\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 161, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/161\hfil Extremal points] {Extremal points for a higher-order fractional boundary-value problem} \author[A. Yang, J. Henderson, C. Nelms Jr. \hfil EJDE-2015/161\hfilneg] {Aijun Yang, Johnny Henderson, Charles Nelms Jr.} \address{Aijun Yang \newline Zhejiang University of Technology, College of Science, Hangzhou 310023, China.\newline Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA} \email{yangaij2004@163.com, Aijun\_Yang@baylor.edu} \address{Johnny Henderson \newline Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA} \email{Johnny\_Henderson@baylor.edu} \address{Charles Nelms Jr. \newline Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA} \email{Charles\_Nelms@baylor.edu} \thanks{Submitted May 19, 2015. Published June 16, 2015.} \subjclass[2010]{26A33, 34B08, 34B40} \keywords{$u_0$-positive operator; fractional boundary value problem; \hfill\break\indent spectral radius} \begin{abstract} The Krein-Rutman theorem is applied to establish the extremal point, $b_0$, for a higher-order Riemann-Liouville fractional equation, $D_{0+}^{\alpha}y+p(t)y = 0$, $0 0$, we investigate the following family of higher-order fractional boundary value problems (BVPs): \begin{gather}\label{e1} D_{0+}^{\alpha}y+p(t)y = 0, \quad 0 0:(b),\text{\eqref{e1}-\eqref{e2} has a nontrivial solution} \}. $$ \end{definition} A cone $\mathcal{P}$ is \emph{solid} if the interior, $\mathcal{P}^{\circ}$, of $\mathcal{P}$, is nonempty. A cone $\mathcal{P}$ is \emph{reproducing} if $\mathcal{B}=\mathcal{P}-\mathcal{P}$; i.e., given $w\in \mathcal{B}$, there exist $u,v\in \mathcal{P}$ such that $w=u-v$. \begin{remark}\label{r1} \rm Krasnosel'skii \cite{M.Krasnoselskii} showed that every solid cone is reproducing. \end{remark} Let $\mathcal{P}$ be a cone in a real Banach space $\mathcal{B}$. For $u,v\in \mathcal{B}$, $u\preceq v$ with respect to $\mathcal{P}$, if $u-v\in \mathcal{P}$. A bounded linear operator $L:\mathcal{B}\to\mathcal{B}$ is said to be \emph{positive} with respect to the cone $\mathcal{P}$ if $L:\mathcal{P}\to\mathcal{P}$. $L:\mathcal{B}\to\mathcal{B}$ is $u_0$-\emph{positive with respect to} $\mathcal{P}$ if there exists $u_0\in \mathcal{P}\backslash \{0\}$ such that for each $u\in \mathcal{P}\backslash \{0\}$, there exist $k_1(u)>0$ and $k_2(u)>0$ such that $k_1u_0\preceq Lu\preceq k_2u_0$ with respect to $\mathcal{P}$. \begin{remark}\label{r2} \rm Throughout this article, let $\mathcal{B}$ be a partially ordered Banach space over $\mathbb{R}$ and $\mathcal{P}$ a cone in the Banach space $\mathcal{B}$. Let $\preceq$ be the partial ordering on the Banach space $\mathcal{B}$ induced by the cone $\mathcal{P}$, and $\leq$, the usual partial ordering on $\mathbb{R}$ induced by $\mathbb{R}^{+}$. Also, $u\preceq v$ will be used in the same way as $v\succeq u$. In addition, We will denote the \emph{spectral radius} of the bounded linear operator $L$ by $r(L)$. \end{remark} The following five results are fundamental to our extremal point results. The first two results are found in Krasnosel'skii's book \cite{M.Krasnoselskii}. The third one is proved in Nussbaum \cite{R.D.Nussbaum}. The last two results are found in \cite{M.Krasnoselskii,Krein-Rutman}. In each of the following theorems, assume that $\mathcal{B}$ is a Banach space and $\mathcal{P}$ is a reproducing cone, and that $L:\mathcal{B}\to\mathcal{B}$ is a compact, linear, and positive operator with respect to $\mathcal{P}$. \begin{theorem}\label{t1} Let $\mathcal{P}\subset\mathcal{B}$ be a solid cone. If $L:\mathcal{B}\to\mathcal{B}$ is a linear operator such that $L:\mathcal{P}\setminus\{0\}\to\mathcal{P}^{\circ}$, then $L$ is $u_0$-positive. \end{theorem} \begin{theorem}\label{t2} Let $L:\mathcal{B}\to\mathcal{B}$ be a compact, $u_0$-positive linear operator. Then $L$ has an essentially unique eigenvector in $\mathcal{P}$, and the corresponding eigenvalue is simple, positive, and larger than the absolute value of any other eigenvalue. \end{theorem} \begin{theorem}\label{t3} Let $L_b$, $\eta\leq b\leq\beta$ be a family of compact, linear operators on Banach space such that the mapping $b\mapsto L_b$ is continuous in the uniform operator topology. Then the mapping $b\mapsto r(L_b)$ is continuous. \end{theorem} \begin{theorem}\label{t4} Assume $r(L)>0$. Then $r(L)$ is an eigenvalue of $L$, and there is a corresponding eigenvalue in $\mathcal{P}$. \end{theorem} \begin{theorem}\label{t5} Suppose there exists $\gamma>0$, $u\in\mathcal{B}$, $-u\notin \mathcal{P}$, such that $\gamma u\preceq Lu$ with respect to $\mathcal{P}$. Then $L$ has an eigenvector in $\mathcal{P}$ which corresponding to an eigenvalue $\lambda$ with $\lambda\geq\gamma$. \end{theorem} \section{Criteria for extremal points} First, we introduce a family of Green's functions for $-D_{0+}^{\alpha}y = 0$ with $n<\alpha\leq n+1$, $n\geq2$, under the boundary conditions\eqref{e2}, can be calculated as \[ G(b;t,s)=\frac{1}{\Gamma(\alpha)b^{\alpha-n}} \begin{cases} t^{\alpha-1}(b-s)^{\alpha-n},& 0 \leq t \leq s \leq b,\\ t^{\alpha-1}(b-s)^{\alpha-n}-b^{\alpha-n}(t-s)^{\alpha-1}, & 0 \leq s < t \leq b. \end{cases} \] Obviously, $G(b;t,s)>0$ and \[ \frac{\partial G(b;t,s)}{\partial b} =\frac{(\alpha-n)t^{\alpha-1}s}{\Gamma(\alpha)b^{\alpha+1-n}(b-s)^{n+1-\alpha}}>0 \] on $(0,b)\times(0,b)$. In particular, we note that $G(b;t,s)=t^{\alpha-n}K(b;t,s)$, where \[ K(b;t,s)=\frac{1}{\Gamma(\alpha)b^{\alpha-n}} \begin{cases} t^{n-1}(b-s)^{\alpha-n},& 0 \leq t \leq s \leq b,\\ t^{n-1}(b-s)^{\alpha-n}-b^{\alpha-n}t^{n-\alpha}(t-s)^{\alpha-1}, & 0 \leq s < t \leq b. \end{cases} \] It is easy to deduce the sign properties of $K$ as: \begin{itemize} \item[(1)] $K(b;t,s)>0$ for $(t,s)\in(0,b]\times(0,b)$. \item[(2)] $K(b;0,s)=0$ for $s\in (0,b)$. \item[(3)] $\frac{\partial^{i} K(b;0,s)}{\partial t^{i}}=0$, $i=1,2,\ldots,n-2$. \item[(4)] $\frac{\partial^{n-1} K(b;0,s)}{\partial t^{n-1}} =\frac{(n-1)!(b-s)^{\alpha-n}}{\Gamma(\alpha)b^{\alpha-n}}>0$ for $s\in(0,b)$. \item[(5)] $\frac{\partial K(b;t,s)}{\partial b} =\frac{\alpha-n}{\Gamma(\alpha)}b^{n-\alpha-1}(b-s)^{\alpha-n-1}st^{n-1}>0$ for $(t,s)\in (0,b)\times(0,b)$. \item[(6)] $\frac{\partial}{\partial b}(\frac{\partial^{n-1} K(b;0,s)}{\partial t^{n-1}})=\frac{(\alpha-n)(n-1)!