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Volume 9, Issue 1

# Berestycki-Lions conditions on ground state solutions for a Nonlinear Schrödinger equation with variable potentials

Sitong Chen
/ Xianhua Tang
• Corresponding author
• School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083, P.R.China
• Email
• Other articles by this author:
Published Online: 2019-06-16 | DOI: https://doi.org/10.1515/anona-2020-0011

## Abstract

This paper is dedicated to studying the nonlinear Schrödinger equations of the form

$−△u+V(x)u=f(u),x∈RN;u∈H1(RN),$

where V ∈ 𝓒1(ℝN, [0, ∞)) satisfies some weak assumptions, and f ∈ 𝓒(ℝ, ℝ) satisfies the general Berestycki-Lions assumptions. By introducing some new tricks, we prove that the above problem admits a ground state solution of Pohožaev type and a least energy solution. These results generalize and improve some ones in [L. Jeanjean, K. Tanka, Indiana Univ. Math. J. 54 (2005), 443-464], [L. Jeanjean, K. Tanka, Proc. Amer. Math. Soc. 131 (2003) 2399-2408], [H. Berestycki, P.L. Lions, Arch. Rational Mech. Anal. 82 (1983) 313-345] and some other related literature. In particular, our assumptions are “almost” necessary when V(x) ≡ V > 0, moreover, our approach could be useful for the study of other problems where radial symmetry of bounded sequence either fails or is not readily available, or where the ground state solutions of the problem at infinity are not sign definite.

MSC 2010: 35J20; 35J65

## 1 Introduction

In this paper, we consider the nonlinear Schrödinger equations of the form:

$−△u+V(x)u=f(u),x∈RN;u∈H1(RN),$(1.1)

where N ≥ 3, V: ℝN → ℝ and f : ℝ → ℝ satisfy the following basic assumptions:

• (V1)

V ∈ 𝓒(ℝN, [0, ∞));

• (V2)

V(x) ≤ V := lim|y|→∞V(y) for all x ∈ ℝN;

• (F1)

f ∈ 𝓒(ℝ, ℝ) and there exists a constant 𝓒0 > 0 such that

$|f(t)|≤C01+|t|2∗−1,∀t∈R;$

• (F2)

f(t) = o(t) as t → 0 and |f(t)| = o (|t|(N+2)/(N−2)) as |t| → +∞.

Clearly, under (V1), (V2), (F1) and (F2), the weak solutions of (1.1) correspond to the critical points of the energy functional defined in H1(ℝN) by

$I(u)=12∫RN|∇u|2+V(x)u2dx−∫RNF(u)dx,$(1.2)

where $\begin{array}{}F\left(t\right):={\int }_{0}^{t}f\left(s\right)\mathrm{d}s\end{array}$.

If the potential V (x) ≡ V, then (1.1) reduces to the following autonomous form:

$−△u+V∞u=f(u),x∈RN;u∈H1(RN),$(1.3)

its energy functional is as follows:

$I∞(u)=12∫RN|∇u|2+V∞u2dx−∫RNF(u)dx.$(1.4)

It is well known that every solution u(x) of (1.3) satisfies the following Pohožaev type identity [9]:

$P∞(u):=N−22∥∇u∥22+NV∞2∥u∥22−N∫RNF(u)dx=0.$(1.5)

Let

$M∞:=u∈H1(RN)∖{0}:P∞(u)=0.$(1.6)

Berestycki-Lions [1] proved that (1.3) has a radially symmetric positive solution provided f satisfies (F1), (F2) and the following two assumptions:

• (F0)

f is odd;

• (F3)

there exists s0 > 0 such that F(s0) > $\begin{array}{}\frac{1}{2}{V}_{\mathrm{\infty }}{s}_{0}^{2}\end{array}$.

To prove the above result, Berestycki-Lions [1] considered the following constrained minimization problem

$min∥∇u∥22:u∈S,$(1.7)

where

$S=u∈H1(RN):∫RNF(u)−12V∞u2dx=1;$(1.8)

they first showed that by the Pólya-Szegö inequality for the Schwarz symmetrization, the minimum can be taken on radial and radially nonincreasing functions. Then they showed the existence of a minimizer ŵH1(ℝN) by the direct method of the calculus of variations. With the Lagrange multiplier Theorem, they concluded that ū(x) := ŵ(x/tŵ) with $\begin{array}{}{t}_{\stackrel{^}{w}}=\sqrt{\frac{N-2}{2N}}\parallel \mathrm{\nabla }\stackrel{^}{w}{\parallel }_{2}\end{array}$ is a least energy solution of (1.3). By noting the one-to-one correspondence between 𝓢 and 𝓜, Jeanjean-Tanaka [6] proved that ū is also a ground state solution of Pohožaev type for (1.3), i.e. ū ∈ 𝓜 and satisfies

$I∞(u¯)=infM∞I∞.$(1.9)

By using a different way, Shatah [12] showed that there exists ũ$\begin{array}{}{\mathcal{M}}_{r}^{\mathrm{\infty }}\end{array}$ such that

$I∞(u~)=infMr∞I∞,$(1.10)

where

$Mr∞:=u∈Hr1(RN)∖{0}:P∞(u)=0$

and

$Hr1(RN)=u∈H1(RN):u is radially symmetric function onRN.$

Obviously, (F1)-(F3) are satisfied by a very wide class of nonlinearities. In particular only conditions on f(t) near 0, ∞ and the point s0 are required. Moreover, in view of [1, 2.2], (F1) is “almost” necessary, and (F2) and (F3) are necessary for the existence of a nontrivial solution of problem (1.3). For more related results under Berestycki-Lions conditions, we refer to [3, 4, 11, 18, 19].

When V(x) ≢ V, the approach used in [1] does not work any more for nonautonomous equation (1.1), since the Schwarz symmetrization can only be applied to autonomous problems. In a different way, Rabinowitz [10] proved that (1.1) has a nontrivial solution if V satisfies (V1) and (V2) and f does (F1), (F2), the Nehari monotonic condition:

• (Ne)

f(t)/|t| is strictly increasing on (− ∞, 0) ∪ (0, ∞);

and the global growth Ambrosetti-Rabinowitz condition:

• (AR)

there exists μ > 2 such that f(t)t ≥ μ F(t) > 0, ∀ t ∈ ℝ ∖ {0}.

(Ne) and (AR) are used to recover the compactness and to get the boundedness of Palais-Smale sequences, respectively, see [17] for more details. By means of Jeanjean’s monotonicity trick, developed in [5], which is a generalization of the Struwe’s one (see [13]), consisting in a suitable approximating method, Jeanjean and Tanaka [7] derived an existence result using two weaker conditions instead of (Ne) and (AR). More precisely, Jeanjean and Tanaka proved that (1.1) has a least energy solution if f satisfies (F1), (F2) and the following nonnegativity condition:

• (NG)

f(t) ≥ 0 for t ≥ 0;

and the superlinear growth condition:

• (SL)

$\begin{array}{}\underset{t\to +\mathrm{\infty }}{lim}\frac{f\left(t\right)}{t}=\mathrm{\infty };\end{array}$

and V does (V1), (V2) and the decay condition:

• (Vd)

V ∈ 𝓒1(ℝN, ℝ) and there exists φL2(ℝN) ∩ W1,∞(ℝN) such that

$|∇V(x)||x|≤[φ(x)]2,∀x∈RN.$

Clearly, (NG) and (SL) are stronger than (F3), moreover, (Vd) puts relatively strict constrains on the decay of |∇ V(x)|. For example, V(x) = a$\begin{array}{}\frac{b}{1+|x{|}^{\alpha }}\end{array}$ does not satisfy (Vd) for a, b > 0 and 0 < αN.

