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Optimal product differentiation in a circular model

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Abstract

Since circular model was introduced in Salop (Bell J Econ 10:141–156, 1979), it has been the workhorse for analyzing spatial competition among differentiated firms. A common assumption in this literature is that firms are evenly spaced on the circle, even when entry is allowed. We characterize conditions for even spacing to be an equilibrium, using a two-stage (location-then-price) circular model with general transport cost function. Under duopoly competition, we characterize a mild sufficient condition—the first derivative of transport cost is concave (together with an assumption governing the transport cost difference to the two firms). If one only considers pure strategy equilibrium in prices, this sufficient condition is weakened to the first derivative of transport cost being \(-\)1-concave. These conditions ensure that firms’ profits are concave in their prices when firms are evenly spaced and that even spacing maximizes profits. Under oligopoly competition (\(N\ge 2\) firms), we characterize a necessary condition for even spacing to be an equilibrium. This necessary condition requires a firm’s profit to be concave in location at the symmetric location. It involves the third derivative of transport cost function, so having convex transport cost in general is neither necessary nor sufficient to determine equilibrium location choice. Our results have implications for studies employing circular models, especially in terms of welfare analysis which depends on firms’ location choices.

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Notes

  1. Another common assumption in the literature is that consumers are uniformly distributed on the circle. We keep this assumption in the paper. It would be interesting to see how nonuniform distribution affects the equilibrium location choice. With non-uniform distribution, a Hotelling model may be more suitable for the analysis (e.g., Anderson et al. 1997; Kim 2009; Loertscher and Muehlheusser 2011) and likely one needs to impose more constraints on the transport cost function.

  2. Our results support de Frutos et al. (2002) who point out the equivalence between games with convex transport costs and games with concave transport costs, using a duopoly circular model.

  3. Simple searches show that Salop (1979) has 671 Web of Science citations and 2192 Google Scholar citations.

  4. Circular models have also been employed to analyze issues such as entry and efficiency (e.g. Bhaskar and To 2004; Liu and Serfes 2005; Matsumura and Okamura 2006).

  5. An exception is Anderson (1986), who analyzes a duopoly model with strictly convex transport cost functions. He focuses on pure strategy price equilibrium, and then characterizes the necessary condition for the existence of a subgame perfect Nash equilibrium.

  6. A subset of the results in this paper, mostly for the case of \(N=2\) firms, are included in Gong and zhang (2011).

  7. On a unit circle, the nearest distance between any two locations on the circle cannot be more than 1 / 2.

  8. An implicit assumption made here is that the shorter routes for marginal consumer at \(x_2\) to travel to firms are clockwise to firm 1 and counterclockwise to firm 2. This clearly holds in a symmetric equilibrium (\(p_1=p_2\)), but may be violated when firms deviate, in which case demand functions are adjusted accordingly in our analysis.

  9. When marginal consumer changes, its transport cost to firm 1 increases and that to firm 2 decreases. As a result, firm 1 can only still \(\frac{1}{2f^{\prime }\left( \frac{a}{2}\right) }\) consumers on the first arc, instead of \(\frac{1}{f^{\prime }\left( \frac{a}{2}\right) }\) consumers.

  10. This is taken from Caplin and Nalebuff (1991) where more details about \(\rho \)-concavity can be found.

  11. If firms’ price supports shrink to individual points, then the mixed strategy equilibrium becomes a pure strategy equilibrium. We verified that our results hold under such degenerate distribution functions so we can consider mixed strategy equilibrium without loss of generality.

  12. The case of \(f(l)=-l^2+2l\) fits in the linear-quadratic case and has been pointed out in earlier studies.

  13. We thank an anonymous referee for suggesting this exercise.

  14. Recall that the equilibrium prices (and thus profits) involve \(f(\cdot )\) under two-plant monopoly, but involve \(f^{\prime } (\cdot )\) under duopoly competition. It is thus unsurprising that the SOC would involve \(f^{\prime \prime }\) under two-plant monopoly but involve \(f^{\prime \prime \prime }\) under duopoly competition.

  15. Since the problem is only because the \(\frac{\mathrm{d}^2 \pi _2^*}{\mathrm{d}a^2}|_{a=1/N}\) expression is lengthy, we can easily check the necessary condition for any specific N and transport cost function f(l).

  16. This proof remains valid if one or both firms’ distribution functions are degenerate, i.e., their price support shrinks to individual points.

  17. Note that \(\underline{p}\), \(\bar{p}\) and \(p^*\) are all functions of a. To ease on notation, we write them as \(\underline{p}\), \(\bar{p}\) and \(p^*\) instead.

  18. Note that the inequality is strict unless \(\underline{p}_i=\bar{p}_i=p^*(a)\), i.e., we are actually considering the case where \(p_1=p_2=p^*(a)\). The same logic can be applied to show that when \(a=1/2\), if there exists any other pure strategy equilibria which have to be asymmetric, they are strictly dominated by the symmetric equilibrium \(p_1=p_2=p^*(a=1/2)\) from firms’ perspective.

  19. The same holds for \(\left. \frac{\mathrm{d}^2\pi _2}{\mathrm{d}a^2}\right| _{a=\frac{1}{N}}\) in the end of step 5 as the general formula applies to the cases of \(N=2\) and \(N=3\) as well.

  20. Details (in Maple programs) are available upon request.

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Acknowledgments

We would like to thank the Editor Giacomo Corneo and two anonymous referees whose comments allowed us to improve the paper significantly. We would also like to thank Greg Burge, George Deltas, Kostas Serfes, John Turner, Hao Wang, Lixin Ye, Artie Zillante, seminar and conference participants at Southwestern University of Finance and Economics, Shanghai University of Finance and Economics, International Economic Association World Congress (2011), International Industrial Organization Conference (2010), Hong Kong Economic Association Conference (2010) and Five-Star Economics Forum (2010) for helpful comments and suggestions.

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Correspondence to Qihong Liu.

