For \(v, w\in C_{1-\alpha}([0,T],E)\), we denote
$$\begin{aligned}{} [v, w] =&\bigl\{ u\in C_{1-\alpha}\bigl([0,T],E\bigr)\mid v(t)\leq u(t)\leq w(t), t\in(0, T], \\ &t^{1-\alpha}v(t)|_{t=0}\leq t^{1-\alpha}u(t)|_{t=0} \leq t^{1-\alpha}w(t)|_{t=0}\bigr\} . \end{aligned}$$
Our main results are as follows.
Theorem 3.1
Let
E
be an ordered Banach space, whose positive cone
P
is normal, \(f:[0, T]\times E\rightarrow E\)
be continuous. Assume that
\(v_{0}, w_{0}\in C_{1-\alpha}([0,T],E)\)
are lower and upper solutions of (1.2) such that (2.2) holds. If the following conditions are satisfied:
-
(H1)
There exists a constant
\(M > 0\)
such that
$$f(t, x_{2})-f(t, x_{1})\geq-M(x_{2}-x_{1}) $$
for
\(\forall t\in[0, T]\)
and
\(v_{0}\leq x_{1}\leq x_{2}\leq w_{0}\).
-
(H2)
There exists a constant
\(K > 0\)
with
$$\frac{2KT^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \biggl(\frac {1}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]}+1 \biggr)< 1 $$
such that
$$\mu \bigl(\bigl\{ f\bigl(t, u_{n}(t)\bigr)+Mu_{n}(t)\bigr\} \bigr)\leq K\mu \bigl(\bigl\{ u_{n}(t)\bigr\} \bigr) $$
for
\(\forall t\in[0, T]\), and a monotonous sequence
\(\{u_{n}\}\subset [v_{0}, w_{0}]\).
Then problem (1.2) has minimal and maximal solutions between
\(v_{0}\)
and
\(w_{0}\), which can be obtained by a monotone iterative procedure starting from
\(v_{0}\)
and
\(w_{0}\), respectively.
Proof
For any \(h\in[v_{0}, w_{0}]\), consider the linear periodic boundary value problem
$$ \textstyle\begin{cases} D^{\alpha}_{0}u(t)+Mu(t)=f(t,h(t))+Mh(t),\quad t\in(0, T], \\ t^{1-\alpha}u(t)|_{t=0}=t^{1-\alpha}u(t)|_{t=T}. \end{cases} $$
(3.1)
By Lemma 2.3, we obtain that problem (3.1) has unique solution u, which can be expressed as follows:
$$\begin{aligned} u(t) =&\frac{T^{1-\alpha}\Gamma(\alpha)t^{\alpha-1}E_{\alpha,\alpha }(-Mt^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int _{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr) \bigl(f\bigl(t,h(s)\bigr) \\ &{}+Mh(s)\bigr)\,ds + \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(t-s)^{\alpha }\bigr) \bigl(f\bigl(t,h(s)\bigr)+Mh(s)\bigr)\,ds \\ :=&Ah(t). \end{aligned}$$
(3.2)
Firstly, we need to show that the operator \(A:[v_{0}, w_{0}]\rightarrow C_{1-\alpha}([0,T],E)\) is well defined, i.e., for \(h\in[v_{0}, w_{0}]\), \(Ah\in C_{1-\alpha}([0,T],E)\). By (H1), for \(h\in[v_{0}, w_{0}]\), we have
$$f\bigl(t,v_{0}(t)\bigr)+Mv_{0}(t)\leq f\bigl(t,h(t) \bigr)+Mh(t)\leq f\bigl(t,w_{0}(t)\bigr)+Mw_{0}(t). $$
We denote
$$F(h) (t)=f\bigl(t,h(t)\bigr)+Mh(t),\quad \forall h\in[v_{0}, w_{0}]. $$
By the normality of the cone P, there exists \(L>0\) such that\(\|F(h)\| _{1-\alpha}\leq L\), that is,
$$ \bigl\Vert f\bigl(t,h(t)\bigr)+Mh(t) \bigr\Vert \leq Lt^{\alpha-1}. $$
(3.3)
By (2.8) and (3.3), we have that
$$\begin{aligned} \bigl\Vert t^{1-\alpha}(Ah) (t) \bigr\Vert =& \biggl\Vert \frac{T^{1-\alpha}\Gamma(\alpha )E_{\alpha,\alpha}(-Mt^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha }(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(T-s)^{\alpha}\bigr)F(h) (s)\,ds \\ &{} +t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t-s)^{\alpha}\bigr)F(h) (s)\,ds \biggr\Vert \\ \leq&\frac{T^{1-\alpha}}{\Gamma(\alpha)[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1} \bigl\Vert F(h) (s) \bigr\Vert \,ds \\ &{} +\frac{t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1} \bigl\Vert F(h) (s) \bigr\Vert \,ds \\ \leq&\frac{T^{1-\alpha}L}{\Gamma(\alpha)[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}s^{\alpha-1}\,ds \\ &{} +\frac{t^{1-\alpha}L}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha -1}s^{\alpha-1}\,ds \\ =&\frac{T^{\alpha}L\Gamma(\alpha)}{\Gamma(2\alpha)[1-\Gamma(\alpha )E_{\alpha,\alpha}(-MT^{\alpha})]}+ \frac{L\Gamma(\alpha)t^{\alpha}}{\Gamma(2\alpha)}. \end{aligned}$$
That is to say, the integral in (3.2) exists and belongs to \(C_{1-\alpha }([0,T],E)\).
