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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 83, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/83\hfil Over-determined systems of PDEs]
{Solutions to over-determined systems of partial differential equations
 related to Hamiltonian stationary Lagrangian surfaces}

\author[B.-Y. Chen\hfil EJDE-2012/83\hfilneg]
{Bang-Yen Chen}

\address{Bang-Yen Chen \newline
Department of Mathematics,
Michigan State University, 619 Red Cedar Road,
East Lansing, Michigan 48824-1027, USA}
\email{bychen@math.msu.edu}

\thanks{Submitted December 1, 2011. Published May 23, 2012.}
\subjclass[2000]{35N05, 35C07, 35C99}
\keywords{Over-determined PDE system;  traveling wave solution;
\hfill\break\indent exact solution; Hamiltonian stationary Lagrangian surfaces}

\begin{abstract}
 This article concerns  the over-determined system  of partial 
 differential equations
 $$
 \Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y=0, \quad
 \frac{f_{y}}{k}=\frac{k_x}{f},\quad
 \Big(\frac{f_y}{k}\Big)_y+\Big(\frac{k_x}{f}\Big)_x=-\varepsilon fk\,.
 $$
 It was shown in \cite[Theorem 8.1]{cgz} that this system
 with $\varepsilon=0$ admits traveling wave solutions as well as
 non-traveling wave solutions.
 In this article we solve completely this system when $\varepsilon\ne 0$.
 Our main result states that this system  admits only traveling wave 
 solutions, whenever $\varepsilon \ne 0$. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

A submanifold $M$ of a K\"ahler manifold $\tilde M$ is called Lagrangian if
the complex structure $J$ of $\tilde M$ interchanges each tangent space
$T_pM$ with the corresponding normal space $T_p^{\perp}M$, $p\in M$
(cf. \cite{book}).

A vector field $X$ on a K\"ahler manifold $\tilde M$ is called Hamiltonian
if ${\mathcal L}_X\omega=f\omega$ for some function $f\in C^\infty(\tilde M)$,
 where $\mathcal L$ is the Lie derivative. Thus, there is a smooth real-valued
function $\varphi$ on $\tilde M$ such that $X=J\tilde\nabla \varphi$, where
 $\tilde\nabla$ is the gradient in  $\tilde M$.
The diffeomorphisms of the flux $\psi_t$ of $X$ satisfy $\psi_t\omega=e^{h_t}\omega$.
Thus they transform Lagrangian submanifolds of $\tilde M$ into Lagrangian
 submanifolds.
A normal vector field $\xi$ to a Lagrangian immersion $\psi:M\to \tilde M$
is called Hamiltonian if $\xi=J\nabla f$, for some $f\in C^\infty(M)$, where
$\nabla f$ is the gradient of $f$.
A Lagrangian submanifold of a K\"ahler manifold is called Hamiltonian stationary
if it is a critical point of the volume under Hamiltonian deformations.

Related to the classification of Hamiltonian stationary Lagrangian surfaces of
constant curvature $\varepsilon$ in a K\"ahler surface of constant holomorphic
sectional curvature $4\varepsilon$  via a construction method introduced by
 Chen,  Dillen,  Verstraelen and  Vrancken in \cite{cdvv}
(see also \cite{c08,cd,cg}), one has to determine  the exact solutions of
 the following overdetermined system of PDEs (see \cite{cgz,DH} for details):
\begin{equation}  \label{1.1}
 \Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y=0, \quad
 \frac{f_{y}}{k}=\frac{k_x}{f},\quad
 \Big(\frac{f_y}{k}\Big)_y+\Big(\frac{k_x}{f}\Big)_x=-\varepsilon fk.
\end{equation}

 This over-determined system was solved completely in \cite{cgz} for the
case $\varepsilon=0$. In particular, it was shown  that
system \eqref{1.1} with $\varepsilon=0$ admits traveling wave solutions
as well as non-traveling wave solutions. More precisely, we have the
following result from \cite[Theorem 8.1]{cgz}.