Appendix: proof of lemmas and propositions

Appendix: proof of lemmas and propositions

Proof of Lemma 1

We first derive the partial derivatives of m and n with respect to \(p_1\) and \(p_2\). From Eq. (1), we have

$$\begin{aligned} V-p_1-f(m)= & {} V-p_2-f(a-m)\\\Rightarrow & {} p_1+f(m)=p_2+f(a-m). \end{aligned}$$

Taking derivative with respect to \(p_1\) at both sides, we can obtain

$$\begin{aligned} 1+f^{\prime }(m)\frac{\partial m}{\partial p_1}=- f^{\prime }(a-m) \frac{\partial m}{\partial p_1}. \end{aligned}$$

This implies that

$$\begin{aligned} \frac{\partial m}{\partial p_1}=-\frac{1}{f^{\prime }(m)+f^{\prime }(a-m)}. \end{aligned}$$

Similarly we can obtain

$$\begin{aligned} {\frac{\partial n}{\partial p_1}}=-\frac{1}{f^{\prime }(n)+f^{\prime }(1-a-n)} \text{ and, } \end{aligned}$$
$$\begin{aligned} \frac{\partial m}{\partial p_2}= -\frac{\partial m}{\partial p_1},\quad \frac{\partial n}{\partial p_2}=- \frac{\partial n}{\partial p_1}. \end{aligned}$$

Firm 1’s first-order conditions implies

$$\begin{aligned} \frac{\partial \pi _1}{\partial p_1}= & {} m+n+(p_1-c)\left( \frac{\partial m}{\partial p_1}+\frac{\partial n}{\partial p_1}\right) =0\nonumber \\\Rightarrow & {} m+n=(p_1-c) \left( \frac{1}{f^{\prime }(m)+f^{\prime }(a-m)} + \frac{1}{f^{\prime }(n)+f^{\prime }(1-a-n)}\right) .\nonumber \\ \end{aligned}$$
(4)

Similarly, from \(\frac{\partial \pi _2}{\partial p_2}=0\), we can obtain

$$\begin{aligned} 1-m-n=(p_2-c) \left( \frac{1}{f^{\prime }(m)+f^{\prime }(a-m)} + \frac{1}{f^{\prime }(n)+f^{\prime }(1-a-n)}\right) . \end{aligned}$$
(5)

Next, we impose the symmetry condition \(p_1=p_2\). Then \(m=a/2\), \(n=(1-a)/2\) and \(m+n=\frac{1}{2}\). Plugging them into Eq. (4) or (5), we can obtain

$$\begin{aligned} p_1=p_2=p^*(a)=c+\frac{1}{\frac{1}{f^{\prime }\left( \frac{a}{2}\right) }+\frac{1}{f^{\prime }\left( \frac{1-a}{2}\right) }}. \end{aligned}$$

Then firms’ profits are

$$\begin{aligned} \pi _i^*(a)=(p^*(a)-c)\times \frac{1}{2}=\frac{1}{\frac{2}{f^{\prime }\left( \frac{a}{2}\right) }+\frac{2}{f^{\prime }\left( \frac{1-a}{2}\right) }}. \end{aligned}$$

\(\square \)

Proof of Proposition 1

Due to symmetry, we fix \(p_2^*=c+\frac{1}{2}f^{\prime }(\frac{1}{4})\) and only check whether firm 1 can increase its profit by deviating \(p_1\). Recall that m and n are firm 1’s market shares on the two arcs respectively. Since \(a=\frac{1}{2}\), the two arcs are symmetric and firm 1’s market share on each arc must be the same (\(m=n\)) for any \(p_1\). Moreover, the marginal consumer on the arc from 0 to \(\frac{1}{2}\) is located at \(x=m\). Therefore

$$\begin{aligned} p_1+f(m)= & {} p_2^*+f(1/2-m)\Rightarrow p_1=p_2^*-f(m)+f(1/2-m)\\\Rightarrow & {} p_1=c+\frac{1}{2}f^{\prime }(1/4)-f(m)+f(1/2-m), \end{aligned}$$

which gives a one-to-one correspondence between \(p_1\) and m.

Firm 1’s deviation profit is

$$\begin{aligned} \pi _1(p_1)=(p_1-c) (m+n)=2m (p_1-c). \end{aligned}$$

Next, we express firm 1’s profit as a function of m (instead of \(p_1\)). Then firm 1’s problem is,

$$\begin{aligned} \max _{m\in (0,\frac{1}{2}]}\; \pi _1 (m) = m \left[ f^{\prime }(1/4)-2f(m)+2f\left( \frac{1}{2}-m\right) \right] . \end{aligned}$$

Take derivative with respect to m, we can obtain

$$\begin{aligned} \pi _1^{\prime }(m)=f^{\prime }(1/4)-2f(m)+2f(1/2-m)-2mf^{\prime }(m)-2mf^{\prime }(1/2-m). \end{aligned}$$

It’s easy to see that \(\pi _1^{\prime }(m=1/4)=0\).

The second derivative is

It can be shown that the following two statements are equivalent: (i) \(\pi _1^{\prime \prime }(m)\le 0\), (ii) \(g(l)\equiv l\left[ f(\frac{1}{2}-l)-f(l)\right] \) is concave in l (i.e., \(g^{\prime \prime }(l)\le 0\)). Note that \(\pi _1^{\prime }(m=1/4)=0\). Therefore, if g(l) is concave on \(l\in (0,\frac{1}{2}]\), then \(\pi _1^{\prime \prime }(m)\le 0\), \(\,\,\,\forall m\in (0,\frac{1}{2}]\). Then \(m=1/4\) is a global maximum and firm 1 has no incentive to deviate. \(\square \)

Proof of Lemma 2

From Lemma 1, we have

$$\begin{aligned} \pi ^*(a)=\frac{1}{\frac{2}{f^{\prime }(\frac{a}{2})}+\frac{2}{f^{\prime }(\frac{1-a}{2})}}=\frac{1}{2}\cdot \frac{1}{\frac{1}{f^{\prime }(\frac{a}{2})}+\frac{1}{f^{\prime }(\frac{1-a}{2})}}. \end{aligned}$$

\(f^{\prime }(l)\) being \(-1\)-concave is equivalent to \(\frac{1}{f^{\prime }(l)}\) being convex. Then

$$\begin{aligned} \frac{1}{f^{\prime }\left( \theta x+ (1-\theta )y\right) } \le \theta \frac{1}{f^{\prime }(x)} + (1-\theta ) \frac{1}{f^{\prime }(y)},\quad \forall \;\; \theta \in [0,1]. \end{aligned}$$

Setting \(\theta =\frac{1}{2}\), \(x=a/2\) and \(y=(1-a)/2\), we can obtain

$$\begin{aligned} \frac{1}{2}\frac{1}{f^{\prime }(\frac{a}{2})}+ \frac{1}{2} \frac{1}{f^{\prime }\left( \frac{1-a}{2}\right) }\ge & {} \frac{1}{f^{\prime }\left( \frac{1}{4}\right) }\\ \Leftrightarrow \frac{1}{\frac{1}{f^{\prime }\left( \frac{a}{2}\right) }+\frac{1}{f^{\prime }\left( \frac{1-a}{2}\right) }}\le & {} \frac{f^{\prime }\left( \frac{1}{4}\right) }{2}. \end{aligned}$$

