# Asymptotic Stability of Solutions to a Nonlinear Urysohn Quadratic Integral Equation

Authors: H. H. G. Hashem; A. R. Al-Rwaily
Journal: International Journal of Analysis
 (2013)

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## Abstract

Here, we prove the existence ofL1-nondecreasing solution to a nonlinear quadratic integral equation of Urysohn type by applying the technique of weak noncompactness. Also, the asymptotic stability of solutions for that quadratic integral equation is studied.


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## Body

## 1. Introduction

Integral equations play an important role in many branches of linear and nonlinear functional analysis and their applications in the theory of elasticity, engineering, mathematical physics, and contact mixed problems, and the theory of integral equations is rapidly developing with the help of several tools of functional analysis, topology, and fixed point theory. For details, we refer to [1–23].Quadratic integral equations often appear in many applications of real world problems, for example, in the theory of radiative transfer, kinetic theory of gases, in the theory of neutron transport, and in the traffic theory (see [12]). The quadratic integral equation can be very often encountered in many applications (see [1, 2, 6–10, 13–26]). However, in most of the previous literature, the main results are realized with the help of the technique associated with the measure of noncompactness. Instead of using the technique of measure of noncompactness, the Tychonoff fixed point theorem is used for some quadratic integral equations [20, 26]. Picard and Adomian decomposition methods are used to compare approximate and exact solutions for quadratic integral equations [13, 19, 22]. Also, nondecreasing solution of a quadratic integral of Urysohn-Stieltjes type is studied in [10].LetL1=L1[0,T] be the class of Lebesgue integrable functions on I=[0,T] with the standard norm.Here, we are concerned with the nonlinear quadratic functional integral equation(1)x(t)=f(t,x(ϕ1(t)))+g(t,x(ϕ2(t)))×∫0α(t)‍u(t,s,x(ϕ3(s)))ds,t∈I,
and we prove the existence of monotonic solutions in L1 by using the technique of measure of noncompactness. The results of this work generalize those obtained in [18]. Finally, the asymptotic stability of solutions for the quadratic integral equation (1) is studied.


## 2. Preliminaries

In this section, we collect some definitions and results needed in our further investigations. Assume that the functionf:I×R→R satisfies Carathèodory condition that is measurable in t for any x and continuous in x for almost all t. Then, to every function x(t) being measurable on the interval I, we may assign the function
(2)(Fx)(t)=f(t,x(t)),t∈I.The operatorF defined in such a way is called the superposition operator. This operator is one of the simplest and most important operators investigated in the nonlinear functional analysis. For this operator, we have the following theorem due to Krasnosel’skii  [3].Theorem 1.
The superposition operatorF maps L1 into itself if and only if
(3)|f(t,x)|≤c(t)+k|x|∀t∈I
and x∈R, where c(t) is a function from L1 and k is a nonnegative constant.Now, letE be a Banach space with zero element θ and X a nonempty bounded subset of E. Moreover denote by Br=B(θ,r) the closed ball in E centered at θ and with radius r. In the sequel, we will need some criteria for compactness in measure; the complete description of compactness in measure was given by Banaś [3], but the following sufficient condition will be more convenient for our purposes (see [3]).Theorem 2.
LetX be a bounded subset of L1. Assume that there is a family of subsets (Ωc)0≤c≤b-a of the interval (a,b) such that measΩc=c for every c∈[0,b-a], and for every x∈X, x(t1)≤x(t2), (t1∈Ωc,t2∉Ωc); then, the set X is compact in measure.The measure of weak noncompactness defined by De Blasi [11, 27] is given by
(4)β(X)=inf(r>0;there exists a weakly compact subsetYofEsuch thatX⊂Y+Kr).The functionβ(X) possesses several useful properties which may be found in [11].The convenient formula for the functionβ(X) in L1 was given by Appell and De Pascale (see [27]) as follows:
(5)β(X)=limϵ→0(supx∈X(sup[∫D‍|x(t)|dt:D⊂[a,b],measD≤ϵ∫D‍]∫D‍)∫D‍),
where the symbol measD stands for Lebesgue measure of the set D.Next, we shall also use the notion of the Hausdorff measure of noncompactnessχ (see [3]) defined by
(6)χ(X)=inf(r>0;there exists a finite subsetYofEsuch thatX⊂Y+Kr).In the case when the setX is compact in measure, the Hausdorff and De Blasi measures of noncompactness will be identical. Namely, we have the following (see [11, 27]).Theorem 3.
LetX be an arbitrary nonempty bounded subset of L1. If X is compact in measure, then β(X)=χ(X).Finally, we will recall the fixed point theorem due to Banaś [5].Theorem 4.
LetQ be a nonempty, bounded, closed, and convex subset of E, and let H:Q→Q be a continuous transformation which is a contraction with respect to the Hausdorff measure of noncompactness χ; that is, there exists a constant α∈[0,1) such that χ(HX)≤αχ(X) for any nonempty subset X of Q. Then, H has at least one fixed point in the set Q.


