\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 113, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/113\hfil The Cauchy problem] {The Cauchy problem and steady state solutions for a nonlocal Cahn-Hilliard equation} \author[Jianlong Han\hfil EJDE-2004/113\hfilneg] {Jianlong Han} \address{Jianlong Han \hfill\break Department of Mathematics, Michigan State University, East Lansing, MI 48824, USA} \email{hanjianl@math.msu.edu} \date{} \thanks{Submitted July 18, 2004. Published September 29, 2004.} \subjclass[2000]{35A05, 35M99} \keywords{Phase transition; long range interaction; steady state solution} \begin{abstract} We study the existence, uniqueness, and continuous dependence on initial data of the solution to the Cauchy problem and steady state solutions of a nonlocal Cahn-Hilliard equation on a bounded domain. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \newtheorem{prop}[theorem]{Proposition} \section{Introduction} We are concerned with two different problems, the first being the Cauchy problem for a nonlocal Cahn-Hilliard equation $$\label{c1.1} \begin{gathered} \frac{\partial u}{\partial t}=\triangle (\varphi(u)-J*u) \quad\mbox{in } \mathbb{R}^n\times (0,T),\\ u(x,0) = u_0(x), \end{gathered}$$ where $\varphi(u) = u+f(u)$, $f$ is bistable (e.g. $f(u)= au(u^2-1)$ for some $a>0$), $*$ is convolution, and $\int_{\mathbb{R}^n} J=1$. The second problem is for the steady state equation $$\label{s1.1} \begin{gathered} \int_{\Omega} J(x-y)dyu(x) -\int_{\Omega} J(x-y)u(y)dy +f(u)=C \quad\mbox{in }\Omega,\\ \int_\Omega u(x)dx=0, \end{gathered}$$ where $\Omega$ is a bounded domain, $C$ is a constant. The case when $\Omega = \mathbb{R}$ or $\mathbb{R}^n$ has been treated in \cite{bates2,bates3a,cham,chen} and references therein. To derive equations (\ref{c1.1}) and (\ref{s1.1}), we consider the free energy $$\label{n1.2} E(u)=C \iint{J(x-y)(u(x)-u(y))^2}dxdy +\int F(u(x))dx,$$ where $C$ is a constant, $F$ is the primitive of $f$, and $u$ represents the concentration of one of the species of a binary material. Following \cite{fife}, we consider the gradient flow for (\ref{n1.2}) in $H^{-1}_0(\Omega)$, where $H^{-1}_0$ is the space of distributions in the dual space of $H^1$ and with mean value zero. We do this since the total energy, $E$, decreases along the trajectories, and the average of $u$ should be conserved. We have $$\label{n1.3a} u_t = -\mathop{\rm grad}{}_{H^{-1}_0} E(u).$$ Since the representative of $\mathop{\rm grad}E(u)$ in $H^{-1}_0$ is $$\mathop{\rm grad}{}_{H^{-1}_0}E(u)=-\triangle(J(x-y)dyu(x) -\int J(x-y)u(y)dy+ f(u)),$$ (\ref{n1.3a}) gives $$\label{n1.3} \frac{\partial u}{\partial t}=\triangle(\int_{\Omega} J(x-y)dyu(x) -\int_{\Omega} J(x-y)u(y)dy+ f(u)).