# Nontrivial Solutions for Time Fractional Nonlinear Schrödinger-Kirchhoff Type Equations

Authors: N. Nyamoradi; Y. Zhou; E. Tayyebi; B. Ahmad; A. Alsaedi
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

We study the existence of solutions for time fractional Schrödinger-Kirchhoff type equation involving left and right Liouville-Weyl fractional derivatives via variational methods.


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## Body

## 1. Introduction

In recent years, there has been a great interest in studying problems involving fractional Schrödinger equations [1–5], Kirchhoff type equations [6–8], fractional Navier-Stokes equations [9, 10], and fractional ordinary differential equations and Hamiltonian systems [11–17], and so forth. For further details and applications, we refer the reader to [18, 19] and the references cited therein.On the other hand, the integer-order Schrödinger-Kirchhoff type equations have also been investigated by many authors; for example, see [20–23]. In fact, Schrödinger-Kirchhoff type equations play an important role in modelling several physical and biological systems. However, to the best of our knowledge, the existence of solutions to the time fractional Schrödinger-Kirchhoff type equations has yet to be addressed.The objective of the present paper is to study time fractional Schrödinger-Kirchhoff type equation of the form(1)a+b∫RDtα-∞ut2dtθ-1D∞αtDtα-∞ut+μVtu=ft,u,t∈R,u∈HαR,where α∈(1/2,1], Dtα-∞ and D∞αt, respectively, denote left and right Liouville-Weyl fractional derivatives of order α on R, a,b>0 are constants, μ>0 is parameter, θ>1, f∈C(R×R,R), and V:R→R+ is a potential function.The rest of the paper is organized as follows. Section2 contains preliminary concepts of fractional calculus and fractional Sobolev space, while some important lemmas, which are needed in the proof of main results, are obtained in Section 3. We present our main results in Section 4.


