In this section, we will present the existence theorem for the abstract fractional differential equation on the half-line. In order to prove our main result, we need the following facts and lemmas.
Lemma 3.1
Let
\(k \in \operatorname{BC}(J,E)\)
and
\(0< \sum_{i=0}^{m} d_{i}<1\). Then, for given
\(u_{0}\in D(A)\), the following linear equation
$$ \left \{ \textstyle\begin{array}{ll} & ^{C}D_{0+}^{\beta}u(t)=Au(t)+k(t),\quad t \in(0,+\infty),\\ & u(0)=u_{0}, \end{array}\displaystyle \right . $$
(3.1)
has a unique mild solution
\(u \in \operatorname{BC}(J,E)\)
such that
$$\begin{aligned}[b] u(t)&= \int_{0}^{t} \int_{0}^{\infty} \biggl[\beta\frac{(t-s)^{\beta-1}}{\theta ^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac{(t-s)^{\beta}}{\theta^{\beta }} \biggr) \biggl( \frac{s^{-\beta}}{\Gamma{(1-\beta)}}u_{0}+k(s) \biggr) \biggr]\,d\theta \,ds \\ &=\frac{\beta}{\Gamma{(1-\beta)}} \int_{0}^{1} \int_{0}^{\infty}\frac{\zeta _{\beta}(\theta)}{\theta^{\beta}}\tau^{-\beta}(1- \tau)^{\beta-1}S \biggl(\frac{t^{\beta}(1-\tau)^{\beta}}{\theta^{\beta}} \biggr)u_{0}\,d\theta\, d \tau +(\Phi k) (t). \end{aligned}$$
(3.2)
Proof
Employing Lemma 2.1, one can rewrite (3.1) as the following equivalent integral equation:
$$ u(t)=u_{0}+\frac{1}{\Gamma{(\beta)}} \int_{0}^{t}(t-s)^{\beta-1}\bigl[Au(s)+k(s) \bigr] \,ds. $$
(3.3)
With an approach similar to [12, 13], by using the Laplace transform, we can get
$$ U(\lambda)={\lambda}^{-1}u_{0}+{ \lambda^{-\beta}}AU(\lambda)+{\lambda ^{-\beta}}K(\lambda),\quad \lambda>0, $$
(3.4)
where \(U(\lambda)\) and \(K(\lambda)\) is the Laplace transform of \(u(t)\) and \(k(t)\), respectively.
Then (3.4) leads to
$$\bigl(\lambda^{\beta}I-A\bigr)U(\lambda)=\lambda^{\beta-1}u_{0}+K( \lambda). $$
In view of (2.5) and Lemma 2.3, we have
$$\begin{aligned} U(\lambda)={}&\bigl(\lambda^{\beta}I-A\bigr)^{-1} \lambda^{\beta-1}u_{0}+\bigl(\lambda ^{\beta}I-A \bigr)^{-1}K(\lambda) \\ ={}&\lambda^{\beta-1} \int_{0}^{\infty}e^{-\lambda^{\beta}s}S(s)u_{0}\,ds+ \int _{0}^{\infty}e^{-\lambda^{\beta}s}S(s)K(\lambda)\,ds \\ ={}&\lambda^{\beta-1} \int_{0}^{\infty} \int_{0}^{\infty}e^{-\lambda s^{1/\beta }\theta}\zeta_{\beta}( \theta)S(s)u_{0}\,d\theta \,ds \\ &+ \int_{0}^{\infty} \int_{0}^{\infty}e^{-\lambda s^{1/\beta}\theta}\zeta _{\beta}( \theta)S(s)K(\lambda)\,d\theta \,ds \\ ={}&\lambda^{\beta-1} \int_{0}^{\infty}e^{-\lambda t} \biggl[ \int_{0}^{\infty }\beta\frac{t^{\beta-1}}{\theta^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac{t^{\beta}}{\theta^{\beta}} \biggr)u_{0}\,d \theta \biggr]\,dt \\ &+ \int_{0}^{\infty}e^{-\lambda t} \biggl[ \int_{0}^{t} \int_{0}^{\infty}\beta \frac{(t-s)^{\beta-1}}{\theta^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac {(t-s)^{\beta}}{\theta^{\beta}} \biggr)k(s)\,d\theta \,ds \biggr]\,dt. \end{aligned}$$
With the help of the convolution theorem, using the inverse Laplace transforms and Lemma 2.5, one can derive that
