Advanced Mathematics for Economics, course 2013-2014Juan Pablo Rincon–Zapatero

Contents

1. Review of Matrices and Determinants 21.1. Square matrices 21.2. Determinants 32. Diagonalization of matrices 51. Introduction 122. Systems of first order difference equations 133. First order linear difference equations 164. Second order linear difference equations 194.1. The nonhomogeneous case 205. Linear systems of difference equations 235.1. Homogeneous systems 235.2. Nonhomogeneous systems 245.3. Stability of linear systems 256. The nonlinear first order equation 296.1. Phase diagrams 311. Introduction. Definitions and classifications of ODEs 332. Elementary integration methods of first order ODEs 352.1. Separable equations 352.2. Exact equations. Integrating factors 362.3. Linear equations 392.4. Phase diagrams 413. Applications 424. Second order linear ODEs 484.1. Stability of second order ODEs with constant coefficients 515. Systems of first order ODEs 525.1. Linear systems 525.2. Nonlinear systems 56

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Topic 1: Matrix diagonalization

1. Review of Matrices and Determinants

Definition 1.1. A matrix is a rectangular array of real numbers

A =

a11 a12 · · · a1ma21 a22 · · · a2m...

.... . .

...an1 an2 · · · anm

.

The matrix is said to be of order n×m if it has n rows and m columns. The set of matricesof 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,...,nj=1,...,m or even A = (aij).

The main or principal, diagonal of a matrix is the diagonal from the upper left– to thelower right–hand corner.

Definition 1.2. The transpose of a matrix A, denoted AT , is the matrix formed by inter-changing the rows and columns of A

AT =

a11 a21 · · · an1a12 a22 · · · an2...

.... . .

...a1m a2m · · · anm

∈Mm×n.

We can define two operations with matrices, sum and multiplication. The main propertiesof these operations as well as transposition are the following. It is assumed that thematrices in each of the following laws are such that the indicated operation can be performedand that α, β ∈ R.

(1) (AT )T = A.(2) (A+B)T = AT +BT .(3) A+B = B + A (commutative law).(4) A+ (B + C) = (A+B) + C (associative law).(5) α(A+B) = αA+ αB.(6) (α + β)A = αA+ βA.(7) Matrix multiplication is not always commutative, i.e., AB 6= BA.(8) A(BC) = (AB)C (associative law).(9) A(B + C) = AB + AC (distributive with respect to addition).

1.1. Square matrices. We are mainly interested in square matrices. A matrix is squareif n = m. The trace of a square matrix A is the sum of its diagonal elements, trace (A) =∑n

i=1 aii.

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Definition 1.3. The identity matrix of order n is

In =

1 0 . . . 00 1 . . . 0...

.... . .

...0 0 . . . 1

.

The square matrix of order n with all its entries null is the null matrix, and will be denotedOn. It holds that InA = AIn = A and OnA = AOn = On.

Definition 1.4. A square matrix A is called regular or invertible if there exists a matrix Bsuch 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 matrixA. Then BA = In and

B = BIn = B(AA−1) = (BA)A−1 = InA−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−1A−1.

1.2. Determinants. To a square matrix A we associate a real number called the determi-nant, |A| or det (A), in the following way.

For a matrix of order 1, A = (a), det (A) = a.

For a matrix of order 2, A =

(a bc d

), det (A) = ad− bc.

For a matrix of order 3

det (A) =

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣ = a11

∣∣∣∣ a22 a23a32 a33

∣∣∣∣− a21 ∣∣∣∣ a12 a13a32 a33

∣∣∣∣+ a31

∣∣∣∣ a12 a13a22 a23

∣∣∣∣ .This is known as the expansion of the determinant by the first column, but it can be donefor any other row or column, giving the same result. Notice the sign (−1)i+j in front of theelement 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 14 3 53 1 3

∣∣∣∣∣∣ = (−1)1+22

∣∣∣∣ 4 53 3

∣∣∣∣+ (−1)2+23

∣∣∣∣ 1 13 3

∣∣∣∣+ (−1)2+31

∣∣∣∣ 1 14 5

∣∣∣∣= −2 · (−3) + 3 · (0)− (1) · 1 = 5

4

For general n the method is the same that for matrices of order 3, expanding the determi-nant by a row or a column and reducing in this way the order of the determinants that mustbe 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 isthe determinant of order n− 1 which results from the deletion of the row i and the columnj 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| = ai1Ai1 + ai2Ai2 + · · ·+ ainAin (by row i)

or, equivalently

|A| = a1jA1j + a2jA2j + · · ·+ anjAnj (by column j).

Example 1.8. Find the value of the determinant∣∣∣∣∣∣∣∣1 2 0 34 7 2 11 3 3 10 2 0 7

∣∣∣∣∣∣∣∣ .Answer: Expanding the determinant by the third column, one gets∣∣∣∣∣∣∣∣

1 2 0 34 7 2 11 3 3 10 2 0 7

∣∣∣∣∣∣∣∣ = (−1)3+22

∣∣∣∣∣∣1 2 31 3 10 2 7

∣∣∣∣∣∣+ (−1)3+33

∣∣∣∣∣∣1 2 34 7 10 2 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|.(4) A matrix A is regular if and only if |A| 6= 0; in this case |A−1| = 1

|A| .

(5) If in a determinant two rows (or columns) are interchanged, the value of the deter-minant is changed in sign.

(6) If two rows (columns) in a determinant are identical, the value of the determinant iszero.

(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) maybe added to the elements of another row (column) without changing the value of thedeterminant.

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.

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2. Diagonalization of matrices

Definition 2.1. Two matrices A and B of order n are similar if there exists a matrix Psuch that

B = P−1AP.

Definition 2.2. A matrix A is diagonalizable if it is similar to a diagonal matrix D, thatis, there exists D diagonal and P invertible such that D = P−1AP .

Of course, D diagonal means that every element out of the diagonal is null

D =

λ1 0 . . . 00 λ2 . . . 0...

.... . .

...0 0 . . . λn

, λ1, . . . , λn ∈ R.

Proposition 2.3. If A is diagonalizable, then for all m ≥ 1

(2.1) Am = PDmP−1,

where

Dm =

λm1 0 . . . 00 λm2 . . . 0...

.... . .

...0 0 . . . λmn

.

Proof. Since A is diagonalizable

Am = (PDP−1)(PDP−1)m· · · (PDP−1)

= PD(P−1P )D · · ·D(P−1P )DP−1

= PDInD · · ·DInDP−1 = PDmP−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 probabilityof doing well is 1/9. Let gt (bt) the probability of good (poor) teaching at day t. Supposethat at time t = 1 the class has been right, that is, g1 = 1, b1 = 0. Which is the probabilitythat the 5th class go fine (bad)?

Answer: The data lead to the following equations that relate the probability of agood/bad class with the performance showed by the teacher the day before

gt+1 =1

2gt +

1

9bt,

bt+1 =1

2gt +

8

9bt.

6

In matrix form (gt+1

bt+1

)=

(12

19

12

89

)(gtbt

).

Obviously (g5b5

)=

(12

19

12

89

)4(g1b1

).

If the matrix were diagonalizable and we could find matrices P and D, then the computationof the 10th power of the matrix would be easy using Proposition 2.3. We will come back tothis example afterwards.

Definition 2.5. Let A be a matrix of order n. We say that λ ∈ R is an eigenvalue of A andthat 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 alleigenvectors of A associated to the same eigenvalue λ, including the null vector, is denotedS(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 eigen-value.

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 linearlyindependent columns or rows (both numbers necessarily coincide). It is also given by theorder 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 homogeneoussystem

(A− λIn)u = 0,

and hence it is a vector subspace, which dimension is

dim S(λ) = n− rank(A− λIn).

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Proof. Suppose that λ ∈ R is an eigenvalue of A. Then the system (A − λIn)u = 0 admitssome non–trivial solution u. Since the system is homogeneous, this implies that the deter-minant of the system is zero, |A − λIn| = 0. The second part about S(λ) follows also fromthe definition of eigenvector, and the fact that the set of solutions of a linear homogenoussystem is a subspace (the sum of two solutions is again a solution, as well as it is the productof a real number by a solution). Finally, the dimension of the space of solutions is given bythe 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 complexroots (not necessarily different, some of the roots may have multiplicity grater than one). Itcould 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

A =

0 −1 01 0 00 0 1

.

