# On the Existence and Stability of Standing Waves for 2-Coupled Nonlinear Fractional Schrödinger System

Authors: Xiuyan Sha; Huanmin Ge; Jie Xin
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

We study a system of 2-coupled nonlinear fractional Schrödinger equations. Firstly, we construct constrained minimization problem to the system. Next, we prove the existence of standing waves for the system by using the concentration-compactness and commutator estimates method. Lastly, we also consider the set of minimizers of the constrained minimization problem. We prove that it is a stable set for initial value of the problem; that is, a solution to the system with initial value which is near the set will remain near it for all time.


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## Body

## 1. Introduction

In the recent years, more and more researchers study the application of fractional calculus and fractional integrodifferential equations in physics and other areas (see [1–4]). The concept of fractional calculus is firstly put forward by Leibniz as a generalization of standard calculus. Afterwards several kinds of definitions of them have been established such as Riemann-Liouville fractional derivative, Caputo fractional derivative, and Weyl fractional derivative (see [5, 6]). We all know that a variety of fractional calculus is brought in inspired by standard calculus, but the calculation laws of fractional calculus are much different (see [5, 7]). To avoid the complicated fractional calculus, in this paper we are only concerned with the fractional Laplacian operator.Here we study the existence and stability of standing waves for the following system of2-coupled nonlinear fractional Schrödinger equations:(1)iψ1t+-Δαψ1=γψ1p-2+ψ1p/2-2ψ2p/2ψ1,x∈RN,iψ2t+-Δαψ2=γψ2p-2+ψ2p/2-2ψ1p/2ψ2,x∈RN,where i2=-1, N⩾2, 2<p<2+4α/N, γ>0, 0<α<1, and (-Δ)α is a pseudo-differential operator which is defined by (-Δ)αψi^ξ=ξ2αψi^(ξ), the Fourier transform Fψi(ξ)=ψ^i(ξ)=(1/(2π)N/2)∫RN‍ψi(x)e-iξ·xdx.It is well known that the coupled nonlinear Schrödinger equations play an important role in describing nonrelativistic quantum mechanical behavior. In particular, the fractional Schrödinger equations can describe better some real physical phenomenon. Schrödinger type equation(s) is first derived by Feynman and Hibbs, applying path integrals over Brownian paths in [8]. In [9–11], Laskin showed the path integral over Lévy-like quantum mechanical paths approaches to a generalization of quantum mechanics. As the path integral over Brownian trajectories, it leads to the standard Schrödinger equations. And as the path integral over Lévy-like quantum mechanical paths, it leads to the fractional Schrödinger equation. Recently, there are some researchers studying fractional Schrödinger equation(s) and its physical applications (see [12–14]). The research of the existence and stability of standing waves for the nonlinear Schrödinger equations arises in various fields of physics such as plasma physics, constructive field theory, and nonlinear optics (see [15–22]). There are many papers about this topic for the standard Schrödinger equations with different nonlinear terms such as [17, 19, 20]. There are some results concerned with the existence and stability of standing waves for the nonlinear fractional Schrödinger equations in [21, 22]. The author [23] obtained the global solution for a class of systems of fractional nonlinear Schrödinger equations with boundary condition. However, there are few results about the standing waves of fractional nonlinear Schrödinger equations (1). In the present paper, we consider the existence and stability of standing waves for the system of 2-coupled nonlinear fractional Schrödinger equations.The rest of this paper is arranged as follows. In Section2, we recall some basic definitions and introduce preparation results. In Section 3, we give the main results and proof of them.


