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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 49, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/49\hfil Positive periodic solutions]
{Positive periodic solutions for the Korteweg-de Vries equation}

\author[S. G. Georgiev\hfil EJDE-2007/49\hfilneg]
{Svetlin Georgiev Georgiev}  % in alphabetical order

\address{Svetlin Georgiev Georgiev \newline 
Department of Differential Equations,
University of Sofia, Sofia, Bulgaria}
\email{sgg2000bg@yahoo.com }


\thanks{Submitted January 18, 2006. Published April 4, 2007.}
\subjclass[2000]{35Q53, 35Q35, 35G25}
\keywords{Nonlinear evolution equation; Kortewg de Vries equation;
\hfill\break\indent  periodic solutions}

\begin{abstract}
 In this paper we prove that the Korteweg-de Vries equation
 $$
 \partial_t u+\partial_x^3 u+u\partial_x u=0
 $$
 has unique positive solution $u(t, x)$ which is $\omega$-periodic
 with respect to the time variable $t$ and
 $u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$,
 $\gamma>0$, $\gamma\notin \{1, 2, \dots\}$, $p>1$, $q\geq 1$,
 $a<b$ are fixed constants, $x\in [a, b]$.
 The period $\omega>0$ is arbitrary chosen and fixed.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction}

In this paper we consider the initial-value problem for the
Korteweg-de Vries equation
\begin{gather}
 \partial_t u+\partial_x^3 u+u\partial_x u=0,\quad t\in \mathbb{R},\quad
x\in [a, b], \label{e1}\\
 \text{$u$  is  periodic  in $t$}, \label{e1'}\\
 u(0, x)=u_0, \quad u_0\in {\dot B}^{\gamma}_{p, q}([a, b]),
\label{e1''}
\end{gather}
where $q\geq 1$, $1<p<\infty$, $\gamma>0$, $\gamma\notin \{1, 2,
\dots\}$. We prove that the \eqref{e1}--\eqref{e1''} has unique
positive solution  in the form $u(t, x)=v(t)q(x)$, which is
continuous $\omega$-periodic with respect to the time variable
$t$. When we say that the solution $u(t, x)$ of the \eqref{e1} is positive we understand: $u(t, x)>0$ for
$t\in \mathbb{R}$, $x\in [a, b]$.
Here the period $\omega>0$ is arbitrary chosen and fixed.

 Bourgain \cite{b1}  consider the initial-value problem
\begin{gather*}
 \partial_t u+\partial_x^3 u+u\partial_x u=0,\\
\text{$u$ is periodic  in $x$},\\
 u(0, x)=u_0.
\end{gather*}
He proved that the above problem is globally well-posed for
$H^s$-data ($s\geq 0$, integer). Bourgain  \cite{b1} used the Fourier
restriction space method, which he introduced.

Here we use the theory of completely continuous vector field
presented by  Krasnosel'skii and  Zabrejko and we prove
that the Korteweg-de Vries \eqref{e1} has unique positive
solution $u(t, x)=v(t)q(x)$, which is continuous $\omega$-periodic
with respect to the time variable $t$ and infinitely
differentiable with respect to the space variable $x\in [a, b]$
and $u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$, $p>1$,
$q\geq 1$, $\gamma>0$, $\gamma\notin \{1, 2, \dots\}$.

To state our main result we use the following hypotheses:
\begin{itemize}
\item[(H1)]  $q\in \mathcal{C}^{\infty}([a, b])$,
 $q(x)>0$ for all $x\in [a, b]$;

\item[(H2)] $q'(x)< 0$, $q'''(x)> 0$ for all $x\in [a, b]$.
\end{itemize}

\begin{theorem} \label{thm1}
 Let $q\geq 1$, $1<p<\infty$, $\gamma>0$,
$\gamma\notin \{1, 2, \dots\}$ be fixed. Then the initial-value
problem \eqref{e1}--\eqref{e1''} has unique positive
solution $u(t, x)=v(t)q(x)$, which is continuous $\omega$-periodic
with respect to the time variable $t$ and infinitely
differentiable with respect to the space variable $x\in [a, b]$,
where $q(x)$ is a fixed function  satisfying {\rm (H1)--(H2)}.
\end{theorem}

This paper is organized as follows: In section 2 we prove that
the \eqref{e1}--\eqref{e1''} has positive solution
$u(t,x)=v(t)q(x)$ which is continuous $\omega$-periodic with respect to
the time variable $t$ and infinitely differentiable with respect
to the space variable $x\in [a, b]$, where $q(x)$ is fixed
function satisfying (H1)--(H2).
In section 3 we prove that the solution obtained in
section 2, is unique.

