# Price Discrimination in Dynamic Cournot Competition

Authors: Wei-li Zhang; Qi-Qing Song; Yi-Rong Jiang
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

This paper introduces a new Cournot duopoly game and gives an applied study for price discrimination in a market by dynamic methods. One of two oligopolies has two different prices for a homogeneous product, while the other charges one kind of price. It is found that there is only one stable equilibrium for the discrete dynamic system, and a corresponding stable condition is given. Using a discriminative price is not always beneficial to a firm in equilibrium. If both oligopolies carry out price discrimination, the market’s average price is lower than when only one oligopoly does it. The results are verified by numerical simulations.


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## Body

## 1. Introduction

Cournot duopoly assumes that there are two oligopolists who compete in a market by offering a homogeneous commodity. This was introduced by Cournot [1] and has become a basic model in modern microeconomics and game theory. Cournot duopoly forms a dynamic as a consequence of oligopolists adjusting their strategies based on the result of the last period.Since Cournot dynamic competition is related to the behavior of consumers, expectations of manufacturer demands, the number of oligopolists, and so on, there are extensive works from different points of view to study Cournot competition. For instance, see the complexity of solutions for Cournot duopoly [2–5] and Cournot games with three oligopolists [6], and see Cournot competition with incomplete information [7–9]. In [10, 11], Guirao and Rubio introduced Cournot-like models by generalizing Cournot duopoly to n players. In another study, stability results were shown for the model with three and four players who competed with their neighbors [12]. For a dynamic duopoly game with heterogeneous expectations for players, stabilities were shown by Agiza and Elsadany [13, 14], Dubiel-Teleszynski [15], and Fanti [16]. Considering bounded rationality, Bischi and Naimzada [17] studied a dynamic duopoly game model where firms adjusted their outputs along the best response direction. Meanwhile, Elsadany gave the necessary and sufficient conditions of the stable region of the Cournot–Nash equilibrium based on relative profit maximization [18].In a static situation, Hazledine extended the homogeneous product Cournot–Nash oligopoly model to allow for price discrimination [19], and further results were shown in [20]. Price discrimination is a widely observed phenomenon in a market. For the traditional classification of price discrimination, see the study by Pigou [21]. Typically, when the good or service does have a price tag, there are consumers who pay this amount and there are other consumers who pay less. For instance, passengers taking the same airline may have different ticket fees, and it is common that a person who books a ticket early will get a lower price. Hotels also charge different prices according to reservation times. Roughly, price discrimination implies that a service is sold at different prices to different groups according to their willingness to pay; i.e., to some extent, price discrimination is driven by heterogeneous consumers or consumers with different demand elasticities. In [19, 20], each airline i(i=1,2,…,n) segments the customers as K groups and sells qik tickets to the group k(≤K). The tickets for the fare class k are sold at price Pk=1-Q1-Q2-⋯-Qk, where Qk is the number of seats sold at price Pk. Following the same demand function, Kutlu gave an analysis of price discrimination in Stackelberg competition with two players [22], in which the leader only uses one kind of price and the follower executes K different prices.In addition, in [19], the average price is independent of the extent of price discrimination. It was generalized to a larger class of demand functions as Pk=α-β(Q1+Q2+⋯+Qk)r by Kutlu [23] and to a situation with asymmetric costs by Mukherjee [24] and Bak and Klecz-Simon [25].Noting that the current dynamic games have no consideration of price discrimination, in the present paper, we examine the equilibrium in a discrete dynamic Cournot game with price discrimination, where players adjust their strategies through local estimates of marginal profits. All equilibrium points of the dynamic system are unstable except for one Nash equilibrium. The sufficient condition to guarantee the local stability of the equilibrium is given. It is shown that, compared with a firm with only one kind of price, a firm does not necessarily benefit more from having two different prices. Numerical examples are presented.The rest of this article is organized as follows. Section2 presents the dynamic model with price discrimination. Section 3 investigates the property of all equilibrium points. Section 4 shows numerical simulation results to verify the main results in Section 3. Section 5 presents our conclusion and discussion.


