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\begin{document}

\title[Turing patterns in general reaction-diffusion systems]{Turing patterns in general reaction-diffusion systems of Brusselator type}


\author{Marius Ghergu}
%    Address of record for the research reported here
%\address{Institute of Mathematics ``Simion Stoilow'' of the Romanian Academy, P.O. BOX 1-764,
%RO-014700 Bucharest, Romania}
%    Current address
\address{School of Mathematical Sciences, University College
Dublin, Belfield, Dublin 4, Ireland}
\email{marius.ghergu@ucd.ie}



\author{Vicen\c tiu R\u adulescu}
\address{Institute of Mathematics "Simion Stoilow" of the Romanian Academy, PO BOX 1-764, RO-014700, Bucharest, Romania}
\address{Department of Mathematics, University of Craiova, 13 A.I. Cuza Street, 200585 Craiova, Romania}
\email{vicentiu.radulescu@imar.ro}












\begin{abstract}
We study the reaction-diffusion system
$$
\left\{\begin{aligned}
&u_t-d_1\Delta u =a-(b+1)u+f(u)v &&\mbox{ in } \Omega\times (0,T),\\
&v_t-d_2\Delta v =bu-f(u)v &&\mbox{ in } \Omega\times (0,T),\\
&u(x,0)=u_0(x), v(x,0)=v_0(x)&& \mbox{ on }\Omega,\\
&\frac{\partial u}{\partial \nu}(x,t)=\frac{\partial u}{\partial \nu}(x,t)=0&&\mbox{ on }\partial\Omega\times(0,T).
\end{aligned}\right.
$$
Here $\o$ is a smooth and bounded domain in $\RR^N$ ($N\geq 1$),
$a,b,d_1,d_2>0$ and $f\in C^1[0,\infty)$ is a non-decreasing
function. The case $f(u)=u^2$ corresponds to the standard
Brusselator model for autocatalytic oscillating chemical reactions.
Our analysis points out the crucial role played by the
nonlinearity $f$ in the existence of Turing patterns. More
precisely we show that if $f$ has a sublinear growth then no
Turing patterns occur, while if $f$ has a superlinear growth then
existence of such patterns is strongly related to the
inter-dependence between the parameters $a,b$ and the diffusion
coefficients $d_1,d_2$.
\end{abstract}

\keywords{Turing patterns, reaction-diffusion system, Brusselator model,
stability, steady-state solutions} 



\subjclass[2000]{Primary 35J25; Secondary 35K50, 47H11, 58C15}


\maketitle


\section{Introduction}

Many physical, chemical, biological, environmental and even
sociological processes are driven by reaction-diffusion systems. These are multi-component models involving two different
mechanisms: on one hand there is diffusion, a random particle
movement, and on the other hand there are chemical, biological or
sociological reactions representing instantaneous interactions,
which depend on the state variables themselves and possibly also
explicitly on the particles' position.




In the early fifties, the British mathematician Alan M. Turing
\cite{turing} proposed a model that accounts for pattern formation
in morphogenesis. Turing \cite{turing} suggested that under
certain conditions, chemicals can react and diffuse in such a way
to produce steady-state heterogeneous spatial patterns of chemical
or morphogen concentrations. He showed that a system of two
reacting and diffusing chemicals could give rise to spatial
patterns from initial near-homogeneity. The idea behind Turing's
model is the existence of a low-range diffusing activator and a
wide-range diffusing inhibitor. The activator production is
inhibited by the presence of inhibitors and enhanced by the
presence of the activator. In contrast, the inhibitor is not
self-enhancing, that is, its production is not linked to the
presence of other inhibitors, but to the presence of activators.
Turing systems show a very rich behavior from the pattern
formation point of view, varying from spots to stripes and from
lamellar to chaotic structures.


Lately, many Turing-type models described by coupled systems of
reaction-diffusion equations have been used for generating
patterns in both organic and inorganic systems.


In this paper we shall be concerned with Turing patterns in a
general Brusselator model for autocatalytic oscillating chemical
reactions. An autocatalytic reaction is one in which a species acts
to increase the rate of its producing reaction. In many
autocatalytic systems complex dynamics are seen, including multiple
steady-states and periodic orbits. The study of oscillating
reactions has only been the subject of interest for the last fifty
years, starting with the  Belousov-Zhabotinsky chemical reactions.

There is now a large number of real and hypothetical systems
that provide insight into the complex behavior of autocatalytic
systems. Among them we mention Brusselator model \cite{prig},
Gray-Scott model \cite{grey1}, Lengyel-Epstein model \cite{leng1},
Oregonator model \cite{fi}, Schnakenberg model \cite{schnak},
Sel'klov model \cite{selkov}.


In this paper we shall consider the system
\neweq{br}
\left\{\begin{aligned}
&u_t-d_1\Delta u =a-(b+1)u+f(u)v &&\mbox{ in } \Omega\times (0,T),\\
&v_t-d_2\Delta v =bu-f(u)v &&\mbox{ in } \Omega\times (0,T),\\
&u(x,0)=u_0(x), v(x,0)=v_0(x)&& \mbox{ on }\Omega,\\
&\frac{\partial u}{\partial \nu}(x,t)=\frac{\partial u}{\partial \nu}(x,t)=0&&\mbox{ on }\partial\Omega\times(0,T),
\end{aligned}\right.
\endeq
where $\o\subset\RR^N$ $(N\geq 1)$ is a smooth and bounded domain,
$a,b,d_1,d_2$ are positive constants and $f\in C^1(0,\infty)\cap
C[0,\infty)$ is non-negative and nondecreasing function such that
$f>0$ in $(0,\infty)$. The initial data $u_0,v_0$ are non-negative continuous functions in $\oo$.
The case $f(s)=s^2$ in system \eq{br} corresponds to the
Brusselator model introduced by Prigogine and Lefever \cite{prig}
in 1968. It consists on the following four intermediate reaction
steps
$$
A\ri X, \qquad B+X\ri Y+D, \qquad 2X+Y\ri 3X, \qquad X\ri E.
$$
The global reaction is  $A+B\ri D+E$ and corresponds to the
transformation of input products $A$ and $B$ into output products
$D$ and $E$. The unknowns $u$, $v$ in system \eq{br} represent the
concentrations of two intermediary reactants having the diffusion
rates $d_1, d_2>0$ while $a$, $b>0$ are fixed concentrations. The
Brusselator system has been extensively investigated in the last
decades from both analytical and numerical point of view (see for
instance \cite{ash,brown,
erneux,ghrn,kang,kol,pena,pengjmaa,twizell,you}).

The analysis in this paper reveals the fact that the dynamics of
the evolution system \eq{br} and its associated steady-state is
strongly related to the behavior of the nonlinearity $f$.
Throughout this paper, we shall assume that $f$ satisfies one of
the following hypotheses:

either $f$ is sublinear, that is,


$ (f1)\;\;\;$ the mapping $(0,\infty)\ni s\longrightarrow
\frac{f(s)}{s}$ is nonincreasing;

of $f$ has a superlinear character, namely,

$ (f2)\;\;\;$ the mapping $(0,\infty)\ni s\longrightarrow
\frac{f(s)}{s}$ is nondecreasing.

