# Opinion dynamics under common influencer assumption
or leadership control

Chiara Cicolani111Dipartimento di Ingegneria e Scienze dell’Informazione e
Matematica
Università dell”Aquila
Via Vetoio, 67100 L’Aquila - Italy
Email<EMAIL_ADDRESS>Badis Ouahab 222 École Militaire
Polytechnique, 16046 Algiers, Algeria
Email<EMAIL_ADDRESS>& Cristina Pignotti333Dipartimento di
Ingegneria e Scienze dell’Informazione e Matematica
Università dell”Aquila
Via Vetoio, 67100 L’Aquila - Italy
Email<EMAIL_ADDRESS>
###### Abstract

We study Hegselmann-Krause type opinion formation models with non-universal
interaction and time-delayed coupling. We assume the presence of a common
influencer between two different agents. Moreover, we explore two cases in
which such an assumption does not hold but leaders with independent opinion
are present. By using careful estimates on the system’s trajectories, we are
able to prove asymptotic convergence to consensus estimates. Some numerical
tests illustrate the theoretical results.

##  1 Introduction

In recent years, multiagent systems have become a highly attractive research
topic due to their numerous applications in various scientific fields. They
naturally appear in biology [6, 9, 22], ecology [44], economics [9, 36],
social sciences [2, 3, 4, 10, 40], physics [25, 45], control theory and
optimization [1, 24, 38, 39, 46], engineering, and robotics [5, 23]. For
additional applications, see also [33, 35, 47].

A significant feature often analyzed is the potential emergence of self-
organization, leading the group’s agents to exhibit globally collective
behaviors. Here, we focus on the renowned Hegselmann-Krause model for opinion
formation, originally proposed in [32]. Since then, several generalizations
have been proposed (see, for instance, [2, 3, 7, 8, 11, 34]).

One of the most natural extensions of the Hegselmann-Krause model involves the
analysis of time-delayed interactions, to account for the time required for
each agent to receive information from other agents or reaction times. Opinion
formation models in the presence of time delay effects have already been
studied by several authors, see, for example, [16, 18, 20, 21, 29, 31, 37,
38]. Regarding the second-order version of the Hegselmann-Krause model, namely
the Cucker-Smale model [22], introduced to describe flocking phenomena,
delayed interactions have also been considered in various works, see, for
example, [12, 13, 14, 15, 17, 19, 26, 30, 41, 42, 43].

In this paper, we focus on studying the convergence to consensus for some
Hegselmann-Krause opinion formation models with time-delayed coupling. We
assume that the time delay between two agents is dependent on the agents
themselves. Specifically, we examine scenarios where interactions are not
universal but where a common influencer always exists between any two
different agents.

Consider a finite set of $N\in\mathbb{N},N\geq 3,$ agents. Let
$x_{i}(t)\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ be the opinion of the i-th
agent at time $t.$ We assume that there is a time lag in the interaction
between the agents, described by positive constants $\tau_{ij},\
i,j=1,\dots,N.$ Moreover, we assume that the interaction is non-universal.
Then, the interaction between the particles of the system is described by the
following Hegselmann-Krause type model:

$\frac{d}{dt}x_{i}(t)=\frac{1}{N-1}\sum_{j\neq
i}\chi_{ij}a_{ij}(t)(x_{j}(t-\tau_{ij})-x_{i}(t)),\quad t>0,\ i=1,...,N,$
(1.1)

with weights $a_{ij}(t)$ of the form

$a_{ij}(t):=\psi_{ij}(x_{i}(t),x_{j}(t-\tau_{ij})),\quad\ i,j=1,...,N,$ (1.2)

where the influence functions $\psi_{ij}:{{\rm I}\kern-1.49994pt{\rm
R}}^{d}\times{{\rm I}\kern-1.49994pt{\rm R}}^{d}\rightarrow{{\rm
I}\kern-1.49994pt{\rm R}}$ are positive, bounded and continuous. We denote

$K_{ij}:=\parallel\psi_{ij}\parallel_{\infty},\quad\forall i,j=1,\dots,N,$

and let

$K:=\max_{i,j=1,\dots,N}K_{ij}.$ (1.3)

The function $\chi_{ij}$ for $i,j=1,\dots,N$ is defined as

$\chi_{ij}=\begin{cases}1\quad\mbox{if $j$ transmits information to $i$,}\\\
0\quad\mbox{otherwise}.\end{cases}$ (1.4)

Moreover, let us assume the following initial conditions:

$x_{i}(t)=x_{i}^{0}(t),\quad i=1,\dots,N,\ t\in[-\tau,0],$ (1.5)

where

$\tau:=\max_{i,j=1,\dots,N}\tau_{ij},$

and $x_{i}^{0}:[-\tau,0]\rightarrow{{\rm I}\kern-1.49994pt{\rm R}}^{d},\
i=1,\dots,N,$ are continuous functions. For existence results about the above
model we refer to the classical books [27, 28]. Here, we are interested in
convergence to consensus results.

We make the following assumption over the structure of the system:

(CI) For all $i,j\in\left\\{1,\dots,N\right\\}$ there exists
$k\in\left\\{1,\dots,N\right\\}$ such that $\chi_{ik}=\chi_{jk}=1$ and
$\tau_{ik}=\tau_{jk}.$ This means that for each pair of agents $(x_{i},x_{j})$
there’s another agent $x_{k}$, that we call common influencer of $x_{i}$ and
$x_{j},$ that transmits information to both them with the same time delay.

Let us define the _diameter function_ as

$d(t):=\max_{i,j=1,\dots,N}|x_{i}(t)-x_{j}(t)|.$

######  Definition 1.1.

We say that a solution converges to consensus if

$\lim_{t\rightarrow\infty}d(t)=0$

After preliminary estimates, we will prove the convergence to consensus for
solutions to system (1.1). As an example falling in the previous setting, we
will consider a model for a population with a leader subset. When the leaders’
numbers is $1$ or $2,$ the common influencer assumption is not satisfied. So,
we will study these cases using different appropriate arguments. In the case
of a unique leader, we will also analyze a control problem (cf. [38, 46]).

The rest of the paper is organized as follows. In Section 2, we introduce some
notations and provide preliminary lemmas. In Section 3, we state and prove the
consensus result under the assumption (CI). Sections 4 and 5 explore different
systems for which the Common Influencer Assumption does not hold. In Section
4, we consider the presence of a leader that influences all other agents but
is not influenced by anyone else, examining both the case of a constant
trajectory for the leader and a controlled trajectory. Section 5 deals with
the scenario involving two leaders. Finally, in Section 6, we present
numerical tests that validate the theoretical results.

##  2 Preliminaries

In this section, we present some preliminary results concerning system (1.1).
We omit the proofs since they are analogous to the ones of Lemma 2.2, Lemma
2.3, and Lemma 2.5 of [20].

######  Lemma 2.1.

Let $\\{x_{i}(t)\\}_{i=1}^{N}$ be a solution to (1.1)-(1.5). Then, for each
$v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ and $T\geq 0$, we have

$\min_{j=1,...,N}\min_{s\in[T-\tau,T]}\langle x_{j}(s),v\rangle\leq\langle
x_{i}(t),v\rangle\leq\max_{j=1,...,N}\max_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,$ (2.6)

for all $t\geq T-\tau,$ $i=1,\dots,N.$

We now introduce the following notation.

######  Definition 2.2.

We define

$D_{0}:=\max_{i,j=1,\dots,N}\max_{s,t\in[-\tau,0]}|x_{i}(s)-x_{j}(t)|,$ (2.7)

and in general for all $n\in\mathbb{N},$

$D_{n}:=\max_{i,j=1,\dots,N}\max_{s,t\in[n\tau-\tau,n\tau]}|x_{i}(s)-x_{j}(t)|.$
(2.8)

Let us denote with $\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}.$

######  Lemma 2.3.

Let $\\{x_{i}(t)\\}_{i=1}^{N}$ be a solution to (1.1)-(1.5). For each
$n\in\mathbb{N}_{0}$ and $\forall\ i,j=1,...,N$ we get

$|x_{i}(t)-x_{j}(t)|\leq D_{n},\quad\forall\ t\geq n\tau-\tau.\ $ (2.9)

######  Remark 2.4.

Let us note that from Lemma 2.3 it follows, in particular,

$|x_{i}(s)-x_{j}(t)|\leq D_{0},\quad\forall s,t\geq-\tau,\ \forall
i,j=1,\dots,N.$

Moreover, it holds that

$D_{n+1}\leq D_{n},\quad\forall n\in\mathbb{N}_{0}.$ (2.10)

As in [20], one can find a bound on $|x_{i}(t)|$ uniform with respect to $t$
and $i=1,...,N$. We have the following lemma (cf. [20, Lemma 2.5]).

######  Lemma 2.5.

Let $\\{x_{i}(t)\\}_{i=1}^{N}$ be a solution to (1.1)-(1.5). For all
$i=1,...,N$ we have

$|x_{i}(t)|\leq M_{0},\quad\forall\ t\geq 0,$ (2.11)

where

$M_{0}:=\max_{i=1,...,N}\max_{s\in[-\tau,0]}|x_{i}(0)|.$

######  Remark 2.6.

From Lemma 2.4, since the influence functions $\psi_{ij}$ are continuous, we
deduce that

$\psi_{ij}(x_{i}(t),x_{j}(t-\tau_{ij}))\geq\psi_{0}:=\min_{i,j=1,\dots,N}\min_{|z_{1}|,|z_{2}|\leq
M_{0}}\psi_{ij}(z_{1},z_{2})>0,$ (2.12)

for each $t\geq 0$ and $i,j=1,...,N$.

Now, we need the following estimates.

