\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 162, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/162\hfil
Uniqueness of a symmetric positive solution]
{Uniqueness of a symmetric positive solution to an ODE system}

\author[O. Lopes\hfil EJDE-2009/162\hfilneg]
{Orlando Lopes}

\dedicatory{In memory of Jack K. Hale (1928--2009)}

\address{Orlando Lopes \newline
IMEUSP- Rua do Matao, 1010, Caixa postal 66281\\
CEP: 05315-970, Sao Paulo, SP, Brazil}
\email{olopes@ime.usp.br}

\thanks{Submitted October 9, 2009. Published December 21, 2009.}
\subjclass[2000]{34A34}
\keywords{Symmetric positive solutions; variational ODE systems}

\begin{abstract}
 In this article, we prove uniqueness of symmetric positive
 solutions of the variational ODE system
 \begin{gather*}
 -w''+a w-wv=0 \\
 -v''+b v -\frac{w^2}{2}=0,
 \end{gather*}
 where $a$ and $b$ are positive constants.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction and Statement of the Result}

In this article, we prove uniqueness of symmetric positive solutions of
the variational ODE system
\begin{equation} \label{e1}
\begin{gathered}
-w''+a w-wv = 0\\
-v''+b v -\frac{w^2}{2}=0
\end{gathered}
\end{equation}
where $a$ and $b$ are positive constants. The solutions under
consideration are defined
for all $x\in \mathbb{R}$ and have finite energy.

To show how \eqref{e1} arises,  we consider  the so-called
$\chi^2$ SHG equations
\begin{equation} \label{e2}
\begin{gathered}
i\frac{\partial w}{\partial t}+r\frac{\partial^2w}{\partial x^2}-\theta w
+w^*v=0\\
i\sigma \frac{\partial v}{\partial t}+s\frac{\partial^2v}{\partial x^2}
-\alpha v+\frac{w^2}{2}=0
\end{gathered}
\end{equation}
where $r,s,\sigma,\theta$ are positive real parameters and $w(x)$
and $v(x)$ are complex functions. This system governs phenomena
in nonlinear optics (see \cite{torner} for instance).

