By using the Nehari manifold, we obtain a positive ground state solution

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

GROUND STATE SOLUTIONS FOR SEMILINEAR ELLIPTIC EQUATIONS WITH ZERO MASS IN R^N

JIU LIU, JIA-FENG LIAO, CHUN-LEI TANG

Abstract. In this article, we study the semilinear elliptic equation

−∆u=|u|^p(x)−2u, x∈R^N u∈D^1,2(R^N), whereN≥3,p(x) =

(p, x∈Ω

2^∗, x6∈Ω, with 2< p <2^∗:= 2N/(N−2), Ω⊂R^N is a bounded set with nonempty interior. By using the Nehari manifold, we obtain a positive ground state solution.

1. Introduction and statement of main result Considering the semilinear elliptic equation

−∆u=|u|^2∗−2u+χΩ(x)(|u|^p−2u− |u|^2∗−2u), x∈R^N

u∈D^1,2(R^N), (1.1)

whereN ≥3, 2< p <2^∗:= 2N/(N−2), Ω⊂R^N is a bounded set with nonempty interior and

χΩ(x) =

(1, x∈Ω

0. x6∈Ω (1.2)

The well-known semilinear elliptic equation with zero mass is

−∆u=|u|^2∗−2u, x∈R^N

u∈D^1,2(R^N), (1.3)

whereN≥3, which has been studied very intensely (see [6, 10, 18]) and the explicit expression of positive solutions was given. Of course, the semilinear elliptic equa- tion with zero mass whose nonlinear term with subcritical growth has also been investigated by many authors, for example [4, 7, 8, 9].

2000Mathematics Subject Classification. 35J20, 35J61.

Key words and phrases. Semilinear elliptic equation; zero mass; Nehari manifold;

ground state solution.

c

2015 Texas State University - San Marcos.

Submitted January 17, 2015. Published April 7, 2015.

1

By a transformation, (1.1) is equivalent to the following elliptic equation with variable exponent

−∆u=|u|^p(x)−2u, x∈R^N

u∈D^1,2(R^N), (1.4)

where

p(x) =

(p, x∈Ω 2^∗. x6∈Ω

The equations with variable exponent appear in various mathematical models, for example: electrorheological fluids [1, 17], nonlinear Darcy’s law in porous medium [5], image processing [12]. Recently, these equations have been investigated by many authors, see for example, [2, 11, 13, 14, 16]. However, they did not consider problem (1.4); thus we study it in this article. Our main result reads as follows.

Theorem 1.1. Assume that N ≥ 3, 2 < p < 2^∗ and χΩ satisfies (1.2). Then equation (1.1)has a positive ground state solution.

Ifp∈C(R^N,[2,2^∗]) andp6≡2^∗, we do not yet know whether equation (1.1) has solution. We shall consider it in the future.

This article is organized as follows. Section 2 contains some preliminaries. Sec- tion 3 gives the proof of theorem 1.1.

2. Preliminaries In what follows, we use the following notation.

•D^1,2(R^N) is the completion ofC₀^∞(R^N) with respect to the norm kuk²_D1,2(R^N)=

Z

R^N

|∇u|²dx.

•L^t(R^N), 2≤t <+∞, denotes a Lebesgue space endowed with the norm

|u|^t_t= Z

R^N

|u|^tdx.

• S denotes the best constant of Sobolev embedding D^1,2(R^N),→ L^2∗(R^N); that is,

S|u|²₂∗≤ kuk²_D1,2(R^N) for allu∈D^1,2(R^N).

•D⁻¹is the dual space ofD^1,2(R^N).

•C,C_i denote various positive constants.

For equation (1.1), the energy functionalI:D^1,2(R^N)→Ris defined by I(u) = 1

2kuk²_D1,2(R^N)− 1 2^∗

Z

R^N

|u|^2∗dx−1 p

Z

R^N

χΩ(x)|u|^pdx + 1

2^∗ Z

R^N

χΩ(x)|u|^2∗dx

= 1

2kuk²_D1,2(R^N)−1 p

Z

Ω

|u|^pdx− 1 2^∗

Z

R^N\Ω

|u|^2∗dx.

The H¨older and Sobolev inequalities imply Z

Ω

|u|^pdx≤ |u|^p₂∗,Ω(meas Ω)^{2∗ −p2∗} ≤ |u|^p₂∗(meas Ω)^{2∗ −p2∗}

≤S^−p2(meas Ω)^{2∗ −p2∗} kuk^p_D1,2(R^N),

(2.1)

where

|u|_s,Ω=Z

Ω

|u|^tdx^1/t

, ∀t∈[1,+∞).

