Soliton Solutions for Quasilinear Schrödinger Equations

Abstract

By using a change of variables, we get new equations, whose respective associated functionals are well defined in 𝐻¹(ℝ^𝑁) and satisfy the geometric hypotheses of the mountain pass theorem. Using this fact, we obtain a nontrivial solution.

1. Introduction

We study the existence of solutions for the following quasilinear Schrödinger equations: −Δ𝑢+𝑉(𝑥)𝑢−Δ1+𝑢^2𝛼/2𝛼𝑢2(1+𝑢²)^(2−𝛼)/2=𝑢^𝑞+𝑢^𝑝,𝑥∈ℝ^𝑁, (1) where 𝑉∈𝐶(ℝ^𝑁,ℝ⁺) is bounded and periodic in each variable of 𝑥_𝑖, 1≤𝑖≤𝑁, 𝑁≥3, 𝑇(𝛼)<𝑞+1<𝑝+1<𝛼2^∗∶=2𝛼𝑁/(𝑁−2), 𝛼≥1, and here 𝑇(𝛼)∶=2𝛼,𝛼₀≤𝛼,2𝛼₀,1<𝛼<𝛼₀,12−4√6,𝛼=1, (2) where 𝛼₀ is defined in Lemma 2. These equations are related to existence of standing wave solutions for quasilinear Schrödinger equations of the form 𝑖𝑧_𝑡=−Δ𝑧+𝑊(𝑥)𝑧−ℎ|𝑧|²𝑧−Δ𝑔|𝑧|²𝑔^′|𝑧|²𝑧,𝑥∈ℝ^𝑁, (3) where 𝑊 is a given potential and 𝑔 and ℎ are real functions. Quasilinear equations such as (3) have been accepted as models of several physical phenomena corresponding to various types of 𝑔. The case of 𝑔(𝑠)=𝑠^𝛼 was used for the superfluid film equation in plasma physics [1]. Besides, (3) also appears in plasma physics and fluid mechanics [2], in dissipative quantum mechanics [3], and in the theory of Heisenberg ferromagnetism and magnons [4, 5]. See also [6, 7] for more physical backgrounds. Equations (3) with 𝛼=1 have been studied extensively recently; see [8, 9]. When 𝑔(𝑠)=(1+𝑠)^𝛼/2, then (3) turn into our equations (1) with ℎ(𝑠)=𝑠^𝑞+𝑠^𝑝. In particular if we let 𝛼=1, that is, 𝑔(𝑠)=(1+𝑠)^1/2, (3) models the self-channeling of a high-power ultrashort laser in matter [10]. In this case, few results are known. In [11], the authors proved global existence and uniqueness of small solutions in transverse space dimensions 2 and 3 and local existence without any smallness condition in transverse space dimension 1. In [12], the authors proved the existence of nontrivial solution. When 𝛼>1, although we do not know the physical background of (3), in a mathematical sense, we give the proof of the existence of nontrivial solution.

For (1), the main difficulty is that the energy functional associated to (1) is not well defined in 𝐻¹(ℝ^𝑁). To overcome this difficulty, enlightened by [8, 9], we give a new change of variables. Then we reduce the quasilinear problem (1) to a semilinear one, which we will prove has a nontrivial solutions.

Our main result is the following.

Theorem 1. Assume that 𝛼≥1 and 𝑇(𝛼)<𝑞+1<𝑝+1<𝛼2^∗. Then (1) has a nontrivial solution.

In this paper, 𝐶 denotes positive (possibly different) constant, 𝐿^𝑝(ℝ^𝑁) denotes the usual Lebesgue space with norm |𝑢|_𝑝=(∫_ℝ𝑁|𝑢|^𝑝𝑑𝑥)^1/𝑝, 1≤𝑝<∞, and 𝐻¹(ℝ^𝑁) denotes the Sobolev space with norm ‖𝑢‖=(∫_ℝ𝑁(|∇𝑢|²+𝑉(𝑥)𝑢²)𝑑𝑥)^1/2.

