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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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Critical elliptic systems involving multiple strongly–coupled Hardy–type terms

Dongsheng Kang
  • Corresponding author
  • School of Mathematics and Statistics, South–Central University for Nationalities, Wuhan 430074, Wuhan, China
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/ Mengru Liu
  • School of Mathematics and Statistics, South–Central University for Nationalities, Wuhan 430074, Wuhan, China
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/ Liangshun Xu
  • School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, Wuhan, China
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Published Online: 2019-07-31 | DOI: https://doi.org/10.1515/anona-2020-0029

Abstract

In this paper, we study the radially–symmetric and strictly–decreasing solutions to a system of critical elliptic equations in RN, which involves multiple critical nonlinearities and strongly–coupled Hardy– type terms. By the ODEs analysis methods, the asymptotic behaviors at the origin and infinity of solutions are proved. It is found that the singularities of u and v in the solution (u, v) are at the same level. Finally, an explicit form of least energy solutions is found under certain assumptions, which has all of the mentioned properties for the radial decreasing solutions.

Keywords: critical elliptic system; radial decreasing solution; asymptotic property; Hardy term

MSC 2010: 35J47; 35J50

1 Introduction

In this paper, we study the following critical elliptic system involving multiple coupled Hardy–type terms:

{Δu-μ1μ+λv|x|2=u2*1+ηa2*uα1vβ,xN\{0},Δv-λu+μ2v|x|2=v2*1+ηβ2*uαvβ1,xN\{0},(u,v)(D1,2(N))2,u,v>0,xN\{0},(1.1)

where D1,2(RN) is the completion of C0(N)with respect to (N|.|2dx)1/2and the parameters satisfy the assumption:

H1N5,1<α,β<2,α+β=2,η>0,μ¯>μ1μ2>0,λ>0,λ2<minμ1μ2,μ¯1μ1μ¯1μ2.

2*:=2NN2is the critical Sobolev exponent and μ¯:=(N22)2is the best constant in the Hardy inequality ([1]):

Nu2|x|2dx1μ¯N|u|2dx,uD1,2(N).

Under the assumption (H1), the matrix E:=(μ1λλμ2)is positive definite and for all (u, v) (D1,2(RN))2 there holds that 0 < у1 < у2 < ¯μ, and furthermore,

y1(u2+v2)μ1μ2+2λuv+μ2v2y2(u2+v2),

where у1, у2, are eigenvalues of the matrix E. According to the Hardy, Sobolev and Young inequalities, the following best constants are well defined ([2, 3, 4, 5]):

S(μ):=infuD1,2(N)\{0}N(|u|2μu2|x|2)dx(N|u|2dx)22,μ<μ¯,S(μ1,μ2,λ):=inf(u,v)D(N)(|u|2+|v|2μ1u2+2λuv+μ2v2|x|2)dx(N(|u|2+|v|2+η|u|α|v|β)dx)22,(1.2)

where D := (D1,2(RN))2 \ {(0, 0)}. Critical elliptic equations and systems involving the Hardy inequality have been studied by many authors (e.g. [2, 5–10] for the problems of single equation, [3], [4], [11, 12, 13, 14] for related elliptic systems). In particular, Terracini proved that S(μ) has the unique positive extremals ([5]):

where

Vμεx:=ε2N2Uμε1x,ε>0,μ0,μ¯,Uμx:=2Nμ¯μμ¯μ¯2xμ¯μ¯μμ¯+xμ¯+μ¯μμ¯μ¯,(1.3)

by which many elliptic problems involving the Hardy inequality have been studied. Furthermore, elliptic systems corresponding to the case λ = 0 in (1.1) have been studied (e.g. [11] and [3] for the existence of solutions, [4], [13] and [14] for the asymptotic properties of solutions, [15] and [16] for the case without Hardy terms).

In this paper, we study the asymptotic behaviors at the origin and infinity of the radially–symmetric and strictly–decreasing solution (u, v) to (1.1) by the ODEs analysis methods. The existence of this kind of solutions to (1.1) can be proved by the arguments similar to those of [3]. Elliptic regularity argument shows that the solution (u, v) to (1.1) satisfies u,vC2(N\{0}).

