\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 114, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/114\hfil Conjugate boundary-value problems] {Solvability of (k,n-k) conjugate boundary-value problems at resonance} \author[W. Jiang, J. Qiu \hfil EJDE-2012/114\hfilneg] {Weihua Jiang, Jiqing Qiu} % in alphabetical order \address{Weihua Jiang \newline College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, China} \email{weihuajiang@hebust.edu.cn} \address{Jiqing Qiu \newline College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, China} \email{qiujiqing@263.net} \thanks{Submitted May 1, 2012. Published July 5, 2012.} \subjclass[2000]{35B34, 34B10, 34B15} \keywords{Resonance; Fredholm operator; boundary-value problem} \begin{abstract} Using the coincidence degree theory due to Mawhin and constructing suitable operators, we prove the existence of solutions for $(k,n-k)$ conjugate boundary-value problems at resonance. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} The existence of solutions for $(k,n-k)$ conjugate boundary-value problems at non-resonance has been studied in many papers (see \cite{1,2,3,4,7,8,11,12,13,14,17,22,26,27,28,31,32,33}). For example, using fixed point theorem in a cone, Jiang \cite{13} obtained the existence of positive solutions for $(k,n-k)$ conjugate boundary-value problem \begin{gather*} (-1)^{n-k}y^{(n)}(t)=f(t,y(t)),\quad 00$, there exists$\Phi_r\in L^\infty[0,1]$such that$|f(t,x_1,x_2,\dots,x_n)|\leq\Phi_r(t)$for all$|x_i|\leq r$,$i=1,2,\dots,n$, a.e.$t\in[0,1]$. \end{itemize} \section{Preliminaries}\label{s2} First, we introduce some notation and state a theorem to be used later. For more details see \cite{24}. Let$X$and$Y$be real Banach spaces and$L:\operatorname{dom}L\subset X\to Y$be a Fredholm operator with index zero,$P:X\to X$,$Q:Y\to Y$be projectors such that $$\operatorname{Im}P=\ker L,\quad \ker Q=\operatorname{Im}L,\quad X=\ker L\oplus \ker P,\quad Y=\operatorname{Im}L\oplus \operatorname{Im}Q.$$ It follows that $$L\big|_{\operatorname{dom}L\cap \ker P}:\operatorname{dom}L\cap \ker P\to \operatorname{Im}L$$ is invertible. We denote the inverse by$K_{P}$. Assume that$\Omega$is an open bounded subset of$X$,$\operatorname{dom}L\cap\overline{\Omega}\neq\emptyset$, the map$N:X\to Y$will be called$L$-compact on$\overline{\Omega}$if$QN(\overline{\Omega})$is bounded and$K_{P}(I-Q)N:\overline{\Omega}\to X$is compact. \begin{theorem}[\cite{24}] \label{thm2.1} Let$L:\operatorname{dom}L\subset X\to Y$be a Fredholm operator of index zero and$N:X\to YL$-compact on$\overline{\Omega}$. Assume that the following conditions are satisfied: \begin{itemize} \item[(1)]$Lx\neq \lambda Nx$for every$(x,\lambda)\in [(\operatorname{dom}L\setminus \ker L)\cap\partial\Omega]\times(0,~1)$; \item[(2)]$Nx\notin \operatorname{Im}L$for every$x\in \ker L\cap\partial\Omega$; \item[(3)]$\deg(QN|_{\ker L},\Omega\cap \ker L,0)\neq 0$, where$Q:Y\to