## 2. Dynamic Cournot Model with Price Discrimination

This section aims to establish a dynamic Cournot model with price discrimination. Suppose that there are two firms in a market and they sell a homogeneous product with a constant marginal costc∈(0,1). Firm 1 divides the consumers into two groups, group 1 and group 2, corresponding to paying high prices and low prices for the product, respectively. Firm 2 thinks that there is only one group and sells the good with a low price. The price function (inverse demand function) of the market is given as below:(1)pk=1-q11,k=1;1-q11-∑i=12qi2,k=2,where pk denotes the price of the good for the group k, qik indicates the sale quantity (strategy) of firm i for group k, and the inverse demand function coincides with that in [19, 20, 23] for two groups. It is clear that p2≤p1. In this situation, it can be assumed that firm 1 has price discrimination by setting a higher price for the same product. For the other situation, where firm 2 sells the good with a high price, the corresponding price function is(2)p′k=1-∑i=12qi1,k=1;1-∑i=12qi1-q12,k=2.Then, firm 2 also has price discrimination. The corresponding results will be discussed in the last section of this paper. According to (1), the profits of firms 1 and 2, Π1 and Π2, can be written as follows:(3)Π1=q111-q11-c+q121-q11-q12-q22-cand(4)Π2=q221-q11-q12-q22-c.The decision of each firm is made in discrete time. For any discrete adjustment process, the firmi adjusts its strategy to qik(t+1) in t period from qik(t). We denote IK as the set {(1,1),(1,2),(2,2)}. The micro-dynamic adjustment equation is given as follows:(5)qikt+1=qikt+αik.qikt.∂Πi∂qikt,i,k∈IK,where αik represents the rate of adjustment and generally takes a number in the interval (0,1). Due to a myopic decision and bounded rationality, in each period, the firm i adjusts its sale quantity along with the fastest direction for increasing its profit. The partial derivative, ∂Πi/∂qik(t), in system (5) is written as follows: (6)∂Π1∂q11t=1-c-2q11t-q12t∂Π1∂q12t=1-c-q11t-2q12t-q22t∂Π2∂q22t=1-c-q11t-q12t-2q22t.To find the stationary points of system (5), let qik(t+1)=qik(t). Then, we have(7)qikt∂Πi∂qikt=0,i,k∈IK.There are three cases for the solutions of (7).Case 1 (qik(t)=0 for all (i,k)∈IK).
Clearly, the solution of (7) is (q¯11,q¯12,q¯22)=(0,0,0).Case 2 (qik(t)≠0 and ∂Πi/∂qik(t)=0 for all (i,k)∈IK).
We have(8)1-c-2q11t-q12t=01-c-q11t-2q12t-q22t=01-c-q11t-q12t-2q22t=0.It follows that the solution of (7) satisfies (9)q¯11,q¯12,q¯22=251-c,151-c,151-c.This is the static Nash equilibrium for the game of the two firms with strategies (q11,q12) and q22. The profits of the two firms at the equilibrium are(10)Π1q¯11,q¯11,q¯21=7521-c2and(11)Π2q¯11,q¯11,q¯21=1521-c2.The average price of the market is(12)pv=7/521-c2+1/521-c24/51-c=251-c.
Then, the market share of firm 1 is three times that of firm 2 at the equilibrium. Moreover, firm 1 benefits much more than firm 2 from selling the good with two different prices.Case 3 (others).
The solution(q¯11,q¯12,q¯22) of (7) consists of the following set: (13)1-c2,0,0,0,1-c2,0,0,0,1-c2,1-c3,1-c3,0,0,1-c3,1-c3,1-c2,0,1-c4.
To analyze the stability of these stationary points, the following two lemmas are needed.Lemma 1.
The spectral radiusρ(A) of a matrix A is not greater than any norm of A.Lemma 2 (see [26]).
Given a polynomial equationax3+bx2+ex+d=0 with variable x, let A=b2-3aeB=be-9ad and C=e2-3bd. If B2-4AC≤0, then the equation has only real roots; if B2-4AC>0, then the equation has complex roots.


