Discussion Paper No.218

Theocharis Problem Reconsidered in
Differentiated Oligopoly
Akio Matsumoto

Ferenc Szidarovszky

Chuo University

University of Pécs
Ｆebruary 2014I

INSTITUTE OF ECONOMIC RESEARCH
Chuo University
Tokyo, Japan

Theocharis Problem Reconsidered inDi¤erentiated Oligopoly
Akio Matsumotoy
Chuo University

Ferenc Szidarovszkyz
University of Pécs

Abstract
Oligopolies with product di¤erentiation are examined with both quantity adjusting and price adjusting …rms. Conditions are given for the positivity of the equilibrium outputs as well as for that of the equilibrium
prices. The stability of the positive equilibria is then investigated, thepositivity and stability conditions are also compared. The dependenceof these conditions on the number of …rms, their quali…cations and thedegree of relation between the goods is discussed. The analysis is given
by assuming both discrete and continuous time scales.

Keywords: Oligopolies with product di¤erentiation, stability, nonnegativity, Cournot and Bertrand competitions

JEL classi…cation: C62, D43

The authors highly appreciate the …nancial supports from the MEXT-Supported Program
for the Strategic Research Foundation at Private Universities 2013-2017, the Japan Society for
the Promotion of Science (Grant-in-Aid for Scientifc Research (C) 24530201 and 25380238)
and Chuo University (Grant for Special Research). The usual disclaimer applies.
y Department of Economics, Chuo University, 742-1, Higashi-Nakano, Tokyo, 192-0393,
Japan. akiom@tamacc.chuo-u.ac.jp
z Department of Applied Mathematics, Universty of Pécs, Pécs, Ifjúság u-6, H-7624, Hungary. szidarka@gmail.com

1

1

Introduction

In his seminal paper, Theocharis (1960) constructs the n-…rm Cournot oligopoly
model with a linear demand function and constant marginal costs and studies
the asymptotic stability of the steady state under discrete-time quantity adjustment. It is then found that stability depends only on the number of …rms; thesteady state is globally stable in the duopoly, marginally stable in the triopoly
and destabilized if the number of …rms becomes more than three. It is also
pointed out that this provocative result (sometimes called as the Theocharisproblem) might be obtained under the limited environment in which the …rms
have full and instantaneous information on the demand function, their own
cost as well as the competitors’costs, the goods are perfectly substitutes (i.e.,
homogenous) and expectations are naively formed. As a natural consequence,
the Theocharis model has been extended in various directions to weaken theserestrictions. Focusing on the type of the adjustment process, McManus andQuandt (1961) and Hahn (1962) prove that the steady state is asymptoticallystable in the continuous-time adjustment process with demand and cost functions having the appropriate slopes. It is also con…rmed in Fisher (1961) that thesteady state can be stabilized when the adjustment coe¢ cient is small and/or
the marginal cost is increasing in the discrete-time adjustment process. Further, models with product di¤erentiation are discussed in Haster (1966), Krelle(1976), Okuguchi (1976), Moorthy (1985) and Friedman (1986) among others.
Models with product di¤erentiations can be considered as special cases of multiproduct oligopolies, which were …rst formulated by Selten (1970) for quantity
strategies and by Eichhorn (1971a, b) for price strategies. Szidarovszky et al.
(1994) examine the stability of equilibria in linear multi-product oligopolies withadaptive adaptive adjustments and derive su¢ cient and necessary stability conditions. A comprehensive summary of the main results on the existence, uniqueness and stability of equilibria in multi-product oligopolies is given in Okguchi
and Szidarivszky (1999) which can be easilty applied to oligopolies with productdi¤erentiation as special cases. Goldberg (1995) presents an application to the
U.S. Automobile Industry. Comparision of Bertrand and Cournot oligopolies
are discussed in Okuguchi (1984), Vives (1984) and Cheng (1985) among others. More recently, taking into account of the non-negativity constraints on thevariables, Cánovas et al. (2008) consider the global dynamics to show that then-…rm model has simple dynamics (i.e., it converges either to a monopoly or a
duopoly or to a two periodic oscillation).
Our purpose of this study is to reconsider the Theocharis problem in differentiated oligopoly, following the spirit of Hadar (1966) in which the steadystate is shown to be stable if the degree of di¤erentiation is weak and the coef…cient matrix of the naive dynamics system has a dominant diagonal under thegeneral forms of the demand and cost functions. In this study, we take away
the assumption of the diagonal dominance but restore the linear structure of
the Theocharis model where demand is linear and marginal costs are constant.
Introducing product di¤erentiation with micro-foundation leads to an environment in which the …rms are also di¤erentiated; each …rm faces its own demand
function and operates production with di¤erent production costs. These two
sorts of di¤erentiation a¤ect dynamics. Further we introduce the price adjustment (i.e, Bertrand competition) in addition to the quantity adjustment (i.e.,
Cournot competition) to investigate the e¤ects caused by the choice of opera2

tional strategies and then address dynamics in the continuous-time scales. In
short, we raise a question whether the Theocharis problem still exist under thosebroader circumstances.
The rest of the paper is organized as follows. Section 2 brie‡y gives a microfoundation of the linear demand functions of the di¤erentiated goods. Section 3
presents an n-…rm linear oligopoly model in Cournot competition and examine
asymptotic dynamics in the extended model. Section 4 considers the Theocharisproblem in Bertrand competitions. Concluding remarks are given in Section 5.

