The influence of leadership and smart level on the strategy choice of the smart logistics platform: a perspective of collaborative innovation participation

Abstract

In a two-tier supply chain consisting of a smart logistics platform and a smart logistics provider, for value-added service innovation in logistics, there are usually two innovation modes: collaborative innovation (CI) in the supply chain and independent innovation by the provider. In this study, we research the strategic choice of the platform about whether to co-innovate with the provider when the platform requires the provider to innovate new value-added services. By building a two-tier supply chain game model, this study focuses on the influence of leadership and smart level on the platform's CI strategy. This study obtains many interesting findings. Firstly, we give the cost thresholds for the platform to choose CI under different relationships, with higher thresholds under the provider-led relationship compared to Nash and platform-led relationships, the platform is most likely to choose CI under the provider-led relationship, while most difficult to choose under Nash relationship. Secondly, we find that under Nash and platform-led relationships, the platform's participation in the CI will lead to an increase in both parties' profits, so that the CI can be achieved, but under the provider-led relationship, there may be a situation where the platform's profits increase and the provider’s profits decrease when the two parties co-innovate, and since under this relationship it is easiest for the platform to choose to participate in the CI, the platform can choose a cost-sharing contract to solve this problem and achieve CI. Thirdly, with the same leadership, the higher the smart level of the provider is, the more incentive the platform is to participate in CI. If the smart level of the provider is very low or the smart level of the platform is very high, the platform will never participate in CI, regardless of the leadership.

Appendices

Appendix 1: Summary of basic model solution

See Table 6.

Table 6 The decisions and profits in different scenario

• (1) The threshold ${N}_{1}$ for the provider to choose collaborative innovation under the Nash relationship

  $$ \begin{aligned} \prod_{S}^{NA} & = \left( {\frac{1}{m} - 1} \right)\frac{{k_{1} }}{2}\left( {e_{1}^{NA} } \right)^{2} - F_{1} \;\;\;\prod^{NO}_{S} = \left( {\frac{1}{m} - 1} \right)\frac{{k_{1} }}{2}(e_{1}^{NO})^{2} - F_{1} \\ e_{1}^{NA} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \;\;\;e_{1}^{NO} = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m)}}} \\ \frac{{e_{1}^{NA} }}{{e_{1}^{NO} }} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{2(1 - m - n)(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \\ \end{aligned} $$

  $\frac{{e_{1}^{NA} }}{{e_{1}^{NO} }} > 1$, we have $\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{2(1 - m - n)(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} > 1$

  $$ \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right) > 1 $$

  When $n > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)$

  Which is $k_{2} < n\left( {\frac{{\left( {1 - \lambda } \right)a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{\lambda k_{1} }}{{\left( {1 - \lambda } \right)m}}} \right)\;\;\;\frac{{e_{1}^{NA} }}{{e_{1}^{NO} }} > 1\;\;\;\frac{{\prod_{S}^{NA} + F_{1} }}{{\prod_{S}^{{NO}} + F_{1} }} > 1$

  So ${N}_{1}=n{\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{\frac{1}{1-m}}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right).$

• (2) The threshold ${N}_{2}$ for the provider to choose collaborative innovation under the platform-led relationship

  $$ \begin{aligned} \prod_{S}^{RL} & = \left( {\frac{1}{m} - 1} \right)\frac{{k_{1} }}{2}\left( {e_{1}^{RL} } \right)^{2} - F_{1} \;\;\;\prod_{S}^{NO} = \left( {\frac{1}{m} - 1} \right)\frac{{k_{1} }}{2}(e_{1}^{NO})^{2} - F_{1} \\ e_{1}^{RL} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )(1 - m)mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \\ \frac{{e_{1}^{RL} }}{{e_{1}^{NO} }} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{2(1 - m - n)(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )(1 - m)mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \\ \end{aligned} $$

  When $n > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )(1 - m)mk_{2} }}{{\lambda k_{1} }}} \right)$, $\frac{{e_{1}^{RL} }}{{e_{1}^{NO} }} > 1\;\;\;\frac{{\prod_{S}^{RL} + F_{1} }}{{\prod_{S}^{{NO}} + F_{1} }} > 1$

  ${k}_{2}<\frac{n}{1-m}{\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{\frac{1}{1-m}}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right)$, so $\frac{{\prod_{S}^{RL} + F_{1} }}{{\prod_{S}^{{NO}} + F_{1} }} > 1$

  So $N_{2} = \frac{n}{1 - m}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{\lambda k_{1} }}{(1 - \lambda )m}} \right)$.

• (3) The threshold ${N}_{3}$ for the provider to choose collaborative innovation under the provider-led relationship

  $$ \begin{aligned} \prod_{S}^{SL} & = \left( {\frac{1 - n}{m} - 1} \right)\frac{{k_{1} }}{2}\left( {e_{1}^{SL} } \right)^{2} - F_{1} \;\;\;\prod_{S}^{NO} = \left( {\frac{1}{m} - 1} \right)\frac{{k_{1} }}{2}(e_{1}^{NO}){2} - F_{1} \\ e_{1}^{SL} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda n(1 - n)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \\ \frac{{e_{1}^{SL} }}{{e_{1}^{NO} }} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{2(1 - m - n)(1 - m)}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda n(1 - n)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \\ \end{aligned} $$

  When $\left( {\frac{1}{1 - n}} \right)^{{\frac{1 - n}{n}}} n > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)$, $\frac{{e_{1}^{SL} }}{{e_{1}^{NO} }} > 1$.

