# Positive Solutions Depending on Parameters for a Nonlinear Fractional System withp-Laplacian Operators

Authors: Chen Yang; Xiaolin Zhu
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

This paper considers a system of fractional differential equations involvingp-Laplacian operators and two parameters D0+α1φp1D0+β1ut+λft,ut,vt=0,0<t<1,D0+α2φp2D0+β2vt+μgt,ut,vt=0,0<t<1,u0=u1=u′0=u′1=0,D0+β1u0=0,D0+β1u1=b1D0+β1uη1,v0=v1=v′0=v′1=0,D0+β2v0=0,D0+β2v1=b2D0+β2vη2, where αi∈1,2, βi∈3,4, D0+αi and D0+βi are the standard Riemann-Liouville derivatives, φpis=spi−2s,pi>1, φpi−1=φqi, 1/pi+1/qi=1,ηi∈0,1,bi∈0,ηi1−αi/pi−1, i=1,2, and f,g∈C0,1×0,+∞×0,+∞,0,+∞ and λ and μ are two positive parameters. We obtain the existence and uniqueness of positive solutions depending on parameters for the system by utilizing a recent fixed point theorem. Furthermore, an example is present to illustrate our main result.


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## Body

## 1. Introduction

During the past several decades, many fractional problems with differential equations have been paid much attention, see [1–10] for example. Also, much attention has been focused on the existence of positive solutions for such equations, see [3–29] and the references therein. As we know, the p-Laplacian operator has very a important position in theoretical research and engineering applications. In 1945, to discuss turbulent flow in a porous medium, a basic mechanical problem, Leibenson [30] introduced a differential equation with a p-Laplacian operator:
(1)φpu′t′=ft,ut.Since then, there are many papers investigating differential equations withp-Laplacian operators. Recently, the study of fractional equations with a p-Laplacian operator has also gained plenty of attention, see [19, 20, 31–40] for instance. In [35], the authors studied a fractional equation with a p-Laplacian operator:
(2)−D0+αφpD0+βut=ft,ut,0<t<1,u0=u1=u′0=u′1=0,D0+βu0=0,D0+βu1=bD0+βuη,where α∈1,2, β∈3,4, D0+α and D0+β denote the Riemann-Liouville derivatives, b∈0,η1−α/p−1, and f∈C0,1×0,+∞,0,+∞. Based on Schauder’s fixed point theorem and by using the upper-lower solution method, they obtained the existence and uniqueness of solutions.Recently, fractional differential systems have been also studied by many people because of their great application value, see [5, 20, 23–30]. So, the results on fractional systems withp-Laplacian operator are many, see [11, 41–45]. For example, Rodica [41] discussed a fractional differential system:
(3)D0+α1φr1D0+β1ut+λft,ut,vt=0,0<t<1,D0+α2φr2D0+β2vt+μgt,ut,vt=0,0<t<1,uj0=0,j=0,⋯,n−2;D0+β1u0=0,D0+p1u1=∑i=1NaiD0+q1uξi,vj0=0,j=0,⋯,m−2;D0+β2v0=0,D0+p2v1=∑i=1MbiD0+q2vηi,where α1,α2∈0,1, β1∈n−1,n, β2∈m−1,m, n,m∈ℕ,n,m≥3, p1,p2,q1,q2∈ℝ, p1∈1,n−2, p2∈1,m−2, q1∈0,p1, q2∈0,p2, ξi,ai∈ℝ for all i=1,⋯,NN∈ℕ, 0<ξ1<⋯<ξN≤1, ηi,bi∈ℝ for all i=1,⋯,MM∈ℕ, 0<η1<⋯<ηM≤1, r1,r2>1, λ,μ>0, and f,g∈C0,1×0,∞×0,∞,0,∞. The existence of solutions was obtained via Guo-Krasnosel’skii’s fixed point theorem.From literature, we see that most results are the existence of solutions, but the uniqueness is scarce. Inspired by [34], we discuss the following system of fractional differential equations withp-Laplacian operators:
(4)D0+α1φp1D0+β1ut+λft,ut,vt=0,0<t<1,D0+α2φp2D0+β2vt+μgt,ut,vt=0,0<t<1,u0=u1=u′0=u′1=0,D0+β1u0=0,D0+β1u1=b1D0+β1uη1,v0=v1=v′0=v′1=0,D0+β2v0=0,D0+β2v1=b2D0+β2vη2,where αi∈1,2, βi∈3,4, D0+αi and D0+βi denote the standard Riemann-Liouville derivatives, φpis=spi−2s,pi>1, φpi−1=φqi, 1/pi+1/qi=1,ηi∈0,1,bi∈0,ηi1−αi/pi−1, i=1,2 and f,g∈C0,1×0,+∞×0,+∞,0,+∞, and λ and μ are two positive parameters. It should be pointed out, in [45], that Hao et al. investigated the existence of solutions for system (4) without considering the uniqueness. They used Guo-Krasnosel’skii’s fixed point theorem to get some existence results for positive solutions under different values of λ and μ. In this paper, based upon a recent fixed point theorem, we aim to present the existence and uniqueness of positive solutions for system (4) depending on fixed positive constants λ and μ. Our results can tell us that the unique positive solution exists in a product set and can be approximated by giving an iterative sequence for any initial point in the product set. Therefore, our result is an extension and improvement of the previous works. At the end, an example is given to illustrate the result.


