# Infinite Existence Solutions of Fractional Systems with Lipschitz Nonlinearity

Authors: Rafik Guefaifia; Salah Mahmoud Boulaaras; Bahri Cherif; Taha Radwan
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

The paper deals with the existence of infinitely many solutions of a class of perturbed nonlinear fractionalp-Laplacian differential systems using one control parameter combined with the variational method.


---

## Body

## 1. Introduction

Fractional differential equations (FDEs) involve fractional derivatives of the formdα/dxαα>0, where is not necessarily an integer. They are generalizations of the ordinary differential equations to a random (noninteger) order. FDEs have attracted considerable interest due to their ability to model complex phenomena in several fields of science, engineering, physics, biology, and economics (see [1–7]). In summary, many improvements have been made in the theory of partial calculus and partial differential equations and partial and ordinary differential equations (see [8–18], [2, 5]). Numerous studies have explored the existence and solutions of different nonlinear elementary and boundary value problems through the use of various nonlinear analysis tools and techniques (see, for example, [7, 19–38]). Some of these ways are the fixed point theorems, critical point theory, monotone iterative methods, coincidence degree theory, and variational methods (see [30]).Motivated by the above, the interest of this paper is the infinite existence solutions of the following fractional system(1)tDTαΦp0Dtαut=λFut,ut,vt+h1u1,a.e.t∈0,T,tDTβΦp0Dtβvt=λFvt,ut,vt+h2u2,a.e.t∈0,T,u0=uT=0,v0=vT=0,where λ is a positive real parameter, α,βϵ0;1, and 0Dtα, tDTα and 0Dtβ, tDTβ are the left and right Riemann-Liouville fractional derivatives of order α,β, respectively, Φps=sp−2s,p>1, H0F:0,T×ℝ2→ℝ, where F·,u,v is continuous in 0,T for any u,v∈ℝ2, Ft,·,· is a C1 function in ℝ2, and Fs is the partial derivative of F with respect to s, and hi:ℝ⟶ℝ are two Lipschitz continuous functions of order p–1 with Lipschitzian constants Li>0 for 1≤i≤2, i.e.,
(2)hix1−hix2≤Lix1−x2p−1.


## 2. Preliminaries

We give some basic lemmas and notations and construct a variational framework in order to apply critical point theory to prove the existence of an infinite number of solutions to the system (1).LetX be a real Banach space, and in addition, let YX denote the class of all functionals
(3)ϕ=X⟶ℝ,,that possess the following property:Ifwn is a sequence in X converge weakly to w∈X with limn→∞ inf ϕwn≤ϕw; thus, wn has a subsequence converge strongly to w.For offer, ifX is uniformly convex and S:0,+∞⟶ℝ is a continuous strictly increasing function, then the functional w⟶Sw belongs toYX.Definition 1 (see Kilbas et al. [4] chapter 2, p. 87).
Letu be a function defined on [a, b]. The right and left Riemann-Liouville fractional derivatives of order>0 for a function u are defined by
(4)Datαut≔dndtnDatα−nut=1Γn−αdndtn∫att−sn−α−1usds,(5)Dabαut≔−1ndndtnDabα−nut=−1nΓn−αdndtn∫abt−sn−α−1usds,for all t∈a,b, provided the right-hand sides are pointwise defined on a,b, where n−1≤α<n and n∈ℕ.
Here,Γα is the standard gamma function given by
(6)Γα≔∫0+∞zα−1e−zdz.
SetACna,b,ℝ the functions space u:a,b⟶ℝ such that
(7)u∈Cn−1a,b,ℝ,with
(8)un−1∈ACna,b,ℝ.
As usual,Cn−1a,b,ℝ denotes the mapping set having n−1 times continuously differentiable on a,b. In particularly, we have
(9)ACa,b,ℝ≔AC1a,b,ℝ.Definition 2 (see [31]).
Let0<α≤1, for 1<p<∞ the derivative fractional space
(10)Eαp=ut∈Lp0,T,ℝ0Dtαut∈Lp0,T,ℝ,u0=uT=0.Thus, for all u∈Eαp, we de ne the norm for Eαp as follows:
(11)uα=∫0Tutpdt+∫0TD0tαutpdt1/p.Lemma 3 (see [3]).
