## Appendix A Mathematical Proofs
### Proof of Theorem 2
First, for each case, from the property of \(g(X)\), we know that \(g(X)<0\) for \(X\in(X_{1},1)\), and hence \(\phi(X)=g(X)\) has no positive solution in \((X_{1},1)\) by (16). (a) If \(\delta_{g}>\delta_{n}\), then we have \(\lim_{X\to 0^{+}}\phi(X)=\infty\). From the properties of \(f(X)\) and the expression of \(\phi(X)\), there exists a unique point \(X^{c}\in(0,X_{1})\subset(0,1)\) satisfying \(f(X^{c})=1\), such that \(\phi(X^{c})=0\) and \(\frac{d\phi}{dX}<0\) for \(X\in(0,X^{c})\) since \[\frac{d\phi_{1}}{dX}=\frac{1}{\delta_{g}-\delta_{n}}\left(\frac{[1-f(X)]X^{-\delta_ {g}}}{f(X)}\right)\frac{1}{\delta_{g}-\delta_{n}}-1\left[-\frac{X^{-\delta_{g}} }{f(X)^{2}}\frac{df}{dX}-\delta_{g}\frac{[1-f(X)]X^{-\delta_{g}-1}}{f(X)}\right]. \tag{100}\]  
Therefore, \(\phi(X)\) is decreasing from infinity to zero in \((0,X^{c}]\) and \(g(X)\) is increasing from zero to infinity in \((0,X_{1})\). Thus, \(\phi(X)=g(X)\) has a unique positive solution in \((0,X^{c})\subset(0,X_{1})\subset(0,1)\). In \((X^{c},X_{1})\), from the monotonicity of \(f(X)\) we have \(f(X)>1\), thus, by (17), \(\phi(X)=g(X)\) has no solution. (b) If \(\delta_{g}<\delta_{n}\), then \(\phi(X)\) is well defined at \(X=0\) and \(\phi(0)=0\). From the properties of \(f(X)\), there exists a unique point \(X^{c}\in(0,X_{1})\subset(0,1)\) such that \(f(X^{c})=1\). In \((X^{c},X_{1})\), from the monotonicity of \(f(X)\) we have \(f(X)>1\); thus, by (17), \(\phi(X)=g(X)\) has no solution. It is easy to check that  
\[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]0  
and \(\frac{d\phi_{1}}{dX}>0\) for \(X\in(0,X^{c})\) since \(\frac{1}{2}<f(X)<1\) and \(f^{\prime}(X)>0\) for \(X\in(0,X^{c})\). Thus, \(\phi_{1}(X)\), and hence \(\phi(X)\), are increasing from zero to infinity in \([0,X^{c})\). Since  
\[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]1  
we have the following three scenarios:  
* When \(\delta_{g}<\delta_{n}<2\delta_{g}\), we can get \(\frac{d\phi_{1}}{dX}\mid_{X=0^{+}}=0\), and hence \[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]2 Notice that \(g(0)=0,g_{+}{}^{\prime}(0)=\frac{e}{2d_{w}}>0\), and \(g(X)\) is increasing from zero to infinity in \([0,X_{1})\), we can conclude that \(\phi(X)=g(X)\) has at least one positive solution in \((0,X^{c})\subset(0,X_{1})\subset(0,1)\). * When \(2\delta_{g}<\delta_{n}\), we can get \(\left.\frac{d\phi_{1}}{dX}\right|_{X=0^{+}}=\infty\). Due to \[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]3 \[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]4 and \(\phi(X)\) is increasing in \((0,X^{c})\), there exists \(\epsilon>0\) such that \[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]5 Let \[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]6Then, if \(\left(\dfrac{\beta_{n}}{\beta_{g}}\right)\dfrac{1}{\delta_{g}-\delta_{n}}\) is large such that \(\left(\dfrac{\beta_{n}}{\beta_{g}}\right)\dfrac{1}{\delta_{g}-\delta_{n}}>M_{e}\), we have \[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]7 Notice that \(\delta_{g}<2\delta_{g}<\delta_{n}\).