Investigating cooperation between competitive manufacturers under the energy performance contracting mechanism

Abstract

In a duopoly market, a superior manufacturer owns more efficient energy-saving technology in production than the inferior manufacturer. To improve the efficiency of energy-saving technology, technical cooperation is a critical option for the inferior manufacturer. However, unaffordable cooperation fees are generally regarded as one of the most significant barriers in technological advancement. To address this issue, this paper investigates cooperation between competitive manufacturers under the energy performance contracting (EPC) mechanism by developing game theoretical models. The findings show that (i) when both manufacturers engage in EPC cooperation, the market equilibrium price decreases and overall market product sales increase. (ii) The EPC cooperation decisions of the two manufacturers depend on the market size and the revenue sharing ratio. (iii) When the market is small, the optimal choice of the inferior manufacturer is to engage in EPC cooperation with the superior manufacturer. (iv) When the superior manufacturer retains part of its technology under EPC, there exists an optimal level of technical cooperation that can yield optimal benefits for both manufacturers. The findings are valuable not only in understanding competitive manufacturers’ optimal cooperation decisions under the EPC mechanism but also for inferior manufacturers seeking cooperation opportunities for technological advancement.

Appendix: proofs

Proof of theorem 1

Firstly, calculate derivatives of Eqs. (2) and (3) with respect to $\mathop q\nolimits_{1}$ and $\mathop q\nolimits_{2}$ respectively, and let $\frac{{\partial \mathop \prod \nolimits_{1} }}{{\partial \mathop q\nolimits_{1} }} = 0$, $\frac{{\partial \mathop \prod \nolimits_{2} }}{{\partial \mathop q\nolimits_{2} }} = 0$.

Then, we can obtain the solutions of Nash equilibrium quantity as $\left\{ \begin{gathered} q_{1}^{*} = \frac{a + ce - 2c\theta e}{3} \hfill \\ q_{2}^{*} = \frac{a + c\theta e - 2ce}{3} \hfill \\ \end{gathered} \right.$. Secondly, substituting the solutions of the equilibrium quantity into the inverse demand function, and we can obtain the equilibrium price $p^{*} = \frac{a + ce + c\theta e}{3}$. Lastly, calculate the second derivatives of Eqs. (2) and (3) with respect to $\mathop q\nolimits_{1}$ and $\mathop q\nolimits_{2}$ respectively, namely, $\frac{{\mathop \partial \nolimits^{{\mathop 2\limits_{{}} }} \mathop \prod \nolimits_{1} }}{{\partial \mathop {\mathop q\nolimits_{1} }\nolimits^{2} }} = - 2 < 0$ and $\frac{{\partial^{2} \mathop \prod \nolimits_{2} }}{{\partial \mathop {\mathop q\nolimits_{2} }\nolimits^{2} }} = - 2 < 0$, which implies that there are the maximum points of $\mathop \prod \nolimits_{1}$ and $\mathop \prod \nolimits_{2}$. Then, substituting the solutions of Nash equilibrium quantity and the equilibrium price into the Eq. (1), and we can obtain the equilibrium profits of two manufacturers as $\left\{ \begin{gathered} \Pi_{1}^{*} = \frac{{(a + ce - 2c\theta e)^{2} }}{9} \hfill \\ \Pi_{2}^{*} = \frac{{(a + c\theta e - 2ce)^{2} }}{9} \hfill \\ \end{gathered} \right.$.

Proof of theorem 2

Similarly, firstly, calculate derivatives of Eqs. (4) and (5) with respect to $\mathop q\nolimits_{1}$ and $\mathop q\nolimits_{2}$ respectively, and let $\frac{{\partial \mathop \prod \nolimits_{1} }}{{\partial \mathop q\nolimits_{1} }} = 0$, $\frac{{\partial \mathop \prod \nolimits_{2} }}{{\partial \mathop q\nolimits_{2} }} = 0$.