\section{Existence of positive periodic solutions}

Here and bellow we will suppose that $q(x)$ is fixed function
satisfying (H1)--(H2). As an example of such function,  we have
$q(x)=2+\sin x$ with $[a, b]=[2\pi/3, 5\pi/6]$.

\begin{proposition} \label{prop2.1}
If for every fixed $x\in [a, b]$,
$u(t, x)=v(t)q(x)$ satisfies
\begin{equation}
u(t, x)=-\int_0^{\omega}
{{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
u^2(t-s, x){{q'(x)}\over {q(x)}}ds,
\label{e2}
\end{equation}
then $u(t, x)=v(t)q(x)$ satisfies the \eqref{e1} for every fixed $x\in [a, b]$.
 Here $v(t)$ is a positive continuous $\omega$-periodic function.
\end{proposition}

\begin{proof} For every fixed $x\in [a, b]$ if $u(t, x)=v(t)q(x)$
is a solution to  \eqref{e2}, we have
\begin{align*}
v(t)q(x)&=-\int_0^{\omega}
{{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
v^2(t-s)q^2(x){{q'(x)}\over {q(x)}}ds\\
&= -\int_0^{\omega}
{{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
v^2(t-s)q(x)q'(x)ds.
\end{align*}
 From here,
$$
v(t)=-\int_0^{\omega}
{{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
v^2(t-s)q'(x)ds;
$$
i.e., for every fixed $x\in [a, b]$, if $u(t, x)=v(t)q(x)$ is a solution to
\eqref{e2} we have
\begin{equation}
v(t)=-{{q'(x)}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}\int_0^{\omega}
e^{-{{q'''(x)}\over {q(x)}}s}v^2(t-s)ds.
\label{star}
\end{equation}
Let us consider the integral
$$
\int_0^{\omega} e^{-{{q'''(x)}\over {q(x)}}s}v^2(t-s)ds.
$$
We make the change of variable $s=t-z$, from where $ds=-dz$ and
\begin{align*}
\int_0^{\omega} e^{-{{q'''(x)}\over {q(x)}}s}v^2(t-s)ds
&= -\int_t^{t-\omega} e^{-{{q'''(x)}\over {q(x)}}(t-z)}v^2(z)dz
\\
&=e^{-{{q'''(x)}\over {q(x)}}t}
\Bigl(\int_0^t e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz-
\int_0^{t-\omega} e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz\Bigr).
\end{align*}
Then the equality \eqref{star} takes the form
$$
v(t)=-{{q'(x)}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
e^{-{{q'''(x)}\over {q(x)}}t}
\Bigl(\int_0^t e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz-
\int_0^{t-\omega} e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz\Bigr).
$$
 From the above equality, for every fixed $x\in [a, b]$, we get
\begin{align*}
v'(t)&=-{{q'(x)}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
e^{-{{q'''(x)}\over {q(x)}}t}\Bigl[-{{q'''(x)}\over {q(x)}}
\Bigl(\int_0^t e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz\\
&\quad-\int_0^{t-\omega} e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz\Bigr)
 + e^{{{q'''(x)}\over {q(x)}}t}v^2(t)-
e^{{{q'''(x)}\over {q(x)}}(t-\omega)}v^2(t-\omega)\Bigr]\\
&={{q'''(x)}\over {q(x)}}
{{q'(x)}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
e^{-{{q'''(x)}\over {q(x)}}t}
\Bigl(\int_0^t e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz-
\int_0^{t-\omega} e^{{{q'''(x)}\over {q(x)}}z}v^2(z)dz\Bigr)\\
&\quad - {{q'(x)}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(1-e^{-{{q'''(x)}\over {q(x)}}\omega}\Bigr)v^2(t)\\
&=-{{q'''(x)}\over {q(x)}}v(t)-q'(x)v^2(t);
\end{align*}
i.e., for every fixed $x\in [a, b]$ we have
$$
v'(t)=-{{q'''(x)}\over {q(x)}}v(t)-q'(x)v^2(t).
$$
Then
\begin{equation}
q(x)v'(t)=-q'''(x)v(t)-q'(x)q(x)v^2(t)
\label{starstar}
\end{equation}
for every fixed $x\in [a, b]$.
Since for every fixed $x\in [a, b]$ we have
\begin{gather*}
u_t=v'(t)q(x),\\
\partial_x^3 u=q'''(x)v(t),\\
u\partial_x u=q'(x)q(x)v^2(t).
\end{gather*}
 From the equality \eqref{starstar} we take
$$
u_t=-\partial_x^3 u-u\partial_x u;
$$
i.e., for every fixed $x\in [a, b]$, if $u(t, x)=v(t)q(x)$
is a solution to the \eqref{e2}, then $u(t, x)$ satisfies the
 Korteweg-de Vries equation \eqref{e1}.
\end{proof}

\begin{proposition} \label{prop2.2}
If for every fixed $x\in [a, b]$, $u(t, x)=v(t)q(x)$
satisfies the Korteweg-de Vries equation \eqref{e1} then
$u(t, x)=v(t)q(x)$ satisfies the integral \eqref{e2}.
Here $v(t)$ is positive continuous $\omega$-periodic function.
\end{proposition}

