\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 118, pp. 1--14.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/118\hfil Ambrosetti-Prodi type results]
{Ambrosetti-Prodi type results in a system of second and
fourth-order ordinary differential equations}

\author[Y. An, J. Feng \hfil EJDE-2008/118\hfilneg]
{Yukun An, Jing Feng}  % in alphabetical order

\address{Yukun An \newline
Department of Mathematics,
Nanjing University of Aeronautics and Astronautics,
Nanjing, 210016,  China}
\email{anyksd@hotmail.com}

\address{Jing Feng \newline
Department of Mathematics,
Nanjing University of Aeronautics and Astronautics,
Nanjing, 210016,  China}
\email{erma19831@sina.com}

\thanks{Submitted March 17, 2008. Published August 25, 2008.}
\subjclass[2000]{34B08, 34B15, 34L30, 47J30}
\keywords{Differential system; Ambrosetti-Prodi type problem;
 subsolution; \hfill\break\indent supersolution; variational method}

\begin{abstract}
 In this paper, by the variational method, we study the
 existence, nonexistence, and multiplicity of solutions
 of an Ambrosetti-Prodi type problem for a system of second
 and fourth order ordinary differential equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

Lazer and McKenna \cite{LM} presented the following (one-dimensional)
mathematical model for the suspension bridge:
 \begin{equation} \label{1}
\begin{gathered}
  y_{tt}+y_{xxxx}+\delta _1 y_t+k(y-z)^+=W(x),
    \quad\text{in } (0,L)\times \mathbb{R}, \\
  z_{tt}-z_{xx}+\delta _2 z_t-k(y-z)^+=h(x,t),
       \quad\text{in } (0,L)\times \mathbb{R}, \\
  y(0,t)=y(L,t)=y_{xx}(0,t)=y_{xx}(L,t)=0,  \quad t\in \mathbb{R}, \\
  z(0,t)=z(L,t)=0,  \quad t\in \mathbb{R}.
\end{gathered}
\end{equation}
Where the variable $z$ measures the displacement from equilibrium of the
cable and the variable $y$ measures the displacement of the road
bed. The constant $k$ is spring constant of the ties.

 When the motion of the cable is ignored, the coupled system (\ref{1})
can be simplified into a single equation which describes the motion of
the road bed of suspension bridge, as follows
\begin{equation} \label{2}
\begin{gathered}
      y_{tt}+y_{xxxx}+\delta y_t+ky^+=W(x,t),
       \quad\text{in } (0,L)\times \mathbb{R}, \\
      y(0,t)=y(L,t)=y_{xx}(0,t)=y_{xx}(L,t)=0,  \quad t\in \mathbb{R}.
\end{gathered}
\end{equation}

  This Problem have been studied by many authors.
In \cite{MW,CJM,CJ1}, the authors, using degree theory and the
variational method,  investigated the multiplicity of some
symmetrical periodic solutions when $\delta =0$ and
$W(x,t)=1+\epsilon h(x,t)$ or $W(x,t)=\alpha\cos x+\beta\cos 2t\cos
x\epsilon $. In \cite{AZ}, the similar results for \eqref{2} are
obtained in case of $\delta \neq 0$ and $W(x,t)=h(x,t)=\alpha \cos
x+\beta \cos{2t}\cos{x}+\gamma \sin{2t}\cos{x}$. Those results give
the conditions impose on the spring constant $k$ which guarantees
the existence of multiple periodic solutions, especially the
sign-changing periodic solutions in the case of $W(x,t)$ is
single-sign. It is notable that the functions $\cos x,
\cos{2t}\cos{x}, \sin{2t}\cos{x}$ are the eigenfunctions of linear
principal operator of \eqref{2} in some function spaces.

 When we consider only the steady state solutions of problem (\ref{1}),
we arrive at the system
\begin{equation} \label{3}
\begin{gathered}
      y_{xxxx}+k(y-z)^+=h_1(x),   \quad\text{in } (0,\pi), \\
      -z_{xx}-k(y-z)^+=h_2(x),    \quad\text{in } (0,\pi), \\
      y(0)=y(\pi)=y_{xx}(0)=y_{xx}(\pi)=0,  \\
      z(0)=z(\pi)=0 .
\end{gathered}
\end{equation}
This problem has little been studied in \cite{DLT,A1}. In \cite{AF,
A2}, the analogous partial differential systems have been considered
when the nonlinearities $k(y-z)^+,-k(y-z)^+$ are replaced by general
$f_1(y,z), f_2(y,z)$. And also, in recently, literature \cite{DM}
studied the  system
\begin{equation} \label{4}
\begin{gathered}
      y_{xx}+k_1y^++\epsilon z^+=\sin x, \quad\text{in } (0,\pi), \\
      z_{xx}+\epsilon y^++k_2 z^+=\sin x, \quad\text{in } (0,\pi), \\
      y(0)=y(\pi)=0,  \\
      z(0)=z(\pi)=0.
\end{gathered}
\end{equation}
Where $u^+=\max\{u,0\}$, the constant $\epsilon$ is small enough
such that the matrix
$$\begin{pmatrix}
k_1 & \epsilon \\
\epsilon & k_2
\end{pmatrix}
$$
is a near-diagonal matrix and the positive numbers
$k_1,k_2$ satisfy
$$
m_1^2<k_1<(m_1+1)^2, \ m_2^2<k_2<(m_2+1)^2 \quad\text{for
  some } m_1,m_2 \in \mathbf{N}.
$$
 This is a first work in the direction of
extending to systems some of well-known results established on
nonlinear equation with an asymmetric nonlinearity. Meanwhile in
\cite{DM} there are two open questions to be interesting:

\subsection*{Question 1} Can one obtain corresponding results if the
second-order differential operator is replaced with a fourth-order
differential operator with corresponding boundary conditions?


\subsection*{Question 2} Can one replace the near-diagonal matrix with
something more general and use information on the eigenvalues of
matrix?

Following the above works and questions, we consider the system
\begin{equation} \label{5}
\begin{gathered}
-u'' =  f_1(x,u,v)+t_1\sin x+h_1(x),   \quad\text{in } (0,\pi)\\
v''''=  f_2(x,u,v)+t_2\sin x+h_2(x),   \quad\text{in } (0,\pi)\\
u(0) =  u(\pi)=0, \\
v(0) =  v(\pi)=v''(0)=v''(\pi)=0,
\end{gathered}
\end{equation}
where $t_1,t_2$ are parameters and $(f_1,f_2):[0,\pi]\times
\mathbb{R}^2\to \mathbb{R}^2$ is asymptotically linear.


