Existence and Global Asymptotic Behavior of Positive Solutions for Superlinear Singular Fractional Boundary Value Problems

Abstract

In this paper, we provide sufficient conditions for the existence, uniqueness and global behavior of a positive continuous solution to some nonlinear Riemann-Liouville fractional boundary value problems. The nonlinearity is allowed to be singular at the boundary. The proofs are based on perturbation techniques after reducing the considered problem to the equivalent Fredholm integral equation of the second kind. Some examples are given to illustrate our main results.

1. Introduction

Nonlinear fractional differential equations have been of great interest for the past three decades. This is due to the intensive development of the theory of fractional calculus and its applications. Apart from diverse areas of pure mathematics, fractional differential equations can be used in the modeling of various fields of science and engineering, including wave and fluid dynamics, mathematical biology, financial systems, structural dynamics, artificial intelligence, etc. (see, for instance [1,2,3,4,5,6] and references therein). A significant characteristic of a fractional order differential operator which distinguishes it from the integer-order differential operator is that it is nonlocal in nature, that is, the future state of a dynamical system or process involving fractional derivative depends on its current state as well as its past states. Therefore, fractional models have become relevant when dealing with phenomena with memory effects instead of relying on ordinary or partial differential equations. The study of fractional differential equations has received a great amount of attention from researchers.

In [7], the authors considered the fractional differential problem:

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)+a\left(\xi \right)g\left(\theta \right)=0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=0,\hfill \end{array}\right.$$

(1)

where a $\in L^{\infty }$ $[0,1]$ and $D^{\alpha }$ is the Riemann-Liouville differential operator of order $\alpha \in (1,2],$ defined by (see, for example [2,4,5]),

$$D^{\alpha }\theta \left(\xi \right)=\left\{\begin{array}{cc}{\left(\frac{d}{d\xi }\right)}^2{I}^{2-\alpha }\theta \left(\xi \right),\hfill & \mathrm{if}1<\alpha <2,\hfill \\ {\theta }^{\prime \prime }\left(\xi \right),\hfill & \mathrm{if}\alpha =2,\hfill \end{array}\right.$$

where for $\beta >0$,

$$I^{\beta }\theta \left(\xi \right)=\frac{1}{\mathrm{\Gamma }\left(\beta \right)}{\int }_0^{\xi }{(\xi -y)}^{\beta -1}\theta \left(y\right)dy.$$

They have proved that, when g satisfies certain growth conditions, problem (1) has at least three positive solutions. The proof is based on the well-known fixed point theorem.

By some fixed-point techniques, the authors [8] established some existence results of positive solutions for:

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)+g(\xi ,\theta \left(\xi \right))=0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=0,\hfill \end{array}\right.$$

(2)

where $1<\alpha \le 2$ and $g:[0,1]\times [0,\infty )\to [0,\infty )$ is continuous satisfying some adequate conditions.

In [9], Cui et al obtained a new uniqueness result for the problem:

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)+g(\xi ,\theta \left(\xi \right))=0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=\theta \left(1\right)=0,\hfill \end{array}\right.$$

(3)

where $1<\alpha \le 2$ and $g\in C\left([0,1]\times [0,\infty ),[0,\infty )\right)$ is satisfying some convenient Lipschitz conditions. They have used some fixed-point theorems.

In [10], the authors consider the second-order noncanonical advanced differential equation

$${\left(A{u}^{\prime }\right)}^{\prime }\left(\xi \right)+\chi \left(\xi \right)u\left(\sigma \left(\xi \right)\right)=0,\xi \ge {\xi }_0,$$

(4)

where $A,\chi \in C([{\xi }_0,\infty ),(0,\infty )),$ with ${\displaystyle \varrho \left({\xi }_0\right):={\int }_{{\xi }_0}^{\infty }\frac{1}{A\left(\xi \right)}d\xi <\infty }$ (i.e., noncanonical form) and $\sigma \in C^1([{\xi }_0,\infty ),\mathbb{R}),$${\sigma }^{\prime }\left(\xi \right)>0,$ $\sigma \left(\xi \right)\ge \xi$ for all $\xi \ge {\xi }_0.$

By transforming the noncanonical differential Equation (4) to an equivalent one in canonical form and applying some known results to the obtained equation in canonical form, the authors established oscillation for the Equation (4) in noncanonical form.

In [11], by using a perturbation argument, some existence results are established for:

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)=\theta \left(\xi \right)g(\xi ,\theta \left(\xi \right)),\hfill & 0<\xi <1,\hfill \\ {\displaystyle \underset{\xi \to 0^{+}}{lim}}{D}^{\alpha -1}\theta \left(\xi \right)=-a,\theta \left(1\right)=b,\hfill \end{array}\right.$$

(5)

where $1<\alpha \le 2, a\ge 0$ and $b\ge 0$ with $a+b>0$ and g satisfying:

(P1)

$g\in C\left((0,1)\times [0,\infty ),[0,\infty )\right)$.

(P2)

There exists a function $q_0\in C\left((0,1)\right)$ satisfying some integrable conditions such that for each $\xi \in \left(0,1\right),$ $\zeta \to q_0\left(\xi \right)\zeta -\zeta g\left(\xi ,\zeta \overline{\omega }\left(\xi \right)\right)$ is nondecreasing on $\left[0,1\right],$

where $\overline{\omega }\left(\xi \right):=\frac{a}{\mathrm{\Gamma }\left(\alpha \right)}{\xi }^{\alpha -2}(1-\xi )+b{\xi }^{\alpha -2},$ for $\xi \in (0,1].$

(P3)

For each $\xi \in \left(0,1\right),$ $\zeta \to \zeta g\left(\xi ,\zeta \right)$ is nondecreasing on $[0,\infty ).$

For further related results, we refer the reader to [12,13,14,15,16,17,18,19,20,21] and the references therein.

Motivated by the approach used in [11], we consider the problem:

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)=\phi (\xi ,\theta \left(\xi \right)),\hfill & 0<\xi <1,\hfill \\ \theta \left(\xi \right)>0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=a>0,\hfill \end{array}\right.$$

(6)

where $1<\alpha \le 2$ and $\phi \in C\left((0,1)\times [0,\infty ),[0,\infty )\right),$ which could be singular at both $\xi =0$ and $\xi =1,$ and satisfying some adequate conditions regarding to the class ${\mathcal{F}}_{\alpha }$ (see, (7) below). Using a perturbation approach, we establish that problem (6) has a unique positive continuous solution and we describe its global behavior.

Notations:

(i)

$$\mathbb{E}:=C\left(\left[0,1\right]\right)\mathrm{and}\mathbb{F}:=C\left((0,1)\right)\cap L^1\left((0,1)\right).$$

(ii)

${\mathfrak{B}}^{+}:=\left\{\mathfrak{q}:\left(0,1\right)\to \left[0,\infty \right),\mathfrak{q}\mathrm{is}\mathrm{a}\mathrm{measurable}\mathrm{function}\right\}$.

(iii)

$C^{+}\left((0,1)\right):=\{\mathfrak{q}\in C\left((0,1)\right)$ with $\mathfrak{q}\ge 0\}.$

(iv)

For $\alpha \in \left(1,2\right],$

$${\mathcal{F}}_{\alpha }:=\left\{\mathfrak{q}\in {\mathfrak{B}}^{+}:{\int }_0^1{\omega }^{\alpha -1}{(1-\omega )}^{\alpha -2}\mathfrak{q}\left(\omega \right)d\omega <\infty \right\}.$$

(7)

(v)

For $\mathfrak{q}\in {\mathfrak{B}}^{+}$,

$${\sigma }_{\mathfrak{q}}:=\underset{\xi ,\zeta \in (0,1)}{sup}{\int }_0^1\frac{G(\xi ,\omega )G(\omega ,\zeta )}{G(\xi ,\zeta )}\mathfrak{q}\left(\omega \right)d\omega ,$$

(8)

where for $\alpha \in (1,2],$ $G(\xi ,\zeta )$ is the Green’s function of the operator $\theta \to -D^{\alpha }\theta ,$ with $\theta \left(0\right)={\theta }^{\prime }\left(1\right)=0.$

Note that (see, Proposition 4) if $\mathfrak{q}\in {\mathcal{F}}_{\alpha },$ then ${\sigma }_{\mathfrak{q}}<\infty$.

