ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012 | 321–345
321
Three firms on a unit disk
market: intermediate product
differentiation
Aljoša Feldin1 Received: 22 October 2012
Accepted: 18 December 2012
Abstract: Irmen and Thisse (1998) demonstrate that two firms competing with multicharacteristic products differentiate them in one characteristic completely, while keeping
them identical in all others. This paper shows that their min-…-min-max differentiation
result is not robust with respect to the number of firms. A market setting that replicates
their result in a duopoly, but fails to do so in a three firm oligopoly is identified. Symmetric pure strategy equilibrium with three firms differentiating their products in two
dimensions, but not completely in either of them, is a novel medium-medium differentiation result.
Keywords: spatial competition, location-price game
JEL classification: L11, L13, R39
Three firms on a unit disk market: intermediate product
differentiation
Models of discrete location choice commonly interpreted as modeling product differentiation, as well, have been explored in a variety of contexts.2 Irmen and Thisse (1998)
provide the most comprehensive study of two firms competing with their products in an
1
University of Ljubljana, Faculty of Economics, Ljubljana, Slovenia, e-mail: [email protected]
ReVelle and Eiselt (2005), and ReVelle, Eiselt, and Daskin (2008) provide a comprehensive survey of the field
and collect an extensive bibliography covering various aspects and problems in this area, respectively. A part
of the literature is interested in the extent of product differentiation that firms should employ, and the number
of product dimensions they should use doing that. Pioneering work with a linear duopoly model by Hotelling
(1929) offered a principle of minimum differentiation. Hotelling’s contribution was revisited 50 years later,
when d’Aspremont et al. (1979) showed that there is no price equilibrium in pure strategies when two firms
are located too closely to each other. Using quadratic instead of linear transportation costs, they find unique
market equilibrium with firms maximizing product differentiation. The direct demand effect that makes
a firm move towards its opponent to capture its demand is followed by the opponent’s price cut. The latter
overrides extra profit gained with new demand from moving towards the opponent. The negative strategic
effect of igniting stronger competition induces firms to differentiate their products as much as they can. As it
was shown later, a maximum differentiation result rests both on the form of consumers’ utility function (e.g.
Economides, 1986), and the uniform distribution of their tastes within the product characteristics space (e.g.
Neven, 1986, Tabuchi & Thisse, 1995).
2
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ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
n-dimensional product space.3 They investigate a unit hyper-cube market that is uniformly populated by consumers and served by two firms. When consumers incur disutility that is quadratic in distance between a product variety that they would prefer the
most and the variety bought, they show that it is always optimal for the two firms to
differentiate their products in one dimension only, and doing so completely. This type of
practice is referred to as min-…-min-max product differentiation.
This paper departs from the Irmen in Thisse model in that it studies a market with three
firms in a two-dimensional product space. We present a unit disk market uniformly
populated by consumers that are served by firms operating one store each. Firms choose
respective store locations in the first stage and compete with prices in the second. There
are some authors that explore how competition between more than two firms affects
product differentiation. Salop (1979) and Economides (1989) look at a circular city model
with location equilibria that place firms equidistantly. Economides (1993) provides price
equilibrium characterizations for every location configuration in a linear city model, and
Brenner (2005) presents location equilibria for different number of firms (from three to
nine) in the same type of the market. Brenner shows that firms depart from a maximum
differentiation result in that the store locations move towards the middle of the market.
We address two questions. First, will more than two firms in a bounded, two dimensional product characteristics space differentiate their products less than completely, as
we might suspect from Brenner (2005)? Brenner shows that a firm with two neighbors
does not cut prices as drastically as in a duopoly case when it is approached by one of
the neighbors. The reason is that it does not want to alter its optimal revenues from the
other side, where it neighbors a firm that did not deviate from an equilibrium position.
Consequently, in equilibrium even the two firms on the two outskirts of the city move towards the center. The direct demand effect outweighs the strategic price effect and extent
of product differentiation is reduced. There are two reinforcing effects facilitating Brenner’s result. First, the area of confrontation between an intrusive firm and its victim is a
single point, a marginal consumer between the two firms. Since the area of confrontation
between the victim and its other side neighbor is of the same size, the incentive to cut
its price to counter the intruder is offset in a large part by a lower price and suboptimal
profits on the other side. Second, the other side neighbor anticipates lower prices; it reduces its price as well. Hence, a reduction of victim’s price does not translate in a sizeable
increase of its demand on the other neighbor’s side and is not profitable. Consequently,
a cut in a victim’s price is not large enough to keep the intrusive firm from moving in to
capture a part of its demand. Consequently, firms locate closer to each other. The role of
the two effects we have described is less obvious in our case of firms competing in two
dimensions. Market configuration may be such that a firm that moves its store does that
in a direction of two and not just one neighbor. The firms’ demands are now delineated
by line segments of consumers that are indifferent between buying from any of the two
Irmen and Thisse are not the first to explore markets where consumers care about more than one product
characteristic. Neven and Thisse (1990) and Tabuchi (1994) were the first to show that in a two dimensional
product characteristics space two firms will never find it optimal to differentiate their products fully. There are
equilibria in which products are completely differentiated along one dimension and identical in the other.
3
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
323
neighboring firms. A confrontation frontier for an intrusive firm may hence be longer
than is a line segment between respective neighbors fighting the intruder. This means
that a price cut by the neighbors does not necessarily be as pronounced as it was in a
one-dimensional case but may still keep the intrusive firm away. As a result, it might be
that firms stay on the outskirts of the market. On the other hand, we might see a freerider effect, meaning that firms under attack would count on each other to counter the
intruder with lower prices. This might lead to a price cut that is insufficient to keep the
intruder on the edge of the market.
Another question is whether firms find it optimal to differentiate their respective products in more than one dimension in the first place. If the market was unbounded and
consumers’ reservation prices finite, the answer is obviously positive. With a bounded
market, the answer is not imminent and might depend on the shape of the market in
general. We expect that firms will find it beneficial to leave the congested competition in
one product characteristic at some point and will choose to differentiate their products
in another one as well. Swann (1990) explores such a process with a simple model and
simulations. Whenever the field of competition becomes too dense at least one firm endogenously finds it optimal to introduce a new product attribute.
We first show that maximum differentiation in one dimension – and no differentiation
in the other (Irmen and Thisse, 1998) – remains optimal in a duopoly. In our setting
this means that the two firms position their stores on the perimeter of the disk, exactly
opposite from each other. We then present two novel results. An oligopoly with three
firms competing on the same market facilitates a pure strategy subgame perfect Nash
equilibrium of our location-price game. The equilibrium has all three firms located at the
same distance from the center of the disk, equidistant from each other. That means that
we observe differentiation in two product characteristics, a result that extends the existent literature. Furthermore, firms do not choose full differentiation, but move inward,
towards the center of the disk considerably. We find medium-medium type of product
differentiation in a setting that yields a min-max differentiation result in a duopoly. This
means that the conjecture based on Brenner (2005), given above, carries over to markets
with more than two competitors and more than just one product characteristic. When
firms are located close to the perimeter of the market, the positive demand effect of a
radial deviation towards the center outweighs the negative strategic effect of rivals decreasing their prices.
Another interesting aspect of the model is that, while in a duopoly a social planner would
have firms differentiating their products less extensively, the result reverses in a threefirm oligopoly. Hence, some cooperative behavior or regulation on product specifications would be beneficial both to firms and to society as a whole.
The organization of the paper is as follows. We set up our model in Section 1, and present
our results for a duopoly and a three-firm oligopoly in Sections 2 and 3, respectively.
Section 4 considers welfare issues, and we make our conclusions in Section 5. Proofs of
Lemmas and Propositions and all necessary derivatives are deferred to the Appendices.
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ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
1. The model
We use the classical spatial model of an oligopoly. Consumers of a total mass π are uniformly distributed on a unit disk.4 Each consumer has a unit demand for a homogeneous
good produced by n firms on the market. We explore configurations in which all firms
are at the same radial distance from the origin, R. Specifically, the firms are located at
Li=(R,φi), i = 1,…, n. Firm 1 is always a counter-clockwise direction neighbor of Firm n,
while Firm n−1 is a clockwise direction neighbor of Firm n. Firms charge pi, i = 1, …,
consumer
at point
x buys at
thepoint
goodxfrom
i, shefrom
derives
utility:
n, per unit of the
good. Ifresiding
a consumer
residing
buysFirm
the good
Firm
i, she
derives utility: u(pi , Li ; x) = v − pi − d 2 ( x, Li ) .
We assume
incur
traveling
that
are utility:
quadratic in the Euclidean
consumer
residing
at point
xconsumers
buys
the
good
from
Firm
i,costs
she
derives
consumer
residing
xthat
buys
thefrom
good
from
i,quadratic
she
derives
We assume
that
consumers
incur
costs
that
arederives
in utility:
the Euclidean disconsumer
residing
at pointatxpoint
buys
good
Firm
i,Firm
she
utility:
2 thetraveling
2
, traveled,
and with cost per unit traveled normalized to one. The surplus v
LiL() x,
u(p
;distance,
Li ) .
u(p
;i ,ix)
=x)
v =−2vpi−−dpand
di(−x,(dx,
tance,
d(x,
),Lpitraveled,
with
i , LiL
i ) .cost per unit traveled normalized to one. The surplus v
u(p
i , Li ; x) = v −
i − d ( x, Li ) .
enjoyed
from
consuming
the
good
assumed
to be large
enough
so
that every consumer buys a
enjoyed from
consuming
theconsumers
good is
assumed
toisbe
large
enough
soquadratic
that
every
consumer
assume
incur
traveling
costs
that
in the
Euclidean
WeWe
assume
thatthat
consumers
incur
traveling
costs
that
are are
quadratic
in by
the
Euclidean
good
from
one
of
the
firms.
Consumers
maximize
their
utility
choosing
the store they buy the
We
assume
that
consumers
incur
traveling
costs
that
are
quadratic
in
the
Euclidean
buys a good
from
one
of
the
firms.
Consumers
maximize
their
utility
by
choosing
the
distance,
,
traveled,
and
with
cost
per
unit
traveled
normalized
to
one.
surplus
d
x,
L
i ) optimally.
distance,
, traveled,
andcost
withper
cost
per
unit traveled
normalized
toThe
one.
TheThe
surplus
v v
d) ,(good
x,
L( i )from
distance,
traveled,
and
with
unit
traveled
normalized
to
one.
surplus
v
d
x,
L
(
i the good from optimally.
store they
buy
enjoyed
from
consuming
good
is assumed
to large
be large
enough
so that
every
consumer
buys
a
enjoyed
from
consuming
theisthe
good
is assumed
to be
enough
so
that
every
consumer
a and
Firms
operate
with
symmetric
constant
marginal
costs
that
we normalize
zero
enjoyed
from
consuming
thethe
good
assumed
to be
large
enough
so
that
every
consumer
buys they
atobuys
good
from
one
of
firms.
Consumers
maximize
their
utility
by
choosing
the
store
buy
the are
good
from
one
of
the
firms.
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maximize
their
utility
by
choosing
the
store
they
buy
the
allowed
to Consumers
operateconstant
one maximize
storemarginal
each.their
We costs
consider
non-cooperative
location-price game
good
from
one of
theoptimally.
firms.
utility
by achoosing
the store
they
Firms
operate
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symmetric
that
we
normalize
totwo-stage
zerobuy
andthe
good
from
good
from
optimally.
of
the
following
form.
good
from
optimally.
are allowed to operate one store each. We consider a non-cooperative two-stage locationFirms
operate
symmetric
constant
marginal
costs
that we
normalize
to zero
and are
Firms
operate
withwith
symmetric
constant
marginal
costs
normalize
zero
price
game
of the
following
form.
Stage
1:marginal
Firms
select
onwe
a unit
disk.
Firms
operate
with
symmetric
constant
costsalocations
that
wethat
normalize
to zerotoand
areand are game
allowed
to
operate
one
store
each.
We
consider
non-cooperative
two-stage
location-price
allowed
to operate
oneeach.
store each.
We consider
a non-cooperative two-stage
location-price
game
allowed
toofoperate
one store
consider
a non-cooperative
game
the
following
form. We
Stage
2: Firms
observe chosentwo-stage
locationslocation-price
and compete in
prices.
of
the
following
form.
ofStage
the following
1: Firms form.
select locations on a unit disk.
Stage
1:i’s
Firms
select
locations
a unit
disk.
Formally,
firm
strategy
space
thedisk.
first
period
is Li = [0,1] × [−π , π ] , i = 1,… , n, and
Stage
1:
Firms
select
locations
on
aon
unit
disk.
Stage
1:
Firms
select
locations
on a in
unit
n
Stage
2: in
Firms
observe
chosen
compete
in prices.
Stage 2: Firms observe
chosen
and
compete
prices.
plocations
Li and
→
ℜ
. We
subgame perfect Nash
each
firm’s
strategy
the
second
stage
isin and
Stage
2:locations
Firms
observe
chosen
locations
compete
in seek
prices.
i :×
i =1and
Stage
2:
Firms
observe
chosen
locations
compete
in +prices.
*
*
*
*
*
*
*
] ×natural
[−π],,πi ]=candidate
, 1,…
i = 1,…
,for
n, equilibrium
and
Formally,
firm
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Lin1 , the
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(period
L2 ,period
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, p= [)0.,1A
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1,… ,, n,
n, and
and, n, and
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firm
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thefirst
first period
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is nLi 2=is[0L,1i ] =
configuration
has
firms
positioned
distance
from
thesubgame
origin, equidistantly
: ×Lequal
Li ℜ
→
seek
perfect
Nash along the
each
firm’s
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in
the
second
stagen pis :atp×inan
i =1→
+ . We
each
firm’s
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in the
second
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. We
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each
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in the
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− π + n2Nash
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*
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the origin, along the
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an
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from
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equidistantly
configuration
haspositioned
firms positioned
at an distance
equal distance
from
the origin,
equidistantly
along the4
*
configuration
has
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at an
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from we
the
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equidistantly
along
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they
occupy.
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locations:
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=firms
(R,equidistantly
),L* L*along
= (R,
− π2 *+ ),
π ),
* 2 L1Two
n πL2 1=2 (R, 4− π −
circle
they
occupy.
we
explore
locations:
=
2− π + n π
circle
we
explore
locations:
L1 = (R, − π + n π ), L2 = (R, − π + n π ), n n
*occupy. Specifically,
*
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*+ –
+ –n2 pthey
), L…
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We
first derive market equilibrium in a duopoly. Here we show that min-max result derived
= (R,
… , and
by Irmen and Thisse (1998) is also2 optimal
strategy in our setting. In our setting that means that
firms
2perimeter
TwoTwo
firms
firms locate their stores 2on the
of the disk, symmetrically across the origin. We search
Two
firms
2. Two firms
We
first
market
equilibrium
a in
duopoly.
Here
we
show
that
min-max
result
derived offfor
anderive
equilibrium
that
is symmetric
firms’
locations
with
1 showing
a possible
Wederive
first
derive
market
equilibrium
in ainduopoly.
Here
we that
show
thatFigure
min-max
result
derived
We first
market
equilibrium
a duopoly.
Here we
show
min-max
result
derived
*
by Irmen
and
Thisse
(1998)
isin
also
optimal
strategy
in
our
setting.
In
our
setting
that
means
that
=
(R,0)
and
L
=
(r,φ).
equilibrium
configuration
L
by
Irmen
and
Thisse
(1998)
is
also
optimal
strategy
in
our
setting.
In
our
setting
that
means
that
1
2
byWe
Irmen
Thisse
(1998)
is alsoonoptimal
our
setting.
In ourthat
setting
that the
means
thatWe search
firstand
derive
market
equilibrium
in strategy
a duopoly.
Here
we
show
min-max
result
firms
locate
their
stores
the
perimeter
ofinthe
disk,
symmetrically
across
origin.
firms
locate
their
stores
on
the
perimeter
of
the
disk,
symmetrically
across
the
origin.
We
search
firms
locate
stores
the
perimeter
theoptimal
symmetrically
across
the1origin.
derived
bytheir
Irmen
andon
Thisse
(1998)
is of
also
strategy
inwith
our
setting.
In ouraWe
setting
for
an equilibrium
that
is symmetric
indisk,
firms’
locations
Figure
showing
asearch
possible offan
equilibrium
that
is symmetric
in firms’
locations
with
Figure
1 showing
possible
forthat
an for
equilibrium
that
is
symmetric
in
showing
a
possible
off- off*firms’ locations with FigureL 1*=(R,
* L1 =
1 the0)disk, symmetrically
means
that
firms
locate
their
stores
on
the
perimeter
of
(R,0)
and
L
=
(r,φ).
equilibrium
configuration
2
B=(1,β)
L2 = (r,φ).
equilibrium configuration
L1 = (R,0)
L2and
= (r,φ).
equilibrium
L1*= (R,0)
across the configuration
origin. We search
for anand
equilibrium
that is symmetric in firms’ locations with
Figure 1 showing a possible off-equilibrium configuration
L1*= (R,0) and L2 = (r,φ).
A=(1,α)
*
{
{
}
{ {
} }
}
L1*=(R, 0)
L =(R, 0)
L1*=(R,1 0)
B=(1,β)
B=(1,β)
B=(1,β)
z α −β
≡
A=(1,α)
A=(1,α)
O
Polar symmetry is used to avoid A=(1,α)
non-differentiable demand functions that arise in a rectangular market
when a line of consumers indifferent between buying fromztwo
≡ α −βneighboring stores touches the corner of a
α −β
market. Moreover, there is no symmetry to exploit
thez ≡
market
with three firms on a square.
z ≡in
α −β
4
O
O
L2=(r, φ)
O
L2=(r, φ)
L2=(r, φ)
L2=(r, φ)
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
325
Figure 1: Configuration of the market, two firms.
Line AB in Figure 1 represents
consumers
are indifferent
between
buying
a product
Line AB
in Figure that
1 represents
consumers
that are
indifferent
between buying a prod
from
either 1ofrepresents
the two
firms
given
their
locations
andbetween
product
prices.
is perpendicuLine AB
in Figure
consumers
are
indifferent
buying
a AB
product
from
either
of that
the
two
firms given
their
locations
and
product
prices. AB is perpendicula
inline
Figure
1 represents
consumers
indifferent
between
buying
product
from eitherLine
of the
two
firms
given
their
locations
and that
product
prices.
AB
is
the
and
L2L. 1It* and
is closer
the
firm
that
setsthat
thesets the higher of
lar
toAB
the
connecting
firms’
locations
L1*are
L2perpendicular
. It to
is closer
toato
the
firm
line
connecting
firms’
locations
*given their locations and
from
either
of
the
two
firms
product
prices.
AB
is
perpendicular
It is closer
to the the
firm
that
sets the
higher
ofat
two toααthe
line connecting
firms’
locations
Lrespective
crosses
theperimeter
perimeter
ofthe
the disk
disk
atthe
angles
and
1 and L*prices.
2. prices.
higher
of the
two respective
ABAB
crosses
of
angles
andβ. Firm 1’s deman
L . disk
It is at
closer
to theand
firm
sets the
higherisofthe
the two
line prices.
connecting
firms’ locations
L1 and
respective
theisarea
perimeter
the
angles
β.that
Firm
demand
α2and
β,α (α−β)/2,
reduced
by
the1’s
area
a triangle
β. FirmAB1’scrosses
demand
thebetween
areaofbetween
and
β, (αα
−β)/2,
reduced
byofthe
area of ABO,
a tri- which is
respective
prices.
AB crosses
the
of of
thea disk
at angles
αwhich
and β.
Firm 1’s demand is the
α − β perimeter
α − βarea
area between
α
and
β,
(α−β)/2,
reduced
by
the
triangle
ABO,
is
angle ABO, which is sin 2 ⋅ cos 2 ==sin(
sin(
firms’ demands:
We set
set zz ≡≡ a
and state
state the
the firms’
αa−–βb))/2.
/ 2 . We
α –− bβ and
area between
α and β, (α−β)/2, reduced by the area of a triangle ABO, which is
α −β
sin α −2 β ⋅ cos
= sin(
. We set z ≡ α − β and state the firms’ demands:
αdemands:
−β 2
α − βα − β ) / 2
sin 2 ⋅ cos 2 = sin(α − β ) / 2 . We set z ≡1α − β and state the firms’ demands:
1
D1 = (1z − sin z ) , and D2 = π − D1 = (2π − z + sin z ) ,
(1
1
2
2 (1)
(1)
D1 = ( z − sin
1 z ) , and D2 = π − D1 = (2π −1 z + sin z ) ,
2D1 = ( z − sin z ) , and D2 = π −2D1 = (2π − z + sin z ) ,
(1)
Firms’ second-stage profits
2
2 given locations chosen in the first stage are:
Firms’ second-stage
profits givenprofits
locations
chosen
in thechosen
first stage
are:first stage are:
Firms’ second-stage
given
locations
in the
p1 ( r , φin
and Π2 ( r , φ ; R ) = p2 ( r , φ ; R ) ·D2 ( r , φ ; R ) . (
Π1 (locations
r , φ ; R ) =chosen
; Rthe
) ·Dfirst
; R ) are:
1 (r , φ
Firms’ second-stage profits given
stage
Π1 ( r , φ ; R ) = p1 ( r , φ ; R ) ·D1 ( r , φ ; R ) and Π2 ( r , φ ; R ) = p2 ( r , φ ; R ) ·D2 ( r , φ ; R ) . (2)
, p 2 ,Πα2 (,r ,βφ ,; R
and
Π1 ( r , φ ; R ) = p1 ( r , φ ; R ) ·DNote
; R ) p1and
) =zp2are
( r ,all
φ ; Rfunctions
) ·D2 ( r , φof
; Rfirms’
) . (2)respective
(2) locations. Sec
1 ( r , φthat
Note that p1 , p 2 , α , β , and
all functions
of firms’
Secondz are
stage
profit
maximization
withrespective
respect to locations.
prices yields
the following result.
Note that p1 , p 2 , α , β , and z are all functions of firms’ respective locations. Secondstage profitNote
maximization
prices
yields the of
following
result.
that p1, p2with
, α , brespect
, and ztoare
all functions
firms’ respective
locations. Second-stage
stage profit maximization
with respect to prices yields the following result.
profit maximization with respect to prices yields the following result.
Lemma 1: When firms choose locations L1* = (R,0) and L2* = (R,π) in the first stage
*
*
1: When
firms
choose
locations
L1* =charge
(R,0)
=and
(R,π)
theπfirst
stagefirst
of the
LemmaLemma
game,
theylocations
both
=Lπ2 ·R
in L
the*insecond.
1: When
When
firms
choose
L* 1=*and
=p(R,0)
(R,0)
(R,
) in
stage
* firms
1:
choose
locations
L
and
L2* 2= =(R,π)
in
thethe
first stage
of of
thethe
Lemma
1
game, theygame,
both charge
p
=
π
·R
in
the
second.
*
*
they
both
charge
p
=
π
·R
in
the
second.
game, they both charge p = π ·R in the second.
The price firms charge inThe
theprice
second-stage
of the
game
increasesofinthe
distance
from thein distance from th
firms charge
in the
second-stage
game increases
origin,
is ininthe
distance
between
their
stores
or their
products.
This
The price
firmsthat
charge
thethat
second-stage
of the
game
increases
in distance
from
theshows
origin,
is in the
distance
between
their
stores
or their
products.
Thisa comshows a common incen
The
price
firms
charge
instores
the second-stage
thedifferentiation.
game
increases
in distance
from the origin,
mon
incentive
dictating
the
need
for product
that is in the
distance
between
their
or their
products.
This
shows
a common
incentive
dictating
the
need
forof
product
differentiation.
thatthe
is need
in thefor
distance
between
their stores or their products. This shows a common incentive
dictating
product
differentiation.
Next, we want to see whether the proposed structure of firms’ first stage locations c
dictating
the
need
for
product
differentiation.
Next,
wetowant
to see supported
whether
the
proposed
of
firms’
locations
can beFirm 1’s location,
Next, we
want
see whether
the proposed
structurestructure
ofWe
firms’
first
stage
can be
that given
in equilibrium.
show
there
isfirst
Rlocations
∈ (stage
0,1] , such
*1’scan
Next,
we
want
to
see
whether
the
proposed
structure
of
firms’
first
stage
locations
be
supported
in
equilibrium.
We
show
there
is
R∈
(0,1],
such
that
given
Firm
location,
*
, such that
givenout
Firm
1’sthis
location,
L1 one;
, Firm
supported in equilibrium. We show
is R ∈
1] optimal.
It turns
that
R equals
firms locate their stores o
2’sthere
location,
L (0, ,is
* * in equilibrium. We* show there2is R ∈ ( 0,1] , such that given Firm 1’s location, L1*, Firm
supported
L
,
Firm
2’s
location,
L
,
is
optimal.
It
turns
out
that
this
R
equals
one;
firms
locate
their
2 out thatof
It turns
this
equals
one;they
firms
their stores
on the
2’s location,1L2 , is optimal.
perimeter
theRdisk.
Also,
willlocate
be positioned
symmetrically
across the origin. To s
*
,
is
optimal.
It
turns
out
that
this
R
equals
one;
firms
locate
their
stores
on
the
2’s
location,
L
stores
on
the
perimeter
of
the
disk.
Also,
they
will
be
positioned
symmetrically
across
2
perimeter of the disk. Also, they we
willlook
be positioned
symmetrically
across in
theradial
origin.
see this,
at respective
effects deviations
andTopolar
directions have on Firm 2’s p
perimeter
of theeffects
disk. Also,
they will
be are:
positioned
the origin.
To see this,
we look
at respective
deviations
in radial
and polarsymmetrically
directions haveacross
on Firm
2’s profits.
These
effects
we
look
at
respective
effects
deviations
in
radial
and
polar
directions
have
on
Firm
2’s
profits.
These effects are:
 ∂D2 ∂D2 d p1 
dΠ 2 ∂Π 2 d p2
These effects are:
(3)
 ∂D ∂D d=p  ⋅ d r + p2 ⋅  ∂r + ∂ p ⋅ d r  , and
dΠ 2 ∂Π 2 d p2
(3) 1
=dΠ 2 ⋅ ∂Π 2+ dp2p⋅2 2 + ∂Dd22 r⋅ ∂D1∂2 p, 2dand
p1 
(3)
dr
∂ p2 = d r ⋅
p2 ⋅  ∂ p1 + d r  ⋅
+∂r
 , and
dr
∂ p2 d r
 ∂r
 ∂D
dΠ ∂ p∂1Π d rd p
∂D d p 
The price firms charge in the second-stage of the game increases in distance from the origin,
that is in the distance between their stores or their products. This shows a common incentive
dictating the need for product differentiation.
Next, we want
of firms’ AND
first BUSINESS
stage locations
be 14 | No. 4 | 2012
326 to see whether the proposed structureECONOMIC
REVIEWcan
| VOL.
supported in equilibrium. We show there is R ∈ (0,1] , such that given Firm 1’s location, L1*, Firm
2’s location, L2*, is optimal. It turns out that this R equals one; firms locate their stores on the
perimeter of thethe
disk.
Also,To
they
bewe
positioned
symmetrically
across
the origin.
To seeand
this,polar direcorigin.
seewill
this,
look at respective
effects
deviations
in radial
we look at respective effects deviations in radial and polar directions have on Firm 2’s profits.
tions have on Firm 2’s profits. These effects are:
These effects are:
 ∂D ∂D d p 
dΠ 2 ∂Π 2 d p2
=
⋅
+ p2 ⋅  2 + 2 ⋅ 1  , and
dr
∂ p2 d r
 ∂r ∂ p1 d r 
 ∂ D2 ∂ D2 d p1 
dΠ 2 ∂Π 2 d p 2
.
=
⋅
+ p 2 ⋅ 
+
⋅
dφ
∂ p2 d φ
∂ p1 d φ 
 ∂φ
(3)
(3)
(4)
(4)
The first term on the right hand side (RHS henceforth) of both (3) and (4) is zero due to
The first term on the right hand side (RHS henceforth) of both (3) and (4) is zero due to
profit maximization with respect to p2 in the second period, so we are left with the direct
maximization
with
respect
to pmove,
in thewhich
second
wesecond
are leftterms
with the direct
2
(demand) effectprofit
and indirect
(strategic)
effect
of each
areperiod,
the firstsoand
(demand)
effect
and
indirect
(strategic)
effect
of
each
move,
which
are
the
in parentheses, respectively. We first derive the demand and strategic effects of a small radial first and secondprofit.
terms in parentheses, respectively. We first derive the demand and strategic effects
move on Firm 2’s
of a small radial move on Firm 2’s profit.
∂D2
∂D2
1
Lemma 2: When L1* = (R,0) and L2* = (R,π), R ∈ (0,1] :∂(a)
D22 ∂p 1 = 2 R , and (c)
D22 ∂r = −1 , ∂(b)
**
**
*
*
= 1 ,, and
WhenLL1 11==(R,0)
(R,0)and
and LL222 == (R,
(R,π),
(a)
(b)
Lemma
= −1,, (b)
R ∈ (0,1]:
(0,1] : (a)
π), R∈
and (c)
Lemma
2: 2:When
∂p11 2 R
∂r
dp1 π 2
(c) π = 2 + R..
dp
11 dr
= + 2 R .3
dr
2 3
Part (a) quantifies Firm 2’s direct gain in demand it captures from its rival by moving
towards its location, while (b) and (c) show that such a move affects Firm 2’s demand
adversely
a rival’s
price
These
are such
that both
would
like to
Partthrough
(a) quantifies
Firm
2’scut.
direct
gaineffects
in demand
it captures
fromfirms
its rival
by moving
Part
(a)
quantifies
Firm
2’s
direct
gain
in
demand
it
captures
from
its
rival
by
moving
locate
their
stores
on
the
perimeter
of
the
disk
when
they
are
separated
by
angle
π
. adversely
towards its location, while (b) and (c) show that such a move affects Firm 2’s demand
towards
its location,
(c) show
move
affects
2’slike
demand
adversely
through
a rival’swhile
price (b)
cut.and
These
effectsthat
aresuch
suchathat
both
firmsFirm
would
to locate
their stores
*
through
a rival’s
pricepair
effects
such
that Lboth
firms
would
likefirms
to locate
their
stores
Lemma
3:perimeter
For any
locations
L1*are
= (R,0)
and
=
(R,
π
),
R∈
(0,1],
find
it
profiton the
ofcut.
theofThese
disk
when
they
are
separated
by
angle
π.
2
onable
the to
perimeter
of the disk
separated by angle π.
move farther
awaywhen
fromthey
the are
origin.
*
3: For
of separating
locations L1the
= (R,0)
and L2is*=exactly
(R,π), R
Finally,Lemma
we argue
thatany
thepair
angle
two firms
π.∈ (0,1] , firms find it
**
**
Lemma
3:
For
any
pair
of
locations
L
=
(R,0)
and
L
=
(R,π),
R
∈
(
0,1] , firms find it
1
1
2
2
profitable to move farther away from the origin.
profitable to move farther away from the origin.
Lemma 4: For any R, φ = π is optimal for Firm 2.
Lemmas
3 andwe
4 prove
1. separating the two firms is exactly π.
Finally,
argueProposition
that the angle
Finally, we argue that the angle separating the two firms is exactly π.
Proposition 1: When in the first stage of the game two firms position their stores on the
perimeter
of the
symmetrically
across for
its Firm
origin;
4: disk
For any
R, φ = π is optimal
2. the equilibrium price they charge
Lemma
4: For any is
R, πφ. =Furthermore,
π is optimal for
Firmfirm’s
2. location is a local best response to
in Lemma
the second-stage
every
their rival’s location.
3 and 4 of
prove
Proposition
Due
to Lemmas
the complexity
the problem
we1.
are not able to show that each firm’s location is
Lemmas 3 and 4 prove Proposition 1.
also a global best response to an opponent’s location analytically. We hence do it numerically. We
fix Firm 1’s
φ1 =of0,the
and
varytwo
Firm
2’s location
across
the disk.
Proposition
1: location
When in at
theR=1
firstand
stage
game
firms
position their
stores
on the
WeProposition
solve
for
the
second-stage
price
equilibrium
for
every
configuration
numerically
and
1:
When
in
the
first
stage
of
the
game
two
firms
position
their
stores
on the
perimeter of the disk symmetrically across its origin; the equilibrium price they charge
in the
perimeter
of
the
disk
symmetrically
across
its
origin;
the
equilibrium
price
they
charge
in
the
calculate
Firm
2’s
profit.
Figure
2
shows
that
the
proposed
location
r=1
and
φ
=
π
is
acsecond-stage is π. Furthermore, every firm’s location is a local best response to2 their rival’s
second-stage
Furthermore,
every firm’s
location
is a local
best response
to attained
their rival’s
tually
Firmis2’sπ.global
best response
to Firm
1’s location
choice.
The profit
there
location.
location.
is π 2/2.
Due to the complexity of the problem we are not able to show that each firm’s location is
Due
the complexity
of the
we arelocation
not ableanalytically.
to show thatWe
eachhence
firm’s
is
also atoglobal
best response
toproblem
an opponent’s
dolocation
it numerically.
also We
a global
best
response
to
an
opponent’s
location
analytically.
We
hence
do
it
numerically.
fix Firm 1’s location at R=1 and φ1 = 0, and vary Firm 2’s location across the disk. We solve
and
varyconfiguration
Firm 2’s location
across the
We solve
We fix
1’s location price
at R=1
and φ11 = 0,for
forFirm
the second-stage
equilibrium
every
numerically
anddisk.
calculate
Firm 2’s
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
327
Figure 2: Firm 2’s profits given the location of its store.
Source: Own calculations.
We now state our first result, which links this work to the existent literature of two firms
competing in a multi-characteristic product space. This gives us a valid reference point
with which to compare our subsequent results.
Result 1: Two firms positioning their stores on the perimeter of the disk, symmetrically
across its origin in the first stage of the game, and setting p=π in the second is a subgame
perfect Nash equilibrium of the game.
This finding is in the spirit of Neven and Thisse (1990), Tabuchi (1994), and Irmen and
Thisse (1998). Two firms offer products that are fully differentiated in one characteristic,
while identical in the other.
3. Three firms
We add another firm to the model and search for symmetric equilibrium. We find that
firms separate their stores in both dimensions, and interestingly, do not choose to locate
them on the perimeter anymore, but relocate them towards the origin noticeably.
Given locations L1* = (R,−π /3), L2* = (R,π /3), and L3 = (r,φ), firms compete in prices, p1, p2,
and p3. A particular market configuration is shown in Figure 3.
328
ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
A=(1,α)
L1*=(R,−π/3)
L2* = (R,π/3)
D = (x,y) = (d·sinδ, d·cosδ)
O
C = (1,γ)
L3 = (r,φ)
B = (1,β)
Figure 3: Configuration of the market, three firms.
Figure 3: Configuration of the market, three firms.
Store locations and prices define the boundaries of market areas covered by firms. There
are three line segments, DA, DB, and DC, representing buyers indifferent between buyStore locations ing
and from
pricesFirms
define1the
market
areas
covered
There areThese segments
andboundaries
2, Firms 2ofand
3, and
Firms
3 andby1,firms.
respectively.
ree line segments,are
DA,needed
DB, and
representing
indifferent
between
fromall join in one point,
inDC,
determining
thebuyers
demand
functions
firms buying
face. They
rms 1 and 2, Firms
and 3, is
and
Firms
3 and
respectively.
These segments
arebe
needed
in Point D = (x, y)
D,2which
due
to the
fact1,that
the delineating
lines must
straight.
termining the demand
functions
firms
face.
They
all
join
in
one
point,
D,
which
is
due
to
represents a consumer indifferent between buying from any of the the
three firms. Points
ct that the delineating
must
y) represents
a consumer
indifferent
A = lines
(1, α),
B =be(1,straight.
β), andPoint
C =D
(1,=γ(x,
) stand
for consumers
on the
perimeter indifferent
tween buying from
any of the
threefrom
firms.respective
Points A =firms.
(1, α),With
B = (1,
and C = (1,
γ) y,
stand
α , βfor
, and γ, which, as
between
buying
theβ),
knowledge
of x,
nsumers on the perimeter indifferent between buying from respective firms. With the
well as p1, p2, and p3, are all functions of r, f, and R, we can write the demand functions
nowledge of x, y, α, β, and γ, 5which, as well as p1 , p 2 , and p 3 , are all functions of r ,φ , and R ,
firms face. Firm 1’s demand
equals the area of the disk between α and γ reduced for the
Firm 1’s
equals
the area
the disk
e can write the demand
functionsbyfirms
face.4 OCD
area covered
triangles
anddemand
ODA. The
other
two of
firms’
demands are obtained
tween α and γ reduced
for
the
area
covered
by
triangles
OCD
and
ODA.
The
other
two firms’
similarly:
mands are obtained similarly:
D1 =
1
(α − γ − d sin(δ − γ ) − d sin(α − δ )) ,
2
D2 =
1
(β − α + d sin(α − δ ) − d sin( β − δ )) , and
2
D3 =
1
(2π − β + γ + d sin( β − δ ) + d sin(δ − γ )) .
2
The sine of the difference rule and definitions of x and y yield:
1
D1 =5 For(αthe
− γderivation
+ x(cosof
α x,
− y,
cos
− y (sin
− proof
sin γ )of) ,Lemma 5.
α,γβ,) and
γ seeαthe
2
(5)
demands are obtained similarly:
D1 =
1
(α − γ − d sin(δ − γ ) − d sin(α − δ )) ,
2
1
(
2
A. FELDIN
ON A
MARKET:
PRODUCT DIFFERENTIATION
D2 =| THREE
β − αFIRMS
+ d sin(
αUNIT
− δ )DISK
− d sin(
β − δINTERMEDIATE
) , and
)
329
1
D3 = (2π − β + γ + d sin( β − δ ) + d sin(δ − γ ) ) .
The sine of2 the difference rule and definitions of x and y yield:
The sine of the difference rule and definitions of x and y yield:
D1 =
1
(α − γ + x(cosα − cos γ ) − y(sin α − sin γ )) ,
2
(5)
(5)
1
(6)
D2 = 1 (β − α + x(cos β − cosα ) − y (sin β − sin α ) ) , and
(6)
D2 = 2 (β − α + x(cos β − cosα ) − y (sin β − sin α ) ) , and
(6)
2 D = 1 (β − α + x(cos β − cosα ) − y (sin β − sin α ) ) , and
(6)
1
1
2
Dcos
−(sin
α +βx(cos
βα−) )cos
α ) − y (sin β − sin
(6)
D2 = (β − α1
+ x(cos
β 12−(cos
αα) +− xy(cos
(sin ββ −−
sin
))()−β, and
(6)
2 =α
D
=
β
−
α
y
−
sin
,
and
(6)α ) ) , and
1
(
)
= 2π − β + γ − x(cos β − cos γ ) + y (sin
− sin γ ) .
(7)
2 β
2D
(7)
D33 = 2 (2π −2 β +2γ − x(cos β − cos γ ) + y (sin
β − sin γ ) ) .
(7)
2 D = 1 (2π − β + γ − x(cos β − cos γ ) + y (sin β − sin γ ) ) .
(7)
1
1 profit functions
3
12are:β − cos γ ) + y (sin
TheDfirms’
(2πγγ ))−)+.β y+(sin
D3β
=−
γ −βx(cos
βγ(7)
−) .cos γ ) + y (sin β(7)
− sin γ ) ) .
(7)
(profit
2π − βfunctions
+ γD− x=(cos
sin
3 =
(
2
π
−
β
+
γ
−
x
(cos
β
−
cos
−
sin
)
The firms’
are:
3
2
2
2
,
Π
(
r
,
φ
;
R
)
=
p
(
r
,
φ
;
R
)
⋅
D
(
r
,
φ
;
R
)
The
The firms’
firms’
profitfunctions
functions
1 profit
1
Πare:
, φare:
; are:
R
) ⋅ D11 (profit
r , φ ; Rfunctions
),
The
firms’
are:
1 ( r , φ ; R ) = p1 ( r
The firms’ profit functions
The firms’
profit functions
are:
=) p⋅ 1D(r2 ,(φr ,;φR;)R⋅ )D,1and
(r , φ ; R) ,
(8)
Π 2 (r , φ ; R)Π=1 (pr2, φ(r;,Rφ); R
, φ ; R) ⋅ D (r , φ ; R) ,1and
φ ;; R
R)) ,= p1 (r , φ ; R) ⋅ D1 (8)
(r , φ ; R) ,
Π 1 (r , φ ;ΠR2) (=r , φp1; (Rr),Π
φ=; (Rpr)2, φ⋅(rD
; R1 ()r=, φp; 1R()r2 ,, φ ; R) ⋅ Π
D1 ((rr ,, φ
1
(8)
. 2 (r , φ ; R) , and
Π 3 ( r , φ ; R )Π=2 (pr3,(φr;, φR;)R=) ⋅pD
, φ; R
; R) ⋅) D
2 (3r(,rφ
, φ2;(Rr ,)φ⋅ ;DR3)( r, ,and
φ ; R )Π. 2 (r , φ ; R ) = p 2 (r , φ(8)
(8)
; R) ⋅ D2 (r , φ ; R) , and
Π 2 (r , φ ;ΠR3)( r=, φp;2R(r)Π
,=φ ;(pRr3),(φ⋅r;D
(8)
,
and
(8)
R
)
=
p
(
r
,
φ
;
R
)
⋅
D
(
r
,
φ
;
R
)
2
r , φ ; R ) =from
p 32 ( rthe
, φ ;system
R ) ⋅ D32(of
r , φnecessary
; R) .
Second-stage optimal prices Π
are3 (derived
conditions obtained
Second-stage
prices
are
system
of necessary
=
r , φ ; R ) ⋅ D3 obtained
(r , φ ; R) .
Π 3 (optimal
r ,φ ; R) =
p 3 ( rΠ
, φ ; (Rderived
) ⋅ D3 ( r=from
, φp; R()rthe
. p 3 (conditions
,. φ ; R ) ⋅Π
D33((rr,,φφ;;RR))solution
from these profits.
The
system
is nonlinear
and
its
analytical
for any possible
3 r ,φ ; R)
3 general
from these profits.
The system
is nonlinear
and
its
general
analytical
solution
for any possible
Second-stage
optimal
prices
are
derived
from
the
system
of
necessary
conditions
obtained
Firm 3’s location,
Lprices
, is therefore
out from
ofSecond-stage
reach.
However,
we are
looking
for symmetric
optimal
prices
are derived
from conditions
thefor
system
of necessary conditions obta
Second-stage
optimal
are optimal
derived
system
offrom
necessary
conditions
obtained
Firm 3’s location,
L33, profits.
is therefore
out
ofprices
reach.
However,
we
are
looking
fornecessary
symmetric
from
these
The
system
isthe
nonlinear
and
itsfrom
general
any
possible
Second-stage
prices
areare
derived
the
system
of
obtained
Second-stage
optimal
derived
the analytical
system
ofsolution
necessary
conditions
obequilibrium,
so
the
derivation
of
optimal
prices
is
straightforward.
from
these
profits.
The
system
is
nonlinear
and
its
general
analytical
for any poss
from these
profits.
The
system
is
nonlinear
and
its
general
analytical
solution
for
any
possible
equilibrium,
so tained
the
of
optimal
prices
isofstraightforward.
Firm
3’s derivation
location,
LThe
therefore
outsystem
reach.
However,
are
looking
for
symmetric
from
these
profits.
system
isThe
nonlinear
and
its generalwe
analytical
solution
for any solution
possible solution
3, is
from
these
profits.
is
nonlinear
and
its
general
analytical
for for symmetric
Firm
3’s
location,
L
,
is
therefore
out
of
reach.
However,
we
are
looking
Firm 3’s location, equilibrium,
L
,
is
therefore
out
of
reach.
However,
we
are
looking
for
symmetric
3
3
so theFirm
of optimal
prices
is straightforward.
Firmany
3’s location,
Lderivation
therefore
out Lof, reach.
However,
wereach.
are looking
for symmetric
3, is 3’s
is
therefore
out
However,
we are looking for
possible
location,
equilibrium,
so3 the
derivation
of of
optimal prices
is straightforward.
equilibrium, so theequilibrium,
derivation
of
optimal
prices
isofstraightforward.
so
the
derivation
optimal
prices
is
straightforward.
three
firms
in
the
first
stage
of
the
game
position
their
stores
R
away from
Lemma 5: When
symmetric
equilibrium,
so
the
derivation
of
optimal
prices
is
straightforward.
Lemma 5: When three firms in the first stage of the game position their stores R away from
the origin, equidistantly from one another, the optimal Nash equilibrium price they charge in the
the origin, equidistantly
onethree
another,
optimal
Nash of
equilibrium
theytheir
charge
in R
theaway from
When
firmsthe
in the
first stage
the game price
position
stores
Lemma 5:from
5:
When
three
firms
in
the
first
stage
of
the
game
position
Lemma
three
firms
in
the
first
stage
of
the
game
position
their
stores
R
away
from
Lemma 5: When
π
R
3
Lemma
5:
When
three
firms
in
the
first
stage
of
the
game
position
their
stores
Rin
away
*
the
origin,
equidistantly
from
one
another,
the
optimal
Nash
equilibrium
price
they
charge
thetheir stores R away
5: When three firms in the first stage of the game position their stores R
away
from
second stage isLemma
pfrom
* = π R 3.
the
origin,
equidistantly
from
one
another,
the
optimal
Nash
equilibrium
the origin,
equidistantly
one
another,
optimal
Nash
equilibrium
price
they
charge
in
the
second
stage
is
.
p
=
from equidistantly
the3 origin,
equidistantly
fromthe
one
another,
the
optimal Nash
the origin,
optimal
Nash
equilibrium
price equilibrium
they charge inprice
the price they charge
3 one another,
3 p* = π Rfrom
second
stage
is
.
πR 3
πR 3
π R3second
3
second
stage is
..
second stage is p*second
.
charge
stage
is p* =
*in the
= they
3
3 stage is p = 3 .
system ofconditions
first-orderderived
conditions
the proof
of Lemma
5 (A.8-A.10) comThe systemThe
of first-order
in thederived
proof ofinLemma
5 (A.8-A.10)
completely
The system of first-order conditions derived in the proof of Lemma 5 (A.8-A.10) completely
pletely
characterizes
in the
second
stageitsgiven
that Firm 3 chooses
characterizes price
competition
in the price
secondcompetition
stage given that
Firm
3 chooses
location
characterizes price
competition
in the
stage given
that
chooses
its location
The
system
of first-order
conditions
in Firm
the *proof
ofaLemma
5 (A.8-A.10)
completely
* 5 second
63such
close
theconditions
proposed
Lderived
. It close
remains
to derived
be
seen
whether
configuration
isproof
.(A.8-A.10)
It remains
to beinseen
whether
such a5 (A.8-A.10) com
itstolocation
reasonably
toproof
the
L53conditions
system
ofproposed
first-order
derived
the
of Lemma
Thereasonably
system ofcharacterizes
first-order
in
the
of
Lemma
completely
reasonably
close
to system
the proposed
L33*.5 ItThe
remains
to derived
be
seen
whether
suchof
aLemma
configuration
is location
price
competition
in
the
second
stage
given
that
Firm
3
chooses
its
The
of
first-order
conditions
in
the
proof
5
(A.8-A.10)
completely
optimal
in the
first
stagein
ofthe
theis
game.
configuration
optimal
in
the
stage
of
the
game.
* 5 first
characterizes
price
competition
in
the
second
stage
given
that
Firm
characterizes
price
competition
second
stage
given
that
Firm
3
chooses
its
location
optimal in reasonably
the first stage
of the
game.
close
to
the
proposedinLthe
It remains
to given
be seen
whether
a configuration
characterizes
price
stage
that
3such
chooses
its location is 3 chooses its location
3 . second
* 5Firm
* 5competition
*
*to be seen whether such a configuration i
reasonably
close
to
the
proposed
L
.
It
remains
5
*
reasonably close
to
the
proposed
L
.
It
remains
to
be
seen
whether
such
a
configuration
is
3
3
L2a*, configuration
and firms
We rewrite
Firm
3’s
profit
function
store positions
being
L1* and
optimal
in the
first
of thewith
game.
reasonably
close
tostage
the
proposed
L3rivals’
. It remains
to be seen
whether
such
is
L2 , L
and
Wefirst
rewrite
Firm
3’s
profit
function
with
store
positions
being
L1 and
* firms *
optimal
in rivals’
the first
stage
of the
game.
optimalcharging
in the
stage
ofin
the
game.
We
rewrite
Firm
profit
function
with
rivals’
store
positions
being
equilibrium
prices
instage
the3’ssecond
stage:
*1 and L*2 , and firms
optimal
the
first
of
the
game.
charging equilibrium
pricesFirm
in the
stage:
We
rewrite
3’ssecond
profit
function
rivals’
store positions
*
* being L1 and L2 , and firms
charging
equilibrium
prices
in thewith
second
stage:
rewrite
Firm
profit
rivals’
being L1* and L2*, and
*
, being
and
firms
We rewrite
function
store
Π 3 (Firm
r ,charging
φ ; R3’s
)We
= profit
prewrite
) ⋅ D3prices
(with
r , φprofit
; rivals’
RinWe
, pthe
, φ ; R ),positions
pstage:
,3’s
φ ; Rbeing
), store
p 3 (Lfunction
r1,positions
φ and
; R ))L.2with
equilibrium
and Lpositions
Firm
3’s
function
with
L1* store
3 (r , φ ; R
1 ( rsecond
2 ( rrivals’
2 , and firms
.
Π
(
r
,
φ
;
R
)
=
p
(
r
,
φ
;
R
)
⋅
D
(
r
,
φ
;
R
,
p
(
r
,
φ
;
R
),
p
(
r
,
φ
;
R
),
p
(
r
,
φ
;
R
))
3
3in the second
3 stage:
1 equilibrium
2
charging
prices in3 the second stage:
charging equilibrium
pricesequilibrium
charging
prices
in the
second stage:
Π 3 ( r , φin
; Rtwo
) = pderivatives:
We are interested
3 ( r , φ ; R ) ⋅ D3 ( r , φ ; R , p1 ( r , φ ; R ), p 2 ( r , φ ; R ), p 3 ( r , φ ; R )) .
two derivatives:;D
Dr 3,.φ( r; ,Rφ),; Rp, p( r1 (, φr ,;φR;))
R ),
Π 3 ( r , φWe
; R ) are
= pinterested
, φ ;(Rr), φ⋅ in
R3),((rrp,,φ2φ(;;rRR, φ), ;=
3 ( rΠ
. p 2 ( r , φ ; R ), p 3 ( r , φ ; R )) .
;DR3 )( r=, φp;3R(,r ,pφ1 ;( rR,)φ⋅Π
pR1 (p),r3,p(φr3;,(φRr ;,),φR;p)R2⋅ ())
3
3
3
are interested
 ∂ Din3 two
dΠ 3 ∂ΠWe
∂ Dderivatives:
∂ D3 d p 2 
3 d p3
3 d p1
 , derivatives:
dΠ 3 = ∂Πin
dare
p derivatives:
∂D
d p interested
We
interested
We
+ p3 ⋅  ∂ Din3intwo
+two
⋅ are
+ ∂ D ⋅ dinp two
and
(9)
We are interested
3 ⋅two
3derivatives:
⋅ are3 interested
+ p ⋅  ∂ r + ∂ pderivatives:
⋅ 1 + ∂ p 32 ⋅ d r2  , and
(9)
d r = ∂ pWe
3 Π d r ∂Π 3 d
1∂ Dd r ∂ D