} {\Gamma(\alpha)}b^{n-\alpha-1}(b-s)^{\alpha-n-1}s>0$ for $s\in(0,b)$. \end{itemize} Next, we consider the Banach space $(\mathcal{B}, \|\cdot \|)$ defined by $$ \mathcal{B}:= \{y: [0,b] \to \mathbb{R}: y=t^{\alpha-n}z,\ z\in C[0,b] \},\quad \|y\| := \sup_{0 \leq t \leq b} |z(t)|=|z|_0. $$ Also, we define a cone $\mathcal{P} \subset \mathcal{B}$ by $$ \mathcal{P} := \{y \in \mathcal{B}: y(t) \geq 0 \text{ on } [0,b]\}. $$ The cone $\mathcal{P}$ is a reproducing cone since if $y\in \mathcal{B}$, $$ y_1(t)=\max\{ 0, y(t)\},\quad y_2(t)=\max\{ 0, -y(t)\}, $$ are in $\mathcal{P}$ and $y=y_1-y_2$. For each $\beta>0$, define the Banach space $$ \mathcal{B}_{\beta}:= \{y: [0,\beta] \to \ \mathbb{R}: y=t^{\alpha-n}z,\, z\in C^{n-1}[0,\beta],\, z^{(i)}(0)=0,\, i=0,1,2,\ldots,n-2 \} $$ with the norm $$ \|y\|_{\beta} := \sup_{0 \leq t \leq \beta} |z^{n-1}(t)|=|z^{n-1}|_0. $$ By this norm, for $y\in \mathcal{B}_{\beta}$, we have \begin{align*} |z(t)| &= \big|\int_0^{t}\int_0^{t_1}\cdots\int_0^{t_{n-2}} z^{(n-1)}(s)ds\cdots dt_{2} dt_1\big|\\ &\leq \frac{t^{n-1}}{(n-1)!}|z^{(n-1)}|_0\\ &= \frac{t^{n-1}}{(n-1)!}\|y\|_{\beta}, \quad t\in[0,\beta]. \end{align*} Then $$ |y(t)|=|t^{\alpha-n}z(t)|\leq \frac{t^{\alpha-1}}{(n-1)!}\|y\|_{\beta}, \quad t\in[0,\beta]. $$ For each $\beta>0$, define the cone $\mathcal{P}_{\beta}\subset \mathcal{B}_{\beta}$ to be $$ \mathcal{P}_{\beta} := \{y \in \mathcal{B}_\beta: y(t) \geq 0 \text{ on } [0,\beta]\}. $$ \begin{lemma}\label{l1} The cone $\mathcal{P}_{\beta}$ is solid in $\mathcal{B}_\beta$ and hence reproducing. \end{lemma} \begin{proof} Define \begin{align*} \Omega_{\beta}=\Big\{&y\in \mathcal{B}_{\beta}: y(t)>0 \text{ for } t\in(0,\beta),\ z^{(n-1)}(0)>0,\\ & z(\beta)>0, \text{ where } y=t^{\alpha-n}z\Big\}. \end{align*} We will show $\Omega_{\beta}\subset\mathcal{P}^{\circ}_{\beta}$. Let $y\in \Omega_{\beta}$. Since $z^{(n-1)}(0)>0$, there exists an $\varepsilon_1>0$ such that $z^{(n-1)}(0)-\varepsilon_1>0$. Since $z\in C^{(n-1)}[0,\beta]$, there exists a $\gamma_1\in (0,\beta)$ such that $z^{(n-1)}(t)>\varepsilon_1$ for $t\in (0,\gamma_1)$. So, \begin{align*} y(t) &= t^{\alpha-n}z(t) \\ &= t^{\alpha-n}\int_0^{t}\int_0^{t_1}\cdots \int_0^{t_{n-2}} z^{(n-1)}(s)ds\cdots dt_{2} dt_1\\ &> \frac{t^{\alpha-1}}{(n-1)!}\varepsilon_1, \quad t\in (0,\gamma_1). \end{align*} Now, since $z(\beta)>0$, there exists an $\varepsilon_2>0$ such that $z(\beta)-\varepsilon_2>0$. Since $z\in C^{n-1}[0,\beta]$, there exists a $\gamma_2\in (0,\beta)$ such that $z(t)>\varepsilon_2$ for $t\in (\gamma_2,\beta)$. Thus $y(t)= t^{\alpha-n}z(t)>\varepsilon_2t^{\alpha-n}$ for all $t\in (\gamma_2,\beta)$. Also, since $y(t)>0$ on $[\gamma_1,\gamma_2]$, there exists an $\varepsilon_3>0$ such that $y(t)>\varepsilon_3$ for all $t\in [\gamma_1,\gamma_2]$. Let $\varepsilon=\min\big\{\frac{\varepsilon_1}{2},\, \frac{(n-1)!\varepsilon_2}{2\beta^{n-1}},\, \frac{(n-1)!\varepsilon_3}{2\beta^{\alpha-1}}\big\}$. Define $B_\varepsilon(y)=\{\hat{y}\in\mathcal{B_{\beta}}: \|y-\hat{y}\|_\beta<\varepsilon\}$. Let $\hat{y}\in B_\varepsilon(y)$, then $\hat{y}=t^{\alpha-n}\hat{z}$, where $\hat{z}\in C^{n-1}[0,\beta]$ with $\hat{z}^{(i)}(0)=0$, $i=0,1,2,\ldots,n-2$. Now, $$ |\hat{y}(t)-y(t)|\leq \frac{t^{\alpha-1}}{(n-1)!}\|\hat{y}-y\|_{\beta} < \frac{t^{\alpha-1}}{(n-1)!}\varepsilon,\quad t\in[0,\beta]. $$ So for $t\in (0,\gamma_1)$, $$ \hat{y}(t) >y(t)-\frac{t^{\alpha-1}}{(n-1)!}\varepsilon >\frac{t^{\alpha-1}}{(n-1)!}\varepsilon_1-\frac{t^{\alpha-1}}{(n-1)!}\varepsilon >\frac{t^{\alpha-1}}{2(n-1)!}\varepsilon_1 >0. $$ For $t\in (\gamma_2,\beta)$, $$ \hat{y}(t) >\varepsilon_2t^{\alpha-n}-\frac{t^{\alpha-1}}{(n-1)!}\varepsilon >\Big(\varepsilon_2-\frac{\beta^{n-1}}{(n-1)!