# The Technique of Measures of Noncompactness in Banach Algebras and Its Applications to Integral Equations

Authors: Józef Banaś; Szymon Dudek
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 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. [1–6]). 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. [7–13], 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 (m) 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 (m).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 byℝ the set of real numbers and put ℝ+=[0,∞). If E is a given real Banach space with the norm ∥·∥ and the zero element θ, then by B(x,r) we denote the closed ball centered at x and with radius r. We will write Br to denote the ball B(θ,r). If X is a subset of E, then the symbols X¯ and ConvX stand for the closure and convex closure of X, respectively. Apart from this the symbol diamX will denote the diameter of a bounded set X while ∥X∥ denotes the norm of X;thatis,∥X∥=sup{∥x∥:x∈X}.Next, let us denote by𝔐E the family of all nonempty and bounded subsets of E and by 𝔑E 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 mappingμ:𝔐E→ℝ+ will be called a measure of noncompactness in E if it satisfies the following conditions.(1°)
The family kerμ={X∈𝔐E:μ(X)=0} is nonempty and kerμ⊂𝔑E.(2°)
X⊂Y⇒μ(X)≤μ(Y).(3°)
μ(X¯)=μ(ConvX)=μ(X).(4°)
μ(λX+(1-λ)Y)≤λμ(X)+(1-λ)μ(Y) for λ∈[0,1].(5°)
If(Xn) is a sequence of closed sets from 𝔐E such that Xn+1⊂Xn for n=1,2,… and if  limn→∞μ(Xn)=0, then the set X∞=⋂n=1∞‍Xn is nonempty.The family kerμ described in (1°) is said to be the kernel of the measure of noncompactnessμ.Observe that the setX∞ from the axiom (5°) is a member of the family kerμ. Indeed, from the inequality μ(X∞)≤μ(Xn) being satisfied for all n=1,2,… we derive that μ(X∞)=0 which means that X∞∈kerμ. This fact will play a key role in our further considerations.In the sequel we will usually assume that the spaceE has the structure of Banach algebra. In such a case we write xy in order to denote the product of elements x,y∈E. Similarly, we will write XY to denote the product of subsets X,Y of E;thatis,XY={xy:x∈X,y∈Y}.Now, we recall a useful concept introduced in [2].Definition 2.
One says that the measure of noncompactnessμ defined on the Banach algebra E satisfies condition (m) if for arbitrary sets X,Y∈𝔐E the following inequality is satisfied:
(1)μ(XY)≤∥X∥μ(Y)+∥Y∥μ(X).
It turns out that the above defined condition (m) 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.
LetΩ be a nonempty subset of a Banach space E, and let F:Ω→E be a continuous operator which transforms bounded subsets of Ω onto bounded ones. One says that F satisfies the Darbo condition with a constant k with respect to a measure of noncompactness μ if μ(FX)≤kμ(X) for each X∈𝔐E such that X⊂Ω. If k<1, then F is called a contraction with respect to μ.Now, assume thatE is a Banach algebra and μ is a measure of noncompactness on E satisfying condition (m). Then we have the following theorem announced above [2].Theorem 4.
Assume thatΩ is nonempty, bounded, closed, and convex subset of the Banach algebra E, and the operators P and T transform continuously the set Ω into E in such a way that P(Ω) and T(Ω) are bounded. Moreover, one assumes that the operator F=P·T transforms Ω into itself. If the operators P and T satisfy on the set Ω the Darbo condition with respect to the measure of noncompactness μ with the constants k1 and k2, respectively, then the operator F satisfies on Ω the Darbo condition with the constant
(2)∥P(Ω)∥k2+∥T(Ω)∥k1.
Particularly, if ∥P(Ω)∥k2+∥T(Ω)∥k1<1, then F is a contraction with respect to the measure of noncompactness μ and has at least one fixed point in the set Ω.Remark 5.
It can be shown [14] that the set Fix F of all fixed points of the operator F on the set Ω is a member of the kernel kerμ.


## 3. Some Measures of Noncompactness in the Banach AlgebraBC(ℝ+)

In this section we present some measures of noncompactness in the Banach algebraBC(ℝ+) consisting of all real functions defined, continuous, and bounded on the half axis R+. The algebra BC(ℝ+) is endowed with the usual supremum norm
(3)∥x∥=sup{|x(t)|:t∈ℝ+}
for x∈BC(ℝ+). Obviously, the multiplication in BC(ℝ+) 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 thatX is an arbitrarily fixed nonempty and bounded subset of the Banach algebra BC(ℝ+);thatis,X∈𝔐BC(ℝ+). Choose arbitrarily ε>0 and T>0. For x∈X denote by ωT(x,ε)the modulus of continuity of the function x on the interval [0,T]; that is,
(4)ωT(x,ε)=sup{|x(t)-x(s)|:t,s∈[0,T],|t-s|≤ε}.
Next, let us put
(5)ωT(X,ε)=sup{ωT(x,ε):x∈X},ω0T(X)=limε→0ωT(X,ε),ω0∞(X)=limT→∞ω0T(X).
Further, define the set quantity a(X) by putting
(6)a(X)=limT→∞{supx∈X{sup{|x(t)-x(s)|:t,s≥T}}}.
Finally, let us put
(7)μa(X)=ω0∞(X)+a(X).
It can be shown [2] that the function μa is the measure of noncompactness in the algebra BC(ℝ+). The kernel kerμa of this measure contains all sets X∈𝔐BC(ℝ+) such that functions belonging to X are locally equicontinuous on ℝ+ and have finite limits at infinity. Moreover, all functions from the set X tend to their limits with the same rate. It can be also shown that the measure of noncompactness μa satisfies condition (m) [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 setX∈𝔐BC(ℝ+) and a number t∈ℝ+. Denote by X(t) the cross-section of the set X at the point t;thatis,X(t)={x(t):x∈X}. Denote by diamX(t) the diameter of X(t). Further, for a fixed T>0 and x∈X denote by dT(x) the so-called modulus of decrease of the function x on the interval [T,∞), which is defined by the formula
(8)dT(x)=sup{|x(t)-x(s)|-[x(t)-x(s)]:T≤s<t}.
Next, let us put
(9)dT(X)=sup{dT(x):x∈X},d∞(X)=limT→∞dT(X).
In a similar way we may define the modulus of increase of function x and the set X (cf.  [2]).Finally, let us define the set quantityμd in the following way:
(10)μd(X)=ω0∞(X)+d∞(X)+limsupt→∞diamX(t).
Linking the facts established in [2, 15] it can be shown that μd is the measure of noncompactness in the algebra BC(ℝ+). The kernel kerμd of this measure consists of all sets X∈𝔐BC(ℝ+) such that functions belonging to X are locally equicontinuous on ℝ+ and the thickness of the bundle X(t) formed by functions from X tends to zero at infinity. Moreover, all functions from X are ultimately nondecreasing on ℝ+ (cf. [16] for details).Now, we show that the measureμd has also an additional property.Theorem 6.