## 2. Preliminaries

Lemma 1 (see [45]).
Assumeα1∈1,2,β1∈3,4,p1>1,1/p1+1/q1=1,η1∈0,1,b1∈0,η11−α1/p1−1. If y∈C0,1, then the unique solution of the following problem:
(5)D0+α1φp1D0+β1ut+yt=0,0<t<1,u0=u1=u′0=u′1=0,D0+β1u0=0,D0+β1u1=b1D0+β1uη1,is
(6)ut=∫01G1t,sφq1∫01H1s,τyτdτds,where
(7)G1t,s=1Γβ1tβ1−21−sβ1−2s−t+β1−21−ts,0≤t≤s≤1,tβ1−21−sβ1−2s−t+β1−21−ts+t−sβ1−1,0≤s≤t≤1,(8)H1t,s=h1t,s+b1p1−1tα1−11−b1p1−1η1α1−1h1η1,s,(9)h1t,s=1Γα1t1−sα1−1,0≤t≤s≤1,t1−sα1−1−t−sα1−1,0≤s≤t≤1.
For convenience, we can easily give the following Lemma by using Lemma1.Lemma 2.
Letα2∈1,2, β2∈3,4, p2>1,1/p2+1/q2=1,η2∈0,1,b2∈0,η21−α2/p2−1. If y∈C0,1, then
(10)D0+α2φp2D0+β2vt+yt=0,0<t<1,v0=v1=v′0=v′1=0,D0+β2v0=0,D0+β2v1=b2D0+β2vη2,has a unique solution
(11)vt=∫01G2t,sφq2∫01H2s,τyτdτds,where
(12)G2t,s=1Γβ2tβ2−21−sβ2−2s−t+β2−21−ts,0≤t≤s≤1,tβ2−21−sβ2−2s−t+β2−21−ts+t−sβ2−1,0≤s≤t≤1,(13)H2t,s=h2t,s+b2p2−1tα2−11−b2p2−1η2α2−1h2η2,s,(14)h2t,s=1Γα2t1−sα2−1,0≤t≤s≤1,t1−sα2−1−t−sα2−1,0≤s≤t≤1.By Lemmas4 and 5 in [45], the following conclusion is clear.Lemma 3.
The functionsGit,s,i=1,2 defined by (7) and (12) have several properties:
(i)
Git,s is continuous on 0,1×0,1 and Git,s>0 for t,s∈0,1×0,1(ii)
βi−2kitlis≤ΓβiGit,s≤Milis,t,s∈0,1×0,1(iii)
βi−2kitlis≤ΓβiGit,s≤Mikis,t,s∈0,1×0,1, where
(15)kit=tβi−21−t2,lis=s21−sβi−2,Mi=maxβi−1,βi−22Suppose thatX,· is a real Banach space with a partial order induced by a cone P⊂X. For any x,y∈X, the notation x~y denotes that there exist λ>0 and μ>0 such that λx≤y≤μx. For h>θ (i.e., h≥θ and h≠θ), define a set Ph=x∈X∣x~h. Evidently, Ph⊂P. For h1,h2∈P with h1,h2≠θ. Suppose h=h1,h2, then h∈P¯≔P×P. If P is normal, then P¯=P,P is normal.Lemma 4 (see [46, 47]).
P¯h=x,y:x∈Ph1,y∈Ph2=Ph1×Ph2.Lemma 5 (see [47]).
LetP be a normal cone in a Banach space X and h=h1,h2∈P×P with h1,h2≠θ. Operators A,B:P×P⟶P are increasing and satisfy the following:
M1 There exist φ1,φ2:0,1⟶0,1 such that
(16)Arx,ry≥φ1rAx,y,Brx,ry≥φ2rBx,y,x,y∈Pwhere φir>r, r∈0,1, i=1,2;
M2 There is e1,e2∈P¯h such that Ae1,e2∈Ph1, Be1,e2∈Ph2.
Then,(a)
A:Ph1×Ph2⟶Ph1, B:Ph1×Ph2⟶Ph2, and exist x1,y1∈Ph1, x2,y2∈Ph2, γ∈0,1 such that γy1,y2≤x1,x2≤y1,y2 andx1≤Ax1,x2≤y1,x2≤Bx1,x2≤y2(b)
for any givenλ,μ>0, the equation x,y=λAx,y,μBx,y has a unique solution xλ,μ∗,yλ,μ∗ in P¯h. Moreover, take any fixed point x0,y0∈P¯h, let(17)xn,yn=λAxn−1,yn−1,μBxn−1,yn−1,n=1,2,⋯then ∥xn−xλ,μ∗∥⟶0, ∥yn−yλ,μ∗∥⟶0, as n⟶∞.


