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.