The following constants and functions are all well defined under (H1):

A=Aμ1,μ2,λ:=2λμ1μ22+4λ2+μ1μ2,lμ1,μ2,λ:=μ¯μ1+μ2+μ1μ22+4λ22,al:=μ¯lμ1,μ2,λ,bl:=μ¯+lμ1,μ2,λ,(1.4)

f(τ):=ηβ2+τατ2βηα2τ2,τ0,g(τ):=λτ2+(μ1μ2)τλ,τ0.(1.5)

Under the assumption (H1), since f(0)=ηβ2>0,f(+)=,forallη(0,),then there exists naturally the smallest positive zero Λ0 of f (e.g. [16]). Furthermore, f (τ) > 0 in [0, Λ0) and a direct calculation shows that g(A) = 0 and g(τ) < 0 for all τ 2 [0,A). To study the explicit form minimizers to S(μ1, μ2, λ), we also define

F(τ):=1+τ2(1+ητβ+τ2)22,τ0,F(τmin):=minτ0F(τ),(1.6)

η:=μ1+2λτmin+μ2(τmin)21+(τmin)2,μ:=μ1+μ2+(μ1μ2)2+4λ22s:=(1+ηα2Aβ)122,(1.7)

where τmin ≥ 0 is a minimal point of F (τ). Direct calculation shows that

2λAλ+λA,μ1+Aλ=μ2+λAμ,μ0,μ¯.(1.8)

Note that Λ0, τmin and μ* depend on η but are independent of λ. However, A and μ* depend on λ but are independent of η.

The main results of this paper are summarized in the following theorems and they are new to the best of our knowledge. It can be checked that the intervals for the parameters are not empty.

Theorem 1.1

Suppose that (H1) holds and (u, v) is a radially–symmetric and strictly–decreasing solution to the problem (1.1). Set r=|x|,xN\{0}.

  • (i)

    Then

limr0+vrur=A,limr0+rurur=limr0+vrvr=al,limr+vrur=A,limr+rurur=limr+vrvr=bl.

  • (ii)

    Assume furthermore that A < Λ0, then

infr>0v(r)u(r)=A,infr>0u(r)u(r)=b(l),supr>0ru(r)u(r)=a(l),

and none of above infimums and supremum can be achieved.

Let (u, v) be a radial decreasing solution to (1.1) and set r = ∣x∣ = et , x ∈ RN \ {0}, t 2 R. By (H1), (1.4) and above Theorem 1.1, we have that λA < ¯μμ1 and the following constant and function are well defined:

T0:=inf{T>0|μ¯μ1λv(et)v(et)>0,t(,T)(T,)},ω(t):=μ¯μ1λv(et)u(et),(,T0)(T0,).

Theorem 1.2

Suppose that (H1) holds with A < Λ0 and (u, v) is a radially–symmetric and strictly–decreasing solution to the problem (1.1). Set r=|x|=et,xN\{0}.Then there exist the constants C1, C2 > 0, such that

limt+eμ¯t+T0tω(s)dsu(et)=C1,limt+e(μ¯+1)t+T0tω(s)ds|u(et)|=C1b(l),limteμ¯t+tT0ω(s)dsu(et)=C2,limte(μ¯+1)t+tT0ω(s)ds|u(et)|=C2a(l),limt+eμ¯t+T0tω(s)dsv(et)=C1A,limt+e(μ¯+1)t+T0tω(s)ds|v(et)|=C1Ab(l),limteμ¯t+tT0ω(s)dsv(et)=C2A,limte(μ¯+1)t+tT0ω(s)ds|v(et)|=C2Aa(l).

In the following theorem, we find the explicit radially–symmetric and strictly–decreasing minimizers to S(μ1, μ2, λ) under certain assumptions, among which there exists an explicit form of least energy solutions to (1.1) satisfying all of the properties mentioned in Theorems 1.1 and 1.2.

Theorem 1.3

Suppose that (H1) holds and Vμε(x)is the minimizer of S(μ) defined as in (1.3). Assume further more that τmin = A. Then

μ=μ,S(μ1,μ2,λ)=F(A)S(μ)=F(A)S(μ),

and S(μ1, μ2, λ) has the radially–symmetric and strictly–decreasing minimizers of the form {C(Vμε,AVμε)C,ε>0},among which the problem (1.1) has the explicit form of least energy solutions

{(sVμε(x),sAVμε(x)),ε>0}.

Corollary 1.4

Suppose that (H1) holds with β<α,ηNN2.min Then there exist μ1, μ2, λ ∈ (0, ¯μ) such that =

A, S(μ1, μ2, λ) = F(A)S(μ*), and thus S(μ1, μ2, λ) has the minimizers of the form {C(Vμε(x),AVμε(x)),C,ε>0}.

