\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 $u(0, x)\in {\dot B}^{\gamma}_{p, q}([a, b])$, $\gamma>0$, $\gamma\notin \{1, 2, \dots\}$, $p>1$, $q\geq 1$, $a0$ 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 $q\geq 1$, $10$, $\gamma\notin \{1, 2, \dots\}$. 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 $q\geq 1$, $10$, $\gamma\notin \{1, 2, \dots\}$ 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 $$u(t, x)=-\int_0^{\omega} {{e^{-{{q'''(x)}\over {q(x)}}s}}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}} u^2(t-s, x){{q'(x)}\over {q(x)}}ds, \label{e2}$$ 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 $$v(t)=-{{q'(x)}\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}}\int_0^{\omega} e^{-{{q'''(x)}\over {q(x)}}s}v^2(t-s)ds. \label{star}$$ 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 $$q(x)v'(t)=-q'''(x)v(t)-q'(x)q(x)v^2(t) \label{starstar}$$ 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 $$v(t)=-q'(x)e^{-{{q'''(x)}\over {q(x)}}t}\int_{-\infty}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds. \label{3star}$$ Now we consider the integral $$\int_{-\infty}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds.$$ We have $$\int_{-\infty}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds= \int_{t-\omega}^t v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds+ \int_{t-2\omega}^{t-\omega} v^2(s)e^{{{q'''(x)}\over {q(x)}}s}ds+ \dots. \label{e3}$$ 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 $$\min_t\chi(u)\geq D_q^{-}{1\over {1-e^{-{{q'''(x)}\over {q(x)}}\omega}}} \Bigl(-{{q'(x)}\over {q(x)}}\int_0^{\omega}u^2(s, x)ds\Bigr). \label{e4}$$ 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 $$r<{{D_q^{-}}\over {{D_q^{+}}^2\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}} \Bigr)\omega}}\Bigl( 1-e^{{ \max_{x\in [a, b]}}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega} \Bigr). \label{e6}$$ 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 $$u(t, x)\leq D_q^{+}\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}} \Bigr){1\over {1-e^{\max_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}} \Bigr)\omega}}}\int_0^{\omega}u^2(t-s, x)ds. \label{e7}$$ 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 $$R>{{D_q^{+}}\over {{D_q^{-}}^2\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}} \Bigr)\omega}}\Bigl( 1-e^{ \min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}}\Bigr)\omega}\Bigr). \label{e8}$$ 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 $$u(t, x)\geq D_q^{-}\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}} \Bigr){1\over {1-e^{\min_{x\in [a, b]}\Bigl(-{{q'(x)}\over {q(x)}} \Bigr)}}}\int_0^{\omega}u^2(t-s, x)ds. \label{e9}$$ 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 $x\in [a, b]$) $$u(t, x)=\int_0^{\omega}K(x, s)u^2(t-s, x)ds,\quad t\in [0, \omega]. \label{e10}$$ The operator $\chi$ (see section 2) we may rewriten in the form $$\chi(u)=\int_0^{\omega}K(x, s)u^2(t-s, x)ds. \label{e11}$$ \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, $$\min_{t\in [0, \omega]}u(t, x)\leq {1\over {m\omega}}. \label{e13}$$ 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}, $$\max_{t\in [0, \omega]}u(t, x)\leq {M\over {m^2\omega}} \label{e14}$$ 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} \begin{thebibliography}{0} \bibitem{b1}J . Bourgain, \emph{Fourier transform restriction phenomena for certain lattice subsets and applications to nonlinear evolution equations, Part II: The KDV-Equation}, Geometric and Functional Analysis, Vol. 3, No. 3 (1993). \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}, Berlin, 1978. \end{thebibliography} \end{document}