Extremal solutions for some periodic fractional differential equations

Abstract

By using the lower and upper solution method, the existence of an iterative solution for a class of fractional periodic boundary value problems,

$$\begin{aligned}& D_{0+}^{\alpha}u(t)=f\bigl(t, u(t)\bigr),\quad t \in(0, h),\\& \lim_{t \to0^{+}}t^{1-\alpha}u(t) = h^{1-\alpha}u(h), \end{aligned}$$

is discussed, where $0< h<+\infty$, $f\in C([0, h]\times R, R)$, $D_{0+}^{\alpha}u (t)$ is the Riemann-Liouville fractional derivative, $0<\alpha< 1$. Different from other well-known results, a new condition on the nonlinear term is given to guarantee the equivalence between the solution of the periodic boundary value problem and the fixed point of the corresponding operator. Moreover, the existence of extremal solutions for the problem is given.

Introduction

Differential equations of fractional order have played a significant role in engineering, science, and pure and applied mathematics in recent years. Some researchers paid attention to the existence results of the solution of the periodic boundary value problem for fractional differential equations, such as [1–17]. Some recent contributions to the theory of fractional differential equations initial value problems can be found in [4, 9].

In [4], by using the fixed point theorem of Schaeffer and the Banach contraction principle, Belmekki et al. obtained the Green’s function and gave some existence results for the nonlinear fractional periodic problem

$$\begin{aligned}& D_{0+}^{\alpha}u (t) -\lambda u(t)= f\bigl(t, u(t)\bigr),\quad t \in(0, 1]\ (0< \alpha< 1),\\& \lim_{t \to0^{+}}t^{1-\alpha}u(t) = u(1), \end{aligned}$$

where $f: [0, 1] \times R \to R$ is continuous and the following assumptions hold:

• (1) there exists a constant $M >0$ such that

  $$\bigl| f(t, u) \bigr|\le M, \quad\mbox{for each } t \in(0, 1), u \in R, $$

• (2) there exists a constant $k > 0$ such that

  $$\bigl|f(t, u) - f(t, v)\bigr| \le k |u-v|, \quad\mbox{for each } t \in(0, 1), u, v \in R. $$

The above conditions (see Lemma 4.2 of [4]) are very strong.

In [13], Wei et al. discussed the properties of the well-known Mittag-Leffler function, and consider the existence and uniqueness of the solution of the periodic boundary value problem for a fractional differential equation involving a Riemann-Liouville fractional derivative

$$\begin{aligned}& D_{0+}^{\alpha}u (t) = f\bigl(t, u(t)\bigr),\quad t \in(0, T)\ (0< \alpha< 1),\\& t^{1-\alpha}u(t)|_{t=0} = t^{1-\alpha}u(t) |_{t=T}, \end{aligned}$$

by using the monotone iterative method. In this result, the bounded demand of f in [13] and the monotone demand of f in [9] were removed. However, the application of Lemma 1.1 in the proof of Theorem 3.1 was not correct, due to $\sigma(\eta)(t) \notin C[0, T]$. In other words, the definition of operator A may be not appropriate. Consequently, while the uniqueness result was correct, the existence of an extremal result was maybe wrong.

In [14], Wei and Dong studied the existence of solutions of the following periodic boundary value problem:

$$\begin{aligned}& D_{0+}^{2\alpha} u (t) = f\bigl(t, u(t), D_{0+}^{\alpha}u(t)\bigr),\quad t \in (0, T)\ (0< \alpha< 1),\\& \lim_{t \to0}t^{1-\alpha}u(t) =\lim_{t \to T} t^{1-\alpha}u(t),\\& \lim_{t \to0}t^{1-\alpha} D_{0+}^{\alpha}u(t) =\lim_{t \to T} t^{1-\alpha}D_{0+}^{\alpha}u(t), \end{aligned}$$

where $D_{0+}^{\alpha}$ is the standard Riemann-Liouville fractional derivative, $D_{0+}^{2\alpha}u = D_{0+}^{\alpha}(D_{0+}^{\alpha}u)$ is the sequential Riemann-Liouville fractional derivative, $0 < T < \infty$, and f defined on $[0, T] \times R^{2}$ is continuous. The methods used in [14] are monotone iterative techniques and the Schauder fixed point theorem under the assumptions that there the upper and lower solutions exist.