By Lemma (2.3), the solution of problem (1.2) is equivalent to the fixed point of the operator A. Now, we complete the proof by four steps.
Step 1. We show that the operator \(A: [v_{0}, w_{0}]\rightarrow C_{1-\alpha}([0,T],E)\) is equicontinuous. For any \(u\in[v_{0}, w_{0}]\) and \(0\leq t_{1}\leq t_{2}\leq T\), we have
$$\begin{aligned}& \bigl\Vert t_{2}^{1-\alpha}Au(t_{2}) - t_{1}^{1-\alpha}Au(t_{1}) \bigr\Vert \\& \quad \leq \biggl\Vert \frac{T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha }(-Mt_{2}^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha })]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha }\bigr)F(u) (s)\,ds \\& \qquad {} +t_{2}^{1-\alpha} \int_{0}^{t} (t_{2}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{2}-s)^{\alpha}\bigr)F(u) (s)\,ds \\& \qquad {} -\frac{T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha}(-Mt_{1}^{\alpha })}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr)F(u) (s)\,ds \\& \qquad {} -t_{1}^{1-\alpha} \int_{0}^{t} (t_{1}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{1}-s)^{\alpha}\bigr)F(u) (s)\,ds \biggr\Vert \\& \quad \leq \frac{T^{1-\alpha}\Gamma(\alpha)|E_{\alpha,\alpha}(-Mt_{2}^{\alpha })-E_{\alpha,\alpha}(-Mt_{1}^{\alpha})|}{[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \\& \qquad {}\times\int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(T-s)^{\alpha}\bigr) \bigl\Vert F(u) (s) \bigr\Vert \,ds \\& \qquad {} + \biggl\Vert t_{2}^{1-\alpha} \int_{0}^{t} (t_{2}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr)F(u) (s)\,ds \\& \qquad {} -t_{1}^{1-\alpha} \int_{0}^{t} (t_{1}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{1}-s)^{\alpha}\bigr)F(u) (s)\,ds \biggr\Vert . \end{aligned}$$
For the first term of the above formula, by (2.8) and (3.3), we have
$$\begin{aligned}& \frac{T^{1-\alpha}\Gamma(\alpha)|E_{\alpha,\alpha}(-Mt_{2}^{\alpha })-E_{\alpha,\alpha}(-Mt_{1}^{\alpha})|}{[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(T-s)^{\alpha}\bigr) \bigl\Vert F(u) (s) \bigr\Vert \,ds \\& \quad \leq \frac{T^{1-\alpha}L}{[1-\Gamma(\alpha)E_{\alpha,\alpha }(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}s^{\alpha-1}\,ds \bigl\vert E_{\alpha ,\alpha}\bigl(-Mt_{2}^{\alpha} \bigr)-E_{\alpha,\alpha}\bigl(-Mt_{1}^{\alpha}\bigr) \bigr\vert \\& \quad = \frac{T^{\alpha}(\Gamma(\alpha))^{2}L}{\Gamma(2\alpha)[1-\Gamma (\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \bigl\vert E_{\alpha,\alpha }\bigl(-Mt_{2}^{\alpha} \bigr)-E_{\alpha,\alpha}\bigl(-Mt_{1}^{\alpha}\bigr) \bigr\vert . \end{aligned}$$
The function \(E_{\alpha,\alpha}(-Mt^{\alpha})\) is continuous, so the previous expression has limit zero as \(| t_{2}-t_{1}| \rightarrow0\).
