On some nonlocal equations with competing coefficients

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J. Math. Anal. Appl. ••• (••••) •••–•••

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

On some nonlocal equations with competing coefficients Tingxi Hu a , Lu Lu b,∗ a

Department of Mathematics, Central China Normal University, Wuhan 430079, PR China School of Statistics and Mathematics, Zhongnan University of Economics and Law, Wuhan 430079, PR China b

a r t i c l e

i n f o

Article history: Received 16 July 2017 Available online xxxx Submitted by G. Chen

a b s t r a c t In this paper we study the following nonlocal problems ⎞ ⎧ ⎛  ⎪ ⎪ ⎪ 2 ⎨ − ⎝a + b |∇u| ⎠ Δu + V (x)u = Q(x)up , x ∈ R3 ,

Keywords: Nonlocal equation Competing coefficients Barycenter map

⎪ ⎪ ⎪ ⎩

R3

u ∈ H (R3 ), u > 0, x ∈ R3 , 1

where the constants a, b > 0, 3 < p < 5, V (x) and Q(x) are two functions  γ in Cloc (R3 ) L∞ (R3 ). By comparing the decay rate of V (x) and Q(x), we first obtain two theorems stating the existence of positive ground states. Under certain assumptions on Q(x), we further prove the existence of positive bound states by using a linking argument with a barycenter map restricted on a Nehari manifold. © 2017 Elsevier Inc. All rights reserved.

1. Introduction In this paper we investigate the existence of positive solutions of the following nonlocal problem with subcritical nonlinearity: ⎞ ⎧ ⎛  ⎪ ⎪ ⎪ ⎨ − ⎝a + b |∇u|2 ⎠ Δu + V (x)u = Q(x)|u|p−1 u, x ∈ R3 , ⎪ ⎪ ⎪ ⎩

R3

(P )

u ∈ H 1 (R3 ), u > 0, x ∈ R3 ,

where a, b > 0 are constants, 3 < p < 5, V (x) and Q(x) are two positive functions such that lim|x|→∞ V (x) = V∞ > 0, lim|x|→∞ Q(x) = Q∞ > 0. * Corresponding author. E-mail addresses: [email protected] (T. Hu), [email protected] (L. Lu). https://doi.org/10.1016/j.jmaa.2017.12.027 0022-247X/© 2017 Elsevier Inc. All rights reserved.

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Problem (P ) is related to the stationary analogue of equation ⎛

⎞ L 2 2 E ∂ u P0 ∂u dx⎠ ∂ u = 0 + ρ 2 −⎝ ∂t h 2L ∂x ∂x2 2

0

proposed by Kirchhoff in [19] as an extension of classical D’Alembert’s wave equation for vibration of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations, so the nonlocal term appears. We refer [5,26] for physical and numerical aspects of Kirchhoff’s model. In the past few years, the following nonlocal problem ⎞ ⎧ ⎛  ⎪ ⎪ ⎪ ⎨ − ⎝a + b |∇u|2 ⎠ Δu + V (x)u = f (x, u), x ∈ R3 , ⎪ ⎪ ⎪ ⎩

(1.1)

R3

u > 0, u ∈ H 1 (R3 ),

has been studied extensively, where V : R3 → R, f ∈ C(R3 × R, R) and a, b > 0 are constants. Various results on the existence of positive solutions, multiple solutions, sign-changing solutions, ground states and semiclassical states have been obtained, see for examples [12,15,16,23,27,30] and the references therein. However, few work concern the effective of the asymptotic behavior at infinity of the potentials. In the present paper, we make some contributions in this direction. Set V (x) = V∞ + λh(x), Q(x) = Q∞ + q(x) in problem (P ), where λ ≥ 0 is a parameter. Then problem (P ) becomes ⎧ 

  ⎪ ⎪ |∇u|2 Δu + V∞ u + λh(x)u = Q∞ + q(x) |u|p−1 u, x ∈ R3 , − a + b ⎨ ⎪ ⎪ ⎩

(Pλ )

R3

u ∈ H 1 (R3 ), u > 0, x ∈ R3 .

γ We assume that h(x) and q(x) belong to Cloc (R3 ) ∩ L∞ (R3 ) for some 0 < γ < 1 such that

(H1) h(x)  0, lim|x|→∞ h(x) = 0; (H2) κ := Q∞ + inf3 q(x) > 0, q(x) ≡ 0, lim|x|→∞ q(x) = 0. x∈R

If q(x) ≥ 0, we consider the existence of ground state of (Pλ ) by comparing the decay rate of h(x) and q(x). On the other hand, if q(x) ≤ 0 and |q(x)| is not large, the existence of bound states of (Pλ ) is also established. Let H 1 (R3 ) be the Sobolev space equipped with the inner product and norm  (u, v) =

 a∇u∇v + V∞ uv dx

1/2

and u = (u, u)

.

R3

We shall seek solutions of (Pλ ) as critical points of the energy functional Iλ : H 1 (R3 ) → R associated with (Pλ ) and given by λ

u

+ 2 2 2

Iλ (u) =

⎛

 h(x)|u|2 dx + R3

b⎝ 4



R3

⎞2 |∇u| dx⎠ − 2

1 p+1

  Q∞ + q(x) |u|p+1 dx. R3

(1.2)

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The functional Iλ is well-defined for every u ∈ H 1 (R3 ) and belongs to C 1 (H 1 , R). A standard elliptic regularity argument and the strong maximum principle, see [14], shows that a nontrivial nonnegative critical point u ∈ H 1 (R3 ) of Iλ is classical positive solution to (Pλ ), and for any ϕ ∈ H 1 (R3 ) we have Iλ (u), ϕ





= (u, ϕ) + λ −



h(x)uϕdx + b

R3



 2

|∇u| dx

R3

Q∞ + q(x) |u|

p−1

∇u∇ϕdx

R3

(1.3)

uϕdx.

R3

In order to obtain a ground state of (Pλ ), we consider the following constraint minimization problem mλ := inf Iλ (u),

(M Pλ )

u∈Nλ

where Nλ denotes the Nehari manifold   Nλ := u ∈ H 1 (R3 ) \ {0} : Iλ (u), u = 0 .

(1.4)

Let {un } ∈ Nλ be a minimizing sequence of problem (M Pλ ). To prove {un } converges strongly, one generally needs some compactness conditions. Indeed, a lack of compactness, due to the invariance of R3 under the action of the noncompact group of translations, prevents a straight application of the variational methods. We should consider the following “limit” functional I∞ : H 1 (R3 ) → R defined by ⎛ I∞ (u) :=

1 b

u 2 + ⎝ 2 4

⎞2



|∇u| dx⎠ − 2

R3

Q∞ p+1

 |u|p+1 dx, R3

and the “limit problem at infinity” m∞ := inf I∞ (u), u∈N∞

(M P∞ )

where N∞ denotes the Nehari manifold for problem (M P∞ ),    N∞ := u ∈ H 1 (R3 ) \ {0} : I∞ (u), u = 0 . By using the well known concentration compactness principle, we can obtain the following lemma, and we give a proof in appendix for completeness. Lemma 1.1. Let (H1), (H2) hold. If the strictly inequality m λ < m∞

(1.5)

is satisfied, then all the minimizing sequences of problem (M Pλ ) are relatively compact, and mλ is achieved. When 3 < p < 5, the ground state energy mλ has a min–max characterization mλ =

inf

max Iλ (tu).