A particular attention will be paid to the steady-states to
\eq{br}, that is, solutions of

\neweq{brs}
\left\{\begin{aligned}
-d_1\Delta u &=a-(b+1)u+f(u)v &&\mbox{ in } \Omega,\\
-d_2\Delta v &=bu-f(u)v &&\mbox{ in } \Omega,\\
\frac{\partial u}{\partial \nu}&=\frac{\partial v}{\partial
\nu}=0&&\mbox{ on }\partial\Omega.
\end{aligned}\right.
\endeq


It is easily seen that there exists a unique uniform steady-state
of \eq{brs}, namely
\neweq{u0v0}
(u,v)=\Big(a,\frac{ab}{f(a)}\Big).
\endeq
In this paper we shall investigate the asymptotic stability of the
above constant solution. In particular, we shall see that if $f$
has a sublinear growth, then the constant solution $(u,v)$ defined by \eq{u0v0} is
uniformly asymptotically stable. Moreover, in the sublinear case on $f$ we prove that \eq{u0v0} 
is the unique solution of system \eq{brs}, so there are no
Turing patterns in this case. In turn, if
$f$ satisfies $(f2)$, the analysis of \eq{brs} is more involved.
The existence of Turing patterns (and implicitly of non-constant
solutions to \eq{brs}) is strongly dependent on the diffusion
coefficients $d_1,d_2$ and on the parameters $a,b$. The most
important issue in the study of steady-state solutions are the
{\it a priori} estimates. Using a similar approach to that in
\cite{ghrn}, we are able to find precise upper and lower bounds
for solutions to \eq{brs} in terms of $a,b,d_1,d_2$ for any
dimension $N\geq 1$. This allows us to extend the study of the
standard Brusselator system started in \cite{brown,
ghrn,pengjmaa}. As a consequence, we are able to provide existence
results in terms of $a$, $b$, $d_1$ and $d_2$ in case where $f$
has a superlinear growth.

The outline of the paper is as follows. Section 2 is devoted to
time-dependent solutions of \eq{br}. The main ingredient in
proving the existence of such solutions if the invariant region
method in the spirit of \cite{smoller} (see also \cite{zheng})
combined with {\it a priori} estimates. Then, uniform stability of
the constant solution \eq{u0v0} is investigated. In Section 3 we
discuss the existence and non-existence of non-constant solutions
to the steady-state system \eq{brs}. Here, we point out the role
played by each parameter $a,b,d_1$ and $d_2>0$.
\section{The evolution system \eqref{br}}


\subsection{Existence of global solutions}



In this part we establish the existence of global solutions to
\eq{br}. Our first result result concerns the case where $f$ is
sublinear.


\begin{them}\label{thme0} Assume that $f$ satisfies $(f1)$ and
$\lim_{s\ri\infty} f(s)/s=0$. Then, for any $a,b,d_1,d_2>0$ and any
non-negative continuous functions $u_0,v_0$, the system \eq{br} has
at least one global solution.
\end{them}
\begin{proof} The proof relies on the invariant region method (see, e.g.,
\cite{smoller,zheng}). To this aim, we rewrite the system \eq{br}
is the vectorial form

\neweq{br1}
{\mathbf w}_t=
\left(\begin{tabular}{ccc}
$d_1$ & $0$\\
$0$&$d_2$
\end{tabular}
  \right)\Delta {\mathbf w}+ F(\mw) \quad\mbox{ in }\Omega\times
  (0,\infty),
\endeq
where ${\mathbf w }=(u,v)^T$ and
$$
F({\mathbf w})  =
\left( \begin{tabular}{cc} $a-(b+1)u+f(u)v$\\
$bu-f(u)v$
\end{tabular}\right).
$$

We claim that the rectangle $\Sigma:=[0,c_1]\times [0,c_2]$ is an
invariant region for \eq{br1} provided $c_1,c_2>0$ are large
enough. In view of $(f1)$ we can choose
$c_1>\max\left\{2a,\|u_0\|_{L^\infty} \right\}$ such that
$$
\frac{(b+1/2)c_1}{f(c_1)}>\|v_0\|_{L^\infty}
$$
and define
$$
c_2:=\frac{(b+1/2)c_1}{f(c_1)}.
$$
We also write $\Sigma$ as
$$
\Sigma=\{ \mw=(u,v)^T\in C(\oo)\cap C(\oo): G_i(\mw)\leq 0, 1\leq
i\leq 4 \},
$$
where
$$
G_1(\mw)=-u,\;\;G_2(\mw)=u-c_1,\;\; G_3(\mw)=-v,\;\; G_4(\mw)=v-c_2.
$$
It is obviously that the initial data $(u_0,v_0)$ belongs to the
interior of $\Sigma$. If $\mw=(u,v)^T\in \partial\Sigma$, by the
definition of $c_1$ and $c_2$  we have
$$
\nabla G_1\cdot F_{\mid_{u=0}} =-a-f(0)v<0,
$$
$$
\nabla G_2\cdot F_{\mid_{u=c_1}}=a-(b+1)c_1+f(c_1)v \leq
a-(b+1)c_1+f(c_1)c_2 =a-\frac{c_1}{2} \leq 0,
$$
$$
\nabla G_3\cdot F_{\mid_{v=0}} =-bu\leq 0,
$$
$$
\nabla G_4\cdot F_{\mid_{v=c_2}} =bu-f(u)v=u\left(b-\frac{f(u)}{u}\right)\leq u\left(b-\frac{f(c_1)}{c_1}c_2 \right)<0.
$$
By Theorem 14.13 in \cite{smoller} it follows that $\Sigma$ is invariant for \eq{br1}. Thus, there exists a global solution of \eq{br1}.
 \qed
\end{proof}


Next, we turn our attention to the case where $f$ is superlinear.
For the standard Brusselator model, that is,  $f(u)=u^2$, the existence of
global solution to \eq{br} was obtained by Rothe \cite{rothebook}.
Here, the existence of global solution to \eq{br} is derived for
more general nonlinearities $f$ under the restriction $d_1=d_2$ and the initial data $u_0$ is strictly positive in $\oo$. 


\begin{them}\label{thme} Assume that $d_1=d_2>0$, the initial data $u_0,v_0$ are continuous function in $\oo$ such 
that $u_0>0$, $v_0\geq 0$ in $\oo$ and the nonlinearity $f$ satisfies 
$(f2)$ and  $\lim_{s\ri 0}f(s)/s=0$.
Then, for any $a,b>0$, the system \eq{br} has a global solution.
\end{them}
\begin{proof}
With the change of variable we can assume $d_1=d_2=1$.
For $\ep>0$ small enough we consider the related problem
\neweq{brUVe}
\left\{\begin{aligned}
&u_t-\Delta u =a-(b+1)u+f(u)v &&\mbox{ in } \Omega\times (0,\infty),\\
&v_t-\Delta v =bu-f(u)v &&\mbox{ in } \Omega\times (0,\infty),\\
&u(x,0)=u_0(x), v(x,0)=v_0(x)+\ep && \mbox{ on }\Omega,\\
&\frac{\partial u}{\partial \nu}(x,t)=\frac{\partial v}{\partial
\nu}(x,t)=0&&\mbox{ on }\partial\Omega\times(0,\infty),
\end{aligned}\right.
\endeq

