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Volume 9, Issue 1

# 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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• Other articles by this author:
/ Mengru Liu
• School of Mathematics and Statistics, South–Central University for Nationalities, Wuhan 430074, Wuhan, China
• Other articles by this author:
/ Liangshun Xu
• School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, Wuhan, China
• Other articles by this author:
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.

MSC 2010: 35J47; 35J50

## 1 Introduction

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

${−Δu-μ1μ+λv|x|2=u2*−1+ηa2*uα−1vβ, x∈ℝN\{0},−Δv-λu+μ2v|x|2=v2*−1+ηβ2*uαvβ−1, x∈ℝN\{0},(u,v)∈(D1,2(ℝN))2, u,v>0,x∈ℝN\{0},$(1.1)

where D1,2(RN) is the completion of ${C}_{0}^{\infty }\left({ℝ}^{N}\right)$with respect to ${\left(\int {}_{{ℝ}^{N}}{|\nabla .|}^{2}\text{d}x\right)}^{1/2}$and the parameters satisfy the assumption:

$H1N≥5,1<α,β<2,α+β=2∗,η>0,μ¯>μ1≥μ2>0,λ>0,λ2

${2}^{*}:=\frac{2N}{N-2}$is the critical Sobolev exponent and $\overline{\mu }:={\left(\frac{N-2}{2}\right)}^{2}$is the best constant in the Hardy inequality ([1]):

$∫ℝNu2|x|2dx≤1μ¯∫ℝN|∇u|2dx, ∀u∈D1,2(ℝN).$

Under the assumption (H1), the matrix $E:=\left(\begin{array}{l}{\mu }_{1\text{\hspace{0.17em}}}\lambda \hfill \\ \lambda \text{\hspace{0.17em}}{\mu }_{2}\hfill \end{array}\right)$is positive definite and for all (u, v) (D1,2(RN))2 there holds that 0 < у1 < у2 < ¯μ, and furthermore,

$y1(u2+v2)≤μ1μ2+2λuv+μ2v2≤y2(u2+v2),$

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]):

$S(μ):=infu∈D1,2(ℝN)\{0}∫ℝN(|∇u|2−μu2|x|2)dx(∫ℝN|u|2⋆dx)22⋆ , μ<μ¯,S(μ1,μ2,λ):=inf(u,v)∈D∫(ℝN)(|∇u|2+|∇v|2−μ1u2+2λuv+μ2v2|x|2)dx(∫ℝN(|u|2⋆+|v|2⋆+η|u|α|v|β)dx)22⋆,$(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

$Vμεx:=ε2−N2Uμε−1x, ∀ε>0,μ∈0,μ¯,Uμx:=2Nμ¯−μμ¯μ¯2xμ¯−μ¯−μμ¯+xμ¯+μ¯−μμ¯−μ¯,$(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 $u,v\in {C}^{2}\left({ℝ}^{N}\\left\{0\right\}\right).$

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

$A=Aμ1,μ2,λ:=2λμ1−μ22+4λ2+μ1−μ2,lμ1,μ2,λ:=μ¯−μ1+μ2+μ1−μ22+4λ22,al:=μ¯−lμ1,μ2,λ, bl:=μ¯+lμ1,μ2,λ,$(1.4)

$f(τ):=ηβ2⋆+τα−τ2−β−ηα2⋆τ2,τ≥0,g(τ):=λτ2+(μ1−μ2)τ−λ,τ≥0.$(1.5)

Under the assumption (H1), since $f\left(0\right)=\frac{\eta \beta }{{2}^{\star }}>0,f\left(+\infty \right)=-\infty ,\text{for}\text{\hspace{0.17em}}\text{all}\eta \in \left(0,\infty \right),$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

$F(τ):=1+τ2(1+ητβ+τ2⋆)22⋆,τ≥0,F(τmin):=minτ≥0F(τ),$(1.6)

$η⋆:=μ1+2λτmin+μ2(τmin)21+(τmin)2,μ⋆:=μ1+μ2+(μ1−μ2)2+4λ22s⋆:=(1+ηα2⋆Aβ)12−2⋆,$(1.7)

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

$2λ≤Aλ+λA,μ1+Aλ=μ2+λAμ⋆,μ⋆∈0,μ¯.$(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 $r=|x|,\text{\hspace{0.17em}}x\in {ℝ}^{N}\\left\{0\right\}.$

