# Existence of Nontrivial Solutions for Fractional Differential Equations with p-Laplacian

Authors: Li Zhang; Fanglei Wang; Yuanfang Ru
Journal: Journal of Function Spaces
 (2019)

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## Abstract

Combining the properties of the Green function with some point theorems, we consider the existence of nontrivial solutions for fractional equations withp-Laplacian operator D0+βϕp[D0+α(ptu′(t))]+f(t,u(t))=0,0<t<1, au(0)-bp0u′(0)=0, and cu(1)+dp1u′(1)=0, D0+αptu′tt=0=0, where a,b,c,d are constants and p(·):[0,1]→(0,+∞) is continuous.


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## Body

## 1. Introduction

Fractional-order models are better than integer order models to describe the real word, which appear frequently in various fields, such as electrical circuits, biology, material, control theory, and physics (see [1–5]). With the rapid development of the theory of fractional differential equations, during the last two decades, the existence of nontrivial solutions of fractional differential equations has been studied by many researchers in nonsingular case as well as singular case. See [6–19]. Usually, the proof is based on either the method of upper and lower solutions, fixed point theorems, alternative principle of Leray-Schauder, topological degree theory, or critical point theory. To our attention, based on a fixed point theorem in cones, K. Lan and W. Lin [20] obtain some new results on existence of multiple positive solutions of systems of nonlinear Caputo fractional differential equations with some of general separated boundary conditions (1)-cDqzit=fit,zt,t∈0,1,αzi0-βzi′0=0,γzi1+δzi′1=0, where z(t)=(z1(t),…,zn(t)), fi:[0,1]×R+n→R+ is continuous on [0,1]×R+n, and cDq is the Caputo differential operator of order q∈(1,2). α,β,γ,δ are positive real numbers. The relations between the linear Caputo fractional differential equations and the corresponding linear Hammerstein integral equations are studied, which show that suitable Lipschitz type conditions are needed when one studies the nonlinear Caputo fractional differential equations.Recently, fractional differential equations withp-Laplacian operator have gained its popularity and importance due to its distinguished applications in numerous diverse fields of science and engineering, such as viscoelasticity mechanics, non-Newtonian mechanics, electrochemistry, fluid mechanics, combustion theory, and material science. There have appeared some results for the existence of solutions or positive solutions of boundary value problems for fractional differential equations with p-Laplacian operator; see [15, 21–26] and the references therein. For example, under different conditions K. Hasib, W. Chen and H. Sun [21] apply some classical fixed-point theorems to study the existence of positive solution for a class of singular fractional differential equations with nonlinear p-Laplacian operator in Caputo sense (2)DβϕpDϵut+Θtφ1t,ut=0,ϕpDϵutit=0=0,i=1,…,n-1uj0=0=u′′1,j=1,…,n,where Dβ,Dϵ is Caputo fractional derivative, n-1<β,ϵ≤n,ϕp(r)=rp-2r is p-Laplacian operator, and Θ(·) is continuous functions. In addition, Hyers-Ulam stability of the proposed problem is also considered.Inspired by the references, based on some fixed point theorems in cones, under different combinations of local superlinearity and local sublinearity of the functionf, we will deal with the existence of nontrivial solutions for a certain p-Laplacian fractional differential equation (3)D0+βϕpD0+αptu′t+ft,ut=0,0<α,β<1,au0-bp0u′0=0,cu1+dp1u′1=0,D0+αptu′tt=0=0, where a,b,c,d are constants with satisfying 0<ad+bc+ac∫01(1/ps)ds<+∞, p(·):[0,1]→R+ is continuous, and ϕp(r)=rp-2r is p-Laplacian operator, where 1/p+1/q=1 and ϕq denotes inverse of p-Laplacian operator. Now we give some notations as follows.Definition 1 (see [2]).
The fractional integral of orderα>0 of a function u:(0,+∞)→R is given by(4)I0+αut=1Γα∫0tt-sα-1usds,provided that the right-hand side integral is pointwise defined on (0,+∞), where Γ(α)=∫0+∞e-ssα-1ds.Definition 2 (see [2]).