Then

$$\begin{aligned} \pi ^*(a)\le \frac{1}{2}\cdot \frac{f^{\prime }\left( \frac{1}{4}\right) }{2}= \frac{1}{4}\cdot f^{\prime }\left( \frac{1}{4}\right) =\pi ^*\left( a=\frac{1}{2}\right) . \end{aligned}$$

Alternatively, it can be shown that \(\frac{\partial \pi ^*(a)}{\partial a}\ge 0\) (\(a\in (0,\frac{1}{2}]\)) if and only if

$$\begin{aligned} f^{\prime }(l)f^{\prime \prime \prime }(l) - 2\left( f^{\prime \prime }(l)\right) ^2 \le 0, \end{aligned}$$

which is equivalent to \(\frac{1}{f^{\prime }(l)}\) being convex. \(\square \)

Proof of Proposition 2

Since \(f^{\prime }(l)\) is concave, it must be \(-1\)-concave. By Lemma 2, \(\pi ^*(a)\) is maximized at \(a=\frac{1}{2}\). This proves (i). Next, we prove (ii) \(\pi ^*(a) \ge \pi _i^{mixed}(a)\), \(\,\,\,\forall a\in (0,\frac{1}{2}]\), \(i=1,2\).

Suppose that in the mixed strategy equilibrium, firm \(i=1,2\) chooses prices according to distribution function \(F_i(p_i)\), \(p_i\in [\underline{p}_i,\bar{p}_i]\).Footnote 16 Recall that \(\tilde{p}_i\) denotes the mixed strategy which firm i plays. Without loss of generality, assume that \(\bar{p}_1\ge \bar{p}_2\). Next, we prove by contradiction that \(\bar{p}_1\) cannot be higher than \(p^*\), which is given in Lemma 1.Footnote 17 Suppose not and \(\bar{p}_1> p^*\). When choosing any price \(p_1\) in the support, firm 1’s expected profit is

$$\begin{aligned} E_{p_2} \pi _1(a,p_1,\tilde{p}_2)=(p_1-c) \int \limits _{\underline{p}_2}^{\bar{p}_2}\left[ m(a,p_1,p_2)+n(a,p_1,p_2)\right] \mathrm{d}F(p_2). \end{aligned}$$

Taking derivative of \(E_{p_2} \pi _1(a,p_1,\tilde{p}_2)\) with respect to \(p_1\), and evaluate it at \(p_1=\bar{p}_1\) (i.e., consider how \(E_{p_2} \pi _1(a,p_1,\tilde{p}_2)\) changes when \(p_1\) decreases slightly), we have

The inequality is because with \(p_1\) at the upper bound of the price support, firm 1’s expected market share must be less than \(\frac{1}{2}\).

Since \(f^{\prime }(l)\) is concave in l, we have

$$\begin{aligned} \theta f^{\prime }(x) + (1-\theta ) f^{\prime }(y) \le f^{\prime }(\theta x+ (1-\theta )y),\forall \theta \in [0,1]. \end{aligned}$$

By setting \(\theta =\frac{1}{2}\), \(x=m\) and \(y=a-m\), we have

$$\begin{aligned} \frac{1}{2}f^{\prime }(m)+\frac{1}{2} f^{\prime }(a-m) \le f^{\prime }\left( \frac{a}{2}\right) \Leftrightarrow -\frac{1}{f^{\prime }(m)+f^{\prime }(a-m)} \le -\frac{1}{2 f^{\prime }\left( \frac{a}{2}\right) }. \end{aligned}$$

Similarly we can show that

$$\begin{aligned} -\frac{1}{f^{\prime }(n)+f^{\prime }(1-a-n)} \le -\frac{1}{2 f^{\prime }\left( \frac{1-a}{2}\right) }. \end{aligned}$$

Then

We have established that if \(\bar{p}_1> p^*\), then \(\frac{\partial E_{p_2} \pi _1(a,p_1,\tilde{p}_2)}{\partial p_1}|_{p_1=\bar{p}_1}<0.\) That is, for \(\varepsilon \rightarrow 0^+\),

$$\begin{aligned} E_{p_2} \pi _1(a,\bar{p}_1,\tilde{p}_2) < E_{p_2} \pi _1(a,\bar{p}_1-\varepsilon ,\tilde{p}_2). \end{aligned}$$

Firm 1 would strictly prefer \(p_1=\bar{p}_1-\varepsilon \) to \(p_1=\bar{p}_1\), contradicting the equilibrium setup that \(\bar{p}_1\) is in the price support. Therefore \(\bar{p}_1>p^*\) can’t happen. This also implies that \(\bar{p}_2\le p^*\).

Compared to the case when \(p_1=p_2=p^*(a)\), in the mixed strategy equilibrium, both firms always choose prices (weakly) below \(p^*\). For each firm, facing a more aggressive competitor, its profit must be lower than \(\pi ^*(a)\). That is, \(\pi _i^{mixed}(a)\le \pi ^*(a)\), \(i=1,2\).Footnote 18 \(\square \)

Proof of Lemma 3

When firms are located symmetrically on the circle, the marginal consumers are defined by

$$\begin{aligned} p_i+f\left( x_i-\frac{i-1}{N}\right)= & {} p_{i+1}+f\left( \frac{i}{N}-x_i\right) ,\quad i=1,\ldots , N-1,\\ p_N+f\left( x_N-\frac{N-1}{N}\right)= & {} p_1+f(1-x_N). \end{aligned}$$

Taking total differentials, we can derive the following partial derivatives,

$$\begin{aligned} \frac{\partial x_i}{\partial p_i}=-\frac{1}{f^{\prime }\left( \frac{i}{N}-x_i\right) +f^{\prime }\left( x_i-\frac{i-1}{N}\right) },\quad i=1,\ldots , N-1, \end{aligned}$$
$$\begin{aligned} \frac{\partial x_i}{\partial p_{i+1}}=-\frac{\partial x_i}{\partial p_i},\quad i=1,\ldots , N-1, \end{aligned}$$
$$\begin{aligned} \frac{\partial x_i}{\partial p_j}=0,\forall j\ne i, i+1, \quad i=1,\ldots , N-1. \end{aligned}$$

The partial derivatives for \(i=N\) can be derived similarly. Firm i’s first-order condition is

$$\begin{aligned} \frac{\partial \pi _i}{\partial p_i}=(x_i-x_{i-1}) + (p_i-c) \left( \frac{\partial x_i}{\partial p_i} -\frac{\partial x_{i-1}}{\partial p_i}\right) =0, \quad i=2,\ldots , N. \end{aligned}$$

Imposing symmetry (\(p_1=p_2=\cdots =p_N\)), we have

$$\begin{aligned} x_i=\frac{i-1}{N}+\frac{1}{2N}\Rightarrow x_i-x_{i-1}=\frac{1}{N},\quad i=2,\ldots , N. \end{aligned}$$

Substituting this back into the first-order conditions, we can obtain

$$\begin{aligned} p_i^*=c+\frac{1}{N}f^{\prime }\left( \frac{1}{2N}\right) ,\quad \pi _i^*=\frac{1}{N^2}f^{\prime }\left( \frac{1}{2N}\right) ,\quad i=1,\ldots ,N. \end{aligned}$$

\(\square \)

Proof of Proposition 4

This proof is quite tedious, requiring us to use implicit function theorem repeated, and to solve for difference equations. We divide the overall proof into the following steps:

  • Step 1: Deriving firms’ first-order conditions in the pricing stage.

