\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 228, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/228\hfil nonlinear surface wind waves model]
{Analysis of a nonlinear surface wind waves model via Lie group method}

\author[B. Gao \hfil EJDE-2016/228\hfilneg]
{Ben Gao}

\address{Ben Gao \newline
College of Mathematics,
Taiyuan University of Technology,
Taiyuan 030024, China}
\email{benzi0116@163.com}


\thanks{Submitted July 21, 2016. Published August 23, 2016.}
\subjclass[2010]{35J05, 35E05, 43A80, 26A18}
\keywords{Nonlinear surface wind waves; symmetry analysis; 
\hfill\break\indent similarity reduction;
power series method}

\begin{abstract}
 This article focuses on two aspects. Firstly, symmetry analysis is performed
 for a nonlinear equation which can model surface wind wave patterns in nature.
 As a byproduct, the similarity reductions and exact solutions of the equation
 are constructed based on the optimal systems.
 Secondly, the explicit solutions are considered by the power series method.
 Moreover, the convergence of the power series solutions are shown.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Manna \cite{ma} studied the surface wind waves qualitative behavior,
represented by the nonlinear equation
\begin{equation} \label{e1}
u_{xt}=-\frac{3g}{hc_0}u-uu_{xx}+(u_x)^2,
 \end{equation}
where $u(x,t)$ is considered as an unidirectional surface
wave propagating in the $x$-direction on a fluid medium
involved in a large scale flow. $h$ is the unperturbed initial depth, $g$ 
is the acceleration of gravity, $c_0$ is the wind velocity and subscripts denote 
partial derivatives. For the sake of providing more information to understand 
\eqref{e1},  some works have been devoted to study \eqref{e1} \cite{ma,yx}. 
The author in \cite{ma} provided some peakon solutions with amplitude, velocity, 
and width in interrelation and static compacton solutions
with amplitude and width in interrelation for \eqref{e1}. 
The exact explicit traveling wave solutions of \eqref{e1} are given by using 
the method of dynamical systems in \cite{yx}. However, as the authors known, 
the Lie group analysis and explicit power series solutions of \eqref{e1} are 
left as open problems.

The application of Lie transformation group theory for the
construction of solutions of nonlinear partial
differential equations (PDEs) is one of the most active fields of
research in the theory of nonlinear PDEs and applications \cite{blu1,bj,mc,olv,ws}. 
The main idea of Lie group method is to transform solutions of a system 
of differential equations to other solutions. Once the
symmetry group of a system of differential equations has be determined, 
one can directly use the defining property of such a group and construct
new solutions to the system from known ones.

The rest of article is arranged as follows: 
Section 2 concentrates on symmetries of \eqref{e1}; 
in Section 3, the similarity reductions for \eqref{e1} are dealt with and 
exact solutions are provided by using Lie group method; 
in Section 4, the explicit solutions for the reduced equations are obtained 
by using the power series method; the last section contains a conclusion
of our work.

\section{Lie point symmetry}

    In this section, we apply Lie point symmetry method for \eqref{e1}, and 
obtain its infinitesimal generators, commutation table of Lie algebra.

First of all, let us consider a one-parameter Lie symmetry group admitted
 by \eqref{e1} with an infinitesimal operator of the form
\begin{equation} \label{e2}
X=\xi\partial_x+\tau\partial_t+\phi\partial_u.
\end{equation}
where $\xi,\tau,\phi$ are functions of $x,t,u$ respectively and are called
infinitesimals of the symmetry group.

