Spatial competition and social welfare considering different feasible location regions

Abstract

This research examines the location-price game of two firms, where consumers with quadratic transportation costs are dispersed along the linear city [0, 1], while firms can select their business sites on a continuous but arbitrarily constrained interval [m, n] or on a discontinuous interval separated by a continuous zoning area, in which business activities are prohibited. We find that at least one subgame perfect equilibrium in pure strategies exists. Multiple equilibria may emerge when the continuous location region is far from consumers with the remote firm earning a zero profit. When the continuous region is relatively small and not too far from consumers, two firms choose to locate separately at the endpoints of the interval. When considering the discontinuous location region, the highest social surplus can be achieved when both firms are located on one side of the zoning area.

Appendix

Proof of Lemma 1

Note that when $\overline{x} = \frac{{b^{2} - a^{2} + p_{2} - p_{1} }}{{2\left( {b - a} \right)}} < 0$, firm 2 owns the whole market. Considering the second stage game, given $p_{1} \ge 0$, it will always be better for firm 2 to increase its price to $p_{2}^{^{\prime}} \left( { > p_{2} \ge 0} \right)$, such that $\overline{x} = \frac{{b^{2} - a^{2} + p_{2} - p_{1} }}{{2\left( {b - a} \right)}} < 0 = \frac{{b^{2} - a^{2} + p_{2}^{^{\prime}} - p_{1} }}{{2\left( {b - a} \right)}}{ } = \overline{x}^{^{\prime}}$. Similarly, when $\overline{x} = \frac{{b^{2} - a^{2} + p_{2} - p_{1} }}{{2\left( {b - a} \right)}} > 1$, firm 1 owns the whole market. Considering the second stage game, given $p_{2} \ge 0,{ }$ it will always be better for firm 1 to increase its price to $p_{1}^{^{\prime}} \left( { > p_{1} \ge 0} \right)$, such that $\overline{x} = \frac{{b^{2} - a^{2} + p_{2} - p_{1} }}{{2\left( {b - a} \right)}} > 1 = \frac{{b^{2} - a^{2} + p_{2} - p_{1}^{^{\prime}} }}{{2\left( {b - a} \right)}}{ } = \overline{x}^{^{\prime}}$. Therefore, in the second stage game, the equilibrium may only exist when $\overline{x} = \frac{{b^{2} - a^{2} + p_{2} - p_{1} }}{{2\left( {b - a} \right)}} \in \left[ {0,{ }1} \right]$.□

Proof of Proposition 1

We only need to prove Proposition 1(i) where $a < b$. (a) When $2 + a + b \le 0 \Rightarrow a + b \le - 2 < 0$, from the discussion above Proposition 1 we know that the equilibrium only exists in Region 1, which is {$p_{1}^{c} = 0$, $p_{2}^{c} = a^{2} - b^{2} > 0$}. (b) Similarly, when $4 - b - a \le 0 \Rightarrow a + b \ge 4 > 2$, the equilibrium only exists in Region 3, which is {$p_{1}^{c} = b^{2} - a^{2} - 2\left( {b - a} \right) > 0$, $p_{2}^{c} = 0$}. (c) It is easy to prove that when $a + b = 1$ (i.e., two firms are symmetric about the city center), {$p_{1}^{c} = \frac{1}{3}\left( {b - a} \right)\left( {2 + a + b} \right) > 0$, $p_{2}^{c} = \frac{1}{3}\left( {b - a} \right)\left( {4 - b - a} \right) > 0$} is the unique equilibrium. When $- 2 < a + b < 1$ (i.e., firm 2 has market advantage), the equilibrium may exist in both Region 1 and 2. Note that considering the outcome {$p_{1} = 0$, $p_{2} = a^{2} - b^{2} \ge 0\}$ in Region 1, given $p_{1} = 0$, if firm 2 sets a higher price $p_{2}^{^{\prime}} = \frac{1}{2}\left( {2\left( {b - a} \right) - b^{2} + a^{2} } \right) > p_{2}$ (note that $p_{2}^{^{\prime}} - p_{2} = \frac{1}{2}\left( {2\left( {b - a} \right) - b^{2} + a^{2} } \right) - \left( {a^{2} - b^{2} } \right) = \frac{1}{2}\left( {b - a} \right)\left( {2 + a + b} \right) > 0$), firm 2 can obtain a higher profit, since $\Pi_{2}^{^{\prime}} = p_{2}^{^{\prime}} \left( {1 - \frac{{b^{2} - a^{2} + p_{2}^{^{\prime}} }}{{2\left( {b - a} \right)}}} \right) = p_{2}^{^{\prime}} \left( {\frac{{2\left( {b - a} \right) - b^{2} + a^{2} - p_{2}^{^{\prime}} }}{{2\left( {b - a} \right)}}} \right) = \frac{1}{8}\left( {b - a} \right)\left( {2 - a - b} \right)^{2} > a^{2} - b^{2} = \Pi_{2}$ (note that $\Pi_{2}^{^{\prime}} - \Pi_{2} = \frac{1}{8}\left( {b - a} \right)\left( {2 - a - b} \right)^{2} - \left( {a^{2} - b^{2} } \right) = \frac{1}{8}\left( {b - a} \right)\left( {2 + a + b} \right)^{2} > 0$). It means that firm 2 has the incentive to deviate from the outcome in Region 1. Therefore, only {$p_{1}^{c} = \frac{1}{3}\left( {b - a} \right)\left( {2 + a + b} \right) > 0$, $p_{2}^{c} = \frac{1}{3}\left( {b - a} \right)\left( {4 - b - a} \right) > 0$} in Region 2 is the equilibrium. Similarly, we can prove that {$p_{1}^{c} = \frac{1}{3}\left( {b - a} \right)\left( {2 + a + b} \right) > 0$, $p_{2}^{c} = \frac{1}{3}\left( {b - a} \right)\left( {4 - b - a} \right) > 0$} is the unique equilibrium when $1 < a + b < 4$ (i.e., firm 1 has market advantage).□