By standard parabolic arguments, there exists a classical solution
$(u^\ep,u^\ep)$ of \eq{brUVe} in a maximal interval
$(0,T^\ep_{max})$. We claim that $T^\ep_{max}=\infty$. First, by
\eq{brUVe} we have that $U^\ep$ satisfies
$$
u_t^\ep-\Delta u^\ep +(b+1)u^\ep\geq a>0 \quad\mbox{
in } \Omega\times (0,T^\ep_{max}).
$$
Since $u_0>0$ in $\oo$, there exists a constant $k>0$ independent
of $\ep$ such that
\neweq{k}
u^\ep\geq k\quad\mbox{ in }\;\o\times (0,T^\ep_{max}).
\endeq
Since $\lim_{s\ri 0}f(s)/s=0$, we can choose $k>0$ small enough
such that
\neweq{vzero}
v_0+1\leq \frac{bk}{f(k)}\quad\mbox{ in }\;\oo.
\endeq
The function $v$ satisfies
\neweq{vg}
\left\{\begin{aligned}
&v_t^\ep-\Delta v^\ep= bu^\ep-f(u^\ep)v^\ep &&\mbox{ in } \Omega\times (0,T^\ep_{max}),\\
&v^\ep(x,0)=v_0(x)+\ep&& \mbox{ on }\Omega,\\
&\frac{\partial v^\ep}{\partial \nu}(x,t)=0&&\mbox{ on
}\partial\Omega\times(0,T^\ep_{max}).
\end{aligned}\right.
\endeq
Using \eq{k} and that fact that $f$ satisfies $(f2)$ we can easily
deduce that the interval $$\Sigma:=[0,bk/f(k)]$$ is an invariant
region for equation \eq{vg}. This means that
\neweq{vv}
v(x,t)\leq \frac{bk}{f(k)}=const.\quad\mbox{ in }\oo\times
(0,T^\ep_{max}).
\endeq
Adding the first two equations in \eq{brUVe} we have
$$
(u^\ep+u^\ep)_t-\Delta(u^\ep+u^\ep)+\frac{1}{d_1}(u^\ep+u^\ep)\leq
a+\frac{bk}{d_1f(k)} \quad\mbox{ in }\o\times (0,T^\ep_{max}).
$$
By maximum principle we deduce that $u^\ep+v^\ep\leq M$ in
$\oo\times (0,T^\ep_{max})$, for some constant $M>0$ independent
of $\ep$. Therefore, for $\ep>0$ small enough, $u^\ep,$ $u^\ep$
satisfy
$$
\ep\leq u^\ep,v^\ep\leq M\quad \mbox{ in }\oo\times
(0,T^\ep_{max}).
$$
This yields $T^\ep_{max}=\infty$, so $u^\ep$ and $v^\ep$ exist
globally. Also by standard parabolic arguments and up to a
subsequence, $u^\ep$ and $v^\ep$ converge to some functions $u$
and $v$ which are global solutions to \eq{br}. This finishes the
proof of Theorem \ref{thme}.\qed
\end{proof}



\subsection{Stability of the uniform steady-state}


The linearization of \eq{br1} at $\mw_0=(a,ab/f(a))^T$ is
\neweq{br2}
{\mathbf w}_t=
\left(\begin{tabular}{ccc}
$d_1$ & $0$\\
$0$&$d_2$
\end{tabular}
  \right)\Delta {\mathbf w}+\nabla F(\mw_0)\cdot
  \mw+{\mathcal{O}}(\|\mw-\mw_0\|^2).
\endeq
Denote by
$$
0=\mu_0<\mu_1< \mu_2<\cdots < \mu_n< \cdots
$$
the eigenvalues of $-\Delta$ with homogeneous Neumann boundary
condition. For any $k\geq 0$ we also denote by $e(\mu_k)$ the multiplicity of $\mu_k$. Consider
\neweq{x}
{\mathbf X}= \left\{ {\bf w}=(u,v)\in C^1(\oo)\times C^1(\oo):
\frac{\partial u}{\partial \nu}=\frac{\partial v}{\partial
\nu}=0\mbox{ on }\partial\o\right\}
\endeq
and decompose
\neweq{xdec}
{\mathbf X}=\bigoplus_{k\geq 0} {\mathbf X}_k,
\endeq
where ${\mathbf X}_k$ denotes the eigenspace corresponding to
$\mu_k$, $k\geq 0$.



\begin{them}\label{tstab}
Assume that
\neweq{m0}
f(a)>\frac{baf'(a)}{f(a)}-b-1
\endeq
and the first eigenvalue $\mu_1$ of the Dirichlet operator subject
to homogeneous Neumann condition satisfies
\neweq{m1}
\mu_1>\frac{1}{d_1}\left(\frac{baf'(a)}{f(a)}-b-1\right)-\frac{f(a)}{d_2}.
\endeq
Then the steady-state $\mw_0$ is uniformly asymptotically stable.
\end{them}
\begin{proof}
Define $\mathcal{L}:{\mathbf X}\rightarrow C(\oo)\times C(\oo)$ by
$$\mathcal{L}=\left(
\begin{tabular}{ccc}
$d_1\Delta+\frac{baf'(a)}{f(a)}-b-1$&$f(a)$\\
$b-\frac{baf'(a)}{f(a)}$& $d_2\Delta -f(a)$
\end{tabular}
\right)$$ Then ${\mathbf X}_k$ is invariant for $\mathcal{L}$ and
$\xi_k$ is an eigenvalue of $\mathcal{L}$ on ${\mathbf X}_k$ if
and only if $\xi$ is an eigenvalue of the matrix
$$A_k=\left(
\begin{tabular}{ccc}
$-d_1\mu_k+\frac{baf'(a)}{f(a)}-b-1$&$f(a)$\\
$b-\frac{baf'(a)}{f(a)}$& $-d_2\mu_k -f(a)$
\end{tabular}
\right).$$
The determinant and trace of $A_k$ are
\neweq{detr}
\begin{aligned}
{\rm det}(A_k)&=\mu_k\left[d_1d_2\mu_k+d_1f(a)-d_2\Big( \frac{baf'(a)}{f(a)}-b-1 \Big)  \right]+f(a), \\
{\rm Tr}(A_k)&=\frac{baf'(a)}{f(a)}-b-1-f(a)-(d_1+d_2)\mu_k.
\end{aligned}
\endeq
Remark that for any $k\geq 0$ we have
$$
{\rm det}(A_k)>0>{\rm Tr}(A_k).
$$
Denote by $\xi_k^+$ and $\xi_k^-$ the two eigenvalues of $A_k$,
$k\geq 0$.