For the rest, by the properties of \(E_{\alpha,\alpha}(x)\) discussed in [16, Proposition 1], we have
$$\begin{aligned}& \biggl\Vert t_{2}^{1-\alpha} \int_{0}^{t_{2}}(t_{2}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{2}-s)^{\alpha}\bigr)F(u) (s)\,ds \\& \qquad {} -t_{1}^{1-\alpha} \int_{0}^{t_{1}} (t_{1}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{1}-s)^{\alpha}\bigr)F(u) (s)\,ds \biggr\Vert \\& \quad \leq \biggl\Vert \int_{0}^{t_{1}} \bigl[t_{2}^{1-\alpha}(t_{2}-s)^{\alpha -1}E_{\alpha,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr) -t_{1}^{1-\alpha}(t_{1}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{1}-s)^{\alpha}\bigr) \bigr] \\& \qquad {}\times F(u) (s)\,ds \biggr\Vert \\& \qquad {} + \biggl\Vert t_{2}^{1-\alpha} \int_{t_{1}}^{t_{2}} (t_{2}-s)^{\alpha -1}E_{\alpha,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr)F(u) (s)\,ds \biggr\Vert \\& \quad \leq L \int_{0}^{t_{1}} \bigl[t_{1}^{1-\alpha}(t_{1}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t_{1}-s)^{\alpha}\bigr) -t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{2}-s)^{\alpha}\bigr) \bigr]s^{\alpha-1}\,ds \\& \qquad {} +L \int_{t_{1}}^{t_{2}}t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \quad = L \int_{0}^{t_{1}}t_{1}^{1-\alpha}(t_{1}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{1}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \qquad {}-L \int_{0}^{t_{2}}t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{2}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \qquad {} +2L \int_{t_{1}}^{t_{2}}t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \quad = L\Gamma(\alpha) \bigl(t_{1}^{1-\alpha}\bigl(I_{0}^{\alpha}t_{1}^{1-\alpha }E_{\alpha,\alpha} \bigl(-Mt_{1}^{\alpha}\bigr)\bigr)- t_{2}^{1-\alpha} \bigl(I_{0}^{\alpha}t_{2}^{1-\alpha}E_{\alpha,\alpha } \bigl(-Mt_{2}^{\alpha}\bigr)\bigr) \bigr) \\& \qquad {} +2L \int_{t_{1}}^{t_{2}}t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \quad = L\Gamma(\alpha) \bigl(t_{1}^{\alpha}E_{\alpha,2\alpha} \bigl(-Mt_{1}^{\alpha}\bigr)- t_{2}^{\alpha}E_{\alpha,2\alpha} \bigl(-Mt_{2}^{\alpha}\bigr) \bigr) \\& \qquad {}+2L \int _{t_{1}}^{t_{2}}t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t_{2}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \quad = -\frac{L\Gamma(\alpha)}{M} \bigl(E_{\alpha,\alpha}\bigl(-Mt_{1}^{\alpha} \bigr)- E_{\alpha,\alpha}\bigl(-Mt_{2}^{\alpha}\bigr) \bigr) \\& \qquad {}+2L \int_{t_{1}}^{t_{2}}t_{2}^{1-\alpha}(t_{2}-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t_{2}-s)^{\alpha}\bigr)s^{\alpha-1}\,ds \\& \quad \leq -\frac{L\Gamma(\alpha)}{M} \bigl(E_{\alpha,\alpha}\bigl(-Mt_{1}^{\alpha} \bigr)- E_{\alpha,\alpha}\bigl(-Mt_{2}^{\alpha}\bigr) \bigr)+ \frac{2Lt_{2}^{1-\alpha }t_{1}^{\alpha-1}}{\Gamma(\alpha)} \int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}\,ds \\& \quad = -\frac{L\Gamma(\alpha)}{M} \bigl(E_{\alpha,\alpha}\bigl(-Mt_{1}^{\alpha} \bigr)- E_{\alpha,\alpha}\bigl(-Mt_{2}^{\alpha}\bigr) \bigr)+ \frac{2Lt_{2}^{1-\alpha }t_{1}^{\alpha-1}}{\Gamma(\alpha)} (t_{2}-t_{1})^{\alpha}. \end{aligned}$$
Obviously, the previous expression also tends to zero for \(| t_{2}-t_{1}|\rightarrow0\). That is to say, \(A:[v_{0}, w_{0}]\rightarrow C_{1-\alpha}([0,T],E)\) is equicontinuous.
