\section{\@startsection{section}{1}{\z@}%
{18\p@ \@plus 6\p@ \@minus 3\p@}%
{9\p@ \@plus 6\p@ \@minus 3\p@}%
{\large\bfseries\boldmath}}
\renewcommand\subsection{\@startsection{subsection}{2}{\z@}%
{12\p@ \@plus 6\p@ \@minus 3\p@}%
{3\p@ \@plus 6\p@ \@minus 3\p@}%
{\bfseries\boldmath}}
\renewcommand\subsubsection{\@startsection{subsubsection}{3}{\z@}%
{12\p@ \@plus 6\p@ \@minus 3\p@}%
{\p@}%
{\bfseries\boldmath}}
\makeatother

\usepackage{microtype}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{corollary}{Corollary}
\newtheorem{proposition}{Proposition}
\newtheorem{conjecture}{Conjecture}
\newtheorem{algorithm}{Algorithm}

\newtheorem{definition}{Definition}
\newtheorem{remark}{Remark}
\newtheorem{example}{Example}
\newtheorem{problem}{Problem}

\numberwithin{equation}{section}
\allowdisplaybreaks
\parindent=18pt
\renewcommand{\baselinestretch}{1.1}

\begin{document}

\title{The Cop Number of Graphs with Forbidden Induced Subgraphs}
\author{Mingrui Liu
\thanks{Yau Mathematical Sciences Center, Tsinghua University, Beijing 100084, China
({\tt lmr16@mails.tsinghua.edu.cn}). }}

\maketitle
\begin{abstract}
\par\vspace{2mm}

In the game of Cops and Robber, a team of cops attempts to capture a robber on a graph $G$. Initially, all cops occupy some vertices in $G$ and the robber occupies another vertex. In each round, a cop can move to one of its neighbors or stay idle, after which the robber does the same. The robber is caught by a cop if the cop lands on the same vertex which is currently occupied by the robber. The minimum number of cops needed to guarantee capture of a robber on $G$ is called the {\em cop number} of $G$, denoted by $c(G)$.

We say a family $\cal F$ of graphs is {\em  cop-bounded} if there is a constant $M$ so that $c(G)\leq M$ for every graph $G\in \cal F$.
Joret, Kamin\'nski, and Theis [Contrib. Discrete Math. 2010] proved that the class of all graphs not containing a graph $H$ as an induced subgraph is cop-bounded if and only if $H$ is a linear forest;
morerover, $C(G)\leq k-2$ if
if $G$ is induced-$P_k$-free for $k\geq 3$.
In this paper, we consider the cop number of a family of graphs forbidding certain two graphs and generalized some previous results.

\noindent{\bfseries Keywords:} Cops and robbers; Cop number; Vertex-pursuit games; Moving target search

\noindent{\bfseries AMS classification:} 05C57
\end{abstract}

\section{Introduction}
The game of $\textit{Cops and Robber}$ is played on a connected graph $G$ by two players: $\cal C$ and $\cal R$. The player $\cal C$ controls $k$ pieces (cops) for some integer $k\geq 1$, and the player $\cal R$ has only one piece (the robber). In the beginning of the game, $\cal C$ put down
$k$ cops on some vertices of $G$ (not necessary distinct). Then $\cal R$ put down the robber on a vertex of $G$. Then, the two players move their pieces alternately. In each of the succeeding rounds, each piece either moves to a vertex adjacent to its current position or stays still.
We say the game is {\em cop-win} if after a finite rounds, a cop and the robber meet at the same vertex (that is, the cops catch the robber). It is {\rm robber-win} if the robber can avoid being caught by cops after arbitrarily many rounds. Both players have complete information, that is, they know the graph and the positions of all the pieces. In this paper, we only consider the game played on finite undirected graphs.

The most important problem we considered in this game is how many cops we need to capture the robber on a given graph. The minimum number of cops guaranteed to capture the robber on the graph $G$ in finite rounds is called the $\textit{cop number}$ of $G$ and denoted by $c(G)$. Nowakowski and Winkler \cite{MR685631} and Quilliot \cite{name1} characterized the class of graphs with cop number $1$. Aigner and Fromme \cite{MR739593}
proved any planar graph $G$ has cop number at most $3$.
Finding a combinatorial characterization of graphs with cop number $k$ (for $k\geq 2$) is a major open problem in this field. Moreover, determining the cop number for a given graph is also a computationally hard problem \cite{MR2757374}. We even don't know the order of the cop number upon the number of vertices of any graph $G$ \cite{MR890640}. For more information about the game of cops and robber we recommend the book by Bonato and Nowakowski \cite{MR2830217} and the survey by Baird and Bonato \cite{MR2980752}.

\begin{definition}
Let $G_1,G_2,\ldots, G_n$ be some graphs, we say a graph $G$ is \{$G_1$,$G_2$,\ldots, $G_n$\}-free if $G$ does not contain any $G_i$ as an induced subgraph. We say $G$ is ($G_1+G_2+\ldots+G_n$)-free if $G$ does not contain the disjoint union of all the graphs $G_1,G_2,\ldots, G_n$ as an induced subgraph.
\end{definition}

\begin{definition}
For a vertex $v$ in a graph $G$, the closed neighbourhood of $v$, denoted by $N[v]$ is the set of all neighbors of $v$, including $v$ itself. The open neighbourhood of $v$, denoted by $N(v)$ is the set of all neighbors of $v$. What's more, $N[v_1,v_2]$ is the union of $N[v_1]$ and $N[v_2]$.
\end{definition}

\begin{definition}
The graph consisting of a path on $k$ vertices is denoted by $P_k$.
\end{definition}

\begin{definition}
For the graph $G$, the subgraph $G[U]$ is the induced subgraph generated by the subset $U$ of the vertices of $G$.
\end{definition}

Joret, Kaminski and Theis \cite{MR2791289} proved that $k-2$ cops can catch the robber in any $P_k$-free graph for every interger $k\geq 3$. In this paper, we prove this theorem with a new method and use this idea to explore the cop number for a graph $G$ with some forbidden induced structures.

\begin{theorem}\label{thm:Pk}\cite{MR2791289}
Let $G$ be a $P_k$-free graph with $k \geq 3$, then $k-2$ cops can catch the robber in $G$. That is, $c(G)\leq k-2$.
\end{theorem}

The idea we use to prove the above theorem can also be used to prove the following two theorems and we can bound the cop number of graphs with some forbidden induced subgraphs with the cop number.

\begin{theorem}\label{thm:two}
If we forbid the following two structures in any graph $G$, then $c(G)\leq l+1$ for $l \geq 1$.
\begin{figure}[hbt]
\begin{center}
\tikzstyle{vertex}=[circle,fill=black,inner sep=1.5pt]
\tikzstyle{new}=[circle,fill=red,inner sep=1.5pt]
\tikzstyle{blue}=[circle,fill=blue, inner sep=1.5pt]
\tikzstyle{green}=[circle,fill=green, inner sep=1.5pt]

\tikzstyle{curly edge}=[]
\tikzstyle{straight edge}=[]

\tikzstyle{straight edge}=[]
\tikzstyle{vertex}=[circle,fill=black,inner sep=1.5pt]

\begin{tikzpicture}[scale=0.8]
	\begin{pgfonlayer}{nodelayer}
		\node [style=vertex] (0) at (0, 0.5) {};
		\node [style=vertex] (1) at (0, -0.5) {};
		\node [style=vertex] (2) at (1, 0) {};
		\node [style=vertex] (3) at (2, 0) {};
		\node [style=vertex] (4) at (3, 0) {};
	\end{pgfonlayer}
	\begin{pgfonlayer}{edgelayer}
		\draw [style=straight edge] (1) to (2);
		\draw [style=straight edge] (0) to (2);
		\draw [style=straight edge] (2) to (3);
		\draw [style=dashed] (3) to (4);
	\end{pgfonlayer}
	\node at (2,-.6) {$\underbrace{\hspace{4 em}}_{l+1 \hbox{ vertices}}$};
\end{tikzpicture}
\hspace{1cm
\begin{tikzpicture}[scale=0.8]
	\begin{pgfonlayer}{nodelayer}
		\node [style=vertex] (0) at (0, 0.5) {};
		\node [style=vertex] (1) at (0, -0.5) {};
		\node [style=vertex] (2) at (1, 0) {};
		\node [style=vertex] (3) at (2, 0) {};
		\node [style=vertex] (4) at (3, 0) {};
	\end{pgfonlayer}
	\begin{pgfonlayer}{edgelayer}
		\draw [style=straight edge] (0) to (1);
		\draw [style=straight edge] (1) to (2);
		\draw [style=straight edge] (0) to (2);
		\draw [style=straight edge] (2) to (3);
		\draw [style=dashed] (3) to (4);
	\end{pgfonlayer}
	\node at (2,-.6) {$\underbrace{\hspace{4 em}}_{l+1 \hbox{ vertices}}$};
\end{tikzpicture}
\hspace{1cm
\label{fig:graphs}
\end{center}
\caption{The structures mentioned in Theorem 2}
\end{figure}
\end{theorem}

\begin{theorem}\label{thm:claw}
Let $G$ be a $\{P_k,K_{1,3}\}$-free graph for $k \geq 5$, then $c(G) \leq k-3$.
\end{theorem}

\noindent\textbf{Remark:} Some mathematicians refer to $K_{1,3}$ as a "claw", we shall use $K_{1,3}$ and "claw" interchangeably.

Next, we will explore the graph $G$ with forbidden induced subgraphs consisted of different paths, we find that we can use fewer cops to catch the robber than we forbid the disjoint union of the same smaller paths.

\begin{theorem}\label{thm:PP}
Let $G$ be a $(P_{i_1}+P_{i_2}+\ldots+P_{i_k})$-free graph for $k\geq 2$, $i_j\geq 1$ and one of $i_j\geq 3$ ($1\leq j \leq k$). Then $c(G)\leq i_1+i_2+\ldots +i_k-2$.
\end{theorem}

\section{Forbidden Induced Structure}
Now we begin to prove \autoref{thm:Pk}, the most important thing to prove this theorem is that similar idea can be used in the proof of other theorems in our paper.
\begin{proof}[Proof of \autoref{thm:Pk}]
In the game of cops and robber, we can define a function $f$, $f$ means the minimal distance between the cop pieces and the robber after each round. Let $t$ be the minimum value of $f$ over all rounds. If $t=0$, it means that at least one cop and the robber are on the same vertex after finite rounds. On the other hand, if $t>0$, it means that all the cop pieces and the robber can never be on the same vertex in the whole game.

Suppose that the cop player can never use $k-2$ cop pieces to capture the robber in the graph $G$. That means $t \geq 1$ in the whole game. We can think in another way that after finite rounds, there exists one cop $c_1$ on the vertex $u$ whose distance is $t$ ($t\geq 1)$ from the robber $r$ on the vertex $v$. We can find a shortest path $P_{t+1}$: $u-u_1-\ldots-u_{t-1}-v$ between $u$ and $v$. Since there is no difference between each piece of the cop player, we can suppose all the cop pieces: $c_1,c_2,\ldots, c_{k-2}$ are on the vertex $u$ without loss of generality.

Now it's the robber's turn to move. In order to keep the minimal distance not decreasing, the robber $r$ must move to a new vertex $v_1$ and the vertex $v_1$ is not adjacent to the vertices $u,u_1,\ldots,u_{t-1}$. That means we can find a shortest path $P_{t+2}$: $u-u_1-\ldots-v-v_1$. Then the cop player will move $k-3$ pieces $c_2,\ldots,c_{k-2}$ to the vertex $u_1$. Since up to now the minimal distance between the cops and the robber is also $t$, the robber must move to another new vertex $v_2$, $v_2$ is not adjacent to the vertices $u,u_1, \ldots, u_{t-1},v$ for the same reason of the first round and we can get a shortest path $P_{t+3}$:$u-u_1-\ldots-v-v_1-v_2$. The cop player will move $k-4$ pieces $c_3,\ldots,c_{k-2}$ to the vertex $u_2$. We use this strategy repeatedly until we don't have cop piece to move to the next vertex, and right now we get a shortest path $P_{t+k-1}$: $u-u_1-\ldots-u_{t-1}-v-v_1-\ldots-v_{k-2}$ and the first $k-2$ vertices are each occupied by one cop.
\begin{figure}[hbt]
\begin{center}

\tikzstyle{straight edge}=[]
\tikzstyle{vertex}=[circle,fill=black,inner sep=1.5pt]

\begin{tikzpicture}[scale=0.8]
	\begin{pgfonlayer}{nodelayer}
		\node [style=vertex, label=above:{$u$},label=below:{$c_1$}] (0) at (0,0) {};
		\node [style=vertex, label=above:{$u_1$},label=below:{$c_2$}] (1) at (1,0) {};
		\node [style=vertex, label=above:{$u_{t-1}$}]  (2) at (2.5,0) {};
		\node [style=vertex, label=above:{$v$}] (3) at (3.5,0) {};
		\node [style=vertex, label=above:{$v_1$}] (4) at (4.5,0) {};
		\node [style=vertex, label=above:{$v_2$}] (5) at (5.5,0) {};
		\node [style=vertex, label=above:{$v_{k-2}$},label=below:{$r$}] (6) at (7,0) {};
	\end{pgfonlayer}
	\begin{pgfonlayer}{edgelayer}
		\draw [style=straight edge] (0) to (1);
		\draw [style=dashed] (1) to (2);
		\draw [style=straight edge] (2) to (3);
		\draw [style=straight edge] (3) to (4);
		\draw [style=straight edge] (4) to (5);
		\draw [style=dashed] (5) to (6);
	\end{pgfonlayer}
\end{tikzpicture}
\end{center}
\label{fig:extendedblockgraph}
\end{figure}

Based on our assumption, $t$ is greater than or equal to $1$, so $t+k-1\geq k$. That means we can get a shortest path $P_k$, however the graph $G$ is $P_k$-free. We draw a contradiction, so we can use $k-2$ cops to catch the robber in the graph $G$.
\end{proof}

The idea we use to prove \autoref{thm:Pk} is very useful to determine the cop number for some graphs with forbidden induced structure. It means that, if we can't use $l$ cops to catch the robber in any graph $G$, we can find an induced path $P_{l+t+1}$: $u_0-\ldots-u_{l+t}$ for $t\geq 1$ in the graph $G$. We can also guarantee the first vertices: $u_0,\ldots, u_{l-1}$ of this path are occupied by the cop $c$ and the last vertex $u_{l+t}$ is occupied by the robber $r$. Next we shall use this idea to prove the following theorems.

\begin{proof}[Proof of \autoref{thm:two}] Suppose not, we can use $l$ cops to form an induced $P_{l+2}$ path: $v_0-v_1-\ldots-v_{l+1}$ and each vertex $v_i$ is occupied by one cop $c_{i+1}$ for $0\leq i \leq l-1$. The robber is on the vertex $v_{l+1}$ by the details in \autoref{thm:Pk}. Now it's the turn for the cop player to move. We let each piece of the cop player move forward to the next vertex along this path. Then the robber cannot stay still, he must move. If he has more than two choices, for example $v_{l+2}, v_{l+2}', \ldots$. Whether or not these vertices are adjacent to each other, the vertices $v_1,\ldots, v_{l+1}$ already form a shortest path $P_{l+1}$. We need to forbid these two structures mentioned in Theorem 2, so one of any two these vertices must be adjacent to one of vertices $v_1,\ldots, v_l$. If either vertex $v_{l+2}$, $v_{l+2}'$, say $v_{l+2}$ is adjacent to any of vertices $v_1, \ldots, v_l$, the robber cannot move to $v_{l+2}$ since $v_1, \ldots, v_l$ are occupied by the cops. There is only one vertex except $v_l$ in the neighbourhood of $v_{l+1}$ which is not adjacent to the vertices $v_1,\ldots, v_l$. That means the robber has only one vertex left to move, we call it $v_{l+2}$. Then the cops will still move forward and the robber will meet the same problem as in the former round. So the robber still has only one choice to move to escape from the capture. However, the graph $G$ is finite and the robber's available vertices will form an induced path or cycle. If it's an induced path, we can use these $l$ cops to catch the robber directly. If it's an induced cycle, we use these $l$ cops to keep the robber moving along this cycle and another cop $c_{l+1}$ to move to any vertex in this cycle and wait for the robber.
So we can use $l+1$ cops to catch the robber in the graph $G$.