######  Lemma 2.7.

Let $\\{x_{i}(t)\\}_{i=1}^{N}$ be a solution to (1.1)-(1.5). For all
$i,j=1,\dots,N,$ unit vector $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ and
$n\in\mathbb{N}_{0}$ we have that

$\langle x_{i}(t)-x_{j}(t),v\rangle\leq e^{-K(t-t_{0})}\langle
x_{i}(t_{0})-x_{j}(t_{0}),v\rangle+(1-e^{-K(t-t_{0})})D_{n},$ (2.13)

for all $t\geq t_{0}\geq n\tau.$ Moreover, for all $n\in\mathbb{N}_{0}$ we get

$D_{n+1}\leq e^{-K\tau}d(n\tau)+(1-e^{-K\tau})D_{n}.$ (2.14)

###### Proof.

Let us fix $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ such that $|v|=1$ and
$n\in\mathbb{N}_{0}.$ We denote by

$M_{n}:=\max_{i=1,...,N}\max_{t\in[n\tau-\tau,n\tau]}\langle
x_{i}(t),v\rangle,$

and

$m_{n}:=\min_{i=1,...,N}\min_{t\in[n\tau-\tau,n\tau]}\langle
x_{i}(t),v\rangle.$

Note that $M_{n}-m_{n}\leq D_{n}.$ Now, fix $i=1,\dots,N$ and $t\geq t_{0}\geq
n\tau.$ Then, by definition of the system (1.1) and applying Lemma 2.1 with
$T=n\tau,$ we have that

$\begin{split}\frac{d}{dt}\langle x_{i}(t),v\rangle&=\frac{1}{N-1}\sum_{j\neq
i}\chi_{ij}a_{ij}(t)\langle x_{j}(t-\tau_{ij})-x_{i}(t),v\rangle\\\
&\leq\frac{1}{N-1}\sum_{j\neq i}a_{ij}(t)(M_{n}-\langle x_{i}(t),v\rangle)\\\
&\leq K(M_{n}-\langle x_{i}(t),v\rangle)\\\ \end{split}$

where we used that $t-\tau_{ij}\geq n\tau-\tau$ and that

$\sum_{j\neq i}\chi_{ij}\leq N-1.$

By applying the Grönwall’s Lemma, we find that

$\langle x_{i}(t),v\rangle\leq e^{-K(t-t_{0})}\langle
x_{i}(t_{0}),v\rangle+(1-e^{-K(t-t_{0})})M_{n}.$ (2.15)

On the other hand, for $i=1,\dots,N$ and $t\geq t_{0}\geq n\tau,$ following a
similar argument, we have that

$\frac{d}{dt}\langle x_{j}(t),v\rangle\geq K(m_{n}-\langle x_{j}(t),v\rangle)$

and, applying again the Grönwall’s Lemma,

$\langle x_{j}(t),v\rangle\geq e^{-K(t-t_{0})}\langle
x_{j}(t_{0}),v\rangle+(1-e^{-K(t-t_{0})})m_{n}.$ (2.16)

From (2.15) and (2.16), we have that

$\begin{split}\langle x_{i}(t)-x_{j}(t),v\rangle&\leq e^{-K(t-t_{0})}\langle
x_{i}(t_{0})-x_{j}(t_{0}),v\rangle+(1-e^{-K(t-t_{0})})(M_{n}-m_{n})\\\ &\leq
e^{-K(t-t_{0})}\langle
x_{i}(t_{0})-x_{j}(t_{0}),v\rangle+(1-e^{-K(t-t_{0})})D_{n}.\end{split}$

So, we have (2.13). The inequality (2.14) follows as in the second part of the
proof of Lemma 2.6 in [20]. ∎

The following result is crucial in order to prove the exponential convergence
to consensus estimate.

######  Lemma 2.8.

Let $\\{x_{i}(t)\\}_{i=1}^{N}$ be a solution to (1.1)-(1.5) and let us assume
(CI). Then, there exists a constant $C\in(0,1)$ such that

$d(n\tau)\leq CD_{n-2},$ (2.17)

for all $n\geq 2$.

###### Proof.

Notice that if $d(n\tau)=0$, the result is trivial. Suppose $d(n\tau)>0$ and
let be $i,j=1,\dots,N$ such that $d(n\tau)=|x_{i}(n\tau)-x_{j}(n\tau)|.$ We
set

$v:=\frac{x_{i}(n\tau)-x_{j}(n\tau)}{|x_{i}(n\tau)-x_{j}(n\tau)|}.$

Then, $d(n\tau)=\langle x_{i}(n\tau)-x_{j}(n\tau),v\rangle$. Consider,

$M_{n-1}:=\max_{i=1,...,N}\max_{s\in[(n-2)\tau,(n-1)\tau]}\langle
x_{i}(s),v\rangle$

and

$m_{n-1}:=\min_{i=1,...,N}\min_{s\in[(n-2)\tau,(n-1)\tau]}\langle
x_{i}(s),v\rangle.$

It holds that $M_{n-1}-m_{n-1}\leq D_{n-1}$.
From (1.1), for $t\in[(n-1)\tau,n\tau],$ we can write

$\begin{split}&\frac{d}{dt}\langle
x_{i}(t)-x_{j}(t),v\rangle=\frac{1}{N-1}\sum_{l:l\neq
i}\chi_{il}a_{il}(t)\langle x_{l}(t-\tau_{il})-x_{i}(t),v\rangle\\\ &\hskip
99.58464pt-\frac{1}{N-1}\sum_{l:l\neq j}\chi_{jl}a_{jl}(t)\langle
x_{l}(t-\tau_{jl})-x_{j}(t),v\rangle\\\ &=\frac{1}{N-1}\sum_{l:l\neq
i}\chi_{il}a_{il}(t)(\langle x_{l}(t-\tau_{il}),v\rangle-M_{n-1})\\\ &\hskip
99.58464pt+\frac{1}{N-1}\sum_{l:l\neq i}\chi_{il}a_{il}(t)(M_{n-1}-\langle
x_{i}(t),v\rangle)\\\ &-\frac{1}{N-1}\sum_{l:l\neq
j}\chi_{jl}a_{jl}(t)(\langle x_{l}(t-\tau_{jl}),v\rangle-m_{n-1})\\\ &\hskip
99.58464pt-\frac{1}{N-1}\sum_{l:l\neq j}\chi_{jl}a_{jl}(t)(m_{n-1}-\langle
x_{j}(t),v\rangle).\end{split}$ (2.18)

Let us define the sums $S_{1}$ and $S_{2}$ and, using Remark 2.6, we have

$\begin{split}S_{1}&:=\frac{1}{N-1}\sum_{l:l\neq i}\chi_{il}a_{il}(t)(\langle
x_{l}(t-\tau_{il}),v\rangle-M_{n-1})\\\ &+\frac{1}{N-1}\sum_{l:l\neq
i}\chi_{il}a_{il}(t)(M_{n-1}-\langle x_{i}(t),v\rangle)\\\
&\leq\frac{\psi_{0}}{N-1}\sum_{l:l\neq i}\chi_{il}(\langle
x_{l}(t-\tau_{il}),v\rangle-M_{n-1})+K(M_{n-1}-\langle
x_{i}(t),v\rangle),\end{split}$ (2.19)

and

$\begin{split}S_{2}&:=\frac{1}{N-1}\sum_{l:l\neq
j}\chi_{jl}a_{jl}(t)(m_{n-1}-\langle x_{l}(t-\tau_{jl}),v\rangle)\\\
&+\frac{1}{N-1}\sum_{l:l\neq j}\chi_{jl}a_{jl}(t)(\langle x_{j}(t),v\rangle-
m_{n-1})\\\ &\leq\frac{\psi_{0}}{N-1}\sum_{l:l\neq j}\chi_{jl}(m_{n-1}-\langle
x_{l}(t-\tau_{jl}),v\rangle)+K(\langle x_{j}(t),v\rangle-m_{n-1}),\\\
\end{split}$ (2.20)

where we used the fact that, being $t\in[(n-1)\tau,n\tau],$ it holds that
$t-\tau_{ij}\in[(n-2)\tau,n\tau],\ \forall ij=1,\dots,N,$ and then

$m_{n-1}\leq\langle x_{l}(t-\tau_{il}),v\rangle\leq M_{n-1},\quad
m_{n-1}\leq\langle x_{l}(t-\tau_{jl}),v\rangle\leq M_{n-1},$

for all $l=1,\dots,N.$ Then, using (2.19) and (2.20) in (2.18), we have that

$\begin{split}&\frac{d}{dt}\langle x_{i}(t)-x_{j}(t),v\rangle\leq
K(M_{n-1}-m_{n-1})-K\langle x_{i}(t)-x_{j}(t),v\rangle\\\
&+\frac{\psi_{0}}{N-1}\sum_{l:l\neq i}\chi_{il}(\langle
x_{l}(t-\tau_{il}),v\rangle-M_{n-1})\\\ &+\frac{\psi_{0}}{N-1}\sum_{l:l\neq
j}\chi_{jl}(m_{n-1}-\langle x_{l}(t-\tau_{jl}),v\rangle).\\\ \end{split}$

Now, we use the assumption (CI). Let $x_{k},$ for some $k\in\\{1,\dots,N\\},$
be a common influencer between $x_{i}$ and $x_{j}.$ Using that, for all
$l=1,\dots,N,$

$\langle x_{l}(t-\tau_{il}),v\rangle-M_{n-1}\leq 0$

and

$m_{n-1}-\langle x_{l}(t-\tau_{jl}),v\rangle\leq 0,$

we find that

$\begin{split}&\frac{d}{dt}\langle x_{i}(t)-x_{j}(t),v\rangle\leq
K(M_{n-1}-m_{n-1})-K\langle x_{i}(t)-x_{j}(t),v\rangle\\\
&+\frac{\psi_{0}}{N-1}(\langle x_{k}(t-\tau_{ik}),v\rangle-
M_{n-1}+m_{n-1}-\langle x_{k}(t-\tau_{jk}),v\rangle)\\\
&=\left(K-\frac{\psi_{0}}{N-1}\right)(M_{n-1}-m_{n-1})-K\langle
x_{i}(t)-x_{j}(t),v\rangle.\end{split}$

Applying the Grönwall’s Lemma in $[(n-1)\tau,t],$ with
$t\in[(n-1)\tau,n\tau],$ we find that