## 3.  Existence Theorem

Let the integral operatorH be defined as
(7)(Hx)(t)=∫0α(t)‍u(t,s,x(s))ds,(Fx)(t)=f(t,x(t)).Then, (1) may be written in operator form as
(8)(Ax)(t)=(Fx(ϕ1))(t)+(Gx(ϕ2))(t)·(Hx(ϕ3))(t),
where (Gx)(t)=g(t,x(t)).Consider the following assumptions.(i)
f,g:I×R→R are functions such that f,g:I×R+→R+. Moreover, the functions f,g satisfy Carathèodory condition (i.e., are measurable in t for all x∈R and continuous in x for all t∈I), and there exist two functions a1,a2∈L1 and constants b1,b2>0 such that
(9)|f(t,x)|≤a1(t)+b1|x|,|g(t,x)|≤a2(t)+b2|x|∀(t,x)∈I×R.
Apart from this, the functions f and g are nondecreasing in both variables.(ii)
u:I×I×R→R is such that u(t,s,x)≥0 for (t,s,x)∈I×I×R+, and u(t,s,x) satisfies Carathéodory condition (i.e., it is measurable in (t,s) for all x∈R and continuous in x for almost all (t,s)∈I×I).(iii)
There exist a positive constantb3, a function a3∈L1, and a measurable (in both variables) function k(t,s)=k:I×I→R+ such that
(10)|u(t,s,x)|≤k(t,s)(a3(t)+b3|x|)∀t,s∈Iand  forx∈R,
and the integral operator K, generated by the function k and defined by
(11)(Kx)(t)=∫0t‍k(t,s)x(s)ds,t∈I,
maps continuously L1 into L∞ on I.(iv)
t→u(t,s,x) is a.e. nondecreasing on I for almost all fixed s∈I and for each x∈R+.(v)
α:I→I is continuous.(vi)
ϕi:I→I, i=1,2,3, are increasing, absolutely continuous functions on I, and there exist positive constants Bi, i=1,2,3, such that ϕi′≥Bi a.e. on I.(vii)
Letd>4Mb2b3B12B2B3(||a1||+M·||a2||||a3||),M=||K||L∞, where d=B1B2B3-b1B2B3-b2MB1B3||a3||-Mb3B1B2||a2||.Now, letr be a positive root of the equation
(12)b2b3B1Mr2-dr+B1B2B3(∥a1∥+M·∥a2∥∥a3∥)=0,
and define the set
(13)Br={x∈L1:∥x∥≤r}.For the existence of at least oneL1-positive solution of the quadratic functional integral equation (1), we have the following theorem.Theorem 5.
Let the assumptions (i)–(vii) be satisfied.
Ifb1B2B3+Mb2B1B3||a3||+rb2b3M<B1B2B3, then the quadratic integral equation (1) has at least one solution x∈L1 which is positive and a.e. nondecreasing on I.Proof.
Take an arbitraryx∈L1; then, we get
(14)|(Ax)(t)|≤|a1(t)|+b1|x(ϕ1(t))|+(a2(t)+b2|x(ϕ2(t))|)×∫0α(t)‍k(t,s)(a3(t)+b3|x(ϕ3(s))|)ds,
which implies that
(15)∥(Ax)(t)∥=∫0T‍|(Ax)(t)|dt≤∫0T‍|a1(t)|dt+b1∫0T‍|x(ϕ1(t))|dt+∫0T‍(a2(t)dt+b2|x(ϕ2(t))|)×∫0α(t)‍k(t,s)(a3(t)+b3|x(ϕ3(s))|)dsdt≤∥a1∥+b1B1∫0T‍|x(ϕ1(t))|·ϕ1′(t)dt+∫0T‍a2(t)∫0α(t)‍k(t,s)a3(s)dsdt+b2∫0T‍|x(ϕ2(t))|∫0α(t)‍k(t,s)a3(s)dsdt+b3×∫0T‍a2(t)∫0α(t)‍k(t,s)|x(ϕ3(s))|dsdt+b2b3∫0T‍|x(ϕ2(t))|∫0α(t)‍k(t,s)|x(ϕ3(s))|dsdt≤∥a1∥+b1B1∫0T‍|x(ϕ1(t))|·ϕ1′(t)dt+∫0T‍a2(t)∫0T‍k(t,s)a3(s)dsdt+b2∫0T‍|x(ϕ2(t))|∫0T‍k(t,s)a3(s)dsdt+b3×∫0T‍a2(t)∫0T‍k(t,s)|x(ϕ3(s))|dsdt+b2b3∫0T‍|x(ϕ2(t))|∫0T‍k(t,s)|x(ϕ3(s))|dsdt≤∥a1∥+b1B1∫ϕ1(0)ϕ1(T)‍|x(ϕ1(t))|·ϕ1′(t)dt+∫0T‍a2(t)∫0T‍k(t,s)a3(s)dsdt+b2B2∫ϕ2(0)ϕ2(T)‍|x(ϕ2(t))|·ϕ2′(t)∫0T‍k(t,s)a3(s)dsdt+b3B3∫0T‍a2(t)∫ϕ3(0)ϕ3(T)‍k(t,s)|x(ϕ3(s))|·ϕ3′(s)dsdt+b2b3B2B3∫ϕ2(0)ϕ2(T)‍|x(ϕ2(t))|·ϕ2′(t)×∫ϕ3(0)ϕ3(T)‍k(t,s)|x(ϕ3(s))|·ϕ3′(s)dsdt≤∥a1∥+b1B1∫0T‍|x(θ)|dθ+M∥a2∥∥a3∥+Mb2∥a3∥B2∫0T‍|x(θ)|dθ+Mb3∥a2∥B3∫0T‍|x(θ)|dθ+Mb2b3B2B3∫0T‍|x(θ)|dθ·∫0T‍|x(θ)|dθ≤∥a1∥+b1B1∥x∥+M∥a2∥∥a3∥+Mb2∥a3∥B2∥x∥+Mb3∥a2∥B3∥x∥+Mb2b3B2B3∥x∥2.