Step 2. We show that \(v_{0}\leq Av_{0}, Aw_{0}\leq w_{o}\), and \(Au_{1}\leq Au_{2}\) for any \(u_{1}, u_{2}\in[v_{0}, w_{0}]\) with \(u_{1}\leq u_{2}\).
Let
$$\sigma(t):= D^{\alpha}v_{0}(t)+Mv_{0}(t), $$
then, by Definition 2.3, we have \(\sigma(t)\leq F(v_{0})(t)\). Hence,
$$\begin{aligned} v_{0}(t) =&\frac{T^{1-\alpha}\Gamma(\alpha)t^{\alpha-1}E_{\alpha,\alpha }(-Mt^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int _{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr)\sigma(s)\,ds \\ &{}+ \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(t-s)^{\alpha}\bigr)\sigma (s)\,ds \\ \leq&\frac{T^{1-\alpha}\Gamma(\alpha)t^{\alpha-1}E_{\alpha,\alpha }(-Mt^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int _{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr)F(v_{0}) (s)\,ds \\ &{}+ \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(t-s)^{\alpha }\bigr)F(v_{0}) (s)\,ds \\ =&Av_{0}(t), \end{aligned}$$
namely \(v_{0}\leq Av_{0}\). Similarly, we can show that \(Aw_{0}\leq w_{0}\). For any \(u_{1}, u_{2}\in[v_{0}, w_{0}]\) with \(u_{1}\leq u_{2}\), by assumption (H1),
$$f\bigl(t,u_{1}(t)\bigr)+Mu_{1}(t)\leq f\bigl(t,u_{2}(t) \bigr)+Mu_{2}(t), $$
which implies that \(Au_{1}\leq Au_{2}\).
Step 3. From Step 2 we know that A maps \([v_{0}, w_{0}]\) into itself, and \(A:[v_{0}, w_{0}]\rightarrow[v_{0}, w_{0}]\) is a continuously increasing operator. We can now define the sequences
$$ v_{n}=Av_{n-1},\qquad w_{n}=Aw_{n-1},\quad n=1,2,\ldots. $$
(3.4)
Then from the monotonicity of A it follows that
$$ v_{0}\leq v_{1}\leq v_{2}\leq\cdots\leq v_{n}\leq\cdots\leq w_{n}\leq\cdots\leq w_{2}\leq w_{1}\leq w_{0}. $$
(3.5)
Obviously, \(\{v_{n}\}, \{w_{n}\}\subset[v_{0}, w_{0}]\) are equicontinuous. Next, we show that \(\{v_{n}\}\) and \(\{w_{n}\}\) are convergent in \(C_{1-\alpha}([0,T],E)\).