d
p
∂
D
d
p
d
p
dr
∂ p 3d r
∂3 r
∂ p d3 +
r ∂3 p⋅ 2 1d +r  3 ⋅
3
2
∂ D3 d(9)p 2 
∂3 D3 ∂∂Π
dΠ 3 ∂Π 3 d p3 d3 Π = ∂∂Π
D3 ⋅ d∂pD3+ pd3p⋅1d1∂ΠD
dD3p 2 ddpp3 ∂ D ∂ dD3p ∂, Dand
3 d p1
 , and
⋅⋅ d, rd1and
++ p∂3 p⋅ 23 ⋅ d r2+ ,(9)and
⋅
+
⋅
(9)
= dΠ ⋅ ∂Π +d rpd33 p=⋅  ∂ p33 ⋅+d∂rD
3 ⋅
⋅Dd3+∂r∂rd3p=p+1∂⋅∂pdp∂13rD
(9)
3 pd3∂r
3d r p
2
∂
r
∂
p
d
r
∂
p 2 (9)
d r 
dr
∂dΠp333 = d∂rΠ 33d⋅rd p33 ∂+∂prp ⋅ d∂∂rD
p1+


3 D3d⋅ r
1

∂
D
∂
d
p
d
p
21 ∂

+
⋅
+
.
(10)
∂
r
p
∂
p
d
r
3
3
2
3

.2
(10)
d φ = ∂ pd3 Π⋅ d φ ∂+Πp3 3 ⋅d p∂ φ + ∂ p 1∂ ⋅Dd φ ∂+D∂1 p 2d⋅pd φ ∂ D
dφ
∂ p3 3d =
φ
∂3φ+ p ∂⋅ p 1 d3 φ+ ∂3 p⋅ 2 d1 φ
3
3 d p2 



⋅
+
⋅
.

d p2 
∂ D3 ∂∂Π
dΠ 3 ∂Π 3 d p3 6dΠ  ∂∂Π
D3 d∂pD3 d3p1d∂ΠD
dD3p 2 ddpp3 ∂ D ∂ dD3p ∂ D3 d p1 ∂ D3 (10)
.
d. φthe
=+of⋅∂(10)
⋅⋅equals
p∂3 p⋅ 23 The
⋅demands
+ would⋅(10)
(10)
= Again,⋅ the first
+
p33term
⋅ 3∂ located
+d its
⋅ both
+∂ φ3top
(10)
dIfφFirm
pin33 RHS
φ3 of
p13 disk,
d φfirst
3and
1 + configuration
2+of
store
at⋅ the
the
the
have changed
and
(9)
zero.
term
in
parentheses

=
⋅
+
p
+
⋅
.
φ∂φand
∂∂(10)
pdbe
ddφφanymore.
∂ dφ φfirst
din
φ parentheses
∂ p 2 d φ 
first
(9)
equals
zero.
The
d φ Again,
∂ p3 dthe
φ quantities
φ RHS
∂in
pof
d3 φ3dwould
p 2not
∂ pterm
3φvalid
1

1 both
 ∂∂defined

Figure
d represents
φ term
pin
d
φ
∂
p
∂
p
in both equations
the
direct
or
demand
effect
of
the
deviation
in
a
respective
variable,
3
1
2
in a respective variable,
in both equations
represents
theterm
direct
or demand
effect
of the
deviation
Again,
the first
in RHS
of both
(9)
and
(10)
equals
zero.
The
first term in parentheses
whilethethefirst
lastterm
two
represent
the
indirect
or(10)
strategic
effects
of
such
aterm
deviation
through
Again,
the
first
term
in
RHS
of
both
(9)
and
equals
zero.
The first term in parenth
Again,
in
RHS
of
both
(9)
and
equals
zero.
The
first
in parentheses
while the last
two
represent
the
indirect
or strategic
effects
of
such
a deviation
through
in
both
equations
represents
the
direct
or
demand
effect
of
the
deviation
in(10)
aterm
variable,
Again,
the
first
term
in
RHS
of
both
(9)
and
(10)
equals
zero.
The
first
in parentheses
prices. We
show
that
there
exists
a distance
from the
origin,
R,
such
that
ifrespective
firms
in
both
equations
represents
the
direct
or
demand
effect
of
the
deviation
in both competitors’
equations represents
the
direct
or
demand
effect
of
the
deviation
in
a
respective
variable,
competitors’
prices.
We
show
that
there
exists
a
distance
from
the
origin,
R,
such
that
if
firms
while
the
last
two
represent
the
indirect
or
strategic
effects
of
such
a
deviation
through
both equations
represents
the direct
or demand
effect of for
the symmetric
deviation inequilibrium
a respective variable, in a respective va
locate
therein
equidistantly
from
one
another,
the
necessary
conditions
while
the
last
two
represent
the indirect
effects
of ifsuch
a deviation through
while the
last two
represent
the prices.
indirect
or show
strategic
effects
ofstrategic
such
a deviation
locate
therecompetitors’
equidistantly
from
one
another,
the
necessary
for
symmetric
equilibrium
We
there
exists
aconditions
distance
from
theororigin,
R, such
that
firms
while
the last
two
represent
thethat
indirect
or
effects
ofthrough
such
astrategic
deviation
through
competitors’
prices.
We
show
that
there
exists
a
distance
from
the
origin,
R, such that if fir
competitors’ prices.
We
show
that
there
exists
a
distance
from
the
origin,
R,
such
that
if
firms
locate
there equidistantly
fromthat
onethere
another,
theanecessary
conditions
for symmetric
equilibrium
competitors’
prices. We show
exists
distance from
the origin,
R, such that
if firms
locate
there
equidistantly
from
one
another,
the
necessary
conditions
for
symmetric
equilib
locate there equidistantly
from
one
another,
the
necessary
conditions
for
symmetric
equilibrium
locate there equidistantly from one another, the necessary conditions for symmetric equilibrium
We rewrite Firm 3’s profit function with rivals’ store positions being L1* and L2*, and firms
charging equilibrium prices in the second stage:
Π 3 ( r , φ ; R ) = p 3 ( r , φ ; R ) ⋅ D3 ( r , φ ; R, p1 ( r , φ ; R ), p 2 ( r , φ ; R ), p 3 ( r , φ ; R )) .
We are interested in two derivatives:
330
ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
 ∂D
dΠ 3 ∂Π 3 d p3
∂ D3 d p1 ∂ D3 d p 2
=
⋅
+ p3 ⋅  3 +
⋅
+
⋅
dr
∂ p3 d r
∂ p1 d r ∂ p 2 d r
 ∂r

 , and

(9)
 ∂D
dΠ 3 ∂Π 3 d p3
∂ D3 d p1 ∂ D3 d p 2 
.
=
⋅
+ p3 ⋅  3 +
⋅
+
⋅
dφ
∂ p3 d φ
∂
φ
∂ p1 d φ ∂ p 2 d φ 

