This paper is concerned with a mixed p-Laplacian boundary value problem involving right-sided and left-sided fractional derivatives and left-sided integral operators with respect to a power function. We prove the uniqueness of positive solutions for the given problem for the cases 1<p≤2 and p>2 by applying an efficient novel approach together with the Banach contraction mapping principle. Estimates for Green's functions appearing in the solution of the problem at hand are also presented. Examples are given to illustrate the obtained results.
Citation: Ahmed Alsaedi, Madeaha Alghanmi, Bashir Ahmad, Boshra Alharbi. Uniqueness results for a mixed p-Laplacian boundary value problem involving fractional derivatives and integrals with respect to a power function[J]. Electronic Research Archive, 2023, 31(1): 367-385. doi: 10.3934/era.2023018
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This paper is concerned with a mixed p-Laplacian boundary value problem involving right-sided and left-sided fractional derivatives and left-sided integral operators with respect to a power function. We prove the uniqueness of positive solutions for the given problem for the cases 1<p≤2 and p>2 by applying an efficient novel approach together with the Banach contraction mapping principle. Estimates for Green's functions appearing in the solution of the problem at hand are also presented. Examples are given to illustrate the obtained results.
Fractional differential equations have received overwhelming interest in the recent years as such equations describe the natural phenomena in a more realistic manner. The p-Laplacian operator is found to be of great help in describing certain problems occurring in mechanics, nonlinear dynamics and many other fields. In consequence, the study of fractional differential equations together with the p-Laplace operator attracted the attention of many researchers. Let us now dwell on some recent works on p-Laplacian fractional boundary value problems.
Liu et al. [1] applied the method of lower and upper solutions to study the existence of solutions for the following problem:
{Dα0+(ϕp(cDβ0+y(x)))=f(x,y(x),cDβ0+y(x)),cDβ0+y(0)=y′(0)=0,y(1)=r1y(η),cDβ0+y(1)=r2cDβ0+y(ξ), |
where 1<α,β≤2,r1,r2≥0,ϕp is the p-Laplacian operator, p>1, Dα0+ is the Riemann-Liouville fractional derivative, and cDβ0+ is the Caputo fractional derivative, f∈C([0,1]×[0,+∞)×(−∞,0],[0,+∞)).
In [2], Bai investigated the existence of positive solutions with the aid of the properties of Green's functions for the following p-Laplacian problem:
{(ϕp(Dα0+y(x)))′+f(x,y(x))=0,y(0)=Dβ0+y(0)=0,cDβ0+y(0)=cDβ0+y(1)=0, |
where 0<β<1,2<α<β+2,Dα0+ and cDβ0+ are the Riemann-Liouville fractional derivative and the Caputo fractional derivative of order α and β respectively, ϕp is the p-Laplacian operator, p>1, and f∈C([0,1]×R,R).
Recently, Wang and Bai [3] discussed the existence and uniqueness of positive solutions to a mixed p-Laplacian fractional boundary value problem given by
{cDγ1−(ϕp(Dδ0+y(t)))=g(t,y(t),cDδ0+y(t)),y(0)=0,y(1)=r1y(μ),Dδ0+y(1)=0,ϕp(Dδ0+y(0))=r2ϕp(Dδ0+y(η)), |
where ϕp(t)=|t|p−2⋅t, 1p+1q=1,p,q>1,0<t<1,1<γ,δ≤2,0<μ,η<1,0≤r1<1μβ−1,0≤r2<1(1−η), cDγ1− is the right Caputo fractional derivative and Dδ0+ is the left Riemann-Liouville fractional derivative and g∈C([0,1]×R2,R). The authors in [3] proved the existence and uniqueness of the solutions to the above problem for 1<p≤2. However, they did not consider the case when p>2.
For some recent results on p-Laplacian boundary value problems, for instance, see the articles [4,5,6,7,8,9,10,11]. The construction of the Green's function together with its properties is a useful tool to investigate the existence of positive solutions to the boundary value problems; for instance, see the text [12].
Let us now review some recent works dealing with a modified form of Caputo and Riemann-Liouville fractional derivatives. In [13], the authors studied the asymptotic stability of solutions of generalized Caputo fractional differential equations. Caputo modification of the generalized fractional derivatives was discussed in [14]. Some existence results for a nonlocal boundary value problem involving generalized Liouville-Caputo derivatives and generalized fractional integral were presented in [15]. The authors in [16] discussed the existence of solutions for generalized fractional differential equations and inclusions equipped with nonlocal generalized fractional integral boundary conditions. In [17], extremal solutions for an integro-initial value problem for generalized Caputo fractional differential equations were obtained. In [18], the authors introduced and studied a new class of coupled systems containing both Caputo and Riemann-Liouville generalized fractional derivatives. For some recent works on the problems involving generalized fractional derivatives, for example, see [19,20,21,22,23,24].
Motivated by the aforementioned studies on boundary value problems involving a p-Laplacian operator and modified versions of fractional derivatives, in this paper, we introduce a new class of mixed p-Laplacian fractional boundary value problems involving right-sided and left-sided fractional derivatives and left-sided integral operators with respect to a power function. In precise terms, we investigate the following problem:
{ρDα1−(ϕp(ρDβ0+y(t)))=ν1f(t,y(t),ρDβ0+y(t))+ν2ρIζ0+g(t,y(t),ρDβ0+y(t)),y(0)=0,y(1)=λ1y(μ),ρDβ0+y(1)=0,ϕp(ρDβ0+y(0))=λ2ϕp(ρDβ0+y(η)), | (1.1) |
where ϕp(t)=|t|p−2⋅t,1p+1q=1,p,q>1,0<t<1,1<α,β≤2,ρ>0,ζ>0,0<μ,η<1,0≤λ1<1μρ(β−1),0≤λ2<1(1−ηρ)α−1, ρDα1− and ρDβ0+ respectively denote the right and left fractional derivatives of orders α and β with respect to a power function (see Definitions 2), ρIζ0+ is the fractional integral operator of order ζ with respect to a power function (see Definitions 1), ν1,ν2∈R and f,g:[0,1]×R2→R are continuous functions.