Thus the functionalI is well defined. By Lemma 3.1 in next section, I is of class C¹(D^1,2(R^N),R) and satisfies

hI⁰(u), vi= Z

R^N

∇u· ∇v dx− Z

R^N

|u|^2∗−2uv dx− Z

R^N

χ_Ω(x)|u|^p−2uv dx +

Z

R^N

χ_Ω(x)|u|^2∗−2uv dx

= Z

R^N

∇u· ∇v dx− Z

Ω

|u|^p−2uv dx− Z

R^N\Ω

|u|^2∗−2uv dx,

(2.2)

for all u, v ∈ D^1,2(R^N). Hence weak solutions of (1.1) correspond to the critical point of the functionalI. Define

N :={u∈D^1,2(R^N)\{0}:J(u) = 0}, m:= inf

u∈NI(u), where

J(u) =kuk²_D1,2(R^N)− Z

Ω

|u|^pdx− Z

R^N\Ω

|u|^2∗dx.

Since all solutions of (1.1) belong to the manifoldN, first we seek for the minimizer uform and then we proveuis a solution of equation (1.1).

3. Proof of Theorem 1.1 The proof relies on the following lemmas.

Lemma 3.1. Assume that N ≥3, 2 < p < 2^∗ and χ_Ω satisfies (1.2). Then the functionalI is of class C¹(D^1,2(R^N),R)and I⁰(·) satisfies (2.2).

Proof. Define

ψ(u) =1 p

Z

Ω

|u|^pdx+ 1 2^∗

Z

R^N\Ω

|u|^2∗dx,

we need only to prove ψ ∈C¹(D^1,2(R^N),R). Let u, h∈ D^1,2(R^N). Given x ∈Ω and 0<|t|<1, by the mean value theorem, there existsλ1∈(0,1) such that

|u+th|^p− |u|^p

t =p|u+λ1th|^p−1|h| ≤p(|u|+|h|)^p−1|h|.

Similarly, givenx∈R^N\Ω and 0<|t|<1, there existsλ₂∈(0,1) such that |u+th|^2∗− |u|^2∗

t = 2^∗|u+λ₂th|^2∗−1|h| ≤2^∗(|u|+|h|)^2∗−1|h|.

The H¨older inequality implies that Z

Ω

(|u|+|h|)^p−1|h|dx≤Z

Ω

(|u|+|h|)^pdx^p−1_p

|h|_p,Ω

≤(|u|p,Ω+|h|p,Ω)^p−1|h|p,Ω<+∞

and Z

R^N\Ω

(|u|+|h|)^2∗−1|h|dx≤Z

R^N\Ω

(|u|+|h|)^2∗dx^{2∗ −1}₂∗

|h|₂∗,R^N\Ω

≤(|u|₂^∗_,RN\Ω+|h|₂^∗_,RN\Ω)^2∗−1|h|₂^∗_,RN\Ω<+∞. It follows from the Lebesgue theorem that

hψ⁰(u), hi= lim

t→0

ψ(u+th)−ψ(u) t

= lim

t→0

Z

Ω

|u+th|^p− |u|^p

pt dx+ lim

t→0

Z

R^N\Ω

|u+th|^2∗− |u|^2∗ 2^∗t dx

= Z

Ω

limt→0

|u+th|^p− |u|^p

pt dx+

Z

R^N\Ω

limt→0

|u+th|^2∗− |u|^2∗ 2^∗t dx

= Z

Ω

|u|^p−2uh dx+ Z

R^N\Ω

|u|^2∗−2uh dx.

Assume that u_n → u in D^1,2(R^N), then u_n → u in L^2∗(R^N) and L^p(Ω). If fol- lows from [19, Theorem A.2 and A.4] that |un|^p−2u_n → |u|^p−2u in L^p−1p (Ω) and

|un|^2∗−2un→ |u|^2∗−2uinL ²

∗

2∗ −1(R^N\Ω). Hence combining the H¨older and Sobolev inequalities, we obtain

kψ⁰(u_n)−ψ⁰(u)kD⁻¹

≤ sup

kϕk_D1,2 (RN)=1,ϕ∈D^1,2(R^N)

Z

Ω

(|un|^p−2un− |u|^p−2u)ϕ dx

+ sup

kϕk_D1,2 (RN)=1,ϕ∈D^1,2(R^N)

Z

R^N\Ω

(|un|^2∗−2un− |u|^2∗−2u)ϕ dx

≤C

|un|^p−2u_n− |u|^p−2u p

p−1,Ω+C

|un|^2∗−2u_n− |u|^2∗−2u ₂∗

2∗ −1,R^N\Ω

=o(1).

