\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 129, 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-2006/129\hfil Existence of positive solutions] {Existence of positive solutions for multi-term non-autonomous fractional differential equations with polynomial coefficients} \author[A. Babakhani, V. Daftardar-Gejji\hfil EJDE-2006/129\hfilneg] {Azizollah Babakhani, Varsha Daftardar-Gejji} % in alphabetical order \address{Azizollah Babakhani \newline Department of Mathematics, University of Mazanderan, Babol, Iran} \email{babakhani@nit.ac.ir} \address{Varsha Daftardar-Gejji \newline Department of Mathematics, University of Pune \\ Ganeshkhind, Pune - 411007, India} \email{vsgejji@math.unipune.ernet.in} \date{} \thanks{Submitted July 27, 2005. Published October 16, 2006.} \subjclass[2000]{26A33, 34B18} \keywords{Riemann-Liouville fractional derivatives and integrals; normal cone; \hfill\break\indent semi-ordered Banach space; completely continuous operator; equicontinuous set} \begin{abstract} In the present paper we discuss the existence of positive solutions in the case of multi-term non-autonomous fractional differential equations with polynomial coefficients; the constant coefficient case has been studied in \cite{bab}. We consider the equation \begin{equation*} \Big(D^{\alpha_n} -\sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}\Big)y = f(x, y). \end{equation*} We state various conditions on $f$ and $p_j$'s under which this equation has: positive solutions, a unique solution which is positive, and a unique solution which may not be positive. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Let $E$ be a real Banach space with a cone $K\subset E$. $K$ introduces a partial order $\leq $ in $E: x \leq y$ if and only if $y - x\in E$. A cone $K$ is said to be normal, if there exists a positive constant $\tau$ such that $\theta \leq f \leq g$ implies $\|f\| \leq \tau \|g\|$, where $\theta$ denotes the zero element of $K$. For $x, y \in E$ the order interval $\langle x, y \rangle$ is define to be \cite{jos}: \[ \langle x , y \rangle = \{ z\in E : x \leq z \leq y \} \] \begin{theorem}[\cite{jos}] \label{thm1.1} Let $K$ be a normal cone in a partially ordered Banach space $E$. Let $F$ be an increasing operator which transforms $\langle x_0, y_0 \rangle$ into itself; i. e., $Fx_0 \geq x_0$ and $Fy_0 \leq y_0$. Assume further that $F$ is compact and continuous. Then $F$ has at least one fixed point $x^* \in \langle x_0, y_0 \rangle$. \end{theorem} \begin{theorem}[Banach fixed point theorem \cite{jos}] \label{thm1.2} Let $K$ be a closed subspace of a Banach space $E$. Let $F$ be a contraction mapping with Lipschitz constant $k<1$ from $K$ to itself. Then $F$ has a unique fixed point $x^*$ in $K$. Moreover if $x_0$ is an arbitrary point in $K$ and $\{x_n\}$ is defined by $x_{n + 1} = Fx_n$, ($n =0, 1, 2, \dots$) then $\lim_{n\to\infty} x_n = x^* \in K$ and $d(x_n, x^*) \leq (k^n /(1 - k))\,d(x_1, x_0)$. \end{theorem} \begin{definition} \label{def1.1} \rm The left sided Riemann-Liouville fractional integral \cite{mil,pod,sam} of order $\alpha$ of a real function $f$ is defined as \begin{equation} I_{a^+}^\alpha y(x) = \frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{y(t)}{(x - t) ^{1 - \alpha}}dt, \quad \alpha > 0,\; x > a. \label{1} \end{equation} \end{definition} \begin{definition} \label{def1.2} \rm The left sided Riemann-Liouville fractional derivative \cite{mil,pod,sam} of order $\alpha$ of a function $f$ is \begin{equation} D_{a^+}^\alpha y(x) = \frac{d^n}{dx^n}\left [I_{a^+}^{n - \alpha}y(x)\right ], \quad n - 1 \leq \alpha < n,\quad n\in\mathbb{N}.\label{2} \end{equation} We denote $D^{\alpha}_{a^+}$ by $D^{\alpha}_{a}y(x)$ and $I_{a^+}^\alpha y(x)$ by $I_{a}^\alpha y(x)$. Also $D^\alpha y(x)$ and $I^\alpha y(x)$ refers to $D^{\alpha}_ {0^+}y(x)$ and $I_{0^+}^\alpha y(x)$, respectively. \end{definition} \begin{proposition} \label{prop1.1} (i) If the fractional derivative $D_a^\alpha y(x)$ is integrable, then \begin{equation} I_a^\alpha(D_a^\beta y(x)) = I_a^{\alpha - \beta}y(x) - \left [I_a^ {1 - \beta}y(x)\right]_{x=a}\frac {(x - a)^{\alpha - 1}}{\Gamma(\alpha)}, \quad 0 < \beta \leq \alpha < 1\,.