A solitary wave is a solution of \eqref{e2} of the form
$$
(w(x)e^{i\gamma t},v(x)e^{2i\gamma t}).
$$
Hence, $(w,v)$ satisfies
\begin{equation} \label{e3}
\begin{gathered}
-rw''+(\theta +\gamma)w-w^*v=0\\
-sv''+(\alpha+2\sigma\gamma) v -\frac{w^2}{2}=0.
\end{gathered}
\end{equation}
The solutions of \eqref{e3} are critical points of $E+\gamma I$
where $E$ and $I$ are the following conserved quantities
for \eqref{e2}
\begin{gather} \label{e4}
E(w,v)=\int_{-\infty}^{+\infty} (r|w'|^2+s|v'|^2+\theta |w|^2
+\alpha|v|^2-{\rm Re}(w^2v^*))\,dx, \\
 \label{e5}
I(w,v)=\int_{-\infty}^{+\infty} (|w|^2+2\sigma |v|^2)\,dx.
\end{gather}
 If $w$ and $v$ are real solutions of \eqref{e3} then it solves
\begin{equation} \label{e6}
\begin{gathered}
-rw''+(\theta +\gamma)w-wv = 0\\
-sv''+(\alpha+2\sigma\gamma) v -\frac{w^2}{2}=0.
\end{gathered}
\end{equation}
Replacing  $(w,v)$ by $(k_1w,k_2v)$ in \eqref{e6}, with $k_2=r$ and
$ k_1^2=rs$,  we get
\begin{gather*}
-w''+\frac{(\theta +\gamma)}{r} w-wv =0\\
-v''+\frac{(\alpha+2\sigma\gamma)}{s} v -\frac{w^2}{2}=0.
\end{gather*}
Therefore, we consider the real variational ODE system
\begin{gather}
-w''+a w-wv=0  \label{e7} \\
-v''+b v -\frac{w^2}{2}=0 \label{e8}
\end{gather}
and we will be interested in  solutions that have finite energy (or equivalently, tend to
zero as $|x|$ tends to infinity). The existence of positive solutions of \eqref{e7}-\eqref{e8} has been
proved in \cite{yew}. Briefly the argument goes as follows. We define
$H=H^1(\mathbb{R})\times H^1(\mathbb{R})$ equipped with the norm
$$
\int_{-\infty}^{+\infty}
(w'^2(x) +v'^2(x) +aw^2(x) +bv^2(x))\, dx.
$$
We consider the functionals
\begin{gather*}
E(w,v)=\int_{-\infty}^{+\infty}(w'^2(x) +v'^2(x)-w^2(x)v(x))\,dx,\\
I(w,v)=\int_{-\infty}^{+\infty} (aw^2(x) +bv^2(x))\, dx.
\end{gather*}
Using the method of concentration-compactness (\cite{lions}),
we minimize $E(w,v)$ under $I(w,v)=1$ in the space $H$.
If we replace $(w(x),v(x))$ by  $(|w(x)|,|v(x)|)$ then $E$
does not increase.  Therefore, any  minimizer is nonnegative
and solves the Euler-Lagrange system
\begin{gather}
-w''+\mu a w-wv=0  \label{e9}\\
-v''+\mu b v -\frac{w^2}{2}=0 \label{e10}
\end{gather}
 with $\mu\geq 0$ (because $(w,v)$ is a minimizer).
On the other hand, it is easy to
see that any solution $(w,v)\in H$ of \eqref{e9}-\eqref{e10}
with $\mu=0$ is the solution identically
zero. Therefore, we must have $\mu >0$. Defining  a new pair
$(k_1w(k_3x), k_2v(k_3 x))$
with $k_3^2=1/\mu,k_1=k_2=1/\mu$, we see that this new pair
satisfies \eqref{e7}-\eqref{e8}.

 In \cite{lopes} the symmetry of any positive solution of
\eqref{e7}-\eqref{e8} has been proved using a result of \cite{busca}.
However, as pointed out in \cite{busca},  their proof
works for $N\geq 2$. Since we are  in dimension one, we
need the following  modified version given in \cite{ikoma}.

 \begin{theorem} \label{thm1.1}
Consider the system
\begin{equation} \label{e11}
\begin{gathered}
w''+f(w,v)=0 \\
v'' +g(w,v)=0
\end{gathered}
\end{equation}
where $f(w,v)$ and $g(w,v)$ are $C^1$ functions satisfying the
conditions:
$$
f(0,0)=0=g(0,0),\quad \frac{\partial f(w,v)}{\partial v},
\frac{\partial g(w,v)}{\partial w}\geq 0.
$$
Suppose that there exist $\epsilon>0$ and $\delta>0$ such that
$w>0$, $v>0$, $w^2+v^2<\epsilon$ imply
$$
\frac{\partial f(w,v)}{\partial w},\frac{\partial g(w,v)}{\partial v}
<-\delta,
\quad 0<\frac{\partial f(w,v)}{\partial v},
\frac{\partial g(w,v)}{\partial w}<\delta.
$$
Then, except for translations, any positive solution of \eqref{e11}
is even and decreasing.
\end{theorem}

We conclude that, except for translations,  any positive solution
of \eqref{e7}-\eqref{e8} is symmetric and decreasing.