Y$is a projection such that$\operatorname{Im}L=\ker Q$. \end{itemize} Then the equation$Lx=Nx$has at least one solution in$\operatorname{dom}L\cap\overline\Omega$. \end{theorem} Take$X=C^{n-1}[0,1]$with norm$\|u\|=\max\{\|u\|_{\infty},\|u'\|_{\infty},\dots,\|u^{(n-1)}\|_{\infty}\}$, where$\|u\|_{\infty}=\max_{t\in[0,1]}|u(t)|$,$Y=L^1[0,1]$with norm$\|x\|_1=\int_0^1|x(t)|dt$. Define the operator$Ly(t)=(-1)^{n-k}y^{(n)}(t)with \begin{align*} \operatorname{dom}L=\{&y\in X: y^{(n)}\in Y,\; y^{(i)}(0)=y^{(j)}(1)=0,\; 0\leq i\leq k-1, \\ & 0\leq j\leq n-k-2,\; y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i)\}. \end{align*} LetN:X\to Y$be defined as $$Ny(t)=f(t,y(t),y'(t),\dots,y^{(n-1)}(t))+\varepsilon(t),\quad t\in [0,1].$$ Then problem \eqref{e1.1}, \eqref{e1.2} becomes$Ly=Ny$. \section{Main results} By Cramer's rule, we can get the following lemmas. \begin{lemma} \label{lem3.1} For given$u\in Y, the system of linear equations \begin{gathered} \frac{x_k}{k!}+\frac{x_{k+1}}{(k+1)!}+\dots+\frac{x_{n-2}}{(n-2)!} +\frac{(-1)^{n-k}}{(n-1)!}\int_0^1(1-s)^{n-1}u(s)ds=0\\ \frac{x_k}{(k-1)!}+\frac{x_{k+1}}{k!}+\dots+\frac{x_{n-2}}{(n-3)!} +\frac{(-1)^{n-k}}{(n-2)!}\int_0^1(1-s)^{n-2}u(s)ds=0\\ \dots\\ \begin{aligned} &\frac{x_k} {[k-(n-k-2)]!}+\frac{x_{k+1}}{[k+1-(n-k-2)]!} +\dots+ \frac{x_{n-2}}{[n-2-(n-k-2)]!}\\ &+ \frac{(-1)^{n-k}}{[n-1-(n-k-2)]!}\int_0^1(1-s)^{k+1}u(s)ds=0 \end{aligned} \end{gathered}\label{e3.1} has an only one solution,(x_k,x_{k+1},\dots,x_{n-2})with \begin{align*} x_m&=\int_0^1\frac{(-1)^{n-k-1}m!}{(m-k)!(k-1)!(n-m-2)!} \sum_{i=0}^{m-k}(-1)^{m-k-i}\frac{C_{m-k}^i}{m-i}\\ &\quad \times \Big[\sum_{j=0}^{n-m-2}(-1)^{j}C_{n-m-2}^j \frac{(1-s)^{n-1-i-j}}{n-1-i-j}\Big]u(s)ds,\quad m=k,k+1,\dots,n-2. \end{align*} \end{lemma} \begin{lemma} \label{lem3.2} The system of linear equations \begin{gathered} \frac{x_k}{k!}+\frac{x_{k+1}}{(k+1)!}+\dots+\frac{x_{n-2}}{(n-2)!}+\frac{1}{(n-1)!}=0 \\ \frac{x_k}{(k-1)!}+\frac{x_{k+1}}{k!}+\dots+\frac{x_{n-2}}{(n-3)!}+\frac{1}{(n-2)!}=0\\ \dots\\ \begin{aligned} &\frac{x_k} {[k-(n-k-2)]!}+\frac{x_{k+1}}{[k+1-(n-k-2)]!} +\dots \\ &+\frac{x_{n-2}}{[n-2-(n-k-2)]!}+ \frac{1}{[n-1-(n-k-2)]!}=0 \end{aligned} \end{gathered} \label{e3.2} has an only one solution,(x_k,x_{k+1},\dots,x_{n-2})with \begin{align*} x_m&=-\frac{m!}{(m-k)!(k-1)!(n-m-2)!}\sum_{i=0}^{m-k}(-1)^{m-k-i} \frac{C_{m-k}^i}{m-i}\\ &\quad \times \Big(\sum_{j=0}^{n-m-2}(-1)^{j}C_{n-m-2}^j\frac{1}{n-1-i-j}\Big), \quad m=k,k+1,\dots,n-2. \end{align*} \end{lemma} Let(B_k(u),B_{k+1}(u),\dots,B_{n-2}(u))$denote the only solution of \eqref{e3.1}, and let$(A_k,A_{k+1},\dots,A_{n-2})$denote the only solution of \eqref{e3.2}, and let$A_{n-1}=1$. In order to obtain our main results, we firstly present and prove the following lemmas. \begin{lemma} \label{lem3.3} Suppose {\rm (H1)} holds, then$L:\operatorname{dom}L\subset X\to Y$is a Fredholm operator of index zero and the linear continuous projector$Q:Y\to Y$can be defined as $$Qu=\frac{1}{1-\sum_{i=1}^{m}\alpha_i\xi_i} \sum_{i=1}^{m}\alpha_i\int_{\xi_i}^1u(s)ds,$$ and the linear operator$K_P:\operatorname{Im}L\to \operatorname{dom}L\cap \ker P$can be written as $$K_Pu=\sum_{i=k}^{n-2}\frac{B_i(u)}{i!