Existence and uniqueness for a kind of nonlocal fractional evolution equations on the unbounded interval

Abstract

By means of a Laplace transform and its inverse transform, we obtain a correct equivalent integral equation for some kind of nonlocal abstract differential equations (fractional order) on the right half-axis. Based on it, the existence result is established by Knasterās theorem, and the uniqueness of the mild solution is obtained using the Banach contraction principle.

1 Introduction

Today, fractional order calculus has a great many of uses in engineering, science, economy, biology, physics and other scientific disciplines (see [1,2,3]). Numerous phenomena and processes in the real world are described by differential equations of fractional order. Given their wide applications, there are numerous scholars focusing on the study of the differential equations of fractional order recently. To acquire more general knowledge of fractional order calculus and equations with fractional derivatives, the reader may refer to [4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28].

Now, nonlocal evolution equations of fractional order have become one of the hot research topics in the field of differential equations with fractional derivatives, which play a role in modeling physics phenomena. For the first time, the existence of mild solutions for the nonlocal problems was discussed in [29]. In [30], Byszewski and Lakshmikantham indicated that as regards describing physical phenomena nonlocal conditions could be more applicable than standard conditions.

In [19], El-Borai obtained the existence results for fractional abstract equations of the following kind:

\begin{aligned} \textstyle\begin{cases} \frac{d^{\beta }z}{d^{\beta }s}=B(s)z(t)+Az(s),\quad s \in [0,a], \\ z(0)=z_{0}, \end{cases}\displaystyle \end{aligned}

where $$0<\beta \leq 1$$, $$a>0$$.

In this paper, the equivalent integral equation in regard to the abstract equations is firstly described by means of some probability densities using Laplace transform and its inverse transform. Since then, many researchers have drawn on El-Boraiās results to investigate fractional evolution equations, such as in [20,21,22,23,24,25,26,27,28].

To the best of our knowledge, fractional evolution equations have attracted more and more attention recently [18,19,20,21,22,23,24,25,26,27,28]. For instance, in [20], Zhou investigated a class of nonlocal evolution equations of fractional order

\begin{aligned} \textstyle\begin{cases} ^{C}D_{0+}^{q}z(s)=f(s,z(s))+Az(s),\quad s \in (0,a], \\ z(0)+g(z)=z_{0}, \end{cases}\displaystyle \end{aligned}

Zhou and Jiao obtained many good conclusions about the existence and uniqueness of mild solution for this equation.

Yet despite all that, a fair number of those papers investigating the mild solutions of evolution equations with fractional derivatives are concerned with bounded intervals, and conclusions on the right half-axis are still rare.

Motivated by [19, 20, 31], we discuss a class of nonlocal functional differential equations of fractional order in a Banach space E

$$\textstyle\begin{cases} {}^{C}D_{0+}^{q}u(t)=f(t,u(m(t)))+Au(t),\quad t \in (0,+\infty ), \\ k(u)+u(0)=u_{0}, \end{cases}$$
(1.1)

where $$0< q < 1$$, $${}^{C}D_{0+}^{q}$$ denotes the fractional derivative in the Caputo sense, $$u_{0}\in E$$, $$\{T(t)\}_{{t\geq 0}}$$ is a $$C_{0}$$ semigroup on Banach E and A is the semigroupās infinitesimal generator, $$m\in C[0,\infty )$$, $$m(t)\geq 0$$ is an increasing function, $$f:[0,+\infty )\times E\rightarrow E$$, and $$k:E\rightarrow E$$ satisfy certain conditions.

Here, entirely different from those already obtained in the previous literature, we obtain a correct equivalent integral equation for the main equation. Applying Knasterās theorem, the existence of positive mild solutions of the main problem (1.1) is given. Then, employing the Banach contraction theorem, the uniqueness of the mild solution is given.

The layout of the rest of the article is listed now. In Sect. 2, fractional derivative, fractional integral and several useful preliminaries are introduced. In Sect. 3, our main conclusions are presented by using Knasterās theorem and Banach contraction theorem, respectively. In Sect. 4, one example is provided to show the application of our main conclusion.

2 Preliminaries

In the following, we assume that E is ordered Banach space. Let $$P\subset E$$ be a cone, which defines a partial order by $$y\leq z$$ if and only if $$z-y \in P$$ on E. We refer the reader to [32, 33] for more details as regards the cone.

Throughout the paper, we set a normal positive cone $$P=\{z\in E \mid z\geq \theta \}$$ with N as its normal constant. Let

$$BC(J,E)= \bigl\{ v(s) \mid v(s) \mbox{ is bounded and continuous on }J \bigr\} , \quad J=[0,+\infty ).$$

Evidently, it is a Banach space equipped with the norm $$\Vert v \Vert _{{b}}=\sup_{{t\in J}}\Vert v(t) \Vert$$. Set

$$P_{{BC}}= \bigl\{ v \mid v(t)\geq \theta , t\in J, v\in BC(J,E) \bigr\} .$$

Obviously, $$P_{{BC}}$$ is a normal cone in $$BC(J,E)$$ with the normal constant N, the same normal to the cone P. Meanwhile, an ordered Banach space $$BC(J,E)$$ is induced by $$P_{{BC}}$$ (for convenience, we also denote the partial order by āā¤ā both on $$BC(J,E)$$ and on E without confusion).