u∈H 1 \{0} t>0

Notice that if q(x) ≥ 0, q(x) ≡ 0, we have m0 < m∞ , hence mλ can be achieved when λ = 0. Clearly, for λ suitably close to zero one expects that the situation does not change, but it is a natural question to wonder

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under which assumptions, when λ increases, the minimum persists and, on the contrary, when it can be lost by the competing effect of two coefficients. If the decay rate of h(x) is faster than that of q(x), we will prove Theorem 1.2. Suppose that conditions (H1), (H2) are satisfied, q(x) ≥ 0, and there exist positive constants α and β, such that (H3)

 R3

√ σ∞ |x|

h(x)eα

dx < +∞ and lim inf q(x)eβ |x|→∞

√ σ∞ |x|

>0

hold, where β < min{2, α}, σ∞ = σ0 V∞ > 0, and σ0 is obtained in Lemma 2.2. Then for all λ ≥ 0, problem (Pλ ) admits a positive ground state. √

Remark 1. The condition (H3) implies that q(x) behave as e−β σ∞ |x| at infinity while the Lebesgue measure √   of the set x ∈ R3 \ BR (0) : h(x) ≥ |x|e−α σ∞ |x| converges to zero as R → ∞, so h(x) decays faster than q(x). On the other hand, if the decay rate of h(x) is slower or equal to that of q(x), and two coefficients additionally satisfy     (H4) Ωh ⊂ Ωq , where Ωh := x ∈ R3 | h(x) = 0 , Ωq := x ∈ R3 | q(x) = 0 , we have the following result. Theorem 1.3. Suppose that conditions (H1), (H2), (H4) are satisfied, q(x) ≥ 0, and there exist positive constants α, β and σ ∈ (0, σ∞ ), such that √ σ|x|

(H5) lim inf h(x)eα |x|→∞

> 0 and

 R3

q(x)eβ

√

σ|x|

dx < +∞

hold, where α ≤ min{p + 1, β} and σ∞ > 0 is defined as in Theorem 1.2. Then there exits a λ∗ > 0, such that problem (Pλ ) admits a positive ground state for all λ < λ∗ , while if λ > λ∗ , problem (Pλ ) has no ground state. √

−α σ|x| Remark 2. From (H5) we obtain that h(x) at infinity, and the Lebesgue measure of the   behave as e √ 3 −β σ|x| set x ∈ R \ BR (0) : q(x) ≥ |x|e converges to zero as R → ∞, so the decay rate of h(x) is slower than that of q(x).

When q(x) ≤ 0, mλ can not be achieved for all λ ≥ 0, which shall be proved in Section 3. We obtain a bound state for problem (Pλ ) by a min–max procedure. By a linking arguments with a barycenter map restricted to the Nehari manifold Nλ , we first get a (PS) sequence {un } with Iλ (un ) → c ∈ (m∞ , 2m∞ ). We then prove that the (PS) sequence is convergent by using a global compactness lemma. Theorem 1.4. Suppose that conditions (H1), (H2) are satisfied, q(x) ≤ 0 and 3−p

(H6) κ > 2 p+1 Q∞ holds, where κ > 0 is defined as in (H2). Then there exists a number λ∗ > 0, such that problem (Pλ ) has a positive solution for all λ ∈ [0, λ∗ ].

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Remark 3. The min–max procedure used here was introduced by Benci–Cerami [4] and Bahri–Li [2] and has been widely applied in different types of elliptic problem, see for example [6,7,10,11,21,25,29,30]. The condition (H6) was first introduced in [10] by Cerami and Vaira. Our main ideas come from two aspects. In [9], Cerami and Pomponio consider the following semilinear elliptic equation with competing coefficients: −Δu + α(x)u = β(x)|u|p−1 u,

x ∈ RN ,

N +2 where N ≥ 2, p > 1, p < N −2 , if N ≥ 3 and functions α(x), β(x) approach their limit at infinity from above. To the best of our knowledge, it is the first work, which concerns on the effective of the competing between coefficients to the existence of ground states using variational methods. For the existence of bound states, we are motivated by Ye [30], in which the existence of positive high energy solutions for a class of Kirchhoff equations was studied. However, we note that the assumptions on the potentials are different between [30] and the present paper. The paper is organized as follows. In Section 2, we present the variational setting and some useful lemma. In Section 3, we show that the map λ → mλ is a nondecreasing, bounded, right continuous map. We then establish the existence and nonexistence of minimizer of problem (M Pλ ) when λ = 0 under the conditions q(x) ≥ 0 and q(x) ≤ 0 respectively. In Section 4, we shall prove Theorem 1.2 and Theorem 1.3. In Section 5, we complete the proof of Theorem 1.4.

2. Variational setting and preliminaries In the following lemma, we summarize some properties of the Nehari manifold Nλ . Lemma 2.1. Suppose that (H1) and (H2) hold. Then for all λ ≥ 0, (1) let u ∈ Nλ , there exist constants η0 > 0 and η1 > 0 independent of λ such that 

u > η0 ,

|u|p+1 dx > η1 ;

(2.1)

R3

(2) for every u ∈ H 1 (R3 )\{0}, there exists a unique tλ (u) ∈ (0, +∞) such that tλ (u)u ∈ Nλ , and Iλ (tλ (u)u) = max Iλ (tu).

(2.2)

t>0

The function tλ (u)u is called the “projection” of u on Nλ ; (3) Nλ is a C 1 regular manifold diffeomorphic to the sphere of H 1 (R3 ); (4) Iλ is bounded form below on Nλ by a positive constant; (5) u is a free critical point of Iλ if and only if u is a critical point of Iλ constrained on Nλ . Proof. We omit details here since the proof is standard (see Lemma 2.1 in [9] for example). 2 We next introduce some results about limit problem (M P∞ ). Obviously, N∞ is a C 1 manifold, and for any given u ∈ H 1 (R3 )\{0}, its projection on N∞ is defined as θ(u)u ∈ N∞ with I∞ (θ(u)u) = max I∞ (tu). t>0

Moreover, m∞ is achieved, see [16]. A critical point of functional I∞ is a weak solution of the autonomous problem:

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−(a + b|∇u|22 )Δu + V∞ u = Q∞ |u|p−1 u,

(P∞ )

u ∈ H 1 (R3 ), x ∈ R3 .

If a = 1, b = 0 in problem (P∞ ), it is well known [13,20,22] that, up to translations, (P∞ ) admits a unique positive radially symmetric solution, which we denote by U = U (|x|). Note also from [13, Prop. 4.1] that U (|x|) has the following exponential decay: lim |Dj U (x)||x|e

√ V∞ |x|

|x|→∞

= dj > 0, j = 0, 1.

(2.3)

Taking the advantage of this fact, Huang et al. in [18] proved the uniqueness result for problem (P∞ ) as follows, also see Lemma 2.3 in [17]. Lemma 2.2. Let u ∈ H 1 (R3 )\{0} be any given positive solution of the problem (P∞ ), then there exist a constant σ0 > 0, independent of u, such that √ u(x) = U ( σ0 x − z)

for some z ∈ R3 .

(2.4)

Remark 4. From Lemma 2.2 and the translation invariant of (P∞ ), all the positive solutions of (P∞ ) are √ minimizers of (M P∞ ). Denote w(x) := U (| σ0 x|) be the positive radially symmetric solution of problem (P∞ ). In view of (2.3), there exists a constant cr > 0, such that lim

|x|→+∞

w(x)|x|e

√ σ∞ |x|

= cr ,

(2.5)

where σ∞ = σ0 V∞ . Note that the uniqueness of positive ground states of (P∞ ) plays an important role in min–max procedure used in Section 5. We close this section by analyze the energy level of the sign-changing solutions to problem (P∞ ). Lemma 2.3. The sign-changing energy level of problem (P∞ ) is strictly larger than 2m∞ , and the least energy level of functional I∞ is isolated. Proof. Let u = u+ − u− solves equation (P∞ ), where u+ (x) = max{u(x), 0} = 0,

u− (x) = max{−u(x), 0} = 0.

