there are two fixed points of
the equation, 0 (extinction of the population) and x0 = K (maximum carrying capacity).
Considering the derivative of f at these two fixed points, we have
r 0
f (0) = 1 + r − 2 x
= 1 + r > 1,
K x=0
r f 0 (K) = 1 + r − 2 x
= 1 − r.
K x=K
Thus, according to Theorem 6.1, 0 is unstable, but K is l.a.e. iff |1 − r| < 1, or 0 < r < 2.
2The
solution is xt = (1 + r)t x0 , why?
31
6.1. Phase diagrams. The stability of a fixed point of the equation
xt+1 = f (xt ),
t = 0, 1, . . . ,
can also be studied by a graphical method based in the phase diagram. This consists in
drawing the graph of the function y = f (x) in the plane xy. Note that a fixed point x0
corresponds to a point (x0 , x0 ) where the graph of y = f (x) intersects the straight line
y = x.
The following figures show possible configurations around a fixed point. The phase diagram is at the left (plane xy), and a solution sequence is shown at the right (plane tx).
Notice that we have drawn the solution trajectory as a continuous curve because it facilitates
visualization, but in fact it is a sequence of discrete points. In Fig. 4, f 0 (x0 ) is positive, and
the sequence x0 , x1 , . . . converges monotonically to x0 , whereas in Fig. 5, f 0 (x0 ) is negative
and we observe a cobweb–like behavior, with the sequence x0 , x1 , . . . converging to x0 but
alternating between values above and below the equilibrium. In Fig. 6, the graph of f near
x0 is too steep for convergence. After many iterations in the diagram, we observe an erratic
behavior of the sequence x0 , x1 , . . .. There is no cyclical patterns and two sequences generated from close initial conditions depart along time at an exponential rate (see Theorem 6.1
above). It is often said that the equation exhibits chaos. Finally, Fig. 7 is the phase diagram
of an equation admitting a cycle of period 3.
Figure 4. x0 stable, f 0 (x0 ) ∈ (0, 1)
32
Figure 5. x0 stable, f 0 (x0 ) ∈ (−1, 0)
Figure 6. x0 unstable, |f 0 (x0 )| > 1
Figure 7. A cycle of period 3
33
Topic 3: Differential Equations
1. Introduction. Definitions and classifications of ODEs
Most often decision agents take optimal actions sequentially and economic variables evolve
along time. Thus it is important to understand the tools of analysis and modeling of dynamical systems. We are looking at functions x : R −→ R or vector functions x : R −→ Rn
described by equations of the form
d
x(t) = f (t, x(t)),
dt
possibly with an initial condition x(t0 ) = x0 .
Objectives:
(1) To find x(t) in closed form or, if this is not possible,
(2) to study qualitative properties of x(t) (e.g. stability).
(3) To apply the above to economic modeling.
Notation:
• x is the independent or unknown variable and t the dependent variable; most often
the variable t is omitted;
dx 0
d
, x (t), x0 , ẋ(t), ẋ, x(1) (t), x(1) .
• x(t) ≡
dt
dt
dk
• Higher order derivatives k x(t) ≡ x(k) (t). Special case k = 2, x00 , ẍ, x(2) .
dt
d
• Other variables are possible, e.g.
y(x), y 0 (x).
dx
Definition 1.1. A one dimensional ordinary differential equation (ODE) of order k is a
relation of the form
(1.1)
x(k) (t) = f (t, x(t), x(1) (t), . . . , x(k−1) (t)).
Note that k is the highest derivative appearing in the equation.
Definition 1.2. A first order system of ordinary differential equations is a relation of the
form
(1.2)
ẋ(t) = f (t, x(t)),
where x = (x1 , . . . , xn )0 , f = (f1 , . . . , fn )0 , xi : R −→ R, fi : Rn+1 −→ R, i = 1, . . . , n.
It is always possible to transform a kth order ODE into a first order system. Let us see
how. Suppose we have the kth order ODE
x(k) (t) = f (t, x(t), x(1) (t), . . . , x(k−1) (t))
34
and consider new functions defined by setting x1 = x, x2 = x0 , . . . , xk = xk−1 . Then the
ODE appears as the following first order system of differential equations:
ẋ1 = x2 ,
ẋ2 = x3 ,
..
.
ẋk = f (t, x1 , x2 , . . . , xk ).
Note that in this case f (t, x) = (x2 , x3 , . . . , xk , f (t, x)). As an example, consider the second
order ODE
ẍ = ẋ2 − 2x − cos t.
Let the new variable y = ẋ, Then the first order system equivalent to the original scalar
equation is
ẋ = y,
ẏ = y 2 − 2x − cos t.
Definition 1.3. A solution of the first order system (1.2) on an interval I ⊆ R is a differentiable function x : I −→ Rn such that x(t) ∈ D for all t ∈ I and ẋ(t) = f (t, x(t)) for all
t ∈ I.
Definition 1.4. An initial value problem or Cauchy problem for a first order system consists
of (1.2) together with a condition x(t0 ) = x0 , where (t0 , x0 ) ∈ D.
Definition 1.5. An initial value problem or Cauchy problem for a kth order ODE consists
of (1.1) together with the conditions
x(t0 ) = x0 ,
x0 (t0 ) = x1 , . . . , x(k−1) (t0 ) = xk−1 ,
where (t0 , x0 , x1 , . . . , xk ) ∈ D.
Thus,
ẋ = t sin x,
x(0) = π
is a Cauchy problem, as well as
ẍ = eẋ − tx,
x(1) = 2,
ẋ(1) = −1.
Under suitable conditions, a Cauchy admits a unique solution.
Definition 1.6.
• The ODE (1.2) is linear if for fixed t, the map x −→ f (t, x) is linear.
• The ODE (1.2) is autonomous if f is independent of t.
Throughout these lecture notes it is assumed the continuity of the functions f and f .
35
2. Elementary integration methods of first order ODEs
Let us look at some particular cases where the scalar first order ODE
(2.1)
ẋ(t) = f (t, x(t)),
t ∈ I ⊆ R,
Ian interval finite or infinite,
can be explicitly solved.
The simplest case is when f is independent of the solution itself, x. That is
ẋ(t) = f (t),
t ∈ I = [a, b] ⊆ R.
Finding x leads to the integration problem.