ThusψisC¹.It is obvious thatI⁰(·) satisfies (2.2). The proof is complete.

Lemma 3.2. Assume that N ≥ 3, 2 < p < 2^∗ and χΩ satisfies (1.2). Then for any u∈D^1,2(R^N)\{0}, there existstu>0 such that tuu∈ N.

Proof. For anyu∈D^1,2(R^N)\{0}, define f(t) :=I(tu) =t²

2kuk²_D1,2(R^N)−t^p p

Z

Ω

|u|^pdx−t^2∗ 2^∗

Z

R^N\Ω

|u|^2∗dx, ∀t∈(0,+∞).

Then one has

f⁰(t)t=hI⁰(tu), tui=t²kuk²_D1,2(R^N)−t^p Z

Ω

|u|^pdx−t^2∗ Z

R^N\Ω

|u|^2∗dx.

Combining 2< p <2^∗, we havef⁰(t)t >0 fort >0 small enough andf⁰(t)t <0 for t >0 large enough. Thus there existstu>0 such thatf⁰(tu)tu=hI⁰(tuu), tuui= 0.

That istuu∈ N. The proof is complete.

Lemma 3.3. Assume thatN ≥3,2< p <2^∗ andχ_Ωsatisfies(1.2). Thenm >0.

Proof. For anyu∈ N, one has kuk²_D1,2(R^N)=

Z

Ω

|u|^pdx+ Z

R^N\Ω

|u|^2∗dx

≤Ckuk^p_D1,2(R^N)+Ckuk²_D^∗1,2(R^N),

which implies that there existsα >0 such that

kuk_D1,2(R^N)≥α, ∀u∈ N. (3.1) Thus for anyu∈ N, we have

I(u) =I(u)−1

phI⁰(u), ui

= 1 2 −1

p

kuk²_D1,2(R^N)+ 1 p− 1

2^∗

Z

R^N\Ω

|u|^2∗dx

≥ 1 2 −1

p

kuk²_D1,2(R^N)

≥ 1 2 −1

p α².

(3.2)

Hencem >0. The proof is complete.

Lemma 3.4. Assume that N ≥ 3, 2 < p < 2^∗ and χΩ satisfies (1.2). Then for any u∈ N,J⁰(u)6= 0.

Proof. By (3.1), for anyu∈ N, one has hJ⁰(u), ui=hJ⁰(u), ui −pJ(u)

= (2−p)kuk²_D1,2(R^N)−(2^∗−p) Z

R^N\Ω

|u|^2∗dx

≤(2−p)kuk²_D1,2(R^N)

≤(2−p)α²<0.

(3.3)

Hence the proof is complete.

Lemma 3.5. Assume thatN ≥3,2< p <2^∗ andχ_Ωsatisfies (1.2). Suppose that u∈ N andI(u) =m. Thenuis a solution of (1.1).

Proof. Assume thatu∈ N andI(u) =m. Then by the Lagrange multiplier rule, there exists λ∈ R such thatI⁰(u) =λJ⁰(u), which implies that 0 = hI⁰(u), ui= λhJ⁰(u), ui. By Lemma 3.4, we obtain λ = 0. Hence I⁰(u) = 0. The proof is

complete.

Lemma 3.6. Assume that N ≥3,2< p <2^∗ and χ_Ω satisfies (1.2). Then there exists a bounded sequence{u_n} ⊂ N satisfyingI(u_n)→mandI⁰(u_n)→0inD⁻¹. Proof. By the Ekeland variational principle in [19], there exist {un} ⊂ N and {λn} ⊂Rsuch thatI(un)→mandI⁰(un)−λnJ⁰(un)→0 inD⁻¹. By (3.2), one has

I(un) =I(un)−1

phI⁰(un), uni ≥1 2 −1

p

kunk²_D1,2(R^N), which implies{un}is bounded in D^1,2(R^N). Then we have

0 =hI⁰(un), uni=λnhJ⁰(un), uni+o(1).

