In this paper, we consider a nonlinear n-term fractional quadratic integral equation. Our investigation is located in the space C(J,R). We prove the existence and uniqueness of the solution for that problem by applying some fixed point theorems. Next, we establish the continuous dependence of the unique solution for that problem on some functions. Finally, we present some particular cases for n-term fractional quadratic integral equation and an example to illustrate our results.
Citation: Hind H. G. Hashem, Asma Al Rwaily. Investigation of the solvability of n- term fractional quadratic integral equation in a Banach algebra[J]. AIMS Mathematics, 2023, 8(2): 2783-2797. doi: 10.3934/math.2023146
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In this paper, we consider a nonlinear n-term fractional quadratic integral equation. Our investigation is located in the space C(J,R). We prove the existence and uniqueness of the solution for that problem by applying some fixed point theorems. Next, we establish the continuous dependence of the unique solution for that problem on some functions. Finally, we present some particular cases for n-term fractional quadratic integral equation and an example to illustrate our results.
Several fixed point problems involving product of operators have been investigated in many literature and monographs, for example [1,2,3,4,5]. However, the problem considered in [6] is more general than those [5,7,8,9,10].
Quadratic integral equations(QIEs) have been investigated from different points of view and using different techniques (see [11,12,13,14,15,16,17,18]). The QIEs can be widely applicable in more applications like the dynamic theory of gases, the theory of radiative exchange, the traffic theory, etc. see [4,6,16,19]. For the case of Banach algebras, many recent references have been appeared, for example [20,21,22]. For solvability of some QIEs on unbounded interval see [19,23,24].
The multi-term fractional quadratic integral equation
x(t)=n∑i=1fi(t,x(t))∫t0(t−s)αi−1Γ(αi)gi(s,x(s))ds,αi>0 | (1.1) |
has been studied in a Banach algebra [25] by using some fixed point theorem [26].
The quadratic integral equation of fractional order
x(t)=k(t,x(η(t)))+f(t,x(μ(t)))(a(t)+∫σ(t)0v(t,s)g(s,x(ν(s)))ds) |
has been investigated in [27] by applying the nonlinear alternative of Leray-Schauder type.
Motivated by these results and by the monographs that studied ϕ-fractional quadratic integral equations, in this article, we focus our attention on a nonlinear multi-term quadratic functional integral equation of fractional order
x(t)=n∑i=1fi(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds),αi>0, | (1.2) |
for all t∈J=[0,1], where ai:J→R,fi:J×R→R,,gi:J×R→R and ηi,ψi:J→J.
By a solution of the quadratic functional integral equation of fractional order (1.2), we mean a function x∈C(J,R) that satisfies Eq (1.2), where C(J,R) stands for the space of continuous real-valued functions on J.
For proving the existence results for the quadratic functional integral equation of fractional Eq (1.2). We recall the following fixed point theorem [26] which enables us to prove the existence theorem for solutions of the functional integral Eq (1.2).
Theorem 1.1. Let n be a positive integer, and C be a nonempty, closed, convex and bounded subset of a Banach Algebra X. Assume that the operators Ai:X→X and Bi:C→X,i=1,2,....,n, satisfy
(a) For each i∈{1,2,....,n},Ai is D−Lipschitzian with a D−function ϕi;
(b) For each i∈{1,2,....,n},Bi is continuous and Bi(C) is precompact;
(c) For each y∈C,x=n∑i=1Aix.Biy implies that x∈C.
Then, the operator equation x=n∑i=1Aix.Bix has a solution provided that
n∑i=1Miϕi(r)<r,∀r>0, |
where Mi=supx∈C||Bix||,i=1,2,....,n.
Equation (1.2) is investigated under the assumptions:
(i∗) gi:J×R→R,i=1,2,....,n satisfy Carathéodory condition (i.e., measurable in t for all x∈R and continuous in x for almost all t∈J) such that:
|gi(t,x)|≤mi(t)+bi|x|,bi≥0,mi∈L1(J,R),i=1,2,....,n∀(t,x)∈J×R |
and ki=supt∈JIβiϕmi(t) for any βi≤αi,i=1,2,....,n such that ki≠0∀i
(ii∗)fi:J×R→R,i=1,2,....,n are continuous and bounded with hi=sup(t,x)∈J×R|fi(t,x)|,i=1,2,....,n.
