Optimal investment and pricing strategies of online–offline model for mobile health provider

Abstract

Due to the lack of medical examinations, in online health platforms (called online channels), a portion of users are unable to be diagnosed. Recently, many mobile health providers have started investing in offline clinics or hospitals (called offline channels) to offer further health services to these customers (i.e., the online–offline model). Providers need to determine the optimal investment and pricing strategies of online and offline channels to maximize profits. In this paper, we first analyze the optimal strategies of the provider in a monopoly case, and find that the properties of optimal online and offline decisions tend to be opposite due to the competitive nature of the two channels. Furthermore, we compare this with the case where the provider invests only in the online channel (i.e., the online-only model) in terms of profitability and show that if the value of the online health (or the offline health) is not high (high enough), then the provider should choose the online–offline model, otherwise she should adopt the online-only model. Finally, we investigate a case of an m-health platform competing with local clinics and find certain interesting insights. In particular, unlike the monopoly case, in the competition case, the platform should operate the offline channel when the sensitivity of patients to the online service quality is at a moderate level, and not otherwise.

Appendix

Proof proposition 1

When ${{u_{1} + \alpha_{1} x^{*} - p_{1} > (u_{2} + \alpha_{2} y^{*} - p_{2} )} \mathord{\left/ {\vphantom {{u_{1} + \alpha_{1} x^{*} - p_{1} > (u_{2} + \alpha_{2} y^{*} - p_{2} )} \gamma }} \right. \kern-0pt} \gamma }$, for any $(x,y)$, it is easy to verify that.

$$p_{1}^{*} = \frac{{u_{1} + \alpha_{1} x}}{2},\quad p_{2}^{*} { = }\frac{{u_{2} + c_{0} + \alpha_{2} y}}{2}$$

So the objective function of profit can be rewritten as:

$\pi (x,y,p_{1}^{*} ,p_{2}^{*} ) = {{\beta \left( {\frac{{u_{1} + \alpha_{1} x}}{2}} \right)^{2} + \left( {\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2}} \right)^{2} (1 - \rho_{0} x)} \mathord{\left/ {\vphantom {{\beta \left( {\frac{{u_{1} + \alpha_{1} x}}{2}} \right)^{2} + \left( {\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2}} \right)^{2} (1 - \rho_{0} x)} \gamma }} \right. \kern-0pt} \gamma } - \frac{{c_{1} x^{2} }}{2} - \frac{{c_{2} y^{2} }}{2} - C$.

Firstly, we check the extreme point of the profit function. Because.

$\frac{{\partial \pi (x,y,p_{1}^{*} ,p_{2}^{*} )}}{\partial x} = {{\beta \alpha_{1} \left( {\frac{{u_{1} + \alpha_{1} x}}{2}} \right) - \rho_{0} \left( {\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2}} \right)^{2} } \mathord{\left/ {\vphantom {{\beta \alpha_{1} \left( {\frac{{u_{1} + \alpha_{1} x}}{2}} \right) - \rho_{0} \left( {\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2}} \right)^{2} } \gamma }} \right. \kern-0pt} \gamma } - c_{1} x$, $\frac{{\partial \pi^{2} (x,y,p_{1}^{*} ,p_{2}^{*} )}}{{\partial x^{2} }} = {{\beta \alpha_{1}^{2} } \mathord{\left/ {\vphantom {{\beta \alpha_{1}^{2} } 2}} \right. \kern-0pt} 2} - c_{1} ,$

according to the hypothesis ${{\beta \alpha_{1}^{2} } \mathord{\left/ {\vphantom {{\beta \alpha_{1}^{2} } 2}} \right. \kern-0pt} 2} - c_{1} { < }\,0$, we know $\pi (x,y,p_{1}^{*} ,p_{2}^{*} )$ is a concave function of $x$, and there is a maximum extreme value point $x^{*}$. We can easily get when $0 < x^{1} { = }$ $\frac{{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} /\gamma - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}$$\le Q$, $x^{*} { = }\frac{{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} /\gamma - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}$. Thus we have

$$\begin{aligned} \pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} ) = & \beta \left( {\frac{{u_{1} + \alpha_{1} \frac{{{{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} } \mathord{\left/ {\vphantom {{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} } \gamma }} \right. \kern-0pt} \gamma } - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}}}{2}} \right)^{2} \\ \quad & + \left( {\frac{{u_{2} - c_{0} + \alpha_{2} y}}{2}} \right)^{2} {{\left( {1 - \rho_{0} \frac{{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} /\gamma - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}} \right)} \mathord{\left/ {\vphantom {{\left( {1 - \rho_{0} \frac{{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} /\gamma - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}} \right)} \gamma }} \right. \kern-0pt} \gamma } \\ \quad & - \, \frac{{c_{1} \left( {\frac{{{{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} } \mathord{\left/ {\vphantom {{\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} } \gamma }} \right. \kern-0pt} \gamma } - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}} \right)^{2} }}{2} - \frac{{c_{2} y^{2} }}{2} - C \\ = & \beta \left( {\frac{{u_{1} }}{2} + \frac{{\alpha_{1} \rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} - 2\beta \alpha_{1}^{2} u_{1} \gamma }}{{4\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right)^{2} \\ \quad & + \left( {\frac{{u_{2} - c_{0} + \alpha_{2} y}}{2}} \right)^{2} \left( {\frac{1}{\gamma } - \frac{{\rho_{0}^{2} (u_{2} + \alpha_{2} y - c_{0} )^{2} - 2\beta \alpha_{1} \gamma u_{1} \rho_{0} }}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right) \\ \quad & - \frac{{c_{1} (\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} - 2\beta \alpha_{1} u_{1} \gamma )^{2} }}{{8\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }} - \frac{{c_{2} y^{2} }}{2} - C \\ \end{aligned}$$

