Mixed duopoly in a Hotelling framework with cubic transportation costs

Abstract

We examine a two-stage location-price model of a mixed duopoly where a private profit-maximizing firm competes with a public welfare-maximizing firm in a Hotelling-type framework. A noteworthy result in this model is that, with quadratic transportation costs, which has become the usual assumption in the literature, the socially optimal locations are obtained in equilibrium. We show here that under the alternative assumption of cubic transportation costs this result no longer holds: equilibrium locations are socially suboptimal. We find that just as in the case of linear transportation costs, previously studied in the literature, for some locations there is price equilibrium in the second stage of the game and for other locations there is not. But, in contrast with such a case, there is a location pair for which there is price equilibrium in the second stage of the game and neither firm has an incentive to marginally change its location. We also find that, in contrast with the case of quadratic transportation costs, this location pair is not socially optimal.

Appendix

1.1 Proof of Proposition 1

Let $p^{*} = \frac{{3\left( {2 - x_{0} - x_{1} } \right)\left( {x_{1} - x_{0} } \right)^{2} }}{4}$ be the candidate equilibrium common price in Eq. (9) and $\Pi_{1}^{*} = \frac{{3\left( {2 - x_{0} - x_{1} } \right)^{2} \left( {x_{1} - x_{0} } \right)^{2} }}{8}$ be the candidate equilibrium firm 1’s profits as given in Eq. (11). We now examine if given $p_{0} = p^{*}$, firm 1 can increase its profits above $\Pi_{1}^{*}$ by choosing a price $p_{1} \ne p^{*}$.

(1) if firm 1 chooses a price in the interval $p_{0} - \left( {x_{1}^{3} - x_{0}^{3} } \right) \le p_{1} \le p_{0} - \left( {x_{1} - x_{0} } \right)^{3} ,$ then $q_{0} \in \left[ {0 ,x_{0} } \right]$ and we have

$$p_{0} + \left( {x_{0} - q_{0} } \right)^{3} = p_{1} + \left( {x_{1} - q_{0} } \right)^{3}$$

(16)

from where

$$\frac{{\partial q_{0} }}{{\partial p_{1} }} = \frac{1}{{3\left( {x_{1} - q_{0} } \right)^{2} - 3\left( {x_{0} - q_{0} } \right)^{2} }}$$

and therefore:

$$\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = \frac{{ - p_{1} }}{{3\left( {x_{1} - q_{0} } \right)^{2} - 3\left( {x_{0} - q_{0} } \right)^{2} }} + 1 - q_{0}$$

(17)

Replacing $p_{1}$ from (16) into (17) and using $p_{0} = p^{*}$ we obtain

$$\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = \frac{{36q_{0}^{2} - 24\left( {x_{0} + x_{1} + 1} \right)q_{0} + 7x_{1}^{2} + 4x_{0} x_{1} + 6x_{1} + x_{0}^{2} + 18x_{0} }}{{12\left( { - 2q_{0} + x_{0} + x_{1} } \right)}}$$

(18)

The denominator in the RHS of (18) is positive because $q_{0} \le x_{0} < x_{1}$. The numerator is also positive because (a) it is decreasing in $q_{0 }$ (its derivative with respect to $q_{0}$ is equal to $72q_{0} - 24\left( {x_{0} + x_{1} + 1} \right) \le 72x_{0} - 24\left( {x_{0} + x_{1} + 1} \right) = - 24\left( {x_{1} - x_{0} } \right) - 24\left( {1 - x_{0} } \right) < 0$) and (b) it is positive when $q_{0} = x_{0}$ (it is then equal to $\left( {x_{1} - x_{0} } \right)\left( { - 13x_{0} + 7x_{1} + 6} \right) > 0).$

