\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 165, 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} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2009/165\hfil An optimal existence theorem] {An optimal existence theorem for positive solutions of a four-point boundary value problem} \author[M. K. Kwong, J. S. W. Wong\hfil EJDE-2009/165\hfilneg] {Man Kam Kwong, James S. W. Wong} % in alphabetical order \address{Man Kam Kwong \newline Department of Applied Mathematics, Hong Kong Polytechnic University, Hong Kong, SAR, China} \email{mankwong@polyu.edu.hk} \address{James S. W. Wong \newline Institute of Mathematical Research, Department of Mathematics, University of Hong Kong Hong Kong, SAR, China} \email{jsww@chinneyhonkwok.com} \thanks{Submitted February 12, 2009. Published December 22, 2009.} \subjclass[2000]{34B10, 34B15, 34B18} \keywords{Four-point boundary value problem; second-order ODE; \hfill\break\indent Krasnoselskii fixed point theorem; mappings on cones} \begin{abstract} We are interested in the existence of positive solutions to a four-point boundary value problem of the differential equation $y''(t) + a(t)f(y(t))=0$ on $[0,1]$. The value of $y$ at $0$ and $1$ are each a multiple of $y(t)$ at an interior point. Many known existence criteria are based on the limiting values of $f(u)/u$ as $u$ approaches $0$ and infinity. In this article we obtain an optimal criterion (thereby improving all existing results of kind mentioned above) by comparing these limiting values to the smallest eigenvalue of the corresponding four-point problem of the associated linear equation. In the simpler case of three-point boundary value problems, the same result has been established in an earlier paper by the first author using the shooting method. The method of proof is based upon a variant of Krasnoselskii's fixed point theorem on cones, the classical Krein-Rutman theorem, and the Gelfand formula relating the spectral radius of a linear operator to its norm. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \section{Introduction} We are interested in the existence of positive solutions of second-order nonlinear differential equations subject to four-point boundary conditions. In an earlier paper \cite{kw2}, we provided improvements of a result by Liu \cite{kw} on the existence of a positive solution for the equation: y''(t) + a(t)f(y(t)) = 0, \quad 00$; \item[(H3)] The following two limits exist: $$f_0 = \lim_{u\to 0} \frac{f(u)}{u} , \quad f_\infty = \lim_{u\to \infty } \frac{f(u)}{u} .$$ \end{itemize} Then \eqref{e} \eqref{bn} has at least one positive solution if either $$f_0 < \Lambda _1, \quad f_\infty > \Lambda _2 , \label{f1}$$ or $$f_0 > \Lambda _2, \quad f_\infty < \Lambda _1 , \label{f2}$$ where \begin{gather*} \Lambda _1 = \Lambda \Big( (1-\alpha +\alpha \xi ) \int_{0}^{1} (1-s)a(s)\,ds \Big) ^{-1} , \\ \Lambda _2 = \Lambda \Big( (1-\alpha +\alpha \xi ) \gamma \eta \int_{\eta }^{1} (1-s)a(s)\,ds \Big) ^{-1} , \end{gather*} and$\gamma =\min\big(\eta ,\beta \eta ,\beta (1-\eta ) /(1-\beta \eta )\big) $. \end{theorem} Clearly$ \Lambda _2>\Lambda _1 $, so there exists a gap between the numbers$ f_0 $and$ f_\infty $in which existence is unknown. If$ \alpha =0 $in \eqref{bn}, then \eqref{e} \eqref{bn} becomes a three-point problem which was studied in an earlier paper by the first named