Jacques I. Mathematics for Economics and Business

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chapter 9

Dynamics

This chapter provides a simple introduction to the dynamics of economic systems. So

far, all of our models have been static. We have implicitly assumed that equilibrium

prices and incomes are somehow attained instantaneously, as if by magic. In

practice, economic variables take time to vary. The incorporation of time into eco-

nomic models is known as dynamics and it enables us to decide whether equilibrium

values are actually achieved and, if so, exactly how individual variables approach

these values. There are two sections that can be read in either order.

In Section 9.1 we consider the case when time, t, is a discrete variable taking whole-

number values. This describes the situation in which variables change only after fixed

periods. For example, the prices of certain agricultural products change from one

season to the next but are fixed throughout each season. We express this time

dependence using an obvious subscript notation. The price during the first period

is denoted by P

, the price during the second period is P

, and so on. Equations

that relate the price in one period, P

, to that of the previous period, P

t−1

, are called

difference equations and a method is described for the solution of such equations.

In Section 9.2 we consider the case when time is a continuous variable taking all pos-

sible values in a certain interval. This describes the situation in which variables

change from one moment to the next. For example, the prices of certain commodi-

ties, such as oil, effectively change instantaneously and are not fixed on a seasonal

basis. We express this time dependence using the usual function notation, P(t). It is

appropriate to model the rate of change of P with respect to t using the derivative,

P′(t). Equations that involve the derivatives of an unknown function are called dif-

ferential equations and a method is described for solving such equations.

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It is not possible in an introductory book such as this to give you more than a flavour

of the mathematics of dynamics. However, in spite of this, we show you how to

solve dynamic systems in both macroeconomics and microeconomics. Also, we hope

that this chapter will encourage you to read other books that describe more

advanced methods and models.

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section 9.1

Difference equations

A difference equation (sometimes called a recurrence relation) is an equation that relates con-

secutive terms of a sequence of numbers. For example, the equation

= 2Y

t−1

describes sequences in which one number is twice its predecessor. There are obviously many

sequences that satisfy this requirement, including

2, 4, 8, 16,...

5, 10, 20, 40,...

−1, −2, −4, −8,...

In order to determine the sequence uniquely, we need to be given some additional information,

such as the ﬁrst term. It is conventional to write the ﬁrst term as Y

, and once this is given a

speciﬁc value all remaining terms are known.

Objectives

At the end of this section you should be able to:

 Find the complementary function of a difference equation.

 Find the particular solution of a difference equation.

 Analyse the stability of economic systems.

 Solve lagged national income determination models.

 Solve single-commodity-market models with lagged supply.

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The following problem gives you an opportunity to solve difference equations for yourself.

Dynamics

554

Example

Write down the ﬁrst four terms of the sequence deﬁned by

= 3 and Y

= 2Y

t−1

and obtain a formula for the general term Y

in terms of t.

Solution

If Y

= 3 then

= 2Y

= 2 × 3 = 6

= 2Y

= 2 × 6 = 12

= 2Y

= 2 × 12 = 24

In order to produce a formula for the general term, we write these as

= 2Y

= 2

× 3

= 2Y

= 2

× 3

= 2Y

= 2

× 3

It is now obvious from this pattern that the general term, given by

= 3(2

)

is the solution of the difference equation

= 2Y

t−1

with initial condition

= 3

Practice Problem

1 Starting with the given initial conditions, write down the first four terms of each of the following

sequences. By expressing these as an appropriate power, write down a formula for the general term,

in terms of t.

(1) (a)

= 3Y

t−1

; Y

= 1 (b) Y

= 3Y

t−1

; Y

= 7 (c) Y

= 3Y

t−1

; Y

= A

(2) (a) Y

= Y

t−1

; Y

= 1 (b) Y

= Y

t−1

; Y

= 7 (c) Y

= Y

t−1

; Y

= A

(3) Y

= bY

t−1

; Y

= A

The result of the last part of Practice Problem 1 shows that the solution of the general equation

= bY

t−1

(1)

with initial condition

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= A

is given by

= A(b

)

Before we can consider the use of difference equations in economic models, we must exam-

ine the solution of more general equations of the form

= bY

t−1

+ c (2)

where the right-hand side now includes a non-zero constant, c. We begin by deﬁning some

terminology. The general solution of equation (2) can be written as the sum of two separate

expressions known as the complementary function (CF) and the particular solution (PS). The

complementary function is the name that we give to the solution of equation (2) when the con-

stant, c, is zero. In this case, equation (2) reduces to equation (1), and so

CF = A(b

)

The particular solution is the name that we give to any solution of equation (2) that we are

clever enough to ‘spot’. This turns out to be rather easier to do than might at ﬁrst appear and

we will see how this can be done in a moment. Finally, once CF and PS have been found, we

can write down the general solution of equation (2) as

= CF + PS = A(b

) + PS

A proof of this result can be found in Practice Problem 8 at the end of this section. The letter

A is no longer equal to the ﬁrst term, Y

, although it can easily be calculated, as the following

example demonstrates.

9.1 • Difference equations

555

Example

Solve the following difference equations with the speciﬁed initial conditions. Comment on the qualitative

behaviour of the solution in each case.