Answer:

A− λI =

−λ −1 01 −λ 00 0 1− λ

; p(λ) = (1− λ)

∣∣∣∣ −λ −11 −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

(A− I3)u =

−1 −1 01 −1 00 0 0

xyz

=

000

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 wouldbe eigenvalues of A in this extended sense.

Example 2.10. Find the eigenvalues and the proper subspaces of

B =

2 1 00 1 −10 2 4

.

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Answer: The eigenvalues are obtained solving∣∣∣∣∣∣2− λ 1 0

0 1− λ −10 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

(B − 3I3)u =

−1 1 00 −2 −10 2 1

xyz

=

000

,

which are x = y and z = −2y, and hence S(3) =< (1, 1,−2) >. To find S(2) we solve thesystem

(B − 2I3)u =

0 1 00 −1 −10 2 2

xyz

=

000

,

from which x = y = 0 and hence S(2) =< (1, 0, 0) >.

Example 2.11. Find the eigenvalues and the proper subspaces of

C =

1 2 00 2 01 1 3

Answer: To compute the eigenvalues we solve the characteristic equation

0 = |C − λI3| =

∣∣∣∣∣∣1− λ 2 0

0 2− λ 01 1 0− λ

∣∣∣∣∣∣=∣∣2− λ∣∣ ∣∣∣∣1− λ 0

1 3− λ

∣∣∣∣ = (2− λ)(1− λ)(3− λ)

So, the eigenvalues are λ1 = 1, λ2 = 2 and λ3 = 3. We now compute the eigenvectors. Theeigenspace S(1) is the solution of the homogeneous linear system whose associated matrix isC−λI3 with λ = 1. That is, S(1) is the solution of the following homogeneous linear system0 2 0

0 2 01 1 2

xyz

=

000

Solving the above system we find that

S(1) = {(−2z, 0, z) : z ∈ R} =< (−2, 0, 1) >

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On the other hand, S(2) is the set of solutions of the homogeneous linear system whoseassociated matrix is C − λI3 with λ = 2. That is, S(2) is the solution of the following−1 2 0

0 0 01 1 1

xyz

=

000

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 matrixis A− λI3 with λ = 3. That is, S(3) is the solution of the following−2 2 0

0 −1 01 1 0

xyz

=

000

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, . . . , λkbe distinct real roots of the characteristic polynomial pA(λ) an let mk be the multiplicityof each λk (Hence mk = 1 if λk is a simple root, mk = 2 if it is double, etc.). Note thatm1 +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 bediagonalizable.

Theorem 2.13. A matrix A is diagonalizable if and only if the two following conditionshold.

(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 eigen-values of A in its main diagonal, with each λj repeated nj times. Moreover, a matrix Psuch that D = P−1AP has as columns independent eigenvectors selected from each propersubspace S(λj), j = 1, . . . , k.

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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

D =

1 0 00 2 00 0 3

, P =

−2 2 00 1 01 −3 1

.

Example 2.16. Returning to Example 2.4, we compute∣∣∣∣ 12− λ 1

912

89− λ

∣∣∣∣ = 0,

or 18λ2 − 25λ+ 7 = 0. We get λ1 = 1 and λ2 = 718

. Now, S(1) is the solution set of(−1

219

12−1

9

)(xy

)=

(00

).

We find y = 92x, so that S(1) =< (2, 9) >. In the same way, S( 7

18) is the solution set of(

19

19

12

12

)(xy

)=

(00

).

We find y = −x, so that S( 718

) =< (1,−1) >. Hence the diagonal matrix is

D =

(1 0

0 718

)and

P =

(2 19 −1

), P−1 =

1

11

(1 19 −2

).

Thus,

An =1

11

(2 19 −1

)(1 0

0 ( 718

)n

)(1 19 −2

).

In particular, for n = 4 we obtain

A4 =

(0.1891 0.1802

0.8111 0.8198

).

Hence (g4b4

)= A4

(g1b1

)=

(0.1891 0.1802

0.8111 0.8198

)(10

)=

(0.18910.8111

).

This means that probability that the 5th class goes right, conditioned to the event that thefirst class was also right is of 0.1891.

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We can wonder what happens in the long run, that is, supposing that the course lastsforever (oh no!). In this case

limn→∞

An = P ( limn→∞

Dn)P−1 = P

(1 0

0 0

)P−1 =

(211

211

911

911

),

to find that the stationary distribution of probabilities is(g∞b∞

)=

(0.18180.8182

).

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Topic 2: Difference Equations

1. Introduction

In this chapter we shall consider systems of equations where each variable has a timeindex t = 0, 1, 2, . . . and variables of different time–periods are connected in a non–trivialway. Such systems are called systems of difference equations and are useful to describedynamical systems with discrete time. The study of dynamics in economics is importantbecause it allows to drop out the (static) assumption that the process of economic adjustmentinevitable leads to an equilibrium. In a dynamic context, this stability property has to bechecked, rather than assumed away.

Let time be a discrete denoted t = 0, 1, . . .. A function X : N −→ Rn that depends onthis 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, wegeneralize 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 theform

(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 initialvalue 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. Obviouslyxt = qtx0. 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;

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Example 1.5. [Fibonacci numbers (1202)] “How many pairs of rabbits will be producedin a year, beginning with a single pair, if every month each pair bears a new pair whichbecomes productive from the second month on?”. With xt denoting the pairs of rabbits inmonth 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 numberof variables. This is the reason we study mainly first order systems. Instead of giving ageneral 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

x2,t+1

)=

(g(x1,t, x2,t)

x1,t

).

If we denote Xt =

(x1,tx2,t

), f(Xt) =

(g(Xt)x1,t

), then the system can be written Xt+1 =

f(Xt).For example, yt+2 = 4yt+1 + y2t + 1 can be reduced to the first order system(

x1,t+1

x2,t+1

)=

(4x1,t + x22,t + 1

x1,t

),

and the Fibonacci equation of Example 1.5 is reduced to(x1,t+1

x2,t+1

)=

(x1,t + x2,t

x1,t

),

For a function f : Rn −→ Rn, we shall use the following notation: f t denotes the t–foldcomposition of f , i.e. f 1 = f , f 2 = f ◦ f and, in general, f t = f ◦ f t−1 for t = 1, 2, . . .. Wealso 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 thatthere exists some subset D such that for any X ∈ D, f(X) ∈ D. Then, given any initialcondition 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. Weare interested what is the behavior of Xt in the future, that is, in the limit

limt→∞

f t(X0).

Generally, we are more interested in this limit that in the analytical expression of Xt. Nev-ertheless, there are some cases where the solution can be found explicitly, so we can studythe above limit behavior quite well. Observe that if the limit exists, limt→∞ f

t(X0) = X0,say, and f is continuous

f(X0) = f( limt→∞

f t(X0)) = limt→∞

f t+1(X0) = X0,

hence the limit X0 is a fixed point of map f . This is the reason fixed points play a distin-guished role in dynamical systems.

Definition 2.3. A point X0 ∈ D is called a fixed point of the autonomous system f if,starting the system from X0, it stays there:

If X0 = X0, then Xt = X0, t = 1, 2, . . . .

Obviously, X0 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; ifq 6= 1, then there exists a unique fixed point: x0 = 0. Notice that the solution xt = qtx0 hasthe following limit (x0 6= 0) depending the value of q.

−1 < q < 1⇒ limt→∞

qtx0 = 0,

q = 1⇒ limt→∞

qtx0 = x0,

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 equationxt+1 = x2t − 6. Then, the fixed points are the solutions of x = x2 − 6, that is, x0 = −2 andx0 = 3.

In the following definitions, ‖X − Y ‖ stands for the Euclidean distance between X andY . For example, if X = (1, 2, 3) and Y = (3, 6, 7), then

‖X − Y ‖ =√

(3− 1)2 + (6− 2)2 + (7− 3)2 =√

36 = 6.

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Definition 2.5.

• A fixed point X0 is called stable if for any close enough initial state X0, the resultingtrajectory {Xt} exists and stays close forever to X0, that is, for any positive real ε,there exists a positive real δ(ε) such that if ‖X0 −X0‖ < δ(ε), then ‖Xt −X0‖ < εfor every t.• A stable fixed point X0 is called locally asymptotically stable (l.a.s.) if the trajectory{Xt} starting from any initial point X0 close to enough to X0, converges to the fixedpoint.• 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 X0 is stable, but not l.a.s., {Xt} need not approach X0.• A g.a.s. fixed point is necessarily unique.• If X0 is l.a.s., then small perturbations around X0 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 iscalled a P–period cycle of system f if a trajectory starting fromX0 goes throughX1, . . . , XP−1and 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

Xt+1 =

(x1,t+1

x2,t+1

)=

(−x2,tx1,t

)≡ f(Xt).