Let0<α≤1 and 1<p<∞. For any u∈Eαp, we have
(12)uLp≤TαΓα+1D0tαuLp.
Also, ifα>p and 1/p+1/q=1, then
(13)u∞≤Tα−1/pΓαα−1q+11/qD0tαuLp.
Under the result of Lemma3, we note that
(14)uLp≤Tα−1/pΓα+1D0tαuLp,for 0<α≤1, and
(15)u∞≤Tα−1/pΓαα−1q+11/qD0tαuLp,for α> p and 1/p+1/q=1.
Under (14), we can see that (11) is equivalent to the following norm:
(16)uα=∫0TD0tαutpdt1/p,∀u∈Eαp.
For0<β≤1, 1<p<∞. Analogous to the space Eαp, we define the fractional derivative space Eαp as
(17)vt∈Lp0,T,ℝ0Dtβvt∈Lp0,T,ℝ,v0=vT=0.Then, for any v∈Eβp, the norm of Eβp is defined by
(18)vβ=∫0Tvtpdt+∫0TD0tβvtpdt1/p,∀v∈Eβp.
Similar with (14) and (15), we get
(19)vLp≤TβΓβ+1D0tαvLp,for 0<β≤1, and
(20)v∞≤Tβ−1/pΓββ−1q+11/qD0tαvLp..
Moreover, if0<β≤1 and 1/p+1/q=1, then, based upon (19), the weighted norm
(21)vβ=∫0TD0tβvtpdt1/p,,is equivalent to (18), for every v∈Eβp.
In the following discussion, for anyu∈Eαp, v∈Eβp denote the space of X=Eαp×Eβp with the norm
(22)u,vX=uαp+vβp1/p,∀u,v∈X,where uα and uβ are defined in (16) and (21), respectively.
Clearly,X is embedded compactly on
(23)C00,T,ℝ×C00,T,ℝ.Lemma 4 (see [33]).
For0<α, β≤1 and 1<p<∞. The derivative fractional space X is a reflexive separable Banach space.Lemma 5.
Assume that1/p<α≤1 and the sequence un converge weakly to u in Eαp, i.e., un⇀u. Then, un converges strongly to u in C0,T,ℝ, i.e., un−u∞⟶0, as n⟶+∞.Definition 6 (see [3]).
We point out to a weak solution to the system (1), for all u,v∈X such that
(24)∫0TΦpD0tαutD0tαxtdt+∫0TΦpD0tαvtD0tβytdt−∫0Th1utxtdt−∫0Th2vtytdt−λ∫0TFut,ut,vtxt+Fvt,ut,vtytdt=0,for all x,y∈X.
We define for allx∈ℝ:
(25)Hix=∫0xhizdz,Θix=∫0THixsdsforalli=1,2,for every t∈0,T.Lemma 7.
Leth1,h2:ℝ→ℝ satisfy (2) and Hix,Θix,i=1,2, defined by (25). Thus, Θu,v:X⟶ℝ defined by
(26)Θu,v=Θ1u+Θ2v=∫0TH1utdt+∫0TH2vtdt,is a Gâteaux function weakly sequentially differentiable over X with
(27)Θ′u,vx,y=∫0Th1utxtdt+∫0Th2vtytdt,forallx,y∈X.Proof.
Assume that(28)un,vn⊂X,un,vn⇀u,vinX,as n→+∞. According to Lemma 5 that un,vn converges uniformly to u,v on 0,T. Then, there exists c1,c2>0 such that un∞≤c1 and vn∞≤c2 for any n∈ℕ.
Then,(29)H1unt−H1ut≤L1∫utuntsp−1ds≤L1puntp+utp≤L1pc1p+ut∞p,(30)H2vnt−H2vt≤L2∫vtvntsp−1ds≤L2pvntp+vtp≤L2pc2p+vt∞p,for any n∈ℕ and t∈0,T. Furthermore, H1unt⟶H1utandH2vnt⟶H2vt at every t∈0,T, and by the Lebesgue Convergence Theorem
(31)Θun,vn=∫0TH1untdt+∫0TH2vntdt⟶∫0TH1utdt+∫0TH2vtdt=Θu,v.
