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\markboth{\hfil Stability of peakons \hfil EJDE--2002/05}
{EJDE--2002/05\hfil Orlando Lopes \hfil}

\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 05, 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} \\
 %
 Stability of peakons for the generalized Camassa-Holm equation
 %
\thanks{ {\em Mathematics Subject Classifications:} 35J20, 49J10.
\hfil\break\indent
{\em Key words:} variational problems, stability of peakons.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted October 25, 2002. Published January 7, 2003.} }
\date{}
%
\author{Orlando Lopes}
\maketitle

\begin{abstract}
 We study the existence of minimizers for a constrained variational
 problems in $H^1(\mathbb{R})$. These minimizers are stable
 waves solutions for the Generalized Camassa-Holm equation,
 and their derivative may have a singularity
 (in which case the travelling wave is called a peakon).
 The existence result is based on a method developed by the same 
 author in a previous work. By giving examples, we show how  
 our method works.
\end{abstract}

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


\section{Introduction}

In this paper of consider the problem
\begin{equation} \label{P}
\begin{gathered}
\mbox{minimize}\quad  V(u)=-\int_{\mathbb{R}} (uu_x^2+F(u)+ku^2)\,dx\\
\mbox{subject to}\quad  I(u)=\int_{\mathbb{R}} (u_x^2+u^2)\,dx=\lambda>0.
\end{gathered}
\end{equation}
This problem appears in connection with the existence and stability of
solitary waves for the generalized Camassa-Holm equation
\begin{equation} \label{1.2}
u_t-u_{txx}=2u_xu_{xx}+uu_{xxx} -[f(u)/2]_x -ku_x.
\end{equation}
In fact, formally, $V(u)$ and $I(u)$ are conserved quantities for \eqref{1.2}
and then the set of the minimizers of the problem above is stable with
respect to \eqref{1.2} (see \cite{c1}) and its elements are solitary waves
(including their translations in the space variable $x$).
Some information about the Cauchy problem for \eqref{1.2} can be found in
\cite{c3}.

 If, for a given $\lambda >0$, the set of the minimizers is finite
(with possibly different multipliers), then each individual solitary wave is
stable. In particular, if, for a given $\lambda >0$, the minimizer is unique
(which is the case if $k=0$ and $F(u)=cu^3, c>0$), then the unique solitary
wave is stable.

The derivative of the solutions of the variational problem above may have a
singularity at some point (which can be taken as $x=0$); it may also be an
everywhere regular solution. The first type will be called a peaked solution
(or peakon) and the later smooth solitary wave.

In the  integrable case $F(u)=u^3$ and $k=0$ problem \eqref{P} has been
studied in \cite{c2} using the concentration-compactness method and some results
about compact sets in some spaces of distributions. The  homogeneity of
$V(u)$  plays an important role in the proof.

In \cite{c3} the local minimization problem has been considered and  an elementary
proof is given. It is based on very special identities and, probably, the
existence of such identities is related to the integrability.

In  the case $F(u)=u^3$ and $k>0$ considered in \cite{c4}, the solutions are regular
solitary waves and their stability is studied using the method of
Grilakkis, Shatah and Strauss \cite{g2}, GSS method for short, for finding local
 minimizers.

In the integrable case $F(u)=u^3$ (with $k=0$) the solitary waves are
$ce^{-|x|}$ and then they belong to the space $W^{1,\infty}(R)$. The reason
for this is that the integrated Euler-Lagrange equation gives the following
equation for the solitary waves:
$$ u_x^2=\frac{\alpha u^2-F(u)}{\alpha-u}
$$
where $\alpha>0$  is the multiplier (velocity). Now for if $F(u)=u^3$
we see that if for some $x$ the denominator $\alpha-u(x)$ vanishes then the
numerator
$\alpha u^2(x)-F(u(x))= \alpha u^2(x)-u^3(x)=u^2(x)(\alpha -u(x))$
 also vanishes. For more general nonlinearities this does not happen
in general; for instance, take $F(u)=cu^3, c\neq 1.$ As a consequence,
close to the point where the solution $u(x)$ is singular, say, $x=0$, $u(x)$
behaves as $x^{2/3}$ and $u_x(x)$ as $x^{-1/3}.$  So, in this case, the best
we can have is that $u\in W_{loc}^{1,p}$ for $1\leq p<3$.

