## 3 Linear Stability Analysis  
### Stable Analysis of the Model  
When all vehicles drive on the road at the same headway and speed, the traffic system is in an equilibrium state. Once the driving state of a vehicle in the fleet changes, an \"interference signal\" is created, leading to a speed fluctuation of the rear vehicles. If this fluctuation weakens over time while spreading until the traffic system regains equilibrium, then the traffic system is considered stable; else, it is considered unstable. Therefore, it is very important to analyze the stability of the car-following model. In previous studies, the Laplace transforms, and the difference equation are usually used to analyze the stability of the car-following model. Due to the complexity of the Laplace transform process, it is more suitable for the stability study of homogeneous traffic flow. Therefore, this paper chooses to use the difference equation to linearly analyze the stability of the extended IDM model. Proceed as follows:  
Step 1. Initial state settings  
Simplifying Formulae (1) and (2) gives:  
\[a_{n}(t)=f(v_{n}(t),s_{n}(t),\ldots s_{n+i-1}(t),\Delta v_{n},\ldots,\Delta v_{ n+i-1},a_{n}(t),\ldots,a_{n+i-1}(t)) \tag{10}\]  
Assuming that the traffic system is in equilibrium in the initial state and all vehicles are traveling at the same headway spacing and velocity, the position of any vehicle in the traffic flow at time \(t\) can be expressed as:  
\[x_{n}^{0}(t)=(N-n)\overline{s}+\overline{v}t\ \ \ \ \ n=1,2,\ldots N \tag{11}\]  
where \(N\) is the total number of vehicles in the fleet. The fluctuation in the steady traffic flow can be expressed as:  
\[y_{n}(t)=ce^{i\alpha_{k}n+zt}=x_{n}(t)-x_{n}^{0}(t),\ y_{n}(t)\to 0 \tag{12}\]The position and headway spacing of the vehicle in the traffic flow can be expressed as:  
\[x_{n}(t)=y_{n}(t)+x_{n}^{0}(t) \tag{13}\]  
\[\Delta x_{n}(t)=\Delta y_{n}(t)+x_{n}^{0}(t) \tag{14}\]  
Taking the first and second derivatives of Formula (13) gives:  
\[v_{n}(t)=y_{n}^{{}^{\prime}}(t)+\overline{x_{n}}(t) \tag{15}\]  
\[a_{n}(t)=\dot{y_{n}^{{}^{\prime}}}(t) \tag{16}\]  
Substituting Formula (16) into Formula (10) gives:  
\[y_{n}^{{}^{\prime\prime}}(t+t_{d})=f(v_{n}(t),s_{n}(t),\ldots s_{n+i-1}(t), \Delta v_{n},\ldots,\Delta v_{n+i-1},a_{n}(t),\ldots,a_{n+i-1}(t)) \tag{17}\]  
Step 2. Linearize and derive the difference equation  
Linearize Formula (17) to obtain:  
\[y_{n}^{{}^{\prime\prime}}(t+t_{d})=f_{n}^{\rm s}\sum\limits_{q=1}^{Q}\varphi_ {q}(t)\big{(}y_{n-q}(t)-y_{n-q+1}(t)\big{)}+f_{n}^{\rm v}y_{n}^{\prime}(t)+f_{ n}^{\rm\Delta v}\rho\sum\limits_{q=1}^{Q}\alpha_{q}\Big{(}y_{n-q+1}^{\prime}(t)-y_{ n-q}^{\prime}(t)\Big{)}+\mu_{n}^{\rm\alpha}\sum\limits_{q=1}^{Q}\beta_{q}y_{n-q}^{{}^{ \prime\prime}}(t) \tag{18}\]  
where: \(f_{n}^{\rm s}=\frac{\partial f_{n}}{\partial\rm s}\geq 0\), \(f_{n}^{\rm v}=\frac{\partial f_{n}}{\partial\rm v}\geq 0\), \(f_{\Delta v}^{\rm s}=\frac{\partial f_{n}}{\partial\rm\Delta v}\leq 0\), \(f_{n}^{\rm a}=\frac{\partial f_{n}}{\partial\rm\alpha}\geq 0\). Rewrite Formula (18) into the following difference equation:  
\[ \begin{array}{l}y_{n}^{{}^{\prime}}(t+2t_{d})-y_{n}^{{}^{\prime}}(t+t_{d}) \\ =t_{d}f_{n}^{\rm s}\sum\limits_{q=1}^{Q}\varphi_{q}\big{(}y_{n-q}(t)-y_{n-q+1} (t)\big{)}+f_{n}^{\rm v}(y_{n}(t+t_{d})-y_{n}(t))\\ +\mu_{n}^{\rm\alpha}\sum\limits_{q=1}^{Q}\beta_{q}\Big{(}y_{n-q}^{{}^{\prime} }(t+t_{d})-y_{n-q}^{{}^{\prime}}(t)\Big{)}\\ +\rho f_{n}^{\rm\Delta v}\sum\limits_{q=1}^{Q}\alpha_{q}\big{(}y_{n-q+1}(t+t_{ d})-y_{n-q+1}(t)-y_{n-q}(t+t_{d})+y_{n-q}(t)\big{)}\end{array}