On the other hand, the second order elliptic systems as follows
\begin{equation}\label{6}
\begin{gathered}
-\Delta u=f_1(u,v)+t_1\varphi_1+h_1(x), \quad\text{in }\Omega , \\
-\Delta v=f_2(u,v)+t_2\varphi_1+h_2(x), \quad\text{in }\Omega , \\
u=v=0, \quad\text{on }\partial \Omega
\end{gathered}
\end{equation}
have been widely studied. Here we mention the papers
\cite{F1,F2,F3,FP} and the references therein. If
$(f_1,f_2):\mathbb{R}^2\to \mathbb{R}^2$ is asymptotically
linear and the asymptotic matrixes at $-\infty$ and $+\infty$ are
$$
\begin{pmatrix}
\underline{a} & \underline{b} \\
\underline{c} & \underline{d}
\end{pmatrix}, \quad
\begin{pmatrix}
\bar{a} & \bar{b} \\
\bar{c} & \bar{d}
\end{pmatrix}
$$
Under some growth conditions on $(f_1,f_2)$,  in those
papers,  the Ambrosetti-Prodi type results for (\ref{6}) have been
given respectively.

We remind that let $g\in C^{\alpha}(\overline{\Omega}\times
\mathbb{R})$ be a given function such that
$$
\limsup_{s\to-\infty}{g(x,s)\over
s}<\lambda_1<\liminf_{s\to+\infty}{g(x,s)\over s}
$$
uniformly in $x\in\Omega$, where $\lambda_1$ is the first eigenvalue
of the Laplacian on a bounded domain $\Omega$ under the Dirichlet
condition and $\varphi_1$ is the associated eigenfunction. The
Ambrosetti-Prodi type result in a Cartesian version states that for
a given $h\in C^{\alpha}(\overline{\Omega})$ there exists a real
number $t_0$ such that the problem
\begin{gather*}
-\Delta u = g(x,u)+t\varphi_1+h,   \quad\text{in }\Omega\\
u=0,  \quad\text{on }\partial\Omega
\end{gather*}
\begin{itemize}
\item[(i)] has no solution if $t>t_0$;

\item[(ii)]has at least two solutions if $t<t_0$.
\end{itemize}
With different variants and formulations this problem has been
extensively studied.

Inspired, we consider the Ambrosetti-Prodi type problem for system
\eqref{5}. This paper is organized as follows: in Section 2, we
prepare the proper variational framework and prove (PS) condition to
the Euler-Lagrange functional associated to our problem. In Section
3, we prove the main theorem. Finally, a piecewise linear problem is
considered as an example in Section 4.

\section{preliminaries}

In this section, we  prepare the proper variational frame work
for \eqref{5}, that is
\begin{gather*}
-u'' =  f_1(x,u,v)+t_1\sin x+h_1(x),   \quad\text{in } (0,\pi)\\
v''''=  f_2(x,u,v)+t_2\sin x+h_2(x),   \quad\text{in } (0,\pi)\\
u(0) =  u(\pi)=0, \\
v(0) =  v(\pi)=v''(0)=v''(\pi)=0.
\end{gather*}
Where $t_1,t_2$ are parameters, $h_1,h_2\in C[0,\pi]$ are fixed
functions with $\int_0^\pi h_1\sin x=\int_0^\pi h_2\sin x=0$.

We shall need some assumptions on the nonlinearities, which are
necessary to settle the existence or not of solutions in the case of
the Ambrosetti-Prodi type problem and to establish (PS) condition.

Let us order $\mathbb{R}^2$ with the order defined by
$$
\xi=(\xi_1,\xi_2)\geq 0\Longleftrightarrow \xi_1,\xi_2\geq 0.
$$
and denote
$W=(u,v)$ and $F(x,W)=(f_1(x,u,v),f_2(x,u,v))$.

We will use the following hypotheses in this article.
\begin{itemize}
\item[(H1)]   $F=(f_1,f_2):[0,\pi]\times \mathbb{R}^2\to
\mathbb{R}^2$ is locally Lipschitzian function respect to $u, v$,
and there exists a function $H:[0,\pi]
\times\mathbb{R}^2\to\mathbb{R}$ such that
$$
\nabla H(x,u,v)=({\partial H\over\partial u}, {\partial
H\over\partial v})=(f_1(x,u,v),f_2(x,u,v)).
$$

\item[(H2)]  For $\xi=(\xi_1,\xi_2)>0$ large enough,
\begin{equation}\label{17}
F(x,\xi)\geq 0.
\end{equation}

\item[(H3)]  $F$ satisfies
\begin{equation}\label{18}
|F(x,\xi)|\leq c(|\xi_1|+|\xi_2|+1), \ \ \forall\xi\in
{\mathbb{R}^2}, \ x \in (0,\pi)
\end{equation}
where $c>0$ is constant.

\item[(H4)]  For $\xi=(\xi_1,\xi_2)\in \mathbb{R}^2$ and $x\in(0,\pi)$
there holds
\begin{equation}\label{9}
F(x,\xi)\geq\underline{A}\xi-ce,
\end{equation}
for some constant $c>0$. Where $e=(1,1)$ and the matrix
$\underline{A}=\begin{pmatrix}
\underline{a} & \underline{b}\\
\underline{c} & \underline{d}
\end{pmatrix}$
satisfies
\begin{gather}\label{11}
\underline{b} ,  \underline{c}\geq 0, \\
\label{12}
(\underline{A}\xi,\xi)\leq\underline\mu|\xi|^2,
\quad\text{for  some }0<\underline\mu<1.
\end{gather}

\item[(H5)]  For $\xi=(\xi_1,\xi_2)\in \mathbb{R}^2$ and $x\in(0,\pi)$
there holds
\begin{equation}\label{10}
F(x,\xi)\geq\overline{A}\xi-ce,
\end{equation}
for some constant $c>0$. Where $e=(1,1)$ and the matrix
$\overline{A}=\begin{pmatrix}
\overline{a} & \overline{b}\\
\overline{c} & \overline{d}
\end{pmatrix}$
satisfies
\begin{gather}\label{13}
\overline{b} ,  \overline{c}\leq 0, \\
\label{14}
(\overline{A}\xi,\xi)\geq\overline\mu|\xi|^2, \quad\text{for  some }
\overline\mu>1.
\end{gather}
\end{itemize}
(If not mentioned, $c$ will always denote a generic positive
constant.)