The Caputo fractional derivative of orderα>0 of a continuous function u:(0,+∞)→R is given by(5)D0+αut=1Γn-α∫0tt-sn-α-1unsds, for n=[α]+1, where [α] is used for the integer part of α, provided that the right-hand side integral is pointwise defined on (0,+∞).Lemma 3 (see [2, Theorem 2.22]).
Letα∈(n-1,n], u∈ACn-1. Then(6)I0+αD0+αut=ut+c0+c1t+c2t2+⋯+cn-1tn-1,where ci=-u(i)(0)/i!, i=1,…,n-1.The paper is organized as follows. In Section2, we give some notations and the Green function is examined whether it is increasing or decreasing and positive or negative function. In Section 3, we will give the main results, which are illustrated by some examples.


## 2. Preliminaries

Lemma 4.
Leth(t)∈AC([0,1]) and 0<α,β<1. Then the solution of the fractional differential equation with p-Laplacian operator (7)D0+βϕpD0+αptu′t+ht=0,au0-bp0u′0=0,cu1+dp1u′1=0,D0+αptu′tt=0=0 can be expressed by(8)ut=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1hωdωdτ,where(9)Gt,τ=1ρΓαb+a∫0t1psdsd1-τα-1+c∫t1s-τα-1psds-Ht,τ,0≤τ≤t≤1;b+a∫0t1psdsd1-τα-1+c∫τ1s-τα-1psds,0≤t≤τ≤1,ρ=ad+bc+ac∫011psds,Ht,τ=ad+c∫t11psds∫τts-τα-1psds.Proof.
Via some computations, from Lemma3 it follows that(10)ptu′t+c1=-I0+αϕqc0+1Γβ∫0tt-sβ-1hsds.Since p(t)>0, we have (11)u′t=-c1pt-1Γαpt∫0tt-sα-1ϕqc0+1Γβ∫0ss-τβ-1hτdτds.Integrating both sides from 0 to t, we can obtain (12)ut=u0-c1∫0t1psds-∫0t1Γαps∫0ss-τα-1ϕqc0+1Γβ∫0ττ-ωβ-1hωdωdτds.Due to D0+αptu′tt=0=0, we get c0=0 and p0u′(0)=-c1. According to the boundary conditions au(0)-bp0u′(0)=0, we have au(0)+bc1=0 and u(0)=-(b/a)c1. Since(13)cu1+dp1u′1=0,p1u′1=-c1-1Γα∫011-sα-1ϕq1Γβ∫0ss-τβ-1hτdτds,u1=u0-c1∫011psds-∫011Γαps∫0ss-τα-1ϕq1Γβ∫0ττ-ωβ-1hωdωdτds,it is clear that(14)0=cu0-c1∫011psds-∫011Γαps∫0ss-τα-1ϕq1Γβ∫0ττ-ωβ-1hωdωdτds+d-c1-1Γα∫011-sα-1ϕq1Γβ∫0ss-τβ-1hτdτds.Since u(0)=-b/ac1, we have (15)c1-bc+ada-c∫011psds=∫01d1-sα-1Γαϕq1Γβ∫0ss-τβ-1hτdτds+∫01∫0scs-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds,which follows that (16)c1=-aρ∫01d1-sα-1Γαϕq1Γβ∫0ss-τβ-1hτdτds+∫01∫0scs-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds.