The measure of noncompactnessμd defined by (10) satisfies condition (m) on the family of all nonempty and bounded subsets X of Banach algebra BC(ℝ+) such that functions belonging to X are nonnegative on ℝ+.Proof.
Observe that it is enough to show the second and third terms of the quantityμd defined by (10) satisfy condition (m). This assertion is a consequence of the fact that the term ω0∞ of the quantity defined by (10) satisfies condition (m) (cf. [2]).
Thus, take setsX,Y∈𝔐BC(ℝ+) and numbers T,s,t∈ℝ+ such that T≤s<t. Moreover, assume that functions belonging to X and Y are nonnegative on the interval ℝ+. Then, fixing arbitrarily x∈X and y∈Y, we obtain
(11)|x(t)y(t)-x(s)y(s)|-[x(t)y(t)-x(s)y(s)]≤|x(t)||y(t)-y(s)|+|y(s)||x(t)-x(s)|-{x(t)|y(t)-y(s)|+y(s)|x(t)-x(s)|}=|x(t)||y(t)-y(s)|+|y(s)||x(t)-x(s)|-{|x(t)||y(t)-y(s)|+|y(s)||x(t)-x(s)|}≤∥x∥·{|y(t)-y(s)|-[y(t)-y(s)]}+∥y∥·{|x(t)-x(s)|-[x(t)-x(s)]}.
Consequently, we get
(12)d∞(XY)≤∥X∥d∞(Y)+∥Y∥d∞(X).
Next, for arbitraryx1,x2∈X, y1,y2∈Y, and t∈ℝ+ we have
(13)|x1(t)y1(t)-x2(t)y2(t)|≤|x1(t)||y1(t)-y2(t)|+|y2(t)||x1(t)-x2(t)|≤∥x1∥·|y1(t)-y2(t)|+∥y2∥·|x1(t)-x2(t)|.
This estimate yields
(14)limsupt→∞diam(XY)(t)≤∥X∥limsupt→∞diamY(t)+∥Y∥limsupt→∞diamX(t).
Hence, in view of the fact that the set quantity ω0∞ satisfies condition (m), we complete the proof.It is worthwhile mentioning that the measure of noncompactnessμd 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 Ω is a nonempty subset of the Banach algebra BC(ℝ+) and F:Ω→BC(ℝ+) is an operator. Consider the operator equation
(15)x(t)=(Fx)(t),t∈ℝ+,
where x∈Ω.Definition 7.
One says that solutions of (15) are asymptotically stable if there exists a ball B(x0,r) in BC(ℝ+) such that B(x0,r)∩Ω≠ϕ, and for each ε>0 there exists T>0 such that |x(t)-y(t)|≤ε for all solutions x,y∈B(x0,r)∩Ω of (15) and for t≥T.Let us pay attention to the fact that if solutions of an operator equation considered in the algebraBC(ℝ+) belong to a bounded subset being a member of the family kerμd, then from the above given description of the kernel kerμd 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 AlgebraBC(ℝ+)

This section is devoted to the study of solvability of a functional integral equation in the Banach algebraBC(ℝ+). 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-calledsuperposition operator. In order to define that operator assume that J is an interval and f:ℝ+×J→ℝ is a given function. Then, to every function x:ℝ+→J we may assign the function Fx defined by the formula
(16)(Fx)(t)=f(t,x(t)),
for t∈ℝ+. The operator F defined in such a way is called the superposition operator generated by the function f(t,x) (cf. [18]).Lemma8 [16] presents a useful property of the superposition operator which is considered in the Banach space B(ℝ+) consisting of all real functions defined and bounded on ℝ+. Obviously, the space B(ℝ+) is equipped with the classical supremum norm. Since B(ℝ+) has the structure of a Banach algebra, we can consider the Banach algebra BC(ℝ+) as a subalgebra of B(ℝ+).Lemma 8.
Assume that the following hypotheses are satisfied.(α)
The functionf is continuous on the set ℝ+×J.(β)
The functiont↦f(t,u) is ultimately nondecreasing uniformly with respect to u belonging to bounded subintervals of J, that is,
(17)limT→∞{sup{|f(t,u)-f(s,u)|-[f(t,u)-f(s,u)]:t>s≥T,u∈J1}}=0for any bounded subintervalJ1⊆J.(γ)
For any fixedt∈ℝ+ the function u↦f(t,u) is nondecreasing on J.(δ)
The functionu↦f(t,u) satisfies a Lipschitz condition; that is, there exists a constant k>0 such that
(18)|f(t,u)-f(t,v)|≤k|u-v|for allt≤0 and all u,v∈J.
Then the inequality(19)d∞(Fx)≤kd∞(x)
holds for any function x∈B(ℝ+), where k is the Lipschitz constant from assumption (δ).Observe that in view of the remark mentioned previously Lemma8 is also valid in the Banach algebra BC(ℝ+).As we announced before, in this section we will study the solvability of the following integral equation:(20)x(t)=(V1x)(t)(V2x)(t),t∈ℝ+,
where Vi are the so-called quadratic Volterra-Hammerstein integral operators defined as follows
(21)(Vix)(t)=pi(t)+fi(t,x(t))×∫0t‍gi(t,s)hi(s,x(s))ds,(i=1,2).
