Differential game analysis of carbon emissions reduction and promotion in a sustainable supply chain considering social preferences

Abstract

We incorporate consumer low-carbon awareness and social preferences, including relationship and status preferences, into a game of emissions reduction and promotion involving one manufacturer and one retailer using a long-term perspective. We investigate the channel members’ decision making and performance under three scenarios, including a decentralized scenario both with and without a cost-sharing contract as well as a centralized scenario. In addition, we examine the effects of some key parameters on the channel members’ decisions and performance. Our study finds that improving consumer low-carbon awareness is beneficial for carbon emissions reduction and both channel members’ utilities. A cost-sharing contract can incentivize the retailer to improve promotion efforts, and the manufacturer’s optimal emissions reduction effort is independent of the cost-sharing contract used. The cost-sharing proportion increases as the manufacturer gives more weight to the relationship and decreases as the retailer gives more weight to the relationship. A cost-sharing contract changes the effect of channel members’ social preferences and marginal profits on their decisions. Most importantly, the supply chain system can achieve Pareto improvement with a cost-sharing contract. If the manufacturer aims to optimize the supply chain’s total profit, then the supply chain system can achieve perfect coordination with a cost-sharing contract.

Appendices

Appendix A: Proof of Proposition 1

Prove: In the decentralized scenario without a cost-sharing contract, the manufacturer’s profit at time $t$ from the long-run and dynamic perspectives is:

$$ \pi_{m}^{N} (\tau ) = \int_{t}^{\infty } {e^{ - \rho t} {\kern 1pt} [b_{m} Q - c_{m} (E_{m} )]{\kern 1pt} {\kern 1pt} } dt . $$

(A1)

Let $v_{m}^{N} { = }\int_{t}^{\infty } {e^{ - \rho (s - t)} \left[ {b_{m} Q - c_{m} (E_{m}^{N} )} \right]} dt$. Then, $\pi_{m}^{N} (\tau ){ = }e^{ - \rho t} v_{m}^{N}$. Similarly, the retailer’s profit at time $t$ from the long-run and dynamic perspectives is $\pi_{r}^{N} (\tau ) = e^{ - \rho t} v_{r}^{N}$ where $v_{r}^{N} { = }\int_{t}^{\infty } {e^{ - \rho (s - t)} \left[ {b_{r} Q - c_{r} (E_{r}^{N} )} \right]} dt$. Based on $u_{r} = \pi_{r} + \theta_{r} \pi_{m}$, the retailer’s objective function satisfies the following HJB function:

$$ \rho v_{r}^{N} (\tau ) + \rho \theta_{r} v_{m}^{N} (\tau ) = \hbox{max} \left\{ \begin{array}{l} b_{r} \left[ {a + \beta E_{r}^{N} (t) + \mu \tau^{N} (t)} \right] - \frac{1}{2}\eta_{r} (E_{r}^{N} (t))^{2} \hfill \\\quad + \theta_{r} \left\{ {b_{m} \left[ {a + \beta E_{r}^{N} (t) + \mu \tau^{N} (t)} \right] - \frac{1}{2}\eta_{m} (E_{m}^{N} (t))^{2} } \right\} \hfill \\ \quad+ \left[ {v_{r} (\tau )^{'} + \theta_{{_{r} }} v_{m} (\tau )^{'} } \right]\left[ {\gamma E_{m}^{N} (t) - \delta \tau^{N} (t)} \right] \hfill \\ \end{array} \right\} $$

(A2)

It is easy to prove that $\rho v_{r}^{N} (\tau ) + \rho \theta_{r} v_{m}^{N} (\tau )$ is concave in $E_{r}^{N}$. Then, the retailer’s promotion effort can be expressed as Equation (A3) based on the first-order condition.

$$ E_{r}^{N} (t) = \frac{{(b_{r} + \theta_{r} b_{m} )\beta }}{{\eta_{r} }} $$

(A3)

Similarly, the manufacturer’s objective function satisfies the following HJB function:

$$ \rho v_{m}^{N} (\tau ) + \rho \theta_{m} v_{r}^{N} (\tau ) = \hbox{max} \left\{ \begin{array}{l} b_{m} [\alpha + \beta E_{r}^{N} (t) + \mu \tau^{N} (t)] - \frac{1}{2}\eta_{m} (E_{m}^{N} (t))^{2} \hfill \\\quad + \theta_{m} \{ b_{r} [\alpha + \beta E_{r}^{N} (t) + \mu \tau^{N} (t)] - \frac{1}{2}\eta_{r} (E_{r}^{N} (t))^{2} \} \hfill \\\quad+ \left[ {v_{m} (\tau )^{'} + \theta_{{_{m} }} v_{r} (\tau )^{'} } \right]\left[ {\gamma E_{m}^{N} (t) - \delta \tau^{N} (t)} \right] \hfill \\ \end{array} \right\}. $$

(A4)

It is easy to prove that $\rho v_{m}^{N} (\tau ) + \rho \theta_{m} v_{r}^{N} (\tau )$ is concave in $E_{m}^{N}$. Then, the manufacturer’s optimal emission reduction effort can be expressed as Equation (A5) based on the first-order condition.

$$ E_{m}^{N} (t) = \frac{{\left[ {v_{m}^{N} (\tau )^{'} + \theta_{m} v_{r}^{N} (\tau )^{'} } \right]\gamma }}{{\eta_{m} }} $$

(A5)

By substituting Equations (A3) and (A5) into Equations (A2) and (A4), the HJB functions of the retailer and manufacturer can be expressed as Equations (A6) and (A7), respectively.

$$ \begin{aligned} \rho v_{r}^{N} (\tau ) + \rho \theta_{r} v_{m}^{N} (\tau ) & = \frac{{(v_{m}^{N} (\tau )^{'} + \theta_{{_{m} }} v_{r}^{N} (\tau )^{'} )\gamma^{2} (2v_{r}^{N} (\tau )^{'} + \theta_{{_{r} }} v_{m}^{N} (\tau )^{'} - \theta_{r} \theta_{m} v_{r}^{N} (\tau )^{'} )}}{{2\eta_{m} }} \\ & \quad + \;\frac{{(\beta^{2} b_{r} + \beta^{2} \theta_{r} b_{m} + 2\alpha \eta_{r} )(b_{r} + \theta_{r} b_{m} )}}{{2\eta_{r} }} \\ &\quad + \left[ {b_{r} \mu + \theta_{r} b_{m} \mu - \left( {v_{r}^{N} (\tau )^{'} + \theta_{{_{r} }} v_{m}^{N} (\tau )^{'} } \right)\delta } \right]\tau^{N} (t) \\ \end{aligned} $$

(A6)

$$ \begin{aligned} \rho v_{m}^{N} (\tau ) + \rho \theta_{m} v_{r}^{N} (\tau ) & = \frac{{\left[ {v_{m}^{N} (\tau )^{'} + \theta_{{_{m} }} v_{r}^{N} (\tau )^{'} } \right]^{2} \gamma^{2} }}{{2\eta_{m} }} + \frac{{(b_{r} - \theta_{r} b_{m} )(b_{r} + \theta_{r} b_{m} )\beta^{2} + 2b_{r} \alpha \eta_{r} }}{{2\eta_{r} }}\theta_{m} \\ & \quad + \left[ {b_{m} \mu + \theta_{m} b_{r} \mu - \left( {v_{m}^{N} (\tau )^{'} + \theta_{{_{m} }} v_{r}^{N} (\tau )^{'} } \right)\delta } \right]\tau^{N} (t) + \frac{{(b_{r} + \theta_{r} b_{m} )\beta^{2} + \alpha \eta_{r} }}{{\eta_{r} }}b_{m} \\ \end{aligned} $$