\begin{proof}
 Let $x\in [a, b]$ is fixed and $u(t, x)=v(t)q(x)$ is a solution to the
Korteweg- de Vries \eqref{e1}, where $v(t)$ is positive continuous
$\omega$-periodic function. Then
$$
v'(t)q(x)=-q'''(x)v(t)-v^2(t)q'(x)q(x).
$$
After we use the definition of the function $q(x)$ (see (H1), (H2))
from the last equation we get
$$
v'(t)=-{{q'''(x)}\over {q(x)}}v(t)-q'(x)v^2(t).
$$
Since $x\in [a, b]$ is fixed, the last equation we may consider as
 ordinary differential equation with respect to the variable $t$. Therefore
\begin{align*}
v(t)&=e^{-\int_0^t{{q'''(x)}\over {q(x)}}ds}
\Bigl(v(0)-\int_0^t q'(x)v^2(s)e^{\int_0^s {{q'''(x)}\over {q(x)}}d\tau}ds\Bigr)\\
&=e^{-{{q'''(x)}\over {q(x)}}t}
\Bigl(v(0)-\int_0^t q'(x)v^2(s)e^{ {{q'''(x)}\over {q(x)}}s}ds\Bigr).
\end{align*}
For $q'''(x)>0$, $q(x)>0$ for $x\in [a, b]$ we have
$\lim_{t\to -\infty}e^{-{{q'''(x)}\over {q(x)}}t}=\infty$.
Therefore,
$$
v(0)=q'(x)\int_0^{-\infty}v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
-q'(x)\int_{-\infty}^0v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds
$$
or
\begin{equation}
v(t)=-q'(x)e^{-{{q'''(x)}\over {q(x)}}t}\int_{-\infty}^t
v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds.
\label{3star}
\end{equation}
Now we consider the integral
$$
\int_{-\infty}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds.
$$
 We have
\begin{equation}
\int_{-\infty}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
\int_{t-\omega}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds+
\int_{t-2\omega}^{t-\omega} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds+
\dots. \label{e3}
\end{equation}
Let
$$
J=\int_{t-\omega}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds.
$$
Let us consider the integral
$$
\int_{t-2\omega}^{t-\omega} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds.
$$
After the change of variable $ s+\omega=\tau$, we obtain
$$
\int_{t-2\omega}^{t-\omega} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
e^{-{{q'''(x)}\over {q(x)}}\omega}
\int_{t-\omega}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
e^{-{{q'''(x)}\over {q(x)}}\omega}J.
$$
In the same way,
$$
\int_{t-3\omega}^{t-2\omega} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
e^{-{{q'''(x)}\over {q(x)}}\omega}
\int_{t-2\omega}^{t-\omega} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
e^{-2{{q'''(x)}\over {q(x)}}\omega}J
$$
and so on and so forth. Then the equality \eqref{e3} takes the form
$$
\int_{-\infty}^{t} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds=
J\Bigl(1+e^{-{{q'''(x)}\over {q(x)}}\omega}+
e^{-2{{q'''(x)}\over {q(x)}}\omega}+\dots\Bigr)
=J{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}},
$$
because ${{q'''(x)}\over {q(x)}}>0$ for every fixed $x\in [a, b]$,
$e^{-{{q'''(x)}\over {q(x)}}\omega}<1$ for every fixed $x\in [a, b]$.
Therefore, from \eqref{3star}, for every fixed $x\in [a, b]$ we get
$$
v(t)=-q'(x)e^{-{{q'''(x)}\over {q(x)}}t}
{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\int_{t-\omega}^tv^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds.
$$
Now we make the change of variable $s-t=\tau$. Then
\begin{align*}
v(t)&=-q'(x)e^{-{{q'''(x)}\over {q(x)}}t}
{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\int_{-\omega}^0 v^2(t+\tau)e^{{{q'''(x)}\over {q(x)}}\tau}
e^{{{q'''(x)}\over {q(x)}}t}d\tau\\
&=-q'(x){1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\int_{-\omega}^0 v^2(t+\tau)e^{{{q'''(x)}\over {q(x)}}\tau}
d\tau.
\end{align*}
Let $\tau=-z$. Then
$$
v(t)=-q'(x) {1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\int_{0}^{\omega} v^2(t-z)e^{-{{q'''(x)}\over {q(x)}}z}dz.
$$
 From where for every fixed $x\in [a, b]$,
$$
u(t, x)=-{{q'(x)}\over {q(x)}}
{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\int_{0}^{\omega} u^2(t-z, x)e^{-{{q'''(x)}\over {q(x)}}z}dz;
$$
i. e., for every fixed $x\in [a, b]$, $u(t, x)$ satisfies \eqref{e2}.
\end{proof}

Let $\mathcal{C}(\omega)$ be the space of the real continuous
$\omega$-periodic functions defined on the whole axis.
With $\mathcal{C}_{+}(\omega)$ we denote the space of the positive
continuous $\omega$-periodic functions defined
on the whole axis.
Let
$$
D_q^{+}=\max_{0\leq s\leq \omega,\, x\in [a, b]}
e^{-{{q'''(x)}\over {q(x)}}s},\quad
D_q^{-}=\min_{0\leq s\leq \omega,\, x\in [a, b]}
e^{-{{q'''(x)}\over {q(x)}}s}.
$$
With $\mathcal{C}_+^{\circ}(\omega)\subset \mathcal{C}_{+}(\omega)$
we denote the cone
$$
\mathcal{C}_+^{\circ}(\omega)=\big\{
x\in \mathcal{C}_+(\omega):\min_tx(t)\geq {{D_q^{-}}\over {D_q^{+}}}
\max_t x(t)\big\}.
$$
For every fixed $x\in [a, b]$ we define the operator
$$
\chi(u)=-{{q'(x)}\over {q(x)}}
\int_0^{\omega}u^2(t-s, x)
{{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}ds,
$$
where $u(t, x)=v(t)q(x)$, $v(t)$ is a positive continuous $\omega$-periodic
function, $q(x)$ is a function  satisfying (H1), (H2).


