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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 140, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/140\hfil Nonexistence of positive solutions]
{Nonexistence of positive solutions for a nonpositone system in a ball}

\author[S. Hakimi \hfil EJDE-2015/140\hfilneg]
{Said Hakimi}

\address{Said Hakimi \newline
Universit\'e Sultan Moulay Slimane,
Facult\'e polydisciplinaire \newline
D\'epartement de Math\'ematiques \\
Mghila B.P. 592, B\'eni Mellal, Morocco}
\email{h\_saidhakimi@yahoo.fr}

\thanks{Submitted October 30, 2014. Published May 21, 2015.}
\subjclass[2010]{35J57, 34B18}
\keywords{Nonpositone system; positive solutions; nonexistence}

\begin{abstract}
 In this article, we prove the nonexistence of positive solutions
 for a nonpositone system in a ball when the nonlinearities may have more
 than one zero.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Reaction-diffusion systems model many phenomena in biology, chemical reaction,
population dynamics etc. A typical example of these models is the
 boundary value problem
\begin{equation}
 \begin{gathered}
 -\Delta u(x)=\lambda f(u(x)),\quad x\in \Omega \\
u(x)=0,\quad x\in \partial \Omega.
\end{gathered} \label{eq0}
\end{equation}
The fact that the reaction term $f$ may be negative at the origin makes
it very challenging problem in showing the positivity of the solution.
In the case of systems, it is even more difficult since we have to
the positivity of every component. In this work we restrict our analysis
 to the system
\begin{equation}
\begin{gathered}
 -\Delta u(x)=\lambda f(v(x)),\quad x\in \Omega \\
-\Delta v(x)=\mu g(u(x)),\quad x\in \Omega \\
u(x)=v(x)=0,\quad x\in \partial \Omega,
\end{gathered} \label{eq1}
\end{equation}
 where $\min(\lambda,\mu) \geq \varepsilon _0>0$, $\Omega=B(0,R) $
is a ball in $\mathbb{R}^N$ with radius $R$, $N\geq2$, $f$ and $g$
are smooth functions that grow at least linearly at infinity.
When $f$ and $g$ are a monotone nondecreasing nonlinearities and have
only one zero, problem \eqref{eq1} has been
studied by Hai, Oruganti and Shivaji \cite{h3} in a ball, and
by Hakimi \cite{h6} in an annulus.

Let $(u,v)$ be a positive solution of \eqref{eq1}.
Then $u, v$ are radial, decreasing and satisfy
\begin{equation}
\begin{gathered}
-(r^{N-1}u')'=\lambda r^{N-1}f(v),\quad 0<r<R\\
-(r^{N-1}v')'=\mu r^{N-1}g(u),\quad 0<r<R\\
u'(0)=v'(0)=0\\
u(R)=v(R)=0.
\end{gathered} \label{eq2}
\end{equation}
In this note, we shall prove that the nonexistence result of positive
solutions of \eqref{eq1} remains valid when $f$ and $g$ have more than
one zero (without loss of generality, we assume that $f$ and $g$ have
exactly three zeros) and are not strictly increasing entirely $[0,+\infty )$;
see \cite[Theorem 1.1]{h3}. To be precise, we shall make the following assumptions
\begin{itemize}
\item[(H1)]  $f, g \in C^1( [ 0,+\infty),\mathbb{R})$ such that $f$ and
$g$ have three zeros $\alpha_1<\alpha_2<\alpha_3$ and
$\beta_1<\beta_2<\beta_3$ respectively with
$f'(\alpha_i)\neq0$, $g'(\beta_i)\neq0$ for all $i \in \{1,2,3\}$.
Moreover, $f'\geq 0$ on $[0,\alpha _{1}] \cup [ \alpha _3,+\infty), g'\geq 0$
on $[ 0,\beta _{1}] \cup [ \beta _3,+\infty)$ and
    $F(\alpha _3)<0$, $G(\beta _3)<0$ where $F(x)=\int_0^xf(t)dt$
and $G(x)=\int_0^xg(t)dt$.

\item[(H2)] $f(0)<0$ and $g(0)<0$.