}t^{i} +\frac{(-1)^{n-k}}{(n-1)!}\int_0^t(t-s)^{n-1}u(s)ds.$$ \end{lemma} \begin{proof} By simple calculations, we obtain that $$\ker L=\big\{y:y=c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^{i}\Big),\; c\in \mathbb{R}\big\}.$$ Define linear operator$P:X\to X$as follows $$Py(t)=\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^{i}\Big)y^{(n-1)}(0).$$ Obviously,$\operatorname{Im}P=\ker L$and$P^2y=Py$. For any$y\in X$, it follows from$y=(y-Py)+Py$that$X=\ker P+\ker L$. By simple calculation, we can get that$\ker L\cap \ker P=\{0\}$. So, we have $$X=\ker L\oplus \ker P.\label{e3.3}$$ We will show that $$\operatorname{Im}L=\big\{u\in Y: \sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds=0 \big\}.$$ In fact, if$u\in \operatorname{Im}L$, there exists$y\in \operatorname{dom}L$such that$u=Ly\in Y$. So, we have $$y=\sum_{i=k}^{n-1}\frac{c_i}{i!}t^{i}+\frac{(-1)^{n-k}}{(n-1)!} \int_0^t(t-s)^{n-1}u(s)ds.$$ Since$\sum_{i=1}^{m}\alpha_i=1$and$y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i)$, we have $$\int_0^1u(s)ds=\sum_{i=1}^{m}\alpha_i\int_0^{\xi_i}u(s)ds;$$ i.e.,$\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds=0$. On the other hand, if$u\in Y$satisfies$\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds=0$, we take $$y=\sum_{i=k}^{n-2}\frac{B_i(u)}{i!}t^{i} +\frac{(-1)^{n-k}}{(n-1)!}\int_0^t(t-s)^{n-1}u(s)ds.$$ Obviously,$Ly=u$and$y^{(n-1)}(1)=\sum_{i=1}^{m}\alpha_iy^{(n-1)}(\xi_i)$. By Lemma \ref{lem3.1}, we obtain that$y\in \operatorname{dom}L$; i.e.,$u\in \operatorname{Im}L$. Define operator$Q:Y\to Y$as follows $$Qu=\frac{1}{1-\sum_{i=1}^m\alpha_i\xi_i} \Big(\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1u(s)ds\Big).$$ Obviously,$Q^2y=Qy$and$\operatorname{Im}L=\ker Q$. For$y\in Y$, set$y=(y-Qy)+Qy$. Then$y-Qy\in \ker Q=\operatorname{Im}L$,$Qy\in \operatorname{Im}Q$. It follows from$\ker Q= \operatorname{Im}L$and$Q^2y=Qy$that$\operatorname{Im}Q\cap \operatorname{Im}L=\{0\}$. So we have $$Y=\operatorname{Im}L\oplus \operatorname{Im}Q.$$ This, together with \eqref{e3.3}, means that$L$is a Fredholm operator of index zero. Define operator$K_P:Y\to X $as follows $$K_Pu=\sum_{i=k}^{n-2}\frac{B_i(u)}{i!}t^{i} +\frac{(-1)^{n-k}}{(n-1)!}\int_0^t(t-s)^{n-1}u(s)ds.$$ Now we show that$K_P(\operatorname{Im}L)\subset \operatorname{dom}L\cap \ker P$. Take$u\in \operatorname{Im}L$. Obviously,$(K_P(u))^{(n-1)}(0)=0$. This implies that$K_P(u)\in \ker P$. It is easy to see that$(K_P(u))^{(i)}(0)=0,~0\leq i\leq k-1$. It follows from Lemma \ref{lem3.1} that$(K_P(u))^{(j)}(1)=0$,$0\leq j\leq n-k-2$. From$u\in \operatorname{Im}L$, we obtain $$(K_P(u))^{(n-1)}(1)=\sum_{i=1}^m\alpha_i(K_P(u))^{(n-1)}(\xi_i).