$$\begin{aligned} u(t) =&\mathcal{L}^{-1}\bigl[\lambda^{\beta-1}\bigr]\ast \biggl[ \int_{0}^{\infty }\beta\frac{t^{\beta-1}}{\theta^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac{t^{\beta}}{\theta^{\beta}} \biggr)u_{0}\,d \theta \biggr] \\ &{}+ \int_{0}^{t} \int_{0}^{\infty}\beta\frac{(t-s)^{\beta-1}}{\theta^{\beta }} \zeta_{\beta}(\theta)S \biggl(\frac{(t-s)^{\beta}}{\theta^{\beta}} \biggr)k(s)\,d\theta \,ds \\ =& \frac{t^{-\beta}}{\Gamma{(1-\beta)}}\ast \biggl[ \int_{0}^{\infty}\beta \frac{t^{\beta-1}}{\theta^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac {t^{\beta}}{\theta^{\beta}} \biggr)u_{0}\,d \theta \biggr] \\ &{}+\beta \int_{0}^{t} \int_{0}^{\infty} \frac{(t-s)^{\beta-1}}{\theta^{\beta }} \zeta_{\beta}(\theta)S \biggl(\frac{(t-s)^{\beta}}{\theta^{\beta}} \biggr)k(s)\,d\theta \,ds \\ =&\frac{\beta}{\Gamma{(1-\beta)}} \int_{0}^{t} \int_{0}^{\infty}s^{-\beta}\frac {(t-s)^{\beta-1}}{\theta^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac {(t-s)^{\beta}}{\theta^{\beta}} \biggr)u_{0}\,d \theta \,ds+(\Phi k) (t) \\ =&\frac{\beta}{\Gamma{(1-\beta)}} \int_{0}^{1} \int_{0}^{\infty}\frac{\zeta _{\beta}(\theta)}{\theta^{\beta}}\tau^{-\beta}(1- \tau)^{\beta-1}S \biggl(\frac{t^{\beta}(1-\tau)^{\beta}}{\theta^{\beta}} \biggr)u_{0}\,d\theta \,d \tau +(\Phi k) (t). \end{aligned}$$
Since
$$\begin{gathered} \frac{\beta}{\Gamma{(1-\beta)}} \int_{0}^{1} \int_{0}^{\infty}\tau^{-\beta }(1- \tau)^{\beta-1}\frac{\zeta_{\beta}(\theta)}{\theta^{\beta}} \biggl\Vert S \biggl(\frac{t^{\beta}(1-\tau)^{\beta}}{\theta^{\beta}} \biggr)u_{0} \biggr\Vert _{{\omega}}\,d\theta \,d\tau \\ \quad\quad{}+\big\| (\Phi k) (t) \big\| _{{\omega}} \\ \quad\leq\frac{\beta}{\Gamma{(1-\beta)}} \int_{0}^{1} \int_{0}^{\infty}\tau ^{-\beta}(1- \tau)^{\beta-1}\frac{\zeta_{\beta}(\theta)}{\theta^{\beta }}e^{-\omega (\frac{t^{\beta}(1-\tau)^{\beta}}{\theta^{\beta}} )} \Vert u_{0} \Vert _{{\omega}}\,d\theta \,d\tau \\ \qquad{}+ \big\| (\Phi k) (t) \big\| _{{\omega}} \\ \quad\leq\frac{\beta}{\Gamma{(1-\beta)}} \Vert u_{0} \Vert _{{\omega}} \int _{0}^{1}\tau^{-\beta}(1- \tau)^{\beta-1} \biggl( \int_{0}^{\infty}\frac{\zeta _{\beta}(\theta)}{\theta^{\beta}}\,d\theta \biggr) \,d \tau+ \big\| (\Phi k) (t) \big\| _{{\omega}} \\ \quad\leq \Vert u_{0} \Vert _{{\omega}}+\frac{1}{\Gamma{(\beta+1)}}\frac { \Vert k \Vert ^{*}_{\omega}}{\omega}. \end{gathered}$$
So, \(u\in \operatorname{BC}(J,E)\). □
Define an operator \(\mathcal{B}_{A}\) on \(\operatorname{BC}(J,E)\) as
$$\begin{aligned} (\mathcal{B}_{A}k) (t) ={}& \int_{0}^{t} \int_{0}^{\infty} \Biggl[\beta\frac{(t-s)^{\beta-1}}{\theta ^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac{(t-s)^{\beta}}{\theta^{\beta }} \biggr) \\ &\times \Biggl(\frac{s^{-\beta}}{\Gamma{(1-\beta)}} \Biggl[(I-\Psi)^{-1}\sum _{i=0}^{m}d_{i}(\Phi k) ( \xi_{i}) \Biggr]+k(s) \Biggr) \Biggr]\,d\theta \,ds, \end{aligned}$$
where \(k \in \operatorname{BC}(J,E)\).