\item[(H3)] There exist two positive real numbers $a_i$ and $b_i$, $i=1, 2$
such that
$$
f(z)\geq a_1z-b_1,\quad g(z)\geq a_2z-b_2,\quad \forall z\geq0.
$$
\end{itemize}

\section{Main result}

Our main result is the following theorem.

\begin{theorem} \label{thm2.1}
Assume that  {\rm (H1)--(H3)} are satisfied. Then
there exists a positive real number $\sigma$ such that
\eqref{eq1} has no positive solution for $\lambda
\mu>\sigma$.
\end{theorem}

\noindent\textbf{Remark.} Existence result for positive
solutions with superlinearities satisfying (H1), (H2),
$\lambda=\mu$ and $\lambda$ small can be found in
\cite{h1, h2}.
For the single equation case, see \cite{a1,c1,h4, h7} for existence
results and \cite{a1,b1,h5} for nonexistence results.

 To prove Theorem \ref{thm2.1}, we need the next three
lemmas. We note that the proofs of the first and the second lemma
are analogous to those of \cite[lemma 2.1, theorem B]{h3}.
On the other hand, the proof of the last is different from that of
\cite[Lemma 2.2]{h3}. This is so because our $f$ and $g$ have no constant
sign in $( \alpha _1,+\infty)$ and $( \beta _1,+\infty)$ respectively.
Here we use ideas adapted from
Hai, Oruganti and Shivaji \cite{h3}.

Let $t_1 \in (0,R)$. We have the following result.

\begin{lemma} \label{lem2.2}
There exists a positive constant $C$ such that
for $\lambda \mu $ large,
\begin{equation*}
u(t_1)+v(t_1)\leq C.
\end{equation*}
\end{lemma}

\begin{proof}
Let $\lambda_1$ be the first eigenvalue of the $-\Delta$ with Dirichlet
boundary conditions.
Multiplying the first equation in \eqref{eq2}
by a positive eigenfunction, say $\phi$ corresponding to
$\lambda _1$, and using (H3) we obtain
\begin{equation*}
-\int_0^{R}(r^{N-1}u')'\phi dr
\geq\int_0^{R}\lambda (a_1v-b_1)\phi r^{N-1}dr;
\end{equation*}
that is,
\begin{equation}
\int_0^{R} \lambda_1 u r^{N-1}\phi dr
\geq\int_0^{R}\lambda (a_1v-b_1)\phi
r^{N-1}dr. \label{eq3}
\end{equation}
Similarly, using the second equation in \eqref{eq2} and (H3), we
obtain
\begin{equation}
\int_0^{R} \lambda_1 v r^{N-1}\phi dr
\geq\int_0^{R}\mu (a_2u-b_2)\phi r^{N-1}dr.
\label{eq4}
\end{equation}
Combining \eqref{eq3} and \eqref{eq4}, we obtain
$$
\int_0^{R}[ \lambda _1-\lambda \mu
\frac{a_1a_2}{\lambda _1}] v\phi r^{N-1}dr\geq
\int_0^{R}\mu [ -\lambda \frac{a_2b_1}{\lambda
_1}-b_2] \phi r^{N-1}dr.
$$
Now, if $\frac{\lambda \mu a_1a_2}2 \geq \lambda _1^2$, then
$$
\int_0^{R}\mu [ -\lambda a_2b_1-b_2\lambda _1]
\phi r^{N-1}dr\leq \int_0^{R}-\frac{\lambda \mu
}2a_1a_2v\phi r^{N-1}dr;
$$
that is,
\begin{equation}
\int_0^{R}\frac{a_1a_2}2v\phi r^{N-1}dr\leq
\int_0^{R}[ a_2b_1+\frac{b_2\lambda _1}{\varepsilon
_0}] \phi r^{N-1}dr, \label{eq5}
\end{equation}
(because $\min(\lambda,\mu) \geq \varepsilon _0$).
Similarly
\begin{equation}
\int_0^{R}\frac{a_1a_2}2u\phi r^{N-1}dr\leq
\int_0^{R}[ a_1b_2+\frac{b_1\lambda _1}{\varepsilon
_0}] \phi r^{N-1}dr. \label{eq6}
\end{equation}
Adding \eqref{eq5} and \eqref{eq6}, we obtain the
inequality
$$
\int_0^{R}(u+v)\phi r^{N-1}dr\leq
\frac{2}{a_1a_2}\int_0^{R}[ a_1b_2+\frac{b_1\lambda
_1}{\varepsilon _0}+a_2b_1+\frac{b_2\lambda _1}{\varepsilon
_0}] \phi r^{N-1}dr.\\
$$
Then
\begin{align*}
(u+v)(t_1)\int_{0}^{t_1}\phi r^{N-1}dr
&\leq  \int_{0}^{t_1}(u+v)\phi r^{N-1}dr\\
&\leq \int_{0}^{R}(u+v)\phi r^{N-1}dr\\
&\leq  \frac{2}{a_1a_2}\int_0^{R}[
a_1b_2+\frac{b_1\lambda _1}{\varepsilon _0}+a_2b_1+\frac{b_2\lambda
_1}{\varepsilon _0}] \phi r^{N-1}dr,
\end{align*}
because $u$ and $v$ are decreasing. The proof is complete.
\end{proof}