The classical infinitesimal Lie invariance criterion for
\eqref{e1} with respect to the operator \eqref{e2} reads as in 
\cite{blu2,olv},
\[
 \operatorname{pr}X^{(2)}\Big[u_{xt}+\frac{3g}{hc_0}u+uu_{xx}-(u_x)^2\Big]=0
\]
for any $u$ solves \eqref{e1}. Here, the symbol
$\operatorname{pr}X^{(2)}$ is the usual $2$ th-order prolongation of the
operator \cite{blu2,olv}, in this situation,
\[
 \operatorname{pr}X^{(2)}=X+\phi_x^{(1)}\frac{\partial}{\partial {u_x}}+\phi_{xx}^{(2)}\frac{\partial}{\partial {u_{xx}}}+\phi_{xt}^{(2)}\frac{\partial}{\partial {u_{xt}}},
\]
where
\begin{gather*}
 \phi_x^{(1)}=D_x\phi-u_xD_x\xi-u_tD_x\tau, \\
 \phi_{xx}^{(2)}=D_x^2(\phi-\xi u_x-\tau u_t)+\xi u_{xxx}+\tau u_{xxt}, \\
 \phi_{xt}^{(2)}=D_xD_t(\phi-\xi u_x-\tau u_t)+\xi u_{xxt}+\tau u_{xtt},
\end{gather*}
and $D_x,D_t$ stand for the total derivative operators, for example,
\[
 D_t=\frac{\partial}{\partial t}+u_t\frac{\partial}{\partial u}
 +u_{tx}\frac{\partial}{\partial u_x}+u_{tt}\frac{\partial}{\partial
 u_t}+\dots.
\]

Substituting $\operatorname{pr}X^{(2)}$ into \eqref{e1} and splitting it
 with respect to the different order derivatives
of $u$, one obtains a system of linear over-determining equations
for the unknown functions $\xi,\tau$ and $\phi$. We can find the following
equations for the symmetry group of \eqref{e1}
\begin{equation} \label{e3}
\begin{gathered}
\xi_x=-\tau_t,\quad  \xi_t=\xi_u=0,\\
\tau_x=\tau_u=0, \quad \tau_{tt}=0,\\
\phi=-2u\tau_t.
\end{gathered}
\end{equation}
Solving above \eqref{e3}, we obtain
\[
 \xi=-c_1x+c_3,\quad \tau=c_1t+c_2, \quad \phi=-2c_1u,
\]
where $c_1,c_2,c_3$ and $c_4$ are arbitrary constants, we find that \eqref{e1} 
admits the operators
\[
X_1=-x\partial_x+t\partial_t-2u\partial_u, \quad
X_2=\partial_t,  \quad
X_3=\partial_x.
\]
It is easy to check that $\{X_1,X_2,X_3\}$ is closed under the Lie bracket. 
In fact, we have
\begin{gather*}
[X_1,X_1]=[X_2,X_2]=[X_3,X_3]=0,\\
[X_1,X_2]=-[X_2,X_1]=-X_2,  [X_1,X_3]=-[X_3,X_1]=X_3,\\
[X_2,X_3]=-[X_3,X_2]=0.
\end{gather*}

Furthermore,  to obtain their adjoint representation, employing the following 
Lie series
\[
 \operatorname{Ad}(\exp (\epsilon X_i))X_j
=X_j-\epsilon[X_i,X_j]+\frac{1}{2}\epsilon^2[X_i,[X_i,X_j]]-\dots,
\]

we can compute the following results
\begin{gather*}
\operatorname{Ad}(\exp (\epsilon X_i))X_i=X_i,\quad i=1,2,3,  \quad
\operatorname{Ad}(\exp (\epsilon X_1))X_2=e^\epsilon X_2, \\
\operatorname{Ad}(\exp (\epsilon X_2))X_1=X_1-\epsilon X_2,\quad
\operatorname{Ad}(\exp (\epsilon X_1))X_3=e^{-\epsilon}X_3,\\
 \operatorname{Ad}(\exp (\epsilon X_3))X_1=X_1+\epsilon X_3, \quad
 \operatorname{Ad}(\exp (\epsilon X_2))X_3=X_3,\\
\operatorname{Ad}(\exp (\epsilon X_3))X_2= X_2,
\end{gather*}
where $\epsilon$ is an arbitrary constant.