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+1 = axt + b, xt ∈ R, a, b ∈ R.

Consider first the case b = 0 (homogeneous case). Then, by Theorem 2.2 the solution isxt = atx0, t = 0, 1, . . .. Consider now the non–homogeneous case, b 6= 0. Let us find thefixed 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

x0 =b

1− a.

Define now yt = xt − x0 and replace xt = yt + x0 into (3.1) to get

yt+1 = ayt,

hence yt = aty0. Returning to the variable xt we find that the solution of the linear equationis

xt = x0 + at(x0 − x0)

=b

1− a+ at

(x0 −

b

1− a

).

Theorem 3.1. In (3.1), the fixed point x0 = b1−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→∞ x

0 +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 propor-tional to national income, and that investment is proportional to the change in income fromperiod 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. Fromthe first and third equation, It = αYt, and so It+1 = αYt+1. Inserting these into the secondequation yields αYt+1 = β(Yt+1 − Yt), or (α− β)Yt+1 = −βYt. Thus,

Yt+1 =β

β − αYt =

(1 +

α

β − α

)Yt, t = 0, 1, 2, . . . .

17

The solution is

Yt =

(1 +

α

β − α

)tY0, t = 0, 1, 2, . . . .

Thus, Y grows at the constant proportional rate g = α/(β − α) each period. Note thatg = (Yt+1 − Yt)/Yt.Example 3.3 (A Cobweb Model). Consider a market model with a single commodity whereproducer’s output decision must be made one period in advance of the actual sale—such asin agricultural production, where planting must precede by an appreciable length of timethe harvesting and sale of the output. Let us assume that the output decision in period t isbased 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, α, β > 0,

Qs,t+1 = −γ + δPt, γ, δ > 0.

By substituting the last two equations into the first the model is reduced to the differenceequation for prices

Pt+1 = − δβPt +

α + γ

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

Pt+1 = P 0 +

(− δβ

)t(P0 − P 0).

Since −δ/β is negative, the solution path is oscillating. It is this fact which gives rise to thecobweb phenomenon. There are three oscillations patterns: it is explosive if δ > β (S steeperthan D), uniform if δ = β, and damped if δ < β (S flatter than D). The three possibilitiesare 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, theinteraction of demand and supply will produce an explosive oscillation as follows: Given aninitial price P0, the quantity supplied in the next period will be Q1 = S(P0). In order toclear the market, the quantity demanded in period 1 must be also Q1, which is possible ifand only if price is set at the level of P1 given by the equation Q1 = D(P1). Now, via the Scurve, the price P1 will lead to Q2 = S(P1) as the quantity supplied in period 2, and to clearthe market, price must be set at the level of P2 according to the demand curve. Repeatingthis 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 + a1xt+1 + a0xt = bt,

where a0 and a1are constants and bt is a given function of t. The associated homogeneousequation is

xt+2 + a1xt+1 + a0xt = 0,

and the associated characteristic equation is

r2 + a1r + a0 = 0.

This quadratic equation has solutions

r1 = −1

2a1 +

1

2

√a21 − 4a0, r2 = −1

2a1 −

1

2

√a21 − 4a0.

There are three different cases depending of the sign of the discriminant a21 − 4a0 of theequation. When it is negative, the solutions are (conjugate) complex numbers. Recall thata 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. Theconjugate of z = a+ib is z = a−ib. Complex numbers can be added, z+z′ = (a+a′)+i(b+b′),and multiplied,

zz′ = (a+ ib)(a′ + ib′) = aa′ + iab′ + ia′b+ i2bb′ = (aa′ − bb′) + i(ab′ + a′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 thefollowing table of trigonometric values

θ sin θ cos θ tan θ

0 0 1 0

π6

12

√32

√33

π3

√32

12

√3

π2

1 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) xt+2 + a1xt+1 + a0xt = 0 (a0 6= 0)

is as follows:

20

(1) If a21 − 4a0 > 0 (the characteristic equation has two distinct real roots),

xt = Art1 +Brt2, r1,2 = −1

2a1 ±

1

2

√a21 − 4a0.

(2) If a21 − 4a0 = 0 (the characteristic equation has one real double roots),

xt = (A+Bt)rt, r = −1

2a1.

(3) If a21 − 4a0 < 0 (the characteristic equation has no real roots),

xt = ρt(A cos θt+B sin θt), ρ =√a0, tan θ = −

√4a0 − a21a1

, 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 aredamped. If ρ > 1, the oscillations are explosive, and in the case ρ = 1, we have undampedoscillations.

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 andr2 = 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. Thegeneral solution is

xt = 3t(A+Bt).

(c) The characteristic equation is r2−2r+4 = 0, with complex solutions r1 = 12(2+√−12) =

(1 + i√

3), r2 = (1 − i√

3). Here ρ = 2 and tan θ = −√12−2 =

√3. this means that θ = π/3.

The general solution is

xt = 2t(A cos

π

3t+B sin

π

3t).

4.1. The nonhomogeneous case. Now consider the nonhomogeneous equation

(4.2) xt+2 + a1xt+1 + a0xt = bt,

and let x∗t be a particular solution. It turns out that solutions of the equation have aninteresting structure, due to the linearity of the equation.

Theorem 4.4. The general solution of the nonhomogeneous equation (4.2) is the sum ofthe general solution of the homogeneous equation (4.1) and a particular solution x∗t of thenonhomogeneous 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 ofthe 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 themethod 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 twounknown parameters A and B:

A+B + 29

= 0−2A+ 2B − 1

3+ 2

9= 1

3

}.

The solution is A = −2/9 and B = 0. Thus, the solution of the nonhomogeneous equationis

xt = −2

9(−2)t − t

2+

2

9.

The method of undetermined coefficients for solving equation (4.2) suppose that a partic-ular solution has the form of the nonhomogeneous term, bt. The method works quite wellwhen 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, withtwo different real roots r1,2 = 2, 3. Its general solution is, therefore, A2t + B3t. To find aparticular 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 = 4,

2D = 1,

−6D + 2E = 0,

−D − 3E + 2F = 3.

It follows that C = 1/2, D = 1/2, E = 3/3 and F = 4. The general solution is

xt = A2t +B3t +1

24t +

1

2t2 +

3

2t+ 4.

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 fort = 0, 1, . . . ,

(i) Yt = Ct + It (income is divided between consumption and investment)

(ii) Ct+1 = aYt + b (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 oflinear difference equations with constant coefficients is given by

x1,t+1 = a11x1,t + · · ·+ a1nxn,t + b1,t...

xn,t+1 = an1x1,t + · · ·+ annxn,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 differentvariables, 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

Xt =

x1,t...xn,t

, A =

a11 . . . a1n...

. . ....

an1 . . . ann

, B =

b1,t...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 = A2X0. Thus, given the initial vector X0, the

solution is

Xt = AtX0.

In the case that A be diagonalizable, P−1AP = D with D diagonal, the expression abovesimplifies to

Xt = PDtP−1X0,

that is easy to compute since D is diagonal.

Example 5.1. Find the general solution of the system(xt+1

yt+1

)=

(4 −12 1

)(xtyt

)

24

Solution: The matrix A =

(4 −12 1

)has characteristic polynomial pA(λ) = λ2−5λ+6,

with roots λ1 = 3 and λ2 = 2. Thus, the matrix is diagonalizable. It is easy to find theeigenspaces

S(3) =< (1, 1) >, S(2) =< (1, 2) > .

Hence, the matrix P is

P =

(1 11 2

), P−1 =

(2 −1−1 1

), D =

(3 00 2

)and the solution

Xt =

(1 11 2

)(3t 00 2t

)(2 −1−1 1

)X0 =

(2 3t − 2t −3t + 2t

2 3t − 2t+1 − 1 −3t + 2t+1

)(x0y0

)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 anon–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 = A2X0 + (A+ In)B,

...

Xt = AXt−1 +B = · · · = AtX0 + (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 = AtX0 + (At − In)(A− In)−1B.

On the other hand, note that the constant solutions of the nonhomogeneous system (or fixedpoints of the system) satisfy

X0 = AX0 +B.

Assuming again that the matrix A− In has inverse, we can solve for X0

(In − A)X0 = B ⇒ X0 = (In − A)−1B.