Let $$\mathcal{S}\subset E$$ be a subset such that

$$\overline{\mathcal{S}}=\{z\in E:y\leq z, \forall y\in \mathcal{S}\}.$$

The point $$z^{*}\in E$$ is called the infimum of $$\overline{\mathcal{S}}$$ if $$z^{*}\in \overline{\mathcal{S}}$$ and for each $$z\in \overline{\mathcal{S}}$$, $$z^{*}\leq z$$.

Similarly, we can define the supremum of $$\overline{\mathcal{S}}$$ by reversing the inequality in the above definition.

Next, we present the theorem established by Knaster [34].

Theorem 2.1

([34])

Let E be a Banach space ordered by partially ordering āāŖÆā. Assume that $$\mathcal{S}\subset E$$ is a subset of E, which has properties as follows: (1) Each nonempty subset of $$\mathcal{S}$$ has a supremum belonging to $$\mathcal{S}$$; (2) The infimum of $$\mathcal{S}$$ is in $$\mathcal{S}$$. Assume that the map $$T:\mathcal{S}\rightarrow \mathcal{S}$$ is increasing, i.e., $$y\preceq z$$ implies that $$Ty\preceq Tz$$ ($$y,z\in \mathcal{S}$$). Then T has a fixed point in $$\mathcal{S}$$.

Next, we list two definitions of fractional integral and fractional derivative and one lemma, which will be used.

Definition 2.1

([1,2,3])

The fractional integral of order $$\alpha >0$$ in the sense of RiemannāLiouville for a function $$g:(0, \infty )\rightarrow \mathbb{R}$$ is defined by

$$I^{\alpha }_{0+}g(t)=D^{-\alpha }_{0+}g(t)= \frac{1}{\varGamma (\alpha )} \int ^{t}_{0}(t-s)^{\alpha -1}g(s)\,ds,$$
(2.1)

provided that the integral above is pointwise defined on $$(0, \infty )$$.

Definition 2.2

([1,2,3])

The fractional derivative of order $$\alpha >0$$ in the sense of Caputo for a function $$g:(0, \infty )\rightarrow \mathbb{R}$$ is defined by

$$^{C}D^{\alpha }_{0+}g(t)=\frac{1}{\varGamma (n-\alpha )} \int ^{t}_{0}(t-s)^{n-\alpha -1} g^{(n)}(s)\,ds,$$
(2.2)

where $$n=[\alpha ]+1$$, provided that the integral above is pointwise defined on $$(0, \infty )$$.

Lemma 2.1

([1, 3])

Let $${}^{C}D^{\alpha }_{0+}g(t) \in L^{1}(0,+\infty )$$, $$\alpha >0$$. Then one has

$$I^{\alpha }_{0+} {^{C}D^{\alpha }_{0+}}g(t)=g(t)+C_{1}+C_{2}t+ \cdots +C_{M}t^{M-1},\quad t>0,$$
(2.3)

for some constants $$C_{i}$$, $$i=1, 2, 3,\ldots ,M$$ā, where $$M=\min \{m \mid m\geq \alpha ,m\textit{ is a integer}\}$$.

Remark 2.1

If the function g appearing in the above lemma and two definitions takes values in the Banach space E, then the integrals here are all to be taken in the sense of Bochner. In addition, if the abstract function g is measurable and its norm is integrable in the sense of Lebesgue, it is Bochner integrable.

Next, we set out to present a series of concepts and results about the semigroups of linear operators. The reader can find more details in [35, 36].

Given a strongly continuous semigroup $$\{T(t)\}_{{t}\geq 0}$$ (i.e. $$C_{0}$$-semigroup), we can define the infinitesimal generator of $$\{T(t)\}_{{t\geq 0}}$$ as

$$Az=\lim_{s\rightarrow 0^{+}}\frac{T(s)z-z}{s}, \quad z\in E.$$

The domain of A is given by

$$D(A)= \biggl\{ z: \lim_{s\rightarrow 0^{+}}\frac{T(s)z-z}{s} \mbox{ exists}, z\in E \biggr\} .$$

Lemma 2.2

([35, 36])

Assume that $$\{T(t)\}_{{t}\geq 0}$$ is a strongly continuous semigroup, then $$\Vert T(t) \Vert \leq C e^{\omega t}$$, where Ļ and $$C\geq 1$$ are two constants.

Lemma 2.3

([35, 36])

Given a strongly continuous semigroup of contractions $$\{T(t)\}_{{t}\geq 0}$$ and a linear operator A, we have the following results: A is the infinitesimal generator of $$\{T(t)\}_{{t}\geq 0}$$ if and only if

1. (1)

$$\overline{D(A)}=E$$ and A is closed.

2. (2)

$$(0,+\infty )\subset \rho (A)$$, and $$\forall \mu >0$$, one has

$$\bigl\Vert R(\mu ,A) \bigr\Vert \leq \frac{1}{\mu },$$

where $$\rho (A)$$ is the resolvent set of A, and

$$R(\mu ,A)z:=(\mu I-A)^{-1}z= \int _{0}^{+\infty }e^{-\mu t}T(t)z\,dt, \quad z\in E.$$

Definition 2.3

([37])

For a given strongly continuous semigroup $$\{T(t)\}_{{t\geq 0}}$$, it is positive on E, if $$\theta \leq T(t)z$$, $$t\geq 0$$, $$z\in E$$.