 Define G∞ (u) := I∞ (u), u . Times u+ on both side of (P∞ ), we get

⎛ ⎝a + b



R3

− 2 ∇u dx + b



⎞    + 2 ∇u dx⎠ ∇u+ 2 dx + V∞ |u+ |2 dx = Q∞ |u+ |p+1 dx,

R3

R3

R3

R3

which implies that G∞ (u+ ) < 0. Then there exists a positive number θ < 1 such that θu+ to be the projection of u+ on N∞ . By maximum principle, θu+ can not be the minimizer of (M P∞ ). Thus, we have  m∞ < I∞ (θu+ ) =  <

1 1 − 2 4 1 1 − 2 4



 θ2 u+ 2 +





u+ 2 +

Q∞ Q∞ − 4 p+1

Q∞ Q∞ − 4 p+1





 |u+ |p+1 dx

θp+1 R3

|u+ |p+1 dx. R3

(2.6)

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By the same argument, we deduce that  m∞ <

1 1 − 2 4



u− 2 +





Q∞ Q∞ − 4 p+1

|u− |p+1 dx.

(2.7)

R3

Therefore, we combine (2.6) and (2.7) to yield that  I∞ (u) =

1 1 − 2 4





Q∞ Q∞ − 4 p+1

u + 2

 |u|p+1 dx > 2m∞ . R3

The proof is finished. 2 3. Properties of the map λ → mλ At the beginning of this section, we shall show that the map λ → mλ is nondecreasing and bounded from above. Lemma 3.1. Let (H1) and (H2) hold. The map λ ∈ [0, +∞) → mλ is nondecreasing, and m λ ≤ m∞

λ ≥ 0.

f or all

(3.1)

¯ > λ ≥ 0. For any u ∈ N ¯ , we have Gλ (u) ≤ G ¯ (u) = 0. Thus, there exists a positive number Proof. Let λ λ λ t ≤ 1 such that tu ∈ Nλ . Then we have  mλ ≤ Iλ (tu) = t2  ≤

1 1 − 2 p+1

1 1 − 2 p+1



⎛



⎝ u 2 + λ



⎞ h(x)|u|2 dx⎠ + t4

R3

⎛ ¯ ⎝ u 2 + λ



⎞

h(x)|u|2 dx⎠ +



R3



b b − 4 p+1

b b − 4 p+1



⎛ ⎝



⎛ ⎝



⎞2 |∇u|2 dx⎠

R3



⎞2

|∇u|2 dx⎠

R3

= Iλ¯ (u). Therefore, by the arbitrariness of u, we conclude that mλ ≤ mλ¯ . To show mλ ≤ m∞ , it suffices to build a sequence {un } ⊂ Nλ , such that lim Iλ (un ) = m∞ . Due to the translation invariant, for any sequence n→∞

{yn } ⊂ R3 with |yn | −−−−→ +∞, we have wn (x) := w(x − yn ) ∈ N∞ and I∞ (wn ) = I∞ (w) = m∞ , where n→∞

w(x) is defined in Remark 4. Set un = tn wn , where {tn } is a sequence of positive number such that tn wn is the projection of wn on Nλ . We deduce from (2.1) that {tn } is bounded away from zero. Besides, note that ⎛ 0 = Gλ (tn wn ) =

t2n wn 2

+

t4n b⎝



⎞2 |∇wn | dx⎠ − tp+1 n Q∞

R3





λh(x + yn )|w|2 dx − tp+1 n

+ t2n

 |wn |p+1 dx

2

R3

R3

q(x + yn )|w|p+1 dx. R3

Since  lim

n→∞ R3

 h(x + yn )|w|2 dx = 0

and

lim

n→∞ R3

q(x + yn )|w|p+1 dx = 0,

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   2    we obtain that t2n w 2 + o(1) + t4n b R3 |∇w|2 = tp+1 Q∞ R3 |w|p+1 dx + o(1) , which implies that n {tn } is bounded. We assert that, up to a subsequence, tn → t as n → ∞. It follows that G∞ (tw) = 0, which yields that t = 1. Therefore,  Iλ (un ) = t2n

1 1 − 2 p+1



⎛ ⎝ w 2 + λ



⎞ h(x + yn )|w|2 dx⎠ + t4n



R3

b b − 4 p+1



⎛ ⎝



⎞2 |∇w|2 dx⎠

R3

−−−−→ I∞ (w) = m∞ . 2 n→∞

¯ ≥ 0, such that m ¯ = m∞ , then for all Lemma 3.2. Suppose that (H1) and (H2) hold. If there exists a λ λ ¯ mλ = m∞ is not attained. λ > λ, ¯ be fixed. By (3.1), we have m ¯ ≤ mλ ≤ m∞ , hence mλ = m∞ . Proof. Let λ > λ λ Next we prove mλ is not attained. Arguing by contradiction, we suppose that there exists u ∈ Nλ such that Iλ (u) = mλ . Further more we can assume that u > 0, since |u| ∈ Nλ and Iλ (u) = Iλ (|u|), then the maximum principle implies that |u| > 0. Set θu is the projection of u to N∞ , where θ < 1. Thus we get a contradiction that m∞ = I∞ (θu) < Iλ (u) = mλ = m∞ . 2 An immediate consequence of Lemma 3.2 is the following corollary. ¯ ≥ 0, such that m ¯ = m∞ , and m ¯ is attained. Corollary 3.3. Let (H1), (H2) hold. There exists at most one λ λ λ The next lemma shows the right continuity of the map λ ∈ [0, +∞) → mλ . Lemma 3.4. Let (H1), (H2) hold. Then the map λ ∈ [0, +∞) → mλ is right continuous. Proof. Let λ ≥ 0 and {λn } be a sequence such that λn  λ as n → ∞. From Lemma 3.1, we know that m λ ≤ mλ n ≤ m∞ . 1) If mλ = m∞ , we have mλn = mλ = m∞ , hence the map is right continuous at λ. 2) If mλ < m∞ , from Lemma 1.1 there exists a positive function u ∈ Nλ such that u is the minimizer of problem (M Pλ ). Let tn u ∈ Nλn be the projection of u on Nλn where tn ∈ (0, +∞). By (2.1) and Gλn (tn u) = 0, we obtain that {tn } is bounded above and bounded away from zero. Up to subsequence, we assert that tn −−−−→ t > 0. We then have Gλ (tu) = 0, which yields that t = 1. Therefore, n→∞

mλ ≤ mλn ≤ Iλn (tn u) = Iλ (u) + o(1) = mλ + o(1), which implies the right continuity of map λ → mλ .