Due to the fact that the terms in the constraint have different behavior
with respect to dilations, the verification of the strict subadditivity of
$V(u)$ for more general nonlinearities could be difficult.

In this paper we show the existence of minimizer for the problem above for
a large class of nonlinearities. We use a method that we have developed and
applied in several problems. A more abstract version of that method is
presented in \cite{l3}. The main assumption that has to be verified  is that
any critical point which is candidate to be minimizer has Morse index
$\geq 1$ (this implies that it is actually equal to 1).
We show that for problem \eqref{P} this condition is met  for very general
nonlinearities $F(u)$. In particular, there is no convexivity assumption
on $F(u)$.

For simplicity we make assumptions to guarantee that the minimization problem
picks positive solutions but we can also consider more general cases.
Under minor changes, we can also study the problem of minimizing $I(u)$
under $V(u)=\lambda$ but, for the sake of tradition,  we prefer to consider
the problem \eqref{P} above because $V(u)/2$ is the hamiltonian and $I(u)$
is the charge (or energy) for the evolution equation \eqref{1.2}.

Note that in the case of peaked solutions, zero belongs to the spectrum
of the linearized equation but it is not an eigenvalue; so, the spectrum
of the linearized equation accumulates at zero (from the right, of course).
This fact seems to exclude the possibility of using the GSS method
for finding peaked local minimizers because the spectral property of
solitons is not satisfied. It seems to exclude also the possibility of
using the implicit function theorem in the functional analytic setting
to show that the solitary wave is isolated modulo translation.
For special $F(u)$, bifurcation diagrams are given in \cite{l2}.

\section{Statement and Proof of the Main Result}

We consider the problem \eqref{P} for $k\geq 0$. The first and second
derivatives of $V$ and $I$ are:
\begin{gather*}
V'(u)h=-\int_{\mathbb{R}} (2uu_xh_x+u_x^2h+f(u)h+2kuh)\,dx\\
V''(u)(h,h)=-\int_{\mathbb{R}}(2uh_x^2+4u_xh_xh+f'(u)h^2+2kh^2)\,dx\\
I'(u)h=\int_{\mathbb{R}}(2u_xh_x+2uh)\,dx\\
I''(u)(h,h)=\int_{\mathbb{R}}(2h_x^2+2h^2)\,dx.
\end{gather*}
The minimization will be done in the space $H^1(R)$ under the following
assumptions:
\begin{itemize}
\item[A1]   $F:\mathbb{R} \to \mathbb{R}$ is a $C^2$ function satisfying the following
two conditions:
\begin{itemize}
\item[i)] $F(u)$ and its derivative $f(u)$
 vanish at $u=0$, and its second derivative satisfies the smallness
 condition, at $u=0$, $ |F''(u)|\leq M|u|^{\beta}$
for some $M>0$, $\beta>0$ and all $|u|\leq 1$.

\item[ii)] for $u\neq 0$, we have $uF(u)>0$.
\end{itemize}

\item[A2] There is an admissible element $u\in H^1(\mathbb{R})$ such that
 $V(u)<-k\lambda$.
\end{itemize}
Note that under Assumption  A1, part i,  the functionals $V$ and $I$ are of class
$C^2$ and their first and second derivative are given by the formulas above
and are uniformly continuous on bounded sets of $H^1(\mathbb{R})$.
Our main result is the following.

\begin{theorem} \label{thm2.1}
Under assumptions A1 and A2  and except for translations in
 the space variable $x$, any   minimizing sequence of the problem
 $\eqref{P}$ is precompact in $H^1(R)$.
Moreover, any minimizer is a positive function decaying exponentially at
infinity.
\end{theorem}


Before proving Theorem 2.1 we make some comments about our assumptions:
Assumption A1 part i simply says that the term $ku^2$ incorporates the
quadratic part.
As far as A2 is concerned, the following statements are true:
\begin{enumerate}
\item If $F(u)\geq k_1 u^{\gamma}$ for $u>0$ and small, where $k_1>0$
and $\gamma<6$, then assumption A2 is satisfied for any $\lambda >0$

\item If $F(u)\leq k_2 u^{\gamma}$ for $u>0$ and small, for some $k_2>0$
and $\gamma >6$, then for $\lambda>0$ and small, the infimum of $V(u)$ on
the admissible set is not achieved

