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Modeling of economic dynamics
1. Motivated examples
1.1 Deposit growth model.
1.2 Samuelson’s model.
1.3 Neo-classical model of economic growth.
1.4 Keynesian model.
2. Difference equations
2.1 Fixed or balance point, example.
2.2 Stable and unstable balance point, example.
2.3 Linear difference systems, example.
2.4 Non-linear difference systems, example.
2.5 Difference equations with delay, example.
2.6 Difference equations with delay and time series.
3. Differential equations
3.1 Equilibrium or balance point, example.
3.2 Stable and unstable balance point, example.
3.3 Linear systems of differential equations, example.
3.4 Non-linear systems of differential equations, example.
4 Chaos in economic systems
4.1 Hyperbolic balance point, example.
4.1 Stable and unstable manifolds, example.
4.2 Homoclinic orbit as a cause of chaos, example.
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4.3 Indestructibility of chaos, example.
5 Computer modeling of economic dynamics
5.1 Symbolic image of dynamical system.
5.2 Construction of symbolic image, example.
5.3 Balance points and periodic orbits on symbolic image, example.
5.4 Matrix of transition and global dynamics, example.
5.5 Attractors and their construction, example.
5.6 Localization of recurrent orbits, example.
5.7 Stochastic Markov chains, example.
5.8 Balance method for stationary distribution.
6 Discussion and propositions.
Тhe lectures can be understood by anyone who has basic knowledge of calculus
and linear algebra.
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1. Motivated examples
1.1 Deposit growth model.
Suppose that we deposit amount X0 in a savings account at an annual the interest
rate of r. Denote Xn amount at the end of n-th year. At the end of the next year the
saving consists of
Xn+1-Xn=rXn.
From this it follows the recurrence relation
Xn+1=(1+r)Xn.
We obtain difference equation. Suppose that interest rate r does not depend on
time. Let us check that solution has the form
Xn=X0(1+r)n .
Substitution this expression in left and right parts of the difference equation gives
correct equality.
Let us consider a continuous case. Denote X(t) amount of deposit at time t. The
saving at the time consists of
( ) ( ) ( )
where a can be treated as the interest rate. We have
( ) ( )
( )
If tends to zero we obtain equality
( )
( )
that contains derivative, i.e. we get differential equation to the deposit. Suppose
that interest rate does not depend on time. Substituting the expression
4
in left and right parts of the differential equation we can verify that the given
function is solution of the equation. If the time unit is one year then between
coefficient a and interest rate r there is the following connection
( ) .
If coefficient a depends on time t, the solution has the form
∫ ( )
.
1.2 Samuelson’s model.
Let us briefly recall the business cycle model of Samuelson (1939).
National income at time t Yt
may be written as the sum of three components: consumption Ct, induced private investment It, and government expenditure Gt. Therefore,
Yt=Ct+It+Gt
The agents consume a constant fraction of their past income
Ct =aYt-1
where 0<a<1 stands for the marginal propensity to consume. Induced private investment is proportional to changes in consumption and thus also to changes in national income
It =b(Ct-Ct-1)=ab(Yt-1-Yt-2) As is well known, ab >0 denotes the relation between the capital stock and output. Since government expenditure is constant G national income may be rewritten as
Yt=G+a(b+1)Yt-1-abYt-2
The recurrence relation in the income variable is a second-order linear difference equation.
1.3 Neo-classical model of economic growth.
Neo-classical model (Swan, 1956) and (Solow, 1956) is based on the following
assumptions.
(i) Labor (L) grows at a constant rate n i.e.
/L=n
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(ii) All saving S = sY are invested in capital (K) by the equality
(s, δ are constant positive fractions)
(iii) Production takes place under constant returns conditions:
Y = F(K, L) = LF(K/L, 1) = Lf(k)
where k = K/L, F is a production function.
As an example we consider the Cobb-Douglas production function
Y=Kα L1-α , (0 < α < 1).
We have
(
)
( )
( )
( )
where δ+n=λ.
These lead to the fundamental dynamic equation
( )
where f(k) is increasing concave differentiable function f" < 0 < f'
For the case of the Cobb-Douglas production function
Y=Kα L1-α, Y/L= Kα L-α=k α=f(k)
and the fundamental growth equation is
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1.4 Keynesian model (1936).
Consider a macro economic model in which velocity of income (Y) rises in response
to excess demand (D — Y).
( )
For a simple closed economy, with investment (I) and government expenditure (G)
both given, demand is consumption (C) plus I plus G.
D=C+I+G
Consumption is an increasing function (assumed linear) of income, i.e.
C = c0 + cY
where c0>0, 0<c<1.
The dynamic model is, for a constant positive k,
( ) ( ) ( )
This is a typical first order linear differential equation of the form
2. Difference equations
2.1 Fixed or balance point, example.
A difference equation expresses the change of economic state as a function of the
current state.
The system of difference equations can be expressed in the vector form
x(t + 1) = f(x(t)) or xn+1=f(xn )
where x ϵ Rd describes the state of economic system, t (or n) is descrete time,
xn+1 is future state of the system and xn is present state of the system.
Orbit or trajectory with initial point x0 is a sequence of points
O(x0 )={…., x-1, x0, x1, x2, …}
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that satisfy the system of difference equations xn+1=f(xn ).
A simple illustration of an orbit is a fixed point
{xn =x0 }.
It describes a state of economic system that doesn’t change in time. The fixed
point is called equilibrium or balance point. Balance point x* is determined as
solution of the equation
x*=f(x*).
Example (The Cobweb Cycle).
Consider the supply S(t) and demand function D(t) for a commodity
( ) ( )
( ) ( )
where α, β, δ, γ are constants and p(t) is price in time t. While demand D(t) is a
function of current price p(t), supply D(t) is a function of price prevailing on the
market at some previous period, p(t-1). If
D(t)=S(t)
then
( ) ( )
( ) ( )
where
.