Remark 1.5

Suppose that (H1) holds. Then the existence of radial decreasing ground state solutions to (1.1) can be proved by the arguments similar to those of [3], where the case λ = 0 was studied and by which S(μ1, μ2) λ, α=β=22,is achieved. Taking in (1.1) μ1 > μ2, η = 2, then a direct calculation shows that 0 < A < 1, Λ0 = 1, and thus A < Λ0 holds naturally.

Remark 1.6

Suppose that (H1) holds with A < Λ0. Taking T > max{|t1|,|t2|}and by Lemma 3.3 of this paper, the radial decreasing solution (u, v) to (1.1) satisfies

limt±ω(t)=l(μ1,μ2,λ)>0,ω(t)<l(μ1,μ2,λ),t,limt+μ¯t+T0tω(s)dsb(l)t=1,limtμ¯t+tT0ω(s)dsa(l)t=1,

which together with Theorem 1.2 reveals that u and v have the singularities of same level at the origin and infinity. When λ = 0 and the other parameters satisfy (H1), then there exists a critical surface II for the positive solution (u, v) to (1.1) (e.g. [14]):

:αμ¯μ1=(2β)μ¯μ2,

such that the singularities of v are different above and below II, which implies that the strongly–coupled critical terms in (1.1) plays a key role for the asymptotic properties of u, v. However, when (H1) holds with λ > 0, then the asymptotic properties of u, v, depend only on the Hardy terms in (1.1). Hence, the methods and conclusions of this paper have crucial differences with those of [14].

Remark 1.7

Suppose that (H1) holds. Then Theorem 1.3 shows that the minimizers of S(μ1, μ2, λ) has an explicit relation with those of S(μ*) if τmin = A. In our following work, we will study singularities of solutions to (1.1) when τmin ≠ A. Then combining

with the conclusions of this paper, the asymptotic properties of the radial decreasing minimizer (U, V) to S(μ1, μ2, λ) will be clear and further studies on (1.1) and related problems can be done, even without the explicit forms of (U, V).

This paper is organized as follows. Some preliminary results are established in Section 2, Theorems 1.1 and 1.2 are proved in Section 3, and Theorem 1.3 is verified in Section 4. For convenience, we always denote positive constants as C and omit dx in integrals if no confusion is caused.

2 Preliminary results

Assume that (H1) holds. Let r = ∣x∣ and (u(r), v(r)) be a radially–symmetric and strictly–decreasing solution to (1.1). Then from (1.1) it follows that

{(rN1u(r))=rN1(μ1ur2+λvr2+u21+ηα2uα1vβ),(rN1v(r))=rN1(λur2+μ2vr2+u21+ηβ2uα1vβ1),(2.1)

which implies that rN−1u'(r) and rN−1v'(r) are strictly decreasing in (0, +∞). Since u'(r), v'(r) ≤ 0, then we obtain that u'(r), v'(r) < 0 in (0, +∞). Set

t=In r,r=et,r(0,=),t,y1(t)=rμ¯u(r),z1(t)=rμ¯+1u(r),y2(t)=rμ¯v(r),z2(t)=rμ¯+1v(r),(2.2)

Since u(r), v(r) ∈ C2(0, +∞), y1(t), y2(t) ∈ C2(R), by (2.1) and (2.2) we have that

{y1=μ¯y1+z1,z1=μ¯z1μ1y1λy2(y121+ηα2y11y2β),y2=μ¯y2+z2,z2=μ¯z2λy1μ2y2(y221+ηβ2y1y2β1),(2.3)

which implies that y1 and y2 satisfy the following ODEs system:

{y1=(μ¯μ1)y1λy2y121ηα2y1α1y2β,y2=(μ¯μ2)y2λy1y221ηβ2y1αy2β1.(2.4)

The complete integral of (2.4) is given by

Vy1,y2,y1,y2:=12y12+y2212μ¯μ1y1212μ¯μ1y22+λy1y2+12y12+y22+ηy1α,y2β.

Set V(t)=V(y1(t),y2(t),y1(t),y2(t)).From (2.4) it follows that V'(t) = 0. Then V(t) is a constant and we set

V(t)K0,t.

Lemma 2.1

Suppose that (H1) holds. Then K0 = 0. Furthermore,

|yi(t)yi(t)|<μ¯μi,t,i=1,2,limt±yi(t)=limt±yi(t)=limt±yi(t)=0,i=1,2,(2.5)

Proof. The arguments are similar to those of Lemmas 2.1 and 2.2 in [14], where the case λ = 0 of (1.1) was studied. The details are omitted for simplicity.