\begin{proposition} \label{prop2.3}
For every fixed $x\in [a, b]$ we have
$\chi:\mathcal{C}_+(\omega)\to \mathcal{C}_{+}^{\circ}(\omega)$.
\end{proposition}

\begin{proof}  Let $x\in [a, b]$ is fixed. Let also
$u(t, x)\in \mathcal{C}_+(\omega)$.
$u(t, x)$ is continuous $\omega$-periodic with respect to the
time variable $t$.
Then
\begin{align*}
\chi(u)&= -{{q'(x)}\over {q(x)}} \int_0^{\omega}u^2(t-s, x)
{{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}ds
\\
&\geq D_q^{-}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(t-s, x)ds\Bigr)\\
&=D_q^{-}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(s, x)ds\Bigr);
\end{align*}
i.e., for every fixed $x\in [a, b]$ we have
$$
\chi(u)\geq
D_q^{-}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(s, x)ds\Bigr).
$$
 From where, for every fixed $x\in [a, b]$, we have
\begin{equation}
\min_t\chi(u)\geq
D_q^{-}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(s, x)ds\Bigr).
\label{e4}
\end{equation}
On the other hand, for every fixed $x\in [a, b]$, we have
$$
\chi(u)\leq
D_q^{+}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(s, x)ds\Bigr).
$$
Therefore, for every fixed $x\in [a, b]$, we have
$$ %5
\max_t\chi(u)\leq
D_q^{+}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}
\Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(s, x)ds\Bigr).
$$
 From this inequality and \eqref{e4},
$$
\min_t\chi(u)\geq {{D_q^-}\over {D_q^+}}\max_t\chi(u)
$$
for every fixed $x\in [a, b]$. Consequently for every fixed
 $x\in [a, b]$ we have
$$
\chi:\mathcal{C}_+(\omega)\to \mathcal{C}_+^{\circ}(\omega).
$$
\end{proof}

 From proposition \ref{prop2.3}, we have that
$\chi:\mathcal{C}_+^{\circ}(\omega)\to \mathcal{C}_{+}^{\circ}(\omega)$,
i.e.
the operator $\chi$ is positive with respect to the cone
$\mathcal{C}_+^{\circ}(\omega)$ for every fixed $x\in [a, b]$.

\begin{proposition} \label{prop2.4}
The operator $\chi$ is completely continuous in the space
$\mathcal{C}(\omega)$ for every fixed $x\in [a, b]$.
\end{proposition}

\begin{proof}  Let $x\in [a, b]$ be fixed.
Let also $u(t, x)\in \mathcal{C}(\omega)$, $\max_{t\in [0, \omega]}|u(t, x)|=r$,  $r>0$.
$u(t, x)$ is continuous $\omega$- periodic with respect to the time
variable $t$. From the definition of the operator $\chi$ we have
$$
|\chi (u)|(t)\leq
\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega r^2
{1\over
{1-e^{\max_{x\in [a, b]}\Bigl(-{{q'''(x)}\over {q(x)}}\Bigr)\omega}}}.
$$
Consequently the functions $\chi(u)(t)$ are uniformly bounded in the
space $\mathcal{C}(\omega)$ for every fixed $x\in [a, b]$.

Let $\epsilon>0$. Then there exists $\delta>0$ such that
$$
-{{q'(x)}\over {q(x)}}{{e^{-{{q'''(x)}\over {q(x)}}s}}\over
{1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}|u^2(t_1-s, x)-u^2(t_2-s, x)|<
{\epsilon\over\omega}
$$
for $|t_1-t_2|<\delta$ and for every $s\in [0, \omega]$, for every fixed
$x\in [a, b]$. Therefore
$$
|\chi(u)(t_1)-\chi(u)(t_2)|<\epsilon
$$
for $|t_1-t_2|<\delta$, for every fixed $x\in [a, b]$.
Then $\chi(u)$ is equicontinuous for every fixed $x\in [a, b]$.
 From the Arzela-Ascoli theorem follows that the set $\{\chi(u)(t)\}$
is compact subset in the space $\mathcal{C}(\omega)$ for every fixed
$x\in [a, b]$. From here and from uniformly bounded of the
functions $\chi(u)(t)$ follows that the operator $\chi$ is
completely continuous in the space $\mathcal{C}(\omega)$ for every fixed
$x\in [a, b]$.
\end{proof}


\begin{proposition} \label{prop2.5}
Let $v(t)$ is continuous $\omega$-periodic
function and $q(x)\in \mathcal{C}^{\infty}([a, b])$. Then for every
$\gamma>0$, $\gamma\notin \{1, 2, \dots\}$, $p>1$, $q\geq 1$ we
have $u(t, x)=v(t)q(x)\in {\dot B}^{\gamma}_{p, q}([a, b])$ for
every $t\in [0, \omega]$.
\end{proposition}