Now we prove the Gâteaux differentiability ofΘ. Assume that u,x∈Eαp and s≠06; thus,
(32)Θ1u+sx−Θ1us−∫0Th1utxtdt≤∫0TH1u+sx−H1us−h1utxtdt=∫0Th1ut+sζtxt−h1utxtdt≤L1x∞ps,where
(33)0<ζt<1,t∈0,T.
Thus(34)Θ1:Eαp⟶ℝ,is a Gâteaux differentiable for all u∈Eαp.
Likewise, we have(35)Θ2:Eβp⟶ℝ,which is a Gâteaux differentiable for all v∈Eαp.
Therefore,(36)Θ:X⟶ℝ,is a Gâteaux differentiable for all u,v∈X with its derivative
(37)Θ′u,vx,y==∫0Th1utxtdt+∫0Th2vtytdt,x,y∈X.
For any three elementsu1,v1, u2,v2, and x,y of X, it is easy to see that
(38)Θ′u1,v1−Θ′u2,v2x,y=∫0Th1u1−h1u2xtdt+∫0Th2v1−h2v2ytdt≤L1∫0Tu1−u2p−1xtdt+L2∫0Tv1−v2p−1ytdt≤L1Tα−1/pΓαα−1q+11/pu1−u2∞p−1xα+L2Tβ−1/pΓββ−1q+11/pv1−v2∞p−1yα,which implies
(39)Θ′u1,v1−Θ′u2,v2X≤T∗u1−u2∞p−1+v1−v2∞p−1,where
(40)T∗≔maxL1Tα−1/pΓαα−1q+11/q,L2Tβ−1/pΓββ−1q+11/q.
Hence,Θ′:X⟶X∗ is a compact operator.Similarly to the proof of Theorem 5.1 of [4], we haveLemma 8 (see [36]).
Let1/p<α, β≤1, and u,v∈X. If u,v is a nontrivial weak solution of problem (1), then u,v is also a nontrivial solution of problem (1).Our analysis is mainly based on the following critical points theorem of Bonanno and Molica Bisci [36], which is a more precise result of Ricceri ([37], Theorem 2.5).Lemma 9 (see [[36], Theorem 2.1]).
LetX be a reflexive real Banach space. Let ϕ,Ψ:X⟶ℝ be two Gâteaux differentiable functionals such that ϕ is sequentially weakly lower semicontinuous, strongly continuous, and coercive and Ψ is sequentially weakly upper semicontinuous. For every r>infXϕ, put
(41)φr=infu∈ϕ−1−∞,rsupv∈ϕ−1−∞,rΨv−Ψur−ϕu,(42)γ≕limr→+∞infφr,δ≕limr→infXϕ+infφr.Then,(1)
Ifγ<+∞ and λ∈0,1/γ, the following alternative holds: either the functional ϕ−λΨ has a global minimum or there exists a sequence un of local minima ϕ−λΨ such that limn→+∞ϕun=+∞(2)
Ifγ<+∞ and λ∈0,1/δ, the following alternative holds: either there exists a global minimum of ϕ or the following alternative holds: either there exists a global minimum of ϕ−λΨ or there exists a sequence {un} of pairwise distinct local minima of ϕ−λΨ, with limn→+1ϕun=infXϕ, which weakly converges to a global minimum of ϕ


## 3. Main Results

Here, we prove our main results.Setting(43)k≔min1−L1TpαΓα+1p,1−L2TpβΓβ+1p,(44)ρ≔max1+L1TpαΓα+1p,1+L2TpβΓβ+1p,(45)M=maxTpα−1Γαpα−1q+1p/q,Tpβ−1Γβpβ−1q+1p/q.For a given constantθ∈1/p,0, set
(46)Pα,θ=1pθTp∫0θTtp1−αdt+∫θT1−θTt1−α−t−θT1−αpdt+∫1−θTTt1−α−t−θT1−α−t−1−θT1−αp,(47)Qβ,θ=1pθTp∫0θTtp1−βdt+∫θT1−θTt1−β−t−θT1−βpdt+∫1−θTTt1−β−t−θT1−β−t−1−θT1−βp.For anyd>0, we denote by Ωd the set
(48)x,y∈ℝ2:1pxp+1pyp≤d.Theorem 10.