$$ In (\ref{n1.2}), making the approximation $$u(x) - u(y)\simeq \nabla u(x)\cdot (x-y),$$ and assuming $J$ to be isotropic, equation (\ref{n1.3a}) leads to $$\label{n1.4} \frac{\partial u}{\partial t}=-\triangle(d\triangle u- f(u)),$$ which is the classical Cahn-Hilliard equation. Equations (\ref{n1.3}) and (\ref{n1.4}) are important in the study of materials science for modelling certain phenomena such as spinodal decomposition, Ostwald ripening, and grain boundary motion. There is a lot of work on equation (\ref{n1.4}) (see for example \cite{Alikakos2, bates1,bates3,cahn,carr,cohen,elliott1,Du,Nic} and references therein). For equation (\ref{n1.3}), there are very few results. In \cite{bates4} and \cite{bates5}, we discussed the Neumann and Dirichlet boundary problems for (\ref{n1.3}). Here, we consider the Cauchy problem, where $\Omega = R^n$ and $\int J=1$. Note that the steady state solutions for (\ref{n1.3}) in a bounded domain with no flux boundary condition satisfy the equation in (\ref{s1.1}) without the constraint. In this paper, we prove the global existence and uniqueness of solutions for equation (\ref{c1.1}). Also we prove the existence of nonconstant solutions for equation (\ref{s1.1}). The techniques used in the proof of the latter result can also be applied to the nonlocal phase field system discussed in \cite{bates6}. We organize this paper as follows. In section 2, we establish the existence, uniqueness and continuous dependence on initial values for classical solutions of equation (\ref{c1.1}). In section 3, we prove that under certain conditions, there exists a discontinuous steady state solution for equation (\ref{s1.1}). \section{The Cauchy problem for the nonlocal Cahn-Hilliard equation} For $T>0$, let $Q_T = \mathbb{R}^n\times(0,T)$. We make the following assumptions: \begin{itemize} \item[(D1)] $f\in C^{2+\beta}(\mathbb{R})$ and $\varphi^{'} (u) \geq c$ for some positive constants $c$ and $\beta$, \item[(D2)] $J\in C^{2+\beta}(\mathbb{R}^n)$, $\triangle J \in L^1(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$, and $\int_{\mathbb{R}^n} J=1$. \end{itemize} First, we prove the uniqueness and continuous dependence of solutions on initial data. We have \begin{prop}\label{cprop1.1} Let $u_i$ $(i=1,2)$ be two solutions of \eqref{c1.1} with initial data $u_{i0}$ $(i=1,2)$. If conditions (D1)--(D2) are satisfied, if $u_i\in C([0,T],L^1(\mathbb{R}^n))\cap L^{\infty}(Q_T)$, and if $u_{i0}\in L^1(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$ $(i=1,2)$, then $$\label{c1.2} \sup_{0\leq t\leq T}\int_{\mathbb{R}^n}|u_1-u_2|dx \leq C(T)\int_{\mathbb{R}^n} |u_{10}-u_{20}|dx$$ for some constant $C(T)$. \end{prop} \begin{proof} For any $\tau\in (0,T)$, and $\psi\in C^{2,1}(Q_\tau)$, with $\psi =0$ for $|x|$ large enough, after multiplying (\ref{c1.1}) by $\psi$, integrating over $[0,\tau]\times \mathbb{R}^n$, we have % %\label{c1.3} \begin{align*} &\int_{\mathbb{R}^n}u_i(x,\tau)\psi(x,\tau)dx\\ & = \int_{\mathbb{R}^n}u_i(x,0)\psi(x,0)dx +\int_0^{\tau}\int_{\mathbb{R}^n}(u_i \psi_{t}+ \varphi(u_i)\triangle \psi) dx\,dt -\int_0^{\tau}\int_{\mathbb{R}^n}\psi \triangle J*u_i dx\,dt. \end{align*} Set $z= u_1-u_2$, $z_0 = u_{10}-u_{20}$, then the above equality gives \label{c1.4} \begin{aligned} \int_{\mathbb{R}^n}z(x,\tau)\psi(x,\tau)dx &= \int_{\mathbb{R}^n}z_0(x)\psi(x,0)dx +\int_0^{\tau}\int_{\mathbb{R}^n}z(x,t)( \psi_{t}+ b(x,t)\triangle \psi)dx\,dt\\ &\quad -\int_0^{\tau}\int_{\mathbb{R}^n}\psi\triangle J*z(x,t) dx\,dt, \end{aligned} where $$\label{c1.5} b(x,t) = \begin{cases} \frac{\varphi(u_1)-\varphi(u_2)}{u_1-u_2} & \mbox{for } u_1 \not= u_2,\\ \varphi'(u_1) & \mbox{for } u_1 = u_2. \end{cases}$$ Let $g(x)\in C_0^{\infty}(\mathbb{R}^n)$ have compact support, $0\leq g(x)\leq 1$, and take $\lambda >0$. We will choose $\psi$, above, to satisfy certain conditions. First, consider the following final value problem on a large ball $B_R(0)$ \label{c1.7} \begin{gathered} \frac{\partial \psi}{\partial t}= -b(x,t) \triangle \psi +\lambda \psi \quad\mbox{for } |x|R-\frac{1}{2},\\ |\nabla \xi_R(x)|,\; |\triangle \xi_R(x)|\leq C \end{gathered} for some constant $C$ which does not depend on $R$. Let $\gamma = \xi_R \psi$, where $\psi$ satisfies (\ref{c1.7}) in $B_R(0)$ and is zero outside. Using $\gamma$ instead of $\psi$ in (\ref{c1.4}), we have \begin{align*} %\label{c1.8} &\int_{\mathbb{R}^n}z(x,\tau)g\xi_Rdx - \int_{\mathbb{R}^n}\xi_R(x)z_0(x)\psi(x,0)dx +\int\int_{Q_{\tau}}(\triangle J*z -\lambda z)\xi_R\psi dx\,dt \\ &= \int\int_{Q_{\tau}}b(x,t)z(x,t)(2\nabla\xi_R\cdot\nabla \psi + \psi \triangle \xi_R)dx\,dt \equiv G(z,R). \end{align*} Since $u_1$ and $u_2$ belong to $L^{\infty}(Q_T)$, and since $b$ is positive, from estimates (\ref{d1})-(\ref{d3}) and (\ref{ch1}), we have \label{c1.9} \begin{aligned} |G(z,R)| &\leq \int_0^{\tau}\int_{B_R\backslash B_{R-1}}(b |u_1-u_2|((2|\nabla \xi_R\|\nabla \psi |+ |\psi \|\triangle \xi_R|)) \\ &\leq C \int_0^{\tau}\int_{B_R\backslash B_{R-1}} b(|u_1|+|u_2|)(|\nabla \psi|+1)dx\,dt\\ &\leq C \int_0^{\tau}\int_{B_R\backslash B_{R-1}} (|u_1|+|u_2|)dx\,dt. \end{aligned} Since $u_1$ and $u_2$ belong to $L^{1}(Q_T)$, letting $R \rightarrow \infty$ we have $G(z,R) \rightarrow 0$. This implies \begin{equation*} %\label{c1.10} \int_{\mathbb{R}^n}z(x,\tau)g(x)dx \leq\int_{\mathbb{R}^n}|z_0(x)|e^{-\lambda\tau}dx +\int_0^{\tau}\int_{\mathbb{R}^n}(|\triangle J*z -\lambda z|e^{\lambda(t-\tau)} dx\,dt. \end{equation*} Letting $\lambda\rightarrow 0$ and $g(x)\rightarrow sign\, z^+(x,\tau)$, we obtain $$\label{c1.11} \int_{\mathbb{R}^n}(u_1-u_2)^+dx \leq\int_{\mathbb{R}^n} |u_{10}-u_{20}|dx + C\int_0^{\tau}\int_{\mathbb{R}^n} |u_1-u_2|dx\,dt.