For simplicity of notation, we denote
$$\sigma:=\frac{1}{\Gamma{(\beta+1)}} \biggl(\frac{1}{1- \sum_{i=0}^{m}d_{i}} \biggr). $$
Lemma 3.2
Let
\(k \in \operatorname{BC}(J,E)\)
and
\(0< \sum_{i=0}^{m} d_{i}<1\). Then, for given
\(u_{0}\in D(A)\), the following linear equation
$$ \left \{ \textstyle\begin{array}{ll} & ^{C}D_{0+}^{\beta}u(t)=Au(t)+k(t),\quad t \in(0,+\infty),\\ & u(0)=\sum_{i=0}^{m}d_{i}u(\xi_{i}), \end{array}\displaystyle \right . $$
(3.5)
has a unique mild solution
\(u \in \operatorname{BC}(J,E)\)
such that
$$\begin{aligned}[b] u(t) ={}&(\mathcal{B}_{A}k) (t) \\ ={}& \int_{0}^{1} \int_{0}^{\infty}\frac{\beta}{\Gamma{(1-\beta)}}\frac{\zeta _{\beta}(\theta)}{\theta^{\beta}} \tau^{-\beta}(1-\tau)^{\beta-1}S \biggl(t^{\beta} \frac{(1-\tau)^{\beta}}{\theta^{\beta}} \biggr) \\ &\times \Biggl[(I-\Psi)^{-1}\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}) \Biggr]\,d\theta \,d\tau +(\Phi k) (t). \end{aligned}$$
(3.6)
Moreover,
$$\|\mathcal{B}_{A}\|^{*}_{\omega}\leq \frac{\sigma}{\omega}. $$
Proof
From Lemma 3.1, we can have
$$u(\xi_{i})= \int_{0}^{\xi_{i}} \int_{0}^{\infty} \biggl[\beta\frac{(\xi_{i}-s)^{\beta -1}}{\theta^{\beta}} \zeta_{\beta}(\theta)S \biggl(\frac{(\xi_{i}-s)^{\beta }}{\theta^{\beta}} \biggr) \biggl( \frac{s^{-\beta}}{\Gamma{(1-\beta )}}u(0)+k(s) \biggr) \biggr]\,d\theta \,ds. $$
By the nonlocal condition of (3.5), we get
$$\begin{aligned} u(0)={}&\frac{\beta}{\Gamma{(1-\beta)}}\sum_{i=0}^{m} d_{i} \int_{0}^{\xi _{i}} \int_{0}^{\infty}\frac{\zeta_{\beta}(\theta)}{\theta^{\beta}}s^{-\beta }( \xi_{i}-s)^{\beta-1}S \biggl(\frac{(\xi_{i}-s)^{\beta}}{\theta^{\beta }} \biggr)u(0)\,d\theta \,ds \\ &+\sum_{i=0}^{m}d_{i}(\Phi k) ( \xi_{i}) \\ ={}& \frac{\beta}{\Gamma{(1-\beta)}}\sum_{i=0}^{m} d_{i} \int_{0}^{1} \int _{0}^{\infty}\frac{\zeta_{\beta}(\theta)}{\theta^{\beta}} \tau^{-\beta }(1-\tau)^{\beta-1}S \biggl(\xi_{i}^{\beta} \frac{(1-\tau)^{\beta}}{\theta ^{\beta}} \biggr)u(0)\,d\theta \,d\tau \\ &+\sum_{i=0}^{m}d_{i}(\Phi k) ( \xi_{i}) \\ ={}&\Psi u(0)+\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}). \end{aligned}$$
Thus,
$$(I-\Psi)u(0)=\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}). $$
So, one can obtain
$$u(0)=(I-\Psi)^{-1}\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}). $$
Then, combined with Lemma 3.1, we can get