2

Consumers

It is assumed that there is a continuum of consumers of the same type and theutility function of the representative consumer is given as
0
1
n
n X
n
n
X
X
X
1@
qi2 + 2qi qj A I;
(1)
U (q; I) =
i qi
2
i=1
i=1 j>ii=1
Pnwhere q = (qi ) is the quantity vector, I = i=1 pi qi with pi being the price ofgood i; i measures the quality of good i and 2 [ 1; 1] measures the degreeof relation between the goods:> 0; < 0 or = 0 imply that the goods
are substitutes, complements or independent. Moreover, the goods are perfect
substitutes if= 1 and perfect complements if= 1: In this study, we
con…ne our analysis to the case in which the goods are imperfect substitutes orcomplements and are not independent, by assuming that j j < 1 and 6= 0:
The linear inverse demand function (or the price function) of good k is
obtained from the …rst-order condition of the interior optimal consumption ofgood k and is given by
pk =

k

qk

n
X

qi for k = 1; 2; :::; n;

(2)

i6=k

where n2 is assumed. More compactly, equation (2) can be written in thevector form: the price vector is a linear function of the output vector:
p=

(3)

Bq;

where p = (pi ); = ( i ) and B = (Bij ) with Bii = 1 and Bij =Since B is invertible1 , solving (3) for q yields the direct demandq = B 1(for i 6= j:
(4)

p);

where the diagonal and the o¤-diagonal elements of B 1 are, respectively,(1

1 + (n 2)and)(1 + (n 1) )(1)(1 + (n1) ):1 The n by n matrix B is invertible if det B = (1)n 1 (1 + (n 1) ) 6= 0: It is clearly
invertible when > 0: In the case of < 0; the inequality constraint 1 + (n 1) > 0 will beassumed in Assumption 2 below and it will guarantee the invertiblity of B:

3

Hence the direct demand of good k, the k th -component of q, is linear in theprices and is given as(1 + (n2) )( k

n
P

pk )

( i

pi )

i6=k

qk =

(1

)(1 + (nfor k = 1; 2; :::; n:

1) )

(5)

For the sake of the later analysis, let us de…ne the admissible region of( ; n) by D(+) or D( ) according to whether the goods are substitutes or complements:
D(+) = f( ; n) j 0 < < 1 and 2 < ngandD( ) = f( ; n) j

3

1<

< 0 and 2 < ng:

Quantity-adjusting …rms

In Cournot competition, …rm k chooses a quantity qk to maximize its pro…t k =(pk ck )qk subject to its price function (2), taking the other …rms’quantities
given. We assume a linear cost function for each …rm, so that the marginal cost
ck is constant and non-negative. To avoid negative optimal production, we also
assume that k ck is positive.
Assumption 1.

ck

0 and

ck > 0 for all k:kAssuming interior maximum and solving its …rst-order condition yield thebest reply of …rm k,
ck

k

qk =

2

2

n
X

qi for k = 1; 2; :::; n:

(6)

i6=k

It can be easily checked that the second-order condition is certainly satis…ed.
The Cournot equilibrium output and price for …rm k are obtained by solving
the following simultaneous equations:
qk +

2

n
X
i6=k

qi =

ck

k

2

for k = 1; 2; :::; n;or in vector form,
B C q = AC ;C
C
C
where AC = (( i ci )=2) and B C = (Bij
) with Bii= 1 and Bij= =2 forC
i 6= j: Since B is invertible, the Cournot output vector is given by

qC = B C

1

AC ;where the diagonal and o¤-diagonal elements of B C
2(2 + (n 2) )and(2)(2 + (n 1) )
4

(2

1

2
)(2 + (nare, respectively,

1) )

:

Hence the Cournot equilibrium output of …rm k is
qkC =

n
P

ck

k

2

(2

)(2 + (n1) ) i=1

and the Cournot equilibrium price of …rm k ispC
k =k + ck

n
P

ck

2

(2

)(2 + (n( i1) ) i=1

ci )

( i

(7)

ci ):

(8)

Subtracting (7) from (8) yields pC
ck = qkC and then by substituting it intokthe pro…t function, the Cournot pro…t becomes
C
k =

3.1

qkC

2

:

(9)

Non-negativity

Before proceeding to the stability analysis, we check the non-negativity conditions for the variables at the Cournot equilibrium point. The relations pC
ck =k
qkC and equation (9) imply that the Cournot price and the Cournot pro…t are
0. We
positive if the Cournot output is positive. So it is su¢ ce to show qkC
start with the case of > 0 in which 2 + (n 1) > 0 always. Equation (7) canbe written as
qkC =
where

ck
) (2 + (nk(2k is de…ned as

1) )

f2

+ (1

k )n

1 Pn
( i ci )
n i=1
:
k =
ck
k

g

(10)