## 3. Main Results

### 3.1. Jacobian Matrix of System (5) at Equilibrium Points

From Section2, we know that the stationary points of system (5) are as follows: (14)E0=0,0,0;E1=1-c2,0,0;E2=0,1-c2,0;E3=0,0,1-c2;E4=1-c3,1-c3,0;E5=0,1-c3,1-c3;E6=1-c2,0,1-c4;E7=21-c5,1-c5,1-c5.System (5) can be rewritten as follows:(15)q11t+1=q11t+α11q11t1-c-2q11t-q12tq12t+1=q12t+α12q12t1-c-q11t-2q12t-q22tq22t+1=q22t+α22q22t1-c-q11t-q12t-2q22t.Then, the Jacobian matrix of system (15) is calculated as(16)Jq11,q12,q22=ξ1-α11q11t0-α12q12tξ2-α12q12t-α22q22t-α22q22tξ3, where ξ1,ξ2, and ξ3 are given below. (17)ξ1=1+α111-c-q12t-4α11q11t,ξ2=1+α121-c-q11t-q22t-4α12q12t,and (18)ξ3=1+α221-c-q11t-q12t-4α22q22t.It is known that if the modulus of all the eigenvalues ofJ at a stationary (equilibrium) point of system (15) is less than 1, then the stationary point is locally asymptotically stable; if there exits an eigenvalue of J at a stationary point such that its modulus is greater than 1, the point is unstable.