author \cite{k0}. Our main result is as follows. \begin{theorem} \label{thmB} Let$ \lambda _1 $be the smallest eigenvalue of the three-point problem u''(t)+\lambda a(t)u(t) = 0, \quad 00$ for $i=1,\dots ,m-2$ satisfying $\sum_{i=1}^{m-2} k_i\xi _i<1$. In both papers \cite{k0} and \cite{kw}, we employed the classical shooting method which resulted in optimal criteria for $f_0$, $f_\infty$ in terms of the smallest eigenvalue of the associated linear problem as given in \eqref{3f1} and \eqref{3f2}. The existence of the smallest eigenvalue for the classical two-point problem (for which the corresponding eigensolution is positive on $(0,1)$), is well-known from the Sturm-Liouville theory. The existence of $\lambda _1$ for the multi-point BVP \eqref{3pl} \eqref{mbn} was proved by the shooting method in \cite{k0}. The purpose of this paper is to prove an analogue of Theorem \ref{thmB} for the four-point problem which includes both Theorems \ref{thmA} and \ref{thmB} as special cases. In fact, a result similar to Theorem \ref{thmB} was also proved in Sun \cite{s} using topological degree theory. However, the proof in this three-point case seems to contain an error where the definition of an infimum of a set of parameters may not exist (see \cite{s}, pp. 1058-1059). In any case, Sun's result is only concerned with symmetric solutions. In Zhang and Sun \cite{zs}, optimal existence theorems given for multi-point boundary value problems were given and proved using topological degree theory. However, the conditions are more restrictive. In the three-point case, their result requires $0\leq \beta <1$ as compared with (H1) of Theorem \ref{thmA}. \section{Main Theorem and Proof} Unlike our earlier result Theorem \ref{thmB} where the shooting method was employed, we use the Krasnoselskii fixed point theorem on cones (see \cite{gl} and \cite{kr}) to prove our main result, similar to that used by Liu in \cite{l} for the three-point case; i.e., BVP \eqref{e} \eqref{bn} with $\alpha =0$. We introduce the operator $A:C[0,1]\to C^2[0,1]$ defined for $y(t)\in C[0,1]$ by $$Ay(t) = \int_{0}^{1} G(t,s)a(s)f(y(s))\,ds, \label{a}$$ where $G(t,s)$ is given in terms of the Green's function $k(t,s)$ for the Dirichlet two-point boundary value problem: $$x''(t)+a(t)f(x(t))=0, \quad x(0)=x(1)=0.$$ Here $$k(t,s) = \begin{cases} t(1-s), & 0\leq t\leq s\leq 1, \\ s(1-t), & 0\leq s\leq t\leq 1, \end{cases}$$ and $$G(t,s) = k(t,s)+l_1(t)k(\xi ,s) + l_2(t) k(\eta ,s), \label{g}$$ where \begin{gather} l_1(t) = \frac{\alpha }{\Lambda } [(\beta -1)t+(1-\beta \eta )], \label{el1} \\ l_2(t) = \frac{\beta }{\Lambda } [(1-\alpha )t+\alpha \xi ]. \label{el2} \end{gather} Define the positive number $\sigma$ by $$\sigma = \min \{ l_1(t)\xi (1-\xi )+l_2(t)\eta (1-\eta ) : 0\leq t\leq 1 \} , \label{s}$$ which is positive because $l_1(t)$ and $l_2(t)$ are strictly positive on $[0,1]$. Indeed by assumption (H1), we easily obtain from (\ref{el1}) and (\ref{el2}) $$\min\{ l_1(t):0\leq t\leq 1\} = \frac{\alpha }{\Lambda } \min (1-\beta \eta ,\beta (1-\eta )) > 0$$ and $$\min\{ l_2(t):0\leq t\leq 1\} = \frac{\beta }{\Lambda } \min (\alpha \xi ,1-\alpha +\alpha \xi ) > 0 .