## 2. Preliminaries

In this section, we firstly introduce the constrained minimization problem in order to study the existence and stability of standing waves for (1). We know that the standing waves of (1) have the special form (ψ1(x,t),ψ2(x,t))=(eiω1tu1(x),eiω2tu2(x)), where ωi∈R(i=1,2). Then we only need to find ω1,ω2,u1(x),u2(x) satisfying the following equations to study the existence and stability of standing waves of (1):(2)-Δαu1-ω1u1=γu1p-2+u1p/2-2u2p/2u1,x∈RN,-Δαu2-ω2u2=γu2p-2+u2p/2-2u1p/2u2,x∈RN,where u1,u2 are complex value functions.By the variational method, we know that in order to study the existence of solution for (2) we need to consider the following constrained minimization problem: (3)Iq≔inf⁡Iu1,u2∣u1,u2∈HαRN×HαRN,Mu1,u2=q,q>0, where (4)Iu1,u2=12∫RN-Δα/2u12+-Δα/2u22dx-∫RNFx,u1,u2dx,Mu1,u2=∫RNu12+u22dx,∂u1Fx,u1,u2=γu1p-2+u1p/2-2u2p/2u1,∂u2Fx,u1,u2=γu2p-2+u2p/2-2u1p/2u2.And Hα(RN) is the fractional order Sobolev space. We denote (5)Gq≔u1,u2∣u1,u2∈HαRN×HαRN,Mu1,u2=q,Iu1,u2=Iqas the set of minimizers of the problem Iq.Next we recall some basic definitions. Define the fractional order Sobolev space(6)HαRN=u∣u∈L2RN;C,F-11+ξ2α/2Fu∈L2RNwith the norm (7)·α,2=F-11+ξ2α/2Fu2.Denote the product space Hα(RN)×Hα(RN)={(u1,u2)∣u1,u2∈Hα(RN)} with the norm (u1,u2)=(u1α,22+u2α,22)1/2. Throughout this paper, we denote the norm of Lperp(RN) by ·p for all 1⩽p⩽∞. Denote the set B(y,r)={x∣x,y∈RN,r>0,x-y<r}. For simplicity and convenience, the letter C will represent a constant, which may be different from one to others. C(·,·), for example, represents the constant C which can be expressed by the parameters appearing in the braces.Lastly, we give the following some lemmas, which will be used in our paper. For more detail, we can see [21, 23, 24].Lemma 1.
If0<α<1, x∈RN, and u∈S, S representing the Schwartz class, then the fractional Laplacian (-Δ)α of u is also expressed for the formula (8)-Δαux=CN,αP.V.∫RNux-uyx-yN+2αdx,where P.V. means the Cauchy principal value on the integral and C(N,α) is some positive normalization constant.Lemma 2.
For0<α<1, there are two properties with Hα(RN):(i)
The norm·α,2 of Hα(RN) is equivalent to ·=F-11+ξαF·2.(ii)
∀u∈Hα(RN), cuWα,2(RN)⩽u⩽CuWα,2(RN),which implies Hα(RN)=Wα,2(RN), where Wα,2(RN) is defined by the method of trace interpolation (see chapter 7 of [24]) and its norm is given by uWα,2(RN). c and C are positive constants.Remark 3.
In the following paper, we denote the norm·α,2 of Hα(RN) by ·.Lemma 4 (see [24]).
For0<α<1, 2⩽p1⩽2N/(N-2α), Hα(RN)↪Lp1(RN) and Wα,2(B(y,r))↪Lp1(B(y,r)). Moreover, if un⇀u in Hα(RN), then un→u in Llocp(RN), and un→u almost everywhere in RN, where 2⩽p<2N/(N-2α).Lemma 5.
If{(u1n,u2n)} is bounded in Hα(RN)×Hα(RN) and limn→+∞⁡supy∈RN⁡∫B(y,r)‍(|u1n|2+|u2n|2)dx=0 for some r>0, then limn→+∞⁡(u1n,u2n)=(0,0) in Lp(RN)×Lp(RN) for 2<p<2N/(N-2α).Remark 6.
The proof of Lemma5 is obtained easily according to Lemma 2.4 (see [21]), so we will omit it.Lemma 7 (commutator estimates).