(A7)

Based on the structures of Equations (A6) and (A7), we assume that the solutions are linear about $\tau$ and:

$$ \left\{ \begin{aligned} v_{m}^{N} (\tau ) = a_{1}^{N} \tau + d_{1}^{N} \hfill \\ v_{r}^{N} (\tau ) = a_{2}^{N} \tau + d_{2}^{N} \hfill \\ \end{aligned} \right. $$

(A8)

where $a_{1}^{N}$, $a_{2}^{N}$, $d_{1}^{N}$ and $d_{2}^{N}$ are constants. By substituting Equation (A8) into (A6) and (A7) and solving the new equations, we have $a_{1}^{N}$, $a_{2}^{N}$, $d_{1}^{N}$ and $d_{2}^{N}$ as follows:

$$ \left\{ \begin{aligned} a_{1}^{N} = \frac{{b_{m} \mu }}{\rho + \delta }{\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} d_{1}^{N} = \frac{{ab_{m} }}{\rho } + \frac{{b_{m} \beta^{2} (b_{r} + b_{m} \theta_{r} )}}{{\eta_{r} \rho }} + \frac{{\gamma^{2} \mu^{2} (b_{m}^{2} - b_{r}^{2} \theta_{m}^{2} )}}{{2\eta_{m} \rho (\rho + \delta )^{2} }}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \hfill \\ a_{2}^{N} = \frac{{b_{r} \mu }}{\rho + \delta },{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} d_{2}^{N} = \frac{{ab_{r} }}{\rho } + \frac{{\beta^{2} (b_{r}^{2} - b_{m}^{2} \theta_{r}^{2} )}}{{2\eta_{r} \rho }} + \frac{{\gamma^{2} \mu^{2} b_{r} (b_{m} + b_{r} \theta_{m} )}}{{\eta_{m} \rho (\rho + \delta )^{2} }} \hfill \\ \end{aligned} \right. $$

(A9)

Then, we obtain the optimal solution of the game as:

$$ \left\{ \begin{array}{l} E_{m}^{N * } (t) = \frac{{(b_{m} + b_{r} \theta_{m} )\gamma \mu }}{{\eta_{m} (\rho + \delta )}} \hfill \\ E_{r}^{N * } (t) = \frac{{(b_{r} + \theta_{r} b_{m} )\beta }}{{\eta_{r} }} \hfill \\ \end{array} \right. $$

(A10)

Moreover, based on $\tau (0) = \tau_{0} \ge 0$, emission reduction effort state trajectories and channel members’ optimal utilities are written as:

$$ \tau^{N} (t) = \omega^{N} + (\tau_{0} - \omega^{N} )e^{ - \delta t} $$

(A11)

$$ \left\{ \begin{aligned} \pi_{m}^{N} (\tau ) = e^{ - \rho t} (a_{1}^{N} \tau + d_{1}^{N} ) \hfill \\ \pi_{r}^{N} (\tau ) = e^{ - \rho t} (a_{2}^{N} \tau + d_{2}^{N} ) \hfill \\ \end{aligned} \right. $$

(A12)

$$ \left\{ \begin{aligned} u_{m}^{N * } = \pi_{m}^{N} + \theta_{m} \pi_{r}^{N} = e^{ - \rho t} [a_{1}^{N} \tau + d_{1}^{N} + \theta_{m} (a_{2}^{N} \tau + d_{2}^{N} )] \hfill \\ u_{r}^{N * } = \pi_{r}^{N} + \theta_{r} \pi_{m}^{N} = e^{ - \rho t} [a_{2}^{N} \tau + d_{2}^{N} + \theta_{r} (a_{1}^{N} \tau + d_{1}^{N} )] \hfill \\ \end{aligned} \right. $$

(A13)

where $\omega^{N} = \frac{{\gamma^{2} \mu (b_{m} + b_{r} \theta_{m} )}}{{\delta \eta_{m} (\rho + \delta )}}$ denotes the manufacturer’s steady-state emission reduction level. In other words, $\omega^{N}$ is the manufacturer’s emission reduction level when $t \to \infty$.

Thus, Proposition 1 is proven.

Appendix B: Proof of Corollary 1

Prove: (i) We have

$$ \pi_{m}^{N} (\tau ) = e^{ - \rho t} \left\{ \begin{array}{l} \frac{{b_{m} \mu }}{\rho + \delta }\frac{{\gamma^{2} \mu (b_{m} + b_{r} \theta_{m} )}}{{\delta \eta_{m} (\rho + \delta )}} + \frac{{b_{m} \mu }}{\rho + \delta }[\tau_{0} - \frac{{\gamma^{2} \mu (b_{m} + b_{r} \theta_{m} )}}{{\delta \eta_{m} (\rho + \delta )}}]e^{ - \delta t} \hfill \\\quad + \frac{{ab_{m} }}{\rho } + \frac{{b_{m} \beta^{2} (b_{r} + b_{m} \theta_{r} )}}{{\eta_{r} \rho }} + \frac{{\gamma^{2} \mu^{2} (b_{m}^{2} - b_{r}^{2} \theta_{m}^{2} )}}{{2\eta_{m} \rho (\rho + \delta )^{2} }} \hfill \\ \end{array} \right\} $$

based on Eqs. (6) and (8). Then, we have $\frac{{\partial \pi_{m}^{N} (\tau )}}{{\partial \theta_{r} }} = e^{ - \rho t} \frac{{\beta^{2} b_{m}^{2} }}{{\eta_{r} \rho }}$ and $\frac{{\partial \pi_{m}^{N} (\tau )}}{{\partial \theta_{m} }} = e^{ - \rho t} \frac{{\gamma^{2} \mu^{2} b_{r} }}{{\eta_{m} (\rho + \delta )^{2} }}[\frac{{b_{m} }}{\delta }(1 - e^{ - \delta t} ) - \frac{{b_{r} \theta_{m} }}{\rho }]$.

Therefore, $\frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{m} }} \ge 0$ if $\frac{{b_{m} }}{{b_{r} }} \ge \frac{{\delta \theta_{m} }}{{\rho (1 - e^{ - \delta t} )}} \Leftrightarrow \theta_{m} \le \frac{{\rho b_{m} }}{{\delta b_{r} }}(1 - e^{ - \delta t} )$; otherwise, $\frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{m} }} < 0$; $\frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{r} }} > 0$.