\end{proof}

The same idea in \autoref{thm:Pk} and \autoref{thm:two} is that we always use some cops to pursue the robber and we find that we can get an induced $P_k$ in the graph $G$ by analyzing the process of chase. In \autoref{thm:two}, we use one more cop to wait for the robber since there may exist an induced cycle by analyzing the trail of the robber's movements. Now we can use what we already know to prove the \autoref{thm:claw}.

\begin{proof}[Proof of \autoref{thm:claw}]
Now, we have $k-3$ cop pieces and we can use them to form an induced path $P_{k-1}$: $v_0-v_1-\ldots-v_{k-2}$, the first $k-3$ vertices are occupied by the cops $c_1,\ldots,c_{k-3}$, the robber $r$ is on the vertex $v_{k-2}$. It's the turn for the cop player to move and all cop pieces will move forward along this path, the vertices $v_1,\ldots,v_{k-3}$ are occupied by the cops as a result. Right now the robber is adjacent to the cop $c_{k-3}$ which is on the vertex $v_{k-3}$. All the existing vertices are occupied by the cops or adjacent to one vertex occupied by the cop, so the robber must choose a new vertex $v_{k-1}$ to escape from the capture. However, the path $P_k$ is forbidden in the graph $G$, the new vertex $v_{k-1}$ must be adjacent to at least one of vertices: $v_0,v_1,\ldots,v_{k-3}$. On the another hand, the robber cannot be adjacent to the vertex occupied by the cop directly. The only choice for the vertex $v_{k-1}$ is to be adjacent to the vertex $v_0$ and we have a $C_k$ cycle. The cop player moves the cop $c_1$ backward to the vertex $v_0$ and the other $k-4$ cops move forward along this path. The robber $r$ is adjacent to the vertices $v_0$ and $v_{k-2}$ occupied by the cops $c_1$ and $c_{k-3}$ respectively, so the robber must move to a new vertex $v_k$. However, the vertices \{$v_{k-2},v_{k-1},v_k,v_0$\} will form an induced "claw". In order not to let this happen, the vertex $v_k$ must be adjacent to the vertex $v_0$ or $v_{k-2}$ since $v_0$ and $v_{k-2}$ are not adjacent ($v_0$ and $v_{k-2}$ are vertices of induced path $P_{k-1}$). However, the cop player has already prepared the cops for the vertices $v_0$ and $v_{k-2}$, so the robber will not escape from capture by the cops. That means we can use $k-3$ cops to catch the robber in $G$.

\end{proof}

If we carefully analyze the proof of the \autoref{thm:claw}, we can find that if we use one fewer cop to catch the robber in the $P_k$-free graphs, it must form an induced cycle $C_k$. Also, if we just keep some cops still and make the other cops move forward in the proof of \autoref{thm:claw}, we can also get an induced cycle $C_j$ for $3\leq j\leq k$. So we can get the following Corollary.

\begin{corollary}\label{cor:cy}
Let $G$ be a $\{P_k,C_l\}$-free graph for $k \ge 5$ and $3 \leq l \leq k$, then $c(G) \leq k-3$.
\end{corollary}

Right now, we can answer the following conjectures made by Sivaraman and Testa \cite{1903.11484} partially.

\begin{conjecture}
Let $G$ be a $P_5$-free graph, then $c(G)\leq 2$.
\end{conjecture}
\begin{conjecture}
Let $G$ be a $2P_2$-free graph, then $c(G)\leq 2$.
\end{conjecture}

$2P_2$-free means ($P_2+P_2$)-free, we shall use the lemma made by Sivaraman and Testa \cite{1903.11484}.

\begin{lemma}\label{lm:3} \cite{1903.11484}
Let $G$ be a $2P_2$-free graph with diameter $3$. Then $c(G)\leq 2$.
\end{lemma}

\begin{theorem}
Let $G$ be a \{$2P_2$, $\overline P_5$\}-free graph, then $c(G)=2$.
\end{theorem}
\begin{proof}
We already know that if $G$ is $2P_2$-free and the diameter of $G$ is 3, we can use 2 cops to capture the robber by Lemma 1. So we just need to consider the case where the diameter of $G$ is $2$. The graph of $\overline P_5$ (complement graph of $P_5$) is like a house and we call it "house" in our proof. The graph $G$ must contain an induced $P_4$ since we already know that we can use two cops to catch the robber in any $P_4$-free graph by \autoref{thm:Pk}. We put two cops on the two internal vertices $u_0$ and $v_0$ of this $P_4$: $u_1-u_0-v_0-v_1$ respectively. The robber $r$ will choose one vertex $z_0$ which is not adjacent to $u_0$ or $v_0$ to stay. Then the vertices except these two vertices in the graph $G$ are divided into four parts: the cop $c_1$'s private neighbourhood (the vertex adjacent to $c_1$ not adjacent to $c_2$) denoted by $U$, the cop $c_2$'s private neighbourhood (the vertex adjacent to $c_2$ not adjacent to $c_1$) denoted by $V$, the common neighbourhood of $c_1$,$c_2$ (the vertex adjacent to both $c_1$ and $c_2$) denoted by $W$ and the vertices not adjacent to $c_1$ or $c_2$ denoted by $Z$. The vertices in $Z$ is an independent set since all the vertices in $Z$ are not adjacent to $P_2$: $u_0-v_0$.
\begin{center}
\begin{tikzpicture}
	\draw  (1,0) circle (2pt) -- (3,0) circle  (2pt);
\filldraw (1,0) -- (0.5,-1.5) circle  (2pt);
\filldraw (1,0) -- (2,-1.5) circle  (2pt);
\filldraw (3,0) -- (2,-1.5) circle  (2pt);
\filldraw (3,0) -- (3.5,-1.5) circle  (2pt);
\draw (1,-1.5) arc (0:360:1cm and 0.5cm) ;
\draw (2.7,-1.5) arc (0:360:0.7cm and 0.5cm) ;
\draw (5,-1.5) arc(0:360:1cm and 0.5cm);
\draw (3,-3) arc(0:360:1cm and 0.5cm);
\draw (2,-3) circle (2pt) node[above]{$r$};
\draw (3,-3) node[right]  {$Z$};
\draw (0.3,-1.5) node[left] {$U$};
\draw (2,-1.5) node[below] {$W$};
\draw (3.7,-1.5) node[right] {$V$};
\draw (1,0) node[above]{$c_1$};
\draw (1,0) node[left]{$u_0$};
\draw (3,0) node[above]{$c_2$};
\draw (3,0) node[right]{$v_0$};

\end{tikzpicture}
\end{center}
\noindent Case 1: If $z_0$ is adjacent to some vertex $w_0$ in $W$, which means the robber is adjacent to one common neighbor of $c_1$ and $c_2$.

We move the cop $c_1$ to the vertex $w_0$, the robber must move and he has three possible choices to move: the sets $U$, $W$, $V$. The parts $W$ and $V$ are controlled by the cop $c_2$ so the robber cannot move to these parts actually. The available choice for the robber is to move to the part $U$ to escape from the capture. For example, the robber moves to one vertex $u'$ in $U$, this vertex must be adjacent to the vertex $w_0$ to prevent from forming an induced "house":\{$v_0$, $u_0$, $w_0$, $u'$, $z_0$\}. So the cop $c_1$ will catch the robber in the next round.\\

\noindent Case 2: If $z_0$ is not adjacent to any vertex in $W$ or $W$ is an empty set.

Since the diameter of $G$ is 2, the vertex $z_0$ must be adjacent to some vertex in $U$ and $V$. These two vertices in $U$ and $V$ are denoted by $u_1$ and $v_1$ respectively for convenience. We move the cop $c_1$ to the vertex $u_1$ and the cop $c_2$ to the vertex $v_1$. The robber must move to escape from the capture. The only choices available for him are the parts $U$ and $V$. Without loss of generality, let the robber move to one vertex $u_2$ in $U$ which is not adjacent to $u_1$. In the next round, let the cop $c_1$ move back to the vertex $u_0$ and the cop $c_2$ stay still. The only available choices for the robber are the parts $Z$ and $V$.

Case 2(i): If the robber moves to some vertex $z_1$ in $Z$. This vertex must be different from the vertex $z_0$ since the cop $c_2$ is still controlling the vertex $z_0$. Actually, the vertex $z_1$ doesn't exist since the vertex \{$z_1-u_2$, $v_0-v_1$\} will form an induced $2P_2$.

Case 2(ii): If the robber moves to some vertex $v_2$ which is not adjacent to the vertex $v_1$ in $V$. Let the cop $c_2$ move back to the vertex $v_0$ and the cop $c_1$ stay still. Then the robber will move back to some vertex $z_2$ in the part $Z$. The vertex $z_2$ must be different from the vertex $z_0$. Suppose not, we can find an induced "house": \{$z_0$, $u_2$, $v_2$, $u_0$, $v_0$\}. The vertex $v_2$ is not adjacent to $z_0$ for the same reason. The vertex $z_2$ must be adjacent to the vertex $v_1$ prevent from forming induced $2K_2$: \{$z_0$-$v_1$, $z_2$-$v_2$\}. The vertex $z_2$ must be adjacent to the vertex $u_2$ to prevent from forming induced $2P_2$: \{$z_2$-$v_1$, $u_0$-$u_2$\}. However, in that way we can get an induced "house": \{$z_2$, $u_2$, $v_2$, $u_0$, $v_0$\}. That means the vertex $z_2$ available for the robber to move actually doesn't exist.

So we can use two cops to catch the robber for each case in \{$2P_2$, $\overline P_5$\}-free graph.

\end{proof}

\noindent\textbf{Remark:} We know that we can use two cops to catch the robber for any \{$2P_2$, $C_i$ ($i=3,4,5$)\}-free graphs, the complement graph $\overline{P_5}$ of $P_5$ contains an induced triangle and $C_4$. So this theorem is more generalized than the theorem for \{$2P_2$, $C_i$ ($i=3,4,5$)\}-free.

Now there is only one theorem left for us, we wouldn't like to show the proof of \autoref{thm:PP} since there are complicated notations in this proof. We only prove two propositions just using the same method in the proof of \autoref{thm:PP}.

\begin{proposition}
Let $G$ be a $(P_1+P_k)$-free graph for $k\geq 3$. Then $c(G)\leq k-1$.
\end{proposition}

\begin{proof}
In the beginning, we put all the $k-1$ cops on any fixed vertex $v_0$. Then the induced subgraph generated by the vertices in $V(G)$ except the vertex $v_0$ and its neighbourhood $N(v_0)$ will form a few components $U_1,U_2,\ldots$. Then robber will choose some vertex $v_1$ in one of these components, we call it $U_1$ since the vertex $v_0$ is occupied by the cops. And also, we know that in the induced subgraph $G[U_1]$ generated by the vertices in the component $U_1$ is $P_k$ forbidden for the reason that all the vertices in $U_1$ is not adjacent to the vertex $v_0$.

We put one cop remaining on the vertex $v_0$ and all the other $k-2$ cops move to the component $U_1$. If the robber wants to move to the other component, he must go through one of the vertices in $N(v_0)$ or $v_0$, that is impossible since the vertex $v_0$ is always occupied by the cop. So the robber will always stay in the component $U_1$, and the induced subgraph $G[U_1]$ is $P_k$-free, we have $k-2$ cops in the component $U_1$ after the cops finish their movements. However, we already knew that $k-2$ cops can catch the robber in any $P_k$-free graphs for $k\geq3$. As a result, we can use $k-1$ cops to catch the robber in any $(P_k,P_1)$-free graphs for $k\geq3$.
\end{proof}

\begin{proposition}
Let $G$ be a $(P_2+P_k)$-free graph for $k\geq3$. Then $c(G)\leq k$.
\end{proposition}
\begin{proof}
In the beginning, we can find two adjacent vertices $v_0,v_1$ and we put one cop on the vertex $v_1$ and other $k-1$ cops on the vertex $v_0$. We also find that all the components for the induced subgraph $G[V(G)\backslash N[v_0,v_1]]$ is $P_k$-free and we keep two cops remaining on the vertex $v_0$ and $v_1$, the other $k-2$ cops will catch the robber eventually.
\end{proof}

\section{Concluding Remarks}
In this paper we find a powerful idea to solve the problem about how many cops we need to capture the robber in some graphs with forbidden induced subgraphs. We also answer the conjecture made by Sivaraman and Testa \cite{1903.11484} partially. The difficulty for proving the cop number of $2P_2$-free graphs is $2$ is that Theorem 1 is right for the condition $k\geq 3$. We cannot use "zero" cops to capture the robber in any $P_2$-free graph.

