## 4 Result and Discussion  
### Stability Analysis  
According to the stability theorem of the differential equations, if the strategy of the coal mining enterprise has a stable state, it needs to meet \(F(p,\,q)=0\). So, we set formula (11) equal to zero to get \(p=0\) or \(p=1\) and \(\omega(q)H+\omega(1-q)I=0\); according to [PERSON]'s expansion, we can get \(q=\left[\frac{I}{I-H}\right]^{\frac{1}{f}}\). Similarly, we set formula (14) equal to 0 to get \(q=0\) or \(q=1\) and \(p=\left[\frac{T}{T-S}\right]^{\frac{1}{f}}\). Therefore, five system evolution equilibrium points of the replicated dynamic equations, composed of (11) and (14), can be obtained: \(E_{1}(0,0)\), \(E_{2}(0,1)\), \(E_{3}(1,0)\), \(E_{4}(1,1)\), if and only if:  
\[E_{5} = \left(\frac{T}{T-S}\right)^{\frac{1}{f}}\in[0,1],\left(\frac{I}{ T-H}\right)^{\frac{1}{f}}\in[0,1],\]  
we get the equilibrium point \(E_{5}\left(\left(\frac{T}{T-S}\right)^{\frac{1}{f}},\left(\frac{I}{I-H}\right) ^{\frac{1}{f}}\right)\). The Jacobian matrix of the system obtained by Equations (11) and (14) is as follows:  
\[J(p,q)=\left[ \begin{array}{ccc}\frac{\partial F_{1}(pA)}{ \partial p}&\frac{\partial F_{1}(pA)}{\partial q}\\ \frac{\partial F_{2}(pA)}{\partial p}&\frac{\partial F_{2}(pA)}{\partial q} \end{array} \right]=\left[ \begin{array}{ccc}(1-2p)(\omega(q)H+\omega(1-q)I)& (p-p^{2})(\frac{\mathrm{d}\omega(q)}{\mathrm{d}q}H+\frac{\mathrm{d}\omega(1-q )}{\mathrm{d}q}I)\\ (q-q^{2})(\frac{\mathrm{d}\omega(p)}{\mathrm{d}p}S+\frac{\mathrm{d}\omega(1-p )}{\mathrm{d}p}T)&(1-2q)(\omega(p)S+\omega(1-p)T)\end{array} \right]\] \[DetI(p,q)=\frac{\partial F_{1}(pA)}{\partial p}\cdot\frac{ \partial F_{2}(pA)}{\partial q}-\frac{\partial F_{1}(pA)}{\partial q}\cdot \frac{\partial F_{2}(pA)}{\partial p}=(1-2p)(\omega(q)H+\omega(1-q)I)(1-2q)\] \[(\omega(p)S+\omega(1-p)T)-(p-p^{2})(\frac{\mathrm{d}\omega(q)}{ \mathrm{d}q}H+\frac{\mathrm{d}\omega(1-q)}{\mathrm{d}q}I)(q-q^{2})(\frac{ \mathrm{d}\omega(p)}{\mathrm{d}p}S+\frac{\mathrm{d}\omega(1-p)}{\mathrm{d}p}T)\]According to the criteria for determining stability specified by [PERSON] in the article, the sign of the determinant and the trace of the matrix _J(p, q)_ can be used to determine the stability of this point. The point in question is only considered a stable node when the determinant of the matrix \(J\) is greater than zero and the trace is less than zero [47]. Since the trace value of the Jacobian matrix of the equilibrium point \(E_{5}\) is zero, in order to further judge the positive and negative properties of the determinant, the following calculations are carried out in this paper. \[ \begin{array}{l}\frac{\mathrm{d}F_{1}(p,q)}{\mathrm{d}q}=(p-p^{2})(\frac{ \mathrm{d}\omega(q)}{\mathrm{d}q}H+\frac{\mathrm{d}\omega(1-q)}{\mathrm{d}q}I) \\ \frac{\mathrm{d}F_{2}(p,q)}{\mathrm{d}p}=(q-q^{2})(\frac{\mathrm{d}\omega(p)}{ \mathrm{d}p}S+\frac{\mathrm{d}\omega(1-p)}{\mathrm{d}p}T)\end{array}  \tag{17}\]  
Since \(0<\left(\frac{T}{T-5}\right)^{\frac{1}{2}}<1,0<\left(\frac{I}{T-H}\right)^{ \frac{1}{2}}<1\), then \(\frac{\mathrm{d}F_{1}(p,q)}{\mathrm{d}q}>0,\frac{\mathrm{d}F_{2}(p,q)}{\mathrm{ d}p}>0\). The value of the determinant of the equilibrium point \(E_{5}\) is less than zero. \(Det(I_{E_{5}})=M=-\frac{\mathrm{d}F_{1}(p,q)}{\mathrm{d}q},\frac{\mathrm{d}F_{2 }(p,q)}{\mathrm{d}p}<0\). Thus, this point is the saddle point. Figure 2: Evolution strategy analysis of system evolution game. det \(J\) sign represents the symbol of the determinant of the matrix \(J\) (_p, q_). tr \(J\) sign represents the sign of the trace of the matrix _J_(_p, q_). Stability is presented in three different situations here. \(\bullet\) represents stable node; \(\bigstar\) represents saddle point; \(\bigstar\) represents instability point. After comprehensively analyzing the values of the remaining four equilibria, this paper finds that the evolutionary strategy of each equilibrium changes when the stability conditions change.