2. The Change of Variables

We note that the solutions of (1) are the critical points of the following functional: 𝐼(𝑢)=12_ℝ^𝑁𝛼1+²𝑢²21+𝑢^22−𝛼|∇𝑢|²+1𝑑𝑥2_ℝ^𝑁𝑉(𝑥)𝑢²−1𝑑𝑥𝑞+1_ℝ^𝑁𝑢^𝑞+11𝑑𝑥−𝑝+1_ℝ^𝑁𝑢^𝑝+1𝑑𝑥. (4) Since the functional 𝐼(𝑢) may not be well defined in the usual Sobolev spaces 𝐻¹(ℝ^𝑁), we make a change of variables as 𝑣=𝐺(𝑢)=^𝑢₀𝑔(𝑡)𝑑𝑡, (5) where 𝑔(𝑡)=1+𝛼²𝑡²/2(1+𝑡²)^2−𝛼. Since 𝑔(𝑡) is monotonous with |𝑡|, the inverse function 𝐺⁻¹(𝑡) of 𝐺(𝑡) exists. Then after the change of variables, 𝐼(𝑢) can be written by 𝐽(𝑣)=12_ℝ^𝑁|∇𝑣|²1𝑑𝑥+2_ℝ^𝑁𝑉(𝑥)𝐺⁻¹(𝑣)²−1𝑑𝑥𝑞+1_ℝ^𝑁𝐺⁻¹(𝑣)^𝑞+1−1𝑑𝑥𝑝+1_ℝ^𝑁𝐺⁻¹(𝑣)^𝑝+1𝑑𝑥. (6) By Lemma 2 listed below, we have lim_𝑡→0𝐺⁻¹(𝑡)/𝑡=1 and lim_𝑡→∞|𝐺⁻¹(𝑡)|^𝛼/𝑡=√2(𝛼>1) or 2/3(𝛼=1), so 𝐽(𝑣) is well defined in 𝐻¹(ℝ^𝑁) and 𝐽(𝑣)∈𝐶¹.

If 𝑢 is a nontrivial solution of (1), then for all 𝜙∈𝐶^∞₀(ℝ^𝑁) it should satisfy _ℝ𝑁𝑔²(𝑢)∇𝑢∇𝜙+𝑔(𝑢)𝑔^′(𝑢)|∇𝑢|²𝜙+𝑉(𝑥)𝑢𝜙−𝑢^𝑞𝜙−𝑢^𝑝𝜙𝑑𝑥=0. (7)

We show that (7) is equivalent to 𝐽^′(𝑣)𝜓=_ℝ𝑁(𝑥)𝐺∇𝑣∇𝜓+𝑉⁻¹(𝑣)𝑔𝐺⁻¹(𝑣)𝜓−𝐺⁻¹(𝑣)^𝑞𝑔𝐺⁻¹(𝑣)𝐺𝜓−⁻¹(𝑣)^𝑝𝑔𝐺⁻¹(𝑣)𝜓𝑑𝑥=0,∀𝜓∈𝐶^∞₀ℝ^𝑁. (8) Indeed, if we choose 𝜙=(1/𝑔(𝑢))𝜓 in (7), then we get (8). On the other hand, since 𝑢=𝐺⁻¹(𝑣), if we let 𝜓=𝑔(𝑢)𝜙 in (8), we get (7). Therefore, in order to find the nontrivial solutions of (1), it suffices to study the existence of the nontrivial solutions of the following equations: (𝑥)𝐺−Δ𝑣=−𝑉⁻¹(𝑣)𝑔𝐺⁻¹(𝑣)+𝐺⁻¹(𝑣)^𝑞𝑔𝐺⁻¹(𝑣)+𝐺⁻¹(𝑣)^𝑝𝑔𝐺⁻¹(𝑣),𝑥∈ℝ^𝑁. (9)

Before we close this section, we give some properties of the change of variables.

