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The dataset generation failed
Error code: DatasetGenerationError
Exception: ArrowInvalid
Message: JSON parse error: Missing a closing quotation mark in string. in row 5
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/packaged_modules/json/json.py", line 145, in _generate_tables
dataset = json.load(f)
File "/usr/local/lib/python3.9/json/__init__.py", line 293, in load
return loads(fp.read(),
File "/usr/local/lib/python3.9/json/__init__.py", line 346, in loads
return _default_decoder.decode(s)
File "/usr/local/lib/python3.9/json/decoder.py", line 340, in decode
raise JSONDecodeError("Extra data", s, end)
json.decoder.JSONDecodeError: Extra data: line 2 column 1 (char 63073)
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1995, in _prepare_split_single
for _, table in generator:
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/packaged_modules/json/json.py", line 148, in _generate_tables
raise e
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/packaged_modules/json/json.py", line 122, in _generate_tables
pa_table = paj.read_json(
File "pyarrow/_json.pyx", line 308, in pyarrow._json.read_json
File "pyarrow/error.pxi", line 154, in pyarrow.lib.pyarrow_internal_check_status
File "pyarrow/error.pxi", line 91, in pyarrow.lib.check_status
pyarrow.lib.ArrowInvalid: JSON parse error: Missing a closing quotation mark in string. in row 5
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1529, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1154, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1027, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1122, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1882, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2038, in _prepare_split_single
raise DatasetGenerationError("An error occurred while generating the dataset") from e
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\section{Introduction}~\label{sec:intro}
Driven by recent advances in computing, communication, and networking technologies, modern engineering systems (e.g., \cite{tac2020wwsc,2014Aunmanned,Lv2020Event}) have gradually shifted their computing and control workload to the cloud, and even edge with data transmitted over wired or wireless networks.
Despite their flexibility, such network-based control systems (a.k.a., networked control systems) are known vulnerable to cyber threats~\cite{2013Attack,Alvaro2009Research}.
In fact, existing works have shown that malicious attacks can severely disrupt the
control performance and even render the system unstable \cite{Jang2014survey}.
Examples of such failures in widely used safety- and security-critical control systems nowadays could put our lives and even national infrastructure at risk \cite{2011Stuxnet}.
Several types of cyberattacks have been studied, including replay attacks \cite{Zhu2017replay,replay2018}, false-data injection attacks \cite{FP-RC-FB:10p ,wu2019Switching}, and Denial-of-Service (DoS) attacks \cite{shi2015jamming,Cetinkaya2019overview,hu2020dos}.~Relative to the others, DoS attacks can cause jamming in communication channels with little knowledge of system dynamics.~They are easy to launch and have received considerable attention \cite{L2010Protection}.~For instance, the work \cite{PersisInput} developed a general DoS framework, under which closed-loop system stability can be preserved via state-feedback control, provided certain DoS attack frequency and duration conditions are met.~This result has been extended in several directions, e.g., via output-feedback control in \cite{FengResilient}, as well as considering multiple output channels in \cite{LuInput}.
All the aforementioned works assumed that the communication channels have an infinite data-rate.
Clearly for real-world engineering systems, this condition is difficult to be met.
Systems with digital communication channels offer a basic paradigm.
The problem of limited bandwidth have been studied by accounting for the effect of quantization.
There is a great deal of research indicating that even without attacks, quantization can compromise system performance \cite{bullo1576851}, which is often addressed by designing suitable encoding schemes and providing enough quantization levels.
To name a few, for stabilization of systems with quantized measurements, \cite{Liberzon2000Quantized} first introduced the so-called ``zooming-in'' and ``zooming-out'' method.
Following this work, a number of stabilization encoding schemes have been designed for systems with quantized output feedback in \cite{Sharon2008Input, WakaikiObserver}, and switched systems in \cite{WakaikiStability, Liberzon2014Finite, YangFeedback, Wakaiki2017Stabilization}.
Recently, a few works have considered these two factors (i.e., quantization and DoS attacks) simultaneously; see \cite{chen2018Event,
feng2020datarate,liu2020datarate, Feng2020multi, 8880482}.
The trade-off between system resilience against DoS attacks and data-rate was analyzed in \cite{feng2020datarate}.
The minimum data-rate for stabilizing a centralized system and a multi-agent system were derived in \cite{liu2020datarate} and \cite{Feng2020multi}, respectively.
Capitalizing on the zooming-in and -out method, the work \cite{8880482} designed a resilient output encoding scheme for systems whose output channel is subject to DoS attacks and limited data-rate.
The goal of this paper is to stabilize systems with both input (controller-to-plant) and output (plant-to-controller) channels subject to DoS attacks and limited bandwidth.
To this aim, the quantizer encoding schemes should be carefully designed.
In the absence of DoS attacks, the work \cite{WakaikiObserver} developed encoding schemes for signals transmitted through both input and output channels.
However, their schemes cannot be applied here, due to the coupling between encoding strategies for different signals in the presence of DoS attacks. To overcome this challenge, we put forth a delicate structure, including a deadbeat controller and a transmission protocol.
Our protocol requires signals transmitted through the input channel at a higher rate than those through the output channel. Precisely, their transmission rate ratio is exactly the controllability index of the system.
Its efficacy is corroborated by the possibility to decouple design of different encoding schemes as, well as, establishing closed-loop stability conditions.
We further apply this structure to stabilize systems with only output channel has network imperfections.
In this scenario, it is proved that the proposed structure can secure the synchronization between encoder and decoder even without acknowledgments (ACKs), which are required by existing works, e.g., \cite{8880482,feng2020datarate}.
In a nutshell, the main contributions of the present work are summarized as follows.
\begin{itemize}
\item[\textbf{c1)}]
To cope with the coupling and synchronization issues, a structure consisting of a deadbeat controller and a transmission protocol for input and output channels, co-designed in terms of the controllability index, is advocated.
\item[\textbf{c2)}]
Under this structure, the input, output, and estimated output encoding schemes can be designed separately to achieve closed-loop stability when both input and output channels are subject to DoS attacks and quantization;
and,
\item[\textbf{c3)}]
When such network phenomena appear only in the output channel, an encoding scheme is designed such that the system can be stabilized through an ACK-free protocol, that is in sharp contrast to existing ACK-based results.
\end{itemize}
\emph{Notation:} Denote the set of integers (real numbers) by $\mathbb{Z}$ ($\mathbb{R}$).
Given $\alpha \in \mathbb{R}$ or $\alpha \in \mathbb{Z}$, let $\mathbb{R}_{>\alpha}$ ($\mathbb{R}_{\ge \alpha}$) or $\mathbb{Z}_{>\alpha}$ ($\mathbb{Z}_{\ge \alpha}$) denote the set of real numbers or integers greater than (greater than or equal to) $\alpha$.
Let $\mathbb{N}$ denote the set of natural numbers and $\mathbb{N}_0 := \mathbb{N} \cup \{0\}$.
For a vector $v = [v_1, v_2, \cdots\!, v_n]^T \in \mathbb{R}^n$, denote its maximum norm by $|v| := \max\{|v_1|, \cdots\!, |v_n|\}$ and the corresponding induced norm of a matrix $M \in \mathbb{R}^{m \times n}$ by $\Vert M\Vert := \sup\{|Mv|:v \in \mathbb{R^n}, |v| = 1\}$.
\section{Preliminaries and Problem Formulation} \label{problem_Formulation}
\subsection{Problem formulation}\label{systemdefination}
In this paper, we study the networked control architecture in Fig. \ref{networkfig}, where a plant is to be stabilized by a remote digital controller over a network subject to DoS attacks.~The plant is described by the following dynamics
\begin{subequations}\label{continuoussystem}
\begin{align}
&\dot{x}(t) = Ax(t) + Bu(t) \label{continuoussysteG_1}\\
&y(t) = Cx(t) \label{continuoussystem_2}
\end{align}
\end{subequations}
where $x(t) \in \mathbb{R}^{n_x}, u(t) \in \mathbb{R}^{n_u}$, and $y(t) \in \mathbb{R}^{n_y}$ are the state, the control input, and the output, respectively.
Here, we consider output signals and control inputs to be transmitted through different channels over a shared network, which are accordingly referred to as output channel and input channel.
Specifically, data transmissions over the output channel occur periodically with interval $\Delta > 0$.
That is, the output encoder samples $y(t)$ and sends its quantized version to the controller every $\Delta$ time.
Likewise, the digital controller generates control signals and transmits their quantized values to the plant periodically with interval $\delta>0$.
At the plant side, the quantized control inputs are first decoded, then pass through a zero-order hold (ZOH) before entering the plant.~To maintain the synchronization between input and output transmissions, we choose $\delta = \Delta/b $ for some $b \in \mathbb{N}$.
For future reference, let
\begin{equation*}
x_{q,k} := x(q\Delta + k\delta),\qquad y_{q,k} := y(q\Delta + k\delta)
\end{equation*}
for every $q \in \mathbb{Z}_{\ge 0}$, and $k = 0, \cdots\!, \frac{\Delta}{b}$, and
\begin{equation}\label{eq:adbd}
A_d := e^{A\delta}, \qquad B_d := \int_{0}^{\delta}{e^{As}B\, ds}.
\end{equation}
Moreover, we use $x_q$ to denote $x_{q,0}$ for simplicity.
\begin{figure}[t]
\centering
\includegraphics[width=8cm]{NetworkedStructure.jpg}\\
\caption{Networked control system with both input (the blue line) and output (the green line) channels subject to DoS attacks.}\label{networkfig}
\centering
\end{figure}
We make the following assumptions on system \eqref{continuoussystem}.
\begin{assumption}[Controllability and observability]\label{as:abca}
The pair $(A,B)$ is controllable, and the pair $(C,A)$ is observable.
\end{assumption}
\begin{assumption}[Initial state bound]\label{x0bound}
An upper bound on the initial state $|x_0|$ is known.
\end{assumption}
\begin{remark}
Thanks to As. \ref{as:abca}, it has been shown in \cite{Kreisselmeier1999On} that if $\delta$ is non-pathological, then $(A_d, B_d)$ in \eqref{eq:adbd} is controllable.
Let $\eta$ denote its controllability index, which can be computed by evaluating
${\rm rank} [B_d, \cdots\!, A_d^{\eta} B_d] = n_x$.
Similarly, $(C, A_d^{\eta})$ is observable.
An upper bound on the initial state in As. \ref{x0bound} can be derived via the zooming-out method; see \cite[Sec. 4]{8880482}.
\end{remark}
\subsection{Denial-of-Service attack}
In Fig. \ref{networkfig}, since both input and output signals are transmitted periodically, we adopt the discrete-time DoS attack model in \cite{8880482}.
Under this model, attacks are launched only at output transmission instants, and each lasts for an output transmission period $\Delta$.
This model is general enough since it only poses requirements on the frequency and duration of DoS attacks.
Here, DoS frequency is the number of DoS \emph{off/on} switches over a fixed time interval, while
DoS duration represents the total number of attacks.
\begin{assumption}[DoS frequency]\label{DoS_frequencyassumption}
There exist constants $\kappa_f \in \mathbb{R}_{\ge 0}$ and $\nu_f \in \mathbb{R}_{\ge 2}$ such that DoS frequency satisfies
\begin{equation}\label{dosfre}
\Phi_f(q) \le \kappa_f + \frac{q}{\nu_f}
\end{equation}
over time interval $[0, q\Delta)$, where $q \in \mathbb{Z}_{\ge 0}$.
\end{assumption}
\begin{assumption}[DoS duration]\label{DoS_durationassumption}
There exist constants $\kappa_d \in \mathbb{R}_{\ge 0}$ and $\nu_d \in \mathbb{Z}_{\ge 1}$ such that DoS duration satisfies
\begin{equation}\label{dosdur}
\Phi_d(q) \le \kappa_d + \frac{q}{\nu_d}
\end{equation}
over time interval $[0, q\Delta)$, where $q \in \mathbb{Z}_{\ge 0}$.
\end{assumption}
\begin{remark}
Given its generality, this attack model has been widely used in e.g., \cite{8880482,Feng2020multi,FengResilient,LuInput,feng2020datarate,PersisInput}.
As pointed out in \cite{Hespanha1999STABILITY}, $\nu_f\Delta$ in As. \ref{DoS_frequencyassumption} can be regarded as the average dwell-time between two consecutive DoS attacks \emph{off/on} switches.
On the other hand, As. \ref{DoS_durationassumption} indicates that, the average duration of DoS attacks does not exceed a proportion $1/\nu_d$ of the time interval.
Constants $\kappa_f$ and $\kappa_d$ are also known as chatter bounds.
Conditions $\nu_d \ge 1$ and $\nu_f \ge 2$ suggest that DoS attacks are not strong enough to prevent all packets from being transmitted, thus rendering it possible for the system to be stabilized by suitable control strategies.
\end{remark}
\section{Networked Phenomena at Both Input and Output Channels}\label{inputoutputsection}
This section aims to design resilient encoding schemes for stabilization of system (\ref{continuoussystem}) via a remote observer-based digital controller over communication channels subject to limited bandwidth and DoS attacks; see Fig. \ref{networkfig}.
To this end, there are three signals that need to be quantized, i.e., the estimated output by observer $\hat{y}_q$, the control input $u_{q,k}$, and the plant output $y_q$, with their quantized values denote by $Q_1(\hat{y}_q)$, $Q_2(u_{q,k})$, and $Q_3(y_q)$, respectively.
In addition, since the input and output channels share a communication network, we assume for simplicity that, once there is a DoS attack,
neither the input nor the output signals will be received, and both of them are set to the default zero.
In this manner, the decoder and encoder at both input and output sides can infer whether there is an attack.
Further, their quantization ranges and centers are identical at every transmission instant.
As a result, they can be synchronized even with an ACK-free protocol.
\subsection{Controller architecture}
To stabilize system (\ref{continuoussystem}), we put forth a two-stage observer-based controller by considering whether there is an attack or not.
Specifically, in the absence of DoS attacks, both $Q_3(y_q)$ and $Q_1(\hat{y}_{q-1,\eta})$ are available at the observer side, so we construct the following controller
\begin{subequations}\label{abdoscontroller}
\begin{align}
&\hat{x}_{q, k+1} \!=\! A_d\hat{x}_{q,k} \!+\! B_du_{q,k},\!\! &k &\le \eta - 1 \label{abdoscontroller_1}\\
&\hat{x}_{q}\!=\! \hat{x}_{q-1,k} \!+\! M\big[Q_3(y_{q}) \!-\! Q_1(\hat{y}_{q-1,k})\big],\!\! &k &= \eta \label{abdoscontroller_2}\\
&\hat{y}_{q,k} \!=\! C \hat{x}_{q,k} \label{abdoscontroller_3}\\
&u_{q,k} \!=\! K \hat{x}_{q,k} \label{abdoscontroller_4}
\end{align}
\end{subequations}
where the initial condition $\hat{x}_0$ is given by $\hat{x}_0 = 0$, and $\delta$ is chosen such that
\begin{equation}\label{delta}
\delta = \frac{\Delta}{\eta}.
\end{equation}
Matrix $M \in \mathbb{R}^{n_x \times n_y}$ can be regarded as an observer gain such that $R := A_d^{\eta}(I - MC)$ is schur stable, which always exists since $(C, A_d^{\eta})$ is observable.
Moreover, $(A_d, B_d)$ is controllable, thus a controller gain matrix $K \in \mathbb{R}^{n_u \times n_x}$ can be designed such that
\begin{equation}\label{dbdb}
\bar{R}^{\eta} = (A_d + B_dK)^{\eta} = 0.
\end{equation}
\begin{remark}\label{remark:dbgain}
Matrix $K$ obeying (\ref{dbdb}) is also known as a class of deadbeat controller gain, since it assigns all the eigenvalues of $A_d + B_d K$ to the origin.
Solution of this eigenstructure assignment problem is non-unique, and can be obtained through several approaches, e.g., \cite{FahmyDead}.
\end{remark}
On the other hand, when there is a DoS attack, none of $Q_1(\hat{y}_q)$, $Q_2(u_{q,k})$, or $Q_3(y_q)$ can be received, thus we simply employ an open-loop controller as follows
\begin{subequations}\label{predoscontroller}
\begin{align}
&\hat{x}_{q,k+1} = A_d \hat{x}_{q,k}\\
&\hat{y}_{q,k} = C \hat{x}_{q,k}\\
&u_{q,k} = 0
\end{align}
\end{subequations}
with the initial estimated state $\hat{x}_0 = 0$.
In addition, to apply the discrete-time signal $Q_2(u_{q,k})$ to the continuous-time system (\ref{continuoussysteG_1}), a ZOH is used, and the control input is given by
\begin{equation*}
u(t) = Q_2(u_{q,k}), \qquad q\Delta + k\delta \le t < q\Delta + (k+1)\delta
\end{equation*}
where $k = 0, \cdots\!, \eta - 1$.
\subsection{Quantizer}
We first design quantizers at the input channel.
According to (\ref{abdoscontroller_1}) and (\ref{abdoscontroller_2}), $u_{q,k}$ is needed for feedback control, whereas $\hat{y}_{q,k}$, resetting the estimated state, is required at each successful transmission instants.
Therefore, the controller sends $u_{q,k}$ and $\hat{y}_{q,\eta}$ to the quantizers periodically at a different rate.
In more precise terms, periods for the former and the latter are $\delta$, and $\eta \delta$, respectively.
Let $E_{1,q} \ge 0$ and $E_{2,q,k} \ge 0$ satisfy
\begin{equation}\label{2E12inequality}
|\hat{y}_{q - 1,\eta}| \le E_{1,q},\ \ \ |u_{q,k}| \le E_{2,q,k}.
\end{equation}
Suppose there are $N_1$ ($N_2$) levels for quantization of $\hat{y}_{q,\eta}$ ($u_{q,k}$).
Partition the hypercubes at the encoders
\begin{equation*}
\begin{split}
\{\hat{y} \in \mathbb{R}^{n_y}: |\hat{y}_{q - 1,\eta}| \le E_{1,q}\},~~
\{u \in \mathbb{R}^{n_u}: |u_{q,k}| \le E_{2,q,k}\}
\end{split}
\end{equation*}
into $N_1^{n_y}$, and $N_2^{n_u}$ equal-sized boxes, respectively.
In addition, each box is represented by a value in $\{1, \cdots\!, N_1^{n_y}\}$, or $\{1, \cdots\!, N_2^{n_u}\}$ following a bijection mapping.
Indices that denote the partitioned boxes containing $\hat{y}_{q, \eta}$ and $u_{q, k}$ are then sent to the decoders.
If $\hat{y}_{q, \eta}$ and $u_{q, k}$ are on the boundary of several boxes, then anyone of them can be chosen.
At the decoders side, $Q_1(\hat{y}_{q, \eta})$ and $Q_2(u_{q, k})$ are recovered from the indices.
This implies that the encoder and its corresponding decoder should share the same quantization ranges and centers.
Since DoS attacks block both input and output signals from transmitting, encoders and decoders at both sides of input and output channels are naturally synchronized.
The quantization errors
of the aforementioned encoding schemes obey
\begin{equation}\label{2E1inequality}
|\hat{y}_{q - 1,\eta} - Q_1(\hat{y}_{q - 1,\eta})| \le \frac{E_{1,q}}{N_1},
\end{equation}
\begin{equation}\label{2E2inequality}
|u_{q,k} - Q_2(u_{q,k})| \le \frac{E_{2,q,k}}{N_2}.
\end{equation}
Since $\hat{x}_0 = 0$, we deduce that $\hat{y}_0 = u_0 = 0$.
Therefore, the initial bounds $E_{1,0}$ and $E_{2,0,0}$ can be set by
\begin{equation*}
E_{1,0} = 0,\ \ \ E_{2,0,0} = 0.
\end{equation*}
Moreover, as for the output $y_q$, choose $E_{3,q} \ge 0$ such that
\begin{equation}\label{2E3inequality}
|y_q - Q_1(\hat{y}_{q - 1,\eta})| \le E_{3,q}.
\end{equation}
Let $N_3$ be the quantization level of $y_q$.
The hypercube
\begin{equation}\label{eq:hypercubeQ3}
\{y \in \mathbb{R}^{n_y} : |y_q - Q_1(\hat{y}_{q - 1,\eta})| \le E_{3, q}\}.
\end{equation}
is partitioned into $N_3^{n_y}$ equal-sized boxes with the center $Q_1(\hat{y}_{q - 1, \eta})$.
Then, following the same procedure as the above two quantizers, $Q_3(y_q)$ is transmitted to the controller every $\Delta$ time.
The quantization error satisfies
\begin{equation*}
|y_q - Q_3(y_q)| \le \frac{E_{3, q}}{N_3}.
\end{equation*}
Define error of the system $e_{q, k} := x_{q,k} - \hat{x}_{q,k}$.
Combining $\hat{x}_0 = 0$ with As. \ref{x0bound}, we deduce that the initial error obeys $|e_0| = |x_0|$.
Thus it suffices to set $E_{3,0} := \Vert C\Vert |x_0|$.
\subsection{Stability analysis}\label{Main_results}
In this subsection, we start by presenting encoding schemes $\{E_{p,q,k}\} (p = 1, 2, 3)$, followed by formal stability conditions.
Design $\{E_{1,q}: q \in \mathbb{Z}_{\ge 1}\}$ such that
\begin{equation}\label{2E1q}
E_{1,q} = E_{1,0},\qquad \forall q \in \mathbb{Z}_{\ge 1}
\end{equation}
and let $\{E_{2,q,k}: q \in \mathbb{Z}_{\ge 1}, k = 0, \cdots\!, \eta - 1\}$ be updated by
\begin{equation}\label{2E2qk}
E_{2,q,k} = \frac{N_3 - 1}{N_3}\left\Vert K\bar{R}^kM\right\Vert E_{3,q}, \qquad {\rm{~if~}}q\Delta = s_r.
\end{equation}
Moreover, the sequence $\{E_{3,q}: q \in \mathbb{Z}_{\ge 1}\}$ is given by
\begin{equation}\label{2E3q}
E_{3,q + 1} :=
\left\{
\begin{aligned}
& \hat{\theta}_a E_{3,q}, & q\Delta \ne s_r\\
& \hat{\theta}_{0} E_{3,q}, & (q-1)\Delta \ne s_r, q\Delta = s_r\\
& \hat{\theta}_{na}E_{3,q}, & (q-1)\Delta = s_r, q\Delta = s_r
\end{aligned}
\right.
\end{equation}
where
\begin{align*}
\hat{\theta}_a & := \Vert A_d^{\eta}\Vert\\
\hat{\theta}_{0} & := a_0\rho + \frac{\Vert C\Vert a_1}{N_3} + \Vert C\Vert a_2\frac{N_3 - 1}{N_2N_3}\\
\hat{\theta}_{na} & := \rho + \frac{\Vert C\Vert a_1}{N_3} + \Vert C\Vert a_2\frac{N_3 - 1}{N_2N_3}
\end{align*}
with positive constants $a_0, a_1, a_2$, and $0 < \rho < 1$ validating the following for all $\ell \ge 1$
\begin{equation}\label{2rho}
\begin{aligned}
\big\Vert R^{\ell}\big\Vert \le a_0\rho^{\ell},~~~~ \big\Vert R^{\ell}A_d^{\eta}M\big\Vert \le a_1\rho^{\ell}\\
\sum_{i = 0}^{\eta - 1}{\big\Vert R^{\ell}A_d^{\eta - i - 1}B_d \big\Vert\big\Vert K\bar{R}^iM\big\Vert}\le a_2\rho^{\ell}.\\
\end{aligned}
\end{equation}
Since $R$ is schur stable, there always exist such constants.
Next, we show that our designed schemes above are resilient to DoS attacks, which is one of our main results too.
\begin{theorem}\label{2convergetheorem}
Consider system (\ref{continuoussystem}) with the observer-based controller in (\ref{abdoscontroller}) and (\ref{predoscontroller}), with $K$ obeying (\ref{dbdb}) and $M$ chosen such that $R$ is schur stable.
Let As. \ref{as:abca}--\ref{DoS_durationassumption} hold.
If i) the input and output transmission periods adhere to (\ref{delta}), ii) the number of quantization levels $N_1 \ge 1$ is odd,
\begin{equation}\label{2Ncondition}
\begin{split}
N_2\! >\! \max\!{\left\{\frac{a_2\Vert C\Vert}{1 - \rho},\, \frac{a_2}{a_1}\right\}},\,{\rm{~and~}}
N_3\! > \!\frac{\Vert C\Vert a_1 - \frac{\Vert C\Vert a_2}{N_2}}{1 - \rho -\frac{\Vert C\Vert a_2}{N_2}}
\end{split}
\end{equation}
and, iii) DoS attacks satisfy
\begin{equation}\label{2doscondition}
\frac{1}{\nu_d} \le \frac{\log{(1/\hat{\theta}_{na})}}{\log{(\hat{\theta}_a/\hat{\theta}_{na})}} - \frac{\log{(\hat{\theta}_0/\hat{\theta}_{na})}}{\log{(\hat{\theta}_a/\hat{\theta}_{na})}}\frac{1}{\nu_f}
\end{equation}
then the system is exponentially stable under the encoding scheme with error bounds $\{E_{p,q,k}:q \in \mathbb{Z}_{\ge 1}, k = 0, \cdots\!, \eta - 1\} (p = 1, 2, 3)$ constructed by the update rule in (\ref{2E1q})-(\ref{2E3q}).