Definition 2.4

([35, 36])

For a given strongly continuous semigroup $$\{T(t)\}_{{t\geq 0}}$$, it is uniformly exponentially stable under the condition that $$\omega _{0}<0$$, where $$\omega _{0}$$ denotes the growth bound of the semigroup, and it is given by

$$\omega _{0}=\inf \bigl\{ \omega \mid \bigl\Vert T(t) \bigr\Vert \leq C e^{\omega t}, t\geq 0, C\geq 1 \bigr\} .$$

By Definition 2.4 and Lemma 2.2, for a uniformly exponentially stable $$C_{0}$$-semigroup $$\{T(t)\}_{{t\geq 0}}$$, $$\Vert T(t) \Vert \leq C e^{\omega t}$$, $$t\geq 0, C\geq 1$$ and $$\omega \in (0,|\omega _{0}|]$$ ($$\omega _{0}$$ is the growth bound of the above semigroup). Next, define a norm on E as follows:

$$\Vert z \Vert _{\omega }=\sup_{s\geq 0}\bigl\Vert e^{\omega s}T(s)z \bigr\Vert .$$

Obviously, one has $$\Vert z \Vert \leq \Vert z \Vert _{\omega }\leq C \Vert z \Vert$$, which implies that $$\Vert \cdot \Vert$$ and $$\Vert \cdot \Vert _{\omega }$$ are equivalent norms. Denote by $$\Vert T(t) \Vert _{\omega }$$ the norm of $$T(t)$$ which is induced by $$\Vert \cdot \Vert _{\omega }$$, then, for each $$t\geq 0$$, we obtain

$$\bigl\Vert T(t) \bigr\Vert _{\omega }\leq e^{-\omega t}.$$
(2.4)

Meanwhile, it is easy to verify that the norm

$$\Vert v \Vert _{{b\omega }}=\sup_{{s\in J}}\bigl\Vert v(s) \bigr\Vert _{{\omega }},\quad v\in BC(J,E),$$

is an equivalent one on $$BC(J,E)$$. Evidently, if $$v(t)\equiv v_{0}, t\in J$$, $$v_{0}\in E$$, then we have

$$\Vert v \Vert _{{b\omega }}=\Vert v_{0} \Vert _{{b\omega }}=\Vert v \Vert _{{\omega }}.$$

Given the stable probability density of one-sided [20, 21, 38]

$$\psi _{q}(\theta )=\frac{1}{\pi }\sum_{k=1}^{\infty } \sin (k\pi q)\frac{\varGamma (kq+1)}{k!}\theta ^{-q k-1},\quad \theta \in (0,+\infty ),$$

where $$0< q<1$$. Noting Remark 2.8 in [21], if $$\beta \in [0,1]$$, we have

$$\int _{0}^{+\infty }\psi _{q}(\theta )\theta ^{-q\beta }\,d\theta =\frac{\varGamma ({1+\beta })}{\varGamma ({1+q\beta })}.$$
(2.5)

From [20, 21, 38], the Laplace transform of $$\psi _{q}(\theta )$$ is

$$\mathcal{L}\bigl[\psi _{q}(\theta )\bigr]= \int _{0}^{\infty }e^{-\lambda \theta }\psi _{q}( \theta )\,d\theta =e^{{-{\lambda }^{q}}},\quad 0< q< 1.$$
(2.6)

In the remaining part, we show that the semigroup $$\{T(t)\}_{{t\geq 0}}$$ is strongly continuous of contractions and is uniformly exponentially stable with the growth bound $$\omega _{0}$$, and $$\omega \in (0,|\omega _{0}|]$$.

Lemma 2.4

Define an operator

\begin{aligned} &(\mathcal{J} h) (t):=\frac{q}{\varGamma {(1-q)}} \int _{0}^{1} \int _{0}^{\infty }\tau ^{-q}(1-\tau )^{q-1}\frac{\psi _{q}(\theta )}{\theta ^{q}}T \biggl( \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta \,d\tau, \\ &\quad h\in BC(J,E). \end{aligned}
(2.7)

