On a Nonlinear Integral Equation with Contractive Perturbation

Abstract

We get an existence result for solutions to a nonlinear integral equation with contractive perturbation by means of Krasnoselskii's fixed point theorem and especially the theory of measure of weak noncompactness.

1. Introduction

The integral equations have many applications in mechanics, physics, engineering, biology, economics, and so on. It is worthwhile mentioning that some problems considered in the theory of abstract differential equations also lead us to integral equations in Banach space, and some foundational work has been done in [1–8].

In this paper we want to study the following integral equation:

(1.1)

in the Banach space .

This equation has been studied when in [9] with and [10] with a perturbation term . Our paper extends their work to more general spaces and some modifications are also given on an error of [10].

Our paper is organized as follows.

In Section 2, some notations and auxiliary results will be given. We will introduce the main tools measure of weak noncompactness and Krasnoselskii's fixed point theorem in Section 3 and Section 4. The main theorem in our paper will be established in Section 5.

2. Preliminaries

First of all, we give out some notations to appear in the following.

denotes the set of real numbers and . Suppose that is a separable locally compact Banach space with norm in the whole paper. (Remark: the locally compactness of will be used in Lemma 2.2). Let be a Lebesgue measurable subset of and denote the Lebesgue measure of .

Let denote the space of all real Lebesgue measurable functions defined on to . forms a Banach space under the norm

(2.1)

for .

Definition 2.1.

A function is said to satisfy Carathéodory conditions if

• (i) is measurable in for any ;

• (ii) is continuous in for almost all .

The following lemma which we will use in the proof of our main theorem explains the structure of functions satisfying Carathéodory conditions with the assumption that the space is separable and locally compact (see [11]).

Lemma 2.2.

Let be a bounded interval and be a functionsatisfying Carathéodory conditions. Then it possesses the Scorza-Dragoni property. That is each , there exists a closed subset of such that and is continuous.

The operator is called superposition operator or Nemytskij operator associated to . The following lemma on superposition operator is important in our theorem (see [12] and also in [13]).

Lemma 2.3.

The superposition operator generated by the function maps continuously the space into itself ( may be unbounded interval) if and only if there exist and a nonnegative constant such that

(2.2)

for all and .

The Volterra operator which is defined by for where is measurable with respect to both variables. If transforms into itself it is then a bounded operator with norm which is majorized by the number

(2.3)

3. Measure of Weak Noncompactness

In this section we will recall the concept of measure of weak noncompactness which is the key point to complete our proof of main theorem in Section 5.

Let be a Banach space. and denote the collections of all nonempty bounded subsets and relatively weak compact subsets, respectively.

Definition 3.1.

A function is said to be a measure of weak noncompactness if it satisfies the following conditions:

• (1) the family Ker  is nonempty and Ker ;

• (2) if , we have ;

• (3) , where denotes the convex closed hull of ;

• (4) for ;

• (5) If is a decreasing sequence, that is, , every is weakly closed,

  and , then is nonempty.

From [14], we see the following measure of weak noncompact:

(3.1)

where denotes the closed ball in centered at 0 with radius .

In [15], Appel and De Pascale gave to the following simple form in space:

(3.2)

for a nonempty and bounded subset of space .

Let

(3.3)

for a nonempty and bounded subset of space .

It is easy to know that is a measure of weak noncompactness in space following the verification in [16].

4. Krasnoselskii's Fixed Point Theorem

The following is the Krasnoselskii's fixed point theorem which will be utilized to obtain the existence of solutions in the next section.

Theorem 4.1.

Let be a closed convex and nonempty subset of a Banach space . Let be two operators such that

• (i) ;

• (ii) is a contraction mapping;

• (iii) is relatively compact and is continuous.

Then there exists such that .

Remark 4.2.

In [9], they proved the existence of solutions by means of Schauder fixed point theorem. With the presence of the Perturbation term in the integral equation, the Schauder fixed point theorem is invalid. To overcome this difficulty we will use the Kransnoselskii's fixed point theorem to obtain the existence of solutions.

Remark 4.3.

We will see in the following section that the important step is the construction of by means of measure of weak noncompactness. This is the biggest difference between our paper from [10].

Remark 4.4.

The Krasnoselskii's fixed point theorem was extended to general case in [17] (see also in [13]). In [10], they used the general Krasnoselskii's fixed point theorem to obtain the existence result. It can be seen in the next section of our paper that the classical Krasnoselskii's fixed point theorem is enough unless we need more general conditions on the perturbation term .

5. Main Theorem and Proof

Our main theorem in this paper is stated as follows.