• (i)

Then

$limr→0+vrur=A, limr→0+ru′rur=limr→0+v′rvr=−al,limr→+∞vrur=A, limr→+∞ru′rur=limr→+∞v′rvr=−bl.$

• (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:

$T0:=inf{T>0|μ¯−μ1−λv(et)v(et)>0,∀t∈(−∞,−T)∪(T,∞)},ω(t):=μ¯−μ1−λv(et)u(et), ∈(−∞,−T0)∪(T0,∞).$

#### 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 $r=|x|={e}^{t},x\in {ℝ}^{N}\\left\{0\right\}.$Then there exist the constants C1, C2 > 0, such that

$limt→+∞ eμ¯t+∫T0tω(s)dsu(et)=C1,limt→+∞ e(μ¯+1)t+∫T0tω(s)ds|u′(et)|=C1b(l),limt→−∞ eμ¯t+∫t−T0ω(s)dsu(et)=C2,limt→−∞ e(μ¯+1)t+∫t−T0ω(s)ds|u′(et)|=C2a(l),limt→+∞ eμ¯t+∫T0tω(s)dsv(et)=C1A,limt→+∞ e(μ¯+1)t+∫T0tω(s)ds|v′(et)|=C1Ab(l),limt→−∞ eμ¯t+∫t−T0ω(s)dsv(et)=C2A,limt→−∞ e(μ¯+1)t+∫t−T0ω(s)ds|v′(et)|=C2Aa(l).$

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 ${V}_{\mu }^{\epsilon }\left(x\right)$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 $\left\{C\left({V}_{{\mu }^{\star }}^{\epsilon },\mathcal{A}{V}_{{\mu }^{\star }}^{\epsilon }\right)C,\epsilon >0\right\},$among which the problem (1.1) has the explicit form of least energy solutions

${(s⋆Vμ⋆ε(x),s⋆AVμ⋆ε(x)),ε>0}.$

#### Corollary 1.4

Suppose that (H1) holds with $\beta <\alpha ,\eta \ge \frac{N}{N-2}.$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 $\left\{C\left({V}_{{\mu }^{\star }}^{\epsilon }\left(x\right),\mathcal{A}{V}_{{\mu }^{\star }}^{\epsilon }\left(x\right)\right),C,\epsilon >$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, $\alpha =\beta =\frac{{2}^{\star }}{2},$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$\left\{|{t}_{1}|,|{t}_{2}|\right\}$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)

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]):

$∏:αμ¯−μ1=(2−β)μ¯−μ2,$

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

${(rN−1u′(r))′=−rN−1(μ1ur2+λvr2+u2⋆−1+ηα2⋆uα−1vβ),(rN−1v′(r))′=−rN−1(λur2+μ2vr2+u2⋆−1+ηβ2⋆uα−1vβ−1),$(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

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

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

${y′1=μ¯y1+z1,z′1=−μ¯z1−μ1y1−λy2−(y12⋆−1+ηα2⋆y1∞−1y2β),y′2=μ¯y2+z2,z′2=−μ¯z2−λy1−μ2y2−(y22⋆−1+ηβ2⋆y1∞y2β−1),$(2.3)

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

${y′′1=(μ¯−μ1)y1−λy2−y12⋆−1−ηα2⋆y1α−1y2β,y′′2=(μ¯−μ2)y2−λy1−y22⋆−1−ηβ2⋆y1αy2β−1.$(2.4)

The complete integral of (2.4) is given by

$Vy1,y2,y′1,y′2:=12y′12+y′22−12μ¯−μ1y12−12μ¯−μ1y22 +λy1y2+12∗y12∗+y22∗+ηy1α,y2β.$

Set $V\left(t\right)=V\left({y}_{1}\left(t\right),{y}_{2}\left(t\right),{{y}^{\prime }}_{1}\left(t\right),{{y}^{\prime }}_{2}\left(t\right)\right).$From (2.4) it follows that V'(t) = 0. Then V(t) is a constant and we set

$V(t)≡K0,∀t∈ℝ.$

#### Lemma 2.1

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

$|y′i(t)yi(t)|<μ¯−μi,∀t∈ℝ, i=1,2,limt→±∞yi(t)=limt→±∞y′i(t)=limt→±∞y′′i(t)=0, i=1,2,$(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