If $\xi_k^+$, $\xi_k^-$ are complex numbers, then by \eq{m1} we
have
$$
{\rm Re }(\xi_k^+)={\rm Re }(\xi_k^-)=\frac{1}{2}{\rm Tr}(A_k)\leq
\frac{1}{2}\left(\frac{baf'(a)}{f(a)}-b-1-f(a)\right)<0.
$$
If $\xi_k^+$, $\xi_k^-$ are real numbers, then by \eq{m1} we have
$$
\begin{aligned}
\xi_k^-\leq \xi_k^+& =\frac{{\rm Tr}(A_k)+\sqrt{{\rm
Tr}^2(A_k)-4{\rm det}(A_k)}}{2}\\
&=\frac{2{\rm det}(A_k)}{{\rm Tr}(A_k)-\sqrt{{\rm Tr}^2(A_k)-4{\rm
det}(A_k)}}\\
&\leq \frac{{\rm det}(A_k)}{{\rm Tr}(A_k)}\\
&<0.
\end{aligned}
$$
Since $\mu_k\ri \infty$ as $k\ri \infty$, from the above estimate
we deduce $\xi_k^+\ri-\infty$ as $k\ri \infty$.

Hence, in both the above cases we can find $\delta>0$ such that
the spectrum of $\mathcal{L}$ lies in the region $\{z\in\CC:{\rm
Re}(z)<-\delta\}$. By Theorem 5.1.1. in \cite{henry} we obtain
that $\mw_0$ is asymptotically uniformly stable for \eq{br1}. This
ends the proof.\qed
\end{proof}

\medskip


If $f$ satisfies $(f1)$ then $\frac{baf'(a)}{f(a)}-b-1<0$ so that
both conditions \eq{m0} and \eq{m1} are satisfied. In this case we
obtain.

\begin{coro}
If $f$ satisfies $(f1)$ then $\mw_0$ is uniformly asymptotically
stable.
\end{coro}
















\subsection{Diffusion-driven instability}

In this part we point out that under certain conditions on the
parameters $a$ and $b$, the uniform steady-state $(u_0,v_0)$
defined by \eq{u0v0} can be linearly stable in the absence of
diffusion but unstable in the presence of diffusion. This is the
well-known phenomenon of {\it diffusion-driven instability}
emphasized by Turing in his pioneering work \cite{turing}.

Let us consider the spatially homogeneous system corresponding to
\eq{br}:


\neweq{brh}
\left\{\begin{aligned}
&\frac{du}{dt}=a-(b+1)u+f(u)v\,, &&t>0,\\
&\frac{dv}{dt}=bu-f(u)v \,,&&t>0,
\end{aligned}\right.
\endeq

We have the following result.

\begin{them}\label{tdriv}
Assume that
\neweq{m01}
f(a)>\frac{baf'(a)}{f(a)}-b-1>0.
\endeq
Then, there exist $d^*,D^*>0$ such that for all
$$
0<d_1<d^*\quad{and}\quad d_2>D^*,
$$
the steady-state $\mw_0=(a,ba/f(a))^T$ is uniformly asymptotically
stable for the system \eq{brh} and instable for the system
\eq{br}, that is, {\it Turing instabilities occur}.
\end{them}
Remark that \eq{m01} does not hold if $f$ satisfies $(f1)$.

\begin{proof} Using the same approach as in Theorem \ref{tstab} we have that $\mw_0$ is uniformly asymptotically stable
for \eq{brh} provided \eq{m01} holds. Also by \eq{m01} we can
choose $D^*>0$ large enough such that
$$
\mu_1D^*\left(\frac{baf'(a)}{f(a)}-b-1  \right)>f(a).
$$
Using \eq{detr}, for all $d_2>D^*$  we have
$$
\lim_{d_1\searrow 0}{\rm det}(A_1)\leq
f(a)-\mu_1D^*\left(\frac{baf'(a)}{f(a)}-b-1 \right)<0.
$$
Therefore we can find $d^*>0$ such that
$$
{\rm det}(A_1)<0\quad\mbox{ for all }0<d_1<d^*, \, d_2>D^*.
$$
This implies that $A_1$, and so the operator $\mathcal{L}$, has at
least one positive eigenvalue. By \cite[Corollary 5.1.1]{henry} it
follows that $\mw_0$ is uniformly asymptotically instable. This
finishes the proof.\qed



\end{proof}



\section{Steady-state solutions}

In this section we shall be concerned with steady-state solutions
to \eq{br}. Basic to our subsequent analysis is the following
result which is due to Lou and Ni (see \cite[Proposition
2.2]{ln96} or  \cite[Lemma 2.1]{ln99}).

\begin{lema}\label{lemani}
Let $g\in C^1(\overline\Omega\times \RR)$.
\begin{enumerate}
\item If $w\in C^2(\Omega)\cap C^1(\overline \Omega)$ satisfies
\[ \Delta w+g(x,w)\geq 0\mbox{\, in \,}\Omega, \qquad  \frac{\partial w}{\partial n}\leq 0 \mbox{\, on \,}\partial\Omega,\]
and $w(x_0)=\max_{\overline\Omega}w$, then $g(x_0,w(x_0))\geq 0$.

\item If $w\in C^2(\Omega)\cap C^1(\overline \Omega)$ satisfies
\[ \Delta w+g(x,w)\leq 0\mbox{\, in \,}\Omega, \qquad  \frac{\partial w}{\partial n}\geq 0 \mbox{\, on \,}\partial\Omega,\]
and $w(x_0)=\min_{\overline\Omega}w$, then $g(x_0,w(x_0))\leq 0$.

\end{enumerate}
\end{lema}


Using the above result, we first derive that if $f$ satisfies $(f1)$ then \eq{brs} has no non-constant solutions. More precisely we have.



\begin{them}\label{nx} Assume that $f$ satisfies $(f1)$. Then, $(u,v)=\big(a,\frac{ab}{f(a)} \big)$ is the unique solution of system \eq{brs}.
\end{them}
\proof
Let $(u,v)$ be a classical solution of
\eq{brs}. Let also $x_1$ (resp. $x_2$) be a maximum
point of $u$ (resp. $v$) and $x_3$ (resp. $x_4$) be a minimum point of $u$
(resp. $v$) in $\oo$. Using Lemma \ref{lemani}(i) in the first equation of
\eq{brs} we have
\begin{equation}\label{x1}
(b+1) u(x_1)\leq a+ f(u(x_1))v(x_1).
\end{equation}
Now, Lemma \ref{lemani}(i) applied to the second equation in \eq{brs} yields
$$
bu(x_2)\geq f(u(x_2))v(x_2),
$$
that is, $v(x_2)\leq
b\frac{u(x_2)}{f(u(x_2))}$. By virtue of $(f1)$ we next derive
\begin{equation}\label{x2}
v(x_1)\leq v(x_2)\leq b\frac{u(x_2)}{f(u(x_2))}\leq b\frac{u(x_1)}{f(u(x_1))}.
\end{equation}
Therefore \eq{x1} and \eq{x2} imply $ (b+1) u(x_1)\leq a+b
u(x_1)$, that is,
\begin{equation}\label{xxs}
u\leq u(x_1)\leq a\quad\mbox{ in }\o.
\end{equation}
On the other hand, Lemma \ref{lemani}(ii) applied to the second equation of \eq{brs} leads us to
$v(x_4)\geq b\frac{u(x_4)}{f(u(x_4))}$. Again by $(f1)$ it follows that
\begin{equation}\label{x2s}
v(x_3)\geq v(x_4)\geq b\frac{u(x_4)}{f(u(x_4))}\geq
b\frac{u(x_3)}{f(u(x_3))}.
\end{equation}
Next, Lemma \ref{lemani}(ii) applied to the first equation in \eq{brs} yields
$$
(b+1) u(x_3)\geq a+f(u(x_3))v(x_3) \geq a+b u(x_3),
$$
which implies
\begin{equation}\label{xxb}
u\geq u(x_3)\geq a\qquad\mbox{ in }\o.
\end{equation}
Now \eq{xxs} and \eq{xxb} produce $u\equiv a$ in $\o$ and by
\eq{brs} we also have $v\equiv ab/f(a)$. This ends the proof. \qed