Motivated by the work of [1, 2, 7, 12, 16], we shall develop a more direct approach (the least energy squeeze approach) to show that (1.1) has a solution ū ∈ 𝓜 such that I(ū) = inf𝓜I under (F1)-(F3), (V1), (V2) and an additional decay condition on V:

• (V3)

V ∈ 𝓒1(ℝN, ℝ) and there exists θ ∈ [0, 1) such that tNV(tx) + ∇ V(tx) ⋅ (tx) + $\begin{array}{}\frac{\theta \left(N-2{\right)}^{3}}{4{t}^{2}|x{|}^{2}}\end{array}$ is nonincreasing on (0, ∞) for all x ∈ ℝN ∖ {0},

where

$M:=u∈H1(RN)∖{0}:P(u)=0$(1.11)

and

$P(u):=N−22∥∇u∥22+12∫RN[NV(x)+∇V(x)⋅x]u2dx−N∫RNF(u)dx$(1.12)

is the Pohožaev functional associated with (1.1) (see [7]).

To prove the above conclusion, we shall divide our arguments into three steps: i). Choosing a minimizing sequence {un} of I on 𝓜, which satisfies

$I(un)→m:=infMI,P(un)=0.$(1.13)

Then showing that {un} is bounded in H1(ℝN). With a concentration-compactness argument, showing that {un} converges to some ūH1(ℝN) ∖ {0} up to translations and extraction of a subsequence. ii). Showing that ū ∈ 𝓜 and I(ū) = inf𝓜I. iii). Showing that ū is a critical point of I. Of them, Step ii) is the most difficult due to lack of global compactness and adequate information on I′(un). Since (1.1) is nonautonomous, the radial compactness does not work for 𝓜. To overcome this difficulty, we establish a crucial inequality related to I(u), I(ut) and 𝓟(u) (the IIP inequality in short, see Lemma 2.2), where ut(x) = u(x/t), it plays an important role in many places of this paper. With the help of the IIP inequality, we then can complete Step ii) by using Lions’ concentration compactness, the least energy squeeze approach and some subtle analysis. In particular, we only use Lions’ concentration compactness in our arguments, the radial and other compactness are not required, see the proofs of Lemmas 2.12 and 2.13. Moreover, such an approach could be useful for the study of other problems where radial symmetry of bounded sequence either fails or is not readily available. In Step iii), usually, one uses the Lagrange multipliers Theorem to show that the minimizer ū is a critical point of I, but it is impossible to verify 𝓟′(u) ≠ 0 for all u ∈ 𝓜 under (V1)-(V3) and (F1)-(F3). To overcome this difficulty, we employ the combination of the IIP inequality and a deformation lemma, see Lemma 2.14. We believe that our approach could be applied to deal with many similar problems, such as quasilinear Schrödinger equations, Choquard equations and fractional Schrödinger equations.

#### Remark 1.1

There are indeed functions which satisfy (V1)-(V3). An example is given by V(x) = V1$\begin{array}{}\frac{A}{|x{|}^{2}+1}\end{array}$, where V1A and 0 < A < (N − 2)3/2(2N + 1) are two positive constants.

We are now in a position to state the main results on ground state solutions of Pohožaev type.

#### Theorem 1.2

Assume that V and f satisfy (V1)-(V3) and (F1)-(F3). Then problem (1.1) has a solution ūH1(ℝN) such that I(ū) = inf𝓜I = infuΛ maxt>0 I(ut) > 0, where

$ut(x):=u(x/t)andΛ=u∈H1(RN):∫RN12V∞u2−F(u)dx<0.$

As a consequence of Theorem 1.2, we can prove the following theorem.

#### Theorem 1.3

Assume that f satisfies (F1)-(F3). Then problem (1.3) has a solution ūH1(ℝN) such that I(ū) = inf𝓜 I = infuΛ maxt>0 I(ut) > 0.

#### Remark 1.4

We point out that, as a consequence of Theorem 1.2, the least energy value m := inf𝓜I has a minimax characterization m = infuΛ maxt>0I(ut) which is much simpler than the usual characterizations related to the Mountain Pass level.

In the second part of the paper, we are interested in the existence of the least energy solutions for (1.1) under (F1)-(F3). In this case, we can replace (V3) by the following weaker decay assumption on ∇ V:

• (V4)

V ∈ 𝓒1(ℝN, ℝ) and there exists θ ∈ [0, 1) such that

$∇V(x)⋅x≤(N−2)2θ2|x|2,∀x∈RN∖{0}.$

As in Jeanjean and Tanaka [6], for λ ∈ [1/2, 1] we consider a family of functionals Iλ : H1(ℝN) → ℝ defined by

$Iλ(u)=12∫RN|∇u|2+V(x)u2dx−λ∫RNF(u)dx.$(1.14)

These functionals have a Mountain Pass geometry, and denoting the corresponding Mountain Pass levels by cλ. orresponding to (1.14), we also let

$Iλ∞(u)=12∫RN|∇u|2+V∞u2dx−λ∫RNF(u)dx.$(1.15)

By Theorem 1.3, for every λ ∈ [1/2, 1], there exists a minimizer $\begin{array}{}{u}_{\lambda }^{\mathrm{\infty }}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{I}_{\lambda }^{\mathrm{\infty }}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}{\mathcal{M}}_{\lambda }^{\mathrm{\infty }}\end{array}$, where

$Mλ∞:=u∈H1(RN)∖{0}:Pλ∞(u)=0$(1.16)

and

$Pλ∞(u)=N−22∥∇u∥22+NV∞2∥u∥22−Nλ∫RNF(u)dx.$(1.17)

Let

$A(u)=12∫RN|∇u|2+V(x)u2dx,B(u)=12∫RNF(u)dx.$

Then Iλ(u) = A(u) − λ B(u). Since B(u) is not sign definite, it prevents us from employing Jeanjean’s monotonicity trick [5]. Thanks to the work of Jeanjean-Toland [8], Iλ still has a bounded (PS)-sequence {un(λ)} ⊂ H1(ℝN) at level cλ for almost every λ ∈ [1/2, 1]. Different from the arguments in the existing literature, by means of $\begin{array}{}{u}_{1}^{\mathrm{\infty }}\end{array}$ and the IIP inequality, we can find λ̄ ∈ [1/2, 1) and then directly prove the following crucial inequality

$cλ(1.18)

which is used to recover the compactness to (PS)-sequence {un(λ)}, see Lemmas 3.5 and 3.7. In particular, it is not required any information on sign of $\begin{array}{}{u}_{1}^{\mathrm{\infty }}\end{array}$ in our arguments. Applying (1.18) and a precise decomposition of bounded (PS)-sequence in [6], we can get a nontrivial critical point uλ of Iλ which possesses energy cλ for almost every λ ∈ (λ̄, 1]. Finally, with a Pohožaev identity we proved that (1.1) admits a least energy solution under (V1), (V2), (V4) and (F1)-(F3). More precisely, we have the following theorem.