\begin{remark} \label{rmk2.1}\rm
 With a simple computation it is easy to show that
\eqref{11}-\eqref{12} and (\ref{13})-(\ref{14}) imply, respectively,
\begin{equation}\label{15}
\begin{gathered}
(1-\underline{a})(1-\underline{d})-\underline{b}\underline{c}>0, \quad
\underline{a},\underline{d}<1, \\
(\underline{A}-I)^{-1}\xi\leq 0, \quad \forall\xi\in \mathbb{R}^2,\;
\xi\geq0,
\end{gathered}
\end{equation}
and
\begin{equation}\label{16}
\begin{gathered}
(1-\overline{a})(1-\overline{d})-\overline{b}\overline{c}>0, \quad
\overline{a},\overline{d}>1,\\
(\overline{A}-I)^{-1}\xi\geq 0, \quad \forall\xi\in \mathbb{R}^2,\;
\xi\geq0,
\end{gathered}
\end{equation}
where $I$ is the identity matrix.
\end{remark}

Let $X=H_0^1(0,\pi)\times(H_0^1(0,\pi)\cap H^2(0,\pi))$
be Hilbert space with the inner product
$$
\langle W,\Psi \rangle=\int_0^\pi(u'\psi_1'+v''\psi_2''), \quad
\forall W=(u,v),\; \Psi=(\psi_1,\psi_2)\in X,
$$
and the corresponding norm
$$
\|W\|_X^2=\int_0^\pi({u'}^2+{v''}^2).
$$

Consider the  second-order ordinary differential eigenvalue
problem
\begin{gather*}
-u''=\lambda u, \quad\text{in } (0,\pi), \\
u(0)=u(\pi)=0,
\end{gather*}
and the  fourth-order ordinary differential eigenvalue
problem
\begin{gather*}
v''''=\lambda v, \quad\text{in } (0,\pi), \\
v(0)=v(\pi)=v''(0)=v''(\pi)=0.
\end{gather*}
It is well known that $\lambda_1=1$ and $\varphi_1=\sin x$ are the
positive first eigenvalue and the associated eigenfunction,
respectively. Hence, it follows from the Poincare inequality that,
for all $W\in X$,
\begin{equation}\label{7}
\int_0^\pi|W|^2\leq\|W\|_X^2.
\end{equation}

A vector $W\in X$ is a weak solution of \eqref{5} if, and only if,
it is a critical point of the associated Euler-Lagrange functional
\begin{equation}\label{8}
J(W)={1\over2}\int_0^\pi({u'}^2+{v''}^2)-\int_0^\pi
H(x,u,v)-\int_0^\pi [(t_1\sin x+h_1)u+(t_2\sin x+h_2)v]
\end{equation}
It is standard to show that the functional $J(W)$ is well defined,
$J(W)\in C^1(X,\mathbb{R})$ and $X\to{\mathbb{R}};\
W\to \int_0^\pi H(x,u,v)+\int_0^\pi [(t_1\sin
x+h_1)u+(t_2\sin x+h_2)v]$ has compact derivative under the
assumptions (H1) and (H3).

\begin{lemma} \label{lem2.2}
Assume that {\rm (H1)-(H5)} hold. Then $J$ satisfies
the (PS) condition.
\end{lemma}

\begin{proof}
 Let $\{W_n=(u_n,v_n)\}\subset X$ be a sequence such
that $|J(W_n)|\leq c$ and $J'(W_n)\to 0$. This implies
\begin{equation}\label{19}
\begin{aligned}
&\Big|\int_0^\pi(u_n'\psi_1'+v_n''\psi_2'')-\int_0^\pi\Big[
(f_1\psi_1+f_2\psi_2)+(t_1\sin x+h_1)\psi_1+(t_2\sin
x+h_2)\psi_2\Big]\Big|\\
& \leq \varepsilon_n\|\Psi\|_X
\end{aligned}
\end{equation}
for all $\Psi=(\psi_1,\psi_2)\in X$, where
$\varepsilon_n\to0(n\to\infty)$. Then by the above
discussion it suffices to prove that $\{W_n\}$ is bounded.

\noindent\textbf{Step 1:} Show the boundedness of $\{W_n^-\}$.
Let $W_n^-=(u_n^-,v_n^-)$, $w^-=\max\{0,-w\}$.
Since $h_1,h_2$ are bounded, there exists $M_1,M_2\geq 0$ such that
\begin{equation}\label{20}
|t_1\sin x+h_1|\leq M_1, \quad  |t_2\sin x+h_2|\leq M_2.
\end{equation}
Moreover, from \eqref{9} and \eqref{11}, we have
\begin{gather*}
f_1(x,u_n,v_n)(-u_n^-)\leq\underline{a}(u_n^-)^2
+\underline{b}u_n^-v_n^-+cu_n^-, \\
f_2(x,u_n,v_n)(-v_n^-)\leq\underline{d}(v_n^-)^2
+\underline{c}u_n^-v_n^-+cv_n^-.
\end{gather*}
Choosing $c>\max\{M_1,M_2\}$ and taking
$\psi_1=-u_n^-,\psi_2=-v_n^-$ in (\ref{19}), then using the above
inequalities and \eqref{12}, we obtain
\begin{align*}
\|W_n^-\|_X^2
& \leq  \int_0^\pi(\underline{A}W_n^-,W_n^-)+
  \int_0^\pi(c{u_n^-}-M_1{u_n^-}+c{v_n^-}-M_2{v_n^-})+c\|W_n^-\|_X\\
& \leq  \underline{\mu}\int_0^\pi|W_n^-|^2+
                       d\int_0^\pi({u_n^-}+{v_n^-})+c\|W_n^-\|_X.
\end{align*}
Where $d\geq\max\{c-M_1,c-M_2\}$ is constant. Using H\"{o}lder
inequality and Poincare inequality, we get
\begin{align*}
\int_0^\pi|u_n^-|\leq c(\int_0^\pi|u_n^-|^2)^{1/2}\leq
c(\int_0^\pi|{u_n^-}'|^2)^{1/2},
\\
\int_0^\pi|v_n^-|\leq c(\int_0^\pi|v_n^-|^2)^{1/2}\leq
c(\int_0^\pi|{v_n^-}''|^2)^{1/2}.
\end{align*}
Then from these two inequalities and (\ref{7}) we have
$$
(1-\underline{\mu})\|W_n^-\|_X^2\leq c\|W_n^-\|_X,
$$
since $0<\underline\mu<1$, $\|W_n^-\|$ is bounded.