The fixed point of the equation gives price P*
such that is not change in time p(t)=p(t-1)=P*
P*=aP*+b
From this it follows
P*=b/(1-a).
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2.2 Stable and unstable balance point, example.
Consider a teeter for children or a pendulum on a rigid suspension
There are two equilibriums: lower equilibrium and upper equilibrium.
It is clear that lower equilibrium is stable and upper equilibrium is unstable.
Suppose that we found a balance state X* in economic system. What happens if
we start a dynamic process not from the balance X*, but from the point x0 closest
to it?
Definition
An equilibrium point X* is stable if every solution that starts at a point x0 close to
it, will stay close to X* at all future time.
More formally, X* is said to be stable if for any ε>0 there exists δ>0 such that if the
|x0-X* |<δ than |xn-X* |<ε for any n>0, xn+1=f(xn ).
A fixed point X* is called asymptotically stable if X* is stable and there exists δ0
such that
if |x0-X*|< δ0 then xn tends to X* if n-->∞.
Example
Consider very simple linear difference equation
xn+1= a xn
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The fixed point is X*=0. It is not hard to establish that solution of the equation has
the form
xn=x0an.
If |a|<1 then xn0 as n∞ . This means that the fixed point X*=0 is
asymptotically stable.
If |a|>1 then |xn| ∞ as n∞. This means that the fixed point X*=0 is unstable.
If |a|=1 then |xn|=|x0| for any n. This means that the fixed point X*=0 is stable,
but it is not asymptotically stable.
Example (continuation of the Cobweb Cycle).
We study the case when the supply Sn and demand function Dn for a commodity
satisfy equalities
Dn =α+βpn,
Sn=γ+δpn-1
From these suppositions we obtain the difference equation for price
pn=apn-1+b,
where a=δ/β, b=(γ-α)/ β.
The equation has a fixed price
P*=b/(1-a).
Let us find conditions for stability of the fixed point P*.
Introduce new variable
x=p - b/(1-a)
that is difference between p and the fixed point P*.
Then from
p= b/(1-a) +x
and the equation
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pn+1=apn+b
it follows equality
b/(1-a) +xn+1=a(b/(1-a) +xn ) +b.
We obtain the difference equation
xn+1=a xn
where a= δ/β >0.
If a<1 (or δ<β) then xn0 and the fixed price P*=b/(1-a) is asymptotically stable,
if a>1 (or δ>β ) the fixed price is unstable.
2.3 Linear difference systems, example.
Let us consider a square matrix A of size nxn. A value λ and vector v are said
to be an eigenvalue and eigenvector of the matrix A if
Av=λv or (A-λE)v=0,
where E is the identity matrix. The equation has solution v it is necessary
det(A-λE)=0.
Eigenvalues are determined from this equation.
Stability theorem.
Let us consider a linear system of difference equations
xn+1=Axn.
1) The fixed point x=0 is asymptotically stable if and only if
module of each eigenvalue of the matrix A
| λ |<1.
2) The fixed point x=0 is stable if
module of each eigenvalue of the matrix A
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| λ |≤1.
3) The fixed point x=0 is unstable if and only if there exists
an eigenvalue λ of the matrix A such that
| λ |>1.
Two-dimension matrix
A=(
), (A-λE)=(
),
det(A-λE) = (a-λ)(d-λ)-cb = λ2-(a+d) λ+ad-cb.
The value
trA=a+d
is called trace of matrix A and
ad-cb=detA
is determinant of A, so eigenvalues is solution of quadratic equation
λ2 - trA λ + detA=0
√( )
.
The value
Δ=(trA)2-4detA
denotes discriminant.
If Δ>0 there exists two different real eigenvalues λ1 and λ2.
If Δ=0 then real eigenvalue λ =trA/2.
If Δ<0 there exists two complex eigenvalues
λ12=trA/2 ± √ √ =α±βi,
where
√ , trA/2= α, √ =β.
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Module of complex number z= α+βi is |z|=√ .
If discriminant Δ<0 then |λ1|=| λ2| and since |detA|= |λ1|| λ2|, the condition
|detA|<1 guarantees asymptotic stability.
Tasks
1. Find a matrix A such that the difference system xn+1=Axn.
is unstable (stable).
2. Construct matrix with complex eigenvalues.
3. Investigate dynamics of system
xn+1=-a xn,
yn+1= bxn,
0<a, b<1; and 0<a<1<b.
4. Investigate dynamics of 2-dimensional system with complex eigenvalues.
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a) are real number, ,
b) are real number, ,
c) are real number, ,
d) are real number, ,
e) α±βi are complex number, ,
f) α±βi are complex number, ,
Example. Samuelson’s model (continuation).
Recall that Samuelson’s model of national income is a second-order linear difference equation
( )
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Where yn+1 is the income in the future period, yn is the income in the current
period and yn-1 is the income the previous period of production. So yn-1 is named
delay. The equation has a unique balance (fixed) point
y*=G/(1-a) that does not change in time. To study stability of the balance point we rewrite the equation as system without delay
{
( )
From the first equation it follows
If it is substituted in the second equation, we get the Samuelson’s model. The
system has fixed point (G/(1-a), G/(1-a)). The matrix of the linear system has the
form
(
( ))
Furthermore, the conditions for stability are determined by eigenvalues of A.
( ) √( ( ))
The condition 0<a<1, 0<ab<1 implies stability of the fixed point. The condition
a(b+1)2<4b generates negative discriminant and the complex eigenvalues that implies cyclical motion around the fixed point.
