a(t). \end{cases} \] Consider the boundary-value problem \[ u''(t)+mu(t)\equiv L_m\,u(t)=q(t), \quad u(0)=u(2\pi),\quad u'(0)=u'(2\pi), \] with $q\in L^1[0,1]$. Since $m<0$, the operator $L_m$ is invertible so that we can write its unique solution $u$ as $u=L_m^{-1}q$, and by the Maximum principle \ref{mp1} we know that if $q(t)\ge0$ then $u(t)\le0$. Since $\alpha(t)=L_m^{-1}a(t)$, $\beta(t)=L_m^{-1}b(t)$ and $b(t)\le p(t,x(t))\le a(t)$, we have \[ \alpha(t)\le L_m^{-1}p(t,x(t))\le \beta(t). \] Let us consider the modified problem \begin{equation} \label{modif} \begin{gathered} \begin{aligned} x''(t)+Mx(t)=(Fx)(t) &\equiv f\left(t,L_m^{-1}p(t,x(t)),p(t,x(t))-mL_m^{-1}p(t,x(t))\right)\\ &\quad +(m+M)p(t,x(t))-m^2L_m^{-1}p(t,x(t)), \end{aligned} \\ x(0)=x(2\pi),\quad x'(0)=x'(2\pi). \end{gathered} \end{equation} Considering the operator $\Phi:C[0,2\pi]\to C[0,2\pi]$ with $\Phi x=L_M^{-1}(Fx)$, since $p(t,x(t))$ and $L_m^{-1}p(t,x(t))$ are bounded and $f$ is a Carath\'eodory function, there exists a $L^1[0,2\pi]$ function $g(t)$ such that $|(Fx)(t)|\le g(t)$ for a.e. $t\in[0,2\pi]$. Therefore, applying Leray-Schauder's fixed point Theorem, we can conclude that $\Phi$ has a fixed point $x(t)$ which is a solution of the modified problem \eqref{modif}. \begin{proposition}\label{prop1} Let $x(t)$ be a solution of the modified problem \eqref{modif}. Assuming (H1), for given lower and upper solutions $\alpha(t)$ and $\beta(t)$, with $\alpha\le \beta$ for all $t\in[0,2\pi]$, we have \[ b(t)\le x(t)\le a(t). \] \end{proposition} \begin{proof} We will only prove that $x(t)\le a(t)$, since the other inequality can be obtained with similar arguments. We have \begin{gather*} a''(t)+Ma(t)\le f\left(t,L_m^{-1}a(t),a(t)-mL_m^{-1}a(t)\right)+(m+M)a(t) -m^2L_m^{-1}a(t), \\ a(0)=a(2\pi),\quad a'(0)\le a'(2\pi). \end{gather*} Setting $y(t)=x(t)-a(t)$, we have $y(0)=y(2\pi)$, $y'(0)\ge y'(2\pi)$. Suppose towards a contradiction that there exists $t_0\in [0,2\pi]$ such that $y(t_0)>0$. If $y(t)>0$ for all $t$, we have $x(t)>a(t)$ and, therefore, $p(t,x(t))=a(t)$, so \begin{align*} x''(t)+Mx(t)=& f\left(t,L_m^{-1}a(t),a(t)-mL_m^{-1}a(t)\right)+(m+M)a(t) -m^2L_m^{-1}a(t) \\ \ge & a''(t)+Ma(t), \end{align*} which is a contradiction, since by the Maximum Principle \ref{mp1} we would have $y(t)\le0$. Otherwise, considering if necessary the periodic extension of $y(t)$, there exists an interval $[p,q]$ with $q-p<2\pi$ such that $y(p)=y(q)=0$, $y'(p)\ge 0 \ge y'(q)$, $y(t)>0$ for $t\in\,]p,q[$, possibly with $p<00$ with $D<4C+1/4$ and $D^2>4C$, such that \begin{equation} f\left(t,u_2,v_1\right)-f\left(t,u_1,v_2\right)\ge -C\left(u_2-u_1\right)-D\left(v_1-v_2\right) \end{equation} for a.e. $t\in[0,2\pi]$, $\alpha(t)\le u_1\le u_2\le\beta(t)$, $v_1\le v_2$. \end{itemize} \begin{remark} \label{rmk3.5} \rm If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in (H1') is equivalent to $\frac{\partial f}{\partial u}\ge -C$ and $\frac{\partial f}{\partial v}\le -D$. \end{remark} Let $0-4C-1/4$, such that \begin{equation} f\left(t,u_1,v_2\right)-f\left(t,u_2,v_1\right)\ge -C\left(u_1-u_2\right)+D\left(v_2-v_1\right) \end{equation} for a.e. $t\in[0,2\pi]$, $\beta(t)\le u_1\le