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Todorova, Tamara

Book Part — Published Version Advanced Differential and Difference Equations

Suggested Citation: Todorova, Tamara (2010) : Advanced Differential and Difference Equations, In: Tamara Todorova, Problems Book to Accompany Mathematics for Economists, ISBN 978-0-470-59181-9, Wiley, Hoboken, pp. 615-701, http://eu.wiley.com/WileyCDA/WileyTitle/productCd-EHEP001511.html

This Version is available at: http://hdl.handle.net/10419/148409

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In this chapter we deal with harder differential and difference equations. We already discussed first- order equations in which a first-order derivative or difference is involved. Some more sophisticated cases are second-, third-, or higher-order differential or difference equations. The chapter is split in two: first we cover more advanced differential equations, and then we turn onto their discrete-time counterpart, higher-order difference equations.

Second-order Differential Equations

Consider the linear differential equation

()nn ( 1) ytutyt()11 () () ... utytutytvtnn () () () () ()

Since it contains the nth derivative ytn () of the function yt(), it is an n -th order differential equation with variable coefficients. It is easy to notice that when only a first-derivative yt() is involved, the equation becomes the special case of a first-order differential equation

dy ut() y vt () dt which we are already familiar with. By analogy with the constant coefficient case, we have the general linear nth order equation

()nn ( 1) ytayt()11 () ... aytaybnn  ()

th ()n where again the n derivative yt() is involved; but this time, the functions uti () and vt()

( in1,2.., ) correspond to the constants ai and b , respectively. Similar to the first-order equation case when only the first derivative yt() is involved, we have the familiar equation dy ay b dt

We found the general solution to this simple first-order differential equation to be b yt() y y  Aeat  cp a b where y  is the particular integral giving the intertemporal equilibrium. This implies that we p a have the simplest possible type of solution for yt(); that is, yt() c where the function y is constant dy in time and the derivative is zero. Consider now the case dt

yt() ayt12  () ayb where the highest derivative is the second-order derivative yt(). If we again assume the simplest possible type, that is, y being a constant, we should have

yt() yt  () 0 615 616 Problems Book to Accompany Mathematics for Economists and the particular integral is b y p  a2  0 a2

Example: Find the particular integral of the equation yt() yt () 2 y 6. Since a2 2 and 6 b 6 , substituting in the expression for the particular integral yields y   3 . p 2

What if a2  0 so the expression for the particular integral is undefined? Then it must be that y is no longer constant. A simple case to consider is yct where again cconst . Then the differential equation becomes

yt() ayt1  () b a2  0

Since yct , it follows that yt() c and yt() 0, which reduces the equation to b yt() a1

We find the particular integral by integrating yt() with respect to t , which gives b ytp  a2  0 a1  0 a1

Given that this time y p is a nonconstant function of time, it constitutes a moving equilibrium.

In the case when aa120 , the second-order differential equation becomes

yt() b

Integrating yt() twice with respect to t gives bt 2 y  aa0 p 2 12

The Complementary Function

In the case of the first-order linear differential equation, its complementary function was the general solution of the homogeneous (reduced) equation yt() ayt () 0, i.e., yt() Aeat . Generally, an expression of the form Aert fits well into complementary functions. One reason why we can apply this exponential term to a second-order differential equation is that the latter is a second-order generalization of the first-order homogeneous equation. If we assume the solution for the function yt() to be of an exponential type yt() Aert , then we have

yt() rAert and yt() rAe2 rt

Substituting these values of the derivatives and the parental function in the homogeneous second-order differential equation yields

2 rt rt rt rAe arAe12 aAe 0

2 which gives rise to the characteristic equation rara 120 and the two characteristic roots aaa2 4 r  112 1,2 2

Chapter 11. Advanced Differential and Difference Equations 617

1 where by Viete’s formula rr12 a 1 and rr12 a 2. These two roots result in two solutions for yt() Aert , respectively

rt1 rt2 yAe11 and yAe22

where A1 and A2 are two arbitrary constants and the complementary function of the nonhomogeneous

(complete) equation is yyyc 12. Three possible situations exist in relation to the characteristic roots r1 and r2 .

Case 1. Distinct real roots

2 If aa12 4 , then both roots r1 and r2 are distinct real numbers and we can write

rt12 rt yyyAeAec 12  1  2 rr12

For particular values of the two constants A1 and A2 implied by some initial conditions of yt() and its derivatives, we can find the general solution to the complete equation as the sum of the complementary function and the particular integral

rt12 rt yt() ycp y  y12 y y p  Ae 1  Ae 2  y p

Example: Solve the differential equation yt() yt () 2 y 6. We already found the particular integral of this nonhomogeneous equation to be y p  3. How to find the complementary function? 2 We see that the equation fits this first case since a1 1, a2  2 and aa12 4 because 18 . Furthermore, the characteristic roots are

11813   r  1,2 22

r1 1 r2 2

rt12 r t tt2 yt() Ae12 Ae  yp Ae 12 Ae  3

To find the particular values of the constants A1 and A2 , we need two initial conditions. Suppose y(0) 10 and y(0) 2 where the initial moment is t  0 . Substituting for t  0 we obtain

02(0) yAeAeAA(0)12  3 12 3 10

Differentiating yt() with respect to t , we get

tt2 yt() Ae12 2 Ae

Then at t  0,

02(0) yAeAeAA(0)12 2  12 2 2 which leads to the system of equations

AA127

AA1222

with solutions A1  4 and A2  3 . Substituting to obtain the definite solution of the second-order differential equation,

1 Named after the French mathematician François Viete (1540-1603).

618 Problems Book to Accompany Mathematics for Economists

yt() 4 ett 3 e2  3

Case 2. Single real root

a If aa2  4 , there is only one root (also called a coincident or repeated real root) r  1 . Then the 12 2 complementary function is

rt yAeAhtc 12() where ht() is a function that cannot be a constant multiple of ert . Therefore, we set ht() tert , and the general solution to the second-order differential equation is

rt rt yt() ycp y  Ae12  Ate  y p

2 Example: Solve the differential equation yt() 2 yt () y 5. We can easily notice that aa12 4 a 2 since a  2 , a 1 and 24(1)2  . Thus, the example is one of a sinlge real root r 1  1 1 2 22 tt and y p  5. Therefore, yt() ycp y  Ae12  Ate  5.

Case 3. Complex roots

2 What if aa12 4 ? Then the roots r1 and r2 contain the square root of a negative number i 1 called an imaginary number. The very roots are called complex numbers as they contain a real part and an imaginary part, for instance, (5  i ), where we already defined i . Complex numbers cannot be ordered along the real line and, therefore, do not belong to the real-number system. They can generally be represented in the form ()mni where m and n are two real numbers. A complex number can be represented graphically in the xy -plane where x is the real-number axis and y is the imaginary-number axis. In this two-dimensional diagram known as the Argand diagram (shown by Figure 1) m is plotted on the horizontal axis and n on the vertical. Thus when n  0 , the complex number does not have an imaginary part and reduces to a real one. When m  0 , it is solely an imaginary number. By Pythagoras theorem the length of the ON line is found as the radius vector R mn22. Imaginary axis

N(m,n)

R  m 2  n 2 n



0 m M Real axis

Figure 1 Argand diagram

2 When aa12 4 , the two roots of the characteristic equation are a pair of conjugate complex numbers: a 4aa 2 rmni where m   1 n  21 and i  1 1,2 2 2

In the complex-root case the complementary function of the differential equation becomes

Chapter 11. Advanced Differential and Difference Equations 619

()mnit () mnit mt nit  nit yAec 12 Ae eAeAe() 12 

Example: Find the roots of the characteristic equation rr2  20. Express the complementary function for this equation. We obtain a pair of conjugate complex numbers for the two characteristic roots. 1717  ri 1,2 222

It can easily be checked that, in accordance with Viete’s formula, rr12  a 1 1 and rr12 a 2 2 . 1 7 Since m  and n  , the complementary function is 2 2 titit77  ye22 AeAe 2 c 12  where the imaginary number i appears in the exponents of the two expressions of the complementary function. To understand such imaginary exponential functions better, we should transform them into circular functions, which requires some discussion of trigonometry.

An Excursion into Trigonometric Functions

In the next part we will briefly revise some basics of trigonometric functions that may be familiar to you from high school. Trigonometric functions are often connected with complex numbers. Given an angle  , as shown in Figure 2 depicting a circle with a radius R , the trigonometric functions are

n m sin  cos  R R where mn, and R happen to be sides of the right-angle triangle OMN . Two more trigonometric functions can be defined on the basis of these two original functions:

sin n cos m tan  cot   cos m sin n Q

N

R

n O  S P m M

T

Figure 2

The angle  is measured in degrees (say, 90 ) or in radians which allow expressing the derivatives of trigonometric functions more easily. The size of the angle  is defined by the PN arc. A complete

620 Problems Book to Accompany Mathematics for Economists circle like PQST involves an angle of 2 radians which is exactly 360 or  180 . Thus, radians transform into degrees according to the following conversion table (see Table 1).

Degrees 0 30 45 60 90 135 180 270 360 Radians 0     3  3 2

6 4 3 2 4 2 sin 0 1 2 3 1 2 0 -1 0

2 2 2 2 cos 1 3 2 1 0 2 -1 0 1  2 2 2 2

Table 1

The sine and the cosine functions are periodic and repeat every 360 . They both fluctuate between 0 and 1 but differ in their peaks as shown on the following two diagrams in Figures 3a and 3b.

1

0   3 2 5 3

-1 2 2 2

sin

(a)

1

0   3 2 5 3

-1 2 2 2

cos

Figure 3 (b)

The sine and cosine functions have the following properties: sin( ) sin cos( ) cos sin22 cos 1 sin(12 ) sin 1 cos 2  cos 1 sin 2 cos(12 ) cos 1 cos 2  sin 1 sin 2

Given that both sin and cos are continuous and smooth, they are differentiable. The derivatives of the functions, applicable to radians only, turn out to be

Chapter 11. Advanced Differential and Difference Equations 621

d sin d cos d tan 1  cos sin  d d d cos2 

In the general case, given that u is a differentiable function of x , the derivatives for sine amd cosine can be written as

dusin du ducos du  cosu sinu dx dx dx dx

Example: Find the second derivative of cos . Applying the formula twice, we get

dd2 cos ( sin ) cos d 2 d

Example: Find the derivative of the trigonometric function sin(5x2  2) . Using the general formula,

dxsin(52  2) 10xx cos(52 2) dx

Transforming Complex Numbers into Trigonometric Functions

As long as the two Cartesian coordinates2 m and n are defined, we can find the angle  and the radius R , also known as polar coordinates. A basic relationship between Cartesian and polar coordinates we obtained previously is R mn22. In the opposite case knowing the values of R and  , we can write mR cos and nR sin . Thus, the pair of conjugate complex numbers mni becomes mniRcos  Ri sin  R (cos  i sin )

By what is known as Euler’s formula for complex numbers, which we will not prove here, eii cos sin and eii cos sin so mniR(cos  i sin )  Re . i

Abraham De Moivre3 discovered further that

k R(cos iRkik sin )k (cos  sin )

This result allows to find a pair of conjugate complex numbers raised to the power k such as ()mni k where by De Moivre’s theorem

()mniRekkikk   R (cossin) k  i k mt nit nit Going back to the complementary function yeAeAec ()12, we let   nt and by Euler’s formula entintnit cos sin and entintnit cos sin and substituting these consequently into the complementary function,

mt mt yc  e A12(cos nt i sin nt ) A (cos nt i sin nt ) e ( B 12 cos nt B sin nt )

where B112AA and B212 ()AAi

2 Named after the talented French mathematician Rene Descartes (1596-1650). 3 Except relating complex numbers to trigonometry, the French mathematician Abraham De Moivre (1667- 1754) is credited for the study of normal distribution and probability theory.

622 Problems Book to Accompany Mathematics for Economists

Example: Find the polar and exponential forms of  22 i . The Cartesian coordinates are m  2 and n  2 . We can find R by the formula Rmn 2222 2. Thus, we can find the sine and cosine functions as

n 2 m 2 sin  and cos  R 2 R 2  But we know that these are the values for  45 . Hence, from the formula 4 mniR(cos  i sin )  Re . i i  222cossin2iie  4 44

Example: Find the complementary function and the particular integral of the differential equation

yt() 2 yt  () 10 y 20

for which the initial conditions are y(0) 3 and y(0) 11. Here we have a1  2 , a2 10 and

b 20 2 b  20, so for the particular integral we get y p  2 . Furthermore, since aa12 4 or a2 10

440 , the characteristic roots are rmni1,2   and a 4aa 2 4(10) 22 6 m 1 1 n 21 3 2 222

mt t yc  e( B12 cos ntBnteB sin ) ( 12 cos3 tB sin3 t )

Thus, the general solution of the differential equation is

t yt() ycp y  e ( B12 cos3 t  B sin3) t  2

To definitize the constants B1 and B2 we must use the initial conditions. Substituting for t  0 in 0 yt(), we obtain yeBB(0) (12 cos0 sin 0) 2 3 where we know that cos0 1 and sin 0 0 , so we have B1 23 or B1 1. Expressing yt() from yt(),

tt yt() e ( B12 cos3 t  B sin3) t  e (  3 B 1 sin3 t  3 B 2 cos3) t and setting t  0 again

00 yeBB(0) (12 cos0  sin 0)  eBB (  3 1 sin 0  3 2 cos0)  BB 12  3  11 or

13B2  11

312B2 

B2  4

So, the differential equation is

yt() et (cos3 t 4sin3) t 2

Dynamic Stability

mt The time path of the complementary function yeBc (cos12 ntB sin) nt depends on the sine and cosine functions as well as on the term emt . Since the period of the trigonometric functions is 2 and

Chapter 11. Advanced Differential and Difference Equations 623 their amplitude is 1, their graphs will repeat their shape every time the expression nt increases by 2 . Alternatively, 2 nt nt2  n t  n 2 The first term in the parentheses B cos nt is a cosine function of t with a period . Similarly, the 1 n second term B2 sin nt has the same period and fluctuates between B2 and B2 . The dynamic stability of the function yt() depends solely on the third term, emt such that for a positive m , as t , the amplitude of (cosB12nt B sin) nt magnifies and causes an explosive fluctuation for yt(). If m  0 , the complementary function has a uniform fluctuation. When m is negative, the time path of the function yt() is dynamically stable (See Figures 4a, 4b, and 4c).

y(t) m  0

Equilibrium level

0 t Explosive fluctuation

(a)

y(t) m  0

0 t Uniform fluctuation

(b) y(t)

m  0

0 t Damped fluctuation

Figure 4 (c)

624 Problems Book to Accompany Mathematics for Economists

For the second-order differential equation

yt() ayt12  () ayb we found that the time path of the parental function yt() in the case of distinct real roots is

rt12 rt yt() Ae12 Ae  yp where A1 and A2 are arbitrary constants. The dynamic stability of the function is insured only if both roots r1 and r2 are negative; so, as t , yt() converges to its equilibrium level y p . If even one of the roots is positive, its exponential term becomes infinitely large, thereby precluding convergence.

rt rt With a single root, the solution is yt() Ae12 Ate yp . Here the necessary and sufficient condition for dynamic stability is that the single root r be negative as t . The second rt multiplicative term, Ate2 , also approaches zero because with a negative exponent the exponential term reaches zero faster than t grows.

mt nit nit Finally, in the case of complex roots, the solution is yt() e ( Ae12 Ae ) yp where a 4aa 2 rmni m   1 and n  21 1,2 2 2 The condition for convergence of the yt() path is m  0 , that is, the real part m of the complex roots to be negative. Then, for the three cases, it is enough to demand that the real part of every characteristic root be negative to ensure dynamic stability of equilibrium.

The Arrow-Pratt Measure of Risk Aversion

Let a person have wealth in the amount w and uw() be the utility function over this wealth. To measure the concavity of the utility function uw() in portfolio choice theory, Kenneth Arrow and John W. Pratt use the so-called Arrow-Pratt measure of relative (or absolute) risk aversion at wealth level w generally given by the expression

uww() E  wu uw() which is nothing but the elasticity of the marginal utility function uw() with respect to the wealth level w . Since we do not want total utility of wealth to be declining, we require uw() 0. If we assume the individual to have a constant relative risk aversion (say, be either risk averse or risk loving or risk neutral), we can adopt a constant elasticity of k . Thus, the expression becomes

uww() k uw() which gives the second-order differential equation

k uw() uw  () 0 w

Furthermore, if we substitute uuww  () for the marginal utility of wealth,

k uu 0 www

we obtain a first-order differential equation in marginal utility uw . Rearranging and solving by the separation of variables method,

Chapter 11. Advanced Differential and Difference Equations 625

u k w  uww

u k w dw dw uww

lnukwcw  ln  and taking the antilog of both sides,

kwcln k uew  eCw

Integrating marginal utility uuww  () further to obtain the total utility function, we have

 C 1k k  wc 1 k 1 u() w u () w dw Cw dw 1 k for  k 1  Cwcln  2

A special case of the measure of risk aversion is the Arrow-Pratt measure of absolute risk aversion representing the percent rate of change of marginal utility of wealth uw() at wealth level w . It is again a measure of the concavity of the total utility function uw() and is given by the expression

uw() aw() uw() thus giving rise to the second-order differential equation uw() awuw ()()  0 which can be solved for a specific function aw().

