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 \(t \in[0, h]\)
$$ \biglf(t, u)  f(t, v)\bigr \le Auv^{r_{1}} + B uv^{r_{2}},\quad u, v \in R. $$
(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
$$\begin{aligned} (T_{\lambda}u) (t) =& \Gamma(\alpha)h^{1\alpha} u(h)t^{\alpha1} E_{\alpha, \alpha}\bigl(\lambda t^{\alpha}\bigr)\\ &{} + \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha} \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds, \end{aligned}$$
where
\(\lambda\geq0\)
is a constant.
Proof
First of all, we show that the operator \(T_{\lambda}\) is well defined. Clearly \(t^{\alpha1} E_{\alpha, \alpha}(\lambda t^{\alpha}) \in C_{1\alpha}[0, h]\), so it is enough to prove that for every \(u \in C_{1\alpha}[0, h] \), the function
$$\int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha} \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds $$
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
$$\int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha} \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds \in C(0, h]. $$
On the other hand, under the condition (S1), we have
$$\biglf(t, u)\bigr \le Au^{r_{1}} + Bu^{r_{2}} + C, $$
where \(C= \max_{t \in[0, h]}f(t, 0)\).
By Lemma 2.1, for \(u \in C_{1\alpha}[0, h]\), we have
$$\begin{aligned} & \biggl\vert t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha } \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds\biggr\vert \\ &\quad \le t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha } \bigl(\lambda( ts)^{\alpha}\bigr)\biglf\bigl(s, u(s)\bigr)+\lambda u(s)\bigr\,ds \\ &\quad \le t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha } \bigl(\lambda( ts)^{\alpha}\bigr) \bigl(A u^{r_{1}} + \lambdau + B u^{r_{2}} + C \bigr)\,ds \\ &\quad \le t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha } \bigl(\lambda( ts)^{\alpha}\bigr) \bigl\{ A s^{(\alpha1)r_{1}} \bigl[s^{1\alpha }\biglu(s)\bigr \bigr]^{r_{1}} \\ &\qquad{} +\lambda s^{\alpha1} s^{1\alpha}\biglu(s)\bigr + B s^{(\alpha 1)r_{2}} \bigl[s^{1\alpha}\biglu(s)\bigr \bigr]^{r_{2}} +C \bigr\} \,ds \\ &\quad \le \frac{A \u\^{r_{1}} t^{1\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (ts)^{\alpha1} s^{(\alpha1)r_{1}} \,ds + \frac{\lambda\u\t^{1\alpha }}{\Gamma(\alpha)} \int_{0}^{t} (ts)^{\alpha1} s^{\alpha1}\,ds \\ &\qquad{} + \frac{B \u\^{r_{2}} t^{1\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (ts)^{\alpha1} s^{(\alpha1)r_{2}} \,ds + \frac{Ct}{\Gamma(\alpha+1)} \\ &\quad \le A \u\^{r_{1}} \frac{\Gamma((\alpha1)r_{1}+1)}{\Gamma((\alpha 1)r_{1}+\alpha+1)} t^{(\alpha1)r_{1} +\alpha+1\alpha} + \lambda\u\ \frac{\Gamma(\alpha)}{\Gamma(2\alpha)} t^{\alpha} \\ &\qquad{} +B \u\^{r_{2}} \frac{\Gamma((\alpha1)r_{2}+1)}{\Gamma((\alpha 1)r_{2}+\alpha+1)} t^{(\alpha1)r_{2} +\alpha+1\alpha} + \frac{Ct}{\Gamma (\alpha+1)} \\ &\quad \le\frac{\Gamma[(\alpha1)r_{1}+1]\cdot A \cdot t^{(\alpha 1)r_{1}+1}}{\Gamma[(\alpha1)r_{1}+\alpha+1]}\u\^{r_{1}} + \lambda\u\ \frac{\Gamma(\alpha)}{\Gamma(2\alpha)} t^{\alpha} \\ &\qquad{} +\frac{\Gamma[(\alpha1)r_{2}+1]\cdot B \cdot t^{(\alpha 1)r_{2}+1}}{\Gamma[(\alpha1)r_{2}+\alpha+1]}\u\^{r_{2}} + \frac{Ct}{\Gamma (\alpha+1)}. \end{aligned}$$