Suppose thatk>0 and (H0) hold. In addition,
(H1) 1/p≤α,β<1
(H2) Ft,x,y≥0 for any t,x,y0,T∈0,+∞0,+∞
(H3) there exists θ∈0,1/p where if we set
(49)A∞=limξ→+∞inf∫0Tsupx+y≤ξFt,x,ydtξp,(50)B∞=limξ→+∞sup∫θT1−θTFt,Γ2−αξ,Γ2−βξdtξp,one has
(51)A∞<k2pMρΔB∞,where
(52)Δ=maxPα,θ,Qβ,θ,and M is given in (45).
Then, for every(53)λ∈ρΔB∞,k2pMA∞.
(1) has an unbounded sequence in X (weak solutions).Proof.
Our goal is to apply a portion (1) of Lemma9 to problem (1). First, by taking
(54)X=Eαp×Eβp,endowed with u,vX similar to what is considered in (22). We define the following functional
(55)Iλu,v=ϕu,v−λΨu,v,for all u,v∈X, where
(56)ϕu,v=1puαp+1pvβp−Θu,v,(57)Ψu,v=∫0TFt,ut,vtdt.
SinceX is embedded compact in
(58)C00,T,ℝ×C00,T,ℝ,it is well known that is a well-defined Gâteaux differentiable functional whose Gâteaux derivative at the point u,v∈X is the functional Ψ′u,v∈X∗, given by
(59)Ψ′u,vx,y=∫0TFut,ut,vtxtdt+∫0TFvt,ut,vtytdt,for every x,y∈X.
We claim that the functionalΨ is a sequentially weakly upper semicontinuous functional on X. Indeed, for fixed u,v∈X, suppose that un,vn⊂X, un,vn⇀u,v in X as n⟶+∞. Then, un,vn converges uniformly to u,v on 0,T. Hence,
(60)limn→+∞supΨun,vn≤∫0Tlimn→+∞supFt,unt,vntdt=∫0TFt,u,t,vtdt=Ψu,v,which implies that it is sequentially weakly upper semicontinuous. Hence, the claim is true.
Concerning the functionalϕ, we can show that what is defined by (56) is a sequentially weakly lower semicontinuous, strongly continuous, and coercive functional on X. In fact since (2) holds for every x1,x2∈R and h10=h20=0, one has hix≤Lixp−1, i=1,2, for all x∈ℝ. It follows from (14), (20), and Lemma 5 that
(61)ϕu,v≥1puαp+1pvβp−∫0TH1utdt−∫0TH2vtdt≥1puαp+1pvβp−L1p∫0Tutpdt−L2p∫0Tvtpdt≥1p−L1TpαpΓα+1puαp+1p−L2TpβpΓβ+1pvβp,for all u,v∈X and similarly
(62)ϕu,v≤1puαp+1pvβp+∫0TH1utdt+∫0TH2vtdt≤1p−L1TpαpΓα+1puαp+1p−L2TpβpΓβ+1pvβp≤ρpuαp+vβp,for all u,v∈X. So ϕ is coercive.
Moreover,ϕ+Θ is a continuous functional on X, and Θ, from Lemma 5, is Gâteaux differentiable sequentially weakly continuous and therefore continuous on X, then ϕ is a continuous functional on X. It is not difficult to verify that the functional is a Gâteaux differentiable functional with the differential
(63)ϕ′u,vx,y=∫0TΦpD0tαutD0tαxtdt+∫0TΦpD0tαvtD0tβytdt−∫0Th1utxtdt−∫0Th2vtytdt.
Furthermore,ϕ is also sequentially weakly lower semicontinuous on X since Θ is sequentially weakly lower semicontinuous, and if un,vn⇀u,v in X then
(64)limn→+∞infϕun,vn=limn→+∞inf1puαp+1pvβp−limn→+∞Θun,vn≥1puαp+1pvβp−Θu,v=ϕu,v.