\begin{proof}
 Here we use the following definition of the
${\dot B}^{\gamma}_{p, q}([a, b])$-norm (see \cite{t1}).
$$
\|u\|_{{\dot B}^{\gamma}_{p, q}([a, b])}^q=
\int_0^1 h^{-1-(\gamma-k)q}
\big\|\Delta_h{{\partial^k}\over {\partial x^k}}u\Bigl|
\Bigl|_{L^p([a, b])}^qdh,
$$
where
$$
\Delta_h u(t, x)=u(t, x+h)-u(t, x),
$$
$k\in\{0, 1, 2, \dots\}$, $\gamma-k=\{\gamma\}$, $\{\gamma\}$ is
the fractional part of $\gamma$, $0<\{\gamma\}<1$. Then, after we
use the middle point theorem we have
\begin{align*}
\|u\|_{{\dot B}^{\gamma}_{p, q}([a, b])}^q
&= \int_0^1 h^{-1-(\gamma-k) q}
\big\|\Delta_h {{\partial^k}\over {\partial x^k}}u\big\|_{L^p([a, b])}^qdh\\
&\leq C_1\int_0^1 h^{-(\gamma-k)q+q-1}\big\|
{{\partial^{k+1}}\over {\partial x^{k+1}}}u\big\|_{L^p[a, b]}^qdh \\
&\leq C_2 \int_0^1 h^{-(\gamma-k)q+q-1}dh<\infty,
\end{align*}
because $q-(\gamma-k)q>0$. Here  $C_1$ and $C_2$ are positive constants.
\end{proof}

 The proof for existence of nontrivial solution to the Korteweg-de Vries
equation, which is positive continuous $\omega$-periodic
with respect to the variable $t$ and positive continuous
with respect to the variable $x$ is based on the theory
of completely continuous vector field presented by  Krasnosel'skii and
 Zabrejko in \cite{k1}. More precisely we will prove that the \eqref{e1}
has nontrivial solution, which is positive continuous $\omega$-periodic
with respect to the variable $t$
and positive continuous with respect to  the variable $x$ after we use the following theorem
which is extracted from \cite{k1}.

\begin{theorem}[\cite{k1}] \label{thm2.1}
Let $Y$ be a real Banach space with a cone $Q$ and
$L:Y\to Y$ be a completely continuous and positive with respect to $Q$
operator. Then the following propositions are valid.
\begin{itemize}
\item[(i)] Let $L(0)=0$. Let also for every sufficiently small $r>0$
there is no $y\in Q$, $\|y\|_Y=r$, with $y\overset{\circ}{\leq}L(y)$.
Then there exists $\mathop{\rm ind}(0,L;Q)=1$.

\item[(ii)] Let for every sufficiently large $R$ there is no $y\in Q$
with $\|y\|_Y=R$ and $L(y)\overset{\circ}{\leq}y$. Then there exists
$\mathop{\rm ind}(\infty, L; Q)=0$.

\item[(iii)] Let $L(0)=0$ and let there exist
$\mathop{\rm ind}(0, L; Q)\ne \mathop{\rm ind}(\infty, L; Q)$.
Then $L$ has nontrivial fixed point in $Q$.
\end{itemize}
\end{theorem}

Here $\mathop{\rm ind}(\cdot ,L;Q)$ denotes an index of a point
with respect to $L$ and $Q$. The symbol
 $\overset{\circ}{\leq}$
denotes the semiordering generated by $Q$.


\begin{theorem} \label{thm2.2}
Let $\gamma>0$, $\gamma\notin \{1, 2, \dots\}$,
$p>1$, $q\geq 1$. Let also $q(x)$ is a function which satisfies
the hypothesis (H1) and (H2). Then the Korteweg- de Vries \eqref{e1}
has a positive solution in the form $u(t, x)=v(t)q(x)$, which
is $\omega$-periodic  with respect to the time variable $t$ and
$u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$.
\end{theorem}

\begin{proof} First we note that $\chi(0)=0$. Also, from
Propositions \ref{prop2.3} and \ref{prop2.4}, we have that the operator $\chi$
is positive and completely continuous with respect to the cone
$\mathcal{C}_+^{\circ}(\omega)$  for every fixed $x\in [a, b]$.
Let $x\in [a, b]$ is fixed.

\noindent(1) Let $r>0$ satisfy the inequality
\begin{equation}
r<{{D_q^{-}}\over {{D_q^{+}}^2\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)\omega}}\Bigl(
1-e^{{ \max_{x\in [a, b]}}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega}
\Bigr).
\label{e6}
\end{equation}
We suppose that there exists $u(t, x)\in \mathcal{C}_+^{\circ}(\omega)$
for which
$$
\max_t u(t, x)=r,\quad u\leq \chi(u),\quad t\in [0, \omega],
$$
for every fixed $x\in [a, b]$.
Then
\begin{equation}
u(t, x)\leq D_q^{+}\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)\omega}}}\int_0^{\omega}u^2(t-s, x)ds.
\label{e7}
\end{equation}
 From the definition of the cone $\mathcal{C}_+^{\circ}(\omega)$ we have for
every fixed $x\in [a, b]$,
$$
u(t, x)\leq\max_t u(t, x)\leq {{D_q^+}\over {D_q^-}}\min_t u(t, x)\leq
{{D_q^+}\over {D_q^-}}\max_t u(t, x)=
r{{D_q^+}\over {D_q^-}}.
$$
 From this and   \eqref{e7}, we have
$$
u(t, x)\leq r{{{D_q^{+}}^2}\over {D_q^-}}\max_{x\in [a, b]}
\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)}}}\int_0^{\omega}u(t-s, x)ds.
$$
Now we integrate the last inequality from 0 to $\omega$ with respect to
the time variable $t$ and we get
$$
\int_0^{\omega}u(s, x)ds\leq \omega r{{{D_q^{+}}^2}\over {D_q^-}}\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)}}}\int_0^{\omega}u(s, x)ds.
$$
 From the last inequality we have
$$
1\leq \omega r{{{D_q^{+}}^2}\over {D_q^-}}\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)}}}
$$
or
$$
r\geq {{D_q^{-}}\over {{D_q^{+}}^2\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)\omega}}\Bigl(
1-e^{ \max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)\omega}\Bigr)
$$
which is a contradiction with \eqref{e6}. Consequently for every enough small
$r>0$ there is no $u(t, x)\in \mathcal{C}_+^{\circ}(\omega)$ such that
$\max_t u(t, x)=r$ for every fixed $x\in [a, b]$, $u(t, x)\leq \chi(u)$
for every fixed $x\in [a, b]$ and $t\in [0, \omega]$. From here and from
Theorem \ref{thm2.1}(i) we get that there exists
$\mathop{\rm ind}(0, \chi;\mathcal{C}_+^{\circ}(\omega))=1$.