From this estimate, we show that the operatorA maps the ball Br into itself with
(16)r=d-d2-4Mb2b3B12B2B3(∥a1∥+M·∥a2∥∥a3∥)2Mb2b3B1,
From assumption (vii) we have(17)0<d2-4Mb2b3B12B2B3(∥a1∥+M·∥a2∥∥a3∥)<d2,
which implies that
(18)0<d2-4Mb2b3B12B2B3(∥a1∥+M·∥a2∥∥a3∥)<d.
Then,d is positive which implies that r is a positive constant.
Now, letQr denote the subset of Br∈L1 consisting of all functions which are positive and a.e. nondecreasing on I.
The setQr is nonempty, bounded, convex, and closed (see [3, page 780]). Moreover, this set is compact in measure (see Lemma  2 in [4, page 63]).
From the assumptions, we deduce that the operatorA maps Qr into itself. Since the operator (Ux)(t)=u(t,s,x) is continuous (Theorem 1 in Section 2), then the operator H is continuous, and, hence, the product G.H is continuous. Also, F is continuous. Thus, the operator A is continuous on Qr.
LetX be a nonempty subset of Qr. Fix ϵ>0, and take a measurable subset D⊂I such that meas D≤ϵ. Then, for any x∈X, using the same reasoning as in [3, 4], we get
(19)∥Ax∥L1(D)=∫D‍|(Ax)(t)|dt≤∫D‍|a1(t)|dt+b1∫D‍|x(ϕ1(t))|dt+∫D‍(a2(t)dt+b2|x(ϕ2(t))|)×∫0α(t)‍k(t,s)(a3(s)+b3|x(ϕ3(s))|)dsdt≤∥a1∥L1(D)+b1B1∫D‍|x(ϕ1(t))|·ϕ1′(t)dt+∫D‍a2(t)∫0T‍k(t,s)a3(s)dsdt+b2∫D‍|x(ϕ2(t))|∫0T‍k(t,s)a3(s)dsdt+Mb3×∫D‍a2(t)∫0T‍|x(ϕ3(s))|dsdt+Mb2b3∫D‍|x(ϕ2(t))|∫0T‍|x(ϕ3(s))|dsdt≤∥a1∥L1(D)+b1B1∫D‍|x(θ)|dθ+M∥a2∥L1(D)∥a3∥L1+Mb2∥a3∥L1B2∫D‍|x(θ)|dθ+Mb3∥a2∥L1(D)B3∫0T‍|x(θ)|dθ+Mb2b3B2B3∫D‍|x(θ)|dθ·∫0T‍|x(θ)|dθ≤∥a1∥L1(D)+b1B1∥x∥L1(D)+M∥a2∥L1(D)∥a3∥L1+Mb2∥a3∥L1B2∥x∥L1(D)+Mb3∥a2∥L1(D)B3∥x∥+Mb2b3B2B3∥x∥L1(D)·∥x∥≤∥a1∥L1(D)+b1B1∥x∥L1(D)+M∥a2∥L1(D)∥a3∥L1+Mb2∥a3∥L1B2∥x∥L1(D)+rMb3∥a2∥L1(D)B3+rMb2b3B2B3∥x∥L1(D).
Since(20)limϵ→0{sup{∫D‍|ai(t)|dt:D⊂I,measD<ϵ}}=0,i=1,2,
we obtain(21)β(Ax(t))≤[b1B1+Mb2∥a3∥L1B2+Mb2b3rB2B3]β(x(t)).
This implies that(22)β(AX)≤[b1B1+Mb2∥a3∥L1B2+Mb2b3rB2B3]β(X),
where β is the De Blasi measure of weak noncompactness.
Keeping in mind Theorem3, we can write (22) in the form
(23)χ(AX)≤[b1B1+Mb2∥a3∥L1B2+Mb2b3rB2B3]χ(X),
where χ is the Hausdorff measure of noncompactness.
Since(b1/B1)+(Mb2||a3||L1/B2)+(Mb2b3r/B2B3)<1, from Theorem 4 follows that A is contraction with respect to the measure of noncompactness χ. Thus, A has at least one fixed point in Qr which is a solution of the quadratic functional integral equation.