The remainder of the paper is arranged as follows. In Section 2, we present the background material related to our problem and prove some important lemmas that play a key role in the forthcoming analysis. Section 3 contains the main results for the given problem. In Section 4, we illustrate our results with the aid of examples. The paper concludes with certain interesting observations.
Let us first recall that the concept of fractional calculus of a function with respect to another function can be found in the books by Samko et al. ([25]; Section 18.2) and Kilbas et al. ([26]; Section 2.5), while an article by Erdelyi [27] contains the first study of fractional integrals with respect to a power function (now mistakenly named after Katugampola).
Definition 2.1. The fractional integral with respect to a power function (tρ/ρ) of order α>0 for a function h∈Xpc(a,b) for −∞<a<t<b<∞, is defined by
(ρIαa+h)(t)=ρ1−αΓ(α)∫tasρ−1(tρ−sρ)1−αh(s)ds,ρ>0, | (2.1) |
where Xpc(a,b) denotes the space of all complex-valued Lebesgue measurable functions ϕ on (a,b) equipped with the norm:
‖ϕ‖Xpc=(∫ba|xcϕ(x)|pdxx)1/p<∞,c∈R,1≤p≤∞. |
Note that the integral in (2.1) is called the left-sided fractional integral. Similarly, we can define right-sided fractional integral ρIαb−f as
(ρIαb−h)(t)=ρ1−αΓ(α)∫btsρ−1(sρ−tρ)1−αh(s)ds. | (2.2) |
Here we mention that the above definitions of fractional integrals follow from the integrals (2.5.1) and (2.5.2) on pages 99–100 in the text [26] by taking the power function g(x)=xρ/ρ.
Definition 2.2. For α>0,n=[α]+1 and ρ>0, the fractional derivatives with respect to a power function (tρ/ρ), associated with the fractional integrals (2.1) and (2.2) are defined, for 0≤a<x<b<∞, by
(ρDαa+g)(t)=(t1−ρddt)n(ρIn−αa+g)(t)=ρα−n+1Γ(n−α)(t1−ρddt)n∫tasρ−1(tρ−sρ)α−n+1g(s)ds, | (2.3) |
and
(ρDαb−g)(t)=(−t1−ρddt)n(ρIn−αb−g)(t)=ρα−n+1Γ(n−α)(−t1−ρddt)n∫btsρ−1(sρ−tρ)α−n+1g(s)ds. | (2.4) |
Note that the above definitions of fractional derivatives follow from the integrals (2.5.17) and (2.5.18) on page 101 in the text [26] by taking the power function g(x)=xρ/ρ.
Lemma 2.1. [28] Let 1<α≤2,ρ>0, u∈Xpc(0,T) and ρI2−αu∈AC2ρ, where AC2ρ([a,b]) denotes the space of absolutely continuous functions possessing the t1−ρddt-derivative defined by
AC2ρ[a,b]={f:[a,b]→R:(t1−ρddtf)∈AC[a,b]}. |
Then the general solution of the fractional differential equation ρDα0+u(t)=0 is
u(t)=c1tρ(α−1)+c2tρ(α−2), |
where ci∈R,i=1,2. Moreover,
(ρIα0+ρDα0+u)(t)=u(t)+c1tρ(α−1)+c2tρ(α−2). |
Lemma 2.2. For any ψ∈C([0,1],R), the integral representation of the solution for the following nonlocal p-Laplacian boundary value problem involving right-sided and left-sided fractional derivatives with respect to a power function:
{ρDα1−(ϕp(ρDβ0+y(t)))=ψ(t),y(0)=0,y(1)=λ1y(μ),ρDβ0+y(1)=0,ϕp(ρDβ0+y(0))=λ2ϕp(ρDβ0+y(η)), | (2.5) |
is given by
y(t)=∫10τρ−1G1(t,τ)ϕq(∫10sρ−1G2(τ,s)ψ(s)ds)dτ, | (2.6) |
where
G1(t,τ)=ρ1−βΓ(β){Λ1[(1−τρ)β−1−λ1(μρ−τρ)β−1]−(tρ−τρ)β−1,0≤τ≤min{t,μ},Λ1(1−τρ)β−1−(tρ−τρ)β−1,μ≤τ≤t,Λ1[(1−τρ)β−1−λ1(μρ−τρ)β−1],t≤τ≤μ,Λ1(1−τρ)β−1,max{t,μ}≤τ≤1, | (2.7) |
and
G2(τ,s)=ρ1−αΓ(α){Λ2sρ(α−1),0≤s≤min{τ,η},Λ2[sρ(α−1)−λ2(sρ−ηρ)α−1],η≤s≤τ,Λ2sρ(α−1)−(sρ−τρ)α−1,τ≤s≤η,Λ2[sρ(α−1)−λ2(sρ−ηρ)α−1]−(sρ−tρ)α−1,max{τ,η}≤s≤1, | (2.8) |
with
Λ1=tρ(β−1)1−λ1μρ(β−1),Λ2=(1−τρ)α−11−λ2(1−ηρ)α−1. |
Proof. Letting −ϕp(ρDβ0+y(t))=H(t), we decompose the mixed boundary value problem (2.5) as
{ρDα1−H(t)=−ψ(t),H(1)=0,H(0)=λ2H(η), | (2.9) |
and
{ρDβ0+y(t)=−ϕq(H(t)),y(0)=0,y(1)=λ1y(μ). | (2.10) |
Solving the equation ρDα1−H(t)=−ψ(t), we get
H(t)=−ρ1−αΓ(α)∫1tsρ−1(sρ−tρ)α−1ψ(s)ds+c0(1−tρ)α−2+c1(1−tρ)α−1, | (2.11) |
where c0 and c1 are arbitrary constants. Using the condition H(1)=0 in (2.11) yields c0=0. Then, inserting (2.11) with c0=0 in the condition: H(0)=λ2H(η), the value of c1 is found to be
c1=ρ1−αΓ(α)(1−λ2(1−ηρ)α−1)(∫10sρα−1ψ(s)ds−λ2∫1ηsρ−1(sρ−tρ)α−1ψ(s)ds). | (2.12) |
So (2.11) becomes
H(t)=−ρ1−αΓ(α)∫1tsρ−1(sρ−tρ)α−1ψ(s)ds+ρ1−α(1−tρ)α−1Γ(α)(1−λ2(1−ηρ)α−1)(∫10sρα−1ψ(s)ds−λ2∫1ηsρ−1(sρ−ηρ)α−1ψ(s)ds)=∫10sρ−1G2(t,s)ψ(s)ds, | (2.13) |
where G2(t,s) is given in (2.8). Applying the integral operator ρIβ0+ on both sides of the differential equation in (2.10), we have
y(t)=−ρ1−βΓ(β)∫t0sρ−1(tρ−sρ)β−1ϕq(H(s))ds+d0tρ(β−2)+d1tρ(β−1), | (2.14) |
where d0 and d1 are arbitrary constants.