\begin{theorem} \label{T:9.1}
 The solutions $\{f,k\}$ of the  over-determined PDE system
\[
\Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y =0,\quad
 \frac{f_y}{k}=\frac{k_x}{f}, \quad
 \Big(\frac{f_y}{k}\Big)_y+\Big(\frac{k_x}{f}\Big)_x=0,
\]
are  the following:
   \begin{gather}
\label{8.1}  f(x,y) =\pm k(x,y)=a e^{b(x+y)}; \\
\label{8.2}  f(x,y) =a me^{b(m^2x+y)},\quad  k(x,y)=\pm a e^{b(m^2x+y)};  \\
\label{8.3}  f(x,y)=\frac{a}{\sqrt{x}}e^{c\arctan \sqrt{-y/x}}, \quad
   k(x,y)=\pm \frac{a}{\sqrt{-y}}e^{c\arctan \sqrt{-y/x}},
  \end{gather}
where $a,b,c,m$ are real numbers with $a,c,m\ne 0$ and $m\ne \pm 1$.
\end{theorem}

The main purpose of  this article is to solve the over-determined system \eqref{1.1}
completely. Our main result states that
  the over-determined $PDE$ system \eqref{1.1} with $\varepsilon\ne 0$ admits
only traveling wave solutions.


\section{Exact solutions of the over-determined system with $\varepsilon =1$}


 \begin{theorem}  \label{thm1}
The solutions $\{f,k\}$ of the over-determined PDE system
\begin{equation} \label{eA}
\Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y =0, \quad
 \frac{f_y}{k}=\frac{k_x}{f}, \quad
 \Big(\frac{f_y}{k}\Big)_y+\Big(\frac{k_x}{f}\Big)_x = -f k,
\end{equation}
are  the traveling wave solutions given by
\begin{equation}\label{2.1}
f= cm \operatorname{sech} \Big(\frac{c(m^2x+y)}{\sqrt{1+m^2}}\Big),\quad
k= c \operatorname{sech}\Big(\frac{c(m^2x+y)}{\sqrt{1+m^2}} \Big),
\end{equation}
where $c$ and $m$ are nonzero real numbers.
\end{theorem}

\begin{proof}
First, let us assume that $f= mk$ for some nonzero real number $m$.
Then the first equation of system \eqref{eA} holds identically.

 If  $\{f,k\}$ satisfies the second equation of system \eqref{eA},
then we have $k_{x}=m^2 k_y$, which implies that
 \begin{equation} \label{2.2}
f= m K(s),\quad k=K(s),\quad s=m^{2}x+y,
\end{equation}
for some function $K$.  By substituting \eqref{2.2} into the third equation
in system \eqref{eA}, we find
\begin{equation} \label{2.3}
(1+m^{2})(K(s)K''(s)-(K')^{2}(s))+{K^{4}(s) }=0.
\end{equation}
Since $K\ne 0$, \eqref{2.3} implies that $K$ is non-constant.
Thus \eqref{2.3} gives
\begin{equation} \label{2.4}
(1+m^{2})\frac{K'{}^{2}}{K^{2}}+ K^{2}=c^{2}
\end{equation}
for some positive real number $c_1$. After solving \eqref{2.4} we conclude that,
up to translations and sign, $K$ is given by
\begin{equation}\label{2.5}
 K=c\operatorname{sech}\Big(\frac{cs}{\sqrt{1+m^2}}\Big).
\end{equation}
 Now, after combining \eqref{2.2} and \eqref{2.5} we obtain the traveling wave
 solutions of the over-determined PDE system given by \eqref{2.1}.