  $$ \begin{aligned} & \frac{{\prod_{S}^{SL} + F_{1} }}{{\prod_{S}^{{NO}} + F_{1} }} = \left( {\frac{1 - n - m}{{1 - m}}} \right)\left( {\frac{{e_{1}^{SL} }}{{e_{1}^{NO} }}} \right)^{2} = \left( {\frac{1 - n - m}{{1 - m}}} \right)\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{(1 - m - n)(1 - m)}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{1}{1 - m - n}}} \left( {\frac{{\lambda n(1 - n)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{1 - m - n}}} \\ & \quad \left( {\frac{1 - m - n}{{1 - m}}} \right)^{{\frac{1 - m - n}{n}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{1 - n}{n}}} n > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right),\;\;\frac{{\prod_{S}^{SL} + F_{1} }}{{\prod_{S}^{{NO}} + F_{1} }} > 1 \\ \end{aligned} $$

  We have:

  $$k_{2} < \left( {\frac{1 - m - n}{{1 - m}}} \right)^{{\frac{1 - m - n}{n}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{1 - n}{n}}} n\left( {\frac{{\left( {1 - \lambda } \right)a^{2} m}}{{2bk_{1} }}} \right)^{{1/\left( {1 - m} \right)}} \left( {\frac{{\lambda k_{1} }}{{\left( {1 - \lambda } \right)m}}} \right),$$

  so

  $$ N_{3} = \left( {\frac{1 - m - n}{{1 - m}}} \right)^{{\frac{1 - m - n}{n}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{1 - n}{n}}} n\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{\lambda k_{1} }}{(1 - \lambda )m}} \right). $$

• (4) The threshold $\overline{N}_{1}$ for the platform to choose collaborative innovation under the Nash relationship

  $$ \begin{aligned} \prod_{R}^{NA} & = \left( {\frac{1}{n} - 1} \right)\frac{{k_{2} }}{2}\left( {e_{2}^{NA} } \right)^{2} - F_{2} \;\;\;e_{2}^{NA} = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{(1 - m)}{{2(1 - m - n)}}}} \\ \prod_{R}^{NA} & = \frac{{\lambda k_{1} }}{2(1 - \lambda )m}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} - F_{2} \\ \prod_{R}^{NA} & = \left( {\frac{1}{n} - 1} \right)\frac{{k_{2} }}{2}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m - n}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{1 - m}{{1 - m - n}}}} - F_{2} \\ \frac{{\prod_{R}^{NA} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} & = \left( {\frac{1}{n} - 1} \right)\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{(1 - m - n)(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{1 - m}{{1 - m - n}}}} \\ & = \left( {\frac{1}{n} - 1} \right)\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{(1 - m - n)(1 - m)}}} \left( {\frac{{\lambda k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{1 - m - n}}} n^{{\frac{1 - m}{{1 - m - n}}}} \\ \end{aligned} $$

  when $n\left( {1 - n} \right)^{{\frac{1 - m - n}{n}}} > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)$, $\frac{{\prod_{R}^{NA} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} > 1$

  That is when ${k}_{2}<n{\left(1-n\right)}^{\frac{1-m-n}{n}}{\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{\frac{1}{1-m}}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right)$, $\frac{{\prod_{R}^{NA} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} > 1$

  So $\overline{N}_{1}$$=n{\left(1-n\right)}^{\frac{1-m-n}{n}}{\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{\frac{1}{1-m}}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right)$.

• (5) The threshold $\overline{N}_{2}$ for the platform to choose collaborative innovation under the platform-led relationship

  $$ \begin{aligned} \prod_{R}^{RL} & = \left( {\frac{1 - m}{n} - 1} \right)\frac{{k_{2} }}{2}\left( ({e_{2}^{RL})^{2} } \right)^{2} - F_{2} \;\;\;e_{2}^{RL} = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )(1 - m)mk_{2} }}} \right)^{{\frac{1 - m}{{2(1 - m - n)}}}} \\ \prod_{R}^{RL} & = \left( {\frac{1 - m - n}{n}} \right)\frac{{k_{2} }}{2}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m - n}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )(1 - m)mk_{2} }}} \right)^{{\frac{1 - m}{{1 - m - n}}}} - F_{2} \\ \prod_{R}^{NO} & = \frac{{\lambda k_{1} }}{2(1 - \lambda )m}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} - F_{2} \\ \frac{{\prod_{R}^{RL} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} & = \left( {\frac{1 - m - n}{n}} \right)\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{(1 - m - n)(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )(1 - m)mk_{2} }}} \right)^{{\frac{1 - m}{{1 - m - n}}}} \\ \frac{{\prod_{R}^{RL} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} & = \left( {\frac{1 - m - n}{n}} \right)\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{(1 - m - n)(1 - m)}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)^{{\frac{ - n}{{1 - m - n}}}} \left( {\frac{n}{1 - m}} \right)^{{\frac{1 - m}{{1 - m - n}}}} \\ \end{aligned} $$

  When $n\left( {1 - m - n} \right)^{{\frac{1 - m - n}{n}}} \left( {\frac{1}{1 - m}} \right)^{{\frac{1 - m}{n}}} > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)$, $\frac{{\prod_{R}^{RL} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} > 1$.