(iii∗)There exist constants Li,i=1,2,....,n satisfying
|fi(t,x)−fi(t,y)|≤Li|x−y|,i=1,2,....,n |
for all t∈J and x,y∈R.
(iv∗)ηi,ψi:J→J,i=1,2,3......,n are continuous functions.
(v∗)ϕ is increasing and absolutely continuous function.
(vi∗) ai:J→R,i=1,...,n are continuous and bound with ki=supt∈J|a(t)|.
Theorem 2.1. Let the assumptions (i∗)-(v∗) be satisfied. Furthermore, if
n∑i=1bihiΓ(αi+1)<1 |
then the quadratic integral Eq (1.2) has at least one solution in C(J,R).
Proof. Set X=C(J,R). Consider the closed ball ¯Br(0) in X centered at origin 0 and of radius r, where r=n∑i=1[hikiΓ(αi−βi+1)+ki][1−n∑i=1bihiΓ(αi+1)]−1>0.
Consider the mapping Ai:X→X and Bi:¯Br(0)→X on C(J,R) defined by:
(Aix)(t)=fi(t,x(η(t))(Bix)(t)=ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds. |
Then the quadratic integral Eq (1.2) can be written in the form:
Tx(t)=n∑i=1Aix(t).Bix(t). | (2.1) |
We shall show that Ai and Bi satisfy all the conditions of Theorem 1.1.
Let us define a subset C of C(J,R) by
C:={x∈C(J,R),||x||≤r}. |
Obviously, C is nonempty, bounded, convex and closed subset of C(J,R).
For all t∈J, since the assumptions (ii*) and (iii*) are satisfied, the mapping fi is well defined and the function Aix is continuous and bounded on J. Again, since each gi is continuous in x and each ψi is continuous function, then the function Bix is also, continuous and bounded by m′i(t)=mi(t)+bir,m′i∈L1(J,R)(in view of assumptions (i*) and (vi*)). Firstly, we show that Ai is Lipschitz on X. Let x,y∈X be arbitrary. Then by assumptions (ii*) and (iii*)
|Aix(t)−Aiy(t)|=|fi(t,x(ηi(t)))−fi(t,y(ηi(t)))|≤Li|x(ηi(t))−y(ηi(t))|≤Li||x−y||,i=1,2,..,n. |
For all t∈J. Taking supremum over t
||Aix−Aiy||≤Li||x−y||,i=1,2,..,n |
for all x,y∈X. This shows that Ai is a Lipschitz mapping on X with the Lipschitz constants Li. Secondly, we show that Bi is continuous and compact operator on ¯Br(0). First we show that each Bi is continuous on ¯Br(0). To do this, let us fix arbitrary ϵ>0 and let {xn} be a sequence of functions in ¯Br(0) converging to x∈¯Br(0). Then we get
|(Bixn)(t)−(Bix)(t)|≤∫t0(ϕ(t)−ϕ(s))αi−1Γ(α)|gi(s,xn(ψi(s)))−gi(s,x(ψi(s)))|ϕ′(s)ds≤∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)[|gi(s,xn(ψi(s)))|+|gi(s,x(ψi(s)))|]ϕ′(s)ds≤2∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)mi(s)ϕ′(s)ds+2bi∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|x(ψi(s))|ϕ′(s)ds≤2Iαi−βiϕIβiϕmi(t)+2bir∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)ϕ′(s)ds≤2ki∫t0(ϕ(t)−ϕ(s))αi−βi−1Γ(αi−βi)ϕ′(s)ds+2bir(ϕ(t)−ϕ(s))αiΓ(αi+1)≤2ki(ϕ(t)−ϕ(s))αi−βiΓ(αi−βi+1)+2bir(ϕ(t)−ϕ(s))αiΓ(αi+1)≤2kiΓ(αi−βi+1)+2birΓ(αi+1)≤ϵfor t∈J. |
Thus
|(Bixn)(t)−(Bix)(t)|≤ϵasn→∞. |
Furthermore, let us assume that t∈J. Then, by Lebesgue dominated convergence theorem, we obtain the estimate:
limn→∞(Bixn)(t)=ai(t)+limn→∞∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds=ai(t)+∫t0ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds=(Bix)(t) |
for all t∈J. Thus, Bixn→Bix as n→∞ uniformly and hence each Bi is a continuous operator on ¯Br(0)into ¯Br(0) has a Cauchy subsequence. Now by (i*) and (vi*)
|Bixn(t)|≤|ai(t)|+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,xn(ψi(s)))|ϕ′(s)ds≤ki+kiΓ(αi−βi+1)+birΓ(αi+1) |
for all t∈J. Then ||Bixn(t)||≤ki+kiΓ(αi−βi+1)+birΓ(αi+1) for all n∈N. This shows that {Bixn} is a uniformly bounded sequence in Bi(¯Br(0)).