Thus, $$\mathcal{J}$$ maps $$BC(J,E)$$ into $$BC(J,E)$$ and

$$\Vert \mathcal{J} h\Vert _{{b\omega }}\leq \Vert h\Vert _{{b\omega }}.$$

In particular, if $$h(t)\equiv x$$, $$t\in J$$, $$x\in E$$, then

$$\Vert \mathcal{J} x\Vert _{{b\omega }}\leq \Vert x\Vert _{{_{\omega }}}.$$

Proof

Since

\begin{aligned} &\bigl\Vert (\mathcal{J} h) (t)\bigr\Vert _{{\omega }} \\ &\quad \leq \frac{q}{\varGamma {(1-q)}} \int _{0}^{1} \int _{0}^{\infty }(1-\tau )^{q-1}\tau ^{-q}\frac{\psi _{q}(\theta )}{\theta ^{q}}\biggl\Vert T \biggl( \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}} \biggr) h(s)\biggr\Vert _{{\omega }}\,d\theta \,d\tau \\ &\quad \leq \frac{q}{\varGamma {(1-q)}} \int _{0}^{1} \int _{0}^{\infty }(1-\tau )^{q-1}\tau ^{-q}\frac{\psi _{q}(\theta )}{\theta ^{q}}\biggl\Vert T \biggl( \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}} \biggr) \biggr\Vert _{{\omega }}\bigl\Vert h(s)\bigr\Vert _{{\omega }}\,d\theta \,d\tau \\ &\quad \leq \frac{q}{\varGamma {(1-q)}} \int _{0}^{1} \int _{0}^{\infty }(1-\tau )^{q-1}\tau ^{-q}\frac{\psi _{q}(\theta )}{\theta ^{q}}e^{-\omega ( \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}} ) }\Vert h\Vert _{{b\omega }}\,d\theta \,d\tau \\ &\quad \leq \frac{q}{\varGamma {(1-q)}}\Vert h\Vert _{{b\omega }} \int _{0}^{1}(1-\tau )^{q-1}\tau ^{-q} \biggl( \int _{0}^{\infty }\frac{\psi _{q}(\theta )}{\theta ^{q}}\,d\theta \biggr)\,d\tau \\ &\quad \leq \Vert h\Vert _{{b\omega }}, \end{aligned}

the proof is finished.āā”

Lemma 2.5

Set

$$(\mathcal{K}h) (t)= \int _{0}^{ t} \int _{0}^{\infty }q\frac{\psi _{q}(\theta )}{\theta ^{q}}(t-s)^{q-1}T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta \,ds,\quad h\in BC(J,E).$$

Thus, $$\mathcal{K}$$ maps $$BC(J,E)$$ into $$BC(J,E)$$ and

$$\bigl\Vert (\mathcal{K}h) (t) \bigr\Vert _{{\omega }} \leq \frac{1}{\varGamma {(q+1)}}\frac{1}{\omega }\Vert h\Vert _{{b\omega }};\qquad \bigl\Vert (\mathcal{K}h) \bigr\Vert _{{b\omega }}\leq \frac{1}{\varGamma {(q+1)}} \frac{1}{\omega }\Vert h\Vert _{{b\omega }}.$$

Proof

Since

\begin{aligned} (\mathcal{K}h) (t)& = \int _{0}^{ t} \int _{0}^{\infty }q\frac{\psi _{q}(\theta )}{\theta ^{q}}(t-s)^{q-1}T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta \,ds \\ & = \int _{0}^{ 1} \int _{0}^{\infty }q\frac{\psi _{q}(\theta )}{\theta ^{q}}(1-\tau )^{q-1}t^{q}T \biggl( \frac{(1-\tau )^{q}t^{q}}{\theta ^{q}} \biggr) h(t\tau )\,d\theta \,d\tau, \end{aligned}

we have

\begin{aligned} \bigl\Vert (\mathcal{K}h) (t) \bigr\Vert _{{\omega }}& \leq \int _{0}^{ 1} \int _{0}^{\infty }q\frac{\psi _{q}(\theta )}{\theta ^{q}}(1-\tau )^{q-1}t^{q}\biggl\Vert T \biggl( \frac{(1-\tau )^{q}t^{q}}{\theta ^{q}} \biggr) \biggr\Vert _{{\omega }}\bigl\Vert h(t\tau ) \bigr\Vert _{{\omega }}\,d\theta \,d\tau \\ & \leq \int _{0}^{ 1} \int _{0}^{\infty }q\frac{\psi _{q}(\theta )}{\theta ^{q}}t^{q}(1- \tau )^{q-1}e^{-\omega \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}}}\bigl\Vert h(t\tau ) \bigr\Vert _{{\omega }}\,d\theta \,d\tau \\ & \leq \frac{1}{\omega }\Vert h\Vert _{{b\omega }} \int _{0}^{\infty } \biggl[ \biggl( \int _{0}^{1} e^{-\omega ( \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}} ) }\,d\biggl( -\omega \frac{t^{q}(1-\tau )^{q}}{\theta ^{q}} \biggr) \biggr) \frac{\psi _{q}(\theta )}{\theta ^{q}} \biggr]\,d\theta \\ & =\frac{1}{\omega }\Vert h\Vert _{{b\omega }} \int _{0}^{\infty } \bigl( 1-e^{-\omega \frac{t^{q}}{\theta ^{q}}} \bigr) \frac{\psi _{q}(\theta )}{\theta ^{q}}\,d\theta \\ & \leq \frac{1}{\omega }\frac{1}{\varGamma {(q+1)}}\Vert h\Vert _{{b\omega }}. \end{aligned}

Therefore,

$$\bigl\Vert (\mathcal{K}h) \bigr\Vert _{{b\omega }}\leq \frac{1}{\omega } \frac{1}{\varGamma {(q+1)}}\Vert h\Vert _{{b\omega }}.$$

āā”