Then, we can obtain the solutions of Nash equilibrium quantity under EPC cooperation as $\left\{ \begin{gathered} \mathop q\nolimits_{{\mathop 1\limits^{{*{*}}} }} = \frac{a - c\theta e - rc\theta e + rce}{3} \hfill \\ \mathop q\nolimits_{{\mathop 2\limits^{{*{*}}} }} = \frac{a - c\theta e + 2rc\theta e - 2rce}{3} \hfill \\ \end{gathered} \right.$. Secondly, substituting the solutions of the equilibrium quantity into the inverse demand function, and we can obtain the equilibrium price $p^{{*{*}}} = \frac{{a + {2}c\theta e - rc\theta e + rce}}{3}$. Lastly, calculate the second derivatives of Eqs. (4) and (5) with respect to $\mathop q\nolimits_{1}$ and $\mathop q\nolimits_{2}$ respectively, namely, $\frac{{\mathop \partial \nolimits^{{\mathop 2\limits_{{}} }} \mathop \prod \nolimits_{1} }}{{\partial \mathop {\mathop q\nolimits_{1} }\nolimits^{2} }} = - 2 < 0$ and $\frac{{\mathop \partial \nolimits^{{\mathop 2\limits_{{}} }} \mathop \prod \nolimits_{2} }}{{\partial \mathop {\mathop q\nolimits_{2} }\nolimits^{2} }} = - 2 < 0$, which implies that there are the maximum points of $\mathop \prod \nolimits_{1}$ and $\mathop \prod \nolimits_{2}$. Then, substituting the solutions of Nash equilibrium quantity and the equilibrium price into Eqs. (4) and (5), and we can obtain the equilibrium profits of two manufacturers as $\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{**} }} = \frac{\begin{gathered} ({2}a^{{2}} + 6a\theta - 3a\theta^{2} + 3ce\theta^{3} + 2c^{2} e^{2} \theta^{2} - 10c^{2} e^{2} r^{2} - 6ce\theta^{2} - 10ac\theta er - 10c^{2} e^{2} r^{2} \theta^{2} + 20c^{2} e^{2} r^{2} \theta \hfill \\ + 10c^{2} e^{2} r\theta^{2} - 10c^{2} e^{2} r\theta + 10cear - 4c\theta ea + 18cer\theta^{2} - 6cer\theta^{3} + 3c\theta e + 6cer - 18c\theta er - 3a) \hfill \\ \end{gathered} }{9} \hfill \\ \mathop \prod \nolimits_{{\mathop 2\limits^{**} }} = \frac{{(a - c\theta e + 2rc\theta e - 2rce)^{2} }}{9} \hfill \\ \end{gathered} \right.$.

Proof of proposition 1

Let $\Delta p = p^{{*{*}}} - p^{*}$, and substitute the solutions of $p^{{*{*}}}$ and $p^{*}$ into this function, and then we obtain $\Delta p = - ce(1 - \theta )({1} - r)/3$.

Because $c > 0,e > 0,0 < \theta < 1,0 < r \le 1$, we have $\Delta p < 0$, so $p^{{*{*}}} < p^{*}$.

Let $\Delta Q = (q_{1}^{{{*}*}} + q_{2}^{{*{*}}} ) - (q_{1}^{*} + q_{2}^{*} )$, and substitute the solutions of $q_{1}^{{{*}*}} ,q_{2}^{{*{*}}} ,q_{1}^{*} ,q_{2}^{*}$ into this function, and then we obtain $\Delta Q = ce(1 - \theta )({1} - r)/3$. Because $c > 0,e > 0,0 < \theta < 1,0 < r \le 1$, we have $\Delta Q > 0$, so $(\mathop q\nolimits_{{\mathop 1\limits^{**} }} + \mathop q\nolimits_{{\mathop 2\limits^{**} }} ) > (\mathop q\nolimits_{{\mathop 1\limits^{*} }} + \mathop q\nolimits_{{\mathop 2\limits^{*} }} )$.

Proof of proposition 2

Based on results in Table 3, the total energy consumption for both manufacturers after EPC cooperation are $\Delta CE^{**} = (q_{1}^{**} + q_{2}^{**} )\theta e = \frac{(2a - 2ce\theta + rce\theta - rce)\theta e}{3}$, and that without EPC cooperation is $\Delta CE^{*} = (\theta q_{1}^{*} + q_{2}^{*} )e = \frac{{(a\theta + 2ce\theta - 2ce\theta^{2} + a - 2ce)e}}{3}$.