Then substitutingc1 and u(0), we obtain(17)ut=bρ∫01d1-sα-1Γαϕq1Γβ∫0ss-τβ-1hτdτds+∫01∫0scs-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds+aρ∫0t1psds∫01d1-sα-1Γαϕq1Γβ∫0ss-τβ-1hτdτds+∫01∫0scs-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds-∫0t∫0ss-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds=b+a∫0t1/psdsρ∫01d1-sα-1Γαϕq1Γβ∫0ss-τβ-1hτdτds+∫01∫0scs-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds-∫0t∫0ss-τα-1Γαpsϕq1Γβ∫0ττ-ωβ-1hωdωdτds=b+a∫0t1/psdsρ∫01d1-τα-1Γαϕq1Γβ∫0ττ-ωβ-1hωdωdτ+∫01∫τ1cs-τα-1Γαpsdsϕq1Γβ∫0ττ-ωβ-1hωdωdτ-∫0t∫τts-τα-1Γαpsdsϕq1Γβ∫0ττ-ωβ-1hωdωdτ=b+a∫0t1/psdsρ∫01d1-τα-1Γα+∫τ1cs-τα-1Γαpsdsϕq1Γβ∫0ττ-ωβ-1hωdωdτ-∫0t∫τts-τα-1Γαpsdsϕq1Γβ∫0ττ-ωβ-1hωdωdτ,which implies the expression of the Green function G(t,s).Lemma 5.
Assume thata,b,c,d>0, and p(t):[0,1]→(0,+∞). The Green function G(t,τ) has the following properties: (i)
G(t,τ)>0, for 0≤t,τ≤1;(ii)
G(t,τ) is an increasing function and maxt∈[0,1]G(t,τ)=G(1,τ);(iii)
For0≤t,τ≤1, there exists a C(t) = (b+a∫0t1/psds)/(b+a∫011/psds) ∈(0,1) such that(18)Gt,τ≥CtG1,τ.Proof.
(i) For0≤t≤τ≤1,(19)Gt,τ=1ρΓαb+a∫0t1psdsd1-τα-1+c∫τ1s-τα-1psds.Since a,b,c,d>0 and p(t)>0, we have ρ=ad+bc+ac∫011/psds>0 and G(t,τ)>0.
For0≤τ≤t≤1, (20)Gt,τ=1ρΓαb+a∫0t1psdsd1-τα-1+∫t1cs-τα-1psds-ad+c∫t11psds∫τts-τα-1psds.Let(21)ht=b+a∫0t1psdsd1-τα-1+∫t1cs-τα-1psds-ad+c∫t11psds∫τts-τα-1psds.Via some computations, we get (22)
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.Let F(t)=-ρt-τα-1 + ad1-τα-1 + ac∫τ1(s-τα-1/ps)ds. Since F′(t) = -(α-1)t-τα-2ρ>0 and F(τ)=ad1-τα-1 + ac∫τ1s-τα-1/p(s)ds>0, we obtain F(t)≥F(τ)>0, which implies that h′(t)=F(t)/p(t)>0 and h(t) is increasing on [τ,t). Furthermore, for 0≤τ≤t≤1, we have (23)Gt,τ=htρΓα≥hτρΓα=b+a∫0τ1psdsd1-τα-1+∫τ1cs-τα-1psds>0.
(ii) For0≤t≤τ≤1, since(24)∂Gt,τ∂t=aρΓαd1-τα-1+c∫τ1s-τα-1psds1pt>0, we obtain that G(t,τ) is increasing on t in (0,1], which implies that G(t,τ)≤G(1,τ).
For0≤τ≤t≤1, from the proof of (i) it follows that G(t,τ) is increasing on t in (0,1) and G(t,τ)≤G(1,τ).
Then from the above discussion, we can obtain the conclusion(25)maxt∈0,1⁡Gt,τ=G1,τ.
(iii) For0≤t≤τ≤1, (26)Gt,τG1,τ=ρΓαb+a∫0t1/psdsd1-τα-1+∫τ1cs-τα-1/psdsρΓαb+a∫011/psdsd1-τα-1+∫τ1cs-τα-1/psds=b+a∫0t1/psdsb+a∫011/psds=Ct.