2

  Let us define λ∗ = sup {λ > 0 : mλ < m∞ } ∪ {0} . We have the following corollary Corollary 3.5. If 0 ≤ λ∗ < +∞, then we have mλ∗ = m∞ . Moreover, λ∗ = min {λ ≥ 0 : mλ = m∞ } . Proof. If λ∗ = +∞, we have mλ < m∞ for all λ ≥ 0. If 0 < λ∗ < +∞, we then have mλ = m∞ for λ > λ∗ , while mλ < m∞ for 0 ≤ λ < λ∗ . If λ∗ = 0, the set {λ > 0 : mλ < m∞ } must be an empty set, thus for all λ > 0, mλ = m∞ . By the right continuity of λ → mλ , the corollary follows. 2

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Lemma 3.6. Suppose that (H1) and (H2) hold, (1) if q(x) ≥ 0, then λ∗ > 0, m0 < m∞ ; (2) if q(x) ≤ 0, then λ∗ = 0, m0 = m∞ and m0 is not attained. Proof. 1) Let w ∈ N∞ be defined in Remark 4. If q(x) ≥ 0, we have  G0 (w) = G∞ (w) −

q(x)|w|p+1 dx < 0, R3

then there exists a t < 1 such that tw ∈ N0 . Therefore,  m0 ≤ I0 (tw) =

1 1 − 2 p+1





tw 2 +

b b − 4 p+1



⎛ ⎝



⎞2 |t∇w|2 dx⎠

R3

< I∞ (w) = m∞ . 2) In view of (3.1), to establish m0 = m∞ , it suffices to prove that m0 ≥ m∞ . For all u ∈ N∞ , let t(u)u ∈ N0 be the projection of u on N0 . If q(x) ≤ 0, we have t(u) ≥ 1 due to  G0 (u) = G∞ (u) −

q(x)|w|p+1 dx ≥ 0. R3

Therefore, we have I∞ (u) ≤ I0 (t(u)u), from which, taking into account that N∞ and N0 are diffeomorphic to the sphere in H 1 (R3 ), we infer that m∞ = inf I∞ (u) ≤ inf I0 (t(u)u) = inf I0 (v) = m0 . u∈N∞

u∈N∞

v∈N0

Finally, we assume, by contradiction, that there exists a positive function v(x) ∈ N0 , such that v(x) is the minimizer of problem (M Pλ ) for λ = 0. From G∞ (v) < G0 (v) = 0, there exists a θ < 1 such that θv ∈ N∞ . Note that m∞ ≤ I∞ (θv) < I0 (v) = m0 , which derives a contradiction. 2 4. Existence of ground states In this section we deal with problem (Pλ ) under the condition q(x) ≥ 0. Mainly, we shall establish the proof of Theorem 1.2 and Theorem 1.3. Before proving Theorem 1.2 and Theorem 1.3, we need the following lemma in [3, Lemma II.2] (also see [2, Proposition 1.2]). Lemma 4.1. Let ϕ ∈ L∞



C(R3 ) and ψ ∈ C(R3 ) are such that, for some α ≥ 0, β ≥ 0, c ∈ R 

ϕ(x)eα|x| |x|β → c ψ(x)eα|x| |x|β dx ≤ ∞,

R3

as |x| → ∞,

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then ⎧ ⎨

⎫ ⎬ ϕ(x + y)ψ(x)dx

⎩

R3

⎭

 eα|x| |x|β → c

e−α(x·y)/|y| ψ(x)dx

as

|y| → ∞.

R3

4.1. The case in which h(x) decays faster than q(x) Proof of Theorem 1.2. In view of Lemma 1.1, in order to prove Theorem 1.2 we need to prove that λ∗ = +∞, which implies that mλ < m∞ for every λ ≥ 0. By Lemma 3.6(1), we already know that m0 < m∞ , so in what follows we can assume λ > 0. Fix a λ > 0, and let w(x) ∈ N∞ be defined in Remark 4. As we proved in Lemma 3.1, for any sequence {yn } ⊂ R3 with |yn | −−−−→ +∞, there exists {tn } ⊂ R with tn −−−−→ 1 such that tn wn (x) = tn w(x − yn ) ∈ n→∞

n→∞

Nλ and mλ ≤ Iλ (tn wn ) → m∞ as n → ∞. Note that Iλ (tn wn ) =

1 1

tn w 2 + 2 2 −

1 p+1



 R3

b λh(x + yn )|tn w|2 dx + |tn ∇w|42 4

 Q∞ + q(x + yn ) |tn w|p+1 dx

R3

λt2 = I∞ (tn w) + n 2



tp+1 h(x + yn )|w| dx − n p+1



2

R3

q(x + yn )|w|p+1 dx. R3

Thus, we get the conclusion if we show that, for n large enough,

λt2n 2

 R3

I∞ (tn w) < I∞ (w) = m∞ ,  tp+1 n 2 h(x + yn )|w| dx < q(x + yn )|w|p+1 dx. p+1

(4.1) (4.2)

R3

Since p > 3, we have I∞ (tn w) − I∞ (w) b 1 1 2 (tn − 1) w 2 + (t4n − 1)|∇w|42 − (tp+1 − 1)|w|p+1 p+1 2 4 p+1 n  (tn − 1)2 = (tn − 1)G∞ (w) + (1 − p) w 2 + (3 − p)b|∇w|42 + o(1) , 2

=

which implies that (4.1) holds. To prove (4.2), we first observe that, by (H3) there exists a positive constant c1 > 0 such that q(x) > √ c1 e−β σ∞ |x| when |x| is large enough and tp+1 n p+1

 p+1

q(x + yn )|w| R3

1 dx > 2(p + 1)

 q(x + yn )|w|p+1 dx B1

1 inf q(x + yn ) > 2(p + 1) x∈B1 >

c1

 B1

2(p +

 |w|p+1 dx

B1

|w|

p+1

dx

√ 1)eβ σ∞

e−β

√

σ∞ |yn |

.

(4.3)

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On the other hand, for β < min{2, α}, we have √ 2 β σ∞ |x|

lim

|x|→+∞

w e



h(x)eβ

√

σ∞ |x|

=

√ 2 β σ∞ |x|

lim

w e

|x|→+∞



R3

lim

|x|→+∞

√ √ σ∞ |x| (β−α) σ∞ |x|

h(x)eα

dx =

=

e

w

2

√ (β−2) σ∞ |x| √ 2 2 σ∞ |x| e |x| e |x|2



R3

√ σ∞ |x|

h(x)eα

dx <

= 0,

dx < +∞.

R3

Then by Lemma 4.1, we deduce that, for n large enough, λt2n 2



√ −β σ∞ |yn |

2

h(x + yn )|w| dx = o(1)e

<

c1

 B1

2(p +

R3

|w|p+1 dx √ 1)eβ σ∞

e−β

√

σ∞ |yn |

.

(4.4)

2

Therefore, we combine (4.3) and (4.4) to yield (4.2). 4.2. The case in which h(x) decays slower than q(x)

Lemma 4.2. Let (H1), (H2), (H4) hold and q(x) ≥ 0. For a sequence {λn } ⊂ R, λn → +∞, let un ∈ Nλn , un > 0, and Iλn (un ) ≤ C. Then {un } is bounded in H 1 (R3 ) and the following relations hold as n → +∞, un → 0 a.e. in R3 \Ωh , and 

 h(x)|un | dx ≤ C,

q(x)|un |p+1 dx → 0.

2

λn R3

R3

Proof. Since  Iλn (un ) =

1 1 − 2 p+1

⎛



⎝ un 2 + λn

⎞



h(x)|un | dx⎠ + 2

R3

we obtain that un is bounded in H 1 (R3 ) and n → ∞. Note that



h(x)|un |2 ≤

R3



 q(x)|un |

p+1

q(x)|un |

p+1

dx =

R3

(H4)

 |∇un |42 ≤ C,

→ 0. Thus, un → 0 a.e. in R3 \Ωh as

q(x)|un |p+1 dx

R3 \Ωh





q(x)|un |p+1 dx +

R3 \BR

b b − 4 p+1



dx ≤

R3 \Ωq

≤

C λn



BR \Ωh



≤ε(R) R3 \BR

q(x)|un |p+1 dx

|un |

p+1



dx + q(x) L∞ (R3 )

|un |p+1 dx.

