\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