\item Assumption  A2 is satisfied if $\lambda$ is large.
In fact, if $u>0$ is any element in $H^1(\mathbb{R})$ then
$ \lim_{t\to +\infty} V(tu)=-\infty$ and this implies the statement.
\end{enumerate}

Next,  we define the even piecewise linear function
\[
u(x)=\begin{cases}
c & \mbox{if }0\leq x\leq a\\
c(1+a/b-x/b) &\mbox{if } a\leq x\leq a+b\\
0 & \mbox{if } a+b \leq x
\end{cases}
\]
Imposing the constraint $I(u)=\lambda$ we obain $c^2= \frac{3\lambda b}{2(3ab+b^2+3)}$.

 Taking $b=1$, $c>0$ and letting
$a\to \infty$ we see that $\lim V(u)=-k\lambda$.
 This shows that assumption
A2 is a necessary condition for Theorem 2.1.
If we take $a=1$ and let $b\to \infty$ then statement 1 follows.
Statement 2 is a consequence of an interpolation inequality.


The Euler-Lagrange equation $V'(u)+\alpha I'(u)=0$ is
\begin{equation}
-2\frac{d}{dx}[(\alpha-u)u_x]-u_x^2+2(\alpha-k) u -f(u)=0
\end{equation}
and in the region where $u(x)\neq \alpha$ we have
\begin{equation}
u_x^2=\frac{(\alpha-k) u^2-F(u)}{\alpha -u(x)}.
\end{equation}
In general, the description of the set of the minimizers for a general
$F(u)$ seems to be a hard task. In section 3 we give example for which
we can give this description.


Next  we state a few lemmata which will be used in the proof of
Theorem \ref{thm2.1}.
The first is taken from \cite[lemma 2.8, part ii]{l3}.

\begin{lemma} \label{lm2.2}
If $u$ is the weak limit of a minimizing sequence  and $u$ satisfies the
Euler-Lagrange equation (2.1), then $(V''(u)+\alpha I''(u))(h,h)\geq 0$
for any $h\in H^1(R)$ such that $I'(u)h=0.$ In particular, there cannot
exist two functions $h_i\in H^1(R),i=1,2$ with disjoint supports such that
$(V''(u)+\alpha I''(u))(h_i,h_i)< 0, i=1,2$.
\end{lemma}

\begin{lemma} \label{lm2.3}
If if $u\in H^1(R),u\not\equiv 0,$ is the weak limit of a minimizing sequence
and $u$ satisfies  the Euler-Lagrange equation (2.1),
then $u(x)>0$, $\alpha -u(x)\geq 0$ everywhere and $\alpha>k$.
In particular, $\alpha >0$.
\end{lemma}

\paragraph{Proof.}
From assumption A1, part ii, and using Fatou's lemma we see that
$$\int_{\mathbb{R}}F(-u_-(x))\,dx=0$$
and this implies that $u(x)\geq 0$ everywhere. The fact that $u(x)>0$
everywhere follows from (2.2).

Now, suppose by contradiction that there is an open interval $J$ where
$\alpha -u(x)<0.$ Then $u$ is a regular solution of (2.1) on $J$ and for
any function $h\in H^1(R)$ with support contained in $J$ we have
\begin{align*}
&V''(u)(h,h)+\alpha I''(u)(h,h)\\
&=\int_{J}(2(\alpha-u)h_x^2 +(2u_{xx}-f'(u(x))+2(\alpha-k))h^2\,dx
\end{align*}
and then we can construct $h_1$ and $h_2$ with disjoint supports contained
in $J$ such that $V''(u)(h_i,h_i)+\alpha I''(u)(h_i,h_i)<0$, $i=1,2,$
but this is impossible because of Lemma \ref{lm2.2}.
If  $\alpha\leq k$ then (2.2)  is violated as $x$ tends to, say, $+\infty$
and this completes the proof. \hfill$\square$


\paragraph{Remark.}
It is easy to construct examples of regular solutions $u(x)$ of (2.1) such
that $u(x)>\alpha$ in some interval. Lemma \ref{lm2.3} says is that such
solutions cannot be local minimizers of problem \eqref{P}.