$eltyi(t)=yi(0)e∫0t(l+y′i(s)yi(s))ds, t∈(0,+∞),e−ltyi(t)=yi(0)e∫t0(l−y′i(s)yi(s))ds, t∈(−∞,0),$(3.1)

Furthermore,

$(eltyi(t))′=eltyi(t)(l+y′i(t)yi(t)), t∈(0,+∞),(eltyi(t))′=e−ltyi(t)(−l+y′i(t)yi(t)),t∈(−∞,0).$

For i = 1, 2, define the functions

$gi(l):=∫0+∞(l+y′i(s)y′i(s))ds, l∈ℝ.$

From Lemma 2.1 it follows that

$−μ¯−μi+y′i(s)yi(s)<0<μ¯−μi+y′i(s)yi(s),s∈ℝ.$(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

$0≤li⋆⋆≤li⋆≤μ¯−μi, i−1,2.$(3.4)

#### Lemma 3.1

Suppose that (H1) holds. Then ${l}_{i}^{\star }={l}_{i}^{\star \star }\ge 0,$i = 1, 2*.

Proof. Assume that ${l}_{i}^{\star \star }<{l}_{i}^{\star }.$Then there exist a, b 2 R such **that and Furthermore, ${l}_{i}^{\star \star }\le b\le a\le {l}_{i}^{\star }$$-\infty <{g}_{i}\left(b\right)<{g}_{i}\left(a\right)<+\infty .$

$gi(a)−gi(b)=∫0+∞(a−b)ds=+∞,$

a contradiction. From (3.4) it follows that ${l}_{i}^{\star }={l}_{i}^{\star \star }\ge 0.$

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

$gi+2(l):=∫−∞0(l−y′i(s)yi(s))ds,i=1,2.$

Arguing as above, the following constants are well defined

$li:=suplgil<+∞=inflgil>−∞,i=3,4.$

Furthermore,

$0≥l1,l3≥μ¯−,μ1, 0≥l2,l4≥μ¯−,μ2.$

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

$∫0t(li±ε+y′i(s)yi(s))ds→±∞ as t→+∞, ε>0, i=1,2,$

which together with (3.1) implies that

$e(li−ε)tyi(t)=yi(o)e∫0t(li−ε+y′i(s)yi(s))ds →0 as t→+∞, i=1,2,e(li+ε)tyi(t)=yi(o)e∫0t(li−ε+y′i(s)yi(s))ds →+∞ as t→+∞, i=1,2. 00,$(3.5)

Consequently,

which implies that

If l1 > l2, then by taking 2ɛ < l1l2 we have that $\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{2}}{{y}_{1}}=+\infty$From (2.4) and the L’Hospital rule it follows that

$limt→+∞|y′1|2y12=limt→+∞y″1y1=limt→+∞(μ¯−μ1−λy2y1−ηα2⋆y1α−2y2β)=−∞,$

a contradiction. If l1 < l2, then $\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{1}}{{y}_{2}}=+\infty$and therefore

$limt→+∞ |y2'|2y22=limt→+∞ y2''y2=limt→+∞ (μ¯−μ2−λy1y2−ηβ2⋆y1αy2β−2)=−∞,$

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

$limt→+∞y2ty1t=limt→−∞y2ty1t=A,limt→+∞y′ityit=−limt→−∞y′ityit=−lμ1,μ2,λ,i=1,2.$

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

$limt→±∞y1α−2y2β=0, limt→±∞y1αy2β−2=0.$(3.6)

• (i)

We claim that either $\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$or $\frac{{y}_{1}\left(t\right)}{{y}_{2}\left(t\right)}$is bounded in $\left(0,\text{\hspace{0.17em}}+\infty \right).$

In fact, if neither $\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$nor $\frac{{y}_{1}\left(t\right)}{{y}_{2}\left(t\right)}$is bounded in $\left(0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\infty \right),$then the continuity implies that $\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$must have sequences of local minimum points $\left\{{\sigma }_{n}\right\}\subset \left(0,\text{\hspace{0.17em}}+\infty \right)$such that

$limn→∞σn=+∞, limn→∞y2(σn)y1(σn)=0, (y2(t)y1(t))′|t=σn=0, (y2(t)y1(t))″|t=σn≥0.$

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

$y2'(σn)y1'(σn)=y2(σn)y1(σn), (y2''(t)y2(t)−y1''(t)y1(t))|t=σn≥0,$

which together with (2.4) imply that

$(μ¯−μ2−λy1y2−y22⋆−2−ηβ2⋆y1αy2β−2)|t=σn≥(μ¯−μ1−λy2y1−y12⋆−2−ηα2⋆y1α−2y2β)|t=σn.$(3.7)