(ii) We have $\pi_{r}^{N} (\tau ) = e^{ - \rho t} \left\{ \begin{aligned} \frac{{\gamma^{2} \mu^{2} b_{r} (b_{m} + b_{r} \theta_{m} )}}{{\eta_{m} (\rho + \delta )^{2} }}[\frac{1}{\delta }(1 - e^{ - \delta t} ) + \frac{1}{\rho }] \hfill \\ + \frac{{b_{r} \mu }}{\rho + \delta }\tau_{0} e^{ - \delta t} + \frac{{ab_{r} }}{\rho } + \frac{{\beta^{2} (b_{r}^{2} - b_{m}^{2} \theta_{r}^{2} )}}{{2\eta_{r} \rho }} \hfill \\ \end{aligned} \right\}$ based on Eqs. (6) and (8). Then,$\frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }} = - e^{ - \rho t} \frac{{\beta^{2} b_{m}^{2} \theta_{r} }}{{\eta_{r} \rho }}$ and $\frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{m} }} = e^{ - \rho t} \frac{{\gamma^{2} \mu^{2} b_{r} b_{r} }}{{\eta_{m} (\rho + \delta )^{2} }}[\frac{1}{\delta }(1 - e^{ - \delta t} ) + \frac{1}{\rho }]$. Thus, we have $\frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{m} }} > 0$; $\frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }} \ge 0$ if $\theta_{r} \le 0$; otherwise, $\frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }} < 0$.

(iii) We have $\frac{{\partial u_{r}^{N * } }}{{\partial \theta_{m} }} = \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{m} }} + \theta_{r} \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{m} }}$ and $\frac{{\partial u_{r}^{N * } }}{{\partial \theta_{r} }} = \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }} + \pi_{m}^{N} + \theta_{r} \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{r} }}$ based on $u_{r}^{N * } = \pi_{r}^{N} + \theta_{r} \pi_{m}^{N}$. Then,$\frac{{\partial u_{r}^{N * } }}{{\partial \theta_{m} }} = \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{m} }} + \theta_{r} \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{m} }} = e^{ - \rho t} \frac{{\gamma^{2} \mu^{2} b_{r} }}{{\eta_{m} (\rho + \delta )^{2} }}\left\{ {\frac{1}{\delta }(1 - e^{ - \delta t} )(b_{r} + b_{m} \theta_{r} ) + \frac{1}{\rho }b_{r} (1 - \theta_{m} \theta_{r} )} \right\} \ge 0$. Thus, we have $\frac{{\partial u_{r}^{N * } }}{{\partial \theta_{m} }} \ge 0$ based on $b_{r} + b_{m} \theta_{r} \ge 0$ and $1 - \theta_{m} \theta_{r} \ge 0$. Similarly, we have$\frac{{\partial u_{r}^{N * } }}{{\partial \theta_{r} }} = \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }} + \pi_{m}^{N} + \theta_{r} \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{r} }} = - e^{ - \rho t} \frac{{\beta^{2} b_{m}^{2} \theta_{r} }}{{\eta_{r} \rho }} + \pi_{m}^{N} + \theta_{r} e^{ - \rho t} \frac{{\beta^{2} b_{m}^{2} }}{{\eta_{r} \rho }} = \pi_{m}^{N} \ge 0$.

We have $\frac{{\partial u_{m}^{N * } }}{{\partial \theta_{m} }} = \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{m} }} + \pi_{r}^{N} + \theta_{m} \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{m} }}$ and $\frac{{\partial u_{m}^{N * } }}{{\partial \theta_{r} }} = \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{r} }} + \theta_{m} \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }}$ based on $u_{m}^{N * } = \pi_{m}^{N} + \theta_{m} \pi_{r}^{N}$. Then,

$$ \begin{aligned} \frac{{\partial u_{m}^{N * } }}{{\partial \theta_{m} }} & = e^{ - \rho t} \frac{{\gamma^{2} \mu^{2} b_{r} }}{{\eta_{m} (\rho + \delta )^{2} }}\left[ {\frac{{b_{m} }}{\delta }(1 - e^{ - \delta t} ) - \frac{{b_{r} \theta_{m} }}{\rho }} \right] \\ &\quad + e^{ - \rho t} \left\{ \begin{aligned} \frac{{\gamma^{2} \mu^{2} b_{r} (b_{m} + b_{r} \theta_{m} )}}{{\eta_{m} (\rho + \delta )^{2} }}\left[ {\frac{1}{\delta }(1 - e^{ - \delta t} ) + \frac{1}{\rho }} \right] \hfill \\ + \frac{{b_{r} \mu \tau_{0} e^{ - \delta t} }}{\rho + \delta } + \frac{{ab_{r} }}{\rho } + \frac{{\beta^{2} (b_{r}^{2} - b_{m}^{2} \theta_{r}^{2} )}}{{2\eta_{r} \rho }} \hfill \\ \end{aligned} \right\} \\ & {\kern 1pt} + \theta_{m} e^{ - \rho t} \frac{{\gamma^{2} \mu^{2} b_{r} b_{r} }}{{\eta_{m} (\rho + \delta )^{2} }}\left[ {\frac{1}{\delta }(1 - e^{ - \delta t} ) + \frac{1}{\rho }} \right] \\ \end{aligned} $$

and $\frac{{\partial u_{m}^{N * } }}{{\partial \theta_{m} }}e^{\rho t} = e^{\rho t} \pi_{r}^{N} (\tau ) + \frac{{\gamma^{2} \mu^{2} b_{r} (b_{m} + b_{r} \theta_{m} )}}{{\eta_{m} (\rho + \delta )^{2} }}\frac{1}{\delta }(1 - e^{ - \delta t} ) \ge 0$. Thus, $\frac{{\partial u_{m}^{N * } }}{{\partial \theta_{m} }} \ge 0$.

In addition, we have $\frac{{\partial u_{m}^{N * } }}{{\partial \theta_{r} }} = \frac{{\partial \pi_{m}^{N} }}{{\partial \theta_{r} }} + \theta_{m} \frac{{\partial \pi_{r}^{N} }}{{\partial \theta_{r} }} = e^{ - \rho t} \frac{{\beta^{2} b_{m}^{2} }}{{\eta_{r} \rho }}(1 - \theta_{m} \theta_{r} )$ based on Corollary 1(i) and (ii). Thus, $\frac{{\partial u_{m}^{N * } }}{{\partial \theta_{r} }} \ge 0$.

(iv) Based on Eq. (8), we have

$$ \frac{{\partial v_{r}^{N} }}{\partial \mu } = \frac{{2\gamma^{2} b_{r} \mu (b_{m} + b_{r} \theta_{m} )(e^{\delta t} \rho - \rho + e^{\delta t} \delta ) + b_{r} (\eta_{m} \tau_{0} \rho^{2} \delta + \eta_{m} \tau_{0} \rho \delta^{2} )}}{{e^{\delta t} \eta_{m} \rho \delta (\rho + \delta )^{2} }} $$

and$\frac{{\partial v_{m}^{N} }}{\partial \mu } = \frac{{b_{m} \eta_{m} \tau_{0} \rho \delta^{2} - \gamma^{2} \mu (b_{m} + b_{r} \theta_{m} )(2b_{m} \rho - 2b_{m} e^{\delta t} \rho - b_{m} e^{\delta t} \delta + b_{r} e^{\delta t} \delta \theta_{m} ) + b_{m} \eta_{m} \tau_{0} \rho^{2} \delta }}{{e^{\delta t} \eta_{m} \rho \delta (\rho + \delta )^{2} }}$.