$$\begin{aligned} u(t)={}& \int_{0}^{1} \int_{0}^{\infty}\frac{\beta}{\Gamma{(1-\beta)}}\frac {\zeta_{\beta}(\theta)}{\theta^{\beta}} \tau^{-\beta}(1-\tau)^{\beta -1}S \biggl(t^{\beta} \frac{(1-\tau)^{\beta}}{\theta^{\beta}} \biggr) \\ &\times \Biggl[(I-\Psi)^{-1}\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}) \Biggr]\,d\theta \,d\tau +(\Phi k) (t). \end{aligned}$$
By virtue of (3.6), one has
$$\begin{aligned} \big\| (\mathcal{B}_{A}k) (t)\big\| _{{\omega}} \leq& \int_{0}^{1} \int_{0}^{\infty}\frac{\beta}{\Gamma{(1-\beta)}}\frac {\zeta_{\beta}(\theta)}{\theta^{\beta}} \tau^{-\beta}(1-\tau)^{\beta -1} \biggl\Vert S \biggl(t^{\beta} \frac{(1-\tau)^{\beta}}{\theta^{\beta}} \biggr) \biggr\Vert _{{\omega}} \\ & \Biggl\Vert (I-\Psi)^{-1}\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}) \Biggr\Vert _{{\omega }}\,d\theta \,d\tau+\big\| (\Phi k) (t)\big\| _{{\omega}} \\ \leq& \int_{0}^{1} \int_{0}^{\infty}\frac{\beta}{\Gamma{(1-\beta)}}\frac {\zeta_{\beta}(\theta)}{\theta^{\beta}} \tau^{-\beta}(1-\tau)^{\beta -1}e^{-\omega\frac{t^{\beta}(1-\tau)^{\beta}}{\theta^{\beta}}} \\ & \bigl\Vert [(I-\Psi)^{-1} \bigr\Vert _{{\omega}} \Biggl\Vert (I-\Psi)^{-1}\sum_{i=0}^{m}d_{i}( \Phi k) (\xi_{i}) \Biggr\Vert _{{\omega}}\,d\theta \,d\tau+\big\| (\Phi k) (t)\big\| _{{\omega}} \\ \leq&\frac{1}{1-\sum_{i=0}^{m} d_{i}} \frac{\sum_{i=0}^{m}d_{i}}{\Gamma {(\beta+1)}}\frac{ \Vert k \Vert ^{*}_{\omega}}{\omega}+\frac{1}{\Gamma {(\beta+1)}} \frac{ \Vert k \Vert ^{*}_{\omega}}{\omega} \\ =&\frac{1}{\Gamma{(\beta+1)}}\frac{1}{1- \sum_{i=0}^{m}d_{i}}\frac{ \Vert k \Vert ^{*}_{\omega}}{\omega} \\ =&\frac{\sigma}{\omega} \Vert k \Vert ^{*}_{\omega}. \end{aligned}$$
Then
$$\big\| (\mathcal{B}_{A}k)\big\| ^{*}_{\omega}\leq \frac{\sigma}{\omega} \Vert k \Vert ^{*}_{\omega}. $$
□
Next, we list some conditions that will be used in our main theorem as follows:
-
\((H1)\)
:
-
There exists a constant \(\mathcal{P}<-\omega_{0}\) such that for
$$\mathcal{F}(v)-\mathcal{F}(u)\leq\mathcal{P}(v-u),\quad \theta\leq u\leq v. $$
-
\((H2)\)
:
-
There exists a constant \(\mathcal{Q}>\max\{-\mathcal {P},\omega_{0}\}\) such that for
$$\mathcal{F}(v)-\mathcal{F}(u)\geq-\mathcal{Q}(v-u),\quad \theta\leq u\leq v. $$
-
\((H3)\)
:
-
$$0< \frac{\mathcal{P}+\mathcal{Q}}{\mathcal{Q}-\omega_{0}}< \frac{1}{\sigma}. $$
Now, it is time for us to state the main result about the existence of positive solutions to problem (1.1) in the following.
