For the convenience of the reader, we give some background material from fractional calculus theory to facilitate the analysis of problem (1.1) and (1.2). These materials can be found in the recent literature, see [7, 8, 26, 27, 30–33].
Definition 2.1[7]
The fractional integral of order \alpha >0 of a function y:(0,+\mathrm{\infty})\to \mathbb{R} is given by
{I}_{0+}^{\alpha}y(t)=\frac{1}{\mathrm{\Gamma}(\alpha )}{\int}_{0}^{t}{(ts)}^{\alpha 1}y(s)\phantom{\rule{0.2em}{0ex}}ds
provided the righthand side is pointwise defined on (0,+\mathrm{\infty}).
Definition 2.2[7]
The fractional derivative of order \alpha >0 of a continuous function y:(0,+\mathrm{\infty})\to \mathbb{R} is given by
{D}_{0+}^{\alpha}y(t)=\frac{1}{\mathrm{\Gamma}(n\alpha )}{\left(\frac{d}{dt}\right)}^{n}{\int}_{0}^{t}\frac{y(s)}{{(ts)}^{\alpha n+1}}\phantom{\rule{0.2em}{0ex}}ds,
where n is the smallest integer greater than or equal to α, provided that the righthand side is pointwise defined on (0,+\mathrm{\infty}).
Lemma 2.1[7]
Let\alpha >0. If we assumeu\in {D}_{0+}^{\alpha}u\in {L}^{1}(0,1), then the fractional differential equation
has
u(t)={c}_{1}{t}^{\alpha 1}+{c}_{2}{t}^{\alpha 2}+\cdots +{c}_{n}{t}^{\alpha n},\phantom{\rule{1em}{0ex}}{c}_{i}\in \mathbb{R},i=1,2,\dots ,n\mathit{\text{as a unique solution}},
where n is the smallest integer greater than or equal to α.
Lemma 2.2[7]
Assume that{D}_{0+}^{\alpha}u\in {L}^{1}(0,1)with a fractional derivative of order\alpha >0. Then
{I}_{0+}^{\alpha}{D}_{0+}^{\alpha}u(t)=u(t)+{c}_{1}{t}^{\alpha 1}+{c}_{2}{t}^{\alpha 2}+\cdots +{c}_{n}{t}^{\alpha n}
for some{c}_{i}\in \mathbb{R}, i=1,2,\dots ,n, where n is the smallest integer greater than or equal to α.
Lemma 2.3[27]
Lety\in C[0,1]and2<\alpha \u2a7d3. Then fractional differential equation boundary value problem
has a unique solution
u(t)={\int}_{0}^{1}G(t,s)y(s)\phantom{\rule{0.2em}{0ex}}ds,
where
G(t,s)=\{\begin{array}{cc}\frac{{t}^{\alpha 1}{(1s)}^{\alpha 2}{(ts)}^{\alpha 1}}{\mathrm{\Gamma}(\alpha )},\hfill & s\u2a7dt,\hfill \\ \frac{{t}^{\alpha 1}{(1s)}^{\alpha 2}}{\mathrm{\Gamma}(\alpha )},\hfill & t\u2a7ds.\hfill \end{array}
(2.3)
Lemma 2.4 Lety\in C[0,1]and2<\alpha \u2a7d3, 1<\beta \u2a7d2. Then the fractional differential equation boundary value problem
has a unique solution
u(t)={\int}_{0}^{1}G(t,s){\varphi}_{q}({\int}_{0}^{1}H(s,\tau )y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds,
where
H(t,s)=\{\begin{array}{cc}\frac{{[t(1s)]}^{\beta 1}{(ts)}^{\beta 1}}{\mathrm{\Gamma}(\beta )},\hfill & s\u2a7dt,\hfill \\ \frac{{[t(1s)]}^{\beta 1}}{\mathrm{\Gamma}(\beta )},\hfill & t\u2a7ds,\hfill \end{array}
(2.6)
G(t,s)is defined as (2.3).
Proof From Lemma 2.2 and 1<\beta \u2a7d2, we have
{I}_{0+}^{\beta}{D}_{0+}^{\beta}\left({\varphi}_{p}({D}_{0+}^{\alpha}u(t))\right)={\varphi}_{p}({D}_{0+}^{\alpha}u(t))+{c}_{1}{t}^{\beta 1}+{c}_{2}{t}^{\beta 2}\phantom{\rule{1em}{0ex}}\text{for some}{c}_{1},{c}_{2}\in \mathbb{R}.