Market Equilibrium with Price Expectations

Very often, market participants base their demand and supply decisions on their expectations about the price and its behavior in the future. Those expectations are often influenced not only by the price prevailing at the moment, but also by the trends in the price movements. We can apply second-order differential equations to establish the time path of market price assuming equilibrium in each moment in time. Let us take, for example, that

qpupvpd     ,0  qps    ,0  dp dp2 where p  and p  . In the context of price trends, a positive p implies that market price dt dt 2 is rising and a positive p shows that it is rising at an increasing rate.

Then, if u  0 , a rising price increases market demand. Buyers, expecting price to rise, would prefer to increase current consumption. An example of such move is the real estate market in Bulgaria. Prior to Bulgaria’s joining the European Union, people expected the prices of houses to continue to rise, so they increased their purchases, thus pushing the prices further up. Conversely, when u  0 , people expect the price trend to reverse and, therefore, they cut back on their purchases in expectation of a lower price in the future. Similarly, the Bulgarian real estate market experienced a slowdown in housing prices after the country was accepted into the EU. Furthermore, the global financial crisis influenced the decisions of house buyers negatively. Expecting real estate prices to fall, they stopped buying, which contributed further to the decline of prices. Thus, this continuous-time model illustrates how people’s expectations of the future shape current prices. In their buying decisions, consumers

626 Problems Book to Accompany Mathematics for Economists may be driven not only by the direction of change in market price, but also by the rate at which this change occurs given here by the parameter v .

Equating market demand with market supply, we get

qqds or

 p up vp p which transforms into vp up  ()  p 

Normalizing this second-order differential equation,

u ()  pp  p  vv v

Note that the present model assumes market clearance at every moment in time. Thus, every price established in the market at any given moment is an equilibrium price, although this market equilibrium may not be the intertemporal equilibrium. You can think of the intertemporal equilibrium price as the normal price that should prevail in the market at any moment. Yet, at different moments the market might clear at too high or too low levels of price. When Bulgaria experiences a real estate bubble, prices of apartments are abnormally high – but demand and supply meet, so the market clears and the price is the market equilibrium price, though not the normal price level or the intertemporal equilibrium. In times of a deep recession, the market still clears – but at very low price levels, much below the intertemporal equilibrium.

For the second-order differential equation that obtains, we have u       a  a  and b  1 v 2 v v

In order to solve the differential equation, we need to find the time path of the price function p()t . The particular integral will give us the intertemporal equilibrium price:

b    pp  a2   

Note that the intertemporal equilibrium price is positive, as it should be, since all the constants in it are positive. Also, because it is constant, it represents a stationary, not a moving equilibrium. To find the complementary function, we discuss three cases.

Case 1. Distinct real roots

2 u     4   vv  

The complementary function for this case is

rt12 r t pc Ae12 Ae

where to definitize the constants A1 and A2 we need some initial conditions for price.

2 1 uu     r1,2  4   2 vv  v  

Thus, the general solution is

Chapter 11. Advanced Differential and Difference Equations 627

   pt() p p  Aert12  Ae r t  cp12  

Case 2. Single real root

2 u     4   vv  

a u The single root is r 1  . Thus, the general solution becomes 22v ut ut    pt() Ae22vv Ate  12  

Case 3. Complex roots

2 u     4   vv  

The characteristic roots are a pair of conjugate complex numbers rmni1,2   where

2 2 a1 u 4aa21 1  u m   and n 4 22v 22vv For the general solution, we have

ut   pt() emt ( Ae nit Ae nit ) e2v ( B cos nt  B sin nt ) 12  1 2    If v  0 , then 4 is always negative, so only the first case of distinct real roots is possible. v 2 u Under the square root we get a number bigger than , which means that at least one characteristic v root is positive. Therefore, the intertemporal equilibrium must be dynamically unstable. If v  0 , all three cases are possible. In the case of distinct real roots both roots will be negative, given u  0 . This 2 u u is because the expression under the square root is definitely smaller than  and the free term is v v positive. Hence, both characteristic roots turn out to be negative. The condition uv,0 also ensures that the single root is negative. In the third case of complex numbers when uv,0 , m turns out to be negative too. Therefore, the dynamic stability of the price function in each case is ensured when both parameters u and v are negative.

The Relationship between Inflation and Unemployment

Let us assume that the rate of inflation is negatively related to the level of unemployment and positively to the expected rate of inflation in a dependence known as the Phillips relation such that

4 p  Uh  ,0  01 h 

4 The original idea underlying the model was expressed by A. W. Phillips in a path-breaking paper titled “The Relationship between Unemployment and the Rate of Change of Money Wage Rates in the United Kingdom, 1861-1957,” Economica, November 1958, pp. 283-299. The expanded version of the Phillips relation incorporates the growth rate of money wage w where the rate of inflation is the difference between the increase in wage and the increase in labor productivity T , that is, p wT  . Thus, inflation would result only when wage increases faster than productivity. Furthermore, wage growth is negatively related to

628 Problems Book to Accompany Mathematics for Economists

p where p  is the rate of growth of the price level (that is, the inflation rate), U is the rate of p unemployment and  denotes the expected rate of inflation. Thus, the expectation of higher inflation shapes the behavior of firms and individuals in a way that stimulates inflation. Expecting prices to rise, they might decide to buy more immediately. As people expect inflation to go down (as a result of appropriate government policies, for example) this brings actual inflation down. This version of the Phillips relation that accounts for the expected rate of inflation is called the expectations-augmented Phillips relation. The adaptive expectations hypothesis further shows how inflationary expectations are formed. The equation

d jp()  01j dt illistrates that when the actual rate of inflation exceeds the expected one, this nurtures people’s d expectations, so  0 . In the opposite case, if the actual inflation is below the expected one, this dt makes people believe that inflation would go down, so  is reduced. If the projected and the real inflation turn out to be equal, people do not expect a change in the level of inflation.

There is also the reverse effect – that of inflation on unemployment. Thus, when inflation is high for too long, for example, this may discourage people from saving, consequently reducing aggregate investment and increasing the rate of unemployment. We can write that

dU km()  p k  0 dt or unemployment increases proportionally with real money where m is the rate of growth of nominal money. Thus, the expression ()mp gives the rate of growth of real money, or the difference between the growth rate of nominal money and the rate of inflation

mp mp   rmp mp where real money is nominal money divided by the average price level in the economy. The model then becomes

p  Uh  ,0  01 h  (expectations-augmented Philips relation) d jp()  01j (adaptive expectations) dt dU km()  p k  0 (monetary policy) dt

We can substitute the first equation into the second, which gives

d jUh()  dt

d jUjh()(1)   dt

Differentiating further with respect to time t,

unemployment and positively to the expected rate of inflation or wUh    where U is the rate of unemployment and  is the expected rate of inflation. If inflationary trends persist long enough, people start forming further inflationary expectations that shape their money-wage demands.

Chapter 11. Advanced Differential and Difference Equations 629

ddUd2  jjh (1)  dt 2 dt dt dU and substituting for , we obtain dt dd2  jkmp ()(1) jh dt 2 dt 1 d where the second equation of the model implies p    . Substituting this last expression for jdt p, we get ddd2 1  2 jkm jh(1) dt jdt dt which is a second-order differential equation in  . Transforming the equation,

dd2  d jkm k  jk  jh(1) dt 2 dt dt dd2 kj(1 h )  jk  jkm   or more simply dt 2 dt

kj(1 h )    jk  jkm where, given the properties of second-order differential equations, we have

akjh1  (1 ) ajk2   bjkm  

The coefficients a1 and a2 are both positive in view of the signs of the parameters. We can immediately find the equilibrium rate of expected inflation to be the particular integral

b  p m a2

Thus, the intertemporal equilibrium of the expected rate of inflation is exactly the rate of growth of nominal money. To find the time path of  we need to find the characteristic roots of the differential equation by the usual formula

aaa2 4 r  112 1,2 2

The time path of  would depend on the particular values of the parameters. Once we find this time path, we migh be able to determine that of unemployment U or the rate of inflation p .

Inflation and Unemployment: An Extended Model

Following Olivier Blanchard’s book Macroeconomics5, we can assume that the rate of change of the inflation rate p is proportional to the difference between the actual unemployment rate U and the natural rate of unemployment Un such that

dp  ()UU    0 dt n

5 Blanchard, Olivier. Macroeconomics. 2nd edition, Prentice Hall, 2000.

630 Problems Book to Accompany Mathematics for Economists

Thus, when UU n , that is, the actual rate of unemployment exceeds the natural rate, the inflation rate decreases and when UU n , the inflation rate increases. The intuitive logic behind this is that in bad economic times when many people are laid off, prices tend to fall. At this point, the actual unemployment would exceed the normal levels. In times of a boom in the business cycle, the rate of actual unemployment would be rather low, but high aggregate demand would push prices up. We also assume that Un is constant and that at any given time the actual unemployment rate U is determined by aggregate demand which, on its own, depends on the real value of money supply given by nominal money supply M divided by the average price level p . Thus, unemployment is negatively related to M real money supply according to the relationship p M U ln ,0  p

By differentiating the first equation with respect to t,

dp2 dU   dt 2 dt dU and the second equation to obtain dt dU d M dln M d ln p ln     (mp  ) dt dt p dt dt where we assume that the growth rate of nominal money supply m is constant. This could be in accordance with government planning or systematic monetary policy. Combining the two results yields

2 dp dU  ()mp  dt 2 dt

2 dp pm dt 2 which is a second-order differential equation in inflation rate p . Solving the differential equation, we have a1  0 , a2   and bm   . Hence, the particular integral is pe  m and the characteristic equation is r 2  0 ri1,2   where m  0 and n  

Thus, the general solution takes the form

0 pt( ) m e B12 cos t  B sin t  m  B 12 cos t  B sin t

Since the real part is zero, the function of inflation rate displays regular oscillations about the rate of growth of money supply, which gives the equilibrium level of inflation. To find the time path of dp unemployment U, we express  : dt dp  Btsin  B cos t dt  12 and substitute it into

Chapter 11. Advanced Differential and Difference Equations 631

1 dp  UU  U  Bsin t  B cos t   nndt 12

UBn 12 sin t   B   cos t 

where the constants B1 and B2 have not been definitized. It follows that the unemployment rate also displays regular oscillations, similar to the inflation rate, but its equilibrium is the natural rate of unemployment. Since the real part is zero, again the time path is neither convergent nor divergent.

Higher-order Differential Equations

Recall the nth -order differential equation with constant coefficients

()nn ( 1) ytayt()11 () ... aytaybnn  ()

To solve such higher-order differential equations, we simply repeat the steps applicable to second- order differential equations. With a simplest constant function assumed, such as yc , all derivatives are zero. The particular integral is b ycp  an  0 an

If an  0 , we try yct such that yt() c ; but all other derivatives are zero, so the equation becomes acbn1  . The particular integral is b ycttp  an  0 an1  0 an1 2 In the case when aann1 0 , the solution must be of the type yct . This produces the derivatives yt() 2 ct and yt() 2 c and the particular integral

22b yctp  t aann1 0 an2  0 an2

The complementary function is the general solution of the homogeneous equation

()nn ( 1) ytayt()11 () ... aytaynn  () 0

If the solution is in the form yAe rt , the derivatives can be written off as yt() rAert , yt() rAe2 rt ,…, ytrAe()nnrt() . This gives rise to the nth -degree characteristic equation with n roots

rt n n1 Ae(...)0 r a11 r ann r a or

nn1 rar11... arann 0

Thus, the complementary function, with all roots real and distinct, can be written as

rt12 rt rtn yAeAecn12 ... Ae

Assuming that the first two roots are repeated, we have rr12 , so the first two terms of the general

rt11 rt solution to the differential equation can be written as Ae12 Ate . If we assume further that the next two roots are complex such that rmni3,4 , the general solution of this differential equation can be written as

rt11 rt mt yt() Ae12 Ate  e ( B 1 cos nt  B 2 sin nt )  yp

632 Problems Book to Accompany Mathematics for Economists

To find the values of the four arbitrary constants, we need four initial conditions.

Example: Solve the differential equation yt() yt  () 3 yt  () 5 y 15. Since the highest derivative is the third-order derivative yt(), this is a third-order differential equation with a 15 particular integral y  3. Its characteristic equation is rr32 350, r which can be p 5 factored out into

(1)(25)0rrr2

with one real root r1 1 and a pair of complex conjugate roots ri1,2  12. The general solution, therefore, is

tt yt() Ae11 e ( B cos2 t  B 2 sin2) t  3

Second-order Difference Equations

We recall that first-order difference equations involve terms like yt1 and yt where the difference in each period is given. Thus knowing some initial value yo we can determine the time path of the y function as the time factor t changes. A simple second-order difference equation is

ybybycttt2112

To find the particular integral in the simplest case, we can take a solution of the form ykt  where in every period y is the constant k :

kbkbkc12  and c ykp  where bb12 1 1bb12

If bb121, we must adopt yktt  . This will result in the terms yktt1  (1) and

yktt2 (2) for the next time periods. Substituting these expressions into the difference equation gives cc k  where bb12 1 (1bbtb12 )  1 2 b 1  2

Hence, the particular integral is c yktp  t where bb12 1 and b1  2 b1  2

Since the particular integral in this second case involves t , it represents a moving equilibrium. If 2 bb121 (so that b1 2 and b2 1), then we try a solution of the type yktt  so the other two c terms are ykt(1)2 and ykt(2)2 . They generate k  and t1 t2 2 c yktt22 where b  2 and b 1 p 2 1 2

The Complementary Function

t With first-order difference equations, we found that the expression yAat  describes well the general solution of such an equation, and we try it to find the complementary function. This implies t1 t2 that yAat1  and yAat2  , which upon substitution in

Chapter 11. Advanced Differential and Difference Equations 633

ybybyttt21120 yields

ttt21 Aa b12 Aa b Aa 0 or

2 abab120 bbb2 4 This characteristic equation has the roots a  11 2. Hence, for the complementary 1,2 2 function, we have three possibilities again:

Case 1. Distinct real roots

2 If bb12 4 , then both roots are real and different, so the complementary function is

tt yyyAaAac 12  1122 

Example: Solve the second-order difference equation yyyttt21 3415. Here we have b1  3 , b2 4, and c 12 . We also note that bb12 1, so

c 15 ytttp 3 b1 232

2 The characteristic equation is aa340 with roots a1 1 and a2  4 . Thus, the general solution is t yyytcp  AA12(4)  3 t

Suppose we are also given that yo  3 and y1 1 for the two periods t  0 and t 1, respectively. Substituting these values for t , we obtain

yAAo 12 3

yA11431 A 2 

Thus, the constants are A1  2 and A2 1 and the final solution is

t ytt 2(4)3 

Case 2. Single real root

b If bb2  4 , there is only one real root a   1 . Then the complementary function is 12 2 tt yyyAaAtac 12  1  2

Example: Solve the second-order difference equation yyyttt21 28. Here we have b1 2 , b2 1, and c  8 . Also, bb121 implies a particular integral of the type c 8 yttt2224 p 22 The characteristic equation is aa2 210 with a single root a 1. The general solution is, 2 therefore, yyytcp  AAtt12 4 .

Case 3. Complex roots

2 When bb12 4 , again a pair of conjugate complex numbers amni1,2   obtains where

634 Problems Book to Accompany Mathematics for Economists

b 4bb 2 m  1 and n  21 2 2 The complementary function is

tt t t yAaAaAmniAmnic 11 2 2  1()()  2 

From the De Moivre’s theorem it follows that ()(cossin)mniRtt ti  t where

bbb224 R mn22 121  b 4 2 m b n b2 Here  is measured in radians and cos 1 and sin  1 1 . Hence, the R 2 b2 R 4b2 complementary function is

ttt yARtitARtitRBc 12(cos sin) (cos sin)(cos 12 tB sin) t where the multiplicative factor Rt substitutes the natural exponential term emt used in differential equations.

Example: Find the general solution to the equation yyyttt21 3914. Here we have b1 3 , b2  9, and c 14 . For the particular integral, c 14 y p 2 1139bb12 

2 This is the case of complex roots since bb12 4 , therefore, Rb 2 93. b 31 b2 93 cos 1   and sin  11  1  2 b2 2(3) 2 44(9)2b2  From the respective values of the two trigonometric functions, we infer that   . Hence, the 3 general solution is

t  yyytcp 3cossin2 B12 tB  t  33

Dynamic Stability

For the second-order difference equation

ybybycttt2112 we have the term at , which may show oscillatory behavior depending on the value of the base a . In the case of two distinct roots, if a1 1 and a2 1, both terms in the complementary function tt yAaAac 11 2 2 will be explosive and the time path of yt is divergent (the time path is oscillatory, if a  0 ). When a1 1 and a2 1, both terms will converge to zero as t  and the time path is convergent. If either a1 or a2 is greater than 1, the time path is divergent. With a single root, the function is dynamically stable if a 1. For the case of complex roots, we found the solution to be

t yRBtp(cos12 tB sin) ty

Chapter 11. Advanced Differential and Difference Equations 635

The parenthesized expression shows a fluctuating path because it contains circular functions. The fluctuation would be a stepped (nonsmooth) fluctuation, rather than oscillation. If R 1, the time path would be dynamically stable. Since R is the absolute value of the conjugate complex roots mni , the condition for convergence is again that the characteristic roots be less than 1. In all cases, the time path of yt will be dynamically stable if the absolute value of every root is less than 1.