Proof of Corollary 1

When $n \le - 1$, we have $a \le b \le n \le - 1 \Rightarrow 2 + a + b \le 0$. From Proposition 1 we know that for any $2 + a + b \le 0$, we have $p_{1}^{c} = 0$ and $\Pi_{1}^{c} = 0$. Therefore, at least one of two firms will have zero profit at the equilibrium. The proof of the case when ${ }m \ge 2$ is similar.□

Proof of Proposition 2

• (i) When $n \le - 1$, if $a < b$, we have $a + b < 2n < - 2$. From Proposition 1 we know that the equilibrium prices for the second stage game are $p_{1} = 0$ and $p_{2} = a^{2} - b^{2}$. In the first stage, for firm 2, we have $D_{2} = 1$ and $\Pi_{2} = a^{2} - b^{2}$. Since $\Pi_{2}$ is a quadratic function and is increasing in $b$ for $b{ } \le n \le - 1 < 0$, we have $b^{c} = n$. For firm 1, any feasible point is an equilibrium location since it will not be better off from deviation (i.e., any deviation will still result in a zero profit). The proof of part (ii) is similar and thus omitted.□

Proof of Proposition 3

Recalling the discussion in Propositions 1 and 2, we need only consider situations in which the second stage equilibrium prices are positive (i.e., when $- 2 < a + b < 4$ and $a < b$). By backward induction, firms’ profits in the first stage can be rewritten as:

$$\Pi_{1} = \frac{1}{18}\left( {b - a} \right)\left( {2 + a + b} \right)^{2} ,\,\,\Pi_{2} = \frac{1}{18}\left( {b - a} \right)\left( {4 - a - b} \right)^{2} .$$

Take the first order derivative of $\Pi_{1}$ and $\Pi_{2}$ w.r.t. $a$ and $b$, respectively. We have

$$\frac{{\partial \Pi_{1} }}{\partial a} = \frac{1}{18}\left( { - 2 - 3a + b} \right)\left( {2 + a + b} \right),\,\,\frac{{\partial \Pi_{2} }}{\partial b} = - \frac{1}{18}\left( {4 + a - 3b} \right)\left( { - 4 + a + b} \right).$$

Solving $\frac{{\partial \Pi_{1} }}{\partial a} = 0$ yields $a_{1} = - 2 - b$ and $a_{2} = \left( { - 2 + b} \right)/3$. Since $- 2 < a + b < 4$, only $a_{2}$ can be the solution. It follows that firm 1’s best response is $a = a_{2} = \left( { - 2 + b} \right)/3$.

Similarly, solving $\frac{{\partial \Pi_{2} }}{\partial b} = 0$, we have $b_{1} = \left( {4 + a} \right)/3$ and $b_{2} = 4 - a$. Since $- 2 < a + b < 4$, only $b_{1}$ can be the solution. It follows that firm 2’s best response is $b = b_{1} = \left( {4 + a} \right)/3$ when $b_{1} < n$, and $b = n$ when $b_{1} \ge n$.

Solving the set of equations $a = \left( { - 2 + b} \right)/3$ and $b = \left( {4 + a} \right)/3$, we have $a = - 1/4$ and ${ }b = 5/4$. Therefore, when $- 1 < n < 5/4$, the unique SPE for the two-stage location-price game is $a^{c} = \left( { - 2 + n} \right)/3$, $b^{c} = n$. From the above discussion, we can readily prove that when $n \ge 5/4$, the equilibrium locations are $a^{c} = - 1/4$,${ }b^{c} = 5/4$, a result the same as that of Lambertini (1994) where firms choose locations on $\left( { - \infty , + \infty } \right)$.

With the equilibriums shown above, we derive the consumer surplus by solving $\mathop \smallint \limits_{0}^{{\overline{x}^{c} }} \left( {v - \left( {x - a^{c} } \right)^{2} - p_{1}^{{\text{c}}} } \right){\text{d}}x + \mathop \smallint \limits_{{\overline{x}^{c} }}^{1} \left( {v - \left( {x - b^{c} } \right)^{2} - p_{2}^{c} } \right){\text{d}}x.$ And the social surplus can be easily derived by solving $\overline{x}^{c} p_{1}^{{\text{c}}} + \left( {1 - \overline{x}^{c} } \right)p_{2}^{c} + \mathop \smallint \limits_{0}^{{\overline{x}^{c} }} \left( {v - \left( {x - a} \right)^{2} - p_{1}^{{\text{c}}} } \right){\text{d}}x + \mathop \smallint \limits_{{\overline{x}^{c} }}^{1} \left( {v - \left( {x - b} \right)^{2} - p_{2}^{c} } \right){\text{d}}x$, where $\overline{x}^{c} = \frac{{{b^{{c}^{2}}} - a^{{c}^{2}} + p_{2}^{c} - p_{1}^{{\text{c}}} }}{{2\left( {b^{c} - a^{c} } \right)}}$.□