Now, assume that there exists $z\geq 0$ ($z\not\equiv 0$) on $\overline{I}$
where
$I=(a,b)$, and a constant $\gamma$ such that
\begin{equation}
-(r^{N-1}z')'\geq \gamma r^{N-1}z\,,\quad r\in I.
\label{eq7}
\end{equation}
Let $\lambda _1=\lambda _1(I)>0$ denote the principal eigenvalue of
\begin{equation}
 \begin{gathered}
-(r^{N-1}\psi')'=\lambda r^{N-1}\psi,\quad r\in (a,b)\\
\psi(a)=0=\psi (b),
\end{gathered}\label{eq8}
\end{equation}
where $0<a<b\leq 1$.

\begin{lemma} \label{lem2.3}
Let \eqref{eq7} hold. Then $\gamma \leq \lambda _1(I)$.
\end{lemma}

\begin{proof}
Multiplying \eqref{eq7} by $\psi$ ($ \psi>0$), an eigenfunction
corresponding to the principal eigenvalue $\lambda _1(I)$, and
 integrating by parts (twice) we obtain
\begin{equation}
\int_a^b[ \gamma -\lambda _1(I)] r^{N-1}z\psi dr
\leq b^{N-1}\psi '(b)z(b)-a^{N-1}\psi'(a)z(a). \label{eq9}
\end{equation}
But $\psi'(b)<0$ and $\psi'(a)>0$. Hence the
right-hand side of \eqref{eq9} is less than or equal to zero.
Then $\gamma\leq\lambda _1(I)$, and the proof is complete.
\end{proof}

Now, we define
\[
t_0=t_1+\frac{R-t_1}3, \quad t_2=t_1+\frac{2(R-t_1)}3.
\]

\begin{lemma}\label{lem2.4}
 For $\lambda \mu $ sufficiently large,
$u(t_2)\leq \beta _3$ or $v(t_2)\leq \alpha _3$.
\end{lemma}

\begin{proof}
We argue by contradiction. Suppose that $u(t_2)>\beta _3$
and $v(t_2)>\alpha _3$.
\smallskip

\noindent\textbf{Case 1:} $u(t_0)>\rho _2$ or $v(t_0)>\rho _1$,
where $\rho _1=\frac{\alpha_3 +\theta_1}2$ and
$\rho _2=\frac{\beta_3 +\theta_2}2$
($\theta_1$ and $\theta_2$ are the greatest zeros of
 $F$ and $G$ respectively.
If $u(t_0)>\rho _2$ then
\[
-(r^{N-1}v')' =\mu r^{N-1}g(u)
\geq \varepsilon _0r^{N-1}g(\rho _2)\quad\text{in }J=(t_1,t_0)
\]
and $v(r)\geq \alpha _3$ on $\bar{J}$.
Let $\omega $ be the unique solution of
\begin{gather*}
-(r^{N-1}\omega')' = \varepsilon _0r^{N-1}g(\rho
_2)\quad \text{in }J\\
\omega =\alpha _3\quad \text{on }\partial J.
\end{gather*}
Then by comparison arguments,
$ v(r)\geq \omega (r)=\varepsilon _0g(\rho _2)\omega _0(r)+\alpha _3$
in $\bar{J}$, where
$\omega _0$ is the unique (positive) solution of
\begin{gather*}
-(r^{N-1}\omega _0')' =r^{N-1}\quad\text{in }J \\
\omega _0 = 0\quad \text{on }\partial J.
\end{gather*}
In particular, there exists $\overline{\alpha }_3>\alpha _3$
($f(\overline{\alpha }_3)\neq 0$) such that
\[
v(t_1+\frac{2(t_0-t_1)}3) \geq \omega (t_1+\frac{2(t_0-t_1)}3)
\geq \overline{\alpha }_3
\]
in $J^{*}=(t_1+\frac{t_0-t_1}3,t_1+\frac{2(t_0-t_1)}3)$.
Then
\begin{align*}
-(r^{N-1}(u-\beta _3)')'
&=\lambda r^{N-1}f(v)\\
&\geq \lambda r^{N-1}f(\overline{\alpha }_3)\\
&\geq (\frac{\lambda f(\overline{\alpha }_3)}C)
r^{N-1}(u-\beta _3)\quad \text{in }J^{*},
\end{align*}
(where $C$ is as in Lemma \ref{lem2.2}).
 Since $u-\beta _3>0$ in $\bar{J}^*$, it follows that
\begin{equation}
\frac{\lambda f(\overline{\alpha }_3)}C\leq \lambda _1(J^{*}),
\label{eq10}
\end{equation}
where $\lambda _1(J^{*})$ is the principal eigenvalue of \eqref{eq8}
(with $(a,b)=J^{*}$).

