## Appendix A  
**Proof.** It is noted that \(\varsigma_{i}(0)>\frac{5}{4}\), which results in \(\varsigma_{i}(0)-\frac{t}{t_{f}}>0\), for \(t\in[0,t_{f})\), and since \(\varsigma_{i}(t_{f})>0\), \(\varsigma_{i}(t)>0\) is obvious. Since \(\varsigma_{i}(t)=-\frac{1}{t_{f}}e^{(1-\frac{t_{f}}{t_{f}-t})}-\frac{\varsigma _{i}(0)t_{f}-t}{(tIf \(t\geq t_{f}\), \(\frac{d^{q}\zeta_{i}(t)}{dt^{q}}\) is continuous, one yields  
\[\lim_{t\to t_{f}^{+}}\frac{d^{q}\zeta_{i}(t)}{dt^{q}}=0,\;q=1,2\cdots \tag{11}\]  
If \(0\leq t<t_{f}\), differentiating (12) with respect to \(t\) yields  
\[\frac{d\zeta_{i}(t)}{dt}=[\frac{t}{t_{f}}-\varsigma_{i}(0)]m(t)-\frac{1}{t_{f} }n(t) \tag{12}\]  
where \(m(t)=e^{\frac{t}{t-t_{f}}}\frac{t_{f}}{(t_{f}-t)^{2}}\) and \(n(t)=e^{\frac{t}{t-t_{f}}}\). Define \(p=\frac{1}{t_{f}-1}\), based on [PERSON] rule, one yields  
\[\lim_{t\to t_{f}^{-}}m(t)=t_{f}\lim_{p\rightarrow+\infty}\frac{p^{2}}{e^{-1+t _{f}p}}=t_{f}\lim_{p\rightarrow+\infty}\frac{2}{t_{f}^{2}e^{-1+t_{f}p}}=0\]  
Similarly, one has \(\lim_{t\to t_{f}^{-}}n(t)=0\), then it is obtained that  
\[\lim_{t\to t_{f}^{-}}\frac{d\varsigma_{i}(t)}{dt}=0 \tag{13}\]  
It follows from (11) and (13) that \(\varsigma_{i}(t)\) is differentiable and \(\frac{d\varsigma_{i}(t)}{dt}\) is continuous.  
Next, differentiating (12) with respect to \(t\) obtains  
\[\begin{split}\frac{d^{2}\varsigma_{i}(t)}{dt^{2}}&= \frac{1}{t_{f}}[m(t)-\frac{dn(t)}{dt}]+[\frac{t}{t_{f}}-\varsigma_{i}(0)] \frac{dm(t)}{dt}\\ &=\frac{2}{t_{f}}m(t)+[\frac{t}{t_{f}}-\varsigma_{i}(0)]\frac{dm( t)}{dt}\end{split} \tag{14}\]  
by noting that \(\frac{dn(t)}{dt}=-m(t)\). Meanwhile, one has \(\frac{dn(t)}{dt}=-m(t)\), then \(\lim_{t\to t_{f}^{-}}\frac{dn(t)}{dt}=0\). Meanwhile, one has  
\[\lim_{t\to t_{f}^{-}}\frac{dm(t)}{dt}=\lim_{t\to t_{f}^{-}}e^{\frac{t}{t-t _{f}}}[\frac{2}{(t_{f}-t)^{3}}-\frac{t_{f}^{2}}{(t_{f}-t)^{4}}]=0\]  
Combined with (11), then it is obtained that  
\[\lim_{t\to t_{f}^{-}}\frac{d^{2}\varsigma_{i}(t)}{dt^{2}}=\lim_{t\to t_{f}^{+ }}\frac{d^{2}\varsigma_{i}(t)}{dt^{2}}=0 \tag{15}\]  
It is verified that \(\varsigma_{i}(t)\) is twice differentiable and \(\frac{d^{2}\varsigma_{i}(t)}{dt^{2}}\) is continuous. Similarly, we can obtain  
\[\frac{d^{q}\varsigma_{i}(t)}{dt^{q}}=\frac{q}{t_{f}}\frac{d^{q-2}m(t)}{dt^{q-2 }}+[\frac{t}{t_{f}}-\varsigma_{i}(0)]\frac{d^{q-1}m(t)}{dt^{q-1}} \tag{16}\]for \(q\geq 2\). It is noted that \(\frac{d^{q-2}m(t)}{dt^{q-2}},\ q\geq 3\) is a polynomial on \(\frac{n(t)}{(t-t_{f})^{2}},c\geq 3\). Owing to the fact that  
\[\lim_{t\to t_{f}^{-}}\frac{e^{t^{-1}_{-f}}}{(t-t_{f})^{c}} =\lim_{t\to t_{f}^{-}}[\frac{\frac{1}{(t-t_{f})^{c}}}{e^{t^{-1}_{f -1}}}]=-\frac{c}{t_{f}}\lim_{t\to t_{f}^{-}}[\frac{\frac{1}{(t-t_{f})^{c-1}}}{e ^{t^{-1}_{f}}}]\] \[=\frac{\prod_{i=0}^{c-3}\left(-c+i\right)}{t_{f}^{c-2}}\lim_{t \to t_{f}^{-}}[\frac{\frac{1}{(t-t_{f})^{2}}}{e^{t^{\frac{1}{f-1}}}}]\] \[=\frac{\prod_{i=0}^{c-3}\left(-c+i\right)}{t_{f}^{c-1}}\lim_{t \to t_{f}^{-}}m(t)=0\] it can be deduced that \[\lim_{t\to t_{f}^{-}}\frac{d^{q}\varsigma_{i}(t)}{dt^{q}}=0,\ q=1,2,\cdots.\] (10) It follows from (11) and (10) that \(\lim_{t\to t_{f}^{-}}\frac{\partial t_{\varsigma_{i}}(t)}{\partial t }=\lim_{t\to t_{f}^{-}}\frac{\partial t_{\varsigma_{i}}(t)}{\partial t }=0\) with \(q\geq 1\). Therefore, \(\varsigma_{i}(t)\) is infinite times differentiable, and the infinite-order partial derivative of \(\varsigma_{i}(t)\) is continuous. In other words, \(\varsigma_{i}(t)\) is a smooth function. The proof is completed.