Proof of Corollary 2

Recall that when $- 1 < n < 5/4$, $a^{c} = \left( { - 2 + n} \right)/3$, $b^{c} = n$, $b^{c} - a^{c} = 2\left( {1 + n} \right)/3$, $p_{1}^{c} = \frac{8}{27}\left( {1 + n} \right)^{2}$, $p_{2}^{c} = \frac{4}{27}\left( {1 + n} \right)\left( {7 - 2n} \right)$, $\Pi_{1}^{c} = \frac{16}{{243}}\left( {1 + n} \right)^{3}$, $\Pi_{2}^{c} = \frac{4}{243}\left( {1 + n} \right)\left( {7 - 2n} \right)^{2}$, and $\Pi_{1}^{c} + \Pi_{2}^{c} = \frac{4}{243}\left( {53 + 33n - 12n^{2} + 8n^{3} } \right)$. It is straightforward that $a^{c}$, $b^{c}$, $b^{c} - a^{c}$, $p_{1}^{c}$, $p_{2}^{c}$, $\Pi_{1}^{c}$ and $\Pi_{1}^{c} + \Pi_{2}^{c}$ all increase in $n$ (note that $\frac{{\partial p_{2}^{c} }}{\partial n} = - \frac{4}{27}\left( { - 5 + 4n} \right) > 0$, and $\frac{{\partial \left( {\Pi_{1}^{c} + \Pi_{2}^{c} } \right)}}{\partial n} = \frac{4}{81}\left( {11 - 8n + 8n^{2} } \right) > 0$). Taking the derivative of $\Pi_{2}^{c}$ with respect to $n$, we have $\frac{{\partial \Pi_{2}^{c} }}{\partial n} = \frac{4}{81}\left( {2n - 1} \right)\left( {2n - 7} \right) > 0$ when $- 1 < n < \frac{1}{2}$, and $\frac{{\partial \Pi_{2}^{c} }}{\partial n} \le 0$ when $1/2 \le n < 5/4$.□

Proof of Corollary 3

For $\left( { - \infty ,n} \right]$ where $n > - 1$, from Proposition 3 and Corollary 2 we know that $\Pi_{2}^{c}$ is maximized when $n = 1/2$ and $\Pi_{2}^{c} = 8/9$, and the equilibrium locations are $a^{c} = - 1/2$, $b^{c} = 1/2$. It is easy to verify that the condition $n = 1/2$ and $m \le - 1/2$ results in the same ontcome. From Proposition 4, when $- 1 < n < 5/4$ and $m \le \left( { - 2 + n} \right)/3$, $\Pi_{2}^{c} = \frac{4}{243}\left( {1 + n} \right)\left( {7 - 2n} \right)^{2} \le 8/9$. When $- 1 < n < 5/4$ and $\left( { - 2 + n} \right)/3 < m < 2$, we have $\Pi_{2}^{c} = \frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2} < \frac{1}{18}\left( {n - \left( { - 2 + n} \right)/3} \right)\left( {4 - \left( { - 2 + n} \right)/3 - n} \right)^{2} = \frac{4}{243}\left( {1 + n} \right)\left( {7 - 2n} \right)^{2} \le 8/9$. When $n \ge 5/4$ and $m \le - 1/4$, $\Pi_{2}^{c} = 3/4 < 8/9$. When $n \ge 5/4$ and $- \frac{1}{4} < m < 3n - 4$, $\Pi_{2}^{c} = \frac{16}{{243}}\left( {2 - m} \right)^{3} < \frac{16}{{243}}\left( {2 - \left( { - 1/4} \right)} \right)^{3} = 3/4 < 8/9$. Finally, when $n \ge 5/4$ and $3n - 4 < m < 2$, $\Pi_{2}^{c} = \frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2} < \frac{1}{18}\left( {n - \left( {3n - 4} \right)} \right)\left( {4 - \left( {3n - 4} \right) - n} \right)^{2} = \frac{16}{9}\left( {2 - n} \right)^{3} \le \frac{16}{9}\left( {2 - 5/4} \right)^{3} = 3/4 < 8/9$. Therefore, Corollary 3 (i) follows. The proof of (ii) and (iii) is similar and thus omitted.□

Proof of Proposition 4

Similar to the proof of Proposition 3.□

Proof of Corollary 4

By solving

$$\overline{x}^{c} p_{1}^{{\text{c}}} + \left( {1 - x^{*} } \right)p_{2}^{c} + \mathop \smallint \limits_{0}^{{\overline{x}^{c} }} \left( {v - \left( {x - a} \right)^{2} - p_{1}^{{\text{c}}} } \right){\text{d}}x + \mathop \smallint \limits_{{\overline{x}^{c} }}^{1} \left( {v - \left( {x - b} \right)^{2} - p_{2}^{c} } \right){\text{d}}x,$$

we can obtain the equilibrium social surplus under different scenarios, shown in Table 3.

Table 3 The social surplus under different regions in Fig. 6

For convenience, here, we define the social surplus in region i under continuous location region as $SW_{i}^{c}$. First, we maximize $SW_{2}^{c} = v - \frac{1}{36}\left( {12 + 5m^{3} - 32n + 28n^{2} - 5n^{3} + m^{2} \left( {8 + 5n} \right) - m\left( {4 + 5n^{2} } \right)} \right)$ to derive the highest social surplus in region ②. By solving $\frac{\partial }{\partial n}\left( {v - \frac{1}{36}\left( {12 + 5m^{3} - 32n + 28n^{2} - 5n^{3} + m^{2} \left( {8 + 5n} \right) - m\left( {4 + 5n^{2} } \right)} \right)} \right) = 0$, we obtain the unique local maximizer in region ②, i.e., $n^{*} = \frac{1}{15}\left( {28 - 5m - 2\sqrt {76 - 70m + 25m^{2} } } \right)$. Substituting $n^{*}$ into the optimization problem and solving the first order condition w.r.t. $m$, we obtain the unique local maximizer in region ②, i.e., $m^{*} = \frac{1}{4}$. Furthermore, we have $n^{*} = \frac{3}{4}$. And the corresponding social surplus is $v - \frac{1}{48}$.

