Infinitely many solutions for a nonlinear Schrödinger equation with general nonlinearity

Abstract

We prove the existence of infinitely many solutions for

$$\begin{aligned} - \Delta u + V(x) u = f(u) \quad \text { in } \mathbb {R}^N, \quad u \in H^1(\mathbb {R}^N), \end{aligned}$$

where V(x) satisfies $\lim _{|x| \rightarrow \infty } V(x) = V_\infty >0$ and some conditions. We require conditions on f(u) only around 0 and at $\infty$.

1 Introduction

In this paper, we consider the following nonlinear Schrödinger equation:

$$\begin{aligned} - \Delta u + V(x) u = f(u) \quad \text { in } \mathbb {R}^N, \quad u \in H^1(\mathbb {R}^N). \end{aligned}$$

(1.1)

Here $N \ge 3$ and we assume that the potential function V(x) satisfies the following:

• (V.1) $V \in C^1(\mathbb {R}^N, (0, \infty ))$.

• (V.2) $\lim _{|x| \rightarrow \infty } V(x) = V_\infty >0$.

• (V.3) There exists $\eta \in L^2(\mathbb {R}^N) \cap W^{1,\infty }(\mathbb {R}^N)$ such that

  $$\begin{aligned} \left( x \cdot \nabla V(x)\right) \le \eta (x)^2 \quad \hbox { for all } x \in \mathbb {R}^N. \end{aligned}$$

• (V.4) There exists $\rho \in (0,1)$ such that, for any $\alpha >0$,

  $$\begin{aligned} \lim _{|x| \rightarrow \infty } \inf _{y \in B(x, \rho |x|)} \left( x \cdot \nabla V(y)\right) e^{\alpha |x|}= \infty . \end{aligned}$$

  Here $B(x,r)=\left\{ y \in \mathbb {R}^N \big | \left| y-x\right| <r\right\}$ for $x \in \mathbb {R}^N$, $r>0$.

We assume that the nonlinearity f(u) satisfies the following:

• (f.1) $f \in C^1(\mathbb {R},\mathbb {R})$ and $f(-s) = -f(s)$.

• (f.2) $f'(0)=0$.

• (f.3) There exists $p \in (1, (N+2)/(N-2))$ such that $\lim _{s \rightarrow \infty } f'(s)/s^{p-1}=0$.

• (f.4) $\lim _{s \rightarrow \infty } f(s)/s = \infty$.

Under the assumptions (V.1)–(V.2), (f.1)–(f.3), the solutions of (1.1) are the critical points of the functional $I \in C^2(H^1(\mathbb {R}^N), \mathbb {R})$ defined by

$$\begin{aligned} I(u)=\frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2+V(x)u^2\, dx-\int _{\mathbb {R}^N}F(u)\, dx, \end{aligned}$$

where $F(u)=\int _0^uf(s)\,ds$.

Many researchers have studied (1.1) under the assumption (V.2) for $f(u)=|u|^{p-1}u$ or more general f(u) (cf. [1,2,3,4,5,6, 11, 13, 14, 16, 17], and their references). When we look for critical points of I(u) by variational approach, we generally need the compactness of Palais–Smale sequences (we denote (PS)-sequences in short) for I(u). If $V(x)\le V_\infty \ (\not \equiv V_\infty )$, then, by the concentration compactness arguments, we obtain the compactness of (PS)-sequences at a mountain pass level (cf. [13, 14, 17]). However, for (PS)-sequences at higher energy levels, it is not easy to get the compactness. To get the compactness, in the case $f(u)=|u|^{p-1}u$, Cerami–Devillanova–Solimini [3] introduced assumptions such as (V.4) (also see Remark 1.2) and balanced sequences which are sequences of solutions of the following equation on a ball:

$$\begin{aligned} -\Delta u+V(x)u=f(u) \quad \hbox { in } B\left( 0,n\right) , \qquad u\in H_0^1\left( B(0,n)\right) . \end{aligned}$$

(1.2)

The balanced sequences are not (PS)-sequences for I(u) but play a similar role to the (PS)-sequences. In fact, the balanced sequences also satisfy concentration-compactness type properties. Moreover, under the assumption (V.4), every balanced sequence is relatively compact (cf. [3, Proposition 2.1]). Consequently, they succeeded to obtain the existence of infinitely many solutions of (1.1).

On the other hand, in the procedure for getting the compactness of (PS)-sequence, in almost all cases, we need the $H^1(\mathbb {R}^N)$-boundedness of (PS)-sequence. For general f(u) as (f.1)–(f.4), it is a problem how to get the bounded (PS)-sequence. As an assumption guaranteeing the boundedness of any (PS)-sequences, the following Ambrosetti-Rabinowitz condition (AR) is well-known.

• (AR) There exists $\mu >2$ such that $\mu F(u)\le f(u)u$ for all $u\in \mathbb {R}$.