(vi)

For $\alpha \in \left(1,2\right]$ and $a>0,$ we let

$$\psi \left(\xi \right):=\frac{a}{\alpha -1}{\xi }^{\alpha -1},\xi \in [0,1],$$

(9)

be the unique solution of

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)=0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=a>0.\hfill \end{array}\right.$$

(10)

Observe that for all $\xi \in [0,1],$ $\psi \left(\xi \right)\in [0,\frac{a}{\alpha -1}].$

To study problem (6), for a given $1<\alpha \le 2$ and $a>0,$ we assume that $\phi$ satisfies:

(C1)

$\phi \in C\left(\right(0,1)\times [0,\infty ),[0,\infty \left)\right)$ with $\phi \left(\xi ,0\right)=0$ for all $\xi \in \left(0,1\right)$.

(C2)

There exists $\overline{\mathfrak{q}}\in {\mathcal{F}}_{\alpha }$ with ${\sigma }_{\overline{\mathfrak{q}}}\le \frac{1}{2}$ such that, for each $\xi \in \left(0,1\right),$ $\zeta \to \overline{\mathfrak{q}}\left(\xi \right)\zeta -\phi \left(\xi ,\zeta \right)$ is nondecreasing on $\left[0,\frac{a}{\alpha -1}\right]$.

(C3)

For each $\xi \in \left(0,1\right),$ $\phi \left(\xi ,{\zeta }_1\right)\le \phi \left(\xi ,{\zeta }_2\right)$, whenever $0\le {\zeta }_1\le {\zeta }_2.$

Remark 1.

Conditions (C1)–(C3) are verified in the particular case $\phi \left(\xi ,\zeta \right)=\lambda \mathfrak{q}\left(\xi \right)p\left(\zeta \right),$ where $\mathfrak{q}\in C\left((0,1)\right)\cap {\mathcal{F}}_{\alpha },$ λ is a small positive parameter and $p\in C\left(\left[0,\infty \right),\left[0,\infty \right)\right)$ is a nondecreasing function with $p\left(0\right)=0$ satisfying: there exists a constant $\eta =\eta (\alpha ,a)>0$ such that

$$p\left(\zeta \right)-p\left(r\right)\le \eta (\zeta -r),\mathrm{for}0\le r\le \zeta \le \frac{a}{\alpha -1}.$$

(11)

Indeed, in this case (C1), (C3) are obviously verified and (C2) holds with $\overline{\mathfrak{q}}\left(\xi \right)=\lambda \eta \mathfrak{q}\left(\xi \right)$ and $\lambda \in [0,\frac{1}{2\eta {\sigma }_{\mathfrak{q}}}).$

As examples of such function $p,$ we quote:

$·p\left(\zeta \right)={\zeta }^{\gamma },$ with $\gamma \ge 1$ (the superlinear case).

$·p\left(\zeta \right)=ln(1+{\zeta }^{\gamma }),$ with $\gamma \ge 1.$

Remark 2.

It is important to observe that conditions of the form (C1)–(C3) are weaker then conditions of the form (P1)–(P3) adopted in [11].

We establish the following main result.

Theorem 1.

Let (C1)–(C3) hold. Then problem (6) possess a unique solution $\theta \in \mathbb{E}$ with the following global behavior

$$(1-{\sigma }_{\overline{\mathfrak{q}}})\psi \left(\xi \right)\le \theta \left(\xi \right)\le \psi \left(\xi \right),forall\xi \in \left[0,1\right].$$

(12)

Corollary 1.

Let $\rho \in C^1\left(\left[0,\infty \right),\left[0,\infty \right)\right),$ be a nondecreasing function with $\rho \left(0\right)=0$ and $\mathfrak{q}\in C\left((0,1)\right)\cap {\mathcal{F}}_{\alpha }.$ There exist ${\lambda }^{*}>0$ such that for $\lambda \in \left[0,{\lambda }^{*}\right)$, the problem

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)=\lambda \mathfrak{q}\left(\xi \right)\rho \left(\theta \left(\xi \right)\right),\hfill & 0<\xi <1,\hfill \\ \theta \left(\xi \right)>0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=a>0,\hfill \end{array}\right.$$

(13)

possess a unique solution $\theta \in \mathbb{E}$ with

$$(1-\lambda \eta {\sigma }_{\mathfrak{q}})\psi \left(\xi \right)\le \theta \left(\xi \right)\le \psi \left(\xi \right),forall\xi \in \left[0,1\right],$$

(14)

where $\eta :=\underset{\zeta \in \left[0,\frac{a}{\alpha -1}\right]}{sup}{\rho }^{\prime }\left(\zeta \right).$

Note that the estimations obtained in (14) are optimal in the sense that if $\lambda =0,$ then $\theta \equiv \psi ,$ the unique solution of problem (10).

In Section 2, some interesting inequalities on $G(\xi ,\zeta )$ are provided. In Section 3, we first show that if $\mathfrak{q}\in C\left((0,1)\right)\cap {\mathcal{F}}_{\alpha }$ satisfying ${\sigma }_{\mathfrak{q}}\le \frac{1}{2},$ then the following fractional boundary value problem:

$$\left\{\begin{array}{cc}-D^{\alpha }\theta \left(\xi \right)+\mathfrak{q}\left(\xi \right)\theta \left(\xi \right)=f\left(\xi \right),\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=0,\hfill \end{array}\right.$$

(15)

has a nonnegative solution provided that $f\ge 0.$ That is problem (15) admits a nonnegative Green’s function ${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right).$ Next, we establish some pertinent properties on ${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right),$ which allow us to prove our main results.

2. Background Materials and Preliminaries

Lemma 1

([2,4,5]). Let $\alpha \in (1,2)$ and $\theta \in \mathbb{F}$. We have

$\left(i\right)$ For $0<\gamma <\alpha ,$ $D^{\gamma }{I}^{\alpha }\theta =I^{\alpha -\gamma }\theta$ and $D^{\alpha }{I}^{\alpha }\theta =\theta$.

$\left(ii\right)$$D^{\alpha }\theta \left(\xi \right)=0$ if and only if $\theta \left(\xi \right)=c_1{\xi }^{\alpha -1}+c_2{\xi }^{\alpha -2},$ where $c_1,c_2\in \mathbb{R}$.

In the next Lemma, we recall the expression of $G.$

Lemma 2

([7]). Let $\alpha \in (1,2]$ and $\varphi \in \mathbb{E},$ then the unique solution of

$$\left\{\begin{array}{cc}-D^{\alpha }\theta \left(\xi \right)=\varphi \left(\xi \right),\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=0,\hfill \end{array}\right.$$

(16)

is given by

$$\theta \left(\xi \right)={\int }_0^{1}G\left(\xi ,\zeta \right)\varphi \left(\zeta \right)d\zeta ,$$

(17)

where for $\left(\xi ,\zeta \right)\in \left[0,1\right]\times \left[0,1\right),$

$$G\left(\xi ,\zeta \right)=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}\left[{\xi }^{\alpha -1}{(1-\zeta )}^{\alpha -2}-{\left({\left(\xi -\zeta \right)}^{+}\right)}^{\alpha -1}\right],$$

(18)

with ${\left(\xi -\zeta \right)}^{+}:=max\left(\xi -\zeta ,0\right)$.

To obtain an immediate idea of the global behavior of $G(\xi ,\zeta ),$ in Figure 1, we have obtained its representation with the contours and some projections for $\alpha =\frac{3}{2}$. In particular, one can see from figure $\left(C\right)$ that $G(\xi ,\zeta )$ is singular at $\zeta =1.$

Figure 1. $G(\xi ,\zeta )$ for $\alpha =3/2$. (a) $G(\xi ,\zeta )$ and contours. (b) Projection on $\xi$z. (c) Projection on $\zeta$z.