2.4 Non-linear difference systems, example.
Let us consider general system of difference equations
x(n+1)=f(x(n)),
where x Rd, f(x) is smooth mapping.
Suppose that x=x* is a fixed (or balace) point, x*=f(x*).
If x lies close to x* we can rewrite the mapping f(x) in the form
f(x)= f(x*)+D(x-x*)+r(x),
where D is matrix and r(x) is a small remaider.
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The matrix D is called Jacobi matrix and has the form
(
),
where derivatives are calculated at x=x*. Since x*=f(x*), the system can be
represented as
x(n+1) x*+D (x(n)-x*) or x(n+1)-x* D (x(n)-x*).
Put y=x-x*, then
y(n+1) D y(n).
Thus the system near fixed point is like as linear system.
Grobman-Hartman Theorem.
Consider system of difference equations
x(n+1)=f(x(n)),
x* is a fixed point. If matrix D has eigenvalues and ,
then the system near the fixed point x* is equivalent to the linear system
y(n+1)=D y(n).
Example
The map f has the form
( ) ( ( ) ( )
) ( ( )
( ))
where a=1.35. The fixed point is (0, 0) and Jacobi matrix at (0,0) has form
(
)
The eigenvalues are solutions of equation det(D- E)=0, or
(1+a- )(1- )-a=0 2-(2+a) +1=0.
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Then
, .
This implies that eigenvalues and . By Grobman-Gartman Theorem
the discrete system
xn+1=xn+yn+axn(1-xn),
yn+1= yn+axn(1-xn)
near the fixed point (0,0) is equivalent to linear system
xn+1=(1+a)xn+yn,
yn+1= axn + yn.
Let us investigate this linear sysem. Eigenvalues
√( )
√
,
.
Eigenvector v1 is determined by equality
Dv1= v1 or (D- E)v1=0.
The system in coordinates v1 = (x, y) has the form
(
) (
) (
)
or
(1+a- )x+y=0
ax+(1- )y=0
Since det(D- E)=0,
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the second equality follows from the first one. Please, check this.
This means that it’s enough to solve the first equation
(1+a- )x+y=0.
There are many solution, one of them is
v1=(1, -1-a)=(1, 0.66)
From equality
Dv1= v1
It follows that vector v=tv1, t 0 is an eigenvector for as well. Thus the
eigenvectors form straight line or linear subspace
L1={ v=tv1, t }
which is invariant to the linear mapping
D: vDv,
that is if v L1 then Dv L1.
Linear system
xn+1=(1+a)xn+yn,
yn+1= axn + yn
is rewrited in the form
(
) (
) (
)
or vn+1=Dvn.
If v0 L1, then all vn L1 for any n.
Moreover
vn=( )n v0=( )n v0.
This means that linear system expands on L1 with coefficient =3.019.
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Because of this the linear line L1 is named unstable subspace and denote Lu .
Simillar, all eigenvectors v2 (to eigenvalue ) form linear subspace Ls which is
called stable subspace. If v0 L2, then all vn L2 for any n and
vn=( )n v0=( )n v0.
The linear system on Ls compresses with coefficient =0.331.
V2=(1, -1-a)=(1, -2.019)
An equilibrium point is called hyperbolic point if .
The origin (0,0) is hyperbolic balace point for the system of difference equations
xn+1= xn+yn+axn(1-xn),
yn+1 =yn+axn(1-xn)).
Moreover, this nonlinear system near (0,0) is equivalent to linear system
(
) (
) (
).
2.5 Difference equations with delay, example.
Consider difference equations of the form
vn+1=f(vn,vn-1,vn-2,…).
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This means that future state vn+1 of economic system depends not only on present
state vn but also on other previous states vn-1, vn-2, … .
Example.
Suppose that vn is volume of production, n is number of periods of production.
We will determine the volume of production by formula
vn+1=vnexp(a(H-vn)),
where a is positive adapt coefficient and H can be treated as maximum volume
that can be sold in one period.
If vn<H, production increases and if vn>H, production decreases.
Let the product have a shelf life equal to three periods of production.
This means that market has fraction of products vn, vn-1, vn-2.
Suppose that during n-th period sale consists of
bvn+ cvn-1+ dvn-2
where b,c,d>0 and b+c+d=1.
Then formula of production take the form
vn+1=vnexp(a(H-(bvn+ cvn-1+ dvn-2))).
We obtane an equation with delay.
Any difference equation with delay can be presanted as system difference
equations without delay.
Consider our example and put
xn+1=yn,
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yn+1=zn,
zn+1= znexp(a(H-(bzn+ cyn+ dxn))).
From the second equation it follows
yn=zn-1.
From the first equation it follows
xn=yn-1 =zn-2.
Finaly we have
zn+1= znexp(a(H-(bzn+ czn-1+ dzn-2))).
2.6 Difference equations with delay and time series.
Takens Theorem. (Takens F. Detecting strange attractor in turbulence. Lect. Notes in
Math., 1981. -- v.898 -- Berlin: Springer -- 89-94.)
Let xn+1=Ф(xn) be a dynamical system, x lies in Rd, and there is an observed value
vn =h(xn) (that is time series {vn}).
Let us construct m-vectors
zn=(vn-m+1, vn-m+2, ... , vn)
then (in typical case) as m>2d there exists mapping
F:RmRm ,
that gives dynamical system
zn+1=F(zn).
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The last equation is a difference equation with delay
vn+1=fm (vn,vn-1,…,vn-m+1).
Theorem say that any time series in the described conditions can be presented as
a difference equations with delay.
Example. The time series of production.
Let us consider a statistic data of production
x0, x1, x2, x3, …, xk.