BR \Ωh

By using Dominated Convergence Theorem and let R → +∞, we complete the proof of Lemma 4.2.

2

From Lemma 3.6(1), we know that λ∗ > 0. Therefore, to establish Theorem 1.3, it suffices to prove that 0 < λ∗ < +∞. On the contrary, we now suppose that λ∗ = +∞, and the rest is to derive a contradiction. For any sequence {λn } with λn −−−−→ +∞, we have mλn < m∞ . Then by Lemma 1.1, there exists a sequence n→∞

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{un } ⊂ H 1 (R3 ), such that un minimizes the problem (M Pλ ) for λ = λn . From Lemma 4.2, it follows that un  u ∈ H 1 (R3 ), u → u a.e. in R3 and u = 0 in R3 \Ωh . Moreover, we have 

 h(x)|un |2 dx ≤ m∞ ,

λn R3

q(x)|un |p+1 dx → 0, R3

as n → ∞. Before proving Theorem 1.3, we first establish several claims. 

h(x)|un |2 dx −−−−→ 0. n→∞  Otherwise, λn R3 h(x)|un |2 dx > c > 0 for n large enough, which implies that

Claim 1: λn

R3



 G∞ (un ) = Gλn (un ) − λn

h(x)|un |2 dx + R3

q(x)|un |p+1 dx < −c/2 < 0, R3

for n large enough. Then there exists θn < 1 such that θn un ∈ N∞ , and {θn } is bounded away form zero. Note that    1 b 1 b 2 −

θn un + − |θn ∇un |42 ≤ I∞ (θn un ) < 2 p+1 4 p+1    1 1 − λn h(x)|un |2 dx + 2 p+1 

m∞

R3


n→∞

Claim 2: u = 0. Indeed, by the boundedness of {un } in H 1 (R3 ), we deduce that  m∞ ≤ I∞ (θn un ) =

1 1 − 2 p+1





un 2 +

b b − 4 p+1

 |∇un |42 + o(1)

≤ mλn + o(1) ≤ m∞ + o(1), which implies that θn un ∈ N∞ is a minimizing sequence of problem (M P∞ ). If u = 0, from the Lemma 4.2 in [16], we see that u = limn→∞ θn un must be a minimizer of problem (M P∞ ), and by maximum principle u > 0 in R3 , which contradicts to u = 0 in R3 \Ωh . As in (2.1), |un |p+1 is bounded from blow, by using Lemma 1.1 in [24], un  0 in H 1 (R3 ) implies that there exist a sequence {yn } ⊂ R3 with |yn | −−−−→ +∞ and positive constant δ such that n→∞

 |un |2 dx > δ > 0.

(4.5)

B1 (yn )

For the sequence {yn } as in (4.5), we set vn (x) = un (x + yn ). Due to translation invariance of problem a.e. (M P∞ ), we see that {θn vn } is also a minimizing sequence of m∞ and θn vn −−−−→ v = 0 by (4.5). With the n→∞

help of the Lemma 4.2 in [16] again, we have vn → v strongly in H 1 (R3 ) and v is a minimizer of problem √ (M P∞ ), so there exists a z ∈ R3 such that v(x) = w(x − z/ σ0 ).

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Claim 3: There exist positive constants R and C, independent of n, such that √

vn (x) < Ce−

for |x| > R

σ|x|

as n → ∞,

(4.6)

where σ < σ∞ and σ, σ∞ > 0 are defined as in Theorem 1.2. Since un minimizes the problem (M Pλ ) for λ = λn , by the definition of vn , it follows that −Δvn +

V∞ λh(x + yn ) Q∞ + q(x + yn ) p vn + vn = v . 2 2 (a + b|∇vn |2 ) (a + b|∇vn |2 ) (a + b|∇vn |22 ) n

2 By virtue of the Schauder interior estimates (see for example [14]), vn → v in Cloc (R3 ). It follows from the strong convergence of {vn } that, for any δ > 0, there exists a large number ρ > 0 such that

 |vn |6 dx ≤ vn − v R3 \Bρ (0) + v R3 \Bρ (0) < δ, R3 \Bρ (0)

when n large enough. Also notice that V∞ + λh(x + yn ) ≥ V∞ > 0, by using Lemma 2.2 in [15], we have vn converge to 0 uniformly at infinity. Notice that lim

|x|→∞

λh(x + yn ) V∞ V∞ + = , (a + b|∇vn |22 ) (a + b|∇vn |22 ) (a + b|∇vn |22 )

and V∞ V∞ = = σ∞ > σ. 2 n→∞ (a + b|∇vn | ) (a + b|∇v|22 ) 2 lim

Then, by applying Lemma 2.11 in [9], there exists a R > 0 such that √

vn (x) ≤ Cn e− where Cn = e

√

σR

σ|x|

for |x| > R

as

n → ∞,

max vn (x). Due to the uniformly boundedness of {vn } in C(B2R ), {Cn } is bounded, then

|x|=R

(4.6) is proved. Claim 4: For n large enough, the inequality 1 2



2 λn h(x + yn )|θn vn | dx − p+1

 q(x + yn )|θn vn |p+1 dx > 0

2

R3

(4.7)

R3

holds true. √ We first note that by (H5), there exists a positive constant c2 > 0 such that h(x) > c2 e−α σ|x| when |x| is large enough. We have λn θn2 2

 h(x + yn )|vn |2 dx > R3

λn θn2 2

 h(x + yn )|vn |2 dx > λn B1

While, for α ≤ min{p + 1, β}, by the decay of vn (4.6),

c2 √ 4eα σ

 B1

√

|v|2 e−α

σ|yn |

dx.

(4.8)

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√ σ|x|

lim |vn |p+1 eα

|x|→∞



q(x)e

= lim

|vn |e



√

|x|→∞

√ α σ|x|

dx =

R3

R3

√

σ|x|

p+1

q(x)eβ σ|x| √ dx ≤ e(β−α) σ|x|

e(α−(p+1))

 q(x)eβ

√

√ σ|x|

σ|x|

≤ C p+1 ,

dx < +∞.

R3

Applying Lemma 4.1, we have ⎛ lim ⎝

⎞

 vnp+1 (x

n→∞

√ σ|yn |

− yn )q(x)dx⎠ eα

 ≤C

p+1

R3

q(x)dx < +∞. R3

It follows from λn −−−−→ +∞ that, for n large enough n→∞

2θnp+1 p+1



√ p+1 α σ

q(x + yn )vnp+1 dx ≤ R3

⎛



e q(x)dx 16C ⎝ c2√  R3 c2 (p + 1) B1 |v|2 dx 4eα σ 

c2 < λn α√σ 4e

√ 2 −α σ|yn |

|v| e

⎞



√ 2 −α σ|yn |

|v| e

dx⎠

B1

(4.9)

dx.