In \cite{lopes} we have also proved the following result.


\begin{theorem} \label{thm1.2}
The linearized operator of \eqref{e7}-\eqref{e8} at any positive
symmetric solution has zero as a simple eigenvalue with odd
eigenfunctions $(w_x,v_x)$  and  it has exactly one negative eigenvalue.
\end{theorem}

The fact that zero is a simple eigenvalue of the linearized operator
is not a proof of uniqueness of symmetric positive solution, but it
may suggest it. Our main result is that this
is indeed the case.

\begin{theorem} \label{thm1.3}
For $a,b>0$, the positive symmetric decreasing solution
of \eqref{e7}-\eqref{e8} is unique.
\end{theorem}

Several interesting numerical experiments concerning system
\eqref{e7}-\eqref{e8} are presented in [6].
They  indicate uniqueness of positive solution
(which is confirmed by Theorem \ref{thm1.3}) and
that \eqref{e7}-\eqref{e8} may have solutions that change sign.


\section{Proof of main result}

First we establish the following abstract uniqueness result.

\begin{theorem} \label{thm2.1}
Let $X$ be a Banach space and  $F: X\times [0,1]\to X$ be a
continuous functions with continuous Frechet derivative with
respect to the first variable. Also assume that
\begin{itemize}
\item[(i)] the set of the solutions $(u,\lambda)$ of $F(u,\lambda)=0$,
$u\in X, \lambda \in [0,1]$ is
precompact;

\item[(ii)] for any solution of $F(u,\lambda)=0$, the derivative
$F_u(u,\lambda)$ is invertible;

\item[(iii)] the equation $F(u,0)=0$ has a unique solution.
\end{itemize}
Then the equation $F(u,\lambda)=0$ has a unique solution for
$\lambda \in [0,1]$.
\end{theorem}

\begin{proof}
 First we claim that there is a $\lambda_0 >0$ such that the solution
of $F(u,\lambda)=0$ is unique for $0\leq  \lambda < \lambda_0$. In
fact, otherwise, there is a sequence $0<\lambda_n \to 0$ such that
$F(u,\lambda_n)=0$ has at least two distinct solution $u_n$ and
$v_n$. In view of assumption (i) and passing to a subsequence if
necessary, we can assume that $u_n$ converges to $u$ and $v_n$
converges to $v$. In view of (iii), we must have $u=v$. However,
by (ii) and the implicit function theorem, in a neighborhood of
$u$, for small $\lambda$, the solution of $F(u,\lambda)=0$ is
unique. This contradiction proves the claim. The same argument
shows that the set $A$ of $\lambda$, $0\leq \lambda \leq 1$, for
which the solution of $F(u,\mu)=0$ is unique for $0\leq \mu \leq
\lambda$ is open. Since by ii) $A$ is clearly closed, $A$ has to
be the whole interval $[0,1]$ and the theorem is proved.
\end{proof}

\subsection*{Remark}
If we take $u\in \mathbb{R}$ and $F(u,\lambda)=u(\lambda
u-1)=\lambda u^2-u$, we have $F_u(u,\lambda)=2\lambda -1$. We see
that, except for assumption i),  all the others are satisfied but
the conclusion of the theorem does not hold. This is so because
there is the branch $u=1/\lambda$ of solutions bifurcating from
infinity.

Theorem \ref{thm1.3} will be a consequence of Theorem \ref{thm2.1}.
To verify all its assumptions, we start with the following result.


\begin{lemma} \label{lem2.1}
The system
\begin{equation} \label{e12}
 \begin{gathered}
-w''+ aw -wv=0 \\
 -v''+ av -\frac{w^2}{2}=0
\end{gathered}
\end{equation}
($a=b$ in \eqref{e7}-\eqref{e8}) has a unique positive  solution
with finite energy.
\end{lemma}

\begin{proof}
Defining $z(x)=w(x)-\sqrt{2}v(x)$, multiplying the second equation by
$\sqrt{2}$  and subtracting we get
$$
-z''+z+ \frac{w}{\sqrt{2}}z=0.
$$
 Multiplying this last equation by $z$ and integrating we get
$$
\int_{-\infty}^{+\infty} (z'^2(x) +z^2(x)+
 \frac{w}{\sqrt{2}} z(x)^2)\,dx=0
$$
and this implies $z\equiv 0$ (because $w$ is a positive).
Therefore, each component of the solution of \eqref{e12} solves
a single second order equation  and this implies uniqueness
and the lemma is proved.
\end{proof}