Our aim is to find parameters of function that gives an opportunity to prognose
the production. We will determine the volume of production by formula
vn+1=vnexp(a+bvn)),
where parameters a and b will be determined by method of least of squares. We
can rewrite the formular in the form
( )
Using the production data
x0, x1, x2, x3, …, xk
we construct other data
y1, y2, y3, …, yk
putting
yn=ln(xn/xn-1)=ln(xn)-ln(xn-1).
The equation to parameters a and b takes the form
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y=a+bx.
The method of least of squares is to find a and b that the expression
∑( )
takes the minimum value. We know the statistic data { (xi, yi) } and the
parameters a and b are determined from the equations
∑( )
( )
∑( )
( )
The first equation is transformed to
∑ ∑
∑
∑
∑
∑
where
∑
∑
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The second equation is transformed to
where is arithmetical mean. We obtain formulas for a and b
3. Differential equations
3.1 Equilibrium or balance point, example.
A differential equation expresses the derivative (velocity) of change of state as a
function of the current state.
The system of differential equations can be expressed in the vector form
( ( )) or ( )
where x ϵ Rd describes the state of economic system, t is time.
Denote by X(x,t) solution of the system of differential equations, with initial
conditions X(x,0)=x.
Trajectory with initial point x0 is the curve
T(x0 ) ( ) } .
A simple illustration of trajectory is an equilibrium or balance point
T(x0 )={x0} or X(x0,t)=x0, as .
Equilibrium or balance of economic system is a state that doesn’t change in time.
Balance point x* is determined as solution of the equation
f(x*)=0.
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Example. (Continuation of Solow model.)
The differential growth equation is
,
The equation has two equilibrium or balance points which are determined by the
equation
-λk+skα=0 or kα(s- λk1-α )=0
k=0 and k1-α=s/λ.
The first balance point k=0 is not interesting for us, because k=0 means that there
is no production.
The second balance is k= k*=(s/λ)1/(1-α) and is very important with practical
point of view.
3.2 Stable and unstable balance point, example.
Definition
An equilibrium point x* is stable if every solution starting at a point x0 close to
x*, will stay close to x* at all future time.
More formally, x* is said to be stable if for any ε>0 there exists δ>0 such that if
the distance
|x0-x* |<δ than |X(x0,t)-x* |<ε for any t>0.
A balance point x* is called asymptotically stable if x* is stable and there exists δ0
such that
if |x0-x*|< δ0 then X(x0,t) tends to x* if t∞.
Example
Consider very simple linear differential equation
The balance point is x*=0. It is not hard to establish that
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X(x,t)=xeat
is a solution of the equation.
If a<0 then xeat 0 as t∞ .
This means that the fixed point x*=0 is asymptotically stable.
If a>0 then |xeat| ∞ as t∞.
This means that the fixed point x*=0 is unstable.
If a=0 then xeat =x for any t .
This means that the fixed point x*=0 is stable, but it is not asymptotically stable.
3.3 Linear systems of differential equations, example.
Let us consider a square matrix A of size nxn.
Eigenvalues λ of A are determined from the equation
det(A-λE)=0,
where E is the identity matrix.
Two-dimension matrix
A=(
), (A-λE)=(
),
det(A-λI) = (a-λ)(d-λ)-cb = λ2-(a+d) λ+ad-cb.
The value trA=a+d is trace of matrix A and ad-cb=detA is determinant of A,
so eigenvalues is solution of quadratic equation
λ2 - trA λ + detA=0
√( )
.
The value
26
Δ=(trA)2-4detA
is discriminant .
If Δ>0 there exists two different real eigenvalues λ1 and λ2.
If Δ=0 then real eigenvalue λ =trA/2.
If Δ<0 there exists two complex eigenvalues
λ12=spA/2 ± √ √
= α±βi,
where √ , trA/2= α, √ =β,
α=Re λ is called real part of λ and β =Im λ is called imaginary part of λ.
Stability theorem.
Let us consider a linear system of differential equations
1) The balance point x=0 is asymptotically stable if and only if real part of each
eigenvalue of the matrix A
Re λ <0.
2) The balance point x=0 is stable if real part of each eigenvalue of the matrix
A
Re λ 0.
3) The balance point x=0 is unstable if and only if there exists an eigenvalue λ
of the matrix A such that
Re λ >0.
Example
Let us consider linear
27
Solution has the form
X(x,y,t)=xCos(t)+ySin(t), X(x,y,0)= x,
Y(x,y,t)=-xSin(t)+yCos(t), Y(x,y,0)=y.
Check, please!
Let us calculate
(X(x,y,t))2+(Y(x,y,t))2=x2+y2.
The value r2=x2+y2 is square of the distance between the point (x,y) and origin of
coordinate (0,0).
This means that the trajectory lies on circle of radius r=√ and center (0,0).
Matrix
(
)
has eigenvalues
, Re =0.
Since Re =0, the balance point (0,0) is stable but not asymptotically stable.
Tasks
1. Find a matrix A such that the system of differential equations
x’=ax+by,
y’=cx+dy.
is unstable (stable).
2. Construct the system of differential equations with complex eigenvalues.
3. Investigate dynamics of system
x’=-a x,
y’= bx,
0<a and 0<b.
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4. Investigate dynamics of 2-dimensional system with complex eigenvalues.
a) are real number, ,
b) are real number, ,
c) are real number, , (
),
d) are real number, , (
),
e) ±βi are complex number (Re =0),
f) α±βi are complex number, Re =α>0.
29
Classification of balance points for linear systems of differential
equations.
The following diagram classifies the balance points of the linear system according to the values of
p = trace(A), q = det(A), and d = disc(A).
The parabola, p2 - 4 q = 0, is the locus of points (p, q) for which d = 0.