}\varepsilon\Big)t^{\alpha-n} >\frac{\varepsilon_2}{2} t^{\alpha-n} >0. $$ Also, \[ \hat{y}(t)>y(t)-\frac{t^{\alpha-1}}{(n-1)!}\varepsilon >\varepsilon_3-\frac{\beta^{\alpha-1}}{(n-1)!}\varepsilon>0 \] for $t\in [\gamma_1,\gamma_2]$. So $\hat{y}\in \mathcal{P}_{\beta}$ and thus $B_\varepsilon(y)\subset\mathcal{P}_{\beta}$. Then $\Omega_{\beta}\subset\mathcal{P}^{\circ}_{\beta}$. \end{proof} Let $N_0y(t)\equiv0$, $t\in [0,b]$, and for each $\beta>0$, define $N_\beta:\mathcal{B}\to\mathcal{B}$ by \begin{equation}\label{e3} N_\beta y(t)= \begin{cases} \int_0^\beta G(\beta;t,s)p(s)y(s)ds,& 0 \leq t \leq \beta,\\[4pt] \int_0^\beta G(\beta;\beta,s)p(s)y(s)ds,& \beta \leq t \leq b. \end{cases} \end{equation} We shall refer to $N_\beta:\mathcal{B}_\beta\to\mathcal{B}_\beta$, where $N_\beta$ is defined by \begin{align*} N_\beta y(t) &= \int_0^\beta G(\beta;t,s)p(s)y(s)ds \\ &= t^{\alpha-n}\int_0^\beta K(\beta;t,s)p(s)y(s)ds,\quad 0 \leq t \leq \beta. \end{align*} By employing the methods used in \cite{Eloe-Jeffrey}, the existence of the extremal point $b_0$ for BVP ($b$),\eqref{e1}-\eqref{e2}, is positive, can be seen from the following theorem. \begin{theorem}\label{t6} Let $\delta>0$ be such that \[ \Big(\frac{1}{\Gamma(\alpha)(\alpha-n+1)} + \frac{2^{n}}{\Gamma(\alpha-n+2)(n-1)!}\Big)P\delta^\alpha=1, \] where $P=\max_{0\leq t\leq\beta}|p(t)|$. Then the BVP($\beta$), \eqref{e1}-\eqref{e2} has a unique solution for $\beta\in (0,\delta)$; in particular, if $\beta\geq\delta$, then $u\equiv 0$ is the only solution of BVP ($\beta$), \eqref{e1}-\eqref{e2}. \end{theorem} \begin{proof} We shall show there exists $\delta>0$ such that for $\beta\in (0,\delta)$, $N_\beta:\mathcal{B}_{\beta}\to\mathcal{B}_\beta$ is a contraction map. Let $y_1, y_2\in\mathcal{B}_\beta$ and consider \begin{align*} (N_\beta y_2-N_\beta y_1)(t) &=t^{\alpha-n} \Big(\int_0^\beta \frac{t^{n-1}(\beta-s)^{\alpha-n}}{\Gamma(\alpha)\beta^{\alpha-n}} p(s)(y_2-y_1)(s)ds \\ &\quad\times \int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)t^{\alpha-n}} p(s)(y_2-y_1)(s)ds\Big). \end{align*} Set $$ z(t) =\int_0^\beta \frac{t^{n-1}(\beta-s)^{\alpha-n}}{\Gamma(\alpha) \beta^{\alpha-n}}p(s)(y_2-y_1)(s)ds -\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)t^{\alpha-n}}p(s)(y_2-y_1)(s)ds. $$ Then, $\|N_\beta y_2-N_\beta y_1\|_{\beta}=|z^{(n-1)}|_0$. For $t\in(0,\beta)$, \begin{align*} &|z^{(n-1)}(t)|\\ & = \Big|\int_0^\beta \frac{(n-1)!(\beta-s)^{\alpha-n}}{\Gamma(\alpha) \beta^{\alpha-n}}p(s)(y_2-y_1)(s)ds \\ &\quad -\frac{1}{\Gamma(\alpha)}\int_0^t \frac{\partial^{n-1}((t-s)^{\alpha-1}t^{n-\alpha})}{\partial t^{n-1}} p(s)(y_2-y_1)(s)ds\Big|\\ &\leq \frac{(n-1)!}{\Gamma(\alpha)\beta^{\alpha-n}}\cdot P\cdot |y_2-y_1|_0\int_0^\beta (\beta-s)^{\alpha-n} ds +\frac{1}{\Gamma(\alpha)}\cdot P\cdot |y_2-y_1|_0\\ &\quad \times\int_0^t\sum_{k=0}^{n-1}C_{n-1}^{k} \frac{\Gamma(\alpha)}{\Gamma(\alpha-k)} \frac{\Gamma(\alpha-n+(n-k-1))}{\Gamma(\alpha-n)}(t-s) ^{\alpha-k-1}t^{n-\alpha-(n-k-1)}ds\\ &\leq \Big(\frac{(n-1)!}{\Gamma(\alpha)(\alpha-n+1)} + \frac{2^{n-1}}{\Gamma(\alpha-n+2)}\Big)P\beta|y_2-y_1|_0\\ &\leq \Big(\frac{1}{\Gamma(\alpha)(\alpha-n+1)} + \frac{2^{n-1}}{\Gamma(\alpha-n+2)(n-1)!