Citation: A. M. A. El-Sayed, Sh. M. Al-Issa. Monotonic solutions for a quadratic integral equation of fractional order[J]. AIMS Mathematics, 2019, 4(3): 821-830. doi: 10.3934/math.2019.3.821
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It is well-known that a useful mathematical tool for physical investigation and description of non-local and anomalous diffusion is fractional calculus, which is that a branch of mathematical analysis dealing with pseudo-differential operators interpreted as integrals and derivatives of non-integer order (see[1,16,21,22]). For example, quadratic integral equations are often applicable in the theory of radiative transfer, of the theory of kinetic gases, the theory of neutron transport and in the theory of traffic. The quadratic integral equation can be very often encountered in many applications (see[6,7,9]).
Recently, the existence of positive monotonic continuous and integrable solutions of the mixed type integral inclusion
x(t)∈p(t)+∫10k(t,s)F1(s,Iβf2(s,x(s))ds,t∈[0,1],β>0 | (1.1) |
has been studied in [13,15] by using Schauder's and nonlinear alternative of Leray-Shauder type fixed-point Theorem. Also, the existence of integrable solution for the nonlinear quadratic integral equation
x(t)=a(t)+g(t,x(t))∫t0k(t,s)f(s,x(s))ds,t∈[0,1] | (1.2) |
has been proven in [12] by using Lusin and Dragoni theorems and applying Schauder Tychonoff fixed-point Theorem.
Here we are concerned with the mixed type nonlinear integral equation of fractional order
x(t)=p(t)+h(t,x(t))∫t0k(t,s)(f1(s,Iαf2(s,x(s)))+g1(s,Iβg2(s,x(s))))ds,t∈[0,1],α,β>0 | (1.3) |
which is more complicated than the equation assumed in [12]. We will use the technique associated with measures of noncompactness to show a global existence theorem for a positive nondecreasing integrable solution of equation (1.3), where the functions f1, f2, g1 and g2 satisfy Carathéodory condition.
Moreover, the existence of at least one integrable solution of the fractional-order quadratic integral equation
x(t)=p(t)+h(t,x(t))Iθ(f1(s,Iαf2(s,x(s))+g1(s,Iβg2(s,x(s))))),t∈[0,1],α,β>0 | (1.4) |
will be studied. Let us mention the result obtained for h(t,x(t))=1 in integral equation (1.4) extend those obtained in the paper [2]. Also, the results concerning the existence of monotonic positive integrable solution of the nonlinear functional equation
x(t)=f1(t,Iαf2(t,x(t))+g1(t,Iβg2(t,x(t)), |
will be given as a special case, which generalized the results proved in [4,14].
In this section, we introduce notations, definitions, and preliminary facts that are used throughout this paper.
Let L1=L1(I) be the class of Lebesgue integrable function on the interval I=[a,b], where 0≤a<b<∞, with the standard norm
‖x‖=∫ba|x(t)|dt. |
Definition 2.1. The Riemann-Liouville fractional integral of the function f(.)∈L1(I) of order α∈R+ is defined by (cf. [18,19,22])
Iαaf(t)=∫ta(t−s)α−1Γ(α)f(s)d(s). |
For the properties of the fractional order integral (see [17,18]).
Definition 2.2. The Caputo fractional derivative Dα of order α∈(a,b] of the absolutely continuous function g is defined as (see [10,19,20,22])
Dαag(t)=I1−αaddtg(t),t∈[a,b]. |
Now, let E denote an arbitrary 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.
The measure of weak noncompactness defined by De Blasi [3,11] is given by
β(X)=inf(r>0;there exists a weakly compact subsetYofEsuch thatX⊂Y+Br). | (2.1) |
The function β(X) possesses several useful properties which may be found in De Blasi's paper [11]. The convenient formula for the function β(X) in L1 was given by Appell and De Pascale (see [3]) as follows:
β(X)=limϵ→0(supx∈X(sup[∫D|x(t)|dt:D⊂[a,b],measD≤ϵ])), | (2.2) |
where the symbol meas D stands for Lebesgue measure of the set D.
Next, we shall also use the notion of the Hausdorff measure of noncompactness χ (see [4]) defined by
χ(X)=inf(r>0;there exist a finite subset Y of E such thatX⊂Y+Br). | (2.3) |
In the case when the set X is compact in measure, the Hausdorff and De Blasi measures of noncompactness will be identical. Namely, we have the following (see [3,11]).
Theorem 2.1. Let X be an arbitrary nonempty bounded subset of L1. If X is compact in measure, then β(X)=χ(X).
Now, we will recall the fixed point theorem due to Banaś [8].
Theorem 2.2. Let Q be a nonempty, bounded, closed, and convex subset of E, and let T: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(TX)≤αχ(X) for any nonempty subset X of Q. Then, T has at least one fixed point in the set Q.