$\pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )$ can be rewritten as a one variable quartic equation of $y$, i.e.,

$$\begin{aligned} f(y) = & \beta \left( {\frac{{u_{1} }}{2} + \frac{{\alpha_{1} \rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} - 2\beta \alpha_{1}^{2} u_{1} \gamma }}{{4\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right)^{2} \\ \quad & + \left( {\frac{{u_{2} - c_{0} + \alpha_{2} y}}{2}} \right)^{2} \left( {\frac{1}{\gamma } - \frac{{\rho_{0}^{2} (u_{2} + \alpha_{2} y - c_{0} )^{2} - 2\beta \alpha_{1} \gamma u_{1} \rho_{0} }}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right) \\ \quad & - \frac{{c_{1} (\rho_{0} (u_{2} + \alpha_{2} y - c_{0} )^{2} - 2\beta \alpha_{1} u_{1} \gamma )^{2} }}{{8\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }} - \frac{{c_{2} y^{2} }}{2} - C \\ \end{aligned}$$

Taking the first and second order derivatives of $f(y)$, we have

$$\begin{aligned} f^{\prime}(y) = & \alpha_{2} \left( {\frac{{\beta \alpha_{1}^{2} \rho_{0}^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }} - \frac{{\rho_{0}^{2} }}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}} - \frac{{c_{1} \rho_{0}^{2} }}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }}} \right)(u_{2} + \alpha_{2} y - c_{0} )^{3} \\ \quad & + \alpha_{2} \left( {\frac{{\beta \alpha_{1} \rho_{0} }}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}(\frac{{u_{1} }}{2} - \frac{{\beta \alpha_{1}^{2} u_{1} }}{{2(\beta \alpha_{1}^{2} - 2c_{1} )}}) + \frac{1}{2\gamma } + \frac{{\beta \alpha_{1} \rho_{0} u_{1} }}{{2\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}} + \frac{{\beta \alpha_{1} \rho_{0} u_{1} c_{1} }}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }}} \right)(u_{2} + \alpha_{2} y - c_{0} ) - c_{2} y \\ = & \alpha_{2} \left( {\frac{{(\beta \alpha_{1}^{2} - 2c_{1} )\rho_{0}^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }} - \frac{{\rho_{0}^{2} }}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right)(u_{2} + \alpha_{2} y - c_{0} )^{3} \\ \quad & + \alpha_{2} \left( {\frac{1}{2\gamma } + \frac{{\beta \alpha_{1} \rho_{0} u_{1} }}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}(1 - \frac{1}{{2(\beta \alpha_{1}^{2} - 2c_{1} )}} + \frac{{c_{1} }}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right)(u_{2} + \alpha_{2} y - c_{0} ) - c_{2} y \\ = & \frac{{ - \alpha_{2} \rho_{0}^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}(u_{2} + \alpha_{2} y - c_{0} )^{3} + \alpha_{2} \left( {\frac{1}{2\gamma } + \frac{{\beta \alpha_{1} \rho_{0} u_{1} }}{{2\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}} \right)(u_{2} + \alpha_{2} y - c_{0} ) - c_{2} y \\ \end{aligned}$$

$$f^{\prime\prime}(y) = = \frac{{ - 3\alpha_{2}^{2} \rho_{0}^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}(u_{2} + \alpha_{2} y - c_{0} )^{2} + \alpha_{2}^{2} (\frac{1}{2\gamma } + \frac{{\beta \alpha_{1} \rho_{0} u_{1} }}{{2\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}) - c_{2} .$$

Note that $p_{2}^{*} - c_{0} { = }\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2} > 0$ (otherwise the profit of the offline clinic must be negative). It is easy to verify whether $f^{\prime\prime}(y) = 0$ has only one root or no root over $y \in (0,\;\infty )$. When $f^{\prime\prime}(y) = 0$ has only one root $\overline{\overline{y}}$ over $(0,\;\infty )$, because ${{\beta \alpha_{1}^{2} } \mathord{\left/ {\vphantom {{\beta \alpha_{1}^{2} } 2}} \right. \kern-0pt} 2} - c_{1} { < }0$, we have $f^{\prime\prime}(y) < 0$ when $y \in (0,\;\overline{\overline{y}})$ and $f^{\prime\prime}(y) > 0$ when $y \in (\overline{\overline{y}},\;\infty )$. If $f^{\prime}(0) > 0$ but $f^{\prime}(\overline{\overline{y}}) < 0$, it is clear that there are two positive roots of $f^{\prime}(y) = 0$. We denote the two roots as $y_{1,2}$ and it can be easily verified that $y_{1}$ and $y_{2}$ are the maximum value point and minimum value point of $f(y)$ respectively. Hence, the optimal offline quality $y^{*} \in \{ 0,\;y_{1} ,\;\xi \}$. If $f^{\prime}(0) \le 0$, $f(y)$ has only a minimum point. Thus, there must be $y^{*} \in \{ 0,\;\xi \}$. If there is no root of $f^{\prime\prime}(y) = 0$ over $y \in (0,\;\infty )$, we can verify that $f^{\prime\prime}(y) > 0$ over $y \in (0,\;\infty )$, which means $f(y)$ is convex function. Hence we have $y^{*} \in \{ 0,\;\xi \}$.