Since $\pi_{m}^{N} (\tau ){ = }e^{ - \rho t} v_{m}^{N}$, $\pi_{r}^{N} (\tau ) = e^{ - \rho t} v_{r}^{N}$, $b_{m} + b_{r} \theta_{m} \ge 0$, $e^{\delta t} \rho - \rho + e^{\delta t} \delta \ge 0$ and $2b_{m} \rho - 2b_{m} e^{\delta t} \rho - b_{m} e^{\delta t} \delta + b_{r} e^{\delta t} \delta \theta_{m} \le 0$, we have $\frac{{\partial \pi_{m}^{N} }}{\partial \mu } \ge 0$ and $\frac{{\partial \pi_{r}^{N} }}{\partial \mu } \ge 0$.

Then, Corollary 1 is proven.

Appendix C: Proof of Proposition 2

Prove: In the decentralized scenario with a cost-sharing contract, the channel members’ profits at time $t$ from the long-run and dynamic perspectives are $\pi_{r}^{Y} = e^{ - \rho t} v_{r}^{Y} (\tau )$ and $\pi_{m}^{Y} = e^{ - \rho t} v_{m}^{Y} (\tau )$ where:

$$ \left\{ \begin{aligned} v_{r}^{Y} (\tau ) = \int_{t}^{\infty } {e^{ - \rho (s - t)} \left[ {b_{r} Q - (1 - X)c_{r} (E_{r}^{Y} )} \right]} dt \hfill \\ v_{m}^{Y} (\tau ) = \int_{t}^{\infty } {e^{ - \rho (s - t)} \left[ {b_{m} Q - c_{m} (E_{m}^{Y} ) - Xc_{r} (E_{r}^{Y} )} \right]} dt \hfill \\ \end{aligned} \right. . $$

(C1)

Then, the retailer’s objective function satisfies the following HJB function:

$$ \rho v_{r}^{Y} (\tau ) + \rho \theta_{r} v_{m}^{Y} (\tau ) = \hbox{max} \left\{ \begin{array}{l} b_{r} \left[ {a + \beta E_{r}^{Y} (t) + \mu \tau^{Y} (t)} \right] - \frac{1}{2}(1 - X)\eta_{r} (E_{r}^{Y} (t))^{2} \hfill \\ \quad + \theta_{r} \{ b_{m} \left[ {a + \beta E_{r}^{Y} (t) + \mu \tau^{Y} (t)} \right] - \frac{1}{2}\eta_{m} (E_{m}^{Y} (t))^{2} \hfill \\ \quad- \frac{1}{2}X\eta_{r} (E_{r}^{Y} (t))^{Y} \} + \left[ {v_{r}^{Y} (\tau )^{'} + \theta_{{_{r} }} v_{m}^{Y} (\tau )^{'} } \right]\left[ {\gamma E_{m}^{Y} (t) - \delta \tau^{Y} (t)} \right] \hfill \\ \end{array} \right\} $$

(C2)

It is easy to prove that $\rho v_{r}^{Y} (\tau ) + \rho \theta_{r} v_{m}^{Y} (\tau )$ is concave in $E_{r}^{Y}$. Then, the retailer’s promotion effort can be expressed as Equation (C3) based on the first-order condition.

$$ E_{r}^{Y} (t) = \frac{{(b_{r} + \theta_{r} b_{m} )\beta }}{{(1 - X + X\theta_{r} )\eta_{r} }} $$

(C3)

Similarly, the manufacturer’s objective function satisfies the following HJB function:

$$ \rho v_{m}^{Y} (\tau ) + \rho \theta_{m} v_{r}^{Y} (\tau ) = \hbox{max} \left\{ \begin{array}{l} b_{m} [\alpha + \beta E_{r}^{Y} (t) + \mu \tau^{Y} (t)] - \frac{1}{2}\eta_{m} (E_{m}^{Y} (t))^{2} - \frac{1}{2}X\eta_{r} (E_{r}^{Y} (t))^{2} \hfill \\ \quad + \left[ {v_{m}^{Y} (\tau )^{'} + \theta_{{_{m} }} v_{r}^{Y} (\tau )^{'} } \right]\left[ {\gamma E_{m}^{Y} (t) - \delta \tau^{Y} (t)} \right] \hfill \\ + \theta_{m} \{ b_{r} [\alpha + \beta E_{r}^{Y} (t) + \mu \tau^{Y} (t)] - \frac{1}{2}(1 - X)\eta_{r} (E_{r}^{Y} (t))^{2} \} \hfill \\ \end{array} \right\} $$

(C4)

It is easy to prove that $\rho v_{m}^{Y} (\tau ) + \rho \theta_{m} v_{r}^{Y} (\tau )$ is concave in $E_{m}^{Y}$ and $X$. Then, the manufacturer’s optimal emission reduction effort can be expressed as Equation (C5) based on the first-order condition.

$$ \left\{ \begin{array}{l} E_{m}^{Y} (t) = \frac{{\left[ {v_{m}^{Y} (\tau )^{'} + \theta_{m} v_{r}^{Y} (\tau )^{'} } \right]\gamma }}{{\eta_{m} }} \hfill \\ X = \frac{{2b_{m} - b_{r} - 3b_{m} \theta_{r} + b_{r} \theta_{m} + 2b_{m} \theta_{m} \theta_{r}^{2} - b_{m} \theta_{m} \theta_{r} }}{{(1 - \theta_{r} )[(b_{r} - b_{m} \theta_{r} - 2b_{r} \theta_{r} )\theta_{m} + 2b_{m} + b_{r} - b_{m} \theta_{r} ]}} \hfill \\ \end{array} \right. $$

(C5)

By substituting Equations (C3) and (C5) into Equations (C2) and (C4), the HJB functions of the retailer and manufacturer can be expressed as Equations (C6) and (C7), respectively.

$$ \begin{aligned} \rho v_{r}^{Y} (\tau ) + \rho \theta_{r} v_{m}^{Y} (\tau ) & = \frac{{\left[ {v_{m}^{Y} (\tau )^{'} + \theta_{m} v_{r}^{Y} (\tau )^{'} } \right]\gamma^{2} \left[ {(2 - \theta_{m} \theta_{r} )v_{r}^{Y} (\tau )^{'} + v_{m}^{Y} (\tau )^{'} \theta_{r} } \right]}}{{2\eta_{m} }} \\ & \quad + \frac{{(\beta^{2} b_{r} + \beta^{2} \theta_{r} b_{m} + 2\alpha \eta_{r} )(b_{r} + \theta_{r} b_{m} )}}{{2\eta_{r} }} \\ &\quad + \left[ {b_{r} \mu + \theta_{r} b_{m} \mu - \left( {v_{r}^{Y} (\tau )^{'} + \theta_{r} v_{m}^{Y} (\tau )^{'} } \right)\delta } \right]\tau^{Y} (t) \\ \end{aligned} $$