Since

$limn→∞y1(σn)=limn→∞y2(σn)=0, limn→∞y1(σn)y2(σn)=+∞,$

by (3.6) and (3.7) we get a contradiction, which implies that either $\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$or $\frac{{y}_{1}\left(t\right)}{{y}_{2}\left(t\right)}$is bounded in (0, +∞).

• (ii)

We claim that if $\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$is bounded, then there exists the limit $\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}=\mathcal{A}.$

In fact, if $\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$doesn’t exist, then the continuity implies that $\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$must have sequences of local minimum points $\left\{{\overline{\sigma }}_{n}\right\}\subset \left(0,\text{\hspace{0.17em}}+\infty \right)$and of local maximum points $\left\{{\overline{\tau }}_{n}\right\}\subset \left(0,+\infty \right),$such that

$limn→∞σ¯n=∞, (y2(t)y1(t))′|t=σ¯n =0, (y2(t)y1(t))″|t=σ¯n≥0,limn→∞τ¯n=∞, (y2(t)y1(t))′|t=τ¯n =0, (y2(t)y1(t))″|t=τ¯n≤0.$

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

$y2′(σ¯n)y1′(σ¯n)=y2(σ¯n)y1(σ¯n), y2′(σ¯n)y1′(τ¯n)=y2(τ¯n)y1(τ¯n),(y2″(t)y2(t)−y1″(t)y1(t))|t=σ¯n≥0, (y2″(t)y2(t)−y1″(t)y1(t))|t=τ¯n≤0.$

For any local minimum points $\left\{{\overline{\sigma }}_{n}\right\}\subset \left(0,+\infty \right)\text{\hspace{0.17em}}\text{of}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)},$since ${\overline{\sigma }}_{n}$is a local maximum point of $\frac{{y}_{1}}{{y}_{2}}-\frac{{y}_{2}}{{y}_{1}},$from (3.6) and (3.7) it follows that

$μ1−μ2+o1≥λy1σ¯ny2σ¯n−y2σ¯ny1σ¯n,as n→∞,μ1−μ2+o1≥λy1ty2t−y2ty1t,as t→+∞.$

which implies that

Similarly, for any local maximum points $\left\{{\overline{\tau }}_{n}\right\}\subset \left(0,\text{\hspace{0.17em}}+\infty \right)\text{of}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)},$since ¯τn is a local minimum point of $\frac{{y}_{1}}{{y}_{2}}-\frac{{y}_{2}}{{y}_{1}}$such that

$μ1−μ2+o(1)≤λ(y1(τ¯n)y2(τ¯n)−y2(τ¯n)y1(τ¯n)), as n→∞,μ1−μ2+o(1)≤λ(y1(t)y2(t)−y2(t)y1(t)), as t→+∞.$

and therefore

Consequently,

$limt→+∞(y1(t)y2(t)−y2(t)y1(t))=μ1−μ2λ,$(3.8)

which implies that

$limt→∞(y1(t)y2(t)+y2(t)y1(t))=(μ1−μ2λ)2+4.$(3.9)

From (3.8) and (3.9) it follows that $\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$exists, a contradiction.

Therefore, there must exist the limit $\mathcal{B}:=\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}.$Since l1 = l2, from (2.4) and the L’Hospital rule it follows that

$B=limt→+∞y2(t)y1(t)=limt→+∞y2′(t)y1′(t)=limt→+∞y2″(t)y1″(t)=(μ¯−μ2)B−λμ¯−μ1−λB,$

which implies that

$B=limt→+∞ y2(t)y1(t)=A2λ(μ1−μ2)2+4 ​​λ​​ 2+(μ1+μ2)>0.$(3.10)

• (iii)

Similarly, if $\frac{{y}_{1}\left(t\right)}{{y}_{2}\left(t\right)}$is bounded, then there exists the limit $\underset{t\to +\infty }{\mathrm{lim}}\frac{{y}_{1}\left(t\right)}{{y}_{2}\left(t\right)}=\frac{1}{\mathcal{A}}.$