\bibliographystyle{abbrv}


\section{Introduction}

The landscape of many vacua%
\footnote{
The vacua here have extended meaning
which indicates the backgrounds in the theory (moduli)
space, or the landscape.
}
is a plausible structure in the fundamental
theory of physical laws in nature.
In particular, this structure is expected
as one of the theoretical ingredients
to understand the observed small cosmological constant
\cite{Wei}.
However, the anthropically allowed region of vacua
in the landscape seems too large to be predictive enough
in the presence of a variety of couplings.
Thus, it is a challenging problem to derive
further physical consequences from the landscape of vacua.

The (non-)presence of inflationary dynamics
is a promising candidate as the first criterion to select realistic vacua
\cite{Iza}.
We can naturally expect that macroscopic universe is realized through
inflation from fundamental-scale physics.
Moreover, under the dynamics of inflation, mediocrity principle
\cite{Vil}
may prefer long-lasting inflations which result in
larger-volume universes where more habitable galaxies are
produced.
In this respect, multiple inflations
\cite{Kaw}
give a remarkable possibility to be considered
\cite{Ibe}.

In a recent article
\cite{Ibe},
we have pointed out that the
inflationary dynamics possess a potential
to select minimal supergravity as a large-cutoff theory,
where the gravitational scale $M_G$ is smaller than the
cutoff scale $M_*$ stemming from the fundamental theory.
Such a large-cutoff supergravity naturally causes
multiple slow-roll inflations, which possibly meet mediocrity principle.

The large-cutoff theory is also attractive from the viewpoint of
particle-physics phenomenology: First of all, the suppression of the
flavor-changing neutral currents is automatic in the large-cutoff
theory, since all of the higher-dimensional operators are suppressed by
the large cutoff $M_*$ except for the genuine gravitational interactions.
In detail,
the large-cutoff supergravity predicts a hierarchical spectrum
\cite{Ibe}
of supersymmetric
particles as $m_0 \gg |M_i|$, where $m_0$ is the universal soft mass for
sfermions and $M_i$ the gaugino masses ($i=1,2,3$).
Thus, the current chargino mass bound suggests heavy sfermions at several TeV.
Such a soft mass parameter belongs to the parabolic
\cite{Ibe}
or hyperbolic
\cite{Cha}
regime allowed for a given $\mu$ parameter.
Indeed the recent detailed analysis
\cite{Ibe}
has confirmed that the region with large sfermion masses
along the small-$\mu$-parameter curve continued from the focus point
\cite{Fen}
is consistent with the electroweak symmetry breaking
\cite{Bar}.
In the region of
heavy sfermions and light gauginos (an order of magnitude lighter than
the sfermions), the constraint
from CP violation is rather weak and even order one CP-violating phases are
allowed for $m_0  \mathop{}_{\textstyle \sim}^{\textstyle >}  10$TeV
\cite{Olive:2005ru}.
Furthermore, the lightest supersymmetric particle
can explain the dark matter density of the present universe in the above
mass region for supersymmetric particles
\cite{Mor}.

In this paper, we discuss a minimal new inflation model
\footnote{We suspect that multiple stages of inflation imply
that the primordial inflation at the last stage tends to be
a new inflation, since it seems naturally
realized with a lower energy scale than that of other types of inflation.
The discussion section includes comments on the case of other inflations.
}
as an example in the framework of the
large-cutoff supergravity with emphasis on baryon asymmetry
generated by leptogenesis
\cite{Fuk}
to complete a model of the large-cutoff hypothesis.\footnote{There are
other new inflation models in the framework of
supergravity\cite{Asaka:1999jb}, although these inflation models can not
explain the observed spectral index in the large-cutoff hypothesis.}
We claim that all the phenomenological requirements from
cosmology and particle physics are satisfied in a certain parameter
region of the large-cutoff theory.

\section{Supergravity new inflation}

We adopt a new inflation model considered in
Ref.\cite{Yan, Izaw}.
As an effective field theory for
an inflaton chiral superfield $\tilde \phi$,
the superpotential is given by
\begin{equation}
 \begin{array}{rr}
 W={\tilde v}^2 {\tilde \phi} -\frac{\tilde g}{n+1}{\tilde \phi}^{n+1},
 &
 \label{inflaton-tilde}
 \end{array}
\end{equation}
for $n \ge 3$ and the K\"ahler potential is given by
\begin{equation}
 K = {\tilde K}\left| {\tilde \phi} \right|^2
 + \frac{\tilde k}{4}\left| {\tilde \phi} \right|^4 + \cdots,
\end{equation}
where we have taken the unit with the reduced Planck scale
$M_{G} \simeq 2.4 \times 10^{18}$GeV equal to
one. The positive parameters $\tilde K$, $\tilde g$, and $\tilde k$
are of orders $1$, $10^{-(n-2)}$, and $10^{-2}$, respectively,
for our large-cutoff hypothesis $M_* \simeq 10 M_G$.
The tiny scale ${\tilde v}^2 >0$ can be generated dynamically
\cite{Hot} and the ellipsis denotes higher-dimensional operators which
may be neglected in the following analysis.

For the canonically normalized field $\phi = \sqrt{\tilde K} {\tilde \phi}$,
the superpotential is given by
\cite{Kumekawa:1994gx}
\begin{equation}
 \begin{array}{rr}
 W=v^2 \phi -\frac{g}{n+1}\phi^{n+1}, &
 \label{inflaton}
 \end{array}
\end{equation}
and the K\"ahler potential is given by
\begin{equation}
 K = \left| \phi \right|^2 + \frac{k}{4}\left| \phi \right|^4 + \cdots,
\end{equation}
where we have defined
\begin{equation}
 {\tilde v}^2=v^2 \sqrt{\tilde K}, \quad
 {\tilde g}=g {\tilde K}^{n+1 \over 2}, \quad
 {\tilde k}=k {\tilde K}^2.
 \label{canonical}
\end{equation}

The effective potential for the lowest component of $\phi$ is given by
\begin{equation}
 V=e^K \left\{ \left(\frac{\partial ^2 K}
 {\partial \phi \partial \phi^\dagger}\right)^{-1}
 \left|D W \right|^2 -3\left|W \right|^2\right\},\label{sugpot}
\end{equation}
where
\begin{equation}
 DW = \frac{\partial W}{\partial \phi}
 + \frac{\partial K}{\partial \phi}W.
\end{equation}
Thus, the potential of the inflaton field
$\varphi=\sqrt{2} \mathrm{Re} \ \phi$ is
approximately given by
\begin{equation}
 V(\varphi) \simeq v^4 - \frac{k}{2}v^4 \varphi^2
 -\frac{g}{2^{\frac{n}{2}-1}}v^2\varphi^n
 +\frac{g^2}{2^n}\varphi^{2n}
\label{potential}
\end{equation}
for the inflationary period near the origin $\varphi > 0$.

The inflationary regime is determined by the slow-roll condition
\cite{Lyt}
\begin{equation}
 \epsilon(\varphi) = \frac{1}{2}\left(\frac{V^\prime(\varphi)}
 {V(\varphi)}\right)^2 \le 1,\quad \left|\eta(\varphi)\right| \le 1,
 \label{slowroll}
\end{equation}
where
\begin{equation}
 \eta(\varphi)=\frac{V^{\prime\prime}(\varphi)}{V(\varphi)}.
\end{equation}
For the potential Eq.(\ref{potential}), we obtain
\begin{eqnarray}
 \epsilon(\varphi) \simeq \frac{1}{2}
 \left(\frac{-k v^4 \varphi-\frac{gn}
 {2^{\frac{n}{2}-1}}v^2\varphi^{n-1}}{v^4}\right)^2, \\
 \eta(\varphi) \simeq \frac{-kv^4 -\frac{g}
 {2^{\frac{n}{2}-1}}n(n-1)v^2\varphi^{n-2}}{v^4}.
\end{eqnarray}
The slow-roll condition Eq.(\ref{slowroll}) is satisfied for
$\varphi\le\varphi_f$ where
\begin{equation}
 \varphi_f \simeq \sqrt{2}\left(\frac{(1-k)v^2}
 {gn(n-1)}\right)^{\frac{1}{n-2}},
\end{equation}
which yields the value of the inflaton field at the end of inflation.

The value $\varphi_{N_e}$ of the inflaton corresponding to the $e$-fold number $N_e$ is
given by
\begin{eqnarray}
 N_e \simeq \int_{\varphi_f}^{\varphi_{N_e}} d\varphi
 \frac{V(\varphi)}{V^\prime(\varphi)}
   \simeq \int_{\varphi_f}^{\varphi_{N_e}}
 d\varphi \frac{v^4}{-kv^4\varphi-\frac{gn}
 {2^{\frac{n}{2}-1}}v^2\varphi^{n-1}}.
\end{eqnarray}
This leads to
\begin{equation}
 \varphi_{N_e}^{n-2} \simeq \frac{kv^2 2^{\frac{n}{2}-1}}{gn}
 \left\{\frac{1+k(n-2)}{1-k}e^{{N_e}k(n-2)}-1\right\}^{-1}.
\end{equation}
Hence the spectral index of the density fluctuations is given by
\cite{Izaw}
\begin{eqnarray}
 n_s &\simeq& 1-6\epsilon(\varphi_{N_e}) + 2 \eta(\varphi_{N_e})\\
  &\simeq& 1 - 2 k \left[1 + \frac{n-1}
 {\left\{1+\frac{k}{1-k}(n-1)\right\}e^{{N_e}k(n-2)}-1}\right].
 \label{spectrum}
\end{eqnarray}
Note that the spectral index does not depend on $v^2$ and $g$
explicitly. We show the $k$ dependence of the spectral index
$n_s$ for $n=4$ and $N_e=45, 50$, $55$
\cite{Yan,Izaw}
in Fig.\ref{spectrum1}.

\begin{figure}
 \begin{center}
 \includegraphics[width=7.5cm]{fig1.eps}
 \caption{The $k$ dependence of the spectral index $n_s$ for $n=4$. The red
 (solid) line corresponds to the $e$-fold number $N_e$=45, the green (dashed)
 line to $N_e$=50, and the blue (dash-dotted)
 line to $N_e$=55. For $k=0$, $n_s \simeq 1-6/(2N_e+3)$.}
 \label{spectrum1}
 \end{center}
\end{figure}

Now we proceed to determine the inflation scale $v$ from the density
fluctuations. The amplitude of primordial density fluctuations is given by
\begin{eqnarray}
 \frac{\delta \rho}{\rho} \simeq \frac{1}
 {5\sqrt{3}\pi}\frac{V^{3 \over 2}(\varphi_{N_0})}{|V^\prime(\varphi_{N_0})|}
                          \simeq \frac{1}{5\sqrt{3}\pi}\frac{v^6}
 {kv^4\varphi_{N_0} + \frac{gnv^2}{2^{\frac{n}{2}-1}}\varphi_{N_0}^{n-1}},
\end{eqnarray}
where $\varphi_{N_0}$ is the value of inflaton field at the epoch of the
present-horizon exit.
Thus we obtain
\begin{eqnarray}
 v^{\frac{2n-6}{n-2}} \simeq \sqrt{2}\frac{V^{3 \over 2}(\varphi_{N_0})}{|V^\prime(\varphi_{N_0})|}\left[\frac{k}{gn}\left\{\frac{1+k(n-2)}{1-k}e^{N_0k(n-2)}-1\right\}^{-1}\right]^{\frac{1}{n-2}}\cr
\quad \times\left[k+k\left\{\frac{1+k(n-2)}{1-k}e^{N_0k(n-2)}-1\right\}^{-1}\right].
\label{v}
\end{eqnarray}
Owing to the COBE normalization
\begin{equation}
 \frac{V^{3 \over 2}(\varphi_{N_0})}{|V^\prime(\varphi_{N_0})|} \simeq 5.3 \times 10^{-4},
\end{equation}
the scale $v$ is expressed as
\begin{equation}
 v \simeq 10^{12} \mathrm{GeV} \times \mathcal{C}(k,N_0) \times \left(\frac{0.1}{g}\right)^{1/2}\label{approv},
\end{equation}
for $n=4$, $N_0 \simeq 50$, and $k\lsim0.01$, where $\mathcal{C}(k,N_0)$ is
a function of order unity.

On the other hand,
the $e$-fold number of the present horizon is also given by
\begin{equation}
 N_0\simeq 67+\frac{1}{3}\ln H +\frac{1}{3}\ln T_{R} \simeq 67
 +\frac{1}{3}\ln \frac{v^2}{\sqrt{3}} + \frac{1}{3}\ln T_R,
\label{efold}
\end{equation}
where $H$ denotes the Hubble scale at the horizon exit
and $T_R$ the reheating temperature.
By means of Eq.(\ref{spectrum}), (\ref{v}), and (\ref{efold}), we can
determine $v$ and $N_0$ from $g$, $k$, and $T_R$. For $n=4$, $g\sim 0.1$,
$k\sim0.01$, and $T_R \sim 10^{5-9}$GeV, the inflation scale $v$ is
given by ${\cal O}(10^{12})$GeV, and the $e$-fold number $N_0$ of the present
horizon is given by $47.6-50.6$.
In Fig.\ref{spectrum3}, we show the $k$ dependence of the spectral index $n_s$
for the reheating temperature $T_R=10^5,10^7$, and $10^9$ GeV.
We conclude that the implication
${\tilde g} \sim {\tilde k} \sim 0.01$
of the large-cutoff hypothesis
\footnote{For instance, Eq.(\ref{canonical}) yields $g=0.1$ and $k=0.01$
for ${\tilde K}=0.5$, ${\tilde g}=0.018$
and ${\tilde k}=0.0025$.}
is consistent with an experimental value $n_s=0.95 \pm 0.02$
\cite{BOOMERANG} of the spectral index for a wide range of the reheating
temperature.\footnote{
The inflaton as a massless scalar field in
the de Sitter background
has quantum fluctuations whose amplitude is given by $\Delta \varphi \sim
H/(2 \pi)$.
Thus the amplitude $\Delta \varphi$ at $\varphi=\varphi_{N_0}$ is given by
\begin{eqnarray}
 \Delta \varphi|_{\varphi_{N_0}} \sim \frac{\sqrt{2\epsilon(\varphi_{N_0})}}{2\pi\sqrt{3}} \frac{V(\varphi_{N_0})^{3 \over 2}}{V'(\varphi_{N_0})}
            \simeq \frac{1}{2\pi\sqrt{3}}\frac{V(\varphi_{N_0})^{3 \over 2}}{V'(\varphi_{N_0})}\left(k+\frac{gn}{2^{\frac{n}{2}-1}}\frac{\varphi_{N_0}^{n-2}}{v^2}\right)\varphi_{N_0}.
\nonumber
\end{eqnarray}
For $n=4, g \sim0.1, k \sim 0.01$, and $T_R\sim 10^{5-9}$GeV,
the fluctuation amplitude $\Delta
\varphi|_{\varphi_{N_0}}$ takes a value of order $10^{-6} \varphi_{N_0}$,
which is much less than the mean-field value $\varphi_{N_0}$ to justify
the above slow-roll analysis.}

\begin{figure}
 \begin{center}
 \includegraphics[width=7.5cm]{fig2.eps}
 \caption{The $k$ dependence of the spectral index $n_s$
 for $n=4$. The shaded regions correspond to
  $T_R=10^5,10^7,10^9$GeV from below, and the lower lines for $g=0.1$ and
  the upper lines for $g=0.01$.}
 \label{spectrum3}
 \end{center}
\end{figure}

\section{The gravitino mass}

In the previous section, we have confirmed that
the new inflation model in the large-cutoff hypothesis
is consistent with the cosmological observations.
In this section, we discuss
the gravitino problem under such an inflationary scenario.

