text:
\begin{definition}[Definition:Subset/Superset]
If $S$ is a [[Definition:Subset|subset]] of $T$, then $T$ is a '''superset''' of $S$.
This can be expressed by the notation $T \supseteq S$.
This can be interpreted as '''$T$ includes $S$''', or (more rarely) '''$T$ contains $S$'''.
Thus $S \subseteq T$ and $T \supseteq S$ mean the same thing.
\end{definition}

config: wiki


TITLE: Is square root of $y^2$ for every $y>0,y\in\mathbb{R}$?
QUESTION [2 upvotes]: I mean I got really confused by this question,  Let's assume that we are considering $y=-2$ so $y^2 = 4$ and so its square root is $2$, but we also have an exponential operation where, $(a^m)^n=a^{mn}$. So by using that when $y=-2, (y^2)^\frac12=y^{\left(2\cdot\frac12\right)}=y=-2$ so, basically $2=-2$? What is wrong here? Will that exponent operation not work in this operation?

REPLY [4 votes]: What's wrong is that the equality$$a^{xy}=(a^x)^y$$doesn't hold without restrictions. It holds when $a>0$ and $x,y\in\mathbb R$, but not in general.


TITLE: Converting a set to a tuple?
QUESTION [3 upvotes]: Okay, so, let's say I have a set:
$\{0,1,2,3\}$
And I want to convert it to a tuple:
$(0,1,2,3)$
How would I do this? Would it be as simple as:
$f(\{0,1,2,3\}) = (0,1,2,3)$
??

REPLY [3 votes]: Converting $\{0,1,2,3\}$ to $(0,1,2,3)$ is not well defined since sets are unordered but tuples are. This means $(3,2,1,0)$ would be an equally valid image for the original set.
A way to avoid this problem is to let each position in the tuple designate a particular element and use either a 0 to denote that element is absent from the set and a 1 to denote that it is present in the set.
For example, if your universal set is $\{0, 1, \dots, 5\}$, then $\{0,1,2,3\}$ would map uniquely to $(1,1,1,1,0,0)$ (if we make the appropriate assumption that position zero in the tuple corresponds to the element $0$ from the set, and so forth).
Defining the function in this way has the added advantage that it is a bijection between the power set of your universal set and the collection of all binary tuples of the appropriate length.


TITLE: How to prove this matrix identity?
QUESTION [0 upvotes]: This was a single step in a derivation, so I'm assuming there is a way to "see" this without writing down the expression for each entry:
$$\sum\limits_{i=1}^n \left(x_i-\frac{1}{n}\sum\limits_{i=1}^n x_i \right) \left(x_i-\frac{1}{n}\sum\limits_{i=1}^n x_i \right)^T = XHX^T$$
where
$$X = [x_1 x_2 \cdots x_n] \quad H=I-n^{-1}\mathbb{1}_{n \times n}$$
where $\mathbb{1}_{n \times n}$ is the $n \times n$ matrix of ones.

REPLY [2 votes]: It's better to use different indices in different sums. Expand
$$\sum\limits_{i=1}^n \left(x_i-\frac{1}{n}\sum\limits_{j=1}^n x_j \right) \left(x_i-\frac{1}{n}\sum\limits_{k=1}^n x_k \right)^T $$
as
$$\sum\limits_{i=1}^n  x_i x_i^T -\frac{1}{n} \sum\limits_{i=1}^n \sum_{k=1}^n x_i x_k^T -\frac{1}{n} \sum\limits_{i=1}^n \sum_{j=1}^n x_j x_i^T +\frac{1}{n^2} \sum\limits_{i=1}^n \sum\limits_{j=1}^n \sum_{k=1}^n x_j x_k^T  $$
and then get rid of extraneous indices:
 $$\sum\limits_{i=1}^n  x_i x_i^T -\frac{2}{n} \sum\limits_{i=1}^n \sum_{k=1}^n x_i x_k^T   +\frac{n}{n^2}   \sum\limits_{j=1}^n \sum_{k=1}^n x_j x_k^T  $$
This further simplifies to
 $$\sum\limits_{i=1}^n  x_i x_i^T -\frac{1}{n} \sum\limits_{i=1}^n \sum_{k=1}^n x_i x_k^T    $$
  where $I-n^{-1}1_{n\times n}$ is clearly seen.


TITLE: Induced orientation on the boundary of a $1$-dimensional manifold
QUESTION [0 upvotes]: Consider a $1$-dimensional smooth compact manifold with boundary $M$. Since $M$ is $1$-dimensional, then $\partial M$ is a $0$-dimensional manifold and it will have an induced orientation. I want to clarify some things about this. An orientation for a $0$-dimensional manifold is simply a function $\epsilon:\partial M\rightarrow \{\pm 1\}$ . Now given an outward-pointing vector field $Y$ of $M$ is the induced orientation $\epsilon$ on $\partial M$ the one such that $\epsilon \cdot Y$ is positively oriented with respect to the orientation of $M$. From the definitions of induced orientation that I found, from the book "Introduction to manifolds" I belive this will be it but I wanted discuss this with more detail.
Any insight is appreciated, thanks in advance.

REPLY [3 votes]: If $Y$ is a (non-vanishing) outward-pointing vector field on $\partial M$, it defines an orientation on $\partial M$ exactly as you outlined: for $p \in \partial M$, define $\varepsilon (p) = 1$ if and only if $(Y(p))$ is a positively oriented basis of $T_pM$ with respect to the orientation of $M$.


text:
\section{Examples}\label{s.ej}

We provide here a list of relevant examples concerning our main result Theorem \ref{mainthm}.
They are based on J. Kwapisz constructions \cite{Kw1}, where axiom A diffeomorphisms on
$\mathbb{T}^2$ having prescribed rational polygons as rotation sets are obtained. The key point in these examples is to construct horseshoe-like invariant sets whose rotation set realizes a prescribed polygon, and then extend the dynamics in the complement avoiding new rotational information. In picture \ref{toro}
we represent this situation, where the horseshoe-like basic piece named by $\Lambda$ contains three points  $a,b,\ c$ having rotation vectors $(0,0),(1,0),\ (1,1)$ respectively for some lift $F$ of $f$, which implies that the triangle generated by these vectors is contained in the possibly bigger convex set $\rho(F)$. As shown in the picture, the basic piece $\Lambda$ is always related to a sink $p$ (and the same holds for some source). Moreover, the example can be constructed
so that $\Lambda$ is the unique saddle basic piece. The disk $D$ in the picture is assumed
to be mapped into its own interior.

\tiny
\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=5cm]{toro.pdf}}
\caption{Representation of Kwapisz's example.}
\label{toro}
\end{center}\end{figure}

\normalsize

\subsection{Non-convex rotation set}

In this paragraph we present an example of a homeomorphism on the double-torus $\Sigma_2$ having a non convex rotation set, in contrast to the Misiurewicz-Ziemian result for $\mathbb{T}^2$. For this, we modify the example above so that the disk $D$ remains invariant. We call this example by $f_1$ and consider it defined on a torus $T$, and consider a copy of its inverse $f_2$ defined on a second torus $T'$. As mentioned above, the maps $f_1$, $f_2$ have unique saddle basic pieces $\Lambda_{f_1},\ \Lambda_{f_2}$ respectively, whose rotation sets equal the rotation set of the respective maps and have non-empty interior.

\smallskip

Then we can consider the connected sum $\Sigma_2$ of $T$ and $T'$ by gluing the boundary of the invariant disks, and the maps $f_1,f_2$ induce a natural homeomorphism $f:\Sigma_2\to\Sigma_2$ represented in the following picture.

\tiny

\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=4cm]{bitoroejemplo1.pdf}}
\caption{An example having non-convex rotation set.}
\label{toro1}
\end{center}\end{figure}

\normalsize

The two connected components that we called $T,T'$ of $\Sigma_2\setminus \partial D$ are
$f$-invariant, and we can name them so that $\rho_{\Lambda_{f_1}}(f)$ is contained in $H_1(T)$ and
$\rho_{\Lambda_{f_2}}(f)$ is contained in $H_1(T')$. These sets, $\rho_{\Lambda_{f_1}}(f)$ and $\rho_{\Lambda_{f_2}}(f)$, are, in light of Lemma \ref{l.conleycasesess}, two convex sets of full dimension in $H_1(T)$ and $H_1(T')$ respectively, which are complementary sub-spaces of dimension two in $H_1(\Sigma_2)$. Furthermore, by construction we have that
$$\rho(f)=\rho_{\Lambda_{f_1}}(f)\cup\rho_{\Lambda_{f_2}}(f),$$
otherwise one could find rotation vectors for the Kwapisz example which are not realized on the saddle basic piece. Thus we conclude that $\rho(f)$ is not convex.

\subsection{An example having a full-dimensional Polyhedra as rotation set}

We now modify our last example to obtain $g\in\textrm{Homeo}_0(\Sigma_2)$ having
a full dimensional rational polyhedra as rotation set. For this consider the map $f$ of the
last example. We can assume that the construction of $f$ implies a decomposition of its non-wandering set given by the two basic pieces $\Lambda_1=\Lambda_{f_1},\Lambda_2=\Lambda_{f_2}$ and a family of sinks
and sources given by contractible fixed points, which in particular have null rotation vectors.

\smallskip

Then composing with a time-one map $g_1$ of a flow supported on an annular neighbourhood of $\partial D$, we can construct a map $g=g_1\circ f\in\textrm{Homeo}_0(\Sigma_2)$ being a fitted axiom A diffeomorphism so that

\begin{itemize}

\item $g(\partial D)\subset T'$,

\item $\Omega(g)=\Omega(f)\setminus \partial D$,

\item $\Lambda_1\prec\Lambda_2$.

\end{itemize}

\tiny

\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=3.5cm]{bitoroejemplo2.pdf}}
\caption{The map $g$ having a full dimensional rational polyhedra as rotation set.}
\label{bitoro2}
\end{center}\end{figure}

\normalsize

The map $g$ is represented in Figure \ref{bitoro2}. Thus in light of Theorem \ref{t.rhorhochain} we have that
$$\rho(g)=\textrm{conv}(\rho_{\Lambda_1}(g)\cup\rho_{\Lambda_2}(g))$$
which is a full dimensional rational polyhedra in $\R^4$.

\subsection{An example concerning the intersection of the convex blocks of $\rho(f)$}

As presented in our main result Theorem \ref{mainthm}, the rotation set of $f\in\mathcal{A}_0(\Sigma_g)$ is given by a union of convex sets
$C_1,\ldots,C_N\subset \R^{2\textrm{g}}$ all of them containing $0$ where $N$ is bounded by $\textrm{c}(\textrm{g})=2^{5\textrm{g}-3}$. Moreover, each $C_i$ can be characterized as the maximal elements with respect to the inclusion in the family of convex sets included in $\rho(f)$. We
call them the basic convex blocks of $\rho(f)$.

Our next example shows that these basic blocks can have a non-trivial intersection, for instance intersecting in a full dimensional set relative to each basic block. We support the example in Figure \ref{bitoro3} where we represent the map $h$ that we want to construct.

\tiny

\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=3.5cm]{bitoroejemplo3.pdf}}
\caption{The map $h$ having non-trivial intersection between blocks.}
\label{bitoro3}
\end{center}\end{figure}
\normalsize

The axiom A diffeomorphism $h$ is intended to have three annular regions $A,B,C$, as shown in the figure, whose images under $h$ are contained in their own interior, defining respectively the attractors $\Omega_A,\Omega_B,\Omega_C$. The two connected
components of $\Sigma_2\setminus(A\cup B\cup C)$, denoted by $S_1,S_2$, contain horseshoe-like basic pieces $\Lambda_1,\Lambda_2$ which carry all the rotational information on each surface given by the convex rotation sets $C_1,C_2$ contained in $V=\textrm{H}_1(S_1)=\textrm{H}_1(S_2)$, which are full dimensional with respect to $V$. Due to the existence of the attracting annular regions one can choose $C_1$ and $C_2$ independently. This last construction can be achieved
by the same techniques employed by Kwapisz for the toral case.

Further, one can manage the construction so that the points of $\Omega_A,\Omega_B,\Omega_C$
which are heteroclinically related to either $\Lambda_1$ or $\Lambda_2$ have null rotation
vector. Thus in light of Theorem \ref{t.rhorhochain} we have that the rotation set of $h$ is given by
$$\rho(h)=C_1\cup C_2 \cup I_A\cup I_B \cup I_C $$
where
\begin{itemize}

\item $C_1$ and $C_2$ are full dimensional convex set of $V$ containing zero, whose interiors intersect,

\item $I_A$, $I_B$ and $I_C$ are intervals contained in $\textrm{H}_1(A),\textrm{H}_1(B),
\ H_1(C)$ respectively, containing zero.

\end{itemize}
We represent this situation in Figure \ref{inter}.
\tiny
\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=6.5cm]{inter.pdf}}
\caption{The rotation set of the map $h$.}
\label{inter2}
\end{center}\end{figure}
\normalsize

\subsection{The number of convex blocks of $\rho(f)$ may be exponential on $g$.} \label{s.ejemploexp}

We present a family of examples of $f\in\mathcal{A}_0(\Sigma_g)$ for $g>1$, satisfying that the number of basic convex blocks of $\rho(f)$ grows exponentially with $g$.

First we need to consider a specific Kwapisz map $h$ on an annulus $A$. Write $\partial A = \gamma^-\cup\gamma^+$  as union of its connected components.

A simple modification of the construction of \cite{Kw1} yields an axiom A diffeomorphism $h$ of $A$ with the following conditions:
\begin{enumerate}
\item $\partial A$ is fixed, $\gamma^+$ is an attractor and $\gamma^-$ a repellor.
\item There is a horseshoe-like basic piece $\Lambda$ so that $\rho_{\Lambda}(h)$ is an interval.
\item $W^s(\Lambda,h)$ accumulates on $\gamma^-$, and $W^u(\Lambda,h)$ accumulates on $\gamma^+$.
\item We may also require that all the non trivial rotation comes from $\Lambda$, and that $0\in \rho(h) = \rho_{\Lambda}(h)$. (This last point is not strictly necessary, but simplifies notations).

\end{enumerate}

Let $k=\left[ \frac{g}{2}\right]$, and consider the surface $\Sigma_g$ as shown in Figure
\ref{exponentialexample1} where for each $i=1,\ldots,k$ we define the essential curves
$\beta_i$, the non-separating annuli $A^0_i$, and $A^1_i$, and the separating annuli $A^*_i$.