Next we consider
\begin{align*}
(r^{N-1}(v-\alpha _3)')'
&= \mu r^{N-1}g(u)\\
&\geq \mu r^{N-1}g(\rho _2)\\
&\geq (\frac{\mu g(\rho _2)}C) r^{N-1}(v-\alpha
_3)\quad \text{in }J.
\end{align*}
Since $v-\alpha _3>0$ in $\bar{J}$, it follows that
\begin{equation}
\frac{\mu g(\rho _2)}C\leq \lambda _1(J), \label{eq11}
\end{equation}
where $\lambda _1(J)$ is the principal eigenvalue of \eqref{eq8}
(with $(a,b)=J$).
Combining \eqref{eq10} and \eqref{eq11}, we obtain
$$
\frac{\lambda \mu f(\overline{\alpha }_3)g(\rho _2)}{C^2}\leq \lambda
_1(J^{*})\lambda _1(J),
$$
but $f(\overline{\alpha }_3)$, $g(\rho _2)$ and $C$ are fixed positive
constants.
This is a contradiction for $\lambda \mu $ large.
A similar contradiction can be reached for the case
$v(t_0)>\rho _1$.
\smallskip

\noindent\textbf{Case 2:} $u(t_0)\leq \rho _2$ and $v(t_0)\leq
\rho _1$.
Then $\beta _3<u\leq \rho _2$ and $\alpha _3<v\leq \rho _1$ in
$J_1=[t_0,t_2]$. Then by the mean value theorem, there exist
$c_1,c_2\in (t_0,t_2)$ such that
$$
|u'(c_2)|\leq \frac{3\rho_2}{R-t_1}, \quad
|v'(c_1)|\leq \frac{3\rho_1}{R-t_1}.
$$
Since $-(r^{N-1}u')'\geq 0$ on $[t_0,t_2)$, it follows that
$$
-r^{N-1}u'(r)\leq -c_2^{N-1}u'(c_2)\quad \text{on }J_2=[t_0,c_2);
$$
thus
\[
| u'(r)|  \leq \frac{c_2^{N-1}}{r^{N-1}}|u'(c_2)|
\leq (\frac{t_2}{t_0}) ^{N-1}\frac{3\rho_2}{R-t_1}\quad
\text{in }J_2.
\]
Similarly, we obtain
$$
| v'(r)| \leq (\frac{t_2}{t_0}) ^{N-1}
\frac{3\rho _1}{R-t_1}\quad \text{in }J_3=[t_0,c_1).
$$
Hence there exists $r_0\in [t_0,R)$ such that
$$
| u'(r_0)| \leq \widetilde{c}, \quad
|v'(r_0)| \leq \widetilde{c},
$$
where
$$
\widetilde{c}=\frac 3{R-t_1}(\frac{t_2}{t_0})
^{N-1}\max (\rho _2,\rho _1).
$$
Now, define the energy function
$$
E(r)=u'(r)v'(r)+\lambda F(v(r))+\mu G(u(r)).
$$
Then
\[
E'(r)=-\frac{2(N-1)}ru'(r)v'(r)\leq 0,
\]
and hence $E\geq 0$ in $[0,R]$, since
$E(R)=u'(R)v'(R)\geq 0$.
However,
\begin{equation}
E(r_0)\leq \widetilde{c}^2+\lambda F(\rho _1)+\mu G(\rho _2),
\label{eq14}
\end{equation}
and $F(\rho _1)<0$ and $G(\rho _2)<0$. Hence $E(r_0)<0$ for
$\lambda \mu$ large which is a contradiction. The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2.1}]
Assume $\lambda \mu $ is large enough so that both
lemmas \ref{lem2.2}, \ref{lem2.4} hold.
We take the case when  $u(t_2)\leq \beta _3$ (we assume that $u(t_2)\leq \beta _{1}$,
unless we can choose $t_{2}^*> t_2$ such that $u(t_{2}^*)\leq \beta _1$). Then
\begin{gather*}
-(r^{N-1}v')' =\mu r^{N-1}g(u)\leq 0\quad \text{in }  J_3=(t_2,R)\\
v(t_2) \leq C,\quad v(R)=0,
\end{gather*}
hence, by comparison arguments, $v(r)\leq \widetilde{\omega }(r)$,
where $\widetilde{\omega }$ is the solution of
\begin{gather*}
-(r^{N-1}\widetilde{\omega }')' =0\quad \text{in }J_3\\
\widetilde{\omega }(t_2)=C,\quad
\widetilde{\omega}(R)=0.
\end{gather*}
However, $\widetilde{\omega }(r)= C\int_r^{R}s^{1-N}ds/\int_{t_2}^{R}s^{1-N}ds$
decreases from $C$ to $0$ on
$[t_2,R]$, hence there exists $r_1\in
(t_2,R)$ (independent of $\lambda \mu $) such that
$\widetilde{\omega }(r_1)=\frac{\alpha _1}2$.