(10)
(10)
Again, the first term in RHS of both (9) and (10) equals zero. The first term in parentheses
Again, the first term in RHS of both (9) and (10) equals zero. The first term in parentheses
in both equations represents the direct or demand effect of the deviation in a respective variable,
in both equations represents the direct or demand effect of the deviation in a respective
while the last two represent the indirect or strategic effects of such a deviation through
variable,
while We
the show
last
two
the
or strategic
of such
deviation
are
satisfied.
The
derivatives
need
to determine
theeffects
demand
andaifstrategic
effects
competitors’
prices.
thatrepresent
there we
exists
a indirect
distance
from
the origin,
R, such
that
firms
through
competitors’
prices.
We
show
that
there
exists
a
distance
from
the
origin,
locate
there equidistantly
from6.one another, the necessary conditions for symmetric equilibriumR,
collected
in
Lemma
The
derivatives
wedetermine
need to determine
demand
strategic
effects
are satisfied.
The
derivatives
we
need
to
the demand
and
strategic
effects inconditions
(9) arein (9) are
such are
thatsatisfied.
if firms
locate
there
equidistantly
from
one the
another,
theand
necessary
collected
in
Lemma
6.
collected
in
Lemma
6.
for symmetric equilibrium are satisfied. The derivatives we need to determine the demand and strategic effects in (9) are collected in Lemma 6.
*
*
∂D
*
in (9) are
4R − 1
3
L1* = (R,−π/3),
and ∂LD33 = (R,π):
,
Lemma 6: When
∂4D
R − 1(a)4 R − 1 = −
*L2 = (R,π/3)
*
* = (R,−π/3),
= (R,−π/3),
(R,π/3)
= (R,π): =(a)− 3 =,,− ∂r ,
6:Lemma
When 6:
LL11*When
L
(R,π/3)
and
LL33** =
=and
(R,π):
(a)
Lemma 6:
= (R,−Lπ1 /3),
L22* == (R,
πL/3)
(R,
πL):3*(a)
Lemma
When
2 =and
2R 3
∂r
2∂Rr 3 2 R 3
dp dp dp
11 dp1 dpdp
−29 + 9π 2 3 −3 πR−22π+ 36
∂D3 ∂D∂3D∂3D3 ∂D∂13D3
3 R− 2π 3 R
(b) =(b) == = =,=and (c) ,, and
..
and=(c)
(c)2 1= −=1 9=+ 29π= −239=−+π9π + 336− π− 2π+ 36
.
and
(c)
.
(b) (b)
drdrdr dr dr 90 − 3π 90
∂p1 ∂p∂1p∂2p1 ∂2pR∂2 p 23 22RR 33 dr
3 − 3π 903 − 3π 3
(
()
()
)
Part (a) considers the direct demand effect. If R > 0.25 Firm 3 would like to position its
store closer to the origin as far as this effect is concerned. This way it captures the oppoPart (a)around
considers
direct
demand
> 0.25
Firm like
3reverses,
would
like
to 3its
position
Part (a)demand
considers
the direct
demand
If effect.
RFor
> 0.25
Firm
3 would
to position
store its store
nents’
thethe
center
of effect.
the disk.
R If
<R
0.25
the
effect
Firm
loses
to the
as far
asdirect
this
effect
is concerned.
way
it captures
the
opponents’
closer to closer
the origin
as origin
far as this
effect
is concerned.
This
wayThis
it captures
theFirm
opponents’
demand
Part
(a)
considers
the
demand
effect.
If
R
>
0.25
3
would
likedemand
to position its stor
demand to competitors and would like to move away from the origin.7
around
the
center
of
the
disk.
For
R
<
0.25
the
effect
reverses,
Firm
3
loses
demand
to
around
the
center
of
the
disk.
For
R
<
0.25
the
effect
reverses,
Firm
3
loses
demand
to
closer to the origin as far as this effect is concerned.
This way it captures the opponents’ demand
6
competitors
and
would
like away
to move
away
the
origin.6
competitors
and would
like
to move
from
the from
origin.
around
the
center
of
the
disk.
For
R
<
0.25
the
effect
Firm 3captured
loses demand to
Part (b) quantifies the positive effect competitors’ prices havereverses,
on the demand
6on the captured
Part
(b)
quantifies
the
positive
effect
competitors’
prices
have
demand
captured
by
Part
(b)
quantifies
the
positive
effect
competitors’
prices
have
on
the
demand
by
competitors
and
would
like
to
move
away
from
the
origin.
by Firm 3.
Firm 3. Firm 3.
Part demanding
(b)most
quantifies
thetask
positive
effect
competitors’
prices
have
on the
demand captured by
demanding
is to evaluate
the
effect
3’s
radial
deviation
on
The most
mostThe
task
evaluate
thethe
effect
Firm
3’s Firm
radial
deviation
has on
The
demanding
taskisistoto
evaluate
effect
Firm
3’s
radial
deviation
has has
on opFirm
3.prices
opponents’
prices
(part
(c)).
In simplify
order
to simplify
a very
complex
exercise
wethe
exploit the
the
opponents’
(part
(c)).(c)).
In order
to
a veryacomplex
exerciseexercise
we
exploit
ponents’
prices
(part
In
order
to simplify
very
complex
we
exploit
symmetry
of
the
problem
extensively.
It
is
clear
that
replies
in
prices
of
both
competitors
symmetry
of
the
problem
extensively.
It
is
clear
that
replies
in
prices
of
both
competitors
must
be must be
symmetry
of thedemanding
problem extensively.
isevaluate
clear thatthe
replies
in prices
of3’s
both
competitors
The
most
task
is Ittothe
effect
Firm
radial
deviation
has on
identical
when
Firm
3
moves
along
vertical
axis.
We
therefore
use
only
necessary
conditions
identical
when
Firm
3
moves
along
the
vertical
axis.
We
therefore
use
only
necessary
conditions
must
be
identical
when
Firm
3
moves
along
the
vertical
axis.
We
therefore
use
only
necopponents’
(part
(c)).
In
order
tofor
simplify
aFirms
very
complex
exercise
weproof
exploit
for
profitprices
maximization
with
to own
for
2prices
and 3 for
(A.9-A.10)
the
of the
for profit
maximization
with
respect
torespect
own
prices
Firms
2 to
and
3 (A.9-A.10)
inFirms
the proof
of
essary
conditions
for
profit
maximization
withprices
respect
own
2inand
3
thebe
problem
extensively.
is,, clear
both
,3 and
areprices
positive
any competitors must b
Itreadily
can
readily that
verified
and
are
for
anyoffor
partsymmetry
(c). Itpart
can(c).
beof
verified
dp 2 dp
drthat
dp1 that
dr , dp12Itdr
dr
dp 3positive
dr in
dr replies
(A.9-A.10) in the proof of part (c). It can be readily verified that dp1/dr, dp2/dr, and dp3 are
identical
when
moves
along
the
axis.
We
therefore
use
only necessary conditions
that
is,
when
Firm
moves
towards
theprices
origin,
prices
as competition
Rthat
is,
when
Firm
33moves
the
origin,
decrease
as
competition
R ∈ [0,1],
∈ [0,1],
positive
for
any
RFirm
[0,1],
that
is,3towards
when
Firm
3 vertical
moves
towards
the decrease
origin,
prices
decrease
∈
toughens,
and
vice
versa.
toughens,
and
vice
versa.
for
profit maximization
with
as competition
toughens, and
vicerespect
versa. to own prices for Firms 2 and 3 (A.9-A.10) in the proof of
, dp 2 (Lemma
, andsome
arewhat
positive
part
(c).
It can
be compared
readily
verified
that
dpfor
dr offer
dr
These
results,
when compared
to
those
a (Lemma
duopoly
2),dp
offer
some
idea
for whatfor any
These
results,
when
to those for
a duopoly
2),
for
1 dr
3idea
follows.
Suppose
all
three
firms
were
located
on
the
perimeter
of
the
disk,
and
Firm
3
follows.
Suppose
all
three
firms
were
located
on
the
perimeter
of
the
disk,
and
Firm
3
These
results,
compared
to 3
those
for atowards
duopoly (Lemma
2), offer
somedecrease
idea for what
thatwhen
is,awhen
Firm
moves
theof
origin,
prices
as affected
competition
R
∈ [0,1],
contemplated
radial
move
towards
the
origin.
Themarginal
line of
of marginal
by
contemplated
a small radial
move
towards
origin.
The
consumers
affected
follows.
Suppose
allsmall
three
firms
werethe
located
on
theline
perimeter
the
disk, consumers
and
Firm by
3
andis vice
versa.
this
move
of
length
two
(BO
and
CO;
see
Figure
3).
The
same
is
true
for
the
duopoly
case.
thistoughens,
move
is
of
length
two
(BO
and
CO;
see
Figure
3).
The
same
is
true
for
the
duopoly
case.
contemplated a small radial move towards the origin. The line of marginal consumers
Hence,
direct
demand
effects
beand
similar
both
cases.
−1
infor
thethe
duopoly
Hence,
direct
effects
beshould
very(BO
similar
in
both
They
are They
−1 inare
the
duopoly
affected
bydemand
this
move
is7 ofshould
length
two
CO;
seecases.
Figure
3).
The
same
is true
These
results,
when
compared
tovery
those
for
ainduopoly
(Lemma
2),
offer
some idea for what
the elasticity
of the
firm’s with
demand with
−0.87
three
firms.7 Furthermore,
Furthermore,
we derivewe
thederive
elasticity
of the firm’s
demand
and −0.87and
with
threewith
firms.
duopoly case.
Hence,alldirect
demand
effectslocated
should be
very
similar
in both
cases.
They
follows.
Suppose
three
firms
were
on
the
perimeter
of
the
disk,
and
Firm 3
to its
radial from
distance
origin
all the
on the perimeter
of the disk.
respect torespect
its radial
distance
the from
originthe
when
all when
the8 firms
arefirms
on theare
perimeter
of the disk.
are −1 in the duopoly
andradial
−0.87 move
with three
firms.the
Furthermore,
weline
derive
the
elasticity
contemplated
a
small
towards
origin.
The
of
marginal
consumers
affected by
Theare
results
for the and
duopoly
and three-firm
oligopoly,
respectively.
At
The results
−0.64are
and−0.64
−0.83and
for−0.83
the duopoly
three-firm
oligopoly,
respectively.
At
current is
demands,
in a(BO
three-firm
case
gainsFigure
relatively
than
in the
duopoly
it
current
a of
firm
in aa firm
three-firm
case
gains
relatively
more than
in the
duopoly
when
it when
thisdemands,
move
length
two
and
CO;
see
3).more
The
same
is true
for
the duopoly
case.
7
Point
D moves
the
vertical
axes
towards
topvery
of the
disk,
so Firm
3 elasticity
gains
someofnew
customers
its along
store
towards
the
origin
<also
0).
We
also
theboth
opponents’
prices
moves
itsmoves
store
towards
the
origin
(∆R
<
0).(∆R
Wethe
derive
thederive
elasticity
of opponents’
prices
Hence,
direct
demand
effects
should
be
similar
in
cases.
They
are
−1
in
the
duopoly
8
from
the
(see
Figure
3).
At
time
line segments
DC8 and
DB
rotate
toward
each other,
7the same
Itin
isthe
0.71
in the
duopoly
and
0.41 demand
in the
withoponents
tothree
the
firm’s
radial
distance
the
is 0.71
duopoly
and
in
the
withand
respect
torespect
the
firm’s
radial
distance
from
the from
origin.
Furthermore,
weItorigin.
derive
the
elasticity
of0.41
firm’s
with
−0.87
with
firms.
which
means
that
Firm
3
loses
some
customers
on
the
outskirts
of
the
market.
total
effect
isthe
negative
oligopoly.
Rivals’
cut response
a firmtowards
movingThe
towards
the
in the first
three-firmthree-firm
oligopoly.
Rivals’ price
cut price
response
to a firm to
moving
the
origin
in origin
the
first
for R < 0.25.
respect
toofits
radial
distance
origin
when
allduopoly.
the
firms
on
perimeter
stage
the
game
will
beinweaker
in the
the three-firm
case
than
in the
duopoly.
Thisthe
is abecause
a of the disk.
stage
of the
game
will
be weaker
thefrom
three-firm
case than
in the
Thisare
is because
8
From
part
(a)
in
Lemmas
2
and
6.
The
results
are
−0.64
and
−0.83
for
the
duopoly
and
three-firm
oligopoly,
respectively.
At
cutofby
the two
notwould
moved
not only
the aggressive
price cut price
by one
theone
twooffirms
thatfirms
havethat
not have
moved
notwould
only affect
theaffect
aggressive
rival, rival,
but the
other
neighbor
as
well.
This
would
provoke
a
response
from
a
peaceful
rival
and
would
but current
the other
neighbor
as
well.
This
would
provoke
a
response
from
a
peaceful
rival
and
would
demands, a firm in a three-firm case gains relatively more than in the duopoly when it
lead profits
lowermade
profits
consumers
affected
by the aggressive
have illustrated
lead to lower
on made
consumers
not affected
by
the aggressive
firm. Wefirm.
have We
illustrated
moves
itsto store
towards
theonorigin
(∆Rnot
< 0).
We also
derive
the elasticity
of opponents’ prices
the incentive
in a three-firm
R = compared
1, when
compared
to the duopoly.
It
the incentive
a firm in aa firm
three-firm
oligopolyoligopoly
has whenhas
R =when
1, when
to the duopoly.
It
8
is 0.71
inathe
duopoly
respect
to
thedemand
firm’s
radial and
distance
thepunished
origin.
gain relatively
more directly
demand
directly
andfrom
will be
byItrelatively
less
severe
a price and 0.41 in th
willwith
gain will
relatively
more
will be
punished
by
relatively
less
severe
price
oligopoly.
Rivals’
price
cut at
response
to=aduopoly
towards
in the first
cut by
rivals
in thestage.
second
If
a firm
residing
at the
R
1firm
in themoving
duopoly
had
an incentive
to
cut three-firm
by rivals
in the
second
If astage.
firm
residing
R = 1 in
had
an incentive
to the origin
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
331
of the firm’s demand with respect to its radial distance from the origin when all the firms
are on the perimeter of the disk. The results are −0.64 and −0.83 for the duopoly and
three-firm oligopoly, respectively. At current demands, a firm in a three-firm case gains
relatively more than in the duopoly when it moves its store towards the origin (∆R < 0).
We also derive the elasticity of opponents’ prices with respect to the firm’s radial distance
from the origin.9 It is 0.71 in the duopoly and 0.41 in the three-firm oligopoly. Rivals’
price cut response to a firm moving towards the origin in the first stage of the game will
be weaker in the three-firm case than in the duopoly. This is because a price cut by one
of the two firms that have not moved would not only affect the aggressive rival, but the
other neighbor as well. This would provoke a response from a peaceful rival and would
lead to lower profits made on consumers not affected by the aggressive firm. We have illustrated the incentive a firm in a three-firm oligopoly has when R = 1, when compared
to the duopoly. It will gain relatively more demand directly and will be punished by relatively
less
severe
a price
rivalswe
inexpect
the second
stage.
If the
a firm
at Rfirms
= 1 in the
move
even
farther
away
fromcut
thebyorigin
this not
to be
caseresiding
with three
duopoly
had
an
incentive
to
move
even
farther
away
from
the
origin
we
expect
this not
anymore. Lemma 7 presents an interior radial distance from the origin for the three firms
to be the
case with three
firmsother,
anymore.
presents
interior
radialindistance
positioned
equidistantly
from each
whichLemma
does not7make
theman
want
to relocate
radial
from the origin for the three firms positioned equidistantly from each other, which does
direction.
not make them want to relocate in radial direction.
72 + 15π 3 − 2π 2
firms are
arelocated
locatedequidistantly,
equidistantly,
and firms
no
(≈ 0.5476) and
288 − 8π 3
firmitfinds
it profitable
to deviate
locally deviate
firmnofinds
profitable
to locally
from it. from it.
Lemma
When R*=
Lemma7:7:When
Proof of Lemma 7 yields another result. Symmetric configuration with maximum distance between three firms, i.e. maximum differentiation, is never optimal.
Proof of Lemma 7 yields another result. Symmetric configuration with maximum distance
between
three1:
firms,
i.e. maximum
differentiation,
is neverof
optimal.
Corollary
The three
firms located
on the perimeter
the disk, equidistantly from each
other, is not a subgame perfect Nash equilibrium of the game.
Corollary 1: The three firms located on the perimeter of the disk, equidistantly from each
Lemma
thatperfect
locating
at equilibrium
the perimeter
of the
disk will still yield the highest posother,
is not5a shows
subgame
Nash
of the
game.
sible prices and profits in a symmetric configuration, but these are not sustainable since
capturing the opponents’ demand around the origin is too tempting; a classic prisoner
dilemma on an oligopoly market.
Lemma 5 shows that locating at the perimeter of the disk will still yield the highest possible
prices
andwe
profits
a symmetric
configuration,
but these
are not sustainable
sincehas
capturing
the
Next,
showinthat,
when firms
are positioned
equidistantly,
none of them
an incenopponents’
demand
around
the
origin
is
too
tempting;
a
classic
prisoner
dilemma
on
an
oligopoly
tive to locally move along a polar direction.
market.
Next, we
that,
are positioned
equidistantly,
φ = πfirms
is locally
optimal for
Firm 3. none of them has an incentive
Lemma
8: show
For any
R,when
to locally move along a polar direction.
Results from Lemmas 5, 7, and 8 are summarized in Proposition 2.
Lemma 8: For any R, φ = π is locally optimal for Firm 3.
Results from Lemmas 5, 7, and 8 are summarized in Proposition 2.
9
From part (c) in Lemmas 2 and 6.
Proposition 2: When in the first stage of the game the three firms locate their stores R*
72 + 15π 3 − 2π 2
(≈ 0.5476) away from origin, equidistantly from each other, the equilibrium
=
288 − 8π 3
*
Lemma 8: For any R, φ = π is locally optimal for Firm 3.
Lemma 8: For any R, φ = π is locally optimal for Firm 3.
332
Results from Lemmas 5, 7, and 8 are summarized in Proposition 2.
Results from Lemmas 5, 7, and 8 are summarized in Proposition 2.
ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
Proposition 2: When in the first stage of the game the three firms locate their stores R*
2
Proposition
the
stage
of the
gamethe
thethree
three firms
firms locate
R*
Proposition
2: When
inπ the
stage
of the
game
locatetheir
theirstores
stores
722:+When
15
3in−first
2π first
(
≈
0
.
5476
)
=
away
from
origin,
equidistantly
from
each
other, the equilibrium
2
72 + 15π 3 − 2π288
−(≈8π0.5476
3 ) away
origin,equidistantly
equidistantlyfrom
from
each
other,
R* =
away from
from origin,
each
other,
thethe
equilibrium
288 − 8π 3
π R* 3
Furthermore, every
price
theycharge
chargeininthe
thesecond
second* stage
stage is
.. Furthermore,
everyfirm’s
firm’slocation is a local best
equilibrium price
they
is
πR 3
3
. Furthermore,
every firm’s location is a local best
price they
charge
in the
secondtostage
is locations.
location
is a local
best
response
rivals’
response to rivals’ locations. 3
response to rivals’ locations.
We lack analytical proof
that
every firm’s
is in
fact location
a globalisbest
reply
to rivals’
We lack
analytical
prooflocation
that every
firm’s
in fact
a global
best reply to rivals’
systemWe
of
first-order
conditions
forprice
second-stage
profit
maximization
with
respect
to We
prices
(A.8- solve the
locations
given
the
price
competition
incompetition
the
second
stage
We
therefore
solve
locations
given
in
the
second
stage
of the
game.
therefore
lack
analytical
proofthe
that
every
firm’s
location
isof
inthe
factgame.
a global
best
reply
to
rivals’
system
of
first-order
conditions
for
second-stage
profit
maximization
with
respect
to
prices
(A.8A.10)
for
an
array
of
Firm
3
s
locations
numerically
to
derive
optimal
profits.
We
find
(see
thelocations
system ofgiven
first-order
conditions
for in
second-stage
profitofmaximization
respect
to the
the price
competition
the second stage
the game. We with
therefore
solve
A.10)
for (A.8-A.10)
an4)array
Firm
numerically
tonumerically
deriveand
optimal
profits.
We
find profits.
(see
Figure
that of
locations
wes locations
propose
are
optimal,
state
nextoptimal
result.
prices
for
an3array
of Firm
3’sglobally
locations
toour
derive
Figure
4) that
locations
wethat
propose
are globally
optimal,
state optimal,
our next result.
We find
(see
Figure 4)
locations
we propose
are and
globally
and state our next
result.
72 + 15π 3 − 2π 2
*
=+ 15π 3 − 2π 2
(≈ 0.5476) ,
Result 2: Three firms positioning their stores at R72
288 − 8π 3 (≈ 0.5476), ,equiThreefirms
firmspositioning
positioning their
their stores
Result2:2:Three
Result
storesatatRR**==
288 − 8π 3
π R* 3
in
the
distantly
fromfrom
eacheach
otherother
in the
firstfirst
stage
of of
thethegame,
equidistantly
in the
stage
game,and
andsetting
setting pπ* R
=* 3
in the
3in the
equidistantly
each other
in the
firstequilibrium
stage of the of
game,
and setting p* =
second, is from
a subgame
perfect
Nash
the game.
3
second, is a subgame perfect Nash equilibrium of the game.
second, is a subgame perfect Nash equilibrium of the game.
There are two novel perspectives on product differentiation to this result. First, in a
settingThere
that are
leads
differentiation
inon
one
characteristic
only in
twotonovel
perspectives
product
differentiation
to duopoly,
this result.which
First, is
in cona setting
sistent
with
the
existing
literature,
the
threedifferentiation
firms
it optimal
to differentiate
theirthe
There
are to
two
novel
perspectives
product
to thiswhich
result.
in a setting
that
leads
differentiation
in oneon
characteristic
onlyfind
in duopoly,
isFirst,
consistent
with
thatproducts
leads
to differentiation
in onefirms
characteristic
only to
indoduopoly,
which
consistent
in two characteristics.
Second,
firms
not differentiate
their products
existing
literature,
the three
find
it optimal
differentiate
theiris products
inwith
two theas
existing
literature,
the
three
firms
it optimal
differentiate
their
products
two could have. In a
characteristics.
Second,
firms
do
not
differentiate
their
products
much asindifferentiation
they
much
as they could
have.
Infind
a setting
thattoyields
familiar
a as
min-max
characteristics.
firms
do
not differentiate
theirofproducts
much
as theya could
have. In a
setting
yields
familiar
a min-max
differentiation
result inasdifferentiation
the
duopoly,
medium-medium
result
inthat
theSecond,
duopoly,
a medium-medium
type
product
is observed.
setting
that
a min-max
differentiation
in the
duopoly,
a medium-medium
type
of
product
differentiation
is observed.
Whenresult
all three
firms
arealocated
on
the perimeter
When
allyields
threefamiliar
firms
are
located
on the perimeter
of
the
disk,
radial
deviation
to- of
type
of product
differentiation
istowards
observed.
all
firms
located
on
theisperimeter
of
the
disk,
radial
deviation
the When
center
of three
theby
disk
by are
a firm
iscut
followed
by
rivals’
price
wards
the acenter
of
the disk
by a firm
is followed
rivals’
price
that
less
severe
the disk,
a radial
deviation
towards
the observed
center of in
theduopoly.
disk by aTherefore,
firm is followed
bydirect
rivals’
price effect
cut
that
is
less
severe
than
the
one
positive
demand
than the one observed in duopoly. Therefore, positive direct demand effect of such a
cut that
is less
severeoutweighs
than the one
in duopoly.
Therefore,
direct
demand
effect
of such
a move
the observed
negative strategic
effect
of rivals’positive
price cuts,
and
firms move
movea outweighs
the negative
strategic
effecteffect
of rivals’
priceprice
cuts,cuts,
andand
firms
move
closer
of such
outweighs
the negative
strategic
of rivals’
firms
move
closermove
together.
together.
closer
together.
We can speculate on whether the min-min-…-min part of the product differentiation result by
We
can
speculate
on
partpart
of the
product
differentiation
result by
Irmen
Thisse (1998)
couldthe
bemin-min-…-min
observed
in a three-firm
market
with
additional
product
We can and
speculate
on whether
whether
the
min-min-…-min
of
the product
differentiation
Irmen
and Thisse (1998)
could
be observed
in a three-firm
market
withinadditional
product
characteristics.
Firms
may
not
want
to
differentiate
their
products
any
additional
result by Irmen and Thisse (1998) could be observed in a three-firm market with adcharacteristics.
Firms
maythey
not find
wantmax-max
to differentiate
their products
in any additional
characteristics,
since
differentiation
to be excessive
in two dimensions,
ditional product
characteristics.
Firms may not want
to
differentiate
their
products in
characteristics,
since
they find
excessive
in
twoproducts
dimensions,
already. This
suggests
thatmax-max
firms maydifferentiation
have no needto
tobe
differentiate
their
in another,
any
additional
characteristics,
since
they
find
max-max
differentiation
to
be
excessive
already.
suggestssince
that firms
have no need
differentiate
their products
in another,
third,This
dimension,
even may
the possibilities
in to
two
were not exhausted.
We therefore
predict
inthat
two
dimensions,
already.
This
suggests
thatwere
firmsnotmay
have
noWe
need
to differentiate
third,
dimension,
since even
the
indifferentiation
two
exhausted.
therefore
the
min-min-…-min
partpossibilities
of a product
result
is going
to
hold
in predict
our setting as
products
another,
dimension,
since This
even
the
possibilities
two
were being
not
thattheir
the
min-min-…-min
part ofthird,
a product
differentiation
result
is going
to
hold
inthe
ourmarket
setting
as
well,
which is in
what
Feldin
(2001)
formally
shows.
hypothesis
rests
onin
exhausted.
We
therefore
predict
thatIf the
min-min-…-min
ofhigher
aonproduct
differentiawell,
which is what
Feldinby(2001)
formally
shows.
This
hypothesis
rests
thepopulation
market
being
uniformly
populated
consumers.
there
were
some
areaspart
with
density there
uniformly
populated
byto
consumers.
If there
were
areas
with
higher
population
density
tion
result
is going
holddifferentiation,
in our
setting
assome
well,
which
istry
what
Feldin
(2001)
formally
would
be obviously
more
since
firms
would
to tailor
their
products
tothere
meet
would
obviously
more
differentiation,
firms
would
try to tailor
their products
to meet
thebetastes
these
different
of since
customers.
shows.
Thisofhypothesis
restsgroups
on the
market
being
uniformly
populated
by consumers.
theIftastes
thesesome
different
of customers.
thereofwere
areasgroups
with higher
population density there would be obviously more
differentiation, since firms would try to tailor their products to meet the tastes of these
different groups of customers.
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
333
Figure 4: Firm 3’s profits with respect to location of its store and given equilibrium locations of opponents.
Source: Own calculations.
4.Welfare analysis
In this section we compare the extent of product differentiation in our competitive market to a social optimum for a duopoly and a three-firm oligopoly. A social planner who
cares about well-being of all agents in the market would simply minimize the traveling
costs borne by buyers. Each of n firms on the market will satisfy π/n of the market demand in a symmetric configuration. Every such piece of the pie can be split into two
halves symmetrically over the line segment connecting the firm’s location and the center
of the disk. One such part of the disk is presented in Figure 5. To find the social optimum, we look at where a particular firm should have been in order to minimize the total
traveling costs that consumers, from such a half of the disk, bear.
Figure 5: Derivation of socially optimal product differentiation.
334
ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
The cost that a consumer residing at (ρ,φ) is faced with when buying from the firm loThecost
costthat
thataaconsumer
consumerresiding
residingatat(ρ,φ)
(ρ,φ)isisfaced
facedwith
withwhen
whenbuying
buyingfrom
fromthe
thefirm
firmlocated
located
The
cated
at R
is:a consumer residing
The
cost
that
at (ρ,φ) is
faced with
when buying
from the
firm located
at
R
is:
at R
R is:
is:
at
c (ρρ,,φφ;;RR))== ρρ222sin
sin22φφ ++((RR−−ρρcos
cosφφ))222 == RR222 −−22ρρRRcos
cosφφ ++ρρ22.2..
cc(( ρ
, φ ; R ) = ρ sin 2 φ
+ ( R − ρ cos φ
) = R − 2 ρR cos φ
+ρ .
Thetotal
totalcost
costborne
borneby
bybuyers
buyersfrom
fromthis
thisportion
portionof
ofthe
thedisk
diskisisthen:
then:
The
The
portion
of the
diskdisk
is then:
The total
totalπcost
costborne
bornebybybuyers
buyersfrom
fromthis
this
portion
of the
is then:
πn 1
πn 1
n 1
CC(((R
RR)))=
==∫∫∫∫ccc(((ρ
ρρ,,φφ,,RR))⋅⋅ρρddρρddφφ..
C
∫0∫0 ,φ , R) ⋅ ρ dρ dφ ..
0 0
0 0
2RR222 ++11 ππ 22
ππ ..
sinπ
Thisintegrates
integratesto:
to:
C(R)
This
integrates
to:C(R)
C(R)===22R
+ 1 ⋅⋅π −−2 RRsin
This
⋅
−
R
sin
This integrates to: C(R) =
4
n
3
n..
44
nn 33
nn
Minimizethis
this
expression
with
respect
toR
Rto
to
getget
thethe
socially
optimal
distance
fromfrom
the the
Minimize
thisexpression
expression
with
respect
to
R get
to
socially
optimal
distance
Minimize
with
respect
to
the
socially
optimal
distance
from
the
Minimize
this for
expression
with
respect
to
R to
get the
socially
optimal
distance
from
the
:
center
of
the
disk
the
firms,
R
S
center
of
the
disk
for
the
firms,
R
:
center of
of the
the disk
disk for
for the
the firms,
firms, R
RS::
S
center
S
2nsin
sinπππn
n .
R S == 22nn sin
R
R SS = 33ππ n ...
3π
Wecompare
comparesocially
socially
optimal
locations
with
competitive
equilibrium
onesderived
derived
in
We
optimal
locations
with
competitive
equilibrium
ones
in
We
compare
sociallyoptimal
optimal
locations
with
competitive
equilibrium
ones derived
in
We
compare
socially
Sections
3and
and344and
in
Table
1. 1. locations with competitive equilibrium ones derived in
Sections
3
in
Table
1.
Sections
4
in
Table
Sections 3 and 4 in Table 1.
Table
Table1:
1:Social
Socialoptimum
optimumvs.
vs.competitive
competitiveequilibrium.
equilibrium.
Table
1:
Social
optimum
vs.
competitive
equilibrium.
Table
1:
Social
optimum
vs.
competitive
equilibrium.
n
2
n
nn3
RS
R*
0.4244
1*
RR**
RRSS
R
R
0.5513
0.5476
S
0.4244
Source:
Own calculations 11
0.4244
2
22
0.4244
1
3
0.5513
0.5476
3
0.5513
0.5476
3
0.5513
0.5476
This is analogous to what Brenner
(2005)
finds. Intense
price competition in a duopoly
Source:Own
Owncalculations
calculations
Source:
drives firms to the edge of theSource:
market,Own
andcalculations
that is too far apart from a social standpoint. In a three-firm oligopoly the situation reverses. Product differentiation becomes
too
when compared
to the social
optimum.
Priceprice
competition
becomes
less intense
Thisweak
analogous
towhat
whatBrenner
Brenner
(2005)
finds.Intense
Intense
price
competition
inaaduopoly
duopoly
This
isisanalogous
analogous
to
(2005)
finds.
competition
in
This
is
to
what
Brenner
(2005)
finds.
Intense
price
competition
in
a duopoly
and
firms
move
in
on
each
other’s
demands.
This
gives
rise
to
an
interesting
drives
firms
to
the
edge
of
the
market,
and
that
is
too
far
apart
from
a
social
standpoint.
Inaa
drives firms
firms to
to the
the edge
edge of
of the
the market,
market, and
and that
that is
is too
too far
far apart
apart from
from aa social
social standpoint.
standpoint. situation.
In
drives
In
a
three-firm
oligopoly
the
situation
reverses.
Product
differentiation
becomes
too
weak
when
It
is
in
the
social
planner’s
interest
to
induce
cooperation
among
firms.
She
might
three-firm oligopoly
oligopoly the
the situation
situation reverses.
reverses. Product
Product differentiation
differentiation becomes
becomes too
too weak
weak when
when pass
three-firm
compared
tothe
thesocial
social
optimum.
Price
competition
becomes
less
intense
and
firms
movehelp
inon
on
a regulation
demanding
thePrice
firms
locate RS becomes
away
from
the
origin.
This
would
the
compared
to
optimum.
competition
less
intense
and
firms
move
in
compared
to
the
social optimum.
competition
becomes
less
intense
and
firms
move
in
on
eachfirms
other’s
demands.
Thisgives
givesPrice
riseto
to
aninteresting
interesting
situation.
It
is
inthe
the
social
planner’s
todemands.
partly resolve
their
prisoner’s
dilemma. situation.
They
would
now
be social
able
toplanner’s
charge higher
each
other’s
This
rise
an
It
is
in
each
other’s
demands.
This gives
rise to
an interesting
It is in the
social planner’s
interest
toinduce
induce
cooperation
among
firms.
She
mightsituation.
pass
demanding
thefirms
firms
prices;
therefore,
they would
befirms.
willing
tomight
accept
such
aregulation
regulation.
interest
to
cooperation
among
firms.
She
might
pass
aaregulation
regulation
demanding
the
interest
to
induce
cooperation
among
She
pass
a
demanding
the
firms
away
from
the
origin.
This
would
help
the
firms
to
partly
resolve
their
prisoner’s
locate
R
S
away from the
the origin.
origin. This
This would
would help
help the
the firms
firms to
to partly
partly resolve
resolve their
their prisoner’s
prisoner’s
locate R
RS away
locate
S Theyfrom
dilemma.
wouldnow
nowbe
beable
ableto
tocharge
chargehigher
higherprices;
prices;therefore,
therefore,
theywould
wouldbe
bewilling
willingto
to
dilemma.
They would
would
they
dilemma.
They
now
be
able
to
charge
higher
prices;
therefore,
they
would
be willing
to
acceptsuch
suchaaregulation.
regulation.
accept
5. such
Concluding
accept
a regulation. remarks
Concludingremarks
remarks
55 Concluding
Concluding
remarksof product differentiation that firms
We have presented some novel5findings
on the extent
Wehave
havepresented
presentedsome
somenovel
novelfindings
findingson
onthe
theextent
extentof
ofproduct
productdifferentiation
differentiationthat
thatfirms
firmsin
in
We
in an
oligopoly
employ.
Contrary
to the
literature
on
duopoly
markets, we
show
We
have
presented
some novel
findings
on on
the
extent of
product
differentiation
that
firmsthat
in
an
oligopoly
employ.
Contrary
to
the
literature
duopoly
markets,
we
show
that
three
firms
an oligopoly
oligopoly
employ.
Contrary
to the
the literature
literature
on duopoly
duopoly
markets, we
we
show
that three
three
firms
three firms
will Contrary
use two product
characteristics
to separate
their
products
from
rivals’.
an
employ.
to
on
markets,
show
that
firms
In a setting that usually yields min-max differentiation we find three firms to utilize a
medium-medium type of differentiation. It remains to be seen how even a larger number
of firms affects product differentiation in multi-characteristic space.
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
335
The demand side of the model is the sole driver of our results. If firms had to cover some
R&D costs to introduce new varieties of products or improve some of the characteristics,
which certainly is the case in reality, differentiation would be even weaker. We therefore
note that firms deciding to introduce new characteristics to their products might want to
be careful as not to engage in excessive differentiation that would not yield the maximum
possible profits. Also, the fact that differentiation possibilities are not exhausted leads us
to think that three firms will not want to use any new product characteristics to differentiate themselves from competitors. The first step towards such a min-…-min-mediummedium differentiation result is presented in Feldin (2001).
References
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A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
337
AppendixA:
A:Proofs
ProofsofofLemmas
Lemmasand
andPropositions
Propositions
Appendix
Appendix A: Proofs of Lemmas and Propositions
Appendix
A: Proofs
Lemmas
Proof of Lemma 1: We first determine
α and
β thatofwe
needand
forPropositions
the demand functions. They
Proof of Lemma 1: We first determine α and β that we need for the demand functions. They
Proof
of Lemma
1:conditions
We
first
αtwo
andof
βconsumers
that
weβ need
forneed
the in
demand
functions.
Proof of
Lemma
1:determine
We
first
determine
α and
that
we
for
the demand
functions.
Appendix
A:
Proofs
Lemmas
and
Propositions
aregiven
givenby
bythe
theindifference
indifference
conditions
for
the
two
residing
inpoints
points
and
B: They
are
for
the
consumers
residing
AAand
B:
Appendix
A: Proofs
of Lemmas
and Propositions
aregiven
given
bythe
theindifference
indifference
conditions
for
the
two
consumers
residing
in
points
A
and
B:
They
are
conditions
for
the
two
consumers
residing
in
points
A
22
2 2by
22
22
Appendix
Proofs
Lemmas
,where
where
α,functions.
sin xx++( (RR−−Proof
cosxx)of) 2Lemma
rsin
sinfirst
sin
+α( r(and
rcos
cos
φ−−and
cosPropositions
xx==α,
β.β.
pp1 1++sin
cos
== pp2 2++1:
( r(We
φφ−−A:
sin
xx) ) +of
cos
xx) ) , for
determine
βφthat
we
need
the2 demand
They
thatφ we
need
the demand
where
=α
α,, β
β..functions. They
sin x of
+ ( Lemma
R − cos x1:
) 2 We
= p 2first
+ ( rdetermine
sin φ − sin α
x ) 2and
+ ( rβcos
− cos
x ) ,,for
and B: p1 + Proof
where
xx =
areProof
given
byLemma
the indifference
conditions
for
the
two
consumers
residing
in points
A and B: (if
Thisequation
equation
simplifies
to
(A.1)
and
by
using
the
fact
that
insymmetric
symmetric
equilibrium
(if
This
simplifies
and
by
using
the
fact
in
equilibrium
of
1:(A.1)
We
first
determine
αand
and
β
that
we
need
for2the
demand
functions.
are
byto
the
indifference
conditions
for
thethat
two
consumers
residing
in
points AThey
and(ifB:
This
equation
simplifies
to
(A.1)
by
the
fact
that
in
symmetric
equilibrium
2 given
2
2 using
,
where
x
=
α,
β. B:
p
+
sin
x
+
(
R
−
cos
x
)
=
p
+
(
r
sin
φ
−
sin
x
)
+
(
r
cos
φ
−
cos
x
)
2indifference conditions
2
2
2
1
2
are
given
by
the
for
the
two
consumers
residing
in
points
A
and
exists)rr=R
=Rand
andφequation
φ=
=
π,
to
(A.2):
exists)
(A.2):
, where x =(if
α, β.
p1π,
+ to
sin
x and
+ ( Rφ−=to
cos
x ) (A.2):
= p 2by
+ (using
r sin φthe
− sin
+ (in
r cos
φ − cos x )equilibrium
This
and
factx )that
symmetric
exists)
r simplifies
=R
π,(A.1)
to
2
2
This
simplifies
using
fact
that
(if
, where
x =equilibrium
α, β. equilibrium
pexists)
x +equation
( R −φequation
cos
)to
=(A.2):
p 2 to
+ A:
((A.1)
r sin
φand
− sin
xand
) 2 +by
( rthe
cos
φPropositions
−the
cosfact
xin) 2symmetric
Proofs
ofbyLemmas
and
1 + sinr =R
and
= πx,Appendix
This
simplifies
to
(A.1)
using
that
inRsymmetric
(if
2
2
2−
2 2r 2 −
.
(A.1)
2
r
sin
φ
⋅
sin
x
+
2
(
r
cos
φ
−
R
)
⋅
cos
x
=
p
p
+
exists)
requation
=R
and
(A.2):
.
(A.1)
2rrsin
sin
φ⋅ ⋅sin
sinxφsimplifies
x+=+2π,2( r(torcos
cos
φ
−
R
)
⋅
cos
x
=
p
−
p
+
r
−
R
.
(A.1)
2This
φ
φ
−
R
)
⋅
cos
x
=
p
−
p
+
r
−
R
2
1
2
(A.1)
and αbyand
using
fact
thatfor
in the
symmetric
(if
2the
1 1need
exists)
r =R and
φ =first
π,toto
(A.2):
Proof
of Lemma
1: We
determine
β that
we
demand equilibrium
functions. They
exists)
r =R
andindifference
φ2=r sin
π, to
(A.1)
φ ⋅(A.2):
sinconditions
x + 2( r cosfor
φ −theR )two
x = px2 =− ppresiding
+ r2 − R2 .2
(A.2)
4 Rconsumers
are
given
by the
A and(A.1)
B:
(A.2)
cosxx−=⋅ =cos
−−44RR⋅ ⋅cos
pp2⋅2)cos
−⋅−cos
pp1 1x =21p−222p−1 p1in+2 points
(A.1)
r − R 2 . (A.2)
2
22 r sin φ ⋅ sin x + 2( r cos2φ − R
,
where
x
=
α,
β.
p1 + sin x + ( R 2−rcos
x
)
=
p
+
(
r
sin
φ
−
sin
x
)
+
(
r
cos
φ
−
cos
x
)
. respect to(A.1)
sinthe
φ ⋅ sin
x2 + 2( r conditions
cos φ −−R4) R⋅for
cos
x x= =pmaximization
r −R
We state
necessary
profit
with
own
prices for
(A.2)
⋅ cos
2p 2− −p1p+
(A.2)
1
This
equation
to (A.1)
and
bymaximization
using
inpsymmetric
equilibrium
(if for
(A.2)
−the
4 Rfact
⋅ costhat
x with
=with
p1
Westate
statethe
the
necessary
conditions
forprofit
profit
maximization
respect
ownprices
prices
for
We
necessary
conditions
for
respect
totoown
2 −
Firm
1 andsimplifies
Firm
2 (from
(2)):
−for
4 Rprofit
⋅ cos xmaximization
= p 2 − p1 with respect to own(A.2)
We
state
the
necessary
conditions
prices
for
exists)
r
=R
and
φ
=
π,
to
(A.2):
Firm11and
andFirm
Firm
2
(from
(2)):
Firm
2
(from
(2)):
We state theWe
necessary
for
profit
with respectwith
to own
prices
for prices for
p1 maximization
1
state
theconditions
necessary
conditions
for profit maximization
respect
to own
Firm 1 and Firm
2 (from
(2)):
((zr−cos
sinφz )−for
+R )profit
⋅ (z pxmaximization
−= cos
z−⋅ p
zp +
) ) r=20with
2 and
We
stateFirm
the
necessary
conditions
respect
to own
prices for
.
(A.1)
Firm
1 Firm
and
2
(from
2and
r sin
φ ⋅ sin2x(2)):
+
2
⋅
cos
p
−
R
1
Firm
(from
(2)):
2
1
2
2
1
pp1 1
pcos
andand
Firm 1 and1Firm
(from
1 cos
sin
) ⋅=z0p0) ) = 0 and
(z(z−−2sin
z1z)1)++((2)):
⋅ ⋅ zzzp)1p+1−−
pz2zpz⋅ −⋅z zcos
z −− sin
p1p1) z=
(A.3)
⋅⋅pcos
p 2 −z ⋅pz1p ) = 0.
(
(
2
π
z )2−
+ ⋅4(R
+ cos
22
22z1 +(zsin
1− z px =
2
and (A.2)
− psin
z )2+
⋅ (z p − cos z ⋅ z p ) ) = 0 and
12
1
2
2
and
1
sin z ) + p ⋅ (z profit
− cosmaximization
z ⋅ z p ) ) = 0 with respect to own prices