Theorem 3.1
Let
\(\{S(t)\}_{{t\geq0}}\)
be a uniformly exponentially stable
\(C_{0}\)-semigroup on a Banach space
E
with the growth bound
\(\omega_{0}\) (\(\omega_{0}<0\)), and
A
is the infinitesimal generator of
\(\{S(t)\} _{{t\geq0}}\). Let
\(0<\sum_{i=0}^{m} d_{i}<1\). Assume that
P
is a positive normal cone on
E
with
N
as its normal constant. Provided that
\(\mathcal{F}:\operatorname{BC}(J,E)\rightarrow \operatorname{BC}(J,E)\)
is continuous and
\(f_{0}(t):=\mathcal{F}(\theta)\geq\theta\)
is bounded on
J. If
\(\mathcal{F}(u) \)
satisfies conditions
\((H1)\), \((H2)\)
and
\((H3)\), then problem (1.1) has a unique positive mild solution in
\(\operatorname{BC}(J,E)\).
Proof
The proof is divided into three parts.
Part 1: Investigate two linear fractional evolution equations.
Consider two abstract fractional differential equations as follows:
$$ \left \{ \textstyle\begin{array}{ll} & ^{C}D_{0+}^{\beta}u(t)=Au(t)+\mathcal{P}u(t)+f_{0}(t),\quad t \in(0,+\infty ),\\ & u(0)= \sum_{i=0}^{m} d_{i} u(\xi_{i}), \end{array}\displaystyle \right . $$
(3.7)
and
$$ \left \{ \textstyle\begin{array}{ll} & ^{C}D_{0+}^{\beta}u(t)+\mathcal{Q}u(t)=Au(t)+h(t), \quad t \in(0,+\infty),\\ & u(0)=\sum_{i=0}^{m} d_{i} u(\xi_{i}), \end{array}\displaystyle \right . $$
(3.8)
where \(h\in \operatorname{BC}(J,E)\) is a given function.
By virtue of the theory of semigroups, it is obvious that \(\{e^{\mathcal {P}t}S(t)\}_{{t\geq0}}\) is a uniformly exponentially stable \(C_{0}\)-semigroup on Banach E generated by \(A+\mathcal{P}I\), and \(A-\mathcal{Q}I\) generates a uniformly exponentially stable \(C_{0}\)-semigroup \(\{e^{\mathcal{Q}t}S(t)\}_{{t\geq0}}\) on Banach E. Both of the semigroups are positive, with the growth bounds \(\mathcal {P}+\omega_{0}\) (\(\mathcal{P}+\omega_{0}<0\)) and \(-\mathcal{Q}+\omega_{0}\) (\(-\mathcal{Q}+\omega_{0}<0\)), respectively.
In view of Lemma 3.2, equation (3.7) has a unique mild solution \(\eta_{0}\in \operatorname{BC}(J,E)\). Moreover, \(\eta_{0}\geq\theta\) due to the fact that \(f_{0}(t)\geq\theta\), \(t\in J\).
In consideration of Lemma 3.2, the unique mild solution of (3.8) can be written as \(u=\mathcal{B}_{{A-\mathcal {Q}I}}h\), where \(\mathcal{B}_{{A-\mathcal{Q}I}}:\operatorname{BC}(J,E)\rightarrow \operatorname{BC}(J,E)\), analogous to the linear bounded operator \(L_{{A}}\), is a positive bounded linear operator with the property that
$$\|\mathcal{B}_{{A-\mathcal{Q}I}}\|^{*}_{\omega}\leq \frac{\sigma }{\mathcal{Q}-\omega_{0}},\quad \omega=\mathcal{Q}-\omega_{0}. $$
In light of the above facts, one can deduce that \(\eta_{0}\) is the mild solution of problem (3.8) for \(h=f_{0}+\mathcal{P}\eta _{0}+\mathcal{Q}\eta_{0}\), so
$$ \eta_{0}=\mathcal{B}_{{A-\mathcal{Q}I}}(f_{0}+ \mathcal{P}\eta_{0}+\mathcal {Q}\eta_{0}). $$
(3.9)
Part 2: Prove the existence of mild solutions for problem (1.1).
