\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 78, pp. 1--8.\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 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/78\hfil Positive solutions for beam equation] {Positive solutions for the beam equation under certain boundary conditions} \author[B. Yang\hfil EJDE-2005/78\hfilneg] {Bo Yang} \address{Bo Yang \hfill\break Department of Mathematics, Kennesaw State University, Kennesaw, GA 30144, USA} \email{byang@kennesaw.edu} \date{} \thanks{Submitted April 29, 2005. Published July 8, 2005.} \subjclass[2000]{34B18} \keywords{Fixed point theorem; cone; nonlinear boundary-value problem; \hfill\break\indent positive solution} \begin{abstract} We consider a boundary-value problem for the beam equation, in which the boundary conditions mean that the beam is embedded at one end and fastened with a sliding clamp at the other end. Some priori estimates to the positive solutions for the boundary-value problem are obtained. Sufficient conditions for the existence and nonexistence of positive solutions for the boundary-value problem are established. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \section{Introduction} In this paper, we consider the fourth order beam equation $$u''''(t) = g(t)f(u(t)) , \quad 0 \le t \le 1, \label{ee}$$ together with boundary conditions $$u(0) = u'(0) = u'(1) = u'''(1) = 0. \label{bb}$$ Throughout this paper, we assume that \begin{itemize} \item[(H1)] $f: [0,\infty) \to [0,\infty)$ is continuous \item[(H2)] $g: [0,1] \to [0,\infty)$ is a continuous function such that $\int_0^1 g(t)\,dt > 0$. \end{itemize} Equation \eqref{ee} and the boundary conditions \eqref{bb} arise from the study of elasticity and have definite physical meanings. Equation \eqref{ee} describes the deflection or deformation of an elastic beam under a certain force. The boundary conditions \eqref{bb} mean that the beam is embedded at the end $t=0$, and fastened with a sliding clamp at the end $t=1$. In 1989, Gupta \cite{Gu1} considered the boundary-value problem \begin{gather} u''''(t)+f(t)u(t)=e(t),\quad 00 $for$ t\in (0,1) $. A beam can have different shapes under different boundary constraints. One of the purposes of this paper is to make some estimates to the shape of the beam under boundary conditions \eqref{bb}. This paper is organized as follows. In Section 2, we give the Green's function for the problem \eqref{ee}-\eqref{bb}, state the Krasnosel'skii's fixed point theorem, and fix some notations. In Section 3, we present some priori estimates to positive solutions to the problem \eqref{ee}-\eqref{bb}. In Sections 4 and 5, we establish some existence and nonexistence results for positive solutions to the problem \eqref{ee}-\eqref{bb}. \section{Preliminaries} The Green's function$ G:[0,1]\times[0,1]\to [0,\infty) $for the problem \eqref{ee}-\eqref{bb} is $$G(t,s) =\begin{cases} \frac{1}{12} t^2 ( 6s - 3s^2 - 2t), & \mbox{if } 0\le t\le