To verify the other  assumptions of Theorem \ref{thm2.1},
we establish a chain of
estimates. Since we wish to find  estimates for solutions
of \eqref{e7}-\eqref{e8} which remain uniform
for $a$ and $b$ in a certain interval, we fix two constants
$0<c_1<c_2$ and we assume
\begin{equation} \label{e13}
c_1\leq a, b \leq c_2.
\end{equation}
In the sequel, $d_i$, $1\leq i \leq $ will indicate constants
depending on $c_1$ and $c_2$ only.
Let $(w(x),v(x))$ be as in Theorem \ref{thm1.3}. Since
\begin{equation} \label{e14}
T(w,v,w',v')\hat{=} -w'^2-v'^2+a w^2+b v^2-w^2v
\end{equation}
is a first integral for \eqref{e7}-\eqref{e8}, we must have
\begin{equation} \label{e15}
 -w'^2(x)-v'^2(x)+a w^2(x)+b v^2(x)-w^2(x)v(x)=0
\end{equation}
for any $x$.

\subsection*{Bound for $v(0)$}
Using the fact the  $(w(x),v(x))$ is symmetric, if we set
$x=0$ in \eqref{e15} we get
\begin{equation} \label{e16}
w^2(0)=\frac{b v^2(0)}{v(0)-a }\cdot
\end{equation}
 In particular $v(0)>a$. Moreover,
$v''(0)\leq 0$ (because $v(x)$ has a maximum at $x=0$) and then
the second  equation \eqref{e8} yields
\begin{equation} \label{e17}
b v(0)\leq \frac{w^2(0)}{2}\,.
\end{equation}
This together with \eqref{e16}  implies
\begin{equation} \label{e18}
 b v(0)\leq \frac{1}{2}\frac{b v^2(0)}{(v(0)-a)}
\end{equation}
and finally
%\label{e}
$v(0)\leq 2a$
because $v(0)>a$.

\subsection*{Bound for $v'(x)$}
Multiplying the second equation  \eqref{e8} by $v'(x)$, then
for $x\geq 0$ we get:
$$
\frac{d}{dx}(-v'(x)^2+b v^2(x))=w^2(x)v'(x) \leq 0.
$$
Therefore $-v'(x)^2+b v^2(x)$ is decreasing and, since it
vanishes at $+\infty$, we get
$$
-v'(x)^2+b v^2(x)\geq 0
$$
and then
\begin{equation} \label{e20}
v'(x)^2\leq b v^2(x)\leq b v^2(0)\leq 4a^2b.
\end{equation}

\subsection*{Bound for $w'(x)$}
We know $w'(x)\leq 0$ and that $w'(x)$  reaches its  minimum
when $w''(x)=0$. By the first
equation  \eqref{e7}, this occurs when $v(x)=a$ and then,
from  \eqref{e15},
$$
w'(x)^2+v'(x)^2=b v^2(x)\leq b v^2(0)\leq 4a^2b.
$$
We conclude
\begin{equation} \label{e21}
|w'(x)|=-w'(x)\leq 2a\sqrt{b}.
\end{equation}