\medskip

When $f$ satisfies $(f2)$ the analysis of the steady-state system
\eq{brs} is more delicate. In some cases, depending of the
parameters $a,b, d_1,d_2$ we obtain the existence of non-constant
solutions to \eq{br}. We start this study with the following
crucial result that provides {\it a priori} estimates for
solutions to \eq{brs}.

\begin{them}\label{e1} Assume that $f$ satisfies $(f2)$. Then, any solution $(u,v)$ of \eq{brs} satisfies
\begin{equation}\label{einf1}
\frac{a}{b+1}\leq u\leq a+\frac{d_2}{d_1}\cdot \frac{ab}{(b+1)f(\frac{a}{b+1})}
\qquad\mbox{ in \,}\o,
\end{equation}
and
\begin{equation}\label{einf2}
\frac{ab}{(b+1)f\big(a+\frac{d_2}{d_1}\cdot \frac{ab}{(b+1)f(\frac{a}{b+1})}\big)}
\leq v\leq \frac{ab}{(b+1)f(\frac{a}{b+1})}
\qquad\mbox{ in \,} \o.
\end{equation}
\end{them}
\proof Consider first a minimum point $x_0\in\oo$ of $u$. By Lemma \ref{lemani}(ii) it follows
$$
a-(b+1)u(x_0)+f(u(x_0))v(x_0)\leq 0
$$
which implies
$u(x_0)\geq a/(b+1)$. Hence
\begin{equation}\label{einf01}
 u\geq \frac{a}{b+1} \qquad\mbox{ in \,}\o.
\end{equation}
At maximum point of $v$ we have $bu-f(u)v\geq 0$, that is, $v\leq
bu/f(u)$. By virtue of $(f2)$ and \eq{einf01} we deduce
\begin{equation}\label{einf02}
 v\leq \frac{ab}{(b+1)f(\frac{a}{b+1})} \qquad\mbox{ in \,}\o.
\end{equation}

Let $w=d_1u+d_2v$. Adding the first two relations in \eq{brs} we have
$$
-\Delta w=a-u \mbox{\, in \,}\o,\qquad  \frac{\partial w}{\partial
\nu}= 0 \mbox{\, on \,}\partial\Omega.
$$
Let now $x_1\in\oo$ be a maximum point of $w$. According to
Lemma \ref{lemani}(i) we have  $a-u(x_1)\geq 0$, that is, $u(x_1)\leq a$. By virtue of \eq{einf02}, for all $x\in\oo$ we have
$$
d_1 u(x)\leq w(x)\leq w(x_1)\leq d_1 a+d_2\cdot \frac{ab}{(b+1)f(\frac{a}{b+1})}\qquad\mbox{ in }\o.
$$
This yields
\begin{equation}\label{ccd}
 u\leq a+\frac{d_2}{d_1}\cdot \frac{ab}{(b+1)f(\frac{a}{b+1})}
\qquad\mbox{ in \,}\o.
\end{equation}
We have proved that $u$ satisfies \eq{einf1}.  Again by Lemma
\ref{lemani}(ii), at minimum points of $v$ we have $bu-f(u)v\leq
0$, which yields $v\geq bu/f(u)$. Combining this inequality with
\eq{ccd} we obtain the first estimate in \eq{einf2}. This
concludes our proof. \qed



From the estimates \eq{einf1}-\eq{einf2} in Theorem \ref{e1} we
derive the following:
\begin{pro}\label{e11}
Assume that $f$ satisfies $(f2)$ and let $a,b,D_1,D_2>0$ be fixed.
Then, there exist two positive constants $C_1,C_2>0$ depending on
$a,b,D_1,D_2$ such that for all
$$
d_1\geq D_1,\quad 0<d_2\leq D_2,
$$
any solution $(u,v)$ of \eq{brs} satisfies
$$
C_1< u, v < C_2\qquad\mbox{ in }\oo.
$$
\end{pro}


Furthermore, by standard elliptic arguments and Theorem \ref{e1}
we now obtain:
\begin{pro}\label{e2}
Assume that $f$ satisfies $(f2)$ and let $a,b, D_1,D_2>0$ be
fixed. Then, for any positive integer $k\geq 1$ there exists a
constant
$$
C=C(a,b,D_1,D_2,k,N,\o)>0
$$
such that for all
$$
d_1\geq D_1,\quad 0<d_2\leq D_2,
$$
any solution $(u,v)$ of \eq{brs} satisfies
$$
\|u\|_{C^k(\oo)}+\|v\|_{C^k(\oo)}\leq C.
$$
In particular, any solution of \eq{brs} belongs to
$C^\infty(\oo)\times C^\infty(\oo)$.
\end{pro}















\begin{them}\label{nex1}
(i) Let $a,b,d_2>0$ be fixed. There exists $D=D(a,b,d_2)>0$ such
that system \eq{brs} has no non-constant solutions for all
$d_1>D$.\\
(ii)  Let $a,d_1,d_2>0$ be fixed. There exists $B=B(a,d_1,d_2)>0$
such that system \eq{brs} has no non-constant solutions for all
$0<b<B$.
\end{them}
\proof (i) Remark first that if $(u,v)$ is a solution of \eq{brs},
then, integrating the two equations in \eq{brs} over $\o$ and
adding them up we have
\begin{equation}\label{ubar}
\int_\o u(x)dx=a|\o|.
\end{equation}