The Fundamental Theorem of Calculus establishes that
Z t
f (t) dt,
x(t) = C +
a
with C a constant. If we want the solution passing through (a, x0 ), then C = x0 . Note that
even in this simple case the solution found can have little practical value, and the study of
the qualitative behavior can be more illuminating. Resorting to numerical approximations
of the solution is another interesting possibility.
2.1. Separable equations.
Definition 2.1. A first order ODE is separable if f (t, x) = g(t)h(x), that is
ẋ(t) = g(t)h(x(t)).
Method of solution: Denoting H(x) and G(t) some antiderivatives of 1/h(x) and g(t) respectively, observe the following steps (notice that H 0 = 1/h and G0 = g):
ẋ
(i) Separation of variables:
= g(t),
h(x)
d
d
H(x(t)) = G(t),
(ii) Chain Rule:
dt
dt
(iii) Integration with respect to t: H(x(t)) = G(t) + C.
The expression obtained defines x implicitly. It is possible to prove that if h(x0 ) 6= 0,
then the solution defined by the implicit expression satisfying x(t0 ) = x0 is unique in a
neighborhood of x0 . The constant C can be determined if an initial condition is fixed.
Example 2.2. To find the solution of the separable ODE ẋ = tx2 , we start with
Z
Z
dx
ẋ
dx
= t ⇒ 2 = t dt ⇒
= t dt.
x2
x
x2
Integrating, we find
t2
−x−1 = + C.
2
Solving for x we get
1
x(t) = − t2
.
+C
2
36
Suppose that we want the solution passing through (0, 1); then
1 = x(0) = −
1
⇒ C = −1,
C
thus
x(t) = − t2
√
2
1
.
−1
The solution exists only in the interval [0, 2).
2.2. Exact equations. Integrating factors. Suppose we have a first order ODE of the
form in the form
P (t, x(t))
(2.2)
ẋ(t) = −
,
Q(t, x(t))
for some functions P , Q, such that Q(t, x) 6= 0 for every point (t, x) in some set D.
dx
This is equivalent to Q(t, x)ẋ = −P (t, x), and interpreting
as a quotient (this has no
dt
sense, of course, but it is useful and it works in this case) we can rewrite the ODE as
(2.3)
P (t, x) dt + Q(t, x) dx = 0.
Consider now a function V of variables (t, x), of class C 2 (the second order partial derivatives exist and are continuous). The differential of V is
∂V
∂V
dt +
dx.
∂t
∂x
Suppose that it is possible to find a function V such that
dV =
∂V
= P.
∂t
∂V
= Q.
∂x
(2.4)
(2.5)
Then, the differential of V
∂V
∂V
dt +
dx = P dt + Q dx = 0
∂t
∂x
along the solutions of the ODE. This means that V is constant. Thus we get that the
solutions of the ODE are given by the implicit equation:
dV =
V (t, x(t)) = C.
This important observation motivates the following definition.
Definition 2.3. The first order ODE (2.2) (or (2.3)) is exact in a neighborhood D of the
point (t0 , x0 ) if Q(t0 , x0 ) 6= 0 and there exists a function V of class C 2 on D satisfying (2.4)
and (2.5).
37
When is there a function V satisfying (2.4) and (2.5)? It both conditions were true, then
(2.6)
∂P
∂ 2V
=
.
∂x∂t
∂x
∂ 2V
∂Q
=
.
∂t∂x
∂t
Since the order of derivation does not matter for a C 2 function,
(2.7)
∂ 2V
∂ 2V
=
,
∂x∂t
∂t∂x
we get the necessary (and sufficient) condition
∂Q
∂P
(t, x) =
(t, x).
∂x
∂t
Theorem 2.4. Assume that P and Q are C 1 in a neighborhood D of the point (t0 , x0 ). The
necessary and sufficient condition for (2.2) (or (2.3)) to be exact in D is
(2.8)
∂Q
∂P
=
∂x
∂t
in D.
Example 2.5. The equation (2t − x2 ) dt + 2tx dx = 0 is not exact, since
∂Q
∂P
= −2x 6= 2x =
.
∂x
∂t
However, the equation
(2t − x2 ) dt − 2tx dx = 0
is exact. Let us solve. Once we determine function V , the problem is finished. To find V ,
we begin with (2.4)
∂V
= P (t, x) = 2t − x2 .
∂t
Integrating with respect to t we get
Z
(2.9)
V (t, x) = (2t − x2 ) dt = t2 − tx2 + ψ(x),
where ψ is a function of x that we must determine using the other condition (2.5), that is,
∂V
= Q(t, x) = −2tx.
∂x
Deriving in (2.9) with respect to x we get
∂V
= −2tx + ψ 0 (x)
∂x
and equating both expressions above
ψ 0 (x) = 0.
38
We choose ψ = 0. Hence, V (t, x) = t2 − tx2 and since the solution satisfies V (t, x(t)) = C,
we have
r
C
2
2
t − tx (t) = C ⇒ x(t) = ± t −
(t 6= 0).
t
If the equation
P (t, x) dt + Q(t, x) dx = 0
is not already exact, we could multiply the equation by a non null function µ(t, x) such that
the equation
µ(t, x)P (t, x) dt + µ(t, x)Q(t, x) dx = 0
be exact. Then µ is called an integrating factor. Unfortunately, to find integrating factors is
difficult, except in the two following cases:
(1) The quotient
∂P
− ∂Q
∂x
∂t
a(t) =
Q
is independent of x. Then
R
µ(t) = e a(t) dt
is an integrating factor.
(2) The quotient
∂Q
− ∂P
∂x
b(x) = ∂t
P
is independent of t. Then
R
µ(x) = e b(x) dx
is an integrating factor.
Example 2.6. The equation
(t2 + x2 ) dt − 2tx dx = 0
(2.10)
is not exact, since ∂P/∂x = 2x 6= −2x = ∂Q/∂t. To find an integrating factor we consider
the two quotients:
∂P
∂Q
−4x
− ∂x = 2
,
∂t
P
t + x2
∂P
∂x
−
Q
∂Q
∂t
=
4x
2
= − , independent of x.
−2tx
t
Hence
R
−2
µ(t) = e− 2/t = e−2 ln t = eln t = t−2
is an integrating factor. We multiply equation (2.10) by µ, transforming the ODE in an
equivalent one
t2 + x2
2x
dt + −
dx = 0,
t2
t
39
which is exact, since
∂P
2x
∂Q
= 2 =
.