  We have ${k}_{2}<n{\left(1-m-n\right)}^{\frac{1-m-n}{n}}{\left(\frac{1}{1-m}\right)}^{\frac{1-m}{n}}$ ${\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{1/(1-m)}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right)$

  So $\overline{N}_{2} = n\left( {1 - m - n} \right)^{{\frac{1 - m - n}{n}}} \left( {\frac{1}{1 - m}} \right)^{{\frac{1 - m}{n}}} \;\;\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{\lambda k_{1} }}{(1 - \lambda )m}} \right)$.

• (6) The threshold $\overline{N}_{3}$ for the platform to choose collaborative innovation under the provider-led relationship

  $$\begin{aligned} \prod_{R}^{S} & = \left( {\frac{1}{n} - 1} \right)\frac{{k_{2} }}{2}\left( {e_{2}^{SL} } \right)^{2} - F_{2} \;\;\;e_{2}^{SL} = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} (1 - n)}}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda n(1 - n)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{1 - m}{{2(1 - m - n)}}}} \\ \prod_{R}^{SL} & = \left( {\frac{1}{n} - 1} \right)\frac{{k_{2} }}{2}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m - n}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{m}{1 - m - n}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{1 - m}{{1 - m - n}}}} - F_{2} \\ \prod_{R}^{NO} & = \frac{{\lambda k_{1} }}{2(1 - \lambda )m}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} - F_{2} \\ \frac{{\prod_{R}^{SL} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} & = \left( {\frac{1}{n} - 1} \right)\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{(1 - m - n)(1 - m)}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)^{{\frac{ - n}{{(1 - m - n)}}}} \left( {\frac{1}{1 - n}} \right)^{{\frac{m}{(1 - m - n)}}} n^{{\frac{1 - m}{{1 - m - n}}}} \\ \end{aligned}$$

  When $n\left( {1 - n} \right)^{{\frac{1 - 2m - n}{n}}} > \left( {\frac{{2bk_{1} }}{{(1 - \lambda )a^{2} m}}} \right)^{{\frac{1}{1 - m}}} \left( {\frac{{(1 - \lambda )mk_{2} }}{{\lambda k_{1} }}} \right)$, $\frac{{\prod_{R}^{SL} + F_{1} }}{{\prod_{R}^{{NO}} + F_{1} }} > 1$

  We have ${k}_{2}<n{\left(1-n\right)}^{\frac{1-2m-n}{n}}{\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{\frac{1}{1-m}}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right)$

  So $\overline{N}_{3}$$=n{\left(1-n\right)}^{\frac{1-2m-n}{n}}{\left(\frac{(1-\lambda ){a}^{2}m}{2b{k}_{1}}\right)}^{\frac{1}{1-m}}\left(\frac{\lambda {k}_{1}}{(1-\lambda )m}\right)$.

• (7) The proof of $\overline{N}_{3} > \overline{N}_{2} > \overline{N}_{1}$.

  The proof of ${\bar{N}}_{2}>{\bar{N}}_{1}$, that is $\frac{{\bar{N}}_{2}}{{\bar{N}}_{1}}>1$:

  ${\left(\frac{1-m-n}{1-n}\right)}^{1-m-n}{\left(\frac{1}{1-m}\right)}^{1-m}>1.$ Set function $C\left(m,n\right)={\left(\frac{1-m-n}{1-n}\right)}^{1-m-n}{\left(\frac{1}{1-m}\right)}^{1-m}$ and

  $$\begin{aligned}D\left(m,n\right)&=\mathrm{ln}C(m,n)\\D\left(m,n\right)&=\left(1-m-n\right)\left[\mathrm{ln}\left(1-\mathrm{m}-\mathrm{n}\right)-\mathrm{ln}\left(1-\mathrm{n}\right)\right]-\left(1-\mathrm{m}\right)\mathrm{ln}(1-\mathrm{m})\\{D}_{m}^{{\prime}}\left(m,n\right)&=\mathrm{ln}\left(1-n\right)-\mathrm{ln}\left(1-m-n\right)+\mathrm{ln}(1-m)\\{D}_{n}^{{\prime}}\left(m,n\right)&=\mathrm{ln}\left(1-n\right)-\mathrm{ln}\left(1-m-n\right)+\left(1-m-n\right)\left(\frac{1}{1-n}-\frac{1}{1-m-n}\right)\\{D}_{m}^{{\prime}}\left(m,n\right)&=0\quad {D}_{n}^{{\prime}}\left(m,n\right)=0 \end{aligned}$$

  $D\left(m,n\right)$ is increasing in all directions, whose first-order derivatives are always greater than 0, ${D}_{m}^{{\prime}}\left(m,n\right)>0 \,and\, {D}_{n}^{{\prime}}\left(m,n\right)>0,$ starting from (0, 0), all directions $D\left(m,n\right)$ is increasing, and the minimum value is obtained at (0, 0), which is 0. As shown in Fig. 9.

  Fig. 9

  

  Graph of multivariate function D(m, n)

  $D(m,n) > D_{\min } = 0$, that is $C(m,n) > 1$, we have ${\left(\frac{1-m-n}{1-n}\right)}^{1-m-n}{\left(\frac{1}{1-m}\right)}^{1-m}>1$, so ${\bar{N}}_{2}>{\bar{N}}_{1}$.

  The proof of ${\bar{N}}_{3}>{\bar{N}}_{1}$, that is $\frac{{\bar{N}}_{3}}{{\bar{N}}_{1}}>1$. Due to 0 < m, n < 1, ${\left(1-n\right)}^{\frac{-m}{n}}>1$ is always true, we have $\frac{{\bar{N}}_{3}}{{\bar{N}}_{1}}>1$.

