Extremal solutions for singular fractional p-Laplacian differential equations with nonlinear boundary conditions

Abstract

In this paper, we establish the existence and uniqueness of extremal solutions for nonlinear boundary value problems of a singular fractional p-Laplacian differential equation involving Riemann-Liouville derivatives. Our results are obtained by constructing monotone iterative sequences of upper and lower solutions and applying the comparison result. At last, we present an example to illustrate the results. The compactness of sequences is proved in the Appendix.

1 Introduction

Fractional differential equations arise in the mathematical modeling of process in physics, chemistry, aerodynamics, polymer rheology, fluid flow phenomena, wave propagation and signal theory, electrical circuits, control theory, viscoelastic materials, and so on. The fractional calculus and its various applications in many fields of science and engineering have gained much attention and developed rapidly. Consequently, fractional differential equations have been of great interest. For details, see [1–8] and the references therein.

The numerical simulation plays an essential role in the analysis of fractional differential equations, and new numerical techniques are being developed; see, for example, [9, 10]. Recently, many research papers have appeared concerning the existence of solutions for the initial and boundary value problems of fractional differential equations; see [11–17]. The monotone iterative technique, combined with the method of upper and lower solutions, is a powerful tool of obtaining the existence of solutions for fractional boundary value problems; see [18–23].

By means of the monotone iterative method, in [24], the following PBVP of fractional differential equation was considered:

$$\left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\alpha}u(t)=f (t,u(t) ),\quad t\in(0,T], \\ t^{1-\alpha}u(t)|_{t=0}=t^{1-\alpha}u(t)|_{t=T}, \end{array}\displaystyle \right . $$

where $D_{0^{+}}^{\alpha}$ is the Riemann-Liouville fractional derivative of order $0<\alpha\le1$. The properties of the well-known Mittag-Leffler function and the existence and uniqueness of solution for this problem were given in [24]. However, fewer papers considered p-Laplacian boundary value problems of fractional order via the upper and lower method and the monotone iterative method; see, for instance, [25–27].

In [28], the authors have discussed the following PBVP of fractional p-Laplacian equation:

$$\left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}u(t)))=f (t,u(t),D_{0^{+}}^{\alpha}u(t) ),\quad t\in[0,T], \\ u(t)|_{t=0}=u(t)|_{t=T}, \qquad D^{\alpha}_{0^{+}}u(t)|_{t=0}=D^{\alpha}_{0^{+}}u(t)|_{t=T}, \end{array}\displaystyle \right . $$

where $0<\alpha, \beta\le1$, $D_{0^{+}}^{\alpha}$ is the Caputo fractional derivative, and $f: [0,T]\times\mathbb{R}^{2} \rightarrow \mathbb{R}$ is a continuous function. By establishing the continuation theorem, which is an extension of the coincidence degree theory for linear differential operators with PBCs, the existence result of solution of the PBVP was stated under the nonlinear growth restriction of f. To the best of our knowledge, the fractional p-Laplacian differential equation with periodic boundary conditions has rarely been considered up to now.

In this paper, we investigate the existence of extremal solutions and uniqueness of solution for singular fractional p-Laplacian differential equation with general nonlinear boundary conditions

$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}u(t)))=f (t,u(t),D_{0^{+}}^{\alpha}u(t) ), \quad t\in(0,T], \\ t^{\frac{1-\beta}{p-1}}D^{\alpha}_{0^{+}}u(t)|_{t=0}=t^{\frac{1-\beta }{p-1}}D^{\alpha}_{0^{+}}u(t)|_{t=T}, \\ g(\tilde{u}(0),\tilde{u}(T))=0, \end{array}\displaystyle \right . $$

(1.1)

where $0<\alpha, \beta\le1$, $1<\alpha+\beta\le2$, $D_{0^{+}}^{\alpha}$ is the Riemann-Liouville fractional derivative of order α, $\phi_{p}(t)=|t|^{p-2}t$ ($p>1$) is the p-Laplacian operator, and $(\phi_{p})^{-1}=\phi_{q}$, $\frac{1}{p}+\frac{1}{q}=1$. Here $f\in C([0,T]\times\mathbb{R}\times\mathbb{R}, \mathbb{R})$, $g\in C(\mathbb{R}\times\mathbb{R},\mathbb{R})$, $\tilde{u}(0)=t^{1-\alpha}u(t)|_{t=0}$, and $\tilde{u}(T)=t^{1-\alpha}u(t)|_{t=T}$.

In the problem (1.1), the boundary condition $g(\tilde{u}(0),\tilde{u}(T))=0$ is a kind of general condition. When $g(x,y)=x\pm y$ or others, this can cover periodic, antiperiodic, or other nonlinear boundary conditions. Moreover, if ${D}_{0^{+}}^{\alpha}u(t)|_{t=0}={D}_{0^{+}}^{\alpha }u(t)|_{t=T}$, then $t^{\frac{1-\beta}{p-1}}{D}_{0^{+}}^{\alpha }u(t)|_{t=0}=t^{\frac{1-\beta}{p-1}}{D}_{0^{+}}^{\alpha}u(t)|_{t=T}$. From this we can see that the boundary conditions in (1.1) are weaker than those in [28]. Thus, our conclusions can be more extensive. Here we not only obtain the existence of extremal solutions, but also the iterative sequences that converge to the extremal solutions.

In the previous related results on boundary value problems for p-Laplacian differential equations by means of the monotone iterative method, the monotone-type conditions for nonlinear terms f with respect to the functions u or their derivatives are usually required. However, in this paper, we only consider the functions $f+M\phi_{p}(D_{0^{+}}^{\alpha}u(t))$, not f, to satisfy the monotone-type conditions (see (H₂)).

The rest of our paper is organized as follows. In Section 2, we provide some preliminaries, the existence results for linear fractional problems with periodic boundary conditions and the comparison result. In Section 3, the existence of extremal solutions and unique solution for (1.1) are established by constructing two well-defined monotone iterative sequences of upper-lower solutions. Finally, an example is given in this section as an application of the theoretical results. Some lengthy proofs of the compactness conclusions used in Theorem 3.1 are settled in the Appendix.

2 Preliminaries and existence results for linear fractional p-Laplacian problems

Let $J=[0,T]$ be a compact interval on the real axis $\mathbb {R}$. It is well known that $C[0,T]$ is a Banach space of continuous functions from $[0,T]$ into $\mathbb{R}$ with the norm $\|u\|_{C}=\max_{t\in[0,T]}|u(t)|$. Denote

$$C_{1-\alpha}[0,T]= \bigl\{ u\in C(0,T]: t^{1-\alpha}u\in C[0,T] \bigr\} , \quad \alpha\in(0,1]. $$

Then $C_{1-\alpha}[0,T]$ is also a Banach space with the norm $\|u\| _{C_{1-\alpha}}=\|t^{1-\alpha}u\|_{C}$ (see Lemma 2.2). It is clear that $C[0,T]:=C_{0}[0,T]\subset C_{1-\alpha}[0,T]\subset C_{1-\beta}[0,T]$ with $\|u\|_{C_{1-\beta}}\le\|u\|_{C_{1-\alpha }}\le\|u\|_{C}$ for $1\ge\alpha\ge\beta>0$ and $C_{1-\alpha}[0,T]\subset L[0,T]$ ($L[0,T]$ is the space of Lebesgue-integrable real functions on $[0,T]$). Denote

$$\begin{aligned} C_{r}^{\alpha}[0,T] =& \bigl\{ u(t)\in C_{1-\alpha}[0,T]: \bigl(D_{0^{+}}^{\alpha }u\bigr) (t)\in C_{r}[0,T] \text{ and} \\ &t^{r}{D}_{0^{+}}^{\alpha}u(t)|_{t=0}=t^{r}{D}_{0^{+}}^{\alpha }u(t)|_{t=T} \bigr\} , \end{aligned}$$

where $r=\frac{1-\beta}{p-1}$, $p>1$, $0<\alpha, \beta\le1$, and $p+\beta>2$.

For convenience, we first present some useful definitions and fundamental facts of fractional calculus theory, some of which can be found in [1, 2].

Definition 2.1

([1])

The Riemann-Liouville fractional integral $I^{\alpha}_{0^{+}}$ and fractional derivative $D^{\alpha}_{0^{+}}$ are defined by

$$I^{\alpha}_{0^{+}}f(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}f(s)\,ds $$

and

$$D^{\alpha}_{0^{+}}f(t)=\frac{1}{\Gamma(n-\alpha)} \biggl(\frac {d}{dt} \biggr)^{n} \int_{0}^{t}(t-s)^{n-\alpha-1}f(s)\,ds= \biggl( \frac {d}{dt} \biggr)^{n}\bigl(I^{n-\alpha}_{0^{+}}f \bigr) (t), $$

where $n-1<\alpha\leq n$, $n \in\mathbb{N}$, provided that the integrals exist.