## 3. Positive Solutions Depending on Parameters

LetX=C0,1, a Banach space with the norm ∥u∥=sup∣ut∣:t∈0,1. We study (4) in the product space X×X. For u,v∈X×X, let ∥u,v∥=max∥u∥,∥v∥. Then, X×X,∥·,·∥ is a Banach space. Let P¯=u,v∈X×X∣ut≥0,vt≥0,t∈0,1, P=u∈X∣ut≥0,t∈0,1, then P¯⊂X×X is a cone and P¯=P×P is normal, and the space X×X has a partial order:
(18)u1,v1≤u2,v2⇔u1t≤u2t,v1t≤v2t,t∈0,1.Lemma 6.
Letft,u,v, gt,u,v be continuous. By using Lemmas 1 and 2 and some results in [45], u,v∈P×P is a positive solution of (4) if and only if u,v∈P×P is a solution of the following equations:
(19)ut=∫01G1t,sφq1∫01H1s,τλfτ,uτ,vτdτds,vt=∫01G2t,sφq2∫01H2s,τμgτ,uτ,vτdτds.Theorem 7.
Letαi∈1,2,βi∈3,4, h1t=tβ1−21−t2, h2t=tβ2−21−t2, t∈0,1. Assume that
H1f,g∈C0,1×0,+∞×0,+∞,0,+∞ and ft,0,0≢0,gt,0,0≢0, t∈0,1
H2f,g are increasing with respect to the second, third variables, i.e., ft,u1,v1≤ft,u2,v2, gt,u1,v1≤gt,u2,v2 for t∈0,1, 0≤u1≤u2,0≤v1≤v2
H3 for r∈0,1, there is ψir:0,1⟶0,1,i=1,2, such that ψir>r1/qi−1 and
(20)ft,ru,rv≥ψ1rft,u,v,gt,ru,rv≥ψ2rgt,u,v,for t∈0,1, u,v∈0,+∞
Then(a)
there areu1,v1∈Ph1, u2,v2∈Ph2, γ∈0,1 such that γv1,v2≤u1,u2≤v1,v2 and(21)u1t≤∫01G1t,sφq1∫01H1s,τfτ,u1τ,u2τdτds≤v1t,t∈0,1,u2t≤∫01G2t,sφq2∫01H2s,τgτ,u1τ,u2τdτds≤v2t,t∈0,1,where Gi,i=1,2, are the Green functions in Lemmas 1 and 2(b)
System (4) has a unique positive solution xλ,μ∗,yλ,μ∗ depending on λ,μ>0 in P¯h, where ht=tβ1−21−t2,tβ2−21−t2, t∈0,1(c)
Take any initial pointu0,v0∈P¯h, let(22)un+1t=λq1−1∫01G1t,sφq1∫01H1s,τfτ,unτ,vnτdτds,n=1,2,⋯,vn+1t=μq2−1∫01G2t,sφq2∫01H2s,τgτ,unτ,vnτdτds,n=1,2,⋯,then unt⟶xλ,μ∗t, vnt⟶yλ,μ∗t as n⟶∞Proof.
We consider three operatorsA,B:P×P⟶X and T:P×P⟶X×X defined by
