\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 155, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/155\hfil Existence and uniqueness] {Existence and uniqueness of a positive solution for a third-order three-point boundary-value problem} \author[A. P. Palamides, N. M. Stavrakakis \hfil EJDE-2010/155\hfilneg] {Alex P. Palamides, Nikolaos M. Stavrakakis} % in alphabetical order \address{Alex P. Palamides \newline Technological Educational Institute of Piraeus, Department Electronic Computer Systems Engineering, Athens, Greece} \email{palamid@teipir.gr} \address{Nikolaos M. Stavrakakis \newline Department of Mathematics, National Technical University, Zografou Campus, 157 80 Athens, Greece} \email{nikolas@central.ntua.gr} \thanks{Submitted May 10, 2010. Published October 28, 2010.} \subjclass[2000]{34B10, 34B18, 34B15, 34G20} \keywords{Three point singular boundary value problem; positive solutions; \hfill\break\indent third order differential equation; existence; uniqueness; fixed points in cones; Green's functions} \begin{abstract} In this work we study a third-order three-point boundary-value problem (BVP). We derive sufficient conditions that guarantee the positivity of the solution of the corresponding linear BVP Then, based on the classical Guo-Krasnosel'skii's fixed point theorem, we obtain positive solutions to the nonlinear BVP. Additional hypotheses guarantee the uniqueness of the solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} In this article, we are concerned with a certain class of third-order differential equations, known as the three-point boundary-value problem (BVP), given by \begin{gathered} u'''(t)=-f(t,u(t)),\quad 0\eta, $G(t,s)=\begin{cases} u_2^{\ast }(t,s),& 0\leq t\leq s \\ v_2^{\ast }(t,s),& s\leq t\leq 1, \end{cases}$ where \begin{gather*} \begin{aligned} u_1^{\ast }(t,s) &=C_0\Big((1-2\eta -2q\eta +s^{2}+2qs)t^{2}-2(s-2s\eta +s^{2}\eta)t\\ &\quad -2q(s-2s\eta +s^{2}\eta )\Big) \end{aligned}\\ v_1^{\ast }(t,s)=C_0\Big((s^{2}+2qs)t^{2} -2(2qs\eta +s^{2}\eta )t -(s^{2}+2qs-4qs\eta -2s^{2}\eta )\Big), \\ u_2^{\ast }(t,s)=C_0\Big( (1+s^{2}-2s)t^{2}-2(\eta -2s\eta +s^{2}\eta )t -2q(\eta -2s\eta +s^{2}\eta )\Big), \\ \begin{aligned} v_2^{\ast }(t,s) &=C_0\Big( (2\eta +2q\eta +s^{2}-2s)t^{2}-2(\eta -s+2qs\eta +s^{2}\eta )t\\ &\quad -2q\eta +s^{2}-4qs\eta -2s^{2}\eta\Big), \end{aligned}\\ C_0=-\frac{1}{2(-1+2\eta +2q\eta )}. \end{gather*} \end{lemma} \begin{proof} To obtain the solution of \eqref{E1}, we proceed by cases on the two branches of the solution, via the above Green's functionG(t,s)$: If$t<\eta , \begin{aligned} u_1(t) &=C_0\Big( \int_0^t\Big( (s^{2}+2qs)t^{2}-2(2qs\eta +s^{2}\eta )t -(s^{2}+2qs-4qs\eta -2s^{2}\eta ) \Big)ds \\ &\quad +\int_{t}^{\eta }\Big( (1-2\eta -2q\eta +s^{2}+2qs)t^{2}-2(s-2s\eta +s^{2}\eta )t\\ &\quad -2q(s-2s\eta +s^{2}\eta )\Big)ds \\ &\quad +\int_{\eta }^{1}\Big( (1+s^{2}-2s)t^{2} -2(\eta -2s\eta +s^{2}\eta )t -2q(\eta -2s\eta +s^{2}\eta )\Big)ds\Big) \end{aligned} \label{14} Hence an easy computation ensures thatu_1(0)-qu_1'(0)=0$and$u_1'(\eta )=0$. For$\eta \leq t\leq 1, \begin{align*} u_2(t) &=C_0\Big( \int_0^{\eta }\Big( (s^{2}+2qs)t^{2}-2(+2qs\eta +s^{2}\eta )t -(s^{2}+2qs-4qs\eta -2s^{2}\eta)\Big)ds\\ &\quad +\int_{\eta }^t\Big( (2\eta +2q\eta +s^{2}-2s)t^{2} -2(\eta -s+2qs\eta +s^{2}\eta )t\\ &\quad -(2q\eta +s^{2}-4qs\eta -2s^{2}\eta)\Big)ds \\ &\quad +\int_{t}^{1}\Big( (1+s^{2}-2s)t^{2}-2(\eta -2s\eta +s^{2}\eta )t -2q(\eta -2s\eta +s^{2}\eta )\Big)ds \Big) \end{align*} By another calculation, we may obtain thatu_2'(\eta )=0$and$u_2(1)=0$. Furthermore, $u_1'''(t)=-1\quad\text{and}\quad u_1(t)=u_2(t)=u(t),\quad 0\leq t\leq 1.$ Hence the obtained function$u(t)$,$0\leq t\leq 1$, is a solution of \eqref{E1}. \end{proof} \begin{lemma}\label{Le00} Assume hypothesis \eqref{H1}. Then the Green function is nonnegative. \end{lemma} \begin{proof} By condition \eqref{H1}, we have$C_0\leq 0$. Then $4s\eta -2s-2s^{2}\eta -t^{2}(2\eta -2s)=-2s(-2\eta +s\eta +1)-t^{2}(2\eta -2s)\leq 0,$ since by the definition of$u_1^{\ast }(t,s)$,$-2\eta +1\geq 0$and$s\leq \eta . Consequently \begin{align*} u_1^{\ast }(t,s) &=C_0\Big( (4s\eta -2s-2s^{2}\eta -t^{2}(2\eta -2s))q\\ \ \ \ &\quad +(t^{2}(s^{2}-2\eta +1)-2t(s-2s\eta +s^{2}\eta )) \Big)\\ &\geq C_0\Big( (4s\eta -2s-2s^{2}\eta -t^{2}(2\eta -2s))(- \frac{1}{2\eta }(2\eta -1))\\ &\quad +(t^{2}(s^{2}-2\eta +1)-2t(s-2s\eta +s^{2}\eta ))\Big). \end{align*} That is, $$u_1^{\ast }(t,s)\geq C_0s(t-1)(t-2\eta +1)\frac{-2\eta +s\eta +1}{\eta }\geq 0. \label{15}$$ Similarly, \begin{align*} u_2^{\ast }(t,s) &=C_0((4\eta s-2\eta s^{2}-2\eta )q+(t^{2}(s^{2}-2s+1)-2t(\eta s^{2}-2\eta s+\eta )) ) \\ & \geq C_0\Big((4\eta s-2\eta s^{2}-2\eta )(-\frac{1}{ 2\eta }(2\eta -1))+(t^{2}(s^{2}-2s+1)\\ &\quad -2t(\eta s^{2}-2\eta s+\eta )) \Big) \end{align*} Hence $$u_2^{\ast }(t,s)\geq C_0(s-1)^{2}( t-1)(t-2\eta +1)\geq 0. \label{16}$$ In the same way, we may verify that $$\begin{gathered} v_1^{\ast }(t,s)\geq C_0s(t-1)(t-2\eta +1)\frac{-2\eta +s\eta +1}{\eta }\geq 0, \\ v_2^{\ast }(t,s)\geq C_0(s-1)^{2}(t-1)(t-2\eta +1)\geq 0. \end{gathered} \label{17}$$ \end{proof} For convenience, we set: Fors \leq \eta$, $\frac{\partial }{\partial t}G(t,s) = 2C_0\times\begin{cases} (s^{2}+2qs+1-2\eta -2q\eta ) t -(s-2s\eta +s^{2}\eta ), & t\leq s \\ (s^{2}+2qs)t-(2q\eta s+s^{2}\eta ), &s\leq t\leq 1 \end{cases}$ and for$s >\eta$, $\frac{\partial }{\partial t}G(t,s) =2C_0\times \begin{cases} (s^{2}-2s+1) t-(\eta s^{2}-2\eta s+\eta), & 0\leq t\leq s \\ (s^{2}-2s+2\eta +2q\eta ) t-(\eta -s+2qs\eta +s^{2}\eta ), & s\leq t\leq 1 \end{cases}$ Moreover, for$s <\eta$, $\frac{\partial ^{2}}{\partial t^{2}}G(t,s) =2C_0\times\begin{cases} s^{2}+2qs+1-2\eta -2q\eta , & 0\leq t\leq s \\ s^{2}+2qs, & s\leq t\leq 1 \end{cases}$ and for$s >\eta$, $\frac{\partial ^{2}}{\partial t^{2}}G(t,s) =2C_0\times\begin{cases} s^{2}-2s+1 , &0\leq t\leq s \\ s^{2}-2s+2\eta +2q\eta , & s\leq t\leq 1. \end{cases}$ Consider the Banach space$C=C([ 0,1] ,\mathbb{R})$of continuous maps, equipped with the standard norm $\|y\|=\max \{ |y(t)|:0\leq t\leq 1\} ,$$0<\theta \leq \eta <1/2and let \begin{align*} K_0=\Big\{& y\in C:y(t)\geq 0,\; t\in [0,1],\; y''(t)\leq 0,\; t\in [\theta ,1-\theta ], \\ &\max_{0\leq t\leq 1}y(t)=y(\eta )\text{ and } y(1)=0\Big\} . \end{align*} It is obvious thatK_0$is a cone in$C$. We define furthermore the subcone $K=\big\{ y\in K_0:\min_{t\in [\theta ,1-\theta ]}y( t)\geq \theta \|y\|\big\}$ \begin{lemma}\label{Le3} For any$y\in K_0$, $\min_{t\in [\theta ,1-\theta ]}y(t)\geq \theta \|y\|=\theta y(\eta ).$ \end{lemma} \begin{proof} Since$y\in K_0$,$y(t)\geq 0$for$0\leq t\leq 1$and moreover it is concave downward on the interval$[\theta ,1-\theta ]$. Thus for any$t_1,t_2\in [\theta ,1-\theta ]$and$\lambda \in [0,1]$, $y(\lambda t_1+(1-\lambda )t_2)\geq \lambda y(t_1)+(1-\lambda )y(t_2).$ Therefore, $y(t)\geq \|y\|\min_{t\in [\theta ,1-\theta ] }\big\{ \frac{t}{\eta },\frac{1-t}{1-\eta }\big\} \geq \|y\|\min_{t\in [\theta ,1-\theta ]}\{ t,1-t\} \geq \theta \|y\|.$ \end{proof} The next result is very useful. \begin{proposition} \label{Pr1} Assume condition \eqref{H1} holds and let$y:[0,1 ]\to [ 0,+\infty ))$be a continuous map. Then the BVP $$\begin{gathered} u'''(t)=-y(t),\quad 0\leq t\leq 1, \\ u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0. \end{gathered} \label{Ey}$$ admits the unique positive solution$u\in K$, where $u(t)=\int_0^{1}G(t,s)y(s)ds.$ \end{proposition} \begin{proof} We notice firstly that$u(t)\geq 0$,$0\leq t\leq 1$. Indeed, this fact follows directly by the nonnegativity of the Green's function (see Lemma \ref{Le00}). On the other hand, for$0\leq t\leq \eta, we have \begin{align*} u'(t)&= \int_0^{1}\frac{\partial }{\partial t}G(t,s)y(s)ds \\ &= \int_0^t2C_0((s^{2}+2qs)t- (\eta s^{2}+2q\eta s))y(s)ds\; \\ &\quad +\int_{t}^{\eta }2C_0\Big( (s^{2}+2qs+1-2\eta -2q\eta ) t-(s-2s\eta +s^{2}\eta ) \Big)y(s)ds \\ &\quad +\int_{\eta }^{1}2C_0\big(( s^{2}-2s+1) t-(\eta s^{2}-2\eta s+\eta ) \big)y(s)ds. \end{align*} Consequently, \begin{align*} u'(\eta )&=\int_0^{\eta }(2C_0(( s^{2}+2qs)\eta -( \eta s^{2}+2q\eta s)))y(s)ds \\ &\quad +\int_{\eta }^{1}( 2C_0((s^{2}-2s+) \eta -(\eta s^{2}-2\eta s+\eta )))y(s)ds \\ &=\int_0^{1}2C_0[(s^{2}+2qs)\eta -( \eta s^{2}+2q\eta s)]y(s)ds \\ & =\int_0^{1}0y(s) ds=0. \end{align*} Similarly, we may prove thatu(0)-qu'(0)=0$and$u(1)=0. Furthermore, \begin{align*} u''(t) &= \int_0^{1}\frac{\partial ^{2}}{ \partial t^{2}}G(t,s)y(s)ds \\ &= \int_0^t2C_0(s^{2}+2qs)y(s)ds \\ &\quad +\int_{t}^{\eta }2C_0(s^{2}+2qs+1-2\eta -2q\eta ) y(s)ds \\ &\quad +\int_{\eta }^{1}2C_0(s^{2}-2s+1)y(s)ds. \end{align*} Hence, recalling thatC_0=1/2(1-2\eta -2q\eta )$, $u'''(t)= 2C_0(t^{2}+2qt)y(t) -2C_0(t^{2}+2qt+1-2\eta -2q\eta )y(t) = -y(t).$ Finally, by the nonnegativity of the solution$u(t)$and the boundary conditions$u(0)-qu'(0)=0$and$u(1)=0$, we may assume that$u''(0) \leq 0$. Otherwise, if$u''(0)>0$, we get$u'(t)>0$, in a right neighborhood of$0$, due to the differential equation$u'''(t)=-y(t)$,$0\leq t\leq 1$, and since$u'(\eta )=0$). Hence there is a$\theta \in [ 0,\eta )$, such that$u''(\theta )=0$. Thus in both the cases, we conclude that $u''(t)\leq 0,\quad \theta \leq t\leq 1-\theta .$ Consequently, in view of Lemma \ref{Le3}, we obtain that$u\in K$. \end{proof} \begin{corollary}\label{Co} Assume that hypotheses of Proposition \ref{Pr1} are satisfied. Consider the BVP $$\begin{gathered} u'''(t)=y(t),\quad 0\leq t\leq 1, \\ u(0)-qu'(0)=0,\quad u'(\eta)=0,\quad u(1)=0. \end{gathered} \label{E1*}$$ Then, the map $u(t)=-\int_0^{1}G(t,s)y(s)ds$ is clearly a non-positive solution of \eqref{E1*}. \end{corollary} \section{Main Results} In this section we prove the existence of at least one positive solution of \eqref{E}. We assume that $$f\in C([0,1]\times [ 0,+\infty ),[0,+\infty)) \label{H2}$$ In view of Proposition \ref{Pr1}, we consider the positive solution$u_1(t)$of \eqref{E1} and set \begin{gather*} A_0 = \max \{ u_1(t):0\leq t\leq 1\} =\max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)ds\Big),\\ B_0 = \max \{ u_1(t):\theta \leq t\leq 1-\theta\} =\max_{\theta \leq t\leq 1-\theta } \Big(\int_{\theta }^{1-\theta}G(t,s)ds\Big). \end{gather*} In view of Lemma \ref{Le00}, we get$A_0\geq B_0>0$. We define the operator $\mathcal{T}u(t)=\int_0^{1}G(t,s)f(s,u(s))ds.