\noindent(2) Let $R>0$ be large enough so that
\begin{equation}
R>{{D_q^{+}}\over {{D_q^{-}}^2\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)\omega}}\Bigl(
1-e^{ \min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega}\Bigr).
\label{e8}
\end{equation}
We suppose that there exists $u(t, x)\in \mathcal{C}_+^{\circ}(\omega)$
for which
$$
\max_t u(t, x)=R,\quad u\geq \chi(u)
$$
for every fixed $x\in [a, b]$ and for every $t\in [0, \omega]$.
Then
\begin{equation}
u(t, x)\geq D_q^{-}\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)}}}\int_0^{\omega}u^2(t-s, x)ds.
\label{e9}
\end{equation}
 From the definition of the cone $\mathcal{C}_+^{\circ}(\omega)$
we have for every fixed $x\in [a, b]$
$$
u(t, x)\geq \min_t u(t, x)\geq {{D_q^-}\over {D_q^+}}\max_t u(t, x)=
R{{D_q^-}\over {D_q^+}}.
$$
Therefore, from  \eqref{e9}, we have
$$
u(t, x)\geq R{{{D_q^{-}}^2}\over {D_q^+}}\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)}}}\int_0^{\omega}u(t-s, x)ds.
$$
Now we integrate the above inequality from 0 to $\omega$ with respect
to  $t$ and obtain
$$
\int_0^{\omega}u(s, x)ds\geq \omega R{{{D_q^{-}}^2}\over {D_q^+}}\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr){1\over {1-e^{\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}
\Bigr)}}}\int_0^{\omega}u(s, x)ds.
$$
 From the above inequality we have
$$
1\geq \omega R{{{D_q^{-}}^2}\over {D_q^+}}\min_{x\in [a, b]}
\Bigl(-{{q'(x)}\over {q(x)}}\Bigr){1\over {1-e^{\min_{x\in [a, b]}
\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)}}}
$$
or
$$
R\leq {{D_q^{+}}\over {{D_q^{-}}^2\min_{x\in [a, b]}
\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega}}
\Bigl(1-e^{ \min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega}\Bigr)
$$
which is a contradiction with \eqref{e8}. Consequently for every enough
large $R>0$ there is no $u(t, x)\in \mathcal{C}_+^{\circ}(\omega)$ such that
$\max_t u(t, x)=R$ for every fixed $x\in [a, b]$, $u(t, x)\geq \chi(u)$
for every fixed $x\in [a, b]$ and $t\in [0, \omega]$.
 From here and from Theorem \ref{thm2.1}(ii) we get that there exists
$\mathop{\rm ind}(\infty, \chi;\mathcal{C}_+^{\circ}(\omega))=0$.

 From (1) and (2) follows that there exist
$$
\mathop{\rm ind}(\infty, \chi;\mathcal{C}_+^{\circ}(\omega))\ne
\mathop{\rm ind}(0, \chi;\mathcal{C}_+^{\circ}(\omega)).
$$
Consequently, from Theorem \ref{thm2.1} (iii), we conclude
that the operator $\chi$ has a nontrivial fixed point in the cone
$\mathcal{C}_+^{\circ}(\omega)$ for every fixed $x\in [a, b]$. Therefore the
Korteweg - de Vries equation \eqref{e1} has positive solution
$u(t, x)=v(t)q(x)$, which is continuous $\omega$-periodic
with respect to the time variable $t$ and from Proposition \ref{prop2.5}
 we have $u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$ for every 
$x\in [a, b]$.
\end{proof}

\section{Uniqueness of the positive periodic solutions}

Here we use the following theorem.

\begin{theorem}[\cite{k1}] \label{thm3.1}
 Let $Q$ is a physical cone in the Banach space $Y$ and the operator
$A: Y\to Q$ is monotonous $u_0$-convex operator ($u_0\in Q$). Let
also for every two solutions $x_1$ and $x_2$ to the equation $x=Ax$ one of the
differences $x_1-x_2$, $x_2-x_1$ is equal to zero or
is inside element for the cone $Q$. Then the equation $x=Ax$ has in the cone $Q$
no more than one nontrivial solution.
\end{theorem}

We say that the operator $A:Y\to Y$, where $Y$ is a Banach space
with a cone $Q$, is {\it monotonous} if:
$y_1\in Y$, $y_2\in Y$, with $y_1\overset{\circ}{\leq} y_2$ then
$Ay_1\overset{\circ}{\leq} Ay_2$.
Here $\overset{\circ}{\leq}$ denotes the semiordering generating by $Q$.