Using (2.14) in the boundary conditions of (2.10), we obtain d0=0 and
d1=ρ1−βΓ(β)(1−λ1μρ(β−1))(∫10sρ−1(1−tρ)α−1ϕq(H(s))ds−λ1∫μ0sρ−1(μρ−sρ)α−1ϕq(H(s))ds). | (2.15) |
Thus, (2.14) takes the form:
y(t)=∫10sρ−1G1(t,s)ϕq(H(s))ds, |
where G1(t,s) and H(.) are respectively given in (2.7) and (2.13).
Lemma 2.3. The functions G1(t,s) and G2(t,s) given in (2.7) and (2.8) respectively, are continuous and possess the following properties:
(i) G1(t,s)>0,G2(t,s)>0, ∀t,s∈(0,1);
(ii) Q1ρ1−βtρ(β−1)sρ(1−sρ)β−1Γ(β)(1−λ1μρ(β−1))≤G1(t,s)≤ρ1−βtρ(β−1)(1−sρ)β−1Γ(β)(1−λ1μρ(β−1)), ∀t,s∈(0,1);
(iii) Q2ρ1−αsρ(α−1)(1−tρ)α−1(1−sρ)Γ(α)(1−λ2(1−ηρ)α−1)≤G2(t,s)≤ρ1−αsρ(α−1)(1−tρ)α−1Γ(α)(1−λ2(1−ηρ)α−1), ∀t,s∈(0,1),
where
0<Q1:=min{1−λ1μρ(β−1),λ1μρ(β−2)(1−μρ),λ1μρ(β−1)}<1, |
and
0<Q2:=min{1−λ2(1−ηρ)α−1,λ2ηρ(1−ηρ)α−2,λ2(1−ηρ)α−1}<1. |
Proof. Let us first prove part (i) with different cases.
Case 1. If 0≤τ≤min{t,μ}<1, then we have τρμρ≥τρ since μρ<1, which implies 1−τρμρ≤1−τρ. Hence, we find that
tρ(β−1)1−λ1μρ(β−1)[(1−τρ)β−1−λ1(μρ−τρ)β−1]=tρ(β−1)1−λ1μρ(β−1)[(1−τρ)β−1−λ1μρ(β−1)(1−τρμρ)β−1]≥tρ(β−1)1−λ1μρ(β−1)[(1−τρ)β−1−λ1μρ(β−1)(1−τρ)β−1]=tρ(β−1)(1−τρ)β−1≥tρ(β−1)(1−τρtρ)β−1=(tρ−τρ)β−1, |
which means that
Λ1[(1−τρ)β−1−λ1(μρ−τρ)β−1]−(tρ−τρ)β−1≥0. |
Case 2. For μ≤τ≤t, we have
tρ(β−1)1−λ1μρ(β−1)(1−τρ)β−1≥tρ(β−1)(1−τρ)β−1≥tρ(β−1)(1−τρtρ)β−1=(tρ−τρ)β−1. |
Thus, Λ1(1−τρ)β−1−(tρ−τρ)β−1≥0.
Case 3. For t≤τ≤μ, we have
(1−τρ)β−1≥λ1μρ(β−1)(1−τρ)β−1≥λ1μρ(β−1)(1−τρμρ)β−1=λ1(μρ−τρ)β−1. |
So, Λ1[(1−τρ)β−1−λ1(μρ−τρ)β−1]≥0.