(a) Y

= 4Y

t−1

+ 21; Y

= 1

(b) Y

= Y

t−1

+ 8; Y

= 2

Solution

(a) The difference equation

= 4Y

t−1

+ 21

is of the standard form

= bY

t−1

+ c

and so can be solved using the complementary function and particular solution. The complementary

function is the general solution of the equation when the constant term on the right-hand side is

replaced by zero: that is, it is the solution of

= 4Y

t−1

which is A(4

The particular solution is any solution of the original equation



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= 4Y

t−1

+ 21

that we are able to ﬁnd. In effect, we need to think of a sequence of numbers, Y

, such that when this is

substituted into

− 4Y

t−1

we obtain the constant value of 21. One obvious sequence likely to work is a constant sequence,

= D

for some number, D. If this is substituted into

= 4Y

t−1

+ 21

we obtain

D = 4D + 21

(Note that Y

= D whatever the value of t so Y

t−1

is also equal to D.) This algebraic equation can be re-

arranged to get

−3D = 21

and so D =−7.

We have therefore shown that the complementary function is given by

CF = A(4

)

and that the particular solution is

PS =−7

Hence

= CF + PS = A(4

) − 7

which is the general solution of the difference equation

= 4Y

t−1

+ 21

To ﬁnd the speciﬁc solution that satisﬁes the initial condition

= 1

we simply put t = 0 in the general solution to get

= A(4

) − 7 = 1

that is,

A − 7 = 1

which gives

A = 8

The solution is

= 8(4

) − 7

A graphical interpretation of this solution is shown in Figure 9.1 where Y

is plotted against t. It is

tempting to join the points up with a smooth curve. However, this does not make sense because t is

allowed to take only whole-number values. Consequently, we join up the points with horizontal lines

to create the ‘staircase’ which more properly reﬂects the fact that t is discrete. Figure 9.1 shows that the

values of Y

increase without bound as t increases. This is also apparent from the formula for Y

because

Dynamics

556

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the numbers 4

get ever larger as t increases. We describe this by saying that the time path diverges uni-

formly or explodes. This sort of behaviour can be expected to occur for any solution

= A(b

) + PS

where b > 1.

(b) The difference equation

= Y

t−1

+ 8

can be solved in a similar way to that of part (a). The complementary function is given by

CF = A

and for a particular solution we try

= D

for some constant D. Substituting this into the difference equation gives

D = D + 8

which has solution D = 12, so

PS = 12

The general solution is therefore

= CF + PS = A

+ 12

9.1 • Difference equations

557

Figure 9.1



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Finally, the speciﬁc value of A can be found from the initial condition

= 2

Setting t = 0 in the general solution gives

2 = A

+ 12 = A + 12

and so A is −10. The solution is

=−10

+ 12

This solution is sketched in Figure 9.2, which shows that the values of Y

increase but eventually settle

down at 12. We describe this by saying that the time path converges uniformly to the value of 12, which

is referred to as the equilibrium value. This behaviour is also apparent from the formula for Y

because

the number (1/3)

gets ever smaller as t increases. In effect, the complementary function disappears,

leaving just the particular solution. The particular solution is the equilibrium value of Y

whereas the

complementary function measures the deviation from the equilibrium which, in this case, converges to

zero as t increases. This sort of behaviour can be expected to occur for any solution

= A(b

) + PS

when 0 < b < 1.

Dynamics

558

Figure 9.2

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9.1 • Difference equations

559

Practice Problem

2 Solve the following difference equations with the specified initial conditions:

(a)

=−

/2Y

t−1

+ 6; Y

= 0 (b) Y

=−2Y

+ 9; Y

= 4

In each case, sketch the corresponding ‘staircase’ diagram and comment on the qualitative behaviour

of the solution as t increases.

The results of the previous example and Practice Problem 2 can be summarized:

If b

1 then Y

displays uniform divergence.

If 0

1 then Y

displays uniform convergence.

−−

0 then Y

displays oscillatory convergence.

If b

<< −−

1 then Y

displays oscillatory divergence.

The remaining possibilities, b = 1, b =−1 and b = 0, are considered in Practice Problem 6 at the

end of this section, which shows that Y

converges when b is 0 but diverges when b is 1 or −1.

We conclude that the solution of the difference equation eventually settles down to an equilib-

rium state only when b lies in the range −1 < b < 1.

If convergence does occur in an economic model, the model is said to be stable. If the vari-

ables diverge, it is said to be unstable.

We now investigate two applications of difference equations taken from macroeconomics

and microeconomics, respectively:

 national income determination

 supply and demand analysis.

We consider each of these in turn.

9.1.1 National income determination

In Section 1.6 we introduced a simple two-sector model with structural equations

Y = C + I

C = aY + b

I = I*

where b and I* denote autonomous consumption and investment, and a is the marginal

propensity to consume, which lies in the range 0 < a < 1. In writing down the equations in this

form, we are implicitly assuming that only one time period is involved, that consumption

depends on national income within this time period and that equilibrium values are attained

instantaneously. In practice, there is a time lag between consumption and national income.

Consumption, C

, in period t depends on national income, Y

t−1

, in the previous period, t − 1.

The corresponding consumption function is given by

= aY

t−1

+ b

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