In view of (2.4), we obtain
{I}_{0+}^{\beta}{D}_{0+}^{\beta}\left({\varphi}_{p}({D}_{0+}^{\alpha}u(t))\right)={I}_{0+}^{\beta}y(t).
Therefore,
{\varphi}_{p}({D}_{0+}^{\alpha}u(t))={I}_{0+}^{\beta}y(t)+{C}_{1}{t}^{\beta 1}+{C}_{2}{t}^{\beta 2}\phantom{\rule{1em}{0ex}}\text{for some}{C}_{1},{C}_{2}\in \mathbb{R},
that is,
{\varphi}_{p}({D}_{0+}^{\alpha}u(t))={\int}_{0}^{t}\frac{{(t\tau )}^{\beta 1}}{\mathrm{\Gamma}(\beta )}y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau +{C}_{1}{t}^{\beta 1}+{C}_{2}{t}^{\beta 2}.
By the boundary conditions {D}_{0+}^{\alpha}u(0)={D}_{0+}^{\alpha}u(1)=0, we have
{C}_{2}=0,\phantom{\rule{2em}{0ex}}{C}_{1}={\int}_{0}^{1}\frac{{(1\tau )}^{\beta 1}}{\mathrm{\Gamma}(\beta )}y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau .
Therefore, the solution u(t) of fractional differential equation boundary value problem (2.4) and (2.5) satisfies
\begin{array}{rcl}{\varphi}_{p}({D}_{0+}^{\alpha}u(t))& =& {\int}_{0}^{t}\frac{{(t\tau )}^{\beta 1}}{\mathrm{\Gamma}(\beta )}y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau {\int}_{0}^{1}\frac{{[t(1\tau )]}^{\beta 1}}{\mathrm{\Gamma}(\beta )}y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau \\ =& {\int}_{0}^{1}H(t,\tau )y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau .\end{array}
Consequently, {D}_{0+}^{\alpha}u(t)+{\varphi}_{q}({\int}_{0}^{1}H(t,\tau )y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )=0. Thus, fractional differential equation boundary value problem (2.4) and (2.5) is equivalent to the following problem:
\begin{array}{c}{D}_{0+}^{\alpha}u(t)+{\varphi}_{q}({\int}_{0}^{1}H(t,\tau )y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )=0,\phantom{\rule{1em}{0ex}}0<t<1,\hfill \\ u(0)={u}^{\prime}(0)={u}^{\prime}(1)=0.\hfill \end{array}
Lemma 2.3 implies that fractional differential equation boundary value problem (2.4) and (2.5) has a unique solution
u(t)={\int}_{0}^{1}G(t,s){\varphi}_{q}({\int}_{0}^{1}H(s,\tau )y(\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds.
The proof is complete. □
Lemma 2.5 Let2<\alpha \u2a7d3, 1<\beta \u2a7d2. The functionsG(t,s)andH(t,s)defined by (2.3) and (2.6), respectively, are continuous on[0,1]\times [0,1]and satisfy

(1)
G(t,s)\u2a7e0, H(t,s)\u2a7e0 for t,s\in [0,1];

(2)
G(t,s)\u2a7dG(1,s), H(t,s)\u2a7dH(s,s) for t,s\in [0,1];

(3)
G(t,s)\u2a7e{t}^{\alpha 1}G(1,s) for t,s\in (0,1);

(4)
there exist two positive functions
{\delta}_{1},{\delta}_{2}\in C[0,1]
such that
Proof Observing the expression of G(t,s) and H(t,s), it is easy to see that G(t,s)\u2a7e0 and H(t,s)\u2a7e0 for s,t\in [0,1].
From Lemma 3.1 in [27] and Lemma 2.4 in [8], we obtain (2) and (3).