Now, we formulate the assumptions under which we will study (20).(i)
pi∈BC(ℝ+) and pi is ultimately nondecreasing; that is, d∞(pi)=0(i=1,2). Moreover, pi(t)≥0 for t∈ℝ+(i=1,2).(ii)
fi:ℝ+×ℝ+→ℝ+ and fi satisfies assumptions of Lemma 8 for J=ℝ+(i=1,2).(iii)
The functionfi satisfies the Lipschitz condition with respect to the second variable; that is, there exists a constant ki>0 such that
(22)|fi(t,x)-fi(t,y)|≤ki|x-y|forx,y∈ℝ+ and for t∈ℝ+(i=1,2).(iv)
The functiongi:ℝ+×ℝ+→ℝ+ is continuous and satisfies the following condition:
(23)limT→∞{sup{∫0t‍{|gi(t,τ)-gi(s,τ)|-[gi(t,τ)-gi(s,τ)]}dτ:T≤s<t{∫0t‍}}}=0,(i=1,2).(v)
The functiont→∫0t‍gi(t,s)ds is bounded on ℝ+ and
(24)limt→∞∫0t‍gi(t,s)ds=0(i=1,2).(vi)
The functionhi:ℝ+×ℝ+→ℝ+ is continuous for i=1,2.(vii)
The functionst→fi(t,0) and t→hi(t,0) are bounded on ℝ+(i=1,2).(viii)
There exists a continuous and nondecreasing functionmi:ℝ+→ℝ+ such that mi(0)=0 and
(25)|hi(t,x)-hi(t,y)|≤mi(|x-y|)forx,y,andt∈ℝ+ and i=1,2.In view of the above assumptions, we may define the following finite constants:(26)F¯i=sup{fi(t,0):t∈ℝ+},H¯i=sup{hi(t,0):t∈ℝ+},G¯i=sup{∫0t‍gi(t,s)ds:t∈ℝ+},fori=1,2.Using these quantities we formulate the last assumption.(ix)
There existS a solutionr0>0 of the inequality
(27)[p+krG¯m1(r)+krG¯H¯+F¯G¯m1(r)+F¯G¯H¯]×[p+krG¯m2(r)+krG¯H¯+F¯G¯m2(r)+F¯G¯H¯]≤rsuch that(28)pkG¯[m1(r0)+m2(r0)+H¯]+2kG¯2[kr0+F¯]×[m1(r0)+H¯][m2(r0)+H¯]<1,wherep=max{∥p1∥,∥p2∥}, F¯=max{F¯1,F¯2}, G¯=max{G¯1,G¯2},
(29)H¯=max{H¯1,H¯2},k=max{k1,k2}.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 x=x(t) in the space BC(ℝ+). Moreover all solutions of (20) are nonnegative, asymptotically stable, and ultimately nondecreasing.Proof.
Consider the subsetΩ of the algebra BC(ℝ+) consisting of all functions being nonnegative on ℝ+. We will consider operators Vi(i=1,2) on the set Ω.
Now, fix an arbitrary functionx∈Ω. Then, from assumptions (i), (ii), and (vi) we deduce that the function Vix is continuous on ℝ+. Moreover, in view of assumptions (i), (ii), (iv), and (vi) we have that the function Vix is nonnegative on the interval ℝ+ for i=1,2.
Further, for arbitrarily fixedt∈ℝ+ we obtain
(30)(Vix)(t)≤pi(t)+fi(t,x(t))∫0t‍gi(t,τ)hi(τ,x(τ))dτ≤pi(t)+[kix(t)+fi(t,0)]×∫0t‍gi(t,τ)[mi(x(τ))+hi(τ,0)]dτ≤∥pi∥+[ki∥x∥+F¯i][mi(∥x∥)+H¯i]G¯i.
Hence we get
(31)∥Vix∥≤∥pi∥+ki∥x∥G¯imi(∥x∥)+ki∥x∥G¯iH¯i+F¯iG¯imi(∥x∥)+F¯iG¯iH¯i,
for i=1,2.
This shows that the functionVix(i=1,2) is bounded on ℝ+, and, consequently, Vix is a member of the set Ω. Consequently we infer that the operator W being the product of the operators V1 and V2 transforms also the set Ω into itself.
Further, taking into account estimate (31) and assumption (ix) we conclude that the operator W is a self-mapping of the set Ωr0 defined in the following way:
(32)Ωr0={x∈BC(ℝ+):0≤x(t)≤r0fort∈ℝ+},
where r0 is the number from assumption (ix).
In the sequel we will work with the measure of noncompactnessμd defined by formula (10).
At the beginning, let us fix nonempty setX⊂Ωr0 and numbers T>0,ε>0. Additionally, let x∈X and t,s∈[0,T] be such that |t-s|≤ε. Without loss of generality we may assume that t<s. Then we obtain
(33)|(Vix)(t)-(Vix)(s)|≤|pi(t)-pi(s)|+|fi(t,x(t))-fi(s,x(s))|×∫0t‍gi(t,τ)hi(τ,x(τ))dτ+|fi(s,x(s))||∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0s‍gi(s,τ),hi(τ,x(τ))dτ|≤|pi(t)-pi(s)|+[|fi(t,x(t))-fi(t,x(s))|+|fi(t,x(s))-fi(s,x(s))|]×∫0t‍gi(t,τ)[hi(τ,x(τ))-hi(τ,0)+hi(τ,0)]dτ+[|fi(s,x(s))-fi(s,0)|+|fi(s,0)|]×∫0t‍|gi(t,τ)-gi(s,τ)|hi(τ,x(τ))dτ≤ωT(pi,ε)+[ki|x(t)-x(s)|+ωr0T(fi,ε)]×∫0t‍gi(t,τ)[mi(|x(τ)|)+H¯i]dτ+[ki|x(s)|+F¯i]∫0T‍ω1T(gi,ε)[mi(|x(τ)|)+H¯i]dτ≤ωT(pi,ε)+[kiωT(x,ε)+ωr0T(fi,ε)][mi(r0)+H¯i]G¯i+[kir0+F¯i]Tω1T(gi,ε)[mi(r0)+H¯i],
where we denoted
(34)ωdT(fi,ε)=sup{|fi(t,y)-fi(s,y)|:t,s∈[0,T],y∈[-d,d],|t-s|≤ε},ω1T(gi,ε)=sup{|gi(t,τ)-gi(s,τ)|:t,s,τ∈[0,T],|t-s|≤ε}.
In view of the uniform continuity of the function fi on the set [0,T]×[-r0,r0] we infer that ωdT(fi,ε)→0 as ε→0. Analogously ω1T(gi,ε)→0 if ε→0 since the function gi is uniformly continuous on the set [0,T]×[0,T]. Hence, in view of estimate (33) we obtain the following evaluation:
(35)ω0∞(ViX)≤kiG¯i[mi(r0)+H¯i]ω0∞(X).
Similarly as above, fort∈ℝ+ and x,y∈X we get
(36)|(Vix)(t)-(Viy)(t)|≤|fi(t,x(t))-fi(t,y(t))|×∫0t‍gi(t,τ)hi(τ,x(τ))dτ+fi(t,y(t))×∫0t‍|gi(t,τ)hi(τ,x(τ))-gi(t,τ)hi(τ,y(τ))|dτ≤ki|x(t)-y(t)|[mi(r0)+H¯i]∫0t‍gi(t,τ)dτ+[ki|y(t)|+F¯i]∫0t‍gi(t,τ)mi(|x(τ)-y(τ)|)dτ≤ki|x(t)-y(t)|[mi(r0)+H¯i]×∫0t‍gi(t,τ)dτ+[kir0+F¯i]mi(r0)∫0t‍gi(t,τ)dτ.
Assumption (v) implies that the right side of the above estimate vanishes at infinity. Hence we obtain
(37)limsupt→∞diam(VX)(t)=0.