Next, let us assume that $v=f(x,y)/k(x,y)$ is a non-constant function.
It follows from the first equation of system \eqref{eA} that
$\frac{\partial v}{\partial y}\ne 0$.  Therefore, after solving  the first
 equation of system \eqref{eA}, we obtain
\begin{equation}  \label{2.6}
 y=-q(v)-xv^2, \quad  f=v k
\end{equation}
for some function $q$. Let us consider the new variables $(u,v)$ with $u=x$
and $v$ being defined by \eqref{2.6}. Then we have
\begin{gather}  \label{2.7}
\frac{\partial x}{\partial u}=1,\quad
\frac{\partial x}{\partial v}=0,\quad
\frac{\partial y}{\partial u}=-v^2,\quad
\frac{\partial y}{\partial v}=-q'(v)-2uv,  \\
 \label{2.8} \frac{\partial u}{\partial x}=1,\quad
\frac{\partial u}{\partial y}=0,\quad
\frac{\partial v}{\partial x}=\frac{-v^2}{q'(v)+2uv},\quad
\frac{\partial v}{\partial y}=\frac{-1}{q'(v)+2uv},
\end{gather}
It follows from \eqref{2.6}, \eqref{2.7} and \eqref{2.8} that
\begin{equation}  \label{2.9}
f_y=-\text{ $ \frac{k+v k_v}{q'(v)+2uv}$},\quad
k_x=k_u- \frac{v^2 k_v}{q'(v)+2uv}.
\end{equation}
By substituting \eqref{2.6}, and \eqref{2.9} into  the second equation
of  \eqref{eA} we obtain
\begin{equation}  \label{2.10}
k_u+\Big(\frac{v }{q'(v)+2uv}\Big)k=0.
\end{equation}
After solving this equation we obtain
\begin{equation} \label{2.11}
f=  \frac{vA(v)}{\sqrt{2 uv+q'(v)}},\quad
k=   \frac{A(v)}{\sqrt{2 uv+q'(v)}}.
\end{equation}
Now, by applying \eqref{2.8} and \eqref{2.11}, we find
\begin{equation}\label{2.12}
\begin{gathered}
f_x=\frac{v^2A(v)(v q''(v)-6uv-4q'(v)) -2v^3A'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}},\\
f_y=\frac{A(v)(v q''(v)-2uv-2q'(v)) -2vA'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}},\\
k_x=\frac{vA(v)(v q''(v)-2uv-2q'(v)) -2v^2A'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}},\\
k_y=\frac{A(v)(2u+q''(v)) -2A'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}}.
\end{gathered}
\end{equation}
After substituting \eqref{2.12} into the last equation in  \eqref{eA} and by
applying \eqref{2.7} and \eqref{2.8}, we obtain a polynomial equation of degree 3
in $u$:
\begin{equation}  \label{2.13}
A^4(v)u^3+B(v)u^2+C(v)u+D(v)=0,
\end{equation}
where $B,C$ and $D$ are functions in $v$. Consequently, we must have
$A(v)=0$ which is a contradiction according to \eqref{2.13}.
 Therefore this case cannot happen.
\end{proof}

\section {Exact solutions of the over-determined system with $\varepsilon =-1$}

\begin{theorem} \label{thm2}
The solutions $\{f,k\}$ of the over-determined PDE system
 \begin{equation} \label{eB}
 \Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y =0, \quad
 \frac{f_y}{k}=\frac{k_x}{f}, \quad
 \Big(\frac{f_y}{k}\Big)_y+\big(\frac{k_x}{f}\big)_x= f k,
\end{equation}
are  the following traveling wave solutions:
\begin{gather}
\label{3.1} f=c m\operatorname{csch}\Big(\frac{c(m^2 x+y)}{\sqrt{1+m^2}} \Big),\quad
 k=c \operatorname{csch}\Big(\frac{c(m^2 x+y)}{\sqrt{1+m^2}} \Big); \\
\label{3.2} f=c m\sec\Big(\frac{c(m^2 x+y)}{\sqrt{1+m^2}}\Big),\quad
k= c \sec\Big(\frac{c(m^2 x+y)}{\sqrt{1+m^2}}\Big);\\
\label{3.3} f=\frac{m\sqrt{1+m^2}}{m^2x+y},\quad
 k=\frac{\sqrt{1+m^2}}{m^2x+y} ,
\end{gather}
where $c$ and $m$ are nonzero real numbers.
 \end{theorem}