Lemma 2.1

([1])

Assume that $f\in C(0,T]\cap L(0,T]$ with a fractional derivative of order α ($0<\alpha\le1$) that belongs to $C(0,T]\cap L(0,T]$. Then

$$I^{\alpha}_{0^{+}} D^{\alpha}_{0^{+}} f(t)=f(t)-ct^{\alpha-1} \quad \textit{for some } c\in\mathbb{R}. $$

Lemma 2.2

$(C_{1-\alpha}[0,T], \|\cdot\|_{C_{1-\alpha}})$ and $(C_{r}^{\alpha}[0,T], \|\cdot\|_{C_{r}^{\alpha}})$ are Banach spaces, where

$$\Vert u\Vert _{C_{1-\alpha}}=\bigl\Vert t^{1-\alpha}u\bigr\Vert _{C}, \qquad \Vert u\Vert _{C_{r}^{\alpha }}=\Vert u\Vert _{C_{1-\alpha}}+\bigl\Vert D_{0^{+}}^{\alpha}u\bigr\Vert _{C_{r}}. $$

Proof

Let $\{u_{n}\}_{n=1}^{\infty}$ be a Cauchy sequence in the space $(C_{1-\alpha}[0,T], \|\cdot\|_{C_{1-\alpha}})$. Then there exist $v_{n}\in C[0,T]$ such that $v_{n}(t)=t^{1-\alpha}u_{n}(t)$, $t\in[0,T]$, and thus $u_{n}(t)=t^{\alpha-1}v_{n}(t)$, $t\in(0,T]$. For any $\varepsilon>0$, there exists $N>0$ such that

$$\|u_{n}-u_{m}\|_{C_{1-\alpha}}=\|v_{n}-v_{m} \|_{C}< \varepsilon,\quad n, m\geq N, $$

which implies that there exists $v(t)\in C[0,T]$ such that $v_{n}(t)\rightarrow v(t)$, $t\in[0,T]$, and so $u_{n}(t)=t^{\alpha-1}v_{n}(t)\rightarrow t^{\alpha-1}v(t)$, $t\in(0,T]$. Let $u(t)=t^{\alpha-1}v(t)$, $t\in(0,T]$. Then $\{t^{1-\alpha}u_{n}(t)\}_{n=1}^{\infty}$ converges uniformly to $t^{1-\alpha}u(t)$, and we can easily find that $u\in C_{1-\alpha}[0,T]$.

Next, we shall prove that $C_{r}^{\alpha}[0,T]$ is a Banach space. It is clear that $\|\cdot\|_{C_{r}^{\alpha}}$ is a norm. Let $\{u_{n}\} _{n=1}^{\infty}$ be a Cauchy sequence in the space $(C_{r}^{\alpha}[0,T], \|\cdot\| _{C_{r}^{\alpha}})$. Evidently, $\{u_{n}\}_{n=1}^{\infty}$ is also a Cauchy sequence in the space $(C_{1-\alpha}[0,T], \|\cdot\| _{C_{1-\alpha}})$; thus, $\lim_{n\rightarrow\infty}t^{1-\alpha}u_{n}(t)=t^{1-\alpha }u(t)$, and $u\in C_{1-\alpha}[0,T]$. Moreover, $\{t^{r}(D_{0^{+}}^{\alpha}u_{n})(t)\}_{n=1}^{\infty}$ converges uniformly to some $w(t)\in C[0,T]$. We need to verify that $w(t)=t^{r}(D_{0^{+}}^{\alpha}u)(t)$, $t\in[0,T]$.

For $\varepsilon=1$, there exists $N>0$ such that $|t^{r}(D_{0^{+}}^{\alpha}u_{n})(t)-w(t)|<1$ for any $t\in[0,T]$ and $n>N$. Denoting

$$M^{*}=\max \Bigl\{ 1+\sup_{t\in[0,T]}\bigl\vert w(t)\bigr\vert , \sup_{t\in[0,T]}\bigl\vert t^{r}\bigl(D_{0^{+}}^{\alpha}u_{i} \bigr) (t)\bigr\vert , i=1,2,\ldots,N \Bigr\} , $$

we have

$$\biggl\vert t^{1-\alpha} \int_{0}^{t}(t-s)^{\alpha-1}s^{-r}s^{r}D_{0^{+}}^{\alpha }u_{n}(s) \,ds\biggr\vert \leq M^{*}t^{1-\alpha} \int_{0}^{t}(t-s)^{\alpha-1}s^{-r}\,ds \leq M^{*}B(\alpha ,1-r)T^{1-r}, $$

where $B(\cdot,\cdot)$ is the Beta function. By Lemma 2.1 we get

$$\begin{aligned} t^{1-\alpha}u_{n}(t)&= t^{1-\alpha}I_{0^{+}}^{\alpha}D_{0^{+}}^{\alpha }u_{n}(t)+c = t^{1-\alpha}\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}D_{0^{+}}^{\alpha}u_{n}(s) \,ds+c \\ &= t^{1-\alpha}\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}s^{-r}s^{r}D_{0^{+}}^{\alpha}u_{n}(s) \,ds+c,\quad t\in[0,T]. \end{aligned}$$

(2.1)

Letting $n\rightarrow\infty$, by the Lebesgue dominated convergence theorem from (2.1) we derive that

$$t^{1-\alpha}u(t)=t^{1-\alpha}\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}s^{-r}w(s) \,ds+c=t^{1-\alpha} I_{0^{+}}^{\alpha}\bigl[t^{-r}w(t) \bigr]+c, \quad t\in[0,T], $$

that is, $u(t)=I_{0^{+}}^{\alpha}[t^{-r}w(t)]+ct^{\alpha-1}$, $t\in (0,T]$, and so $w(t)=t^{r}D_{0^{+}}^{\alpha}u(t)$, $t\in(0,T]$. Obviously, $t^{r}{D}_{0^{+}}^{\alpha}u(t)|_{t=0}=t^{r}{D}_{0^{+}}^{\alpha }u(t)|_{t=T}$; hence, $\|u_{n}-u\|_{C_{r}^{\alpha}}\rightarrow0$, and $u\in C_{r}^{\alpha}$. The proof of the lemma is complete. □

Lemma 2.3

([24], Lemma 1.1)

Assume that $0<\beta\leq1$, $M>0$ is a constant, $u(t)\in C_{1-\beta }[0,T]$, and $h(t)\in C_{1-\beta}[0,T]$. Then the linear fractional periodic boundary value problem

$$\left \{ \textstyle\begin{array}{l} D^{\beta}_{0^{+}}u(t)+Mu(t)=h(t), \quad t\in(0,T], \\ t^{1-\beta}u(t)|_{t=0}=t^{1-\beta}u(t)|_{t=T}, \end{array}\displaystyle \right . $$

has the following integral representation of the solution:

$$\begin{aligned} u(t) =& \frac{\Gamma(\beta)T^{1-\beta}t^{\beta-1}E_{\beta,\beta }(-Mt^{\beta})}{1-\Gamma(\beta)E_{\beta,\beta}(-MT^{\beta})} \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)h(s)\,ds \\ &{}+ \int_{0}^{t}(t-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(t-s)^{\beta}\bigr)h(s)\,ds, \end{aligned}$$

where $E_{\beta,\beta}(x)=\sum_{k=0}^{\infty}\frac{x^{k}}{\Gamma (k\beta+\beta)}$ is the Mittag-Leffler function; see [1, 15].

Remark 2.1

Note that $E_{\beta,\beta}(x)>0$ for all $x\in\mathbb{R}$ and $E_{\beta,\beta}(x)<\frac{1}{\Gamma(\beta)}$ for $x<0$ (see [24], Lemma 2.2), so we know that $1-\Gamma(\beta)E_{\beta,\beta }(-MT^{\beta})>0$.

Lemma 2.4

Assume that $0<\alpha, \beta\leq1$, $M>0$ is a constant, $k\in \mathbb{R}$, $u(t)\in C_{r}^{\alpha}[0,T]$, and $\eta(t)\in C_{1-\beta}[0,T]$. Then the linear fractional periodic boundary value problem

$$ \left \{ \textstyle\begin{array}{l} D^{\beta}_{0^{+}}(\phi_{p}(D^{\alpha}_{0^{+}}u(t)))+M\phi_{p}(D^{\alpha}_{0^{+}}u(t))=\eta(t), \quad t\in(0, T], \\ t^{r}D^{\alpha}_{0^{+}}u(t)|_{t=0}=t^{r}D^{\alpha}_{0^{+}}u(t)|_{t=T}, \qquad \tilde{u}(0)=k, \end{array}\displaystyle \right . $$

(2.2)

has a unique solution of the following integral form:

$$\begin{aligned} u(t) = &kt^{\alpha-1}+\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}\phi_{q} \biggl[\frac{\Gamma(\beta)T^{1-\beta }s^{\beta-1}E_{\beta,\beta}(-Ms^{\beta})}{1-\Gamma(\beta)E_{\beta ,\beta}(-MT^{\beta})} \\ &{}\times\int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta (s)\,ds \\ &{}+ \int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta} \bigl(-M(s-\tau )^{\beta}\bigr)\eta(\tau)\,d\tau\biggr]. \end{aligned}$$

(2.3)

Proof

Let $v(t)=\phi_{p}(D^{\alpha}_{0^{+}}u(t))$. Then $\phi_{p}(t^{r}D^{\alpha}_{0^{+}}u(t))=t^{1-\beta}v(t)$ for $0< t\leq T$. Thus, problem (2.2) is changed to the following fractional periodic boundary problem:

$$\left \{ \textstyle\begin{array}{l} D^{\beta}_{0^{+}}v(t)+Mv(t)=\eta(t), \quad t\in(0,T], \\ t^{1-\beta}v(t)|_{t=0}=t^{1-\beta}v(t)|_{t=T}. \end{array}\displaystyle \right . $$