∂x
t
∂t
Now we compute V using (2.4)
hence
∂V
t2 + x2
=P =
= 1 + x2 t−2 ,
2
∂t
t
Z
V (t, x) = (1 + x2 t−2 ) dt = t − x2 t−1 + ψ(x).
Deriving with respect to x we have
∂V
= −2xt−1 + ψ 0 (x).
∂x
On the other hand, by (2.5)
∂V
x
= Q = −2 .
∂x
t
We obtain that ψ = 0, and the solution is given by
t − x(t)2 t−1 = C.
2.3. Linear equations.
Definition 2.7. The first order ODE
ẋ(t) + a(t)x(t) = b(t)
is called linear. Here, a(t) and b(t) are given functions.
R
To solve the linear equation we proceed as follows. Let µ(t) = e
equation by µ(t) so that
(ẋ + a(t)x)µ(t) = b(t)µ(t).
Notice that µ̇(t) = a(t)µ(t) thus,
a(t) dt
(ẋ(t) + a(t)x(t))µ(t) = ẋ(t)µ(t) + x(t)a(t)µ(t) = ẋ(t)µ(t) + x(t)µ̇(t) =
and multiply the
d
(x(t)µ(t)).
dt
Hence integrating
Z
Z
Z
d
(x(t)µ(t)) dt = b(t)µ(t) dt ⇒ x(t)µ(t) = b(t)µ(t) dt.
dt
Solving for x(t) we find
Z
1
(2.11)
x(t) =
b(t)µ(t) dt.
µ(t)
Recall that the integral symbol means a primitive plus an arbitrary constant. Since for any
constant t0
Z
t
b(s)µ(s) ds
t0
40
is a primitive of b(t)µ(t), we can write
1
x(t) =
µ(t)
Z
t
b(s)µ(s) ds + C ,
t0
from which we can identify the constant C if we look for the solution satisfying x(t0 ) = x0 :
1
C ⇒ C = x0 µ(t0 ),
x0 =
µ(t0 )
Z t
1
b(s)µ(s) ds + x0 µ(t0 ) .
(2.12)
x(t) =
µ(t)
t0
Hence we have proved the following result.
Theorem 2.8. The unique solution of ẋ(t) + a(t)x(t) = b(t) passing through (t0 , x0 ) is given
by (2.12).
Of course, it is not needed to remember the formula. We only need to understand the
method used to find it.
Example 2.9. Solve the Cauchy problem t2 ẋ + tx = 1, t > 0, x(1) = 2.
Solution: First divide by the coefficient of ẋ to have the standard form of the ODE
x
1
ẋ + = 2 .
t
t
R
We identify here a(t) = −1/t and b(t) = 1/t2 . Since a(t) dt = ln t, we have µ(t) = t. Using
(2.11) we get
Z
Z
1
1
1
1
1
x(t) =
t dt =
dt = (ln t + C).
2
t
t
t
t
t
This is the general solution. The individual solution passing through (1, 2) gives 2 = C,
hence x(t) = 1t (ln t + 2).
Example 2.10. Solve the linear equation ẋ + ax = b with initial value x(t0 ) = x0 , where
a 6= 0 and b are constants.
R
= eat and from (2.11)
Z
b at
b
−at
at
−at
−at
be dt = e
x(t) = e
e +C =
+ Ce
.
a
a
Solution: Here µ(t) = e
(2.13)
a dt
Imposing x(t0 ) = x0 it is possible to determine the constant C as follows
b
b
−at0
x0 =
+ Ce
⇒ C = x0 −
eat0
a
a
and plugging this value of C into the (2.13)
b
b
x(t) = + x0 −
e−a(t−t0 ) .
a
a
41
For instance, the solution of the equation ẋ + 2x = 10 with x(0) = −1 is
x(t) = 5 − 6e−2t .
2.4. Phase diagrams. The phase diagram of the autonomous equation ẋ = f (x) consists in
a drawing of the graph of function f in the plane (x, ẋ). The zeroes of f correspond to steady
states, stationary points or equilibrium points of the equation, that is, constant solutions of
the autonomous ODE.
Definition 2.11 (Stationary points). A stationary point of the autonomous ODE ẋ = f (x)
is any constant x0 satisfying f (x0 ) = 0.
Stationary points are important in the study of the behavior of the dynamics. Analyzing
the graph of f , one obtains information on whether the solutions are increasing or decreasing.
If f > 0 in an interval, then x(t) increases in this interval, which can be indicated by an
arrow of motion pointing to the right. Similarly, if f < 0, then x(t) decreases in this interval
and the arrow that describes the motion of x points to the left.
For scalar ODEs, the sign of the f near a stationary point (if any) gives important information on the behavior of the solution near that point. For systems the situation is more
complicated, and will be explored in next sections.
For now, we center on the scalar case, f : D −→ R.
As remarked above, a stationary point is a solution of the ODE, hence if we know that
some uniqueness of solutions criterium holds, then no solution can cross through x0 . In the
scalar case we have the following, where we are assuming that the stationary point x0 is
isolated:
• f > 0 on (a, x0 ) and f > 0 on (x0 , b). The solution x converges to x0 from initial
conditions a < x0 < x0 and diverges of x0 from b > x0 > x0 (unstable solution);
• f > 0 on (a, x0 ) and f < 0 on (x0 , b). The solution x converges to x0 from every
initial condition a < x0 < b (stable solution);
• f < 0 on (a, x0 ) and f > 0 on (x0 , b). The solution x diverges of x0 from every initial
condition a < x0 < b (stable solution);
• f < 0 on (a, x0 ) and f < 0 on (x0 , b). The solution x diverges of x0 from initial
conditions a < x0 < x0 and converges to x0 from b > x0 > x0 (unstable solution).
We can resume the above in the following: a stationary state x0 is locally asymptotically
stable if and only if there exists δ > 0 such that for all x ∈ (x0 − δ, x0 + δ), x 6= x0 we have
(x − x0 )f (x) < 0
and it is unstable in the other case:
(x − x0 )f (x) < 0.
42
Example 2.12. The ODE ẋ = f (x) = x3 − 2x2 − 5x + 6 has three equilibrium points
f (x) = 0: x01 = −2, x02 = 1 and x30 = 3. The function in negative in (−∞, −2), positive in
(−2, 1), negative in (1, 3) and positive in (3, ∞). Hence,
x01 = −2 is unstable;
x02 = 1 is locally asymptotically stable;
x30 = 3 is unstable.