  • Step 2: Taking total differentials using firms’ first-order conditions.

  • Step 3: Calculating \(\left. \frac{\mathrm{d}x_i}{\mathrm{d}a}\right| _{a=1/N}\) and \(\left. \frac{\mathrm{d}^2 x_i}{\mathrm{d}a^2}\right| _{a=1/N}\), \(i=1,\ldots ,N\).

  • Step 4: Solving for \(\left. \frac{\mathrm{d}p_i}{\mathrm{d}a}\right| _{a=1/N}\), \(i=1,\ldots ,N\).

Step 1. Derive firms’ first-order conditions

Note that firm \(i\ne 2\) is located at \(\frac{i-1}{N}\), while firm 2 is located at a around \(\frac{1}{N}\).

The marginal consumers are defined by

$$\begin{aligned} p_1+f(x_1)=p_2+f(a-x_1), \end{aligned}$$
$$\begin{aligned} p_2+f(x_2-a)=p_3+f\left( \frac{2}{N}-x_2\right) , \end{aligned}$$
$$\begin{aligned} p_i+f\left( x_i-\frac{i-1}{N}\right) =p_{i+1}+f\left( \frac{i}{N}-x_i\right) ,\quad i=3,\ldots , N-1, \end{aligned}$$
$$\begin{aligned} p_N+f\left( x_N-\frac{N-1}{N}\right) =p_1+f(1-x_N). \end{aligned}$$

Taking total differentials,, we can derive the following partial derivatives,

$$\begin{aligned} \frac{\partial x_1}{\partial a}=\frac{f^{\prime }(a-x_1)}{f^{\prime }(x_1)+f^{\prime }(a-x_1)},\quad \frac{\partial x_2}{\partial a}=\frac{f^{\prime }(x_2-a)}{f^{\prime }(x_2-a)+f^{\prime }\left( \frac{2}{N}-x_2\right) },\quad \frac{\partial x_i}{\partial a}=0, i\ne 1,2, \end{aligned}$$
$$\begin{aligned} \frac{\partial x_1}{\partial p_1}=-\frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)},\quad \frac{\partial x_1}{\partial p_2}=-\frac{\partial x_1}{\partial p_1}, \end{aligned}$$
$$\begin{aligned} \frac{\partial x_2}{\partial p_2}=-\frac{1}{f^{\prime }(x_2-a)+f^{\prime }\left( \frac{2}{N}-x_2\right) },\quad \frac{\partial x_2}{\partial p_3}=-\frac{\partial x_2}{\partial p_2}, \end{aligned}$$
$$\begin{aligned} \frac{\partial x_i}{\partial p_i}=-\frac{1}{f^{\prime }\left( \frac{i}{N}-x_i\right) +f^{\prime }\left( x_i-\frac{i-1}{N}\right) },\quad \frac{\partial x_i}{\partial p_{i+1}}=-\frac{\partial x_i}{\partial p_i}, i=3,\ldots ,N-1, \end{aligned}$$
$$\begin{aligned} \frac{\partial x_N}{\partial p_N}=-\frac{1}{f^{\prime }(1-x_N)+f^{\prime }\left( x_N-\frac{N-1}{N}\right) },\quad \frac{\partial x_N}{\partial p_1}=-\frac{\partial x_N}{\partial p_N}. \end{aligned}$$

Firms’ profits are

$$\begin{aligned} \pi _1=p_1 [1+x_1(a,p_1,p_2)-x_N(p_1,p_N)], \end{aligned}$$
$$\begin{aligned} \pi _2=p_2 [x_2(a,p_2,p_3)-x_1(a,p_1,p_2)], \end{aligned}$$
$$\begin{aligned} \pi _3=p_3 [x_3(p_3,p_4)-x_2(a,p_2,p_3)], \end{aligned}$$
$$\begin{aligned} \pi _i=p_i [x_i(p_i,p_{i+1})-x_{i-1}(p_{i-1},p_i)], i=4,\ldots , N. \end{aligned}$$

Setting \(\frac{\partial \pi _i}{\partial p_i}=0\), \(i=1,\ldots ,N\), the first-order conditions are

$$\begin{aligned} 1+x_1-x_N=p_1\left[ \frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)} + \frac{1}{f^{\prime }(1-x_N)+f^{\prime }\left( x_N-\frac{N-1}{N}\right) }\right] , \end{aligned}$$
(6)
$$\begin{aligned} x_2-x_1=p_2\left[ \frac{1}{f^{\prime }(x_2-a)+f^{\prime }\left( \frac{2}{N}-x_2\right) }+ \frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)} + \right] , \end{aligned}$$
(7)
$$\begin{aligned} x_3-x_2=p_3\left[ \frac{1}{f^{\prime }\left( x_3-\frac{2}{N}\right) + f^{\prime }\left( \frac{3}{N}-x_3\right) }+ \frac{1}{f^{\prime }(x_2-a)+f^{\prime }\left( \frac{2}{N}-x_2\right) }\right] , \end{aligned}$$
(8)
$$\begin{aligned} x_i-x_{i-1}=p_i\left[ \frac{1}{f^{\prime }\left( x_i-\frac{i-1}{N}\right) + f^{\prime }\left( \frac{i}{N}-x_i\right) }+ \frac{1}{f^{\prime }\left( x_{i-1}-\frac{i-2}{N}\right) + f^{\prime }\left( \frac{i-1}{N}-x_{i-1}\right) }\right] ,\quad i=4,\ldots ,N.\nonumber \\ \end{aligned}$$
(9)

Step 2. Taking total differentials using firms’ first-order conditions

Next, we will utilize firms’ first-order conditions (Eqs. 69), starting with firm 1’s.