\tiny
\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=5cm]{exexampletop.pdf}}
\caption{Decomposition of $\Sigma_g$ that will be used for our examples.}
\label{exponentialexample1}
\end{center}\end{figure}
\normalsize

Let us define $f_0\in\textrm{Homeo}_0(\Sigma)$ so that:

\begin{itemize}

\item $f_0$ is a copy of $h$ on each annuli $A^0_i$, $A^1_i$, and $A^*_i$ for $i=1,\ldots,k$, where the boundary components that appear on the left are identified with $\gamma^-$, and those on the right with $\gamma^+$.

\item In the complement of these annuli $f_0$ is given by a Morse-Smale
diffeomorphism $h_0$ satisfying the following properties:

\begin{itemize}
\item The non-wandering set of $h_0$ consists of finitely many irrotational fixed points.

\item For every $i=1,\ldots,k$, the curve $\beta_i$ is mapped towards its right side.

\item For every $i=1,\ldots,k$ and $j=0,1$ there is a point $x_{i,j}$ whose
$\alpha$-limit is contained in the right boundary component of $A_i^j$ and whose $\omega$-limit
is contained in the left boundary component of $A_i^*$.

\item For every $i=1,\ldots,k-1$ and $j=0,1$ there is a point $z_{i,j}$
whose $\alpha$-limit is contained in the right boundary component of $A_i^*$ and whose
$\omega$-limit is contained in the left boundary component of $A_{i+1}^j$.

\end{itemize}
\end{itemize}

Notice that the last two properties of $h_0$ imply that the boundary components of the annuli $A_i^j$, $A_i^*$ are not attractors nor repellors for $f_0$, so one can actually assume that they attract from their left side and repel from their right side.

Under these circumstances, one can modify the map $f_0$ by composing
with a time-one map $g_1$ of a flow whose support is contained in annular neighbourhoods of $\partial A^0_i$, $\partial A^1_i$, and $\partial A^*_i$ for $i=0,\ldots,k$, so that each of these boundary curves is mapped to its right. We assume that the support of $g_1$ is disjoint from the curves $\beta_i$, and intersects $\Omega(f_0)$ only on $\bigcup_{i=1}^k(\partial A^0_i\cup \partial A^1_i\cup \partial A^*_i) $. This way
we obtain $f=g_1\circ f_0 \in\mathcal{A}_0(\Sigma_g)$ represented in the next Figure,
and having the following features.

\tiny
\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=5cm]{exponentialexample.pdf}}
\caption{The map $f\in\mathcal{A}_0(\Sigma_g)$. The small arrows indicate
where $f$ maps the free curves given by $\partial A_i^j,\partial A_i^*$, and $\beta_i$.}
\label{exponentialexample}
\end{center}\end{figure}
\normalsize

\begin{enumerate}
\item $\Omega(f) = \Omega(f_0) \setminus \bigcup_{i=1}^k(\partial A^0_i\cup \partial A^1_i \cup \partial A^*_i)$.

\item $\rho_{\Lambda^j_i}(f)=\rho_{\Lambda^j_i }(f_0)$ is an interval containing $0$ in $H_1(A^j_i)\cong\R$, for $i=1,\ldots,k$, $j=0,1$.

\item $\Lambda_i^j\prec\Lambda_i^*$ for each $i=1,\ldots,k$, $j=0,1$.

\item $\Lambda_i^*\prec \Lambda_{i+1}^j$ for each $i=1,\ldots,k-1$, $j=0,1$.

\item For each $i=1,\ldots,k$, the basic pieces $\Lambda_{i}^0$ and $\Lambda_i^1$ are not
heteroclinically related. In order to ensure this point, one uses the fact that $f$
sends every curve $\beta_i$ to its right.

\end{enumerate}

The last three properties say that the graph $\Gamma$
given in Figure \ref{graph} is a full subgraph of $\mathcal{G}_f$, i.e. all the heteroclinical relations among the pieces $\Lambda^0_i$, $\Lambda^1_i$ and $\Lambda^*_i$, (for $i=1,\ldots,k$), arise from following the arrows of the directed graph $\Gamma$.

\tiny
\begin{figure}[ht]\begin{center}

\centerline{\includegraphics[height=3cm]{graphexpex.pdf}}
\caption{The graph $\Gamma$.}
\label{graph}
\end{center}\end{figure}
\normalsize

Since the basic pieces of $f$ that are not in $\Gamma$ have trivial rotation, and induce no other heteroclinical connections among the pieces in $\Gamma$, we see that $$\rho(f) = \bigcup_{\mathcal{C}}\rho_{\mathcal{C}}(f)$$ for $\mathcal{C}$ ranging over the maximal chains of $\Gamma$. We can index such chains by all the functions $$\xi:\{1,\ldots,k\}\to\{0,1\}, \quad \text{where}\quad \mathcal{C}(\xi) = (\Lambda_1^{\xi(1)}\prec\Lambda_1^{*}\prec\cdots\prec\Lambda_k^{\xi(k)}\prec\Lambda_k^{*} )$$
Let $C_{\xi} = \rho_{\mathcal{C}(\xi)}(f)$. For $i=1,\ldots,k$, notice that $H_1(A_i^*)=\{0\}$ and choose generators $[\alpha_i^j]$ of  $H_1(A_i^j) \cong\R$ for $j=0,1$. Notice from Figure \ref{exponentialexample1} that $\{[\alpha_i^j]:i=1,\ldots,k;j=0,1\}$ is a linearly independent set of $H_1(\Sigma_g;\R)$, so we get that

\begin{itemize}
\item by item (2) above, $C_{\xi}$ is a $k$-dimensional simplex in the subspace $V_{\xi}$ (of dimension $k$) spanned by $\{[\alpha_i^{\xi(i)}]:i=1,\ldots,k\}$,

\item and if $\xi\neq \xi'$, then  $V_{\xi}$ and $V_{\xi'}$ intersect in a subspace of dimension at most $k-1$.
\end{itemize}

Thus the sets $C_{\xi}$ for $\xi:\{1,\ldots,k\}\to\{0,1\}$ are the basic convex blocks of $\rho(f)$, and there are $2^k \sim 2^{g/2}$ of them, which shows the desired property for this family of examples.

config: arxiv


TITLE: Finding the minimum amount?
QUESTION [0 upvotes]: So let's say a company sells life insurance policies, which require customers to pay an annual fee. Assume that if the customer happens to die in that year, the company does not charge the customer's beneficiaries the fee and will also pay out the amount that the policy is worth. If the average customer has a 2.5% chance of dying in the next year and the policy is worth $100,000 then what is the minimum amount the company should charge for the yearly payment so that on average they won't lose any money?
I'm having trouble starting out this question. I've learned about things like how to calculate profit however those questions give me the mean and standard deviation and this one doesn't.

REPLY [1 votes]: Hint:  The company has a $2.5\%$ chance of losing $\$100,000$ and a $97.5\%$ chance of collecting the fee $f$.  You want the expectation to be zero.

REPLY [0 votes]: If 2.5% of people will not be paying the fee, and will be receiving $100 000, then we have 97.5% of people who will be paying.
What we need is that the money payed out equals the money taken in :
100000 * 0.025 = 0.975 * p

Solving for the premium (p) we get :
2500         = 0.975p
2564.1025641 = p

The company should therefore charge a minimum of 2564.11 (rounding up) to not lose money.


TITLE: A special class of commutative rings
QUESTION [2 upvotes]: Can we characterize all commutative finite rings the sum of all elements is zero? For instance, all odd characteristic finite commutative rings have this property (easy). GF(2) does not.

REPLY [6 votes]: Since you're only using addition, you might as well ask this for abelian groups instead of rings.  The sum of all the elements of a finite abelian group is equal to the sum of elements of order $2$ (since the others cancel their inverses), which (together with $0$) form a subgroup which is a vector space over $\mathbb{F}_2$.  The sum of all elements of a finite-dimensional vector space $V$ over $\mathbb{F}_2$ is $0$ iff the dimension is not equal to $1$ (there are various ways you can see this; for instance, the sum must be invariant under all invertible linear maps $V\to V$, and we can map any nonzero element of $V$ to any other by an invertible linear map, so if the sum is nonzero there can only be one nonzero element).
So we conclude that the sum of the elements in a finite abelian group is nonzero iff it has exactly one element of order $2$, in which case the sum is that element of order $2$.  In terms of the classification of finite abelian groups, this means that when you write the group as a direct sum of cyclic groups, there is exactly one summand whose order is even.


TITLE: Proof by contradiction involving set theory
QUESTION [0 upvotes]: Using a contradiction, prove the following:
If $S\cap T = \emptyset$ and $S\cup T = T$, then $S = \emptyset$.

So far, I've written the definitions of the intersection and union, and I've assumed that $S\neq\emptyset$, but I'm not sure how to derive the contradiction.
Any advice would be greatly appreciated.

REPLY [1 votes]: Without contradiction:
$$
S=S\cap(S\cup T)=S\cap T=\emptyset
$$
The first equality follows from $S\subseteq S\cup T$.
If you really need a proof by contradiction, then the path is almost obvious: negating $S=\emptyset$ means there exists $x\in S$. Since $T=S\cup T$ and $x\in S$, we get $x\in T$. But then $x\in S\cap T$: contradiction.


text:
\begin{document}

\begin{frontmatter}

\title{ Regularized finite difference methods for the logarithmic Klein-Gordon equation }

\author[nudt,nus]{Jingye Yan}
\ead{yanjingye0205@163.com}
\author[nudt]{Hong Zhang}
\ead{zhanghnudt@163.com}

\author[nudt]{Xu Qian\corref{cor1}}
\ead{qianxu@nudt.edu.cn}
\author[nudt,hpc]{Songhe Song}
\ead{shsong@nudt.edu.cn}

\cortext[cor1]{Corresponding author.}

\address[nudt]{Department of Mathematics, College of Arts and Sciences, National University of Defense Technology, Changsha 410073, China}
\address[nus]{Department of Mathematics, National University of Singapore, Singapore 119076}

\address[hpc]{State Key Laboratory of High Performance Computing, National University of Defense Technology, Changsha 410073, China}

\begin{abstract}
We propose and analyze two regularized finite difference methods for the logarithmic Klein-Gordon equation (LogKGE). Due to the blowup phenomena caused by the logarithmic nonlinearity of the LogKGE, it is difficult to construct numerical schemes and establish their error bounds. In order to avoid singularity, we present a regularized logarithmic Klein-Gordon equation (RLogKGE) with a small regularized parameter $0<\varepsilon\ll1$. Besides, two finite difference methods are adopted to solve the regularized logarithmic Klein-Gordon equation (RLogKGE) and rigorous error bounds are estimated in terms of the mesh size $h$, time step $\tau$, and the small regularized parameter $\varepsilon$. Finally, numerical experiments are carried out to verify our error estimates of the two numerical methods and the convergence results from the LogKGE to the RLogKGE with the linear convergence order $O(\varepsilon)$.
\end{abstract}

\begin{keyword}
logarithmic Klein-Gordon equation; regularized logarithmic Klein-Gordon equation; finite difference method; error estimate; convergence order
\end{keyword}

\end{frontmatter}

\section{Introduction}

The logarithmic  Klein-Gordon equation (LogKGE) known as the relativistic version of the logarithmic Schr\"{o}dinger equation \cite{bartkowski2008one} has been introduced in the quantum field theory by Rosen \cite{rosen1969dilatation} and has the form
\begin{align}\label{LogKGE}
\left\{
\begin{aligned}
&u_{tt}(\mathbf{x},t)-\Delta u(\mathbf{x},t)+u(\mathbf{x},t)+\lambda u(\mathbf{x},t)\ln(|u(\mathbf{x},t)|^{2})=0, ~~\mathbf{x}\in\mathbb{R}^{d}, t>0,\\
&u(\mathbf{x},0)=\phi(\mathbf{x}),~~~\partial_{t}u(\mathbf{x},0)=\gamma(\mathbf{x}),~~~\mathbf{x}\in\mathbb{R}^{d},
\end{aligned}
\right.
\end{align}
where $\mathbf{x}=(x_{1},\ldots,x_{d})^{\mathrm{T}}\in \mathbb{R}^{d}, (d =1, 2, 3)$ is the spatial coordinate, $t$ is time, $u:=u(\mathbf{x},t)$ is a real-valued scalar field, the parameter $\lambda$ measures the force of the nonlinear interaction.
 This kind of nonlinearity frequently appears in inflation cosmology and supersymmetric field theories \cite{rosen1969dilatation,enqvist1998q,linde1992strings}.

 The LogKGE (\ref{LogKGE}) has been used to describe the spinless particle \cite{sakurai1967advanced} in optics, nuclear physics and geophysics \cite{buljan2003incoherent,de2003logarithmic,hefter1985application,krolikowski2000unified}. Assume that $u(\cdot,t)\in H^{1}(\mathbb{R}^{d})$ and $\partial_{t}u(\cdot,t)\in L^{2}(\mathbb{R}^{d})$,
the LogKGE (\ref{LogKGE}) admits the energy conservation law \cite{masmoudi2002nonlinear,machihara2001nonrelativistic}, which is defined as:
\begin{align}
E(t) =\int_{\Omega}\left[ (u_{t}(\mathbf{x},t))^{2}+(\nabla u(\mathbf{x},t))^{2}+(1-\lambda)u^{2}(\mathbf{x},t)+\lambda u^{2}(\mathbf{x},t) \ln \left(|u(\mathbf{x},t)|^{2} \right) \right] \mathrm{d} x \equiv E(0). \label{energy}
\end{align}

The Klein-Gordon equation with logarithmic potentials posses some special analytical solutions in quantum mechanics \cite{maslov1990pulsons,bialynicki1979gaussons,koutvitsky2006instability}, and the existence of classical solutions and weak solutions have been investigated in \cite{bartkowski2008one,gorka2009logarithmic}. In the paper \cite{bialynicki1979gaussons}, the author studies the solutions named Gaussons which represent solutions of Gaussian shape  \cite{wazwaz2016gaussian}. Besides, the interaction of Gaussons has been introduced in \cite{makhankov1981interaction}. For the nonlinear Klein-Gordon equation (NKGE) and the oscillatory NKGE,  various analysis and numerical results have been represented in literature.
Along the mathematical front, the derivation, Cauchy problem, well-posedness and dynamical properties have been proposed in \cite{bainov1995nonexistence,brenner1981global,ibrahim2006global,kosecki1992unit,simon1993cauchy} and the references therein.
Along the numerical aspects, a surge of efficient and accurate numerical methods have been proposed and analyzed for the nonlinear Klein-Gordon equation (NKGE) and the oscillatory NKGE in the literature. For example, the standard finite difference time domain (FDTD) methods including energy conservative /semi-implicit /explicit finite difference time domain methods \cite{bao2012analysis,chang1991conservative,duncan1997sympletic,zhang2005convergence,bao2019long}, multiscale time integrator Fourier pseudospectral (MWI-FP) method \cite{bao2014uniformly}, finite element method \cite{cao1993fourier}, exponential wave integrator Fourier pseudospectral (EWI-FP) method \cite{bao2012analysis,bao2013exponential}, asymptotic preserving (AP) \cite{faou2014asymptotic} method, ect.  For numerical comparisons of different numerical methods of the NKGE and the oscillatory NKGE, we refer to \cite{bao2012analysis,jimenez1990analysis,pascual1995numerical,bao2019comparison}. However, due to the singularity of the logarithmic nonlinearity at the origin, these methods can not be applied to the LogKGE (\ref{LogKGE}) equation directly.

