Stability of Cournot duopoly games with isoelastic demands and quadratic costs

Let us consider a market served by two firms producing homogeneous products. We use $q_{i}(t)$ to denote the output of firm $i$ at period $t$. Moreover, the cost function of firm $i$ is supposed to be quadratic, i.e., $C_{i}(q_{i})=c_{i}q_{i}^{2}$ with $c_{i}>0$. At each period $t$, firm $i$ first estimates the possible price $p_{i}^{e}(t)$ of the product, then the expected profit of firm $i$ would be

In order to maximize the expected profit, at period $t$ each firm would decide the quantity of the output by solving

Furthermore, assume that the demand function of the market is isoelastic, which is founded on the hypothesis that the consumers have the Cobb-Douglas utility function. Hence, the real (not expected) price of the product should be

where $Q=q_{1}+q_{2}$ is the total supply. Four types of players with distinct rationality degrees are involved in this draft, which are described as follows one by one.

A rational player not only knows clearly the form of the price function, but also has complete information of the decision of its rival. If firm $i$ is a rational player, at period $t+1$ we have

where $q_{-i}^{e}(t+1)$ is the expectation of the output of the rival. Due to the assumption of complete information, which means that $q_{-i}^{e}(t+1)=q_{-i}(t+1)$, it is acquired that the expected profit of firm $i$ would be

The first order condition for profit maximization gives rise to a cubic polynomial equation. To be exact, the condition for the reaction function of firm $i$ would be

which is simply denoted as $F_{i}(q_{i}(t+1),q_{-i}(t+1))=0$ in the sequel. The player could maximize its profit by solving the above equation. It is easy to verify that there exists only one positive solution for $q_{i}(t+1)$ if solving (1), but the expression could be quite complex.

where

For simplicity, we temporarily denote (2) by

where $R_{i}$ is called the best response function of firm $i$. It is evident that if the two firms in the market are both rational players, the equilibrium (the best decision of output) would be arrived in a shot and there are no dynamics in the system. In order to tackle this problem, Puu introduced the bounded rational player in [9].

A boundedly rational player knows the form of the price function, but do not know the rival’s decision of the production. If firm $i$ is a boundedly rational player, then it naively expects its competitor to produce the same quantity of output as the last period, i.e., $q_{-i}^{e}(t+1)=q_{-i}(t)$. Thus,

Then the best response for firm $i$ would be $q_{i}(t+1)=R_{i}(q_{-i}(t))$. By now, the following two dynamic duopoly models could be introduced.

The stability analysis of Model LL was first given in [2], which is restated here.

In [8], the first author of this paper proposed an approach of systematically computing semi-algebraic systems, which could also be used in this work to prove similar theoretical results as Proposition 1. To illustrate the basic idea of this approach, we reprove the above proposition in a computational style in the sequel.

In order to acquire the equilibria, we set $q_{1}(t+1)=q_{1}(t)=q_{1}$ and $q_{2}(t+1)=q_{2}(t)=q_{2}$ in (7). Moreover, we focus only on such equilibria that $q_{1},q_{2}>0$, then the following system is obtained.

where $c_{1}>0$, $c_{2}>0$ are obvious according to the economic constraints of the parameters. The problem of counting non-vanishing equilibria is equivalent to counting real roots of system (12). Using our approach proposed in [8], this problem can be solved in 4 steps as follows.

Step 1. The solutions of the equation part of (12) are equivalent to zeros of

Using the method of triangular decomposition¹¹1The method of triangular decomposition can be viewed as an extension of the method of Gaussian elimination. The main ideas of both are to transform a system into a triangular form. However, the triangular decomposition method is for polynomial systems, while the Gaussian elimination method is for linear systems. Refer to [11, 7, 6, 10] for more details., we decompose $P$ into two triangular forms:

and

According to the properties of triangular decomposition, the zero set of $P$ is the same as the union of zeros of $T_{1}$ and $T_{2}$. Of them, only $T_{2}$ should be considered as $T_{1}$ is corresponding to the original equilibrium $(0,0)$. Furthermore, one may observe that the second polynomial in $T_{2}$ has degree $1$ with respect to $q_{2}$, thus $q_{2}$ could be represented as

Step 2. Substitute (14) into the inequality $q_{2}>0$ in (12), we obtain

or equivalently

Thus, system (12) is transformed to a univariate system

Step 3. Define the border polynomial of system (15) to be

where $\operatorname{discr}(T)$ represents the discriminant of $T$, $\operatorname{res}(T,Q)$ stands for the resultant of $T$ with respect to $Q$, with

It is noted that $A_{0}$ is the leading coefficient of $T$ with respect to $q_{1}$, i.e., the coefficient of $q_{1}^{4}$. We have

and its squarefree part is

It is proved in [8] that $BP_{LL}=0$, or equivalently $SP_{LL}=0$, divides the parameter space into several connected regions (see Figure 1), and on each of them the number of distinct real solutions is invariant.

Step 4. From each region of our concerned parameter set with $c_{1},c_{2}>0$, select²²2For a simple border polynomial as (17), the selection of sample points could be done by hand. Generally, however, the selection could be automated by using, e.g., the partial cylindrical algebraic decomposition (PCAD) method [4]. one sample point:

At each sample point, count³³3The first thought is to count real solutions by solving the system numerically. However, numerical methods have several shortcomings: first, the numerical computation may encounter the problem of instability, which could make the results completely useless; second, floating-point errors may cause the problem that it is extremely hard to distinguish between real solutions and complex solutions with tiny imaginary parts; third, most numerical algorithms only search for a single equilibrium and are nearly infeasible for multiple equilibria. Thus we herein need symbolic methods, e.g., [12], which can be used to count the real solutions exactly. the number of distinct real solutions of system (15). We obtain that there exists exactly one real solution for each sample point. This means that system (15), or equivalently (12), has one real solution for any feasible parameter value. Hence, we conclude that the dynamic system (7) has one unique equilibrium with $q_{1},q_{2}>0$ for any $c_{1},c_{2}>0$.

It is worth noting that all computations involved in the above 4 steps are symbolic and rigorous. This is to say that our computational approach permit us to acquire analytical results, thus could be used to discover and prove theorems of economic models involving polynomials.

In order to investigate the local stability of this unique equilibrium, the Jacobian matrix

plays an ambitious role. We use $\operatorname{Det}(J)$ and $\operatorname{Tr}(J)$ to denote the determinant and the trace of $J$, respectively. According to the Jury’s criterion [5], an equilibrium $E$ is locally stable provided that

Combine these inequalities with system (12), transform the resulting system to a univariate system, and then compute its border polynomial likewise. It is obtained that the squarefree part of the border polynomial is

Accordingly, the selected sample points might be

The parameter plane of Model LL is depicted in Figure 2, where $SP^{*}_{LL}=0$ is represented with blue lines, and the selected sample points are marked in red. It could be verified that the system (12)+(18) has exactly one real solution at any of these sample points, which means the unique non-vanishing equilibrium of (7) is locally stable for all parameter values that satisfy $c_{1},c_{2}>0$.

Indeed, the above proposition has been rigorously proved in Appendix of [2] by explicitly solving the non-vanishing equilibrium. It is lucky that the closed-form solution of the non-vanishing equilibrium exists, which is

Then substituting the solution into the stability condition by Jury’s criterion, it is not hard to verify that all the inequalities in the condition are satisfied. In the next section, it will be found that the verification of the stability through substituting the solution might be impractical for some models. In comparison, our approach do not require to explicitly compute the solution of the considered equilibrium. If the closed-form solution has a super complicated expression or even if there exist no closed-form solutions, our approach could still work.