It is easy to show that the critical points of the functionalIλ and the weak solutions of the problem (1) are the same, and by Lemma 9, we prove our result. According to
(65)u∞=max0,Tutandu∞=max0,Tut,taking (13) and (20) into account, one has
(66)maxrt∈0,Tutp≤Muαpandmaxt∈0,Tvtp≤Mvβpfor every u,v∈X.
Hence,(67)maxt∈0,Tutp+vtp≤Muαp+vβp.
So, for everyr>0, from the definition of and by using (61), one has
(68)ϕ−1−∞,r≔u,v∈X:ϕu,v≤r⊆u,v∈X:1puαp+1pvβp≤rk⊆u,v∈X:Γαpα−1q+1p/qTpα−1u∞p+Γβpβ−1q+1p/qTpβ−1u∞p≤rk⊆u,v∈X:1pup+1pvp≤Mrk,forallt∈0,T.
Set(69)φr=infu,v∈ϕ−1−∞,rsupx,y∈ϕ−1−∞,rΨx,y−Ψu,vr−ϕu,v.
Note thatϕ0,0=0, and from the condition (H1), Ψ0,0≥0. Hence, for every r>0,
(70)φr=infu,v∈ϕ−1−∞,rsupx,y∈ϕ−1−∞,rΨx,y−Ψu,vr−ϕu,v≤supx,y∈ϕ−1−∞,rΨx,yr,and it follows from (68) that
(71)φr≤1rsupΩMr/k∫0TFt,u,vdt,where
(72)ΩMrk=u,v∈X:1putp+1pvtp≤Mrk,∀t∈0,T.
Letξn be a sequence of positive numbers such that ξn⟶+∞ and
(73)limξ→+∞inf∫0Tsupx+y≤ξFt,x,ydtξp=A∞<+∞.
Putrn=k/p2pMξnp for all n∈ℕ. Let u,v∈ϕ11−∞,rn, by (68) one has
(74)1putp+1pvtp≤Mkrn,∀t∈0,T,which implies
(75)ut≤pMkrnpandvt≤pMkrnp.
Hence, forn large enough rn>1(76)ut+vt≤2pMkrnp=ξn.
Thus, for alln∈ℕ,
(77)φrn=p2pMkξnpsupu,v∈X:ut+vt<ξn,∀t∈0,T∫0TFt,ut,vtdt≤p2pMk.∫0Tsupx+y<ξnFt,x,ydtξnp.
Let(78)γ≔limr→+∞infφr.
Then,(79)γ≤limn→+∞infφrn≤p2pMk⋅limn→+∞∫0Tsupx+y<ξnFt,x,ydtξnp=p2pMkA∞<+∞.
Hence,Λ⊑0,1/γ.
Forλ∈Λ, we shall show that the functional Iλ is unbounded from below.
Indeed, sinceB∞/ρΔ>1/λ, we can choose a sequence ηn of positive numbers and ε>0 such that ηn⟶+∞ and
(80)1λ<ε<1ρΔ⋅∫θT1−θTFt,Γ2−αηn,Γ2−βηndtηnp,for n large enough.
For alln∈ℕ, and 0,1/p define ωnt=ω1,nt,ω2,nt by setting
(81)ω1,nt=Γ2−αθTt,t∈0,θT,Γ2−αηn,t∈θT,1−θT,Γ2−αηnθTT−t,t∈1−θT,T,(82)ω2,nt=Γ2−βηnθTt,t∈0,θT,Γ2−αβηn,t∈θT,1−θT,Γ2−βηnθTT−t,t∈1−θT,T.
Clearlyωi,n0=ωi,nT=0 and ωi,n∈Lp0,T for i=1,2. A direct calculation shows that
(83)0Dtαω1,nt=ηnθTt1−α,t∈0,θT,ηnθTt1−α−t−θT1−α,t∈θT,1−θT,ηnθTt1−α−t−θT1−α−t−1−θT1−α,t∈1−θT,T,(84)0Dtβω2,nt=ηnθTt1−β,t∈0,θT,ηnθTt1−β−t−θT1−β,t∈θT,1−θT,ηnθTt1−β−t−θT1−β−t−1−θT1−β,t∈1−θT,T.
