Beatrice Venturi1 Economic Faculty STABILITY AND DINAMICAL SYSTEMS prof. Beatrice Venturi.

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Beatrice Venturi 1

Economic Faculty

STABILITY AND DINAMICAL SYSTEMS

prof. Beatrice Venturi

mathematics for economics Beatrice Venturi

1.STABILITY AND DINAMICAL

SYSTEMS We consider a differential equation:

)((*) xfx

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

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

Market Price

)]()([ padt

)()( apadt

( )d s

dpa Q Q

Dynamics Market Price

The equilibrium Point

costante)( tp

)( pfdt

dp 0)( pf

0)]()([ pa

Dynamics Market Price

,))0(()(

akdove

pepptp kt

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

(unstable)

The Time Path of the Market Price

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:

Mathematics for Economics Beatrice Venturi

A Market Model with Time Expectation:

Let the demand and supply functions be:

40)(222

5)(3 tPQs

A Market Model with Time Expectation

45)(522

In equilibrium we have

tCetP )(

tt eCdt

PdandeC

We adopt the trial solution:

In the first we find the solution of the homogenous equation

tt eCdt

PdandeC

We get:

0)52( 2 teC

The characteristic equation

We have two different rootsiandi 2121 21

the general solution of its reduced homogeneous equation is

tectectP tt 2sin2cos)( 21

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

12)0( P

1)0(' PThe solution became

92sin22cos3)( tetetP tt

The equilibrium points of the system

))(),((

))(),(()1(

xyxyfdx

Are the solutions :

0))(),((

0))(),(()2(

)()((*)

tdytcxdt

tbytaxdt

The linear case

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. (*)

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

Knot and Focus The stable case

Knot and Focus The unstable case’

Some ExamplesCase a)λ1= 1 e λ2 = 3

)()(2)1(

txtxdt

Case b) λ1= -3 e λ2 = -1

)()(2)2(

txtxdt

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

)()(2)3(

txtxdt

System of LINEAR Ordinary Differential Equations

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

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:

Examples

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

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:

This equation is called the eigenvalues equation.

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):

Example

01det)det(

0)2)(1(

We consider

)()()1( 212

xfxyadx

We get the system:

)()()()(

xfxyxaxyadx

12 xaxaA

The Characteristic Equation

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

0)()1( 212

0)(23)3(2

)(3)(2

txtxdt

it’s equivalent to :

EXAMPLE

Eigenvalues

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

The solutions

are the eigenvalues of the matrix A.

)()(2)()3(

txtxtxdt

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

0)()(2)()4(

txtxtx

),()(3

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

212212

212111

xxgtxtxtxdt

xxftxtxtxdt

)0,0(),( 21 xx

1(),( 21 xx

Jacobian Matrix

121 ),(

221),(

Jacobian Matrix

01)0,0(J

01)det( AI

Jacobian Matrix

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

Stability and Dynamical Systems

Given the non linear system:

)()()4(

txtxdt

01)()(

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

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

LOTKA-VOLTERRAPrey – Predator Model

The Lotka-Volterra Equations,

We shall consider an ecologic

system

PREy PREDATOR

)()()(

)()((*)

tdxtxtcxdt

txbxtaxdt

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

Eigenvalues

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

The solutions

are the eigenvalues of the matrix A.

TrJ = a11+ a22

a11 a12

a21 a22 J =

THE TRACE

THE DETERMINANT

Det J = a11 a22 – a12 a21

The equilibrium solutions x = 0 y = 0 Unstable

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

22112 /xxx

)( dxx

cxxxx ||ln||ln 1122

||ln||ln),( 112221 xxxxxxF

1 2 3 4 5 6 7

Cycles

Beatrice Venturi1 Economic Faculty STABILITY AND DINAMICAL SYSTEMS prof. Beatrice Venturi.

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