# Ordered-Bipartite Consensus of Multi-Agent Systems under Finite Time Control
## 3 Main Results  
This section will give rigorous proofs for the _MAS_ (1) controlled by (8) to reach _GC_. Theorem 6: _For the MAS_ (8)_, consider the unsigned directed graph \(\mathcal{G}\) as its network topology. If \(\mathcal{G}\) satisfies Assumptions 1-3, then the MAS is divided into \(p\) groups to achieve finite time agreement in an orderly manner._  
Proof: Note that \(\mathcal{G}\) is a graph with an acyclic partition \(\{\mathcal{V}_{1},\mathcal{V}_{2},\ldots,\mathcal{V}_{p}\}\), which means that information is transmitted from \(\mathcal{V}_{i}\) to \(\mathcal{V}_{j}\) only when \(j>i\). Therefore, based on this feature, the proof process in this part is divided into three parts. Firstly, it is proved that the first group \(\mathcal{G}_{1}\) achieves agreement in finite time. This is immediately followed by proving that both the first group \(\mathcal{G}_{1}\) and the second group \(\mathcal{G}_{2}\) achieve finite time agreement. Finally, the result of the second step of the proof is extended to \(\mathcal{G}_{2},\ldots,\mathcal{G}_{p}\). Now, for simplicity, denote that  
\[\left\{ \begin{aligned} x^{[i]}&=(x_{n_{i-1}+1},x_{n_{ i-1}+2},\ldots,x_{n_{i-1}+k_{i}})^{T}\\ y^{[i]}&=(y_{n_{i}+1},y_{n_{i-1}+2},\ldots,y_{n_{i-1}+k_ {i}})^{T},\\ u^{[i]}&=(u_{n_{i-1}+1},u_{n_{i-1}+3},\ldots,u_{n_{i-1}+k_ {i}})^{T}\end{aligned} \right. \tag{9}\]  
where \(n_{j}=\sum\limits_{\ell=1}^{j}k_{\ell}\) and \(k_{\ell}\) is the cardinality of \(\mathcal{V}_{\ell}\). Consider first the trivial case of \(p=1\). From Assumption 2, \(\mathcal{G}\) has a directed spanning tree, which means that the first group system solves the consensus problem in finite time. The next discussion is on the case of \(p=2\)--that is, to show that the agents in \(\mathcal{G}_{1}\) and \(\mathcal{G}_{2}\) reach consensus in finite time, which is a key step in the whole proof process. In this case, \(L= \begin{pmatrix}L_{11}&0\\ L_{21}&L_{22}\end{pmatrix} \) and \(L_{22}=L(A_{11}),L_{22}=L(A_{22})\). Let \(y=Lx\); it is not difficult to obtain \(\dot{y}=-Lu\). Thus, the control law (8) has that  
\[\dot{y}=-L\,\Big{(}\beta sig^{\alpha}(y)+\gamma\,y\Big{)}. \tag{10}\]  
By Lemma 2, the orthogonal space of the column vector space of vector \(y\) is \(\mathbb{R}\pi_{1}\bigoplus\mathbb{R}\pi_{2}\). Thus, there exists a positive column vector \(\omega\,\in\,\mathbb{R}\pi_{1}\bigoplus\mathbb{R}\pi_{2}\) such that \(\omega\,\perp\,y\). For ease of presentation, let \(\omega\) be denoted as \((\omega_{1},\ldots,\omega_{n_{2}})\). Now, consider the Lyapunov candidate  
\[ \begin{split} V_{2}(t)=& y^{T}\operatorname{diag}( \omega_{1},\ldots,\omega_{n_{2}})\,\Big{(}\frac{\beta}{1+\alpha}sig^{\alpha}( y)+\frac{\gamma}{2}y\Big{)}\\ &=\sum\limits_{j=1}^{n_{2}}\omega_{j}\Big{(}\frac{\beta}{1+ \alpha}|y_{j}|^{1+\alpha}+\frac{\gamma}{2}y_{j}^{2}\Big{)}\,\geq\,0\,,\end{split}  \tag{11}\]for which differentiation at \(t\) yields  
\[ \begin{split}\hat{V}_{2}(t)=&\sum_{j=1}^{n_{2}}\omega_{ j}\Big{(}\beta\text{sig}^{\alpha}(y_{j})+\gamma\,y_{j}\Big{)}\,\dot{y}_{j}\\ =&\Big{(}\beta\text{sig}^{\alpha}(y)+\gamma\,y\Big{)} \,\text{diag}(\omega_{1},\dots,\omega_{n_{2}})\,\dot{y}\\ =&-\Big{(}\beta\text{sig}^{\alpha}(y)+\gamma\,y\Big{)} ^{T}\,\text{diag}(\omega_{1},\dots,\omega_{n_{2}})\,L\,\Big{(}\beta\text{sig}^ {\alpha}(y)+\gamma\,y\Big{)}\\ \leq& 0\,.\end{split}  \tag{12}\]  
Since \(\text{diag}(\omega_{1},\dots,\omega_{n_{2}})\) is positive definite, \(V_{2}=0\) implies that \(y=0\) due to the sign of the values of \(\alpha,\beta,\gamma\). Therefore, it is reasonable to assume that \(V_{2}\
eq 0\); further, the following inequalities follow from Lemma 4,  
\[