Abstract

We study the solvability of some nonlinear functional integral equations in the Banach algebra of real functions defined, continuous, and bounded on the real half axis. We apply the technique of measures of noncompactness in order to obtain existence results for equations in question. Additionally, that technique allows us to obtain some characterization of considered integral equations. An example illustrating the obtained results is also included.

1. Introduction

The purpose of the paper is to study the solvability of some functional integral equations in the Banach algebra consisting of real, continuous, and bounded functions defined on an unbounded interval. Equations of such a kind are recently often investigated in the mathematical literature (cf. [16]). It is worthwhile mentioning that some functional integral equations of that type find very interesting applications to describe real world problems which appeared in engineering, mathematical physics, radiative transfer, kinetic theory of gases, and so on (cf. [713], e.g.).

Functional integral equations considered in Banach algebras have rather complicated form, and the study of such equations requires the use of sophisticated tools. In the approach applied in this paper we will use the technique associated with measures of noncompactness and some fixed point theorems [14]. Such a direction of investigations has been initiated in the paper [2], where the authors introduced the so-called condition () related to the operation of multiplication in an algebra and playing a crucial role in the use of the technique of measures of noncompactness in Banach algebras setting. The usefulness of such an approach has been presented in [2], where the solvability of some class of functional integral equations was proved with help of some measures of noncompactness satisfying the mentioned condition ().

This paper is an extension and continuation of the paper [2]. Here we are going to unify the approach with the use of the technique of measures of noncompactness to some general type of functional integral equations in the Banach algebra described above. Measures of noncompactness used here allow us not only to obtain the existence of solutions of functional integral equations but also to characterize those solutions in terms of stability, asymptotic stability, and ultimate monotonicity, for example.

Let us notice that such an approach to the theory of functional integral equations in Banach algebras is rather new and it was not exploited up to now except from the paper [2] initiating this direction of investigation.

2. Notation, Definitions, and Auxiliary Results

This section is devoted to presenting auxiliary facts which will be used throughout the paper. At the beginning we introduce some notation.

Denote by the set of real numbers and put . If is a given real Banach space with the norm and the zero element , then by we denote the closed ball centered at and with radius . We will write to denote the ball . If is a subset of , then the symbols and Conv stand for the closure and convex closure of , respectively. Apart from this the symbol diam will denote the diameter of a bounded set while denotes the norm of .

Next, let us denote by the family of all nonempty and bounded subsets of and by its subfamily consisting all relatively compact sets.

In what follows we will accept the following definition of the concept of a measure of noncompactness [14].

Definition 1. A mapping will be called a measure of noncompactness in if it satisfies the following conditions. The family ker is nonempty and ker. . . for . If is a sequence of closed sets from such that for and if , then the set is nonempty.

The family ker described in is said to be the kernel of the measure of noncompactness .

Observe that the set from the axiom is a member of the family ker. Indeed, from the inequality being satisfied for all we derive that which means that . This fact will play a key role in our further considerations.

In the sequel we will usually assume that the space has the structure of Banach algebra. In such a case we write in order to denote the product of elements . Similarly, we will write to denote the product of subsets of .

Now, we recall a useful concept introduced in [2].

Definition 2. One says that the measure of noncompactness defined on the Banach algebra satisfies condition () if for arbitrary sets the following inequality is satisfied:
It turns out that the above defined condition () is very convenient in considerations connected with the use of the technique of measures of noncompactness in Banach algebras. Apart from this the majority of measures of noncompactness satisfy this condition [2]. We recall some details in the next section.

Now, we are going to formulate a fixed point theorem for operators acting in a Banach algebra and satisfying some conditions expressed with help of a measure of noncompactness. To this end we first recall a concept parallel to the concept of Lipschitz continuity (cf. [14]).

Definition 3. Let be a nonempty subset of a Banach space , and let be a continuous operator which transforms bounded subsets of onto bounded ones. One says that satisfies the Darbo condition with a constant with respect to a measure of noncompactness if for each such that . If , then is called a contraction with respect to .

Now, assume that is a Banach algebra and is a measure of noncompactness on satisfying condition (). Then we have the following theorem announced above [2].