## 2. Preliminaries

In this section, we recall important definitions and concepts of fractional calculus and then prove certain results about fractional Sobolev spaceHα(R) related to our study of the problem at hand.Definition 1 (see [24]).
The left and right Liouville-Weyl fractional integrals of orderα∈(0,1) on R are defined by(2)Ixα-∞ϕx=1Γα∫-∞xx-ξα-1ϕξdξ,I∞αxϕx=1Γα∫x∞ξ-xα-1ϕξdξ,respectively, where x∈R.
The left and right Liouville-Weyl fractional derivatives of orderα∈(0,1) on R are defined by(3)Dxα-∞ϕx=ddxIx1-α-∞ϕx,D∞αxϕx=-ddxI∞1-αxϕx,respectively, where x∈R.
The definitions (3) may be written in an alternative form as follows:(4)Dxα-∞ϕx=αΓ1-α∫0∞ϕx-ϕx-ξξα+1dξ,D∞αxϕx=αΓ1-α∫0∞ϕx-ϕx+ξξα+1dξ.Also, we define the Fourier transformF(u)(ξ) of u(x) as (5)Fuξ=∫-∞∞e-ix·ξuxdx.For anyα>0, we define the seminorm and norm, respectively, as [16](6)uI-∞α=Dxα-∞uL2,uI-∞α=uL22+uI-∞α21/2,and let the space I-∞α(R) denote the completion of C0∞(R) with respect to the norm ·I-∞α.Next, for0<α<1, we give the relationship between classical fractional Sobolev space Hα(R) and I-∞α(R), where Hα(R) is defined by (7)HαR=C0∞R¯·α, with the norm(8)uα=uL22+uα21/2,and seminorm (9)uα=ξαFuL2. Observe that the spaces Hα(R) and I-∞α(R) are equal and have equivalent norms (see [16]).Therefore, we define(10)HαR=u∈L2R∣ξαFu∈L2R.Let(11)Xα=u∈HαR∣∫RDtα-∞ut2+ut2dt<∞. The space Xα is a reflexive and separable Hilbert space with the inner product(12)u,vXα=∫RDtα-∞ut·Dtα-∞vt+utvtdtand the corresponding norm (13)uXα2=u,uXα. Define the space (14)Xμα=u∈Xα:∫RμVtu2dt<+∞, with the norm (15)uXμα=∫Raθ-1Dtα-∞ut2dt+∫RμVtu2dt1/2.Lemma 2.
(
X
μ
α
,
·
X
μ
α
) is a uniformly convex Banach space.Proof.
X
μ
α is obviously Banach space. Now, we can prove that (Xμα,·Xμα) is uniformly convex. To this end, let 0<ε<2 and u,v∈Xμα with uXλα=vXμα=1 and u-vXμα≥ε. Using the following inequality:(16)a+b22+a-b22≤12a2+b2,∀a,b∈R, we get(17)u+v2Xμα2+u-v2Xμα2=∫Raθ-1Dtα-∞u+v2t2dt+∫RμVtu+v22dt+∫Raθ-1Dtα-∞u-v2t2dt+∫RμVtu-v22dt≤12∫Raθ-1Dtα-∞ut2dt+∫Raθ-1Dtα-∞vt2dt+∫RμVtu2dt+∫RμVtv2dt=12uXμα2+vXμα2=1,which implies that (u+v)/2Xμα2≤1-ε/2. Hence, taking δ=δ(ε) such that 1-ε/2=1-δ, we have (u+v)/2Xμα2≤1-δ. Therefore, (Xμα,·Xμα) is uniformly convex.In the sequel, we need the following assumptions.(V1)
V(t)∈C(R,R), V0≔inft∈R⁡V(t)>0;(V2)
there existsr>0 such that, for any M>0, (18)meas⁡t∈y-r,y+r:Vt≤M⟶0asy⟶∞;(V3)
there existsl0>0 such that ∫t≥l0Vt-1dt<∞;(F1)
f∈C(R×R,R) and there exist constants c0,c1,…,cl>0 and qj∈(2,2θ) such that (19)ft,u≤c0u+∑j=1lcjuqj-1,∀t,u∈R×R;(F2)
f(t,u)=o(u) as u→0 uniformly in t∈RN;(F3)
there existλ∈(2θ,∞) such that (20)λFt,u≤ft,uu,∀t∈R,u∈R;(F4)
F(t,u)/u2θ→+∞ as u→+∞ uniformly in t∈R;(F5)
f(t,-u)=-f(t,u) for all (t,u)∈R×R;(F6)
f∈C(R×R,R) and there exists 1<p<2 such that (21)ft,u≤up-1,∀t,u∈R×R;(F7)
there existσ1>0, 0<σ2<1/8D22 (D2 is defined in Remark 6), 1≤γ<2, and small constants 0<r0<r1 such that (22)σ1uγ<Ft,u≤σ2u2,r0≤u≤r1,a.e.t∈R.Lemma 3.