As a result, the changes of total energy consumption after EPC cooperation are $\Delta CE^{**} - \Delta CE^{*} = \frac{{e(a\theta + rce\theta^{2} - rce\theta - 2ce\theta - a + 2ce)}}{3}$. When $r < \frac{2ce - a}{{ce\theta }}$, $\Delta CE^{**} - \Delta CE^{*} > 0$ and total energy consumption increases after EPC cooperation; when $r > \frac{2ce - a}{{ce\theta }}$, that reversely decreases.

Proof of proposition 3 and Proof of proposition 4

Let $\Delta \mathop \prod \nolimits_{1} = \mathop \prod \nolimits_{{\mathop 1\limits^{**} }} - \mathop \prod \nolimits_{{\mathop 1\limits^{*} }}$, and substitute the equilibrium solutions in Table 1 into this equation, and we can obtain:

$\Delta \mathop \prod \nolimits_{1} = \frac{{1}}{18}(1 - \theta )(10c^{2} e^{2} r^{2} \theta - 10c^{2} e^{2} r^{2} - 10c^{2} e^{2} r\theta + 6c^{2} e^{2} \theta + 6cer\theta^{2} + 10acer - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} - 4ace + 6cer + ce\theta + 3a\theta - 3a)$ Because $c > 0,e > 0,0 < \theta < 1,0 < r \le 1$, we have $\frac{{1}}{{{18}}}(1 - \theta ) > 0$,the sign of $\Delta \mathop \prod \nolimits_{1}$ depends on $a,r$:

• (i) when $r \ge 0.4$,we have:$\left\{ \begin{gathered} \Delta \mathop \prod \nolimits_{1} \ge 0,\forall :a \ge \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )} \hfill \\ \Delta \mathop \prod \nolimits_{1} < 0,\forall :a < \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )} \hfill \\ \end{gathered} \right.$

• (ii) (ii)when $r < 0.4$,we have:

  $\left\{ \begin{gathered} \Delta \mathop \prod \nolimits_{1} \ge 0,\forall :a \le \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )} \hfill \\ \Delta \mathop \prod \nolimits_{1} < 0,\forall :a > \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )} \hfill \\ \end{gathered} \right.$ Correspondingly, we have:

$\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{{*{*}}} }} \ge \mathop \prod \nolimits_{{\mathop 1\limits^{*} }} ,\forall :a \ge \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )},0.4 \le r \le 1 \hfill \\ \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \cup \forall :a \le \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )},0 < r < 0.4 \hfill \\ \mathop \prod \nolimits_{{\mathop 1\limits^{{*{*}}} }} < \mathop \prod \nolimits_{{\mathop 1\limits^{*} }} ,\forall :a < \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )},0.4 \le r \le 1 \hfill \\ \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \mathop {}\limits^{{}} \cup \forall :a > \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )},0 < r < 0.4 \hfill \\ \end{gathered} \right.$ Let $\Delta \mathop \prod \nolimits_{{2}} = \mathop \prod \nolimits_{{\mathop {2}\limits^{**} }} - \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }}$, and substitute the equilibrium solutions in Tab.1 into this equation, and we can obtain:$\Delta \mathop \prod \nolimits_{{2}} = \frac{{4}}{{9}}(1 - \theta )(1 - r)ce(cer\theta - cer - ce + a)$. Because $c > 0,e > 0,0 < \theta < 1,0 < r \le 1$, we have $\frac{{4}}{{9}}(1 - \theta )(1 - r)ce > 0$,the sign of $\Delta \mathop \prod \nolimits_{{2}}$ depends on $a$:$\left\{ \begin{gathered} \Delta \mathop \prod \nolimits_{{2}} \ge 0,\forall :a \ge ce + cer - c\theta er \hfill \\ \Delta \mathop \prod \nolimits_{{2}} < 0,\forall :a < ce + cer - c\theta er \hfill \\ \end{gathered} \right.$.

