\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 36, pp. 1--14.\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-2013/36\hfil Positive solutions] {Positive solutions for a $2n$th-order $p$-Laplacian boundary value problem involving all derivatives} \author[Y. Ding, J. Xu, X. Zhang\hfil EJDE-2013/36\hfilneg] {Youzheng Ding, Jiafa Xu, Xiaoyan Zhang} % in alphabetical order \address{Youzheng Ding \newline School of Mathematics, Shandong University, Jinan 250100, Shandong, China} \email{dingyouzheng@139.com} \address{Jiafa Xu \newline School of Mathematics, Shandong University, Jinan 250100, Shandong, China} \email{xujiafa292@sina.com} \address{Xiaoyan Zhang \newline School of Mathematics, Shandong University, Jinan 250100, Shandong, China} \email{zxysd@sdu.edu.cn} \thanks{Submitted September 10, 2012. Published January 30, 2013.} \subjclass[2000]{34B18, 45J05, 47H11} \keywords{Integro-ordinary differential equation; a priori estimate; index; \hfill\break\indent fixed point; positive solution} \begin{abstract} In this work, we are mainly concerned with the positive solutions for the $2n$th-order $p$-Laplacian boundary-value problem \begin{gather*} -(((-1)^{n-1}x^{(2n-1)})^{p-1})' =f(t,x,x',\ldots,(-1)^{n-1}x^{(2n-2)},(-1)^{n-1}x^{(2n-1)}),\\ x^{(2i)}(0)=x^{(2i+1)}(1)=0,\quad (i=0,1,\ldots,n-1), \end{gather*} where $n\ge 1$ and $f\in C([0,1]\times \mathbb{R}_+^{2n}, \mathbb{R}_+)(\mathbb{R}_+:=[0,\infty))$. To overcome the difficulty resulting from all derivatives, we first convert the above problem into a boundary value problem for an associated second order integro-ordinary differential equation with $p$-Laplacian operator. Then, by virtue of the classic fixed point index theory, combined with a priori estimates of positive solutions, we establish some results on the existence and multiplicity of positive solutions for the above problem. Furthermore, our nonlinear term $f$ is allowed to grow superlinearly and sublinearly. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} %\newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \section{Introduction} In this paper, we investigate the existence and multiplicity of positive solutions for the following $2n$th-order $p$-Laplacian boundary value problem involving all derivatives $$\label{Problem} \begin{gathered} -(((-1)^{n-1}x^{(2n-1)})^{p-1})' =f(t,x,x',\ldots,(-1)^{n-1}x^{(2n-2)},(-1)^{n-1}x^{(2n-1)}),\\ x^{(2i)}(0)=x^{(2i+1)}(1)=0,\quad (i=0,1,\ldots,n-1), \end{gathered}$$ where $f\in C([0,1]\times \mathbb{R}_+^{2n},\mathbb{R}_+)$. Here, by a positive solution \eqref{Problem} we mean a function $u\in C^{2n}[0,1]$ that solves \eqref{Problem} and satisfies $u(t)> 0$ for all $t\in (0,1]$. We are here interested in the case where $f$ depends explicitly on all derivatives. When $f$ involves all even derivatives explicitly, many researchers \cite{AOS,Bai-Ge,Davis,YuhongMa,Wei3} study the Lidstone boundary value problem $$\label{Lidstone-1} \begin{gathered} (-1)^nu^{(2n)}=f(t,u,-u'',\ldots,(-1)^{n-1}u^{(2n-2)}), \quad n\geq 2,\\ u^{(2i)}(0)=u^{(2i)}(1)=0,\quad i=0,1,2,\ldots,n-1. \end{gathered}$$ Yang \cite{Yang1} considered the existence and uniqueness of positive solutions for the following generalized Lidstone boundary value problem $$\label{Lidstone-2} \begin{gathered} (-1)^nu^{(2n)}=f(t,u,-u'',\ldots,(-1)^{n-1}u^{(2n-2)}),\\ \alpha_0u^{(2i)}(0)-\beta_0u^{(2i+1)}(0)=0, \alpha_1u^{(2i)}(1)-\beta_1u^{(2i+1)}(1)=0, \; i=0,1,2,\ldots,n-1, \end{gathered}$$ where $\alpha_j\ge 0$, $\beta_j\ge 0$ $(j=0,1)$ and $\alpha_0\alpha_1+\alpha_0\beta_1+\alpha_1\beta_0>0$. In view of the symmetry, the results in \cite{AOS,Bai-Ge,Davis,YuhongMa,Wei3,Yang1} demonstrate that problems \eqref{Lidstone-1} and \eqref{Lidstone-2} are essentially identical with second-order Dirichlet problem and Sturm-Liouville problem (the case $n=1$), respectively. Yang, O'Regan