## 4. Asymptotic Stability of the Quadratic Integral Equation

We shall show that the solution of the quadratic integral equation (1) is asymptotically stable on ℝ+.Definition 6.
The functionx is said to be asymptotically stable solution of (1) if for any ϵ>0 there exists T′=T′(ϵ)>0 such that for every t≥T′ and for every other solution y of (1),
(24)|x(t)-y(t)|≤ϵ.Proof.
Letr be defined by (16), and consider the following assumptions.(*)
There exist constantsl1 and l2 satisfying that
(25)|f(t,x)-f(t,y)|≤l1|x-y|,|g(t,x)-g(t,y)|≤l2|x-y|,∀t∈I,x,y∈R+.(**)
2MB1B2B3||a2||||a3||+2rb2MB1B3||a2||+2rMb3B1B2||a2||+2Mr2b2b3B1<ϵ(B1B2B3-   l1B2B3-Ml2B1B3||a3||-Mrl2b3B1).
For solutionsx=x(t) and y=y(t) of (1) in Br, by the assumptions (*) and (**), we deduce that
(26)|x(t)-y(t)|=|(Ax)(t)-(Ay)(t)|≤|f(t,x(ϕ1(t)))-f(t,y(ϕ1(t)))|+|g(t,x(ϕ2(t)))-g(t,y(ϕ2(t)))|×∫0α(t)‍|u(t,s,y(ϕ3(s)))|ds+|g(t,y(ϕ2(t)))|×∫0α(t)‍|u(t,s,y(ϕ3(s)))-u(t,s,x(ϕ3(s)))|ds≤l1|x(ϕ1(t))-y(ϕ1(t))|+l2|x(ϕ2(t))-y(ϕ2(t))|×∫0T‍k(t,s)(a3(s)+b3|y(ϕ3(s))|)ds+2(a2(t)+b2|y(ϕ2(t))|)×∫0T‍k(t,s)(a3(s)+b3|x(ϕ3(s))|)ds;
for any t∈I, using (**), we have
(27)∥x-y∥=∫0T‍|x(t)-y(t)|dt≤l1B1∥x-y∥+Ml2B2∥x-y∥×[∥a3∥+b3rB3]+2M∥a2∥∥a3∥+2rb2M∥a3∥B2+2Mr2b2b3B2B3+2Mrb3∥a2∥B3.
Then,
(28)∥x-y∥≤[2M∥a2∥∥a3∥+2rb2M∥a3∥B2+2Mr2b2b3B2B3+2Mrb3∥a2∥B3]·[1-l1B1-Mrb3l2B2B3-l2M||a3||B2]-1≤ϵ,t∈I.
That is, the solutionx=x(t) of (1) is asymptotically stable on R+. This completes the proof.