When $x^{1} < 0$ or $x^{1} > \xi$, $x^{*} \in \{ 0,\;\xi \}$. Then we get the profit function.

$\pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} ) = \beta \left( {\frac{{u_{1} + \alpha_{1} x^{*} }}{2}} \right)^{2} + \left( {\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2}} \right)^{2} {{(1 - \rho_{0} x^{*} )} \mathord{\left/ {\vphantom {{(1 - \rho_{0} x^{*} )} \gamma }} \right. \kern-0pt} \gamma } - \frac{{c_{1} x^{*2} }}{2} - \frac{{c_{2} y^{2} }}{2} - C$.

Taking the first and second derivatives of $\pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )$ respectively, we have

$$\begin{gathered} \frac{{\partial \pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )}}{\partial y} = \alpha_{2} \left( {\frac{{u_{2} + \alpha_{2} y - c_{0} }}{2}} \right){{(1 - \rho_{0} x^{*} )} \mathord{\left/ {\vphantom {{(1 - \rho_{0} x^{*} )} \gamma }} \right. \kern-0pt} \gamma } - c_{2} y, \hfill \\ \frac{{\partial^{2} \pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )}}{{\partial y^{2} }} = \frac{{\alpha_{2}^{2} }}{2}{{(1 - \rho_{0} x^{*} )} \mathord{\left/ {\vphantom {{(1 - \rho_{0} x^{*} )} \gamma }} \right. \kern-0pt} \gamma } - c_{2} . \hfill \\ \end{gathered}$$

If $\alpha_{2}^{2} (1 - \rho_{0} x^{*} ){/(2}\gamma {)} - c_{2} \ge 0$, $\pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )$ is convex function over $y \in (0,\;\infty )$, and hence we have $y^{*} \in \{ 0,\;\xi \}$. If $\alpha_{2}^{2} {{(1 - \rho_{0} x^{*} )} \mathord{\left/ {\vphantom {{(1 - \rho_{0} x^{*} )} {{2}\gamma }}} \right. \kern-0pt} {{2}\gamma }} - c_{2} < 0$, $\pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )$ is concave function over $y \in (0,\;\infty )$. Letting $\frac{{\partial \pi (x^{*} ,y,p_{1}^{*} ,p_{2}^{*} )}}{\partial y} = 0$, we have $\dot{y} = \frac{{\alpha_{2} (u_{2} - c_{0} ){{(1 - \rho_{0} x^{*} )} \mathord{\left/ {\vphantom {{(1 - \rho_{0} x^{*} )} \gamma }} \right. \kern-0pt} \gamma }}}{{2c_{2} - \alpha_{2}^{2} {{(1 - \rho_{0} x^{*} )} \mathord{\left/ {\vphantom {{(1 - \rho_{0} x^{*} )} \gamma }} \right. \kern-0pt} \gamma }}}$. Thus we know that $y^{*} \in \{ 0,\;\dot{y},\;\xi \}$.

Proof of proposition 2

If $u_{1} + \alpha_{1} x^{*} - p_{1} { = }{{(u_{2} + \alpha_{2} y^{*} - p_{2} )} \mathord{\left/ {\vphantom {{(u_{2} + \alpha_{2} y^{*} - p_{2} )} \gamma }} \right. \kern-0pt} \gamma }$, we can rewrite it as $p_{1} { = }\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{p}_{1} { = }u_{1} + \alpha_{1} x - {{(u_{2} + \alpha_{2} y - p_{2} )} \mathord{\left/ {\vphantom {{(u_{2} + \alpha_{2} y - p_{2} )} \gamma }} \right. \kern-0pt} \gamma }$. Thus the optimal problem is rewritten as.

$$\begin{aligned} {\text{Max }}\pi (x,y,\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{p}_{1} ,p_{2} ) = & \beta (u_{1} + \alpha_{1} x - {{(u_{2} + \alpha_{2} y - p_{2} )} \mathord{\left/ {\vphantom {{(u_{2} + \alpha_{2} y - p_{2} )} \gamma }} \right. \kern-0pt} \gamma }){{(u_{2} + \alpha_{2} y - p_{2} )} \mathord{\left/ {\vphantom {{(u_{2} + \alpha_{2} y - p_{2} )} \gamma }} \right. \kern-0pt} \gamma }) \\ \quad & + (p_{2} - c_{0} )(1 - \rho_{0} x)(u_{2} + \alpha_{2} y - p_{2} ){/}\gamma - \frac{{c_{1} x^{2} }}{2} - \frac{{c_{2} y^{2} }}{2} - C \\ \end{aligned}$$

$${\text{s.t}}{.}\quad 0 \le x,y \le \xi .$$

Firstly, for any ($x$, $y$), since

$$\frac{\partial \pi }{{\partial p_{2} }} = 2\beta (u_{2} + \alpha_{2} y - p_{2} ){/}\gamma^{2} - \beta (u_{1} + \alpha_{1} x){/}\gamma { + }(1 - \rho_{0} x)(u_{2} + \alpha_{2} y - p_{2} ){/}\gamma - (p_{2} - c_{0} )(1 - \rho_{0} x){/}\gamma$$