Assume that (V1) holds. Then the embeddingsXμα↪Xα↪L2(R) are continuous. In particular, there exists a constant C2>0 such that(23)uL2R≤C2uXμα∀u∈Xμα.Moreover, if (V1) and (V2) hold, then the embedding Xμα↪L2(R) is compact.Proof.
Clearly, the chain of embeddingsXμα↪Xα↪L2(R) is continuous and consequently one can obtain (23). Also in view of (V1), (V2), and following the method of proof similar to that of Lemma  2.2 in [15], the embedding Xμα↪L2(R) is compact.Lemma 4.
Letα>1/2. Then Hα(R)⊂C(R) and there exists a constant C=Cα such that(24)supx∈R⁡ux≤CuXμα.Proof.
The proof is similar to that of Theorem  2.1 in [16], so we omit it.Also by Lemma4, there is a constant Cα>0 such that(25)u∞≤CαuXμα.Remark 5.
Ifu∈Hα(R) with 1/2<α<1, then it follows by Lemma 4 that u∈Lq(R) for all q∈[2,∞) as (26)∫Ruxqdx≤u∞q-2uL2R2.Remark 6.
From Remark5 and Lemma 3, it is easy to verify that the imbedding of Xμα in Lq(R) is also compact for q∈(2,∞). Hence, for all 2≤q<∞, the imbedding of Xμα in Lq(R) is continuous and compact, which together with Lemma 4 implies that there exists Dq>0 such that (27)uLqR≤DquXμα.Lemma 7.
Assume that (V1) and (V3) hold. Then the embeddingXμα↪Lp(R) is continuous and compact for p∈[1,+∞).Proof.
By (V3) and Hölder’s inequality, we have(28)∫t≥l0utdt≤∫t≥l0Vtut2dt1/2∫t≥l0Vt-1/2dt1/2≤c1uXμα, for some positive constant c1. So Lemma 4 implies that (29)u1=∫-l0l0utdt+∫t≥l0utdt≤2l0u∞+c1uXμα≤c2uXμα, for some positive constant c2. Hence, by Remark 6, we can get continuous embeddings Xμα into Lp(R) for p∈[1,+∞). Now, we will show that the embedding is compact for p∈[1,+∞). Let un⊂Xμα such that un⇀0 and M>0 such that uXμα≤M. In view of (V3), given ε>0, for l>0 large enough, one can obtain (30)∫t≥l0Vt-1/2dt<ε2M2.Then, (31)∫t≥lutdt≤∫t≥lVtut2dt1/2∫t≥lVt-1/2dt1/2≤ε2MuXμα≤ε2. On the other hand, by Sobolev’s theorem (see, e.g., [25]) which implies that un→0 uniformly on [-l,l], there is n0 such that ∫-ll‍utdt<ε/2 for all n≥n0. Thus un→0 in L1(R). So, for 1<p<∞, we have (32)∫Ruxpdx≤u∞p-1∫Rutdt≤c3u1⟶0, and consequently, un→0 in Lp(R) for p∈[1,+∞).Definition 8.
LetX be a Banach space, I∈C1(X,R). One says that I satisfies the Palais-Smale (PS) condition if any sequence (un)∈X for which I(un) is bounded and I′(un)→0 as n→∞ possesses a convergent subsequence.In order to establish the main results, we need the following known Theorems.Theorem 9 (see [26, Theorem  2.2]).
LetX be a real Banach space and I∈C1(X,R) satisfies (PS) condition. Suppose I(0)=0 and (i)
there are constantsρ,α>0 such that I∂Bρ(0)≥0;(ii)
there is ane∈X∖Bρ(0)¯ such that I(e)≤0.Then I possesses a critical value c≥α. Moreover c can be characterized as (33)c=infγ∈Γ⁡maxs∈0,1⁡Iγs, where (34)Γ=γ∈C0,1,X:γ0=0,γ1=e.Theorem 10 (see [26, Theorem  9.12]).
LetX be an infinite dimensional Banach space and let I∈C1(X,R) be even, satisfying (PS) condition, and I(0)=0. If X=Y⊕Z, where Y is finite dimensional and I satisfies the following conditions:(I1)
there exist constantsρ,α>0 such that I∂Bρ∩Z≥α;(I2)
for any finite dimensional subspaceX~⊂X, there is R=R(X~)>0 such that I(u)≤0 on X~∖BR,then I possesses an unbounded sequence of critical values.