s\le 1, \\[3pt] \frac{1}{12} s^2 ( 6t -3t^2 - 2s), & \mbox{if } 0\le s\le t\le 1. \end{cases}$$ Then problem \eqref{ee}-\eqref{bb} is equivalent to the integral equation $$u(t) = \int_0^1 G(t,s) g(s) f(u(s))\,ds,\quad 0\leq t\leq 1. \label{int}$$ It is easy to verify that$G$is a continuous function, and$G(t,s)>0$if$t,s\in(0,1)$. We will need the following fixed point theorem, which is due to Krasnosel'skii \cite{K}, to prove some of our results. \begin{theorem} \label{thmK} Let$ (X,\|\cdot\|) $be Banach space over the reals, and let$P\subset X $be a cone in$ X $. Let$ H_1 $and$ H_2 $be real numbers such that$ H_2>H_1>0 $, and let $$\Omega_i = \{ v\in X\ |\ \|v\| < H_i\}, \quad i=1,2.$$ If$L:P\cap(\overline{\Omega_2}-\Omega_1 )\to P$is a completely continuous operator such that, either \begin{itemize} \item[(K1)]$ \|Lv\|\leq\|v\| $if$ v\in P\cap\partial \Omega_1$, and$ \|Lv\|\geq\|v\| $if$ v\in P\cap\partial \Omega_2$, or \item[(K2)]$ \|Lv\|\geq\|v\| $if$ v\in P\cap\partial \Omega_1 $, and$ \|Lv\|\leq\|v\| $if$ v \in P\cap\partial \Omega_2 $. \end{itemize} Then$ L $has a fixed point in$ P\cap(\overline{\Omega_2}-\Omega_1) $. \end{theorem} For the rest of this paper, we let$ X = C[0,1] $be with norm $$\|v\|=\max_{t\in [0,1]} |v(t)|,\quad \forall v\in X.$$ Clearly$X$is a Banach space. We define$Y=\{v\in X\,|\,v(t)\ge 0\mbox{ for }0\le t\le 1\}$, and define the operator$ T : Y \to X $by $$(Tu)(t) = \int_0^1 G(t,s)g(s)f(u(s))\,ds,\quad 0\leq t\leq 1. \label{op}$$ It is clear that if (H1) and (H2) hold, then$ T: Y\to Y $is a completely continuous operator. We also define the constants \begin{gather*} F_0 = \limsup_{x\to 0^+}\frac{f(x)}{x}, \quad f_0=\liminf_{x\to 0^+}\frac{f(x)}{x}, \\ F_{\infty} = \limsup_{x\to +\infty}\frac{f(x)}{x}, \quad f_{\infty}=\liminf_{x\to +\infty}\frac{f(x)}{x}. \end{gather*} These constants, which are associated with the function$f$, will be used in Sections 4 and 5. \section{Estimates for Positive Solutions} In this section, we shall give some estimates for positive solutions of the problem \eqref{ee}-\eqref{bb}. To this purpose, we define the functions$ a:[0,1]\to [0,1] $,$ b:[0,1]\to [0,1] $, and$ c:[0,1]\to [0,1] $by $$a(t) = 3t^2 - 2t^3, \quad b(t) = 2t-t^2 ,\quad c(t) = 4t^2 - 4t^3 + t^4.$$ It is easy to see that$ b(t)\ge c(t)\ge a(t)\ge t^2 $for$0\le t\le 1$. \begin{lemma} \label{lem3.1} If$ u \in C^4[0,1] $satisfies the boundary conditions \eqref{bb}, and $$u''''(t)\ge 0\quad\mbox{for}\quad 0\le t\le 1, \label{2101}$$ then $$u'''(t)\le 0,\quad u'(t)\ge 0, \quad u(t)\ge 0 \quad\mbox{for}\quad 0\le t\le 1. \label{2102}$$ \end{lemma} \begin{proof} Note that \eqref{2101} implies that$u'''$is nondecreasing. Since$ u'''(1)=0$, we have$u'''(t)\le 0$on$ [0,1]$, which means that$u'$is concave downward on$[0,1]$. Since$u'(0)=u'(1)=0$, we have$u'(t)\ge 0 $on$[0,1]$. Since$u(0)=0$, we have$u(t)\ge 0 $on$[0,1]$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.2} If$ u \in