(7)

$$\frac{{\partial^{2} \pi (x,y,\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{p}_{1} ,p_{2} )}}{{\partial p_{2}^{2} }} = - 2\beta {/}\gamma^{2} - {{2(1 - \rho_{0} x)} \mathord{\left/ {\vphantom {{2(1 - \rho_{0} x)} \gamma }} \right. \kern-0pt} \gamma } < 0.$$

It is clear that $p_{2}^{*}$ is determined by $\frac{\partial \pi }{{\partial p_{2} }} = 0$. Because $0 \le x,y \le \xi$, $x^{*}$ and $y^{*}$ can be got from $\frac{\partial \pi }{{\partial x}} = 0$ and $\frac{\partial \pi }{{\partial y}} = 0$ respectively, or at the endpoints $0,\;\xi$. Now, we discuss the following cases:

Case (1): $x^{*}$ and $y^{*}$ are satisfied $\frac{\partial \pi }{{\partial x}} = 0$ and $\frac{\partial \pi }{{\partial y}} = 0$ respectively, i.e.,

$$\frac{\partial \pi }{{\partial x}}{ = }\beta \alpha_{1} (u_{2} + \alpha_{2} y - p_{2}^{*} ){/}\gamma - \rho_{0} (p_{2}^{*} - c_{0} )(u_{2} + \alpha_{2} y - p_{2}^{*} ){/}\gamma - c_{1} x = 0,$$

(8)

$$\frac{\partial \pi }{{\partial y}} = - 2\beta \alpha_{2} (u_{2} - p_{2}^{*} ){/}\gamma^{2} - 2\beta \alpha_{2}^{2} y{/}\gamma^{2} - c_{2} y{ + }\beta \alpha_{2} (u_{1} + \alpha_{1} x){/}\gamma { + }\alpha_{2} (p_{2}^{*} - c_{0} )(1 - \rho_{0} x){/}\gamma { = }0.$$

(9)

We can derive $x^{*}$ and $y^{*}$ from the simultaneous Eqs. (7) (8) (9), i.e.,

$\begin{gathered} x^{*} = \frac{{\beta \alpha_{1} - \rho_{0} (p_{2}^{*} - c_{0} )}}{{c_{1} \gamma }}\frac{{(u_{2} - p_{2}^{*} )c_{1} c_{2} \gamma^{2} /\alpha_{2}^{2} { + }\beta u_{1} c_{1} \gamma + c_{1} \gamma (p_{2}^{*} - c_{0} )}}{{2\beta c_{1} + c_{1} c_{2} \gamma^{2} /\alpha_{2}^{2} - (\beta \alpha_{1} - \rho_{0} (p_{2}^{*} - c_{0} ))^{2} }}, \hfill \\ y^{*} = \frac{{ - 2\beta c_{1} (u_{2} - p_{2}^{*} ) + \beta u_{1} c_{1} \gamma + c_{1} \gamma (p_{2}^{*} - c_{0} ) + (\beta \alpha_{1} - \rho_{0} (p_{2}^{*} - c_{0} ))^{2} (u_{2} - p_{2}^{*} )}}{{2\beta \alpha_{2} c_{1} + c_{1} c_{2} \gamma^{2} /\alpha_{2} - \alpha_{2} (\beta \alpha_{1} - \rho_{0} (p_{2}^{*} - c_{0} ))^{2} }}; \hfill \\ \end{gathered}$, where $p_{2}^{*}$ is one root of the equation $p_{2} = \frac{{ - c_{2} \gamma y^{*} }}{{\alpha_{2} (1 - \rho_{0} x^{*} )}} + u_{2} + \alpha_{2} y^{*}$.

Case (2): $x^{*}$ is satisfied $\frac{\partial \pi }{{\partial x}} = 0$ and $y^{*} \in \{ 0,\xi \}$. Then we can get

$$\begin{aligned} \frac{\partial \pi }{{\partial x}}{ = } & \beta \alpha_{1} (u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma - \rho_{0} (p_{2} - c_{0} )(u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma - c_{1} x = 0, \\ \frac{\partial \pi }{{\partial p_{2} }} = & 2\beta (u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma^{2} - (u_{1} + \alpha_{1} x){/}\gamma \\ \quad & { + }(1 - \rho_{0} x)(u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma - (p_{2} - c_{0} )(1 - \rho_{0} x){/}\gamma \\ \end{aligned}$$

Solving the equations, we have

$$x^{*} = \frac{{\beta \alpha_{1} (u_{2} + \alpha_{2} y^{*} - p_{2} ) - \rho_{0} (p_{2} - c_{0} )(u_{2} + \alpha_{2} y^{*} - p_{2} )}}{{\gamma c_{1} }},$$

$p_{2}^{*} { = }\frac{{{2}\beta (u_{2} + \alpha_{2} y^{*} ){ + }\gamma (1 - \rho_{0} x)(u_{2} + \alpha_{2} y^{*} { + }c_{0} ) - \gamma (u_{1} + \alpha_{1} x^{*} )}}{{2\beta { + }2\gamma (1 - \rho_{0} x^{*} )}}$, where $y^{*} \in \{ 0,\xi \}$.

