\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small 2003 Colloquium on Differential Equations and Applications, Maracaibo, Venezuela.\newline {\em Electronic Journal of Differential Equations}, Conference 13, 2005, pp. 29-34.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \setcounter{page}{29} \begin{document} \title[\hfilneg EJDE/Conf/13 \hfil Critical points of the steady state] {Critical points of the steady state of a Fokker-Planck equation} \author[J. Gu\'{\i}\~{n}ez, R. Quintero, A. D. Rueda \hfil EJDE/Conf/13 \hfilneg] {Jorge Gu\'{\i}\~{n}ez, Robert Quintero, Angel D. Rueda} % in alphabetical order \address{Centro de Investigaci\'{o}n de Matem\'{a}tica Aplicada (C.I.M.A.)\\ Facultad de Ingenier\'{\i}a\\ Universidad del Zulia\\ Apartado 10482, Maracaibo, Venezuela} \email[J. Gu\'{\i}\~{n}ez]{jguinez@luz.edu.ve} \email[R. Quintero]{rquintero@luz.edu.ve} \email[A. D. Rueda]{ad-rueda@cantv.net} \date{} \thanks{Supported by CONDES University of Zulia} \subjclass[2000]{58J60, 37C20} \keywords{Almost gradient vector fields} \begin{abstract} In this paper we consider a set of vector fields over the torus for which we can associate a positive function $v_{\epsilon }$ which define for some of them in a solution of the Fokker-Planck equation with $\epsilon $ diffusion: $$ \epsilon \Delta v_{\epsilon }-\mathop{\rm div}(v_{\epsilon }X)=0\,. $$ Within this class of vector fields we prove that $X$ is a gradient vector field if and only if at least one of the critical points of $v_{\epsilon }$ is a stationary point of $X$, for an $\epsilon >0$. In particular we show a vector field which is stable in the sense of Zeeman but structurally unstable in the Andronov-Pontriaguin sense. A generalization of some results to other kind of compact manifolds is made. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}[theorem]{Definition} \newtheorem{lemma}[theorem]{Lemma} \section{Vector fields in covering spaces} Let $\pi :\widetilde{M}\to M$ be a covering space of a Riemannian and oriented manifold M. In $\widetilde{M}$ there exists one and only one Riemannian structure such that \begin{equation*} d\pi _{y}:T_{y}(\widetilde{M})\to T_{\pi (y)}(M) \end{equation*} is an isometry for all $y\in \widetilde{M}$. Then it is able to associate to every $C^{r}$ vector field $X$ in $M$, another $C^{r}$ vector field $\widetilde{X}$ in $\widetilde{M}$ in the following way: \begin{equation*} \widetilde{X}(y)=(d(\pi )(y))^{-1}(X(\pi (y))). \end{equation*} It is easy to verify the following theorem: \begin{theorem} \label{thm1} \begin{gather} ( \widetilde{X+Y}) =\widetilde{X}+\widetilde{Y}m \\ % 1 \widetilde{\nabla f}=\nabla (\pi \circ f) \\ %2 \mathop{\rm div}(\widetilde{X}(y)=\mathop{\rm div}(X)(\pi (y)) %3 \end{gather} \end{theorem} \noindent\textbf{Definition} %1 A vector field $X$ is called almost