#### Theorem 1.5

Assume that V and f satisfy (V1), (V2), (V4) and (F1)-(F3). Then problem (1.1) has a least energy solution.

#### Remark 1.6

Relative to (Vd), there seem to be more functions satisfying (V4). For example, it is easy to verify that V(x) = a$\begin{array}{}\frac{b}{1+|x{|}^{\alpha }}\end{array}$ satisfies (V4) for α ≥ 2, a > 0 and

$4b<(N−2)2ifα=2;(α−2)(α−2)/α(α+2)(α+2)/α2αb<(N−2)2,ifα>2.$

However, it does not satisfy (Vd) when 2 ≤ αN.

Applying Theorem 1.5 to the following perturbed problem:

$−△u+[V∞−εh(x)]u=f(u),x∈RN;u∈H1(RN),$(1.19)

where V is a positive constant and the function h ∈ 𝓒1(ℝN, ℝ) verifies:

• (H1)

h(x) ≥ 0 for all x ∈ ℝN and lim|x|→∞h(x) = 0;

• (H2)

supx∈ℝN [−|x|2(∇ h(x), x)] < ∞.

Then we have the following corollary.

#### Corollary 1.7

Assume that h and f satisfy (H1), (H2) and (F1)-(F3). Then there exists a constant ε0 > 0 such that problem (1.19) has a least energy solution ūεH1(ℝN) ∖ {0} for all 0 < εε0.

Classically, in order to show the existence of solutions for (1.1), one compares the critical level of I with the one of I (i.e. the energy functional corresponds to the problem at infinity). To this end, it is necessary to establish a strict inequality similar to

$maxt∈[0,1]I(y0(t))

for some path y0 ∈ 𝓒([0, 1], H1(ℝN)). Clearly, y0(t) > 0 is a natural requirement under (V1) and (V2). But we only need y0(t) ≠ 0 in our arguments. Therefore, our approach could be useful for the study of other problems where paths or the ground state solutions of the problem at infinity are not sign definite.

Throughout the paper we make use of the following notations:

• H1(ℝN) denotes the usual Sobolev space equipped with the inner product and norm

$(u,v)=∫RN(∇u⋅∇v+uv)dx,∥u∥=(u,u)1/2,∀u,v∈H1(RN);$

• Ls(ℝN) (1 ≤ s < ∞) denotes the Lebesgue space with the norm ∥us = (∫N|u|sdx)1/s;

• For any uH1(ℝN) ∖ {0}, ut(x) := u(x/t) for t > 0;

• For any x ∈ ℝN and r > 0, Br(x) := {y ∈ ℝN : |yx| < r};

• C1, C2, ⋅ s denote positive constants possibly different in different places.

The rest of the paper is organized as follows. In Section 2, we give some preliminaries, and give the proof of Theorem 1.2. Section 3 is devoted to finding a least energy solution for (1.1) and Theorem 1.5 will be proved in this section.

## 2 Ground state solutions for (1.1)

In this section, we give the proofs of Theorem 1.2. To this end, we give some useful lemmas. Since V(x) ≡ V satisfies (V1)-(V3), thus all conclusions on I are also true for I. For (1.3), we always assume that V > 0. By a simple calculation, we can verify Lemma 2.1.

#### Lemma 2.1

The following inequality holds:

$g(t):=2−NtN−2+(N−2)tN>g(1)=0,∀t∈[0,1)∪(1,+∞).$(2.1)

Moreover (V3) implies the following inequality holds:

$NtNV(x)−V(tx)+tN−1∇V(x)⋅x≥−(N−2)2θ2−NtN−2+(N−2)tN4|x|2,∀t≥0,x∈RN∖{0}.$(2.2)

#### Lemma 2.2

Assume that (V1), (V3), (F1) and (F2) hold. Then

$I(u)≥I(ut)+1−tNNP(u)+(1−θ)2−NtN−2+(N−2)tN2N∥∇u∥22,∀u∈H1(RN),t>0.$(2.3)

#### Proof

According to Hardy inequality, we have

$∥∇u∥22≥(N−2)24∫RNu2|x|2dx,∀u∈H1(RN).$(2.4)

Note that

$I(ut)=tN−22∥∇u∥22+tN2∫RNV(tx)u2dx−tN∫RNF(u)dx.$(2.5)

Thus, by (1.2), (1.12), (2.1), (2.2), (2.4) and (2.5), one has

$I(u)−I(ut)=1−tN−22∥∇u∥22+12∫RNV(x)−tNV(tx)u2dx−1−tN∫RNF(u)dx=1−tNNN−22∥∇u∥22+12∫RN[NV(x)+∇V(x)⋅x]u2dx−N∫RNF(u)dx+2−NtN−2+(N−2)tN2N∥∇u∥22+12∫RNtN[V(x)−V(tx)]−1−tNN∇V(x)⋅xu2dx≥1−tNNP(u)+(1−θ)2−NtN−2+(N−2)tN2N∥∇u∥22,∀u∈H1(RN),t>0.$

This shows that (2.3) holds. □

From Lemma 2.2, we have the following two corollaries.

#### Corollary 2.3

Assume that (F1) and (F2) hold. Then

$I∞(u)=I∞(ut)+1−tNNP∞(u)+2−NtN−2+(N−2)tN2N∥∇u∥22,∀u∈H1(RN),t>0.$(2.6)

#### Corollary 2.4

Assume that (V1), (V3), (F1) and (F2) hold. Then for u ∈ 𝓜

$I(u)=maxt>0I(ut).$(2.7)

#### Lemma 2.5

Assume that (V1)-(V3) hold. Then there exist two constants y1, y2 > 0 such that

$y1∥u∥2≤(N−2)∥∇u∥22+∫RNNV(x)+∇V(x)⋅xu2dx≤y2∥u∥2,∀u∈H1(RN).$(2.8)

#### Proof

Let t = 0 and t → ∞ in (2.2), respectively, and using (V2), one has

$−(N−2)3θ4|x|2+NV∞≤NV(x)+∇V(x)⋅x≤NV∞+(N−2)2θ2|x|2,∀x∈RN∖{0}.$(2.9)

Thus it follows from (2.4) and (2.9) that

$(N−2)∥∇u∥22+∫RNNV(x)+∇V(x)⋅xu2dx≤(N−2+2θ)∥∇u∥22+NV∞∥u∥22≤[N−2+2θ+NV∞]∥u∥2:=y2∥u∥2,∀u∈H1(RN)$(2.10)

and

$(N−2)∥∇u∥22+∫RNNV(x)+∇V(x)⋅xu2dx≥(1−θ)(N−2)∥∇u∥22+NV∞∥u∥22≥min(1−θ)(N−2),NV∞∥u∥2:=y1∥u∥2,∀u∈H1(RN).$(2.11)

Both (2.10) and (2.11) imply that (2.8) holds. □

To show 𝓜 ≠ ∅, we define a set Λ as follows:

$Λ=u∈H1(RN):∫RN12V∞u2−F(u)dx<0.$(2.12)

#### Lemma 2.6

Assume that (V1)-(V3) and (F1)-(F3) hold. Then Λ ≠∅ and

$u∈H1(RN)∖{0}:P∞(u)≤0orP(u)≤0⊂Λ.$(2.13)

#### Proof

In view of the proof of [1, Theorem 2], (F3) implies Λ ≠∅. Next, we have two cases to distinguish:

1. uH1(ℝN) ∖ {0} and 𝓟(u) ≤ 0, then (1.5) implies uΛ.