and Agarwal \cite{Yang4} studied the existence and multiplicity of positive solutions for the second-order boundary value problem depending on the first-order derivative $u'$ $$\label{yang4} \begin{gathered} u''+f(t,u,u')=0,\\ u(0)=u'(1)=0. \end{gathered}$$ In order to overcome the difficulty resulting from the first-order derivative, they imposed the Bernstein-Nagumo condition \cite{SB,MN} on the nonlinear term $f$ to establish several existence theorems for \eqref{yang4}. Yang and O'Regan \cite{Yang2} studied the existence, multiplicity and uniqueness of positive solutions for the $2n$th-order boundary value problem involving all derivatives of odd orders $$\label{yang} \begin{gathered} (-1)^nu^{(2n)}=f(t,u,u',-u''',\ldots,(-1)^{n-1}u^{(2n-1)}),\\ u^{(2i)}(0)=u^{(2i+1)}(1)=0,\quad i=0,1,2,\ldots,n-1, \end{gathered}$$ where $n\ge 2$ and $f\in C([0,1]\times \mathbb{R}_+^{n+1}, \mathbb{R}_+)$ depends on $u$ and all derivatives of odd orders. As application, they utilized their results to discuss the positive symmetric solutions for a Lidostone problem involving an open question posed by Eloe \cite{Eloe}. Yang \cite{Yang5} discussed a $2n$th-order ordinary differential equation involving all derivatives, and the results improved and extended the corresponding ones in \cite{Yang1,Yang4,Yang2}. Equations of the $p$-Laplacian form occur in the study of non-Newtonian fluid theory and the turbulent flow of a gas in a porous medium. Since 1980s, there exist a very large number of papers devoted to the existence of solutions for differential equations with $p$-Laplacian, for instance, see \cite{Graef,Ke,Li,Lakmeche,Wang,Jiafa,Wei,Yang3,Yang6,Zhang} and the references therein. Yang and his coauthors \cite{Jiafa,Yang3,Yang6} studied some boundary value problems with the $p$-Laplacian operator. Yang and O'Regan \cite{Yang3} studied the existence and multiplicity of positive solutions for the focal problem involving both the $p$-Laplacian and the first order derivative $$\label{JiafaXu} \begin{gathered} ((u')^{p-1})'+f(t,u,u')=0,\quad t\in (0,1),\\ u(0)=u'(1)=0, \end{gathered}$$ where $p>1$ and $f\in C([0,1]\times \mathbb{R}_+^2,\mathbb{R}_+)$. Moreover, they applied their main results obtained here to establish the existence of positive symmetric solutions to the Dirichlet problem $$\begin{gathered} (|u'|^{p-2}u')'+f(u,u')=0,\quad t\in (-1,0)\cup (0,1),\\ u(-1)=u(1)=0. \end{gathered}$$ However, the existence of positive solutions for $p$-Laplacian equation with the nonlinear term involving the derivatives, such as Lidstone problem, has not been extensively studied yet. To the best of our knowledge, only \cite{Guo-Ge,Su-Wei,Zhao-Ge} is devoted to this direction. Guo and Ge \cite{Guo-Ge} considered the following boundary-value problem $$\label{GG} \begin{gathered} (\Phi(y^{(2n-1)}))'=f(t,y,y'',\ldots,y^{(2n-2)}),\quad 0\le t\le 1,\\ y^{(2i)}(0)=y^{(2i)}(1)=0,\quad i=0,1,2,\ldots,n-1, \end{gathered}$$ where $f\in C([0,1]\times \mathbb{R}^n, \mathbb{R}) (\mathbb{R}:=(-\infty,+\infty))$. Some growth conditions are imposed on $f$ which yield the existence of at least two symmetric positive solutions by using a fixed point theorem in cones. An interesting feature in \cite{Guo-Ge} is that the nonlinearity $f$ may be sign-changing. Motivated by the works mentioned above, in particular \cite{Jiafa,Yang1,Yang4,Yang2,Yang5,Yang3,Yang6}, in this work, we discuss the existence and multiplicity of positive solutions for \eqref{Problem}. To overcome the difficulty resulting from all derivatives, we first transform \eqref{Problem} into a boundary value problem for an associated second order integro-ordinary differential equation. Then, we will use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some properties of concave functions, properties including Jensen's inequalities and our inequality \eqref{le4} below. The results obtained here improve some existing results in the literature. \section{Preliminaries} Let $E:=C^1[0,1]$, $\|u\|:=\max\{\|u\|_0,\|u'\|_0\}$, where $\|u\|_0:=\max_{t\in [0,1]}|u(t)|$. Furthermore, let $P:=\{u\in E: u(t)\geq 0,u^\prime(t)\geq 0, \forall t\in [0,1]\}$. Then $E$ is a real Banach space and $P$ a cone on $E$. For any positive integer $i\ge 2$, we denote $k_1(t,s):=\min\{t,s\}, \quad k_i(t,s):=\int_0^1k_{i-1}(t,\tau)k_1(\tau,s)\,{\rm d} \tau, \forall t,s\in [0,1].