If0<α<1, f,g∈S, the Schwartz class, then (9)-Δα/2fg-f-Δα/2g2⩽C∇fp¯1-Δα-1/2gq¯1+-Δα/2fp¯2gq¯2,where p¯2,q¯1∈[2,+∞) satisfying 1/p¯1+1/q¯1=1/p¯2+1/q¯2=1/2.In [23], the authors have obtained the existence and uniqueness of the global solution of a class of systems for fractional nonlinear Schrödinger equations with periodic boundary condition by using the Faedo-Galërkin method. In the proof of Theorem 4.1 in the references, only needing to let the period 2πei→+∞, we can obtain the following theorem.Theorem 8.
Let0<α<1. If Ψ0=(Ψ10,Ψ20)∈Hα(RN)×Hα(RN) then there exists a global weak solution Ψ(x,t)=(Ψ1(x,t),Ψ2(x,t))∈C(R,Hα(RN)×Hα(RN)) to the initial value problem of coupled nonlinear fractional Schrödinger equations (1) with initial date Ψ10=Ψ1(x,0),Ψ20=Ψ2(x,0).


## 3. Uniform Estimates of the Solution

In the section, we consider coupled nonlinear equations (2). Since the scaling uiγ=γ1/(p-2)ui(x)(i=1,2), we can transform (2) to the following equations:(10)-Δαu1γ-ω1u1γ=u1γp-2+u1γp/2-2u2γp/2u1γ,x∈RN,-Δαu2γ-ω2u2γ=u2γp-2+u1γp/2u2γp/2-2u2γ,x∈RN.So we only consider coupled nonlinear equations (10). So we have (11)Fx,u1,u2=1pu1p+u2p+2pu1p/2u2p/2.For any q>0, let I~q≔Iq and G~≔Gq.Under the case we have the following main results.Theorem 9.
Assuming{(u1n,u2n)} is a minimizing sequence of problem Iq~, then there exists a sequence {yn}∈RN such that {(u1n(·+yn),u2n(·+yn)} contains a convergent subsequence in Hα(RN)×Hα(RN). In particular, there exists a minimizer for problem (I~q)(i.e.,G~q≠ϕ), and (12)limn→+∞⁡infg1,g2∈G~⁡u1n-g1+u2n-g2=0.Theorem 10.
The setG~q is Hα(RN)×Hα(RN)-stable with respect to (1); that is, ∀ε>0, there exists δ>0 such that if (u1,u2)∈C([0,T),Hα(RN)×Hα(RN)) is a solution of (1) with initial value (u10,u20) satisfying (13)infg1,g2∈G~q⁡u10-g1+u20-g2<δ,then, ∀t∈[0,T), (14)infg1,g2∈G~q⁡u1t,·-g1+u2t,·-g2<ε.Before we give the proof of Theorems9 and 10, we firstly prove the following some lemmas.Lemma 11.
For anyq>0, one has the following:(i)
-∞<I~q<0,(ii)
every minimizing sequence problemI~q is bounded in Hα(RN)×Hα(RN). And for sufficiently large n, there exists a positive constant δ such that |u1n|p+|u2n|p⩾δ.Proof.
(i) For given(u1,u2)∈Hα(RN)×Hα(RN), |u1|22+|u2|22=q, |u1|pp+|u2|qq≠0. And set uiθ(x)=θ1/2ui(θ1/Nx)(i=1,2); then |u1θ|22+|u2θ|22=|u1|22+|u2|22=q.
By Lemma1, we have that (15)I~u1θ,u2θ=12θ2α/N∫RN-Δα/2u12+-Δα/2u22dx-1pθp/2-1∫RNu1p+u2pdx-2pθp/2-1∫RNu1p/2u2p/2dx⩽12θ2α/N∫RN-Δα/2u12+-Δα/2u22dx-1pθp/2-1∫RNu1p+u1pdx.By 0<p/2-1<2α/N, there exists I~(u1θ,u2θ)<0 taking sufficiently small θ. Therefore, we obtain that I~⩽I~(u1θ,u2θ)<0.