Now, we proceed to show that it is also equicontinuous. Let t1,t2∈J (without loss of generality assume that t1<t2), then we have
|Bixn(t2)−Bixn(t1)|≤|ai(t2)−ai(t1)|+|∫t20(ϕ(t2)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds−∫t10(ϕ(t1)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds|≤|ai(t2)−ai(t1)|+|∫t10(ϕ(t2)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds+∫t2t1(ϕ(t2)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds−∫t10(ϕ(t1)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds|≤|ai(t2)−ai(t1)|+|∫t10(ϕ(t2)−ϕ(s))αi−1−(ϕ(t1)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds+∫t2t1(ϕ(t2)−ϕ(s))αi−1Γ(αi)gi(s,xn(ψi(s)))ϕ′(s)ds|≤|ai(t2)−ai(t1)|+∫t2t1(ϕ(t2)−ϕ(s))αi−1Γ(αi)|gi(s,xn(ψi(s)))|ϕ′(s)ds+∫t10(ϕ(t2)−ϕ(s))αi−1−(ϕ(t1)−ϕ(s))αi−1Γ(αi)|gi(s,xn(ψi(s)))|ϕ′(s)ds. |
Therefore,
|Bixn(t2)−Bixn(t1)|≤|ai(t2)−ai(t1)|+∫t2t1(ϕ(t2)−ϕ(s))αi−1Γ(αi)[mi(t)+bir]ϕ′(s)ds+∫t10(ϕ(t1)−ϕ(s))αi−1−(ϕ(t1)−ϕ(s))αi−1Γ(αi)[mi(t)+bir]ϕ′(s)ds≤|ai(t2)−ai(t1)|+ki∫t2t1(ϕ(t2)−ϕ(s))αi−βi−1Γ(αi−βi)ϕ′(s)ds+bir∫t2t1(ϕ(t2)−ϕ(s))αi−1Γ(αi)ϕ′(s)ds≤|ai(t2)−ai(t1)|+ki(ϕ(t2)−ϕ(t1))αi−βiΓ(αi−βi+1)+bir(ϕ(t2)−ϕ(t1))αiΓ(αi+1). |
Using the uniform continuity of the functions ai and ϕ on J, we obtain
|Bixn(t2)−Bixn(t1)|→0ast2→t1. |
As a consequence, |Bixn(t2)−Bixn(t1)|→0ast2→t1. This shows that {Bixn} is an equicontinuous sequence in X. Now an application of Arzela-Ascoli [28] theorem yields that {Bixn} has a uniformly convergent subsequence on the the compact subset J of R. without loss of generality, call the subsequence it self. We show that {Bixn} is Cauchy sequence in X. Now |Bixn(t)−Bix(t)|→0asn→∞ for all t∈J. Then for given ϵ>0 there exists an n0∈N such that for all m,n≥n0. Therefore (or m,n≥n0), we have
|Bixm(t)−Bixn(t)|≤supt∈J|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)[gi(s,xm(ψi(s)))−gi(s,xn(ψi(s)))]ϕ′(s)ds|≤supt∈J∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,xm(ψi(s)))−gi(s,xn(ψi(s)))|ϕ′(s)ds<ϵ. |
This shows that {Bixn}⊂F(¯Br(0))⊂X is Cauchy sequence. Since X is complete, {Bixn} converges to a point in X. Hence Bi(¯Br(0)) is relatively compact and consequently Bi is a continuous and compact operator on ¯Br(0).
