For convenience of the reader, we present here some necessary definitions.
Definition 2.1
[24]
Let X be the Banach space with norm \(\|\cdot\|\). A nonempty closed set \(K\subset X\) is called a cone if K satisfies the following conditions: (i) if \(x,y\in K\), then \(x+y\in K\); (ii) if \(x\in K\), then \(\mu x\in K\), for any \(\mu\geq0\); (iii) if \(0\neq x\in K\), then \(-x\notin K\). Let \(x_{1}, x_{2} \in X\). We write \(x_{1}\preceq x_{2}\), if \(x_{2}-x_{1}\in K\). The cone K is called normal, if there exists \(\rho>0\) such that
$$\|x_{1}+x_{2}\|\geq\rho,\quad \forall x_{1}, x_{2} \in K, \|x_{1}\|=\|x_{2}\|=1. $$
We call the set \([x_{1},x_{2}]=\{x \in X, x_{1}\preceq x\preceq x_{2}\}\) an order interval in X. The operator \(T:[x_{1},x_{2}]\rightarrow X\) is called increasing if \(T\bar{x}\preceq T\tilde{x}\) for any \(\bar{x}, \tilde{x}\in[x_{1},x_{2}]\) with \(\bar{x}\preceq \tilde{x} \).
Definition 2.2
[3]
The Riemann-Liouville fractional integral of order \(\alpha>0\) of a function \(y:(0,\infty)\rightarrow R\) is given by
$$I_{0+}^{\alpha}y(t)=\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds, $$
provided the right side is pointwise defined on \([0,\infty)\).
Definition 2.3
[3]
The Riemann-Liouville fractional derivative of order \(\alpha>0\) of a function \(y:(0,\infty)\rightarrow R\) is given by
$$D_{0+}^{\alpha}y(t)=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}} \int _{0}^{t}(t-s)^{n-\alpha-1}y(s)\,ds, $$
where \(n=[\alpha]+1\), \([\alpha]\) denotes the integer part of number α, provided the right side is pointwise defined on \([0,\infty)\).
From the definitions of Riemann-Liouville’s derivative, we can obtain the statements.
Lemma 2.4
[3]
Let
\(\alpha>0\), if we assume
\(u\in C(0,1)\cap L^{1}(0,1)\), then the fractional differential equation
$$D_{0+}^{\alpha}u(t)=0 $$
has
\(u(t)=C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+\cdots+C_{N}t^{\alpha-N}\), from some
\(C_{i}\in R\), \(i=1,2,\ldots,N\), as a unique solution, where
N
is the smallest integer greater than or equal to
α.
Lemma 2.5
[3]
Let
\(\alpha>0\), if we assume
\(u\in C(0,1)\cap L^{1}(0,1)\), then the fractional differential equation
$$I_{0+}^{\alpha}D_{0+}^{\alpha}u(t)=u(t)+C_{1}t^{\alpha-1}+C_{2}t^{\alpha -2}+ \cdots+C_{N}t^{\alpha-N}, $$
from some
\(C_{i}\in R\), \(i=1,2,\ldots,N\), where
N
is the smallest integer greater than or equal to
α.
