# Dynamic Investigations of Shared Bicycle Operators' Competition Based on Profit Maximization
## 3 Model Description
### Stability Analysis of Cournot Competition  
Let \(x_{1}^{C}(t+1)=x_{1}^{C}(t)\); by solving nonlinear algebraic systems (15)  
\[ \begin{cases}v_{1}[(1-k-\beta_{1})-2x_{1}^{C}-gx_{2}^{C}-\mu_{1}-\mu_{2}D]=0\\ v_{2}[(1-k-\beta_{2}+S)-2x_{2}^{C}-gx_{1}^{C}-\mu_{1}-\mu_{2}D]=0\end{cases}  \tag{15}\]  
we can obtain four equilibrium points of the system (14): \(E_{1}=(0,0)\), \(E_{2}=\bigg{(}0,\dfrac{1}{2}A_{2}\bigg{)}\), \(E_{3}=\bigg{(}\dfrac{1}{2}A_{1},0\bigg{)}\) and \(E_{4}=\bigg{(}\dfrac{1}{4-g^{2}}(2A_{1}-gA_{2}),\dfrac{1}{4-g^{2}}(2A_{2}-gA_ {1})\bigg{)}\). Where  
\[A_{1}=1-k-\beta_{1}-\mu_{1}-\mu_{2}D>0,\]  
\[A_{2}=1-k-\beta_{2}-\mu_{1}-\mu_{2}D+S>0.\]  
It is evident that the aforementioned system (14) possesses two stable equilibrium points, \(E_{4}\) and the other two equilibrium points are all unstable. The Jacobian matrix of system (14) is  
\[J\Big{(}x_{1}^{C},x_{2}^{C}\Big{)}= \begin{bmatrix}1+v_{1}\big{(}A_{1}-4x_{1}^{ C}-gx_{2}^{C}\big{)}&-gv_{1}x_{1}^{C}\\ -gv_{2}x_{2}^{C}&1+v_{2}\big{(}A_{2}-4x_{2}^{C}-gx_{1}^{C}\big{)}\end{bmatrix} \]  
**Theorem 1**. : _The equilibrium point \(E_{1}\) is an unstable equilibrium point._  
Proof. : The Jacobian matrix of the system (14) at equilibrium point \(E_{1}\) is  
\[J(E_{1})= \begin{bmatrix}1+v_{1}A_{1}&0\\ 0&1+v_{2}A_{2}\end{bmatrix} \]  
and the corresponding eigenvalues are  
\[\lambda_{1}=(1+v_{1}A_{1})>0,\lambda_{2}=(1+v_{2}A_{2})>0,\]  
therefore, \(E_{1}\) is an unstable equilibrium point. **Theorem 2**. : _The equilibrium point \(E_{2}\) is a stable equilibrium point if_  
\[ \begin{cases}1+v_{1}(A_{1}-0.5 gA_{2})<0,\\ 1-v_{2}A_{2}<0.\end{cases}  \tag{16}\]Proof. : The Jacobian matrix of the system (14) at equilibrium point \(E_{2}\) is  
\[J(E_{2})= \begin{bmatrix}1+v_{1}(A_{1}-0.5 gA_{2})&0\\ -0.5 gv_{2}A_{2}&1-v_{2}A_{2}\end{bmatrix} \]  
The corresponding eigenvalues are \(\lambda_{1}=1+v_{1}(A_{1}-0.5 gA_{2})\), \(\lambda_{2}=1-v_{2}A_{2}\). When \(|\lambda_{1}|<0\) and \(|\lambda_{2}|<0\), the Equation (16) can be obtained. Theorem 3.2: _The equilibrium point \(E_{3}\) is an unstable equilibrium point._  
Proof. : The Jacobian matrix of the system (14) at equilibrium point \(E_{3}\) is  
\[J(E_{3})= \begin{bmatrix}1-v_{1}A_{1}&-0.5 gv_{1}A_{1}\\ 0&1+v_{2}(A_{2}-0.5 gA_{1})\end{bmatrix} \]  
The corresponding eigenvalues are \(\lambda_{1}=1-v_{1}A_{1}\) and \(\lambda_{2}=1+v_{2}(A_{2}-0.5 gA_{1})\). Since \(A_{2}>0.5 gA_{1}\), we have\(|\lambda_{1}|<1\), \(|\lambda_{2}|>1\). Therefore, \(E_{3}\) is a stable equilibrium point. Theorem 3.3: _The equilibrium point \(E_{4}\) is a stable equilibrium point if_  
\[ \begin{cases}v_{1}v_{2}<\dfrac{g^{2}+4}{2g(A_{1}^{2}+A_{2}^{2})},\\ A_{1}v_{1}+A_{2}v_{2}>\dfrac{g}{2}(A_{1}v_{2}+A_{2}v_{1}).\end{cases}  \tag{17}\]  
Proof. : The Jacobian matrix of the system (14) at equilibrium point \(E_{4}\) is  
\[J(E_{4})= \begin{bmatrix}1+\dfrac{2v_{1}}{4-g^{2}}(gA_{2}-2A_{1})&\dfrac{gv_{1 }}{4-g^{2}}(gA_{2}-2A_{1})\\ \dfrac{gv_{2}}{4-g^{2}}(gA_{1}-2A_{2})&1+\dfrac{2v_{2}}{4-g^{2}}(gA_{1}-2A_{2} )\end{bmatrix} \]  
Let \(Tr(J)\) denote the trace of the Jacobian matrix \(J(E_{4})\) and \(Der(J)\) denote the determinant of the Jacobian matrix \(J(E_{4})\), then  
\[ \begin{cases}Tr(J)=2+\dfrac{2}{4-g^{2}}[v_{1}(gA_{2}-2A_{1})+v_{2}(gA_{1}-2A_{ 2})]\\ Der(J)=1+\dfrac{1}{4-g^{2}}[(g^{2}+4)A_{1}A_{2}-2g(A_{1}^{2}+A_{2}^{2})v_{1}v_{ 2}+2g(A_{1}v_{2}+A_{2}v_{1})-4(A_{1}v_{1}+A_{2}v_{2})]\end{cases} \]  
Based on the Jury condition [34; 35], the local stability of the equilibrium point must meet the following criteria:  
\[ \begin{cases}1+Tr(J)+Der(J)>0\\ 1-Tr(J)+Der(J)>0\\ 1-Der(J)>0\end{cases}  \tag{18}\]  
By substituting the expressions of \(Tr(J)\) and \(Der(J)\) into Equation (18), stability condition (17) for equilibrium point \(E_{4}\) can be obtained.