Using firm 1’s FOC

Let \(\Delta _{1N}\equiv \frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)} + \frac{1}{f^{\prime }(1-x_N)+f^{\prime }\left( x_N-\frac{N-1}{N}\right) }\). Then

Next, we calculate \(\frac{\mathrm{d}^2 \Delta _{1N}}{\mathrm{d}a^2}\) and evaluate it at \(a=\frac{1}{N}\),

Taking total differential using firm 1’s FOC (Eq. 6), we have

$$\begin{aligned} \frac{\mathrm{d}x_1}{\mathrm{d}a}-\frac{\mathrm{d}x_N}{\mathrm{d}a}=\frac{\mathrm{d}p_1}{\mathrm{d}a}\Delta _{1N}+p_1\frac{d\Delta _{1N}}{\mathrm{d}a}. \end{aligned}$$

Taking total differential one more time, we have

$$\begin{aligned} \frac{\mathrm{d}^2 x_1}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_N}{\mathrm{d}a^2}=\frac{\mathrm{d}^2 p_1}{\mathrm{d}a^2}\Delta _{1N} +2\frac{\mathrm{d}p_1}{\mathrm{d}a} \frac{d\Delta _{1N}}{\mathrm{d}a}+p_1\frac{\mathrm{d}^2\Delta _{1N}}{\mathrm{d}a^2}. \end{aligned}$$

Evaluating at \(a=\frac{1}{N}\), we have

$$\begin{aligned} p_1=\frac{A_1}{N},\quad \Delta _{1N}=\frac{1}{A_1},\quad \frac{d\Delta _{1N}}{\mathrm{d}a}=-\frac{A_2}{4(A_1)^2}. \end{aligned}$$

Then

$$\begin{aligned} \frac{\mathrm{d}x_1}{\mathrm{d}a}-\frac{\mathrm{d}x_N}{\mathrm{d}a}=\frac{1}{A_1}\frac{\mathrm{d}p_1}{\mathrm{d}a}-\frac{A_2}{4NA_1}. \end{aligned}$$
(10)
(11)

Using firm 2’s FOC

Let \(\Delta _{21}=\frac{1}{f^{\prime }(x_2-a)+f^{\prime }\left( \frac{2}{N}-x_2\right) }+ \frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)}\). Then

$$\begin{aligned} \frac{d\Delta _{21}}{\mathrm{d}a}= & {} -\frac{f^{\prime \prime }(x_2-a)\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}-1\right) -f^{\prime \prime }\left( \frac{2}{N}-x_2\right) \frac{\mathrm{d}x_2}{\mathrm{d}a}}{[f^{\prime }(x_2-a)+f^{\prime }\left( \frac{2}{N}-x_2\right) ]^2}\nonumber \\&-\frac{f^{\prime \prime }(x_1)\frac{\mathrm{d}x_1}{\mathrm{d}a} +f^{\prime \prime }(a-x_1)\left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) }{\left[ f^{\prime }(x_1)+f^{\prime }(a-x_1)\right] ^2}. \end{aligned}$$

Next, we calculate \(\frac{\mathrm{d}^2 \Delta _{21}}{\mathrm{d}a^2}\) and evaluate it at \(a=\frac{1}{N}\),

$$\begin{aligned} \left. \frac{\mathrm{d}^2 \Delta _{21}}{\mathrm{d}a^2}\right| _{a_1=1/N}= & {} -\frac{A_3\left[ \left( \frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2 + \left( 1-\frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2\right] 4(A_1)^2 -A_2 2(2A_1)A_2}{16(A_1)^4} \\&\;- \frac{A_3\left[ \left( \frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2 + \left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2\right] 4(A_1)^2 -A_2 2(2A_1)A_2}{16(A_1)^4} \\= & {} -\frac{A_3}{4(A_1)^2}\left[ \left( \frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2 +\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2 + \left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2+ \left( 1-\frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2\right] \\&+\frac{(A_2)^2}{2(A_1)^3}. \\ \end{aligned}$$

Taking total differential using firm 2’s FOC, we can obtain

$$\begin{aligned} \frac{\mathrm{d}x_2}{\mathrm{d}a}-\frac{\mathrm{d}x_1}{\mathrm{d}a}=\frac{\mathrm{d}p_2}{\mathrm{d}a}\Delta _{21}+p_2\frac{d\Delta _{21}}{\mathrm{d}a}. \end{aligned}$$

Taking total differential one more time, we have

$$\begin{aligned} \frac{\mathrm{d}^2 x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}=\frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}\Delta _{21} +2\frac{\mathrm{d}p_2}{\mathrm{d}a} \frac{d\Delta _{21}}{\mathrm{d}a}+p_2\frac{\mathrm{d}^2\Delta _{21}}{\mathrm{d}a^2}. \end{aligned}$$

Evaluating at \(a=\frac{1}{N}\), we have

$$\begin{aligned} \Delta _{21}=\frac{1}{A_1},\quad \frac{d\Delta _{21}}{\mathrm{d}a}=0. \end{aligned}$$

Then

$$\begin{aligned} \frac{\mathrm{d}x_2}{\mathrm{d}a}-\frac{\mathrm{d}x_1}{\mathrm{d}a}=\frac{1}{A_1}\frac{\mathrm{d}p_2}{\mathrm{d}a}. \end{aligned}$$
(12)
$$\begin{aligned} \frac{\mathrm{d}^2 x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}= & {} \frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}\frac{1}{A_1}+p_2\left[ -\frac{A_3}{4(A_1)^2}\left[ \left( \frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2 +\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2 \right. \right. \nonumber \\&\quad \left. \left. + \left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2+ \left( 1-\frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2\right] +\frac{(A_2)^2}{2(A_1)^3}\right] . \end{aligned}$$
(13)

Using firm 3’s FOC

Following similar steps, we can obtain

$$\begin{aligned} {\frac{\mathrm{d}x_3}{\mathrm{d}a}}-\frac{\mathrm{d}x_2}{\mathrm{d}a}=\frac{1}{A_1}\frac{\mathrm{d}p_3}{\mathrm{d}a}+ \frac{A_2}{4N A_1}, \end{aligned}$$
(14)
$$\begin{aligned} \frac{\mathrm{d}^2 x_3}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_2}{\mathrm{d}a^2}= & {} \frac{\mathrm{d}^2 p_3}{\mathrm{d}a^2}\frac{1}{A_1} + \frac{\mathrm{d}p_3}{\mathrm{d}a}\frac{A_2}{2(A_1)^2}\nonumber \\&+\; p_3\left[ -\frac{A_3}{2(A_1)^2} \left( \frac{\mathrm{d}x_3}{\mathrm{d}a}\right) ^2 -\frac{A_1A_3\left[ \left( \frac{\mathrm{d}x_2}{\mathrm{d}a}-1\right) ^2 +\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2\right] - (A_2)^2}{4(A_1)^3}\right] \!.\nonumber \\ \end{aligned}$$
(15)