\subsection*{Bound for $w(0)$}
Suppose $w(0)=M$ and $w(x_0)=M/2$ for some $x_0>0$.
Since
$$
w(0)-w(x_0)=-  \int_0^{x_0} w'(s)\, ds,
$$
then, in view of \eqref{e21}, we have
$ \frac{M}{2}\leq 2a\sqrt{b} x_0$ and this
implies
\begin{equation} \label{e22}
x_0\geq  \frac{M}{4a\sqrt{b}}.
\end{equation}
Moreover,  the solution of the linear equation
\begin{equation} \label{e23}
-v''(x)+b v(x)=h(x)
\end{equation}
 is given by
\begin{equation} \label{e24}
v(x)= \frac{1}{2\sqrt{b}} \int_{-\infty}^{+\infty}
 e^{-\sqrt{b}|x-y|}h(y)\,dy,
\end{equation}
and then, the second equation  \eqref{e8} and \eqref{e22}  give
\begin{align*}
v(0)
&=\frac{1}{4\sqrt{b}}\int_{-\infty}^{+\infty}
  e^{-\sqrt{b}|y|}w^2(y) \,dy\\
&= \frac{1}{2\sqrt{b}}\int_0^{+\infty} e^{-\sqrt{b}y}w^2(y) \,dy \\
&\geq \frac{1}{2\sqrt{b}}\int_0^{x_0} e^{-\sqrt{b}y}w^2(y) \,dy \\
& \geq \frac{M^2}{8\sqrt{b}}\int_0^{x_0}e^{-\sqrt{b}y}\,dy\\
&=\frac{M^2}{8b}(1-e^{-\sqrt{b} x_0})\\
&\geq \frac{M^2}{8b}
 (1-e^{- \frac{M}{4a}}).
\end{align*}
Therefore,
$$
2a \geq v(0) \geq \frac{M^2}{8b}(1-e^{- \frac{M}{4a}})
$$
and this gives  that $M=w(0)\leq d_1$, for some constant $d_1$.
In  view of \eqref{e16},  this gives also that
$v(0)\geq d_2>a$, for some constant $d_2$, and also gives a lower
bound for $w(0)\geq d_3$.



\subsection*{Bound for the length of the interval for which
$v(x) \geq a$}
By the first equation in \eqref{e7} and the previous estimates
for $v(0)$ and $w(0)$, we have
$w''(0)\leq -d_4<0$ and  $|w'''(x)|\leq d_5$.
Defining  $X=-\frac{w''(0)}{2d_5}$
then, for $0\leq x \leq X$ we have
\[
w''(x)-w''(0)=\int_0^x w'''(s)\,ds\leq d_5 X=-w''(0)/2,
\]
 and then
$w''(x)\leq w''(0)/2\leq -d_4/2$ for $0\leq x \leq X$.
Moreover,
\[
w'(X)=w'(0)+\int_0^X w''(s)\,ds \leq \int_0^X
\frac{w''(0)}{2} \,ds=X\frac{w''(0)}{2}
=- \frac{w''(0)^2}{4d_5}\leq -d_6.
\]
Since, by \eqref{e7},  $w''(x)\leq 0$ whenever $v(x) \geq a$,
we have $w'(x) \leq -d_6$ whenever
$v(x) \geq a$  and $x\geq X$. Furthermore,
$$
-w(0)\leq -w(X)\leq w(x)-w(X)= \int_X^xw'(s)\, ds\leq -d_6(x-X).
$$
Therefore,  defining $X_1=w(0)/d_6+X$, we see that we
must have  $v(X_1)\leq a$.

\subsection*{Estimate for the time $v(x)$ stays close (and less)
than $a$}

Let $x_0\leq X_1$ be such that $v(x_0)=a$ and  let $d_7>0$ and
$d_8<a$ be such that
$$
(a-v)w^2+b v^2\geq d_7^2
$$
whenever $d_8\leq v \leq a, w\leq d_1$. Then, if $d_8\leq v(x)
\leq a$ for $x_0\leq x \leq x_0+X_2$, by \eqref{e15}  we have
$-w'(x)-v'(x) \geq d_7$ and then
$$
w(x_0)+v(x_0)\geq -w(x)+w(x_0)-v(x)+v(x_0)\geq d_7X_2
$$
and this gives a uniform  upper bound for $X_2$.