B1

Therefore, we combine (4.8) and (4.9) to yield (4.7). Completion for the Proof of Theorem 1.3. We calculate from (4.7) that m∞ ≤ I∞ (θn un ) < I∞ (θn un ) +  =  ≤

θp+1 θn2 − n 2 p+1 1 1 − 2 p+1





θn2 2

⎛

 λn h(x)|un |2 dx − R3

⎝ un 2 + λn

⎛ ⎝ un 2 + λn

⎞



h(x)|un |2 dx⎠ +

R3



⎞

h(x)|un | dx⎠ + 2

R3

θnp+1 p+1 



 q(x)|un |p+1 dx R3

θp+1 θn4 − n 4 p+1

1 1 − 4 p+1

 b|∇un |42

 b|∇un |42

= Iλn (un ) = mλn , which however derives a contradiction. This therefore complete the proof of Theorem 1.3. 2 5. Existence of bound states From Lemma 3.6, when q(x) ≤ 0, mλ = m∞ and mλ can not be achieved for all λ ≥ 0. In this section, we shall prove Theorem 1.4 and find a bound state of problem (Pλ ) by min–max procedure. We shall first give the representation theorem of the (PS) sequences. Lemma 5.1. Suppose (H1), (H2) hold. Let {un } be a (P S) sequence of Iλ constrained on Nλ , that is un ∈ Nλ , Iλ (un ) ≤ C and ∇|Nλ Iλ (un ) −−−−→ 0 in H 1 (R3 ). Then, up to a subsequence, either un is strongly convergent n→∞

in H 1 (R3 ), or there exist a function u ¯ ∈ H 1 (R3 ), an integer k > 0, k functions {Wj }, Wj = 0 for 1 ≤ j ≤ k j 3 and k sequences {yn } ⊂ R for 1 ≤ j ≤ k, such that |ynj | → +∞,

|ynj − yni | → +∞, i = j,

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and ⎞ ⎧ ⎛ k ⎪  ⎪ p−1 ⎪ ⎪ b|∇Wj |22 ⎠ Δ¯ ¯ + λh(x)¯ u = (Q∞ + q(x))|¯ u| u ¯, u + V∞ u − ⎝a + b|∇¯ u|22 + ⎪ ⎪ ⎨ j=1   ⎪ k ⎪  ⎪ ⎪ 2 2 ⎪ u|2 + b|∇Wi |2 ΔWj + V∞ Wj = Q∞ |Wj |p−1 Wj , ⎪ ⎩ − a + b|∇¯

(5.1)

j = 1, 2 . . . , k,

i=1

un (x) −

k 

Wj (x − ynj ) → u ¯(x)

in H 1 (R3 ),

(5.2)

j=1

lim Iλ (un ) ≥ I˜λ (¯ u) +

n→∞

k 

I˜∞ (Wj ),

(5.3)

j=1

where 1 λ u 2 + I˜λ (¯ u) = ¯ 4 4



 h(x)|¯ u|2 + R3

1 I˜∞ (Wj ) = Wj 2 + 4



1 1 − 4 p+1

Q∞ Q∞ − 4 p+1



 (Q∞ + q(x))|¯ u|p+1 , R3

|Wj |p+1 p+1 .

Proof. The proof of Lemma 5.1 is similar to that of Lemma 3.4 in [23]. The main ingredients are Lions Lemma and Brezis–Lieb Lemma [28]. We give a proof in appendix. 2 Remark 5. By Lemma 2.1, ∇|Nλ Iλ (un ) → 0 implies that Iλ (un ) → 0, see [1, Sect. 6.3]. Remark 6. Note that Wj = 0 is a solution of equation ⎧   2 p−1 ⎪ ⎨ − aj + b|∇Wj |2 ΔWj + V∞ Wj = Q∞ |Wj | Wj ,  u|22 + b|∇Wi |22 ≥ a. ⎪ ⎩ aj := a + b|∇¯ i=j

By the similar argument to (2.6), if k ≥ 2, we have I˜∞ (Wj ) > m∞ . Since q(x) ≤ 0, if k ≥ 1 and u ¯ = 0, we also have I˜λ (¯ u) > mλ = m∞ and I˜∞ (Wj ) > m∞ . Lemma 5.2. Suppose that (H1), (H2) hold and q(x) ≤ 0. Then Iλ satisfies the (P S)c condition for all c ∈ (m∞ , 2m∞ ). Proof. Let {un } be a (P S) sequence of Iλ constrained on Nλ at level c, with c ∈ (m∞ , 2m∞ ). We argue by contradiction. Once we assume that un is not strongly convergent, by Lemma 5.1, and then there exist u ¯ and k functions {Wj }, Wj = 0 1 ≤ j ≤ k such that (5.1)–(5.3) hold. By Remark 6, both k ≥ 2 and u ¯ = 0 imply that c = lim Iλ (un ) = I˜λ (¯ u) + n→∞

k 

I˜∞ (Wj ) > 2m∞ .

j=1

Thus, we have un (x) = W1 (x − yn1 ) + o(1) in H 1 (R3 ), and W1 is a nontrivial solution of problem (P∞ ). Due to Lemma 2.3, we obtain that either I˜∞ (W1 ) = I∞ (W1 ) = m∞ or I˜∞ (W1 ) = I∞ (W1 ) > 2m∞ , which

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however contradicts I˜∞ (W1 ) = lim Iλ (un ) = c ∈ (m∞ , 2m∞ ). n→∞

This completes the proof of Lemma 5.2. 2 Now, we need to build the min–max scheme with the help of the barycenter map which is given in [8]. The barycenter map β : H 1 (R3 ) \ {0} → R3 of a function u satisfies following properties: (B1) β(u) is continuous in H 1 (R3 )\{0}; (B2) if u is radial function, β(u) = 0 ∈ R3 ; (B3) for all t = 0 and for all u ∈ H 1 (R3 )\{0}, β(tu) = β(u); (B4) given z ∈ R3 and setting uz (x) = u(x − z), β(uz ) = β(u) + z. Set   Nλ0 := u ∈ Nλ β(u) = 0 ∈ R3 . Then Nλ0 = ∅. Indeed, taking into account that Nλ and N∞ are diffeomorphic to the sphere in H 1 (R3 ), there exists a t > 0 such that tw ∈ Nλ in which w is defined in Remark 4. Then, by (B3) we have tw ∈ Nλ0 . Hence,   m0 := inf Iλ (u) u ∈ Nλ0 is well defined. Lemma 5.3. Let (H1), (H2) hold and q(x) ≤ 0. Then m0 > m∞ . Proof. Clearly m0 ≥ m∞ . Suppose that, by contradiction, m0 = m∞ . Let {un } be a minimizing sequence of m0 . By the Ekeland’s variational principle in [28], there exists another sequence {vn } ⊂ Nλ0 such that Iλ (vn ) → m∞ , ∇|Nλ Iλ (vn ) → 0 in H 1 (R3 ) and un − vn → 0 as n → ∞. Note that m∞ can not be achieved, and apply Lemma 5.1 to {vn }, then there exist a function W1 and a sequence yn1 ∈ R3 with |yn1 | −−−−→ +∞ such that n→∞

vn (x) = W1 (x − yn1 ) + o(1),

(5.4)

√ where W1 is a minimizer of problem (M P∞ ). Set W1 (x) = w(x −z/ σ 0 ) for some z ∈ R3 , where w is defined in Remark 4. Computing the barycenter of both terms in (5.4) and using the properties of barycenter map, we get √ √ β(W1 (x − yn1 ) + o(1)) = β(w(x − yn1 − z/ σ 0 )) + o(1) = yn1 + z/ σ 0 + o(1) and β(vn (x)) = β(un (x) + o(1)) = β(un (x)) + o(1) −−−−→ 0 ∈ R3 , n→∞

√

which leads to a contradiction because |yn1 + z/ σ 0 + o(1)| −−−−→ +∞. n→∞

Now we define the operator Γ : R3 → Nλ as Γ[y](x) = ty w(x − y), where w is defined in Remark 4 and ty is chosen such that Γ[y] ∈ Nλ .