$\begin{split}\langle x_{i}(t)-x_{j}(t),v\rangle&\leq
e^{-K(t-n\tau+\tau)}\langle x_{i}(n\tau-\tau)-x_{j}(n\tau-\tau),v\rangle\\\
&+\left(1-\frac{\psi_{0}}{K(N-1)}\right)(M_{n-1}-m_{n-1})(1-e^{-K(t-n\tau+\tau)}).\end{split}$

Since this is valid $\forall\ t\in[(n-1)\tau,n\tau],$ let consider $t=n\tau.$
Then,

$\begin{split}d(n&\tau)\leq e^{-K\tau}\langle
x_{i}(n\tau-\tau)-x_{j}(n\tau-\tau),v\rangle\\\
&\quad\quad+\left(1-\frac{\psi_{0}}{K(N-1)}\right)(M_{n-1}-m_{n-1})(1-e^{-K\tau})\\\
&\leq e^{-K\tau}|x_{i}(n\tau-\tau)-x_{j}(n\tau-\tau)||v|\\\
&+\left(1-\frac{\psi_{0}}{K(N-1)}\right)(M_{n-1}-m_{n-1})(1-e^{-K\tau})\\\
&\leq
D_{n-1}\left[e^{-K\tau}+1-\frac{\psi_{0}}{K(N-1)}(1-e^{-K\tau})\right]\\\
&\leq D_{n-2}\left[1-\frac{\psi_{0}}{K(N-1)}(1-e^{-K\tau})\right].\end{split}$
(2.21)

Therefore, (2.17) follows with

$C:=1-\frac{\psi_{0}}{K(N-1)}(1-e^{-K\tau}).$ (2.22)

∎

##  3 Consensus under Common Influencer assumption

We are ready to give our convergence to consensus estimate.

######  Theorem 3.1.

Assume (CI). Then, every solution $\\{x_{i}(t)\\}_{i=1}^{N}$ to (1.1)-(1.5)
satisfies the exponential decay estimate

$d(t)\leq{D_{0}}e^{-\gamma(t-2\tau)}\ \mbox{for all}\ t\geq 0,$ (3.23)

for a suitable positive constant $\gamma.$

###### Proof.

Let $\left\\{x_{i}(t)\right\\}_{i=1}^{N}$ be the solution of (1.1)-(1.5). We
claim that

$D_{n+1}\leq\tilde{C}D_{n-2},\ \forall n\geq 2,$ (3.24)

for some constant $\tilde{C}\in(0,1).$ Using (2.10),(2.14) and (2.17) we get

$\begin{split}D_{n+1}&\leq e^{-K\tau}d(n\tau)+(1-e^{-K\tau})D_{n}\\\ &\leq
e^{-K\tau}CD_{n-2}+(1-e^{-K\tau})D_{n}\\\ &\leq
e^{-K\tau}CD_{n-2}+(1-e^{-K\tau})D_{n-2}\\\
&=\left(1-e^{-K\tau}(1-C)\right)D_{n-2},\end{split}$

where $C$ is defined in (2.22). So, setting

$\tilde{C}:=1-e^{-K\tau}(1-C),$

we have the claim. This implies that

$D_{3n}\leq\tilde{C}^{n}D_{0},\ \forall n\geq 1.$ (3.25)

From (3.25), we have that

$D_{3n}\leq e^{-3n\gamma\tau}D_{0},\ \forall n\in\mathbb{N}_{0}.$

where

$\gamma:=\frac{1}{3\tau}\ln\left(\frac{1}{\tilde{C}}\right).$

Now, fix $i,j=1,\dots,N$ and $t\geq 0.$ Then one can find
$t\in[3n\tau-\tau,3n\tau+2\tau]$ for some $n\in\mathbb{N}_{0}.$ Therefore,
using Lemma 2.3, we find that

$|x_{i}(t)-x_{j}(t)|\leq D_{3n}\leq e^{-3n\gamma\tau}D_{0}.$

Thus, being $t\leq 3n\tau+2\tau,$ we get

$|x_{i}(t)-x_{j}(t)|\leq e^{-\gamma t}e^{2\gamma\tau}D_{0},$

and, finally, we find that

$d(t)\leq e^{-\gamma(t-2\tau)}D_{0},\quad\forall t\geq 0.$

So, (3.23) is proved. ∎

As a particular case of such a model, we can consider a system in which there
is a (eventually small) group of $m$ leaders, $m\in\mathbb{N}.$ These agents
influence all other agents in the population, but they are influenced only by
the other leaders. Let $y_{i}(t)\in{{\rm I}\kern-1.49994pt{\rm
R}}^{d},i=1,\dots,m,$ be the opinion of the i-th leader at time $t$ and
$x_{i}(t)\in{{\rm I}\kern-1.49994pt{\rm R}}^{d},\ i=1,\dots,N$ be the opinion
of the i-th non-leader at time $t.$ We assume a non-universal interaction
between the non-leaders. We still consider a time delay in the interaction
between the agents to appear as a time needed to discuss and make a decision.
Then, the opinions of the population evolve following the Hegselmann-Krause
opinion formation model:

$\begin{array}[]{l}\displaystyle{\frac{d}{dt}y_{i}(t)=\frac{1}{m-1}\sum_{j\neq
i}a_{ij}(t)(y_{j}(t-\tilde{\tau}_{j})-y_{i}(t)),\quad t>0,\ i=1,...,m,}\\\
\displaystyle{\frac{d}{dt}x_{i}(t)=\frac{1}{N+m-1}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)(x_{j}(t-\tau_{ij})-x_{i}(t))}\\\ \displaystyle{\hskip
85.35826pt+\frac{1}{N+m-1}\sum_{j=1}^{m}c_{ij}(t)(y_{j}(t-\tilde{\tau}_{j})-x_{i}(t)),}\\\
\displaystyle{\hskip 199.16928ptt>0,\ i=1,...,N,}\end{array}$ (3.26)

with the interaction weights $a_{ij}(t),b_{ij}(t),c_{ij}(t)\ t\geq 0,$ defined
analogously to (1.2), namely

$a_{ij}(t):=\tilde{\psi}_{ij}(y_{i}(t),y_{j}(t-\tilde{\tau}_{j})),\quad\
i,j=1,...,m,$ (3.27) $b_{ij}(t):=\psi_{ij}(x_{i}(t),x_{j}(t-\tau_{ij})),\quad\
i,j=1,...,N,\hskip 48.36958pt$ (3.28)
$c_{ij}(t):=\psi^{*}_{ij}(x_{i}(t),y_{j}(t-\tilde{\tau}_{j})),\quad\
i=1,\dots,m,\ j=1,...,N,$ (3.29)

being the influence functions $\psi,\tilde{\psi},\psi^{*}$ positive,
continuous and bounded. Let us assume the initial conditions

$y_{i}(t)=y_{i}^{0}(t),\quad i=1,\dots,m,\ t\in[-\tau,0],$ (3.30)

and

$x_{i}(t)=x_{i}^{0}(t),\quad i=1,\dots,N,\ t\in[-\tau,0],$ (3.31)

where

$\tau=\max\left\\{\max_{i,j=1,\dots,N}{\tau_{ij},\max_{j=1,\dots,m}\tilde{\tau}_{j}}\right\\},$

and $y_{i}^{0}(\cdot),i=1,\dots,m,$ and $x_{i}^{0}(\cdot),i=1,\dots N,$ are
continuous functions.

######  Remark 3.2.

Note that system (3.26) satisfies the assumption (CI) if $m\geq 3.$ Then, in
that case, the agents exponentially convergence to an asymptotic consensus.

The special cases with only one or two leaders deserve different analyses that
will be developed in the next two sections.

##  4 A HK-model with one leader

Here, we will focus on the case in which only one leader, influencing the
other agents but not influenced by anyone, is present in the agents’ group.
First, we consider the leader to have a fixed opinion; then, we will deal with
a controlled leader.

###  4.1 A unique leader with constant trajectory

In this case, the system reads as:

$\begin{split}&\frac{d}{dt}y_{0}(t)=0,\quad t>0,\\\
&\frac{d}{dt}x_{i}(t)=\frac{1}{N}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)(x_{j}(t-\tau_{ij})-x_{i}(t))+\frac{c_{i0}(t)}{N}(y^{0}-x_{i}(t)),\\\
&\hskip 199.16928pt\quad t>0,\ i=1,...,N,\end{split}$ (4.32)

where $b_{ij}(t),$ $i,j=1,\dots,N,$ are defined as in (3.28) and
$c_{0i}(t)=\phi_{0i}(x_{i}(t),y^{0}),$ with $\phi_{0i}:{{\rm
I}\kern-1.49994pt{\rm R}}^{d}\times{{\rm I}\kern-1.49994pt{\rm
R}}^{d}\rightarrow{{\rm I}\kern-1.49994pt{\rm R}}$ continuous, positive and
bounded. Let us assume the initial conditions

$y_{0}(0)=y^{0}\quad\mbox{\rm and}\quad x_{i}(t)=x_{i}^{0}(t),\quad
t\in[-\tau,0],\ \forall\ i=1,\dots,N,$ (4.33)

being, as before, $\tau=\max_{i,j=1,\dots,N}\tau_{ij}.$ We want to study the
convergence to consensus of system (4.32). The _diameter function_ for
solutions to (4.32) can be written as

$d(t):=\max\left\\{\max_{i=1,\dots,N}|x_{i}(t)-y^{0}|,\max_{i,j=1,\dots,N}|x_{i}(t)-x_{j}(t)|\right\\}.$

Note that, in this case, assumption (CI) is not satisfied. Indeed, for each
$i=1,\dots,N,$ the pair $(x_{i},y_{0})$ does not admit a common influencer.
Let us define the following quantities:

$M_{T}:=\max\left\\{\max_{j=1,...,N}\max_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,\langle y^{0},v\rangle\right\\},$ (4.34)

and

$m_{T}:=\min\left\\{\min_{j=1,...,N}\min_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,\langle y^{0},v\rangle\right\\}.$ (4.35)

Moreover, let us define

$\tilde{K}:=\max\left\\{\max_{i,j=1,\dots,N}\|\psi_{ij}\|_{\infty},\max_{i=1,\dots,N}\|\phi_{0i}\|_{\infty}\right\\}.$
(4.36)

Similarly to the previous case, we can state the following results.