Based on the adjoint representation of the infinitesimal operators, we
obtain the optimal systems of \eqref{e1} as follows,
\[
\{X_1,X_2,X_3,X_3+aX_2\},
\]
where $a$ is an arbitrary constant.

\begin{remark} \label{rmk2.1} \rm
The optimal systems of \eqref{e1} can also be obtained using the results
of the paper \cite{1}.
\end{remark}

\section{Similarity reductions and exact solutions}

    The symmetry group properties are very useful for the construction
of invariant solutions of the differential equation under study.
Next, We will consider the following similarity
reductions and group-invariant solutions for \eqref{e1} based on the optimal system.
\smallskip

\noindent\textbf{Case 1:} Reduction by $X_1$.
Integrating the characteristic
equation for $X_1$, we get similarity variables
\[
 z=tx, \quad p=\frac{u}{x^2},
\]
and the group-invariant solution is $p=f(z)$, that is,
\begin{equation} \label{e4}
 u=x^2f(tx).
\end{equation}
 Substituting this expression into \eqref{e1}, we obtain
\begin{equation} \label{e5}
\frac{3g}{hc_0}f-2f^2+3f'-z^2f'^2+zf''+z^2ff''=0
 \end{equation}
where $f'=\frac{df}{dz}$.
\smallskip

\noindent\textbf{Case 2:} Reduction by $X_2$.
Similarly, we have $u=f(z)$ in which $z=x$. Substituting it into \eqref{e1}, 
we obtain
\begin{equation} \label{e6}
\frac{3g}{hc_0}f-f'^2+ff''=0
\end{equation}
where $f'=\frac{df}{dz}$.
\smallskip

\noindent\textbf{Case 3:} For generator $X_3$, we have $z=t,u=f(z)$. 
The corresponding reduction equation is
\begin{equation} \label{e7}
\frac{3g}{hc_0}f=0.
\end{equation}
Therefore, \eqref{e1} has a solution $u=0$. Obviously, the solution 
is not meaningful.
\smallskip

\noindent\textbf{Case 4:} For generator $X_3+aX_2$, we have $z=-ax+t,u=f(z)$. 
The corresponding reduction equation is
\begin{equation} \label{e8}
\frac{3g}{hc_0}f-a^2f'^2-af''+a^2ff''=0
 \end{equation}
where $f'=\frac{df}{dz}$.

\section{Explicit power series solutions}

    In Section 4, we obtained the reduced equations by using symmetry analysis.
 The power series can be used to treat differential equations, 
including many complicated differential equations with nonconstant coefficients 
\cite{hh}. In this section, we solve the nonlinear
ODEs \eqref{e5}, \eqref{e6}, and \eqref{e8} by the power series method.
\subsection{Explicit solutions to \eqref{e5}}
Now, we seek a solution of \eqref{e5} in a power series of the form
\begin{equation} \label{e9}
 f(z)=\sum_{n=0}^\infty p_nz^n,
\end{equation}
where the coefficients $p_n$ $(n=0,1,2,\dots)$ are constants to be determined.
Substituting \eqref{e9} into \eqref{e5}, we have
\begin{equation}
\begin{aligned}
&\frac{3g}{hc_0}\sum_{n=0}^\infty p_nz^n-2
 \sum_{n=0}^\infty\sum_{k=0}^np_kp_{n-k}z^n+3\sum_{n=0}^\infty(n+1) p_{n+1}z^n\\
&-z^2\sum_{n=0}^\infty\sum_{k=0}^n(k+1)(n+1-k)p_{k+1}p_{n+1-k}z^n
 +z\sum_{n=0}^\infty(n+1)(n+2)p_{n+2}z^n\\
&+z^2\sum_{n=0}^\infty\sum_{k=0}^n(k+1)(k+2)p_{n-k}p_{k+2}z^n=0.
\end{aligned}
\end{equation}
From this equality, comparing coefficients, we have
\begin{equation} \label{e11}
\begin{gathered}
p_1=\frac{1}{3}p_0(2p_0-\frac{3g}{hc_0}),\\
p_2=\frac{1}{8}p_1(4p_0-\frac{3g}{hc_0}).
\end{gathered}
\end{equation}
Generally, for $n\geq 2$, we have
\begin{equation} \label{e12}
\begin{aligned}
 p_{n+1}&=\frac{1}{(n+1)(n+3)}\Big\{2p_{n-1}p_1
 -\big(\frac{3g}{hc_0}-2p_0\big)p_n\\
&\quad +\sum_{k=0}^{n-2}[2p_kp_{n-k}+(k+1)(n-1-k)p_{k+1}p_{n-1-k}\\
&\quad -(k+1)(k+2)p_{n-2-k}p_{k+2}]\Big\}.
\end{aligned}
\end{equation}