For0≤τ≤t≤1, (27)Gt,τG1,τ=ρΓα​b+a∫0t1/psdsd1-τα-1+c∫t1s-τα-1/psdsρΓαd1-τα-1b+a∫011/psds-ad∫τ1s-τα-1/psds-ad+c∫t11/psds∫τts-τα-1/psdsρΓαd1-τα-1b+a∫011/psds-ad∫τ1s-τα-1/psds≥d1-τα-1b+a∫0t1/psds+cb+a∫0t1/psds∫t1s-τα-1/psdsd1-τα-1b+a∫011/psds-ad+c∫t11/psds∫τts-τα-1/psdsd1-τα-1b+a∫011/psds.Let(28)Kt=cb+a∫0t1psds∫t1s-τα-1psds-ad+c∫t11psds∫τts-τα-1psds,K1t=ac∫τ1s-τα-1psds-t-τα-1ρ.On one hand, since (29)K′t=acpt∫t1s-τα-1psds-cb+a∫0t1psdst-τα-1pt+acpt∫τts-τα-1psds-ad+c∫t11psdst-τα-1pt=1ptac∫τ1s-τα-1psds-t-τα-1bc+ad+ac∫011psds=1ptac∫τ1s-τα-1psds-t-τα-1ρ,we have K′(t)=K1(t)/p(t). On the other hand, it is not hard to obtain that K1′(t)=-(α-1)t-τα-2ρ>0. Then K1(t) is increasing on t, which implies that (30)K1t≥K1τ=ac∫τ1s-τα-1psds>0. Hence, we get K′(t)>0. Furthermore, K(t) is increasing on t and (31)Kt≥Kτ=cb+a∫0τ1psds∫τ1s-τα-1psds>0.Then, we have (32)Gt,τG1,τ≥db+a∫0t1/psds1-τα-1+Ktdb+a∫011/psds1-τα-1≥db+a∫0t1/psds1-τα-1db+a∫011/psds1-τα-1=b+a∫0t1/psdsb+a∫011/psds=Ct.
Therefore, from the above discussion, we get the conclusionG(t,τ)≥C(t)G(1,τ),t∈(0,1) and it is clear to see that 0<C(t)<1.At the end of this section, we give some notations and crucial lemmas.LetE=C[0,1] be the Banach space and it endowed with the norm u=maxt∈0,1ut:u∈E. Define a subcone K as (33)K=u∈E:ut≥Ctu,t∈0,1.For any given r>0, let(34)Ωr=u∈K:u<r,∂Ωr=u∈K:u=r.Lemma 6 (see [27]).
LetE be a Banach space, E1 a closed, convex subset of E, Ω an open subset of E1, and 0∈Ω. Suppose that T:Ω¯→E1 is completely continuous. Then either (i)
T has a fixed point in Ω¯, or(ii)
there are anu∈∂Ω (the boundary of Ω in E1) and λ∈(0,1) with u=λTu.Lemma 7 (see [27]).
LetE be a Banach space and K⊂E be a cone in E. Assume Ω1,Ω2 are open subsets of E with 0∈Ω1, Ω1¯⊂Ω2, and let T:K∩(Ω2¯∖Ω1)→K be a completely continuous operator such that either (i)
Tu≤u, u∈K∩∂Ω1 and Tu≥u, u∈K∩∂Ω2; or(ii)
Tu≥u, u∈∂Ω1 and Tu≤u, u∈K∩∂Ω2.ThenT has a fixed point in K∩(Ω2¯∖Ω1).Lemma 8 (see [21]).
Letϕp be a p-Laplacian operator. Then(i)
If1<p≤2,x1x2>0 and x1,x2≥λ>0, then(35)ϕpx1-ϕpx2≤p-1λp-2x1-x2.(ii)
Ifp>2, and x1,x2≤λ∗, then(36)ϕpx1-ϕpx2≤p-1λ∗p-2x1-x2.


## 3. Existence Results

For convenience, the following assumptions hold throughout this paper:(A1)
f:(0,1)×(0,+∞)→R is continuous;A1′
f:(0,1)×(0,+∞)→[0,+∞) is continuous;(A2)
there exists positive constantsμ1,μ2 and k∈[0,1] such that f satisfies(37)ft,ut≤ϕpμ1utk+μ2;(A3)
there exists a positive constantL such that for all u,v∈V,(38)ft,u-ft,v≤Lut-vt.In addition, let(39)ϖ0=∫01τβG1,τdτ,ω¯=∫01G1,τϕq1Γβ∫0ττ-ωβ-1dωdτ,max0≤t≤1⁡ft,ut=μ¯.Theorem 9.