By Lemma 2.3 we get

$$\begin{aligned} v(t) = &\frac{\Gamma(\beta)T^{1-\beta}t^{\beta-1}E_{\beta,\beta }(-Mt^{\beta})}{1-\Gamma(\beta)E_{\beta,\beta}(-MT^{\beta})} \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta(s)\,ds \\ &{}+ \int_{0}^{t}(t-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(t-s)^{\beta}\bigr)\eta(s)\,ds. \end{aligned}$$

(2.4)

Hence, $v(t)\in C_{1-\beta}[0,T]$, and

$$\begin{aligned} D^{\alpha}_{0^{+}}u(t) = &\phi_{q}\biggl[ \frac{\Gamma(\beta)T^{1-\beta }t^{\beta-1}E_{\beta,\beta}(-Mt^{\beta})}{1-\Gamma(\beta)E_{\beta ,\beta}(-MT^{\beta})} \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta (s)\,ds \\ &{}+ \int_{0}^{t}(t-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(t-s)^{\beta}\bigr)\eta(s)\,ds\biggr]. \end{aligned}$$

(2.5)

Since $v(t)\in C(0,T]\cap L(0,T]$, we have $D_{0^{+}}^{\alpha}u(t)\in C(0,T]\cap L(0,T]$. By Lemma 2.1 we arrive at

$$\begin{aligned} u(t) = &ct^{\alpha-1}+\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}\phi_{q} \biggl[\frac{\Gamma(\beta)T^{1-\beta }s^{\beta-1}E_{\beta,\beta}(-Ms^{\beta})}{1-\Gamma(\beta)E_{\beta ,\beta}(-MT^{\beta})} \\ &{}\times \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta (s)\,ds \\ &{}+ \int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta} \bigl(-M(s-\tau )^{\beta}\bigr)\eta(\tau)\,d\tau\biggr]\,ds. \end{aligned}$$

In view of $\tilde{u}(0)=k$, we find $c=k$ and

$$\begin{aligned} u(t) = &kt^{\alpha-1}+\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1} \phi_{q}\biggl[\frac{\Gamma(\beta)T^{1-\beta}s^{\beta-1}E_{\beta ,\beta}(-Ms^{\beta})}{1-\Gamma(\beta)E_{\beta,\beta}(-MT^{\beta})} \\ &{}\times\int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta (s)\,ds \\ &{}+ \int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta} \bigl(-M(s-\tau )^{\beta}\bigr)\eta(\tau)\,d\tau\biggr]\,ds. \end{aligned}$$

(2.6)

Conversely, it is obvious that $u(t)\in C_{1-\alpha}[0,T]$ and $\tilde{u}(0)=k$. Note that $D_{0^{+}}^{\alpha}t^{\alpha-1}=0$ and $D_{0^{+}}^{\alpha}I^{\alpha}u=u$ for all $u\in C(0,T]\cap L(0,T]$. Differentiating (2.6) with order α, we get (2.5). Since $\eta(t)\in C_{1-\beta}[0,T]$, we have $\phi _{p}(D_{0^{+}}^{\alpha}u(t))\in C_{1-\beta}[0,T]$ and $D_{0^{+}}^{\alpha}u(t)\in C_{r}[0,T]$. By (2.4) we see that

$$\begin{aligned} t^{1-\beta}v(t) = &\frac{\Gamma(\beta)T^{1-\beta}E_{\beta,\beta }(-Mt^{\beta})}{1-\Gamma(\beta)E_{\beta,\beta}(-MT^{\beta})} \\ &{}\times\int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta(s)\,ds \\ &{}+t^{1-\beta} \int_{0}^{t}(t-s)^{\beta-1}E_{\beta,\beta } \bigl(-M(t-s)^{\beta}\bigr)\eta(s)\,ds \end{aligned}$$

and

$$\begin{aligned} t^{1-\beta}v(t)|_{t=0} =&t^{1-\beta}v(t)|_{t=T} \\ =& \frac{T^{1-\beta }}{1-\Gamma(\beta)E_{\beta,\beta}(-MT^{\beta})} \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta(s)\,ds. \end{aligned}$$

Thus, $t^{r}D_{0^{+}}^{\alpha}u(t)|_{t=0}=t^{r}D_{0^{+}}^{\alpha}u(t)|_{t=T}$. Differentiating (2.4) with order β, by Lemma 2.3 we obtain

$$D_{0^{+}}^{\beta}\bigl(\phi_{p}\bigl(D_{0^{+}}^{\alpha}u(t) \bigr)\bigr)+M\phi _{p}\bigl(D_{0^{+}}^{\alpha}u(t)\bigr)= \eta(t). $$

This completes the proof. □

Lemma 2.5

(Comparison result)

If $u(t)\in C_{r}^{\alpha}[0,T]$ and satisfies

$$\left \{ \textstyle\begin{array}{l} D^{\beta}_{0^{+}}(\phi_{p}(D^{\alpha}_{0^{+}}u(t)))+M\phi_{p}(D^{\alpha}_{0^{+}}u(t))\geq0, \quad t\in(0, T], \\ t^{r}D^{\alpha}_{0^{+}}u(t)|_{t=0}=t^{r}D^{\alpha}_{0^{+}}u(t)|_{t=T}, \\ \tilde{u}(0)\geq0, \end{array}\displaystyle \right . $$

where $M>0$ is a constant, then $D^{\alpha}_{0^{+}}u(t)\ge0$ and $u(t)\ge0$ for $t\in(0,T]$.

Proof

Let $w(t)=\phi_{p}(D^{\alpha}_{0^{+}}u(t))$. Then $w(t)\in C_{1-\beta}[0,T]$ and satisfies

$$\left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}w(t)+Mw(t)\geq0,\quad t\in(0,T], \\ t^{1-\beta}w(t)|_{t=0}= t^{1-\beta}w(t)|_{t=T}, \end{array}\displaystyle \right . $$

and hence $w(t)\ge0$ for $t\in(0,T]$ by Lemma 2.3 and Remark 2.1. Since $\phi_{p}(x)$ is nondecreasing, $u(t)\in C_{r}^{\alpha}[0,T]$ satisfies

$$\left \{ \textstyle\begin{array}{l} D^{\alpha}_{0^{+}}u(t)\ge0, \quad t\in(0,T], \\ \tilde{u}(0)\geq0, \end{array}\displaystyle \right . $$

and so we get $u(t)\ge0$, $t\in(0,T]$, by (2.5) and (2.6). This lemma is complete. □

Remark 2.2

In fact, from the above proof, we can see that Lemma 2.5 unifies and includes two separate comparison results, which are applied to the next Theorem 3.1 directly.

3 Main results

We first introduce the definition of a pair of lower and upper solutions for using the monotone iterative method.

Definition 3.1

A function $u(t)\in C_{r}^{\alpha }[0,T]$ is called a lower solution of problem (1.1) if it satisfies

$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}u(t)))\le f(t,u(t),D_{0^{+}}^{\alpha}u(t)), \quad t\in(0,T], \\ t^{r}{D}_{0^{+}}^{\alpha}u(t)|_{t=0}= t^{r}{D}_{0^{+}}^{\alpha}u(t)|_{t=T},\qquad g(\tilde{u}(0),\tilde {u}(T))\geq0. \end{array}\displaystyle \right . $$

(3.1)

A function $v(t)\in C_{r}^{\alpha}[0,T]$ is called an upper solution of problem (1.1) if it satisfies

$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}v(t)))\ge f(t,v(t),D_{0^{+}}^{\alpha}v(t)), \quad t\in(0,T], \\ t^{r}{D}_{0^{+}}^{\alpha}v(t)|_{t=0}= t^{r}{D}_{0^{+}}^{\alpha}v(t)|_{t=T}, \qquad g(\tilde{v}(0),\tilde {v}(T))\leq0. \end{array}\displaystyle \right . $$

(3.2)

For our main results, we need the following assumptions.

(H₁):

Assume that $u_{0}, v_{0}\in C_{r}^{\alpha}[0,T]$ are lower and upper solutions of problem (1.1), respectively, and $u_{0}(t)\le v_{0}(t)$, $t\in(0,T]$.

(H₂):

There exists a constant $M>0$ such that

$$f\bigl(t,u(t),D_{0^{+}}^{\alpha}u(t)\bigr)-f\bigl(t,v(t),D_{0^{+}}^{\alpha}v(t) \bigr)\le M \bigl[\phi_{p}\bigl(D_{0^{+}}^{\alpha}v(t) \bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha }u(t)\bigr) \bigr] $$

for $u_{0}(t)\le u(t)\le v(t)\le v_{0}(t)$, $D_{0^{+}}^{\alpha}u_{0}(t)\le D_{0^{+}}^{\alpha}u(t)\le D_{0^{+}}^{\alpha}v(t)\le D_{0^{+}}^{\alpha }v_{0}(t)$, $t\in(0,T]$.

(H₃):

There exist constants $\lambda>0$ and $\mu\ge0$ such that

$$g(x_{1},y_{1})-g(x_{2},y_{2})\le \lambda(x_{2}-x_{1})-\mu(y_{2}-y_{1}) $$

for $\tilde{u}_{0}(0)\le x_{1}\le x_{2}\le\tilde{v}_{0}(0)$ and $\tilde{u}_{0}(T)\le y_{1}\le y_{2}\le\tilde{v}_{0}(T)$.