$$ Interchanging $u_1$ and $u_2$ yields $%\label{c1.12} \int_{\mathbb{R}^n}|u_1-u_2|dx \leq\int_{\mathbb{R}^n} |u_{10}-u_{20}|dx + C\int_0^{\tau}\int_{\mathbb{R}^n} |u_1-u_2|dx\,dt.$ Inequality (\ref{c1.2}) follows from the above inequality and Gronwall's inequality. \end{proof} Next we prove the existence of a solution to (\ref{c1.1}). \begin{theorem}\label{cthm1.2} For any $T>0$, if $u_0(x)\in C^{2+\beta}_0(\mathbb{R}^n)$, and if $\varphi$ and $J$ satisfy assumptions $(D_1)-(D_2)$, then there exists a unique solution of (\ref{c1.1}) which belongs to $C^{2+\beta,\frac{2+\beta}{2}}(Q_T)\cap L^1(Q_T)\cap L^{\infty}(Q_T)$. \end{theorem} \begin{proof} Since $u_0(x) =0$ for $|x|$ large enough, we consider $$\label{c1.13} \begin{gathered} \frac{\partial u}{\partial t}=\triangle (\varphi(u)-J*u) \quad\mbox{in } B_R(0)\times (0,T),\\ u(x,t) = 0 \quad\mbox{on } \partial{B_R(0)}\times (0,T),\\ u(x,0) = u_0(x). \end{gathered}$$ From \cite[Theorem 2.4]{bates5}, there exists a unique solution $u(x,t)\in C^{2+\beta,\frac{2+\beta}{2}}(B_R(0)\times (0,T))$ of (\ref{c1.13}). Let $u(x,t)= v e^ t$ in (\ref{c1.13}), we have $$\label{c1.14} \begin{split} e^{t}v_t + v e^{ t}= \varphi'(u)e^{ t}\triangle v +\varphi''(u)|\nabla v|^2e^{2 t} -e^{ t} \triangle J*v . \end{split}$$ Multiplying (\ref{c1.14}) by $v$ and using $v\triangle v =\frac{1}{2}\triangle v^2 -|\nabla v|^2$, we obtain $$\label{c1.15} \begin{split} \frac{1}{2}(v^2)_t + v^2 &= \frac{1}{2}\varphi'(u)\triangle v^2+ \frac{1}{2} \varphi''(u)\nabla v\cdot\nabla v^2 e^{ t} -\varphi'(u)|\nabla v|^2-v \triangle (J*v). \end{split}$$ If there exists $(P_0, t_0)\in B_R(0)\times (0,T]$ such that $v^2(P_0,t_0) =\max v^2$, then we have $\triangle v^2(P_0, t_0)\leq 0$, $\nabla v^2(P_0, t_0) =0$, $\nabla v(P_0,t_0)=0$, $v^2_t(P_0, t_0) \geq 0$, and (\ref{c1.15}) yields $$\label{c1.16} v^2(P_0,t_0)\leq -\int_{B_R}\triangle J(P_0-y)v(y,t_0)dy v(P_0,t_0).$$ This yields $$\label{c1.17} \max|v|\leq M\int_{B_R}|v(y,t_0)|dy$$ for some constant $M$ which does not depend on $R$. Since $u = 0$ is also a solution of (\ref{c1.13}) with initial data $u_0=0$, by \cite[Theorem 2.5]{bates5}, we have $$\label{c1.18} \int_{B_R}|u(x,t)-0|dx \leq C(T)\int_{B_R}|u_0-0|dx$$ for some constant $C(T)$ which does not depend on $R$. Inequalities (\ref{c1.17}) and (\ref{c1.18}) imply $$\label{c1.19} \max|v|\leq C(T)\int_{B_R}|u_0|dx.