The Multiplier-Accelerator Model

The multiplier-accelerator model shows the interaction between aggregate investment and output. Usually, in the presence of positive exogenous shocks, increased investment has a multiplying effect on GDP by the amount of the investment multiplier, but the increase in GDP makes firms believe that demand for their goods has increased. This stimulates them to invest more in capital stock, a process known as the accelerator. Thus, investment stimulates GDP through the multiplier process while GDP further pushes up investment through the accelerator process in an interactive way. Of course, a downturn in the economy would have an effect opposite to the multiplier-accelerator process or would force the economy to contract. The model was first advanced by Paul Samuelson, who extended the Keynesian national-income model of the investment multiplier by the accelerator principle.6 The model assumes the following three equations:

YCIGttto

CYtt  1 01

ICCttt()1   0

People spend based on income earned in the previous period where  shows the share of income that is consumed, that is, the marginal propensity to consume. Furthermore, investment is positively related to the increase in aggregate consumption CCCttt11() showing here the accelerator effect. That is, based on increased consumption, firms expect demand for their product to rise and, hence, decide to increase investment. Note also that the parameter  is called an accelerator coefficient and is greater than zero. Substituting the respective terms for Ct in the last equation, we obtain

IYYttt()12 and substitute this new result and the second equation into the first one:

YY  Y   YG  tt112 t t o

YYYGttto21 (1 )  

The parameters in this second-order difference equation are b1   (1 ), b2  , and cG o . We can easily find the particular integral as

c GGoo Yp   11(1)1bb12    

Since  is less than 1, we could expect a meaningful intertemporal equilibrium for national income 1 that is positively related to exogenous government spending. Furthermore, is the value of the 1  multiplier. The characteristic equation of the model is  (1 )22 (1  )  4  aa2 (1 )   0 where a  1,2 2

6 Samuelson, Paul A. “Interaction between the Multiplier Analysis and the Principle of Acceleration.” Review of Economic Statistics, May, 1939, pp. 75-78; reprinted in American Economic Association, Readings in Business Cycle Theory, Richard D. Irwin, Inc., Homewood, Ill., 1944, pp. 261-269

636 Problems Book to Accompany Mathematics for Economists where we know the two roots satisfy the conditions

aa12 (1  ) and aa12 

Thus,

(1aa12 )(1  )  1 ( aaaa 1212  )   1 (1  )   1

which implies that 0(1)(1)1aa12  . Furthermore, for the complementary function, there are 2 three possible cases depending on whether bb12 4 or not. This first case is equivalent to

 22(1 ) 4 4   (1 ) 2

In this first case of distinct real roots since aa12 0 and aa12  0 , both roots are positive. This precludes oscillation, and convergence would depend on whether a1 and a2 are smaller or bigger than 1. Several cases might be considered, but the legitimate ones are presented in Table 2. Similar is  (1  ) the case of a single real root a  , which is positive again. Oscillation is excluded and the 2 dynamic stability of national income depends on whether a is smaller or bigger than 1. In the case of conjugate complex roots the presence of Rb2  determines stepped fluctuation. If R 1, the fluctuation would be narrowed down, while for R 1 we would have explosive growth. These conclusions are summarized in the table below

Case Subcase Time path of Yt Dynamic stability 1. Distinct real roots 4 01aa Nonoscillatory Convergence   12 (1  ) 2 4 1aa Nonoscillatory Divergence   12 (1  ) 2 2. Single real root 4 01a Nonoscillatory Convergence   (1  ) 2 4 a 1 Nonoscillatory Divergence   (1 ) 2 3. Complex roots 4 R 1 Stepped Convergence   (1 ) 2 fluctuation 4 R 1 Stepped Divergence   (1 ) 2 fluctuation

Table 2

In conclusion, the time path of national income is convergent only if  1 in all cases. Furthermore, the model shows that it is possible for national income to have cyclical fluctuations endogenously without any external shocks present, but merely due to the interactive play between the multiplier and the accelerator process.

Chapter 11. Advanced Differential and Difference Equations 637

Inflation and Unemployment in Discrete Time

Recall that the model of the augmented Philips curve showing the relationship between inflation and unemployment in continuous time, that is, with differential equations, took the form:

p  Uh  ,0  01 h  d jp()  01 j  dt dU km()  p k  0 dt

We can transform this model in a discrete-time form so it becomes

pttt Uh  ,0  01 h 

tt1 jp() tt  01 j 

UUkmptt11() t k  0

In solving the discrete-time model, we notice the difference terms for expected inflation  t and unemployment Ut that obtain in the second and the third equation. To take advantage of these differences, we can further express the difference for actual inflation, which is

pttpp1  t where we substitute pUhttt111  

pptt111  Uh t  t  Uh tt  ()() UUh t  11  t   tt  and substituting the last two equations of the model into this expression,

ppkmphjptt11 ()()   t   tt 

(1kp )ttt1 (1 hjp ) hj km 

Since the  t term appears in the difference equation for price, we can express it from the first equation of the model as hptt  U t and substitute:

(1kp )tttt1 (1 hjp ) jp j jU  km

(1kp )ttt1 (1 j hjp ) jU kmj 

We still have one term Ut to get rid of. In order to do that, we can extend the upper equation by one period so that it becomes

(1kp )ttt211 (1 j hjp ) jU kmj 

From the monetary policy equation in the original model, we know that the difference for unemployment is UUkmptt11(), t and we use this when subtracting the last two equations:

(1kp )ttttt21 (1 j hj 1 kp ) (1 j hjp ) j ( U 1 U ) 0

After substituting for the difference of unemployment and some further transformations, we reach the following second-order difference equation in p :

(1kp )ttttt21 (1 j hj 1 kp ) (1 j hjp ) j ( U 1 U ) 0

Here the parameters are

638 Problems Book to Accompany Mathematics for Economists

1(1)(1)hj j  k (1 jhj ) jkm b  b  and c   1 1 k 2 1 k 1 k

The particular integral can be immediately found as

cjkm  pp pm    1bb12 jk

The characteristic roots must satisfy the conditions

1 hj (1jhj ) aa b 1 j aa b 12 11 k 12 2 1 k

From the values of the respective parameters, we can conclude that

aa120 aa12(0,1) and (1 aa12 )(1 ) 0

Hence, a1 and a2 both are positive fractions and the time path of inflation p is convergent and 2 2 nonoscillatory with distinct or repeated roots – that is, when bb12 4 . If bb12 4 , we have complex roots where R  b2 . Since b2 itself is a positive fraction, so must be R which renders the time path of p convergent in the form of stepped fluctuation.

We can as well study the behavior of unemployment U in time. From the last equation of the model,

UUkmptt11() t

But from the expectations-augmented Philips curve equation, we know that pUhttt111   . Substituting this expression in the difference equation for unemployment gives

UUkmUhtt111()   t  t

Extending this by one time period,

UUtt21 kmUh()   t  2 t  2 and subtracting the last two equations, we obtain

(1kU )ttttt2121 (2 kU ) U kh ( )

From the adaptive-expectations equation we know that

tt21jp() tt  11 

So, substituting this difference term for expected inflation results in

(1kU )ttttt21 (2 kU ) U khjp (  11 ) where pUhttt111  

(1kU )tttttt21  (2 kU ) U khj ( U  111 h ) or

(1kU )tttt21 (2 k kjhU ) U khj  khjh ( 1)   1

From the original difference equation for unemployment, we express the term t1

UUtt1  mUh  tt11  k

(1kU ) tt1 U km  k  t1  hk Substituting this finally gives a second-order difference equation solely in U .

Chapter 11. Advanced Differential and Difference Equations 639

(1kU ) tt1 U km  k (1kU ) (2 k kjhU ) U khj khjh ( 1) ttt21 hk

(1k ) U (2 k kjh ) U j ( h 1)(1  k ) U (1 j jh ) U tttt211 khj jk(1)() h m

(1kU )ttt21 1 hj (1 k )(1 j ) U (1 j jhU ) kj h ( h 1) m h  1(1)(1)hj k  j(1jjh ) kj  (1) h m  UUU ttt21111kkk

The parameters here are

1(1)(1)hj k  j (1jjh ) kj (1) h m  b  b  and c  1 1 k 2 1 k 1 k

It can be checked that the intertemporal equilibrium level of unemployment is

chm (1) UUp   1bb12 

Since in a state of general equilibrium the equilibrium rate of inflation was found to be exactly the growth rate of money m , the last equation could be written as   (1)hp U   

Therefore, the equilibrium inflation and unemployment must be negatively related in the long run, a relationship which we previously denoted as the long-run Phillips curve. In the special case of h 1, the p term will drop out of the equation and the unemployment rate will become a constant. This will give rise to a vertical long-run Phillips curve with the unemployment rate plotted on the horizontal axis. This fixed value of unemployment which represents the natural rate of unemployment teaches economic policy-makers that inflation and unemployment may be unrelated in the long run. If h 1, the coefficient of p will be negative and the long-run Phillips curve will be negatively sloped. Thus, the value of the h parameter determines the trade-off between inflation and unemployment. By definition this parameter measures the degree to which expectations form actual inflation, that is, the interrelationship between expected and actual inflation. Thus, as stronger expectations of higher inflation form in the nation and penetrate the wage structure of the economy, there will be little interdependence between inflation and unemployment and little the government can do. With lower inflationary expectations, the potential for government policies to take advantage of the trade-off between inflation and unemployment increases.

Higher-order Difference Equations

To find the particular integral of the nth order difference equation with constant coefficients and a constant term,

ybytn11 tn... bybyc n 11 t n t

2 we again try solutions ykt  , yktt  , yktt  , etc. As to the complementary function, an n th- degree characteristic equation obtains

nn1 aba11... babnn 0

with n characteristic roots ai ( in1,2,..., ). If some of the roots are repeated (for instance, the first two), and the next two are complex numbers, the general solution would be

640 Problems Book to Accompany Mathematics for Economists

ttt yAaAtaRBtp11 2 1 (cos 1 tB  2 sin) ty 

At least n initial conditions are necessary to find the values of the n arbitrary constants.

Example: Find the general solution of the third-order difference equation

511 yyy y3 ttt3214216 t

Trying a solution of the type ykt  , we find the particular integral to be

c 3 3(16) y 16 p 1162081bbb 51 1   1231 4216

The characteristic equation is

511 aaa320 4216 1 Factoring out the term a  , we transform this cubic equation into 4 2 11 1 1 aa0 with roots aa12  and a3  . Hence, 42 2 4 ttt 111   yAt 12 At   A 3  16 224  

Since aaa123,, 1, yt converges to the stationary intertemporal equilibrium of 16.

Problems

1. The logistic model of population growth (also known as the Verhulst model) assumes that the growth rate of a population decreases as this population grows in size. Similar to Malthusian growth, it assumes that there are limits to the increase of human population. This might be due to a decline in the arable or other land as a fixed input, the depletion of nonrenewable resources, crowding, and the eventual spreading of epidemics that could reduce the human population considerably. Thus, if y is the cumulative world population and its growth rate is y , according to the model

yaby  ab,0

Find the time path of human population using the logistic model.

Solution:

We can rewrite the equation as

1 dy aby or ydt 1 a dy bdt ybay

Integrating both sides of the equation,

Chapter 11. Advanced Differential and Difference Equations 641

y ln bt c bay and taking the antilog of both sides,

y y  eecbt or  Aebt bay bay Abebt which transforms into y  . To definitize the A constant, we set t  0 1 Aaebt Ab y(0) y(0)  which gives A  1 Aa bay (0)

Substituting A and transforming further gives the definite solution of population

by(0) ebt yt() bay(0) aye (0) bt

2. The population of a country grows according to the logistic equation yaby   where the rate of change of the population with time is dy dt and ab,0 . Find the equilibrium size of this population – that is, the one for which the rate of change is zero.

Solution:

In order for the population to be in equilibrium we need to have yt() 0. From the solution obtained by(0) ebt previously, we have yt() , so differentiating with respect to t , bay(0) aye (0) bt

b222 y(0) ebt b ay (0) ay (0) e bt  ab y (0)2 e bt   yt()2 bt bay(0) aye (0) b22222 y(0) ebt b ay (0)  ab  y (0)22 e bt  ab  y (0) e bt 2 bt bay(0) aye (0) by2 (0) ebt  b ay (0) 2 0 bt bay(0) aye (0) b which implies that the expression bay (0) in the numerator should be 0. Thus, we obtain y(0)  a for the initial condition in equilibrium. Substituting to find that equilibrium value, be2 bt b yt() ab() b bebt a

Thus, we have found that the equilibrium population size is ba. Depending on whether the initial population is less than or greater than ba, there will be growth or decline in the population according to the equation of logistic growth.

642 Problems Book to Accompany Mathematics for Economists

3. For the logistic growth of the population discussed in the previous problem, prove that the maximum rate of growth occurs when the population is equal to half its equilibrium size, that is, when the population is ba2 .

Solution:

In order for the growth rate to be maximum we need to find the maximum of dy dt . This means that dy2 we can express and set it equal to zero. From the logistic equation, we have dt 2 dy yb() ay dt by(0) ebt or given the solution for y  bay(0) aye (0) bt

bt 2 by(0) ebt aby (0) e bt by(0) e b aby (0)    yt()bt b bt 2 bay(0) aye (0) bay  (0) aye (0)  bt  bay(0) aye (0) 

Differentiating further to find yt(),

by(0) b2  aby (0) bebt b ay (0) ay (0) e bt 2 ay (0) e bt     yt()3 bt bay(0) aye (0)

3 bt bt by(0) e b ay (0) b ay (0) ay (0) e 3 0 bt bay(0) aye (0) which implies that the numerator would be zero when bay(0) 0 or bay(0) aye (0)bt  0

But the first case implies that y  0, which cannot give the condition for maximum growth. We have also discussed this case. Thus, we review the second, which gives

b y(0)  ae(1 bt )

Substituting the initial value in the expression for yt(),

be22bt be bt b yt() bt bt bt bbe abe2 2a aeb(1 ) bt bt 11ee which proves that the maximum growth rate of the population is achieved when population is ba2 .

4. Imagine that the Isle of Timbuktu has an initial population of 100,000 and an equilibrium population of 1 million. The population is known to have a logistic growth pattern. Statisticians count that at the end of one year the population doubles from the initial moment, that is, there are 200,000 citizens of Timbuktu. Determine the time path of Timbuktu’s population. What is the time at which the population is increasing most rapidly?

Chapter 11. Advanced Differential and Difference Equations 643

Solution: dy From the general result for the logistic equation of the type yb(), ay the time path of the dt by(0) ebt function is given by yt() . Here we have bay(0) aye (0) bt b y(0) 100,000 y 1,000,000 y(1) 200,000 a by(0) ebt yt() b ayye(0) (0) bt a

Substituting in the equation,

1,000,000(100,000)eebt 106 bt yt() (1,000,000 100,000 100,000eebt ) 9  bt

To find b, we use that y(1) 200,000 106 eb y(1) 200,000 which transforms into 9  eb

59eebb 49eb  eb  2.25 b  ln 2.25

Hence, the time path for the Timbuktu population is

106ln2.25et yt() 9  et ln 2.25 b Maximum growth would be achieved when y  or y  500,000 . Substituting in the obtained 2a function, 106ln2.25et 500,000  9  et ln 2.25 2et ln 2.25 1  9  et ln 2.25 92eettln 2.25 ln 2.25 et ln 2.25  9

Taking the log of both sides, t ln 2.25 ln9 ln 9 t 2.7 years ln 9 ln 4

The population will grow most rapidly after 2.7 years.

5. A textile factory has 300 workers, all of whom are vulnerable to the Brisbane flu virus. An epidemic is known to have spread out where the number of infected workers is I . The rate of change with respect to time of the number of infected workers is proportional both to the number of infected and the number of uninfected, that is, 300  I , according to the equation

644 Problems Book to Accompany Mathematics for Economists

dI  kI(300 I ) dt where k is a constant of proportionality. Find the number of infected people at time t days, if at time t  0 one worker becomes infected. If k  0.01 , find the value of the rate of new cases It() after 3 days, that is, I(3) .

Solution:

The equation can be rewritten as dI IkkI(300 ) dt which implies bk 300 and ak . We also know that I(0) 1. Substituting in the general solution, bI(0) ebt It() baI(0) aIe (0) bt

300ke300kt 300 e 300 kt It() 300kkke300kt 299  e 300 kt

Setting k  0.01 , we have

300e3t It() 299  e3t

Differentiating to find It(),

336ttt 300 3eee (299 ) 3 900(299)ee33tt 269,100 It() (299eee32ttt ) (299 32 ) (299 32 )

After 3 days, the rate of new cases is

269,100e9 It() (299 e92 )

6. The growth rate of a certain population depends on the supply of food which changes seasonally. The growth of the population is given by the equation

dp  cp()cos t t dt where c is a positive constant. Solve this simple model of seasonal population growth in terms of an initial population p(0) . Analyze the time path of the population function.

Solution:

We can rearrange the equation in the form

1 dp  ctcos pdt

Integrating both sides with respect to t ,

Chapter 11. Advanced Differential and Difference Equations 645

1 dp dt ccos tdt pdt dtsin lnp ctc sin since  cost 1 dt Taking the antilog of both sides,

pt() ectsin ec1 Ae ct sin

To definitize A, we set t  0:

p(0) Aecsin 0 Ae 0 A where sin 0 0 which gives the definite solution for population:

pt() p (0) ectsin

Since sint is a circular function, it fluctuates around -1 and 1. Hence, the population function would fluctuate between p(0)ec and p(0)ec . Furthermore, the equilibrium value of population p is dp obtained when the rate of change is zero. This would be, if dt dp cp()cos t t 0 or cost  0 dt  3  3 This can occur at t  or t  when sin 1 and sin  1 . Therefore, the population 2 2 2 2 function takes values p()tpe (0) c and p()tpe (0) c in a stationary state. Thus, it is oscillating between these equilibrium values.