(11)

The constant term k of the price function is the maximum price of the productqk and thought to be a proxy of the product quality (that is, the high-priceproduct is usually highly-quali…ed) and thus k ck could measure the net
quality. k is the ratio of the average net quality over the individual net qualityof …rm k. When k < 1; …rm k is called higher-quali…ed as its individual net
quality is larger than the average net quality. On the other hand, when k > 1;
…rm k is called lower-quali…ed as the individual net quality is less than the
average net quality. Equation (10) implies that qkC > 0 if k
1. A di¤erent
form of equation (10) is
qkC =and leads to qkC > 0 if

( k ck )( k(2
) (2 + (n

1)
1) )

2
( k

1)

n

(12)

k > 1 and

n<

2
( k

1)

:

(13)

We call the locus of ( ; n) satisfying relation (13) with equality the zero-output
curve.

5

We now turn to the case of< 0 in which the sign of 2 + (n 1) is
ambiguous. Equation (7) indicates qkC > 0 if 2 + (n 1) > 0. Positiveity of theoutput does not depend on whether the …rm is higher- or lower-quali…ed if
2

n<1

:

We now suppose the opposite situation where 2 + (n 1) < 0. Then equation(10) and k
1 together imply qkC < 0: The lower-quali…ed …rms temporarilychooses zero-production and will exit the market if nothing is changed. Declining
the number of …rms could reverse the inequality of the condition, 2+(n 1) < 0;and in consequence, the remaining …rms produce positive output. On the other
hand, equation (12) implies that qkC > 0 if k < 1 and
2
( k

n>

1)

:

(14)

Solving 2 + (n 1) = 0 for n and subtracting the resultant expression fromthe right hand side of (14) present
2
( k

1

1)

2

=

(2
(

) k
)(1

k)

> 0:

This inequality implies that the zero-output curve with < 0 is in the region of
2 + (n 1) < 0: Hence the non-negativity conditions for Cournot output aresummarized as follows:
Lemma 1 In the case of> 0; Cournot output is positive if

either the …rm is higher-quali…ed (i.e.,or
it is lower-quali…ed (i.e.,
while in the case ofk > 1) and n <k < 1)

2( k< 0; it is positive if
either n < 12orthe …rm is lower-quali…ed (i.e.,

3.2

1)

k < 1) and n >

2
( k

1)

:

Stability

We now turn attention to the stability of the Cournot output. Best response
dynamics is assumed with naive expectations, when each …rm believes thatthe other …rms remain unchanged with their outputs from the previous period.
By assuming discrete time scales, the best response (6) gives rise to the timeinvariant linear dynamic systemqk (t + 1) =

ck

k

2

2

n
X

6

i6=k

qi (t); k = 1; 2; :::; n:(15)

Substituting qk (t + 1) into the price function (2) of …rm k yields the pricedynamic equation associated with the output dynamics:
pk (t + 1) =knXqk (t + 1)

qi (t + 1); k = 1; 2; :::; n:

(16)

i6=k

Equations (15) and (16) imply that the price dynamics is essentially the sameas the quantity dynamics: In other words, the Cournot price is stable (resp.
unstable) if the Cournot output is stable (resp. unstable). Therefore it isenough for our purpose to draw attention only to the stability of the Cournot
output.
The coe¢ cient matrix of system (15) is
1
0
0
B
2
2 C
C
B
0
C
B
JC = B 2
2 C:
C
B
A
@
0
2
2
The corresponding characteristic equation reads
Ij = ( 1)n

jJ C

n 1

+

2

(n

1)

= 0;

2

which indicates that there are n 1 identical eigenvalues and one di¤erent
eigenvalue. Without loss of generality, the …rst n 1 eigenvalues are assumed
to be identical,
C
1 =C
2 = ::: =

C
n 1 =

2

and

(n

C
n =

1)
2

:

Since j j < 1 is assumed; the …rst n 1 eigenvalues are less than unity in absolutevalue. It depends on the absolute value of Cn whether the Cournot output isCstable or not. It is clear that n < 1 for n = 2 and 3: Solving C
n < 1 forn > 3 presents the stability conditions of the Cournot output2 :
n<1+

2

if

> 0 and n < 1

2

if

< 0:

(17)

We can now summarize these stability results as follows.
Lemma 2 In Cournot competition with discrete time scales, the Cournot output
is stable, marginally stable or unstable according to the number of the …rms isless than, equal to or greater than 1 + j 2 j .
2 If the Jacobian of the price function satis…es the diagonally dominant condition (i.e.,P
jdpk =dqk j > i6=k jdpk =dqi j), then we haven<1+

under whichCn1< 1: However, we proceed our analysis without this strong assumption.