(C6)

$$ \begin{aligned} \rho v_{m}^{Y} (\tau ) + \rho \theta_{m} v_{r}^{Y} (\tau ) & = [b_{m} \mu + \theta_{m} b_{r} \mu - \left( {v_{m} (\tau )^{'} + \theta_{{_{m} }} v_{r} (\tau )^{'} } \right)\delta ]\tau (t) \\ &\quad + b_{m} \left[ {\alpha + \frac{{(b_{r} + \theta_{r} b_{m} )\beta^{2} }}{{(1 - X + X/\theta_{m} )\eta_{r} }}} \right] + \frac{{[v_{m} (\tau )^{'} + \theta_{{_{m} }} v_{r} (\tau )^{'} ]^{2} \gamma^{2} }}{{2\eta_{m} }} \\ &\quad + \theta_{m} \left\{ {b_{r} \left[ {\alpha + \frac{{(b_{r} + \theta_{r} b_{m} )\beta^{2} }}{{(1 - X + X/\theta_{m} )\eta_{r} }}} \right] - \frac{{(b_{r} + \theta_{r} b_{m} )^{2} \beta^{2} }}{{2(1 - X + X/\theta_{m} )\eta_{r} }}} \right\} \\ \end{aligned} $$

(C7)

Based on the structures of Equations (C6) and (C7), we assume that the solutions are linear about $\tau$ and:

$$ \left\{ \begin{aligned} v_{m}^{Y} (\tau ) = a_{1}^{Y} \tau + d_{1}^{Y} \hfill \\ v_{r}^{Y} (\tau ) = a_{2}^{Y} \tau + d_{2}^{Y} \hfill \\ \end{aligned} \right. $$

(C8)

where $a_{1}^{Y}$, $a_{2}^{Y}$, $d_{1}^{Y}$ and $d_{2}^{Y}$ are constants. By substituting Equation (C8) into (C6) and (C7) and solving the new equations, we have $a_{1}^{Y}$, $a_{2}^{Y}$, $d_{1}^{Y}$ and $d_{2}^{Y}$ as follows:

$$ \left\{ \begin{array}{l} a_{1}^{Y} = \frac{{b_{m} \mu }}{\rho + \delta },{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} a_{2}^{Y} = \frac{{b_{r} \mu }}{\rho + \delta } \hfill \\ d_{1}^{Y} = \frac{{\beta^{2} \theta_{m} (b_{r} + b_{m} \theta_{r} )\left[ {(1 - X + X/\theta_{m} )\eta_{r} (b_{r} + b_{m} \theta_{r} ) + (1 - X + X\theta_{r} )\eta_{r} (b_{m} \theta_{r} - b_{r} )} \right]}}{{2[(1 - X) + X\theta_{r} ]^{2} \eta_{r}^{2} \rho (\theta_{m} \theta_{r} - 1)}} \hfill \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} + \frac{{\alpha b_{m} }}{\rho } + \frac{{\gamma^{2} \mu^{2} (b_{m}^{2} - b_{r}^{2} \theta_{m}^{2} )}}{{2\eta_{m} \rho (\rho + \delta )^{2} }}\left[ {b_{r}^{2} + \frac{{b_{m} \beta^{2} (b_{r} + b_{m} \theta_{r} )}}{{(1 - X + X\theta_{r} )\eta_{r} \rho (1 - \theta_{m} \theta_{r} )}}} \right] \hfill \\ d_{2}^{Y} = \frac{{\beta^{2} \theta_{m} \theta_{r} \left[ {(1 - X + X/\theta_{m} )\eta_{r} (b_{m} \theta_{r} + b_{r} )^{2} - 2b_{r} (1 - X + X\theta_{r} )\eta_{r} (b_{r} + b_{m} \theta_{r} )} \right]}}{{2(1 - X + X\theta_{r} )^{2} \eta_{r}^{2} \rho (1 - \theta_{m} \theta_{r} )}} \hfill \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} + \frac{{\alpha b_{r} }}{\rho } + \frac{{(b_{r}^{2} - b_{m}^{2} \theta_{r}^{2} )\beta^{2} }}{{2(1 - X + X\theta_{r} )\eta_{r} \rho (1 - \theta_{m} \theta_{r} )}} + \frac{{b_{r} \gamma^{2} (b_{m} + b_{r} \theta_{m} )}}{{\eta_{m} \rho (\rho + \delta )^{2} }} \hfill \\ \end{array} \right. $$

(C9)

Then, the optimal solution of the game is:

$$ \left\{ \begin{array}{l} E_{m}^{Y * } (t) = \frac{{(b_{m} + b_{r} \theta_{m} )\gamma \mu }}{{\eta_{m} (\rho + \delta )}} \hfill \\ E_{r}^{Y * } (t) = \frac{{(b_{m} + b_{r} \theta_{m} )(1 - \theta_{r} )\beta }}{{2(1 - \theta_{m} \theta_{r} )\eta_{r} }} + \frac{{(b_{m} + b_{r} )\beta }}{{2\eta_{r} }} \hfill \\ X^{ * } (t) = \frac{{2b_{m} - b_{r} - 3b_{m} \theta_{r} + b_{r} \theta_{m} + 2b_{m} \theta_{m} \theta_{r}^{2} - b_{m} \theta_{m} \theta_{r} }}{{(1 - \theta_{r} )[(b_{r} - b_{m} \theta_{r} - 2b_{r} \theta_{r} )\theta_{m} + 2b_{m} + b_{r} - b_{m} \theta_{r} ]}} \hfill \\ \end{array} \right. $$

(C10)

Moreover, based on $\tau (0) = \tau_{0} \ge 0$, we have the emission reduction effort state trajectories and the channel members’ profits and optimal utilities as:

$$ \tau^{Y} (t) = \omega^{Y} + (\tau_{0} - \omega^{Y} )e^{ - \delta t} $$

(C11)

$$ \left\{ \begin{aligned} \pi_{m}^{Y} (\tau ) = e^{ - \rho t} (a_{1}^{Y} \tau + d_{1}^{Y} ) \hfill \\ \pi_{r}^{Y} (\tau ) = e^{ - \rho t} (a_{2}^{Y} \tau + d_{2}^{Y} ) \hfill \\ \end{aligned} \right. $$

(C12)

$$ \left\{ \begin{aligned} u_{m}^{Y * } (t) = \pi_{m}^{Y} + \theta_{m} \pi_{r}^{Y} = e^{ - \rho t} [a_{1}^{Y} \tau + d_{1}^{Y} + \theta_{m} (a_{2}^{Y} \tau + d_{2}^{Y} )] \hfill \\ u_{r}^{Y * } (t) = \pi_{r}^{Y} + \theta_{r} \pi_{m}^{Y} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = e^{ - \rho t} [a_{2}^{Y} \tau + d_{2}^{Y} + \theta_{r} (a_{1}^{Y} \tau + d_{1}^{Y} )] \hfill \\ \end{aligned} \right. $$

(C13)

where $\omega^{Y} = \frac{{\gamma^{2} \mu (b_{m} + b_{r} \theta_{m} )}}{{\delta \eta_{m} (\rho + \delta )}}$ refers to the manufacturer’s steady-state emission reduction level in the case with a cost-sharing contract. In other words, $\omega^{Y}$ is the manufacturer’s emission reduction level when $t \to \infty$.

Thus, Proposition 2 is proven.