\label{3} \end{equation} (ii) If $y$ is continuous on $[a,b]$, then $D_a^\alpha y(x)$ is integrable, $ I^{1-\beta}y(x)|_{x=a} = 0$ and \begin{equation} I_a^\alpha\left (D_a^\beta y(x)\right ) = I_a^{\alpha - \beta}y(x),\quad 0 < \beta \leq \alpha < 1.\label{4} \end{equation} \end{proposition} \begin{proof} For (i), we refer the reader to \cite{pod}. For (ii), let $M = \max _{_{a \leq x \leq b}} y(x)$ then, using \eqref{2} we get \[ \big|\int_{a}^{x}D_a^\alpha y(t)\,dt\big| \leq \frac{M}{\Gamma(1 - \alpha)}\int_{a}^{x}(x - t)^{- \alpha}dt = \frac {M (x - a)^{1 - \alpha}}{\Gamma(2 - \alpha)}, \] so $D_a^\alpha y(t)$ is integrable. On the other hand \[ |I_a^{1 - \beta}y(x)|_{x=a} \leq \frac{M}{\Gamma(1 - \beta)} \Big[\int_{a}^{x}(x- t)^{- \beta}dt\Big]_{x=a} = \frac {M}{\Gamma(2 -\beta)}\big[(x - a)^{1 - \beta}\big]_{x=a} = 0, \] and hence (\ref{3}) reduces to $I_a^\alpha(D_a^\beta y(x) ) = I_a^{\alpha - \beta}y(x)$. \end{proof} \begin{proposition} \label{prop1.2} Let $y$ be continuous on $[0, \lambda]$, $ \lambda > 0$ and $n$ be a non negative integer, then \begin{equation} I^\alpha(x^n y(x)) = \sum_{k = 0}^{n}\binom{-\alpha}{k} \left [D^k x^n\right ]\left [I^{\alpha + k} y(x)\right ] = \sum_{k = 0}^{n}\binom{-\alpha}{k}\frac{n ! x^{n - k}} {(n - k) !}I^{\alpha + k} y(x), \label{5} \end{equation} where \begin{equation} \binom{-\alpha}{k} = (-1)^k\frac{\Gamma(\alpha + 1)}{n! \Gamma(\alpha)} = (-1)^k\binom{\alpha + k - 1}{k} = \frac{\Gamma(1 - \alpha)}{\Gamma(k + 1)\Gamma(1 - \alpha - k)}.\label{6} \end{equation} \end{proposition} The proof of the above proposition can be found in \cite[p. 53]{mil}. \begin{corollary} \label{coro1.1} Let $y\in C[0, \lambda]$, $\lambda > 0$ and $p_j(x) = \sum_{k = 0}^{N_j} a_{jk} x^k$, $N_j\in \mathbb{N}\cup \{0\}$, $j = 1, 2, \dots, n$. Then \begin{equation} I^\alpha\Big(\sum_{j = 1}^{n}p_{j}(x) y(x)\Big) = \sum_{j = 1}^{n}\sum_ {k = 0}^{N_j}\sum_{r = 0}^{k}a_{jk}\binom{-\alpha_n }{r}\frac{k! x^{k - r}} {(k - r)!} [I^{\alpha + r}y(x)].\label{7} \end{equation} \end{corollary} \begin{proof} Using $I^\alpha(x^n y(x)) = \sum_{k = 0}^{n}\binom{-\alpha}{k} [D^k x^n][I^{\alpha + k} y(x)]$ and $D^r( x^k) = \frac{k! x^{k-r}}{(k-r)!} $ we have \begin{equation} \begin{aligned} I^\alpha\Big(\sum_{j = 1}^{n}p_{j}(x) y(x)\Big) &= \sum_{j =1}^{n}I^\alpha(p_j(x)y(x) ) \\ &= \sum_{j = 1}^{n}\sum_ {k= 0}^{N_j}a_{jk} I^\alpha (x^k y(x)) \\ &= \sum_{j = 1}^{n}\sum_ {k = 0}^{N_j}a_{jk}\Big[\sum_{r = 0}^{k} \binom{-\alpha_n}{r}(D^r x^k)I^{\alpha +r} y(x)\Big] \\ &= \sum_{j = 1}^{n}\sum_ {k = 0}^{N_j}\sum_{r =0}^{k}a_{jk} \binom{-\alpha_n }{r}\frac{k! x^{k - r}} {(k -r)!} [I^{\alpha + r}y(x)]. \end{aligned}\label{8} \end{equation} \end{proof} \section{Existence of Positive Solutions} In this section we discuss conditions under which the following fractional initial-value probelm has a positive solution. \begin{equation} \Big(D^{\alpha_n} - \sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}\Big)y = f(x, y),\quad y(0) = 0,\quad 0 \leq x \leq \lambda, \lambda > 0, \label{9} \end{equation} where $0 < \alpha_1 < \alpha_2 < \dots < \alpha_n < 1$; $p_j(x) = \sum_{k = 0}^{N_j} a_{jk} x^k, p^{(2m)}_j(x) \geq 0$, $p^{(2m + 1)}_j(x) \leq 0$, $m = 0, 1, \dots, [\frac{N_j}{2}]$, $j = 1, 2, \dots, n - 1$, $D^{\alpha_j}$ is the standard Riemann-Liouville fractional derivative and $f :[0, 1]\times [0, +\infty)\to [0, +\infty)$ a given continuous function. Let us denote by $Y = C[0, \lambda]$, the Banach space of all continuous real functions on $[0, \lambda]$ endowed with the sup norm and $K$ be the cone: \[ K = \{y \in Y : y(x) \geq 0, 0 \leq x \leq \lambda\}. \] \begin{definition} \label{def2.1} \rm By a solution of \eqref{9}, we mean a continuous function $y \in C[0, \lambda]$, that satisfies \eqref{9}. \end{definition} We remark that in \cite{sam} the initial-value problems \begin{gather*} D_0^\alpha y(x) = f(x,y),\quad 0 <\alpha < 1,\\ I_0^{1-\alpha }y(x)|_{x=0}=b, \quad0 < \alpha< 1, \end{gather*} are studied where $D_0^\alpha$ denotes the Riemann-Liouville derivative and the underlying space of functions is $C(0, \lambda )$. However, in the present paper we are dealing with the space of functions $C[0, \lambda ]$. For $y(x) \in C[0, \lambda ]$, always $I_0^{1-\alpha }y(x)|_{x=0}=0$, and is not free data. \begin{lemma} \label{lem2.1} The fractional initial-value problem \eqref{9} is equivalent to the Volterra integral equation \begin{equation} y(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n }{r} \frac{k! x^{k - r} }{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}f(x, y(x)).\label{10} \end{equation} \end{lemma} \begin{proof} Suppose $y(x)$ satisfies \eqref{9}, then \[ I^{\alpha_n}\big[\big(D^{\alpha_n} - \sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}\big)y \big] = I^{\alpha_n}f(x, y). \] Proposition \ref{prop1.1} (ii) yields $I^{1-\alpha_n}y(x)\big|_{x=0}=0$, $I^{1 - \alpha_n -\alpha_{n-j}}y(x)\big|_{x=0} = 0$, hence $I^{\alpha_n } (D^{\alpha_n} y(x)) = y(x)$ and using (\ref{7}) we obtain the integral equation \eqref{10}. Conversely, let $y(x)$ satisfy the integral equation \eqref{10}. Then \begin{align*} &\sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n }{r} \frac{k! x^{k - r} }{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}f(x, y(x)) \\ &=\sum_{j=1}^{n-1}\sum_{k = 0}^{N_j} a_{jk} \Big[ \sum_{r=0}^{k} \binom{-\alpha_n }{r} D^r x^k I^{\alpha_n - \alpha_{n - j} + r}y(x)\Big] + I^{\alpha_n}f(x, y(x)) \\ &= \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j} a_{jk} I^{\alpha_n} (x^k D^{\alpha_{n-j}} y(x)) + I^{\alpha_n}f(x, y(x))\\ & = \sum_{j=1}^{n-1}I^{\alpha_n}(p_j(x) D^{\alpha_{n-j}} y(x)) + I^{\alpha_n}f(x, y(x)) \\ &= I^{\alpha_n}(\sum_{j=1}^{n-1}[p_j(x) D^{\alpha_{n-j}} y(x) )] + f(x, y(x))) = y(x) \end{align*} But $y(x) = I^{\alpha_n}(D^{\alpha_n} y(x))$, hence $y(x)$ satisfies \eqref{9}, and $y(0)=0$. \end{proof} \begin{lemma} \label{lem2.2} If $p_j^{(2m)}(x) \geq 0$ and $p_j^{(2m + 1)}(x)\leq 0$ for $m = 0, 1, \dots, [\frac{N_j}{2}]$ where $N_j = \deg(p_j)$, $j = 1, 2, \dots , n - 1$ and $p_j$'s are as in \eqref{9}, then $F$ defined as \begin{equation} Fy(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}f(x, y(x)),\label{11} \end{equation} is from $K$ to itself. \end{lemma} \begin{proof} The right-hand side of \eqref{11} can be expressed as: \begin{align*} Fy(x) &= \sum_{j=1}^{n-1}\binom{-\alpha_n}{0} \Big(\sum_{k = 0}^{N_j} a_{jk} x^k\Big) I^{\alpha_n - \alpha_{n - j}}y(x)\\ &\quad + \sum_{j=1}^{n-1}\binom{-\alpha_n}{1} \Big(\sum_{k = 1}^{N_j} ka_{jk} x^{k - 1}\Big) I^{\alpha_n - \alpha_ {n - j} + 1}y(x)\\ &\quad + \sum_{j=1}^{n-1}\binom{-\alpha_n}{2} \Big(\sum_{k = 2}^{N_j} k(k - 1)a_{jk} x^{k - 2}\Big)I^{\alpha_n - \alpha_ {n - j} + 2}y(x) + \dots \\ &\quad + \sum_{j=1}^{n-1}\binom{-\alpha_n}{N_j} (N_j! a_{j N_j}) I^{\alpha_n - \alpha_{n - j} + N_j}y(x) \\ &= \sum_{j=1}^{n-1} \sum_{k=0}^{N_j}\binom{-\alpha_n}{k} p_{j}^{(k)}(x) I^{\alpha_n - \alpha_{n - j} + k }y(x) \end{align*} In view of (\ref{6}), we have \[ \binom{-\alpha_n}{2m}> 0,\quad \binom{-\alpha_n}{2m + 1} < 0,\quad m \in \mathbb{N}. \] Then by assumptions on $p^{(k)}_j(x)$, $k = 0, 1, \dots, N_j$, $j = 1, 2, \dots, n - 1$ we get $Fy(x)\in K$. \end{proof} Furthermore, it is easy to show the following result. \begin{lemma} \label{lem2.3} The operator $F : K \to K$ defined in Lemma \ref{lem2.2} is completely continuous. \end{lemma} \begin{lemma} \label{lem2.4} Let $M \subset K$ be bounded; i.e. there exists a positive constant $l$ such that $\|y\| \leq l$, for all $y \in M$. Then $\overline{F(M)}$ is compact. \end{lemma} \begin{proof} Let $L = \max \{1 + f(x, y) : 0 \leq x \leq 1, 0 \leq y \leq l\}$. For $y \in M$, we have \[ |F(y(x))| \leq \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \Big|a_{jk} \binom{-\alpha_n}{r}\Big| \frac{k! x^{\alpha_n - \alpha_{n - j} + k}}{(k - r)! \Gamma (\alpha_n - \alpha_{n - j} + r + 1)} + \frac{L x^{\alpha_n}}{\Gamma (\alpha_n + 1)}. \] Hence \[ \|Fu\| \leq \Big[\sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \Big|a_{jk} \binom{-\alpha_n}{r}\Big| \frac{k!