Hence $v(r_1)\leq \alpha _1/2$, and
\begin{align*}
-(r^{N-1}(\beta _3-u)')'
&= -\lambda r^{N-1}f(v)\\
&\geq -\lambda r^{N-1}f(\frac{\alpha _1}2)\\
&\geq \lambda \big(-f(\frac{\alpha _1}2)\big) r^{N-1}\frac{\beta
_3-u}{\beta _3}\quad \text{in }J_4=(r_1,R).
\end{align*}
Since $\beta _3-u>0$ in $\bar{J}_4$, we have
\begin{equation}
\frac{\lambda \widetilde{K}_1}{\beta _3}\leq \lambda _1(J_4),
\label{eq15}
\end{equation}
where $\widetilde{K}_1=-f(\alpha _1/2)$ and
$\lambda_1(J_4)$ is the principal eigenvalue of \eqref{eq8}
(with $(a,b)=J_4$).
Similarly, there exists $r_2\in (r_1,R)$
(independent of $\lambda \mu $) such that
$$
v(r_2)<\frac{\alpha _1}2.
$$
Hence
\begin{gather*}
-(r^{N-1}u')' = \mu r^{N-1}f(v)\leq 0\quad \text{in }  J_5=(r_2,R)\\
u(r_2) \leq C,\quad u(R)=0,
\end{gather*}
then, by comparison arguments we obtain
$$
u(r)\leq \omega _1(r)
=\frac C{\int_{r_2}^{R}s^{1-N}ds}\int_r^{R}s^{1-N}ds;
$$
which satisfies
\begin{gather*}
-(r^{N-1}\omega _1')'=0,\quad \text{in }J_5,\\
\omega_1(r_2)=C,\quad \omega _1(R)=0.
\end{gather*}
Arguing as before there exists $r_3\in (r_2,R)$
(independent of $\lambda \mu $) such that
$$
u(r_3)\leq \omega _1(r_3)\leq \frac{\beta _1}2<C.
$$
Hence
\begin{align*}
-(r^{N-1}(\alpha _3-v)')'
&= -\mu r^{N-1}g(u)\\
&\geq -\mu r^{N-1}g(\frac{\beta _1}2)\\
&\geq \mu \big(-g(\frac{\beta _1}2)\big) r^{N-1}
\frac{\alpha _3-v}{\alpha _3}\quad \text{on }J_6=(r_3,R).
\end{align*}
Since $\alpha _3-v>0$ in $\bar{J}_6$, it follows that
\begin{equation}
\frac{\mu \widetilde{K}_2}{\alpha _3}\leq \lambda_1(J_6),\label{eq16}
\end{equation}
where $\widetilde{K}_2=-g(\frac{\beta _1}2)$ and
$\lambda_1(J_6)$ is the principal eigenvalue of \eqref{eq8}
(with $(a,b)=J_6$).
Combining \eqref{eq15} and \eqref{eq16}, we obtain
$$
\frac{\lambda \mu \widetilde{K}_1\widetilde{K}_2}{\alpha _3\beta
_3}\leq \lambda _1(J_4)\lambda _1(J_6),
$$
which is a contradiction to $\lambda \mu $ being large.

A similar contradiction can be reached for the case
$v(t_2)\leq \alpha_3$. Hence Theorem \ref{thm2.1} is proven.
\end{proof}

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\end{document}