3. Applications
Example 3.1 (Walras adjustment mechanism). Economic models often analyze rates of
change of economic variables. In equilibrium analysis the rate of change of the market price
for commodity x depends on excess demand E (demand quantity minus the supply quantity,
E = D − S)
(3.1)
ṗ(t) = E(p(t)),
where p is the price. This is a first order differential equation, called the Walrasian price
adjustment mechanism. Note that E(p) > 0 implies that p rises, and E(p) < 0 that p falls.
Suppose that D(p) = b − ap and S(p) = β + αp, with a, b, α, β > 0, with b > β; then
ṗ = b − β − (a + α)p.
This is a linear ODE with constant coefficients. The solution is
b−β
p(t) = p(0)e−(a+α)t +
(1 − e−(a+α)t )
a
+
α
b−β
b−β
= p(0) −
.
e−(a+α)t +
a+α
α+a
The solution tends to the equilibrium solution p0 =
b−β
α+a
> 0.
Example 3.2 (An asset pricing model). Let p(t) denote the price of an equity that pays
dividend D(t) dt, and let r denote the yield on a risk freeR bond. Consider an interval of time
τ
[t, τ ]. The total cash flow of the asset in interval [t, τ ] is t D(s) ds, and the capital gain in p
is p(τ ) − p(t). By a non–arbitrage condition, the cash flow plus capital gains must be equal
to earnings of keeping the asset in the bank account, hence
Z
t
τ
D(s) ds + p(τ ) − p(t) = p(t)er(τ −t) − p(t).
43
Dividing by (τ − t), taking limits as τ → t, assuming D is (right) continuous and applying
L’Hospital rule, we have
Rτ
D(s) ds
0
D(τ )
lim t
= = lim
= D(t),
(L’Hospital rule).
τ →t
τ −t
0 τ →t 1
p(τ ) − p(t)
= ṗ(t),
τ →t
τ −t
lim
(Definition of derivative).
er(τ −t) − 1
0
rer(τ −t)
= = lim
= r,
τ →t
τ −t
0 τ →t
1
Then we get the linear ODE
lim
(3.2)
(L’Hospital rule).
⇒
D(t) + ṗ(t) = rp(t)
ṗ(t) − rp(t) = D(t),
which is the fundamental pricing equation.
Given dividends D(t), the price of the asset isR driven by ODE (3.2). It is a linear equation
that can be solved using (2.12) with µ(t) = e− r dt = e−rt to give
Z t
rt
−rs
p(t) = e
−D(s)e ds + p(0) .
0
Here, p(0) is the current price of the asset, and solving for it we find
Z t
−rt
D(s)e−rs ds.
p(0) = e p(t) +
0
Notice that we find that the price of the equity at time 0 equals the present value of future
dividends only if
lim e−rt p(t) = 0.
t→∞
Supposing that this holds (the non–bubble condition), then price of the asset today is
Z ∞
D(s)e−rs ds,
p(0) =
0
that is, the fundamental value of the equity equals the discounted sum of all future dividends
from t = 0 onwards.
Some examples: Which is the price of an asset that pays the constant amount of 1 dt euros
perpetually? It is
Z ∞
1
1
p(0) =
e−rs ds = lim (1 − e−rt ) = .
r t→∞
r
0
Which is the price of an asset that pays 1 dt euro up to t < 10 and then 2 dt euros forever
if the risk–free rate is r = 0.025? It is
Z 10
Z ∞
−0.025s
p(0) =
e
ds + 2
e−0.025s ds
0
10
= 40(1 − e−0.25 ) + 80 lim (e−0.25 − e−0.025t )
= 40(1 − e
−0.25
t→∞
−0.25
) + 80e
= 40(1 + e−0.25 ) = 71.152 euros.
44
Example 3.3 (Malthus’ model). The British economist Thomas Malthus (1766–1834) observed that many biological populations increase at a rate proportional to the population,
P , that is,
(3.3)
Ṗ (t) = rP (t),
where the constant of proportionality r is called the rate of growth (r > 0) or decline (r < 0).
The mathematical model with r > 0 predicts that the population will grow exponentially
for all time. Malthus was led to this formulation by inspecting the census records of the
United States, which showed a doubling of population every 50 years. Since the means of
subsistence were found to increase in arithmetic progression, he argued that the earth could
not feed the human population. This point of view had a major impact on social philosophy
in the 19th Century. It is immediate to see that the solution is
P (t) = P (0)ert ,
which is unique given the initial condition P (0). The solution shows exponential growth
r > 0.
As an example, consider the population of the United States in 1800, that was recorded as
5.3 million. Taking r = 0.03 (which is a good approximation of the true rate of growth for
years around 1800) we get P (t) = 5.3e0.03t million for the population in year 1800+t. For 1850
it predicts P (50) = 23.75 million, whereas the actual population was 23.19. However, for
1900 it gives P (100) = 106.45, but the actual population was 76.21. The model approximate
the data for years near the initial one, but the accuracy of the approximation diminishes
over time because the increase in population is not proportional to the population.
Example 3.4 (Verhulst’ model). The Belgian mathematician P.F. Verhulst (1804–1849)
observed that limitations on space, food supply or other resources will reduce the growth
rate, precluding exponential growth. He modified Eq. (3.3) replacing the constant r by a
function r(P )
Ṗ (t) = r(P (t))P (t).
Verhulst supposed that r(P ) = r − mP , where r and m are constants. Then, the ODE is
(3.4)
Ṗ (t) = rP (t) − mP 2 (t),
that is also known as the logistic equation. It can be rewritten as
P (t)
Ṗ (t) = r 1 −
P (t),
K
with K = r/m. The constant r is called the intrinsic growth rate, and K is the saturation
level or environmental carrying capacity.
The logistic equation is separable and can be integrated explicitly from the identity
Z
Z
dP
= r dt.
P
P (1 − K
)
Noticing that
1
P (1 −
P
)
K
=
A
B
+
P
P
1− K
45
gives A = 1 and B = 1/K we have
(3.5)
P = rt + C.
ln K −P
Imposing that P (0) = P0 is the initial population, the constant C is given by
P0 .
C = ln K − P0 Plugging this value of C into (3.5) and taking the exponential on both sides we get
P P0 rt
K − P = e K − P0 .
It is possible to show that if P0 < K, then P (t) < K and that if P0 > K, then P (t) > K for
all t, hence eliminating the absolute value on both sides and solving for P (t) we get
P (t) =
KP0
.