This fact helps us prove thatVi is continuous on the set Ωr0. Indeed, let us fix ε>0 and take x,y∈Ωr0 such that ∥x-y∥≤ε. In view of (37) we know that there exists a number T>0 such that for arbitrary t≥T we get |(Vix)(t)-(Viy)(t)|≤ε. Then, if we take t∈[0,T] we obtain the following estimate:
(38)|(Vix)(t)-(Viy)(t)|≤|fi(t,x(t))-fi(t,y(t))|∫0t‍gi(t,τ)hi(τ,x(τ))dτ+fi(t,y(t))∫0t‍gi(t,τ)|hi(τ,x(τ))-hi(τ,y(τ))|dτ≤ki|x(t)-y(t)|[mi(∥x∥)+H¯i]∫0t‍gi(t,τ)dτ+[ki|y(t)|+F¯i]∫0t‍gi(t,τ)mi(|x(τ)-y(τ)|)dτ≤kiε[mi(r0)+H¯i]G¯i+[kir0+F¯i]mi(ε)G¯i.
Observe that continuity of the function mi yields that for each T>0 the expression on the right hand side of estimate (38) can be sufficiently small.
In what follows let us fixT>0 and t>s≥T. Then, if x∈Ωr0, we derive the following estimates:
(39)|(Vix)(t)-(Vix)(s)|-[(Vix)(t)-(Vix)(s)]≤|pi(t)-pi(s)|+|fi(t,x(t))∫0t‍gi(t,τ)hi(τ,x(τ))dτ-fi(s,x(s))∫0s‍gi(s,τ)hi(τ,x(τ))dτ|-[pi(t)-pi(s)]-[fi(t,x(t))∫0t‍gi(t,τ)hi(τ,x(τ))dτ-fi(s,x(s))∫0s‍gi(s,τ)hi(τ,x(τ))dτ]≤dT(pi)+|fi(t,x(t))-fi(s,x(s))|×∫0t‍gi(t,τ)hi(τ,x(τ))dτ+fi(s,x(s))|∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0s‍gi(s,τ)hi(τ,x(τ))dτ|-[fi(t,x(t))-fi(s,x(s))]∫0t‍gi(t,τ)hi(τ,x(τ))dτ-fi(s,x(s))[∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0s‍gi(s,τ)hi(τ,x(τ))dτ]≤dT(pi)+{|fi(t,x(t))-fi(s,x(s))|-[fi(t,x(t))-fi(s,x(s))]}×[mi(∥x∥)+H¯i]∫0t‍gi(t,τ)dτ+fi(s,x(s)){|∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0s‍gi(s,τ)hi(τ,x(τ))dτ|-[∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0s‍gi(s,τ)hi(τ,x(τ))dτ]}.
Observe that in virtue of imposed assumptions we have
(40)|∫0t‍gi(t,τ)hi(τ,x(τ))-∫0s‍gi(s,τ)hi(τ,x(τ))dτ|-[∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0s‍gi(s,τ)hi(τ,x(τ))dτ]≤|∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0t‍gi(s,τ)hi(τ,x(τ))dτ|+∫st‍gi(s,τ)hi(τ,x(τ))dτ-[∫0t‍gi(t,τ)hi(τ,x(τ))dτ-∫0t‍gi(s,τ)hi(τ,x(τ))dτ]-∫st‍gi(s,τ)hi(τ,x(τ))dτ≤[mi(∥x∥)+H¯i]×∫0t‍{|gi(t,τ)-gi(s,τ)|-[gi(t,τ)-gi(s,τ)]}dτ.
Linking the above estimate and (39), we obtain
(41)sup{|(Vix)(t)-(Vix)(s)|-[(Vix)(t)-(Vix)(s)]:t>s≥T}≤dT(pi)+dT(fi)G¯i[mi(r0)+H¯i]+[kir0+F¯i][mi(r0)+H¯i]dTc(gi),
where we denoted
(42)dT(fi)=sup{|fi(t,x(t))-fi(s,x(s))|-[fi(t,x(t))-fi(s,x(s))]:t>s≥T},dTc(gi)=sup{∫0t‍{|gi(t,τ)-gi(s,τ)|-[gi(t,τ)-gi(s,τ)]}:t>s≥T{∫0t‍}}.
In view of assumptions (i), (iv) and Lemma 8, if T tends to infinity, then the above obtained estimate allows us to deduce the following inequality:
(43)d∞(ViX)≤kiG¯i[mi(r0)+H¯i]d∞(X).
Now, linking (35), (37) and (43) we obtain
(44)μd(ViX)≤kiG¯i[mi(r0)+H¯i]μd(X)
for i∈{1,2}.
Hence, if we use Theorem4, we obtain that the operator W=V1V2 is a contraction with respect to the measure of noncompactness μd and fulfills the Darbo condition with the below indicated constant
(45)L=pkG¯[m1(r0)+m2(r0)+H¯]+2kG¯2[kr0+F¯][m1(r0)+H¯][m2(r0)+H¯].
Taking into account the second part of assumption (ix) we have additionally that L<1.
Finally, invoking Theorem4 we deduce that the operator W has at least one fixed point x=x(t) in the set Ωr0. Obviously, the function x is a solution of (20). From Remark 5 we conclude that x is asymptotically stable and ultimately nondecreasing. Obviously, x is nonnegative on ℝ+.
The proof is complete.


## 5. The Existence of Solutions of a Quadratic Fractional Integral Equation in the Banach AlgebraBC(ℝ+)

In this section we will investigate the existence of solutions of the quadratic fractional integral equation having the form(46)x(t)=(U1x)(t)(U2x)(t),
where
(47)(Uix)(t)=mi(t)+fi(t,x(t))∫0t‍vi(t,s,x(s))(t-s)αids
for t∈ℝ+ and i=1,2. Here we assume that αi∈(0,1) is a fixed number for i=1,2.Our investigations will be conducted, similarly as previously, in the Banach algebraBC(ℝ+).For further purposes we define a few operators on the spaceBC(ℝ+) by putting
(48)(Fix)(t)=fi(t,x(t)),(Vix)(t)=∫0t‍vi(t,s,x(s))(t-s)αids
for i=1,2. Obviously, we have
(49)(Uix)(t)=mi(t)+(Fix)(t)(Vix)(t)
for i=1,2 and for t∈ℝ+.Now, we are going to formulate assumptions imposed on functions involved in (46).(i)
The functionmi is nonnegative, bounded, continuous, and ultimately nondecreasing (i=1,2).(ii)
The functionfi:ℝ+×ℝ+→ℝ+ satisfies the conditions (α)–(γ) of Lemma 8 for i=1,2.(iii)
The functionsfi(i=1,2) satisfy the Lipschitz condition with respect to the second variable; that is, there exists a constant ki>0 such that
(50)|fi(t,x)-fi(t,y)|≤ki|x-y|
for, t∈ℝ+(i=1,2).(iv)
vi:ℝ+×ℝ+×ℝ→ℝ is a continuous function such that vi:ℝ+×ℝ+×ℝ+→ℝ+ (i=1,2).(v)
There exist a continuous and nondecreasing functionGi:ℝ+→ℝ+ and a bounded and continuous function gi:ℝ+×ℝ+→ℝ+ such that vi(t,s,x)=gi(t,s)Gi(|x|) for t,s∈ℝ+, x∈ℝ, and i=1,2.(vi)
The functiont↦∫0t‍(gi(t,s)/(t-s)αi)ds is bounded on ℝ+ and
(51)limt→∞∫0t‍gi(t,s)(t-s)αids=0,
for i=1,2.(vii)
The functiongi (i=1,2) satisfies the following condition (i=1,2):
(52)limT→∞{sup{∫0t‍{|gi(t,τ)(t-τ)αi-gi(s,τ)(s-τ)αi|-[gi(t,τ)(t-τ)αi-gi(s,τ)(s-τ)αi]}dτ:T≤s<t{∫0t‍}}}=0.
