# Positive Solution for the Integral and Infinite Point Boundary Value Problem for Fractional-Order Differential Equation Involving a Generalizedϕ-Laplacian Operator

Authors: Nadir Benkaci-Ali
Journal: Abstract and Applied Analysis
 (2020)

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## Abstract

In this paper, we establish the existence of nontrivial positive solution to the following integral-infinite point boundary-value problem involvingϕ-Laplacian operator D0+αϕx,D0+βux+fx,ux=0,x∈0,1,D0+σu0=D0+βu0=0,u1=∫01gtutdt+∑n=1n=+∞αnuηn, where ϕ:0,1×R→R is a continuous function and D0+p is the Riemann-Liouville derivative for p∈α,β,σ. By using some properties of fixed point index, we obtain the existence results and give an example at last.


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## Body

## 1. Introduction

Our aim in this article is to study the existence of a nontrivial positive solution to the following integral and infinite point boundary-value problem involving a two-dimensionalϕ-Laplacian operator
(1)D0+αϕx,D0+βux+fx,ux=0,x∈0,1,D0+σu0=D0+βu0=0,u1=∫01gtutdt+∑n=1n=+∞αnuηn,where D0+p is the Riemann-Liouville derivative for p∈α,β,σ, 0<α≤1≤β≤2 and
(2)σ=0if1<β≤2,1ifβ=1,and αn,ηn∈0,1 for n≥1 such that
(3)∑n=1n=+∞αn<∞.Throughout this paper, we assume that the following conditions are satisfied;(A1)η0=0<ηn<ηn+1<1 for n∈N with
(4)limn→+∞ηn=η≤1.(A2)f:0,1×R+→R+ is continuous and g:0,1→R+ is an integrable function.(A3)∫01tβ−1gtdt+∑n=1n=+∞αnηnβ−1<1.(A4)ϕ:0,1×R→R is continuous and for t∈0,1, the function ϕt,. is odd and increasing, ϕ−1t,. is the inverse function of ϕt,. denoted by ψt,. where ψ:0,1×R→R is continuous.(A5) There existp+,p−∈ℝ with p+≥p−>1 such that
(5)ϕ−x≤ϕ.,x≤ϕ+xfort,x∈0,1×ℝ,with
(6)ϕ−x=ϕp+xifx∈0,1∪−∞,−1,ϕp−xifx∈−1,0∪1,+∞,and
(7)ϕ+x=ϕp−xifx∈0,1∪−∞,−1,ϕp+xifx∈−1,0∪1,+∞.Boundary value problems involving apt-Laplacian operator have attracted a great deal of attention in the last ten years (see [1] [2–9]). At the same time, boundary value problems with fractional-order differential equations involving pt-Laplacian are of great importance and are an interesting class of problems. Such kind of BVPs in Banach space has been studied by many authors, see, for example [10–13] and the references therein. Noting that the generalized ϕ-Laplacian operator can turn into the well-known pt-Laplacian operator when we replace ϕ by ϕptx=xpt−2x, so our results extend and enrich some existing papers.By using the homotopy deformation property of the fixed point index, our paper aims at investigating the existence of at least one positive solution for bvp 1.The paper is organized as follows. In the first section, we recall some lemmas giving fixed point index calculations. In the second section, we present a fixed point formulation for bvp (1), and we close this section by some lemmas making use of homotopical arguments. After that, we give our main results and their proofs and we end by giving as an example, a problem involving a sum of manypt-Laplacian operators.


## 2. Preliminaries

For the sake of completeness, let us recall some basic facts needed in this paper. LetE be a real Banach space equipped with its norm noted ., LE is the set of all linear continuous mapping from E into E . For L∈LE, rL=limn→∞∥Ln∥1/n denotes the spectral radius of L. A nonempty closed convex subset K of E is said to be a cone if K∩−K=0 and tK⊂K for all t≥0.LetK be a cone in E. A cone K induces a partial ordering “≤”, defined so that x≤y if and only if y−x∈K.K is said to be normal if there exists a positive constant N such that for all u,v∈K,
(8)u≤vimplies∥u∥≤N∥v∥.L∈LE is said to be positive in K if LK⊂K, it is said to be strongly positive in K if intK≠∅ and LK0⊂intK, and it is said to be K-normal if for all u,v∈K,
(9)u≤vimpliesLu≤Lv.LetE be a real Banach space and let K be a cone.LetR>0, B0,R be the ball of radius R in E and A:KR→K a completely continuous mapping, where KR=B0,R∩K. We will use the following lemmas concerning computations of the fixed point index, i, for a compact map A (see [14]).Lemma 1.
If∥Ax∥<∥x∥ for all x∈∂B0,R∩K, then
(10)iA,KR,K=1.Lemma 2.
If∥Ax∥>∥x∥ for all x∈∂B0,R∩K, then
(11)iA,KR,K=0.Lemma 3.
IfAx≥x for all x∈∂B0,R∩K, then
(12)iA,KR,K=1.Lemma 4.
IfAx≤x for all x∈∂B0,R∩K, then
(13)iA,KR,K=0.Lemma 5.
IfAx≠λx for all x∈∂B0,R∩K and λ>1, then
(14)iA,KR,K=1.


