stability and dynamical systems -...

Beatrice Venturi 1

STABILITY AND

DINAMICAL SYSTEMS

Lesson # 4

prof. Beatrice Venturi

PhD in Economics and Business

Course:

Quantitative Methods

1.STABILITY AND DINAMICAL SYSTEMS

• We consider a differential equation:

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)((*) 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


(*)

Market Price

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)]()([ padt

dp

)()( apadt

dp

( )d s

dpa Q Q

dt

Dynamics Market Price

The Equilibrium Point

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costante)(tp

)( pfdt

dp0)( pf

0)]()([ pa

)(

)(p

Dynamics Market Price

)(

,))0(()(

akdove

pepptp kt

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The general solution with k>0 (k<0) converges to

(diverges from) equilibrium asintotically stable

(unstable)

The Time Path of the Market Price

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)(xdt

df

x

)(xfx


• 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:

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atxt

)(lim

A Market Model with Time

Expectation

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:

Let the demand and supply functions be:

40)(222

2

tPdt

dP

dt

PdQd

5)(3 tPQs


Expectation

45)(522

2

tPdt

dP

dt

Pd

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In equilibrium we have

sD QQ


Expectation

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tCetP )(

tt eCdt

PdandeC

dt

dP 2

2

2

We adopt the trial solution:

In the first we find the solution of the homogenous equation

tt eCdt

PdandeC

dt

dP 2

2

2


Expectation

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We get:

0)52( 2teC

The characteristic equation

0522


Expectation

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We have two different roots

iandi 2121 21

the general solution of its reduced

homogeneous equation is

tectectP tt 2sin2cos)( 21


Expectation

95/45)(tP

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The intertemporal equilibrium is given by the

particular integral

92sin2cos)( 21 tectectP tt


Expectation

• With the following initial conditions

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12)0(P

1)0('P

The solution became

92sin22cos3)( tetetP tt

The equilibrium points of the system

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))(),((

))(),((

)1(

2122

2111

xyxyfdx

dy

xyxyfdx

dy

STABILITY AND DINAMICAL

SYSTEMS

STABILITY AND DINAMICAL SYSTEMS

• Are the solutions :

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0))(),((

0))(),(()2(

212

211

xyxyf

xyxyf

The Linear Case

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)()(

)()(

(*)

tdytcxdt

dy

tbytaxdt

dx

We remember that

x'' = ax' + bcx + bdy

• by = 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 Examples Case a)λ1= 1 e λ2 = 3

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)(2)(

)()(2

)1(

212

211

txtxdt

dx

txtxdt

dx

Case b) λ1= -3 e λ2 = -1

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)(2)(

)()(2

)2(

212

211

txtxdt

dx

txtxdt

dx

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

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)(2)(

)()(2

)3(

212

211

txtxdt

dx

txtxdt

dx

System of LINEAR Ordinary Differential Equations

• Where A is the matrix associeted to the coefficients of the linear system of ODE ‘s:

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)()(

)()(

2221

1211

xaxa

xaxaA


• Definition of Matrix

• A 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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20

01A

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• Examples

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)(2

)(

)1(

22

11

txdt

dx

txdt

dx


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t

t

ectx

ectx

2

22

11

)(

)(


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)(2

)(

)2(

22

11

txdt

dx

txdt

dx


20

01A

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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.

http://en.wikipedia.org/wiki/File:Eigenvalue_equation.svg


• 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:

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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):

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• Example

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020

01det)det( IA

0)2)(1(

• We consider

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)()()1( 212

2

xfxyadx

yda

dx

yd


• We get the system:

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)sin(

)2(

pydt

dp

ydt

dp



• The equilibrium solutions are

• P1(0,0) and P2(0, ).

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)()(

10

12 xaxaA

The Characteristic Equation

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0)()(

1det

)det(

12 xaxa

IA


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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


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Eigenvalues

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

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The solutions

are the eigenvalues of the matrix A.


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)(3

1)()(

)()(2)(

)3(

2212

2111

txtxtxdt

dx

txtxtxdt

dx


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

:

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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

Jacobian Matrix

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21

2

1

1

21 ),(

x

g

x

g

x

f

x

f

xxJ

3

1

221

),(12

12

21xx

xx

xxJ

Jacobian Matrix

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3

10

01

)0,0(J

3

10

01

)det( AI

Jacobian Matrix

??

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

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Stability and Dynamical Systems

.

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01

dt

dx02

dt

dx


• Given the non linear system:

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1)()(

)()(

)4(

2

2

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)()(

2

2

2

2

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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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)

61

LOTKA-VOLTERRA

Prey – Predator Model

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The Lotka-Volterra Equations,

63

We shall consider an ecologic system

PREy PREDATOR

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Predator-Prey cycles

1

1

dFa b S

F dtdS

c d FS dt

Rate of growth of Fish

Food supply Interactions with Sharks

Rate of growth of Sharks

Rate of death in absence of Fish to eat

Interactions with Fish

• Generates a cycle: – Lots of fish—>lots of interactions with Sharks—>rapid growth of Sharks—>Fall in Fish numbers—>less interactions with Sharks —>Fall in Shark numbers—>Lots of fish again...

dFa F b S F

dtdS

c S d F Sdt

Linear bits unstable

near equilibrium

Nonlinear bits stabilise far from

equilibrium

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),()()()(

),()()()(

(*)

SFgtStdFtcSdt

dS

SFftStbFtaFdt

dF

The Model

Steady State Solutions

a F –b F S=0

d F S– c S=0

The Jacobian Matrix

J =

∂f/∂F ∂f/∂S

∂g/∂F ∂g/∂S

Eigenvalues

p( λ) = λ2 − Tr J λ + det J = 0

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69

TrJ = a11+ a22

Det J = a11 a22 – a12 a21

a11 a12

a21 a22 J =

THE TRACE & THE DETERMINANT

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70

The equilibrium solutions

F = 0 S = 0 Unstable

F = c/d S = a/b Stable Center

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Fish Cycles

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0 200 400 600 8001000

900

1000

1100

Time

Fis

h

Sharks Cycles

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0 200 400 600 800100098

99

100

101

102

Time

Shar

ks


98 99 100 101 102

900

950

1000

1050

1100

Sharks

Fis

h

Equilibrium here, but

system will never reach it

Linear forces push away

Nonlinear forces push back in

System cycles indefinitely


0 200 400 600 8001000

900

1000

1100

Time

Fis

h

0 200 400 600 800100098

99

100

101

102

Time

Shar

ks

98 99 100 101 102

900

950

1000

1050

1100

Sharks

Fis

h

Equilibrium here, but system will never reach it

Linear forces push away

Nonlinear forces push back in

System cycles indefinitely

75

bFSaF

cSdFS

dt

dF

dt

dS/

cFcdFbSSa ||ln||ln

||ln||ln),( 21 FcdFbSSaxxHBeatrice Venturi

76

1 2 3 4 5 6 7

1

2

3

4

Cycles

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77 Beatrice Venturi

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