\end{theorem}
We begin proving Thm. \ref{2convergetheorem} by giving a lemma demonstrating that the update rules in \eqref{2E1q}-\eqref{2E3q} satisfy (\ref{2E12inequality}) and (\ref{2E3inequality}).
\begin{lemma}\label{lem:lem1}
Consider system (\ref{continuoussystem}) with the controller in (\ref{abdoscontroller}) and (\ref{predoscontroller}), where $K$ obeys (\ref{dbdb}).
Let As. \ref{as:abca}--\ref{DoS_durationassumption} hold.
If $\{E_{p,q,k}:q \in \mathbb{Z}_{\ge 1}, k = 0, \cdots\!, \eta - 1\} (p = 1, 2, 3)$ obey (\ref{2E1q})-(\ref{2E3q}), then (\ref{2E12inequality}) and (\ref{2E3inequality}) hold true for all $q \in \mathbb{Z}_{\ge 1}$.
\end{lemma}
\begin{proof}
Encoding schemes for systems with quantized inputs and outputs in the absence of DoS attacks have been discussed in \cite{WakaikiObserver}.
However, their methods cannot be directly applied due to the DoS-induced coupling between these schemes.
This challenge is addressed through our carefully designed controller structure in (\ref{abdoscontroller})-(\ref{dbdb}).
According to (\ref{abdoscontroller}) and (\ref{dbdb}),
\begin{equation*}
\hat{y}_{q - 1, \eta} = C\hat{x}_{q - 1, \eta} = C\bar{R}^{\eta}\hat{x}_{q - 1} = 0
\end{equation*}
holds true irrespective of DoS attacks,
which implies $E_{1,q} \ge |\hat{y}_{q - 1, \eta}| = E_{1,0}$ for all $q \in \mathbb{Z}_{\ge 1}$, so $E_{1,q}$ remains unchanged.
This result further indicates that $Q_1(\hat{y}_{q - 1, \eta}) = 0$.
Hence, it follows from (\ref{eq:hypercubeQ3}) that the quantization center of $Q_3(y_q)$ is at the origin.
On the other hand, if no DoS attacks occur within $[q_1\Delta, (q_1 + 1)\Delta)$, then
\begin{equation}\label{2hatxqk}
\begin{aligned}
\hat{x}_{q_1, k} =&\ (A_d + B_dK)^{k }\hat{x}_{q_1} \\
= &\ \bar{R}^{k}(\hat{x}_{q_1 - 1, \eta} + M[Q_3(y_{q_1}) - Q_1(\hat{y}_{q_1 - 1, \eta})])\\
= &\ \bar{R}^k M[Q_3(y_{q_1}) - Q_1(\hat{y}_{q_1 - 1, \eta})]
\end{aligned}
\end{equation}
hence $u_{q_1,k}$ can be expressed by $Q_3(y_{q_1}) - Q_1(\hat{y}_{q_1 - 1, \eta})$.
In addition, since
\begin{equation*}
|Q_3(y_q) - Q_1(\hat{y}_{q-1, \eta})| \le \frac{N_3 - 1}{N_3}E_{3,q}
\end{equation*}
it follows that, in the absence of DoS attacks, $u_{q,k}$ satisfies
\begin{equation}\label{2uqk}
|u_{q,k}| \le \frac{N_3 - 1}{N_3}\big\Vert K\bar{R}^kM\big\Vert E_{3,q} =: E_{2,q,k}
\end{equation}
for all $q \ge 1, k = 0, \cdots\!, \eta -1$.
When DoS attacks occur, the plant cannot receive inputs from controller.
In other words, $E_{2,q,k}$ only depends on the latest $E_{3,q}$, thus $E_{2,q,k}$ can remain unchanged during DoS attacks.
Following the definitions of $E_{1,q}$ and $E_{2,q,k}$, we are able to design sequence $\{E_{3,q}: q\in \mathbb{Z}_{\ge 1}\}$.
First, in the absence of DoS attacks, the error just before each transmission instant, $e_{q - 1, \eta} = x_q - \hat{x}_{q - 1,\eta}$, satisfies
\begin{align*}
e_{q - 1, \eta} =&\ A_d^{\eta}(I - MC)e_{q - 1} - A_d^{\eta}M[Q_3(y_q)- y_q]\\
& + \sum_{i = 0}^{\eta - 1}{A_d^{\eta - i - 1}B_d[Q_2(u_{q-1,i}) - u_{q- 1,i}]}\\
& - A_d^{\eta}\big[\hat{y}_{q - 1, \eta} - Q_1(\hat{y}_{q - 1, \eta})\big]
\end{align*}
which implies that $e_{q - 1, \eta}$ generally relies on $\hat{y}_{q-1, \eta}, u_{q,k}$, and itself, thus introducing coupling in $E_{3,q}$ design.
Here, this issue is addressed by (\ref{dbdb}).
To see this, recalling (\ref{2uqk}), $\hat{y}_{q - 1, \eta} = 0$, and $Q_1(\hat{y}_{q - 1, \eta}) = 0$, we have that
\begin{align}\label{2error}
&e_{q + {\ell} - 1, \eta} =R^{\ell} e_{q - 1} + \sum_{j = 0}^{{\ell} - 1}R^jA_d^{\eta}M(Q_3(y_{q - j})- y_{q - j})\nonumber\\
&+ \sum_{j = 0}^{{\ell} - 1}R^j\sum_{i = 0}^{\eta \!-\! 1}A_d^{\eta \!-\! i - 1}B_d \big[Q_2(u_{q+\ell \!-\!j \!-\!1,i}) \!-\! u_{q+{\ell}-j-1,i}\big].
\end{align}
Define $E_{3,q}$ as follows
\begin{align*}
E_{3,q+{\ell}} :=&\ a_0\rho^{\ell}E_{3,q} \\
& + \sum_{i = 0}^{{\ell} - 1}\left(\frac{(N_3 - 1)a_2\Vert C\Vert }{N_2N_3} + \frac{\Vert C\Vert a_1}{N_3}\right)\rho^{i}E_{3,q-i}.
\end{align*}
Hence, combining (\ref{2uqk}) with (\ref{2error}) yields
\begin{align}\label{eq:yqy}
|y_{q + 1} - Q_1(\hat{y}_{q, \eta})| & \le |y_{q + 1} - \hat{y}_{q, \eta}| + |\hat{y}_{q, \eta} - Q_1(\hat{y}_{q, \eta})|\nonumber\\
& \le \Vert C\Vert |x_{q + 1} - \hat{x}_{q, \eta}|\nonumber\\
& \le \hat{\theta}_{na} E_{3,q}
\le E_{3,q + 1}.
\end{align}
Moreover, since both the input and output channels are blocked in the presence of DoS attacks, and $\hat{y}_{q,\eta} = 0$, due to the property of $\bar{R}$, it follows that
\begin{equation*}
|y_{q + 1} - Q_1(\hat{y}_{q, \eta})| \le \hat{\theta}_{a} E_{3,q} \le E_{3,q + 1}
\end{equation*}
and we complete the proof.
\end{proof}
Next, we establish upper bounds on the sequences $\{E_{p,q,k}:q \in \mathbb{Z}_{\ge 1}, k = 0, \cdots\!, \eta - 1\} (p = 1, 2, 3)$, whose existence will imply the boundness of state trajectory.
\begin{lemma}\label{lem:eqbound}
Consider system (\ref{continuoussystem}) with controller in (\ref{abdoscontroller}) and (\ref{predoscontroller}), where $K$ satisfies (\ref{dbdb}) and $M$ is chosen such that $R$ is schur stable.
Let the assumptions and conditions in Thm. \ref{2convergetheorem} hold.
If further $\{E_{p,q,k}:q \in \mathbb{Z}_{\ge 1}, k = 0, \cdots\!, \eta - 1\} (p = 1, 2, 3)$ obey (\ref{2E1q})-(\ref{2E3q}), there exist $\Omega_1 \ge 1$ and $\gamma \in (0,1)$ such that
\begin{equation}\label{Eqomegagamma}
E_{3, q} \le \Omega_1 \gamma^q |x_0| ,\qquad \forall k \in \mathbb{Z}_{\ge 1}
\end{equation}
and $E_{1, q} = E_{1, 0}$, and $E_{2, q, k} \le \Omega_2\gamma^q|x_0|$.
\end{lemma}
\begin{proof}
Using (\ref{2E1q}), $E_{1, q}$ remains unchanged within the considered interval, therefore, $E_{1, q} = E_{1, 0}$ holds for all $q \in \mathbb{Z}_{\ge 1}$.
The proof for $E_{3, q} \le \Omega \gamma^q |x_0|, \forall q \in \mathbb{Z}_{\ge 1}$ follows directly from that of Lemma
3.9 in \cite{8880482},
where $\Omega_1 := \frac{\hat{\theta}_0^{\Pi_f + 1}\hat{\theta}_a^{\Pi_d}}{\hat{\theta}_{na}^{\Pi_f + \Pi_d + 1}} \big(\hat{\theta}_{na}\big(\frac{\hat{\theta}_0}{\hat{\theta}_{na}}\big)^{\nu_f}\big(\frac{\hat{\theta}_a}{\hat{\theta}_{na}}\big)^{\nu_d}\big)$.
Moreover, applying (\ref{2uqk}), $E_{2, q, k} \le \Omega_2\gamma^q |x_0|$ can be verified with $\Omega_2 := \frac{N_3 - 1}{N_3}\Vert K\bar{R}^{\eta - 1}\Vert\Omega_1$.
\end{proof}
We are now in a position to prove Thm. \ref{2convergetheorem}.
\begin{proof}
[Proof of Theorem \ref{2convergetheorem}]
We first establish the bound of the state $x$ at the transmission instants, i.e., $|x(q\Delta)|$, then derive its bound at the sampling instants, i.e., $|x(q\Delta + k\delta)|$.
Finally, combining these two bounds to yield bound $|x(t)|$ in the considered horizon.
First, according to (\ref{continuoussystem}), (\ref{abdoscontroller}), and (\ref{predoscontroller}), one has
\begin{align}\label{1xqeta2}
x_{q, \eta} =&~ \bar{R}^{\eta}x_{q,k} + \sum_{i = 0}^{\eta - 1}\bar{R}^iB_dK(x_{q, \eta - i - 1} - \hat{x}_{q, \eta - i - 1}) \nonumber\\
& + \sum_{i = 0}^{\eta - 1}\bar{R}^iB_d(Q_2(u_{q, \eta - i - 1}) - K \hat{x}_{q, \eta - i - 1})
\end{align}
and
\begin{align}\label{eq:vertx}
\vert x_{q, \eta}\vert \le&~ \Vert \bar{R}^{\eta}\Vert |x_{q}| + \sum_{i = 0}^{\eta - 1}\Vert\bar{R}^iB_dK\Vert |(x_{q, \eta - i - 1} - \hat{x}_{q, \eta - i - 1})|\nonumber\\
& + \sum_{i = 0}^{\eta - 1}\Vert \bar{R}^iB_d\Vert|(Q_2(u_{q, \eta - i - 1}) - K \hat{x}_{q, \eta - i - 1})|.
\end{align}
Since (\ref{eq:yqy}), it follows that
\begin{equation}\label{eq:xqx}
\Vert x_{q} - \hat{x}_{q - 1, \eta}\Vert \le \frac{E_{3,q}}{\Vert C\Vert}.
\end{equation}
Noticing that $\bar{R}^{\eta} = 0$, substituting (\ref{2E2qk}) and (\ref{eq:xqx}) into (\ref{eq:vertx}),
\begin{align}
\vert x_{q, \eta}\vert\le&\ \sum_{i = 0}^{\eta - 1} \frac{N_3 - 1}{N_2 N_3}\Vert \bar{R}^i B_d\Vert\Vert K\bar{R}^{\eta - i - 1}M\Vert E_{3,q}\nonumber\\
& + \sum_{i = 0}^{\eta - 1} \frac{\Vert \bar{R}^i B_d K A_d^{\eta - i - 1}\Vert}{\Vert C\Vert} E_{3,q}\nonumber\\
\le&\ \Omega_x \Omega_1 \gamma^q |x_0|
\end{align}
where $\Omega_x := \sum_{i = 0}^{\eta - 1}\big\{ \frac{N_3 - 1}{N_2 N_3}\Vert \bar{R}^i B_d\Vert\Vert K\bar{R}^{\eta - i - 1}M\Vert + \frac{\Vert \bar{R}^i B_d K A_d^{\eta - i - 1}\Vert}{\Vert C\Vert}\big\}$, and the last inequality holds due to (\ref{Eqomegagamma}).
Since $x_{q, k+1} = A_dx_{q,k} + B_d Q_2(u_{q,k})$, we have that
\begin{align}\label{eq:xql}
\vert x_{q, {\ell}}\vert\le &\ \Vert\bar{R}^{{\ell}}\Vert |x_{q}| + \sum_{i = 0}^{{\ell} - 1} \frac{\Vert \bar{R}^i B_d K A_d^{{\ell} - i - 1}\Vert}{\Vert C\Vert}E_{3,q}\nonumber\\
&+ \sum_{i = 0}^{{\ell} - 1} \frac{N_3 - 1}{N_2 N_3}\Vert \bar{R}^i B_d\Vert\Vert K\bar{R}^{{\ell} - i - 1}M\Vert E_{3,q}\\
\le&\ \Omega_x \Omega_1 \gamma^q |x_0| + \Omega_3 \gamma^q |x_0|\nonumber \le \bar{\Omega}_x \gamma^q |x_0|
\end{align}
where $\Omega_3 := \Omega_1\max_{{\ell}} \sum_{i = 0}^{{\ell}}\big\{ \frac{N_3 - 1}{N_2 N_3}\Vert \bar{R}^i B_d\Vert\Vert K\bar{R}^{{\ell} - 1}M\Vert + \frac{\Vert \bar{R}^i B_d K A_d^{\eta - i - 1}\Vert}{\Vert C\Vert}\big\}, {\ell} \in \{1, \cdots\!, \eta - 1\}$, and $\bar{\Omega}_x := \Omega_3 + \Vert\bar{R}^{{\ell}}\Vert\Omega_x$.
Finally, abiding by (\ref{continuoussystem}), $x(t)$ satisfies
\begin{align*}
x(t) = e^{A(t - q\Delta - k\delta)} + \int_{q\Delta + k\delta}^te^{As}BQ_2(u_{q, k}) \,ds
\end{align*}
for all $t \in [q\Delta + k\delta, q\Delta + (k+1)\delta)$.
Combining Lem. \ref{lem:eqbound} and (\ref{eq:xql}), it follows that
\begin{align*}
\vert x(t) \vert \le \big(\Vert A_d\Vert \bar{\Omega}_x + \frac{N_2 + 1}{N_2}\Vert B_d\Vert\Omega_2\big)\gamma^q|x_0|\le \tilde{\Omega}_x e^{-\sigma t}|x_0|
\end{align*}
where $\sigma := \frac{1}{\eta \delta}\log \frac{1}{\gamma}$ and $\tilde{\Omega}_x := \Vert A_d\Vert \bar{\Omega}_x + \frac{N_2 + 1}{N_2}\Vert B_d\Vert\Omega_2$.
This implies exponential convergence of the state.
\end{proof}
\begin{remark}\label{2dbdbremark}
Leveraging the same technique as in Rmk. \ref{remark:dbgain}, one can also design $M$ to nullify $R^{\mu} = 0$, where $\mu$ is the observability index of $(C, A_d^{\eta})$.
A direct benefit from using the deadbeat observer gain is that the encoding schemes can be simplified, since $R^{\ell} = 0$ holds for all $\ell \ge \mu$.
However, the results in \cite{WakaikiObserver} indicate that despite exhibiting faster convergence and fewer quantization levels, due to the deadbeat property of matrices $R$ and $\bar{R}$, the quantization step size $E_{p,q}/N_p$ is large, which leads to large quantization errors.
Moreover, it was shown in \cite{8880482} that if the quantization step size $E_{p,q}/N_p$ grows slower during DoS attacks, then the overshoot from an attack is smaller, and the level of system robustness is stronger.
Therefore, instead of a deadbeat observer gain, a general one that can make $R$ schur stable is employed in the present work.
\end{remark}
\section{Network Phenomena at Output Channel}\label{outputsection}
In this section, we consider stabilizing linear systems over a communication network, where only the output channel is subject to DoS attacks, i.e., the input channel is assumed ideal; see Fig. \ref{siglenetworkfig}.
The transmission policy in the previous section is considered here; that is, the digital controller receives quantized output $Q(y_q)$ from the plant with period $\Delta$ and generates control input $u_{q,k}$ with period $\delta$.
Notice that the decoder can recover the correct quantized value from the index sent by the encoder only if they share the same quantization ranges and centers.
It is thus necessary to ensure that the encoder and the decoder are \emph{synchronized} before designing encoding schemes.
A direct way to maintain synchronization is through using an ACK-based protocol; see Fig. \ref{tcpfig}, which has been adopted in previous studies, such as, \cite{feng2020datarate, 8880482}.
Nevertheless, in real-time applications, protocols without ACKs, e.g., UDP, are often preferred since the resulting implementation is simpler as well as saves the additional energy required for sending ACKs \cite{HongUDP}.
Hence, in the following, we first show that method for stabilizing systems with ACK-based protocols can no longer be used under ACK-free protocols.
Then, we demonstrate that our proposed methods can inform the encoder of DoS attacks from zero inputs, thus the decoder and the encoder can be synchronized even without ACKs.
\begin{figure}
\centering
\includegraphics[width=6cm]{NetworkedStructure0.jpg}\\
\caption{Closed-loop system
with an ACK-free protocol.}\label{siglenetworkfig}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=6cm]{TCP.jpg}\\
\caption{Closed-loop system
with an ACK-based protocol. The black dashed line represents the ACKs sent from the decoder to the encoder.}\label{tcpfig}
\centering
\end{figure}
\subsection{Controller under an acknowledgment-based protocol}
Recall that $\{s_r\}_{r\in \mathbb{N}_0}$ collects the sequence of successful transmission instants.
Let $\delta = \Delta$, and choose $K$ such that $\bar{R} = A_d + B_dK$ is schur stable.
We consider an observer-based controller described by
\begin{subequations}\label{eq:tcpcontroller}
\begin{align}
&\hat{x}_{q+1} = A_d\hat{x}_q + B_du_q + L(Q(y_q) - \hat{y}_q), & q\Delta = s_r \label{eq:tcpcontroller_1}\\
&\hat{x}_{q+1} = A_d\hat{x}_q + B_du_q, & q\Delta \ne s_r \label{eq:tcpcontroller_2}\\
&\hat{y}_q = C\hat{x}_q \label{eq:tcpcontroller_3}\\
&u_q = K\hat{x}_q \label{eq:tcpcontroller_4}
\end{align}
\end{subequations}
where $\hat{x}_{q}\! \in\! \mathbb{R}^{n_x}, \hat{y}_{q} \!\in\! \mathbb{R}^{n_y}$, and $Q(y_{q}) \!\in\! \mathbb{R}^{n_y}$ are the estimated state, the estimated output, and the quantized output, respectively.
The initial condition is set to be $\hat{x}_0 = 0$.
Since the input channel is ideal, it follows that
\begin{equation*}
u(t) = u_q, \quad q \Delta \le t < (q + 1)\Delta, \quad q \in \mathbb{Z}_{\ge 0}.
\end{equation*}
To design an encoding scheme such that the output $y_q$ can be quantized without saturation, an error bound between the estimated output and the actual output, i.e., $| e_q| := |x_q - \hat{x}_q| \le E_q$, should be derived.
Based on (\ref{continuoussystem_2}) and (\ref{eq:tcpcontroller_2}), it can be deduced that
\begin{equation}\label{eq:tcperror}
|y_q - \hat{y}_q| = |C(x_q - \hat{x}_q)| = |Ce_q| \le \Vert C \Vert E_q.
\end{equation}
Let $N$ denote the number of quantization levels of $y_q$.
Similar to the previous section, we partition the hypercube $ \{ y \in \mathbb{R} ^{n_y} : | y_q - \hat{y}_q| \le \Vert C \Vert E_q\}$
into $N^{n_y}$ equal-sized boxes.
The quantization error obeys $ |Q(y_q) - y_q| \le \frac{\Vert C\Vert}{N}E_q$.
According to As. \ref{x0bound}, the initial value $E_0$ is given by
\begin{equation}\label{eq:tcpinitial}
|e_0| = |x_0| \stackrel{\triangle}{=} E_0.
\end{equation}
Sequence $\{E_q, q \in \mathbb{Z}_{\ge 1}\}$ will be specified latter.
Notice that the hypercube center is $\hat{y}_q$, which is generated by the predictor-based observer in (\ref{eq:tcpcontroller}).
Therefore, this predictor should also be equipped at the encoder side.
Under ACK-based protocol, the decoder sends ACKs to the encoder without delay at successful transmission instants; and when the encoder does not receive the ACKs, it infers that there is a DoS attack.
In this manner, synchronization between these two predictors is ensured, which consequently implies that the quantization ranges and the centers at the encoder are identical to that of the decoder.
Before giving stability condition for ACK-based protocol case, we present an output encoding scheme.
Let
\begin{equation}\label{eq:tcperrorbound}
E_{q + 1} :=
\left\{
\begin{aligned}
& \theta_a E_{q}, & q\Delta \ne s_r\\
& \theta_0 E_{q}, & (q-1)\Delta \ne s_r, q\Delta = s_r\\
& \theta_{na}E_{q}, & (q-1)\Delta = s_r, q\Delta = s_r
\end{aligned}
\right.
\end{equation}
with
\begin{subequations}\label{eq:tcpencode}
\begin{align}
\theta_{a} &:=\left\|A_d \right\| \label{eq:tcpencode1}\\
\theta_{0} &:=H_{0} \rho+\frac{H_1\left\|C \right\|}{N} \label{eq:tcpencode2}\\
\theta_{na} &:=\rho+\frac{H_1 \left\|C \right\|}{N} \label{eq:tcpencode3}
\end{align}
\end{subequations}
where constants $H_0$, $H_1$, and $0 < \rho < 1$ satisfy
\begin{equation*}
\Vert (A_d - LC)^\ell \Vert \le H_0 \rho^{\ell},\quad \Vert (A_d - LC)^\ell L\Vert \le H_1 \rho^{\ell}.
\end{equation*}
\begin{theorem}\label{th:outputtheorem1}
Consider system (\ref{continuoussystem}) with controller (\ref{eq:tcpcontroller}), where $M$ and $K$ are chosen such that $A_d - LC$ and $A_d + B_d K$ are schur stable.
Under As. \ref{as:abca}--\ref{DoS_durationassumption}, if i) the quantization levels
\begin{equation}\label{2Ntcpcondition}
\begin{split}
N > \frac{H_1\Vert C\Vert}{1 - \rho}
\end{split}
\end{equation}
and, ii) DoS attacks satisfy
\begin{equation}\label{2dostcpcondition}
\frac{1}{\nu_d} \le \frac{\log{(1/{\theta}_{na})}}{\log{({\theta}_a/{\theta}_{na})}} - \frac{\log{({\theta}_0/{\theta}_{na})}}{\log{(\theta_a/{\theta}_{na})}}\frac{1}{\nu_f}
\end{equation}
then the system is exponentially stable under the encoding scheme with error bounds $\{E_{q}:q \in \mathbb{Z}_{\ge 1}\}$ constructed by the update rule (\ref{eq:tcperrorbound}).
\end{theorem}
The proof is similar to that of \cite[Thm. 3.4]{8880482} and is thus omitted here due to space limitations.
\subsection{Controller under an acknowledgment-free protocol}\label{1Acontrollersection}
In this subsection, we show that the aforementioned controller and encoding scheme cannot stabilize the system when the ACK-based protocol is replaced by an ACK-free protocol.
This is because synchronization between the encoder and decoder is no longer guaranteed.
To see this, consider controller (\ref{eq:tcpcontroller}) with the encoding scheme in (\ref{eq:tcperrorbound}) employing an ACK-free protocol.
In this setting, predictors at the encoder and decoder sides may become asynchronized, since no matter whether DoS attacks happen or not, the decoder does not send ACKs to the encoder.