Next, we show that AixBiy=x⇒x∈¯Br(0) for all y∈¯Br(0). Then,
|x(t)|≤|Aix(t)||Biy(t)|≤|fi(t,x(ηi(t)))|[Iαi|gi(t,x(ψi(t)))|]≤hi(ki+kiΓ(αi−βi+1)+birΓ(αi+1))≤hi(ki+kiΓ(αi−βi+1)+birΓ(αi+1))≤rfor all t∈J. |
Taking the supremum over t. we obtain ||x||≤r for all y∈¯Br(0),r=n∑i=1[hikiΓ(αi−βi+1)+ki][1−n∑i=1bihiΓ(αi+1)]−1. Hence hypothesis (c) of Theorem 1.1 holds. Here,
Mi=||Bix||≤ki+Iαi|gi(t,x(ψi(t)))|≤ki+kiΓ(αi−βi+1)+birΓ(αi+1). |
Therefore we can get, for every x∈C we have
|(Tx)(t)|≤n∑i=1[kihi+hikiΓ(αi−βi+1)+bihirΓ(αi+1)]=r. |
Then, Tx∈C and hence TC⊂C.
Since all conditions of Theorem 1.1 are satisfied, then the operator T=n∑i=1Ai.Bi has a fixed point in C.
In this section, we shall demonstrate some characteristics for the solutions of the n-term quadratic integral Eq (1.2).
In aim of proving the uniqueness of the solution of (1.2), we replace assumption (i∗) by the following assumption:
gi:J×R→R,i=1,2,....,n satisfy Carathéodory condition (i.e., measurable in t for all x∈R and continuous in x for almost all t∈J) and
|gi(t,x)−gi(t,y)|≤ci|x−y|,i=1,2,....,n∀(t,x)∈J×R |
and ρi=supt∈J|gi(t,0)| for any i=1,2,....,n such that ki≠0∀i.
Theorem 3.1. Let the assumptions (i) and (ii∗)–(v∗) be satisfied. Moreover, if
n∑i=1(Li(ki+ρi+cirΓ(αi+1))+hiciΓ(αi+1))<1 |
then the quadratic integral Eq (1.2) has a unique solution in C(J,R).
Proof. From the assumption (i∗) we have
|gi(s,x(s))|−|gi(s,0)|≤|gi(s,x(s))−gi(s,0)|≤ci|x||gi(s,x(s))|≤ci|x|+|gi(s,0)||gi(s,x(s))|≤ci|x|+ρi. |
Let x1,x2 be two solutions of the integral Eq (1.2), then
|x1(t)−x2(t)|=|n∑i=1fi(t,x1(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x1(ψi(s)))ϕ′(s)ds)−n∑i=1fi(t,x2(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x2(ψi(s)))ϕ′(s)ds)+n∑i=1fi(t,x2(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x1(ψi(s)))ϕ′(s)ds)−n∑i=1fi(t,x2(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x1(ψi(s)))ϕ′(s)ds)|≤n∑i=1|fi(t,x1(ηi(t)))−fi(t,x2(ηi(t)))||ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x1(ψi(s)))ϕ′(s)ds|+n∑i=1|fi(t,x2(ηi(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x1(ψi(s)))−gi(s,x2(ψi(s)))|ϕ′(s)ds≤n∑i=1Li|x1(ηi(t))−x2(ηi(t))|(ki+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x1(ψi(s)))|ϕ′(s)ds)+n∑i=1hi∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)ci|x1(ψi(s))−x2(ψi(s))|ϕ′(s)ds≤n∑i=1Li|x1(ηi(t))−x2(ηi(t))|(ki+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)(ρi+ci|x1(ψi(s))|)ϕ′(s)ds)+n∑i=1hi∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)cisups∈J|x1(s)−x2(s)|ϕ′(s)ds≤n∑i=1Li|x1(t)−x2(t)|(ki+ρi+cirΓ(αi+1))+n∑i=1hici||x1−x2||Γ(αi+1)≤n∑i=1Li||x1−x2||(ki+ρi+cirΓ(αi+1))+n∑i=1hici||x1−x2||Γ(αi+1)≤n∑i=1(Li(ki+ρi+cirΓ(αi+1))+hiciΓ(αi+1))||x1−x2||. |
Then
(1−n∑i=1(Li(ki+ρi+cirΓ(αi+1))+hiciΓ(αi+1)))||x1−x2||≤0. |
Since n∑i=1(Li(ki+ρi+cirΓ(αi+1))+hiciΓ(αi+1))<1, then x1(t)=x2(t).