In this paper, we will focus our attention on the following problem:

$$\begin{aligned}& D_{0+}^{\alpha}u(t)=f\bigl(t, u(t)\bigr),\quad t \in(0, h), \end{aligned}$$

(1.1)

$$\begin{aligned}& \lim_{t \to0^{+}}t^{1-\alpha}u(t) =h^{1-\alpha} u(h), \end{aligned}$$

(1.2)

where $f\in C([0, h]\times R, R)$, $D_{0+}^{\alpha}u (t)$ is the Riemann-Liouville fractional derivative, $0<\alpha< 1$. The existence of the solution is obtained by the use of the upper and lower solution method which has been used by authors to deal with the fractional initial value problems [2].

The remainder of this paper is as follows. In Section 2, we recall some notions and the theory of the fractional calculus. Section 3 is devoted to the study of the existence of a solution utilizing the method of upper and lower solutions. The existence of extremal solutions is given. An example is given to illustrate the main result.

Preliminaries

Given $0 \le a < b <+\infty$ and $r>0$, define

$$C_{r}[a, b]=\bigl\{ u \mid u \in C(a, b], (t-a)^{r} u(t) \in C[a, b]\bigr\} . $$

Clearly, $C_{r}[a, b]$ is a linear space with the normal multiplication and addition. Given $u \in C_{r}[a, b]$, define

$$\|u\| = \max_{t \in[a, b]}(t-a)^{r}\bigl|u(t)\bigr|, $$

then $(C_{r}[a, b], \|\cdot\|)$ is a Banach space.

Lemma 2.1

([13])

For $0 < \alpha\le1$, $\lambda\ge0$, the Mittag-Leffler type function $E_{\alpha, \alpha}(-\lambda t^{\alpha})$ satisfies

$$0 \le E_{\alpha, \alpha}\bigl(-\lambda t^{\alpha}\bigr) < \frac{1}{\Gamma(\alpha )},\quad t \in(0, \infty). $$

Lemma 2.2

The linear periodic problem

$$\begin{aligned}& D_{0+}^{\alpha}u(t) + \lambda u(t) = q(t), \end{aligned}$$

(2.1)

$$\begin{aligned}& \lim_{t \to0^{+}}t^{1-\alpha}u(t) = h^{1-\alpha} u(h), \end{aligned}$$

(2.2)

where $\lambda\ge0$ is a constant and $q \in L(0, h)$, has the following integral representation of the solution:

$$u(t) = \Gamma(\alpha) u(h) t^{\alpha-1}E_{\alpha, \alpha}\bigl(-\lambda t^{\alpha}\bigr) + \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha, \alpha} \bigl(-\lambda (t-s)^{\alpha}\bigr)q(s)\,ds. $$

Proof

According to [8], for every initial condition

$$\lim_{t \to0+}t^{1-\alpha}u(t) = u_{0} $$

the unique solution of equation (2.1) is given by

$$u(t) = \Gamma(\alpha)u_{0} t^{\alpha-1}E_{\alpha, \alpha}\bigl(- \lambda t^{\alpha}\bigr) + \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha, \alpha} \bigl(-\lambda (t-s)^{\alpha}\bigr)q(s)\,ds. $$

Specially, choose $u_{0}$ as

$$u_{0} = \frac{h^{1-\alpha}\int_{0}^{h} (h-s)^{\alpha-1} E_{\alpha, \alpha }(-\lambda(h-s)^{\alpha})q(s)\,ds}{ 1- \Gamma(\alpha)E_{\alpha, \alpha}(-\lambda h^{\alpha})}, $$

then $u(t)$ satisfies the periodic boundary condition (2.2). That is to say that the linear periodic problem (2.1), (2.2) has the following integral representation of the solution:

$$u(t) = \Gamma(\alpha) h^{1-\alpha}u(h) t^{\alpha-1}E_{\alpha, \alpha }\bigl(- \lambda t^{\alpha}\bigr) + \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha, \alpha } \bigl(-\lambda(t-s)^{\alpha}\bigr)q(s)\,ds. $$