When a DoS attack occurs, the predictor at the controller side switches to (\ref{eq:tcpcontroller_2}), whereas the predictor at the encoder side sticks to (\ref{eq:tcpcontroller_1}).
Moreover, the update rule of sequence $E_q$ at the decoder switches to (\ref{eq:tcpencode1}), while adhering to (\ref{eq:tcpencode2})-(\ref{eq:tcpencode3}) at the encoder.
As a result, their quantization ranges and centers may deviate, and the correct output value cannot be recovered by the decoder.
We prove that even if one DoS attack occurs (i.e., decoder and encoder are asynchronized for only one transmission period), the state may diverge eventually.
To distinguish between predictors at the encoder and decoder, let $\hat{x}_q$, $\hat{y}_q$, and $\hat{Q}(y_q)$ denote the estimated state, estimated output, and quantized output at the controller side, and $\tilde{x}_q, \tilde{y}_q$, and $Q(y_q)$ denote their counterparts at the encoder side.
In addition, let $u_q$ stand for the input sent by the controller, and $\tilde{u}_q$ the estimated input generated by the predictor at the encoder side.
Predictor at the controller side can be expressed by
\begin{subequations}\label{eq:tcpdecoder}
\begin{align}
&\hat{x}_{q+1} = A_d\hat{x}_q + B_du_q + L(\hat{Q}(y_q) - \hat{y}_q), & q\Delta = s_r \label{eq:tcpdecoder_1}\\
&\hat{x}_{q+1} = A_d\hat{x}_q + B_du_q, & q\Delta \ne s_r \label{eq:tcpdecoder_2}\\
&\hat{y}_q = C\hat{x}_q \label{eq:tcpdecoder_3}\\
&u_q = K\hat{x}_q \label{eq:tcpdecoder_4}
\end{align}
\end{subequations}
and predictor at the encoder side is described by
\begin{subequations}\label{eq:tcpencoder}
\begin{align}
&\tilde{x}_{q+1} = A_d\tilde{x}_q + B_d\tilde{u}_q + L({Q}(y_q) - \tilde{y}_q)\label{eq:tcpencoder_1}\\
&\tilde{y}_q = C\tilde{x}_q \label{eq:tcpencoder_2}\\
&\tilde{u}_q = K\tilde{x}_q
\end{align}
\end{subequations}
where $q \in \mathbb{Z}_{\ge 0}$.
Similarly, let $E_{d,q}$, and $E_{e,q}$ denote the error bound at the decoder, and the encoder side, respectively
\begin{equation*}
\begin{aligned}
&E_{d,q + 1} :=
\left\{
\begin{aligned}
& \theta_a E_{d,q}, & q\Delta \ne s_r\\
& \theta_0 E_{d,q}, & (q-1)\Delta \ne s_r, q\Delta = s_r\\
& \theta_{na}E_{d,q}, & (q-1)\Delta = s_r, q\Delta = s_r
\end{aligned}
\right.\\
&E_{e,q + 1} :=
\left\{
\begin{aligned}
& \theta_0 E_{e,q}, & q\Delta = 0\\
& \theta_{na}E_{e,q}, & q\Delta > 0
\end{aligned}
\right.
\end{aligned}
\end{equation*}
where $\theta_{0}$, $\theta_{a}$, and $\theta_{na}$ are defined in (\ref{eq:tcpencode}).
Accordingly, the errors at the encoder and decoder sides are $e_{e,q} := x_q - \tilde{x}_q$, and $e_{d,q} := x_q - \hat{x}_q$.
Moreover, the quantized outputs in (\ref{eq:tcpdecoder_2}) and (\ref{eq:tcpencoder_1}) are
\begin{align}\label{hatQ}
\hat{Q}(y_q) = \hat{y}_q + Q^{i}_q\frac{\Vert C\Vert E_{d,q}}{N}\\
Q(y_q) = \tilde{y}_q + Q^{i}_q\frac{\Vert C\Vert E_{e,q}}{N}
\end{align}
where $Q_q^{i}$ denotes the quantization index transmitted from the encoder to the decoder.
Suppose that a DoS attack is launched at $q_a\Delta$ and no attacks happen before or after $q_a\Delta$. It follows that $\hat{Q}(y_q) = Q(y_q)$ for all $q \le q_a$, and
\begin{subequations}\label{tildex-x}
\begin{align}
&\hat{x}_{q_a} = \tilde{x}_{q_a} \\
&\hat{x}_{q_a\!+\!1} = (A_d \!+\! B_d K)\hat{x}_{q_a} \\
&\tilde{x}_{q_a \!+\! 1} = (A_d \!+\! B_d K) \tilde{x}_{q_a} \!+\! L Q^{i}_{q_a}\frac{\Vert C\Vert E_{e,q_a}}{N} \\
&\hat{x}_{q_a+2} = (A_d \!+\! B_d K)^2 \hat{x}_{q_a} \!+\! L Q^{i}_{q_a\!+\!1}\frac{\Vert C\Vert E_{d,q_a \!+\! 1}}{N}\\
\begin{split}
&\tilde{x}_{q_a + 2} =
(A_d \!+\! B_d K)^2 \tilde{x}_{q_a} \!+\! L Q^{i}_{q_a+1}\frac{\Vert C\Vert E_{e,q_a + 1}}{N}\\
&~~~~~~~~~~ \!+\! (A_d \!+\! B_d K)L Q^{i}_{q_a}\frac{\Vert C\Vert E_{e,q_a}}{N} \\
\end{split}\\
& \cdots \nonumber
\end{align}
\end{subequations}
Notice that the quantizer operates normally without saturation only if $E_{e,q} \ge |e_{e,q}| = |x_{q} - \tilde{x}_q|$ and $E_{d,q} \ge |e_{d,q}| = |x_{q} - \hat{x}_q|$ hold for all $q \in \mathbb{Z}_{\ge 1}$.
If the quantizer saturates, the error between the actual output and the quantized output maybe large, which consequently renders the system unstable.
In the following, we assume that the quantizer is not saturated; that is $E_{e,q} \ge |e_{e,q}|$ and $E_{d,q} \ge |e_{d,q}|$ for all $q \in \mathbb{Z}_{\ge 1}$, and reach a contradiction.
Since $E_{e,q + 1} = \theta_{na}E_{e,q}, q > 0$, and $\theta_{na} <1$, sequence $\{|x_q - \tilde{x}_q|\}$ is decreasing.
Let $\Pi_{L} := A_d - LC$ and $\Pi_{K} := A_d + B_dK$.
Combining (\ref{hatQ}) and (\ref{tildex-x}) yields
\begin{align*}
|x_{q_a + 1} - \tilde{x}_{q_a + 1}| & = |\Pi_{L} (x_{q_a} - \tilde{x}_{q_a}) - L(Q(y_{q_a}) - y_{q_a})|\\
& \le E_{e, q_a + 1} \stackrel{\triangle}{=} \tilde{E}_{e,q_a + 1}.
\end{align*}
Likewise,
\begin{align*}
&~|x_{q_a + 2}- \tilde{x}_{q_a + 2}| \\
=&\ \big|\Pi_{L} (x_{q_a + 1} - \tilde{x}_{q_a + 1}) - L(Q(y_{q_a + 1}) - y_{q_a + 1})\\
& - BKLQ^{i}_{q_a}\frac{\Vert CR^{-1}\Vert E_{e,q_a}}{N}\big|\\
\le &\ E_{q_a + 2} + \frac{1}{\theta_{na}^2}\frac{\Vert BKLQ^{i}_{q_a}\Vert\Vert CR^{-1}\Vert}{N}E_{e,q_a+2}\\
\stackrel{\triangle}{=} &\ \tilde{E}_{e,q_a + 2}.
\end{align*}
Iteratively, for $\ell \ge 3$, it follows that
\begin{align*}
&~ |x_{q_a + \ell} - \tilde{x}_{q_a + \ell}|\\
\le&\ E_{e, q_a + \ell} + \frac{1}{\theta_{na}^\ell}\frac{\Vert B_dK\Pi_{K}^{\ell - 1}LQ^{i}_{q_a}\Vert\Vert C\Vert E_{e, q_a}}{N}\\
& + \frac{1}{\theta_{na}^\ell}\frac{\Vert B_dK\Pi_{K}^{\ell - 2}LQ^{i}_{q_a + 1}\Vert\Vert C\Vert(\theta_{a} - \theta_{na}) E_{e, q_a}}{N}\\
& + \sum_{i = 0}^{\ell - 3}\frac{1}{\theta_{na}^{i + 3}}\frac{\Vert B_dK\Pi_{K}^{i}LQ^{i}_{q_a + \ell - i - 1}\Vert\Vert CR^{-1}\Vert}{N}
\\
&~~ \times (\theta_0\theta_a - \theta_{na}^2) E_{e, q_a}\\
\stackrel{\triangle}{=}& ~\tilde{E}_{e,q_a + \ell}.
\end{align*}
Since ${1}/{\theta_{na}}>1$, $\{\tilde{E}_{e,q}\}$ is an increasing sequence, which contradicts the assumption that $\{|x_q - \tilde{x}_q|\}$ is a decreasing sequence.
Therefore, it can be concluded that without ACKs, predictors at the encoder and controller sides may get asynchronized even if there is a single DoS attack.
This causes mismatches on their quantization centers and ranges, and there exists $\hat{q} \ge q_a$ such that $E_{e, q} < |x_{q} - \tilde{x}_{q}|$ holds for all $q \ge \hat{q}$, and the state diverges eventually.
We have just shown that the synchronization between decoder and encoder is essential.
However, ACK-based protocol is not the only way to achieve this goal.
In the absence of ACKs, this challenge can be overcome by using a deadbeat controller, and the prove will be given in the following.
Let the number of the quantization level $N$ to be even.
We adopt the same quantizer as in (\ref{eq:tcperror})-(\ref{eq:tcpinitial}), with $\hat{x}_q$, $e_q$, and $\hat{y}_q$ replaced by $\hat{x}_{q-1, \eta}$, $e_{q-1, \eta}$, and $\hat{y}_{q - 1, \eta}$, respectively.
The observer-based controller is employed only at the decoder side
\begin{subequations}\label{controller1}
\begin{align}
&\hat{x}_{q, k+1} = A_d\hat{x}_{q,k} + B_du_{q,k}, & k&\le \eta - 1 \label{controller1_1}\\
&\hat{x}_{q} = \hat{x}_{q-1, \eta} + M_q[Q(y_{q}) - \hat{y}_{q-1, \eta}], & k &= \eta \label{controller1_2}\\
&\hat{y}_{q,k} = C \hat{x}_{q,k} \label{controller1_3}\\
&u_{q,k} = K \hat{x}_{q,k} \label{controller1_4}
\end{align}
\end{subequations}
Thanks to the ideal input channel,
\begin{equation*}\label{ut=uk}
u(t) = u_{q,k}, \qquad q\Delta + k\delta \le t < q\Delta +(k+1)\delta
\end{equation*}
for every $q \in \mathbb{Z}_{\ge 0}$, and $k = 0, \cdots\!, \eta - 1$.
Consider an arbitrary transmission interval $[q \Delta, (q + 1)\Delta)$.
From the property (\ref{dbdb}), one gets that $\hat{x}_{q,\eta} = (A_d + B_d K)^{\eta}\hat{x}_{q} = 0$, and $\hat{y}_{q, \eta} = C\hat{x}_{q, \eta} = 0$.
It is thus sufficient to choose the quantization center to be the origin, and predictor (\ref{controller1}) is not needed at the encoder side.
This saves computational resources.
If an attack is launched at $(q + 1)\Delta$, the decoder is not going to receive the quantized output $Q(y_{q + 1})$, and instead it will use a default zero.
Then, it follows from (\ref{controller1_1})-(\ref{controller1_2}) that $\hat{x}_{q + 1} = \hat{x}_{q,\eta} = 0$, and $u_{q+1} = K\hat{x}_{q + 1} = 0$.
On the other hand, in the absence of DoS attacks, since the quantization center is zero and $N$ is even, the quantized value is nonzero.
Therefore, the decoder receives a quantized output $Q(y_{q + 1}) \ne 0$.
As a result, $\hat{x}_{q + 1} = \hat{x}_{q, \eta} + M(Q(y_{q+1}) - \hat{y}_{q, \eta})= MQ(y_{q + 1}) \ne 0$, and $u_{q+1} = K\hat{x}_{q + 1} \ne 0$.
This suggests that the encoder can infer whether there is an attack or not from the input signals, thus its quantization ranges can be updated following the same scheme with the decoder.
We have secured synchronization between the encoder and decoder.
Now, what is left behind is the system stability analysis.
Recall that $A_d^{\eta}(I - MC)$ is schur stable, there exist constants $G_0, G_1$, and $0 <\rho <1$ such that
\begin{equation}\label{m0m1}
\begin{aligned}
\big\Vert R^{\ell}\big\Vert \le G_0 \rho^{\ell},\quad
\big\Vert {R}^{\ell}A_d^{\eta}M\big\Vert \le G_1 \rho^{\ell}.
\end{aligned}
\end{equation}
Define constants
\begin{align*}
\tilde{\theta}_{a} := \Vert A_d^{\eta}\Vert,~~
\tilde{\theta}_{0} := G_0\rho + \frac{G_1\Vert C\Vert}{N},~~
\tilde{\theta}_{na} := \rho + \frac{G_1\Vert C\Vert}{N}
\end{align*}
and the error bound $\{E_q:q \in \mathbb{Z}_{\ge 1}\}$ is updated by
\begin{equation}\label{errorboundEq}
E_{q + 1} :=
\left\{
\begin{aligned}
& \tilde{\theta}_a E_q, & q\Delta \ne s_r\\
& \tilde{\theta}_0 E_q, & (q-1)\Delta \ne s_r, q\Delta = s_r\\
& \tilde{\theta}_{na}E_q, & (q-1)\Delta = s_r, q\Delta = s_r
\end{aligned}
\right..
\end{equation}
The following result is an extension of Thm. \ref{th:outputtheorem1} under an ACK-free protocol, whose proof follows from that of Thm. \ref{th:outputtheorem1}.
\begin{theorem}\label{1convergetheorem}
Consider system (\ref{continuoussystem}) equipped with controller in (\ref{controller1}), where $M$ and $K$ are chosen such that $R$ is schur stable and (\ref{dbdb}) is met.
Let As. \ref{as:abca}--\ref{DoS_durationassumption} hold.
If i) the output and input transmission periods satisfy (\ref{delta}), ii) the quantization levels $N$ is even, and obey
\begin{equation}\label{Ncondition}
N > \frac{G_1\Vert C\Vert}{1 - \rho}
\end{equation}
and, iii) DoS attacks satisfy
\begin{equation}\label{1doscondition}
\frac{1}{\nu_d} \le \frac{\log{(1/\tilde{\theta}_{na})}}{\log{(\tilde{\theta}_a/\tilde{\theta}_{na})}} - \frac{\log{(\tilde{\theta}_0/\tilde{\theta}_{na})}}{\log{(\tilde{\theta}_a/\tilde{\theta}_{na})}}\frac{1}{\nu_f}
\end{equation}
then the system is exponentially stable under the encoding scheme with error bound $\{E_q:q \in \mathbb{Z}_{\ge 1}\}$ constructed by (\ref{errorboundEq}).
\end{theorem}
\begin{figure}[b]
\centering
\includegraphics[width=9cm]{doubledblqnormxbd.pdf}\\
\caption{Maximum norm of state $x$ and its estimate $\hat{x}$ with controller (\ref{abdoscontroller}).}\label{figdoubledblqnormx}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=9cm]{doubledblqnormE3bd.pdf}\\
\caption{Relationship between normalized quantization range $E_{3,k}/N_3$ and actual error $|y_q - Q_1(\hat{y}_q)|$ with controller (\ref{abdoscontroller}).}\label{figdoubledblqnormE}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=9cm]{doubledblqnormE2bd.pdf}\\
\caption{Normalized quantization range $E_{2,q,k}/N_2$ with controller (\ref{abdoscontroller}).}\label{figdoubledblqnormE2}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=9cm]{logdblq.pdf}\\
\caption{Normalized quantization ranges $E_{3,q}/N_3$ from using general observer gain and deadbeat observer gain in log space.}\label{figlogdblq}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=9cm]{doubledbdbnormxbd.pdf}\\
\caption{Maximum norm of state $x$ and its estimate $\hat{x}$ with controller (\ref{abdoscontroller}) and deadbeat observer gain.}\label{figdoubledbdbnormx}
\centering
\end{figure}
\section{Numerical Example}
\label{simulation}
A linearized model of the unstable batch reactor in \cite{8880482} is given by $\dot{x}(t) = A x(t) + B u(t)$ and $y = C x(t)$, where
\begin{align*}
A :=
\left[
\begin{matrix}
1.38 & -0.2077 & 6.715 & -5.676\\
-0.5814 & -4.29 & 0 & 0.675\\
1.067 & 4.273 & -6.654 & 5.893\\
0.048 & 4.273 & -1.343 & -2.104
\end{matrix}
\right],\\
B := \left[
\begin{matrix}
0 & 0\\
5.679 & 0\\
1.136 & -3.146\\
1.136 & 0
\end{matrix}
\right],
C :=
\left[
\begin{matrix}
1 & 0 & 1 & -1\\
0 & 1 & 0 & 0
\end{matrix}
\right].
\end{align*}
This system $(A, B, C)$ is observable and controllable with $\eta =\mu = 2$.
Let the output transmission period $\Delta = 0.2$, so $\delta = \Delta/\eta = 0.1$.
Choosing matrix $K$, such that (\ref{dbdb}) is met, i.e.,
\begin{align*}
& K :=
\left[
\begin{matrix}
1.0106 & -1.5661 & 0.0385 & -4.0366\\
8.1074 & -0.0347 & 4.3337 &- 3.6241
\end{matrix}
\right].
\end{align*}
Calculating the gain of the steady-state Kalman filter
\begin{align*}
M :=
\left[
\begin{matrix}
0.5534 & -0.0249\\
-0.0287 & 0.0396\\
0.1489 & 0.0892\\
0.0810 & 0.0931
\end{matrix}
\right].
\end{align*}
We first present the time responses when both input and output channels suffer from the network phenomena.
Applying Thm. \ref{2convergetheorem}, when both the quantization levels $N_2$ and $N_3$ go to infinity, the duration bound $1/\nu_d$ and the frequency bound $1/\nu_f$ of DoS attacks approach to the line $\frac{1}{\nu_d} \approx -0.5544\frac{1}{\nu_f} + 0.2707$.
According to (\ref{2doscondition}), if
$\frac{1}{\nu_d} < -2.0380\frac{1}{\nu_f} + 0.2269$,
then the closed-loop system with encoding schemes (\ref{2E1q})-(\ref{2E3q}) is stabilized.
Over a simulation horizon of $160$s ($800$ time-step), DoS attacks (the gray shades) are generated randomly with $\Phi_d = 47$ and $\Phi_f = 44$.
Setting $\kappa_d = 3, \nu_d = 18, \kappa_f = 2, \nu_f = 19$, condition (\ref{2doscondition}) holds, i.e., $1/\nu_d = 0.056 < 0.119$.
Figs. \ref{figdoubledblqnormx} and \ref{figdoubledblqnormE} illustrate the time response in this situation.
Since the condition in Thm. \ref{2convergetheorem} is satisfied, the maximum norm of the state converges, and the bound $E_{3,q}$ exponentially decreases.
Fig. \ref{figdoubledblqnormE} depicts that $E_{3,q}$ shares the same trend with $|y_q - Q_1(\hat{y}_{q})|$, and Fig. \ref{figdoubledblqnormE2} demonstrates the evolution of the quantization step size $E_{2,q,k}/N_2$, which jumps up and down within an output transmission period, and decreases in general.
Difference between the trend of $E_{3,q}/N_3$ and $E_{2,q,k}/N_2$ lies in the property of $\Vert R \Vert$ and $\Vert \bar{R}\Vert$.
Fig. \ref{figlogdblq} compares the quantization step size $E_{3,q}/N_3$ of a general observer gain (blue line), such that $R$ is schur stable, and the deadbeat observer gain (dot marked green line), namely $R^{\mu} = 0$.
This panel illustrates that although $E_{3,q}$ responds faster under deadbeat observer, the large quantization step size results in large overshoot of the state; see Fig. \ref{figdoubledbdbnormx}, which confirms Rmk. \ref{2dbdbremark}.
\begin{figure}
\centering
\includegraphics[width=9cm]{siglenormx.pdf}\\
\caption{Maximum norm of state $x$ and its estimate $\hat{x}$ with controller (\ref{controller1}).}\label{figsiglenormx}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=9cm]{sigleerrorE.pdf}\\
\caption{Relationship between normalized quantization range $E_q/N$ and actual error $|e_q|$ with controller (\ref{controller1}).}\label{figsigleerrorE}
\centering
\end{figure}
\begin{figure}
\centering
\includegraphics[width=9cm]{sigleu.pdf}\\
\caption{Input signal $u_{q,k}$ with controller (\ref{controller1}).}\label{figsigleu}
\centering
\end{figure}
Next, consider network phenomena only at output channel.
From (\ref{Ncondition}), the quantization levels satisfies $N > 6.957$, also since $N$ is even, we set $N = 100$.
Over a simulation horizon of $60$s ($300$ time-step), generating DoS attacks randomly with $\Phi_d = 27$ and $\Phi_f = 25$.
Setting $\kappa_d = 1, \nu_d = 11, \kappa_f = 1, \nu_f = 11$, so condition (\ref{1doscondition}) is met with $1/\nu_d = 0.01 < 0.198$, and convergence of the state is presented in Figs. \ref{figsiglenormx} and \ref{figsigleerrorE}.
Further, Fig. \ref{figsigleu} shows that when a DoS attack happens, the control input is set to zero immediately, which verifies the effectiveness of our method.
\section{Conclusions}\label{conclusion}
This paper considered the problem of stabilizing networked control systems in the presence of DoS attacks and limited data rates. To overcome the network-induced challenges, a structure consisting of a deadbeat controller and a transmission protocol which are carefully co-designed based on the system controllability index, was proposed to address the network-induced challenges.
Specifically, when both input and output channels are subject to the network phenomena, it was shown that the proposed structure can decouple and thus allow for separate design of encoding schemes for the input, output, and estimated output signals. Furthermore, easy-to-check conditions were derived such that exponential stability of the closed-loop system under this structure is ensured. On the other hand, when only the output channel is subject to the network phenomena, the proposed structure was shown able to guarantee synchronization between the encoder and decoder under an ACK-free protocol.
Finally, a numerical example was presented to verify the effectiveness of our approach as well as the correctness of our theory.
Future developments will focus on generalizing the results to more general systems and controllers under ACK-free protocols.
\bibliographystyle{IEEEtran}
|
{
"timestamp": "2021-03-23T01:36:44",
"yymm": "2103",
"arxiv_id": "2103.11862",
"language": "en",
"url": "https://arxiv.org/abs/2103.11862"
}
|
\section{Introduction}
We present a simple experiment that allows us to demonstrate graphically that the intensity of sound waves is proportional to the square of their amplitude, a result that is theoretically analysed in any introductory wave course but rarely demonstrated empirically.\\
To achieve our goal, we use an audio signal generator that, when connected to a loudspeaker, produces sine waves that can be easily observed and measured using an oscilloscope. The measurements made with these instruments allow us to create a plot of amplitude versus sound intensity level, which verifies the mathematical relationship between amplitude and intensity mentioned above. Among the experimental errors, the plot obtained is in excellent agreement with what is theoretically expected.
\section{Theoretical framework}
It is well known that the intensity level ($\beta$) of a sound wave is related to its intensity $I$ (in the international system it is measured in $W\cdot m^{-2}$) as [1]:
\begin{equation}
\beta = 10 log \left( \frac{I}{I_{0}} \right),
\end{equation}
where the factor 10 is due to the fact that the intensity level is expressed in decibels (one tenth of a bel) and the $I_{0}$ value corresponds to the lowest intensity that a human ear in good condition can detect ($10^{-12} W\cdot m^{-2}$).\\
\begin{figure}
\centering
\includegraphics[width=0.3\textwidth]{Fig1}
\caption{Expected theoretical plot of $\beta$ vs. $log(A)$.}
\end{figure}
As demonstrated in any general physics textbook, the intensity $I$ of a wave is proportional to the square of the amplitude $A$ [2]:
\begin{equation}
I = c_{0} A^{2},
\end{equation}
where $c_{0}$ is a constant. This means that the intensity of a wave increases faster than its amplitude. Starting from Eq. (2), Eq. (1) can be rewritten as:
\begin{equation}
\beta = 10 log \left( \frac{c_{0} A^{2}}{c_{1}} \right)
\end{equation}
or,
\begin{equation}
\beta = 20 log \left( A \right) + c_{2},
\end{equation}
where $c_{1}$ and $c_{2}$ are constants. This equation shows us that if the intensity sound level, $\beta$ (dependent variable), is plotted versus the logarithm of its amplitude, $log (A)$ (independent variable), a line is obtained with a slope of 20, as illustrated in Fig. 1.\\
Using an audio signal generator to produce sound through a speaker, an oscilloscope to visualise and measure the amplitude of the emitted wave, and a sound level meter to measure the intensity level, it is possible to reproduce the plot in Fig. 1, which allows us to demonstrate experimentally the relationship between intensity and amplitude given by Eqs. (2) and (3). Fig. 2 schematises the experimental setup.
\begin{figure}[h]
\centering
\includegraphics[width=0.8\textwidth]{Fig2}
\caption{Schematic representation of the experimental setup.}
\end{figure}
\section{Materials list}
Audio signal generator EICO 377.\\
Oscilloscope HAMEG HM 203-7.\\
Digital multimeter with a sound level meter MASTECH MS 8209.\\
Loudspeakers $3 W – 8 \Omega$.\\
Connection cables.\\
Coaxial cable.