Theorem 4. Assume that is nonempty, bounded, closed, and convex subset of the Banach algebra , and the operators and transform continuously the set into in such a way that and are bounded. Moreover, one assumes that the operator transforms into itself. If the operators and satisfy on the set the Darbo condition with respect to the measure of noncompactness with the constants and , respectively, then the operator satisfies on the Darbo condition with the constant Particularly, if , then is a contraction with respect to the measure of noncompactness and has at least one fixed point in the set .

Remark 5. It can be shown [14] that the set Fix of all fixed points of the operator on the set is a member of the kernel ker.

3. Some Measures of Noncompactness in the Banach Algebra ()

In this section we present some measures of noncompactness in the Banach algebra consisting of all real functions defined, continuous, and bounded on the half axis . The algebra is endowed with the usual supremum norm for . Obviously, the multiplication in is understood as the usual product of real functions. Let us mention that measures of noncompactness, which we intend to present here, were considered in details in [2].

In what follows let us assume that is an arbitrarily fixed nonempty and bounded subset of the Banach algebra . Choose arbitrarily and . For denote by the modulus of continuity of the function on the interval ; that is, Next, let us put Further, define the set quantity by putting Finally, let us put It can be shown [2] that the function is the measure of noncompactness in the algebra . The kernel ker of this measure contains all sets such that functions belonging to are locally equicontinuous on and have finite limits at infinity. Moreover, all functions from the set tend to their limits with the same rate. It can be also shown that the measure of noncompactness satisfies condition () [2].

In our considerations we will also use another measure of noncompactness which is defined below.

In order to define this measure, similarly as above, fix a set and a number . Denote by the cross-section of the set at the point . Denote by diam the diameter of . Further, for a fixed and denote by the so-called modulus of decrease of the function on the interval , which is defined by the formula Next, let us put In a similar way we may define the modulus of increase of function and the set (cf. [2]).

Finally, let us define the set quantity in the following way: Linking the facts established in [2, 15] it can be shown that is the measure of noncompactness in the algebra . The kernel ker of this measure consists of all sets such that functions belonging to are locally equicontinuous on and the thickness of the bundle formed by functions from tends to zero at infinity. Moreover, all functions from are ultimately nondecreasing on (cf. [16] for details).

Now, we show that the measure has also an additional property.

Theorem 6. The measure of noncompactness defined by (10) satisfies condition on the family of all nonempty and bounded subsets of Banach algebra such that functions belonging to are nonnegative on .

Proof. Observe that it is enough to show the second and third terms of the quantity defined by (10) satisfy condition . This assertion is a consequence of the fact that the term of the quantity defined by (10) satisfies condition () (cf. [2]).
Thus, take sets and numbers such that . Moreover, assume that functions belonging to and are nonnegative on the interval . Then, fixing arbitrarily and , we obtain Consequently, we get
Next, for arbitrary , , and we have This estimate yields Hence, in view of the fact that the set quantity satisfies condition (), we complete the proof.

It is worthwhile mentioning that the measure of noncompactness defined by formula (10) allows us to characterize solutions of considered operator equations in terms of the concept of asymptotic stability.

To formulate precisely that concept (cf. [17]) assume that is a nonempty subset of the Banach algebra and is an operator. Consider the operator equation where .

Definition 7. One says that solutions of (15) are asymptotically stable if there exists a ball in such that , and for each there exists such that for all solutions of (15) and for .

Let us pay attention to the fact that if solutions of an operator equation considered in the algebra belong to a bounded subset being a member of the family ker, then from the above given description of the kernel ker we infer that those solutions are asymptotically stable in the sense of Definition 7 (cf. also Remark 5).

4. Existence of Asymptotically Stable and Ultimately Nondecreasing Solutions of a Functional Integral Equation in the Banach Algebra ()

This section is devoted to the study of solvability of a functional integral equation in the Banach algebra . Apart from the existence of solutions of the equation in question we obtain also some characterization of those solutions expressed in terms of asymptotic stability and ultimate monotonicity. Characterizations of such type are possible due to the technique of measures of noncompactness. Obviously, we will apply measures of noncompactness described in the preceding section.

In our considerations we will often use the so-called superposition operator. In order to define that operator assume that is an interval and is a given function. Then, to every function we may assign the function defined by the formula for . The operator defined in such a way is called the superposition operator generated by the function (cf. [18]).