$$ So,$K_P(u)\in \operatorname{dom}L$. Now we prove that$K_P$is the inverse of$L|_{\operatorname{dom}L\cap \ker P}$. Obviously,$LK_Pu=u$, for$u\in \operatorname{Im}L$. On the other hand, for$y\in \operatorname{dom}L\cap \ker P, we have \begin{align*} K_PLy(t) &=\sum_{i=k}^{n-2}\frac{B_i(Ly)}{i!}t^{i}+\frac{(-1)^{n-k}}{(n-1)!} \int_0^t(t-s)^{n-1}(-1)^{n-k}y^{(n)}(s)ds\\ &=\sum_{i=k}^{n-2}\big(\frac{B_i(Ly)-y^{(i)}(0)}{i!}\big)t^i+y(t). \end{align*} Sincey$and$K_PLy\in \operatorname{dom}L$, we have$(K_PLy)^{(j)}(1)=y^{(j)}(1)=0,~0\leq j\leq n-k-2$. This means that$(B_k(Ly)-y^{(k)}(0),~B_{k+1}(Ly)-y^{(k+1)}(0),\dots,B_{n-2}(Ly)-y^{(n-2)}(0))$is the only zero solution of the system of linear equations \begin{gather*} \frac{x_k}{k!}+\frac{x_{k+1}}{(k+1)!}+\dots+\frac{x_{n-2}}{(n-2)!}=0\\ \frac{x_k}{(k-1)!}+\frac{x_{k+1}}{k!}+\dots+\frac{x_{n-2}}{(n-3)!}=0\\ \dots\\ \frac{x_k} {[k-(n-k-2)]!}+\frac{x_{k+1}}{[k+1-(n-k-2)]!} +\dots+ \frac{x_{n-2}}{[n-2-(n-k-2)]!}=0. \end{gather*} So, we have$K_PLy=y$, for$y\in \operatorname{dom}L\cap \ker P$. Thus,$K_P=(L|_{\operatorname{dom}L\cap \ker P})^{-1}$. The proof is complete. \end{proof} \begin{lemma} \label{lem3.4} Assume$\Omega\subset X$is an open bounded subset and$\operatorname{dom}L\cap \overline{\Omega}\neq \emptyset$, then$N$is$L$-compact on$\overline{\Omega}$. \end{lemma} \begin{proof} Obviously,$QN(\overline{\Omega})$is bounded. Now we will show that$K_P(I-Q)N:\overline{\Omega}\to X$is compact. It follows from (H2) that there exists constant$M_0>0$such that$|(I-Q)Ny|\leq M_0$; a.e.,$t\in [0,1]$,$y\in \overline{\Omega}$. Thus,$K_P(I-Q)N(\overline{\Omega})$is bounded. By (H2) and Lebesgue Dominated Convergence theorem, we get that$K_P(I-Q)N:\overline{\Omega}\to X$is continuous. Since$\big\{\int_0^t(t-s)^j(I-Q)Ny(s)ds,~y\in \overline{\Omega}\big\}$,$j=0,1\dots,n-1$are equi-continuous, and$t^j$,$j=0,1\dots,n-1$are uniformly continuous on$[0,1]$, using Ascoli-Arzela theorem, we obtain that$K_P(I-Q)N:\overline{\Omega}\to X$is compact. The proof is complete. \end{proof} To obtain our main results, we need the following conditions. \begin{itemize} \item[(H3)] There exists a constant$M>0$such that if$|y^{(n-1)}(t)|>M$,$t\in [\xi_m,1]$then $$\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1\big[f(s,y(s),y'(s), \dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds\neq 0.$$ \item[(H4)] There exist functions$g,h,\psi_i\in L^1[0,1]$,$i=1,2,\dots,n$, with$\sum_{i=1}^{n}\|\psi_i\|_1<1/2$,$\theta\in[0,1)$, some$1\leq j\leq n$such that $$|f(t,x_1,x_2,\dots,x_n)|\leq g(t)+\sum_{i=1}^{n}\psi_i(t)|x_i|+h(t)|x_j|^\theta.$$ \item[(H5)] There exists a constant$c_0>0$such that, if$|c|>c_0$, one of the following two conditions holds \begin{gather} c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big), c \Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots, c\Big)+\varepsilon(s)\Big]ds>0,\label{e3.4} \\ c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big), c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots, c\Big)+\varepsilon(s)\Big]ds<0.