# Multiple Positive Solutions for Quadratic Integral Equations of Fractional Order

Authors: Hui-Sheng Ding; Man-Man Liu; Juan J. Nieto
Journal: Journal of Function Spaces
 (2017)

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## Abstract

In this paper, the existence of multiple positive solutions for a class of quadratic integral equation of fractional order is obtained, by utilizing Avery-Henderson and Leggett-Williams multiple fixed point theorems on cones. An example is given to illustrate the applicability of our results. We believe that this is a first result concerning the existence of multiple solutions for such quadratic integral equation of fractional order.


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## Body

## 1. Introduction and Preliminaries

Recently, there has been great interest for many authors to study quadratic functional integral equations, which has become one of the most attractive and interesting research areas of integral equations and functional integral equations. There is large literature on this topic. We refer the reader to [1–6] for some of very recent results. In fact, as noted in some earlier literature (see, e.g., [5] and references therein), the nonlinear quadratic functional integral equations have been applied to, for example, the theory of radiative transfer, kinetic theory of gases, the theory of neutron transport, the traffic theory, plasma physics, and numerous branches of mathematical physics.On the other hand, due to the fact that fractional differential and integral equations have recently been extensively applied in various areas of engineering, mathematics, physics and bioengineering, and other applied sciences, there has been a significant development in fractional integral equations in recent years. Many authors especially have focused on the existence and qualitative properties of solutions for quadratic integral equations of fractional order such as(1)xt=ft,xt+gt,xt∫0tt-sα-1Γαhs,xsds,α>0,and several important variants of (1), and obtain substantial results on this topic (cf. [1–3, 5–9]). However, to the best of our knowledge, it seems that there are no results concerning the existence of multiple solutions for (1) and its variants. That is the main goal of this work. It seems that this is a first result concerning the existence of multiple solutions for such quadratic integral equation of fractional order.Stimulated by the above works, we aim to investigate the existence ofmultiple positive solutions for the following fractional order quadratic integral equation:(2)xt=ft,xλt+gt,xμt∫0tt-sα-1Γαhs,xνsds,t∈0,1,α>0.See Section 2 for the hypotheses on the involved functions.Throughout the rest of this paper, if there is no special statement, we denote byR the set of real numbers, and by Lip(R×R,R+) the set of all functions f:R×R→R+ satisfying that there exists a constant Lf>0 such that (3)ft,x-ft,y≤Lfx-y,t∈R,x,y∈R.Next, let us recall some notations about cones and two fixed point theorem. For more details, we refer the reader to [10, 11].LetX be a real Banach space. A closed convex set K in X is called a cone if the following conditions are satisfied:(i)
Ifx∈K, then λx∈K for any λ≥0;(ii)
Ifx∈K and -x∈K, then x=0. A nonnegative continuous functional ψ is said to be a concave on K if ψ is continuous and (4)ψμx+1-μy≥μψx+1-μψy,x,y∈K,μ∈0,1.Letting c1,c2,c3 be three positive constants and ϕ be a nonnegative continuous functional on K, we denote (5)Kc1=y∈K:y<c1,Kc1¯=y∈K:y≤c1,Kϕ,c1≔x∈K:ϕx<c1,Kϕ,c1¯≔x∈K:ϕx≤c1,∂Kϕ,c1≔x∈K:ϕx=c1,Kϕ,c2,c3=y∈K:c2≤ϕy,y≤c3. In addition, we say that ϕ is increasing on K if ϕ(x)≥ϕ(y) for all x,y∈K with x-y∈K.The following two theorems are the well-known Avery-Henderson multiple fixed point theorem and Leggett-Williams multiple fixed point theorem, respectively.Lemma 1 (see [10]).
LetK be a cone in a real Banach space X, α and φ be two increasing, nonnegative, and continuous functionals on K, and ρ be a nonnegative continuous functional on K with ρ(0)=0 such that, for some c>0 and M>0, (6)φx≤ρx≤αx,x≤Mφx,x∈Kφ,c¯.Moreover, suppose that there exists a completely continuous operator Φ:K(φ,c)¯→K and 0<a<b<c such that (7)ρλx≤λρx,0≤λ≤1,x∈∂Kρ,b,and(i)
φ(Φx)>c, for all x∈∂K(φ,c);(ii)
ρ(Φx)<b, for all x∈∂K(ρ,b);(iii)
K(α,a)≠∅, and α(Φx)>a, for all x∈∂K(α,a).Then Φ has at least two fixed points x1,x2 belonging to K(φ,c)¯ such that (8)a<αx1,ρx1<b,b<ρx2,φx2<c.Lemma 2 (see [11]).
LetK be a cone in a real Banach space X,  c4 be a positive constant, Φ:Kc4¯→Kc4¯ be a completely continuous mapping, and ψ be a concave nonnegative continuous functional on K with ψ(u)≤u for all u∈Kc4¯. Suppose that there exist three constants c1,c2,c3 with 0<c1<c2<c3≤c4 such that (i)
{u∈K(ψ,c2,c3):ψ(u)>c2}≠⌀, and ψ(Φu)>c2 for all u∈K(ψ,c2,c3);(ii)
Φu<c1 for all u∈Kc1¯;(iii)
ψ(Φu)>c2 for all u∈K(ψ,c2,c4) with Φu>c3.Then Φ has at least three fixed points u1,u2,u3 in Kc4¯. Furthermore, u1<c1<u2, and ψ(u2)<c2<ψ(u3).