Appendix D: Proof of Corollary 2

Prove: (i) We have $\frac{\partial X(t)}{{\partial \theta_{r} }} = - \frac{{(b_{m} \theta_{r} + b_{r} )(1 - \theta_{m} )\left[ {3(b_{m} + b_{r} \theta_{m} )(1 - \theta_{r} ) + (1 - \theta_{m} \theta_{r} )(b_{m} + b_{r} )} \right]}}{{[2b_{m} + b_{r} - \theta_{r} (b_{m} \theta_{m} + b_{m} + 2b_{r} \theta_{m} ) + b_{r} \theta_{m} ]^{2} (\theta_{r} - 1)^{2} }}$ and $\frac{\partial X(t)}{{\partial \theta_{m} }} = \frac{{2(b_{r} + b_{m} \theta_{r} )^{2} }}{{[2b_{m} + b_{r} - b_{m} \theta_{r} - \theta_{m} (b_{m} \theta_{r} - b_{r} + 2b_{r} \theta_{r} )]^{2} }}$ from Eq. (13). Since $\theta_{m} - 1 \le 0$,$b_{r} + b_{m} \theta_{r} \ge 0$ and $b_{m} + b_{r} \theta_{m} \ge 0$, we have $\frac{\partial X}{{\partial \theta_{m} }} \ge 0$ and $\frac{\partial X}{{\partial \theta_{r} }} \le 0$. Similarly, we have $\frac{{\partial E_{r}^{Y * } (t)}}{\partial X} > 0$ from $E_{r}^{Y * } (t) = \frac{{(b_{r} + \theta_{r} b_{m} )\beta }}{{(1 - X + X\theta_{r} )\eta_{r} }}$.

(ii) We have $\frac{{\partial \pi_{m}^{Y} (t)}}{{\partial \theta_{m} }} = - \theta_{m} \left[ {\frac{{\beta^{2} (1 - \theta_{r} )(b_{r} + b_{m} \theta_{r} )^{2} }}{{4\eta_{r} \rho (1 - \theta_{m} \theta_{r} )^{3} }} + \frac{{b_{r}^{2} \gamma^{2} \mu^{2} }}{{\eta_{m} \rho (\rho + \delta )^{2} }}} \right]e^{ - \rho t}$ from Eq. (15). Since $1 - \theta_{r} \ge 0$ and $1 - \theta_{m} \theta_{r} \ge 0$, $\frac{{\partial \pi_{m}^{Y} (t)}}{{\partial \theta_{m} }} \ge$ if $\theta_{m} \le 0$; otherwise, $\frac{{\partial \pi_{m}^{Y} (t)}}{{\partial \theta_{m} }} < 0$. Similarly, $\frac{{\partial \pi_{r}^{Y} (t)}}{{\partial \theta_{m} }} = e^{ - \rho t} (\frac{{\beta^{2} (\theta_{r} - 1)(b_{r} + b_{m} \theta_{r} )^{2} }}{{4\eta_{r} \rho (\theta_{m} \theta_{r} - 1)^{3} }} + \frac{{b_{r}^{2} \gamma^{2} \mu^{2} (e^{\delta t} \rho - \rho + e^{\delta t} \delta )}}{{e^{\delta t} \eta_{m} \rho \delta (\rho + \delta )^{2} }}) \ge 0$ based on $e^{\delta t} \rho - \rho \ge 0$.

(iii) Based on Eq. (16), we have$\frac{{\partial u_{r}^{Y} (t)}}{{\partial \theta_{m} }} = \frac{{\beta^{2} (1 - \theta_{r} )(b_{r} + b_{m} \theta_{r} )^{2} }}{{2e^{\rho t} \eta_{r} \rho (X\theta_{r} - X + 1)^{2} }}\frac{dX}{{d\theta_{m} }} - \frac{{b_{r} \gamma^{2} \mu^{2} \left[ {\rho (b_{r} + b_{m} \theta_{r} )(1 - e^{\delta t} ) + b_{r} e^{\delta t} \delta (\theta_{r} \theta_{m} - 1)} \right]}}{{e^{(\delta + \rho )t} \eta_{m} \rho \delta (\delta + \rho )^{2} }}$. Based on $b_{r} + b_{m} \theta_{r} \ge 0$ and $\frac{dX(t)}{{d\theta_{m} }} \ge 0$, we have $\frac{{\partial u_{r}^{Y} (t)}}{{\partial \theta_{m} }} \ge 0$.

(iv) Based on Eq. (14), we have$\frac{{\partial \pi_{r}^{Y} (t)}}{\partial \mu } = \frac{{b_{r} [2\mu \gamma^{2} (b_{m} + b_{r} \theta_{m} )(e^{\delta t} \rho - \rho + e^{\delta t} \delta ) + \eta_{m} \tau_{0} \rho^{2} \delta + \eta_{m} \tau_{0} \rho \delta^{2} ]}}{{e^{\delta t} \eta_{m} \rho \delta (\rho + \delta )^{2} }}$ and

$\frac{{\partial \pi_{m}^{Y} (t)}}{\partial \mu } = \frac{{\mu \gamma^{2} [2b_{m}^{2} \rho (e^{\delta t} - 1) + 2b_{m} b_{r} \rho \theta_{m} (e^{\delta t} - 1) + b_{r}^{2} \delta e^{\delta t} (1 - \theta_{m}^{2} )] + b_{m} \eta_{m} \tau_{0} \rho \delta (\rho + \delta )}}{{e^{(\delta - \rho )t} \eta_{m} \rho \delta (\rho + \delta )^{2} }}$.

Since $e^{\delta t} \rho - \rho + e^{\delta t} \delta \ge 0$ and $b_{m} + b_{r} \theta_{m} \ge 0$, $\frac{{\partial \pi_{m}^{Y} (t)}}{\partial \mu } \ge 0$ and $\frac{{\partial \pi_{r}^{Y} (t)}}{\partial \mu } \ge 0$.

Then, Corollary 2 is proven.

Appendix E: Proof of Proposition 3

Prove: In the centralized scenario, the supply chain’s profit at time $t$ from the long-run and dynamic perspectives is $\prod (\tau ,t){ = }e^{ - \rho t} v^{C}$ where:

$$ v^{C} { = }\int_{t}^{\infty } {e^{ - \rho (s - t)} {\kern 1pt} {\kern 1pt} [b_{m} Q + b_{r} Q - c_{m} (E_{m} ) - c_{r} (E_{r} )]} {\kern 1pt} {\kern 1pt} dt . $$

(E1)

Then, the supply chain’s objective function satisfies the following HJB function:

$$ \rho v^{C} (\tau ) = \hbox{max} \left\{ \begin{aligned} b_{m} [\alpha + \beta E_{r} (t) + \mu \tau (t)] - \frac{1}{2}\eta_{m} E_{m}^{2} (t) - \frac{1}{2}\eta_{r} E_{r}^{2} (t) \hfill \\ +\, b_{r} [\alpha + \beta E_{r} (t) + \mu \tau (t)] + v(\tau )^{'} [\gamma E_{m} (t) - \delta \tau (t)] \hfill \\ \end{aligned} \right\} $$

(E2)

It is easy to prove that $\rho v^{C} (\tau )$ is concave in $E_{r}$ and $E_{m}$. Then, the solution can be expressed as Equation (E3) based on the first-order condition.