## 3. Related Lemmas

LetN:E→E be an operator and K be a cone of a real Banach space E, and consider the partial ordering “≤” in E, defined so that x≤y if and only if y−x∈K.Letρ∈K∗, and consider the following cone.P=Kρ=u∈K:u≥uρ and the positive value
(15)λ0K=infΛ−K,where
(16)Λ−K=λ≥0:thereexistsu∈K∩∂B0,1suchthatNu≤λu.Remark 6.
It is clear that IfN is completely continuous, then from Lemmas 4 and 5, there exist λ≥1 such that λ∈Λ−K.Lemma 7.
Assume thatN:E→E is increasing, positively 1-homogeneous, and completely continuous, such that NK0⊂K\0.
If there existρ∈K∗ such that NK⊂P=Kρ, then
(17)λ0K=λ0P>0.Proof. In first, we claimλ0K=λ0P.Letλ≥0,u∈K∩∂B0,1 such that Nu≤λu . Since NK⊂P, N is strictly increasing and positively 1-homogenuous, we have
(18)NNuNu≤λNuNu,then
(19)Λ−K⊂Λ−P,with P⊂K we deduce
(20)Λ−P=Λ−K,and so
(21)λ0K=λ0P.Now, we show that(22)λ0P>0.In the contrary, we assume that there existλn∈R+ and un∈P∩∂B0,1n with limn→∞λn=0. Such that
(23)Nun≤λnun.Forn∈N,
(24)λnun≥Nun≥Nunρ=Nρ,and so
(25)λnun−Nρ∈K.Then,limn→∞λnun−Nρ=−Nρ∈K, and we obtain
(26)Nρ=0,which is a contradiction.Remark 8.
IfK is a normal cone in a Banach space E, with the constant of normality n=1 (i.e, u≥v if u≥v≥0), then
(27)λ0K≥Nρ.Since forλ∈ΛP,u∈P∩B0,1(28)λu≥Nu≥Nuρ=Nρ.In the following lemma, we assume thatN0:E→E is a positively 1-homogeneous and completely continuous operator, and N:E→E is a completely continuous, increasing and positively 1-homogeneous operator, such that
(29)NK\0⊂P\0,where P=Kρ, ρ∈K∗, and K is a normal cone in a Banach space E, with the constant of normality n=1.Lemma 9.
LetQ,Q0,G2:K→K be continuous mappings with
(30)limu→+∞Quu<+∞andlimu→+∞G2uu=0≤limu→+∞Q0uu<+∞,such that
(31)NQu−G2u≤N0Q0u,foru∈K.Suppose that there existλ1∈ℝ+ and G1:K→K with
(32)limu→+∞G1uu=0,such that
(33)Qu≥λ1u−G1u,foru∈K.If(34)λ1>λ0−1K,then there exist R1>0 such that for all R≥R1(35)iN0Q0,KR,K=iNQ,PR,P.Moreover, if(36)λ1>Nρ−1,then there exist R2>0 such that for all R≥R2(37)iN0Q0,KR,K=0.Proof. In first, we show that there existR1>0 such that for all R≥R1(38)iNQ,KR,K=iN0Q0,KR,K.We consider the homotopy(39)Ht,u=tN0Q0u+1−tNQu.We show that there existR1>0 such that for all R≥R1 the equation Ht,u=u has not solutions in 0,1×K∩∂B0,R. In the contrary, we assume that for all n∈N, there exist Rn≥n and tn,un∈0,1×K∩∂B0,Rn such that
(40)un=Htn,un=tnN0Q0un+1−tnNQun.By dividing the above equation byun, we obtain
(41)vn=unun=tnN0Q0unun+1−tnNQunun.From (5),
(42)limn→∞Q0unun<∞andlimn→∞Qunun<∞,then the sequences Q0un/unn,Qun/unn are bounded, and we deduce from the compactness of N and N0, that vnn admits a convergent subsequence also denoted by vnn.Letv=limn→∞vn∈K∩∂B0,1 and t=limn→∞tn.By using the conditions (31) and (33), it follows from 12 that for all n∈N(43)vn≥NQunun−tnG2unun≥Nλ1vn−G1unun−tnG2unun.With the fact that(44)limn→+∞G1unun=limn→+∞G2unun=0,we have