$ Obviously, BVP \eqref{E} has a solution$u=u(t)$, if and only if$u$is a fixed point of$T$. Moreover, recalling that the operator$\mathcal{T}:K\to C([0,1])$is called \textit{completely continuous}, if it is continuous and maps bounded sets into precompact sets we state the next well-known result \cite{Su}. \begin{proposition} Assume that \eqref{H1}-\eqref{H2} hold. Then$\mathcal{T}:K\to K$is a completely continuous operator. \end{proposition} \begin{proof} It is sufficient to show that$\mathcal{T}(K)\subset K$. This is easily derived from Lemma \ref{Le3} and Proposition \ref{Pr1}, due to assumption \eqref{H1} and the definition of the cone$K$. \end{proof} We will employ the following fixed point theorem due to Krasnosel'skii \cite{Kr}. \begin{theorem} \label{Th1} Let$E$be a Banach space,$K\subseteq E$be a cone and suppose that$\Omega _1$,$\Omega _2$are bounded open balls of$E$centered at the origin with$\overline{\Omega }_1\subset \Omega _2$. Furthermore, suppose that$\mathcal{T}:K\cap (\overline{\Omega }_2 \setminus \Omega_1)\to K$is a completely continuous operator such that either :$\| \mathcal{T}u\| \leq \| u\| $,$u\in K\cap \partial \Omega _1$and$\| \mathcal{T}u\| \geq \| u\| $,$u\in K\cap \partial \Omega _2$; or$\| \mathcal{T}u\| \geq \| u\| $,$u\in K\cap \partial \Omega _1$and$\| \mathcal{T}u\| \leq \|u\|$,$u\in K\cap \partial \Omega _2$holds. Then$\mathcal{T}$admits a fixed point in$K\cap (\overline{\Omega }_2\setminus \Omega _1)$. \end{theorem} Now we are ready to formulate and prove our main result. \begin{theorem} \label{Th2} Assume that \eqref{H1}-\eqref{H2} hold and there exist positive constants$r\neq R$such that \begin{gather} |f(t,x)|\leq \frac{r}{A_0},\quad (t,x)\in [0,1]\times [0,r]; \label{A1}\\ | f(t,x)|\geq \frac{R}{B_0},\quad (t,x)\in [0,1] \times [ \theta R,R]. \label{A2} \end{gather} Then the boundary value problem \eqref{E} admits a positive solution$u=u(t)$,$0\leq t\leq 1$, such that $\min \{ r,R\} \leq \|u\|\leq \max \{ r,R\} .$ Moreover, the obtained solution$u=u(t)$,$0\leq t\leq 1$is concave downward. \end{theorem} \begin{proof} Assuming first that$rR. We consider the open balls \Omega _1=\{ u\in C([0,1]): \|u\|0 such that \frac{f(t,x)}{x}\leq \frac{1}{A_0}, for all (t,x)\in [0,1] \times [0,r] and this yields assumption \eqref{A1} of previous Theorem \ref{Th2}. Similarly by the superlinearity at +\infty , we get an R>r such that \frac{f(t,x)}{x}\geq \frac{1}{\theta B_0}, for all (t,x)\in [0,1]\times [ \theta R,R]. Hence Theorem \ref{Th2} is applicable. On the other hand, when the nonlinearity is sublinear, we examine the following cases: (a) If f is bounded, say by M>0, we may choose any R\geq A_0M and then we obtain \begin{aligned} \| \mathcal{T}u\| &\leq \max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\ &\leq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)Mds\Big) \\ &=MA_0\leq R=\| u\| , \end{aligned} \label{33} for u\in K with \|u\|=R. (b) If f is unbounded, let also an R be large enough such that \[ \frac{|f(t,R)|}{R}\leq \frac{1}{A_0}\quad \text{and}\quad |f(t,u)|\leq |f(t,R)|,\quad (t,u)\in [0,1]\times [0,R]. Therefore, $|f(t,u)|\leq |f(t,R)|\leq \frac{R}{A_0},\quad (t,u)\in [0,1]\times [0,R].