2. uH1(ℝN) ∖ {0} and 𝓟(u) ≤ 0, then it follows from (1.12), (2.4) and (2.9) that

$N∫RN12V∞u2−F(u)dx=P(u)−N−22∥∇u∥22−N2∫RN(V(x)−V∞)+∇V(x)⋅xNu2dx≤−N−22∥∇u∥22+(N−2)3θ8∫RNu2|x|2dx≤−(1−θ)(N−2)2∥∇u∥22<0,$

which implies uΛ. □

#### Lemma 2.7

Assume that (V1)-(V3) and (F1)-(F3) hold. Then for any uΛ, there exists a unique tu > 0 such that utu ∈ 𝓜.

#### Proof

Let uΛ be fixed and define a function ζ(t) := I(ut) on (0, ∞). Clearly, by (1.12) and (2.5), we have

$ζ′(t)=0⇔N−22tN−2∥∇u∥22+tN2∫RN[NV(tx)+∇V(tx)⋅(tx)]u2dx−NtN∫RNF(u)dx=0⇔P(ut)=0⇔ut∈M.$(2.14)

It is easy to verify, using (V1), (V2), (F1), (2.5) and the definition of Λ, that limt→0 ζ(t) = 0, ζ(t) > 0 for t > 0 small and ζ(t) < 0 for t large. Therefore maxt∈[0, ∞)ζ(t) is achieved at some tu > 0 so that ζ′(tu) = 0 and utu ∈ 𝓜.

Next we claim that tu is unique for any uΛ. In fact, for any given uΛ, let t1, t2 > 0 such that ut1, ut2 ∈ 𝓜. Then 𝓟 (ut1) = 𝓟 (ut2) = 0. Jointly with (2.3), we have

$Iut1≥Iut2+t1N−t2NNt1NPut1+(1−θ)2t1N−Nt12t2N−2+(N−2)t2N2Nt1N∥∇ut1∥22=Iut2+(1−θ)2t1N−Nt12t2N−2+(N−2)t2N2Nt12∥∇u∥22$(2.15)

and

$Iut2≥Iut1+t2N−t1NNt2NPut2+(1−θ)2t2N−Nt22t1N−2+(N−2)t1N2Nt2N∥∇ut2∥22=Iut1+(1−θ)2t2N−Nt22t1N−2+(N−2)t1N2Nt22∥∇u∥22.$(2.16)

(2.15) and (2.16) imply t1 = t2. Therefore, tu > 0 is unique for any uΛ. □

#### Corollary 2.8

Assume that (F1)-(F3) hold. Then for any uΛ, there exists a unique tu > 0 such that utu ∈ 𝓜.

From Corollary 2.4, Lemma 2.6 and Lemma 2.7, we have 𝓜 ≠ ∅ and the following lemma.

#### Lemma 2.9

Assume that (V1)-(V3) and (F1)-(F3) hold. Then

$infu∈MI(u):=m=infu∈Λmaxt>0I(ut).$

The following lemma is a known result which can be proved by a standard argument (see [14, 15]).

#### Lemma 2.10

Assume that (V1), (V2), (F1) and (F2) hold. If unū in H1(ℝN), then

$I(un)=I(u¯)+I(un−u¯)+o(1)$(2.17)

and

$P(un)=P(u¯)+P(un−u¯)+o(1).$(2.18)

#### Lemma 2.11

Assume that (V1)-(V3) and (F1)-(F3) hold. Then

1. there exists ρ0 > 0 such thatu∥ ≥ ρ0, ∀ u ∈ 𝓜;

2. m = infu∈𝓜 I(u) > 0.

#### Proof

1. Since 𝓟(u) = 0, ∀ u ∈ 𝓜, by (F1), (F2), (1.12), (2.8) and Sobolev embedding theorem, one has

$y12∥u∥2≤N−22∥∇u∥22+12∫RN[NV(x)+∇V(x)⋅x]u2dx=N∫RNF(u)dx≤y14∥u∥2+C1∥u∥2∗,$(2.19)

which implies

$∥u∥≥ρ0:=y14C1(N−2)/4,∀u∈M.$(2.20)

2. For uH1(ℝN), by the Sobolev inequality, one has $\begin{array}{}S\parallel u{\parallel }_{{2}^{\ast }}^{2}\le \parallel \mathrm{\nabla }u{\parallel }_{2}^{2}\end{array}$. By (V2), there exists R > 0 such that V(x) ≥ $\begin{array}{}\frac{{V}_{\mathrm{\infty }}}{2}\end{array}$ for |x| ≥ R\$. It follows from (F1) and (F2) that there exists C2 > 0 such that

$|F(t)|≤14minSR2ωN2/N,V∞|t|2+C2|t|2∗,∀t∈R,$(2.21)

where ωN denote the volume of the unit ball of ℝN. For u ∈ 𝓜, let

$tu=(N−2)SN/(N−2)4NC21/2∥∇u∥2−2/(N−2).$

Making use of the Hölder inequality and the Sobolev embedding theorem, we get

$∫|tux|(2.22)

Then from (2.3), (2.5), (2.21), (2.22) and the Sobolev embedding theorem, we have

$I(u)≥Iutu=tuN−22∥∇u∥22+tuN2∫RNV(tux)u2dx−tuN∫RNF(u)dx≥tuN−24∥∇u∥22+S4R2ωN2/NtuN∫|tux|

This shows that m = infu∈𝓜I(u) > 0. □

#### Lemma 2.12

Assume that (V1)-(V3) and (F1)-(F3) hold. Then mm = infu∈ 𝓜 I(u).

#### Proof

Arguing indirectly, we assume that m > m. Let ε := mm. Then there exists $\begin{array}{}{u}_{\epsilon }^{\mathrm{\infty }}\end{array}$ such that

$uε∞∈M∞andm∞+ε2>I∞(uε∞)$(2.23)

In view of Lemmas 2.6 and 2.7, there exists tε > 0 such that $\begin{array}{}\left({u}_{\epsilon }^{\mathrm{\infty }}{\right)}_{{t}_{\epsilon }}\end{array}$ ∈ 𝓜. Thus, it follows from (V1), (V2), (1.2), (1.4), (2.6) and (2.23) that

$m∞ε2>I∞(uε∞)≥I∞(uε∞)tε≥I(uε∞)tε≥m.$

This contradiction shows the conclusion of Lemma 2.12 is true. □

#### Lemma 2.13

Assume that (V1)-(V3) and (F1)-(F3) hold. Then m is achieved.