}\Big)P\beta^{\alpha}\|y_2-y_1\|_\beta, \end{align*} where $\frac{\partial^{n-1}((t-s)^{\alpha-1}t^{n-\alpha})}{\partial t^{n-1}}$ is calculated by Leibniz rule formula for derivative, and $\sum_{k=0}^{n-1}C_{n-1}^{k}=2^{n-1}$ due to binomial formula $(a+b)^{n-1}=\sum_{k=0}^{n-1}C_{n-1}^{k}a^{k}b^{n-1-k}$ with $a=b=1$. Thus, if $$ \Big(\frac{1}{\Gamma(\alpha)(\alpha-n+1)} + \frac{2^{n-1}}{\Gamma(\alpha-n+2)(n-1)!}\Big)P\beta^{\alpha}<1, $$ then $N_{\beta}$ is a contraction map. Choose $\delta>0$ such that $\Big(\frac{1}{\Gamma(\alpha)(\alpha-n+1)} + \frac{2^{n}}{\Gamma(\alpha-n+2)(n-1)!}\Big)P\delta^\alpha=1$ and the proof is complete. \end{proof} \begin{lemma}\label{l2} For each $\beta>0$, $N_\beta$ is positive with respect to $\mathcal{P}$ and $\mathcal{P}_\beta$. In addition, $N_\beta:\mathcal{P}_\beta\setminus \{0\}\to\mathcal{P}_\beta^{\circ}$. \end{lemma} \begin{proof} The positivity of $N_\beta$ with respect to $\mathcal{P}$ and $\mathcal{P}_\beta$ is an easy consequence of the sign properties of Green's function $G$ and the kernel $K$. Now, we shall show that $N_\beta:\mathcal{P}_\beta\setminus \{0\}\to\mathcal{P}_\beta^{\circ}$. From Lemma \ref{l1}, we have $\Omega_{\beta}\subset\mathcal{P}^{\circ}_{\beta}$. Next, we prove $N_\beta:\mathcal{P}_\beta\setminus \{0\}\to\Omega_{\beta}$. Let $y\in\mathcal{P}_\beta\setminus \{0\}$, then there exists $[\gamma_1,\gamma_2]\subset[0,\beta]$ such that $p(t)>0$ and $y(t)>0$ for all $t\in[\gamma_1,\gamma_2]$. So \begin{align*} N_\beta y(t)& = \int_0^\beta G(\beta;t,s)p(s)y(s)ds\\ &\geq \int_{\gamma_1}^{\gamma_2} G(\beta;t,s)p(s)y(s)ds > 0,\quad \text{for all } t\in (0,\beta). \end{align*} Note $z(t)=\int_0^\beta K(\beta;t,s)p(s)y(s)ds$, we have \begin{gather*} z(\beta) = \int_0^\beta K(\beta;\beta,s)p(s)y(s)ds \geq \int_{\gamma_1}^{\gamma_2} K(\beta;\beta,s)p(s)y(s)ds > 0, \\ z^{(n-1)}(0)=\int_0^\beta \frac{\partial^{n-1} K(\beta;0,s)}{\partial t^{n-1}}p(s)y(s)ds>0. \end{gather*} Thus, $N_\beta y\in \Omega_\beta$ and $N_\beta:\mathcal{P}_\beta\setminus \{0\}\to\mathcal{P}_\beta^{\circ}$. \end{proof} \begin{remark}\label{r3} \rm According to Theorem \ref{t1}, $N_\beta$ is $u_0$-positive with respect to $\mathcal{P}_\beta$. \end{remark} \begin{lemma}\label{l3} The mapping $\beta\mapsto r(N_\beta)$ with $N_\beta$ defined on $\mathcal{B}$ for each $\beta\in(0,b]$ is continuous. \end{lemma} \begin{proof} We shall prove that the mapping $\beta\mapsto N_\beta$ is continuous in the uniform operator topology with $N_\beta$ defined on $\mathcal{B}$ for each $\beta\in(0,b]$. Since $p(t)$ is continuous on $[0,\infty)$, the linear operator $N_\beta$ defined on $\mathcal{B}$ can be proved to be compact as in \cite{P.W.Eloe}. Now, let $f:(0,b]\to\{N_\beta\}|_{0}^b$ be given by $f(\beta)=N_\beta$. Assume $y=t^{\alpha-n}z\in\mathcal{B}$ with $\|y\|=1$. Note $P=\max_{0\leq t\leq b}|p(t)|$. Let $0<\gamma_1<\gamma_2\leq b$. Then \begin{align*} &\|f(\gamma_2)-f(\gamma_1)\|\\ &= \|N_{\gamma_2}-N_{\gamma_1}\|\\ &= \sup_{\|y\|=1}\|N_{\gamma_2}y-N_{\gamma_1}y\|\\ &= \sup_{\|y\|=1}\sup_{0\leq t\leq b} \Big|\int_0^{\gamma_2} K(\gamma_2;t,s)p(s)y(s)ds -\int_0^{\gamma_1} K(\gamma_1;t,s)p(s)y(s)ds\Big|. \end{align*} Since $K(\beta;t,s)$ is continuous for each $\beta\in (0,b]$, for $\varepsilon>0$, there exists $\delta>0$ such that $|K(\gamma_2;t,s)-K(\gamma_1;t,s)|<\frac{\varepsilon}{2bP}$ whenever $|\gamma_2-\gamma_1|<\delta$. \smallskip \noindent\textbf{Case (i)} $t\leq \gamma_1$. Let $\sup_{0\leq t\leq \gamma_1,\gamma_1 \leq