Lemma 2. For all 𝑡>0, one has the following: (1)lim_𝑡→0(𝐺⁻¹(𝑡)/𝑡)=1, (2)(i) if 𝛼>1 then lim_𝑡→∞(|𝐺⁻¹(𝑡)|^𝛼/𝑡)=√2, and (ii) if 𝛼=1 then lim_𝑡→∞(|𝐺⁻¹(𝑡)|/𝑡)=2/3, (3)|𝐺⁻¹(𝑡)|≤𝑡, (4)(i) if 𝛼₀≤𝛼 then 𝑡𝑔^′(𝑡)/𝑔(𝑡)≤𝛼−1, (ii) if 1<𝛼≤𝛼₀ then 𝑡𝑔^′(𝑡)/𝑔(𝑡)≤𝛼₀−1, and (iii) if 𝛼=1 then 𝑡𝑔^′(𝑡)/𝑔(𝑡)≤5−2√6, where 𝛼₀≈1.36 is a real root of the equation 𝛼³−4𝛼²+8𝛼−6=0.

Proof. (1) We easily get lim_𝑡→0(𝐺⁻¹(𝑡)/𝑡)=(𝐺⁻¹(𝑡))^′|_𝑡=0=1/𝑔(𝐺⁻¹(0))=1.
For (2) if 𝛼>1, since 𝑔(𝑡)=1+𝛼²𝑡²/2(1+𝑡²)^2−𝛼=1+(𝛼²𝑡²/2(1+𝑡²))(1+𝑡²)^𝛼−1, so 𝑔(𝑡)∼(𝛼²/2)𝑡^2(𝛼−1)=(𝛼/√2)𝑡^𝛼−1 as 𝑡→∞, then 𝐺(𝑡)=∫^𝑡₀𝑔(𝑠)𝑑𝑠∼(1/√2)𝑡^𝛼 as 𝑡→∞. Since 𝐺⁻¹(𝑡) is the inverse of 𝐺(𝑡), so 𝐺⁻¹(𝑡)∼(√2𝑡)^1/𝛼 as 𝑡→∞, thus we have lim_𝑡→∞(|𝐺⁻¹(𝑡)|^𝛼/𝑡)=√2. When 𝛼=1, the result is obvious since 𝑔(𝑡) is an increasing bounded function.
For (3), since [𝐺⁻¹(𝑡)−(1/𝑔(0))𝑡]^′=1/𝑔(𝐺⁻¹(𝑡))−1/𝑔(0)≤0, so 𝐺⁻¹(𝑡)≤(1/𝑔(0))𝑡=𝑡, which proves (3).
Now we prove (4), since (𝑡/𝑔(𝑡))𝑔^′(𝑡)=𝑡²/2(1+𝑡²)²𝑔²(𝑡)=𝑡²/(2+5𝑡²+3𝑡⁴)=1/(2/𝑡²+5+3𝑡²)≤5−2√6, which is (iii). To prove (i), that is, 𝛼²(2−𝛼)𝑡²≤2(𝛼−1)1+𝑡^23−𝛼, (10) we set 𝑗(𝑡)=2(𝛼−1)(1+𝑡²)^3−𝛼−𝛼²(2−𝛼)𝑡², so 𝑗^′(𝑡)=2𝑡[2(𝛼−1)(3−𝛼)(1+𝑡²)^2−𝛼−𝛼²(2−𝛼)]∶=2𝑡𝑘_𝛼(𝑡), where 𝑘_𝛼(𝑡)=2(𝛼−1)(3−𝛼)(1+𝑡²)^2−𝛼−𝛼²(2−𝛼). Then 𝑘^′_𝛼(𝑡)=4(𝛼−1)(3−𝛼)(2−𝛼)𝑡(1+𝑡²)^1−𝛼. If 