In view of the above assumptions we may define the following constants (i=1,2):
(53)F¯i=sup{|fi(t,0)|:t∈ℝ+},G¯i=sup{∫0t‍gi(t,s)(t-s)αids:t∈ℝ+},g¯i=sup{gi(t,s):t,s∈ℝ+},F¯=max{F¯1,F¯2},k=max{k1,k2},m=max{∥m1∥,∥m2∥}.
The last assumption has the form 0.(viii)
There exists a solutionr0>0 of the inequality
(54)[m+kG¯1rG1(r)+F¯G¯1G1(r)]×[m+kG¯2rG2(r)+F¯G¯2G2(r)]≤r
such that
(55)mk(G¯1G1(r0)+G¯2G2(r0))+2kF¯G¯1G1(r0)G¯2G2(r0)+2k2r0G¯1G1(r0)G¯2G2(r0)<1.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 x=x(t) in the space BC(ℝ+) which is nonnegative, asymptotically stable, and ultimately nondecreasing.Proof.
Similarly as in the proof of Theorem9 let us consider the subset Ω of the Banach algebra BC(ℝ+) which consists of all functions being nonnegative on ℝ+. Further, choose arbitrary function x∈Ω. Then, applying assumptions (i), (ii), (iv), and (v) we infer that the function Uix is nonnegative on ℝ+ (i=1,2).
Next, fort∈ℝ+, in view of (47) and the imposed assumptions, we get
(56)(Uix)(t)≤mi(t)+[kix(t)+fi(t,0)]∫0t‍vi(t,s,x(s))(t-s)αids≤mi(t)+[kix(t)+fi(t,0)]Gi(∥x∥)∫0t‍gi(t,s)(t-s)αids≤∥mi∥+kiG¯i∥x∥Gi(∥x∥)+F¯G¯iGi(∥x∥),
for i=1,2.
This estimate yields that the functionUix is bounded on ℝ+ (i=1,2).
Furthermore, let us observe that in view of the properties of the superposition operator [18] and assumption (ii) we derive that the function Fix is continuous on ℝ+ (i=1,2). Thus, in order to show that Uix is continuous on the interval ℝ+, it is sufficient to show that the function Vix is continuous on ℝ+.
To this end fixT>0 and ε>0. Next, choose arbitrarily t,s∈[0,T] such that |t-s|≤ε. Without loss of generality we may assume that s<t. Then we have
(57)|(Vix)(t)-(Vix)(s)|≤|∫0t‍vi(t,τ,x(τ))(t-τ)αidτ-∫0t‍vi(s,τ,x(τ))(t-τ)αidτ|+|∫0t‍vi(s,τ,x(τ))(t-τ)αidτ-∫0s‍vi(s,τ,x(τ))(t-τ)αidτ|+|∫0s‍vi(s,τ,x(τ))(t-τ)αidτ-∫0s‍vi(s,τ,x(τ))(s-τ)αidτ|≤∫0t‍|vi(t,τ,x(τ))-vi(s,τ,x(τ))|(t-τ)αidτ+∫st‍vi(s,τ,x(τ))(t-τ)αidτ+∫0s‍vi(s,τ,x(τ))[1(s-τ)αi-1(t-τ)αi]dτ≤ω∥x∥T(vi,ε)∫0t‍1(t-τ)αidτ+Gi(∥x∥)g¯i∫st‍1(t-τ)αidτ+Gi(∥x∥)g¯i∫0s‍[1(s-τ)αi-1(t-τ)αi]dτ≤ω∥x∥T(vi,ε)t1-αi1-αi+Gi(∥x∥)g¯i(t-s)1-αi1-αi+Gi(∥x∥)g¯i[s1-αi1-αi-t1-αi1-αi+(t-s)1-αi1-αi]≤ω∥x∥T(vi,ε)T1-αi1-αi+2Gi(∥x∥)g¯iε1-αi1-αi,
where we denoted
(58)ωdT(vi,ε)=sup{|vi(t,τ,x)-vi(s,τ,x)|:t,s,τ∈[0,T],|t-s|≤ε,x∈[-d,d]}.
From the above estimate and the fact that functionvi is uniformly continuous on the set [0,T]×[0,T]×[-∥x∥,∥x∥] we infer that the function Vix is continuous on the real half axis ℝ+.
Gathering the above established facts and estimate (57) we conclude that the operator Ui (i=1,2) transforms the set Ω into itself.
Apart from this, in view of (56) and assumption (vii) we infer that there exists a number r0>0 such that the operator S=U1U2 transforms into itself the set Ωr0 defined in the following way:
(59)Ωr0={x∈BC(ℝ+):0≤x(t)≤r0fort∈ℝ+}.
Moreover, the following inequality is satisfied:
(60)∥UiΩr0∥≤m+kiG¯r0Gi(r0)+F¯G¯Gi(r0).
In the sequel we will work with the measure of noncompactnessμd. Thus, let us fix a nonempty subset X of the set Ωr0 and choose arbitrary numbers T>0 and ε>0. Then, for x∈X and for t,s∈[0,T] such that |t-s|≤ε and t≥s we have
(61)|(Uix)(t)-(Uix)(s)|≤ωT(mi,ε)+|(Fix)(t)(Vix)(t)-(Fix)(s)(Vix)(s)|≤ωT(mi,ε)+|(Fix)(t)(Vix)(t)-(Fix)(s)(Vix)(t)|+|(Fix)(s)(Vix)(t)-(Fix)(s)(Vix)(s)|≤ωT(mi,ε)+|(Fix)(t)-(Fix)(s)||(Vix)(t)|+|(Fix)(s)||(Vix)(t)-(Vix)(s)|.