In order to avoid blowup of the LogKGE (\ref{LogKGE}), i.e., $\log |u|\rightarrow -\infty, |u|\rightarrow 0$, we consider a regularized logarithmic  Klein-Gordon equation (RLogKGE) with a small regularized parameter $0<\varepsilon \ll 1$,
\begin{align}\label{RLogKGE}
\left\{
\begin{aligned}
&u^{\varepsilon}_{tt}(\mathbf{x},t)-\Delta u^{\varepsilon}(\mathbf{x},t)+u^{\varepsilon}(\mathbf{x},t)+\lambda u^{\varepsilon}(\mathbf{x},t)\ln\left(\varepsilon^{2}+ \left( u^{\varepsilon}(\mathbf{x},t) \right)^{2} \right)=0, ~~\mathbf{x}\in\mathbb{R}^{d}, t>0,\\
&u^{\varepsilon}(\mathbf{x},0)=\phi(\mathbf{x}),~~~\partial_{t}u^{\varepsilon}(\mathbf{x},0)=\gamma(\mathbf{x}),~~~\mathbf{x}\in\mathbb{R}^{d}.
\end{aligned}
\right.
\end{align}
The above RLogKGE (\ref{RLogKGE}) is time symmetric or time reversible, i.e., they are invarient if interchanging $n+1\leftrightarrow n-1$ and $\tau \leftrightarrow -\tau$.
\begin{rem}
The Cauchy problem of the LogKGE (\ref{LogKGE}) and the RLogKGE (\ref{RLogKGE}), the convergence estimate between the regularized model (\ref{RLogKGE}) and the LogKGE (\ref{LogKGE}) will be represented in another paper.
\end{rem}
\begin{thm}
Assume $u^{\varepsilon}(\cdot,t)\in H^{1}(\mathbb{R}^{d})$ and $\partial_{t}u^{\varepsilon}(\cdot,t)\in L^{2}(\mathbb{R}^{d})$,
the RLogKGE (\ref{RLogKGE}) conserves energy conservation law
, which is defined as:
\begin{align}
E^{\varepsilon}(t) =\int_{\Omega}\left[ (u^{\varepsilon}_{t}(\mathbf{x},t))^{2}+(\nabla u^{\varepsilon}(\mathbf{x},t))^{2}+(u^{\varepsilon}(\mathbf{x},t))^{2}+\lambda F_{\varepsilon} \left( (u^{\varepsilon}(\mathbf{x},t))^{2} \right) \right] \mathrm{d} x \equiv E^{\varepsilon}(0), \label{energyr}
\end{align}
where $F_{\varepsilon}(\rho)=\int_{0}^{\rho} \ln (\varepsilon^{2}+s) \mathrm{d} s= \rho\ln (\varepsilon^{2}+\rho)+\varepsilon^{2}\ln (1+\frac{\rho}{\varepsilon^{2}})- \rho,   \quad \rho=(u^{\varepsilon}(\mathbf{x},t))^{2}.$
\end{thm}
\begin{proof}
\begin{align}
\begin{split}
\frac{d}{dt}E^{\varepsilon}(t)&=2\int_{\Omega}\left[ u^{\varepsilon}_{t}\cdot u^{\varepsilon}_{tt}+\nabla u^{\varepsilon}\cdot \nabla u^{\varepsilon}_{t}+u^{\varepsilon}u^{\varepsilon}_{t}+\lambda F^{'}_{\varepsilon}\left((u^{\varepsilon})^{2}\right)\cdot u^{\varepsilon}\cdot u^{\varepsilon}_{t}\right](\mathbf{x},t) \mathrm{d} x\\
&=2\int_{\Omega} \left[ u^{\varepsilon}_{t} \left( u^{\varepsilon}_{tt}-\Delta u^{\varepsilon}+u^{\varepsilon}+\lambda u^{\varepsilon}\ln\left(\varepsilon^{2}+( u^{\varepsilon})^{2} \right) \right) \right] (\mathbf{x},t) \mathrm{d} x=0.
\end{split}
\end{align}
This ends the proof.
\end{proof}

The main purpose of this work is to analyze two FDTD schemes for the RLogKGE (\ref{RLogKGE}) and study the efficiency, accuracy between the LogKGE (\ref{LogKGE}) and the RLogKGE (\ref{RLogKGE}) as well as their numerical simulations.

The rest of this paper is organized as follows. In Section 2, a semi-implicit and an explicit FDTD schemes are proposed for the RLogKGE (\ref{RLogKGE}). Besides, we analyze the stability and solvability of the two schemes. The details of error analysis are established in Section 3. Section 4 is devoted to verifying our error etimates using the numerical experiments. At last, some concluding remarks are drawn in Section 5. Throughout this paper, we denote $p\lesssim q$ to represent that there exists a generic
constant $C$ which is independent of $\tau,h,\varepsilon$, such that $|p|\leq Cq$.

\section{FDTD methods and their stability}

In this section, we construct two FDTD schemes to approximate the RLogKGE (\ref{RLogKGE}) and study their stability, solvability and analyze their error estimates. For simplicity of notations, we set $\lambda=1$ and only make analysis and construct numerical schemes in one dimensional space ($d=1$) for the RLogKGE (\ref{RLogKGE}). When $d=1$, we truncate the RLogKGE (\ref{RLogKGE}) with periodic boundary conditions
\begin{align}\label{RLogKGE1}
\left\{
\begin{aligned}
&u^{\varepsilon}_{tt}(x,t)-\Delta u^{\varepsilon}(x,t)+u^{\varepsilon}(x,t)+ u^{\varepsilon}(x,t)\ln\left(\varepsilon^{2}+ \left( u^{\varepsilon}(x,t) \right)^{2} \right)=0, ~~x\in\Omega=(a,b), ~~t>0,\\
&u^{\varepsilon}(x,0)=\phi(x),~~~\partial_{t}u^{\varepsilon}(x,0)=\gamma(x),~~~x\in\overline{\Omega}=[a,b].
\end{aligned}
\right.
\end{align}

\subsection{FDTD methods}

Choose time step $\tau:=\Delta t$ and time steps $t_{n}:=n\tau, n=0,1,2,\ldots;$  let the mesh size $h:=\frac{b-a}{N}$ with $N$ being a positive integer and denote the grid points as $x_{j}:=a+jh,j=0,1,\ldots,N$.
Define the index sets as:
\begin{align}
\mathcal{T}_{N}=\left\{j|j=0,1,2,\ldots,N-1\right\},\quad \mathcal{T}_{N}^{0}=\left\{j|j=0,1,2,\ldots,N\right\}.
\end{align}
Assume $u^{\varepsilon,n}_{j}, u^{n}_{j}$ are the approximations of the exact solution $u^{\varepsilon}(x_{j},t_{n})$ and $ u(x_{j},t_{n})$, $j\in \mathcal{T}_{N}^{0}$ and $n\geq 0$. Define $u^{\varepsilon,n}=(u^{\varepsilon,n}_{0},u^{\varepsilon,n}_{1},\ldots,u^{\varepsilon,n}_{N})^{\mathrm{T}}, u^{n}=(u^{n}_{0},u^{n}_{1},\ldots,u^{n}_{N})^{\mathrm{T}} \in \mathbb{C}^{N+1} $ as the numerical solutions vector at time $t=t_{n}$. The followings are the finite difference operators:
\begin{align*}
&\delta^{+}_{t}u^{n}_{j}=\frac{u^{n+1}_{j}-u^{n}_{j}}{\tau},~~~~~\delta_{t}^{-}u^{n}_{j}=\frac{u^{n}_{j}-u^{n-1}_{j}}{\tau},~~~~\delta^{2}_{t}u^{n}_{j}=\frac{u^{n+1}_{j}-2u^{n}_{j}+u^{n-1}_{j}}{\tau^{2}},\\
&\delta^{+}_{x}u^{n}_{j}=\frac{u^{n}_{j+1}-u^{n}_{j}}{h},~~~~~~\delta^{-}_{x}u^{n}_{j}=\frac{u^{n}_{j}-u^{n}_{j-1}}{h},~~~~~\delta^{2}_{x}u^{n}_{j}=\frac{u^{n}_{j+1}-2u^{n}_{j}+u^{n}_{j-1}}{h^{2}}.\\

\end{align*}
We denote a space of grid functions
\begin{align}
X_{N}=\left\{ u| u=(u_{0},u_{1},u_{2},\ldots,u_{N})^{\mathrm{T}},u_{0}=u_{N},u_{-1}=u_{N-1}\right\}\subseteq \mathbb{C}^{N+1}.
\end{align}
We define the standard discrete $l^{2}$, semi-$H^{1}$ and $l^{\infty}$ norms and inner product over $X_{N}$ as follows
\begin{align}
\|u\|_{l^{2}}^{2}=(u,u)=h\sum \limits^{N-1}_{j=0}|u_{j}|^{2},~~\|\delta_{x}^{+}u\|^{2}_{l^{2}}=h\sum \limits^{N-1}_{j=0}|\delta_{x}^{+}u_{j}|^{2},~~\|u\|_{l^{\infty}}=\sup\limits_{0\leq j\leq N-1}|u_{j}|,~~(u,v)=h\sum \limits^{N-1}_{j=0}u_{j}v_{j},
\end{align}
where $u,v\in X_{N}$, and $(\delta^{2}_{x}u,v)=-(\delta^{+}_{x}u,\delta^{+}_{x}v)=(u,\delta^{2}_{x}v)$.
In the following, we introduce two frequently used FDTD methods for the RLogKGE (\ref{RLogKGE}):\bigskip

$\mathbf{I}$. Semi-implicit finite difference (SIFD) scheme
\begin{align}\label{SIFD2}
\delta_{t}^{2}u^{\varepsilon,n}_{j}-\frac{1}{2}\delta_{x}^{2} (u^{\varepsilon,n+1}_{j}+u^{\varepsilon,n-1}_{j})+\frac{1}{2} (u^{\varepsilon,n+1}_{j}+u^{\varepsilon,n-1}_{j})+u^{\varepsilon,n}_{j}f_{\varepsilon}\left( (u^{\varepsilon,n}_{j})^{2} \right)=0,~~n\geq1;
\end{align}

$\mathbf{II}$. Explicit finite difference (EFD) scheme
\begin{align}\label{EFD}
\delta_{t}^{2}u^{\varepsilon,n}_{j}-\delta_{x}^{2} u^{\varepsilon,n}_{j}+u^{\varepsilon,n}_{j}+ u^{\varepsilon,n}_{j}f_{\varepsilon}\left( (u^{\varepsilon,n}_{j})^{2} \right)=0,~~n\geq1;
\end{align}
where, $f_{\varepsilon}(\rho)=\ln (\varepsilon^{2}+\rho)$.
 The initial and boundary conditions are discretized as
\begin{align}\label{initialvalue1}
u^{\varepsilon,n+1}_{0}=u^{\varepsilon,n+1}_{N},u^{\varepsilon,n+1}_{-1}=u^{\varepsilon,n+1}_{N-1},~~n\geq0,~~u^{\varepsilon,0}_{j}=\phi(x_{j}),~~j\in \mathcal{T}_{N}^{0}.
\end{align}
Using the Taylor expansion we can get the first step solution $u_{j}^{\varepsilon,1}$,
\begin{align}\label{initialvalue}
u_{j}^{\varepsilon,1}=\phi(x_{j})+\tau\gamma(x_{j})+\frac{\tau^{2}}{2}\left[\delta^{2}_{x}\phi(x_{j})-\phi(x_{j})- \phi(x_{j})\ln(\varepsilon^{2}+(\phi(x_{j}))^{2}) \right], ~~j\in \mathcal{T}_{N}^{0}.
\end{align}
It is easy to prove that the above FDTD schemes are all time symmetric or time reversible.

\subsection{Stability analysis}

Let $0<T<T_{\max}$ with $T_{\max}$ being the maximum existence time. Define
\begin{align}
\sigma_{\max}:=\max\{|\ln(\varepsilon^{2})|,\left| \ln(\varepsilon^{2}+\|u^{\varepsilon,n}\|^{2}_{l^{\infty}}) \right|\},~~0\leq n\leq\frac{T}{\tau}-1.
\end{align}

According to the von Neumann linear stability analysis, we can get the following stability results for the FDTD schemes.
\begin{thm}\label{stability}
For the above FDTD schemes applied to the RLogKGE (\ref{RLogKGE}) up to $t=T$, we have:

(i)~When $-1\leq\sigma_{\max}\leq 1$, the SIFD scheme (\ref{SIFD2}) is unconditionally stable; and when $\sigma_{\max}> 1$, it is conditionally stable under the stability condition
\begin{align}\label{sifd2stability}
\tau\leq \frac{2}{\sqrt{\sigma_{\max}-1}}.
\end{align}

(ii)~The EFD scheme (\ref{EFD}) is conditionally stable under the stability condition
\begin{align}\label{efdstability}
\tau\leq \frac{2h}{\sqrt{(\sigma_{\max}+1)h^{2}+4}}.
\end{align}
\end{thm}
\begin{proof}
Substituting
\begin{align}
u^{\varepsilon,n-1}_{j}=\sum_{l}\hat{U}_{l}e^{2ijl\pi/N},~~u^{\varepsilon,n}_{j}=\sum_{l}\xi_{l}\hat{U}_{l}e^{2ijl\pi/N},~~u^{\varepsilon,n+1}_{j}=\sum_{l}\xi_{l}^{2}\hat{U}_{l}e^{2ijl\pi/N},~~
\end{align}
into (\ref{SIFD2})-(\ref{EFD}), where $\xi_{l}$ is the amplification factor of the $l$th mode in phase space, we can get the characteristic equation with the following structure
\begin{align}
\xi_{l}^{2}-2 \theta_{l} \xi_{l}+1=0, \quad l=-\frac{N}{2}, \ldots, \frac{N}{2}-1,
\end{align}
where $\theta_{l}$ is invarient with different methods. By the above equation, we get $\xi_{l}=\theta_{l}\pm \sqrt{\theta_{l}^{2}-1}$. The stability of numerical schemes amounts to
\begin{align}
\left|\xi_{l}\right| \leq 1 \Longleftrightarrow\left|\theta_{l}\right| \leq 1, \quad l=-\frac{N}{2}, \ldots, \frac{N}{2}-1.
\end{align}
Denote $s_{l}=\frac{2}{h} \sin \left(\frac{l \pi}{N}\right), \quad l=-\frac{N}{2}, \ldots, \frac{N}{2}-1$, we have
\begin{align}\label{stability1}
0\leq s_{l}^{2}\leq \frac{4}{h^{2}}.
\end{align}
Firstly, we prove linear stability. We assume $f_{\varepsilon}\left((u^{\varepsilon})^{2}\right)=\alpha$, and $\alpha$ is a constant satisfying $\alpha>-1$, then (\ref{SIFD2}) and (\ref{EFD}) are linear.