\section{Graphic relation between amplitude and sound intensity level}
Before starting with the description of the experiment, it is important to note that for its correct performance, we recommend reducing the background noise to a minimum, ensuring that it is uniform during the experiment. For this reason, we also do not recommend taking measurements outdoors.\\
\begin{figure}[h]
\centering
\includegraphics[width=0.8\textwidth]{Fig3}
\caption{Experimental setup.}
\end{figure}
To carry out the experiment, a sinusoidal signal generator was connected to a speaker, which reproduces the pure sound emitted by the generator. In contrast, an oscilloscope was used to measure the amplitude of the generated wave, which is expressed in millivolts ($mV$). Finally, very close to the speaker, a sound level meter was installed to measure the intensity level due, on the one hand, to the background noise, and on the other, to the generated wave. Frequencies were established with the generator scale but were also measured with the oscilloscope.\\
\begin{figure}[h]
\centering
\includegraphics[width=0.9\textwidth]{Fig4}
\caption{Sound intensity level vs. amplitude plots for three sets of measurements.}
\end{figure}
The experiment was carried out in a laboratory in the absence of students, with a background intensity level of $33.3 dB$. Nine amplitude values were measured, which varied between 10 and 300 $mV$. \\
The experiment was tested with different frequencies and it was decided to perform it with those that gave the best results, i.e., those occurring in a range between 1000 and 2000 $Hz$. Higher frequencies (over 2000 $Hz$ and up to 5000 $Hz$) or lower (from 500 to 1000 $Hz$) caused a deviation from the linear relationship between sound intensity level and amplitude at the ends of the plotted line, and also generated a greater distortion of the sine wave shape, which is a consequence of the corresponding increase of amplitude.\\
The measurements at the intensity level ranged from 40 dB to a little less than $90 dB$, and the amplitude variations were in a range between 10 and 300 $mV$; therefore, both the amplitude variation range and the range intensity level were significant. Fig. 3 shows the experimental setup.\\
Table 1 summarises the measurements for frequencies of 1000, 1500 and 2000 $Hz$. For each of these frequencies, the table shows the amplitude $A$ (in $mV$), the sound intensity level $\beta$ (in $dB$) and the logarithm of the amplitude $log (A)$. \\
The plots for the three sets of measurements are shown in Fig. 4. As can be seen, the linear trend is satisfactory. Furthermore, in all three cases, the experimental slope ($19.8\pm0.5$, $21,5\pm0.8$, and $20,1\pm0.5$) is very close to the expected theoretical value (20). \\
\begin{table}[htbp]
\begin{center}
\caption{Summary of amplitude and intensity level measurements.}
\begin{tabular}{l l l l l l l l l}
\toprule
\multicolumn{3}{l}{First set of measures} &
\multicolumn{3}{l}{Second set of measures} &
\multicolumn{3}{l}{Third set of measures} \\
\multicolumn{3}{l}{(1000 Hz)} &
\multicolumn{3}{l}{(1500 Hz)} &
\multicolumn{3}{l}{(2000 Hz)} \\
\midrule
$A (mV)$ & $\beta (dB)$ & $log (A)$ & $A (mV)$ & $\beta (dB)$ & $log (A)$ & $A (mV)$ & $\beta (dB)$ & $log (A)$ \\
$\pm 5\%$ & $\pm 0.3\%$ & & $\pm 5\%$ & $\pm 0.3\%$ & & $\pm 5\%$ & $\pm 0.3\%$ & \\
\midrule
10 & 59.8 & 1.000 & 10 & 51.3 & 1.000 & 10 & 59.1 & 1.000 \\
\midrule
20 & 66.0 & 1.301 & 20 & 60.1 & 1.301 & 20 & 66.3 & 1.301 \\
\midrule
30 & 70.3 & 1.477 & 30 & 64.3 & 1.477 & 30 & 69.7 & 1.477 \\
\midrule
40 & 73.3 & 1.602 & 40 & 67.4 & 1.602 & 40 & 72.5 & 1.602 \\
\midrule
60 & 76.5 & 1.778 & 60 & 70.5 & 1.778 & 60 & 76.3 & 1.778 \\
\midrule
100 & 81.0 & 2.000 & 100 & 75.3 & 2.000 & 100 & 80.7 & 2.000 \\
\midrule
150 & 84.1 & 2.176 & 150 & 78.5 & 2.176 & 150 & 83.8 & 2.176 \\
\midrule
200 & 86.3 & 2.301 & 200 & 80.9 & 2.301 & 200 & 85.9 & 2.301 \\
\midrule
300 & 88.7 & 2.477 & 300 & 83.7 & 2.477 & 300 & 88.8 & 2.477 \\
\bottomrule
\end{tabular}
\label{Summary of amplitude and intensity level measurements}
\end{center}
\end{table}
The uncertainties were estimated using the least squares method. For this reason we have included the $R$ squared coefficient in the graphs. The uncertainties are basically associated with the amplitude of the wave displayed by the oscilloscope. We estimate its value between 3 and 5\% depending on the scale of the instrument (considering half of the smallest division of the oscilloscope). The uncertainty associated with the sound level meter is lower than that of the oscilloscope (except for very small amplitudes), and is in the range $0.1 - 0.2 dB$, approximately 0.3\% (on the order of 10 times lower than the uncertainty associated with the amplitude). Therefore, it is feasible to neglect this last value.\\
It was not possible to extend the measurements beyond the range $10 - 300 mV$. The reason is that for values less than $10 mV$ the uncertainty in the oscilloscope reading increases due to the widening of the wave (see Fig. 5, left), while for values greater than $300 mV$ the wave deforms significantly (see Fig. 5, right).
\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{Fig5}
\caption{ Distortion in the waveform displayed by the oscilloscope.}
\end{figure}
\section{Comments and conclusions}
Firstly, it seems remarkable to us that with accessible materials and with a simple experimental setup, it is possible to establish the relationship between intensity and amplitude, a relationship that is generally only addressed theoretically in wave courses. In particular, the excellent agreement between the expected and obtained slopes is gratifying. Another aspect that should be highlighted is that, together with the graphic representation of the results, the experiment allows us to visualise the waves and their amplitude variations in the oscilloscope, and in parallel it makes it possible to measure the intensity level and perceive it with the ear, that is, it makes it possible to appreciate the concepts intuitively and concretely.\\
In relation to the uncertainties associated with the experiment, we raise three aspects. The first is that the amplitude is visualised at a guess on the oscilloscope screen, which introduces uncertainty in the value of the amplitude. The second aspect is that the intensity level measurement with the sound level meter is affected by background noise that cannot be controlled or eliminated, which generates fluctuations in the intensity level. Finally, with respect to Fig. 4, it is important to note that the difference between the intercept of the first graph (40.7) with respect to the other two (31.8 and 40), is due to the point of cut with the vertical axis, which is determined by the sensitivity of the speaker and cannot be controlled. In relation to this issue, it is important to note that to carry out the experiment we used a low-cost speaker that was extracted from a broken radio. Taking into account all the antecedents, the estimated fluctuations for the slope are of the order of 5\%, and since its value is 20, the estimated variation is of the order of unity.\\
It is important to bear in mind that once the experiment has been set up, the measurements and their respective graphic representation require a time of $\sim 15 min$, meaning it is perfectly feasible to carry out this experimental activity in a demonstrative way in a classroom (with silence from the students).\\
A key aspect that we believe is important for the teacher to discuss with their students is that although a sound wave is generated by pressure variations in the air, and therefore the amplitude of a wave of this type should be associated with said variations, in the experiment, we use an amplitude scale expressed in $mV$. As we know, the reason is that the potential difference applied to the speaker was measured, and therefore the corresponding amplitude is expressed in $mV$. However, said voltage is related to the current in the speaker, which in turn is related to the amplitude of the movement generated in the speaker and the pressure differences of the sound wave.\\
In summary, considering the simplicity of the presented experiment and the accuracy of the results obtained, we hope that this work constitutes a contribution to the work carried out by teachers who teach wave courses, both at school and university level.
\section*{Acknowledgments}
I would like to thank to Daniela Balieiro for their valuable comments in the writing of this paper.
\section*{References}
[1] R.A. Serway, J.W. Jewett, Physics for Scientists and Engineers with Modern Physics, 8 ed., (Thomson, Cengage Learning), 2010, p. 480.
\vspace{2mm}
[2] P.A. Tipler, Physics for Scientists and Engineers, 5 ed., (W. H. Freeman and Company, New York, 2004), p. 476.
\end{document}
|
{
"timestamp": "2021-03-23T01:35:35",
"yymm": "2103",
"arxiv_id": "2103.11822",
"language": "en",
"url": "https://arxiv.org/abs/2103.11822"
}
|
\section{Introduction and preliminaries}
Let $\mathbb Z$, $\mathbb N$, $\mathbb N_{0}$ and $\mathbb C$ denote the set of integers, positive integers, nonnegative integers and complex numbers, respectively. The well-known polylogarithm function is defined as
$$
Li_{p}(x):=\sum_{n=1}^\infty \frac{x^n}{n^p}\quad (\lvert x \lvert \leq 1,\quad p \in \mathbb N_{0})\,.
$$
Note that when $p=1$, $-Li_{1}(x)$ is the logarithm function $\log(1-x)$. Furthermore, $Li_{n}(1)=\zeta(n)$, where $\zeta(s)$ denotes the Riemann zeta function which is defined as $\zeta(s):=\sum_{n=1}^\infty n^{-s}$. The famous generalized harmonic numbers of order $m$ is defined by the partial sum of the Riemann Zeta function $\zeta(m)$ as:
$$
H_n^{(m)}:=\sum_{j=1}^n \frac{1}{j^m} \quad (n, m \in \mathbb N)\,.
$$
Before going further, we introduce some notations. Let $p \in \mathbb N$ and $m \in \mathbb N_{0}$, define
\begin{align*}
&J_{0}(m,p):=\int_{0}^{1}x^{m} Li_{p}(x)\mathrm{d}x\,.
\end{align*}
Let $p, m \in \mathbb N_{0}$ and $0 \leq x \leq 1$, define
\begin{align*}
&J_{0}(m,p,x):=\int_{0}^{x}t^{m} Li_{p}(t)\mathrm{d}t\,,
\quad J_{1}(m,p,x):=\int_{0}^{x}\log^{m}(t) Li_{p}(t)\mathrm{d}t\,.
\end{align*}
Let $p, q \in \mathbb N$ and $m \in \mathbb Z$ with $m \geq -2$, define
\begin{align*}
&J(m,p,q):=\int_{0}^{1}x^{m} Li_{p}(x)Li_{q}(x)\mathrm{d}x\,.
\end{align*}
Let $p, q \in \mathbb N_{0}$ with $p+q \geq 1$, $r \in \mathbb N$, define
\begin{align*}
&K(r,p,q):=\int_{0}^{1} \frac{\log^{r}(x) Li_{p}(x)Li_{q}(x)}{x}\mathrm{d}x\,.
\end{align*}
Freitas \cite{Freitas} showed that integrals $J_{0}(m,p)$, $J(m,p,q)$ and $K(r,p,q)$ satisfy the following recurrence relations:
\begin{align*}
&J_{0}(m,q)=\frac{\zeta(q)}{m+1}-\frac{1}{m+1}J_{0}(m,q-1)\quad (q \geq 2, m \geq 0)\,,\\
&J(m,p,q)=\frac{\zeta(p)\zeta(q)}{m+1}-\frac{1}{m+1}\bigg(J(m,p-1,q)+J(m,p,q-1)\bigg)\\
&(p, q\geq 2, m \in \mathbb N_{0}\cup\{-2\})\,,\\
&K(r,p,q)=-\frac{1}{r+1}\bigg(K(r+1,p-1,q)+K(r+1,p,q-1)\bigg)\quad (p, q, r \in \mathbb N)\,.
\end{align*}
From this Freitas proved that integrals $K(r,p,q)$ with $p+q+r$ even and $J(m,p,q)$ could be reduced to zeta values. Note that the proof was constructive, Freitas didn't give explicit evaluations for these integrals. On the contrary, Freitas \cite{Freitas} gave explicit evaluations for $J(-1,p,q)$ and $K(r,0,q)$ with $r+q$ even.
An anonymous reviewer told the author that Sofo \cite{Sofo2016,Sofo2018,Sofo2020}and Xu \cite{xu1,xu2,xu3} had made many progresses in the area of integrals involving polylogarithm functions, which the author was initially unaware of. It is known to all that polylogarithmic functions are intrinsically connected with sums of harmonic numbers. For instance, Sofo \cite{Sofo2016} developed closed form representations for infinite series containing generalized harmonic numbers of type
$$
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n^{(3)}}{n^{p}\binom{n+k}{k}}\quad (p=0, 1)\,.
$$
The author \cite{Lirusensen} gave closed form representations for generalized hyperharmonic number sums with reciprocal binomial coefficients, which greatly extend Sofo's result. Sofo \cite{Sofo2016} also obtained explicit evaluations for some integrals involving polylogarithm functions. Motivated by the work of Freitas \cite{Freitas}, Sofo \cite{Sofo2018} investigated the representations of integrals of polylogarithms with negative argument of the type
$$
\int_{0}^{1}x^{m} Li_{p}(-x)Li_{q}(-x)\mathrm{d}x\,
$$
for $m \ge -2$, and for integers $p$ and $q$. For $m=-2, -1, 0$, Sofo also gave explicit
representations of the integral in terms of Euler sums and for $m \ge 0$, Sofo obtained a
recurrence relation for the integral. As a more general consideration, Sofo \cite{Sofo2020} considered integrals of polylogarithms with alternating argument of the type
$$
\int_{0}^{1}x^{m} Li_{p}(x)Li_{q}(-x)\mathrm{d}x\,
$$
for integers $p$ and $q$. Similarly, for $m=-2, -1, 0$, Sofo gave explicit representations of the integral in terms of Euler sums. Some more integrals involving polylogarithms were obtained.
Xu \cite{xu1} showed that quadratic Euler sums of the form
\begin{align*}
\sum_{n=1}^\infty \frac{H_n H_n^{(m)}}{n^{p}}\quad(m+p\leq 8)\,,
\end{align*}
and some integrals of polylogarithm functions of the form
\begin{align*}
\int_{0}^{1} \frac{Li_{r}(x) Li_{p}(x) Li_{q}(x)}{x}\mathrm{d}x\quad(r+p+q\leq 8)\,
\end{align*}
can be written in terms of Riemann zeta values. It is interesting that integrals of polylogarithm functions can be related to multiple zeta (star) values. By using integrals of polylogarithm functions, Xu \cite{xu2} gave explicit expressions for some restricted multiple zeta (star) values. Some of lemmas used by Xu \cite{xu2} were also re-discovered by the author in different forms. Furthermore, by using the iterated integral representation of multiple polylogarithm functions, Xu \cite{xu3} proved some conjectures proposed by J. M. Borwein, D. M. Bradley and D. J. Broadhurst \cite{Borwein}. Xu also obtained numerous formulas for alternating multiple zeta values.
In this paper, we mainly give explicit expressions for integrals of types $J_{0}(m,p,x)$, $J_{1}(m,p,x)$, $J(m,p,q)$ and $K(r,p,q)$. In addition, some more explicit formulas for integrals involving the logarithm function of types
\begin{align*}
\int_{0}^{x}\frac{\log^{m}(1-t)}{t^{n}}\mathrm{d}t\,,\quad \int_{0}^{x}\frac{\log^{m}(1+t)}{t^{n}}\mathrm{d}t\,,\quad
\int_{0}^{x}\frac{\log^{m}(t)}{(1-t)^{n}}\mathrm{d}t\,\quad (m, n \in \mathbb N, m\geq n)
\end{align*}
will also be derived.
\section{Integrals involving logarithm function}
De Doelder \cite{Doelder} used the integral $\int_{0}^{x}\frac{\log^{2}(1-t)}{t}\mathrm{d}t$ to evaluate infinite series of type $\sum_{n=1}^\infty \frac{H_{n}}{n^{2}}x^{n}$. As a natural consideration, The author \cite{LIRUSEN} gave explicit evaluations for infinite series involving generalized (alternating) harmonic numbers of types
$\sum_{n=1}^\infty \frac{H_{n}}{n^{3}}x^{n}$,
$\sum_{n=1}^\infty \frac{H_{n}}{n^{3}}(-x)^{n}$,
$\sum_{n=1}^\infty \frac{H_{n}^{(2)}}{n}x^{n}$,
$\sum_{n=1}^\infty \frac{H_{n}^{(2)}}{n}(-x)^{n}$,
$\sum_{n=1}^\infty \frac{H_{n}^{(2)}}{n^{2}}x^{n}$,
$\sum_{n=1}^\infty \frac{H_{n}^{(2)}}{n^{2}}(-x)^{n}$,
$\sum_{n=1}^\infty \frac{\overline{H}_{n}}{n}x^{n}$,
$\sum_{n=1}^\infty \frac{\overline{H}_{n}^{(2)}}{n}x^{n}$,
$\sum_{n=1}^\infty \frac{\overline{H}_{n}^{(2)}}{n}(-x)^{n}$ in terms of polylogarithm functions. However, it seems difficult to give explicit expressions for infinite series of types
$\sum_{n=1}^\infty \frac{H_{n}}{n^{4}}x^{n}$ and
$\sum_{n=1}^\infty \frac{H_{n}^{(2)}}{n^{3}}x^{n}$, since the integrals $\int_{0}^{x}\frac{\log^{2}(t)\log^{2}(1-t)}{t}\mathrm{d}t$ and $\int_{0}^{x}\frac{\log^{2}(t)Li_{2}(t)}{1-t}\mathrm{d}t$ are not known to be related to the polylogarithm functions, even with the help of a mathematical package. It is interesting to evaluate similar type integrals, e.g., $\int_{0}^{x}\frac{\log^{m}(1-t)}{t^{n}}\mathrm{d}t$. Before going further, We introduce some notations.
\begin{Definition}\label{def1}
For $m, n \in \mathbb N$ with $m \geq n$ and $0 \leq x \leq 1$, define the quantities $A(m,n,x)$, $B(m,n,x)$ and $C(m,n,x)$ as
\begin{align*}
&A(m,n,x):=\int_{0}^{x}\frac{\log^{m}(1-t)}{t^{n}}\mathrm{d}t\,,\quad
B(m,n,x):=\int_{0}^{x}\frac{\log^{m}(1+t)}{t^{n}}\mathrm{d}t\,,\\
&C(m,n,x):=\int_{0}^{x}\frac{\log^{m}(t)}{(1-t)^{n}}\mathrm{d}t\,.
\end{align*}
\end{Definition}
\begin{Lem}\label{lem9}
Let $m \in \mathbb N$ and $0 \leq x \leq 1$, then we have
\begin{align*}
&\quad A(m,1,x)\\
&=\log(x)\log^{m}(1-x)+\sum_{k=0}^{m-2}(-1)^{k}(m-k)_{k+1}\log^{m-k-1}(1-x)Li_{k+2}(1-x)\\
&\quad +(-1)^{m-1}m!Li_{m+1}(1-x)+(-1)^{m}m!\zeta(m+1)\,.
\end{align*}
In particular, we have $A(m,1,1)=(-1)^{m}m!\zeta(m+1)$.
\end{Lem}
\begin{proof}
From the definition of $A(m,1,x)$, by using integration by parts, we can write
\begin{align*}
&\quad A(m,1,x)\\
&=\log(x)\log^{m}(1-x)-m \int_{1}^{1-x}\frac{\log(1-t)\log^{m-1}(t)}{t}\mathrm{d}t\\
&=\log(x)\log^{m}(1-x)+m \log^{m-1}(1-x)Li_{2}(1-x)\\
&\quad -m(m-1)\int_{1}^{1-x}\frac{Li_{2}(t)\log^{m-2}(t)}{t}\mathrm{d}t\\
&=\log(x)\log^{m}(1-x)+\sum_{k=0}^{m-2}(-1)^{k}(m-k)_{k+1}\log^{m-k-1}(1-x)Li_{k+2}(1-x)\\
&\quad +(-1)^{m-1}m!Li_{m+1}(1-x)+(-1)^{m}m!Li_{m+1}(1)\,.
\end{align*}
\end{proof}
\begin{Lem}[\cite{xu2}]\label{lem10}
Let $m \in \mathbb N$ and $x \geq 0$, then we have
\begin{align*}
B(m,1,x)
&=\log(x)\log^{m}(1+x)-\frac{m}{m+1}\log^{m+1}(1+x)+m!\zeta(m+1)\\
&\quad -\sum_{i=1}^{m}\binom{m}{i}i!\log^{m-i}(1+x)Li_{i+1}(\frac{1}{1+x})\,.
\end{align*}
In particular, we have
$$
B(m,1,1)
=-\frac{m}{m+1}\log^{m+1}(2)+m!\zeta(m+1)-\sum_{i=1}^{m}\binom{m}{i}i!\log^{m-i}(2)Li_{i+1}(\frac{1}{2})\,.
$$
\end{Lem}
\begin{proof}
From the definition of $B(m,1,x)$, by using integration by parts, we can write
\begin{align*}
&\quad B(m,1,x)\\
&=\log(x)\log^{m}(1+x)-m \int_{0}^{x}\frac{\log(t)\log^{m-1}(1+t)}{1+t}\mathrm{d}t\\
&=\log(x)\log^{m}(1+x)-m \int_{0}^{x}\log^{m-1}(1+t)\bigg(\frac{\mathrm{d}Li_{2}(\frac{1}{1+t})}{\mathrm{d}t}
+\frac{\log(1+t)}{1+t}\bigg)\mathrm{d}t\\
&=\log(x)\log^{m}(1+x)-\frac{m}{m+1}\log^{m+1}(1+x)-m\log^{m-1}(1+x)Li_{2}(\frac{1}{1+x})\\
&\quad +m(m-1) \int_{0}^{x}\frac{Li_{2}(\frac{1}{1+t})\log^{m-2}(1+t)}{1+t}\mathrm{d}t\\
&=\log(x)\log^{m}(1+x)-\frac{m}{m+1}\log^{m+1}(1+x)+m!Li_{m+1}(1)\\
&\quad -\sum_{i=1}^{m}\binom{m}{i}i!\log^{m-i}(1+x)Li_{i+1}(\frac{1}{1+x})\,.
\end{align*}
\end{proof}
\begin{Prop}\label{prop}
Let $m \in \mathbb N$ and $0 \leq x \leq 1$, then we have
\begin{align*}
&\quad C(m,1,x)\\
&=-\log(1-x)\log^{m}(x)+m\sum_{i=2}^{m+1}(-1)^{i-1}\binom{m-1}{i-2}(i-2)!\log^{m+1-i}(x)Li_{i}(x)\,.
\end{align*}
In particular, we have $C(m,1,1)=(-1)^{m}m!\zeta(m+1)$.
\end{Prop}
\begin{proof}
From the definition of $C(m,1,x)$, by using integration by parts, we can write
\begin{align*}
&\quad C(m,1,x)\\
&=-\log(1-x)\log^{m}(x)+m \int_{0}^{x}\frac{\log(1-t)\log^{m-1}(t)}{t}\mathrm{d}t\\
&=-\log(1-x)\log^{m}(x)-m \log^{m-1}(x)Li_{2}(x)+m(m-1)\int_{0}^{x}\frac{Li_{2}(t)\log^{m-2}(t)}{t}\mathrm{d}t\\
&=-\log(1-x)\log^{m}(x)+m\sum_{i=2}^{m+1}(-1)^{i-1}\binom{m-1}{i-2}(i-2)!\log^{m+1-i}(x)Li_{i}(x)\,.