$$ \left\{ \begin{aligned} E_{m}^{C} (t) = v^{C} (\tau )^{'} \frac{\gamma }{{\eta_{m} }} \hfill \\ E_{r}^{C} (t) = \frac{{(b_{r} + b_{m} )\beta }}{{\eta_{r} }} \hfill \\ \end{aligned} \right.$$

(E3)

By substituting Equation (E3) into Equation (E2), we have:

$$ \begin{aligned} \rho v^{C} (\tau ) & = (b_{m} \mu + b_{r} \mu - (v^{C} )^{'} \delta )\tau^{C} (t) - \frac{1}{2}\frac{{(b_{r} + b_{m} )^{2} \beta^{2} }}{{\eta_{r} }} \\ & \quad {\kern 1pt} + \;(b_{r} + b_{m} )\left( {\alpha + \frac{{(b_{r} + b_{m} )\beta^{2} }}{{\eta_{r} }}} \right) + \frac{{(v^{C} )^{2} \gamma^{2} }}{{2\eta_{m} }} \\ \end{aligned} $$

(E4)

Based on the structure of Equation (E4), we assume that the solutions are linear about $\tau$ and:

$$ v^{C} (\tau ) = a^{C} \tau + d^{C} $$

(E5)

where $a^{C}$ and $d^{C}$ are constants. By substituting Equation (E5) into Equation (E4), the HJB function of the supply chain is written as Equation (E6).

$$ \begin{aligned} \rho (a^{C} \tau + d^{C} ) = (b_{m} \mu + b_{r} \mu - a^{C} \delta )\tau (t) - \frac{{(b_{r} + b_{m} )^{2} \beta^{2} }}{{2\eta_{r} }} \hfill \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} + (b_{r} + b_{m} )\frac{{\alpha \eta_{r} + (b_{r} + b_{m} )\beta^{2} }}{{\eta_{r} }} + \frac{{(a^{C} )^{2} \gamma^{2} }}{{2\eta_{m} }} \hfill \\ \end{aligned} $$

(E6)

Then, we have:

$$ \left\{ \begin{array}{l} \rho a^{C} = b_{m} \mu + b_{r} \mu - a^{C} \delta \hfill \\ \rho d^{C} = - \frac{1}{2}\frac{{(b_{r} + b_{m} )^{2} \beta^{2} }}{{\eta_{r} }} + (b_{r} + b_{m} )(\alpha + \frac{{(b_{r} + b_{m} )\beta^{2} }}{{\eta_{r} }}) + \frac{{(a^{C} )^{2} \gamma^{2} }}{{2\eta_{m} }} \hfill \\ \end{array} \right. $$

(E7)

By solving Equation (E7), we have:

$$ \left\{ \begin{array}{l} a^{C} = \frac{{(b_{r} + b_{m} )\mu }}{\rho + \delta } \hfill \\ d^{C} = \frac{{\alpha (b_{r} + b_{m} )}}{\rho } + \frac{{(b_{r} + b_{m} )^{2} \beta^{2} }}{{2\eta_{r} \rho }} + \frac{{\gamma^{2} \mu^{2} (b_{m} + b_{r} )^{2} }}{{2\eta_{m} \rho (\rho + \delta )^{2} }} \hfill \\ \end{array} \right. $$

(E10)

Then, the optimal solution of the game is:

$$ \left\{ \begin{array}{l} E_{m}^{C} (t) = \frac{{(b_{m} + b_{r} )\gamma \mu }}{{\eta_{m} (\rho + \delta )}} \hfill \\ E_{r}^{C} (t) = \frac{{(b_{r} + b_{m} )\beta }}{{\eta_{r} }} \hfill \\ \end{array} \right. $$

(E11)

Moreover, based on $\tau (0) = \tau_{0} \ge 0$, the emission reduction effort state trajectories and the channel’s optimal profit are:

$$ \tau^{C} (t) = \omega^{C} + (\tau_{0} - \omega^{C} )e^{ - \delta t} $$

(E12)

$$ \prod^{C * } = e^{ - \rho t} (a^{C} \tau + d^{C} ) $$

(E13)

where $\omega^{C} = \frac{{\gamma^{2} \mu (b_{m} + b_{r} )}}{{\delta \eta_{m} (\rho + \delta )}}$ denotes the manufacturer’s steady-state emission reduction level in the centralized scenario. In other words, $\omega^{C}$ is the manufacturer’s emission reduction level when $t \to \infty$.

Thus, Proposition 3 is proven.

Appendix F: Proof of Corollary 4

Prove: From $E_{r}^{N * } (t) = \frac{{(b_{r} + \theta_{r} b_{m} )\beta }}{{\eta_{r} }}$ and $E_{r}^{Y * } (t) = \frac{{(b_{m} + b_{r} \theta_{m} )(1 - \theta_{r} )\beta }}{{2(1 - \theta_{m} \theta_{r} )\eta_{r} }} + \frac{{(b_{m} + b_{r} )\beta }}{{2\eta_{r} }}$, we have $\frac{{\partial E_{m}^{j * } }}{{\partial b_{m} }} > 0$; $\frac{{\partial E_{m}^{j * } }}{{\partial \theta_{m} }} > 0$; $\frac{{\partial E_{m}^{j * } }}{{\partial b_{r} }} \ge 0$ if $\theta_{m} \ge 0$; otherwise, $\frac{{\partial E_{m}^{j * } }}{{\partial b_{r} }} < 0$ and $\frac{{\partial E_{r}^{N * } }}{{\partial b_{r} }} > 0$ easily, where $j = N,Y$.

In addition, $\frac{{\partial E_{r}^{Y * } (t)}}{{\partial b_{r} }} = \frac{{(\theta_{m} - 2\theta_{m} \theta_{r} + 1)\beta }}{{2(1 - \theta_{m} \theta_{r} )\eta_{r} }}$. When $\theta_{m} \ge 0$, $\theta_{m} - 2\theta_{m} \theta_{r} + 1 \ge 0$. Then, $\frac{{\partial E_{r}^{Y * } (t)}}{{\partial b_{r} }} \ge 0$. When $\theta_{m} < 0$,$\theta_{m} - 2\theta_{m} \theta_{r} + 1 \ge 0$ if $\theta_{r} \ge \frac{{\theta_{m} + 1}}{{2\theta_{m} }}$. Then,$\frac{{\partial E_{r}^{Y * } (t)}}{{\partial b_{r} }} \ge 0$ if $\theta_{m} < 0$ and $\theta_{r} \ge \frac{{\theta_{m} + 1}}{{2\theta_{m} }}$;$\frac{{\partial E_{r}^{Y * } (t)}}{{\partial b_{r} }} < 0$ if $\theta_{m} < 0$ and $\theta_{r} < \frac{{\theta_{m} + 1}}{{2\theta_{m} }}$. Similarly, $\frac{{\partial E_{r}^{Y * } (t)}}{{\partial b_{m} }} = \frac{{(2 - \theta_{r} - \theta_{m} \theta_{r} )\beta }}{{2(1 - \theta_{m} \theta_{r} )\eta_{r} }} > 0$.