1 state the necessary
1We
2 2sin z ) + 2for
for
(A.3)
(2zπz)2)+−(+zzp−+pconditions
(cos
) = 0. it is enough to use only
−pbe
zpzp zsymmetric,
⋅0
z0.p .hence,
(A.3)
(A.3)
sin
+⋅cos
zp2p2 z==
(2(2ππ−−Firms’
zz++sin
−−zzp2p222 +
⋅+⋅zcos
behavior
in⋅ ⋅prices
will
first equation in
2
(A.3)
Firm221 and Firm12 2(from1(2)):
2π − z + psin
(
)
(
)
(A.3)
2
z
+
⋅
−
z
+
cos
z
⋅
z
=
0
.
p
p
(A.3), the (necessary
condition
1. Since
z =π, we get:(A.3)
2+ cos
(
2π − z2+ sin
z ) + 2 ⋅for
− zFirm
z ⋅ z p )in= equilibrium
0.
p
Firms’ behavior
in prices will
hence, it is enough to use only first equation in
p21 be symmetric,
1will
2
Firms’behavior
behavior
prices
bezsymmetric,
symmetric,
it0is
enough
toenough
use
only
first
equation
Firms’
ininnecessary
prices
be
hence,
enough
use
only
first
equation
inin in
and it to
(will
)be
z − sin
+prices
⋅ (zwill
cos
zsymmetric,
=
πhence,
+⋅ z2ppin1) )zit
=is0enough
.hence,
p −1.
p is
be
is
to use
only
first equation
Firms’
in behavior
prices
willin
symmetric,
hence,
it
use
only
equation
(A.3),behavior
the Firms’
for
Since
equilibrium
zto=π,
we
get:first
2 condition
2 Firm
Firms’
behavior
in
prices
will
be
symmetric,
hence,
it
is
enough
to
use
only
first
equation
in
(A.3),the
thenecessary
necessary
condition
for
Firm
1.Since
Since
inequilibrium
equilibrium
=π,we
(A.3),
condition
for
Firm
in1.
zz=π,
(A.3),
the
necessary
condition
for
Firm
1.are
Since
in equilibrium
zwe
=π,get:
we get:(B.1) completes
,get:
in (A.3),
the
necessary
condition
Firm
in
zwe
=πget:
p1.for
1derivatives
All
collected
inz Appendix
π +section
+Since
2 pSince
0=
. equilibrium
(A.3)
1zz⋅pin
(2π −condition
(
)
z + sinwe
z ) +need
⋅in
− this
z p1.
cos
z p=equilibrium
0.
(A.3), the necessary
for2 Firm
=π, we get:B. Using
2
2
the proof.
Q.E.D.
ππ++22pp1 z1 zp1p1==00.π. + 2 p1 z p = 0.
All derivatives we need in this section
in Appendix B. Using (B.1) completes
π + 2 p1are
z p collected
= 0.
Firms’ behavior in prices will be symmetric, hence, it is enough to use only first equation in
the proof.
Q.E.D.
derivatives
we
need are
in
this
section are
collected
in B.
Appendix
B.
Using
(B.1) completes
Allderivatives
derivatives
weAll
need
inthis
thissection
section
are
collected
Appendix
B.
Using
(B.1)
completes
All
we
need
in
collected
ininAppendix
(B.1)
completes
(A.3),All
thederivatives
necessary
condition
forthis
Firm
1. Since
equilibrium
z =π, we
get:
we
need
in
section
arein
collected
in Appendix
B.Using
Using
(B.1)
completes
the
proof.
Q.E.D.
Proof
of
Lemma
2:
Since
firms
will
charge
symmetric
prices:
α
=
π/2,
β
=
−π/2,
z = π.
All
derivatives
we
need
in
this
section
are
collected
in
Appendix
B.
Using
(B.1)
completes
theproof.
proof. the proof.
Q.E.D.and
the
Q.E.D.
Q.E.D.
π
+
2
p
z
=
0
.
(a)
In
equilibrium
line
AB
crosses
the
origin,
therefore,
any
firm’s
infinitesimal
radial
move
1
p
the proof.
Q.E.D.
Proof
of Lemma
2: Since
firms
will
chargeinsymmetric
prices: hence,
α = π/2,
β = −π/2,
and by
z =dr.
π.
towards
origin,
−dr,
yields
new
customers
the
area
2×dr/2,
raising
demand
All(a)
derivatives
we need
thiscrosses
sectionthe
areorigin,
collected
in Appendix
B. Using
(B.1) completes
In Proof
equilibrium
lineinAB
therefore,
any firm’s
infinitesimal
radial
move
ofSince
Lemma
2:will
Since
firmssymmetric
will charge
symmetric
prices:
α
=Q.E.D.
π/2,
Proof
of
Lemma
2:
firms
charge
prices:
α
=
π
/2,
β
=
−
π
/2,
and β
z and
== π−π/2,
.z = π.and z = π.
the
proof.
(b)
Using
(1)
and
derivative
(B.1)
the
partial
derivative
in
question
is:
Proof
of
Lemma
2:
Since
firms
will
charge
symmetric
prices:
α
=
π/2,
β
=
−π/2,
towards
origin,
−dr,
yields
new
customers
in
the
area
2×dr/2,
hence,
raising
demand
by
dr.
Proof
of
Lemma
2:
Since
firms
will
charge
symmetric
prices:
α
=
π/2,
β
=
−π/2,
and
=π.π. move
Proof of Lemma 2:(a)Since
firms will
symmetric
prices:
α = any
π/2,firm’s
β = −π/2,
and zz=radial
In equilibrium
linecharge
AB crosses
the origin,
therefore,
infinitesimal
(a) In equilibrium
line AB crosses the origin,1therefore, any firm’s infinitesimal radial move
∂
D
1
(a)InInequilibrium
equilibrium
line
AB
crosses
origin,
therefore,
any
firm’s
infinitesimal
radialmove
move
towards
yields
new
in
area
2×dr/2,
hence,
demand
by dr.
(a)
line
crosses
the
origin,
therefore,
firm’s
infinitesimal
radial
2 origin,
Using
(1)
and
derivative
the
derivative
in
question
is: raisingradial
(a) In (b)
equilibrium
line
ABnew
crosses
thecustomers
origin,
any
firm’s
infinitesimal
=AB
⋅ z(B.1)
= therefore,
(−
z−dr,
cos
zthe
. theany
towards
origin,
−dr,
yields
customers
inppartial
p1 +
p1 ) = − z
1 the area 2×dr/2, hence, raising demand by dr.
∂
p
2
2
R
Proof
of
Lemma
2:
Since
firms
will
charge
symmetric
prices:
α
=
π/2,
β
=
−π/2,
and
z
=
π.
towards
origin,
−dr,
yields
new
customers
in
the
area
2×dr/2,
hence,
raising
demand
by
dr.
towards origin, −dr,
new
customers
in the
area
2×dr/2,
hence,
demand
by dr.
move
towards
origin,
−dr,derivative
yields
new
customers
in derivative
the
arearaising
2×dr/2,
hence,
(b)11 Using
(1)
and
(B.1)
partial
in question
is:raising
∂Using
Dyields
1 the
2
(b)In
partial
derivative
question
is:
(a)
equilibrium
origin,
anyinfirm’s
infinitesimal
radial move
=by(1)
−and
+derivative
= −the
(dr.
z p1 line
cosAB
z ⋅crosses
z p1 )(B.1)
zthe
.therefore,
demand
p1 =
′
total derivative
of2derivative
any
variable
xinwith
respectis:
to
r by x for the length of this
∂origin,
p1 (c)
2Denote
R area
(b)Using
Using
(1)and
and
(B.1)
thepartial
partial
derivative
question
is:demand
(b)
(1)
derivative
the
in
question
∂derivative
D
1 the(B.1)
12×dr/2,
towards
customers
raising
by dr.
2
∂Dappendix.
1 −dr,
1the
= yields
⋅ z p1 ) =in−
(− z new
. fromhence,
The
derivative
inz question
isz obtained
the system of first-order conditions (A.3)
2
p1 + cos
p1 =
=
−
+
⋅
=
−
=
(
z
cos
z
z
)
z
.
∂
p
2
2
R
p
p
p
1derivative
1 (B.1)
1partial
′ forpair
(1)
and
(B.1)
the
derivative
ininrespect
question
is:
1toand
Denote
the
totalequilibrium
derivative
any
x with
r by
the of
length
of thisthe
Using
derivative
the
partial
derivative
question
is:
we
used
derive
prices.
Since
this
system
must to
hold
forxany
locations,
1 (b)∂p(c)
∂∂DD2 2 1(b)Using
11of
2 (1)
2 Rvariable
1
−inzzpquestion
(−zzptotal
cos
== (appendix.
−
⋅ ⋅zzp1p)1 )==−with
== is
zzderivative
.to. obtained
The
from
the
system
of
first-order
conditions
(A.3)
p1++cos
p1respect
′
differentiation
r
yields
.
The
total
derivatives
are
(Note:
sin
z
=
0 and cos
p
1
1
1
x with respect to r by x ′ for the length
of this
∂D2 1(c) Denote the total derivative
1 of any variable
∂∂pp1 1 22we(c)
2of2RRany
′ pair
x
Denote
the
total
derivative
variable
x
with
respect
to
r
by
for
the
length
of
this
used
to
derive
equilibrium
prices.
Since
this
system
must
hold
for
any
of
locations,
the
=
−
+
⋅
=
−
=
(
z
cos
z
z
)
z
.
.
p
p
p
z
=
−1
in
equilibrium):
1 The derivative
1
1
appendix.
in
question
is
obtained
from
the
system
of
first-order
conditions
(A.3)
∂p1differentiation
2 derivative
2 R p ′ .from
appendix.
The
question
obtained
the system
of first-order
conditions
(A.3)
total
withinrespect
to risyields
The total
derivatives
are (Note:
sin z = 0 and
cos
used
to derive
equilibrium
prices.
Since
system
must
hold
for
pair
of locations,
x′ ′pair
(c)Denote
Denote
thewe
total
derivative
of
any
variable
x1with
with
to
r=by
by
forany
the
length
ofthis
this the
x
(c)
the
total
derivative
of
any
variable
x
respect
to
for
the
length
of
and
.
′ +this
′respect
′prfor
′
′
′
z
+
p
⋅
z
+
p
⋅
z
=
0
z
p
⋅
z
+
p
⋅
z
0
we
used
to
derive
equilibrium
prices.
Since
this
system
must
hold
any
of
locations,
the
1
p
p
2
p
′
(c)
Denote
the
total
derivative
ofrespect
xxwith
respect
to
by xx′ for
forthe
the
length
z=
−1
in equilibrium):
(c)
Denote
the
total
derivative
ofany
anyvariable
variable
with
respect
to rr derivatives
by
length
ofof
thisz = 0 and cos
′
total
differentiation
with
to
r
yields
.
The
total
are
(Note:
sin
p
1
appendix.The
The
derivative
question
obtained
from
the
system
offirst-order
first-order
conditions
(A.3)
appendix.
derivative
ininquestion
isisobtained
from
the
system
of
first-order
(A.3)
total
differentiation
withderivative
respect
to
risyields
.isThe
total
derivatives
are (Note:
sin
z =(A.3)
0 and
cos
p1′ from
this
appendix.
The
in
question
obtained
from
system
ofconditions
first-order
appendix.
The
derivative
obtained
the
system
ofinthe
conditions
The
needed
above
system
are
collected
It
becomes:
and
. (B.1)-(B.3).
′ in question
z ′ + prices.
pprices.
pSince
⋅ z ′pin=this
0this
z ′ + p2′ ⋅must
zmust
⋅hold
z ′p =for
0for
zequilibrium
= −1
inderivatives
equilibrium):
1 ⋅ z p + Since
p + phold
weused
usedtotoderive
derive
system
any
pair
of
locations,
we
equilibrium
system
any
pair
of
locations,
the
conditions
we usedprices.
to derive
equilibrium
Sincefor
this
system
hold for
we
derive(A.3)
equilibrium
Since
this systemprices.
must hold
any
pair ofmust
locations,
the the
z = used
−1
intoequilibrium):
′
′
′
′
,
and
,
−
4
p
+
2
p
+
4
R
+
π
=
0
2
p
−
4
p
−
4
R
+
π
=
0
1
2
1
2
′
′
′
total
differentiation
with
respect
to
r
yields
.
The
total
derivatives
are
(Note:
sin
z
=
andcos
cos
p
′
and
total differentiation
with
to
rp ′yields
total
(Note:
= 00cos
and
total any
differentiation
withneeded
to
r+yields
′ derivatives
′r yields
total
sin
= 0z total
and
p=1 0. The
′ +the
pair
ofrespect
locations,
total
differentiation
.zThe
zrespect
⋅in
z above
pp⋅ 1system
z1.p The
zwith
+ derivatives
p′respect
⋅ z in +(B.1)-(B.3).
pto
⋅ are
zare
=(Note:
0 . Itp
The
derivatives
are
collected
becomes:
1 sin
z ′ + p1′ ⋅ z p + p1 ⋅ z ′pp = 0 and
z ′ + p2′ ⋅ z p + p2⋅ z ′p p = 0 . p
derivatives
are (Note: sin z = 0 and cos z = −1 in equilibrium):
z = −1
in− equilibrium):
−1ininequilibrium):
equilibrium):
zz==−1
′
4 p1′The
+ 2 pderivatives
0 , and
2 p1′ − 4system
p′ − 4 Rare
+ πcollected
= 0 , in (B.1)-(B.3). It becomes:
needed
in above
2 + 4R + π =
The derivatives
needed in above system
are 2collected in (B.1)-(B.3). It becomes:
z ′ + p1′ ⋅ z p + p ⋅ z ′p = 0 and z ′ + p2′ ⋅ z p + p ⋅ z ′p = 0 .
and= z0z′ ,′++pand
p′ ′⋅ ⋅zz +′+pp⋅ ⋅zz′′ ′ ==00.. .
zz′ ′++pp1′ 1′⋅ ⋅−zzp41p1p+1′++pp2⋅ ⋅pzz2′p1′p+1=4=0R0+and
and
− 4 p1′ + 2 p2′ + 4 R + π = 0 , πand
2 p1′ −2 24 pp22′2pp2−1 4−R4 +pp2π2p−2 =40R,+ π = 0 ,
The derivatives needed in above system are collected in (B.1)-(B.3). It becomes:
1
1
((
))
1
1
2
2
1
((
1
))
1
2
1
2
2
2
2
2
1
1
1
1
2
2
1
1
1
1
1
1
1
2
2
1
1
1
1
2
2
1
2
2
1
2
2
2
2
Thederivatives
derivatives′ needed
neededininabove
abovesystem
systemare
arecollected
collectedinin(B.1)-(B.3).
(B.1)-(B.3). ItItbecomes:
becomes:
The
− 4 p + 2 p′ + 4 R + π = 0 , and 2 p′ − 4 p′ − 4 R + π = 0 ,
1
2
1
2
and 22pp1′ 1′−−44pp2′ 2′−−44RR++ππ==00, ,
−−44pp1′ 1′++22pp2′ 2′++44RR++ππ==00, , and
∂p1∂p1 2 2
p1
p1
p1
p1
p1
p1
2 R2 R
′ for
x ′ xfor
(c)(c)
Denote
Denote
thethe
total
total
derivative
derivative
of any
of any
variable
variable
x with
x with
respect
respect
to rtoby
r by
thethe
length
length
of this
of this
appendix.
appendix.
TheThe
derivative
derivative
in question
in question
is obtained
is obtained
from
from
thethe
system
system
of first-order
of first-order
conditions
conditions
(A.3)
(A.3)
wewe
used
used
to derive
to derive
equilibrium
equilibrium
prices.
prices.
Since
Since
thisthis
system
system
must
must
hold
hold
forfor
anyany
pairpair
of locations,
of locations,
thethe
′
′
total
total
differentiation
differentiation
with
with
respect
respect
to
r
to
yields
r
yields
.
The
.
The
total
total
derivatives
derivatives
are
are
(Note:
(Note:
sin
sin
z
=
z
0
=
and
0
cos
p
p
1 ECONOMIC
1
338
AND BUSINESS REVIEW | VOL. 14 | No. and
4 cos
| 2012
z =z−1
= −1
in equilibrium):
in equilibrium):
z ′ +z ′p+1′ ⋅pz1′ p⋅ z+p p+⋅ zp′p⋅ z=′p 0π
=and
0 and
z ′ +z ′p+2′ ⋅pz2′p⋅ z+p pπ
+⋅ zp′p⋅ z=′p 0=. 0 .
2
π system
π 2
2 andarep 2′collected
2 in. (B.1)-(B.3). It becomes:
p
=
with
the
The derivatives
needed in
p11′′ above
= 2+
+3R
R and p 2′ =
=2−
−3R
R.
with
the unique
unique solution
solution
TheThe
derivatives
derivatives
needed
needed
in above
in above
system
system
are
collected
collected
in3π(B.1)-(B.3).
in (B.1)-(B.3).
It becomes:
It becomes:
π
2 are
2
2
3
2
′
′
p
=
+
R
p
=
− 22 R .
with
the
unique
solution
and
π
π
2
1
2
Q.E.D.
π
π
2
′
′
p
=
+
R
p
=
−
R
and
.
with
the
unique
solution
Q.E.D.
2
3
2
3
′ 2′ the
,and
and
− 4−p14′ +pwith
+
p2′4+unique
R4+Rπ+=πsolution
0=, 02,and
4−pπ42′ R−
p′42and
−R4+Rpπ+22′ =π=0=
p1π1′2 =p12′22−p1′+
12+p2
22,, 0−, 3 R .
′
+ pR
= 33and
−2 pR2′ .= π
with the unique
solution
Q.E.D.
′ and
with the
uniquepsolution
1 =
π +p222′ R
π − 22 R .3
1 =π
π
2
3
2
3
R2..
Q.E.D.
with the
the unique
unique solution
solution pp′1′ == 2ππ++ 32R
π− 3 2R
2R and pp2′2′ == 2π−
Q.E.D.
with
1 ′ ′ ==2 ++
3 RRand
2and−33−put
with
the
uniquesolution
solution
and
.. . in parentheses in
RR
and
and pp2′ 2′22==
with
withthe
the
unique
solution
Proof
of
Lemma
3:
Use
from
Lemma
them
of
Q.E.D. Q.E.D.
1 2
3
2
Proof
of unique
Lemma
3:
Use the
thepp1results
results
from
Lemma
and
put
in RHS
RHSQ.E.D.
of (3)
(3)
22 33
22 33 them 2in parentheses
Q.E.D.
π
Q.E.D.
π
2
Proof
of
Lemma
3:
Use
the
results
from
Lemma
2
and
put
them
in
parentheses
in
RHS
of (3)
−
+
to
get
the
sign
of
the
effect
a
radial
move
has
on
Firm
2’s
profits:
>0,
.
∀R
∈
(
0
.
1
]
Q.E.D.
Q.E.D.
to
getProof
the
sign
the
effect
radial
has
on Firm
>0,
. RHS
∀R ∈
0in
.1in
]RHS
of Lemma
3: Use
the
results
from
Lemma
22’s
and22profits:
put
in+parentheses
in(RHS
of (3)ofof(3)
Proof
ofof
3:3:aUse
the
results
from
Lemma
andthem
put−them
them
inparentheses
parentheses
(3)
3
4
R
Proof
ofLemma
Lemma
Use
themove
results
from
Lemma
and
put
π
2in
3
4
R
Proof of Lemma
3: Lemma
Use
results
from
Lemma
2 and
parentheses
in+RHS
of in
(3)
to get of
the
signthe
of 3:
the
effect
aresults
radial
move
hasput
on them
Firm
2’s
profits:
>0,
.
∀R
∈ (0of.1](3)
Proof
Use
the
from
Lemma
2 andin
put
them in−parentheses
RHS
π
2
π
2
3 ++ 4 R >0,
Proof
of
Lemma
3:
Use
theaaresults
results
from
Lemma
and22’s
put
them 2in
in−−
parentheses
in∀R
RHS
(3)
to
get
the
sign
of
the
effect
aradial
radialmove
move
hasononFirm
2’sprofits:
profits:
toProof
get
the
sign
ofof
the
effect
has
profits:
∀R
(of
1.] ..
to
get
the
sign
the
effect
radial
move
has
∈∈(Q.E.D.
0of
.01.](3)
π
Q.E.D.
of
Lemma
3:
Use
the
from
Lemma
22Firm
and
put
them
parentheses
RHS
π (04.1R] .>0,in
+put
to get the sign
of
thesign
effect
radial
move
has
on
Firm
profits:
>0,
4 R ∀R ∈ (in
Proof
Proof
of
ofLemma
Lemma
3:3:Use
Use
the
the
results
results
from
from
Lemma
Lemma
22−and
and
putthem
them
parentheses
parentheses
RHS
of(3)
(3)
− 22 ∀R
+inin3ππ3∈
to get
the
of athe
effect
a radial
move
has2’s
on
Firm 2’s
profits:
>0,
. of
0in.1RHS
]Q.E.D.
3profits:
4 R − 3 + 4 R >0, ∀R ∈ (0.1] . Q.E.D.
to get
get the
the sign
sign of
of the
the effect
effect aa radial
radial move
move has
has on
on Firm
Firm 2’s
2’s profits:
− 3 +224 Rπ>0,
to
π ∀R ∈ (0.1] . Q.E.D.
−
totoget
get
the
thesign
sign
of
ofthe
theeffect
effectademand
aradial
radialmove
movehas
hasany
on
onFirm
Firm2’s
2’saprofits:
profits:direction
>0,
∀R
∀R
∈∈((0the
0.1.1]]. . Q.E.D.
3− 4++R is >0,
Proof
move
in
since
Q.E.D.
Q.E.D.
Proof
of
Lemma
4:4:The
The
effect
ofofany
move
in in
a polar
polar
direction
isR zero,
zero,
since
the
Proofof
ofLemma
Lemma4:
Thedemand
demandeffect
effectof
any
move
a polar
direction
is
zero,
since
the
33 44R
Q.E.D.
new
of
indifferent
consumers
(line
must
through
origin
(firms
will
charge
Q.E.D.
new line
line
ofline
indifferent
consumers
(line AB)
AB)
mustAB)
pass
through
the
origindirection
(firms
will
charge
of indifferent
Lemma
4: The
demand
effect
ofpass
any
move
inthe
athrough
polar
is zero,
since
the
newProof
of
consumers
(line
must
pass
the origin
(firms
will
Q.E.D.
Q.E.D.
identical
prices).
The
demands
firms
face
are
unaffected
by
such
move,
but,
any
such
identical
prices).
The
demands
firms
face
areAB)
unaffected
bythrough
such
move,
but,
any
such
move
Proof
Lemma
The
demand
effect
of
in
direction
isis
zero,
since
the
new
lineidentical
indifferent
consumers
(line
must
pass
the origin
(firms
willmove
charge
Proof
ofofLemma
4:4:The
demand
effect
of
any
move
in aaaapolar
polar
direction
zero,
since
the
charge
prices).
The
demands
firms
face
are
unaffected
by
such
a
move,
but,
any
Proof that
of Lemma
4: Lemma
The
demand
effect
of any
move
in
a more
polar
direction
isdirection
zero,
since
the
brings
both
firms
closer
together
makes
them
competitive
and
lowers
the
price
they
Proof
of
4:
The
demand
effect
of
any
move
in through
a polar
is(firms
zero,
since
the
that
brings
both
firms
closer
together
makes
them
more
competitive
and
lowers
theany
price
they
new
line
of
indifferent
consumers
(line
AB)
must
pass
the
will
charge
identical
prices).
The
demands
firms
face
are
unaffected
bythem
such
a origin
move,
but,
such
move
new
line
ofLemma
indifferent
consumers
(line
AB)
must
through
the
origin
(firms
will
charge
such
move
that
brings
both
firms
closer
together
makes
more
competitive
and
lowProof
of
4:
The
demand
effect
of
any
move
in
a
polar
direction
is
zero,
since
the
new line ofcharge.
indifferent
consumers
(line
AB) must
pass
through
thethrough
origin
(firms
will
charge
result
follows.
Proof
Lemma
4:
The
demand
effect
of
any
move
in a competitive
polar
direction
islowers
zero,
since
themove
new
lineThe
ofofindifferent
consumers
(line
AB)
must
pass
origin
(firms
will
charge
charge.
The
result
follows.
identical
prices).
The
demands
firms
face
arethem
by
such
aamove,
but,
any
that
brings
both
firms
closer
together
makes
more
and
thesuch
price
they
identical
prices).
The
demands
firms
face
are
unaffected
bythe
such
move,
but,
any
such
move
new
line
of
indifferent
consumers
(line
AB)
must
pass
through
the
origin
(firms
will
charge
ers
the
price
they
charge.
The
result
follows.
Q.E.D.
Proof
Proof
of
of
Lemma
Lemma
4:
4:
The
The
demand
demand
effect
effect
of
of
any
any
move
move
in
in
a
a
polar
polar
direction
direction
is
is
zero,
zero,
since
since
the
the
identical prices).
The
demands
firms
face
are
unaffected
by
such
a
move,
but,
any
such
move
Q.E.D.
new
line
of
indifferent
consumers
(line
AB)
must
pass
through
the
origin
(firms
will
charge
identical
prices).
The
demands
firms
face
are
unaffected
by
such
a
move,
but,
any
such
move
Q.E.D.
that
brings
both
firms
closer
together
makes
competitive
and
lowers
the
price
they
charge.
The
result
follows.
that
brings
both
firms
closer
together
makes
them
more
competitive
and
lowers
the
price
they
identical
prices).
The
demands
firms
face
arethem
unaffected
by
such
move,
but,
any
such
move
that bringsnew
bothcharge.
firms
closer
together
makes
them
more
competitive
and
lowers
the lowers
price
they
new
line
line
of
ofboth
indifferent
indifferent
consumers
consumers
(line
(line
AB)
AB)
must
mustmore
pass
passthrough
through
the
origin
origin
(firms
(firms
will
will
charge
charge
identical
prices).
The
demands
firms
face
are
unaffected
by
such
aathe
move,
but,
any
move
that
brings
firms
closer
together
makes
competitive
and
thesuch
price
they
The
result
follows.
Q.E.D.
charge.
The
result
follows.
that
brings
both
firms
closer
together
makes
them
more
competitive
and
lowers
the
price
they
charge. The
result
follows.
Proof
ofresult
Lemma
5: First
we
characterize
x,more
y, αcompetitive
, βby
,byand
γ,aawhich
define
the
firms’
deidentical
identical
prices).
prices).
The
The
demands
demands
firms
firms
face
faceare
are
unaffected
unaffected
such
such
move,
move,
but,
but,the
any
any
such
such
move
move
that
brings
both
firms
closer
together
makes
them
and
lowers
price
they
charge.
The
follows.
Q.E.D.
Q.E.D.
charge.
Theof
result
follows.
Proof
Lemma
5:
First
we
characterize
x,
y,
α,
β,
and
γ,
which
define
the
firms’
demands.
Q.E.D. that
thatbrings
brings
both
both
firms
firms
closer
closer
together
together
makes
makes
them
them
more
more
competitive
competitive
and
and
lowers
lowers
the
the
price
price
they
they
mands.
Coordinates
of
point
D,
x
and
y
solve
the
system:
charge.
The
result
follows.
Q.E.D.
Proof of Lemma 5: First we characterize x, y, α, β, and γ, which define the firms’ demands.
Q.E.D.
Coordinates
of
point
D,
yy solve
the
charge.
charge.The
The
result
result
follows.
follows.
Q.E.D.
Proof
of
Lemma
5: First
we characterize
Coordinates
of
point
D, 2xx and
and
solve
the system:
system: x, y, α, β, and2 γ, which define2 the firms’ demands.
π
π
π
2 +
Q.E.D.
Q.E.D.
p
+
(
R
sin(
−
)
−
x
)
(
R
cos(
−
)
−
yy)) 22the
= ppsystem:
xx)) 2 γ,+
((which
R
cos
−
yy)) 2,,,the firms’ demands.
π
π
Proof
of
Lemma
5:
First
we
characterize
x,(( R
y,sin
α, ππβ,
Coordinates
of
D,
x
and
y
solve
1 Proof
2 +
3 ) point
3characterize
3 −
p
+
(
R
sin(
−
−
x
)
+
(
R
cos(
−
)
−
R
sin
−and
+the
R firms’
cos π33define
−demands.
of
Lemma
5:
First
we
x,
y,
α,
β,
and
γ,
which
define
the firms’
demands.
1
2 +
3 we characterize x,
3 y, α, β,=and
3 define
Proof of Lemma
5:
First
γ,
which
Proof of Lemma
5:
characterize
2we
2 y, α, β, and γ, πwhich2 define the
2 demands.
π First
π thex,
π firms’
Coordinates
of
point
D,
x
and
y
solve
system:
p
+
(
R
sin(
−25:
)
−
x
)
+
(
R
cos(
−
)
−
y
)
=
p
+
(
R
sin
−
x
)
+
(
R
cos
−
y
)
,
2
2
2
Proof
of
Lemma
First
we
characterize
x,
y,
α,
β,
and
γ,
which
define
the
firms’
demands.
π
π
Coordinates
of
point
D,
x
and
y
solve
the
system:
1
2
3 +First
3R cos πdefine
Coordinates
ofProof
point
x φpoint
and
solve
the
25:
+ ((of
rD,sin
sin
− xxy))D,
cos
φsystem:
−
=π3ppsystem:
+x,
(R
Ry,sin
− and
x)) 2 +
+γ,((πwhich
− yy)) 2,,,theπ3 firms’
Lemma
we
characterize
α,π3β,−
demands.
Coordinates
of
x−((rrand
solve
π+
2−+
pp33π+
φpoint
−
−
yy)) 2 the
=
sin
R cos
p+1r +
sin(
−
xcos
)2 2 +yyφ((R
Rcos(
cos(
−π3 )system:
)22π ==
p232+
+ ((xR
sin
xx))22332++−((R
cos π2−−yy) )2 2, ,
2 y(
πD,
2( R
2solve
Coordinates
of
the
3 )−2x
3π2π2−
πand
2 sin
p
(
R
sin(
)
x
)
−
y
)
p
R
sin
−
R
πR−cos(
πp −
πand
π cos3the
2)) −+we
2R
2thefirms’
π
π
p1 + (Coordinates
R sin(
−
)
−
x
)
+
(
−
)
=
+
(
−
x
)
+
(
R
cos
−
y
)
of
point
D,
x
and
y
solve
the
system:
,
Proof
Proof
of
of
Lemma
Lemma
5:
5:
First
First
we
characterize
characterize
x,
x,
y,
y,
α,
α,
β,
β,
and
γ,
γ,
which
which
define
define
firms’
demands.
demands.
1
2
3
3
3
3
p1 3+ (pR3 sin(
) − x) 2)3+ (+R(cos(
−φ3 )−
sin3π3−−xx))32 ++((RRcos
cos3π3 −− yy)) 2,,
2 −yy
+that
(r−sin
rwe
cosproceed
) ) 2==pp232β.
++((RRsolves:
sin
π3 φ −=x
π
We
and
p1 ++note
(R
Rsin(
sin(
− π3sinα
) −−D,
xD,
)2xxxx++2and
(R
Rcos(
cos(
− π3 ) − y2)2with
= pp2β.++It
(R
Rsolves:
sinππ3 −− xx))22 ++ ((R
Rcos
cosππ3 −− yy))22 ,,
We
note
that
sinα
=
and
we
proceed
with
It
Coordinates
Coordinates
of
of
point
point
and
y
y
solve
solve
the
the
system:
system:
p
(
−
)
x
)
(
−
)
−
y
)
=
(
sin
p32 + that
(r sin
φ 2−ax2)= 2x+and
(r cos
φ3 proceed
−
y)2 = pwith
+ (R
sin
y) 2,
1 note
3 sin
3−
π33π2− x) 2 2+ ( R cos 3π
2 2+
We
we
It
solves:
2 y )2 π=− p
π.sin
p+xWe
(+note
r(sin
φπ xφ−
x+)y2(2)r+cos
rxcos
φywe
(R
−π−+πx()yR) +
,
p3 + (rpsin
φp −sin
)+
−
pand
()−
R
(βR
cos
,222(+Rπ(cos
2
πx3)solves:
ππ sin
sinα
=((=(R
β.−
It
rRsin
φr−cos
−πthat
φ
=proceed
+p2 =x2(=2)R+with
cos
−Rcos
ycos
)3 22−ππ,π−y−)cos
2−+
32 sin
−
sin
)))22−22x+
φ
−
cos
))−2ppy3=
((+2R
−
β
)
π3sin
π3((R
π3 (R
+3((((Rφ
sin(
sin(
−3β
x+)(()(rrr+cos
+cos
R
cos(
cos(
−−32β
−
y2))2+
ppsin
++sin
R
sin
−
−
x
x
)
+
yy))22β
, , )) 2 ..
3
2
3)x)−
3))
3
3
3
p33 +
+ppp((p1rr331+
sin
φ
−
sin
β
+
cos
φ
−
cos
=
p
+
R
sin
−
sin
β
)
+
(
R
cos
−
cos
β
2β
2
2
+
r
sin
φ
−
φ
−
y
)
=
(
R
sin
−
x
)
+
(
R
cos
−
y
)
,
π3
π3
2
2
3
3
note
= x φand
+We
(r sin
φ −that
x) sinα
+ (2r cos
− ywe
) =proceed
p2 +2( Rwith
sin 3β.−Itx)solves:
+π ( R cos 3 −
2 y) ,
2
π
that
sinα
we
proceed
Itsin
solves:
We note that
sinα
=that
xφand
we
with
solves:
p3We
+note
(γrnote
sin
−sinα
sin
)x =and
+x(rand
cos
φ −β.
) =with
(R
We
=proceed
we
proceed
with
β.p2It+β.
solves:
22β
2cos
2 It β
232 − sin β ) ππ+ ( R cos
22 3 − cos β ) ..
ππ
And,
solves:
2((rand
2++((R
2 −−yy)) , , π
2
ppp333+
+
(
(
r
r
sin
sin
φ
φ
−
−
x
x
)
)
+
+
r
cos
cos
φ
φ
−
−
y
y
)
)
=
=
p
p
R
sin
sin
−
−
x
x
)
)
+
+
(
(
R
R
cos
cos
π
And,
γ
solves:
We
note
that
sinα
=
x
we
proceed
with
β.
It
solves:
2)2 = β.
3 sin − sin β ) 23+
3 ( R2 cos − cos β ) .2
(2r sin
φ −sinα
sin β=2)x and
+ (r2we
cosproceed
φ − cos2 βwith
p2It+solves:
(3R
We
that
2π
3 +
3π
π
p3 + (r sin φpp3−p++sin
βsin
) sin
r−sin
cos
β(cos
p−2−cos
+cos
βr (sin
)R2πsin
+−(sin
Rπ3 cos
2) p3=−
(note
rAnd,
sin
r) cos
βsin
psin
−βsin
+β()Rπ.cos
r sin
φ− +ππ−φ(−γsin
βγφ) )22β−22+)+cos
(r(+Rcos
φ −=−ππφcos
β()R
sin
) 2 2−+βcos
()R cos
− cos
βγ cos
))222 .. β ) .
solves:
3((+
2R(+
3 −
2 = 2 p+ (+
2 )=
R
γ
π3 φ −3sin γ 32) 2 + (r cosπ3φ − cos
1
3
And,
γ
solves:
3
3
2) +=
2 )= with
2) ++( R
p
+
(
R
sin
− 3−
−
sin
γ
+
(
R
cos
−we
−
cos
γ
=
p
+
(
r
sin
−
sin
γ
(
r
cos
−
cos
γ
+
(
r
sin
φ
sin
β
)
(
r
cos
φ
−
cos
β
)
p
+
(
R
sin
−
sin
β
)
cos
cos
β
)2 .
π3 φ
π3φ −
We
We
note
note
that
that
sinα
sinα
=
x
x
and
and
we
proceed
proceed
with
β.
β.
It
It
solves:
solves:
1
3
3
2
3
p + (r sin
φ − γsinsolves:
β ) + (r cos
φ − cos β
) = p2 +
And,
2
2 ( R sin 3 − sin β ) + ( R2cos 3 − cos β ) .
2
π
π
And, γ 3solves:
p1And,
+four
(γRsolves:
sin
− 3 − sin γ ) + ( R cos− 3 − cos γ ) = p3 + (r sin φ − sin γ ) + (r cosφ − cos γ ) .
γ equations
solves:
And,
Last
and
lead
to:
22 simplify
2
2
2
2
22 2
ππ
ππ
2
2
2
π
π
Last
four
equations
simplify
and
lead
to:
pp33++And,
φφ
−−sin
sin
) ++γ()(rrcos
cos
φφ2cos
−−cos
cos
) ==γp)p22+=+(p(RR
sin
sin 3−−sin
β)) γ++)((R+Rcos
cos
ββ)γ) ). ...
solves:
p((r1rsin
+sin
(γγR
sin
− β
−β)sin
+
(R
− 3β
−β)cos
φsin
−βsin
(cos
r2 cos
φ−cos
−
cos
33 −
2solves:
3 + (2r3sin
π 2
2 − sin
2 − cos
2
π + (r2sin φ
π (πR3cos
p1 + ( R sinp− 1π3p+−And,
sin
γLast
)sin
γ−)−cos
(rφcos
andcos
to:
R
−−3sin
−equations
sin
(simplify
R
= p γ+)(r+sin
− φsin
γ ) 2 γ+)(r. cosφ − cos
γ )2 .
R(sin
− 3+four
γ )−22γ3+)−(cos
R+ppcos
)lead
pππ31=−− pcos
1(+
2 γ=) p
33 − γ
3 + (3r sin φ − sin γ ) 2 + (r cosφ − cos γ ) 2 .
2 −
π
p
2
2
2
2
p
+
(
R
sin
−
−
sin
γ
)
+
(
R
cos
−
−
cos
γ
)
=
p
+
(
r
sin
φ
−
sin
γ
)
+
(
r
cos
φ
−
cos
γ
)
.
2
1
,
(A.4)
x
=
sin
α
=
π
π
And,
γγ− solves:
3
3solves:
Last
four
simplify
and
p11 +And,
( R sin
−xsin
γ ) α+=( R2cos
− 33 −
γ )lead
= pto:
cosγ ) .
, cos
=equations
sin
3 + (r sin φ − sin γ ) + (r cosφ −(A.4)
3 simplify
R
3
pto:
−and
pand
Last
four
equations
simplify
Last four
equations
and
lead
1 lead
Last
four
equations
simplify
lead
to:2to:
Last
four
equations
lead
to:
2simplify
R
32and
,
(A.4)22
x
=
sin
α
=
2
2
2
2
2
π
π
π
π
Last
four
equations
simplify
lead
to:
p−−and
− p3lead
pp11++Last
((RRsin
sin
−− 33equations
−−sin
sinγγ)) +simplify
+((RRcos
cos
cos
)) == pp33++((rrsin
sinφφ−−sin
sinγγ)) ++((rrcos
cosφφ−−cos
cos
γγ)) . .
1cos
four
to:
2R
2 γγ
33−−
(A.4)
x p=p11sin
αp2 2+
=−p2r2p2and
p32 −−
2−−p2 R, 2 ) − 2r sin φ ( p − p )
R
3
(
2
p
−
p
2
1
1 −2
R α3 (=2xp=3 x−
p
−
p
+
2
r
R
)
−
2
r
sin
φ
(
p
−
p
)
2
1
,
(A.4)
= sin
,
1
2
2
1
yyx =
,
(A.5)
, 3 2,
(A.4)(A.4) (A.4)
sin
=α
sin=
αpp=2 −− 2ppR
1and
= four
,
(A.5)
2
Last
Last
fourRequations
equations
simplify
and
lead
lead
to:
to:
2(Rsin
,
(A.4)
x 3==
sin
α=
1r
2simplify
R122−R3
(
2
cos
φ
−
R
)
2 p3α
p
+
2
r
−
2
R
)
−
2
r
sin
φ
(
p
−
p
)
3
2
R
3
3 −=2pR
2
2
1
,
(A.4)
x
r cos2φ − R )2
y = R 3 (2 p − p 2−R3p(23+
,
(A.5)
2
r
−
2
R
)
−
2
r
sin
φ
(
p
−
p
)
R22−R
32
12
23p2p3 ( 2r cos
2
p
p
−
φ
−
R
)
2 (p
2 1 , 2
2
2
2
2
1
1
y p=φR
(A.5)
R 3 (2(2pr3 sin
−R
−
p
+
2
r
−
2
R
)
−
2
r
sin
φ
−
p
)
(2R
p−α
r+2 2−φ
2−R−R
−22)rβ
sin
)rr 2p−
13−
)sin
sin
2
cos
=
p
,,
(A.6)
(A.4)
(A.4)
xxp2==3−sin
α−β
p=1=pp+
y=
,(3 p−
21 −
(y2r=ysin
+−2 ((+
2prr22cos
cos
R222))) R
cos
=1 rφ
psin
−2φp
p−( 2pp+
+
−1 )R
R, 2(A.5)
(A.6)
(,r,222−rR
)2
(A.5)(A.5) (A.5)
R= φ33−((223Rpp(3323−−p)3sin
p11 −−β
−φ
Rcos
−φ 22−rrβ−R
sin
2+R3−R22R(R
R
pR22φ2+
rr232r)33−
)−
sin
φφ((3pp22 −−2pp11)) ,,,
2
2
2
R
y
=
(A.5)
3 3 (p2
1 r cos
2
cos
φ
−
R
)
(
2
r
sin
φ
−
R
3
)
sin
β
+
(
2
r
cos
φ
−
R
)
cos
β
=
p
−
p
+
r
−
R
,
= φ + R 3 ) sin γ2+R(223R
(A.5)(A.6)
3 (cos
2r−φcos
φR)−γR=) p − p 3 + r 22,2− R 22 .
− cos
((y2
φ
R
(A.7)
2r
2sin
((2cos
222r(rrr22cos
R
)R−γ2)2cos
2rr sin
sin
+((2R
)−sin
γ)−
+R
(β2+3rr++
cos
φ−cos
−2φ2R
RR
=
pβ33φ2φ−
(A.7)(A.6)
(R
2rφ3sin
φpp−3−R
3−
φ−))22cos
−
=((ppp1132−+−−rppp2)2−)+Rr 2 .2− R 2 ,
R
3
2
p
p
p
p
−
)
)
−
r
r
sin
sin
3(32r cos
11 φ −
22R) cos β = p − p + r − R
22 , 11
2R2(A.6)
2 2
(2r sin φ − y(Ry2=r=3sin
)
sin
β
+
φ
−
R
3
)
sin
β
+
(
2
r
cos
φ
−
R
)
cos
β
=
p
−
p
+
r
−
,
(A.6)
3
2
,
(A.5)
(A.5)
(
2
r
sin
φ
+
R
3
)
sin
γ
+
(
2
r
cos
φ
−
γ
=
p
−
p
−
R
.
(A.7)
2+,+r
2−
3
2
(
2
r
sin
φ
−
R
3
)
sin
β
+
(
2
r
cos
φ
−
R
)
cos
β
=
p
−
p
r
R
,
(A.6) (A.6)
3
1
The
system
profit
maximization
(8),
which
sin
Rof
sin ββ22++RR((22conditions
cos
φcos
Rφfor
cos
− pp32 ++ rr2 −−derived
R2 ,,, 2 from
(A.6)
2R
The
system
of+33first-order
first-order
conditions
for
profit
derived
from
(8), (A.7)
which we
we
3−
3r+r3(cos
(222rrrφcos
R
R
)))ββcos
((22rr sin
φφsin
−− R
))sin
−− φR
))−−−
cos
==γ2maximization
pp=
(A.6)
(
2
r
φ
R
3
)
sin
γ
(
cos
φ
R
p
−
p
+
r
−
R
.
3
32 2
12
mainly
need
for
future
reference,
is:
2
(2r sinmainly
φ + (R2rneed
3The
) sin
γ R+future
(32r)of
cos
φ +−(R
) cos
γφ =
p)3 cos
− pfor
rp3−−Rpmaximization
. + r 2 − R22 .(A.7)
for
reference,
is:
system
first-order
conditions
profit
derived
from
(8),
which
we
2
sin
φ
+
sin
γ
2
r
cos
−
R
γ
=
(A.7)
1 +
1
2sin
r sin+φR+ R
3 ) γsin+γ(2+r (cos
2r φcos
φR)−cos
R)γcos
γp3=−pp31 −+ pr 21 +−2rR
(A.7)
2 2−
22 .
((222rr(rsin
−is:
.RRR
(A.7)
(mainly
sinφφneed
φsystem
RRfor
)sin
sin
sin
+
rcos
cos
−R
−R
))cos
cos
=
−−
p
p
+
+
r
r
−
−
The
first-order
conditions
for
derived
from
(8),
which
we
,
,
(A.6)
(A.6)
future
+−−R
3333)))of
sin
γββ++reference,
(2((2r2rcos
φφ−φ
)Rcos
γβ
==βprofit
p=3 p−p33pmaximization
+
r
−
R
.
(A.7)
.
(A.7)
1 22
The systemmainly
of first-order
conditions
for profit
derived
from
(8), which
we which we
The
system
conditions
for profit
maximization
derived
from from
(8),
needof
forfirst-order
reference,
is: maximization
The
system
offuture
first-order
conditions
for profit
maximization
derived
(8), which
The
system
of
first-order
conditions
for
profit
maximization
derived
from (8),
(8), which
which
we we
22 derived
22
mainly need
for
future
reference,
is:
The
system
of
first-order
conditions
for
profit
maximization
from
we
mainly
need
for
future
reference,
is:
(
(
2
2
r
r
sin
sin
φ
φ
+
+
R
R
3
3
)
)
sin
sin
γ
γ
+
+
(
(
2
2
r
r
cos
cos
φ
φ
−
−
R
R
)
)
cos
cos
γ
γ
=
=
p
p
−
−
p
p
+
+
r
r
−
−
R
R
.
.
(A.7)
(A.7)
3
3
1
1
mainly
need
for
future
reference,
is:
mainly need
need for
for future
future reference,
reference, is:
is:
mainly
1
1
1
1
2
2
2
2
The
Thesystem
systemof
offirst-order
first-orderconditions
conditionsfor
forprofit
profitmaximization
maximizationderived
derivedfrom
from(8),
(8),which
whichwe
we
mainly
mainlyneed
needfor
forfuture
futurereference,
reference,is:
is:
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
339
The system of first-order conditions for profit maximization derived from (8), which we
need
forγγfuture
reference,
cos
(sin
) )+p+1p(α
− cos
αα−−−γγmainly
ααcos
cos
)γ)−)−−
yy(sin
−−sin
) γ+
x x(cos
ααα−−−
ααα
γ γis:
γ− γ +
α−
γ ) γ γ) )
p +
x(cos
(cos
−sin
sin
p11 1α
xp x(cos
(cos
γ++−+xxγ(cos
pp−
11p1 p1 p11p1 p1 p11p1
(
+
xα(cos
−γ cos
) y−(sin
yα(sin
−γ sin
pp1α
(cos
cos
αcos
γy (sin
α
γp) (+α
αγp−1 −γ+γ xp1++(cos
α−−γcos
1−
p1α
−
+
x
(cos
−
cos
)
−
−
sin
)
+
−
cos
)) γγ, )) , ,(A.8)
α
γγα
(A.8)
1
p
p
p
−
+
x
(cos
−
cos
)
−
y
(sin
−
sin
)
+
p
−
+
x
(cos
−
cos
α
α
γ
α
γ
α
γ
α
1
1
1
1
1
1
(
+ xpα
(cos
(sin
−γ−sin
sin
+)α−yppp1(cos
x−p1cos
α+−α
γ(cos
αγγ−γ⋅ ⋅γcos
αα−
α
γ⋅xpα1pp1 +−p(cos
α⋅ γ)−γγ)pcos
p1 −
cos
)γ⋅pγ−pp)1γγy)−)(sin
− sin
)−sin
+sin
+α
−⋅γγcos
)) )0=γ,=0) 0 (A.8)
α
γ(sin
γ(cos
α(cos
, (A.8)
(A.8)
x(sin
1−
p1α
−−x−
αxα
α
sin
−−−yypyy1α
(sin
(cos
α
⋅⋅α
−−−−sin
α
γγγpγ))1)1 γ(−
⋅
cos
γ
=
p1 −
p(sin
x(sin
⋅
α
sin
γ
α
)
−
⋅
α
cos
γ
⋅
γ
1 y(y
1+
1p
1p (sin
1p x − (cos
1cos
p
p
p
−
+
(cos
−
cos
)
(sin
−
sin
+
p
−
α
γ
α
α
α
γ
α
ppx
p
p
,
(A.8)
p
p
1
1
1
1
1
1
1
1
1
, (A.8)
1y
1 p1− cos γ ⋅ γ 1 ) ), = 0 (A.8)
− xα(sin
α ⋅−1αsin
γ py1 ) −(sin
(sin
α −γ sin
γy (cos
)1 − ypα
αp1 ⋅−αcos
1(cos
p1 −
p1α
p1 γ ⋅ γ p ) ) =p10
−
⋅⋅α
γγ sin
⋅⋅ γγ pγγ11 ))⋅⋅1−
−
sin
)sin
−
⋅⋅α
pp11 ⋅−
p
p
−− xxx(sin
(sin
α
α
sin
−
y
(sin
α
−
sin
γ
)
−
y
(cos
α
α
−
cos
γ
⋅
γ
)
=
0
,
(A.8)(A.8)
,
1
1
1
1
1
1
)
−
x
(sin
α
α
−
sin
γ
)
−
y
(sin
α
−
γ
)
−
y
(cos
α
⋅
α
−
cos
γ
⋅
γ
)
=
0
p (sin
(sin
⋅αα
p1 γ ⋅ γ pp1 ) − py
p1α − sin γ ) − y (cos α ⋅ α pp1 − cos
p1 γ ⋅ γ pp1 ) ) =p10
1 p11
p11 ⋅−
1
1
1
− xα(sin
αsin
p1 − sin γ ⋅ γ p1 ) − y p1 (sin α − sin γ ) − y (cos α ⋅ α p1 − cos γ ⋅ γ p1 ) ) = 0
((
))
(
(( ( ( ((
(
(
(( ( ( ((
(
cos
yy(sin
x++px(cos
βββ−−−
α )α)−)−−
ββ−
α )α+) )+p+2pp2β
α
β −ββcos
α )αα) )
x(cos
cos
−α
ββ−−−ααβα++−+xα
2 β
p βp−
p2αpp+
x(cos
(cos
cos
−
sin
xpp2x2p(cos
(cos
β−sin
2p −
22p 2
+
xβ(cos
−ααcos
) y(sin
−(sin
yβ(sin
−αα
sin
(coscos
−αcos
cos
) (A.9)
β
βsin
α
β−−cos
2α
2p β
p2p −+
++−xxα(cos
−− cos
)) −−α
yy(sin
−− sin
)) ++α
pp))2 ++ββ2ppp2222 p−−2β
xxp2p222++(cos
)) α
ββ −− α
, , (A.9)
2 β −
(cos
cos
(sin
sin
(cos
−
cos
α
ββ
α
ββ(sin
α
α
α
22−+
+
x
(cos
−
cos
)
−
y
−
sin
x
(cos
−
cos
β
β
α
β
α
β
α
β
α,, )(A.9)
p
p
2 β p22 −
−
+
x
(cos
−
cos
)
−
y
(sin
−
sin
)
+
p
+
x
(cos
−
cos
)
β
α
α
α
α
β
α
(A.9)(A.9)
2 ⋅ β p 2p2 − cos
p2β
p2 α ⋅ α
,, (A.9)
β
⋅
β
−
sin
α
⋅
α
)
−
y
(sin
β
−
sin
α
)
−
y
(cos
)
=
0
p
p
2
(sin
2
2
2
−−xx(sin
β
⋅
β
−
sin
α
⋅
α
)
−
y
(sin
β
−
sin
α
)
−
y
(cos
β
⋅
β
−
cos
α
⋅
α
)
=
0
+β
(cos
−
cos
)
−
−
sin
)
+
p
−
x
(cos
−
)
β x−(sin
α
β
β
α
β
α
β
α
pppx
p2ppα
py2y
p⋅2 βp+
p2⋅ α
p(sin
pβ
pcos
x
(sin
β
⋅
β
−
sin
α
⋅
α
)
−
β
−
sin
α
)
−
y
(cos
β
−
cos
α
)
=
0
,
(A.9)
222p
2pα
2p y(sin
2
2
p
p
p
p
p
2
, (A.9)
−
⋅
β
−
sin
α
⋅
)
−
(sin
β
−
sin
α
)
−
y
(cos
β
⋅
−
cos
α
⋅
α
)
=
0
p
p
2
2
2
2
2 sin
2− py
2 p β − sin α ) − y (cos 2β ⋅ β 2 − cos
2p 2 α ⋅ α
2p 0 , (A.9)
p
−− xx(sin
β
⋅
β
−
α
⋅
α
)
(sin
)
=
2
2
2
2
2
pp22 )
pp22−(sin
pp22 ⋅−
pp22 )
⋅⋅ βββpp222 ⋅−−βsin
α
⋅
α
−
y
β
−
sin
α
)
−
y
(cos
β
⋅
β
cos
α
⋅
α
=
0
,
(A.9)
− xββ(sin
−
sin
α
⋅
α
)
y
(sin
β
−
sin
α
)
−
y
(cos
β
β
−
cos
α
⋅
α
)
=
0
− x(sin
(sin
sin
α
⋅
α
)
−
y
(sin
β
−
sin
α
)
−
y
(cos
β
⋅
β
−
cos
α
⋅
α
)
=
0
p2
p2
p22 ⋅ α p2 )p22− y p2 (sin β − sin α )
p2
p22 ⋅ α p2 ) = 0
−βββx++(sin
β(cos
−
α
cos
α
π
γγ γ−−βp−x2x⋅(cos
γγ)γ)+
γ )γ γ+) )+p+3pp−3 −
βy−(cos
γβ
β −ββcos
γcos
−
y (sin
+
) γ γ) )
p2 β
p)
p2ββ−
p−
pβ
ββ−sin
22π
−−
(cos
−cos
cos
(sin
−sin
sin
+2pγ⋅ γβp−pγ33ppx2−p−−3x−(cos
(cos
cos
2++yy
2 −
pβ
px
π
β
β
2
+
x
−
cos
(sin
−
sin
x
3−
3p +3+
3p (cos
3
p
p
π
β
γ
β
γ
β
γ
β
γ
2
−
+
−
x
(cos
−
cos
)
+
y
(sin
−
sin
)
+
p
(cos
−
cos
)
3
3
3
β
γγ )) +
22π
−
)) +
y (sin
−
−
)) . γ ).
ππ −
βπ +
ββ
ββ(sin
ββ33pp333−+
β(cos
−−2β
+−γγγβ−
−+xxγ(cos
(cos
− cos
cos
(sin
−− sin
sin
−− pβ
+ γ3γγpp3pp333+−
−−γx3xxpp3pp333 −(cos
(cos
− cos
cos
β −γγ cos
β−
γyppp(cos
β −γγγcos
xβ
(cos
)(sin
+ yβ
)333 +−
x3p3 β
)+y++yγγyy(sin
sin
) +++) +
) . γ ) ..
3γ
p3 −
+2+
xx(sin
−−−
sin
γγ−γ⋅ γ⋅cos
γ−γ )sin
β3p⋅3β
β+β
⋅(cos
=−))0cos
(sin
β−β⋅⋅ββ⋅β−β+ppp3x3γ⋅(cos
sin
γ⋅βpγ3pp)3−⋅γγ)+cos
β−β
−sin
sin
γsin
y++
(cos
⋅βp⋅β3pβ⋅33pp−pβ33+cos
−γ−cos
γxγpγ⋅pp33γ⋅3 )γ⋅β(cos
===0−00cos
β
γ
β
β
2βπxβ+(sin
−
x
(cos
+
y
(sin
)
p
−
p3yp) (sin
+
. . . (A.10)
p
x
(sin
−
sin
)
+
(sin
−
sin
γ
)
+
y
(cos
β
cos
p
+
β
−
sin
)
+
y
(sin
β
−
sin
γ
)
y
(cos
−
cos
γ
)
3
3
p
p
3
3p−
3p3 p β − sin γ ) + y (cos β ⋅ β
3pcos
3p
p3 γ ⋅ γ p 3)p3+ py
+
x
(sin
β
⋅
β
(sin
−
γ
⋅
γ
)
=
3 sin
3p3
3p30
3
3
p
p
p
p
++ xx(sin
⋅⋅ βββp333 ⋅−−βsin
γγ sin
⋅⋅ γγ pγ333 ))⋅ γ++ pyy)p333+(sin
−− sin
)) ++ γyy(cos
⋅⋅ βββp333 ⋅−−βcos
γ ⋅⋅ γγ pγ333 ))⋅ γ==p 00) = 0 .
.
+ xββ(sin
y p3ββ(sin
β −γγ sin
)(cos
+ yββ
(cos
(sin
(sin
sin
p3 −
p3 −γcos
pγ3 ⋅ γ 3 )p3+
p3 ⋅ βcos
pγ3 ⋅ γ 3 ) = 0
+ x(sin
βp3 ⋅ βsin
−
sin
y
(sin
β
−
sin
γ
)
+
y
(cos
β
−
cos
(A.10)
p3
p3
p3
p3
p3
(A.10)
(A.10)
(A.10)
(A.10)
(A.10)
(A.10)
We
the
assumed
symmetry
ofofof
equilibrium
configuration
to to
find
the
optimal
prices.
WeWe
We use
use
the
assumed
symmetry
equilibrium
configuration
to
find
the
optimal
prices.
(A.10)
We
use
the
assumed
symmetry
equilibrium
configuration
find
the
optimal
prices.
We
use
the
assumed
symmetry
of
equilibrium
configuration
findoptimal
theoptimal
optimal
prices.
Weuse
usethe
theassumed
assumed
symmetry
of
equilibrium
configuration
totofind
the
prices.
We
We
symmetry
of
equilibrium
configuration
to
find
the
prices.
We
We
use
the
assumed
symmetry
of
equilibrium
configuration
to
find
the
optimal
prices.
We
first
note
that
in
equilibrium,
xsymmetry
==0,0,yofy= equilibrium
0,xof
α=equilibrium
=0,=0,y0,
β= β
=
2π/3,
and
γ=
−2π/3,
which
immediately
We
use
the
assumed
configuration
toand
find
the
optimal
prices.
We
We
first
note
that
in
equilibrium,
0,
α
=
0,
β
=
2
π
/3,
γ
=
−2
π
/3,
which
imWe
use
the
assumed
symmetry
configuration
to
find
the
optimal
prices.
We
first
note
that
in
equilibrium,
x
=
0,
α
=
2π/3,
and
γ
=
−2π/3,
which
immediately
first
note
that
in assumed
equilibrium,
x 0,
= =0,
yofα
= equilibrium
0,
α= β=0,=
0,β2π/3,
β==configuration
2π/3,
and
γ==to
−2π/3,
which
immediately
first
note
that
in
equilibrium,
x
=
y
=
0,
α
2π/3,
and
γ
−2π/3,
which
immediately
We
use
the
symmetry
find
the
optimal
prices.
We
first
note
that
in
equilibrium,
x
=
0,
y
0,
=
0,
and
γ
=
−2π/3,
which
immediately
first
note
that
in
equilibrium,
xx =
0,
yy= =
0,
α
=
0,
ββ
=
2π/3,
and
γγand
=
−2π/3,
which
immediately
simplifies
(A.10):
first
note
that
in
equilibrium,
x
0,
y
=
0,
α
=
0,
β
=
2π/3,
γ
=
−2π/3,
which
immediately
mediately
simplifies
(A.10):
first
note
that
in
equilibrium,
=
0,
=
0,
α
=
0,
=
2π/3,
and
=
−2π/3,
which
immediately
simplifies
(A.10):
simplifies
(A.10):
simplifies
(A.10):
first
note
that
in
equilibrium,
x
=
0,
y
=
0,
α
=
0,
β
=
2π/3,
and
γ
=
−2π/3,
which
immediately
simplifies
(A.10):
simplifies
(A.10):
simplifies
(A.10):
simplifies
2ππ (A.10):
simplifies
(A.10):
2π
+
p
−
β p3 + γ p3 + y p3 3 = 0 .
π
2
222π
3
−β
+2+ppπp3p33+3−−−p−βββ3β
yyp+p3 y33p 3==03=0. 0=.. .0 .
p
+γ γp++3γγ+py+
23ππ +
pp33p+
+
++3+
+
p
βγγγpp33ppp333p+33p+
0.
π
2
3
3
+
p
−
+ pyy3 pp333+3p3y33p3 == 0033 .. =
3
333 3 3 + pβ3pp3333−
3+ γ 3 p3+ y 3
−
β
= 0.
3
p3
p3
3 symmetry
γp3 p3 = −
β p3 . Remaining two partial derivatives are in (B.8) and
Next,
yields
3
γ=−p=3−
=β− βRemaining
. Remaining
two
partial
derivatives
are
inin
(B.8)
and
Next,
symmetry
yields
...pRemaining
two
partial
derivatives
are
inin
(B.8)
and
Next,
symmetry
yields
γγ γpγ33pp33p=
β
two
partial
derivatives
are
in
(B.8)
and
Next,
symmetry
yields
Remaining
two
derivatives
are
and
Next,
symmetry
yields
Next,
symmetry
yields
two
partial
derivatives
are
(B.8)
and
3Remaining
β−pβpp...33pβ
Remaining
two
partial
derivatives
are
in
(B.8)
and
Next,
symmetry
yields
3γ −
3
twopartial
partial
derivatives
are
in(B.8)
(B.8)
and
Next,
symmetry
yields
p3 =
γ
=
−
Remaining
two
partial
derivatives
are
in
(B.8)
and
Next,
symmetry
yields
p3 β=pp3333−
p3 . Remaining
(B.14).
We
get:
p
3
γ
=
−
β
.
Remaining
two
partial
derivatives
are
in
(B.8)
and
Next,
symmetry
yields
(B.14).
We
get:
(B.14).
We
get:
p
p
(B.14).
We
get:
3
3
(B.14).
We
(B.14).
Weget:
get:get:
(B.14).
We
get:
(B.14).
We
(B.14).
 get:
2π We get:
2
1

