Solutions to Fractional Differential Equations with Nonlocal Initial Condition in Banach Spaces

Abstract

A new existence and uniqueness theorem is given for solutions to differential equations involving the Caputo fractional derivative with nonlocal initial condition in Banach spaces. An application is also given.

1. Introduction

Fractional differential equations have played a significant role in physics, mechanics, chemistry, engineering, and so forth. In recent years, there are many papers dealing with the existence of solutions to various fractional differential equations; see, for example, [1–6].

In this paper, we discuss the existence of solutions to the nonlocal Cauchy problem for the following fractional differential equations in a Banach space :

(11)

where is the standard Caputo's derivative of order , and is a given -valued function.

2. Basic Lemmas

Let be a real Banach space, and the zero element of . Denote by the Banach space of all continuous functions with norm . Let be the Banach space of measurable functions which are Lebesgue integrable, equipped with the norm . Let , function is called a solution of (1.1) if it satisfies (1.1).

Recall the following defenition

Definition 2.1.

Let be a bounded subset of a Banach space . The Kuratowski measure of noncompactness of is defined as

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Clearly, . For details on properties of the measure, the reader is referred to [2].

Definition 2.2 (see [7, 8]).

The fractional integral of order with the lower limit for a function is defined as

(22)

where is the gamma function.

Definition 2.3 (see [7, 8]).

Caputo's derivative of order with the lower limit for a function can be written as

(23)

Remark.

Caputo's derivative of a constant is equal to .

Lemma (see [7]).

Let . Then we have

(24)

Lemma (see [7]).

Let and . Then

(25)

Lemma(see [9]).

If is bounded and equicontinuous, then

(a);

(b) where

Lemma.

(26)

where , .

Proof.

A direct computation shows

(27)

and

(28)

3. Main Results

, and there exist such that for and each

For any and , is relatively compact in , where and

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Lemma.

If holds, then the problem (1.1) is equivalent to the following equation:

(32)

Proof.

By Lemma 2.6 and (1.1), we have

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Therefore,

(34)

So,

(35)

and then

(36)

Conversely, if is a solution of (3.2), then for every , according to Remark 2.4 and Lemma 2.5, we have

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It is obvious that This completes the proof.

Theorem.

If (H₁) and (H₂) hold, then the initial value problem (1.1) has at least one solution.

Proof.

Define operator , by

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Clearly, the fixed points of the operator are solutions of problem (1.1).

It is obvious that is closed, bounded, and convex.

Step 1.

We prove that is continuous.

Let

(39)

Then and For each ,

(310)

It is clear that

(311)

It follows from (3.11) and the dominated convergence theorem that

(312)

Step 2.

We prove that .

Let . Then for each , we have

(313)

Step 3.

We prove that is equicontinuous.

Let , and . We deduce that

(314)

As , the right-hand side of the above inequality tends to zero.

Step 4.

We prove that is relatively compact.

Let be arbitrarily given. Using the formula

(315)

for and , we obtain

(316)

It follows from (3.16) that This, together with Lemma 2.7, yields that

(317)

From (3.17), we see that is relatively compact. Hence, is completely continuous. Finally, the Schauder fixed point theorem guarantees that has a fixed point in .

Theorem.

Besides the hypotheses of Theorem 3.2, we suppose that there exists a constant such that

(318)

(319)

where

(320)

Then, the solution of (1.1) is unique in .

Proof.

From Theorem 3.2, we know that there exists at least one solution in . We suppose to the contrary that there exist two different solutions and in . It follows from (3.8) that

(321)

Therefore, we get

(322)

By (3.18), we obtain So, the two solutions are identical in .

4. Example

Let

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with the norm Consider the following nonlocal Cauchy problem for the following fractional differential equation in :

(42)

Conclusion 4.

Problem (4.2) has only one solution on

Proof.

Write

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Then it is clear that

(44)

So, is satisfied.

In the same way as in Example in [9], we can prove that is relatively compact in .

By a direct computation, we get

(45)

Hence, condition is also satisfied.

Moreover, we have

(46)

so

(47)

Clearly,

(48)

Therefore, . Thus, our conclusion follows from Theorem 3.3.