  The proof of ${\bar{N}}_{3}>{\bar{N}}_{2}$, that is $\frac{{\overline{N}_{3} }}{{\overline{N}_{2} }} > 1$:

  $${\left(\frac{1-n}{1-m-n}\right)}^{1-m-n}{(1-n)}^{-m}{\left(1-m\right)}^{1-m}>1$$

  Set function $g\left(m,n\right)={\left(\frac{1-n}{1-m-n}\right)}^{1-m-n}{(1-n)}^{-m}{\left(1-m\right)}^{1-m}$ and $f\left(m,n\right)=\mathrm{ln}g(m,n)$

  $$\begin{aligned}f\left(m,n\right)&=\left(1-m-n\right)\left[\mathrm{ln}\left(1-n\right)-\mathrm{ln}\left(1-m-n\right)\right]-m\mathrm{ln}\left(1-n\right)+\left(1-m\right)\mathrm{ln}(1-m)\\{f}_{m}^{{\prime}}\left(m,n\right)&=-2\mathrm{ln}\left(1-n\right)+\mathrm{ln}\left(1-m-n\right)-\mathrm{ln}(1-m)\\{f}_{n}^{{\prime}}\left(m,n\right)&=-\mathrm{ln}\left(1-n\right)+\mathrm{ln}\left(1-m-n\right)+\left(1-m-n\right)\left(-\frac{1}{1-n}+\frac{1}{1-m-n}\right)+\frac{m}{1-n}\end{aligned}$$

  ${f}_{m}^{{\prime}}\left(m,n\right)=0$ ${f}_{n}^{{\prime}}\left(m,n\right)=0$ we have obtained the Stagnation point $\left({m}_{0},{n}_{0}\right),$ where ${m}_{0}=\frac{1-{n}_{0}}{2-{n}_{0}}$, ${n}_{0}$ is the solution of $ln\frac{1-n}{2-n}+\frac{2}{2-n}=0$, we can know function $G\left(n\right)=ln\frac{1-n}{2-n}+\frac{2}{2-n}$ is decreasing in (0, 1), we get ${n}_{0}\approx 0.75$ ${m}_{0}\approx 0.2$

  $$\begin{aligned}&{f}_{mm}^{{\prime\prime} }\left(m,n\right)=\frac{-n}{(1-m)(1-m-n)}<0\quad {f}_{nn}^{{\prime\prime} }\left(m,n\right)=\frac{m(1-2m-n)}{{(1-n)}^{2}(1-m-n)}\\&{f}_{mn}^{{\prime\prime} }\left(m,n\right)=\frac{1-2m-n}{(1-n)(1-m-n)}\\&{f}_{mm}^{{\prime\prime} }\left(m,n\right) {f}_{nn}^{{\prime\prime} }\left(m,n\right)-{{[f}_{mn}^{{\prime\prime} }\left(m,n\right)]}^{2}=\frac{-(1-2m)(1-2m-n)}{{\left(1-n\right)}^{2}(1-m)(1-m-n)}>0\end{aligned}$$

  It can be seen that $f\left(m,n\right)$ takes the maximum value at $({m}_{0},{n}_{0})$, which is 0.179.

  Through the partial derivative function ${f}_{mn}^{{\prime\prime} }\left(m,n\right)=\frac{1-2m-n}{(1-n)(1-m-n)}$, we can see that in the interval (0, 1), since 0 < m + n < 1 and 0 < m, n < 1, no matter what value n takes, f(m) has a maximum value; no matter what value m takes, f(n) has a maximum value. It can be seen that the points (0, 0), (0, 1), (1, 0) and (1, 1) can be regarded as the minimum value in the interval, and we find f(0, 0) = 0, f(0, 1) = 0.

  Considering $\left(m,n\right)=\mathrm{ln}g(m,n)$, we see g_min = 1 and g_max = $g\left({m}_{0},{n}_{0}\right)$ = 1.196, we find $1<g\left(m,n\right)<1.196$ is always true.

  So ${\left(\frac{1-n}{1-m-n}\right)}^{1-m-n}{(1-n)}^{-m}{\left(1-m\right)}^{1-m}>1$

  So ${\bar{N}}_{3}>{\bar{N}}_{2}$.

Appendix 2: Proof of Propositions 4, 5 and Lemma 2

• (1) The proof of Proposition 4

  When $m>\frac{2b{k}_{1}}{(1-\lambda ){a}^{2}}$:

  $${\mathrm{Y}}\left(\mathrm{m}\right)=\mathrm{ln}{\bar{N}}_{1}=\mathrm{ln}n+\frac{1-m-n}{n}\mathrm{ln}\left(1-n\right)+\frac{1}{1-m}\mathrm{ln}\left(\frac{(1-\lambda ){a}^{2}}{2b{k}_{1}}\right)+\mathrm{ln}\left(\frac{\lambda {k}_{1}}{(1-\lambda )}\right)+\frac{m}{1-m}\mathrm{ln}m$$

  ${Y}_{m}^{{\prime}}=\frac{n[\mathrm{ln}\left((1-\lambda )m{a}^{2}\right)-\mathrm{ln}\left(2b{k}_{1}\right)+1-m]-{\left(1-m\right)}^{2}\mathrm{ln}\left(1-n\right)}{{n(1-m)}^{2}}$ > 0 is always true.