(45)v≥λ1Nv,and so
(46)λ1−1∈Λ−K,where
(47)Λ−K=λ≥0;thereexistsu∈K∩∂B0,1suchthatNu≤λu.Then(48)λ1−1≥λ0K,which contradicts (10).Then, there existR1>0 such that for all R≥R1 the equation Ht,u=u has not the solutions in
(49)0,1×K∩∂B0,R,and by invariance property of fixed point index, we deduce that for all R≥R1.(50)iNQ,KR,K=iN0Q0,KR,K.By the fact thatNQK⊂P, we have from the excision property of the fixed point index that
(51)iNQ,KR,K=iNQ,PR,P.Then(52)iN0Q0,KR,K=iNQ,PR,P.Now, we assume that the condition (36) holds.By using Lemma4, we prove that there exists R0>0 such that, for all R≥R0(53)iNQ,PR,P=0.In the contrary, we assume that for alln∈N, there exist Rn≥n and un∈P∩∂B0,Rn such that
(54)un≥NQun.By the condition (33), we have
(55)vn=unun≥NQunun−tnG2unun≥Nλ1vn−G1unun−tnG2unun,with
(56)limn→∞G1unun=limn→∞G2unun=0.As(57)un≥ρun,we have
(58)vn≥Nλ1ρ−G1unun−tnG2unun.Set(59)An=Nλ1ρ−G1unun−tnG2unun−Nλ1ρ.We have(60)limn→∞An=0,vn−An≥Nλ1ρ≥0.SinceK is normal with the constant of normality N=1, then for n∈N,(61)vn−An≥Nλ1ρ,and so
(62)1=limn→∞vn=limn→∞vn−An≥λ1Nρ,then
(63)1≥λ1Nρ,which contradicts (10).Consequently, forR≥R2=maxR1,R0,
(64)iN0Q0,KR,K=iNQ,PR,P=0.Definition 10.
[10, 11] The Riemann-Liouville fractional integral of order p>0 of f∈L1a,b,ℝ+ is defined by
(65)Ia+pfx=1Γp∫axx−tp−1ftdt,where Γ is the gamma function.Definition 11.
[10, 11] The Riemann-Liouville fractional derivative of order p≥0 of a function f is defined by
(66)Da+pfx=dndxnIa+n−pfx,n=α+1,where n is the integer part of α.Remark 12.
Ifp∈ℕ, then
(67)Da+pf=δpδxpf,and for p=1,
(68)Ia+1fx=∫axftdt.Lemma 13.
[15] Let p>0, and let ut be an integrable function in a,b.
(69)Ia+pDa+pux=ux+c1x−ap−1+c2x−ap−2⋯+cnx−ap−n,where ck∈ℝ, k∈1,2,⋯,n, n=α+1 and n is the integer part of α.Lemma 14.
Leth∈L0,1, 0<α≤1≤β≤2 and
(70)σ=0if1<β≤2,1ifβ=1.Then the unique solution of(71)D0+αϕx,D0+βux+hx=0,x∈0,1,D0+σu0=D0+βu0=0,u1=∫01gtutdt+∑n=1n=+∞αnuηn,is given by
(72)ux=∫01Gx,tHtdt,with
(73)Ht=ψt,1Γα∫0tt−sα−1hsds,and
(74)Gx,t=1Γβxβ−1Gmt−x−tβ−1if0≤t<minx,η,xβ−1Gηt−x−tβ−1ifη≤t≤x,xβ−1Gmtifx≤t<η,xβ−1Gηtift≥maxx,ηwith
(75)η=limn→∞ηn,,and m∈ℕ∗ such that
(76)ηm−1≤t≤ηm,where
(77)Gmt=μt−∑n≥mαnηn−tβ−11−L,Gηt=μt1−L,μt=1−tβ−1−∫t1s−tβ−1gsds,with
(78)L=∑n≥1αnηnβ−1+∫01sβ−1gsds<1.Proof. By Lemma13, equation
(79)D0+αϕx,D0+βux+hx=0,gives
(80)ϕx,D0+βux=−I0+αhx+c1xα−1if0<α<1,−I0+1hx+c1+c2x−1ifα=1.SinceD0+βu0=0, we have that c1=c2=0 and
(81)D0+βut=−Ht,with
(82)Ht=ψt,1Γα∫0tt−sα−1hsds.And also from Lemma13, we have