\noindent\textbf{Step 2:} Show the boundedness of $\{W_n\}$.
Suppose by contradiction that $\{W_n\}$ is unbounded, then there
exists a subsequence (still denote $\{W_n\}$) such that
$\|W_n\|_X\to\infty \quad\text{as } n\to\infty$.
Setting $V_n=(x_n,y_n)=W_n/\|W_n\|_X$, then $\|V_n\|_X=1$ and there
exists a subsequence such that
\begin{gather}\label{21}
V_n \rightharpoonup V_0=(x_0,y_0), \quad \text{in } X, \\
\label{b1}
V_n \to V_0, \quad\text{in } L^2(0,\pi)\times L^2(0,\pi), \\
V_n \to V_0, \quad\text{a.e.  in } (0,\pi), \notag \\
\label{b2}
\text{with } |x_n(x)|, |y_n(x)|\leq h(x)\in L^2,\; x\in (0,\pi).
\end{gather}
By step 1 we may assume that $ V_n^- \to 0$ in $ L^2\times
L^2$ and $V_n^- \to 0$ a.e.in $ (0,\pi)$. Clearly, $V_0 \geq
0$. Denote
\begin{align*}
G_n(x)&=(g_n^1(x),g_n^2(x))\\
&={(f_1(x,W_n(x))+t_1\sin x+h_1,
f_2(x,W_n(x))+t_2\sin x+h_2)\over\|W_n\|_X}.
\end{align*}
We claim that
\begin{equation}\label{22}
G_n\to\gamma=(\gamma_1,\gamma_2)\geq 0 \quad\text{in }
L^2\times L^2.
\end{equation}
In fact, let $A_n=\{x\in(0,\pi);u_n(x)\leq 0 \ and \ v_n(x)\leq 0\}$
and let $\chi_n$ denotes its characteristic function, then
$G_n=\chi_n G_n+(1-\chi_n)G_n$. By (H3), (\ref{b1}), (\ref{b2})
and using the Lebesgue Dominated Convergence Theorem, we get
$$
\chi_n{F(x,W_n)\over\|W_n\|_X}\to 0 \quad\text{in }L^2\times L^2.
$$
Moreover, from (\ref{20}) we have
$$
\chi_n{(t_1\sin x+h_1,t_2\sin x+h_2)\over\|W_n\|_X}\to 0
\quad\text{in }L^2\times L^2.
$$
Hence $\chi_n G_n\to 0$ in $L^2\times L^2$. With the same
reasoning $(1-\chi_n)G_n\to\gamma'=(\gamma'_1,\gamma'_2)$
in $L^2\times L^2$. Therefore, we only need to prove that
$\gamma'\geq0$.

(i) If $u_n(x)\geq 0$ and $v_n(x)\leq 0$, since $\overline{a}>1$,
from \eqref{10} we have
$$
(1-\chi_n)g_n^1(x)+\overline{b}(y_n^-(x))+{c\over\|W_n\|_X}-
(1-\chi_n){t_1\sin x+h_1\over\|W_n\|_X}
\geq \overline{a}x_n^+(x)
\geq 0
$$
and from \eqref{9} and \eqref{11}, we obtain
$$
(1-\chi_n)g_n^2(x)+\underline{d}(y_n^-(x))+{c\over\|W_n\|_X}-
(1-\chi_n){t_2\sin x+h_2\over\|W_n\|_X} \geq \underline{c}x_n^+(x)
\geq 0
$$
Since $ V_n^- \to 0$ in $ L^2\times L^2$ and
\begin{gather*}
(1-\chi_n)g_n^1(x)+\overline{b}(y_n^-(x))+{c\over\|W_n\|_X}-
(1-\chi_n){t_1\sin x+h_1\over\|W_n\|_X}\to\gamma'_1,
\\
(1-\chi_n)g_n^2(x)+\underline{d}(y_n^-(x))+{c\over\|W_n\|_X}-
(1-\chi_n){t_2\sin x+h_2\over\|W_n\|_X}\to\gamma'_2
\end{gather*}
we get $\gamma'\geq0$.

(ii) If $u_n(x)\leq 0$ and $v_n(x)\geq 0$, we can handle in the same
way to obtain that $\gamma'\geq0$.

(iii) If $u_n(x)\geq 0$ and $v_n(x)\geq 0$,  the assertion
$\gamma'\geq0$ can be inferred from (H2).

Now dividing (\ref{19}) by $\|W_n\|_X$, using (\ref{21}), (\ref{22})
and passing to the limit we obtain
\begin{equation}\label{23}
\int_0^\pi (x_0'\psi_1'+y_0''\psi_2'')=\int_0^\pi
(\gamma_1\psi_1+\gamma_2\psi_2), \quad
\forall\Psi=(\psi_1,\psi_2)\in X.
\end{equation}
 From \eqref{10} we have
$$
{(f_1(x,W_n(x))+t_1\sin x+h_1,
f_2(x,W_n(x))+t_2\sin
x+h_2)\over\|W_n\|_X}\geq\overline{A}V_n-{ce\over\|W_n\|_X}.
$$
Passing to the limit in this inequality we get
\begin{equation}\label{24}
\gamma\geq\overline{A}V_0.
\end{equation}

Taking $\psi_1=\sin x,\psi_2=0$ and then $\psi_1=0$, $\psi_2=\sin x$ in
(\ref{23}) and using (\ref{24}), it is achieved that
\begin{equation}\label{25}
(\overline{A}-I)
\begin{pmatrix}
\int_0^\pi x_0\sin x\\
\int_0^\pi y_0\sin x
\end{pmatrix} \leq 0.
\end{equation}
 From Remark \ref{rmk2.1}, applying $(\overline{A}-I)^{-1}$ to (\ref{25}) we
get $(\int_0^\pi x_0\sin x,\int_0^\pi y_0\sin x)\leq 0$. Hence
$x_0=y_0=0$ a.e. So, from (\ref{23}), $\int_0^\pi(\gamma,\Psi)=0$
and taking $\Psi>0$ we have $\gamma=0$.

Finally, consider $\psi_1=x_n,\psi_2=y_n$ in (\ref{19}). Dividing
the resulting expression by $\|W_n\|_X$, and passing to the limit we
obtain $1\leq 0$, that is impossible.
\end{proof}

\begin{lemma} \label{lem2.3}
Suppose {\rm (H5)} hold. Then
\begin{equation}\label{26}
\lim_{s\to+\infty}J(s\sin x,s\sin x)=-\infty.
\end{equation}
\end{lemma}