In the region of the (q, p) plain enclosed by the parabola, d = p2 - 4 q < 0. Consequently, the eigenvalues 1 and 2 are complex numbers. If p is not zero,
complex eigenvalues result in spiral trajectories. If p = 0, eigenvalues are purely
imaginary and trajectories enclose the fixed point in ellipses or circles.
If p > 0 and q > 0 (the positive quadrant of the (q, p) plain), both eigenvalues
have positive real parts (or are purely positive numbers), and the fixed point is
unstable. If p > 0, and q = 0, one eigenvalue is positive and the other is zero: the
trajectories are unstable lines.
Conversely, if p < 0 and q > 0 (the negative quadrant), both eigenvalues have
negative real parts (or are purely negative numbers), and the fixed point is stable.
If p < 0, and q = 0, one eigenvalue is negative and the other is zero: the
trajectories are stable lines.
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However, if q < 0, one eigenvalue is a positive number and the other is negative:
trajectories follow the saddle point pattern.
3.4 Non-linear systems of differential equations, example.
Let us consider general system of differential equations
( )
where x Rd, f(x) is smooth mapping.
Suppose that x=x* is an equilibrium (or balace) point,
f(x*)=0.
If x lies close to x* we can rewrite the mapping f(x) in the form
f(x)= f(x*)+D(x-x*)+r(x),
where D is matrix and r(x) is a small remaider.
The matrix D is Jacobi matrix and has the form
(
),
where derivatives are calculated at x=x*. Since f(x*)=0, the system can be
represented as
( )
Put y=x-x*, then
Thus the system near fixed point is like as linear system.
Grobman-Hartman Theorem.
Consider a system of differential equations
( )
31
x* is an equilibrium point. If matrix D has eigenvalues ,
then the system near the equlibrium point x* is equivalent to the linear system
Example. (Continuation of Solow model.)
The differential growth equation is
,
The equation has two equilibrium or balance points which are determined by the
equation
-λk+skα=0.
The equation has solutions
k=0 and k1-α=s/λ.
The second balance is k= k*=(s/λ)1/(1-α) and is very important. Its stability is
determined by the derivative of the right side of the equation
and as
k=k*=(s/λ)1/(1-α)
the value
-λ+sαkα-1 = - λ(1-α)<0.
This means that the balance k* is stable and any trajectory tends to k*.
4 Chaos in economic systems
4.1 Hyperbolic balance point, example.
Consider a system of difference equations
x(n+1)=f(x(n)),
32
where x lies in R2. Suppose that a point x* is fixed point, i.e. x*=f(x*).
Recall that the point is hyperbolic if the eigenvalues of the matrix
(
( ))
satisfy inequality . There exist stable Ls and unstable Lu lines (or
subspaces) such that
DLs= Ls , DLu= Lu,
matrix D expands on Lu and D contracts on Ls.
Theorem. Let x* is a hyperbolic fixed point of difference system
x(n+1)=f(x(n)),
x R2 then the set of points
Ws={x: fn (x) x* for n }
is a curve that is tangent to stable subspace Ls and the set of points
Wu={x: f -n (x) x* for n }
is a curve that is tangent to unstable subspace Lu.
Ws and Wu are called stable and unstable manifolds.
Algorithm for construction Wu:
Let L is a small segment that intersects Ws then
the iterates fn(L) converge to the unstable manifolds Wu as n .
Example
Consider the mapping
f(x,y)=( 1.1x-0.1y sin(x), 0.7y+0.5x2)
(0, 0) is a fixed point
33
D=(
),
The segment L starts at the point (0.31, 0.31) and finishes at the point (-0.3,
0.03). Iterations fn(L) have the form:
Example (Line3)
Let us continue to study hyperbolic point of the mapping
f(x,y)=(x+y+ax(1-x), y+ax(1-x)), a=1.35
Unstable manifold of this mapping has very complicate behavior
Algorithm for construction Ws:
Let L be a small segment that intersects Wu.
34
The iterates f-n(L) converge to the unstable manifolds Ws as n .
Conctrustion inverse mapping
Let the mapping f has form
f(x,y)=(f1(x,y), f2(x,y))
Problem is to construct f-1.
Set
X =f1(x,y),
Y= f2(x,y).
Solve this system wth respect to x and y. We get equalities
x=g1(X,Y),
y=g2(X,Y).
Then the mapping G(X,Y)=(g1(X,Y), g2(X,Y)) is inverse to initial one.
Example
Let
f(x,y)=(x+y+ax(1-x), y+ax(1-x)),
X= x+y+ax(1-x),
Y= y+ax(1-x),
We have
X= x+Y x=X-Y,
Y= y+ax(1-x) y=Y- ax(1-x)=Y-a(X-Y)(1-(X-Y)).
Thus the mapping
G(X,Y)=(X-Y, Y-a(X-Y)(1-(X-Y))
is inverse mapping to f(x,y)=(x+y+ax(1-x), y+ax(1-x)).
35
It can be used to construction stable manifold Ws
4.2 Homoclinic orbit as a cause of chaos, example.
Homoclinic point
Let x* is hyperbolic fixed point, a poit x0 is called homoclinic if
fn(x0) x* as n+ and n-
This means x0 lies in the intersection Ws and Wu. Orbit of the homoclinic poit
T={x=fn(x0), n=0, 1, 2, 3, …}
is called homoclinic orbit, it starts and finishis at the hyperbolic point x*.
Existence homoclinic point generates nontrivial dynamics near the orbit T.
Dynamics around a homoclinic orbit is very complicated, hard to predict and has
stochastic character.
Such behavior is called CHAOS.
Let x* is honoclinic poit or intersection point Ws and Wu
. This inresection is called
transversal if angle between Ws and Wu
greater than zero.