$ Define $(B_iu)(t):=\int_0^1 k_i(t,s)u(s)\,{\rm d} s,\quad h_i(t,s):=\partial k_i(t,s)/\partial t, \quad i=1,2,\ldots,$ Then $((B_iu)(t))':=\int_0^1 h_i(t,s)u(s)\,{\rm d} s,\quad i=1,2,\ldots,$ and $B_i,B_i': E\to E$ are completely continuous linear operators and $B_i,B_i'$ are also positive operators. Let $(-1)^{n-1}x^{(2n-2)}:=u$, it is easy to see that \eqref{Problem} is equivalent to the following system of integro-ordinary differential equations $$\label{integro-diff} \begin{gathered} -((u')^{p-1})'=f(t,(B_{n-1}u)(t),((B_{n-1}u)(t))',\ldots,u,u'),\\ u(0)=u'(1)=0. \end{gathered}$$ Furthermore, the above system can be written in the form $$\label{integral} u(t)=\int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau)) \,{\rm d} \tau\Big)^{\frac{1}{p-1}}\,{\rm d} s.$$ Denote by $(Au)(t):=\int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau)) \,{\rm d} \tau\Big)^{\frac{1}{p-1}}\,{\rm d} s.$ Hence, $f\in C([0,1]\times \mathbb{R}_+^{2n},\mathbb{R}_+)$ implies that $A:P\to P$ is a completely continuous operator, and the existence of positive solutions for \eqref{integro-diff} is equivalent to that of positive fixed points of $A$. \begin{lemma} \label{lem2.1} Let $\kappa:=1-2/e$ and $\psi(t):=te^t, t\in [0,1]$. Then $\psi(t)$ is nonnegative on $[0,1]$ and $$\label{Gineq} \kappa \psi(s)\le \int_0^1 k_1(t,s)\psi(t)\,{\rm d} t\le \psi(s).$$ \end{lemma} \begin{lemma} \label{lem2.2} Let $u$ is concave, increasing and nonnegative on $[0,1]$. Then $$\label{le4} \int_0^1u(t)\psi(t)\,{\rm d} t\ge \kappa e\|u\|.$$ \end{lemma} \begin{proof} The concavity of $u$ and $\max_{t\in [0,1]}u(t)=u(1)=\|u\|$ imply $\int_0^1 u(t)\psi(t)\,{\rm d} t =\int_0^1 u(t\cdot 1+(1-t)\cdot 0)\psi(t)\,{\rm d} t\ge u(1)\int_0^1t\psi(t)\,{\rm d} t= \kappa e\|u\|.$ This completes the proof. \end{proof} \begin{lemma}[\cite{Yang2}] \label{lem2.3} Let $u\in P$ and $q>0$. Then $$\label{hkeq} \int_0^1 \Big[(B_{n-1}u^q)(t)+2\sum_{i=0}^{n-2}((B_{n-1-i}u^q)(t))'\Big] \psi(t)\,{\rm d} t=\int_0^1 u^q(t)\psi(t)\,{\rm d} t.$$ \end{lemma} \begin{lemma}[\cite{GuoDajun}] \label{lem2.4} Let $\Omega \subset E$ be a bounded open set and $A:\overline{\Omega}\cap P\to P$ is a completely continuous operator. If there exists $v_0\in P\setminus\{0\}$ such that $v-Av\neq \lambda v_0$ for all $v\in \partial\Omega\cap P$ and $\lambda \geq 0$, then $i(A,\Omega\cap P,P)=0$, where $i$ is the fixed point index on $P$. \end{lemma} \begin{lemma}[\cite{GuoDajun}] \label{lem2.5} Let $\Omega \subset E$ be a bounded open set with $0\in \Omega$. Suppose $A:\overline{\Omega}\cap P\to P$ is a completely continuous operator. If $v\neq \lambda Av$ for all $v\in \partial\Omega \cap P$ and $0\leq \lambda\leq 1$, then $i(A,\Omega\cap P,P)=1$. \end{lemma} \begin{lemma}[Jensen's inequalities] \label{lem2.6} Let $\theta>0$ and $\varphi\in C([0,1],\mathbb R^+)$. Then \begin{gather*} \Big(\int_0^1 \varphi (t)\,{\rm d} t\Big)^\theta \leq \int_0^1 (\varphi(t))^\theta\,{\rm d} t, \quad \text{if }\theta\geq 1,\\ \Big(\int_0^1 \varphi (t)\,{\rm d} t\Big)^\theta \geq \int_0^1 (\varphi(t))^\theta\,{\rm d} t,\quad \text{if } 0<\theta\leq 1. \end{gather*} \end{lemma} \section{Main results} For brevity, we define $y=(y_1,y_2,\ldots,y_{2n-1},y_{2n})\in \mathbb R^{2n}_+$, $\gamma_p:=\max\{1,2^{p-2}\}$, $p_*=\min\{1,p-1\}$, $p^*=\max\{1,p-1\}$, $\mathscr{K}_i:=\max_{t,s\in [0,1]}k_i(t,s)>0$, $\mathscr{H}_i:=\max_{t,s\in [0,1]}h_i(t,s)>0$, \begin{gather*} \beta_p:=\Big\{2^{p_*-1}\kappa\Big[(n-1) \Big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p_*-1}+1\Big]\Big\} ^{1-p/p_*}, \\ \alpha_p:=\Big\{2^{p^*-1}\Big[(n-1) \Big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p^*-1}+1\Big]\Big\} ^{1-p/p^*}. \end{gather*} \begin{itemize} \item[(H1)] $f\in C([0,1]\times \mathbb{R}_+^{2n},\mathbb{R}_+)$. \item[(H2)] There exist $a_1>\beta_p$ and $c>0$ such that $f(t,y)\ge a_1\Big(\sum_{i=1}^{n-1}(y_{2i-1}+2(n-i)y_{2i})+y_{2n-1}\Big)^{p-1}-c, \quad\text{for all }y\in \mathbb R^{2n}_+\text{ and }t\in [0,1].$ \item[(H3)] For any $M_0>0$ there is a function $\Phi_{M_0}\in C(\mathbb{R}_+,\mathbb{R_+})$ such that \begin{gather*} f(t,y)\le \Phi_{M_0}(y_{2n}^{p-1}), \forall (t,y)\in [0,1]\times [0,{M_0}]^{2n-1}\times \mathbb{R}_+,\\ \int_\delta^\infty\frac{\,{\rm d} \xi}{\Phi_{M_0}(\xi)}=\infty\quad \text{for any }\delta>0. \end{gather*} \item[(H4)] There exist $b_1\in (0,\alpha_p)$ and $r>0$ such that $f(t,y)\le b_1\Big(\sum_{i=1}^{n-1}(y_{2i-1}+2(n-i)y_{2i})+y_{2n-1}\Big)^{p-1} \quad \text{for all y\in [0,r]^{2n} and }t\in [0,1].$ \item[(H5)] There exist $a_2>\beta_p$ and $r>0$ such that $f(t,y)\ge a_2\Big(\sum_{i=1}^{n-1}(y_{2i-1}+2(n-i)y_{2i})+y_{2n-1}\Big)^{p-1} \text{ for all y\in [0,r]^{2n} and }t\in [0,1].$ \item[(H6)] There exist $b_2\in (0,\alpha_p)$ and $c>0$ such that $f(t,y)\le b_2\Big(\sum_{i=1}^{n-1}(y_{2i-1}+2(n-i)y_{2i})+y_{2n-1}\Big)^{p-1}+c \text{ for all y\in \mathbb R^{2n}_+ and }t\in [0,1].$ \item[(H7)] $f$ is increasing in $y$ and there is a constant $\omega > 0$ such that $\int_0^1 f^{p^*/(p-1)}(s,\omega,\ldots,\omega)\,{\rm d} s<\omega.$ \end{itemize} \begin{remark} \label{rmk3.1} \rm A function $f$ is said to be increasing in $y$ if $f(t, x) \le f(t, y)$ holds for every pair $x, y \in\mathbb{R}^{2n}_+$ with $x \le y$, where the partial ordering $\le$ in $\mathbb{R}^{2n}_+$ is understood componentwise.\end{remark} \begin{theorem} \label{thm3.2} If {\rm (H1)--(H4)} hold, then \eqref{Problem} has at least one positive solution. \end{theorem} \begin{proof} Let $\mathscr{M}_1: =\{u\in P: u=Au+\lambda \varphi, \text{ for some } \lambda\ge 0\},$ where $\varphi(t):=te^{-t}$. Clearly, $\varphi(t)$ is nonnegative and concave on $[0, 1]$. We claim $\mathscr{M}_1$ is bounded. We first establish the a priori bound of $\|u\|_0$ for $\mathscr{M}_1$. Indeed, $u\in\mathscr{M}_1$ implies $u$ is concave (by the concavity of $A$ and $\varphi$) and $u(t) \ge (Au)(t)$. By definition we obtain $$u(t)\ge \int_0^t \Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))', \ldots,u(\tau),u'(\tau))\,{\rm d} \tau\Big)^{\frac{1}{p-1}}\,{\rm d} s$$ for all $u\in\mathscr{M}_1$. Note that $p_*, p_*/p-1\in [0,1]$. Now, by (H2), we find $$\Big[a_1\Big(\sum_{i=1}^{n-1}(y_{2i-1}+2(n-i)y_{2i})+y_{2n-1} \Big)^{p-1}\Big]^{\frac{p_*}{p-1}} \le (f(t,y)+c)^{\frac{p_*}{p-1}}\le f^{\frac{p_*}{p-1}}(t,y)+c^{\frac{p_*}{p-1}}.$$ Combining this and Jensen's inequality, we obtain \label{A-1} \begin{aligned} & u^{p_*}(t)\\ & \ge \Big[\int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau))\,{\rm d} \tau\Big)^{\frac{1}{p-1}}\,{\rm d} s\Big]^{p_*}\\ & \ge \int_0^t\int_s^1 f^{\frac{p_*}{p-1}}(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau), u'(\tau))\,{\rm d} \tau\,{\rm d} s\\ & =\int_0^1 k_1(t,s)f^{\frac{p_*}{p-1}}(s,(B_{n-1}u)(s),((B_{n-1}u)(s))', \ldots,u(s),u'(s))\,{\rm d} s\\ &\ge a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)\Big[\sum_{i=1}^{n-1}\left((B_iu)(s)+2(n-i)((B_iu)(s))'\right)+u(s) \Big]^{p_*}\,{\rm d} s-\frac{c^{\frac{p_*}{p-1}}}{2}\\ &\ge 2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)\Big[ \sum_{i=1}^{n-1}\Big((B_iu)(s)+2(n-i)((B_iu)(s))'\Big)\Big]^{p_*}\,{\rm d} s\\ &\quad +2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)u^{p_*}(s)\,{\rm