\begin{proof} From \eqref{10} we have
\begin{gather}\label{27}
H(x,u,v)\geq{\overline{a}\over2}u^2+\overline{b}uv-cu+H(x,0,v)\quad
\text{as } u\geq0,\forall v, \\
\label{28}
H(x,u,v)\geq{\overline{d}\over2}v^2+\overline{c}uv-cv+H(x,u,0)\quad
\text{as } v\geq0,\forall u.
\end{gather}
Adding (\ref{27}), (\ref{28}) and using them again,
\begin{align*}
2H(x,u,v) & \geq
{\overline{a}\over2}u^2+(\overline{b}+\overline{c})uv+{\overline{d}\over2}v^2-cu-cv+H(x,0,v)+H(x,u,0)\\
& \geq
\overline{a}u^2+(\overline{b}+\overline{c})uv+\overline{d}v^2-2cu-2cv+2H(x,0,0)\\
& \geq
\overline{a}u^2+(\overline{b}+\overline{c})uv+\overline{d}v^2-2cu-2cv+2c,
\quad\text{for }u,v\geq0.
\end{align*}
Then by (\ref{14}) we have
\begin{equation}\label{29}
H(x,W)\geq{\overline{\mu}\over2}|W|^2-cu-cv+c.
\end{equation}
Taking $W=(s\sin x,s\sin x)$, where $s>0$, from (\ref{20}) and
(\ref{29}) we get
\begin{align*}
J(s\sin x,s\sin x)
& \leq  {\pi s^2\over2}(1-\overline{\mu})+
   (c+M_1)\int_0^\pi s\sin x+(c+M_2)\int_0^\pi s\sin x-c\\
& \leq  {\pi s^2\over2}(1-\overline{\mu})+cs-c
\end{align*}
since $\overline{\mu}>1$, (\ref{26}) holds.
\end{proof}

\section{The Ambrosetti-Prodi type result}

In this section, we  state and prove the Ambrosetti-Prodi type
result for system \eqref{5}. We need the following concepts.

\begin{definition} \label{def3.1} \rm
(1) We say that a vector function
$W\in X$ is a weak subsolution of \eqref{5} if
$$
J'(W)(\Psi)\leq 0, \quad \forall  \Psi \in X, \; \Psi\geq 0.
$$

(2) $W=(u,v)\in C^2\times C^4$ is a  subsolution (classical)
of \eqref{5} if
\begin{gather*}
-u'' \leq  f_1(x,u,v)+t_1\sin x+h_1,   \quad\text{in } (0,\pi),\\
v''''\leq  f_2(x,u,v)+t_2\sin x+h_2,   \quad\text{in } (0,\pi),\\
u(0) =  u(\pi)=0, \\
v(0) =  v(\pi)=v''(0)=v''(\pi)=0.
\end{gather*}

(3) Weak supersolutions and supersolutions (classical) are
defined likewise by reversing the above inequalities.
\end{definition}

We can easily show that each a subsolution or a supersolution of
\eqref{5} is indeed also a weak subsolution or a weak supersolution,
respectively.

For to present the subsolution and supersolution for \eqref{5}, we
firstly show a maximum principle as follows.

\begin{lemma} \label{lem3.2}
Let $A$ be a matrix-function with entries in
$C[0,\pi]$ satisfy \eqref{11} and \eqref{12}. If $W=(u,v)\in X$ is
such that
\begin{equation}\label{31}
\int_0^\pi(u'\psi_1'+v''\psi_2'')\geq\int_0^\pi (AW,\Psi),\quad \forall
\Psi=(\psi_1, \psi_2)\in X,
\end{equation}
then $W\geq 0$.
\end{lemma}

\begin{proof}
Let $\Psi=W^-=(u^-, v^-)$ in (\ref{31}), by \eqref{11}
and \eqref{12}, we obtain
\begin{align*}
\int_0^\pi(|{u^-}'|^2+|{v^-}''|^2)
&\leq\int_0^\pi(A{W^-},{W^-})-\int_0^\pi(A{W^+},{W^-})\\
&\leq\underline{\mu}\int_0^\pi|{W^-}|^2
  \leq\underline{\mu}\|{W^-}\|_X^2.
\end{align*}
Therefore, ${W^-}=0$, i.e. $W\geq0$.
\end{proof}

\begin{remark} \label{3.3} \rm
In the classical sense, \eqref{11} and \eqref{12}
are also sufficient conditions for having a maximum principle for
the problem
\begin{gather*}
-u'' =  \underline{a}u+\underline{b}v+g_1(x),   \quad\text{in } (0,\pi),\\
v''''=  \underline{c}u+\underline{d}v+g_2(x),   \quad\text{in } (0,\pi),\\
u(0) =  u(\pi)=0, \\
v(0) =  v(\pi)=v''(0)=v''(\pi)=0.
\end{gather*}
This is, $W=(u,v)\geq0$ if $g_1\geq0,g_2\geq0$.
\end{remark}

\begin{lemma} \label{lem3.4}
Assume condition {\rm (H4)}, i.e. \eqref{9},
\eqref{11} and \eqref{12} hold. Then, for all
$t=(t_1,t_2)\in\mathbb{R}^2$, system \eqref{5} has a subsolution
$W_t$ such that, if $W^t$ is any supersolution we have
\begin{equation}\label{35}
W_t\leq W^t \quad\text{in } (0,\pi).
\end{equation}
\end{lemma}

\begin{proof}
We consider the  system
\begin{equation} \label{36}
\begin{gathered}
-u'' =  \underline{a}u+\underline{b}v-c+t_1\sin x+h_1,   \quad\text{in } (0,\pi),\\
v''''=  \underline{c}u+\underline{d}v-c+t_2\sin x+h_2,   \quad\text{in } (0,\pi),\\
u(0) =  u(\pi)=0, \\
v(0) =  v(\pi)=v''(0)=v''(\pi)=0,
\end{gathered}
\end{equation}
where $c$ is the constant in \eqref{9} and \eqref{10}. From the
hypotheses on $\underline{A}$ and $h_1,\ h_2$, (\ref{36}) has a
unique solution $W_t\in C^2\times C^4$. Then, using \eqref{9} we
conclude that $W_t$ is in fact a subsolution of \eqref{5}.

Finally, suppose that $W^t$ is any supersolution of \eqref{5}, from
\eqref{9} and applying Lemma \ref{lem3.2} directly we can get the assertion
(\ref{35}).
\end{proof}

\begin{lemma} \label{lem3.5}
 Suppose {\rm (H1)} holds and $(h_1,h_2)\in C[0,\pi]\times C[0,\pi]$.
Then there exists
$t^0\in\mathbb{R}^2$ such that, for all $t\leq t^0$, system
\eqref{5} has a supersolution $W^t$.
\end{lemma}