By the Hölder inequality and the Sobolev inequality, we can derive(16)uip⩽ui21-λui2N/N-2αλ⩽Cui21-λuiλ,i=1,2,where p(1-λ)/2+λp/(2N/(N-2α))=1. Because of the Young inequality and 0<λp<2, for ε>0, from the above inequality we can derive that(17)uipp⩽εui2+Cεui22p1-λ/2-λp,i=1,2.So for (u1,u2)∈Hα(RN)×Hα(RN), setting |ui|22=qi(i=1,2), q1+q2=q by the Young inequality and (17), we have that (18)I~u1,u2⩾12u12+u22-12q-1p∫RNu1p+u2pdx-2p∫RNu1p/2u1p/2dx⩾12u12+u22-12q-εpu12+u22-Cεpu122p1-λ/2-λp+u222p1-λ/2-λp-1p∫RNu1p+u2pdx⩾12u12+u22-12q-2εpu12+u22-2Cεpq1p1-λ/2-λp+q2p1-λ/2-λp.Therefore for sufficiently small ε, we have that (19)I~u1,u2⩾-12q-Cε,pq1p1-λ/2-λp+q2p1-λ/2-λp>-∞.So -∞<I~q<0.
(ii) Let{(u1n,u2n)} be minimizing sequence for problem (I~q).
By (16), we have that (20)12u1n2+u2n2⩽I~u1n,u2n+q2+1p∫RNu1np+u2npdx+2p∫RNu1p/2u2p/2dx⩽I~u1n,u2n+q2+Cu1n2p1-λu1npλ+Cu2n2p1-λu2npλ⩽I~u1n,u2n+q2+Cq1p1-λ/2u1npλ+q2p1-λ/2u2npλ.By 0<λp<2, we get that u1n2+u2n2<C; that is, {(u1n,u2n)} is bounded in Hα(RN)×Hα(RN).
Using an argument by contradiction, we prove the second part of (ii). Assumingδ does not exist such that |u1n|p+|u2n|p⩾δ>0, then (21)limn→+∞⁡inf⁡u1np+u2np⩽0.Now we have that (22)limn→+∞⁡I~u1n,u2n=limn→+∞⁡∫RN-Δα/2u1n2+-Δα/2u2n2dx-1p∫RNu1np+u2npdx-2p∫RNu1np/2u2np/2dx⩾-2plimn→+∞⁡∫RNu1np+u2npdx⩾0,which contradicts -∞<I~q<0. Therefore for sufficiently large n, there exists a positive constant δ such that u1np+u2np⩾δ. We have completed the proof of the lemma.Now, for minimizing sequence{(u1n,u2n)} of problem (I~q) we introduce the Lévy concentration function Qn(r)=supy∈RN⁡∫By,r‍u1n2+u2n2dx. Apparently, each function Qn is nondecreasing on (0,+∞). Using Helly’s selection theorem, {Qn} has a convergent subsequence {Qm}, such that limm→+∞⁡Qm(r)=Q(r), for all r>0, where Q(r) is a nondecreasing function. By 0⩽Qm(r)⩽q, there exists β(0⩽β⩽q) satisfying limγ→+∞⁡Q(r)=β.Next, we will proveβ=q.Lemma 12.
Assumingq>0 and {(u1n,u2n)} is a minimizing sequence for (I~q), then β>0.Proof.
By the definition ofβ, we know β⩾0. Using an argument by contradiction, assume β=0; then there exist nk and r0, satisfying (23)limnk→+∞⁡supy∈RN⁡∫By,r0u1nk2+u2nk2dx⟶0.By Lemma 5, we have u1nk,u2nk→(0,0) in Lp(RN)×Lp(RN). According to Lemma 11(ii), (u1nk,u2nk)↛(0,0) in Lp(RN)×Lp(RN). They are contradictory. Therefore, we obtain β>0.Lemma 13.
Forγ1,γ2>0, then I~γ1+γ2<I~γ1+I~γ2.Proof.