\end{align*}
\end{proof}
Now we develop explicit expressions for $A(m,n,x)$, $B(m,n,x)$ and $C(m,n,x)$.
\begin{theorem}\label{maintheorem5}
Let $m, n \in \mathbb N$ with $m \geq n \geq 2$ and $0 < x \leq 1$, then we have
\begin{align*}
&\quad A(m,n,x)\\
&=\sum_{y=0}^{n-2}\binom{m}{y}y!(-1)^{y+1}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\cdot\frac{\log^{m-y}(1-x)}{x^{i_{y}-1}}\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y}y!(-1)^{y}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\log^{m-y}(1-x)\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y+1}(y+1)!(-1)^{y+1}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\\
&\qquad \times A(m-y-1,1,x)\,,
\end{align*}
and
\begin{align*}
&\quad C(m,n,x)\\
&=\frac{(-1)^{m}m!}{n-1}\sum_{y=0}^{n-2}\zeta(m-y)
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
+\sum_{y=0}^{n-2}\binom{m}{y}y!(-1)^{y}\\
&\quad \times \sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\bigg(\frac{\log^{m-y}(x)}{(1-x)^{i_{y}-1}}-\log^{m-y}(x)\bigg)\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y+1}(y+1)!(-1)^{y}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\\
&\qquad \times A(m-y-1,1,1-x)\,,
\end{align*}
where $A(m-y-1,1,x)$ and $A(m-y-1,1,1-x)$ are given in Lemma \ref{lem9}. In particular, we have
\begin{align*}
A(m,n,1)=C(m,n,1)=\frac{(-1)^{m}m!}{n-1}\sum_{y=0}^{n-2}\zeta(m-y)
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\,.
\end{align*}
\end{theorem}
\begin{proof}
From the definition of $A(m,n,x)$, when $n \geq 2$, by using integration by parts, we can write
\begin{align*}
&\quad A(m,n,x)\\
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1-x)}{x^{n-1}}
-\frac{m}{n-1}\int_{0}^{x}\frac{\log^{m-1}(1-t)}{t^{n-1}(1-t)}\mathrm{d}t\\
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1-x)}{x^{n-1}}-\frac{m}{n-1}\sum_{i=1}^{n-1}\int_{0}^{x}\frac{\log^{m-1}(1-t)}{t^{i}}\mathrm{d}t\\
&\quad -\frac{m}{n-1}\int_{0}^{x}\frac{\log^{m-1}(1-t)}{1-t}\mathrm{d}t\\
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1-x)}{x^{n-1}}+\frac{1}{n-1}\log^{m}(1-x)
-\frac{m}{n-1}\sum_{i=1}^{n-1}A(m-1,i,x)\,,
\end{align*}
successive application of the above relation $n-2$ times, we can obtain that
\begin{align*}
&\quad A(m,n,x)\\
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1-x)}{x^{n-1}}+\frac{1}{n-1}\log^{m}(1-x)-\frac{m}{n-1}A(m-1,1,x)\\
&\quad +\frac{m}{n-1}\sum_{i_{1}=2}^{n-1}\frac{1}{i_{1}-1}\cdot\frac{\log^{m-1}(1-x)}{x^{i_{1}-1}}
-\frac{m}{n-1}\sum_{i_{1}=2}^{n-1}\frac{1}{i_{1}-1}\log^{m-1}(1-x)\\
&\quad +\frac{m}{n-1}\sum_{i_{1}=2}^{n-1}\frac{m}{i_{1}-1}A(m-2,1,x)
+\frac{m}{n-1}\sum_{i_{1}=2}^{n-1}\frac{m}{i_{1}-1}\sum_{i_{2}=2}^{i_{1}-1}A(m-2,i_{2},x)\\
&=\sum_{y=0}^{n-2}\binom{m}{y}y!(-1)^{y+1}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\cdot\frac{\log^{m-y}(1-x)}{x^{i_{y}-1}}\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y}y!(-1)^{y}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\log^{m-y}(1-x)\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y+1}(y+1)!(-1)^{y+1}\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\\
&\qquad \times A(m-y-1,1,x)\,.
\end{align*}
Note that $C(m,n,x)=A(m,n,1)-A(m,n,1-x)$, thus we get the desired result.
\end{proof}
\begin{theorem}\label{maintheorem6}
Let $m, n \in \mathbb N$ with $m \geq n \geq 2$ and $x > 0$, then we have
\begin{align*}
&\quad B(m,n,x)\\
&=\sum_{y=0}^{n-2}\binom{m}{y}y!\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\cdot\frac{(-1)^{n+i_{y}+y+1}\log^{m-y}(1+x)}{x^{i_{y}-1}}\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y}y!\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}
\cdot(-1)^{n+y+1}\log^{m-y}(1+x)\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y+1}(y+1)!\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\\
&\qquad \times (-1)^{n+y}B(m-y-1,1,x)\,,
\end{align*}
where $B(m-y-1,1,x)$ are given in Lemma \ref{lem10}. In particular, we have
\begin{align*}
B(m,n,1)
&=\sum_{y=0}^{n-2}\binom{m}{y}y!\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\\
&\qquad \times\log^{m-y}(2)(-1)^{n+y+1}((-1)^{i_{y}}+1)\\
&\quad +\sum_{y=0}^{n-2}\binom{m}{y+1}(y+1)!\sum_{i_{0}=n}^{n}\frac{1}{i_{0}-1}
\sum_{i_{1}=2}^{i_{0}-1}\frac{1}{i_{1}-1}\cdots\sum_{i_{y}=2}^{i_{y-1}-1}\frac{1}{i_{y}-1}\\
&\qquad \times (-1)^{n+y}\bigg\{-\frac{m-y-1}{m-y}\log^{m-y}(2)+(m-y-1)!\zeta(m-y)\\
&\qquad -\sum_{i=1}^{m-y-1}\binom{m-y-1}{i}i!\log^{m-y-1-i}(2)Li_{i+1}(\frac{1}{2})\bigg\}\,.
\end{align*}
\end{theorem}
\begin{proof}
From the definition of $B(m,n,x)$, when $n \geq 2$, by using integration by parts, we can write
\begin{align*}
B(m,n,x)
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1+x)}{x^{n-1}}
+\frac{m}{n-1}\int_{0}^{x}\frac{\log^{m-1}(1+t)}{t^{n-1}(1+t)}\mathrm{d}t\\
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1+x)}{x^{n-1}}
+\frac{m}{n-1}\sum_{i=1}^{n-1}(-1)^{n-1-i}\int_{0}^{x}\frac{\log^{m-1}(1+t)}{t^{i}}\mathrm{d}t\\
&\quad +\frac{m}{n-1}\int_{0}^{x}\frac{(-1)^{n-1}\log^{m-1}(1+t)}{1+t}\mathrm{d}t\\
&=-\frac{1}{n-1}\cdot\frac{\log^{m}(1+x)}{x^{n-1}}+\frac{(-1)^{n-1}}{n-1}\log^{m}(1+x)\\
&\quad +\frac{m}{n-1}\sum_{i=1}^{n-1}(-1)^{n-1-i}B(m-1,i,x)\,,
\end{align*}
similar with Theorem \ref{maintheorem5}, successive application of the above relation $n-2$ times, we get the desired result.
\end{proof}
\section{Integrals involving polylogarithms}
In this section, we develop explicit expressions for integrals of types $J_{0}(m,p,x)$, $J_{1}(m,p,x)$, $J(m,p,q)$ and $K(r,p,q)$. Before going further, we introduce some notations and lemmata. Following Flajolet-Salvy's paper \cite{Flajolet}, we write the classical linear Euler sums as
$$
S_{p,q}^{+,+}:=\sum_{n=1}^\infty \frac{H_n^{(p)}}{{n}^{q}}\,.
$$
\begin{Lem}[\cite{xu2,Lirusensen}]\label{lem1}
Let $n, m \in \mathbb N_{0}$ and $x \geq 0$, defining
\begin{align*}
L(n, m, x):=\int_{0}^{x}y^{n}\log^{m}(y) \mathrm{d}y\,,
\end{align*}
then we have
\begin{align*}
L(n, m, x)=\frac{x^{n+1}}{n+1}\sum_{j=0}^{m}\frac{(m+1-j)_{j}}{(n+1)^j}(-1)^{j}\log^{m-j}(x)\,,
\end{align*}
where $(t)_{n}=t(t+1)\cdots(t+n-1)$ is the Pochhammer symbol. In particular, we have $L(n, m, 1)=\frac{m!(-1)^m}{(n+1)^{m+1}}$.
\end{Lem}
\begin{Lem}[\cite{Lirusensen}]\label{lem2}
Let $n, m \in \mathbb N_{0}$ and $0 \leq x \leq 1$, defining
\begin{align*}
M(n, m, x):=\int_{x}^{1}y^{n}\log^{m} (1-y) \mathrm{d}y\,,
\end{align*}
then we have
\begin{align*}
M(n, m, x)=\sum_{j=0}^{n}\binom{n}{j}(-1)^{j}\frac{(1-x)^{j+1}}{j+1}
\sum_{i=0}^{m}\frac{(m+1-i)_{i}}{(j+1)^i}(-1)^{i}\log^{m-i} (1-x)\,.
\end{align*}
In particular, we have $M(n, m, 0)=(-1)^{m}m!\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^{j}}{(j+1)^{m+1}}$.
\end{Lem}
\begin{Lem}\label{lem3}
Let $n, m \in \mathbb N_{0}$ and $0 \leq x \leq 1$, then we can obtain that
\begin{align*}
&\int_{0}^{x}y^{n}\log^{m} (1-y) \mathrm{d}y\\
&=(-1)^{m}m!\sum_{j=0}^{n}\binom{n}{j}\frac{(-1)^{j}}{(j+1)^{m+1}}\\
&\quad-\sum_{j=0}^{n}\binom{n}{j}(-1)^{j}\frac{(1-x)^{j+1}}{j+1}
\sum_{i=0}^{m}\frac{(m+1-i)_{i}}{(j+1)^i}(-1)^{i}\log^{m-i} (1-x)\,.
\end{align*}
\begin{proof}
Note that
$$
\int_{0}^{x}y^{n}\log^{m} (1-y) \mathrm{d}y=M(n, m, 0)-M(n, m, x)\,,
$$
with the help of Lemma \ref{lem2}, we get the desired result.
\end{proof}
\end{Lem}
\begin{theorem}\label{maintheorem1}
Let $p \in \mathbb N$ and $m \in \mathbb N_{0}$, then we have
\begin{align*}
&J_{0}(m,p,x)\\
&=\sum_{j=2}^{p}\frac{(-1)^{p-j}}{(m+1)^{p+1-j}}x^{m+1}Li_{j}(x)
+\frac{(-1)^{p-1}}{(m+1)^{p-1}}\bigg(\sum_{j=0}^{m}\binom{m}{j}\frac{(-1)^{j}}{(j+1)^{2}}\\
&\quad+\sum_{j=0}^{m}\binom{m}{j}(-1)^{j}\frac{(1-x)^{j+1}}{j+1}
\sum_{i=0}^{1}\frac{(2-i)_{i}}{(j+1)^i}(-1)^{i}\log^{1-i} (1-x)\bigg)\,.
\end{align*}
In particular, $J_{0}(m,p)$ can be reduced to zeta values and harmonic numbers:
$$
J_{0}(m,p)=J_{0}(m,p,1)
=\sum_{j=2}^{p}\frac{(-1)^{p-j}}{(m+1)^{p+1-j}}\zeta(j)+\frac{(-1)^{p-1}}{(m+1)^{p}}H_{m+1}\,.
$$
Note that we have used the fact \cite{Lirusensen}
$$
H_{m+1}=(m+1)\sum_{j=0}^{m}\binom{m}{j}\frac{(-1)^{j}}{(j+1)^{2}}\,.
$$
\end{theorem}
\begin{proof}
By using integration by parts, we have
\begin{align*}
&\quad J_{0}(m,p,x)\\
&=\frac{1}{m+1}x^{m+1} Li_{p}(x)-\frac{1}{m+1}\int_{0}^{x}t^{m} Li_{p-1}(t)\mathrm{d}t\\
&=\frac{1}{m+1}x^{m+1} Li_{p}(x)-\frac{1}{m+1}J_{0}(m,p-1,x)\\
&=\frac{1}{m+1}x^{m+1} Li_{p}(x)-\frac{1}{(m+1)^{2}}x^{m+1} Li_{p-1}(x)+\frac{1}{(m+1)^{2}}J_{0}(m,p-2,x)\\
&=\sum_{j=2}^{p}\frac{(-1)^{p-j}}{(m+1)^{p+1-j}}x^{m+1}Li_{j}(x)+\frac{(-1)^{p-1}}{(m+1)^{p-1}}J_{0}(m,1,x)\,.
\end{align*}
Note that
$$
J_{0}(m,1,x)=-\int_{0}^{x}t^{m} \log(1-t)\mathrm{d}t\,,
$$
with the help of Lemma \ref{lem3}, we get the desired result.
\end{proof}
\begin{Lem}\label{lem4}
Let $m \in \mathbb N_{0}$, then we have
\begin{align*}
J_{1}(m,0,x)
&=x\sum_{j=0}^{m}(m+1-j)_{j}(-1)^{j+1}\log^{m-j}(x)-\log(1-x)\log^{m}(x)\\
&\quad+m\sum_{j=2}^{m+1}(-1)^{j-1}\binom{m-1}{j-2}(j-2)!Li_{j}(x)\log^{m+1-j}(x)\,.
\end{align*}
\begin{proof}
Note that
\begin{align*}
J_{1}(m,0,x)
&=-\int_{0}^{x}\log^{m}(t)\mathrm{d}t+\int_{0}^{x}\frac{\log^{m}(t)}{1-t}\mathrm{d}t\,.
\end{align*}
with the help of Lemma \ref{lem1} and Proposition \ref{prop}, we get the desired result.
\end{proof}
\end{Lem}
\begin{theorem}\label{maintheorem2}
Let $p \in \mathbb N$ and $m \in \mathbb N_{0}$, then we have
\begin{align*}
&J_{1}(m,p,x)\\
&=\sum_{y=1}^{p}\sum_{i_{1}=0}^{m-1}\cdots\sum_{i_{y}=0}^{m-i_{1}-\cdots-i_{y-1}-1}
m(m-1)\cdots(m-i_{1}-\cdots-i_{y}+1)\\
&\qquad \times(-1)^{i_{1}+\cdots+i_{y}+y-1} x Li_{p-y+1}(x)\log^{m-i_{1}-\cdots-i_{y}}(x)\\
&\quad +\sum_{y=1}^{p}\sum_{i_{1}=0}^{m-1}\cdots\sum_{i_{y-1}=0}^{m-i_{1}-\cdots-i_{y-2}-1}
m!(-1)^{i_{1}+\cdots+i_{y-1}+y-1}J_{1}(0,p-y+1,x)\\
&\quad +\sum_{i_{1}=0}^{m-1}\cdots\sum_{i_{p}=0}^{m-i_{1}-\cdots-i_{p-1}-1}
m(m-1)\cdots(m-i_{1}-\cdots-i_{p}+1)\\
&\qquad \times(-1)^{i_{1}+\cdots+i_{p}+p}J_{1}(m-i_{1}-\cdots-i_{p},0,x)\,,
\end{align*}
where $J_{1}(0,p-y+1,x)=J_{0}(0,p-y+1,x)$ are given Theorem \ref{maintheorem1} and $J_{1}(m-i_{1}-\cdots-i_{p},0,x)$ are given in Lemma \ref{lem4}.
\end{theorem}
\begin{proof}
By using integration by parts, we can obtain that
\begin{align*}
&\quad J_{1}(m,p,x)\\
&=x Li_{p}(x)\log^{m}(x)-\int_{0}^{x}\log^{m}(t) Li_{p-1}(t)\mathrm{d}t-m \int_{0}^{x}\log^{m-1}(t) Li_{p}(t)\mathrm{d}t\\
&=x Li_{p}(x)\log^{m}(x)-J_{1}(m,p-1,x)-m J_{1}(m-1,p,x)\\
&=x Li_{p}(x)\log^{m}(x)-m x Li_{p}(x)\log^{m-1}(x)-J_{1}(m,p-1,x)\\
&\quad +m J_{1}(m-1,p-1,x)+m(m-1) J_{1}(m-2,p,x)\\
&=\sum_{i=0}^{m-1}\binom{m}{i}i!(-1)^{i} x Li_{p}(x)\log^{m-i}(x)+(-1)^{m}m!J_{1}(0,p,x)\\
&\quad +\sum_{i=0}^{m-1}\binom{m}{i}i!(-1)^{i+1} J_{1}(m-i,p-1,x)\\
&=\sum_{i_{1}=0}^{m-1}\binom{m}{i_{1}}i_{1}!(-1)^{i_{1}} x Li_{p}(x)\log^{m-i_{1}}(x)+(-1)^{m}m!J_{1}(0,p,x)\\
&\quad +\sum_{i_{1}=0}^{m-1}\binom{m}{i_{1}}i_{1}!(-1)^{i_{1}+1}\bigg\{(-1)^{m-i_{1}}(m-i_{1})!J_{1}(0,p-1,x)\\
&\quad +\sum_{i_{2}=0}^{m-i_{1}-1}
\binom{m-i_{1}}{i_{2}}i_{2}!(-1)^{i_{2}} x Li_{p-1}(x)\log^{m-i_{1}-i_{2}}(x)\\
&\quad +\sum_{i_{2}=0}^{m-i_{1}-1}
\binom{m-i_{1}}{i_{2}}i_{2}!(-1)^{i_{2}+1} J_{1}(m-i_{1}-i_{2},p-2,x)\bigg\}\\
&=\sum_{y=1}^{p}\sum_{i_{1}=0}^{m-1}\cdots\sum_{i_{y}=0}^{m-i_{1}-\cdots-i_{y-1}-1}
m(m-1)\cdots(m-i_{1}-\cdots-i_{y}+1)\\
&\qquad \times(-1)^{i_{1}+\cdots+i_{y}+y-1} x Li_{p-y+1}(x)\log^{m-i_{1}-\cdots-i_{y}}(x)\\
&\quad +\sum_{y=1}^{p}\sum_{i_{1}=0}^{m-1}\cdots\sum_{i_{y-1}=0}^{m-i_{1}-\cdots-i_{y-2}-1}
m!(-1)^{i_{1}+\cdots+i_{y-1}+y-1}J_{1}(0,p-y+1,x)\\
&\quad +\sum_{i_{1}=0}^{m-1}\cdots\sum_{i_{p}=0}^{m-i_{1}-\cdots-i_{p-1}-1}
m(m-1)\cdots(m-i_{1}-\cdots-i_{p}+1)\\
&\qquad \times(-1)^{i_{1}+\cdots+i_{p}+p}J_{1}(m-i_{1}-\cdots-i_{p},0,x)\,.
\end{align*}
\end{proof}
\begin{Lem}[{{Abel's lemma on summation by parts} \cite{Abel,Chu}}]\label{lem5}
Let $\{f_k\}$ and $\{g_k\}$ be two sequences, and define the forward difference and backward difference, respectively, as
$$
\Delta\tau_k=\tau_{k+1}-\tau_k\quad\hbox{and}\quad \nabla\tau_k=\tau_k-\tau_{k-1}\,,
$$
then, there holds the relation:
\begin{align*}
\sum_{k=1}^\infty f_k\nabla g_k=\lim_{n\to\infty}f_n g_n-f_1 g_0-\sum_{k=1}^\infty g_k \Delta f_k \,.
\end{align*}
\end{Lem}
We now provide a criterion concerning the exchange of summation and integral for improper integrals.
\begin{Lem}\label{exchange}
Given a series of functions $\sum_{n=1}^\infty u_{n}(x), a\leq x \leq b$ with $u_{n}(x)\geq 0$, $\sum_{n=1}^\infty u_{n}(b)=\infty$ and $\sum_{n=1}^\infty u_{n}(x)$ converges for $a \leq x < b$. Suppose $u_{n}(x)$ is integrable (Riemann integrable or integrable as an improper integral) on $[a, b]$, the improper integral $\int_{a}^{b}\sum_{n=1}^\infty u_{n}(x)\mathrm{d}x$ converges, and $\sum_{n=1}^\infty u_{n}(x)$ internally closed uniform converges on $[a, b)$, i.e. for any $a\leq c < d <b$, $\sum_{n=1}^\infty u_{n}(x)$ converges uniformly on $[c, d]$, then we can exchange summation and integral, i.e.
$$
\int_{a}^{b}\sum_{n=1}^\infty u_{n}(x)\mathrm{d}x=\sum_{n=1}^\infty \int_{a}^{b} u_{n}(x)\mathrm{d}x\,.
$$
\end{Lem}
\begin{proof}
Since the improper integral $\int_{a}^{b}\sum_{n=1}^\infty u_{n}(x)\mathrm{d}x$ converges, then for any $\epsilon >0$, there exists $\delta >0$, s.t. $\lvert \int_{b-\delta}^{b}\sum_{n=1}^\infty u_{n}(x)\mathrm{d}x \rvert\leq\frac{\epsilon}{3}$. It is not hard to see that we can choose $\delta >0$ such that $b-\delta-a>0$. Note that, $\sum_{n=1}^\infty u_{n}(x)$ converges uniformly on $[a, b-\delta]$, then for fixed $\epsilon, \delta >0$, there exists $N > 0$, for any $M \geq N$, we have
$$
\left\lvert \sum_{n=M+1}^\infty u_{n}(x) \right\rvert\leq\frac{\epsilon}{3(b-\delta-a)}\quad x\in[a, b-\delta]\,.
$$
Thus we can obtain that
\begin{align*}
&\quad \left\lvert \sum_{n=1}^{M} \int_{a}^{b}u_{n}(x)\mathrm{d}x-\int_{a}^{b} \sum_{n=1}^\infty u_{n}(x)\mathrm{d}x \right\rvert\\
&\leq\left\lvert \int_{a}^{b-\delta}\sum_{n=1}^{M} u_{n}(x)-\sum_{n=1}^{\infty} u_{n}(x)\mathrm{d}x\right\rvert+\left\lvert\int_{b-\delta}^{b} \sum_{n=1}^{M} u_{n}(x)\mathrm{d}x \right\rvert+\left\lvert\int_{b-\delta}^{b} \sum_{n=1}^{\infty} u_{n}(x)\mathrm{d}x \right\rvert\\
&\leq(b-\delta-a)\frac{\epsilon}{3(b-\delta-a)}+\frac{\epsilon}{3}+\frac{\epsilon}{3}\\
&=\epsilon\,,
\end{align*}
which completes the proof.
\end{proof}
\begin{Lem}\label{lem6}
Let $m \in \mathbb N_{0}$ and $p \in \mathbb N$, then we have
\begin{align*}
&\quad J(m,p,1)\\
&=\sum_{j=2}^{p}(-1)^{p-j}\zeta(j)\bigg(\sum_{i=2}^{p+1-j}\frac{-1}{(m+1)^{p+2-j-i}}\zeta(i)
+\sum_{i=1}^{p+1-j}\frac{1}{(m+1)^{p+2-j-i}}H_{m+1}^{(i)}\bigg)\\
&\quad +(-1)^{p-1}\bigg\{\sum_{i=2}^{p}\frac{-1}{(m+1)^{p-i+1}}\bigg((1+\frac{i}{2})\zeta(i+1)
-\frac{1}{2}\sum_{k=1}^{i-2}\zeta(k+1)\zeta(i-k)\\
&\qquad -\sum_{n=1}^{m+1}\frac{H_{n}}{n^{i}}\bigg)+\frac{1}{(m+1)^{p}}
\bigg(H_{m+1}^{2}+\sum_{b=0}^{m}\frac{H_{m+1}-H_{b}}{m+1-b}\bigg)\bigg\}\,.
\end{align*}
\end{Lem}
\begin{proof}
From the definition of $J(m,p,1)$, we can write
\begin{align*}
J(m,p,1)
&=\sum_{n=1}^\infty \frac{1}{n} \int_{0}^{1}x^{m+n} Li_{p}(x)\mathrm{d}x\\
&=\sum_{j=2}^{p}(-1)^{p-j}\zeta(j)\sum_{n=1}^\infty \frac{1}{n(m+n+1)^{p+1-j}}+\sum_{n=1}^\infty \frac{(-1)^{p-1}H_{m+n+1}}{n(m+n+1)^{p}}\,.