3 Asymptotic properties

Assume that (H1) holds and the functions y1(t), y2(t), are defined as in (2.2). For all l ∈ R, i = 1, 2, a direct calculation shows that

eltyi(t)=yi(0)e0t(l+yi(s)yi(s))ds,t(0,+),eltyi(t)=yi(0)et0(lyi(s)yi(s))ds,t(,0),(3.1)

Furthermore,

(eltyi(t))=eltyi(t)(l+yi(t)yi(t)),t(0,+),(eltyi(t))=eltyi(t)(l+yi(t)yi(t)),t(,0).

For i = 1, 2, define the functions

gi(l):=0+(l+yi(s)yi(s))ds,l.

From Lemma 2.1 it follows that

μ¯μi+yi(s)yi(s)<0<μ¯μi+yi(s)yi(s),s.(3.2)

For any ɛ > 0, from (3.2) it follows that

gi(μ¯μi+ε)=+,gi(o)=(3.3)

Therefore the following constants are well defined:

li:=sup{l|gi(l)<+|},ll:=inf{l|gi(l)>}.

Since ɛ > 0 is arbitrary, from (3.3) it follows that

0liliμ¯μi,i1,2.(3.4)

Lemma 3.1

Suppose that (H1) holds. Then li=li0,i = 1, 2*.

Proof. Assume that li<li.Then there exist a, b 2 R such **that and Furthermore, libali<gi(b)<gi(a)<+.

gi(a)gi(b)=0+(ab)ds=+,

a contradiction. From (3.4) it follows that li=li0.

According to Lemma 3.1, the following constants are well defined:

li:=sup{l|gi(l)<+}=inf{l|gi(l)>},i=1,2.

Similarly, we define the functions

gi+2(l):=0(lyi(s)yi(s))ds,i=1,2.

Arguing as above, the following constants are well defined

li:=suplgil<+=inflgil>,i=3,4.

Furthermore,

0l1,l3μ¯,μ1,0l2,l4μ¯,μ2.

Therefore, we can assume that l ≥ 0 when studying li , 1 ≤ i ≤ 4.

Lemma 3.2

Suppose that (H1) holds. Then l1 = l2, l3 = l4.

Proof. For all ɛ > 0, from the definitions of li it follows that

0t(li±ε+yi(s)yi(s))ds±ast+,ε>0,i=1,2,

which together with (3.1) implies that

e(liε)tyi(t)=yi(o)e0t(liε+yi(s)yi(s))ds0ast+,i=1,2,e(li+ε)tyi(t)=yi(o)e0t(liε+yi(s)yi(s))ds+ast+,i=1,2.0<e(liε)tyi(t)<1<e(li+ε)tyi(t)ast+,i=1,2,e(l1l22ε)t<y2y1<e(l1l2+2ε)tast+,ε>0,(3.5)

Consequently,

which implies that

If l1 > l2, then by taking 2ɛ < l1l2 we have that limt+y2y1=+From (2.4) and the L’Hospital rule it follows that

limt+|y1|2y12=limt+y1y1=limt+(μ¯μ1λy2y1ηα2y1α2y2β)=,

a contradiction. If l1 < l2, then limt+y1y2=+and therefore

limt+|y2'|2y22=limt+y2''y2=limt+(μ¯μ2λy1y2ηβ2y1αy2β2)=,

a contradiction.

Therefore l1 = l2 must hold and similarly l3 = l4 also holds.

Lemma 3.3

Suppose that (H1) holds. Let A and l(μ1, μ2, λ) be defined as in (1.4). Then

limt+y2ty1t=limty2ty1t=A,limt+yityit=limtyityit=lμ1,μ2,λ,i=1,2.

Proof. From (3.5) and Lemma 3.2 it follows that

limt±y1α2y2β=0,limt±y1αy2β2=0.(3.6)

  • (i)

    We claim that either y2(t)y1(t)or y1(t)y2(t)is bounded in (0,+).

In fact, if neither y2(t)y1(t)nor y1(t)y2(t)is bounded in (0,+),then the continuity implies that y2(t)y1(t)must have sequences of local minimum points {σn}(0,+)such that

limnσn=+,limny2(σn)y1(σn)=0,(y2(t)y1(t))|t=σn=0,(y2(t)y1(t))|t=σn0.