#### Proof

In view of Lemmas 2.6, 2.7 and 2.11, we have 𝓜 ≠ ∅ and m > 0. Let {un} ⊂ 𝓜 be such that I(un) → m. Since 𝓟(un) = 0, then it follows from (2.3) with t → 0 that

$m+o(1)=I(un)≥1−θN∥∇un∥22.$(2.24)

This shows that {∥∇ un2} is bounded. Next, we prove that {∥un∥} is also bounded. By (F1), (F2), (1.12), (2.8) and the Sobolev embedding theorem, one has

$y1∥un∥2≤(N−2)∥∇un∥22+∫RN[NV(x)+∇V(x)⋅x]un2dx=2N∫RNF(un)dx≤y12∥un∥2+C4∥un∥2∗2∗≤y12∥un∥2+C4S−2∗/2∥∇un∥22∗.$(2.25)

This shows that {un} is bounded in H1(ℝN). Passing to a subsequence, we have unū in H1(ℝN). Then unū in $\begin{array}{}{L}_{\mathrm{l}\mathrm{o}\mathrm{c}}^{s}\end{array}$(ℝN) for 2 ≤ s < 2* and unū a.e. in ℝN. There are two possible cases: i). ū = 0 and ii). ū ≠ 0.

• Case i)

ū = 0, i.e. un ⇀ 0 in H1(ℝN). Then un → 0 in $\begin{array}{}{L}_{\mathrm{l}\mathrm{o}\mathrm{c}}^{s}\end{array}$(ℝN) for 2 ≤ s < 2* and un → 0 a.e. in ℝN. By (V2) and (2.9), it is easy to show that

$limn→∞∫RN[V∞−V(x)]un2dx=limn→∞∫RN∇V(x)⋅xun2dx=0.$(2.26)

From (1.2), (1.4), (1.5), (1.12) and (2.26), one can get

$I∞(un)→m,P∞(un)→0.$(2.27)

From Lemma 2.11 (i), (1.5) and (2.27), one has

$min{N−2,NV∞}ρ02≤min{N−2,NV∞}∥un∥2≤(N−2)∥∇un∥22+NV∞∥un∥22=2N∫RNF(un)dx+o(1).$(2.28)

Using (F1), (F2), (2.28) and Lions’ concentration compactness principle [20, Lemma 1.21], we can prove that there exist δ > 0 and a sequence {yn} ⊂ ℝN such that ∫B1(yn) |un|2 dx > δ. Let ûn(x) = un(x + yn). Then we have ∥ûn∥ = ∥un∥ and

$P∞(u^n)=o(1),I∞(u^n)→m,∫B1(0)|u^n|2dx>δ.$(2.29)

Therefore, there exists ûH1(ℝN) ∖ {0} such that, passing to a subsequence,

$u^n⇀u^,inH1(RN);u^n→u^,inLlocs(RN),∀s∈[1,2∗);u^n→u^,a.e. onRN.$(2.30)

Let wn = ûnû. Then (2.30) and Lemma 2.10 yield

$I∞(u^n)=I∞(u^)+I∞(wn)+o(1)$(2.31)

and

$P∞(u^n)=P∞(u^)+P∞(wn)+o(1).$(2.32)

Moreover,

$1N∥∇wn∥22=m−1N∥∇u^∥22+o(1),P∞(wn)=−P∞(u^)+o(1).$(2.33)

If there exists a subsequence {wni} of {wn} such that wni = 0, then going to this subsequence, we have

$I∞(u^)=m,P∞(u^)=0.$(2.34)

Next, we assume that wn ≠ 0. We claim that 𝓟(û) ≤ 0. Otherwise, if 𝓟(û) > 0, then (2.33) implies 𝓟(wn) < 0 for large n. In view of Lemma 2.6 and Corollary 2.8, there exists tn > 0 such that (wn)tt ∈ 𝓜. From (1.4), (1.5), (2.6) and (2.33), we obtain

$m−1N∥∇u^∥22+o(1)=1N∥∇wn∥22=I∞(wn)−1NP∞(wn)≥I∞(wn)tn−tnNNP∞(wn)≥m∞−tnNNP∞(wn)≥m∞,$

which implies 𝓟(û) ≤ 0 due to ∥∇ û2 > 0. Since û ≠ 0 and 𝓟(û) ≤ 0, in view of Lemma 2.6 and Corollary 2.8, there exists > 0 such that û ∈ 𝓜. From (1.4), (1.5), (2.6), (2.29) and the weak semicontinuity of norm, one has

$m=limn→∞I∞(u^n)−1NP∞(u^n)=1Nlimn→∞∥∇u^n∥22≥1N∥∇u^∥22=I∞(u^)−1NP∞(u^)≥I∞u^t^−t^NNP∞(u^)≥m∞−t^NNP∞(u^)≥m−t^NNP∞(u^)≥m,$

which implies (2.34) holds also. In view of Lemmas 2.6 and 2.7, there exists > 0 such that û ∈ 𝓜, moreover, it follows from (V2), (1.2), (1.4), (2.34) and Corollary 2.3 that

$m≤I(u^t~)≤I∞(u^t~)≤I∞(u^)=m.$

This shows that m is achieved at û ∈ 𝓜.

• Case ii)

ū ≠ 0. Let vn = unū. Then Lemma 2.10 yields

$I(un)=I(u¯)+I(vn)+o(1)$(2.35)

and

$P(un)=P(u¯)+P(vn)+o(1).$(2.36)

Set

$Ψ(u)=1N∥∇u∥22−12N∫RN(∇V(x),x)u2dx.$(2.37)

Then it follows from (2.2) with t = 0 and (2.4) that

$Ψ(u)≥1−θN∥∇u∥22,∀u∈H1(RN).$(2.38)

Since I(un) → m and 𝓟(un) = 0, then it follows from (1.2), (1.12), (2.35), (2.36) and (2.37) that

$Ψ(vn)=m−Ψ(u¯)+o(1),P(vn)=−P(u¯)+o(1).$(2.39)

If there exists a subsequence {vnn} of {vn} such that vnn = 0, then going to this subsequence, we have

$I(u¯)=m,P(u¯)=0,$(2.40)

which implies the conclusion of Lemma 2.13 holds. Next, we assume that vn ≠ 0. We claim that 𝓟(ū) ≤ 0. Otherwise 𝓟(ū) > 0, then (2.39) implies 𝓟(vn) < 0 for large n. In view of Lemmas 2.6 and 2.7, there exists tn > 0 such that (vn)tt ∈ 𝓜. From (1.2), (1.12), (2.3) and (2.39), we obtain

$m−Ψ(u¯)+o(1)=Ψ(vn)=I(vn)−1NP(vn)≥I(vn)tn−tnNNP(vn)≥m−tnNNP(vn)≥m,$

which implies 𝓟(ū) ≤ 0 due to Ψ(ū) > 0. Since ū ≠ 0 and 𝓟(ū) ≤ 0, in view of Lemmas 2.6 and 2.7, there exists > 0 such that ū ∈ 𝓜. From (1.2), (1.12), (2.3), (2.37), (2.38) and the weak semicontinuity of norm, one has

$m=limn→∞I(un)−1NP(un)=limn→∞Ψ(un)≥Ψ(u¯)=I(u¯)−1NP(u¯)≥Iu¯t¯−t¯NNP(u¯)≥m−t¯NNP(u¯)≥m,$

which implies (2.40) also holds. □

#### Lemma 2.14

Assume that (V1)-(V3) and (F1)-(F3) hold. If ū ∈ 𝓜 and I(ū) = m, then ū is a critical point of I.