As considered in
Ref.\cite{Yan},
we assume that the positive energy  $\Lambda^4_{\rm{SUSY}}$
of the SUSY breaking
is dominantly canceled out by the negative energy at the inflaton potential minimum.
Namely we impose
\begin{equation}
 \Lambda^4_{\rm{SUSY}}-3|W(\phi_0)|^2=0,
\end{equation}
where $\phi_0$ is the minimum point of $\phi$ in Eq.(\ref{sugpot}).

Then we obtain the gravitino mass as
\begin{equation}
 m_{3/2} \simeq \frac{\Lambda^2_{\rm{SUSY}}}{\sqrt{3}}=W(\phi_0).
\end{equation}
The value of $\phi_0$ is approximately given by
\begin{equation}
 \phi_0 \simeq \left(\frac{v^2}{g}\right)^{\frac{1}{n}}.
 \label{phivev}
\end{equation}
Consequently the gravitino mass is given by
\begin{equation}
 m_{3/2} \simeq  \frac{n v^2}{n+1}\left(\frac{v^2}{g}\right)^{\frac{1}{n}}
         \simeq  9 \mathrm{TeV} \times \left(\frac{0.1}{g}\right)^{3 \over 2}.
 \label{gmass}
\end{equation}
The second equality holds for $n=4$,
where we have used Eq.(\ref{approv}) and omitted the
weak dependence on $k$ and $T_R$.

More precisely, by means of Eq.(\ref{v}) and (\ref{efold}),
the gravitino mass can be
expressed as a function of $g$, $k$, and $T_R$,
although the dependence on
$T_R$ is very weak, as can be seen from Eq.(\ref{v}) and (\ref{efold}).
The result is shown in
Fig.\ref{gravitino}.
For $g  \mathop{}_{\textstyle \sim}^{\textstyle <}  0.2$ and $k  \mathop{}_{\textstyle \sim}^{\textstyle <}  0.035$,
the gravitino mass is larger than $4$TeV,
which may avoid the gravitino overproduction
for a reheating temperature $T_R \sim 10^{6-7}$GeV
\cite{Koh}.

\begin{figure
 \begin{center}
 \includegraphics[width=7.5cm]{fig3.eps}
 \caption{The contours of the gravitino mass for $n=4$
 and $T_R=4 \times 10^6$GeV. The dependence on
 the reheating temperature is very weak.}
 \label{gravitino}
 \end{center}
\end{figure}

In contrast,
the sfermion soft mass is given as $m_0 \simeq m_{3/2}$ if no $D$-term contributes to the SUSY breaking.
Thus, $m_0 < 10$TeV implies $g  \mathop{}_{\textstyle \sim}^{\textstyle >}  0.07$ for $k  \mathop{}_{\textstyle \sim}^{\textstyle <}  0.035$.

\section{Reheating for baryogenesis}
\label{se:baryon}
Now we are ready to consider the baryon asymmetry
in the present new inflation model with the large cutoff.

We assume the baryon asymmetry is generated by leptogenesis
\cite{Fuk}
through non-thermal production of
right-handed neutrinos, as investigated in
Ref.\cite{Mor,Fuk},
which provides a numerical estimate
\begin{equation}
 \frac{n_B}{s} \simeq 8.2 \times 10^{-11} \left(\frac{T_R}{10^6\mathrm{GeV}}\right)\left(\frac{2m_N}{m_\phi}\right)\left(\frac{m_{\nu_3}}{{0.05\mathrm{eV}}}\right)\frac{1}{\sin^2 \beta}\delta_{eff}.
 \label{lepto}
\end{equation}
Here $m_N$, $m_\phi$, and $m_{\nu_3}$ are the masses of the right-handed
neutrino $N$, the inflaton $\phi$ and the heaviest (active) neutrino, respectively.
The phase $\delta_{eff}$ is the effective CP phase defined in
Ref.\cite{Fuk} and $\tan \beta$ is the ratio of the vacuum
expectation value of up- and down-type Higgs bosons in the MSSM.
The reheating temperature is given by
\begin{equation}
 T_R \simeq \left(\frac{10}{g_*\pi^2} \Gamma_\phi^2\right)^{\frac{1}{4}},
\label{tr}
\end{equation}
where $\Gamma_\phi$ is the decay width of the inflaton and $g_*$ is the effective number of massless degrees of
freedom to be taken as $g_*=228.75$ numerically.
Note that the inflaton mass
\begin{equation}
 m_\phi \simeq n v^2 \left({v^2 \over g}\right)^{-{1 \over n}}
\label{mphi}
\end{equation}
in our new inflation model
also weakly depends on the $k$ and the reheating
temperature $T_R$, as is the case for the gravitino mass in
Eq.(\ref{gmass}).

Let us introduce the following superpotential interaction
as the dominant source of the $N$ production:
\footnote{The inflaton
field is also expected to decay through couplings
with light fields $\psi_i$ in the K\"ahler potential
such as $\sum_i c_i|\phi|^2 |\psi_i|^2$.
However, the decay width $\Gamma \sim \sum_i |c_i|^2\phi_0^2m_\phi^3$
through these couplings is so small that
we neglect such contributions.}
\begin{equation}
 \delta W = \frac{h}{2(n-1)}\phi^{n-1}N^2,
 \label{hNN}
\end{equation}
where $h$ is a positive parameter of the order of the inflaton
self-coupling $g$.
\footnote{Here, we assign the same charge for $\phi$ and
$N$ under $Z_{2n}$ R-symmetry, while we
assign the matter parity $+$ for
$\phi$ and $-$ for $N$.
Hence we expect the presence of
such operators as $\phi^{n-3}N^{4}$ in addition to Eq.(\ref{hNN}).
We do not include such operators since
the operator Eq.(\ref{hNN}) with the smallest number of $N$
dominates the reheating and leptogenesis.}
The coupling Eq.(\ref{hNN}) gives a decay width
\begin{equation}
  \Gamma_\phi \simeq \frac{|h|^2}{16\pi}\phi_0^{2(n-2)}m_\phi.
\label{gammaphi}
\end{equation}
From this decay width the reheating temperature after inflation for
$n=4$ is given by
\footnote{The cross term between $\phi^{n-1}N^2$ and
$v^2\phi$ in the superpotential
gives a comparable decay width.
We neglect this contribution since it does not essentially affect
our conclusions.}
\begin{equation}
T_R \simeq 2.6\times 10^6 \mathrm{GeV} \left(\frac{h}{0.1}\right)\left(\frac{0.1}{g}\right)^{5/4},
\end{equation}
where we have omitted the weak dependence on $k$ in Eq.(\ref{approv}).
Therefore, the reheating temperature $T_R \sim 10^{6-7}$ GeV is typical
in this model. As mentioned above, this reheating temperature is low
enough to avoid the gravitino overproduction.

Note that the operator Eq.(\ref{hNN}) also gives the
Majorana mass to the neutrino:
\begin{equation}
 m_N=\frac{h}{n-1} \phi_0^{n-1}\simeq \frac{h}{n-1}\left(\frac{v^2}{g}\right)^{1-\frac{1}{n}}.
\end{equation}
Thus the mass inequality $2m_N<m_\phi$, namely,
\begin{equation}
 h <\frac{n(n-1)}{2}g,
\label{tiny}
\end{equation}
is satisfied with a typical parameter set $g \sim h$.
This is appropriate for the non-thermal production of neutrinos
which leads to the non-thermal leptogenesis.

Based on the above setup, we now
estimate the baryon asymmetry due to the decay of inflaton
\footnote{
In our setup, we also have an additional contribution to the baryon asymmetry and the gravitino abundance.
However, as we see in Appendix, this contribution
is small in typical parameter region so that we neglect this
contribution in the following analysis.}
in our model as a function of the couplings $g,k$, and the
reheating temperature $T_R$.
The baryon asymmetry is determined
by four independent parameters $g,k,v$, and $h$.
In terms of the observed density fluctuations,
we can represent $v$ with the other parameters.
We further use the reheating temperature as an input parameter
instead of $h$ by means of Eq.(\ref{tr}), (\ref{mphi}) and (\ref{gammaphi}):
\begin{equation}
 h\simeq \sqrt{\frac{16\pi}{m_\phi}}\left(\frac{g_*\pi^2}{10M_G^2}\right)^\frac{1}{4} \left(\frac{g}{v^2}\right)^{\frac{n-2}{n}} T_R.
\label{htr}
\end{equation}
Then the baryon asymmetry ${n_B}/{s}$ is given in terms of $g$, $k$, and
$T_R$ by
\begin{equation}
  \frac{n_B}{s} \simeq 8.2 \times 10^{-11} \left(\frac{T_R}{10^6\mathrm{GeV}}\right)\left(\frac{2h}{n(n-1)g}\right)\left(\frac{m_{\nu_3}}{{0.05\mathrm{eV}}}\right)\frac{1}{\sin^2 \beta}\delta_{eff},
\label{tocompare}
\end{equation}
where $h(g,k,T_R)$ is given by Eq.(\ref{htr}) with $v$ determined by
Eq.(\ref{v}) and (\ref{efold}).

\begin{figure}[tbp]
 \begin{center}
 \includegraphics[width=7cm]{fig4.eps}
  \caption{The contours of
 $\left(\frac{n_B}{s}\right)/\left(\frac{n_B}{s}\right)_{0}$ for $n=4$, $T_R=4 \times 10^6$GeV, $\delta_{eff}=1$, $\sin \beta=1$ are plotted in red (solid)
 lines. The
 blue (dashed-dotted) lines correspond to the contours of gravitino mass.}
 \label{bar}
 \end{center}
\end{figure}

In Fig.\ref{bar},
we plot the contours of $m_{3/2}$ and $(\frac{n_B}{s})/(\frac{n_B}{s})_0$ for
$T_R=4\times 10^6$GeV, $m_{\nu_3}=0.05$eV, $\delta_{eff}=1$, $\sin \beta=1$, where
$(n_B/s)_0$ is the baryon asymmetry of the universe suggested by WMAP
\cite{WMAP}:
\begin{equation}
 \left(\frac{n_B}{s}\right)_{0} \simeq 8.7 \times 10^{-11}.
\end{equation}
We note that the baryon asymmetry and the gravitino mass
for different reheating
temperatures can also be seen from Fig.\ref{bar}: The baryon asymmetry is
proportional to the square of the reheating temperature $T_R^2$, since the
coupling $h$ is approximately proportional to the reheating temperature.
As for the gravitino mass, its value is almost independent of $T_R$,
since $v$ is almost independent of $T_R$.

This figure shows that the sufficient baryon asymmetry is produced in
a typical parameter region of the large-cutoff hypothesis:
$k\sim0.01$, $g\sim0.01-0.1$,
and $T_R\sim10^6$GeV, which turns out to be low enough to avoid the gravitino
overproduction. Thus it is revealed that the large-cutoff hypothesis is also
consistent with the observed baryon asymmetry.

\section{Discussion}\label{se:discussion}

We have studied the large-cutoff hypothesis from the viewpoint of
cosmology.
We first confirmed
that the spectral index in the new inflation
model has an upper bound $n_s \mathop{}_{\textstyle \sim}^{\textstyle <}  0.95$ (see Ref.\cite{Izaw})
and the large-cutoff hypothesis implies its boundary value,
which remarkably
agrees with the present experimental suggestion $n_s=0.95 \pm
0.02$ \cite{BOOMERANG}.
Secondly, we found a concrete setup where the sufficient baryon
asymmetry can be produced via non-thermal leptogenesis with the reheating
temperature low enough to avoid the gravitino overproduction in a typical
parameter region of large-cutoff hypothesis.

We again emphasize
that the large cut-off hypothesis has several advantages
from the viewpoint of particle-physics phenomenology.
It solves the FCNC problem
and produces the mass spectrum $m_0\sim10 m_{1/2}\sim 10\mu$,
which yields the correct electroweak symmetry breaking
\cite{Ibe}.
Furthermore,
the spectrum realized in the large-cutoff hypothesis
accommodates the
appropriate amount of the dark matter density
\cite{Mor}.

We also mention CP violations in the visible-sector
supersymmetric standard model
as a sensitive low-energy probe of the supersymmetry breaking.
Phases of the theory would be limited severely if the scalar masses were
to be less than the TeV scale.
In contrast, for $m_{0}\sim10$TeV, such a constraint
is far milder, with the very heavy scalar masses
expected to be realized in the large-cutoff hypothesis
from the viewpoint of electroweak symmetry breaking and dark matter,
as mentioned above.

The heavy scalar masses are remarkably consistent
with the cosmological constraint, as we saw in this paper.
Thus we conclude that the large-cutoff theory with the supergravity new inflation
and non-thermal leptogenesis is consistent with all the phenomenological requirements from cosmology and particle physics.