Program Line2
36
Let x0 is honoclinic poit or intersection point Ws and Wu. This inresection is called
transversal if angle between Ws and Wu
greater than zero.
Smale Theorem
S.Smale showed that near to transversal homoclinic orbit there exists a collection
of orbits which can be coded by all possible binary sequences. In particular,
periodic sequences correspond to periodic orbit of the same period. This implies
the existence of periodic orbits of any large period near homoclinic orbit.
Example
Let us continue to study hyperbolic point of the mapping
f(x,y)=(x+y+ax(1-x), y+ax(1-x)), a=1.35
We can get coordinates of the transversal homoclinic point by computer
x = 0.009848517670, y = -0.019195942408
Smale theorem guarantees existence of chaos in this case.
Problem
Let us investigate a system of differnce equations
(xn+1, yn+1)=(f1(xn, yn), f2(xn, yn))
where f and f2 is the form
f1(x,y)=x+y+ax(1-x2),
f1(x,y)=y+ ax(1-x2),
37
where a=0.4. The fixed point is (0,0) and Jacobi matrix at (0,0) has form
(
)
The eigenvalues are solutions of equation
(1+a- )(1- )-a=0 or 2-(2+a) +1=0.
Then
, .
This implies that eigenvalues and . By Grobman-Gartman Theorem
the discrete system
x(n+1)=x(n)+y(n)+ax(n)(1-x(n)2),
y(n+1)= y(n)+ax(n)(1-x(n)2)
near the fixed point (0,0) is equivalent to linear system
x(n+1)=(1+a)x(n)+y(n),
y(n+1)= a x(n) + y(n).
Let us investigate the linear sysem.
√( )
√
,
.
Eigenvector v1 is determined by equality
Dv1= v1 or (D- E)v1=0
(
) (
) (
)
(1+a- )x+y=0
38
ax+(1- )y=0
Since
det(D- E)=0,
the second equality follows from the first one. This means that it’s enough to solve
the first equation
(1+a- )x+y=0.
There are many solution, one of them is
v1=(1, -1-a)=(1, 0.46…)
We can get coordinates of the transversal homoclinic point by computer
Smale theorem guarantees existence of chaos in this case.
x = -0.085871691404, y = 0.017047586207
4.3 Indestructibility of chaos, example.
Let a system have a hyperbolic fixed point x* and transversal homoclinic point X*.
Suppose that the system is perturbed by small perturbation. If the perturbation is
sufficiently small then the perturbed system has hyperbolic fixed point x** close
to x* and transversal homoclinic point X** close X*. Smale theorem guarantees
existence of chaos in this case.
Example
Let us consider a system of differnce equations generated by the mapping
39
f1(x,y)=x+y+ax(1-x2),
f1(x,y)=y+ ax(1-x2),
with the perturbed parameter a=0.5.
5 Computer modeling of economic dynamics
5.1 Symbolic image of dynamical system.
Let an economic system have n states {i, j, q, … }. For each state i it is fixed
admissible transition from i to some states {q, l, …} in unit of time.
Consider a directed graph G that has n vertices corresponding to the states
{i, j, q, … } and has edges {iq} corresponding to the fixed admissible passes.
The constructed graph G is called symbolic image of the dynamic system.
40
An infinite in both directions sequence {zk} of vertices is called an admissible path
(or simply a path) if for each k the graph G contains the directed edge
zk zk+1.
A path {zk} is said to be p-periodic if
zk=zk+p
for each k. An admissible path describes dynamics of the economic system.
5.2 Construction of symbolic image, example.
There are many methods for constructing a symbolic image. In any case
preliminary information is needed. It may be expert estimates, the information
obtained from the reports, mathematical calculations and models.
Example.
Let us consider dynamical system generated by the forced Duffing system
( )
in the domain M=[-2,2]x[-2,2]. The system is -periodic in t.
Construct a covering of the domain M. The covering consists of the boxes M(i) of
the size 0.25x0.25. So we have 16x 16=256 cells. Each cell is identified with the
state of the system.
41
The numbering of the sells starts from the left-upper corner and finishes to the
right-down corner. Let m(i) be a finite set of points in M(i). The placement of
points may be systematic or random.
Let (X(x,y,t),Y(x,y,t)) be solution of the system and mapping
f(x,y)= (X(x,y, ),Y(x,y, ))
be the shift along the trajectories on the period T= .
The image of points f(m(i)) is an approximation of the image f(M(i)).
We can check by computer the inclusions
f(x,y) M(j), (x,y) m(i)$
and if the inclusions hold we fix the edge
ij.
Let us consider the described construction and take as example cell M(87), see
picture. The image of the cell M(87) is shown in Figure
The image
f(M(87))
intersects the cells
M(213), M(214), M(229,) and M(230).
42
So we have the edges
87213, 87214, 87229, and 87230
in the graph G
The same way we can construct each edge and, hence, the symbolic image.
Let us consider a discrete dynamical system generated by a system of difference
equation
xn+1=f(xn)
where x lies in a domain M Rd. Let
C={M(1),...,M(n)}
be a finite covering of the domain M by closed cells. We identify the cell M(i) with
the state i. We will say that there exists (admissible) transition from i to j if the
image f(M(i)) intersects the cell M(j)
( ( )) ⋂ ( )
The existence of the edge guarantees the existence of a point x in the cell
M(i) such that its image f(x) lies in M(j). In other words, the edge is a trace
of the matching x f(x), where x M(i), f(x) M(j).
Let G be a directed graph with vertices {i} corresponding to cells {M(i)}. Two
vertices i and j are connected by the directed edge i j if there is admissible
transition . The graph G is called the symbolic image of f with respect to the
covering C. A symbolic image is a geometric tool to describe the quantization
process. We can consider the symbolic image as a finite approximation of the
mapping $f$.