Let \(G(u)=\mathcal{F}(u)+\mathcal{Q}u\). It is clear that \(G(\theta )=\mathcal{F}(\theta)=f_{0}\geq\theta\) and \(G:\operatorname{BC}(J,E)\rightarrow \operatorname{BC}(J,E)\) is continuous as a result of the continuity of \(\mathcal{F}\) and conditions \((H1)\) and \((H2)\).
In view of \((H2)\), one can get that
$$G(v)-G(u)=\mathcal{F}(v)+\mathcal{Q}v-\mathcal{F}(u)-\mathcal {Q}u= \mathcal{F}(v)-\mathcal{F}(u)+\mathcal{Q}(v-u)\geq\theta,\quad \theta \leq u \leq v, $$
which shows that G is an increasing operator on the positive cone \(P^{*}\).
Let \(\mathcal{M}=\mathcal{B}_{{A-\mathcal{Q}I}}\circ G\). It is easy to see that the fixed point of the composition operator \(\mathcal{M}\) is the mild solution of problem (1.1). Below we are going to prove that the operator \(\mathcal{M}\) has at least one fixed point by the monotone iterative method.
Now, we define two sequences
$$ \eta_{n}=\mathcal{M}(\eta_{n-1}),\quad n=1,2,3, \ldots, $$
(3.10)
and
$$ \varpi_{0}=\theta,\qquad \varpi_{n}=\mathcal{M}( \varpi_{n-1}),\quad n=1,2,3,\ldots. $$
(3.11)
By \((H1)\), we have
$$ \mathcal{F}(\eta_{0})-\mathcal{F}(\theta)\leq \mathcal{P}\eta_{0}, $$
that is,
$$ \mathcal{F}(\eta_{0})\leq\mathcal{P}\eta_{0}+f_{0}. $$
Thus, we obtain
$$ \theta\leq G(\theta)\leq G(\eta_{0})=\mathcal{F}( \eta_{0})+\mathcal{Q}\eta _{0}\leq\mathcal{P} \eta_{0}+\mathcal{Q}\eta_{0}+f_{0}. $$
Combined with (3.9) and the positivity of \(\mathcal {B}_{{A-\mathcal{Q}I}}\), we get
$$ \theta\leq(\mathcal{B}_{{A-\mathcal{Q}I}}\circ G) (\theta)=\mathcal {M}(\theta)\leq (\mathcal{B}_{{A-\mathcal{Q}I}}\circ G) (\eta_{0})=\mathcal{M}(\eta _{0})\leq\mathcal{B}_{{A-\mathcal{Q}I}}( \mathcal{P}\eta_{0}+ \mathcal {Q}\eta_{0}+f_{0})=\eta_{0}, $$
that is,
$$ \theta=\varpi_{0}\leq\eta_{1}\leq \eta_{0}. $$
(3.12)
Since \(\mathcal{M}\) is an increasing operator on the order interval \([\theta,\eta_{0}]\), by the definition of \(\mathcal{M}\) and (3.12), we can obtain two sequences \(\{\eta_{n}\}\) and \(\{\varpi_{n}\}\) (\(n=0,1,2,3,\ldots \)) satisfying
$$ \theta=\varpi_{0}\leq\varpi_{1}\leq\varpi_{2}\leq \cdots\leq\varpi_{n}\leq \cdots\leq\eta_{n} \leq\cdots\leq \eta_{2} \leq\eta_{1} \leq\eta_{0}. $$
Then, from \((H1)\), one can get that
$$\begin{aligned} \theta&\leq \eta_{n}-\varpi_{n}= \mathcal{M}( \eta_{n-1})-\mathcal {M}(\varpi_{n-1}) =(\mathcal{B}_{{A-\mathcal{Q}I}} \circ G) (\eta_{n-1})-(\mathcal {B}_{{A-\mathcal{Q}I}}\circ G) ( \varpi_{n-1}) \\ &= \mathcal{B}_{{A-\mathcal{Q}I}} \bigl[f(\cdot,\eta_{n-1})+\mathcal {Q} \eta_{n-1}-f(\cdot,\varpi_{n-1})-\mathcal{Q} \varpi_{n-1} \bigr] \\ &\leq(\mathcal{P}+\mathcal{Q})\mathcal{B}_{{A-\mathcal{Q}I}} (\eta _{n-1}-\varpi_{n-1} ). \end{aligned}$$