Case 4. For max{t,μ}≤τ≤1, it is obvious that Λ1(1−τρ)β−1≥0. Consequently, we get G1(t,τ)≥0. By a similar argument, one can show that G2(t,s)≥0. In order to establish (ii), let Γ(β)ρβ−1(1−λ1μρ(β−1))G1(t,s)=g1(t,s). Then, for 0≤s≤min{t,μ}, we have
g1(t,s)=tρ(β−1)(1−sρ)β−1−λ1tρ(β−1)(μρ−sρ)β−1−(tρ−sρ)β−1(1−λ1μρ(β−1))=tρ(β−1)(1−sρ)β−1−λ1tρ(β−1)μρ(β−1)(1−sρμρ)β−1−tρ(β−1)(1−sρtρ)β−1(1−λ1μρ(β−1))≥tρ(β−1)(1−sρ)β−1−λ1tρ(β−1)μρ(β−1)(1−sρμρ)β−1−tρ(β−1)(1−sρ)β−1(1−λ1μρ(β−1))=λ1μρ(β−1)tρ(β−1)(−(1−sρμρ)β−1+(1−sρ)β−1)≥λ1μρ(β−1)tρ(β−1)(1−sρ)β−2(−(1−sρμρ)+(1−sρ))≥λ1μρ(β−1)tρ(β−1)(1−sρ)β−1sρ(1−μρμρ)≥Q1tρ(β−1)sρ(1−sρ)β−1. |
For μ≤s≤t, let Γ(β)ρβ−1(1−λ1μρ(β−1))G1(t,s)=g2(t,s). Then, we get
g2(t,s)=tρ(β−1)(1−sρ)β−1−(tρ−sρ)β−1(1−λ1μρ(β−1))=tρ(β−1)(1−sρ)β−1−tρ(β−1)(1−sρtρ)β−1(1−λ1μρ(β−1))≥tρ(β−1)(1−sρ)β−1−tρ(β−1)(1−sρ)β−1(1−λ1μρ(β−1))=λ1μρ(β−1)tρ(β−1)(1−sρ)β−1≥Q1tρ(β−1)sρ(1−sρ)β−1. |
For t≤s≤μ, let Γ(β)ρβ−1(1−λ1μρ(β−1))G1(t,s)=g3(t,s). Then, we obtain
g3(t,s)=tρ(β−1)(1−sρ)β−1−λ1tρ(β−1)(μρ−sρ)β−1=tρ(β−1)(1−sρ)β−1−λ1tρ(β−1)μρ(β−1)(1−sρμρ)β−1≥tρ(β−1)(1−sρ)β−1−λ1tρ(β−1)μρ(β−1)(1−sρ)β−1=tρ(β−1)(1−λ1μρ(β−1))(1−sρ)β−1≥Q1tρ(β−1)sρ(1−sρ)β−1. |
Lastly, when max{t,μ}≤s≤1, it is clear that
Γ(β)ρβ−1(1−λ1μρ(β−1))G1(t,s)=tρ(β−1)(1−sρ)β−1≥Q1tρ(β−1)sρ(1−sρ)β−1. |
Thus,
G1(t,s)≥Q1ρ1−βtρ(β−1)sρ(1−sρ)β−1Γ(β)(1−λ1μρ(β−1)). |
On the other hand, it easy to show that
G1(t,s)≤ρ1−βtρ(β−1)(1−sρ)β−1Γ(β)(1−λ1μβ−1),∀(t,s)∈(0,1)×(0,1). |
Now, for (iii), consider Γ(α)ρα−1(1−λ2(1−ηρ)α−1)G2(t,s). If 0≤s≤min{t,η}, then we find that
(1−tρ)α−1sρ(α−1)≥Q2(1−tρ)α−1sρ(α−1)(1−sρ). |
When t≤s≤η, let Γ(α)ρα−1(1−λ2(1−ηρ))G2(t,s)=h1(s,t); then we have
h1(t,s)=(1−tρ)α−1sρ(α−1)−(sρ−tρ)α−1(1−λ2(1−ηρ)α−1)=(1−tρ)α−1sρ(α−1)−sρ(α−1)(1−tρsρ)α−1(1−λ2(1−ηρ)α−1)≥sρ(α−1)(1−tρ)α−1(1−1+λ2(1−ηρ)α−1)≥sρ(α−1)(1−tρ)α−1λ2(1−ηρ)α−1≥Q2sρ(α−1)(1−tρ)α−1(1−sρ). |
When η≤s≤t, let Γ(α)ρα−1(1−λ2(1−ηρ))G2(t,s)=h2(t,s); then we have
h2(t,s)=(1−tρ)α−1sρ(α−1)−λ2(1−tρ)α−1(sρ−ηρ)α−1=(1−tρ)α−1sρ(α−1)−λ2(1−tρ)α−1sρ(α−1)(1−ηρsρ)α−1≥sρ(α−1)(1−tρ)α−1(1−λ2(1−ηρ)α−1)≥Q2sρ(α−1)(1−tρ)α−1(1−sρ). |
When max{η,t}≤s≤1, let Γ(α)ρα−1(1−λ2(1−ηρ))G2(t,s)=h3(t,s). Then, we obtain
h3(t,s)=(1−tρ)α−1sρ(α−1)−λ2(1−tρ)α−1(sρ−ηρ)α−1−(sρ−tρ)α−1(1−λ2(1−ηρ)α−1)=(1−tρ)α−1sρ(α−1)−λ2(1−tρ)α−1sρ(α−1)(1−ηρsρ)α−1−sρ(α−1)(1−tρsρ)α−1(1−λ2(1−ηρ)α−1)≥sρ(α−1)(1−tρ)α−1(1−λ2(1−ηρsρ)α−1−1+λ2(1−ηρ)α−1)≥λ2sρ(α−1)(1−tρ)α−1(1−ηρ)α−2((1−ηρ)−(1−ηρsρ))≥λ2sρ(α−1)(1−tρ)α−1(1−ηρ)α−2ηρs−ρ(1−sρ))=λ2(1−ηρ)α−2ηρsρ(α−2)(1−tρ)α−1(1−sρ)≥Q2sρ(α−1)(1−tρ)α−1(1−sρ). |
Hence,
G2(t,s)≥Q2ρ1−αsρ(α−1)(1−tρ)α−1(1−sρ)Γ(α)(1−λ2(1−ηρ)α−1). |
On the other hand, it is easy to show that
G2(t,s)≤ρ1−αsρ(α−1)(1−tρ)α−1Γ(α)(1−λ2(1−ηρ)β−1),∀(t,s)∈(0,1)×(0,1). |
Thus, the proof is completed.
Remark 2.1. When ρ→1, (ii) is similar to the result presented in Theorem 1 in [29].