u_2\le\alpha(t)$, $v_1\le v_2$. \end{itemize} \begin{remark} \label{rmk4.1} \rm If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in (H2) is equivalent to $\frac{\partial f}{\partial u}\le -C$ and $\frac{\partial f}{\partial v}\ge D$. \end{remark} Let $M<0$ and $0 0$. If $y(t)>0$ for all $t$, we have (noting that $b(t)-m\beta(t)\ge a(t)-m\alpha(t)$) \begin{align*} y''(t)+My(t)\ge&f(t,\beta(t),b(t)-m\beta(t))-f(t,\alpha(t),a(t)-m\alpha(t))\\ &+(m+M)y(t)-m^2(\beta(t)-\alpha(t))\ge0, \end{align*} and this is a contradiction, since by the Maximum Principle \ref{mp1} we would have $y(t)\le0$. Otherwise, considering if necessary the periodic extension of $y(t)$, there exists an interval $[p,q]$ with $q-p<2\pi$ such that $y(p)=y(q)=0$, $y'(p)\ge 0 \ge y'(q)$, $y(t)>0$ for $t\in\,]p,q[$, possibly with $p<0 a(t). \end{cases} \] Consider the boundary-value problem \[ u''(t)+mu(t)\equiv L_m\,u(t)=q(t), \quad u(0)=u(2\pi),\quad u'(0)=u'(2\pi), \] with $q\in L^1[0,1]$. Since $m<1$ , the operator $L_m$ is invertible so that we can write its unique solution $u$ as $u=L_m^{-1}q$, and by the Maximum principle \ref{mp2} we know that if $q(t)\ge0$, then $u(t)\ge0$. Since $\alpha(t)=L_m^{-1}a(t)$, $\beta(t)=L_m^{-1}b(t)$ and $b(t)\le p(t,x(t))\le a(t)$, we have \[ \beta(t)\le L_m^{-1}p(t,x(t))\le \alpha(t). \] Let us consider the modified problem \begin{equation} \label{modif2} \begin{gathered} \begin{aligned} x''(t)+Mx(t)=(Fx)(t)&\equiv f\left(t,L_m^{-1}p(t,x(t)),p(t,x(t)) -mL_m^{-1}p(t,x(t))\right)\\ &\quad +(m+M)p(t,x(t))-m^2L_m^{-1}p(t,x(t)), \end{aligned}\\ x(0)=x(2\pi),\quad x'(0)=x'(2\pi). \end{gathered} \end{equation} Considering the operator $\Phi:C[0,2\pi]\to C[0,2\pi]$ with $\Phi x=L_M^{-1}(Fx)$ since $p(t,x(t))$ and $L_m^{-1}p(t,x(t))$ are bounded and $f$ is a Carathéodory function, there exists a $L^1[0,2\pi]$ function $g(t)$ such that $|(Fx)(t)|\le g(t)$ for a.e. $t\in[0,2\pi]$. Therefore, applying Leray-Schauder's fixed point Theorem, we can conclude that $\Phi$ has a fixed point $x(t)$ which is a solution of the modified problem \eqref{modif2}. \begin{proposition} \label{prop4.3} Let $x(t)$ be a solution of the modified problem \eqref{modif2}. Assuming (H2), for given lower and upper solutions $\alpha$ and $\beta$, with $\alpha\le \beta$, we have \[ b(t)\le x(t)\le a(t). \] \end{proposition} The proof of the above proposition is similar to the one of proposition \ref{prop1}. \begin{theorem} \label{thm4.4} Assuming (H2), for given lower and upper solutions $\alpha$ and $\beta$, with $\beta\le \alpha$, the boundary value problem \eqref{eq}--\eqref{bcper} has a solution $u(t)\in W^{4,1}(0,2\pi)$ such that $\beta\le u\le\alpha$. \end{theorem} The proof of the above theorem is similar to the one of theorem \ref{teo1}. We can reach a similar conclusion, assuming the following hypothesis \begin{itemize} \item[(H2')]there exist constants $C,D$ with $C<0$ and $D<4C+1/4$ such that \begin{equation} f\left(t,u_1,v_1\right)-f\left(t,u_2,v_2\right)\ge -C\left(u_1-u_2\right)-D\left(v_1-v_2\right) \end{equation} for a.e. $t\in[0,2\pi]$, $\beta(t)\le u_1\le u_2\le\alpha(t)$, $v_1\le v_2$. \end{itemize} \begin{remark} \label{rmk4.5} \rm