######  Lemma 4.1.

Let $\Big{(}\\{x_{i}(t)\\}_{i=1}^{N},y_{0}\Big{)}$ be a solution to
(4.32)-(4.33). Then, for each $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ and
$T\geq 0$, we have

$m_{T}\leq\langle x_{i}(t),v\rangle\leq M_{T},$ (4.37)

for all $t\geq T-\tau,$ $i=1,\dots,N.$

######  Remark 4.2.

Notice that, from the definition (4.34) and (4.35), it is trivial that

$m_{T}\leq\langle y^{0},v\rangle\leq M_{T}.$

Let us specify the notation in Definition 2.2 for this case.

######  Definition 4.3.

We define

$\begin{array}[]{l}\displaystyle{D_{0}:=\max\Big{\\{}\max_{i,j=1,\dots,N}\max_{s,t\in[-\tau,0]}|x_{i}(s)-x_{j}(t)|,}\\\
\displaystyle{\hskip
113.81102pt\max_{i=1,\dots,N}\max_{s\in[-\tau,0]}|x_{i}(s)-y^{0}|\Big{\\}},}\end{array}$
(4.38)

and in general for all $n\in\mathbb{N},$

$\begin{array}[]{l}\displaystyle{D_{n}:=\max\Big{\\{}\max_{i,j=1,\dots,N}\max_{s,t\in[n\tau-\tau,n\tau]}|x_{i}(s)-x_{j}(t)|,}\\\
\displaystyle{\hskip
113.81102pt\max_{i=1,\dots,N}\max_{s\in[n\tau-\tau,n\tau]}|x_{i}(s)-y^{0}|\Big{\\}}}.\end{array}$
(4.39)

As before, we need preliminary estimates in order to prove the consensus
result.

######  Lemma 4.4.

Let $\Big{(}\\{x_{i}(t)\\}_{i=1}^{N},y_{0}\Big{)}$ be a solution to
(4.32)-(4.33). For each $n\in\mathbb{N}_{0}$ and $\forall\ i,j=1,...,N,$ we
get

$|x_{i}(t)-x_{j}(t)|\leq D_{n},\quad\forall\ t\geq n\tau-\tau,\ $ (4.40)

and, similarly, $\forall i=\dots,N,$

$|x_{i}(t)-y^{0}|\leq D_{n},\quad\forall\ t\geq n\tau-\tau.\ $ (4.41)

Let us note that, as in Remark 2.4, Lemma 4.4 implies that
$\\{D_{n}\\}_{n\in\mathbb{N}_{0}}$ is a non-increasing sequence.

######  Lemma 4.5.

Let $\Big{(}\\{x_{i}(t)\\}_{i=1}^{N},y_{0}\Big{)}$ be a solution to
(4.32)-(4.33). For all $i=1,...,N,$ we have

$|x_{i}(t)|\leq C_{0},\quad\forall\ t\geq 0,$ (4.42)

and, in particular,

$|y^{0}|\leq C_{0},\quad\forall\ t\geq 0,$ (4.43)

where

$C_{0}:=\max\left\\{\max_{i=1,...,N}\max_{s\in[-\tau,0]}|x_{i}(0)|,|y^{0}|\right\\}.$

######  Remark 4.6.

As in Remark 2.6, from Lemma 4.5, since the influence functions $\psi_{ij}$
and $\phi_{0i}$ are continuous and positive, we deduce that

$\psi_{ij}(x_{i}(t),x_{j}(t-\tau_{ij}))\geq\psi_{0}:=\min_{i,j=1,\dots,N}\min_{|z_{1}|,|z_{2}|\leq
C_{0}}\psi_{ij}(z_{1},z_{2})>0,$ (4.44)

for each $t\geq 0$ and $i,j=1,...,N,$ and

$\phi_{0i}(x_{i}(t),y^{0})\geq\phi_{0}:=\min_{i=1,\dots,N}\min_{|z_{1}|,|z_{2}|\leq
C_{0}}\phi_{0i}(z_{1},z_{2})>0,$ (4.45)

for each $t\geq 0$ and $i=1,...,N$.

Let us denote

$\tilde{\psi}_{0}:=\min\\{\psi_{0},\phi_{0}\\}.$ (4.46)

######  Lemma 4.7.

Let $\Big{(}\\{x_{i}(t)\\}_{i=1}^{N},y_{0}\Big{)}$ be a solution to
(4.32)-(4.33). For all $i,j=1,\dots,N,$ unit vector $v\in{{\rm
I}\kern-1.49994pt{\rm R}}^{d}$ and $n\in\mathbb{N}_{0}$ we have that

$\langle x_{i}(t)-x_{j}(t),v\rangle\leq e^{-\tilde{K}(t-t_{0})}\langle
x_{i}(t_{0})-x_{j}(t_{0}),v\rangle+(1-e^{-\tilde{K}(t-t_{0})})D_{n},$ (4.47)

for all $t\geq t_{0}\geq n\tau,\ i,j=1,\dots,N,$ and

$\langle x_{i}(t)-y^{0},v\rangle\leq e^{-\tilde{K}(t-t_{0})}\langle
x_{i}(t_{0})-y^{0},v\rangle+(1-e^{-\tilde{K}(t-t_{0})})D_{n},$ (4.48)

for all $t\geq t_{0}\geq n\tau.$ Moreover, (2.14) holds true for all
$n\in\mathbb{N}_{0}.$

###### Proof.

The estimate (4.47) can be proved analogously to (2.13). So, let us prove
(4.48). Let us fix $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ such that $|v|=1$
and $n\in\mathbb{N}_{0}.$ We denote by

$M_{n}:=\max\left\\{\max_{i=1,...,N}\max_{t\in[n\tau-\tau,n\tau]}\langle
x_{i}(t),v\rangle,\langle y^{0},v\rangle\right\\}$

and

$m_{n}:=\min\left\\{\min_{i=1,...,N}\min_{t\in[n\tau-\tau,n\tau]}\langle
x_{i}(t),v\rangle,\langle y^{0},v\rangle\right\\}.$

Note that $M_{n}-m_{n}\leq D_{n}.$ Now, fix $i=1,\dots,N$ and $t\geq t_{0}\geq
n\tau.$ Then, as before, we have that

$\begin{split}\frac{d}{dt}\langle x_{i}(t),v\rangle&=\frac{1}{N}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)\langle
x_{j}(t-\tau_{ij})-x_{i}(t),v\rangle+\frac{c_{i0}(t)}{N}\langle
y^{0}-x_{i}(t),v\rangle\\\ &\leq\frac{1}{N}(N-1)\tilde{K}\left(M_{n}-\langle
x_{i}(t),v\rangle\right)+\frac{\tilde{K}}{N}\left(M_{n}-\langle
x_{i}(t),v\rangle\right)\\\ &\leq\tilde{K}(M_{n}-\langle
x_{i}(t),v\rangle).\end{split}$

By applying the Grönwall’s Lemma, we find that

$\langle x_{i}(t),v\rangle\leq e^{-\tilde{K}(t-t_{0})}\langle
x_{i}(t_{0}),v\rangle+(1-e^{-\tilde{K}(t-t_{0})})M_{n}.$ (4.49)

On the other hand, notice that holds true that

$\frac{d}{dt}\langle y^{0},v\rangle=0\geq\tilde{K}(m_{n}-\langle
y^{0},v\rangle).$

Then, applying again the Grönwall’s Lemma,

$\langle y^{0},v\rangle\geq e^{-\tilde{K}(t-t_{0})}\langle
y^{0},v\rangle+(1-e^{-\tilde{K}(t-t_{0})})m_{n}.$ (4.50)

From (4.49) and (4.50), we have find (4.48). ∎

Finally, we have the following result.

######  Lemma 4.8.

Let $\Big{(}\\{x_{i}(t)\\}_{i=1}^{N},y_{0}\Big{)}$ be a solution to
(4.32)-(4.33). There exists a constant $C\in(0,1)$ such that

$d(n\tau)\leq CD_{n-2},$ (4.51)

for all $n\geq 2$.

###### Proof.

Let us assume that $d(n\tau)=|x_{i}(n\tau)-y^{0}|$ for a given $i=1,\dots,N.$
We set

$v:=\frac{x_{i}(n\tau)-y^{0}}{|x_{i}(n\tau)-y^{0}|}.$

Then, $d(n\tau)=\langle x_{i}(n\tau)-y^{0},v\rangle$. Consider,

$M_{n-1}:=\max\left\\{\max_{i=1,...,N}\max_{s\in[(n-2)\tau,(n-1)\tau]}\langle
x_{i}(s),v\rangle,\langle y^{0},v\rangle\right\\}$

and

$m_{n-1}:=\min\left\\{\min_{i=1,...,N}\min_{s\in[(n-2)\tau,(n-1)\tau]}\langle
x_{i}(s),v\rangle,\langle y^{0},v\rangle\right\\}.$

It holds that $M_{n-1}-m_{n-1}\leq D_{n-1}$.
From (4.32), for $t\in[(n-1)\tau,n\tau],$ we can write

$\begin{split}&\frac{d}{dt}\langle
x_{i}(t)-y^{0},v\rangle=\frac{1}{N}\sum_{j\neq i}\chi_{ij}b_{ij}(t)\langle
x_{j}(t-\tau_{ij})-x_{i}(t),v\rangle+\frac{c_{i0}(t)}{N}\langle
y^{0}-x_{i}(t),v\rangle\\\ &=\frac{1}{N}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)\left(\langle x_{j}(t-\tau_{ij}),v\rangle-
M_{n-1}\right)+\frac{1}{N}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)\left(M_{n-1}-x_{i}(t),v\rangle\right)\\\ &\hskip
99.58464pt+\frac{c_{i0}(t)}{N}\left(\langle y^{0},v\rangle-
M_{n-1}\right)+\frac{c_{i0}(t)}{N}\left(M_{n-1}-x_{i}(t),v\rangle\right)\\\
&\leq\frac{N-1}{N}\tilde{K}(M_{n-1}-\langle
x_{i}(t),v\rangle)+\frac{\tilde{K}}{N}(M_{n-1}-\langle
x_{i}(t),v\rangle)-\frac{\tilde{\psi_{0}}}{N}(\langle y^{0},v\rangle-
M_{n-1}).\end{split}$