P (0) + (K − P0 )e−rt
Notice that limt→∞ P (t) = K if r > 0.
Turning back to the example above about United States population, suppose that K = 300
(this is close to the actual population of year 2009, and thus a very modest level for the
carrying capacity) and r = 0.03 (a good estimation of the intrinsic growth rate around 1800,
but far away from the actual intrinsic growth rate in year 2009). Recall that the initial data
from year 1800 was 5.3. Using this we find P (50) = 22.38 and P (100) = 79.61, whereas the
actual population in 1900 was 79.61.
In Figure 8 it is represented P/K for a population model driven by the logistic equation
with r = 0.71 (for illustrative purposes), for several initial conditions. The thicker line is the
solution with P (0) = 0.25K.
Example 3.5 (Population with a threshold). Suppose now that when the population of
a species falls below a certain level, the species cannot sustain itself, but otherwise, the
population follows logistic growth. To describe this situation, we can consider the differential
equation (we omit t)
P
P
1−
P,
(3.6)
Ṗ = −r 1 −
A
B
where 0 < A < B. The constant A is called the threshold of the population (we will see why
afterwards) and B is now the carrying capacity. There are three stationary points, P = 0,
P = A and P = B, corresponding to the equilibrium solutions P1 (t) = 0, P2 (t) = A and
P3 (t) = B, respectively. From Figure 9, it is clear that P 0 > 0 for A < P < B. The reverse
is true for y < A or y > B. Consequently, the equilibrium solution P1 (t) and P3 (t) are
asymptotically stable, and the solution P2 (t) is unstable. Notice that the population goes to
extinction when P (0) < A, so we call A the threshold of the population.
46
1.75
1.50
P/K
1.25
1.00
0.75
0.50
0.25
0
1
2
3
4
t
5
6
7
8
Figure 8. P/K versus t for the logistic equation with r = 0.71.
In Figure 10 we graph several solutions to Ṗ = −0.25P (1 − P )(1 − P/3) using different
values for the initial population P0 .When 0 < P (0) < 1, limt→∞ P (t) = 0, and if 1 < P (0) <
3 or P (0) > 3, then limt→∞ P (t) = 3.
f (P )
O
<<
A
>>
>>
B
<<
P
Figure 9. Phase space in the population model with threshold.
Example 3.6 (The Solow model). The dynamic economic model of Solow (1956) marked
the beginning of modern growth theory. It is based on the following assumptions.
(1) Labour, L, growths at a constant rate n, i.e. L̇/L = n;
(2) All saving S = sY are invested in capital formation, I = K̇ + δK, where Y denotes
income, K capital and δ, s ∈ (0, 1] (δ is capital depreciation):
sY = K̇ + δK.
47
5
4.5
4
3.5
B
y
3
2.5
2
1.5
A
1
0.5
0
0
5
10
15
20
25
x
Figure 10. Several solutions in the population model with threshold.
(3) The production function F (L, K) depends on labor L and capital K, and shows
constant returns, F (λL, λK) = λF (K, L). A typical example is the Coob–Douglas
production function F (K, L) = ALα K 1−α , α ∈ [0, 1]. Observe that taking λ = L
K
K
Y = F (K, L) = F L , L = LF
, 1 = Lf (k),
L
L
K
where k = K
and
f
(k)
=
F
,
1
.
L
L
The fundamental dynamic equation of this growth model is obtained as follows:
dK
K̇L − K L̇
K̇ K L̇
=
=
−
2
dt L
L
L
LL
sLf (k) − δK
− kn = sf (k) − (δ + n)k.
=
L
k̇ =
Thus, we have obtained that the per–capita capital moves according to the ODE
k̇ = sf (k) − (δ + n)k.
Common assumptions are that the function f is increasing and strictly concave and that
there exists a maximal productive stock of capital, km , that is, f (k) < k for k > km and
f (k) < k for k < km .
There are two steady states k̇ = 0, k = 0 and k = ke satisfying sf (ke ) = (δ + n)ke . The
steady state 0 is unstable and ke is stable. This can be seen in the figure below (λ = δ + n).
48
λk
sf (k)
k̇
0
ke
>>
>>
<<
Figure 11. Phase diagram in the model of Solow.
4. Second order linear ODEs
Definition 4.1. A second order linear ODEs is of the form
(4.1)
ẍ + a1 (t)ẋ + a0 (t)x = b(t),
where a1 , a0 and b are given functions. In the case that a1 and a0 are constant, then the
ODE is called of constant coefficients (even if b is not constant). In the case b = 0, the ODE
is called homogeneous.
Definition 4.2. The general solution of (4.1) is the set of all its solutions; a particular
solution is any element of this set.
The space of solutions of the homogeneous ODE has the structure of a vector subspace.
Proposition 4.3. If x1 and x2 are solutions of the homogeneous ODE, then for any constants
C1 , C2 , x(t) = C1 x1 (t) + C2 x2 (t) is also a solution.
Theorem 4.4. The general solution of the complete ODE (4.1) is the sum of the general
solution of the homogeneous equation, xh , and a particular solution, xp :
x(t) = xh (t) + xp (t).
Next, we give a result that shows how to find the general solution xh for the equation with
constant coefficients
(4.2)
ẍ + a1 ẋ + a0 x = 0,
a1 , a0
constant.
Definition 4.5. The characteristic equation of (4.2) is
r2 + a1 r + a0 = 0.
Theorem 4.6. Let r1 , r2 be the solutions (real or complex) of the characteristic equation.
Then, the general solution of the homogeneous equation is of one of the following forms:
49
(1) r1 and r2 are both real and distinct,
xh (t) = C1 er1 t + C2 er2 t .
(2) r1 = r2 = r is real and of multiplicity two,
xh (t) = C1 ert + C2 tert .
(3) r1 , r2 are complex conjugates, r1,2 = a ± ib,
xh (t) = eat (C1 cos bt + C2 sin bt).
Example 4.7. Find the general solution of the following homogeneous equations.
(1) ẍ − x = 0; since r2 − 1 = 0 is the characteristic equation,
xh (t) = C1 et + C2 e−t .
(2) ẍ − 4ẋ + 4x = 0; since r2 − 4r − 4 = (r − 2)2 = 0 is the characteristic equation,
xh (t) = C1 e2t + C2 te2t .