25

Then, collecting all the above observations, we can write the solution of the nonhomoge-neous system in a nice form as

(5.1) Xt = AtX0 − (At − In)X0 = X0 + At(X0 −X0).

Theorem 5.2. Suppose that |A−In| 6= 0. Then, the general solution of the nonhomogeneoussystem is given in Eqn. (5.1). Moreover, when A is diagonalizable, the above expression maybe written as

(5.2) Xt = X0 + PDtP−1(X0 −X0),

where P−1AP = D with D diagonal.

Proof. Eqn. (5.2) easily follows from Eqn. (5.1), taking into account that At = PDtP−1. �

Example 5.3. Find the general solution of the system(xt+1

yt+1

)=

(4 −12 1

)(xtyt

)+

(1−1

)

Solution: The fixed point X∗ is given by

(I3 − A)−1B =

(−3 1−2 0

)−1(1−1

)=

1

2

(0 −12 −3

)(1−1

)=

(1/25/2

)By the example above we already know the general solution of the homogeneous system.The general solution of the nonhomogeneous system is then(

xtyt

)=

(2 3t − 2t −3t + 2t

2 3t − 2t+1 − 1 −3t + 2t+1

)(x0 − 1/2y0 − 5/2

)+

(1/25/2

).

5.3. Stability of linear systems. We study here the stability properties of a first ordersystem Xt+1 = AXt +B where |In − A| 6= 0.

For the following theorem, recall that for a complex number z = α + βi, the modulus isρ =

√α2 + β2. 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 lessthan 1. In this case, any trajectory converges to X∗ = (In − A)−1B as t→∞.

We can give an idea of the proof of the above theorem in the case where the matrix A isdiagonalizable. As we have shown above, the solution of the nonhomogeneous system in thiscase is

Xt = X0 + PDtP−1(X0 −X0),

where

D =

λt1 0 . . . 00 λt2 . . . 0...

.... . .

...0 0 . . . λtn

,

26

and λ1, . . . , λn are the real roots (possibly repeated) of pA(λ). Since |λj| < 1 for all j, thediagonal elements of Dt tends to 0 as t goes to ∞, since λtj ≤ |λj|t → 0. Hence

limt→∞

Xt = X0.

Example 5.5. Study the stability of the system

xt+1 = xt −1

2yt + 1,

yt+1 = xt − 1.

Solution: The matrix of the system is

(1 −1/21 0

), with characteristic equation λ2 −

λ+1/2 = 0. The (complex) roots are λ1,2 = 1/2±i/2. Both have modulus ρ =√

1/4 + 1/4 =

1/√

2 < 1, hence the system is g.a.s. and the limit of any trajectory is the equilibrium point,

X0 =

(1− 1 0− (−1/2)0− 1 1− 0

)−1(1−1

)=

(32

).

Example 5.6. Study the stability of the system

xt+1 = −xt + yt,

yt+1 = −xt/2− yt/2.

Solution: The matrix of the system is

(1 3

1/2 1/2

), with characteristic equation λ2−

(3/2)λ−1 = 0. The roots are λ1 = 2 and λ2 = −1/2. The system is not g.a.s. However, thereare initial conditions X0 such that the trajectory converges to the fixed point X0 = (0, 0).This can be seen once we find the solution

Xt = PDtP−1X0.

The eigenspaces are S(2) =< (3, 1) > and S(−1/2) =< (2,−1) >, thus

P =

(3 21 −1

), P−1 =

(1/5 2/51/5 −3/5

).

The solution is (xt+1

yt+1

)=

(2t 3

5(x0 + 2y0) + 21−t 1

5(x0 − 3y0)

2t 15(x0 + 2y0)− 2−t 1

5(x0 − 3y0)

).

If the initial conditions are linked by the relation x0 +2y0 = 0, then the solution convergesto (0, 0). For this reason, the line x + 2y = 0 is called the stable manifold. Notice that thestable 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 investigateunder what conditions a given adjustment process converges to the Nash equilibrium of theCournot game.

Consider a Cournot duopoly in which two firms produce the same product and face con-stant marginal costs c1 > 0 and c2 > 0. The market price Pt is a function of the totalquantity 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 theproduction level of the other firm, qj. At time t, firm i’s profit is

πi = qiP − ciqi.As it is well–known, taking ∂πi/∂qi = 0 we obtain the best response of firm i, which dependson the output of firm j as follows1

br1 = a1 − q2/2, br2 = a2 − q1/2,

where ai =α− ci

2β, i = 1, 2. We suppose that a1 > a2/2 and that a2 > a1/2 in order to have

positive quantities in equilibrium, as will be seen below.The Nash equilibrium of the game, (qN1 , q

N2 ), 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 againstitself. This means that the Nash equilibrium of the static game solves

qN1 = br1(qN2 ),

qN2 = br2(qN1 ).

In this case

qN1 = a1 − qN2 /2,qN2 = a2 − qN2 /2.

Solving, we have

qN1 =4

3

(a2 −

a12

),

qN2 =4

3

(a1 −

a22

),

which are both positive by assumption. As a specific example, suppose for a moment that

the game is symmetric, with c1 = c2 = c. Then, a1 = a2 =α− c

2βand the Nash equilibrium

is the output

qN1 =α− c

3β,

qN2 =α− c

3β.

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 thegame as follows. Suppose that each firm does not choose its Nash output instantaneously, butthey adjust gradually its output qi towards its best response bri at each time t as indicatedbelow

(5.3)

{q1,t+1 = q1,t + d1(br1,t − q1,t) = q1,t + d1(a1 − 1

2q2,t − q1,t),

q2,t+1 = q1,t + d2(br2,t − q2,t) = q2,t + d2(a2 − 12q1,t − q2,t),

where d1 and d2 are positive constants. The objective is to study whether this tattonementprocess converges to the Nash equilibrium.

To simplify notation, let us rename x = q1 and y = q2. Then, rearranging terms in thesystem (5.3) above, it can be rewritten as{

xt+1 = (1− d1)xt − d12yt + d1a1,

yt+1 = (1− d2)yt − d22xt + d2a2.

It is easy to find the equilibrium points by solving the system{x = (1− d1)x− d1

2y + d1a1,

y = (1− d2)y − d22x+ d2a2.

The only solution is precisely the Nash equilibrium,

(xN , yN) =

(4

3

(a2 −

a12

),4

3

(a1 −

a22

)).

Under what conditions this progressive adjustment of the produced output does converge tothe Nash equilibrium? According to the theory, it depends on the module of the eigenvaluesbeing smaller than 1. Let us find the eigenvalues of the system. The matrix of the system is(

1− d1 −d12

−d22

1− d2

)To simplify matters, let us suppose that the adjustment parameter is the same for bothplayers, d1 = d2 = d. The eigenvalues of the matrix are

λ1 = 1− d

2, λ2 = 1− 3d

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 andsufficient condition for convergence to the Nash equilibrium of the one shot game from anyinitial 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 differenceequation

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 issaid to be of class C1 in an open interval, if f ′ exists and it is continuous in that interval. Forexample, the functions x2, cosx or ex are C1 is the whole real line, but |x| is not differentiableat 0, so is not C1 in any open interval that contains 0.

Theorem 6.1. Let x0 ∈ I a fixed point of f , and suppose that f is C1 in an open intervalaround x0, Iδ = (x0 − δ, x0 + δ).

(1) If |f ′(x0)| < 1, then x0 is locally asymptotically stable;(2) If |f ′(x0)| > 1, then x0 is unstable.

Proof. Since f ′ is continuous in Iδ and f ′(x0) < 1, there exists some open interval Iδ =(x0 − δ, x0 + δ) and a positive number k < 1 such that |f ′(x)| ≤ k for any x ∈ Iδ.

(1) By the mean value theorem (also called Theorem of Lagrange), there exists some cbetween x0 and x0 such that

f(x0)− f(x0) = f ′(c)(x0 − x0),or

x1 − x0 = f ′(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 theequality above we get

(6.1) |x1 − x0| = |f ′(c)||x0 − x0| ≤ k|x0 − x0|.Also note that |x1 − x0| ≤ kδ < δ, thus x1 ∈ Iδ. Reasoning as above, one gets

|x2 − x0| = |f(x1)− x0| = |f ′(c)||x1 − x0| ≤ k|x1 − x0| ≤ k2|x0 − x0|.where c is a number between x1 and x0 that belongs to Iδ (and thus |f ′(c)| ≤ k).Continuing in this fashion we get after t steps

|xt − x0| ≤ kt|x0 − x0| → 0, as t→∞.So xt converges to the fixed point x0 as t→∞, and x0 is l.a.s.