### 3.2. Stability Results for Stationary Points

Theorem 3.
The stationary pointEi of system (15) is unstable for each i=0,1,…,6.Proof.
(i) At the stationary pointE0=(0,0,0), the Jacobian matrix becomes(19)JE0=1+α111-c0001+α121-c0001+α221-c.Clearly, three eigenvalues of J(E0) are λ1=1+α11(1-c),λ2=1+α12(1-c), and λ3=1+α22(1-c). Since α11,α12,α22∈(0,1), and c∈(0,1), we have λi>1 for each i=1,2,3. Then, the point E0 is unstable.
(ii) For the equilibrium pointE1=(1/2(1-c),0,0), the Jacobian matrix is(20)JE1=1-α111-c,-12α111-c001+12α121-c0001+12α221-c.The determinant of the matrix λI-J(E1) (I is the 3×3 identity matrix) follows with (21)λI-JE1=λ-1-α111-c12α111-c00λ-1+12α121-c000λ-1+12α221-c. Three eigenvalues of J(E1) are λ1,λ2, and λ3, with λ1=1-α11(1-c),λ2=1+1/2α12(1-c), and λ3=1+1/2α22(1-c). The eigenvalues λ2 and λ3 are greater than 1. Hence, the equilibrium point is also unstable. Similarly, we can obtain unstable results for E2 and E3.
(iii)J at the equilibrium point E5 becomes(22)JE5=1+23α111-c00-13α121-c1-23α121-c-13α121-c-13α221-c-13α221-c1-23α221-c. It is easy to check that one of the eigenvalues of J(E5) is λ1=1+2/3α11(1-c)>1 by noting that c∈(0,1). Hence, the equilibrium point is also unstable. Similarly, we can get that E4 is unstable.
(iv)J at the point E6 is(23)JE6=1-α111-c-12α111-c001+14α121-c,0-14α221-c-14α221-c1-12α221-c. The determinant of the matrix λI-J(E6) is (24)λI-JE6=λ-1-α111-c12α111-c00λ-1+14α121-c014α221-c14α221-cλ-1-12α221-c.Obviously, we can check that the three eigenvalues of J(E6) are λ1=1-α11(1-c),λ2=1+1/4α12(1-c), and λ3=1-1/2α22(1-c). Then, λ2>1; hence, E6 is unstable.
From the above (i), (ii), (iii), and (iv), the equilibrium solutionsEi are unstable for each i=0,1,…,6. These complete the proof.Theorem 4.
There existsα>0 such that the spectral radii of J(E7) and ρ(J(E7)) satisfy ρ(J(E7))<1 if(25)α11,α12,α22∈Ωα=α11,α12,α22∣0<α11,α12,α22<α,B2-4AC≤0, where A=b2-3e,B=be-9d, and C=e2-3bd with b=4α11+2α12+2α22, e=6α11α12+8α11α22+3α12α22, and d=10α11α12α22.Proof.
The Jacobian matrixJ at the stationary point E7=(2/5(1-c),1/5(1-c),1/5(1-c)) is given below:(26)JE7=1-45α111-c-25α111-c0-15α121-c1-25α121-c-15α121-c-15α221-c-15α221-c1-25α221-c.The matrix norm of J(E7) is written as(27)JE7∞=max⁡1-45α111-c+-25α111-c+0,-15α121-c+1-25α121-c+-15α121-c,-15α221-c+-15α221-c+1-25α221-c.Then, it holds that (28)JE7∞=max⁡1-25α111-c,1,1=1.By Lemma 1, the spectral radius ρJ7 is not greater than the norm J(E7)∞; that is, ρJ7≤1. Hence, the modulus of each eigenvalue of J(E7) is no more than 1.
Next, we check the eigenvalues ofJ(E7). Consider the determinant of the matrix λI-J(E7),(29)λI-JE7=λ-1+45α111-c25α111-c015α121-cλ-1+25α121-c15α121-c15α221-c15α221-cλ-1+25α221-c. Let λ-1=(1-c)/5t. Then, the polynomial eigenvalue of J(E7) can be written as(30)λI-JE7=1-c353t+4α112α110α12t+2α12α12α22α22t+2α22=0. This is equivalent to(31)t+4α112α110α12t+2α12α12α22α22t+2α22=0.Furthermore, the left-hand side of (31) is as follows:(32)t+4α112α110α12t+2α12α12α22α22t+2α22=t+4α11t+2α12α12α22t+2α22-2α11α12α12α22t+2α22=t+4α11t+2α12t+2α22-α12α22-2α11α12t+2α22-α12α22=t3+4α11+2α12+2α22t2+6α11α12+8α11α22+3α12α22t+10α11α12α22.Let a=1, b=4α11+2α12+2α22, e=6α11α12+8α11α22+3α12α22, and d=10α11α12α22. Then, from (32), (31) can be written as(33)at3+bt2+et+d=0.
The next steps, (i) and (ii), show that the eigenvalues are not1 and -1.
(i) Suppose thatλ=1; i.e., t=0. Equation (33) is (34)d=10α11α12α22.This contradicts α12α11α22>0. Thus, λ≠1.
(ii) Assuming thatλ=-1, we have t=-10/1-c. In this case, (33) is(35)1001-cb-101-c2e+d1-c3=1000.When α12,α11, and α22 take the value sufficiently near 0, the left-hand side of (35) is less than 1000. Therefore, there is a small enough α>0 (for instance, α can take 1) such that the value on the left-hand side of (35) is smaller than 1000 if α12,α11,α22<α. This contradicts (35). Therefore, we have λ≠-1.
LetA=b2-3e, B=be-9d, and C=e2-3bd. From Lemma 2, it can be deduced that when Δ=B2-4AC≤0, (33) has only real roots. Noting (i) and (ii), we can conclude that for each (α11,α12,α22)∈Ωα, it holds that ρ(J(E7))<1.Noting thatαik∈(0,1) for any (i,k)∈IK by the settings in Section 2, from the proof of Theorem 4, we can take α=1 in Theorem 4. Then, the following result related to the stability of the equilibrium point E7 can be obtained.Theorem 5.
The stationary pointE7 is locally asymptotically stable for system (15) if (α11,α12,α22)∈Ω1.