Lemma 2.6
Assume that
\(y\in C(0,1)\cap L^{1}(0,1)\), α, β
are two positive constants with
\(\alpha-\beta\geq0\). Then
$$D_{0+}^{\beta}I_{0+}^{\alpha}y(t)=I_{0+}^{\alpha-\beta}y(t). $$
Proof
Let \(h(t)=\int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds\). Then by Definition 2.3,
$$D_{0+}^{\beta}h(t)=\frac{1}{\Gamma(n-\beta)}\frac{d^{n}}{dt^{n}} \int _{0}^{t}(t-\tau)^{n-1-\beta}h(\tau)\,d\tau. $$
On the other hand, let \(\frac{\tau-s}{t-s}=u\), then
$$\begin{aligned} \int_{0}^{t}(t-\tau)^{n-1-\beta}h(\tau)\,d\tau =& \int_{0}^{t}(t-\tau )^{n-1-\beta} \biggl( \int_{0}^{\tau}(\tau-s)^{\alpha-1}y(s)\,ds \biggr) \,d\tau \\ =& \int_{0}^{t}ds \int_{s}^{t}(t-\tau)^{n-1-\beta}( \tau-s)^{\alpha -1}y(s)\,d\tau \\ =& \int_{0}^{t}(t-s)^{\alpha-\beta+(n-1)}y(s)\,ds \int_{0}^{1}u^{\alpha -1}(1-u)^{(n-\beta)-1}\,du \\ =&B(\alpha,n-\beta) \int_{0}^{t}(t-s)^{\alpha-\beta+(n-1)}y(s)\,ds. \end{aligned}$$
Therefore, we have
$$\begin{aligned} D_{0+}^{\beta}h(t) =&\frac{B(\alpha,n-\beta)}{\Gamma(n-\beta)}\frac {d^{n}}{dt^{n}} \int_{0}^{t}(t-s)^{\alpha-\beta+(n-1)}y(s)\,ds \\ =&\frac{B(\alpha,n-\beta)}{\Gamma(n-\beta)} \int_{0}^{t}\bigl[\alpha-\beta +(n-1)\bigr] \bigl[ \alpha-\beta+(n-2)\bigr]\cdots[\alpha-\beta] \\ &{}\times(t-s)^{\alpha-\beta -1}y(s)\,ds \\ =&\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)} \int_{0}^{t}(t-s)^{\alpha -\beta-1}y(s)\,ds. \end{aligned}$$
□
In the remainder of the paper, we always assume that \(2<\alpha\leq3\) and \(0\leq\beta\leq1\), so we also easily get \(0<\alpha-\beta-1\leq2\).
Lemma 2.7
Let
\(y\in C(0,1)\cap L^{1}[0,1]\)
and
\(\delta:=\int_{0}^{1}t^{\alpha-\beta -1}\,dA(t)<1\), the unique solution of
$$ \left \{ \textstyle\begin{array}{l} D_{0+}^{\alpha}u(t)+y(t)=0,\quad t\in(0,1), \\ u ( 0 ) =u' ( 0 ) =0,\qquad D_{0+}^{\beta}u(1)=\int _{0}^{1}D_{0+}^{\beta}u(t)\,dA(t) \end{array}\displaystyle \right . $$
(2.1)
is
\(u(t)=\int_{0}^{1}G(t,s)y(s)\,ds\), in which
$$G(t,s)=K(t,s)+\frac{t^{\alpha-1}}{1-\delta} \int_{0}^{1}H(t,s)\,dA(t), $$
where
$$ K(t,s)=\frac{1}{\Gamma(\alpha)}\left \{ \textstyle\begin{array}{l@{\quad}l} t^{\alpha-1}(1-s)^{\alpha-\beta-1}-(t-s)^{\alpha-1}, & 0\leq s\leq t\leq1, \\ t^{\alpha-1}(1-s)^{\alpha-\beta-1},&0\leq t\leq s\leq1 \end{array}\displaystyle \right . $$
and
$$ H(t,s)=\frac{1}{\Gamma(\alpha)}\left \{ \textstyle\begin{array}{l@{\quad}l} {[t(1-s) ]}^{\alpha-\beta-1}-(t-s)^{\alpha-\beta-1}, & 0\leq s\leq t\leq1, \\ {[t(1-s) ]}^{\alpha-\beta-1},&0\leq t\leq s\leq1. \end{array}\displaystyle \right . $$
Proof
As deduced from Lemma 2.5, we have
$$u(t)=-I_{0+}^{\alpha}y(t)+C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+C_{3}t^{\alpha-3}. $$