𝛼≤2 or 𝛼≥3, we get 𝑘^′_𝛼(𝑡)≥0, so 𝑘_𝛼(𝑡)≥𝑘_𝛼(0). We notice that 𝑘_𝛼(0)=𝛼³−4𝛼²+8𝛼−6 and 𝑘_𝛼(0) is an increasing function with respect to 𝛼. By Cardano's formula for cubic equations, we know that 𝑘_𝛼(0) has one real root and two complex roots. If we set 𝛼₀≈1.36 to be the real root of 𝑘_𝛼(0), then 𝑘_𝛼(𝑡)≥𝑘_𝛼(0)≥0 as 𝛼≥𝛼₀. So 𝑗^′(𝑡)=2𝑡𝑘_𝛼(𝑡)≥0. That is 𝑗(𝑡) is a increasing function, so 𝑗(𝑡)≥𝑗(0)=2(𝛼−1)>0 as 𝛼>1. If 2<𝛼<3, we get 𝑘^′_𝛼(𝑡)<0, so 𝑘_𝛼(𝑡) is a decreasing function, but in this case lim_𝑡→∞𝑘_𝛼(𝑡)=(𝛼−2)𝛼²>0, so 𝑘_𝛼(𝑡)≥0 for all 𝑡>0. Thus we have the same result as 𝛼≤2 or 𝛼≥3, which proves (i). For (ii), by the definition of 𝛼₀, we have 𝛼²₀𝑡²1+𝛼₀−1𝑡²2(1+𝑡²)^3−𝛼0+𝛼²₀𝑡²1+𝑡²≤𝛼₀−1, (11) so 𝛼²₀𝑡²1+𝛼₀−1𝑡²≤2𝛼₀−11+𝑡^23−𝛼0+𝛼²₀𝛼₀−1𝑡²1+𝑡². (12) We add (𝛼₀−1)(𝛼²−𝛼²₀)𝑡²(1+𝑡²) to both sides of (12), where 𝛼<𝛼₀. Then 𝛼²₀𝑡²+𝛼²₀𝛼₀−1𝑡⁴+𝛼₀−1𝛼²−𝛼²₀𝑡²+𝑡⁴≤2𝛼₀−11+𝑡^23−𝛼0+𝛼₀−1𝛼²𝑡²1+𝑡²,∴𝛼²₀+𝛼₀−1𝛼²−𝛼²₀𝑡²+𝛼₀−1𝛼²𝑡⁴≤𝛼₀−121+𝑡^23−𝛼+𝛼²𝑡²1+𝑡². (13) We notice that 𝛼²₀+(𝛼₀−1)(𝛼²−𝛼²₀)≥𝛼². In fact, 𝛼²₀+𝛼₀𝛼²−𝛼³₀−𝛼²+𝛼²₀≥𝛼²⇔2(𝛼²₀−𝛼²)+𝛼₀(𝛼²−𝛼²₀)≥0⇔(𝛼²₀−𝛼²)(2−𝛼₀)>0, and the last inequality is obvious. So 𝛼²𝑡²+𝛼₀−1𝛼²𝑡⁴≤𝛼₀−121+𝑡^23−𝛼+𝛼²𝑡²1+𝑡²∴𝛼²𝑡²+(𝛼−1)𝛼²𝑡⁴≤𝛼₀−121+𝑡^23−𝛼+𝛼²𝑡²1+𝑡², (14)(15) which implies that 𝑡𝑔^′(𝑡)/𝑔(𝑡)≤𝛼₀−1.