Combining (3.3), we obtainλn→0. For anyϕ∈D^1,2(R^N) withkϕk_D1,2(R^N)= 1, it follows from (2.1), the H¨older and Sobolev inequalities that

Z

Ω

|un|^p−2unϕ dx ≤Z

Ω

|un|^pp−1_p Z

Ω

|ϕ|^p1_p

≤

S^−p2(meas Ω)^{2∗ −p2∗} kunk^p_D1,2(

R^N)

^p−1_p

|ϕ|^p₂∗(meas Ω)^{2∗ −p2∗ 1}_p

≤Ckunk^p−1_D1,2(R^N)kϕk_D1,2(R^N)≤C.

Thus combining the H¨older and Sobolev inequalities, we have kJ⁰(un)k_D⁻¹ = sup

kϕk_D1,2 (RN)=1,ϕ∈D^1,2(R^N)

|hJ⁰(un), ϕi|

= sup

kϕk_D1,2 (RN)=1,ϕ∈D^1,2(R^N)

2

Z

R^N

∇un· ∇ϕ dx−p Z

Ω

|un|^p−2unϕ dx

−2^∗ Z

R^N\Ω

|un|^2∗−2unϕ dx

≤ sup

kϕk_D1,2 (RN)=1,ϕ∈D^1,2(R^N)

[2kunk_D1,2(R^N)kϕk_D1,2(R^N)+C +Ckunk²_D^∗_1,2⁻¹₍

R^N)kϕk_D1,2(R^N)]≤C.

Hence we obtain

kI⁰(u_n)k_D−1 ≤ kI⁰(u_n)−λ_nJ⁰(u_n)k_D−1+|λ_n|kJ⁰(u_n)k_D−1 =o(1).

The proof is complete.

If p(x) ≡ 2^∗, equation (1.4) reduces to (1.3). It is well known that (1.3) has ground state solution

v(x) = C_N

(1 +|x|²)^N−22 , (3.4)

whereCN := [N(N−2)]^N−24 andvsatisfies Z

R^N

|∇v|²dx= Z

R^N

|v|^2∗dx=S^N/2. Let the energy functional of (1.3) be

I∞(u) =1

2kuk²_D1,2(R^N)− 1 2^∗

Z

R^N

|u|^2∗dx.

Then we have

I_∞(v) =I_∞(v)− 1

2^∗hI_∞⁰ (v), vi= 1 N

Z

R^N

|∇v|²dx= 1 NS^N/2. Now for the energym, we make the following estimation.

Lemma 3.7. Assume that N ≥ 3, 2 < p < 2^∗ and χΩ satisfies (1.2). Then m < _N¹S^N/2.

Proof. Inspired by the idea in [3, 15]. For ground state solutionvof equation (1.3), we define vn(x) := v(x+xn), where xn := (0,0, . . . ,0, n). Thus kvnk_D1,2(R^N) = kvk_D1,2(R^N) =S^N4 and then v_n * u in D^1,2(R^N), v_n →u in L^p_loc(R^N), v_n(x)→ u(x) a.e. in R^N. Since for anyx∈R^N, vn(x)→0, u= 0. By Lemma 3.2, there existstn∈(0,+∞) such thattnvn∈ N. Then one has

kv_nk²_D1,2(R^N)=t^p−2_n Z

Ω

|v_n|^pdx+t²_n^∗−2 Z

R^N\Ω

|v_n|^2∗dx. (3.5)

By (2.1) and (3.5), one has

kvk²_D1,2(R^N)=kvnk²_D1,2(R^N)

≤C(t^p−2_n kvnk^p_D1,2(

R^N)+t²_n^∗−2kvnk²_D^∗1,2(R^N))

=C(t^p−2_n kvk^p_D1,2(

R^N)+t²_n^∗−2kvk²_D^∗1,2(R^N)),

which indicates thattn cannot appraoch zero, that istn≥t0for somet0>0. Since Ω is bounded, there exists R >0 such that Ω⊂BR:={x∈R^N :|x|< R}. Since fornlarge enough,

Z

|x−x_n|<R

1

(1 +|x|²)^N dx≤ Z

|x−x_n|<R

2^N

n^2N dx= 2^N

n^2N measB_R=o(1), (3.6) we have

Z

R^N

|v|^2∗dx= Z

R^N

|vn|^2∗dx

≥ Z

R^N\Ω

|vn|^2∗dx

≥ Z

|x|≥R

|vn|^2∗dx

=C_N^2∗ Z

|x|≥R

1

(1 +|x+xn|²)^N dx

=C_N^2∗ Z

|x−xn|≥R

1

(1 +|x|²)^N dx

=C_N^2∗ Z

R^N

1

(1 +|x|²)^N dx−C_N^2∗ Z

|x−xn|<R

1

(1 +|x|²)^N dx

= Z

R^N

|v|^2∗dx+o(1).