Then,

$\begin{split}\frac{d}{dt}\langle
x_{i}(t)-y^{0},v\rangle&\leq\tilde{K}(M_{n-1}-\langle
x_{i}(t),v\rangle)+\frac{\tilde{\psi_{0}}}{N}(\langle y^{0},v\rangle-
M_{n-1})\\\ &\hskip 36.98866pt+c_{i0}(t)(\langle y^{0},v\rangle-
m_{n-1}+m_{n-1}-\langle y^{0},v\rangle)\\\ &\leq\tilde{K}(M_{n-1}-\langle
x_{i}(t),v\rangle)+\frac{\tilde{\psi_{0}}}{N}(\langle y^{0},v\rangle-
M_{n-1})\\\ &\hskip 36.98866pt+\tilde{K}(\langle y^{0},v\rangle-
m_{n-1})+\tilde{\psi_{0}}(m_{n-1}-\langle y^{0},v\rangle)\\\
&\leq\tilde{K}(M_{n-1}-\langle
x_{i}(t),v\rangle)+\frac{\tilde{\psi_{0}}}{N}(\langle y^{0},v\rangle-
M_{n-1})\\\ &\hskip 36.98866pt+\tilde{K}(\langle y^{0},v\rangle-
m_{n-1})+\frac{\tilde{\psi_{0}}}{N}(m_{n-1}-\langle y^{0},v\rangle)\\\
&\leq\left(\tilde{K}-\frac{\tilde{\psi_{0}}}{N}\right)(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-y^{0},v\rangle,\end{split}$ (4.52)

where we used the fact that $m_{n-1}-\langle y^{0},v\rangle\leq 0.$
Notice that, repeating the argument in the proof of Lemma 2.8, from the
estimate above, applying the Grönwall’s Lemma, we can obtain a similar
estimate as (2.21). Finally, assuming that
$d(n\tau)=|x_{i}(n\tau)-x_{j}(n\tau)|,$ for some $i,j=1,\dots,N,$ and using
similar steps as above, we can find that

$\frac{d}{dt}\langle
x_{i}(t)-x_{j}(t),v\rangle\leq\left(\tilde{K}-\frac{\tilde{\psi_{0}}}{N}\right)(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-x_{j}(t),v\rangle.$

Therefore, we have the result with

$C:=1-\frac{\tilde{\psi_{0}}}{\tilde{K}N}(1-e^{-\tilde{K}\tau}).$ (4.53)

∎

Then, the exponential consensus result follows arguing as in previous section.

######  Theorem 4.9.

Every solution $\Big{(}\\{x_{i}(t)\\}_{i=1}^{N},y_{0}\Big{)}$ to (4.32)-(4.33)
satisfies the exponential decay estimate

$d(t)\leq{D_{0}}e^{-\tilde{\gamma}(t-2\tau)},\quad\forall\ t\geq 0,$ (4.54)

for a suitable positive constant $\tilde{\gamma}.$

###  4.2 A unique leader with controlled trajectory

Consider now a HK-model with a unique leader following a controlled
trajectory:

$\begin{split}&\frac{d}{dt}y_{0}(t)=u(t)\in{{\rm I}\kern-1.49994pt{\rm
R}}^{d},\quad t>0,\\\ &\frac{d}{dt}x_{i}(t)=\frac{1}{N}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)(x_{j}(t-\tau_{ij})-x_{i}(t))+\frac{c_{i0}(t)}{N}(y_{0}(t-\tau_{i0})-x_{i}(t)),\\\
&\hskip 199.16928pt\quad t>0,\ i=1,...,N,\end{split}$ (4.55)

where $b_{ij}(t),$ $i,j=1,\dots,N,$ are defined as in (3.28) and
$c_{0i}(t)=\phi_{0i}(x_{i}(t),y_{0}(t)),$ with $\phi_{0i}:{{\rm
I}\kern-1.49994pt{\rm R}}^{d}\times{{\rm I}\kern-1.49994pt{\rm
R}}^{d}\rightarrow{{\rm I}\kern-1.49994pt{\rm R}}$ continuous, positive and
bounded, as before. Let us assume the initial conditions

$y_{0}(0)=y^{0}(t)\quad\mbox{\rm and}\quad x_{i}(t)=x_{i}^{0}(t),\quad
t\in[-\tau,0],\ \forall\ i=1,\dots,N,$ (4.56)

being, as before,
$\tau=\max\left\\{\max_{i,j=1,\dots,N}\tau_{ij},\max_{i=1,\dots,N}\tau_{i0}\right\\}.$
Here, we want to study the convergence to any consensus state for solutions to
system (4.55). In this case, the definition of the _diameter function_ can be
written as

$d(t):=\max\left\\{\max_{i=1,\dots,N}|x_{i}(t)-y_{0}(t)|,\max_{i,j=1,\dots,N}|x_{i}(t)-x_{j}(t)|\right\\}.$

Note that, also in this configuration, assumption (CI) is not satisfied.
Consider, again, the appropriate definitions of $m_{T}$ and $M_{T}$:

$M_{T}:=\max\left\\{\max_{j=1,...,N}\max_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,\max_{s\in[T-\tau,T]}\langle y_{0}(s),v\rangle\right\\},$
$m_{T}:=\min\left\\{\min_{j=1,...,N}\min_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,\min_{s\in[T-\tau,T]}\langle y_{0}(s),v\rangle\right\\}.$

and let $\tilde{K}$ be the constant defined in (4.36).

######  Definition 4.10.

Fix a constant $M>0.$ We say that a ${\mathcal{L}^{1}}$-measurable strategy
$u(\cdot)$ is admissible if $\|u\|_{\infty}\leq M.$

We will prove that, given any state $\bar{x}\in{{\rm I}\kern-1.49994pt{\rm
R}}^{d},$ there exists an admissible strategy that steers each agent to
$\bar{x}.$ Firstly, in the following lemma we prove that there exists an
admissible strategy that steers the leader to $\bar{x}\in{{\rm
I}\kern-1.49994pt{\rm R}}^{d}$ and the other agents stay close enough to
$\bar{x}.$

######  Lemma 4.11.

Let us consider the system (4.55)-(4.56) and let $\bar{x}\in{{\rm
I}\kern-1.49994pt{\rm R}}^{d}.$ Then, there exists an admissible control
strategy $u:[0,+\infty)\rightarrow{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ such
that $y_{0}$ reaches $\bar{x}$ in finite time, i.e. there exists $t_{0}>0$
such that $y_{0}(t)=\bar{x}$ for all $t\geq t_{0}.$ Moreover, if we denote

$R:=\max\left\\{\max_{i=1,\dots,N}\max_{s\in[-\tau,0]}|x_{i}(s)-\bar{x}|,\max_{s\in[-\tau,0]}|y_{0}(s)-\bar{x}|\right\\},$
(4.57)

then

$\max_{i=1,\dots,N}|x_{i}(t)-\bar{x}|\leq R,$ (4.58)

for all $t\geq 0.$

###### Proof.

Let us define for $t\geq 0$ the admissible control

$u(t):=\begin{cases}M\frac{\bar{x}-y^{0}}{|\bar{x}-y^{0}|},\quad\ \ \mbox{ if
}y_{0}(t)\neq\bar{x},\\\ 0,\hskip 56.9055pt\mbox{ if
}y_{0}(t)=\bar{x},\end{cases}$ (4.59)

where $y^{0}=y_{0}(0).$ Consider the unit vector $v\in{{\rm
I}\kern-1.49994pt{\rm R}}^{d}$ defined as

$v:=\frac{y^{0}-\bar{x}}{|y^{0}-\bar{x}|},$

so that $\langle y^{0}-\bar{x},v\rangle=|y^{0}-\bar{x}|.$ Therefore, from the
definition of the system (4.55) and (4.59), we have for $t\geq 0$ that

$\frac{d}{dt}\langle
y_{0}(t)-\bar{x},v\rangle=\frac{M}{|\bar{x}-y^{0}|}\langle\bar{x}-y^{0},v\rangle=-\frac{M}{|\bar{x}-y^{0}|}|\bar{x}-y^{0}|=-M.$

Therefore,

$\langle
y_{0}(t)-\bar{x},v\rangle\leq|y_{0}(t)-\bar{x}|\leq|y^{0}-\bar{x}|-Mt,\quad
t\in[0,t_{0}],$

where

$t_{0}:=\frac{|y^{0}-\bar{x}|}{M}.$

It is immediate to see that

$\langle y_{0}(t)-\bar{x},v\rangle=|y_{0}(t)-\bar{x}|,$

then the first part of the lemma is proven. Notice now that by definition
$|y^{0}-\bar{x}|\leq R,$ then

$|y_{0}(t)-\bar{x}|\leq R,$

for all $t\geq 0.$ Now, let $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ be a
unit vector and let us define the set

$\mathcal{K}_{\epsilon}:=\left\\{t\in[0,+\infty)\Big{|}\max_{i=1,\dots,N}\langle
x_{i}(s)-\bar{x},v\rangle<R+\epsilon,\ \forall s\in[0,t)\right\\}.$

By continuity, $\mathcal{K}\neq\emptyset.$ Let
$S:=\sup\mathcal{K}_{\epsilon}.$ We claim that $S=+\infty.$ In order to prove
it, suppose by contraddiction that $S<+\infty.$ Then,

$\lim_{t\rightarrow S^{-}}\max_{i=1,\dots,N}\langle
x_{i}(s)-\bar{x},v\rangle=R+\epsilon.$