Theorem 3.1

Suppose that $f\in C([0,T]\times\mathbb{R}\times\mathbb {R}, \mathbb{R})$, $g\in C(\mathbb{R}\times\mathbb{R},\mathbb{R})$, and (H₁), (H₂), and (H₃) hold. Then there exist sequences $\{ u_{n}(t)\}, \{v_{n}(t)\}\subset C_{r}^{\alpha}[0,T]$ such that $\lim_{n\to\infty}u_{n}=x$, $\lim_{n\to\infty }v_{n}=y$ on $(0,T]$ and x, y are minimal and maximal solutions on the interval $[u_{0},v_{0}]$ of problem (1.1), respectively, where

$$[u_{0}, v_{0}]= \bigl\{ u\in C_{r}^{\alpha}[0,T]: u_{0}(t)\le u(t)\le v_{0}(t), t\in(0,T], \tilde{u}_{0}(0)\le\tilde{u}(0)\le\tilde{v}_{0}(0) \bigr\} , $$

that is, for any solution $u\in[u_{0},v_{0}]$,

$$u_{0}\le u_{1}\le\cdots\le u_{n}\le\cdots\le x \le u\le y\le\cdots\le v_{n}\le\cdots\le v_{1}\le v_{0}. $$

Moreover, we have

$$\begin{aligned} D_{0^{+}}^{\alpha}u_{0} \le& D_{0^{+}}^{\alpha}u_{1} \le\cdots\le D_{0^{+}}^{\alpha }u_{n}\le\cdots\le D_{0^{+}}^{\alpha}x \\ \le& D_{0^{+}}^{\alpha}u\le D_{0^{+}}^{\alpha}y\le\cdots\le D_{0^{+}}^{\alpha}v_{n} \le\cdots\le D_{0^{+}}^{\alpha}v_{1}\le D_{0^{+}}^{\alpha}v_{0}. \end{aligned}$$

Proof

Let $F(u(t)):=f(t,u(t),D_{0^{+}}^{\alpha}u(t))$. For $n=1,2,\ldots$ , we define

$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}u_{n}(t)))+M\phi _{p}(D_{0^{+}}^{\alpha}u_{n}(t)) \\ \quad =F(u_{n-1}(t))+M\phi_{p}(D_{0^{+}}^{\alpha }u_{n-1}(t)), \quad t\in (0,T], \\ t^{r}D_{0^{+}}^{\alpha}u_{n}(t)|_{t=0}=t^{r}D_{0^{+}}^{\alpha }u_{n}(t)|_{t=T}, \\ \tilde{u}_{n}(0)=\tilde{u}_{n-1}(0)+\frac{1}{\lambda }g(\tilde{u}_{n-1}(0),\tilde{u}_{n-1}(T)), \end{array}\displaystyle \right . $$

(3.3)

and

$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha}v_{n}(t)))+M\phi _{p}(D_{0^{+}}^{\alpha}v_{n}(t)) \\ \quad =F(v_{n-1}(t))+M\phi_{p}(D_{0^{+}}^{\alpha}v_{n-1}(t)),\quad t\in(0,T], \\ t^{r}D_{0^{+}}^{\alpha}v_{n}(t)|_{t=0}=t^{r}D_{0^{+}}^{\alpha }v_{n}(t)|_{t=T}, \\ \tilde{v}_{n}(0)=\tilde{v}_{n-1}(0)+\frac{1}{\lambda }g(\tilde{v}_{n-1}(0),\tilde{v}_{n-1}(T)). \end{array}\displaystyle \right . $$

(3.4)

Since $u_{0}, v_{0}\in C_{r}^{\alpha}[0,T]$, we know that $D_{0^{+}}^{\alpha }u_{0}(t), D_{0^{+}}^{\alpha}v_{0}(t)\in C_{r}[0,T]$, and so $F(u_{0}(t))+\phi_{p}(D_{0^{+}}^{\alpha}u_{0}(t)), F(v_{0}(t))+\phi _{p}(D_{0^{+}}^{\alpha}v_{0}(t))\in C_{1-\beta}[0,T]$. In view of Lemma 2.4, the functions $u_{1}$ and $v_{1}$ are well defined in the space $C_{r}^{\alpha}[0,T]$. By induction, we can infer that $u_{n}$ and $v_{n}$ are well defined in the space $C_{r}^{\alpha}[0,T]$.

First, we prove that $u_{0}(t)\le u_{1}(t)\le v_{1}(t)\le v_{0}(t)$, $t\in (0,T]$, and $D_{0^{+}}^{\alpha}u_{0}(t)\le D_{0^{+}}^{\alpha}u_{1}(t)\le D_{0^{+}}^{\alpha}v_{1}(t)\le D_{0^{+}}^{\alpha}v_{0}(t)$, $t\in(0,T]$. Let $\delta(t):=\phi_{p}(D_{0^{+}}^{\alpha}u_{1}(t))-\phi_{p}(D_{0^{+}}^{\alpha}u_{0}(t))$. The definition of $u_{1}$ and the assumption that $u_{0}$ is a lower solution imply that

$$ D_{0^{+}}^{\beta}\delta(t)+M\delta(t)=F\bigl(u_{0}(t) \bigr)-D_{0^{+}}^{\beta}\bigl(\phi _{p} \bigl(D_{0^{+}}^{\alpha}u_{0}(t)\bigr)\bigr)\ge0 $$

and $t^{1-\beta}\delta(t)|_{t=0}=t^{1-\beta}\delta(t)|_{t=T}$, $\tilde{u}_{1}(0)-\tilde{u}_{0}(0)=\frac{1}{\lambda }g(\tilde{u}_{0}(0),\tilde{u}_{0}(T))\ge 0$. Thus, we have $D_{0^{+}}^{\alpha}u_{0}(t)\le D_{0^{+}}^{\alpha}u_{1}(t)$ and $u_{1}(t)\ge u_{0}(t)$, $t \in(0,T]$ by Lemma 2.5.

Using a similar method, we can show that $v_{1}(t)\le v_{0}(t)$ and $D_{0^{+}}^{\alpha}v_{1}(t)\le D_{0^{+}}^{\alpha}v_{0}(t)$ for all $t\in(0,T]$. Now, we put $\xi(t)=\phi_{p}(D_{0^{+}}^{\alpha}v_{1}(t))-\phi _{p}(D_{0^{+}}^{\alpha}u_{1}(t))$. From (3.3), (3.4), and (H₂) we get

$$ D_{0^{+}}^{\beta}\xi(t)+M\xi(t)=F \bigl(v_{0}(t)\bigr)-F\bigl(u_{0}(t)\bigr)+M \bigl[\phi _{p}\bigl(D_{0^{+}}^{\alpha}v_{0}(t)\bigr)- \phi_{p}\bigl(D_{0^{+}}^{\alpha}u_{0}(t)\bigr) \bigr]\ge0 $$

(3.5)

and

$$ t^{1-\beta}\xi(t)|_{t=0}=t^{1-\beta} \xi(t)|_{t=T}. $$

(3.6)

We find, by (H₃) and (H₁), that

$$\begin{aligned} \begin{aligned}[b] \tilde{v}_{1}(0)-\tilde{u}_{1}(0)&= \tilde{v}_{0}(0)+\frac {1}{\lambda}g\bigl(\tilde{v}_{0}(0), \tilde{v}_{0}(T)\bigr) - \biggl[\tilde{u}_{0}(0)+ \frac{1}{\lambda}g\bigl(\tilde {u}_{0}(0),\tilde{u}_{0}(T) \bigr) \biggr] \\ &=\frac{1}{\lambda} \bigl[\lambda\bigl(\tilde{v}_{0}(0)-\tilde {u}_{0}(0)\bigr)+g\bigl(\tilde{v}_{0}(0), \tilde{v}_{0}(T)\bigr)-g\bigl(\tilde {u}_{0}(0), \tilde{u}_{0}(T)\bigr) \bigr] \\ &\ge\frac{\mu}{\lambda} \bigl(\tilde{v_{0}}(T)-\tilde {u}_{0}(T) \bigr)\ge0. \end{aligned} \end{aligned}$$

(3.7)

It follows from (3.5)-(3.7) and Lemma 2.5 that $D_{0^{+}}^{\alpha}v_{1}(t)\ge D_{0^{+}}^{\alpha}u_{1}(t)$ and $v_{1}(t)\ge u_{1}(t)$, $t\in(0,T]$.

Next, we show that $u_{1}$ and $v_{1}$ are lower and upper solutions of problem (1.1), respectively. From (3.3) and assumptions (H₂) and (H₃) we have

$$\begin{aligned} D_{0^{+}}^{\beta}\bigl(\phi_{p}\bigl(D_{0^{+}}^{\alpha }u_{1}(t) \bigr)\bigr) =&F\bigl(u_{0}(t)\bigr)-F\bigl(u_{1}(t)\bigr)+F \bigl(u_{1}(t)\bigr) \\ &{}-M \bigl[\phi_{p}\bigl(D_{0^{+}}^{\alpha }u_{1}(t) \bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha}u_{0}(t) \bigr) \bigr] \\ \le& F\bigl(u_{1}(t)\bigr) \end{aligned}$$

and

$$\begin{aligned} 0&=g\bigl(\tilde{u}_{0}(0),\tilde{u}_{0}(T)\bigr)-g\bigl( \tilde {u}_{1}(0),\tilde{u}_{1}(T)\bigr)+g\bigl( \tilde{u}_{1}(0),\tilde {u}_{1}(T)\bigr)-\lambda \bigl[ \tilde{u}_{1}(0)-\tilde{u}_{0}(0)\bigr] \\ &\le g\bigl(\tilde{u}_{1}(0),\tilde{u}_{1}(T)\bigr)-\mu \bigl(\tilde {u}_{1}(T)-\tilde{u}_{0}(T)\bigr). \end{aligned}$$

Since $\tilde{u}_{1}(T)\ge\tilde{u}_{0}(T)$, the last inequality implies $g(\tilde{u}_{1}(0),\tilde{u}_{1}(T))\ge0$. This proves that $u_{1}$ is a lower solution of problem (1.1). In the same way, we can show that $v_{1}$ is an upper solution of (1.1).