Lemma 2.6

The evolution equation of fractional order

\begin{aligned} \textstyle\begin{cases} ^{C}D_{0+}^{q}u(t)=h(t)+Au(t), \quad t \in (0,+\infty ), \\ u(0)=u_{0}, \end{cases}\displaystyle \end{aligned}
(2.8)

where $$h \in BC(J,E)$$ and $$u_{0}\in D(A)$$, has a unique solution in $$BC(J,E)$$

\begin{aligned} u(t) =&(\mathcal{J} u_{0}) (t)+(\mathcal{K}h) (t) \\ =& \int _{0}^{1} \int _{0}^{\infty }\frac{q}{\varGamma {(1-q)}} \frac{\psi _{q}(\theta )}{\theta ^{q}}(1-\tau )^{q-1}\tau ^{-q}T \biggl( \frac{(1-\tau )^{q}t^{q}}{\theta ^{q}} \biggr) u_{0}\,d\theta \,d\tau \\ &+ \int _{0}^{ t} \int _{0}^{\infty }q\frac{\psi _{q}(\theta )}{\theta ^{q}}(t-s)^{q-1}T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta \,ds. \end{aligned}
(2.9)

Proof

By Lemma 2.1, Definition 2.1, and Definition 2.2, (2.8) is equivalent with the following integral equation:

$$u(t)=u_{0}+ \int _{0}^{t}\frac{1}{\varGamma {(q)}}(t-s)^{q-1} \bigl[Au(s)+h(s)\bigr]\,ds.$$
(2.10)

By a similar method used in [18, 19], after Laplace transformation of the above equation, we can get

$$U(\lambda )=\frac{1}{\lambda }u_{0}+ \frac{1}{\lambda ^{q}}AU(\lambda )+\frac{1}{\lambda ^{q}}H(\lambda ),\quad \lambda >0,$$
(2.11)

where $$H(\lambda )$$ and $$U(\lambda )$$ are the Laplace transform of $$h(t)$$ and $$u(t)$$, respectively.

Then one has

$$\bigl(\lambda ^{q}I-A\bigr)U(\lambda )=\lambda ^{q-1}u_{0}+H( \lambda ).$$

From Lemma 2.3 and (2.6), we obtain

\begin{aligned} U(\lambda ) =&\bigl(\lambda ^{q}I-A\bigr)^{-1}\lambda ^{q-1}u_{0}+\bigl(\lambda ^{q}I-A \bigr)^{-1}H(\lambda ) \\ =&\lambda ^{q-1} \int _{0}^{\infty }e^{-\lambda ^{q}s}T(s)u_{0}\,ds+ \int _{0}^{\infty }e^{-\lambda ^{q}s}T(s)H(\lambda )\,ds \\ =&\lambda ^{q-1} \int _{0}^{\infty } \int _{0}^{\infty }e^{-\lambda s^{1/q}\theta }\psi _{q}( \theta )T(s)u_{0}\,d\theta \,ds \\ &+ \int _{0}^{\infty } \int _{0}^{\infty }e^{-\lambda s^{1/q}\theta }\psi _{q}( \theta )T(s)H(\lambda )\,d\theta \,ds \\ =&\lambda ^{q-1} \int _{0}^{\infty } \biggl[ \int _{0}^{\infty }q \frac{t^{q-1}}{\theta ^{q}}\psi _{q}(\theta )T \biggl( \frac{t^{q}}{\theta ^{q}} \biggr) u_{0}\,d\theta \biggr] e^{-\lambda t}\,dt \\ &+ \int _{0}^{\infty } \biggl[ \int _{0}^{t} \int _{0}^{\infty } \psi _{q}(\theta ) \frac{(t-s)^{q-1}}{\theta ^{q}} T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta ds \biggr] q e^{-\lambda t}\,dt. \end{aligned}

Taking the inverse Laplace transforms on the above equations, according to the convolution theorem and Lemma 2.5, we have

\begin{aligned} u(t)&=\mathcal{L}^{-1}\bigl[\lambda ^{q-1}\bigr]\ast \mathcal{L}^{-1} \biggl[ \int _{0}^{\infty } \biggl[ \int _{0}^{\infty }q\psi _{q}(\theta ) \frac{t^{q-1}}{\theta ^{q}}T \biggl( \frac{t^{q}}{\theta ^{q}} \biggr) u_{0}\,d\theta \biggr] e^{-\lambda t}\,dt \biggr] \\ &\quad {}+\mathcal{L}^{-1} \biggl[ \int _{0}^{\infty }e^{-\lambda t} \biggl( \int _{0}^{t} \int _{0}^{\infty }q \frac{(t-s)^{q-1}}{\theta ^{q}}\psi _{q}(\theta )T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta \,ds \biggr)\,dt \biggr] \\ &=\frac{t^{-q}}{\varGamma {(1-q)}}\ast \biggl[ \int _{0}^{\infty }q \frac{t^{q-1}}{\theta ^{q}}\psi _{q}(\theta )T \biggl( \frac{t^{q}}{\theta ^{q}} \biggr) u_{0}\,d\theta \biggr] \\ &\quad {}+ \int _{0}^{t} \int _{0}^{\infty }q \psi _{q}(\theta ) \frac{(t-s)^{q-1}}{\theta ^{q}}T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) h(s)\,d\theta \,ds \\ &=\frac{q}{\varGamma {(1-q)}} \int _{0}^{t} \int _{0}^{\infty }s^{-q}\psi _{q}( \theta )\frac{(t-s)^{q-1}}{\theta ^{q}}T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) u_{0}\,d\theta \,ds+(\mathcal{K}h) (t) \\ &=(\mathcal{J} u_{0}) (t)+(\mathcal{K}h) (t). \end{aligned}