Firstly, we discuss the continuous dependence of the unique solution of the Eq (1.2) on the delays functions ηi and ψi.
Definition 1. The solutions of the quadratic functional integral Eq (1.2) depends continuously on the delay functions ηi and ψi if ∀ϵ>0,∃δ,σ>0, such that
|ηi(t)−η∗i(t)|≤δ and|ψi(t)−ψ∗i(t)|≤σ⇒‖x−x∗‖≤ϵ,i=1,2,...,n. |
Theorem 3.2. Let the assumptions of Theorem 3.1 be satisfied, then the solution of the functional integral Eq (1.2) depends continuously on the delay functions ηi and ψi.
Proof. Let δ>0 and σ>0 be given such that |ηi(t)−η∗i(t)|≤δ and |ψi(t)−ψ∗i(t)|≤σ ∀t∈J, then for
x∗(t)=n∑i=1fi(t,x∗(η∗i(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x∗(ψ∗i(s)))ϕ′(s)ds),t∈J,αi>0, |
|x(t)−x∗(t)|=|n∑i=1fi(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds)−n∑i=1fi(t,x∗(η∗i(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x∗(ψ∗i(s)))ϕ′(s)ds)+n∑i=1fi(t,x(η∗i(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds)−n∑i=1fi(t,x(η∗i(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds)+n∑i=1fi(t,x(η∗i(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x∗(ψ∗i(s)))ϕ′(s)ds)−n∑i=1fi(t,x(η∗i(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x∗(ψ∗i(s)))ϕ′(s)ds)|≤n∑i=1|fi(t,x(ηi(t)))−fi(t,x(η∗i(t)))|(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψi(s)))|ϕ′(s)ds)+n∑i=1|fi(t,x(η∗i(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψi(s)))−gi(s,x∗(ψ∗i(s)))|ϕ′(s)ds+n∑i=1|fi(t,x(η∗i(t)))−fi(t,x∗(η∗i(t)))|.(|ai(t)|+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x∗(ψ∗i(s)))|ϕ′(s)ds) |
≤n∑i=1Li|x(ηi(t))−x(η∗i(t))|(|ai(t)|+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)(ρi+ci|x(ψi(s))|)ϕ′(s)ds)+n∑i=1|fi(t,x(η∗i(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψi(s)))−gi(s,x(ψ∗i(s)))|ϕ′(s)ds+n∑i=1|fi(t,x(η∗i(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψ∗i(s)))−gi(s,x∗(ψ∗i(s)))|ϕ′(s)ds+n∑i=1Li|x(η∗i(t))−x∗(η∗i(t))|(ki+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)(ρi+ci|x∗(ψ∗i(s))|)ϕ′(s)ds)≤n∑i=1Li|ηi(t)−η∗i(t)|(ki+ρi+cirΓ(αi+1))+n∑i=1hiciΓ(αi+1)supt∈J|ψi(t)−ψ∗i(t)|+n∑i=1hiciΓ(αi+1)supt∈J|x(t)−x∗(t)|+n∑i=1Lisupt∈J|x(t)−x∗(t)|(ki+ρi+cirΓ(αi+1))≤n∑i=1Liδ(ki+ρi+cirΓ(αi+1))+n∑i=1hiciΓ(αi+1)σ+n∑i=1(Liki+hici+Li(ρi+cir)Γ(αi+1))||x−x∗||. |
Then
[1−n∑i=1(kiLi+hici+Li(ρi+cir)Γ(αi+1))]||x−x∗||≤n∑i=1(Liδ(ki+ρi+cirΓ(αi+1))+hiciΓ(αi+1)σ)||x−x∗||≤n∑i=1(Liδ(ki+ρi+cirΓ(αi+1))+hiciΓ(αi+1)σ).[1−n∑i=1(Liki+hici+Li(ρi+cir)Γ(αi+1))]−1 |
Next, we investigate the continuous dependence of the unique solution of the Eq (1.2) on the functions fi and gi.