It would appear natural that this approximation describes dynamics more precise
if the mesh of the covering is smaller. By investigating the symbolic image we can
analyze the evolution of a system. It is easily to note that there is a
correspondence between orbits of a system and the paths on G. The investigation
of the symbolic image permits to get valuable information about the global
structure of economic system.
43
5.3 Balance points and periodic orbits on symbolic image, example.
Suppose the economic system has fixed (or balance) point x* that lies in M(i).
Since the balance point does not change in time the point x* lies in the
intersection
( ( )) ( )
That implies existence transition
ii.
Suppose the system has periodic regime
i1, i2, i3, …, ip,
that implies existence transitions
i1i2 i3 … ip i1
and periodic path {i1, i2, i3, …, ip} on the symbolic image G.
Task
Find
1) balance states,
2) periodic paths.
How many difference periodic paths?
44
5.4 Matrix of transition and global dynamics, example.
The directed graph G is uniquely determined by its nxn (adjacency) matrix of
transitions
( )
where if and only if there is the directed edge i j, otherwise .
It should be remarked that if there is mapping f:MM we can consider the
matrix of transitions independently of the symbolic image by putting
if ( ( )) ( )
if ( ( )) ( )
Let
( )
be the square the transition matrix, where
∑
and upper script 2 stands for an index (not for power).
Clearly,
45
if and only if
and ,
otherwise
So if and only if there exists the path
ikj
from i to j through k. Then the sum
∑
is the number of all admissible paths of length 2 from i to j.
In the similar way one can verify that the element
of the power
( )
is the number of all admissible paths of length p from i to j.
A vertex of the symbolic image is called recurrent if there is a periodic path passing
through it. The set of recurrent vertices is denoted by RV. Two recurrent vertices i
and j are called equivalent if there is a periodic path containing i and j. According
to the definition, the set of recurrent vertices RV decomposes into classes {Hk} of
equivalent recurrent vertices. In the graph theory the classes Hk are called a
strongly connected components of the graph G.
Task
Determine recurrent and non-recurrent vertices.
Find classes of equivalent recurrent vertices.
46
Proposition.
The vertices of a symbolic image G can be renumbered so that the equivalent
recurrent vertices are numbered with consecutive integers and the transition matrix has the form
(
)
where the elements under the diagonal blocks are zeros,
each diagonal block corresponds to either a class of equivalent recurrent
vertices Hk or a non-recurrent vertex. In the last case coincides with a single
zero.
The described renumbering is not uniquely defined.
Task
Determined described renumbering for the graph considered above.
5.5 Attractors and their construction, example.
Consider discrete dynamical system generated by system of difference equations
xn+1=f(xn).
47
The set S is called invariant if from an inclusion it follows the inclusions
( ) and ( ) . This means that the orbit (or trajectory) of x has to be
in S. So invariant set is an union of orbits.
An invariant set A is called attractor if there exists a neighborhood U(A) of A such
that iteration of closure lies in :
( )
and positive iterations ( ) tend to A:
⋂
( )
Asymptotically stable balance point is an example of attractor.
The set
( ) ( )
is called the domain of attraction or the basin of A.
Suppose that we construct a symbolic image by method described above and
numbering of vertices is according to the Proposition. Transition matrix has the
form
(
)
where the elements under the diagonal blocks are zeros, each diagonal block
corresponds to either a class of equivalent recurrent vertices Hk or a non-
recurrent vertex. In the last case coincides with zero.
Under this numbering there is no admissible path from a class Hp to a class Hq if
q<p:
48
As the numbering of rows of the transition matrix coincides with the numbering
of the vertices, we have to move in the same class or to go down, as the following
scheme shows.
(
)
Hence we cannot get higher than Hk. Consider a collection B of vertices of G. We say that B is an attractor on G if one can get into the A, but one cannot get out. Thus, each class Hk generates an attractor consisting of the vertices of numbers (indexes) more or equal to the numbers of the ones from Hk.
Proposition
Suppose G be symbolic image of the mapping f constructed by method described
above and B is an attractor on G. Then the union of cells
⋃ ( )
is a neighborhood such that
( )
and
⋂
( )
is an attractor of the mapping f.
According to this Proposition we can determine attractors by symbolic image.
Task.
Find attractor of the symbolic image
49
Strange attractors.
There are many attractors with so nontrivial structure that they are called "strange
attractors". By this is meant that the attractor is chaotic, i.e., it has a dense orbit
and a nontrivial fractal structure. Henon explored numerically the mapping
f(x,y)=(1-ax2+y, bx).
Henon's computation represents numerical evidence of the existence of
a strange attractor for the case a=1.4 and b=0.3. The Henon attractor has the form
Ikeda Attractor.
The Ikeda system of difference equations has the form
( )
( )
where r=2, a=-0.9, b=0.9, (
). The system has a
chaotic attractor of the form
50
5.6 Localization of recurrent orbits, example.
Consider a system of difference equations
xn+1=f(xn)
where x lies in Rd. For a given an infinite in both directions sequence
is called an -orbit (a pseudo-trajectory or a pseudo-orbit) of f if for any k
the distance
( ( ) )
Example
On the plane R2 consider a map of the form
f(x,y)=(y, 0.05(1-x2)y-x).
Let us check that the sequence
x1=(2,0), x2=(0,-2), x3=(-2,0), x4=(0,2), xk+4=xk
forms a 4-periodic -orbit for any >0.1. In fact,
f(2,0)=(0,-2), & (f(x1),x2)=0,
f(0,-2)=(-2,-0.1), & (f(x2),x3)=0.1,
f(-2,0)=(0,2), & (f(x3),x4)=0,
51
f(0,2)=(2,0.1), & (f(x4),x1)=0.1.