In the following, we consider the existence of {\delta}_{1}(s) and {\delta}_{2}(s). Firstly, for given s\in (0,1), G(t,s) is increasing with respect to t for t\in (0,1). Consequently, setting
{g}_{1}(t,s)=\frac{{t}^{\alpha 1}{(1s)}^{\alpha 2}{(ts)}^{\alpha 1}}{\mathrm{\Gamma}(\alpha )},\phantom{\rule{2em}{0ex}}{g}_{2}(t,s)=\frac{{t}^{\alpha 1}{(1s)}^{\alpha 2}}{\mathrm{\Gamma}(\alpha )},
we have
\begin{array}{rcl}\underset{1/4\u2a7dt\u2a7d3/4}{min}G(t,s)& =& \{\begin{array}{cc}{g}_{1}(\frac{1}{4},s),\hfill & s\in (0,\frac{1}{4}],\hfill \\ {g}_{2}(\frac{1}{4},s),\hfill & s\in [\frac{1}{4},1),\hfill \end{array}\\ =& \{\begin{array}{cc}\frac{1}{\mathrm{\Gamma}(\alpha )}[{(\frac{1}{4})}^{\alpha 1}{(1s)}^{\alpha 2}{(\frac{1}{4}s)}^{\alpha 1}],\hfill & s\in (0,\frac{1}{4}],\hfill \\ \frac{1}{\mathrm{\Gamma}(\alpha )}{(\frac{1}{4})}^{\alpha 1}{(1s)}^{\alpha 2},\hfill & s\in [\frac{1}{4},1).\hfill \end{array}\end{array}
Secondly, with the use of the monotonicity of G(t,s), we have
\underset{0\u2a7dt\u2a7d1}{max}G(t,s)=G(1,s)=\frac{1}{\mathrm{\Gamma}(\alpha )}[{(1s)}^{\alpha 2}{(1s)}^{\alpha 1}].
Thus, setting
{\delta}_{1}(s)=\{\begin{array}{cc}\frac{{(\frac{1}{4})}^{\alpha 1}{(1s)}^{\alpha 2}{(\frac{1}{4}s)}^{\alpha 1}}{{(1s)}^{\alpha 2}{(1s)}^{\alpha 1}},\hfill & s\in (0,\frac{1}{4}],\hfill \\ \frac{{(\frac{1}{4})}^{\alpha 1}{(1s)}^{\alpha 2}}{{(1s)}^{\alpha 2}{(1s)}^{\alpha 1}},\hfill & s\in [\frac{1}{4},1),\hfill \end{array}
then (2.7) holds.
Similar to Lemma 2.4 in [8], we choose
{\delta}_{2}(s)=\{\begin{array}{cc}\frac{{[\frac{3}{4}(1s)]}^{\beta 1}{(\frac{3}{4}s)}^{\beta 1}}{{[s(1s)]}^{\beta 1}},\hfill & s\in (0,r],\hfill \\ {(\frac{1}{4s})}^{\beta 1},\hfill & s\in [r,1).\hfill \end{array}
The proof is complete. □
Lemma 2.6 Let2<\alpha \u2a7d3. Ify(t)\in C[0,1]andy(t)\u2a7e0, then fractional differential equation boundary value problem (2.1) and (2.2) has a unique solutionu(t)\u2a7e0, t\in [0,1].
Proof From Lemma 2.3, the fractional differential equation boundary value problem (2.1) and (2.2) has a unique solution
u(t)={\int}_{0}^{1}G(t,s)y(s)\phantom{\rule{0.2em}{0ex}}ds.
In view of Lemma 2.5, we know G(t,s) is continuous on [0,1]\times [0,1] and G(t,s)\u2a7e0 for t,s\in [0,1]. If y(t)\in C[0,1] and y(t)\u2a7e0, we obtain u(t)\u2a7e0. The proof is complete. □
Let {E}_{0}=\{u:u\in {C}^{3}[0,1],{\varphi}_{p}({D}_{0+}^{\alpha}u)\in {C}^{2}[0,1]\}. Now, we introduce definitions about the upper and lower solutions of fractional differential equation boundary value problem (1.1) and (1.2).