For given(u1,u2)∈Hα(RN)×Hα(RN), u122+u222=β>0, let uiθ(x)=θγ1ui(θγ2x)(i=1,2), where θ>0, γ1=2α/(4α-p-2N),γ2=(p-2)/(4α-p-2N). Then u1θ22+u2θ22=θβ and (24)I~u1θ,u2θ=θ2αp-Np-2/4α-p-2N12∫RN-Δα/2u12+-Δα/2u22dx-1p∫RNu1p+u2pdx-2p∫RNu1p/2u2p/2dx,where we apply Lemma 1.
Now, we can derive that(25)I~θβ=infu1,u2∈HαRN×HαRN,u122+u222=θβ⁡12∫RN-Δα/2u12+-Δα/2u22dx-1p∫RNu1p+u2pdx-2p∫RNu1p/2u2p/2dx=θ2αp-Np-2/4α-p-2Ninfu1,u2∈HαRN×HαRN,u122+u222=β⁡12∫RN-Δα/2u12+-Δα/2u22dx-1p∫RNu1p+u2pdx-2p∫RNu1p/2u2p/2dx=θ2αp-Np-2/4α-p-2NI~β.Therefore by Lemma 11(ii) and (2αp-N(p-2))/(4α-p-2N)>1, we have that (26)I~γ1+γ2=γ1+γ22αp-Np-2/4α-p-2NI~1<γ12αp-Np-2/4α-p-2N+γ22αp-Np-2/4α-p-2NI~1=I~γ1+I~γ2.Lemma 14.
Assumingq>0 and 0<β<q, then I~q⩾I~β+I~q-β.Proof.
For given minimizing sequence{(u1n,u2n)} for problem (I~q), r>0, define Qn(r)=supy∈RN⁡∫B(y,r)‍(|u1n|2+|u2n|2)dx. Then Qn is nondecreasing on (0,+∞); moreover for all r>0, {Qn} has subsequences denoted again by {Qn}, existing nondecreasing function Q(r) such that limn→+∞⁡Qn(r)=Q(r),limr→+∞⁡Q(r)=β. Therefore ∀ε>0,∃rε,nε such that r>rε,n>nε satisfy(27)β-ε2<Qr⩽Q2r⩽β,β-3ε4<Qnr⩽Qn2r⩽β+3ε4.Apparently, there exists {yn}⊂RN, satisfying (28)∫Byn,ru1n2+u2n2dx>β-ε,∫Byn,2ru1n2+u2n2dx<β+ε,where because of definition of Qn and (27).
Settingvin=Ψir(x-yn)uin,win=Ψ~ir(x-yn)uin,i=1,2, they satisfy that Ψir(x)=Ψi(x/r),Ψ~ir(x)=Ψ~i(x/r); ∀x∈B~(0,1),Ψi(x)=1; for |x|⩾2,Ψi(x)=0;Ψi(x)∈C0∞(B(0,2)), Ψ~i∈C∞(RN); and ∀x∈RN,Ψi2(x)+Ψ~i2(x)=1. Then we have that(29)β-ε<∫RNv1n2+v2n2dx<β+ε,q-β-ε<∫RNw1n2+w2n2dx<q-β+ε.Therefore (29) imply that there exist (μ1n,μ2n) and (ν1n,ν2n), |μin-1|<ε,|νin-1|<ε,i=1,2, satisfying μ1nv1n22+μ2nv2n22=β, ν1nw1n22+ν2nw2n22=q-β. Then we have that(30)I~β⩽I~μ1nv1n,μ2nv2n⩽I~v1n,v2n+Cε,I~q-β⩽I~ν1nw1n,ν2nw2n⩽I~w1n,w2n+Cε.