$$ Since $u_0\in L^1(\mathbb{R}^n)$, we have $$\label{c1.20} \max|v|\leq B(T)$$ for some constant $B(T)$ which does not depend on $R$. This yields $$\label{c1.21} \max|u|\leq B(T)e^T$$ for some constant $B(T)$ which does not depend on $R$. We have proved the solution of (\ref{c1.13}) is uniformly bounded, i.e., $\max_{B_R\times [0,T]}|u(x,t)|\leq C$ for any $R>0$, where $C$ does not depend on $R$. A similar argument to that in the proof in \cite[Theorem 2.2]{bates5} yields $$\label{c1.22} \|u_R\|_{2+\beta} \leq C(K,T)$$ for any $R>K\equiv constant$, where $u_R$ is a solution of (\ref{c1.13}) in $B_R\times (0,T)$ and $C(K,T)$ is a constant which does not depend on $R$ ($\|\cdot\|_{2+\beta}$ is a $H\ddot{o}lder$ norm defined in \cite{Ladyzenskaja}). By employing the usual diagonal process, we can choose a sequence $\{R_i\}$ such that $u_{R_i}$, $Du_{R_i}$, and $D^2u_{R_i}$ converge to $u$, $Du$, and $D^2u$ pointwise, and $u$ satisfies equation (\ref{c1.1}). From (\ref{c1.18}) and (\ref{c1.21}), we also have $u\in L^1(Q_T)\cap L^{\infty}(Q_T)$. Uniqueness follows from Proposition \ref{cprop1.1}. \end{proof} \section{Steady state solutions for the nonlocal Cahn-Hilliard equation} In this part, we study equation (\ref{s1.1}). \begin{prop}\label{sprop1.1} Suppose $\Omega \subset \mathbb{R}^n$ is a closed and bounded set, $J(x)\geq 0$ and is continuous on $\mathbb{R}^n$, $supp J\supset B_{\delta} (0)$ for some positive constant $\delta$, and $f$ is nondecreasing. Then the only continuous solution of equation (\ref{s1.1}) is zero. \end{prop} \begin{proof} Without loss of generality, we assume that $f(0)= 0$. If $f(0)\neq 0$, we may use $f(u) - f(0)$ instead of $f(u)$ in (\ref{s1.1}). \noindent Case 1: $C\leq 0$ in equation (\ref{s1.1}). If the conclusion is not true, since $\int udx = 0$, and u is continuous on $\Omega$, there exists $P_0\in \Omega$ such that $u(P_0) = \max u(x)>0$. Let $$A=\{y\in \Omega : u(y) = \max u(x)\}.$$ We claim: There exist $P_0\in \partial A$ and $r>0$ such that $K:=(\Omega\setminus A)\cap B_r(P_0)$ has positive measure. If this is not true, we have $meas(\Omega\setminus A)=0$. This and $u(x)=\max u$ on $A$ imply $\int_\Omega u= \int_A u >0$. This contradicts $\int_\Omega u =0$. Since $\mathop{\rm supp}J \supset B_\delta(0)$ implies $\mathop{\rm supp}J(P_0-\cdot)\supset B_\delta(P_0)$, choosing $r_1 = \min\{\delta, r\}$ gives \begin{gather}\label{1} \mathop{\rm meas}(K\cap B_{r_1}(P_0))>0, \\ \label{2} J(P_0-y)>0\quad\mbox{on } K\cap B_{r_1}(P_0),\\ \label{3} u(P_0)-u(y)>0\quad \mbox{on } K\cap B_{r_1}(P_0). \end{gather} Inequalities (\ref{1})-(\ref{3}) imply \begin{equation*} %\label{4} \int_{\Omega}J(P_0-y)(u(P_0)-u(y))dy\geq \int_{K\cap B_{r_1}(P_0)}J(P_0-y)(u(P_0)-u(y))dy>0. \end{equation*} This and $f(u(P_0))\geq 0$ imply $$\label{s1.2} \int_{\Omega}J(P_0-y)u(P_0)dy -\int_{\Omega} J(P_0-y)u(y)dy+ f(u(P_0))>0,$$ contradicting (\ref{s1.1}). \noindent Case 2: $C>0$ in (\ref{s1.1}). In this case, taking $P_0$ such that $u(P_0)= \min u<0$ leads to a contradiction in a similar way. \end{proof} If $f'(u)$ changes sign, we make the following assumptions: \begin{itemize} \item[(E1)] $\Omega = (-1,1)$ if $\dim\Omega =1$, $\Omega = (-1,1)\times \Omega'$ if $\dim\Omega >1$. \item[(E2)] $J(x) = J(|x|)$, $J(x)\geq 0$, and \begin{align*} M \geq\sup_{x\in\Omega}\int_{\Omega}J(x-y)dy \geq \inf_{x\in\Omega}\int_{\Omega}J(x-y)dy \geq m>0 \end{align*} for positive constants $M$ and $m$. \item[(E3)] $f\in C^1(R)$ is odd, $f(1) =0$, there exist $\delta>0$ and $a\in (0,1)$ such that $f'(u)\geq \delta$ on $[a,\infty)$, and $f(-a)\geq (1+a)M$. \item[(E4)] $C=0$ in (\ref{s1.1}). \end{itemize} \begin{remark}\label{r3.2}\rm Condition (E3) implies that $f(-1)=0$, $f'(u)\geq \delta$ on $(-\infty,-a]$, and $-f(a)\geq (1+a)M$. \end{remark} Let $j(x)=\int_{\Omega} J(x-y)dy$. From $(E_2)$, we have $$\label{s1} m\leq j(x)\leq M.$$ Dividing (\ref{s1.1}) by $j(x)$, we consider $$\label{s1.3} \begin{gathered} u(x)-\frac{1}{j(x)}\int_{\Omega} J(x-y)u(y)dy +\frac{f(u(x))}{j(x)}=0,\\ \int_\Omega u(x)dx=0. \end{gathered}$$ \begin{theorem}\label{sthm1.3} If assumptions $(E_1)-(E_4)$ are satisfied, then there exists a solution of equation (\ref{s1.3}) such that $$\label{s1.4} u(x)\begin{cases} \geq a & \text{for x\in M_1\equiv (0,1)\times \Omega'},\\ \leq -a &\text{for x\in M_2\equiv(-1,0)\times \Omega'}. \end{cases}$$ Moreover, we have $$\label{ss1.4} -1\leq u(x)\leq 1.$$ \end{theorem} \begin{proof} Following \cite{bates2}, we let $B=\{u\in L^{\infty}(\Omega): u(-x_1,x')=-u(x_1,x'),\, \,u(x)\in[a,1] \text{ for x\in M_1}\}.$ The definition of $B$ implies that $u(x)\in [-1,-a]$ for $x\in M_2$. Define $$Tu(x)=u(x)+h[\frac{1}{j(x)}\int_{\Omega}J(x-y)u(y)dy-u(x) -\frac{1}{j(x)} f(u(x))].$$ We want to show $T:B\rightarrow B$ is a contraction map if $h$ is small enough. In fact, since $j(x) = \int_{\Omega}J(x-y)dy$, with assumption (E2), we have $j(-x_1,x') =j(x_1,x')$. And if $u(x)\in B$, we have \label{s1.5} \begin{aligned} &T(u(-x_1,x'))\\ &=u(-x_1,x')+\frac{h}{j(-x_1,x')}\int_{-1}^1 \int_{\Omega'}J(-x_1-y_1,x'-y')u(y_1,y')dy_1dy'\\ &\quad -hu(-x_1,x')+\frac{h}{j(-x_1,x')}f(u(-x_1,x'))\\ &=-u(x_1,x')-\frac{h}{j(x_1,x')}\int_{-1}^1\int_{\Omega'} J(-x_1+z_1,x'-y')u(z_1,y')dz_1dy'\\ &\quad +hu(x_1,x')-\frac{h}{j(x_1,x')}f(u(x_1,x'))\\ &=-u(x_1,x')-\frac{h}{j(x_1,x')}\int_{-1}^1\int_{\Omega'}J(x_1-z_1,x'-y')u(z_1,y')dz_1dy'\\ &\quad +hu(x_1,x') -\frac{h}{j(x_1,x')} f(u(x_1,x')) \\ &=-(u(x_1,x')+\frac{h}{j(x_1,x')}\int_{-1}^1\int_{\Omega'}J(x_1-z_1,x'-y')u(z_1,y')dz_1dy' \\ &\quad -hu(x_1,x')+\frac{h}{j(x_1,x')} f(u(x_1,x')))\\ &=-T(u(x_1,x')). \end{aligned} Choose $h$ small enough such that $$\label{s1.6} h\frac{1}{j(x)} f'(u)<1-h$$ for $u\in [-1,-a]\cup [a,1]$ and $x\in \Omega$. This implies that $u-h[u+\frac{1}{j(x)} f(u)]$ is increasing in $u$ on $[a,1]$. Since $u(y)\geq