Consequently, the solution of (2.1) is
$$u(t)=-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}y(s) \,ds+C_{1}t^{\alpha-1}+C_{2}t^{\alpha-2}+C_{3}t^{\alpha-3}. $$
By \(u(0)=u'(0)=0\), there is \(C_{2}=C_{3}=0\). Moreover, \(D_{0+}^{\beta}u(1)=\int_{0}^{1}D_{0+}^{\beta}u(t)\,dA(t)\) and Lemma 2.6 yield
$$C_{1}=\frac{1}{\Gamma(\alpha)(1-\delta)} \biggl[ \int_{0}^{1}(1-s)^{\alpha -\beta-1}y(s)\,ds- \int_{0}^{1} \biggl( \int_{0}^{t}(t-s)^{\alpha-\beta -1}y(s)\,ds \biggr) \,dA(t) \biggr]. $$
Therefore, the solution of (2.1) is
$$\begin{aligned} u(t) =&-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds+ \frac {t^{\alpha-1}}{\Gamma(\alpha)(1-\delta)} \\ &{}\times \biggl[ \int_{0}^{1}(1-s)^{\alpha-\beta-1}y(s)\,ds- \int_{0}^{1} \biggl( \int _{0}^{t}(t-s)^{\alpha-\beta-1}y(s)\,ds \biggr) \,dA(t) \biggr] \\ =&-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha-1}y(s)\,ds+ \frac {t^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-\beta -1}y(s)\,ds \\ &{}+\frac{t^{\alpha-1}}{\Gamma(\alpha)(1-\delta)} \biggl( \int_{0}^{1} \biggl[ \int_{0}^{1}t^{\alpha-\beta-1}(1-s)^{\alpha-\beta-1}y(s) \,ds \biggr]\,dA(t) \\ &{}- \int_{0}^{1} \biggl[ \int_{0}^{t}(t-s)^{\alpha-\beta-1}y(s)\,ds \biggr] \,dA(t) \biggr) \\ =& \int_{0}^{1} \biggl[K(t,s)+\frac{t^{\alpha-1}}{1-\delta} \int _{0}^{1}H(t,s)\,dA(t) \biggr]y(s)\,ds. \end{aligned}$$
□
Lemma 2.8
For any
\(t,s\in[0,1]\), \(H(t,s)\)
has the following property:
$$\min \{\alpha-\beta-1,1 \}t^{\alpha-\beta -1}(1-t) (1-s)^{\alpha-\beta-1}s\leq\Gamma( \alpha)H(t,s)\leq\max \{\alpha-\beta-1,1 \}(1-s)^{\alpha-\beta-1}. $$
Proof
Case 1: \(0<\alpha-\beta-1\leq1\). If \(s\leq t\), then
$$\begin{aligned} \begin{aligned} \Gamma(\alpha)H(t,s)&=\bigl[t(1-s)\bigr]^{\alpha-\beta-1}-(t-s)^{\alpha-\beta -1} \\ &= (\alpha-\beta-1) \int_{t-s}^{t(1-s)}x^{\alpha-\beta-2}\, dx \\ &\geq(\alpha-\beta-1)\bigl[t(1-s)\bigr]^{\alpha-\beta-2}\bigl[t(1-s)-(t-s)\bigr] \\ &\geq (\alpha-\beta-1)\bigl[t(1-s)\bigr]^{\alpha-\beta-1}s(1-t) \\ &= (\alpha-\beta-1)t^{\alpha-\beta-1}(1-t) (1-s)^{\alpha-\beta-1}s \end{aligned} \end{aligned}$$
and
$$\begin{aligned} \Gamma(\alpha)H(t,s) =&\bigl[t(1-s)\bigr]^{\alpha-\beta-1}-(t-s)^{\alpha-\beta -1} \\ =& \bigl[t(1-s)\bigr]^{\alpha-\beta-2} \biggl[t(1-s)-(t-s)\cdot\frac{(t-s)^{\alpha -\beta-2}}{[t(1-s)]^{\alpha-\beta-2}} \biggr] \\ =&\bigl[t(1-s)\bigr]^{\alpha-\beta-2} \biggl[t(1-s)-(t-s)\cdot\frac{(1-\frac {s}{t})^{\alpha-\beta-2}}{(1-s)^{\alpha-\beta-2}} \biggr] \\ \leq& \bigl[t(1-s)\bigr]^{\alpha-\beta-2}\bigl[t(1-s)-(t-s)\bigr] \\ \leq&t^{\alpha-\beta-2}(1-t) (1-s)^{\alpha-\beta-2}s \\ \leq&s^{\alpha-\beta-2}(1-s) (1-s)^{\alpha-\beta-2}s \\ =&s^{\alpha-\beta-1}(1-s)^{\alpha-\beta-1}\leq(1-s)^{\alpha-\beta-1}. \end{aligned}$$