Lemma 8 [16] presents a useful property of the superposition operator which is considered in the Banach space consisting of all real functions defined and bounded on . Obviously, the space is equipped with the classical supremum norm. Since has the structure of a Banach algebra, we can consider the Banach algebra as a subalgebra of .

Lemma 8. Assume that the following hypotheses are satisfied. The function is continuous on the set . The function is ultimately nondecreasing uniformly with respect to belonging to bounded subintervals of , that is, for any bounded subinterval . For any fixed the function is nondecreasing on . The function satisfies a Lipschitz condition; that is, there exists a constant such that for all and all .
Then the inequality holds for any function , where is the Lipschitz constant from assumption .

Observe that in view of the remark mentioned previously Lemma 8 is also valid in the Banach algebra .

As we announced before, in this section we will study the solvability of the following integral equation: where are the so-called quadratic Volterra-Hammerstein integral operators defined as follows Now, we formulate the assumptions under which we will study (20).(i) and is ultimately nondecreasing; that is, . Moreover, for .(ii) and satisfies assumptions of Lemma 8 for .(iii) The function satisfies the Lipschitz condition with respect to the second variable; that is, there exists a constant such that for and for .(iv) The function is continuous and satisfies the following condition: (v) The function is bounded on and (vi) The function is continuous for .(vii) The functions and are bounded on .(viii) There exists a continuous and nondecreasing function such that and for and .In view of the above assumptions, we may define the following finite constants: for . Using these quantities we formulate the last assumption.(ix) There existS a solution of the inequality such that where , , ,

Now we formulate the main existence result for the functional integral equation (20).

Theorem 9. Under assumptions (i)–(ix) (20) has at least one solution in the space . Moreover all solutions of (20) are nonnegative, asymptotically stable, and ultimately nondecreasing.

Proof. Consider the subset of the algebra consisting of all functions being nonnegative on . We will consider operators on the set .
Now, fix an arbitrary function . Then, from assumptions (i), (ii), and (vi) we deduce that the function is continuous on . Moreover, in view of assumptions (i), (ii), (iv), and (vi) we have that the function is nonnegative on the interval for .
Further, for arbitrarily fixed we obtain Hence we get for .
This shows that the function is bounded on , and, consequently, is a member of the set . Consequently we infer that the operator being the product of the operators and transforms also the set into itself.
Further, taking into account estimate (31) and assumption (ix) we conclude that the operator is a self-mapping of the set defined in the following way: where is the number from assumption (ix).
In the sequel we will work with the measure of noncompactness defined by formula (10).
At the beginning, let us fix nonempty set and numbers . Additionally, let and be such that . Without loss of generality we may assume that . Then we obtain where we denoted In view of the uniform continuity of the function on the set we infer that as . Analogously if since the function is uniformly continuous on the set . Hence, in view of estimate (33) we obtain the following evaluation:
Similarly as above, for and we get Assumption (v) implies that the right side of the above estimate vanishes at infinity. Hence we obtain
This fact helps us prove that is continuous on the set . Indeed, let us fix and take such that . In view of (37) we know that there exists a number such that for arbitrary we get . Then, if we take we obtain the following estimate: Observe that continuity of the function yields that for each the expression on the right hand side of estimate (38) can be sufficiently small.
In what follows let us fix and . Then, if , we derive the following estimates: Observe that in virtue of imposed assumptions we have Linking the above estimate and (39), we obtain where we denoted In view of assumptions (i), (iv) and Lemma 8, if tends to infinity, then the above obtained estimate allows us to deduce the following inequality: Now, linking (35), (37) and (43) we obtain for .
Hence, if we use Theorem 4, we obtain that the operator is a contraction with respect to the measure of noncompactness and fulfills the Darbo condition with the below indicated constant Taking into account the second part of assumption (ix) we have additionally that .
Finally, invoking Theorem 4 we deduce that the operator has at least one fixed point in the set . Obviously, the function is a solution of (20). From Remark 5 we conclude that is asymptotically stable and ultimately nondecreasing. Obviously, is nonnegative on .
The proof is complete.

5. The Existence of Solutions of a Quadratic Fractional Integral Equation in the Banach Algebra

In this section we will investigate the existence of solutions of the quadratic fractional integral equation having the form where for and . Here we assume that is a fixed number for .