The proof is complete. □

Lemma 2.3

([18])

Suppose that E is an ordered Banach space, $x_{0}, y_{0} \in E$, $x_{0} \le y_{0}$, $D=[x_{0}, y_{0}]$, $T: D \to E$ is an increasing completely continuous operator and $x_{0} \le Tx_{0}$, $y_{0} \ge Ty_{0}$. Then the operator T has a minimal fixed point $x^{*}$ and a maximal fixed point $y^{*}$. If we let

$$x_{n} = Tx_{n-1},\qquad y_{n} = Ty_{n-1},\quad n=1, 2, 3, \ldots, $$

then

$$\begin{aligned}& x_{0} \le x_{1} \le x_{2} \le\cdots\le x_{n} \le\cdots\le y_{n} \le\cdots \le y_{2} \le y_{1} \le y_{0},\\& x_{n} \to x^{*},\qquad y_{n} \to y^{*}. \end{aligned}$$

Definition 2.1

A function $v(t) \in C_{1-\alpha}[0, h]$ is called a lower solution of problem (1.1), (1.2), if it satisfies

$$\begin{aligned}& D_{0+}^{\alpha}v(t) \le f\bigl(t, v(t)\bigr),\quad t \in(0, h), \end{aligned}$$

(2.3)

$$\begin{aligned}& \lim_{t \to0^{+}}t^{1-\alpha}v(t) \le h^{1-\alpha} v(h). \end{aligned}$$

(2.4)

Definition 2.2

A function $w(t) \in C_{1-\alpha}[0, h]$ is called an upper solution of problem (1.1), (1.2), if it satisfies

$$\begin{aligned}& D_{0+}^{\alpha}w(t) \ge f\bigl(t, w(t)\bigr),\quad t \in(0, h), \end{aligned}$$

(2.5)

$$\begin{aligned}& \lim_{t \to0^{+}}t^{1-\alpha}w(t) \ge h^{1-\alpha} w(h). \end{aligned}$$

(2.6)

The main results

The following assumptions will be used in this section:

• (S1) $f: [0, h] \times R \to R$ is continuous and there exist constants $A, B \ge0$ and $0 < r_{1} \le1 < r_{2}<1/(1-\alpha)$ such that for $t \in[0, h]$

  $$ \bigl|f(t, u) - f(t, v)\bigr| \le A|u-v|^{r_{1}} + B |u-v|^{r_{2}},\quad u, v \in R. $$

  (3.1)

Theorem 3.1

Suppose (S1) holds. Then u solves problem (1.1), (1.2) if and only if it is a fixed point of the operator $T_{\lambda}: C_{1-\alpha} [0, h] \to C_{1-\alpha} [0, h]$ defined by

$$\begin{aligned} (T_{\lambda}u) (t) =& \Gamma(\alpha)h^{1-\alpha} u(h)t^{\alpha-1} E_{\alpha, \alpha}\bigl(-\lambda t^{\alpha}\bigr)\\ &{} + \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha} \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds, \end{aligned}$$

where $\lambda\geq0$ is a constant.

Proof

First of all, we show that the operator $T_{\lambda}$ is well defined. Clearly $t^{\alpha-1} E_{\alpha, \alpha}(-\lambda t^{\alpha}) \in C_{1-\alpha}[0, h]$, so it is enough to prove that for every $u \in C_{1-\alpha}[0, h]$, the function

$$\int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha} \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds $$

belongs to $C_{1-\alpha}[0, h]$. Taking into account that f is continuous on $[0, h] \times R$, for $u \in C_{1-\alpha}[0, h]$, we have

$$\int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha} \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds \in C(0, h]. $$

On the other hand, under the condition (S1), we have

$$\bigl|f(t, u)\bigr| \le A|u|^{r_{1}} + B|u|^{r_{2}} + C, $$

where $C= \max_{t \in[0, h]}f(t, 0)$.