## 3. Some Lemmas

Recall thatu∈Xμα is said to be a weak solution of problem (1) if(35)a+b∫RDtα-∞ut2dtθ-1∫RDtα-∞ut·Dtα-∞φtdt+∫RμVtutφtdt=∫Rft,utφtdt,∀φ∈Xμα,and the energy functional Iμ,θ:Xμα→R is given by the formula(36)Iμ,θu=12bθa+b∫RDtα-∞ut2dtθ+12∫RμVxut2dt-∫RFt,utdt,where F(x,u)=∫0u‍f(t,s)ds.In view of assumptions (V1) and (F1), the functionalIμ,θ is of class C1(Xμα,R) and by similar method in Theorem  4.1 in [27] and the definition of Gâteaux derivative, one can get(37)Iμ,θ′u,φ=a+b∫RDtα-∞ut2dtθ-1∫RDtα-∞ut·Dtα-∞φtdt+∫Rμtutφtdt-∫Rft,utφtdt,∀u,φ∈Xμα.Lemma 11.
Assume that (V) and (F1)–(F3) hold. ThenIμ,θ satisfies the (PS) condition.Proof.
Letunn∈N⊂Xμα be a sequence such that Iμ,θunn∈N is bounded and Iμ,θ′(un)→0 as n→∞. Then there exits D>0 such that Iμ,θ′un,un≤DuXμα and Iμ,θun≤D. So, by (F3), (23), and the fact that λ>2θ>1, we get (38)λD+DuXμα≥λIμ,θun-Iμ,θ′un,un=λ2bθa+b∫RDtα-∞unt2dtθ+λ2∫RμVtunt2dt-λ∫RFt,untdt-a+b∫RDtα-∞unt2dtθ-1∫RDtα-∞unt2dt-∫RμVtunt2dt+∫Rft,untuntdt=aλ2bθa+b∫RDtα-∞unt2dtθ-1+λ-2θ2θa+b∫RDtα-∞unt2dtθ-1∫RDtα-∞unt2dt+λ-22∫RμVtunt2dt+∫Rft,untunt-λFt,untdt≥λ-2θ2θaθ-1∫RDtα-∞unt2dt+λ-22∫RμVtunt2dt≥λ-2θ2θuXμα2. Hence, unn∈N is bounded in Xμα.
So, passing onto subsequence if necessary, thanks to Lemma3, we have(39)un⇀u,weakly  inXμα,un⟶u,strongly  a.e.  inR,un⟶u,strongly  a.e.  inLsRN,2≤s<+∞,(40)∫RDtα-∞unt2dt⟶ρ1≥0,∫RμVtun2dt⟶ρ2≥0.We will prove that (41)∫RDtα-∞ut2dt=ρ1,∫RμVtu2dx=ρ2. Let φ∈Xμα be fixed and denote by Bφ the linear functional on Xμα defined by(42)Bφv≔∫RDtα-∞φt·Dtα-∞vtdt, and set(43)Δαu≔∫RDtα-∞ut2dt, for all v∈Xμα. In view of the Hölder inequality and definition of Bφ, we have(44)Iμ,θ′un-Iμ,θ′u,un-u=a+bΔαunθ-1Bunun-u-a+bΔαunθ-1Buun-u+∫RμVtun-uun-udt-∫Rft,un-ft,uun-udt≥a+bΔαunθ-1Δαun-a+bΔαunθ-1Δαunθ-1/2Δαun1/2+a+bΔαuθ-1Δαu-a+bΔαuθ-1Δαuθ-1/2Δαun1/2+∫RμVtun2dt-∫RμVtun2dt1/2∫RμVtu2dt1/2+∫RμVtu2dt-∫RμVtu2dt1/2∫RμVtun2dt1/2-∫Rft,un-ft,uun-udt=a+bΔαunθ-1Δαunθ-1/2Δαun1/2-Δαu1/2+a+bΔαuθ-1Δαuθ-1/2Δαu1/2-Δαun1/2+∫RμVtun2dt1/2∫RμVtun2dt1/2-∫RμVtu2dt1/2+∫RμVtu2dt1/2∫RμVtu2dt1/2-∫RμVtun2dt1/2-∫Rft,un-ft,uun-udt=Δαun1/2-Δαu1/2a+bΔαunθ-1Δαunθ-1/2-a+bΔαuθ-1Δαuθ-1/2+∫RμVtun2dt1/2-∫RμVtu2dt1/22-∫Rft,un-ft,uun-udt. Since un⇀u in Xμα and Iμ,θ′(un)→0 as n→∞ in (Xμα)∗, therefore Iμ,θ′(un)-Iμ,θ′(u),un-u→0 as n→∞. Now, using (F1) and Hölder inequality, we obtain (45)∫Rft,un-ft,uun-udt≤∫Rc0un+u+∑j=1lcjunqj-1+uqj-1un-udx≤c0unL2R+uL2Run-uL2R+∑j=1lcjunLqjRqj-1+unLqjRqj-1un-uLqjR, which, in view of (39), yields(46)limn→∞⁡∫Rft,un-ft,uun-udt=0.Since un→u a.e. in R, it follows by Fatou’s lemma that(47)Δαu≤liminfn→∞⁡Δαun=ρ1,∫RμVtu2dt≤liminfn→∞⁡∫RμVtun2dt=ρ2.Noting that