Our investigations will be conducted, similarly as previously, in the Banach algebra .

For further purposes we define a few operators on the space by putting for . Obviously, we have for and for .

Now, we are going to formulate assumptions imposed on functions involved in (46).(i)The function is nonnegative, bounded, continuous, and ultimately nondecreasing ().(ii)The function satisfies the conditions of Lemma 8 for .(iii)The functions satisfy the Lipschitz condition with respect to the second variable; that is, there exists a constant such that for, .(iv) is a continuous function such that ().(v)There exist a continuous and nondecreasing function and a bounded and continuous function such that for , , and .(vi)The function is bounded on and for .(vii)The function () satisfies the following condition (): In view of the above assumptions we may define the following constants (): The last assumption has the form 0. (viii)There exists a solution of the inequality such that

Now we are prepared to formulate and prove the existence result concerning the functional integral equation (46).

Theorem 10. Under assumptions (i)–(viii) (46) has at least one solution in the space which is nonnegative, asymptotically stable, and ultimately nondecreasing.

Proof. Similarly as in the proof of Theorem 9 let us consider the subset of the Banach algebra which consists of all functions being nonnegative on . Further, choose arbitrary function . Then, applying assumptions (i), (ii), (iv), and (v) we infer that the function is nonnegative on ().
Next, for , in view of (47) and the imposed assumptions, we get for .
This estimate yields that the function is bounded on ().
Furthermore, let us observe that in view of the properties of the superposition operator [18] and assumption (ii) we derive that the function is continuous on (). Thus, in order to show that is continuous on the interval , it is sufficient to show that the function is continuous on .
To this end fix and . Next, choose arbitrarily such that . Without loss of generality we may assume that . Then we have where we denoted
From the above estimate and the fact that function is uniformly continuous on the set we infer that the function is continuous on the real half axis .
Gathering the above established facts and estimate (57) we conclude that the operator () transforms the set into itself.
Apart from this, in view of (56) and assumption (vii) we infer that there exists a number such that the operator transforms into itself the set defined in the following way: Moreover, the following inequality is satisfied:
In the sequel we will work with the measure of noncompactness . Thus, let us fix a nonempty subset of the set and choose arbitrary numbers and . Then, for and for such that and we have In the similar way we obtain the estimate where we denoted Moreover, we derive the following evaluations: which hold for an arbitrary .
Further, linking (61), (62) with the above obtained evaluation, we arrive at the following estimate: Observe that the terms and tend to zero as since the functions and are uniformly continuous on the set and , respectively. Hence we obtain and consequently
In what follows, let us choose arbitrarily and . Then, based on our assumptions, we obtain Hence, keeping in mind assumption (vi), we derive the following equality:
Now, we show that is continuous on the set . To this end fix and take such that . In view of (69) we know that we may find a number such that for arbitrary we get . On the other hand, if we take , we derive the following estimate: where we denoted In view of the uniform continuity of the function on the set we have that as . This yields that we can find such that last term in the above estimate is sufficiently small for and .
Next, fix arbitrarily , and choose such that . Then, for an arbitrary we obtain On the other hand we get Now, taking into account assumptions (i), (v), and (vii) and estimate (72), we obtain for . Hence, in view of Lemma 8, we derive the following inequality: (). Further, combining the above inequality and (62), (67), and (72), we obtain for .
Next, applying Theorem 4 we derive that the operator is a contraction with respect to the measure of noncompactness with the constant given by the formula Observe that assumption (viii) implies that . Thus, in view of Theorem 4 we infer that the operator has at least one fixed point belonging to the set . Moreover, in view of Remark 5 we conclude that is nonnegative on , asymptotically stable, and ultimately nondecreasing.
This completes the proof.

Now we provide an example illustrating Theorem 10.

Example 11. Consider the quadratic fractional integral equation having form of (46) with the operators , defined by the following formulas for .

Observe that in this case the functions involved in (46) have the form Moreover, , .

