\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): $$\label{1.1} \Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y=0, \quad \frac{f_{y}}{k}=\frac{k_x}{f},\quad \Big(\frac{f_y}{k}\Big)_y+\Big(\frac{k_x}{f}\Big)_x=-\varepsilon fk.$$ 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 $$\label{eA} \Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y =0, \quad \frac{f_y}{k}=\frac{k_x}{f}, \quad \Big(\frac{f_y}{k}\Big)_y+\Big(\frac{k_x}{f}\Big)_x = -f k,$$ are the traveling wave solutions given by $$\label{2.1} f= cm \operatorname{sech} \Big(\frac{c(m^2x+y)}{\sqrt{1+m^2}}\Big),\quad k= c \operatorname{sech}\Big(\frac{c(m^2x+y)}{\sqrt{1+m^2}} \Big),$$ 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 $$\label{2.2} f= m K(s),\quad k=K(s),\quad s=m^{2}x+y,$$ for some function $K$. By substituting \eqref{2.2} into the third equation in system \eqref{eA}, we find $$\label{2.3} (1+m^{2})(K(s)K''(s)-(K')^{2}(s))+{K^{4}(s) }=0.$$ Since $K\ne 0$, \eqref{2.3} implies that $K$ is non-constant. Thus \eqref{2.3} gives $$\label{2.4} (1+m^{2})\frac{K'{}^{2}}{K^{2}}+ K^{2}=c^{2}$$ for some positive real number $c_1$. After solving \eqref{2.4} we conclude that, up to translations and sign, $K$ is given by $$\label{2.5} K=c\operatorname{sech}\Big(\frac{cs}{\sqrt{1+m^2}}\Big).$$ 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 $$\label{2.6} y=-q(v)-xv^2, \quad f=v k$$ 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 $$\label{2.9} f_y=-\text{ \frac{k+v k_v}{q'(v)+2uv}},\quad k_x=k_u- \frac{v^2 k_v}{q'(v)+2uv}.$$ By substituting \eqref{2.6}, and \eqref{2.9} into the second equation of \eqref{eA} we obtain $$\label{2.10} k_u+\Big(\frac{v }{q'(v)+2uv}\Big)k=0.$$ After solving this equation we obtain $$\label{2.11} f= \frac{vA(v)}{\sqrt{2 uv+q'(v)}},\quad k= \frac{A(v)}{\sqrt{2 uv+q'(v)}}.$$ Now, by applying \eqref{2.8} and \eqref{2.11}, we find $$\label{2.12} \begin{gathered} f_x=\frac{v^2A(v)(v q''(v)-6uv-4q'(v)) -2v^3A'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}},\\ f_y=\frac{A(v)(v q''(v)-2uv-2q'(v)) -2vA'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}},\\ k_x=\frac{vA(v)(v q''(v)-2uv-2q'(v)) -2v^2A'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}},\\ k_y=\frac{A(v)(2u+q''(v)) -2A'(v)(2uv+q'(v))}{2 (2uv+q'(v))^{5/2}}. \end{gathered}$$ 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 $u$: $$\label{2.13} A^4(v)u^3+B(v)u^2+C(v)u+D(v)=0,$$ 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 $$\label{eB} \Big(\frac{k}{f}\Big)_x+\Big(\frac{f}{k}\Big)_y =0, \quad \frac{f_y}{k}=\frac{k_x}{f}, \quad \Big(\frac{f_y}{k}\Big)_y+\big(\frac{k_x}{f}\big)_x= f k,$$ 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 $$\label{3.4} f=m K(s),\quad k=K(s),\quad s=m^{2}x+y,$$ for some function $K$. By substituting \eqref{2.2} into the third equation in system \eqref{eB}, we find $$\label{3.5} (1+m^{2})(K(s)K''(s)-K'{}^{2}(s))=K^{4}(s).$$ Since $K\ne 0$, \eqref{3.5} implies that $K$ is non-constant. Thus \eqref{2.3} gives $$\label{3.6} (1+m^{2})\frac{K'{}^{2}}{K^{2}}- K^{2}=c_1$$ 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 $$\label{3.7} (1+m^{2})\frac{K'{}^{2}}{K^{2}}- K^{2}=c^2.$$ After solving \eqref{3.7} we conclude that, up to translations and sign, $K$ is given by $$\label{3.8} K=c \operatorname{csch}\Big(\frac{cs}{\sqrt{1+m^2}} \Big).$$ 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 $$\label{3.9} (1+m^{2})\frac{(K')^{2}}{K^{2}}- K^{2}=-c^2.$$ After solving \eqref{3.9} we conclude that, up to translations and sign, $K$ is given by $$\label{3.10} K=c \sec\big(\frac{cs}{\sqrt{1+m^2}}\big).$$ 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 $$\label{13.8} (1+m^{2}){K'{}^{2}}= K^{4}.$$ After solving \eqref{13.8} we conclude that, up to translations and sign, $K$ is given by $$\label{33.7} K=\frac{\sqrt{1+m^3}}{m^2x+y},$$ 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 $$\label{9.1} g(X,Y)=f^2_1\cdot g_1({\pi_1}_*X,{\pi_1}_*Y)+\dots + f^2_m\cdot g_m({\pi_m}_*X,{\pi_m}_*Y).$$ Let $N^{n-\ell}( \varepsilon)$ be an $(n-\ell)$-dimensional real space form of constant curvature $\varepsilon$. For \$\ell