Π(s)=(a+bs)θ-1sθ-1/2 is a nondecreasing function for s≥0, we get (48)ρ11/2-Δαu1/2a+bρ1θ-1ρ1θ-1/2-a+bΔαuθ-1Δαuθ-1/2,ρ21/2-∫RμVtu2dt1/22≥0.Now, in view of Iμ,θ′(un)-Iμ,θ′(u),un-u→0 as n→∞, (46), and (47), one has(49)0≥liminfn→∞⁡Δαun1/2-Δαu1/2a+bΔαunθ-1Δαunθ-1/2-a+bΔαuθ-1Δαuθ-1/2+∫RμVtun2dt1/2-∫RμVtu2dt1/22-∫Rft,un-ft,uun-udt≥limn→∞⁡Δαun1/2-Δαu1/2a+bΔαunθ-1Δαunθ-1/2-a+bΔαuθ-1Δαuθ-1/2+limn→∞⁡∫RμVtun2dt1/2-∫RμVtu2dt1/22-limn→∞⁡∫Rft,un-ft,uun-udt≥ρ11/2-Δαu1/2a+bρ1θ-1ρ1θ-1/2-a+bΔαuθ-1Δαuθ-1/2+ρ21/2-∫RμVtu2dt1/22.Then, from (48)-(49), we get (50)Δαu=∫RDtα-∞ut2dt=ρ1,∫RμVtu2dt=ρ2. Hence, we obtain unXμα→uXμα. As Xμα is a reflexive Banach space (see Lemma 2), it is isomorphic to a locally uniformly convex space. So the weak convergence and norm convergence imply strong convergence. This completes the proof.Letej be a total orthonormal basis of L2(R) and define Xj=Rej, j∈N, (51)Yk=⊕j=1kXj,Zk=⊕j=k+1∞Xj,k∈N.Lemma 12.
Assume that (V1) holds. Then, for2<p<+∞, (52)βk≔supu∈Zk,uXμα=1⁡uLpR⟶0,k⟶∞.Proof.
The proof is similar to that of Lemma  3.8 in [28]. So it is omitted.In view of Lemma12, we can choose an integer k≥1 such that(53)∫Ru2dt≤12c0∫RaDtα-∞ut2dt+∫RμVtu2dt∀u∈Zm∩Xμα,where c1 is a constant given in condition (F1). Let (54)Rt=1,t>r,0,t≤r, and set Y=1-Ru:u∈Xμα,1-Ru∈Yk and Z=1-Ru:u∈Xμα,1-Ru∈Zk+Rv:v∈Xμα. Hence Y and Z are subspaces of Xμα, and Xμα=Y⊕Z.Lemma 13.
Suppose that (V1), (V2), and (F1) are satisfied. Then there exist constantsϱ,β>0 such that Iμ,θ∂Bϱ∩Z≥α.Proof.
In view of (V2), (53), and definition of the space Z, we have(55)uL2R2=∫t<rut2dt+∫t≥rut2dt≤12c0uXμα2+1μω∫t∈R,Vt>ωμVtut2dt≤12c0uXμα2+1μωuXμα2∀u∈Z.Therefore, from (23), (55), and (F1) and for large enough value of μ, we get (56)Iμ,θu=12bθa+bΔαuθ+12∫RμVtu2dt-∫RFt,udt≥aθ-12Δαu+12∫RμVtu2dt-∫RFt,udt≥12uXμα2-c02uL2R2-∑j=1lcjqjuLqjRqj≥14uXμα2-c0μω2uXμα2-∑j=1lcjDqjqjqjuXμαqj≥18uXμα2-∑j=1lcjDqjqjqjuXμαqj.Since 2<qj(j=1,…,l), there exist constants ϱ,β>0 such that Iμ,θ∂Bϱ∩Z≥β.Lemma 14.
Assume that (F1) and (F4) are satisfied. Then, for any finite dimensional subspaceX~μα⊂Xμα, there is R=R(X~μα)>0 such that Iμ,θ(u)≤0 on X~μα∖BR.Proof.
Since all the norms in the finite dimensional space are equivalent, there exists a constantΥ such that (57)uL2θR≥ΥuXμα,∀u∈X~μα. From (F1) and (F4), for any L>bθ-1/2θΥ2θaθ(θ-1), there exists a constant CL>0 such that (58)Ft,u≥Lu2θ-CLu2,∀t,u∈R×R. Thus (59)Iμ,θu=12bθa+bΔαuθ+12∫RμVtu2dt-∫RFt,udt≤12bθa+baθ-1uXμα2θ+12uXμα2+CLuL2R2-LuL2θR2θ≤12bθa+baθ-1uXμα2θ+12+CLD22uXμα2-LΥ2θuXμα2θ for all u∈X~μα. Consequently, there is a large R>0 such that Iμ,θ(u)≤0 on X~μα∖BR. Therefore, the proof is completed.