(83)ux=−I0+βHx+d1xβ−1+d2xβ−2if1≤β<2,−I0+βHx+d1x+d2+d3x−1ifβ=2.Ifβ≠1, then the condition u0=0 leads d2=d3=0, and if β=1, the equation D0+1u=δu/δx=−H leads
(84)ux=−I0+1Hx+d1,with
(85)d1=u0.Then(86)ux=−I0+βHx+d1xβ−1,forβ∈1,2.In addition, from equation(87)u1=∑n≥1αnuηn+∫01gsusds,we deduce that
(88)Γβ1−Ld1=−∑n≥1αn∫0ηnηn−tβ−1Htdt−∫01gt∫0tt−sβ−1Hsdsdt+∫011−tβ−1Htdt,with L=∑n≥1αnηnβ−1+∫01sβ−1gsds . The Fubini’s theorem gives
(89)∫01Ht∫t1s−tβ−1gsdsdt=∫01Ht∫t1s−tβ−1gsdsdt.Then(90)ux=1ΓβC.xβ−11−L−∫0xx−tβ−1Htdt,where
(91)C=∫011−tβ−1Htdt−∫01Ht∫t1t−sβ−1gsdsdt−∑n≥1αn∫0ηnηn−tβ−1Htdt.Consequently, the solution of (19) is
(92)ux=∫01Gx,tHtdt,with
(93)Gx,t=1Γβxβ−1Gmt−x−tβ−1if0≤t<minx,η,xβ−1Gηt−x−tβ−1ifη≤t≤x,xβ−1Gmtifx≤t<η,xβ−1Gηtift≥maxx,η,with
(94)η=limn→∞ηn,,and m∈N∗ such that
(95)ηm−1≤t≤ηm,where
(96)Gmt=μt−∑n≥mαnηn−tβ−11−L,Gηt=μt1−L,μt=1−tβ−1−∫t1s−tβ−1gsds.This finishes the proof.Lemma 15.
Forx,t∈0,1, we have
(97)h1txβ≤Gx,t≤h2txβ−1,where
(98)h1t=1−tβ−1∫0tsβ−1gsdsΓβ1−L,h2t=μtΓβ1−L,with
(99)L=∫01sβ−1gsds+∑n≥1αnηnβ−1.Proof. It is clear that the right hand inequality(100)Gx,t≤h2txβ−1,is obvious.Now, we show that(101)Gx,t≥h1txβ,where
(102)h1t=1−tβ−1∫0tsβ−1gsdsΓβ1−L,with
(103)L=∫01sβ−1gsds+∑n≥1αnηnβ−1.Letx,t∈0,1.Forn∈N∗, as t≥tηn and t≥tx, we have
(104)ηn−tβ−1≤ηnβ−11−tβ−1,x−tβ−1≤xβ−11−tβ−1,and t≥ts gives
(105)∫t1s−tβ−1gsds≤1−tβ−1∫t1sβ−1gsds.Fort,x∈0,1, we have
(106)Gx,t≥xβ−11−tβ−1Γβ1−∫t1sβ−1gsds−∑n≥1αnηnβ−11−L−1,and with xβ−1≥xβ leads
(107)Gx,t≥xβ1−tβ−1Γβ∫01sβ−1gsds−∫t1sβ−1gsds1−L,then
(108)Gx,t≥xβh1t.Remark 16.
The functionG:0,1×0,1→ℝ defined by (20) is continuous, and from Lemma 15, we have Gx,t≥0.According to Lemma14, u is solution of 1 if and only if u is a fixed point of the operator
(109)Tux=∫01Gx,tψt,1Γα∫0tt−sα−1fs,usdsdt.T can be written as
(110)T=N0Q0,where
(111)N0ux=∫01Gx,tutdtQ0ut=ψt,1Γα∫0tt−sα−1fs,usds.Remark 17.
LetE=C0,1 be the Banach space equipped with the sup-norm u=supx∈0,1ux, and the cone
(112)K=E+=u∈E;u≥0.We have from 2 of the condition (A5) that(113)ψ+x≤ψ.,x≤ψ−xfort∈0,1,where ψ−,ψ+ are the inverse functions of ϕ−,ϕ+, respectively, defined by
(114)ψ−x=ψp+xifx∈0,1∪−∞,−1,ψp−xifx∈−1,0∪1,+∞,and
(115)ψ+x=ψp−xifx∈0,1∪−∞,−1,ψp+xifx∈−1,0∪1,+∞.where ψ+ is the inverse function of ϕ+ defined in the condition (A5).Remark 18.
There existc,e>0 such that for all t,x∈0,1×ℝ+,
(116)ψp−x+e≥ψt,x≥ψp+x−c.Remark 19.
By Lemma15 and Remark 18, we have for u∈K(117)N0Q0u≥NQu−G2u,where
(118)Nux=xβ∫01h1tψp+1Γα∫0tt−sα−1ϕp+usdsdt,and
(119)Qux=ψp+fx,ux,with
(120)G2ux=c.xβ∫01h1tdt,Moreover, the linear operatorN0:E→E is compact, and N:E→E is completely continuous, increasing, positively 1- homogeneous and verifying
(121)NK\0⊂P\0,where
(122)P=Kρ=u∈K;u≥ρu,with
(123)ρx=xβ∈K∗.