From (H2), Lemma 2.1, and Lemma 2.2, for any \(t\in[0, T]\), we have that
$$\begin{aligned}& t^{1-\alpha}\mu\bigl\{ v_{n}(t)\bigr\} \\& \quad = t^{1-\alpha}\mu\bigl\{ Av_{n-1}(t)\bigr\} \\& \quad = t^{1-\alpha}\mu \biggl( \biggl\{ \frac{T^{1-\alpha}\Gamma(\alpha)t^{\alpha -1}E_{\alpha,\alpha}(-Mt^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha }(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(T-s)^{\alpha}\bigr)F(v_{n-1}) (s)\,ds \\& \qquad {} + \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(t-s)^{\alpha }\bigr)F(v_{n-1}) (s)\,ds \biggr\} \biggr) \\& \quad \leq \mu \biggl( \biggl\{ \frac{T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha }(-Mt^{\alpha})}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int _{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha }\bigr)F(v_{n-1}) (s)\,ds \biggr\} \biggr) \\& \qquad {} +t^{1-\alpha}\mu \biggl( \biggl\{ \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha ,\alpha} \bigl(-M(t-s)^{\alpha}\bigr)F(v_{n-1}) (s)\,ds \biggr\} \biggr) \\& \quad \leq \frac{2T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha}(-Mt^{\alpha })}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr)\mu \bigl(\bigl\{ F(v_{n-1}) (s)\bigr\} \bigr)\,ds \\& \qquad {} +2t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t-s)^{\alpha}\bigr)\mu \bigl(\bigl\{ F(v_{n-1}) (s)\bigr\} \bigr)\,ds \\& \quad \leq \frac{2KT^{1-\alpha}E_{\alpha,\alpha}(-Mt^{\alpha})}{[1-\Gamma (\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}\mu \bigl(\bigl\{ v_{n-1}(s)\bigr\} \bigr)\,ds \\& \qquad {} +\frac{2Kt^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha-1}\mu \bigl(\bigl\{ v_{n-1}(s)\bigr\} \bigr)\,ds \\& \quad \leq \frac{2KT^{1-\alpha}\Gamma(\alpha)}{[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}s^{\alpha-1}\,ds \mu \bigl(\{v_{n-1}\} \bigr) \\& \qquad {} +\frac{2Kt^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha -1}s^{\alpha-1}\,ds \mu \bigl(\{v_{n-1}\} \bigr) \\& \quad = \frac{2KT^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \biggl(\frac {1}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]}+1 \biggr)\mu \bigl(\{ v_{n}\} \bigr). \end{aligned}$$
Since \(\{v_{n}\}\) is equicontinuous, using Lemma (2.1), we have
$$\mu \bigl(\{v_{n}\} \bigr)=\max_{t\in[0,T]}\bigl\{ t^{1-\alpha}\mu\big(\bigl\{ v_{n}(t)\bigr\} \big) \bigr\} \leq \frac{2KT^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \biggl(\frac {1}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]}+1 \biggr)\mu \bigl(\{ v_{n} \} \bigr). $$
While
$$\frac{2KT^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \biggl(\frac {1}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]}+1 \biggr)< 1, $$
hence \(\mu (\{v_{n}\} )=0\). So, \(\{v_{n}\}\) are relatively compact in \(C_{1-\alpha}([0,T],E)\). Hence, \(\{v_{n}\}\) has a convergent subsequence in \(C_{1-\alpha}([0,T],E)\). Combining this with the monotonicity (3.5), we easily prove that \(\{v_{n}\}\) itself is convergent in \(C_{1-\alpha }([0,T],E)\).
Using a similar argument to that for \(\{w_{n}\}\), we can prove that \(\{ w_{n}\}\) is also convergent in \(C_{1-\alpha}([0,T],E)\). Then there are \(\underline{u}, \bar{u} \in C_{1-\alpha}([0,T],E)\) such that
$$\underline{u}=\lim_{n\rightarrow\infty}v_{n},\qquad \bar{u}=\lim _{n\rightarrow\infty}w_{n}. $$
Letting \(n\rightarrow\infty\) in (3.4), we see that
$$\underline{u}=A\underline{u},\qquad \bar{u}=A\bar{u}. $$
Therefore, \(\underline{u}, \bar{u} \in C_{1-\alpha}([0,T],E)\) are fixed points of A.
Step 4. We prove the minimal and maximal property of \(\underline{u}\), ū. Assume that ũ is a fixed point of A in \([v_{0}, w_{0}]\), then we have
$$\begin{aligned}& v_{0}(t)\leq\tilde{u}(t)\leq w_{0}(t),\quad t\in(0,T], \\& t^{1-\alpha}v_{0}(t)|_{t=0}\leq t^{1-\alpha} \tilde{u}(t)|_{t=0}\leq t^{1-\alpha}w_{0}(t)|_{t=0}. \end{aligned}$$
By the monotonicity of A, it is easy to see that
$$\begin{aligned}& v_{1}(t)= (Av_{0}) (t)\leq(A\tilde{u}) (t)=\tilde{u}(t) \leq(Aw_{0}) (t)= w_{1}(t), \quad t\in(0,T], \\& t^{1-\alpha}v_{1}(t)|_{t=0}\leq t^{1-\alpha} \tilde{u}(t)|_{t=0}\leq t^{1-\alpha}w_{1}(t)|_{t=0}. \end{aligned}$$
Furthermore, we have
$$ v_{n}\leq\tilde{u}\leq w_{n},\quad n=1,2, \ldots. $$
(3.6)
Letting \(n\rightarrow\infty\) in (3.6), we obtain \(\underline{u}\leq \tilde{u}\leq\bar{u}\). So \(\underline{u}\), ū are the minimal and maximal fixed points of A in \([v_{0}, w_{0}]\), and therefore, they are the minimal and maximal solutions of problem (1.2) in \([v_{0}, w_{0}]\), respectively.