The next lemma tells us that any solution of the Euler-Lagrange equation (2.1)
which is candidate to be a minimizer of problem \eqref{P} has Morse index
$\geq 1.$ This information will be crucial for applying our method.

\begin{lemma}. \label{lm2.4}
If  $u\in H^1(R),u\not\equiv 0,$ is the weak limit of a minimizing sequence
and $u$ satisfies  (2.1) then there is an element $h\in H^1(R)$ such that
\begin{equation}
V''(u)(h,h)+\alpha I''(u)(h,h)=
\int_{\mathbb{R}}[2(\alpha-u)h_x^2-4u_xhh_x+(2\alpha-2k-f'(u))h^2]\,dx<0.
\end{equation}
\end{lemma}

\paragraph{Proof.}
The proof will be divided in three cases. Note that according to the previous lemma, for such element $u$ we must have $\alpha >k$ and $\alpha -u(x)\geq 0.$ In all cases we assume that a translation have been made so that $u(x)$ is an even function.

\noindent {\bf First Case:} $u$ is a regular solution.
In this case for $h=u_x$ we have $V''(u)(h,h)+\alpha I''(u)(h,h)=0$ because $u_x$ satisfies the linearized equation. Since $u_x$ changes sign the first case follows from the maximum principle.
To handle the singular case we define the even function
\[
h_{\epsilon}(x)=\begin{cases}
u_x(x) & \mbox{if }x\leq -\epsilon \\
u_x(-\epsilon) &\mbox{if } -\epsilon\leq x\leq \epsilon\\
-u_x(x) & \mbox{if }\epsilon \leq x
\end{cases}
\]
where $\epsilon >0$ will be small. The function $h_{\epsilon}$ is an element
of $H^1(\mathbb{R})$ because $u_x$ is odd. Taking in account that $u_x$ and $-u_x$
satisfy the linearized equation in its  region of regularity, an integration
by parts gives
\begin{align*}
&[V''(u)+\alpha I''(u)](h_{\epsilon},h_{\epsilon})\\
&= 2[2(\alpha-u)u_xu_{xx}-2u_x^3](-\epsilon)+u_x^2(-\epsilon)\int_{-\epsilon}^
{\epsilon} (2(\alpha-k) -f'(u(x))\,dx\\
&=2\big[-u_x^3+(2(\alpha-k) u-f(u))u_x\big]
(-\epsilon)+u_x^2(-\epsilon)\int_{-\epsilon}^{\epsilon} (2(\alpha-k)
 -f'(u(x))\,dx.
\end{align*}

\noindent {\bf Second Case:} $u$ is peaked but $u\in W^{1,\infty}(R)$.
In this case, at the singular point $x=0$ we must have $u(0)=\alpha$. So, $u(x)>0$ everywhere and increasing for $x<0.$ From the integrated  Euler-Lagrange equation (2.2) we see
that we also must have $(\alpha-k) u(0)^2-F(u(0))=0$ and then as a consequence of l'Hopital rule  we get $u_x(0)^2=-[2(\alpha-k) u(0)-f(u(0))]$  and this implies that
$$\lim_{\epsilon \to  0} V''(u)((h_{\epsilon},h_{\epsilon})=-4\lim_{x\to 0_-} u^3_x(x)<0$$
and then $h_{\epsilon}$ verifies (2.3) for $\epsilon$ small.

\noindent {\bf Third Case:} $u$ is peaked and $u\not\in W^{1,\infty}(R)$.
In this case $(\alpha-k) u(0)^2-F(u(0))\neq 0$, hence
$$\lim_{x\to 0_-} u_x(x)=+\infty$$
and then
$$\lim_{\epsilon \to 0} V''(u)((h_{\epsilon},h_{\epsilon})=-\infty$$
because the term $-u_x^3$ dominates and Lemma \ref{lm2.4} is proved in all cases.
\hfill$\square$ \smallskip

 As a preparation for the proof of Theorem \ref{thm2.1}, we start with a lemma
that prevents vanishing. This lemma is due to P. Lions \cite{l1} and it is
related to the so called Lieb's lemma \cite{b1} (see also \cite[lemma 2.3]{l3}).

\begin{lemma} \label{lm2.5}
Let $u_n$ be a bounded sequence in $H^1(R)$ such that the sequence $u(x+c_n)$
converges to zero weakly in $H^1(R)$ for any sequence $c_n$ of elements of
$R$. Then $u_n$ converges to zero strongly in $L^p(R)$ for any $p$ in the
interval $2<p\leq \infty$.
\end{lemma}