(i) For the SIFD scheme (\ref{SIFD2}), we have
\begin{align}
\theta_{l}=\frac{2- \alpha\tau^{2}}{2 +\tau^{2}\left(s_{l}^{2}+1\right)} , \quad l=-\frac{N}{2}, \ldots, \frac{N}{2}-1.
\end{align}
When $-1\leq\alpha\leq 1$, it implies that $|\theta_{l}| \leq 1$ and the SIFD scheme (\ref{SIFD2}) is unconditional stable. On the other hand, when $1<\alpha$, we have
\begin{align}

2-\alpha\tau^{2}\geq-2-\tau^{2},
\end{align}
it implies that, when $\tau\leq \frac{2}{\sqrt{\alpha-1}}$, the SIFD scheme (\ref{SIFD2}) is stable.

And when SIFD scheme is nonlinear, with the same method we can get
stability condition is
\begin{align}
\tau\leq \frac{2}{\sqrt{\sigma_{\max}-1}}.
\end{align}

(ii) For the EFD scheme (\ref{EFD}), we have
\begin{align}
\theta_{l}=\frac{2- \tau^{2}(1+\alpha+s_{l}^{2})}{2} , \quad l=-\frac{N}{2}, \ldots, \frac{N}{2}-1.
\end{align}
By (\ref{stability1}), we get
\begin{align}
\tau^{2}(\alpha+1+s_{l}^{2})\leq \tau^{2}(\alpha+1+\frac{4}{h^{2}})\leq 4, \Rightarrow |\theta_{l}|<1.
\end{align}
It implies that, when $\tau\leq \frac{2h}{\sqrt{(\alpha+1)h^{2}+4}}$, the EFD scheme (\ref{EFD}) is stable. Besides, when the EFD scheme (\ref{EFD}) is nonlinear, the stability condition is
\begin{align}
\tau\leq \frac{2h}{\sqrt{(\sigma_{\max}+1)h^{2}+4}}.
\end{align}
\end{proof}

\begin{rem}
Since the scheme SIFD (\ref{SIFD2}) is linear, and the coefficient matrix is strictly diagonal, it is easy to conclude that the SIFD (\ref{SIFD2}) is solvable. In addition, (\ref{EFD}) is explicit, so it is evident that there exists a unique solution.
\end{rem}

\section{Error esitimates}

\subsection{Main results}

Motivated by the analytical results in \cite{bao2019long,bao2012analysis,bao2019error,bao2018uniform,bao2018uniforme}, we will establish the error estimates of the FDTD schemes.

Assume that the solution $u^{\varepsilon}$ is smooth enough over $\Omega_{T}:\Omega\times [0,T],$ i.e.
\begin{align}
(A)~~~~~u^{\varepsilon}\in C\left( [0,T]; H^{5}(\Omega)\right)\cap C^{2}\left( [0,T]; H^{4}(\Omega)\right)\cap C^{4}\left( [0,T]; H^{2}(\Omega)\right),
\end{align}
and there exsit $\varepsilon_{0} >0$ and $C_{0}>0$ independent of $\varepsilon$ such that
\begin{align}
\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{5}(\Omega)\right)}+\left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{4}(\Omega)\right)}+\left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{2}(\Omega)\right)} \leq C_{0},
\end{align}
is valid uniformly in $0\leq \varepsilon \leq\varepsilon_{0}$.

Denote $\Lambda=\|u^{\varepsilon}(x,t)\|_{L^{\infty}(\Omega_{T})}$ and the grid `error' function $e^{\varepsilon,n} \in X_{N}~(n\geq0)$ as
\begin{align}
e^{\varepsilon,n}_{j}=u^{\varepsilon}(x_{j},t_{n})-u^{\varepsilon,n}_{j},~~j\in \mathcal{T}^{0}_{N},~~n=0,1,2,\ldots,
\end{align}
where $u^{\varepsilon}$ and $u^{\varepsilon,n}_{j}$ are the exact solution and numerical approximation of (\ref{RLogKGE1}) respectively.
\begin{thm}\label{SIFD2err}
Assume $\tau\lesssim h$ and under the assumption (A), there exist $h_{0}>0,\tau_{0}>0$  sufficiently small and independent of $\varepsilon,$ for any $0<\varepsilon \ll1$, when $0<h\leq h_{0}$ and $0<\tau\leq \tau_{0}$ and under the stability condition (\ref{sifd2stability}),  the SIFD (\ref{SIFD2}) with (\ref{initialvalue1}) and (\ref{initialvalue}) satisfies the following error estimates
\begin{align}\label{SIFD2error}

\|\delta_{x}^{+}e^{\varepsilon,n}\|_{l^{2}} + \|e^{\varepsilon,n}\|_{l^{2}}\lesssim e^{\frac{T}{2}(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2}),~~\|u^{\varepsilon,n}\|_{l^{\infty}}\leq \Lambda+1.
\end{align}
\end{thm}
\begin{thm}\label{EFDerr}
Assume $\tau \leq \frac{1}{2}\min\{1,h\}$ and under the assumption (A), there exist $h_{0}>0,\tau_{0}>0$  sufficiently small and independent of $\varepsilon,$  for any $0<\varepsilon \ll1$, when $0<h\leq h_{0}$ and $0<\tau\leq \tau_{0}$  and under the stability condition (\ref{efdstability}), the EFD (\ref{EFD}) with (\ref{initialvalue1}) and (\ref{initialvalue}) satisfies the error estimates
\begin{align}\label{EFDerror}

\|\delta_{x}^{+}e^{\varepsilon,n}\|_{l^{2}} + \|e^{\varepsilon,n}\|_{l^{2}}\lesssim e^{\frac{T}{2}(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2}),~~\|u^{\varepsilon,n}\|_{l^{\infty}}\leq \Lambda+1.
\end{align}
\end{thm}
\begin{rem} \cite{bao2012analysis,bao2019long}
Extending to $2$ and $3$ dimensions, the above Theorems are still valid under the conditions $0<h\lesssim \sqrt{C_{d}(h)},~0<\tau\lesssim \sqrt{C_{d}(h)}$. Besides, the inverse inequality becomes

\begin{align}
\|u^{\varepsilon,n}\|_{l^{\infty}}\lesssim \frac{1}{C_{d}(h)}\left(\|\delta_{x}^{+}u^{\varepsilon,n}\|_{l^{2}} + \|u^{\varepsilon,n}\|_{l^{2}} \right),
\end{align}
where $C_{d}(h)=1/|\ln h|$ when $d=2$ and when $d=3$, $C_{d}(h)=h^{1/2}$.
\end{rem}

\subsection{Proof of Theorem \ref{SIFD2err} for the SIFD}

Define the local trunction error for the SIFD (\ref{SIFD2}) as
\begin{align}\label{trunerre}
\begin{split}
\xi^{0}_{j}:=&\delta_{t}^{+}u^{\varepsilon}(x_{j},0)-\gamma(x_{j})-\frac{\tau}{2}\left[ \delta^{2}_{x}\phi(x_{j})-\phi(x_{j})- \phi(x_{j})\ln(\varepsilon^{2}+(\phi(x_{j}))^{2})\right],\\
\xi^{\varepsilon,n}_{j}:=&\delta_{t}^{2}u^{\varepsilon}(x_{j},t_{n})-\frac{1}{2}\delta_{x}^{2} \left( u^{\varepsilon}(x_{j},t_{n+1})+u^{\varepsilon}(x_{j},t_{n-1}) \right)+\frac{1}{2} \left( u^{\varepsilon}(x_{j},t_{n+1})+u^{\varepsilon}(x_{j},t_{n-1}) \right)\\
&+u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right),~~j\in \mathcal{T}_{N},~~1\leq n\leq \frac{T}{\tau}-1,
\end{split}
\end{align}
then we have the following bounds for the local trunction error.
\begin{lem}\label{trunerrlem}
Under the assumption (A), we have
\begin{align}
&\|\xi^{\varepsilon,0}\|_{H^{1}}\lesssim h^{2}+\tau^{2},\\
&\|\xi^{\varepsilon, n}\|_{l^{2}}\lesssim h^{2}+\tau^{2},\\
&\|\delta^{+}_{x}\xi^{\varepsilon, n}\|_{l^{2}}\lesssim h^{2}+\tau^{2},~~1\leq n \leq \frac{T}{\tau}-1.
\end{align}
\end{lem}
\begin{proof}
By (\ref{initialvalue}) and the Taylor expansion, it leads to
\begin{align}
|\xi^{\varepsilon,0}_{j}|\leq\frac{\tau^{2}}{6}\|\partial_{t}^{3}u\|_{L^{\infty}(0,T;L^{\infty}(\Omega))}+\frac{\tau h}{6}\|\partial_{x}^{3}\phi\|_{L^{\infty}(\Omega)}\lesssim h^{2}+\tau^{2}.
\end{align}
Similarly, we have
\begin{align}
|\delta_{x}^{+}\xi^{\varepsilon,0}_{j}|\leq\frac{\tau^{2}}{6}\|\partial_{ttt}u\|_{L^{\infty}(0,T;H^{1}(\Omega))}+\frac{\tau h}{6}\|\partial_{x}^{4}\phi\|_{L^{\infty}(\Omega)}\lesssim h^{2}+\tau^{2},~~j\in \mathcal{T}_{N}.
\end{align}
Therefore
\begin{align}
\|\xi^{\varepsilon,0}\|_{H^{1}}\lesssim h^{2}+\tau^{2}.
\end{align}
Noting that
\begin{align}
\begin{split}
\xi^{\varepsilon,n}_{j}:=&\delta_{t}^{2}u^{\varepsilon}(x_{j},t_{n})-\frac{1}{2}\delta_{x}^{2} \left( u^{\varepsilon}(x_{j},t_{n+1})+u^{\varepsilon}(x_{j},t_{n-1}) \right)+\frac{1}{2} \left( u^{\varepsilon}(x_{j},t_{n+1})+u^{\varepsilon}(x_{j},t_{n-1}) \right)\\
&+ u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right)\\
&-[ \partial_{tt}u^{\varepsilon}(x_{j},t_{n})-\partial_{xx} u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right) ]\\
=&\left[ \delta_{t}^{2}u^{\varepsilon}(x_{j},t_{n})-\partial_{tt}u^{\varepsilon}(x_{j},t_{n})\right] - \left[\frac{1}{2}\delta_{x}^{2} \left( u^{\varepsilon}(x_{j},t_{n+1})+u^{\varepsilon}(x_{j},t_{n-1}) \right)-\partial_{xx}u^{\varepsilon}(x_{j},t_{n}) \right]\\
&+\frac{1}{2} \left( u^{\varepsilon}(x_{j},t_{n+1})+u^{\varepsilon}(x_{j},t_{n-1}) \right) -u^{\varepsilon}(x_{j},t_{n}).
\end{split}
\end{align}
Taking the Taylor expansion, we obtain
\begin{align}
\xi^{\varepsilon,n}_{j}=\frac{\tau^{2}}{12}\alpha^{\varepsilon,n}_{j}+\frac{\tau^{2}}{2}\beta^{\varepsilon,n}_{j}+\frac{h^{2}}{12}\eta^{\varepsilon,n}_{j}+\frac{\tau^{2}}{2}\phi^{\varepsilon,n}_{j},
\end{align}
where
\begin{align}
&\alpha^{\varepsilon,n}_{j}=\int^{1}_{-1}(1-|s|)^{3}\partial^{4}_{t} u^{\varepsilon}(x_{j},t_{n}+s\tau)ds,~~\beta^{\varepsilon,n}_{j}=\int^{1}_{-1}(1-|s|)\partial^{2}_{t} u^{\varepsilon}_{xx}(x_{j},t_{n}+s\tau)ds,\\
&\eta^{\varepsilon,n}_{j}=\int^{1}_{-1}(1-|s|)^{3}\left( \partial^{4}_{x} u^{\varepsilon}(x_{j}+sh,t_{n+1})+ \partial^{4}_{x} u^{\varepsilon}(x_{j}+sh,t_{n-1}) \right) ds,\\
&\phi^{\varepsilon,n}_{j}=\int^{1}_{-1}(1-|s|)\partial^{2}_{t} u^{\varepsilon}(x_{j},t_{n}+s\tau)ds.
\end{align}
Applying Cauchy-Schwarz inequality, we obtain
\begin{align}
\begin{split}
\|\alpha^{\varepsilon,n}\|^{2}_{l^{2}}&=h\sum\limits^{N-1}_{j=1}|\alpha^{\varepsilon,n}_{j}|^{2}\leq h\int^{1}_{-1}(1-|s|)^{6}ds\sum\limits^{N-1}_{j=1}\int^{1}_{-1}|\partial^{4}_{t} u^{\varepsilon}(x_{j},t_{n}+s\tau)|^{2}ds\\
&=\frac{2}{7}\left[ \int^{1}_{-1}\|\partial^{4}_{t} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|^{2}_{L^{2}(\Omega)}ds- \int^{1}_{-1}\sum\limits^{N-1}_{j=0} \int^{x_{j+1}}_{x_{j}}\int^{\omega}_{x_{j}}\partial_{x} |\partial^{4}_{t} u^{\varepsilon}(\hat{x},t_{n}+s\tau)|^{2}d\hat{x}d\omega ds \right]\\
&\leq \frac{2}{7}\int^{1}_{-1}\left[ \|\partial^{4}_{t} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|^{2}_{L^{2}(\Omega)} +2h\|\partial^{4}_{t} u^{\varepsilon}_{x}(\cdot,t_{n}+s\tau)\|_{L^{2}(\Omega)} \|\partial^{4}_{t} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|_{L^{2}(\Omega)} \right] ds\\
&\leq \max_{0\leq t\leq T}\left( \|\partial^{4}_{t} u^{\varepsilon}\|_{L^{2}(\Omega)}+h \|\partial^{4}_{t} u_{x}^{\varepsilon}\|_{L^{2}(\Omega)} \right)^{2},
\end{split}
\end{align}
which implies that when $h\leq 1,$
\begin{align}
\left\|\alpha^{\varepsilon, n}\right\|_{l^{2}} \leq \left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{1}(\Omega)\right)} .
\end{align}
Similarly, we can get
\begin{align}
&\|\beta^{\varepsilon, n}\|_{l^{2}} \leq 2\left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{3}(\Omega)\right)},\\
&\left\|\phi^{\varepsilon, n}\right\|_{l^{2}} \leq \left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{1}(\Omega)\right)} .
\end{align}
On the other hand, it can be estimated that
\begin{align}
\begin{split}
\|\eta^{\varepsilon, n}\|^{2}_{l^{2}} & \leq h \int_{-1}^{1}(1-|s|)^{6} d s \sum_{j=1}^{N-1} \int_{-1}^{1}\left|\partial_{x}^{4} u^{\varepsilon}\left(x_{j}+s h, t_{n+1}\right)+\partial_{x}^{4} u^{\varepsilon}\left(x_{j}+s h, t_{n-1}\right)\right|^{2} d s \\
& \leq \frac{4 h}{7} \sum_{j=1}^{N-1} \int_{-1}^{1}\left(\left|\partial_{x}^{4} u^{\varepsilon}\left(x_{j}+s h, t_{n+1}\right)\right|^{2}+\left|\partial_{x}^{4} u^{\varepsilon}\left(x_{j}+s h, t_{n-1}\right)\right|^{2}\right) d s \\
& \leq \frac{8}{7}\left(\left\|\partial_{x}^{4} u^{\varepsilon}\left(\cdot, t_{n-1}\right)\right\|_{L^{2}(\Omega)}^{2}+\left\|\partial_{x}^{4} u^{\varepsilon}\left(\cdot, t_{n+1}\right)\right\|_{L^{2}(\Omega)}^{2}\right) \\
& \leq 4\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{4}(\Omega)\right)}^{2},
\end{split}
\end{align}
which yields $\|\eta^{\varepsilon, n}\|_{l^{2}}\leq 2\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{4}(\Omega)\right)}$.
Therefore, according to the assumption (A), we get
\begin{align}
\begin{split}
\left\|\xi^{\varepsilon, n}\right\|_{l^{2}} \leq &\frac{\tau^{2}}{12}\left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{1}(\Omega)\right)}+\tau^{2}\left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{3}(\Omega)\right)}+\frac{h^{2}}{6}\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{4}(\Omega)\right)}\\
&+\frac{\tau^{2}}{2}\left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{1}(\Omega)\right)} \\
\lesssim &\tau^{2}+h^{2}.
\end{split}
\end{align}
Using the same approach, we can get
\begin{align}
\begin{split}
\|\delta^{+}_{x}\xi^{\varepsilon, n}\|_{l^{2}}\leq& \frac{\tau^{2}}{12}\left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{2}(\Omega)\right)}+\tau^{2}\left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{4}(\Omega)\right)}+\frac{h^{2}}{6}\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{5}(\Omega)\right)}\\
&+\frac{\tau^{2}}{2}\left\|\partial_{t}^{2} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{2}(\Omega)\right)}  \\
\lesssim & \tau^{2}+h^{2}.
\end{split}
\end{align}
This ends the proof.
\end{proof}