3. Mountain Pass Geometry

In this section, we establish the geometric hypotheses of the mountain pass theorem.

Lemma 3. There exist 𝜌₀,𝑎₀>0 such that 𝐽(𝑣)≥𝑎₀ for all ‖𝑣‖=𝜌₀.

Proof. Let 𝑄()1𝑥,𝑡∶=−2𝑉(𝑥)𝐺⁻¹(𝑡)²+1𝐺𝑞+1⁻¹(𝑡)^𝑞+1+1𝐺𝑝+1⁻¹(𝑡)^𝑝+1. (16) Then, by Lemma 2 and 𝑝+1<𝛼2^∗, we have lim_𝑡→0𝑄(𝑥,𝑡)𝑡²=lim_𝑡→0−12𝑉(𝑥)𝐺⁻¹(𝑡)𝑡²+1𝐺𝑞+1⁻¹(𝑡)𝑡²𝐺⁻¹(𝑡)^𝑞−1+1𝐺𝑝+1⁻¹(𝑡)𝑡²𝐺⁻¹(𝑡)^𝑝−11=−2𝑉(𝑥),lim_𝑡→∞𝑄(𝑥,𝑡)𝑡^2∗=lim_𝑡→∞−12𝑉(𝑥𝐺)⁻¹(𝑡)^𝛼𝑡^2/𝛼1𝑡^{2∗−2/𝛼}+1𝐺𝑞+1⁻¹(𝑡)^𝛼𝑡^{(𝑞+1)/𝛼}1𝑡^{2∗−(𝑞+1)/𝛼}+1𝐺𝑝+1⁻¹(𝑡)^𝛼𝑡^{(𝑝+1)/𝛼}1𝑡^{2∗−(𝑝+1)/𝛼}=0. (17) Thus, for 𝜖>0 sufficiently small, there exists a constant 𝐶_𝜖>0 such that 𝑄()≤−1𝑥,𝑡2𝑉(𝑥)+𝜖𝑡²+𝐶_𝜖|𝑡|^2∗. (18) Then, we have 𝐽(𝑣)=12_ℝ^𝑁|∇𝑣|²1𝑑𝑥+2_ℝ^𝑁𝑉(𝑥)𝐺⁻¹(𝑣)²−1𝑑𝑥𝑞+1_ℝ^𝑁𝐺⁻¹(𝑣)^𝑞+11𝑑𝑥−×𝑝+1_ℝ𝑁𝐺⁻¹(𝑣)^𝑝+1≥1𝑑𝑥2_ℝ^𝑁|∇𝑣|²1𝑑𝑥+2_ℝ^𝑁𝑉(𝑥)𝑣²×𝑑𝑥−𝜖_ℝ𝑁𝑣²𝑑𝑥−𝐶_𝜖ℝ𝑁𝑣^2∗‖𝑣‖𝑑𝑥≥𝐶²−𝐶‖𝑣‖^2∗. (19) Thus, by choosing 𝜌₀ small, we get the result when ‖𝑣‖=𝜌₀.

Lemma 4. There exists 𝑣∈𝐻¹(ℝ^𝑁) such that 𝐽(𝑣)<0.

Proof. Given 𝜙∈𝐶^∞₀(ℝ^𝑁,[0,1]) with supp𝜙∶=𝐵₁, we will prove that 𝐽(𝑠𝜙)→−∞ as 𝑠→∞, which will prove the result if we take 𝑣=𝑠𝜙 with 𝑠 large enough. By the proof of Lemma 2, we have 𝐺⁻¹(𝑡)≥𝐶𝑡^1/𝛼 as 𝑡≥1, so 𝐽≤1𝑠𝜙2𝑠²_ℝ𝑁∇𝜙²1𝑑𝑥+2𝑠²×_ℝ𝑁𝑉(𝑥)𝜙²𝑑𝑥−𝑠^{(𝑝+1)/𝛼}×_{{|𝑠𝜙|≥1}}𝜙^{(𝑝+1)/𝛼}𝑑𝑥⟶−∞, (20) as 𝑠→∞. Thus, we get the result.

4. Existence

In consequence of Lemmas 3 and 4 of the Ambrosetti-Rabinowitz mountain pass Theorem [13], see also [14–16], for the constant 𝑐=inf_𝛾∈Γsup_[]𝑡∈0,1𝐽𝛾(𝑡)>0, (21) where Γ={𝛾∈𝐶([0,1],𝐻¹(ℝ^𝑁))∶𝛾(0)=0,𝛾(1)≠0,𝐽(𝛾(1))<0}, and there exists a Palais-Smale sequence at level 𝑐; that is, 𝐽(𝑣_𝑛)→𝑐 and 𝐽^′(𝑣_𝑛)→0 as 𝑛→∞.

Lemma 5. The Palais-Smale sequence {𝑣_𝑛} for 𝐽 is bounded in 𝐻¹(ℝ^𝑁).