Proposition 1.

Let $\alpha \in (1,2]$ be fixed. Then

(i) 

For $\left(\xi ,\zeta \right)\in \left[0,1\right]\times \left[0,1\right),$

$$\frac{\alpha -1}{\mathrm{\Gamma }\left(\alpha \right)}{H}_0\left(\xi ,\zeta \right)\le G(\xi ,\zeta )\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{H}_0\left(\xi ,\zeta \right),$$

(19)

where $H_0\left(\xi ,\zeta \right)={\xi }^{\alpha -2}{(1-\zeta )}^{\alpha -2}min\left(\xi ,\zeta \right)$.

(ii) 

For $\left(\xi ,\zeta \right)\in \left[0,1\right]\times \left[0,1\right),$

$$\frac{\alpha -1}{\mathrm{\Gamma }\left(\alpha \right)}{\xi }^{\alpha -1}\zeta {(1-\zeta )}^{\alpha -2}\le G(\xi ,\zeta )\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\xi }^{\alpha -1}{(1-\zeta )}^{\alpha -2}.$$

(20)

Proof. 

(i)

From (18), for $\left(\xi ,\zeta \right)\in (0,1]\times \left[0,1\right),$ we have

$$G(\xi ,\zeta )=\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\xi }^{\alpha -1}{(1-\zeta )}^{\alpha -2}\left(1-(1-\zeta ){\left(\frac{{(\xi -\zeta )}^{+}}{\xi (1-\zeta )}\right)}^{\alpha -1}\right).$$

(21)

Using this fact and that for $\gamma ,\nu >0$ and $d,\zeta \in [0,1],$

$$min(1,\frac{\nu }{\gamma })(1-d{\zeta }^{\gamma })\le 1-d{\zeta }^{\nu }\le max(1,\frac{\nu }{\gamma })(1-d{\zeta }^{\gamma }),$$

(22)

we deduce the required result with $\nu =\alpha -1,$ $\gamma =1$ and $d=(1-\zeta )$.

(ii)

The results follows from the previous estimates and the fact that

$$\xi \zeta \le min(\xi ,\zeta )\le \xi ,\mathrm{for}\xi ,\zeta \in [0,1].$$

□

For $f\in {\mathfrak{B}}^{+},$ we let

$$Vf\left(\xi \right):={\int }_0^{1}G\left(\xi ,\zeta \right)f\left(\zeta \right)d\zeta ,\mathrm{for}\xi \in \left[0,1\right].$$

(23)

The next Corollary follows immediately from (18) and Proposition 1 (ii).

Corollary 2.

Let $f\in$ ${\mathfrak{B}}^{+}$ with ${\int }_0^1{(1-\zeta )}^{\alpha -2}f\left(\zeta \right)d\zeta <\infty .$ Then $\xi \to Vf\left(\xi \right)$ belongs to $\mathbb{E}.$

The next Proposition is important in the rest of the paper. It improves Lemma 2.

Proposition 2.

Let $1<\alpha <2$ and f $\in C\left(\right(0,1\left)\right).$ Assume that the function $\zeta \to {\left(1-\zeta \right)}^{\alpha -2}f\left(\zeta \right)$ belongs to $\mathbb{F}$, then $Vf$ is the unique solution in $\mathbb{E}$ of (16).

Proof. 

Since by Corollary 2, $Vf$ belongs to $\mathbb{E},$ then $I^{2-\alpha }\left(V\left|f\right|\right)$ becomes finite on $\left[0,1\right].$ Hence

$$\begin{array}{ccc}\hfill I^{2-\alpha }\left(Vf\right)\left(\xi \right)& =& \frac{1}{\mathrm{\Gamma }\left(2-\alpha \right)}{\int }_0^{\xi }{(\xi -\zeta )}^{\alpha -1}Vf\left(\zeta \right)d\zeta \hfill \\ & =& \frac{1}{\mathrm{\Gamma }\left(2-\alpha \right)}{\int }_0^1\left({\int }_0^{\xi }{(\xi -\zeta )}^{\alpha -1}G\left(\zeta ,\omega \right)d\zeta \right)f\left(\omega \right)d\omega \hfill \\ & =& {\int }_0^1\mathcal{K}\left(\xi ,\omega \right)f\left(\omega \right)d\omega ,\hfill \end{array}$$

with $\mathcal{K}\left(\xi ,\omega \right):=\frac{1}{\mathrm{\Gamma }\left(2-\alpha \right)}{\int }_0^{\xi }{(\xi -\zeta )}^{\alpha -1}G\left(\zeta ,\omega \right)d\zeta$.

By simple computation, we obtain

$$\mathcal{K}\left(\xi ,\omega \right)=\xi {\left(1-\omega \right)}^{\alpha -2}-{(\xi -\omega )}^{+}.$$

Thus

$$I^{2-\alpha }\left(Vf\right)\left(\xi \right)=\xi {\int }_0^1{(1-\omega )}^{\alpha -2}f\left(\omega \right)d\omega -\xi {\int }_0^{\xi }f\left(\omega \right)d\omega +{\int }_0^{\xi }\omega f\left(\omega \right)d\omega .$$

(24)

Differentiate twice (24), we obtain

$$D^{\alpha }\left(Vf\right)\left(\xi \right):=\frac{d^2}{d{\xi }^2}\left(I^{2-\alpha }\left(Vf\right)\right)\left(\xi \right)=-f\left(\xi \right).$$

Since $Vf$ $\in \mathbb{E},$ then ${\displaystyle Vf\left(0\right)={\int }_0^{1}G\left(0,\zeta \right)f\left(\zeta \right)d\zeta =0.}$ Combining (18) and (22), we find that

$$\underset{\xi \to 1}{lim}\frac{G\left(\xi ,\zeta \right)-G\left(1,\zeta \right)}{\xi -1}=0\mathrm{and}\left|\frac{G\left(\xi ,\zeta \right)-G\left(1,\zeta \right)}{\xi -1}\right|\le \frac{2}{\mathrm{\Gamma }\left(\alpha \right)}{(1-\zeta )}^{\alpha -2},\mathrm{for}\xi \in [0,1).$$

Hence, we deduce by the Lebesgue dominated convergence theorem, that ${\left(Vf\right)}^{\prime }\left(1\right)=0$.

Finally, by Lemma 1, we obtain the uniqueness. □

Proposition 3.

For each $\xi ,\omega ,\zeta \in \left(0,1\right),$

$$\frac{G\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)}{G\left(\xi ,\zeta \right)}\le \frac{1}{\left(\alpha -1\right)\mathrm{\Gamma }\left(\alpha \right)}{\omega }^{\alpha -1}{(1-\omega )}^{\alpha -2}.$$

(25)

Proof. 

Due to Proposition 1 (i), for each $\xi ,\omega ,\zeta \in \left(0,1\right)$,

$$\frac{G\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)}{G\left(\xi ,\zeta \right)}\le \frac{1}{\left(\alpha -1\right)\mathrm{\Gamma }\left(\alpha \right)}{\omega }^{\alpha -2}{(1-\omega )}^{\alpha -2}\frac{min\left(\xi ,\omega \right)min\left(\omega ,\zeta \right)}{min\left(\xi ,\zeta \right)}.$$

Hence, we obtain inequality (25) by observing that

$$\frac{min\left(\xi ,\omega \right)min\left(\omega ,\zeta \right)}{min\left(\xi ,\zeta \right)}\le \omega .$$

□

Proposition 4.