\subsection*{Exponential decay for $w(x)$ and for $v(x)$}
Since
$$
\frac{d}{dx}( -w'(x)^2+a w^2(x) -w^2(x) v(x))=-w^2(x) v'(x) \geq 0
$$
the function $-w'(x)^2+(a-v(x)) w^2(x) $ is increasing and
then $-w'(x)^2+(a-v(x)) w^2(x)\leq 0 $ for all $x\geq 0$ because
it vanishes at infinity.  Now, for $x\geq
X_3\hat{=}X_1 + X_2$ we have $-w'(x)^2 +d_8w(x)^2 \leq 0$ and then
$ w'(x)+d_9w(x) \leq 0$ and then $ \frac{d}{dx} e^{d_9x} w(x) \leq 0$
and finally, $w(x)\leq e^{-d_9(x-X_3)}w(X_3) $ for $ x\geq X_3$ and
this implies
\begin{equation} \label{e25}
w(x)\leq d_{10}e^{-d_9x}, \quad x\geq 0.
\end{equation}
 From the second equation \eqref{e8} we get
$$
v(x)= \frac{1}{4\sqrt{b}} \int_{-\infty}^{+\infty}
e^{-\sqrt{b}|x-y|}w^2(y)\,dy
$$
and this together with \eqref{e25} and elementary calculation
gives a similar exponential decay
\begin{equation} \label{e26}
v(x)\leq d_{11}e^{-d_{12}x}, \quad x\geq 0
\end{equation}
for $v(x)$.


\begin{proof}[Proof of Theorem \ref{thm1.3}]
Using \eqref{e24} to invert  the linear operators $-w''+aw$ and
$-v''+bv$, we see that system \eqref{e7}-\eqref{e8} can be
written as
\begin{equation} \label{e27}
\begin{gathered} w(x)
= \frac{1}{2\sqrt{a}} \int_{-\infty}^{+\infty}
e^{-\sqrt{a}|x-y|}w(y)v(y)\,dy\\
 v(x)= \frac{1}{4\sqrt{b}} \int_{-\infty}^{+\infty}
e^{-\sqrt{b}|x-y|}w^2(y)\,dy.
\end{gathered}
\end{equation}
 Defining $u$ as the pair $(w,v)$, system \eqref{e27} can be viewed
as the equation
\begin{equation} \label{e28}
F(u,\lambda)=0
\end{equation}
where $\lambda$, say, is $b$, with $a$ kept fixed.
We denote by $H^1_{ev}\subset H^1(\mathbb{R})$ the subspace of
the even functions. If we take
$X=H^1_{ev}\times H^1_{ev}$, then $F:X\to X$ is a well defined
very smooth function. In view of Theorem \ref{thm1.2}, assumption ii) of
Theorem \ref{thm2.1} is satisfied because $X$ consists of even functions.
Uniqueness for $\lambda=a$ is given by Lemma \ref{lem2.1}.
To verify  assumption (i) of Theorem \ref{thm2.1},  we recall that a subset $K$
of $X$ is precompact if and only if the following conditions are
satisfied:
\begin{itemize}
\item[(1)] for each $n$ the restriction of the functions of $K$
to the interval $[-n,n]$ is
precompact;

\item[(2)] for every $\epsilon >0$, there is an $x(\epsilon)>0$ such
that for all $u\in K$ we have
$$
\int _{|x| \geq x(\epsilon)} (|u'|^2(x) + |u(x)|^2)\, dx <\epsilon.
$$
\end{itemize}
To verify these conditions we first notice that we have obtained
uniform bound for the $H^1(\mathbb{R})$ norm of the solution $(w,v)$
of \eqref{e7}-\eqref{e8}. This implies uniform bound for
the $H^2$ norm of such solutions and this verifies condition
(1) for precompactness. The
uniform exponential decay \eqref{e25} and \eqref{e26} for $w(x)$ and
$v(x)$  together with \eqref{e14} gives
the uniform exponential decay also for the derivatives.
This implies that condition (2)
for precompactness is satisfies; therefore, Theorem \ref{thm1.3}
 is proved.
\end{proof}

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\end{document}