Correspondingly, we have: $\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop {2}\limits^{{{*}*}} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :a \ge ce + cer - c\theta er \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{{*{*}}} }} < \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :a < ce + cer - c\theta er \hfill \\ \end{gathered} \right.$.

We assume that the lower limit of the market is $\underline{a} > (2 - \theta )ce$.Thus, let $a_{{1}} = \frac{{(10r^{2} c^{2} e^{2} \theta - 10r^{2} c^{2} e^{2} - 10rc^{2} e^{2} \theta + 6c^{2} e^{2} \theta + 6rce\theta^{2} - 2c^{2} e^{2} - 12cer\theta - 3ce\theta^{2} + 6cer + 3ce\theta )}}{(4ce - 10rce + 3 - 3\theta )}$;$a_{{2}} = ce + cer - c\theta er$; $\underline{a} = (2 - \theta )ce$.

Let $\Delta {1} = \underline{a} - a_{{1}}$, we have $\Delta {1} = \frac{2ce(1 - r)(1 - \theta )(5rce - 5ce + 3\theta - 3)}{{(4ce - 10rce + 3 - 3\theta )}}$. Since $c > 0,e > 0,0 < \theta < 1,0 < r \le 1$, the sign of $\Delta {1}$ depends on $c$ and $r$:

(i)when $r \ge 0.4$:

(a) when $c \le {3}(1 - \theta )/e({10}r - 4)$, we have $\Delta {1} \ge 0$, so $\underline{a} \ge a_{{1}}$;

(b) when $c > {3}(1 - \theta )/e({10}r - 4)$, we have $\Delta {1} < 0$, so $\underline{a} < a_{{1}}$;

(ii) when $r < 0.4$:

we have $c > {3}(1 - \theta )/e({10}r - 4)$. Because ${3}(1 - \theta )/e({10}r - 4) < 0,c > 0$, we have $\Delta {1} \ge 0$, so $\underline{a} \ge a_{{1}}$;

Let $\Delta {2} = \underline{a} - a_{{2}}$, we have $\Delta {2} = ({1} - r)(1 - \theta )ce > 0$, so $\underline{a} \ge a_{{2}}$;

Let $\Delta {3} = a_{{2}} - a_{{1}}$, we have $\Delta {3} = 3ce(1 - \theta )({1} - r)(2ce - \theta + 1)/(4ce - 10rce + 3 - 3\theta )$, the sign of $\Delta {3}$ depends on $c$ and $r$:

(i)when $r \ge 0.4$:

(a) when $c \le {3}(1 - \theta )/e({10}r - 4)$, $\Delta {3} \ge 0$, so $a_{{2}} \ge a_{{1}}$;

(b) when $c > {3}(1 - \theta )/e({10}r - 4)$, $\Delta {3} < 0$, so $a_{{2}} < a_{{1}}$;

(ii) when $r < 0.4$:

we have $c > {3}(1 - \theta )/e({10}r - 4)$, Because ${3}(1 - \theta )/e({10}r - 4) < 0,c > 0$, we have $\Delta {3} \ge 0$, so $a_{{2}} \ge a_{{1}}$;

Let $c_{1} = {3}(1 - \theta )/e({10}r - 4)$, and the relationship of $a_{{1}} ,a_{{2}} ,\underline{a}$ depends on $c$, that is $\left\{ \begin{gathered} a_{{1}} \le a_{{2}} \le \underline{a} ,\forall :0 < c \le c_{1} \hfill \\ a_{{2}} \le \underline{a} < a_{{1}} ,\forall :c > c_{{1}} \hfill \\ \end{gathered} \right.$.

In summary, the relationship between market size and energy prices is shown as follows:

According to the above figure, the comparative results of the two manufacturers’ profits with or without EPC cooperation can be divided into two situations:

(i) when $0 < c \le c_{1}$,$a_{{1}} \le a_{{2}} \le \underline{a}$, the comparative results of the two manufacturers’ profits are shown as follows:

Because $a > \underline{a}$, we can obtain $\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{{{*}*}} }} \ge \mathop \prod \nolimits_{{\mathop 1\limits^{*} }} \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{**} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} \hfill \\ \end{gathered} \right.,\forall :0 < c \le c_{1} ,0 < r \le 1$.