Our next lemma is also taken from \cite[lemmas 2.6 and 2.7]{l3}. It says that
a minimizing sequence satisfies an approximate Euler-Lagrange equation.

\begin{lemma} \label{lm2.6}
If $u_n$ is a minimizing sequence of the problem \eqref{P} then there are sequences $\alpha_n$ $\gamma_n$ of real numbers, $\alpha_n$ bounded and $\gamma_n$ tending to zero, and a sequence $\phi_n$ of unitary elements of $H^1(R)$ such that for any $h\in H^1(R)$ we have
\begin{equation}
\begin{aligned}
& V'(u_n)h+\alpha_nI'(u_n)h+\gamma_n \langle\phi_n,h\rangle\\
&= \int_{\mathbb{R}}[2(\alpha _n-u_n)u_{n,x}h_x-u_{n,x}^2h +2(\alpha_n-k)u_nh\\
&\quad -f(u_n)h +2\gamma_n \phi_{n,x}h_x+2\gamma_n\phi_nh]\,dx=0.
\end{aligned}
\end{equation}
\end{lemma}


\paragraph{Proof of Theorem \ref{thm2.1}}
Let $u_n$ be a minimizing sequence for problem \eqref{P}. We first note that
there is a sequence of real numbers $c_n$ such that the sequence
$u_n(\cdot+c_n)$ has a subsequence converging weakly in $H^1(R)$ to a
nonzero function. In fact, otherwise, according to Lemma \ref{lm2.5}, $u_n$
converges to zero strongly in $L_{\infty}(R)$ and then
$$
\lim_{n\to \infty} V(u_n)=-k\lambda+\lim_{n\to \infty}
\int_{\mathbb{R}}(-u_nu_{n,x}^2-F(u_n)+ku_{n,x}^2)\,dx \geq -k\lambda
$$
and this contradicts assumption  A2.

Passing to a subsequence, for which we keep the same notation, we can assume
that $\alpha_n$ converges to $\alpha$ and that $u_n$ converges weakly in
$H^1(R)$ to a nonzero element $u\in H^1(R).$ Passing (2.4) to the limit we
also have
\begin{equation} V'(u)h+\alpha I'(u)h=
\int_{\mathbb{R}}(2(\alpha-u) u_xh_x -u_x^2h+2(\alpha-k)uh-f(u)h)\,dx=0
\end{equation}
for any $h\in H^1(R)$.

 In view of the assumptions we have made and the lemmata we have proven,
we can use \cite[Theorem 2.1]{l3} to conclude that $u_n$ converges to $u$ in
$L_p(R)$, $2<p\leq \infty$.

 We can improve this convergence in the following way: taking $h=u_n-u$
in (2.4) and (2.5), subtracting one from the other and using the convergence
we have already proved we get
$$
\lim_{n\to \infty} \int_{\mathbb{R}}[2(\alpha-u)(u_{n,x}-u_x)^2
+2(\alpha-k)(u_n-u)^2]\,dx=0.
$$
From Lemma \ref{lm2.3}, we know that $\alpha >k$ and $\alpha-u(x)\geq 0$ everywhere
and then $u_n$ converges to $u$ also in $L_2(R)$ and
\begin{equation}
\lim_{n\to \infty} \int_{\mathbb{R}}(\alpha-u)(u_{n,x}-u_x)^2\,dx=0.
\end{equation}
This equation allows us to conclude that $u_n$ converges to
$u$ in the $H^1$ norm but only outside the points where $u(x)\neq \alpha$. If we assume that $u(0)=\alpha$, it remains to show that
\begin{equation}
\lim_{n\to \infty}\int_{-1}^1u_{n,x}^2\,dx=\int_{-1}^1u_x^2\,dx.
\end{equation}
To accomplish this, we consider a smooth function $\psi$ such that
$$
\psi(x)=\begin{cases} 1 &\mbox{for }  -1\leq x \leq 1\\
  0 &\mbox{for }  |x|\geq 2.
\end{cases}
$$
 With $J=[-2,-1]\cup [1,2]$ and $h=\psi$ in (2.4) we obtain
\begin{align*}
\int_{-1}^1u_{n,x}^2\,dx
=&\int_{-1}^1(-f(u_n)-2ku_n+2\alpha u_n+\gamma_n\phi_n)\,dx\\
&+\int_J(2(\alpha_n-u_n)u_{n,x}\psi_x-u_{n,x}^2\psi\\
&-f(u_n)\psi+(2\alpha-2k)u_n\psi+2\gamma_n\phi_{n,x}\psi_x
+2\gamma_n\phi_n\psi)\,dx.
\end{align*}
Taking the limit in the above equation, using the convergence we have already
proved, and taking $h=\psi$ in (2.5), we see that (2.7) holds and
Theorem \ref{thm2.1} is proved. \hfill$\square$ \smallskip