a$ for $y\in M_1$, and $u(y)\geq -1$ for $y\in M_2$, we have for $x\in M_1$ \begin{align*} % \label{s1.7} &Tu(x) \\ &=h\frac{1}{j(x)}\int_\Omega J(x-y)u(y)dy+u-h[u+\frac{1}{j(x)}f(u)]\\ &\geq h\frac{1}{j(x)}\int_\Omega J (x-y)u(y)dy +a-ha-h\frac{1}{j(x)} f(a)\\ &=h\frac{1}{j(x)}\int_{M_1} J(x-y)u(y)dy + h\frac{1}{j(x)}\int_{M_2} J(x-y)u(y)dy+a-ha-h\frac{1}{j(x)}f(a)\\ &\geq ha\frac{1}{j(x)}\int_{M_1} J(x-y)dy -h\frac{1}{j(x)}\int_{M_2} J(x-y)dy +a-ha-h\frac{1}{j(x)}f(a)\\ &=a-ha\frac{1}{j(x)}\int_{M_2} J(x-y)dy -h\frac{1}{j(x)}\int_{M_2} J(x-y)dy-\frac{h}{j(x)}f(a)\\ &\geq a-\frac{h}{j(x)}[(1+a)\int_{M_2} J(x-y)dy+f(a)] \geq a \end{align*} by (E3). Also, \label{s1.8} \begin{aligned} Tu(x) &=h\frac{1}{j(x)}\int_\Omega J(x-y)u(y)dy+u-h[u+\frac{1}{j(x)}f(u)]\\ &\leq h\frac{1}{j(x)}\int_\Omega J(x-y)u(y)dy + 1-h-h\frac{1}{j(x)} f(1) \leq 1 \end{aligned} for $x\in M_1$. Estimates (\ref{s1.5})-(\ref{s1.8}) imply that $T$ maps $B$ to $B$. For $u, v \in B$, choosing $h$ small enough so that $0<1-h(1+\delta\frac{1}{M})<1$, we have % \label{s1.9} \begin{align*} &\|Tu-Tv\|_{\infty}\\ &=\|(u-v)+\frac{h}{j(x)}\int_{\Omega} J(x-y)(u(y)-v(y))dy \\ &\quad - h(u(x)-v(x)) -\frac{h}{j(x)}(f(u)-f(v))\|_{\infty}\\ &=\|(1-h-\frac{hf'(\theta u+(1-\theta)v)}{j(x)})(u-v)+\frac{h}{j(x)}\int_{\Omega} J(x-y)(u(y)-v(y))dy \|_{\infty} \\ &\leq (1-h(1+\delta\frac{1}{M}))\|u-v\|_\infty+h\|(u-v)\|_\infty\\ &\leq (1-h\delta\frac{1}{M})\|u-v\|_\infty, \end{align*} where $\theta(x)\in (0,1)$ for all $x\in \Omega$. Here we used (E3) and the fact that for any $x\in \Omega$ either $u(x),\,v(x)\geq a$ or $u(x),\, v(x)\leq -a$. Therefore, $T$ is a contraction map from $B$ to $B$. There exists a unique fixed point $u(x)$ such that $Tu =u$. Estimates (\ref{ss1.4}) follows from the definition of B. \end{proof} \begin{remark}\rm If we just consider the solution to \label{new1} \int_{\Omega} J(x-y)u(x)dy -\int_{\Omega} J(x-y)u(y)dy + f(u)=0\quad\mbox{in } \Omega without the condition $\int_{\Omega} udx =0$, then the conditions that $f$ is odd and $J(x) = J(|x|)$ are not necessary. In this case, we can use a similar method to that in \cite{bates2} to prove the existence of a discontinuous solution under conditions (E2), (E3)', and (E4), where \begin{itemize} \item[(E3)'] $f\in C^1(R)$, $f(-1)=f(c)=f(1) =0$ for $c\in (-1,1)$, there exist $\delta>0$, $a\in (0,1)$, $b\in (-1,0)$ such that $f'(x)\geq \delta$ on $[a,\infty)\cup (-\infty, b)$, $f(a)\leq -(1+a)M$, and $f(b)\geq (1+b)M$, where $M$ is defined in (E2). \end{itemize} \end{remark} \subsection*{Acknowledgement} The author would like to thank Professor Peter W. Bates for his helpful discussions. \begin{thebibliography}{00} \bibitem{Alikakos2} N. D. Alikakos, P. W. 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