(B.14).
We
2
π
+
p

gives the result.
Q.E.D.
222πππ
111−1 311 =30, =which
22 2−
2π++2pπp3 +−−−p3 22−
0which
, which
gives
the
result.
Q.E.D.
2
−
3
=00=,0,, 0which
, ,which
gives
the
result.
Q.E.D.
3
R
23π
2
1

−
3
=
which
gives
the
result.
Q.E.D.

2
3
R



gives
the
result.
Q.E.D.
3

+
p

−
−
3
gives
the
result.
Q.E.D.
3

+
p
−
−
3
=
which
gives
the
result.
Q.E.D.
3


3
0 , which
the result.
Q.E.D.
2 333RR− 331R =30, =which
givesgives
the result.
Q.E.D.
3333 +233pπ33++−pp2232R
2R−−R3233R3−
3
R
3
R
−
3
=
0
,
which
gives
the
result.
Q.E.D.

R


3
R
3 3  23 R 23R 3R 3R 

 2R 3
Proof of Lemma 6: (a) Partially differentiate (7) with respect to r to get:10
Proof of Lemma 6: (a) Partially differentiate (7) with respect to r to get: 99 9 9
99
(a) Partially
differentiate
(7)
with
respect
to
r to
Proof
of Lemma
6:Partially
99get:
(a)
differentiate
(7)
with
respect
to
rrto
to
Proof
of
6:
(a)
Partially
differentiate
(7)
with
respect
r rget:
to
get:
Proof
ofofLemma
Lemma
6:6:
(a)
differentiate
(7)
with
respect
to
get:
Proof
Lemma
6:
(a)
Partially
differentiate
(7)
with
respect
get:
Proof
Lemma
1of
(a)
Partially
differentiate
(7)
with
respect
to
rtoto
get: 99
Proof
of Lemma
6:Partially
(a)
Partially
differentiate
(7)
with
respect
to to
r to
to
get:
Proof
of
Lemma
6:
1
(
)
Dr =Proof
x
(cos
cos
)
y
(sin
sin
)
−
β
+
γ
−
β
−
γ
+
β
−
γ
.
(a)
Partially
differentiate
(7)
with
respect
to
r
to
get:
of
Lemma
6:
.
r
D11211r 1(=
xr β
(cos
y rβ(sin
γ xrr (cos
−(cos
β −γ cos
γy )ry(sin
+ (sin
−γ sin
γ )) .
1 r(−+
D
))γγ+
−
−
DD
+βγγ+γγrrrγrr+
β−−
−cos
β
−ββ−sin
rrrr =
D
xxrrr xr(cos
cos
sin
))γ)))γ...)γ))).).) .
==D
−−−
++
−−r−γx−
++))+
β(sin
−−βsin
γγ )sin
r(cos
(
cos
=2r (((−=
−ββ2β1βrrβ
β
−βcos
(
x r ββ
(cos
−r (cos
−γγcos
γyy+r)rr (sin
+yrr r y(sin
−sin
rr−
D
cos
)
(sin
sin
−
r β
rr +
rβ
r r=D
r
r
2
22r2= 2 (− β r + γ r − x r (cos β − cos γ ) + y r (sin β − sin γ ) ) .
Again,
symmetry,
and
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
andpartials
partials
(B.4),
(B.9),
and
(B.15)
give:
2symmetry,
Again,
and
partials
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
and
partials
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
and
partials
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
and
partials
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
and
partials
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
and
partials
(B.4),
(B.9),
and
(B.15)
give:
Again,
symmetry,
and
partials
(B.4),
(B.9),
and
(B.15)
give:




1
2
R
1
2
1
4
R
2
4
R
4
R
−
−
−1
Again, symmetry,
(B.15)
 − 14(B.4),
R −42R(B.9),
4 R give:
21R−−21and
2 partials
4 R=
− 24 Rand
. −1
Dr = 11  − 12 22R
3
=
−
−
1
−
−−R4411−R



R
1
2
1
4
R
2
4
R
4
R
−
−
D
2
3
=

−
−

=

−
−

=
−




1
2
R
1
2
1
4
R
2
4
R
4
−
−




1
2
R
1
2
1
4
R
2
4
R
−
−
42R4−R32−−−22 R4 R43R == −−24R 4−3R1R...−1−11 .
2r 1=−−−21222−R22R−31−−−1−−3212−3332===3211=−−
D
rr =
D
=
DD
=

−
−

=
−
D


=
−
−
−


1
4
R
2
4
R
4
−
2
2
r
2
3
=

−
−

=

−
−

=
−
. −.31. .
2
3
D
=
−
=
−
−
=
−
−
2
R
3
2
3
2
R
3
2
R




222r = 2 22−
rr r D
RR2 3323R 3333 − 3 3222= 222R
3 3 223R
332 R
 3 22=R
3
R
R
R333 .
r2
2R−2Rto
2−2RR
2−2R332R
2 2(a),