## 4. Existence of Weak Solutions

In this section, we present our main results.Theorem 15.
Assume that (V1), (V2), (F1), (F3), (F4), and (F5) hold. Then problem (1) has infinitely many nontrivial weak solutions whenever μ>0 is sufficiently large.Proof.
We know thatIμ,θ(0)=0, and it is even by (F5). Let X=Xμα and Y and Z be as defined in Section 2. By Lemmas 11, 13, and 14, it follows that Iμ,θ satisfies all the condition of the Theorem 10. Therefore, problem (1) has infinitely many nontrivial weak solutions whenever μ>0 is sufficiently large.Theorem 16.
Assume that (V1), (V2), (F1), (F2), (F3), and (F4) hold. Then problem (1) has at least one nontrivial weak solution when μ>0.Proof.
We complete the proof in three steps.
Step  1. Clearly Iμ,θ(0)=0 and Iμ,θ∈C1(Xμα,R) satisfies the (PS) condition by Lemma 11.
Step  2. It will be shown that there exist constants ϱ,β>0 such that Iμ,θ satisfies condition (i) of Theorem 9. For any ε>0, by (F1) and (F2), there exists a constant cε>0 such that(60)Ft,u≤ϵ2u2+∑j=1lcjεqjuqj.Thus, by (23) and (60), for small ρ>0, we get (61)Iμ,θu=12bθa+bΔαuθ+12∫RμVtu2dt-∫RFt,udt≥aθ-12Δαu+12∫RμVtu2dt-∫RFt,udt≥12uXμα2-εD22uXμα2-∑j=1lcjεqjDqjqjuXμαqj≥181-εD22ϱ2, for all u∈B¯ϱ, where Bϱ=u∈Xμα:uXμα<ϱ. So it suffices to choose ε=1/2D22 so that (62)Iμ,θ∂Bϱ≥116ϱ2≔β>0.
Step  3. It remains to prove that there exists an e∈Xμα such that uXμα>ϱ and Iμ,θ(e)≤0, where ρ is defined in Step  2. Let us consider (63)Iμ,θσu=12bθa+bσ2Δαuθ+σ22∫RμVtu2dt-∫RFt,σudt, for all σ∈R. Take 0≠u∈Xμα. By (F1) and (F4), for any κ>bθ-1(Δα(u))θ/2θ∫R‍|u|2θdt, there is a constant Cκ>0 such that (64)Ft,u≥κu2θ-Cκu2. So we have (65)Iμ,θσu≤12bθa+bσ2Δαuθ+σ22∫RμVtu2dt+Cκσ2∫Ru2dt-κσ2θ∫Ru2θdt⟶-∞, as σ→+∞. Thus, there is a point e∈Xμα∖B¯ϱ such that Iμ,θ(e)≤0. By Theorem 9, Iμ,θ possesses a critical value c≥α>0 given by (66)c=infγ∈Γ⁡maxs∈0,1⁡Iμ,θγs, where (67)Γ=γ∈C0,1,X:γ0=0,γ1=e. Hence there is u∈Xμα such that Iμ,θ(u)=c and Iμ,θ′(u)=0; that is, problem (1) has a nontrivial weak solution in Xμα.Theorem 17.
Assume that (V1), (V3), (F5), (F6), and (F7) hold. Then problem (1) has infinitely many nontrivial weak solutions for μ>0.Proof.
One can obtain the proof by employing the method of proof for Theorem15 and using Lemma 7.



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