In the similar way we obtain the estimate
(62)|(Fix)(t)-(Fix)(s)|≤|fi(t,x(t))-fi(t,x(s))|+|fi(t,x(s))-fi(s,x(s))|≤ki|x(t)-x(s)|+|fi(t,x(s))-fi(s,x(s))|≤kiωT(x,ε)+ω∥x∥T(fi,ε),
where we denoted
(63)ωdT(fi,ε)=sup{|fi(t,x)-fi(s,x)|:t,s∈[0,T],|t-s|≤ε,x∈[-d,d]}.
Moreover, we derive the following evaluations:
(64)|(Vix)(t)|≤Gi(∥x∥)∫0t‍gi(t,τ)(t-τ)αidτ≤Gi(∥x∥)G¯i,|(Fix)(s)|≤ki|x(s)|+|fi(s,0)|≤kir0+F¯i,
which hold for an arbitrary t,s∈ℝ+.
Further, linking (61), (62) with the above obtained evaluation, we arrive at the following estimate:
(65)|(Uix)(t)-(Uix)(s)|≤ωT(mi,ε)+[kiωT(x,ε)+ω∥x∥T(fi,ε)]Gi(∥x∥)G¯i+[kir0+F¯i][ω∥x∥T(ui,ε)T1-αi1-αi+2Gi(∥x∥)g¯iε1-αi1-αi].
Observe that the terms ω∥x∥T(fi,ε) and ω∥x∥T(ui,ε) tend to zero as ε→0 since the functions fi and ui are uniformly continuous on the set [0,T]×[-∥x∥,∥x∥] and [0,T]×[0,T]×[-∥x∥,∥x∥], respectively. Hence we obtain
(66)ω0T(UiX)≤kiG¯iGi(r0)ω0T(X)
and consequently
(67)ω0∞(UiX)≤kiG¯iGi(r0)ω0∞(X).
In what follows, let us choose arbitrarilyx,y∈X and t∈ℝ+. Then, based on our assumptions, we obtain
(68)|(Uix)(t)-(Uiy)(t)|≤|fi(t,x(t))-fi(t,y(t))|∫0t‍ui(t,τ,x(τ))(t-τ)αidτ+fi(t,y(t))∫0t‍|ui(t,τ,x(τ))-ui(t,τ,y(τ))|(t-τ)αidτ≤ki|x(t)-y(t)|Gi(∥x∥)∫0t‍gi(t,τ)(t-τ)αidτ+[kiy(t)+fi(t,0)]×∫0t‍gi(t,τ)[Gi(x(τ))-Gi(y(τ))](t-τ)αidτ≤ki|x(t)-y(t)|Gi(r0)∫0t‍gi(t,τ)(t-τ)αidτ+[kir0+F¯i]2Gi(r0)∫0t‍gi(t,τ)(t-τ)αidτ.
Hence, keeping in mind assumption (vi), we derive the following equality:
(69)limsupt→∞diam(UiX)(t)=0.
Now, we show thatUi is continuous on the set Ωr0. To this end fix ε>0 and take x,y∈Ωr0 such that ∥x-y∥≤ε. In view of (69) we know that we may find a number T>0 such that for arbitrary t≥T we get |(Uix)(t)-(Uiy)(t)|≤ε. On the other hand, if we take t∈[0,T], we derive the following estimate:
(70)|(Uix)(t)-(Uiy)(t)|≤|fi(t,x(t))-fi(t,y(t))|∫0t‍ui(t,τ,x(τ))(t-τ)αidτ+fi(t,x(t))∫0t‍|ui(t,τ,x(τ))-ui(t,τ,y(τ))|(t-τ)αidτ≤ki|x(t)-y(t)|Gi(∥x∥)∫0t‍gi(t,τ)(t-τ)αidτ+[ki|y(t)|+fi(t,0)]ωr0T(ui,ε)∫0t‍dτ(t-τ)αi≤kεG¯iGi(r0)+(kr0+F¯)T1-αi1-αiωr0T(ui,ε),
where we denoted
(71)ωdT(ui,ε)=sup{|ui(t,s,x)-ui(t,s,y)|:t,s∈[0,T],x,y∈[-d,d],|x-y|≤ε}.
In view of the uniform continuity of the function ui on the set [0,T]×[0,T]×[-r0,r0] we have that ωr0T(v,ε)→0 as ε→0. This yields that we can find T>0 such that last term in the above estimate is sufficiently small for t≥T and i=1,2.
Next, fix arbitrarilyT>0, and choose t,s such that t>s≥T. Then, for an arbitrary x∈X we obtain
(72)|(Uix)(t)-(Uix)(s)|-[(Uix)(t)-(Uix)(s)]≤|mi(t)-mi(s)|-[mi(t)-mi(s)]+|(Fix)(t)(Vix)(t)-(Fix)(s)(Vix)(t)|+|(Fix)(s)(Vix)(t)-(Fix)(s)(Vix)(s)|-[(Fix)(t)(Vix)(t)-(Fix)(s)(Vix)(t)]-[(Fix)(s)(Vix)(t)-(Fix)(s)(Vix)(s)]≤dT(mi)+dT(Fix)(Vix)(t)+(Fix)(s){|(Vix)(t)-(Vix)(s)|-[(Vix)(t)-(Vix)(s)]}.
On the other hand we get
(73)|(Vix)(t)-(Vix)(s)|-[(Vix)(t)-(Vix)(s)]≤|∫0t‍gi(t,τ)Gi(x(τ))(t-τ)αidτ-∫0t‍gi(s,τ)Gi(x(τ))(s-τ)αidτ|+|∫st‍gi(s,τ)Gi(x(τ))(s-τ)αidτ|-[∫0t‍gi(t,τ)Gi(x(τ))(t-τ)αidτ-∫0t‍gi(s,τ)Gi(x(τ))(s-τ)αidτ]-∫st‍gi(s,τ)Gi(x(τ))(s-τ)αidτ≤∫0t‍|gi(t,τ)Gi(x(τ))(t-τ)αidτ-gi(s,τ)Gi(x(τ))(s-τ)αi|dτ-∫0t‍[gi(t,τ)Gi(x(τ))(t-τ)αidτ-gi(s,τ)Gi(x(τ))(s-τ)αi]dτ≤Gi(∥x∥)∫0t‍{|gi(t,τ)(t-τ)αi-gi(s,τ)(s-τ)αi|-[gi(t,τ)(t-τ)αi-gi(s,τ)(s-τ)αi]}dτ.
Now, taking into account assumptions (i), (v), and (vii) and estimate (72), we obtain
(74)d∞(Uix)≤d∞(Fix)Gi(r0)G¯i,
for i=1,2. Hence, in view of Lemma 8, we derive the following inequality:
(75)d∞(Uix)≤kiG¯iGi(r0)d∞(x)
(i=1,2). Further, combining the above inequality and (62), (67), and (72), we obtain
(76)μd(UiX)≤kiG¯iGi(r0)μd(X)
for i=1,2.