7. If the demand and supply functions for a commodity are given by qpd     and qnts   sin , use the model of market price dynamics to determine p()t and analyze its behavior as t increases.

Solution:

In accordance with the model,

dp jq() q dt ds

Substituting for the demand and supply functions,

dp jpnt(sin)  dt

dp jp j(sin) nt dt

sinnt jt  sin nt pt() p (0) e   As we would normally expect  and the adjustment coefficient j to be positive, the time path of market price is convergent with time. The equilibrium value of price is

 sin nt p  

646 Problems Book to Accompany Mathematics for Economists

Since the equilibrium price contains a circular function, the values of which fluctuate between -1 and       1, the price is oscillating between two stationary values, and .  

8. Consider the following demand and supply functions for a commodity:

qpppd 22 3 4   qps 32  with initial conditions p(0) 7 and p(0) 4 . Find the time path of market price p()t and determine whether price converges to its intertemporal equilibrium.

Solution:

The concrete values of the parameters are

  22   3   3   2 u  4 v 1

Equating the two market forces,

22 3p 4pp  3 2 p

ppp45  25

We can find the intertemporal equilibrium given by the particular integral either directly from the differential equation or by the formula

  22 3 p  5 p 32

Since v  0 , the only feasible case is that of distinct real roots. Solving for the characteristic roots by the formula 2 111uu   r1,2  4    4  16  4(3  2)  (4  36)  5, 1 222vv  v   

Thus, the general solution is

5tt pt() pcp p  Ae12  Ae  5

To definitize the constants A1 and A2 , we use the initial conditions:

00 pAeAe(0)12 5 7 or AA12  2

Differentiating to find the first derivative of price,

5tt p()tAeAe 5 12 or pAA(0) 512 4

which gives AA121, and the definite solution is pt() e5tt e 5

Since one of the characteristic roots is positive ( r1  5 ), the intertemporal equilibrium of 5 is dynamically unstable.

9. The following market model is given:

Chapter 11. Advanced Differential and Difference Equations 647

qpppd 38 3 6  2 qps 10  5 where p(0) 11 and p(0) 5. Find the time path of market price p()t assuming that the market clears at every point of time.

Solution:

We can solve again by substituting the values for the different parameters, but one other way is to solve the characteristic equation.

38 3p 6pp  2  10 5 p

268ppp  48 ppp34  24

a1 3 a2 4 b 24

The intertemporal equilibrium is given by the particular integral:

b 24 pp  6 a2 4

Here u  0 and v  0 , so only the case of distinct real roots is possible. Solving for the characteristic roots by the formula

2 rara120 or rr2 340

aaa2 4 11 r 112(3 9 16) (3 5) 4, 1 1,2 22 2 which produces one positive root so the intertemporal equilibrium of 6 for price is dynamically unstable. The general solution becomes

4tt pt() Ae12 Ae  6

To find the constants A1 and A2 , we set t  0 .

00 pAeAe(0)12 6 11 or AA12  5

4tt p()tAeAe 4 12 or 45AA12 

so A1  2 and A2  3 . The definite solution is

pt() 2 e4tt 3 e 6

10. For the following demand and supply functions

qpppd 83   qps 2  where p(0) 8 and p(0) 4 , express the general and definite solution for price p()t assuming market equilibrium at every moment.

648 Problems Book to Accompany Mathematics for Economists

Solution:

83p pp   2 p

ppp3210 

a1  3 a2  2 b 10 u  3 v  1

b 10 pp 5 a2 2

Here uv,0 , so the intertemporal equilibrium of 5 must be dynamically stable and the time path of price should be convergent. The characteristic equation is rr2 320

aaa2 4 11 r 112(3 9 8) (3 1) 2,1 1,2 22 2

As we expected, both characteristic roots are negative, which ensures the dynamic stability of the time path of price. Solving further,

tt2 pt() Ae12 Ae  5

To find the constants A1 and A2 , we set t  0 .

00 pAeAe(0)12 5 8 or AA12  3

tt p()tAeAe12  2 or AA1224

which results in A1  2 and A2 1 . Thus, the definite solution is

pt() 2 ett e2 5

11. If the time path of price stems from the equation ppp 4412, what is the general solution for p()t ? Is the time path of price likely to be convergent or divergent? Assume that p(0) 4 and p(0) 1 to find the definite solution.

Solution:

We can immediately determine the parameters:

a1  4 a2  4 b 12

b 12 pp 3 a2 4

We must have u 4 and v 1 in order for the right signs of the other parameters to obtain (check why with some hypothetical demand and supply functions). Then, the intertemporal equilibrium of 3 must be dynamically stable. The characteristic equation is rr2 440

aaa2 4 r 112(2 4 4) 2 1,2 2

Chapter 11. Advanced Differential and Difference Equations 649

We get case 2 of a single real root. Since this root of -2 is negative, this ensures the dynamic stability for the price function. Therefore, the general solution is

22tt pt() Ae12 Ate  3

To find the constants A1 and A2 , we set t  0 .

0 pAe(0)1 3 4 or A1 1

22tt  t p()tAeAeAte 212   2 2 or 21AA12

so A1 1 and A2  3 . Therefore, the definite solution is

pt() e22tt 3 te  3

12. Consider the following demand and supply functions:

qpppd 20 5 6   qps 74 

Assume market equilibrium at any point in time and find the time path of p()t .

Solution:

Equating demand and supply,

ppp6927  b pp 3 a2 Here u 6 and v 1; therefore, the intertemporal equilibrium of 3 must be dynamically stable. The characteristic equation is rr2 690 which can be written as (3)0r  2  or r 3 . Since it is a negative single root we get, we can be sure that the time path of the price function is convergent. The general solution is

33tt pt() Ae12 Ate  3

13. In the following market model,

qpppd 20 3 4   qps 52  where p(0) 7 and p(0) 1, express the general and definite solution for price p()t assuming market equilibrium at every moment.

Solution:

20 3p 4pp  5 2 p

ppp4525  b 25 pp 5 a2 5

Here u 4 and v 1, so all the three cases are feasible. The intertemporal equilibrium is dynamically stable. The characteristic equation is

650 Problems Book to Accompany Mathematics for Economists rr2 450

ri1,2 2452      12  

Here the roots are a pair of conjugate complex numbers of the type rmni1,2   where the real part is m 2 and the imaginary is n 1 . As we expected, a negative real part ensures the dynamic stability of the time path of price. Using the formula for complex numbers, we can write the general solution

2t pt() e ( B12 cos t B sin) t 5

To definitize the constants, we set t  0 .

0 peBB(0) (12 cos0 sin0) 5 7 cos0 1 sin 0 0 or B1  2

22tt p()teBtBteBtBt 2 (12 cos  sin)  (  12 sin  cos)

00 peBBeBB(0) 2 (12 cos0  sin 0)  (  12 sin 0  cos0)  2 BB 12   1

which gives B1  2 and B2  5

pt() e2t (2cos t 5sin) t 5 2 With circular functions in the solution, the time path is periodic fluctuation with a period or 2 . n The price performs a full cycle every time t increases by 2 . Since m  2 , the fluctuation is damped, and price converges to the intertemporal equilibrium of 5 in a cyclical and damped way.

14. Given the demand and supply functions on a market for a commodity

qpppd 11 3 qppps 44    5 where p(0) 7 and p(0) 5, determine whether price fluctuates in time. Is the fluctuation explosive or damped? Assume the market clears at every moment.

Solution:

11p pp  3  4 4 ppp    5

22515ppp 

pp  2.5 p  7.5

7.5 p 3 p 2.5

The characteristic equation is rr2 2.5  0

1113 rii1110(13) 1,2 2222  1 3 The roots are a pair of complex numbers with a real part m  and an imaginary one n  . Hence, 2 2 the general solution is

Chapter 11. Advanced Differential and Difference Equations 651

t 2 33tt pt() e B12 cos B sin 3 22

To definitize the constants, we set t  0 .

0 peBB(0) (12 cos0 sin0) 3 7 cos0 1 sin 0 0 or B1  4

t 2 ett33t 2 33BB12 3 t 3 t pt() B12 cos B sin e  sin cos 222 2222

0 e 0 33BB12 BB 12 3 pBBe(0)12 cos0 sin 0 sin 0 cos0 5 22222

BB12310 which results in B1  4 and B2  2 t 33tt pt() e2  4cos 2sin 3 22 22(3) The price function fluctuates periodically with a period 3 . The price performs a full n 2 1 cycle every time t increases by 3 . Since the real part is positive or m  , the fluctuation is 2 explosive, and price diverges from its intertemporal equilibrium of 3 cyclically.

15. Consider an expanded market equilibrium model that takes into account both buyers’ and sellers’ expectations of price change such that

qpupvpd   11  ,0  qpupvps   22   ,0 

Assuming the market is always in equilibrium, express the time path of price. Find also its intertemporal equilibrium and determine how it is influenced by the expectations of market participants. Under what circumstances could a single real root and periodic fluctuation be ruled out?

Solution:

Equating demand and supply,

 p up11  vp  p up 2   vp 2

()()()vvp12 uup 12    p

Let vv12 v and uu12 u. Normalizing the equation, we get

u ()  pp  p  vv v

As with the simple market equilibrium model, the intertemporal equilibrium is

   p  p   

We can see that it does not contain any of the coefficients uvii, ( i 1, 2 ) that reflect the expectations of the market participants. Hence, the intertemporal equilibrium price does not depend on the expectations of either market group, but depends solely on current price. This is quite logical; the time path shows a changing equilibrium price at any moment in time. This dynamically changing equilibrium depends on people’s expectations. The intertemporal equilibrium price, though, does not

652 Problems Book to Accompany Mathematics for Economists depend on trends or short-term expectations and is the normal-level pice throughout a long period of u    time. Since a  and a  , the characteristic roots are 1 v 2 v 2 1 uu     r1,2  4   2 vv  v  

The roots are identical with the simple model but note that here u and v are products of both the demand and the supply functions. Thus, the general solution in the three known cases is

Case 1. Distinct real roots

2 u     4   vv      pt() p p  Aert12  Ae r t  cp12  

Case 2. Single real root

2 u     4   vv   a u The single root is r 1  . 22v ut ut    pt() Ae22vv Ate  12  

Case 3. Complex roots

2 u     4   vv  

The characteristic roots are a pair of conjugate complex numbers and 2 u 1  u m  and n 4  2v 2 vv ut     pt() e2v ( B cos nt B sin nt ) 12      To rule out the last two cases, we should have v  0 ; that is, vv12  0 or vv12 . Then 4 v is always negative, so cases 2 and 3 are impossible.

16. With the price of real estate rising at an increasing rate in Bulgaria ( pp,0  ) at the time of the real estate boom, the construction business felt stimulated to supply many more new buildings. Expecting prices to rise further and profits to grow higher, builders built more and more intensively. The model, with consumer expectations ignored, becomes

qpd   ,0  qpupvps      ,0  uv,0

Assuming the market is always in equilibrium, express the time path of price. Find also its intertemporal equilibrium. Which of the three familiar cases are possible and dynamically stable?

Chapter 11. Advanced Differential and Difference Equations 653

Solution:

Assuming equilibrium on the housing market,

 p pupvp vp up  ()  p 

u ()  pp  p  vv v

As with the simple market equilibrium model, the intertemporal equilibrium is

   p  p   

u which again is independent of the expectations of the Bulgarian builders. Since a  and 1 v    a  , the characteristic roots are 2 v 2 1 uu     r1,2  4   2 vv  v  

Case 1. Distinct real roots

2 u      4   vv      pt() p p  Aert12  Ae r t  cp12  

Case 2. Single real root

2 u      4   vv   a u The root is r 1  . 22v ut ut    pt() Ae22vv Ate  12  

Case 3. Complex roots

2 u      4   vv   ut     The solution is pt() e2v ( B cos nt B sin nt ) 12  

In view of the positive u and v , all the three cases are possible. Furthermore, in case 1 both characteristic roots are negative. (Can you tell why?) So is the single root in the second case. Lastly, in case 3 the real part of the characteristic roots h is negative. This is sufficient for the time path of price to be dynamically stable in all the three cases. Hence, the condition uv,0 denoting the optimistic expectations of the Bulgarian builders guarantees the dynamic stability of price of real estate in the Bulgarian construction market.

654 Problems Book to Accompany Mathematics for Economists

17. The national income of a country is changing according to the equation Yt() 8 Yt () 32. Find the time path of national income and its intertemporal equilibrium, if it exists. Say whether income is converging to this equibrium. Assume initial conditions of Y (0) 5 and Y(0)  12 . Express the national income in the third period.

Solution:

This is a second-order differential equation for which

a1  8 a2  0 and b  32

Using the formula for the particular integral, we have

b 32 Ytttp  4 a1 8

Since Yp is a function of t , this is not a stationary, but a moving equilibrium. Furthermore, since every next moment of time t is bigger, the equilibrium value is expected to grow. The characteristic equation is

2 rar1 0 rr()0 a1

which gives the characteristic roots r1  0 , ra21 8 . Thus, the general solution for national income is

rt12 r t (0)tt 8 8 t Yt() Ae12 Ae  Yp Ae 1 Ae 2  4 t A 12 Ae 4 t

To specify the constants,

8(0) YAAe(0)12  4(0)  AA 12  5

8t Yt() 8 Ae2  4 0 YAeA(0)822  48   412  or A2  2 and A1  3

The definite solution is

Yt() 3 2 e8t  4 t

Since the nonzero characteristic root is negative, national income converges to its intertemporal equilibrium – but it is a moving equilibrium. Furthermore, the income of the nation is increasing with time starting from a positive initial level of 5. In the third period, the income in this growing economy is Ye(3) 3 28(3)  4(3)  15  2 e 24

18. The increase in the rate of change of national income of a country is given by Yt() 14. Find the time path of national income and its intertemporal equilibrium, if it exists. Assume that in the initial moment t  0 the national income is 15. How is the income of this country changing with time? Check the first and the second derivative of the national income function.

Solution:

This is a second-order differential equation for which

a1  0 a2  0 and b 14

Chapter 11. Advanced Differential and Difference Equations 655

Therefore, using the formula for the particular integral, we have

b 14 Yt222 tt 7 p 22

Since Yp is a function of t , this is not a stationary but a moving equilibrium, and it is rapidly increasing with time. Since rr120 , the general solution becomes

rt12 r t (0)tt (0) 2 2 Yt() Ae12 Ae  Yp Ae 1 Ae 2  7 t A 12 A 7 t

In the initial moment, Y (0) 15 . Hence,

YAA(0)12 7(0)  AA 12  15

We can write the national income function as

Yt() 15 7 t2

The national income takes a higher value in every next period. Thus, in the first period it is Y (1) 22 , while in the second it is Y (2) 43. From the definite solution, we can easily check the two consecutive derivatives Yt() 14 t and Yt() 14, which prove our calculations correct.

19. Let uw() be the utility function over wealth w . At any wealth level w , the Arrow-Pratt measure uw() of absolute risk aversion is the percent rate of change of marginal utility u at w so it equals  . uw() Assume a utility function that has a constant absolute risk aversion a . Find the general form of that utility function. Check that, indeed, it has a constant absolute risk aversion. What is the condition for marginal utility u to be positive?

Solution:

To find the particular utility function we have to solve the second-order differential equation

uw() a or uw() uw() auw  () 0

Here we have aa1  , a2  0, and b  0 . Hence, the particular integral is u p  0 and the 2 characteristic equation is rar0 which gives roots ra1,2  0, . Thus, the general solution is

aw uw() A12 Ae

To check for absolute risk aversion, we differentiate sequentially:

aw 2 aw uw() aAe2 and uw() aAe2

Taking the ratio of the two derivatives with a minus, we get a constant absolute risk aversion of exactly a :

2 aw uw() aAe2 aw a uw() aA2 e

656 Problems Book to Accompany Mathematics for Economists

aw In order for marginal utility to be positive for a positive total utility uw()  aAe2 0 implies that the constant A2 be negative.

20. In relation to the previous problem, assume now that the absolute risk aversion function is non- constant taking the specific form aw() bw where b is a positive parameter. Thus, when the individual’s wealth is increasing his risk aversion increases as well at the constant rate b . Find the marginal utility of wealth uw() and set the condition for it to be positive.

Solution:

uw() aw() bw uw() uw() bwuw  () 0

which is a second-order differential equation in marginal utility of wealth uw(). Setting uuww  (),

ubwuww 0 and resorting to separation of variables

u w bw uw

u w dw bwdw uw bw2 ln uc  w 2 and taking the antilog of both sides

bw22 bw  22c ueeCew  and uw  0 implies C  0 bw2  Integrating marginal utility u would give the total utility function uw() Ce2 dw. w 

21. The consumption function of an individual grows with time according to the equation Ct() 9 Ct  () 14 C 42. Find the time path of this person’s consumption given that C(0) 6 and C(0) 16 . What is the amount of his consumption in the second period?

Solution:

For the particular integral, we have

42 C 3 p 14 The characteristic roots are

9 81  4(14) 95 r 2, 7 1,2 22 Since both characteristic roots are negative, this implies dynamic stability for the time path of the consumer. For the general solution

rt12 r t 27tt Ct() Ae12 Ae  Cp Ae 1 Ae 2  3

Chapter 11. Advanced Differential and Difference Equations 657

To specify the constants,

oo CAeAeAA(0)12 3 12 3 6 AA12  3

27tt Ct() 2 Ae12  7 Ae 00 CAeAeAA(0) 212  7  2 12  7  16

which gives A1 1 and A2  2 . The definite solution is

Ct() e27tt 2 e  3

The consumption of the individual converges to the intertemporal equilibrium of 3. In the second period the consumption is

Ce(2)2(2) 2 e 7(2)  3 ee  4 2 14 3

22. The following differential equation gives the aggregate savings of a country St() 6 St  () 5 S 10. Establish the time path of aggregate savings for S(0)  4 and S(0) 2 . Is it dynamically stable? How do savings change from their initial level?

