economic faculty

Beatrice Venturi 1

Economic Faculty

STABILITY AND DINAMICAL SYSTEMS

prof. Beatrice Venturi

mathematics for economics Beatrice Venturi

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1.STABILITY AND DINAMICAL

SYSTEMS We consider a differential equation:

)((*) xfx

dt

d

with f a function independent of time t , represents a dynamical system .(*)


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a = is an equilibrium point of our system

x(t) = a is a constant value.such that

f(a)=0 The equilibrium points of our system are the

solutions of the equation

f(x) = 0

1.STABILITY AND DINAMICAL SYSTEMS

(*)


4

Market Price

)]()([ padt

dp

)()( apadt

dp

( )d s

dpa Q Q

dt


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Dynamics Market Price

The equilibrium Point

)(

)(

p

costante)( tp

)( pfdt

dp 0)( pf

0)]()([ pa


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Dynamics Market Price

)(

,))0(()(

akdove

pepptp kt

The general solution with k>0 (k<0) converges to (diverges from) equilibrium asintotically stable

(unstable)


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The Time Path of the Market Price


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Given

)(xdt

df

x

)(xfx


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Let B be an open set and a Є B, a = is a stable equilibrium point if for any

x(t) starting in B result:

atxt

)(lim

Mathematics for Economics Beatrice Venturi

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A Market Model with Time Expectation:

Let the demand and supply functions be:

40)(222

2

tPdt

dP

dt

PdQd

5)(3 tPQs

A Market Model with Time Expectation


11

45)(522

2

tPdt

dP

dt

Pd

In equilibrium we have

sD QQ


12


tCetP )(

tt eCdt

PdandeC

dt

dP 22

2

We adopt the trial solution:

In the first we find the solution of the homogenous equation

tt eCdt

PdandeC

dt

dP 22

2


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We get:

0)52( 2 teC

The characteristic equation

0522


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We have two different rootsiandi 2121 21

the general solution of its reduced homogeneous equation is

tectectP tt 2sin2cos)( 21



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95/45)( tP

The intertemporal equilibrium is given by the particular integral

92sin2cos)( 21 tectectP tt

A Market Model with Time ExpectationWith the following initial conditions


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12)0( P

1)0(' PThe solution became

92sin22cos3)( tetetP tt


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The equilibrium points of the system

))(),((

))(),(()1(

2122

2111

xyxyfdx

dy

xyxyfdx

dy



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Are the solutions :

0))(),((

0))(),(()2(

212

211

xyxyf

xyxyf


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)()(

)()((*)

tdytcxdt

dy

tbytaxdt

dx

The linear case


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We remember that

x'' = ax' + bcx + bdyby = x' − ax

x'' = (a + d)x' + (bc − ad)x x(t) is the solution (we assume z=x)

z'' − (a + d)z' + (ad − bc)z = 0. (*)


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The Characteristic Equation

If x(t), y(t) are solution of the linear system then x(t) and y(t) are solutions

of the equations (*).

The characteristic equation of (*) is

p(λ) = λ2 − (a + d)λ + (ad − bc) = 0


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Knot and Focus The stable case


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Knot and Focus The unstable case’


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Some ExamplesCase a)λ1= 1 e λ2 = 3

)(2)(

)()(2)1(

212

211

txtxdt

dx

txtxdt

dx


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Case b) λ1= -3 e λ2 = -1

)(2)(

)()(2)2(

212

211

txtxdt

dx

txtxdt

dx


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Case c) Complex roots λ1 =2+i and λ2 = 2-i,

)(2)(

)()(2)3(

212

211

txtxdt

dx

txtxdt

dx


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System of LINEAR Ordinary Differential Equations

Where A is the matrix associeted to the coefficients of the system:

)()(

)()(

2221

1211

xaxa

xaxaA


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Definition of MatrixA matrix is a collection of numbers

arranged into a fixed number of rows and columns. Usually the numbers are real numbers. Here is an example of a matrix with two rows and two columns:


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t

t

ectx

ectx2

22

11

)(

)(



30

20

01A


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Examples

)(2

)()1(

22

11

txdt

dx

txdt

dx


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)(2

)()2(

22

11

txdt

dx

txdt

dx



33

20

01A

Eigenvectors and Eigenvalues of a Matrix

The eigenvectors of a square matrix are the non-zero vectors that after being multiplied by the matrix, remain parellel to the original vector.

mathematics for economist Beatrice Venturi

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Matrix A acts by stretching the vector x, not changing its direction, so x is an eigenvector of A. The vector x is an eigenvector of the matrix A with eigenvalue λ (lambda) if the following equation holds:

xAx


This equation is called the eigenvalues equation.


