Nonlocal Cauchy Problem for Nonautonomous Fractional Evolution Equations

Abstract

We investigate the mild solutions of a nonlocal Cauchy problem for nonautonomous fractional evolution equations , , in Banach spaces, where , . New results are obtained by using Sadovskii's fixed point theorem and the Banach contraction mapping principle. An example is also given.

1. Introduction

During the past decades, the fractional differential equations have been proved to be valuable tools in the investigation of many phenomena in engineering and physics; they attracted many researchers (cf., e.g., [1–9]). On the other hand, the autonomous and nonautonomous evolution equations and related topics were studied in, for example, [6, 7, 10–20], and the nonlocal Cauchy problem was considered in, for example, [2, 5, 18, 21–26].

In this paper, we consider the following nonlocal Cauchy problem for nonautonomous fractional evolution equations

(1.1)

in Banach spaces, where , . The terms , are defined by

(1.2)

the positive functions are continuous on and

(1.3)

Let us assume that and is a family linear closed operator defined in a Banach space . The fractional order integral of the function is understood here in the Riemann-Liouville sense, that is,

(1.4)

In this paper, we denote that is a positive constant and assume that a family of closed linear satisfying

(A1) the domain of is dense in the Banach space and independent of ,

(A2) the operator exists in for any with and

(1.5)

(A3) There exists constant and such that

(1.6)

Under condition (A2), each operator , generates an analytic semigroup ,, and there exists a constant such that

(1.7)

where , , ([11]).

We study the existence of mild solution of (1.1) and obtain the existence theorem based on the measures of noncompactness. An example is given to show an application of the abstract results.

2. Preliminaries

Throughout this work, we set . We denote by a Banach space, the space of all linear and bounded operators on , and the space of all -valued continuous functions on .

Lemma 2.1 (see [9]).

• (1) .

• (2) For , we have

  

  (2.1)

where is a Beta function.

Definition 2.2.

Let be a bounded set of seminormed linear space . The Kuratowski's measure of noncompactness (for brevity, -measure) of is defined as

(2.2)

From the definition, we can get some properties of -measure immediately, see ([27]).

Lemma 2.3 (see [27]).

Let and be bounded sets of . Then

• (1) , if .

• (2) , where denotes the closure of .

• (3) if and only if is precompact.

• (4) , .

• (5) .

• (6) , where .

• (7) , for any .

For we define

(2.3)

for , where .

The following lemma will be needed.

Lemma 2.4 (see [27]).

If is a bounded, equicontinuous set, then

• (1) .

• (2) , for .

Lemma 2.5 (see [28]).

If and there exists a such that

(2.4)

then is integrable and

(2.5)

We need to use the following Sadovskii's fixed point theorem.

Definition 2.6 (see [29]).

Let be an operator in Banach space . If is continuous and takes bounded, sets into bounded sets, and for every bounded set of with , then is said to be a condensing operator on .

Lemma 2.7 (Sadovskii's fixed point theorem [29]).

Let be a condensing operator on Banach space . If for a convex, closed, and bounded set of , then has a fixed point in .

According to [4], a mild solution of (1.1) can be defined as follows.

Definition 2.8.

A function satisfying the equation

(2.6)

is called a mild solution of (1.1), where

(2.7)

and is a probability density function defined on such that its Laplace transform is given by

(2.8)

where

(2.9)

To our purpose, the following conclusions will be needed. For the proofs refer to [4].

Lemma 2.9 (see [4]).

The operator-valued functions and are continuous in uniform topology in the variables , , where , , for any . Clearly,

(2.10)

Moreover, we have

(2.11)

Remark 2.10.

From the proof of Theorem 2.5 in [4], we can see

• (1) .

• (2) For , is uniformly continuous in the norm of and

  

  (2.12)

3. Existence of Solution

Assume that

(B1) satisfies is measurable for all , and is continuous for a.e , and there exist a positive function and a continuous nondecreasing function such that

(3.1)

and set .

(B2) For any bounded sets , and ,

(3.2)

where is a nonnegative function, and ,

(3.3)

(B3) is continuous and there exists

(3.4)

such that

(3.5)

(B4) The functions and satisfy the following condition:

(3.6)

where , and .

Theorem 3.1.

Suppose that (B1)–(B4) are satisfied, and if , then (1.1) has a mild solution on .

Proof.

Define the operator by

(3.7)

Then we proceed in five steps.

Step 1.

We show that is continuous.

Let be a sequence that as . Since satisfies (B1), we have

(3.8)

Then

(3.9)

According to the condition (A2), (2.12), and the continuity of , we have

(3.10)

Noting that in , there exists such that for sufficiently large. Therefore, we have

(3.11)

Using (2.10) and by means of the Lebesgue dominated convergence theorem, we obtain

(3.12)

Similarly, by (2.10) and (2.11), we have

(3.13)

Therefore, we deduce that

(3.14)

Step 2.

We show that maps bounded sets of into bounded sets in .

For any , we set . Now, for , by (B1), we can see

(3.15)

Based on (2.12), we denote that , we have

(3.16)

Then for any , by (A2), (2.10), (2.11), and Lemma 2.1, we have

(3.17)

where .

By means of the Hölder inequality, we have

(3.18)

Thus

(3.19)

This means .

Step 3.

We show that there exists such that .

Suppose the contrary, that for every , there exists and , such that . However, on the other hand

(3.20)

we have

(3.21)

Dividing both sides by and taking the lower limit as , we obtain

(3.22)

which contradicts (B4).

Step 4.

Denote

(3.23)

where

(3.24)

We show that is equicontinuous.

Let and . Then

(3.25)

where

(3.26)

It follows from Lemma 2.9, (B1), and (3.20) that .

For , from (2.10), (3.20), and (B1), we have

(3.27)

Similarly, by (2.10), (2.11), (B1), and Lemma 2.1, we have

(3.28)

Step 5.

We show that for every bounded set . For any , we can take a sequence such that

(3.29)

(cf. [30]). So it follows from Lemmas 2.3–2.5, 2.9, (2) in Remark 2.10, and (B2) that

(3.30)

Since is arbitrary, we can obtain

(3.31)

In summary, we have proven that has a fixed point . Consequently, (1.1) has at least one mild solution.

Our next result is based on the Banach's fixed point theorem.

(G1) There exists a positive function and a constant such that

(3.32)

(G2) There exists a constant such that the function defined by

(3.33)

Theorem 3.2.

Assume that (G1), (G2) are satisfied, then (1.1) has a unique mild solution.

Proof.

Let be defined as in Theorem 3.1. For any , we have

(3.34)

Thus, from (A2), (2.10), (2.11), Lemma 2.1, we have

(3.35)

We get

(3.36)

By the Banach contraction mapping principle, has a unique fixed point, which is a mild solution of (1.1).

4. An Example

To illustrate the usefulness of our main result, we consider the following fractional differential equation:

(4.1)

where , , , , is continuous function and is uniformly Hölder continuous in , that is, there exist and such that

(4.2)

Let and define by

(4.3)

Then generates an analytic semigroup .

For , , we set

(4.4)

where

(4.5)

Moreover, we can get

(4.6)

for any . Then the above equation (4.1) can be written in the abstract form as (1.1). On the other hand,

(4.7)

where , satisfying (B1). For any ,

(4.8)

Therefore, for any bounded sets , we have

(4.9)

Moreover,

(4.10)

Similarly, we obtain

(4.11)

Suppose further that

• (1) ,

• (2) .

Then (4.1) has a mild solution by Theorem 3.1.