In view of this equality, we can obtain all the coefficients $p_i(i\geq3)$ 
of the power series \eqref{e9}, e.g.,
\begin{equation} \label{e13}
p_3=\frac{1}{15}(3p_1^2-\frac{3g}{hc_0}p_2+2p_0p_2).
\end{equation}

Therefore, for arbitrary chosen constant number $p_0$, the other terms of the 
sequence $\{p_n\}_{n=0}^\infty$\\can be determined by
\eqref{e11} and \eqref{e12}.
 This implies that for \eqref{e5}, there is a power series solution \eqref{e9}
with the coefficients constructed by \eqref{e11} and \eqref{e12}.

Now we show that the convergence of the power series
solution \eqref{e9} of \eqref{e5}. In fact, from \eqref{e12}, we have
\[
|p_{n+1}|{\leq}M[|p_{n-1}|+|p_n|+\sum_{k=0}^{n-2}(|p_k||p_{n-k}|
+|p_{k+1}||p_{n-1-k}|+|p_{n-2-k}||p_{k+2}|)],
\]
for  $n=2,3,\dots$,
where  $M=\max \{|p_1|,|\frac{3g}{hc_0}-2p_0|,1\}$.

Now, we define a power series  $R=R(z)=\sum_{n=0}^\infty r_nz^n$
by 
\[
r_i=|p_i|,  \quad i=0,1,2,
\]
 and
\[
r_{n+1}=M[r_{n-1}+r_n+\sum_{k=0}^{n-2}(r_kr_{n-k}+r_{k+1}r_{n-1-k}
+r_{n-2-k}r_{k+2})]
\]
where  $n=2,3,\dots$. Then, it is easily seen that
\[
 |p_n|{\leq}r_n,  \quad n=0,1,2,\dots.
\]
Thus, the series  $R=R(z)=\sum_{n=0}^\infty r_nz^n$ is a majorant series 
of \eqref{e9}. Next, we show that the series $R = R(z)$ has a positive radius of
convergence.
\begin{align*}
 R(z)&=r_0+r_1z+r_2z^2+\sum_{n=2}^\infty r_{n+1}z^{n+1} \\
&=r_0+r_1z+r_2z^2+M\Big(\sum_{n=2}^\infty r_{n-1}z^{n+1}
 +\sum_{n=2}^\infty r_nz^{n+1}\\
&\quad +\sum_{n=2}^\infty\sum_{k=0}^{n-2} r_kr_{n-k}z^{n+1}
 +\sum_{n=2}^\infty \sum_{k=0}^{n-2} r_{k+1}r_{n-1-k}z^{n+1} \\
&\quad +\sum_{n=2}^\infty \sum_{k=0}^{n-2} r_{n-2-k}r_{k+2}z^{n+1}\Big)\\
&=r_0+r_1z+r_2z^2+M[z^2(R-r_0)+z(R-r_0-r_1z)\\
&\quad +zR(R-r_0-r_1z)+z(R-r_0)^2   +zR(R-r_0-r_1z)].
\end{align*}
Consider now the implicit functional equation with respect to the independent 
variable $z$,
\begin{align*}
F(z,R)&=R-r_0-r_1z-r_2z^2-M[z^2(R-r_0)+z(R-r_0-r_1z)\\
&\quad +zR(R-r_0-r_1z)  +z(R-r_0)^2+zR(R-r_0-r_1z)].
\end{align*}
Since $F$ is analytic in the neighborhood of $(0,r_0)$ and 
$ F(0,r_0) = 0, F_R(0,r_0) = 1\neq 0$.
if we choose the parameter $r_0 = |p_0|$ properly. By the implicit function
theorem \cite{wr}, we see that $R=R(z)$ is analytic in a neighborhood of 
the point $(0,r_0)$ and with the positive radius. 
This implies that the power series \eqref{e9} converges in a neighborhood
of the point $(0,r_0)$. This completes the proof.