(23)Au,vt=∫01G1t,sφq1∫01H1s,τfτ,uτ,vτdτds,Bu,vt=∫01G2t,sφq2∫01H2s,τgτ,uτ,vτdτds,Tu,vt=λ~Au,vt,μ~Bu,vt,where λ~≔λq1−1, μ~≔μq2−1, Gi,Hi, and i=1,2 are defined by (7) and (12). From Lemma 3 and H1, it is clear that A,B:P¯⟶P and T:P¯⟶P¯. From our above discussion, we can easily claim that u,v∈P¯ is a solution of system (4) if and only if u,v∈P¯ is a fixed point of operator T. Next, we only need to prove that all assumptions of Lemma 5 are satisfied for operators A,B.We first show thatA,B are increasing. To do this, for ui,vi∈P,i=1,2, with u1≤u2, v1≤v2, one has u1t≤u2t, v1t≤v2t, t∈0,1 and by H2 and Lemma 3,
(24)Au1,v1t=∫01G1t,sφq1∫01H1s,τfτ,u1τ,v1τdτds≤≤∫01G1t,sφq1∫01H1s,τfτ,u2τ,v2τdτds=Au2,v2t,Bu1,v1t=∫01G2t,sφq2∫01H2s,τgτ,u1τ,v1τdτds≤∫01G2t,sφq1∫01H2s,τgτ,u2τ,v2τdτds=Bu2,v2t.That is,Au1,v1≤Au2,v2 and Bu1,v1≤Bu2,v2.Second, we indicate thatA,B satisfy condition M1 of Lemma 5. Let Ψ1r=φq1ψ1r, Ψ2r=φq2ψ2r. Then, for r∈0,1, by H2, we have
(25)Ψ1r=ψ1rq1−1>r1/q1−1q1−1=r.Similarly,Ψ2r>r. For r∈0,1 and u,v∈P, by H3, we obtain
(26)Aru,rvt=∫01G1t,sφq1∫01H1s,τfτ,ruτ,rvτdτds≥∫01G1t,sφq1∫01H1s,τψ1rfτ,uτ,vτdτds=φq1ψ1r∫01G1t,sφq1∫01H1s,τfτ,uτ,vτdτds=Ψ1rAu,vt,Bru,rvt=∫01G2t,sφq2∫01H2s,τgτ,ruτ,rvτdτds≥∫01G2t,sφq2∫01H2s,τψ2rgτ,uτ,vτdτds=φq2ψ2r∫01G2t,sφq2∫01H2s,τgτ,uτ,vτdτds=Ψ2rBu,vt.That is,Aru,rv≥Ψ1rAu,v, Bru,rv≥Ψ2rBu,v for r∈0,1, u,v∈P.Seth=h1,h2, where h1t=tβ1−21−t2, h2t=tβ2−21−t2, t∈0,1. Then, h1,h2∈P¯h. Now, we prove that Ah1,h2∈Ph1, Bh1,h2∈Ph2. In view of H2 and Lemma 3, for t∈0,1, we have
(27)Ah1,h2t=∫01G1t,sφq1∫01H1s,τfτ,h1τ,h2τdτds≥∫01β1−2k1tl1sΓβ1φq1∫01H1s,τfτ,τβ1−21−τ2,τβ2−21−τ2dτds≥β1−2k1tΓβ1∫01l1sφq1∫01H1s,τfτ,0,0dτds=β1−2Γβ1h1t∫01l1sφq1∫01H1s,τfτ,0,0dτds,Ah1,h2t=∫01G1t,sφq1∫01H1s,τfτ,h1τ,h2τdτds≤∫01M1k1tΓβ1φq1∫01H1s,τfτ,τβ1−21−τ2,τβ2−21−τ2dτds≤M1k1tΓβ1∫01φq1∫01H1s,τfτ,1,1dτds=M1Γβ1h1t∫01φq1∫01H1s,τfτ,1,1dτds.Noting thatH1 and H2 guarantee that ft,1,1≥ft,0,0≥0 and ft,0,0≡0, t∈0,1. Because l1s=s21−sβ1−2≤1, s∈0,1, then
