\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 238, 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} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/238\hfil Patterns on surfaces of revolution] {Patterns on surfaces of revolution in a diffusion problem with variable diffusivity} \author[A. S. do Nascimento, M. S\^onego \hfil EJDE-2014/238\hfilneg] {Arnaldo Simal do Nascimento, Maicon S\^onego} % in alphabetical order \address{Arnaldo S. do Nascimento \newline Universidade Federal de S\~ao Carlos - DM, S. Carlos, S. P., Brasil} \email{arnaldon@dm.ufscar.br} \address{Maicon S\^onego \newline Universidade Federal de Itajub\'a - IMC, Itajub\'a, M.G. Brasil} \email{mcn.sonego@unifei.edu.br} \thanks{Submitted January 16, 2014. Published November 13, 2014.} \subjclass[2000]{35K57, 35B36, 35R01, 35B25, 35B35, 34K20, 58J32} \keywords{Patterns; diffusion; surface of revolution; stability} \begin{abstract} In this article we study the existence of non-constant stable stationary solutions to the the diffusion equation $u_t=\operatorname{div}(a \nabla u)+f(u)$ on a surface of revolution whose border is supplied with zero Neumann boundary condition. Sufficient conditions on the geometry of the surface and on the diffusivity function $a$ are given for the existence of a function $f$ such the problem possesses such solutions. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction}\label{I} The main goal in this article is to find sufficient conditions for the existence of nonconstant stable stationary solutions (herein referred to as {\it patterns}, for short) to the diffusion problem \begin{equation}\label{N} \begin{gathered} \frac{\partial u}{\partial t} = \operatorname{div}_g[a(x)\nabla_g u] + f(u), \quad (t,x) \in \mathbb{R}^{+}\times \mathcal{D} \\ \frac{\partial u}{\partial\nu}=0,\quad x \in \partial \mathcal{D} \end{gathered} \end{equation} where $a \in C^2(\mathcal{D})$ is positive in $\overline{\mathcal{D}}$, $\mathcal{D} \subset \mathbb{R}^3$ is a surface of revolution with border $\partial \mathcal{D}$, metric $g$ and $\nu$ the outer co-normal vector to $\partial \mathcal{D}$. This work should be seen as an attempt to understand the role played by the diffusivity function $a$, the geometry of the surface $\mathcal{D}$ and the reaction term $f$ (a source or sink, depending on its sign) in existence of patterns to \eqref{N}. Typically this kind of problem appears as a mathematical model in many distinct areas and, roughly speaking, a solution models the time evolution of the concentration of a diffusing substance in a heterogeneous medium whose diffusivity is given by a positive function $a$, under the effect of the term $f$. Usually the diffusivity is a property of the material of the surface. The question of existence and nonexistence of patterns for scalar diffusion equations of the type considered here seems to have been first addressed in \cite{M,CH} in bounded domains of $\mathbb{R}^n$ when diffusivity is constant; indeed it is proved that no pattern exists if the domain is convex regardless of the function $f$ and in \cite{M} a non-convex domain and a source term $f$ for which there exists a nonconstant stable stationary solution are provided. In the case when $a$ is a constant function, non-existence of patterns to the diffusion equation appearing in \eqref{N} on a Riemannian manifold without boundary with non-negative Ricci curvature was proved in \cite{AA}, thus generalizing a similar result for surface of revolution