Finally we comment on other types of inflations.
The presence of the large cutoff seems advantageous for
other inflationary models such as hybrid inflation and chaotic inflation.
In particular, large-field inflations imply the presence
of a larger scale
(see Ref.\cite{Feldstein:2005bm})
than the reduced Planck scale.
In fact, we suspect that multiple inflations may be so generic
as to include various types of inflations as their components,
whose slow-roll conditions are realized by the large-cutoff mechanism
\cite{Ibe}.

\section*{Acknowledgments}
M.~I. thanks the Japan Society for the Promotion of Science
for financial support. This work is partially supported by Grand-in-Aid Scientific Research (s) 14102004.


\section{Introduction}

Reaching consensus is an important construct in distributed
coordination of multi-agent systems \cite{mesbahi2010graph,olfati2007consensus,zhang2018fully,degroot1974reaching}.
Although the consensus problem has been extensively investigated in the literature,
it has often been assumed that the network has scalar-weighted edges;
extensions of the scalar weights to matrix-valued weights
has become relevant in order to characterize interdependencies
among multi-dimensional states of neighboring agents.
Recently, a broader category of networks referred to as matrix-weighted
networks has been introduced to address such interdependencies~\cite{trinh2018matrix,sun2018dimensional}.
In fact, matrix-weighted networks arise in scenarios such as graph
effective resistance examined in the context of
distributed control and estimation \cite{barooah2006graph,tuna2017observability},
logical inter-dependencies amongst topics in opinion evolution \cite{friedkin2016network,ye2018continuous},
bearing-based formation control \cite{zhao2015translational}, dynamics of an array
of coupled LC oscillators \cite{tuna2019synchronization}, as well
as consensus and synchronization on matrix-weighted networks \cite{trinh2018matrix,tuna2016synchronization,pan2018bipartite}.

For matrix-weighted networks, network connectivity does not translates to
achieving consensus. To this end, properties of weight matrices play an important role in characterizing
consensus. For instance, positive
definiteness and positive semi-definiteness of weight matrices have been
employed to provide consensus conditions in \cite{trinh2018matrix};
negative definiteness and negative semi-definiteness of weight matrices
are further introduced in \cite{pan2018bipartite,su2019bipartite}.
In the meantime, the notion of network connectivity
can be further extended for matrix-valued networks. For instance, one
can identify edges with positive/negative definite matrices as ``strong'' connections; whereas
an edge weighted by positive/negative semi-definite matrices can be considered
as a ``weak'' connection \cite{trinh2017theory}.

To the best of our knowledge, conditions under which consensus
can be achieved for time-varying matrix-weighted networks have
not been developed in the literature; this is in contrast with conditions
that have been examined for scalar-weighted networks \cite{cao2008reaching,olfati2004consensus,ren2005consensus,Jadbabaie2003,cao2011necessary,moreau2005stability,meng2018uniform,meng2014modulus,proskurnikov2014consensus}.
In this paper, we provide necessary and/or sufficient conditions for
achieving consensus on matrix-weighed time-varying  networks. Under
mild assumptions on the switching pattern for such networks,
necessary and/or sufficient conditions for which average consensus
is achieved are provided in terms of the null space of the matrix-valued
Laplacian of the associated integral networks. In particular, for periodic
matrix-weighted time-varying networks with period $T>0$, a necessary
and sufficient condition for average consensus is obtained; we
further show that from a graph-theoretic perspective, when the integral
network over time span $[0,T)$ has a positive spanning tree, then
average consensus is achieved. Simulation results are provided to demonstrate
the theoretical analysis.

The remainder of this paper is organized as follows. Preliminaries
are introduced in \S\ref{sec:Preliminaries}. The problem formulation
is provided in \S\ref{sec:Problem-Formulation}, followed
by the consensus conditions in \S\ref{sec:Consensus-on-General}
and \S\ref{sec:Consensus-on-Periodic}, respectively. A simulation example is
presented in \S\ref{sec:Simulation-Results} followed by concluding
remarks in \S~\ref{sec:Conclusion}.

\section{Preliminaries \label{sec:Preliminaries}}

Let $\mathbb{R}$, $\mathbb{N}$ and $\mathbb{Z}_{+}$ be the set
of real numbers, natural numbers and positive integers, respectively.
For $n\in\mathbb{Z}_{+}$, denote $\underline{n}=\left\{ 1,2,\ldots,n\right\}$.
A symmetric matrix $M\in\mathbb{R}^{n\times n}$ is positive definite,
denoted by $M\succ0$, if $\boldsymbol{z}^{\top}M\boldsymbol{z}>0$
for all $\boldsymbol{z}\in\mathbb{\mathbb{R}}^{n}$ and $\boldsymbol{z\not}=\boldsymbol{0}$
and is positive semi-definite, denoted by $M\succeq0$, if $\boldsymbol{z}^{\top}M\boldsymbol{z}\ge0$
for all $\boldsymbol{z}\in\mathbb{\mathbb{R}}^{n}$.
The null space of a matrix $M\in\mathbb{R}^{n\times n}$ is denoted by $\text{{\bf null}}(M)=\left\{ \boldsymbol{z}\in\mathbb{R}^{n}|M\boldsymbol{z}=\boldsymbol{0}\right\} $.
\begin{lem}
\label{lem:Rayleigh Theorem}\cite{horn2012matrix} Let $M\in\mathbb{R}^{n\times n}$
be symmetric with eigenvalues $\lambda_{1}\leq\cdots\leq\lambda_{n}$.
Let $\boldsymbol{x}_{i_{1}},\cdots,\boldsymbol{x}_{i_{k}}$ be mutually
orthonormal vectors such that $M\boldsymbol{x}_{i_{p}}=\lambda_{i_{p}}\boldsymbol{x}_{i_{p}}$,
where $i_{p}\in\mathbb{Z}_{+}$, $p\in\underline{k}$ and $1\leq i_{1}<\cdots<i_{k}\leq n$.
Then
\[
\lambda_{i_{1}}=\underset{\{\boldsymbol{x}\not={\bf 0},\boldsymbol{x}\in S_k\}}{\text{{\bf min}}}\frac{\boldsymbol{x}^{\top}M\boldsymbol{x}}{\boldsymbol{x}^{\top}\boldsymbol{x}},
\]
and
\[
\lambda_{i_{k}}=\underset{\{\boldsymbol{x}\not={\bf 0},\boldsymbol{x}\in S_k\}}{\text{{\bf max}}}\frac{\boldsymbol{x}^{\top}M\boldsymbol{x}}{\boldsymbol{x}^{\top}\boldsymbol{x}},
\]
where $S_k=\text{{\bf span}}\{\boldsymbol{x}_{i_{1}},\cdots,\boldsymbol{x}_{i_{k}}\}$.
\end{lem}

\section{Problem Formulation \label{sec:Problem-Formulation}}

Consider a multi-agent system consisting of $n>1$ ($n\in\mathbb{Z}_{+}$)
agents whose interaction network is characterized by a matrix-weighted
time-varying graph $\mathcal{G}(t)=(\mathcal{V},\mathcal{E}(t),A(t))$, where $t$ refers to the time index. The node and edge sets of
$\mathcal{G}$ are denoted by $\mathcal{V}=\left\{ 1,2,\ldots,n\right\} $
and $\mathcal{E}(t)\subseteq\mathcal{V}\times\mathcal{V}$, respectively.
The weight on the edge $(i,j)\in\mathcal{E}(t)$ is encoded by the symmetric
matrix $A_{ij}(t)\in\mathbb{R}^{d\times d}$ such that $A_{ij}(t)\succeq0$
or $A_{ij}(t)\succ0$, and $A_{ij}(t)=0_{d\times d}$ for $(i,j)\not\in\mathcal{E}(t)$.\textbf{
}Thereby, the matrix-weighted adjacency matrix $A(t)=[A_{ij}(t)]\in\mathbb{R}^{dn\times dn}$
is a block matrix such that the block located in its $i$-th row and
the $j$-th column is $A_{ij}(t)$. It is assumed that $A_{ij}(t)=A_{ji}(t)$
for all $i\not\not=j\in\mathcal{V}$ and $A_{ii}(t)=0_{d\times d}$
for all $i\in\mathcal{V}$.

Denote the state of an agent $i\in\mathcal{V}$ as $\boldsymbol{x}_{i}(t)=[x_{i1}(t),x_{i2}(t),\ldots,x_{id}(t)]^{\top}\in\mathbb{R}^{d}$ evolving according to the protocol,
\begin{equation}
\dot{\boldsymbol{x}}_{i}(t)=-\sum_{j\in\mathcal{N}_{i}(t)}A_{ij}(t)(\boldsymbol{x}_{i}(t)-\boldsymbol{x}_{j}(t)),\thinspace i\in\mathcal{V},\label{equ:matrix-consensus-protocol}
\end{equation}
where $\mathcal{N}_{i}(t)=\left\{ j\in\mathcal{V}\,|\,(i,j)\in\mathcal{E}(t)\right\} $
denotes the neighbor set of agent $i\in\mathcal{V}$ at time $t$.
Note that protocol \eqref{equ:matrix-consensus-protocol} degenerates
into the scalar-weighted case when $A_{ij}(t)=a_{ij}(t)I$, where $a_{ij}(t)\in\mathbb{R}$
and $I$ denotes the $d\times d$ identity matrix.

Let $C(t)=\text{{\bf diag}}\left\{ C_{1}(t),C_{1}(t),\cdots,C_{n}(t)\right\} \in\mathbb{R}^{dn}$
be the matrix-valued degree matrix of $\mathcal{G}(t)$,
where $C_{i}(t)=\sum_{j\in\mathcal{N}_{i}}A_{ij}(t)\in\mathbb{R}^{d\times d}$.
The matrix-valued Laplacian is subsequently defined as $L(t)=C(t)-A(t)$.
The dynamics of the overall multi-agent system now admits the form,
\begin{equation}
\dot{\boldsymbol{x}}(t)=-L(t)\boldsymbol{x}(t),\label{equ:matrix-consensus-overall}
\end{equation}
where $\boldsymbol{x}(t)=[\boldsymbol{x}_{1}^{\top}(t),\boldsymbol{x}_{2}^{\top}(t),\ldots,\boldsymbol{x}_{n}^{\top}(t)]^{\top}\in\mathbb{R}^{dn}$.
\begin{defn}
Let $\boldsymbol{x}_{f}=\boldsymbol{1_{n}}\otimes(\frac{1}{n}\sum_{i=1}^{n}\boldsymbol{x_{i}}(0))$.
Then the multi-agent system \eqref{equ:matrix-consensus-overall}
admits an average consensus solution if $\lim{}_{t\rightarrow\infty}\boldsymbol{x}_{i}(t)=\lim{}_{t\rightarrow\infty}\boldsymbol{x}_{j}(t)=\boldsymbol{x}_{f}$
for all $i,j\in\mathcal{V}$.
\end{defn}

This work aims to investigate the necessary and/or sufficient
conditions under which the multi-agent system \eqref{equ:matrix-consensus-overall}
admits an average consensus solution. It is well-known that network
connectivity plays a central role in determining consensus for scalar-weighted
networks \cite{olfati2004consensus}.
However, as we shall show subsequently, definiteness
of the weight matrices is also a crucial factor in examining consensus for a matrix-weighted
networks in addition to its connectivity.
First, we shall recall a few facts on network connectivity. In graph theory, network
connectivity captures how a pair of nodes in the network can be ``connected'' by traversing
a sequence of consecutive edges called paths. A path of $\mathcal{G}(t)$ is a sequence
of edges of the form $(i_{1},i_{2}),(i_{2},i_{3}),\ldots,(i_{p-1},i_{p})$,
where nodes $i_{1},i_{2},\ldots,i_{p}\in\mathcal{V}$ are distinct;
in this case we say that node $i_{p}$ is reachable from $i_{1}$.
The graph $\mathcal{G}(t)$ is connected if any two distinct nodes in $\mathcal{G}(t)$
are reachable from each other. A tree is a connected graph with $n\ge2$
nodes and $n-1$ edges where $n\in\mathbb{Z}_{+}$.
For matrix weighted graphs, we adopt the following terminology.
An edge $(i,j)\in\mathcal{E}(t)$ is positive definite or positive semi-definite
if the associated weight matrix $A_{ij}(t)$ is positive definite
or positive semi-definite, respectively. A positive path in $\mathcal{G}(t)$
is a path such that every edge on this path is positive definite.
A tree in $\mathcal{G}(t)$ is a positive tree if every edge contained
in this tree is positive definite. A positive spanning tree of $\mathcal{G}(t)$
is a positive tree containing all nodes in $\mathcal{G}(t)$.
\section{Consensus on General Matrix-weighted Time-varying Networks \label{sec:Consensus-on-General}}

In order to analyze multi-agent systems of the form~\eqref{equ:matrix-consensus-overall},
we adopt the following assumption on the matrix-weighted time-varying
network~\cite{olfati2004consensus,ren2005consensus,cao2011necessary,meng2018uniform}.

\textbf{Assumption 1. }There exists a sequence $\{t_{k}|k\in\mathbb{N}\}$
such that $\lim_{k\rightarrow\infty}t_{k}=\infty$ and $\triangle t_{k}=t_{k+1}-t_{k}\in[\alpha,\beta]$
for all $k\in\mathbb{N}$, where $\beta>\alpha>0$, $t_{0}=0$, and
$\mathcal{G}(t)$ is time-invariant for $t\in[t_{k},t_{k+1})$ for
all $k\in\mathbb{N}$.