Let $1<\alpha \le 2$ and $\mathfrak{q}\in {\mathcal{F}}_{\alpha },$ then

(i) 

$${\sigma }_{\mathfrak{q}}\le \frac{1}{\left(\alpha -1\right)\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1{\omega }^{\alpha -1}{(1-\omega )}^{\alpha -2}\mathfrak{q}\left(\omega \right)d\omega <\infty .$$

(26)

(ii) 

$$V\left(\mathfrak{q}\psi \right)\left(\xi \right)\le {\sigma }_{\mathfrak{q}}\psi \left(\xi \right),\mathrm{for}\xi \in \left[0,1\right],$$

(27)

where V is given in (23).

Proof. 

Let $1<\alpha \le 2$ and $\mathfrak{q}\in {\mathcal{F}}_{\alpha }.$ Inequality (26) follows immediately from (8) and (25). To prove inequality (27), first we observe that, for each $\xi ,\zeta \in (0,1],$

$$\underset{\omega \to 1}{lim}\frac{G\left(\zeta ,\omega \right)}{G\left(\xi ,\omega \right)}=\frac{\psi \left(\zeta \right)}{\psi \left(\xi \right)}.$$

Hence by Fatou’s lemma and (8), we obtain

$$\begin{array}{ccc}\hfill \frac{1}{\psi \left(\xi \right)}V\left(\mathfrak{q}\psi \right)\left(\xi \right)& =& {\int }_0^{1}G\left(\xi ,\zeta \right)\frac{\psi \left(\zeta \right)}{\psi \left(\xi \right)}\mathfrak{q}\left(\zeta \right)d\zeta \hfill \\ & \le & \underset{\omega \to 1}{liminf}{\int }_0^{1}G\left(\xi ,\zeta \right)\frac{G\left(\zeta ,\omega \right)}{G\left(\xi ,\omega \right)}\mathfrak{q}\left(\zeta \right)d\zeta \le {\sigma }_{\mathfrak{q}}.\hfill \end{array}$$

□

3. Proofs of the Existence Results

3.1. Properties of ${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)$

Our first goal is to show that if $\mathfrak{q}\in C\left((0,1)\right)\cap {\mathcal{F}}_{\alpha }$ satisfying ${\sigma }_{\mathfrak{q}}\le \frac{1}{2},$ the operator $\theta \to -D^{\alpha }\theta +\mathfrak{q}\left(\xi \right)\theta$ subjected to $\theta \left(0\right)=0,$ ${\theta }^{\prime }\left(1\right)=0$ admits a nonnegative Green’s function ${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)$. Next, we will study properties of the kernel $V_{\mathfrak{q}}$ defined by

$$V_{\mathfrak{q}}f\left(\xi \right):={\int }_0^1{\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)f\left(\zeta \right)d\zeta ,\mathrm{where}f\in {\mathfrak{B}}^{+}\mathrm{and}\xi \in \left[0,1\right].$$

(28)

This will allow us to prove our main results.

For $\mathfrak{q}\in$ ${\mathcal{F}}_{\alpha }$ and $\left(\xi ,\zeta \right)\in \left[0,1\right]\times \left[0,1\right),$ set $G_0\left(\xi ,\zeta \right)=G\left(\xi ,\zeta \right)$ and

$$G_k\left(\xi ,\zeta \right):={\int }_0^{1}G\left(\xi ,\omega \right)G_{k-1}\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega ,k\ge 1.$$

(29)

Define ${\mathcal{H}}_{\mathfrak{q}}$ on $\left[0,1\right]\times \left[0,1\right)$ by

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)=\underset{k=0}{\sum ^{\infty }}{\left(-1\right)}^k{G}_k\left(\xi ,\zeta \right).$$

(30)

Lemma 3.

Consider $\mathfrak{q}\in$ ${\mathcal{F}}_{\alpha }$ with ${\sigma }_{\mathfrak{q}}<1$. For all $k\in \mathbb{N}$ and $\left(\xi ,\zeta \right)\in \left[0,1\right]\times \left[0,1\right),$

(i) 

$$0\le G_k\left(\xi ,\zeta \right)\le {\sigma }_{\mathfrak{q}}^{k}G\left(\xi ,\zeta \right).$$

(31)

Hence, ${\mathcal{H}}_{\mathfrak{q}}$ is well defined in $\left[0,1\right]\times \left[0,1\right).$

(ii) 

$$A_k{\xi }^{\alpha -1}\zeta {\left(1-\zeta \right)}^{\alpha -2}\le G_k\left(\xi ,\zeta \right)\le B_k{\xi }^{\alpha -1}{\left(1-\zeta \right)}^{\alpha -2},$$

(32)

where

$$A_k:={\left(\frac{\alpha -1}{\mathrm{\Gamma }\left(\alpha \right)}\right)}^{k+1}{\left({\int }_0^1{\omega }^{\alpha }{(1-\omega )}^{\alpha -2}\mathfrak{q}\left(\omega \right)d\omega \right)}^k,$$

and

$$B_k:=\frac{1}{{\left(\mathrm{\Gamma }\left(\alpha \right)\right)}^{k+1}}{\left({\int }_0^1{\omega }^{\alpha -1}{(1-\omega )}^{\alpha -2}\mathfrak{q}\left(\omega \right)d\omega \right)}^k.$$

(iii) 

${\displaystyle G_{k+1}\left(\xi ,\zeta \right)={\int }_0^1{G}_k\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega .}$

(iv) 

${\displaystyle {\int }_0^1{\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega ={\int }_0^{1}G\left(\xi ,\omega \right){\mathcal{H}}_{\mathfrak{q}}\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega .}$

Proof. 

Using (29) and (8), we obtain inequalities in $\left(i\right)$ by simple induction.

Inequality (32) holds by induction and Proposition 1 (ii).

(iii) 

The assertion is valid for $k=0$. Assume that

$$G_k\left(\xi ,\zeta \right)={\int }_0^1{G}_{k-1}\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega .$$

(33)

Therefore, by using (29) and Fubini-Tonelli’s theorem, we deduce that

$$\begin{array}{ccc}\hfill G_{k+1}\left(\xi ,\zeta \right)& =& {\int }_0^{1}G\left(\xi ,\omega \right)\left({\int }_0^1{G}_{k-1}\left(\omega ,\sigma \right)G\left(\sigma ,\zeta \right)\mathfrak{q}\left(\sigma \right)d\sigma \right)\mathfrak{q}\left(\omega \right)d\omega \hfill \\ & =& {\int }_0^1\left({\int }_0^{1}G\left(\xi ,\omega \right)G_{k-1}\left(\omega ,\sigma \right)\mathfrak{q}\left(\omega \right)d\omega \right)G\left(\sigma ,\zeta \right)\mathfrak{q}\left(\sigma \right)d\sigma \hfill \\ & =& {\int }_0^1{G}_k\left(\xi ,\sigma \right)G\left(\sigma ,\zeta \right)\mathfrak{q}\left(\sigma \right)d\sigma .\hfill \end{array}$$

(iv) 

Let $k\in \mathbb{N},$ $\xi \in \left[0,1\right]$ and $\omega ,\zeta \in \left[0,1\right).$ From part (i), we get

$$0\le G_k\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)\le {\sigma }_{\mathfrak{q}}^{k}G\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right).$$

So, the series ${\displaystyle \sum _{k\ge 0}}{\int }_0^1{G}_k\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega$ converges.

Hence, we deduce by the dominated convergence theorem and Lemma 3 (iii), that

$$\begin{array}{ccc}\hfill {\int }_0^1{\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega & =& \underset{k=0}{\sum ^{\infty }}{\int }_0^1{\left(-1\right)}^k{G}_k\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega \hfill \\ & =& \underset{k=0}{\sum ^{\infty }}{\int }_0^1{\left(-1\right)}^{k}G\left(\xi ,\omega \right)G_k\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega \hfill \\ & =& {\int }_0^{1}G\left(\xi ,\omega \right){\mathcal{H}}_{\mathfrak{q}}\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega .\hfill \end{array}$$

□

Proposition 5.