Case (3): $x^{*} = {0}$ while $y^{*}$ is at the point $\{ 0,\xi \}$ or satisfied $\frac{\partial \pi }{{\partial y}} = 0$:

• (a) $y^{*}$ is satisfied $\frac{\partial \pi }{{\partial y}} = 0$. This is equivalent to.

$\frac{\partial \pi }{{\partial y}} = = - 2\beta c_{2} y(1 - 5\gamma ){ + }\beta \alpha_{2} u_{1} + \alpha_{2} (u_{2} + \alpha_{2} y - c_{0} {) = }0$, and.

$\frac{\partial \pi }{{\partial p_{2} }}{ = 2}\beta (u_{2} + \alpha_{2} y - p_{2} ){/}\gamma^{2} - u_{1} {/}\gamma + (u_{2} + \alpha_{2} y - p_{2} ){/}\gamma - (p_{2} - c_{0} ){/}\gamma = 0$.

Solving the above equations, we have.

$y^{*} = \frac{{\beta \alpha_{2} u_{1} - \alpha_{2} (c_{0} - u_{2} )}}{{2\beta c_{2} (1 + 5\gamma ) - \alpha_{2}^{2} }}$, $p_{2}^{*} = u_{2} - \frac{{(c_{2} \gamma - \alpha_{2}^{2} {)(}\beta u_{1} - (c_{0} - u_{2} ))}}{{2\beta c_{2} (1 + 5\gamma ) - \alpha_{2}^{2} }}$.

• (b) $y^{*} \in \{ 0,\xi \}$. From $\frac{\partial \pi }{{\partial p_{2} }}{ = 2}\beta (u_{2} + \alpha_{2} y - p_{2} ){/}\gamma^{2} - u_{1} {/}\gamma + (u_{2} + \alpha_{2} y - p_{2} ){/}\gamma - (p_{2} - c_{0} ){/}\gamma = 0$, we have.

$p_{2}^{*} = \frac{{(u_{2} + \alpha_{2} y^{*} )({2}\beta + \gamma ) - (u_{1} - c_{0} )\gamma }}{{{2}\beta + 2\gamma }}$.

Case (4): $x^{*} = \xi$ while $y^{*}$ is at the point $\{ 0,\xi \}$ or satisfied $\frac{\partial \pi }{{\partial y}} = 0$.

• (a) $y^{*}$ is satisfied $\frac{\partial \pi }{{\partial y}} = 0$. Then we can get.

$$\begin{gathered} \frac{\partial \pi }{{\partial y}} = - 2\beta \alpha_{2} (u_{2} - p_{2} ){/}\gamma^{2} - 2\beta \alpha_{2}^{2} y{/}\gamma^{2} - c_{2} y{ + }\beta \alpha_{2} (u_{1} + \alpha_{1} \xi ){/}\gamma { + }\alpha_{2} (p_{2} - c_{0} )(1 - \rho_{0} \xi ){/}\gamma = 0 \hfill \\ \frac{\partial \pi }{{\partial p_{2} }} = 2\beta (u_{2} + \alpha_{2} y - p_{2} ){/}\gamma^{2} - (u_{1} + \alpha_{1} \xi ){/}\gamma { + }(1 - \rho_{0} \xi )(u_{2} + \alpha_{2} y - p_{2} ){/}\gamma - (p_{2} - c_{0} )(1 - \rho_{0} \xi ){/}\gamma = 0 \hfill \\ \end{gathered}$$

Solving the above equations, we have

$$y^{*} = \frac{{\beta \alpha_{2} (u_{1} + \alpha_{1} \xi ) + (\alpha_{2} u_{2} - \alpha_{2} c_{0} )(1 - \rho_{0} \xi )}}{{{{2c_{2} \gamma + 2\beta c_{2} } \mathord{\left/ {\vphantom {{2c_{2} \gamma + 2\beta c_{2} } {(1 - \rho_{0} \xi ) - \alpha_{2}^{2} (1 - \rho_{0} \xi )}}} \right. \kern-0pt} {(1 - \rho_{0} \xi ) - \alpha_{2}^{2} (1 - \rho_{0} \xi )}}}},\quad p_{2}^{*} = \frac{{ - c_{2} \gamma y^{*} }}{{\alpha_{2} (1 - \rho_{0} \xi )}} + u_{2} + \alpha_{2} y^{*} .$$