s\leq\gamma_2}|K(\gamma_2;t,s)|\leq K_1$. Choose $\delta=\frac{\varepsilon}{2K_1P}$. Then \begin{align*} &\Big|\int_0^{\gamma_2} K(\gamma_2;t,s)p(s)y(s)ds -\int_0^{\gamma_1} K(\gamma_1;t,s)p(s)y(s)ds\Big|\\ &\leq \int_0^{\gamma_1}| K(\gamma_2;t,s) -K(\gamma_1;t,s)|p(s)y(s)ds+\int_{\gamma_1}^{\gamma_2} K(\gamma_2;t,s)p(s)y(s)ds\\ &\leq \frac{\varepsilon}{2bP}\cdot\gamma_1\cdot P\cdot 1 +K_1\cdot P\cdot1\cdot|\gamma_2-\gamma_1| < \varepsilon. \end{align*} \smallskip \noindent\textbf{Case (ii)} $\gamma_1\leq t\leq \gamma_2$. Let $\sup_{\gamma_1\leq t\leq\gamma_2,0\leq s\leq \gamma_1} \big|\frac{\partial K(\gamma_2;t,s)}{\partial t}\big|\leq K_2$ and \[ \sup_{t,s\in[\gamma_1,\gamma_2]}|K(\gamma_2;t,s)|\leq K_3. \] Choose $\delta=\frac{\varepsilon}{2(K_2b+K_3)P}$. Then \begin{align*} &\Big|\int_0^{\gamma_2} K(\gamma_2;t,s)p(s)y(s)ds -\int_0^{\gamma_1} K(\gamma_1;\gamma_1,s)p(s)y(s)ds\Big|\\ &\leq \int_0^{\gamma_1}| K(\gamma_2;t,s)-K(\gamma_1;\gamma_1,s)|p(s)y(s)ds+\int_{\gamma_1}^{\gamma_2} K(\gamma_2;t,s)p(s)y(s)ds\\ &\leq \int_0^{\gamma_1}| K(\gamma_2;t,s)-K(\gamma_2;\gamma_1,s)|p(s)y(s)ds\\ &\quad +\int_0^{\gamma_1}| K(\gamma_2;\gamma_1,s)-K(\gamma_1;\gamma_1,s)|p(s)y(s)ds +\int_{\gamma_1}^{\gamma_2} K(\gamma_2;t,s)p(s)y(s)ds\\ &\leq \int_0^{\gamma_1}\Big| \frac{\partial K(\gamma_2;\xi_{t,\gamma_1},s)}{\partial t}\Big|(t-\gamma_1)p(s)y(s)ds +\frac{\varepsilon}{2bP}\cdot\gamma_1\cdot P\cdot1 +K_3\cdot P\cdot1\cdot|\gamma_2-\gamma_1|\\ &\leq (K_2\gamma_1+K_3)P|\gamma_2-\gamma_1|+\frac{\varepsilon}{2} < \varepsilon. \end{align*} \smallskip \noindent\textbf{Case (iii)} $t\geq \gamma_2$. The similar technique is used in Case (ii), so we omit it here. From above discussion we can see that $\beta\mapsto N_\beta$ is continuous in the uniform operator topology. Therefore, the mapping $\beta\mapsto r(N_{\beta})$ is continuous due to Theorem \ref{t3}. \end{proof} \begin{theorem}\label{t7} For $0<\beta\leq b$, $r(N_\beta)$ is strictly increasing as a function of $\beta$. \end{theorem} \begin{proof} Let $\lambda>0$ and $y\in \mathcal{P}_\beta\setminus\{0\}$. Theorem \ref{t2} implies that $N_\beta y(t)=\lambda y(t)$ for $t\in [0,\beta]$. Let $y(t)=y(\beta)$ for $t>\beta$. Then, for $t\in [0,b]$, $N_\beta y(t)=\lambda y(t)$, and $r(N_\beta)\geq\lambda>0$, i.e., $r(N_\beta)>0$. Next, let $0<\beta_1<\beta_2\leq b$. Since $r(N_{\beta_1})>0$, by Theorem \ref{t4}, there exists $y\in\mathcal{P}_{\beta_1}$ such that $N_{\beta_1} y=r(N_{\beta_1}) y$. Let $u_1=N_{\beta_1} y$ and $u_2=N_{\beta_2} y$. Then for $t\in [0,\beta_1]$, we claim that $u_2-u_1\in \mathcal{P}_{\beta_1}^{\circ}$. In fact, by noting $(u_2-u_1)(t)=t^{\alpha-2}z_{12}(t)$, we have \begin{align*} z_{12}(t) &= \int_0^{\beta_2} K(\beta_2;t,s)p(s)y(s)ds -\int_0^{\beta_1} K(\beta_1;t,s)p(s)y(s)ds\\ &= \int_0^{\beta_1} [K(\beta_2;t,s)-K(\beta_1;t,s)]p(s)y(s)ds +\int_{\beta_1}^{\beta_2} K(\beta_2;t,s)p(s)y(\beta_1)ds. \end{align*} Since $y\in\mathcal{P}_{\beta_1}\setminus \{0\}$ and $p(t)$ does not vanish identically on any compact subinterval $[0,\beta_1]\subset[0,b]$, it follows that $z_{12}(t)>0$ as $K(\beta_2;t,s)>K(\beta_1;t,s)$. So, $u_2(t)>u_1(t)$ on $(0,\beta_1)$. In view of $\frac{\partial^{i}K(\beta;0,s)}{\partial t^{i}}=0$, for $\beta\in(0,b]$ and $s\in [0,b]$, $i=0,1,2,\ldots,n-2$, we have $$ z_{12}^{(i)}(0)=\int_0^{\beta_2} \frac{\partial^{i}K(\beta_2;0,s)}{\partial