Let $\mathfrak{q}\in$ ${\mathcal{F}}_{\alpha }$ with ${\sigma }_{\mathfrak{q}}<1.$ The function $\left(\xi ,\zeta \right)\to {\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)$ is in $C\left(\left[0,1\right]\times \left[0,1\right)\right)$.

Proof. 

Let $A\in \left(0,1\right)$ be fixed. We claim that the function $\left(\xi ,\zeta \right)\to {\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\in C\left(\left[0,1\right]\times \left[0,A\right]\right).$

To this end, we need first to prove that the function $\left(\xi ,\zeta \right)\to G_k\left(\xi ,\zeta \right)\in C(\left[0,1\right]\times \left[0,A\right])$ for all $k\ge 0$.

We proceed by induction. It is clear that $\left(\xi ,\zeta \right)\to G_0\left(\xi ,\zeta \right)\in C(\left[0,1\right]\times \left[0,A\right]).$

Assume that $\left(\xi ,\zeta \right)\to G_k\left(\xi ,\zeta \right)\in C(\left[0,1\right]\times \left[0,A\right]),$ for some $k\ge 0.$

Let $\omega \in (0,1)$ be fixed. By using (31) and (20) we obtain for $(\xi ,\zeta )\in \left[0,1\right]\times \left[0,A\right],$

$$G\left(\xi ,\omega \right)G_k\left(\omega ,\zeta \right)\le \frac{{\sigma }_{\mathfrak{q}}^k}{{\left(\mathrm{\Gamma }\left(\alpha \right)\right)}^2}{\left(1-A\right)}^{\alpha -2}{\omega }^{\alpha -1}{\left(1-\omega \right)}^{\alpha -2}.$$

(34)

Hence from (29), (34) and the dominated convergence theorem, we conclude that the function $\left(\xi ,\zeta \right)\to G_{k+1}\left(\xi ,\zeta \right)$ belongs to $C\left(\left[0,1\right]\times \left[0,A\right]\right).$

On the other hand, since for all $(\xi ,\zeta )\in \left[0,1\right]\times \left[0,A\right],$

$$G_k\left(\xi ,\zeta \right)\le \frac{{\sigma }_{\mathfrak{q}}^k}{\mathrm{\Gamma }\left(\alpha \right)}{\left(1-A\right)}^{\alpha -2},$$

the series $\left(\xi ,\zeta \right)\to {\displaystyle \sum _{k=0}^{\infty }}{\left(-1\right)}^k{G}_k\left(\xi ,\zeta \right)$ becomes uniformly convergent on $\left[0,1\right]\times \left[0,A\right].$

Hence, ${\mathcal{H}}_{\mathfrak{q}}$ is continuous on $\left[0,1\right]\times \left[0,1\right).$ □

Lemma 4.

Consider $\mathfrak{q}\in$ ${\mathcal{F}}_{\alpha }$ with ${\sigma }_{\mathfrak{q}}\le \frac{1}{2}.$ Then

$$\left(1-{\sigma }_{\mathfrak{q}}\right)G\left(\xi ,\zeta \right)\le {\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\le G\left(\xi ,\zeta \right),on(\xi ,\zeta )\in \left[0,1\right]\times \left[0,1\right).$$

(35)

In particular, ${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\ge 0.$

Proof. 

From Lemma 3 (i), we obtain

$$\left|{\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\right|\le \underset{k=0}{\sum ^{\infty }}{\left({\sigma }_{\mathfrak{q}}\right)}^{k}G\left(\xi ,\zeta \right)\le \frac{1}{1-{\sigma }_{\mathfrak{q}}}G\left(\xi ,\zeta \right).$$

(36)

By writing

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)=G\left(\xi ,\zeta \right)-\underset{k=0}{\sum ^{\infty }}{\left(-1\right)}^k{G}_{k+1}\left(\xi ,\zeta \right),$$

(37)

we conclude from (29) that

$$\begin{array}{ccc}\hfill {\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)& =& G(\xi ,\zeta )-\underset{k=0}{\sum ^{\infty }}{\left(-1\right)}^k{\int }_0^{1}G\left(\xi ,\omega \right)G_k\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega \hfill \\ & =& G(\xi ,\zeta )-{\int }_0^{1}G\left(\xi ,\omega \right){\mathcal{H}}_{\mathfrak{q}}\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)d\omega .\hfill \end{array}$$

Namely,

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)=G(\xi ,\zeta )-V\left(\mathfrak{q}{\mathcal{H}}_{\mathfrak{q}}\left(.,\zeta \right)\right)\left(\xi \right).$$

(38)

From (36) and Lemma 3 (i), we derive that

$$V\left(\mathfrak{q}{\mathcal{H}}_{\mathfrak{q}}\left(.,\zeta \right)\right)\left(\xi \right)\le \frac{1}{1-{\sigma }_{\mathfrak{q}}}V\left(\mathfrak{q}G\left(.,\zeta \right)\right)\left(\xi \right)=\frac{1}{1-{\sigma }_{\mathfrak{q}}}{G}_1(\xi ,\zeta )\le \frac{{\sigma }_{\mathfrak{q}}}{1-{\sigma }_{\mathfrak{q}}}G(\xi ,\zeta ).$$

Therefore, by (38),

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\ge G(\xi ,\zeta )-\frac{{\sigma }_{\mathfrak{q}}}{1-{\sigma }_{\mathfrak{q}}}G(\xi ,\zeta )=\frac{1-2{\sigma }_{\mathfrak{q}}}{1-{\sigma }_{\mathfrak{q}}}G(\xi ,\zeta )\ge 0.$$

So ${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\le G(\xi ,\zeta ).$ Now, by using (38) and (31), we obtain

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\ge G(\xi ,\zeta )-V\left(\mathfrak{q}G\left(.,\zeta \right)\right)\left(\xi \right)=G(\xi ,\zeta )-G_1(\xi ,\zeta )\ge \left(1-{\sigma }_{\mathfrak{q}}\right)G(\xi ,\zeta ).$$

Finally from (35) and (18), we conclude that

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\ge \left(1-{\sigma }_{\mathfrak{q}}\right)G(\xi ,\zeta )\ge 0.$$

□

Corollary 3.

Consider $\mathfrak{q}\in {\mathcal{F}}_{\alpha }$ with ${\sigma }_{\mathfrak{q}}\le \frac{1}{2}.$ Let $f\in {\mathfrak{B}}^{+}$ satisfying ${\int }_0^1{(1-\zeta )}^{\alpha -2}f\left(\zeta \right)d\zeta <\infty .$ Then

$$\xi \to V_{\mathfrak{q}}f\left(\xi \right)\in \mathbb{E},$$

where $V_{\mathfrak{q}}$ is given in (28).

Proof. 

Let $\zeta \in \left[0,1\right)$ be fixed. Since the function $\xi \to {\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\in \mathbb{E}$ and by Lemma 4 and Proposition 1$\left(ii\right),$

$${\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\le G(\xi ,\zeta )\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{(1-\zeta )}^{\alpha -2},\mathrm{for}\mathrm{all}\xi \in [0,1],$$

we deduce by the dominated convergence theorem that $V_{\mathfrak{q}}f$ is a continuous function on $\left[0,1\right].$ □

Lemma 5.