Subtracting (\ref{SIFD2}) from  (\ref{trunerre}), the error $e^{\varepsilon,n}_{j}$ satisfies
\begin{subequations}\label{SIFD2error}
\begin{align}
&\delta_{t}^{2}e^{\varepsilon,n}_{j}-\frac{1}{2}\delta_{x}^{2} (e^{\varepsilon,n+1}_{j}+e^{\varepsilon,n-1}_{j})+\frac{1}{2} (e^{\varepsilon,n+1}_{j}+e^{\varepsilon,n-1}_{j})=\xi^{\varepsilon,n}_{j} - \zeta^{\varepsilon,n}_{j},\label{SIFD2errora}\\
&e^{\varepsilon,n}_{0}=e^{\varepsilon,n}_{N},~~e^{\varepsilon,n}_{-1}=e^{\varepsilon,n}_{N-1},~~n=0,1,\ldots\label{SIFD2errorb}\\
&e^{\varepsilon,0}_{j}=0,~~e^{\varepsilon,1}_{j}=\tau\xi^{0}_{j},~~j\in \mathcal{T}_{N},\label{SIFD2errorc}
\end{align}
\end{subequations}
where
\begin{align}
\zeta^{\varepsilon,n}_{j}=u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right)-u^{\varepsilon,n}_{j}f_{\varepsilon}\left( ( u^{\varepsilon,n}_{j})^{2} \right).
\end{align}
We define the ``energy'' for the error vector $e^{\varepsilon,n}~(n=0,1,\ldots)$ as
\begin{align}
E^{n}_{e}:=&\|\delta_{t}^{+}e^{\varepsilon,n}\|^{2}_{l^{2}}+\frac{1}{2}\left( \|\delta_{x}^{+}e^{\varepsilon,n+1}\|^{2}_{l^{2}}+ \|\delta_{x}^{+}e^{\varepsilon,n}\|^{2}_{l^{2}} \right)+\frac{1}{2}\left( \|e^{\varepsilon,n+1}\|^{2}_{l^{2}}+\|e^{\varepsilon,n}\|^{2}_{l^{2}} \right).
\end{align}
Besides, we can get that
\begin{align}\label{E0}
E^{0}_{e}:=&\|\xi^{0}\|^{2}_{l^{2}}+\frac{\tau^{2}}{2}\|\delta_{x}^{+}\xi^{0}\|^{2}_{l^{2}}+\frac{\tau^{2}}{2} \|\xi^{0}\|^{2}_{l^{2}}\lesssim (\tau^{2}+h^{2})^{2}.
\end{align}
\begin{proof} ($\mathbf{Proof~of~Theorem~{\ref{SIFD2err}}}$)
When $k=1$,
under the assumption (A), by Lemma \ref{trunerrlem} we can conclude the errors of the first step discretization (\ref{initialvalue})
\begin{align}
e^{\varepsilon,0}=0,~~\|e^{\varepsilon,1}_{j}\|_{H^{1}}\lesssim\tau^{2}+h^{2},
\end{align}
for sufficiently small $0<\tau<\tau_{1}$ and $0<h<h_{1}$. So it is true for $k=0, 1$.

Assume (\ref{SIFD2error}) is valid for $k\leq n\leq \frac{T}{\tau}-1$. Next, we need to verify (\ref{SIFD2error}) is true for $k=n+1$.
Denote
\begin{align}
\zeta^{\varepsilon,m}_{j}=u^{\varepsilon}(x_{j},t_{m})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{m}))^{2} \right)-u^{\varepsilon,m}_{j}f_{\varepsilon}\left( ( u^{\varepsilon,m}_{j})^{2} \right).
\end{align}
When $|u^{\varepsilon,m}_{j}|\leq|u^{\varepsilon}(x_{j},t_{m})|$, we get
\begin{align}
\begin{split}
|\zeta^{\varepsilon,m}_{j}|=&|u^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon,m}_{j} )^{2} \right)-u^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} \right) \\
&+ u^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} \right)
-u^{\varepsilon}(x_{j},t_{m})\ln \left( {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} \right) |\\
=&\left| e^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} \right)+ u^{\varepsilon,m}_{j} \ln \left(1+ \frac{ ( u^{\varepsilon,m}_{j} )^{2}-(u^{\varepsilon}(x_{j},t_{m}))^{2} }{{\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}) )^{2}} \right) \right|\\
\leq & \left| e^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} \right)\right|+ \left|u^{\varepsilon}(x_{j},t_{m}) \frac{ ( u^{\varepsilon,m}_{j} )^{2}- \left(u^{\varepsilon}(x_{j},t_{m})\right)^{2} }{( u^{\varepsilon}(x_{j},t_{m}) )^{2}}\right|\\
\leq & | e^{\varepsilon,m}_{j}|\max\{\ln(\frac{1}{\varepsilon^{2}}),|\ln(\Lambda^{2}+{\varepsilon^{2}})|\}+2| e^{\varepsilon,m}_{j}|\\
=&| e^{\varepsilon,m}_{j}|\left( \max\{\ln(\frac{1}{\varepsilon^{2}}),|\ln(\Lambda^{2}+{\varepsilon^{2}})|\}+2\right).
\end{split}
\end{align}
In addition, when $|u^{\varepsilon,m}_{j}|\geq|u^{\varepsilon}(x_{j},t_{m})|$, we obtain
\begin{align}
\begin{split}
|\zeta^{\varepsilon,m}_{j}|=&|u^{\varepsilon}(x_{j},t_{m})\ln \left( {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} \right)-
u^{\varepsilon}(x_{j},t_{m})\ln \left( {\varepsilon^{2}}+( u^{\varepsilon,m}_{j} )^{2} \right)\\
&+u^{\varepsilon}(x_{j},t_{m})\ln \left( {\varepsilon^{2}}+( u^{\varepsilon,m}_{j} )^{2} \right)
- u^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon,m}_{j} )^{2} \right) |\\
=&\left| e^{\varepsilon,m}_{j}\ln \left( {\varepsilon^{2}}+( u^{\varepsilon,m}_{j})^{2} \right)+ u^{\varepsilon}(x_{j},t_{m}) \ln \left(\frac{ {\varepsilon^{2}}+( u^{\varepsilon}(x_{j},t_{m}))^{2} }{{\varepsilon^{2}}+( u^{\varepsilon,m}_{j} )^{2}} \right) \right|\\
\leq & | e^{\varepsilon,m}_{j}|\max\{\ln(\frac{1}{\varepsilon^{2}}),|\ln((1+\Lambda)^{2}+{\varepsilon^{2}})|\}+2| e^{\varepsilon,m}_{j}|\\
=&| e^{\varepsilon,m}_{j}|\left( \max\{\ln(\frac{1}{\varepsilon^{2}}),|\ln((1+\Lambda)^{2}+{\varepsilon^{2}})|\}+2\right),
\end{split}
\end{align}
where we use the assumption $\|u^{\varepsilon,m}\|_{l^{\infty}}\leq 1+\Lambda$ above for $m\leq n$. Since $\varepsilon$ is sufficiently small, we have
\begin{align}\label{nonlinearerror}
\|\zeta^{\varepsilon,m}\|^{2}_{l^{2}}\lesssim \| e^{\varepsilon,m}\|^{2}_{l^{2}}(\ln\varepsilon^{2})^{2}.
\end{align}

Multiplying both sides of (\ref{SIFD2errora}) by $h(e^{\varepsilon,m+1}-e^{\varepsilon,m-1})$, then summing up for $j\in \mathcal{T}_{N}$. And by Young's inequality, Lemma \ref{trunerrlem}, and (\ref{nonlinearerror}) we can obtain
\begin{align}
\begin{split}
E^{m}_{e}-E^{m-1}_{e}&=h\sum\limits_{j=0}^{N-1}(\xi^{\varepsilon,m}_{j}-\zeta^{\varepsilon,m}_{j})(e^{\varepsilon,m+1}_{j}-e^{\varepsilon,m-1}_{j})\\
&\leq h\sum\limits_{j=0}^{N-1}\left( |\xi^{\varepsilon,m}_{j}|+|\zeta^{\varepsilon,m}_{j}|\right)\left|e^{\varepsilon,m+1}_{j}-e^{\varepsilon,m-1}_{j} \right|\\

&\leq\tau \left( \|\xi^{\varepsilon,m}\|^{2}_{l^{2}}+\|\zeta^{\varepsilon,m}\|^{2}_{l^{2}}+\left\|\delta^{+}_{t}e^{\varepsilon,m}\|^{2}_{l^{2}}+\|\delta^{+}_{t}e^{\varepsilon,m-1} \right\|^{2}_{l^{2}}\right)\\
&\lesssim \tau\left((\tau^{2}+h^{2})^{2} + (\ln\varepsilon^{2})^{2}\| e^{\varepsilon,m}\|^{2}_{l^{2}}+E^{m}_{e}+E^{m-1}_{e}\right),~~1\leq m\leq \frac{T}{\tau}-1.
\end{split}
\end{align}
 Therefore, there exists a constant $\tau_{2}>0$ sufficiently small and independent of $\varepsilon$ and $h$, such that when $0<\tau<\tau_{2}$, we get
 \begin{align}
 E^{m}_{e}-E^{m-1}_{e}
 &\lesssim \tau\left((\tau^{2}+h^{2})^{2} + (\ln\varepsilon^{2})^{2}\| e^{\varepsilon,m}\|^{2}_{l^{2}}+E^{m-1}_{e}\right),~~1\leq m\leq
 \frac{T}{\tau}-1.
 \end{align}
Summing above the inequalities up to $n$, and noticing (\ref{E0}), the following holds
\begin{align}

&E^{n}_{e}\lesssim   (\tau^{2}+h^{2})^{2}+ \tau(\ln\varepsilon^{2})^{2}\sum\limits_{m=0}^{n-1}E^{m}_{e},~~1\leq n\leq \frac{T}{\tau}-1.
\end{align}
By applying the discrete Gronwall's inequality \cite{holte2009discrete}, we have
\begin{align}