7

Lemma 1 is concerned with the non-negativity condition for the Cournot
output and so is Lemma 2 with the stability condition. Combining these lemmas
provides the condition for which the Cournot output is positive and stable. We
now investigate the following e¤ects on stability of the positive Cournot output;
the di¤ erentiation e¤ ect caused by a change in , the heterogeneous e¤ ect caused
by a change in k and the time scale e¤ ect caused by changing the time scale
to continuous from discrete.

3.3

Di¤erentiation E¤ect

To see the di¤erentiation e¤ect on stability, we assume away the heterogeneityof the …rms for a while only for analytical simplicity by assuming that k =and ck = c for all k: This simpli…cation leads to k = 1 for all k. Due toequation (10), qkC > 0 for all k if > 0 or if < 0 and 2 + (n 1) > 0: The
positive-sloping black curve in Figures 1(A) and the negative-sloping black curvein Figure 1(B) are the corresponding C
n = 1 or marginal-stability curves. The
yellow regions under the black curves are the stability regions. Thus, the positive
Cournot output is always stable if n = 2 or n = 3 and the goods are neither
perfect substitutes ( = 1) nor perfect complements ( = 1). The second
result implies that the Theocharis problem does not arise when the goods aredi¤erentiated (i.e., j j 6= 1) in the triopoly market in which equation (15) exhibit
marginal stability when j j = 1. For n 4; stability depends on the degree ofthe di¤erentiation. Solving C
n = 1 gives the threshold value of ,
C
n

=

2
n

1

:

For example, when n increases to 6 from 4 with an increment of unity, thethreshold value decreases to 2=3; 1=2 and 2=5, and Cournot output is stable if
j j<

2
1
2
when n = 4; j j < when n = 5 and j j < when n = 6325

and becomes unstable when the inequality is reversed. In short, Cournot output
for n
4 can be stable if the goods are strongly di¤erentiated in the sensethat the degree of di¤erentiation in absolute value is less than 2=(n 1). A
larger number of n could be a destabilizer in the sense that the stability region
shrinks as n increases. Nevertheless, the di¤erentiation partially settles the
Theocharis problem. Further, observing Figures 1(A) and 1(B), we see that thedi¤erentiation e¤ect is symmetric, that is, one …gure is the mirror image of theother.
Lemma 3 If k = 1 for all k, then the Cournot output is positive and stable if
the degree of di¤ erentiation satis…es
j j<

3.4

2
n

1

:

Heterogeneity E¤ect

Since k = 1 for all k is a special case, we examine the heterogeneous e¤ect of> 0; Cournot output is always positive if k < 1:
k 6= 1 on stability. For
8

The stability condition of the positive Cournot output produced by the higherquali…ed …rm is the same as the one with k = 1,
n<1+

2

:

Graphically the Cournot output is positive and stable in the yellow region of
Figure 1(B) and unstable in the white region. The heterogeneity among thehigher-quali…ed …rms does not a¤ect stability.
When …rm k is lower-quali…ed (i.e., k > 1), the stability region is accordingly modi…ed. The zero-output curve and the marginal-stability curve yield
2
( k

1)

1+

2

=

2+
( k 1)

4
2+

k

(18)

where the …rst factor on the right hand side is positive for k > 1 and thesecond factor is negative if k > 2 and positive if 1 < k < 4=3. We distinguishthree cases, Case I where k > 2; Case II where k < 4=3 and Case III where
4=3 < k < 2: In Case I, the non-negativity condition is stronger than thestability condition. Stability of the Cournot output can be examined with theconsideration of the non-negativity condition. The negative-sloping blue curvein Figure 1(B) is the zero-output curve with k = 2. The Cournot output ispositive and stable in the yellow region under the zero-output curve. It is still
stable in the yellow region between the zero-output curve and the marginalstability curve but it is negative there. Notice that as k becomes larger than2; the blue curve shifts downward making the non-negativity region smaller.
The heterogeneity k > 2 can be a destabilizer.
Lemma 4 If

k > 2; then the Cournot output is positive and stable if

n<

2
( k

1)

:

We draw attention to Case II. The zero-output curve with k = 4=3 is in red
and located above the marginal-stability (black) curve in Figure 1(B). It canbe veri…ed that the zero-output curve shifts upward as k decreases. Thus for
1 < k < 4=3; equation (18) implies that the stability condition is stronger than
the non-negativity condition. In Figure 1(B), the Cournot output is positiveand stable in the yellow region below the black curve. The stability conditionsof the higher-quali…ed …rm (i.e., k < 1) and the lower-quali…ed …rms (i.e.,k > 1) are the same. However the optimal behavior is di¤erent when thestability condition is violated. The Cournot output of the lower-quali…ed …rm
is negative in the white region above the red curve so that the …rm chooses to
produce nothing and may exit the market. Further, it is positive and unstablein the white region between the red curve and the black curve. As is seen above,
the higher-quali…ed …rm produces positive output which is unstable, implyingthat the …rm may stay in the market even if the stability condition is violated.
Notice that changing k does not a¤ect the stability condition and thus has noheterogeneous e¤ect.
Lemma 5 If 1 <k < 4=3; then the Cournot output is positive and stable if

n<1+

9

2

:

In Case III, the zero-output curve intersects the marginal-stability curve at
the threshold value k determined as
k =

2(2

k)
k

where k ! 1 as k ! 4=3 and k ! 0 as k ! 2: For > k ; the Cournot
output satisfying n < 1 + 2= is positive and stable and for < k ; the Cournot
output satisfying n < (2)=[ ( k1)] is positive and stable. In Figure
1(B), the negative-sloping green curve is the zero-output curve with k = 3=2,
intersects the black curve at the point of k = 2=3 and n = 4 and passes throughthe point of = 1 and n = 2. The alternative expression of the zero-output curve
is
2
0
n =
1 + ( k 1)nwhere, given

k = 3=2,0
2 = 1 for n = 2;20
for n = 4:4 =40
for n = 3 and
3 =
5

3

Thereby the Cournot output is always positive and stable for any 2 (0; 1) inthe duopoly market (i.e., n = 2). It is positive and stable for < 03 in thetriopoly market (i.e., n = 3). For > 03 it is stable but negative. So …rm kchooses to produce nothing. For n = 4; notice that 4 = 04 : Cournot output ispositive and stable for < 04 and chosen to be zero for > 04 ; leading to the…rm’s exit from the market. For n = 5; Cournot output is positive and stablefor < 5 ; positive and unstable for 5 < < 05 ; negative and unstable for> 05 : We then have the hybrid result that the Cournot output is positive andstable for < min( n ; 0n ) and is zero for > 0k :Lemma 6 If 4=3 <k< 2; then the Cournot output is positive and stable ifn < min 1 +

2

;

2
( k

:

1)

In the case of < 0; the situation is much simpler since the non-negativity
condition is identical with the stability condition. In Figure 1(A), the positive
sloping black curve is not only the zero-output curve but also the marginalstability curve. The positive Cournot output is stable in the yellow region below
the black curve. The heterogenous e¤ect among the …rms (i.e., k 6= 1) provides
a sharp contrast in whether the goods are substitutes or complements. In summary, the heterogeneity a¤ects stability of positive Cournot output by shifting
the zero-output curve when > 0 and has no e¤ect on stability when < 0.
Therefore we can say that the heterogeneity partially settles the Theocharisproblem only when the goods are substitutes. Lemmas 1-6 leads to the main
result obtained under the Cournot competition
Theorem 1 In the case of > 0; the positive Cournot output is stable if oneof the following exclusive conditions holds:(i)k < 4=3 and n < 1 +

10

2

,

(ii)

4
<
3

k < 2 and n < min

1+

and
(iii)
while in the case of

2

2
( k

k > 2 and n <

;

2
( k

1)

1)

< 0; it is stable, irrespective of the value of
if n < 1(A) Complements:

2

<0

k,

:

(B) Substitutes:>0

Figure 1. Cournot competition: stability and non-negativity region

3.5

Time Scale E¤ect

If continuous time scales are assumed by replacing q(t + 1)
system (15) is modi…ed as follows:
q_k (t) =

ck

k

2

2

X

qi (t)

q(t) by q(t);_thenqk (t):

(19)

i6=k

Notice that the steady state of the continuous system is the same at the one ofthe discrete system. Its Jacobian matrix is
0
1
1
B
2
2 C
B
C
1
B
C
JC I = B 2
2 C:
B
C
@
A
1
2
2
The corresponding characteristic equation has the form,
jJ C

I

Ij = ( 1)n

n 1

+1

2
11

+1+

(n

1)
2

=0

implying that the eigenvalues are
1+
If

2

and

(n

1

> 0; then both are negative and if

1)
2

:

< 0; then they are negative if2

n<1

which is the same stability condition as in the discrete case. We then summarize
these results as follows:
Theorem 2 In Cournot competition with continuous time scales, the Cournot
output is stable for any n
2 if the goods are substitutes and it is stable forn < 1 2 if the goods are complements.
Comparing Theorem 1 with Theorem 2 reveals that the time-scale e¤ect
solves the Theocharis problem since a change of the time scales completely
stabilizes the Cournot point in the D(+) plane. On the other hand the scalechange does not change the stability condition at all if< 0. The timesacle e¤ect exhibits a sharp di¤erence, depending on whether the goods aresubstitutes or complements.