It follows that $\Pi_{1}$ is increasing over the whole range $p_{0} - \left( {x_{1}^{3} - x_{0}^{3} } \right) \le p_{1} \le p_{0} - \left( {x_{1} - x_{0} } \right)^{3}$. Let $\Pi_{1}^{a}$ be firm 1’s profits when $p_{1} = p_{0} - \left( {x_{1} - x_{0} } \right)^{3}$-which implies $q_{0} = x_{0}$- and $p_{0} = p^{*}$. Then

$$\Pi_{1}^{a} = \left( {\frac{{3\left( {2 - x_{1} - x_{0} } \right)\left( {x_{1} - x_{0} } \right)^{2} }}{4} - \left( {x_{1} - x_{0} } \right)^{3} } \right)\left( {1 - x_{0} } \right)$$

If we substract from these profits the candidate equilibrium profits $\Pi_{1}^{*}$ we obtain, after some simplifications,

$$\Pi_{1}^{a} - \Pi_{1}^{*} = \frac{{\left( {x_{1} - x_{0} } \right)^{3} \left( { - 3x_{1} + 5x_{0} - 2} \right)}}{8} < 0$$

which implies that firm 1 will not deviate to such a price, and will thus neither deviate to any price $p_{1}$ with $p_{0} - \left( {x_{1}^{3} - x_{0}^{3} } \right) \le p_{1} \le p_{0} - \left( {x_{1} - x_{0} } \right)^{3}$.

(2) If firm 1 chooses a price in the interval $p_{0} - \left( {x_{1} - x_{0} } \right)^{3} \le p_{1} \le p_{0} + \left( {x_{1} - x_{0} } \right)^{3}$, then $q_{0} \in \left[ {x_{0} ,x_{1} } \right]$ and we will have:

$$p_{0} + \left( {q_{0} - x_{0} } \right)^{3} = p_{1} + \left( {x_{1} - q_{0} } \right)^{3}$$

(19)

and thus

$$\frac{{\partial q_{0} }}{{\partial p_{1} }} = \frac{1}{{3\left( {q_{0} - x_{0} } \right)^{2} + 3\left( {x_{1} - q_{0} } \right)^{2} }}$$

Therefore:

$$\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = \frac{{ - p_{1} }}{{3\left( {q_{0} - x_{0} } \right)^{2} + 3\left( {x_{1} - q_{0} } \right)^{2} }} + 1 - q_{0}$$

(20)

Replacing $p_{1}$ from (19) into (20) and using $p_{0} = p^{*}$ we obtain, after some simplifications:

$$\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = \frac{{\left[ {2q_{0} - x_{0} - x_{1} } \right]\left[ {16q_{0}^{2} - \left( {10\left( {x_{0} + x_{1} } \right) + 12} \right)q_{0} + \left( {7x_{1} - 10x_{0} + 6} \right)x_{1} + 7x_{0}^{2} + 6x_{0} } \right]}}{{ - 12\left( {q_{0} - x_{0} } \right)^{2} - 12\left( {x_{1} - q_{0} } \right)^{2} }}$$

Therefore, $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = 0$ if $q_{0} = \frac{{x_{0} + x_{1} }}{2}$, which implies $p_{1} = p_{0} = p^{*} .$ Notice also that, from (20):

$$\frac{{\partial^{2} \Pi_{1} }}{{\partial p_{1}^{2} }} = \frac{{ - \left( {3\left( {q_{0} - x_{0} } \right)^{2} + 3\left( {x_{1} - q_{0} } \right)^{2} } \right) + 6p_{1} \left( {\left( {q_{0} - x_{0} } \right) - \left( {x_{1} - q_{0} } \right)} \right)\frac{{\partial q_{0} }}{{\partial p_{1} }}}}{{\left( {3\left( {q_{0} - x_{0} } \right)^{2} + 3\left( {x_{1} - q_{0} } \right)^{2} } \right)^{2} }} - \frac{{\partial q_{0} }}{{\partial p_{1} }}$$

and that this second derivative evaluated at $p_{1} = p_{0} = p^{*}$ (and thus at $q_{0} = \frac{{x_{0} + x_{1} }}{2}$) is negative:

$$\frac{{\partial^{2} \Pi_{1} }}{{\partial p_{1}^{2} }} = \frac{ - 4}{{3\left( {x_{1} - x_{0} } \right)^{2} }} < 0$$