## 3.1. Jacobian Matrix of System (5) at Equilibrium Points

From Section2, we know that the stationary points of system (5) are as follows: (14)E0=0,0,0;E1=1-c2,0,0;E2=0,1-c2,0;E3=0,0,1-c2;E4=1-c3,1-c3,0;E5=0,1-c3,1-c3;E6=1-c2,0,1-c4;E7=21-c5,1-c5,1-c5.System (5) can be rewritten as follows:(15)q11t+1=q11t+α11q11t1-c-2q11t-q12tq12t+1=q12t+α12q12t1-c-q11t-2q12t-q22tq22t+1=q22t+α22q22t1-c-q11t-q12t-2q22t.Then, the Jacobian matrix of system (15) is calculated as(16)Jq11,q12,q22=ξ1-α11q11t0-α12q12tξ2-α12q12t-α22q22t-α22q22tξ3, where ξ1,ξ2, and ξ3 are given below. (17)ξ1=1+α111-c-q12t-4α11q11t,ξ2=1+α121-c-q11t-q22t-4α12q12t,and (18)ξ3=1+α221-c-q11t-q12t-4α22q22t.It is known that if the modulus of all the eigenvalues ofJ at a stationary (equilibrium) point of system (15) is less than 1, then the stationary point is locally asymptotically stable; if there exits an eigenvalue of J at a stationary point such that its modulus is greater than 1, the point is unstable.