(3) ẍ + x = 0; since r2 + 1 = 0 is the characteristic equation, that has roots ±i (a = 0,
b = 1),
xh (t) = C1 cos t + C2 sin t.
Now, to obtain the general solution of the complete equation, we need to give methods
to obtain particular solutions. This is possible only in some limited cases that are described
next. Hence, consider the equation with constant coefficients
ẍ + a1 ẍ + a0 x = b(t),
where b(t) is:
-: A polynomial P (t) = bn tn + · · · b1 t + b0 of degree n = 0, 1, . . .;
-: An exponential beat ;
-: Trigonometric b1 cos at + b2 sin at;
-: Product and sums of the above (e.g. (t2 − t + 1)e−t + 2 sin t).
Then, the a particular solution of the complete equation is of the form
-: xp (t) = Bn tn + · · · B1 t + B0 of degree n = 0, 1, . . .;
-: xp (t) = Beat ;
-: xp (t) = B1 cos at + B2 sin at;
-: Product and sums of the above (e.g. (B2 t2 + B1 t + B0 )e−t + D1 sin t + D2 sin t),
respectively.
The procedure to find xp is to substitute the guessed form for xp (depending of the structure
of b(t)) on the equation, and then to match coefficients to obtain a linear systems for the
unknown constants B0 , . . . Bn .
Example 4.8. Find particular solutions of the following equations.
50
(1) ẍ − x = 2et/2 + e−t/2 . We guess
xp (t) = B1 et/2 + B2 e−t/2
and put into the equation (after obtaining ẋp and ẍp ) to have
B1 t/2 B2 −t/2
e +
e
− B1 et/2 − B2 e−t/2 = 2et/2 + e−t/2 .
4
4
Then, B1 = −8/3 and B2 = −4/3.
(2) ẍ − 4ẋ + 4x = t2 − t. We guess
xp (t) = B2 t2 + B1 t + B0 .
We find ẋp = 2B2 t + B1 and ẍp = 2B2 and substituting into the equation we obtain
2B2 − 8B2 t − 4B1 + 4B2 t2 + 4B1 t + 4B0 = t2 − t.
Hence, it must be

4B2 = 1 
−8B2 + 4B1 = −1 .
2B2 − 4B1 + 4B0 = 0 
Solving, we get B0 = 1/8, B1 = 1/4 and B2 = 1/4.
(3) ẍ + x = te−t − 2. We guess
xp (t) = (B1 t + B0 )e−t + C.
We find ẋp = B1 e−t − (B1 t + B0 )e−t and ẍp = −B1 e−t + (B1 t + B0 )e−t − B1 e−t and
substituting into the equation we get
−2B1 e−t + B0 e−t + B1 te−t + B1 te−t + B0 e−t + C = te−t − 2.
Hence,

C = −2 
2B1 = 1
−2B1 + 2B0 = 0 
implies C = −2, B1 = 1/2 and B0 = 1/2.
Definition 4.9. The Cauchy problem of the equation (4.1) consists in finding a solution
satisfying the initial conditions
x(t0 ) = x0 ,
ẋ(t0 ) = x1 .
Example 4.10. Solve the Cauchy problem
ẍ − 4ẋ + 4x = t2 − t,
x(0) = 1,
ẋ(0) = 0.
As we know from examples above, the general solution of the complete equation is
1
1
1
x(t) = C1 e2t + C2 te2t + t2 + t + .
8
4
4
We need to compute the derivative to use the condition imposed in ẋ(0).
t 1
ẋ(t) = 2e2t (C1 + C2 t) + C2 e2t + + .
2 4
51
Then
x(0) = 1 = C1 + 81
ẋ(0) = 0 = 2C1 + C2 +
1
4
.
This linear system can be solved to obtain C1 = 7/8 and C2 = −2. The solution of the
Cauchy problem is thus
7
1
1
1
x(t) =
− 2t e2t + t2 + t + .
8
8
4
4
4.1. Stability of second order ODEs with constant coefficients. Consider the equation
ẍ + a1 ẋ + a0 x = b,
where a1 , a0 and b are constant with b 6= 0. Then x0 = b/a0 is an equilibrium state, since
x(t) = b/a0 is a constant solution of the equation.
For the next result, recall that for a complex number z = α + iβ, α is the real part, and
that the real part of a real number is the number itself.
Proposition 4.11. The equilibrium point x0 = b/a0 is globally asymptotically stable iff both
roots of the characteristic equation r2 + a1 r + a0 = 0 have negative real parts.
Proof. We have two cases to consider:
(1) The roots are real. Then the general solution is
C1 er1 t + C2 er2 t +
b
a0
if r1 6= r2 or
b
a0
if r1 = r2 = r. In both cases the exponential tends to 0 as t → ∞ iff the exponent is
negative. Moreover,
t
1
1
lim tert = lim −rt = lim
=
= 0.
−rt
t→∞
t→∞ e
t→∞ −re
∞
Then, in both cases any solution tends to b/a0 iff r1 < 0 and r2 < 0.
(2) The roots are complex, α + iβ. Then the general solution is
C1 ert + C2 tert +
eαt (C1 cos βt + C2 sin bt) +
b
,
a0
that tends to b/a0 iff a < 0.
The next proposition gives an immediate criterium to check the sign of the real part of
the roots of the characteristic equation.
Proposition 4.12. The equilibrium point x0 = b/a0 is globally asymptotically stable iff
a0 > 0 and a1 > 0.
52
Proof. Suppose that r1 and r2 are the roots of the equation (real or complex). Then
r2 + a1 r + a0 = (r − r1 )(r − r2 ) = r2 − (r1 + r2 )r + r1 r2 ,
from which we find
a1 = −(r1 + r2 ),
a0 = r1 r2 .
(4.3)
We have two cases to consider:
(1) The roots are real. Then, if b/a0 is g.a.s., both roots are negative, thus from (4.3) both
a1 and a0 are positive. Reciprocally, if both a1 and a0 are positive, then a0 = r1 r2 > 0
implies r1 , r2 > 0 or r1 , r2 < 0. But we know a1 = −(r1 + r2 ) > 0, thus r1 + r2 < 0,
and since both roots have the same sign, both are negative.
(2) The roots are complex. Then, if b/a0 is g.a.s., the real part α < 0. Hence a0 =
−2α > 0 and
a1 = (α + iβ)(α − iβ) = α2 + β 2 > 0.