t^{i}}p(s)y(s)ds -\int_0^{\beta_1} \frac{\partial^{i}K(\beta_1;0,s)}{\partial t^{i}}p(s)y(s)ds=0, $$ for $i=0,1,2,\ldots,n-2$. Since $\frac{\partial K(b;0,s)}{\partial t}>0$ and $\frac{\partial}{\partial b}\Big(\frac{\partial K(b;0,s)}{\partial t}\Big)>0$ for $s\in(0,b)$, we can get \begin{align*} z^{(n-1)}_{12}(0) &= \int_0^{\beta_2} \frac{\partial^{n-1} K(\beta_2;0,s)}{\partial t^{n-1}}p(s)y(s)ds -\int_0^{\beta_1} \frac{\partial^{n-1} K(\beta_1;0,s)}{\partial t^{n-1}}p(s)y(s)ds\\ &= \int_0^{\beta_1} \Big[\frac{\partial^{n-1} K(\beta_2;0,s)}{\partial t^{n-1}} -\frac{\partial^{n-1} K(\beta_1;0,s)}{\partial t^{n-1}}\Big]p(s)y(s)ds\\ &\quad+\int_{\beta_1}^{\beta_2} \frac{\partial^{n-1} K(\beta_2;0,s)}{\partial t^{n-1}}p(s)y(\beta_1)ds >0. \end{align*} Also, \begin{align*} z_{12}(\beta_1) &= \int_0^{\beta_2} K(\beta_2;\beta_1,s)p(s)y(s)ds-\int_0^{\beta_1} K(\beta_1;\beta_1,s)p(s)y(s)ds\\ &= \int_0^{\beta_1} [K(\beta_2;\beta_1,s)-K(\beta_1;\beta_1,s)]p(s)y(s)ds\\ &\quad +\int_{\beta_1}^{\beta_2} K(\beta_2;\beta_1,s)p(s)y(\beta_1)ds >0, \end{align*} due to $\frac{\partial K(\beta;t,s)}{\partial b}>0$ for $\beta\in (0,b)$ and $K(\beta_2;\beta_1,s)>0$ on $(\beta_1,\beta_2)$. Thus, the restriction of $u_2-u_1$ to $[0,\beta_1]$ belongs to $\Omega_{\beta_1}\subset \mathcal{P}_{\beta_1}^{\circ}$. So there exists $\delta>0$ such that $u_2-u_1\succeq \delta y$ with respect to $\mathcal{P}_{\beta_1}$. Let $u_1(t)=u_1(\beta_1)$ for $t>\beta_1$. In view of $u_2\in \mathcal{P}_{{\beta_2}}$, it follows that $u_2-u_1\succeq \delta y$ with respect to $\mathcal{P}_{\beta_2}$. Thus, $$ u_2\succeq u_1+\delta y=r(N_{\beta_1})y+\delta y=(r(N_{\beta_1})+\delta)y, $$ i.e., $N_{\beta_2}y\succeq (r(N_{\beta_1})+\delta)y$. So by Theorem \ref{t5}, $$ r(N_{\beta_2})\geq r(N_{\beta_1})+\delta > r(N_{\beta_1}). $$ Hence, $r(N_{\beta})$ is strictly increasing for $0<\beta\leq b$. \end{proof} \begin{theorem}\label{t8} The following three statements are equivalent: \begin{itemize} \item[(i)] $b_0$ is the first extremal point of the BVP (b), \eqref{e1}-\eqref{e2}; \item[(ii)] there exists a nontrivial solution $y$ of the BVP($b_0$), \eqref{e1}-\eqref{e2} such that $y\in \mathcal{P}_{b_0}$; \item[(iii)] $r(N_{b_0})=1$. \end{itemize} \end{theorem} \begin{proof} (iii) $\Rightarrow$ (ii) is an immediate consequence of Theorem \ref{t4}. Next, we prove (ii) $\Rightarrow$ (i). Let $y\in \mathcal{P}_{b_0}\setminus\{0\}$ satisfy BVP($b_0$),\eqref{e1}-\eqref{e2} for $0\leq t\leq b_0$. Extend $y(t)=y(b_0)$ for $t>b_0$. For $N_{b_0}y(t)=y(t)$, we have $r(N_{b_0})\geq1$. If $r(N_{b_0})=1$, then by Theorem \ref{t7} that $r(N_{\beta})1$. Let $v\in \mathcal{P}_{b_0}\setminus\{0\}$ such that $N_{b_0}v=r(N_{b_0})v$. From Lemma \ref{l2}, we know that the restriction of $v$ to $[0,b_0]$ belongs to $\mathcal{P}_{b_0}^\circ$. Thus, there exists $\delta >0$ such that $y\succeq \delta v$ with respect to $\mathcal{P}_{b_0}$, $0\leq t\leq b_0$. Extend $v(t)=v(b_0)$ for $t>b_0$. Then $y\succeq \delta v$ with respect to $\mathcal{P}$. Assume $\delta$ is maximal such that the inequality $y\succeq \delta v$ holds. Then, $$ y=N_{b_0}y\succeq N_{b_0}(\delta v)=\delta N_{b_0}v=\delta r(N_{b_0})v. $$ Since $r(N_{b_0})>1$, $\delta r(N_{b_0})>\delta$. But this contradicts the assumption that $\delta$ is the maximal value satisfying the inequality $y\succeq \delta v$. So $r(N_{b_0})=1$. Finally, to prove (i) $\Rightarrow$ (iii) observe that $\lim_{b\to0^+}r(N_b)=0$. If $b_0$ is the first extremal point of BVP ($b$),\eqref{e1}-\eqref{e2}, then $r(N_{b_0})\geq 1$. If $r(N_{b_0})> 1$, then by the continuity of $r$ about $b$, there exists $\beta_0\in (0,b_0)$ such that $r(N_{\beta_0})=1$, and for this $\beta_0$, the BVP ($\beta_0$),\eqref{e1}-\eqref{e2} has a nontrivial solution, which is a contradiction. \end{proof} \section{A nonlinear problem} Consider a BVP for a nonlinear fractional differential equation of the form \begin{equation}\label{e4} D_{0+}^{\alpha}y+f(t,y) = 0, \quad 0 1$ such that $Nw=\mu w$, and $Nu=u$ implies $u\notin\mathcal{P}$. Further, there exists $\rho>0$ such that, if $ u=\frac{1}{\lambda}Mu$, $u\in \mathcal{P}$ and $\|u\|=\rho$, then $\lambda\leq 1$. \end{itemize} Then the equation $u=Mu$ has a solution $u\in \mathcal{P}\setminus\{0\}$. \end{theorem} Now, we shall use this theorem and the main conclusions of Section 3 to prove the following result. \begin{theorem}\label{t10} Suppose that $b_0$ is the first extremal point of BVP \eqref{e5}-\eqref{e2}. For each $\beta>b_0$ assume the property: \begin{itemize} \item[(H1)] There exists $\rho(\beta)>0$ such that if $y(t)$ is a nontrivial solution of the BVP \begin{equation}\label{e6} D_{0+}^{\alpha}y+\frac{1}{\lambda}f(t,y) = 0, \quad 0 b_0$, let $N_\beta:\mathcal{B}\to\mathcal{B}$ be defined by \eqref{e3}, where $ p(t)\equiv \frac{\partial f(t,0)}{\partial y}$. Define the nonlinear operator $M_\beta:\mathcal{B}\to\mathcal{B}$ by \[ M_\beta y(t)=\begin{cases} \int_0^\beta G(\beta;t,s)f(s,y(s))ds,& 0 \leq t \leq \beta,\\[4pt] \int_0^\beta G(\beta;\beta,s)f(s,y(s))ds,& \beta \leq t \leq b. \end{cases} \] The differentiability of $f$ with respect to $y$ is sufficient to argue that $M_\beta$ is Fr$\acute{e}$chet differentiable at $y=0$ since \begin{align*} &\big|\int_0^\beta G(\beta;t,s)[f(s,y(s))-p(s)y(s)]ds\big|\\ &= \big|\int_0^\beta G(\beta;t,s)[f_y(s,\tilde{y}(s))-p(s)]y(s)ds\big|\\ &\leq Q\beta\|y\|\int_0^\beta |f_y(s,\tilde{y}(s))-p(s)|ds, \end{align*} where $0\leq \tilde{y}(t)\leq y(t)$ for $t\in[0,\beta]$ and $Q=\sup_{t,s\in [0,b]}|G(\beta;t,s)|$. Moreover, $M_\beta'(0)=N_\beta$. By Theorems \ref{t7} and \ref{t8}, it follows that $r(N_{b_0})=1$ and $r(N_\beta)>1$ if $\beta >b_0$. Moreover, since $b_0$ is the first extremal point of the BVP ($b$),\eqref{e5}-\eqref{e2}, it also follows from Theorem \ref{t8} that if $N_\beta y=y$ and $y$ is nontrivial for $\beta >b_0$, then $y\notin \mathcal{P}$. So, for $\beta >b_0$, we can apply property (H1) to check the condition (A1) in Theorem \ref{t9}. Then we obtain the existence of a $y\in \mathcal{P}\setminus\{0\}$ such that $y=N_\beta y$ and the proof is complete. \end{proof} \begin{remark}\label{r4} \rm Condition \eqref{e6} may always be satisfied when $f(t,y)$ is sublinear for large $|y|$, in the case when $\alpha=2$ we can refer the readers to see \cite{Schmitt-Smith}. \end{remark} \subsection*{Acknowledgments} A. Yang was supported by Project 201408330015 from the China Scholarship Council, and by the NNSF of China (61273016). This research was carried out while A. Yang was a Visiting Research Professor at Baylor University. \begin{thebibliography}{99} \bibitem{W.Coppel} W. 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