## 3.2. Stability Results for Stationary Points

Theorem 3.
The stationary pointEi of system (15) is unstable for each i=0,1,…,6.Proof.
(i) At the stationary pointE0=(0,0,0), the Jacobian matrix becomes(19)JE0=1+α111-c0001+α121-c0001+α221-c.Clearly, three eigenvalues of J(E0) are λ1=1+α11(1-c),λ2=1+α12(1-c), and λ3=1+α22(1-c). Since α11,α12,α22∈(0,1), and c∈(0,1), we have λi>1 for each i=1,2,3. Then, the point E0 is unstable.
(ii) For the equilibrium pointE1=(1/2(1-c),0,0), the Jacobian matrix is(20)JE1=1-α111-c,-12α111-c001+12α121-c0001+12α221-c.The determinant of the matrix λI-J(E1) (I is the 3×3 identity matrix) follows with (21)λI-JE1=λ-1-α111-c12α111-c00λ-1+12α121-c000λ-1+12α221-c. Three eigenvalues of J(E1) are λ1,λ2, and λ3, with λ1=1-α11(1-c),λ2=1+1/2α12(1-c), and λ3=1+1/2α22(1-c). The eigenvalues λ2 and λ3 are greater than 1. Hence, the equilibrium point is also unstable. Similarly, we can obtain unstable results for E2 and E3.
(iii)J at the equilibrium point E5 becomes(22)JE5=1+23α111-c00-13α121-c1-23α121-c-13α121-c-13α221-c-13α221-c1-23α221-c. It is easy to check that one of the eigenvalues of J(E5) is λ1=1+2/3α11(1-c)>1 by noting that c∈(0,1). Hence, the equilibrium point is also unstable. Similarly, we can get that E4 is unstable.
(iv)J at the point E6 is(23)JE6=1-α111-c-12α111-c001+14α121-c,0-14α221-c-14α221-c1-12α221-c. The determinant of the matrix λI-J(E6) is (24)λI-JE6=λ-1-α111-c12α111-c00λ-1+14α121-c014α221-c14α221-cλ-1-12α221-c.Obviously, we can check that the three eigenvalues of J(E6) are λ1=1-α11(1-c),λ2=1+1/4α12(1-c), and λ3=1-1/2α22(1-c). Then, λ2>1; hence, E6 is unstable.
From the above (i), (ii), (iii), and (iv), the equilibrium solutionsEi are unstable for each i=0,1,…,6. These complete the proof.Theorem 4.
There existsα>0 such that the spectral radii of J(E7) and ρ(J(E7)) satisfy ρ(J(E7))<1 if(25)α11,α12,α22∈Ωα=α11,α12,α22∣0<α11,α12,α22<α,B2-4AC≤0, where A=b2-3e,B=be-9d, and C=e2-3bd with b=4α11+2α12+2α22, e=6α11α12+8α11α22+3α12α22, and d=10α11α12α22.Proof.
The Jacobian matrixJ at the stationary point E7=(2/5(1-c),1/5(1-c),1/5(1-c)) is given below:(26)JE7=1-45α111-c-25α111-c0-15α121-c1-25α121-c-15α121-c-15α221-c-15α221-c1-25α221-c.The matrix norm of J(E7) is written as(27)JE7∞=max⁡1-45α111-c+-25α111-c+0,-15α121-c+1-25α121-c+-15α121-c,-15α221-c+-15α221-c+1-25α221-c.Then, it holds that (28)JE7∞=max⁡1-25α111-c,1,1=1.By Lemma 1, the spectral radius ρJ7 is not greater than the norm J(E7)∞; that is, ρJ7≤1. Hence, the modulus of each eigenvalue of J(E7) is no more than 1.
Next, we check the eigenvalues ofJ(E7). Consider the determinant of the matrix λI-J(E7),(29)λI-JE7=λ-1+45α111-c25α111-c015α121-cλ-1+25α121-c15α121-c15α221-c15α221-cλ-1+25α221-c. Let λ-1=(1-c)/5t. Then, the polynomial eigenvalue of J(E7) can be written as(30)λI-JE7=1-c353t+4α112α110α12t+2α12α12α22α22t+2α22=0. This is equivalent to(31)t+4α112α110α12t+2α12α12α22α22t+2α22=0.Furthermore, the left-hand side of (31) is as follows:(32)t+4α112α110α12t+2α12α12α22α22t+2α22=t+4α11t+2α12α12α22t+2α22-2α11α12α12α22t+2α22=t+4α11t+2α12t+2α22-α12α22-2α11α12t+2α22-α12α22=t3+4α11+2α12+2α22t2+6α11α12+8α11α22+3α12α22t+10α11α12α22.Let a=1, b=4α11+2α12+2α22, e=6α11α12+8α11α22+3α12α22, and d=10α11α12α22. Then, from (32), (31) can be written as(33)at3+bt2+et+d=0.
The next steps, (i) and (ii), show that the eigenvalues are not1 and -1.
(i) Suppose thatλ=1; i.e., t=0. Equation (33) is (34)d=10α11α12α22.This contradicts α12α11α22>0. Thus, λ≠1.
(ii) Assuming thatλ=-1, we have t=-10/1-c. In this case, (33) is(35)1001-cb-101-c2e+d1-c3=1000.When α12,α11, and α22 take the value sufficiently near 0, the left-hand side of (35) is less than 1000. Therefore, there is a small enough α>0 (for instance, α can take 1) such that the value on the left-hand side of (35) is smaller than 1000 if α12,α11,α22<α. This contradicts (35). Therefore, we have λ≠-1.
LetA=b2-3e, B=be-9d, and C=e2-3bd. From Lemma 2, it can be deduced that when Δ=B2-4AC≤0, (33) has only real roots. Noting (i) and (ii), we can conclude that for each (α11,α12,α22)∈Ωα, it holds that ρ(J(E7))<1.Noting thatαik∈(0,1) for any (i,k)∈IK by the settings in Section 2, from the proof of Theorem 4, we can take α=1 in Theorem 4. Then, the following result related to the stability of the equilibrium point E7 can be obtained.Theorem 5.
The stationary pointE7 is locally asymptotically stable for system (15) if (α11,α12,α22)∈Ω1.