(28)∫01φq1∫01H1s,τfτ,1,1dτds≥∫01l1sφq1∫01H1s,τfτ,0,0dτds>0.So, we have(29)m1≔β1−2Γβ1∫01l1sφq1∫01H1s,τfτ,0,0dτds>0,m2≔M1Γβ1∫01φq1∫01H1s,τfτ,1,1dτds>0.By the definition ofM1, it is clear that M1≥β1−2; then, we have m1≤m2, so m1h1t≤Ah1,h2t≤m2h1t, t∈0,1, that is, Ah1,h2∈Ph1. Similarly, from Lemma 3 and H1−H2, we get Bh1,h2∈Ph2.Now, by Lemma5, we obtain the following conclusions:
(1)
We can findu1,v1∈Ph1, u2,v2∈Ph2, γ∈0,1 such that γv1,v2≤u1,u2≤v1,v2 and u1≤Au1,u2≤v1,u2≤Bu1,u2≤v2, that is,
(30)u1t≤∫01G1t,sφq1∫01H1s,τfτ,u1τ,u2τdτds≤v1t,t∈0,1,u2t≤∫01G2t,sφq2∫01H2s,τgτ,u1τ,u2τdτds≤v2t,t∈0,1(2)
The operator equationu,v=λ~Au,v,μ~Bu,v has a unique solution uλ,μ∗,vλ,μ∗ depending on λ,μ>0 in P¯h, where λ~=λq1−1, μ~=μq2−1. That is, u∗,v∗=Tu∗,v∗. So, system (4) has a unique positive solution uλ,μ∗,vλ,μ∗ in P¯h(3)
Taking any initial pointu0,v0∈P¯h, let
(31)un+1t=λq1−1∫01G1t,sφq1∫01H1s,τfτ,unτ,vnτdτds,n=1,2,⋯,vn+1t=μq2−1∫01G2t,sφq2∫01H2s,τgτ,unτ,vnτdτds,n=1,2,⋯,then unt⟶uλ,μ∗t, vnt⟶vλ,μ∗t as n⟶∞Takingλ=μ=1, we have the following conclusion.Corollary 8.
Letαi∈1,2,βi∈3,4, h1t=tβ1−21−t2, and h2t=tβ2−21−t2, t∈0,1. Assume that H1, H2, and H3 hold. Then, the following system:
(32)D0+α1φp1D0+β1ut+ft,ut,vt=0,0<t<1,D0+α2φp2D0+β2vt+gt,ut,vt=0,0<t<1,u0=u1=u′0=u′1=0,D0+β1u0=0,D0+β1u1=b1D0+β1uη1,v0=v1=v′0=v′1=0,D0+β2v0=0,D0+β2v1=b2D0+β2vη2,has a unique positive solution x∗,y∗ in P¯h. In addition, take any given point u0,v0∈P¯h, construct
(33)un+1t=∫01G1t,sφq1∫01H1s,τfτ,unτ,vnτdτds,n=1,2,⋯,vn+1t=∫01G2t,sφq2∫01H2s,τgτ,unτ,vnτdτds,n=1,2,⋯,then unt⟶x∗t, vnt⟶y∗t as n⟶∞.