$ Consequently, \begin{align*} \| \mathcal{T}u\| &=\max_{0\leq t\leq 1}|\int_0^{1}G(t,s)f(s,u(s))ds| \\ &\leq \max_{0\leq t\leq 1}\Big(\int_0^{1}G(t,s)\frac{R}{ A_0}ds\Big) \\ &\leq \frac{R}{A_0}A_0=\| u\| \end{align*} foru\in K$with$\|u\|=R$. Moreover, by the sublinearity of$f$at$u=0$, there exists an$r0,\quad u''(0)\leq 0,\quad u'(1)<0, \quad u''(1)\leq 0. \] Furthermore the map $u''(t)$, $0\leq t\leq 1$ is non-increasing. \end{remark} \begin{corollary} \label{Co0} Under the assumptions of Theorem \ref{Th2} or Corollary \ref{Co1}, the BVP \begin{gathered} u'''(t)=f(t,u(t)),\quad 00,\quad u'(1)>0,\quad u''(1) \geq 0. \] \end{proof} \begin{corollary}\label{Co2} Under the assumptions of Theorem \ref{Th2} or Corollary\ \ref{Co1}, BVP \eqref{34} admits a negative, concave upward solution $u=u(t)$, $0\leq t\leq 1$. \end{corollary} \begin{proof} Let $u=u_1(t)$, $0\leq t\leq 1$ be a solution of BVP \eqref{E}. Then the function $u=-u_1(t),\quad 0\leq t\leq 1$ is obviously the desired solution of \eqref{34}. We notice that $u'(0)<0,\quad u''(0)>0,\quad u'(1)>0,\quad u''(1) \geq 0.$ Moreover, the map $u''(t)$, $0\leq t\leq 1$ is nondecreasing. \end{proof} \begin{corollary} \label{Co3} Under the assumptions of Theorem \ref{Th2} or Corollary\ \ref{Co1}, BVP \eqref{34} admits a positive and concave downward solution $u=u(t)$, $0\leq t\leq 1$. \end{corollary} \begin{proof} Obviously the desired solution is given by $u(t)=\int_0^{1}[-G(t,s)]f(s,u(s))ds.$ Here also the map $u''(t)$, $0\leq t\leq 1$ is nondecreasing. \end{proof} \begin{example} \label{exa1} \rm Consider the boundary value problem \begin{gather*} u'''(t)=-\sqrt[3]{u(t)+(u(t))^{2}},\quad 0\frac{1}{6}\eta (\eta -1)^{2} \frac{2q+\eta +q\eta }{2\eta +2q\eta -1}. \] \end{theorem} \begin{proof} Let $w_{i}(t)$, $i=1,2$ be two solutions of BVP ( \eqref{E}, Since the Green's function $G(t,s)$ is positive, we have \begin{align*} w_2(t)-w_1(t) &=\int_0^{1}G(t,s) [f(s,w_2(s))-f(s,w_1(s))]ds \\ &\leq \int_0^{1}G(t,s)|f(s,w_2(s))-f(s,w_1(s))|ds \\ &\leq \int_0^{1}G(t,s)L| w_2(s)-w_1(s)|ds \\ &\leq \int_0^{1}G(t,s) L\|w_2-w_1\|ds\\ &=L\|w_2-w_1\|\int_0^{1}G(t,s)ds \\ &=L\|w_2-w_1\|u_1(t),\quad 0\leq t\leq 1, \end{align*} where $u_1(\ t)$, is the unique positive solution of BVP \eqref{E1}. 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