Then, through mathematical induction, we have
$$\theta\leq \eta_{n}-\varpi_{n} \leq(\mathcal{P}+ \mathcal{Q})^{n}\mathcal {B}_{{A-\mathcal{Q}I}}^{n} ( \eta_{0}-\varpi_{0} )=(\mathcal {P}+\mathcal{Q})^{n} \mathcal{B}_{{A-\mathcal{Q}I}}^{n} (\eta_{0} ). $$
As the cone \(P^{*}\) is normal with the normal constant N, in view of \((H3)\), we can get
$$ \begin{aligned}[b]\|\eta_{n}-\varpi_{n}\|^{*}_{\omega} &\leq N(\mathcal{P}+\mathcal{Q})^{n} \bigl\Vert \mathcal{B}_{{A-\mathcal{Q}I}}^{n} (\eta_{0} ) \bigr\Vert ^{*}_{\omega} \leq N( \mathcal{P}+\mathcal{Q})^{n} \bigl\Vert \mathcal{B}_{{A-\mathcal {Q}I}}^{n} \bigr\Vert ^{*}_{\omega}\|\eta_{0} \|^{*}_{\omega} \\ &\leq N(\mathcal{P}+\mathcal{Q})^{n} \bigl( \Vert \mathcal {B}_{{A-\mathcal{Q}I}} \Vert ^{*} \bigr)^{n}\| \eta_{0}\|^{*}_{\omega } \\ &\leq N(\mathcal{P}+\mathcal{K}_{2})^{n} \biggl( \frac{\sigma}{\mathcal {Q}-\omega_{0}} \biggr)^{n}\|\eta_{0}\|^{*}_{\omega} \\ &= N \biggl(\frac{\sigma(\mathcal{P}+\mathcal{Q})}{\mathcal{Q}-\omega _{0}} \biggr)^{n}\|\eta_{0} \|^{*}_{\omega}\rightarrow0,\quad n\rightarrow+\infty. \end{aligned}$$
(3.13)
Analogous to the nested interval method, by virtue of (3.13), we can prove that there exists a unique mild solution \(u^{*}\in\bigcap_{n=1}^{\infty}[\varpi_{n},\eta_{n}]\) such that
$$u^{*}=\lim_{n\rightarrow\infty}\eta_{n}=\lim_{n\rightarrow\infty} \varpi_{n}. $$
By taking limit of n tending to +∞ on both of (3.10) and (3.11), one can obtain
$$u^{*}=\mathcal{M}\bigl(u^{*}\bigr), $$
which shows that \(u^{*}\) is the fixed point of \(\mathcal{M}\). Therefore, \(u^{*}\) is a mild positive solution of problem (1.1).
Part 3: Certify the uniqueness of the mild solution for problem (1.1).
In this part, we will discuss the uniqueness of the mild solution for problem (1.1) by reduction to absurdity.
Suppose that \(u_{1}^{*}\) and \(u_{2}^{*}\) are two different positive mild solution for the fractional evolution equation (1.1), that is, \(\|u_{1}^{*}-u_{2}^{*}\|^{*}_{\omega}> 0\).
By the same steps as above, replacing \(\eta_{0}\) by \(u_{1}^{*}\) and \(u_{2}^{*}\) in (3.10), respectively, then one can get that
$$u_{i}^{*}=\mathcal{M}\bigl(u_{i}^{*}\bigr),\qquad \big\| u_{i}^{*}-\varpi_{n}\big\| ^{*}\rightarrow0\quad (n \rightarrow\infty), \qquad\eta_{n}=u_{i}^{*}, \quad n\in\mathbb{N}, i=1,2. $$
Hence,
$$0< \big\| u_{1}^{*}-u_{2}^{*}\big\| ^{*}_{\omega}\leq \big\| u_{1}^{*}-\varpi_{n}\big\| ^{*}_{\omega}+\big\| u_{2}^{*}-\varpi_{n}\big\| ^{*}_{\omega} \rightarrow0,\quad n\rightarrow\infty, $$
which is a contradiction.
Therefore, problem (1.1) has a unique positive solution. The proof is completed. □