\end{align*}
For the first part, by using fraction expansion, we have
\begin{align*}
&\quad \sum_{n=1}^\infty \frac{1}{n(m+n+1)^{p+1-j}}\\
&=\sum_{n=1}^\infty \bigg(\sum_{i=2}^{p+1-j}\frac{-1}{(m+1)^{p+2-j-i}}\cdot\frac{1}{(n+m+1)^{i}}
+\frac{1}{(m+1)^{p-j}}\cdot\frac{1}{n(n+m+1)}\bigg)\\
&=\sum_{i=2}^{p+1-j}\frac{-1}{(m+1)^{p+2-j-i}}\bigg(\zeta(i)-H_{m+1}^{(i)}\bigg)
+\frac{1}{(m+1)^{p-j+1}}H_{m+1}\\
&=\sum_{i=2}^{p+1-j}\frac{-1}{(m+1)^{p+2-j-i}}\zeta(i)
+\sum_{i=1}^{p+1-j}\frac{1}{(m+1)^{p+2-j-i}}H_{m+1}^{(i)}\,.
\end{align*}
For the second part, by using fraction expansion, we have
\begin{align*}
&\quad \sum_{n=1}^\infty \frac{H_{m+n+1}}{n(m+n+1)^{p}}\\
&=\sum_{n=1}^\infty H_{m+n+1}\bigg(\sum_{i=2}^{p}\frac{-1}{(m+1)^{p-i+1}}\cdot\frac{1}{(n+m+1)^{i}}
+\frac{1}{(m+1)^{p-1}}\cdot\frac{1}{n(n+m+1)}\bigg)\\
&=\sum_{i=2}^{p}\frac{-1}{(m+1)^{p-i+1}}\bigg(\sum_{n=1}^\infty \frac{H_{n}}{n^{i}}-\sum_{n=1}^{m+1} \frac{H_{n}}{n^{i}}\bigg)+\frac{1}{(m+1)^{p-1}}\sum_{n=1}^\infty\frac{H_{m+n+1}}{n(n+m+1)}\,.
\end{align*}
Note that
\begin{align*}
S_{1,i}^{+,+}=\sum_{n=1}^\infty \frac{H_{n}}{n^{i}}=(1+\frac{i}{2})\zeta(i+1)
-\frac{1}{2}\sum_{k=1}^{i-2}\zeta(k+1)\zeta(i-k)\,.\quad \cite{Flajolet}
\end{align*}
Set
$$
f_{n}:=H_{n+m+1}\quad\hbox{and}\quad
g_{n}:=\frac{1}{n+1}+\cdots+\frac{1}{n+m+1}\,,
$$
by using Lemma \ref{lem5}, we have
\begin{align*}
&\quad -\sum_{n=1}^\infty \frac{(m+1) H_{n+m+1}}{n(n+m+1)}\\
&=\sum_{n=1}^\infty H_{n+m+1}\left(\bigg(\frac{1}{n+1}+\cdots+\frac{1}{n+m+1}\bigg)-\bigg(\frac{1}{n}+\cdots+\frac{1}{n+m}\bigg)\right)\\
&=-H_{m+2}\bigg(\frac{1}{1}+\cdots+\frac{1}{1+m}\bigg)-\sum_{n=1}^\infty \bigg(\frac{1}{n+1}+\cdots+\frac{1}{n+m+1}\bigg)\frac{1}{n+m+2}\\
&=-H_{m+1}^{2}-\sum_{n=1}^\infty \sum_{b=0}^{m}\frac{1}{n+b}\cdot\frac{1}{n+m+1}\\
&=-H_{m+1}^{2}-\sum_{b=0}^{m}\frac{1}{m+1-b}\bigg(H_{m+1}-H_{b}\bigg)\,.
\end{align*}
Combining the above results, we get the desired result.
\end{proof}
\begin{Remark}
In the proof of the above theorem, we exchange the order of summation and integration, i.e.
\begin{align*}
\int_{0}^{1}x^{m}\sum_{n=1}^\infty \frac{x^{n}}{n} Li_{p}(x)\mathrm{d}x=\sum_{n=1}^\infty \frac{1}{n} \int_{0}^{1}x^{m+n} Li_{p}(x)\mathrm{d}x\,.
\end{align*}
To verify this, we only need to note that for any $0<\delta<1$, $\frac{x^{m+n}}{n} Li_{p}(x)$ is monotonic increasing on the interval $[0, 1-\delta]$ and the series $\sum_{n=1}^\infty \frac{x^{m+n}}{n} Li_{p}(x)$ converges when $x=1-\delta$. With the help of Lemma \ref{exchange}, we get the desired result. The other cases can be checked in a similar manner.
\end{Remark}
Now we provide another formula for $J(m,p,1)$.
\begin{Lem}\label{lem7}
Let $m \in \mathbb N_{0}$ and $p \in \mathbb N$, then we have
\begin{align*}
J(m,p,1)
&=\sum_{i=2}^{p}\frac{(-1)^{p-i}}{(m+1)^{p-i+1}}\bigg((1+\frac{i}{2})\zeta(i+1)
-\frac{1}{2}\sum_{k=1}^{i-2}\zeta(k+1)\zeta(i-k)\\
&\quad +\sum_{k=2}^{i}(-1)^{i-k}\zeta(k)H_{m+1}^{(i-k+1)}
+\sum_{j=1}^{m+1}\frac{(-1)^{i-1}}{j^{i}}H_{j}\bigg)\\
&\quad +\frac{(-1)^{p-1}}{(m+1)^{p}}
\bigg(H_{m+1}^{2}+\sum_{b=0}^{m}\frac{H_{m+1}-H_{b}}{m+1-b}\bigg)\,.
\end{align*}
\end{Lem}
\begin{proof}
From the definition of $J(m,p,1)$, we can write
\begin{align*}
J(m,p,1)
&=-\sum_{n=1}^\infty \frac{1}{n^{p}} \int_{0}^{1}x^{m+n} \log(1-x)\mathrm{d}x\\
&=\sum_{n=1}^\infty \frac{1}{n^{p}} \frac{H_{m+n+1}}{m+n+1}\\
&=\sum_{n=1}^\infty H_{m+n+1}\bigg(\sum_{i=2}^{p}\frac{(-1)^{p-i}}{(m+1)^{p-i+1}}\cdot\frac{1}{n^{i}}
+\frac{(-1)^{p-1}}{(m+1)^{p-1}}\cdot\frac{1}{n(n+m+1)}\bigg)\,.
\end{align*}
For the first part, by using fraction expansion, we have
\begin{align*}
&\quad \sum_{n=1}^\infty \frac{H_{m+n+1}}{n^{i}}\\
&=\sum_{n=1}^\infty \frac{H_{n}}{n^{i}}
+\sum_{j=1}^{m+1}\sum_{n=1}^\infty \frac{1}{n^{i}(n+j)}\\
&=\sum_{n=1}^\infty \frac{H_{n}}{n^{i}}+\sum_{j=1}^{m+1}\sum_{n=1}^\infty \bigg(\sum_{k=2}^{i}\frac{(-1)^{i-k}}{j^{i-k+1}}\cdot\frac{1}{n^{k}}
+\frac{(-1)^{i-1}}{j^{i-1}}\cdot\frac{1}{n(n+j)}\bigg)\\
&=\sum_{n=1}^\infty \frac{H_{n}}{n^{i}}+\sum_{k=2}^{i}(-1)^{i-k}\zeta(k)H_{m+1}^{(i-k+1)}
+\sum_{j=1}^{m+1}\frac{(-1)^{i-1}}{j^{i}}H_{j}\,.
\end{align*}
For the second part, from the previous lemma we know that
\begin{align*}
&\quad \sum_{n=1}^\infty \frac{(m+1) H_{n+m+1}}{n(n+m+1)}
=H_{m+1}^{2}+\sum_{b=0}^{m}\frac{1}{m+1-b}\bigg(H_{m+1}-H_{b}\bigg)\,.
\end{align*}
Combining the above results, we get the desired result.
\end{proof}
When $p=1$, we have
$$
J(m,1,1)=\frac{1}{m+1}
\bigg(H_{m+1}^{2}+\sum_{b=0}^{m}\frac{H_{m+1}-H_{b}}{m+1-b}\bigg)\,.
$$
It is known that \cite{Devoto}
$$
J(m,1,1)=\frac{2}{m+1}\bigg(H_{m+1}^{(2)}+\sum_{k=1}^{m}\frac{H_{k}}{k+1}\bigg)\,,
$$
thus we have the following proposition:
\begin{Prop}
\begin{align*}
H_{m+1}^{(2)}
&=\frac{1}{2}\bigg(\sum_{j=0}^{m}\binom{m+1}{j+1}\frac{(-1)^{j}}{j+1}\bigg)^{2}
+\frac{1}{2}\sum_{b=0}^{m}\frac{1}{m+1-b}\bigg\{\sum_{j=0}^{m}\binom{m+1}{j+1}\frac{(-1)^{j}}{j+1}\\
&\quad -\sum_{j=0}^{b-1}\binom{b}{j+1}\frac{(-1)^{j}}{j+1}\bigg\}
-\sum_{k=1}^{m}\frac{1}{k+1}\sum_{j=0}^{k-1}\binom{k}{j+1}\frac{(-1)^{j}}{j+1}\,.
\end{align*}
\end{Prop}
Now we give explicit expression for $J(-2,p,1)$.
\begin{Lem}\label{lem8}
Let $p \in \mathbb N$, then we have
\begin{align*}
J(-2,p,1)
&=\zeta(p+1)+\zeta(2)+\sum_{j=2}^{p}\zeta(j)\bigg(1+\sum_{i=2}^{p+1-j}(-1)^{1+i}\zeta(i)\bigg)\\
&\quad +\sum_{i=2}^{p}(-1)^{1+i}\bigg((1+\frac{i}{2})\zeta(i+1)
-\frac{1}{2}\sum_{k=1}^{i-2}\zeta(k+1)\zeta(i-k)\bigg)\\
&=2\zeta(2)-\sum_{i=2}^{p}\frac{i}{2}\zeta(i+1)+\frac{1}{2}\sum_{i=3}^{p}\sum_{k=1}^{i-2}\zeta(k+1)\zeta(i-k)\,.
\end{align*}
\end{Lem}
\begin{proof}
From the definition of $J(-2,p,1)$, we can write
\begin{align*}
J(-2,p,1)
&=\sum_{n=1}^\infty \frac{1}{n} \int_{0}^{1}x^{n-2} Li_{p}(x)\mathrm{d}x\\
&=\zeta(p+1)+\sum_{n=1}^\infty \frac{1}{n+1}\bigg(\sum_{j=2}^{p}\frac{(-1)^{p-j}}{n^{p+1-j}}\zeta(j)
+\frac{(-1)^{p-1}}{n^{p}}H_{n}\bigg)\\
&=\zeta(p+1)+\sum_{j=2}^{p}(-1)^{p-j}\zeta(j)\bigg(\sum_{i=2}^{p+1-j}(-1)^{p+1-j-i}\zeta(i)+(-1)^{p-j}\bigg)\\
&\quad +(-1)^{p-1}\sum_{n=1}^\infty H_{n}\bigg(\sum_{i=2}^{p}(-1)^{p-i}\frac{1}{n^{i}}+(-1)^{p-1}\frac{1}{n(n+1)}\bigg)\,.
\end{align*}
It is known that \cite{Sofo}
$$
\sum_{n=1}^\infty \frac{H_{n}}{n(n+1)}=\zeta(2)\,,
$$
thus we get the first result.
On the contrary,
\begin{align*}
J(-2,p,1)
&=-\sum_{n=1}^\infty \frac{1}{n^{p}} \int_{0}^{1}x^{n-2} \log(1-x)\mathrm{d}x\\
&=\zeta(2)+\sum_{n=2}^\infty H_{n-1}\bigg(\sum_{i=2}^{p}\frac{-1}{n^{i}}+\frac{1}{n(n-1)H_{n}}\bigg)\\
&=\zeta(2)+\sum_{n=1}^\infty \frac{H_{n}}{n(n+1)}-\sum_{i=2}^{p}\sum_{n=1}^\infty \frac{H_{n}}{n^{i}}
+\sum_{i=2}^{p}\zeta(i+1)\,,
\end{align*}
thus we get the second result.
\end{proof}
Now we derive explicit expression for $J(m,p,q)$.
\begin{theorem}\label{maintheorem3}
Let $p, q \in \mathbb N$ with $p \geq q$ and $m \in \mathbb N_{0}\cup\{-2\}$, then we have
\begin{align*}
&\quad J(m,p,q)\\
&=\sum_{x=1}^{q-1}\sum_{i_{1}=0}^{p-2}\frac{(-1)^{i_{1}+1}}{(m+1)^{i_{1}+1}}\cdots
\sum_{i_{x-1}=0}^{p-i_{1}-\cdots-i_{x-2}-2}\frac{(-1)^{i_{x-1}+1}}{(m+1)^{i_{x-1}+1}}\\
&\quad \times \sum_{i_{x}=0}^{p-i_{1}-\cdots-i_{x-1}-2}\frac{(-1)^{i_{x}}}{(m+1)^{i_{x}+1}}
\zeta(p-i_{1}-\cdots-i_{x})\zeta(q-x+1)\\
&\quad +\sum_{x=1}^{q-1}\frac{(-1)^{p-2+x}}{(m+1)^{p-2+x}}J(m,1,q-x+1)
\sum_{i_{1}=0}^{p-2}\cdots\sum_{i_{x-1}=0}^{p-i_{1}-\cdots-i_{x-2}-2}1\\
&\quad +\sum_{i_{1}=0}^{p-2}\frac{(-1)^{i_{1}+1}}{(m+1)^{i_{1}+1}}\cdots
\sum_{i_{q-1}=0}^{p-i_{1}-\cdots-i_{q-2}-2}\frac{(-1)^{i_{q-1}+1}}{(m+1)^{i_{q-1}+1}}
J(m,p-i_{1}-\cdots-i_{q-1},1)\,.
\end{align*}
where $K(m+p+x-1,0,q-x+1)$ and $K(m+i_{1}+\cdots+i_{q},p-i_{1}-\cdots-i_{q}+q,0)$ are given in Lemmata \ref{lem6}, \ref{lem7} and \ref{lem8}. Therefore $J(m,p,q)$ can be reduced to zeta values and generalized harmonic numbers.
\end{theorem}
\begin{proof}
It is known that \cite{Freitas}
$$
J(m,p,q)=\frac{\zeta(p)\zeta(q)}{m+1}-\frac{1}{m+1}\bigg(J(m,p-1,q)+J(m,p,q-1)\bigg)\,,
$$
successive application of the above relation $p-1$ times, we can obtain the following recurrence relation:
\begin{align*}
J(m,p,q)
&=\sum_{i_{1}=0}^{p-2}\frac{(-1)^{i_{1}}}{(m+1)^{i_{1}+1}}
\bigg(\zeta(p-i_{1})\zeta(q)-J(m,p-i_{1},q-1)\bigg)\\
&\quad +\frac{(-1)^{p-1}}{(m+1)^{p-1}}J(m,1,q)\,.
\end{align*}
Successive application of the new relation gives
\begin{align*}
&\quad J(m,p,q)\\
&=\sum_{x=1}^{q-1}\sum_{i_{1}=0}^{p-2}\frac{(-1)^{i_{1}+1}}{(m+1)^{i_{1}+1}}\cdots
\sum_{i_{x-1}=0}^{p-i_{1}-\cdots-i_{x-2}-2}\frac{(-1)^{i_{x-1}+1}}{(m+1)^{i_{x-1}+1}}\\
&\quad \times \sum_{i_{x}=0}^{p-i_{1}-\cdots-i_{x-1}-2}\frac{(-1)^{i_{x}}}{(m+1)^{i_{x}+1}}
\zeta(p-i_{1}-\cdots-i_{x})\zeta(q-x+1)\\
&\quad +\sum_{x=1}^{q-1}\frac{(-1)^{p-2+x}}{(m+1)^{p-2+x}}J(m,1,q-x+1)
\sum_{i_{1}=0}^{p-2}\cdots\sum_{i_{x-1}=0}^{p-i_{1}-\cdots-i_{x-2}-2}1\\
&\quad +\sum_{i_{1}=0}^{p-2}\frac{(-1)^{i_{1}+1}}{(m+1)^{i_{1}+1}}\cdots
\sum_{i_{q-1}=0}^{p-i_{1}-\cdots-i_{q-2}-2}\frac{(-1)^{i_{q-1}+1}}{(m+1)^{i_{q-1}+1}}
J(m,p-i_{1}-\cdots-i_{q-1},1)\,.
\end{align*}
\end{proof}
Freitas \cite{Freitas} gave the following recurrence relation for $K(r,0,q)$: For $r\geq 1$, $q\geq 2$, one has
\begin{align*}
K(r,0,q)
=(-1)^{r+q}\frac{r!}{(q-1)!}K(q-1,0,r+1)+(-1)^{r}r!\bigg(\zeta(r+1)\zeta(r+q+1)\bigg)\,.
\end{align*}
From this, Freitas showed that $K(r,p,q)$ could be reduced to zeta values as $p+q+r$ even. We now provide an explicit formula for $K(r,p,q)$.
\begin{theorem}\label{maintheorem4}
Let $p, q, m \in \mathbb N$ with $p \geq q$, then we have
\begin{align*}
&\quad K(m,p,q)\\
&=\sum_{x=1}^{q}K(m+p+x-1,0,q-x+1)\frac{(-1)^{p+x-1}}{(m+1)_{p+x-1}}\sum_{i_{1}=1}^{p}\cdots
\sum_{i_{x-1}=1}^{p-i_{1}-\cdots-i_{x-2}+x-2}1\\
&\quad +\sum_{i_{1}=1}^{p}\frac{(-1)^{i_{1}}}{(m+1)_{i_{1}}}\cdots\sum_{i_{q}=1}^{p-i_{1}-\cdots-i_{q-1}+q-1}
\frac{(-1)^{i_{q}}}{(m+i_{1}+\cdots+i_{q-1}+1)_{i_{q}}}\\
&\quad \times K(m+i_{1}+\cdots+i_{q},p-i_{1}-\cdots-i_{q}+q,0)\,.
\end{align*}
Therefore when $m+p+q$ is even, $K(m,p,q)$ can be reduced to zeta values and generalized harmonic numbers.
\end{theorem}
\begin{proof}
It is known that \cite{Freitas}
$$
K(m,p,q)=-\frac{1}{m+1}\bigg(K(m+1,p-1,q)+K(m+1,p,q-1)\bigg)\,,
$$
successive application of the above relation $p-1$ times, we can obtain the following recurrence relation:
\begin{align*}
K(m,p,q)
=\sum_{i_{1}=1}^{p}\frac{(-1)^{i_{1}}}{(m+1)_{i_{1}}}K(m+i_{1},p-i_{1}+1,q-1)
+\frac{(-1)^{p}}{(m+1)^{p}}K(m+p,0,q)\,.
\end{align*}
Successive application of the new relation gives
\begin{align*}
&\quad K(m,p,q)\\
&=\sum_{x=1}^{q}K(m+p+x-1,0,q-x+1)\frac{(-1)^{p+x-1}}{(m+1)_{p+x-1}}\sum_{i_{1}=1}^{p}\cdots
\sum_{i_{x-1}=1}^{p-i_{1}-\cdots-i_{x-2}+x-2}1\\
&\quad +\sum_{i_{1}=1}^{p}\frac{(-1)^{i_{1}}}{(m+1)_{i_{1}}}\cdots\sum_{i_{q}=1}^{p-i_{1}-\cdots-i_{q-1}+q-1}
\frac{(-1)^{i_{q}}}{(m+i_{1}+\cdots+i_{q-1}+1)_{i_{q}}}\\
&\quad \times K(m+i_{1}+\cdots+i_{q},p-i_{1}-\cdots-i_{q}+q,0)\,.
\end{align*}
Note that
\begin{align*}
K(m,0,q)=m!(-1)^{m}\bigg(S_{q,m+1}^{+,+}-\zeta(m+q+1)\bigg)\,,\quad \cite{Freitas}
\end{align*}
and
\begin{align*}
S_{p,q}^{+,+}
&=\zeta(p+q)\bigg(\frac{1}{2}-\frac{(-1)^{p}}{2}\binom{p+q-1}{p}-\frac{(-1)^{p}}{2}\binom{p+q-1}{q}\bigg)\\
&\quad +\frac{1-(-1)^{p}}{2}\zeta(p)\zeta(q)+(-1)^{p}\sum_{k=1}^{\lfloor \frac{p}{2} \rfloor}\binom{p+q-2k-1}{q-1}\zeta(2k)\zeta(m-2k)\\
&\quad +(-1)^{p}\sum_{k=1}^{\lfloor \frac{q}{2} \rfloor}\binom{p+q-2k-1}{p-1}\zeta(2k)\zeta(m-2k)\,,\quad (p+q \quad \hbox{odd})\cite{Flajolet}
\end{align*}
where $\zeta(1)$ should be interpreted as $0$ whenever it occurs and $\lfloor x \rfloor$ denotes the floor function. Combining the above results, we get the desired result.
\end{proof}
\section{Acknowledgements}
The author is grateful to the referee for her/his useful comments and suggestions. The author is also grateful to Dr. Wanfeng Liang and Dr. Ke Wang for some useful discussions.
|
{
"timestamp": "2021-03-23T01:36:08",
"yymm": "2103",
"arxiv_id": "2103.11839",
"language": "en",
"url": "https://arxiv.org/abs/2103.11839"
}
|
\section{Introduction}\label{sec:intro}
Erd\H{o}s in~\cite{Erd88} wrote about a problem he proposed with Ne\v{s}et\v{r}il:
\begin{quote}
``One could perhaps try to determine the smallest integer $h_t(\Delta)$ so that every $G$ of $h_t(\Delta)$ edges each vertex of which has degree $\le \Delta$ contains two edges so that the shortest path joining these edges has length $\ge t$ \dots {\em This problem seems to be interesting only if there is a nice expression for $h_t(\Delta)$}.''
\end{quote}
Equivalently, $h_t(\Delta)-1$ is the largest number of edges inducing a graph of maximum degree $\Delta$ whose line graph has diameter at most $t$.
Alternatively, one could consider this an edge version of the (old, well-studied, and exceptionally difficult) degree--diameter problem, cf.~\cite{B78}.
It is easy to see that $h_t(\Delta)$ is at most $2\Delta^t$ always, but one might imagine it to be smaller.
For instance, the $t=1$ case is easy and $h_1(\Delta)=\Delta+1$.
For $t=2$, it was independently proposed by Erd\H{o}s and Ne\v{s}et\v{r}il~\cite{Erd88} and Bermond, Bond, Paoli and Peyrat~\cite{BPPT83} that
$h_2(\Delta) \le 5\Delta^2/4+1$, there being equality for even $\Delta$. This was confirmed by Chung, Gy\'arf\'as, Tuza and Trotter~\cite{CGTT90}.
For the case $t=3$, we suggest the following as a ``nice expression''.
\begin{conjecture}\label{conj:CJK}
$h_3(\Delta)\le \Delta^3-\Delta^2+\Delta+2$, with equality if $\Delta$ is one more than a prime power.
\end{conjecture}
\noindent
As to the hypothetical sharpness of this conjecture, first consider the point--line incidence graphs of finite projective planes of prime power order $q$. Writing $\Delta=q+1$, such graphs are bipartite, $\Delta$-regular, and of girth $6$; their line graphs have diameter $3$; and they have $\Delta^3-\Delta^2+\Delta$ edges. At the expense of bipartiteness and $\Delta$-regularity, one can improve on the number of edges in this construction by {\em one} by subdividing one edge, which yields the expression in Conjecture~\ref{conj:CJK}.
We remark that for multigraphs instead of simple graphs, one can further increase the number of edges by $\left \lfloor\frac \Delta2 \right \rfloor-1,$ by deleting some arbitrary vertex $v$ and replacing it with a multiedge of multiplicity $\left \lfloor\frac \Delta2 \right \rfloor,$ whose endvertices are connected with $\left \lfloor\frac \Delta2 \right \rfloor$ and $\left \lceil \frac \Delta2 \right \rceil$ of the original $\Delta$ neighbours of $v.$
This last remark contrasts to what we know for multigraphs in the case $t=2$, cf.~\cite{CaKa19,JKP19}.
Through a brief case analysis, we have confirmed Conjecture~\ref{conj:CJK} in the case $\Delta=3$.
\begin{thr}\label{thm:cubic}
The line graph of any (multi)graph of maximum degree $3$ with at least $23$ edges has diameter greater than $3$.
That is, $h_3(3) = 23$.
\end{thr}
For larger fixed $t$, although we are slightly less confident as to what a ``nice expression'' for $h_t(\Delta)$ might be, we believe that $h_t(\Delta)=(1+o(1))\Delta^t$ holds for infinitely many $\Delta$.
We contend that this naturally divides into two distinct challenges, the former of which appears to be more difficult than the latter.
\begin{conjecture}\label{conj:lower}
For any $\varepsilon>0$, $h_t(\Delta) \ge (1-\varepsilon)\Delta^t$ for infinitely many $\Delta$.
\end{conjecture}
\begin{conjecture}\label{conj:upper}
For $t\ne 2$ and any $\varepsilon>0$, $h_t(\Delta) \le (1+\varepsilon)\Delta^t$ for all large enough $\Delta$.