If \(s\geq t\), then
$$ \Gamma(\alpha)H(t,s)=\bigl[t(1-s)\bigr]^{\alpha-\beta-1}\leq s^{\alpha-\beta -1}(1-s)^{\alpha-\beta-1} \leq(1-s)^{\alpha-\beta-1} $$
and
$$\begin{aligned} \Gamma(\alpha)H(t,s) =&\bigl[t(1-s)\bigr]^{\alpha-\beta-1}\geq t^{\alpha-\beta -1}(1-t) (1-s)^{\alpha-\beta-1}s \\ \geq& (\alpha-\beta-1)t^{\alpha-\beta-1}(1-t) (1-s)^{\alpha-\beta-1}s. \end{aligned}$$
Case 2: \(1<\alpha-\beta-1\leq2\). We have
$$(1-t)t^{\alpha-\beta-1}s(1-s)^{\alpha-\beta-1}\leq\Gamma(\alpha )H(t,s)\leq(\alpha- \beta-1) (1-s)^{\alpha-\beta-1}. $$
The proof is similar to Lemma 2.3 in [16], so it is omitted. □
Lemma 2.9
The function
\(K(t,s)\)
satisfies:
-
(i)
\(\Gamma(\alpha)K(t,s)\leq t^{\alpha-1}(1-s)^{\alpha-\beta-1}\), \(t,s\in[0,1]\);
-
(ii)
\(\Gamma(\alpha)K(t,s)\geq t^{\alpha-1}\beta s(1-s)^{\alpha-\beta -1}\), \(t,s\in[0,1]\).
Proof
Since (i) holds obviously, we only show that (ii) is true. Here we need the fact \(1-(1-s)^{\beta}\geq\beta s\), \(0<\beta\leq1\), \(s\in[0,1]\). In fact,
$$\bigl[1-(1-s)^{\beta}-\beta s \bigr]'=\beta \bigl[(1-s)^{\beta-1}-1 \bigr]\geq0, $$
which implies \(1-(1-s)^{\beta}-\beta s\) is nondecreasing in \([0,s]\), so \(1-(1-s)^{\beta}\geq\beta s\).
If \(s\leq t\), then
$$\begin{aligned} \Gamma(\alpha)K(t,s) =&t^{\alpha-1}(1-s)^{\alpha-\beta-1}-(t-s)^{\alpha -1} \\ \geq& t^{\alpha-1}(1-s)^{\alpha-\beta-1}-t^{\alpha-1}(1-s)^{\alpha-1} \\ =& t^{\alpha-1}(1-s)^{\alpha-\beta-1} \bigl[1-(1-s)^{\beta} \bigr] \\ \geq&t^{\alpha-1}(1-s)^{\alpha-\beta-1}\beta s. \end{aligned}$$
If \(t\leq s\), then
$$ \Gamma(\alpha)K(t,s)=t^{\alpha-1}(1-s)^{\alpha-\beta-1}\geq t^{\alpha -1}(1-s)^{\alpha-\beta-1}\beta s. $$
□
Lemma 2.10
Let
$$\Phi_{1}(s):=\frac{1}{\Gamma(\alpha)} \biggl[\beta+\frac{\min \{\alpha -\beta-1,1 \}\int_{0}^{1}t^{\alpha-\beta-1}(1-t)\,dA(t)}{1-\delta } \biggr]s(1-s)^{\alpha-\beta-1} $$
and
$$\Phi_{2}(s):=\frac{1}{\Gamma(\alpha)} \biggl[1+\frac{\max \{\alpha-\beta -1,1 \}\int_{0}^{1}\,dA(t)}{1-\delta} \biggr](1-s)^{\alpha-\beta-1}. $$
Then the function
\(G(t,s)\)
has the following property:
$$t^{\alpha-1}\Phi_{1} (s)\leq G(t,s)\leq t^{\alpha-1} \Phi_{2} (s),\quad t,s\in[0,1]. $$
Proof
For any \(t,s\in[0,1]\), it follows from Lemmas 2.8 and 2.9 that
$$\begin{aligned} G(t,s) =&K(t,s)+\frac{t^{\alpha-1}\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta} \\ \leq&\frac{1}{\Gamma(\alpha)}t^{\alpha-1}(1-s)^{\alpha-\beta-1}+\frac {t^{\alpha-1}\max \{\alpha-\beta-1,1 \}\int _{0}^{1}dA(t)}{\Gamma(\alpha) (1-\delta )}(1-s)^{\alpha-\beta -1} \\ \leq&t^{\alpha-1}\frac{1}{\Gamma(\alpha)} \biggl[1+\frac{\max \{ \alpha-\beta-1,1 \}\int_{0}^{1}dA(t)}{ (1-\delta )} \biggr](1-s)^{\alpha-\beta-1} \\ =&t^{\alpha-1}\Phi_{2}(s). \end{aligned}$$