}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r + 1)} + \frac{L}{\Gamma(\alpha_n + 1)}\Big]\zeta, \] where $\zeta = \max \{\lambda^{\alpha_n}, \lambda^{\alpha_n - \alpha_ {n - 1}}, \lambda^{\alpha_n - \alpha_{n - 1} + b}\}$ and $b = \max\{N_1, N_2, \dots, N_{n - 1}\}$. Hence $F(M)$ is bounded. Let $y\in M, x_1, x_2 \in [0, \lambda], x_1 < x_2$ then for given $\epsilon > 0$, choose \begin{equation} \delta = \min\Big\{\Big[\frac{\epsilon C(j, k, r)}{2}\Big]^{1/(\alpha_n - \alpha_{n - j} + r)}\quad \Big[\frac{\epsilon \Gamma(\alpha_n + 1)}{4 \|f\|_ \infty}\Big]^{1/ \alpha_n}\Big\},\label{12} \end{equation} where $j = 1, 2, \dots, n - 1$, $k = 0, 1, \dots, N_j$, $r = 0, 1, \dots, k$, \[ C(j, k, r) = \frac{(k - r)!}{\sum_{i = 1}^{n - 1}(N_i + 1)(N_i + 2)} \times \frac{\Gamma(\alpha_n - \alpha_{n - j} + r + 1)}{\left|a_{jk} \binom{-\alpha_n}{r}\right| l \eta k!} \] and $\eta = \max\{1, \lambda^{N_j}, j = 1, 2, \dots, n - 1\}$. If $\vert x_1 - x_2 \vert < \delta$, \begin{align*} &| Fy(x_1) - Fy(x_2)| \\ &\leq \Big| \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk}\binom{-\alpha_n}{r}| k! x_1^{k - r}}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r)}\\ &\quad \times\Big[\int_{0}^{x_1}\Big(\frac{y(t)}{(x_1 - t)^{ \alpha_{n - j} - \alpha_n - r + 1}} - \frac{y(t)}{(x_2 - t)^{\alpha_{n - j} - \alpha_n - r + 1}}\Big)dt \\ &\quad - \int_{x_1}^{x_2}\frac{dt}{(x_2 - t)^{\alpha_{n - j} -\alpha_n - r + 1}}\Big] \\ &\quad + \frac{1}{\Gamma(\alpha_n)}\int_{0}^{x_1}\big((x_1 - t)^{\alpha_n - 1} - (x_2 - t)^{\alpha_n - 1}\big) f(t, y(t))dt \\ &\quad - \frac{1}{\Gamma(\alpha_n)}\int_{x_1}^{x_2}(x_2 - t)^{\alpha_n - 1} f(t, y(t))dt \Big| \\ &\leq \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk}\binom{-\alpha_n}{r}| l k! \eta}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r)}(x_2 - x_1)^ {\alpha_n - \alpha_{n - j} + r} + \frac{2L (x_2 - x_1)^{\alpha_n}}{\Gamma (\alpha_n + 1)} \\ &= \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk} \binom{-\alpha_n}{r}| l k! \eta}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r)}\delta^ {\alpha_n - \alpha_{n - j} + r} + \frac{2L \delta^{\alpha_n}}{\Gamma (\alpha_n + 1)}\\ &\leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Hence $F(M)$ is equicontinuous and Arzela-Ascoli theorem implies that $\overline{F(M)}$ is compact. \end{proof} \begin{theorem} \label{thm2.1} Consider the fractional differential equation \begin{equation} \Big(D^{\alpha_n} - \sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}\Big)y = g(y),\quad y(0) = 0,\quad 0 \leq x \leq \lambda, \; \lambda > 0,\label{13} \end{equation} where $0 < \alpha_1 < \alpha_2 < \dots < \alpha_n < 1$, $p_j(x) = \sum_{k=0}^{N_j}a_{j k}x^{k}$, $N_j\in\mathbb{N}\cup \{0\}$, $j = 1, 2, \dots, n - 1$, $g :[0, 1]\times [0, +\infty) \to [0, +\infty)$ satisfies the Lipschitz condition with constant $L$ and $g(0) <\infty$. If $p^{(2m)}_j(x) \geq 0$, $p^{(2m + 1)}_j(x) \leq 0$, $m = 0, 1, \dots, [\frac{N_j}{2}]$, then \eqref{13} has a positive solution. \end{theorem} \begin{proof} In view Lemma \ref{lem2.1}, (\ref{13}) is equivalent to the integral equation \[ y(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}g(y). \] In view of Lemma \ref{lem2.3}, $y(x) \in K$. Let $T : K \to K$ be defined as \begin{equation} Ay(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r} }{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}g(y). \nonumber \end{equation} $A$ is completely continuous by Lemma \ref{lem2.2}. \noindent Case (i) $g(0) \neq 0$. Let \[ B(r) = \Big\{y(x) \in C[0, \delta]: y(x) \geq 0\,\; \|y - \frac{g(0) x^{\alpha_n}}{\Gamma(1 + \alpha_n)}\| \leq r \Big\}, \] be a convex bounded and closed subset of the Banach space $C[0, \delta]$ where \[ \delta < \min \Big\{\lambda, \; \Big(\frac{r\Gamma(\alpha_n + 1)} {2 E g(0)}\Big)^{1/\alpha_n},\; \big(\frac{1}{2E}\big)^{1/\alpha_n} \Big\}, \] where \[ E = \xi \Big(\sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac {|a_{jk} \binom{-\alpha_n}{r}| k!