C^4[0,1] $satisfies \eqref{bb} and \eqref{2101}, then $$u(t)\ge a(t)u(1) \quad \mbox{ for }\quad 0\le t\le 1. \label{2201}$$ \end{lemma} \begin{proof} If$ u \in C^4[0,1] $satisfies \eqref{bb} and \eqref{2101}, then$ u(0)=0 $,$ u(1)\ge 0 $, and$ u'(t) \ge 0 $for$ 0\le t\le 1 $. If$ u(1)=0 $, then$ u(t)\equiv 0 $, and it is easy to see that \eqref{2201} is true in this case. Now we prove \eqref{2201} when$ u(1)>0 $. Without loss of generality, we assume that$ u(1)=1 $. If we define $$h(t) = u(t) -a(t)u(1) = u(t) - (3t^2-2t^3),\quad 0\le t\le 1,$$ then \begin{equation*} \begin{gathered} h'(t)= u'(t)-(6t-6t^2), \quad h''(t) = u''(t)-(6-12t), \\ h'''(t) = u'''(t)+12, \end{gathered} \end{equation*} % $$h''''(t) = u''''(t)\geq 0,\quad 0\le t \le 1. \label{2202}$$ To prove the lemma, it suffices to show that$ h(t)\ge 0 $on$ [0,1] $. It is easy to see that$ h(0) = h(1) = 0$. By mean value theorem, there exists$ r_1\in (0,1) $such that$ h'(r_1)=0 $. It is also easy to see that$ h'(0) = h'(1) = 0$. Since$ h'(0) = h'(r_1) = h'(1) = 0 $, there exist$ r_2\in (0,r_1)$and$ t_2\in (r_1,1) $such that$ h''(r_2) = h''(t_2) = 0 $. Note that \eqref{2202} implies that$ h''$is concave upward. Since$ h''(r_2) = h''(t_2) = 0 $, we have $$h''(t)\geq 0 \ \ {\rm on}\ \ (0,r_2),\ \ h''(t)\leq 0 \ \ {\rm on}\ \ (r_2,t_2),\ \ {\rm and}\ \ h''(t)\geq 0 \ \ {\rm on}\ \ (t_2,1).$$ These inequalities, together with the fact that$ h'(0)=h'(r_1) = h'(1)=0 $, imply that $$h'(t)\geq 0 \text{ on } (0,r_1),\quad h'(t)\leq 0 \text{ on } (r_1,1).$$ Since$ h(0)=h(1)=0 $, we have$h(t)\geq 0$on$(0,1)$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.3} If$ u \in C^4[0,1] $satisfies \eqref{bb} and \eqref{2101}, then $$u(t) \leq u(1)b(t) \quad \mbox{ for }\quad t\in [0,1]. \label{2301}$$ \end{lemma} \begin{proof} Without loss of generality, we assume that$ u(1)=1$. If we define $$h(t)= b(t) u(1) -u(t) = 2t -t^2 - u(t),\quad 0\le t\le 1,$$ then \begin{equation*} h'(t)= 2-2t - u'(t),\quad h''(t) = -2 - u''(t), \end{equation*} $$h'''(t) = - u'''(t),\quad 0\le t \le 1. \label{2302}$$ It is easy to see that$ h(0) = h(1) = h'(1) = 0$. By mean value theorem, because$ h(0) = h(1) = 0 $, there exists$ r_1\in (0,1) $such that$ h'(r_1)=0 $. We see from \eqref{2302} and \eqref{2102} that$h'''(t)\ge 0 $on$[0,1]$, which implies that$ h'$is concave upward on$ [0,1]$. Since$ h'(r_1) = h'(1) = 0 $, we have $$h'(t)\geq 0 \text{ on } (0,r_1),\quad h'(t)\leq 0 \text{ on } (r_1,1).$$ Since$ h(0) = h(1) = 0 $, we have$h(t)\ge 0$on$(0,1)$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.4} If$ u \in C^4[0,1] $satisfies \eqref{bb} and \eqref{2101}, and$ u''''(t) $is nondecreasing on$ [0,1] $, then $$u(t) \leq u(1)c(t) \quad \mbox{ for }\quad t\in [0,1]. \label{2401}$$ \end{lemma} \begin{proof} Without loss of generality, we assume that$ u(1)=1 $. If we define $$h(t) = c(t)u(1)-u(t) = 4t^2-4t^3+t^4 - u(t),\quad 0\le t\le 1,$$ then \begin{equation*} \begin{gathered} h'(t) = 