Solution:

For the intertemporal equilibrium of savings, we have

10 S 2 p 5

The characteristic roots are

r1,2 395325,1      

Since both characteristic roots are negative, the time path of savings is dynamically stable and converges to the equilibrium of 2. The general solution is

5tt St() Ae12 Ae 2

To specify the constants,

00 St() Ae12 Ae 2 A 12 A 2 4 AA12 6

For the first derivative, we have

5tt St() 5 Ae12  Ae

00 SAeAeAA(0) 512   5 12   2

which gives A1 1 and A2 7 . Therefore, the savings function is

St() e5tt 7 e  2

We can check the parental function in the initial moment

See(0)00 7  2 1 7 2 4 which is the initial value of savings. Since this initial value is negative, we can conclude that the nation borrows in the beginning. Since the time path is convergent, as time passes the savings become positive and converge to the equilibrium level.

658 Problems Book to Accompany Mathematics for Economists

23. The aggregate savings of a nation change according to the equation St() 6 St () 9 S 36. Find the time path of the aggregate savings function if S(0)  2 and S(0) 16 . What is the intertemporal equilibrium for savings, and is the nation moving to it or divergining from it? Interpret the results economically. Differentiate the definite solution to check the validity of your calculations.

Solution:

For the intertemporal equilibrium of savings, we have 36 S 4 p 9

The characteristic roots are r1,2 3993    . We obtain a single root that is negative, implying dynamic stability for the time path of savings. Thus, the nation is moving in the direction of this intertemporal equilibrium for savings. The general solution can be expressed as

rt rt33 t t S() t Ae12 Ate  Sp Ae 1 Ate 2  4

To specify the constants,

00 SAeAe(0)12 (0)  4 2 A1  6

Furthermore, for the first derivative from the general solution, we have

33tt  3 t St() 3 Ae12  Ae  3 Ate 2

00 SAeAeAA(0) 312   0  3 12   16

which gives A1 6 and A2 2 . The definite solution becomes

St() 6 e33tt  2 te  4

We can check the parental function and its first derivative using this definite solution.

See(0)600  2(0)4642     which is the initial level of savings. Note that since aggregate savings are initially negative, the nation must be a net borrower in the beginning. With the passage of time, savings become positive and converge to the equilibrium level of 4. For the first derivative from the definite solution, we have

Stetee() 1833333ttttt 6 2 16 ete 6 Setting t  0 ,

See(0) 1600 6(0)  16

Checking for S(0) and S(0), we get exactly the values given, which proves our computations correct.

24. Let the demand and supply functions be the well-known

qpupvpd     ,0  qps    ,0 

It is given that the market does not always clear but adjusts according to the relationship dp jq() q for j  0 where the change in the price depends on the level of excess demand by dt ds the amount of a positive adjustment coefficient j . Find and compare the intertemporal equilibrium price and the market-clearing equilibrium price. State the conditions for dynamic stability.

Chapter 11. Advanced Differential and Difference Equations 659

Solution:

Writing the differential equation,

pjpupvpp() 

pjjpjupjvp()()  

jvp(1)( ju p   j  ) p  j ( )

Normalizing,

(1)(ju   ) (  ) ppp  jv v v (1)uj (  ) ( ) ppp  vv v    We can easily find the intertemporal equilibrium by the well-known formula p  . The p    intertemporal equilibrium is identical to that under market clearance in every instant. Whether the market is continually in equilibrium or not, the intertemporal equilibrium stays the same. Furthermore, we see that the intertemporal equilibrium, unlike the market-clearing one, does not depend on (1)uj ()   expectation coefficients u and v . Here we have a  and a  . There are three 1 v 2 v possibilities, as with the simple market equilibrium model:

Case 1. Distinct real roots 2 uj1   4 vv

The characteristic roots are 2 11uj uj 1   r1,2   4 2 vv v 

The general solution is    pt() p p  Aert12  Ae r t  cp12  

Case 2. Single real root

2 uj1   4 vv

a (1)uj r 1  22v

The general solution is

(1)ujt (1) ujt    pt() Ae22vv Ate  12  

Case 3. Complex roots

2 uj1   4 vv

660 Problems Book to Accompany Mathematics for Economists

The characteristic roots are the complex numbers for which 2 2 a1 (1)uj 4aa21 11uj m   and n 4 22v 22vv For the general solution,

(1)ujt     pt() e2v ( B cos nt B sin nt ) 12   1 The condition for dynamic stability is v  0 and u  . All three cases considered are possible. j Under the above conditions, all roots, distinct or repeated, are negative. In the case of complex numbers, m is negative too. Therefore, the dynamic stability of the price function is ensured when 1 v  0 and v  . j

25. The following three equations of the Phillips relation model are given:

pU 32  d 1 ()p  dt 2 dU ()mp  dt

Find the time path of the expected rate of inflation and determine whether it converges to its intertemporal equilibrium. Express also the time paths of the price level and the unemployment rate. Discuss the relationship between inflation and unemployment in the short and in the long run.

Solution:

We have the following values of the parameters: 1   2 j  h 1 and k 1 2

Using the formulas for the coefficients aa12, and b , we obtain, respectively, 1 akjh (1 ) 2(1) (0)  2 1 2 1 ajk2  (2)(1)  1 bjkmm  2

The intertemporal equilibrium of the expected rate of inflation is  p  m . For the characteristic roots, aaa2 4 12 r 112 244 1 1,2 22 2 or we have the single-root case, so the time path of expected inflation can be written as

tt  ()tAeAtem12 

Since the root is negative, the time path of expected inflation is convergent to the growth rate of nominal money. Knowing the time path of  , we can find that of p . From the relationship 1 d p  we get jdt

Chapter 11. Advanced Differential and Difference Equations 661

d p 2  dt

Differentiating  ,

ttt  ()tAeAteAe12   2 and substituting in the expression,

ttttt  tt  t p2(AeAteAeAeAtemAeAteAem12   2 )  12   12   2 2 

tt (2AAeAtem21 )  2 

Similar to the expected rate of inflation  , the actual rate of inflation p also converges to the intertemporal equilibrium m . In fact, in an intertemporal equilibrium, the expected rate of inflation would equal the actual one, so there will be no change in expected inflation. From the first equation of the given model, we can also determine how unemployment changes with time. It is

33pp  U  222 Hence, the time path of the rate of unemployment is

13tt  tt  t U() t ( Ae12 Ate m Ae 12 Ate 2 Ae 2 m ) or 22 133 Ut() (2 Aettt 2 Ate 2 Ae ) Ae  ttt Ate Ae 22212 2 122

3 ()AAeAtett   12 2 2 which is also a convergent time path. Expected and real inflation both converge to the intertemporal equilibrium given by the growth rate of nominal money. If this growth rate changes, we will have a moving equilibrium, so the monetary policy of the government will affect the inflation levels. Note, though, that the unemployment rate does not depend on m . In this example, it converges to the 3 constant . This constant value is known as the natural rate of unemployment. The independence of 2 unemployment of any equilibrium rate of inflation is reflected by a vertical line known as the long-run Phillips curve. We would normally expect the short-run Phillips curve to be negatively sloped; that is, we would expect higher inflation to be accompanied by a low level of unemployment and vice versa. This short-term negative relationship can easily be deduced from the first equation of the model. The intuitive explanation is that in the business cycle at the time of a boom the economy is nearly at its full-employment level (not accounting for frictional unemployment). At that point, aggregate demand grows substantially exerting an inflationary pressure on prices. Conversely, when in a recession a lot of people are unemployed, aggregate demand falls, pushing prices down to their normal levels.

26. Consider the Phillips relation model

pU 53  d 2 ()p  dt 3 dU ()mp  dt

Find the time path of the expected rate of inflation, the actual rate of inflation, and the unemployment rate.

662 Problems Book to Accompany Mathematics for Economists

Solution:

For the parameters, we have 2   3 j  h 1 and k 1 3

For the coefficients aa12, and b, we have 2 2 akjh1  (1 ) 3(1) (0)  3 ajk2   (3)(1) 2 bjkmm   2 3 3

The intertemporal equilibrium of the expected rate of inflation is  p  m . For the characteristic roots, aaa2 4 131 r 112 394(2) 2,1 1,2 22 2 which are two distinct roots, so the solution is

tt2  ()tAeAem12 

The time path of expected inflation is convergent to the intertemporal equilibrium. Knowing the time 1 d path of  , we can find that of p . From the relationship p    , we get jdt 3 d p  2 dt

Differentiating  ,

tt2  ()tAeAe12  2 and substituting in the expression,

31tttt22  tt 2 p(2)Ae12  Ae  Ae 12  Ae  m Ae 12  2 Ae  m 22

Both the expected rate of inflation  and the actual rate of inflation p converge to the intertemporal equilibrium m , so they tend to be equal. From the first equation of the given model, we can also determine unemployment. It is

pU 53  55pp  U  333

Hence, the time path of the rate of unemployment is

11tt22  t  t 515  tt 2 U() t Ae12 Ae m Ae 1 2 Ae 2 m Ae 12 Ae 32 323

The unemployment rate converges to the natural rate of unemployment, which in this example 5 happens to be . 3

27. Let the three equations of the Phillips relation model be

pU 3  d 1 ()p  dt 4

Chapter 11. Advanced Differential and Difference Equations 663

dU 1 ()mp  dt 2

Find the time path of the expected rate of inflation, the real rate of inflation, and the unemployment rate. Do the functions converge to their equilibrium; and if so, how?

Solution:

We have the following values of the parameters: 1 1  1 j  h 1 and k  4 2 Consequently, 1 1 m akjh1  (1 ) ajk2    bjkm    2 8 8

Thus, the intertemporal equilibrium is  p  m . 2 aaa 4 11 14111 11 rii112       1,2  2 2 2 48 2 22 44 1 1 The roots are a pair of complex numbers with a real part, m   , and an imaginary one, n  . 4 4 Hence, the general solution for the expected rate of inflation is

t  4 tt  ()te B12 cos B sin m 44 1 d Knowing the time path of  , we can find that of p . From the relationship p    , we get jdt d p 4  dt t  t 4  ett4 BB12 tt  ()tBBe12 cos  sin   sin  cos 444 4444

Substituting for  and  , we obtain for the time path of actual inflation

t  4 tt p()te B21 cos B sin m 44

And for the unemployment rate,

Up  3 t  4 tt Ut() e ( B12 B )cos  ( B 12 B )sin 3 44 22(4)  All the functions fluctuate periodically with a period 4 . They perform a full cycle n 1 1 every time t increases by 4 . Since the real part is negative or m   , the fluctuation is damped 4 and the time path is dynamically stable. Both  and p converge cyclically to the intertemporal equilibrium equal to the monetary-policy parameter m , the rate of growth of nominal money. This is a moving equilibrium, since this rate of growth would change upon the discretion of the government. The rate of unemployment also fluctuates in a dynamically stable way, but around the natural rate of unemployment, which is equal to 3 here.

664 Problems Book to Accompany Mathematics for Economists

28. In the model given in the previous problem, assume that all coefficients are the same, except 1 h  . Find  ()t , p()t , and Ut() and analyze their time paths. Find also the intertemporal equilibria 3 of the variables. What do you notice about the intertemporal equilibrium of the unemployment rate? What can you conclude about the long-term Phillips curve for the new value of the h parameter?

Solution:

The model now becomes

1 pU 3  3 d 1 ()p  dt 4 dU 1 ()mp  dt 2

Rewriting the parameters, 1 1 1  1 j  h  and k  4 3 2 Hence, we obtain 11 1 112 1 m akjh1  (1  )  1   ajk2    bjkm    24 3 263 8 8

Thus, the intertemporal equilibrium for expected inflation is  p  m . aaa2 4 12 4412 2 1 2 rii112       1,2  22398236312 1 2 These are complex roots with m  and n  . The general solution for the expected rate of 3 12 inflation is t  22tt  ()te3 B cos B sin m 12 12 12 1 d From the relationship p  , we have jdt d p 4  dt t  t ett3 22 22BB 2 t 2 t  ()tB cos  B sin  e3 12 sin  cos 12  3121212121212

Substituting for  and  , we obtain the time path of actual inflation:

tt  422222ettett33 pt() B12 cos  B sin   B 12 sin  B cos  3121231212  t  22tt eB3 cos B sin m 12 12 12

Chapter 11. Advanced Differential and Difference Equations 665

tt  ettett332222 2 p()tB cos  B sin   BB sin  cos  m 12 12 3 12 12 3 12 12 tt  (2 1)33 2tt (2 1) 2 p()tBBeBBem (12 ) cos  ( 12 ) sin  312312

Hence, the equilibrium values of both expected and actual inflation are equal to the nominal money growth m . Since both variables are circular functions, they must be fluctuating in time. With the real part of the roots negative we have damped fluctuation, which means they converge to this equilibrium level. For the unemployment rate, 1 Up  3 3 Substituting the already obtained expressions,

t t 122(21)2 ttm   t Ut() e3  B cos B sin ( B B ) e3 cos 312123312 12 12  t  (2 1)3 2t ()sin3BBe12  m 312

tt  222ttmBBBB1122(21)() (21)()  BB 12 Ut() e33 cos  e sin  3 12 3 3 12 3 3 3 tt 12ttm 12 2 Ut() e33 cos (2 2) B ( 2 1) B e sin ( 2 1) B (2 2) B 3  31212 312 1 2 3

Unemployment is also a circular function that fluctuates in a damped way so it converges to its intertemporal equilibrium. Note, though, that now this equilibrium depends on m since it is 2m U 3  3 Hence, the equilibrium level of unemployment depends on the monetary (that is, inflationary) policy of the government and the long-run Phillips curve is no longer vertical. Since p  m , the inflation rate is negatively related to unemployment in equilibrium. A higher rate of actual inflation is accompanied by a lower unemployment rate. A lower rate of actual inflation may come at the expense of greater unemployment. We obtain this negative correlationship between inflation and unemployment in the long run, particularly when the coefficient h in the first equation of the model is chosen to be less than 1. Therefore, the generally vertical shape of the long-run Phillips curve is contingent on the 1 special value of this parameter. Since here h  , we have a negatively sloped long Phillips curve as 3 the result for the equilibrium unemployment rate show.

29. Solve the following Phillips relation model:

11 pU  54 d 1 ()p  dt 3 dU 3 ()mp  dt 4

Find  ()t , p()t , and Ut() and their intertemporal equilibria. What is the slope of the long-term Phillips curve for the specific value of the h parameter?

666 Problems Book to Accompany Mathematics for Economists

Solution:

The parameters are 1 1 3  1 j  h  k  3 4 4

For the coefficients a1, a2 and b, 31 1 31 13 1 m akjh1  (1 ) 1 1 ajk2    bjkm  43 4 44 34 4 4

The intertemporal equilibrium of the expected rate of inflation is  p  m .

aaa2 4 141 r 112 11 1,2  22 42 Since this is a single real root, we have tt  22  ()tAeAtem12  1 d From the relationship p  , jdt d p 3  dt

Differentiating  ,

tt 11  ()tAeAteAe22   t 22122 and substituting in the expression,

ttttt  3322222 pAe12212  Ate 3 Ae  Ae  Ate  m  22

ttt  11222 Ae122  Ate 3 Ae  m 22

For unemployment,

  41p U  45

Hence, the time path of the rate of unemployment is

tt t t t 11    U() t Ae22 Ate m 2 Ae 2 2 Ate 2 12 Ae 2 4 m or 12 1 2 2 45 tt t 11  Ut() 3 Ae22 3 Ate 12 Ae 2 3 m 12 2 45 ttt 33 13m Ut() Ae222 Ate  3 Ae  44122 54

13m U  54 Expected and real inflation both converge to the intertemporal equilibrium given by the growth rate of nominal money m . The unemployment rate also has a dynamically stable time path, but here the equilibrium is a moving one and depends on government monetary policy. Furthermore, we see that

Chapter 11. Advanced Differential and Difference Equations 667 the intertemporal equilibrium unemployment and inflation are negatively related. Therefore, the long- 3 run Phillips curve is not vertical, but is negatively sloped with a slope of  . This is because, as we 4 1 can notice from the model, h  . 4

30. Consider the simple inflation-unemployment model

p  TUh  ,0  01 h  d jp()  01j dt in which the third equation is dropped as unemployment U is assumed to be exogenous. What is the differential equation in  that obtains? Solve it to find the time path of  as well as its intertemporal equilibrium. How does this new equilibrium differ from the one for actual inflation rate p ?

Solution:

Substituting the first equation into the second,

d jTUh()  dt d jh(1 )  j (  TU ) dt which is a first-order differential equation in  . It is easy to solve using the definite solution formula

bbat ()te (0)  where aj (1 h ) and bj ()  T U aa

()()TUjht(1 )   TU ()te (0)  11hh

Thus, the intertemporal equilibrium of the expected rate of inflation is ()TU   1 h To find the equilibrium value of actual inflation rate, we substitute for  :

hTU() pTU   or 1 h () TU p  1 h

In intertemporal equilibrium, actual and expected inflation rates are equal. This result obtains alternatively from the second equation of the model by which the change in expected inflation is the difference between actual and projected inflation. Since in intertemporal equilibrium it could be expected that  will be constant, then   p . Given that the parameter h is assumed to be less than 1 (for h 1 the equilibrium value is undefined), the expected rate of inflation has a dynamically stable time path and converges to this equilibrium level. Furthermore, since the denominator of p is positive, if the numerator is positive, then the equilibrium value is positive and there is indeed inflation. If, however, the numerator turns out to be negative, there is a fall in the average price level – that is, deflation. For instance, when labor productivity T is sufficiently high, we may expect a fall in the price level. Also, in times of a recession or depression we could have a decline in the general price level again, this time driven by a sizeable level of unemployment due to massive layoffs and limited aggregate demand.