• (b) $y^{*} \in \{ 0,\xi \}$. From $\frac{\partial \pi }{{\partial p_{2} }} = - (u_{1} + \alpha_{1} \xi ){/}\gamma + 2(u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma^{2} + (1 - \rho_{0} \xi )(u_{2} + \alpha_{2} y^{*} - 2p_{2} + c_{0} ){/}\gamma = 0$, we have

$$p_{2}^{*} = \frac{{2\beta (u_{2} + \alpha_{2} y^{*} ) - \gamma (u_{1} + \alpha_{1} \xi ){ + }(1 - \rho_{0} \xi )(u_{2} + \alpha_{2} y^{*} + c_{0} )\gamma }}{{2\beta + 2\gamma (1 - \rho_{0} \xi )}}.$$

Proof of proposition 3

When $u_{1} + \alpha_{1} x^{*} - p_{1} > {{(u_{2} + \alpha_{2} y^{*} - p_{2} )} \mathord{\left/ {\vphantom {{(u_{2} + \alpha_{2} y^{*} - p_{2} )} \gamma }} \right. \kern-0pt} \gamma }$ and $y^{*} { = }y_{1}$, we first get.

$\frac{{\partial f^{\prime}(y^{*} )}}{{\partial u_{1} }} = \beta \alpha_{1} \alpha_{2} \rho_{0} (\beta \alpha_{1}^{2} - 2c_{1} )\frac{{(u_{2} + \alpha_{2} y^{*} - c_{0} )}}{{4\gamma (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }} + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{{\partial u_{1} }} = 0$.

Because $y_{1}$ is the maxima point, we have $\frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }} < 0$. Note that $n_{2}^{*} = {{(u_{2} + \alpha_{2} y^{*} - p_{2}^{*} )} \mathord{\left/ {\vphantom {{(u_{2} + \alpha_{2} y^{*} - p_{2}^{*} )} \gamma }} \right. \kern-0pt} \gamma } > 0$ and $p_{2}^{*} { = }\frac{{u_{2} + c_{0} + \alpha_{2} y^{*} }}{2}$, we have $u_{2} - c_{0} + \alpha_{2} y^{*} > 0$. Due to the basic assumption $\beta \alpha_{1}^{2} - 2c_{1} { < }0$. Hence we have $\frac{{\partial y^{*} }}{{\partial u_{1} }} < 0$.

Similarly, because $\frac{{\partial f^{\prime}(y^{*} )}}{{\partial c_{2} }} = - y^{*} + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{{\partial c_{2} }} = 0$, we can similarly have $\frac{{\partial y^{*} }}{{\partial c_{2} }} < 0$.

Further, we can obtain the following properties. From.

$$\frac{{\partial f^{\prime}(y^{*} )}}{{\partial c_{0} }} = \frac{{3\alpha_{2} \rho_{0}^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}(u_{2} - c_{0} + \alpha_{2} y^{*} )^{2} + \frac{{\beta \alpha_{1} \alpha_{2} u_{1} \rho_{0} }}{{2\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}} + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{{\partial c_{0} }} = 0,$$

we can easily verify that $\frac{{\partial y^{*} }}{{\partial c_{0} }} < 0$. Similarly, from.

$$\frac{{\partial f^{\prime}(y^{*} )}}{{\partial u_{2} }} = \frac{{ - 3\alpha_{2} \rho_{0}^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}(u_{2} + \alpha_{2} y - c_{0} )^{2} - \frac{{\beta \alpha_{1} \alpha_{2} u_{1} \rho_{0} }}{{4\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}} + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{{\partial u_{2} }} = 0,$$

we can easily verify that $\frac{{\partial y^{*} }}{{\partial u_{2} }} > 0$.

And from

$$\frac{{\partial f^{\prime}(y^{*} )}}{{\partial \rho_{0} }} = \frac{{ - \alpha_{2} \rho_{0} }}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}}(u_{2} + \alpha_{2} y - c_{0} )^{3} - \beta \alpha_{1} \alpha_{2} u_{1} (\beta \alpha_{1}^{2} - 2c_{1} ))\frac{{(u_{2} + \alpha_{2} y - c_{0} )}}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )^{2} }} + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{{\partial \rho_{0} }} = 0,$$

we have $\frac{{\partial y^{*} }}{{\partial \rho_{0} }} > 0$.

And from

$$\begin{aligned} \frac{{\partial f^{\prime}(y^{*} )}}{\partial \gamma } = & \frac{{\alpha_{2} \rho_{0}^{2} }}{{2\gamma^{3} (\beta \alpha_{1}^{2} - 2c_{1} )}}(u_{2} + \alpha_{2} y^{*} - c_{0} )^{3} \\ \quad & + \frac{{(\beta \alpha_{1} \alpha_{2} u_{1} \rho_{0} - \alpha_{2} (\beta \alpha_{1}^{2} - 2c_{1} ))(u_{2} - c_{0} + \alpha_{2} y^{*} )}}{{2\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}} \\ \quad & + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{\partial \gamma } = 0, \\ \end{aligned}$$

we have $\frac{{\partial y^{*} }}{\partial \gamma } < 0$.