## 5. Applications

As particular cases of Theorem5, we can obtain theorems on the existence of positive and a.e. nondecreasing solutions belonging to the space L1(I) of the following quadratic integral equations.(1)
Ifα(t)=1, then we obtain the quadratic integral equation
(29)x(t)=f(t,x(ϕ1(t)))+g(t,x(ϕ2(t)))×∫01‍u(t,s,x(ϕ3(s)))ds,t∈I.(2)
Ifα(t)=1 and f(t,x)=a(t), then we obtain the quadratic integral equation
(30)x(t)=a(t)+g(t,x(ϕ2(t)))∫01‍u(t,s,x(ϕ3(s)))ds,t∈I.(3)
Iff(t,x)=a(t), u(t,s,x)=h(t,x), α(t)=1, and g(t,s)=1, then we obtain the quadratic integral equation
(31)x(t)=a(t)+∫01‍h(s,x(ϕ3(s)))ds,t∈I,
which was proved by Banaś in [4].(4)
Ifg(t,x)=0, then we obtain the functional equation
(32)x(t)=f(t,x(ϕ1(t))),t∈I,
which is the same results proved by Banaś in [3].(5)
Iff(t,x)=a(t), and u(t,s,x)=k(t,s)h(t,x), then we obtain the quadratic integral equation
(33)x(t)=a(t)+g(t,x(t))∫01‍k(t,s)h(s,x(ϕ3(s)))ds,t∈I,
which is the same result proved in [16].(6)
Iff(t,x)=a(t), u(t,s,x)=h(t,x), and α(t)=t, then we obtain the quadratic integral equation
(34)x(t)=a(t)+g(t,x(t))∫0t‍h(s,x(ϕ3(s)))ds,t∈I,
which is the same result proved in [18].Example 7.
Let us consider the quadratic integral equation of Urysohn type having the form(35)x(t)=a(t)+x(t)∫01‍tt+su(t,s,x(s))ds,t∈[0,1].
This equation represents the Hammerstein counterpart of the famous Chandrasekhar quadratic integral equation which has numerous application (cf. [1, 2, 6, 24]). It arose originally in connection with scattering through a homogeneous semi-infinite plane atmosphere [24].
In casea(t)=1 and u(t,s,x(s))=λϕ(s)x(s),λ is a positive constant. Then, (35) has the form
(36)x(t)=1+λx(t)∫01‍tϕ(s)t+sx(s)ds.
In order to apply our results, we have to impose an additional condition that the so-called “characteristic” functionϕ is continuous on I.
In this case,r=(1-1-4λk1)/2λk, and the assumption (vii) may be reduced to 4λk1≤1 where sups∈Iϕ(s)=k1.Example 8.
Consider the following quadratic functional integral equation:(37)x(t)=16x(t)+[t+13+tx(sin(t2+3t))]×∫0t‍[1+13+sx(sin(s2+4s))]dst∈[0,1].
Taking(38)f(t,x)=16x(t),g(t,x)=1+13+tx,u(t,s,x)=t+13+tx,
then we can easily deduce that (i)
|u(t,s,x)|≤1+(1/4)|x| and |g(t,x)|≤t+(1/4)|x|(i.e.,α(t)=t, a1(t)=0, a2(t)=t, and a3(t)=1 which implies that ||a1||=0,||a2||=||a3||=1, and b1=1/6, b2=b3=1/4);(ii)
ϕ1(t)=t, ϕ2(t)=sin(t2+3t), and ϕ3(t)=sin(t2+4t), and then ϕ1′(t)=1, ϕ2′(t)=(2t+3)cos(t2+3t)>2, and ϕ3′(t)=(2t+4)cos(t2+4t)>3(i.e.,B1=1/2, B2=2, B3=3, and M=1).
Now, we will calculater.
Then,r=2.30228>0 and (b1/B1)+(Mb2||a3||L1/B2)+(Mb2b3r/B2B3)=0.48231546246258117<1.
Thus, all the assumptions of Theorem5 are satisfied; so, the quadratic functional integral equation (37) possesses at least one solution being positive, a.e. nondecreasing, and integrable in [0,1].



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