  Y(m) increases monotonically, indicating that ${\bar{N}}_{1}$ also increases with the increase of m.

  $$\begin{aligned}{\mathrm{W}}\left(\mathrm{m}\right)&=\mathrm{ln}{\bar{N}}_{2}=\mathrm{ln}n+\frac{1-m-n}{n}\mathrm{ln}\left(1-m-n\right)+\frac{1-m}{n}\mathrm{ln}\left(\frac{1}{1-m}\right)\\&\quad +\frac{1}{1-m}\mathrm{ln}\left(\frac{(1-\lambda ){a}^{2}}{2b{k}_{1}}\right)+\mathrm{ln}\left(\frac{\lambda {k}_{1}}{(1-\lambda )}\right)+\frac{m}{1-m}\mathrm{ln}m\end{aligned}$$

  ${W}_{m}^{{\prime}}=\frac{n[\mathrm{ln}\left((1-\lambda )m{a}^{2}\right)-\mathrm{ln}\left(2b{k}_{1}\right)+1-m]+{\left(1-m\right)}^{2}[\mathrm{ln}\left(1-m\right)-\mathrm{ln}(1-m-n)]}{{n(1-m)}^{2}}$ > 0 is always true.

  $\mathrm{W}\left(\mathrm{m}\right)$ increases monotonically, indicating that ${\bar{N}}_{2}$ also increases with the increase of m.

  $${\mathrm{V}\left(\mathrm{m}\right)=\mathrm{ln}\bar{N}}_{1}=\mathrm{ln}n+\frac{1-2m-n}{n}\mathrm{ln}\left(1-n\right)+\frac{1}{1-m}\mathrm{ln}\left(\frac{(1-\lambda ){a}^{2}}{2b{k}_{1}}\right)+\mathrm{ln}\left(\frac{\lambda {k}_{1}}{(1-\lambda )}\right)+\frac{m}{1-m}\mathrm{ln}m$$

  ${V}_{m}^{{\prime}}=\frac{n[\mathrm{ln}\left((1-\lambda )m{a}^{2}\right)-\mathrm{ln}\left(2b{k}_{1}\right)+1-m]-2{\left(1-m\right)}^{2}\mathrm{ln}\left(1-n\right)}{{n(1-m)}^{2}}$ > 0 is always true.

  $\mathrm{V}\left(\mathrm{m}\right)$ increases monotonically, indicating that ${\bar{N}}_{3}$ also increases with the increase of m.

• (2) The proof of Proposition 5 and Lemma 2

  $${\mathrm{S}\left(\mathrm{n}\right)=\mathrm{ln}\bar{N}}_{1}=\mathrm{ln}n+\frac{1-m-n}{n}\mathrm{ln}\left(1-n\right)+\frac{1}{1-m}\mathrm{ln}\left(\frac{(1-\lambda ){a}^{2}}{2b{k}_{1}}\right)+\mathrm{ln}\left(\frac{\lambda {k}_{1}}{(1-\lambda )}\right)+\frac{m}{1-m}\mathrm{ln}m$$

  Due to 0 < n < 1, we see ln(1 − n) < 0. ${S}_{n}^{{\prime}}=\frac{mn-\left(1-n\right)\left(1-m\right)\mathrm{ln}(1-n)}{{n}^{2}(1-n)}$ > 0 is always true.

  $\mathrm{S}\left(\mathrm{n}\right)$ increases monotonically, indicating that ${\bar{N}}_{1}$ also increases with the increase of n.

  $$\begin{aligned}{\mathrm{R}}\left(\mathrm{n}\right)&=\mathrm{ln}{\bar{N}}_{2}=\mathrm{ln}n+\frac{1-m-n}{n}\mathrm{ln}\left(1-m-n\right)+\frac{1-m}{n}\mathrm{ln}\frac{1}{1-m}\\&\quad+\frac{1}{1-m}\mathrm{ln}\left(\frac{(1-\lambda ){a}^{2}}{2b{k}_{1}}\right)+\mathrm{ln}\left(\frac{\lambda {k}_{1}}{(1-\lambda )}\right)+\frac{m}{1-m}\mathrm{ln}m\end{aligned}$$

  ${R}_{n}^{{\prime}}=\frac{[\mathrm{ln}\left(1-m\right)-\mathrm{ln}(1-m-n)](1-m)}{{n}^{2}}$ > 0 is always true.

  $\mathrm{R}\left(\mathrm{n}\right)$ increases monotonically, indicating that ${\bar{N}}_{2}$ also increases with the increase of n.

  $$\begin{aligned}{\mathrm{P}}\left(\mathrm{n}\right)&=\mathrm{ln}{\bar{N}}_{3}=\mathrm{ln}n+\frac{1-2m-n}{n}\mathrm{ln}\left(1-n\right)+\frac{1}{1-m}\mathrm{ln}\left(\frac{(1-\lambda ){a}^{2}}{2b{k}_{1}}\right)+\mathrm{ln}\left(\frac{\lambda {k}_{1}}{(1-\lambda )}\right)+\frac{m}{1-m}\mathrm{ln}m\\{P}_{n}^{{\prime}}&=\frac{2mn-\left(1-n\right)\left(1-2m\right)\mathrm{ln}(1-n)}{{n}^{2}(1-n)}\end{aligned}$$

  When m < 0.5, ${P}_{n}^{{\prime}}=\frac{2mn-\left(1-n\right)\left(1-2m\right)\mathrm{ln}(1-n)}{{n}^{2}(1-n)}>0$ is always true.