Since

$$\bigl\Vert (\mathcal{K}h) (t)+(\mathcal{J} u_{0}) (t) \bigr\Vert _{{\omega }}\leq \bigl\Vert (\mathcal{K}h) (t) \bigr\Vert _{{\omega }}+ \bigl\Vert (\mathcal{J} u_{0}) (t) \bigr\Vert _{{\omega }}\leq \Vert u_{0} \Vert _{{\omega }}+\frac{1}{\omega } \frac{1}{\varGamma {(q+1)}}\Vert h\Vert _{{b\omega }},$$

we have

$$\Vert \mathcal{J} u_{0}+\mathcal{K}h \Vert _{{b\omega }}\leq \Vert u_{0} \Vert _{{\omega }}+\frac{1}{\omega } \frac{1}{\varGamma {(q+1)}}\Vert h\Vert _{{b\omega }}.$$

Hence, $$u\in BC(J,E)$$.āā”

In consequence, it leads to the conclusion.

Lemma 2.7

The main equation (1.1) can be written in the equivalent form

\begin{aligned} u(t)=Tu(t):= \bigl[\mathcal{J}\bigl(u_{0}-k(u)\bigr) \bigr](t)+( \mathcal{K}f) (t). \end{aligned}
(2.12)

It is obvious that problem (1.1) has a mild solution if T has a fixed point.

3 Main results

In the part, we give the main conclusions about the existence of mild solutions for the main equation on the right half-axis.

First, we give the result about the existence of positive mild solutions for problem (1.1).

Theorem 3.1

Assume that the normal cone P is positive in the Banach space E with the normal constant N. We assume that semigroup $$\{T(t)\}_{{t\geq 0}}$$ is strongly continuous and is uniformly exponentially stable with the growth bound $$\omega _{0} (\omega _{0}<0)$$, and the infinitesimal generator of $$\{T(t)\}_{{t\geq 0}}$$ is the operator A. Let $$m\in C[0,\infty )$$ and $$m(t)$$ is increasing and non-negative. Provided that $$u_{0}\in P, u_{0}\neq \theta$$, $$f(t,u):J\times E\longrightarrow E$$, $$f_{0}(t):=f(t,\theta )\geq \theta$$ is bounded on J, $$k(u)$$ maps E into E, both $$f(t,u)$$ and $$k(u)$$ are continuous. If $$k(u)$$ and $$f(t,u)$$ satisfy:

1. (F1)

for any $$t\in J$$,

$$f(t,z)\geq f(t,y), \quad \forall \theta \leq y\leq z.$$
2. (F2)

For a given $$\omega _{1}\in (0,\vert \omega _{0} \vert ]$$, there exists a positive number R such that

$$\Vert u_{0} \Vert _{{b}}+\frac{1}{\omega _{1}}\frac{1}{\varGamma {(q+1)}} R_{f}\leq R,$$

where

$$R_{f}=\sup_{s\in J, 0\leq \Vert z \Vert _{{b}}\leq R}\bigl\Vert f(s,z) \bigr\Vert _{{b}}< +\infty .$$
3. (F3)
$$\theta \leq k(y)\leq k(x)\leq u_{0}, \quad \forall \theta \leq x\leq y.$$

Thus, there exists a positive mild solution for Eq. (1.1) in $$BC(J,E)$$.

Proof

Define a nonempty subset of $$BC(J,E)$$

$$\mathcal{D}:=\bigl\{ u\in P_{{BC}}:\Vert u \Vert _{{b\omega }}\leq R\bigr\} .$$

By (F1), $$\forall u\in \mathcal{D}$$, one has

\begin{aligned} \theta &\leq Tu(t) \end{aligned}
(3.1)
\begin{aligned} &=(\mathcal{K}f) (t)+ \bigl[\mathcal{J}\bigl(u_{0}-k(u)\bigr) \bigr](t) \\ &\leq (\mathcal{K}f) (t)+ \bigl[\mathcal{J}(u_{0}) \bigr](t) \\ &=q \int _{0}^{ t} \int _{0}^{\infty }\frac{\psi _{q}(\theta )}{\theta ^{q}}(t-s)^{q-1}T \biggl( \frac{(t-s)^{q}}{\theta ^{q}} \biggr) f(s,u\bigl(m(s)\bigr)\,d\theta \,ds \\ &\quad {}+ \int _{0}^{1} \int _{0}^{\infty }\frac{q}{\varGamma {(1-q)}} \frac{\psi _{q}(\theta )}{\theta ^{q}}(1-\tau )^{q-1}\tau ^{-q}T \biggl( \frac{(1-\tau )^{q}t^{q}}{\theta ^{q}} \biggr) u_{0}\,d\theta \,d\tau . \end{aligned}
(3.2)

Then, combined with Lemma 2.4 and Lemma 2.5, one has

\begin{aligned} \bigl\Vert Tu(t) \bigr\Vert _{{b}} \leq &\frac{1}{\omega } \frac{1}{\varGamma {(q+1)}}R_{f}+\Vert u_{0} \Vert _{{b}}. \end{aligned}

Thus,

\begin{aligned} \Vert Tu \Vert _{{b\omega }} \leq &\Vert u_{0} \Vert _{{b}}+\frac{1}{\omega }\frac{1}{\varGamma {(q+1)}} R_{f}. \end{aligned}

From condition (F2), one can get

\begin{aligned} \Vert Tu \Vert _{{b\omega }} \leq &\Vert u_{0} \Vert _{{b}}+\frac{1}{\omega _{1}}\frac{1}{\varGamma {(q+1)}} R_{f} \leq R, \end{aligned}

which shows that $$T(\mathcal{D})\subseteq \mathcal{D}$$.