Using mathematical induction, we have

$$\begin{aligned}& u_{0}(t)\le u_{1}(t)\le\cdots \le u_{n}(t)\le u_{n+1}(t)\le v_{n+1}(t)\le v_{n}(t)\le\cdots\le v_{1}(t)\le v_{0}(t), \\& D_{0^{+}}^{\alpha}u_{0}\le D_{0^{+}}^{\alpha}u_{1} \le\cdots\le D_{0^{+}}^{\alpha }u_{n}\le D_{0^{+}}^{\alpha}u_{n+1} \\& \hphantom{D_{0^{+}}^{\alpha}u_{0}}\le D_{0^{+}}^{\alpha}v_{n+1} \le D_{0^{+}}^{\alpha}v_{n}\le\cdots\le D_{0^{+}}^{\alpha}v_{1} \le D_{0^{+}}^{\alpha}v_{0} \end{aligned}$$

(3.8)

for $t\in(0,T]$ and $n=1,2,3,\ldots$ .

The sequences $\{t^{1-\alpha}u_{n}\}$ and $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ are uniformly bounded and equicontinuous (see Lemma A.1 in the Appendix). Similarly, we can prove that the sequences $\{t^{1-\alpha}v_{n}\}$ and $\{t^{r}D_{0^{+}}^{\alpha}v_{n}\}$ are uniformly bounded and equicontinuous. The Arzelà-Ascoli theorem guarantees that $\{t^{1-\alpha}u_{n}\}$ and $\{t^{1-\alpha}v_{n}\}$ converge to $t^{1-\alpha}x(t)$ and $t^{1-\alpha}y(t)$ uniformly on $[0,T]$, respectively, and $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ and $\{t^{r}D_{0^{+}}^{\alpha}v_{n}\}$ converge to $\{t^{r}D_{0^{+}}^{\alpha }x(t)\}$ and $\{t^{r}D_{0^{+}}^{\alpha}y(t)\}$ uniformly on $[0,T]$, respectively. Therefore, $\|u_{n}-x\|_{C_{r}^{\alpha}}\rightarrow0$, $\|v_{n}-y\| _{C_{r}^{\alpha}}\rightarrow0$ ($n\rightarrow\infty$).

By the integral representation (2.3) for the linear fractional problem, the solution $u_{n}(t)$ of problem (3.3) can be expressed as

$$\begin{aligned} u_{n}(t) = &t^{\alpha-1} \biggl[\tilde{u}_{n-1}(0)+ \frac{1}{\lambda }g\bigl(\tilde{u}_{n-1}(0),\tilde{u}_{n-1}(T) \bigr) \biggr] \\ &{}+\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}\phi_{q} \biggl[M_{0}s^{\beta-1}E_{\beta,\beta}\bigl(-Ms^{\beta}\bigr) \\ &{}\times\int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta _{n-1}(s)\,ds \\ &{}+ \int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta} \bigl(-M(s-\tau )^{\beta}\bigr)\eta_{n-1}(\tau)\,d\tau\biggr], \quad t \in(0,T], \end{aligned}$$

where $\eta_{n-1}(s)=F(u_{n-1}(s))+M\phi_{p}(D^{\alpha}_{0^{+}}u_{n-1}(s))$ and

$$ M_{0}:=\frac{\Gamma(\beta)T^{1-\beta}}{1-\Gamma(\beta)E_{\beta ,\beta}(-MT^{\beta})}. $$

(3.9)

By the assumption on f, applying the dominated convergence theorem, we get that $x(t)$ satisfies the following integral equation:

$$\begin{aligned} x(t) = &t^{\alpha-1}\tilde{x}(0)+\frac{1}{\Gamma(\alpha)} \int _{0}^{t}(t-s)^{\alpha-1}\phi_{q} \biggl[M_{0}s^{\beta-1}E_{\beta,\beta }\bigl(-Ms^{\beta}\bigr) \\ &{}\times\int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta (s)\,ds \\ &{}+ \int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta} \bigl(-M(s-\tau )^{\beta}\bigr)\eta(\tau)\,d\tau\biggr],\quad t \in(0,T], \end{aligned}$$

where $\eta(s)=F(x(s))+M\phi_{p}(D^{\alpha}_{0^{+}}x(s))$. By Lemma 2.4 we have that $x(t)$ is a solution of problem (1.1). Meanwhile, $y(t)$ is also a solution of problem (1.1) and satisfies $u_{0}\le x\le y\le v_{0}$ on $(0,T]$.

To prove that $x(t)$ and $y(t)$ are extremal solutions of (1.1), let $u\in[u_{0},v_{0}]$ be any solution of problem (1.1). We suppose that $u_{n}\le u\le v_{n}$, $t \in(0,T]$, for some n. Let $\zeta(t)=\phi_{p}(D_{0^{+}}^{\alpha}u(t))-\phi_{p}(D_{0^{+}}^{\alpha }u_{n+1}(t))$, $\eta(t)=\phi_{p}(D_{0^{+}}^{\alpha}v_{n+1}(t))-\phi _{p}(D_{0^{+}}^{\alpha}u(t))$. Thus, by condition (H₂) we have

$$D_{0^{+}}^{\beta}\zeta(t)+M\zeta(t)=F\bigl(u(t)\bigr)-F \bigl(u_{n}(t)\bigr)+M\bigl[\phi _{p}\bigl(D_{0^{+}}^{\alpha}u \bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha}u_{n} \bigr)\bigr]\ge0 $$

and

$$D_{0^{+}}^{\beta}\eta(t)+M\eta(t)=F\bigl(v_{n}(t) \bigr)-F\bigl(u(t)\bigr)+M\bigl[\phi _{p}\bigl(D_{0^{+}}^{\alpha}v_{n} \bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha}v\bigr)\bigr]\ge0. $$

Moreover, from condition (H₃) we find

$$\begin{aligned} \tilde{u}(0)-\tilde{u}_{n+1}(0) &=\frac{1}{\lambda} \bigl[\lambda \tilde{u}(0)+g\bigl(\tilde {u}(0),\tilde{u}(T)\bigr)-\bigl(\lambda \tilde{u}_{n}(0)+g\bigl(\tilde {u}_{n}(0),\tilde {u}_{n}(T)\bigr)\bigr) \bigr] \\ &\ge\frac{\mu}{\lambda} \bigl(\tilde{u}(T)-\tilde {u}_{n}(T) \bigr)\ge0 \end{aligned}$$

and

$$\begin{aligned} \tilde{v}_{n+1}(0)-\tilde{u}(0) &=\frac{1}{\lambda} \bigl[\lambda \tilde{v}_{n}(0)+g\bigl(\tilde {u}(0),\tilde{u}(T)\bigr)-\bigl(\lambda \tilde{u}(0)+g\bigl(\tilde {u}_{n}(0),\tilde{u}_{n+1}(T)\bigr) \bigr) \bigr] \\ &\ge\frac{\mu}{\lambda} \bigl(\tilde{v}_{n+1}(T)-\tilde {u}(T) \bigr) \ge0. \end{aligned}$$

These inequalities and Lemma 2.5 imply that $D_{0^{+}}^{\alpha}u_{n+1}(t)\le D_{0^{+}}^{\alpha}u(t)\le D_{0^{+}}^{\alpha}v_{n+1}(t)$ and $u_{n+1}(t)\le u(t)\le v_{n+1}(t)$, $t\in(0,T]$, so by induction $x(t)\le u(t)\le y(t)$ and $D_{0^{+}}^{\alpha}x\le D_{0^{+}}^{\alpha }u\le D_{0^{+}}^{\alpha}y$ on $(0,T]$ by taking the limits as $n\to \infty$. This finishes the proof. □

Remark 3.1

In Definition 3.1, we also can use $g(\tilde{u}(0),\tilde{u}(T))\le0$ instead of $g(\tilde{u}(0),\tilde{u}(T))\geq0$ to define the lower solution of problem (1.1) and use $g(\tilde{v}(0),\tilde{v}(T))\geq0$ instead of $g(\tilde {v}(0),\tilde{v}(T))\leq0$ to define the upper solution of problem (1.1), with the remaining conditions unchanged. However, the conclusions of Theorem 3.1 hold under assumptions (H₁), (H₂), and

($\mathrm{H}_{3}'$):

there exist constants $\lambda'>0$, $\mu' \ge0$ such that

$$g(x_{1},y_{1})-g(x_{2},y_{2})\ge- \lambda'(x_{2}-x_{1})+\mu'(y_{2}-y_{1}) $$

for $\tilde{u}_{0}(0)\le x_{1}\le x_{2}\le\tilde{v}_{0}(0)$ and $\tilde{u}_{0}(T)\le y_{1}\le y_{2}\le\tilde{v}_{0}(T)$. Meanwhile, in the proof, we need to transform the definitions of $\tilde{u}_{n}(0)$ and $\tilde{v}_{n}(0)$ in (3.3) and (3.4) into the forms

$$\tilde{u}_{n}(0)=\tilde{u}_{n-1}(0)-\frac{1}{\lambda '}g\bigl( \tilde{u}_{n-1}(0),\tilde{u}_{n-1}(T)\bigr), \qquad \tilde{v}_{n}(0)=\tilde{v}_{n-1}(0)-\frac{1}{\lambda '}g\bigl( \tilde{v}_{n-1}(0),\tilde{v}_{n-1}(T)\bigr) $$

and make the corresponding modification in view of ($\mathrm{H}_{3}'$).