(2) Now suppose that |f ′(x0)| > 1. Again by continuity of f ′, there exists δ > 0 andK > 1 such that |f ′(x)| > K for any x ∈ Iδ. By equation (6.1) one has

|x1 − x0| = |f ′(c)||x0 − x0| > K|x0 − x0|and after t steps

|xt − x0| > Kt|x0 − x0|.Since Kt 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 ′(x)| < 1 for every point x ∈ I, then the fixed point x0 is globallyasymptotically stable.

Example 6.3 (Population growth models). In the Malthus model of population growth itis postulated that a given population x grows at constant rate r,

xt+1 − xtxt

= r, or xt+1 = (1 + r)xt.

This is a linear equation and the population grows unboundedly if the per capita growth rater is positive2. This is not realistic for large t. When the population is small, there are ampleenvironmental resources to support a high birth rate, but for later times, as the populationgrows, there is a higher death rate as individuals compete for space and food. Thus, thegrowth rate should be decreasing as the population increases. The simplest case is to take alinearly decreasing per capita rate, that is

r(

1− xtK

),

where K is the carrying capacity. This modification is known as the Verhulst’ law. Thenthe population evolves as

xt+1 = xt

(1 + r − r

Kxt

),

which is not linear. The function f is quadratic, f(x) = x(1 + r − rx/K). In Fig. 6.3 isdepicted 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 ofthe equation, 0 (extinction of the population) and x0 = K (maximum carrying capacity).Considering the derivative of f at these two fixed points, we have

f ′(0) = 1 + r − 2r

Kx∣∣∣x=0

= 1 + r > 1,

f ′(K) = 1 + r − 2r

Kx∣∣∣x=K

= 1− r.

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)tx0, 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 indrawing 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 liney = x.

The following figures show possible configurations around a fixed point. The phase di-agram 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 facilitatesvisualization, but in fact it is a sequence of discrete points. In Fig. 4, f ′(x0) is positive, andthe sequence x0, x1, . . . converges monotonically to x0, whereas in Fig. 5, f ′(x0) is negativeand we observe a cobweb–like behavior, with the sequence x0, x1, . . . converging to x0 butalternating between values above and below the equilibrium. In Fig. 6, the graph of f nearx0 is too steep for convergence. After many iterations in the diagram, we observe an erraticbehavior of the sequence x0, x1, . . .. There is no cyclical patterns and two sequences gener-ated from close initial conditions depart along time at an exponential rate (see Theorem 6.1above). It is often said that the equation exhibits chaos. Finally, Fig. 7 is the phase diagramof an equation admitting a cycle of period 3.

Figure 4. x0 stable, f ′(x0) ∈ (0, 1)

32

Figure 5. x0 stable, f ′(x0) ∈ (−1, 0)

Figure 6. x0 unstable, |f ′(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 evolvealong time. Thus it is important to understand the tools of analysis and modeling of dy-namical systems. We are looking at functions x : R −→ R or vector functions x : R −→ Rn

described by equations of the form

d

dtx(t) = f(t, x(t)),

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 oftenthe variable t is omitted;

• d

dtx(t) ≡ dx

dt, x′(t), x′, x(t), x, x(1)(t), x(1).

• Higher order derivativesdk

dtkx(t) ≡ x(k)(t). Special case k = 2, x′′, x, x(2).

• Other variables are possible, e.g.d

dxy(x), y′(x).

Definition 1.1. A one dimensional ordinary differential equation (ODE) of order k is arelation 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 theform

(1.2) x(t) = f(t,x(t)),

where x = (x1, . . . , xn)′, f = (f1, . . . , fn)′, 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 seehow. 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 = x′, . . . , xk = xk−1. Then theODE appears as the following first order system of differential equations:

x1 = x2,

x2 = x3,...

xk = f(t, x1, x2, . . . , xk).

Note that in this case f(t,x) = (x2, x3, . . . , xk, f(t,x)). As an example, consider the secondorder ODE

x = x2 − 2x− cos t.

Let the new variable y = x, Then the first order system equivalent to the original scalarequation is

x = y,

y = y2 − 2x− cos t.

Definition 1.3. A solution of the first order system (1.2) on an interval I ⊆ R is a differ-entiable function x : I −→ Rn such that x(t) ∈ D for all t ∈ I and x(t) = f(t,x(t)) for allt ∈ I.

Definition 1.4. An initial value problem or Cauchy problem for a first order system consistsof (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 consistsof (1.1) together with the conditions

x(t0) = x0, x′(t0) = x1, . . . , x(k−1)(t0) = xk−1,

where (t0, x0, x1, . . . , xk) ∈ D.

Thus,

x = t sinx, x(0) = π

is a Cauchy problem, as well as

x = ex − tx, x(1) = 2, x(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) x(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

x(t) = f(t), t ∈ I = [a, b] ⊆ R.Finding x leads to the integration problem.

The Fundamental Theorem of Calculus establishes that

x(t) = C +

∫ t

a

f(t) dt,

with C a constant. If we want the solution passing through (a, x0), then C = x0. Note thateven in this simple case the solution found can have little practical value, and the study ofthe qualitative behavior can be more illuminating. Resorting to numerical approximationsof 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

x(t) = g(t)h(x(t)).

Method of solution: Denoting H(x) and G(t) some antiderivatives of 1/h(x) and g(t) respec-tively, observe the following steps (notice that H ′ = 1/h and G′ = g):

(i) Separation of variables:x

h(x)= g(t),

(ii) Chain Rule:d

dtH(x(t)) =

d

dtG(t),

(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 aneighborhood 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 x = tx2, we start with

x

x2= t⇒ dx

x2= t dt⇒

∫dx

x2=

∫t dt.

Integrating, we find

−x−1 =t2

2+ C.

Solving for x we get

x(t) = − 1t2

2+ C

.

36

Suppose that we want the solution passing through (0, 1); then

1 = x(0) = − 1

C⇒ C = −1,

thus

x(t) = − 1t2

2− 1

.

The solution exists only in the interval [0,√

2).

2.2. Exact equations. Integrating factors. Suppose we have a first order ODE of theform in the form

(2.2) x(t) = −P (t, x(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.

This is equivalent to Q(t, x)x = −P (t, x), and interpretingdx

dtas a quotient (this has no

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 C2 (the second order partial deriva-tives exist and are continuous). The differential of V is

dV =∂V

∂tdt+

∂V

∂xdx.

Suppose that it is possible to find a function V such that

∂V

∂t= P.(2.4)

∂V

∂x= Q.(2.5)

Then, the differential of V

dV =∂V

∂tdt+

∂V

∂xdx = P dt+Qdx = 0

along the solutions of the ODE. This means that V is constant. Thus we get that thesolutions of the ODE are given by the implicit equation:

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 thepoint (t0, x0) if Q(t0, x0) 6= 0 and there exists a function V of class C2 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

∂2V

∂x∂t=∂P

∂x.(2.6)

∂2V

∂t∂x=∂Q

∂t.(2.7)

Since the order of derivation does not matter for a C2 function,

∂2V

∂x∂t=

∂2V

∂t∂x,

we get the necessary (and sufficient) condition

∂P

∂x(t, x) =

∂Q

∂t(t, x).

Theorem 2.4. Assume that P and Q are C1 in a neighborhood D of the point (t0, x0). Thenecessary and sufficient condition for (2.2) (or (2.3)) to be exact in D is

(2.8)∂P

∂x=∂Q

∂t

in D.

Example 2.5. The equation (2t− x2) dt+ 2tx dx = 0 is not exact, since

∂P

∂x= −2x 6= 2x =

∂Q

∂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

∂t= P (t, x) = 2t− x2.

Integrating with respect to t we get

(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

∂x= Q(t, x) = −2tx.

Deriving in (2.9) with respect to x we get

∂V

∂x= −2tx+ ψ′(x)

and equating both expressions above

ψ′(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

t2 − tx2(t) = C ⇒ x(t) = ±√t− C

t(t 6= 0).

If the equationP (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 thatthe 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 isdifficult, except in the two following cases:

(1) The quotient

a(t) =∂P∂x− ∂Q

∂t

Qis independent of x. Then

µ(t) = e∫a(t) dt

is an integrating factor.(2) The quotient

b(x) =∂Q∂t− ∂P

∂x

Pis independent of t. Then

µ(x) = e∫b(x) dx

is an integrating factor.