By Lemma 2.1, for $u \in C_{1-\alpha}[0, h]$, we have

$$\begin{aligned} & \biggl\vert t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha } \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds\biggr\vert \\ &\quad \le t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha } \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl|f\bigl(s, u(s)\bigr)+\lambda u(s)\bigr|\,ds \\ &\quad \le t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha } \bigl(-\lambda( t-s)^{\alpha}\bigr) \bigl(A |u|^{r_{1}} + \lambda|u| + B |u|^{r_{2}} + C \bigr)\,ds \\ &\quad \le t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha } \bigl(-\lambda( t-s)^{\alpha}\bigr) \bigl\{ A s^{(\alpha-1)r_{1}} \bigl[s^{1-\alpha }\bigl|u(s)\bigr| \bigr]^{r_{1}} \\ &\qquad{} +\lambda s^{\alpha-1} s^{1-\alpha}\bigl|u(s)\bigr| + B s^{(\alpha -1)r_{2}} \bigl[s^{1-\alpha}\bigl|u(s)\bigr| \bigr]^{r_{2}} +C \bigr\} \,ds \\ &\quad \le \frac{A \|u\|^{r_{1}} t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1} s^{(\alpha-1)r_{1}} \,ds + \frac{\lambda\|u\|t^{1-\alpha }}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1} s^{\alpha-1}\,ds \\ &\qquad{} + \frac{B \|u\|^{r_{2}} t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1} s^{(\alpha-1)r_{2}} \,ds + \frac{Ct}{\Gamma(\alpha+1)} \\ &\quad \le A \|u\|^{r_{1}} \frac{\Gamma((\alpha-1)r_{1}+1)}{\Gamma((\alpha -1)r_{1}+\alpha+1)} t^{(\alpha-1)r_{1} +\alpha+1-\alpha} + \lambda\|u\| \frac{\Gamma(\alpha)}{\Gamma(2\alpha)} t^{\alpha} \\ &\qquad{} +B \|u\|^{r_{2}} \frac{\Gamma((\alpha-1)r_{2}+1)}{\Gamma((\alpha -1)r_{2}+\alpha+1)} t^{(\alpha-1)r_{2} +\alpha+1-\alpha} + \frac{Ct}{\Gamma (\alpha+1)} \\ &\quad \le\frac{\Gamma[(\alpha-1)r_{1}+1]\cdot A \cdot t^{(\alpha -1)r_{1}+1}}{\Gamma[(\alpha-1)r_{1}+\alpha+1]}\|u\|^{r_{1}} + \lambda\|u\| \frac{\Gamma(\alpha)}{\Gamma(2\alpha)} t^{\alpha} \\ &\qquad{} +\frac{\Gamma[(\alpha-1)r_{2}+1]\cdot B \cdot t^{(\alpha -1)r_{2}+1}}{\Gamma[(\alpha-1)r_{2}+\alpha+1]}\|u\|^{r_{2}} + \frac{Ct}{\Gamma (\alpha+1)}. \end{aligned}$$

That is to say that

$$\int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha} \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds \in C_{1-\alpha}[0, h]. $$

The above inequalities and the assumption $0< r_{1} \le1 < r_{2} < 1/(1-\alpha)$ imply that

$$\lim_{ t\to0+} t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha } \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds=0. $$

Combining with the fact that $\lim_{t \to0+}E_{\alpha, \alpha }(-\lambda t^{\alpha}) = E_{\alpha, \alpha}(0)=1/\Gamma(\alpha)$ yields

$$\lim_{t \to0+} t^{1-\alpha}(T_{\lambda}u)(t) = h^{1-\alpha}u(h). $$

The above arguments combined with Lemma 2.2 imply that the fixed point of the operator $T_{\lambda}$ solves the periodic boundary value problem (1.1), (1.2), and vice versa. The proof is complete. □

In the following, we consider the compactness of the set of the space $C_{r}[0, h]$.

Let $F \subset C_{r}[0, h]$ and $E= \{g(t)= t^{r} h(t) \mid h(t)\in F\}$, then $E \subset C[0, h]$. It is clear that F is a bounded set of $C_{r}[0, h]$ if and only if E is a bounded set of $C[0, h]$.

Therefore, to prove that $F \subset C_{r}[0, h]$ is a compact set, it is enough to prove that $E \subset C[0, h]$ is a bounded and equicontinuous set.