\begin{proof}
Let $\overline{u},\overline{v}$ be the solution of the system
\begin{equation} \label{36'}
\begin{gathered}
-\overline{u}'' =  f_1(x,0,0)+h_1(x),   \quad\text{in } (0,\pi),\\
\overline{v}''''=  f_2(x,0,0)+h_2(x),   \quad\text{in } (0,\pi),\\
u(0) =  u(\pi)=0, \\
v(0) =  v(\pi)=v''(0)=v''(\pi)=0.
\end{gathered}
\end{equation}
Due to the locally Lipschitzian condition on $f_1,f_2$, it is
possible to choose $t^0=(t_1^0,t_2^0)<0$ such that
\begin{gather*}
f_1(x,\overline{u},\overline{v})-f_1(x,0,0)+t_1^0\sin x\leq 0,\\
f_2(x,\overline{u},\overline{v})-f_2(x,0,0)+t_2^0\sin x\leq 0.
\end{gather*}
Hence, from these inequalities and the system (\ref{36'}), for all
$t\leq t^0$, $W^{t^0}=(\overline{u},\overline{v})$ is a supersolution
for \eqref{5}.
\end{proof}

\begin{lemma} \label{lem3.6}
Let {\rm (H4), (H5)} hold. Then for a given
$h_1,h_2$, there exists an unbounded domain $\Re$ in the plane such
that if $t\in\Re$, system \eqref{5} has no supersolution.
\end{lemma}

\begin{proof} Suppose $W=(u,v)$ is a supersolution for \eqref{5}.
Multiplying both equations of this system by $\sin x$, integration
them by parts and using \eqref{9}, \eqref{10} we deduce that
\begin{gather} \label{37}
(\underline{A}-I) \begin{pmatrix}
\rho_1\\
\rho_2
\end{pmatrix}
\leq \frac{\pi}{2}
\begin{pmatrix}
-s_1\\
-s_2
\end{pmatrix}, \\
 \label{38}
(\overline{A}-I)\begin{pmatrix}
\rho_1\\
\rho_2
\end{pmatrix}
\leq \frac{\pi}{2}\begin{pmatrix}
-s_1\\
-s_2
\end{pmatrix}.
\end{gather}
Where $\rho_1=\int_0^\pi u\sin x, \rho_2=\int_0^\pi v\sin x,
s_1=t_1-c, s_2=t_2-c$ and $c$ is the constant in \eqref{9} and
\eqref{10}. From remark \ref{rmk2.1}, applying $(\underline{A}-I)^{-1}$ and
$(\overline{A}-I)^{-1}$ to (\ref{37}) and (\ref{38}), respectively,
we obtain that
\begin{itemize}
\item[(i)] If $\rho_1\leq0$, then
$s_2\leq{\underline{d}-1\over\underline{b}}s_1$  when
$\underline{b}\neq0$,  or $ s_1\leq0$  when
$\underline{b}=0$.

\item[(ii)] If $\rho_1\geq0$, then
$s_2\leq{\overline{d}-1\over\overline{b}}s_1$ when
$\overline{b}\neq0$,  or $s_1\leq0$  when
$\overline{b}=0$.
\end{itemize}
Therefore, independently of the sign of $\rho_1$, the pair
$(s_1,s_2)$ is in a region composed of the union of two half-planes
passing through the origin, each of them bounded above by a
straight-line of negative or infinity slope. $\Re$ is the complement
of this region in the original variables $t_1$ and $t_2$.
\end{proof}


Now, we are at a position  to prove the Ambrosetti-Prodi
type result for system \eqref{5}.


\begin{theorem} \label{thm3.7}
Suppose that conditions {\rm (H1)--(H5)} are
satisfied and that there exists a  matrix
\[
A(x)=\begin{pmatrix}
a(x) & b(x)\\
c(x) & d(x)
\end{pmatrix},
\]
with $b(x),c(x)\geq 0$ (cooperativeness condition on $A(x)$) satisfies
\eqref{12} such that
\begin{equation}\label{30}
F(x,\xi)-F(x,\eta)\geq A(x)(\xi-\eta), \quad\text{for }
 \xi,\eta\in \mathbb{R}^2,\; \xi\geq\eta.
\end{equation}
Then there exists a continuous curve $\Gamma$ splitting
$\mathbb{R}^2$ into two unbounded
components $N$ and $E$ such that:
\begin{itemize}
\item[(1)] for each $t=(t_1,t_2)\in N$, \eqref{5} has no solution;
\item[(2)] for each $t=(t_1,t_2)\in E$, \eqref{5} has at least two solutions.
\end{itemize}
\end{theorem}

\begin{proof}  For each $\theta\in \mathbb{R}$, define
$$
L_\theta=\{(t_1,t_2)\in {\mathbb{R}}^2; t_2+\theta=t_1\},
$$
and $R(\theta)=\{t_1\in {\mathbb{R}}; \eqref{5}$ has a supersolution
with $t\in L_\theta$ for some $t_2\in \mathbb{R}\}$.

Lemmas \ref{lem3.5} and \ref{lem3.6} allows us to define the continuous curve
$$
\Gamma(\theta)=(\sup R(\theta),\sup R(\theta)-\theta),
$$
which splits the plane into two disjoints unbounded domains $N$ and
$E$ such that for all $t\in N$ no supersolution exists for
\eqref{5}, while for all $t\in E$ \eqref{5} has a supersolution.

Obviously, for all $t\in N$, no solution exists for \eqref{5},
result (1) is proved.

To prove result (2), now we use the abstract
variational theorems to find the solutions of \eqref{5} when
$t\in E$.
We write
\begin{align*}
&\langle J'(W),\Psi \rangle\\
&= \langle W,\Psi
\rangle-\int_0^\pi[(f_1(x,u,v)+t_1\sin
x+h_1)\psi_1+(f_2(x,u,v)+t_2\sin x+h_2)\psi_2].
\end{align*}
Given $t\in E$ there exists a supersolution $W^t=(u^t,v^t)$ and a
subsolution $W_t=(u_t,v_t)$ of \eqref{5} such that
$W_t\leq W^t$ in $(0,\pi)$.
Let
$$
M=[W_t,W^t]=\{W\in X;W_t\leq W\leq W^t\},
$$
since $W_t,W^t\in L^{\infty}$ by assumption, also
$M\subset L^{\infty}$ and
$H(x,W(x))+(t_1\sin x+h_1)u+(t_2\sin x+h_2)v\leq c$
for all $W\in M$ and almost every $x\in(0,\pi)$.

Clearly, $M$ is a closed and convex subset of $X$, hence weakly
closed. Since $M$ is essentially bounded,
$J(W)\geq{1\over2}\|W\|_X^2-c$ is coercive on $M$. On the other
hand, if $W_n\rightharpoonup W$ weakly in $X$, where $W_n,W\in M$,
we may assume that $W_n\to W$ pointwise almost everywhere;
moreover, $|H(x,W_n)+(t_1\sin x+h_1)u_n+(t_2\sin x+h_2)v_n|\leq c$
uniformly, using Lebesgue Dominated Convergence Theorem, we have
\begin{align*}
&\int_0^\pi H(x,W_n)+\int_0^\pi [(t_1\sin x+h_1)u_n+(t_2\sin
x+h_2)v_n] \\
&\to  \int_0^\pi H(x,W)+\int_0^\pi [(t_1\sin x+h_1)u+(t_2\sin
x+h_2)v].
\end{align*}
Hence $J$ is weakly lower semi-continuous on $M$. Then we can use
\cite[Theorem 1.2]{MS} to find a vector function
$W_0=(u_0,v_0)\in X$ such that $W_0\in M$ is the infimum of the
functional $J$ restricted to $M$.