Therefore, $p_{1} = p^{*}$ is a local maximum, and it is also the only value where $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }}$ vanishes (in the segment $q_{0} \in \left[ {x_{0} ,x_{1} } \right]$) unless the following equation has roots in $q_{0} \in \left[ {x_{0} ,x_{1} } \right]:$

$$16q_{0}^{2} - \left( {10\left( {x_{0} + x_{1} } \right) + 12} \right)q_{0} + \left( {7x_{1} - 10x_{0} + 6} \right)x_{1} + 7x_{0}^{2} + 6x_{0} = 0$$

(21)

Equation (21) is a second degree equation in $q_{0}$ with discriminant equal to

$$12\left( { - 29x_{1}^{2} + \left( {70x_{0} - 12} \right)x_{1} - 29x_{0}^{2} - 12x_{0} + 12} \right)$$

(22)

If the discriminant in (22) is negative $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }}$ will only vanish at $q_{0} = \frac{{x_{0} + x_{1} }}{2}$ (which corresponds to $p_{1} = p^{*}$) in the segment $q_{0} \in \left[ {x_{0} ,x_{1} } \right].$ If this discriminat is positive, then Eq. (21) will have two roots:

$$q_{0}^{H} = \frac{{5x_{0} + 5x_{1} + 6 + \sqrt 3 \sqrt { - 29x_{1}^{2} + \left( {70x_{0} - 12} \right)x_{1} - 29x_{0}^{2} - 12x_{0} + 12} }}{16}$$

(23)

$$q_{0}^{L} = \frac{{5x_{0} + 5x_{1} + 6 - \sqrt 3 \sqrt { - 29x_{1}^{2} + \left( {70x_{0} - 12} \right)x_{1} - 29x_{0}^{2} - 12x_{0} + 12} }}{16}$$

(24)

But, $q_{0}^{H} > x_{1}$, since the positiveness of the discriminant in (22) implies that (a) $q_{0}^{H} > \frac{{5x_{0} + 5x_{1} + 6}}{11}$, and (b) $x_{1} < \frac{{5x_{0} + 6}}{11}$, because this discriminat is strictly decreasing in $x_{1}$ and it is negative when $x_{1} = \frac{{5x_{0} + 6}}{11}$. These two facts imply $q_{0}^{H} - x_{1} \ge \frac{{5x_{0} + 5x_{1} + 6}}{11} - x_{1} = \frac{{5x_{0} - 11x_{1} + 6}}{11} > 0$. Thus, $q_{0}^{L}$ is the only possible root additional to $q_{0} = \frac{{\left( {x_{0} + x_{1} } \right)}}{2}$ in the range $q_{0} \in \left[ {x_{0} ,x_{1} } \right]$, which cannot therefore be a local maximum (since $p_{1} = p^{*}$ is a local maximum and $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }}$ only vanishes at $p^{*}$ and $p^{L}$ (associated to $q_{0}^{L}$) then $\Pi_{1}$ is strictly decreasing in $p_{1} \in \left( {p^{*} , p^{L} } \right)$ if $p^{*} < p^{L}$ and, similarly, it is strictly increasing in $p_{1} \in \left( {p^{L} , p^{*} } \right)$ if $p^{L} < p^{*}$).

Since we proved above that firm 1 will not deviate to a price $p_{1}$ such that $q_{0} = x_{0}$, the only price that remains to be considered in this interval is $p_{1} = p_{0} + \left( {x_{1} - x_{0} } \right)^{3}$ which corresponds to $q_{0} = x_{1}$. If $p_{0} = p^{*}$ and firm 1 chooses such a price, firm 1’s profits will be equal to

$$\Pi_{1}^{b} = \left( {\frac{{3\left( {2 - x_{0} - x_{1} } \right)\left( {x_{1} - x_{0} } \right)^{2} }}{4} + \left( {x_{1} - x_{0} } \right)^{3} } \right) \left( {1 - x_{1} } \right)$$