}{(k - r)!\Gamma(\alpha_n - \alpha_{n - j} + r + 1)}\Big) + \frac{L}{\Gamma (\alpha_n + 1)}, \] and \[ \xi = \max\left\{x^{\alpha_n}, x^{\alpha_n - \alpha_{n - 1} + \rho}: 0 \leq x \leq \delta\right\},\quad \rho = \max\{N_1, \dots, N_{n - 1}\}. \] Note that, for all $y \in B(r)$, \begin{align*} &\big|Ay(x) - \frac{g(0) x^{\alpha_n}}{\Gamma(1 + \alpha_n)} \big|\\ &\leq \|u\|\Big[\sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac {|a_{jk} \binom{-\alpha_n}{r}| k! x^{\alpha_n - \alpha_{n - j} + k}}{(k - r)!\Gamma(\alpha_n - \alpha_{n - j} + r + 1)} + \frac{L x^{\alpha_n}}{\Gamma (\alpha_n + 1)}\Big] \\ &\leq \|u\|\Big[\sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac {|a_{jk} \binom{-\alpha_n}{r}| k!}{(k - r)!\Gamma(\alpha_n - \alpha_{n - j)} + r + 1} + \frac{L} {\Gamma (\alpha_n + 1)}\Big]\xi. \end{align*} Since \[ \| u \| \leq \frac{g(0)}{\Gamma(1 + \alpha_n)}x^{\alpha_n} + r \leq \frac{g(0)}{\Gamma(1 + \alpha_n)}\delta^{\alpha_n} + r, \] we have \[ \Big|Ay(x) - \frac{g(0)}{\Gamma(1 + \alpha_n)} x^{\alpha_n} \Big| \leq E \Big(\frac{g(0)}{\Gamma(\alpha_n + 1)} \delta^{\alpha_n + r} \Big) \leq \frac{r}{2} + \frac{r}{2} = r. \] So we have $A (B(r)) \subseteq B(r)$. It can be seen that $A {(B(r))}$ is equicontinuous (the proof is similar to the proof of Lemma \ref{lem2.4}). Let $\{y_n\}$ be a bounded sequence in $B(r)$. Then $\{A(y_n)\} \subset T(B(r))$. Hence $\{A(y_n)\}$ is equicontinuous. Since $y_n \in C[a, b]$, Arzela-Ascoli theorem \cite{ali,gol} implies that $\{A(y_n)\}$ has a convergent subsequence. Therefore $A : B(r) \to B(r)$ is compact. Hence by Schauder fixed point theorem \cite{jos} it has a fixed point, which is a positive solution of(\ref{13}). \end{proof} A similar proof can be given for the case $g(0) = 0$. \begin{example} \label{exa2.1}\rm Consider the equation \[ D^{\alpha_3}y(x) - (x^2 - 3x +2)D^{\alpha_2}y(x) - (1-x)D^{\alpha_1}y(x) =\frac{1+y}{1+y^2}, \] $y(0) = 0$, $0 \leq x \leq 1, 0 < \alpha_1 < \alpha_2 < \alpha_3 < 1$. Note that $p_1(x) = x^2 - 3x +2$, $p_2(x) = 1 - x$ and $g(y) = \frac{1+y}{1+y^2}$ satisfy the conditions required in Theorem \ref{thm2.1}, hence this equation has a positive solution. \end{example} \begin{theorem} \label{thm2.2} Let $f :[0, \lambda] \times [0, \infty) \to [0, \infty)$ be continuous and $f(x,.)$ be increasing for each $x\in [0,\lambda]$. Assume there exist $v_0, w_0$ satisfying $\mathcal{L}(D)v_0 \leq f(x, v_0)$, $\mathcal{L}(D)w_0 \geq f(x, w_0)$ and $0 \leq v_0(x) \leq w_0(x)$, $0 \leq x \leq 1$, where $\mathcal{L}(D) = D^{\alpha_n} - \sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}$. Then \eqref{9} has a positive solution. \end{theorem} \begin{proof} We need to consider the fixed point of the operator $F$. Let $y_1, y_2 \in K$, $y_1 \leq y_2$, then \[ Fy_1 = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y_1 + I^{\alpha_n}f(x, y_1(x)) \leq Fy_2, \] as $f$ is nondecreasing. Hence $F$ is an increasing operator. Assuming $Fv_0 \geq v_0, Fw_0 \leq w_0$, implies that $F :\langle v_0, w_0\rangle\to \langle v_0, w_0 \rangle$ is compact operator in view of Lemma \ref{lem2.3} and completely continuous in view of Lemma \ref{lem2.2}. Since $K$ is a normal cone and $F$ is compact continuous, by Theorem \ref{thm1.1} $F$ has a fixed point $u^* \in \langle v_0, w_0\rangle$, which is the required positive solution. \end{proof} \begin{example} \label{exa2.2} \rm Consider the equation \[ D^{1/2}y(x) - (\Gamma(\frac{3}{2}))^{-1}\Gamma(\frac{7}{4})xD^{1/4}y(x) =(\Gamma(\frac{3}{2}))^{-1}f(x,y), \] where $0 \leq x \leq 1$, $0 \leq y \leq +\infty$, $f(x,y)=\mu(x^{1/2}- x^{\frac{7}{4}})e^{2y-x}$ and $0 < \mu\leq 1$ which is equivalent to equation \[ \Gamma(\frac{3}{2})D^{1/2}y(x) - \Gamma(\frac{7}{4})xD^{1/4}y(x)=f(x,y). \] If we let $v_0 = 0$, $w_0 = \frac{1}{2}x$, then $0 \leq v_0 \leq w_0$, $\mathcal{L}(D)v_0=0$, $\mathcal{L}(D)w_0=x^{1/2} -x^{\frac{7}{4}}$, $\mathcal{L}(D)v_0 \leq f(x, 0)$ and $\mathcal{L}(D)w_0 \geq f(x, \frac{1}{2}x)$. Then this equation has a positive solution . \end{example} \begin{theorem} \label{thm2.3} Let $f :[0, \lambda] \times [0, \infty) \to [0, \infty)$ be continuous and $f(x, .)