Using firm i’s FOC, \(i=4,\ldots ,N\)

Following similar steps, we have

$$\begin{aligned} \frac{\mathrm{d}x_i}{\mathrm{d}a}-\frac{\mathrm{d}x_{i-1}}{\mathrm{d}a}=\frac{1}{A_1}\frac{\mathrm{d}p_i}{\mathrm{d}a}, \quad i=4,\ldots ,N. \end{aligned}$$
(16)
$$\begin{aligned} \frac{\mathrm{d}^2 x_i}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_{i-1}}{\mathrm{d}a^2}=\frac{\mathrm{d}^2 p_i}{\mathrm{d}a^2}\frac{1}{A_1} - p_i\left[ \left( \frac{\mathrm{d}x_i}{\mathrm{d}a}\right) ^2 +\left( \frac{\mathrm{d}x_{i-1}}{\mathrm{d}a}\right) ^2\right] \frac{A_3 }{2(A_1)^2}. \end{aligned}$$
(17)

Step 3: Calculating \(\left. \frac{\mathrm{d}x_i}{\mathrm{d}a}\right| _{a=1/N}\) and \(\left. \frac{\mathrm{d}^2 x_i}{\mathrm{d}a^2}\right| _{a=1/N}\), \(i=1,\ldots ,N\).

Note that

$$\begin{aligned} \frac{\mathrm{d}x_1}{\mathrm{d}a}= & {} \frac{\partial x_1}{\partial a} + \frac{\partial x_1}{\partial p_1}\frac{\mathrm{d}p_1}{\mathrm{d}a} + \frac{\partial x_1}{\partial p_2}\frac{\mathrm{d}p_2}{\mathrm{d}a}\\= & {} \frac{f^{\prime }(a-x_1)}{f^{\prime }(x_1)+f^{\prime }(a-x_1)} -\frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)}\frac{\mathrm{d}p_1}{\mathrm{d}a}\\&+\; \frac{1}{f^{\prime }(x_1)+f^{\prime }(a-x_1)}\frac{\mathrm{d}p_2}{\mathrm{d}a}. \end{aligned}$$

Next, we calculate \(\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}\) and evaluate it at \(a=\frac{1}{N}\) to obtain

$$\begin{aligned} {\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}}= \frac{A_2}{2A_1}\left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) - \frac{A_2}{4 A_1}+ \frac{A_2}{4(A_1)^2}\frac{\mathrm{d}p_1}{\mathrm{d}a} + \frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_2}{\mathrm{d}a^2}-\frac{\mathrm{d}^2p_1}{\mathrm{d}a^2}\right) . \end{aligned}$$

We then evaluate \(\frac{\mathrm{d}x_1}{\mathrm{d}a}\) at \(a=\frac{1}{N}\) to obtain

$$\begin{aligned} \frac{\mathrm{d}x_1}{\mathrm{d}a}=\frac{1}{2} + \frac{1}{2A_1}\left( \frac{\mathrm{d}p_2}{\mathrm{d}a}-\frac{\mathrm{d}p_1}{\mathrm{d}a}\right) . \end{aligned}$$

Similarly, it can be shown that

$$\begin{aligned} \frac{\mathrm{d}x_2}{\mathrm{d}a}=\frac{1}{2} + \frac{1}{2A_1}\left( \frac{\mathrm{d}p_3}{\mathrm{d}a}-\frac{\mathrm{d}p_2}{\mathrm{d}a}\right) , \end{aligned}$$
$$\begin{aligned} \frac{\mathrm{d}x_i}{\mathrm{d}a}=\frac{1}{2A_1}\left( \frac{\mathrm{d}p_{i+1}}{\mathrm{d}a}-\frac{\mathrm{d}p_i}{\mathrm{d}a}\right) ,\quad i=3,\ldots , N-1, \end{aligned}$$
$$\begin{aligned} \frac{\mathrm{d}x_N}{\mathrm{d}a}=\frac{1}{2A_1}\left( \frac{\mathrm{d}p_1}{\mathrm{d}a}-\frac{\mathrm{d}p_N}{\mathrm{d}a}\right) . \end{aligned}$$

and

$$\begin{aligned} \frac{\mathrm{d}^2x_2}{\mathrm{d}a^2}=\frac{A_2}{2A_1}\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}-1\right) + \frac{A_2}{4 A_1} + \frac{A_2}{4(A_1)^2}\frac{\mathrm{d}p_3}{\mathrm{d}a} +\frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_3}{\mathrm{d}a^2}-\frac{\mathrm{d}^2p_2}{\mathrm{d}a^2}\right) , \end{aligned}$$
$$\begin{aligned} \frac{\mathrm{d}^2x_i}{\mathrm{d}a^2}=\frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_{i+1}}{\mathrm{d}a^2}-\frac{\mathrm{d}^2 p_i}{\mathrm{d}a^2}\right) ,\quad i=3,\ldots ,N-1, \end{aligned}$$
$$\begin{aligned} \frac{\mathrm{d}^2x_N}{\mathrm{d}a^2}=\frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_1}{\mathrm{d}a^2}-\frac{\mathrm{d}^2 p_N}{\mathrm{d}a^2}\right) . \end{aligned}$$

Step 4: Solving for \(\left. \frac{\mathrm{d}p_i}{\mathrm{d}a}\right| _{a=1/N}\), \(i=1,\ldots ,N\).

Substituting the above \(\frac{\mathrm{d}x_i}{\mathrm{d}a}\) expressions into Eqs. (10), (12), (14) and (16), we can obtain

$$\begin{aligned} \frac{1}{2}+\frac{1}{2A_1}\left( \frac{\mathrm{d}p_2}{\mathrm{d}a}-4\frac{\mathrm{d}p_1}{\mathrm{d}a}+\frac{\mathrm{d}p_N}{\mathrm{d}a}\right) =-\frac{A_2}{4NA_1}, \end{aligned}$$
(18)
$$\begin{aligned} \frac{1}{2A_1}\left( \frac{\mathrm{d}p_3}{\mathrm{d}a}-4\frac{\mathrm{d}p_2}{\mathrm{d}a}+\frac{\mathrm{d}p_1}{\mathrm{d}a}\right) =0, \end{aligned}$$
$$\begin{aligned} -\frac{1}{2}+\frac{1}{2A_1}\left( \frac{\mathrm{d}p_4}{\mathrm{d}a}-4\frac{\mathrm{d}p_3}{\mathrm{d}a}+\frac{\mathrm{d}p_2}{\mathrm{d}a}\right) =\frac{A_2}{4NA_1}, \end{aligned}$$
$$\begin{aligned} \frac{1}{2A_1}\left( \frac{\mathrm{d}p_{i+1}}{\mathrm{d}a}-4\frac{\mathrm{d}p_i}{\mathrm{d}a} +\frac{\mathrm{d}p_{i-1}}{\mathrm{d}a}\right) =0, \quad i=4,\ldots , N-1, \end{aligned}$$
(19)
$$\begin{aligned} \frac{1}{2A_1}\left( \frac{\mathrm{d}p_1}{\mathrm{d}a}-4\frac{\mathrm{d}p_N}{\mathrm{d}a}+\frac{\mathrm{d}p_{N-1}}{\mathrm{d}a}\right) =0. \end{aligned}$$