Subtracting the candidate equilibrium profits $\Pi_{1}^{*}$ from $\Pi_{1}^{b}$ we obtain:

$$gb\left( {x_{0} ,x_{1} } \right) = \Pi_{1}^{b} - \Pi_{1}^{*} = \left( {\frac{{3\left( {2 - x_{0} - x_{1} } \right)\left( {x_{1} - x_{0} } \right)^{2} }}{4} + \left( {x_{1} - x_{0} } \right)^{3} } \right)\left( {1 - x_{1} } \right) - \frac{{3\left( {2 - x_{0} - x_{1} } \right)^{2} \left( {x_{1} - x_{0} } \right)^{2} }}{8}$$

with

$$\frac{{\partial gb\left( {x_{0} ,x_{1} } \right)}}{{\partial x_{1} }} = \frac{{ - \left( {x_{1} - x_{0} } \right)^{2} \left( {10x_{1} - 7x_{0} - 3} \right)}}{4}$$

For any given $x_{0}$,$gb\left( {x_{0} ,x_{0} } \right) = 0$, $gb\left( {x_{0} ,x_{1} } \right)$ reaches a maximum at $x_{1} = \frac{{\left( {7x_{0} + 3} \right)}}{10}$ with $gb\left( {x_{0} ,\frac{{\left( {7x_{0} + 3} \right)}}{10}} \right) = \frac{{27\left( {1 - x_{0} } \right)^{4} }}{16,000} > 0$, and $gb\left( {x_{0} ,1} \right) = \frac{{ - 3\left( {1 - x_{0} } \right)^{4} }}{8} < 0$. Since $gb\left( {x_{0} ,x_{1} } \right)$ is strictly increasing in $x_{1}$ for $x_{0} < x_{1} < \frac{{\left( {7x_{0} + 3} \right)}}{10}$, strictly decreasing in $x_{1}$ for $\frac{{\left( {7x_{0} + 3} \right)}}{10} < x_{1} < 1$, and $gb\left( {x_{0} ,\frac{{\left( {3x_{0} + 2} \right)}}{5}} \right) = 0$, it follows that, when $\frac{{3x_{0} + 2}}{5} \le x_{1} \le 1$, $gb\left( {x_{0} ,x_{1} } \right) \le 0,$ and firm 1 does not find this deviation profitable, while for $x_{1} < \frac{{3x_{0} + 2}}{5}$, $gb\left( {x_{0} ,x_{1} } \right) > 0$, firm 1 does deviate from the candidate equilibrium price and there does not exist an equilibrium in the second stage of the game.

(3) if firm 1 chooses a price in the interval $p_{0} + \left( {x_{1} - x_{0} } \right)^{3} \le p_{1} \le p_{0} + \left( {1 - x_{0} } \right)^{3} - \left( {1 - x_{1} } \right)^{3}$, then $q_{0} \in \left[ {x_{1} ,1 } \right]$ and we have

$$p_{0} + \left( {q_{0} - x_{0} } \right)^{3} = p_{1} + \left( {q_{0} - x_{1} } \right)^{3}$$

(25)

from where

$$\frac{{\partial q_{0} }}{{\partial p_{1} }} = \frac{1}{{3\left( {q_{0} - x_{0} } \right)^{2} - 3\left( {q_{0} - x_{1} } \right)^{2} }}$$

and therefore:

$$\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = \frac{{ - p_{1} }}{{3\left( {q_{0} - x_{0} } \right)^{2} - 3\left( {q_{0} - x_{1} } \right)^{2} }} + 1 - q_{0}$$

(26)

Replacing $p_{1}$ from (25) into (26) and using $p_{0} = p^{*}$ we obtain

$$\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = \frac{{ - 36q_{0}^{2} + 24\left( {x_{0} + x_{1} + 1} \right)q_{0} - x_{1}^{2} - 4x_{0} x_{1} - 18x_{1} - 7x_{0}^{2} - 6x_{0} }}{{12\left( {2q_{0} - x_{0} - x_{1} } \right)}}$$