Example 2.6. The equation

(2.10) (t2 + x2) dt− 2tx dx = 0

is not exact, since ∂P/∂x = 2x 6= −2x = ∂Q/∂t. To find an integrating factor we considerthe two quotients:

∂Q

∂t−

∂P∂x

P=−4x

t2 + x2,

∂P∂x− ∂Q

∂t

Q=

4x

−2tx= −2

t, independent of x.

Henceµ(t) = e−

∫2/t = e−2 ln t = eln t

−2

= t−2

is an integrating factor. We multiply equation (2.10) by µ, transforming the ODE in anequivalent one

t2 + x2

t2dt+

(−2x

t

)dx = 0,

39

which is exact, since∂P

∂x=

2x

t2=∂Q

∂t.

Now we compute V using (2.4)

∂V

∂t= P =

t2 + x2

t2= 1 + x2t−2,

hence

V (t, x) =

∫(1 + x2t−2) dt = t− x2t−1 + ψ(x).

Deriving with respect to x we have

∂V

∂x= −2xt−1 + ψ′(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)2t−1 = C.

2.3. Linear equations.

Definition 2.7. The first order ODE

x(t) + a(t)x(t) = b(t)

is called linear. Here, a(t) and b(t) are given functions.

To solve the linear equation we proceed as follows. Let µ(t) = e∫a(t) dt and multiply the

equation by µ(t) so that(x+ a(t)x)µ(t) = b(t)µ(t).

Notice that µ(t) = a(t)µ(t) thus,

(x(t) + a(t)x(t))µ(t) = x(t)µ(t) + x(t)a(t)µ(t) = x(t)µ(t) + x(t)µ(t) =d

dt(x(t)µ(t)).

Hence integrating∫d

dt(x(t)µ(t)) dt =

∫b(t)µ(t) dt ⇒ x(t)µ(t) =

∫b(t)µ(t) dt.

Solving for x(t) we find

(2.11) x(t) =1

µ(t)

∫b(t)µ(t) dt.

Recall that the integral symbol means a primitive plus an arbitrary constant. Since for anyconstant t0 ∫ t

t0

b(s)µ(s) ds

40

is a primitive of b(t)µ(t), we can write

x(t) =1

µ(t)

(∫ t

t0

b(s)µ(s) ds+ C

),

from which we can identify the constant C if we look for the solution satisfying x(t0) = x0:

x0 =1

µ(t0)C ⇒ C = x0µ(t0),

(2.12) x(t) =1

µ(t)

(∫ t

t0

b(s)µ(s) ds+ x0µ(t0)

).

Hence we have proved the following result.

Theorem 2.8. The unique solution of x(t) +a(t)x(t) = b(t) passing through (t0, x0) is givenby (2.12).

Of course, it is not needed to remember the formula. We only need to understand themethod used to find it.

Example 2.9. Solve the Cauchy problem t2x+ tx = 1, t > 0, x(1) = 2.

Solution: First divide by the coefficient of x to have the standard form of the ODE

x+x

t=

1

t2.

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

x(t) =1

t

∫1

t2t dt =

1

t

∫1

tdt =

1

t(ln t+ C).

This is the general solution. The individual solution passing through (1, 2) gives 2 = C,hence x(t) = 1

t(ln t+ 2).

Example 2.10. Solve the linear equation x + ax = b with initial value x(t0) = x0, wherea 6= 0 and b are constants.

Solution: Here µ(t) = e∫a dt = eat and from (2.11)

(2.13) x(t) = e−at∫beat dt = e−at

(b

aeat + C

)=

(b

a+ Ce−at

).

Imposing x(t0) = x0 it is possible to determine the constant C as follows

x0 =

(b

a+ Ce−at0

)⇒ C =

(x0 −

b

a

)eat0

and plugging this value of C into the (2.13)

x(t) =b

a+

(x0 −

b

a

)e−a(t−t0).

41

For instance, the solution of the equation x+ 2x = 10 with x(0) = −1 is

x(t) = 5− 6e−2t.

2.4. Phase diagrams. The phase diagram of the autonomous equation x = f(x) consists ina drawing of the graph of function f in the plane (x, x). The zeroes of f correspond to steadystates, stationary points or equilibrium points of the equation, that is, constant solutions ofthe autonomous ODE.

Definition 2.11 (Stationary points). A stationary point of the autonomous ODE x = f(x)is any constant x0 satisfying f(x0) = 0.

Stationary points are important in the study of the behavior of the dynamics. Analyzingthe 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 anarrow of motion pointing to the right. Similarly, if f < 0, then x(t) decreases in this intervaland 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 infor-mation on the behavior of the solution near that point. For systems the situation is morecomplicated, 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 thescalar case we have the following, where we are assuming that the stationary point x0 isisolated:

• f > 0 on (a, x0) and f > 0 on (x0, b). The solution x converges to x0 from initialconditions 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 everyinitial 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 initialcondition a < x0 < b (stable solution);

• f < 0 on (a, x0) and f < 0 on (x0, b). The solution x diverges of x0 from initialconditions 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 asymptoticallystable 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 x = f(x) = x3 − 2x2 − 5x + 6 has three equilibrium pointsf(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 ofchange of economic variables. In equilibrium analysis the rate of change of the market pricefor commodity x depends on excess demand E (demand quantity minus the supply quantity,E = D − S)

(3.1) p(t) = E(p(t)),

where p is the price. This is a first order differential equation, called the Walrasian priceadjustment 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

p = b− β − (a+ α)p.

This is a linear ODE with constant coefficients. The solution is

p(t) = p(0)e−(a+α)t +b− βa+ α

(1− e−(a+α)t)

=

(p(0)− b− β

a+ α

)e−(a+α)t +

b− βα + a

.

The solution tends to the equilibrium solution p0 = b−βα+a

> 0.

Example 3.2 (An asset pricing model). Let p(t) denote the price of an equity that paysdividend D(t) dt, and let r denote the yield on a risk free bond. Consider an interval of time[t, τ ]. The total cash flow of the asset in interval [t, τ ] is

∫ τtD(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 equalto earnings of keeping the asset in the bank account, hence∫ τ

t

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 applyingL’Hospital rule, we have

limτ→t

∫ τtD(s) ds

τ − t=

0

0= lim

τ→t

D(τ)

1= D(t), (L’Hospital rule).

limτ→t

p(τ)− p(t)τ − t

= p(t), (Definition of derivative).

limτ→t

er(τ−t) − 1

τ − t=

0

0= lim

τ→t

rer(τ−t)

1= r, (L’Hospital rule).

Then we get the linear ODE

(3.2) D(t) + p(t) = rp(t) ⇒ p(t)− rp(t) = D(t),

which is the fundamental pricing equation.Given dividends D(t), the price of the asset is 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

p(t) = ert(∫ t

0

−D(s)e−rs ds+ p(0)

).

Here, p(0) is the current price of the asset, and solving for it we find

p(0) = e−rtp(t) +

∫ t

0

D(s)e−rs ds.

Notice that we find that the price of the equity at time 0 equals the present value of futuredividends only if

limt→∞

e−rtp(t) = 0.

Supposing that this holds (the non–bubble condition), then price of the asset today is

p(0) =

∫ ∞0

D(s)e−rs ds,

that is, the fundamental value of the equity equals the discounted sum of all future dividendsfrom t = 0 onwards.

Some examples: Which is the price of an asset that pays the constant amount of 1 dt eurosperpetually? It is

p(0) =

∫ ∞0

e−rs ds =1

rlimt→∞

(1− e−rt) =1

r.

Which is the price of an asset that pays 1 dt euro up to t < 10 and then 2 dt euros foreverif the risk–free rate is r = 0.025? It is

p(0) =

∫ 10

0

e−0.025s ds+ 2

∫ ∞10

e−0.025s ds

= 40(1− e−0.25) + 80 limt→∞

(e−0.25 − e−0.025t)

= 40(1− e−0.25) + 80e−0.25 = 40(1 + e−0.25) = 71.152 euros.

44

Example 3.3 (Malthus’ model). The British economist Thomas Malthus (1766–1834) ob-served that many biological populations increase at a rate proportional to the population,P , that is,

(3.3) P (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 exponentiallyfor all time. Malthus was led to this formulation by inspecting the census records of theUnited States, which showed a doubling of population every 50 years. Since the means ofsubsistence were found to increase in arithmetic progression, he argued that the earth couldnot feed the human population. This point of view had a major impact on social philosophyin 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 growthr > 0.