Lemma 2.4. Let ψ∈C([0,1],R) and
ω(τ)=−ϕq(∫10sρ−1G2(τ,s)ψ(s)ds),y(t)=−∫10τρ−1G1(t,τ)ω(τ)dτ. |
Then, for t,τ,s∈[0,1] and 1p+1q=1,p,q≥1, the following results hold:
(i)‖ω‖≤Ωq−11‖ψ‖q−1,‖y‖≤Ω2Ωq−11‖ψ‖q−1, where
Ω1:=1ραΓ(α+1)(1−λ2(1−ηρ)α−1),Ω2:=1ρβΓ(β+1)(1−λ1μρ(β−1)); |
(ii) ω(τ)≤−mq−1Ωq−13(1−τρ)(α−1)(q−1),y(t)≥mq−1Ωq−13Ω4tρ(β−1), for ψ(t)≥m>0 and ∀t∈[0,1], where
Ω3:=Q2ραΓ(α+2)(1−λ2(1−ηρ)α−1),Ω4:=Q1ρβΓ(β)(β+(α−1)(q−1))(β+(α−1)(q−1)+1)(1−λ1μρ(β−1)). |
Proof. By Lemma 2.3, we have
Q1tρ(β−1)ρβΓ(β+2)(1−λ1μρ(β−1))≤∫10τρ−1G1(t,τ)dτ≤tρ(β−1)ρβΓ(β+1)(1−λ1μρ(β−1)), |
and
Q2(1−τρ)α−1ραΓ(α+2)(1−λ2(1−ηρ)α−1)≤∫10sρ−1G2(τ,s)ds≤(1−τρ)α−1ραΓ(α+1)(1−λ2(1−ηρ)α−1). |
Consequently,
|ω(τ)|=|−ϕq(∫10sρ−1G2(τ,s)ψ(s)ds)|=ϕq(∫10sρ−1G2(τ,s)|ψ(s)|ds)≤ϕq(∫10sρ−1G2(τ,s)ds‖ψ‖)≤(1Γ(α+1)ρα(1−λ2(1−ηρ)α−1)))q−1‖ψ‖q−1=Ωq−11‖ψ‖q−1. |
Thus, ‖ω‖≤Ωq−11‖ψ‖q−1. Similarly, we have that ‖y‖≤Ω2Ωq−11‖ψ‖q−1. This establishes (i).
For (ii), we have
ω(τ)=−ϕq(∫10sρ−1G2(τ,s)ψ(s)ds)≤−ϕq(∫10sρ−1G2(τ,s)mds)≤−(Q2(1−τρ)α−1ραΓ(α+2)(1−λ2(1−ηρ)α−1))q−1mq−1=−Ωq−13mq−1(1−τρ)(α−1)(q−1). |
Likewise, we have that y(t)≥mq−1Ωq−13Ω4tρ(β−1). This completes the proof.
The following lemma describes some properties of the p-Laplace operator which can easily be proved by using the mean value theorem when the function ϕp(k)=|k|p−2k is differentiable at all k except k=0 and |∂ϕ(k)∂k| is bounded by (p−1)max|k|p−2 for p>2 and bounded by (p−1)min|k|p−2 for 1<p<2.
Lemma 2.5. (see (2.1) and (2.2) on page 3268 in [30]) The following relations hold for the p-Laplace operator:
(i) For 1<p≤2, |k1|,|k2|≥Δ1>0 and k1k2>0, |ϕp(k2)−ϕp(k1)|≤(p−1)Δp−21|k2−k1|;
(ii) For p>2, |k1|,|k2|≤Δ2 and k1k2>0, |ϕp(k2)−ϕp(k1)|≤(p−1)Δp−22|k2−k1|.
In this section, we discuss the existence and uniqueness of the solutions to the problem (1.1). For a given number M>0, let us consider the following set
ΥM={(t,y,ω):0≤t≤1,‖y‖≤Ω2Ωq−11Mq−1,‖ω‖≤Ωq−11Mq−1}; |
and denote by Γ[O,M] a closed ball in the space of the continuous function C[0,1].
Theorem 3.1. Assume that 1<p≤2 and there exist positive constants M1,M2,C1,C2,κ1andκ2 such that
(A1) |f(t,y,ω)|≤M1,|g(t,y,ω)|≤M2for(t,y,ω)∈ΥM;
(A2) |f(t,y1,ω1)−f(t,y2,ω2)|≤C1|y2−y1|+C2|ω2−ω1|,for(t,yi,ωi)∈ΥM,i=1,2;
(A3) |g(t,y1,ω1)−g(t,y2,ω2)|≤κ1|y2−y1|+κ2|ω2−ω1|,for(t,yi,ωi)∈ΥM,i=1,2;
(A4) L1:=(q−1)Mq−2Ωq−11{|ν1|(C1Ω2+C2)+|ν2|ρζΓ(ζ+1)(κ1Ω2+κ2)}<1.
Then the mixed boundary value problem (1.1) has a unique solution satisfying the following inequalities:
|y(t)|≤Ω2Ωq−11Mq−1,|ρDβ0+y(t)|≤Ωq−11Mq−1,for allt∈[0,1], | (3.1) |
where M≥|ν1|M1+|ν2|M2ρζΓ(ζ+1).
Proof. Define an operator G:C[0,1]→C[0,1] by
(Gψ)(t)=ν1f(t,∫10τρ−1G1(t,τ)ϕq(∫10sρ−1G2(τ,s)ψ(s)ds)dτ,−ϕq(∫10sρ−1G2(τ,s)ψ(s)ds))+ν2∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)g(r,∫10τρ−1G1(r,τ)ϕq(∫10sρ−1G2(τ,s)ψ(s)ds)dτ,−ϕq(∫10sρ−1G2(r,s)ψ(s)ds))dr. | (3.2) |
Observe that the continuity of G1(t,τ), G2(τ,s), f(t,y,ω) and g(t,y,ω) leads to that of the operator G. Moreover, if y(t) is a solution to the problem (1.1), then ψ(t)=ρDα1−(ϕp(ρDβ0+y(x))) is the fixed point of the operator G. Conversely, if ψ(t) is a fixed point of the operator G, then
y(t)=∫10τρ−1G1(t,τ)ϕq(∫10sρ−1G2(τ,s)ψ(s)ds)dτ, |
is a solution to the problem (1.1).