## 4. An Example

Considering the following system:(34)D0+3/2φ3/2D0+7/2ut+λu1/3+v1/3+2t2=0,0<t<1,D0+3/2φ3/2D0+7/2vt+μu1/4+v1/4+3t3=0,0<t<1,u0=u1=u′0=u′1=0,D0+7/2u0=0,D0+7/2u1=b1D0+7/2u1/2,v0=v1=v′0=v′1=0,D0+7/2v0=0,D0+7/2v1=b2D0+7/2v1/3,whereα1=α2=3/2, β1=β2=7/2, λ,μ>0, η1=1/2, η2=1/3, b1=1/2, b2=1/3, p1=p2=3/2, q1=q2=3 and
(35)ft,u,v=u1/3+v1/3+2t2,gt,u,v=u1/4+v1/4+3t3.Obviously,f,g∈C0,1×0,+∞×0,+∞,0,+∞) and ft,0,0=2t2≡0,gt,0,0=3t3≡0.Note thatx1/3 and x1/4 are increasing in 0,+∞, it implies that ft,u,v, gt,u,v are increasing with respect to the second and third variables for t∈0,1. Moreover, set ψ1r=r1/3, ψ2r=r1/4, r∈0,1. Then, ψ1r, ψ2r∈0,1, ψ1r=r1/3>r1/2=r1/q1−1, ψ2r=r1/4>r1/2=r1/q2−1, and
(36)ft,ru,rv=r1/3u1/3+v1/3+2t2≥r1/3u1/3+v1/3+r1/32t2=r1/3ft,u,v=ψ1rft,u,v,gt,ru,rv=r1/4u1/4+v1/4+3t3≥r1/4u1/4+v1/4+r1/43t3=r1/4gt,u,v=ψ2rgt,u,v,for t∈0,1, u,v∈0,+∞. Hence, all conditions of Theorem 7 are satisfied. Then, Theorem 7 shows that system (34) has a unique positive solution uλ,μ∗,vλ,μ∗ in P¯h, where ht=t3/21−t2,t3/21−t2, t∈0,1, and taking any given point u0,v0∈P¯h, let
(37)un+1t=λ2∫01G1t,s∫01H1s,τunτ1/3+vnτ1/3+2τ2dτ2ds,n=1,2,⋯,vn+1t=μ2∫01G2t,s∫01H2s,τunτ1/4+vnτ1/4+3τ3dτ2ds,n=1,2,⋯,then unt⟶uλ,μ∗t, vnt⟶vλ,μ∗t as n⟶∞, where
(38)G1t,s=G2t,s=1Γ7/2t3/21−s3/2s−t+3/21−ts,0≤t≤s≤1,t3/21−s3/2s−t+3/21−ts+t−s5/2,0≤s≤t≤1,H1t,s=h1t,s+2t1/2h112,s,h1t,s=1Γ3/2t1−s1/2,0≤t≤s≤1,t1−s1/2−t−s1/2,0≤s≤t≤1,H2t,s=h2t,s+32t1/2h113,s,h2t,s=1Γ3/2t1−s1/2,0≤t≤s≤1,t1−s1/2−t−s1/2,0≤s≤t≤1.



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