As we have pointed out in the introduction, in the case of a peaked solution,
zero belongs to the spectrum of the linearized  operator
$$L(h)=-2\frac{d}{dx}[(\alpha-u)h_x-u_xh]-2u_xh_x+2(\alpha-k)h-f'(u)h$$
but is not an eigenvalue.
Zero belongs to the spectrum of $L$ because, due to translation invariance,
a critical point of $V(u)+\alpha I(u)$ is not isolated. Furthermore,
if $L(h)=0$ then for $x<0$ and for $x>0$, $h$ has to a multiple
(with possible different factors) of $u_x.$ Since $h$ has to be continuous,
it remains to consider the case $u_x$ is bounded and has a nonzero limit
at $x=0$. But, in this case,
$$u_{xx}=\frac{u_x^2+\alpha u-f(u)}{\alpha -u}$$
does not belong to $L^2(\mathbb{R})$ because the numerator has a nonzero limit,
equal to $ 3\lim_{x\to 0} u_x^2(x)/2$, and the denominator behaves as $x$
for $x=0$.

\section {Examples}

First we use a trick presented in \cite{c4}: To give a formula for
$I(u(\alpha))$ as a function of the multiplier $\alpha$.

We start by defining the functions $F_1(u)=F(u)/u^2$,
$u_1(\alpha)$ as the first zero (if any) of $\alpha -k -F_1(u)$,
and $u_0(\alpha)=\min(u_1(\alpha),\alpha)$.
Note that $u_0(\alpha)=\alpha$ for peaked solutions and
$u_0(\alpha)=u_1(\alpha)$ for regular solitary waves.

For $x<0$ the solution $u(x)$ is increasing and satisfies
$0\leq u(x)\leq u_0(\alpha)$. Then
$$ u_x=\frac{u\sqrt{\alpha-k-F_1(u)}}{\sqrt{\alpha-u}}u
$$
and then
$$  u_x^2+u^2=\frac{(\alpha-k)u^2-F(u)}{\alpha -u}+u^2
  =\frac{u[(2\alpha -k)-u-F_1(u)]}{\sqrt{(\alpha-u)(\alpha-k-F_1(u)}}u_x\,.
$$
This implies
\begin{eqnarray}
I(\alpha)=I(u(\alpha))&=&2\int_{-\infty}^{0}
\frac{u[(2\alpha -k)-u-F_1(u)]}{\sqrt{(\alpha-u)(\alpha-k-F_1(u))}}u_x\,dx\nonumber\\
&=&2\int_0^{u_0(\alpha)}\frac{u[(2\alpha -k)-u-F_1(u)]}
{\sqrt{(\alpha-u)(\alpha-k-F_1(u))}}\,du.
\end{eqnarray}
As a first example we take  $F(u)=pu^3+ku^2$ with $p,k>0$.
According to Theorem \ref{thm2.1} for any $\lambda >0$ problem (1.1) has a
 global minimizer. In this section we study the function $I(\alpha)$ as a
function of the multiplier.