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xAx


The eigenvalues of A are precisely the solutions λ to the equation:

Here det is the determinant of matrix formed by

A - λI ( where I is the 2×2 identity matrix). This equation is called the characteristic equation

(or, less often, the secular equation) of A. For example, if A is the following matrix (a so-called diagonal matrix):


37


38


Example

020

01det)det(

IA

0)2)(1(


39

We consider

)()()1( 212

2

xfxyadx

yda

dx

yd



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We get the system:

)()()()(

)()2(

21122

21

xfxyxaxyadx

dy

xydx

dy



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)()(

10

12 xaxaA


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0)()(

1det

)det(

12

xaxa

IA

The Characteristic Equation


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The Characteristic Equation of the matrix A is the same of the equation (1)

0)()1( 212

2

xyadx

yda

dx

yd


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0)(23)3(2

2

txdt

xd

dt

xd

)(3)(2

)()4(

212

21

txtxdt

dx

txdt

dx

it’s equivalent to :

EXAMPLE


45



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Eigenvalues

p( λ) = λ2 − (a + d) λ + (ad − bc) = 0

The solutions

are the eigenvalues of the matrix A.


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)(3

1)()(

)()(2)()3(

2212

2111

txtxtxdt

dx

txtxtxdt

dx


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Solving this system we find the equilibrium point of the non-linear system (3):

:

0)(3

1)()(

0)()(2)()4(

221

211

txtxtx

txtxtx


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),()(3

1)()(

),()()(2)()3(

212212

212111

xxgtxtxtxdt

dx

xxftxtxtxdt

dx


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)0,0(),( 21 xx

)2

1,

3

1(),( 21 xx


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Jacobian Matrix

21

2

1

121 ),(

x

g

x

g

x

f

x

f

xxJ

3

1

221),(

12

12

21xx

xxxxJ


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Jacobian Matrix

3

10

01)0,0(J

3

10

01)det( AI


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Jacobian Matrix

??

??)2/1,3/1(J


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Stability and Dynamical Systems

.01

dt

dx02

dt

dx


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Given the non linear system:

1)()(

)()()4(

22

12

211

txtxdt

dx

txtxdt

dx


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01 dt

dx

)()(

0)()(

12

21

txtx

txtx


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02 dt

dx

1)()(

01)()(

22

22

1

1

txtx

txtx


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f(x)=(x^2)-1

f(x)=x

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5 3 3.5 4 4.5

-4

-3

-2

-1

1

2

3

4

x

f(x)


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20

01A

11 22

t

t

ectx

ectx2

22

11

)(

)(


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f(x)=e^x

f(x)=e^(-2x)

-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5 3 3.5 4 4.5

-4

-3

-2

-1

1

2

3

4

x

f(x)

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LOTKA-VOLTERRAPrey – Predator Model

The Lotka-Volterra Equations,

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We shall consider an ecologic

system

PREy PREDATOR


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)()()(

)()((*)

2212

2111

tdxtxtcxdt

dx

txbxtaxdt

dx

The Model

Steady State Solutions

a x1-bx1x2=0c x1x2– d x2=0

a prey growth rate; d mortality rate

The Jacobian Matrix

J= a11 a12

a21 a22


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Eigenvalues

p( λ) = λ2 − (a + d) λ + (ad − bc) = 0

The solutions

are the eigenvalues of the matrix A.

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TrJ = a11+ a22

a11 a12

a21 a22 J =

THE TRACE

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THE DETERMINANT

Det J = a11 a22 – a12 a21

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The equilibrium solutions x = 0 y = 0 Unstable

x = d/g y = a/b Stable center

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211

22112 /xxx

xxx

dt

dx

dt

dx

22

)( dxx

1

1

)( dxx

cxxxx ||ln||ln 1122

||ln||ln),( 112221 xxxxxxF

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1 2 3 4 5 6 7

1

2

3

4

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