As an example, consider the population of the United States in 1800, that was recorded as5.3 million. Taking r = 0.03 (which is a good approximation of the true rate of growth foryears around 1800) we get P (t) = 5.3e0.03t million for the population in year 1800+t. For 1850it predicts P (50) = 23.75 million, whereas the actual population was 23.19. However, for1900 it gives P (100) = 106.45, but the actual population was 76.21. The model approximatethe data for years near the initial one, but the accuracy of the approximation diminishesover 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 growthrate, precluding exponential growth. He modified Eq. (3.3) replacing the constant r by afunction r(P )

P (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) P (t) = rP (t)−mP 2(t),

that is also known as the logistic equation. It can be rewritten as

P (t) = r

(1− P (t)

K

)P (t),

with K = r/m. The constant r is called the intrinsic growth rate, and K is the saturationlevel or environmental carrying capacity.

The logistic equation is separable and can be integrated explicitly from the identity∫dP

P (1− PK

)=

∫r dt.

Noticing that1

P (1− PK

)=A

P+

B

1− PK

45

gives A = 1 and B = 1/K we have

(3.5) ln

∣∣∣∣ P

K − P

∣∣∣∣ = rt+ C.

Imposing that P (0) = P0 is the initial population, the constant C is given by

C = ln

∣∣∣∣ P0

K − P0

∣∣∣∣.Plugging this value of C into (3.5) and taking the exponential on both sides we get∣∣∣∣ P

K − P

∣∣∣∣ = ert∣∣∣∣ P0

K − P0

∣∣∣∣ .It is possible to show that if P0 < K, then P (t) < K and that if P0 > K, then P (t) > K forall 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 thecarrying 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 datafrom year 1800 was 5.3. Using this we find P (50) = 22.38 and P (100) = 79.61, whereas theactual population in 1900 was 79.61.

In Figure 8 it is represented P/K for a population model driven by the logistic equationwith r = 0.71 (for illustrative purposes), for several initial conditions. The thicker line is thesolution with P (0) = 0.25K.

Example 3.5 (Population with a threshold). Suppose now that when the population ofa species falls below a certain level, the species cannot sustain itself, but otherwise, thepopulation follows logistic growth. To describe this situation, we can consider the differentialequation (we omit t)

(3.6) P = −r(

1− P

A

)(1− P

B

)P,

where 0 < A < B. The constant A is called the threshold of the population (we will see whyafterwards) 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 andP3(t) = B, respectively. From Figure 9, it is clear that P ′ > 0 for A < P < B. The reverseis true for y < A or y > B. Consequently, the equilibrium solution P1(t) and P3(t) areasymptotically stable, and the solution P2(t) is unstable. Notice that the population goes toextinction when P (0) < A, so we call A the threshold of the population.

46

0 1 2 3 4 5 6 7 8

0.25

0.50

0.75

1.00

1.25

1.50

1.75

t

P/K

Figure 8. P/K versus t for the logistic equation with r = 0.71.

In Figure 10 we graph several solutions to P = −0.25P (1 − P )(1 − P/3) using differentvalues 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.

O>> >><< <<

f(P )

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) markedthe beginning of modern growth theory. It is based on the following assumptions.

(1) Labour, L, growths at a constant rate n, i.e. L/L = n;(2) All saving S = sY are invested in capital formation, I = K + δK, where Y denotes

income, K capital and δ, s ∈ (0, 1] (δ is capital depreciation):

sY = K + δK.

47

0 5 10 15 20 250

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x

y

A

B

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 showsconstant returns, F (λL, λK) = λF (K,L). A typical example is the Coob–Douglasproduction function F (K,L) = ALαK1−α, α ∈ [0, 1]. Observe that taking λ = L

Y = F (K,L) = F

(LK

L,L

)= LF

(K

L, 1

)= Lf(k),

where k = KL

and f(k) = F(KL, 1).

The fundamental dynamic equation of this growth model is obtained as follows:

k =d

dt

K

L=KL−KL

L2=K

L− K

L

L

L

=sLf(k)− δK

L− kn = sf(k)− (δ + n)k.

Thus, we have obtained that the per–capita capital moves according to the ODE

k = sf(k)− (δ + n)k.

Common assumptions are that the function f is increasing and strictly concave and thatthere exists a maximal productive stock of capital, km, that is, f(k) < k for k > km andf(k) < k for k < km.

There are two steady states k = 0, k = 0 and k = ke satisfying sf(ke) = (δ + n)ke. Thesteady state 0 is unstable and ke is stable. This can be seen in the figure below (λ = δ+ n).

48

0

>>

k

ke

<<>>

sf(k)

λk

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) x+ a1(t)x+ a0(t)x = b(t),

where a1, a0 and b are given functions. In the case that a1 and a0 are constant, then theODE is called of constant coefficients (even if b is not constant). In the case b = 0, the ODEis called homogeneous.

Definition 4.2. The general solution of (4.1) is the set of all its solutions; a particularsolution 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 constantsC1, C2, x(t) = C1x1(t) + C2x2(t) is also a solution.

Theorem 4.4. The general solution of the complete ODE (4.1) is the sum of the generalsolution 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 withconstant coefficients

(4.2) x+ a1x+ a0x = 0, a1, a0 constant.

Definition 4.5. The characteristic equation of (4.2) is

r2 + a1r + 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) = C1er1t + C2e

r2t.

(2) r1 = r2 = r is real and of multiplicity two,

xh(t) = C1ert + C2te

rt.

(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) x− x = 0; since r2 − 1 = 0 is the characteristic equation,

xh(t) = C1et + C2e

−t.

(2) x− 4x+ 4x = 0; since r2 − 4r − 4 = (r − 2)2 = 0 is the characteristic equation,

xh(t) = C1e2t + C2te

2t.

(3) x + 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 methodsto obtain particular solutions. This is possible only in some limited cases that are describednext. Hence, consider the equation with constant coefficients

x+ a1x+ a0x = b(t),

where b(t) is:

-: A polynomial P (t) = bntn + · · · b1t+ 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) = Bntn + · · ·B1t+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. (B2t

2 +B1t+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 theunknown constants B0, . . . Bn.

Example 4.8. Find particular solutions of the following equations.

50

(1) x− x = 2et/2 + e−t/2. We guess

xp(t) = B1et/2 +B2e

−t/2

and put into the equation (after obtaining xp and xp) to have

B1

4et/2 +

B2

4e−t/2 −B1e

t/2 −B2e−t/2 = 2et/2 + e−t/2.

Then, B1 = −8/3 and B2 = −4/3.(2) x− 4x+ 4x = t2 − t. We guess

xp(t) = B2t2 +B1t+B0.

We find xp = 2B2t+B1 and xp = 2B2 and substituting into the equation we obtain

2B2 − 8B2t− 4B1 + 4B2t2 + 4B1t+ 4B0 = t2 − t.

Hence, it must be4B2 = 1

−8B2 + 4B1 = −12B2 − 4B1 + 4B0 = 0

.

Solving, we get B0 = 1/8, B1 = 1/4 and B2 = 1/4.(3) x+ x = te−t − 2. We guess

xp(t) = (B1t+B0)e−t + C.

We find xp = B1e−t − (B1t+ B0)e

−t and xp = −B1e−t + (B1t+ B0)e

−t − B1e−t and

substituting into the equation we get

−2B1e−t +B0e

−t +B1te−t +B1te

−t +B0e−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 solutionsatisfying the initial conditions

x(t0) = x0, x(t0) = x1.

Example 4.10. Solve the Cauchy problem

x− 4x+ 4x = t2 − t, x(0) = 1, x(0) = 0.

As we know from examples above, the general solution of the complete equation is

x(t) = C1e2t + C2te

2t +1

8t2 +

1

4t+

1

4.

We need to compute the derivative to use the condition imposed in x(0).

x(t) = 2e2t(C1 + C2t) + C2e2t +

t

2+

1

4.

51

Thenx(0) = 1 = C1 + 1

8x(0) = 0 = 2C1 + C2 + 1

4

}.

This linear system can be solved to obtain C1 = 7/8 and C2 = −2. The solution of theCauchy problem is thus

x(t) =

(7

8− 2t

)e2t +

1

8t2 +

1

4t+

1

4.

4.1. Stability of second order ODEs with constant coefficients. Consider the equa-tion

x+ a1x+ a0x = b,

where a1, a0 and b are constant with b 6= 0. Then x0 = b/a0 is an equilibrium state, sincex(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, andthat 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 bothroots of the characteristic equation r2 + a1r + a0 = 0 have negative real parts.