We can easily derive the maximum social surplus with no constraint (denoted by the superscript “*”), i.e., $SW_{i}^{{c}^{*}} = \left\{ \begin{gathered} v - \frac{13}{{48}},i = 1 \hfill \\ v - \frac{{595 - 29\sqrt {145} }}{1440},i = 3,4 \hfill \\ v - \frac{1}{12},i = 5,6 \hfill \\ \end{gathered} \right.$. It can be easily verified that $SW_{i}^{{c}^{*}} < SW_{2}^{c}$. That is, the highest social surplus is achieved in region ②.□

Proof of Proposition 5

First consider that two firms locate in different intervals of the discontinuous line, i.e., $a \le m < 1/2 < 1 - m \le b$. In this case, from Lemma 1 and Proposition 1, we know that in equilibrium firms will always choose locations such that $- 2 < a + b < 4$ and thus be able to set positive prices in the second stage (note that $- 2 < a + b < 4$ can be satisfied when $a + b = 1$). Recall the proof of Proposition 3. By backward induction, firms’ profits in the first stage can be rewritten as:

$$\Pi_{1} = \frac{1}{18}\left( {b - a} \right)\left( {2 + a + b} \right)^{2} ,\,\,\Pi_{2} = \frac{1}{18}\left( {b - a} \right)\left( {4 - a - b} \right)^{2} .$$

Solving $\frac{{\partial \Pi_{1} }}{\partial a} = 0$ yields $a_{1} = - 2 - b < a_{2} = \left( { - 2 + b} \right)/3$. As $a$ increases from $a_{1}$, $\Pi_{1}$ first increases, and then decreases, achieving the maximum at $a = a_{2}$. Under the condition $a \le m < 1/2 < 1 - m \le b$, only when $m \ge a \ge \left( { - 2 + b} \right)/3 \ge \left( { - 2 + 1 - m} \right)/3 = \left( { - 1 - m} \right)/3$, i.e., $- 1/4 \le m < 1/2$, is firm 1’s best response $a = a_{2} = \left( { - 2 + b} \right)/3$. We can readily verify that firm 2’s best response is $b = b_{1} = \left( {4 + a} \right)/3$. As a result, the equilibrium locations are $a^{d} = - 1/4$,${ }b^{d} = 5/4$, and the corresponding prices and profits are $p_{1}^{d} = p_{2}^{d} = 3/2$, $\Pi_{1}^{d} = \Pi_{2}^{d} = 3/4$. When $m < - 1/4$, firms’ best responses are $a = m$ and $b = 1 - m$, respectively. Therefore, the equilibrium locations are $a^{d} = m$,${ }b^{d} = 1 - m$, and the corresponding prices and profits are $p_{1}^{d} = p_{2}^{d} = 1 - 2m$, $\Pi_{1}^{d} = \Pi_{2}^{d} = \left( {1 - 2m} \right)/2$.

When two firms locate in the same interval, without loss of generality, assume $a \le {\text{b}} \le m < 1/2$. When $- 1 < m < 1/2$, this is the same case as Proposition 3 (i) with the constraint that $- 1 < n < 1/2$. Thus the equilibrium outcomes are $a^{d} = \left( { - 2 + m} \right)/3$, $b^{d} = m$, $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$, and $\Pi_{1}^{d} = \frac{16}{{243}}\left( {1 + m} \right)^{3}$, $\Pi_{2}^{d} = \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2}$. When $m \le - 1$, Proposition 2 (i) we have {$a^{d} = a$, $b^{d} = n$; $p_{1}^{d} = 0$, $p_{2}^{d} = a^{2} - n^{2}$}, where $m \le a \le n$. The corresponding profits are $\Pi_{1}^{d} = 0$, $\Pi_{2}^{d} = a^{2} - n^{2}$.

Comparing the above two cases, when $- 1 < m < - 1/4$, $\Pi_{1}^{d} = \left( {1 - 2m} \right)/2 > \frac{16}{{243}}\left( {1 + m} \right)^{3}$ and $\Pi_{2}^{d} = \left( {1 - 2m} \right)/2 > \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2}$. When $- 1/4 \le m < 1/2$, $\Pi_{1}^{d} = 3/4 > \frac{16}{{243}}\left( {1 + m} \right)^{3}$ and $\Pi_{2}^{d} = 3/4 > \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2}$. Thus, both firms will be better off if they locate separately at the endpoints of the zoning area. When $m \le - 1$, one of the firms always has zero profit and it has an incentive to relocate to the opposite side. Therefore, the firms prefer to locate in different intervals of the discontinuous line. The derivation for social welfare will be shown in the proof of Proposition 6.□

Proof of Corollary 5

Note that when $m = 1 - n \le - 1/4$, $p_{1}^{d} = p_{2}^{d} = 1 - 2m$ and the corresponding profits are $\Pi_{1}^{d} = \Pi_{2}^{d} = \left( {1 - 2m} \right)/2$. It is easy to verified that $p_{i}^{d}$ and $\Pi_{i}^{d}$ $\left( {i = 1,2} \right){ }$ are increasing as $m$ decreases. Note that the social surplus is equal to $v - \left( {\frac{1}{12} - \frac{m}{2} + m^{2} } \right)$, which is increasing in $m$ when $m \le 1/4$. Thus, when $m = 1 - n \le - 1/4$, the social surplus is decreasing as $m$ decreases.□

Proof of Proposition 6

(i) According to the above discussion, we can get (i) straightforwardly.