If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in (H2') is equivalent to $\frac{\partial f}{\partial u}\le -C$ and $\frac{\partial f}{\partial v}\le -D$. \end{remark} Let $m<0$ and $00$, with $D>-C-1$, $D^2>4C$ , and \begin{equation} f\left(t,u_2,v_2\right)-f\left(t,u_1,v_1\right)\ge -C\left(u_2-u_1\right)+D\left(v_2-v_1\right) \end{equation} for a.e. $t\in[0,2\pi]$, $\alpha(t)\le u_1\le u_2\le\beta(t)$, $v_1\le v_2$. \end{itemize} \begin{remark} \label{rmk5.1} \rm If $f(t,u,v)$ is a $C^1$ function in $(u,v)$, the inequality in (H3) is equivalent to $\frac{\partial f}{\partial u}\ge -C$ and $\frac{\partial f}{\partial v}\ge D$. \end{remark} Let $m<0$ and $M<1$ be the two roots of the equation $x^2+Dx+C=0$ (note that $C=mM,\,D=-(m+M)$). Defining $a(t)=\alpha''(t)+m\alpha(t)$ and $b(t)=\beta''(t)+m\beta(t)$, we have the following result. \begin{proposition}\label{ab5} If $f$ is a $L^1$-Carath\'eodory function satisfying (H3) for $\alpha(t),\beta(t)$ lower and upper solutions such that $\alpha(t)\le\beta(t)$, then $b(t)\le a(t)$. \end{proposition} \begin{proof} Setting $y(t)=b(t)-a(t)$, we have $y(0)\le0$ and $y(\pi)\le0$. Suppose towards a contradiction that there exists $t_0\in [0,\pi]$ such that $y(t_0)>0$. The result follows using similar arguments to the ones of Proposition \ref{ab1} (using Maximum principle \ref{mp3} instead). \end{proof} Let \[ p(t,x)= \begin{cases} b(t), & x a(t). \end{cases} \] Consider the boundary value problem \[ u''(t)+mu(t)\equiv L_m\,u(t)=q(t), \quad u(0)=0,\;\quad u(\pi)=0, \] with $q\in L^1[0,1]$. Since $m<0$, the operator $L_m$ is invertible so that we can write its unique solution $u$ as $u=L_m^{-1}q$. Let us define $\tilde{a}(t)$ such that $\tilde{a}''+m\tilde{a}=0$, $\tilde{a}(0)=\alpha(0)$, $\tilde{a}(\pi)=\alpha(\pi)$, and $\tilde{b}(t)$ such that $\tilde{b}''+m\tilde{b}=0$, $\tilde{b}(0)=\beta(0)$, $\tilde{b}(\pi)=\beta(\pi)$. It is obvious that $\tilde{a}(t)\le0$ and $\tilde{b}(t)\ge0$ for all $t\in [0,\pi]$. We have $\alpha(t)=L_m^{-1}a(t)+\tilde{a}(t)$ and $\beta(t)=L_m^{-1}b(t)+\tilde{b}(t)$, then, by the Maximum principle \ref{mp3}, \[ \alpha(t)\le L_m^{-1}p(t,x(t))\le \beta(t). \] Proceeding in a similar way as in the previous cases, we can reach an analogue conclusion: \begin{theorem} Assuming (H3), for given lower and upper solutions $\alpha$ and $\beta$, with $\alpha\le \beta$, the boundary value problem \eqref{eq2}--\eqref{bcss} has a solution $u(t)\in W^{4,1}(0,\pi)$ such that $\alpha\le u\le\beta$. \end{theorem} Let us now consider a hypothesis somehow different from the ones considered above. Suppose that \begin{itemize} \item[(H4)] There exist constants $C,D$ with $C>0$, $00$ and $D-m<1$. Defining $a(t)=\alpha''(t)+m\alpha(t)$ and $b(t)=\beta''(t)+m\beta(t)$, we have the following result. \begin{proposition}\label{ab6} If $f$ is a $L^1$-Carath\'eodory function satisfying (H4) for $\alpha(t),\beta(t)$ lower and upper solutions such that $\alpha(t)\le\beta(t)$, then $b(t)\le a(t)$. \end{proposition} \begin{proof} Setting $y(t)=b(t)-a(t)$, we have $y(0)\le0$, $y(\pi)\le0$ and \begin{align*} y''(t)=&\beta^{(4)}(t)-\alpha^{(4)}(t)+m(\beta''(t)-\alpha''(t)) +m^2(\beta(t)-\alpha(t))-m^2(\beta(t)-\alpha(t))\\ \ge&f(t,\beta(t),b(t)-m\beta(t))-f(t,\alpha(t),a(t)-m\alpha(t))+my(t) -m^2(\beta(t)-\alpha(t)) \\ \ge&C(\beta(t)-\alpha(t))-D\left|y(t)-m(\beta(t)-\alpha(t))\right| +my(t)-m^2(\beta(t)-\alpha(t)) \\ \ge&(C+Dm-m^2)(\beta(t)-\alpha(t))-D\left|y(t)\right|+my(t). \end{align*} To apply Maximum principle \ref{mp4}, we rewrite the previous inequality as \[ y''(t)+(D-m)y^+(t)+(D+m)y^-(t)\ge0 \] and conclude that $y(t)\le 0$. \end{proof} Proceeding in a similar way as above, we can reach an analogue conclusion: \begin{theorem} \label{thm5.6} Assuming (H4), for given lower and upper solutions $\alpha$ and $\beta$, with $\alpha\le \beta$, the boundary value problem \eqref{eq2}--\eqref{bcss} has a solution $u(t)\in W^{4,1}(0,\pi)$ such that $\alpha\le u\le\beta$. \end{theorem} \subsection*{Acknowledgments} This work was supported by grant POCTI/Mat/57258/2004 from Funda\c{c}\~ao para a Ci\^encia e Tecnologia, and by grant POCTI-ISFL-1-209 from Centro de Matem\'atica e Aplica\c{c}\~{o}es Fundamentais. The authors would like to thank the anonymous referee for his/her careful reading of the original manuscript. \begin{thebibliography}{00} \bibitem {BW02} Zhanbing Bai and Haiyan Wang; \emph{On positive solutions of some nonlinear fourth order beam equations}, J. Math. Anal. Appl. \textbf{270} (2002) 357-368. \bibitem {C94} A. Cabada; \emph{The method of lower and upper solutions for nth order periodic boundary value problems}, J. Appl. Math. Stochastic Anal. \textbf{7}, No. 1, 33-47 (1994). \bibitem {CCS06} A. Cabada, J. A. Cid and L. Sanchez; \emph{Positivity and lower and upper solutions for fourth order boundary value problems}, to appear in Nonlinear Analysis TMA. \bibitem{GWJH06} H. Gao, S. Weng, D. Jiang, X. Hou; \emph{On second order periodic boundary value problems with upper and lower solutions in the reversed order}, Electronic Journal of Diff. Eq., \textbf{25} (2006) 1--8. \bibitem{HS06} P. Habets, L. Sanchez; \emph{A monotone method for fourth order boundary value problems involving a factorizable linear operator}, Portugaliae Mathematica \textbf{64} (2007), 255-279. \bibitem{dJwGaW02} Daqing Jiang, Wenjie Gao and Aying Wan; \emph{A monotone method for constructing extremal solutions to fourth-order periodic boundary value problems}, Appl. Math. Comput. \textbf{132} (2002), 411-421. \bibitem{Li} Yongxiang Li; \emph{Positive solutions of fourth order boundary value problems with two parameters}, J. Math. Anal. Appl. \textbf{281} (2003), 477-484. \bibitem{L03} Yongxiang Li; \emph{Positive solutions of fourth-order periodic boundary value problems}, Nonlinear Analysis \textbf{54} (2003), 1069-1078. \bibitem{LiuLi05} Xi-Lan Liu and Wan-Tong Li; \emph{Positive solutions of the nonlinear fourth order beam equation with three parameters}, J. Math. Anal. Appl. \textbf{303} (2005), 150-163. \bibitem {R02} B. P. Rynne; \emph{Infinitely many solutions of superlinear fourth order boundary value problems}, Top. Meth. Nonl. Anal. \textbf{19} (2002), 303-312. \bibitem {Y04} Qingliu Yao; \emph{On the positive solutions of a nonlinear fourth order boundary value problem with two parameters}, Applicable Analysis \textbf{83} (2004), 97-107. \end{thebibliography} \end{document}