d} s-\frac{c^{\frac{p_*}{p-1}}}{2}\\ &= 2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)\Big[\sum_{i=1}^{n-1}\int_0^1 (k_i(s,\tau)+2(n-i)h_i(s,\tau))u(\tau)\,{\rm d} \tau\Big]^{p_*}\,{\rm d} s\\ &\quad +2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)u^{p_*}(s)\,{\rm d} s-\frac{c^{\frac{p_*}{p-1}}}{2}\\ & = 2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)\Big[\int_0^1 \frac{\sum_{i=1}^{n-1}(k_i(s,\tau)+2(n-i)h_i(s,\tau))}{\sum_{i=1}^{n-1} (\mathscr{K}_i+\mathscr{H}_i)}\sum_{i=1}^{n-1}(\mathscr{K}_i\\ &\quad +\mathscr{H}_i) u(\tau)\,{\rm d} \tau\Big]^{p_*}\,{\rm d} s +2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)u^{p_*}(s)\,{\rm d} s-\frac{c^{\frac{p_*}{p-1}}}{2}\\ & \ge \Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p_*-1} a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)\\ &\quad\times \Big[\sum_{i=1}^{n-1}\int_0^1 (k_i(s,\tau)+2(n-i)h_i(s,\tau)) u^{p_*}(\tau)\,{\rm d} \tau\Big]\,{\rm d} s\\ &\quad +2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)u^{p_*}(s)\,{\rm d} s-\frac{c^{\frac{p_*}{p-1}}}{2}\\ &=\Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p_*-1} a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)\\ &\quad\times \Big[\sum_{i=1}^{n-1}((B_iu^{p_*})(s)+2(n-i)((B_iu^{p_*})(s))') \Big]\,{\rm d} s\\ &\quad +2^{p_*-1}a_1^{\frac{p_*}{p-1}} \int_0^1 k_1(t,s)u^{p_*}(s)\,{\rm d} s-\frac{c^{\frac{p_*}{p-1}}}{2}. \end{aligned} Multiply both sides of the above expression by $\psi(t)$ and integrate over [0,1] and use \eqref{Gineq} and \eqref{hkeq} to obtain \begin{aligned} & \int_0^1 \psi(t)u^{p_*}(t)\,{\rm d} t\\ &\ge \Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p_*-1} a_1^{\frac{p_*}{p-1}}\kappa \int_0^1 \psi(t)\Big[\sum_{i=1}^{n-1}\big((B_iu^{p_*})(t)\\ &\quad +2(n-i)((B_iu^{p_*})(t))'\big) \Big]\,{\rm d} t +2^{p_*-1}a_1^{\frac{p_*}{p-1}}\kappa \int_0^1 \psi(t)u^{p_*}(t)\,{\rm d} t-\frac{c^{\frac{p_*}{p-1}}}{2} \\ & =2^{p_*-1}a_1^{\frac{p_*}{p-1}}\kappa\Big[(n-1) \Big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p_*-1}+1\Big] \int_0^1 \psi(t)u^{p_*}(t)\,{\rm d} t-\frac{c^{\frac{p_*}{p-1}}}{2}. \end{aligned} Therefore, $\int_0^1 \psi(t)u^{p_*}(t)\,{\rm d} t \le \frac{c^{\frac{p_*}{p-1}}}{2^{p_*}a_1^{\frac{p_*}{p-1}}\kappa \Big[(n-1) \Big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p_*-1}+1 \Big]-2}:=\mathscr{N}_1.$ Recall that every $u \in \mathscr{M}_1$ is concave and increasing on $[0,1]$. So is $u^{p_*}$ with $p_*\in (0,1]$. Now Lemma \ref{lem2.2} yields $$\|u\|_0\le (\kappa e)^{-1/p_*}\mathscr{N}_1^{1/p_*}$$ for all $u\in\mathscr{M}_1$, which implies the a priori bound of $\|u\|_0$ for $\mathscr{M}_1$, as claimed. It follows, from the boundedness of $\|u\|_0$ for $\mathscr{M}_1$, that there is $\lambda_0 > 0$ such that $\lambda\le \lambda_0$ for all $\lambda\in \Lambda$, where $\Lambda:=\{\lambda\ge 0: \text{ there exists } u\in \mathscr{M}_1 \text{ such that } u=Au+\lambda \varphi\}.$ If $u\in \mathscr{M}_1$, then $u'(t)=\Big(\int_t^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))', \ldots,u(\tau),u'(\tau))\,{\rm d} \tau\Big)^{\frac{1}{p-1}}+\lambda(1-t)e^{-t}$ for some $\lambda\ge 0$, and by (H3), \begin{align*} (u')^{p-1}(t) & \le \gamma_p\int_t^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))', \ldots,u(\tau),u'(\tau))\,{\rm d} \tau+\gamma_p\lambda_0^{p-1}\\ & \le \gamma_p\int_t^1 \Phi_{M_0}((u')^{p-1}(\tau))\,{\rm d} \tau+\gamma_p\lambda_0^{p-1}. \end{align*} Let $v(t):= (u')^{p-1}(t)$. Then $v(t)\in C([0,1], \mathbb{R}_+)$ and $v(1)=0$. Moreover, $v(t)\le \gamma_p\int_t^1 \Phi_{M_0}(v(\tau))\,{\rm d} \tau+\gamma_p\lambda_0^{p-1}.$ Let $F(t):= \int_t^1 \Phi_{M_0}(v(\tau))\,{\rm d} \tau$. Then $-F'(t)=\Phi_{M_0}(v(t))\le \Phi_{M_0}(\gamma_pF(t)+\gamma_p\lambda_0^{p-1}).$ Therefore, $\int_{\gamma_p\lambda_0^{p-1}}^{v(t)}\frac{\,{\rm d} \xi}{\Phi_{M_0}(\xi)}\le \int_{\gamma_p\lambda_0^{p-1}}^{\gamma_pF(t) +\gamma_p\lambda_0^{p-1}}\frac{\,{\rm d} \xi}{\Phi_{M_0}(\xi)}\le \gamma_p(1-t).