We say that the operator $A:Y\to Y$, $Y$ is a Banach space
with a cone $Q$, $A:Q\to Q$,
is a {\it $u_0$-convex} operator ($u_0\in Q$) if
 for every $x\in Q$, $x\ne 0$, then
$$
\alpha(x)u_0\leq Ax\leq \beta(x)u_0,
$$
where $\alpha(x)>0$, $\beta(x)>0$; and for every $x\in Q$ for which
$$
\alpha_1(x)u_0\leq Ax\leq \beta_1(x)u_0
$$
($\alpha_1(x)>0$, $\beta_1(x)>0$) we have
$$
A(\lambda x)\leq [1-\eta(x, \lambda)]\lambda Ax, \quad 0<\lambda<1,
$$
where $\eta(x, \lambda)>0$.

Here and bellow we suppose that $q(x)$ is the function
satisfying the conditions in Theorem \ref{thm2.2}.
Let
$$
K(x, s)=-{{q'(x)}\over {q(x)}}{{e^{-{{q'''(x)}\over {q(x)}}s}}\over
{1-e^{-{{q'''(x)}\over {q(x)}}\omega}}},\quad x\in [a, b], s\in [0, \omega].
$$
 From the above assumptions follows that there exist constants $m>0$, $M>0$
such that
$$
m\leq K(x, s)\leq M, \quad \forall x\in [a, b],\quad \forall s\in [0, \omega].
$$
For instance
\begin{gather*}
m=\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)
{{e^{-\max_{x\in [a, b]}{{q'''(x)}\over {q(x)}}\omega}}\over
{1-e^{-\max_{x\in [a, b]}{{q'''(x)}\over {q(x)}}\omega}}},\\
M=\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)
{1\over
{1-e^{-\min_{x\in [a, b]}{{q'''(x)}\over {q(x)}}\omega}}}.
\end{gather*}
Now we consider the integral equation (for a fixed $x\in [a, b]$)
\begin{equation}
u(t, x)=\int_0^{\omega}K(x, s)u^2(t-s, x)ds,\quad t\in [0, \omega].
\label{e10}
\end{equation}
The operator $\chi$ (see section 2) we may rewriten in the form
\begin{equation}
\chi(u)=\int_0^{\omega}K(x, s)u^2(t-s, x)ds.
\label{e11}
\end{equation}

\begin{theorem} \label{thm3.2}
Let $\gamma>0$, $\gamma\notin \{1, 2, \dots\}$,
$p>1$, $q\geq 1$. Let also
$$
{{M^2}\over {m^2}}-{{m^2}\over {M^2}}<{1\over 2}.
$$
Then  \eqref{e1} has a unique positive solution $u(t, x)=v(t)q(x)$
which is continuous $\omega$-periodic with respect to the
time variable $t$ and $u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$.
\end{theorem}

\begin{proof} From Theorem \ref{thm2.2} follows that the problem
\eqref{e1}--\eqref{e1''} has positive solution $u(t, x)=v(t)q(x)$. Let
$x\in [a, b]$ is fixed. Let also $T\subset \mathcal{C}_+^{\circ}(\omega)$
is the set
$$
T=\big\{u(t, x)\in \mathcal{C}_+^{\circ}(\omega),\quad {m\over {M^2\omega}}\leq u(t, x)\leq {M\over {m^2\omega}},
\forall t\in [0, \omega]\big\}.
$$
If $u(t, x)$ is positive solution to  \eqref{e1}, which is $\omega$-periodic
with respect to the time variable $t$ then $u(t, x)\in T$. Indeed,
for every fixed $x\in [a, b]$ we have
$$
u(t, x)= \chi(u)\leq \Bigl(\max_{t\in [0, \omega]}u(t, x)\Bigr)^2 M\omega
$$
for every $t\in [0, \omega]$. From where,
$$
\max_{t\in [0, \omega]}u(t, x)\leq \Bigl(\max_{t\in [0, \omega]}u(t, x)\Bigr)^2 M\omega
$$
or
$\max_{t\in [0, \omega]}u(t, x)\geq 1/M\omega$
for every fixed $x\in [a, b]$. On the other hand from  proposition \ref{prop2.3},
we have
$$
u(t, x)\geq {m\over M}\max_{t\in [0, \omega]}u(t, x)\geq {m\over {M^2\omega}}
\quad \forall t\in [0, \omega],
\label{e12}
$$
for every fixed $x\in [a, b]$.
Also, for every fixed $x\in [a, b]$
$$
u(t, x)=\chi(u)\geq m\omega\Bigl(\min_{t\in [0, \omega]}u(t, x)\Bigr)^2,\quad
\forall t\in [0, \omega].
$$
 From the above inequality,
\begin{equation}
\min_{t\in [0, \omega]}u(t, x)\leq {1\over {m\omega}}.
\label{e13}
\end{equation}
Since $u(t, x)\in \mathcal{C}_+^{\circ}(\omega)$, we have
$$
\min_{t\in [0, \omega]}u(t, x)\geq {m\over M}\max_{t\in [0, \omega]}u(t, x)
$$
for every fixed $x\in [a, b]$. From the above inequality and \eqref{e13},
\begin{equation}
\max_{t\in [0, \omega]}u(t, x)\leq {M\over {m^2\omega}}
\label{e14}
\end{equation}
for every fixed $x\in [a, b]$. From \eqref{e12} and \eqref{e14}
it  follows that $u(t, x)\in T$
for every $t\in [0, \omega]$ and for every fixed $x\in [a, b]$.