Proof. We have two cases to consider:

(1) The roots are real. Then the general solution is

C1er1t + C2e

r2t +b

a0

if r1 6= r2 or

C1ert + C2te

rt +b

a0if r1 = r2 = r. In both cases the exponential tends to 0 as t→∞ iff the exponent isnegative. Moreover,

limt→∞

tert = limt→∞

t

e−rt= lim

t→∞

1

−re−rt=

1

∞= 0.

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

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 ofthe roots of the characteristic equation.

Proposition 4.12. The equilibrium point x0 = b/a0 is globally asymptotically stable iffa0 > 0 and a1 > 0.

52

Proof. Suppose that r1 and r2 are the roots of the equation (real or complex). Then

r2 + a1r + a0 = (r − r1)(r − r2) = r2 − (r1 + r2)r + r1r2,

from which we find

(4.3)a1 = −(r1 + r2),

a0 = r1r2.

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) botha1 and a0 are positive. Reciprocally, if both a1 and a0 are positive, then a0 = r1r2 > 0implies 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 + β2 > 0.

Reciprocally, if both a1 and a0 is positive (as we have just seen, when the there arecomplex roots, a1 is necessarily positive), then α = −a0/2 < 0.

�

Example 4.13. Study the asymptotic properties of the equation

x+ (4− a2)x+ (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 conditionshold 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

X(t) = AX(t) +B,

where

X(t) =

x1(t)...

xn(t)

, A =

a11 . . . a1n...

. . ....

an1 . . . ann

, B(t) =

b1...bn

.

The unknowns are the functions x1(t), . . . , xn(t). We are interested only in studying thestability properties of the equilibrium points.

Definition 5.1. A constant vector X0 is an equilibrium point iff AX0 +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 byX0 = −A−1B.

Most often, it is better finding X0 by solving directly the algebraic system than using theinverse matrix.

We center on the two dimensional case n = 2, that is, in systems of the form

(5.1)

{x = a11x+ a12y + b1,

y = a21x+ a22y + b2.

where ∣∣∣∣ a11 a12a21 a22

∣∣∣∣ 6= 0.

Let λ1, λ2 be the roots (real or complex) of the characteristic polynomial of A, that is, ofthe 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) = C1eλ1tv1 + C2e

λ2tv2 +X0.

(2) λ1 = λ2 = λ.(a) A is diagonalizable. Let v1,v2 ∈ S(λ) two independent eigenvectors. Then, the

general solution is

X(t) = eλt(C1v1 + C2v2) +X0.

(b) A is not diagonalizable. Let v ∈ S(λ) the only independent eigenvector corre-sponding to λ, and let w another vector satisfying

(A− λI2)w = v.

Then, the general solution is

X(t) = eλt(C1v + C2w + C2tv) +X0.

(3) λ1 = α + iβ, λ2 = α − iβ, with β 6= 0. Then, there are non trivial vectors v and wsuch that the general solution is3

X(t) = eαtC1(w cos βt− v sin βt) + eαtC2(w sin βt+ v cos βt) +X0.

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(

(2A− αI2)2 + β2I2)v = 0, w =

1

β(2A− αI2)v.

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 con-

verges to X0. We say that X0 is a saddle point. Initial conditions X0 = (x0, y0)from which the corresponding solution converges form the stable manifold, andit 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 systemis in fact the trivial system x = 0, y = 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 βtand w sin βt+ v cos βt are periodic functions with period 2π/β(a) The real part α = 0. The solution oscillates around X0 with constant amplitude.

It is said that X0 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

X0, 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 eigenval-ues of A have non–negative real parts; it is g.a.s. if and only if the eigenvalues have negativereal parts.

Example 5.3. Determine the behavior of solutions near the origin for the system

x = 3x− 2y,

y = 2x− 2y.

Find the general solution.

Solution: The coefficient matrix

A =

(3 −22 −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

2 −2− λ

)(v1v2

)=

(00

).

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(x(t)y(t)

)= C1e

−t(

12

)+ C2e

2t

(21

).

Example 5.4. Determine the behavior of solutions near the origin for the system

X(t) =

(3 b1 1

)X(t).

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 thatb ≥ −1 gives λ1 > 0 for all b, hence the origin is unstable for all b. However, λ2 < 0 forb > 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

X(t) =

(−a −11 −a

)X(t).

Solution: The characteristic equation is

λ2 + 2aλ+ a2 + 1 = 0,

with solutions

λ1,2 =1

2(−2a±

√4a2 − 4a2 − 4) = −a± i.

The real part is α = −a, thus the origin is a g.a.s. spiral for a > 0, a center for a = 0 and itis an unstable spiral for a > 0.

56

5.2. Nonlinear systems. Consider the two dimensional nonlinear system

(5.2)x = P (x, y),

y = Q(x, y).

We want to study the stability properties of the equilibrium points of the system. Wewill use for this a technique that consists in substituting the non–linear system for anotherthat is linear, and that constitutes a local approximation near the equilibrium point of theoriginal system. Then, we will apply, if possible, Theorem 5.2.

The linearization of the non–linear system around the equilibrium point (x0, y0) is thelinear system

(5.3)

u =∂P

∂x(x0, y0)u+

∂P

∂y(x0, y0) v,

v =∂Q

∂x(x0, y0)u+

∂Q

∂y(x0, y0) v.

We call the matrix

A(x0, y0) =

(Px(x

0, y0) Py(x0, y0)

Qx(x0, y0) Qy(x

0, y0)

)the Jacobian matrix of the system (5.2).

Example 5.6. The system

x = y − x,

y = −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 pointscan be computed as follows. The partial derivatives are

Px(x, y) = −1, Py(x, y) = 1,

Qx(x, y) =40x

(x2 + 4)2, Qy(x, y) = −1.

Hence, the Jacobian matrices are

A(0, 0) =

(−1 10 −1

), A(1, 1) =

(−1 185−1

), A(4, 4) =

(−1 125−1

),

57

and the associated linear systems are{u = −u+ v

v = u− v,

u = −u+ v

v =8

5u− v

,

u = −u+ v

v =2

5u− v

,

respectively. Hence, the linearization depends on the equilibrium point.

Theorem 5.7. Let (x0, y0) an isolated equilibrium point for the nonlinear system (5.2) andlet A = A(x0, y0) be the Jacobian matrix for linearization (5.3), with |A| 6= 0. Then (x0, y0) isan equilibrium point of the same type as the origin (0, 0) for the linearization in the followingcases.

(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 forthe nonlinear system near the equilibrium points mirrors of the linearization in the non–exceptional cases.

Example 5.8. The nonlinear system

x = −y − x3, y = x,

has Jacobian matrix

A =

(0 −11 0

)at the origin (0, 0), which has imaginary eigenvalues ±i, and hence (0, 0) is a center for thelinearization. This is the exceptional case in the theorem, thus we cannot assure that thenonlinear system had a center at (0, 0) based in the associated linear system. However, it ispossible 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), thenit is locally asymptotically stable for (5.2).

Example 5.10. Consider the nonlinear system

x = 2x+ 3y + xy, y = −x+ y − 2xy3,

which has an isolated critical point at (0, 0). The Jacobian matrix at the origin is

A =

(−2 3−1 1

)and it has eigenvalues −1

2± (√32

)i. Thus the linearization has an asymptotically stable spiralat (0, 0), and thus the nonlinear system has an asymptotically stable spiral at (0, 0).

58

Example 5.11. The system {x = x(ρ1 − κ1x)

y = y(ρ2 − κ2y)

models two populations governed by the logistic equation that does not interact to eachother. Here ρi is the growth rate and ρi/κi is the saturation level. When both speciesare present, they compete for a limited amount of available food. To capture the effect ofcompetition, we modify the growth rate factor by −α1y and −α2x respectively, where αi isa measure of the effect of one of the species on the other. The system modifies to{

x = x(ρ1 − κ1x− α1y)

y = y(ρ2 − κ2y − α2x).

Suppose that ρ1 = 1, ρ2 = 0.75, κ1 = κ2 = 1, α1 = 1 and α2 = 0.5. There are fourequilibrium 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 ofthe system is

A = A(x, y) =

(−2x− y −x−0.5y 0.75− 2y

).

We analyze the qualitative behavior of the system around the equilibrium points. Recallthat 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 andthe 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 byv2 = (4,−5). Every solution with the initial condition not in the stable manifolddepart 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 pointof the linear and the nonlinear system.

We conclude that for every initial condition with positive population for both species thedynamical process converges to the coexistence equilibrium.