&E^{n}_{e}\lesssim e^{T(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2})^{2},~~1\leq n\leq \frac{T}{\tau}-1.
\end{align}
Recalling $\|\delta_{x}^{+}e^{\varepsilon,n+1}\|^{2}_{l^{2}} + \|e^{\varepsilon,n+1}\|^{2}_{l^{2}}\leq 2E^{n}_{e}$ when $0<\varepsilon \ll1$, we can get the error estimate
\begin{align}
\|\delta_{x}^{+}e^{\varepsilon,n+1}\|^{2}_{l^{2}} + \|e^{\varepsilon,n+1}\|^{2}_{l^{2}}\lesssim
&e^{T(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2})^{2}.
\end{align}
By Sobolev inequality, we obtain
\begin{align}
\|e^{\varepsilon,n}\|^{2}_{l^{\infty}}\leq\|\delta_{x}^{+}e^{\varepsilon,n}\|^{2}_{l^{2}} + \|e^{\varepsilon,n}\|^{2}_{l^{2}}\lesssim
&e^{T(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2})^{2}.
\end{align}
Therefore, there exist $\tau_{3}>0,h_{2}>0$ sufficiently small. When $0<h<h_{2},0<\tau<\tau_{3}$, applying the triangle inequality, it implies that
\begin{align}
\|u^{\varepsilon,n}\|_{l^{\infty}}\leq \|u^{\varepsilon}(\cdot,t_{n})\|_{L^{\infty}(\Omega)}+\|e^{\varepsilon,n}\|_{l^{\infty}}\leq \Lambda +1.
\end{align}
We complete the proof by choosing $h_{0}=\min\{h_{1},h_{2}\},~\tau_{0}=\min\{\tau_{1},\tau_{2},\tau_{3}\}$.
\end{proof}

\subsection{The proof of Theorem for EFD}

Define the local trunction error for the EFD (\ref{EFD}) as
\begin{align}\label{EFDtrunerre}
\begin{split}
\xi^{0}_{j}:=&\delta_{t}^{+}u^{\varepsilon}(x_{j},0)-\gamma(x_{j})-\frac{\tau}{2}\left[ \delta^{2}_{x}\phi(x_{j})-\phi(x_{j})- \phi(x_{j})\ln(\varepsilon^{2}+(\phi(x_{j}))^{2})\right],~~j\in \mathcal{T}_{N},\\
\xi^{\varepsilon,n}_{j}:=&\delta_{t}^{2}u^{\varepsilon}(x_{j},t_{n})-\delta_{x}^{2} u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right),~~1\leq n\leq \frac{T}{\tau}-1,
\end{split}
\end{align}
then we have the following bounds for the local trunction error.
\begin{lem}\label{EFDtrunerrlem}
Under the assumption (A), we have
\begin{align}
&\|\xi^{\varepsilon,0}\|_{H^{1}}\lesssim h^{2}+\tau^{2},\\
&\|\xi^{\varepsilon, n}\|_{l^{2}}\lesssim h^{2}+\tau^{2},\\
&\|\delta^{+}_{x}\xi^{\varepsilon, n}\|_{l^{2}}\lesssim h^{2}+\tau^{2},~~1\leq n \leq \frac{T}{\tau}-1.
\end{align}
\end{lem}
\begin{proof}
According to the Lemma \ref{trunerrlem}, we have
\begin{align}
\|\xi^{\varepsilon,0}\|_{H^{1}}\lesssim h^{2}+\tau^{2}.
\end{align}
Noting that
\begin{align}
\begin{split}
\xi^{\varepsilon,n}_{j}:=&\delta_{t}^{2}u^{\varepsilon}(x_{j},t_{n})-\delta_{x}^{2} u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right)\\

&-[ \partial_{tt}u^{\varepsilon}(x_{j},t_{n})-\partial_{xx} u^{\varepsilon}(x_{j},t_{n}) + u^{\varepsilon}(x_{j},t_{n})+ u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right) ]\\
=&\left[ \delta_{t}^{2}u^{\varepsilon}(x_{j},t_{n})-\partial_{tt}u^{\varepsilon}(x_{j},t_{n})\right] - \left[\delta_{x}^{2} u^{\varepsilon}(x_{j},t_{n})-\partial_{xx} u^{\varepsilon}(x_{j},t_{n})\right].
\end{split}
\end{align}
Taking the Taylor expansion, we obtain
\begin{align}
\xi^{\varepsilon,n}_{j}=\frac{\tau^{2}}{12}\alpha^{\varepsilon,n}_{j}+\frac{h^{2}}{6}\beta^{\varepsilon,n}_{j},
\end{align}
where
\begin{align}
&\alpha^{\varepsilon,n}_{j}=\int^{1}_{-1}(1-|s|)^{3}\partial^{4}_{t} u^{\varepsilon}(x_{j},t_{n}+s\tau)ds,~~\beta^{\varepsilon,n}_{j}=\int^{1}_{-1}(1-|s|)^{3}\partial^{3}_{x} u^{\varepsilon}(x_{j}+sh,t_{n})ds.
\end{align}
By Cauchy-Schwarz inequality, we obtain
\begin{align}
\begin{split}
\|\alpha^{\varepsilon,n}\|^{2}_{l^{2}}&=h\sum\limits^{N-1}_{j=1}|\alpha^{\varepsilon,n}_{j}|^{2}\leq h\int^{1}_{-1}(1-|s|)^{6}ds\sum\limits^{N-1}_{j=1}\int^{1}_{-1}|\partial^{4}_{t} u^{\varepsilon}(x_{j},t_{n}+s\tau)|^{2}ds\\
&=\frac{2}{7}\left[ \int^{1}_{-1}\|\partial^{4}_{t} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|^{2}_{L^{2}(\Omega)}ds- \int^{1}_{-1}\sum\limits^{N-1}_{j=0} \int^{x_{j+1}}_{x_{j}}\int^{\omega}_{x_{j}}\partial_{x} |\partial^{4}_{t} u^{\varepsilon}(\hat{x},t_{n}+s\tau)|^{2}d\hat{x}d\omega ds \right]\\
&\leq \frac{2}{7}\int^{1}_{-1}\left[ \|\partial^{4}_{t} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|^{2}_{L^{2}(\Omega)} +2h\|\partial^{4}_{t}\partial_{x} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|_{L^{2}(\Omega)} \|\partial^{4}_{t} u^{\varepsilon}(\cdot,t_{n}+s\tau)\|_{L^{2}(\Omega)} \right] ds\\
&\leq \max_{0\leq t\leq T}\left( \|\partial^{4}_{t} u^{\varepsilon}\|_{L^{2}(\Omega)}+h \|\partial^{4}_{t}\partial_{x} u^{\varepsilon}\|_{L^{2}(\Omega)} \right)^{2},
\end{split}
\end{align}
which implies that when $h\leq 1,$
\begin{align}
\left\|\alpha^{\varepsilon, n}\right\|_{l^{2}} \leq \left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{1}(\Omega)\right)}.
\end{align}
On the other hand, it can be estimated that
\begin{align}
\begin{split}
\|\beta^{\varepsilon, n}\|^{2}_{l^{2}} & \leq h \int_{-1}^{1}(1-|s|)^{6} d s \sum_{j=1}^{N-1} \int_{-1}^{1}\left|\partial_{x}^{3} u^{\varepsilon}\left(x_{j}+s h, t_{n}\right)\right|^{2} d s \\
& \leq \frac{2 h}{7} \sum_{j=1}^{N-1} \int_{-1}^{1}\left|\partial_{x}^{3} u^{\varepsilon}\left(x_{j}+s h, t_{n}\right)\right|^{2} d s \\
& \leq \frac{4}{7}\left(\left\|\partial_{x}^{3} u^{\varepsilon}\left(\cdot, t_{n}\right)\right\|_{L^{2}(\Omega)}^{2}\right) \\
& \leq \left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{3}(\Omega)\right)}^{2},
\end{split}
\end{align}
which yields that $\|\beta^{\varepsilon, n}\|_{l^{2}}\leq \left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{3}(\Omega)\right)}$.
Therefore, according to assumption (A), we get
\begin{align}
\begin{split}
\left\|\xi^{\varepsilon, n}\right\|_{l^{2}} & \leq \frac{\tau^{2}}{12}\left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{1}(\Omega)\right)}+\frac{h^{2}}{6}\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{3}(\Omega)\right)} \\
& \lesssim \tau^{2}+h^{2}.
\end{split}
\end{align}
With the same method, we have
\begin{align}
\begin{split}
\|\delta^{+}_{x}\xi^{\varepsilon, n}\|_{l^{2}}& \leq \frac{\tau^{2}}{12}\left\|\partial_{t}^{4} u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{2}(\Omega)\right)}+\frac{h^{2}}{6}\left\|u^{\varepsilon}\right\|_{L^{\infty}\left(0, T ; H^{4}(\Omega)\right)} \\
& \lesssim \tau^{2}+h^{2}.
\end{split}
\end{align}
This completes the proof.
\end{proof}

Subtracting (\ref{EFD}) from  (\ref{EFDtrunerre}), the error $e^{\varepsilon,n}_{j}$ satisfies
\begin{subequations}\label{EFDerror}
\begin{align}
&\delta_{t}^{2}e^{\varepsilon,n}_{j}-\delta_{x}^{2} e^{\varepsilon,n}_{j}+ e^{\varepsilon,n}_{j}=\xi^{\varepsilon,n}_{j} - \zeta^{\varepsilon,n}_{j},\label{EFDerrora}\\
&e^{\varepsilon,n}_{0}=e^{\varepsilon,n}_{N},~~e^{\varepsilon,n}_{-1}=e^{\varepsilon,n}_{N-1},~~n=0,1,\ldots\label{EFDerrorb}\\
&e^{\varepsilon,0}_{j}=0,~~e^{\varepsilon,1}_{j}=\tau\xi^{0}_{j},~~j\in \mathcal{T}_{N},\label{EFDerrorc}
\end{align}
\end{subequations}
where
\begin{align}
\zeta^{\varepsilon,n}_{j}=u^{\varepsilon}(x_{j},t_{n})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{n}))^{2} \right)-u^{\varepsilon,n}_{j}f_{\varepsilon}\left( ( u^{\varepsilon,n}_{j})^{2} \right).
\end{align}
We define the ``energy'' for the error vector $e^{\varepsilon,n}~(n=0,1,\ldots)$ as
\begin{align}
E^{n}_{e}:=&(1-\frac{\tau^{2}}{2}-\frac{\tau^{2}}{h^{2}})\|\delta_{t}^{+}e^{\varepsilon,n}\|^{2}_{l^{2}}+\frac{1}{2}\left( \|e^{\varepsilon,n}\|^{2}_{l^{2}}+\|e^{\varepsilon,n+1}\|^{2}_{l^{2}} \right)+ \frac{1}{2h}\sum\limits_{j=0}^{N-1}\left[ (e^{\varepsilon,n+1}_{j+1}-e^{\varepsilon,n}_{j})^{2} + (e^{\varepsilon,n}_{j+1}-e^{\varepsilon,n+1}_{j})^{2}\right].
\end{align}
Besides, we can get that
\begin{align}\label{EFDE0}
 E^{0}_{e}:=&(1-\frac{\tau^{2}}{2}-\frac{\tau^{2}}{h^{2}})\|\delta_{t}^{+}e^{\varepsilon,0}\|^{2}_{l^{2}}+(\frac{1}{2}+\frac{1}{h^{2}}) \|e^{\varepsilon,1}\|^{2}_{l^{2}}=\|\xi^{0}\|^{2}_{l^{2}}\lesssim (\tau^{2}+h^{2})^{2}
 .
\end{align}
\begin{proof} ($\mathbf{Proof~of~Theorem~{\ref{EFDerr}}}$)
When $m=1$,
under assumption (A), by Lemma \ref{trunerrlem} we can conclude the first step errors of the discretization (\ref{initialvalue})
\begin{align}
e^{\varepsilon,0}=0,~~\|e^{\varepsilon,1}_{j}\|_{H^{1}}\lesssim\tau^{2}+h^{2},
\end{align}
for sufficiently small $0<\tau<\tau_{1}$ and $0<h<h_{1}$. So it is true for $m=0, 1$.
Assume (\ref{SIFD2error}) is valid for $m\leq n\leq \frac{T}{\tau}-1$. Next, we need to verify (\ref{SIFD2error}) is true for $m=n+1$.
Denote
\begin{align}
\zeta^{\varepsilon,m}_{j}=u^{\varepsilon}(x_{j},t_{m})f_{\varepsilon}\left( ( u^{\varepsilon}(x_{j},t_{m}))^{2} \right)-u^{\varepsilon,m}_{j}f_{\varepsilon}\left( ( u^{\varepsilon,m}_{j})^{2} \right).
\end{align}
With the same method in Theorem \ref{SIFD2err}, we have
\begin{align}\label{EFDnonlinearerror}
\|\zeta^{\varepsilon,m}\|^{2}_{l^{2}}\lesssim \| e^{\varepsilon,m}\|^{2}_{l^{2}}(\ln\varepsilon^{2})^{2}.
\end{align}
Besides, under the assumption $\tau \leq \frac{1}{2}\min\{1,h\}$, we have $1-\frac{\tau^{2}}{2}-\frac{\tau^{2}}{h^{2}}\geq \frac{1}{4}>0$.
By
\begin{align}
\left\|\delta_{x}^{+} e^{m+1}\right\|_{l^{2}}^{2}=\frac{1}{h} \sum_{j=0}^{N-1}\left(e_{j+1}^{m+1}-e_{j}^{m}-\tau \delta_{t}^{+} e_{j}^{m}\right)^{2} \leq \frac{2}{h} \sum_{j=0}^{N-1}\left(e_{j+1}^{m+1}-e_{j}^{m}\right)^{2}+\frac{2 \tau^{2}}{h^{2}}\left\|\delta_{t}^{+} e^{m}\right\|_{l^{2}}^{2},
\end{align}
we have
\begin{align}
E^{m}_{e} \geq \frac{1}{4}\left\|\delta_{x}^{+} e^{m+1}\right\|_{l^{2}}^{2}+\frac{1}{2}\left(\left\|e^{m}\right\|_{l^{2}}^{2}+\left\|e^{m+1}\right\|_{l^{2}}^{2}\right), \quad 1 \leq m \leq n-1.
\end{align}
Similar to the proof of Theorem \ref{SIFD2err}, there exists a $\tau_{2}>0$ sufficiently small, when $0<\tau\leq\tau_{2}$, we get
\begin{align}
E^{m}_{e}\lesssim
&e^{T(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2})^{2}, \quad 1 \leq m \leq n-1.
\end{align}
Therefore, we can get the $(n+1)th$ error estimate
\begin{align}
\|\delta_{x}^{+}e^{\varepsilon,n+1}\|^{2}_{l^{2}} + \|e^{\varepsilon,n+1}\|^{2}_{l^{2}}\lesssim
&e^{T(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2})^{2}.
\end{align}
By Sobolev inequality, we obtain
\begin{align}
\|e^{\varepsilon,n}\|^{2}_{l^{\infty}}\leq\|\delta_{x}^{+}e^{\varepsilon,n}\|^{2}_{l^{2}} + \|e^{\varepsilon,n}\|^{2}_{l^{2}}\lesssim
&e^{T(\ln(\varepsilon^{2}))^{2}}(\tau^{2}+h^{2})^{2}.
\end{align}
Applying the triangle inequality, it implies that
\begin{align}
\|u^{\varepsilon,n}\|_{l^{\infty}}\leq \|u^{\varepsilon}(\cdot,t_{n})\|_{L^{\infty}(\Omega)}+\|e^{\varepsilon,n}\|_{l^{\infty}}\leq \Lambda +1.
\end{align}
This ends the proof  by choosing $h_{0}=h_{1},\tau_{0}=\min\{\tau_{1},\tau_{2}\}$.
\end{proof}
\section{Numerical results}\label{conclusion}
In this section, we represent some numerical experiments of the EFD (\ref{EFD}) scheme to quantify the error bounds. Since the results of the SIFD (\ref{SIFD2}) are similar to the EFD (\ref{EFD}), we omit the details here for brevity. Here we take $d=1, \lambda=1$ and we define the error functions as:
\begin{align}
&\hat{e}^{\varepsilon}(t_{n}):=u(\cdot,t_{n})-u^{\varepsilon}(\cdot,t_{n}),
\quad e^{\varepsilon}(t_{n}):=u^{\varepsilon}(\cdot,t_{n})-u^{\varepsilon,n}, \quad \tilde{e}^{\varepsilon}(t_{n}):=u(\cdot,t_{n})-u^{\varepsilon,n}.
\end{align}
Besides, we denote the error functions:
\begin{align}
&e^{\varepsilon}_{\infty}(t_{n}):=\|u^{\varepsilon}(\cdot,t_{n})-u^{\varepsilon,n}\|_{l^{\infty}}, \quad  e^{\varepsilon}_{2}(t_{n}):=\|u^{\varepsilon}(\cdot,t_{n})-u^{\varepsilon,n}\|_{l^{2}}, \\
&e^{\varepsilon}_{H^{1}}(t_{n}):=\sqrt{(e^{\varepsilon}_{2}(t_{n}))^{2}+\|\delta^{+}_{x}(u^{\varepsilon}(\cdot,t_{n})-u^{\varepsilon,n})\|^{2}_{l^{2}}}.
\end{align}
Here $u, u^{\varepsilon}$ are the exact solutions of the LogKGE (\ref{LogKGE}) and the RLogKGE (\ref{RLogKGE}), $u^{n}, u^{\varepsilon,n}$ are the numerical solutions of the LogKGE (\ref{LogKGE}) and the RLogKGE (\ref{RLogKGE}).