And from

$$\begin{aligned} \frac{{\partial f^{\prime}(y^{*} )}}{{\partial \alpha_{2} }} = & \frac{{ - \rho_{0}^{2} (u_{2} - c_{0} + \alpha_{2} y^{*} )^{3} - 3\rho_{0}^{2} \alpha_{2} y^{*} (u_{2} - c_{0} + \alpha_{2} y^{*} )^{2} }}{{4\gamma^{2} (\beta \alpha_{1}^{2} - 2c_{1} )}} \\ \quad & - \frac{{(u_{2} - c_{0} + \alpha_{2} y^{*} )(\beta \alpha_{1} u_{1} \rho_{0} - (\beta \alpha_{1}^{2} - 2c_{1} ))}}{{2\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}} \\ \quad & + \frac{{\alpha_{2}^{2} (\beta \alpha_{1} u_{1} \rho_{0} - (\beta \alpha_{1}^{2} - 2c_{1} ))}}{{4\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}} + \frac{{\partial f^{\prime\prime}(y^{*} )}}{{\partial y^{*} }}\frac{{\partial y^{*} }}{{\partial \alpha_{2} }} = 0, \\ \end{aligned}$$

we can similarly verify that $\frac{{\partial y^{*} }}{{\partial \alpha_{2} }} > 0$.

Proof of proposition 4

When $u_{1} + \alpha_{1} x^{*} - p_{1} > (u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma$ and $y^{*} { = }y_{1}$, taking the derivative of $x^{*} { = }\frac{{{{\rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )^{2} } \mathord{\left/ {\vphantom {{\rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )^{2} } \gamma }} \right. \kern-0pt} \gamma } - 2\beta \alpha_{1} u_{1} }}{{2\beta \alpha_{1}^{2} - 4c_{1} }}$ with respect to $u_{1}$, we can obtain $\frac{{\partial x^{*} }}{{\partial u_{1} }}{ = }{{\left( {{{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )} \mathord{\left/ {\vphantom {{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )} \gamma }} \right. \kern-0pt} \gamma }\frac{{\partial y^{*} }}{{\partial u_{1} }} - \beta \alpha_{1} } \right)} \mathord{\left/ {\vphantom {{\left( {{{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )} \mathord{\left/ {\vphantom {{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )} \gamma }} \right. \kern-0pt} \gamma }\frac{{\partial y^{*} }}{{\partial u_{1} }} - \beta \alpha_{1} } \right)} {(\beta \alpha_{1}^{2} - 2c_{1} )}}} \right. \kern-0pt} {(\beta \alpha_{1}^{2} - 2c_{1} )}}$. Note that from the proof of Proposition 3 we have $\frac{{u_{2} - c_{0} + \alpha_{2} y^{*} }}{2} > 0$ and $\frac{{\partial y^{*} }}{{\partial u_{1} }} < 0$. Hence we have $\frac{{\partial x^{*} }}{{\partial u_{1} }} > 0$.

Taking the derivative of $x^{*}$ with respect to $c_{2}$, we have $\frac{{\partial x^{*} }}{{\partial c_{2} }} = \frac{{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )}}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}\frac{{\partial y^{*} }}{{\partial c_{2} }}$. Because $\frac{{\partial y^{*} }}{{\partial c_{2} }} < 0$, we can obtain that $\frac{{\partial x^{*} }}{{\partial c_{2} }} > 0$.

Take the derivative of $x^{*}$ with respect to $c_{0}$ and we can have $\frac{{\partial x^{*} }}{{\partial c_{0} }} = \frac{{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )}}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}\left( {\alpha_{2} \frac{{\partial y^{*} }}{{\partial c_{0} }} - 1} \right)$. Because $\frac{{\partial y^{*} }}{{\partial c_{0} }} < 0$, it’s easy to verify that $\frac{{\partial x^{*} }}{{\partial c_{0} }} > 0$.

Similarly, as $\frac{{\partial x^{*} }}{{\partial u_{2} }} = \frac{{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )}}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}\left( {1 + \alpha_{2} \frac{{\partial y^{*} }}{{\partial u_{2} }}} \right)$, and $\frac{{\partial y^{*} }}{{\partial u_{2} }} > 0$, we can obtain $\frac{{\partial x^{*} }}{{\partial u_{2} }} < 0$. Because $\frac{{\partial x^{*} }}{{\partial \rho_{0} }} = \frac{1}{{\gamma (2\beta \alpha_{1}^{2} - 4c_{1} )}}\left( {(u_{2} + \alpha_{2} y^{*} - c_{0} )^{2} + 2\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )\frac{{\partial y_{1}^{*} }}{{\partial \rho_{0} }}} \right)$, and $\frac{{\partial y^{*} }}{{\partial \rho_{0} }} > 0$, we can obtain $\frac{{\partial x^{*} }}{{\partial \rho_{0} }} < 0$. As $\frac{{\partial x^{*} }}{\partial \gamma } = \frac{1}{{2\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}\left( { - \frac{{\rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )^{2} }}{\gamma } + 2\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )\frac{{\partial y^{*} }}{\partial \gamma }} \right)$, and $\frac{{\partial y^{*} }}{\partial \gamma } < 0$, we can obtain $\frac{{\partial x^{*} }}{\partial \gamma } > 0$. As $\frac{{\partial x^{*} }}{{\partial \alpha_{2} }} = \frac{{\alpha_{2} \rho_{0} (u_{2} + \alpha_{2} y^{*} - c_{0} )}}{{\gamma (\beta \alpha_{1}^{2} - 2c_{1} )}}\left( {y^{*} + \alpha_{2} \frac{{\partial y^{*} }}{{\partial \alpha_{2} }}} \right)$, and $\frac{{\partial y^{*} }}{{\partial \alpha_{2} }} > 0$, we can obtain $\frac{{\partial x^{*} }}{{\partial \alpha_{2} }} < 0$.