From condition (F1) and (F3), it is obvious that T maps $$\mathcal{D}$$ into $$\mathcal{D}$$ and it is also an increasing mapping. Then, applying Theorem 2.1, the map T possesses a fixed point in the nonempty set $$\mathcal{D}$$. Thus, Eq. (1.1) has a positive mild solution in $$BC(J,E)$$.āā”

Corollary 3.1

Assume that the normal cone P is positive in the Banach space E with the normal constant N. Provided that semigroup $$\{T(t)\}_{{t\geq 0}}$$ is strongly continuous and is uniformly exponentially stable with the growth bound $$\omega _{0}$$ ($$\omega _{0}<0$$), and the infinitesimal generator of $$\{T(t)\}_{{t\geq 0}}$$ is the operator A. Let $$m\in C[0,\infty )$$ and $$m(t)$$ be increasing and non-negative. Provided that $$u_{0}\in P, u_{0}\neq \theta$$, $$f(t,u):J\times E\longrightarrow E$$ is continuous and bounded, $$f_{0}(t):=f(t,\theta )\geq \theta$$, and $$k(u): E\longrightarrow E$$ is continuous. If $$f(t,u)$$ and $$k(u)$$ satisfy the conditions $$(F1)$$ and $$(F3)$$, respectively, there exists a positive mild solution for Eq. (1.1) in $$BC(J,E)$$.

Proof

Since $$f(t,u)$$ is bounded and continuous on $$J\times E$$, condition (F2) is fulfilled, then Theorem 3.1 implies that there is a positive mild solution of Eq. (1.1) in $$BC(J,E)$$.āā”

Theorem 3.2

Provided that $$\{T(t)\}_{{t\geq 0}}$$ (the semigroup of operators on Banach space E) is strongly continuous and is uniformly exponentially stable with the growth bound $$\omega _{0}$$ ($$\omega _{0}<0$$), and the infinitesimal generator of $$\{T(t)\}_{{t\geq 0}}$$ is the operator A. Assume that $$m(t)\geq 0$$ and $$m\in C[0,\infty )$$. If $$f(t,u)$$ and $$k(u)$$ satisfy:

1. (H1)

For a given $$\omega ^{*}\in (0,\vert \omega _{0} \vert ]$$, we have two positive numbers $$\mathcal{C}_{f}$$ and $$\mathcal{C}_{k}$$ satisfying $$\mathcal{M}<\omega ^{*}/\chi$$ such that

$$\mathcal{C}_{f} \Vert z_{1}-z_{2} \Vert _{{\omega ^{*}}}\geq \bigl\Vert f(t,z_{1})-f(t,z_{2}) \bigr\Vert _{{\omega ^{*}}},\quad \forall z_{1}, z_{2}\in E,$$

and

$$\mathcal{C}_{k} \Vert z_{1}-z_{2} \Vert _{{\omega ^{*}}}\geq \bigl\Vert k(z_{1})-k(z_{2}) \bigr\Vert _{{\omega ^{*}}}, \quad \forall z_{1}, z_{2}\in E.$$
2. (H2)
$$\mathcal{L}_{kf}:=\mathcal{C}_{k}+\frac{1}{\omega ^{*}} \frac{\mathcal{C}_{f}}{\varGamma {(q+1)}}< 1.$$

Thus, one has a unique mild solution for Eq. (1.1) in $$BC(J,E)$$.

Proof

By condition (H1), Lemma 2.4, Lemma 2.5, and Lemma 2.7, āv, $$u\in BC(J,E)$$, one has