Definition 2.3[26]
A function \eta (t) is called an upper solution of fractional differential equation boundary value problem (1.1) and (1.2) if \eta (t)\in {E}_{0} and \eta (t) satisfies
\begin{array}{c}{D}_{0+}^{\beta}\left({\varphi}_{p}({D}_{0+}^{\alpha}\eta (t))\right)\u2a7ef(t,\eta (t)),\phantom{\rule{1em}{0ex}}0<t<1,2<\alpha \u2a7d3,1<\beta \u2a7d2,\hfill \\ \eta (0)\u2a7e0,\phantom{\rule{2em}{0ex}}{\eta}^{\prime}(0)\u2a7e0,\phantom{\rule{2em}{0ex}}{\eta}^{\prime}(1)\u2a7e0,\hfill \\ {D}_{0+}^{\alpha}\eta (0)\u2a7d0,\phantom{\rule{2em}{0ex}}{D}_{0+}^{\alpha}\eta (1)\u2a7d0.\hfill \end{array}
Definition 2.4[26]
A function \xi (t) is called a lower solution of fractional differential equation boundary value problem (1.1) and (1.2) if \xi (t)\in {E}_{0} and \xi (t) satisfies
\begin{array}{c}{D}_{0+}^{\beta}\left({\varphi}_{p}({D}_{0+}^{\alpha}\xi (t))\right)\u2a7df(t,\xi (t)),\phantom{\rule{1em}{0ex}}0<t<1,2<\alpha \u2a7d3,1<\beta \u2a7d2,\hfill \\ \xi (0)\u2a7d0,\phantom{\rule{2em}{0ex}}{\xi}^{\prime}(0)\u2a7d0,\phantom{\rule{2em}{0ex}}{\xi}^{\prime}(1)\u2a7d0,\hfill \\ {D}_{0+}^{\alpha}\xi (0)\u2a7e0,\phantom{\rule{2em}{0ex}}{D}_{0+}^{\alpha}\xi (1)\u2a7e0.\hfill \end{array}
Definition 2.5[8]
The map θ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E provided that \theta :P\to [0,+\mathrm{\infty}) is continuous and
\theta (tx+(1t)y)\u2a7et\theta (x)+(1t)\theta (y)
for all x,y\in P and 0\u2a7dt\u2a7d1.
Lemma 2.7[8]
Let E be a Banach space, P\subseteq Ebe a cone, and{\mathrm{\Omega}}_{1}, {\mathrm{\Omega}}_{2}be two bounded open balls of E centered at the origin with{\overline{\mathrm{\Omega}}}_{1}\subset {\mathrm{\Omega}}_{2}. Suppose that\mathcal{A}:P\cap ({\overline{\mathrm{\Omega}}}_{2}\setminus {\mathrm{\Omega}}_{1})\to Pis a completely continuous operator such that either

(i)
\parallel \mathcal{A}x\parallel \u2a7d\parallel x\parallel, x\in P\cap \partial {\mathrm{\Omega}}_{1} and \parallel \mathcal{A}x\parallel \u2a7e\parallel x\parallel, x\in P\cap \partial {\mathrm{\Omega}}_{2} or

(ii)
\parallel \mathcal{A}x\parallel \u2a7e\parallel x\parallel, x\in P\cap \partial {\mathrm{\Omega}}_{1} and \parallel \mathcal{A}x\parallel \u2a7d\parallel x\parallel, x\in P\cap \partial {\mathrm{\Omega}}_{2}
holds. Then\mathcal{A}has a fixed point inP\cap ({\overline{\mathrm{\Omega}}}_{2}\setminus {\mathrm{\Omega}}_{1}).
Let a,b,c>0 be constants, {P}_{c}=\{u\in P:\parallel u\parallel <c\}, P(\theta ,b,d)=\{u\in P:b\u2a7d\theta (u),\parallel u\parallel \u2a7dd\}.
Lemma 2.8[8]
Let P be a cone in a real Banach space E, {P}_{c}=\{x\in P\mid \parallel x\parallel \u2a7dc\}, θ be a nonnegative continuous concave functional on P such that\theta (x)\u2a7d\parallel x\parallelfor allx\in {\overline{P}}_{c}, andP(\theta ,b,d)=\{x\in P\mid b\u2a7d\theta (x),\parallel x\parallel \u2a7dd\}. Suppose\mathcal{B}:{\overline{P}}_{c}\to {\overline{P}}_{c}is completely continuous and there exist constants0<a<b<d\u2a7dcsuch that
(C1) \{x\in P(\theta ,b,d)\mid \theta (x)>b\}\ne \mathrm{\varnothing}and\theta (\mathcal{B}x)>bforx\in P(\theta ,b,d);
(C2) \parallel \mathcal{B}x\parallel <aforx\u2a7da;
(C3) \theta (\mathcal{B}x)>bforx\in P(\theta ,b,c)with\parallel \mathcal{B}x\parallel >d.