4

Price-adjusting …rms

In Bertrand competition, …rm k chooses the price of good k to maximize thepro…t k = (pk ck )qk subject to its direct demand (5), taking the other …rms’
prices given. Solving the …rst-order condition yields the best reply of …rm k,
pk =

n
P

k + ck2

2[1 + (n2) ] i6=k( ipi ); for k = 1; 2; :::; n:(20)

The second-order condition for an interior optimum solution is
@2 k
=
@p2k

2 (1 + (n 2) )< 0;
(1)(1 + (n 1) )(21)

where the direction of inequality depends on the parameter con…guration.3 For( ; n) 2 D(+) ; we see that (21) is always satis…ed. On the other hand, for ( ; n) 2
D( ) ; we need an additional condition to ful…ll the second-order condition. Since
1 + (n1) < 1 + (n2) ;for < 0, the required condition is either 0 < 1 + (n 1) or 1 + (n 2) < 0.
Since the …rst inequality is the stability condition as well as the non-negativity
condition under the Cournot competition, we take it even under the Bertrand
competition:
Assumption 2. 1 + (n1) > 0 for< 0.

3 Note that inequality (21) is always ful…lled for n = 2.

12

The Bertrand equilibrium prices are obtained by solving the simultaneous
equationspk

2[1 + (n

n
P

pi =

2) ] i6=knP

k + ck2

2[1 + (ni2) ] i6=k

for k = 1; 2; :::; n with unknowns pk : In vector form,
B B p = AB ;kwith AB = ( k +c2

2[1+(n 2) ]2[1+(n 2) ] for i 6= j. Since B

nPi ), B

B

B
B
B= (Bij) where Bii= 1 and Bij
=

i6=k

B

is invertible, the solution is1

p = BBAB ;where the diagonal and o¤-diagonal elements of B B

1

are, respectively,

2 (1 + (n 2) )
2) )
and
:
3) )
(2 + (n 3) )(2 + (2n 3) )2(1 + (n 2) )(2 + (n(2 + (n 3) )(2 + (2nHence, the Bertrand equilibrium price and output of …rm k are given by(2+(n 3) )[(1+(n 1) )(pBk =k +ck )ck ](1+(n 2) )

(

i

ci )

i=1

(2+(2n 3) )(2+(n 3) )

and
qkB =

1 + (n 2)(pB)(1 + (n 1) ) k(1(2+(n 3) )(1+(n 1) )(

k

ck )

(1+(n 2) )

(23)
n
P

(

i=1

ck =

(22)

ck )

with
pB
k

n
P

(2+(2n 3) )(2+(n 3) )

i

ci )

:

(24)

Due to (23), the Bertrand pro…t of …rm k becomes
(1
B
k =

4.1

)(1 + (n 1) ) B 2(qk ) :
1 + (n 2)(25)

Non-negativity and Stability

Equation (23) implies that the Bertrand output is positive if pB
ck is positive.
k
qkB 0 leads to pB
>
0
but
not
vice
versa.
So
we
con…neourattention
to thekconditions for qkB
0: Equation (24) implies that pB
cisalwayspositiveifkk< 0 and Assumption 2 holds or it is nonnegative if > 0 andz B ( ; n)
where

(26)

k

k is de…ned by (11) andz B ( ; n) =(2 + (n 3) )(1 + (n(1 + (n 2) )n
13

1) )

:

(27)

Since it can be con…rmed that z B ( ; n) > 1 for 0 < < 1 and 2
n < 1;4
the Bertrand price is positive if …rm k is higher-quali…ed or if the …rms arehomogeneous. For k > 1; it is positive if n n+ whereq
3 4 + k (21) + (1 2 )2 ( 2k + 1) + 2 k (1 2 2 )n+ =2( k 1)is a positive solution of z B ( ; n) =Lemma 7 In the case ofk : In short, we have the following:> 0; the Baretrand price is positive if

either the …rm is higher-quali…ed or the …rms are all homogenousorthe …rm is lower-quali…ed and nwhile in the case oftion 2.

n+< 0; the Bertrand price is always positive under Assump-

In examining stability of the Bertrand price with discrete time scales, we
assume best response dynamics with naive expectations on price formation andobtain the following system of time-invariant di¤erence equations:pk (t + 1) =nP

k + ck2

2[1 + (n2) ] i6=k[ i

pi (t)] for k = 1; 2; :::; n: (28)

Similarly to the Cournot competition, we can also obtain the output di¤erence
equations under Bertrand competition by substituting pk (t + 1) into the direct
demand function (5):
(1 + (n2) )( knP

pk (t + 1))( i

pi (t + 1))

i6=kqk (t + 1) =(1)(1 + (n

:

1) )

(29)

It is clear from (28) and (29) that the output dynamics is synchronized with theprice dynamics. We focus on price dynamics henceforth. The coe¢ cient matrixof the price adjusting system (28) is0102[1 + (n 2) ]2[1 + (n 2) ] C
B
B
C
B
C
0
C:
2[1
+
(n
2)
]
2[1
+
(n
2)
]
JB = B
B
C
B
C
@
A
0
2[1 + (n 2) ] 2[1 + (n 2) ]This matrix has the same structure as J C . So replacing(n 2) ] yields the eigenvalues of J B ,B
1 =B
2 = ::: =Bn 1 =4 (2 + (n3) )(1 + (n

1) )

and

2[1 + (n(1 + (n2) ]and2) )n = (1

< 1:

14

B
n =

of

C
k by

=2[1 +(n 1):
2[1 + (n 2) ])(2 + (2n3) ) > 0 for n

1

When

> 0 and nBk2; we have

< 1 for k = 1; 2; :::; n: That is, theBertrand price is globally asymptotically stable in D(+) :5 Neither product differentiation nor homogeneity a¤ect the stability of the Bertrand price if thegoods are substitute. When < 0 and Assumption 2 is given, 0 < B
k < 1
holds for k = 1; 2; :::; n 1: The value of B
n is clearly negative and the stability
condition B
1 can be rewritten as
n >
n<

5
3

2
:
3

(30)

Under Assumption 2, the Bertrand price is stable if the following condition
holds,6
5
2
1
n < min
; 1
(31)
3 3
The stability of the Bertrand price is summarized as follows:
Lemma 8 In Bertrand competition with discrete time scales, (i) the Bertrand
price is stable for any n 2 if the goods are substitutes whereas it is stable if
the goods are complements, Assumption 2 holds andn < min

5
3

2
; 1
3

1

:

Lemma 7 is concerned with the non-negativity condition of the Bertrand
price and Lemma 8 is concerned with the stability condition. Thus combining
these lemmas gives the following result:
Theorem 3 In the case of> 0; the Bertrand price is positive and stable ifeither the …rm is higher-quali…ed or homogenousorthe …rm is lower-quali…ed and nwhile in the case ofn+< 0; it is positive and stable if Assumption 2 holds andn < min

5
3

2
; 1
3

1

:

We graphically con…rm Theorem 3. In Figure 2(A) in which < 0, thepositive-sloping red and blue curves correspond, respectively, to the marginalstability curve and the SOC (i.e., the second-order condition) curve under which
Assumption 2 holds. Thus the Bertrand price is positive and stable in the yellow
region. It is checked that the two curves intersects for = 1=2 and n = 3: Thus
the Bertrand price is always stable in the duopoly and stable if > 1=2 in
5 Okuguchi (1987) has already shown the same result with a more general demand function.
6
P If the Jacobian of the demand function is diagonally dominant (i.e., jdqk =dpk j >i6=k jdqi =dpk j), then we have
3
1
n<
2
2
which is stronger than the price-stability condition and thus leads to the stability of Bertrand
price. However we do not assume this strong condition in what follows.

15

the triopoly. For ndetermined by

4; it is stable if>B
n where the threshold value

B
n is

2
:
5 3n
It is also checked that the Bartrand price is unstable but satis…es Assumption
2 if 1=(1 n) << B
n while it is unstable and violates Assumption 2 if
< 1=(1 n).
In the case of > 0; the Bertrand price is stable but its positivity is sensitive
to the value of k . Indeed, it is positive and stable if the …rm is higher-quali…edor the …rms are all homogenous. On the other hand, if the …rm is lower-quali…ed,then zero-output curve on which n = n+ divides the D+ plane and the Bertrand
price is positive and stable only in the region below the curve. In Figure 2(B),three curves with k = 3=2; k = 5=2 and k = 7=2 are illustrated as threenegative-sloping curves. Increasing the value of k shifts the zero-output curve
leftward. Thus these colored regions give the stability region of the positive
Bertrand price when k = 3=2: The union of the blue and dark-orange regions
is the stability region when k = 5=2 and increasing k to 7=2 takes away
the blue region, limiting the stability region to the dark-orange region. The
stronger heterogeneity makes the stability region smaller through shifting thezero-output curve leftward. In the n-poly model, the Bertrand price is positiveand stable for < k where k solves z B ( ; n) = k : As seen in Figure 2(B),the threshold values of for n = 2 are
p
p
p
2 <
1 + 3 ' 0:732; 52 < 2 + 6 ' 0:449 and 72 < 3 + 11 ' 0:317:
3
Bn =It is also seen that the larger number of the …rms is an destabilizing factor as
is decreasing in n.
k(A) Complements:<0(B) Substitutes:>0

Figure 2. Bertrand competition: stability and non-negativity region
If continuous time scales are considered, then the best response dynamics
has the Jacobian J B I with eigenvalues1

2 (1 + (n

2) )

and

16

1+

(n 1)
:
2(1 + (n 2) )If> 0, then these eigenvalues are negative. If
n<

3
2

1

< 0; then they are negative if:

However this inequality always holds under Assumption 2 and thus is ine¤ective.
In summary, a change of the time scales does not a¤ect the stability conditions
regardless of whether the goods are substitutes or complements:
Theorem 4 Under Bertrand competition with continuous time scales, the Bertrand
point is stable for any n if the goods are substitutes and for ( ; n) satisfying equationrelation (31) if the goods are complements and Assumption 2 holds.

