## 3 Modeling Methodology
### Solvability Analysis of the Proposed CLC Control Strategy
 \tag{47}\]  
Let \(a_{M}(k)=a_{min}\), and if the speed difference between vehicles C and M satisfies \(v_{C}(k)-v_{M}(k)\geq a_{min}dt/2\) at the same time, the following inequality will be held after multi-step amplification. \[ \begin{array}{ll}S_{CM}(k+1)-&S_{safe,CM}(k+1)LPFA(k+1)\\ &\geq[v_{C}(k)-v_{M}(k)]dt-a_{M}(k)dt^{2}/2=0\end{array}  \tag{48}\]  
Hence, when \(y_{M}(k)>d/2\), the longitudinal distance constraint \(S_{CM}(k+1)\geq S_{safe,CM}(k+1)LPFA(k+1)\) will be satisfied at \(k+1\) step. To sum up, the analysis, if \(v_{C}(k)-v_{M}(k)\geq a_{min}dt/2\) is satisfied, \(S_{CM}(k+1)\geq S_{safe,CM}(k+1)LPFA(k+1)\) will hold, given that \(S_{CM}(k)\geq S_{safe,CM}(k)LPFA(k)\). The proof of Lemma 1 is closed. \(\square\)  
**Lemma 2**. : _Given \(S_{BA}(k)\geq S_{safe,BA}(k)LPFA(k)\) is satisfied at \(k\) step, a feasible solution \(a_{A}(k)=a_{min}\) can make \(S_{BA}(k+1)\geq S_{safe,BA}(k+1)LPFA(k+1)\) hold, if the speed difference between vehicles \(B\) and \(A\) satisfies \(v_{B}(k)-v_{A}(k)\geq a_{min}dt/2\) for any step \(k\) (\(k=1,\ 2,\ \ldots,\ N\))._  
Proof. : The proof process of lemma 2 is similar to that of Lemma 1 and will not be repeated here. \(\square\)  
**Lemma 3**. : _Given \(S_{BM}(k)\geq S_{safe,BM}(k)LPFB(k)\) is satisfied at \(k\) step, a feasible solution \(a_{M}(k)=a_{min}\) can make \(S_{BM}(k+1)\geq S_{safe,BM}(k+1)LPFB(k+1)\) hold, if the speed difference between vehicles \(B\) and \(M\) satisfies \(v_{B}(k)-v_{M}(k)\geq a_{min}dt/2+d_{0}L_{3}\Delta y_{M_{min}}\) for any step \(k\) (\(k=1,\ 2,\ \ldots,\ N\))._  
Proof. : Since \(LPFB(*)\) is also a segmented function, it needs to be discussed separately. When \(0\leq y_{M}(k)\leq d/2\), the following relationship exists. \[S_{BM}(k)\geq S_{safe,BM}(k)LPFB(k)=d_{0}L_{3}y_{M}(k) \tag{49}\]  
Equation (49) can be rewritten in the following form. \[S_{BM}(k)-S_{safe,BM}(k)LPFB(k)=S_{BM}(k)-d_{0}L_{3}y_{M}(k)\geq 0 \tag{50}\]  
According to the mathematical expression of \(LPFB(*)\), the following Equation (51) can be obtained:  
\[S_{safe,BM}(k+1)LPFB(k+1)=d_{0}L_{3}y_{M}(k+1)=d_{0}L_{3}[y_{M}(k)+\Delta y_{M }(k)] \tag{51}\]  
By substituting (4) and (51), the following Equation (52) can be obtained. \[ \begin{array}{ll}S_{BM}(k+1)-&S_{safe,BM}(k+1)LPFB(k+1)\\ &=S_{BM}(k+1)-d_{0}L_{3}[y_{M}(k)+\Delta y_{M}(k)]\\ &=S_{BM}(k)+[v_{B}(k)-v_{M}(k)]dt-a_{M}(k)dt^{2}/2\\ &-d_{0}L_{3}[y_{M}(k)+\Delta y_{M}(k)]\\ &\geq[v_{B}(k)-v_{M}(k)]dt-a_{M}(k)dt^{2}/2-d_{0}L_{3}\Delta y_{M}(k)\end{array}  \tag{52}\]  
Let \(a_{M}(k)=a_{min}\) and \(\Delta y_{M}(k)=\Delta y_{M_{min}}\), where \(\Delta y_{M_{min}}\) is the minimal single-step lateral movement distance, and if the speed difference between vehicles B and M satisfies \(v_{B}(k)-v_{M}(k)\geq a_{min}dt/2+d_{0}L_{3}\Delta y_{M_{min}}\) at the same time, the following inequality (53) will be held after the multi-step amplification. \[ \begin{array}{ll}S_{BM}(k+1)&-S_{safe,BM}(k+1)LPFB(k+1)\\ &\geq[v_{B}(k)-v_{M}(k)]dt-a_{M}(k)dt^{2}/2-d_{0}L_{3}\Delta y_{M}(k)\\ &=a_{min}dt^{2}/2+d_{0}L_{3}\Delta y_{M_{min}}-a_{min}dt^{2}/2-d_{0}L_{3} \Delta y_{M_{min}}=0\end{array}  \tag{53}\]  
Thus, when \(0\leq y_{M}(k)\leq d/2\), the longitudinal distance constraint \(S_{BM}(k+1)\geq S_{safe,BM}(k+1)LPFB(k+1)\) will be satisfied at \(k+1\) step. When \(y_{M}(k)>d/2\), the following relationship exists. \[S_{BM}(k)\geq S_{safe,BM}(k)LPFB(k)=d_{0}L_{4} \tag{54}\]  
Equation (54) also can be rewritten as following Equation (55). \[S_{BM}(k)-S_{safe,BM}(k)LPFB(k)=S_{BM}(k)-d_{0}L_{4}\geq 0 \tag{55}\]When \(y_{M}(k)>d/2\), by substituting (4) and (55), the following inequality (56) can be obtained. \[ \begin{array}{ll}S_{BM}(k+1)-&S_{safe,BM}(k+1)LPFB(k+1)=S_{BM}(k+1)-d_{0}L_{4}\\ &=S_{BM}(k)+[v_{B}(k)-v_{M}(k)]dt-a_{M}(k)dt^{2}/2-d_{0}L_{4}\\ &\geq[v_{B}(k)-v_{M}(k)]dt-a_{M}(k)dt^{2}/2\end{array}