\begin{proof}
First, let us assume that $f=mk$ for some nonzero real number $m$.
Then the first equation of system \eqref{eB} holds identically.
As in the previous section, we obtain from the second equation of
system \eqref{eB} that
\begin{equation} \label{3.4}
f=m K(s),\quad k=K(s),\quad s=m^{2}x+y,
\end{equation}
for some function $K$.  By substituting \eqref{2.2} into the third equation
in system \eqref{eB}, we find
\begin{equation} \label{3.5}
(1+m^{2})(K(s)K''(s)-K'{}^{2}(s))=K^{4}(s).
\end{equation}
Since $K\ne 0$, \eqref{3.5} implies that $K$ is non-constant.
Thus \eqref{2.3} gives
\begin{equation} \label{3.6}
 (1+m^{2})\frac{K'{}^{2}}{K^{2}}- K^{2}=c_1
\end{equation}
for some real number $c_1$.

 If $c_1>0$, we put $c_1=c^2$ with $c\ne 0$. Then \eqref{3.6} becomes
\begin{equation} \label{3.7}
 (1+m^{2})\frac{K'{}^{2}}{K^{2}}- K^{2}=c^2.
\end{equation}
 After solving \eqref{3.7} we conclude that, up to translations and sign, $K$
is given by
\begin{equation} \label{3.8}
K=c \operatorname{csch}\Big(\frac{cs}{\sqrt{1+m^2}} \Big).
\end{equation}
Now, after combining \eqref{3.4} and \eqref{3.8} we obtain the traveling
 wave solutions  \eqref{3.1}.


If $c_1<0$, we put $c_1=-c^2$ with $c\ne 0$. Then \eqref{3.6} becomes
\begin{equation} \label{3.9}
(1+m^{2})\frac{(K')^{2}}{K^{2}}- K^{2}=-c^2.
\end{equation}
 After solving \eqref{3.9} we conclude that, up to translations and sign,
$K$ is given by
\begin{equation} \label{3.10}
K=c \sec\big(\frac{cs}{\sqrt{1+m^2}}\big).
\end{equation}
 By combining \eqref{3.4} and \eqref{3.10} we obtain the traveling wave
 solutions of the over-determined PDE system given by \eqref{3.2}.

If $c_1=0$,  \eqref{3.6} becomes
\begin{equation} \label{13.8}
(1+m^{2}){K'{}^{2}}= K^{4}.
\end{equation}
After solving \eqref{13.8} we conclude that, up to translations and sign,
$K$ is given by
\begin{equation} \label{33.7}
 K=\frac{\sqrt{1+m^3}}{m^2x+y},
\end{equation}
which yields solutions \eqref{3.3}.

Finally, by applying a argument  similar to the one given in section 2,
we conclude that the remaining case is impossible.
\end{proof}


\section{Applications to Hamiltonian-stationary Lagrangian surfaces}

Let $(M_j,g_j),j=1,\dots,m$, be Riemannian manifolds, $f_i$ a positive
function on $ M_1\times\dots\times M_m$ and $\pi_i: M_1\times\dots\times M_m\to M_i$
the $i$-th canonical projection for $i=1,\dots,m.$ The \emph{twisted  product}
$$
{}_{f_1}M_1\times\dots\times_{f_m}M_m
$$
is the product manifold $M_1\times\dots\times M_m$ equipped with the twisted
product metric $g$ defined by
\begin{equation} \label{9.1}
g(X,Y)=f^2_1\cdot g_1({\pi_1}_*X,{\pi_1}_*Y)+\dots +
f^2_m\cdot g_m({\pi_m}_*X,{\pi_m}_*Y).
\end{equation}
Let $N^{n-\ell}( \varepsilon)$ be an $(n-\ell)$-dimensional real space form
of constant  curvature $ \varepsilon$. For $\ell<n-1$ we consider
the following twisted product:
\begin{equation} \label{9.2}
{}_{f_1}I_1\times\dots\times_{f_\ell}I_\ell\times_1
N^{n-\ell}( \varepsilon)
\end{equation}
 with twisted product metric given by
\begin{equation} \label{9.3}
g=f^2_1dx_1^2+\dots+f^2_\ell dx_\ell^2+g_0,
\end{equation}
where $g_0$ is the canonical metric of $N^{n-\ell}( \varepsilon)$ and
$I_1,\dots,I_\ell$ are open intervals.
When $\ell=n-1$, we shall replace $N^{n-1}(\varepsilon)$ by an open interval.
If the twisted product  is a real-space-form $M^n( \varepsilon)$, it is called
 a \emph{twisted product decomposition} of  $M^n( \varepsilon)$ (cf. \cite{cdvv}).
 We denote such a decomposition by ${\mathcal TP}_{f_1\dots f_\ell}^n( \varepsilon)$.