## 2. Main Results

Firstly, we list some assumptions:(H1)
f,g,h∈C([0,1]×R+,R+) and f,g∈Lip(R×R,R+).(H2)
λ,μ,ν∈C1([0,1],R+). Moreover, λ′(t),μ′(t)>0 and ν′(t)>1.(H3)
inft∈[0,1],x≥0f(t,x)>0, supt∈[0,1],x≥0f(t,x)<+∞, supt∈[0,1],x≥0g(t,x)<+∞, and (9)limr→+∞⁡supt∈0,1,x∈0,r⁡ht,xr=0.(H4)
There exist0<b<c such that (10)infx≥0⁡f1,x+infx≥0⁡g1,xinfν-11≤s≤1⁡hs,c·1-ν-11ααΓα>c,supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,b⁡ht,xαΓα<b.By a solution of (2) we mean a function x∈C[0,T] satisfying the equation, where (11)T=max⁡maxt∈0,1⁡λt,maxt∈0,1⁡μt,maxt∈0,1⁡νt.Now, we are ready to present our main result.Theorem 3.
Let (H1)–(H4) hold. Then, there existsL∗>0 such that (2) has at least two nonnegative solutions provided that Lf<1 and Lg<L∗.Proof.
By using (H3), we can chooseσ∈(0,1) satisfying(12)inft∈0,1,x≥0⁡ft,x≥σsupt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,x·supt∈0,1,x∈0,c/σ⁡ht,xαΓα.Let (13)K=x∈C0,T:xt=x1for  everyt∈1,T,mint∈0,1⁡xt≥σx. It is not difficult to verify that K is a cone in C[0,T]. Let (14)φu=u1,ρu=αu=u=maxt∈0,T⁡ut=maxt∈0,1⁡ut,u∈K. Obviously, φ,  ρ, and α are increasing, nonnegative, and continuous functionals on K with ρ(0)=0. Moreover, we have (15)u≤1σmint∈0,1⁡ut≤σ-1φu,ρλu=λρu,for all u∈K and 0≤λ≤1. We divide the remaining proof by three steps.
Step  1. Let (16)Ω=x∈C0,T:xt=x1for  everyt∈1,T.For every y∈K(φ,c)¯, define an operator Ay on Ω by (17)Ayxt=ft,xλt+gt,xμt∫0tt-sα-1Γαhs,yνsds,x∈Ω,t∈0,1,Ayx1,x∈Ω,t∈1,T. It is easy to see that Ay(Ω)⊂Ω.
Forx,z∈Ω and t∈[0,1], there holds (18)Ayxt-Ayzt≤ft,xλt-ft,zλt+gt,xμt-gt,zμt·∫0tt-sα-1Γαhs,yνsds≤Lfx-z+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓαx-z, which yields that (19)Ayx-Ayz=maxt∈0,T⁡Ayxt-Ayzt=maxt∈0,1⁡Ayxt-Ayzt≤Lfx-z+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓαx-z. Let (20)L∗=αΓα1-Lfsupt∈0,1,x∈0,c/σ⁡ht,x. Then, noting that Lf<1,  Ay has a unique fixed point xy∈Ω provided that Lg<L∗.
Step  2. Now, define an operator A on K(φ,c)¯ by (21)Ayt=xyt=ft,xyλt+gt,xyμt∫0tt-sα-1Γαhs,yνsds,y∈Kφ,c¯,t∈0,1,Ay1,y∈Kφ,c¯,t∈1,T. Noting that, for every t∈[0,1] and y∈K(φ,c)¯, by (12), we have (22)Ayt≥inft∈0,1,x≥0⁡ft,x≥σsupt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,x·supt∈0,1,x∈0,c/σ⁡ht,xαΓα≥σAy, which yields that AK(φ,c)¯⊂K.