Furthermore,(85)∫0T0Dtαω1,ntpdt=∫0θT+∫θT1−θT+∫1−θTT0Dtαω1,ntpdt=ηnPθTp∫0θTtp1−αdt+∫θT1−θTt1−α−t−θT1−αpdt+∫1−θTTt1−α−t−θT1−α−t−1−θT1−αp=pPα,θηnp,(86)∫0T0Dtβω2,ntpdt=∫0θT+∫θT1−θT+∫1−θTT0Dtβω1,ntpdt=ηnPθTp∫0θTtp1−βdt+∫θT1−θTt1−β−t−θT1−βpdt+∫1−θTTt1−β−t−θT1−β−t−1−θT1−βp=pQβ,θηnp.
Thus,ωn∈X, and
(87)ω1,ntp=∫0T0Dtαω1,ntpdt=pPα,θηnp,ω2,ntp=∫0T0Dtαω1,ntpdt=pQβ,θηnp.
This and (61) imply that
(88)Φω1,n,ω2,n=1pω1,ntp+1pω2,ntp−Θω1,n,ω2,n≤ρpω1,ntp+ω2,ntp=ρPα,θ+Qβ,θηnp≤ρΔηnp.
From (H2), we have
(89)Ψω1,n,ω2,n=∫0θT+∫θT1−θT+∫1−θTTFt,ω1,n,ω2,ndt≥∫θT1−θTFt,ω1,n,ω2,ndt=∫θT1−θTFt,Γ2−αηn,Γ2−βηndt.
According to (80), (88), and (89), we have
(90)Iλω1,n,ω2,n=ϕω1,n,ω2,n−λΨω1,n,ω2,n≤ρPα,θ+Qβ,θηnp−λ∫θT1−θTFt,Γ2−αηn,Γ2−βηndt≤ρΔ1−λεηnp,for n large enough. Taking into account the choice of ε, the above inequality shows that
(91)limn→+∞Iλω1,n,ω1,n=−∞,which implies that the functional Iλ is unbounded from below and the claim follows.
By using the case (1) of Lemma9, the functional Iλ has a sequence un,vn of critical points such that
(92)Φun,vn⟶+∞.
From (22) and (61), we get
(93)un,vnX≥ppΦun,vnρ,which implies un,vnX⟶+∞ and the proof of Theorem 10 is complete.Theorem 11.
Assume thatk>0 and (H0) and (H2) hold. Furthermore, (H4) Ft,0,0=0 for all t∈0,T.
(H5) There exists θ∈0,1/p such that, if we put
(94)A0=limξ→0+inf∫0Tsupx+y≤ξFt,x,ydtξp,(95)B0=limξ→0+sup∫θT1−θTFt,Γ2−αξ,Γ2−βξdtξp,one has
(96)A0<k2pMρΔB0,where Δ=maxPα,θ,Qβ,θ and M is given in (45).
Then, for every(97)λ∈Λ′≔ρΔB0,k2pMA0,
(1) has a sequence un,vn of weak solutions such that un,vn⇀0,0.Proof.
Our goal is to apply part (2) of Lemma9 to ϕ and Ψ defined in (48) and (51), respectively.
As it has been pointed out before, the functionalsϕ and Ψ satisfy the assumption regularity required in Lemma 9.
SinceFt,0,0=0 for all t∈0,T, then
(98)minu,v∈Xϕu,v=ϕ0,0=0.
Letξn be a sequence of positive numbers such that ξn⟶0 and
(99)limn→+∞∫0Tsupx+y≤ξnFt,x,ydtξnp=A0<+∞.
Settingrn=k/p2pMξnP for all n∈ℕ, and working as in the proof of Theorem 10, we can show that
(100)δ=limr→infXΦinfφr≤p2pMk⋅limn→+∞∫0Tsupx+y≤ξnFt,x,ydtξnp=p2pMkA0,and so Λ′⊂0,1/δ.