Next, we need to show that the operator G maps Γ[O,M] into itself. Let ψ∈Γ[O,M]; then, by Lemma 2.3, we have
|ω(t)|≤Ωq−11Mq−1,|y(t)|≤Ω2Ωq−11Mq−1. |
Consequently, for any t∈[0,1], there is (t,y(t),ω(t))∈ΥM. So, from (A1), we have
|(Gψ)(t)|=|ν1f(t,y(t),ω(t))+ν2ρIζ0+g(t,y(t),ω(t))|≤|ν1|M1+|ν2|M2Γ(ζ+1)≤M. |
Therefore, (Gψ)(t)∈Γ[O,M]. Thus, the operator G maps Γ[O,M] into itself. Now, we show that the operator G:Γ[O,M]→Γ[O,M] is a contraction. From (A2),(A3), Lemma 2.3, Lemma 2.4 and (ii) of Lemma 2.5, there is Δ2:=Ω1M≥|∫10sρ−1G2(τ,s)ψ(s)ds| for each ψ1(t),ψ2(t)∈Γ[O,M], and 1<p≤2 (that is, q≥2). Thus, we obtain
|(Gψ2)(t)−(Gψ1)(t)|≤|ν1||f(t,y2(t),ω2(t))−f(t,y1(t),ω1(t))|+|ν2||ρIζ0+g(t,y2(t),ω2(t))−ρIζ0+g(t,y1(t),ω1(t))|≤|ν1|{C1|y2(t)−y1(t)|+C2|ω2(t)−ω1(t)|}+|ν2|{κ1ρIζ0+|y2(t)−y1(t)|+κ1ρIζ0+|ω2(t)−ω1(t)|}=|ν1|{C1|∫10τρ−1G1(t,τ)[ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)]dτ|+C2|ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)|}+|ν2|{κ1∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)|∫10τρ−1G1(r,τ)[ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)]dτ|dr+κ2∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)|ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)|dτ}≤|ν1|{C1(q−1)(MΩ1)q−2[∫10τρ−1G1(t,τ)|∫10sρ−1G2(τ,s)(ψ2(s)−ψ1(s))ds|dτ]+C2(q−1)(MΩ1)q−2|∫10G2(τ,s)(ψ2(s)−ψ1(s))ds|}+|ν2|{κ1(q−1)(MΩ1)q−2[∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)∫10τρ−1G1(r,τ)××|∫10sρ−1G2(τ,s)(ψ2(s)−ψ1(s))ds|dτdr]+κ2(q−1)(MΩ1)q−2∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)|∫10G2(τ,s)(ψ2(s)−ψ1(s))ds|dτ}≤|ν1|{C1(q−1)(MΩ1)q−2[∫10τρ−1G1(t,τ)|∫10sρ−1G2(τ,s)ds|dτ‖ψ2−ψ1‖]+C2(q−1)(MΩ1)q−2[∫10sρ−1G2(τ,s)ds‖ψ2−ψ1‖]}+|ν2|{κ1(q−1)(MΩ1)q−2[∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)∫10τρ−1G1(r,τ)××|∫10sρ−1G2(τ,s)ds|dτdr‖ψ2−ψ1‖]+κ2(MΩ1)q−2[∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)∫10sρ−1G2(τ,s)ds‖ψ2−ψ1‖dτ]}≤|ν1|(q−1)Mq−2Ωq−11(C1Ω2+C2)‖ψ2−ψ1‖+|ν2|(q−1)Mq−2Ωq−11Γ(ζ+1)(κ1Ω2+κ2)‖ψ2−ψ1‖=(q−1)Mq−2Ωq−11{|ν1|(C1Ω2+C2)+|ν2|Γ(ζ+1)(κ1Ω2+κ2)}‖ψ2−ψ1‖=L1‖ψ2−ψ1‖, |
which, on taking the norm for [0,1], yields
‖(Gψ2)−(Gψ1)‖≤L1‖ψ2−ψ1‖. |
Since L1<1 by (A4), the operator G is a contraction. So, we deduce by Banach's contraction mapping principle that ψ(t) is the unique fixed point of the operator G. Hence, there exists a unique solution to the mixed boundary value problem (1.1) satisfying (3.1). The proof is completed.
In the following result, we consider a special case of ΥM. With the aid of Lemma 2.4, for m>0 with M>m, we define the sets
Υ+M={(t,y,ω):0≤t≤1,Ω4Ωq−13mq−1tρ(β−1)≤y(t)≤Ω2Ωq−11Mq−1,−Ωq−11Mq−1≤ω(t)≤−Ωq−13mq−1(1−tρ)(α−1)(q−1),} |
and
Γm={ψ(t)∈C[0,1]:m<ψ(t)<M}. |
The following theorem is concerned with the existence of a unique solution to the problem (1.1) when p>2.
Theorem 3.2. Assume that p>2 and there exist positive numbers m1,m2,M1,M2,C1andC2 such that
(B1) m1≤f(t,y,ω)≤M1andm2≤g(t,y,ω)≤M2,for(t,y,ω)∈Υ+M;
(B2) |f(t,y1,ω1)−f(t,y2,ω2)|≤C1|y2−y1|+C2|ω2−ω2|,for(t,yi,ωi)∈Υ+M,i=1,2;
(B3) |g(t,y1,ω1)−g(t,y2,ω2)|≤κ1|y2−y1|+κ2|ω2−ω1|,for(t,yi,ωi)∈Υ+M,i=1,2;
(B4) L2:=(q−1)mq−2Ωq−23Ω1{|ν1|(C1Ω2+C2)+|ν2|ρζΓ(ζ+1)(κ1Ω2+κ2)}<1.