2

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Note that, since q(x) ≤ 0 



Gλ (w(x − y)) = G∞ (w) + λ

h(x + y)|w|2 dx −

R3

q(x + y)|w|p+1 dx > 0, R3

which implies that ty > 1 for all y ∈ R3 . On the other hand, note 

ty w + λ 2

 2

h(x + y)|ty w| dx +

|ty ∇w|42

(Q∞ + q(x + y))|ty w|p+1 dx ≥ κ|ty w|p+1 p+1 .

=

R3

R3

Thus, we deduce that ty is bounded. Lemma 5.4. Let (H1), (H2) hold and q(x) ≤ 0. Then

lim

|y|→+∞

Iλ (Γ[y]) = m∞ .

Proof. Let t = lim sup|y|→+∞ ty , we have Gλ (Γ[y]) → G∞ (tw) as |y| → ∞, which implies ty → t = 1 as |y| → ∞. Thus, we deduce that  Iλ (Γ[y]) =

1 1 − 2 p+1

⎛



⎝ ty w 2 + λ



⎞ h(x + y)|ty w|2 ⎠ +



R3

→ I∞ (w) = m∞

as

b b − 4 p+1

 |ty ∇w|42

|y| → ∞. 2

Lemma 5.5. If q(x) ≤ 0 and (H6) hold, there exists a λ∗ > 0 such that for every λ ∈ [0, λ∗ ], y ∈ R3 .

Iλ (Γ[y]) < 2m∞ Proof. Since Γ[y] ∈ Nλ , we have  t2y w 2

+

t2y

2

λh(x + y)|w| dx +

t4y b|∇w|42

=

tp+1 y

R3



 Q∞ + q(x + y) |w|p+1 dx.

R3

Using p > 3 and ty > 1, we deduce that tp−3 y



w 2 + λ R3 h(x + y)|w|2 dx + b|∇w|42  < (Q∞ + q(x + y))|w|p+1 dx R3 <

=

w 2 + λ h(x) L∞ (R3 ) |w|22 + b|∇w|42 κ|w|p+1 p+1 λ h(x) L∞ (R3 ) |U |22 Q∞ + . κ κ|U |p+1 p+1

Note that t2y t2y Iλ (Γ[y]) = w 2 + λ 4 4  + tp+1 y

 h(x + y)|w|2 dx

R3

1 1 − 4 p+1

 R3

 Q∞ + q(x + y) |w|p+1 dx

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   1 1 λ 1 p+1 2 2

w + h(x) L∞ (R3 ) |w|2 + − Q∞ |w|p+1 < 4 4 4 p+1 p+1   p−3   λ h(x) L∞ (R3 ) |U |22 Q∞ λ 2 ∞ 3 < +

h(x)

m := e(λ). + |w| ∞ L (R ) 2 κ 4 κ|U |p+1 p+1 

tp+1 y

p−3 p+1

Since κ1 < 2Q∞ , we have e(0) < 2m∞ . Observing that e(λ) is monotone increasing on [0, +∞). Let λ∗ be the positive root of the equation e(λ) = 2m∞ , then for all λ ∈ [0, λ∗ ], Iλ (Γ[y]) < e(λ) ≤ 2m∞ .

2

Proof of Theorem 1.4. By Lemma 5.3 and Lemma 5.4, there exists ρ¯ > 0 such that for all ρ > ρ¯ m∞ < max Iλ (Γ[y]) < m0 . |y|=ρ

(5.5)

In order to apply the Linking Theorem in [28], we take ¯ρ¯), M = Γ(B

S = Nλ0 = {u ∈ Nλ |β(u) = 0}.

We claim that (a) ∂M ∩ S = ∅. (b) τ (M ) ∩ S = ∅, for all τ ∈ T = {τ ∈ C(M, Nλ )|τ|∂M = id}. ¯ρ¯ such that u = Γ[¯ To prove (a), we notice that for every u ∈ ∂M , there exists y¯ ∈ ∂ B y ]. By using the properties of barycenter map we have β(u) = β(Γ[¯ y ]) = β(ty w(x − y¯)) = β(w) + y¯ = y¯ = 0. To verify (b), we consider τ ∈ T and define ¯ρ¯ → R3 , Ψ:B

Ψ(y) = β ◦ τ ◦ Γ[y].

Ψ is a continuous function, and for all |y| = ρ¯, we see Γ[y] ∈ ∂M . Hence, τ ◦ Γ[y] = Γ[y], which implies that Ψ(y) = y. We claim that there exists y˜ ∈ Bρ¯ such that Ψ(˜ y ) = 0, which means that τ (Γ[˜ y ]) ∈ S. Otherwise Ψ(y) ρ¯ · |Ψ(y)| must be a retraction from Bρ¯ to ∂Bρ¯, which contradicts to the Brower Fixed Point Theorem. Therefore, τ (M ) ∩ S = ∅. Then (5.5) can be written as m0 = inf Iλ > max Iλ . Let us define S

∂M

m := inf max Iλ (τ (u)). τ ∈T u∈M

Then by (5.5) and (b) we have m ≥ m0 > m∞ . Moreover, taking τ = id and using Lemma 5.5, we deduce that m < 2m∞ for all λ ∈ [0, λ∗ ], where λ∗ is defined in Lemma 5.5. By Lemma 5.2 and the Linking Theorem, we conclude that m is a critical value and problem (Pλ ) has a nontrivial solution. In order to get a positive solution to problem (Pλ ), it suffices to repeat the whole procedure to the functional   1 λ b 1 4 Iλ+ (u) = u 2 + h(x)|u|2 + |∇u|2 − (Q∞ + q(x))(u+ )p+1 . 2 2 4 p+1 R3

R3

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Therefore, we can get a solution u = 0 of problem   − a + b|∇u|22 Δu + V∞ u + λh(x)u = (Q∞ + q(x))(u+ )p . Thus u− = 0, and by maximum principle we conclude that u > 0. This completes the proof of Theorem 1.4. 2 Acknowledgments The authors thank Professor Yinbin Deng and Professor Shuangjie Peng very much for stimulating discussions and helpful suggestions on the present paper. This work is supported by NSFC Grant No. 11601523 and NSFC Grant No. 11561024. Appendix Proof of Lemma 1.1. Let {un } ⊂ Nλ be a minimizing sequence of problem (M Pλ ) with Iλ (un ) → mλ < m∞ as n → ∞. Obviously, {un } is bounded in H 1 (R3 ). Assume that un  u in H 1 (R3 ), un → u in Lqloc (R3 ) for 1 ≤ q < 6, un → u a.e. in R3 . We first prove that u = 0. Otherwise, un → 0 in Lqloc (R3 ) for 1 ≤ q < 6. We compute that 

 h(x)|un |2 dx = R3

 h(x)|un |2 dx +

BR



≤ |h|∞

h(x)|un |2 dx

R3 \BR

|un |2 dx + ε(R)|un |22 ,

BR

where ε(R) → 0 as R → ∞. Hence  q(x)|un |p+1 dx = o(1). Then R3

 R3

h(x)|un |2 dx = o(1) as n → ∞. By a similar argument we have

G∞ (un ) = un 2 + 4b|∇un |42 − Q∞ |un |p+1 p+1   2 = Gλ (un ) − λ h(x)|un | dx + q(x)|un |p+1 dx = o(1). R3

(A.1)

R3

Set θn un ∈ N∞ for some θn > 0. Since N∞ is bounded away from 0, there exists η˜0 > 0, independent of λ, such that θn un ≥ η˜0 . On the other hand {un } is bounded, so θn →  0. Note that G∞ (θn un ) = θn2 un 2 + θn4 b|∇un |42 − θnp+1 Q∞ |un |p+1 p+1 = 0.