Theorem 3.2

Suppose (S1) holds. Then the operator $T_{\lambda}: C_{1-\alpha}[0, h] \to C_{1-\alpha}[0, h]$ is completely continuous.

Proof

Given $u_{n} \to u \in C_{1-\alpha}[0, h]$, with the definition of $T_{\lambda}$, the condition (S1), and Lemma 2.1, one has

$$\begin{aligned} &\|T_{\lambda}u_{n} -T_{\lambda}u\| \\ &\quad= \bigl\| t^{1-\alpha}(T_{\lambda}u_{n} - T_{\lambda}u) \bigr\| _{\infty}\\ &\quad= \max_{0 \le t \le h} \biggl\{ \bigl\vert \Gamma( \alpha)h^{1-\alpha }E_{\alpha, \alpha}\bigl(-\lambda t^{\alpha}\bigr) \bigl[u_{n}(h)-u(h)\bigr]\bigr\vert \\ &\qquad{}+ \biggl\vert t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha, \alpha } \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[f(s, u_{n})-f(s, u)+ \lambda(u_{n}-u)\bigr]\,ds\biggr\vert \biggr\} \\ &\quad\le\frac{1}{\Gamma(\alpha)} \max_{0 \le t \le h} t^{1-\alpha} \int _{0}^{t} (t-s)^{\alpha-1} \bigl[A|u_{n}-u|^{r_{1}} + B|u_{n}-u|^{r_{2}} + \lambda |u_{n}-u|\bigr]\,ds \\ &\qquad{} +\|u_{n} -u\| \\ &\quad\le\frac{1}{\Gamma(\alpha)} \biggl[A \max_{0 \le t \le h} t^{1-\alpha } \int_{0}^{t} (t-s)^{\alpha-1}\cdot s^{-r_{1}(1-\alpha)} \cdot s^{r_{1}(1-\alpha)}\cdot|u_{n}-u|^{r_{1}} \,ds \\ &\qquad{} + \lambda\max_{0 \le t \le h} t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha -1}\cdot s^{-(1-\alpha)} \cdot s^{(1-\alpha)}\cdot|u_{n}-u|\,ds \\ &\qquad{} + B \max_{0 \le t \le h} t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha -1}\cdot s^{-r_{2}(1-\alpha)} \cdot s^{r_{2}(1-\alpha)}\cdot |u_{n}-u|^{r_{2}} \,ds \biggr] +\|u_{n} -u\| \\ &\quad\le\frac{1}{\Gamma(\alpha)} \biggl[A \|u_{n}-u\|^{r_{1}}\max _{0 \le t \le h} t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}\cdot s^{-r_{1}(1-\alpha)}\,ds \\ &\qquad{} + \lambda\|u_{n}-u\|\max_{0 \le t \le h} t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}\cdot s^{-(1-\alpha)}\,ds \\ &\qquad{} + B \|u_{n}-u\|^{r_{2}}\max_{0 \le t \le h} t^{1-\alpha} \int _{0}^{t} (t-s)^{\alpha-1}\cdot s^{-r_{2}(1-\alpha)}\,ds \biggr] +\|u_{n} -u\| \\ &\quad\le \frac{A\|u_{n}-u\|^{r_{1}} \Gamma[1-r_{1}(1-\alpha)]}{\Gamma [1-r_{1}(1-\alpha) +\alpha]} h^{1-r_{1}(1-\alpha)} + \frac{\lambda\|u_{n}-u\| \Gamma[\alpha]}{\Gamma[2\alpha]} h^{\alpha} \\ &\qquad{} +\frac{B\|u_{n}-u\|^{r_{2}} \Gamma[1-r_{2}(1-\alpha)]}{\Gamma [1-r_{2}(1-\alpha) +\alpha]} h^{1-r_{2}(1-\alpha)} +\|u_{n} -u\| \\ &\quad \to0\quad (n \to\infty). \end{aligned}$$

That is to say that $T_{\lambda}$ is continuous.

Suppose that $F \subset C_{1-\alpha}[0, h]$ is a bounded set and there is a positive constant M such that $\|u\| \le M$ for $u \in F$. The proof process of Theorem 3.1 shows that $T_{\lambda}(F) \subset C_{1-\alpha}[0, h]$ is bounded.

