## 2 Problem Formulation
### Control Problem Formulation
\right], \tag{17}\]  
where \(\dot{e}_{ik,I1}=e^{q_{21}/p_{21}}_{ik,D_{11}}\), \(e_{ik,I2}=e^{q_{21}/p_{22}}_{ik,D_{12}}\). \(\gamma_{ik,1}\), \(\gamma_{ik,2}\), \(\lambda_{ik,1}\), \(\lambda_{ik,2}\) are all positive constants. \(p_{kj}>q_{Kj}\), here \(K\), \(j=1\), \(2\), \(p_{Kj}\) and \(q_{Kj}\) are all odd positive constants. **Theorem 1. **_Considering the derivative and integral terminal sliding mode surface \(S_{ik}(t)\) with the fractional function, the state error of formation system can reach the equilibrium point e = 0 at the limited time_  
\[T=\max_{K=1,2}\Biggl{(}\frac{|e_{D1K}(0)|^{1-q_{2K}/q_{2K}}}{\kappa_{K}(1-q_{2 K}/q_{2K})}+\frac{|e_{D0K}(t_{1K})|^{1-q_{1K}/q_{1K}}}{\beta_{K}(1-q_{1K}/q_{1K})} \Biggr{)}. \tag{18}\]  
_Here, \(t_{1K}\) (\(K=1\), 2) is the reaching time of terminal slide mode \(e_{D1K}\)._  
**Proof. ** From (17), the sliding mode \(s_{ik}\) starts on \(t=0\). Then, the equations \(e_{ik,D_{11}}=-\lambda_{ik,1}e_{ik,I1}\) and \(e_{ik,D_{12}}=-\lambda_{ik,2}e_{ik,I2}\) can always hold true by control design. Subsequently, substituting \(e_{ik,D_{11}}=-\lambda_{ik,1}e_{ik,I1}\), \(e_{ik,D_{12}}=-\lambda_{ik,2}e_{ik,I2}\) into \(\dot{e}_{ik,I1}=e^{q_{21}/p_{21}}_{ik,D_{11}}\) and \(e_{ik,I2}=e^{q_{22}/p_{22}}_{ik,D_{12}}\), respectively, yields  
\[\dot{e}_{ik,I1}(t)=-\lambda^{q_{21}/p_{21}}_{ik,I1}e^{q_{21}/p_{21}}_{ik,I1}, \ \ \dot{e}_{ik,I2}(t)=-\lambda^{q_{22}/p_{22}}_{ik,I2}e^{q_{22}/p_{22}}_{ik,I2}. \tag{19}\]  
\(\square\)  
The converge time of sliding mode \(e_{ik,I1}\) and \(e_{ik,I2}\) can obtained by solving (19). \[t_{11}=\frac{|e_{ik,D11}(0)|^{1-q_{21}/p_{21}}}{\lambda_{ik,1}(1-q_{21}/p_{21} )},\ \ \ t_{12}=\frac{|e_{ik,D12}(0)|^{1-q_{22}/p_{22}}}{\lambda_{ik,2}(1-q_{22}/p_{22})} \tag{20}\]In sliding mode \(s_{ik}=0\), \(e_{ik,D_{11}}=-\lambda_{ik,1}e_{ik,I1}\), and \(e_{ik,D_{12}}=-\lambda_{ik,2}e_{ik,I2}\) can always hold true. Therefore, the reaching time of \(e_{ik,D_{11}}\) and \(e_{ik,D_{12}}\) are the same as the convergence time of \(e_{ik,I1}\) and \(e_{ik,I2}\), respectively. When \(e_{ik,D_{11}}=e_{ik,D_{12}}=0\), \(e_{ik,I1}\) and \(e_{ik,I2}\) will converge to zero successfully. At \(t=t_{1K}\) (\(K\) = 1, 2), the \(e_{ik,I1}\) and \(e_{ik,I2}\) are formulated by  
\[\dot{e}_{ik,D01}(t_{11})=-\gamma_{ik,1}^{q_{11}/p_{11}}e_{ik,D01}^{q_{11}/p_{11 }}(t_{11}),\ \ \ \dot{e}_{ik,D02}(t_{12})=-\gamma_{ik,2}^{q_{12}/p_{12}}e_{ik,D02}^{q_{12}/p_{12 }}(t_{12}). \tag{21}\]  
Solving (21), the \(\dot{e}_{ik,D01}\), \(\dot{e}_{ik,D02}\) from \(\dot{e}_{ik,D01}(t_{11})\), \(\dot{e}_{ik,D02}(t_{12})\) to \(\dot{e}_{ik,D01}=\dot{e}_{ik,D01}=0\) will spend the time  
\[t_{01}=\frac{\left|e_{ik,D01}(0)\right|^{1-q_{11}/p_{11}}}{\gamma_{ik,1}(1-q_{ 11}/p_{11})},\ \ \ t_{02}=\frac{\left|e_{ik,D02}(0)\right|^{1-q_{12}/p_{12}}}{\gamma_{ik,2}(1-q_ {12}/p_{12})}. \tag{22}\]  
Since the sliding mode \(s_{ik}=0\) consists of the derivative term and integral term, the time \(T_{K}\) spending from \(s_{ik,K}=0\) (\(K\) = 1, 2) to \(e_{ik,K}=0\) (\(K\) = 1, 2) is the summation of the two terms. Since the fact that each sliding mode \(s_{ik,1}\), \(s_{ik,2}\) is independent, the time spent for equilibrium point is the max of the \(T_{K}\). **Hypothesis 2**. : \(\parallel d_{ik}\parallel_{\infty}\leq\overline{d}_{ik}\)_. It means that the unmatched uncertainties of formation system (13) have a boundary._  
Differentiating the sliding function \(s_{ik}\) with the respect to time \(t\), and substituting (13) into the derivative of \(s_{ik}\) can get  
\[\dot{s}_{ik}=diag\bigg{[}\frac{p_{11}}{q_{11}}\dot{e}_{D01}^{(p_{11}/q_{11}-1 )},\frac{p_{12}}{q_{12}}\dot{e}_{D02}^{(p_{12}/q_{12}-1)}\bigg{]}\Big{[}d_{ik} +H(x_{ik},d_{ik})u_{k}-\ddot{y}_{ik}^{d}\Big{]}+\phi_{ik}, \tag{23}\]  
where \(y_{ik}^{d}=\left[ \begin{array}{cc}y_{ik,1}^{d}&y_{ik,2}^{d}\end{array} \right] ^{T}\) denotes the desired distance and angle between the leader robot and follower robots. \(\phi_{ik}=\left[ \begin{array}{cc}\phi_{ik,1}&\phi_{ik,2}\end{array} \right] ^{T}\).