gradient respect to the projection $\pi $, if and only if $\widetilde{X}$ is a gradient in $M$. This set will be denoted by $V_{ag}(\pi )$. Particular we can write \begin{equation*} \mathop{\rm grad}(M)=V_{ag}(1_{M}) \end{equation*} There are non trivial projections for which is true the preceding statement, so we have the following theorem. \begin{theorem} \label{thm2} If $\pi :\widetilde{M}\to M$ is a finite covering and $M$ is compact, then $V_{ag}$ is the set of gradient vector fields in $M$. \end{theorem} \begin{proof} Let $X$ be a vector field in $M$ and let \begin{equation*} X=\nabla f+W \end{equation*} be its Hodge's decomposition. If $\widetilde{X}$ is gradient, then there exists a $C^{\infty }$ function $g$ such that: \begin{equation*} \nabla g=\widetilde{X}=\nabla (f\circ \pi )+\widetilde{W}. \end{equation*} Then by Theorem \ref{thm1} it follows that $\widetilde{W}$ is a gradient vector field in $\widetilde{M}$ and we can write $\widetilde{W}=\nabla h$. Finally from Theorem \ref{thm1} we get, $\mathop{\rm div}(\widetilde{W})=\mathop{\rm div}(W)\circ \pi =0$. Thus $\nabla h=0$ and by compactness of $\widetilde{M}$ it follows $h$ to be a constant. So $W=0$ and $W=0$. \end{proof} \section{Vector Fields in $T_{n}$} Let $\pi :\mathbb{R}^{n}\to T_{n}=\mathbb{R}^{n}/\mathbb{Z}^{n}$ be the universal covering space of the torus $T_{n}$. So there exists a Riemannian structure in $T_{n}$ such that $\widetilde{T}_{n}=\mathbb{R}^{n}$, where $\mathbb{R}^{n}$ is considered with the usual Riemannian structure. It is easy to realize that $V_{ag}(\pi )$ is different from $\mathop{\rm grad}(M)$. More precisely we have the following statement. \begin{theorem} \label{thm3} $X\in V_{ag}(\pi )$ if and only if $X$ is in the form $X=\nabla f+\lambda $, where $\lambda \in \mathbb{R}^{n}$. \end{theorem} \begin{proof} Let $X=\nabla f+W$ be the Hodge's decomposition of $X$. Then $\widetilde{X}\in V_{ag}(\pi )$ implies \begin{equation*} \widetilde{X}=\nabla g, \end{equation*} but \begin{equation*} \widetilde{X}=\widetilde{\nabla f}+\widetilde{W}=\nabla (f\circ \pi )+% \widetilde{W} \end{equation*} and $\widetilde{W}=\nabla h$ with $h=g-f\circ \pi $. By the periodicity of $\widetilde{W}$ it follows that $\| \widetilde{W}\| \leq K$. Then \begin{equation} %4 | h(x)| \leq K\| x\| \quad \forall x\in \mathbb{R}^{n}. \end{equation} Because $W$ has divergence zero so it does $\widetilde{W}$, and $h$ is an harmonic function in whole $\mathbb{R}^{n}$. From the estimate (4) $h$ is a linear function; i.e., \begin{equation} %5 h(x)=a+\lambda \cdot x \end{equation} with $\lambda \in \mathbb{R}^{n}$ so $\nabla h=\widetilde{W}=\lambda $, consequently $W=\lambda $. \end{proof} \section{The function $v_{\epsilon}$} For the rest of this article, we consider $X$ in $T_{n}$ to be of the form $X=\nabla f+\lambda $. Let