  When m > 0.5, due to 1 − m − n > 0, so 0 < n < 0.5.

  $y(n)=2mn-\left(1-n\right)\left(1-2m\right)\mathrm{ln}(1-n)$ increases monotonically, and ${y}_{min}=y\left(0\right)=0$, so $y\left(n\right)>0$ is always true, so ${P}_{n}^{{\prime}}=\frac{2mn-\left(1-n\right)\left(1-2m\right)\mathrm{ln}(1-n)}{{n}^{2}(1-n)}>0$ is always true.

  $\mathrm{P}\left(\mathrm{n}\right)$ increases monotonically, indicating that ${\bar{N}}_{3}$ also increases with the increase of n.

• (3) Graphs of multivariate functions

  See Figs. 10, 11 and 12.

  Fig. 10

  

  Graph of multivariate function $k_{2} = \overline{N}_{1} (m,n)$

  Fig. 11

  

  Graph of multivariate function $k_{2} = \overline{N}_{2} (m,n)$

  Fig. 12

  

  Graph of multivariate function $k_{2} = \overline{N}_{3} (m,n)$

Appendix 3: The results of the cost-sharing contract model

See Table 7.

Table 7 The decisions and profits in different scenario

When both parties innovate together:

$$ \begin{aligned} \prod_{S} & = (1 - \lambda )(ae_{1}^{m} e_{2}^{n} - bp)p - \frac{1}{2}(1 - c)k_{1} e_{1}^{2} - F_{1} \\ \prod_{R} & = \lambda (ae_{1}^{m} e_{2}^{n} - bp)p - \frac{1}{2}ck_{1} e_{1}^{2} - \frac{1}{2}k_{2} e_{2}^{2} - F_{2} \\ \end{aligned} $$

For the convenience of calculation, the platform’s profit can be written as:

$$ \prod_{R} = \left[ {\lambda c(ae_{1}^{m} e_{2}^{n} - bp)p - \frac{1}{2}ck_{1} e_{1}^{2} } \right] + \left[ {\lambda (1 - c)(ae_{1}^{m} e_{2}^{n} - bp)p - \frac{1}{2}k_{2} e_{2}^{2} } \right] - F_{2} $$

When the provider innovates independently:

$$ \begin{aligned} \prod_{S} & = (1 - \lambda )(ae_{1}^{m} - bp)p - \frac{1}{2}k_{1} e_{1}^{2} - F_{1} \\ \prod_{R} & = \lambda (ae_{1}^{m} - bp)p - F_{2} \\ \end{aligned} $$

Appendix 4: Proof of Proposition 6

When $\prod_{S}^{SL} > \prod_{S}^{NO}$ and $\prod_{R}^{SL} > \prod_{R}^{NO}$, both parties will choose collaborative innovation. For the provider:

$$ \begin{aligned} e_{1}^{NO} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m)}}} \;\;\prod_{S}^{NO} = \left( {\frac{1}{m} - 1} \right)\frac{{k_{1} }}{2}\left( {e_{1}^{NO} } \right)^{2} - F_{1} \\ e_{1}^{SL} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2b(1 - c)k_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda n(1 - c)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \left( {1 - n} \right)^{{\frac{n - 1}{{2(1 - m - n)}}}} \\ \prod_{S}^{SL} & = \left( {\frac{1 - n}{m} - 1} \right)\frac{{(1 - c)k_{1} }}{2}\left( {e_{1}^{SL} } \right)^{2} - F_{1} \\ \end{aligned} $$

If $\prod_{S}^{SL} > \prod_{S}^{NO}$, we obtain:

$$ \left( {\frac{1 - m - n}{m}} \right)(1 - n)^{{\frac{n - 1}{{1 - m - n}}}} \frac{{(1 - c)k_{1} }}{2}\left( {\frac{{(1 - \lambda )a^{2} m)}}{{2b(1 - c)k_{1} }}} \right)^{{\frac{1}{1 - m - n}}} \left( {\frac{{\lambda n(1 - c)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{1 - m - n}}} > \left( {\frac{1 - m}{m}} \right)\frac{{k_{1} }}{2}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{1 - m}}} $$

After simplification:

$$ c > 1 - \left( {\frac{1 - m - n}{{1 - m}}} \right)^{{\frac{1 - m - n}{m}}} \left( {1 - n} \right)^{{\frac{n - 1}{m}}} \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{m(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{\frac{n}{m}} $$

Let $V_{1} = 1 - \left( {\frac{1 - m - n}{{1 - m}}} \right)^{{\frac{1 - m - n}{m}}} \left( {1 - n} \right)^{{\frac{n - 1}{m}}} \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{n}{m(1 - m)}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{\frac{n}{m}}.$