Suppose that(A1), (A2), and (A3) hold and p>2. Then problem (3) has a unique solution if (40)Lq-1λ∗q-2Γβ+1ϖ0<1and2-qq-1q-1/2-qL1/2-qϖ1/2-q+μ¯≤0.Proof.
By Lemma4, (3) is equivalent to the following integral equation: (41)ut=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ.Define an operator T:C[0,1]→C[0,1] by(42)Tut=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ.From (A1) and (A2), the operator T is well defined.
LetBr0=u∈C0,1:u≤r0 with r0=q-1(q-1)/(2-q)L(q-1)/(2-q)ϖ1/(2-q). Then we have (43)Tu=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,uω-fω,0+fω,0dωdτ≤∫01G1,τϕqLr0+μ¯1Γβ∫0ττ-ωβ-1dωdτ=ϕqLr0+μ¯∫01G1,τϕq1Γβ∫0ττ-ωβ-1dωdτ=ϖLr0+μ¯q-1. Let g(r)=Lr-ϖ-1/(q-1)r1/(q-1)+μ¯, and then we have (44)gr0=Lq-1q-1/2-qLq-1/2-qϖ1/2-q-ϖ-1/q-1q-1q-1/2-qLq-1/2-qϖ1/2-q1/q-1+μ¯=q-1q-1/2-qL1/2-qϖ1/2-q-q-11/2-qL1/2-qϖ1/2-q+μ¯=q-11/2-qL1/2-qϖ1/2-qq-1-1-1+μ¯=2-qq-1q-1/2-qL1/2-qϖ1/2-q+μ¯≤0, which implies that Tu=ϖ(Lr0+μ¯)q-1≤r0. Therefore, we proved that T:Br0→Br0.
For anyu,v∈E, by Lemma 8, we have (45)Tu-Tv=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdω-ϕq1Γβ∫0ττ-ωβ-1fω,vωdωdτ≤∫01G1,τq-1λ∗q-21Γβ∫0ττ-ωβ-1fω,uωdω-1Γβ∫0ττ-ωβ-1fω,vωdωdτ≤q-1λ∗q-2Γβ∫01G1,τ∫0ττ-ωβ-1fω,uω-fω,vωdωdτ≤Lq-1λ∗q-2Γβ+1∫01τβG1,τdτu-v≤Lq-1λ∗q-2Γβ+1ϖ0u-v.Since Lq-1λ∗q-2/Γβ+1ϖ0<1, from Banach’s contraction mapping principle it follows that there exists a unique fixed point for the operator T, which corresponds to the unique solution for problem (3).Lemma 10.
Assume that(A1′) and (A2) hold. Then the operator T:K→K is completely continuous.Proof.
For anyu∈K, according to Lemma 5, we can get (46)Tut=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ,and (47)Tut=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≥Ct∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ. So we have(48)Tut≥CtTu,t∈0,1. This implies T:K→K.
GivenR>r>0, now we show that T is completely continuous on Ω(R)¯∖Ω(r).
Firstly, we will show thatT is continuous, and we only need to prove that T(un)-T(u)→0 for any un→u as n→∞. It is clear that (49)Tun-Tu=max0≤t≤1⁡∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,unωdωdτ-∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,unωdω-ϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ.From the continuity of f, ϕq(·), we have(50)ϕq1Γβ∫0ττ-ωβ-1fω,unωdω→ϕq1Γβ∫0ττ-ωβ-1fω,uωdωas n→∞. Therefore, T(un)-T(u)→0, as n→∞, and T is continuous.
Next, we show that the operatorT is uniformly bounded. By (A2), we get (51)Tu=max0≤t≤1∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1ϕpμ1utk+μ2dωdτ=∫01G1,τϕq1Γβ∫0ττ-ωβ-1dωdτμ1utk+μ2≤ω¯μ1utk+μ2, which implies that the operator T is uniformly bounded on Ω(R)¯∖Ω(r).
