## Appendix A Mathematical Proofs
### Proof of Theorem 2  
**Proof. ** Recall  
\[\psi(X)=-X^{2}\big{(}ed_{g}+2d_{g}-2d_{w}\big{)}+2X\big{(}d_{g}-2d_{w}\big{)}+ 2d_{w}.\] (A5)  
When \(ed_{g}+2d_{g}-2d_{w}\
eq 0\), i.e., \(\frac{d_{g}}{d_{w}}\
eq\frac{2}{2+e}\), \(\psi(X)\) has two real zeros:  
\[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}, \text{ and }X_{2}=\frac{d_{g}-2d_{w}-\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}.\]  
If \(\frac{d_{g}}{d_{w}}=\frac{2}{2+e}\), which implies that \(d_{g}-2d_{w}<0\), \(\psi(X)\) has the unique positive solution \(\frac{d_{w}}{2d_{w}-d_{g}}\). We still denote it as \(X_{1}\), i.e., \(X_{1}=\frac{d_{w}}{2d_{w}-d_{g}}=\frac{2+e}{2+e}(<1)\). We claim that \(X_{1}\) is the unique zero of \(\psi(X)\) in the interval \((0,1)\). In fact, if \(ed_{g}+2d_{g}-2d_{w}>0\), i.e., \(1>\frac{d_{g}}{d_{w}}>\frac{2}{2+e}\) (Note: we always assume that \(d_{g}<d_{w}\)), then \(X_{1}>0>X_{2}\). Note that \(\frac{d_{g}}{d_{w}}>\frac{2}{2+e}\) implies that  
\[\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}-(e+1)d_{g}<d_{g}\bigg{(}\sqrt{e(2+e)+1}-(e+1) \bigg{)}=0.\] (A6)  
Thus, we have by (A6) that  
\[X_{1}=\frac{d_{g}-2d_{w}+\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}}{ed_{g}+2d_{g}-2d_{w}}= 1+\frac{\sqrt{2 ed_{g}d_{w}+d_{g}^{2}}-(e+1)d_{g}}{ed_{g}+2d_{g}-2d_{w}}<1.\]  
It follows that \(X_{2}<0<X_{1}<1\). If \(ed_{g}+2d_{g}-2d_{w}<0\), i.e., \(\frac{d_{g}}{d_{w}}<\frac{2}{2+e}\), then similarly by the opposite inequality of (A6), we have \(0<X_{1}<1<X_{2}\). Noticing that \(\psi(1)=-ed_{g}<0\) when \(\frac{d_{g}}{d_{w}}\
eq\frac{2}{2+e}\) and \(\psi(1)=2d_{g}-2d_{w}<0\) when \(\frac{d_{g}}{d_{w}}=\frac{2}{2+e}\), and \(\psi(0)=2d_{w}>0\) and \(\psi(X_{1})=0\), we know that \(X=X_{1}\) is a vertical asymptote of \(g(X)=\frac{X(1-X)e}{\psi(X)}\) in the interval of \((0,1)\). Since for \(x\in(0,X_{1})\cup(X_{1},1)\),  
\[\frac{dg}{dX} = \frac{e\Big{(}ed_{g}X^{2}+2d_{w}(1-X)^{2}\Big{)}}{\big{(}-X^{2}( ed_{g}+2d_{g}-2d_{w})+2X(d_{g}-2d_{w})+2d_{w}\big{)}^{2}}>0,\]  
we know that the function \(g(X)\) has the following properties:  
* \(g(0)=g(1)=0\), \(g(X)\) is increasing in the both intervals of \((0,X_{1})\) and \((X_{1},1)\), and \(g(x)>0\) for \(X\in(0,X_{1})\) and \(g(x)<0\) for \(X\in(X_{1},1)\);
* \(\lim_{X\to X_{1}^{+}}g(X)=-\infty\), \(\lim_{X\to X_{1}^{-}}g(X)=\infty\). In the meantime, \(X=X_{1}\) is also the vertical asymptote of \(f(X)=\frac{ed_{g}X^{2}}{2\varphi(X)}+\frac{1}{2}\) in the interval of \((0,1)\) and for \(x\in(0,X_{1})\cup(X_{1},1)\),  
\[\frac{df}{dX}=\frac{ed_{g}X\big{(}d_{g}X+2d_{w}(1-X)\big{)}}{\big{(}-X^{2}(ed_{ g}+2d_{g}-2d_{w})+2X(d_{g}-2d_{w})+2d_{w}\big{)}^{2}}>0.\]  
So, similar to the above arguments, \(f(X)\) has the following properties:  
* \(f(0)=1/2>0\), \(f(1)=0\), \(f(X)\) is increasing in the both intervals of \((0,X_{1})\) and \((X_{1},1)\), and \(f(x)>0\) for \(X\in(0,X_{1})\) and \(f(x)<0\) for \(X\in(X_{1},1)\);
* \(\lim_{X\to X_{1}^{+}}f(X)=-\infty\), \(\lim_{X\to X_{1}^{-}}f(X)=\infty\). If \(\delta_{g}=\delta_{n}\), then from (16) and (17), we have \(N=g(x)\) and  
\[\frac{\beta_{n}}{\beta_{g}X^{\delta_{g}}+\beta_{n}}=f(X). \tag{18}\]  
From the property of \(f(x)\), we know that \(f(x)<0\) for \(X\in(X_{1},1)\), so (18) has no solution in \((X_{1},1)\), which implies that system (9) and (10) has no interior equilibrium. In \((0,X_{1})\), (18) has a unique solution \(X^{*}\) which gives the interior equilibrium \((N^{*},N_{g}^{*})\) of (9) and (10) with \(N^{*}=g(X^{*})\) and \(N_{g}^{*}=X^{*}N^{*}\). Now, we set \(\delta_{g}\
eq\delta_{n}\) and  
\[\phi_{1}(X)=\left(\frac{[1-f(X)]X^{-\delta_{g}}}{f(X)}\right)\frac{1}{\delta_ {g}-\delta_{n}}\,.\]  
Then  
\[\phi(X)=\left(\frac{\beta_{n}}{\beta_{g}}\right)\frac{1}{\delta_{g}-\delta_{n }}\,\phi_{1}(X).\]  
We consider the solution of \(\phi(X)=g(X)\) in the interval \((0,1)\) for two cases (a) \(\delta_{g}>\delta_{n}\) and (b) \(\delta_{g}<\delta_{n}\).