Consider $\mathfrak{q}\in {\mathcal{F}}_{\alpha }$ with ${\sigma }_{\mathfrak{q}}\le \frac{1}{2}.$ Let $f\in {\mathfrak{B}}^{+},$ then the following resolvent equation hold:

$$Vf=V_{\mathfrak{q}}f+V_{\mathfrak{q}}\left(\mathfrak{q}Vf\right)=V_{\mathfrak{q}}f+V\left(\mathfrak{q}{V}_{\mathfrak{q}}f\right).$$

(39)

Furthermore, if $V\left(\mathfrak{q}f\right)<\infty$, then $V_{\mathfrak{q}}\left(\mathfrak{q}f\right)<\infty$ and we have

$$\left(I-V_{\mathfrak{q}}\left(\mathfrak{q}.\right)\right)\left(I+V\left(\mathfrak{q}.\right)\right)f=\left(I+V\left(\mathfrak{q}.\right)\right)\left(I-V_{\mathfrak{q}}\left(\mathfrak{q}.\right)\right)f=f,$$

(40)

where $V\left(\mathfrak{q}.\right)$ and $V_{\mathfrak{q}}\left(\mathfrak{q}.\right)$ are the operators defined on ${\mathfrak{B}}^{+}$ by

$$V\left(\mathfrak{q}.\right)\left(f\right)\left(\xi \right):=V\left(\mathfrak{q}f\right)\left(\xi \right)={\int }_0^{1}G\left(\xi ,\zeta \right)\mathfrak{q}\left(\zeta \right)f\left(\zeta \right)d\zeta ,\mathrm{for}\xi \in \left[0,1\right],$$

(41)

and

$$V_{\mathfrak{q}}\left(\mathfrak{q}.\right)\left(f\right)\left(\xi \right):=V_{\mathfrak{q}}\left(\mathfrak{q}f\right)\left(\xi \right)={\int }_0^1{\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)\mathfrak{q}\left(\zeta \right)f\left(\zeta \right)d\zeta ,\mathrm{for}\xi \in \left[0,1\right].$$

(42)

Proof. 

Let $\left(\xi ,\zeta \right)\in \left[0,1\right]\times \left[0,1\right),$ then by (38),

$$G(\xi ,\zeta )={\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)+V\left(\mathfrak{q}{\mathcal{H}}_{\mathfrak{q}}\left(.,\zeta \right)\right)\left(\xi \right).$$

Therefore

$$\begin{array}{ccc}\hfill Vf\left(\xi \right)& =& {\int }_0^1\left({\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\zeta \right)+V\left(\mathfrak{q}{\mathcal{H}}_{\mathfrak{q}}\left(.,\zeta \right)\right)\left(\xi \right)\right)f\left(\zeta \right)d\zeta \hfill \\ & =& V_{\mathfrak{q}}f\left(\xi \right)+V\left(\mathfrak{q}{V}_{\mathfrak{q}}f\right)\left(\xi \right).\hfill \end{array}$$

Hence by Lemma 3 (iv), we obtain

$${\int }_0^1{\int }_0^1{\mathcal{H}}_{\mathfrak{q}}\left(\xi ,\omega \right)G\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)f\left(\zeta \right)d\omega d\zeta ={\int }_0^1{\int }_0^{1}G\left(\xi ,\omega \right){\mathcal{H}}_{\mathfrak{q}}\left(\omega ,\zeta \right)\mathfrak{q}\left(\omega \right)f\left(\zeta \right)d\omega d\zeta .$$

Namely,

$$V_{\mathfrak{q}}\left(\mathfrak{q}Vf\right)\left(\xi \right)=V\left(\mathfrak{q}{V}_{\mathfrak{q}}f\right)\left(\xi \right),$$

from which, we obtain

$$Vf=V_{\mathfrak{q}}f+V\left(\mathfrak{q}{V}_{\mathfrak{q}}f\right)=V_{\mathfrak{q}}f+V_{\mathfrak{q}}\left(\mathfrak{q}Vf\right).$$

Furthermore, if $V\left(\mathfrak{q}f\right)<\infty$, then from (39), we deduce that $V_{\mathfrak{q}}\left(\mathfrak{q}f\right)\le V\left(\mathfrak{q}f\right)<\infty$ and

$$\begin{array}{ccc}\hfill \left(I-V_{\mathfrak{q}}\left(\mathfrak{q}.\right)\right)\left(I+V\left(\mathfrak{q}.\right)\right)f& =& \left(I-V_{\mathfrak{q}}\left(\mathfrak{q}.\right)\right)(f+V\left(\mathfrak{q}f\right))\hfill \\ & =& f+V\left(\mathfrak{q}f\right)-V_{\mathfrak{q}}\left(\mathfrak{q}f\right)-V_{\mathfrak{q}}\left(\mathfrak{q}V\left(\mathfrak{q}f\right)\right)\hfill \\ & =& f.\hfill \end{array}$$

Similarly, we obtain

$$\left(I+V\left(\mathfrak{q}.\right)\right)\left(I-V_{\mathfrak{q}}\left(\mathfrak{q}.\right)\right)f=f.$$

□

Proposition 6.

Let $\mathfrak{q}$ belonging to $C\left(\left(0,1\right)\right)\cap {\mathcal{F}}_{\alpha }$ such that ${\sigma }_{\mathfrak{q}}\le \frac{1}{2}.$ For any $f\in {\mathfrak{B}}^{+}$ such that $\zeta \to {\left(1-\zeta \right)}^{\alpha -2}f\left(\zeta \right)\in \mathbb{F},$ the function $\theta :=V_{\mathfrak{q}}f$ is the unique nonnegative continuous solution to problem (15) satisfying

$$\left(1-{\sigma }_{\mathfrak{q}}\right)Vf\le \theta \le Vf.$$

(43)

Proof. 

From Corollary 3, it is clear that the function $\xi \to \mathfrak{q}\left(\xi \right)V_{\mathfrak{q}}f\left(\xi \right)$ is in $C^{+}\left(\left(0,1\right)\right).$ By using (39) and Proposition 1 (iii), there exists $m\ge 0$ such that

$$V_{\mathfrak{q}}f\left(\xi \right)\le Vf\left(\xi \right)\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1{\xi }^{\alpha -1}{(1-\zeta )}^{\alpha -2}f\left(\zeta \right)d\zeta \equiv m{\xi }^{\alpha -1}.$$

(44)

This implies that

$${\int }_0^1{(1-\zeta )}^{\alpha -2}\mathfrak{q}\left(\zeta \right)V_{\mathfrak{q}}f\left(\zeta \right)d\zeta \le m{\int }_0^1{\zeta }^{\alpha -1}{(1-\zeta )}^{\alpha -2}\mathfrak{q}\left(\zeta \right)d\zeta <\infty .$$

Due to Proposition 2, the function $\theta =V_{\mathfrak{q}}f=Vf-V\left(\mathfrak{q}{V}_{\mathfrak{q}}f\right)$ belongs to $\mathbb{E}$ and verify

$$\left\{\begin{array}{cc}-D^{\alpha }\theta \left(\xi \right)=f\left(\xi \right)-\mathfrak{q}\left(\xi \right)\theta \left(\xi \right),\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=0.\hfill \end{array}\right.$$

That is $\theta =V_{\mathfrak{q}}f$ is a solution to problem (15).

Inequalities (43) hold trivially by integrating (35).

Next, to establish the uniqueness, we consider $h\in \mathbb{E}$ be another solution to problem (15) satisfying $h\le Vf.$

From (43) and (44), we conclude that the function $\zeta \to {\left(1-\zeta \right)}^{\alpha -2}\mathfrak{q}\left(\zeta \right)h\left(\zeta \right)$ is in $\mathbb{F}.$

Therefore, by Proposition 2, $\overline{h}:=h+V\left(\mathfrak{q}h\right)$ satisfies

$$\left\{\begin{array}{cc}-D^{\alpha }\overline{h}\left(\xi \right)=f\left(\xi \right),\hfill & 0<\xi <1,\hfill \\ \overline{h}\left(0\right)=0,{\overline{h}}^{\prime }\left(1\right)=0.\hfill \end{array}\right.$$

So

$$\overline{h}:=h+V\left(\mathfrak{q}h\right)=Vf,$$

from which, we deduce that

$$\left(I+V\left(\mathfrak{q}.\right)\right)\left({\left(h-\theta \right)}^{+}\right)=\left(I+V\left(\mathfrak{q}.\right)\right)\left({\left(h-\theta \right)}^{-}\right),$$

where ${\left(h-\theta \right)}^{+}=max\left(h-\theta ,0\right)$ and ${\left(h-\theta \right)}^{-}=max\left(\theta -h,0\right).$

Since $\left|h\left(\zeta \right)-\theta \left(\zeta \right)\right|\le 2Vf\left(\zeta \right)\le 2m{\zeta }^{\alpha -1}$, we conclude by Proposition 1 (ii) that

$$V\left(\mathfrak{q}\left|h-\theta \right|\right)\left(\xi \right)\le \frac{2m}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1{\zeta }^{\alpha -1}{(1-\zeta )}^{\alpha -2}\mathfrak{q}\left(\zeta \right)d\zeta <\infty .$$

Hence, from (40), we obtain that $\theta =h.$ □

3.2. Proofs of the Existence of Solutions

Proof of Theorem 1. 