We omit the proof details of the equicontinuity of $T(F)$ here and refer the reader to [2] for a similar details. The proof is complete. □

Theorem 3.3

Assume (S1) hold and $v, w \in C_{1-\alpha}[0, h]$ are lower and upper solutions of problem (1.1), (1.2), respectively, such that

$$ v(t) \le w(t),\quad 0\le t\leq h. $$

(3.2)

Moreover, $f: [0, h] \times R \to R$ satisfies

$$ f(t, x) - f(t, y) + \lambda(x-y) \ge0, \quad\textit{for } v \le y \le x \le w. $$

(3.3)

Then the fractional periodic boundary value problem (1.1), (1.2) has a minimal solution $x^{*}$ and a maximal solution $y^{*}$ such that

$$x^{*} = \lim_{n \to\infty} T_{\lambda}^{n}v,\qquad y^{*} = \lim_{n \to\infty} T_{\lambda}^{n}w. $$

Proof

Clearly, if the functions v, w are lower and upper solutions (or strict) of problem (1.1), (1.2), then there are $v \le T_{\lambda}v$, $w \ge T_{\lambda}w$ (or the inequality is strict). In fact, by the definition of the lower solution, there exist $q(t) \ge0$ and $\epsilon\ge0$ such that

$$\begin{aligned}& D_{0+}^{\alpha}v(t) = f\bigl(t, v(t)\bigr) -q(t),\quad t \in(0, h),\\& \lim_{t \to0^{+}}t^{1-\alpha} v(t) =h^{1-\alpha}v(h) - \epsilon. \end{aligned}$$

By the use of Theorem 3.1 and Lemma 2.1, one has

$$\begin{aligned} v(t) =& \Gamma(\alpha) \bigl(h^{1-\alpha}v(h) -\epsilon \bigr)t^{\alpha-1} E_{\alpha, \alpha}\bigl(-\lambda t^{\alpha}\bigr) \\ &{} + \int_{0}^{t} (t-s)^{\alpha-1} E_{\alpha, \alpha} \bigl(-\lambda( t-s)^{\alpha}\bigr)\bigl[ f\bigl(s, v(s)\bigr)+\lambda v(s)-q(s)\bigr]\,ds \\ \le&(T_{\lambda}v) (t). \end{aligned}$$

Similarly, we have $w \ge T_{\lambda}w$.

By condition (3.3) and Theorem 3.2, the operator $T_{\lambda}: C_{1-\alpha }[0, h]\to C_{1-\alpha}[0, h]$ is an increasing completely continuous operator. Setting $D:= [v, w]$, by the use of Lemma 2.3, the existence of $x^{*}$, $y^{*}$ is obtained. The proof is complete. □

Remark 3.1

The main result is a consequence of the classical monotone iterative technique [19, 20]. However, the periodic condition is not the same.

Example 3.1

Consider the following periodic fractional boundary value problem:

$$\begin{aligned}& D_{0+}^{\alpha}u(t)=f\bigl(t, u(t)\bigr),\quad t \in(0, h), \end{aligned}$$

(3.4)

$$\begin{aligned}& \lim_{t \to0^{+}}t^{1-\alpha}u(t) = h^{1-\alpha}u(h), \end{aligned}$$

(3.5)

where $\alpha=0.3$, $h=0.7$, $f(t,u)=\frac{t}{10} [1+u(t)]$. Obviously, the function $f(t, u)$ satisfies condition (3.3) and (S1), $f(t,0)\geq0$, and $f(t,0) \not\equiv0$ for $t \in[0, h]$. Thus, $v(t) \equiv0$ is a lower solution of problem (3.4), (3.5). Choose $u(t) = 2t^{\alpha-1}\operatorname{Cos}[2t] +t^{\alpha}$, one can check that $u\in C_{1-\alpha}[0, h]$ is an upper solution of problem (3.4), (3.5), and $v(t) \leq u(t)$ for $t \in[0, h]$. By the use of Theorem 3.3, problem (3.4), (3.5) has at least one solution.