Then the mixed boundary value problem (1.1) has a unique solution satisfying the following inequalities
Ω4Ωq−13mq−1tρ(β−1)≤y(t)≤Ω2Ωq−11Mq−1,for allt∈[0,1],−Ωq−11Mq−1≤ρDβ0+y(t)≤−Ωq−13mq−1(1−tρ)(α−1)(q−1),for allt∈[0,1], | (3.3) |
where M≥|ν1|M1+|ν2|M2ρζΓ(ζ+1) and 0<m≤ν1m1+ν2m2ρζΓ(ζ+1).
Proof. As argued in the proof of the last theorem, the operator G defined by (3.2) is continuous and G maps any ψ∈Γm into itself.
Now, from (B2),(B3), Lemma 2.3, Lemma 2.4, and (i) of Lemma 2.5, there exists Δ1:=mΩ3(1−τρ)α−1≤|∫10sρ−1G2(τ,s)ψ(s)ds|. Then, for all ψ1,ψ2∈Γm, and p>2 (that is, 1<q≤2), we find that
|(Gψ2)(t)−(Gψ1)(t)|≤|ν1||f(t,y2(t),ω2(t))−f(t,y1(t),ω1(t))|+|ν2||ρIζ0+g(t,y2(t),ω2(t))−ρIζ0+g(t,y1(t),ω1(t))|≤|ν1|{C1|y2(t)−y1(t)|+C2|ω2(t)−ω1(t)|}+|ν2|{κ1ρIζ0+|y2(t)−y1(t)|+κ1ρIζ0+|ω2(t)−ω1(t)|}=|ν1|{C1|∫10τρ−1G1(t,τ)[ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)]dτ|+C2|ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)|}+|ν2|{κ1∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)|∫10τρ−1G1(r,τ)[ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)]dτ|dr+κ2∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)|ϕq(∫10sρ−1G2(τ,s)ψ2(s)ds)−ϕq(∫10sρ−1G2(τ,s)ψ1(s)ds)|dτ}≤|ν1|{C1(q−1)∫10τρ−1G1(t,τ)mq−2Ωq−23(1−τρ)(α−1)(q−2)|∫10sρ−1G2(τ,s)(ψ2(s)−ψ1(s))ds|dτ+C2(q−1)mq−2Ωq−23(1−τρ)(α−1)(q−2)|∫10sρ−1G2(τ,s)(ψ2(s)−ψ1(s))ds|}+|ν2|{κ1(q−1)∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)∫10τρ−1G1(r,τ)mq−2Ωq−23(1−τρ)(α−1)(q−2)××|∫10sρ−1G2(τ,s)(ψ2(s)−ψ1(s))ds|dτdr+κ2(q−1)mq−2Ωq−23ρ∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)(1−τρ)(α−1)(q−2)××|∫10sρ−1G2(τ,s)(ψ2(s)−ψ1(s))ds|dτ}≤|ν1|{C1(q−1)mq−2Ωq−23[∫10τρ−1G1(t,τ)(1−τρ)(α−1)(q−2)|∫10sρ−1G2(τ,s)ds|dτ‖ψ2−ψ1‖]+C2(q−1)mq−2Ωq−23(1−τρ)(α−1)(q−2)[∫10sρ−1G2(τ,s)ds‖ψ2−ψ1‖]}+|ν2|{κ1(q−1)mq−2Ωq−23[∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)∫10τρ−1G1(r,τ)(1−τρ)(α−1)(q−2)××|∫10sρ−1G2(τ,s)ds|dτdr‖ψ2−ψ1‖]+κ2(q−1)mq−2Ωq−23[∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)(1−τρ)(α−1)(q−2)∫10sρ−1G2(τ,s)dsdτ‖ψ2−ψ1‖]}≤|ν1|{C1(q−1)mq−2Ωq−23[∫10τρ−1G1(t,τ)(1−τρ)(α−1)(q−2)Ω1(1−τρ)α−1dτ‖ψ2−ψ1‖]+C2(q−1)mq−2Ωq−23(1−τρ)(α−1)(q−2)[Ω1(1−τρ)α−1‖ψ2−ψ1‖]}+|ν2|{κ1(q−1)mq−2Ωq−23[∫t0rρ−1(tρ−rρ)ζ−1ρζ−1Γ(ζ)∫10τρ−1G1(r,τ)(1−τρ)(α−1)(q−2)××Ω1(1−τρ)α−1dτdr‖ψ2−ψ1‖]+κ2(q−1)mq−2Ωq−23[∫t0τρ−1(tρ−τρ)ζ−1ρζ−1Γ(ζ)(1−τρ)(α−1)(q−2)Ω1(1−τρ)α−1dτ‖ψ2−ψ1‖]}≤(q−1)mq−2Ωq−23Ω1{|ν1|(C1Ω2+C2)+|ν2|ρζΓ(ζ+1)(κ1Ω2+κ2)}‖ψ2−ψ1‖=L2‖ψ2−ψ1‖, |
which, after taking the norm for t∈[0,1], takes the form:
‖(Gψ2)−(Gψ1)‖≤L2‖ψ2−ψ1‖, |
with L2<1 by the condition (B4). Consequently, the operator G is a contraction. Hence, by Banach's contraction mapping principle, ψ(t) is the unique fixed point of the operator G. Therefore, there exists a unique solution to the problem (1.1) satisfying (3.3). This finishes the proof.
Consider the following problem
{1/2D5/31−(ϕp(1/2D3/20+y(x)))=34f(t,y(t),1/2D3/20+y(t))−231/2I1/40+g(t,y(t),1/2D3/20+y(t)),y(0)=0,y(1)=12y(3/4),1/2D3/20+y(1)=0,ϕp(1/2D3/20+y(0))=ϕp(1/2D3/20+y(1/2)), | (4.1) |
where α=5/3,β=3/2,ρ=1/2,λ1=1/2,λ2=1,μ=3/4,η=1/2,ν1=3/4andν2=−2/3, and p, f(t,y,ω) and g(t,y,ω) will be fixed later.