## 4. Main Results

Set(124)λ=ϕp+∫01h2tψp+tαΓα+1dt−1.Theorem 20.
Assume that there existr0>0,r1>0 and
(125)γ>ϕp+Nρ−1,such that
(126)ft,x<λϕp+x,fort,x∈0,1×0,r0,and
(127)ft,x≥γϕp+x,fort,x∈0,1×r1,+∞,with
(128)limx→+∞supt∈0,1ft,xϕp−x<∞,then problem 1 has at least one nontrivial positive solution.Proof. In first, we show thatiT,Kr,K=1 where
(129)r=minr0,1,ψp+Γα+1λ.From (25), we have
(130)ft,x<λϕp+x,t,x∈0,1×0,r.Foru∈∂B0,r∩K,
(131)Tux=∫01Gt,xψt,1Γα∫0tt−sα−1fs,usdsdt≤∫01xβ−1h2tψt,1Γα∫0tt−sα−1fs,usdsdt≤∫01h2tψ−1Γα∫0tt−sα−1fs,usdsdt<∫01h2tψ−1Γα∫0tt−sα−1λϕp+usdsdt<∫01h2tψ−1Γα+1tαλϕ−rdt,and from the definition of the constant r, we have that
(132)1Γα+1tαλϕp+r≤1,then
(133)Tux<∫01h2tψp+1Γα+1tαλϕp+rdt.Then,(134)Tu<u.By Lemma1,
(135)iT,Kr,K=1.Now, by using Lemma9, we show that there exists R>0 such that
(136)iT,KR,K=0.In first, we have from Remark19 that
(137)T=N0Q0≥NQ−G2,where
(138)G2u=c.Aslimu→+∞G2u/u=0, then the condition (31) of Lemma 9 is satisfied.Now, we have from (26), for x≥r1(139)ψp+ft,x≥λ1x,with
(140)λ1=ψp+γ>Nρ−1.Then, there existsd∈R such that
(141)ψp+ft,x≥λ1x−d,forx≥0.and set
(142)G1u=d.We have foru∈K(143)Qut≥λ1ut−G1ut,with
(144)limu→∞G1uu=0.Moreover, from Remark18, for u∈K,
(145)Qut=ψp+ft,ut≤ψp−ft,ut+e+c,fort∈0,1,and
(146)Q0ut=ψt,1Γα∫0tt−sα−1fs,usds≤ψp−t,1Γα∫0tt−sα−1fs,usds+e.Then, from (27), we have
(147)limu→+∞Quu<∞andlimu→+∞Q0uu<∞.By Lemma9, there exist R>r0 such that
(148)iN0Q0,KR,K=0.Consequently,T=N0Q0 has at least one fixed point u in K∩B¯0,R\B0,r, which is a nontrivial positive solution for problem 1.Set(149)λ2=ϕp−∫01h2tψp−tαΓα+1dt−1,and
(150)N2ux=xβ∫01h1tψp−1Γα∫0tt−sα−1ϕp−usdsdt.Theorem 21.
Assume that there existr2>0,r3>0 and
(151)γ>ϕp−N2ρ−1,such that
(152)ft,x<λ2ϕp−x,fort,x∈0,1×r2,+∞,and
(153)ft,x≥γϕp−x,fort,x∈0,1×0,r3,then problem 1 has at least one nontrivial positive solution.Proof. In first, by using Lemma3, we show that there exists R≥r2 such that iT,PR,P=1. In the contrary, we assume that there exists a sequence unn in P with
(154)limn→∞un=∞,such that