Hence, the power series
solution \eqref{e9} for \eqref{e5} is an analytic solution. 
The power series solution of \eqref{e5} can be written as
\begin{align*}
f(z)&=p_0+p_1z+p_2z^2+\sum_{n=2}^\infty p_{n+1}z^{n+1}\\
&=p_0+\frac{1}{3}p_0(2p_0-\frac{3g}{hc_0})z+\frac{1}{8}p_1(4p_0-\frac{3g}{hc_0})z^2
+\sum_{n=2}^\infty\frac{1}{(n+1)(n+3)}\{2p_{n-1}p_1\\
&\quad -(\frac{3g}{hc_0}-2p_0)p_n+\sum_{k=0}^{n-2}
[2p_kp_{n-k}+(k+1)(n-1-k)p_{k+1}p_{n-1-k}\\
&\quad-(k+1)(k+2)p_{n-2-k}p_{k+2}]\}z^{n+1}.
\end{align*}
Furthermore, the explicit power series solution of \eqref{e1} is
\begin{align*}
 u(x,t)
&=p_0x^2+p_1tx^3+p_2t^2x^4+\sum_{n=2}^\infty p_{n+1}t^{n+1}x^{n+3}\\
&=p_0x^2+\frac{1}{3}p_0(2p_0-\frac{3g}{hc_0})tx^3
 +\frac{1}{8}p_1(4p_0-\frac{3g}{hc_0})t^2x^4 \\
&\quad +\sum_{n=2}^\infty\frac{1}{(n+1)(n+3)}\{2p_{n-1}p_1\\
&\quad -(\frac{3g}{hc_0}-2p_0)p_n+\sum_{k=0}^{n-2}[2p_kp_{n-k}
 +(k+1)(n-1-k)p_{k+1}p_{n-1-k}\\
&\quad -(k+1)(k+2)p_{n-2-k}p_{k+2}]\}t^{n+1}x^{n+3},
\end{align*}
where $p_0$ is an arbitrary constant, the other coefficients $p_n(n\geq1)$ 
depend on \eqref{e11} and \eqref{e12} completely.

\subsection{Explicit solutions to \eqref{e6}}

    Similarly, we seek a solution of \eqref{e6} in a power series of 
the form \eqref{e9}. Substituting it into \eqref{e6}, and comparing
coefficients, we have
\begin{equation} \label{e14}
 p_2=\frac{1}{2p_0}(p_1^2-\frac{3g}{hc_0}p_0).
\end{equation}
Generally, for $n\geq 1$, we have
\begin{equation} \label{e15}
\begin{aligned}
p_{n+2}&=\frac{1}{(n+1)(n+2)p_0}\Big\{\sum_{k=0}^{n-1}(k+1)[(n+1-k)p_{k+1}
 p_{n+1-k} \\
&\quad -(k+2)p_{n-k}p_{k+2}] +(n+1)p_1p_{n+1}-\frac{3g}{hc_0}p_n\Big\}.
\end{aligned}
\end{equation}

In view of \eqref{e14} and \eqref{e15}, we can obtain all the coefficients 
$p_i(i\geq3)$ of the power series \eqref{e9}, e.g.,
\[
p_3=\frac{p_1}{6p_0}(2p_2-\frac{3g}{hc_0}),  \quad
p_4=\frac{p_2}{12p_0}(2p_2-\frac{3g}{hc_0}).
\]

Therefore, for arbitrary chosen constant numbers $p_0\neq 0$ and $p_1$, 
the other terms of the sequence $\{p_n\}_{n=0}^\infty$ can be determined by
\eqref{e14} and \eqref{e15}. This implies that for \eqref{e6}, there is a
 power series solution \eqref{e9}
with the coefficients constructed by \eqref{e14} and \eqref{e15}.