Let \(y_i\equiv \frac{\mathrm{d}p_i}{\mathrm{d}a}\), \(i=1,\ldots , N\). Equation (19) is a homogeneous linear second-order difference equation. The general solution takes the form of

$$\begin{aligned} y_i=c_1\left( 2+\sqrt{3}\right) ^{i-3} +c_2\left( 2-\sqrt{3}\right) ^{i-3},\quad i=3,\ldots ,N. \end{aligned}$$
(20)

Then

$$\begin{aligned} y_N=c_1\left( 2+\sqrt{3}\right) ^{N-3} +c_2\left( 2-\sqrt{3}\right) ^{N-3}. \end{aligned}$$

Using the initial conditions \(y_i(i=3)=y_3\) and \(y_i(i=4)=y_4\), we can recover

$$\begin{aligned} c_1=\left( \frac{1}{2}-\frac{\sqrt{3}}{3}\right) y_3+\frac{\sqrt{3}}{6}y_4,\quad c_2=\left( \frac{1}{2}+\frac{\sqrt{3}}{3}\right) y_3-\frac{\sqrt{3}}{6}y_4. \end{aligned}$$

Due to symmetry, we have \(y_2=0\), \(y_1=-y_3\) and \(y_N=-y_4\). Using the symmetry condition, Eq. (18) is an equation of \(c_1\) and \(c_2\). Moreover, the symmetry condition \(y_N=-y_4\) is another equation of \(c_1\) and \(c_2\). Combining these two equations, we can obtain

$$\begin{aligned} c_1= \frac{1}{2} \cdot \frac{\left( 2 A1+\frac{A2}{N}\right) \left[ 2 \sqrt{3}-3+\left( 26 \sqrt{3} +45\right) (2-\sqrt{3})^N\right] }{\left[ 24-\left( 7 \sqrt{3}+12\right) (2-\sqrt{3})^N+\left( 7 \sqrt{3}-12\right) (2+\sqrt{3})^N\right] }, \end{aligned}$$
(21)
$$\begin{aligned} c_2= \frac{1}{2} \cdot \frac{\left( 2 A1+\frac{A2}{N}\right) \left[ -2 \sqrt{3}-3+\left( 45- 26 \sqrt{3}\right) (2+\sqrt{3})^N\right] }{\left[ 24-\left( 7 \sqrt{3}+12\right) (2-\sqrt{3})^N+\left( 7 \sqrt{3}-12\right) (2+\sqrt{3})^N\right] }. \end{aligned}$$
(22)

Substituting them into Eq. (20), we can then obtain all \(y_i\), \(i=3,\ldots ,N\). While this formula is derived for \(N\ge 4\), it can be verified that it leads to the correct results for \(N=2,3\) as well.Footnote 19 In particular, when \(N=3\), \(\left. \frac{\mathrm{d}p_3}{\mathrm{d}a}\right| _{a=1/3}=-\frac{1}{5}\left( A_1+\frac{A_2}{2N}\right) \). When \(N=2\), the above formula leads to \(\left. \frac{\mathrm{d}p_1}{\mathrm{d}a}\right| _{a=1/2}=\left. \frac{\mathrm{d}p_2}{\mathrm{d}a}\right| _{a=1/2}=0\). \(\square \)

Proof of Proposition 5

Note that at \(a=1/N\), \(\frac{\mathrm{d}p_2}{\mathrm{d}a}=\frac{\mathrm{d}\pi _2}{\mathrm{d}a}=0\). A necessary condition for firm 2 not to have an incentive to deviate locally is \(\left. \frac{\mathrm{d}^2\pi _2}{\mathrm{d}a^2}\right| _{a=\frac{1}{N}}\le 0\). Next, we derive the expression for \(\left. \frac{\mathrm{d}^2\pi _2}{\mathrm{d}a^2}\right| _{a=\frac{1}{N}}\).

Using the expressions of \(\frac{\mathrm{d}p_i}{\mathrm{d}a}\), we can obtain \(\frac{\mathrm{d}x_i}{\mathrm{d}a}\) (see step 3 in Proof of Proposition 4). Next, substituting \(\frac{\mathrm{d}p_i}{\mathrm{d}a}\), \(\frac{\mathrm{d}x_i}{\mathrm{d}a}\) and the \(\frac{\mathrm{d}^2 x_i}{\mathrm{d}a^2}\) expressions into Eqs. (11), (13), (15) and (17), we can obtain (all evaluated at \(a=\frac{1}{N}\))