When $L(t)=L$ for all $t\in[0,\infty)$, then \eqref{equ:matrix-consensus-overall}
encodes the consensus protocol on a time-invariant network. The
following observation characterizes the structure of the null space of matrix-valued
Laplacian $L$ on time-invariant networks, that in turn, can determine the steady-state
of the network~\eqref{equ:matrix-consensus-overall}.
\begin{lem}
\label{lem:1}\cite{trinh2018matrix} Let $\mathcal{G}=(\mathcal{V},\mathcal{E},A)$
be a matrix-weighted time-invariant network with matrix-valued Laplacian
$L$. Then $L\succeq0$ and $\text{{\bf null}}(L)=\text{{\bf span}}\left\{ \mathcal{R},\mathcal{H}\right\}$,
where $\mathcal{R}=\text{{\bf range}}\{\boldsymbol{1}\otimes I_{d}\}$
and
\begin{align*}
\mathcal{H=}\{&[\boldsymbol{v}_{1}^{T},\boldsymbol{v}_{2}^{T},\cdots,\boldsymbol{v}_{n}^{T}]^{T}\in\mathbb{R}^{dn}\mid\\
 & (\boldsymbol{v}_{i}-\boldsymbol{v}_{j})\in\text{{\bf null}}(A_{ij}),\,(i,j)\in\mathcal{E}\}.
\end{align*}
\end{lem}
Note that the null space of a matrix-valued Laplacian is not only determined
by the network connectivity, but also by the properties of weight
matrices; this is distinct from the scalar-weighted networks.
For matrix-weighted time-invariant networks, a condition under
which the multi-agent system \eqref{equ:matrix-consensus-overall}
achieves consensus is provided in the following lemma.
\begin{lem}
\label{lem:2}\cite{trinh2018matrix} Let $\mathcal{G}=(\mathcal{V},\mathcal{E},A)$
be a matrix-weighted time-invariant network with matrix-valued Laplacian
$L$. Then the multi-agent system \eqref{equ:matrix-consensus-overall}
admits an average consensus if and only if $\text{{\bf null}}(L)=\mathcal{R}$.
\end{lem}
\begin{defn}
Define the consensus subspace of the multi-agent system \eqref{equ:matrix-consensus-overall}
as $\mathcal{R}=\text{{\bf range}}\{\boldsymbol{1}\otimes I_{d}\}$.
\end{defn}
\begin{lem}
\label{lem:3}\cite{trinh2018matrix} Let $\mathcal{G}=(\mathcal{V},\mathcal{E},A)$
be a matrix-weighted time-invariant network. If $\mathcal{G}$ has
a positive spanning tree $\mathcal{T}$, then the network
\eqref{equ:matrix-consensus-overall} admits an average consensus.
\end{lem}
In order to characterize the related properties of the time-varying
networks $\mathcal{G}(t)$ over a given time span, we introduce the
notion of matrix-weighted integral network; this notion proves
crucial in characterizing algebraic and graph-theoretic conditions for reaching
consensus on matrix-weighted time-varying networks.
\begin{defn}
\label{def:integral graph}Let $\mathcal{G}(t)=(\mathcal{V},\mathcal{E}(t),A(t))$
be a matrix-weighted time-varying network. Then the matrix-weighted
integral network of $\mathcal{G}(t)$ over time span $[\tau_{1},\tau_{2})\subseteq[0,\infty)$
is defined as $\widetilde{\mathcal{G}}_{[\tau_{1},\tau_{2})}=(\mathcal{V},\widetilde{\mathcal{E}},\widetilde{A})$,
where
\[
\widetilde{A}=\frac{1}{\tau_{2}-\tau_{1}}\intop_{\tau_{1}}^{\tau_{2}}A(t)dt
\]
and
\[
\widetilde{\mathcal{E}}=\left\{ (i,j)\in\mathcal{V}\times\mathcal{V}\mid\intop_{\tau_{1}}^{\tau_{2}}A_{ij}(t)dt\succ0\; \text{or}\; \intop_{\tau_{1}}^{\tau_{2}}A_{ij}(t)dt\succeq0\right\} .
\]
\end{defn}
According to Definition \ref{def:integral graph}, denote by $\widetilde{C}$
as the matrix-weighted degree matrix of $\widetilde{\mathcal{G}}_{[\tau_{1},\tau_{2})}$, that is,
$\widetilde{C}=\frac{1}{\tau_{2}-\tau_{1}}\intop_{\tau_{1}}^{\tau_{2}}C(t)dt$.
Denote the matrix-valued Laplacian of $\widetilde{\mathcal{G}}_{[\tau_{1},\tau_{2})}$
as $\widetilde{L}_{[\tau_{1},\tau_{2})}$. Thus,

\begin{align*}
\widetilde{L}_{[\tau_{1},\tau_{2})} & =\widetilde{C}-\widetilde{A}=\frac{1}{\tau_{2}-\tau_{1}}\intop_{\tau_{1}}^{\tau_{2}}L(t).
\end{align*}

According to Assumption 1, we denote $\mathcal{G}(t)$ on dwell
time $t\in[t_{k},t_{k+1})$ as $\mathcal{G}_{[t_{k},t_{k+1})}(t)=\mathcal{G}^{k}$
and denote the associated matrix-valued Laplacian as $L^{k}$, where
$k\in\mathbb{N}$. The following lemma reveals the connection between
the null space of the matrix-valued Laplacian of a sequence of matrix-weighted
networks and that of the corresponding integral network.
\begin{lem}
\label{thm:null space equality} Let $\mathcal{G}(t)$ be a matrix-weighted
time-varying network satisfying \textbf{Assumption 1. }Then $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})=\mathcal{R}$
if and only if
\[
\underset{i\in\underline{k^{\prime\prime}-k^{\prime}}}{\bigcap}\text{{\bf null}}(L^{k^{\prime}+i-1})=\mathcal{R},
\]
where $k^{\prime}<k^{\prime\prime}\in\mathbb{N}$.
\end{lem}
\begin{IEEEproof}
(Necessity) From the definition of matrix-valued Laplacian, one has
$\mathcal{R}\subseteq\underset{i\in\underline{k^{\prime\prime}-k^{\prime}}}{\bigcap}\text{{\bf null}}(L^{k^{\prime}+i-1})$.

Assume that $\underset{i\in\underline{k^{\prime\prime}-k^{\prime}}}{\bigcap}\text{{\bf null}}(L^{k^{\prime}+i-1})\neq\mathcal{R}$;
then there exists an $\boldsymbol{\eta}\notin\mathcal{R}$ such that
$L^{k^{\prime}+i-1}\boldsymbol{\eta}=\boldsymbol{0}$ for all $i\in\underline{k^{\prime\prime}-k^{\prime}}$,
which would imply,
\begin{align*}
\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})}\boldsymbol{\eta} & =\left(\frac{1}{\triangle t_{k^{\prime}}}\intop_{t_{k^{\prime}}}^{t_{k^{\prime\prime}}}L(t)dt\right)\boldsymbol{\eta}\\
 & =\frac{1}{\triangle t_{k^{\prime}}}\sum_{i=1}^{k^{\prime\prime}-k^{\prime}}L^{k^{\prime}+i-1}(t_{k^{\prime}+i}-t_{k^{\prime}+i-1})\boldsymbol{\eta}\\
 & =\boldsymbol{0},
\end{align*}
contradicting the fact that $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})=\mathcal{R}$.
Therefore, $\underset{i\in\underline{k^{\prime\prime}-k^{\prime}}}{\bigcap}\text{{\bf null}}(L^{k^{\prime}+i-1})=\mathcal{R}$.

(Sufficiency) Assume that $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})\neq\mathcal{R}$;
then there exists $\boldsymbol{\eta}\notin\mathcal{R}$ such that
$\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})}\boldsymbol{\eta}=\boldsymbol{0}$.
Hence, $\boldsymbol{\eta}^{\top}\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})}\boldsymbol{\eta}=0$,
implying that,
\begin{align*}
 & \frac{1}{\triangle t_{k^{\prime}}}\boldsymbol{\eta}^{\top}\left(\intop_{t_{k^{\prime}}}^{t_{k^{\prime\prime}}}L(t)dt\right)\boldsymbol{\eta}\\
 & =\frac{1}{\triangle t_{k^{\prime}}}\sum_{i=1}^{k^{\prime\prime}-k^{\prime}}\boldsymbol{\eta}^{\top}L^{k^{\prime}+i-1}(t_{k^{\prime}+i}-t_{k^{\prime}+i-1})\boldsymbol{\eta}\\
 & =0.
\end{align*}
Due to the fact that $L^{k^{\prime}+i-1}$ is positive semi-definite
for all $i\in\underline{k^{\prime\prime}-k^{\prime}}$, $\boldsymbol{\eta}^{\top}L^{k^{\prime}+i-1}\boldsymbol{\eta}=0$,
which would imply that $L^{k^{\prime}+i-1}\boldsymbol{\eta}=\boldsymbol{0}$;
this on the other hand, contradicts the premise $\underset{i\in\underline{k^{\prime\prime}-k^{\prime}}}{\bigcap}\text{{\bf null}}(L^{k^{\prime}+i-1})=\mathcal{R}$.
Thus $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})=\mathcal{R}$.
\end{IEEEproof}
In order to link the state evolution of the multi-agent system \eqref{equ:matrix-consensus-overall}
and the null space of the integral of matrix-weighted time-varying
networks, we need to employ the state transition matrix. Denote $\varPhi(k^{\prime},k^{\prime\prime})=e^{-L^{k^{\prime\prime}-1}\triangle t_{k^{\prime\prime}-1}}\cdots e^{-L^{k^{\prime}}\triangle t_{k^{\prime}}}$.
Then $\boldsymbol{x}(t_{k^{\prime\prime}})=\varPhi(k^{\prime},k^{\prime\prime})\boldsymbol{x}(t_{k^{\prime}})$,
where $k^{\prime}<k^{\prime\prime}\in\mathbb{N}$. Note that the matrix-valued
Laplacian $L$ has at least $d$ zero eigenvalues. Let $\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{dn}$
be the eigenvalues of $L$. Then we have $0=\lambda_{1}=\cdots=\lambda_{d}\leq\lambda_{d+1}\le\cdots\leq\lambda_{dn}$.
Denote by $\beta_{1}\geq\beta_{2}\geq\cdots\geq\beta_{dn}$ as the
eigenvalues of $e^{-Lt}$; then $\beta_{i}(e^{-Lt})=e^{-\lambda_{i}(L)t}$,
i.e., $1=\beta_{1}=\cdots=\beta_{d}\geq\beta_{d+1}\geq\cdots\geq\beta_{dn}$.
In the meantime, the eigenvector corresponding to the eigenvalue
$\beta_{i}(e^{-Lt})$ is equal to that corresponding to $\lambda_{i}(L)$.
Consider the symmetric matrix $\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime})$
which has at least $d$ eigenvalues at $1$. Let $\mu_{j}$ be the
eigenvalues of $\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime})$,
where $j\in\underline{dn}$ such that $\mu_{1}=\cdots=\mu_{d}=1$
and $\mu_{d+1}\geq\mu_{d+2}\geq\cdots\geq\mu_{dn}$. The following
lemma provides the relationship between the null space of the matrix-valued
Laplacian of $\widetilde{\mathcal{G}}_{[t_{k^{\prime}},t_{k^{\prime\prime}})}$
and the eigenvalue $\mu_{d+1}$ of $\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime})$. This relationship will prove useful in the proof of our main theorem.
\begin{lem}
Let $\mathcal{G}(t)$ be a matrix-weighted time-varying network satisfying
\textbf{Assumption 1. }Then $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})=\mathcal{R}$
if and only if
\[
\mu_{d+1}(\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime}))<1,
\]
where $k^{\prime}<k^{\prime\prime}\in\mathbb{N}$.
\end{lem}
\begin{IEEEproof}
(Sufficiency) Assume that $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})\neq\mathcal{R}$;
then according to Lemma \ref{thm:null space equality}, there exists
an $\boldsymbol{\eta}\notin\mathcal{R}$ such that $L^{k^{\prime}+i-1}\boldsymbol{\eta}=\boldsymbol{0}$
for all $i\in\underline{k^{\prime\prime}-k^{\prime}}$. Thus one can
obtain $e^{-L^{k^{\prime}+i-1}t}\boldsymbol{\eta}=\boldsymbol{\eta}$
for all $i\in\underline{k^{\prime\prime}-k^{\prime}}$ and $\varPhi(k^{\prime},k^{\prime\prime})\boldsymbol{\eta}=\boldsymbol{\eta}$.
According to the Lemma \ref{lem:Rayleigh Theorem}, one has
\begin{align*}
 & \mu_{d+1}(\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime}))\\
\geq & \frac{\boldsymbol{\eta}^{\top}\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime})\boldsymbol{\eta}}{\boldsymbol{\eta}^{\top}\boldsymbol{\eta}}\\
= & 1,
\end{align*}
contradicting,
\[
\mu_{d+1}(\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime}))<1.
\]
Therefore $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})=\mathcal{R}$
holds.

(Necessity) Assume that $\mu_{d+1}(\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime}))\geq1$.
Again, according to Lemma \ref{lem:Rayleigh Theorem}, there exists
a $\boldsymbol{\eta}\notin\mathcal{R}$ and $\boldsymbol{\eta}\neq\boldsymbol{0}$
such that
\begin{align*}
\mu_{d+1}(\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime}))
& = \frac{\boldsymbol{\eta}^{\top}\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime})\boldsymbol{\eta}}{\boldsymbol{\eta}^{\top}\boldsymbol{\eta}}\\
&\geq 1.
\end{align*}
Thus,
\[
\parallel\boldsymbol{\eta}\parallel\leq\parallel\varPhi(k^{\prime},k^{\prime\prime})\boldsymbol{\eta}\parallel.
\]