That is to say that
$$\int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha} \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds \in C_{1\alpha}[0, h]. $$
The above inequalities and the assumption \(0< r_{1} \le1 < r_{2} < 1/(1\alpha)\) imply that
$$\lim_{ t\to0+} t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha } \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, u(s)\bigr)+\lambda u(s) \bigr]\,ds=0. $$
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
$$\begin{aligned} &\T_{\lambda}u_{n} T_{\lambda}u\ \\ &\quad= \bigl\ t^{1\alpha}(T_{\lambda}u_{n}  T_{\lambda}u) \bigr\ _{\infty}\\ &\quad= \max_{0 \le t \le h} \biggl\{ \bigl\vert \Gamma( \alpha)h^{1\alpha }E_{\alpha, \alpha}\bigl(\lambda t^{\alpha}\bigr) \bigl[u_{n}(h)u(h)\bigr]\bigr\vert \\ &\qquad{}+ \biggl\vert t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1}E_{\alpha, \alpha } \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[f(s, u_{n})f(s, u)+ \lambda(u_{n}u)\bigr]\,ds\biggr\vert \biggr\} \\ &\quad\le\frac{1}{\Gamma(\alpha)} \max_{0 \le t \le h} t^{1\alpha} \int _{0}^{t} (ts)^{\alpha1} \bigl[Au_{n}u^{r_{1}} + Bu_{n}u^{r_{2}} + \lambda u_{n}u\bigr]\,ds \\ &\qquad{} +\u_{n} u\ \\ &\quad\le\frac{1}{\Gamma(\alpha)} \biggl[A \max_{0 \le t \le h} t^{1\alpha } \int_{0}^{t} (ts)^{\alpha1}\cdot s^{r_{1}(1\alpha)} \cdot s^{r_{1}(1\alpha)}\cdotu_{n}u^{r_{1}} \,ds \\ &\qquad{} + \lambda\max_{0 \le t \le h} t^{1\alpha} \int_{0}^{t} (ts)^{\alpha 1}\cdot s^{(1\alpha)} \cdot s^{(1\alpha)}\cdotu_{n}u\,ds \\ &\qquad{} + B \max_{0 \le t \le h} t^{1\alpha} \int_{0}^{t} (ts)^{\alpha 1}\cdot s^{r_{2}(1\alpha)} \cdot s^{r_{2}(1\alpha)}\cdot u_{n}u^{r_{2}} \,ds \biggr] +\u_{n} u\ \\ &\quad\le\frac{1}{\Gamma(\alpha)} \biggl[A \u_{n}u\^{r_{1}}\max _{0 \le t \le h} t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1}\cdot s^{r_{1}(1\alpha)}\,ds \\ &\qquad{} + \lambda\u_{n}u\\max_{0 \le t \le h} t^{1\alpha} \int_{0}^{t} (ts)^{\alpha1}\cdot s^{(1\alpha)}\,ds \\ &\qquad{} + B \u_{n}u\^{r_{2}}\max_{0 \le t \le h} t^{1\alpha} \int _{0}^{t} (ts)^{\alpha1}\cdot s^{r_{2}(1\alpha)}\,ds \biggr] +\u_{n} u\ \\ &\quad\le \frac{A\u_{n}u\^{r_{1}} \Gamma[1r_{1}(1\alpha)]}{\Gamma [1r_{1}(1\alpha) +\alpha]} h^{1r_{1}(1\alpha)} + \frac{\lambda\u_{n}u\ \Gamma[\alpha]}{\Gamma[2\alpha]} h^{\alpha} \\ &\qquad{} +\frac{B\u_{n}u\^{r_{2}} \Gamma[1r_{2}(1\alpha)]}{\Gamma [1r_{2}(1\alpha) +\alpha]} h^{1r_{2}(1\alpha)} +\u_{n} u\ \\ &\quad \to0\quad (n \to\infty). \end{aligned}$$
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
$$ f(t, x)  f(t, y) + \lambda(xy) \ge0, \quad\textit{for } v \le y \le x \le w. $$
(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
$$\begin{aligned} v(t) =& \Gamma(\alpha) \bigl(h^{1\alpha}v(h) \epsilon \bigr)t^{\alpha1} E_{\alpha, \alpha}\bigl(\lambda t^{\alpha}\bigr) \\ &{} + \int_{0}^{t} (ts)^{\alpha1} E_{\alpha, \alpha} \bigl(\lambda( ts)^{\alpha}\bigr)\bigl[ f\bigl(s, v(s)\bigr)+\lambda v(s)q(s)\bigr]\,ds \\ \le&(T_{\lambda}v) (t). \end{aligned}$$
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 \(u(t) = 2t^{\alpha1}\operatorname{Cos}[2t] +t^{\alpha}\), 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.