Consider $t\in[0,S).$ Then, from (4.55), we have that

$\begin{split}\frac{d}{dt}\langle
x_{i}(t)-\bar{x},v\rangle&=\frac{1}{N}\sum_{j\neq i}\chi_{ij}b_{ij}(t)\langle
x_{j}(t-\tau_{ij})-x_{i}(t),v\rangle\\\ &\hskip
71.13188pt+\frac{c_{i0}(t)}{N}\langle y_{0}(t-\tau_{i0})-x_{i}(t),v\rangle\\\
&=\frac{1}{N}\sum_{j\neq i}\chi_{ij}b_{ij}(t)\left(\langle
x_{j}(t-\tau_{ij})-\bar{x},v\rangle-\langle
x_{i}(t)-\bar{x},v\rangle\right)\\\ &\hskip
71.13188pt+\frac{c_{i0}(t)}{N}\left(\langle
y_{0}(t-\tau_{i0})-\bar{x},v\rangle-\langle
x_{i}(t)-\bar{x},v\rangle\right)\\\ &\leq\tilde{K}(R+\epsilon-\langle
x_{i}(t)-\bar{x},v\rangle),\end{split}$

where we used the fact that, if $t\in[0,S),$ then $t-\tau_{ij}\in[-\tau,S),$
for all $i=1,\dots,N,\ j=0,\dots,N,$ and, by definition of
$\mathcal{K}_{\epsilon}$ we have

$\langle x_{j}(t-\tau_{ij})-\bar{x},v\rangle\leq R+\epsilon,$

and

$\langle y_{0}(t-\tau_{i0})-\bar{x},v\rangle\leq R+\epsilon.$

Using the Grönwall inequality, we get

$\begin{split}\langle x_{i}(t)-\bar{x},v\rangle&\leq e^{-\tilde{K}t}\langle
x_{i}(0)-\bar{x},v\rangle+(R+\epsilon)(1-e^{-\tilde{K}t})\\\ &\leq
e^{-\tilde{K}t}R+R+\epsilon-e^{-\tilde{K}t}R-\epsilon e^{-\tilde{K}t}\\\
&=R+\epsilon-\epsilon e^{-\tilde{K}t}\\\ &\leq R+\epsilon-\epsilon
e^{-\tilde{K}S}<R+\epsilon.\end{split}$

Therefore, sending $t\rightarrow S^{-},$ we finally find

$\lim_{t\rightarrow S^{-}}\max_{i=1,\dots,N}\langle
x_{i}(t)-\bar{x},v\rangle<R+\epsilon,$

that gives the contradiction. Thus, $S=+\infty$ and

$\max_{i=1,\dots,N}\langle x_{i}(t)-\bar{x},v\rangle<R+\epsilon,$

for all $t\geq 0$ and $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}.$ Because of
the arbitrarily of $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ we can choose

$v:=\frac{x_{i}(t)-\bar{x}}{|x_{i}(t)-\bar{x}|},$

so that $\langle x_{i}(t)-\bar{x},v\rangle=|x_{i}(t)-\bar{x}|.$ Hence, we have
that

$\max_{i=1,\dots,N}|x_{i}(t)-\bar{x}|<R+\epsilon.$

Finally, sending $\epsilon\rightarrow 0,$ we have the result. ∎

We are now able to prove the following result.

######  Theorem 4.12.

Let us consider the system (4.55)-(4.56) and let $\bar{x}\in{{\rm
I}\kern-1.49994pt{\rm R}}^{d}.$ Then, there exists an admissible control
strategy $u:[0,+\infty)\rightarrow{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ such
that the leader $y_{0}$ reaches $\bar{x}$ in finite time, and

$d(t)\leq{C^{*}}e^{-{\gamma}^{*}t},\quad\forall\ t\geq t_{0},$ (4.60)

for suitable positive constants ${\gamma}^{*}$ and $C^{*}.$

###### Proof.

Notice that with the preliminary lemma we prove that the leader reaches in a
finite time $t_{0}$ the state $\bar{x}\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$
and stay there for all $t\geq,t_{0},$ and the other agents maintain the
distance from $\bar{x}$ less or equal than $R.$ Now, in order to prove the
consensus result one can apply the results in Section 4.1, since

$\frac{d}{dt}y_{0}(t)=0,$

for all $t\geq t_{0}.$ ∎

######  Remark 4.13.

Note that the above result extends the one in [46], that proved the consensus
in the absence of time delays, and the one in [38], considering delay effects,
since now we do not need any restrictions on the time delay size. We point
that [38, 46] assume the coefficients $b_{ij}(t)$ compactly supported while
here they are always positive. However, it is easy to see from the proof that
our result is valid also in the case $b_{ij}(t)$ compactly supported. What we
need is, indeed, that the leader’s influence is always positive only, exactly
as in [38, 46].

##  5 A HK-model with two leaders

Now, we want to analyze the case of a Hegselmann-Krause model in the presence
of two leaders. The system reads as follows:

$\begin{array}[]{l}\displaystyle{\frac{d}{dt}y_{i}(t)=a_{ij}(t)(y_{j}(t-\tilde{\tau}_{j})-y_{i}(t)),\quad
t>0,\ i=1,2,}\\\ \displaystyle{\frac{d}{dt}x_{i}(t)=\frac{1}{N+1}\sum_{j\neq
i}\chi_{ij}b_{ij}(t)(x_{j}(t-\tau_{ij})-x_{i}(t))}\\\ \displaystyle{\hskip
36.98866pt+\frac{1}{N+1}\sum_{j=1}^{2}c_{ij}(t)(y_{j}(t-\tilde{\tau}_{j})-x_{i}(t)),\quad
t>0,\ i=1,...,N,}\end{array}$ (5.61)

with the interaction weights $a_{ij}(t),b_{ij}(t),c_{ij}(t),\ t\geq 0,$
defined as in (3.27), (3.28), (3.29).
Also in this case the assumption (CI) is not satisfied. Indeed, the two
leaders do not admit a common influencer. Let us assume the initial conditions

$\begin{split}&y_{i}(t)=y_{i}^{0}(t),\ \quad t\in[-\tau,0],\ i=1,2,\\\
&x_{i}(t)=x_{i}^{0}(t),\quad t\in[-\tau,0],\ \forall\ i=1,\dots,N,\end{split}$
(5.62)

being
$\tau=\max\left\\{\tilde{\tau}_{1},\tilde{\tau}_{2},\max_{i,j=1,\dots,N}\tau_{ij}\right\\}.$
We want to study the convergence to consensus of system (5.61). The _diameter
function_ can be written, in this case, as

$d(t):=\max\left\\{\max_{\begin{subarray}{c}i=1,\dots,N\\\
j=1,2\end{subarray}}|x_{i}(t)-y_{i}(t)|,\max_{i,j=1,\dots,N}|x_{i}(t)-x_{j}(t)|,|y_{1}(t)-y_{2}(t)|\right\\}.$

Again, let us define the following quantities:

$M_{T}:=\max\left\\{\max_{j=1,...,N}\max_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,\max_{i=1,2}\max_{t\in[-\tau,0]}\langle
y_{i}(t),v\rangle\right\\},$

and

$m_{T}:=\min\left\\{\min_{j=1,...,N}\min_{s\in[T-\tau,T]}\langle
x_{j}(s),v\rangle,\min_{i=1,2}\min_{t\in[-\tau,0]}\langle
y_{i}(t),v\rangle\right\\},$

and let

$\tilde{K}:=\max\left\\{\max_{i=1,2}\|\tilde{\psi}_{ij}\|_{\infty},\max_{i,j=1,\dots,N}\|\psi_{ij}\|_{\infty},\max_{i=1,...,N}\|\psi^{*}_{i1}\|_{\infty},\max_{i=1,...,N}\|\psi^{*}_{i2}\|_{\infty}\right\\}.$

Analogously to the one leader case, we can prove the following preliminary
lemmas.

######  Lemma 5.1.

Let $\left(\\{x_{i}(t)\\}_{i=1}^{N},\\{y_{j}(t)\\}_{j=1}^{2}\right)$ be a
solution to (5.61)-(5.62). Then, for each $v\in{{\rm I}\kern-1.49994pt{\rm
R}}^{d}$ and $T\geq 0$, we have

$m_{T}\leq\langle x_{i}(t),v\rangle\leq M_{T},$ (5.63)

and

$m_{T}\leq\langle y_{j}(t),v\rangle\leq M_{T},$ (5.64)

for all $t\geq T-\tau,$ $i=1,\dots,N$ and $j=1,2.$

Let us introduce the appropriate notation:

######  Definition 5.2.

We define

$\begin{split}&D_{0}:=\max\Big{\\{}\max_{i,j=1,\dots,N}\max_{s,t\in[-\tau,0]}|x_{i}(s)-x_{j}(t)|,\max_{\begin{subarray}{c}i=1,\dots,N\\\
j=1,2\end{subarray}}\max_{s,t\in[-\tau,0]}|x_{i}(s)-y_{j}(t)|,\\\ &\hskip
156.49014pt\max_{s,t\in[-\tau,0]}|y_{1}(s)-y_{2}(t)|\Big{\\}},\end{split}$
(5.65)

and in general for all $n\in\mathbb{N},$

$\begin{split}&D_{n}:=\max\Big{\\{}\max_{i,j=1,\dots,N}\max_{s,t\in[n\tau-\tau,n\tau]}|x_{i}(s)-x_{j}(t)|,\\\
&\hskip 56.9055pt\max_{\begin{subarray}{c}i=1,\dots,N\\\
j=1,2\end{subarray}}\max_{s,t\in[n\tau-\tau,n\tau]}|x_{i}(s)-y_{j}(t)|,\max_{s,t\in[n\tau-\tau,n\tau]}|y_{1}(s)-y_{2}(t)|\Big{\\}}.\end{split}$
(5.66)

######  Lemma 5.3.

Let $\left(\\{x_{i}(t)\\}_{i=1}^{N},\\{y_{j}(t)\\}_{j=1}^{2}\right)$ be a
solution to (5.61)-(5.62). For each $n\in\mathbb{N}_{0},$ we get

$\begin{split}&|x_{i}(t)-x_{j}(t)|\leq D_{n},\quad\forall\ i,j=1,...,N,\\\
&|x_{i}(t)-y_{j}(t)|\leq D_{n},\quad\forall\ i=1,\dots,N,\ j=1,2,\\\
&|y_{1}(t)-y_{2}(t)|\leq D_{n},\end{split}$ (5.67)

for all $t\geq n\tau-\tau.$

Again, as a consequence of Lemma 5.3 we have that
$\\{D_{n}\\}_{n\in\mathbb{N}_{0}}$ is a non-increasing sequence.