$$ Using (\ref{s}) in (\ref{g}), we obtain $$G(t,s) \geq \sigma s(1-s), \quad 0\leq s,t\leq 1. \label{gs}$$ Now let $$\frac{M_0}{2} = \max (\alpha (1-\beta \eta ), \alpha \beta (1-\eta ), \alpha \beta \xi , \beta (1-\alpha +\alpha \xi ) ). \label{m0}$$ From the definition of $k(t,s)$ it is easy to see that $$k(t,s), \, k(\xi ,s), \, k(\eta ,s) \leq s(1-s) \label{gs2}$$ where $0\leq t,s\leq 1$ and $0<\xi \leq \eta <1$. Upon combining (\ref{m0}) and (\ref{gs2}), we have $$G(\tau ,s) \leq (1+M_0\Lambda ^{-1})^{-1}s(1-s), \quad 0\leq \tau ,s\leq 1. \label{gs3}$$ Using (\ref{gs3}) and the definition of $A$ given by \eqref{a}, we find that $$Ay(t) \geq \sigma (1+M_0\Lambda ^{-1})^{-1}Ay(\tau ). \label{a2}$$ Define $c_0=\sigma (1+M_0\Lambda ^{-1})^{-1}$ and the positive cone $$K = \{ y(t)\in C[0,1] : y(t)\geq c_0\|y\| \} ,$$ where $\|y\|=\max\{ |y(t)|:0\leq t\geq 1\}$ is the supremum norm of $C[0,1]$. By (\ref{a2}), it is clear that $A(K)\subseteq K$. We now apply the Krasnoselskii fixed point theorem on cones given in the form by Kwong \cite{k1} as follows: \begin{theorem}[Krasnoselskii-Petryshyn-Benjamin] \label{thmKPB} Let $A:K\to K$ be a completely continuous operator, where $K$ is a cone on $C[0,1]$. Suppose that \begin{itemize} \item[(i)] there exists $p\in K$ such that $x-Ax\neq \mu p$, for all $\mu \geq 0$ and all $x\in K_a=\{ x\in K:\|x\|=a\}$; and \item[(ii)] (Leary-Schauder condition) given any $\mu \in[0,1]$, $x\neq \mu Ax$ for all $x\in K_b=\{ x\in K:\|x\|=b\}$, where $b>a>0$. \end{itemize} Then $A$ has a non-zero fixed point $x^*\in K$ satisfying $a\leq \|x^*\|\leq b$. \end{theorem} We are now ready to prove our main result: \begin{theorem} \label{thm1} Let $\lambda _1$ be the smallest eigenvalue of the linear four-point boundary value problem \eqref{3pl} \eqref{bn}. If $f(y)$ satisfies \eqref{3f1} or \eqref{3f2}, then \eqref{e} \eqref{bn} has at least one positive solution. \end{theorem} Consider at first the case when $f_\infty <\lambda _10$ is given in terms of (\ref{s}) and (\ref{m0}) by $c_0=\sigma (1+M_0\Lambda ^{-1})^{-1}$. From the definition \eqref{a} of the operator $A$ and by (\ref{a2}), we know that $A:K\to K$. It is now a standard argument to prove that $A$ is completely continuous. Next we need to verify conditions (i) and (ii) in Theorem \ref{thmKPB} to establish that $A$ has a fixed point $\hat y\in K$ which is clearly non-zero since $\hat y(t)\geq c_0\|\hat y\|$, $c_0>0$, and by the definition of $K$, the function $\hat y(t)$ is positive. It is also easy to verify that the fixed point $\hat y$ satisfies the four-point boundary condition \eqref{bn}. Now consider the linear operator $L$ defined by setting $f(y)\equiv y$ in \eqref{a}, namely $$L y(t) = \int_{0}^{1} G(t,s)a(s)y(s)\,ds .$$ Clearly $L$ maps the cone $K$ into itself. We quote the famous Krein-Rutman theorem. \begin{theorem}[Krein-Rutman \cite{z}] \label{thmKR} Let $L:K\to K$ be a linear, completely continuous operator which maps a cone $K$ in a Banach space $X$ into itself. Then the equation $Ly=\lambda y$ has a smallest positive eigenvalue $\lambda _1>0$ which satisfies $\lambda _1r(L)=1$ where $r(L)$ denotes the spectral radius of the operator $L$. \end{theorem} \begin{proof} Assume that $f_\infty <\lambda _1\lambda _1$, there exists $a>0$ sufficiently small such that $f(u)\geq \lambda _1u$ for all $u\in K_a$. Suppose that condition (i) is false, then there exists $u_0\in K_a$ such that $\|u_0\|=a$ and $$u_0 = Au_0 + \mu _0 p \label{u0}$$ for some $\mu _0\geq 0$. From (\ref{u0}), it is clear that $\mu _0>0$, for otherwise $u_0$ is a fixed point of $A$. Since $A$ is positive; i.e., $Ay(t)\geq 0$ for all $y\geq 0$, we have $u_0\geq \mu _0p$. Let $$\mu ^* = \sup \{ \mu : u_0\geq \mu p \} .