R
32 R3 33 3 as
2R
2
R
3
2
R
3
2R
R3232R
3
3
  2 22R

 2part

(b) We2proceed
similarly
get:
3in
R to3 get:2 R 3 
2R 3
 in part
 (a),
(b)proceed
We proceed
similarly
as
(b)
We
similarly
as
in
part
(a),
to
get:
(b)
proceed
similarly
as
in
part
(a),
to
get:
(b)
We
proceed
similarly
as
in
part
(a),
to
get:
(b)
We
proceed
similarly
as
in
part
(a),
to
get:
(b)
We
proceed
similarly
as
in
part
(a),
to
get:
(b)
We
proceed
similarly
as
in
part
(a),
to
get:
(b)We
We
proceed
similarly
as
in
part
(a),
to
get:
1
partγ )(a),
D p2 (b)
= 11We
− β1proceed
+ γ similarly
− x (cosas
β in
− cos
+ ytop2 get:
(sin β − sin γ ) .
D p2112 1−
= β1 p2−+βγp2p2+−γ xp2p2−(cos
x p2 β
(cos
β −γ cos
γy ) +(sin
y p2 β(sin
β −γ sin
D
=
−
cos
)
+
− sin
)) .. γγ )) ..
p
p
p
p
pp22+(sin
D
β
γ
x
(cos
β
−
cos
γ
)
+
y
ββ(sin
22−+
22+−
22 −
=
β
γ
x
(cos
β
−
cos
γ
)
−γsin
1
==2p2 −
2
p
p
p
−
β
+
γ
−
x
(cos
β
−
cos
γ
)
+
y
(sin
sin
γ sin
)γ γ..) γ).). .
DDppp22222p ==D
−
β
+
γ
−
x
(cos
β
−
cos
γ
)
+
y
(sin
β−−βsin
−ββ−sin
2pp2 (cos
p2pγ2pp2 −
p2px
p2
p(sin
p2py
−
β
β
−
cos
γ
)
+
y
2
pp2p2−
p22+
p
2p β+
2
2
p
p
2
2
2
2
2
D
=
+
γ
−
x
(cos
β
−
cos
γ
)
y
(sin
−
sin
2
2
2
2
2
p2
2p222 22we note
2
CD is unaffected by Firm 2’s price
From (A.7)
thatp2γ p2 =p2 0 (the indifferencepline
= 0 indifference
(the indifference
lineisCD
is unaffected
by Firm
2’s price
From
(A.7)
we note
γthat
=γ 00p2 (the
line
CD
unaffected
by
Firm
2’s
price
From
(A.7)
we
note
that
γthat
line
CD
is
unaffected
by
Firm
2’s
price
From
(A.7)
we
note
that
=(the
0 indifference
(the
indifference
line
CD
is
unaffected
byFirm
Firm
2’s
price
From
(A.7)
we
note
γpppγ2222pp2==γγand
=0p=20(the
From
(A.7)
we
note
that
indifference
line
CD
is
unaffected
by
2’s
price
(the
indifference
line
CD
is
unaffected
by
Firm
2’s
price
From
(A.7)
we
note
that
γ
(the
indifference
line
CD
is
unaffected
by
Firm
2’s
price
From
(A.7)
we
note
that
change).
With
partials
(B.4),
(B.7)
(B.14)
we
obtain:
0
(the
indifference
line
CD
is
unaffected
by
Firm
2’s
From
(A.7)
we
note
that
0 (the
indifference
line CD is unaffected by Firm 2’s price
price
FromWith
(A.7)
we
note
that2p(B.7)
change).
partials
(B.4),
(B.14)
we obtain:
2 p2 =and
change).
With
partials
(B.4),
(B.7)
and
(B.14)
we
obtain:
change).
With
partials
(B.4),
(B.7)
and
(B.14)
we
obtain:
change).
With
partials
(B.4),
(B.7)
and
(B.14)
we
obtain:
change).
With
partials
(B.4),
(B.7)
and
(B.14)
we
obtain:
With
partials
(B.4),
(B.7)
and
(B.14)
we
obtain:
change).
With
partials
(B.4),
(B.7)
and
(B.14)
we
obtain:
change).
With
partials
(B.14)
we
obtain:
 (B.7)
1 With
1 partials
1(B.4),
1and
change).
(B.4),
(B.7)
and
(B.14)
we
obtain:
D p2 = 11  111 +1 11 31 =  11 .1
11 +1 61R1+ 331==32R=113 ..1 .
D
=D p2112 1==2 R
D
1 631R3 ==3  = 21R..1. 3 .
2111323R++1++
D
3R++
DDppppp22222 ==D
==22p2 =22 R
3 322=R
. .
D
R
R= 332 R
2R 3233R 66636R
2p 2
2p22 2 22R
R6RR66R
33 33
R  22R2RR32 R
  2 R 2 R
3 3
(
((( ( ((
(
(
(((( ((
(
)
))
) ))
))
)
)) ))))
)
))) ) ) ))
)
)
)) ) ) ))
)
10
We denote partial derivatives with the variable with respect to which we differentiate in subscript and omit
the index of the firm, since we always work with Firm 3.
340
ECONOMIC AND BUSINESS REVIEW | VOL. 14 | No. 4 | 2012
(c) The reaction in first two firms’ prices to a radial move by Firm 3 will be identical;
(c)
The we
reaction
in first twoFirm
firms’1’sprices
to a radial
move
by Firm 3 will
identical;
hence,
can disregard
behavior.
Totally
differentiate
(A.9)beand
(A.10)hence,
with
(c) disregard
The reaction
in 1’s
firstbehavior.
two firms’
pricesdifferentiate
to a radial move
by
Firm
3 will
berespect
identical;
we can
Firm
Totally
(A.9)
and
(A.10)
with
to rhence,
to hence,
(c)
(c)
The
The
reaction
reaction
in
in
first
first
two
two
firms’
firms’
prices
prices
to
to
a
a
radial
radial
move
move
by
by
Firm
Firm
3
3
will
will
be
be
identical;
identical;
hence,
respect
to
r
to
get:
we can
disregard
Firm
1’s
behavior.
Totally
differentiate
(A.9)
and
(A.10)
with
respect
to rhence,
to
(c)(c)
The
inin
first
two
firms’
prices
totoa radial
by
Firm
3 3will
identical;
Thereaction
reaction
first
two
firms’
prices
a radialmove
move
by
Firm
willbe
be
identical;
hence,
get:
(c)
The
reaction
in
first
two
firms’
prices
to
aadifferentiate
radial
move
by
Firm
3
will
be
identical;
hence,
we
we
can
can
disregard
disregard
Firm
Firm
1’s
1’s
behavior.
behavior.
Totally
Totally
differentiate
(A.9)
(A.9)
and
and
(A.10)
(A.10)
with
with
respect
respect
to
to
r
r
to
to
(c)
The
reaction
in
first
two
firms’
prices
to
radial
move
by
Firm
3
will
be
identical;
hence,
get:
we
Firm
Totally
(A.9)
and
(A.10)
with
to
wecan
candisregard
disregard
Firm
1’sbehavior.
behavior.
Totally
differentiate
(A.9)
and
(A.10)
with
respectto
torrhence,
r to
(c)
The
reaction
in1’s
first
two
firms’
pricesdifferentiate
to
a radial move
by
Firm
3 will
berespect
identical;
we
can
disregard
Firm
1’s
behavior.
Totally
differentiate
(A.9)
and
(A.10)
with
respect
to
to
get:
get:
′
′
′
′
−
+
−
−
−
β
α
x
(cos
β
cos
α
)
y
(sin
β
sin
α
)
(c)
The
reaction
in
first
two
firms’
prices
to
a
radial
move
by
Firm
3
will
be
identical;
hence,
we
can
disregard
Firm
1’s
behavior.
Totally
differentiate
(A.9)
and
(A.10)
with
respect
to
r
to
get:
weget:
can
Firm
1’s
behavior.
Totally
differentiate
(A.9) and (A.10) with respect to r to
− α ′ + x ′(cos
β ′ disregard
β−
cos
α ) − y ′(sin
β − sin
α)
get:
we can
disregard
Firm
respect to r to
get:
′′ ++
′(cos
′ − α with
′
−−
−−Totally
−− )sin
βββ′′′p+−−
α
αα
x′+β
(cos
β
βbehavior.
cos
cos
α
))(sin
(sin
βdifferentiate
β
sinβα
α−)) sin(A.9)
−
−− ysin
x−1’s
(cos
β−−α
cos
αyyβ
)′′(sin
(sin
α ) + and
p β(A.10)
get: β+ ′βp−′2′ α
p2x
p2cos
p2 α
2 x ′(cos
′
α
)
y
′ +−′(cos
−βα
−(cos
−psin
xα′(cos
β
cos
α
y ′(sin
β− ysin
α )β − sin α ) + p 22 β ′pp22 − α ′pp22
x−p2cos
β−)−−ycos
αβ
)−
(sin
get: β+ ′′p−2α
′
′
p+2 x
p2 +
β
α
)
(sin
α
)
,,
2
′
′
′
− cos
− y ′β(sin
β+ ′x−′ α++(cos
(cos
)(cos
β′ −−α
sin
α⋅yy)αpp ′)(sin
−−+β
+(cos
+
−−
−− )sin
−−α
pp+2′2′x−
αα
β ⋅−−
cos
cos
)) α
sin
βα
α′′
−
− xxα
(sin
sin
−− sin
xβ′β(sin
β+α
⋅αpβ))p2++β−′ppsin
α′p′p2′2⋅ α
pp22 cos
22−β
,
−β−ycos
xβββ′α
(cos
βα
αppp222)β(cos
(sin
β −)−α
α(sin
) 22β(sin
p2 p)p22
+β
px−′2′pp2αβ
xp−pp22))x2cos
αβ
)α
ysin
(sin
α
′ ′′ pβ+
′
′
+
−
+
−
−
−
+
−
α
(cos
cos
β
sin
α
)
α
p2cos
py
p⋅2α
2p 2 −′pβ2sin
2 p−
2α
′
′
′
′
′
′
p
p
p
p
p
2
(cos
β
−
α
−
x
(sin
β
⋅
β
−
sin
⋅
α
)
−
x
(sin
β
⋅
β
)
−
+
−
−
−
β
α
x
(cos
β
cos
α
)
y
(sin
β
sin
α
)
′
+
+
−
+
−
p
β
α
x
(cos
β
cos
α
)
y
(sin
β
sin
α
)
p
β
α
2
2
2
2p
p2β − cos α ) − y p2 (sin
′ppβ22 −2 α
′ppsin
p
p
2
+− py ′22′p2++β(sin
+2 α
−
+ ⋅pβ22βpβ2β⋅⋅ −
α−ppsin
x )cos
β
sin
α x)x′′(sin
2β
2 (cos
2
, , ,,
px22′′ −
pcos
p′′2−
′
′
x
(cos
(cos
−
−
α
α
)
)
−
−
x
x
(sin
(sin
β
β
⋅
⋅
β
β
−
sin
sin
α
α
⋅
⋅
α
α
)
)
−
−
(sin
β
−
−
sin
α
α
⋅
⋅
α
α
)
)
2 β
2−
2⋅ α
′
′
′
′
′
β
y
(cos
β
⋅
β
−
cos
α
⋅
α
)
−
y
(cos
β
cos
α
)
=
0
+
−
+
−
−
+
−
p
β
α
x
(cos
β
cos
α
)
y
(sin
β
sin
α
)
p
β
α
p
p
p
p
p
p
p
p
22β ⋅ β ′ − sin
22 α p
,
′β′) ′−) −xsin
′(sin
+
xpp2x2 β
(sin
⋅⋅αα⋅ α
p2 22β β
p2α α
2 ) 22
2 −
′α
(cos
−2 +
cos
) y−
(sin
β⋅ α
⋅ )β−−′ sin
−y sin
x(cos
(sin
−
sin
⋅α
p p)
′⋅ 22α
+ +xp′p2px2 ′(cos
− α−pcos
+β⋅⋅ ⋅β
x)pp))α22−−
(cos
cos
)β
p⋅ 2pβp2β−2−′ppsin
αα
pβ
p2 −
(sin
αα
β
cos
α(sin
cos
= 0 ,,
+
α
xxpp22 (cos
(sin
⋅⋅ α
))) −
xxy′′′(sin
β
sin
α
α
p2α
2 p 2 β −psin
p)
2
2
+− xxy′′′pp′2p2222−(cos
(cos
β(sin
− cos
cos
αsin
)−
−α
(sin
β(cos
⋅β
β β′′β
−⋅⋅sin
α
α ′′′α
−−⋅⋅α
(sin
β′′β(cos
⋅β
ββ2ppβ222pβ2p−
−⋅⋅ ββ
sin
α−−αp⋅⋅cos
α⋅ αpp222α
)p2 2)⋅⋅α
p2−
′
′
′
′
′
′
−
y
y
(sin
β
β
−
−
sin
α
)
)
−
y
y
(cos
β
β
−
−
cos
cos
α
α
)
)
−
−
y
y
(cos
cos
α
α
)
)
′
′
+
x
(cos
β
−
cos
α
)
−
x
(sin
β
⋅
β
−
sin
α
⋅
α
)
−
x
(sin
β
⋅
β
−
sin
α
⋅
α
)
p
p
p
p
p
p
p
p
2 α ⋅ αp
−
(sin
sin
αα
αα
α)⋅ α
−) −
(cos
cos
= 2=02 ,0== 00
+22ββ
−cos
γ ′ 2−2ββxβ−−′(cos
β)α)−)−cos
)(cos
y ′β⋅(sin
sin
2p )) )
′ cos
−cos
sin
−yyxpypp2 2γ(cos
−sin
y ′(cos
⋅ppβ2 −−2 sin
− 2cos
′′)) ′−
+ −yxβ′′pp′y2 +′(cos
⋅β
β⋅ ′β′ −−β
α
⋅⋅γα
xy′′(sin
ββ ⋅β⋅⋅ββ
α ⋅α
α ⋅pα
p2 (sin
p(sin
−
sin
− cos
β
α
2) + β
− sin
γ ′ −β
− yyβ′′′ppp′2222 +(sin
(sin
β x−
−′(cos
sin α
αβ)) −
y pp22 2γ(cos
(cos
βy ′⋅⋅(sin
β ′′′ −
−βcos
cos
α ⋅⋅γα
α) ′′′)) −
− yy ′′′(cos
(cos β
β ⋅β
βppp2222 p−
− cos
cos α
α ⋅⋅ α
α 2pp222 ))p2 =
=0
0
−
ypp3′2−−(sin
β
−′γ′sin
α
)x−p y(cos
(cos
β
⋅ +β
−′′(sin
cos
α
⋅−αsin
)β−γγ−))ysin
(cos
β+⋅ βp3p2 −−βcos
α
⋅ α′p p2 ) = 0
′
′
′
′
′
+
+
−
−
−
−
+
−
β
β
γ
γ
β
β
γ
γ
β
β
x
x
(cos
(cos
cos
cos
)
)
y
y
(sin
sin
+
+
−
−
+
+
β
cos
γ
)
y
(sin
γ
)
γ
p
p
p
p
p
2
3 x−
3 α
3cos
3 γ
3 α ⋅ α3
′
′
′
′
′
′
−+ −
−
−
+
−
βyp′p′β2+(sin
γ
β
γ
β
(cos
)
y
(sin
sin
)
β
sin
)
−
y
(cos
β
⋅
β
−
cos
α
⋅
α
)
−
y
(cos
β
⋅
β
−
cos
)
=
0
′
′
′
′
+
−
−
+
−
γ
β
γ
β
γ
x
(cos
cos
)
y
(sin
sin
)
p
p
p
′
′
−γβ′ −
+′(cos
− cos γ )β+−ysin
− sin γ ) + p3 2 − β p3 + γ p3 2
γ p3 −β x−p3cos
(cos
(sin
2
p3 x
−
+
γγ yy)) β(sin
))β(cos
−β
++
+ βyy ′′′β⋅(sin
−γγsin
β ′′3 +
γpp′′3′3′−
βsin
(cos
cos
(sin
−−xx+−
+−γβ
−(cos
−x pxxγγγp(sin
−−
−β)sin
+sin
β′β
γγx)p−
(cos
β
cos
cos
)γ)p3+⋅+
(sin
ββ −
sin
γγpp)) −+−
pβp33′γ−−+⋅ βγβγ′pp′p′33 )++ γγ ′p′p33 ...
′
′
′
(cos
β
cos
β
−
)
x
(sin
⋅
β
pp33 +
p+
p33β
pp3+
′
− +βxp′3p′p3+−
−
−
+
−
γ
β
β
γ
(cos
)
y
(sin
sin
)
33 cos
3
+
−
+
−
+
β
γ
cos
γ
)
y
(sin
β
sin
γ
3
′ β′ pβ−3βpx+3−′γ(cos
+cos
−′⋅(sin
− sin
+p33p−−3 sin
−
cos
)−+ysin
(sin
pγ3 p γ
px
pβ ′⋅ γ+ ′p3γ3)′p
3 cos
′) +
′(sin
3 γ−
px3 p(cos
pγ
(cos
βycos
βγγ′ ))−γβ+
sin
γppy3 ⋅(sin
− βxp
γ(sin
) +β−
+
−
−3β−x)−+
+
(cos
β
β βx−
sin
γβ))γ⋅)+
β
3 )
3 γp′
+
−
+
+β β
+sin
p′33p′′′3+
(cos
cos
(sin
ppβ333⋅⋅ −
β ′′pp′−γ−333 psin
γ p3 3 )) . . ..
3 +γ p
3 − xp
pβ
p33 y
′′′))′sin
−
xβ
xβ′p′ppp3p3β
(cos
(cos
β
−
cos
cos
γγp))3 +
+β
xxpp−
(sin
(sin
β′β−−
⋅⋅sin
ββ ′yy′ −
−pγp33sin
sin
γγ)+β
⋅⋅γγx−
++ xxγγβ′′(sin
(sin
β
βsin
3β
3 −
′
(sin
−
sin
γ
)
+
(cos
β
⋅
β
cos
⋅
γ
+
y
(cos
β
⋅
β
⋅γγγγ′ppp333)3⋅⋅)γγ p=
+−
−
+
−
−
+
−
+
−
py′3px3−
γ
x
(cos
β
cos
γ
)
(sin
β
sin
)
p
β
pcos
p33 pγ γ⋅+
p33 0 .
3
3
′
′
p
3
p
p
−
x
(cos
−
cos
γ
)
+
x
(sin
β
⋅
β
γ
⋅
γ
)
(sin
⋅
β
−
γ
3
3
3 +
3cos
3+
3γ
3 γ
′
′
′
′
(cos
β
−
γ
)
x
(sin
β
⋅
β
−
sin
⋅
γ
)
+
x
(sin
β
⋅
β
−
sin
⋅pγ′ p3)) =
p
p
p
′
+
−
−
−
+
−
+
−
+
p
β
γ
x
(cos
β
cos
γ
)
y
(sin
β
sin
γ
)
p
β
3
3
3
′
′
′
′
p(sin
p3
sin
cosγγγp3 ⋅⋅⋅γγγ′′)))+
+ xxy′′(sin
(cosβ
β⋅⋅ ⋅β
βp p33p−
cos
0 ..
pβ
p3 γ
pγ3 γ⋅ γ⋅γγ3pp3)
−
−
γγ ))) ++
(sin
β
⋅⋅ β
−
sin
3
3 sin
− xxy′′′pp3p333+(cos
(cos
β3(sin
− cos
cos
+p3γyγxx)pp)333++(cos
(sin
β(cos
β ′β
−
sin
+
(sin
β
β
3 − sin γ ⋅ γ p
3 )3
p
p
′
′
′
′
′
′
′
′
+
y
y
(sin
β
β
−
−
sin
sin
y
y
(cos
β
⋅
⋅
β
β
−
−
cos
cos
γ
γ
⋅
⋅
γ
γ
)
)
+
+
y
y
(cos
(cos
β
β
⋅
⋅
β
β
−
−
cos
cos
γ
γ
⋅
⋅
γ
γ
)). == 00
3
3
′
′
′
−
x
(cos
β
−
cos
γ
)
+
x
(sin
β
⋅
β
−
sin
γ
⋅
γ
)
+
x
(sin
β
⋅
β
−
sin
γ
⋅
γ
)
p
p
p
p
p
p
p
p
3
3
3
3
3
3
3
3
′
′
′
′
p
p
p
p
+
y
(sin
β
−
sin
γ
)
+
y
(cos
β
⋅
β
−
cos
γ
⋅
γ
)
+
y
(cos
β
⋅
β
−
cos
γ
⋅
γ
)
=
0
Next,
we assume
equality
part
and
′(sin
−cos
sin
γ)+)+first
+yxppy33 (cos
(cos
β
−
⋅′′γ)) +′+)ofx+y′′present
y ′(cos
⋅pp3β3 −p−3 sin
− cos
⋅3pγ)3 )p3 )= =0from
0 (B.5)
′−
− +xy′′′ppy333 (cos
β β−− sin
γthe
β β⋅⋅ β
β⋅ from
−′ cos
sincos
γγ ⋅γ⋅γγ(c)
(sin
ββ⋅β⋅βLemma
γγ⋅ γ⋅ observe
p3 (sin
p(sin
′
p
+
β
γ
)
β
(cos
β
cos
γ
3
3
3
3
′
′
′
Next,
we
assume
the
first
equality
from
part
(c)
of
present
Lemma
and
observe
from
(B.5)
p
p
p
p
+ y ′p33 (sin
βx ′−=sin
) + αy p′33 =(cos
⋅ βalso
− cos
γ the
⋅ γ ′)second
+ y ′(cos
β ⋅of
β p(B.5),
3 − cos γ ⋅ γ p3 ) = 0
0 γγand
0. β
and (A.4)
that
We
use
part
which
thatfrom
the
total
3 − cos
3says
′
+ y p3Next,
(sin
βwe
− sin
) + first
ythe
(cos
β
⋅
β
−
cos
γ
⋅
γ
)
+
y
(cos
β
⋅
β
γ
⋅
γ
)
=
0
Next,
we
assume
the
equality
from
part
(c)
of
present
Lemma
and
observe
(B.5)
Next,
we
assume
assume
the
first
first
equality
equality
from
from
part
part
(c)
(c)
of
of
present
present
Lemma
Lemma
and
and
observe
observe
from
from
(B.5)
(B.5)
p
p
p
3=
3 − cos
3says
′
′
′
′
′
′
x
=
0
α
0
and
(A.4)
that
and
.
We
also
use
the
second
part
of
(B.5),
which
that
the
total
+
y
(sin
β
−
sin
γ
)
+
y
(cos
β
⋅
β
−
cos
γ
⋅
γ
)
+
y
(cos
β
⋅
β
γ
⋅
γ
)
=
0
Next,
we
assume
the
first
equality
from
part
(c)
of
present
Lemma
and
observe
from
(B.5)
Next,
we
assume
the
first
equality
from
part
(c)
of
present
Lemma
observe
from
(B.5)
p3 with
p3 partial
p3 (A.5) and
pof
3
derivative
respect
to
r
of
derivative
of
x,
and,
hence
by
also
α,
with
respect
Next,
we assume
assume
the
first
equality
from
part
(c)
of
present
Lemma
and
observe
from
(B.5)
′
′
′
′
x
x
=
=
0
0
α
α
=
=
0
0
and
and
(A.4)
(A.4)
that
that
and
and
.
.
We
We
also
also
use
use
the
the
second
second
part
part
of
of
(B.5),
(B.5),
which
which
says
says
that
that
the
the
total
total
and
(A.4)
that
and
.
We
also
use
the
second
part
of
(B.5),
which
says
that
the
Next,
we
the
first
equality
from
part
(c)
of
present
Lemma
and
observe
from
(B.5)
′x=′ =0 0and
′ =′partial
xrespect
α
and
(A.4)
that
also
use
the
second
part
which
says
that
derivative
with
to is
r of
derivative
of
x,in
and,
hence
by
(A.5)
also
of
α,symmetry
with
respect
α
=00 ..0We
and
(A.4)
that
and
. again
We
also
use
second
partof
of(B.5),
(B.5),
which
says
thatthe
thetotal
Next,
we
assume
the
first
equality
from
part
(c)
of
present
Lemma
and
observe
from
(B.5)
to
any
of the
three
prices
0.
Note
that
αthe
=the
0second
equilibrium,
use
(B.4)
and
oftotal
′′ =
′′to
xxwith
0
α
=
and
(A.4)
that
and
We
also
use
part
of
(B.5),
which
says
that
the
total
derivative
derivative
with
respect
respect
to
r
r
of
of
partial
partial
derivative
derivative
of
x,
x,
and,
and,
hence
hence
by
by
(A.5)
(A.5)
also
also
of
of
α,
α,
with
with
respect
respect
total
derivative
with
respect
to
r
of
partial
derivative
of
x,
and,
hence
by
(A.5)
also
of
α
,
Next,
we
assume
the
first
equality
from
part
(c)
of
present
Lemma
and
observe
from
(B.5)
=
0
α
=
0
and
(A.4)
that
and
.
We
also
use
the
second
part
of
(B.5),
which
says
that
the
total
derivative
with
respect
tothe
rαof
derivative
of
hence
also
of
respect
toderivative
any
of the
three
is
Note
again
that
=of
0x,
in
equilibrium,
use
(B.4)
and
symmetry
of
′ respect
′=
xsimplify
=prices
0 and
0partial
and
(A.4)
that
. We
also
useαthe
second
part
ofby
(B.5),
which
says
that
the
total
with
to
r0.
ofpartial
derivative
x,and,
and,
hence
by(A.5)
(A.5)
also
ofα,
α,with
with
respect
the
problem
to
above
system:
derivative
with
respect
to
r of
partial
derivative
of
x,
hence
also
of
α,
with
respect
′ =three
′=
xsimplify
0any
and
(A.4)
that
and
. We
also
use
the
second
part
ofby
which
says
thatsymmetry
the
total
to
toof
any
any
of
of
the
the
three
prices
prices
is
is0three
0.
0.
Note
Note
again
again
that
that
α
αinand,
==again
00 in
in
equilibrium,
equilibrium,
use
use
(B.4)
(B.4)
and
and
symmetry
of
with
respect
to
of
the
prices
isα0.
that
α(A.5)
= 0 (B.4)
in
equilibrium,
use
(B.4)
derivative
with
respect
to
of
partial
derivative
of
and,
hence
by(B.5),
(A.5)
also
of
α,symmetry
with
respect
the
problem
to
the
above
system:
to
any
the
three
prices
isrrαis
0.
Note
again
that
=Note
0x,
equilibrium,
use
and
ofof of
derivative
with
respect
to
of
partial
derivative
of
x,
and,
hence
by
(A.5)
also
of
α,
with
respect
to
any
of
the
three
prices
0.
Note
again
that
α
=
0
in
equilibrium,
use
(B.4)
and
symmetry
to
any
of
the
three
prices
is
0.
Note
again
that
α
=
0
in
equilibrium,
use
(B.4)
and
symmetry
of
the
the
problem
problem
to
to
simplify
simplify
the
the
above
above
system:
system:
derivative
with
respect
to
r
of
partial
derivative
of
x,
and,
hence
by
(A.5)
also
of
α,
with
respect
and
symmetry
of
the
problem
to
simplify
the
above
system:
to
any
of
the
three
prices
is
0.
Note
again
that
α
=
0
in
equilibrium,
use
(B.4)
and
symmetry
of
′
′
′
βtosimplify
− prices
3 y +the
p 20.
2above
β p2 system:
−again
2α p2 that
− 3xαp2 =− 0 in
3 yequilibrium,
the
problem
to
above
p2
problem
system:
tothe
any
of the2
three
isthe
use (B.4) and symmetry of
the
problem
above
′ −simplify
βsimplify
3 y ′ +the
2Note
β p2 system:
−again
2α p2 that
− 3xαp2 =− 0 in
3 yequilibrium,
to any
of the2to
three
prices
isp 2′0.
Note
use (B.4) and symmetry of
the
problem
to
simplify
the
above
system:
p2
′
′
′
′
′
′
the problem 2+toβpsimplify
the
above
system:
2
2
β
β
−
−
3
3
y
y
+
+
p
p
2
2
β
β
−
−
2
2
α
α
−
−
3
3
x
x
−
−
3
3
y
y
py
22− − 2
′ −−ppy232 ′y3−pyβ p2 pp−22 2α p2 = 0
3px2′pabove
ββ2′pβ
3pα
22 ′
′β−2′ β− ′p32 y3−′y+the
− 3−yppxp2322pxβ
−
the problem 2toβ2simplify
system:
′
′
p2
pp22 +
+
−
2
α
22
2
2
2
p
p
p
p
−2 β ′3
3 yy−′′ +
+ 3p
px2′′ p2
2ββ ′p−
−2 2
23α
αy ′pp22 +−
−2 3
3yxx pβ
−
3′yy−p β2 − 2α p2 = 0
′2 − y3
2+βp′′′ −
−
2 −
2′ 22β
2 y
3β3′xxpp−222pp22−ββ32′′α
−−′pp22 3+
3−yyy3′p′p2x2p2p+
− 22α
α=pp022 == 00
2 βp 2−++β p′pp32 2y2−′ββ+′p′p232px−−
−y−pp22 yββ3′′y′−−−pp222βyy′p′ 2−−−ββ2pα
p −
p+
2′ β
2 ′−
+
y
β
′
′
2
β
−
3
y
+
p
2
β
2
α
−
3
x
′
′
′
′
′p32−
+
p
2
β
−
3
x
β
−
3
y
+
y
β
−
=
p
p
p
p
pβ
′
′
′
′
− 2β ++
y pyy3′′pp22+p+
p −p2 p2β
+3′′y−
yppβ′p23 p−
−−
yα222p3α
y ′β0 = 0
2 p′ −
p2+
p22β
2 33
′′p−
′′ −
p3 y2 β+′ 2pp− 32xβ2pp223β
β
2 y
2 2α
2 = 0
p3 y
pβ
β
−
3
β
−
y
−
β
2
= y0′β p3 = 0 ..
2 ++py−
22
2 −
2 ′−
′
′
p
p
p
p
p
− 2β ′ +++ p3p y22′ββ+′pp′ 22p33′−−2 − 323xβ
+
3
y
β
+
3
y
−
y
2 β′ −
2 + y 2 β′ −
2 − 2α
2
′
′
p
p
p
p
p
x
3
y
y
−
β
=
0 ′′ −−p3yy′′ββ == 00 ..
3
3
3
3
3
p
p 3y
′′ +2′ β33′pp2yy′−
′′ ++ p3p3′x3′ pp−−2 β22β′β−pp33 ++
′′p33 pp−−2 ′yyp=p33 ββ
−
p′ −
′ −′p′p333 β+
y pp++222α
βpβ
yββ
0′β′ = 0
pp33..
3 −
′β−+2′2++ββnote
′pβ−3 x2′2+
′ppy223 ′3−
−−
2β2next
3p+
y3 y+that
p2 3pby
−3′ −
2(A.5)
β2pβ23 p+α+p3 y33=pyy3 x′pp3322p+yy++ppp3cos
y+3equilibrium.
− y2−ypα
βp2 β−′ −yThe
′
=
in
We
+
p
−
2
+
y
p
3
p
p
p
p3= =
− 22β
βnext
+ note
3 yy ′′ +
+that
p3′′ by
−2
2(A.5)
β p3 +
+
32 yy=p3 x 3+
+2 p
pcos
−2
2αβ
β=′′p3 x+
+3 2 in
3 yyequilibrium.
−3 yy p3 β
β3 ′′ −
− yyThe
βy ppβ33remaining
0 ..0 . derivatives
3
′′ +
′′pp33 −
′′β
−
3
p
−
β
3
−
3
=
0
α
remaining
derivatives
We
3
p
p
p
p
p
p
p
p
3
3
3
3
3
′ + in
′ +note
′β p3The
2 p − 2β ′
2 3 y′
− 2βWe
3 y(B.7),
p3′(B.8),
−that
2βby
+ 32 y(B.12),
+(B.14),
− equilibrium.
y p3 β ′and
− y(B.18).
= 0Applying
.
α
α
=pcos
xx(B.13),
cos
α
==in
p(B.11),
p3 pp +=
in
in
The
remaining
remaining
derivatives
derivatives
We
next
next
note
that
by
(A.5)
are collected
all
3 (A.5)
pp222
pequilibrium.
p′2p23 (B.17),
′next
′+
′pp33=xα
′ − The
′β
−We
2βnext
+ note
3note
y(B.7),
p3′(B.8),
−by
2(A.5)
β(A.5)
+α α
3 y(B.12),
−cos
βcos
+(B.14),
3xxin
yin
− yequilibrium.
y(B.18).
=remaining
0Applying
.
x2p2+
α
=
derivatives
We
that
by
p(B.11),
p3=
p3equilibrium.
p3 β and
p3remaining
We
next
note
that
by
(A.5)
The
derivax
α
x
equilibrium.
The
remaining
derivatives
that
p2 p=
p
3
are
collected
in
(B.13),
(B.17),
all
2
2
p2cosα = x pin
α
=
equilibrium.
The
remaining
derivatives
next
note
that
by
(A.5)
2 x
2
theseWe
we
get
the
final
form
of
the
system:
p
p
p
α(B.11),
x p22 (B.12),
cosα =(B.13),
x p22 in(B.14),
equilibrium.
The
derivatives
We
next
note in
that
by (A.5)
are
are
collected
in (B.7),
(B.7),
(B.8),
(B.8),
(B.11),
(B.12),
(B.17),
(B.17),
and
andremaining
(B.18).
(B.18).
Applying
Applying
all
p22 =
α(B.8),
= (B.12),
x(B.11),
α =(B.13),
x p2(B.14),
in (B.14),
equilibrium.
The
remaining
derivatives
We
next
note
that
by
tives
arethe
collected
in(A.5)
(B.7),
(B.12),
(B.13),
(B.14),
(B.17),
and
(B.18).
Applying
these
wecollected
get
final
form
of
the
system:
are
collected
in
(B.8),
(B.11),
(B.13),
(B.17),
(B.18).
Applying
allall all
p2 (B.12),
p2 cos
collected
in(B.7),
(B.7),
(B.11),
(B.13),
(B.17),and
and
(B.18).
Applying
α
= system:
xsystem:
α = x(B.14),
The
remaining
derivatives
We
next
note
thatfinal
by(B.8),
(A.5)
arearecollected
collected
in
(B.7),
(B.8),
(B.11),
(B.12),
(B.13),
(B.14),
(B.17),
and
(B.18).
Applying
all
these
these
we
we
get
get
the
the
final
form
form
of
of
the
p2the
p2 cos
p2 in equilibrium.
are
in
(B.7),
(B.8),
(B.11),
(B.12),
(B.13),
(B.14),
(B.17),
and
(B.18).
Applying
all
−54
pget
+18
pfinal
+ 36R
− 9of+(B.11),
πform
3 (system:
p2′ the
− p′system:
− 2R + 3(B.14),
′ final
these
wewe
the
form
system:
all
these
the
final
of
) == 00 ,, (B.17), and (B.18). Applying all
these
the
form
ofthe
2′get
3get
are
collected
inwe
(B.7),
(B.8),
these
we
the
form
the
−54
pfinal
− 9of
πthesystem:
3 ( (B.12),
p2′ − p33′ −(B.13),
2R + 3(B.14),
) + 33) ==(B.17),
are collected
in (B.7),
(B.8),
(B.11),
(B.12),
(B.13),
and (B.18). Applying all
these
we pget
get
the
final
form
of+
the
system:
2′ +18
3′ + 36R
these we get
the
final
form
of
the
system:
−54
−54
p
p
+18
+18
p
p
+
+
36R
36R
−
−
9
9
+
+
π
π
3
3
p
p
−
−
p
p
−
−
2R
2R
′
′
′
′
′
′
′
′
(−pp3′2′2p−+′ 2R
22pfinal
33 − 9of
33 + 3) =+
,,0 , 00,,
−54
p
+18
+
36R
+
π
3
p
−
0
′
′
′
(
these18
we
get
the
form
the
system:
−54
p
+18
p
+
36R
−
9
+
π
3
p
−
2R
+
3
=
′
′
′
p
−
36
p
−
36R
+
9
+
π
3
−
p
+
2R
+
3
=
0
′
′
′
2
3
2
(
)
(
2
2p ′ +18
−54
36R
−54
+336R
36R
−−9999++
++ππ
ππ 3333( −( ppp22′′22 −
−+2 pppp33′′33′ −
−+3 2R
2R
+++ 3333)) =
=== 0000,,,,
18
−22′′ +18
36
p33pp′p33−′′′ +
+−
2R +
−54p2′pp18
36R
2R
(
)
2 +18
3 +p
2′ (−
3′2′2′ −
18
p
p
−
−
36
36
p
−
−
36R
36R
+
+
9
9
+
+
π
π
3
3
−
−
p
p
+
+
p
p
+
+
2R
2R
′
′
′
′
′
′
2
2
3
3
3
3
−54
p
+18
p
+
36R
−
9
+
π
3
p
−
p
−
2R
+
3
′
′
′
′
(
)
with
18
p2′pa−22′ solution:
36
p3′p−33′ 36R
+ 9+ +9 π+ π 3 (3−( −22p+2′ +33p+3′ + 2R + 3=)++=033,)0=,= 00,,
18
− 36
− 36R
18
pp2′′ a−
36
pp3′′ −
36R
+ 9 + π 3 ( − p2′ + p3′ + 2R +
3) = 0 ,,
with
solution:
18
−
36
−
36R
2
3
18 p2′with
−solution:
36 apa3′solution:
− 36R ++299 ++ ππ 33 (−− pp2′2′ ++ pp33′′ ++ 2R
2R ++ 33) == 00,,
with
solution:
with
a
+
36
−
2π
3
R
−9
+
9π
3
−
π
18with
p2′ a−solution:
p3′ − 36R +29 + π 3 ( − p2′ + p3′ + 2R + 3) = 0 ,
a36solution:
with
+ 9π
3 − π + 36
with
a−9
solution:
p
=
′
22 − 2π 3 R , and
2
with
with
asolution:
solution:
36
36 −− 2π
2π ,33and
RR
−9
−9 ++39π
9π
332 3π
−−2 ππ36
p2′ = aa−9
90
−
3++− 2π
with
solution:
+
+
9π
−
π
2
− 2π 33 3R
3 π−−2π3π
,, and
and
pp2′2′−9
==
+ +36
36336
− 2π
2π
RR
−9
++
9π9π 33 90
−
+
−
3
R
+
9π
−
π 2 90
,
and
p2′p=2′ =−9
, and
−36
−33π
3π
33 3 R ,, and
290
−
π
+
−
2π
+
9π
3
and
= −9
2
90
−
3π
pp22′′ =
72−
4π 33RR,, and
18 ++12π
−2+−
3π36
3−2π
9π 3390
−−90
π−π3π
and
pp2′ ′ == −9
− 3372
390−−
π3π
Q.E.D.
22 − 4π 3 R, .and
p2′3 = 18 +12π
90
−
3π
3
−
−
72
72 −−34π
4πR .33 RR
18
18
+12π
+12π
3
3
−
−
π
π
2
Q.E.D.
p3′ = 1818
2 72
90
−
3π
3
90
−
3π
3
−
−
4π
+12π
3
−
π
2π − 72 − 4π 3 R
+12π
3
−
2
.
.
Q.E.D.
Q.E.D.
p
p
=
=
′
′
3−
− 72
72
− 4π
4π 33 R
R.
1833 +12π
+12π 3390
−π
π−23π
−
18
−
Q.E.D.
p3′p=3′ =
.
Q.E.D.
90
90
−
−
3π
3π
3
3
−
72
−
4π
3
R
18
+12π
3
−
π
.
Q.E.D.
pp3′′ =
2
90
−
3π
3
.
Q.E.D.
=
−−3π72 3− 4π 3 R
−90
π− 3π
p33′ = 18 +12π 390
− 3π 33results from... Lemma 6 to calculate the expression inQ.E.D.
Q.E.D.
Q.E.D.
p3′ = of Lemma 90
Proof
7:
parentheses
90Collect
− 3π 3
90Collect
−
3π 3results from Lemma 6 to calculate the expression in parentheses
of Lemma
7:
of (9)Proof
andProof
equate
it
with
zero:
of
ofitLemma
Lemma
7:
7: Collect
Collect
results
results from
from Lemma
Lemma
66 to
to calculate
calculateexpression
the
the expression
expression
in
in parentheses
parentheses
of (9)
andProof
equate
with
zero:
Proof
of
7:7:Collect
results
6 6totocalculate
parentheses
Proof
ofLemma
Lemma
Collect
resultsfrom
fromLemma
Lemma
calculatethe
expressionin
parentheses
Proof
of
Lemma
7:ititCollect
Collect
results
from
Lemma
6 to
to calculate
calculate
thetheexpression
expression
ininparentheses
parentheses
of
of
(9)
(9)
and
and
equate
equate
with
with
zero:
zero:
Proof
of
Lemma
7:
results
from
Lemma
6
the
in
ofof(9)(9)
and
it itwith
zero:
andequate
with
zero: results from Lemma 6 to calculate the expression in parentheses
ofequate
Lemma
7: Collect
of (9)
(9)Proof
and
equate
it with
with
zero:
Proof
of Lemma
7: Collect
results from Lemma 6 to calculate the expression in parentheses
of
and
equate
it
zero:
of (9) and equate it with zero:
of (9) and equate it with zero:
((
(( ( (
((
))
)) )
))
((
)( ( (
(
((
((
(( ( (
((
))
)) )
))
(
)(( ( (
((
(
((
(( (
((
((
((
(( ( (
((
(
(( (
((
(
((
(( ( (
(( (
(
(( ( (
((
))
)) ) )
))
))
)) ) )
))
))
(( ) ) ) ))))
)) (
(
)))) )) )) )
(
)) ((
( )
)
))))
)) ) (() ( ( ((
)) ) )
)) ((
))
))
)) ) )
)) )
)
)) ) )
))
A. FELDIN | THREE FIRMS ON A UNIT DISK MARKET: INTERMEDIATE PRODUCT DIFFERENTIATION
341
Proof of Lemma 7: Collect results from Lemma 6 to calculate the expression in parentheses of (9) and equate it with zero:
−9 + 9π 3 − π 2 + 36 − 2π 3 R*
4R* −1
2
− ** + * ⋅ −9 + 9π 3 − π 2 + 36 − 2π 3 R* = 0 ..
2
*
4R −1
* + 9π
3 **+−1
2R22 3 2⋅ −9 4R
+−9
36
− 2π
3π=R20**+. 36 − 2π 3 R*
−9
−3π
π36
+ 9π
+π
−36
2π
3 3R−
+−9
9π
3 90
− π33−22−
2 3
4R
−1
4R** −1
− 2R
+
−
2π
3
+
9π
*
−1
4R
2
− 3 *+ 2R+* 3 ⋅* − ⋅ *
= 0 ..
+90 −* 3π ⋅ 3
= 0.
− 2R−
=R0 . =
⋅ result.
0
Simple
algebra
the
3 +yields
90
− 3π
3
2R
2R
3
3
90
2R
3 * 2R
3
90
−
3π
3
2R* 2R
* 3 2R
2
*− 3π 3
2
*
2
*
3 yields
3
90
−
3π
3
2R −9
** *algebra
Simple
the
result.
− 2π33 3RR R
−9
+ 9π33−−3ππ− π++ +36
−−2π
++9π
3636
2π
9π
−1
2 −9
4R
22yields
−1
4R4R−1
Simple
algebra
the
result.
However,
we
have
a firm’s
Simple
algebra
thealgebra
result.really
Simple
yieldsfound
the result.
+establish
⋅that
Simple
algebra
yields
the result.
−− −Simple
+to
⋅⋅yields
== 0local
+ algebra
0=..0 . best response to the rival
** *
** * yields the result.
3to2R
−condition
3π33found
3 also.
2R 33 3thethat
locations
we33must
check
second
order
The local
second
order
derivative
ofrival
Firm
90
−−really
3π
2R
9090
3π
2R2R
2R
However,
establish
we
have
a
firm’s
best
response
to the
However,
toestablish
establish
thatwe
wehave
really
found
a afirm’s
local
best
torival
therival
rival
However,
to establish
that
we
have
really
found
firm’s
local
best
response
to
the
However,
tohave
establish
that
we
have
really
found
a response
firm’s
best
response to the rival
However,
to
that
really
found
a firm’s
local
best
response
tolocal
the
3’s
profit
with
respect
to rthe
is
derived
from
(9):
However,
to
establish
that
we
have
really
found
a
firm’s
local
best
response
to
the
rival
locations
we
must
check
second
order
condition
also.
The
second
order
derivative
of
Firm
Simple
algebra
yields
the
result.
Simple
algebra
yields
the
result.
Simple
algebra
yields
the
result.
locations
wemust
must
check
the
second
order
condition
also.
The
second
order
derivative
ofderivative of Firm
locations
we
check
the
second
order
condition
also.
The
second
order
derivative
of
Firm
locations
we
must
check
the
second
order
condition
also.
The
second
order
locations
we
must
check
the
second
order
condition
also.
The
second
order
derivative
of
Firm
locations
we
must to
check
the second
order
3’s2 profit
with
respect
r is derived
from
(9): condition also. The second order derivative of Firm



∂
Π
∂
∂
∂
d Π
d
p
dp
D
D
D
d
p
d
p
d
3’s
profit
with
respect
to
r
is
derived
from
(9):
3’s
profit
with
respect
to
r
is
derived
from
(9):
Firm
3’s
profit
with
respect
to
r
is
derived
from
(9):
3’s
profit
with
respect
to
r
is
derived
from
(9):
3 However,
3
3 really
3 found
3
1 aa firm’s
2local
establish
that
we
have
really
found
a firm’s
best
response
rival
However,
to
establish
we
have
response
to
the
rival
However,
to3to
establish
that
we
have
local
best
response
to to
thethe
rival
3’s
to
r3is
from
(9):
 with
 +that
 best
=profit
⋅ respect
⋅ derived
+really
⋅found
+ firm’s
⋅ local
must
 ∂∂D
 order
∂Π
∂also.
dd22locations
ddcheck
pr32 the
dp
D
D
ddsecond
pr2second
dwewe
∂
∂
∂
p
dr
r
p
r
p
rΠ2d3 2 Πdr
must
the
condition
The
order
derivative
of
Firm
locations
we
second
order
condition
The
derivative
of
Firm
locations
must
check
the
second
order
condition
also.
second
order
derivative
of
Firm
3 check
3second
3 order
3 d p
3The
1 also.
3
1
2







∂
Π
∂
∂
∂
d
p
dp
D
D
D


d
p
d
p


 +3 3dp
 ∂dΠ3 ⋅ d 3dp3 Π
d∂3pD33 ∂⋅ d dp
d Π2d3 2=
D33 ∂+D
DD3 3∂⋅ D
d 3p∂2Dd3 p 2dp1 ∂ D3 d p 2
⋅  ∂3 Π
3d 
3p1 3 + ∂1∂
3d dp

Π
∂
Π
d
p
D
d
p
d







=
⋅
+
⋅
+
⋅
+
⋅




=
⋅
+
⋅
+
⋅
+
⋅
3’s
profit
with
respect
to
r
is
derived
from
(9):
3’s
profit
with
respect
to
r
is
derived
from
(9):
=
⋅
+
⋅
+
⋅
+
⋅
3’s
profit
with
respect
to
r
is
derived
from
(9):
3
3
3
3
3
3
3
1
2
dr
dr ′  ∂⋅r ∂ p+1 d′ r⋅ ∂ p+2 d r⋅  
dr
2dr ∂ p3
∂ p ′ ⋅ d2 r  +dr
dr
dr
rd ddp r∂r ∂p 2r∂∂∂Dppd2∂rdpd2d1 prr d r ∂ D ∂ pd2 2 pd r 
d r 2 dd rr 2dr= ∂dr
 ∂∂ pr13 ∂∂ rrd∂∂rpD13∂∂ ppd11dr
p∂3D
∂ 3pd33 rd rd ∂r Ddr
3  drd p
3
22 2





′
′
′
+
⋅
+
⋅
+
⋅
⋅ 2 22 
p


∂33D3 ∂∂D
dd dΠ
D
Π3Π
Π∂3Π
∂D
dppd33p3 ′ dp
dpdp
D
D∂33D3dd ppd11 p′1′ 2∂∂+
D∂33D3dd3 2p⋅pd22p22 21 ′+
d3 d∂∂Π
33 ′ 3∂
3 3 d
3 3d
′
′
′
′












∂
∂
∂
D
D
D
D
D




dd rp 2 
d
p
d
p
d
p



=
⋅
+
⋅
+
⋅
+
⋅
=
⋅
+
⋅
+
⋅
+
⋅
=
⋅
+
⋅
+
⋅
+
⋅




∂
∂
∂
∂
∂
p
r
p
r
p
d
r
p
r
2
3
2
r32 3p1d1 2+pd∂1D
13 ∂D⋅3d∂∂p′r∂r1∂1D
D
 3 ∂D
D
dr+3p∂∂1∂pp∂D23p3 ∂∂dD
p 2233 ∂∂⋅ D
d∂32Dp32dd22 2dpp22 p1 ∂ D3 d 2 p 2 
dD
p ⋅∂∂pp∂33D
rrd++3r′∂+Ddr
dr
dr
dD⋅⋅r3rd3drp′ ⋅2∂⋅∂ddd++pp∂ppp2∂221pD2+13dd∂∂⋅⋅rr∂dD
dd rrd22r 2++dr
2d
1 D
3 ∂pdr



11




∂
D
d
p
p





+⋅ p3 ⋅∂pr33 d∂dD
⋅
+
⋅
+
⋅




+
⋅
+
⋅
+
⋅
+
⋅
⋅
+
+
⋅
p33dr
3
3
3
3
3
2
1
2
1
∂
∂
∂
∂
p
p
d
r
p
d
r
p
d r⋅2  2 2
2
+ p3 ∂⋅ r ∂ r  ∂+p1 ∂3p1d r⋅∂ rd r∂+p2 ∂∂ pp2 d r⋅ d r∂ p+1 ∂∂pddp1rr⋅2d r 2∂2d p+
drd2 r  ∂ p 2 d r 2 
1d
   ∂′′r ′ 1 ∂ ′′p1′  d r 2∂ ′′p 21′  d r 1 ∂ p1222 2d r r2 ∂∂ ppd∂2222rp(A.11)
r  (A.11)
2 

