\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 36, 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 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/36\hfil Positive solutions for boundary-value problems] {Positive solutions for boundary-value problems of nonlinear fractional differential equations} \author[S. Zhang\hfil EJDE-2006/36\hfilneg] {Shuqin Zhang} \address{Shuqin Zhang \hfill\break Department of Mathematics, University of Mining and Technology, Beijing, 100083 China} \email{zhangshuqin@tsinghua.org.cn} \date{} \thanks{Submitted November 30, 2005. Published March 21, 2006.} \subjclass[2000]{34B15} \keywords{Caputo's fractional derivative; fractional differential equation; \hfill\break\indent boundary-value problem; positive solution; fractional Green's function; fixed-point theorem} \begin{abstract} In this paper, we consider the existence and multiplicity of positive solutions for the nonlinear fractional differential equation boundary-value problem \begin{gather*} \mathbf{D}_{0+}^\alpha u(t)=f(t,u(t)),\quad 00 $$where s>0, is called Riemann-Liouville fractional integral of order s. \smallskip \noindent\textbf{Definition.} \cite[page 36-37]{a} For a function f(x) given in the interval [0,\infty), the expression$$ D_{0+}^sf(x)=\frac 1{\Gamma(n-s)}(\frac d{dx})^n \int_0^x\frac{f(t)}{(x-t)^{s-n+1}}dt where  n=[s]+1, [s] denotes the integer part of number s, is called the Riemann-Liouville fractional derivative of order s. As examples, for \mu>-1, we have \begin{gather*} \mathbf{D}_{0+}^\alpha x^\mu=\mu(\mu-1)\dots (\mu-n+1) \frac{\Gamma(1+\mu-n)}{\Gamma(1+\mu-\alpha)}x^{\mu-\alpha} \\ D_{0+}^\alpha x^\mu=\frac{\Gamma(1+\mu-n)}{\Gamma(1+\mu-\alpha)}x^{\mu-\alpha} \end{gather*} where n=[\alpha]+1. From the definition of Caputo's derivative and Remark \ref{rmk2.1}, we can obtain the statement. \begin{lemma} \label{lem2.1} Let \alpha>0, then the differential equation $\mathbf{D}_{0+}^\alpha u(t)=0$ has solutions u(t)=c_0+c_1t+c_2t^2+\dots+c_nt^{n-1}, c_i\in \mathbb{R}, i=0,1,\dots,n, n=[\alpha]+1. \end{lemma} From the lemma above, we deduce the following statement. \begin{lemma} \label{lem2.2} Let \alpha>0, then \begin{align*} I_{0+}^\alpha \mathbf{D}_{0+}^\alpha u(t)=u(t)+c_0+c_1t+c_2t^2+\dots+c_nt^{n-1} \end{align*} for some c_i\in \mathbb{R}, i=0,1,\dots,n, n=[\alpha]+1. \end{lemma} The following theorems will play major role in our next analysis. \begin{lemma}[\cite{g}] \label{lem2.3} Let X be a Banach space, and let P\subset X be a cone in X. Assume \Omega_1, \Omega_2 are open subsets of X with 0\in \Omega_1\subset {\overline{\Omega}_1}\subset{\Omega _2}, and let S:P\to P be a completely continuous operator such that, either \begin{enumerate} \item \|Sw\|\leq \|w\|, w\in P\cap {\partial \Omega_1},  \|Sw\|\geq \|w\|, w\in P\cap {\partial \Omega_2}, or \item \|Sw\|\geq \|w\|, w\in P\cap {\partial \Omega_1}, \|Sw\|\leq \|w\| w\in P\cap {\partial \Omega_2} \end{enumerate} Then S has a fixed point in P\cap {\overline{\Omega}_2\backslash{\Omega_1}}. \end{lemma} \noindent\textbf{Definition} %2.4 A map \delta is said to be a nonnegative continuous concave functional on K if \delta : K\to [0,+\infty) is continuous and \delta(tx+(1-t)y)\geq t\delta(x)+(1-t)\delta(y) $$for all x, y\in K and 0\leq t\leq 1. And let$$ K(\delta ,a,b)=\{u\in K|a\leq \delta (u),\|u\|\leq b\} $$\begin{lemma}[\cite{f}] \label{lem2.4} Let K be a cone and K_c=\{y\in K|\|y\|\leq c\}, and A:\overline{K_c}\to \overline{K_c} be completely continuous and \alpha  be a nonnegative continuous concave function on K such that \alpha(y)\leq\|y\|, for all y\in\overline{K_c}. Suppose there exist 0b\}\neq\emptyset and \alpha(Ay)>b, for all y\in K\{\alpha,b,d\}, \item[(C2)] \|Ay\|b, for y\in K\{\alpha,b,c\} with \|Ay\|>d. \end{itemize} Then A has at least three fixed points y_1,y_2,y_3 satisfying$$ \|y_1\|a\quad \mbox{with}\quad \alpha(y_3)0$,$u\in L(0,1)(see \cite{a}), we have $u'(t)=\frac 1{\Gamma(\alpha-1)}\int_0^t(t-s)^{\alpha-2}h(s)ds-c_2$ As boundary conditions for problem \eqref{e4}, we have \begin{gather*} -c_1-c_2=0\\ -c_1-2c_2=-I_{0+}^\alpha h(1)-I_{0+}^{\alpha-1}h(1); \end{gather*} that is, \begin{gather*} c_1=-I_{0+}^\alpha h(1)-I_{0+}^{\alpha-1}h(1)\\ c_2=I_{0+}^\alpha h(1)+I_{0+}^{\alpha-1}h(1) \end{gather*} Therefore, the unique solution of \eqref{e4} is \begin{align*} &u(t)=\frac 1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}h(s)ds+ \frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha -1}h(s)ds\\ &\quad +\frac{1}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha -2}h(s)ds-\frac{t}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha -1}h(s)ds\\ &\quad -\frac{t}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha -2}h(s)ds\\ &= \int_0^t(\frac{(1-s)^{\alpha-1}(1-t)+(t-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}(1-t)}{\Gamma(\alpha-1)})h(s)ds\\ &\quad +\int_t^1 (\frac{(1-s)^{\alpha-1}(1-t)}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}(1-t)}{\Gamma(\alpha-1)})h(s)ds\\ &=\int_0^1G(t,s)h(s)ds \end{align*} which completes the proof. \end{proof} \begin{lemma} \label{lem3.2} Leth(t)\in{C[0,1]}$be a given function, then function$G(t,s)$defined by \eqref{e6} has the following properties: \begin{itemize} \item[(R1)]$G(t,s)\in C([0,1]\times [0,1))$, and$G(t,s)>0$for$t,s\in (0,1)$; \item[(R2)] There exists a positive function$\gamma\in C(0,1)$such that $$\label{e7} \begin {gathered} \min_{1/4\leq t\leq 3/4}G(t,s)\geq \gamma(s)M(s), \quad\ s\in(0,1) \\ \max_{0\leq t\leq 1}G(t,s)\leq M(s), \end{gathered}$$ where $$\label{e8} M(s)=\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)},\quad s\in [0,1)$$ \end{itemize} \end{lemma} \begin{proof} From the expression of$G(t,s)$, it is obvious that$G(t,s)\in C([0,1]\times[0,1))$and$G(t,s)\geq 0$for$s, t\in (0,1)$. Next, we will prove (R2). From the definition of$G(t,s)$, we can known that, for given$s\in (0,1)$,$G(t,s)$is decreasing with respect to$t$for$t\leq s$, we let \begin{gather*} g_1(t,s)=\frac{(1-t)(1-s)^{\alpha -1}+(t-s)^{\alpha-1}}{\Gamma(\alpha)}+\frac{(1-t)(1-s)^{\alpha-2}}{ \Gamma(\alpha-1)},\quad s\leq t \\ g_2(t,s)=\frac{(1-t)(1-s)^{\alpha -1}}{\Gamma(\alpha)}+\frac{(1-t)(1-s)^{\alpha-2}}{ \Gamma(\alpha-1)},\quad t\leq s \end{gather*} That is,$g_1(t,s)$is a continuous function for$\frac 14\leq t\leq \frac 34$, and$g_2(t,s)$is decreasing with respect to$t$. Hence, we have \begin{gather*} g_1(t,s)\geq \frac{ (1-s)^{\alpha}}{4\Gamma(\alpha)}+ \frac{ (1-s)^{\alpha-2}}{4\Gamma(\alpha-1)},\quad \mbox{for$1/4\leq t\leq 3/4} \\ \max_{0\leq t\leq 1}g_1(t,s)\leq\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)} \\ \min_{1/4\leq t\leq 3/4}g_2(t,s)= g_2(\frac 34,s)=\frac{(1-s)^{\alpha-1}}{4\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{4\Gamma(\alpha-1)} \\ \begin{aligned} \max_{0\leq t\leq 1}g_2(t,s)= g_2(0,s)&=\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\\ &<\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)} \end{aligned} \end{gather*} Thus, we have \begin{gather} \label{e9} \min_{1/4\leq t\leq 3/4}G(t,s)\geq m(s)=\frac{(1-s)^{\alpha-1}}{4\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{4\Gamma(\alpha-1)},\quad s\in[0,1) \\ \label{e10} \max_{0\leq t\leq 1}G(t,s)\leq M(s)=\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)},\quad s\in [0,1) \end{gather} Let $$\label{e11} \gamma(s)= m(s)/M(s)=\frac 14 \frac{(1-s)^{\alpha-1}+(\alpha-1)(1-s)^{\alpha-2}}{2(1-s)^{\alpha-1}+ (\alpha-1)(1-s)^{\alpha-2}},\quad s\in (0,1)$$ It is obviously that\gamma(s)\in C((0,1), (0,+\infty))$. The proof is completed. \end{proof} \begin{remark} \label{rmk3.1} \rm From the definition of function$\gamma(s)$, we see that$\gamma(s)\geq \frac 18$. \end{remark} Let$E=C[0,1]$be endowed with the ordering$u\leq v$if$u(t)\leq v(t)$for all$t\in[0,1]$, and the maximum norm,$\|u\|=\max_{0\leq t\leq 1}|u(t)|$, Define the cone$K\subset E$by $K=\{u\in E|u(t)\geq 0, \min_{1/4\leq t\leq 3/4}\geq \frac 18\|u\|\}$ and the nonnegative continuous concave functional$\varphi$on the cone$K$by $\varphi(u)=\min_{1/4\leq t\leq 3/4}|u(t)|$ \begin{lemma} \label{lem3.3} Assume that$f(t,u)$is continuous on$[0,1]\times [0,\infty)$. A function$u\in K$is a solution of boundary-value problem \eqref{e1} if and only if it is a solution of the integral equation \eqref{e5}. \end{lemma} \begin{proof} Let$u\in K$be a solution of boundary-value problem \eqref{e1}. Applying the operator$I_{0+}^\alpha$to both sides of equation of problem \eqref{e1}, we have $u(t)=c_1+c_2t+I_{0+}^\alpha f(t,u(t))$ for some$c_1,c_2\in \mathbb{R}$. By the same methods as obtaining the Green's function of problem \eqref{e1} (Lemma \ref{lem3.1}), by boundary value conditions of problem \eqref{e1}, we can calculate out constants$c_1$and$c_2$, so $u(t)=\int_0^1G(t,s)f(s,u(s))ds$ From Lemma \ref{lem3.2} and Remark \ref{rmk3.1}, we obtain that$\int_0^1 G(t,s) f(s,u(s))ds \in K$. Hence,$u$is also a solution of integral equation \eqref{e5}. Let$u\in K$be a solution of integral equation \eqref{e5}. If we denote the right-hand side of integral equation \eqref{e5} by$w(t)$, then, applying Caputo's fractional operator to both sides of integral equation \eqref{e5}, by the Definition of function$G(t,s), since \begin{align*} w(t)&=\int_0^1G(t,s)f(s,u(s))ds\\ &=\frac 1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}f(s,u(s))ds+ \frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha -1}f(s,u(s))ds\\ &\quad +\frac{1}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha -2}f(s,u(s))ds-\frac{t}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha -1}f(s,u(s))ds\\ &\quad -\frac{t}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha -2}f(s,u(s))ds\,. \end{align*} Therefore, \begin{align*} w'(t)&=\frac d{dt}I_{0+}^\alpha f(t,u(t))-I_{0+}^\alpha f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))\\ &= D_{0+}^1I_{0+}^\alpha f(t,u(t))-I_{0+}^\alpha f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))\\ &= D_{0+}^1I_{0+}^1I_{0+}^{\alpha-1} f(t,u(t))-I_{0+}^\alpha f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))\\ &= I_{0+}^{\alpha-1} f(t,u(t))-I_{0+}^\alpha f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1)) \end{align*} and \begin{gather*} w''(t)=D_{0+}^1I_{0+}^{\alpha-1} f(t,u(t))=D_{0+}^{2-\alpha}f(t,u(t)) \\ \mathbf{D}_{0+}^\alpha w(t)=I_{0+}^{2-\alpha}w''(t) =I_{0+}^{2-\alpha}D_{0+}^{2-\alpha}f(t,u(t)) =f(t,u(t)) \end{gather*} here, use the relationI_{0+}^sI_{0+}^tg(t)=I_{0+}^{s+t}g(t)$,$D_{0+}^sI_{0+}^sg(t)=g(t)$,$s>0$,$t>0$,$g\in L(0,1)$and$I_{0+}^s D_{0+}^s g(t)=g(t)$,$s>0$,$g\in C[0,1]$(see \cite{a}), where$D_{0+}^s$is Riemann-Liouville fractional derivative. That is,$\mathbf{D}_{0+}^\alpha u(t)=f(t,u(t))$. On the other hand, one has \begin{gather*} u(0) =\int_0^1(\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})f(s,u(s))ds \\ u'(0) =-\int_0^1(\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+ \frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})f(s,u(s))ds \\ u(1) =\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds \\ u'(1) =-\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds\,. \end{gather*} We obtain $u(0)+u'(0)=0, u(1)+u'(1)=0$ which implies that$u\in K$is a solution of problem \eqref{e1}. \end{proof} \begin{lemma} \label{lem3.4} Assume that$f(t,u)$is continuous on$[0,1]\times [0,\infty)$, and define the operator$T:K\to E$by $Tu(t)=\int_0^1G(t,s)f(s,u(s))ds$ Then$T:K\to K$is completely continuous. \end{lemma} \begin{proof} Firstly, we prove that$T:K\to K$. In view of the expression of$G(t,s)$, it is clear that,$Tu(t)\geq 0$,$t\in [0,1]$,$Tu(t)$is continuous for$u\in K$. And that, for$u\in K, by means of the Lemma \ref{lem3.1} and Remark \ref{rmk3.1}, we have $\min_{1/4\leq t\leq 3/4}Tu(t)=\min_{1/4\leq t\leq 3/4}\int_0^1G(t,s)f(s,u(s))ds\geq\frac 18\int_0^1M(s)f(s,u(s))ds$ On the other hand, \begin{align*} \|Tu\|=\max_{0\leq t\leq 1}|Tu(t)|\leq\int_0^1M(s)f(s,u(s))ds\,. \end{align*} Thus, we obtain $\min_{1/4\leq t\leq 3/4}Tu(t)\geq\frac 18\|Tu\|$ which impliesT:K\to K$. Let$P\subset K$be bounded, i.e. there exists a positive constant$L>0$such that$\|u\|\leq L$, for all$u\in P$. Let$M=\max_{0\leq t\leq 1, 0\leq u\leq L}|f(t,u)|+1$, then for$u\in P$, from the Lemma \ref{lem3.1}, one has $|Tu(t)|\leq \int_0^1|G(t,s)f(t,u(s))|ds\leq M\int_0^1M(s)ds$ Hence,$T(P)$is bounded. For all$\varepsilon>0$, each$u\in P$,$t_1, t_2\in [0,1]$,$t_10, N_2>0$, such that for all$0<\varepsilon<(2\int_0^1M(s)ds)^{-1}$and$\rho>\frac {64}{\int_{1/4}^{3/4}M(s)ds}>0$. Then \begin{gather*} f(t,u(t))\leq \varepsilon u^{\mu_1}, \quad \mbox{for$t\in [0,1],u\geq N_1$}\\ f(t,u(t))>\rho u^{\nu_1}, \quad \mbox{for$t\in [0,1],0\leq u\leq N_2} \end{gather*} So we have $$f(t,u(t))\leq \varepsilon u^{\mu_1}+c, \quad \mbox{for t\in[0,1],u\in [0,+\infty)}$$ where $$c=\max_{0\leq t\leq 1 ,0\leq u\leq N_1}|f(t,u(t))|+1$$ Let \Omega_1=\{u\in K; \|u\|\{1,2c\int_0^1M(s)ds\}. For u\in \partial{\Omega_1}, from the Lemma \ref{lem3.2}, we have \begin{align*} |T u(t)|&=\int_0^1G(t,s)f(s,u(s))ds\\ &\leq\int_0^1M(s)(\varepsilon |u|^{\mu_1}+c)ds\\ &\leq \varepsilon R_1^{\mu_1}\int_0^1M(s)+c\int_0^1M(s)ds\\ &\leq \frac {R_1}2+\frac {R_1}2=R_1\,, \end{align*} \|Tu\|\leq R_1=\|u\|. Let \Omega_2=\{u\in K; \|u\|\frac \rho 8\int_{1/4}^{3/4}M(s)u(s)^{\nu_1} ds\\ &\geq \frac \rho {64}\int_{1/4}^{3/4}M(s)\|u\|^{\nu_1} ds\\ &=\frac \rho {64}\int_{1/4}^{3/4}M(s)R_2R_2^{\nu_1-1} ds\\ &\geq \frac \rho {64}\int_{1/4}^{3/4}M(s)R_2 ds\\ &> R_2=\|u\| \end{align*} so\|Tu\|\geq R_2=\|u\|$. Then Lemma \ref{lem2.3} implies that operator$T$has one fixed point$u^*(t)\in {\overline {\Omega_1}\backslash {\Omega_2}}$. Then$u^*(t)$is one positive solution of problem \eqref{e1}. For condition (H1'), we can obtain the result in a similarly way. Now, we give a briefly description. Assume that (H1') holds, thus, there exist$M_1>0, M_2>0$, such that for all$0< \varepsilon<(\int_0^1M(s)ds)^{-1}$and$\lambda>(\frac{\int_{1/4}^{3/4}M(s)ds}{64})^{-1}>0$, we have have \begin{gather*} f(t,u(t))>\lambda u^{\mu_2}, \quad \mbox{for$t\in [0,1],u\geq M_1$} \\ f(t,u(t))\leq\varepsilon u^{\nu_2}, \quad \mbox{for$t\in [0,1],0\leq u\leq M_2$} \end{gather*} Let$$\Omega_1=\{u\in K; \|u\|\{1,8M_1\}$, $0M_1$. thus, from the Lemma \ref{lem3.2}, we have \begin{align*} |T u(t)|&\geq\int_{1/4}^{3/4}G(t,s)f(s,u(s))ds\\ &\geq \frac \lambda {64}\int_{1/4}^{3/4}M(s)\|u\|^{\mu_2} ds\\ &\geq \frac \lambda {64}\int_{1/4}^{3/4}M(s)\|u\| ds\\ &> R_1=\|u\| \end{align*} for $u\in \partial{ \Omega_2}$, we obtain $|T u(t)|\leq\int_0^1M(s)\varepsilon \|u\|^{\nu_2}ds \leq \varepsilon R_2\int_0^1M(s)ds \leq R_2$ Thence, the Lemma \ref{lem2.3} implies that operator $T$ has one fixed point $u^*(t)\in {\overline {\Omega_1}\backslash {\Omega_2}}$, then $u^*(t)$ is a positive solution of problem \eqref{e1}. \end{proof} Let $M=(\int_0^1M(s)ds)^{-1},\quad N=(\int_{1/4}^{3/4}\gamma(s)M(s)ds)^{-1}$ \begin{theorem} \label{thm3.2} Assume that $f(t,u)$ is continuous on $[0,1]\times [0,\infty)$, and there exist constants $0b$, thus, $\{u\in K(\delta,b,c)|\delta(u)>b\}\neq\emptyset$. Thence, if $u\in K(\delta,b,c)$, then $b\leq u(t)\leq c$ for $1/4\leq t\leq 3/4$, by assumption $(H_3)$, we have $f(t,u)\geq Nb$, for $\frac 14\leq t\leq\frac 34$, so, by the Lemma \ref{lem3.2}, there has $\delta(Tu)=\min_{1/4\leq t\leq 3/4}|Tu(t)|> \int_{1/4}^{3/4}\gamma(s)M(s)Nbds=b$ By Lemma \ref{lem2.4}, problem \eqref{e1} has at least three positive solutions $u_1, u_2, u_3$ with the required conditions; which completes the proof. \end{proof} \begin{thebibliography}{99} \bibitem{d} O. P. Agrawal, \emph{Formulation of Euler-Larange equations for fractional variational problems}, J. Math. Anal. Appl. 272, 368-379, 2002. \bibitem{c} D. Delbosco and L.Rodino, \emph{Existence and Uniqueness for a Nonlinear Fractional Differential Equation}, J. Math. Appl, 204, 609-625, 1996. \bibitem{g} Krasnosel'skii M. A., \emph{Positive solutions of operator equations}, Noordhoff Gronigen, Netherland, 1964. \bibitem{f} R. W. Leggett, L. R. 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Appl. 278, 1, 136-148, 2003. \end{thebibliography} \section*{Addendum posted on November 9, 2009.} The definition $n=[\alpha]+1$ in Lemmas \ref{lem2.1} and \ref{lem2.2} is incomplete. It should be $n=\begin{cases} [\alpha]+1 &\text{if } n \not\in \{0,1,2,\dots\}\\ \alpha &\text{if } n \in \{0,1,2,\dots\}\,. \end{cases}$ The author wants to thank Yige Zhao, Shurong Sun, Zhenlai Han, and Meng Zhang (at the University of Jinan) for pointing out this misprint. \end{document}