Next, applying Theorem4 we derive that the operator S=U1U2 is a contraction with respect to the measure of noncompactness μd with the constant L given by the formula
(77)L=mk(G¯1G1(r0)+G¯2G2(r0))+2kF¯G¯1G1(r0)G¯2G2(r0)+2k2r0G¯1G1(r0)G¯2G2(r0).
Observe that assumption (viii) implies that L<1. Thus, in view of Theorem 4 we infer that the operator S has at least one fixed point x=x(t) belonging to the set Ωr0. Moreover, in view of Remark 5 we conclude that x is nonnegative on ℝ+, asymptotically stable, and ultimately nondecreasing.
This completes the proof.Now we provide an example illustrating Theorem10.Example 11.
Consider the quadratic fractional integral equation having form of (46) with the operators U1, U2 defined by the following formulas
(78)(U1x)(t)=t3t+1+arctan(t2+x(t))∫0t‍e-(t+s)|x(t)|(t-s)1/3ds,(U2x)(t)=1-e-t4+12ln(x(t)+1)×∫0t‍8x4(t)5(t+s+2)3(t-s)1/5ds
for t∈ℝ+.Observe that in this case the functions involved in (46) have the form
(79)m1(t)=t3t+1,m2(t)=1-e-t4,f1(t,x)=arctan(t2+x),f2(t,x)=12ln(x+1),v1(t,s,x)=e-(t+s)|x|,v2(t,s,x)=8x45(t+s+2)3.
Moreover, α1=1/3, α2=1/5.It is easy to check that for the above functions there are satisfied assumptions of Theorem10. Indeed, we have that the function mi=mi(t) is nonnegative, bounded, p and continuous on ℝ+ (i=1,2). Since m1 and m2 are increasing on ℝ+ we derive that they are also ultimately nondecreasing on ℝ+. Moreover, ∥m1∥=1/3 and ∥m2∥=1/4. Thus, there is satisfied assumption (i). Further notice that the functions fi(i=1,2) transform continuously the set ℝ+×ℝ+ into ℝ+. Moreover, f1 is nondecreasing with respect to both variables and satisfies the Lipschitz condition (with respect to the second variable) with the constant k1=1. Similarly, the function f2=f2(t,x) is increasing with respect to x and satisfies the Lipschitz condition with the constant k2=1/2. Apart from this it is easily seen that F¯1=π/2, F¯2=0.Summing up, we see that functionsf1 and f2 satisfy assumptions (ii) and (iii).Next, let us note that the functionvi(t,s,x) is continuous on the set ℝ+×ℝ+×ℝ and transforms the set ℝ+×ℝ+×ℝ+ into ℝ+ for i=1,2. Apart from this the function vi can be represented in the form vi(t,s,x)=gi(t,s)Gi(|x|) (i=1,2), where g1(t,s)=e-(t+s),G1(x)=|x|,g2(t,s)=8/5(t+s+2)3, and G2(x)=x4. It is easily seen that assumptions (iv) and (v) are satisfied for the functions v1 and v2.Further on, we have(80)∫0t‍g1(t,s)(t-s)α1ds=∫0t‍e-t-s(t-s)1/3ds≤e-t∫0t‍ds(t-s)1/3=32e-tt2/3.
Hence we see that
(81)limt→∞∫0t‍g1(t,s)(t-s)α1ds=0.
Moreover, we get
(82)∫0t‍g2(t,s)(t-s)α2ds=∫0t‍85(t+s+2)3(t-s)1/5ds≤85(t+2)3∫0t‍ds(t-s)1/5=2t4/5(t+2)3.
Thus, we have
(83)limt→∞∫0t‍g2(t,s)(t-s)α2ds=0.
This shows that assumption (vi) is satisfied.In order to show that the functiongi(t,s) satisfies assumption (vii) let us fix arbitrarily T>0. Then, for T≤s<t we obtain
(84)∫0t‍{|g1(t,τ)(t-τ)α1-g1(s,τ)(s-τ)α1|-[g1(t,τ)(t-τ)α1-g1(s,τ)(s-τ)α1]}dτ=∫0t‍{|e-(t+τ)(t-τ)1/3-e-(s+τ)(s-τ)1/3|-[e-(t+τ)(t-τ)1/3-e-(s+τ)(s-τ)1/3]}dτ=2∫0t‍[e-(s+τ)(s-τ)1/3-e-(t+τ)(t-τ)1/3]dτ≤2e-s∫0t‍dτ(s-τ)1/3-2e-2t∫0t‍dτ(t-τ)1/3=3e-s(s23-(s-t)23)-3e-2tt23≤3e-ss23.
In the similar way, we get
(85)∫0t‍{|g2(t,τ)(t-τ)α2-g2(s,τ)(s-τ)α2|-[g2(t,τ)(t-τ)α2-g2(s,τ)(s-τ)α2]}dτ=∫0t‍{|85(t+τ+2)3(t-τ)1/5-85(s+τ+2)3(s-τ)1/5|-[85(t+τ+2)3(t-τ)1/5-85(s+τ+2)3(s-τ)1/5]}dτ=2∫0t‍[85(s+τ+2)3(s-τ)1/5-85(t+τ+2)3(t-τ)1/5]dτ≤165(s+2)3∫0t‍dτ(s-τ)1/5-165(2t+2)3∫0t‍dτ(t-τ)1/5=4(s+2)3(s45-(s-t)45)-4(2t+2)3t45≤4(s+2)3s45.
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 thatm=max{∥m1∥,∥m2∥}=1/3, k=max{k1,k2}=1, and F¯=max{F¯1,F¯2}=π/2. Thus, the first inequality from assumption (viii) has the form
(86)[13+G¯1(rr+π2r)][13+G¯2(r5+π2r4)]≤r.
It can be shown that the number r0=1/2 is a solution of the above inequality such that it satisfies also the second inequality from assumption (viii).Applying Theorem10 we infer that the quadratic fractional integral equation considered in this example has a solution belonging to the set
(87)Ω1/2={x∈BC(ℝ+):0≤x(t)≤12fort∈ℝ+},
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 AlgebraBC(ℝ+)

In this final section we are going to investigate the following integral equation of mixed type:(88)x(t)=(Vx)(t)(Ux)(t),t∈ℝ+,
where V denotes the nonlinear Volterra integral operator of the form
(89)(Vx)(t)=p1(t)+f1(t,x(t))∫0t‍v(t,s,x(s))ds,
while U is the Urysohn integral operator having the form
(90)(Ux)(t)=p2(t)+f2(t,x(t))∫0∞‍u(t,s,x(s))ds.
