\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2004(2004), No. 23, 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 (login: ftp)} \thanks{\copyright 2004 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2004/23\hfil Existence of positive solutions] {Existence of positive solutions for fractional differential equations with Riemann-Liouville left-hand and right-hand fractional derivatives} \author[Shuqin Zhang\hfil EJDE-2004/23\hfilneg] {Shuqin Zhang} \address{Shuqin Zhang \hfill\break Department of Mathematical Sciences, Tsinghua University,\\ Beijing, 100084 China} \email{szhang@math.tsinghua.edu.cn\quad shuqinzhang\_1971@hotmail.com} \date{} \thanks{Submitted June 15, 2003. Published February 16, 2004.} \thanks{Supported by the Postdoctoral Foundation of China.} \subjclass[2000]{34B15} \keywords{Existence of positive solution, fractional differential equation, \hfill\break\indent fixed point, cone} \begin{abstract} Combining properties of Riemann-Liouville fractional calculus and fixed point theorems, we obtain three existence results of one positive solution and of multiple positive solutions for initial value problems with fractional differential equations. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{prop}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Let $s$ be a real number and $n = [s]+1$ where $[s]$ the integer part of $s$. For a function $f:[a,b] \to \mathbb{R}$ the expressions \begin{gather*} D_{a+}^sf(x)=\frac 1{\Gamma(n-s)}(\frac d{dx})^n \int_a^x\frac{f(t)}{(x-t)^{s-n+1}}dt\\ D_{b-}^sf(x)=\frac {(-1)^n}{\Gamma(n-s)}(\frac d{dx})^n \int_x^b\frac{f(t)}{(t-x)^{s-n+1}}dt \end{gather*} are called, respectively, the Riemann-Liouville left-hand and right-hand fractional derivative of order $s$. If $s$ is an integer, the derivative of order $s$ is understood in the sense of usual differentiation: $$D_{a+}^s=(\frac d{dx})^s,\quad D_{b-}^s=(-1)^s(\frac d{dx})^s\quad s=1,2,3,\dots$$ Here we consider the initial-value problem \label{e1} \begin{aligned} D_{1-}^\alpha D_{0+}^\delta u(t) =g(t,u(t))\quad 0< t< 1 , 0<\alpha,\delta<1\\ t^{1-\delta}u(t)\big|_{t=0}=a\geq 0,\quad (1-t)^{1-\alpha}D_{0+}^\delta u(t)|_{t=0}=b\geq 0, \end{aligned} where $D_{1-}^\alpha,D_{0+}^\delta$ are the Riemann-Liouville right-hand and left-hand fractional derivatives. For $x >0$, the expressions \begin{gather*} I_{a+}^sf(x)= \frac 1{\Gamma(s)}\int_a^x\frac{f(t)}{(x-t)^{1-s}}dt,\quad x>a,\\ I_{b-}^sf(x)= \frac 1{\Gamma(s)}\int_x^b\frac{f(t)}{(t-x)^{1-s}}dt,\quad x0$then$D_{a+}^sI_{a+}^sf(x)=f(x)$for any$f\in L_1(a,b)$, while $$\label{e2} I_{a+}^sD_{a+}^sf(x)=f(x)$$ is satisfied for$f\in I_{a+}^s(L_1(a,b))$with $I_{a+}^s(L_1(a,b))=\{g(x): g(x)=I_{a+}^s\varphi(x),\quad \varphi\in L_1(a,b)\}$ If$f, D_{a+}^s f\in L_1(a,b)$, then \eqref{e2} is not true in general. However $$I_{a+}^sD_{a+}^sf(x)=f(x)-\Sigma_{k=0}^{n-1}\frac{(x-a)^{s-k-1}}{\Gamma(s-k)} f_{n-s}^{(s-k-1)}(a)$$ where$n=[s]+1, f_{n-s}(x)=I_{a+}^{n-s}f(x)$. In particular for$0b\}\neq\emptyset$and$\alpha(Ay)>b$, for all$y\in K\{\alpha,b,d\}$, \item[(C2)]$\|Ay\|b$, for$y\in K\{\alpha,b,c\}$with$\|Ay\|>d$, \end{itemize} then$A$has at least three fixed points$y_1,y_2,y_3satisfying \|y_1\|a \quad \mbox{with}\quad\alpha(y_3)0\}. For u \in M, \begin{align*} &t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\ &=|a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}} {\Gamma(\delta)}\int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &+\quad \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta) \Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} g(\tau,u(\tau))d\tau\,ds|\\ &\leq a+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\ &+\quad \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta) \Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} |g(\tau,u(\tau))|d\tau\,ds\\ &\leq a+\frac{b}{\Gamma(1+\delta)}+\frac{L}{\Gamma(\delta)\Gamma(\alpha)} \int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} d\tau\,ds\\ &= a+\frac{b}{\Gamma(1+\delta)}+\frac {L}{\Gamma(\delta)\Gamma(1+\alpha)} \int_0^t(t-s)^{\delta-1}(1-s)^\alpha ds\\ &\leq a+\frac{b}{\Gamma(1+\delta)}+\frac{L}{\Gamma(\delta) \Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}ds\\ &= a+\frac{b}{\Gamma(1+\delta)}+\frac {L}{\Gamma(1+\delta)\Gamma(1+\alpha)}t^\delta\\ &\leq a+\frac{b}{\Gamma(1+\delta)}+\frac {L}{\Gamma(1+\delta)\Gamma(1+\alpha)} \end{align*} where L=\max_{0\leq t\leq 1,\|u\|\leq l}|g(t,u(t))|+1\,. $$So T(M) is bounded. Let us see that \overline {T(M)} is equicontinuous. For u\in M, \varepsilon>0, t_1,t_2\in[0,1],t_10, such that for 0< \varepsilon<\frac {\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)} {4\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))},$$ g(t,u(t))\leq \varepsilon |u|^\mu, \quad \mbox{fort\in [0,1],|u|\geq N$} $$So we have$$g(t,u(t))\leq \varepsilon |u|^\mu+c, \quad \mbox{for$t\in[0,1],u\in [0,+\infty)} $$where$$c=\max_{0\leq t\leq 1,|u|\leq N}|g(t,u(t))|+1. Let \Omega_1=\{u\in K; \|u\|\{1,4a,\frac {4b}{\Gamma(1+\delta)},\frac {4c}{\Gamma(1+\delta)\Gamma(1+\alpha)}\}, for u\in \partial{\Omega_1}, we have \begin{align*} &t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\ &=|a(1-t)^{1-\alpha}+\frac{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &+\quad \frac{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)} \int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}g(\tau,u(\tau))d\tau\,ds|\\ &\leq a+\frac{bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\ &\quad +\frac{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)} \int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1}|g(\tau,u(\tau))|d\tau\,ds\\ &\leq a+\frac b{\Gamma(1+\delta)}+\frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)} \int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} (\varepsilon |u(\tau)|^\mu+c)d\tau\,ds\\ &\leq \frac{t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)} \int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} (\varepsilon \tau^{\mu(\delta-1)}(1-\tau)^{\mu(\alpha-1)}\|u\|^\mu+c)d\tau\,ds\\ &+a+\frac{b}{\Gamma(1+\delta)}\\ &\leq\frac{t^{1-\delta}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu}{\Gamma(\delta)\Gamma(\alpha)} \int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} s^{\mu(\delta-1)}(1-\tau)^{\mu(\alpha-1)} d\tau\,ds\\ &+\quad \frac c{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1} \int_s^1(\tau-s)^{\alpha-1}d\tau\,ds+a+\frac{b}{\Gamma(1+\delta)}\\ &=\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\alpha+ \mu(\alpha-1))}\int_0^t(t-s)^{\delta-1} s^{\mu(\delta-1)}(1-s)^{\mu(\alpha-1)+\alpha} ds\\ &\quad +\frac c{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}(1-s)^\alpha ds +a+\frac{b}{\Gamma(1+\delta)} \end{align*} If \mu(\alpha-1)+\alpha<0, then the first equality of above becomes \begin{align*} &\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\alpha+\mu(\alpha-1))}\int_0^t(t-s)^{\delta-1} s^{\mu(\delta-1)}(1-s)^{\mu(\alpha-1)+\alpha} ds \\ &\leq \frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\mu(\alpha-1)+\alpha)}\int_0^t(t-s)^{\delta-1} s^{\mu(\delta-1)}(1-t)^{\mu(\alpha-1)+\alpha} ds \\ &= \frac {t^{1-\delta}(1-t)^{1+\mu(\alpha-1)}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\mu(\alpha-1)+\alpha)} \int_0^t(t-s)^{\delta-1} s^{\mu(\delta-1)} ds \\ &=\frac {t^{1+\mu(\delta-1)}(1-t)^{1+\mu(\alpha-1)}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\\ &\leq \frac {\varepsilon R_1^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\,. \end{align*} If \mu(\alpha-1)+\alpha\geq 0, then the first equality of above becomes \begin{align*} &\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\alpha+\mu(\alpha-1))}\int_0^t(t-s)^{\delta-1} s^{\mu(\delta-1)}(1-s)^{\mu(\alpha-1)+\alpha} ds \\ &\leq\frac {t^{1-\delta}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))}{\Gamma(\delta)\Gamma(1+\mu(\alpha-1)+\alpha)}\int_0^t(t-s)^{\delta-1} s^{\mu(\delta-1)} ds \\ &=\frac {t^{1+\mu(\delta-1)}(1-t)^{1-\alpha}\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\\ &\leq \frac {\varepsilon R_1^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\,. \end{align*} Therefore, we obtain \begin{align*} t^{1-\delta}(1-t)^{1-\alpha}|T u(t)| &\leq\frac {\varepsilon \|u\|^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\\ &\quad +\frac c{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1}ds+a +\frac{b}{\Gamma(1+\delta)}\\ &\leq \frac {\varepsilon R_1^\mu\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))}{\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\\ &\quad +\frac c{\Gamma(1+\delta)\Gamma(1+\alpha)}t^\delta+a +\frac{b}{\Gamma(1+\delta)}\\ &\leq \frac {\varepsilon R_1\Gamma(1+\mu(\alpha-1))\Gamma(1+\mu(\delta-1))} {\Gamma(1+\mu(\alpha-1)+\alpha) \Gamma(1+\mu(\delta-1)+\delta)}\\ &\quad +\frac c{\Gamma(1+\delta)\Gamma(1+\alpha)}+a+\frac{b}{\Gamma(1+\delta)}\\ &\leq \frac {R_1}4+\frac {R_1}4+\frac {R_1}4+\frac {R_1}4 =R_1 \end{align*} and \|Tu\|\leq R_1=\|u\|. Taking \Omega_2=\{u\in K; \|u\|0 and $1\leq \lambda<\min\{\frac \alpha {1-\alpha},\frac 1{1-\delta}\}$ with $\alpha\geq\frac 12$ or $1\leq\lambda<\min\{\frac 1{1-\alpha},\frac 1{1-\delta}\}$ with $0<\alpha\leq\frac 12$ such that $$\label{h3} g(t,u(t))\leq c_1+c_2|u(t)|^\lambda, \quad \mbox{for all t\in[0,1], u\in [0,+\infty)}$$ then problem \eqref{e1} has at least one solution. \end{theorem} \begin{proof} As in Theorem \ref{theo1}, we only need to consider existence of fixed point of operator $T$. By Lemma \ref{lm1}, $T$ is a completely continuous operator. We will make use of the Schauder Fixed Point Theorem to prove this theorem. Let $0(\frac 12)^{-1-\delta}(\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac 12)^\delta} {\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)\Gamma(1+\alpha)}{\delta}$ for $0\leq t\leq 1$, $\overline b\leq u\leq 2\overline b$, \end{itemize} then the initial-value problem \eqref{e1} has three positive solutions $u_1,u_2,u_3$ satisfying \label{e6} \|u_1\|<\overline a,\quad \overline b\overline a \quad\mbox{with}\quad f(u_3)<\overline b \end{theorem} \begin{proof} We apply Theorem \ref{theoB}. Since $g$ is continuous, by Lemma \ref{lm1}, operator $T$ is completely continuous. Now we choose $u\in \overline {K_c}$, then $\|u\|\leq c$, and $g(t,u(t))<(c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta) \Gamma(1+\alpha)$ for $t\in[0,1]$ by (H5), so we have \begin{align*} &t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\ &=|a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &\quad + \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} g(\tau,u(\tau))d\tau\,ds|\\ &\leq