\label{e3.5} \end{gather} \end{itemize} \begin{lemma} \label{lem3.5} Assume {\rm (H1)--(H4)}. Then the set $$\Omega_1=\big\{y\in \operatorname{dom}L\setminus \ker L: Ly=\lambda Ny,\;\lambda\in (0,1)\big\}$$ is bounded. \end{lemma} \begin{proof} Take$y\in \Omega_1$. Since$Ny\in \operatorname{Im}L$, we have $$\sum_{i=1}^{m}\alpha_i\int _{\xi_i}^1\left[f(s,y(s),y'(s), \dots,y^{(n-1)}(s))+\varepsilon(s)\right]ds=0.\label{e3.6}$$ Since$Ly=\lambda Ny$and$y\in \operatorname{dom}L$, it follows that $$y(t)=\sum_{i=k}^{n-1}\frac{c_i}{i!}t^{i}+\frac{(-1)^{n-k}}{(n-1)!} \lambda\int_0^t(t-s)^{n-1}\left[f(s,y(s),y'(s),\dots,y^{(n-1)}(s)) +\varepsilon(s)\right]ds,\label{e3.7}$$ where$c_i$,$i=k,k+1,\dots,n-1satisfy \begin{gather*} \sum_{i=k}^{n-1}\frac{c_i}{i!} =-\frac{(-1)^{n-k}}{(n-1)!}\lambda\int_0^1(1-s)^{n-1} \big[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds\\ \sum_{i=k}^{n-1}\frac{c_i}{(i-1)!}=- \frac{(-1)^{n-k}}{(n-2)!}\lambda\int_0^1(1-s)^{n-2} \big[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds\\ \dots\\ \begin{aligned} \sum_{i=k}^{n-1}\frac{c_i}{[i-(n-k-2)]!} &=-\frac{(-1)^{n-k}}{[n-1-(n-k-2)]!}\lambda\int_0^1(1-s)^{k+1}\\ &\quad \times \big[f(s,y(s),y'(s),\dots,y^{(n-1)}(s))+\varepsilon(s)\big]ds. \end{aligned} \end{gather*} It follows fromy^{(i)}(0)=y^{(j)}(1)=0$,$0\leq i\leq k-1$,$0\leq j\leq n-k-2$that there exists at least one point$\delta_i\in [0,1]$such that$y^{(i)}(\delta_i)=0$,$i=0,1,\dots,n-2$. So, we have $$y^{(i)}(t)=\int_{\delta_i}^ty^{(i+1)}(s)ds,\quad i=0,1,\dots,n-2.$$ Therefore, $$\|y^{(i)}\|_\infty\leq \|y^{(i+1)}\|_1\leq \|y^{(i+1)}\|_\infty,\quad i=0,1,\dots,n-2.\label{e3.8}$$ By \eqref{e3.6} and (H3), there exists$t_0\in [\xi_m,1]$such that$|y^{(n-1)}(t_0)|\leq M. This, together with \eqref{e3.7}, implies $$|c_{n-1}|\leq M+\int_0^1\left|f(s,y(s),y'(s),\dots,y^{(n-1)}(s))\right|ds +\|\varepsilon\|_1.\label{e3.9}$$ It follows from \eqref{e3.7}-\eqref{e3.9} and (H4) that \begin{align*} \|y^{(n-1)}\|_\infty &\leq M+2\int_0^1\big|f(s,y(s),y'(s),\dots,y^{(n-1)}(s))\big|ds +2\|\varepsilon\|_1\\ &\leq M+2[\|g\|_1+\sum_{i=1}^{n}\|\psi_i\|_1\|y^{(i-1)}\|_\infty +\|h\|_1\|y^{(j-1)}\|_\infty^\theta]+2\|\varepsilon\|_1\\ &\leq M+2\|g\|_1+2\sum_{i=1}^{n}\|\psi_i\|_1\|y^{(n-1)}\|_\infty +2\|h\|_1\|y^{(n-1)}\|_\infty^\theta+2\|\varepsilon\|_1. \end{align*} So, we obtain $$\|y^{(n-1)}\|_\infty \leq\frac{M+2\|g\|_1+2\|\varepsilon\|_1}{1-2\sum_{i=1}^{n}\|\psi_i\|_1}+ \frac{2\|h\|_1}{1-2\sum_{i=1}^{n}\|\psi_i\|_1}\|y^{(n-1)}\|_\infty^\theta.$$ Then\theta\in [0,1)$implies that$\{\|y^{(n-1)}\|_\infty|: y\in \Omega_1\}$is bounded. Considering of \eqref{e3.8}, we obtain that$\Omega_1$is bounded. \end{proof} \begin{lemma} \label{lem3.6} Assume {\rm (H1), (H2), (H5)}. Then the set $$\Omega_2=\{y:y\in \ker L,\;Ny\in \operatorname{Im}L\}$$ is bounded. \end{lemma} \begin{proof} Take$y\in \Omega_2$, then$y(t)= c\big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^i\big)$,$c\in \mathbb{R}$and$Ny\in \operatorname{Im}L$. So, we have $$c \sum_{i=1}^m\alpha_i\int_{\xi_i}^1\Big[f\Big(s,c\Big(\sum_{i=k}^{n-1} \frac{A_i}{i!