(A.2)

By Lemma 2.1(1), there exists η1 > 0 such that |un |p+1 p+1 > η1 hence   θnp+1 Q∞ η1 < max θn2 un 2 , θn4 b|∇un |42 , which implies {θn } is bounded. We combine (A.1) and (A.2) to yield that (θn2 − θn4 ) un 2 = (θnp+1 − θn4 )Q∞ |un |p+1 p+1 + o(1). Since {un } is bounded from above and bounded away from 0, we know that θn −−−−→ θ = 1. Then we have n→∞

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⎧ ⎫  ⎨ ⎬ p−1 p−3

un 2 + λ h(x)|un |2 dx + b|∇un |42 mλ ← Iλ (un ) = ⎭ 4(p + 1) 2(p + 1) ⎩ R3

=

p−1 p−3

θn un 2 + b|θn ∇un |42 + o(1) 2(p + 1) 4(p + 1)

= I∞ (θn un ) + o(1) ≥ m∞ + o(1), which is a contradiction. Thus we have u = 0. Next we prove Gλ (u) ≥ 0. Set vn = un − u  0 in H 1 (R3 ). By Brezis–Lieb Lemma, we have 

 h(x)|un |2 dx = R3



 h(x)|vn |2 dx +

h(x)u2 dx + R3

R3 \BR

BR



h(x)|vn |2 dx

h(x)u2 dx + o(1).

= R3

Similarly, 





(Q∞ + q(x)) |un |p+1 dx = R3

(Q∞ + q(x)) |u|p+1 dx + R3

Q∞ |vn |p+1 dx + o(1). R3

Then 1 λ Iλ (un ) = un 2 + 4 4 =

1 λ

u 2 + 4 4



 h(x)|un | dx + 2

R3





h(x)|u|2 dx + R3

1 + vn 2 + 4



1 1 − 4 p+1

1 1 − 4 p+1

1 1 − 4 p+1







(Q∞ + q(x))|un |p+1 dx R3

(Q∞ + q(x))|u|p+1 dx

(A.3)

R3

Q∞ |vn |p+1 dx + o(1) R3

Set tu ∈ Nλ , and we claim that t ≥ 1. Otherwise, t < 1 and 1 λ mλ ≤ Iλ (tu) = tu 2 + 4 4 1 λ < u 2 + 4 4





 2

h(x)|tu| dx + R3



2

h(x)|u| dx + R3

1 1 − 4 p+1

1 1 − 4 p+1



 (Q∞ + q(x))|tu|p+1 dx R3

(A.4) p+1

(Q∞ + q(x))|u|

dx.

R3

Combine with (A.3) and (A.4) we get a contradiction. At last we prove vn → 0 as n → ∞. We claim that lim inf G∞ (vn ) ≥ 0. Otherwise, up to subsequence, there exists c < 0 with G∞ (vn ) < c. Set θn vn ∈ N∞ , we have θn < 1 for all n. Then mλ < m∞ ≤ I∞ (θn vn ) =

<

1

vn 2 + 4



≤ mλ + o(1),

1

θn vn 2 + 4 1 1 − 4 p+1





1 1 − 4 p+1

 Q∞ |θn vn |p+1 dx R3

Q∞ |vn |p+1 dx R3

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which is a contradiction. Hence we have   2 2 4 0 = Gλ (un ) = un + λ h(x)|un | dx + b|∇un |2 − (Q∞ + q(x))|un |p+1 dx R3

 = u 2 + λ



h(x)|u|2 dx + b|∇u|42 −

R3

(Q∞ + q(x))|u|p+1 dx R3



+ vn 2 + b|∇vn |42 −

R3

Q∞ |vn |p+1 dx + 2b|∇u|22 |∇vn |22 R3

= Gλ (u) + G∞ (vn ) + 2b|∇u|22 |∇vn |22 . Then Gλ (u) = 0, limn→∞ G∞ (vn ) = limn→∞ |∇vn |2 = 0 and V∞ |vn |22 + o(1) = vn 22 + b|∇vn |22 + o(1) = Q∞ |vn |p+1 p+1 ≤ Q∞ S −

3p−3 4

(5−p)/2

|vn |2

(3p−3)/2

|∇vn |2

,

which implies |vn |2 −−−−→ 0. Thus un → u in H 1 (R3 ) and u is the minimizer of problem (M Pλ ). n→∞

2

Proof of Lemma 5.2. Let un  u ¯ ∈ H 1 , un → u ¯ a.e. in R3 , un → u ¯ ∈ Lqloc (R3 ), q ∈ [1, 6). Set u ˜1n = un − u ¯. By Remark 5 and Brezis–Lieb Lemma we have − (a + bA) Δ¯ u + (V∞ + λh(x))¯ u = (Q∞ + q(x))|¯ u|p−1 u ¯ − (a +

bA) Δ˜ u1n

+

V∞ u ˜1n

=

Q∞ |˜ u1n |p−1 u ˜1n

+ o(1)

where A = limn→∞ |∇un |22 = |∇¯ u|22 + |∇˜ u1n |22 + o(1). Define δ := lim sup

in H −1 ,

in H

−1



n→∞ y∈R3 BR (y)

,

(A.5) (A.6)

|˜ u1n |2 dx. If δ = 0, by Lions’

Lemma we have u ˜1n → 0 in Lp+1 (R3 ). Then by (A.6) we have ˜ u1n 2 ≤ Q∞ |˜ u1n |p+1 p+1 → 0, which implies that 1 1 3 1 un → u ¯ in H . If δ > 0, there exists a sequence {yn } ⊂ R such that |yn | → ∞ and u ˜1n (x + yn1 )  W1 = 0 ˜2n := u ˜1n (x + yn1 ) − W1 , we have in H 1 . Set u |∇˜ u1n |22 = |∇˜ u2n |22 + |∇W1 |22 + o(1). Similarly, W1 and u ˜2n solve equation − (a + bA) ΔW1 + V∞ W1 = Q∞ |W1 |p−1 W1 + o(1) − (a +

bA) Δ˜ u2n

+

V∞ u ˜2n

=

Q∞ |˜ u2n |p−1 u ˜2n

+ o(1)

in H −1 , in H

−1

,

(A.7) (A.8)

where A = |∇¯ u|22 + |∇W1 |22 + limn→∞ |∇˜ u2n |22 . By iterating the above procedure, we construct sequences j {Wj } and {yn }. However, the iteration must terminate at some finite index k. Indeed, Wj satisfies − (¯ a + b|∇Wj |) ΔWj + V∞ Wj = Q∞ Wjp , where a ¯ = a + |∇¯ u|22 +

 i
b|∇Wi |22 , hence I˜∞ (Wj ) > m∞ . On the other hand,

b u|22 |∇˜ Iλ (un ) = I˜λ (un ) = Iλ (¯ u) + I˜∞ (˜ u1n ) + |∇¯ u1n |22 4 b u2n |22 |∇W1 |22 . ≥ Iλ (¯ u) + I˜∞ (W1 ) + |∇˜ 4

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It follows from Iλ (un ) < C that iteration must terminate at finite step. Thus, un (x) − u ¯(x) ∈ H 1 . The proof is finished. 2

k j=1

Wj (x − ynj ) →

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