Then for all n ∈ N, a direct calculation shows that

y2'(σn)y1'(σn)=y2(σn)y1(σn),(y2''(t)y2(t)y1''(t)y1(t))|t=σn0,

which together with (2.4) imply that

(μ¯μ2λy1y2y222ηβ2y1αy2β2)|t=σn(μ¯μ1λy2y1y122ηα2y1α2y2β)|t=σn.(3.7)

Since

limny1(σn)=limny2(σn)=0,limny1(σn)y2(σn)=+,

by (3.6) and (3.7) we get a contradiction, which implies that either y2(t)y1(t)or y1(t)y2(t)is bounded in (0, +∞).

  • (ii)

    We claim that if y2(t)y1(t)is bounded, then there exists the limit limt+y2(t)y1(t)=A.

In fact, if limt+y2(t)y1(t)doesn’t exist, then the continuity implies that y2(t)y1(t)must have sequences of local minimum points {σ¯n}(0,+)and of local maximum points {τ¯n}(0,+),such that

limnσ¯n=,(y2(t)y1(t))|t=σ¯n=0,(y2(t)y1(t))|t=σ¯n0,limnτ¯n=,(y2(t)y1(t))|t=τ¯n=0,(y2(t)y1(t))|t=τ¯n0.

Then for all n ∈ N, a direct calculation shows that

y2(σ¯n)y1(σ¯n)=y2(σ¯n)y1(σ¯n),y2(σ¯n)y1(τ¯n)=y2(τ¯n)y1(τ¯n),(y2(t)y2(t)y1(t)y1(t))|t=σ¯n0,(y2(t)y2(t)y1(t)y1(t))|t=τ¯n0.

For any local minimum points {σ¯n}(0,+)ofy2(t)y1(t),since σ¯nis a local maximum point of y1y2y2y1,from (3.6) and (3.7) it follows that

μ1μ2+o1λy1σ¯ny2σ¯ny2σ¯ny1σ¯n,asn,μ1μ2+o1λy1ty2ty2ty1t,ast+.

which implies that

Similarly, for any local maximum points {τ¯n}(0,+)ofy2(t)y1(t),since ¯τn is a local minimum point of y1y2y2y1such that

μ1μ2+o(1)λ(y1(τ¯n)y2(τ¯n)y2(τ¯n)y1(τ¯n)),asn,μ1μ2+o(1)λ(y1(t)y2(t)y2(t)y1(t)),ast+.

and therefore

Consequently,

limt+(y1(t)y2(t)y2(t)y1(t))=μ1μ2λ,(3.8)

which implies that

limt(y1(t)y2(t)+y2(t)y1(t))=(μ1μ2λ)2+4.(3.9)

From (3.8) and (3.9) it follows that limt+y2(t)y1(t)exists, a contradiction.

Therefore, there must exist the limit B:=limt+y2(t)y1(t).Since l1 = l2, from (2.4) and the L’Hospital rule it follows that

B=limt+y2(t)y1(t)=limt+y2(t)y1(t)=limt+y2(t)y1(t)=(μ¯μ2)Bλμ¯μ1λB,

which implies that

B=limt+y2(t)y1(t)=A2λ(μ1μ2)2+4λ2+(μ1+μ2)>0.(3.10)

  • (iii)

    Similarly, if y1(t)y2(t)is bounded, then there exists the limit limt+y1(t)y2(t)=1A.

From (i)–(iii) it follows that (3.10) holds under (H1). Arguing as above, under (H1) we also have that

limty2(t)y1(t)=A(3.11)

Let l(μ1, μ2, λ) be defined as in (1.4). Then a direct calculation shows that

0<l(μ1,μ2,λ)μ¯μi,i=1,3,l(μ1,μ2,λ)=μ¯μ1λA=μ¯μ2λA.(3.12)

From (2.4), (3.6), (3.11), (3.12) and the L’Hospital rule it follows that

limt±|y1|2y12=limt±y1y1=limt±(μ¯μ1λy2y1)=l(μ1,μ2,λ)2,limt±|y2|2y22=limt±y2y2=limt±(μ¯μ2λy1y2)=l(μ1,μ2,λ).(3.13)

By (3.13) we deduce that y1,y2>0as ∣t∣ large enough, which together with (2.5) implies that y1,y2<0as t+and y1y2>0as t ! −∞. Consequently,

limt+yi(t)yi(t)=limt+yi(t)yi(t)=l(μ1,μ2,λ),i=1,2.(3.14)

The proof is complete.