\begin{aligned} &\bigl\Vert Tv(t)-Tu(t) \bigr\Vert _{{\omega ^{*}}} \\ &\quad = \bigl\Vert \bigl[\mathcal{J}\bigl(u_{0}-k(v)\bigr) \bigr](t)+ \mathcal{K}f\bigl(t,v\bigl(m(t)\bigr)\bigr)- \bigl[\mathcal{J} \bigl(u_{0}-k(u)\bigr) \bigr](t)-\mathcal{K}f\bigl(t,u\bigl(m(t)\bigr) \bigr) \bigr\Vert _{{\omega ^{*}}} \\ &\quad =\bigl\Vert \mathcal{K} \bigl[f\bigl(t,v\bigl(m(t)\bigr)\bigr)-f\bigl(t,u \bigl(m(t)\bigr)\bigr) \bigr]\bigr\Vert _{{\omega ^{*}}}+ \bigl(\mathcal{J} \bigl[k(u)-k(v) \bigr] \bigr) (t) \\ &\quad \leq \bigl\Vert \mathcal{K} \bigl[f\bigl(t,v\bigl(m(t)\bigr)\bigr)-f \bigl(t,u\bigl(m(t)\bigr)\bigr) \bigr]\bigr\Vert _{{\omega ^{*}}}+\bigl\Vert \bigl(\mathcal{J} \bigl[k(u)-k(v) \bigr] \bigr) (t)\bigr\Vert _{{\omega ^{*}}} \\ &\quad \leq \frac{1}{\varGamma {(q+1)}}\frac{1}{\omega ^{*}} \bigl\Vert f\bigl(t,v\bigl(m(t) \bigr)\bigr)-f\bigl(t,u\bigl(m(t)\bigr)\bigr) \bigr\Vert _{{\omega ^{*}}}+ \bigl\Vert k\bigl(u(t)\bigr)-k\bigl(v(t)\bigr) \bigr\Vert _{{\omega ^{*}}} \\ &\quad \leq \frac{\mathcal{C}_{f}}{\varGamma {(q+1)}}\frac{1}{\omega ^{*}} \bigl\Vert v\bigl(m(t)\bigr)-u \bigl(m(t)\bigr) \bigr\Vert _{{\omega ^{*}}}+\mathcal{C}_{k} \bigl\Vert u(t)-v(t) \bigr\Vert _{{\omega ^{*}}} \\ &\quad \leq \biggl(\mathcal{C}_{k}+\frac{1}{\omega ^{*}}\frac{\mathcal{C}_{f}}{\varGamma {(q+1)}} \biggr) \Vert v-u \Vert _{{b\omega ^{*}}}. \end{aligned}

Therefore,

$$\mathcal{L}_{kf} \Vert v-u \Vert _{{b\omega ^{*}}} \geq \Vert Tv-Tu \Vert _{{b\omega ^{*}}}.$$

Then, according to condition (H2), Eq. (1.1) has a unique mild solution.āā”

4 Examples

In order to certify the effectiveness of the main conclusion, an example is given in the following. Consider the following partial differential equation of fractional order.

Example 4.1

$$\textstyle\begin{cases} \partial ^{q}_{t}z(t,x)=F(t,m(z(t,x)))+\partial ^{2}_{x}z(t,x),&t \in [0,+\infty ), \\ z(0,x)+k(z(t,x))=z_{0},& x\in [0,\pi ], \\ z(t,0)=z(t,\pi )=0,& t \in [0,+\infty ), \end{cases}$$
(4.1)

where $$\partial ^{q}_{t}$$ ($$q\in (0,1)$$) is fractional partial derivative of fractional order in the Caputo sense.

Set $$E=L^{2}([0,\pi ],\mathbb{R})$$ and $$Az=\partial ^{2}_{x}z$$, in view of [39], we can conclude that A is a linear operator mapping $$D(A)$$ into E, and the domain $$D(A)= \{v \mid v(0)=v(\pi )=0, v'\in E, v\in E \}$$. In addition, the operator A generates a semigroup $$\{T(t)\}_{{t\geq 0}}$$, which is strongly continuous and uniformly exponentially stable. Denote $$\omega _{0}$$ as the growth bound of $$\{T(t)\}_{{t\geq 0}}$$, then we can get $$\omega _{0}\leq -1$$.

Set $$v(t)=z(t,\cdot )$$, $$v_{0}=z_{0}$$, $$g(t,m(v(t)))=F(t,z(t,\cdot ))$$. At this point, the above problem can be written as

\begin{aligned} \textstyle\begin{cases} ^{C}D_{0+}^{q}v(t)=f(t,m(v(t)))+Av(t),\quad t \in (0,+\infty ), \\ v(0)+k(v)=v_{0}. \end{cases}\displaystyle \end{aligned}
(4.2)

Consider the function

$$g(t,v)=\frac{2+t^{2}}{1+t^{2}} \biggl( 2-\frac{1}{1+v^{2}} \biggr)$$

and

$$k(v)=1+\frac{1}{4+\arctan v}.$$

Take $$q=1/2$$, $$v_{0}=3$$ and $$m(v)=v^{2}$$.

Then the conditions of Theorem 3.1 are all fulfilled. Therefore, there one has a positive mild solution for Eq. (4.1).

5 Conclusion

Here, using Laplace transform and its inverse transform, we obtain a correct equivalent integral equation for some kind of nonlocal abstract differential equations (fractional order) on the right half-axis, which is different from those given in the existing literature. According to this equivalent integral equation as obtained above, we investigate a kind of abstract fractional order differential equations. By using Knasterās theorem, we get the existence of positive mild solution for the main equation, and the uniqueness of a mild solution by the Banach contraction theorem.

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Acknowledgements

The authors are grateful to the referees for their careful reading of the manuscript. The authors thank the help from the editor.

Not applicable.

Funding

This research is supported by the National Natural Science Foundation of China (No. 71672195, 11601508), and supported by Science Fund for Creative Research Groups of the National Natural Science Foundation of China (No. 71221061).

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Correspondence to Yanping Gong.

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Lv, Z., Gong, Y. & Chen, Y. Existence and uniqueness for a kind of nonlocal fractional evolution equations on the unbounded interval. Bound Value Probl 2018, 176 (2018). https://doi.org/10.1186/s13661-018-1095-7