$\mathbf{Example~1}$. The initial datum is taken as $\phi(x)=e^{-\frac{k^{2}x^{2}}{2(c^{2}-k^{2})}},\gamma(x)=\frac{ckx}{c^2-k^2}e^{-\frac{(kx)^{2}}{2(c^{2}-k^{2})}}$, and the Gaussian solitary wave solution is
\begin{align}
u(x,t)=e^{-\frac{(kx-ct)^{2}}{2(c^{2}-k^{2})}},
\end{align}
where $c = 2, k=1$. The RLogKGE (\ref{RLogKGE}) is simulated on the domain $\Omega=[-16,16]$. The `exact' solution $u^{\varepsilon}$ is obtained numerically by the EFD (\ref{EFD}) scheme with $\varepsilon=10^{-7}$.

$\mathbf{Example~2}$. We take the initial value as $\phi(x)=\frac{2}{e^{-x^{2}}+e^{x^{2}}},\gamma(x)=0$.  The computation domain is chosen as $\Omega=[-16,16]$ with periodic boundary conditions. Since the analytical solution is not available in this example. The `exact' solution $u^{\varepsilon}$ is obtained by the EFD (\ref{EFD}) with a small mesh size $h=2^{-10}$, and time step $\tau=0.01\times2^{-9}$. In addition, the `exact' solution $u$ is approximated by $u^{\varepsilon}$ with $\varepsilon=10^{-7}$.

\subsection{Convergence of the regularized model}

Here we test the order of accuracy of the regularized model, that is the convergence rate between the solutions of the RLogKGE (\ref{RLogKGE}) and the LogKGE (\ref{LogKGE}). Figure \ref{fig:regexhat} represents $\|\hat{e}^{\varepsilon}\|_{l^{2}}, \|\hat{e}^{\varepsilon}\|_{l^{\infty}},\|\hat{e}^{\varepsilon}\|_{H^{1}}$ with the scheme EFD (\ref{EFD}) for Example~1 and Example~2. The errors are displayed at $T=0.5$.

From Figure \ref{fig:regexhat}, we can observe that the solutions of the RLogKGE (\ref{RLogKGE}) are linearly convergent to the LogKGE (\ref{LogKGE}) with regard to $\varepsilon$, and the convergence rate is $O(\varepsilon)$ in the $l^{2}$-norm, $l^{\infty}$-norm, $H^{1}$-norm.
\begin{center}
\begin{figure}[htbp]
\centering
\subfigure{\includegraphics[width=.43\textwidth, height=0.38\textwidth]{logkg_efd_order_ex_hat_time_epsilon.eps}}~~~~~~~
\subfigure{\includegraphics[width=.43\textwidth, height=0.38\textwidth]{logkg_efd_order_eref_hat_time05_epsilon_case1.eps}}
\caption{The errors $\hat{e}^{\varepsilon}(0.5)$ in three different norms with the scheme  EFD $(\ref{EFD})$ for Example~1 (left) and Example~2 (right).}
\label{fig:regexhat}
\end{figure}
\end{center}

\subsection{Convergence of FDTD to the RLogKGE}

Then we check the convergence patterns of the finite difference scheme: EFD (\ref{EFD}) to the RLogKGE (\ref{RLogKGE})  for various mesh size $h$, time step $\tau$ under any fixed
parameter $0<\varepsilon\ll1$ for Example~1 and Example~2.

Firstly, we perform test on the temporal errors with the EFD (\ref{EFD}) in the $l^{2}$-norm, $l^{\infty}$-norm, $H^{1}$-norm at $T=1$, depicted in Figure \ref{fig:efdcase2error}. Due to the stability condition of the EFD (\ref{EFD}), we set $0< \tau <\min\{\frac{1}{2}, \frac{h}{2}\}$, varying the mesh size and time step simultaneously as $\tau_{j}=0.01\times2^{-j}, h_{j}=2^{-j}$ for $j=1,\ldots,7$.

Secondly, for spatial accuracy of the EFD (\ref{EFD}) at $T=1$, we set time step $\tau=\tau_{e}=0.01\times{2^{-9}}$, such that the errors from the time discretization are ignored and solve the RLogKGE (\ref{RLogKGE}) with the FDTD schemes versus mesh size $h$. The results are displayed in Figure \ref{fig:regefdcase2spaceerr}. Figure \ref{fig:regefdcase2spaceerr} depict $\|e^{\varepsilon}\|_{l^{2}}, \|e^{\varepsilon}\|_{l^{\infty}},\|e^{\varepsilon}\|_{H^{1}}$ with different $h$ of the scheme EFD (\ref{EFD}) for Example~1 and Example~2.

From Figure \ref{fig:regefdcase2spaceerr}, we can make the observations:
the scheme EFD (\ref{EFD}) are uniformly second order accurate for the RLogKGE (\ref{RLogKGE}) for any $0<\varepsilon\ll 1$ which demonstrate the Theorem \ref{EFDerr}.

\begin{center}
\begin{figure}[htbp]
\centering
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_ex_time_t1_l2.eps}}~~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_ex_time_t1_linf.eps}}~~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_ex_time_t1_H.eps}}
\caption{The temporal errors $e^{\varepsilon}(1)$ in three different norms for Example~1.}
\label{fig:regefderr}
\end{figure}
\end{center}
\begin{center}
\begin{figure}[htbp]
\centering
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_time_epsilon_t1_l2_case1.eps}}~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_time_epsilon_t1_linf_case1.eps}}~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_time_epsilon_t1_H_case1.eps}}
\caption{The temporal errors $e^{\varepsilon}(1)$ in three different norms for Example~2.}
\label{fig:efdcase2error}
\end{figure}
\end{center}

\begin{center}
\begin{figure}[htbp]
\centering
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_ex_space_t1_l2.eps}}~~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_ex_space_t1_linf.eps}}~~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_ex_space_t1_H.eps}}
\caption{The spatial errors $e^{\varepsilon}(1)$ in three different norms for Example~1.}
\label{fig:regefdspaceerr}
\end{figure}
\end{center}

\begin{center}
\begin{figure}[htbp]
\centering
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_space_case1_l2.eps}}~~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_space_case1_linf.eps}}~~~
\subfigure{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_space_case1_H.eps}}
\caption{The spatial errors $e^{\varepsilon}(1)$ in three different norms for Example~2.}
\label{fig:regefdcase2spaceerr}
\end{figure}
\end{center}

\subsection{Convergence of FDTD to the LogKGE}

We check the convergence rates of the finite difference schemes: SIFD (\ref{SIFD2}) and EFD (\ref{EFD}) to the LogKGE (\ref{LogKGE}) for Example~1. Tables \ref{SIFD2He} and \ref{EFDHe} display $l^{2}$-norm, $l^{\infty}$-norm, $H^{1}$-norm of $\tilde{e}^{\varepsilon}(1)$, respectively, for various mesh size $h$, time step $\tau$ and parameter $\varepsilon$.

\begin{table}
\centering
\caption{The convergence of the SIFD (\ref{SIFD2}) scheme to the LogKGE (\ref{LogKGE}) with different $\tau,h,\varepsilon$}

\begin{tabular}{cccccccccccc}
\toprule
$\|\tilde{e}^{\varepsilon} (1)\|_{l^{\infty}} $ &$  h=0.1 $ & $ h/2   $& $h/2^{2}   $ & $h/2^{3}    $& $h/2^{4}   $ &$h/2^{5}   $& \\
$ $ &$\tau=0.1$ & $\tau/2 $& $\tau/2^{2}$ & $\tau/2^{3} $& $\tau/2^{4}$ &$\tau/2^{5}$&\\
\midrule
$\varepsilon=10^{-3} $&$\mathbf{4.03E}$-$\mathbf{03}$ &1.03E-03& 7.34E-04 &7.66E-04 & 7.74E-04 &7.76E-04\\
 rate   &--& $1.97$ & 0.48 & -0.06 &-0.02&0.00\\\hline
$\varepsilon/4$ &4.03E-03 &$\mathbf{1.03E}$-$\mathbf{03}$& 2.59E-04 &2.24E-04 & 2.28E-04 &2.29E-04\\
 rate   &--& $\mathbf{1.97}$ & $1.99$ & $0.21$ &-0.03&-0.01\\\hline
$\varepsilon/4^{2}$&4.03E-03 &1.03E-03& $\mathbf{2.59E}$-$\mathbf{04}$ &6.52E-05& 6.70E-05 &6.75E-05\\
 rate   &--& $1.97$ & $\mathbf{1.99}$ & $1.99$ &-0.04&-0.01\\\hline
$\varepsilon/4^{3}$ &4.03E-03 &1.03E-03& 2.59E-04 &$\mathbf{6.50E}$-$\mathbf{05}$ & 1.96E-05 &1.98E-05\\
 rate   &--& $1.97$ & 1.99 & $\mathbf{1.99}$ &$1.73$&-0.01\\\hline
$\varepsilon/4^{4}$ &4.03E-03 &1.03E-03& 2.59E-04  &6.50E-05 & $\mathbf{1.63E}$-$\mathbf{05}$ &5.81E-06\\
 rate   &--& $1.97$ & 1.99 & $1.99$ &$\mathbf{2.00}$&1.49\\
\toprule
$\|\tilde{e}^{\varepsilon}(1) \|_{l^{2}} $ &$  h=0.1 $ & $ h/2   $& $h/2^{2}   $ & $h/2^{3}    $& $h/2^{4}   $ &$h/2^{5}   $& \\
$ $ &$\tau=0.1$ & $\tau/2 $& $\tau/2^{2}$ & $\tau/2^{3} $& $\tau/2^{4}$ &$\tau/2^{5}$&\\
\midrule
$\varepsilon=10^{-3} $&$\mathbf{7.72E}$-$\mathbf{03}$ &2.23E-03& 1.30E-03 &1.24E-03 & 1.25E-03 &1.25E-03\\
 rate   &--& $1.80$ & 0.78 & $0.06$ &-0.01&0.00\\\hline
$\varepsilon/4$ &7.73E-03 &$\mathbf{1.99E}$-$\mathbf{03}$& 5.93E-04 &3.69E-04 & 3.56E-04 &3.56E-04\\
 rate   &--& $\mathbf{1.96}$ & $1.75$ & $0.68$ &0.05&0.00\\\hline
$\varepsilon/4^{2}$&7.74E-03 &1.98E-03& $\mathbf{5.07E}$-$\mathbf{04}$ &1.59E-04 & 1.05E-04 &1.02E-04\\
 rate   &--& $1.97$ & $\mathbf{1.97}$ & $1.68$ &0.59&0.05\\\hline
$\varepsilon/4^{3}$ &7.74E-03 &1.98E-03& 5.02E-04 &$\mathbf{1.29E}$-$\mathbf{04}$ & 4.25E-05 &3.01E-05\\
 rate   &--& $1.96$ & 1.98 & $\mathbf{1.96}$ &$1.60$&0.50\\\hline
$\varepsilon/4^{4}$ &7.74E-03 &1.98E-03& 5.01E-04 &1.26E-04 & $\mathbf{3.26E}$-$\mathbf{05}$ &1.14E-05\\
 rate   &--& $1.96$ & 1.98 & $1.99$ &$\mathbf{1.95}$&1.51\\
\toprule
$\|\tilde{e}^{\varepsilon}(1) \|_{H^{1}}$ &$  h=0.1 $ & $ h/2   $& $h/2^{2}   $ & $h/2^{3}    $& $h/2^{4}   $ &$h/2^{5}   $& \\
$ $ &$\tau=0.1$ & $\tau/2 $& $\tau/2^{2}$ & $\tau/2^{3} $& $\tau/2^{4}$ &$\tau/2^{5}$&\\
\midrule
$\varepsilon=10^{-3} $&$\mathbf{1.08E}$-$\mathbf{02}$ &3.07E-03& 1.68E-03 &1.59E-03 & 1.59E-03 &1.59E-03\\
 rate   &--& $1.81$ & 0.87 & $0.09$ &0.00&0.00\\\hline
$\varepsilon/4$ &1.08E-02 &$\mathbf{2.79E}$-$\mathbf{03}$& 8.26E-04 &4.91E-04 & 4.65E-04 &4.64E-04\\
 rate   &--& $\mathbf{1.95}$ & $1.76$ & $0.75$ &0.08&0.00\\\hline
$\varepsilon/4^{2}$&1.08E-02 &2.77E-03& $\mathbf{7.10E}$-$\mathbf{04}$ &2.21E-04 & 1.42E-04 &1.36E-04\\
 rate   &--& $1.97$ & $\mathbf{1.96}$ & $1.68$ &0.64&0.07\\\hline
$\varepsilon/4^{3}$ &1.08E-02 &2.76E-03& 6.99E-04 &$\mathbf{1.80E}$-$\mathbf{04}$ & 5.92E-05 &4.10E-05\\
 rate   &--& $1.97$ & 1.98 & $\mathbf{1.96}$ &$1.60$&0.53\\\hline
$\varepsilon/4^{4}$ &1.08E-02 &2.76E-03& 6.98E-04 &1.76E-04 & $\mathbf{4.54E}$-$\mathbf{05}$ &1.60E-05\\
 rate   &--& $1.97$ & 1.99 & $1.99$ &$\mathbf{1.95}$&1.51\\
\bottomrule
\end{tabular}
\label{SIFD2He}