4.2

Strategy E¤ect

We can consider the strategy e¤ect by comparing Figure 1 with Figure 2. We
start with the case of < 0: it is easily veri…ed that
max

5
3

2
; 1
3

1

2

<1

:

This inequality implies that the marginal stability curve under the Cournot
competition is located above the partition curve under the Bertrand competition. In other words, the Cournot output is more stable than the Bertrand pricein the sense that the stability region for the positive Cournot output is larger
than the stability region for the positive Bertrand price.
On the other hand, in the case of > 0; the crucial factor is whether the…rm is higher- or lower-quali…ed. For the higher-quali…ed …rm, the Bertrand
price is always positive and stable while the Cournot output is positive andstable if n < 1 + 2= : Changing the strategy to price-adjustment from quantityadjustment works for stabilization. When the …rm is lower-quali…ed, its zerooutput curve is obtained from inequality (13) as
n = f( ;k ) :=

2
( k

(32)

1)

under the Cournot competition and n = n+ under the Bertrand competition.
We abbreviate the former to the C-curve and the latter to the B-curve. Thensolving f ( ; k ) = n+ for present the following; the C-curve is above, respectively below, the C-curve according tok <2; respectively

k >

2

:

The case 2 > k (> 1) is divided into two sub cases according to the relative
location of the C-curve with respect to the marginal-stability curve, 1 < k <4=3 and 4=3 < k < 2: The …rst subcase is depicted by Figure 3(A) in whichk = 1:3 is taken and the second case by Figure 3(B) with k = 3=2. In both…gures, the red curve is described by n = f ( ; k ); the green curve by n = n+
and the black curve by the marginal-stability condition, n = 1 + 2= : As shownin Figure 2(B), the C-curve is above the black curve implying that the Cournot
output is stable in the union of the yellow and blue regions. The B-curve crosses
the marginal stability curve from below, adding the red region to the yellow
17

region and subtracting the blue region from it to construct the stability regionof the Bertrand price. In the second subcase, the C-curve as well as the B-curve
shifts down by increasing the value of k : As a result, as already shown in Figure2(B), the stability region of the Cournot output is the union of the yellow, blue
and meshed-yellow regions. The stability region of the Bertrand price is theunion of the yellow and red regions. Since it may not be determined which islarger, the red region and the blue region, the strategy e¤ect caused by changing
the quantity adjustment to the price adjustment on stability is ambiguous whenk < 2.

(A) 1 <

k < 4=3(B) 4=3 <k <2Figure 3. Stability regions under Cournot and Bertrand competitionFor k
2; the solution of the simultaneous equations n = f ( ; k ) andn = n+ is = 2= k and n = 2: As far as n 2 is assumed, the C-curve is always
located above the B-curve, implying that the Cournot output is more stable
than the Bertrand price in the sense that the stability region under the Cournot
competition is larger than the stability region under the Bertrand competition.
Summarizing the results, we have
Theorem 5 In the discrete time framework, when the goods are substitutes,
the Bertrand competition is more stable for k < 1, the Cournot competition ismore stable for k > 2; it is ambiguous which is stronger for 1 < k < 2
The Cournot output and Bertrand price are always stable in the continuoustime scale if the goods are substitutes. On the other hand, if the goods arecomplements, the changing of time scales do not a¤ect the stability conditions
in both Cournot and Bertrand competition. So we have the following:
Theorem 6 In the continuous time framework, both of the Cournot and Bertrand
competitions are stable if the goods are substitutes and Cournot competition ismore stable than Bertrand competition if the goods are complements.

18

5Concluding Remarks

Discrete and continuous linear oligopolies are examined with di¤erentiated products. Conditions are derived for the positivity of the equilibrium quantities aswell as for that of the equilibrium prices. Then conditions are given for thestability of the positive equilibria. These conditions are analyzed in detail when
their dependence on the number of …rms (n) on the degree of di¤erentiation of
the products ( ); on the heterogeneity of the …rms ( k ) and on the selection of
the time scales is examined. In the case of Cournot competition our …ndings canbe summarized as follows. With discrete time scales the positive equilibrium ofa …rm is stable if the number of …rms is less than a threshold that depends on
both and k . If the time scale is continuous, then the positive equilibrium isalways stable if the products are substitutes and it is stable if the products arecomplements and the number of …rms is below a certain threshold. In the caseof price adjusting …rms and discrete time scales, we have the following facts. If
the products are substitutes, then the positive equilibrium price is stable if either the …rm is higher quali…ed ( k < 1). If the products are complements, thenthe positive equilibrium price is stable if the number of …rms is below a given
threshold. With continuous time scales the positive equilibrium price is stable
for any number of …rms if the products are substitutes and with a bounded
number of …rms if the products are complements. In comparing the stability
conditions we can conclude as follows: In the discrete time framework when
the goods are substitutes, the Bertrand competition is more stable for k < 1;
the Cournot competition is more stable for k > 2; however for 1 < k < 2;
no clear comparison is possible. In the continuous time framework both the
Cournot and Bertrand equilibria are stable if the goods are substitutes andCournot competition is more stable than Bertrand competition if the goods arecomplements.
Since all models resulted in linear dynamics, stability of the equilibria means
global asymptotical stability. Our next research project will be to analyze nonlinear oligopolies and investigate their local and global stability.

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21