Now fixλ as in the conclusion, then
(101)1λ<1ρΔlimξ→0+sup∫θT1−θTFt,Γ2−αξ,Γ2−βξdtξp,and there exist a sequence τn of positive numbers and a constant ε1 such that τn≤1/n and
(102)limn→+∞∫θT1−θTFt,Γ2−ατn,Γ2−βτndtτnp=limξ→0+sup∫θT1−θTFt,Γ2−αξ,Γ2−βξdtξp,and in addition
(103)1λ<ε1<1ρΔlimn→+∞∫θT1−θTFt,Γ2−ατn,Γ2−βτndtτnp.
For alln∈ℕ, and θ∈0,1/p define ωnt=ω1,nt,ω2,nt by setting
(104)ω1,nt=Γ2−ατnθTt,t∈0,θT,Γ2−ατn,t∈θT,1−θT,Γ2−ατnθTT−t,t∈1−θT,T,(105)ω2,nt=Γ2−βτnθTt,t∈0,θT,Γ2−βτn,t∈θT,1−θT,Γ2−βτnθTT−t,t∈1−θT,T.
Clearlyωi,n0=ωi,nT=0 for i=1,2,andωn converges strongly to 0,0 in X.
By the same argument as in Theorem10, we have
(106)Iλω1,n,ω2,n−λΨω1,n,ω2,n≤ρPα,θ+Qβ,θτnp−λ∫θT1−θTFt,Γ2−ατn,Γ2−βτndt≤ρΔ1−λε1τnp<0=Iλ0,0,for n large enough. This together with the fact that ωnX=ω1,n,ω2,nX⟶0 shows that Iλ has no local minimum at zero, and the claim follows.
The alternative of Lemma9 case (2) ensures the existence of sequence un,vn of pairwise distinct local minima of Iλ which weakly converges to 0,0. This completes the proof of Theorem 11.Finally, we present an example to illustrate our main results.Example 12.
Consider the following fractional differential system:(107)Dt10,6Φ3D0t0,6ut=λFut,ut,vt+sinu122,a.e.t∈0,T,Dt10,75Φ3D0t0,75ut=λFvt,ut,vt+arctanu232,a.e.t∈0,T,u0=u1=0,v0=v1=0,where T=1,α=0,6,β=0,75,andh1u1=sinu1/22,h2u2=arctanu2/32. Moreover, for all t,u,v∈0,1×ℝ2 put
(108)Ft,ut,vt=1+t2Hu,v,where
(109)Hu,v=ξn+13exp−11−u−0.8873ξn+12+v−0.9064ξn+12,u,v∈Ω,0,u,v∈ℝ2\Ω,where
(110)Ω=∪n≥1u,v:u−0.8873ξn+12+v−0.9064ξn+1<1,and ξ1=1,ξn+1=nξn4/3+1 for all n∈ℕ.Clearly,h1,h2:ℝ⟶ℝ are two Lipschitz continuous functions of order 2 with Lipschitzian constants L1=1/2,L2=1/3andh10=h20=0,Ft,0,0=0 for all t∈0,1. With the aid of direct computation we have that
(111)M≈1.8925,k≈0.2991,ρ≈1.7009.Letθ=1/3, then we have
(112)Pα,13=P0,6,13≈0.3366,Qβ,13=Q0,75,13≈0.3745.Then,Δ≈0.3745. Thus, all conditions of Theorem 10 are satisfied.In fact, the conditions (H0), (H1), and (H2) hold. For all n∈ℕ.Restriction ofHu,v on Ω attains its maximum in 0.8873ξn+1,0.9064ξn+1 and
(113)H0.8873ξn+1,0.9064ξn+1=ξn+13exp−1.In addition,(114)supu+v≤0,8873ξn+1−1Hu,v=ξn3exp−1,and so
(115)B∞=limsupn→+∞∫1/32/3H0.8873ξn+1,0.9064ξn+1dtξn2+1=limn→+∞ξn+13exp−1ξn2+1=+∞,(116)A∞=limn→+∞inf∫01supu+v≤0.8873ξn+1−1Hu,vdt0.8873ξn+12=limn→+∞ξn3exp−10.8873ξn+12=0<k2pMρΔB0,which implies that the condition (H3) holds. Hence, owing to Theorem 10, for each λ∈0;+∞, the coupled system (107) has an unbounded sequence of weak solutions.



---

*Source: 6679101-2020-12-17.xml*