Reciprocally, if both a1 and a0 is positive (as we have just seen, when the there are
complex roots, a1 is necessarily positive), then α = −a0 /2 < 0.
Example 4.13. Study the asymptotic properties of the equation
ẍ + (4 − a2 )ẋ + (a + 1)x = 1,
a 6= −1.
Solution: The equilibrium 1/(a +1) is g.a.s. iff 4− a2 > 0 and a > −1. These conditions
hold iff |a| < 2 and a > −1, thereby the equilibrium is g.a.s. iff a ∈ (−1, 2).
5. Systems of first order ODEs
5.1. Linear systems. Consider the n–dimensional linear system of constant coefficients
Ẋ(t) = AX(t) + B,
where

x1 (t)
X(t) =  ...  ,
xn (t)


a11 . . . a1n
A =  ... . . . ...  ,
an1 . . . ann


b1
B(t) =  ...  .
bn

The unknowns are the functions x1 (t), . . . , xn (t). We are interested only in studying the
stability properties of the equilibrium points.
Definition 5.1. A constant vector X 0 is an equilibrium point iff AX 0 + B = 0.
53
To assure that only one equilibrium exists, we will impose in the following the condition
|A| =
6 0.
Then the equilibrium is given by
X 0 = −A−1 B.
Most often, it is better finding X 0 by solving directly the algebraic system than using the
inverse matrix.
We center on the two dimensional case n = 2, that is, in systems of the form
ẋ = a11 x + a12 y + b1 ,
(5.1)
ẏ = a21 x + a22 y + b2 .
where
a11 a12 a21 a22 6= 0.
Let λ1 , λ2 be the roots (real or complex) of the characteristic polynomial of A, that is, of
the equation
pA (λ) = |A − λI2 | = 0.
We have the following cases:
(1) λ1 6= λ2 are real (A is diagonalizable). Let v1 ∈ S(λ1 ) and v2 ∈ S(λ2 ) be eigenvectors.
Then, the general solution is
X(t) = C1 eλ1 t v1 + C2 eλ2 t v2 + X 0 .
(2) λ1 = λ2 = λ.
(a) A is diagonalizable. Let v1 , v2 ∈ S(λ) two independent eigenvectors. Then, the
general solution is
X(t) = eλt (C1 v1 + C2 v2 ) + X 0 .
(b) A is not diagonalizable. Let v ∈ S(λ) the only independent eigenvector corresponding to λ, and let w another vector satisfying
(A − λI2 )w = v.
Then, the general solution is
X(t) = eλt (C1 v + C2 w + C2 tv) + X 0 .
(3) λ1 = α + iβ, λ2 = α − iβ, with β 6= 0. Then, there are non trivial vectors v and w
such that the general solution is3
X(t) = eαt C1 (w cos βt − v sin βt) + eαt C2 (w sin βt + v cos βt) + X 0 .
3The
vectors can be obtained as solutions to the (complex) linear system (A − (α + iβ)I2 )(v + iw) = 0.
We are not interested in how to find these vectors, but it can be proved that they satisfy
1
(2A − αI2 )2 + β 2 I2 v = 0, w = (2A − αI2 )v.
β
54
Using this we can deduce the asymptotic behavior of the solutions.
(1) λ1 6= λ2 are real.
(a) λ1 , λ2 < 0. The equilibrium is globally asymptotically stable. It is called an
stable node.
(b) λ1 < 0 < λ2 . The equilibrium is unstable, but solutions for which C2 = 0 converges to X 0 . We say that X 0 is a saddle point. Initial conditions X0 = (x0 , y0 )
from which the corresponding solution converges form the stable manifold, and
it is given by the eigenspace S(λ1 ).
(c) λ1 , λ2 > 0. The equilibrium is unstable. It is called an unstable node.
(2) λ1 = λ2 = λ. The equilibrium is g.a.s. iff λ < 0. In this case it is called an improper
stable node. In the case λ > 0 the system is unstable. In the case λ = 0, the system
is in fact the trivial system ẋ = 0, ẏ = 0, which is not interesting.
(3) λ1 = α+iβ, λ2 = α−iβ, with β 6= 0. Notice that the two functions w cos βt−v sin βt
and w sin βt + v cos βt are periodic functions with period 2π/β
(a) The real part α = 0. The solution oscillates around X 0 with constant amplitude.
It is said that X 0 is a center. It is stable, but not g.a.s.
(b) The real part α < 0. The solution oscillates with a decreasing amplitude towards
X 0 , hence it is g.a.s. and it is called an spiral point.
Summarizing:
Theorem 5.2. The equilibrium point of the system (5.1) is stable if and only if the eigenvalues of A have non–negative real parts; it is g.a.s. if and only if the eigenvalues have negative
real parts.
Example 5.3. Determine the behavior of solutions near the origin for the system
ẋ = 3x − 2y,
ẏ = 2x − 2y.
Find the general solution.
Solution: The coefficient matrix
A=
3 −2
2 −2
has characteristic equation
3−λ
−2
2
−2 − λ
= λ2 − λ − 2,
and therefore the eigenvalues are −1 and 2. Therefore the origin is an unstable saddle point.
To find the solution, we need the eigenvectors associated to the eigenvalues, and they are
found by solving the homogeneous system
3−λ
−2
v1
0
=
.
2
−2 − λ
v2
0
55
For λ = −1
4v1 − 2v2 = 0,
2v1 − v2 = 0.
and an eigenvector associated to λ1 = −1 is v1 = (1, 2). When λ = 2
v1 − 2v2 = 0,
2v1 − 4v2 = 0,
which gives v2 = (2, 1). The general solution to the system is
2
1
x(t)
2t
−t
.
+ C2 e
= C1 e
1
2
y(t)
Example 5.4. Determine the behavior of solutions near the origin for the system
3 b
Ẋ(t) =
X(t).
1 1
Solution: The characteristic equation is
λ2 − 4λ + (3 − b) = 0.
The solutions are
λ1 = 2 +
√
1 + b,
λ2 = 2 −
√
1 + b.
Note that b < −1 implies that λ1 , λ2 are both complex, with real part α = 2 > 0, and that
b ≥ −1 gives λ1 > 0 for all b, hence the origin is unstable for all b. However, λ2 < 0 for
b > 3, hence for these values of b the origin is an unstable saddle point.
Example 5.5. Determine the behavior of solutions near the origin for the system
−a −1
Ẋ(t) =
X(t).