668 Problems Book to Accompany Mathematics for Economists

31. For the general inflation-unemployment model

p  TUh  ,0  01 h  d jp()  01j dt dU km()  p k  0 dt we expressed a second-order differential equation in the variable  . Write the model alternatively in the form of a second-order differential equation in U . Prove that the coefficients a1 and a2 are the same but bkjT (1 hm )  .

Solution:

Since both  and p are endogenous and functions of time, in order to solve for Ut() we need to drop both of these variables from the equation for U . Expressing  from the first equation,

p  TU    h and substituting it in the second equation,

dpTUj  jp () hpp  T U dt h h

Rewriting the third equation,

dU km()(  p  km    T  U  h ) dt and differentiating both sides with respect to t,

dU2 dU d khk  dt 2 dt dt d Substituting for , dt dU2 dU jkh khppTU ()     dt 2 dt h

dU2 dU kjkhpjkTjkU(1)(  ) dt 2 dt

Finally from the last equation of the model, we have for p 1 dU pm kdt which we substitute in the second-order differential equation

d2 U dU1 dU kjkhm(1) jkTjkU ( ) dt 2 dt k dt

dU2 dU kj(1 h )  jkUjkTm   (1 h ) dt 2 dt

Comparing this result to the second-order differential equation for  (),t we see that, indeed, the coefficients a1 and a2 are the same or akjh1   (1 ) and ajk2   . Just for b we get

Chapter 11. Advanced Differential and Difference Equations 669 bjkTm (1 h )

Thus, the equilibrium value U should be given by the particular integral

bTmh (1  ) U p  a2 

Solving the second-order differential equation for U would yield a general solution that should be the same as the time path of U obtainable through  . Furthermore, the intertemporal equilibrium for the unemployment rate should be the same for a particular model.

32. For the model given in problem 28, find the general solution for the time path and equilibrium value of Ut(). Compare the results to the ones obtained previously.

Solution:

The model is 1 pU 3  3 d 1 ()p  dt 4 dU 1 ()mp  dt 2

So, for the parameters, we have 1 1 1  T 3  1 j  h  and k  4 3 2 Hence, we obtain 11 1 112 1 akjh1  (1  )  1   ajk2    24 3 263 8 1112m  bjkTm (1 h ) 3 m  1 3  8383 

Thus, the intertemporal equilibrium for the rate of unemployment is

2m U 3  p 3 1 which is exactly the intertemporal equilibrium value we obtained previously. Again, since h  , we 3 get a negative relationship between unemployment and equilibrium inflation rate m . This results in a negatively sloped long-run Phillips curve. For the characteristic roots, we have

aaa2 4 12 4412 2 1 2 rii112       1,2  22398236312 1 2 These are complex roots with m  and n  . Hence, the general solution for unemployment is 3 12 t  222ttm Ut() e3 B cos B sin 3  34 12 12 3

This is quite similar to the function obtained previously, but this time with different coefficients:

670 Problems Book to Accompany Mathematics for Economists

tt  222ttmBBBB1122(21)() (21)()  BB 12 Ut() e33 cos  e sin  3 12 3 3 12 3 3 3   BB3 4

33. Consider the extended inflation-unemployment model in which the rate of change of the inflation rate is a decreasing function not only of the level of the unemployment rate, but aso of its rate of change (an example of the so called hysteresis system). Thus, the infation-unemployment model is dp dU  ()UU   ,0  dtn dt M U ln ,0  p Solve for p and U and find their equilibrium values.

Solution:

Substituting for U,

dp2  M d M 2   ln Un      ln dt p dt p and differentiating with respect to t,

2 dp d ()mp  () mp  dt 2 dt 2 dp dp ()mp  dt 2 dt 2 dp dp pm dt 2 dt

Again, nominal money supply m is a stationary value for inflation rate p . Here we have a1   , a2  , and bm   . Hence, the particular integral is pe  m and the characteristic roots are

aaa2 4 22 4 r 112 1,2 22

Thus the general solution for inflation would depend on the values of the characteristic roots where if 2  4 , we have real roots such that

rt12 r t p()tmAeAe12 

Since the constants  and  are positive, the roots (or their real part) turn out to be negative, and the equilibrium is dynamically stable. For the unemployment rate, we know that

1 dp UU  n  dt which again gives the natural rate of unemployment as the equilibrium rate for U . The general solution for unemployment by differentiation of the inflation rate 1 UU AreArert12  rt n  11 2 2

Chapter 11. Advanced Differential and Difference Equations 671

34. In the extended inflation-unemployment model, assume that the rate of change of the inflation rate is a decreasing function of the level of unemployment but the unemployment rate itself is a decreasing M M function of both real money supply and the inflation rate p . An increase in p , given , p p increases aggregate demand and, therefore, lowers unemployment. This results in the following infation-unemployment model: dp  ()UU    0 dt n M Up ln  ,, 0 p Solve for p and U and analyze their time paths.

Solution:

Substituting for U,

dp2  M 2   ln  p Un dt p and differentiating with respect to t,

2 dp dp ()mp  dt 2 dt 2 dp dp pm dt 2 dt

Again, nominal money supply m is a stationary value for inflation rate p . Here we have a1  , a2  , and bm   . Hence, the particular integral is pe  m and the characteristic roots are

aaa2 4 22 4 r 112 1,2 22

Thus, the general solution for inflation would depend on the values of the characteristic roots. If it happens that 2  4 , we have real roots . If 2  4 , then we get complex roots for the time path of inflation. In all cases, though, we know that this time path is unstable since the parameters  and  are positive and the real part of the characteristic roots is also positive.

p()tmAeAert12  r t 12 From the expression for the unemployment rate, we obtain 1 dp UU  n  dt which again gives the natural rate of unemployment as the equilibrium rate for U . The general solution for unemployment by differentiation of the inflation rate 1 UU AreArert12  rt n  11 2 2

35. The following model is given where the Phillips relation involves aggregate output Y rather than unemployment. Thus, when actual output exceeds the potential one, actual inflation exceeds the expected one.

p ()YY    0 (Phillips relation)

672 Problems Book to Accompany Mathematics for Economists

d jp()  01j (adaptive expectations) dt M lnYr ( ) ,0  (LM schedule) p It is still assumed that expectations are adaptive. Finally, the LM schedule gives the relationship M between real money supply , where M is nominal money supply and aggregate output is Y . The p government increases money supply as aggregate output expands or as people expect inflation to decline, with r being the real interest rate. Solve for expected inflation  .

Solution:

We substitute the first equation into the second:

d jYY()  dt d  jY() Y dt

Differentiating with respect to t,

ddY2   j dt 2 dt and differentiating the third equation also with respect to t,

dY d mp  which gives dt dt

dY1  d dY mp and substituting for in the equation for expected inflation, dt  dt dt

dj2  d mp dt 2  dt 1 d And finally, substituting p  for p from the last equation, jdt djd2 1 d 2 m  dt  jdt dt

Rearranging,

ddjj2   (1jm )   or dt 2  dt 

 jj (1jm )       j The parameters in this second-order differential equation are aj(1 ), a  , and 1  2   j b bm  . Thus,  p m . The characteristic roots for  are  a2

Chapter 11. Advanced Differential and Difference Equations 673

22(1 jj ) 4 (1 j )   2 (1jjj ) 22 (1  )  4 r  1,2 22 where the time path of  would depend on the particular values of the parameters.

36. For the national-income model in the previous problem, find the intertemporal equilibrium value of aggregate output Y and study its behavior with time. If instead of the absolute value of output, Y denotes the natural log of output, how does the result change for Y ?

Solution:

To trace the time path of aggregate output, we use the equations obtained in the previous example:

dY1  d mp dt  dt and we differentiate once again with respect to t :

22 2 dY1  dm dp d ddY 22 where 2   j dt dt dt dt dt dt

2 d Y1  dm dp dY  j dt 2  dt dt dt

dp d dY d and from   and   jY() Y we have dt dt dt dt

dp dY  jY() Y dt dt

Substituting in the equation for aggregate output,

2 d Y1  dm dY dY jY jY  j dt 2  dt dt dt

Rearranging,

2 dY(1 j ) dY  j 1 dm YmjY   where we set m   dt 2 dt dt

(1 jj )  1 YYYmjY    Thus, 

mjYm YY p jj

We can conclude that the intertemporal equilibrium value of national income is positively related to the full-employment (potential) output level Y . Note, however, that the two equilibria are not quite the same. One is a static equilibrium, while the other is a moving equilibrium. Since money growth rate is presumed to change with time, Yp is a moving equilibrium. The output levels are interrelated, though. Thus, when the government increases nominal money supply at a constant rate m (so that m   0 ), the intertemporal equilibrium is exactly equal to the potential output. Also, intertemporal equilibrium is negatively related to the adjustment coefficient j measuring the discrepancy between

674 Problems Book to Accompany Mathematics for Economists real and expected inflation as well as  showing the effect an expanding national output has on prices. The differential equation in Y is quite similar to that for expected inflation, which implies that expected inflation and aggregate output have similar time paths. The characteristic roots, hence, are the same as those for expected inflation.

With a log function of aggregate output, the differential equation would be

(1 jj )  1 (lnYYYmjY ) (ln )  (ln )   ln  where (lnYY )   is the rate of growth of aggregate output and (lnYY )  is its rate of change. Hence, the particular integral is

m (lnYY ) ln p  j m Going a step further and differentiating with respect to t , we obtain Y   . Alternatively, we can p  j solve the differential equation

(1 jj )  1 YYYmjY(ln )  ln 

Differentiating once more transforms the equation into a second-order differential one in the growth rate of national income:

(1 jjm )   YYY   Thus, again,   m Y   p  j which shows that at the optimum the government should determine the rate of growth of money supply in pace with the growth rate of equilibrium aggregate output.

37. In the national-income model in problem 35, assume that aggregate output is exogenously determined and find the values of actual and expected inflation.

Solution:

The model thus becomes

p ()YYo    0 (Phillips relation) d jp()  01j (adaptive expectations) dt M lnYr ( ) ,0  (LM schedule) p o

We substitute the first equation into the second:

d jYY()  dt o d  jY() Y dt o

From the third equation,

Chapter 11. Advanced Differential and Difference Equations 675

d mp  jYY()o  and, hence, for p, dt

pmjYY ()o 

From the first equation, we express  :

p()YYoo   m  (1)() j  YY 

We see that in equilibrium, that is, when YYo  , both actual and expected inflation would be equal to the growth rate of nominal money supply or

p m

Furthermore, both types of inflation are nurtured by an aggregate output that grows much above the full-employment level. This causes an overheated economy and an inflationary spiral. A downturn in the economic cycle and a fall of actual output below the normal level leads to low inflation or even deflation.

38. From the national-income model in problem 35, express the intertemporal equilibrium and characteristic roots for actual inflation p .

Solution:

From the first two equations, we have

d  jY() Y dt

Differentiating the first equation with respect to t,

dp d dY dY   jY()  Y  dt dt dt dt and differentiating once again,

dp22 dY dY  j dt22dt dt

From the last equation of the model,

dY d ddpdY  mp  and substituting  , dt dt dt dt dt

dY dp dY mp    dt dt dt

dY dp mp()    dt dt

dY1  dp mp and differentiating this once again with respect to t, dt()  dt

dY221  dm dp d p dm     22 where we set  m dt() dt dt dt dt

676 Problems Book to Accompany Mathematics for Economists

dY dY2 Substituting and in the equation for inflation p , dt dt 2

dp2  j dp dp dp2  2 mp m  2 dt()jdtjdt ()   dt

2 2 dp j dp dp ()jm m p  (1) j dt2 ()jjjdtj ()()   ()  dt 2

2 dp dp ()()(1)jjmmjpj           dt 2 dt

2 dp dp ()(1)()j    j    jp   jm  m dt 2 dt

(1)jjjmm ( ) ppp   jjj  

Therefore,

()jm m m  pm  jj m  In equilibrium  pm such that when money supply is growing at a constant rate, the j optimum is  pm . The characteristic roots for p are (1)jjj 22 (1)  4   () jj ()j 2 ()     r  1,2 2 (1)(1)4(jjjjjjjj22     )(1)(1)4(  22    )  2(jj  ) 2(    )

39. When analyzing the quality level s()t of the items produced in a firm the statisticians found that quality varies according to the differential equation st() 4 st  () 6 st  () 4 s 20. If the intertemporal equilibrium represents the desired quality standard or optimal quality level, find that standard and conclude whether, with time, the firm converges to or diverges from this standard.

Solution: 20 This is a third-order differential equation with a particular integral y   5 that is the optimal p 4 quality the firm strives to achieve. The characteristic equation is rrr324640

This transforms into

(2)(22)0rrr2

Thus, we have a real root, r1 2, and a pair of complex roots, ri2,3  1, where m 1 and n 1 . The general solution is

Chapter 11. Advanced Differential and Difference Equations 677

2tt st() Ae112 e ( B cos t  B sin) t  5

Since both the real root and the real part of the complex roots are negative, the time path of quality is convergent to the equilibrium level of 5. Although the time path is fluctuating around this quality standard, given the circular functions it contains, the firm is reaching the desired level with time. Note that due to the fluctuation, at times the firm might produce below the optimum but also at a quality higher than the required standard. It might be optimal for the firm to produce just at the standard rather than provide too high or too low quality.

40. In studying the dynamics of the value of stock at the stock exchange, the stockbrokers found that the value changes according to the differential equation pt() 4 pt  () 5 pt  () 2 p 12. Find the time path of the value p()t . Is it dynamically stable?

Solution: 12 This is a third-order differential equation with a particular integral p  6 . The characteristic 2 equation is rrr324520 which can conveniently be transformed into

(2)(21)0rrr2 or

(2)(1)0rr2

Thus, we have two real roots, r1 1 and r2  2 , one of which is repeated. The general solution is

ttt2 pt() Ae12 Ate  Ae 3  6

Since both roots are positive and the terms grow infinitely as t , the market price of stock is not dynamically stable.

41. Given the multiplier-accelerator model, determine the time path of national income if the accelerator is   0.8 and the marginal propensity to consume is   0.6 .

Solution: 4 Comparing  and would allow us to determine which particular case we are dealing with. (1  ) 2 44(0.8) 0.98 (1 )22 (1 0.8) 4 Thus, we have   and  1, which is a time path characterized by damped stepped (1  ) 2 fluctuation. Thus, the time path of national income, given the values of the parameters, is dynamically stable.

42. In the multiplier-accelerator model, it is given that the accelerator is   0.3, while the simple investment multiplier is 4. Express and analyze the time path of national income.

Solution:

We need to find the marginal propensity to consume. From the multiplier-accelerator model, we know that the simple multiplier gives the intertemporal equilibrium of national income

678 Problems Book to Accompany Mathematics for Economists

G Y  o p 1  1 where the multiplier is  4 . We can deduce that   0.75 . Hence, the intertemporal 1  equilibrium of national income is

YGpo 4

where Go is exogenously determined. From the model, we can also express the roots. Comparing  4 and , we establish (1  ) 2 44(3) 0.75  (1 )22 (1 3)  (1  ) 0.75(1 3) Therefore, we have the single real root a  1.5 . The function of the national 22 income becomes

tt t t YYYtcp  AaAtaY12   p  A 1(1.5)  At 2 (1.5)  4 G o

Since  3(0.75) 2.25 1 and a 1, the time path of national income is nonoscillatory and divergent from the intertemporal equilibrium.

43. If the accelerator is   0.6 and the investment multiplier as defined by Samuelson’s multiplier- accelerator model is 2.5, determine the time path of national income.

Solution:

According to the model, the intertemporal equilibrium is given by the particular integral G YGo 2.5 po1  1 where the multiplier is  2.5 . Thus, we have   0.6 . 1  44(0.6)2.4 0.9375 0.6 or (1 )22 (1 0.6) 2.56

4   and  1 (1 ) 2

This is the subcase of complex roots where the function of national income demonstrates damped stepped fluctuation to the equilibrium level of 2.5Go . Alternatively, we could also check that R  0.6 1, so again the time path is dynamically stable. We can further write the general solution of national income as

tt YRBtBtc (12 cos sin ) (0.6) BtBt 12 cos  sin  where b 0.6(1.6) b2 cos 1   0.8 and sin  11  1 0.82  0.6 2 b2 2(0.6) 4b2

The general solution is

t YBtBtGto(0.6) 12 cos sin 2.5

Chapter 11. Advanced Differential and Difference Equations 679

44. Suppose a particlar nation’s marginal propensity to consume is   0.9 . The nation’s government wants to predict the effect of different accelerator values given by 1  0.2, 2  0.4, and 3  0.5, respectively. Using the framework of the multiplier-accelerator model, help the government analyze the time path of national income. What do you observe about higher values of the accelerator?