Proof. Since {𝑣_𝑛}⊂𝐻¹(ℝ^𝑁) satisfies 𝐽𝑣_𝑛=12_ℝ^𝑁∇𝑣_𝑛²1𝑑𝑥+2_ℝ^𝑁𝑉(𝑥)𝐺⁻¹𝑣_𝑛²−1𝑑𝑥𝑝+1_ℝ^𝑁𝐺⁻¹𝑣_𝑛^𝑝+1−1𝑑𝑥𝑞+1_ℝ^𝑁𝐺⁻¹𝑣_𝑛^𝑞+1𝑑𝑥=𝑐+𝑜(1) (22) and for any 𝜓∈𝐶^∞₀(ℝ^𝑁), 𝐽^′𝑣_𝑛𝜓=_ℝ𝑁∇𝑣_𝑛(𝑥)𝐺∇𝜓+𝑉⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛𝜓−𝐺⁻¹𝑣_𝑛^𝑞𝑔𝐺⁻¹𝑣_𝑛𝐺𝜓−⁻¹𝑣_𝑛^𝑝𝑔𝐺⁻¹𝑣_𝑛𝜓(1𝜓.𝑑𝑥=𝑜) (23) Now, we consider the function 𝐺⁻¹(𝑣_𝑛)𝑔(𝐺⁻¹(𝑣_𝑛)). Note by Lemma 2 that ∇𝐺⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛=𝐺1+⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛𝑔^′𝐺⁻¹𝑣_𝑛∇𝑣_𝑛≤12𝑇(𝛼)∇𝑣_𝑛. (24) Combining Lemma 2, we have 𝐺⁻¹(𝑣_𝑛)𝑔(𝐺⁻¹(𝑣_𝑛))∈𝐻¹(ℝ^𝑁). Thus, since 𝐶^∞₀(ℝ^𝑁) is dense in 𝐻¹(ℝ^𝑁), by choosing 𝜓=𝐺⁻¹(𝑣_𝑛)𝑔(𝐺⁻¹(𝑣_𝑛)) in (23), we deduce that 𝑜(1)𝑣_𝑛=𝐽^′𝑣_𝑛𝐺⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛=_ℝ𝑁𝐺1+⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛𝑔^′𝐺⁻¹𝑣_𝑛∇𝑣_𝑛²+𝑉(𝑥)𝐺⁻¹𝑣_𝑛²−𝐺⁻¹𝑣_𝑛^𝑞+1−𝐺⁻¹𝑣_𝑛^𝑝+1≤𝑑𝑥_ℝ𝑁12𝑇(𝛼)∇𝑣_𝑛²+𝑉(𝑥)𝐺⁻¹𝑣_𝑛²−𝐺⁻¹𝑣_𝑛^𝑞+1−𝐺⁻¹𝑣_𝑛^𝑝+1𝑑𝑥. (25) Therefore, by (22) and (25), we have 𝑞+1𝐽𝑣_𝑛−𝐽^′𝑣_𝑛𝐺⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛≥𝑞+12−12𝑇(𝛼)×_ℝ𝑁∇𝑣_𝑛²𝑑𝑥+𝑞−12_ℝ^𝑁𝑉(𝑥)𝐺⁻¹𝑣_𝑛²+𝑑𝑥1−𝑞+1𝑝+1_ℝ^𝑁𝐺⁻¹𝑣_𝑛^𝑝+1≥𝑑𝑥𝑞+12−12𝑇(𝛼)_ℝ𝑁∇𝑣_𝑛²𝑑𝑥+𝑞−12×_ℝ𝑁𝑉(𝑥)𝐺⁻¹𝑣_𝑛²𝑑𝑥. (26) Combining (22) and (26), we get ∫_ℝ𝑁|𝐺⁻¹(𝑣_𝑛)|^𝑝+1𝑑𝑥 is bounded. To verify that {𝑣_𝑛} is bounded in 𝐻¹(ℝ^𝑁) we start splitting _ℝ𝑁𝑉(𝑥)𝑣²_𝑛𝑑𝑥=_{{𝑥∶|𝑣𝑛(𝑥)|>1}}𝑉(𝑥)𝑣²_𝑛+𝑑𝑥_{{𝑥∶|𝑣𝑛(𝑥)|≤1}}𝑉(𝑥)𝑣²_𝑛𝑑𝑥. (27) By the proof of Lemma 2, we have 𝐺⁻¹(𝑡)≥𝐶𝑡^1/𝛼, for all 𝑡>1 and 𝐺(1)=∫¹₀𝑔(𝑡)𝑑𝑡≥∫¹₀𝑑𝑡=1. Therefore _{{𝑥∶|𝑣𝑛(𝑥)|>1}}𝑉(𝑥)𝑣²_𝑛𝑑𝑥≤𝐶_{{𝑥∶|𝑣𝑛(𝑥)|>1}}𝐺⁻¹𝑣_𝑛^2𝛼𝑑𝑥≤𝐶_{{𝑥∶|𝑣𝑛(𝑥)|>1}}𝐺⁻¹𝑣_𝑛^𝑝+1𝑑𝑥≤𝐶. (28) Since 𝑔(𝑡) is increasing and 𝐺(𝑡)=∫^𝑡₀𝑔(𝑠)𝑑𝑠≤𝑔(𝑡)𝑡, we have _{{𝑥∶|𝑣𝑛(𝑥)|≤1}}𝑉(𝑥)𝐺⁻¹𝑣_𝑛²≥1𝑑𝑥𝑔²𝐺⁻¹(1)^{{𝑥∶|𝑣}_𝑛^(𝑥)|≤1}𝑉(𝑥)𝑣²_𝑛𝑑𝑥. (29) Hence {𝑣_𝑛} is bounded in 𝐻¹(ℝ^𝑁), and this proves Lemma 5.