######  Lemma 5.4.

Let $\left(\\{x_{i}(t)\\}_{i=1}^{N},\\{y_{j}(t)\\}_{j=1}^{2}\right)$ be a
solution to (5.61)-(5.62). Then, for all $i=1,...,N$ and $j=1,2,$ we have

$\begin{split}&|x_{i}(t)|\leq C_{0},\\\ &|y_{j}(t)|\leq C_{0}\end{split}$
(5.68)

for all $t\geq 0,$ where

$C_{0}:=\max\left\\{\max_{i=1,...,N}\max_{s\in[-\tau,0]}|x_{i}(0)|,\max_{i=1,2}\max_{s\in[-\tau,0]}|y_{i}(0)|\right\\}.$

From Lemma 5.4 we deduce again a positive lower bound, $\tilde{\psi}_{0},$ as
in (2.12), for the interaction rates. Moreover, we can prove the following
estimates.

######  Lemma 5.5.

Let $\left(\\{x_{i}(t)\\}_{i=1}^{N},\\{y_{j}(t)\\}_{j=1}^{2}\right)$ be a
solution to (5.61)-(5.62). For any unit vector $v\in{{\rm
I}\kern-1.49994pt{\rm R}}^{d}$ and $n\in\mathbb{N}_{0}$ we have that

$\begin{split}&\langle x_{i}(t)-x_{j}(t),v\rangle\leq
e^{-\tilde{K}(t-t_{0})}\langle
x_{i}(t_{0})-x_{j}(t_{0}),v\rangle+(1-e^{-\tilde{K}(t-t_{0})})D_{n},\\\
&\hskip 170.71652pt\quad i,j=1,\dots,N\\\ &\langle
x_{i}(t)-y_{j}(t),v\rangle\leq e^{-\tilde{K}(t-t_{0})}\langle
x_{i}(t_{0})-y_{j}(t_{0}),v\rangle+(1-e^{-\tilde{K}(t-t_{0})})D_{n},\\\
&\hskip 159.3356pt\quad i=1,\dots,N,\ j=1,2,\\\ &\langle
y_{1}(t)-y_{2}(t),v\rangle\leq e^{-\tilde{K}(t-t_{0})}\langle
y_{1}(t_{0})-y_{2}(t_{0}),v\rangle+(1-e^{-\tilde{K}(t-t_{0})})D_{n},\end{split}$
(5.69)

for all $t\geq t_{0}\geq n\tau.$ Moreover, (2.14) holds true for all
$n\in\mathbb{N}_{0}.$

###### Proof.

Notice that the first two inequalities in (5.69) can be obtained arguing as in
Section 2. For the last inequality in (5.69), we want to emphasize that the
equations for the leaders $y_{1}(t)$ and $y_{2}(t)$ are independent from the
other agents $\\{x_{i}(t)\\}_{i=1}^{N}.$ Therefore, it represents an
Hegselmann-Krause $2\times 2$ system itself. So, the estimate follows as in
[20]. ∎

######  Lemma 5.6.

Let $\left(\\{x_{i}(t)\\}_{i=1}^{N},\\{y_{j}(t)\\}_{j=1}^{2}\right)$ be a
solution to (5.61)-(5.62). Then, there exists a constant $C\in(0,1)$ such that

$d(n\tau)\leq CD_{n-2},$ (5.70)

for all $n\geq 2$.

###### Proof.

Let us suppose, as before, without loss of generality that $d(n\tau)>0$. Let
us assume that $d(n\tau)=|x_{i}(n\tau)-y_{j}(n\tau)|,$ for some $i=1,\dots,N$
and $j=1,2.$ In particular, suppose $d(n\tau)=|x_{i}(n\tau)-y_{1}(n\tau)|.$
Let us define the unit vector $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ as

$v:=\frac{x_{i}(n\tau)-y_{1}(n\tau)}{|x_{i}(n\tau)-y_{1}(n\tau)|}.$

Then we have that $d(n\tau)=\langle x_{i}(n\tau)-y_{1}(n\tau),v\rangle.$
Consider, as before,

$M_{n-1}=\max\left\\{\max_{l=1,...,N}\max_{s\in[(n-2)\tau,(n-1)\tau]}\langle
x_{l}(s),v\rangle,\max_{l=1,2}\max_{s\in[(n-2)\tau,(n-1)\tau]}\langle
y_{l}(s),v\rangle\right\\}$

and

$m_{n-1}=\min\left\\{\min_{l=1,...,N}\min_{s\in[(n-2)\tau,(n-1)\tau]}\langle
x_{l}(s),v\rangle,\min_{l=1,2}\min_{s\in[(n-2)\tau,(n-1)\tau]}\langle
y_{l}(s),v\rangle\right\\}.$

Again, we have that $M_{n-1}-m_{n-1}\leq D_{n-1}$. For $t\in[(n-1)\tau,n\tau]$
we have that

$\begin{split}&\frac{d}{dt}\langle
x_{i}(t)-y_{1}(t),v\rangle=\frac{1}{N+1}\sum_{k\neq
i}\chi_{ik}b_{ik}(t)\langle x_{k}(t-\tau_{ik})-x_{i}(t),v\rangle\\\ &\hskip
17.07182pt+\frac{1}{N+1}\sum_{k=1}^{2}c_{ik}(t)\langle
y_{k}(t-\tilde{\tau}_{k})-x_{i}(t),v\rangle-a_{12}(t)\langle
y_{2}(t-\tilde{\tau}_{2})-y_{1}(t),v\rangle\\\ &=\frac{1}{N+1}\sum_{k\neq
i}\chi_{ik}b_{ik}(t)(\langle x_{k}(t-\tau_{ik}),v\rangle-
M_{n-1}+M_{n-1}-\langle x_{i}(t),v\rangle)\\\ &\hskip
28.45274pt+\frac{1}{N+1}\sum_{k=1}^{2}c_{ik}(t)(\langle
y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1}+M_{n-1}-\langle
x_{i}(t),v\rangle)\\\ &\hskip 28.45274pt-a_{12}(t)(\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle-m_{n-1}+m_{n-1}-\langle
y_{1}(t),v\rangle)\\\ &=:S_{1}+S_{2}.\end{split}$ (5.71)

Since $t\in[(n-1)\tau,n\tau]$, then
$t-\tilde{\tau}_{k},t-\tau_{ij}\in[(n-2)\tau,n\tau]$ for all $k=1,2,$
$i,j=1,\dots,N.$ Then, we can write

$\begin{split}&S_{1}=\frac{1}{N+1}\sum_{k\neq i}\chi_{ik}b_{ik}(t)(\langle
x_{k}(t-\tau_{ik}),v\rangle-M_{n-1})\\\ &\hskip
22.76228pt+\frac{1}{N+1}\sum_{k\neq i}\chi_{ik}b_{ik}(t)(M_{n-1}-\langle
x_{i}(t),v\rangle)\\\ &\hskip
22.76228pt+\frac{1}{N+1}\sum_{k=1}^{2}c_{ik}(t)(\langle
y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1})\\\ &\hskip
22.76228pt+\frac{1}{N+1}\sum_{k=1}^{2}c_{ik}(t)(M_{n-1}-\langle
x_{i}(t),v\rangle).\end{split}$ (5.72)

So, we find that

$\begin{split}&S_{1}\leq\frac{\tilde{\psi}_{0}}{N+1}\sum_{k\neq
i}\chi_{ik}(\langle x_{k}(t-\tau_{ik}),v\rangle-
M_{n-1})+\frac{N-1}{N+1}\tilde{K}(M_{n-1}-\langle x_{i}(t),v\rangle)\\\
&\hskip 22.76228pt+\frac{\tilde{\psi}_{0}}{N+1}\sum_{k=1}^{2}(\langle
y_{k}(t-\tilde{\tau}_{k}),v\rangle-
M_{n-1})+\frac{2}{N+1}\tilde{K}(M_{n-1}-\langle x_{i}(t),v\rangle)\\\
&=\tilde{K}(M_{n-1}-\langle
x_{i}(t),v\rangle)+\frac{\tilde{\psi}_{0}}{N+1}\sum_{k\neq i}\chi_{ik}(\langle
x_{k}(t-\tau_{ik}),v\rangle-M_{n-1})\\\ &\hskip
22.76228pt+\frac{\tilde{\psi}_{0}}{N+1}\sum_{k=1}^{2}(\langle
y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1}).\end{split}$ (5.73)

Analogously, we can estimate

$\displaystyle\begin{aligned} &S_{2}:=a_{12}(t)(m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle+a_{12}(t)(\langle y_{1}(t),v\rangle-
m_{n-1})\cr&\leq\tilde{\psi}_{0}(m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle+\tilde{K}(\langle y_{1}(t),v\rangle-
m_{n-1}).\end{aligned}$ (5.74)

Putting (5.73) and (5.74) in (5.71), we can write

$\displaystyle\begin{aligned} &\frac{d}{dt}\langle
x_{i}(t)-y_{1}(t),v\rangle\leq\tilde{K}(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-y_{1}(t),v\rangle\cr&+\frac{\tilde{\psi}_{0}}{N+1}\sum_{k\neq
i}\chi_{ik}(\langle x_{k}(t-\tau_{ik}),v\rangle-
M_{n-1})+\frac{\tilde{\psi}_{0}}{N+1}\sum_{k=1}^{2}(\langle
y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1})\cr&\hskip
142.26378pt+\tilde{\psi}_{0}(m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle\cr&\leq\tilde{K}(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-y_{1}(t),v\rangle\cr&\hskip
28.45274pt+\frac{\tilde{\psi}_{0}}{N+1}\sum_{k=1}^{2}(\langle
y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1})+\tilde{\psi}_{0}(m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle).\end{aligned}$ (5.75)

Notice that, since $\langle y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1}\leq 0$
for $k=1,2$ and $t\in[(n-1)\tau,n\tau],$ we have that

$\sum_{k=1}^{2}(\langle y_{k}(t-\tilde{\tau}_{k}),v\rangle-M_{n-1})\leq\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle-M_{n-1}.$