For the platform:

$$ \begin{aligned} e_{1}^{NO} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{1}{2(1 - m)}}} \;\;\;\prod_{R}^{NO} = \frac{\lambda }{(1 - \lambda )m}\frac{{k_{1} }}{2}\left( {e_{1}^{NO} } \right)^{2} - F_{2} \\ e_{2}^{SL} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2b(1 - c)k_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda n(1 - c)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{1 - m}{{2(1 - m - n)}}}} (1 - n)^{{\frac{ - m}{{2(1 - m - n)}}}} \\ e_{1}^{SL} & = \left( {\frac{{(1 - \lambda )a^{2} m}}{{2b(1 - c)k_{1} }}} \right)^{{\frac{1}{2(1 - m - n)}}} \left( {\frac{{\lambda n(1 - c)k_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{n}{2(1 - m - n)}}} \left( {1 - n} \right)^{{\frac{n - 1}{{2(1 - m - n)}}}} \\ \prod_{R}^{SL} & = \left( {\frac{\lambda (1 - n)(1 - c)}{{(1 - \lambda )m}} - 1} \right)\frac{{ck_{1} }}{2}\left( {e_{1}^{SL} } \right)^{2} + \left( {\frac{1 - c}{n} - 1} \right)\frac{{k_{2} }}{2}\left( {e_{2}^{SL} } \right)^{2} - F_{2} \\ \end{aligned} $$

If $\prod_{R}^{SL} > \prod_{R}^{NO}$, we obtain:

$$ \begin{aligned} & \left( {\frac{\lambda }{(1 - \lambda )m} - \frac{1}{(1 - c)(1 - n)}} \right)\frac{{ck_{1} }}{2} + \left( {\frac{1 - c - n}{n}} \right)\frac{{k_{2} }}{2}\left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right) \\ & \quad > \frac{\lambda }{(1 - \lambda )m}\frac{{k_{1} }}{2}\left( {\frac{{(1 - \lambda )a^{2} m}}{{2bk_{1} }}} \right)^{{\frac{ - n}{{(1 - m)(1 - m - n)}}}} [(1 - c)(1 - n)]^{{\frac{m}{1 - m - n}}} \left( {\frac{{\lambda nk_{1} }}{{(1 - \lambda )mk_{2} }}} \right)^{{\frac{ - n}{{1 - m - n}}}} \\ \end{aligned} $$

Let c₁ be the smallest solution of the inequality and c₂ be the largest solution of the inequality, where $c_{1} < c < c_{2}$. To sum up, under the provider-led relationship, to ensure that both parties make more profits due to collaborative innovation, the value range of the cost coefficient c that the platform needs to share with the provider is:

$$ \max \{ V_{1} ,c_{1} \} < c < \min \{ c_{2} ,1\} . $$

Appendix 5: Solving process of extended model

Without loss of generality, we set b₁ = b₂ = b.

The results of independent innovation under the Nash relationship:

$$\begin{aligned} e_{1}^{{\underline{NA} }}& = \left( {\frac{{2a^{2} m[3b + \left( {1 - \lambda } \right)b]^{2} - 3b^{2} \left( {1 - \lambda )^{2} } \right]}}{{bk_{1} [3b - \left( {1 - \lambda } \right)b]^{2} }}} \right)^{{\frac{1}{{2\left( {1 - m} \right)}}}} \\{\varvec{p}}_{\mathbf{1}}^{{\underline{{{\varvec{NA}}}} }} &= \frac{{{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} \left[ {\mathbf{3}{\varvec{b}} + {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}{{{\varvec{b}}\left[ {\mathbf{3}{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}{\varvec{p}}_{\mathbf{2}}^{{\underline{{{\varvec{NA}}}} }} = \frac{{\mathbf{3}{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} }}{{\mathbf{3}{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)}}\end{aligned} $$

The results of collaborative innovation under the Nash relationship: $e_{1}^{NA} = \eta_{1}^{{\frac{1}{2(1 - m - n)}}} \eta_{2}^{{\frac{1 - n}{{2(1 - m - n)}}}} \,e_{2}^{NA} = \eta_{1}^{{\frac{1}{2(1 - m - n)}}} \eta_{2}^{{\frac{m}{2(1 - m - n)}}}$. Note:

$$\begin{aligned} {{\varvec{\eta}}}_{\mathbf{1}}&=\left(\frac{\mathbf{18}{{\varvec{a}}}^{\mathbf{2}}{\varvec{b}}{\varvec{n}}{\varvec{\lambda}}}{{{\varvec{k}}}_{\mathbf{2}}[\mathbf{3}{\varvec{b}}-(\mathbf{1}-{\varvec{\lambda}}){\varvec{b}}{]}^{\mathbf{2}}}\right)\quad {{\varvec{\eta}}}_{\mathbf{2}}=\left(\frac{{{\varvec{k}}}_{\mathbf{2}}{\varvec{m}}\{[\mathbf{3}{\varvec{b}}+(\mathbf{1}-{\varvec{\lambda}}){\varvec{b}}{]}^{\mathbf{2}}-\mathbf{3}{{\varvec{b}}}^{\mathbf{2}}(\mathbf{1}-{\varvec{\lambda}}{)}^{\mathbf{2}}}{\mathbf{9}{{\varvec{b}}}^{\mathbf{2}}{{\varvec{k}}}_{\mathbf{1}}{\varvec{n}}{\varvec{\lambda}}}\right) \\{{\varvec{p}}}_{\mathbf{1}}^{{\varvec{N}}{\varvec{A}}}&=\frac{{\varvec{a}}{{\varvec{e}}}_{\mathbf{1}}^{{\varvec{m}}}[\mathbf{3}{\varvec{b}}+{\varvec{b}}(\mathbf{1}-{\varvec{\lambda}})]}{{\varvec{b}}[\mathbf{3}{\varvec{b}}-{\varvec{b}}(\mathbf{1}-{\varvec{\lambda}})]}\quad {{\varvec{p}}}_{\mathbf{2}}^{{\varvec{N}}{\varvec{A}}}=\frac{\mathbf{3}{\varvec{a}}{{\varvec{e}}}_{\mathbf{1}}^{{\varvec{m}}}}{\mathbf{3}{\varvec{b}}-{\varvec{b}}(\mathbf{1}-{\varvec{\lambda}})}.\end{aligned}$$

