\documentclass[twoside]{article} \pagestyle{myheadings} \markboth{\hfil The nonlocal bistable equation \hfil EJDE--2002/02} {EJDE--2002/02\hfil Adam J. J. Chmaj \& Xiaofeng Ren \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 02, pp. 1--12. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % The nonlocal bistable equation: Stationary solutions on a bounded interval % \thanks{ {\em Mathematics Subject Classifications:} 45G10. \hfil\break\indent {\em Key words:} local minimizers, monotone solutions. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted July 18, 2001. Published January 2, 2002. \hfil\break\indent A.J.J. Chmaj was supported by a Marie Curie Fellowship of the European Community \hfil\break\indent IHP programme under contract number HPMFCT-2000-00465 and in part by \hfil\break\indent NSF grant DMS-0096182. \hfil\break\indent X. Ren was supported in part by NSF grant DMS-9703727. } } \date{} % \author{ Adam J. J. Chmaj \& Xiaofeng Ren } \maketitle \begin{abstract} We discuss instability and existence issues for the nonlocal bistable equation. This model arises as the Euler-Lagrange equation of a nonlocal, van der Waals type functional. Taking the viewpoint of the calculus of variations, we prove that for a class of nonlocalities this functional does not admit nonconstant $C^1$ local minimizers. By taking variations along non-smooth paths, we give examples of nonlocalities for which the functional does not admit local minimizers having a finite number of discontinuities. We also construct monotone solutions and give a criterion for nonexistence of nonconstant solutions. \end{abstract} \newtheorem{theorem}{Theorem}[section] \renewcommand{\theequation}{\thesection.\arabic{equation}} \catcode`@=11 \@addtoreset{equation}{section} \catcode`@=12 \section{Introduction} We study the semilinear integral equation (the nonlocal bistable equation) \begin{equation} -J[u]+ju+f(u)=0, \label{eq} \end{equation} on the interval $(0,1)$, where \[ J[u](x)=\int_{0}^{1} J(x,y)u(y)dy, \quad j(x)=\int_{0}^{1} J(x,y)dy. \] We assume $J \in W^{1,1}$ is symmetric, in the sense $J(x,y)=J(y,x)$ for $x,y \in (0,1)$, and $f \in C^1$ is a bistable function with three zeros: $-1, a\in (-1,1),1$, with $f'(\pm 1)>0$ and $f'(a)<0$. Equation (\ref{eq}) (which has no boundary conditions) arises as the Euler-Lagrange equation of the functional (defined on $L^2 (0,1)$) \begin{equation} I(u)=\frac{1}{4} \int_{0}^{1} \int_{0}^{1} J(x,y)(u(x)-u(y))^2 dxdy +\int_{0}^{1} W(u(x))\,dx , \label{f} \end{equation} where $W$ is a double-well function. Depending on the type of problem studied, one can impose the mass constraint $\int_0^1 u(x)dx=m$ on (\ref{f}). This was done in the seminal work of van der Waals in 1892 \cite{w}, who simplified (\ref{f}) by expanding the nonlocal part in power series and considering the first order approximation \begin{equation} I^{{\rm loc}} (u)=\frac{1}{2} \int_0^1 |u'(x)|^2 dx+\int_0^1 W(u(x)) dx. \label{fl} \end{equation} The gradient flow of (\ref{fl}) without the mass constraint: \begin{equation} u_t =u_{xx} -f(u),\quad u'(0)=u'(1)=0 , \label{ac} \end{equation} is sometimes referred to as the Ginzburg-Landau or Allen-Cahn equation. (\ref{f}) can be viewed as a model for materials whose constitutive relations are nonlocal (see \cite{rt,rw} for other examples). Namely, if $u$ denotes the general phase