\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 266, pp. 1--21.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/266\hfil Symmetric positive solutions] {Symmetric positive solutions for $\phi$-Laplacian boundary-value problems with integral \\ boundary conditions} \author[W. Yang \hfil EJDE-2013/266\hfilneg] {Wengui Yang} % in alphabetical order \address{Wengui Yang\newline Ministry of Public Education, Sanmenxia Polytechnic, Sanmenxia 472000, China} \email{wgyang0617@yahoo.com} \thanks{Submitted June 17, 2013. Published November 30, 2013.} \subjclass[2000]{34B15, 34B18} \keywords{Boundary value problems; integral boundary conditions; \hfill\break\indent symmetric positive solutions; $\phi$-Laplacian operator; fixed point theorem} \begin{abstract} In this article, we study the existence, multiplicity, and nonexistence of symmetric positive solutions for a class of four-order integral boundary value problems with $\phi$-Laplacian operator. The arguments mainly rely on the Guo-Krasnosel'skii fixed point theorem of cone expansion and compression of norm type and Leggett-Williams fixed point theorem. Finally, some examples are presented to illustrate the main results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction}\label{sect1} In this article, we consider the following $\phi$-Laplacian boundary-value problems (BVP) with integral boundary conditions $$\label{e1.1} \begin{gathered} \big[q(t)[\phi((p(t)u'(t))')]'\big]'=w(t)f(t,u(t)),\quad t\in[a,b],\\ \alpha u(a)-\beta p(a)u'(a)=\int_a^bg(s)u(s)ds,\\ \alpha u(b)+\beta p(b)u'(b)=\int_a^bg(s)u(s)ds,\\ \gamma \phi((p(a)u'(a))')-\delta q(a)[\phi((p(a)u'(a))')]' =\int_a^bh(s)\phi((p(s)u'(s))')ds,\\ \gamma \phi((p(b)u'(b))')+\delta q(b)[\phi((p(b)u'(b))')]' =\int_a^bh(s)\phi((p(s)u'(s))')ds. \end{gathered}$$ We use the following assumptions throughout this article: \begin{itemize} \item[(H0)] $\phi$ is an odd, increasing homeomorphism from $\mathbb{R}$ onto $\mathbb{R}$ and there exist two increasing homeomorphisms $\psi_1$ and $\psi_2$ of $(0,\infty)$ onto $(0,\infty)$ such that $\psi_1(u)\phi(v)\leq\phi(uv)\leq\psi_2(u)\phi(v)$ for all $u,v>0$. Moreover, $\phi,\phi^{-1}\in C^1(\mathbb{R})$, where $\phi^{-1}$ denotes the inverse of $\phi$. \item[(H1)] $\alpha>0$, $\gamma>0$, and $\beta\geq0$, $\delta\geq0$. \item[(H2)] $p(t),q(t):[a,b]\to (0,\infty)$ are continuous functions, and $p(t),q(t)$ are symmetric on $[a,b]$. \item[(H3)] $w\in L^1[a,b]$ is nonnegative, symmetric on $[a,b]$ and $w(t)\not\equiv0$ on any subinterval of $[a,b]$. \item[(H4)] $f:[a,b]\times[0,\infty)\to [0,\infty)$ is continuous and $f(\cdot,u)$ is symmetric on $[a,b]$ for all $u\in[0,\infty)$. \item[(H5)] $g,h\in C([a,b],[0,\infty))$ are symmetric functions on $[a,b]$, and $\mu\in[0,\alpha)$, $\nu\in[0,\gamma)$, where $\mu=\int_a^bg(s)ds, \quad \nu=\int_a^bh(s)ds.$ \end{itemize} Boundary-value problems with integral boundary conditions have gained considerable popularity and importance due to their application in different areas of applied mathematics and physics and so on. They include two, three, multi-point and nonlocal boundary-value problems as special cases. There have been some papers dealing with the existence and multiplicity of solutions or positive solutions for such problems by the use of some well-known fixed point theorems and upper and lower solutions method. For some recent developments on the subject, see \cite{3,2,4,6,1,9,7,5,8} and the references therein. Recently, many researchers have extensively studied the existence, multiplicity and nonexistence of symmetric positive solutions of boundary-value problems by using fixed point theorem; i.e., fixed point theorem of cone expansion and compression of norm type, fixed point theory in cones and Leggett-Williams fixed point theorem. To identify a few, we refer the reader to \cite{13,16,10,11,17,14,12,15} and the references therein. In particular, we would like to mention some results of Feng \cite{18}, Ma \cite{19}, Xu \cite{20}, Luo and Luo \cite{21}, and Zhang et al.