found in \cite{RW}. For bounded domains in $\mathbb{R}^N$ the question of how the diffusivity function can give rise to patterns, or not, has been considered by some authors. For the diffusion equation considered in one-dimensional domains, i.e. when $\mathcal{M}$ is an interval, subjected to zero Neumann boundary condition, a sufficient condition for nonexistence of patterns was found to be $a''<0$ in \cite{CH2} and $(\sqrt{a})''<0$ in \cite{Y}. Such result in domains with dimension $N \geq 2$ still remains an open problem. When the domain is an interval, existence of a diffusivity function $a$ which gives rise to patterns to \eqref{N} was addressed in \cite{FH,HR}. These results were generalized to two-dimensional domains in \cite{A2}, using a $\Gamma-$convergence approach, and for any dimension in \cite{A1}, employing a variational approach to dynamical systems. Let us now state our main result. To this end consider a smooth curve $C$ in $\mathbb{R}^3$ parametrized by $ (\psi(s),0,\chi(s))$, $s \in [l_1,l_2] $ and the borderless surface of revolution $\mathcal{M}$ generated by $C$. We consider $\mathcal{D}\subset\mathcal{M}$ a surface of revolution with boundary. Moreover, we suppose that the diffusivity function does not depend on the angular variable $\theta$, so that, abusing notation, we set $a(x(s, \theta))=a(s)$, $s \in [l_1,l_2]$. We prove that if $a \in C^2([l_1, l_2])$, $a>0$ in $[l_1, l_2]$ and \begin{equation}\label{CA} \Big(\frac{(a\psi)'}{\psi}\Big)'(s_0)>0 \end{equation} for some $s_0\in (0,l)\subset [l_1,l_2]$ and if $a(\cdot)$ satisfies \begin{equation}\label{CB} a'(s)\geq 0\text{ in } (0,s_0),\quad a'(s)\leq 0\quad \text{in } (s_0,l), \end{equation} then there exists $f\in C^1(\mathbb{R})$ such that problem \eqref{N} admits a nonconstant stable stationary solution. Our proof stems from \cite{BPT} which by its turn was inspired in \cite{Y}. In \cite{Y} the problem is considered in a interval and it is proved that if $a''(s_0)\geq 0$ for some $s_0$ in this interval then there exists $f$ such that the corresponding problem possesses patterns. In \cite{BPT} problem \eqref{N} is considered with constant diffusivity and the sufficient condition found for existence of a function $f$ such that \eqref{N} possesses patterns is $({\psi'}/{\psi})'(s_0)>0$ for some $s_0\in (0,l)$. Note that our result generalizes \cite{Y} and \cite{BPT} and although the mathematical procedure used in the proof is basically the same, the problem of diffusion on surfaces made of inhomogeneous material - and therefore with variable diffusivity - is a basic one in Mathematical Physics and therefore worthy been studied. Regarding problem \eqref{N}, it was proved in \cite{ARMA} that if $$ -\big(\frac{\psi'}{\psi}\big)'(s)\geq\frac{(a'\psi)'}{2(a\psi)}(s) $$ for all $s\in (0,l)$, then \eqref{N} has no pattern. As expected, our conditions (see \eqref{CA} and \eqref{CB}) for existence of patterns implies that $$ -\big(\frac{\psi'}{\psi}\big)'(s_0)<\frac{(a'\psi)'}{2(a\psi)}(s_0). $$ The expression $-\big(\frac{\psi'}{\psi}\big)'(s)$ has a interesting geometrical meaning. Note that the Gaussian curvature of $\mathcal{M}$ is given by $$ K(s)=\frac{-\psi''(s)}{\psi(s)}\quad (s\in (0,l)) $$ whereas $$ K_g(s)=\frac{\pm\psi'}{\psi}\ \ (s\in (0,l)) $$ represents the geodesic curvature of the parallel circles $s=\text{constant}$ on $\mathcal{M}$. Here the sign depends on the orientation of the parametrization. Therefore $$ K(s)+[K_g(s)]^2=-\big(\frac{\psi'}{\psi}\big)'(s). $$ \section{Preliminaries} We begin with some definitions and known results from Differential Geometry which will be used in the following sections. \subsection{Surface of revolution}\label{subsec1} Consider $\mathcal{M}=(\mathcal{M}, g)$ a $n$-dimensional Riemannian manifold with a metric in local coordinates $x=(x^1, x^2, \dots, x^n)$ given by (using Einstein summation convention) $$ ds^2=g_{ij}dx^i dx^j,\;\;\;(g^{ij})=(g_{ij}^{-1}),\quad |g|=\det(g_{ij}). $$ Given a smooth vector field $X$ on $\mathcal{M}$, the divergence operator of $X$ is defined as $$ \operatorname{div}_g X=\frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^i}(\sqrt{|g|}X^i) $$ and the Riemannian gradient, denoted by $\nabla_g$, of a sufficiently smooth real function $\phi$ defined on $\mathcal{M}$, as the vector field $$ (\nabla_g \phi)^i=g^{ij}\partial_j \phi. $$ We shall see how the operator $\operatorname{div}_g(a(x)\nabla_g u)$ can be expressed for the particular case where $\mathcal{M}$ is a surface of revolution. Let $C$ be the curve of $\mathbb{R}^3$ parametrized by \begin{gather*} x_1=\psi(s)\\ x_2=0\quad (s\in I:=[l_1,l_2])\\ x_3=\chi(s) \end{gather*} where $\psi,\chi\in C^2(I)$, $\psi>0$ in $(l_1,l_2)$ and $(\psi')^2+(\chi')^2=1$ in $I$. Moreover, \begin{gather}\label{boundary} \psi(l_1)=\psi(l_2)=0, \\ \label{boundary2} \psi'(l_1)=-\psi'(l_2)=1. \end{gather} Let $\mathcal{M}$ be the surface of revolution parametrized by \begin{equation}\label{sur} \begin{gathered} x_1=\psi(s)\cos(\theta)\\ x_2=\psi(s)\sin(\theta)\quad (s,\theta)\in [l_1,l_2]\times [0,2\pi).\\ x_3=\chi(s) \end{gathered} \end{equation} Setting $x^1=s, x^2=\theta$ then a surface of revolution in $\mathbb{R}^3$ with the above parametrization is a 2-dimensional Riemannian manifold with metric $$ g=d s^2+\psi^2(s)d\theta^2. $$ By \eqref{boundary} and \eqref{boundary2} $\mathcal{M}$ has no boundary and we always assume that $\mathcal{M}$ and the Riemannian metric $g$ on it are smooth (see \cite{B}, for instance). The area element on $\mathcal{M}$ is $d\sigma=\psi d\theta d s$ and the gradient of $u$ with respect to the metric $g$ is given by $$ \nabla_gu=\Big(\partial_s u,\frac{1}{\psi^2}\,\partial_{\theta}u\Big). $$ Now, consider $\mathcal{D}\subset\mathcal{M}$ a surface of revolution with boundary, i.e., $\mathcal{D}$ is delimited by two circles $C_0$ and $C_l$, $ l_1<00$. Thus there holds $\nu=\frac{\partial}{\partial s}$ on $C_{l}$ and $\nu=-\frac{\partial}{\partial s}$ on $C_{0}$. Although the diffusivity function $a$ may depend on $(s, \theta)$ throughout this work we suppose, for the sake of simplicity, that it depends just on the variable $s$. Thus abusing notation, for simplicity sake, we set \begin{equation}\label{a} a(x(s, \theta ))=a(s),\quad \text{for } x=(\psi(s)\cos(\theta),\psi(s)\sin(\theta),\chi(s))\in \mathcal{D} \end{equation} and therefore \begin{equation}\label{ab} \operatorname{div}_g(a(x)\nabla_g u) =au_{ss}+\frac{(\psi a)_s}{\psi}u_s+\frac{a}{\psi^2}u_{\theta\theta}. \end{equation} Hence throughout the text, problem \eqref{N} on $\mathcal{D}$ reduces to \begin{equation}\label{parabolica2} \begin{gathered} u_t=au_{ss}+\frac{(\psi a)_s}{\psi}u_s+\frac{a}{\psi^2}u_{\theta\theta}+f(u),\quad (s,\theta)\in (0,l) \times [0,2\pi)\\ u'(0)=u'(l)=0. \end{gathered} \end{equation} \subsection{Stability analysis} By a stationary solution of problem \eqref{N} we mean a solution to the problem \begin{equation}\label{est} \begin{gathered} \operatorname{div}_g(a(x)\nabla_g