Thus one has Z

R^N\Ω

|vn|^2∗dx= Z

R^N

|v|^2∗dx+o(1) =S^N/2+o(1). (3.7) It follows from (3.5) that

kvnk²_D1,2(R^N)

Z

R^N\Ω

|vn|^2∗dx−1

≥t²_n^∗−2, which implies

lim sup

n→∞

t²_n^∗−2≤ kvk²_D1,2(R^N)S^−N2 = 1.

Thus up to a subsequence, one hastn→T ∈(t0,1]. Notice that Z

Ω

|vn|^pdx=o(1).

By (3.5) and (3.7), one hasS^N/2=T^2∗−2S^N/2. ThusT = 1. From (3.7) it follows that

Z

Ω

|vn|^2∗dx=o(1).

We claim that

t²_n^∗ 2^∗

R

Ω|vn|^2∗dx

t^p_n p

R

Ω|vn|^pdx →0.

Indeed, by (3.4) and (3.6), fornlarge enough, one has Z

Ω

|vn|^2∗dx≤C_N^2∗ Z

B_R

1

(1 +|x+xn|²)^N dx

=C_N^2∗ Z

|x−xn|<R

1

(1 +|x|²)^N dx

= C_N^2∗2^N

n^2N measBR.

Since the interior of Ω is nonempty, there exist z₀ ∈ R^N and r > 0 such that B_r(z₀) :={x∈R^N :|x−z₀|< r} ⊂Ω. Thus for nlarge enough, one has

Z

Ω

|vn|^pdx≥C_N^p Z

Br(z0)

1

(1 +|x+x_n|²)^(N−2)p2 dx

≥C_N^p Z

B_r(z₀)

1 2^(N−2)p2 n^(N−2)p

dx

=C_N^p2^(2−N)p2

n^(N−2)p measBr. Then we obtain

R

Ω|vn|^2∗dx R

Ω|vn|^pdx ≤C_N⁰ 1

n^{2N−(N−2)p} =o(1),

sincep < _N^2N₋₂. Combining tn → 1, we implies the claim holds. By calculations, one has

1−t²_n 2 1−t²_n^∗

2^∗

=2^∗(1−t²_n) 2(1−t²_n^∗) →1.

Recall thattnvn∈ N. Hence fornlarge enough, one has m≤I(tnvn)

= t²_n

2kvnk²_D1,2(R^N)−t^p_n p

Z

Ω

|vn|^pdx−t²_n^∗ 2^∗

Z

R^N\Ω

|vn|^2∗dx

= t²_n

2kvnk²_D1,2(R^N)−t²_n^∗ 2^∗

Z

R^N

|vn|^2∗dx−t^p_n p

Z

Ω

|vn|^pdx+t²_n^∗ 2^∗

Z

Ω

|vn|^2∗dx

=I∞(vn)−1−t²_n

2 kvnk²_D1,2(R^N)+1−t²_n^∗ 2^∗

Z

R^N

|vn|^2∗dx

−t^p_n p

Z

Ω

|vn|^pdx+t²_n^∗ 2^∗

Z

Ω

|vn|^2∗dx

=S^N/2+1−t²_n^∗

2^∗ −1−t²_n 2

S^N/2−t^p_n p

Z

Ω

|vn|^pdx+t²_n^∗ 2^∗

Z

Ω

|vn|^2∗dx

< S^N/2.

The proof is complete.

Lemma 3.8. Assume that N ≥ 3, 2 < p < 2^∗ and χΩ satisfies (1.2). Suppose that the sequence {un} ⊂ N is bounded in D^1,2(R^N) and satisfies I(un) →m <

1

NS^N/2 andI⁰(un)→0 inD⁻¹. Then there existsu∈D^1,2(R^N)such that up to a subsequence,u_n→uinD^1,2(R^N).