Using Lemma7, we derive that(31)-Δα/2Ψirx-ynuin-Ψirx-yn-Δα/2uin2⩽C∇Ψir∞-Δα-1/2uin2+-Δα/2Ψirx-ynp~uinp,where 1/p~+1/p=1/2,i=1,2. By ∇Ψir∞=(1/r)∇Ψi∞⩽C/r, the Sobolev inequality, Lemma 11, and (31), we obtain that (32)12∫RN-Δα/2Ψirx-ynuin2dx⩽12∫RNΨir2x-yn-Δα/2uin2dx+Cε,i=1,2, for sufficiently large r. Similarly, for sufficiently large r we have that (33)12∫RN-Δα/2Ψ~irx-ynuin2dx⩽12∫RNΨ~ir2x-yn-Δα/2uin2dx+Cε,i=1,2. So we have that(34)12∫RN-Δα/2Ψ1rx-ynu1n2+-Δα/2Ψ2rx-ynu2n2dx+12∫RN-Δα/2Ψ~1rx-ynu1n2+-Δα/2Ψ~2rx-ynu2n2dx⩽12∫RN-Δα/2u1n2+-Δα/2u2n2dx+Cε.
By Lemma11, (16), (27), 2<p<2+4α/N, and the definitions of Ψir, Ψ~ir(i=1,2), we can derive that (35)-1p∫RNΨ1rx-ynu1np+Ψ2rx-ynu2npdx+1p∫RNu1np+u2npdx-1p∫RNΨ~1rx-ynu1np+Ψ~2rx-ynu2npdx=-1p∫RNΨ1rx-ynu1np+Ψ2rx-ynu2npdx+1p∫RNΨ1r2x-yn+Ψ~1r2x-ynu1npdx+1p∫RNΨ2r2x-yn+Ψ~2r2x-ynu2npdx-1p∫RNΨ~1rx-ynu1np+Ψ~2rx-ynu2npdx=1p∫RN-Ψ1rp+Ψ1r2x-ynu1nxpdx+1p∫RN-Ψ2rp+Ψ2r2x-ynu2nxpdx+1p∫RN-Ψ~1rp+Ψ~1r2x-ynu1nxpdx+1p∫RN-Ψ~2rp+Ψ~2r2x-ynu2nxpdx⩽4p∫Ωu1nxp+u2nxpdx⩽Cpu1nL2Ωp1-λu1nWα,2Ωλp+u2nL2Ωp1-λu2nWα,2Ωλp,-∫RNΨ1rx-ynu1np/2Ψ2rx-ynu2np/2dx-∫RNΨ~1rx-ynu1np/2Ψ~2rx-ynu2np/2dx+∫RNu1np/2u2np/2dx=∫RN1-Ψ1rx-ynp/2Ψ2rx-ynp/2-Ψ~1rx-ynp/2Ψ~2rx-ynp/2u1p/2u2p/2dx⩽3p∫Ωu1np/2u2np/2dx⩽32p∫Ωu1npdx+∫Ωu2npdx⩽Cpu1nL2Ωp1-λu1nWα,2Ωλp+u2nL2Ωp1-λu2nWα,2Ωλp⩽Cε,where Ω=B(yn,2r)-B(yn,r) and λ=N(p-2)/2αp. Therefore by the two inequations, we obtain that(36)I~v1n,v2n+I~w1n,w2n⩽12∫RN-Δα/2u1n2+-Δα/2u2n2dx-1p∫RNu1np+u2npdx-2p∫RNu1np/2u2np/2dx+Cε=I~u1n,u2n+Cε.By (30) and (36), we obtain (37)I~β+I~q-β-3Cε⩽I~v1n,v2n+I~w1n,w2n-Cε⩽I~u1n,u2n⟶I~q.So, we prove that I~q⩾I~β+I~q-β.Obviously, using Lemmas11–14, we obtain β=q.Proof of Theorem9.