In the sequel, we will need some criteria for compactness in measure; the complete description of compactness in measure was given in Banaś [4], but the following sufficient condition will be more convenient for our purposes (see [4]).
Theorem 2.3. Let X be a bounded subset of L1. Assume that there is a family of measurable subsets (Ωc)0≤c≤b−a of the interval (a,b) such that meas Ωc=c. If 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.
To facilitate our discussion, let us first state the following assumptions:
(1) the function p:[0,1]→R+ is integrable and nondecreasing on [0,1].
(2) h:[0,1]×R+→R+, satisfies Carathéodory condition i.e., h is measurable in t for any x∈R+ and continuous in x for almost all t∈[0,1]. There exists a function m(t)∈L1 such that
|h(t,x)|≤m(t). |
Moreover it is nondecreasing in the two arguments.
(3) fi:[0,1]×R+→R+, and gi:[0,1]×R+→R+,i=1,2 satisfy Carathéodory condition i.e., fi,gi are measurable in t for any x∈R+ and continuous in x for almost all t∈[0,1].
There exist four functions t→ai(t),t→bi(t),t→ci(t),andt→di(t). such that
|fi(t,x)|≤ai(t)+bi(t)|x|,i=1,2∀t∈[0,1]andx∈R |
and
|gi(t,x)|≤ci(t)+di(t)|x|,i=1,2∀t∈[0,1]andx∈R |
where ai(.),ci(.)∈L1, and bi(.), di(.) are measurable and bounded.
(4) k:[0,1]×R+→R+ is measurable with respect to both variables and the integral operator K defined by
(Kx)(t)=∫t0k(t,s)x(s)ds,t∈[0,1] |
map nondecreasing positive function L1 into itself and such that
∫t0k(t,s)m(t)dt<M,s∈[0,1]. |
Moreover, it is nondecreasing in the first argument.
For the existence of at least one nondecreasing L1−positive solution of a mixed type integral equation (1.3) we have the following theorem.
Theorem 3.1. Let the assumptions (1)–(4) be satisfied and assume that Mb1b2Γ(α+1)+Md1d2Γ(β+1)<1, then equation (1.3) has at least one solution x∈L1 which is nondecreasing on the interval [0,1].
Proof. Firstly, for t1,t2∈[0,1], t1<t2 and x(t1)≤x(t2), we have
x(t1)=p(t1)+h(t1,x(t1))∫t0k(t1,s)(f1(s,Iαf2(s,x(s)))+g1(s,Iβg2(s,x(s))))ds≤P(t2)+h(t2,x(t2))∫t0k(t2,s)(f1(s,Iαf2(s,x(s)))+g1(s,Iβg2(s,x(s))))ds=x(t2). |
This implies that, if the solution of the integral equation (1.3) exists, then it is nondecreasing on [0,1].
Let the operator T be defined by the formula
(Tx)(t)=p(t)+h(t,x(t))∫t0k(t,s)(f1(s,Iαf2(s,x(s)))+g1(s,Iβg2(s,x(s))))ds |