#### Proof

Assume that I′(ū) ≠ 0. Then there exist δ > 0 and ϱ > 0 such that

$∥u−u¯∥≤3δ⇒∥I′(u)∥≥ϱ.$(2.41)

First, we prove that

$limt→1u¯t−u¯=0.$(2.42)

Arguing by contradiction, suppose that there exist ε0 > 0 and a sequence {tn} such that

$limn→∞tn=1,u¯tn−u¯2≥ε0.$(2.43)

Since ūH1(ℝN), there exist U ∈ 𝓒0(ℝN, ℝN) and v ∈ 𝓒0(ℝN, ℝ) such that

$∫RN|∇u¯−U|2<ε020,∫RN|u¯−v|2<ε020.$(2.44)

From (2.43) and (2.44), one has

$∇u¯tn−∇u¯22=∫RN∇u¯tn−∇u¯2dx≤2∫RN∇u¯tn−U2dx+2∫RN|∇u¯−U|2dx=2∫RNtn−1∇u¯tn−1x−U(x)2dx+2∫RN|∇u¯−U|2dx≤6tn−2∫RNUtn−1x−U(x)2dx+6|tn−1−1|2∫RN|U|2dx+(1+3tnN−2)ε010=25ε0+o(1)$(2.45)

and

$u¯tn−u¯22=∫RNu¯tn−u¯2dx≤2∫RNu¯tn−v2dx+2∫RN|u¯−v|2dx≤4∫RNvtn−1x−v(x)2dx+(1+2tnN)ε010=310ε0+o(1).$(2.46)

Combining (2.45) with (2.46), one has

$u¯tn−u¯2=∇(u¯tn)−∇u¯22+u¯tn−u¯22≤710ε0+o(1).$(2.47)

(2.47) contradicts with (2.43). Therefore, (2.42) holds. Thus, there exists δ1 ∈ (0, 1/4) such that

$|t−1|<δ1⇒u¯t−u¯<δ.$(2.48)

In view of Lemma 2.2, one has

$Iu¯t≤I(u¯)−(1−θ)2−NtN−2+(N−2)tN2N∥∇u¯∥22=m−(1−θ)g(t)2N∥∇u¯∥22,∀t>0.$(2.49)

It follows from (1.12), (2.4) and (2.9) that there exist T1 ∈ (0, 1) and T2 ∈ (1, ∞) such that

$Pu¯T1>0,Pu¯T2<0.$(2.50)

Let $\begin{array}{}\epsilon :=min\left\{\left(1-\theta \right)\mathfrak{g}\left({T}_{1}\right)\parallel \mathrm{\nabla }\overline{u}{\parallel }_{2}^{2}/5N,\left(1-\theta \right)\mathfrak{g}\left({T}_{2}\right)\parallel \mathrm{\nabla }\overline{u}{\parallel }_{2}^{2}/5N,1,\varrho \delta /8\right\}\end{array}$ and S := B(ū, δ). Then [20, Lemma 2.3] yields a deformation η ∈ 𝓒([0, 1] × H1(ℝN), H1(ℝN)) such that

1. η(1, u) = u if I(u) < m - 2ε or I(u) > m + 2ε;

2. η (1, Im+εB(ū, δ)) ⊂ Imε;

3. I(η(1, u)) ≤ I(u), ∀ uH1(ℝN);

4. η(1, u) is a homeomorphism of H1(ℝN).

By Corollary 2.4, I (ūt) ≤ I(ū) = m for t > 0, then it follows from (2.48) and ii) that

$Iη1,u¯t≤m−ε,∀t>0,|t−1|<δ1.$(2.51)

On the other hand, by iii) and (2.49), one has

$Iη1,u¯t≤Iu¯t≤m−(1−θ)g(t)2N∥∇u¯∥22≤m−(1−θ)δ22N∥∇u¯∥22,∀t>0,|t−1|≥δ1,$(2.52)

where

$δ2:=min{g(1−δ1),g(1+δ1)}>0.$

Combining (2.51) with (2.52), we have

$maxt∈[T1,T2]Iη1,u¯t(2.53)

Define Ψ0(t) := 𝓟 (η (1, ūt)) for t > 0. It follows from (2.49) and i) that η(1, ūt) = ūt for t = T1 and t = T2, which, together with (2.50), implies

$Ψ0(T1)=Pu¯T1>0,Ψ0(T2)=Pu¯T1<0.$

Since Ψ0(t) is continuous on (0, ∞), then we have that η (1, ūt) ∩ 𝓜 ≠ ∅ for some t0 ∈ [T1, T2], contradicting to the definition of m. □

#### Proof of Theorem 1.2

In view of Lemmas 2.9, 2.13 and 2.14, there exists ū ∈ 𝓜 such that

$I(u¯)=m=infu∈Λmaxt>0I(ut),I′(u¯)=0.$

This shows that ū is a ground state solution of Pohožaev type for (1.1). □

## 3 The least energy solutions for (1.1)

In this section, we give the proof of Theorem 1.5.

#### Proposition 3.1

[8] Let X be a Banach space and let J ⊂ ℝ+ be an interval, and

$Φλ(u)=A(u)−λB(u),∀λ∈J,$

be a family of 𝓒1-functional on X such that

1. either A(u) → +∞ or B(u) → +∞, asu∥ → ∞;

2. B maps every bounded set of X into a set ofbounded below;

3. there are two points v1, v2 in X such that

$c~λ:=infy∈Γ~maxt∈[0,1]Φλ(y(t))>max{Φλ(v1),Φλ(v2)},$(3.1)

where

$Γ~=y∈C([0,1],X):y(0)=v1,y(1)=v2.$

Then, for almost every λJ, there exists a sequence {un(λ)} such that

1. {un(λ)} is bounded in X;

2. Φλ(un(λ)) → cλ;

3. $\begin{array}{}{\mathit{\Phi }}_{\lambda }^{\prime }\left({u}_{n}\left(\lambda \right)\right)\end{array}$ → 0 in X*, where X* is the dual of X.

#### Lemma 3.2

[7] Assume that (V1), (V2), (F1) and (F2) hold. Let u be a critical point of Iλ in H1(ℝN), then we have the following Pohožaev type identity

$Pλ(u):=N−22∥∇u∥22+12∫RNNV(x)+∇V(x)⋅xu2dx−Nλ∫RNF(u)dx=0.$(3.2)

By Corollary 2.3, we have the following lemma.