Let $1<\alpha \le 2,$ $a>0$ be fixed and set $M=\frac{a}{\alpha -1}.$ By (C2), there exists a function $\overline{\mathfrak{q}}\in {\mathcal{F}}_{\alpha }$ with ${\sigma }_{\overline{\mathfrak{q}}}\le \frac{1}{2}$ such that, for each $\xi \in \left(0,1\right),$

$$\phi \left(\xi ,\zeta \right)-\phi \left(\xi ,r\right)\le \overline{\mathfrak{q}}\left(\xi \right)(\zeta -r),\mathrm{for}\mathrm{all}0\le r\le \zeta \le M.$$

(45)

In particular, from (C1), we obtain

$$0\le \phi \left(\xi ,\zeta \right)\le \overline{\mathfrak{q}}\left(\xi \right)\zeta ,\mathrm{for}\mathrm{all}0\le \zeta \le M.$$

(46)

Let

$$\Omega :=\left\{\theta \in {\mathfrak{B}}^{+}:\left(1-{\sigma }_{\overline{\mathfrak{q}}}\right)\psi \le \theta \le \psi \right\}.$$

Since, for all $\xi \in [0,1],$ $\psi \left(\xi \right)\in [0,M],$ we deduce from (46), that

$$0\le \phi \left(\xi ,\theta \right)\le \overline{\mathfrak{q}}\left(\xi \right)\theta ,\mathrm{for}\theta \in \Omega .$$

(47)

Define the operator T on $\Omega$ by

$$T\theta =\psi -V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\psi \right)+V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\theta -\phi \left(.,\theta \right)\right).$$

(48)

From (39) and (27), we obtain

$$V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\psi \right)\le V\left(\overline{\mathfrak{q}}\psi \right)\le {\sigma }_{\overline{\mathfrak{q}}}\psi \le \psi ,$$

(49)

which implies that

$$\left\{\begin{array}{c}T\theta \ge \psi -V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\psi \right)\ge \left(1-{\sigma }_{\overline{\mathfrak{q}}}\right)\psi ,\hfill \\ \mathrm{and}\hfill \\ T\theta \le \psi -V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\psi \right)+V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\theta \right)\le \psi .\hfill \end{array}\right.$$

(50)

That is $T\left(\Omega \right)\subset \Omega .$

From (45), we derive that T is a nondecreasing operator on $\Omega .$

Let $\left\{{\theta }_k\right\}$ defined by ${\theta }_0=\left(1-{\sigma }_{\overline{\mathfrak{q}}}\right)\psi$ and ${\theta }_{k+1}=T{\theta }_k,$ for $k\in \mathbb{N}.$

Since $T\left(\Omega \right)\subset \Omega$ and T is nondecreasing, we obtain

$$\left(1-{\sigma }_{\overline{\mathfrak{q}}}\right)\psi ={\theta }_0\le {\theta }_1\le ...\le {\theta }_k\le {\theta }_{k+1}\le \psi .$$

Hence by the dominated convergence theorem and (C1)–(C2), the sequence $\left\{{\theta }_k\right\}$ converges to a function $\theta \in \Omega$ satisfying

$$\theta =\psi -V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\psi \right)+V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\theta -\phi \left(.,\theta \right)\right).$$

(51)

So

$$\theta -V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\theta \right)=\psi -V_{\overline{\mathfrak{q}}}\left(\overline{\mathfrak{q}}\psi \right)-V_{\overline{\mathfrak{q}}}\left(\phi \left(.,\theta \right)\right).$$

(52)

Since by (49), we have $V\left(\overline{\mathfrak{q}}\psi \right)\le {\sigma }_{\overline{\mathfrak{q}}}\psi \le \psi <\infty ,$ then by applying $\left(I+V\left(\mathfrak{q}.\right)\right)$ on (52) and using (39) and (40), we deduce that

$$\theta =\psi -V\left(\phi \left(.,\theta \right)\right).$$

(53)

We claim that $\theta$ is a solution of (6). Indeed, from (47), we have

$$\phi \left(\zeta ,\theta \left(\zeta \right)\right)\le \overline{\mathfrak{q}}\left(\zeta \right)\theta \left(\zeta \right)\le \overline{\mathfrak{q}}\left(\zeta \right)\psi \left(\zeta \right)=M{\zeta }^{\alpha -1}\overline{\mathfrak{q}}\left(\zeta \right).$$

(54)

This implies that

$${\int }_0^1{(1-\zeta )}^{\alpha -2}\phi \left(\zeta ,\theta \left(\zeta \right)\right)d\zeta \le M{\int }_0^1{\zeta }^{\alpha -1}{(1-\zeta )}^{\alpha -2}\overline{\mathfrak{q}}\left(\zeta \right)d\zeta <\infty .$$

(55)

Hence, by Corollary 2, $V\left(\phi \left(.,\theta \right)\right)$ belongs to $\mathbb{E}$ and from (53), $\theta$ becomes in $\mathbb{E}$.

Since the function $\zeta \to {(1-\zeta )}^{\alpha -2}\phi \left(\zeta ,\theta \left(\zeta \right)\right)$ belongs to $\mathbb{F}$, then by Proposition 2 and (10), we conclude that $\theta$ is a solution to problem (6) satisfying (12). Finally, by using (C3) and following the steps as in the proof of Proposition 6, we obtain the uniqueness. □

Remark 3.

Assume that (C1)–(C2) hold. Let Λ be the nonempty, closed, bounded and convex set given by

$$\mathrm{\Lambda }:=\left\{\theta \in \mathbb{E}:\left(1-{\sigma }_{\overline{\mathfrak{q}}}\right)\psi \le \theta \le \psi \right\},$$

where $\overline{\mathfrak{q}}$ is as in the proof of Theorem 1. Define the operator T on Λ by (48), then T is a compact operator mapping Λ to itself.

Indeed, following the proof of Theorem 1, one can see that the family $T\left(\mathrm{\Lambda }\right)$ is equicontinuous in $[0,1].$ In particular, for all $\theta \in \mathrm{\Lambda },$ $T\theta \in \mathbb{E}$ and $T\left(\mathrm{\Lambda }\right)\subset \mathrm{\Lambda }.$ Moreover, the family $\left\{T\theta \right(\xi ),$ $\theta \in \mathrm{\Lambda }\}$ is uniformly bounded in $[0,1].$ It follows by Ascoli’s theorem that $T\left(\mathrm{\Lambda }\right)$ is relatively compact in $\mathbb{E}.$ To prove the continuity of T in $\mathrm{\Lambda },$ we consider a sequence $\left\{{\theta }_k\right\}$ in Λ which converges uniformly to a function $\theta \in \mathrm{\Lambda }.$ Then from (35) we have

$$\begin{array}{ccc}\hfill \left|T\left({\theta }_k\right)\left(\xi \right)-T\left(\theta \right)\left(\xi \right)\right|& \le & {\int }_0^1{\mathcal{H}}_{{}_{\overline{\mathfrak{q}}}}\left(\xi ,\zeta \right)\overline{\mathfrak{q}}\left(\zeta \right)\left|{\theta }_k\left(\zeta \right)-\theta \left(\zeta \right)\right|d\zeta \hfill \\ & & +{\int }_0^1{\mathcal{H}}_{{}_{\overline{\mathfrak{q}}}}\left(\xi ,\zeta \right)\left|\phi \left(\zeta ,\theta \left(\zeta \right)\right)-\phi \left(\zeta ,{\theta }_k\left(\zeta \right)\right)\right|d\zeta \hfill \\ & \le & \underset{\xi \in [0,1]}{sup}\left|{\theta }_k\left(\zeta \right)-\theta \left(\zeta \right)\right|{\int }_0^{1}G\left(\xi ,\zeta \right)\overline{\mathfrak{q}}\left(\zeta \right)d\zeta \hfill \\ & & +{\int }_0^{1}G\left(\xi ,\zeta \right)\left|\phi \left(\zeta ,\theta \left(\zeta \right)\right)-\phi \left(\zeta ,{\theta }_k\left(\zeta \right)\right)\right|d\zeta .\hfill \end{array}$$