The denominator of $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }}$ is positive because $x_{0} < x_{1} \le q_{0}$. $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }}$ will vanish when the numerator does so, which yields a quadratic equation in $q_{0}$ with discriminant equal to

$$144 \left( {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} \right)$$

(27)

The discriminant in (27) is strictly decreasing in $x_{1}$ and it is equal to zero when $x_{1} = x_{1}^{R}$, with

$$x_{1}^{R} = \frac{{\left( {\sqrt {13} - 2} \right)x_{0} + 5 - \sqrt {13} }}{3} \in \left( {\frac{{3x_{0} + 2}}{5},1} \right)$$

Therefore,

(a) when $x_{1} > x_{1}^{R}$, the discriminant in (27) is negative and $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }}$ never vanishes (it is negative). It then suffices to consider the price deviation associated to $q_{0} = x_{1}$ (which yields profits $\Pi_{1}^{b}$). Since $x_{1}^{R} > \frac{{3x_{0} + 2}}{5}$, we have that $x_{1} > \frac{{3x_{0} + 2}}{5}$ and thus $gb\left( {x_{0} ,x_{1} } \right) < 0$, $\Pi_{1}^{b} < \Pi_{1}^{*}$ and there is equilibrium.

(b) when $x_{1} \le x_{1}^{R}$ the discriminant in (27) is positive and $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = 0$ has the following roots:

$$\begin{aligned} q_{0}^{c} & = \frac{{2x_{1} + 2x_{0} + 2 + \sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} }}{6} \\ q_{0}^{d} & = \frac{{2x_{1} + 2x_{0} + 2 - \sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} }}{6} \\ \end{aligned}$$

Now, from (26)

$$\frac{{\partial^{2} \Pi_{1} }}{{\partial p_{1}^{2} }} = \frac{{ - 3\left( {q_{0} - x_{0} } \right)^{2} + 3\left( {q_{0} - x_{1} } \right)^{2} + p_{1} \left( {6\left( {q_{0} - x_{0} } \right) - 6\left( {q_{0} - x_{1} } \right)} \right)\frac{{\partial q_{0} }}{{\partial p_{1} }}}}{{\left( {3\left( {q_{0} - x_{0} } \right)^{2} - 3\left( {q_{0} - x_{1} } \right)^{2} } \right)^{2} }} - \frac{{\partial q_{0} }}{{\partial p_{1} }}$$

(28)

Also, from (26), $\frac{{\partial \Pi_{1} }}{{\partial p_{1} }} = 0$ implies that

$$p_{1} = \left( {1 - q_{0} } \right)\left( {3\left( {q_{0} - x_{0} } \right)^{2} - 3\left( {q_{0} - x_{1} } \right)^{2} } \right)$$

(29)

Replacing $p_{1}$ from (29) and $q_{0}$ with $q_{0}^{c}$ in (28) we get:

$$\frac{{\partial^{2} \Pi_{1} }}{{\partial p_{1}^{2} }} = \frac{{ - 3\sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} }}{{\left( {x_{1} - x_{0} } \right)\left( {\sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} - x_{0} - x_{1} + 2} \right)^{2} }}$$

Similarly, when $q_{0} = q_{0}^{d}$ we get:

$$\frac{{\partial^{2} \Pi_{1} }}{{\partial p_{1}^{2} }} = \frac{{3\sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} }}{{\left( {x_{1} - x_{0} } \right)\left( {\sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} + x_{0} + x_{1} - 2} \right)^{2} }}$$

Therefore $\Pi_{1}$ reaches a local maximum at $q_{0}^{c}$ and a local minimum at $q_{0}^{d}$.