The results of independent innovation under the provider-led relationship:

$$\begin{aligned} e_{1}^{{\underline{SL} }} &= \left( {\frac{{a^{2} m\left[ {9b + 8b\left( {1 - \lambda } \right)} \right]}}{{2bk_{1} \left[ {2b - b\left( {1 - \lambda } \right)} \right]}}} \right)^{{\frac{1}{{2\left( {1 - m} \right)}}}} \\{\varvec{p}}_{\mathbf{1}}^{{\underline{{{\varvec{SL}}}} }} &= \frac{{{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} \left[ {\mathbf{3}{\varvec{b}} + {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}{{{\varvec{b}}\left[ {\mathbf{2}{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}\quad {\varvec{p}}_{\mathbf{2}}^{{\underline{{{\varvec{SL}}}} }} = \frac{{\mathbf{5}{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} }}{{\mathbf{2}\left[ {\mathbf{2}{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}} \end{aligned}$$

The results of collaborative innovation under the provider-led relationship: $e_{1}^{SL} = \eta_{3}^{{\frac{n}{2(1 - m - n)}}} \eta_{4}^{{\frac{1 - n}{{2(1 - m - n)}}}} \,e_{2}^{SL} = \eta_{3}^{{\frac{1 - m}{{2(1 - m - n)}}}} \eta_{4}^{{\frac{m}{2(1 - m - n)}}}$. Note:

$$\begin{aligned} {\varvec{\eta}}_{\mathbf{3}} &= \left( {\frac{{\mathbf{25}\varvec{\lambda a}^{\mathbf{2}} {\varvec{bn}}}}{{\mathbf{2}{\varvec{k}}_{\mathbf{2}} [\left( {\mathbf{1} - {\varvec{\lambda}}} \right){\varvec{b}} - 2{\varvec{b}}]^{\mathbf{2}} }}} \right) \quad \varvec{ \eta }_{\mathbf{4}} = \left( {\frac{{\mathbf{4}{\varvec{m}}\left[ {\left( {\mathbf{1} - {\varvec{\lambda}}} \right){\varvec{b}} + \frac{\mathbf{9}}{\mathbf{8}}{\varvec{b}}} \right]{\varvec{a}}^{\mathbf{2}} }}{{\left[ {\mathbf{2}{\varvec{b}}_{\mathbf{2}} - \left( {\mathbf{1} - {\varvec{\lambda}}} \right){\varvec{b}}} \right]\left( {\mathbf{1} - {\varvec{n}}} \right){\varvec{bk}}_{\mathbf{1}} }}} \right)\\{\varvec{p}}_{\mathbf{1}}^{{{\varvec{SL}}}} &= \frac{{{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} {\varvec{e}}_{\mathbf{2}}^{{\varvec{n}}} \left[ {\mathbf{3}{\varvec{b}} + {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}{{{\varvec{b}}\left[ {\mathbf{2}{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}\quad {\varvec{p}}_{\mathbf{2}}^{{{\varvec{SL}}}} = \frac{{\mathbf{5}{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} {\varvec{e}}_{\mathbf{2}}^{{\varvec{n}}} }}{{\mathbf{2}\left[ {\mathbf{2}{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}} \end{aligned}$$

The results of independent innovation under the platform-led relationship:

$$\begin{aligned} e_{1}^{{\underline{RL} }} &= \left( {\frac{{a^{2} m\left[ {25b + 23b\left( {1 - \lambda } \right)} \right]}}{{8bk_{1} \left[ {b - b\left( {1 - \lambda } \right)} \right]}}} \right)^{{\frac{1}{{2\left( {1 - m} \right)}}}}\\ {\varvec{p}}_{\mathbf{1}}^{{\underline{{{\varvec{RL}}}} }} &= \frac{{{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} \left[ {\mathbf{5}{\varvec{b}} + {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}{{\mathbf{4}{\varvec{b}}_{\mathbf{1}} \left[ {{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}}\quad {\varvec{p}}_{\mathbf{2}}^{{\underline{{{\varvec{RL}}}} }} = \frac{{\mathbf{3}{\varvec{ae}}_{\mathbf{1}}^{{\varvec{m}}} }}{{\mathbf{2}\left[ {{\varvec{b}} - {\varvec{b}}\left( {\mathbf{1} - {\varvec{\lambda}}} \right)} \right]}} \end{aligned}$$

The results of collaborative innovation under the platform-led relationship: $e_{1}^{RL} = \eta_{5}^{{\frac{n}{2(1 - m - n)}}} \eta_{6}^{{\frac{1 - n}{{2(1 - m - n)}}}}$ $e_{2}^{RL} = \eta_{5}^{{\frac{1 - m}{{2(1 - m - n)}}}} \eta_{6}^{{\frac{m}{2(1 - m - n)}}}$. Note:

$$ \eta_{5} = \left( {\frac{{9\lambda na^{2} }}{{4k_{2} \left( {1 - m} \right)\left[ {b - b\left( {1 - \lambda } \right)} \right]}}} \right). $$