For convenience we change notation and replace $\alpha$ by $y$.
Defining the function $ R(x)=a+bx+cx^2$
with $c>0$, and the indefinite integral
$$ I_m(x)=\int \frac{x^m}{\sqrt{R(x)}}\,dx,
$$
then according to \cite[pages 81 and 83]{g1}, we have
\begin{gather*}
I_0(x)=\frac{1}{\sqrt{c}} \log (2\sqrt{cR(x)}+2cx+b),\\
I_1(x)=\frac{\sqrt{R(x)}}{c}-\frac{b}{2c}I_0(x),\\
I_2(x)=\big(\frac{x}{2c}-\frac{3b}{4c^2}\big)\sqrt{R(x)}
+\big(\frac{3b^2}{8c^2}-\frac{a}{2c}\big)I_0(x).
\end{gather*}
Using (3.9) we get
$$
I(y)/2=\int_0^{u_0(y)}\frac{(2y-k)u-(p+1)u^2}{\sqrt{(y-u)(y-k-pu)}}\,dx
=(2y-k)I_1-(p+1)I_2.
$$
Next we define the following functions and compute their derivatives:
\begin{gather*}
f(y)=(-kp+3k-3p^2y+2py-3y)/4p^2,\quad f'(y)=( - 3p^2 + 2p - 3)/(4p^2),\\
r(y)=(p-1)y+k,\quad  r'(y)=(p-1),\\
t(y)=\sqrt{y(y-k)},\quad  t'(y)=(2y-k)/(2t(y)),\\
s(y)= 2\sqrt{p}t(y)+k-(p+1)y, \quad s'(y)=\sqrt{p}(2y-k)/t(y)-(p+1),\\
\overline{s}(y)=2\sqrt{p}t(y) -k+(p+1)y,\quad
\overline{s}'(y)= (2y-k)/2t(y)+(p+1),\\
q(y)=(kp-3k-3p^2y+3y)r(y)/(8p^2),\quad q'(y)=(1-p)(3p^2y+pk+3k-3y),\\
h(y)=8(1-p)y^2+4k(p-3)y+(p+3)k^2.
\end{gather*}
Then
$I(y)/2=-t(y)f(y)+q(y)I_0(y) $ and
$$ I'(y)/2=-\frac{(2y-k)}{t(y)}f(y)-t(y)f'(y)+q'(y)I_0(y)+q(y)I'_0(y).
$$
In order to analyze the behavior of $I'(y)$ we define the functions
$$
g(y)=\big(\frac{2y-k}{t(y)} f(y)+t(y)f'(y)-q(y)I'_0(y)\big)/q'(y)
$$
and $w(y)=g(y)-I_0(y)$.

To give the formula for $I_0(y)$ we have to consider two cases.

\noindent {\bf First case (smooth solitary wave)} $k\leq y \leq k/(1-p)$:
In this case, $u_0(y)=(y-k)/p$ and
$$
I_0(y)=\frac{1}{\sqrt{p}}\log(-r(y)/s(y))
=\frac{1}{\sqrt{p}}\log(\overline{s}(y)/r(y))
$$
because $s(y)\overline{s}(y)=-r^2(y)$.
Defining $y_0=\frac{k(p+3)}{3(1-p)(1+p)}$, we have $k< y_0<k/(1-p)$.
 Now we collect the following elementary facts: Regarding $q$, we have
$q'(y)>0$ for $k<y<y_0$, $q'(y_0)=0$ and $q'(y)<0$ for $y_0<y<k/(1-p)$.
Regarding $g$, we have
$$g(y)=\frac{-2(3p^2-2p+3)t(y)}{q'(y)}
$$
then $g(k)=0$,  $g(y)<0$  for $k<y<y_0$,
$\lim_{y\to y^-_0}g(y)=-\infty$, $\lim_{y\to y^+_0}g(y)=+\infty$
and $g(y)>0$ for $y_0<y\leq k/(1-p)$.
Regarding $I_0$, we have
 $I_0(k)=0$ and $\lim_{y\to (k/(1-p))^-}I_0(y)=+\infty$.
Regarding $w$, we have
$w(k)=0$, $\lim_{y\to y^-_0}w(y)=-\infty$,
$\lim_{y\to y^+_0}w(y)=+\infty$,
$\lim_{y\to (k/(1-p))^-}w(y)=-\infty$, and
$$ w'(y)=\frac{-4p^2(p-1)h(y)}{r(y)q'^2(y)}<0.
$$
This last inequality holds because the interval $[k,k/(1-p)]$
is interior to the interval whose ends are the roots of the quadratic
equation $h(y)=0$.

Putting the above  facts together we see that there is a point $y_1$,
$y_0<y_1<k/(1-p)$ such that $I'(y)>0$ for $ k<y<y_1$, $I'(y_1)=0$ and
$I'(y)<0$ for $ y_1<y<k/(1-p)$.