\end{conjecture}
\noindent
With respect to Conjecture~\ref{conj:lower}, we mentioned earlier how it is known to hold for $t\in\{1,2,3\}$. For $t\in\{4,6\}$, it holds due to the point--line incidence graphs of, respectively, a symplectic quadrangle with parameters $(\Delta-1,\Delta-1)$ and a split Cayley hexagon with parameters $(\Delta-1,\Delta-1)$ when $\Delta-1=q$ is a prime power. For all other values of $t$ the conjecture remains open.
Conjecture~\ref{conj:lower} may be viewed as the direct edge analogue of an old conjecture of Bollob\'as~\cite{B78}. That conjecture asserts, for any positive integer $t$ and any $\varepsilon>0$, that there is a graph of maximum degree $\Delta$ with at least $(1-\varepsilon)\Delta^t$ vertices of diameter at most $t$ for infinitely many $\Delta$.
The current status of Conjecture~\ref{conj:lower} is essentially the same as for Bollob\'as's conjecture: it is unknown if there is an absolute constant $c>0$ such that $h_t(\Delta) \ge c \Delta^t$ for all $t$ and infinitely many $\Delta$.
For large $t$ the best constructions we are aware of are (ones derived from) the best constructions for Bollob\'as's conjecture.
\begin{prop}\label{prop:CaGo05}
There is $t_0$ such that $h_t(\Delta) \ge 0.629^t \Delta^t$ for $t\ge t_0$ and infinitely many $\Delta$.
\end{prop}
\begin{proof}
Canale and G\'omez~\cite{CaGo05} proved the existence of graphs of maximum degree $\Delta$, of diameter $t'$, and with more than $(0.6291\Delta)^{t'}$ vertices, for $t'$ large enough and infinitely many $\Delta$. Consider this construction for $t'=t-1$ and each valid $\Delta$.
Now in an iterative process arbitrarily add edges between vertices of degree less than $\Delta$.
Note that as long as there are at least $\Delta+1$ such vertices, then for every one there is at least one other to which it is not adjacent.
Thus by the end of this process, at most $\Delta$ vertices have degree smaller than $\Delta$, and so the resulting graph has at least $\frac 12 ( (0.6291\Delta)^{t'}\Delta - \Delta^2 )$ edges, which is greater than $(0.629\Delta)^{t}$ for $t$ sufficiently large.
Furthermore since the graph has diameter at most $t-1$, its line graph has diameter at most $t$.
\end{proof}
\noindent
By the above argument (which was noted in~\cite{DeSl19eurocomb}), the truth of Bollob\'as's conjecture would imply a slightly weaker form of Conjecture~\ref{conj:lower}, that is, with a leading asymptotic factor of $1/2$.
As far as we are aware, a reverse implication, i.e.~from Conjecture~\ref{conj:lower} to some form of Bollob\'as's conjecture is not known.
Our main result is partial progress towards Conjecture~\ref{conj:upper} (and thus Conjecture~\ref{conj:CJK}).
\begin{thr}\label{thm:mainht}
$h_t(\Delta) \le \frac{3}{2}\Delta^t+1$.
\end{thr}
\noindent
Theorem~\ref{thm:mainht} is a result/proof valid for all $t\ge1$, but as we already mentioned there are better, sharp determinations for $t\in\{1,2\}$.
We have also settled Conjecture~\ref{conj:upper} in the special case of graphs containing no cycle $C_{2t+1}$ of length $2t+1$ as a subgraph.
\begin{thr}\label{thm:2t+1ht}
The line graph of any $C_{2t+1}$-free graph of maximum degree $\Delta$ with at least $\Delta^t$ edges has diameter greater than $t$.
\end{thr}
\noindent
For $t\in\{1,2,3,4,6\}$, this last statement is asymptotically sharp (and in its more precise formulation the result is in fact exactly sharp) due to the point--line incidence graphs of generalised polygons.
The cases $t\in\{3,4,6\}$ are perhaps most enticing in Conjecture~\ref{conj:upper}, and that is why we highlighted the case $t=3$ first in Conjecture~\ref{conj:CJK}.
In order to discuss one consequence of our work, we can reframe the problem of estimating $h_t(\Delta)$ in stronger terms. Let us write $L(G)$ for the line graph of $G$ and $H^t$ for the $t$-th power of $H$ (where we join pairs of distinct vertices at distance at most $t$ in $H$). Then the problem of Erd\H{o}s and Ne\v{s}et\v{r}il framed at the beginning of the paper is equivalent to seeking optimal bounds on $|L(G)|$ subject to $G$ having maximum degree $\Delta$ and $L(G)^t$ inducing a clique.
Letting $\omega(H)$ denote the clique number of $H$, our main results are proven in terms of bounds on the {\em distance-$t$ edge-clique number} $\omega(L(G)^t)$ for graphs $G$ of maximum degree $\Delta$. In particular, we prove Theorem~\ref{thm:mainht} by showing the following stronger form.
\begin{thr}\label{thm:mainclique}
For any graph $G$ of maximum degree $\Delta$, it holds that $\omega(L(G)^t) \le \frac32\Delta^t$.
\end{thr}
\noindent
We should remark that D\c{e}bski and \'Sleszy\'nska-Nowak~\cite{DeSl19eurocomb} announced a bound of roughly $\frac74\Delta^t$.
Note that the bound in Theorem~\ref{thm:mainclique} can be improved in the cases $t\in\{1,2\}$: $\omega(L(G)) \le \Delta+1$ is trivially true, while $\omega(L(G)^2) \le \frac43\Delta^2$ is a recent result of Faron and Postle~\cite{FaPo19}.
We also have a bound on $\omega(L(G)^t)$ analogous to Theorem~\ref{thm:2t+1ht}, a result stated and shown in Section~\ref{sec:2t+1Free}.
A special motivation for us is a further strengthened form of the problem. In particular, there has been considerable interest in $\chi(L(G)^t)$ (where $\chi(H)$ denotes the chromatic number of $H$), especially for $G$ of bounded maximum degree. For $t=1$, this is the usual chromatic index of $G$; for $t=2$, it is known as the {\em strong chromatic index} of $G$, and is associated with a more famous problem of Erd\H{o}s and Ne\v{s}et\v{r}il~\cite{Erd88}; for $t>2$, the parameter is referred to as the {\em distance-$t$ chromatic index}, with the study of bounded degree graphs initiated in~\cite{KaMa12}.
We note that the output of Theorem~\ref{thm:mainclique} may be directly used as input to a recent result~\cite{HJK21} related to Reed's conjecture~\cite{Ree98} to bound $\chi(L(G)^t)$. This yields the following.
\begin{cor}\label{cor:fromreed}
There is some $\Delta_0$ such that, for any graph $G$ of maximum degree $\Delta\ge \Delta_0$, it holds that $\chi(L(G)^t) < 1.941\Delta^t$.
\end{cor}
\begin{proof}
By Theorem~\ref{thm:mainclique} and~\cite[Thm.~1.6]{HJK21}, $\chi(L(G)^t) \le \left \lceil0.881(\Delta(L(G)^t)+1)+0.119\omega(L(G)^t\right \rceil \le \left \lceil0.881(2\Delta^t+1)+0.119\cdot1.5\Delta^t\right \rceil < 1.941\Delta^t$ provided $\Delta$ is taken large enough.
\end{proof}
\noindent
For $t=1$, Vizing's theorem states that $\chi(L(G)) \le \Delta+1$.
For $t=2$, the current best bound on the strong chromatic index~\cite{HJK21} is $\chi(L(G)^2) \le 1.772\Delta^2$ for all sufficiently large $\Delta$.
For $t>2$, note for comparison with Corollary~\ref{cor:fromreed} that the local edge density estimates for $L(G)^t$ proved in~\cite{KaKa14} combined with the most up-to-date colouring bounds for graphs of bounded local edge density~\cite{HJK21} yields only a bound of $1.999\Delta^t$ for all large enough $\Delta$.
We must say though that, for the best upper bounds on $\chi(L(G)^t)$, $t>2$, rather than bounding $\omega(L(G)^t)$ it looks more promising to pursue optimal bounds for the local edge density of $L(G)^t$, particularly for $t\in\{3,4,6\}$. We have left this to future study.
\subsection{Terminology and notation}
For a graph $G=(V,E)$, we denote the $i^{th}$ neighbourhood of a vertex $v$ by $N_i(v)$, that is, $N_i(v)=\{u \in V \mid d(u,v)=i\}$, where $d(u,v)$ denotes the distance between $u$ and $v$ in $G$.
Similarly, we define $N_i(e)$ as the set of vertices at distance $i$ from an endpoint of $e$.
Let $T_{k,\Delta}$ denote a tree rooted at $v$ of height $k$ (i.e.~the leafs are exactly $N_k(v)$) such that all non-leaf vertices have degree $\Delta$.
Let $T^1_{k,\Delta}$ be one of the $\Delta$ subtrees starting at $v$, i.e.~a subtree rooted at $v$ of height $k$ such that $v$ has degree $1$, such that $N_k(v)$ only contains leaves and all non-leaf vertices have degree $\Delta$.
\section{A bound on $\omega(L(G)^t)$ for $C_{2t+1}$-free $G$}\label{sec:2t+1Free}
In this section, we prove the following theorem.
\begin{thr}\label{thm:L^tforC_{2t+1}-free}
Let $t \ge 2$ be an integer. Let $G$ be a $C_{2t+1}$-free graph with maximum degree $\Delta.$
Then $\omega(L(G)^t) \le \lvert E(T_{t,\Delta}) \rvert$.
When $t \in \{2,3,4,6\}$ equality can occur for infinitely many $\Delta$.
\end{thr}
\noindent
Since $\lvert E(T_{t,\Delta}) \rvert \le \Delta^t$, the expression is at most the bound desired for Conjecture~\ref{conj:upper}, and thus this implies Theorem~\ref{thm:2t+1ht}. In fact, the expression matches the order of the point--line incidence graphs of generalised polygons when $t \in \{2,3,4,6\}$, which are the examples for which equality holds.
On the other hand, by subdividing one edge of any these constructions, one can see in the cases $t\in\{2,3,4,6\}$ that the result fails if we omit the condition of $C_{2t+1}$-freeness.
We note that Theorem~\ref{thm:L^tforC_{2t+1}-free} is a generalisation of a result in~\cite{CKP20} which was specific to the case $t=2$.
It is also a stronger form of a result announced in~\cite{DeSl19eurocomb} for bipartite graphs.
One might wonder about excluding other cycle lengths, particularly even ones. Implicitly this was already studied in~\cite{KaPi18}, in that the local sparsity estimations there imply the following statement: for any $t\ge 2$ and even $\ell\ge 2t$, $\omega(L(G)^t) = o(\Delta^t)$ for any $C_\ell$-free graph of maximum degree $\Delta$.
Similarly, it would be natural to pursue a similar bound as in Theorem~\ref{thm:L^tforC_{2t+1}-free} but for an excluded odd cycle length (greater than $2t+1$), which was done for $t=2$ in~\cite{CKP20}.
The bound in Theorem~\ref{thm:L^tforC_{2t+1}-free} is a corollary of the following proposition.
\begin{prop}\label{sub}
For fixed $\Delta$ and $t$, let $G$ be a $C_{2t+1}$-free graph with maximum degree $\Delta$ and $H\subseteq G$ be a subgraph of $G$ with maximum degree $\Delta_H$.
Let $v$ be a vertex with degree $d_H(v)=\Delta_H=j$ and let $u_1,u_2, \ldots, u_j$ be its neighbours. Suppose that in $L(G)^t$, every edge of $H$ is adjacent to $vu_i$ for every $1 \le i \le j$.
Then $\lvert E(H) \rvert \le \lvert E(T_{t,\Delta}) \rvert$.
\end{prop}
\begin{proof}
For fixed $\Delta$, let $H$ and $G$ be graphs satisfying all conditions, such that $\lvert E(H) \rvert$ is maximised. This can be done since $\lvert E(H) \rvert$ is upper bounded by say $j\Delta ^t$.
With respect to the graph $G$, we write $N_i=N_i(v)$ for $0 \le i \le t+1.$
We start proving a claim that makes work easier afterwards.
\begin{claim}\label{cl:no_edge_inNi}
For any $1 \le i \le t$, $H$ does not contain any edge between two vertices of $N_i$.
\end{claim}
\begin{claimproof}
Suppose it is not true for some $i \le t-1$.
Take an edge $yz \in E(H)$ with $y,z \in N_i$.
Construct the graph $H'$ with $V(H')=V(H) \cup \{y',z'\}$ and $E(H')=E(H) \setminus yz \cup \{yy',zz'\}$, where $y'$ and $z'$ are new vertices, and let $G'$ be the corresponding modification of $G$.
Then $H' \subseteq G'$ also satisfies all conditions in Proposition~\ref{sub} and $\lvert E(H') \rvert=\lvert E(H) \rvert+1,$ contradictory with the choice of $H$.
Next, suppose there is an edge $yz \in E(H)$ with $y,z \in N_t$.
Take a shortest path from $u_1$ to $yz$, which is wlog a path $P_y$ from $u_1$ to $y$. Note that a shortest path $P_z$ from $v$ to $z$ will intersect $P_y$ since $G$ is $C_{2t+1}$-free. Let $w$ be the vertex in $V(P_y)\cap V(P_z)$ that minimises $d_G(w,z)$ and assume $w \in N_m,$ i.e.~$m=d_G(v,w)$ is the distance from $v$ to $w$.
The condition $w \in V(P_y)\cap V(P_z)$ ensures that $d_G(w,z)=d_G(w,y)$.
Furthermore note that $y$ and $z$ are interchangeable at this point, as both are at the same distance from $u_1$.
If $d_G(w,u_i)=m-1$ for every $1 \le i \le j$, we can remove $yz$ again and add two edges $yy'$ and $zz'$ to get a graph $H'$ satisfying all conditions, leading to a contradiction again. This is the blue scenario illustrated in Figure~\ref{fig:CloseLookAtG&H}.
In the other case there is some $1 <s \le j$ such that $d_G(w,u_s)>m-1$. Since $d_G(u_s,yz)=t-1$, wlog $d_G(u_s,z)=t-1$, there is a shortest path from $u_s$ to $z$ which is disjoint from the previously selected shortest path $P_y$ between $u_1$ and $y$. Hence together with the edges $u_1v, vu_s$ and $yz$, this forms a $C_{2t+1}$ in $G$, which again is a contradiction. This is sketched as the red scenario in Figure~\ref{fig:CloseLookAtG&H}.
\end{claimproof}
\begin{figure}[h]
\centering
\begin{tikzpicture}[line cap=round,line join=roundx,x=1.3cm,y=1.3cm]
\clip(-1,-0.7) rectangle (11,2.4);
\draw [line width=1.1pt] (0,0.) -- (2.,2);
\draw [line width=1.1pt] (0,0.) -- (2.,1.5);
\draw [line width=1.1pt] (0,0.) -- (2.,0.);
\draw [dashed, line width=1.1pt] (0,0.) -- (2.,-0.5);
\draw [dotted, line width=0.8pt] (2,0.) -- (2.,1.5);
\draw[line width=1.1pt] (2,2)--(10,2) ;
\draw[line width=1.1pt] (10,1)--(10,2) ;
\draw[line width=1.1pt] (6,2)--(10,1) ;
\draw[line width=1.1pt] (0,0)--(2,0.8) ;
\draw[ultra thick, color=red] (7,1.75)--(2,0.8) ;
\draw[ultra thick, color=blue] (6,2)--(2,0.8) ;
\draw[ultra thick, color=blue] (6,2)--(2,1.5) ;
\draw[ultra thick, color=blue] (6,2)--(2,0.) ;
\draw [fill=xdxdff] (0,0.) circle (2.5pt);
\draw [fill=xdxdff] (2.,0.) circle (2.5pt);
\draw [fill=xdxdff] (2.,1.5) circle (2.5pt);
\draw [fill=xdxdff] (2.,-.5) circle (2.5pt);
\draw [fill=xdxdff] (2.,2.) circle (2.5pt);
\draw [fill=xdxdff] (2,0.8) circle (2.5pt);
\draw [fill=xdxdff] (10.,2.) circle (2.5pt);
\draw [fill=xdxdff] (10.,1) circle (2.5pt);
\draw [fill=xdxdff] (6.,2.) circle (2.5pt);
\coordinate [label=right:$y$] (A) at (10.05,2);
\coordinate [label=right:$z$] (A) at (10.05,1);
\coordinate [label=above:$w$] (A) at (6,2.05);
\coordinate [label=left:$v$] (A) at (-0.1,0);
\coordinate [label=above left:$u_{1}$] (A) at (1.9,2.05);
\coordinate [label=above:$u_{2}$] (A) at (2,1.5);
\coordinate [label=left :$u_{s}$] (A) at (1.95,0.9);
\coordinate [label=above left:$u_{j}$] (A) at (1.9,0.05);
\end{tikzpicture}
\caption{Sketch of two scenarios (red and blue) referred to in the proof of Claim~\ref{cl:no_edge_inNi}.}
\label{fig:CloseLookAtG&H}
\end{figure}
For every $1 \le m \le t+1$, let $A_m$ be the set of all vertices $x$ in $N_m$ such that $d_G(v,x)=d_G(u_i,x)+1=m$ for at least one index $1 \le i \le j$ and let $R_m=N_m \backslash A_m.$
Also let $A_0=\{v\}.$
Let $A=\cup_{i=0}^{t+1} A_i$ and $R=\cup_{i=0}^{t+1} R_i$.
We observe that a vertex in $R_{t+1}$ cannot be an endvertex of an edge of $H$. Indeed, assuming the contrary, the other endvertex of such an edge would be at distance $t-1$ from every $u_i$, $1 \le i \le j$ and in particular would belong to $A_t$, leading to a contradiction.
Also we observe that there are no edges in $H$ between $R_t$ and $A$. By definition of $R_t$, any vertex $r \in R_t$ has no neighbour in $A_i$ where $i<t$, nor does it have a neighbour in $A_t$ by Claim~\ref{cl:no_edge_inNi}. To end with, an edge with endvertices in $R_t$ and $A_{t+1}$ is not connected to any edge $vu_s$, $1 \le s \le j$, in $L(G)^t$.
As a consequence, the number of edges of $H$ which contain at least one vertex of $R$ can be upper bounded by $ |R_1| \cdot \left(|E(T_{t,\Delta}^{1})| -1\right)$, which equals \begin{equation}\label{eq:EH_R}
(\deg(v)-\Delta_H)\left(\frac1{\Delta}\lvert E(T_{t,\Delta}) \rvert -1 \right).
\end{equation}
All other edges of $H$ are in the induced subgraph $H[A]$. We will now compute a bound on the number of those remaining edges.
We start with defining a weight function $w$ on the vertices $x$ in $A$ which will turn out to be useful.
For every $x \in A_m$ where $1 \le m \le t$, we define $w(x)$ to be equal to the number of paths (in $G$) of length $m-1$ between $x$ and $A_1$.
Note that by definition $w(x)$ is at least equal to the number of vertices $u_i \in A_1$ with $d_G(x,u_i)=m-1$ and by definition of $A_m$ this implies $w(x) \ge 1.$
An equivalent recursive definition of $w$ is the following: we let $w(u_i)=1$ for any $u_i \in A_1$ and for every vertex $x \in A_m$ where $m\ge 2$, we let $$w(x)= \sum_{y \in A_{m-1} \colon yx \in E(G)} w(y).$$
We observe by induction that
\begin{equation}\label{eq:sum_w}
\sum_{ x \in A_m} w(x) \le j(\Delta-1)^{m-1}
\end{equation} for every $1 \le m \le t.$ For $m=1$ this is by definition of $A_1$ and $j$.
For $m\ge 2$, we have by induction that
\begin{align*}
\sum_{ x \in A_m} w(x) &= \sum_{ x \in A_m} \sum_{y \in A_{m-1} \colon yx \in E(G)} w(y)\\
&=\sum_{ y \in A_{m-1}} \sum_{x \in A_m \colon yx \in E(G)} w(y)\\
&\le \sum_{ y \in A_{m-1}} (\Delta-1) w(y)\\
&\le j(\Delta-1)^{m-1}.
\end{align*}
Let $A^{'}_t=\{a \in A_t\mid w(a)<j\}$ and $A^*_t=\{a \in A_t\mid w(a) \ge j\}.$ We first count the edges that are incident to some fixed $a\in A^{'}_t$.
Note that $H$ contains no edges between $a$ and $A_{t+1}$ since for such an edge we would need that $a$ is connected by a path of length $t-1$ to every $u_i, 1 \le i \le j$ and thus in particular we would have $w(a) \ge j.$ By Claim~\ref{cl:no_edge_inNi} we also know that $a$ cannot be incident with an other vertex in $A_t$.
So we only need to count the edges in $H$ between $a$ and $A_{t-1}$, and by definition of the weight function, this is bounded by $w(a).$
On the other hand, for every $a \in A^*_t$ there are at most $\Delta_H=j \le w(a)$ edges in $E(H)$ incident to $a$.
Having proven that for every $a \in A_t$ there are at most $w(a)$ edges in $H[A]$ incident with $a$, we conclude (remembering~\eqref{eq:sum_w}) that there are at most $\sum_{ x \in A_t} w(x) \le j(\Delta-1)^{t-1}$ edges in $E(H[A])$ having a vertex in $A_{t}.$
Also we have for every $1 \le m \le t-1$ that the number of edges between $A_{m-1}$ and $A_m$ is bounded by $j(\Delta-1)^{m-1}.$
Hence $$\lvert E(H[A]) \rvert \le \sum_{m=1}^t j(\Delta-1)^{m-1}=\frac{\Delta_H}{\Delta}\lvert E(T_{t,\Delta}) \rvert.$$
Together with~\eqref{eq:EH_R} on the number of edges that intersect $R$, this gives the result as $\deg(v)\le \Delta$ by definition.
\end{proof}
An inspection of the proof yields that the extremal graphs $H$ for Proposition~\ref{sub} satisfy Claim~\ref{cl:no_edge_inNi}, $R=\emptyset$ and for every $x \in A_m$ where $0 \le m \le t-1$, there are exactly $\Delta -1$ edges towards $A_{m+1}.$
Hence such an extremal graph $H$ is exactly $T_{t,\Delta}$ where possibly some of its leaves are identified as one (as long as the maximum degree is still $\Delta$). Let us call such a graph a \emph{quasi-$T_{t,\Delta}$}.\\
Next, we discuss some properties that should be satisfied by any graph that attains the bound of Theorem~\ref{thm:L^tforC_{2t+1}-free} (provided such a graph exists for the given values of $t$ and $\Delta$!).
Let $H \subseteq G$ be a graph such that $E(H)$ is a clique in $L(G)^t$ and $v$ be a vertex of maximum degree in $H$, which maximises $|E(H)|$ among all choices for $G$ and $H$.
Let $N_H(v)=\{u_1, \ldots, u_j\}$. Then in particular, in $L(G)^t$ every edge of $H$ is adjacent to every edge $vu_i$, for all $1\le i \le j$.
So by Proposition~\ref{sub}, for every vertex $v$ of degree $\Delta$
we observe locally a quasi-$T_{t, \Delta}$ again, and in particular every neighbour of such a $v$ has degree $\Delta$ (for $t \ge 2$).
So $H$ is $\Delta$-regular and in particular a connected component of $G$. So it is not a tree and hence has some girth.
The girth is at least $2t$ (as for every vertex we locally have a quasi-$T_{t, \Delta}$), but it cannot be $2t+1$ since $G$ is $C_{2t+1}$-free and it cannot be $2t+2$ or more since $E(H)$ is a clique in $L(G)^t$.
Also we observe that for every $a \in A_t^*$ the condition that $w(a)\ge j$ implies that it has $\Delta$ neighbours in $A_{t-1}$ as these all have a weight function equal to $1$ and so it has no neighbours in $A_{t+1}.$
Hence $H$ is a $\Delta$-regular graph with girth $2t$ and diameter $t$.
In particular they need to be Moore graphs and consequently by~\cite{Sing66, BI73} the extremal graphs are polygons when $t \ge 3.$
\section{A general bound on $\omega(L(G)^t)$}\label{sec:general}
When $H\subseteq G$ is a graph whose edges form a clique in $L(G)^t$, it implies in particular that all edges adjacent to a specific vertex $v$ are at distance at most $t-1$ from all other edges.
As $\lvert E(T_{t,\Delta}) \rvert \le \Delta^t$, the following proposition implies Theorem~\ref{thm:mainclique}.
\begin{prop}\label{substrong}
For fixed $\Delta$ and $t$, let $G$ be a graph with maximum degree $\Delta$ and
$H\subseteq G$ be a subgraph of $G$ with maximum degree $\Delta_H$.