\begin{lema}\label{limit0}
Let $a,b,d_2>0$ be fixed and let
$\{\delta_n\}\subset(0,\infty)$ be such that $\delta_n\rightarrow
\infty$ as $n\rightarrow \infty$. If $(u_n,v_n)$ is a solution of
\eq{brs} with $d_1=\delta_n$ then
\begin{equation}\label{conv1}
(u_n,v_n)\rightarrow \left(a,\frac{ab}{f(a)}\right)\qquad\mbox{ in }C^2(\oo)\times C^2(\oo)
\mbox{ as } n\rightarrow \infty.
\end{equation}
\end{lema}
\proof  By Proposition \ref{e2} the sequence $\{(u_n,v_n)\}$ is
bounded in $C^3(\oo)\times C^3(\oo)$. Hence, passing to a
subsequence if necessary, $\{(u_n,v_n)\}$ converges in
$C^2(\oo)\times C^2(\oo)$ to some $(u,v)\in C^2(\oo)\times
C^2(\oo)$. We divide by $\delta_n$ in the corresponding equation
to $u_n$ and then we pass to the limit with $n\ri\infty$. We
obtain that $(u,v)$ satisfies
\neweq{brsz}
\left\{\begin{aligned}
-\Delta u &=0 &&\mbox{ in } \Omega,\\
-d_2\Delta v &=bu-f(u)v &&\mbox{ in } \Omega,\\
\frac{\partial u}{\partial \nu}&=\frac{\partial v}{\partial \nu}=0&&\mbox{ on }\partial\Omega.
\end{aligned}\right.
\endeq
Also, $u_n$ and $u$ satisfy \eq{ubar}. Now, the first equation in
\eq{brsz} together with $\partial u/\partial \nu=0$ on
$\partial\Omega$ implies that $u$ is constant. Combining this fact
with \eq{ubar} it follows that $u\equiv a$. Thus, from \eq{brsz},
$v$ satisfies
$$
-d_2\Delta v =ab-f(a)v \mbox{ in } \Omega, \quad  \frac{\partial v}{\partial \nu}=0\;\mbox{ on }\partial\Omega.
$$
Multiplying the above equality with $ab-f(a)v$ and then integrating over $\o$ we obtain
$$
0\leq \frac{d_2}{f(a)}\int_\o |\nabla (ab-f(a)v)|^2dx=-\int_\o (ab-f(a)v)^2 dx\leq 0.
$$
Hence $v\equiv \frac{ab}{f(a)}$ and the proof follows.
\qed

\medskip


We first introduce the function spaces
$$
H^2_{\rm n}(\o)=\left\{w\in W^{2,2}(\o): \frac{\partial
w}{\partial \nu}=0\right\},\qquad L^2_0(\o)=\left\{w\in L^2(\o):
\int_\o w=0\right\}.
$$



Thus, letting $w=u-a$,
by \eq{ubar} and the standard elliptic regularity, system \eq{brs} is
equivalent to
\neweq{brss}
\left\{\begin{aligned}
-\Delta w &=\delta(a-(b+1)(w+a)+f(w+a)v) &&\mbox{ in } \Omega,\\
-d_2\Delta v &=b(w+a)-f(w+a)v &&\mbox{ in } \Omega,\\
w\in & H^2_{\rm n}(\o)\cap L^2_0(\o),\;v\in H^2_{\rm n}(\o),
\end{aligned}\right.
\endeq
where $\delta=1/d_1$. Define
$$
{\mathcal F}:\RR\times (H^2_{\rm n}(\o)\cap L^2_0(\o))\times
H^2_{\rm n}(\o)\rightarrow L^2_0(\o)\times L^2(\o),
$$
by
$$
{\mathcal F}(\delta,w,v)=\left( \begin{tabular}{cc} $\Delta w +\delta {\mathcal P} (a-(b+1)(w+a)+f(w+a)v)$\\
\\
$d_2\Delta v+ b(w+a)-f(w+a)v$
\end{tabular}
\right),
$$
where ${\mathcal P}:L^2(\o)\rightarrow L^2_0(\o)$ is the
projection operator from $L^2(\o)$ onto $L^2_0(\o)$, namely,
$$
{\mathcal P}(z)=z-\frac{1}{|\o|}\int_\o z(x)dx,\qquad\mbox{ for all
}z\in L^2(\o).
$$
Now \eq{brss} is equivalent to
\begin{equation}\label{fi}
{\mathcal F}(\delta,w,v)={\bf 0}. \end{equation} Indeed, if
${\mathcal F}(\delta,w,v)={\bf 0}$, then
$$
d_2\Delta v+b(w+a)-f(w+a)v=0\mbox{ in
}\o,\qquad v\in H^2_{\rm n}(\o).
$$
It is easy to see that the above relations imply $b(w+a)-f(w+a)v\in
L^2_0(\o)$. Since $w\in L_0^2(\o)$, this yields
$$
a-(b+1)(w+a)+f(w+a)v\in L^2_0(\o),
$$
so that
$$
{\mathcal P}(a-(b+1)(w+a)+f(w+a)v)=a-(b+1)(w+a)+f(w+a)v.
$$
Therefore \eq{brss} is satisfied.



With the same method as in the proof of Lemma \ref{limit0} we have
that the equation ${\mathcal F}(0,w,v)={\bf 0}$ has the unique
solution $(w,v)=(0,ab/f(a))$. Next it is easy to see that
$$
D_{(w,v)}{\mathcal F}(0,0,ab/f(a)):(H^2_{\rm n}(\o)\cap L^2_0(\o))\times
H^2_{\rm n}(\o)\rightarrow L^2_0(\o)\times L^2(\o),
$$
is given by
$$
D_{(w,v)}{\mathcal F}(0,0,ab/f(a))=
\left( \begin{tabular}{cc} $\Delta$ & $0$\\
\\
$b\frac{f(a)-af'(a)}{f(a)}$& $d_2\Delta -f(a)$
\end{tabular}
\right).
$$
Thus $D_{(w,v)}{\mathcal F}(0,0,ab/f(a))$ is invertible and we are in
the frame of the Implicit Function Theorem. It follows that there
exists $\delta_0,r>0$ such that $(0,0,ab/f(a))$ is the
unique solution of
$${\mathcal F}
(\delta,w,v)={\bf 0}\qquad\mbox{ in }[0,\delta_0]\times B_r\left(0,\frac{ab}{f(a)}\right),
$$
where $B_r(0,\frac{ab}{f(a)})$ denotes the open ball in $(H^2_{\rm n}(\o)\cap
L^2_0(\o))\times H^2_{\rm n}(\o)$ centered at $(0,ab/f(a))$ and having
the radius $r>0$.

Let now $\{\delta_n\}$ be a sequence of positive real numbers such that
$\delta_n\ri\infty$ as $n\ri\infty$  and let $(u_n,v_n)$ be an
arbitrary solution of \eq{brs} for $a,b,d_2$ fixed and $d_1=\delta_n$.
Letting $w_n=u_n-a$, it follows that
$$
{\mathcal F}\Big( \frac{1}{\delta_n}, w_n, v_n\Big)={\bf 0}.
$$
According to Lemma \ref{limit0}  we have
$$
(w_n,v_n)\rightarrow \left(0,\frac{ab}{f(a)}\right)\quad \mbox{ in } C^2(\oo)\times
C^2(\oo)\mbox{ as } n\ri\infty.
$$
This means that for $n\geq 1$ large enough there holds
$(1/\delta_n,w_n,v_n)\in (0,\delta_0)\times
B_r(0,\frac{ab}{f(a)})$ which yields
$(w_n,v_n)=(0,\frac{ab}{f(a)})$. Hence, for $d_1=1/\delta_n$ small
enough, system \eq{brs} has only the constant solution
$(a,\frac{ab}{f(a)})$. The proof of (ii) is similar.
\qed