Theorem 3.2

The assumptions of Theorem  3.1 hold, and there exists a constant $N>0$ such that

$$ N \bigl[\phi_{p}\bigl(D_{0^{+}}^{\alpha}v(t) \bigr)-\phi_{p}\bigl(D_{0^{+}}^{\alpha }u(t)\bigr) \bigr] \le f\bigl(t,u(t),D_{0^{+}}^{\alpha }u(t)\bigr)-f\bigl(t,v(t),D_{0^{+}}^{\alpha}v(t) \bigr) $$

(3.10)

for $u_{0}(t)\le u(t)\le v(t)\le v_{0}(t)$, $D_{0^{+}}^{\alpha}u_{0}(t)\le D_{0^{+}}^{\alpha}u(t)\le D_{0^{+}}^{\alpha}v(t)\le D_{0^{+}}^{\alpha}v_{0}(t)$, $t\in(0,T]$, and $\tilde {u}_{0}(0)=\tilde{v}_{0}(0)$. Then problem (1.1) has a unique solution in the order interval $[u_{0}, v_{0}]$.

Proof

By Theorem 3.1 we see that $x(t)$ and $y(t)$ are extremal solutions and $x(t)\le y(t)$, $t\in(0,T]$. In order to prove that $x(t)\ge y(t)$, $t\in(0,T]$, we let $w(t)=\phi_{p}(D_{0^{+}}^{\alpha}x(t))-\phi_{p}(D_{0^{+}}^{\alpha}y(t))$, $t\in(0,T]$. From (3.10) we arrive at

$$\left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}w(t)=F(x(t))-F(y(t))\ge N[\phi_{p}(D_{0^{+}}^{\alpha}y(t))-\phi_{p}(D_{0^{+}}^{\alpha }x(t))]=-Nw(t), \\ t^{1-\beta}w(t)|_{t=0}=t^{1-\beta}w(t)|_{t=T}. \end{array}\displaystyle \right . $$

Then $w(t)\ge0$, $t\in(0,T]$, that is, $D_{0^{+}}^{\alpha}x(t)\ge D_{0^{+}}^{\alpha}y(t)$, $t\in(0,T]$. Also, by (3.8), since $\tilde{u}_{0}(0)=\tilde{v}_{0}(0)$, we have $\tilde{x}(0)=\tilde{y}(0)$. Therefore, Lemma 2.5 implies $x(t)\ge y(t)$, $t\in(0,T]$. Thus, we obtain $x=y$. The proof is complete. □

Example 3.1

Consider the following fractional periodic boundary value problem:

$$ \left \{ \textstyle\begin{array}{l} D_{0^{+}}^{\beta}(\phi_{p}(D_{0^{+}}^{\alpha }u(t)))=t^{1/2}(1-t)-2[D_{0^{+}}^{\alpha}u(t)]^{2}+u(t), \quad t\in(0,1], \\ t^{1/6}{D}_{0^{+}}^{\alpha}u(t)|_{t=0}=t^{1/6}{D}_{0^{+}}^{\alpha }u(t)|_{t=1}, \\ \tilde{u}(0) (\frac{\Gamma(5/6)}{2\Gamma(4/3)}-\tilde {u}(1) )=0, \end{array}\displaystyle \right . $$

(3.11)

where $\alpha=1/2$, $\beta=2/3$, $p=3$, $T=1$, and $f(t,u,D_{0^{+}}^{\alpha}u)=t^{1/2}(1-t)-2[D_{0^{+}}^{\alpha }u(t)]^{2}+u(t)$, $g(x,y)=x (\frac{\Gamma(5/6)}{2\Gamma (4/3)}-y )$. Set

$$u_{0}(t)\equiv0, \qquad v_{0}(t)=\frac{\Gamma(5/6)}{\Gamma(4/3)}t^{1/3}, \quad t\in[0,1]. $$

It is easy to verify that $D_{0^{+}}^{1/2}u_{0}(t)\equiv0$ and $D_{0^{+}}^{1/2}v_{0}(t)=t^{-1/6}$ for $t\in(0,1]$ and

$$\begin{aligned}& t^{1/6}D_{0^{+}}^{1/2}u_{0}(t)|_{t=0}=0=t^{1/6}D_{0^{+}}^{1/2}u_{0}(t)|_{t=1}, \qquad t^{1/6}D_{0^{+}}^{1/2}v_{0}(t)|_{t=0}=1=t^{1/6}D_{0^{+}}^{1/2}v_{0}(t)|_{t=1}, \\& D_{0^{+}}^{2/3}\bigl(\phi_{3}\bigl(D_{0^{+}}^{1/2}u_{0}(t) \bigr)\bigr)\equiv0\le f\bigl(t,u_{0},D_{0^{+}}^{1/2}u_{0} \bigr)=t^{1/2}(1-t), \\& D_{0^{+}}^{2/3}\bigl(\phi _{3}\bigl(D_{0^{+}}^{1/2}v_{0}(t) \bigr)\bigr)=D_{0^{+}}^{2/3}\bigl(t^{-1/3}\bigr)=0\ge f \bigl(t,v_{0},D_{0^{+}}^{1/2}v_{0} \bigr) \\& \hphantom{D_{0^{+}}^{2/3}\bigl(\phi _{3}\bigl(D_{0^{+}}^{1/2}v_{0}(t) \bigr)\bigr)}=t^{1/2}(1-t)-2t^{-1/3}+\frac{\Gamma (5/6)}{\Gamma(4/3)}t^{1/3}, \\& g\bigl(\tilde{u}_{0}(0),\tilde{u}_{0}(1)\bigr)=0, \qquad g \bigl(\tilde {v}_{0}(0),\tilde{v}_{0}(1)\bigr)=0. \end{aligned}$$

These show that $u_{0}$ and $v_{0}$ are the lower and upper solutions of (3.11), respectively, and $u_{0}(t)\le v_{0}(t)$ on $[0,1]$.

For $u_{0} \le u\le v\le v_{0}$, we have $\phi_{3}(D_{0^{+}}^{1/2}v)-\phi _{3}(D_{0^{+}}^{1/2}u)=(D_{0^{+}}^{1/2}v)^{2}-(D_{0^{+}}^{1/2}u)^{2}$ and

$$f\bigl(t,u,D_{0^{+}}^{1/2}u\bigr)+2\phi _{3} \bigl(D_{0^{+}}^{1/2}u\bigr)-\bigl[f\bigl(t,v,D_{0^{+}}^{1/2}v \bigr)+2\phi _{3}\bigl(D_{0^{+}}^{1/2}v\bigr)\bigr]=u-v \le0. $$

Thus, $f(t,u, D_{0^{+}}^{1/2}u)-f(t,v, D_{0^{+}}^{1/2}v)\le M[\phi _{3}(D_{0^{+}}^{1/2}v)-\phi_{3}(D_{0^{+}}^{1/2}u)]$, where $M=2$.

In addition, $\frac{\partial g(x,y)}{\partial x}= \frac{\Gamma(5/6)}{2\Gamma(4/3)}-y\geq-\frac{\Gamma(5/6)}{2\Gamma (4/3)}$, $\frac{\partial g(x,y)}{\partial y}=-x$ for $\tilde{u}_{0}(0)\le x\le\tilde{v}_{0}(0)$, $y\in[\tilde {u}_{0}(1),\tilde{v}_{0}(1)]=[0,\frac{\Gamma(5/6)}{\Gamma(4/3)}]$. Therefore, $g(u_{1},v_{1})-g(u_{2},v_{2})\le\frac{\Gamma(5/6)}{2\Gamma(4/3)}(u_{2}-u_{1})$ for $\tilde{u}_{0}(0)\le u_{1}\le u_{2}\le \tilde{v}_{0}(0)$, $\tilde{u}_{0}(1)\le v_{1}\le v_{2}\le\tilde{v}_{0}(1)$. Hence, conditions (H₁), (H₂), and (H₃) are satisfied. There exist two monotone iterative sequences $\{u_{k}\}$ and $\{v_{k}\}$ that converge uniformly to the minimal and maximal solutions of fractional periodic boundary problem (3.11) in $[u_{0},v_{0}]$ by Theorem 3.1.

Appendix

Lemma A.1

The sequences $\{t^{1-\alpha}u_{n}\}$ and $\{t^{r}D^{\alpha}u_{n}\}$ are uniformly bounded and equicontinuous in $C[0,T]$, where $u_{n}$ is defined by (3.3) in Theorem  3.1.

