Abstract

Using the classical Schauder fixed point theorem, we prove the existence of solutions of a quadratic integral equation of Fredholm type with supremum in the space of functions satisfying the Hölder condition.

1. Introduction

Quadratic integral equations arise naturally in applications of real-world problems. For example, problems in the theory of radiative transfer in the theory of neutron transport and in the kinetic theory of gases lead to the quadratic equation where is a continuous function defined on the interval (see [1–4]).

Equations of this type have been studied by several authors [5–11].

The aim of this paper is to investigate the existence of solutions of the following quadratic integral equation of Fredholm type with supremum:

Differential and integral equations with supremum are adequate models of real-world problems, in which the present state depends significantly on this maximum value on a past time interval [12]. Equations of such kind have been studied in some papers appearing in the literature (see [13–20]).

Our solutions are placed in the space of functions satisfying the Hölder condition. A sufficient condition for the relative compactness in these spaces and the classical Schauder fixed point theorem are the main tools used in our study.

2. Preliminaries

Our starting point in this section is to introduce the space of functions satisfying the Hölder condition and some properties in this space. These properties appear in [21].

Let be a closed interval in ; by we denote the space of the continuous functions on equipped with the norm .

For fixed , by , we will denote the space of the real functions defined on and satisfying the Hölder condition, that is, those functions for which there exists a constant such that for all .

It is easily seen that forms a linear subspace of .

In what follows, for , by , we will denote the least possible constant for which inequality (3) is satisfied. More “precisely,” we put The space with can be normed under the following norm: for .

It is proved in [21] that with is a Banach space.

Now, we recollect some results about the spaces with which appear in [21].

Lemma 1. For with the following inequality is satisfied:

Lemma 2. For , one has Moreover, for the following inequality holds:

The following sufficient condition for relative compactness in the spaces with appears in Example 6 of [21].

Theorem 3. Suppose that and that is a bounded subset in this means that there exists a constant such that for any and for any . Then is a relatively compact subset of .

Remark 4. Suppose that and by we denote the ball centered at and radius in the space ; that is, . Then is compact in the space .

Proof. In fact, by Theorem 3, since is a bounded subset in , is a relatively compact subset of .
Suppose that and with . This means that for we can find such that or, equivalently This implies that .
Moreover, if in (10) we put , then we get The last inequality implies that Therefore, for any and any and taking into account (10) and (12), we have Consequently,
Next, we will prove that .
In fact, since , we have that and, accordingly, Letting in the above inequality and taking into account (14), we deduce that Hence, we get and this means that .
This proves that is a closed subset of . Thus, is a compact subset of .
This finishes the proof.

Next, we recall some results appearing in [18].

In what follows, we consider .

Lemma 5. Let be a continuous and nondecreasing function and . Let be the function defined by Then .

Under the above assumption we have the following.

Lemma 6. Let be a sequence in and such that in . Then

Finally, we recall Schauder’s fixed point theorem.

Theorem 7. Let be a nonempty, convex, and compact subset of a Banach space and let be a continuity mapping. Then has at least one fixed point in .

3. Main Result

In this section we will investigate the solvability of the integral equation (2) in the Hölder spaces.

We will formulate the following assumptions:(i) ();(ii) is a continuous function such that it satisfies the Hölder condition with exponent with respect to the first variable; that is, there exists a constant such that for any ;(iii) is a continuous and nondecreasing function;(iv)the following inequality is satisfied: where the constant is defined by and whose existence is guaranteed by virtue of (ii).

Theorem 8. Under assumptions , (2) has at least one solution belonging to the space , where is arbitrarily fixed number satisfying .

Proof. Let us consider the operator defined on the space by for .
First, we will prove that transforms the space into itself.
In fact, we take an arbitrary function and such that . Then, in view of assumptions (i) and (ii), we get By Lemma 1, , and, since , from the last inequality it follows that Therefore, where we have used the fact that .
This shows that the operator transforms into itself.
On the other hand, the inequality is satisfied by the number which are positive by virtue of (iv), and, consequently, from (27) we infer that transforms the ball into itself, for any .
Therefore, , where .
By Theorem 3 and Remark 4, is a compact subset in the space for any .
Next, we will prove that the operator is continuous on , where in we consider the induced norm by , where .
To this end fix and . Suppose that and , where .
Then, for any with , we have On the other hand, we have From (30) and (31), we infer that This shows that the operator is continuous at the point , where is arbitrary.
Finally, applying Theorem 7, we deduce that desired result.
Finally, we illustrate our results by presenting an example.
Example 9. Let us consider the following quadratic integral equation: where and , and are positive constants.
Notice that (33) is a particular case of (2), where and .
It is easily seen that and, consequently, .
Moreover, using the inequality proved in [11], we have for any . Therefore, assumption (ii) of Theorem 8 is satisfied, since for any where .
Moreover, we have that Since , the function is continuous and increasing on and for all . Thus, assumption (iii) of Theorem 8 is satisfied.
In our case, the constant is given by In our case, the inequality appearing in assumption (iv) of Theorem 8 takes the form It is easily seen that the above inequality is satisfied when, for example, and .
Therefore, using Theorem 8, we infer that (33) for and has at least one solution in the space with .