Now we give the completion of the proof of Theorem 1.

Proof. First, we will prove that 𝐽^′(𝑣)=0. That is, 𝑣 is a weak solution of (9). To prove this, it suffices to show that 𝐽^′(𝑣)𝜓=_ℝ𝑁(𝑥)𝐺∇𝑣∇𝜓+𝑉⁻¹(𝑣)𝑔𝐺⁻¹(𝑣)𝜓−𝐺⁻¹(𝑣)^𝑞𝑔𝐺⁻¹(𝑣)𝐺𝜓−⁻¹(𝑣)^𝑝𝑔𝐺⁻¹(𝑣)𝜓𝑑𝑥=0,∀𝜓∈𝐶^∞₀ℝ^𝑁. (30) From Lemma 5, {𝑣_𝑛} is a bounded Palais-Smale sequence, and there exists 𝑣∈𝐻¹(ℝ^𝑁) such that 𝑣_𝑛⇀𝑣 weakly in 𝐻¹(ℝ^𝑁). By the Lebesgue dominated theorem, we have 𝐽^′𝑣_𝑛𝜓−𝐽^′(𝑣)𝜓=_ℝ𝑁∇𝑣_𝑛+−∇𝑣∇𝜓𝑑𝑥_ℝ𝑁𝑉(𝑥𝐺)⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛−𝐺⁻¹(𝑣)𝑔𝐺⁻¹(𝑣)−𝜓𝑑𝑥_ℝ𝑁𝐺⁻¹𝑣_𝑛^𝑞𝑔𝐺⁻¹𝑣_𝑛−𝐺⁻¹(𝑣)^𝑞𝑔𝐺⁻¹(𝑣)−𝜓𝑑𝑥_ℝ𝑁𝐺⁻¹𝑣_𝑛^𝑝𝑔𝐺⁻¹𝑣_𝑛−𝐺⁻¹(𝑣)^𝑝𝑔𝐺⁻¹(𝑣)𝜓𝑑𝑥⟶0. (31) Hence, 𝐽^′(𝑣)=0. That is, 𝑣 is a weak solution of (1).
Next, in order to complete the proof of Theorem 1, we must show that 𝑣 is nontrivial. By contradiction, we assume 𝑣=0. To prove this, we claim that, for all 𝑅>0, lim_𝑛→∞sup_{𝑦∈ℝ𝑁𝐵𝑅(𝑦)}𝑣²_𝑛𝑑𝑥=0 (32) cannot occur. Suppose by contradiction that (32) occurs; that is, {𝑣_𝑛} vanishes. Then by the Lions compactness lemma [16], 𝑣_𝑛→0 in 𝐿^𝑟(ℝ^𝑁) for any 𝑟∈(2,2^∗). By the proof of Lemma 2, we get 𝐺⁻¹(𝑡)∶(√2𝑡)^1/𝛼 as 𝑡→∞, so there exists a suitable constant 𝐶 such that 𝐺⁻¹(𝑡)≤𝐶𝑡^1/𝛼. In addition, since 𝐺(𝑡)≤𝑔(𝑡)𝑡, we have