Finally, we show that the operatorT is equicontinuous. Before that, we will proceed with ∂G(t,τ)/∂t being bounded.
For0≤t≤τ≤1, (52)∂Gt,τ∂t=aρΓαptd1-τα-1+c∫τ1s-τα-1psds. Since the function p(·), d1-τα-1 and ∫τ1(s-τα-1/ps)ds are bounded, there exists a positive constant M1 such that ∂Gt,τ/∂t≤M1. In a similar way, for 0≤τ≤t≤1, there exists a positive constant M2 such that (53)∂Gt,τ∂t=1ρΓαpt-ρt-τα-1+ad1-τα-1+ac∫τ1s-τα-1psds≤M2. Choosing M¯=max⁡M1,M2, we can obtain ∂Gt,τ/∂t≤M¯.
Finally, for anyt1,t2∈[0,1] with t1<t2, there exists a ξ∈(t1,t2) such that (54)Tut2-Tut1=∫01Gt2,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ-∫01Gt1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ=∫01Gt2,τ-Gt1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ=∫01∂Gt,τ∂tt=ξt2-t1ϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01M¯t2-t1μ1utk+μ2ϕq1Γβ∫0ττ-ωβ-1dωdτ≤M¯μ1utk+μ2∫01ϕq1Γβ∫0ττ-ωβ-1dωdτt2-t1. Thus, the operator T is equicontinuous. According to Arzela-Ascoli theorem, T:Ω(R)¯∖Ω(r)→E is compact.Theorem 11.
Suppose that(A1) and (A2) hold. In addition,(C1)
there exists a continuous functiong:R+→R+ such that(55)ft,ut≤ϕpgu,for  anyt∈0,1,u∈R;(C2)
there exists a constantR such that R/ϖg(R)>1.
Then problem (3) has at least one solution.Proof.
Now we show the (ii) of Lemma6 does not hold. If u is a solution of (3), then, for λ∈(0,1), we have (56)u=λTu=λ∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤λ∫01G1,τϕq1Γβ∫0ττ-ωβ-1ϕpgudωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1dωϕpϕqgudτ≤guϖ. Let BR=u∈E:u<R. From the above inequality and (C2), it yields a contradiction. Therefore, the operator T has a fixed point in BR.Forr>0, define the following functions:(57)fMt,r=maxft,ut∣Ctr≤u≤r,fmt,r=min⁡ft,ut∣Ctr≤u≤r.Theorem 12.
Suppose that(A1′) and (A2) hold. In addition, there exist r0,R0∈R+ such that one of the following conditions satisfied: (B1)
(58)r0≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fmω,r0dωdτ<+∞and(59)∫01G1,τϕq1Γβ∫0ττ-ωβ-1fMω,r0dωdτ≤R0;or(B2)
(60)∫01G1,τϕq1Γβ∫0ττ-ωβ-1fMω,x0dωdτ<r0and(61)R0≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fMω,r0dωdτ<+∞.Then problem (3) has a positive solution u0∈K such that r0≤u0≤R0.Proof.
We only verify the case(B1). On one hand, for any u∈∂Ω(r0), we have C(t)r0≤u≤r0 and (62)Tu=max0≤t≤1∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≥max0≤t≤1∫01CtG1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ=max0≤t≤1⁡Ct∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≥∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≥∫01G1,τϕq1Γβ∫0ττ-ωβ-1fmω,r0dωdτ≥r0=u.Thus, T(u)≥u, for any u∈∂Ω(r0).
On the other hand, for anyu∈∂Ω(R0), we have C(t)R0≤u≤R0, t∈[0,1] and (63)Tu=max0≤t≤1∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1fMω,R0dωdτ≤R0=u. Then, T(u)≥u, for any u∈∂Ω(R0).
Therefore, By Lemma7, the operator T has a fixed point u0∈Ω(R0)¯∖Ω(r0) with r0≤u0≤R0.Theorem 13.