$ Hence there is $\mathscr{N}_2 > 0$ such that$\|(u')^{p-1}\|_0=\|v\|_0=v(0)\le \mathscr{N}_2.$ Let $\mathscr{N}_3:=\max\{(\kappa e)^{-1/p_*}\mathscr{N}_1^{1/p_*},\mathscr{N}_2^{1/p-1}\}$. Then $\|u\|\le\mathscr{N}_3, \quad \forall u\in \mathscr{M}_1.$ This proves the boundedness of $\mathscr{M}_1$. As a result of this, for every $R>\mathscr{N}_3$, we have $u-Au\not=\lambda \psi,\quad \forall u\in\partial B_R\cap P, \; \lambda\geq 0.$ Now by Lemma \ref{lem2.4}, we obtain $$\label{fi-1} i(A,B_R\cap P,P)=0.$$ Let $\mathscr{M}_2:=\{u\in \overline{B}_r\cap P:u=\lambda Au \text{ for some } \lambda\in [0,1]\}.$ We shall prove $\mathscr{M}_2= \{0\}$. Indeed, if $u\in \mathscr{M}_2$, we have for any $u\in \overline{B}_r\cap P$ $$u(t)\le \int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau))\,{\rm d} \tau \Big)^{\frac{1}{p-1}}\,{\rm d} s.$$ Notice that $p^*, p^*/p-1\ge 1$. Now, similar to \eqref{A-1}, by Jensen's inequality and (H4), we obtain \label{A-2} \begin{aligned} &u^{p^*}(t)\\ & \le \Big[\int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau)) \,{\rm d} \tau\Big)^{\frac{1}{p-1}}\,{\rm d} s\Big]^{p^*}\\ & \le \int_0^1 k_1(t,s)f^{p^*/(p-1)}(s,(B_{n-1}u)(s),((B_{n-1}u)(s))', \ldots,u(s),u'(s))\,{\rm d} s\\ & \le b_1^{p^*/(p-1)} \int_0^1 k_1(t,s)\Big[\sum_{i=1}^{n-1}\Big((B_iu)(s)+2(n-i)((B_iu)(s))'\Big)+u(s) \Big]^{p^*}\,{\rm d} s\\ &\le 2^{p^*-1}b_1^{p^*/(p-1)} \int_0^1 k_1(t,s)\Big[\sum_{i=1}^{n-1}\int_0^1 (k_i(s,\tau)+2(n-i)h_i(s,\tau))u(\tau)\,{\rm d} \tau\Big]^{p^*}\,{\rm d} s\\ &\quad +2^{p^*-1}b_1^{p^*/(p-1)} \int_0^1 k_1(t,s)u^{p^*}(s)\,{\rm d} s\\ & = 2^{p^*-1}b_1^{p^*/(p-1)} \int_0^1 k_1(t,s) \Big[\int_0^1 \frac{\sum_{i=1}^{n-1}(k_i(s,\tau)+2(n-i)h_i(s,\tau))} {\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)}\\ &\quad\times \sum_{i=1}^{n-1} (\mathscr{K}_i+\mathscr{H}_i) u(\tau)\,{\rm d} \tau\Big]^{p^*}\,{\rm d} s +2^{p^*-1}b_1^{p^*/(p-1)} \int_0^1 k_1(t,s)u^{p^*}(s)\,{\rm d} s\\ & \le \Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p^*-1} b_1^{p^*/(p-1)} \int_0^1 k_1(t,s) \Big[\sum_{i=1}^{n-1}((B_iu^{p^*})(s)\\ &\quad +2(n-i)((B_iu^{p^*})(s))') \Big]\,{\rm d} s +2^{p^*-1}b_1^{p^*/(p-1)} \int_0^1 k_1(t,s)u^{p^*}(s)\,{\rm d} s. \end{aligned} Multiply both sides of the above expression by $\psi(t)$ and integrate over $[0,1]$ and use \eqref{Gineq} and \eqref{hkeq} to obtain \begin{aligned} & \int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t\\ &\le \Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p^*-1} b_1^{p^*/(p-1)} \int_0^1 \psi(t)\Big[\sum_{i=1}^{n-1} \big((B_iu^{p^*})(t)\\ &\quad +2(n-i)((B_iu^{p^*})(t))'\big)\Big]\,{\rm d} t +2^{p^*-1}b_1^{p^*/(p-1)} \int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t\\ & =2^{p^*-1}b_1^{p^*/(p-1)}\Big[(n-1) \Big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p^*-1}+1\Big] \int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t. \end{aligned} Therefore, $\int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t=0$, whence $u(t)\equiv 0, \forall u\in\mathscr{M}_2$. As a result, $\mathscr{M}_2= \{0\}$, as claimed. Consequently, $u\neq \lambda Au, \quad \forall u\in \partial B_r\cap P, \;\lambda\in [0,1].$ Now Lemma \ref{lem2.5} yields $$\label{index11} i(A,B_r\cap P,P)=1.$$ Combining this with \eqref{fi-1} gives $i(A,(B_R\backslash \overline{B}_r)\cap P,P)=0-1=-1.$ Hence the operator $A$ has at least one fixed point on $(B_R\setminus \overline{B}_r)\cap P$ and therefore \eqref{Problem} has at least one positive solution. This completes the proof. \end{proof} \begin{theorem} \label{thm3.3} If {\rm (H1), (H5), (H6)} are satisfied, then \eqref{Problem} has at least one positive solution. \end{theorem} \begin{proof} Let $\mathscr{M}_3:=\{u\in \overline{B}_r\cap P: u=Au+\lambda \psi\ \text{for some}\ \lambda\ge 0\}.