Let x∈L1, then by assumptions (1)–(4) we find that
|(Tx)(t)|=|p(t)|+|h(t,x(t))|∫t0k(t,s)|f1(s,Iαf2(s,x(s)))+g1(s,Iβg2(s,x(s)))|ds≤|p(t)|+m(t)∫t0k(t,s)(|f1(s,Iαf2(s,x(s)))|+|g1(s,Iβg2(s,x(s)))|)ds‖Tx‖=∫10|(Tx)(t)|dt≤∫10|p(t)|dt+∫10m(t)∫t0k(t,s)|f1(s,∫s0(s−τ)α−1Γ(α)f2(τ,x(τ))dτ|dsdt+∫10m(t)∫t0k(t,s)|g1(s,∫s0(s−τ)β−1Γ(β)g2(τ,x(τ))dτ|dsdt≤‖p‖+∫10m(t)∫t0k(t,s)(a1(s)+b1(s)|∫s0(s−τ)α−1Γ(α)f2(τ,x(τ))dτ|)dsdt+∫10m(t)∫t0k(t,s)(c1(s)+d1(s)|∫s0(s−τ)β−1Γ(β)f2(τ,x(τ))dτ|)dsdt≤‖p‖+∫10∫1sk(t,s)m(t)dt(a1(s)+b1(s)|∫s0(s−τ)α−1Γ(α)f2(τ,x(τ))dτ|)dsdt+∫10∫1sk(t,s)m(t)dt(c1(s)+d1(s)|∫s0(s−τ)β−1Γ(β)f2(τ,x(τ))dτ|)dsdt≤‖p‖+M∫10|a1(s)|ds+M∫10|b1(s)|∫s0(s−τ)β−1Γ(α)|f2(τ,x(τ))|dτds+M∫10|c1(s)|ds+M∫10|d1(s)|∫s0(s−τ)β−1Γ(β)|f2(τ,x(τ))|dτds≤‖p‖+M‖a1‖+Mb1∫10∫s0(s−τ)α−1Γ(α)[a2(τ)+b2(τ)|x(τ)|]dτds+M‖c1‖+Md1∫10∫s0(s−τ)β−1Γ(β)[c2(τ)+d2(τ)|x(τ)|]dτds≤‖p‖+M‖a1‖+Mb1∫10∫1τ(s−τ)α−1Γ(α)a2(τ)dsdτ+Mb1∫10∫1τ(s−τ)α−1Γ(α)|b2(τ)||x(τ)|dsdτ+M‖c1‖+Md1∫10∫1τ(s−τ)β−1Γ(β)c2(τ)dsdτ+Md1∫10∫1τ(s−τ)β−1Γ(β)|d2(τ)||x(τ)|dsdτ≤‖p‖+M‖a1‖+Mb1∫10a2(τ)∫1τ(s−τ)α−1Γ(α)dsdτ+Mb1b2∫10|x(τ)|(1−τ)αΓ(α+1)dτ+M‖c1‖+Md1∫10c2(τ)∫1τ(s−τ)β−1Γ(β)dsdτ+Md1d2∫10|x(τ)|(1−τ)βΓ(β+1)dτ≤‖p‖+M‖a1‖+Mb1∫10a2(τ)(1−τ)αΓ(α+1)dτ+Mb1b2Γ(α+1)∫10|x(τ)|dτ+M‖c1‖+Md1∫10c2(τ)(1−τ)βΓ(β+1)dτ+Md1d2Γ(β+1)∫10|x(τ)|dτ≤‖p‖+M‖a1‖+Mb1Γ(α+1)∫10|a2(τ)|dτ+Mb1b2‖x‖Γ(α+1)+M‖c1‖+Md1Γ(β+1)∫10|c2(τ)|dτ+Md1d2‖x‖Γ(β+1)≤‖p‖+M‖a1‖+Mb1‖a2‖Γ(α+1)+Mb1b2‖x‖Γ(α+1)+Md1‖c2‖Γ(β+1)+Md1d2‖x‖Γ(β+1),≤‖p‖+M‖a1‖+M‖c1‖+Mb1‖a2‖+Mb1b2‖x‖Γ(α+1)+Md1‖c2‖+Md1d2‖x‖Γ(β+1) |
which gives
‖Tx‖≤‖p‖+M‖a1‖+M‖c1‖+Mb1‖a2‖+Mb1b2‖x‖Γ(α+1)+Md1‖c2‖+Md1d2‖x‖Γ(β+1) | (3.1) |
and proves that Tx∈L1. Moreover, the estimate (3.1) shows that the operator T maps the ball Br into itself, where
r=[‖p‖+M‖a1‖+M‖c1‖+Mb1‖a2‖Γ(α+1)+Md1‖c2‖Γ(β+1)][1−(Mb1b2Γ(α+1)+Md1d2Γ(β+1))]−1. |
Let Qr⊂Br consisting of all positive and nondecreasing functions on I. Clearly Qr is nonempty, bounded, closed and convex (see Banas [4], pp. 780). Now Qr is a bounded subset of L1 consisting of all positive and nondecreasing functions on [0,1], then Theorem 2.3 shows that Qr is compact in measure (see Lemma 2 in [5] pp. 63).
Thus the operator T maps Qr into itself, by using assumptions (1)–(4), the operator T is continuous on Qr, and the operator T transforms a positive and nondecreasing function into the function of the same type (see [5,23]).