Therefore, we have that

$\begin{split}&\frac{d}{dt}\langle
x_{i}(t)-y_{1}(t),v\rangle\leq\tilde{K}(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-y_{1}(t),v\rangle\\\ &+\frac{\tilde{\psi}_{0}}{N+1}(\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle-M_{n-1}+m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle)\\\
&=\left(\tilde{K}-\frac{\tilde{\psi}_{0}}{N+1}\right)(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-y_{1}(t),v\rangle,\end{split}$ (5.76)

where we used that $\tilde{\psi}_{0}(m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle)\leq\frac{\tilde{\psi}_{0}}{N+1}(m_{n-1}-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle)$.
Applying now the Grönwall inequality over $[(n-1)\tau,t],$ with
$t\in((n-1)\tau,n\tau],$ we have

$\begin{split}&\langle x_{i}(t)-y_{1}(t),v\rangle\leq
e^{-\tilde{K}(t-(n-1)\tau)}\langle
x_{i}((n-1)\tau)-y_{1}((n-1)\tau),v\rangle\\\ &\hskip
28.45274pt+\left(1-\frac{\tilde{\psi}_{0}}{\tilde{K}(N+1)}\right)(M_{n-1}-m_{n-1})(1-e^{-\tilde{K}(t-(n-1)\tau)}).\end{split}$
(5.77)

For $t=n\tau,$ we have

$\begin{split}&d(n\tau)\leq e^{-\tilde{K}\tau}\langle
x_{i}((n-1)\tau)-y_{1}((n-1)\tau),v\rangle\\\ &\hskip
34.14322pt+\left(1-\frac{\tilde{\psi}_{0}}{\tilde{K}(N+1)}\right)(M_{n-1}-m_{n-1})(1-e^{-\tilde{K}\tau})\\\
&\hskip
28.45274pt\leq\left(1-\frac{\tilde{\psi}_{0}}{\tilde{K}(N+1)}(1-e^{-\tilde{K}\tau})\right)D_{n-2},\end{split}$
(5.78)

where we used that $M_{n-1}-m_{n-1}\leq D_{n-1}$ and the monotonicity property
of $D_{n}.$
Notice that if $d(n\tau)=|x_{i}(n\tau)-x_{j}(n\tau)|,$ for some fixed
$i,j=1,\dots,N,$ using a similar argument, we can find again that

$\frac{d}{dt}\langle
x_{i}(t)-x_{j}(t),v\rangle\leq\left(\tilde{K}-\frac{\tilde{\psi}_{0}}{N+1}\right)(M_{n-1}-m_{n-1})-\tilde{K}\langle
x_{i}(t)-x_{j}(t),v\rangle,$

for $t\in[(n-1)\tau,n\tau].$
Finally, assume that $d(n\tau)=|y_{1}(n\tau)-y_{2}(n\tau)|.$ Let us define
again an unit vector $v\in{{\rm I}\kern-1.49994pt{\rm R}}^{d}$ as

$v:=\frac{y_{1}(n\tau)-y_{2}(n\tau)}{|y_{1}(n\tau)-y_{2}(n\tau)|}.$

Then, we can write $d(n\tau)=\langle y_{1}(n\tau)-y_{2}(n\tau),v\rangle.$
Consider $t\in[(n-2)\tau,n\tau].$ Let us distinguish two cases.
Case 1. There exists $t_{0}\in[(n-2)\tau,n\tau]$ such that $\langle
y_{1}(t_{0})-y_{2}(t_{0}),v\rangle<0.$ Applying Lemma 5.5, we get

$\begin{split}d(n\tau)&\leq e^{-\tilde{K}(n\tau-t_{0})}\langle
y_{1}(t_{0})-y_{2}(t_{0}),v\rangle+(1-e^{-\tilde{K}(n\tau-t_{0})})D_{n-2}\\\
&\leq(1-e^{-\tilde{K}(n\tau-t_{0})})D_{n-2}\\\
&\leq(1-e^{-2\tilde{K}\tau})D_{n-2},\end{split}$ (5.79)

and we have the statement.
Case 2. Assume that $\langle y_{1}(t)-y_{2}(t),v\rangle\geq 0,\ \forall
t\in[(n-2)\tau,n\tau].$ Consider $t\in[(n-1)\tau,n\tau].$ From (5.61) we have
that

$\begin{split}\frac{d}{dt}\langle y_{1}(t)-y_{2}(t)&,v\rangle=a_{12}(t)\langle
y_{2}(t-\tilde{\tau}_{2})-y_{1}(t),v\rangle-a_{21}(t)\langle
y_{1}(t-\tilde{\tau}_{1})-y_{2}(t),v\rangle\\\ &=a_{12}(t)\left(\langle
y_{2}(t-\tilde{\tau}_{2},v\rangle-
M_{n-1}\right)+a_{12}(t)\left(M_{n-1}-\langle y_{1}(t),v\rangle\right)\\\
&+a_{21}(t)\left(m_{n-1}-\langle
y_{1}(t-\tilde{\tau}_{1}),v\rangle\right)+a_{21}(t)\left(\langle
y_{2}(t),v\rangle-m_{n-1}\right)\\\ &\leq\tilde{\psi}_{0}\left(\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle-
M_{n-1}\right)+\tilde{K}\left(M_{n-1}-\langle y_{1}(t),v\rangle\right)\\\
&+\tilde{\psi}_{0}\left(m_{n-1}-\langle
y_{1}(t-\tilde{\tau}_{1}),v\rangle\right)+\tilde{K}\left(\langle
y_{2}(t),v\rangle-m_{n-1}\right).\end{split}$

Then,

$\begin{split}\frac{d}{dt}\langle
y_{1}(t)-y_{2}(t),v\rangle&=\left(\tilde{K}-\tilde{\psi}_{0}\right)(M_{n-1}-m_{n-1})-\tilde{K}\langle
y_{1}(t)-y_{2}(t),v\rangle\\\ &-\tilde{\psi}_{0}(\langle
y_{1}(t-\tilde{\tau}_{1}),v\rangle-\langle
y_{2}(t-\tilde{\tau}_{2}),v\rangle)\\\
&\leq\left(\tilde{K}-\tilde{\psi}_{0}\right)(M_{n-1}-m_{n-1})-\tilde{K}\langle
y_{1}(t)-y_{2}(t),v\rangle,\end{split}$

where, since $t-\tilde{\tau}_{1},t-\tilde{\tau}_{2}\geq(n-2)\tau,$ we use the
initial assumption. Therefore, applying the Grönwall inequality we find that,
for $t=n\tau$,

$d(n\tau)\leq\left(1-\frac{\tilde{\psi}_{0}}{\tilde{K}}(1-e^{-\tilde{K}\tau})\right)D_{n-2}.$

Thus, the result follows for a suitable constant $C.$ ∎

Again, from this result we can deduce the exponential consensus estimate for
the case of two leaders.

######  Theorem 5.7.

Every solution
$\left(\\{x_{i}(t)\\}_{i=1}^{N},\\{y_{j}(t)\\}_{j=1}^{2}\right)$ to
(5.61)-(5.62) satisfies the exponential decay estimate

$d(t)\leq D(0)e^{-\bar{\gamma}(t-2\tau)}\ \mbox{for all}\ t\geq 0,$

for a suitable positive constant $\bar{\gamma}.$

##  6 Numerical tests

In this section, we present some numerical simulations illustrating the
theoretical results.
In Figure 1, we consider the case with one leader with constant trajectory. We
consider the weight functions $b_{ij}(t)$ defined by

$\psi(r,r^{\prime}):=\tilde{\psi}(|r-r^{\prime}|),\quad
r,r^{\prime}\in[0,+\infty),$

for all $i=1,\dots,N.$ Meanwhile, the weight functions $c_{i0}(t)$ are assumed
to be the same for all $i=1,\dots,N$. In particular, we assume

$\begin{split}&\tilde{\psi}(r):=e^{-(r-1)^{2}},\ r\in[0,+\infty),\\\
&c_{i0}(t):=\frac{\tilde{K}}{N},\ \forall\ i\in\\{1,\dots,N\\},\\\
\end{split}$ (6.80)

with $\tilde{K}$ is a positive constant. Moreover, for simplicity, we assume a
constant time delay $\tau_{ij}=5,$ for all $i,j=1,\dots,N.$

Figure 1: HK-model with a unique leader with constant trajectory

In Figure 2, we present the case with one leader with a controlled trajectory.
Here, we use the weight functions defined in (6.80) and a constant time delay
$\tau_{ij}=1,$ for all $i=1,\dots,N,\ j=0,\dots,N.$

Figure 2: HK-model with a unique leader with control

In Figure 3, we consider the case with two leaders. Again, we choose the
functions $a_{ij}(t)$ for $i,j=1,2,$ and $b_{ij}(t)$ for $i,j=1,\dots,N,$
equal to the function $\tilde{\psi}(r)$ in (6.80). Meanwhile, the function
$c_{ij}(t)$ is supposed constant and equal to $\frac{\tilde{K}}{N+1}.$ The
time delays $\tilde{\tau}_{j}$ for $j=1,2$ is constant and
$\tilde{\tau}_{j}=5.$

Figure 3: HK-model with two leaders

For simplicity, we assumed $\chi_{ij}=1$ in all the figures above. Finally, in
Figure 4, we plot the general case in which the Common Influencer assumption
(CI) holds. As above, the function $a_{ij}(t)=\tilde{\psi}(r)$ and the time
delays $\tau_{ij}=5$ for all $i,j=1,\dots,N.$

Figure 4: HK-model under common influencer assumption

Acknowledgements. C. Cicolani and C. Pignotti are members of Gruppo Nazionale
per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of
the Istituto Nazionale di Alta Matematica (INdAM). C. Pignotti is partially
supported by PRIN 2022 (2022238YY5) Optimal control problems: analysis,
approximation and applications, PRIN-PNRR 2022 (P20225SP98) Some mathematical
approaches to climate change and its impacts, and by INdAM GNAMPA Project
Modelli alle derivate parziali per interazioni multiagente non simmetriche
(CUP E53C23001670001).

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