$ We claim $\mathscr{M}_3\subset\{0\}$. Indeed, if $u\in\mathscr{M}_3$, then we have $u\geq Au$ by definition. That is, $$u(t)\ge \int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau), ((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau))\,{\rm d} \tau\Big)^{\frac{1}{p-1}} \,{\rm d} s.$$ Similar to $\mathscr{M}_2= \{0\}$, we can also obtain $\mathscr{M}_3\subset\{0\}$. As a result of this, we have $u-Au\not=\lambda \psi, \forall u\in\partial B_r\cap P, \lambda\geq 0.$ Now Lemma \ref{lem2.4} gives $$\label{theorem2index0} i(A,B_r\cap P,P)=0.$$ Let $\mathscr{M}_4:=\{u\in P:u=\lambda Au \text{ for some } \lambda\in [0,1]\}.$ We assert $\mathscr{M}_4$ is bounded. We first establish the a priori bound of $\|u\|_0$ for $\mathscr{M}_4$. Indeed, if $u\in \mathscr{M}_4$, then $u$ is concave and $u\leq Au$, which can be written in the form $$u(t)\le \int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau),((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau)) \,{\rm d} \tau\Big)^{\frac{1}{p-1}}\,{\rm d} s,$$ for all $u\in \mathscr{M}_4$. Note that $p^*, p^*/p-1\ge 1$. Now by (H6) and Jensen's inequality, we obtain \label{A-4} \begin{aligned} u^{p^*}(t) & \le \Big[\int_0^t\Big(\int_s^1 f(\tau,(B_{n-1}u)(\tau), ((B_{n-1}u)(\tau))',\ldots,u(\tau),u'(\tau))\,{\rm d} \tau\Big)^{\frac{1}{p-1}} \,{\rm d} s\Big]^{p^*}\\ & \le \int_0^1 k_1(t,s)f^{p^*/(p-1)}(s,(B_{n-1}u)(s), ((B_{n-1}u)(s))',\ldots,u(s),u'(s))\,{\rm d} s\\ & \le \int_0^1 k_1(t,s)\Big\{b_2\Big[\sum_{i=1}^{n-1} \big((B_iu)(s)+2(n-i)((B_iu)(s))'\big) +u(s)\Big]^{p-1}\\ &\quad +c\Big\}^{p^*/(p-1)}\,{\rm d} s\\ &\le b_3^{p^*/(p-1)} \int_0^1 k_1(t,s)\Big[\sum_{i=1}^{n-1}\left((B_iu)(s)+2(n-i)((B_iu)(s))'\right)+u(s) \Big]^{p^*}\,{\rm d} s\\ &\quad +\frac{c_1^{p^*/(p-1)}}{2}\\ &\le 2^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 k_1(t,s) \Big[\sum_{i=1}^{n-1}\left((B_iu)(s)+2(n-i)((B_iu)(s))'\right)\Big]^{p^*} \,{\rm d} s\\ &\quad +2^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 k_1(t,s)u^{p^*}(s)\,{\rm d} s+\frac{c_1^{p^*/(p-1)}}{2}\\ &= 2^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 k_1(t,s) \Big[\int_0^1 \frac{\sum_{i=1}^{n-1}(k_i(s,\tau)+2(n-i)h_i(s,\tau))} {\sum_{i=1}^{n-1} (\mathscr{K}_i+\mathscr{H}_i)}\\ &\quad\times \sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i) u(\tau)\,{\rm d} \tau\Big]^{p^*}\,{\rm d} s +2^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 k_1(t,s)u^{p^*}(s)\,{\rm d} s\\ &\quad +\frac{c_1^{p^*/(p-1)}}{2}\\ & \le \Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i) \Big)^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 k_1(t,s) \Big[\sum_{i=1}^{n-1}((B_iu^{p^*})(s)\\ &\quad +2(n-i)((B_iu^{p^*})(s))') \Big]\,{\rm d} s +2^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 k_1(t,s)u^{p_*}(s)\,{\rm d} s\\ &\quad +\frac{c_1^{p^*/(p-1)}}{2}, \end{aligned} for all $u\in \mathscr{M}_4$, $b_3\in (b_2,\alpha_p)$ and $c_1>0$ being chosen so that $(b_2z+c)^{p^*/(p-1)}\le b_3^{p^*/(p-1)} z^{p^*/(p-1)}+c_1^{p^*/(p-1)}, \forall z\geq 0.$ Multiply both sides of \eqref{A-4} by $\psi(t)$ and integrate over $[0,1]$ and use \eqref{Gineq} and \eqref{hkeq} to obtain \begin{aligned} & \int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t\\ &\le \Big(2\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p^*-1} b_3^{p^*/(p-1)} \int_0^1 \psi(t)\Big[\sum_{i=1}^{n-1}\big((B_iu^{p^*})(t)\\ &\quad +2(n-i) ((B_iu^{p^*})(t))'\big) \Big]\,{\rm d} t +2^{p^*-1}b_3^{p^*/(p-1)} \int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t+\frac{c_1^{p^*/(p-1)}}{2}\\ & =2^{p^*-1}b_3^{p^*/(p-1)}\Big[(n-1) \Big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\Big)^{p^*-1}+1\Big] \int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t\\ &\quad +\frac{c_1^{p^*/(p-1)}}{2}. \end{aligned} Therefore, $\int_0^1 \psi(t)u^{p^*}(t)\,{\rm d} t \le \frac{c_1^{p^*/(p-1)}}{2-2^{p^*}b_3^{p^*/(p-1)} \big[(n-1) \big(\sum_{i=1}^{n-1}(\mathscr{K}_i+\mathscr{H}_i)\big)^{p^*-1}+1\big]} :=\mathscr{N}_4.