This completes the proof of Theorem 3.1. □
Remark 3.1
When \(E=\mathbb{R}\), we do not need condition (H3), \(\{v_{n}\}\) and \(\{w_{n}\}\) defined in (3.4) are convergent in \(C_{1-\alpha}([0,T],\mathbb{R})\) automatically. Therefore, Theorem 3.1 improves the main results in [13].
Next, we discuss the uniqueness of the solution to problem (1.2) in \([v_{o}, w_{o}]\).
Theorem 3.2
Let
E
be an ordered Banach space, whose positive cone
P
is normal, \(f:[0, T]\times E\rightarrow E\)
be continuous. Assume that
\(v_{0}, w_{0}\in C_{1-\alpha}([0,T],E)\)
are lower and upper solutions of (1.2) such that (2.2) holds. If conditions (H1), (H2) and the following condition are satisfied:
-
(H3)
There exists a constant
\(C > 0\)
with
$$\frac{N(M+C)T^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \biggl(\frac {1}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]}+1 \biggr)< 1 $$
such that
$$f(t, x_{2})-f(t, x_{1})\leq C(x_{2}-x_{1}), $$
for
\(\forall t\in[0, T]\), and
\(v_{0}\leq x_{1}\leq x_{2}\leq w_{0}\), where
N
is a normal constant.
Then problem (1.2) has a unique solution between
\(v_{0}\)
and
\(w_{0}\), which can be obtained by a monotone iterative procedure starting from
\(v_{0}\)
or
\(w_{0}\).
Proof
From the proof of Theorem 3.1, we know that the iterative sequences \(\{v_{n}\}\) and \(\{w_{n}\}\) defined by (3.4) satisfy (3.5). Now, we show that there exists a unique \(u^{\star} \in C_{1-\alpha }([0,T],E)\) such that \(u^{\star}=Au^{\star} \). For \(t\in[0, T]\), by (H3), we have
$$\begin{aligned}& t^{1-\alpha} \bigl(w_{n}(t) - v_{n}(t) \bigr) \\ & \quad =t^{1-\alpha} \bigl(Aw_{n}(t)-Av_{n}(t) \bigr) \\ & \quad = \frac{T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha}(-Mt^{\alpha })}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr)F(w_{n-1}) (s)\,ds \\ & \qquad {} +t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t-s)^{\alpha}\bigr)F(w_{n-1}) (s)\,ds \\ & \qquad {} -\frac{T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha}(-Mt^{\alpha })}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha}\bigr)F(v_{n-1}) (s)\,ds \\ & \qquad {} -t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t-s)^{\alpha}\bigr)F(v_{n-1}) (s)\,ds \\ & \quad = \frac{T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha}(-Mt^{\alpha })}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \\ & \qquad {}\times\int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha }\bigr) \bigl(F(w_{n-1}) (s)-F(v_{n-1}) (s)\bigr)\,ds \\ & \qquad {} +t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t-s)^{\alpha}\bigr) \bigl(F(w_{n-1}) (s)-F(v_{n-1}) (s)\bigr)\,ds \\ & \quad \leq \frac{(M+C)T^{1-\alpha}\Gamma(\alpha)E_{\alpha,\alpha}(-Mt^{\alpha })}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]} \\ & \qquad {}\times\int_{0}^{T} (T-s)^{\alpha-1}E_{\alpha,\alpha} \bigl(-M(T-s)^{\alpha }\bigr) \bigl(w_{n-1}(s)-v_{n-1}(s) \bigr)\,ds \\ & \qquad {}+(M+C)t^{1-\alpha} \int_{0}^{t} (t-s)^{\alpha-1}E_{\alpha,\alpha } \bigl(-M(t-s)^{\alpha}\bigr) \bigl(w_{n-1}(s)-v_{n-1}(s) \bigr)\,ds \\ & \quad \leq \frac{(M+C)T^{1-\alpha}}{\Gamma(\alpha)[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}s^{\alpha -1} \bigl[s^{1-\alpha}\bigl(w_{n-1}(s)-v_{n-1}(s)\bigr)\bigr]\,ds \\ & \qquad {} +\frac{(M+C)t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha -1}s^{\alpha-1} \bigl[s^{1-\alpha}\bigl(w_{n-1}(s)-v_{n-1}(s)\bigr)\bigr] \,ds. \end{aligned}$$
From the normality of the cone P, it follows that
$$\begin{aligned}& \bigl\Vert t^{1-\alpha} \bigl(w_{n}(t)-v_{n}(t) \bigr) \bigr\Vert \\& \quad \leq \frac{N(M+C)T^{1-\alpha}}{\Gamma(\alpha)[1-\Gamma(\alpha )E_{\alpha,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}s^{\alpha -1} \bigl\Vert s^{1-\alpha}\bigl(w_{n-1}(s)-v_{n-1}(s)\bigr) \bigr\Vert \,ds \\& \qquad {} +\frac{N(M+C)t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha -1}s^{\alpha-1} \bigl\Vert s^{1-\alpha}\bigl(w_{n-1}(s)-v_{n-1}(s)\bigr) \bigr\Vert \,ds \\& \quad \leq \frac{N(M+C)T^{1-\alpha}}{\Gamma(\alpha)[1-\Gamma(\alpha)E_{\alpha ,\alpha}(-MT^{\alpha})]} \int_{0}^{T} (T-s)^{\alpha-1}s^{\alpha-1}\,ds \Vert w_{n-1}-v_{n-1} \Vert _{1-\alpha} \\& \qquad {} +\frac{N(M+C)t^{1-\alpha}}{\Gamma(\alpha)} \int_{0}^{t} (t-s)^{\alpha -1}s^{\alpha-1}\,ds \Vert w_{n-1}-v_{n-1} \Vert _{1-\alpha} \\& \quad = \frac{N(M+C)T^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)[1-\Gamma(\alpha )E_{\alpha,\alpha}(-MT^{\alpha})]} \Vert w_{n-1}-v_{n-1} \Vert _{1-\alpha} \\& \qquad {}+\frac{N(M+C)t^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \Vert w_{n-1}-v_{n-1} \Vert _{1-\alpha} \\& \quad \leq \frac{N(M+C)T^{\alpha}\Gamma(\alpha)}{\Gamma(2\alpha)} \biggl(\frac {1}{[1-\Gamma(\alpha)E_{\alpha,\alpha}(-MT^{\alpha})]}+1 \biggr) \Vert w_{n-1}-v_{n-1} \Vert _{1-\alpha}. \end{aligned}$$
Therefore,
$$\|w_{n}-v_{n}\|_{1-\alpha}\leq\frac{N(M+C)T^{\alpha}\Gamma(\alpha )}{\Gamma(2\alpha)} \biggl( \frac{1}{[1-\Gamma(\alpha)E_{\alpha,\alpha }(-MT^{\alpha})]}+1 \biggr)\|w_{n-1}-v_{n-1}\|_{1-\alpha}. $$
Again using the above inequality, we get
$$\|w_{n}-v_{n}\|_{1-\alpha}\leq \biggl[\frac{N(M+C)T^{\alpha}\Gamma(\alpha )}{\Gamma(2\alpha)} \biggl(\frac{1}{[1-\Gamma(\alpha)E_{\alpha,\alpha }(-MT^{\alpha})]}+1 \biggr) \biggr]^{n}\|w_{0}-v_{0} \|_{1-\alpha}, $$
which implies that for \(n\rightarrow\infty\) we have \(\|w_{n}-v_{n}\| _{1-\alpha}\rightarrow0\). Then there exists a unique \(u^{\star} \in C_{1-\alpha}([0,T],E)\) such that
$$\lim_{n\rightarrow\infty}w_{n}=\lim_{n\rightarrow\infty}v_{n}=u^{\star}. $$
So let \(n\rightarrow\infty\) in (3.4), we have \(u^{\star}=Au^{\star}\), which means that \(u^{\star}\) is a unique solution of problem (1.2).
This completes the proof of Theorem 3.2. □