Proof

We first show that $\{t^{1-\alpha}u_{n}\}$ are uniformly bounded in $C[0,T]$. Since $u_{0}, v_{0}\in C_{r}^{\alpha}[0,T]$, we have $\phi_{p}(D^{\alpha}_{0^{+}}v_{0}(t))\in C_{1-\beta}[0,T]$, that is, $t^{1-\beta} \phi_{p}(D^{\alpha}_{0^{+}}v_{0}(t))\in C[0,T]$. Thus, there exists a constant $\gamma_{1}>0$ such that

$$\bigl\Vert t^{1-\beta} \phi_{p}\bigl(D^{\alpha}_{0^{+}}v_{0} \bigr)\bigr\Vert _{C}\leq\gamma_{1}, \quad t\in[0,T], $$

which is equivalent to

$$ \bigl\vert \phi_{p}\bigl(D^{\alpha}_{0^{+}}v_{0}(t) \bigr)\bigr\vert \leq\gamma_{1}t^{\beta-1}, \quad t\in(0,T]. $$

(A.1)

Let

$$ \eta_{n-1}(t)=F\bigl(u_{n-1}(t)\bigr)+M \phi_{p}\bigl(D^{\alpha}_{0^{+}}u_{n-1}(t)\bigr), \quad t\in(0,T]. $$

(A.2)

By condition (H₂) and (A.1) we get

$$\eta_{n-1}(t)\leq F\bigl(v_{0}(t)\bigr)+M\phi_{p} \bigl(D^{\alpha}_{0^{+}}v_{0}(t)\bigr)\leq F \bigl(v_{0}(t)\bigr)+M\gamma_{1}t^{\beta-1}, \quad t \in(0,T]. $$

Hence,

$$ \Vert \eta_{n-1}\Vert _{C_{1-\beta}}\leq\bigl\Vert F(v_{0})\bigr\Vert _{C}T+M\gamma_{1}=:\gamma _{2},\quad t\in[0,T]. $$

(A.3)

Let

$$\begin{aligned} x_{n-1}(s) = &M_{0}s^{\beta-1}E_{\beta,\beta} \bigl(-Ms^{\beta}\bigr) \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta _{n-1}(s)\,ds \\ &{}+ \int_{0}^{s}(s-\tau)^{\beta-1}E_{\beta,\beta} \bigl(-M(s-\tau)^{\beta}\bigr)\eta_{n-1}(\tau)\,d\tau, \end{aligned}$$

(A.4)

where $M_{0}$ is defined in (3.9). Then $x_{n-1}\in C_{1-\beta }[0,T]$. Noting that $E_{\beta,\beta}(x)<\frac{1}{\Gamma(\beta)}$ for $x<0$, by (A.4) and (A.3) we have

$$\begin{aligned} \bigl\vert x_{n-1}(s)\bigr\vert \leq &M_{0}s^{\beta-1}\frac{1}{\Gamma^{2}(\beta)} \int_{0}^{T}(T-s)^{\beta-1}s^{\beta-1}\| \eta_{n-1}\|_{C_{1-\beta }}\,ds \\ &{}+\frac{1}{\Gamma(\beta)} \int_{0}^{s}(s-\tau)^{\beta-1}\tau ^{\beta-1}\|\eta_{n-1}\|_{C_{1-\beta}}\,d\tau \\ \leq & M_{0}s^{\beta-1}\frac{\gamma_{2}}{\Gamma^{2}(\beta)}B(\beta , \beta)T^{2\beta-1}+\frac{\gamma_{2}}{\Gamma(\beta)}B(\beta,\beta )s^{2\beta-1}, \quad s \in(0,T], \end{aligned}$$

(A.5)

which yields

$$\bigl\vert s^{1-\beta}x_{n-1}(s)\bigr\vert \leq M_{0}\frac{\gamma_{2}}{\Gamma^{2}(\beta )}B(\beta,\beta)T^{2\beta-1}+ \frac{\gamma_{2}}{\Gamma(\beta )}B(\beta,\beta)T^{\beta},\quad s\in(0,T]. $$

Thus, for $s\in(0,T]$, we get

$$\begin{aligned} s^{r}\phi_{q}\bigl(x_{n-1}(s)\bigr) =& \phi_{q}\bigl(s^{1-\beta}x_{n-1}(s)\bigr) \\ \leq& \phi_{q} \biggl(M_{0}\frac{\gamma_{2}}{\Gamma^{2}(\beta)}B(\beta,\beta )T^{2\beta-1}+\frac{\gamma_{2}}{\Gamma(\beta)}B(\beta,\beta )T^{\beta} \biggr)=:C. \end{aligned}$$

(A.6)

This implies that $\phi_{q}(x_{n-1}(s))=D^{\alpha}_{0^{+}}u_{n}(s)$ is bounded in $C_{r}[0,T]$. From (3.3) and Lemma 2.4 we find

$$ t^{1-\alpha}u_{n}(t)=\tilde{u}_{n-1}(0)+ \frac{1}{\lambda}g\bigl(\tilde {u}_{n-1}(0),\tilde{u}_{n-1}(T) \bigr) +\frac{t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}\phi _{q} \bigl(x_{n-1}(s)\bigr)\,ds. $$

(A.7)

Using (A.7), (A.6), and condition (H₃), we get

$$\begin{aligned} \bigl\vert t^{1-\alpha}u_{n}(t)\bigr\vert &= \biggl\vert \tilde{u}_{n-1}(0)+\frac{1}{\lambda}g\bigl(\tilde {u}_{n-1}(0), \tilde{u}_{n-1}(T)\bigr)\biggr\vert +\biggl\vert \frac{t^{1-\alpha }}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}\phi _{q} \bigl(x_{n-1}(s)\bigr)\,ds\biggr\vert \\ &\leq \biggl\vert \tilde{v}_{0}(0)+\frac{1}{\lambda}g\bigl(\tilde {v}_{0}(0),\tilde{v}_{0}(T)\bigr)\biggr\vert + \frac{C t^{1-\alpha}}{\Gamma(\alpha )} \int_{0}^{t}(t-s)^{\alpha-1}s^{-r}\,ds \\ &\leq \biggl\vert \tilde{v}_{0}(0)+\frac{1}{\lambda}g\bigl(\tilde {v}_{0}(0),\tilde{v}_{0}(T)\bigr)\biggr\vert + \frac{CB(\alpha,1-r)}{\Gamma(\alpha )}T^{1-r}, \quad t\in[0,T]. \end{aligned}$$

Hence, $\{t^{1-\alpha}u_{n}\}$ are uniformly bounded in $C[0,T]$.

Next, we prove that $\{t^{1-\alpha}u_{n}\}$ are equicontinuous in $C[0,T]$. Suppose $0< t_{1}\leq t_{2}\leq T$. From (A.7) and (A.6) we have

$$\begin{aligned}& \bigl\vert t_{2}^{1-\alpha}u_{n}(t_{2})-t_{1}^{1-\alpha}u_{n}(t_{1}) \bigr\vert \\& \quad = \biggl\vert \frac{t_{2}^{1-\alpha}}{\Gamma(\alpha)} \int _{0}^{t_{2}}(t_{2}-s)^{\alpha-1} \phi_{q}\bigl(x_{n-1}(s)\bigr)\,ds -\frac{t_{1}^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t_{1}}(t_{1}-s)^{\alpha -1} \phi_{q}\bigl(x_{n-1}(s)\bigr)\,ds\biggr\vert \\& \quad \leq \biggl\vert \frac{t_{2}^{1-\alpha}}{\Gamma(\alpha)} \int _{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1} \phi_{q}\bigl(x_{n-1}(s)\bigr)\,ds\biggr\vert \\& \qquad {}+\biggl\vert \frac{t_{2}^{1-\alpha}}{\Gamma(\alpha)} \int _{0}^{t_{1}}\bigl[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1} \bigr]\phi _{q}\bigl(x_{n-1}(s)\bigr)\,ds\biggr\vert \\& \qquad {} +\biggl\vert \frac{t_{2}^{1-\alpha}-t_{1}^{1-\alpha}}{\Gamma(\alpha)} \int _{0}^{t_{1}}(t_{1}-s)^{\alpha-1} \phi_{q}\bigl(x_{n-1}(s)\bigr)\,ds\biggr\vert \\& \quad \leq \frac{Ct_{2}^{1-\alpha}}{\Gamma(\alpha)} \int _{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}s^{-r} \,ds +\frac{Ct_{2}^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t_{1}} \bigl\vert \bigl[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1} \bigr]s^{-r}\bigr\vert \,ds \\& \qquad {} +\frac{C(t_{2}^{1-\alpha}-t_{1}^{1-\alpha})}{\Gamma(\alpha)} \int _{0}^{t_{1}}(t_{1}-s)^{\alpha-1}s^{-r} \,ds \\& \quad =: \mathrm{I}+\mathrm{II}+\mathrm{III}. \end{aligned}$$

For part I, we have

$$\int_{t_{1}}^{t_{2}}t_{2}^{1-\alpha} (t_{2}-s)^{\alpha-1}s^{-r}\,ds \leq t_{2}^{1-\alpha}t_{2}^{\alpha-r}B( \alpha,1-r)=t_{2}^{1-r}B(\alpha ,1-r)< \infty. $$

By the absolute continuity of the integral, we have that I can be sufficiently small when $t_{1}$ is sufficiently close to $t_{2}$. For part II, we have

$$\begin{aligned} \mathrm{II}&=\frac{C}{\Gamma(\alpha)} \biggl[t_{2}^{1-\alpha} \int _{0}^{t_{1}}(t_{1}-s)^{\alpha-1}s^{-r} \,ds-t_{2}^{1-\alpha} \int _{0}^{t_{1}}(t_{2}-s)^{\alpha-1}s^{-r} \,ds \biggr] \\ &=\frac{C}{\Gamma(\alpha)} \biggl[t_{2}^{1-\alpha} \int _{0}^{t_{1}}(t_{1}-s)^{\alpha-1}s^{-r} \,ds-t_{2}^{1-\alpha} \biggl( \int _{0}^{t_{2}}(t_{2}-s)^{\alpha-1}s^{-r} \,ds- \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha -1}s^{-r} \,ds \biggr) \biggr] \\ &=\frac{C}{\Gamma(\alpha)} \biggl( \biggl(\frac{t_{2}}{t_{1}} \biggr)^{1-\alpha} \cdot t_{1}^{\alpha}-t_{2}^{\alpha}\biggr)B( \alpha,1-r)+t_{2}^{1-\alpha} \int _{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}s^{-r} \,ds. \end{aligned}$$

(A.8)

It is easy to see that as $t_{1}$ approaches $t_{2}$, II goes to zero.