a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &\quad + \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} |g(\tau,u(\tau))|d\tau\,ds\\ &< a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &\quad +\frac {t^{1-\delta}(1-t)^{1-\alpha}(c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1} \int_s^1(\tau-s)^{\alpha-1} d\tau\,ds\\ &=a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &\quad+ \frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1} (c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)(1-s)^\alpha ds\\ &\leq a+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-t)^{\alpha-1}ds\\ &\quad+\frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(1+\alpha)}\int_0^t(t-s)^{\delta-1} (c-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)ds\\ &\leq a+\frac {bt}{\Gamma(1+\delta)}+t(1-t)^{1-\alpha} (c-a-\frac b{\Gamma(1+\delta)})\\ &\leq a+\frac {b}{\Gamma(1+\delta)}+c-a-\frac b{\Gamma(1+\delta)} =c \end{align*} That is $\|Tu\|\leq c$. On the other hand, $f(u)\leq\|u\|$ for $u\in K_c$, and $T:K_c\rightarrow K_c$ is completely continuous by the above deduction and Lemma \ref{lm1}. Similarly, if $u\in \overline {K_{\overline a}}$, then $\|u\|\leq \overline a$, and $g(t,u(t))<(\overline a-a-\frac b{\Gamma(1+\delta)})\Gamma(1+\delta)\Gamma(1+\alpha)$ for each $t\in[0,1]$ by (H4), we obtain \begin{align*} &t^{1-\delta}(1-t)^{1-\alpha}|T u(t)|\\ &=|a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &\quad+\frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} g(\tau,u(\tau))d\tau\,ds|\\ &\leq a(1-t)^{1-\alpha}+\frac {bt^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)} \int_0^t(t-s)^{\delta-1}(1-s)^{\alpha-1}ds\\ &\quad+\frac {t^{1-\delta}(1-t)^{1-\alpha}}{\Gamma(\delta)\Gamma(\alpha)}\int_0^t(t-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} |g(\tau,u(\tau))|d\tau\,ds\\ &\overline b$, so$\{u\in K(f,\overline b,2\overline b)|f(u)>\overline b\}\neq \emptyset$. In addition, if we choose$u\in K(f,\overline b,2\overline b)$, then we have$\overline b\frac {(\frac 12)^{-1-\delta}(\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac 12)^\delta} {\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta) \Gamma(1+\alpha)}{\delta}}{\Gamma(\delta)\Gamma(\alpha)}\\ &\quad\times \int_0^{\frac12}(\frac 12-s)^{\delta-1}\int_s^1(\tau-s)^{\alpha-1} d\tau\,ds +a(\frac 12)^{\delta-1}+\frac b{\Gamma(\delta)}\int_0^{\frac 12} (\frac 12-s)^{\delta-1}ds\\ &=\frac {(\frac 12)^{-1-\delta}(\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac 12)^\delta} {\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)}{\delta}}{\Gamma(\delta)}\int_0^{\frac 12}(\frac 12-s)^{\delta-1} (1-s)^\alpha ds\\ &\quad+a(\frac 12)^{\delta-1}+\frac {b(\frac 12)^\delta}{\Gamma(1+\delta)}\\ &\geq \frac {(\frac 12)^{-1-\delta}(\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac 12)^\delta} {\Gamma(1+\delta)})\frac{(1+\delta)\Gamma(1+\delta)}{\delta}}{\Gamma(\delta)}\int_0^{\frac 12}(\frac 12-s)^{\delta-1}(1-s)ds\\ &\quad+a(\frac 12)^{\delta-1}+\frac {b(\frac12)^\delta}{\Gamma(1+\delta)}\\ &> \frac {(\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac 12)^\delta}{\Gamma(1+\delta)})(1+\delta)}\delta- \frac {\overline b-a(\frac 12)^{\delta-1}-\frac {b(\frac12)^\delta} {\Gamma(1+\delta)}}\delta\\ &\quad+a(\frac 12)^{\delta-1}+\frac {b(\frac 12)^\delta}{\Gamma(1+\delta)} =\overline b\,. \end{align*} So $f(Tu)>\overline b$ for $u\in K(f,\overline b,2\overline b)$ which proves the condition (C1) of Theorem \ref{theoB}. 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