Let $\boldsymbol{\eta}_{k^{\prime}}=\boldsymbol{\eta}$ and $\boldsymbol{\eta}_{k^{\prime}+i}=e^{-L^{k^{\prime}+i-1}\triangle t_{k^{\prime}+i-1}}\boldsymbol{\eta}_{k^{\prime}+i-1}$
for $i\in\underline{k^{\prime\prime}-k^{\prime}}$. Due to the fact
$\lambda_{j}(e^{-L^{k^{\prime}+i-1}\triangle t_{k^{\prime}+i-1}})\leq1$
for $j\in\underline{dn}$ and $\boldsymbol{\eta}\notin\mathcal{R}$,
then
\[
\parallel e^{-L^{k^{\prime}+i-1}\triangle t_{k^{\prime}+i-1}}\boldsymbol{\eta}\parallel\leq\parallel\boldsymbol{\eta}\parallel,
\]
which implies that,
\begin{align*}
\parallel\boldsymbol{\eta}\parallel & \leq\parallel\varPhi(k^{\prime},k^{\prime\prime})\boldsymbol{\eta}\parallel\\
 & =\parallel\boldsymbol{\eta}_{k^{\prime\prime}}\parallel\leq\parallel\boldsymbol{\eta}_{k^{\prime\prime}-1}\parallel\leq\ldots\leq\parallel\boldsymbol{\eta}_{k^{\prime}}\parallel\\
 & =\parallel\boldsymbol{\eta}\parallel.
\end{align*}
Hence, $\parallel e^{-L^{k^{\prime}+i-1}\triangle t_{k^{\prime}+i-1}}\boldsymbol{\eta}_{k^{\prime}+i-1}\parallel=\parallel\boldsymbol{\eta}_{k^{\prime}+i-1}\parallel$
for $i\in\underline{k^{\prime\prime}-k^{\prime}}$. Then, one can
further derive $L^{k^{\prime}+i-1}\boldsymbol{\eta}_{k^{\prime}+i-1}=\boldsymbol{0}$;
thus $\boldsymbol{\eta}_{k^{\prime}+i-1}\in\text{{\bf ker}}(L^{k^{\prime}+i-1})$.
Note that since,
\begin{align*}
\parallel\boldsymbol{\eta}_{k^{\prime}+i}-\boldsymbol{\eta}_{k^{\prime}+i-1}\parallel & =\parallel e^{-L^{k^{\prime}+i-1}\triangle t_{k^{\prime}+i-1}}\boldsymbol{\eta}_{k^{\prime}+i-1}-\boldsymbol{\eta}_{k^{\prime}+i-1}\parallel\\
 & =\parallel\sum_{t=1}^{\infty}\frac{1}{t!}(-L^{k^{\prime}+i-1}\triangle t_{k^{\prime}+i-1})^{t}\boldsymbol{\eta}_{k^{\prime}+i-1}\parallel\\
 & =0,
\end{align*}
one can further obtain $\boldsymbol{\eta}_{k^{\prime}+i-1}=\boldsymbol{\eta}_{k^{\prime}+i}$
for $\forall i\in\underline{k^{\prime\prime}-k^{\prime}}$, which
implies that $\boldsymbol{\eta}\in\underset{i\in\underline{k^{\prime\prime}-k^{\prime}}}{\cap}\text{{\bf ker}}(L^{k^{\prime}+i-1})$
and $\text{{\bf null}}(\widetilde{L}_{[t_{k^{\prime}},t_{k^{\prime\prime}})})\neq\mathcal{R}$.
This is a contradiction however. As such $\mu_{d+1}(\varPhi(k^{\prime},k^{\prime\prime})^{\top}\varPhi(k^{\prime},k^{\prime\prime}))<1$.
\end{IEEEproof}
\begin{thm}
\label{thm:consensus theorem-necessary}Let $\mathcal{G}(t)$ be a
matrix-weighted time-varying network satisfying \textbf{Assumption
1.} If the multi-agent network \eqref{equ:matrix-consensus-overall}
admits an average consensus, then there exists a subsequence of $\{t_{k}|k\in\mathbb{N}\}$
denoted by $\{t_{k_{l}}|l\in\mathbb{N}\}$, such that the null space
of the matrix-valued Laplacian of $\widetilde{\mathcal{G}}_{[t_{k_{l}},t_{k_{l+1}})}(t)$
is $\mathcal{R}$, namely, $\text{{\bf null}}(\widetilde{L}_{[t_{k_{l}},t_{k_{l+1}})})=\mathcal{R}$
for all $l\in\mathbb{N}$, where $\triangle t_{k_{l}}=t_{k_{l+1}}-t_{k_{l}}<\infty$
and $t_{k_{0}}=t_{0}$.
\end{thm}
\begin{IEEEproof}
Assume that there does not exist a subsequence $\{t_{k_{l}}|l\in\mathbb{N}\}$
such that $\text{{\bf null}}(\widetilde{L}_{[t_{k_{l}},t_{k_{l+1}})})=\mathcal{R}$
for all $l\in\mathbb{N}$, which implies that there exists $k^{*}\in\mathbb{N}$
such that $\text{{\bf null}}(\widetilde{L}_{[t_{k^{*}},\infty)})\neq\mathcal{R}$.
Then $\underset{k\geq k^{*},k\in\mathbb{N}}{\bigcap}\text{{\bf null}}(L^{k})\neq\mathcal{R}$.
Denote $\boldsymbol{\eta}\notin\mathcal{R}$ and $\boldsymbol{\eta}\in\underset{k\geq k^{*},k\in\mathbb{N}}{\bigcap}\text{{\bf null}}(L^{k})$.
Then $L^{k}\boldsymbol{\eta}=\boldsymbol{0}$ for all $k\geq k^{*},k\in\mathbb{N}$.
One can choose a suitable $\boldsymbol{x}(0)$ such that $\boldsymbol{x}(t_{k^{*}})=\boldsymbol{\eta}$;
then $\underset{t\rightarrow\infty}{\text{{\bf lim}}}\boldsymbol{x}(t)=\boldsymbol{\eta}$,
establishing a contradiction to the fact that the multi-agent network
\eqref{equ:matrix-consensus-overall} admits an average consensus.
Thus, there exists a subsequence $\{t_{k_{l}}|l\in\mathbb{N}\}$
such that $\text{{\bf null}}(\widetilde{L}_{[t_{k_{l}},t_{k_{l+1}})})=\mathcal{R}$
for all $l\in\mathbb{N}$.
\end{IEEEproof}
\begin{rem}
Although the existence of a subsequence of $\{t_{k}|k\in\mathbb{N}\}$
denoted by $\{t_{k_{l}}|l\in\mathbb{N}\}$ such that $\text{{\bf null}}(\widetilde{L}_{[t_{k_{l}},t_{k_{l+1}})})=\mathcal{R}$,
for all $l\in\mathbb{N}$ is a necessary condition for an average
consensus, it is not sufficient. To see this fact, we choose, for
instance, the multi-agent system $\dot{\boldsymbol{x}}(t)=-\frac{1}{t^{2}}L\boldsymbol{x}(t)$,
where $L$ is the matrix-valued Laplacian of a time-invariant matrix-weighted
network for which $\text{{\bf null}}(L)=\mathcal{R}$. Now consider the
underlying matrix-weighted time-varying network corresponding to the
Laplacian matrix $\frac{1}{t^{2}}L$. Then for the arbitrary subsequence
$\{t_{k_{l}}|l\in\mathbb{N}\}$ of $\{t_{k}|k\in\mathbb{N}\}$, one
always has $\text{{\bf null}}(\widetilde{L}_{[t_{k_{l}},t_{k_{l+1}})})=\mathcal{R}$
for all $l\in\mathbb{N}$. However, the solution to the above system
is $\boldsymbol{x}(t)=e^{\frac{L}{t}}e^{-L}\boldsymbol{x}(0)$, and
$\text{{\bf lim}}_{t\rightarrow\infty}\boldsymbol{x}(t)=e^{-L}\boldsymbol{x}(0)$.
Therefore, an average consensus cannot be achieved in this example.
Thus, we need additional conditions in order to guarantee average consensus
for \eqref{equ:matrix-consensus-overall}. These observations motivate the following
result.
\end{rem}
\begin{thm}
\label{thm:consensus theorem}Let ${\normalcolor \mathcal{G}(t)}$
be a matrix-weighted time-varying network satisfying \textbf{Assumption
1}; furthermore, suppose there exists a subsequence of $\{t_{k}|k\in\mathbb{N}\}$, denoted
by $\{t_{k_{l}}|l\in\mathbb{N}\}$, such that $\text{{\bf null}}(\widetilde{L}_{[t_{k_{l}},t_{k_{l+1}})})=\mathcal{R}$
for all $l\in\mathbb{N}$, where $\triangle t_{k_{l}}=t_{k_{l+1}}-t_{k_{l}}<\infty$
and $t_{k_{0}}=t_{0}$. If there exists a scalar $0<q<1$
such that $\mu_{d+1}(\varPhi(t_{k_{l}},t_{k_{l+1}})^{\top}\varPhi(t_{k_{l}},t_{k_{l+1}}))\leq q$
for all $l\in\mathbb{N}$, then the multi-agent network \eqref{equ:matrix-consensus-overall}
admits an average consensus.
\end{thm}
\begin{IEEEproof}
Let $\boldsymbol{\omega}(t)=\boldsymbol{x}(t)-\boldsymbol{x}_{f}$.
Then $\dot{\boldsymbol{\omega}}(t)=-L(t)\boldsymbol{\omega}(t)$.
Choose $\boldsymbol{\omega}(0)\notin\mathcal{R}$ and observe that,
\begin{align*}
 & \mu_{d+1}(\varPhi(t_{k_{0}},t_{k_{1}})^{\top}\varPhi(t_{k_{0}},t_{k_{1}}))\\
\geq & \frac{\boldsymbol{\omega}(0)^{\top}(\varPhi(t_{k_{0}},t_{k_{1}})^{\top}\varPhi(t_{k_{0}},t_{k_{1}}))\boldsymbol{\omega}(0)}{\boldsymbol{\omega}(0)^{\top}\boldsymbol{\omega}(0)}\\
= & \frac{\boldsymbol{\omega}(t_{k_{1}})^{\top}\boldsymbol{\omega}(t_{k_{1}})}{\boldsymbol{\omega}(0)^{\top}\boldsymbol{\omega}(0)},
\end{align*}
implying that,
\[
\parallel\boldsymbol{\omega}(t_{k_{1}})\parallel\leq\mu_{d+1}(\varPhi(t_{k_{0}},t_{k_{1}})^{\top}\varPhi(t_{k_{0}},t_{k_{1}}))^{\frac{1}{2}}\parallel\boldsymbol{\omega}(0)\parallel.
\]
Therefore,
\begin{align*}
\parallel\boldsymbol{\omega}(t_{k_{l+1}})\parallel & \leq\mu_{d+1}(\varPhi(t_{k_{l}},t_{k_{l+1}})^{\top}\varPhi(t_{k_{l}},t_{k_{l+1}}))^{\frac{1}{2}}\parallel\boldsymbol{\omega}(t_{k_{l}})\parallel\\
 & \leq\mu_{d+1}(\varPhi(t_{k_{l}},t_{k_{l+1}})^{\top}\varPhi(t_{k_{l}},t_{k_{l+1}}))^{\frac{1}{2}}\\
{\color{red}} & \thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\vdots\\
{\color{red}} & \mu_{d+1}(\varPhi(t_{k_{0}},t_{k_{1}})^{\top}\varPhi(t_{k_{0}},t_{k_{1}}))^{\frac{1}{2}}\parallel\boldsymbol{\omega}(0)\parallel\\
 & \leq q^{\frac{1}{2}(l+1)}\parallel\boldsymbol{\omega}(0)\parallel.
\end{align*}
Let
\[
V(t)=\boldsymbol{\omega}(t)^{\top}\boldsymbol{\omega}(t)=\parallel\boldsymbol{\omega}(t)\parallel^{2};
\]
then
\[
\dot{V}(t)=2\boldsymbol{\omega}(t)^{\top}(-L(t))\boldsymbol{\omega}(t)\leq0.
\]
 Thus
\[
\parallel\boldsymbol{\omega}(t)\parallel\leq\parallel\boldsymbol{\omega}(t_{k_{l+1}})\parallel\leq q^{\frac{1}{2}(l+1)}\parallel\boldsymbol{\omega}(0)\parallel,
\]
for $\forall t\in[t_{k_{l+1}},\infty)$. Note that $0<q<1$, and hence,
\[
{\displaystyle \lim_{t\rightarrow\infty}}\parallel\boldsymbol{\omega}(t)\parallel=0.
\]
As such, the multi-agent network \eqref{equ:matrix-consensus-overall}
achieves average consensus.
\end{IEEEproof}

\section{Consensus on Periodic Matrix-weighted Time-varying Networks \label{sec:Consensus-on-Periodic}}

In the subsequent discussion, we consider a special class of time-varying
networks, where $\mathcal{G}(t)$ is periodic. The periodic network
$\mathcal{G}(t)$ is formally characterized by the following assumption.

\textbf{Assumption 2. } There exists a $T>0$ such that $\mathcal{G}(t+T)=\mathcal{G}(t)$
for any $t\geq0$. Moreover, there exists a time sequence $\left\{ t_{k}|k\in\mathbb{N}\right\}$
satisfying $\triangle t_{k}=t_{k+1}-t_{k}>\alpha$ for all $k\in\mathbb{N}$,
where $\alpha>0$, and there exists $m>2$ ($m\in\mathbb{N}$) partitions
for each time span $[lT,(l+1)T)$ for which,
\[
lT=t_{lm}<t_{lm+1}<\cdots<t_{(l+1)m}=(l+1)T,\thinspace l\in\mathbb{N},
\]
and $\mathcal{G}(t)$ is time-invariant for $t\in[t_{k},t_{k+1})$,
where $k\in\mathbb{N}$.

Under Assumption 2, we now proceed to provide the algebraic and graph-theoretic
conditions under which the multi-agent system \eqref{equ:matrix-consensus-overall}
admits average consensus.
\begin{thm}
\label{thm:consensus theorem-periodic}Let $\mathcal{G}(t)$ be a
periodic matrix-weighted time-varying network satisfying \textbf{Assumption
2}. Then the multi-agent network \eqref{equ:matrix-consensus-overall}
admits average consensus if and only if,
\[
\text{{\bf null}}(\widetilde{L}_{[0,T)})=\mathcal{R}.
\]
\end{thm}
\begin{IEEEproof}
(Necessity) Assume that $\text{{\bf null}}(\widetilde{L}_{[0,T)})\neq\mathcal{R}$;
then there exists a $\boldsymbol{\eta}\notin\mathcal{R}$ such that
$L^{i-1}\boldsymbol{\eta}=\boldsymbol{0}$ for all $i\in\underline{m}$.
Let $\boldsymbol{x}(0)=\boldsymbol{\eta}$. Thereby, we can obtain $\boldsymbol{x}(t)=\boldsymbol{\eta}$
for all $t>0$, contradicting the fact that the multi-agent
network \eqref{equ:matrix-consensus-overall} admits average consensus.