Theorem 5.1.

Suppose that the following assumptions are satisfied.

(H1) The functions satisfy Carathéodory conditions, and there exist constants and functions such that

(5.1)

for and .

(H2) Then function satisfies Carathéodory conditons, and the linear Volterra integral operator defined by

(5.2)

transforms the space into itself.

(H3) The function is measurable in t and continuous in x and y for almost all t. And there exist two positive constants and a function such that

(5.3)

for and . Additionally, the function satisfies the following Lipschitz condition for almost all :

(5.4)

(H4) The function such that where is an arbitrary subset of , and is bounded by for all .

(H5) , where denotes the norm of the linear Volterra operator .

(H6) .

Then the integral equation (1.1) has at least one solution .

Proof.

Equation (1.1) may be written in the following form:

(5.5)

where is the linear Volterra integraloperator and is the superposition operator generated by the function .

The proof will be given in six steps.

Step 1.

There exists such that , where is a ball centered zero element with radius in .

Let and be arbitrary functions in with to be determined later. In view of our assumptions we get

(5.6)

We then derive that by taking

(5.7)

where by assumption (H5).

Step 2.

for allbounded subset of .

Take a arbitrary numbers and such that .

For any , we have

(5.8)

It follows that by definition (3.2).

For and any , we have

(5.9)

and then by definition (3.3).

From above, we then obtain for all bounded subset of .

Step 3.

We will construct a nonempty closed convex weakly compact set in on which we will apply fixed point theorem to prove the existence of solutions.

Let where is defined in Step 1, and so on. We then get a decreasing sequence , that is, for . Obviously all sets belonging to this sequence are closed and convex, so weakly closed. By the fact proved in Step 2 that for all bounded subset of , we have

(5.10)

which yields that .

Denote , and then . By the definition of measure of weak noncompact we know that is nonempty. Moreover, .

is just nonempty closed convex weakly compact set which we need in the following steps.

Step 4.

is relatively compact in , where is just the set constructed in Step 3.

Let be arbitrary sequence. Since , , , the following inequality is satisfied:

(5.11)

Considering the function on and on , we can find a closed subset of interval , such that , and such that and is continuous. Especially is uniformly continuous.

Let us take arbitrarily and assume without loss of generality. For an arbitrary fixed and denoting we obtain:

(5.12)

where denotes the modulusof continuity of the function on the set and . The last inequality of (5.12) is obtained since , where is just the one in the Step 1.

Taking into account the fact that the , we infer that the terms of the numerical sequence are arbitrarily small provided that the number is small enough.

Since is also arbitrarily small provided that the number is small enough, the right of (5.12) then tends to zero independent of as tends to zero. We then have is equicontinuous in the space .

On the other hand,

(5.13)

From above, we then obtain that is equibounded in the space .

By assumption (H1),we have the operator is continuous. So forms a relatively compact set in the space .

Further observe that the above result does not depend on the choice of . Thus we can construct a sequence of closed subsets of the interval such that as and such that the sequence is relatively compact in every space . Passing to subsequence if necessary we can assume that is a cauchy sequence in each space .

Observe the fact , then . By the definition (3.2), let us choose a number such that for each closed subset of the interval provided that we have

(5.14)

for any , where .

By the fact that is a cauchy sequence in each space , we can choose a natural number such that and , and for arbitrary natural number the following inequality holds:

(5.15)

for any .

Combining (5.11), (5.14) and (5.15), we get

(5.16)

which means that is a cauchy sequence in the space . Hence we conclude that is relatively compact in .

Step 5.

The operator is a contraction mapping:

(5.17)

where we have made a transformation in the above process. Since by assumption (H6), we then get the fact that the operator is a contraction mapping.

Step 6.

We now check out that the conditions needed in Krasnoselskii's fixed point theorem are fulfilled.

• (1) From Step 3, we know that , where is the set constructed in Step 3.

• (2) From Step 5, we know that is a contraction mapping.

• (3) From the Step 4 and assumptions (H1),   (H2), is relatively compact and is continuous.

We apply Theorem 4.1, and then obtain that (1.1) has at least one solution in .

Remark 5.2.

When , in [10] they said is weakly sequence compact in their Step 1 of main proof. From our proof, we know that their proof is not precise, since in Step 4, one of the crucial conditions to prove the relatively compactness of is that is weakly compact. We can only obtain that is weakly sequence compact as a mapping from to which is the weakly compact set defined in Step 3. The construction of set overcomes the fault in [10], and we obtain the existence result finally.