(ii) Next, we consider the cases when $m + n > 1$ and $m < 5/4$ (note that $n > 1/2$ since $1 < m + n < 2n$).

(ii)(a) When $m \le - 1$ and $1 < m + n\left\langle {4 \Rightarrow n} \right\rangle 1 - m \ge 2$, recalling Proposition 2, we know that when two firms locate at the same sides of the location region ($a \le b \le m$ or $n \le a \le {\text{b}}$), at least one firm will get a zero profit. Therefore, two firms will locate at different sides of the zoning area where both firms can get positive profits, and the unique SPE is {$a^{d} = m$, $b^{d} = n$; $p_{1}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {2 + m + n} \right) > 0$, $p_{2}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {4 - m - n} \right) > 0$} (Proposition 1(i)(c)).

(ii)(b) When $m \le - 1$ and $m + n \ge 4$ and $m < n$, at least one firm will locate at the end point of the left side of the zoning area which is more close to consumers, while the other firm will always get a zero profit wherever it locates. Next, we want to prove that {$a^{d} = m$, $b^{d} = b \ge n$; $p_{1}^{d} = b^{2} - m^{2} - 2\left( {b - m} \right)$, $p_{2}^{d} = 0$} and {$a^{d} = a \le m$, $b^{d} = m$; $p_{1}^{d} = 0$, $p_{2}^{d} = a^{2} - m^{2}$} are all SPEs. To prove {$a^{d} = m$, $b^{d} = b \ge n$; $p_{1}^{d} = b^{2} - m^{2} - 2\left( {b - m} \right)$, $p_{2}^{d} = 0$} is an SPE, we only need to show that in the first stage, given $b^{d} = b > n$, firm 1 has no incentive to locate at $a^{\prime} = n$. The profit of firm 1 when locating at $a^{\prime} = n$ is $\Pi_{1}^{^{\prime}} = b^{2} - n^{2} - 2\left( {b - n} \right) < b^{2} - m^{2} - 2\left( {b - m} \right) = \Pi_{1}^{d}$. Similarly, to prove {$a^{d} = a \le m$, $b^{d} = m$; $p_{1}^{d} = 0$, $p_{2}^{d} = a^{2} - m^{2}$} is an SPE, we only need to show that in the first stage, given $a^{d} = a < m$, firm 2 has no incentive to locate at $b^{^{\prime}} = n$. When $4 - n \le a < m$, $\Pi_{2}^{^{\prime}} = 0 < a^{2} - m^{2} = \Pi_{2}^{d}$; when $a \le - 2 - n$, $\Pi_{2}^{^{\prime}} = a^{2} - n^{2} < a^{2} - m^{2} = \Pi_{2}^{d}$; when $- 2 - n < a < 4 - n$, $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - a} \right)\left( {4 - a - n} \right)^{2} < a^{2} - m^{2} = \Pi_{2}^{d}$.

(ii)(c) When $- 1 < m \le - 1/4$ and $1 < m + n < 4$ and $m < n$, there are two possible SPEs, {$a^{d} = m$, $b^{d} = n$; $p_{1}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {2 + m + n} \right)$, $p_{2}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {4 - m - n} \right)$} and {$a^{d} = \frac{ - 2 + m}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$}. The condition under which {$a^{d} = m$, $b^{d} = n$; $p_{1}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {2 + m + n} \right)$, $p_{2}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {4 - m - n} \right)$} is an SPE is that given $a^{d} = m$, firm 2 has no incentive to locate at the left side of firm 1 ($b^{\prime} = \frac{ - 2 + m}{3}$), i.e., $\Pi_{2}^{^{\prime}} = \frac{16}{{243}}\left( {1 + m} \right)^{3} \le \frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2} = \Pi_{2}^{d}$ (We could easily prove that under this condition given $b^{d} = n$, firm 1 has no incentive to locate at the right side of firm 2: if $n \ge 2$, $\Pi_{1}^{^{\prime}} = 0 < \frac{1}{18}\left( {n - m} \right)\left( {2 + m + n} \right)^{2} = \Pi_{1}^{d}$; if $n < 2$, $\Pi_{1}^{^{\prime}} = \frac{16}{{243}}\left( {2 - n} \right)^{3} < \frac{1}{18}\left( {n - m} \right)\left( {2 + m + n} \right)^{2} = \Pi_{1}^{d}$.). Next, we consider the solution {$a^{d} = \frac{m - 2}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$}. Given $b^{d} = m$, from the discussion above we can see that when $\frac{16}{{243}}\left( {1 + m} \right)^{3} \ge \frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2}$, firm 1 has no incentive to locate at $n$. We only need to ensure that given $a^{d} = \frac{ - 2 + m}{3}$, firm 2 has no incentive to locate at $b^{\prime} = n$. By calculation we can prove that $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - \frac{ - 2 + m}{3}} \right)\left( {4 - \frac{ - 2 + m}{3} - n} \right)^{2} < \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2} = \Pi_{2}^{d}$.