Lemma 3.4

Suppose that (H1) holds and A < Λ0. Then

infty2(t)y1(t)=A,supty1(t)y1(t)=l(μ1,μ2,λ),infty1(t)y1(t)=l(μ1,μ2,λ).

Proof. If infty2(t)y1(t)can’t be achieved at any finite point, then from Lemma 3.3 it follows that infty2(t)y1(t)=AIf the infimum is achieved at a finite point t3,then

0<infty2(t)y1(t)=y2(t3)y1(t3)A,(y2y1)|t=t3=[y2y1(y2y1y1y1)]t=t3=0,(y2y1)|t=t3=[y2y1(y2y2yy1)]t=t30.(3.15)

Let the functions f (τ) and g(τ) be defined as in (1.5) and Λ0 be the smallest positive zero of f (τ). Then f (τ) > 0 for all τ 2 [0, Λ0), g(A) = 0 and g(τ) < 0 for all τ ∈ [0,A). Since A < Λ0, from (2.4) and (3.15) it follows that

0g(y2(t3)y1(t3))f(y2(t3)y1(t3))(y2(t3)y1(t3))β1y1(t3)22>0,

a contradiction, which together with (3.10) implies that infty2(t)y1(t)can’t be achieved at any finite point and furthermore,

limxy2(t)y1(t)=infty2(t)y1(t)=A.(3.16)

Consider the supremum of|y1(t)y1(t)|in R. If the supremum can’t be achieved at any finite point, then (3.16) implies that supt|y1(t)y1(t)|=l(μ1,μ2,λ).If the supremum is achieved at finite point t4 2 R, then

Furthermore,

|y1(t4)y1(t4)|=supt|y1(t)y1(t)|l(μ1,μ2,λ).(y1(t)y1(t))|t=t4=(y1(t)y1(t))|y1(t)|2|y1(t)|2|t=t4=0.(3.17)

Since infty2(t)y1(t)can’t be achieved at any finite point, from (2.4), (3.16) and (3.17) it follows that

|y1|2y12|t=t4=[(μ¯μ1)λy2y1y122ηα2y1α2y2β]t=t4μ¯μ1λy2(t4)y1(t4)<μ¯μ1λA=l(μ1,μ2,λ)2,

a contradiction, which implies that supt|y1(t)y1(t)|can’t be achieved at any finite points. Then from (3.14) and (3.16) it follows that

limt+y1(t)y1(t)=infty1(t)y1(t)=μ¯μ1λA=l(μ1,μ2,λ),limty1(t)y1(t)=supty1(t)y1(t)=μ¯μ1λA=l(μ1,μ2,λ).(3.18)

The proof is complete.

Lemma 3.5

Suppose that (H1) holds and A < Λ0. Then li = l(μ1, μ2, λ), 1 ≤ i ≤ 4.

Proof. For any ɛ > 0, by the definition of l1 we have that

0+((l1+εl(μ1,μ2,λ))+l(μ1,μ2,λ)+y1y1)ds=0+(l1+ε+y1y1)ds =+.

From Lemma 3.4 it follows that

l1+εl(μ1,μ2,λ)0,ε>0,

that is, l1l(μ1, μ2, λ) ≥ 0. For any ɛ > 0, from Lemma 3.4 it follows that

g1(l(μ1,μ2,λ)+ε)=+.

Then l1l(μ1, μ2, λ). Therefore, l1 = l(μ1, μ2, λ), which together with Lemma 3.2 implies that l1 = l2 = l(μ1, μ2, ).

Under the assumption (H1), arguing similarly as above we also have that

limty1y1=limty2y2=l(μ1,μ2,λ),l3=l4=l(μ1,μ2λ).

The proof is complete.

Lemma 3.6

Suppose that (H1) holds with A < Λ0 and let t1, t2, be defined as in Theorem 1.1. Then there exist the positive constants C1, > 0, C2 such that

limt+eT0tμ¯μ1λy2sy1sdsy1t=C1,limtetT0μ¯μ1λy2sy1sdsy1t=C2,limt+eT0tμ¯μ1λy2sy1sdsy2t=C1A,limtetT0μ¯μ1λy2sy1sdsy2t=C2A.

Proof. We only prove the first equality. The second one can be verified similarly and the last two can be concluded by the first two equalities and (3.10).