To see that $W_0$ is a weak solution of \eqref{5}, for
$\varphi=(\varphi_1,\varphi_2)\in C_0^{\infty}(0,\pi)$ and
$\varepsilon>0$ let
\begin{gather*}
u_{\varepsilon}=\min\{u^t,\max\{u_t,u_0+\varepsilon\varphi_1\}\}
=u_0+\varepsilon\varphi_1-\varphi_1^{\varepsilon}+\varphi_{1\varepsilon}
\\
v_{\varepsilon}=\min\{v^t,\max\{v_t,v_0+\varepsilon\varphi_2\}\}
=v_0+\varepsilon\varphi_2-\varphi_2^{\varepsilon}+\varphi_{2\varepsilon}
\end{gather*}
with
\begin{gather*}
\varphi_1^{\varepsilon}=\max\{0,u_0+\varepsilon\varphi_1-u^t\}\geq0, \\
\varphi_2^{\varepsilon}=\max\{0,v_0+\varepsilon\varphi_2-v^t\}\geq0,
\end{gather*}
and
\begin{gather*}
\varphi_{1\varepsilon}=-\min\{0,u_0+\varepsilon\varphi_1-u_t\}\geq0,\\
\varphi_{2\varepsilon}=-\min\{0,v_0+\varepsilon\varphi_2-v_t\}\geq0.
\end{gather*}
Note that $W_{\varepsilon}=(u_{\varepsilon},v_{\varepsilon})\in M$
and
$\varphi^{\varepsilon}=(\varphi_1^{\varepsilon},\varphi_2^{\varepsilon}),
\varphi_{\varepsilon}=(\varphi_{1\varepsilon},\varphi_{2\varepsilon})\in
X\cap L^{\infty}(0,\pi)$.

The functional $J$  is differentiable in direction $W_{\varepsilon}-W_0$.
Since $W_0$ minimizes $J$ in $M$ we have
$$
0\leq\langle W_{\varepsilon}-W_0,J'(W_0)\rangle= \varepsilon\langle
\varphi,J'(W_0)\rangle-\langle \varphi^{\varepsilon},J'(W_0)\rangle+
\langle \varphi_{\varepsilon},J'(W_0)\rangle,
$$
so that
$$
\langle \varphi,J'(W_0)\rangle \geq {1\over\varepsilon}[\langle
\varphi^{\varepsilon},J'(W_0)\rangle-\langle
\varphi_{\varepsilon},J'(W_0)\rangle].
$$
Now, from $W^t$ is a supersolution to \eqref{5}, we get
\begin{align*}
&\langle \varphi^{\varepsilon},J'(W_0)\rangle\\
&= \langle \varphi^{\varepsilon},J'(W^t)\rangle+
       \langle \varphi^{\varepsilon},J'(W_0)-J'(W^t)\rangle\\
&\geq  \langle \varphi^{\varepsilon},J'(W_0)-J'(W^t)\rangle\\
&=  \int_{\Omega}[(u_0-u^t)'(u_0+\varepsilon\varphi_1-u^t)'+
                        (v_0-v^t)''(v_0+\varepsilon\varphi_2-v^t)'']\\
&\quad  -\int_{\Omega}[f_1(x,W_0)-f_1(x,W^t)](u_0+\varepsilon\varphi_1-u^t)\\
&\quad  -\int_{\Omega}[f_2(x,W_0)-f_2(x,W^t)](v_0+\varepsilon\varphi_2-v^t)\\
&\geq  \varepsilon\int_{\Omega}[(u_0-u^t)'\varphi_1'+(v_0-v^t)''\varphi_2'']\\
&\quad  -\varepsilon\int_{\Omega}|f_1(x,W_0)-f_1(x,W^t)||\varphi_1|
            -\varepsilon\int_{\Omega}|f_2(x,W_0)-f_2(x,W^t)||\varphi_2|
\end{align*}
where $\Omega=\{x\in(0,\pi);W_0(x)+\varepsilon\varphi(x)\geq
W^t(x)>W_0(x)\}$. Note that meas$(\Omega)\to 0$ as
$\varepsilon \to 0$. Hence by absolute continuity of the
Lebesgue integral we obtain that
$$
\langle \varphi^{\varepsilon},J'(W_0)\rangle \geq o(\varepsilon)
$$
where $o(\varepsilon)/\varepsilon\to 0$ as
$\varepsilon \to 0$. Similarly, we conclude that
$\langle \varphi_{\varepsilon},J'(W_0)\rangle \leq o(\varepsilon)$;
thus
$$
\langle \varphi,J'(W_0)\rangle \geq 0
$$
for all $\varphi \in C_0^{\infty}(0,\pi)$. Reversing the sign of
$\varphi$ and since $C_0^{\infty}(0,\pi)$ is dense in $X$ we finally
get that $J'(W_0)=0$, i.e. $W_0$ is a weak solution to \eqref{5}.
Then using (\ref{30}) and a Maximum Principle Lemma \ref{lem3.2},  we claim
that $W_0$ is a local minimum of $J$.