(ii)(d) When $- 1 < m \le - 1/4$ and $m < n$ and $m + n \ge 4 \Rightarrow n \ge 4 - m \ge 17/4$, the only possible SPE is the one where both firms locate at the left side of the zoning area and both firms get a positive profit. To prove that {$a^{d} = \frac{ - 2 + m}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$} is the unique SPE, we only need to show that given $a^{d} = \frac{ - 2 + m}{3}$, firm 2 has no incentive to locate at $b^{\prime} = n$. When $\frac{ - 2 + m}{3} + n \ge 4$, $\Pi_{2}^{^{\prime}} = 0 < \Pi_{2}^{d}$; when $\frac{ - 2 + m}{3} + n < 4$, by calculation we can prove that $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - \frac{ - 2 + m}{3}} \right)\left( {4 - \frac{ - 2 + m}{3} - n} \right)^{2} < \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2} = \Pi_{2}^{d}$.

(ii)(e) When $- 1/4 < m < 5/4$ and $1 < m + n < 4$ and $m < n$ and $m \le \frac{n - 2}{3}$, the possible SPEs are {$a^{d} = m$, $b^{d} = n$; $p_{1}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {2 + m + n} \right)$, $p_{2}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {4 - m - n} \right)$} and {$a^{d} = \frac{ - 2 + m}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$}. The condition under which {$a^{d} = m$, $b^{d} = n$; $p_{1}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {2 + m + n} \right)$, $p_{2}^{d} = \frac{1}{3}\left( {n - m} \right)\left( {4 - m - n} \right)$} is an SPE is that given $a^{d} = m$, if firm 2 has no incentive to locate at $b^{^{\prime}} = \frac{m - 2}{3}$, i.e., $\Pi_{2}^{d} = \frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2} \ge \frac{16}{{243}}\left( {1 + m} \right)^{3} = \Pi_{2}^{^{\prime}}$ (We could easily prove that under this condition given $b^{d} = n$, firm 1 has no incentive to locate at the right side of firm 2). The condition under which {$a^{d} = \frac{ - 2 + m}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$} is an SPE is that given $a^{d} = \frac{ - 2 + m}{3}$, firm 2 has no incentive to locate at $b^{^{\prime}} = n$, i.e., $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - \frac{m - 2}{3}} \right)\left( {4 - \frac{m - 2}{3} - n} \right)^{2} \le \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2} = \Pi_{2}^{d}$ and given $b^{d} = n$, firm 1 has no incentive to locate at the right side of firm 2, i.e., $a^{^{\prime}} = n$, i.e., $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2} \le \frac{16}{{243}}\left( {1 + m} \right)^{3} = \Pi_{2}^{d}$ (Note that $\frac{4 + m}{3} < n$, so the best response of firm 1 is $a^{\prime} = n$.). In addition, we can prove that $\frac{1}{18}\left( {n - \frac{m - 2}{3}} \right)\left( {4 - \frac{m - 2}{3} - n} \right)^{2} \le \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2}$ when $\frac{1}{18}\left( {n - m} \right)\left( {4 - m - n} \right)^{2} \le \frac{16}{{243}}\left( {1 + m} \right)^{3}$.

(ii)(f) When $- 1/4 < m < 5/4$ and $1 < m + n < 4$ and $m < n$ and $n > 5/4$ and $m > \frac{n - 2}{3}$, the possible SPEs are {$a^{d} = \frac{ - 2 + n}{3}$, $b^{d} = n$; $p_{1}^{d} = \frac{8}{27}\left( {1 + n} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + n} \right)\left( {7 - 2n} \right)$} and {$a^{d} = \frac{m - 2}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$}. The condition under which {$a^{d} = \frac{n - 2}{3}$, $b^{d} = n$; $p_{1}^{d} = \frac{8}{27}\left( {1 + n} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + n} \right)\left( {7 - 2n} \right)$} is an SPE is that given $a^{d} = \frac{n - 2}{3}$, firm 2 has no incentive to locate at the left side of firm 1 $\left( {b^{^{\prime}} = \frac{{\frac{n - 2}{3} - 2}}{3}} \right)$, i.e., $\Pi_{2}^{^{\prime}} = \frac{16}{{243}}\left( {1 + \frac{n - 2}{3}} \right)^{3} \le \frac{4}{243}\left( {1 + n} \right)\left( {7 - 2n} \right)^{2} = \Pi_{2}^{d}$ (which yields $n \le \frac{191 - 27\sqrt 3 }{{52}}$), and firm 2 has no incentive to locate at the right side of firm 1 ($b^{^{\prime\prime}} = m$), i.e., $\Pi_{2}^{{^{\prime\prime}}} = \frac{1}{18}\left( {m - \frac{n - 2}{3}} \right)\left( {4 - \frac{n - 2}{3} - m} \right)^{2} \le \frac{4}{243}\left( {1 + n} \right)\left( {7 - 2n} \right)^{2} = \Pi_{2}^{d}$ (Note that given $b^{d} = n$, firm 1 has no incentive to locate at $a^{^{\prime}} = \frac{4 + n}{3}$ since $1 < m + n < 4 \Rightarrow \Pi_{1}^{^{\prime}} = \frac{16}{{243}}\left( {2 - m} \right)^{3} \le \frac{16}{{243}}\left( {1 + n} \right)^{3} = \Pi_{1}^{d}$). Next, we derive the condition under which {$a^{d} = \frac{m - 2}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$} is an SPE. Given $a^{d} = \frac{m - 2}{3}$, the condition under which firm 2 has no incentive to locate at $b^{\prime} = n$ is $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - \frac{m - 2}{3}} \right)\left( {4 - \frac{m - 2}{3} - n} \right)^{2} \le \frac{4}{234}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2} = \Pi_{2}^{d} { }\left( {{\text{which yields }}m \ge \frac{1}{2}} \right)$. And given $b^{d} = m$, the condition under which firm 1 has no incentive to locate at right side of firm 2 is that when $\frac{4 + m}{3} > n$, $a^{\prime} = \frac{4 + m}{3}$, $\Pi_{1}^{d} = \frac{16}{{243}}\left( {1 + m} \right)^{3} \ge \frac{16}{{243}}\left( {2 - m} \right)^{3} = \Pi_{1}^{^{\prime}}$ and when $\frac{4 + m}{3} \le n$, $a^{\prime\prime} = n$, $\Pi_{1}^{d} = \frac{16}{{243}}\left( {1 + m} \right)^{3} \ge \frac{1}{18}\left( {n - m} \right)\left( {4 - n - m} \right)^{2} = \Pi_{1}^{{^{\prime\prime}}}$.