On the other hand, for any \(t,s\in[0,1]\), we have
$$\begin{aligned} G(t,s) =&K(t,s)+\frac{t^{\alpha-1}\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta} \\ \geq&\frac{1}{\Gamma(\alpha)}t^{\alpha-1}\beta s(1-s)^{\alpha-\beta -1}+ \frac{t^{\alpha-1}\min \{\alpha-\beta-1,1 \}\int _{0}^{1}t^{\alpha-\beta-1}(1-t)\,dA(t)}{\Gamma(\alpha) (1-\delta)} \\ &{}\times s(1-s)^{\alpha-\beta-1} \\ =&t^{\alpha-1}\frac{1}{\Gamma(\alpha)} \biggl[\beta+\frac{\min \{ \alpha-\beta-1,1 \}\int_{0}^{1}t^{\alpha-\beta -1}(1-t)\,dA(t)}{1-\delta} \biggr]s(1-s)^{\alpha-\beta-1} \\ =&t^{\alpha-1}\Phi_{1}(s). \end{aligned}$$
□
Lemma 2.11
The function
\(G(t,s)\)
is continuous and satisfies
$$\begin{aligned} \bigl\vert G(t_{2},s)-G(t_{1},s)\bigr\vert &\leq\max _{0\leq s\leq1} \bigl\vert G(t_{2},s)-G(t_{1},s) \bigr\vert \\ &\leq(\alpha-1) \biggl[\frac{2}{\Gamma(\alpha )}+\frac{\max_{0\leq s\leq1}\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta } \biggr](t_{2}-t_{1}), \end{aligned}$$
for
\(0\leq t_{1}\leq t_{2}\leq1\).
Proof
For \(0\leq t_{1}\leq t_{2}\leq1\), we have
$$\begin{aligned} \bigl\vert G(t_{2},s)-G(t_{1},s)\bigr\vert =&\biggl\vert K(t_{2},s)-K(t_{1},s)+\frac{ (t_{2}^{\alpha-1}-t_{1}^{\alpha-1} )\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta }\biggr\vert \\ =&\biggl\vert \frac{1}{\Gamma(\alpha)} \bigl(t_{2}^{\alpha-1}-t_{1}^{\alpha -1} \bigr) (1-s)^{\alpha-\beta-1}+\frac{1}{\Gamma(\alpha)} \bigl[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1} \bigr] \\ &{}+\frac{ (t_{2}^{\alpha -1}-t_{1}^{\alpha-1} )\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta}\biggr\vert \\ \leq& \biggl(\frac{1}{\Gamma(\alpha)}\bigl\vert t_{2}^{\alpha-1}-t_{1}^{\alpha -1} \bigr\vert +\frac{1}{\Gamma(\alpha)}\bigl\vert (t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha-1} \bigr\vert \\ &{}+\frac{\vert t_{2}^{\alpha-1}-t_{1}^{\alpha -1}\vert \max_{0\leq s\leq1}\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta } \biggr). \end{aligned}$$
Note that, applying the mean value theorem, we arrive at \(t_{2}^{\alpha -1}-t_{1}^{\alpha-1}<(\alpha-1)(t_{2}-t_{1})\) and \((t_{2}-s)^{\alpha -1}-(t_{1}-s)^{\alpha-1}<(\alpha-1)(t_{2}-t_{1})\), which implies that
$$\begin{aligned} \bigl\vert G(t_{2},s)-G(t_{1},s)\bigr\vert &\leq\max _{0\leq s\leq1} \bigl\vert G(t_{2},s)-G(t_{1},s) \bigr\vert \\ &\leq(\alpha-1) \biggl[\frac{2}{\Gamma(\alpha)}+\frac{\max_{0\leq s\leq1}\int_{0}^{1}H(t,s)\,dA(t)}{1-\delta} \biggr](t_{2}-t_{1}), \end{aligned}$$
for \(0\leq t_{1}\leq t_{2}\leq1\). □