us consider \begin{equation} %6 v_{\epsilon }(x)=\int_{T_{n}}\exp \big( \frac{h(x,z)}{\epsilon }\big) dz \end{equation} where \begin{gather} g(x)=f(x)+\lambda \cdot x \,,\\ %7 h(x,z)=g(x)-g(x+z) %8 \end{gather} \begin{lemma} \label{lem1} $X$ is a gradient if $\lambda =0$ and we have \begin{equation*} v_{\epsilon }=L(\epsilon )\exp ( f/\epsilon ) \end{equation*} \end{lemma} \begin{proof} If $\lambda =0$, we have $$ v_{\epsilon }=\exp (\frac{f}{\epsilon }) \int_{T_{n}} \exp ( \frac{-f(x+z)}{\epsilon }) dz =\exp (\frac{f}{\epsilon }) \int_{T_{n}}\exp (\frac{-f(z)}{\epsilon }) dz\,. $$ \end{proof} \noindent\textbf{Definition.}% 2. $X$ will be called without coupling, if \begin{equation*} \big( (i\neq j)\text{ and }( \frac{\partial X_{i}}{\partial x_{j}}\neq 0) \big) \Rightarrow \lambda _{i}=0. \end{equation*} \begin{theorem}\label{thm4} Let $X$ be a vector field without coupling. Then $v_{\epsilon }$ is a solution of the Fokker-Planck equation \begin{equation} \label{e9} \epsilon \Delta v-\mathop{\rm div}(vX)=0. \end{equation} \end{theorem} \begin{proof} Let $I$ be the set of indices for which $\lambda _{i}\neq 0$. Then the $i$-component of the vector field $X$ is \begin{equation*} X_{i}=f_{i}'(x_{i})+\lambda _{i} \end{equation*} with $f_{i}$ a function in the variable $x_{i}$. Therefore, $X=\nabla f+\lambda$ with \begin{equation*} f=\sum (f_{i}(x_{i}))+p(x) \end{equation*} where $p(x)$ is a periodic function and \begin{equation*} \frac{\partial }{\partial x_{i}}(p(x))=0,\quad i\in I. \end{equation*} Then \begin{equation*} h(x,z)=\sum_{i\epsilon I}h_{i}(x_{i},z_{i})+p(x)-p(x+z). \end{equation*} Because $X$ is without coupling, applying Lemma \ref{lem1} to $\nabla p$, \begin{equation} % 10 v_{\epsilon } =\int_{T_{n}}\exp \big( \frac{h(x,z)}{\epsilon } \big) dz=K\big( \prod_{i\epsilon I}v_{\epsilon }^{i}(x_{i})\big) \exp \big( \frac{p(x)}{\epsilon }\big)\,, \end{equation} where \begin{equation*} v_{\epsilon }^{i}(x_{i})=\int_{0}^{1}\exp \big( \frac{h_{i}(x_{i},z_{i})}{\epsilon }\big) dz_{i} \end{equation*} is associated with the vector field $\nabla f_{i}+\lambda _{i}$. If $i\in I$ it follows that \begin{equation*} \epsilon (\nabla v_{\epsilon })_{i}=( X_{i}(x_{i})v_{\epsilon }^{i}-\epsilon R_{i}) \prod_{k\epsilon (I-\{i\})}v_{\epsilon }^{k}\exp \big( \frac{p(x)}{\epsilon }\big) \end{equation*} where \begin{equation*} R_{i}=\int_{0}^{1}\frac{X_{i}(x_{i}+z_{i})}{\epsilon }\exp \big( \frac{f_{i}(x_{i})-f(x_{i}+z_{i})-\lambda _{i}z_{i}}{\epsilon }\big) dz_{i} -\exp \big( -\frac{-\lambda _{i}}{\epsilon }\big) +1. \end{equation*} For $i\notin I$, \begin{equation*} (\epsilon \nabla v_{\epsilon })_{i}=(\nabla p(x))_{i}v_{\epsilon }=X_{i}v_{\epsilon } \end{equation*}% thus $v_{\epsilon }$ is solution of \eqref{e9}. \end{proof} \section{Dynamics and Steady State} We begin this section with