Definition 2. The solutions of the quadratic functional integral Eq (1.2) depends continuously on the functions fi and gi if ∀ϵ>0,∃δ,σ>0, such that
|fi(t,x)−f∗i(t,x)|≤δ and|gi(t,x)−g∗i(t,x)|≤σ⇒‖x−x∗‖≤ϵ,i=1,2,...,n. |
Theorem 3.3. Let the assumptions of Theorem 3.1 be satisfied, then the solution of the functional integral equation (1.2) depends continuously on the functions fi and gi.
Proof. Let δ>0 and σ>0 be given such that |fi(t,x)−f∗i(t,x)|≤δ and |gi(t,x)−g∗i(t,x)|≤σ ∀t∈J, then for
x∗(t)=n∑i=1f∗i(t,x∗(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)g∗i(s,x∗(ψi(s)))ϕ′(s)ds),t∈J,αi>0, |
|x(t)−x∗(t)|=|n∑i=1fi(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds)−n∑i=1f∗i(t,x∗(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)g∗i(s,x∗(ψi(s)))ϕ′(s)ds)+n∑i=1f∗i(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds)−n∑i=1f∗i(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds)+n∑i=1f∗i(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)g∗i(s,x∗(ψi(s)))ϕ′(s)ds)−n∑i=1f∗i(t,x(ηi(t)))(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)g∗i(s,x∗(ψi(s)))ϕ′(s)ds)|≤n∑i=1|fi(t,x(ηi(t)))−f∗i(t,x(ηi(t)))|(ai(t)+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψi(s)))|ϕ′(s)ds)+n∑i=1|f∗i(t,x(ηi(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψi(s)))−g∗i(s,x∗(ψi(s)))|ϕ′(s)ds+n∑i=1|f∗i(t,x(ηi(t)))−f∗i(t,x∗(ηi(t)))|(|ai(t)|+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|g∗i(s,x∗(ψi(s)))|ϕ′(s)ds)≤n∑i=1δ(|ai(t)|+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)(ρi+ci|x(ψi(s))|)ϕ′(s)ds)+n∑i=1|f∗i(t,x(ηi(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|gi(s,x(ψi(s)))−g∗i(s,x(ψi(s)))|ϕ′(s)ds+n∑i=1|f∗i(t,x(ηi(t)))|∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)|g∗i(s,x(ψi(s)))−g∗i(s,x∗(ψi(s)))|ϕ′(s)ds+n∑i=1Li|x(ηi(t))−x∗(ηi(t))|(|ai(t)|+∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)(ρi+ci|x∗(ψi(s))|)ϕ′(s)ds)≤n∑i=1δ(ki+ρi+cirΓ(αi+1))+n∑i=1hiσΓ(αi+1)+n∑i=1hi∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)ci|x(ψi(s))−x∗(ψi(s))|ϕ′(s)ds+n∑i=1Li|x(ηi(t))−x∗(ηi(t))|(ki+ρi+cirΓ(αi+1)) |
≤n∑i=1δ(ki+ρi+cirΓ(αi+1))+n∑i=1hiσΓ(αi+1)+n∑i=1hiciΓ(αi+1)||x−x∗||+n∑i=1Li||x−x∗||(ki+ρi+cirΓ(αi+1)). |
Then
[1−n∑i=1(Liki+hici+Li(ρi+cir)Γ(αi+1))]||x−x∗||≤n∑i=1[δ(ki+ρi+cirΓ(αi+1))+hiσΓ(αi+1)]||x−x∗||≤n∑i=1[δ(ki+ρi+cirΓ(αi+1))+hiσΓ(αi+1)].[1−n∑i=1(Liki+hici+Li(ρi+cir)Γ(αi+1))]−1 |
● In case ai=0, the operators Bi have the following form
(Bix)(t)=∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds. |
Then we get the n-term quadratic integral equation of fractional order
x(t)=n∑i=1fi(t,x(ηi(t))).∫t0(ϕ(t)−ϕ(s))αi−1Γ(αi)gi(s,x(ψi(s)))ϕ′(s)ds,t∈J. | (4.1) |
By a simple calculation we can verify that the operators Bi satisfy the assumptions of Theorem 1.1 and hence the quadratic integral Eq (4.1) has a solution in C. For ϕ(t)=t in Eq (4.1), we obtain the quadratic equation which is studied in [25].