It is easy to check that for the above functions there are satisfied assumptions of Theorem 10. Indeed, we have that the function is nonnegative, bounded, and continuous on (). Since and are increasing on we derive that they are also ultimately nondecreasing on . Moreover, and . Thus, there is satisfied assumption (i). Further notice that the functions ) transform continuously the set into . Moreover, is nondecreasing with respect to both variables and satisfies the Lipschitz condition (with respect to the second variable) with the constant . Similarly, the function is increasing with respect to and satisfies the Lipschitz condition with the constant . Apart from this it is easily seen that , .

Summing up, we see that functions and satisfy assumptions (ii) and (iii).

Next, let us note that the function is continuous on the set and transforms the set into for . Apart from this the function can be represented in the form (), where , and . It is easily seen that assumptions (iv) and (v) are satisfied for the functions and .

Further on, we have Hence we see that Moreover, we get Thus, we have This shows that assumption (vi) is satisfied.

In order to show that the function satisfies assumption (vii) let us fix arbitrarily . Then, for we obtain In the similar way, we get In view of the above obtained estimates we conclude that assumption (vii) is also satisfied.

Finally, let us notice that taking into account the above established facts we have that , , and . Thus, the first inequality from assumption (viii) has the form It can be shown that the number is a solution of the above inequality such that it satisfies also the second inequality from assumption (viii).

Applying Theorem 10 we infer that the quadratic fractional integral equation considered in this example has a solution belonging to the set which is asymptotically stable and ultimately nondecreasing.

6. The Existence of Solutions Having Limits at Infinity of an Integral Equation of Mixed Type in the Banach Algebra ()

In this final section we are going to investigate the following integral equation of mixed type: where denotes the nonlinear Volterra integral operator of the form while is the Urysohn integral operator having the form Equation (88) will be considered in the Banach algebra . The existence result concerning (88), which we intend to present here, creates an extension of a result obtained in [2].

In our considerations we will use the measure of noncompactness defined in Section 3. The use of that measure enables us to obtain a result on the existence of solutions of (88) having (finite) limits at infinity.

In what follows we will study (88) under the below formulated assumptions.(i) and as .(ii) is continuous and such that as , for .(iii)The functions satisfy the Lipschitz condition with respect to the second variable; that is, there exists a constant such that for and for .(iv) is continuous and there exist a continuous, function and a continuous and nondecreasing function such that for all and .(v) is continuous and there exist a continuous, function and a continuous and nondecreasing function such that for all and .(vi)The function is bounded on .(vii)For each the function is integrable on and the function is bounded on .(viii)The improper integral is uniformly convergent with respect to ; that is, (ix)There exists a positive solution of the inequality such that where the constants involved in the above inequalities are defined as follows: It is worthwhile mentioning that a result obtained in the paper [2] asserts that under assumptions (i)–(ix) (88) has at least one solution in the Banach algebra such that as . We generalize that result showing the existence of solutions of (88) which have finite limits at infinity. To this end we will need the following additional hypotheses. (x)The following conditions hold for .(xi) is bounded function, and for each there exists a finite limit .

Remark 12. Observe that assuming additionally hypothesis (xi) we can dispense with a certain part of assumption (ii).

Now, we can formulate the main result of this section.

Theorem 13. Under assumptions (i)–(xi) (88) has at least one solution belonging to the Banach algebra and such that there exists a finite limit .

Proof. Observe that based on the paper [2] we infer that operators and defined earlier transform the Banach algebra into itself. Moreover, recalling [2] again, we have In virtue of the above estimates and assumption (ix) we deduce that there exists a number such that the operator maps the ball into itself. Moreover, from (99) we derive the following estimates: Further, let us take an arbitrary nonempty subset of the ball . Then, using some estimates proved in [2] we have Now, let us fix and take arbitrary numbers . Then we obtain Hence, we derive the following inequality: Combining the above inequality with assumptions (i), (vi), (x), and (xi), we obtain Similarly we can show, based on assumptions (i), (ii), (v), (vii), (viii), (x), and (xi), that the following inequality holds: Further, let us observe that from (101) and (105) we derive In the same way, linking (102) and (106), we get
Finally, taking into account estimates (100), (107), and (108), assumption (ix), and Theorem 4 we deduce that operator has at least one fixed point in the ball being a subset of the Banach algebra . It is clear that the function is a solution of (88). Moreover, from Remark 5 and the description of the kernel ker we infer that has a finite limit at infinity.
The proof is complete.