## 2 Materials and Methods
### Models and Solutions
#### 2.3.1 Subsidy-Free Model
 \tag{8}\]  
Substitute Equation (8) into Equation (4) to obtain the optimal market price \(p^{*}=\frac{ck^{2}-bc\mu r-3a\mu r}{k^{2}-4b\mu r}\), then the optimal profits of supplier, distributor, and supply chain can be calculated according to Equations (1)-(3):  
\[U_{0}^{*}(\pi_{S})=\frac{4b[(bc-a)\mu r]^{2}-9\mu rk^{2}(bc-a)^{2}}{2(k^{2}-4b \mu r)^{2}} \tag{9}\]  
\[U_{0}^{*}(\pi_{R})=\frac{(a-bc)\mu}{(k^{2}-4b\mu r)^{2}}\bigg{[}ar\Big{(}b\mu r -2k^{2}\Big{)}+\frac{1}{2}k^{2}(1-r)(bc-a)\bigg{]} \tag{10}\]  
\[U_{0}^{*}(\pi_{T})=\frac{(bc-a)^{2}\mu}{(k^{2}-4b\mu r)^{2}}\bigg{[}3b\mu r^{ 2}-\frac{1}{2}k^{2}(12r+1)\bigg{]} \tag{11}\]  
**Proposition 1**. : _Under the subsidy-free condition, \(\theta\) is inversely proportional to the invested cost of ecological conservation and restoration, and the utilities of the distributor and supply chain are proportionate to the invested cost of ecological conservation and restoration; while the cost-sharing contract coefficient is constant, \(w\), \(p\) and the supplier's utility will show a trend of decreasing first and then increasing._  
Proof.Based on the given condition \(\theta>0\) and \(b\mu r>\frac{k^{2}}{4}\), it can be concluded that \(k^{2}-4b\mu r<0\); and from given \(\theta^{*}=\frac{k(bc-a)}{k^{2}-4b\mu r}>0\), it can also be concluded that \(bc-a<0\), so \(\frac{\partial\theta}{\partial\theta}=\frac{4b\mu k(bc-a)}{(k2-4b\mu r)^{2}}<0\). \(\square\)  
Because \(\frac{\partial p}{\partial r}=\frac{-\mu(bc+3a)}{k^{2}-4b\mu r}+\frac{(ck^{2} -bc\mu r-3a\mu r)(-4b\mu)}{(k^{2}-4b\mu r)^{2}}\frac{-\mu[(bc+3a)k^{2}+8b\mu r (bc-3a)]}{(k^{2}-4b\mu r)^{2}}\), so when \(0<r<\frac{(bc+3a)k^{2}}{8b\mu(3a-bc)}\), it obtains \((bc+3a)k^{2}+8b\mu r(bc-3a)>0\) and \(\frac{\partial p}{\partial r}<0\), and when \(\frac{(bc+3a)k^{2}}{8b\mu(3a-bc)}<r<1\), it obtains \((bc+3a)k^{2}+8b\mu r(bc-3a)<0\) and \(\frac{\partial p}{\partial r}>0\). Similarly, when \(0<\delta<\frac{(bc+3a)k^{2}}{8b\mu(3a-bc)-(bc+3a)k^{2}}\), \(\frac{\partial p}{\partial r}<0\), and when \(\frac{(bc+3a)k^{2}}{8b\mu(3a-bc)-(bc+3a)k^{2}}<\delta<1\), \(\frac{\partial p}{\partial r}>0\). Because \(\frac{\partial\theta}{\partial r}=\frac{-2\mu r[(a+3 bc)k^{2}-8b\mu r(a+bc)]}{(k^ {2}-4b\mu r)^{2}}\), so when \(0<r<\frac{(a+3 bc)k^{2}}{8b\mu(a+bc)}\), it obtains \((a+3 bc)k^{2}-8b\mu r(a+bc)>0\) and \(\frac{\partial\theta}{\partial r}<0\), and when \(\frac{(a+3 bc)k^{2}}{8b\mu(a+bc)}<r<1\), it obtains\((a+3 bc)k^{2}-8b\mu r(a+bc)<0\) and \(\frac{3\mu r}{\partial r}>0\). Similarly, when \(0<\delta<\frac{(a+3 bc)k^{2}}{8b\mu(a+bc)-(a+3 bc)k^{2}}\), \(\frac{\partial w}{\partial r}<0\); and when \(\frac{(a+3 bc)k^{2}}{8b\mu(a+bc)-(a+3 bc)k^{2}}<\delta<1\), \(\frac{\partial w}{\partial r}>0\). Because \(\frac{\partial U_{0}(\tau_{S})}{\partial r}=\frac{-2(k-a)^{2}ik^{2}[28b\mu r -9k^{2}]}{(k^{2}-4b\mu r)^{3}}\), so when \(\frac{k^{2}}{4b\mu}<r<\frac{9k^{2}}{28b\mu}\), it obtains \(28b\mu r-9k^{2}<0\) and \(\frac{\partial U_{0}(\tau_{S})}{\partial r}<0\), and when \(\frac{9k^{2}}{28b\mu}<r<1\), it obtains \(28b\mu r-9k^{2}>0\) and \(\frac{\partial U_{0}(\tau_{S})}{\partial r}>0\). Because \(\frac{\partial U_{0}(\tau_{S})}{\partial r}=\frac{-2\mu k^{2}[\frac{2 ab\mu r +b^{2}cr+2b(a-bc)}{(k^{2}-4b\mu r)^{3}}}{\text{, so }\frac{\partial U_{0}(\tau_{R})}{\partial r}>0\text{. Similarly, }\frac{ \partial U_{0}(\tau_{T})}{\partial r}>0\).