Suppose by contradiction that $W_0$ is not a local minimum, then for
every $\varepsilon>0$ there is
$\widetilde{W_\varepsilon}\in\overline{B_\varepsilon(W_0)}$ (a ball
of radius $\varepsilon$ around $W_0\in X$) such that
$J(\widetilde{W_\varepsilon})<J(W_0)$. We know that
$\overline{B_\varepsilon(W_0)}$ is weaker sequentially compact in
$X$ and $J$ is weakly lower semi-continuous, therefore there is
$\widehat{W_\varepsilon}\in\overline{B_\varepsilon(W_0)}$ such that
$$
J(\widehat{W_\varepsilon})=\inf_{\overline{B_\varepsilon(W_0)}}J
\leq J(\widetilde{W_\varepsilon})<J(W_0),
$$
and
$\langle J'(\widehat{W_\varepsilon}),
\widehat{W_\varepsilon}-W_0\rangle \leq 0$,
or
$$
J'(\widehat{W_\varepsilon})=\lambda_\varepsilon(\widehat{W_\varepsilon}-W_0)
 \quad\text{with } \lambda_\varepsilon\leq 0,
$$
namely
\begin{equation}\label{w1}
\begin{aligned}
&\int_0^\pi(\widehat{u_\varepsilon}'\psi_1'+\widehat{v_\varepsilon}''\psi_2'')
-\int_0^\pi[f_1(x,\widehat{u_\varepsilon},\widehat{v_\varepsilon})\psi_1+
f_2(x,\widehat{u_\varepsilon},\widehat{v_\varepsilon})\psi_2]
\\
&-\int_0^\pi[(t_1\sin x+h_1)\psi_1+(t_2\sin x+h_2)\psi_2]\\
&=\lambda_\varepsilon[(\widehat{u_\varepsilon}-u_0)
\psi_1+(\widehat{v_\varepsilon}-v_0)\psi_2].
\end{aligned}
\end{equation}
On the other hand, from Definition \ref{def3.1} we have
\begin{equation}\label{w2}
\begin{aligned}
&\int_0^\pi(u_t'\psi_1'+v_t''\psi_2'')-\int_0^\pi[f_1(x,u_t,v_t)\psi_1+
f_2(x,u_t,v_t)\psi_2]\\
&-\int_0^\pi[(t_1\sin x+h_1)\psi_1+(t_2\sin x+h_2)\psi_2] \leq 0,
\end{aligned}
\end{equation}
and
\begin{equation}\label{w3}
\begin{aligned}
&\int_0^\pi({u^t}'\psi_1'+{v^t}''\psi_2'')-\int_0^\pi[f_1(x,{u^t},{v^t})\psi_1
+f_2(x,{u^t},{v^t})\psi_2] \\
&-\int_0^\pi[(t_1\sin x+h_1)\psi_1+(t_2\sin x+h_2)\psi_2] \geq 0.
\end{aligned}
\end{equation}
 From (\ref{w1})--(\ref{w2}), we obtain
\begin{align*}
&\int_0^\pi[(\widehat{u_\varepsilon}'-u_t')\psi_1'+(\widehat{v_\varepsilon}''-v_t'')\psi_2'']\\
&-\int_0^\pi[(f_1(x,\widehat{W_\varepsilon})-f_1(x,W_t))\psi_1+(f_2(x,\widehat{W_\varepsilon})-
f_2(x,W_t))\psi_2]\\
&\geq\lambda_\varepsilon[(\widehat{u_\varepsilon}-u_t+u_t-u_0)\psi_1+(\widehat{v_\varepsilon}-v_t+v_t-v_0)\psi_2].
\end{align*}
This implies
\begin{gather*}
-(\widehat{u_\varepsilon}-u_t)''\geq
f_1(x,\widehat{W_\varepsilon})-f_1(x,W_t)+
\lambda_\varepsilon(\widehat{u_\varepsilon}-u_t)+\lambda_\varepsilon(u_t-u_0),
\\
(\widehat{v_\varepsilon}-v_t)^{(4)}\geq
f_2(x,\widehat{W_\varepsilon})-f_2(x,W_t)+
\lambda_\varepsilon(\widehat{v_\varepsilon}-v_t)
+\lambda_\varepsilon(v_t-v_0).
\end{gather*}
Then from (\ref{30}) we obtain
\[
\begin{pmatrix}
-(\widehat{u_\varepsilon}-u_t)''\\
(\widehat{v_\varepsilon}-v_t)^{(4)}
\end{pmatrix}
 \geq
A(x)(\widehat{W_\varepsilon}-W_t)+\lambda_\varepsilon(\widehat{W_\varepsilon}
-W_t),
\]
note that $\lambda_\varepsilon\leq 0$, and by using Lemma \ref{lem3.2}
 we obtain
$$
\widehat{W_\varepsilon}-W_t\geq 0,\quad\text{or}\quad
W_t\leq\widehat{W_\varepsilon}.
$$
Similarly, from (\ref{w3})$-$(\ref{w1}), we can obtain
$$
\widehat{W_\varepsilon}\leq W^t.
$$
Which contradicts
$J(W_0)=\inf_M J(W)$.

Finally, since $J$ is not bounded from below, a weaker form of the
Mountain Pass Theorem can be used to find another solution
$W_1\not=W_0$ of \eqref{5}. Then  result (2) is
proved.
\end{proof}

\section{Example: A piecewise linear problem}

Consider the  system
\begin{equation} \label{40}
\begin{gathered}
      -u''=k_1u^++\epsilon v^++t_1\sin x+h_1(x),
       \quad\text{in } (0,\pi), \\
      v^{(4)}=\epsilon u^++k_2 v^++t_2\sin x+h_2(x),
       \quad\text{in } (0,\pi), \\
      u(0)=u(\pi)=0,  \\
      v(0)=v(\pi)=v''(0)=v''(\pi)=0.
\end{gathered}
\end{equation}
Where  $\epsilon$ and $k_1,k_2$ are constants, $t_1,t_2$ are
parameters and $h_1,h_2\in C[0,\pi]$ are fixed functions with
$\int_0^\pi h_1\sin x=\int_0^\pi h_2\sin x=0$. This  problem is
similar to system (\ref{4}).


\begin{theorem} \label{thm4.1}
Suppose that $k_1>1$, $k_2>1$ and $\epsilon\geq0$.
Then there exists a curve $\Gamma$ splitting
$\mathbb{R}^2$ into two unbounded
components $N$ and $E$ such that:
\begin{itemize}
\item[(1)]  for each $t=(t_1,t_2)\in N$, \eqref{40} has no solution;
\item[(2)]  for each $t=(t_1,t_2)\in E$, \eqref{40} has at least two
solutions.
\end{itemize}
\end{theorem}

\begin{proof}  Let
\[
\overline{A}=\begin{pmatrix}
k_1 & 0\\
0 & k_2\end{pmatrix}, \quad
\underline{A}=\begin{pmatrix}
0 & 0\\
0 & 0 \end{pmatrix}.
\]
Then we can easily verify that the conditions of Theorem \ref{thm3.7} hold
and therefore the results are follow.
\end{proof}

\begin{remark} \label{4.2} \rm (1)  Denote by $\mu_i$ ($i=1,2$) the
eigenvalues of matrix
\[
A=\begin{pmatrix}
k_1 & \epsilon\\
\epsilon & k_2 \end{pmatrix}
\]
and let $\mu_1\leq\mu_2$. It can be shown that $\mu_2>1$ since
$k_1>1$ and $k_2>1$.

 (2)  This result gives a partial answer
to Question 1 and Question 2 that were posted in \cite{DM} and stated
in Section 1.
\end{remark}

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\end{document}