∂∂D














D∂33D3dd dpp22p2  
D∂33D3 dd ppd11 p1 ∂∂D
D∂33D3 dd ppd22p 2 ∂∂D
D∂33D3dd dpp11 p1 ∂∂D
D∂33D3 +++∂∂D



*









+
⋅
⋅
+
⋅
+
⋅
+
⋅
p
+
⋅
⋅
+
⋅
+
⋅
+
⋅
p
+
⋅
⋅
+
⋅
+
⋅
+
⋅
p
(A.11)
3both
33 
2
22 2 behavior
22 due











∂∂rrterms




At r = R
in
the
first
row
of
(A.11)
are
zero.
First
one
to
optimizing
(A.11)
(A.11)
pp∂11 p1  dd rrd r ∂∂pp∂22p2  dd rrd r ∂∂ pp∂11 p1 dd rrd r ∂∂ pp∂22p 2 dd rrd(A.11)
r (A.11)
 behav*   ∂ r   ∂
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both
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(A.11)
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rr =
=R
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rof
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due
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the
second
one
due
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in the
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the
one
due
to
the
expression
in
parentheses
being
* stage
theses
being
zero
at
=
R
(F.O.C.).
It
to
evaluate
the
sign
of
the
expression
in
second
row
of
(A.11).
We
prove
a
series
of
claims.
stage It
ofremains
the game,
the second
one
to the
expression
parentheses
evaluate
the sign
of due
the big
expression
in in
parentheses
in being
the
zeroinatthe
r =second
R*** (F.O.C.).
* to and
*R** (F.O.C.).
It
remains
to
evaluate
the
sign
of
the
big
expression
in
parentheses
in
the
zero
at
r
=
(F.O.C.).
It
remains
to
evaluate
the
sign
of
the
big
expression
zero
at
r
=
R
(F.O.C.).
It
remains
to
evaluate
the
sign
of
the
big
expression
in
parentheses
in
the
zerozero
at
r
=
R
=of=Rboth
terms
in
the
row
(A.11)
zero.
First
due
optimizing
behavior
At
rrow
terms
in
the
first
row
of
(A.11)
are
zero.
First
one
due
to
optimizing
behavior
AtAt
r=
=r RrR
both
terms
inprove
the
first
row
ofof
(A.11)
areare
zero.
First
one
due
to to
optimizing
behavior
parentheses
in
the
second
row
ofto
(A.11).
Wethe
prove
aofseries
ofone
claims.
(F.O.C.).
It
remains
evaluate
sign
the
big
expression
in parentheses
inintheparentheses in t
at
Rboth
second
(A.11).
We
afirst
series
of
claims.
′second
second
row
of
(A.11).
We
prove
a
series
of
claims.
row
of
(A.11).
We
prove
a
series
of
claims.
second
row
of
(A.11).
We
prove
a
series
of
claims.
in
the
second
stage
of
the
game,
and
the
second
one
due
to
the
expression
in
parentheses
being
in
second
stage
of
and
second
in the
the
second
the game,
game,
and the
the
second
one due
due to
to the
the expression
expression in
in parentheses
parentheses being
being
∂of
D3(A.11).
stage
of the
second
row
We prove
a series
of one
claims.
*
Claim
1:
.It
(F.O.C.).
 ′ <ItIt0 remains
(F.O.C.).
remains
evaluate
sign
expression
parentheses
zero
=**R(F.O.C.).
to
evaluate
the
sign
of
the
big
expression
in
parentheses
in
the
zero
at
remains
to to
evaluate
thethe
sign
ofof
thethe
bigbig
expression
in in
parentheses
in in
thethe
zero
at rat
r ==r R
R
′
′
∂
D
′


∂
r
3
 ∂D ∂D< 0′ ..
∂
D


Claim
1:
Claim
1:
3
3
second
row
of
(A.11).
We
prove
a
series
of
claims.
3
second
row
of
(A.11).
We
prove
a
series
of
claims.
second
row
of
(A.11).
We
prove
a
series
of
claims.
Claim
1:∂r ∂D<3 0Claim
Claim
1:  1:
. < 0 . 1: 
 < 0.
Claim
 < 0of. Lemma
We go
back
rthe
 to obtain the relevant part of the partial derivative of
 ∂r 6(a)
∂∂′′rr proof
 ∂to
′ 

∂3symmetry
∂∂D
D
the
3 proof of the
interest.
Using
problem
and
the fact
weofget:
x r ≡ 0part
We
go
back
to
the
Lemma
6(a)
to obtain
thethat
relevant
the partial
derivative
of
3D
00<proof
Claim
1:
.of
Claim
..0proof
We
go1:
of
to
the
relevant
of
the
partial
Claim
1:back
totothe
<proof
 back
the
We
go
to<
the
ofLemma
Lemma
6(a)
toofobtain
obtain
the
relevant
part
ofpartial
the
partial
derivative
of
We
goLemma
back
to 6(a)
the 6(a)
proof
Lemma
6(a)
topart
obtain
the
relevant
part
ofdethe
derivativ
We
go
back
to obtain
the relevant
ofpart
the
derivative
of partial
We
go
back
to
the
proof
of
Lemma
6(a)
to
obtain
the
relevant
part
of
the
partial
derivative
of
∂
r
interest.
Using
the
symmetry
of
the
problem
and
the
fact
that
we
get:
x
≡
0
∂
r
∂
r
.
Hence,
Dr =interest.
−
β
+
y
sin
β






r
r
r interest.
we
get:
rivative
of
Using
the
symmetry
ofthe
the
problem
and
the
fact
that
Using
the
symmetry
ofproblem
the symmetry
problem
and
the
fact that
we
get:
x
≡
0
interest.
Using
the
of
the
problem
and
the
fact
that
we
get:
x
≡
0
interest.
Using
the
symmetry
of
the
and
fact
that
we
get:
x
≡
0
r
symmetry of the problem and the fact that
x rr ≡ 0 we get: r
. the
Hence,
D =interest.
−We
βgo
yUsing
sin βthe
r +
rback
.=.proof
Hence,
of
Lemma
6(a)
obtain
relevant
part
partial
derivative
back
the
of
Lemma
6(a)
to
obtain
the
relevant
part
of
the
partial
derivative
of
We
the
proof
of
Lemma
6(a)
to to
obtain
thethe
relevant
part
ofof
thethe
partial
derivative
ofof
Hence,
D
=go
−+go
βback
+sinto
ytorβto
sin
βproof
.
Hence,
D
−
β
+
y
sin
β
.
Hence,
Drr =We
−
β
y
3
1
r
r
r
r
r
r
r
′
−=β−rβ+′ +
y r sin yβ′ .−Hence,
(DDr r =)Using
yof
β ′the
.problem
rthe
r
r ⋅of
interest.
Using
the
symmetry
problem
and
the
fact
that
we
get:
x
≡
0
interest.
the
symmetry
the
and
the
fact
that
we
get:
x
≡
0
interest.
Using
symmetry
of
the
problem
and
the
fact
that
we
get:
x
≡
0
rr r
23 ′ 3 21 1 ′
′ sin
3
1
)r′(′rβD+==+r −y−)+y′′rββr=sin
D
y. rHence,
− 1 y′r 1⋅ β .
3.β
r +
=βrβ−
y
Hence,
D
−−
.
Hence,
DrrD==r((D
sin
β
3
′
′
′
′
′
′
−
β
+
y
−
y
⋅
β
.
′
′
′
(
)
D
=
−
β
+
y
−
y
⋅
β
.
r
)
+
y
−
y
⋅
β
.
.
r
r
′ . 2 r (B.16)
Ther total
(Drr ) derivatives
=r − β 2rr′2+ r2are
y2r′r in
−r (B.9),
2 y r ⋅r β(B.10),
2 and (B.18). Putting them all together yields:
2 2 2
The total
derivatives
and (B.18). Putting them all together yields:
1 (B.9),p(B.10),
33 3 are
11 in
′ (B.10),
′ (B.16)
2 R in
− 1(B.9),
The
total
are
in⋅′2derivatives
(B.9),
(B.16)
and
(B.18).
Putting
them
all together
yields:
2−r′R
2in
The
total
are
(B.10),
(B.16)
andthem
(B.18).
Putting
them
all together yie
′derivatives
′ .(B.10),
The
total
derivatives
are
(B.9),
and (B.18).
Putting
them
all together
yields:
′ +++
′β..(B.9),
+1 yyr′r′3y−
)The
D
ββ
yyare
⋅⋅1βrβin
)′′′r ==)=′ −−−=total
D(rrD
−′ −
3 − p
2 +(B.16)
rderivatives
r+
rβ
ry
(B.10),
(B.16)
and
(B.18).
Putting
together
yields:
(((D
−2 r 2are
. and
The
derivatives
in (B.9),
(B.10),
(B.16)
(B.18).
Putting
themallall
together
2
r ) total
2
2
2
′
′
p
p
2
R
1
−
+
−
R
+
2
1
3
2
1
2
2
9
R
2
3
′
4 23R + 1
−+33p2′2′R2+−21R.1−21 p3′ − p 2′ + 2 R − 1
3 21 22 R
p13′3+2−1pp2p′3′R2′2−
(DDThe
yields:
r )′ = −′ 2 R +21R−+ 13 2 3′ +2
p32(B.16)
1(B.18).
+ 2+R
2 (B.10),
(total
)′derivatives
derivatives
Putting
together
yields:
D=total
=R
− 3 − 2are
+ 1(B.10),
. (B.18).
The
total
in
(B.9),
and
Putting
them
all
together
yields:
derivatives
(B.9),
(B.10),
(B.16)
and
(B.18).
Putting
them
allall
together
yields:
(−−are
)in+in
D9are
=
−(B.9),
−3R (B.16)
.them
(The
. −and
2
3
r−
rR
2
r )( D
4
)
=
−
+
2 33 2 R 2from
r
2 3 . 6(c)2yields
2Rusing
34 R
R 32 9*R2 99R
23R9 R3Lemma
R 3 (Dr )′ ≈ −1.3167 .
R 443at
4this
Evaluating
2 3result
R R3= R 2and
2
R
3
−11− 1
RR2+R
+11+ 1 33 32*2 2 11 221 p2p3′3′p−−3′ p−p2′2′p++2′ 22+RR2−R
′
Evaluating
this at
R−= R* and
using
result from
. 6(c) yields (Dr )′ ≈ −′1.3167 .
(Evaluating
+R
(DD(rrD)Evaluating
)′′r =)=′ −−=22−this
+*+and
... Lemma
′
* Lemma
′ −3−R
′using
this
at
R
=and
result
from
6(c)
yields
(
D )′ ≈.3167
−1yields
.3167
Evaluating
this
at
R
=
R
and
using
result
from
Lemma
at
=
R
result
from
Lemma
6(c)
yields
. .(. Dr ) ≈ −1.3167 .
(
)
D
*22using
2
9
R
2
3
2
9
R
3
2
9
R
3
r
R
2
R
3
4
R
2
R
3
4
3
R
2
R
3
4
3
Evaluating
and
using result from Lemma 6(c) yields (D≈rr −) 16(c)
≈ −1.3167
 ∂D3 this at R = R
 ∂D
3 
 ′ < 0 and 
′ < 0.
Claim 2: 
′ *** * ∂D
′
 ∂pD23 ′
p13 ∂′D ′′′

′


∂
D
 ∂Lemma

D 3result


D3  6(c)
∂

and
Evaluating
at
=RRand
from
Lemma
6(c)
yields
1.3167
Evaluating
at
using
Lemma
yields
.. ..
((DD(rrD))′′r ≈)≈′ −−≈11−..3167
Dthis
∂D
Evaluating
this
atatR
==R
R
and
using
from
Lemma
6(c)
yields
3167
Claim
2:  ∂this
and
. from
<
0result
and
using
from
6(c)
yields
Evaluating
this
R0=R
3R
3 ∂
0result
3using
∂D3<<3result
 D




Claim
and
.
<
0
<
0
 2:

Claim
2:
and
.
<
0
<
0
Claim
2:
and
.
<
0
3 
∂
p
∂
p
 < 0 . 6(b)and

1 ∂∂pp Lemma
 ∂pfrom
 < 0 and
2 ∂ p derivative
Claim
2:
p
∂
∂
p
We take
the
partial
proceed
similarly
as
in
previous
claim:


2
1




 1
 2 ∂′p′ 2 ′ 
 1 ∂′p′21 ′
 .derivative
Lemma 6(b) and proceed similarly as in previous claim:
D
D








∂
D
∂
D
∂
∂
∂
D
∂
D
D p2 Claim
=
−
β
+
y
sin
β
So,
3
3
3
3
3
3
We
take
the
partial
from
p22:
<derivative
<
partial
 p2partial
 the
Claim
2:
and
. 6(b) 6(b)
Claim
2:take
0and
and
Claim
2:
<
00<We
<
00<...0Lemma
We
the partial
derivative
from
and proceed
similarly
as in previous
claim:
take
derivative
6(b)
and
proceed
similarly
as in previous clai
We
take
the
from
Lemma
andfrom
proceed
similarly
as in
claim:
take
Lemma
6(b)
andLemma
proceed
similarly
asprevious
in previous
claim:
∂2p2partial
∂∂ppthe
∂∂pp1∂1 p1 from
D p2 = − βWe
β . So,derivative



p2 +y p2 2sin
D
=′p2−+β yp2 p2+sin
y p2D
sin
β
.
So,
=
−
β
+
y
sin
β
.
So,
D p2 D
= p−2 β=
β
.
So,
p
p
p
3 2 β . So,
12
2
− β 2 ′+ y+p2 sin
2 ) = −pβ
(We
DpWe
y ′pderivative
− y pfrom
⋅ from
β ′ .Lemma
p2take
ppartial
take
Lemma
6(b)
and
proceed
similarly
previous
claim:
We
the
6(b)
and
proceed
similarly
as
in
previous
claim:
take
thethe
partial
derivative
Lemma
6(b)
and
proceed
similarly
asas
in in
previous
claim:
2 partial
2
3derivative
1 2from
2
2
′
′p32So,
1
(
)
D−−
=+p+−′y+
βpp′py2sin
+sin
ySo,
− 1′ y p11
⋅ β′. 3
3.β
′
=
−
β
D
=
β
DppD
=
β
y
sin
β
.
So,
.
pβ
2
2
p
p
p
p
3
′
′
′
(D pp2 2−))′β==′2p22−−+2ββ ′pp22++y(′pD22 p−2yy)′pp22 =y−−p−22β⋅ yyβ′ppp2′2 .+⋅⋅ ββ ′ .. y ′p2 − y p2 ⋅ β ′ .
2
22 (D
p2 ()D22=
2
2
22
2 22 2 2 2
′′ ′ ′′ ′ 33 3′′ ′ 11 1
((DD(ppD22 ))p2 =)= −−=ββ−ppβ22 +p+2 +22 2yypp2y2 −p−2 2−2 yy2pp2y2 ⋅p⋅β2β⋅′′β.. ′ .
(
(( ((
((
)
)) )) (
))
)
2 R2 + 1 1 2 3p3′ 2− p 2′ 1+22 Rp3′−−1 p 2′ + 2 R − 1
+
.
+−
.
2R 3
2 49RR 3 2 32 9 R 2 R2 33
(Dr )′ = − 2 R(D+r 1)′−= −3
4R 3
′
*
Evaluating
at using
R = R*result
and using
result from
(Dr )′. ≈ −1.3167 .
Evaluating this
at R = Rthis
and
from Lemma
6(c)Lemma
yields (6(c)
Dr ) yields
≈ −1.3167
342
′
′
ECONOMIC
AND BUSINESS REVIEW
′
′
 ∂D3  ∂D3 
 ∂D3 
 ∂D3 




Claim
2:
and
.
<
0
<
0
 < 0and 
 < 0.
Claim 2: 