(155)Tun≥un.From (28), there exist ε>0 and b∈R such that
(156)ft,x≤λ2−εϕp−x+b,t,x∈0,1×0,+∞.Then, forn∈N(157)un≤Tunx=∫01Gt,xψt,1Γα∫0tt−sα−1fs,unsdsdt≤∫01xβ−1h2tψt,1Γα∫0tt−sα−1fs,unsdsdt≤∫01h2tψp−1Γα∫0tt−sα−1fs,unsdsdt+e∫01h2tdt≤∫01h2tψp−1Γα∫0tt−sα−1λ−εϕp−uns+bdsdt+e∫01h2tdt≤unψp−λ−ε∫01h2tψp−1Γα∫0tt−sα−11+rndsdt+e∫01h2tdt,where
(158)rn=bϕp−unλ−ε.Then,(159)1≤ψp−λ−ε∫01h2tψp−1Γα∫0tt−sα−11+rndsdt,with
(160)limn→∞rn=0=limn→∞e∫01h2tdtun,it follows the following contradiction
(161)1≤ψp−λ−εψp−λ−1<1.Then, there existsR≥r2 such that
(162)iT,PR,P=1.Now, we prove thatiT,Pr3,P=0.Letu∈P∩∂B0,r0, with
(163)r0=min1,r3,ψp−Γα+1γ.We haveu≥ρu, and from (29)
(164)Tu1≥∫01h1tψt,1Γα∫0tt−sα−1γϕp−usdsdt≥∫01h1tψp−1Γα∫0tt−sα−1γϕp−ρsudsdt.Then,(165)Tu≥ψp−γN2ρu>u.From Lemma1, we have
(166)iT,Pr0,P=0.Consequently,T=N0Q0 has at least one fixed point u in K∩B¯0,R\B0,r0, which is a nontrivial positive solution for problem 1.Example 22.
We consider the followingp1x,p2x,⋯,pnx-Laplacian boundary value problem
(167)∑k=1k=ND0+αϕpkxx,D0+βux+hx,ux=0,x∈0,1,D0+σu0=D0+βu0=0,u1=∫01gtutdt+∑n=1n=+∞αnuηn,where ϕpkt is the pkt-Laplacian operator defined in 0,1×ℝ as
(168)ϕpktt,x=xpkt−2.x,fork∈1,2,⋯,N,N∈ℕ∗,with
(169)pkt∈C10,1,1,+∞.We consider the problem 1 withft,x=ht,x/N and ϕt,x=1/N∑k=1k=Nϕpktt,x. We assume that the conditions (A1), (A2), and (A3) are satisfied, and ϕ verifies (A4) and (A5) with
(170)p+=maxpkt,t∈0,1,k∈1,2,..,N,and
(171)p−=minpkt,t∈0,1,k∈1,2,..,N.Set(172)λ=ϕp+∫01h2tψp+tαΓα+1dt−1,λ2=ϕp−∫01h2tψp−tαΓα+1dt−1,and
(173)γ0=maxϕp+Nρ−1,ϕp−N2ρ−1.We deduce from Theorems20 and 21 that, if there exist R0>0,R1>0 and γ>γ0 such that h verifies one of the following conditions;(H1)(174)ht,x<Nλϕp+x,fort,x∈0,1×0,R0.ht,x≥Nγϕp+x,fort,x∈0,1×R1,+∞,and
(175)limx→+∞supt∈0,1ht,xϕp−x<∞,or(H2)(176)ht,x<Nλ2ϕp−x,fort,x∈0,1×R1,+∞,and
(177)ht,x≥Nγϕp−x,fort,x∈0,1×0,R0,then problem (30) has at least one nontrivial positive solution.



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*Source: 2127071-2020-08-28.xml*