## 3 Mathematical Analysis  
Based on the formulation of Model (9) and (10), we observe that if \(N=0\), then the system automatically has \(N_{g}=N_{w}=0\) as \(N=N_{g}+N_{w}\). At the same time, the expression of the metabolic cost \(C_{w}\) and \(C_{g}\) is not well defined at \(N=0\). For convenience, by using the compactness concept, we define the following:  
\[\frac{dN}{dt}\Big{|}_{N=0}=0\text{ and }\frac{dN_{g}}{dt}\Big{|}_{N=0}=0. \tag{11}\]  
Thus, we provide the following basic properties of the system (9) and (10):  
Theorem 3.1: _The system (9) and (10) is positive invariant and bounded in \(\mathbb{R}^{+2}\)._  
Notes: Theorem 3.1 implies that our proposed system (9) and (10) is well-defined biologically. The first step to study the system (9) and (10) is to solve its steady states, i.e., equilibrium. According to (11), we can conclude that \(N=N_{g}=0\) is always an equilibrium of the system (9) and (10). Moreover, this equilibrium is unstable. Let us look at its interior equilibrium \((N^{*},N_{g}^{*})\). From the biological point of view, at the equilibrium the relative cost do not change and also the relative cost of workers is larger than that for gamergates. For convenience, let us define the _reproductive ratio_\(X=\frac{N_{g}}{N}\). Then by letting \(\frac{dN}{dt}=\frac{dN_{g}}{dt}=0\), we obtain the following two equations:  
\[0 = eN_{g}\frac{\beta_{n}N^{\delta_{n}}}{\beta_{g}N_{g}^{\delta_{g} }+\beta_{n}N^{\delta_{n}}}-d_{g}N_{g}N-d_{w}(N-N_{g})N \tag{12}\] \[0 = (N-N_{g})\frac{\beta_{n}N^{\delta_{n}}-\beta_{g}N_{g}^{\delta_{g }}}{\beta_{n}N^{\delta_{n}}+\beta_{g}N_{g}^{\delta_{g}}}-d_{g}N_{g}N \tag{13}\]  
Suppose \(N_{g},N\
eq 0\). Dividing (12) by \(N^{2}\) and letting \(X=\frac{N_{g}}{N}\), we have the following equation:  
\[0=eX\frac{\beta_{n}N^{\delta_{n}-1}}{\beta_{g}(NX)^{\delta_{g}}+\beta_{n}N^{ \delta_{n}}}-d_{g}X-d_{w}+d_{w}X\]  
Simplifying it further, we obtain this expression:  
\[\frac{\beta_{n}N^{\delta_{n}-1}}{\beta_{g}(NX)^{\delta_{g}}+\beta_{n}N^{ \delta_{n}}}=\frac{d_{g}X+d_{w}(1-X)}{eX}=\frac{d_{g}-d_{w}}{e}+\frac{d_{w}}{ eX}\]  
which gives  
\[\frac{\beta_{n}N^{\delta_{n}}}{\beta_{g}(NX)^{\delta_{g}}+\beta_{n}N^{\delta _{n}}}=N\bigg{[}\frac{d_{g}-d_{w}}{e}+\frac{d_{w}}{eX}\bigg{]}. \tag{14}\]  
Notice that  
\[\frac{\beta_{n}N^{\delta_{n}}-\beta_{g}N_{g}^{\delta_{g}}}{\beta_{n}N^{ \delta_{n}}+\beta_{g}N_{g}^{\delta_{g}}}=\frac{2\beta_{n}N^{\delta_{n}}}{ \beta_{n}N^{\delta_{n}}+\beta_{g}N_{g}^{\delta_{g}}}-1.\]  
Based on (13), then we obtain the following equation:  
\[\frac{2\beta_{n}N^{\delta_{n}}}{\beta_{n}N^{\delta_{n}}+\beta_{g}N_{g}^{\delta _{g}}}-1=\frac{d_{g}NX}{1-X}\]which gives  
\[\frac{\beta_{n}N^{\beta_{n}}}{\beta_{g}(NX)^{\delta_{g}}+\beta_{n}N^{\beta_{n}}}= \frac{d_{g}NX}{2(1-X)}+1/2. \tag{15}\]  
Therefore, from (14), we have the following  
\[N\bigg{[}\frac{d_{g}-d_{w}}{e}+\frac{d_{w}}{eX}\bigg{]}=\frac{d_{g}NX}{2(1-X)}+ 1/2.\]  
It follows  
\[0 = eN_{g}\frac{\beta_{n}N^{\delta_{n}}}{\beta_{g}N_{g}^{\delta_{g} }+\beta_{n}N^{\delta_{n}}}-d_{g}N_{g}N-d_{w}(N-N_{g})N \tag{12}\]0  
Substituting \(X=\frac{N_{g}}{N}\) and \(N=g(X)\) of (16) to the Equation (15), we obtain the following after some arrangement  
\[ \begin{array}{ll}\frac{\beta_{n}}{\beta_{g}N^{\beta_{g}-d_{n}}X^{\beta_{g}}+ \beta_{n}}&=\frac{d_{g}NX}{2(1-X)}+1/2\\ &=\frac{ed_{g}X^{2}}{2[-X^{2}(ed_{g}+2d_{g}-2d_{w})+2X(d_{g}-2d_{w})+2d_{w}]} +1/2\\ &=:f(X),\end{array}  \tag{17}\]  
which gives  
\[N^{\delta_{g}-\delta_{n}}=\frac{\beta_{n}[1-f(X)]X^{-\delta_{g}}}{\beta_{g}f( X)},\]  
or equivalently,  
\[N=\phi(X):=\left(\frac{\beta_{n}[1-f(X)]X^{-\delta_{g}}}{\beta_{g}f(X)}\right)^{ \frac{1}{g-\beta_{n}}}\]  
provided that \(\delta_{g}\
eq\delta_{n}\). Thus, when \(\delta_{g}\
eq\delta_{n}\), the positive equilibrium \((N^{*},N^{*}_{g})\) occurs if and only if \(X^{*}=\frac{N^{*}_{g}}{N^{*}}\in(0,1)\) solves the equation \(\phi(X)=g(X)\) subject to \(g(X)>0\) and \(0<f(X)<1\).