#### Lemma 3.3

Assume that (F1) and (F2) hold. Then

$Iλ∞(u)=Iλ∞ut+1−tNNPλ∞(u)+2−NtN−2+(N−2)tN2N∥∇u∥22,∀u∈H1(RN),t>0,λ≥0.$(3.3)

In view of Theorem 1.3, $\begin{array}{}{I}_{1}^{\mathrm{\infty }}\end{array}$ = I has a minimizer $\begin{array}{}{u}_{1}^{\mathrm{\infty }}\ne 0\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}{\mathcal{M}}_{1}^{\mathrm{\infty }}={\mathcal{M}}^{\mathrm{\infty }}\end{array}$, i.e.

$u1∞∈M1∞,(I1∞)′(u1∞)=0andm1∞=I1∞(u1∞),$(3.4)

where $\begin{array}{}{m}_{\lambda }^{\mathrm{\infty }}\end{array}$ is defined by (1.18). Since (1.3) is autonomous, V ∈ 𝓒(ℝN, ℝ) and V(x) ≤ V but V(x)≢ V, then there exist ∈ ℝN and > 0 such that

$V∞−V(x)>0,|u1∞(x)|>0a.e.|x−x¯|≤r¯.$(3.5)

#### Lemma 3.4

Assume that (V1), (V2) and (F1)-(F3) hold. Then

1. there exists T > 0 independent of λ such that $\begin{array}{}{I}_{\lambda }\left(\left({u}_{1}^{\mathrm{\infty }}{\right)}_{T}\right)\end{array}$ < 0 for all λ ∈ [0.5, 1];

2. there exists a positive constant κ0 independent of λ such that for all λ ∈ [0.5, 1],

$cλ:=infy∈Γmaxt∈[0,1]Iλ(y(t))≥κ0>maxIλ(0),Iλ(u1∞)T,$

where

$Γ=y∈C([0,1],H1(RN)):y(0)=0,y(1)=(u1∞)T;$

3. cλ is bounded for λ ∈ [0.5, 1];

4. $\begin{array}{}{m}_{\lambda }^{\mathrm{\infty }}\end{array}$ is non-increasing on λ ∈ [0.5, 1];

5. lim supλλ0 cλcλ0 for λ0 ∈ (0.5, 1].

Since $\begin{array}{}{m}_{\lambda }^{\mathrm{\infty }}={I}_{\lambda }^{\mathrm{\infty }}\left({u}_{\lambda }^{\mathrm{\infty }}\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{\int }_{{\mathbb{R}}^{N}}F\left({u}_{\lambda }^{\mathrm{\infty }}\right)\mathrm{d}x\end{array}$ > 0, then the proofs of (i)-(iv) in Lemma 3.4 is standard, (v) can be proved similar to [5, Lemma 2.3], so we omit it.

#### Lemma 3.5

Assume that (V1), (V2) and (F1)-(F3) hold. Then there exists λ̄ ∈ [1/2, 1) such that cλ < $\begin{array}{}{m}_{\lambda }^{\mathrm{\infty }}\end{array}$ for λ ∈ (λ̄, 1].

#### Proof

It is easy to see that $\begin{array}{}{I}_{\lambda }\left(\left({u}_{1}^{\mathrm{\infty }}{\right)}_{t}\right)\end{array}$ is continuous on t ∈ (0, ∞). Hence for any λ ∈ [1/2, 1], we can choose tλ ∈ (0, T) such that $\begin{array}{}{I}_{\lambda }\left(\left({u}_{1}^{\mathrm{\infty }}{\right)}_{{t}_{\lambda }}\right)=\underset{t\in \left[0,T\right]}{max}{I}_{\lambda }\left(\left({u}_{1}^{\mathrm{\infty }}{\right)}_{t}\right)\end{array}$. Setting

$y0(t)=(u1∞)(tT),fort>0,0,fort=0.$

Then y0Γ defined by Lemma 3.4 (ii). Moreover

$Iλ(u1∞)tλ=maxt∈[0,1]Iλy0(t)≥cλ.$(3.6)

Let

$ζ0:=min{3r¯/8(1+|x¯|),1/4}.$(3.7)

Then it follows from (3.5) and (3.7) that

$|x−x¯|≤r¯2ands∈[1−ζ0,1+ζ0]⇒|sx−x¯|≤r¯.$(3.8)

Since $\begin{array}{}{\mathcal{P}}^{\mathrm{\infty }}\left({u}_{1}^{\mathrm{\infty }}\right)\end{array}$ = 0, then $\begin{array}{}{\int }_{{\mathbb{R}}^{N}}F\left({u}_{1}^{\mathrm{\infty }}\right)\mathrm{d}x\end{array}$ > 0. Let

$λ¯:=max12,1−(1−ζ0)Nmins∈[1−ζ0,1+ζ0]∫RNV∞−V(sx)|u1∞|2dxTN∫RNF(u1∞)dx,1−min{g(1−ζ0),g(1+ζ0)}∥∇u1∞∥22NTN∫RNF(u1∞)dx.$(3.9)

Then it follows from (2.1), (3.5) and (3.8) that 1/2 ≤ λ̄ < 1. We have two cases to distinguish:

• Case i)

tλ ∈ [1 − ζ0, 1 + ζ0]. From (1.14), (1.15), (3.3)-(3.6), (3.8), (3.9) and Lemma 3.4 (iv), we have

$mλ∞≥m1∞=I1∞(u1∞)≥I1∞(u1∞)tλ=Iλ(u1∞)tλ−(1−λ)tλN2∫RNF(u1∞)dx+tλN2∫RN[V∞−V(tλx)]|u1∞|2dx≥cλ−(1−λ)T0N2∫RNF(u1∞)dx+(1−ζ0)N2mins∈[1−ζ0,1+ζ0]∫RNV∞−V(sx)|u1∞|2dx>cλ,∀λ∈(λ¯,1].$

• Case ii)

tλ ∈ (0, 1 − ζ0) ∪ (1 + ζ0, T). From (V2), (1.14), (1.15), (2.1), (3.3), (3.4), (3.6), (3.9) and Lemma 3.4 (iv), we have

$mλ∞≥m1∞=I1∞(u1∞)≥I1∞(u1∞)tλ+g(tλ)∥∇u1∞∥222N=Iλ(u1∞)tλ−(1−λ)tλN2∫RNF(u1∞)dx+tλN2∫RN[V∞−V(tλx)]|u1∞|2dx+g(tλ)∥∇u1∞∥222N≥cλ−(1−λ)TN2∫RNF(u1∞)dx+min{g(1−ζ0),g(1+ζ0)}∥∇u1∞∥222N>cλ,∀λ∈(λ¯,1].$

In both cases, we obtain that cλ < $\begin{array}{}{m}_{\lambda }^{\mathrm{\infty }}\end{array}$ for λ ∈ (λ̄, 1]. □

#### Lemma 3.6

[7] Assume that (V1), (V2) and (F1)-(F3) hold. Let {un} be a bounded (PS) sequence for Iλ, for λ ∈ [1/2, 1]. Then there exists a subsequence of {un}, still denoted by {un}, an integer l ∈ ℕ ∪ {0}, a sequence $\begin{array}{}\left\{{y}_{n}^{k}\right\}\end{array}$ and wkH1(ℝ3) for 1 ≤ kl, such that

1. unu0 with $\begin{array}{}{I}_{\lambda }^{\prime }\left({u}_{0}\right)\end{array}$ = 0;

2. wk ≠ 0 and $\begin{array}{}\left({I}_{\lambda }^{\mathrm{\infty }}{\right)}^{\prime }\end{array}$ (wk) = 0 for 1 ≤ kl;

3. $\begin{array}{}∥{u}_{n}-{u}_{0}-\sum _{k=1}^{l}{w}^{k}\left(\cdot +{y}_{n}^{k}\right)∥\to 0;\end{array}$

4. Iλ(un) → Iλ(u0) + $\begin{array}{}\sum _{i=1}^{l}{I}_{\lambda }^{\mathrm{\infty }}\left({w}^{i}\right)\end{array}$;

where we agree that in the case l = 0 the above holds without wk.