\end{table}

\begin{table}
\centering
\caption{The convergence of the EFD (\ref{EFD}) scheme to the LogKGE (\ref{LogKGE}) with different $\tau,h,\varepsilon$}

\begin{tabular}{cccccccccccc}
\toprule
$\|\tilde{e}^{\varepsilon} (1)\|_{l^{\infty}} $ &$  h=0.1 $ & $ h/2   $& $h/2^{2}   $ & $h/2^{3}    $& $h/2^{4}   $ &$h/2^{5}   $& \\
$ $ &$\tau=0.1$ & $\tau/2 $& $\tau/2^{2}$ & $\tau/2^{3} $& $\tau/2^{4}$ &$\tau/2^{5}$&\\
\midrule
$\varepsilon=10^{-3} $&$\mathbf{1.63E}$-$\mathbf{03}$ &6.76E-04& 7.43E-04 &7.68E-04 & 7.75E-04 &7.76E-03\\
 rate   &--& $1.27$ & -0.14 & -0.05 &-0.01&0.00\\\hline
$\varepsilon/4$ &1.70E-03 &$\mathbf{4.31E}$-$\mathbf{04}$& 2.14E-04 &2.25E-04 & 2.28E-04 &2.29E-04\\
 rate   &--& $\mathbf{1.98}$ & $1.01$ & -0.08 &-0.02&0.00\\\hline
$\varepsilon/4^{2}$&1.71E-03 &4.37E-04& $\mathbf{1.10E}$-$\mathbf{04}$ &6.58E-05 & 6.72E-05 &6.76E-05\\
 rate   &--& $1.97$ & $\mathbf{1.99}$ & $0.73$ &-0.03&-0.01\\\hline
$\varepsilon/4^{3}$ &1.71E-03 &4.37E-03& 1.10E-04 &$\mathbf{2.76E}$-$\mathbf{04}$ & 1.97E-05 &1.99E-05\\
 rate   &--& $1.96$ & 1.99 & $\mathbf{2.00}$ &$0.48$&-0.01\\\hline
$\varepsilon/4^{4}$ &1.71E-03 &4.37E-04& 1.10E-04 &2.76E-05 & $\mathbf{6.90E}$-$\mathbf{06}$ &5.82E-06\\
 rate   &--& $1.96$ & 1.99 & $2.00$ &$\mathbf{2.00}$&0.25\\
 \toprule
 $\|\tilde{e}^{\varepsilon} (1) \|_{l^{2}}$ &$  h=0.1 $ & $ h/2   $& $h/2^{2}   $ & $h/2^{3}    $& $h/2^{4}   $ &$h/2^{5}   $& \\
$ $ &$\tau=0.1$ & $\tau/2 $& $\tau/2^{2}$ & $\tau/2^{3} $& $\tau/2^{4}$ &$\tau/2^{5}$&\\
\midrule
$\varepsilon=10^{-3} $&$\mathbf{3.73E}$-$\mathbf{03}$ &1.43E-03& 1.23E-03 &1.24E-03 & 1.25E-03 &1.25E-03\\
 rate   &--& $1.39$ & 0.21 & -0.01 &-0.01&0.00\\\hline
$\varepsilon/4$ &3.71E-03 &$\mathbf{9.72E}$-$\mathbf{04}$& 4.06E-04 &3.55E-04 & 3.56E-04 &3.56E-04\\
 rate   &--& $\mathbf{1.93}$ & $1.26$ & $0.19$ &0.00&0.00\\\hline
$\varepsilon/4^{2}$&3.72E-03 &9.43E-04& $\mathbf{2.52E}$-$\mathbf{04}$ &1.15E-04 & 1.02E-04 &1.02E-04\\
 rate   &--& $1.98$ & $\mathbf{1.90}$ & $1.13$ &0.17&0.00\\\hline
$\varepsilon/4^{3}$ &3.72E-03 &9.42E-04& 2.38E-04 &$\mathbf{6.53E}$-$\mathbf{05}$ & 3.23E-05 &2.93E-05\\
 rate   &--& $1.98$ & 1.99 & $\mathbf{1.86}$ &$1.01$&0.14\\\hline
$\varepsilon/4^{4}$ &3.72E-03 &9.43E-04& 2.37E-04 &5.98E-05 & $\mathbf{1.69E}$-$\mathbf{05}$ &9.10E-06\\
 rate   &--& $1.98$ & 1.99 & $1.99$ &$\mathbf{1.82}$&0.89\\
 \toprule
$\|\tilde{e}^{\varepsilon} (1)\|_{H^{1}} $ &$  h=0.1 $ & $ h/2   $& $h/2^{2}   $ & $h/2^{3}    $& $h/2^{4}   $ &$h/2^{5}   $& \\
$ $ &$\tau=0.1$ & $\tau/2 $& $\tau/2^{2}$ & $\tau/2^{3} $& $\tau/2^{4}$ &$\tau/2^{5}$&\\
\midrule
$\varepsilon=10^{-3} $&5.05E-03 &1.92E-03& 1.59E-03 &1.59E-03 & 1.59E-03 &1.59E-03\\
 rate   &--& $1.39$ & 0.27 & $0.01$ &0.00&0.00\\\hline
$\varepsilon/4$ &$\mathbf{4.93E}$-$\mathbf{03 } $&1.31E-03& 5.51E-04&4.68E-04 &4.64E-04 & 4.64E-04 \\
 rate   &--& $1.91$ & $1.25$ & $0.23$ &0.01&0.00\\\hline
$\varepsilon/4^{2}$&4.92E-03 &$\mathbf{1.25E}$-$\mathbf{03}$& 3.39E-04 &1.56E-04 & 1.37E-04 &1.35E-04\\
 rate   &--& $\mathbf{1.98}$ & $1.88$ & $1.12$ &0.19&0.01\\\hline
$\varepsilon/4^{3}$ &4.92E-03 &1.24E-03& $\mathbf{3.14E}$-$\mathbf{04}$ &8.76E-05 & 4.12E-05 &3.98E-05\\
 rate   &--& $1.99$ & $\mathbf{1.98}$ & 1.84 &$0.99$&0.15\\\hline
$\varepsilon/4^{4}$ &4.92E-03 &1.24E-03& 3.11E-04 &$\mathbf{7.87E}$-$\mathbf{05}$ & 2.27E-05 &1.25E-05\\
 rate   &--& $1.99$ & 1.99 & $\mathbf{1.98}$ &$1.78$&0.86\\\hline
 $\varepsilon/4^{5}$ &4.92E-03 &1.24E-03& 3.11E-04 &7.79E-05 & $\mathbf{1.96E}$-$\mathbf{05}$ &5.92E-06\\
 rate   &--& $1.99$ & 2.00 & $2.00$ &$\mathbf{1.98}$&1.74\\
\bottomrule
\end{tabular}
\label{EFDHe}
\end{table}

\subsection{The evolution of the solution}

Figure \ref{fig:efdcase2ut1} represents the numerical solutions of the EFD (\ref{EFD}) at three different time $T=1,5,9$ for Example~2. We take the step size as $\tau=0.01\times2^{-7}$, and the mesh size as $h=2^{-7}$ at the large domain $[-16,16]$. From Figure \ref{fig:efdcase2ut1}, we can see that the numerical solutions of those two schemes are very close different $\varepsilon$ at fixed times. Besides the number of wave crests increase over time. We can conclude the two discritization schemes are stable under the stability conditions.

\begin{center}
\begin{figure}[htbp]
\centering
\subfigure[$T=1$]{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_u_t1_case1.eps}}~~
\subfigure[$T=5$]{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_u_t5_case1.eps}}~~
\subfigure[$T=9$]{\includegraphics[width=.33\textwidth, height=0.30\textwidth]{logkg_efd_order_eref_u_t9_case1.eps}}
\caption{The numerical solution $u^{\varepsilon}$ in three different time for the scheme EFD $(\ref{EFD})$ .}
\label{fig:efdcase2ut1}
\end{figure}
\end{center}

\section{Conclusions}\label{conclusion}
To avoid the singularity of the LogKGE (\ref{LogKGE}) at the origin, we proposed the RLogKGE (\ref{RLogKGE}) with a small regularized parameter $0<\varepsilon\ll1$. Two finite difference methods: SIFD, EFD were proposed and analyzed theoretically for the RLogKGE, which showed that the orders of accuracy are all second in both space and time. Besides, The numerical results demonstrated that the solutions of the RLogKGE (\ref{RLogKGE}) are linearly convergent to the LogKGE (\ref{LogKGE}) at $O(\varepsilon)$. In addition, the error bounds of FDTD methods to the LogKGE (\ref{LogKGE}) were numerically investigated and depended on $\tau,h,\varepsilon$.
\section*{Acknowledgments}

This work is supported by the National Natural Science Foundation of China (Grant No. 11971481,11901\\577), the Natural Science Foundation of Hunan (Grant No.S2017JJQNJJ0764,  S2020JJQNJJ1615), the Basic Research Foundation of National Numerical Wind Tunnel Project (No. NNW2018-ZT4A08). Research Fund of NUDT (Grand No. ZK17-03-27,ZK19-37), and the fund from Hunan Provincial Key Laboratory of Mathematical Modeling and Analysis in Engineering (Grand No.2018MMAEZD004).

\section*{References}
\bibliographystyle{elsarticle-num}
\bibliography{rlogkg_sifd}

\end{document}


TITLE: Solving for a vector that is orthogonal to two other vectors.
QUESTION [1 upvotes]: Suppose $[a, b, c]^T$ is orthogonal to both $[3, 4, -2]^T$ and $[4, 2, -3]^T$ and $a, b, c$ are not all zero. Which of the following is true?
A. $\frac{a}{b} = -5/7$
B. $\frac{b}{c} = -1$
C. $\frac{c}{a} = -1/6$
D. $\frac{a}{c} = 4/5$
E. There is not solution.
I know the solution is D, but I am having trouble coming to that answer. I tried making a system where $3a + 4b -2c = 0$ and $4a + 2b -3c = 0$ because the dot product must be zero. However, that only gives me the solution to $a$, $b$. How do I get $c$?

REPLY [3 votes]: If the equations are
$$ 3a + 4b -2c = 0\ \text{ (I)}$$
$$ 4a + 2b -3c = 0\ \text{ (II)}$$
then notice that if we multiply equation (II) by $-2$ and add it to the equation (I), we get
$$-5a+4c = 0$$
which gives $\dfrac{a}{c} = \dfrac{4}{5}$.


TITLE: Is the quasi-equational theory of groups the same as cancellative semigroups?
QUESTION [3 upvotes]: Consider the class of groups in the signature {*}. Is the quasi-equational theory of that class axiomatized by the associative law and the left and right cancellative laws?

REPLY [7 votes]: No.
Every group satisfies
$$
(xy\approx x'y')\wedge (zy\approx z'y')\wedge (zw\approx z'w')\to (xw\approx x'w')
$$
but this quasi-identity is not derivable from associativity + cancellativity. You can find this in Maltsev's papers Uber die Einbettug von assoziativen Systemen in Gruppen parts I and II. In fact, Maltsev proves that the quasi-equational theory of groups in the signature $\{*\}$ is not finitely axiomatizable relative to the theory axiomatized by the associative law and the cancellation laws.


TITLE: Determine the number of equivalence relations on the set {1, 2, 3, 4}
QUESTION [14 upvotes]: this was a question listed on my last "proofs and conjectures" midterm. It is similar to my previous post however this asks a different question which is throwing me off.

What is the number of equivalence relations on the set $\{1, 2, 3, 4\}\,$?

Do I simply list all equivalence relations and then count them?
This seems rather tedious.
Is there a general rule I can use as a shortcut to determine the number?
I was wondering if anybody could help guide me or list the solution, as I'm expected to know the material for the final.
Note: standard proof procedure is required in ALL my solutions for this course so I may come back and ask to clarify if a particular solution isn't specific enough.

REPLY [0 votes]: Number of equivalence relations or number of partitions is given by
$$S(n,k)\;=\;S(n-1,k-1)+kS(n-1,k),$$ where $n$ is the number of elements in a set and $k$ is the number of elements in a subset of partition,
with initial condition $S(n,1)=S(n,n)=1.$

e.g., $A=\{1,2,3,4\}$ (here $n=4$):
$k=4:\quad \quad S(4,4)=1.$
for $k=3:$
$$S(4,3)=S(3,2)+3*S(3,3)
      =S(2,1)+2*S(2,2)+3*S(3,3)
      =1+2+3=6.$$
for $k=2:$
$$S(4,2)=S(3,1)+2*S(3,2)
      =S(3,1)+2*[S(2,2)+2*S(2,2)]
      =1+2*[1+2*1]
      =7.$$
for $k=1:\quad \quad S(4,1)=1.$
So, the total number of equivalence relations $$= S(4,4)+S(4,3)+S(4,2)+S(4,1)=1+6+7+1=15.$$