\subsection{Explicit solutions to \eqref{e8}}

    Now, we seek a solution of \eqref{e8} in a power series of the form \eqref{e9}. 
Substituting \eqref{e9} into \eqref{e8}, and comparing
coefficients, we have
\begin{equation} \label{e16}
p_2=\frac{1}{2a(ap_0-1)}(a^2p_1^2-\frac{3g}{hc_0}p_0).
\end{equation}
Generally, for $n\geq 1$, we have
\begin{equation} \label{e17}
\begin{aligned}
p_{n+2}&=\frac{1}{(n+1)(n+2)a(ap_0-1)}\\
&\quad\times \Big\{a^2\sum_{k=0}^{n-1}(k+1)[(n+1-k)p_{k+1}p_{n+1-k}
-(k+2)p_{n-k}p_{k+2}]\\
&\quad +a^2(n+1)p_1p_{n+1}-\frac{3g}{hc_0}p_n\Big\}.
\end{aligned}
\end{equation}
In view of \eqref{e16} and \eqref{e17}, we can obtain all the coefficients 
$p_n(n\geq3)$ of the power series \eqref{e9}, e.g.,
\[
p_3=\frac{p_1}{6a(ap_0-1)}(2a^2p_2-\frac{3g}{hc_0}), \quad
p_4=\frac{p_2}{12a(ap_0-1)}(2a^2p_2-\frac{3g}{hc_0}).
\]

Thus, for arbitrary chosen constant numbers $a\neq0,p_0\neq\frac{1}{a}$ and 
$p_1$, the other terms of the sequence $\{p_n\}_{n=0}^\infty$ can be constructed by
\eqref{e16} and \eqref{e17}. This implies that for \eqref{e8}, there is a 
power series solution \eqref{e9}
with the coefficients determined by \eqref{e16} and \eqref{e17}.

Furthermore, the explicit power series solution of \eqref{e1} is
\begin{align*}
u(x,t)&=p_0+p_1(-ax+t)+p_2(-ax+t)^2+\sum_{n=1}^\infty p_{n+2}(-ax+t)^{n+2}\\
&=p_0+p_1(-ax+t)+\frac{1}{2a(ap_0-1)}(a^2p_1^2-\frac{3g}{hc_0}p_0)(-ax+t)^2\\
&\quad +\sum_{n=1}^\infty\frac{1}{(n+1)(n+2)a(ap_0-1)}
\Big\{a^2\sum_{k=0}^{n-1}(k+1)[(n+1-k)p_{k+1}p_{n+1-k}\\
&\quad -(k+2)p_{n-k}p_{k+2}]+a^2(n+1)p_1p_{n+1}
 -\frac{3g}{hc_0}p_n\Big\}(-ax+t)^{n+2},
\end{align*}
where $a\neq0,p_0\neq\frac{1}{a}$ and $p_1$ are arbitrary constants, 
the other coefficients $p_n(n\geq2)$ depend on \eqref{e16} and \eqref{e17} completely.


\begin{remark} \label{rmk4.1} \rm
The proofs of convergence of the power series solutions to \eqref{e6}, \eqref{e8} 
are similar to the one of \eqref{e5}. The details are omitted here. 
Furthermore, such power series solutions can greatly enrich the solutions
 of the \eqref{e1} and converge quickly, so it is convenient for computations
 in both theory and applications.
\end{remark}

\begin{remark} \label{rmk4.2} \rm
The solutions of equations \eqref{e6} and \eqref{e8} (obviously, it is sufficiently
to consider only the case $a=1$) in the explicit form can also be found without
using the power series method (see the handbook \cite{2}) and read as
\[
 f(z)=f(x)=be^{cx}-\frac{d}{c^2}+\frac{d^2}{4bc^4}e^{-cx},
\]
and
\[
 f(z)=be^{cz}-(\frac{d}{c^2}-1)+\frac{d}{4bc^2}(\frac{d}{c^2}-1)e^{-cz} (\mbox{for} a=1),
\]
where $d = \frac{3g}{hc_0}$, $b\neq 0$, $c \neq 0$.
\end{remark}

\subsection*{Conclusions}
    In this article, We present Lie approach for a nonlinear surface wind waves 
model. As a byproduct, new invariant solutions are constructed based on the 
optimal systems which may exert potential applications
about the problems of physical phenomena. Moreover, we apply the power series 
method to obtain the explicit solutions of \eqref{e1}. 
It can be seen that the Lie symmetry analysis and the power series method are 
very efficient to research the explicit solutions of PDEs.

\subsection*{Acknowledgments}
This research was supported by the Natural Science Foundation of Shanxi
(No. 2014021010-1).


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\end{thebibliography}

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