(Sufficiency) Let $\boldsymbol{\omega}(t)=\boldsymbol{x}(t)-\boldsymbol{x}_{f}$;
then we have $\dot{\boldsymbol{\omega}}(t)=-L(t)\boldsymbol{\omega}(t)$.
Denote
\[
\varPhi(0,T)=e^{-L^{m-1}\triangle t_{m-1}}\cdots e^{-L^{0}\triangle t_{0}},
\]
and choose $\boldsymbol{\omega}(0)\notin\mathcal{R}$. Then,
\begin{align*}
\mu_{d+1}(\varPhi(0,T)^{\top}\varPhi(0,T)) & \geq\frac{\boldsymbol{\omega}(0)^{\top}(\varPhi(0,T)^{\top}\varPhi(0,T))\boldsymbol{\omega}(0)}{\boldsymbol{\omega}(0)^{\top}\boldsymbol{\omega}(0)}\\
 & =\frac{\boldsymbol{\omega}(T)^{\top}\boldsymbol{\omega}(T)}{\boldsymbol{\omega}(0)^{\top}\boldsymbol{\omega}(0)},
\end{align*}
implying that,
\[
\boldsymbol{\omega}(T)^{\top}\boldsymbol{\omega}(T)\leq\mu_{d+1}(\varPhi(0,T)^{\top}\varPhi(0,T))\boldsymbol{\omega}(0)^{\top}\boldsymbol{\omega}(0).
\]
Therefore
\[
\parallel\boldsymbol{\omega}(T)\parallel\leq\mu_{d+1}(\varPhi(0,T)^{\top}\varPhi(0,T))^{\frac{1}{2}}\parallel\boldsymbol{\omega}(0)\parallel,
\]
implying that,
\[
\parallel\boldsymbol{\omega}(kT)\parallel\leq\mu_{d+1}(\varPhi(0,T)^{\top}\varPhi(0,T))^{\frac{1}{2}k}\parallel\boldsymbol{\omega}(0)\parallel.
\]
Hence, one has
\[
\parallel\boldsymbol{\omega}(t)\parallel\leq\parallel\boldsymbol{\omega}(kT)\parallel\leq\mu_{d+1}(\varPhi(0,T)^{\top}\varPhi(0,T))^{\frac{1}{2}k}\parallel\boldsymbol{\omega}(0)\parallel,
\]
for $t\in[kT,(k+1)T)$; then $\text{{\bf lim}}_{t\rightarrow\infty}\parallel\boldsymbol{\omega}(t)\parallel=0$.
Therefore, the multi-agent network \eqref{equ:matrix-consensus-overall}
admits average consensus.
\end{IEEEproof}
Theorem \ref{thm:consensus theorem-periodic} provides
an algebraic condition for reaching consensus for periodic matrix-weighted
time-varying networks using the structure of the null space
of the matrix-valued Laplacian matrix of the corresponding integral
network. An analogous graph theoretic condition is as follows.
\begin{thm}
\label{thm:graph condition-integral graph}Let $\mathcal{G}(t)$ be
a periodic matrix-weighted time-varying network satisfying \textbf{Assumption
2}. If the integral graph of $\mathcal{G}(t)$ over time span $[0,T)$
has a positive spanning tree, then the multi-agent network \eqref{equ:matrix-consensus-overall}
admits average consensus.
\end{thm}
\begin{IEEEproof}
Let $\widetilde{\mathcal{G}}_{[0,T)}$ be the integral network of
$\mathcal{G}(t)$ over time span $[0,T)$. If $\widetilde{\mathcal{G}}_{[0,T)}$
has a positive spanning tree, from Lemma \ref{lem:2} and Lemma \ref{lem:3},
one has $\text{{\bf null}}(\widetilde{L}_{[0,T)})=\mathcal{R}$, where
$\widetilde{L}_{[0,T)}$ is the matrix-valued Laplacian matrix of $\widetilde{\mathcal{G}}_{[0,T)}$.
Theorem \ref{thm:consensus theorem-periodic}, now implies that
the multi-agent network \eqref{equ:matrix-consensus-overall}
admits average consensus.
\end{IEEEproof}

\section{Simulation Results \label{sec:Simulation-Results}}

Consider a sequence of matrix-weighted networks, consisting
of (the same) four agents, and the topologies of the networks are as $\mathcal{G}_{1},\mathcal{G}_{2}$
and $\mathcal{G}_{3}$, as shown in Figure \ref{fig:example-network}.
Note that $n=4$ and $d=2$ in this example.
\begin{figure}[H]
\begin{centering}
\begin{tikzpicture}[scale=0.6]

    \node (n1) at (1 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {1};
	\node (n2) at (0,-1.5) [circle,circular drop shadow,fill=black!20,draw] {2};
    \node (n3) at (2,-1.5) [circle,circular drop shadow,fill=black!20,draw] {3};
    \node (n4) at (3 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {4};

    \node (G2) at (1.2,-2.5) {\large{$\mathcal{G}_1$}};

	\draw[-, ultra thick, color=black!70] [-] (n1) -- (n2);
	\draw [-, ultra thick, dashed, color=black!70] (n2) -- (n3);

\end{tikzpicture}\,\,\,\,\,\,\,\,\,\,\,\,\begin{tikzpicture}[scale=0.6]

    \node (n1) at (1 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {1};
	\node (n2) at (0,-1.5) [circle,circular drop shadow,fill=black!20,draw] {2};
    \node (n3) at (2,-1.5) [circle,circular drop shadow,fill=black!20,draw] {3};
    \node (n4) at (3 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {4};
    \node (G2) at (1.2,-2.5) {\large{$\mathcal{G}_2$}};

	\draw[-, ultra thick, color=black!70] (n2) -- (n4);
	\draw[-, ultra thick, dashed, color=black!70] (n4) -- (n3);

\end{tikzpicture}\,\,\,\,\,\,\,\,\,\,\,\,\begin{tikzpicture}[scale=0.6]

    \node (n1) at (1 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {1};
	\node (n2) at (0,-1.5) [circle,circular drop shadow,fill=black!20,draw] {2};
    \node (n3) at (2,-1.5) [circle,circular drop shadow,fill=black!20,draw] {3};
    \node (n4) at (3 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {4};

    \node (G2) at (1.2,-2.5) {\large{$\mathcal{G}_3$}};

	\draw[-, ultra thick, color=black!70] (n2) -- (n3);

\end{tikzpicture}
\par\end{centering}
\caption{Three matrix-weighted networks $\mathcal{G}_{1}$, $\mathcal{G}_{2}$,
and $\mathcal{G}_{3}$. \textcolor{black}{Those edges weighted by
positive definite matrices are illustrated by solid lines and edges
weighted by positive semi-definite matrices are illustrated by dotted
lines.}}
\label{fig:example-network}
\end{figure}

The matrix-valued edge weights for each network are,
\[
A_{12}(\mathcal{G}_{1})=\left[\begin{array}{cc}
1 & 1\\
1 & 2
\end{array}\right],\thinspace\thinspace A_{23}(\mathcal{G}_{1})=\left[\begin{array}{cc}
1 & 1\\
1 & 1
\end{array}\right],
\]
\[
A_{24}(\mathcal{G}_{2})=\left[\begin{array}{cc}
1 & 0\\
0 & 2
\end{array}\right],\thinspace\thinspace A_{34}(\mathcal{G}_{2})=\left[\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right],
\]
and
\[
A_{23}(\mathcal{G}_{3})=\left[\begin{array}{cc}
1 & -1\\
-1 & 2
\end{array}\right],
\]
respectively. The matrix-valued Laplacian matrices corresponding to
above three networks are,

\[
L(\mathcal{G}_{1})=\left[\begin{array}{cccccccc}
1 & 1 & -1 & -1 & 0 & 0 & 0 & 0\\
1 & 2 & -1 & -2 & 0 & 0 & 0 & 0\\
-1 & -1 & 2 & 2 & -1 & -1 & 0 & 0\\
-1 & -2 & 2 & 3 & -1 & -1 & 0 & 0\\
0 & 0 & -1 & -1 & 1 & 1 & 0 & 0\\
0 & 0 & -1 & -1 & 1 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0
\end{array}\right],
\]

\[
L(\mathcal{G}_{2})=\left[\begin{array}{cccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & -1 & 0\\
0 & 0 & 0 & 2 & 0 & 0 & 0 & -2\\
0 & 0 & 0 & 0 & 1 & 0 & -1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & -1 & 0 & -1 & 0 & 2 & 0\\
0 & 0 & 0 & -2 & 0 & 0 & 0 & 2
\end{array}\right],
\]
and

\[
L(\mathcal{G}_{3})=\left[\begin{array}{cccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & -1 & -1 & 1 & 0 & 0\\
0 & 0 & -1 & 2 & 1 & -2 & 0 & 0\\
0 & 0 & -1 & 1 & 1 & -1 & 0 & 0\\
0 & 0 & 1 & -2 & -1 & 2 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0
\end{array}\right],
\]
respectively.

\begin{figure}[tbh]
\begin{centering}
\begin{tikzpicture}[scale=1]

    \node (G1) at (0,0) [] {\large{$\mathcal{G}_1$}};
    \node (G2) at (2.5,0) [] {\large{$\mathcal{G}_2$}};
    \node (G3) at (5,0) [] {\large{$\mathcal{G}_3$}};

\path[]
	(G1) [->,  thick] edge node[above] {$ 2\Delta t$} (G2);

\path[]
	(G2) [->,  thick]  edge node[above] {$3\Delta t$} (G3);

\path[]
	(G1) [<-,  thick, bend left=50] edge node[above] {$\Delta t$} (G3);

\end{tikzpicture}
\par\end{centering}
\caption{Switching sequence amongst networks $\mathcal{G}_{1}$, $\mathcal{G}_{2}$,
and $\mathcal{G}_{3}$.}
\label{fig:switching-network}
\end{figure}

Consider a time sequence $\{t_{k}\thinspace|\thinspace k\in\mathbb{N}\}$
such that $t_{k}=k\Delta t$ where $\Delta t>0$. The evolution is
initiated from network $\mathcal{G}_{1}$ (i.e., ${\normalcolor \mathcal{G}(0)=\mathcal{G}_{1}}$)
with $\boldsymbol{x}_{1}(0)=[0.6787,\thinspace0.7577]^{\top}$, $\boldsymbol{x}_{2}(0)=[0.7431,\thinspace0.3922]^{\top}$,
$\boldsymbol{x}_{3}(0)=[0.6555,\thinspace0.1712]^{\top}$ and $\boldsymbol{x}_{4}(0)=[0.7060,\thinspace0.0318]^{\top}$.
The switching among networks $\mathcal{G}_{1},\mathcal{G}_{2}$ and
$\mathcal{G}_{3}$ satisfies $\{t_{k}\thinspace|\thinspace k\in\mathbb{N}\}$,
\[
{\normalcolor \mathcal{G}(t)}=\begin{cases}
\begin{array}{c}
\mathcal{G}_{1},\\
\mathcal{G}_{2},\\
\mathcal{G}_{3},
\end{array} & \begin{array}{c}
t\in[t_{6l},t_{6l+2}),\\
t\in[t_{6l+2},t_{6l+5}),\\
t\in[t_{6l+5},t_{6(l+1)}),
\end{array}\end{cases}
\]
where $l\in\mathbb{N}$. The network switching process is demonstrated
in Figure \ref{fig:switching-network}. Examine the dimension of the
null space of $L(\mathcal{G}_{1})$, $L(\mathcal{G}_{2})$ and $L(\mathcal{G}_{3})$,
respectively. We have $\text{{\bf null}}(L(\mathcal{G}_{1}))\not=\mathcal{R}$,
$\text{{\bf null}}(L(\mathcal{G}_{2}))\not=\mathcal{R}$ and $\text{{\bf null}}(L(\mathcal{G}_{3}))\not=\mathcal{R}$.
However, note that from Figure \ref{fig:integral-network},
the integral graph of $L(\mathcal{G}_{1})$, $L(\mathcal{G}_{2})$
and $L(\mathcal{G}_{3})$ over time span $[t_{6l},t_{6(l+1)})$, where
$l\in\mathbb{N}$, denoted by $\widetilde{\mathcal{G}}$, has a positive
spanning tree $\mathcal{T}(\widetilde{\mathcal{G}})$. Therefore,
according to Theorem \ref{thm:graph condition-integral graph}, the
multi-agent system \eqref{equ:matrix-consensus-overall} admits an
average consensus solution at $[0.6958,\thinspace0.3382]^{\top}$;
see Figure \ref{fig:trajectory}.

\begin{figure}[tbh]
\begin{centering}
\begin{tikzpicture}[scale=0.7]

    \node (n1) at (1 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {1};
	\node (n2) at (0,-1.5) [circle,circular drop shadow,fill=black!20,draw] {2};
    \node (n3) at (2,-1.5) [circle,circular drop shadow,fill=black!20,draw] {3};
    \node (n4) at (3 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {4};

    \node (G2) at (1.2,-2.5) {{$\widetilde{\mathcal{G}}$}};

	\draw[-, ultra thick, color=black!70] (n1) -- (n2);
	\draw[-, ultra thick, color=black!70] (n2) -- (n3);
    \draw[-, ultra thick, color=black!70] (n2) -- (n4);
    \draw[-, ultra thick, dashed, color=black!70] (n3) -- (n4);

\end{tikzpicture}\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{tikzpicture}[scale=0.7]

    \node (n1) at (1 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {1};
	\node (n2) at (0,-1.5) [circle,circular drop shadow,fill=black!20,draw] {2};
    \node (n3) at (2,-1.5) [circle,circular drop shadow,fill=black!20,draw] {3};
    \node (n4) at (3 ,0.3) [circle,circular drop shadow,fill=black!20,draw] {4};

    \node (G2) at (1.2,-2.5) {{$\mathcal{T}(\widetilde{\mathcal{G}}$})};

	\draw[-, ultra thick, color=black!70] (n1) -- (n2);
	\draw[-, ultra thick, color=black!70] (n2) -- (n3);
    \draw[-, ultra thick, color=black!70] (n2) -- (n4);

\end{tikzpicture}
\par\end{centering}
\caption{The integral graph $\mathcal{G}(t)$ over time span $[t_{6l},t_{6(l+1)})$
where $l\in\mathbb{N}$ (left) and the associated positive spanning
tree $\mathcal{T}(\widetilde{\mathcal{G}})$ (right).}
\label{fig:integral-network}
\end{figure}

\begin{figure}[tbh]
\begin{centering}
\includegraphics[width=9cm]{figures/trajectory}
\par\end{centering}
\caption{State evolution in the multi-agent system \eqref{equ:matrix-consensus-overall}.}
\label{fig:trajectory}
\end{figure}

\section{Conclusion \label{sec:Conclusion}}

This paper examines consensus problems on matrix-weighted time-varying
networks. For such networks, necessary
and/or sufficient conditions for reaching average consensus are provided.
Furthermore, for matrix-weighted
periodic time-varying networks, necessary and sufficient algebraic
and graph theoretic conditions are obtained for reaching consensus.

\bibliographystyle{IEEEtran}
\phantomsection\addcontentsline{toc}{section}{\refname}