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

$limt→−∞ y2(t)y1(t)=A$(3.11)

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

$0(3.12)

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

$limt→±∞|y1′|2y12=limt→±∞y1″y1=limt→±∞(μ¯−μ1−λy2y1)=l(μ1,μ2,λ)2,limt→±∞|y2′|2y22=limt→±∞y2″y2=limt→±∞(μ¯−μ2−λy1y2)=l(μ1,μ2,λ).$(3.13)

By (3.13) we deduce that ${y}_{1}^{″},{y}_{2}^{″}>0$as ∣t∣ large enough, which together with (2.5) implies that ${y}_{1}^{\prime },{y}_{2}^{\prime }<0$as $t\to +\infty$and ${y}_{1}^{\prime }{y}_{2}^{\prime }>0$as 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

$inft∈ℝy2(t)y1(t)=A, supt∈ℝy1′(t)y1(t)=l(μ1,μ2,λ), inft∈ℝy1′(t)y1(t)=−l(μ1,μ2,λ).$

Proof. If $\underset{t\in ℝ}{\mathrm{inf}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$can’t be achieved at any finite point, then from Lemma 3.3 it follows that $\underset{t\in ℝ}{\mathrm{inf}}\text{ }\text{ }\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}=\mathcal{A}$If the infimum is achieved at a finite point ${t}_{3}\in ℝ,$then

$0(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

$0≥g(y2(t3)y1(t3))≥f(y2(t3)y1(t3))(y2(t3)y1(t3))β−1y1(t3)2⋆−2>0,$

a contradiction, which together with (3.10) implies that $\underset{t\in ℝ}{\mathrm{inf}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$can’t be achieved at any finite point and furthermore,

$limx→−∞y2(t)y1(t)=inft∈ℝy2(t)y1(t)=A.$(3.16)

Consider the supremum of$|\frac{{{y}^{\prime }}_{1}\left(t\right)}{{y}_{1}\left(t\right)}|$in R. If the supremum can’t be achieved at any finite point, then (3.16) implies that $\underset{t\in ℝ}{\mathrm{sup}}|\frac{{{y}^{\prime }}_{1}\left(t\right)}{{y}_{1}\left(t\right)}|=l\left({\mu }_{1},{\mu }_{2},\lambda \right).$If the supremum is achieved at finite point t4 2 R, then

Furthermore,

$|y′1(t4)y1(t4)|=supt∈ℝ|y′1(t)y1(t)|≥l(μ1,μ2,λ).(y′1(t)y1(t))′|t=t4=(y″1(t)y1(t))−|y′1(t)|2|y1(t)|2|t=t4=0.$(3.17)

Since $\underset{t\in ℝ}{\mathrm{inf}}\frac{{y}_{2}\left(t\right)}{{y}_{1}\left(t\right)}$can’t be achieved at any finite point, from (2.4), (3.16) and (3.17) it follows that

$|y′1|2y12|t=t4=[(μ¯−μ1)−λy2y1−y12⋆−2−ηα2⋆y1α−2y2β]t=t4 ≤μ¯−μ1−λy2(t4)y1(t4) <μ¯−μ1−λA=l(μ1,μ2,λ)2,$

a contradiction, which implies that $\underset{t\in ℝ}{\mathrm{sup}}|\frac{{{y}^{\prime }}_{1}\left(t\right)}{{y}_{1}\left(t\right)}|$can’t be achieved at any finite points. Then from (3.14) and (3.16) it follows that

$limt→+ ∞y′1(t)y1(t)=inft∈ℝy′1(t)y1(t)=−μ¯−μ1−λA=−l(μ1,μ2,λ),limt→− ∞y′1(t)y1(t)=supt∈ℝy′1(t)y1(t)=−μ¯−μ1−λA=l(μ1,μ2,λ).$(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

$∫0+ ∞((l1+ε−l(μ1,μ2,λ))+l(μ1,μ2,λ)+y′1y1) ds=∫0+ ∞(l1+ε+y′1y1) ds =+∞.$

From Lemma 3.4 it follows that

$l1+ε−l(μ1,μ2,λ)≥0, ∀ε>0,$

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

$limt→−∞y′1y1=limt→−∞y′2y2=l(μ1,μ2,λ), l3=l4=l(μ1,μ2λ).$

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

$limt→+∞e∫T0tμ¯−μ1−λy2sy1sdsy1t=C1,limt→−∞e∫t−T0μ¯−μ1−λy2sy1sdsy1t=C2,limt→+∞e∫T0tμ¯−μ1−λy2sy1sdsy2t=C1A,limt→−∞e∫t−T0μ¯−μ1−λy2sy1sdsy2t=C2A.$