We recall the following result from \cite[Theorem 3.2]{cgz} (see also \cite{DH}).

\begin{theorem} \label{T:3.2}
Let $f,k$ be a pair of positive functions  satisfying PDE system \eqref{eA}.
Then, up to rigid motions of $\tilde M^2(4\varepsilon)$, there is a unique
$H$-stationary Lagrangian  isometric immersion:
\begin{equation} \label{3.12}
 L_{f,k}: {\mathcal TP}_{f^2k^2}^2( \varepsilon) \to \tilde M^2(4\varepsilon)
\end{equation}
whose second fundamental form  satisfies
\begin{equation} \label{3.13}
h\Big(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_1}\Big)
= J \frac{\partial}{\partial x_1},\quad
 h\Big(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}\Big)= 0,\quad
 h\Big(\frac{\partial}{\partial x_2},\frac{\partial}{\partial x_2}\Big)
= J\frac{\partial}{\partial x_2}.
\end{equation}
\end{theorem}
If the two twistor functions $f^2$ and $ k^2$ are equal  and if they satisfy
PDE system \eqref{1.1},  then the corresponding Hamiltonian-stationary adapted
Lagrangian immersion of  $ {\mathcal TP}_{f^2 k^2}^2(\varepsilon)$ is
said to be \emph{of type I}.
If the two twistor functions $f^2$ and $ k^2$ are unequal,  then the corresponding
Hamiltonian-stationary adapted Lagrangian immersion  is said to be \emph{of type II}.

 By applying Theorem \ref{thm1} and results of \cite[Section 5]{cgz}, we can determine
 all type II adapted Hamiltonian stationary Lagrangian surfaces in the
complex projective plane $CP^{2}(4)$ of constant holomorphic sectional
curvature 4. In fact,  by combining Theorem \ref{thm1} and \cite[Section 5]{cgz}
we have the following.

\begin{corollary}  \label{coro1}
A type II adapted Hamiltonian-stationary Lagrangian surface in  $CP^{2}(4)$ is
 congruent to $\pi\circ L$, where $\pi:S^{5}(1)
\to CP^{2}(4)$ is the Hopf fibration and $L$ is given by
\begin{align*}
L(x,y)
&=\frac{\operatorname{sech} \big( \frac{m^2x+y}{\sqrt{1+m^2}} \big)}
{\sqrt{2+m^2}}
\bigg(\frac{2 m \sqrt{2+m^2}}{\sqrt{1+5m^2}} e^{i(x+y)/2}
\sin \Big( \frac{\sqrt{1+5m^2}}{2\sqrt{1+m^2}}(x-y)\Big),\\
&\quad e^{i(x+y)/2}  \Big[ \sqrt{1+ m^2}
\cos\Big(\frac{\sqrt{1+5m^2}}{2\sqrt{1+m^2}}(x-y)\Big)\\
&\quad -  i(1-m^2) {\sqrt{1+5m^2}} \sin
\Big(\frac{\sqrt{1+5m^2}}{2\sqrt{1+m^2}} (x-y)\Big) \Big], \\
&\quad \frac{1}{\sqrt{2}} \sqrt{1+\cosh \big(\frac{2m^2x+2y}{\sqrt{1+m^2}} \big)}
\Big(1-i \sqrt{1+m^2} \tanh \Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big)\Big)
\bigg)
\end{align*}
for some positive number $m\ne 1$.
\end{corollary}