8t -12t^2 +4t^3 - u'(t), \quad h''(t) = 8-24t +12t^2 - u''(t), \\ h'''(t) = -24 +24 t - u'''(t), \end{gathered} \end{equation*} $$h''''(t) = 24 - u''''(t),\quad 0\le t \le 1. \label{2402}$$ It is easily seen that$ h(0) = h(1) = h'(0) = h'(1)=0 $. By the mean value theorem, because$ h(0) = h(1) = 0 $, there exists$ r_1 \in (0,1) $such that$ h'(r_1)=0 $. Since$ h'(0) = h'(r_1) = h'(1) = 0 $, there exist$ r_2 \in (0,r_1) $and$ t_2\in(r_1,1) $such that$ h''(r_2) = h''(t_2) = 0 $. As a consequence, there exists$ r_3\in(r_2,t_2) $such that$ h'''(r_3)=0 $. Note that$ u'''' $is nondecreasing by assumption. It follows from \eqref{2402} that$ h'''(t) $is concave downward. It is easy to see that$ h'''(1)=0 $. Because$ h'''(r_3) = h'''(1) = 0 $, we have $$h'''(t) \le 0 \mbox{ on } (0,r_3),\quad \mbox{and}\quad h'''(t) \ge 0 \mbox{ on } (r_3,1).$$ Since$ h''(r_2) = h''(t_2) = 0 $, we have $$h''(t) \ge 0 \mbox{ on } (0,r_2), \quad h''(t) \le 0 \mbox{ on } (r_2,t_2), \quad h''(t) \ge 0 \mbox{ on } (t_2,1).$$ Because$ h'(0) = h'(r_1) = h'(1) = 0 $, we have $$h'(t) \ge 0 \mbox{ for } t\in (0,r_1), \quad h'(t) \le 0 \mbox{ for } t\in (r_1,1).$$ Hence,$h(t) \ge 0 $for$ 0\le t\le 1$. The proof is complete. \end{proof} \begin{theorem} \label{thm3.5} Suppose that (H1) and (H2) hold. If$ u(t) $is a nonnegative solution to the problem \eqref{ee}-\eqref{bb}, then$ u(t) $satisfies \eqref{2102}, \eqref{2201}, and \eqref{2301}. \end{theorem} \begin{proof} If$ u(t) $is a nonnegative solution to the problem \eqref{ee}-\eqref{bb}, then$ u(t) $satisfies the boundary conditions \eqref{bb}, and $$u''''(t)=g(t)f(u(t))\ge 0,\quad 0\le t\le 1.$$ Now Theorem \ref{thm3.5} follows directly from Lemmas \ref{lem3.1}, \ref{lem3.2}, and \ref{lem3.3}. The proof is complete. \end{proof} \begin{theorem} \label{thm3.6} Suppose that (H1), (H2), and the following condition hold. \begin{itemize} \item[(H3)] Both$ f $and$ g $are nondecreasing functions. \end{itemize} If$ u(t) $is a nonnegative solution to the problem \eqref{ee}-\eqref{bb}, then$ u(t) $satisfies \eqref{2102}, \eqref{2201}, and \eqref{2401}. \end{theorem} \begin{proof} By Theorem \ref{thm3.5},$ u(t) $satisfies \eqref{2102} and \eqref{2201}. Therefore$ u(t) $is nondecreasing on$[0,1] $. It is obvious that$ u''''(t) = g(t)f(u(t)) \ge 0 $. By (H3), we have that$ u''''(t) = g(t)f(u(t)) $is nondecreasing on the interval$ [0,1] $. It follows directly from Lemma \ref{lem3.4} that$ u(t) satisfies \eqref{2401}. The proof is complete. \end{proof} \section{Existence and Nonexistence Results} First, we define some important constants: $$A = \int_0^1 G(1,s)g(s)a(s)\,ds, \quad B = \int_0^1 G(1,s)g(s)b(s)\,ds.