Now from (47), we have

$$\left|\phi \left(\zeta ,\theta \left(\zeta \right)\right)-\phi \left(\zeta ,{\theta }_k\left(\zeta \right)\right)\right|\le 2\overline{\mathfrak{q}}\left(\zeta \right)\psi \left(\zeta \right)\le \left(\frac{2a}{\alpha -1}\right)\overline{\mathfrak{q}}\left(\zeta \right).$$

So we deduce from (C1)–(C2), (20) and the dominated convergence theorem that

$$\forall \xi \in [0,1],T\left({\theta }_k\right)\left(\xi \right)\to T\left(\theta \right)\left(\xi \right)\mathrm{as}k\to \infty .$$

Since $T\left(\mathrm{\Lambda }\right)$ is relatively compact in $\mathbb{E},$ we have the uniform convergence, namely

$$\underset{\xi \in [0,1]}{sup}\left|T{\theta }_k\left(\zeta \right)-T\theta \left(\zeta \right)\right|\to 0ask\to \infty .$$

Hence T is a compact operator mapping Λ to itself. Therefore by the Schauder fixed-point theorem, there exists a function $\theta \in \mathrm{\Lambda }$ satisfying (51), which is a solution of (6).

Proof of Corollary 1. 

Let $\phi \left(\xi ,\zeta \right)=\lambda \mathfrak{q}\left(\xi \right)\rho \left(\zeta \right),$ with $\mathfrak{q}$ and $\rho$ as in Corollary 1.

Assumptions (C1) and (C3) are trivially satisfied.

Let $\eta :={\displaystyle \underset{\zeta \in [0,\frac{a}{\alpha -1}]}{sup}}{\rho }^{\prime }\left(\zeta \right)$ and set ${\lambda }^{*}:=\frac{1}{2\eta {\sigma }_{\mathfrak{q}}}.$ Then, for $\lambda \in \left[0,{\lambda }^{*}\right),$ hypotheses (C2) is satisfied with $\overline{\mathfrak{q}}\left(\xi \right):=\lambda \eta \mathfrak{q}\left(\xi \right).$

Hence by applying Theorem 1, we deduce the result. □

Example 1.

Let $1<\alpha \le 2$ and $a>0$ be fixed. Let $\gamma <\alpha$ and $\delta <\alpha -1.$ Put $\epsilon :=\frac{\mathcal{B}(\alpha -\gamma ,\alpha -\delta -1)}{(\alpha -1)\mathrm{\Gamma }\left(\alpha \right)},$ where $\mathcal{B}$ is the Beta function.

Then for $\lambda \in [0,{\displaystyle \frac{1}{2\epsilon }}),$ the singular problem

$$\left\{\begin{array}{cc}{D}^{\alpha }\theta \left(\xi \right)=\frac{\lambda }{{\xi }^{\gamma }{(1-\xi )}^{\delta }}ln\left(1+\theta \left(\xi \right)\right),\hfill & 0<\xi <1,\hfill \\ \theta \left(\xi \right)>0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=a>0,\hfill \end{array}\right.$$

possess a unique solution $\theta \in \mathbb{E}$ satisfying

$$(1-\lambda \epsilon )\psi \left(\xi \right)\le \theta \left(\xi \right)\le \psi \left(\xi \right),forall\xi \in \left[0,1\right],$$

where $\psi \left(\xi \right)=\frac{a}{\alpha -1}{\xi }^{\alpha -1}.$

Indeed, we may apply Corollary 1 with $\mathfrak{q}\left(\xi \right)=\frac{1}{{\xi }^{\gamma }{(1-\xi )}^{\delta }}\in C\left((0,1)\right)\cap {\mathcal{F}}_{\alpha }$ (which could be singular at both $\xi =0$ and $\xi =1$) and $\rho \left(\zeta \right)=ln(1+\zeta ).$

In this case, we have $\eta :={\displaystyle \underset{\zeta \in \left[0,\frac{a}{\alpha -1}\right]}{sup}}{\rho }^{\prime }\left(\zeta \right)=1$ and from (25), we obtain ${\sigma }_{\mathfrak{q}}\le \epsilon .$ Therefore, ${\lambda }^{*}:=\frac{1}{2\eta {\sigma }_{\mathfrak{q}}}\ge {\displaystyle \frac{1}{2\epsilon }}.$

Example 2.

Let $\alpha =\frac{3}{2},$ $\gamma \ge 1$ and $a>0$. Consider $\mathfrak{q}\in \mathbb{E}$ and put $\epsilon :={\displaystyle \frac{a^{1-\gamma }}{\gamma 2^{\gamma +1}\sqrt{\pi }{∥\mathfrak{q}∥}_{\infty }}},$ where ${∥\mathfrak{q}∥}_{\infty }:={\displaystyle \underset{\xi \in [0,1]}{sup}}\mathfrak{q}\left(\xi \right).$

Then for $\lambda \in [0,{\displaystyle \frac{1}{2\epsilon }}),$ the problem

$$\left\{\begin{array}{cc}{D}^{\frac{3}{2}}\theta \left(\xi \right)=\lambda \mathfrak{q}\left(\xi \right){\theta }^{\gamma }\left(\xi \right),\hfill & 0<\xi <1,\hfill \\ \theta \left(\xi \right)>0,\hfill & 0<\xi <1,\hfill \\ \theta \left(0\right)=0,{\theta }^{\prime }\left(1\right)=a>0,\hfill \end{array}\right.$$

possess a unique solution $\theta \in \mathbb{E}$ satisfying

$$2a(1-\lambda \epsilon )\sqrt{\xi }\le \theta \left(\xi \right)\le 2a\sqrt{\xi },forall\xi \in \left[0,1\right].$$

Indeed, we may again apply Corollary 1 with $\mathfrak{q}\in C^{+}\left(\left[0,1\right]\right)\subset {\mathcal{F}}_{\alpha },$ and $\rho \left(\zeta \right)={\zeta }^{\gamma }.$

In this case, we have $\eta :={\displaystyle \underset{\zeta \in \left[0,2a\right]}{sup}}{\rho }^{\prime }\left(\zeta \right)=\gamma {\left(2a\right)}^{\gamma -1}$ and from (25), we obtain ${\sigma }_{\mathfrak{q}}\le 2\sqrt{\pi }{∥\mathfrak{q}∥}_{\infty }.$ Therefore, ${\lambda }^{*}:=\frac{1}{2\eta {\sigma }_{\mathfrak{q}}}\ge {\displaystyle \frac{1}{2\epsilon }}.$

4. Conclusions

Under mild assumptions, the existence and uniqueness of a positive continuous solution to the nonlinear fractional boundary value problem (6) has been studied. Sharp estimates on the solution are obtained. The proofs are based on perturbation techniques, which consist, for a convenient function $\mathfrak{q}$, at the construction of the Green’s function of the operator $u\to -D^{\alpha }u+\mathfrak{q}\left(\xi \right)u$ subject to the boundary condition $u\left(0\right)=0,$ $u^{\prime }\left(1\right)=0$. This allows us to prove the resolvent Equation (39) which plays an important role for the existence. It will be interesting to investigate, by using this approach, fractional boundary value problems for higher dimensions.