We can consider the transition from the point (-2,-0.1) to the point (-2,0) and from the point (2,0.1) to the point (2,0) as jumps or corrections of the value 0.1. So we
have a 4-periodic -orbit for any >0.1.
A point x is called chain recurrent if x is -periodic for each , i.e., there
exists a periodic -trajectory passing through x.
A chain recurrent set, denoted Q, is the set of all the chain recurrent points.
It is known, that the chain recurrent set Q is invariant, closed, and contains
periodic, homoclinic, nonwandering and other singular trajectories. It should be remarked that if a chain recurrent point is not periodic and then there exists as
small as one likes perturbation of f for which this point is periodic. One may say
that a chain recurrent point may become periodic under a small perturbation of
the map f.
Let us consider a symbolic image G of the mapping f with respect to a covering C. Recall that a vertex of the symbolic image is called recurrent if a periodic path passes through it.
Denote by P(d) the union of the cells M(i) for which the vertex i is recurrent, i.e.,
( ) {⋃ ( ) }
where by d is the largest diameter of the cells M(i).
Theorem.
1. The set P(d) is a closed neighborhood of the chain recurrent set
52
( )
2. The chain recurrent set Q is limit of the sets P(d) as diameter d tends to 0
( )
Example. Let us consider the Van-der-Pol system
( )
It is well known that the chain recurrent set of the Van der-Pol system consists of an equilibrium point (0,0) and a periodic orbit. The system has been studied numerically in the square M=[-3,5;3,5]x[-3,5;3,5]. The initial covering consists of 49 cells, which are 1x1 squares. The subsequent subdivisions cells are into 4 equal squares. The picture presents the neighborhoods of the chain recurrent set. According to Theorem these neighborhoods tend to the equilibrium point and the periodic orbit.
A computer program realizing the algorithm described above has been implemented in St. Petersburg Technical University, 1991 by A. Moiseev.
5.7 Stochastic Markov chains, example.
Stochastic Markov chain is defined by a collection of states {i=1, 2, … n} and
probabilities Pij of transitions from the state i to the state j. The matrix of the
transition probabilities
53
( )
is a stochastic matrix if
∑
for each i. The latter condition means that the probability of transition from state i
to some state (possibly the same) is equal to 1. So stochastic Markov chain is a
symbolic image with probability of transition over each directed edge i j. In
addition, for each state, the sum of the probabilities of all outputs is equal to one.
Task.
Determine stochastic Markov chain on the given symbolic image.
Stochastic Markov chain can be obtained from a symbolic image. Let
( )
be transition matrix of a symbolic image,
if the edge ij exists else
By setting
∑
we get a stochastic matrix. In this case, all the outputs of the state are equivalent.
However, if some passages are more preferred that they should be more
probability. Thus, the Markov chain can be viewed as a generalization of the
54
symbolic image. Markov chains are used to study dynamics when we know state
of economic system with some probability.
Suppose we know that initial state of system is distributed with probability
P0=(P1, P2, …, Pn)
to states {1, 2, …, n}, ∑ . The matrix of the transition probabilities is
( )
During the unit of time, the system switches to a new state that has a distribution
of the form
( )
After k units of time the system has a distribution
( )
We obtain a limit distribution when k tends to infinity. The limit distribution is
called stationary distribution and it is great importance for the practice. Stationary
distribution is determined by the equation
( )
Example.
Consider symbolic image with two vertices and transition matrix of the form
(
)
The corresponding stochastic matrix is
(
)
The stationary distribution is
(
)
55
Please, check this.
Task.
Show graph G of the symbolic image described above.
5.8 Balance method for stationary distribution.
Many economic problems are reduced to the study of flows in graphs. Let G be a
directed graph. A distribution {mij} on the edges {i j} is called flow on G if
∑
∑
∑
The last property may be treated as a Kirchhoff law:
the flow incoming in the vertex i ∑ equals
the flow outgoing from i ∑ .
The second property is normalization.
In the graph theory the described distribution is named a closed or invariant flow.
We define the probability of the vertex i as
∑
∑
In this case we get
∑
∑
Each flow {mij} on a graph G generates the stochastic Markov chain such that the
states {i} are vertices
56
and the transition probability ij is defined as
The constructed stochastic matrix
(
)
has a stationary distribution of the form
( )
It turns out that the inverse is true: for any stochastic matrix
( )
and its stationary distribution p=(pi) there exists a flow
( )
with the distribution on vertices
It's enough to put
It follows from above that the flow technology on graph is equivalent to the
stochastic matrix method.
Let G be a graph with n vertices. A matrix {mij} is a flow on G if the following
conditions are satisfied:
∑
∑
∑
57
The task on computation of flow on graph may be considered as a specific
transport task. Leningrad architect G.V. Sheleikhovsky solved such a task in the
fifties of last century using the balance method. Starting from an arbitrary
distribution he recalculated the distribution sequentially such that only one
equality be satisfied and other ones have escaped the attention. By repeating such
balance in cycle he obtained the sequence of distributions which converges rapidly
to the desired solution.
Example.
Consider the graph G
For G a stochastic matrix and stationary distribution are constructed by the
Sheleikhovsky method of balance. The original matrix has been the transition
matrix of the symbolic image. The stochastic matrix and stationary distribution has
the form
(
)
(
)
Task.
Please, check the equality
58
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W.A, BROCK , A.G. MALLIARIS, DIFFERENTIAL EQUATIONS, STABILITY AND CHAOS
IN DYNAMIC ECONOMICS . Amsterdam, The Netherlands, 1996.
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Sargent, Thomas J. Dynamic macroeconomic theory. United States of America
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WEI-BIN ZHANG, Discrete Dynamical Systems, Bifurcations and Chaos in Economics
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