Set H1(s):=y1(s)y1(s).From (2.4) and Lemma 3.4 it follows that

l(μ1,μ2,λ)±H1(s)>0,s.

Note that A < Λ0 and

limt+(μ¯μ1λy2(t)y1(t))=l(μ1,μ2,λ)2>0.(3.19)

Arguing as in the proof of Lemma 3.4, we have that

μ¯μ1λy2(t)y1(t)>0,t(T0,+).

According to (2.4) and by direct calculation we have that

H1(s)=μ¯μ1λy2(s)y1(s)H12(s)y122ηα2y1α2y2β,

which implies that

μ¯μ1λy2(s)y1(s)+H1(s)=H1(s)+y122+ηα2y1α2y2βμ¯μ1λy2(s)y1(s)H1(s),t>T0.(3.20)

Define

I:=T0+(μ¯μ1λy2(s)y1(s)+H1(s))ds,I1:=T0+H1(s)μ¯μ1λy2(s)y1(s)+H1(s)ds,I2:=T0+y122+ηα2y1α2y2βμ¯μ1λy2(s)y1(s)+H1(s)ds.

We claim that the integral I:=T0+(μ¯μ1λy2(s)y1(s)+H1(s))ds converges.

In fact, from (3.18) and (3.19) follows that

limt+μ¯μ1λy2(s)y1(s)=l(μ1,μ2,λ),limt+H1(s)=l(μ1,μ2,λ).(3.21)

Since A < Λ0, arguing as in the proof of Lemma 3.4 we have that y2(s)y1(s)and H 1(s) are strictly decreasing as s ! +∞. Taking ¯τ > T0 large enough we have that H1(s)<0for all s > ¯τ and

|τ¯+H1(s)μ¯μ1λy2(s)y1(s)H1(s)ds|=|τ¯+H1(s)l(μ1,μ2,λ)H1(s)l(μ1,μ2,λ)H1(s)μ¯μ1λy2(s)y1(s)H1(s)ds|C|τ¯+H1(s)l(μ1,μ2,λ)H1(s)ds|=Cln2l(μ1,μ2,λ)|l(μ1,μ2,λ)H1(τ¯)|<+.(3.22)

Then the integral I1 converges. Furthermore, (3.21) implies that there exists R > 0 such that

Therefore,

μ¯μ1λy2(s)y1(s)H1(s)>l(μ1,μ2,λ),s>R.0<I2:=R+2y122+ηαy1α2y2βμ¯μ1λy2(s)y1(s)H1(s)ds<1l(μ1,μ2,λ)R+(2y122+ηαy1α2y2β)ds.(3.23)

For any ɛ > 0, t > 0, from (3.5) and Lemma 3.5 it follows that

R+(y122+y1α2y2β)ds<C(ε)R+e((22)l(μ1,λ,μ2)(2+2)ε)sds.

By taking ε0+we have that

((22)l(μ1,μ2,λ)(2+2)ε)<0.

Then the integral T0+(2y122+ηαy1α2y2β)ds converges, which together with (3.23) implies that the integralI2 converges.

By (3.22) and (3.23) we have that the integral I = I1 + I2 converges. Furthermore,

limt+eT0tμ¯μ1λy2(s)y1(s)dsy1(t)=y1(T0)limt+eT0t(μ¯μ1λy2(s)y1(s)+H1(s))ds=y1(T0)eT0+(μ¯μ1λy2(s)y1(s)+H1(s))ds=C1,

where C1:=y1(T0)eT0+(μ¯μ1λy2(s)y1(s)+H1(s))ds>0.

Proof of Theorem 1.1

he results follow directly from (2.2), (2.3), Lemma 3.3 and Lemma 3.4.

Proof of Theorem 1.2

The asymptotic properties of u(r) and v(r) at the origin and infinity follow from (2.2) and Lemma 3.6, and the asymptotic properties of u'(r) and v'(r) follow from (2.2), (2.3), (3.14) and Lemma 3.6.

4 Explicit form solutions

In this section, we study the explicit form of radially–symmetric and strictly–decreasing minimizers to S(μ1, μ2, λ), among which there exists an explicit form of least energy solutions to (1.1), satisfying all of the properties in Theorems 1.1 and 1.2. For convenience we set k(τ) := −f (τ), τ > 0, where f (τ) is defined as in (1.5).