$$ We also define \begin{align*} P = \big\{& v\in X : v(1)\geq 0,\ v(t)\mbox{ is nondecreasing on }[0,1],\\ & a(t)v(1)\leq v(t)\leq b(t) v(1) \mbox{ on } [0,1]\big\}. \end{align*} Clearly P $is a positive cone in$ X $. It is obvious that if$u\in P$, then$u(1)=\|u\|$. We see from Theorem \ref{thm3.5} that if$ u(t) $is a nonnegative solution to the problem \eqref{ee}-\eqref{bb}, then$ u\in P $. In a similar fashion to Theorem \ref{thm3.5}, we can show that$T(P)\subset P$. To find a positive solution to the problem \eqref{ee}-\eqref{bb}, we need only to find a fixed point$ u $of$ T $such that$ u\in P $and$ u(1)=\|u\|>0 $. The next two theorems provide sufficient conditions for the existence of at least one positive solution for the problem \eqref{ee}-\eqref{bb}. \begin{theorem} \label{thm4.1} Suppose that (H1) and (H2) hold. If$ B F_0 < 1 < A f_\infty $, then problem \eqref{ee}-\eqref{bb} has at least one positive solution. \end{theorem} \begin{proof} First, we choose$ \varepsilon >0 $such that$ (F_0+\varepsilon) B\leq 1 $. By the definition of$ F_0 $, there exists$ H_1 >0 $such that$ f(x)\leq (F_0+\varepsilon)x$for$ 0 < x\leq H_1 $. Now for each$ u\in P $with$ \|u\|=H_1 , we have \begin{align*} (Tu)(1) & = \int_0^1 G(1,s) g(s) f(u(s))\,ds\\ &\leq \int_0^1 G(1,s) g(s) (F_0+\varepsilon)u(s)\,ds\\ & \leq (F_0+\varepsilon)\|u\| \int_0^1 G(1,s) g(s) b(s)\,ds\\ & = (F_0+\varepsilon)\|u\| B \leq \|u\|, \end{align*} which means \|Tu\|\leq \|u\| $. Thus, if we let$ \Omega_1=\{ u\in X\ |\ \|u\|0 $and$ \tau \in (0,1/4) $such that $$\int_\tau^{1} G(1,s) g(s) a(s)\,ds \cdot(f_\infty-\delta) \ge 1.$$ There exists$ H_3 >2H_1 $such that$f(x)\geq (f_\infty-\delta)x \ \mbox{for}\ x\geq H_3$. Let$ H_2=H_3/{\tau^2} $. If$ u\in P $such that$ \|u\| = H_2 $, then for each$ t\in[\tau,1] $, we have $$u(t)\ge H_2 a(t)\ge H_2 t^2 \ge H_2 \tau^2 \ge H_3 .$$ Therefore, for each$ u\in P $with$ \|u\| = H_2 , we have \begin{align*} (Tu)(1) & = \int_0^{1} G(1,s) g(s) f(u(s))\,ds\\ &\geq \int_\tau^{1} G(1,s) g(s) f(u(s))\,ds\\ & \geq \int_\tau^{1} G(1,s) g(s) (f_\infty-\delta)u(s)\,ds\\ & \geq \int_\tau^{1} G(1,s) g(s) a(s)\, ds\cdot (f_\infty-\delta)\|u\| \geq \|u\|, \end{align*} which means \|Tu\|\geq \|u\| $. Thus, if we let$ \Omega_2=\{u\in X\ |\ \|u\|0 $, then the problem \eqref{ee}-\eqref{bb} has no positive solutions. \end{theorem} \begin{proof} Assume the contrary that$ u(t) $is a positive solution of the problem \eqref{ee}-\eqref{bb}. Then$u\in P $,$u(t)>0 $for$ 0 x $for all$x>0$, then the problem \eqref{ee}-\eqref{bb} has no positive solutions. \end{theorem} \section{More Existence and Nonexistence Results} In this section, we define a new constant $$C = \int_0^1 G(1,s)g(s)c(s)\,ds,$$ and define the positive cone$ Q $of$ X by \begin{align*} Q = \big\{& v\in X : v(1)\geq 0,\ v(t)\mbox{ is nondecreasing on } [0,1], \\ & a(t)v(1)\leq v(t)\leq c(t) v(1)\mbox{ on } [0,1]\big\}. \end{align*} It is obvious that ifu\in Q$, then$u(1)=\|u\|$. We see from Theorem \ref{thm3.6} that if (H1), (H2), and (H3) hold, and$ u(t) $is a nonnegative solution to the problem \eqref{ee}-\eqref{bb}, then$ u\in Q $. In a similar fashion to Theorem \ref{thm3.6}, we can show that if (H1), (H2), and (H3) hold, then$T(Q)\subset