\subsection{Existence results}
Let ${\mathbf X}$ be the space defined in \eq{x} and let
$$
{\mathbf X}^+=\{ (u,v)\in X\,:\, u,v>0\mbox{ in }C(\oo)\}.
$$
We write the system \eq{brs} in the form
\neweq{eqw}
-\Delta {\bf w}={\mathcal G}({\bf w})\,, \qquad {\bf w}\in
{\mathbf X}^+,
\endeq
where
$$
{\mathcal G}({\bf w})  =
\left( \begin{tabular}{cc} $\frac{1}{d_1}(a-(b+1)u+f(u)v)$\\
\\
$\frac{1}{d_2}(bu-f(u)v)$
\end{tabular}
\right).
$$
It is  more convenient to write \eq{eqw} in the form
\begin{equation}\label{ect0}
{\mathcal F}({\bf w})={\bf 0}\,, \qquad {\bf w}\in {\mathbf X}^+,
\end{equation}
where
\neweq{ect}
{\mathcal F}({\bf w})={\bf w}-({\bf I}-\Delta )^{-1}({\mathcal
G}({\bf w})+{\bf w})\,, \qquad {\bf w}\in {\mathbf X}^+.
\endeq

Let ${\bf w}_0=(a,ab/f(a))^T$ be the uniform steady state solution of \eq{brs}. Then


$$
\nabla {\mathcal F}({\bf w}_0)={\bf I}-({\bf I}-\Delta )^{-1}({\bf
I}+A),
$$
where
$$
{A}:= \nabla{\mathcal G}({\bf w}_0)=
\left( \begin{tabular}{cc} $\frac{1}{d_1}\Big( b \frac{af'(a)-f(a)}{f(a)}-1\Big)$ & $ \frac{f(a)}{d_1}$\\
\\
$-\frac{b}{d_2} \frac{af'(a)-f(a)}{f(a)}$& $ -\frac{f(a)}{d_2}$
\end{tabular}
\right).
$$

If $\nabla {\mathcal F}({\bf w}_0)$ is invertible, by
\cite[Theorem 2.8.1]{nir} the index of ${\mathcal F}$ at ${\bf
w}_0$ is given by
\begin{equation}\label{index1}
\mbox{index}({\mathcal F},{\bf w}_0)=(-1)^{\gamma},
\end{equation}
where $\gamma$ denotes the number of the negative eigenvalues of
$\nabla {\mathcal F}({\bf w}_0)$. On the other hand, using the
decomposition \eq{xdec} we have that ${\bf X}_i$ is an invariant
space under $\nabla {\mathcal F}({\bf w}_0)$ and $\xi\in\RR$ is an
eigenvalue of $\nabla {\mathcal F}({\bf w}_0)$ in ${\bf X}_i$ if
and only if $\xi$ is an eigenvalue of $(\mu_i+1)^{-1}(\mu_i{\bf
I}-{ A})$. Therefore, $\nabla {\mathcal F}({\bf w}_0)$ is
invertible if and only if for any $i\geq 0$ the matrix $(\mu_i{\bf
I}-{ A})$ is invertible.


Let us define
\neweq{hh}
H(a,b,d_1,d_2,\mu)=\det(\mu {\bf I}-A).
\endeq
Then, if $(\mu_i{\bf I}-{ A})$ is invertible for any $i\geq 0$, with the same arguments as in \cite{pengshi} we have
\neweq{eqqqq}
\gamma=\sum_{\substack{ i\geq 0,\\ H(a,b,d_1,d_2,\mu_i)<0}}\!\!\!\!\!\!\!\!\!e(\mu_i). \endeq A
straightforward computation yields
$$
H(a,b,d_1,d_2,\mu)=\mu^2-\Big(\frac{abf'(a)-(b+1)f(a)}{d_1 f(a)} -\frac{f(a)}{d_2}
\Big)\mu+\frac{f(a)}{d_1d_2}.
$$

If
\neweq{mdd}
b\frac{af'(a)-f(a)}{f(a)}>\Big(1+\sqrt{\frac{d_1}{d_2}f(a)}\;    \Big)^2,
\endeq
then the equation $H(\mu)=0$ has two positive solutions $\mu^{\pm}(a,b,d_1,d_2)$ given by
$$
\mu^{\pm}(a,b,d_1,d_2)=\frac{1}{2}\Big(\theta(a,b,d_1,d_2)\pm\sqrt{\theta(a,b,d_1,d_2)^2-4f(a)/(d_1d_2)}\;\Big),
$$
where
$$
\theta(a,b,d_1,d_2)=\frac{abf'(a)-(b+1)f(a)}{d_1 f(a)} -\frac{f(a)}{d_2}.
$$




With the same method as in \cite{pengshi} (see also
\cite{ghrn,pengjmaa}) we have the following result.

\begin{them}\label{tex}
Assume that  condition \eq{mdd} holds and there exist $i> j\geq 0$
such that
\begin{enumerate}
\item[(i)] $\mu_i<\mu^+(a,b,d_1,d_2)<\mu_{i+1}$ and  $\mu_j<\mu^-(a,b,d_1,d_2)<\mu_{j+1}$;
\item[(ii)] $\sum_{k=j+1}^{i}e_k$ is odd.
\end{enumerate}
Then \eq{brs} has at least one non-constant solution.
\end{them}

\begin{proof} The proof uses some topological degree arguments (see \cite{brezisaf,brezish}).
By Theorem \ref{nex1}(i) we can fix $D>d_1$ such
that 
\begin{enumerate}
\item[(a)] system \eq{brs} with diffusion coefficients $D$ and $d_2$ has no non-constant solutions; 
\item[(b)] $H(a,b,D,d_2,\mu)>0$ for all $\mu\geq 0$.
\end{enumerate}
Further, by Proposition \ref{e11} one can find $C_1,C_2>0$ depending
on $a,b,d_1,d_2$ such that for any $d\geq d_1$, any solution $(u,v)$ of \eq{brs} with diffusion 
coefficients $d$ and $d_2$ satisfies
$$
C_1<u,v< C_2\quad\mbox{ in }\oo.
$$
Set
$$
{\mathcal M}=\{(u,v)\in C(\oo)\times C(\oo) : C_1<u,v<C_2\mbox{ in
}\oo  \},
$$
and define
$$
\Psi: [0,1]\times {\mathcal M}\ri C(\oo)\times C(\oo),
$$
by
$$
\Psi(t,{\bf w})=(-\Delta+{\bf I})^{-1}\left( \begin{tabular}{cc} $u+\Big(\frac{1-t}{D}+\frac{t}{d_1}\Big)(a-(b+1)u+f(u)v)$\\
\\
$v+\frac{1}{d_2}(bu-f(u)v)$
\end{tabular}
\right).
$$
It is easy to see that solving \eq{brs} is equivalent to find a
fixed point of $\Psi(1,\cdot)$ in ${\mathcal M}$. 
Further, from the definition of ${\mathcal M}$ and Proposition \ref{e11}, we have that $\Psi(t,\cdot)$ has no fixed points in 
$\partial {\mathcal M}$ for all $0\leq t\leq 1$. Therefore, the Leray-Schauder 
topological degree deg$({\bf I}-\Psi(t,\cdot),{\mathcal M},0)$ is well defined.