~\cite{22}, Zhang and Ge \cite{22A}. Feng \cite{18} considered the following second-order nonlinear ordinary differential equation with integral boundary conditions \begin{gather*} (g(t)u'(t))'+w(t)f(t,u(t))=0,\quad 00$,$g\in C^1([0,1],(0,\infty))$is symmetric on$[0,1]$,$w\in L^p[0,1]$for some$1\leq p\leq+\infty$, and is symmetric on$[0,1]$,$f:[0,1]\times[0,+\infty)$is continuous,$f(1-t,u)=f(t,u)$for all$(t,u)\in[0,1]\times[0,+\infty)$, and$h\in L^p[0,1]$is nonnegative and symmetric on$[0,1]$. The author investigated the existence of at least one symmetric positive solution by applying the theory of fixed point index in cones. Ma \cite{19} obtained the existence of at least one symmetric positive solution for fourth-order boundary-value problem with integral boundary conditions by using the fixed point index in cones. \label{e1.2} \begin{gathered} u''''(t)=w(t)f(u(t)),\quad 01$, and $f(t,u(t),u'(t))=f(t,u(t))$. Zhang and Ge \cite{22A} considered the following second-order nonlinear ordinary differential equation with integral boundary conditions \begin{gather*} (q(t)u'''(t))'=w(t)f(t,u(t)),\quad 00$,$q\in C^1([0,1],(0,\infty))$is symmetric on$[0,1]$,$w\in L^p[0,1]$for some$1\leq p\leq+\infty$, and is symmetric on$[0,1]$,$f:[0,1]\times[0,+\infty)$is continuous,$f(1-t,u)=f(t,u)$for all$(t,u)\in[0,1]\times[0,+\infty)$, and$g,h\in L^p[0,1]$is nonnegative and symmetric on$[0,1]$. The authors obtained the existence of at least one symmetric positive solution based upon a specially constructed cone and the fixed point theory in a cone. Motivated greatly by the above mentioned works, we establish sufficient conditions for the existence, multiplicity and nonexistence of symmetric positive solutions of BVP \eqref{e1.1} by applying the Guo-Krasnosel'skii fixed point theorem of cone expansion and compression of norm type and Leggett-Williams fixed point theorem. Our paper improves and generalizes the results of mentioned results to some degree. At the end of this paper, some examples are presented to illustrate the main results. \section{Preliminaries}\label{sect2} We recall that the function$u$is said to be concave on$[a,b]$, if$u(\lambda t_2+(1-\lambda)t_1)\geq\lambda u(t_1)+(1-\lambda)u(t_1)$,$t_1,t_2\in[a,b]$,$\lambda\in[0,1]$and the function$u$is said to be symmetric on$[a,b]$, if$u(b-t+a)=u(t)$,$t\in[a,b]$. A function$u$is called a symmetric positive solution of \eqref{e1.1} provided$u$is symmetric and positive on$[a,b]$, and satisfies the differential equation and the boundary value conditions in \eqref{e1.1}. To prove the main results in this article, we will use the following lemmas. \begin{lemma}\label{L2.1} Assume {\rm (H0)--(H2)} hold and$\mu\neq\alpha$. Then for any$v\in C[a,b]$, the BVP $$\label{e2.1} \begin{gathered} (p(t)u'(t))'=\phi^{-1}(v(t)),\quad t\in[a,b],\\ \alpha u(a)-\beta p(a)u'(a)=\int_a^bg(s)u(s)ds,\\ \alpha u(b)+\beta p(b)u'(b)=\int_a^bg(s)u(s)ds, \end{gathered}$$ has a unique solution$u$and$u$can be expressed in the form $$\label{e2.2} u(t)=-\int_a^bH_1(t,s)\phi^{-1}(v(s))ds,$$ where \begin{gather}\label{e2.3} H_1(t,s)=G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau,\\ \label{e2.4} G_1(t,s)=\frac{1}{\Delta_1} \begin{cases} \big(\beta+\alpha\int_a^s\frac{d\tau}{p(\tau)}\big) \big(\beta+\alpha\int_t^b\frac{d\tau}{p(\tau)}\big), &a\leq s\leq t\leq b,\$3pt] \big(\beta+\alpha\int_a^t\frac{d\tau}{p(\tau)}\big) \big(\beta+\alpha\int_s^b\frac{d\tau}{p(\tau)}\big), &a\leq t\leq s\leq b. \end{cases} \end{gather} Here \Delta_1=\alpha\Big(2\beta+\alpha\int_a^b\frac{ds}{p(s)}\Big), \quad \mu=\int_a^bg(s)ds. \end{lemma} \begin{proof} First suppose that u is a solution of problem \eqref{e2.1}. It is easy to see by integration of both sides of \eqref{e2.1} on [a,t] that \[ p(t)u'(t)=p(a)u'(a)+\int_a^t\phi^{-1}(v(s))ds.$ Letting$A=p(a)u'(a), then we have $u'(t)=\frac{A}{p(t)}+\frac{1}{p(t)}\int_a^t\phi^{-1}(v(s))ds.$ Integrating again, we can get $u(t)=u(a)+A\int_a^t\frac{1}{p(s)}ds+\int_a^t\frac{1}{p(s)} \int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds.$ By the boundary condition, we obtain \begin{gather*} \alpha u(a)-\beta A=\int_a^bg(s)u(s)ds,\\ \begin{aligned} \alpha u(a)+\Big(\alpha\int_a^b\frac{ds}{p(s)}+\beta\Big)A &= \int_a^bg(s)u(s)ds-\beta\int_a^bg(s)\phi^{-1}(v(s))ds \\ &\quad -\alpha\int_a^b\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds. \end{aligned} \end{gather*} Then \begin{align*} A=-\frac{1}{2\beta+\alpha\int_a^b\frac{ds}{p(s)}} \Big(\alpha\int_a^b\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds +\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big), \end{align*} and \begin{align*} u(a)&=-\frac{1}{\Delta_1}\Big(\alpha\int_a^b\frac{1}{p(s)} \int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds +\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big)\\ &\quad +\frac{1}{\alpha}\int_a^bg(s)u(s)ds. \end{align*} Thus \begin{align*} u(t)&=\frac{1}{\alpha}\int_a^bg(s)u(s)ds +\int_a^t\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds\\ &\quad -\frac{1}{\Delta_1}\Big(\alpha\int_a^b\frac{1}{p(s)} \int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds +\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big)\\ &\quad -\frac{\int_a^t\frac{1}{p(s)}ds}{2\beta+\alpha\int_a^b\frac{ds}{p(s)}} \Big(\alpha\int_a^b\frac{1}{p(s)}\int_a^s\phi^{-1}(v(\tau))\,d\tau\,ds +\beta\int_a^bg(s)\phi^{-1}(v(s))ds\Big)\\ &=\frac{1}{\alpha}\int_a^bg(s)u(s)ds-\int_a^bG_1(t,s)\phi^{-1}(v(s))ds, \end{align*} whereG_1(t,s)$is defined by \eqref{e2.4}. Multiplying the above equation with$g(t)and integrating it again, we obtain $\int_a^bg(s)u(s)ds=-\frac{\alpha}{\alpha -\int_a^bg(s)ds}\int_a^bG_1(t,s)\phi^{-1}(v(s))ds.