field characterizing the state of a material, and if the energy density of the continuum is postulated to be $ e= -J[u]u/2+ j u^2 /2 +W(u)$, then the total free energy can be written as \begin{equation} I(u)=\int_{0}^{1} e(x)dx=\int_{0}^{1} \bigl( -\frac{1}{2} J[u]u+ \frac{j u^2}{2} +W(u) \bigr) dx. \label{f2} \end{equation} (\ref{f}) can also be derived from elementary statistical mechanics. In \cite{bc1} and \cite{bc2} it was shown that (\ref{f}) arises as the Helmholtz free energy of an Ising-like spin system with long range interactions. In particular, in this approach $J$ can change sign and $W$ does not have to be balanced. An infinite lattice system similar to (\ref{eq}) was studied in \cite{bc2,cbc}. Results on solutions of the whole line version of (\ref{eq}) (i.e., $-J*u+ju+f(u)=0$) can be found in e.g., \cite{bfrw,c,cr1,bc1,cr2}. An interesting discovery was made in those papers: the existence of discontinuous solutions when $ju+f(u)$ is not monotone. In \cite{cr2} and \cite{cr3} we studied (\ref{eq}) for $ju+f(u)$ monotone and established, using singular perturbation techniques, another interesting phenomenon: the {\em existence} of nonconstant local minimizers of $I$ for a class of sign-changing interaction kernels $J$. By comparison, the local functional (\ref{fl}) does not admit nonconstant local minimizers, the stationary solutions of (\ref{ac}) are metastable, and the evolution of (\ref{ac}) is through slow motion \cite{fh,cp}. In this paper we show that for $ju+f(u)$ monotone and a different wide class of $J$'s which are nonnegative and translationally invariant (i.e., $J(x,y)=J(x-y)$), nonconstant local minimizers do {\em not} exist. In the case $ju+f(u)$ non-monotone (where solutions are in general discontinuous), we show in some examples that variations along non-smooth paths lead to a similar nonexistence result. Using an iteration method, we construct monotone solutions of (\ref{eq}) under certain assumptions. To our knowledge, it is the first nonperturbative existence result for a class of nonlocal equations such as (\ref{eq}). Finally, we give a criterion for nonexistence of stationary solutions of (\ref{eq}). \section{Nonexistence of local minimizers} (\ref{eq}) has a rich structure of solutions, whose properties in general depend both on the nonlocality $J$ and on the nonlinearity $f$. In \cite{cr2} and \cite{cr3} we showed using the $\Gamma$-convergence method that by taking $J$ and $W$ having particular forms: \begin{equation} J(x,y)=\frac{1}{\epsilon} J^s \Bigl( \frac{x-y}{\epsilon} \Bigr) - \epsilon J^l (x,y), \quad W = W_0 +\epsilon W_1 , \label{sp} \end{equation} with $J^s \geq 0$, $\frac{js^2}{2}+W(s)$ convex in $s$, $W_0 (-1)=W_0 (1)$ and $|W_1 (1)-W_1 (-1) |< 2 \int_0^1 J^l (0,y)dy$, there exist nonconstant local minimizers of $I$ for $\epsilon >0$ small enough. Observe that $J$ in (\ref{sp}) changes sign for $\epsilon >0$ small enough. The properties of these minimizers are as follows. Briefly speaking, if $I_\epsilon$ is the energy (\ref{f}) corresponding to (\ref{sp}), then $\frac{1}{\epsilon} I_\epsilon$ $\Gamma$-converges to $I_0$, defined by \begin{equation} I_0 (u)= \left\{ \begin{array}{ll} c_0 \frac{||Du||(0,1)}{2} + I^l (u) & \rm{if} ~u \in BV((0,1), \{ -1,1 \} ) , \\ \infty & \rm{otherwise} \end{array} \right. \label{bv} \end{equation} where $I^l (u)=-\frac{1}{4} \int_0^1 \int_0^1 J^l (x,y) (u(x)-u(y))^2 dxdy +\int_0^1 W_1 (u(x))dx$ and\break $c_0 \frac{1}{2} (||Du||(0,1))$ is equal to a constant multiplied by the number of jumps $u$ has. If an isolated local minimizer of $I_0$ exists in the space of step functions having a fixed number of jumps, then $\frac{1}{\epsilon}I_\epsilon$ also has a local minimizer, which is $C^1$ and $L^2$-close to the BV one. The layers $\xi_1 ,\ldots, \xi_n$ of the minimizers of $I_0$ are determined from the system \begin{equation} J^l [u](\xi_i )=-\frac{1}{2} f_1 (r)dr ,\quad i=1,\ldots,n, \label{red} \end{equation} where $f_1 =W_1 '$. (\ref{red}) is in general difficult to solve. In \cite{cr2} we considered $J^l$ to be the Green's function of the linear differential equation $-\gamma^2 v''+v=u$, $v'(0)=v'(1)=0$, i.e., \begin{equation} J^l(x,y)=\frac{1}{\gamma(e^{\frac{1}{\gamma}}- e^{-\frac{1}{\gamma}})} \Bigl[ \mbox{cosh} \Bigl( \frac{x+y-1}{\gamma} \Bigr) + \mbox{cosh} \Bigl( \frac{|x-y|-1}{\gamma} \Bigr) \Bigr], \ \gamma>0 . \label{G} \end{equation} In this case, (\ref{red}) can be written as a system of ODE's, and we showed that for every $n$, there exists a unique solution of (\ref{red}) such that $u(\xi_1 +)=-1$, and it is an isolated local minimum of (\ref{bv}). In \cite{cr3}, we considered more general $J^l (x,y)=J^l (x-y)$. This is a more complicated case, as (\ref{red}) becomes progressively more difficult to solve with an increasing number of layers. On an interesting note, we found that if $J^l$ changes sign, then in the class of critical points $u$ of (\ref{bv}) having one jump at $\xi$ such that $u(\xi +)=-1$, there are in general two local minima $\xi^1$, $\xi^3$, and a local maximum $\xi^2$ between them, i.e., $\xi^1 < \xi^2 < \xi^3$. We do not know if $\xi^2$ can be perturbed for small $\epsilon >0$ to, say, a mountain pass solution of $\frac{1}{\epsilon} I_\epsilon$. We now show that for a class of nonnegative $J$'s there are no local minimizers of $I$. \begin{theorem} Let $J(x,y)=J(x-y)$ and $j(\frac 12 )+f'(s) >0$ for $s \in [-1,1]$. Let $J' (x)\leq 0$ for $x\in (0,1)$ and $J(1) \geq 0$. Then any nonconstant solution of (\ref{eq}) is unstable, in the sense that it is not a local minimum of $I$. \label{t-non} \end{theorem} \paragraph{Proof.} With the regularity of $J$ and $f$ we can compute the second variation of $I$. For every $w, \phi \in L^2(0,1)$ \begin{equation} \frac{d^2 I(w+\epsilon \phi)}{d\epsilon^2}\Big|_{\epsilon =0}=\int_0^1 [-J[\phi]\phi + j \phi^2 +f'(w) \phi^2] dx. \label{sv} \end{equation} It suffices to show that the right side of (\ref{sv}) is $<0$ for $w=u$ and a particular choice of $\phi$. Let us assume that $u$ is a critical point of $I$, i.e., a solution of (\ref{eq}). Then (\ref{eq}) can be rewritten as \[ u= (j \cdot +f(\cdot))^{-1}(J[u]). \] The regularity of $J$ and $f$ imply that $u$ is in $C^1$. Differentiating (\ref{eq}) with respect to $x$, we deduce \[ -J[u']+j u'+ f'(u) u' = J(x-1)(u(x)-u(1))-J(x)(u(x)-u(0)). \] We now choose $\phi =u'$. Multiplying the last equation by $u'$ and integrating over $(0,1)$, we obtain \[ \frac{d^2 I(u+\epsilon u')}{d\epsilon^2}\Big|_{\epsilon =0}=\int_0^1[ J(x-1)(u(x)-u(1))- J(x)(u(x)-u(0))]u'(x) dx. \] We break up this expression into four integrals and integrate by parts each one of them to get \[ \int_0^1 J(x-1)u(x)u'(x)dx= \frac{1}{2} [J(0)u(1)^2 -J(1)u(0)^2 ] - \frac{1}{2} \int_0^1 J'(x-1) u(x)^2 dx , \] \[ \int_0^1 J(x-1) u(1)u'(x)dx = J(0)u(1)^2 -J(1)u(1)u(0)- \int_0^1 J'(x-1) u(1)u(x)dx , \] \[ \int_0^1 J(x)u(x)u'(x)dx = \frac{1}{2} [J(1)u(1)^2 -J(0)u(0)^2 ] - \frac{1}{2} \int_0^1 J'(x) u(x)^2 dx , \] \[ \int_0^1 J(x)u(0)u'(x)dx = J(1)u(0)u(1)-J(0)u(0)^2 -\int_0^1 J'(x)u(0)u(x)dx .