## 4. Numerical Simulations

From Theorem4, there is a range Ωα for the adjustment rate (α11,α12,α22) that guarantees the stability of system (15) near the stationary point E7, where(36)Ωα=α11,α12,α22∣0<α11,α12,α22<α,Δ≤0.Figure1 shows the simulation result for the value of Δ=B2-4AC when (α11,α12,α22)∈Ω1. + symbols represent the cases for Δ≤0; ⋆ symbols denote cases for Δ>0. It can be seen that when α12<α11, there is Δ=B2-4AC≤0. Then, there are many cases such that (α11,α12,α22)∈Ω1, which means that system (15) is locally asymptotically stable at stationary point E7 for parameters with a large range of values.Figure 1
Blue + signs denote cases forΔ=B2-4AC≤0; black ⋆ signs indicate cases for Δ=B2-4AC>0. Here, A,B, and C are from Theorem 4.Numerical solutions for system (15) are shown in Figure 2, where we set the following parameters: c=0.075, (α11,α12,α22)=(0.01,0.02,0.01)∈Ω1.Figure 2
Subgraphs (a), (b), and (c) show the solutions of (15) with an initial value near E0, E2, and E6, respectively. Subgraph (d) displays the solution of (15) with an initial value near E7. It can be observed that E7 is stable and Ei is unstable for i=0,2,6. Subgraph titles clarify the initial points of system (15). The parameters are set as follows: c=0.075,α11=0.01,α12=0.02, and α22=0.01.
(a) (b) (c) (d)As shown in Figures2(a), 2(b), and 2(c), when the initial points are near the corresponding equilibrium points E0=(0,0,0), E1=(1/2(1-c),0,0), E5=(0,1/3(1-c),1/3(1-c)), the solutions deviate from the equilibrium points. This is consistent with Theorem 3. For E7=(2/5(1-c),1/5(1-c),1/5(1-c)), the solution tends to the equilibrium E7, as seen in Figure 2(d), which verifies Theorems 4 and 5.


## 5. Conclusion and Discussion

We assume that one firm sells the product with a certain price while another firm sells the same good for two different prices. This paper contributes to applied research on price discrimination by introducing a new dynamic Cournot competition model. For the dynamic system, each firm makes a decision with bounded rationality in each discrete period to maximize its profit. The stability analysis shows that only one static Nash equilibrium is stable for the dynamic in a large parameter range. The seven other stationary points of the dynamic system are all unstable. At the stable Nash equilibrium, the firm with two prices benefits greatly due to having a major market share; the firm with only a low price has one-third of the market share of the other firm.If firm 2 only charges a high pricep′1 in (2), then, eight stationary points (q¯11,q¯12,q¯21) can be obtained for the corresponding dynamic system as follows: (37)E0′=E0;E1′=E1;E2′=E2;E3′=E3;E4′=E4;E5′=1-c3,0,1-c3;E6′=0,1-c4,1-c2;E7′=1-c5,1-c5,21-c5. Similar to the results presented in Sections 2 and 3, E7′ is the only stable equilibrium point for the dynamic system. The profits of the two firms at E7′ are(38)Π1′E7′=3521-c2 and(39)Π2′E7′=4521-c2. Then, although each firm has the same market share 2/5(1-c) at the equilibrium E7′, firm 1 cannot benefit more than firm 2. The average price is as follows:(40)pv′=3/521-c2+4/521-c24/51-c=7201-c. Comparing (10) and (11) with Π1′(E7′) and Π2′(E7′), we see that having two different prices is not always beneficial to a firm. From pv in (12) and the above pv′, it holds that pv′<pv. Therefore, if both firms have a high (discriminative) price for the homogeneous product, the market’s average price is lower than when only one firm has a higher price.



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*Source: 9231582-2019-06-25.xml*