Next, let us show thatA:K(φ,c)¯→K is completely continuous. Let yn→y in K(φ,c)¯. For t∈[0,1], we have (23)Aynt-Ayt=ft,Aynλt-ft,Ayλt+gt,Aynμt∫0tt-sα-1Γαhs,ynνsds-gt,Ayμt∫0tt-sα-1Γαhs,yνsds≤LfAynλt-Ayλt+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓαAynμt-Ayμt+supt∈0,1,x≥0gt,x∫0tt-sα-1Γαhs,ynνs-hs,yνsds≤Lf+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓαAyn-Ay+supt∈0,1,x≥0⁡gt,x∫0tt-sα-1Γαhs,ynνs-hs,yνsds, which yields that(24)Ayn-Ay=maxt∈0,T⁡Aynt-Ayt=maxt∈0,1⁡Aynt-Ayt≤αΓαsupt∈0,1,x≥0⁡gt,x·supt∈0,1⁡∫0tt-sα-1/Γαhs,ynνs-hs,yνsdsαΓα-αΓαLf-Lgsupt∈0,1,x∈0,c/σ⁡ht,x,where (25)αΓα>αΓαLf+Lgsupt∈0,1,x∈0,c/σ⁡ht,x since Lg<L∗. Combining (24) with the fact h is uniformly continuous on [0,1]×[0,c/σ] and yn(t) uniformly converges to y(t) on [0,T], we conclude that Ayn→Ay in K. It suffices to show AK(φ,c)¯ is precompact. It follows from the above proof that {(Ay)(t):y∈K(φ,c)¯} is uniformly bounded on [0,T]. For every y∈K(φ,c)¯ and t1,t2∈[0,1] with t1≤t2, we have (26)Ayt1-Ayt2≤ft1,Ayλt1-ft2,Ayλt2+gt1,Ayμt1∫0t1t1-sα-1Γαhs,yνsds-gt2,Ayμt2∫0t2t2-sα-1Γαhs,yνsds≤ft1,Ayλt1-ft2,Ayλt1+LfAyλt1-Ayλt2+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓαAyμt1-Ayμt2+supt∈0,1,x∈0,c/σ⁡ht,xαΓαgt1,Ayμt1-gt2,Ayμt1+supt∈0,1,x≥0⁡gt,x·supt∈0,1,x∈0,c/σ⁡ht,x∫0t1t1-sα-1Γαds-∫0t2t2-sα-1Γαds. On the other hand, we have (27)∫0t1t1-sα-1Γαds-∫0t2t2-sα-1Γαds≤∫0t1t1-sα-1Γα-t2-sα-1Γαds+∫t1t2t2-sα-1Γαds≤1Γα∫0t1sα-1-t2-t1+sα-1ds+1αΓαt2-t1α. By Lebesgue’s dominated convergence theorem, as t2-t1→0, (28)∫0t1sα-1-t2-t1+sα-1ds≤∫01sα-1-t2-t1+sα-1ds⟶0. Combining this with the fact that f,g are uniformly continuous on compact sets, we conclude that, for every ε>0, there exists δ′>0 such that, for all t1,t2∈[0,1] with t1≤t2 and |t1-t2|<δ′, and y∈K(φ,c)¯, there hold (29)ft1,Ayλt1-ft2,Ayλt1<ε3,supt∈0,1,x∈0,c/σ⁡ht,xαΓαgt1,Ayμt1-gt2,Ayμt1<ε3,supt∈0,1,x≥0⁡gt,x·supt∈0,1,x∈0,c/σ⁡ht,x∫0t1t1-sα-1Γαds-∫0t2t2-sα-1Γαds<ε3. Thus, we conclude that, for every ε>0, (30)Ayt1-Ayt2≤LfAyλt1-Ayλt2+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓαAyμt1-Ayμt2+ε holds for all t1,t2∈[0,1] with t1≤t2 and |t1-t2|<δ′, and y∈K(φ,c)¯, which yields that, for all t1,t2∈[0,1] with t1≤t2 and |t1-t2|<δ′, there holds (31)supy∈Kφ,c¯⁡Ayt1-Ayt2≤Lf·supy∈Kφ,c¯⁡Ayλt1-Ayλt2+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓα·supy∈Kφ,c¯⁡Ayμt1-Ayμt2+ε. Then, we have (32)limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayt1-Ayt2≤Lf·limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayλt1-Ayλt2+Lg·supt∈0,1,x∈0,c/σ⁡ht,xαΓα·limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayμt1-Ayμt2+ε, where (33)limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayt1-Ayt2≔infδ>0⁡sup0≤t1≤t2≤t1+δ≤1⁡supy∈Kφ,c¯⁡Ayt1-Ayt2,limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayλt1-Ayλt2,limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayμt1-Ayμt2are similarly defined. Noting that λ′(t),μ′(t)>0 for all t∈[0,1], we conclude (34)limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayt1-Ayt2=limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayλt1-Ayλt2=limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayμt1-Ayμt2, and so we have (35)limsupt2-t1→0⁡supy∈Kφ,c¯⁡Ayt1-Ayt2≤αΓαεαΓα-αΓαLf-Lgsupt∈0,1,x∈0,c/σ⁡ht,x. Thus, {(Ay)(t):y∈K(φ,c)¯} is equicontinuous on [0,1]. Then, it is not difficult to obtain that {(Ay)(t):y∈K(φ,c)¯} is equicontinuous on [0,T]. This proves that AK(φ,c)¯ is precompact, and thus A:K(φ,c)¯→K is completely continuous.
Step  3. It remains to verify the assumptions (i)–(iii) of Lemma 1. For every y∈∂K(φ,c), noting that y(1)=c and y(t)=c for all t∈[1,T], by (H4), we have (36)φAy=Ay1=f1,Ayλ1+g1,Ayμ1∫011-sα-1Γαhs,yνsds≥infx≥0⁡f1,x+infx≥0⁡g1,x∫ν-1111-sα-1Γαhs,cds≥infx≥0⁡f1,x+infx≥0⁡g1,xinfν-11≤s≤1⁡hs,c·1-ν-11ααΓα>c. For every y∈∂K(ρ,b), noting that y(t)≤b for all t∈[0,T], again by (H4), we have (37)ρAy=supt∈0,1⁡ft,Ayλt+gt,Ayμt∫0tt-sα-1Γαhs,yνsds≤supt∈0,1⁡supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,b⁡ht,x∫0tt-sα-1Γαds≤supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,b⁡ht,xαΓα<b. In addition, letting (38)a=12min⁡inft∈0,1,x≥0⁡ft,x,b, it is easy to see that the assumption (iii) of Lemma 1 holds.
Then, by Lemma1, we conclude that A has at least two fixed points x1,x2∈K(φ,c)¯. Thus, (39)xit=ft,xiλt+gt,xiμt∫0tt-sα-1Γαhs,xiνsds,t∈0,1,i=1,2; that is, x1,x2 are two nonnegative solutions for (2).Remark 4.
Due to the influence of fractional termt-sα-1 in (2), it becomes more difficult to verify the assumptions of Lemma 1. In addition, we have tried to consider the case of ν′(t)≤1, where it seems that the approach used here can not be applied to study (2). We leave it for further research.By using Lemma2, we can obtain the existence of three nonnegative solutions to (2).Corollary 5.
Let (H1)–(H4) hold and the first inequality in (H4) is strengthened to(40)infx≥0⁡f1,x+infx≥0⁡g1,xinfν-11≤s≤1,x≥c⁡hs,x·1-ν-11ααΓα>c.Then, there exists L∗>0 such that (2) has at least three nonnegative solutions provided that Lf<1 and Lg<L∗.Proof.