Proof of proposition 5

When $u_{1} + \alpha_{1} x^{*} - p_{1} > (u_{2} + \alpha_{2} y^{*} - p_{2} ){/}\gamma$ and $y^{*} { = }y_{1}$, taking the derivative of $p_{1}^{*} = \frac{{u_{1} }}{2} + \frac{1}{2}\alpha_{1} x^{*}$ with respect to $u_{1}$, we can obtain $\frac{{\partial p_{1}^{*} }}{{\partial u_{1} }} = \frac{1}{2} + \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{{\partial u_{1} }}$. Due to $\frac{{\partial x^{*} }}{{\partial u_{1} }} > 0$, we have $\frac{{\partial p_{1}^{*} }}{{\partial u_{1} }} > 0$. Then taking the derivative of $p_{2}^{*} { = }\frac{{u_{2} + c_{0} + \alpha_{2} y_{1} }}{2}$ with respect to $u_{1}$, we have $\frac{{\partial p_{2}^{*} }}{{\partial u_{1} }} = \frac{{\alpha_{2} }}{2}\frac{{\partial y^{*} }}{{\partial u_{1} }}$. Due to $\frac{{\partial y^{*} }}{{\partial u_{1} }} < 0$, we have $\frac{{\partial p_{2}^{*} }}{{\partial u_{1} }} < 0$.

Similarly, as $\frac{{\partial p_{1}^{*} }}{{\partial c_{2} }} = \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{{\partial c_{2} }}$ and $\frac{{\partial x^{*} }}{{\partial c_{2} }}$ > 0, we can obtain $\frac{{\partial p_{1}^{*} }}{{\partial c_{2} }} > 0$. As $\frac{{\partial p_{2}^{*} }}{{\partial c_{2} }} = \frac{{\alpha_{2} }}{2}\frac{{\partial y^{*} }}{{\partial c_{2} }}$ < 0, we can obtain $\frac{{\partial p_{2}^{*} }}{{\partial c_{2} }} < 0$.

Further, we can have the following properties. Taking the derivative of $p_{1}^{*} = \frac{{u_{1} }}{2} + \frac{1}{2}\alpha_{1} x^{*}$ with respect to $c_{0}$, we can obtain $\frac{{\partial p_{1}^{*} }}{{\partial c_{0} }} = \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{{\partial c_{0} }}$. Due to $\frac{{\partial x^{*} }}{{\partial c_{0} }} > 0$, we have $\frac{{\partial p_{1}^{*} }}{{\partial c_{0} }} > 0$. Similarly, because $\frac{{\partial y^{*} }}{{\partial c_{0} }} < 0$, $\frac{{\partial x^{*} }}{\partial \gamma } > 0$, $\frac{{\partial y^{*} }}{\partial \gamma } > 0$, $\frac{{\partial x^{*} }}{{\partial u_{2} }} < 0$, $\frac{{\partial x^{*} }}{{\partial \rho_{0} }} < 0$, $\frac{{\partial y^{*} }}{{\partial \rho_{0} }} > 0$, $\frac{{\partial x^{*} }}{{\partial \alpha_{2} }} < 0$, $\frac{{\partial y^{*} }}{{\partial \alpha_{2} }} > 0$, we can obtain that $\frac{{\partial p_{2}^{*} }}{{\partial c_{0} }} = \frac{{\alpha_{2} }}{2}\frac{{\partial y^{*} }}{{\partial c_{0} }} < 0$,$\frac{{\partial p_{1}^{*} }}{\partial \gamma } = \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{\partial \gamma } > 0$, $\frac{{\partial p_{2}^{*} }}{\partial \gamma } = \frac{{\alpha_{2} }}{2}\frac{{\partial y^{*} }}{\partial \gamma } > 0$, $\frac{{\partial p_{1}^{*} }}{{\partial u_{2} }} = \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{{\partial u_{2} }} < 0$, $\frac{{\partial p_{2}^{*} }}{{\partial u_{2} }} = \frac{1}{2} + \frac{{\alpha_{2} }}{2}\frac{{\partial y^{*} }}{{\partial u_{2} }} > 0$, $\frac{{\partial p_{1}^{*} }}{{\partial \rho_{0} }} = \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{{\partial \rho_{0} }} < 0$, $\frac{{\partial p_{2}^{*} }}{{\partial \rho_{0} }} = \frac{{\alpha_{2} }}{2}\frac{{\partial y^{*} }}{{\partial \rho_{0} }} > 0$, $\frac{{\partial p_{1}^{*} }}{{\partial \alpha_{2} }} = \frac{1}{2}\alpha_{1} \frac{{\partial x^{*} }}{{\partial \alpha_{2} }} < 0$, $\frac{{\partial p_{2}^{*} }}{{\partial \alpha _{2} }} = \frac{1}{2}\left( {y_{1} + \alpha _{2} \frac{{\partial y^{*} }}{{\partial \alpha _{2} }}} \right){\text{ > }}0$.