Solution:

We know that the intertemporal equilibrium of national income is independent of the value of the accelerator, since

GG YGoo 10 where the multiplier is 10. For   0.2, we have po110.9 1

4 4(0.2) 0.8 4 0.556 0.9 or   (1 )22 (1 0.2) 1.44 (1  ) 2

Furthermore, 1 0.2(0.9) 0.18 1 . The two results show that the time path of national income is nonoscillatory and convergent (subcase 1 of the distinct-root case). For 2  0.4, we obtain

4 4(0.4) 0.16 4 0.816 0.9 so again   and    0.4(0.9) 0.36 1. (1 )22 (1 0.4) 1.96 (1 ) 2 2

Finally for 3  0.5, 44(0.5)2 4 0.889 0.9 or   (1 )22 (1 0.5) 2.25 (1 ) 2

Again, 3 0.5(0.9) 0.45 1

All these fall within subcase 1 of the distinct-root case, which means the time path of national income is nonoscillatory and convergent. However, with the increase in the accelerator value we see that there is greater tendency for divergence. It could be checked that for greater values of the accelerator (in any of the three cases) the time path becomes divergent. The simple logic is that when  1, we already have instability. For the particular value of   0.9, we obtain that 0.9  1 or  1.111 is a condition for the dynamic instability of national income.

45. For the multiplier-accelerator model, find the characteristic roots and determine the time path of national income if the accelerator is   3 and the marginal propensity to consume is   0.8 .

Solution:

The particular integral gives the intertemporal equilibrium of national income

GG YGoo 5 po110.8

From the model, we find the roots

(1 ) 22 (1   )  4  0.8(1  3) 0.8 22 (1  3)  4(3)0.8 a1,2   22 3.2 10.24 9.6 3.2  0.8 2;1.2 22

The general solution, therefore, is

680 Problems Book to Accompany Mathematics for Economists

tt YYYtcp  A12(1.2)  A (2)  5 G o 4 Comparing  and , we find (1  ) 2 44(3) 0.75 0.8 (1 )22 (1 3) 4 Since  3(0.8) 2.4 1 and   , the time path of national income is nonoscillatory and (1  ) 2 divergent. This is the divergent subcase of the first case of distinct real roots, both of which are greater than one, since 11.22.

46. Consider a different form of the multiplier-accelerator model given by

YCIttt

CYtt  1 01   0

IYYttt ()12  ,0  where present aggregate investment depends on the increase in national income from the previous period such that the accelerator is still  . Solve for national income.

Solution:

Substituting the last two equations into the first, we obtain

YYtt 112  YY tt 

YYY()    ttt12 Extrapolating this equation by two time periods gives the more convenient form

YYYttt21()   

We have b1 ()  , b2   , and c     . The intertemporal equilibrium national income is

c   YYp    11bb12   1   1 Note that again is the value of the multiplier. Furthermore,  and  , being autonomous 1  consumption and autonomous investment, respectively, influence equilibrium national income positively. The larger the two types of autonomous spending, the greater the value of the multiplier is.

For the characteristic roots, we obtain aa12   and aa12 

(1aa12 )(1  )  1 ( aaaa 1212  )   1    1  so

0(1)(1)1aa12  

Similar to the standard model, we could have several possibilities where the condition for dynamic stability again is for both roots to be fractions. Then aa12  1, or, the requirement is for the accelerator to be less than 1. In the case of repeated and complex roots, we get that  1 which is the same condition.

47. Given the specific multiplier-accelerator model

YCIttt

Chapter 11. Advanced Differential and Difference Equations 681

3 CY10 tt4 1 1 IYY5(  ) ttt4 12 where the accelerator is 14, solve for national income finding its intertemporal equilibrium and time path. Analyze the behavior of national income in time.

Solution:

Substituting into the first equation, 311 YYYY10  5  tt444112 tt 1 YY Y 15 tt124 t Transforming into a more convenient form, 1 YY Y 15 tt214 t 1 Here we have b 1, b  and c 15 . Thus, the intertemporal equilibrium national income is 1 2 4 c 15 YY  60 p 1bb 1 1211 4 Solving for the characteristic roots, 4(1) 2 11 bbb 4 1 a 11 2 4 1,2 222 which is a case of a real, single root. Hence, the general solution for the time path of national income, given the assumptions of the model, is

tt YYYtcp  A12(0.5)  At (0.5)  60

Since the root is less than 1, we conclude that the time path of national income is dynamically stable; that is, as t , national income would tend to approach the equilibrium value of 60.

48. Solve the standard Phillips curve model for expected inflation  . Assume that

pttt Uh  ,0  01 h 

tt1 jp() tt  01 j 

UUkmptt11() t k  0

Solution:

This time, we use the difference for  t and substitute for pt :

 tt1 jUh()  ttt  

ttt1 (1jjh )  j  jU

Extending this by one time period,

ttt211(1jjh )  j  jU

Subtracting the last two equations gives a difference term for unemployment U :

682 Problems Book to Accompany Mathematics for Economists

ttttt21(2jjh )  (1 jjh )  jU (  1  U )

 tttt21(2jjh )  (1 jjh )  jkmp (   1 )

But from the second equation of the model, we also have

jptt12 (1 j ) t  1

And substituting this expression in the equation finally gives a second-order difference equation solely in  :

 ttttt21(2jjh )  (1 jjh )  jkmk   21  k (1  j )

(1kjhjkjjhjkm )ttt21 1 (1 )(1  )  (1 )   or

1(1)(1)jh j  k (1jjh ) jkm  ttt21111kkk

But this equation is absolutely identical to the one for actual inflation p .

(1kp )ttttt21 (1 j hj 1 kp ) (1 j hjp ) j ( U 1 U ) 0

c Hence, the equilibrium value for expected inflation is    m , and all other conclusions 1aa12 relevant to actual inflation relate also to expected inflation.

49. Consider the standard inflation-unemployment model in discrete time. Assume that the change in unemployment depends on inflation rate from the previous period such that UUkmptt1 () t. Express equilibrium inflation rate.

Solution:

The model becomes

pttt Uh  ,0  01 h 

 tt1 jp() tt  01 j 

UUkmptt1 () t k  0

Finding the difference for actual inflation,

pttpp1  t where pUhttt111  

pptt111 ()() UUh tt   tt  and substituting the last two equations of the model into this expression,

ppkmphjptt1  ()()   t  tt 

pttt1 (1kjhp  )  kmhj  where hp tt   U t and substituting further,

ptttt1 (1kjhp  )  kmjpj    jU or

pttt1 (1kjhjp   )  kmj   jU and extending this by one time period,

pttt211(1kjhjp   )  kmj   jU

Chapter 11. Advanced Differential and Difference Equations 683

Subtracting the last two equations and substituting for the difference term in inflation finally gives

pttttt21(2kjhjp  )  (1 kjhjp )  jkUU (  1  )

ptttt21(2kjhjp  )  (1 kjhjpjkmp )  (   )

ptt21(2kjhjp  )  (1 kjhjjkpjkm ) t  

For the equilibrium actual inflation rate, we get cjkm  p m 1121bb12  kjhj   kjhjjk 

This result is consistent with the previous findings of the model according to which the rate of growth of nominal money supply gives the intertemporal equilibrium for actual inflation rate. Note that this result obtains whether the increase in unemployment is assumed to depend on inflation in the current or previous periods.

50. Assume that the inflation-unemployment model is

pUhttt   1 ,0  01 h 

tt1 jp() tt  01 j 

UUkmptt11() t k  0 so that actual inflation in the present period depends on people’s expectations of inflation in the previous period. Write the difference equation for actual inflation p . What is the order of the equation that obtains? Using the particular integral, find the intertemporal equilibrium of actual inflation rate.

Solution:

Alternatively, the model can be written in the form

pttt11 Uh 

 tt21jp() tt  11 

UUtt21 kmp() t  2

Extending the equation for actual inflation by one more period,

pUhttt221  

pttt11 Uh  and expressing the difference,

pptt21 ()() UUh t  2 t  1  tt  1  and substituting the last two equations of the model into this expression,

pptt21 kmphjp()()   t  2   tt 

p  U Expressing   tt11 from the first equation, t h

pptt21 kmpjhpp()(   t  2   tt   1 U t  1 )

(1kp )ttt21 (1 jp ) jhp km j jU t  1

(1k ) pttt11 (1 j ) p jhp km j jU t and subtracting both sides of the last two equations to find the difference again,

(1kp )tttttt21 (1 j 1 kp ) ( jh 1 jp ) jhp  11 jU ( U )

Substituting for the difference of unemployment,

684 Problems Book to Accompany Mathematics for Economists

(1kp )ttttt21 (2 j kp ) ( jh 1 jp ) jhp  11 j km ( p )

(1kp )tttt211 (2 j k j kp ) ( jh 1 jp ) jhp j  km

Normalizing the equation and extending it by one time period finally gives

(2jkjk  ) ( jhj  1 ) jhjkm pppptttt321 1111kkkk

We see that this is a third-order difference equation in p , which could solved given specific values of the parameters. Using the steps for finding the particular integral of such a third-order difference equation, we get cjkm  p m 1bbb123 12kjkjkjhjjh  1 (1 k )  1 k Our findings again are consistent with the standard inflation-unemployment model where the rate of growth of nominal money supply gives the intertemporal equilibrium of actual inflation rate. This result obtains whether actual inflation is assumed to depend on past or on present expectations.

51. For the model in the previous problem, assume that the change in unemployment results from inflation in the previous, not in the current, period, such that UUkmptt1  ()  t. Write again the difference equation for actual inflation p . What is the order of the equation that obtains? Given this new assumption, express intertemporal equilibrium actual inflation rate.

Solution:

Alternatively, the model can be written in the form

pttt11 Uh 

 tt21jp() tt  11 

UUtt21 kmp() t  1

Extending the equation for actual inflation by one more period,

pUhttt221  

pttt11 Uh  and expressing the difference,

pptt21 ()() UUh t  2 t  1  tt  1  and substituting the last two equations of the model into this expression,

ptt11  U  pptt21 kmphjp()()   t  1   tt  where  t  h

pptt21 kmpjhpp()(   t  1   tt   1 U t  1 )

ptttt21(1kjp  )  jhpkmj    jU  1

ptttt11(1kjpjhp  )   kmj   jU

Subtracting,

ptt21(2kjp  )  ( jh  1 kjpjhp )  tttt   11  jUU (  )

ptt21(2kjp  )  ( jh  1 kjpjhpjkmp )  ttt   1  (  )

ptt21(2kjp  )  ( jh  1 kjjkpjhpjkm )  tt   1  

Chapter 11. Advanced Differential and Difference Equations 685

Extending the equation by one time period,

ptt32(2kjp  )  ( jh  1 kjjkp )  tt  1  jhpjkm  

Again, we have a third-order difference equation in p , which could be solved following the steps of higher-order difference equations. For the equilibrium value of inflation, we use the particular integral cjkm  p m 1121bbb123  kjjh  kjjkjh 

The result again is consistent with the one obtained previously. Whether actual inflation is assumed to depend on past or present expectations, or unemployment depends on previous or current inflation, the equilibrium value for actual inflation does not change.

52. Given the third-order difference equation for p in the previous problem, assume the equation 111 takes the specific form ppptt321 t p  t16 . Using the steps of third-order difference 3412 equations, find the general solution for inflation p . What is its intertemporal equilibrium, and does inflation converge to or diverge from it?

Solution: 1 1 1 Since we have b  , b  and b  , the particular integral is 1 3 2 4 3 12 c 16 16(12) p 32 112431bbb 11 1  1231 3412 The characteristic equation is

111 aaa320 3412

2 1 Factoring out the term a  gives 4

2211 1 aa a 0 or 43 4

111 aaa0 223 1 1 1 Thus, the characteristic roots are a  , a  , and a  . The general solution of actual 1 2 2 2 3 3 inflation, therefore, can be written as

ttt 111   pAt 123 A   A  32 223  

We can further definitize the arbitrary constants, if we are given some initial conditions. Since all the three roots have absolute values smaller than 1, the time path of inflation is convergent to the intertemporal equilibrium of 32.

53. Consider a simplified inflation-unemployment model where the unemployment rate is assumed to be exogenous:

ptot Uh  ,0  01 h 

686 Problems Book to Accompany Mathematics for Economists

tt1 jp() tt  01 j 

Find and determine the time path of actual inflation rate p . What is the intertemporal equilibrium value of inflation? What kind of difference equation obtains?

Solution:

Using the difference

pttpp1  t where we have pUhtot11  

pphtt11() tt  and substituting the difference for expected inflation,

pphjptt1 ()  tt 

From the first equation of the model, we also have hp tt   U o, and substituting further,

ptt1 phjpjpj tt  ()  U o

ptto1 (1jjhpj )  (  U ) which is a first-order difference equation solely in p . We know from before that the general solution for the time path of actual inflation can by found by the formula

cct ppto()  b where bjjh (1 ) and cj ()   Uo 11bb

Substituting,

jU()oot jU ()  ppto(1  jjh ) 11jjh  11  jjh 

()UUoot ()  ppto(1  jjh ) 11hh

  Uo where the equilibrium value for the inflation rate is p  . Although the result is different 1 h from the one for the expanded model, we still get an inverse relationship between inflation and unemployment, which illustrates the negatively sloped long-run Phillips curve. Analyzing the time path further, we find out that the term 1 jjh is always less than 1, since 0,1 jh . Therefore, the time path is convergent. Since this term is also positive, it follows that the time path is nonoscillatory.

54. For the simplified model in the previous problem, obtain the time path of expected inflation  . How does it differ from the one for real inflation p ? Solution:

We can solve easily taking the difference for expected inflation from the second equation

 tt1 jp() tt  and substituting the term pt ,

 tt1 jUh()  ott  

Chapter 11. Advanced Differential and Difference Equations 687

tto1 (1jh  j )  j (  U ) which is again a first-order difference equation in  . Thus, the general solution is

()UUoot ()  to(1 jjh ) 11hh   U where the equilibrium expected inflation rate is   o . This is the same as the value obtained 1 h for actual inflation. Logically, in a state of equilibrium the value of the two inflation rates should be equal. Similar to actual inflation, expected inflation is convergent and nonoscillatory.

55. Consider the simplified inflation-unemployment model in which unemployment is assumed to be exogenous. Imagine that there is a time lag in the way people’s inflationary expectations form actual inflation. In other words, actual inflation in the present period depends on expected inflation from the previous period. Hence, the model is

ptot1  Uh  ,0  01 h 

 tt1 jp() tt  01 j 

Solve for actual inflation rate p . What is its intertemporal equilibrium value, and how does it depend on expectations? Analyze also its time path.

Solution:

pttpp21  t where we have ptot1   Uh and pUhtot21  

pphtt21()() tt  1  jhp  tt 

From the first equation of the model, we also have hp tt  1  U o, and substituting further,

ptt21phjpjpj  tt  1 ()  U o

pttto21(1jp )  hjp  j (  U )

This is a second-order difference equation where the equilibrium value for the inflation rate is

c   Uo p  11bb12  h

This is a result we obtained previously with unemployment again considered exogenous. It shows that equilibrium actual inflation is unaffected by people’s expectations. Whether those expectations were formed in the previous or the current period, the equilibrium level of inflation stays the same.

bbb2 4 1(1)4jjjh 2  a 11 2 1,2 22 and we have different outcomes depending on whether (1jjh )2 4 . We also know that the characteristic roots must satisfy the conditions aa12  b 1 1  j and aa12 b 2 jh. Then it must be that

(1aa12 )(1 ) 1 ( aaaajjhjh 1212 ) 1 1 (1 ) 0

Since 0,1jh , we conclude that

aa120 aa12 0 and (1 aa12 )(1 ) 0

688 Problems Book to Accompany Mathematics for Economists

so one of the roots a2 is positive and the other a1 is negative where the two must be fractions, but a2 prevails over a1; that is, 1 aa21. Since the two roots are fractions smaller than 1, the time path of actual inflation p must be convergent.

56. Consider the extended inflation-unemployment model, dealt with previously, in its continuous- time form

dp  ()UU    0 dt n dU  ()mp    0 dt

where U is the rate of actual unemployment while Un is a fixed, natural rate of unemployment. Convert the model in a discrete-time form and solve for the time path of inflation p .

Solution:

From the first equation of the model, by further differentiation we obtained dp2 dU   dt 2 dt

In discrete time this should involve a second difference of price on the left side, or

2 pt()()()()2pppppppppp  t  tt121121  t t  ttt  tt

The equation in its discrete form becomes

pttttt212()pp UU  1 where from the second equation of the model we have in discrete time

UUtt1  () mp t

Thus, the new model becomes

pttttt212()pp UU  1

UUtt1  () mp t

Substituting the difference term for unemployment gives a second-order difference equation in p :

pttt212()pp mp t or

ptt212(1)ppm  t m The equilibrium value for p is pm . This result is consistent with our previous 121 findings. For the characteristic roots, we get bbb2 4 244(1) 22 i  ai11 2 1  1,2 222 which turn out to be complex numbers, so the time path of the inflation rate must involve stepped fluctuation. Since Rb2 (1 ) where both  and  are positive constants, it must be that R 1. Hence, the fluctuating path of inflation, given the assumptions of the model, must be explosive.

Chapter 11. Advanced Differential and Difference Equations 689

57. For the discrete time model in the previous problem, assume that the difference for unemployment is given by UUtt11 () mp t, that is the increase in unemployment depends on inflation in the present, not in the previous period.