 ∂p 2  ∂p1 
 ∂p1 
 ∂p 2 
| VOL. 14 | No. 4 | 2012
We take
takethe
the
partial
derivative
from
Lemma
6(b)
and
proceed
similarly
as inasprevious
We
takederivative
the
partialfrom
derivative
from
6(b)
and
proceed
in
previous claim:
We
partial
Lemma
6(b)Lemma
and proceed
similarly
as similarly
in previous
claim:
− ββp2 .+So,
y p2 sin β .. So,
So,
D p2 claim:
= − β p2 D
+ py2 p=2 sin
(D )′ = − β(′D + )′23= y−′β ′− 12+ y23 ⋅yβ′ ′..− 12 y
p2
p2 p2
p2 p2
p2
p2
p2
⋅ β′.
Derivatives
needed
are in
(B.7),(B.17),
(B.11),and
(B.17),
andWe
(B.18).
Derivatives
neededare
are
(B.7),
(B.11),
(B.17),
and
(B.18).
We
get: We get:
Derivatives
needed
inin(B.7),
(B.11),
(B.18).
get:
Derivatives
needed
areare
in (B.7),
(B.11),
(B.17),
andand
(B.18).
WeWe
get:get:
Derivatives
needed
in (B.7),
(B.11),
(B.17),
(B.18).
−1
31 11 p(B.17),
′1− 1p ′ +p23′ R− (B.18).
−p12′ + 2 RWe
1′
31 1
′
Derivatives
get:
′
.
R1−. 1
1 =1−−are32in 3(B.7),
1 −2 1− 1(B.11),
1 21⋅ −p3′3 p−3′ p−2′2 ⋅p+and
D p2 ′ =′ −D pneeded
2 2+R2−
2 2
.
.
D
=
−
−
−
⋅
.
D
=
−
−
−
⋅
2
2
6
R
96R (B.17),
2 R 3 We get:
4 R2in93
p2 p2
R 2 2 2(B.11),
R 2 23 are
2
R
3
4needed
Derivatives
(B.7),
and
(B.18).
2
2
6
R
9
R
2
2
6
R
9
R
R
2
R
3
4
3
R
2
R
3
4
3
′ 2R − 1
p3′ − pare
1
3 1
1 1 needed
′
2 +in
(B.7),
(B.17),
and
needed
are
(B.17),
and
(B.18).
We get:
. (B.11),
D p2 Derivatives
=− 2
− *Derivatives
−in *(B.7),
⋅ (B.11),
′ We get:
′ (B.18).
2
′
′
p
p
2 Rfrom
−
+Lemma
− 1 6(c)
1 at3 R this
3*at
1
1
1
and
using
result
Lemma
6(c)(Dyields
R
=
R
)
≈ .−Due
0.8121
′ Evaluating
2
2
6
R
and
using
result
from
yields
to . Due to
Evaluating
this
=
R
)
9
R
3
2
′ ≈′ (−D0.p8121
R
2
R
3
4
*
*
p
2
2
.
D
=
−
−
−
⋅
using
result
from
Lemma
6(c)
yields
. Due
to to
Evaluating
this
atatR
(D(D) ≈) −≈0−.8121
and
using
result
from
Lemma
6(c)
yields
.. Due
Due
Evaluating
this
atR=R=R=RRand
and
using
result
from
Lemma
6(c)
yields
Evaluating
this
0.8121
p2
′R 2 3 12 9DR 2 3′ ′=21−6 R 11′ 1 −2 Rp33′ 3− 1p2′ +− 21R 1− 1⋅ p3′ − p 2′ p2+ p22 R − 1 .
4
.
D
= − itit must
⋅
′
to the
also
be
p2 )′be
thesymmetry
symmetry
(2D0−R.−.p8121
pit
2
the symmetry
(−DDalso
≈R
8121
2 must also
2) ≈ −0.8121
′−
2 be
p2
1 2 6from
and
6(c)
Evaluating
atalso
=must
R3* be
1 using
2 6 Ryields (2DRp2 )3 ≈ −0.8121 . Due to
9R
9≈
3R 2Lemma
thethe
symmetry
it2 this
must
be
(
)
−≈40result
symmetry
it must
(
)
D
0.8121
2
R
3
4RRalso
p
p
1
1
′
using result from Lemma 6(c) yields (D p2 ) ≈ −0.8121 . Due to
Evaluating this
at R =2 R* and
2
d pbe2 (dD22 pp12)′* ≈ −d02.8121
p1
′
p22 also
the symmetry
itdd 2must
using
result
Lemma
6(c)from
yields
Due
Evaluating
atEvaluating
R
and using
result
Lemma
this
at R
= R*from
(Dp22 )′6(c)
≈ −0yields
.8121 .(D
≈ −0.8121 . Due to
<d 0=dp12R
dp22 p3:
and
Claim
<2 this
0 and
<and
Claim
3:
Claim
3:
p 2 ) to
1′ p
1 0 .. 2 < 0 .
2
<
0
<
0
<
0
<
0
and
.
Claim
3: 3:
and
.
Claim
the symmetry
itdr
must
also
be
(
)
D
≈
−
0
.
8121
dr
dr
dr
p
2
2
2
2
1
2 dr
2 dr
′
ddr
p 2 it must
ddr
pbe
the symmetry
be (D p1 ) ≈ −0.8121
the symmetry
also
1 (Ditp must
)′ ≈ −also
0.8121
< 0this
<we
0 .11 would
andclaim
Claim
3: To
we
would
have
to solve
the(A.8)-(A.10)
system (A.8)-(A.10)
2 prove
2 2analytically
To prove
this
claim
analytically
have
to
solve
the
system
without without
2
dthis
p 2claim
ddrp1 wewe
To
prove
analytically
would
have
to to
solve
thethe
system
(A.8)-(A.10)
without
To
prove
claim
analytically
we
would
have
to solve
solve
the
system
(A.8)-(A.10)
without
drthis
To
prove
this
claim
analytically
would
have
system
(A.8)-(A.10)
without
′
′
<
0
<
0
′
2
2
and
.
Claim
3:
asserting
r
=
R
to
get
,
and
as
functions
of
r.
The
dependence
of
relevant
derivatives
p
p
p
2
2
asserting r = R to 2get dp1′′p, p ′2′ , and
as
functions
of dr. The
dependence of relevant derivatives
12
3
32 ,d
d
p
p
p
′
′
′
asserting
r =rR=rdr
to
get
,p2 1′p,p2Claim
and
as
functions
of
r.
The
dependence
of
relevant
derivatives
p1get
p
asserting
R
to
get
and
as
functions
of
r.
The
dependence
of
relevant
derivatives
p
2
1
p
1
dr
ʹ,
p
ʹ
and
p
ʹ
as
functions
of
r.
The
dependence
of
relevant
derivaasserting
=
R
to
, 2and
,2, 3:3we33would
<
0
<
0
and
.
<
<
0
To
prove
this
claim
analytically
have
to
solve
the
system
(A.8)-(A.10)
without
10
.
Claim
3:
2 is
r from Appendix
Bthat
shows
that
solution
out of
reach.
We as
proceed
as follows.
2
2 2 such analytical
on r tives
fromonon
Appendix
B
shows
such
analytical
solution
outsolution
of is
reach.
We
proceed
follows.
dr
dr
dras
from
Appendix
Bthat
shows
that
such
analytical
is
out
ofproceed
reach.
pro- *
on on
r To
from
Appendix
shows
such
analytical
solution
is dependence
out
of of
reach.
We
proceed
as2We
follows.
r from
B
analytical
solution
is out
reach.
We
as
′ analytically
′ that
* follows.
′such
asserting
rthe
= Appendix
Rrsystem
to get
, shows
and
functions
of
r.dr
The
of
relevant
derivatives
psystem
pfirst-order
pfirst-order
prove
this
claim
would
have
to
solve
the
system
(A.8)-(A.10)
without
1the
2 , of
3we
We
solve
theBof
the
conditions
(A.8)-(A.10)
Firms
1 and
R away
We solve
conditions
(A.8)-(A.10)
for
Firmsfor
1 and
2 located
Rlocated
* away
*
WeWe
solve
of
the
first-order
conditions
(A.8)-(A.10)
for
Firms
1(A.8)-(A.10)
and
2solve
located
Rresults
ceed
as
follows.
We
solve
the
of3’s
the
first-order
conditions
for
Firms
away
solve
first-order
conditions
(A.8)-(A.10)
for
Firms
1toand
2 located
Raway
* have
*we
′of
′Toand
prove
claim
analytically
would
the
system
(A.8)-(A.10)
without
′ this
asserting
rthe
=Tothe
Rsystem
tosystem
get
as
functions
of
r.
The
dependence
of
relevant
derivatives
pvarying
pthe
psystem
prove
this
analytically
we
would
have
to
solve
the
system
(A.8)-(A.10)
without
,
numerically.
The
are:
the
origin,
while
varying
Firm
location,
r,
around
R
1 ,claim
2 , that
33’s
on
r from
Appendix
B
shows
such
analytical
solution
is
out
of
reach.
We
proceed
as
follows.
,
numerically.
The
results
are:
from
the from
origin,
while
Firm
location,
r,
around
R
*
*
* varying
*
,Ras
numerically.
The
results
are:
from
the
origin,
while
varying
Firm
3’s
location,
r,paround
R
,Firm
numerically.
The
results
from
the
origin,
while
Firm
3’s
location,
r,
around
away
from
the
origin,
while
varying
3’s
location,
r,
around
R
,
1
and
2
located
R
*are:
′
′
′
asserting
r
=
R
to
get
,
and
functions
of
r.
The
dependence
of
relevant
derivatives
p
p
′
′
′
asserting
r
=
R
to
get
,
and
as
functions
of
r.
The
dependence
of
relevant
derivatives
p
p
p
1
2
1
2
3
3
Wer solve
systemBof*shows
the first-order
conditions
(A.8)-(A.10)
1 and 2 located R away
on
from the
Appendix
that
such
outfor
ofFirms
,) = 0.is
,reach. We proceed as follows.
R results
)==R
p*, 2−(*r0analytical
R *) −= 00..solution
005
, 991583
p1 ( r = R **p1−(*r0The
.=005
) −= 0p.005
( rare:
.=005
991583
numerically.
numerically.
The2 results
are:
fromsolve
varying
r,
R,*, for
,such
pthe
(prorigin,
R=system
0−.005
)the
=) p=
(pshows
r2Firm
−conditions
0−location,
.005
)B=
.991583
R−while
0of
.005
(=r R
=*3’s
R
0.005
)0=(A.8)-(A.10)
0around
.991583
away
We
and
located
R* reach.
r(=rfrom
Appendix
B22from
that
such
analytical
isFirms
out of1solution
reach.
We
proceed
as follows.
on
rfirst-order
Appendix
shows
thatsolution
analytical
is out
of
We proceed as follows.
1on
1the
*
*
*
*
* while
*of
,
numerically.
The
results
are:
from the
origin,
varying
Firm
3’s
location,
r,
around
R
We
solve
the
system
of
the
first-order
conditions
(A.8)-(A.10)
for
Firms
1
and 2 located R* away
away
We
solve
the
system
the
first-order
conditions
(A.8)-(A.10)
for
Firms
1
and
2
located
R
,
and
*
*
r .p=
R )==Rpp* )2( r(=
R −) 0=.005
0,.993237
and
=R
R *p)−
0=.R993237
1 (=
*0
*r=
2 ( r) =
,
ppp111(((prrr =
005
)
=
0
.
991583
*
*
,
and
=
R
)
=
p
(
r
=
R
)
=
0
.
993237
2
,
and
(
r
=
R
)
=
p
(
r
=
R
)
=
0
.
993237
2 2while
1
from varying
the origin,
while
varying
3’s location,
r, around The
R , numerically.
results are: The results are:
from
the origin,
Firm
3’s
location,Firm
r, around
R , numerically.
pp1 ((rr == RR***p−+(0r0..005
)*) =+= 0pp.*2005
( rr =)==RR*p* −+(0r0..005
)*) =+= 000...991583
.
=005
R
R
005
) = 0,..994882
005
994882
* )1 =
* 2 ** =
* p
2)(=
*
,
and
pp111((prr =(=r R
(
r
=
R
0
.
993237
*
*
. .− 0.,005) = 0.991583 ,
R=(R
+ 0+.005
) =) p=2 p()r12=((=rrpR
0==.994882
02 .005
005
994882
==2 (RR
00..005
p0)2.=
( r0=.991583
R
r+=0−+.R005
−)0=).)005
1 p
1* r = R − 0.005
, and representing
p1 ( rare
= Rinterested
) =are
p 2 (interested
rin= two
R * ) differences
= in
0.993237
We
two
differences
the first-order
derivatives
*
*
We
representing
the
first-order
derivatives
of p2 withof p2 with
* 0.*005
* thethe
p1 We
( rare
=are
R
+*=0.R
005
) =two
r=(*=rdifferences
) = 0.R
994882
* in
We
interested
differences
representing
first-order
derivatives
of of
p2 pwith
inp(two
first-order
derivatives
2 (p
2 with
, and
=*and
p representing
) = 0..993237
pat
rinterested
RR
)*R=+ R
0) .=993237
1 * +0.0025:
2 ( r =, and
respect
r at) =R*p−0.0025
+0.0025:
1*(R
2 rR
respect
to
r
−0.0025
and
*to
*
*
p1 (to
r =rtoat
Rr R
+ −0.0025
0.005
) = and
p 2and
(R
r =+0.0025:
R+0.0025:
+ 0.005) = 0.994882 .
respect
respect
at
R
−0.0025
R
We are interested
differences
representing
the first-order
derivatives of p2 with
*
* two
−*3
* in
*
.
r.=005
.005
=1.−−654
p33)2−=(3r×0=.10
Rthe
+ first-order
0..005) = 0.994882
)in
p** 2p−)(*1r=
)0×)=
(p*r)2*(−
=*r R
.−005
p=2R(R*)r −==+01R.0005
+representing
.10
005
994882
p ( rare
=pR1interested
p= (R+r 0=
Rtwo
0*(differences
654
We
derivatives
of p2 with
respect
−0.0025
and
R
+0.0025:
We
interested
two
differences
representing
the
first-order
derivatives
of p2 with
p 22 to
(pare
r2 r(=rat
R=*RR
)−
=
R
−
0
.
005
)
=
1
.
654
×
10
) p−22 *(prin
(
r
=
R
−
0
.
005
)
=
1
.
654
×
10
2
*
*
*
respect
to
r
at
R
−0.0025
and
R
+0.0025:
*
*
−
3
* are
−3 differences
We
interested
representing
the first-order
derivatives of p2 with
respect
to r=We
atR*pR
−0.0025
Rr two
+0.0025:
representing
derivatives
of p 2 with
.the first-order
−.−645
33 .−3 × 10
(*r0p.interested
=005
R )=+
0p*.005
)are
−Rp**differences
R )××in
=10
1two
+−
−and
(in
=
) (=
645
*r 1=
ppp222 (((prrr =(=
R=* )R2+
)R
...654
*1=
*) −)Rp−22 −
*=)1
. R.*+0.0025:
0+.005
(pr20to
Rr= at
)2 =
645
×10
10
20(.r
005
(=.and
r005
1.645
×and
10
2 rR
respect
R
−0.0025
respect
R
+0.0025:
*to r at R −0.0025
*
−
3
p 2 ( rsee
= Rthat
) −in
pvicinity
r = in
R of
−R
0*.*005
)of
= R1*.654
10
We
see
vicinity
the ×first-order
derivative
rivals’
prices
withtorespect
2 (that
*
−3
We
the
first-order
derivative
of rivals’ofprices
with
respect
r is to r is
*.
3with
p 2We
( rsee
=see
R
+in
0.R
005
)−−ppof
=R
R*=**−the
)first-order
=*.)005
1−.645
* vicinity
3 of of
We
that
vicinity
R
the
derivative
respect
to rtoisr is proof of
that
in
R
first-order
with
respect
2r(prof
(
r
R
p)order
(×r110
Rderivative
0.−005
)rivals’
= 1rivals’
.be
654prices
×prices
10 −completes
p
(
r
=
)
(
=
0
.=
654
×−10
decreasing
in r,the
hence,
derivative
must
negative.
This
completes
2 theorder
2=
decreasing
in2*r, hence,
derivative
This
the proofthe
of
2second
*second
−3must be negative.
. must
p 2 ( r =inR r,
005)the
− the
psecond
r =* Rorder
)order
= 1derivative
.645
× 10 must
decreasing
be be
negative.
This
completes
thethe
proof
of of
decreasing
in+hence,
r,0.hence,
derivative
negative.
This
completes
proof
2 (second
Claim
We
that3.in vicinity
of
R the first-order
derivative−3 of* rivals’ prices−with
respect to r is
Claim
3.see
*
3
*
*
− p 2×(10
r = R. ) = 1.645 × 10 .
Claim
3. 3. p 2 ( r = R + 0.005p)2−(*rp=2 (R
Claim
r =+R0.)005
= 1).645
decreasing
inthat
r, hence,
the second
derivative
mustinbeLemma
This
the
* order
We
inin
vicinity
ofthat
first-order
derivative
ofnegative.
rivals’
prices
with
respect
r to
is *r,ofis
* toproof
We see
see
vicinity
ofRRthe
first-order
derivative
of rivals’
prices
with
respect
and
p 2′ presented
(c)
arecompletes
positive
when
while
One
can
verify
′1′ presented
p1′′ and
ppthe
in
Lemma
positive
when
r=R
One
canthat
verify
that
* , while
* r=R
* 6 (c) are 6
2
*
′
′
′
pvicinity
and
p
presented
in
Lemma
6
(c)
are
positive
when
r=R
,
while
One
can
verify
thatthat
Claim
3.
p
and
p
presented
in
Lemma
6
(c)
are
positive
when
r=R
,
while
One
can
verify
decreasing
in
r,
hence,
the
second
order
derivative
must
be
negative.
This
completes
the
proof
of r is
We
see
that
in
of
R
the
first-order
derivative
of
rivals’
prices
with
respect
to
We
see
that
in
vicinity
of
R
the
first-order
derivative
of
rivals’
prices
with respect to r is
1 the
2 2
1
decreasing
in
r, hence,
second
order
derivative
must
be
negative.
This
completes
the
and
are
also
positive.
Using
the
three
claims
above, we
Lemma
6(b)
says
that
∂
D
∂
p
∂
D
∂
p
and
are
also
positive.
Using
the
three
claims
above,
we
Lemma
6(b)
says
that
∂
D
∂
p
∂
D
∂
p
3
1
3
2
3
1 and
3 order
2 are
Claim
3.
decreasing
in
r,
hence,
the
second
order
derivative
must
be
negative.
This
completes
the proof of
*
decreasing
in
r,
hence,
the
second
derivative
must
be
negative.
This
completes
the
proof
also
positive.
Using
the
three
claims
above,
we
Lemma
6(b)
says
that
∂Dp∂3′D∂and
∂D∂3 D∂3p 2∂pin2 Lemma
are also 6positive.
Using the
three claims
above, we of
Lemma
6(b)
saysthat
that
3 p1∂p1p ′and
(c) in
arethe
positive
when
, while
One
can
verify
proof
ofthat
Claim
3.
1
2 presented
see
every
single
summand
the big parentheses
second
row r=R
ofis(A.11)
is negative.
see that
every
single
summand
in the
biginparentheses
in the second
row
of (A.11)
negative.
Claim
3.
Claim
3.
*
seesee
that
every
single
summand
parentheses
in the
second
row
of when
(A.11)
is negative.
that
every
single
summand
in
in6 the
second
row
of
(A.11)
is, while
negative.
′ the
p ′ andinpthe
in Lemma
(c)
are
positive
r=R
One
can
verify
2 that
Lemma
6(b)
∂big
D3big
∂pparentheses
2 presented
2 are also positive. Using the three claims above, we
dthat
Π 3∂D31* ∂p1 and
d 22 Π2says
*
*
3
* positive
′
′
p
and
p
presented
in
Lemma
6
(c)
are
whentor=R*, while
One
can
verify
that
′
′
d
Π
<
0
d
Π
p
and
p
presented
in
Lemma
6
(c)
are
positive
when
r=R
, while
One
can
verify
that
Hence,
at
r
=
R
.
We
have
found
a
local
maximum
of
Firm
3’s
profits
One
can
verify
that
p
ʹ
and
p
ʹ
presented
in
Lemma
6
(c)
are
positive
when
r=R
,with
while
<
0
Hence,
at2summand
r∂=DR3 *.∂1 We
have
found
ofUsing
Firm
3’s
profits
with
respect
torespect
are also
positive.
the
three
claims
above,
we
Lemma
says
p*1 inand
D
∂p 2a local
1 maximum
3 that
1 the
2parentheses
see
that 6(b)
every
inmaximum
the2 second
row
of3’s
(A.11)
is
negative.
3 found
2 ∂big
2 3 single
<
0
<
0
Hence,
at
r
=
R
.
We
have
found
a
local
maximum
of
Firm
3’s
profits
with
respect
to
Hence,
at
r
=
R
.
We
have
a
local
of
Firm
profits
with
respect
to
dr
dr
26(b) saysLemma
2 2dr6(b)
andthat
are
also
Using
the
three
claims
above,
weclaims above, we
Lemma
that ∂D
and
also
positive.
Using
the three
∂psays
∂D3∂D∂3p2∂pare
∂positive.
D
p 2 are
Lemma
says
that
and
also
positive.
Using
the
three
claims
1 the
3 ∂row
see
summand
in6(b)
in
second
of
(A.11)
is Q.E.D.
negative.
3 the 1big parentheses
dr.dr
Π 3 single
Q.E.D.
*
r. that every
0
Hence,
at
r
=
R
.
We
have
found
a
local
maximum
of
Firm
3’s
profits
with
respect
to
2 2we<see
that
every
single
summand
in
the
big
parentheses
in
the
second
row
of
r. r. above,
Q.E.D.
Q.E.D.
see* summand
that every in
single
summand
in theinbig
second
row of (A.11) is negative.
see
the big
parentheses
theparentheses
second rowinofthe
(A.11)
is negative.
d that
Π every single
2
Hence, dr 2 3 < 0
found a local maximum
of Firm 3’s profits with respect to
2 at r = R . We have
d
Π
*
d
Π
* at r = R . We have found a local
3
r. (A.11)
Q.E.D.
*
3
dr is negative.
maximum
< 0going
Hence,
atfound
r =toRshow
. We
have
aoflocal
ofwith
Firmrespect
3’s profits
0Hence,
Hence,
at rare
=8:Rgoing
. We
local
maximum
Firm
3’s
to with respect to
2 of<
2 are
Lemma
We
that
thefound
expression
inmaximum
the profits
parentheses
Proof
of Proof
Lemma
8:
We
tohave
show
that athe
expression
in the
parentheses
in (10)
isin (10) is
r. Proof
Q.E.D.
drgoing
dr
of
Lemma
8:
We
are
going
to
show
that
the
expression
in
the
parentheses
in
(10)
is is
Proof
of
Lemma
8:
We
are
to
show
that
the
expression
in
the
parentheses
inQ.E.D.
(10)
of
Firm
3’s
profits
with
respect
to
r.
same of
number
of customers
by a move
zero.
it isr.that
obvious
∂φ = the
0 , since
samethe
number
customers
won by awon
move
zero. First,
it isFirst,
obvious
∂D that
∂φ =∂D0 , since
Q.E.D.
r.First,
thethe
same
number
of of
customers
won
by by
a move
zero.
First,
it isit obvious
thatthat
∂D∂33D∂3φ ∂=φ 0=3, 0since
, since
same
number
customers
won
aQ.E.D.
move
zero.
is obvious
Proof
ofdirection
Lemma
8:
Weone
arefrom
going
show
the
expression
in one.
the parentheses in (10) is
along
polar direction
onetoneighbor
lost
to the
other
along
polar
from
neighbor
is lostthat
to is
the
other
one.
along
polar
from
one
is lost
tothethe
other
one.
along
polar
direction
from
neighbor
is lost
tothe
the
other
one.of
same
number
customers
by
abymove
zero.
First,
itdirection
iswe
obvious
that
∂one
Dneighbor
∂φ d=to
Proof
of
Lemma
8:
We
are
going
expression
the becomes
parentheses
in (10)
is aby such a
Second,
we
claim
φ0 show
=, since
− dpthat
Clearly,
the in
neighbor
whowon
becomes
closer
3 dp
Second,
claim
dp
neighbor
who
closer
such
2 dφ . the
1 dφ = −1dp 2 dφ . Clearly,
Second,
we
claim
dp
dProof
φDWe
=φ ∂−=of
dp
d2, φsince
.φClearly,
thegoing
neighbor
who
becomes
closer
bymove
such
a ais
Second,
we
claim
dp
d
−
dp
d
.
Clearly,
the
neighbor
who
becomes
closer
by
such
1
2
1
the
Lemma
8:
We
are
to
show
that
the
expression
in
the
in (10) is
the
same
number
of
customers
won
by
a
zero.
First,
it
is
obvious
that
∂
φ
=
0
Proof
of
Lemma
8:
are
going
to
show
that
the
expression
in
parentheses
in
(10)
along
direction
oneaggressive
neighbor
isand
lost
toprice
the other
one.and
3 and reduces
move
becomes
more
reduces
its price
vice
the neighbor
that
isparentheses
now
move polar
becomes
more from
aggressive
its
and
vice
versa,
theversa,
neighbor
that is now
move
becomes
more
aggressive
and
reduces
its
price
and
vice
versa,
the
neighbor
that
is
now
move
becomes
more
aggressive
and
reduces
its
price
and
vice
versa,
the
neighbor
that
is
now
,
since
same
number
of
customers
won
by
a move
zero.
First,
it
is
obvious
that
∂
D
∂
φ
=
0
,
since
the
same
number
of
customers
won
by
a
move
zero.
First,
it
is
obvious
that
∂
D
∂
φ
=
0
Second,
weaway
claim
dp1one
dφ neighbor
= − dp
. Clearly,
the3 neighbor
who
becomes
closer
such
a
along
polar
direction
from
isφraises
lost
to
the
other
one.
3 d
farther
becomes
less
aggressive
and
raises
its
price.
The magnitude
ofderivatives
thebytwo
derivatives
farther
away
becomes
less
aggressive
its
price.
The
magnitude
of the two
2and
farther
away
becomes
less
aggressive
and
raises
its
price.
The
magnitude
of
the
two
derivatives
farther
away
becomes
less
aggressive
and
raises
its
price.
The
magnitude
of
the
two
derivatives
along
polar
direction
from
one
neighbor
is
lost
to
the
other
one.
along
polar
direction
from
one
neighbor
is
lost
to
the
other
one.
wemore
claim
dp1 dφ =and
− dpreduces
who
such a
moveSecond,
becomes
aggressive
its price the
andneighbor
vice versa,
thebecomes
neighborcloser
that isbynow
2 dφ . Clearly,
((
(
(
( )) ) ( )
)
)
( )
( )
One can verify that p1′ and p 2′ presented in Lemma 6 (c) are positive when r=R*, while
p1′ and
in Lemma 6 (c) are positive when r=R*, while
One
can verify
that that
Lemma
6(b) says
∂D3 p∂2′ ppresented
1 and ∂D3 ∂p 2 are also positive. Using the three claims above, we
also positive.
Using
the row
threeofclaims
we
Lemmasee
6(b)
thatsingle
∂D3 summand
∂p1 and ∂in
D3the
∂pbig
2 are
thatsays
every
parentheses
in the
second
(A.11)above,
is negative.
2
see that every single
summand
in
the
big
parentheses
in
the
second
row
of
(A.11)
is
negative.
d Π3
*
Hence,
d 2 Π 3 dr 2 < 0 at* r = R . We have found a local maximum of Firm 3’s profits with respect to
A. FELDIN |<THREE
UNIT DISK
MARKET:
PRODUCT
0 at rFIRMS
343
Hence,
= R . ON
WeA have
found
a localINTERMEDIATE
maximum of Firm
3’s DIFFERENTIATION
profits with respect to
2
r. dr
Q.E.D.
r.
Q.E.D.
ProofProof
of Lemma
8: We8:are
to show
that the
in the
parentheses
in in
(10)
is is
of Lemma
Wegoing
are going
to show
thatexpression
the expression
in the
parentheses
(10)
Proof
Lemma
We arethat
going
show
that
the
expression
in
the
parentheses
in
(10)
is
zero.
First,
ititisis8:
obvious
,
since
the
same
number
of
customers
won
by
a
,
since
the
same
number
of
customers
won
by
a
move
zero.of
First,
obvious
that
∂Dto
∂
φ
=
0
3
, since
same
number
of
customers
zero. move
First,
is obvious
thatfrom
∂D3 one
∂from
φ =neighbor
0one
along
polar
direction
neighbor
lost
to the
other
one. won by a move
alongit
polar
direction
isthe
lost
toisthe
other
one.
along polar Second,
directionwe
from
onedp
neighbor
is lost to the other one.
claim
1 dφ = − dp 2 dφ . Clearly, the neighbor who becomes closer by such a
Second,
we
dp1 aggressive
dφ = − dp 2 and
dφ .reduces
.Clearly,
Clearly,itsthe
the
neighbor
who
becomes
closer
such
aa
Second,
we claim
claimmore
neighbor
who
becomes
closerby
by
such
move becomes
price
and vice
versa,
the neighbor
that
is now
becomes
more
aggressive
and reduces
itsand
price
and
vicemagnitude
versa,
theofneighbor
that is
movemove
becomes
morebecomes
aggressive
reduces
its price
versa,
the
neighbor
that
is now
farther
away
lessand
aggressive
and
raises
its vice
price.
The
the two
derivatives
farther
away
becomes
aggressive
raises its
price.
Theitsmagnitude
the two derivatives
now
farther
awayless
becomes
less and
aggressive
and
raises
price. Theofmagnitude
of the two
derivatives is the same due to symmetry of the proposed configuration. The result follows.
Q.E.D.
isis
the
same
due
to
symmetry
of
the
proposed
configuration.
configuration.
The
The
result
follows.
follows.
344
ECONOMIC
AND BUSINESS
REVIEW
| VOL. 14 | No. 4 | 2012
isthe
thesame
samedue
dueto
tosymmetry
symmetryof
ofthe
theproposed
proposed
configuration.
Theresult
result
follows.
is
the
same
due
to
symmetry
of
the
proposed
configuration.
The
result
follows.
is
symmetry
isthe
thesame
samedue
dueto
toQ.E.D.
symmetryof
ofthe
theproposed
proposedconfiguration.
configuration.The
Theresult
resultfollows.
follows.
Q.E.D.
is
the
same
due
to
symmetry
of
the
proposed
configuration.
The
result
follows.
Q.E.D. of the proposed configuration. The result follows.
is the same due to symmetry
Q.E.D.
Q.E.D.
Q.E.D.
Q.E.D.
Q.E.D.
Appendix B: Derivatives
Appendix
Appendix
B:
Derivatives
AppendixB:
B:Derivatives
Derivatives
Appendix
B:
Derivatives
Appendix
B:Derivatives
Derivatives
Appendix
B:
We We
collect
all
the
derivatives
needed
in
preceding
Appendix
in
subsections
B.1 and
andB.2
B.2for
Appendix
B:
Derivatives
Appendix
B:
Derivatives
We
collect
collect
all
all
the
the
derivatives
derivatives
needed
needed
in
in
preceding
preceding
Appendix
Appendix
in
subsections
B.1
We collect
collect
allthree-firm
the derivatives
derivatives
needed
in preceding
preceding
Appendixin
insubsections
subsectionsB.1
B.1and
and B.2
B.2for
for
We
all
the
needed
in
Appendix
in
subsections
B.1
and
B.2
for
for
duopoly
and
oligopoly
markets,
respectively.
We
collect
all
the
derivatives
needed
in
preceding
Appendix
in
subsections
B.1
and
B.2
for
We
collect
all
the
derivatives
needed
in
preceding
Appendix
in
subsections
B.1
and
B.2
for
duopoly
duopoly
and
and
three-firm
oligopoly
oligopoly
markets,
markets,
respectively.
Wecollect
collect
allthe
thederivatives
derivatives
needed
inrespectively.
precedingAppendix
Appendixininsubsections
subsectionsB.1
B.1and
andB.2
B.2for
for
duopoly
andthree-firm
three-firm
oligopolyneeded
markets,
respectively.
We
all
in
preceding
duopoly
and
three-firm
oligopoly
markets,
respectively.
duopoly
duopolyand
andthree-firm
three-firmoligopoly
oligopolymarkets,
markets,respectively.
respectively.
duopoly
and
three-firm
oligopoly
markets,
respectively.
duopoly
and
three-firm
oligopoly
markets,
respectively.
B.1
B.1
Two
Two
firms
firms
B.1
Two
firms
B.1 Two
Two firms
firms
B.1
B.1Two
Twofirms
firms
B.1
B.1
Two
firms
B.1
Two
firms
In
In
equilibrium
equilibrium
rr===R,
R,
=π,
and
and
αα
π/2.
In equilibrium
equilibrium
R,φφφπ=π,
=π,
and
α===
=π/2.
π/2.
In
R,
and
π/2.
In equilibrium
r =rrrrrR,
φR,
,=π,
and
αα
=αα
Inequilibrium
equilibrium
=R,
R,=φφφ
φ=π,
=π,
and
απ
π/2.
In
===
and
===/2.
π/2.
In
equilibrium
=π,
and
π/2.
In equilibrium r = R, φ =π, and 1α11 = π/2.
111
111
===−−− 11
From
From
(A.2):
and
(B.1)
(B.1)
1
1 and
αpp1ppp111 ====−−−−αα
αppp2pp222 ====−−−−22((rr++RR11)1
=−−−−zzzzppp2pp222 ====−−−−2211R11R
From(A.2):
(A.2):αα
and zzzzpp1ppp111 ===
(B.1)
1
α
α
=
−
From
(A.2):
and
(B.1)
⋅
⋅
α
α
)
sin
sin
4
4
R
R
1R) ⋅ sin α ===−−−14 R and
α p111 ===−−−α
α p 2 ==−−
From
(B.1)
and
(B.1)
From
(A.2):α
From(A.2):
(A.2):
and zzzppp1p1111 ===−−−zzzpppp22222 ===−−−12 R
(B.1)
From
(A.2):
and
(B.1)
⋅
α
R
)
sin
4
R
2
R
ααp1pppp1111=
−ααp2pppp222222==−− 2222((((rrrr+++
=
−
p
p
=
−
=
−
z
z
From
(A.2):
and
(B.1)
222RRR
+RRR)))⋅⋅⋅sin
sinα
p1 p11
p2 p22
αα 444R4RRR
sin
⋅
α
2(2r(r¸+¸2+
R
)
sin
2
R
111
sin
111
111
¸222sin
sinαα
α
1
sin
α
and
From
From
(B.1):
12222
sin
1 2222 ==== 111
12222 and
α 2222 ====
222sin
111
=−−−
−zzzz′p′p′′p2pp222 ===
=44R1R
and zzzz′p′p′1′ppp111 ===
From(B.1):
(B.1):αα
α′p′p′1p′pp111 ====−−−−αα
α′p′p′p′2pp222 ====44((rr++¸¸¸¸R
222 α
1R
sin
α
2
and
From
(B.1):
α
α
⋅
⋅
α
α
+
+
R
)
)
sin
sin
2
2
(
(
r
r
R
R
)
)
8
8
R
R
′
′
′
′
′
′
′
′
z
−
z
From
(B.1):
22α⋅ sin 222 α == 2( r1 + R ) 222 == 1
22 and
α
−
α
z
=
−
z
=
and
From
(B.1):
α
=
−
α
=
1
11 =
22 =
p
p
¸
2
sin
11 =
22 =
p
p
p
p
p
p
222
2
2
1
2
1
2
4
+
4
(
r
R
)
8
R
p
p
p
p
p
p
p
p
′
′
22 ⋅ sin22 α = 2( r + R )22 = 8 R22 and
1 1 = − z 2 2 = 4 R22
1 = −α ′2 2 =
z
and
From
(B.1):
From
(B.1):αα′p ′pp111=
+
4
(
r
R
)
′
′
′
p
p
p
z
=
−
z
=
and
From
(B.1):
−
α
=
=
=
2
2
2
2
44RR22 (B.2)
44((rr++RR))2 ⋅⋅⋅sin
α 22((rr++RR))2 88RR2
sin
p1 p11
p2 p22
p2 p22
2 2α
1
sin
44RR2
αα 2(2r(r++RR) 2) 88RR2
4(4r(r++RR) 2) ⋅ sin
(B.2)
(B.2)
(B.2)
(B.2)
(B.2)
(B.2)
(B.2)
(B.2)
ppp2′2′2′ −−−ppp1′11′′ +++222RRR
ppp2′2′2′ −−−ppp1′11′′ +++222rrr ppp2′2′2′ −−−ppp1′11′′ +++222RRR
From
(A.1):ααα′′′=== ppp′2′222′ −−− ppp′1111′′ +++222rrr === ppp′2′222′ −−− ppp′1111′′ +++222RR
and
(B.3)
From
From
(A.1):
and
and
(B.3)
(B.3)
R
−ppp′′1111′′ ++++2222RR
R
From(A.1):
(A.1):
and zzz′′′=== pp′pp2′′2′222′ −−−
(B.3)
α
From
(A.1):
and
(B.3)
′1 22RRR and
−pRRR
+22rαα
p′)1))′1⋅⋅+⋅sin
r ==
sin
11R+ 2 R
=p2 22 − p221p22′R1R
α′′′′===
=p2222p2′((((2r2′2r2rr−+++
=pp2′ 22′22−−p441p44′R1R
11R++
From(A.1):
(A.1): α
and zzzz′′′′===
(B.3)
From
(B.3)
α
+
sin
1 11)
α
=
From
(A.1):
and
(B.3)
R
α
⋅
R
sin
R
From (A.1): α ′ = 22((rr++RR))⋅⋅sin
and z ′ =
(B.3)
α=
sinα
22RR
44RR
⋅ sin
22RR
αα
2(2r(r++RR) ⋅)sin
44RR
B.2
Three
firms
B.2
B.2
Three
Three
firms
firms
B.2 Three
Three firms
firms
B.2
B.2
B.2Three
Threefirms
firms
B.2
Three
firms
B.2
Three
firms
111
(B.4)
(B.4)
From
From
(A.4):
(A.4):
(B.4)
From
(A.4):
=−−−
−xxxxppp2pp222 ===
=−−−
− 1111 ,,,, xxxxppp3pp333 ===
=000,,, xxxrrr ===0000,,,,
(B.4)
From (A.4):
(A.4): xxxxpp1ppp111 ===
(B.4)
From
=000,,,
(B.4)
From(A.4):
(A.4): xxxppp11111 ===−−−xxxppp22222 ===−−−222R
1RR 333,,, xxxppp33333 ===0000,,,, xxxxrrrrr ===
(B.4)
From
(B.4)
From
(A.4):
3, x pp = 0, xr rr= 0,
(B.4)
From (A.4): x p1 pp11= − x p2pp22= − 222RR
RR 333 p3 33
2p2R
pp2′2′2′ −3−−ppp1′11′′
(B.5)
(B.5)
From
From
(A.4)
and
(B.4):
−−−− pppp1′′11111′′ ,,,, xxxx′p′p′1′pp111 ===
=xxxx′p′p′′p2pp222 ===
=xxx′p′p′′p3pp333 ===
=000,,,,,
(B.5)
From(A.4)
(A.4)
and
(B.4):
From
(A.4)and
and(B.4):
(B.4):xxxxx′′′′′′=====ppp′pp22′22′2′2222′R
(B.5)
(B.5)
From
(A.4)
and
(B.4):
′
′
′
R
3
3
′
′
11 , , xx′p′pp
x
,
(B.5)
From
(A.4)
and
(B.4):
p
−
x
=
=
x
x
0
,
(B.5)
From
(A.4)
and
(B.4):
ppp222 == xx
ppp333 ==00
p1111 =
2
p
p
p
2
1
2
R
3
′
′
′
′
2
3
x
=
,
x
=
x
=
x
=
0
1
2
3
(B.5)
From(A.4)
(A.4)and
and(B.4):
(B.4):x ′ = 22RR 33 , x ′ pp1= x ′ pp2= x ′ pp3= 0 , ,
(B.5)
From
2RR 33
p1 1
p2 2
p3 3
222R
3 φφ
−−−RRR 333+++
22rrrsin
sin
111
111
φ
sin
yyypp1p1 === −−−RR
From
From
(A.5):
(B.6)
(B.6)
sin
11 ,,,
R 3333++++2222rrrrsin
sinφφφφ ====22((22rr1111++RR)) ====6611R
From(A.5):
(A.5):
(B.6)
11 =
p
From
(A.5):
(B.6)
−
R
sin
y
From
(A.5):
(B.6)
φφφ−φ−−RRR))) === 2(21r + R ) ===16RR,,,,,
RRRR 333(3((222+rrrcos
From
(B.6)
From(A.5):
(A.5): yyypppp1p1111 ===22−
(B.6)
2cos
r sin
2
cos
From
(A.5):
(B.6)
φ
−
2
R
3
(
2
r
cos
R
)
2
(
2
r
R
)
6
R
+
p
= 22((22rr++RR))= 66RR,
From (A.5): y p1 p11= 22RR 33((22rrcos
(B.6)
φ
−
cos
R
)
φ
−
R
)
cosφφ−−RR) ) 2(22(2r r++RR) ) 66RR
22RR 33(2(2r rcos
sin
111
111
−−−RRR 333−−−222rrrsin
sinφφφ ==
R
3
sin
−
−
yyyppp22 === −−RR 33−−22
=== 11
(B.7)
(B.7)
From
From
(A.5):
2rrrrsin
sinφφφ
1
1 ,,,
φ ==
111
(B.7)
From(A.5):
(A.5):
p22 = 22R
R
3
2
sin
−
−
y
=
(B.7)
From
(A.5):
From
(A.5):
(B.7)
2
2
(
(
2
2
r
r
R
R
)
)
6
6
+
+
p
RRR 333(3((222−rrrcos
φφφ−φ−−RRR))) === 2(21r + R) ===161RRR,,,,,
(B.7)
From
(B.7)
From(A.5):
(A.5): yyypppp22222 === −
2cos
r sin
cos
(B.7)
From
(A.5):
2
(
2
r
R
)
6
R
+
y p2pp22= 2222RR
=
=
,
(B.7)
From
(A.5):
3
(
2
r
cos
R
)
φ
−
222(((222rrr+++RRR))) 666RRR
3
(
2
r
cos
R
)
φ
−
R
3
(
2
r
cos
R
)
φ
−
cosφφ−−RR) ) 2(2r + R) 6 R
22RR 33(2(2r rcos
222RRR 333
11
11
2
R
=
=
===−−− 1111 ===−−− 1111 ,,,
y
y
From
From
(A.5):
(A.5):
(B.8)
(B.8)
222RRR 3333
From (A.5):
(A.5): yyppp3pp333 ==
(B.8)
1
1RR,,,,
=
−
=
−
From
(B.8)
+
+
2
2
r
r
R
R
3
3
From
(A.5):
(B.8)
2
2
R
R
3
3
(
(
2
r
r
cos
cos
R
R
)
)
φ
φ
−
−
From
(B.8)
From(A.5):
(A.5): yyypppp33333 === 2 R 32(R
(B.8)
3 φ − R) ===−−− 21r + R ===−−−13R
2 r cos
From
(A.5):
,
(B.8)
+
2
r
R
3
R
=
−
=
−
From (A.5): y p3pp33= 222RR
,
(B.8)
3
(
2
r
cos
R
)
φ
−
+
2
r
R
3
R
+
2
r
R
3
R
3
(
2
r
cos
R
)
φ
−
R
3
(
2
r
cos
R
)
φ
−
+
2
r
R
3
R
cosφφ−−RR) ) 2r + R
3R
22RR 33(2(2r rcos
222rrr
222
RRR 333⋅⋅⋅444rrr
=
=
=
=
−
−
=
=
−
−
y
y
(B.9)
(B.9)
From
From
(A.5):
(A.5):
3
⋅
4
2
R
r
r
RRR 333⋅⋅⋅444rrr
rrrRR =
2+
=
=−
−22rr22+
=−
−322322,,,,
(B.9)
From (A.5):
(A.5): yyrrrrr =
=
(B.9)
From
φ
φ
2
2
3
3
(
(
2
2
cos
cos
−
−
)
)
R
R
r
r
R
R
=
=
−
=
−
y
(B.9)
From
(A.5):
=
=
−
=
−
y
,,,,
(B.9)
From
(A.5):
From
(A.5):
(B.9)
3
⋅
4
2
2
R
r
r
r
2
3
r
+
R
r rr = 2 R 3 ( 2 r cos φ − R ) = − 2
=
−
y
(B.9)
From
(A.5):
3
r
+
R
(B.9)
From (A.5): y r rr= 222RR
φφ−−−RR
33(((222rrrcos
cos
))= − 22rr++RR= − 33,
φ
3
)
cos
R
R
2
3
r
+
R
φ
2
3
(
2
cos
−
)
R
r
R
2r + R
3
2 R22((232r(r2++r R
cos
− R)
RR)))−−φ
−4444rrrr == −− 2222 ,,
(B.10)
(B.10)
From
From
(B.9):
(B.9):
yyyr′r′r′ ===−−−2222((((2222rrrr+++
R
))−−
R
)
4
r
2
R
4
r
2
+
−
,
(B.10)
From
(B.9):
=
−
2
2
′
)−))−)4222224r r ===−−−9299R
2RR,,,
(B.10)
From
(B.9):
(B.10)
From
yyyyr′′rrr′rr ====−−−−2(22((2r((22r2+rrr++
(B.10)
From(B.9):
(B.9):
R++R)RRR
2
2
From
(B.9):
,
,
(B.10)
From
(B.9):
=
−
9
R
R
(B.10)(B.10)
From (B.9): y r′ rr= −
((((2222rrrr++++RR
R)2)))22 = − 999RRR,
9R
( 2r + R )
R 33 ⋅⋅ 22cos
cosφφ
1
R
−−11
= − 1 2 ,,,
=
From (B.6)
(B.6)
and
(B.7):
From
(B.6) and
and(B.7):
(B.7): yy′′p1 == yy′′p2 ==
2 = −
From
2 =
p1
p2
2
(
2
r
R
)
9
R
+
2
2
R
3
(
2
r
cos
R
)
φ
−
( 2r + R )
9R 2
2 R 3 (2r cos φ − R )
(B.11)
(B.11)
(B.11)
cosφφ
22cos
22
22
From (B.8):
(B.8):
From
From
(B.8): yy′p′p33 == −− (2r cos φ − R)22 == (2r + R)22 == 9 R22 ,,,
(2r cos φ − R )
( 2r + R )
9R
(B.12)
(B.12)
(B.12)
R 3 ( 2 p ′ − p ′ − p ′ + 4r ) 2 p ′ − p ′ − p ′ + 4r
2 p′ − p′ − p′ + 4R
(B.13)
From (A.5):
(A.5): yy′′ == R 3 (2 p3′3 − p1′1 − p 2′2 + 4r ) == 2 p3′3 − p1′1 − p 2′2 + 4r == −− 2 p3′3 − p1′1 − p 2′2 + 4 R ,, (B.13)
From
cosφφ −− R
R))
R
R 33((22rr cos
cosφφ −− R
R))
22((22rr cos
66R
22R
11
ββ pp22 == −−ββ pp33 == −− (2r sin φ − R 3 ) cos β − (2r cos φ − R) sin β
RRR 333⋅ 2⋅⋅2cos
1 1
φ φφ
− 1−−11
2cos
cos
= −= −− 2 1, 2 ,, (B.11)
==
From (B.6)
(B.6)
(B.11)
From
(B.6) and
and (B.7):
(B.7): yyy′p′p1′pp111 ===yyy′p′2p′pp222===
(B.11)
From
and
(B.7):
2 222 =
2 222 =
22
(
2
r
R
)
9
R
+
cos
222RRRR33(32(3(r22⋅rcos
φ φφ−φ −R
++1RR))= − 199RR
r2⋅cos
cos
− )RR))= ((22−rr1−
′
′
y
y
=
=
,
(B.11)
From
(B.6)
and
(B.7):
R
3
2
cos
1
φ
p1 A UNIT
A. FELDIN
| THREE
FIRMS y
ON
DISK MARKET: INTERMEDIATE
′ = yp2′p2 =
=9−R 2 2 , (B.11) 345
From
(B.6)
and (B.7):
(2r PRODUCT
+ R) 2 2 DIFFERENTIATION
2 R 3 (22r cos φ − R2) 2 2 =
2p221cos
cosφφφ
2r cos φ − R
2)
(
2
r
R
)
9 R (B.12)
+
2
R
3
(
2
cos
2
2
From (B.8):
(B.8):
===
= == 2 ,2 ,,
(B.12)
From
(B.8): yyy′′pp′pp3333 ===−−−
(B.12)
From
φφφ−
(((222rrrcos
)R2))222 9 R
φ−−RRR) )2)222 (2((2r2rr+2++RR
2cos
cos
299RR22
cos
,
(B.12)
From (B.8): y ′p3 = −
2 cos φ 2 =
2 2 =
2
(B.12)
From (B.8): y ′p3 = −(2r cos φ − R) 2 =(2r + R) 2 =9 R 2 2 ,
RRR 3(332(((2r22pcos
2 2)p23′pp−′33′ −−p91′R
+R
p3′33′ p−−1′ −pp1′11′p−−2′ +
R , (B.13)
+ 44, R
pp3′3′33′−−φ−p−
pp1′1′11′−R−−)ppp2′ 2′2′2+++4(24r4)rrr))=
pp−1′11′ −p−2′pp+2′2′2 4++r44=rr −=2−p223′ p−
pp2′2′24+R
3
′
y
=
(B.13)
(A.5):
′
y
=
=
From
(A.5):
From(A.5):
(A.5): y ′ =
, (B.13)
= ′
=− ′
, (B.13)
From
′cos
′
′
′
′
′
φ
2
(
r
cos
R
)
6
R
−
R 2232RR
(R2 p3333′((2(−22rrrpcos
p
4
r
)
2
p
p
p
4
r
2
p
p
p
4
R
−
+
−
−
+
−
−
+
φ
2
(
2
r
cos
R
)
6
R
−
φ
R
)
−
′ φφ2−−′ RR))
(B.13)
From (A.5): y ′ = R 3 (2 p3′ − p1cos
23p23′(2−r1cos
p1′ −φ2p−2′ R+)4=r − 23 p3′ −1 p16′ R
−2 p 2′ + 4, R
1 − p 2 + 4r )=
= 2(2r cos φ − R)
=−
, (B.13)
From (A.5): y ′ = 2 R 3 (2r cos φ − R)
6R
12(112r cos φ − R)
6R
φ
2
R
3
(
2
r
cos
R
)
−
βββppp22 ===−−−βββp3pp3 ===−−−
p22
p33
β1ββ− −(−2((r22cos
φ−
β ββ
((2(22rrrsin
sin
sinφφφ−−−RRR 3 3)3cos
cos
cos
sin
φφ R
−−)RR
sin
))cos
))sin
rr cos
1
From
(A.6):
, ,
(B.14)
From(A.6):
(A.6): ββ p2 == −−ββp3 ==−−
(B.14)
From
(A.6):
(B.14)
From
(B.14)
(2r sin φ111− R 3 ) cos β − (21r11cos φ − R) sin
1 11β ,,
p2
p3
===−−− (23 r sin φ − R 33 )=cos
− −−β − (2r cos
= φ−
−−−R) sin ,β
From (A.6):
(B.14)
=
=
=
=
From (A.6):
,
(B.14)
R 333+ (21r + R) 333
3 (r1+ R)
2 R1 3
= − RR23222 ++ ((221rr ++ RR))23 222= − 33((rr1++ RR=)) − 22RR1 33
= −R 2 3+ (2r + R) 2 3 = − 3 (r + R) = 2−R 3
β ββ 3 (r + R )
2r − 1
2R − 1
+ φRφφ)cos
R 2222r+
rcos
cos
cos
rr−−(−2222cos
cos
2
= == 23R 322rr −−11 3 = == 22RR ,−−11,,,
(A.6):
From
(A.6):
From(A.6):
(A.6): βββrrrr ===
From
3332(r
333 2 R −
2
r
2
cos
cos
−
φ
β
1
1 3
−
R
2
r
R
)
+
+
(((222rrrsin
R
3
)
cos
(
2
r
cos
R
)
sin
φ
−
β
−
φ
−
β
R
(
2
r
R
)
+
+
sin
R
3
)
cos
(
2
r
cos
R
)
sin
23R
φ
−
β
−
φ
−
β
sin φ − R 23r)−cos
2r cos
β −φ(cos
From (A.6): β =
2 cos
β φ − R) sin β= R23 222 +2(r2r−+1 R23) =222 22R
R ,−31
= 2 +3(2r + R) 2 3 2=R (B.15)
,
From (A.6): βrr = (2r sin φ − R 3 ) cos β − (2r cos φ − R) sin β R
3
(2r sin φ − R 3 ) cos β − (2r cos φ − R) sin β R 2 + (2r + R) 2
2 R(B.15)
3(B.15)
(B.15)
(B.15)
′
(B.15)
 222rrr−−−111 ′′ 222 333⋅ 2⋅⋅2R2RR−−−
(2((R
− 1−−)11))3 33 2 R22+RR1++11
2
R
2
R
= ==
, ,,,
(B.16)
(B.15):
(B.16)
From
(B.15):
(B.16)
From
From(B.15):
(B.15): βββr′r′rr′ === 2r − 1 ′′===2 3 ⋅ 2 R12
(B.16)
2 222
R(2RR − 1) 3 24 R4+R31 33
−12
33((rr++RR) )
(B.16)
From (B.15): β ′ =  32(rr −+1R) = 2 3 ⋅ 2 R12−R2(2 R − 1) 3= 42R
R +, 1
=4 R 3 ,
(B.16)
From (B.15): βrr′ = 3 (r + R)   =
12 R 2
12
R
3
(
r
R
)
4
R
3
+
′
φφφcos
βββ+++RRR3 3sin
β ββ
φ sin
β −ββ(−2−r((2cos
φ −φφR−−) cos
β ⋅ βββ′⋅⋅ ββ′′
⋅ β⋅⋅ ββ−′′ −
222sin
2−cos

φ
r
R
sin
cos
3
sin
2
cos
sin
2
cos
)
cos
φ
r
R
sin
cos
sin
2
cos
sin
cos
)
cos
βββ′p′′pp2 2 ===βββ′p′3′pp3 ===
p22
p33
φ(2sin
β φ− −(2Rr )cos
φβ−2R )222cos β ⋅ β ′
⋅ 3β)′cos
− 2βcos
2 sin φ cos β(+2rRsin3φsin
RβRR
sin
R
sin3φφ−sin
cos
cos
sinφ2ββ− R ) cos β ⋅, β ′ ,
β−β
φφ− −−
rr cos
R
−−
−−φr((22cos
((2+2rrRsin
33β))′cos
))cos
sin
′p2 = β ′p3 = 2 sin φ cos β
β
β
β
r
⋅
−
2
cos
sin
(
2
(B.14):
From
(B.14):
From (B.14): β ′p = β ′p =
,
2
2
3
φ
β
φ
β
r
R
r
R
−
−
−
(
2
sin
3
)
cos
(
2
cos
)
sin
333
1r11sin φ − R 3 ) cos β − (2r cos φ − R) sin
From
(B.14):
,,
From (B.14):
β
(
2
=
=
=
=
=
=
From (B.14):
,
33(r(r++
3RR) 2)222 4 R 12 222 3 33
= 3(r +3R)2 = 44RR
1
=3(r + R) 2 =4 R 2 2 3
(B.17)
(B.17)
3(r + R)
(B.17)
4R 3
(B.17)
(B.17)
From (A.6):
(A.6):
From
(A.6):
From
(B.17)
From (A.6): p ′′′ − p ′′′ + 2r − 2 cos φ cos β
′
′
′
′
p
−
p
+
2
r
−
1
p
−
p
+
2
R
−
1
′
′
′
′
′
′
′
′
φ
β
p
−
p
+
2
r
−
2
cos
cos
p
−
p
+
2
r
−
1
p
−
p
+
2
R
−
1
3
2
3
2
3
2
φ
β
p
−
p
+
2
r
−
2
cos
cos
p
−
p
+
2
r
−
1
p
−
p
+
2
R
−
1
From
(A.6):
From
222
222
222
= (A.6): 333
= == 333
= == 333
. ..
ββ′′ ==
−R p332′3)))+cos
2rββ
−β−2−−(cos
cos
r)RR
−))1 p3′ − 2pR2′ 22+RR
φφφp−−−3′ R
φφφ−β−−RRR
β ββ p3′ −3p(332′r ((+rr 2R
((222rrrsin
sin
2(22rrrφcos
) sin
32 R33− 1
sin
cos
cos
)
sin
+
(
R
cos
(
cos
)
sin
+
β′ =
= p ′ − p ′ + 2r −=1 p3′ − p 2′ + 2 R −
.1
p ′ − p ′ + 2r − 2 cos φ cos β
..
β ′ = (2r sin φ −3 R 32) cos β − (2r cos φ − R) sin β = 3 3 (r +2 R)
= 2R 3
(B.18)
(B.18)
(2r sin φ − R 3 ) cos β − (2r cos φ − R) sin β
3 (r + R)
2R 3
(B.18)
(B.18)
(B.18)
(B.18)
((
((
) )
) )