(ii)(g) When $- 1/4 < m < 5/4$ and $m + n \ge 4 \Rightarrow n \ge 4 - m \ge 11/4$, the only possible SPE is the one where both firms locate at the left side of the zoning area and both firms get positive profit. To prove that {$a^{d} = \frac{ - 2 + m}{3}$, $b^{d} = m$; $p_{1}^{d} = \frac{8}{27}\left( {1 + m} \right)^{2}$, $p_{2}^{d} = \frac{4}{27}\left( {1 + m} \right)\left( {7 - 2m} \right)$} is the unique SPE, we only need to show firm 2 has no incentive to locate at $b^{\prime} = n$. When $\frac{ - 2 + m}{3} + n \ge 4$, $\Pi_{2}^{^{\prime}} = 0 < \Pi_{2}^{d}$; when $\frac{ - 2 + m}{3} + n < 4$, by calculation we can prove that $\Pi_{2}^{^{\prime}} = \frac{1}{18}\left( {n - \frac{ - 2 + m}{3}} \right)\left( {4 - \frac{ - 2 + m}{3} - n} \right)^{2} < \frac{4}{243}\left( {1 + m} \right)\left( {7 - 2m} \right)^{2} = \Pi_{2}^{d}$.

With the equilibriums shown above, the social surplus can be easily derived by solving $\overline{x}^{d} p_{1}^{d} + \left( {1 - \overline{x}^{d} } \right)p_{2}^{{\text{d}}} + \mathop \smallint \limits_{0}^{{\overline{x}^{d} }} \left( {v - \left( {x - a} \right)^{2} - p_{1}^{{\text{d}}} } \right){\text{d}}x + \mathop \smallint \limits_{{\overline{x}^{d} }}^{1} \left( {v - \left( {x - b} \right)^{2} - p_{2}^{{\text{d}}} } \right){\text{d}}x$.□

Proof of Corollary 6

Here, we only proof for the region satisfying $m + n > 1$ and $m < n$. We can follow a similar proofs procedure when $m + n < 1$ and $m < n$. By solving

$$\overline{x}^{d} p_{1}^{d} + \left( {1 - x^{*} } \right)p_{2}^{d} + \mathop \smallint \limits_{0}^{{\overline{x}^{d} }} \left( {v - \left( {x - a} \right)^{2} - p_{1}^{d} } \right){\text{d}}x + \mathop \smallint \limits_{{\overline{x}^{d} }}^{1} \left( {v - \left( {x - b} \right)^{2} - p_{2}^{d} } \right){\text{d}}x,$$

we can obtain the equilibrium social surplus under different scenarios, as shown in Table 4.

Table 4 The social surplus under different regions in Fig. 8

For convenience, here, we define the social surplus in region i under discontinuous location region as $SW_{i}^{d}$. First, we maximize $SW_{3}^{d} = v - \frac{1}{243}\left( {113 - 219m + 195m^{2} - 40m^{3} } \right)$ to derive the highest social surplus in region ③. By solving $\frac{\partial }{\partial m}\left( {v - \frac{1}{243}\left( {113 - 219m + 195m^{2} - 40m^{3} } \right)} \right) = 0$, we obtain the unique maximizer in region ③, i.e., $m^{*} = \frac{1}{40}\left( {65 - 3\sqrt {145} } \right)$. Substituting $m^{*}$ into the above optimization problem, we can obtain corresponding social surplus, i.e., $v - \frac{{595 - 29\sqrt {145} }}{1440}$. Consider the constraints of region ③, the highest social surplus can be reached only if $n \ge \frac{1}{440}\left( {1090 + 18\sqrt {145} - 9\sqrt {5610 - 310\sqrt {145} } } \right) \approx 2.0837$, which can be derived by simplifying $\frac{1}{18}\left( {m^{*} - \frac{n - 2}{3}} \right)\left( {4 - \frac{n - 2}{3} - m^{*} } \right)^{2} \le \frac{4}{243}\left( {1 + n} \right)\left( {7 - 2n} \right)^{2}$.

We next compare the highest social surplus in region ③ with those in other regions.

• 1. We can easily verify that $SW_{3}^{d} = v - \frac{{595 - 29\sqrt {145} }}{1440} > v - \frac{13}{{48}} = SW_{1}^{d}$.