(ii) when $c > c_{{1}}$, $a_{{2}} \le \underline{a} < a_{{1}}$:

(a) when $0.4 < r \le 1$, the comparative results of the two manufacturers’ profits are shown as follows:

Because $a > \underline{a}$, we can obtain $\left\{ \begin{gathered} \left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{{*{*}}} }} < \mathop \prod \nolimits_{{\mathop 1\limits^{*} }} \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{{*{*}}} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} \hfill \\ \end{gathered} \right.,\forall :\underline{a} < a \le a_{1} ,c > c_{{1}} ,0.4 \le r \le 1 \hfill \\ \left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{{{*}*}} }} \ge \mathop \prod \nolimits_{{\mathop 1\limits^{*} }} \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{{*{*}}} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} \hfill \\ \end{gathered} \right.,\forall :a > a_{{1}} ,c > c_{{1}} ,0.4 \le r \le 1 \hfill \\ \end{gathered} \right.$.

(b) when ${0} < r < {0}{\text{.4}}$, the comparative results of the two manufacturers’ profits are shown as follows:

Because $a > \underline{a}$, we can obtain $\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{**} }} < \mathop \prod \nolimits_{{\mathop 1\limits^{*} }} \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{**} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} \hfill \\ \end{gathered} \right.,\forall :c > c_{{1}} ,{0} < r < {0}{\text{.4}}$.

In summary, the comparative results of the two manufacturers’ profits are shown as follows:

Because $\Delta \mathop \prod \nolimits_{{2}} = \frac{{4}}{{9}}(1 - \theta )(1 - r)ce(cer\theta - cer - ce + a)$, and $\frac{{4}}{{9}}(1 - \theta )(1 - r)ce > 0$, so the sign of $\Delta \mathop \prod \nolimits_{{2}}$ depends on $(cer\theta - cer - ce + a)$. Let $(cer\theta - cer - ce + a){ = 0}$, we have $r_{1} = \frac{a - ce}{{(1 - \theta )ce}}$. Then, the sign of $\Delta \mathop \prod \nolimits_{{2}}$ depends on $a$ and $r$:$\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop {2}\limits^{{*{*}}} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :r < r_{1} \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{{*{*}}} }} < \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :r > r_{1} \hfill \\ \end{gathered} \right.$.

when $(\theta - 2r\theta + 2r)ce < a < (2 - \theta )ce$, we have:

(i) if $\frac{1}{2} < r < 1$, we have $\mathop r\nolimits_{{1}} > \frac{1}{2}$, and $\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop {2}\limits^{{*{*}}} }} \ge \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :\frac{1}{2} < r \le r_{1} \hfill \\ \mathop \prod \nolimits_{{\mathop {2}\limits^{{{*}*}} }} < \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :r_{1} < r < 1 \hfill \\ \end{gathered} \right.$;

(ii) if $0 < r \le \frac{1}{2}$, we have $\mathop r\nolimits_{{1}} < 0$, and $\mathop \prod \nolimits_{{\mathop {2}\limits^{**} }} < \mathop \prod \nolimits_{{\mathop {2}\limits^{*} }} ,\forall :0 < r \le \frac{1}{2}$.

Proof of proposition 5

Calculate derivative of Eq. (6) with respect to $\mathop q\nolimits_{1}$ and $\theta$ respectively, and let $\frac{{\partial \mathop \prod \nolimits_{1} }}{{\partial \mathop q\nolimits_{1} }} = 0$, $\frac{{\partial \mathop \prod \nolimits_{1} }}{\partial \theta } = 0$.