\] By completing the squares in the multiples of $J'(x)$ and $J'(x-1)$ we finally get \begin{eqnarray} \frac{d^2 I(u+\epsilon u')}{d\epsilon^2}\Big|_{\epsilon =0} \equiv R(u') &=&-J(1) (u(1)-u(0))^2 \nonumber \\ &&- \frac{1}{2} \int_0^1 J'(x-1)[u(x)-u(1)]^2 dx \nonumber \\ &&+ \frac{1}{2} \int_0^1 J'(x) [u(x)-u(0)]^2 dx \leq 0. \label{ko} \end{eqnarray} We now show that actually $R(u')<0$. Assume that $R(u')=0$. Then also $R(|u'|)=0$ ($\int_0^1 \int_0^1 J(x-y)|u'(x)||u'(y)| dxdy \geq \int_0^1 \int_0^1 J(x-y)u'(x)u'(y)dxdy$). Consider the variational problem $\inf_{||\phi ||_2 =1} R(\phi )$. If this inf is less than $0$, then there exists some $\phi_0$ for which $R(\phi_0 )<0$, thus $u$ is unstable. If it is equal to $0$, then since it is achieved at $\phi_1 \equiv |u'|/||u'||_2$ we have \[ -J[|u'|]+j|u'|+f'(u)|u'|=\lambda |u'| .\] If $|u'|=0$ at some $x_0$, then $-J[|u'|](x_0 )+j(x_0 ) |u'(x_0 )|=0$, which implies that $|u'|\equiv 0$ (an inductive argument is used if $|\rm{supp}J|<2$), a contradiction. So $|u'|>0$. But then it is easily seen that $R(u')<0$. \hfill \fbox{}\smallskip The following example \cite{cr3} shows that condition $J\geq 0$ might be relaxed somewhat. \begin{theorem} If in Theorem \ref{t-non} we set $J(x) = b-m|x|$, $b,m>0$, and $b\geq 3m/4$, then any nonconstant solution of (\ref{eq}) is unstable. \end{theorem} \paragraph{Proof.} Let $u$ be a solution of (\ref{eq}). In (\ref{ko}) we break up $-(b-m)(u(1)-u(0))^2$ and use $\frac{m}{4}(u(1)-u(0))^2$ together with the other two terms in (\ref{ko}) to complete the square. We get \begin{eqnarray*} \frac{d^2 I(u+\epsilon u')}{d\epsilon^2}\Big|_{\epsilon =0} &=& -\bigl( b-\frac{3m}{4} \bigr) [u(1)-u(0)]^2 \\ & & -m \int_{0}^{1} \bigl[ u(x)-\frac{1}{2} (u(1)+u(0)) \bigr]^2 dx <0. \end{eqnarray*} Note that $J$ changes sign on $(-1,1)$ if $\frac{3m}{4} \leq b 0$ on $[-1,1] \setminus [s_1 ,s_2 ]$, $c+f'(s)<0$ on $(s_1 ,s_2 )$. Then any nonconstant solution of (\ref{eq}) with a finite number of discontinuities is unstable, in the sense it is not a local minimum of $I$. \end{theorem} \paragraph{Proof.} Let $u$ be a solution of (\ref{eq}). Note that $g(u)= \int_{0}^{1} u(y)dy=const$, thus $u$ is a step function. Assume that $u(x)=u_i$ on $(\xi_i ,\xi_{i+1})$, $1 \leq i \leq k-1$. For small $\epsilon >0$, extend $u$ to $(-\epsilon ,0)$ by setting $u(x) =u(0)$, so that $u(x-\epsilon )$ is defined on $(0,1)$. It now suffices to note that the second directional derivative along the continuous but non-smooth path $u(\cdot -\epsilon )$ is negative: \[ \frac{d^2 I(u(\cdot -\epsilon ))}{d\epsilon^2}\Big|_{\epsilon =0} =- \bigl( \sum_{i=1}^{n} (-1)^{i+1} u_i \bigr)^2 <0 . \] \section{Variations along non-smooth paths} In this section we further examine the role played by non-smooth paths of variations when studying the functional $I$. We show that when $ju+f(u)$ is not monotone, or in other words $\frac{ju^2}{2} + W(u)$ in (\ref{f2}) is not convex, variations along non-smooth paths select some special discontinuous