Let $v$ be a vertex with degree $d_H(v)=\Delta_H=j$ and let $u_1,u_2, \ldots, u_j$ be its neighbours. Suppose that in $L(G)^t$, every edge of $H$ is adjacent to $vu_i$ for every $1 \le i \le j$.
Then $\lvert E(H) \rvert \le \frac{3}{2}\lvert E(T_{t,\Delta}) \rvert$.
\end{prop}
\begin{proof}
We do this analogously to the proof of Proposition~\ref{sub}.
For fixed $\Delta$, let $H$ and $G$ be graphs satisfying all conditions, such that $\lvert E(H) \rvert$ is maximized (which is again possible since $j\Delta ^t$ is an upper bound for $\lvert E(H) \rvert$).
It suffices to show that $\lvert E(H) \rvert \le \frac{3}{2} \lvert E(T_{t,\Delta}) \rvert$.
By the proof of Claim~\ref{cl:no_edge_inNi} we know that for any $1 \le i \le t-1$, the set $N_i$ does not induce any edges of $H$ (but this is not necessarily true anymore for $N_t$).
Define $A_m, R_m$, the weight function $w$, $A^{'}_t$ and $A_t^*$ as has been done in the proof of Proposition~\ref{sub}.
As before, the number of edges that (are not induced by $N_t$ and) use at least one vertex of $R$ is bounded by~\eqref{eq:EH_R}. Also, we again have for every $1 \le m \le t-1$ that the number of edges between $A_{m-1}$ and $A_m$ is bounded by $j(\Delta-1)^{m-1}.$ Furthermore, $R_t$ does not induce any edge of $H$, because such an edge would be at distance larger than $t$ from $vu_1$. Thus the number of edges of $H$ that are either disjoint from $A_t$, or join $A_t$ and $R \backslash R_{t}$, is at most
\begin{equation}\label{eq:EH_bla}
(\Delta-j)\left(\frac1{\Delta}\lvert E(T_{t,\Delta}) \rvert -1 \right) + \sum_{m=1}^{t-1} j(\Delta-1)^{m-1}.
\end{equation}
We will derive that the remaining edges of $H$ (which all intersect $A_t$) can be bounded by a linear combination of the weight functions $w(a)$ of the vertices $a\in A_t$.\\
For every $a \in A_t^*$ there are at most $j \le w(a)$ edges in $E(H)$ having $a$ as one of its endvertices. So let us now focus on the edges that intersect $A^{'}_t$.
We observe that there are no edges in $H$ between any $a \in A^{'}_t$ and $r \in R_t$, because there is some $u_i$ such that $d(a,u_i) \ge t$, which implies that $vu_i$ and $ar$ would be at distance larger than $t$. For the same reason $H$ has no edges between $A^{'}_t$ and $A_{t+1}$.\\
Finally, we want to count the edges between $A_{t-1}$ and $A^{'}_t$, as well as those that are induced by $A^{'}_t$.
We will prove that their number is bounded by $\frac 32 \sum_{a \in A^{'}_t} w(a).$
For that, we need the following technical claim.
\begin{claim}\label{claim:technical}
Let $j$ be fixed and assume $j>x \ge m>0$ and $j>y \ge n>0$ with $x+y \ge j$. Then
$$\frac{ \frac32 x -m}{j-m}+\frac{ \frac32 y -n}{j-n} \ge 1.$$
Equality occurs if and only $m=n=x=y=\frac j2.$
\end{claim}
\begin{claimproof}
Multiplying both sides with the positive factor $2(j-m)(j-n)$, we need to prove that $3(x+y)j-3xn-3ym+2mn \ge 2j^2$.
For fixed $j,x$ and $y$ the left hand side is minimal when $m=x$ and $n=y$. This reduces to proving that $3(x+y)j -4xy \geq 2j^2.$
But this is true since
\begin{align*}
3(x+y)j -4xy-2j^2 &= 0.25 j^2 - (x+y-1.5j)^2 + (x-y)^2\\
&=(2j-(x+y))\cdot(x+y-j)+(x-y)^2\\&\ge 0,
\end{align*}
as $j \le x+y <2j$, i.e.~$|x+y-1.5j| \le 0.5j$.
\end{claimproof}
For every $a\in A_t^{'}$, let $m(a)$ denote the number of neighbours (in $H$) of $a$ in $A_{t-1}$ and let $q(a)$ denote the number of neighbours (in $H$) of $a$ in $A^{'}_t$.
Furthermore, we define $f(a)=\frac{ \frac32 w(a) -m(a)}{j-m(a)}$.
Suppose $H$ contains an edge $e$ between two vertices $a_1, a_2 \in A^{'}_t$. Then $w(a_1)+w(a_2) \ge j$, since $a_1a_2$ must be within distance $t-1$ of each of $vu_1,vu_2,\ldots,vu_j$. Hence by Claim~\ref{claim:technical} (applied with $m=m(a_1)$, $n=m(a_2)$, $x=w(a_1)$ and $y=w(a_2)$), we obtain that $f(a_1)+f(a_2)\geq 1$ for every edge $a_1a_2$ of $H[A^{'}_t]$.
From this it follows that $|E(H[A^{'}_t])| \leq \sum_{a_1a_2 \in E(H[A^{'}_t])} f(a_1)+f(a_2)$. The right hand side can further be rewritten as $ \sum_{a \in A^{'}_t} q(a) \cdot f(a)$. Since every vertex $a \in A^{'}_t$ has $q(a) \leq j-m(a)$ neighbours in $A^{'}_t$ and has $m(a)$ neighbours in $A_{t-1}$, we conclude that
the number of edges of $H$ that are either induced by $A^{'}_t$ or join $A^{'}_t$ and $A_{t-1}$ is at most
\begin{align*}
\sum_{a \in A^{'}_t} \left((j-m(a)) \cdot f(a) + m(a) \right) &= \sum_{a \in A^{'}_t} \frac{3}{2} w(a).
\end{align*}
Thus the number of edges in $E(H)$ using
at least one vertex in $A_t$ is bounded by $$\sum_{ a \in A^{'}_t} \frac32 w(a) +\sum_{ a \in A^{*}_t} w(a) \le \sum_{ x \in A_t} \frac32 w(x),$$
which (see the derivation of~\eqref{eq:sum_w}) is at most $\frac{3}{2} j (\Delta-1)^{t-1}$.
Summing this and~\eqref{eq:EH_bla}, we conclude that $H$ has fewer than $(\Delta-j)\left(\frac1{\Delta}\lvert E(T_{t,\Delta}) \rvert -1 \right) + \frac{3}{2}\frac{j}{\Delta} |E(T_{t,\Delta})| \leq \frac{3}{2} |E(T_{t,\Delta})|$ edges.
\end{proof}
Note that the exact maximum in Proposition~\ref{substrong} is $\sum_{m=1}^{t-1} \Delta( \Delta-1)^{m-1}+\frac32 \Delta( \Delta-1)^{t-1}$ and this can be attained when $\Delta$ is even.
For example when $t=2$, the following example in Figure~\ref{fig:beatC5} shows that the blow-up of a $C_5$ is not extremal anymore when only taking into account the weaker conditions from Proposition~\ref{substrong}.
\begin{figure}[h]
\centering
\scalebox{0.9}{
\rotatebox{90}{
\begin{tikzpicture}[line cap=round,line join=roundx,x=1.3cm,y=1.3cm]
\clip(1.4,-2.1) rectangle (3.6,2.5);
\draw [line width=1.1pt] (1.5,0.) -- (2.,2);
\draw [line width=1.1pt] (1.5,0.) -- (2.,1);
\draw [line width=1.1pt] (1.5,0.) -- (2.,-1);
\draw [line width=1.1pt] (1.5,0.) -- (2.,-2);
\draw [line width=1.1pt](3,-0.5)--(2.5,0.5);
\draw [line width=1.1pt](3,-0.5)--(3.5,0.5);
\draw [line width=1.1pt](2.5,-0.5)--(3,0.5);
\draw [line width=1.1pt](2.5,-0.5)--(3.5,0.5);
\draw [line width=1.1pt](3.5,-0.5)--(3,0.5);
\draw [line width=1.1pt](3.5,-0.5)--(2.5,0.5);
\draw [line width=1.1pt](2.,2)--(2.5,0.5);
\draw [line width=1.1pt](2.,2)--(3.5,0.5);
\draw [line width=1.1pt](2.,2)--(3,0.5);
\draw [line width=1.1pt](2.,1)--(3.5,0.5);
\draw [line width=1.1pt](2.,1)--(3,0.5);
\draw [line width=1.1pt](2.,1)--(2.5,0.5);
\draw [line width=1.1pt](3,-0.5)--(2.,-1);
\draw [line width=1.1pt](3.5,-0.5)--(2.,-1);
\draw [line width=1.1pt](2.5,-0.5)--(2.,-1);
\draw [line width=1.1pt](2.5,-0.5)--(2.,-2);
\draw [line width=1.1pt](3,-0.5)--(2.,-2);
\draw [line width=1.1pt](3.5,-0.5)--(2.,-2);
\draw [fill=red] (1.5,0.) circle (3pt);
\draw [fill=xdxdff] (2.,1) circle (2.5pt);
\draw [fill=xdxdff] (2.,-2.) circle (2.5pt);
\draw [fill=xdxdff] (2.,-1) circle (2.5pt);
\draw [fill=xdxdff] (2.,2.) circle (2.5pt);
\draw [fill=xdxdff] (3,-0.5) circle (2.5pt);
\draw [fill=xdxdff] (2.5,-0.5) circle (2.5pt);
\draw [fill=xdxdff] (3.5,-0.5) circle (2.5pt);
\draw [fill=xdxdff] (3,0.5) circle (2.5pt);
\draw [fill=xdxdff] (2.5,0.5) circle (2.5pt);
\draw [fill=xdxdff] (3.5,0.5) circle (2.5pt);
\end{tikzpicture}}}
\caption{An extremal graph for Proposition~\ref{substrong} for $\Delta=4$ and $t=2$.}
\label{fig:beatC5}
\end{figure}
\section{Determination of $h_3(3)$}\label{sec:subcubic}
\begin{proof}[Proof of Theorem~\ref{thm:cubic}]
Let $G=(V,E)$ be a graph of maximum degree $3$ such that the line graph $L(G)$ of $G$ has diameter at most $3$, i.e.~$L(G)^3$ is a clique. If we can show that $G$ must have at most $22$ edges, then the result is proven. Suppose to the contrary that $|E|\ge23$. The proof proceeds through a series of claims that establish structural properties of $G$.
In each claim, we will estimate $|E|$ by performing a breadth-first search rooted at some specified edge $e$ up to distance $3$. To avoid repetition, let us set out the notation we use each time. We write $e=uv$. Let $u_0$ and $u_1$ be the two neighbours of $u$ other than $v$ (if $u$ has degree $3$). For $i\in\{0,1\}$, let $u_{i0}$ and $u_{i1}$ be the two neighbours of $u_i$ other than $u$ (if $u_i$ has degree $3$). For $i,j\in\{0,1\}$, let $u_{ij0}$ and $u_{ij1}$ be the two neighbours of $u_{ij}$ other than $u_{i}$ (if $u_{ij}$ has degree $3$). Similarly, define $v_i$, $v_{ij}$, $v_{ijk}$ for $i,j,k\in\{0,1\}$.
\begin{claim}\label{clm:triangle}
$G$ contains no triangle, loop or multi-edge.
\end{claim}
\begin{claimproof}
These $3$ cases are straightforwardly bounded by the breadth-first search.
If the edge $e$ is in a triangle, $|E|=|N_{L(G)^3}[e]| \le |E(K_3)|+2\cdot|E(T^1_{3,3})|+|E(T^1_{2,3})|= 3+2\cdot7+3= 20$.
Analogously, if the edge $e$ is a loop, one obtains $|E|=|N_{L(G)^3}[e]| \le 1+7=8$.
If the edge $e$ has a parallel edge then $|E|=|N_{L(G)^3}[e]| \le 2+2\cdot7= 16$.
\end{claimproof}
\begin{claim}\label{clm:regular}
$G$ is $3$-regular, and so $|E|$ is divisible by $3$.
\end{claim}
\begin{claimproof}
If not, say, $v$ has degree at most $2$, then, say, $v_1$, $v_{1j}$, $v_{1jk}$ are undefined, and so $|E|=|N_{L(G)^3}[e]| \le 1+3\cdot7=22$, a contradiction.
\end{claimproof}
\begin{claim}\label{clm:C4}
$G$ contains no $4$-cycle.
\end{claim}
\begin{claimproof}
If the edge $e$ is in a $4$-cycle, then without loss of generality suppose $u_1 =v_{00}$, $v_0=u_{11}$, and so on. Already $|E|=|N_{L(G)^3}[e]| \le 4+2\cdot7+2\cdot3=24$ and by Claim~\ref{clm:regular} we have a contradiction if we can show that $|E|$ is $1$ lower. So we may assume that $u_0$, $u_1$, $v_0$, $v_1$, $u_{00}$, $u_{01}$, $u_{10}$, $v_{01}$, $v_{10}$, $v_{11}$ are all distinct vertices and that the vertices $u_{00k}$, $u_{01k}$, $u_{10k}$, $v_{01k}$, $v_{10k}$, $v_{11k}$ (possibly not all distinct) are all at distance exactly $3$ from $e$.
Consider the edges $u_{00}u_{000}$, $u_{00}u_{001}$, $u_{01}u_{010}$ and $u_{01}u_{011}$. They are within distance $3$ (in $L(G)$) from $vv_0$, so $u_{000}$, $u_{001}$, $u_{010}$ and $u_{011}$ all need to be adjacent to $v_{01}$, leading to a contradiction as $\deg{v_{01}}\le 3.$
\end{claimproof}
\begin{claim}\label{clm:C5}
$G$ contains no $5$-cycle.
\end{claim}
\begin{claimproof}
If the edge $e$ is in a $5$-cycle, then without loss of generality suppose $u_{11} =v_{00}$, $u_{111}=v_{000}$ and so on. Already $|E|=|N_{L(G)^3}[e]| \le 5+2\cdot 7+ 2\cdot 3+1= 26$.
Since $26 \ge |E|\geq 23$ and $G$ is $3$-regular by Claim~\ref{clm:regular}, it follows that $|E|=24$ and $|V|= \frac{2 |E|}{3} = 16$.
Note first that $N_2(e)=\{u_{00},u_{01},u_{10},u_{11},v_{01},v_{10},v_{11}\}$ are all distinct vertices or else $|E|$ is already at most $23$.
Thus $|N_3(e)|=16-13=3$ and so (again using $3$-regularity, and also the fact that $N_3(e)$ must be an independent set) there are exactly $2$ edges in the subgraph induced by $N_2(e)$.
We divide our considerations into two cases. First, we assume $u_{11}$ has a neighbour in $N_2(e).$ By Claim~\ref{clm:triangle}, without loss of generality we can assume that this neighbour is $u_{00}$ and thus $u_{111} =v_{000}=u_{00}$, $u_{11}=u_{001}$ and so on.
Since $N_2(e)$ induces two edges, we can assume that $v_{100}, v_{110} \in N_3(e)$.
Note that the edge $u_{00}u_{11}$ is within distance $3$ (in $L(G)$) of both $v_{10}v_{100}$ and $v_{11}v_{110}$.
It cannot be that $u_{000}$ is equal to $v_{10}$ or $v_{11}$ or else one of the edges $v_{10}v_{100}$ and $v_{11}v_{110}$ remains too far from $u_{00}u_{11}$ (taking Claim~\ref{clm:C4} into account).
At this point, we note that $v_{10}$ and $v_{11}$ both need to be adjacent to $u_{000}$, creating a $C_4$ and hence leading to a contradiction.
Second, since we are not in the first case, $u_{111}=v_{000}$ must be at distance exactly $3$ from $e$.
The vertex $u_{111}$ must have all of its $3$ neighbours in $N_2(e)$, one of which is $u_{11}$. Keeping in mind that there is no four-cycle,
we can therefore assume without loss of generality that $u_{111}=u_{000}=v_{111}$.
Let us consider as a subcase the possibility that $u_{01}$ and $v_{11}$ are adjacent (the case $v_{10}$ and $u_{00}$ being adjacent, is done in exactly the same way), say, $u_{010}=v_{11}$. Note that the edge $u_{01}v_{11}$ is within distance $3$ (in $L(G)$) of both $u_1u_{10}$ and $v_0v_{01}$. Since $v_{11}$ has all its neighbours already fixed (and keeping in mind that $N_2(e)$ induces only one edge other than $u_{01}v_{11}$), it can only be that $u_{01},u_{10}$ and $v_{01}$ have a common neighbour in $N_3(e)$.
So without loss of generality, $u_{011}=u_{100}=v_{010}$. But now, with only the free placement of $u_{001}$, the only possibility to have $u_{00}u_{000}$ within distance $3$ (in $L(G)$) of both edges $u_{10}u_{100}$ and $v_{01}v_{010}$, is if $u_{001}$ is equal to $u_{10}$ or $v_{01}$.
But then we have already determined the two edges induced by $N_2(e)$, none of which is incident to $v_{10}$, so that both $v_{100}$ and $v_{101}$ must be in $N_3(e)$, leading to the contradiction that $|N_3(e)|=|\left\{u_{100},u_{111},v_{100},v_{101} \right\}| \geq 4$.
We have thus shown that $u_{01}v_{11}$ and $v_{10}u_{00}$ are not present as an edge.
Let $i\in \left\{0,1\right\}$. The vertex $u_{01}$ is not adjacent to any vertex in $\{u_{00},u_1,v_0,v_{11}\}$ and so $u_{01i}$ has to be adjacent to one of them to ensure that $u_{11}u_{111}$ is within distance $3$ (in $L(G)$) of $u_{01}u_{01i}$
This implies $u_{01i}$ has to be equal to $v_{110}$, $u_{10}$, $v_{01}$ or $v_{10}$
(taking Claims~\ref{clm:triangle} and~\ref{clm:C4} into account).
Incidentally, $u_{01i}$ can also not be equal to $v_{10}$, because in order for $u_{01}v_{10}$ to be within distance $3$ of $u_{01}u_{01i}$, we would need an edge between $\left\{u_{01},v_{10} \right\}$ and $\left\{u_{00},v_{11}\right\}$, which would either create a triangle or an edge that we already showed to be not present.
So $u_{01i}$ has to be equal to $v_{110}$, $u_{10}$ or $v_{01}$, and symmetrically $v_{10i}$ equals $u_{001}, u_{10}$ or $v_{01}$, for all $i\in \left\{0,1\right\}$.
As there are only two edges in the graph induced by $N_2(e)$, and both $u_{01}$ and $v_{10}$ are an endvertex of one of them,
we may conclude without loss of generality that $u_{010}=v_{110}$ and $v_{101}=u_{001}$.
Note that the edge $v_{11}v_{110}$ is within distance $3$ (in $L(G)$) of $uu_1$, and so $u_{10}v_{110}$ must be an edge. But then the distance between $u_{01}v_{110}$ or $u_{10}v_{110}$ and $vv_0$ is at least $4$, a contradiction.
\end{claimproof}
By the above claims, it only remains to consider $G$ being $3$-regular and of girth at least $6$. Let $e\in E$ be arbitrary. Then we have $|E|=|N_{L(G)^3}[e]| \le 29$ and by Claim~\ref{clm:regular} we have a contradiction if we can show that $|E|$ is $6$ lower.
Since $|E|\geq 23$ and $G$ is $3$-regular, we know $|V|=\left \lceil \frac{2 |E|}{3} \right \rceil \geq 16$, hence there are at least $16 - (2+4+8) = 2$ vertices at distance $3$ from $e$.
Let $x$ and $y$ be vertices at distance $3$ from $e$. We may assume without loss of generality that $x$ is adjacent to $u_{00}$, $u_{10}$, and $v_{00}$. Since the edge $vv_1$ is within distance $3$ (in $L(G)$) of both edges $u_{00}x$ and $u_{10}x$, it follows (without loss of generality) that $u_{00}v_{10}$ and $u_{10}v_{11}$ are edges.
Since $y$ must satisfy similar constraints as $x$, and it cannot be adjacent to $u_{00}$ nor to $u_{10}$, there will be at least three edges between vertices in $N_2(e)$ and so similarly as before, we know that $|V|=\left \lfloor\frac{2|E|}{3}\right \rfloor \leq \left \lfloor\frac{2 \cdot (29-3)}{3}\right \rfloor =17$. Because every $3$-regular graph has an even number of vertices, it follows that $|V|=16$, so that in fact $x$ and $y$ are the only vertices in $N_{3}(e)$. From this and $3$-regularity, we conclude that the subgraph induced by $N_2(e)$ must have exactly $5$ edges.
Since every edge between $2$ vertices in $N_2(e)$ will be between some $u_{ij}$ and a $v_{k\ell}$ and $G$ is $3$-regular, we know that $y$ is adjacent to exactly one of $u_{01}$ and $u_{11}$, wlog $u_{11}$.
Similarly the two neighbours of $y$ of the form $v_{ij}$ are not a neighbour of $x$ and so $N(y)$ and $N(x)$ are disjoint.
In particular $y$ cannot be adjacent to $v_{00}$ and so it has to be adjacent to $v_{01}.$
The last neighbour of $y$ is either $v_{10}$ or $v_{11}$. If it is $v_{11}$, then to ensure that $uu_0$ is within distance $3$ of both $yv_{01}$ and $yv_{11}$, we would need $u_{01}$ to be a neighbour of both $v_{10}$ and $v_{11}$, creating a four-cycle; contradiction. Thus the neighbours of $y$ must be $u_{11}, v_{01}$ and $v_{10}$.
So to ensure this, $u_{01}v_{01}$ is an edge as well.
We have now determined the whole graph, apart from two edges between $\left\{u_{01},u_{11}\right\}$ and $\left\{v_{00},v_{11} \right\}$. However, the edge $v_{00}u_{11}$ would create the five-cycle $xu_{10}u_1u_{11}v_{00}$, while the edge $v_{00}u_{01}$ would yield the four-cycle $u_{01}v_{00}v_{0}v_{01}$. So in both cases, we get a contradiction, from which we conclude.
\end{proof}
A brief inspection of the proof in Claim~\ref{clm:regular} yields that the extremal graph has exactly one vertex of degree $2$ and $14$ vertices of degree $3$.
Let the vertex of degree $2$ be $w$. Let its two neighbours be $u$ and $v$ and then $u_i,v_i,u_{ij},v_{ij}$ for $i,j \in \{0,1\}$ are defined as before.
Noting that every $u_{ij}$ has two neighbours of the form $v_{0k}$ and $v_{1 \ell}$ where $k, \ell \in \{0,1\}$, one can check that there is a unique extremal example with respect to Theorem~\ref{thm:cubic}, namely, the point--line incidence graph of the Fano plane, in which exactly one edge is subdivided.
\subsection*{Acknowledgement}
We are grateful to the anonymous referees for their helpful comments and suggestions.
\bibliographystyle{abbrv
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{
"timestamp": "2021-12-13T02:14:03",
"yymm": "2103",
"arxiv_id": "2103.11898",
"language": "en",
"url": "https://arxiv.org/abs/2103.11898"
}
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"\\section{Introduction}\n\nThe extraction of meaningful intermediate features in classification pro(...TRUNCATED)
| {"timestamp":"2021-06-14T02:18:47","yymm":"2103","arxiv_id":"2103.11888","language":"en","url":"http(...TRUNCATED)
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"\\section{SignalBackgrounds} \n{\\it Neutrino magnetic moment signals and relevant backgrounds} - S(...TRUNCATED)
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"\\section{Introduction}\n\nPlease follow the steps outlined below when submitting your manuscript t(...TRUNCATED)
| {"timestamp":"2021-03-23T01:35:54","yymm":"2103","arxiv_id":"2103.11833","language":"en","url":"http(...TRUNCATED)
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"\\section{Introduction}\\label{sec1}\r\n\r\nThroughout this paper we work over a fixed algebraicall(...TRUNCATED)
| {"timestamp":"2021-03-23T01:35:15","yymm":"2103","arxiv_id":"2103.11813","language":"en","url":"http(...TRUNCATED)
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"\\section{Introduction}\n}{}\n\n\nThe off-resonant interface of light with atomic ensembles has bee(...TRUNCATED)
| {"timestamp":"2021-03-23T01:32:21","yymm":"2103","arxiv_id":"2103.11729","language":"en","url":"http(...TRUNCATED)
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"\\section{Introduction}\n\nRecent advances in generative adversarial networks (GANs) \\cite{goodfel(...TRUNCATED)
| {"timestamp":"2021-03-23T01:38:28","yymm":"2103","arxiv_id":"2103.11897","language":"en","url":"http(...TRUNCATED)
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