Let $u_1$ and $u_2$ be two solutions to the integral equation \eqref{e10}.
Let $y=u_1-u_2$. We suppose that $y$ changes its sign. Then for every
positive constants $c$ we have
$$
\|y-c\|\geq {1\over 2}\|y\|.
$$
(because $y$ changes your sign)
We note that in our case $\|y\|=\max_{t\in [0, \omega]}|y|$ for every
fixed $x\in [a, b]$, $y\in \mathcal{C}(\omega)$.
Let
$$
b_1=2{{m^2}\over {M^2\omega}},\quad b_2=2{{M^2}\over {m^2\omega}}.
$$
In particular we have
$$
\big\|y-{{b_1+b_2}\over 2}\int_0^{\omega}y(s)ds\big\|
\geq {1\over 2}\|y\|
\label{e15}
$$
for every fixed $x\in [a, b]$. Also, we have
$$
y(t, x)=\int_0^{\omega}K(x, s)(u_1^2(t-s, x)-u_2^2(t-s, x))ds
=2\int_0^{\omega}K(x, s)z(s)y(s)ds
$$
for every fixed $x\in [a, b]$. In the last equality we use the middle
point theorem.
Here
$$
\min\{u_1, u_2\}\leq z\leq \max\{u_1, u_2\}.
$$
 From where it follows that $z\in T$ for every fixed $x\in [a, b]$. Then
\begin{gather*}
2K(x, s)z(s)\geq 2m{m\over {M^2\omega}}=b_1,\\
2K(x, s)z(s)\leq 2M{M\over {m^2\omega}}=b_2.
\end{gather*}
Consequently
$$
\Bigl| 2K(x, s)z(s)-{{b_1+b_2}\over 2}\Bigr|\leq {{b_2-b_1}\over 2}
$$
for every fixed $x\in [a, b]$. On the other hand
\begin{align*}
\Bigl| y(t)-{{b_1+b_2}\over 2}\int_0^{\omega}y(s)ds\Bigr|
&=\Bigl| 2\int_0^{\omega} K(x, s)z(s)y(s)ds-{{b_1+b_2}\over 2}
\int_0^{\omega}y(s)ds\Bigr|\\
&=\Bigl|\int_0^{\omega}\Bigl(2K(x, s)z(s)-{{b_1+b_2}\over 2}\Bigr)y(s)ds
\Bigr|\\
&\leq \int_0^{\omega}\Bigl|2K(x, s)z(s)-{{b_1+b_2}\over 2}\Bigr| |y(s)|ds \\
&\leq {{b_2-b_1}\over 2}\int_0^{\omega}|y(s)|ds\leq {{b_2-b_1}\over 2}\|y\|
\omega
\end{align*}
for every fixed $x\in [a, b]$. From where,
$$
\big\|y-{{b_1+b_2}\over 2}\int_0^{\omega}y(s)ds\big\|\leq
{{b_2-b_1}\over 2}\|y\|\omega
$$
for every fixed $x\in [a, b]$. Now we use the inequality \eqref{e15}
 and we get
$$
{1\over 2}\|y\|\leq {{b_2-b_1}\over 2}\|y\|\omega
$$
or
$$
1\leq (b_2-b_1)\omega=2\Bigl({{M^2}\over {m^2\omega}}-{{m^2}\over {M^2\omega}}
\Bigr)\omega,
$$
from where,
$$
{1\over 2}\leq {{M^2}\over {m^2}}-{{m^2}\over {M^2}},
$$
which is a contradiction with the conditions of the theorem \ref{thm3.2}.
Consequently, if $u_1$ and $u_2$ are two solutions to the integral equation
$u=\chi(u)$ we have $u_1\equiv u_2$ or $u_1-u_2$ or $u_2-u_1$ is inside element
for the cone $\mathcal{C}_+^{\circ}(\omega)$. Now we will show that the
operator $\chi$ is 1-convex operator with respect to the cone
$\mathcal{C}_+^{\circ}(\omega)$. First we note that
$1\in \mathcal{C}_+^{\circ}(\omega)$.
Let $\eta(x, \lambda)=1-\lambda$, $\lambda\in (0, 1)$. Then we have
$$
\chi(\lambda u)=\lambda^2\chi(u)=(1-\eta(x, \lambda))\lambda\chi(u).
$$
Consequently the operator $\chi$ is 1-convex operator with respect to the cone
$\mathcal{C}_+^{\circ}(\omega)$.

 From here and from Theorems \ref{thm2.2}, \ref{thm3.1}, it follows that the
Korteveg-de Vries \eqref{e1} has unique positive solution
$u(t, x)=v(t)q(x)$, which is $\omega$-periodic with respect to
the time variable $t$ and $u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$.
\end{proof}

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\bibitem{k1}  M. A. Krasnosel'skii and P. P. Zabrejko;
\emph{Geometrical Methods of Nonlinear Analysis} (in Russian),
Nauka, Moscow, 1975.

\bibitem{t1} H. Triebel,
\emph{Interpolation theory, function spaces, differential operators},
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\end{thebibliography}
\end{document}