Proof of Theorem 1.3

Suppose that (H1) holds with τmin = A. We first investigate the functions F (τ) and k(τ). A direct calculation shows that

Note that

F(τ)=2τβ1k(τ)(1+ητβ+τ2)22+1,τ>0.limτ0+F(τ)=limτ+F(τ)=1,k(τ)<0asτ0+,k(τ)>0asτ+.(4.1)

Then minτ0F(τ)must be achieved at finite τmin > 0 and from (4.1) it follows that

F(τmin)=0,k(τmin)=0,0<F(τmin)<1.

For all w ∈ D1, 2(RN) \ {0}, testing the second Rayleigh quotient in (1.2) by (w, τminw), we have that

S(μ1,μ2,λ)F(τmin)N(|w|2μw2|x|2)(N|w|2)22,

which together with (1.2) implies that

S(μ1,μ2,λ)F(τmin)S(μ).(4.2)

Let {(un , vn)} ⊂ D be a minimizing sequence of S(μ1, μ2, λ) and define zn = snvn, where

which implies that

sn=((N|vn|2)1N|un|2)12,N|zn|2=N|un|2.(4.3)

By the Young inequality and (4.3) we have that that

Nunαznβα2Nun2+β2Nzn2=Nun2=Nzn2.(4.4)

Then from (1.2), (1.6), (1.8) and (4.4) it follows that

RNun2+vn2μ1un2+2λunvn+μ2v2x2RNun2+ηunαvnβ+vn222RNun2+vn2μ1+Aλun2x2μ2+λAvn2x21+ηsnβ+sn2RNun222RNun2μun2x21+ηsnβ+sn2RNun222+sn2RNzn2μzn2x21+ηsnβ+sn2RNzn222Fsn1SμFτminSμ.

Taking nwe have that

S(μ1,μ2,λ)F(τmin)S(μ).(4.5)

A direct calculation shows that

τmin=Aμ=μ.(4.6)

Then from (4.2), (4.5) and (4.6) it follows that

S(μ1,μ2,λ)=F(A)S(μ)=F(A)S(μ),

which implies that S(μ1, μ2, λ) has the minimizers of the form:

{C(Vμε(x),AVμε(x)),C,ε>0}.

Since k(τmin) = k(A) = 0, a direct calculation shows that the problem (1.1) has the explicit form of least energy solutions

{(sVμε(x),sAVμε(x)),ε>0},

that is, (uε,vε):=(sVμε(x),sAVμε(x))is a solution to (1.1) such that

N(|uε|2+|vε|2μ1uε2+2λuεvε+μ2vε2|x|2)=N(uε2+vε2+ηuεαvεβ)

=(F(A)S(μ))N2=S(μ1,μ2,λ)N2,

where s* is defined as in (1.7). By (3.18) we have that l(μ1,μ2,λ)=μ¯μ1λAand therefore the solution (sVμε(x),sAVμε(x))satisfies all of the properties mentioned in Theorems 1.1 and 1.2 with T0 = 0.

The proof is complete.

Proof of Corollary 1.4

Since τmin depends on , β, η, and A depends on μ1, μ2, λ, then τmin is independent of A. Obviously, by (4.1) a sufficient condition to ensure τmin = A is

k(A)=0,k(τ)<0in(0,A),k(τ)>0in(A,+).

In the following discussion, we only consider the case k(τ) > 0 in [1, + ∞).

Suppose that 1<β<α<2,ηNN2.Since

then we have that

k(τ)=τ1β(2β+2ηα2τβατ22),k(1)=η2(αβ)>0,2β>0,β>22,2ηα2α,k(τ)>0,k(τ)>k(1)>0,τ(1,+).(4.7)

Since k(τ)<0asτ0+,from (4.1) and (4.7) it follows that minτ0F (τ) must be achieved at finite τmin 2 (0, 1). Noting that 0 < A ≤ 1, A 0 as λ → 0 and A = 1 as μ1 = μ2, and A(μ1, μ2, λ) is a continuous function, there must exist certain μ1, μ2, λ 2 (0, ¯μ), such that A(μ1, μ2, λ) = τmin. Then the desired result follows directly from Theorem 1.3.

Acknowledgement

The authors acknowledge the anonymous referee for carefully reading this paper and making many important comments. This work is supported by the Fundamental Research Funds for the Central Universities of China, South– Central University for Nationalities (No. CZT18008).

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About the article

Received: 2018-11-25

Accepted: 2019-03-29

Published Online: 2019-07-31

Published in Print: 2019-03-01


Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 866–881, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2020-0029.

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© 2020 D. Kang et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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