● Taking n=1 we obtain the quadratic integral equation
x(t)=f(t,x(ψ1(t)))[a(t)+∫t0(ϕ(t)−ϕ(s))α−1Γ(α)g(s,x(ψ2(s)))ϕ′(s)ds] | (4.2) |
and putting ϕ(t)=t, then we have the same result studied in [27].
● Letting αi→1,i=1,2. Then we have the n- term quadratic integral equation
x(t)=n∑i=1fi(t,x(ηi(t)))(ai(t)+∫t0gi(s,x(ψi(s)))ds),t∈J |
and for n=2, we have
x(t)=f1(t,x(η1(t)))(a1(t)+∫t0g1(s,x(ψ1(s)))ds)+f2(t,x(η2(t)))(a2(t)+∫t0g2(s,x(ψ2(s)))ds). |
Therefore we obtain the same result obtained in [29] when ai=0,i=1,2 and ηi(t)=ψi(t)=t.
● Letting n=2 we obtain the two term quadratic integral equation
x(t)=f1(t,x(η1(t)))∫t0(ϕ(t)−ϕ(s))α1−1Γ(α1)g1(s,x(ψ1(s)))ϕ′(s)ds+f2(t,x(η2(t)))∫t0(ϕ(t)−ϕ(s))α2−1Γ(α2)g2(s,x(ψ2(s)))ϕ′(s)ds,t∈J. | (4.3) |
● Taking ϕ(t)=t,η1(t)=η2(t)=t,ψ1(t)=ψ2(t)=t and f1(t,x)=1 in (4.3), then we get the hybrid differential equation of fractional order
RDα2(x(t)−Iα1g1(t,x(t))f2(t,x(t)))=g2(t,x(t)),t∈J, |
where RDα2 is the Riemann-Liouville fractional derivative of order α2∈(0,1).
● Taking α2→1,g1(t,x)=0,ϕ(t)=t,η1(t)=η2(t)=t,ψ1(t)=ψ2(t)=t and f1(t,x)=1 in (4.3) we have the initial value problem
(x(t)f2(t,x(t)))′=g2(t,x(t)),t∈J,x(0)=x0∈R. |
Example 1. For n=2, consider the nonlinear quadratic integral equation
x(t)=f1(t,x(t))(2t+sin(t)10+∫t0(t−s)12−1Γ(12)g1(s,x(s))ds)+f2(t,x(t))(1+∫t0(t−s)14−1Γ(14)g2(s,x(s))ds),t∈[0,1], |
where
f1(t,x(t))=1+cos(t)x(t)1+t,f2(t,x(t))=√t+3+|x(t)|g1(t,x(t))=sin(t)+3|x(t)|10(0.1+|x(t)|),g2(t,x(t))=1+2t5+12.|x(t)|1+|x(t)|a1(t)=2t+sin(t)10,a2(t)=1. |
We can easily verify that the functions f1,f2,g1 and g2 satisfy all assumptions of Theorem 2.1. Then h1=h2=1,b1=12,b2=310, which implies that b1h1Γ(α1+1)+b2h2Γ(α2+1)=0.3Γ(12+1)+0.5Γ(14+1)=0.898437<1. Moreover, we get r>0.
Many previous papers have discussed quadratic integral equations of fractional order by using different techniques in several classes of functions [3,18,30,31,32,33,34].
We have investigated existence and uniqueness of the solutions for a multi-term ϕ-quadratic integral equation of fractional order in C(J,R). The giving problem is converted into an analogous fixed point problem which is solved using typical functional analysis tools to prove our results. As a pursuit of this, sufficient conditions are given for the existence of solutions to that singular quadratic integral equation. Next, the continuous dependence of the solution on some functions has been proved. Finally, some particular cases, remarks and example to validate our results.
The researchers would like to thank the Deanship of Scientific Research, Qassim University for funding the publication of this project. The authors are thankful to the referees for remarks and suggestions for the improvement of this paper.
All authors declare no conflicts of interest in this paper.
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