Similarly, by applying Theorem \ref{thm2} and results of \cite[Section 7]{cgz}
we can determine all type II adapted Hamiltonian-stationary Lagrangian
surfaces in the complex hyperbolic plane $CH^{2}(-4)$ of constant holomorphic
sectional curvature $-4$. More precisely, we have the following result.


\begin{corollary}  \label{coro2}
A type II adapted Hamiltonian-stationary Lagrangian surface in  $CH^{2}(-4)$
is  congruent to $\pi\circ L$, where $\pi:H^{5}_1(-1)
\to CH^{2}(-4)$ denotes the Hopf fibration and $L(x,y)$ is given by one
of the following five immersions:\\
(a)
\[ 
L= \Big(1-\frac{ i(1+m^2)}{m^2x+y} ,\,
    \frac{ m\sqrt{1+m^2}}{m^2x+y}  e^{ix} ,\,
    \frac{ \sqrt{1+m^2}}{m^2x+y} e^{iy}\Big);
\]
(b)
\[ 
 L=\operatorname{sech} \Big(\frac{x+3y}{2\sqrt{3}} \Big)
 \Big(\frac{x-y+4i}{2}e^{i(x+y)/2} ,\,
  \frac{x-y}{2} e^{i(x+y)/2},\,
  \sqrt{3}+2i \tan \Big(\frac{x+3y}{2\sqrt{3}} \Big) \Big);
\]
(c)
\begin{align*} 
L&= \Big( \frac{ \sqrt{3m^4+2m^2-1} \cosh (\alpha (x-y))+  i(m^2-1)
 \sinh( \alpha (x-y))} { m \sqrt{3m^2-1} e^{-i(x+y)/2}}
 \sec \Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big),\\
& \quad
\frac{2m e^{i(x+y)/2}}{ \sqrt{3m^2-1} }
\sec \Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big)
 \sinh ( \alpha (x-y)),  \frac{1}{m}+ \frac{i\sqrt{1+m^2}}{m}
 \tan\Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big)   \Big);
\end{align*}
(d)
\begin{align*}
  L&= \Big( \frac{ \sqrt{1-2m^2-3m^4} \cos (\beta (x-y))
+  i(1-m^2)\sin(\beta (x-y))}
{m \sqrt{1-3m^2} e^{-i(x+y)/2} }
\sec \Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big),
\\
&\quad
\frac{2m e^{i(x+y)/2}}{\sqrt{1-3m^2}}
 \sec \Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big) \sinh (\beta (x-y)),
 \frac{1}{m}+ \frac{i\sqrt{1+m^2}}{m}
 \tan\Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big)   \Big);
\end{align*}
(e)
\begin{align*}
  L&= \frac{1}{\sqrt{2+m^2}}
\operatorname{csch} \Big(\frac{m^2x+y}{\sqrt{1+m^2}}\Big)
\bigg(\sinh\Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big)
-i\sqrt{1+m^2}\cosh\Big(\frac{m^2x+y}{\sqrt{1+m^2}} \Big),\\
&\quad  e^{i(x+y)/2} \Big\{\sqrt{1+m^2}
 \cos \Big(\frac{\sqrt{1+5m^2}}{2\sqrt{1+m^2}}(x-y)\Big)\\
&\quad +\frac{i(m^2-1)}{\sqrt{1+5m^2}}
 \sin \Big(\frac{\sqrt{1+5m^2}}{2\sqrt{1+m^2}}(x-y)\Big)\Big\},\\
&\quad  \frac{2m\sqrt{2+m^2}}{\sqrt{1+5m^2}} e^{i(x+y)/2}
 \sin \Big(\frac{\sqrt{1+5m^2}}{2\sqrt{1+m^2}}(x-y)\Big)\bigg),
\end{align*}
 where $\alpha$ and $\beta$ are constants given by
$$
\alpha =\frac{\sqrt{3m^2-1}}{2\sqrt{1+m^2}},\quad
\beta =\frac{\sqrt{1-3m^2}}{2\sqrt{1+m^2}}.
$$
\end{corollary}

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