\noindent {\bf Second Case (peakon):} $k/(1-p)\leq y$.
In this case, $u_0(y)=y$ and
$$
I_0(y)=\frac{1}{\sqrt{p}}\log(r(y)/s(y))=\frac{1}{\sqrt{p}}
\log(-\overline{s}(y)/r(y))\,.
$$
As elementary facts we have: Regarding $q$:
 $q'(y)<0$. Regarding, $g$ we have
 $g(y)>0$ and
 $$\lim_{y\to+\infty}g(y)= \frac{2(3p^2-2p+3)}{3(1-p)^2(p+1)}\,.
$$
Regarding $I_0$, we have:
$\lim_{y\to (k/(1-p))^+}I_0(y)=+\infty$ and
$$\lim_{y\to +\infty} I_0(y)=\frac{1}{\sqrt{p}}
\log\Big(\frac{(1-p)}{p-2\sqrt{p}+1}\Big)\,.
$$
Regarding $w$, we have:
$$ \lim_{y \to +\infty}w(y)= \frac{2(3p^2-2p+3)}{3(1-p)^2(p+1)}
 -\frac{1}{\sqrt{p}}\log\Big(\frac{(1-p)}{p-2\sqrt{p}+1}\Big)>0,
$$
this inequality because by defining
$$
\phi(p)=\sqrt{p}\Big(\frac{2(3p^2-2p+3)}{3(1-p)^2(p+1)}
-\frac{1}{\sqrt{p}}\log\Big(\frac{(1-p)}{p-2\sqrt{p}+1}\Big)\Big)\,,
$$
we have $\phi(0)=0$ and $\phi'(p)=\frac{-32p^2}{3\sqrt{p}(p+1)^3(p-1)^3}>0$.
Also, there exists $y_2>k/(1-p)$ such that
$w'(y)=\frac{-4p^2(p-1)h(y)}{r(y)q'^2(y)}>0$ for
$k/(1-p)<y<y_2$, $w(y_2)=0$, and $w'(y)<0$ for $y_2<y$.
From the inequalities above, we see that there is a point $y_3>y_2$ such
that $I'(y)<0$ for $k/(1-p)<y<y_3$, $I'(y_3)=0$ and $I'(y)>0$ for $y_3<y$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.8\textwidth]{fig1.eps}
\caption{Behavior of $I(y)$ for  $y<k/(1-p)$}
\end{center}
\end{figure}

Figure 1 shows the qualitative behavior of $I(y)$. For  $y<k/(1-p)$ we are
in the region of solitary waves.
The region AB corresponds to global minimizers whose existence is given by
Theorem \ref{thm2.1}. Numerical experiments indicate that the branch of global
minimizer goes up to the point C which is in the level of the transition
point E. The branch ABCDE, except for the point D, falls in the GSS method.
So, from A to D the solitary waves are stable (as local minimizers) and from
D to E they are linearly unstable.

For $y>k/(1-p)$ we are in the region of peakons. There the GSS method as it
is proved  in \cite{g2} does not apply because the spectrum of the linearized
operator does not have the spectral property of solitons.

The upper branch starting at the point H corresponds to global minimizers
(hence stable peakons) whose existence is given by Theorem \ref{thm2.1}.
As in the region of solitary waves, numerical experiments indicate that the
branch of global minimizers goes further to the point G which is in the
level of the transition point E. Probably the branch EF corresponds to
unstable peakons and FG to stable ones (local minimizers) but so far there
are no results that support this conclusion.

We can also consider the case $p\geq 1$ and $k>0$. In this case
$u_0(y)=(y-k)/p$ (solitary wave) and elementary calculations give the
following facts:
$I_0(y)=\frac{1}{\sqrt{p}}\log(\overline{s}(y)/r(y))$ is regular everywhere,
$q'(y)>0$, and $w'(y)>0$.
Then $I'(y)>0$ for $y>k$  So, these solitary waves are stable global
minimizers. The fact that they are local minimizers follows from the
GSS method. The case $p=1$ has been considered in \cite{c4}; but their trick to
verify that $I'(y)>0$ does not seem to work in the case $p>1$.

When $k=0$ it is easy to see that $I(y)=yI(1)$ and then our minimization
result gives the existence of a stable solitary wave if $p>1$ and of a
stable peakon if $p\leq 1$ (the case $p=1$ has been considered in \cite{c2}).


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\end{thebibliography}

\noindent\textsc{Orlando Lopes}\\
Departamento de Matematica-IMECC-UNICAMP- \\
C.P. 6065 Campinas, SP 13083-859, Brasil \\
e-mail: lopes@ime.unicamp.br

\end{document}