$ This, together with Jensen's inequality and $\psi(t)/e\in [0,1]$ (Note that $p^*\ge 1$), leads to $$e\int_0^1 u(t)\frac{\psi (t)}{e} \,{\rm d} t \le e\Big(\int_0^1 u^{p^*}(t)\big(\frac{\psi(t)}{e}\big)^{p^*} \,{\rm d} t\Big)^{1/p^*} \le e^{\frac{p^*-1}{p^*}} \mathscr{N}_4^{1/p^*}$$ for all $u\in\mathscr{M}_4$. From Lemma \ref{lem2.2}, we find $\|u\|_0\le \kappa^{-1} e^{-1/p^*}\mathscr{N}_4^{1/p^*} :=\mathscr{N}_5, \forall u\in \mathscr{M}_4,$ which implies the a priori bound of $\|u\|_0$ for $\mathscr{M}_4$, as claimed. Furthermore, for any positive integer $i\ge 1$, this estimate leads to $\|(B_iu)\|_0=(B_iu)(1)\le \mathscr{N}_5, \quad \forall u\in\mathscr{M}_4$ and for each positive integer $i\ge 2$, we see $\|(B_iu)'\|_0=(B_iu)'(0)=\int_0^1 (B_{i-1}u)(t)\,{\rm d} t\le \mathscr{N}_5, \forall u\in\mathscr{M}_4.$ Moreover, for $i=1$, we have $\|(B_1u)'\|_0=(B_1u)'(0)=\int_0^1 u(t)\,{\rm d} t \le \mathscr{N}_5, \forall u\in\mathscr{M}_4.$ Combining these and (H6), we have \begin{gather*} -((u')^{p-1})'\le f(t,(B_{n-1}u)(t),((B_{n-1}u)(t))',\ldots,u,u'),\\ u(0)=u'(1)=0. \end{gather*} Let $(u')^{p-1}(t):=w'(t)$. Then $w\in C([0,1], \mathbb{R}_+)$ and $u'(1)=0$ implies $w'(1)=0$. Therefore, $-w''(t)\le n^2 \mathscr N_5, \forall u\in\mathscr{M}_4,$ so that $\|w'\|_0=w'(0)\le n^2 \mathscr N_5.$ Consequently, $\|u'\|_0=\|w'\|_0^{1/p-1}\le (n^2 \mathscr N_5)^{1/p-1},\quad \forall u\in\mathscr{M}_4.$ Let $\mathscr{N}_6:=\max\{\mathscr{N}_5,(n^2 \mathscr N_5)^{1/p-1}\}$. Then $\|u\|\le\mathscr{N}_6, \quad \forall u\in \mathscr{M}_4.$ This proves the boundedness of $\mathscr{M}_4$. As a result of this, for every $R>\mathscr{N}_6$, we have $u\neq \lambda Au, \quad \forall u\in \partial B_R\cap P, \lambda\in [0,1].$ Now Lemma \ref{lem2.5} yields $$\label{index10} i(A,B_R\cap P,P)=1.$$ Combining this with \eqref{theorem2index0} gives $i(A,(B_R\backslash \overline{B}_r)\cap P,P)=1-0=1.$ Hence the operator $A$ has at least one fixed point on $(B_R\setminus \overline{B}_r)\cap P$ and therefore \eqref{Problem} has at least one positive solution. This completes the proof. \end{proof} \begin{theorem} \label{thm3.4} If {\rm (H1)--(H3), (H6), (H7)} are satisfied. Then \eqref{Problem} has at least two positive solutions. \end{theorem} \begin{proof} By (H2), (H3), and (H6), we know that \eqref{fi-1} and \eqref{theorem2index0} hold. Note we may choose $R >\omega > r$ in \eqref{fi-1} and \eqref{theorem2index0} (see the proofs of Theorems \ref{thm3.2} and \ref{thm3.3}). By (H7) and Jensen's inequality, we have that for all $u \in \partial B_\omega \cap P$, \begin{aligned} &[(Au)(t)]^{p^*}\\ & \le \int_0^1 k_1(t,s)f^{p^*/(p-1)}(s,(B_{n-1}u)(s),((B_{n-1}u)(s))',\ldots,u(s),u'(s))\,{\rm d} s\\ & \le \int_0^1 f^{p^*/(p-1)}(s,(B_{n-1}u)(s),((B_{n-1}u)(s))',\ldots,u(s),u'(s))\,{\rm d} s <\omega \end{aligned} and \begin{aligned} &[((Au)(t))']^{p^*}\\ & =\left(\int_t^1 f(s,(B_{n-1}u)(s),((B_{n-1}u)(s))',\ldots,u(s),u'(s))\,{\rm d} s\right)^{p^*/(p-1)} \\ &\le \int_0^1 f^{p^*/(p-1)}(s,(B_{n-1}u)(s),((B_{n-1}u)(s))',\ldots,u(s),u'(s))\,{\rm d} s<\omega. \end{aligned} Thus we obtain $\|Au\|<\omega=\|u\|,\quad \forall u \in \partial B_\omega \cap P,$ This implies $u\neq \lambda Au, \quad \forall u\in \partial B_\omega\cap P, \;\lambda\in [0,1].$ Now Lemma \ref{lem2.5} yields $$i(A,B_\omega\cap P,P)=1.$$ Combining this with \eqref{fi-1} and \eqref{theorem2index0} gives $i(A,(B_R\backslash \overline{B}_\omega)\cap P,P)=0-1=-1,\quad i(A,(B_\omega\backslash \overline{B}_r)\cap P,P)=1-0=1.$ Hence the operator $A$ has at least two fixed points, with one on $(B_R\setminus \overline{B}_\omega)\cap P$ and the other on $(B_\omega\setminus \overline{B}_r)\cap P$. 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