}s^i\Big),c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big), \dots,c\Big) +\varepsilon(s)\Big]ds=0.$$ By (H5), we obtain that$|c|\leq c_0$. So,$\Omega_2$is bounded. \end{proof} \begin{lemma} \label{lem3.7} Assume {\rm (H1), (H2), (H5)}. Then the set $$\Omega_3=\{y\in \ker L:\lambda Jy+(1-\lambda)\theta QNy=0,\,\lambda\in [0,1]\}$$ is bounded, where$J:\ker L\to \operatorname{Im}Q$is a linear isomorphism given by $$J\Big(c\sum_{i=k}^{n-1}\frac{A_i}{i!}t^i\Big) = \frac{c}{1-\sum_{i=1}^{m}\alpha_i\xi_i} ,\quad c \in \mathbb{R}$$ and$\theta=\begin{cases} 1 & \text{if \eqref{e3.4} holds},\\ -1,&\text{if \eqref{e3.5} holds}. \end{cases}$. \end{lemma} \begin{proof} For$y\in \Omega_3$, we get$y=c\big(\sum_{i=k}^{n-1}\frac{A_i}{i!}t^i\big)$with $$\lambda c+(1-\lambda)\theta \sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big), c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots,c\Big) +\varepsilon(s)\Big]ds=0.$$ If$\lambda=0$, by (H5), we get$|c|\leq c_0$. If$\lambda=1$,$c=0$. For$\lambda\in (0,1)$, if$|c|\geq c_0, then \begin{align*} \lambda c^2&=-(1-\lambda)\theta c\sum_{i=1}^m\alpha_i\int_{\xi_i}^1 \Big[f\Big(s,c\Big(\sum_{i=k}^{n-1}\frac{A_i}{i!}s^i\Big), c\Big(\sum_{i=k}^{n-1}\frac{A_i}{(i-1)!}s^{i-1}\Big),\dots, c\Big) \\ &\quad +\varepsilon(s)\Big]ds<0. \end{align*} This is a contradiction. So,\Omega_3$is bounded. \end{proof} \begin{theorem} \label{thm3.1} Assume {\rm (H1)--(H5)} Then problem \eqref{e1.1}--\eqref{e1.2} has at least one solution in$X$. \end{theorem} \begin{proof} Let$\Omega\supset\cup_{i=1}^3\overline{\Omega_i}\cup\{0\}$be a bounded open subset of$X$. It follows from Lemma \ref{lem3.4} that$N$is$L-$compact on$\overline{\Omega}$. By Lemmas \ref{lem3.5} and \ref{lem3.6}, we obtain: (1)$Ly\neq \lambda Ny$for every$(y,\lambda)\in [(\operatorname{dom}L\setminus \ker L)\cap\partial\Omega]\times(0,~1)$; and (2)$Ny\notin \operatorname{Im}L$for every$y\in \ker L\cap\partial\Omega$. We need to prove only (3)$\deg(QN|_{\ker L},~\Omega\cap \ker L,~0)\neq0$. To do this, we take $$H(y,\lambda)=\lambda Jy+\theta(1-\lambda)QNy.$$ According to Lemma \ref{lem3.7}, we know$H(y,\lambda)\neq 0$for$y\in \partial\Omega\cap \ker L. By the homotopy of degree, we obtain \begin{align*} \deg(QN|_{\ker L}, \Omega \cap \ker L,0) &=\deg(\theta H(\cdot,0),\Omega \cap \ker L,0)\\ &=\deg(\theta H(\cdot,1),\Omega \cap \ker L,0)\\ &=\deg(\theta J, \Omega \cap \ker L,0)\neq 0. \end{align*} By Theorem \ref{thm2.1}, we obtain thatLy=Ny$has at least one solution in$\operatorname{dom}L\cap\overline{\Omega}$; i.e., \eqref{e1.1}-\eqref{e1.2} has at least one solution in$X$. 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