From the above parts; we know thatβ=q, that is, there exists a subsequence {(u1m,u2m)} of {(u1n,u2n)} satisfying (38)limr→+∞⁡limm→+∞⁡supy∈RN⁡∫By,ru1m2+u2m2dx=β=q.Therefore, for all k∈N, there exist rk>0,Mk∈N, such that, ∀m>Mk,∃ym∈RN, having (39)∫Bym,rku1m2+u2m2dx>q-1k.For the above inequality, with a Cantor diagonalization argument, using Lemma 11(ii), u1m22+u2m22=q, we know that there exists (g1,g2)∈Hα(RN)×Hα(RN) satisfying ∫RN‍(|g1|2+|g2|2)dx=q, such that going if necessary to exist a subsequence {(u1j(·+yj),u2j(·+yj))} of {(u1m(·+ym),u2m(·+ym))} satisfying (40)u1j·+yj,u2j·+yj⇀g1,g2inHαRN×HαRN,u1j·+yj,u2j·+yj⟶g1,g2inL2RN×L2RN.Then by the interpolation theorem, we obtain that (41)u1j·+yj,u2j·+yj⟶g1,g2inLpRN×LpRN.And by the weak low semicontinuity of the norm in Hα(RN), we also obtain that (42)u1j·+yj,u2j·+yj⟶g1,g2inHαRN×HαRN,and (g1,g2) is a minimizer of problem (I~q).
Using an argument by contradiction, we prove(43)limn→+∞⁡infg1,g2∈G~q⁡u1n-g1+u2n-g2=0.Assume that there exist ε0>0 and a subsequence {(u1nj,u2nj)} of {(u1n,u2n)} satisfying (44)infg1,g2∈G~q⁡u1nj-g1+u2nj-g2⩾ε0>0.By the above proof, we know that there exist a subsequence {(u1j,u2j)} of {(u1nj,u2nj)}, {yj}∈RN, and (g1,g2)∈G~q satisfying (45)u1j·+yj,u2j·+yj⟶g1,g2inHαRN×HαRN.So we derive (46)limj→+∞⁡u1j·+yj-g1+u2j·+yj-g2=0,where since (g1(·-yj),g2(·-yj))∈G~q, we have that (47)limj→+∞⁡infg1,g2∈G~q⁡u1j·+yj-g1+u2j·+yj-g2=0.Then there exists a contradiction. We completed the proof of Theorem 9.Proof of Theorem10.
Now using an argument by contradiction, we complete the proof of Theorem10. Assume that the set G~q is not Hα(RN)×Hα(RN)-stable. Then there exist ε0>0, (u1k0,u2k0)⊂Hα(RN)×Hα(RN), and tk∈[0,T) such that(48)infg1,g2∈G~q⁡u1k0-g1+u2k0-g2⩽1k,k∈N,and for (u1k,u2k)∈C([0,T),Hα(RN)×Hα(RN)) satisfying (1) with initial value (u1k0,u2k0), we have(49)infg1,g2∈G~q⁡u1ktk,·-g1+u2ktk,·-g2⩾ε0.By (48), we know that (50)limk→+∞⁡I~u1k0,u2k0=I~q,limk→+∞⁡u1k022+u2k022=q.Then there exist {μ1k},{μ2k}∈R satisfying limk→+∞⁡μ1k=1 and limk→+∞⁡μ2k=1 such that {(μ1ku1k0,μ2ku2k0)} is a minimizing sequence of the problem (I~q). Since I~(u1k0,u2k0)=I~(u1k(tk),u2k(tk)), |u1k0|22=|u1k(tk)|22 and |u2k0|22=|u2k(tk)|22, we easily derive that {(μ1ku1k(tm),μ2ku2k(tm))} is a minimizing sequence of the problem (I~q). By Theorem 9, there exists {(g1km,g2km)}⊂G~q, for sufficiently large km satisfying (51)μ1kmu1kmtkm-g1km+μ2kmu2kmtkm-g2km<ε02.Therefore by Lemma 11(ii), (49), limk→+∞⁡μ1k=1, limk→+∞⁡μ2k=1, and the above inequality, for sufficiently large km, we derive that (52)ε0⩽u1kmtkm-g1km+u2kmtkm-g2km⩽u1kmtkm-μ1kmu1kmtkm+μ1kmu1kmtkm-g1km+u2kmtkm-μ2kmu2kmtkm+μ2kmu2kmtkm-g2km⩽3ε04.Then obtaining a contradiction, the set G~q is Hα(RN)×Hα(RN)-stable with respect to (1). The proof is completed.



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*Source: 427487-2015-12-27.xml*