$$\begin{aligned}&\left( \frac{\mathrm{d}^2p_2}{\mathrm{d}a^2}-4\frac{\mathrm{d}^2 p_1}{\mathrm{d}a^2}+\frac{\mathrm{d}^2p_N}{\mathrm{d}a^2}\right) \nonumber \\&\quad = -A_2\left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) +\frac{A_2}{2} -\frac{3A_2}{2A_1}\frac{\mathrm{d}p_1}{\mathrm{d}a}\nonumber \\&\qquad -\; p_1\left[ \frac{A_1A_3\left[ \left( \frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2 + \left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2\right] -(A_2)^2}{2(A_1)^2} + \frac{A_3}{A_1}\left( \frac{\mathrm{d}x_N}{\mathrm{d}a}\right) ^2\right] , \end{aligned}$$
(23)
$$\begin{aligned}&\frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_3}{\mathrm{d}a^2}-4\frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}+\frac{\mathrm{d}^2p_1}{\mathrm{d}a^2}\right) \nonumber \\&\quad = -\frac{A_2}{2A_1}\left( \frac{\mathrm{d}x_1}{\mathrm{d}a}+\frac{\mathrm{d}x_2}{\mathrm{d}a}-2\right) -\frac{A_2}{2A_1} +\frac{A_2}{2(A_1)^2}\frac{\mathrm{d}p_1}{\mathrm{d}a}\nonumber \\&\qquad +\; p_2\left[ -\frac{A_3}{4(A_1)^2}\left[ \left( \frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2 +\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2 + \left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) ^2+ \left( 1-\frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2\right] +\frac{(A_2)^2}{2(A_1)^3}\right] ,\nonumber \\ \end{aligned}$$
(24)
$$\begin{aligned}&\frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_4}{\mathrm{d}a^2}-4\frac{\mathrm{d}^2 p_3}{\mathrm{d}a^2}+\frac{\mathrm{d}^2p_2}{\mathrm{d}a^2}\right) \nonumber \\&\quad = \frac{A_2}{2A_1}\left( 1-\frac{\mathrm{d}x_1}{\mathrm{d}a}\right) -\frac{A_2}{4A_1} -\frac{3A_2}{4(A_1)^2}\frac{\mathrm{d}p_1}{\mathrm{d}a}\nonumber \\&\qquad + p_3\left[ -\frac{A_3}{2(A_1)^2}\left( \frac{\mathrm{d}x_3}{\mathrm{d}a}\right) ^2 -\frac{A_1A_3\left[ \left( \frac{\mathrm{d}x_2}{\mathrm{d}a}-1\right) ^2 +\left( \frac{\mathrm{d}x_2}{\mathrm{d}a}\right) ^2\right] - (A_2)^2}{4(A_1)^3}\right] ,\nonumber \\ \end{aligned}$$
(25)
$$\begin{aligned}&\left( \frac{\mathrm{d}^2p_{i+1}}{\mathrm{d}a^2}-4\frac{\mathrm{d}^2 p_i}{\mathrm{d}a^2}+\frac{\mathrm{d}^2p_{i-1}}{\mathrm{d}a^2}\right) \nonumber \\&\quad = - \frac{A_3}{N}\left[ \left( \frac{\mathrm{d}x_i}{\mathrm{d}a}\right) ^2 +\left( \frac{\mathrm{d}x_{i-1}}{\mathrm{d}a}\right) ^2\right] ,\quad i=4,\ldots ,N-1, \end{aligned}$$
(26)
$$\begin{aligned} \frac{1}{2A_1}\left( \frac{\mathrm{d}^2p_1}{\mathrm{d}a^2}-4\frac{\mathrm{d}^2 p_N}{\mathrm{d}a^2}+\frac{\mathrm{d}^2p_{N-1}}{\mathrm{d}a^2}\right) = - p_N\left[ \left( \frac{\mathrm{d}x_N}{\mathrm{d}a}\right) ^2 +\left( \frac{\mathrm{d}x_{N-1}}{\mathrm{d}a}\right) ^2\right] \frac{A_3 }{2(A_1)^2}.\nonumber \\ \end{aligned}$$
(27)

Equation (25) is symmetric to Eq. (23). Equation (26) is a non-homogeneous linear second-order difference equation. The general solution takes the form of

$$\begin{aligned} \frac{\mathrm{d}^2 p_i}{\mathrm{d}a^2}=k_1\left( 2+\sqrt{3}\right) ^{i-3} +k_2\left( 2-\sqrt{3}\right) ^{i-3}+sol_p,\quad i=3,\ldots ,N, \end{aligned}$$
(28)

where \(sol_p\) is any particular solution to the difference equation. We guess and then verify the following particular solution

$$\begin{aligned} sol_p=\theta _0+\theta _1(7+4\sqrt{3})^i+\theta _2(7-4\sqrt{3})^i, \end{aligned}$$

where \(\theta _0=-\frac{A_3c_1c_2}{N(A_1)^2}\), \(\theta _1 = \frac{A_3 (c_1)^2(-1351+780\sqrt{3})}{5NA1^2}\), \(\theta _2= -\frac{A_3(c_2)^2(780\sqrt{3}+1351)}{5N(A_1)^2}\), and \(c_1\) and \(c_2\) come from the \(\frac{\mathrm{d}p_i}{\mathrm{d}a}\) expression earlier. Using initial conditions \(\frac{\mathrm{d}^2 p_i}{\mathrm{d}a^2} (i=3)=\frac{\mathrm{d}^2 p_3}{\mathrm{d}a^2}\) and \(\frac{\mathrm{d}^2 p_i}{\mathrm{d}a^2} (i=4)=\frac{\mathrm{d}^2 p_4}{\mathrm{d}a^2}\), \(k_1\) and \(k_2\) then can be expressed in \(\frac{\mathrm{d}^2 p_3}{\mathrm{d}a^2}\) and \(\frac{\mathrm{d}^2 p_4}{\mathrm{d}a^2}\). Different from \(\frac{\mathrm{d}p_2}{\mathrm{d}a}=0\), we do not have \(\frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}=0\). So we use Eqs. (23), (24) and (27) to solve for \(\frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}\), \(\frac{\mathrm{d}^2 p_3}{\mathrm{d}a^2}\) and \(\frac{\mathrm{d}^2 p_4}{\mathrm{d}a^2}\).

With \(\pi _2=p_2(x_2-x_1)\), we have

$$\begin{aligned} \frac{\mathrm{d}^2 \pi _2}{\mathrm{d}a^2}= & {} \frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}(x_2-x_1)+ 2p_2\left( \frac{\mathrm{d}x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}x_1}{\mathrm{d}a^2}\right) + p_2\left( \frac{\mathrm{d}^2x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}\right) \\= & {} \frac{1}{N} \frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}+ 2p_2\left( \frac{\mathrm{d}x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}x_1}{\mathrm{d}a^2}\right) + p_2\left( \frac{\mathrm{d}^2x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}\right) \\= & {} {\frac{1}{N} \frac{\mathrm{d}^2 p_2}{\mathrm{d}a^2}+ p_2\left( \frac{\mathrm{d}^2x_2}{\mathrm{d}a^2}-\frac{\mathrm{d}^2x_1}{\mathrm{d}a^2}\right) }. \end{aligned}$$

Using the expressions of \(\frac{\mathrm{d}p_i}{\mathrm{d}a}\), \(\frac{\mathrm{d}x_i}{\mathrm{d}a}\), \(\frac{\mathrm{d}^2p_i}{\mathrm{d}a^2}\) and \(\frac{\mathrm{d}^2 x_i}{\mathrm{d}a^2}\), we can then obtain the expression of \(\left. \frac{\mathrm{d}^2\pi _2}{\mathrm{d}a^2}\right| _{a=\frac{1}{N}}\). It involves the third derivative of f(l) and is too lengthy to report. To obtain cleaner results, we consider specific classes of transport cost functions including polynomials, exponential and logarithmic functions. It is then straightforward to verify the results in Proposition 5.Footnote 20 \(\square \)

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Gong, Q., Liu, Q. & Zhang, Y. Optimal product differentiation in a circular model. J Econ 119, 219–252 (2016). https://doi.org/10.1007/s00712-016-0489-1

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