Q$. \begin{theorem} \label{thm5.1} Suppose that (H1), (H2), and (H3) hold. If either$ C F_0 < 1 < A f_\infty $or$ C F_{\infty} < 1 < A f_0 $, then problem \eqref{ee}-\eqref{bb} has at least one positive solution. \end{theorem} The proof of the above theorem is omitted, because it is very similar to that of Theorem \ref{thm4.1}. The only difference is that we use the positive cone$ Q $, instead of$ P $, in the proof of Theorem \ref{thm5.1}. \begin{theorem} \label{thm5.2} Suppose (H1), (H2), and (H3) hold. If$ C f(x)0$, then problem \eqref{ee}-\eqref{bb} has no positive solutions. \end{theorem} The proof of the above theorem is quite similar to that of Theorem \ref{thm4.3} and therefore omitted. \begin{example} \label{ex5.3} \rm Consider the beam equation $$u''''(t) = \lambda (t + 2t^2) u(t)(1+2u(t))/(1+u(t)),\quad 0\le t\le 1, \label{eee}$$ where$\lambda>0$is a parameter, together with the boundary conditions \eqref{bb}. In this example,$g(t)=t + 2t^2$and$f(u)=\lambda u(1+2u)/(1+u)$. It is easy to see that$ f_0 = F_0 =\lambda $,$ f_\infty = F_\infty = 2\lambda $, and $$\lambda x < f(x) < 2\lambda x \quad \mbox{for}\quad x > 0.$$ Calculations indicate that$ A = 47/756$,$B = 43/630$, and$C = 1937/30240$. By Theorem \ref{thm4.1}, if $$8.04 \approx1/(2A) < \lambda < 1/B\approx 14.56 ,$$ then the problem \eqref{eee}-\eqref{bb} has at least one positive solution. From Theorems \ref{thm4.3} and \ref{thm4.4} we see that if $$\lambda \le 1/(2B) \approx 7.326 \quad \mbox{or}\quad \lambda \ge 1/A\approx 16.085,$$ then the problem \eqref{eee}-\eqref{bb} has no positive solutions. Note that$ g(t) $is increasing on$[0,1]$, and$ f(u) $is increasing on$[0,+\infty)$. Therefore Theorems \ref{thm5.1} and \ref{thm5.2} apply. From Theorem \ref{thm5.1} we see that if $$8.04 \approx1/(2A) < \lambda < 1/C\approx 15.612 ,$$ then the problem \eqref{eee}-\eqref{bb} has at least one positive solution. From Theorem \ref{thm5.2} we see that if $$\lambda \le 1/(2C) \approx 7.806,$$ then the problem \eqref{eee}-\eqref{bb} has no positive solutions. This example shows that our existence and nonexistence results are quite sharp indeed. \end{example} \subsection*{Acknowledgment} The author is grateful to the anonymous referee for his/her valuable comments and suggestions. \begin{thebibliography}{99} \bibitem{A} R. P. Agarwal; \emph{On fourth-order boundary-value problems arising in beam analysis, Differential Integral Equations}, 2 (1989), 91--110. \bibitem{AOL} P. R. Agarwal, D. O'Regan, and V. Lakshmikantham; \emph{Singular$(p,n-p)$focal and$(n,p)\$ higher order boundary-value problems}, Nonlinear Anal., 42 (2000), no. 2, Ser. A: Theory Methods, 215--228. \bibitem{AD} D. R. Anderson and J. M. Davis; \emph{Multiple solutions and eigenvalues for third-order right focal boundary-value problems}, J. Math. Anal. Appl., 267 (2002), no. 1, 135--157. \bibitem{BW} Z. Bai and H. 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