$ increasing for each $x \in [0, \lambda]$. If $0 < \lim_{y \to +\infty}f(x, y) < +\infty$ for each $x \in [0, \lambda]$ then \eqref{9} has a positive solution. \end{theorem} \begin{proof} There exist positive constants $N, R$ such that $f(x, y) \leq N$, for all $x \in [0, \lambda]$, and all $y \geq R$. Let $C = \max \{f(x, y) | 0 \leq y \leq \lambda, 0 \leq y \leq R\}$. Then we have $f \leq N + C$, for all $y \geq 0$. Now we consider the equation, \[ \Big(D^{\alpha_n} - \sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}\Big)w(y) = N + C ,\quad w(0) = 0 ,\quad 0 < x < \lambda. \] Using Lemma \ref{lem2.1}, the above equation is equivalent to the integral equation \[ w(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}(N + C). \] This integral equation has a positive solution $w(x)$ in view of Theorem \ref{thm2.2}. Also \[ w(x) \geq \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r}\frac{k! x^{k - r}} {(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}f(x, w(x)) = Fw(x). \] Now for $v(x) \equiv 0, F(v(x)) = I^{\alpha_n}f(x, v(x)) \geq v(x)$. Hence in view of Theorem \ref{thm2.2}, the result follows. \end{proof} It is easy to prove the following existence theorem using Theorems \ref{thm2.2} and \ref{thm2.3}. \begin{theorem} \label{thm2.4} Let $f :[0, \lambda] \times [0, \infty) \to [0, \infty)$ be continuous and $f(x, .)$ increasing for each $x\in [0, \lambda]$. If \[ 0 \leq \lim_{ y \to \infty} \max_{ 0 \leq x \leq \lambda} \frac{f(x, y)}{y} < +\infty. \] Then\eqref{9} has a positive solution. \end{theorem} \begin{example} \label{exa2.3} (1) $f(x, y) = x(1 +e^{-y})^{-1}$, satisfies the condition required in Theorem \ref{thm2.3}. \noindent (2) $f(x, y) = x \ln(1 + y)$ satisfies the conditions required in Theorem \ref{thm2.4}. \end{example} \section{Uniqueness and existence of solutions} In this section we give conditions on $f$ and $p_j$'s, which render unique positive solution to \eqref{9}. \begin{theorem} \label{thm3.1} Let $f :[0, \lambda] \times [0, \infty) \to [0, \infty)$ be continuous and Lipschitz with respect to the second variable with constant $L$. If (i) $p^{(2m)}_j(x) \geq 0$ and $p^{(2m + 1)}_j(x) \leq 0$, $m = 0, 1, \dots, [\frac{N_j}{2}]$, $N_j = \deg (p_j)$ $j = 1, 2, \dots, n - 1$; and (ii) \[0 < \frac{L \lambda^{\alpha_n}}{\Gamma(\alpha_n + 1)} + \sum_{j = 1}^{n - 1} \sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk}\binom{-\alpha_n}{r}|k! \lambda^{\alpha_n - \alpha_{n - j} + k}}{(k - r)! \Gamma(\alpha_n - \alpha_ {n - j} + r + 1)} < 1, \] then \eqref{9} has unique solution which is positive. \end{theorem} \begin{proof} As pointed out in the preceding section, \eqref{9} is equivalent to \eqref{10}. For $y_1, y_2 \in K$ we have \begin{align*} &\big|F(y_1(x)) - F(y_2(x))\big|\\ & \leq \sum_{j = 1}^{n - 1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk} \binom{-\alpha_n}{r}k! x^{k -r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}|y_1(x) -y_2(x)| + L I^{\alpha_n}|y_1(x) -y_2(x)| \\ &\leq \| y_1(x) - y_2(x)\| \Big[\frac{L x^{\alpha_n}}{\Gamma(\alpha_n + 1)} + \sum_{j = 1}^{n - 1} \sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk}\binom{-\alpha_n}{r}| k! x^{\alpha_n - \alpha_{n - j} + k}} {(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r + 1)}\Big], \end{align*} where $F$ is given in (\ref{11}). Hence \begin{align*} & \|Fy_1 - Fy_2 \|\\ &\leq \Big[\frac{L \lambda^{\alpha_n}}{\Gamma(\alpha_n + 1)} + \sum_{j = 1}^{n - 1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk} \binom{-\alpha_n}{r}|k! \lambda^{\alpha_n - \alpha_{n - j} + k}}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r +1)}\Big] \| y_1(x) - y_2(x)\|. \end{align*} In view of Theorem \ref{thm1.2}, $F$ has unique fixed point in $K$, which is the unique positive solution of \eqref{9}. \end{proof} In the following, we omit the condition on $p_j(x)'$s and study the equation \begin{equation} \Big(D^\alpha_n - \sum_{j = 1}^{n - 1} p_j(x)D^{\alpha_{n - j}}\Big)y = f(x, y),\quad y(0) = 0,\quad 0 \leq x \leq \lambda,\; \lambda > 0, \label{15} \end{equation} where $0 < \alpha_1 < \alpha_2 < \dots < \alpha_n < 1$, $p_j(x) = \sum_{k=0}^{N_j}a_{j k}x^{k}, N_j\in\mathbb{N}\cup \{0\}$, $j = 1, 2, \dots, n - 1$. Using Banach fixed point theorem for $F:C[0, \lambda]\to C[0, \lambda]$ we obtain the following result. \begin{example} \label{exa3.1} \rm Consider the equation \begin{equation} \Big(D^{1/2} - \frac{1}{60}(A - x)(B - x)D^{1/4} - \frac{1}{40}(C - x) D^{1/6} - MD^{1/8}\Big)y = L y + e^x,\label{16} \end{equation} where $y(0) = 0$, $0 \leq x \leq 1$, $A \geq 1$ and $B \geq 1$. (\ref{16}) is equivalent to the integral equation \[ y(x) = \sum_{j=1}^{3}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-1/2}{r} \frac{k!}{(k - r)!} x^{k - r} I^{\frac{1}{2} - \alpha_{n - j} + r}y(x) + I^{1/2}(Ly + e^x). \] Here $p_1(x) = \sum_{k = 0}^{2}a_{1k}x^k = \frac{1}{60}[x^2 - (A + B)x + AB]$, hence $N_1 = 2$; $a_{10} = \frac{1}{60}AB$, $a_{11}= \frac{-1}{60} (A + B)$, $ a_{12} =\frac{1}{60}$, $p_2(x) = \sum_{k = 0}^ {1}a_{2k}x^k = \frac{1}{40}(C - x)$, so $N_2 = 1$; $a_{20} = \frac{1}{40}C$, $a_{21} = \frac{-1}{40}$, $p_3(x) = \sum_{k = 0}^{0}a_{3k}x^k = M$, so $N_3 = 0$, $a_{30} = M$. Hence \begin{align*} y(x) &= a_{10}\binom{-1/2}{0}I^{\frac{1}{2} - \frac{1}{4}}y + a_{11}\Big[\binom{-1/2}{0}xI^{\frac{1}{2} - \frac{1}{4}}y + \binom{-1/2}{1}I^{\frac{1}{2} - \frac{1}{4} + 1}y\Big] \\ &\quad + a_{12}\Big[\binom{-1/2}{0}x^2I^{\frac{1}{2} - \frac{1}{4}}y + 2\binom{-1/2}{1}xI^{\frac{1}{2} - \frac{1}{4} + 1}y + 2\binom{-1/2}{2}I^{\frac{1}{2} - \frac{1}{4} + 2}y \Big] \\ &\quad+ a_{20}\binom{2\frac{-1}{2}}{0}I^{\frac{1}{2} - \frac{1}{6}}y + a_{21}\Big[\binom{-1/2}{0}xI^{\frac{1}{2} - \frac{1}{6}}y + \binom{-1/2}{1}I^{\frac{1}{2} - \frac{1}{6} + 1}y\Big] \\ &\quad + a_{30}\binom{-1/2}{0}I^{\frac{1}{2} - \frac{1}{8}}y + LI^{\frac{1} {2}}y + I^{1/2}e^x \\ &= \frac{AB}{60}I^{\frac{1}{2} - \frac{1}{4}}y - \frac{A + B}{60}\Big[xI^ {\frac{1}{2} - \frac{1}{4}}y -\frac{1}{2} I^{\frac{1}{2} - \frac{1}{4} + 1} y\Big] \\ &\quad + \frac{1}{60}\Big[x^2I^{\frac{1}{2} - \frac{1}{4}}y - xI^{\frac{1}{2} - \frac{1}{4} + 1}y + \frac{3}{4}I^{\frac{1}{2} - \frac{1}{4} + 2}y + \Big] \\ &\quad + \frac{C}{40}I^{\frac{1}{2} - \frac{1}{6}}y - \frac{1} {40}\Big[xI^{\frac{1}{2} - \frac{1}{6}}y - \frac{1}{2}I^{\frac{1}{2} - \frac{1}{6} + 1}y\Big] + MI^{\frac{1}{2} - \frac{1}{8}}y + LI^{1/2}y + I^{1/2}e^x. \end{align*} If $1 \leq A \leq 3$, $1 \leq B \leq 3$, $0 < M \leq \frac{1}{40}$ and $0 < L \leq \frac{1}{4}$ in the above equation satisfy the conditions required in Theorem \ref{thm3.1}. The iterated sequence is \[ y_1(x) = I^{1/2}e^x = x^{1/2}E_{1, \frac{3}{4}}(x) \] \begin{align*} y_2(x) & =\Big[\frac{AB}{60}I^{1/4} - \frac{A + B}{60}\big(xI^{\frac{1} {4}} -\frac{1}{2} I^{5/4}\big) + \frac{1}{60}\big(x^2I^{1/4} - xI^{\frac{3}{4}} + \frac{3}{4} I^{9/4} \big)\\ &\quad + \frac{C}{40}I^{1/3} - \frac{1}{40} \big(xI^{1/3}- \frac{1}{2}I^{4/3}\big) + MI^{3/8} + LI^{1/2}\Big]y_1 + y_1, \end{align*} and \begin{align*} y_{n + 1}(x) &= \sum_{k = 0}^{n} \Big[\frac{AB}{60}I^{1/4} - \frac{A + B} {60}\big(xI^{1/4} - \frac{1}{2} I^{5/4}\big) + \frac{1}{60}\big(x^2I^{1/4} - xI^{5/4} + \frac{3}{4} I^{9/4} \big) \\ &\quad + \frac{C}{40}I^{1/3} - \frac{1}{40}\big(xI^ {\frac{1}{3}} - \frac{1}{2}I^{4/3}\big) + MI^{3/8} + LI^{1/2}\Big]^{n - k}y_1, \end{align*} $n = 1, 2, 3, \dots$, where $I^\alpha y_1 = x^{\alpha + \frac{1}{2}}E_{1, \alpha + \frac{3}{4}}(x)$, $ \alpha > 0$. $y(x) = \lim_{n \to \infty} y_n(x)$ is the unique positive solution. \end{example} \begin{theorem} \label{thm3.2} Let $f :[0, \lambda] \times [0, \infty) \to [0, \infty)$ be continuous and Lipschitz with respect to the second variable with constant $L$. Let $a_{jk}$'s satisfy \[ 0 < \frac{L \lambda^{\alpha_n}}{\Gamma(\alpha_n + 1)} + \sum_{j = 1}^{n - 1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk} \binom{-\alpha_n}{r}|k! \lambda^{\alpha_n - \alpha_{n - j} + k}}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r + 1)} < 1. \] Then \eqref{15} has unique solution, which may not necessarily be positive. \end{theorem} \begin{proof} Using Lemma \ref{lem2.1}, \eqref{15} is equivalent to the integral equation \[ y(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}f(x, y(x)). \] We define an operator $F : C[0, \lambda] \to C[0, \lambda]$ as \[ Fy(x) = \sum_{j=1}^{n-1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\alpha_n}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\alpha_n - \alpha_{n - j} + r}y(x) + I^{\alpha_n}f(x, y(x)), \] For $y_1, y_2 \in C[0, \lambda]$, \begin{align*} &\|Fy_1 - Fy_2\| \\ &\leq \Big[\frac{L \lambda^{\alpha_n}}{\Gamma(\alpha_n + 1)} + \sum_{j = 1}^{n - 1}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} \frac{|a_{jk} \binom{-\alpha_n}{r}|k! \lambda^{\alpha_n - \alpha_{n - j} + k}}{(k - r)! \Gamma(\alpha_n - \alpha_{n - j} + r +1)}\Big] \| y_1(x) - y_2(x)\|. \end{align*} Hence in view of Theorem \ref{thm1.2}, $F$ will have unique fixed point in $C[0,\lambda]$, which is the unique solution of (\ref{15}). This solution is not necessarily positive one. \end{proof} \begin{example} \label{exa3.2} \rm Consider the euation \begin{equation} \big(D^{1/2} - ax^2 D^{1/4} - bx D^{1/6} - cD^{\frac{1}{8}}\big)y = L y + e^x,\quad y(0) = 0, \; 0 \leq x \leq 1. \label{17} \end{equation} This equation is equivalent to the integral equation \[ y(x) = \sum_{j=1}^{3}\sum_{k = 0}^{N_j}\sum_{r = 0}^{k} a_{jk} \binom{-\frac{1}{2}}{r} \frac{k! x^{k - r}}{(k - r)!} I^{\frac{1}{2} - \alpha_{n - j} + r}y(x) + I^{1/2}(Ly + e^x). \] Here $p_1(x) = \sum_{k = 0}^{2}a_{1k}x^k = ax^2$, then $N_1 = 2$, $a_{10} = a_{11} = 0$, $a_{12} = a$, $p_2(x) = \sum_{k = 0}^{1}a_{2k}x^k = bx$, then $N_2 = 1$, $a_{20} = a_{21} = b$, and $p_3(x) = \sum_{k = 0}^{0}a_{3k}x^k = c$, then $N_3 = 0$, $a_{30} = c$. Hence \begin{align*} y(x) &= a_{10}\binom{-1/2}{0} I^{\frac{1}{2} - \frac{1}{4}}y + a_{11}\Big[\binom{-1/2}{0}xI^{\frac{1}{2} - \frac{1}{4}}y + \binom{-1/2}{1}I^{\frac{1}{2} - \frac{1}{4} + 1}y\Big] \\ &\quad+ a_{12}\Big[\binom{-1/2}{0}x^2I^{\frac{1}{2} - \frac{1} {4}}y + 2\binom{-1/2}{1}xI^{\frac{1}{2} - \frac{1}{4} + 1}y + 2\binom{-1/2}{2}I^{\frac{1}{2} - \frac{1}{4} + 2}y + \Big] \\ &\quad+ a_{20}\binom{2\frac{-1}{2}}{0}I^{\frac{1}{2} - \frac{1}{6}}y + a_{21}\Big[\binom{-1/2}{0}xI^{\frac{1}{2} - \frac{1}{6}}y + \binom{-1/2}{1}I^{\frac{1}{2} - \frac{1}{6} + 1}y\Big] \\ &\quad + a_{30}\binom{-1/2}{0}I^{\frac{1}{2} - \frac{1}{8}}y + LI^{\frac{1} {2}}y + I^{1/2}e^x. \end{align*} In view of (\ref{6}) and that $\Gamma(\frac{1}{2}) = \sqrt{\pi}$, $\Gamma(\frac{-1}{2}) = -2\sqrt{\pi}$ and $\Gamma(\frac{-3}{2}) = \frac {4\sqrt{\pi}}{3}$ we obtain \begin{align*} y(x) &= a\Big[x^2I^{1/4}y(x) - I^{5/4}y(x) + \frac{3}{4}I^ {9/4}y(x)\Big] + b\Big[xI^{1/3}y(x) - \frac{1}{2}I^{4/3}y(x) \Big]\\ &\quad + c I^{3/8}y(x) + LI^{1/2}y(x) + I^{1/2}e^x. \end{align*} If $|a| \leq \frac{3}{5}$, $|b| \leq \frac{2}{5}$, $|c| \leq \frac{1}{5}$, $0 < L \leq \frac{4}{5}$ in the above equation satisfy the conditions required in Theorem \ref{thm3.2}. The iterated sequence is \begin{gather*} y_1(x) = I^{1/2}e^x = x^{1/2}E_{1, \frac{3}{4}}(x), \\ y_2(x) = \Big[a(x^2I^{1/4} - I^{5/4} + \frac{3}{4}I^ {\frac{9}{4}}) + b(xI^{1/3} - \frac{1}{2}I^{4/3} ) + cI^{3/8} + LI^{1/2}\Big]y_1 + y_1, \\ y_{n + 1} = \sum_{k = 0}^{n}\Big[a\big(x^2I^{1/4} - I^{5/4} + \frac{3}{4}I^{9/4}\big) + b\big(xI^{1/3} - \frac{1}{2} I^{4/3} \big) + cI^{3/8} + LI^{1/2}\Big]^{n - k} y_1, \end{gather*} for $n = 1, 2, 3, \dots$, where $I^\alpha y_1 = x^{\alpha + \frac{1}{2}}E_{1, \alpha + \frac{3}{4}}(x) , \alpha > 0$. $y(x) = \lim_{n \to \infty} y_n(x)$ is the unique solution, which may not be positive. \end{example} \medskip V. Daftardar-Gejji acknowledges University Grants Commission, N. Delhi, India for the support through Major Research Project. \begin{thebibliography}{99} \bibitem{ali} C. D. Aliprantis, O. Burkinshaw; \emph{Principles of Real Analysis}, (Second Edition), Academic Press, New York, 1990. \bibitem{bab} A. Babakhani, V. 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