solutions of (\ref{eq}). We consider the following example. Let $J$ be as in (\ref{G}) with $\gamma =1$. Let $f$ be the piecewise linear function \begin{equation} f(u)= \left \{ \begin{array}{ll} u+1 & u<0 \\ u-1 & u>0 . \end{array} \right. \end{equation} Since the solutions we will consider jump across value $0$, the discontinuity of $f$ at $0$ does not really violate the requirement $f\in C^1$. We may modify $f$ to make it smooth after we have found jump solutions. The equation (\ref{eq}) can be written as a system \begin{equation} \begin{array}{c} -v'' + v = u \\ -v+u+ u \pm 1=0 \\ v'(0)=v'(1)=0 \end{array} \label{syst} \end{equation} Denote the set of discontinuities of a solution by $\xi_1$, $\xi_2$, \dots, $\xi_k$. Assume $u<0$ on $(0,\xi_1)$, $u>0$ on $(\xi_1,\xi_2)$, etc. Denote the set of all such vectors $\xi_i$ by $A_k^-$, where $-$ refers to the fact that $u<0$ on $(0, \xi_1)$. Let \begin{equation} u=\frac{v-(\pm 1)}{2} \label{ubyv} \end{equation} Substitute the second equation of (\ref{syst}) into the first to obtain \begin{equation} -v''+\frac{v}{2} = -\frac{\pm 1}{2}, \ v'(0)=v'(1)=0 \label{vchi} \end{equation} Let $G$ be Green's function of this ODE. With the $\xi_i$'s fixed we solve the last equation to find $v$ and then $u$ by (\ref{ubyv}). This way we have obtained a discontinuous solution of (\ref{eq}). The energy of this solution can be written as \begin{eqnarray} \nonumber I(u) & = & \int_0^1 [-\frac{1}{2} J[u] u + \frac{j}{2} u^2 + W(u)] \,dx \\ \nonumber &=& \int_0^1 [ -\frac{1}{2} u(v-u) +W(u)]\,dx \\ \nonumber &=& \int_0^2 [-\frac{1}{2}u(u \pm 1) + \frac{1}{2} (u \pm 1)^2] \,dx \\ \nonumber &=& \int_0^1 \frac{1}{2}(1 \pm u)\,dx \\ &=& \frac{1}{4}+\frac{1}{4} [ \int_0^{\xi_1} v \,dx - \int_{\xi_1}^{\xi_2} v\,dx + \int_{\xi_2}^{\xi_3}v\,dx- \dots] \label{iv} \end{eqnarray} Now we treat the $\xi_i$'s as variables and obtain a family of variations of $I$. Here the variations are taken along discontinuous solutions of (\ref{eq}). This family is continuous but not $C^1$ under the $L^2$-norm. We differentiate $I$ with respect to $\xi_i$. For instance \begin{eqnarray*} \frac{\partial I(u)}{\partial \xi_1} & = & \frac{1}{2} v(\xi_1) + \frac{1}{4} [ \int_0^{\xi_1} \frac{\partial v}{\partial \xi_1} \,dx - \int_{\xi_1}^{\xi_2} \frac{\partial v}{\partial \xi_1} \,dx + \int_{\xi_2}^{\xi_3} \frac{\partial v}{\partial \xi_1} \,dx - \dots] \\ &=& \frac{1}{2} v(\xi_1) + \frac{1}{4} [- \int_0^{\xi_1} G(x,\xi_1) \,dx + \int_{\xi_1}^{\xi_2} G(x,\xi_1) \,dx + \dots] \\ &=& \frac{1}{2} v(\xi_1) +\frac{1}{2} v(\xi_1) \\ &=& v(\xi_1) \end{eqnarray*} since \begin{eqnarray*} \frac{\partial v}{\partial \xi_1} & = & \frac{\partial}{\partial \xi_1} [ \int_0^{\xi_1}G(x,y)(-\frac{1}{2})\,dy + \int_{\xi_1}^{\xi_2} G(x,y)\frac{1}{2}\,dy + \dots ] \\ & = & -G(x,\xi_1), \end{eqnarray*} and in a similar way \begin{equation} \frac{\partial I(u)}{\partial \xi_i} = (-1)^{i+1} v(\xi_i). \end{equation} Let us now look for solutions of $I$ with $k$ jump discontinuous points that are stationary with respect to these non-smooth paths of variations. Set $\frac{\partial I(u)}{\partial \xi_i}=0$, we conclude that \begin{equation} v(\xi_i)=0, \ i=1,2,\dots,k. \label{key} \end{equation} Two conclusions are drawn from (\ref{key}). First, according to (\ref{ubyv}) at $\xi_1$ $u$ must jump from $-1/2$ to $1/2$, and in general at $\xi_i$ $u$ jumps from $(-1)^{i+1}/2$ to $(-1)^i/2$. The two numbers $-1/2$ and $1/2$ are precisely the two global minima of the function $\frac{u^2}{2} + W(u)$ that appears in (\ref{f2}). Also see \cite{rwm} for the role played by these two numbers. The second conclusion is that the $\xi_i$'s are equally distributed. This is because we may solve the equation (\ref{vchi}) on each $(\xi_i, \xi_{i+1})$ with the boundary conditions (\ref{key}). Then the continuity of $v'$ across the $\xi_i$'s requires that the sub-intervals (with the exception of $(0,\xi_1)$ and $(\xi_k, 1)$) all have the same length. The two end intervals have half the length. Such a solution $u$ is unique in $A_k^-$. Let us now compute the energy of this particular $u$. Because of (\ref{iv}) and $\xi_1= 1/(2k)$ we deduce \[ I(u) = \frac{1}{4}- 2k v'(\frac{1}{\xi_1}). \] Because $v$ satisfies (\ref{vchi}) and $v'(0)=v(\xi_1)=0$, we solve the ODE to find \[ v(x)= -\frac{\sinh \frac{x}{\sqrt{2}} } {\cosh \frac{1}{2k\sqrt{2}}}. \] So \begin{equation} I(u)=\frac{1}{4} - \frac{k}{2\sqrt{2}} \tanh \frac{1}{2k \sqrt{2}}, \label{ik} \end{equation} which is increasing in $k$. A similar computation in $A_k^+$ shows that the critical point $u$ in $A_k^+$ has the same energy. We remark that $u$ a local maximum of $I$ in $A_k^-$ with respect to $\xi_i$. Suppose this is not true. Since there is only one critical point of $I$ in $A_k^-$, the maximum of $I$ with respect to $\xi_i$ must be achieved at some $\overline{u}$ on the boundary of the domain of $\xi_i$, which is identified by the union of all $A_m^{\pm}$ with $m (x-\frac 12 ) f'(0), \quad {\rm for } x \in (\frac 12 ,1] \label{J} \end{equation} and \begin{equation} J(\frac 12) < -f'(0). \label{J1} \end{equation} Note that the slightly weaker $J(\frac 12) \leq -f'(0)$ follows from (\ref{J}). Also, note that (\ref{J}) guarantees that $J$ is not constant. Otherwise, if $J=c$, then (\ref{J}) is equivalent to $f'(0)<-c$, which violates $j(\frac 12 )+f'(s)=c+f'(s) \geq 0$. For $J=c$ and $j(\frac 12) + f'(s) \geq 0$ on $[-1,1]$, it can be easily determined that (\ref{eq}) has only constant solutions, whose values are the zeros of $f$. Conditions (\ref{J}) and (\ref{J1}) relate the nonlocal effect with $f'(0)$. As an example, let us choose $J(x)=b-m|x|$, $b,m>0$. $j(\frac 12 )+f'(s) \geq 0$ is equivalent to $m\leq 4 (b+f'(s))$. (\ref{J}) and (\ref{J1}) are then satisfied if and only if $m > 2 (b+f'(0))$. With these assumptions we have the following. \begin{theorem} There exists an increasing solution $U$ of (\ref{eq}), such that $U(x)=-U(1-x)$ (i.e., $U(\cdot +\frac 12 )$) is odd. Moreover, with $U'$ denoting the pointwise derivative of $U$ and assuming $J(\frac 12 ) 0$ for $s \in [-1,1]$, $U' >0$. \item In the case $j(\frac 12) + f'(s) > 0$ for $s \in [-1,-u_0 ) \cup (u_0 ,1]$ and $j(\frac 12) + f'(s) < 0$ for $s \in (-u_0 ,u_0 )$, $U(\frac 12 \pm )=\pm u_0$ and $U'(x)>0$ for $x \in (0,\frac 12) \cup (\frac 12 ,1)$. \item In the case $j(\frac 12 ) + f'(s) > 0$ for $s \in [-1,0) \cup (0,1]$ and $j(\frac 12 ) + f'(0)=0$, $U'(0\pm )=+ \infty$ and $U'(x)>0$ for $x \in (0,\frac 12 ) \cup (\frac 12 , 1)$. \end{enumerate} \label{ex} \end{theorem} \paragraph{Proof.