$ It follows that \begin{align*} u(t)&=-\int_a^bG_1(t,s)\phi^{-1}(v(s))ds-\frac{1}{\alpha-\mu}\int_a^bG_1(t,s)\phi^{-1}(v(s))ds\\ &=-\int_a^bH_1(t,s)\phi^{-1}(v(s))ds, \end{align*} whereH_1(t,s)$is defined in \eqref{e2.3}. The proof is complete. \end{proof} \begin{lemma}\label{L2.2} Assume (H1)-(H4) hold and$\nu\neq\gamma$. Then for any$u\in C[a,b]$, the BVP \begin{equation*} \begin{gathered} (q(t)v'(t))'=w(t)f(t,u(t)),\quad t\in[a,b],\\ \gamma v(a)-\delta q(a)v'(a)=\int_a^bh(s)v(s)ds,\quad \gamma v(b)+\delta q(b)v'(b)=\int_a^bh(s)v(s)ds, \end{gathered} \end{equation*} has a unique solution$v$and$vcan be expressed in the form $v(t)=-\int_a^bH_2(t,s)w(s)f(s,u(s))ds,$ where \begin{gather}\label{e2.5} H_2(t,s)=G_2(t,s)+\frac{1}{\gamma-\nu}\int_a^bG_2(s,\tau)h(\tau)d\tau, \\ \label{e2.6} G_2(t,s)=\frac{1}{\Delta_2}\begin{cases} \big(\delta+\gamma\int_a^s\frac{d\tau}{q(\tau)}\big) \big(\delta+\gamma\int_t^b\frac{d\tau}{q(\tau)}\big), &a\leq s\leq t\leq b,\3pt] \big(\delta+\gamma\int_a^t\frac{d\tau}{q(\tau)}\big) \big(\delta+\gamma\int_s^b\frac{d\tau}{q(\tau)}\big), &a\leq t\leq s\leq b. \end{cases} \end{gather} Here \Delta_2=\gamma\Big(2\delta+\gamma\int_a^b\frac{ds}{q(s)}\Big), \quad\nu=\int_a^bh(s)ds. \end{lemma} The above lemma can be proved in a way similar to Lemma \ref{L2.1}. \begin{lemma}\label{L2.3} Assume {\rm (H5)} holds. Then for any t,s\in[a,b], the following results are true. \begin{itemize} \item[(i)] G_1(t,s)\geq0, H_1(t,s)\geq0, G_2(t,s)\geq0, H_2(t,s)\geq0; \item[(ii)] \frac{\beta^2}{\Delta_1}\leq G_1(t,s)\leq G_1(s,s) \leq\frac{\Lambda_1}{\Delta_1}, \frac{\beta^2\sigma_1}{\Delta_1} \leq H_1(t,s)\leq H_1(s,s)\leq\frac{\sigma_1\Lambda_1}{\Delta_1}, \frac{\delta^2}{\Delta_2}\leq G_2(t,s)\leq G_2(s,s) \leq\frac{\Lambda_2}{\Delta_2}, \frac{\delta^2\sigma_2}{\Delta_2} \leq H_2(t,s)\leq H_2(s,s)\leq\frac{\sigma_2\Lambda_2}{\Delta_2} with \begin{gather*} \Lambda_1=\Big(\beta+\alpha\int_a^b\frac{d\tau}{p(\tau)}\Big)^2,\quad \sigma_1=\frac{\alpha}{\alpha-\mu},\\ \Lambda_2=\Big(\delta+\gamma\int_a^b\frac{d\tau}{q(\tau)}\Big)^2,\quad \sigma_2=\frac{\gamma}{\gamma-\nu}; \end{gather*} \item[(iii)] G_1(b-t+a,b-s+a)=G_1(t,s), H_1(b-t+a,b-s+a)=H_1(t,s), G_2(b-t+a,b-s+a)=G_2(t,s), H_2(b-t+a,b-s+a)=H_2(t,s); \item[(iv)] \rho_1G_1(s,s)\leq H_1(t,s)\leq\sigma_1 G_1(s,s), \rho_2G_2(s,s)\leq H_2(t,s)\leq\sigma_2 G_2(s,s) with \rho_1=\frac{\Delta_1}{(\alpha-\mu)\Lambda_1}\int_a^bG_1(s,s)g(s)ds,\quad \rho_2=\frac{\Delta_2}{(\gamma-\nu)\Lambda_2}\int_a^bG_1(s,s)h(s)ds; \end{itemize} where H_1(t,s), G_1(t,s), H_2(t,s) and G_2(t,s) are defined by \eqref{e2.3}--\eqref{e2.6}, respectively. \end{lemma} \begin{proof} By simple computations, we have (i) and (ii). Firstly, we prove that (iii) holds. If a\leq t\leq s\leq b, then b-t+a\geq b-s+a. In view of \eqref{e2.4} and the assumption (H2), we get \begin{align*} &G_1(b-t+a,b-s+a)\\ &=\frac{1}{\Delta_1}\Big(\beta+\alpha\int_a^{b-s+a}\frac{d\tau}{p(\tau)}\Big) \Big(\beta+\alpha\int_{b-t+a}^b\frac{d\tau}{p(\tau)}\Big)\\ &= \frac{1}{\Delta_1}\Big(\beta+\alpha\int_b^s\frac{d(b-\tau+a)}{p(b-\tau+a)} \Big)\Big(\beta+\alpha\int_t^a\frac{d(b-\tau+a)}{p(b-\tau+a)}\Big)\\ &= \frac{1}{\Delta_1}\Big(\beta+\alpha\int_s^b\frac{d\tau}{p(\tau)}\Big) \Big(\beta+\alpha\int_a^t\frac{d\tau}{p(\tau)}\Big)\\ &=G_1(t,s), \quad a\leq t\leq s\leq b. \end{align*} Similarly, we can prove that G_1(b-t+a,b-s+a)=G_1(t,s), a\leq s\leq t\leq b. So, we have G_1(b-t+a,b-s+a)=G_1(t,s), for any t,s\in[a,b]. By \eqref{e2.3} and (H5), we have \begin{align*} &H_1(b-t+a,b-s+a)\\ &= G_1(b-t+a,b-s+a)+\frac{1}{\alpha-\mu}\int_a^bG_1(b-s+a,\tau)g(\tau)d\tau\\ &= G_1(t,s)+\frac{1}{\alpha-\mu}\int_b^aG_1(b-s+a,b-\tau+a) g(b-\tau+a)d(b-a\tau+a)\\ &= G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau=H_1(t,s). \end{align*} Similarly, we can prove that G_2(b-t+a,b-s+a)=G_2(t,s), H_2(b-t+a,b-s+a)=H_2(t,s). Next, we prove that (iv). First, we prove that \frac{\Delta_1}{\Lambda_1}G_1(t,t)G_1(s,s)\leq G_1(t,s) and \frac{\Delta_2}{\Lambda_2}G_2(t,t)G_2(s,s)\leq G_2(t,s). If a\leq t\leq s\leq b, then \[ \frac{G_1(t,s)}{G_1(t,t)G_1(s,s)} =\frac{\Delta_1}{\big(\beta+\alpha\int_a^s\frac{d\tau}{p(\tau)}\big) \big(\beta+\alpha\int_t^b\frac{d\tau}{p(\tau)}\big)} \geq\frac{\Delta_1}{\Lambda_1}. Similarly, we