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\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} \\
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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{}
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\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]
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\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