Then ℬ has at least three fixed points{x}_{1}, {x}_{2}, and{x}_{3}with
\parallel {x}_{1}\parallel <a,\phantom{\rule{2em}{0ex}}b<\theta ({x}_{2}),\phantom{\rule{2em}{0ex}}a<\parallel {x}_{3}\parallel \phantom{\rule{1em}{0ex}}\mathit{\text{with}}\phantom{\rule{1em}{0ex}}\theta ({x}_{3})<b.
Let E=C[0,1] be endowed with \parallel u\parallel ={max}_{0\u2a7dt\u2a7d1}u(t). Define the cone P\subset E by
P=\{u\in E\mid u(t)\u2a7e0\}.
Let the nonnegative continuous concave functional θ on the cone P be defined by
\theta (u)=\underset{1/4\u2a7dt\u2a7d3/4}{min}u(t).
Lemma 2.9
Let
T:P\to E
be the operator defined by
Tu(t):={\int}_{0}^{1}G(t,s){\varphi}_{q}({\int}_{0}^{1}H(s,\tau )f(\tau ,u(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds.
ThenT:P\to Pis completely continuous.
Proof Let u\in P, in view of the nonnegativeness and continuity of G(t,s), H(t,s), and f(t,u(t)), we have T:P\to P is continuous.
Let \mathrm{\Omega}\subset P be bounded, i.e., there exists a positive constant M>0 such that \parallel u\parallel \u2a7dM for all u\in \mathrm{\Omega}. Let L={max}_{0\u2a7dt\u2a7d1,0\u2a7du\u2a7dM}f(t,u)+1, then for u\in \mathrm{\Omega}, we have
\begin{array}{rcl}Tu(t)& =& {\int}_{0}^{1}G(t,s){\varphi}_{q}({\int}_{0}^{1}H(s,\tau )f(\tau ,u(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \u2a7d& {L}^{q1}{\int}_{0}^{1}G(t,s){\varphi}_{q}({\int}_{0}^{1}H(s,\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ \u2a7d& {L}^{q1}{\int}_{0}^{1}G(1,s){\varphi}_{q}({\int}_{0}^{1}H(\tau ,\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ <& +\mathrm{\infty}.\end{array}
Hence, T(\mathrm{\Omega}) is uniformly bounded.
On the other hand, since G(t,s) is continuous on [0,1]\times [0,1], it is uniformly continuous on [0,1]\times [0,1]. Thus, for fixed s\in [0,1] and for any \epsilon >0, there exists a constant \delta >0 such that any {t}_{1},{t}_{2}\in [0,1] and {t}_{1}{t}_{2}<\delta,
G({t}_{1},s)G({t}_{2},s)<\frac{\epsilon}{{L}^{q1}{\varphi}_{q}({\int}_{0}^{1}H(\tau ,\tau )\phantom{\rule{0.2em}{0ex}}d\tau )}.
Then, for all u\in \mathrm{\Omega},
\begin{array}{c}Tu({t}_{2})Tu({t}_{1})\hfill \\ \phantom{\rule{1em}{0ex}}\u2a7d{\int}_{0}^{1}G({t}_{2},s)G({t}_{1},s){\varphi}_{q}({\int}_{0}^{1}H(s,\tau )f(\tau ,u(\tau ))\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\hfill \\ \phantom{\rule{1em}{0ex}}\u2a7d{L}^{q1}{\int}_{0}^{1}G({t}_{2},s)G({t}_{1},s){\varphi}_{q}({\int}_{0}^{1}H(\tau ,\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\hfill \\ \phantom{\rule{1em}{0ex}}={L}^{q1}{\varphi}_{q}({\int}_{0}^{1}H(\tau ,\tau )\phantom{\rule{0.2em}{0ex}}d\tau ){\int}_{0}^{1}G({t}_{2},s)G({t}_{1},s)\phantom{\rule{0.2em}{0ex}}ds\hfill \\ \phantom{\rule{1em}{0ex}}<\epsilon ,\hfill \end{array}
that is to say, T(\mathrm{\Omega}) is equicontinuous. By the ArzelaAscoli theorem, we have T:P\to P is completely continuous. The proof is complete. □