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 ${H}_{1}\left(s\right):=\frac{{{y}^{\prime }}_{1}\left(s\right)}{{y}_{1}\left(s\right)}.$From (2.4) and Lemma 3.4 it follows that

$l(μ1,μ2,λ)±H1(s)>0, ∀s ∈ ℝ.$

Note that A < Λ0 and

$limt→+∞(μ¯−μ1−λy2(t)y1(t))=l(μ1,μ2,λ)2>0.$(3.19)

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

$μ¯−μ1−λy2(t)y1(t)>0, ∀t ∈ (T0,+∞).$

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

$H′1(s)=μ¯−μ1−λy2(s)y1(s)−H12(s)−y12⋆−2−ηα2⋆y1α−2y2β,$

which implies that

$μ¯−μ1−λy2(s)y1(s)+H1(s)=H′1(s)+y12⋆−2+ηα2⋆y1α−2y2βμ¯−μ1−λy2(s)y1(s)−H1(s), ∀t>T0.$(3.20)

Define

$I:=∫T0+∞(μ¯−μ1−λy2(s)y1(s)+H1(s)) ds,I1:=∫T0+∞H′1(s)μ¯−μ1−λy2(s)y1(s)+H1(s) ds,I2:=∫T0+∞y12⋆−2+ηα2⋆y1α−2y2βμ¯−μ1−λy2(s)y1(s)+H1(s) ds.$

We claim that the integral $I\text{\hspace{0.17em}}:={\int }_{{T}_{0}}^{+\infty }\left(\sqrt{\overline{\mu }-{\mu }_{1}-\lambda \frac{{y}_{2}\left(s\right)}{{y}_{1}\left(s\right)}}+{H}_{1}\left(s\right)\right)$ds converges.

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

$limt→+∞μ¯−μ1−λy2(s)y1(s)=l(μ1,μ2,λ), limt→+∞H1(s)=−l(μ1,μ2,λ).$(3.21)

Since A < Λ0, arguing as in the proof of Lemma 3.4 we have that $\frac{{y}_{2}\left(s\right)}{{y}_{1}\left(s\right)}$and H 1(s) are strictly decreasing as s ! +∞. Taking ¯τ > T0 large enough we have that ${{H}^{\prime }}_{1}\left(s\right)<0$for all s > ¯τ and

$|∫τ¯+∞H′1(s)μ¯−μ1−λy2(s)y1(s)−H1(s)ds|=|∫τ¯+∞H′1(s)l(μ1,μ2,λ)−H1(s) l(μ1,μ2,λ)−H1(s)μ¯−μ1−λy2(s)y1(s)−H1(s) ds|≤C|∫τ¯+∞H′1(s)l(μ1,μ2,λ)−H1(s) ds|=Cln2l(μ1,μ2,λ)|l(μ1,μ2,λ)−H1(τ¯)|<+∞.$(3.22)

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

Therefore,

$μ¯−μ1−λy2(s)y1(s)−H1(s)>l(μ1,μ2,λ), ∀s>R.0(3.23)

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

$∫R+∞(y12⋆−2+y1α−2y2β)ds

By taking $\epsilon \to {0}^{+}$we have that

$−((2⋆−2)l(μ1,μ2,λ)−(2⋆+2)ε)<0.$

Then the integral ${\int }_{{T}_{0}}^{+\infty }\left({2}^{\star }{y}_{1}^{{2}^{\star }-2}+\eta \alpha {y}_{1}^{\alpha -2}{y}_{2}^{\beta }\right)$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,

$limt→+∞ e∫T0tμ¯−μ1−λy2(s)y1(s)dsy1(t) =y1(T0) limt→+∞ e∫T0t(μ¯−μ1−λy2(s)y1(s)+H1(s))ds =y1(T0) e∫T0+∞(μ¯−μ1−λy2(s)y1(s)+H1(s))ds=C1,$

where ${C}_{1}:={y}_{1}\left({T}_{0}\right){e}^{{\int }_{{T}_{0}}^{+\infty }\left(\sqrt{\overline{\mu }-{\mu }_{1}-\lambda \frac{{y}_{2}\left(s\right)}{{y}_{1}\left(s\right)}+}{H}_{1}\left(s\right)\right)\text{\hspace{0.17em}}\text{d}s}>0.$