• 2. In region ②, it can be verified that $\frac{{\partial SW_{2}^{{\text{d}}} }}{\partial m} > 0$ always holds. Hence, the highest social surplus is achieved when $m$ takes the maximum value. We have the local highest welfare follows $SW_{2}^{d} = \left\{ \begin{gathered} v - \frac{7}{3},if\,n > 5 \hfill \\ v - \frac{49329\sqrt 3 - 52414}{{52728}},if\,2 + \frac{3}{52}\left( {29 - 9\sqrt 3 } \right) < n \le 5 \hfill \\ v - \frac{49329\sqrt 3 - 52414}{{52728}},if\,\frac{5}{4} < n \le 2 + \frac{3}{52}\left( {29 - 9\sqrt 3 } \right) \hfill \\ \end{gathered} \right.$. Finally, we obtain the highest value (i.e., $v - \frac{49329\sqrt 3 - 52414}{{52728}}$) when $m = \frac{1}{52}\left( {29 - 9\sqrt 3 } \right)$ and $n = 2 + \frac{3}{52}\left( {29 - 9\sqrt 3 } \right)$. It can be verified that $SW_{3}^{d} = v - \frac{{595 - 29\sqrt {145} }}{1440} > v - \frac{49329\sqrt 3 - 52414}{{52728}} = SW_{2}^{d}$.

• 3. Note that region ④ is limited to $\frac{5}{4} \le n \le 2 + \frac{3}{52}\left( {29 - 9\sqrt 3 } \right)$. It can be verified that for $\frac{5}{4} \le n \le \frac{1}{40}\left( {65 + 3\sqrt {145} } \right)$ and $\frac{{\partial SW_{4}^{d} }}{\partial n} < 0$$\frac{{\partial SW_{4}^{d} }}{\partial n} > 0$ for $\frac{1}{40}\left( {65 + 3\sqrt {145} } \right) < n \le 2 + \frac{3}{52}\left( {29 - 9\sqrt 3 } \right)$. By comparing the social surplus when $n = \frac{5}{4}$ with that when $n = 2 + \frac{3}{52}\left( {29 - 9\sqrt 3 } \right)$, we can obtain the highest social surplus in region ④, i.e.,$v - \frac{13}{{48}}$. It can be verified that $SW_{3}^{d} = v - \frac{{595 - 29\sqrt {145} }}{1440} > v - \frac{13}{{48}} = SW_{4}^{d}$.

• 4. In region ⑤, there exist two equilibrium, i.e., ${\mathbb{E}}_{2}$ and ${\mathbb{E}}_{3}$, which correspond with $SW_{3}^{d}$ and $SW_{4}^{d}$ respectively. Under ${\mathbb{E}}_{2}$(${\mathbb{E}}_{3}$), the profits of firm 1 and firm 2 are $\frac{16}{{243}}\left( {1 + m} \right)^{3}$ $\left( {\frac{16}{{243}}\left( {1 + n} \right)^{3} } \right)$ and $\frac{4}{243}\left( {7 - 2m} \right)^{2} \left( {1 + m} \right)$ $\left( {\frac{4}{243}\left( {7 - 2n} \right)^{2} \left( {1 + n} \right)} \right)$, respectively. With further comparison, we obtain the highest value of $SW_{4}^{d}$ in region ⑤, i.e.,$v-\frac{13}{48}$, when$n=\frac{5}{4}$. Similarly, the lowest value of $SW_{3}^{d}$, which is equal to $v-\frac{13}{48}$, is obtained when m takes the endpoint value, i.e., $\frac{5}{4}$ or $\frac{1}{52}\left( {29 - 9\sqrt 3 } \right)$. Thus, $SW_{4}^{d} \le SW_{3}^{d}$ always holds in region ⑤. Hence, in region ⑤, the highest social surplus is reached when $SW_{3}^{d}$ reaches $v - \frac{{595 - 29\sqrt {145} }}{1440}$, wherein $m = \frac{1}{40}\left( {65 - 3\sqrt {145} } \right)$ and $n \ge 1.5464$.

• 5. Note that region ⑥ is limited to $m \le - 1$ and it can be verified that $\frac{{\partial SW_{6}^{d} }}{\partial n} > 0$ for $m \le - 1$. Hence, we can obtain the highest local social surplus $SW_{4}^{d} = v - \frac{7}{3}$ when $m = - 1$. Again, we show that $SW_{4}^{d} = v - \frac{7}{3} < v - \frac{{595 - 29\sqrt {145} }}{1440} = SW_{3}^{d}$.

According to the above analysis, we show that when $m + n > 1$ and $m < n$, the global highest social surplus $\left( {{\text{i}}.{\text{e}}.,\,\,v - \frac{{595 - 29\sqrt {145} }}{1440}} \right)$ is obtain when $m = \frac{1}{40}\left( {65 - 3\sqrt {145} } \right)$ and $n \ge \frac{1}{440}\left( {1090 + 18\sqrt {145} - 9\sqrt {5610 - 310\sqrt {145} } } \right) \approx 2.0837$ in region ③. In region ⑤, the highest social welfare is reached only at equilibrium ${\mathbb{E}}_{2}$, wherein $m = \frac{1}{40}\left( {65 - 3\sqrt {145} } \right)$ and $n \ge 1.5464$. When $m + n < 1$ and $n > - 1/4$, the results can be similarly derived. □

Table 5 The equilibrium results and their corresponding constraints on $m$ and $n$ under different location regions $\left[ {m,n} \right]$ ($m < n$)

Table 6 The equilibrium results and their corresponding constraints on $m$ and $n$ under different zoning area $\left( {m,n} \right)$ when $m + n > 1$ and $m < n$