1 −a
Solution: The characteristic equation is
λ2 + 2aλ + a2 + 1 = 0,
with solutions
√
1
λ1,2 = (−2a ± 4a2 − 4a2 − 4) = −a ± i.
2
The real part is α = −a, thus the origin is a g.a.s. spiral for a > 0, a center for a = 0 and it
is an unstable spiral for a > 0.
56
5.2. Nonlinear systems. Consider the two dimensional nonlinear system
ẋ = P (x, y),
(5.2)
ẏ = Q(x, y).
We want to study the stability properties of the equilibrium points of the system. We
will use for this a technique that consists in substituting the non–linear system for another
that is linear, and that constitutes a local approximation near the equilibrium point of the
original system. Then, we will apply, if possible, Theorem 5.2.
The linearization of the non–linear system around the equilibrium point (x0 , y 0 ) is the
linear system
u̇ =
∂P 0 0
∂P 0 0
(x , y ) u +
(x , y ) v,
∂x
∂y
v̇ =
∂Q 0 0
∂Q 0 0
(x , y ) u +
(x , y ) v.
∂x
∂y
(5.3)
We call the matrix
0
0
A(x , y ) =
Px (x0 , y 0 ) Py (x0 , y 0 )
Qx (x0 , y 0 ) Qy (x0 , y 0 )
the Jacobian matrix of the system (5.2).
Example 5.6. The system
ẋ = y − x,
ẏ = −y +
5x2
4 + x2
has three equilibrium points, that is, there are three solutions of the equations
0 = y − x,
0 = −y +
5x2
,
4 + x2
(0, 0), (1, 1) and (4, 4). The linearization of the system about the differen equilibrium points
can be computed as follows. The partial derivatives are
Px (x, y) = −1,
40x
,
Qx (x, y) = 2
(x + 4)2
Py (x, y) = 1,
Qy (x, y) = −1.
Hence, the Jacobian matrices are
−1 1
−1 1
, A(1, 1) =
A(0, 0) =
,
8
0 −1
−1
5
A(4, 4) =
−1
2
5
1
−1
,
57
and the associated linear systems are


(
 u̇ = −u + v
 u̇ = −u + v
u̇ = −u + v
,
,
,
 v̇ = 8 u − v
 v̇ = 2 u − v
v̇ = u − v
5
5
respectively. Hence, the linearization depends on the equilibrium point.
Theorem 5.7. Let (x0 , y 0 ) an isolated equilibrium point for the nonlinear system (5.2) and
let A = A(x0 , y 0 ) be the Jacobian matrix for linearization (5.3), with |A| =
6 0. Then (x0 , y 0 ) is
an equilibrium point of the same type as the origin (0, 0) for the linearization in the following
cases.
(1) The eigenvalues of A are real, either equal or distinct, and have the same sign (node).
(2) The eigenvalues of A are real and have opposite signs (saddle).
(3) The eigenvalues of A are complex, but not purely imaginary (spiral)
Therefore, the exceptional case is when the linearization has a center. The structure for
the nonlinear system near the equilibrium points mirrors of the linearization in the non–
exceptional cases.
Example 5.8. The nonlinear system
ẋ = −y − x3 ,
ẏ = x,
has Jacobian matrix
A=
0 −1
1 0
at the origin (0, 0), which has imaginary eigenvalues ±i, and hence (0, 0) is a center for the
linearization. This is the exceptional case in the theorem, thus we cannot assure that the
nonlinear system had a center at (0, 0) based in the associated linear system. However, it is
possible to show that (0, 0) is an asymptotically stable spiral for the non–linear system.
Theorem 5.9. If (0, 0) is a globally asymptotically stable equilibrium point for (5.3), then
it is locally asymptotically stable for (5.2).
Example 5.10. Consider the nonlinear system
ẋ = 2x + 3y + xy,
ẏ = −x + y − 2xy 3 ,
which has an isolated critical point at (0, 0). The Jacobian matrix at the origin is
−2 3
A=
−1 1
√
and it has eigenvalues − 21 ± ( 23 )i. Thus the linearization has an asymptotically stable spiral
at (0, 0), and thus the nonlinear system has an asymptotically stable spiral at (0, 0).
58
Example 5.11. The system
(
ẋ = x(ρ1 − κ1 x)
ẏ = y(ρ2 − κ2 y)
models two populations governed by the logistic equation that does not interact to each
other. Here ρi is the growth rate and ρi /κi is the saturation level. When both species
are present, they compete for a limited amount of available food. To capture the effect of
competition, we modify the growth rate factor by −α1 y and −α2 x respectively, where αi is
a measure of the effect of one of the species on the other. The system modifies to
(
ẋ = x(ρ1 − κ1 x − α1 y)
.
ẏ = y(ρ2 − κ2 y − α2 x)
Suppose that ρ1 = 1, ρ2 = 0.75, κ1 = κ2 = 1, α1 = 1 and α2 = 0.5. There are four
equilibrium points: (0, 0) (extinction of both species), (0, 0.75) (extinction of population x),
(1, 0) (extinction of population y), and (0.5, 0.5) (long–term survival of both species).
Let us study the stability properties of the equilibrium points. The Jacobian matrix of
the system is
−2x − y
−x
.
A = A(x, y) =
−0.5y 0.75 − 2y
We analyze the qualitative behavior of the system around the equilibrium points. Recall
that σ(A) denotes the set of eigenvalues of A.
• (0, 0). σ(A) = {1, 0.75}, thus the origin is an unstable node of both the linear and
the nonlinear system.
• (1, 0). σ(A) = {−1, 0.25} and it is a saddle point. The stable manifold is the line
trough (1, 0) in the direction v1 = (1, 0) and the unstable manifold is generated by
v2 = (4, −5). Every solution with the initial condition not in the stable manifold
depart from (1, 0).
• (0, 0.75). σ(A) = {0.25, −0.75}, hence again it is a saddle point. The stable manifold
is generated by v1 = (0, 1) and
√ the unstable manifold by v2 = (8, −3).
• (0.5, 0.5). σ(A) = {−0.5 ± 2/4}, the eigenvalues are negative, thus this is a stable
node. All trajectories near (0.5, 0.5) converge asymptotically to the equilibrium point
of the linear and the nonlinear system.
We conclude that for every initial condition with positive population for both species the
dynamical process converges to the coexistence equilibrium.