Proof. Since {un} ⊂ D^1,2(R^N) is a bounded, up to a subsequence, there exists u∈D^1,2(R^N) such thatu_n* uinD^1,2(R^N),u_n→uinL^p(Ω) andu_n(x)→u(x) a.e. inR^N. For anyv∈D^1,2(R^N), byI⁰(u_n)→0 inD⁻¹, one has

0 =hI⁰(u_n), vi+o(1)

= Z

R^N

∇un· ∇v dx− Z

Ω

|un|^p−2unv dx− Z

R^N\Ω

|un|^2∗−2unv dx+o(1)

= Z

R^N

∇u· ∇v dx− Z

Ω

|u|^p−2uv dx− Z

R^N\Ω

|u|^2∗−2uv dx

=hI⁰(u), vi.

Thus we have

I(u) =I(u)−1

phI⁰(u), ui

= 1 2 −1

p

kuk²_D1,2(R^N)+ 1 p− 1

2^∗

Z

R^N\Ω

|u|^2∗dx

≥ 1 2 −1

p

kuk²_D1,2(R^N)≥0.

Definevn=un−u. Thus one has

kunk²_D1,2(R^N)=kvnk²_D1,2(R^N)+kuk²_D1,2(R^N)+o(1).

The Brezis-Lieb lemma implies Z

Ω

|un|^pdx= Z

Ω

|vn|^pdx+ Z

Ω

|u|^pdx+o(1) = Z

Ω

|u|^pdx+o(1)

and Z

R^N\Ω

|un|^2∗dx= Z

R^N\Ω

|vn|^2∗dx+ Z

R^N\Ω

|u|^2∗dx+o(1).

Combining this withI(un)→m, we obtain m= 1

2kvnk²_D1,2(R^N)− 1 2^∗

Z

R^N\Ω

|vn|^2∗dx+I(u) +o(1)

≥ 1

2kvnk²_D1,2(R^N)− 1 2^∗

Z

R^N\Ω

|vn|^2∗dx+o(1).

(3.8)

It follows fromhI⁰(un), uni= 0 andI⁰(u) = 0 that 0 =kvnk²_D1,2(R^N)−

Z

R^N\Ω

|vn|^2∗dx+hI⁰(u), ui+o(1)

=kvnk²_D1,2(R^N)− Z

R^N\Ω

|vn|^2∗dx+o(1).

Up to a subsequence, we assume that kv_nk²_D1,2(R^N)+o(1) =b=

Z

R^N\Ω

|v_n|^2∗dx+o(1).

Thus we have

Sb^2/2∗ =SZ

R^N\Ω

|vn|^2∗dx2/2^∗

+o(1)

≤SZ

R^N

|vn|^2∗dx^2/2∗ +o(1)

≤ kv_nk²_D1,2(R^N)+o(1) =b.

Assume thatb6= 0. Then one hasb≥S^N/2. From (3.8), we obtain m≥ 1

2− 1 2^∗

b≥ 1 NS^N/2,

which is a contradiction. Henceb= 0, and the proof is complete.

Proof of Theorem 1.1. By Lemmas 3.3, 3.6 and 3.7, there exists a bounded sequence {un} ⊂ N satisfying I(un) →m ∈ (0,_N¹S^N/2) andI⁰(un)→ 0 in D⁻¹. Lemma 3.8 implies that there existsu∈D^1,2(R^N) such that up to a subsequence,un→u in D^1,2(R^N). ThenI(u) =mand J(u) = 0. That is,m is achieved by a function u ∈ D^1,2(R^N). Since I(|u|) = I(u) and J(|u|) = J(u), we can assume that u is nonnegative. Lemma 3.5 implies thatu∈D^1,2(R^N) is a solution of equation (1.1).

It follows from the definition ofmthatu∈D^1,2(R^N) is a ground state solution of equation (1.1). It follows from the strongly maximum principle that u > 0. This

completes the proof.

Acknowledgments. The authors would like to thank Professor C. O. Alves for his suggestioins. This research was supported by the National Natural Science Foundation of China (No. 11471267), by the Fundamental Research Funds for the Central Universities (No. XDJK2015D015), and by the Natural Science Foundation of Education of Guizhou Province (No. LKZS[2014]22)

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Jiu Liu

School of Mathematics and Statistics, Southwest University, Chongqing 400715, China E-mail address:[email protected]

Jia-Feng Liao

School of Mathematics and Statistics, Southwest University, Chongqing 400715, China.

School of Mathematics and Computational Science, Zunyi Normal College, Zunyi, Guizhou 563002, China

E-mail address:[email protected]

Chun-Lei Tang (corresponding author)

School of Mathematics and Statistics, Southwest University, Chongqing 400715, China E-mail address:[email protected], Phone +86 23 68253135, Fax +86 23 68253135