For part III, we have

$$\mathrm{III} \leq \frac{C(t_{2}^{1-\alpha}-t_{1}^{1-\alpha})}{\Gamma(\alpha )t_{1}^{1-\alpha}}t_{1}^{1-r}B( \alpha,1-r). $$

Combining the results of I, II, and III, we have that $|t_{2}^{1-\alpha}u_{n}(t_{2})-t_{1}^{1-\alpha}u_{n}(t_{1})|\rightarrow0$ as $t_{1}\rightarrow t_{2}$.

When $t_{1}=0\le t_{2}\le T$, from (A.7) and (A.6) we have

$$\begin{aligned} \bigl\vert t_{2}^{1-\alpha}u_{n}(t_{2})- \tilde{u}_{n}(0)\bigr\vert &=\biggl\vert \frac {t_{2}^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t_{2}}(t_{2}-s)^{\alpha-1}\phi _{q}\bigl(x_{n-1}(s)\bigr)\,ds\biggr\vert \\ &\leq\frac{C}{\Gamma(\alpha)} \int_{0}^{t_{2}}t_{2}^{1-\alpha }(t_{2}-s)^{\alpha-1}s^{-r} \,ds \\ &=\frac{C}{\Gamma(\alpha)}t_{2}^{1-r}B(\alpha,1-r) \\ &\rightarrow0 \quad \mbox{as } t_{2}\rightarrow0. \end{aligned}$$

This shows that $\{t^{1-\alpha}u_{n}\}$ are equicontinuous in $C[0,T]$.

In the following, we will check that $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ are relatively compact in $C[0,T]$.

First, we prove that $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ are uniformly bounded in $C[0,T]$. By (3.3) and (2.5) we get

$$\begin{aligned} D^{\alpha}_{0^{+}}u_{n}(t) =& \phi_{q}\biggl[M_{0}t^{\beta-1}E_{\beta,\beta } \bigl(-Mt^{\beta}\bigr) \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(T-s)^{\beta}\bigr)\eta _{n-1}(s)\,ds \\ &{}+ \int_{0}^{t}(t-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(t-s)^{\beta}\bigr)\eta _{n-1}(s)\,ds\biggr], \end{aligned}$$

(A.9)

where $\eta_{n-1}$ is defined in (A.3). By (A.9) and the definition of $x_{n-1}$ in (A.4) and (A.6), we still have

$$\bigl\vert t^{r}D_{0^{+}}^{\alpha}u_{n}(t) \bigr\vert =\bigl\vert t^{r}\phi_{q}\bigl(x_{n-1}(t) \bigr)\bigr\vert =\bigl\vert \phi _{q}\bigl(t^{1-\beta}x_{n-1}(t) \bigr)\bigr\vert \leq C, \quad t\in(0,T]. $$

Therefore, $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ are uniformly bounded in $C[0,T]$.

Second, we prove that $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ are equicontinuous in $C[0,T]$. Since $\phi_{p}(t^{r}D^{\alpha}_{0^{+}}u(t))=t^{1-\beta}\phi_{p}(D^{\alpha}_{0^{+}}u(t))$, we need to deal with the equicontinuity of $\{t^{1-\beta}\phi_{p}(D_{0^{+}}^{\alpha}u_{n})\}$ in $C[0,T]$. Choosing $0< t_{1}\leq t_{2}\leq T$, by (A.9) and (A.3) we have

$$\begin{aligned}& \bigl\vert t_{2}^{1-\beta}\phi_{p} \bigl(D_{0^{+}}^{\alpha}u_{n}(t_{2}) \bigr)-t_{1}^{1-\beta }\phi_{p}\bigl(D_{0^{+}}^{\alpha}u_{n}(t_{1})\bigr)\bigr\vert \\& \quad = \biggl\vert M_{0} \bigl[E_{\beta,\beta} \bigl(-Mt_{2}^{\beta}\bigr)-E_{\beta,\beta }\bigl(-Mt_{1}^{\beta}\bigr) \bigr] \int_{0}^{T}(T-s)^{\beta-1}E_{\beta,\beta } \bigl(-M(T-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds \\& \qquad {} +t_{2}^{1-\beta} \int_{0}^{t_{2}}(t_{2}-s)^{\beta-1}E_{\beta,\beta } \bigl(-M(t_{2}-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds \\& \qquad {}-t_{1}^{1-\beta} \int_{0}^{t_{1}}(t_{1}-s)^{\beta-1}E_{\beta,\beta } \bigl(-M(t_{1}-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds\biggr\vert \\& \quad \leq M_{0}\bigl\vert E_{\beta,\beta}\bigl(-Mt_{2}^{\beta}\bigr)-E_{\beta,\beta }\bigl(-Mt_{1}^{\beta}\bigr)\bigr\vert \int_{0}^{T}(T-s)^{\beta-1} \frac{1}{\Gamma(\beta )}s^{\beta-1}\|\eta_{n-1}\|_{C_{1-\beta}}\,ds \\& \qquad {} +\biggl\vert t_{2}^{1-\beta} \int_{0}^{t_{2}}(t_{2}-s)^{\beta-1}E_{\beta,\beta } \bigl(-M(t_{2}-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds \\& \qquad {}-t_{1}^{1-\beta} \int_{0}^{t_{1}}(t_{1}-s)^{\beta-1}E_{\beta,\beta } \bigl(-M(t_{1}-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds\biggr\vert \\& \quad \leq t_{2}^{1-\beta}\biggl\vert \int_{0}^{t_{1}} \bigl[(t_{2}-s)^{\beta -1}E_{\beta,\beta} \bigl(-M(t_{2}-s)^{\beta}\bigr)-(t_{1}-s)^{\beta-1}E_{\beta ,\beta} \bigl(-M(t_{1}-s)^{\beta}\bigr) \bigr]\eta_{n-1}(s)\,ds \biggr\vert \\& \qquad {} +\biggl\vert \bigl(t_{2}^{1-\beta}-t_{1}^{1-\beta} \bigr) \int_{0}^{t_{1}}(t_{1}-s)^{\beta -1}E_{\beta,\beta} \bigl(-M(t_{1}-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds\biggr\vert \\& \qquad {} +\biggl\vert t_{2}^{1-\beta} \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\beta-1}E_{\beta ,\beta} \bigl(-M(t_{2}-s)^{\beta}\bigr)\eta_{n-1}(s)\,ds\biggr\vert \\& \qquad {}+C^{*}\bigl\vert E_{\beta,\beta}\bigl(-Mt_{2}^{\beta}\bigr)-E_{\beta,\beta }\bigl(-Mt_{1}^{\beta}\bigr)\bigr\vert \\& \quad =: \mathrm{I}'+\mathrm{II}'+ \mathrm{III}'+\mathrm{IV}', \end{aligned}$$

where

$$C^{*}=M_{0}\frac{\gamma_{2}}{\Gamma(\beta)}T^{2\beta-1}B(\beta,\beta). $$

It is easy to verify that $\mathrm{II}'$, $\mathrm{III}'$, and $\mathrm{IV}'$ go to zero as $t_{1}\to t_{2}$. In the following, we only consider $\mathrm{I}'$:

$$\begin{aligned} \begin{aligned} \mathrm{I}'\leq{}& \gamma_{2}t_{2}^{1-\beta} \biggl\vert \int _{0}^{t_{1}}\bigl[(t_{2}-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(t_{2}-s)^{\beta}\bigr)-(t_{1}-s)^{\beta-1}E_{\beta,\beta} \bigl(-M(t_{1}-s)^{\beta}\bigr)\bigr]s^{\beta -1}\,ds\biggr\vert \\ \leq{}& \gamma_{2}t_{2}^{1-\beta} \int_{0}^{t_{1}}(t_{1}-s)^{\beta-1} \bigl\vert E_{\beta,\beta}\bigl(-M(t_{2}-s)^{\beta}\bigr)-E_{\beta,\beta}\bigl(-M(t_{1}-s)^{\beta}\bigr)\bigr\vert s^{\beta-1}\,ds \\ &{} +\frac{\gamma_{2}}{\Gamma(\beta)}t_{2}^{1-\beta} \int _{0}^{t_{1}}\bigl((t_{1}-s)^{\beta-1}-(t_{2}-s)^{\beta-1} \bigr)s^{\beta-1}\,ds. \end{aligned} \end{aligned}$$

By the continuity of the Mittag-Leffler function and (A.8) we have that $\mathrm{I}'$ goes to zero as $t_{1}\to t_{2}$. It is easy to verify that the equicontinuity of $\{t^{r}D_{0^{+}}^{\alpha}u_{n}\}$ is true for $t_{1}=0$ by (A.9) and similar estimates. This completes the proof of the lemma. □