Solution:

In this new version, the model becomes

p 2()pp UU ttttt21  1 UUtt11 () mp t

Substituting again, the difference term for unemployment results in

pttt212()pp mp t  1

pttt21(2 )pp   m m The equilibrium value for p is pm . Again, the intertemporal equilibrium of 12  1 inflation is the growth rate of nominal money supply. The characteristic roots are

bbb2 4 2(2)42(4)  2       a 11 2  1,2 22 2

By analyzing the roots further, we find

aa12 b 12  and aa12 b 2 1. Then

(1aa12 )(1  )  1 ( aaaa 1212  )   1 2   1  0

Since both  and  are positive constants, one possibility is for both roots to be negative where one is a fraction. (Can you see why the two roots cannot both be fractions?) From the second equation, we also see that one root is reciprocal of the other. Therefore, we conclude that

aa12,0 a1 1 and a2 1

Since the absolute value of one of the roots turns out to be greater than 1, the time path of inflation is divergent and nonoscillatory.

58. The extended inflation-unemployment model in its continuous-time form is

dp dU  ()UU   ,0  dtn dt dU  ()mp    0 dt

where U is the rate of actual unemployment and Un is the natural rate of unemployment. Convert the model in a discrete-time form and solve for the time path of inflation p . Solution:

From the first equation of the model, by further differentiation we have dp22 dU dU    dt22dt dt

In discrete time, this should involve a second difference of price on the left side and a second difference of the rate of unemployment on the right side:

2 pt()()()()2pppppppppp  t  tt121121  t t  ttt  tt

690 Problems Book to Accompany Mathematics for Economists

2 UUUUUUUUUUUt()( t tt121121 )( t t )( ttt ) 2 tt

The equation in its discrete form becomes

p 2()(2)pp UU U UU tttttt21  1  2 tt  1 where from the second equation of the model we have in discrete time

UUtt1  () mp t and also

UUUppttttt212()   1

Therefore, the equation for inflation becomes

pttt212()()pp mp t pp tt  1 

ptt21(2 )ppm  (1  )  t 

m The equilibrium value for p is pm , which we have obtained 12  1    previously. Analyzing the characteristic roots,

aa12 b 12  and aa12 b 21   and, therefore,

(1aa12 )(1  )  1 ( aaaa 1212  )   1 2   1    0

The last result implies that the characteristic roots can both be bigger than 1 or smaller than 1. This means that a convergent time path for inflation is not impossible. The condition 01 1 ensures the dynamic stability of inflation.

59. For the discrete-time model in the previous problem, assume the difference

UUtt11 () mp t where the change in unemployment depends on current inflation. How do the results differ from those in the previous problem?

Solution:

The equation of inflation is still

p 2()(2)pp UU U UU tttttt21  1  2 tt  1 where UUtt11 () mp t and also

UUUppttttt212()   1

Substituting in the first equation,

pttt212()()pp mp t  1 pp tt  1 

pttt21(2  )ppm  (1  )  

m The equilibrium value for p is pm , which we have obtained 12    1  previously. For the characteristic roots, we have aa12  b 1 2   and aa12 b 2 1  .

(1aa12 )(1  )  1 ( aaaa 1212  )   1 2    1    0

The last result again shows that a convergent time path for inflation is not impossible. However, this depends on the exact values of the parameters. In this sense, the results are similar to those in the previous model. Furthermore, we see that 1  could be less than 1, given the positive values of the parameters, which also allows for convergence.

Chapter 11. Advanced Differential and Difference Equations 691

60. If the extended inflation-unemployment model in its continuous-time form is

dp  ()UU    0 dt n dU dp ()mp   ,0  dt dt modify the model in a discrete-time form and analyze the characteristic roots for inflation p . How does inflation in the present or previous period affect unemployment?

Solution:

From the first equation of the model, by further differentiation we obtained dp2 dU   dt 2 dt

In discrete time, this should involve a second difference of price on the left side, or

2 pt()()()()2pppppppppp  t  tt121121  t t  ttt  tt

The equation in its discrete form becomes

pttttt212()pp UU  1

The second equation translates into

UUtt11 ()() mp t  p  tt p 

Thus, the new model becomes

pttttt212()pp UU  1

UUtt11 ()() mp t  p  tt p 

Substituting the difference term for unemployment gives a second-order difference equation in p :

pttt212()()pp mp t pp tt  1  or

ptt21(2 )ppm  (1  )  t 

m The equilibrium value for p is pm . For the characteristic roots, we 12  1    have aa12 b 12  and aa12 b 21   .

(1aa12 )(1  )  1 ( aaaa 1212  )   1 2   1    0

Here, since 1 cannot be between 0 and 1, the roots cannot both be fractions. Therefore, the time path of inflation would not be dynamically stable. Note that, if a different assumption is made about unemployment, such as UUtt111 ()() mp t  p  tt p , the equation becomes

pttt212()()pp mp t  1 pp tt  1  or

pttt21(2  )ppm  (1  )  

m The intertemporal equilibrium for p is p  m . For characteristic roots, 12    1  we have aa12 b 12  and aa12 b 21  .

(1aa12 )(1  )  1 ( aaaa 1212  )   1 2    1    0

692 Problems Book to Accompany Mathematics for Economists

Here since 1 cannot be between 0 and 1, the roots cannot both be fractions. Therefore, the time path of inflation would not be dynamically stable.

61. Transform the national-income model presented in problem 35 from continuous into discrete time. Solve for aggregate output Y, and find its intertemporal equilibrium.

Solution:

We recall that in its continuous form the model is

p ()YY    0 (Phillips relation) d jp()  01j (adaptive expectations) dt dY d mp  ,0  (LM schedule) dt dt

Solution:

In discrete time, the model can be written as

ptt()YY t 

 tt1 jp() tt  mpttttt ()() Y11  Y   

From the first two equations,

tt1 j()()pjYY tt    t 

We directly substitute the difference for  t in the third equation:

mpttt()() Y1  Y  jYY t 

We have one pt to get rid of in order to obtain a difference equation in Y solely. Extending the first equation by one time period gives

ptt11()YY t  1 

ptt()YY t 

and expressing the difference pt ,

ptttt11pYYjYYYY ()()() tt  1    t   tt  1 

Furthermore, from the equation for aggregate output,

mpttt()() Y1  Y  jYY t  mpttt121()() Y  Y  jYY t  1  where we subtract the two equations

ptt1121pYYYjYY (2 ttt   )  ( tt  )

But for the difference pt , we already have

ptt11pjYYYY()() t  tt 

Thus, equating the two,

Chapter 11. Advanced Differential and Difference Equations 693

jY()()(2tttttttt Y Y1121  Y  Y  Y  Y )  jY (  Y ) which is a second-order differential equation solely in Y . Rearranging and normalizing leads to

(2jj ) (   jjY ) YY Y  tt21  t

Intertemporal equilibrium output is

cjY YYp   1bb12 2 jj  j  1  The intertemporal equilibrium is exactly equal to the full-employment output, which represents a static equilibrium.

62. Convert the national-income model given in problem 35 from continuous into discrete time and solve for actual inflation p . Prove that its time path is the same as that for aggregate output Y .

Solution:

We convert the model into

ptt()YY t 

tt1 jp() tt  mpttttt ()() Y11  Y   

Extending the first equation by one time period gives

ptt11()YY t  1 

ptt()YY t 

and expressing the difference pt ,

ptttt11pYYjpYY ()()() tt  1    tttt      1 

From the last equation of the model, mpttt ()() Y1  Y  jp  tt 

and expressing the term (),YYtt1  1 YYtt1  mpjp  t  ()  tt   Substituting this term in the difference equation for inflation,   j ppjptt1 ()()()  tt   mp   t p  tt   ()jj  ptt1 ppmp()()  tt     t 

Extending this by one time period,

()jj  ptt21ppmp()()  tt  11     t  1  and subtracting the two equations,

694 Problems Book to Accompany Mathematics for Economists

() jj  2()()ptt12pp tpp  tttt   11 pp  tt  1  

()jj (  j22  j )  2()()()ptt12pp tpp  tt   1  p tt  pp  tt  1   1 where ptt()()pp t1  t  mp  t jj

Substituting the last term in the equation for inflation,

()jj  j 2()()()ptt12pp tppjpp  tt   1   tt  1  mp   t  Rearranging, ()jj  2()()ptt12pp tpp  tt   1  mp   t  (2)(jjjjm    )  pptt21p  t   

The coefficients b1 and b2 are the same as those in the differential equation from aggregate output Y obtained in the previous problem. This indicates that the two variables have the same time path. Furthermore, for the equilibrium level of inflation, cjm  p m 1bb12 jjj2    1 

63. In the discrete-time national-income model, assume that the government sets the growth of money supply based on the increase in national income from the previous period. Thus, the model becomes

ptt()YY t 

tt1 jp() tt  mpttttt ()() YY 11   

Solve the model for expected inflation  .

Solution:

We rewrite the last equation mpttttt11 ()() Y  Y   21  

From the first two equations,

tt1 j()()pjYY tt    t 

Extending this equation by a period and subtracting the two equations,

tt21jY() t  1  Y

tt1 j()YY t 

ttttt212()jY  1 Y

and substituting the term YYtt1  in the equation for expected inflation,  mptttttt12121(2   )(  )  j

Chapter 11. Advanced Differential and Difference Equations 695

From the second equation, we express pt : 1(1)j  1(1)j  ptt 1  t which corresponds to ptt12  t 1 and substituting again jj jj 11j   m ttttttt212121 (2  )( ) jj j

 jm ttttttt212121 (1) j   (   2   )  j (    )

Rearranging and normalizing,

()(1)2jjttt21      j    jm

(1)2jj  jm  ttt21()()() jjj    

Hence, the equilibrium value for expected inflation is cjm    m 1bb12 jj2   ()1j    j

64. A market equilibrium model is given such that the demand and supply functions for a commodity are d qptt65 s qptt415  1

Furthermore, it is known that the increase in market price from one period to another depends on 1 excess demand by the adjustment coefficient 15 such that p pqq()ds  . Solve the model tt1 5 tt with the help of a second-order difference equation. Find the general solution and intertemporal equilibrium for pt . Then, using the steps of the Cobweb model, form a first-order difference equation and solve again for market equilibrium. Compare the time paths and equilibrium values of price pt using the two approaches.

Solution:

Substituting the demand and the supply function in the last equation, 1 p ppp(6  5  4 15 ) tt115 t t

ptt11ppp23 tt

pptt1132 which transforms into

pptt2 32 where b1  0 b2  3

Finding intertemporal equilibrium price, 21 p  13 2

For the characteristic roots we have a2  30 and ai 3; or, we have the complex-root case. This implies that the time path of price would be stepped fluctuation and divergent from the

696 Problems Book to Accompany Mathematics for Economists equilibrium of 12 since the absolute value of both roots is greater than 1, that is, 31 . The general solution for price could be written as tt1 pAi33  Ai  t 12 2 Alternatively, we can use the steps of the Cobweb model according to which at any point in time the market clears. This is different from the previous approach where we assumed that the market may not always be in equilibrium and actually diverges from this equilibrium as t . Therefore,

ds qqtt

65ptt 415p 1

pptt321

This equation resembles the one previously obtained; but unlike it, it is a first-order difference equation. Solving through the well-known method,

22t ppto 3 13 13

11t ppto3 22

Again, an equilibrium value of 12 is obtained, and the time path is oscillatory and divergent since 30 and 31 . Thus, the results of the two approaches are pretty similar – in both cases, price diverges from the intertemporal equilibrium of 12, although in the first case the fluctuation is stepped and in the second it is ordinary oscillation.

65. In a market for a given commodity, supply depends on price in the previous period but also on the increase in price from the previous to the current period. Thus, as producers see price rising, they feel stimulated to supply more. The demand and supply functions are

d qptt  s qppptttt 11  ()  ds ptt1 pjqq() tt  where all parameters are positive. Solve the model with the help of a second-order difference equation. Then assume the market is constantly in equilibrium, and solve using a first-order difference equation. Compare the two equilibrium values.

Solution:

Substituting the demand and the supply function in the last equation,

ptt111pj()  p t  p t  p tt  p

pjpjpjttt111(  )  (  )  (  )

Extending further by one time period,

pjpjpjttt211(  )  (  )  (  )

Finding equilibrium price,

Chapter 11. Advanced Differential and Difference Equations 697

j()  p  11(  )jj  (  ) 

Assuming the market always clears, we equate demand and supply:

 pttttppp11() 

()() pptt1

Normalizing the equation,

()  pp tt 1  t    ppto      ()1 ()1       t      ppto   where p      

Thus, the two equilibrium values are the same.

66. Assume a market for a commodity in which producers continuously follow price trends and their decision to supply presently depends on price levels in two consecutive previous periods. Thus, the model becomes d qptt  s qppttt 12  

Assume the market clears at any point in time and all parameters are positive. Find the intertemporal equilibrium price and quantity.

Solution:

Equating demand with supply,

 ptttpp12

 pptt12 p t  

Normalizing and extending further by two time periods,

  ppp ttt21 

Finding equilibrium price,

()  p      1  From the demand function,

 ()()   qpd        

698 Problems Book to Accompany Mathematics for Economists

67. Assume a market for a commodity in which consumers expect price to rise in the future so their present demand is positively related to price in the future period. Thus, the demand and supply functions are

d qppttt  1 s qppttt   1

Assume the market always clears and all parameters are positive. Find the intertemporal equilibrium price and quantity. How does a positive or a negative value of the expectations coefficient  affect the value of equilibrium price p ?

Solution:

Equating demand with supply,

 pttppp11   t t

 pppttt11()    

Normalizing and extending by one time period,

()  ppp ttt21 

Finding equilibrium price,

()   p           1  From the demand function,

()()()()      qpppd     ()      ()()  

To have a meaningful price, we need   . Note that if   0 , as the model assumes, consumers expect price to rise in the future, so they buy more presently. Thus, the equilibrium price indeed turns out to be higher than if consumer expectations of price are not taken into account. However, if hypothetically,   0 and, hence, consumers predict that price would fall in the future period, this would reduce their present demand for the good. With  negative, the equilibrium price of the commodity falls lower. Therefore, we can conclude that consumers’ expectations do in fact shape market price and its movement.

68. In a given market, producers continuously follow price trends and base their decisions on price in two consecutive previous periods. Consumers, on the other hand, are influenced by current price but also by the increase in the price level. Thus, the model is

d qppptttt  () 1 s qppttt 12  

Assume that the market clears at any point in time and that all parameters are positive. Find the intertemporal equilibrium price and check how it depends on the expectations coefficient  .

Solution:

Equating demand with supply,

Chapter 11. Advanced Differential and Difference Equations 699

 pttt()pp 112   p t p t

()() pppttt12  

Normalizing and extending further by two time periods,

   ppp ttt21 

Finding equilibrium price, ()  p           ()1  

We obtain that intertemporal equilibrium price is independent of the expectations coefficient  according to the assumptions of the model.

69. Recall the nonlinear Cobweb model discussed in chapter 10. Consumers are negatively influenced by current price. Producers, on the other hand, make their decisions based on price in the current and the previous two periods.

d  qptt22  ,0  s  qppptttt221   ,0 

Solve the model for intertemporal equilibrium price and quantity. Express the characteristic roots. What is the condition for a single root to exist? Is dynamic stability of market price plausible?

Solution:

ds In equilibrium qqtt22 and

 ptttt221 ppp

Taking the natural log of both sides,

ln  lnptttt221 ln lnppp ln ln

(1)ln pppttt11 lnlnlnln

Normalizing,  1ln lnppp ln ln ttt21111

Setting yptt ln ,

 1ln yyy ttt21111

ln  ln y  or  1 2   (1)1   1 ln  ln p  2  

Taking the antilog of both sides,

700 Problems Book to Accompany Mathematics for Economists

1 ln 1 2  22ln    pe() e   From the demand function, we express equilibrium quantity

  2 2   22   qp   

If the market cleared at any point in time, we would have

()pp () 2

  ()p 2  and  1  2  p    which is the equilibrium value obtained previously. The characteristic roots are 2 4   1(1)(1)22 4(1) a  1,2 22(1)  There will be a single real root, if the parameters are such that  21 . Furthermore,  1 aa b and aa b so aa (0,1) 12 1  1 12 2  1 12 11 (1aa )(1  )  1 ( aaaa  )   1   1  1 12 121211   1

Since the parameters  and  are positive by definition, we have (1 aa12 )(1 ) 1 and both a1 and a2 are negative (why?), we may have a1 1 and a2 1. Therefore, dynamic stability for market price is plausible.

70. Assume that the nonlinear Cobweb model describes the electricity sector. Consumers are negatively influenced by rises in electricity price in three consecutive periods. They have switched to alternative energy sources in the previous periods, so they are most elastic at present. The electrical company, on the other hand, made a huge investment in period t, so supply was most elastic then. The company is stimulated by a higher price in each period.

 d  pt2 qt2  ,0  ptt1 p s  qppptttt221   ,0 

Solve the model for intertemporal equilibrium price and quantity. Express the characteristic roots. What is the condition for complex roots to exist?

Solution:

ds In equilibrium, qqtt22 and   pt2    pttt21pp pptt1

Chapter 11. Advanced Differential and Difference Equations 701

Taking the natural log of both sides,

ln  lnpttt21 lnpp ln ln ln p tt  21 ln p ln p t

(  1)lnpptt21 2ln ( 1)ln p t ln ln

Normalizing,

21ln   lnppp ln ln ttt21111

Setting yptt ln ,

21ln   yyy ttt21111

ln  ln y  or 21 4   (1)1   1 ln  ln p  4  

Taking the antilog of both sides,

1 ln 1 4  44ln    pe() e   From the demand function, we express equilibrium quantity

 2  22 4  2  44   qp   

If the market was constantly in equilibrium, we would have

()pp22 () 

  ()p 4  and  1  4  p    which confirms our previous result. For the characteristic roots, 24(1)    1(1)(1)2 24(1)(1) a  1,2 22(1) 

For complex roots to obtain, we need (1)(1)4   .