$$ So $u_0\geq \mu ^*p\geq \mu _0p$. Note that $Lu_0\geq L(\mu ^*p)=\mu ^*Lp$ and $Au_0\geq \lambda _1Lu_0$. Thus \label{u01} \begin{aligned} u_0 &= Au_0 + \mu _0 p \\ &\geq \lambda _1Lu_0 + \mu _0p \\ &\geq \lambda _1\mu ^*Lp + \mu _0p \\ &= \mu ^*(\lambda _1Lp) + \mu _0p \\ &= (\mu ^* + \mu _0) p. \end{aligned} Since $\mu _0>0$, we have $u_0\geq (\mu ^*+\mu _0)p$, contradicting the definition of $\mu ^*$. So condition (i) holds. We now turn to condition (ii). From $f_\infty <\lambda _1$, there exist $\varepsilon >0$ and $b$ sufficiently large such that $f(u)\leq (\lambda _1-\varepsilon )u$ for all $\|u\|\geq b$. Let $u_1\in K_b$; i.e., $\|u_1\|=b$, and $u_1=\mu Au_1$ for all $\mu \in [0,1]$. Note that $$u_1 = \mu A u_1 \leq (\lambda _1-\varepsilon )\mu Lu_1 = \sigma Lu_1, \quad 0<\sigma <(\lambda _1-\varepsilon )\mu .\label{u1}$$ Since $L$ is linear, we can prove by induction from (\ref{u1}) that $u_1\leq \sigma ^nL^nu_1$ for $n=1,2,\dots$ from which it follows that $$\|L^n\| \geq ~f \|L^nu_1\| \|u_1\| = \sigma ^{-n} . \label{Ln}$$ Using the Gelfand formula for the spectral radius $r(L)$ (see \cite{z}), we obtain from (\ref{Ln}) $$r(L) = \lambda _1^{-1} = \lim_{n\to \infty } (\|L^n\|)^{1/n} \geq \sigma ^{-1} \geq (\lambda _1-\varepsilon )^{-1} . \label{rL}$$ Since $\varepsilon >0$, (\ref{rL}) gives the desired contradiction. Hence, we conclude that for $x\in K_b$, $x\neq \mu Ax$ for all $\mu \in[0,1]$ and that condition (ii) holds. The existence of a non-zero fixed point of $A$ now follows from Theorem \ref{thmKPB} cited above. In the superlinear case; i.e., $f_0<\lambda _10$. Since $Ay(t)\geq 0$ for all $y(t)\geq 0$, and, in particular, $Au_2\geq 0$, so $u_2\geq \mu _2p$, $\mu _2>0$. Define $\hat\mu =\sup\{ \mu :u_2\geq \mu p\}$ which exists and $\hat\mu \geq u_2$. Note that $Au_2\geq \lambda _1Lu_2$ and $Lu_2\geq L(\hat\mu p)=\hat\mu Lp$. Using the fact that $p=\lambda _1Lp$, we observe that \label{u2} \begin{aligned} u_2 &= Au_2 + \mu _2 p \\ &\geq \lambda _1Lu_2 + \mu _2p \\ &\geq \lambda _1 \hat \mu Lp + \mu _2p \\ &= (\hat \mu + \mu _2)p . \end{aligned} Since $\mu _2>0$, (\ref{u2}) shows that $\hat\mu$ does not exist. Hence, condition (i) is satisfied. Next we show that condition (ii) holds for $u\in K_a$ where $a$ satisfies $00$ sufficiently small, there exists $a$, $01$. Since $Ay(t)\leq \lambda _1Ly(t)$ for all $y\in K_a$, we have $Av_0=\bar\mu v_0\leq \lambda _1Lv_0$. By the linearity of $L$, we can prove by induction that $\lambda _1^nLv_0\geq \mu ^{-n}v_0$, for $n=2,3,\dots$, so $$\|L^n\| \geq \frac{\|L^nv_0\|}{\|v_0\|} = \big( \frac{\bar\mu }{\lambda _1} \big) ^n . \label{Ln2}$$ Again by the spectral radius formula, we have from (\ref{Ln2}) $$r(L) = \lambda _1^{-1} = \lim_{n\to \infty } (\|L^n\|)^{1/n} \geq \frac{\bar\mu }{\lambda _1} > \frac{1}{\lambda _1} ,$$ which gives the desired contradiction. Therefore condition (ii) of Theorem \ref{thmKPB} is also true. This implies the existence of a non-zero fixed point of $A$ and completes the proof of the theorem. \end{proof} \section{Examples and Remarks} We give two examples to illustrate the usefulness of our main Theorem. \begin{example} \label{exa1} \rm Consider