Equation (88) will be considered in the Banach algebra BC(ℝ+). 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 noncompactnessμa 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)
pi∈BC(ℝ+) and pi(t)→0 as t→∞(i=1,2).(ii)
fi:ℝ+×ℝ→ℝ is continuous and such that fi(t,0)→0 as t→∞, for i=1,2.(iii)
The functionsfi(i=1,2) satisfy the Lipschitz condition with respect to the second variable; that is, there exists a constant ki>0 such that
(91)|fi(t,x)-fi(t,y)|≤ki|x-y|
for x,y∈ℝ and for t∈ℝ+(i=1,2).(iv)
v:ℝ+×ℝ+×ℝ→ℝ is continuous and there exist a continuous, function g:ℝ+×ℝ+→ℝ+ and a continuous and nondecreasing function G:ℝ+→ℝ+ such that
(92)|v(t,s,x)|≤g(t,s)G(|x|)
for all t,s∈ℝ+ and x∈ℝ.(v)
u:ℝ+×ℝ+×ℝ→ℝ is continuous and there exist a continuous, function h:ℝ+×ℝ+→ℝ+ and a continuous and nondecreasing function H:ℝ+→ℝ+ such that
(93)|u(t,s,x)|≤h(t,s)H(|x|)
for all t,s∈ℝ+ and x∈ℝ.(vi)
The functiont→∫0t‍g(t,s)ds is bounded on ℝ+.(vii)
For eacht∈ℝ+ the function s→h(t,s) is integrable on ℝ+ and the function t→∫0∞‍h(t,s)ds is bounded on ℝ+.(viii)
The improper integral∫0∞‍h(t,s)ds is uniformly convergent with respect to ℝ+; that is,
(94)limT→∞{supt∈ℝ+∫T∞‍h(t,s)ds}=0.(ix)
There exists a positive solutionr0 of the inequality
(95)[p+kG¯rG(r)+F¯G¯G(r)]×[p+kH¯rH(r)+F¯H¯H(r)]≤r
such that
(96)pk(G¯G(r0)+H¯H(r0))+2kF¯G¯H¯G(r0)H(r0)+2k2r0G¯H¯G(r0)H(r0)<1,
where the constants involved in the above inequalities are defined as follows:
(97)G¯=sup{∫0t‍g(t,s)ds:t∈ℝ+},H¯=sup{∫0∞‍h(t,s)ds:t∈ℝ+},F¯i=sup{|fi(t,0)|:t∈ℝ+}fori=1,2,p=max{∥p1∥,∥p2∥},F¯=max{F¯1,F¯2},k=max{k1,k2}.
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 BC(ℝ+) such that x(t)→0 as t→∞. 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(98)limt→∞∫0t‍gi(t,s)ds=0,limt→∞∫0∞‍hi(t,s)ds=0
for i=1,2.(xi)
fi is bounded function, and for each x∈ℝ there exists a finite limit limt→∞fi(t,x)(i=1,2).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 x=x(t) belonging to the Banach algebra BC(ℝ+) and such that there exists a finite limit limt→∞x(t).Proof.
Observe that based on the paper [2] we infer that operators V and U defined earlier transform the Banach algebra BC(ℝ+) into itself. Moreover, recalling [2] again, we have
(99)∥Vx∥≤p+kG¯∥x∥G(∥x∥)+F¯G¯G(∥x∥),∥Ux∥≤p+kH¯∥x∥H(∥x∥)+F¯H¯H(∥x∥).
In virtue of the above estimates and assumption (ix) we deduce that there exists a number r0>0 such that the operator W=VU maps the ball Br0 into itself. Moreover, from (99) we derive the following estimates:
(100)∥VBr0∥≤p+kG¯r0G(r0)+F¯G¯G(r0),∥UBr0∥≤p+kH¯r0H(r0)+F¯H¯H(r0).
Further, let us take an arbitrary nonempty subset X of the ball Br0. Then, using some estimates proved in [2] we have
(101)ω0∞(VX)≤kG¯G(r0)ω0∞(X),(102)ω0∞(UX)≤kH¯H(r0)ω0∞(X).
Now, let us fix T>0 and take arbitrary numbers t,s≥T. Then we obtain
(103)|(Vx)(t)-(Vx)(s)|≤|p1(t)|+|p1(s)|+|f1(t,x(t))-f1(s,x(s))|×∫0t‍|v(t,τ,x(τ))|dτ+|f1(s,x(s))|×|∫0t‍v(t,τ,x(τ))dτ-∫0s‍v(s,τ,x(τ))dτ|≤|p1(t)|+|p1(s)|+[k1|x(t)-x(s)|+|f1(t,x(s))-f1(s,x(s))|]×∫0t‍|v(t,τ,x(τ))|dτ+[k1|x(s)|+f1(s,0)]×[∫0t‍|v(t,τ,x(τ))|dτ+∫0s‍|v(s,τ,x(τ))|dτ]≤|p1(t)|+|p1(s)|+[k1|x(t)-x(s)|+|f1(t,x(s))-f1(s,x(s))|]×G(∥x∥)∫0t‍g(t,τ)dτ+[k1∥x∥+|f1(s,0)|]G(∥x∥)×[∫0t‍g(t,τ)dτ+∫0s‍g(s,τ)dτ].
Hence, we derive the following inequality:
(104)|(Vx)(t)-(Vx)(s)|≤sup{|p1(t)|+|p1(s)|:t,s≥T}+k1G¯G(r0)sup{|x(t)-x(s)|:t,s≥T}+G¯G(r0)sup{|f1(t,x(s))-f1(s,x(s))|:t,s≥T}+[k1r0+F1¯]G(r0)sup{∫0t‍g(t,τ)dτ+∫0s‍g(s,τ)dτ:t,s≥T}.
Combining the above inequality with assumptions (i), (vi), (x), and (xi), we obtain
(105)a(VX)≤kG¯G(r0)a(X).
Similarly we can show, based on assumptions (i), (ii), (v), (vii), (viii), (x), and (xi), that the following inequality holds:
(106)a(UX)≤kH¯H(r0)a(X).
Further, let us observe that from (101) and (105) we derive
(107)μa(VX)≤kG¯G(r0)μa(X).
In the same way, linking (102) and (106), we get
(108)μa(UX)≤kH¯H(r0)μa(X).
Finally, taking into account estimates (100), (107), and (108), assumption (ix), and Theorem 4 we deduce that operator W=VU has at least one fixed point x=x(t) in the ball Br0 being a subset of the Banach algebra BC(ℝ+). It is clear that the function x is a solution of (88). Moreover, from Remark 5 and the description of the kernel kerμa we infer that x has a finite limit at infinity.
The proof is complete.



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*Source: 537897-2013-04-27.xml*