We have $\frac{{\partial E_{r}^{Y * } (t)}}{{\partial \theta_{m} }} = \frac{\beta }{{2\eta_{r} }}\frac{{(1 - \theta_{r} )(b_{r} + b_{m} \theta_{r} )}}{{(1 - \theta_{m} \theta_{r} )^{2} }}$,$\frac{{\partial E_{r}^{Y * } (t)}}{{\partial \theta_{r} }} = \frac{{(b_{m} + b_{r} \theta_{m} )\beta }}{{2\eta_{r} }}\frac{{ - 1 + \theta_{m} }}{{(1 - \theta_{m} \theta_{r} )^{2} }}$ and $\frac{{\partial E_{r}^{N * } }}{{\partial \theta_{r} }} = \frac{{b_{m} \beta }}{{\eta_{r} }}$. Thus, we have $\frac{{\partial E_{r}^{N * } }}{{\partial \theta_{r} }} > 0$;$\frac{{\partial E_{r}^{Y * } (t)}}{{\partial \theta_{r} }} \le 0$; $\frac{{\partial E_{r}^{N * } (t)}}{{\partial \theta_{m} }} = 0$ and $\frac{{\partial E_{r}^{Y * } (t)}}{{\partial \theta_{m} }} \ge 0$ based on $b_{r} + b_{m} \theta_{r} \ge 0$ and $b_{m} + b_{r} \theta_{m}$.

Thus, Corollary 4 is proven.

Appendix G: Proof of Corollary 5

Prove: (i) Based on Eqs. (5), (13) and (18), we easily have $E_{m}^{N * } (t) = E_{m}^{Y * } (t) \le E_{m}^{C * } (t)$. Similarly, we have $E_{r}^{C * } (t) - E_{r}^{Y * } (t) = \frac{\beta }{{2\eta_{r} }}\frac{{(b_{r} + b_{m} \theta_{r} )(1 - \theta_{m} )}}{{(1 - \theta_{m} \theta_{r} )}} \ge 0$ and $E_{r}^{N * } (t) \le E_{r}^{Y * } (t)$ based on $0 < 1 - X + X\theta_{r} \le 1$ and $b_{r} + \theta_{r} b_{m} \ge 0$. Thus, $E_{r}^{N * } (t) \le E_{r}^{Y * } (t) \le E_{r}^{C * } (t)$.

(ii) From Propositions 1, 2 and 3, we have $u_{r}^{Y} (t) - u_{r}^{N} (t) = \frac{{X\beta^{2} (1 - \theta_{r} )(b_{r} + b_{m} \theta_{r} )^{2} }}{{2e^{\rho t} \eta_{r} \rho (X\theta_{r} - X + 1)}} \ge 0,$$u_{m}^{N * } (t) = e^{ - \rho t} \left[ {(b_{m} + \theta_{m} b_{r} )(a + \beta E_{r}^{N * } + \mu \tau^{N * } (t)) - C_{m} (E_{m}^{N * } ) - \theta_{m} C_{r} (E_{r}^{N * } )} \right]$ and$u_{m}^{Y * } (t) = e^{ - \rho t} \big[ (b_{m} + \theta_{m} b_{r} )(a + \beta E_{r}^{Y * } + \mu \tau^{Y * } (t)) - C_{m} (E_{m}^{Y * } ) - X^{ * } C_{r} (E_{r}^{Y * } ) - \theta_{m} (1 - X^{ * } )C_{r} (E_{r}^{Y * } ) \big].$

Since $(E_{m}^{Y * } ,X^{ * } )$ is the manufacturer’s optimal decision made in the decentralized scenario with a cost-sharing contract,$\begin{aligned} u_{m}^{Y * } (t) = e^{ - \rho t} \left[ {(b_{m} + \theta_{m} b_{r} )(a + \beta E_{r}^{Y * } + \mu \tau^{Y * } (t)) - C_{m} (E_{m}^{Y * } ) - X^{ * } C_{r} (E_{r}^{Y * } ) - \theta_{m} (1 - X^{ * } )C_{r} (E_{r}^{Y * } )} \right] \hfill \\ \ge u_{m}^{Y \prime } (t){ = }u_{m}^{Y} (E_{m}^{Y * } { = }E_{m}^{N * } ,X^{ * } { = }0) = e^{ - \rho t} \left[ {(b_{m} + \theta_{m} b_{r} )(a + \beta E_{r}^{Y * } + \mu \tau^{Y} (t)) - C_{m} (E_{m}^{N * } ) - \theta_{m} C_{r} (E_{r}^{Y * } )} \right] \hfill \\ \end{aligned}$. Based on $E_{m}^{N * } (t) = E_{m}^{Y * } (t)$ and $\tau^{N} (t) = \tau^{Y} (t)$, we have$\begin{aligned} u_{m}^{Y \prime } (t) - u_{m}^{N * } (t) & = e^{ - \rho t} \left[ {(b_{m} + \theta_{m} b_{r} )\beta (E_{r}^{Y * } - E_{r}^{N * } ) - \theta_{m} C_{r} (E_{r}^{Y * } ) + \theta_{m} C_{r} (E_{r}^{N * } )} \right] \\ {\kern 1pt} & = e^{ - \rho t} (E_{r}^{Y * } - E_{r}^{N * } )\left[ {(b_{m} + \theta_{m} b_{r} )\beta - \frac{{\theta_{m} }}{2}\eta_{r} (E_{r}^{N * } + E_{r}^{Y * } )} \right] \\ \end{aligned}$.

Then, $u_{m}^{Y \prime } (t) - u_{m}^{N * } (t) > 0$ if $\theta_{m} < 0$; otherwise, we have$\begin{aligned} u_{m}^{Y'} (t) - u_{m}^{N * } (t) \ge e^{ - \rho t} (E_{r}^{Y * } - E_{r}^{N * } )\left[ {(b_{m} + \theta_{m} b_{r} )\beta - \eta_{r} \theta_{m} E_{r}^{Y * } } \right] \hfill \\ = [b_{r} \theta_{m} (1 - \theta_{m} ) + b_{m} (2 - 2\theta_{m} + \theta_{m} \theta_{m} \theta_{r} - \theta_{m} \theta_{r} )]\frac{{e^{ - \rho t} \beta (E_{r}^{Y * } - E_{r}^{N * } )}}{{2(1 - \theta_{m} \theta_{r} )}} \hfill \\ \end{aligned}$ based on $E_{r}^{N * } \le E_{r}^{Y * }$.

Let $f(\theta_{r} ) = 2 - 2\theta_{m} + \theta_{m}^{2} \theta_{r} - \theta_{m} \theta_{r}$. $f^{\prime}(\theta_{r} ) = \theta_{m}^{2} - \theta_{m} \le 0$ given that $\theta_{m} \ge 0$. Then, $\hbox{min} f(\theta_{r} ) = f(\theta_{r} = 1) = \theta_{m}^{2} - 3\theta_{m} + 2$. We thus easily prove that $\hbox{min} f(\theta_{r} ) \ge 0$ when $0 \le \theta_{m} \le 1$. Therefore, when $\theta_{m} \ge 0$, we have $u_{m}^{Y'} (t) - u_{m}^{N * } (t) \ge 0$. Thus, we prove that $u_{m}^{Y*} (t) \ge u_{m}^{N*} (t)$.

(iii) From Propositions 1, 2 and 3, we have Corollary 5(iii).

Thus, Corollary 5 is proven.