the three-point boundary value problem \begin{gather} y''(t) + \frac{7y^2(t)+y(t)}{1+y(t)} = 0 \label{x1} \\ y(0)=0, \quad y(1) = \frac{1}{2} y\big( \frac{1}{2} \big) . \label{x1b} \end{gather} In Liu \cite{l}, it was proved that \eqref{e} \eqref{bn} with $\alpha =0$ has a positive solution if $f_0<\overline \Lambda _1$ and $f_\infty >\overline \Lambda _2$ where \begin{gather*} \overline \Lambda _1 = (1-\beta \eta ) \Big( \int_{0}^{1} (1-s)a(s)\,ds \Big) ^{-1} , \\ \overline \Lambda _2 = (1-\beta \eta ) \Big( \mu \eta \int_{0}^{1} (1-s)a(s)\,ds \Big) ^{-1} , \end{gather*} with $\mu =\min(\eta ,\beta \eta ,\beta (1-\eta )(1-\beta \eta ))$. In case of the specific equation we have $\overline \Lambda _1=3/2$ and $\overline \Lambda _2=48$. Now consider the linear boundary value problem $y''+\lambda y=0$, subject to the three-point boundary condition (\ref{x1b}). It is easy to determine the smallest positive eigenvalue $\lambda _1$ by equating $\sin\sqrt{\lambda }/2=2\sin\sqrt{\lambda }$, yielding $\lambda _1=6.917$. Since $1=f_0<\lambda _10 \label{x2} subject to the boundary conditions $$y(0)= \frac{1}{2} y\big( \frac{1}{3} \big) , \quad y(1) = \frac{1}{3} y\big( \frac{1}{2} \big) . \label{x2b}$$ Here$ f_0=a/(1+b) $and$ f_\infty =a $. Also$ \Lambda =19/36 $,$ \Lambda _1=19/12 $and$ \Lambda _2=76 $. Since$ a,b>0 $, in order to apply Theorem \ref{thmA}, we require$ a>76 $and$ b>\frac{12a}{19} -1 $. This gives existence of positive solution only when$ a>76 $and$ b>47 $. We can compute the smallest positive eigenvalue of the linear BVP$ y''+\lambda y=0 $, subject to the four-point boundary condition (\ref{x2b}) numerically. The answer is$ \lambda _1=5.3163775 $and the corresponding eigenfunction is$ \sin\{ (2.3057271)t+0.4971368\} $. By our main theorem, we obtain the existence of positive solution to (\ref{x2}) (\ref{x2b}) if$ a $and$ b $satisfy $$\frac{a}{1+b} < \lambda _1 < a. \label{ab}$$ Condition (\ref{ab}) gives a much greater range in$ a $and$ b$. If$ a=6 $, we only need$ b>0.128588 $. By comparison, if$ a>76 $, we require$ b>13.2954 $which is better than$ b>47 $as required by the estimate (\ref{f1}) given in Theorem \ref{thmA}. \end{example} We close our discussion with a few remarks relating our work to others in the existing literature. \begin{remark} \label{rmk1} \rm Our main result in this paper is closely related to our two earlier papers [6], [7], which included extensive references on the subject matter. Therefore we shall not reproduce here. \end{remark} \begin{remark} \label{rmk2} \rm Boundary condition \eqref{bn} is sometimes referred as separated (2;2) four point boundary conditions. It should be distinguished from the (1;3) four point boundary condition such as $$y(0) = 0, \alpha _1y(\xi _1)+\alpha _2y(\xi _2)+\alpha _3y(\xi _3) = 0,$$ where$ \alpha _1 $,$ \alpha _2 $, and$ \alpha _3 $are real constants and$ 0<\xi _1<\xi _2<\xi _3<1 $. \end{remark} \begin{remark} \label{rmk3} \rm In \cite{kw2}, we showed that condition (H3), namely,$ \Lambda =\alpha \xi (1-\beta )+(1-\alpha )(1-\beta \eta )>0 $is a necessary condition for the existence of a positive solution. Previously, (H3) has always been assumed as a sufficient condition. When$ \Lambda =0 $, the BVP \eqref{e} \eqref{bn} is said to be at resonance in the sense that the associated linear homogeneous boundary value problem$ x''(t)=0 $,$ 0