some definitions: \noindent\textbf{Definition} Let $X$ be a vector field in $T_{n}$ and let $% u_{\epsilon }=\sum_{0}^{\infty }\frac{F_{i}}{\epsilon ^{i}}$ be a series with a positive ratio of convergence. Suppose that $u$, is a solution of \eqref{e9}. We will denote: \begin{gather} C_{\epsilon }=\{x\in M:\nabla u_{\epsilon }=0\} \\ %11 E(X)=\{x\in M:X(x)=0\} \\ %12 D(X)=\{ x\in M:\det \big( \frac{\partial X_{i}}{\partial x_{j}}\big) =0\} %13 \end{gather} In \cite{g1} and \cite{g2}, we have such series on $T_{n}$ and $S_{n}$. \begin{theorem} \label{thm5} Consider $X\in V_{ag}(T_{n})$ such that \begin{itemize} \item[(i)] There exists a convergent series $u_{\epsilon}=\sum_{i=0}^{\infty }\frac{F_{i}}{\epsilon ^{i}}$ solving \eqref{e9} for $\frac{1}{\epsilon }\leq r,r>0$ \item[(ii)] There exists an infinity set $S\subset \lbrack r_{1},r]$, $r1>0$ and a point $x$ in $T_{n}$ such that \begin{equation} %14 x\in C_{\epsilon }\cap E(X)\quad \forall \frac{1}{\epsilon }\in S \end{equation} \end{itemize} Then $X$ is a gradient vector field. \end{theorem} \begin{proof} Because $X\in V_{ag}(T_{n})$, by Theorem \ref{thm3} we can write \begin{gather} X=\nabla f+\lambda\,,\\ %15 \nabla u_{\epsilon }=0=\sum_{i=0}^{\infty }\nabla F_{i}(x) \big( \frac{1}{\epsilon }\big) ^{i}\quad \forall \epsilon \in S. \end{gather} Then $\nabla F_{i}(x)=0$, for every $i$. In particular $\nabla F_{1}(x)=\nabla f(x)=0$ and by (15) $\lambda =0$. \end{proof} \begin{theorem} \label{thm6} Let $X$ be a vector field without coupling. Then the following statements are equivalent. \begin{itemize} \item[(i)] There exists $\epsilon$ such that $C_{\epsilon }\cap E(X)\neq \emptyset$ \item[(ii)] For all $\epsilon$, $ C_{\epsilon }\cap E(X)\neq \emptyset$ \item[(iii)] $X$ is gradient. \end{itemize} \end{theorem} \begin{proof} For $x\in C_{\epsilon }\cap E(X)$ and $i\in I$ we have a contradiction: $$ 0=(v_{\epsilon }^{i})'(x)=e^{\frac{\lambda _{i}}{\epsilon }}-1 +\frac{\lambda _{i}}{\epsilon }v_{\epsilon }(x)\,. $$ wich completes the proof.\end{proof} \noindent\textbf{Remark.} The main idea here is that for non-gradient cases critical points of a steady state are different from stationary points of the vector field. This fact enable us to find a vector field $X$ with an associated $u_{\epsilon }$ which has not generated critical points, even when $X$ has degenerated stationary points. \begin{lemma} \label{lem2} Let's suppose that $X=\nabla f+\lambda $ is without coupling and let $I_{+}$ be the set of index such that $\lambda _{i}>0$ and let $I_{-}$ be the set of index such that $\lambda _{i}<0$. Then \begin{equation*} C_{\epsilon }\subset \cap _{i\epsilon I_{+}}X_{i}^{-1}((0,+\infty ))\cap _{i\epsilon I_{-}}X_{i}^{-1}((-\infty ,0)) \end{equation*} \end{lemma} \begin{proof} For a such $f$ we can write \begin{equation*} f=\sum (f_{i}(x_{i}))+p(x) \end{equation*}% where $p(x)$ is not depending of $x_{i}$ for $i\in I=I_{-}\cup I_{+}$ and \begin{equation} %17 v_{\epsilon }=K\big( \prod_{i\epsilon I}v_{\epsilon }^{i}(x_{i})\big) \exp \big( \frac{p(x)}{\epsilon }\big) \end{equation} where \begin{equation} %18 v_{\epsilon }^{i}=\int_{0}^{1}\exp \big( \frac{f_{i}(x_{i}) -f(x_{i}+z_{i})-\lambda _{i}z_{i}}{\epsilon }\big) dz_{i} \end{equation} So for every $i\in I$, we have \begin{equation} \label{e19} \frac{\partial v_{\epsilon }}{\partial x_{i}} =\big( \frac{X_{i}(x_{i})v_{\epsilon }^{i}}{\epsilon }-R_{i}\big) \prod_{k\epsilon (I-\{i\})}v_{\epsilon }^{k} \exp \big( \frac{p(x)}{\epsilon }\big)\,, \end{equation} where $R_{i}=-\exp (-\lambda _{i}/\epsilon) +1$. So if $x\in C_{\epsilon }$, \begin{equation} \label{e20} \frac{X_{i}(x_{i})}{\epsilon }v_{\epsilon }^{i} =-\exp \big( -\frac{\lambda_{i}}{\epsilon }\big) +1,\quad i\in I \end{equation} Then for $i\in I_{+}$ \ we have $X_{i}(x_{i})>0$ and for $i\in I_{-}$ we have $X_{i}(x_{i})<0$. \end{proof} \begin{lemma} \label{lem3} Under the hypothesis of Lemma \ref{lem2}, the set of degenerated critical points of $u_{\epsilon }$ is a subset of \[ % 21 D_{1}(X) =\cup _{i\epsilon I_{+}}[D(X_{i})\cap (X_{i}^{-1}(0,+\infty ))] \cup _{i\epsilon I_{-}}[D(X_{i})\cap (X_{i}^{-1}(-\infty ,0))]\cup D(\nabla _{p}) \] Here $x\in D(X_{i})$, means $X'(x_{i})=0$ and $x\in D(\nabla _{p})$ means $\det( \frac{\partial ^{2}p}{\partial x_{i}x_{j}}) =0$. \end{lemma} \begin{proof} With the notation of Lemma \ref{lem2} and by \eqref{e19} and \eqref{e20}, for every $x\in C_{\epsilon }$, \begin{gather} \frac{\partial ^{2}u_{\epsilon }}{\partial x_{i}^{2}}(x)=X'(x_{i})u_{\epsilon }, \quad i\in I\\ \frac{\partial ^{2}u_{\epsilon }}{\partial x_{i}\partial x_{j}}(x)=0,\quad i,j\in I,i\neq j \\ % 22 \frac{\partial ^{2}u_{\epsilon }}{\partial x_{i}\partial x_{j}}(x) =\frac{\partial ^{2}p}{\partial x_{i}\partial x_{j}}(x)u_{\epsilon },\quad i,j\in I' %23 \end{gather} where $I'=\{1,2,\dots,n\}-I$. So \begin{equation} \det \big( \frac{\partial ^{2}u_{\epsilon }}{\partial x_{i}\partial x_{j}} (x)\big) =\big( \prod_{i\epsilon I}X_{i}'(x_{i})\big) \big( \det \big( \frac{\partial ^{2}p}{\partial x_{i}\partial x_{j}} (x)\big) _{j,i\epsilon I'}\big) (u_{\epsilon }(x))^{n} \end{equation} If $x$ is a degenerated critical point of $u_{\epsilon }$, by Lemma \ref{lem2}, we get $x\in D_{1}(X)$. \end{proof} \begin{theorem} \label{thm7} Let $X=\nabla f+\lambda $ be a vector field without coupling and suppose \begin{equation*} I_{+}=\{i:\lambda _{i}>0\},\quad I_{-}=\{i:\lambda _{i}<0\},\quad I=I_{+}\cup I_{-},\quad k=\mathop{\rm card}(I) \end{equation*} Let also suppose: \begin{itemize} \item[(i)] For every $i\in I_{+}$ the set $D_{1}(X_{i})=D(X_{i})\cap X_{i}^{-1}(0,+\infty )$ is finite and for $x_{i}\in D_{1}(X_{i})$ \ there exits $z_{i}\in (0,1)$ such that $f(x_{i})-f(x_{i}+z_{i})-\lambda _{i}z_{i}>0$. \item[(ii)] For $i\in I_{-}$ the set $D_{1}(X_{i})=D(X_{i})\cap X_{i}^{-1}(-\infty ,0)$ is finite and for every $z_{i}\in (0,1)$ we have $f(x_{i} )-f(x_{i}+z_{i})-\lambda _{i}z_{i}\leq 0$. \item[(iii)] Considering $p$ as a function in $T_{n-k}$ do not has critical points which are degenerated. \end{itemize} Then there exists $\epsilon _{0}>0$ such that $u_{\epsilon }$ does not have degenerated critical points for $0<\epsilon <\epsilon _{0}$. \end{theorem} \begin{proof} Suppose there exists a sequence of values $\epsilon_{n}$ with $\epsilon _{n}>0$ and $\lim_{n\to \infty }\epsilon _{n}=0$ and a sequence of point $x_{n}$ in such way that $x_{n}$ is a critical degenerated point of $u_{\epsilon _{n}}$. Then by the proceeding Lemma and under conditions (i), (ii) and (iii) we can find a sequence of $(\epsilon_{n_{k}})$ such that $\lim_{k\to \infty }x_{n_{k}}=x$ with $x_{i}\in D_{1}(X_{i)}$ for some index $i\in I$. Clearly $(x_{n_{k}})_{i}=x_{i}$ for $k>k_{0}$ because $D_{1}(X_{i})$ is finite set. Then for that index $i$, it follows: \begin{equation} X_{i}(x_{i})u_{\epsilon _{n_{k}}}^{i}(x_{i})=\epsilon _{n_{k}} \big( -\exp\big( \frac{\lambda _{i}}{\epsilon _{n_{k}}}\big) +1\big) \end{equation} then for (i) or (ii) we have a contradiction when $k\to \infty $. \end{proof} \subsection*{Example } Consider the vector field \[ X(x)=\begin{cases} \alpha \exp\big(-\frac{1}{\sin (2\pi x)}\big) & 0\leq x\leq 1/2,\\ -\beta \exp\big(-\frac{1}{\sin (2\pi x)}\big) & 1/2\leq x\leq 1 \end{cases} \] It is a $C^{\infty }$ vector field on $T_{1}$. We put \begin{gather*} H=\int_{0}^{1/2}\exp \big( -\frac{1}{sen2\pi x}\big) ,\\ X=\nabla f+\lambda , \\ \begin{aligned} h(x,z) &= f(x_{i})-f(x+z)-\lambda z \\ &= \int_{0}^{x}X(t)dt-\int_{0}^{x+z}X(t)dt \end{aligned} \end{gather*} Then we have \begin{equation*} h(1/4,3/4)=(\beta -\frac{\alpha }{2})H,\quad \lambda =(\alpha -\beta )H \end{equation*} Then if $\alpha >\beta >\frac{\alpha }{2}$, $\lambda =(\alpha -\beta )>0$, $D_{1}(X)=\{1/4\}$ and by theorem \ref{thm7}, we have $\epsilon _{0}>0$ such that $u_{\epsilon }$ does not have degenerated critical points. In this case, $v_{\epsilon }$ is a Morse function for $\epsilon <\epsilon_{0}$ and $X$ is Zeeman Stable vector field \cite{z1}. \begin{thebibliography}{0} \bibitem{g1} J. Gu\'{\i}\~{n}ez, R. Quintero, A. D. Rueda; \emph{Calculating steady state for a Fokker-Planck equation}. Act. Math. Hungar. 91 (2001), No. 4, 311-323. \bibitem{g2} J. Gu\'{\i}\~{n}ez and A. D. Rueda; \emph{Steady state for a Fokker-Planck equation on $S_{n}$}. Act. Math. Hungar. Vol. 94 (2002), No. 3, 211-221. \bibitem{z1} E. C. Zeeman; \emph{Stability of dinamical systems}, Nonlinearity, Vol. 1 (1988), No. 1, 115--155. \end{thebibliography} \end{document}