Let $\Pi_{1}^{c}$ be firm 1’s profits when $p_{1}$ is such that $q_{0} = q_{0}^{c}$,

$$\Pi_{1}^{c} = \left( {\frac{{3\left( {2 - x_{0} - x_{1} } \right)\left( {x_{1} - x_{0} } \right)^{2} }}{4} + \left( {q_{0}^{c} - x_{0} } \right)^{3} - \left( {q_{0}^{c} - x_{1} } \right)^{3} } \right)\left( {1 - q_{0}^{c} } \right)$$

and let $gc\left( {x_{0} ,x_{1} } \right) = \Pi_{1}^{c} - \Pi_{1}^{*}$. $gc\left( {x_{0} ,x_{1} } \right)$ is the increase in profits when firm 1 chooses $p_{1}$ such that $q_{0} = q_{0}^{c}$ instead of the candidate equilibrium price $p^{*}$.

Since we know that for $x_{1} < \frac{{3x_{0} + 2}}{5}$ there is no equilibrium (because $\Pi_{1}^{b} > \Pi_{1}^{*}$), while for $x_{1} > x_{1}^{R}$ there is equilibrium, we will focus our attention on the behaviour of $gc\left( {x_{0} ,x_{1} } \right)$ in the interval $x_{1} \in \left[ {\frac{{3x_{0} + 2}}{5},x_{1}^{R} } \right]$.

We have that $gc\left( {x_{0} ,\frac{{3x_{0} + 2}}{5}} \right) > 0$, $gc\left( {x_{0} ,x_{1}^{R} } \right) < 0$,and we will now prove that $\frac{{\partial gc\left( {x_{0} ,x_{1} } \right)}}{{\partial x_{1} }} < 0$ for $x_{1} \in \left[ {\frac{{3x_{0} + 2}}{5},x_{1}^{R} } \right]$. This implies that there exists $x_{1}^{*} \in \left( {\frac{{3x_{0} + 2}}{5},x_{1}^{R} } \right)$ such that $\Pi_{1}^{c} = \Pi_{1}^{*}$ if $x_{1} = x_{1}^{*}$ and $\Pi_{1}^{c} > \Pi_{1}^{*}$$\left( {\Pi_{1}^{c} < \Pi_{1}^{*} } \right)$ if $x_{1} < x_{1}^{*}$$\left( {x_{1} > x_{1}^{*} } \right)$. This in turn implies that there does not exist (there exists) equilibrium if $x_{1} < x_{1}^{*}$$\left( {x_{1} > x_{1}^{*} } \right).$

We have:

$$\frac{{\partial gc\left( {x_{0} ,x_{1} } \right)}}{{\partial x_{1} }} = \frac{{f\left( {x_{0} ,x_{1} } \right)\sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} + h\left( {x_{0} ,x_{1} } \right)}}{{36\sqrt {3x_{1}^{2} + \left( {4x_{0} - 10} \right)x_{1} - 3x_{0}^{2} + 2x_{0} + 4} }}$$

with

$$f\left( {x_{0} ,x_{1} } \right) = - \left( {34x_{1}^{3} + \left( { - 12x_{0} - 90} \right)x_{1}^{2} + \left( { - 18x_{0}^{2} + 60x_{0} + 60} \right)x_{1} + 4x_{0}^{3} + 6x_{0}^{2} - 36x_{0} - 8} \right)$$

and

$$\begin{aligned} h\left( {x_{0} ,x_{1} } \right) & = 36x_{1}^{4} + \left( {51x_{0} - 195} \right)x_{1}^{3} + \left( { - 59x_{0}^{2} - 35x_{0} + 310} \right)x_{1}^{2} \\ & \quad + \,\left( { - 39x_{0}^{3} + 235x_{0}^{2} - 200x_{0} - 140} \right)x_{1} + 27x_{0}^{4} - 69x_{0}^{3} \\ & \quad - \,14x_{0}^{2} + 76x_{0} + 16 \\ \end{aligned}$$

Now, $f\left( {x_{0} ,x_{1} } \right) < 0$ because, as can be easily checked, $\frac{{\partial^{2} f}}{{\partial x_{1}^{2} }} > 0$, $\frac{\partial f}{{\partial x_{1} }} < 0$ when $x_{1} = \frac{{3x_{0} + 2}}{5}$, $\frac{\partial f}{{\partial x_{1} }} > 0$ when $x_{1} = x_{1}^{R}$ and $f < 0$ when both $x_{1} = \frac{{3x_{0} + 2}}{5}$ and $x_{1} = x_{1}^{R}$.