Using \eq{ect} we have ${\bf I}-\Psi(1,\cdot)={\mathcal F}$. Thus, if \eq{brs}
has no other solutions except the constant one ${\bf w}_0$, then
by \eq{index1} and \eq{eqqqq} we have
\neweq{ssaaa}
{\rm deg}({\bf I}-\Psi(1,\cdot),{\mathcal M},0)={\rm
index}({\mathcal F},{\bf w}_0)=(-1)^{\sum_{k=j+1}^{i}e(\mu_k)}=-1.
\endeq
On the other hand, from the invariance of the Leray-Schauder
degree at the homotopy we deduce
\neweq{oqq}
{\rm deg}({\bf I}-\Psi(1,\cdot),{\mathcal M},0)={\rm deg}({\bf
I}-\Psi(0,\cdot),{\mathcal M},0).
\endeq
Remark that by our choice of $D$, we have that ${\bf w}_0$ is
the only fixed point of $\Psi(0,\cdot)$. Furthermore by (b) above
we have
\neweq{ssaa}
{\rm deg}({\bf I}-\Psi(0,\cdot),{\mathcal M},0)={\rm index}({\bf
I}-\Psi(\cdot, 0),{\bf w}_0)=1.
\endeq

Now, from \eq{ssaaa}-\eq{ssaa} we reach a contradiction.
Therefore, there exists a non-constant solution of \eq{brs}. This
ends the proof. \qed
\end{proof}










































\begin{coro}\label{corod1} Let $a,b,d_2>0$ be fixed. Assume that
\neweq{ba}
abf'(a)>(b+1)f(a)
\endeq
and all the eigenvalues $\mu_i$ have odd multiplicity.
Then, there exists a sequence of intervals $\{(k_n,K_n)\}$ with $0<k_n<K_n<k_{n-1}\ri 0$ (as $n\ri \infty$) 
such that the steady-state system \eq{brs} has at least one non-constant solution for all $d_1\in\bigcup_{n\geq 1} (k_n,K_n)$.
\end{coro}
\begin{proof}
In view of \eq{ba}, condition \eq{mdd} holds for small values of $d_1>0$. Also for $a,b,d_2>0$ fixed we have
$$
\mu^-(a,b,d_1,d_2)\ri \frac{f(a)^2}{d_2(abf'(a)-(b+1)f(a))}\; \quad\mbox{ as }d_1\ri 0.
$$

$$
\mu^+(a,b,d_1,d_2)\ri \infty \; \quad\mbox{ as }d_1\ri 0.
$$


Therefore we can find a sequence of intervals $\{(k_n,K_n)\}_{n}$ such that
\neweq{dddd}
\sum_{\substack{ i\geq 0,\\ \mu^-(a,b,d_1,d_2)<\mu_i<\mu^+(a,b,d_1,d_2)}}\!\!\!\!e(\mu_i)\;\;\mbox{ is odd}
\endeq
for all $d_1\in\bigcup_{n\geq 1} (k_n,K_n)$.Therefore, conditions (i)-(ii) in Theorem \ref{tex} are fulfilled.
\end{proof}\qed



\begin{coro}\label{corod2}Let $a,b,d_1>0$ be fixed. Assume that \eq{ba} holds and
\neweq{emm}
\sum_{\substack{ i\geq 0,\\ 0<\mu_i< 
\frac{abf'(a)-(b+1)f(a)}{d_1 f(a)}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!e(\mu_i)\;\;\mbox{ is odd}.
\endeq
Then there exists $D>0$ such that the steady-state system \eq{brs} has at least one non-constant solution for any $d_2>D$.
\end{coro}
\begin{proof}
By virtue of \eq{ba}, for any $d_2>0$  large enough condition \eq{mdd} holds. Also for any $a,b,d_1$ fixed we have
$$
0<\mu^-(a,b,d_1,d_2)<\mu^+(a,b,d_1,d_2)<\frac{abf'(a)-(b+1)f(a)}{d_1 f(a)}
$$
and
$$
\mu^-(a,b,d_1,d_2)\ri 0\;,\;\;\mu^+(a,b,d_1,d_2)\ri \frac{abf'(a)-(b+1)f(a)}{d_1 f(a)} \; \quad\mbox{ as }d_2\ri \infty.
$$
Therefore, for $d_2>0$ large, condition \eq{emm} implies (i)-(ii) in Theorem \ref{tex}. This concludes the proof.
\end{proof}\qed




\begin{coro}\label{corob} Let $a,d_1,d_2>0$ be fixed. Assume that
$af'(a)>f(a)$ and all the eigenvalues $\mu_i$ have odd multiplicity.
Then, there exists a sequence of intervals $\{(b_n,B_n)\}$ with $0<b_n<B_n<b_{n+1}\ri \infty$ (as $n\ri \infty$) 
such that the steady-state system \eq{brs} has at least one non-constant solution for all $b\in\cup_{n\geq 1} (b_n,B_n)$.
\end{coro}
\begin{proof} We proceed similarly. Since $af'(a)>f(a)$, for large values of $b$
condition \eq{mdd} is fulfilled.  Also for $a,d_1,d_2>0$ fixed we have
$$
\mu^-(a,b,d_1,d_2)\ri 0\;,\;\;\mu^+(a,b,d_1,d_2)\ri \infty \; \quad\mbox{ as }b\ri \infty.
$$
Hence, we can find a sequence of non-overlapping intervals $\{(b_n,B_n)\}$ such that $b_n\ri \infty$ as $n\ri\infty$ and \eq{dddd} holds for all  $b\in\bigcup_{n\geq 1} (b_n,B_n)$.
\end{proof}\qed




If $f(s)=s^m$, $m>1$, then condition \eq{ba} is independent of $a$. We obtain


\begin{coro}\label{coroa} Let $f(s)=s^m$, $m>1$. Assume that $b(m-1)>1$ and
\neweq{emma}
\sum_{\substack{ i\geq 0,\\ 0<\mu_i< (b(m-1)-1)/d_1}}\!\!\!\!\!\!\!\!\!\!\!e(\mu_i)\;\;\mbox{ is odd}.
\endeq
Then there exists $A>0$ such that the steady-state system \eq{brs} has at least one non-constant solution for any $0<a<A$.
\end{coro}
\begin{proof} It is easy to see that \eq{mdd} holds for small values of $a>0$. As before
$$
0<\mu^-(a,b,d_1,d_2)<\mu^+(a,b,d_1,d_2)<\frac{b(m-1)-1}{d_1}
$$
and
$$
\mu^-(a,b,d_1,d_2)\ri 0\;,\;\;\mu^+(a,b,d_1,d_2)\ri \frac{b(m-1)-1}{d_1} \; \quad\mbox{ as } a\ri 0.
$$
Therefore, for $a>0$ small, condition \eq{emma} implies (i)-(ii) in Theorem \ref{tex}. This ends the proof.
\end{proof}\qed


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