#### 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

$F′(τ)=2τβ−1k(τ)(1+ητβ+τ2⋆)22⋆+1, τ>0. limτ→0+F(τ)=limτ→+∞F(τ)=1,k(τ)< 0 as τ→0+, k(τ)> 0 as τ→+∞.$(4.1)

Then $\underset{\tau \ge 0}{\mathrm{min}}F\left(\tau \right)$must be achieved at finite τmin > 0 and from (4.1) it follows that

$F′(τmin)=0, k(τmin)=0, 0

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

$S(μ1,μ2,λ)≤F(τmin)∫ℝN(|∇w|2−μ⋆w2|x|2)(∫ℝN|w|2⋆)22⋆,$

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

$sn=((∫ℝN|vn|2⋆)−1∫ℝN|un|2⋆)12⋆, ∫ℝN|zn|2⋆=∫ℝN|un|2⋆.$(4.3)

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

$∫ℝNunαznβ≤α2∗∫ℝNun2∗+β2∗∫ℝNzn2∗=∫ℝNun2∗=∫ℝNzn2∗.$(4.4)

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

$∫RN∇un2+∇vn2−μ1un2+2λunvn+μ2v2x2∫RN∇un2⋆+ηunαvnβ+vn2⋆22⋆≥∫RN∇un2+∇vn2−μ1+Aλun2x2−μ2+λAvn2x21+ηsn−β+sn−2⋆∫RNun2⋆22⋆≥∫RN∇un2−μ⋆un2x21+ηsn−β+sn−2⋆∫RNun2⋆22⋆+sn−2∫RN∇zn2−μ⋆zn2x21+ηsn−β+sn−2⋆∫RNzn2⋆22⋆≥Fsn−1Sμ⋆≥FτminSμ⋆.$

Taking $n\to \infty$we have that

$S(μ1,μ2,λ)≥F(τmin)S(μ⋆).$(4.5)

A direct calculation shows that

$τmin=A⇔μ⋆=μ⋆.$(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:

${C(Vμ⋆ε(x),AVμ⋆ε(x)), C,ε>0}.$

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

${(s⋆Vμ⋆ε(x),s⋆AVμ⋆ε(x)), ε>0},$

that is, $\left({u}_{\epsilon },{v}_{\epsilon }\right):=\left({s}^{\star }{V}_{{\mu }^{\star }}^{\epsilon }\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{s}^{\star }\mathcal{A}{V}_{{\mu }^{\star }}^{\epsilon }\left(x\right)\right)$is a solution to (1.1) such that

$∫ℝN(|∇uε|2+|∇vε|2−μ1uε2+2λuεvε+μ2vε2|x|2)=∫ℝN(uε2⋆+vε2⋆+ηuεαvεβ)$

$=(F(A)S(μ⋆))N2=S(μ1,μ2,λ)N2,$

where s* is defined as in (1.7). By (3.18) we have that $l\left({\mu }_{1},{\mu }_{2},\lambda \right)=\sqrt{\overline{\mu }-{\mu }_{1}-\lambda \mathcal{A}}$and therefore the solution $\left({s}^{\star }{V}_{{\mu }^{\star }}^{\epsilon }\left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{s}^{\star }\mathcal{A}{V}_{{\mu }^{\star }}^{\epsilon }\left(x\right)\right)$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

$k(A)=0, k(τ)<0 in (0,A), k(τ)>0 in (A,+∞).$

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

Suppose that $1<\beta <\alpha <2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\eta \text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{N}{N-2}.\text{\hspace{0.17em}}$Since

then we have that

$k′(τ)=τ1−β(2−β+2ηα2⋆τβ−ατ2⋆−2),k(1)=η2⋆(α−β)>0,2−β>0, β>2⋆−2, 2ηα2⋆≥α,k′(τ)>0, k(τ)>k(1)>0, ∀τ ∈(1,+∞).$(4.7)

Since $k\left(\tau \right)<0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\tau \to {\text{0}}^{+},$from (4.1) and (4.7) it follows that $\underset{\tau \ge 0}{\mathrm{min}}$F (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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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,

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