lim_{𝑛→∞ℝ𝑁}𝐺⁻¹𝑣_𝑛^𝑝𝑔𝐺⁻¹𝑣_𝑛𝑣_𝑛𝑑𝑥≤lim_{𝑛→∞ℝ𝑁}𝐺⁻¹𝑣_𝑛^𝑝+1𝑑𝑥≤lim_𝑛→∞𝐶_ℝ𝑁𝑣_𝑛^{(𝑝+1)/𝛼}𝑑𝑥=0,lim_{𝑛→∞ℝ𝑁}𝐺⁻¹𝑣_𝑛^𝑞+1𝑑𝑥=lim_{𝑛→∞ℝ𝑁}𝐺⁻¹𝑣_𝑛^𝑝+1𝑑𝑥=0, (33) and lim_𝑛→∞∫_ℝ𝑁(|𝐺⁻¹(𝑣_𝑛)|^𝑞/𝑔(𝐺⁻¹(𝑣_𝑛)))𝑣_𝑛𝑑𝑥=0 is obvious since 𝑞<𝑝, which implies that 0=lim_𝑛→∞𝐽^′𝑣_𝑛𝑣_𝑛=lim_{𝑛→∞ℝ𝑁}∇𝑣_𝑛²(𝑥)𝐺+𝑉⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛𝑣_𝑛−𝐺⁻¹𝑣_𝑛^𝑞𝑔𝐺⁻¹𝑣_𝑛𝑣_𝑛−𝐺⁻¹𝑣_𝑛^𝑝𝑔𝐺⁻¹𝑣_𝑛𝑣_𝑛𝑑𝑥=lim_{𝑛→∞ℝ𝑁}∇𝑣_𝑛²(𝑥)𝐺+𝑉⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛𝑣_𝑛𝑑𝑥. (34) Then, lim_{𝑛→∞ℝ𝑁}∇𝑣_𝑛²𝑑𝑥=0,lim_{𝑛→∞ℝ𝑁}𝑉(𝑥)𝐺⁻¹𝑣_𝑛𝑔𝐺⁻¹𝑣_𝑛𝑣_𝑛𝑑𝑥=0. (35)(36) On the other hand, by (25), we have lim_{𝑛→∞ℝ𝑁}𝑉(𝑥)𝐺⁻¹𝑣_𝑛²𝑑𝑥=0. (37) Combining (35) and (37), we get a contradiction since 𝐽(𝑣_𝑛)→𝑐>0. Thus, {𝑣_𝑛} does not vanish and there exist 𝑘,𝑅>0 and {𝑦_𝑛}⊂ℝ^𝑁 such that lim_{𝑛→∞𝐵𝑅(𝑦𝑛)}𝑣²_𝑛𝑑𝑥≥𝑘>0. (38) Define 𝑣_𝑛̃(𝑥)=𝑣_𝑛(𝑥+𝑦_𝑛). We may assume that the components of {𝑦_𝑛} are integer multiples of the periods of 𝑉(𝑥). Since {𝑣_𝑛} is a Palais-Smale sequence for 𝐽 and 𝑉(𝑥) is periodic in 𝑥_𝑖, 1≤𝑖≤𝑁, {𝑣_𝑛̃} is also a Palais-Smale sequence for 𝐽 with 𝐽^′(𝑣)̃=0 if 𝑣_𝑛̃⇀𝑣̃ in 𝐻¹(ℝ^𝑁). Since {𝑣_𝑛̃} does not vanish, we have that 𝑣̃≠0 is a nontrivial solution of (9).

Acknowledgment

The research is supported by NSF of China (11201154).