Suppose that(A1′) holds. In addition D1
limu→0+ft,u/u1/q-1=0;(D2)
there exists a constantK>0 such that f(t,u)≤K;(D3)
there existsR¯>0 and θ∈(0,1/2) such that (64)minϑR¯≤u≤R¯⁡ft,u>σR¯1/q-1,
where(65)0<ϑ=minθ≤t≤1-θ⁡Ct<1,σ=minθ≤t≤1-θ⁡Ct∫θ1-θG1,τϕq1Γβ∫0ττ-ωβ-1dωdτ-1.Then problem (3) has at least two solutions.Proof.
Sincelimu→0+⁡ft,u/u1/q-1=0, there exist ε>0 and r>0 such that f(t,u)<εu1/(q-1), for 0≤u≤r,t∈[0,1], where ε satisfies εq-1ϖ<1. For u∈∂Ω(r)=u∈E:u<r, we have (66)Tu=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕq1Γβ∫0ττ-ωβ-1εu1/q-1dωdτ≤εq-1u∫01G1,τϕq1Γβ∫0ττ-ωβ-1dωdτ≤εq-1ϖu<u.Choosing R>Kq-1ϖ. For u∈∂Ω(R), we have (67)Tu=∫01Gt,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≤∫01G1,τϕqK1Γβ∫0ττ-ωβ-1dωdτ=Kq-1ϖ<R=u.For any u∈∂Ω¯(R¯), choosing t0∈(θ,1-θ), we have u(t0)∈[ϑR¯,R¯]. Furthermore, we have (68)Tut0=∫01Gt0,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≥∫01Ct0G1,τϕq1Γβ∫0ττ-ωβ-1fω,uωdωdτ≥∫θ1-θCt0G1,τϕq1Γβ∫0ττ-ωβ-1minϑR¯≤u≤R¯⁡fω,uωdωdτ≥minθ≤t≤1-θCt∫θ1-θG1,τϕq1Γβ∫0ττ-ωβ-1dωϕqσR¯1/q-1dτ=δ-1δR¯=R¯=u.By Lemma 7, problem (3) has at least two positive solution r≤u1(t)≤R¯ and R¯≤u2(t)≤R.At the end of this section, we give some examples to illustrate our main results.Example 1.
Let us consider the problem(69)D0+βϕpD0+αptu′t+ϕpsin⁡tarctan⁡u1/3+cos⁡u1/2+3=0,au0-bp0u′0=0,cu1+dp1u′1=0,D0+αptu′tt=0=0. It is clear to see that ft,u=ϕpsin⁡tarctan⁡u1/3+cos⁡u1/2+3≤ϕpu1/3+u1/2+3=ϕpgu, for t∈[0,1], u∈R. In addition, there exists a sufficiently large R>0, we have R/ϖg(R)>1. Therefore, problem (69) has at least one solution by Theorem 11.Example 2.
Let us consider the problem(70)D0+βϕpD0+αptu′t+σpp-1+1pϑpe-pϑsin⁡t+2e-uup=0,au0-bp0u′0=0,cu1+dp1u′1=0,D0+αptu′tt=0=0. Since f(t,u)=σpp-1+1/(pϑ)pe-pϑsin⁡t+2e-uup, we have f(t,u)/up-1 = σpp-1+1/(pϑ)pe-pϑ(sin⁡t+2)ue-u→0 as u→0+, and (D2) hold. By some calculations, we get f′(t,u) = (σpp-1+1)/(pϑ)pe-p-1ϑ(sin⁡t+2)e-uup-1(1/p-u), and it is clear to see that f′(t,u)>0, for u∈(0,1/p), and f′(t,u)<0, for u∈(1/p,+∞). Let R¯=1/p, and then for any u∈(ϑ1/p,1/p), we have minu∈ϑ1/p,1/pf(t,u) = (σpp-1+1/(pϑ)pe-pϑ)(sin⁡t+2)(pϑ)pe-1/pϑ ≥ (pσ)p-1+1 ≥ (R¯ϑ)p-1, and (D3) holds. Therefore, problem (70) has at least two positive solutions u(t) by Theorem 13.



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*Source: 3486410-2019-01-01.xml*