From the given data, we have
G1(t,s)≈√2Γ(3/2){(1.87022)t1/4[(1−s1/2)1/2−1/2((3/4)1/2−s1/2)1/2]−(t1/2−s1/2)1/2,0≤s≤min{t,3/4};(1.87022)t1/4(1−s1/2)1/2−(t1/2−s1/2)1/2,3/4≤s≤t;(1.87022)t1/4[(1−s1/2)1/2−1/2((3/4)1/2−s1/2)1/2],t≤s≤3/4;(1.87022)t1/4(1−s1/2)1/2,max{t,3/4}≤s≤1, |
and
G2(t,s)≈22/3Γ(5/3){1.78902(1−t1/2)2/3s1/3,0≤s≤min{t,1/2};1.78902(1−t1/2)2/3[s1/3−(s1/2−(1/2)1/2)2/3],1/2≤s≤t;1.78902(1−t1/2)2/3s1/3−(s1/2−t1/2)2/3,t≤s≤1/2;1.78902(1−t1/2)2/3[s1/3−(s1/2−(1/2)1/2)2/3]−(s1/2−t1/2)2/3,max{t,1/2}≤s≤1, |
which satisfy the properties expressed in Lemma 2.3. Moreover, for t,s∈(0,1), we have G1(t,s)>0,G2(t,s)>0 and
√2Γ(3/2)(0.8702161)t1/4s1/2(1−s1/2)1/2≤G1(t,s)≤√2Γ(3/2)(1.87022)t1/4(1−s1/2)1/2, |
22/3Γ(5/3)(0.7890202)s1/3(1−t1/2)2/3(1−s1/2)≤G2(t,s)≤22/3Γ(5/3)(1.78902)s1/3(1−t1/2)2/3, |
with Q1≈0.465302 and Q2≈0.4410348.
For illustrating Theorem 3.1, let us take
f(t,u,ω)=e−t36√900+t(y2+|ω|2+cost), | (4.2) |
g(t,u,ω)=tan−1y+ω2(t+25)2, | (4.3) |
and p=3/2 (that is, q=3). Using the given values, it is found that Ω1=3.7750084,Ω2=3.9792441,Ω3=0.6243413andΩ4=0.2557131. Also, M satisfies the following relations:
2.977481684M4+0.1319508185M2+0.9259259259≤M1,0.2513274123+0.3249313918M4≤M2,3/4M1+2/3M21/21/4Γ(1/4+1)≤M, |
with ‖y‖≤56.706968M2 and ‖ω‖≤14.250688M2. Choosing M1=0.3,M2=0.08 and M=0.4, it can easily be verified that the functions f(t,y,ω) and g(t,y,ω) given by (4.2) and (4.3) respectively, satisfy the condition (A1). Furthermore, on the domain:
Υ0.4:={(t,y,ω),0≤t≤1,|y|≤9.073114878,|ω|≤2.280110144}, |
we find that
|fy|=|2e−ty36√900+t|≤0.01680206459,|fω|=|e−t72√900+t|≤0.001055606548,|gy|=|1(1+y2)(t+25)2|≤0.0016,|gω|=|2ω(t+25)2|≤0.00729635. |
Obviously the conditions (A2) and (A3) are satisfied with C1=0.01680206459,C2=0.001055606548,κ1=0.0016andκ2=0.00729635246. Also, L1=0.7169476783<1. Thus all of the conditions of Theorem 3.1 are satisfied and hence the problem (4.1) has a unique solution on Υ0.4.
We illustrate Theorem 3.2 by choosing
f(t,y,ω)=1120(y+ω2+5(t+3)), | (4.4) |
g(t,y,ω)=e−t3√t+900(y+3ω2/4) | (4.5) |
and p=4 (that is, q=4/3). Here the values of Ω1,Ω2andΩ3 are the same as those found in the first example and Ω4=0.592399. Letting m=0.05andM=0.4, as argued in the first example, we find that the functions f(t,y,ω) and g(t,y,ω) given by (4.4) and (4.5) respectively, satisfy the condition (B1) in the following domain: Υ+0.4:={(t,y,ω),0≤t≤1,0.1865284535t1/4≤y(t)≤4.565200828,−1.147253270≤ω≤−0.3148695697(1−√t)2/9}.
Moreover, the conditions (B2) and (B3) hold true with C1=1/120,C2=0.01912088784,κ1=1/90,κ2=0.005247826157 and L2=0.3153939012<1. Thus all of the conditions of Theorem 3.2 are satisfied and hence the problem (4.1) has a unique solution on Υ+0.4.
In this paper, we have investigated the criteria for ensuring the uniqueness of positive solutions for a class of fractional integro-differential equations with a p-Laplacian operator, complemented with nonlocal boundary conditions involving fractional derivatives and the p-Laplacian operator. Using a method employed in [31] together with the properties of the associated Green's functions established for the given problem, we proved two uniqueness results for the cases 1<p≤2 and p>2, respectively. Illustrative examples demonstrating application of the obtained results are presented. It is worthwhile to note that our results are new in the given configuration and enrich the literature on p-Laplacian fractional boundary value problems involving right-sided and left-sided fractional derivative operators, as well as left-sided fractional integral operators with respect to the power function.
The Deanship of Scientific Research (DSR) at King Abdulaziz University (KAU), Jeddah, Saudi Arabia has funded this project, under grant no. (KEP-PhD: 35-130-1443). The authors thank the Editor for indicating the correct terminology and references for the concepts of fractional calculus used in this paper. The authors also thank the reviewers for their constructive remarks on their work.
The authors declare that there is no conflict of interest.
[1] |
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