can prove that $\frac{G_1(t,s)}{G_1(t,t)G_1(s,s)}\geq\frac{\Delta_1}{\Lambda_1},$ fora\leq s\leq t\leq b$. So, we have$\frac{\Delta_1}{\Lambda_1}G_1(t,t)G_1(s,s)\leq G_1(t,s)$, for any$t,s\in[a,b]. From \eqref{e2.3}, then we obtain \begin{align*} H_1(t,s) &= G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau\\ &\leq G_1(s,s)+\frac{1}{\alpha-\mu}\int_a^bg(s)dsG_1(s,s)\\ &=\frac{\alpha}{\alpha-\mu}G_1(s,s)=\sigma_1G_1(s,s). \end{align*} On the other hand, we have \begin{align*} H_1(t,s) &= G_1(t,s)+\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau\\ &\geq\frac{1}{\alpha-\mu}\int_a^bG_1(s,\tau)g(\tau)d\tau\\ &\geq \frac{\Delta_1}{(\alpha-\mu)\Lambda_1} \int_a^bG_1(\tau,\tau)g(\tau)d\tau G_1(s,s)=\rho_1G_1(s,s). \end{align*} Hence we get\rho_1G_1(s,s)\leq H_1(t,s)\leq\sigma_1 G_1(s,s)$. Similarly, we can prove that$\rho_2G_2(s,s)\leq H_2(t,s)\leq\sigma_2 G_2(s,s)$. \end{proof} From Lemmas \ref{L2.1} and \ref{L2.2}, we have the following result. \begin{lemma}\label{L2.4} Assume {\rm (H0)--(H5)} hold. If$u$is a solution of \eqref{e1.1}, then $u(t)=\int_a^bH_1(t,s)\phi^{-1}\Big(\int_a^bH_2(s,\tau)w(\tau) f(\tau,u(\tau))d\tau\Big)ds.$ \end{lemma} Now, we let$\mathbb{C}=C[a,b]$, then$\mathbb{C}$is a Banach space with$\|u\|=\max_{t\in[a,b]}|u(t)|$, and define a cone$\mathbb{P}$by$\mathbb{P}=\{x\in\mathbb{C} : u(t)\geq0$,$u''(t)\leq0$,$u(t)$is a symmetric and concave function on$[a,b]$and$u(t)\geq\omega\|u\|\}$, where$\omega$is defined as in Lemma \ref{L2.6}. Also, define, for$00$the following set$Q(\chi,d)=\{u\in Q|\chi(u)0$be constants with$Q$and$\kappa$as defined above, we note $$Q_r=\{u\in Q:\|u\|b\} \neq\emptyset and \kappa(Tu)>b, for all u\in Q(\kappa,b,d). \item[(ii)] \|Tu\|\leq a, for all \|u\|\leq a. \item[(iii)] \kappa(Tu)\geq b, for all u\in Q(\kappa,b,c) with \|Tu\|>d. \end{itemize} Then T has at least three fixed points u_1,u_2 and u_3 satisfying$$ \|u_1\|\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and \\ $f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$\\ (particularly, $f_0=\infty$ and $f^\infty=0$). \item[(A3)] $f^0<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and \\ $f_\infty>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f^0=0$ and $f_\infty=\infty$). \end{itemize} Then the BVP \eqref{e1.1} has at least one symmetric positive solution. \end{theorem} \begin{proof} Let the operator $T$ be defined by \eqref{e2.9}. (A1) For $u\in\mathbb{P}\cap\partial\Omega_r$, we have $u\in[0,r]$, which implies $f(t,u)\geq\phi\big(\frac{r}{k_1}\big)$. Hence for $t\in[a,b]$, by Lemma \ref{L2.3}, we obtain \begin{align*} u(t) &\geq \rho_1\int_a^b G_1(s,s)\phi^{-1} \Big(\rho_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\phi\big(\frac{r}{k_1}\big)\Big)ds\\ &\geq \rho_1\int_a^b G_1(s,s)ds\psi_2^{-1} \Big(\rho_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\Big)\frac{r}{k_1}=r=\|u\|, \end{align*} which implies that for $u\in\mathbb{P}\cap\partial\Omega_r$, $$\label{e3.1} \|Tu\|\geq\|u\|.$$ Next, for $u\in\mathbb{P}\cap\partial\Omega_R$, we have $u\in[0,R]$, which implies $f(t,u)\leq\phi\big(\frac{R}{k_2}\big)$. Hence for $t\in[a,b]$, by Lemma \ref{L2.3}, we obtain \begin{align*} u(t) &\leq \sigma_1\int_a^b G_1(s,s)\phi^{-1} \Big(\sigma_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\phi\big(\frac{R}{k_2}\big)\Big)ds\\ &\leq \sigma_1\int_a^b G_1(s,s)ds\psi_1^{-1} \Big(\sigma_2\int_a^b G_2(\tau,\tau)w(\tau)d\tau\Big)\frac{R}{k_2}=R=\|u\|, \end{align*} which implies that for $u\in\mathbb{P}\cap\partial\Omega_R$ $$\label{e3.2} \|Tu\|\leq\|u\|.$$ (A2) At first, in view of $f_0>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},$ there exists $r>0$ such that $f(t,u)\geq(f_0-\varepsilon_1)\phi(\|u\|)$, for $t\in[a,b]$, $\|u\|\in[0,r]$, where $\varepsilon\geq0$ satisfies $$\rho_1\int_a^bG_1(s,s)ds\psi_2^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau(f_0-\varepsilon_1)\Big)\geq1.$$ Then, for $t\in[a,b]$, $u\in\mathbb{P}\cap\partial\Omega_r$, which implies $\|u\|\leq r$, we have \begin{align*} (Tu)(t) &\geq \int_a^bH_1(t,s)\phi^{-1} \Big(\int_a^bH_2(s,\tau)w(\tau)(f_0-\varepsilon_1)\phi(\|u\|)d\tau\Big)ds\\ &\geq \rho_1\int_a^bG_1(s,s)\phi^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau(f_0-\varepsilon_1)\phi(\|u\|)\Big)ds\\ &\geq \rho_1\int_a^bG_1(s,s)ds\psi_2^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau(f_0-\varepsilon_1)\Big)\|u\| \geq\|u\|, \end{align*} which implies that for $u\in\mathbb{P}\cap\partial\Omega_r$, $$\label{e3.3} \|Tu\|\geq\|u\|.