Similarly, $h\left( {x_{0} ,x_{1} } \right) < 0$ because $\frac{{\partial^{2} h}}{{\partial x_{1}^{2} }} > 0$, $\frac{\partial h}{{\partial x_{1} }} > 0$ when $x_{1} = \frac{{3x_{0} + 2}}{5}$ and $h = 0$ when $x_{1} = x_{1}^{R}$.

1.2 Proof of socially optimal locations in Proposition 2

Since demand is inelastic, the socially optimal locations are those that minimize transportation costs as given in (5). To find them, remember first that, as shown above, $q_{0} = \frac{{x_{0} + x_{1} }}{2}$ minimizes transportation costs for any locations $x_{0} ,x_{1}$. Replacing this value back in (5) yields TC as a function of $x_{0}$ and $x_{1}$ as given in (10). The first-order conditions to minimize TC in (10) with respect to both $x_{0}$ and $x_{1}$ are:

$$\frac{\partial TC}{{\partial x_{0} }} = x_{0}^{3} - \left( {\frac{{x_{1} - x_{0} }}{2}} \right)^{3} = 0$$

$$\frac{\partial TC}{{\partial x_{1} }} = \left( {\frac{{x_{1} - x_{0} }}{2}} \right)^{3} - \left( {1 - x_{1} } \right)^{3} = 0$$

from where we can obtain, respectively, $x_{0} = \frac{{x_{1} }}{3}$ and $x_{0} = 3x_{1} - 2$ and, therefore, $x_{0} = \frac{1}{4}$, $x_{1} = \frac{3}{4}$. The second order conditions are also satisfied.

1.3 Firm 0’s demand $q_{0}$ written as an explicit function of $p_{0}$ and $p_{1}$

If $- \left( {x_{1}^{3} - x_{0}^{3} } \right) \le p_{1} - p_{0} \le - \left( {x_{1} - x_{0} } \right)^{3}$ then:

$$q_{0} = \frac{{x_{0} + x_{1} }}{2} - \sqrt {\frac{{4\left( {p_{0} - p_{1} } \right) - \left( {x_{1} - x_{0} } \right)^{3} }}{{12\left( {x_{1} - x_{0} } \right)}} }$$

If $- \left( {x_{1} - x_{0} } \right)^{3} \le p_{1} - p_{0} \le \left( {x_{1} - x_{0} } \right)^{3}$ then:

$$\begin{aligned} q_{0} & = \frac{{\left( {x_{0} + x_{1} } \right)}}{2} + \left( {\frac{{\sqrt {4\left( {p_{1} - p_{0} } \right)^{2} + \left( {x_{1} - x_{0} } \right)^{6} } }}{8} - \frac{{p_{0} - p_{1} }}{4}} \right)^{1/3} \\ & \quad - \,\frac{{\left( {x_{1} - x_{0} } \right)^{2} }}{{4\left( {\frac{{\sqrt {4\left( {p_{1} - p_{0} } \right)^{2} + \left( {x_{1} - x_{0} } \right)^{6} } }}{8} - \frac{{p_{0} - p_{1} }}{4}} \right)^{1/3} }} \\ \end{aligned}$$

If $\left( {x_{1} - x_{0} } \right)^{3} \le p_{1} - p_{0} \le \left( {1 - x_{0} } \right)^{3} - \left( {1 - x_{1} } \right)^{3}$ then:

$$q_{0} = \frac{{x_{0} + x_{1} }}{2} + \sqrt {\frac{{4\left( {p_{1} - p_{0} } \right) - \left( {x_{1} - x_{0} } \right)^{3} }}{{12\left( {x_{1} - x_{0} } \right)}} }$$