#### Lemma 3.7

Assume that (V1), (V2), (V4) and (F1)-(F3) hold. Then for almost every λ ∈ (λ̄, 1], there exists uλH1(ℝN) ∖ {0} such that

$Iλ′(uλ)=0,Iλ(uλ)=cλ.$(3.10)

#### Proof

Under (V1), (V2) and (F1)-(F3), Lemma 3.4 implies that Iλ(u) satisfies the assumptions of Proposition 3.1 with X = H1(ℝN), J = [λ̄, 1] and Φλ = Iλ. So for almost every λ ∈ (λ̄, 1], there exists a bounded sequence {un(λ)} ⊂ H1(ℝN) (for simplicity, we denote the sequence by {un} instead of {un(λ)}) such that

$Iλ(un)→cλ>0,Iλ′(un)→0.$(3.11)

By Lemmas 3.2 and 3.6, there exist a subsequence of {un}, still denoted by {un}, and uλH1(ℝN), an integer l ∈ ℕ ∪ {0}, and w1, …, wlH1(ℝN) ∖ {0} such that

$un⇀uλinH1(RN),Iλ′(uλ)=0,$(3.12)

$(Iλ∞)′(wk)=0,Iλ∞(wk)≥mλ∞,1≤k≤l$(3.13)

and

$cλ=Iλ(uλ)+∑k=1lIλ∞(wk).$(3.14)

Since $\begin{array}{}{I}_{\lambda }^{\prime }\left({u}_{\lambda }\right)\end{array}$ = 0, then it follows from Lemma 3.2 that

$Pλ(uλ)=N−22∥∇uλ∥22+12∫RNNV(x)+∇V(x)⋅xuλ2dx−Nλ∫RNF(uλ)dx=0.$(3.15)

Since ∥un∥ ↛ 0, we deduce from (3.13) and (3.14) that if uλ = 0 then l ≥ 1 and

$cλ=Iλ(uλ)+∑k=1lIλ∞(wk)≥mλ∞,$

which contradicts with Lemma 3.5. Thus uλ ≠ 0. It follows from (1.14), (2.4), (3.15) and (V4) that

$Iλ(uλ)=Iλ(uλ)−1NPλ(uλ)=1N∥∇uλ∥22−12N∫RN(∇V(x),x)uλ2dx≥1−θN∥∇uλ∥22>0.$(3.16)

From (3.14) and (3.16), one has

$cλ=Iλ(uλ)+∑k=1lIλ∞(wk)>lmλ∞.$(3.17)

By Lemma 3.5, we have cλ < $\begin{array}{}{m}_{\lambda }^{\mathrm{\infty }}\end{array}$ for λ ∈ (λ̄, 1], which, together with (3.17), implies that l = 0 and Iλ(uλ) = cλ. □

#### Lemma 3.8

Assume that (V1), (V2), (V4) and (F1)-(F3) hold. Then there exists ūH1(ℝN) ∖ {0} such that

$I′(u¯)=0,0(3.18)

#### Proof

In view of Lemma 3.7, there exist two sequences {λn} ⊂ [λ̄, 1] and {uλn} ⊂ H1(ℝN) ∖ {0}, denoted by {un}, such that

$λn→1,cλn→c∗,Iλn′(un)=0,0(3.19)

Then it follows from Lemma 3.2 that

$Pλn(un):=N−22∥∇un∥22+12∫RNNV(x)+∇V(x)⋅xun2dx−Nλn∫RNF(un)dx=0.$(3.20)

From (V4), (1.14), (2.4), (3.19), (3.20) and Lemma 3.4 (iii), one has

$C5≥cλn=Iλn(un)−1NPλn(un)=1N∥∇un∥22−12N∫RN∇V(x)⋅xun2dx≥1−θN∥∇un∥22.$(3.21)

This shows that {∥∇ un2} is bounded. Next, we demonstrate that {un} is bounded in H1(ℝN). According to (V1) and (V2), it is easy to show that there exists a constant y3 > 0 such that

$∫RN|∇u|2+V(x)u2dx≥y3∥u∥2,∀u∈H1(RN).$(3.22)

From (F1), (F2), (1.14), (3.19), (3.21), (3.22), Lemma 3.4 (iii) and the Sobolev embedding theorem, we have

$y3∥un∥2≤∫RN|∇un|2+V(x)un2dx=2cλn+2λn∫RNF(un)dx≤2C5+y32∥un∥2+C6∥un∥2∗2∗≤2C5+y32∥un∥2+C6S−2∗/2∥∇un∥22∗.$

Hence, {un} is bounded in H1(ℝN). In view of Lemma 3.4 (v), we have limn→∞ cλn = c*c1. Hence, it follows from (1.2), (1.14) and (3.19) that

$I(un)→c∗,I′(un)→0.$(3.23)

This shows that {un} satisfy (3.11) with cλ = c*. In view of the proof of Lemma 3.7, we can show that there exists ūH1(ℝN) ∖ {0} such that (3.18) holds. □

#### Proof of Theorem 1.5

Let

$K:=u∈H1(RN)∖{0}:I′(u)=0,m^:=infu∈KI(u).$

Then Lemma 3.8 shows that 𝓚 ≠ ∅ and c1. For any u ∈ 𝓚, Lemma 3.2 implies 𝓟(u) = 𝓟1(u) = 0. Hence it follows from (3.16) that I(u) = I1(u) > 0, and so ≥ 0. Let {un} ⊂ 𝓚 such that

$I′(un)=0,I(un)→m^.$(3.24)

In view of Lemma 3.5, c1 < $\begin{array}{}{m}_{1}^{\mathrm{\infty }}\end{array}$. By a similar argument as in the proof of Lemma 3.7, we can prove that there exists ūH1(ℝN) ∖ {0} such that

$I′(u¯)=0,I(u¯)=m^.$(3.25)

This shows that ū is a nontrivial least energy solution of (1.1). □

## Acknowledgement

This work was partially supported by the National Natural Science Foundation of China (11571370).

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Accepted: 2019-01-07

Published Online: 2019-06-16

Published in Print: 2019-03-01

Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 496–515, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496,

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