Solving this simultaneous equation, we can obtain $\left\{ \begin{gathered} \mathop q\nolimits_{{\mathop 1\limits^{{**{*}}} }} = \frac{{a - c\theta e - \mathop q\nolimits_{2} }}{2} \hfill \\ \mathop \delta \nolimits^{{*{*}*}} = 1 - rce \hfill \\ \end{gathered} \right.$. Because $\frac{{\mathop \partial \nolimits^{2} \mathop \prod \nolimits_{1} }}{{\partial \mathop {\mathop q\nolimits_{1} }\nolimits^{2} }} = - 2$, $\frac{{\mathop \partial \nolimits^{2} \mathop \prod \nolimits_{1} }}{{\partial \mathop \theta \nolimits^{2} }} = - q_{2}$ and $\frac{{\mathop \partial \nolimits^{2} \mathop \prod \nolimits_{1} }}{{\partial \mathop q\nolimits_{1} \partial \theta }} = \frac{{\mathop \partial \nolimits^{2} \mathop \prod \nolimits_{1} }}{{\partial \theta \partial \mathop q\nolimits_{1} }} = 0$, the corresponding Hessian Matrix is $H = \left[ {\begin{array}{*{20}c} { - 2} & 0 \\ 0 & { - q_{2} } \\ \end{array} } \right]$. Solving this Hessian Matrix, we can obtain the first-order principal minor $- 2 < 0$, and the second-order principal minor $2q_{2} > 0$. Thus, we have $H < 0$, and the function $\mathop \prod \nolimits_{1}$ has the maximum value at $(\mathop q\nolimits_{{\mathop 1\limits^{{{**}*}} }} ,\mathop \delta \nolimits_{{\mathop {}\limits^{{{**}*}} }} )$.

Proof of proposition 6

Calculate derivative of Eq. (7) with respect to $\mathop q\nolimits_{2}$, and let $\frac{{\partial \mathop \prod \nolimits_{2} }}{{\partial \mathop q\nolimits_{2} }} = 0$.

And then we can get $\mathop q\nolimits_{{\mathop 2\limits^{***} }} = \frac{{a + (\delta r - \delta - r)ce - \mathop q\nolimits_{1} }}{2}$. Since $\frac{{\mathop \partial \nolimits^{2} \mathop \prod \nolimits_{2} }}{{\partial \mathop {\mathop q\nolimits_{2} }\nolimits^{2} }} = - 2 < 0$, $\mathop \prod \nolimits_{2}$ has the maximum value at $\mathop q\nolimits_{{\mathop 2\limits^{***} }}$. Solving the first derivation of Eqs. (6) and (7) simultaneously, we can obtain the optimal solutions as $\left\{ \begin{gathered} \mathop q\nolimits_{{\mathop 1\limits^{***} }} = \frac{{a - 2c\theta e + c^{2} e^{2} r^{2} - c^{2} e^{2} r + ce}}{3} \hfill \\ \mathop q\nolimits_{{\mathop 2\limits^{***} }} = \frac{{a + c\theta e - 2c^{2} e^{2} r^{2} + 2c^{2} e^{2} r - 2ce}}{3} \hfill \\ \delta^{***} = 1 - rce \hfill \\ \end{gathered} \right.$. Then, substituting those optimal solutions into the inverse demand function, we can obtain the optimal price $\mathop p\nolimits_{{\mathop {}\limits^{***} }} = \frac{{a + c\theta e + c^{2} e^{2} r^{2} - c^{2} e^{2} r + ce}}{3}$. Substituting those optimal solutions into the Eqs. (6) and (7), we can obtain the optimal profits of two manufacturers as $\left\{ \begin{gathered} \mathop \prod \nolimits_{{\mathop 1\limits^{***} }} = \frac{\begin{gathered} (2a^{2} + 2c^{2} e^{2} + 4ace + 8c^{2} e^{2} \theta^{2} - 4c^{3} e^{3} r - 4c^{4} e^{4} r^{4} - 2c^{3} e^{3} r^{2} + 2c^{4} e^{4} r^{3} \hfill \\ + 2c^{4} e^{4} r^{2} - 5c^{3} e^{3} r^{2} \theta + 8c^{3} e^{3} r\theta + 7ac^{2} e^{2} r^{2} - 4ac^{2} e^{2} r - 8ac\theta e - 8c^{2} e^{2} \theta ) \hfill \\ \end{gathered} }{18} \hfill \\ \mathop \prod \nolimits_{{\mathop 2\limits^{***} }} = \frac{{(a + c\theta e - 2c^{2} e^{2} r^{2} + 2c^{2} e^{2} r - 2ce)^{2} }}{9} \hfill \\ \end{gathered} \right.$