u)+f(u)=0,\quad x\in\mathcal{D}\\ \frac{\partial u}{\partial\nu}=0,\quad x\in\partial\mathcal{D} \end{gathered} \end{equation} or equivalently, in our setting, a solution to \eqref{parabolica2} which does not depend on time. A stationary solution $U$ of \eqref{parabolica2} is called {\it stable} (in the sense of Lyapunov) if for every $\epsilon>0$ there exists $\delta>0$ such that $\|u(\cdot, t)-U\|_{\infty}<\epsilon$ for all $t>0$, whenever $\|u_0-U\|_{\infty}<\delta$, where $\|\cdot\|_{\infty}$ stands for the norm of the space $L^\infty(\mathcal{D})$. If there exists $\delta_1>0$ such that $\|u_0-U\|_{\infty}<\delta_1$ implies that $\|u(\cdot, t)-U\|_{\infty}\to 0$, as $t\to \infty$, then $U$ is called {\it asymptotically stable}. We say that $U$ is {\it unstable} if it is not stable. Regarding the eigenvalue problem for the linearized problem \begin{equation}\label{lin} \begin{gathered} \operatorname{div}_g(a\nabla_g\phi)+f'(U)\phi+\lambda\phi=0\quad \text{in } \mathcal{D}\\ \frac{\partial\phi}{\partial\nu}=0\quad \text{in } \partial\mathcal{D} \end{gathered} \end{equation} the first eigenvalue $\lambda_1$ is given by \begin{equation}\label{eigen} \lambda_1=\min\{R_U(\phi):\phi\in H^1(\mathcal{D}), |\phi|_{L^2(\mathcal{D})}=1\} \end{equation} where $$ R_U(\phi)=\int_{\mathcal{D}}\{a|\nabla_g\phi|^2-f'(U)\phi^2 \}d\sigma. $$ It is well known that if $\lambda_1>0$ then $U$ is asymptotically stable and if $\lambda_1<0$ then $U$ is unstable. If $\lambda_1=0$ then stability or instability can occur. \section{Existence of patterns} Next theorem is the main result of this work. \begin{theorem}\label{tt} If for some $s_0\in (0,l)$ it holds that \begin{equation}\label{he} \Big(\frac{(a\psi)'}{\psi}\Big)'(s_0)>0 \end{equation} and if $a$ satisfies \begin{equation}\label{new} a'(s)\geq 0\text{ in } (0,s_0)\quad\text{and}\quad a'(s)\leq 0\text{ in } (s_0,l), \end{equation} then there exists $f\in C^1(\mathbb{R})$ such that problem \eqref{N} admits a nonconstant asymptotically stable solution. \end{theorem} We start with a lemma concerning stationary solutions of \eqref{N}. This result was observed in \cite{RW} for $a\equiv 1$ and for convenience of the reader we will prove it in our case. \begin{lemma}\label{W} Every stationary solution $u$ of problem \eqref{N} on $\mathcal{D}$, which depends on the angular variable $\theta$, is unstable. \end{lemma} \begin{proof} By \eqref{ab} $u$ satisfies the equation $$ au_{ss}+\frac{(\psi a)_s}{\psi}u_s+\frac{a}{\psi^2}u_{\theta\theta}+f(u)=0. $$ As the function $a$ does not depend on $\theta$ we have that $u_{\theta}$ is an eigenfunction of \eqref{lin} with corresponding eigenvalue $\lambda=0$. Since $u_{\theta}$ must change sign it cannot be the eigenfunction corresponding to the lowest eigenvalue. Hence $\lambda_1 < 0$. \end{proof} \begin{lemma}\label{ban} Let $v$ be a solution of problem \eqref{N}. Let there exist $w\in C^2(\mathcal{D})\cap C^1(\overline{\mathcal{D}})$ such that $w\geq 0$, $w$ not identically zero on $\overline{\mathcal{D}}$ and \begin{equation} \begin{gathered} \operatorname{div}_g(a\nabla_g w)+f'(v)w\leq 0,\quad \mathcal{D}\\ \frac{\partial w}{\partial\nu}>0,\quad \partial \mathcal{D}. \end{gathered} \end{equation} Then $v$ is asymptotically stable. \end{lemma} \begin{proof} Let $\lambda_1$ be the smallest eigenvalue of the linearized problem \begin{equation} \begin{gathered} \operatorname{div}_g(a\nabla_g\phi)+f'(v)\phi+\lambda\phi=0,\quad \mathcal{D}\\ \frac{\partial\phi}{\partial\nu}=0,\quad \partial \mathcal{D} \end{gathered} \end{equation} and let $\phi_1$ be the corresponding eigenfunction. As $\phi_1>0$ in $\overline{\mathcal{D}}$ (see \cite{SMO}), we have \begin{align*} 0 &\geq \int_{\mathcal{D}}\phi_1[\operatorname{div}_g(a\nabla_g w)+f'(v)w]d\sigma\\ &=- \int_{\mathcal{D}}\nabla_g\phi_1(a\nabla_g w)d\sigma+ \int_{\partial \mathcal{D}}\phi_1a\frac{\partial w}{\partial\nu}d\gamma+ \int_{\mathcal{D}}\phi_1f'(v)wd\sigma\\ &= \int_{\mathcal{D}}w\big[\operatorname{div}_g(a\nabla_g\phi_1)+f'(v)\phi_1\big] d\sigma+ \int_{\partial \mathcal{D}}a \big[\phi_1\frac{\partial w}{\partial\nu}-w\frac{\partial\phi_1}{\partial\nu}\big] d\gamma\\ &=-\lambda_1 \int_{\mathcal{D}}w\phi_1d\sigma+ \int_{\partial \mathcal{D}}a\phi_1 \frac{\partial w}{\partial\nu}d\gamma\\ &>-\lambda_1 \int_{\mathcal{D}}w\phi_1d\sigma. \end{align*} It follows that $\lambda_1>0$ and then $v$ is asymptotically stable. \end{proof} The following steps are essential for the proof of Theorem \ref{tt}. Using \eqref{he} and the regularity of $\psi$ and $a$, we can find a neighborhood $V$ of $s_0$ such that \begin{equation} \Big(\frac{(a\psi)'}{\psi}\Big)'(s)>0,\quad \forall s\in V. \end{equation} Consider four points $R_1$, $R_2$, $R_3$, $R_4$ in $V$ such that $$ R_10,\quad \forall s\in [R_1,R_4]. \end{equation} Let $z_1=z_1(s)$ be the solution of the Cauchy problem \begin{equation}\label{ez1} \begin{gathered} \Big[\frac{(a\psi z)'}{\psi}\Big]'-Bz=0\quad \text{in } [0,R_2)\\ z(0)=0,\quad z'(0)=1 \end{gathered} \end{equation} where \begin{equation}\label{db} B>\overline{B}:=\max_{[0,l]}\Big|\Big[\frac{(a\psi)'}{\psi}\Big]'\Big|. \end{equation} Analogously let $z_2=z_2(s)$ be the solution of the Cauchy problem \begin{equation}\label{ez2} \begin{gathered} \Big[\frac{(a\psi z)'}{\psi}\Big]'-Bz=0\quad \text{in } (R_3,l]\\ z(l)=0,\quad z'(l)=-1. \end{gathered} \end{equation} We shall write $z_i(s)=z_i(s,B)$ $(i=1,2)$ to indicate the dependence of the solution on the parameter $B$. \begin{lemma}\label{tr} The solution $z_1$ of problem \eqref{ez1} has the following properties: \begin{itemize} \item[(i)] $z_1>0$ in $(0,R_2)$; \item[(ii)] %\label{2i} $z_1(\cdot,B)$ is increasing in $[0,R_2)$ for any $B>\overline{B}$; \item[(iii)] %\label{3i} $z_1(s,\cdot)$ is increasing on $(\overline{B},\infty)$ for any $s\in (0,R_2)$; \item[(iv)] $ \lim_{B\to \infty}z_1(s,B)=\infty$ for any $s\in (0,R_2)$. \end{itemize} Similarly, for $z_2$ we have that: \begin{itemize} \item[(i')] $z_2>0$ in $(R_3,l)$; \item[(ii')] %\label{2i1} $z_2(\cdot,B)$ is decreasing in $(R_3,l]$ for any $B>\overline{B}$; \item[(iii')] $z_2(s,\cdot)$ is decreasing in $(\overline{B},\infty)$ for any $s\in (R_3,l)$; \item[(iv')] $ \lim_{B\to \infty}z_2(s,B)=\infty$ for any $s\in (R_3,l)$. \end{itemize} \end{lemma} \begin{proof} (i) Recalling that $z_1(0)=0$ and $ z_1'(0)=1$, let us assume that $z_1$ vanishes in $(0,R_2)$ and let $s_1\in (0,R_2)$ be the first root of $z_1$ in $(0,R_2)$. Then $$ z_1(s_1)=0\text{ and } z_1(s)>0,\quad \forall s\in (0,s_1). $$ Then for some $s_2\in (0,s_1)$ $$ z_1(s_2)=\max_{[0,s_1]}z_1>0, $$ i.e., $z_1'(s_2)=0$ and $z_1''(s_2)\leq 0$. It follows that \begin{align*} &\big[\frac{(a\psi z_1)'}{\psi}\big]'(s_2)-Bz_1(s_2)\\ &=\big[\big(\frac{(a\psi)'}{\psi}\big)'z_1+\frac{(a\psi)'}{\psi}z_1'\big](s_2) + (a'z_1'+az_1'')(s_2)-Bz_1(s_2)\\ &=(az_1'')(s_2)+z_1(s_2)\big[\big(\frac{(a\psi)'}{\psi}\big)'(s_2)-B\big]\\ &\stackrel{(*)}{<} 0, \end{align*} what contradicts the definition of $z_1$. Note that in $(*)$ the fact that $a(s)>0$ in $I$ and \eqref{db} were used. (ii) Suppose by contradiction that thee exists $s_1\in (0,R_2)$ such that $$ z_1'(s)>0,\quad \forall s\in (0,s_1)\text{ and } z_1'(s_1)=0. $$ Then $z_1''(s_1)\leq 0$. On the other hand, $$ z_1''(s_1)=-\big(\frac{z_1}{a}\big)(s_1) \big[\big(\frac{(a\psi)'}{\psi}\big)'(s_1)-B\big]>0, $$ since $a(s_1)>0$ and $z_1(s_1)>0$ (from (i)). Therefore $z_1$ is increasing in $(0,R_2)$. (iii) Let $B_1>B_2\geq\overline{B}$ and note that \begin{equation}\label{ez} \begin{gathered} \Big[\frac{(a\psi z_1(s,B_1))'}{\psi}\Big]'-B_2z_1(s,B_1)\geq 0\quad (0,R_2)\\ z_1(0,B_1)=0,\quad z_1'(0,B_1)=1. \end{gathered} \end{equation} The inequality in the above problem occurs by $$ \Big[\frac{(a\psi z_1(s,B_1))'}{\psi}\Big]'-B_2z_1(s,B_1) =(B_1-B_2)z_1(s,B_1)\geq 0, $$ for all $s\in (0,R_2)$. Now, as $z_1(s, B_2)$ satisfies \begin{equation} \begin{gathered} \Big[\frac{(a\psi z_1(s,B_2))'}{\psi}\Big]'-B_2z_1(s,B_2)= 0\quad (0,R_2)\\ z_1(0,B_2)=0,\quad z_1'(0,B_2)=1 \end{gathered} \end{equation} following the procedure used to prove Theorem 13 in Chapter 1 of \cite{PW}, we can prove that $$ z_1(s,B_2)\leq z_1(s,B_1), \quad \forall s\in (0,R_2). $$ (iv) Fix any $B_1>\overline{B}$. By integrating the equation \eqref{ez1} and remembering that $z_1$ is a solution to (3.24), we get for any $B\geq B_1$, $$ (a\psi z_1(\eta,B))'=\psi \int_0^{\eta}Bz_1(t,B)d t+\psi c_1. $$ Integrating again $$ a\psi z_1(s,B)=B \int_0^s\psi(\eta) \int_0^{\eta}z_1(t,B)d td\eta +c_1 \int_0^s\psi(\eta)d\eta+c_2, $$ where $c_1$ and $c_2$ are constants independent of $B$. As $\psi>0$ and $a>0$ then using (iii) we obtain $$ z_1(s,B)\geq\frac{1}{a\psi}\Big[B \int_0^s\psi(\eta) \int_0^{\eta}z_1(t,B_1)d td\eta +c_1 \int_0^s\psi(\eta)d\eta+c_2\Big]. $$ The claim follows by letting $B\to \infty$. The proof for $z_2$ is analogous. \end{proof} Now, we define the function $z:[0,l]\to \mathbb{R}$, \begin{equation}\label{z0} z(s):=\begin{cases} z_1(s),&\text{if } s\in [0,R_2)\\ z_3(s),&\text{if } s\in [R_2,R_3]\\ z_2(s),&\text{if } s\in (R_3,l] \end{cases} \end{equation} where $z_3$ is a positive smooth function such that $z$ is smooth at the points $s=R_2$ and $s=R_3$. Under these conditions, by Lemma \ref{tr} (ii) and (ii') we have that $z_3'(R_2)>0$ and $z_3'(R_3)<0$. Then there exist $\overline{s}\in [R_2,R_3]$ such that $z_3'(\overline{s})=0$, and therefore we take $z_3$ such that $s_0$ (recall that $s_0\in [R_2,R_3]$) is the only critical point of $z_3$. Consequently $z_3''(s_0)<0$. We take $z_3$ in this way to ensure that the function $a'z_3'$ is positive in $[R_2,R_3]$. Thus by Lemma \ref{tr} and by the assumptions on $a(\cdot)$ (see \eqref{new}) we have \begin{equation}\label{eqm} a'z'(s)\geq 0,\quad \forall s\in (0,l). \end{equation} Furthermore, the function $z$ is smooth in $[0,l]$, $z>0$ in $(0,l)$ and $z(0)=z(l)=0$. \begin{lemma}\label{im} Let the function $z$ be defined by \eqref{z0}. Then there exists $f\in C^1(\mathbb{R})$ such that the function \begin{equation}\label{FZ} Z(s):= \int_0^sz(t)d t\quad (s\in [0,l]) \end{equation} is a stationary nonconstant solution of problem \eqref{N}. \end{lemma} \begin{proof} Since $z>0$ in $(0,l)$ we have that $u=Z(s)$ is increasing in $(0,l)$. Hence we can define the inverse function $X(u)=Z^{-1}(u)$. Put \begin{equation}\label{ef} f(u):= \begin{cases} -Bu-a(0)&\text{if } u\leq 0\\[4pt] - \frac{\frac{d}{d u}\left\{(a\psi)[X(u)]z[X(u)]\right\}}{\psi[X(u)] \frac{d}{d u}[X(u)]}&\text{ if } 00$; i.e., $f(u)$ is smooth in $(0,Z(l))$. Therefore we have to prove that $f$ is smooth at $u=0$ and $u=Z(l)$. The smoothness at $u=0$ will follow if we show that $$ f(u)=-Bu-a(0), $$ for all $u\in (0,Z(R_2))$. Integrating \eqref{ez1} in $(0,s)$, for any fixed $s$ in $(0,R_2)$, we obtain $$ \frac{(a\psi z)'}{\psi}(s)- \frac{(a\psi z)'}{\psi}(0)=B \int_0^sz(t)d t, $$ i.e. \begin{equation}\label{n1} \frac{(a\psi z)'}{\psi}(s)=BZ(s)+a(0). \end{equation} By the definition of $f$, \begin{equation}\label{n2} f(Z(s))=-\frac{(a\psi z)'}{\psi}(s) \end{equation} for any $s\in (0,l)$. Hence by \eqref{n1} and \eqref{n2} $$ f(Z(s))=-BZ(s)-a(0), $$ for any $s\in (0,R_2]$. Since $Z$ is increasing, $f(u)$ can be written as $$ f(u)=-Bu-a(0), $$ for any $u\in (0,Z(R_2))$. Similarly it is seen that $$ f(u)=-Bu+BZ(l)+a(l) $$ for any $u\in [Z(R_3),Z(l))$. Therefore, $f\in C^1(\mathbb{R})$. Now we prove that $Z(\cdot)$ is a stationary solution to problem \eqref{N} with $f$ defined by \eqref{ef}. For all $s\in (0,l)$, $Z(s)\in (0, Z(l))$, thus $$ f(Z(s))=- \frac{(a\psi z)'}{\psi}(s)=- \frac{(a\psi Z')'}{\psi}(s); $$ i.e., $$ \frac{(a\psi Z')'}{\psi}(s) + f(Z(s))=0\ \ s\in (0,l), $$ and further $$ Z'(0)=z(0)=0\quad \text{and}\quad Z'(l)=z(l)=0. $$ Moreover, $Z(\cdot)$ is nonconstant since $Z'=z>0$ in $(0,l)$. \end{proof} \begin{proof}[Proof of Theorem \ref{tt}] Let $z$ be the function defined by \eqref{z0} and $m_1>0$ and $m_2>0$ constants to be chosen later. Define \begin{equation} w(s):=\begin{cases} z(s)-m_1z(R_1)(s-R_2)^3 &\text{if } s\in [0,R_2)\\ z(s) &\text{if } s\in [R_2, R_3]\\ z(s)+m_2z(R_4)(s-R_3)^3 &\text{if } s\in (R_3,l]. \end{cases} \end{equation} Note that $w>0$ in [0,l], and $w$ depends on the parameter $B$, because $z$ depends on $B$ (see \eqref{ez1}, \eqref{ez2} and \eqref{z0}). We have (see \eqref{chi}) \begin{equation}\label{par} \frac{\partial w}{\partial\nu}(0)=-w'(0), \quad \frac{\partial w}{\partial\nu}(l)=w'(l). \end{equation} To use Lemma \ref{ban}, we prove the following claim. \noindent\textbf{Claim:} Let $Z$ be the stationary solution of \eqref{N} defined by \eqref{FZ}. Then there exist $m_1>0$, $m_2>0$ and $B>0$ satisfying \eqref{db} such that \begin{equation} \begin{gathered} \operatorname{div}_g(a\nabla_g w)+f'(Z)w\leq 0, \quad \mathcal{D}\\ \frac{\partial w}{\partial\nu}>0, \quad \partial \mathcal{D}. \end{gathered} \end{equation} The above problem can be written as \begin{equation}\label{ete} \begin{gathered} \frac{(a\psi w')'}{\psi}+f'(Z)w\leq 0,\quad (0,l)\\ w'(0)<0,\quad w'(l)>0. \end{gathered} \end{equation} Now we partition the interval $(0,l)$ as follows: $$ (0,l)=(0,R_2)\cup [R_2,R_3]\cup (R_3,l). $$ For any $s\in (0,R_2)\cup (R_3,l)$ we have that \begin{equation} f'(Z(s))=-B. \end{equation} By \eqref{ez1} and \eqref{ez2}, \begin{equation}\label{ig} \frac{(a\psi z')'}{\psi}-Bz=-\big[ \frac{(a\psi)'}{\psi}\big]'z-a'z', \end{equation} for all $s\in (0,R_2)\cup (R_3,l)$. Thus, for any $s\in (0,R_2)$, \begin{align*} &\frac{(a\psi w')'}{\psi}+f'(Z)w\\ &= \frac{(a\psi w')'}{\psi}-Bw\\ &= \frac{(a\psi z'-3a\psi m_1z(R_1)(s-R_2)^2)'}{\psi}-Bz + Bz(R_1)m_1(s-R_2)^3\\ &= \frac{(a\psi z')'}{\psi}-Bz-3 \frac{(a\psi)'}{\psi}m_1z(R_1)(s-R_2)^2 - 6am_1z(R_1)(s-R_2)\\ &\quad +Bz(R_1)m_1(s-R_2)^3\\ &\stackrel{(*)}{=}-[ \frac{(a\psi)'}{\psi}]'z-a'z' +m_1z(R_1)(R_2-s)\big[6a +3 \frac{(a\psi)'}{\psi}(s-R_2)-B(s-R_2)^2\big]. \end{align*} In (*) we used \eqref{ig}. Recall that $(0,R_2)=(0,R_1)\cup [R_1,R_2)$, $a'z'\geq 0$ in $(0,l)$ (see \eqref{eqm}) and that $z=z_1$ in $(0,R_2)$. Then $\bullet$ in $(0,R_1)$, \begin{align*} -\big[\big( \frac{(a\psi)'}{\psi}\big)'z\big](s)-a'z'(s) &\leq -\big[\big( \frac{(a\psi)'}{\psi}\big)'z\big](s)\\ &\leq \big|\big( \frac{(a\psi)'}{\psi}\big)'z\big|(s)\\ &\leq \overline{B}z(R_1), \end{align*} with $\overline{B}$ defined by \eqref{db}, and $\bullet$ in $[R_1,R_2)$, $$ -\big[\big( \frac{(a\psi)'}{\psi}\big)'z\big](s)-a'z'(s) \leq-\hat{B}z(s)\leq-\hat{B}z(R_1) $$ where $$ \hat{B}:=\min_{[R_1,R_2]}\big[ \frac{(a\psi)'}{\psi}\big]', $$ and $\hat{B}>0$ due to \eqref{con}. By the above remarks, in $(0,R_1)$ it holds that \begin{equation}\label{et1} \frac{(a\psi w')'}{\psi}+f'(Z)w\leq z(R_1) \big[\overline{B}+m_1R_2\big(3(CR_2+2a(s_0))-B(R_1-R_2)^2)\big)\big], \end{equation} where $$ C:=\max_{[0,l]}| \frac{(a\psi)'}{\psi}|. $$ Similarly in $[R_1, R_2)$, \begin{equation}\label{et2} \frac{(a\psi w')'}{\psi}+f'(Z)w\leq z(R_1) \big[-\hat{B}+m_1R_2(6a(s_0)+3CR_2)\big]. \end{equation} Now, if $$ m_1:= \frac{\hat{B}}{3R_2(CR_2+2a(s_0))} $$ a simple calculation shows that the right-sides of \eqref{et1} and \eqref{et2} are both negative if $$ B\geq\max\big\{\overline{B},3\big(1+ \frac{\overline{B}}{\hat{B}}\big) \big(\frac{CR_2+2a(s_0)}{(R_1-R_2)^2}\big)\big\}. $$ It follows that \begin{equation} \frac{(a\psi w')'}{\psi}+f'(Z)w<0\quad \text{in } (0,R_2]. \end{equation} Similarly we obtain \begin{equation} \frac{(a\psi w')'}{\psi}+f'(Z)w<0\quad \text{in } (R_3,l). \end{equation} Now consider the interval $[R_2,R_3]$. Since $Z$ is a stationary solution to problem \eqref{N}, in $[R_2,R_3]$ we have $$ aZ''+ \frac{(a\psi)'}{\psi}Z'+f(Z)=0. $$ Differentiating the above equation and remembering that $Z'=z$, we have $$ az''+ \frac{(a\psi)'}{\psi}z'+f'(Z)z=-[ \frac{(a\psi)'}{\psi}]'z-a'z', $$ as $w=z$ in $[R_2,R_3]$ $$ \frac{(a\psi w')'}{\psi}+f'(Z)w=aw''+ \frac{(a\psi)'}{\psi}w'+f'(Z)w =-[ \frac{(a\psi)'}{\psi}]'z-a'z'<0. $$ The above inequality holds because $z>0$ in $[R_2,R_3]$, and using \eqref{con} and \eqref{eqm}. Thus we conclude the first inequality in \eqref{ete}. It remains to prove that $w'(0)<0$ and $w'(l)>0$. Note that by Lemma \ref{tr} (iv) and (iv') we can take $B$ sufficiently large such that $$ z(R_1)=z_1(R_1,B)> \frac{1}{3m_1R_2^2}, \quad z(R_4)=z_2(R_4,B)> \frac{1}{3m_2(l-R_3)^2} $$ thus implying that $$ w'(0)=1-3m_1z(R_1)R_2^2<0, \quad w'(l)=-1+3m_2z(R_4)(l-R_3)^2>0. $$ Therefore our {\it Claim} is proved. At last, by Lemmas \ref{ban} and \ref{im}, we have that $Z$ is a stable nonconstant stationary solution of problem \eqref{N} with $f$ given by \eqref{ef}. \end{proof} We finish this work by giving a simple example of a surface and a diffusivity function which satisfy the hypotheses of Theorem \ref{tt}. Consider the domain of revolution $\mathcal{D}$ obtained from the parametrization $$ \psi(s)=\frac{s^2}{4}+\frac{1}{2}\, ,\quad \chi(s)=\frac{s}{4}\sqrt{4-s^2}+\arcsin(\frac{s}{2}), $$ $s\in (0,1)$, which resembles a frustum of a hyperboloid. Also consider $$ a(s)=-s^2+s+3; $$ i.e., $a(x)=-4\sqrt{x_1^2+x_2^2}+\sqrt{4\sqrt{x_1^2+x_2^2}-2}+5$, $x\in\mathcal{D}$. A simple computation shows that the hypotheses \eqref{CA} and \eqref{CB} are satisfied by taking $s_0=1/2$. Hence by Theorem \ref{tt} we conclude that there is $f$ such that the problem \eqref{N} possesses patterns. \begin{remark}\rm Theorem \ref{tt} should hold when $\mathcal{D}$ is a surface of revolution without border. 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