In what follows we show that the operator T is a contraction with respect to the measure of weak noncompactness β. To do this let us fix ϵ>0 and X⊂Qr. Further, take a measurable subset D⊂[0,1] such that measD≤ϵ, then for any x∈X by our assumptions and using the same reasoning as in [4,5] we obtain
‖Tx‖L1(D)=∫D|(Tx)(t)|dt≤∫D|p(t)|dt+∫Dm(t)∫t0k(t,s)[a1(s)+b1(s)|∫s0(s−τ)α−1Γ(α)f2(τ,x(τ))dτ|dsdt+∫Dm(t)∫t0k(t,s)[c1(s)+d1(s)|∫s0(s−τ)β−1Γ(β)g2(τ,x(τ))dτ|dsdt≤‖p‖D+M‖a1‖D+Mb1∫Da2(τ)(1−τ)βΓ(β+1)dτ+Mb1b2Γ(β+1)∫D|x(τ)|dτ+M‖c1‖+Mb1∫Da2(τ)(1−τ)αΓ(α+1)dτ+Mb1b2Γ(α+1)∫D|x(τ)|dτ≤‖p‖D+M‖a1‖D+Mb1Γ(α+1)∫D|a2(τ)|dτ++Mb1b2Γ(α+1)∫D|x(τ)|dτ+M‖c1‖+Md1Γ(β+1)∫D|c2(τ)|dτ+Md1d2Γ(β+1)∫D|x(τ)|dτ,≤‖p‖D+M‖a1‖D+Mb1‖a2‖DΓ(α+1)+Mb1b2Γ(α+1)‖x‖D+M‖c1‖D+Md1‖c2‖DΓ(β+1)+Md1d2Γ(β+1)‖x‖D‖Tx‖L1(D)≤‖p‖D+M‖a1‖D+M‖c1‖D+Mb1‖a2‖DΓ(α+1)+Md1‖c2‖DΓ(β+1)+(Mb1b2Γ(α+1)+Md1d2Γ(β+1))‖x‖D. |
But
limϵ→0{sup{∫D|p(t)|dt:D⊂I,meas.D<ϵ}}=0 |
limϵ→0{sup{∫D|ai(t)|dt:D⊂I,meas.D<ϵ}}=0i=1,2 |
and
limϵ→0{sup{∫D|ci(t)|dt:D⊂I,meas.D<ϵ}}=0i=1,2. |
We obtain
β(Tx(t))≤(Mb1b2Γ(α+1)+Md1d2Γ(β+1))β(x(t)) |
and
β(TX)≤(Mb1b2Γ(α+1)+Md1d2Γ(β+1))β(X) | (3.2) |
where β is the De Blasi measure of weak noncompactness. Keeping in mind Theorem 2.1, we can write (3.2) in the form
χ(TX)≤(Mb1b2Γ(α+1)+Md1d2Γ(β+1))χ(X) |
where χ is the Hausdorff measure of noncompactness. Since (Mb1b2Γ(α+1)+Md1d2Γ(β+1))<1, it follows, from Theorems 2.2, that T is a contraction with respect to the measure of noncompactness χ and has at least one fixed point in Qr which proves that the nonlinear quadratic functional integral equation (1.3) has at least one positive nondecreasing solution x∈L1[0,1].
As particular cases of Theorem 3.1, we can obtain theorems on the existence of a positive and nondecreasing solutions belonging to the space L1.
Theorem 4.1. Let the assumptions of Theorem 3.1 be satisfied with k(t,s)=(t−s)θ−1Γ(θ), then the fractional-order quadratic integral equation
x(t)=p(t)+h(t,x(t))Iθ(f1(s,Iαf2(s,x(s))+g1(s,Iβg2(s,x(s))))),t∈[0,1],α,β>0 | (4.1) |
has at least one positive nondecreasing solution x∈L1.
Proof. From the properties of fractional order integral operator, we deduce that the operator
(Kx)(t)=∫t0(t−s)θ−1Γ(θ)x(s)ds,θ∈[0,1] |
satisfy the assumption (4) in Theorem 3.1, and the result follows from the results of Theorem 3.1.
Corollary 4.1.1. Under the assumptions of Theorem 3.1, with p(t)=0, h(t,x(t))=1 and letting θ→0 the nonlinear functional equation
x(t)=f1(t,Iαf2(t,x(t))+g1(t,Iβg2(t,x(t)) | (4.2) |
has at least one positive nondecreasing solution x∈L1.
Proof. Fractional-order quadratic integral equation (4.1) will be the functional equation (4.2) and the result follows from Theorem 4.1.
Finally, for the existence of a monotonic positive integrable solution of the nonlinear functional differential equation of fractional order
Dθx(t)=f1(t,Iαf2(t,x(t))+g1(t,Iβg2(t,x(t)),t∈(0,1]andI1−θx(t)|t=0=p | (5.1) |
where Dθ is the Riemann-Liouville fractional order derivative, we have the following theorem.
Theorem 5.1. Under the assumptions of Theorem 3.1, with p(t)=ptθ−1Γ(θ) and h(t,x(t))=1, the Cauchy type problem (5.1) has at least one positive nondecreasing integrable solution.
Proof. Integrating Cauchy problem (5.1) we obtain the integral equation
x(t)=ptθ−1Γ(θ)+∫t0(t−s)θ−1Γ(θ)[f1(s,Iαf2(s,x(s))+g1(s,Iβg2(s,x(s))]ds | (5.2) |
which, by Theorem 3.1, has the desired solution, operating by Dθ on equation (5.2) we obtain the problem (5.1). So the equivalence between problem (5.1) and integral equation (5.2) is proven and then the results follow from Theorem 3.1.
The authors thank the reviewers for their useful remarks on our paper.
All authors declare no conflicts of interest in this paper.
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