} $U$ is constructed from an iteration scheme. Let \[ u_0 (x)= \left\{ \begin{array}{ll} -1, & x\in (0, \frac 12 ) \\ 1, & x\in (\frac 12 , 1) \end{array} \right. . \] Define the sequence $\{ u_n \}$, $n=0,1,2, \dots , $ by \begin{equation} J[u_n ](x) + (j(\frac 12 )-j(x))u_n (x)=g(u_{n+1}(x)) , \label{iter} \end{equation} where $g(s)\equiv j(\frac 12 ) s+f(s)$. Note that since $J$ is decreasing on $(0,1)$, $j(\frac 12 )-j(x) \geq 0$ for $x \in (0,1)$. First, $u_n (x)=-u_n (1-x)$ and $J$ even imply \[ J[u_n ](x)= -\int_{0}^{1} J(x-y) u_n (1-y)dy = - \int_{0}^{1} J(1-x-y) u_n (y) = -J[u_n ](1-x). \] Since $g$ is odd, it then easily follows that $u_{n+1} (x)=- u_{n+1} (1-x)$ as well. We show by induction that $u_{n} '(x) \geq 0$ for $x \in (0,\frac 12 )\cup (\frac 12 ,1)$ and $n \geq 1$, where $u_n '$ denotes the pointwise derivative of $u_n$. First assume that $g'(s)>0$ for $s \in [-1,0) \cup (0,1]$. Inverting (\ref{iter}), we see that $u_{n+1}$ is $C^1$ on $(0,\frac 12 ) \cup (\frac 12 ,1)$. Thus we can differentiate (\ref{iter}) on $(0,\frac 12 )\cup (\frac 12 ,1)$ to get: \begin{eqnarray} \lefteqn{-J(x-1) u_n (1) +J(x) u_n (0)+ J(x-\frac 12 ) (u_n (\frac 12 +)-u_n (\frac 12 -)) }\nonumber \\ \lefteqn{ +\int_{0}^{1} J(x-y)u_n '(y)dy + (J(x-1)-J(x)) u_n (x) + (j(\frac 12 )-j(x)) u_n '(x) }\nonumber \\ & = & g'(u_{n+1}(x)) u_{n+1} '(x). \hspace{7cm} \label{est} \end{eqnarray} Since $J$ is decreasing, the integral appearing on the left side of (\ref{est}) can be estimated as follows: \begin{equation} \int_{0}^{1} J(x-y)u_n '(y)dy \geq \int_{0}^{x} J(x) u_n '(y)dy + \int_{x}^{1} J(1-x) u_n '(y)dy . \label{ineq} \end{equation} With this estimate, (\ref{est}) becomes \begin{eqnarray*} \lefteqn{ [J(x-\frac 12 )-J(\frac 12 +|x-\frac 12 |)] [u_n (\frac 12 +)-u_n (\frac 12 -)] + (j(\frac 12 )-j(x)) u_n '(x) }\\ & \leq & g'(u_{n+1}(x)) u_{n+1} '(x) . \hspace{65mm} \end{eqnarray*} Since $u_n '(x) \geq 0$ and $J$ is decreasing on $(0,1)$, also $u_{n+1}'(x) \geq 0$ for $x \in (0,\frac 12 )\cup (\frac 12 ,1)$. We now show that (\ref{iter}) has the following monotonicity property. If $v$ and $w$ are two functions such that $v(x)=-v(1-x)$ and $w(x)=-w(1-x)$, with $v(x) \leq w(x)$ for $\frac 12 \leq x \leq 1$, and $\tilde{v}$ and $\tilde{w}$ are defined by (\ref{iter}), namely $J[v](x) + (j(\frac 12 )-j(x))v(x)=g(\tilde{v}(x))$ and similarly for $\tilde{w}$, then $\tilde{v} \leq \tilde{w}$ for $\frac 12 \leq x \leq 1$. First note that if $u$ is such that $u(x)=-u(1-x)$ and $u(x) \geq 0$ for $x \in (\frac 12 ,1)$, then $J$ being even and decreasing on $(0,1)$ implies that \begin{equation} J[u](x)=\int_{\frac 12}^{1} [J(x-y)-J(1-x-y)]u(y)dy \geq 0 . \label{diff} \end{equation} For $x \in (\frac 12 ,1)$, (\ref{iter}), (\ref{diff}) and $v(x) \leq w(x)$ imply that $0 \geq g(\tilde{v})- g(\tilde{w})=g'(c(x) ) (\tilde{v} -\tilde{w})$ for some function $c(x)$, thus since $g'>0$ we get $\tilde{v} \leq \tilde{w}$. To show by induction that $\{ u_n (x) \}$ is nonincreasing in $n$ for $x\in (\frac 12 ,1)$, it now suffices to note by direct computation that $u_0 (x)$ is a supersolution of (\ref{iter}) on $(\frac 12 ,1)$: \[ J[u_0 ](x) + (j(\frac 12 )-j(x))u_0 (x) \leq g(u_0 (x)). \] We can now set $U(x)\equiv \lim_{n \to \infty} u_n (x)$. Clearly, $U$ solves (\ref{eq}). To guarantee that $U$ is not the constant solution $0$, we show that $\underline{u} (x) \equiv m(x-\frac{1}{2})$ is a subsolution of (\ref{iter}) on $(\frac 12 ,1)$: \[ J[\underline{u} ](x)-j(x)\underline{u} =m \int_{x}^{x-1} tJ(t)dt \geq m (x-\frac 12 )(f'(0)+ \epsilon )\geq f(m(x-\frac 12 )) =f (\underline{u} ), \] for $m$ and $\epsilon >0$ small enough, where we used (\ref{J}) and (\ref{J1}). In the case $j(\frac 12) + f'(s) > 0$ for $s \in [-1,-u_0 )\cup (u_0 ,1]$ and $j(\frac 12) +f'(s) < 0$ for $s \in (-u_0 ,u_0 )$, the construction is similar, except that $\underline{u}$ is now taken to be $\underline{u} (x)=H(x-\frac 12 ) u_0$, where $H$ is the Heaviside function. That $\underline{u}$ is indeed a subsolution on $(\frac 12 ,1)$, follows from (\ref{diff}) and $g(u_0 )=0$. We now show that $U'>0$. First, in the case $j(\frac 12 )+f'(s)>0$ for $s\in [-1,1]$, as we discussed before, $u$ is in $C^1$. Using (\ref{est}) and (\ref{ineq}), we see that $J(\frac 12 )0$. The other two cases are discussed in a similar way. In particular, the regularity of $J[U]$ implies that $U(\frac 12 \pm )=\pm u_0$. \hfill\fbox{}\smallskip Note that for the whole line version of (\ref{eq}), the increasing solution $\hat{U}$ of $-J*u+ju+f(u)=0$ has the property $\hat{U}'>0$ under the milder assumption $J\geq 0$ \cite{bfrw}. One cannot expect the same property to hold in Theorem \ref{ex} (recall that the discontinuous increasing solution for $J=c$ is piecewise constant, as was discussed before). For $J\geq 0$, the $C^1$ solutions constructed in Theorem \ref{ex} are unstable, by Theorem \ref{t-non}. We do not know if they are unique in the class of increasing functions. Recall that for the scaled local (\ref{ac}) equation $-\epsilon^2 u''+f(u)=0$, $u'(0)=u'(1)=0$, where $f(u)=u^3 -u$, for $\epsilon_{n+1} \leq \epsilon \leq \epsilon_n$ there exist $n$ solutions (such that $u(0)<0$), where $\epsilon_i =\sqrt{f'(0)}/2\pi i$. The existence of similar nonmonotone solutions of (\ref{eq}) is left as an open problem. As was noted before, for $J=c$ and $c+f'(s) \geq 0$, (\ref{eq}) has only constant solutions. This nonexistence result can be improved in the following way. \begin{theorem} Let $J(x,y) \geq 0$ and $j(x)+f'(s)>0$ for $s\in [-1,1]$, where $j(x)=\int_0^1 J(x,y)dy$. Then, assuming \begin{equation} \max_{x\in [0,1]} \int_0^1 |x-y| | J_x (x,y)| dy < \min_{s \in [-1,1]} \min_{x\in [0,1]} ~ [j(x)+f'(s)], \label{nonex} \end{equation} there are no nonconstant solutions of (\ref{eq}). \end{theorem} \paragraph{Proof.} First, note that from the comparison principle $J(x,y) \geq 0$ implies that any solution $u$ of (\ref{eq}) is such that $|u|\leq 1$. Also, $j(x)+f'(s)>0$ implies that $u$ is $C^1$. We write (\ref{eq}) as $J[u]=ju+f(u)$, then differentiate it and estimate both sides in the following way: \begin{eqnarray*} \lefteqn{ \bigl[ \max_{x\in [0,1]} u'(x) \bigr] \int_0^1 |x-y| |J_x (x,y)| dy }\\ & \geq& \int_0^1 J_x (x,y) [\int_0^1 (y-x) u'(x-t(x-y))dt ]dy\\ & =& \int_0^1 J_x (x,y) (u(y)-u(x))dy\\ &=& (j(x)+f'(u))u'(x) . \end{eqnarray*} We take the maximum of both sides of this inequality to get \[ \bigl[ \max_{x\in [0,1]} u'(x) \bigr] \int_0^1 |x-y| |J_x (x,y)| dy \geq \min_{s \in [-1,1]} \min_{x\in [0,1]} ~ [j(x)+f'(s)] \max_{x\in [0,1]} u'(x) . \] If $u$ is nonconstant, we divide both sides by $\max_{x\in [0,1]} u'(x)$ to get a contradiction. \hfill \fbox{} To illustrate this theorem, we again choose $J(x)=b-m|x|$, $b,m>0$ (as was already noted, $j(\frac 12 )+f'(s)>0$ is equivalent to $m<4 (b+f'(s))$. Then condition (\ref{nonex}) is equivalent to $m