By using (H3), we can chooseρ∈(0,1) such that(41)inft∈0,1,x≥0⁡ft,x≥ρsupt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,x·supt∈0,1,x∈0,c/ρ⁡ht,xαΓα,(42)supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,c/ρ⁡ht,xαΓα≤cρ.Let T be the same as in Theorem 3 and (43)K=x∈C0,T:xt=x1for  everyt∈1,T,mint∈0,1⁡xt≥ρx. Moreover, denote Ω and Ay (y∈Kc4¯) be the same as in Theorem 3, and (44)c1=b,c2=c,c3=c4=cρ. Then, by a similar proof to Step 1 of Theorem 3, we can prove that Ay has a unique fixed point xy∈Ω provided that Lg<L∗, where (45)L∗=αΓα1-Lfsupt∈0,1,x∈0,c/ρ⁡ht,x.
Now, we can define(46)Ayt=xyt=ft,xyλt+gt,xyμt∫0tt-sα-1Γαhs,yνsds,y∈Kc4¯,t∈0,1,Ay1,y∈Kc4¯,t∈1,T. Analogously to the proof of Theorem 3, we can also show that A:Kc4¯→K is completely continuous. Let (47)ψu=u1,u∈K. Obviously, ψ is a concave nonnegative continuous functional on K and ψ(u)≤∥u∥. For every y∈Kc4¯, by (42), we have (48)Ay=supt∈0,1⁡ft,xyλt+gt,xyμt∫0tt-sα-1Γαhs,yνsds≤supt∈0,1⁡supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,c/ρ⁡ht,x∫0tt-sα-1Γαds≤supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,c/ρ⁡ht,xαΓα<cρ=c4, which means that A maps Kc4¯ to Kc4¯. Similarly, by (H4), we have (49)Ay≤supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,b⁡ht,xαΓα<b=c1,y∈Kc1¯;that is, the condition (ii) of Lemma 2 holds. Moreover, the condition (iii) of Lemma 2 follows from the definition of K and c3. It remains to show that the condition (i) of Lemma 2 holds. For every y∈K(ψ,c2,c3), noting that y(1)≥c2=c and thus y(t)≥c for all t∈[1,T], by (40), we have (50)ψAy=Ay1=f1,xyλ1+g1,xyμ1∫011-sα-1Γαhs,yνsds≥infx≥0⁡f1,x+infx≥0⁡g1,x∫ν-1111-sα-1Γαinfx≥c⁡hs,xds≥infx≥0⁡f1,x+infx≥0⁡g1,xinfν-11≤s≤1,x≥c⁡hs,x·1-ν-11ααΓα>c.
Then, by Lemma2, A has at least three fixed points in Kc4¯, and thus (2) has at least three nonnegative solutions.Next, we present an example to show the applicability of our results.Example 6.
Fort∈[0,1] and x≥0, let α=1/2: (51)λt=μt=t,νt=2t,ft,x=1+14sin2⁡tx,gt,x=1+βe-t2x2,ht,x=x,0≤x≤10,2γ-110x+2-2γ10,10<x<20,γx,x≥20, where β,γ>0 are two constants.
It is not difficult to show that (H1) holds withLf≤1/2 and Lg≤β2e-1/2. Moreover, it is easy to verify that (H2) and (H3) hold. Letting b=10 and c=20, we have (52)infx≥0⁡f1,x+infx≥0⁡g1,xinfν-11≤s≤1,x≥c⁡hs,x·1-ν-11ααΓα≥1+γ20·2π=1+γ40π>20=c, provided that γ>19π/40, and (53)supt∈0,1,x≥0⁡ft,x+supt∈0,1,x≥0⁡gt,xsupt∈0,1,x∈0,b⁡ht,xαΓα≤54+1+β10π/2<10=b, provided that β<1. Then, by Corollary 5, (2) has at least three nonnegative solutions provided that γ>19π/40 and β is sufficiently small.



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