$$ Next, turning to $f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},$ there exists $\overline{R}>0$ large enough such that $f(t,u)\leq(f^\infty+\varepsilon_2)\phi(\|u\|)$, for $t\in[a,b]$, $\|u\|\in(\overline{R},\infty)$, where $\varepsilon_2>0$ satisfies $$\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}-f^\infty-\varepsilon_2>0.$$ Set $M=\max_{\|u\|\leq\overline{R},t\in[a,b]}f(t,u)$. Then $f(t,u)\leq M+(f^\infty+\varepsilon_2)\phi(\|u\|)$. Choose \begin{align*} R>&\max\Big\{r,\overline{R},\phi^{-1}\Big(M\Big[\psi_1 \Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big)\\ &\times\Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1} -f^\infty-\varepsilon_2\Big]\Big)\Big\}. \end{align*} Hence for $u\in\mathbb{P}\cap\partial\Omega_R$, we have \begin{align*} (Tu)(t) &\leq \sigma_1\int_a^bG_1(s,s)\phi^{-1} \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)(M+(f^\infty+\varepsilon_2) \phi(\|u\|))d\tau\Big)ds\\ &\leq \sigma_1\int_a^bG_1(s,s)ds\phi^{-1} \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau \Big(\frac{M}{\phi(R)}+f^\infty+\varepsilon_2\Big)\phi(R)\Big)\\ &\leq \sigma_1\int_a^bG_1(s,s)ds\psi_1^{-1} \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau \Big(\frac{M}{\phi(R)}+f^\infty+\varepsilon_2\Big)\Big)R\\ &\leq R=\|u\|, \end{align*} which implies that for $u\in\mathbb{P}\cap\partial\Omega_R$, $$\label{e3.4} \|Tu\|\leq\|u\|.$$ (A3) Considering $f^0<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1},$ there exists $r>0$ such that $f(t,u)\leq\eta\phi(u)$, for any $u\in[0,r]$, $t\in[a,b]$, where $$\eta\leq\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.$$ Then if $\Omega_r$ is the ball in $\mathbb{C}$ centered at the origin with radius $r$ and if $u\in\mathbb{P}\cap\partial\Omega_r$, then we have \begin{align*} \|Tu\| &= \max_{t\in[a,b]}\int_a^bH_1(t,s)\phi^{-1} \Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\ &\leq \sigma_1\int_a^bG_1(s,s)\phi^{-1} \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)\eta\phi(u(\tau))d\tau\Big)ds\\ &\leq \sigma_1\int_a^bG_1(s,s)ds\phi^{-1} \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)\eta\phi(r)d\tau\Big)\\ &\leq r\sigma_1\int_a^bG_1(s,s)ds\psi_1^{-1} \Big(\eta\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)\leq r=\|u\|, \end{align*} which implies that for $u\in\mathbb{P}\cap\partial\Omega_r$ $$\label{e3.5} \|Tu\|\leq\|u\|.$$ On the other hand, we use the assumption $$f_\infty>\psi_2 \Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.$$ Then there exists $R>0$ large enough such that $f(t,u)\geq\varrho\phi(u)$ for any $u\in[R,\infty)$, $t\in[a,b]$, where $\varrho\geq\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.$ If we define $\Omega_R=\{u\in\mathbb{C}:\|u\|\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and $f_\infty>\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f_0=f_\infty=\infty$). \item[(ii)] There exists $c>0$ satisfying $f(t,u)<\phi\big(\frac{c}{k_2}\big)$, $(t,b)\in[a,b]\times[0,c]$. \end{itemize} \item[(A5)] \begin{itemize} \item[(i)] $f^0<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and $f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f^0=f^\infty=0$). \item[(ii)] There exists $c>0$ satisfying $f(t,u)>\phi\big(\frac{c}{k_1}\big)$, $(t,b)\in[a,b]\times[0,c]$. \end{itemize} \end{itemize} Then \eqref{e1.1} has at least two symmetric positive solutions $u_1(t)$ and $u_2(t)$, which satisfy 0<\|u_1\|\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}, it follows from the proof of \eqref{e3.3} that we can choose r with 0\psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}, \] then as in the proof of \eqref{e3.6}, we can choose R with cy\}\neq\emptyset. Moreover, if u\in Q(\varphi,y,\frac{y}{\omega}), then \varphi(u)\geq y, so y\leq\|u\|\leq\frac{y}{\omega}. By the definition of \varphi and (A7), we obtain \begin{align*} \|Tu\| &= \min_{t\in[a,b]}(Tu)(t)\\ &=\min_{t\in[a,b]}\int_a^bH_1(t,s)\phi^{-1} \Big(\int_a^bH_2(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds\\ &\geq \rho_1\int_a^bG_1(s,s)\phi^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)\phi\big(\frac{y}{k_1}\big)d\tau\Big)ds\\ &\geq \frac{y\rho_1}{k_1}\int_a^bG_1(s,s)ds\psi_2^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)=y. \end{align*} Therefore, condition (i) of Lemma \ref{L2.10} is satisfied. Finally, we address condition (\rm iii) of Lemma \ref{L2.10}. For this we choose u\in Q(\varphi,y,z) with \|Tu\|>\frac{y}{\omega}. Then from Lemma \ref{L2.6}, we deduce \varphi(Tu)=\min_{t\in[a,b]}(Tu)(t)\geq\omega\|Tu\|>y. $$Hence, (\rm iii) of Lemma \ref{L2.10} holds. By Lemma \ref{L2.10}, then we obtain that \eqref{e1.1} has at least three symmetric positive solutions u_1(t), u_2(t) and u_3(t) such that$$ \|u_1\|\phi\big(\frac{\|u\|}{k_1}\big) for all $t\in[a,b], u\in[0,\infty)$. \item[(A10)] $f(t,u)<\phi\big(\frac{\|u\|}{k_2}\big)$ for all $t\in[a,b], u\in[0,\infty)$. \end{itemize} Then \eqref{e1.1} has no positive solution. \end{theorem} \begin{proof} Assume $u(t)$ is a positive solution of \eqref{e1.1}, we have \begin{align*} \|u\|&= \|Tu\|\geq(Tu)(t)\\ >&\rho_1\int_a^bG_1(s,s)\phi^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)\phi\big(\frac{\|u\|}{k_1}\big)d\tau \Big)ds\\ &\geq \frac{\|u\|\rho_1}{k_1}\int_a^bG_1(s,s)ds\psi_2^{-1} \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)=\|u\|. \end{align*} which is a contradiction. So, due to (A9), equation \eqref{e1.1} has no positive solution. Similarly, due to (A9), we obtain that \eqref{e1.1} has no positive solution. \end{proof} \section{Further remarks}\label{sect4} Consider the fourth-order integral BVP with with $\phi$-Laplacian operator $$\label{e4.1} \begin{gathered} \big[q(t)\phi(p(t)u''(t))\big]''=w(t)f(t,u(t)),\quad t\in[a,b],\\ \alpha u(a)-\beta u'(a)=\int_a^bg(s)u(s)ds,\quad \alpha u(b)+\beta u'(b)=\int_a^bg(s)u(s)ds,\\ \gamma q(a)\phi(p(a)u''(a))-\delta [q(a)\phi(p(a)u''(a))]' =\int_a^bh(s)q(s)\phi(p(s)u''(s))ds,\\ \gamma q(b)\phi(p(b)u''(b))+\delta [q(b)\phi(p(b)u''(b))]' =\int_a^bh(s)q(s)\phi(p(s)u''(s))ds. \end{gathered}$$ By methods analogous to the ones abov, we have the following results. \begin{lemma}\label{L4.1} Assume {\rm (H0)--(H2)} hold and $\mu\neq\alpha$. Then for any $v\in C[a,b]$, the BVP $$\label{e4.2} \begin{gathered} q(t)\phi(p(t)u''(t))=v(t),\quad t\in[a,b],\\ \alpha u(a)-\beta u'(a)=\int_a^bg(s)u(s)ds,\quad \alpha u(b)+\beta u'(b)=\int_a^bg(s)u(s)ds, \end{gathered}$$ has a unique solution $u$ and $u$ can be expressed in the form $$\label{e4.3} u(t)=-\int_a^bH_1^\ast(t,s)\frac{1}{p(s)}\phi^{-1} \Big(\frac{v(s)}{q(s)}\Big)ds,$$ where \begin{gather}\label{e4.4} H_1^\ast(t,s)=G_1^\ast(t,s)+\frac{1}{\alpha-\mu} \int_a^bG_1^\ast(s,\tau)g(\tau)d\tau, \\ \label{e4.5} G_1^\ast(t,s)=\frac{1}{\Delta_1^\ast} \begin{cases} [\beta+\alpha(s-a)][\beta+\alpha(b-t)], &a\leq s\leq t\leq b,\\ [\beta+\alpha(t-a)][\beta+\alpha(b-s)], &a\leq t\leq s\leq b. \end{cases} \end{gather} Here $$\Delta_1^\ast=\alpha[2\beta+\alpha(b-a)], \quad\mu=\int_a^bg(s)ds.$$ \end{lemma} \begin{lemma}\label{L4.2} Assume {\rm (H1)--(H4)} hold and $\nu\neq\gamma$. Then for any $u\in C[a,b]$, the BVP \begin{gather*} v''(t)=w(t)f(t,u(t)),\quad t\in[a,b],\\ \gamma v(a)-\delta v'(a)=\int_a^bh(s)v(s)ds,\quad \gamma v(b)+\delta v'(b)=\int_a^bh(s)v(s)ds, \end{gather*} has a unique solution $v$ that can be expressed as $v(t)=-\int_a^bH_2^\ast(t,s)w(s)f(s,u(s))ds,$ where \begin{gather*}\label{e4.6} H_2^\ast(t,s)=G_2^\ast(t,s) +\frac{1}{\gamma-\nu}\int_a^bG_2^\ast(s,\tau)h(\tau)d\tau, \\ \label{e4.7} G_2^\ast(t,s)=\frac{1}{\Delta_2^\ast} \begin{cases} [\delta+\gamma(s-a)][\delta+\gamma(b-t)], &a\leq s\leq t\leq b,\\ [\delta+\gamma(t-a)][\delta+\gamma(b-s)], &a\leq t\leq s\leq b. \end{cases} \end{gather*} Here $$\Delta_2^\ast=\gamma[2\delta+\gamma(b-a)], \quad\nu=\int_a^bh(s)ds.$$ \end{lemma} \begin{lemma}\label{L4.3} Assume {\rm (H5)} holds. Then for any $t,s\in[a,b]$, the following results are true. \begin{itemize} \item[(i)] $G_1^\ast(t,s)\geq0$, $H_1^\ast(t,s)\geq0$, $G_2^\ast(t,s)\geq0$, $H_2^\ast(t,s)\geq0$; \item[(ii)] $\frac{\beta^2}{\Delta_1^\ast}\leq G_1^\ast(t,s) \leq G_1^\ast(s,s)\leq\frac{\Lambda_1^\ast}{\Delta_1^\ast}$, $\frac{\beta^2\sigma_1}{\Delta_1^\ast}\leq H_1^\ast(t,s) \leq H_1^\ast(s,s)\leq\frac{\sigma_1\Lambda_1^\ast}{\Delta_1^\ast}$, $\frac{\delta^2}{\Delta_2^\ast}\leq G_2^\ast(t,s) \leq G_2^\ast(s,s)\leq\frac{\Lambda_2^\ast}{\Delta_2^\ast}$, $\frac{\delta^2\sigma_2}{\Delta_2^\ast}\leq H_2^\ast(t,s) \leq H_2^\ast(s,s)\leq\frac{\sigma_2\Lambda_2^\ast}{\Delta_2^\ast}$ with $$\Lambda_1^\ast=[\beta+\alpha(b-a)]^2,\quad \sigma_1 =\frac{\alpha}{\alpha-\mu},\quad \Lambda_2^\ast=[\delta+\gamma(b-a)]^2,\quad \sigma_2=\frac{\gamma}{\gamma-\nu};$$ \item[(iii)] $G_1^\ast(b-t+a,b-s+a)=G_1^\ast(t,s)$, $H_1^\ast(b-t+a,b-s+a)=H_1^\ast(t,s)$, $G_2^\ast(b-t+a,b-s+a)=G_2^\ast(t,s)$, $H_2^\ast(b-t+a,b-s+a)=H_2^\ast(t,s)$; \item[(iv)] $\rho_1G_1^\ast(s,s)\leq H_1^\ast(t,s)\leq\sigma_1 G_1^\ast(s,s)$, $\rho_2G_2^\ast(s,s)\leq H_2^\ast(t,s)\leq\sigma_2 G_2^\ast(s,s)$ with $$\rho_1^\ast=\frac{\Delta_1^\ast}{(\alpha-\mu)\Lambda_1^\ast} \int_a^bG_1^\ast(s,s)g(s)ds,\quad \rho_2^\ast =\frac{\Delta_2^\ast}{(\gamma-\nu)\Lambda_2^\ast}\int_a^bG_1^\ast(s,s)h(s)ds;$$ \end{itemize} where $H_1^\ast(t,s)$, $G_1^\ast(t,s)$, $H_2^\ast(t,s)$ and $G_2^\ast(t,s)$ are defined by \eqref{e4.4}--\eqref{e4.7}, respectively. \end{lemma} \begin{lemma}\label{L4.4} Assume {\rm (H0)--(H5)} hold. If $u$ is a solution of \eqref{e4.1}, then $u(t)=\int_a^bH_1^\ast(t,s)\phi^{-1} \Big(\int_a^bH_2^\ast(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds.$ \end{lemma} \begin{lemma}\label{L4.5} Assume {\rm (H0)-(H5)} hold. the solution $u(t)$ of \eqref{e4.1} is positive and symmetric on $[a,b]$ and $\min_{t\in[a,b]}u(t)\geq\omega^\ast\|u\|$, where $\omega^\ast=\frac{\rho_1^\ast\psi_2^{-1}(\rho_2^\ast)}{\sigma_1\psi_1^{-1} (\sigma_2)}$. \end{lemma} Let $\mathbb{P}^\ast=\{x\in\mathbb{C} : u(t)\geq0$, $u''(t)\leq0$, $u(t)$ is a symmetric and concave function on $[a,b]$ and $u(t)\geq\omega^\ast\|u\|\}$. We introduce the integral operator $T^\ast:\mathbb{C}\to \mathbb{C}$ by $$\label{e4.8} (T^\ast u)(t)=\int_a^bH_1^\ast(t,s)\phi^{-1} \Big(\int_a^bH_2^\ast(s,\tau)w(\tau)f(\tau,u(\tau))d\tau\Big)ds.$$ \begin{lemma}\label{L4.6} Assume {\rm (H0)--(H5)} hold. Then $u\in\mathbb{C}$ is a solution of \eqref{e1.1} if and only if $u$ is a fixed point of the operator $T^\ast$. \end{lemma} \begin{lemma}\label{L4.7} Assume {\rm (H0)-(H5)} hold. Then $T^\ast:\mathbb{P}\to \mathbb{P}$ is a completely continuous operator. \end{lemma} Now we need to introduce new notation. \begin{gather*} k_1^\ast=\rho_1^\ast\int_a^b G_1^\ast(s,s)ds\psi_2^{-1} \Big(\rho_2^\ast\int_a^b G_2^\ast(\tau,\tau)w(\tau)d\tau\Big),\\ k_2^\ast=\sigma_1\int_a^b G_1^\ast(s,s)ds\psi_1^{-1} \Big(\sigma_2\int_a^b G_2^\ast(\tau,\tau)w(\tau)d\tau\Big). \end{gather*} \begin{theorem}\label{th4.1} Assume {\rm (H0)--(H5)} hold. Furthermore, suppose one of the following conditions are satisfied. \begin{itemize} \item[(A1*)] There exist two constants $r$ and $R$ with $0\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and\\ $f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f_0=\infty$ and $f^\infty=0$). \item[(A3*)] $f^0<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and\\ $f_\infty>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f^0=0$ and $f_\infty=\infty$). \end{itemize} Then \eqref{e4.1} has at least one symmetric positive solution. \end{theorem} \begin{theorem}\label{th4.2} Assume {\rm (H0)--(H5)} hold. Furthermore, suppose one of the following conditions are satisfied. \begin{itemize} \item[(A4*)] \begin{itemize} \item[(i)] $f_0>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and \\ $f_\infty>\psi_2\Big((\rho_1^\ast\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\rho_2^\ast\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f_0=f_\infty=\infty$). \item[(ii)] There exists $c>0$ satisfying $f(t,u)<\phi\big(\frac{c}{k_2^\ast}\big)$, $(t,b)\in[a,b]\times[0,c]$. \end{itemize} \item[(A5*)] \begin{itemize} \item[(i)] $f^0<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ and\\ $f^\infty<\psi_1\Big((\sigma_1\int_a^bG_1^\ast(s,s)ds)^{-1}\Big) \Big(\sigma_2\int_a^bG_2^\ast(\tau,\tau)w(\tau)d\tau\Big)^{-1}$ \\ (particularly, $f^0=f^\infty=0$). \item[(ii)] There exists $c>0$ satisfying $f(t,u)>\phi\Big(\frac{c}{k_1^\ast}\Big)$, $(t,b)\in[a,b]\times[0,c]$. \end{itemize} \end{itemize} Then \eqref{e4.1} has at least two symmetric positive solutions $u_1(t)$ and $u_2(t)$, which satisfy $0<\|u_1\|\phi\big(\frac{\|u\|}{k_1^\ast}\big)$ for all $t\in[a,b], u\in[0,\infty)$. \item[(A10*)] $f(t,u)<\phi\big(\frac{\|u\|}{k_2^\ast}\big)$ for all $t\in[a,b], u\in[0,\infty)$. \end{itemize} Then \eqref{e4.1} has no positive solution. \end{theorem} \begin{remark} \label{re4.1} \rm $\phi$ defined as in (H0) generalizes the projection $\varphi:\mathbb{R}\to \mathbb{R}$ which is an increasing homeomorphism and homomorphism with $\varphi(0)=0$. A projection $\varphi:\mathbb{R}\to \mathbb{R}$, which generates the $p$-Laplacian operator $\varphi_p(u)=|u|^{p-2}u$ for $p>1$, is called an increasing homeomorphism and homomorphism if the following conditions are satisfied: \begin{itemize} \item[(i)] If $x\leq y$, then $\varphi(x)\leq\varphi(y)$, for all $x,y\in\mathbb{R}$; \item[(ii)] $\varphi$ is a continuous bijection and its inverse mapping is also continuous; \item[(iii)] $\varphi(xy)=\varphi(x)\varphi(y)$, for all $x,y\in\mathbb{R}$. If $\phi$ is replaced by $\varphi$, so all of our results also hold. \end{itemize} \end{remark} \section{Examples}\label{sect5} \begin{example} \label{ex5.1} \rm Consider the following fourth-order integral BVP with $\phi$-Laplacian operator $$\label{e5.1} \begin{gathered} \Big[\frac{1}{1+\sin(\pi t)}\Big[\phi\Big(\big(\frac{u'(t)}{1+\sin(\pi t)} \big)'\Big)\Big]'\Big]'=w(t)f(t,u(t)),\quad t\in[0,1],\\ u(0)-u'(0)=\frac{1}{2}\int_0^1u(s)ds,\quad u(1)+u'(1)=\frac{1}{2}\int_0^1u(s)ds,\\ \phi((u'(0))')-[\phi((u'(0))')]' =\frac{1}{2}\int_0^1\phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds,\\ \phi((u'(1))')+[\phi((u'(1))')]' =\frac{1}{2}\int_0^1\phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds, \end{gathered}$$ where $\phi(u)=|u|u$, $w(t)=1/100$ and $f(t,u)=(1+\sin(\pi t))(1+u)$ for $(t,u)\in[0,1]\times[0,\infty)$. Let $\psi_1(u)=\psi_2(u)=u^2$, $u>0$. Then, by calculations we obtain that $\mu=\nu=1/2$, $\Delta=\Delta_1=\Delta_2=3+\frac{2}{\pi}$, \begin{align*} &G_1(t,s)=G_2(t,s)\\ &=\frac{1}{\Delta} \begin{cases} \big(1+\frac{1}{\pi}+s-\frac{1}{\pi}\cos(\pi s)\big) \big(2+\frac{1}{\pi}-t+\frac{1}{\pi}\cos(\pi t)\big), &0\leq s\leq t\leq 1,\\ \big(1+\frac{1}{\pi}+t-\frac{1}{\pi}\cos(\pi t)\big) \big(2+\frac{1}{\pi}-s+\frac{1}{\pi}\cos(\pi s)\big), &0\leq t\leq s\leq 1, \end{cases} \end{align*} \begin{gather*} \rho_1=\rho_2 =\frac{(2+3\pi)(-24+3\pi+18\pi^2+13\pi^3)}{24\pi^2(1+\pi)^2}\approx1.59199, \quad \sigma_1=\sigma_2=2,\\ \omega\approx0.710176,\quad k_1=\frac{(2+3\pi)^\frac{3}{2}(-24+3\pi+18\pi^2+13\pi^3)^3} {17280\pi^\frac{15}{2}(1+\pi)^3}\approx1.06639,\\ k_2=\frac{1}{10\pi^\frac{3}{2}} \Big(\frac{13}{3}-\frac{8}{\pi^3}+\frac{1}{\pi^2}+\frac{6}{\pi}\Big) \sqrt{\frac{1}{3}(-24+3\pi+18\pi^2+13\pi^3)}\approx1.50159. \end{gather*} Clearly, the conditions (H0)--(H5) hold. Next, we prove that the condition (A1) of Theorem \ref{th3.1} is satisfied. In fact, choosing $r=1$ and $R=6$, we have $r<\frac{k_1}{k_2}R$, $\phi\big(\frac{r}{k_1}\big)\approx0.879356$, $\phi\big(\frac{R}{k_2}\big)\approx15.9661$. For $(t,u)\in[0,1]\times[0,1]$, then $f(t,u)=(1+\sin(\pi t))(1+u)\geq1>0.879356 \approx\phi\big(\frac{r}{k_1}\big)$. For $(t,u)\in[0,1]\times[0,10]$, then $f(t,u)=(1+\sin(\pi t))(1+u)\leq14<15.9661 \approx\phi\big(\frac{R}{k_2}\big)$. Hence, by (A1) of Theorem \ref{th3.1}, then BVP \eqref{e5.1} has at least one symmetric positive solution. \end{example} \begin{example} \label{ex5.2} \rm Consider the following fourth-order integral BVP with $\phi$-Laplacian operator $$\label{e5.2} \begin{gathered} \Big[\frac{1}{1+t(1-t)}\Big[\phi\Big(\big(\frac{u'(t)}{1+t(1-t)}\big)'\Big) \Big]'\Big]' =w(t)f(t,u(t)),\quad t\in[0,1],\\ u(0)-u'(0)=\int_0^1su(s)ds,\quad u(1)+u'(1)=\int_0^1su(s)ds,\\ \phi((u'(0))')-[\phi((u'(0))')]'= \int_0^1s \phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds,\\ \phi((u'(1))')+[\phi((u'(1))')]' =\int_0^1s\phi\Big(\big(\frac{u'(s)}{1+\sin(\pi s)}\big)'\Big)ds, \end{gathered}$$ where $\phi(u)=u$, $w(t)=t^2(1-t)^2$ and $f(t,u)=\big(\big(t-\frac{1}{2}\big)^2+1\big)\big(\frac{1}{2}+u\big) \big(\frac{1}{4}+u^2\big)$ for $(t,u)\in[0,1]\times[0,\infty)$. Let $\psi_1(u)=\psi_2(u)=u$, $u>0$. Then, by calculations we obtain that $\mu=\nu=1/2$, $\Delta=\Delta_1=\Delta_2=19/6$, \begin{align*} &G_1(t,s)=G_2(t,s)\\ &=\frac{1}{\Delta} \begin{cases} \big(1-\frac{s}{6}(2s^2-3s-6)\big) \big(\frac{13}{6}+\frac{t}{6}(2t^2-3t-6)\big), &0\leq s\leq t\leq 1,\\ \big(1-\frac{t}{6}(2t^2-3t-6)\big) \big(\frac{13}{6}+\frac{s}{6}(2s^2-3s-6)\big), &0\leq t\leq s\leq 1, \end{cases} \end{align*} \begin{gather*} \rho_1=\rho_2=\frac{1745663}{1171170}\approx1.49053, \quad \sigma_1=\sigma_2=2,\quad \omega=\frac{3047339309569}{5486556675600}\approx0.555419,\\ k_1=\frac{263200328806935685377871}{711426606363060588000000}\approx0.369961, \quad k_2=\frac{86370535759}{129667230000}\approx0.666094. \end{gather*} Clearly, the conditions (H0)-(H5) hold. Next, we prove that the condition (A4) of Theorem \ref{th3.2} is satisfied. $$f_0=f_\infty=\infty>2.70299\approx \psi_2\Big((\rho_1\int_a^bG_1(s,s)ds)^{-1}\Big) \Big(\rho_2\int_a^bG_2(\tau,\tau)w(\tau)d\tau\Big)^{-1}.$$ On the other hand, choosing $c=\frac{1}{2}$, for $(t,u)\in[0,1]\times[0,c]$, we have $f(t,u)\leq f(0,c)=0.625<0.750645\approx\phi\big(\frac{c}{k_2}\big)$. Hence, by (A4) of Theorem \ref{th3.2}, BVP \eqref{e5.2} has at least two symmetric positive solutions $u_1$ and $u_2$ satisfying $0<\|u_1\|<\frac{1}{2}<\|u_2\|$. \end{example} \begin{example} \label{ex5.3} \rm Consider the BVP \eqref{ex5.2} with \label{e5.3} f(t,u)= \begin{cases} \frac{1}{t(1-t)+79u}+8u^3, &0\leq t\leq 1,\; u\leq1,\\ \frac{u}{t(1-t)+79u}+u+7, &0\leq t\leq 1,\; 15.40598\approx\phi\big(\frac{y}{k_1}\big), \] for all $t\in[0,1]$, and $y\leq u\leq \frac{y}{\omega}$. $f(t,u)\leq\frac{1}{79}+\frac{8}{8^3}\approx0.0282832<0.187661 \approx\phi\big(\frac{x}{k_2}\big),$ for all $t\in[0,1]$, and $0\leq u\leq x$. Thus, according to Theorem \ref{th3.3}, BVP \eqref{e5.2} with \eqref{e5.3} has at least three positive solutions $u_1$, $u_2$, and $u_3$ satisfying $$\|u_1\|<\frac{1}{8},\quad 2<\varphi(u_2), \quad \frac{1}{8}<\|u_3\|,\quad \varphi(u_3)<2.$$ \end{example} \begin{thebibliography}{00} \bibitem{13} R. I. Avery, J. 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