There are many researches to obtain the bounded (PS)-sequence under weaker conditions than (AR) (cf. [9,10,11,12, 16, 19], and their references). In our knowledge, one of the weakest assumptions is (V.3) and (f.4) which were introduced by Jeanjean–Tanaka [11]. By the monotonicity trick, they obtained a (PS)-sequence which is a sequence of solutions of

$$\begin{aligned} -\Delta u+V(x)u=\lambda _n f(u) \quad \hbox { in } \mathbb {R}^N, \qquad u\in H^1(\mathbb {R}^N), \end{aligned}$$

(1.3)

where $\lambda _n \rightarrow 1-0$ $(n\rightarrow \infty )$. To get the $H^1(\mathbb {R}^N)$-boundedness of this (PS)-sequence, they used (V.3), (f.4), and the Pohozaev identity for the solutions of (1.3). Consequently, for $N\ge 2$, they obtained a positive solution of (1.1) under the conditions (V.1)–(V.3), $V(x)\le V_\infty \ (\not \equiv V_\infty )$, and

(f.1$'$):

$f \in C([0,\infty ),\mathbb {R})$.

(f.2$'$):

$\lim _{s \rightarrow +0} f(s)/s=0$.

(f.3$'$):

There exists $p \in (1, (N+2)/(N-2))$ if $N\ge 3$, $p\in (1,\infty )$ if $N=2$ such that $\lim _{s \rightarrow \infty } f(s)/s^p=0$.

(f.4):

$\lim _{s \rightarrow \infty } f(s)/s = \infty$.

The main result of this paper is the following theorem which is considered as a development from [3] and [11].

Theorem 1.1

Assume $N\ge 3$, (V.1)–(V.4) and (f.1)–(f.4). Then (1.1) have infinitely many solutions.

Remark 1.2

If V(x) satisfies the following (i)–(ii), V(x) satisfies (V.3)–(V.4).

• (i) There exist $c_0, c_1, r_1>0$ and $\ell _0\ge \ell _1>N$ such that

  $$\begin{aligned} c_0|x|^{-\ell _0}\le \left( x\cdot \nabla V(x)\right) \le c_1|x|^{-\ell _1} \quad \hbox { for all } |x|\ge r_1. \end{aligned}$$

• (ii) There exist $c_2, r_2>0$ such that

  $$\begin{aligned} \left| \left( \frac{\xi }{|\xi |}\cdot \nabla V(x)\right) \right| \le c_2\left( \frac{x}{|x|}\cdot \nabla V(x)\right) \hbox { for all } |x|\ge r_2 \hbox { and } \xi \hbox { with } (\xi \cdot x)=0. \end{aligned}$$

The radial functions V always satisfy (ii). Cerami–Devillanova–Solimini [3] assumed (ii) and

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\left( x\cdot \nabla V(x)\right) e^{\alpha |x|}=\infty \quad \text { for any } \alpha >0 \end{aligned}$$

(1.4)

instead of (V.4). (V.4) follows from (ii) and (1.4). In fact, for $y\in B(x,\rho |x|)$ ($\rho \in (0,1)$), we set $\xi =x-\frac{(y\cdot x)}{|y|^2}y$. Then $x=\frac{(y\cdot x)}{|y|^2}y+\xi \in (\text {span}\{y\})\oplus (\text {span}\{y\})^{\perp }$. Moreover, since $|y-x|\le \rho |x|$ and $(1-\rho )|x| \le |y| \le (1+\rho )|x|$, we have

$$\begin{aligned} \frac{1}{1+\rho }|y|^2\le (y\cdot x)\le \frac{1}{1-\rho }|y|^2, \qquad |\xi |\le \frac{\rho }{1-\rho }|y|. \end{aligned}$$

From (ii), for $y\in B(x,\rho |x|)$ and $|y|\ge r_2$,

$$\begin{aligned} \begin{aligned} \left( x\cdot \nabla V(y)\right)&= \frac{(y\cdot x)}{|y|^2}\left( y\cdot \nabla V(y)\right) +|\xi |\left( \frac{\xi }{|\xi |}\cdot \nabla V(y)\right) \\&\ge \frac{(y\cdot x)}{|y|^2}\left( y\cdot \nabla V(y)\right) -c_2|\xi | \left( \frac{y}{|y|}\cdot \nabla V(y)\right) \\&\ge \left( \frac{1}{1+\rho }-c_2\frac{\rho }{1-\rho }\right) \left( y \cdot \nabla V(y)\right) . \end{aligned} \end{aligned}$$

Thus, choosing $\rho \in (0,1)$ such that $\frac{1}{1+\rho }-c_2\frac{\rho }{1-\rho }>0$, from (1.4), we obtain (V.4).

Here we emphasize that we can not prove Theorem 1.1 by only combining the methods of [3] and [11]. In fact, if we use a balanced sequence which is a sequence of solutions of (1.2), then we don’t know the boundedness of that sequence. On the other hand, we can not obtain infinitely many solutions of (1.3) because of the compactness problem. Therefore the sequences of solutions of (1.2) or (1.3) are not proper to show Theorem 1.1. From those reasons, we need introduce another sequence which satisfies both boundedness and compactness. Just to state only conclusions, this sequence is obtained as a sequence of solutions of

$$\begin{aligned} - \Delta u+V(x)u=\lambda _n f(u)+\left( \lambda _n-1\right) g(u) \quad \text { in } B(0,n), \qquad u\in H_0^1\left( B(0,n)\right) , \end{aligned}$$

(1.5)

where $\lambda _n\rightarrow 1+0$ $(n\rightarrow \infty )$. Here g(u) is an auxiliary function which is defined in Sect. 2. We obtain a solution of (1.5) as a critical point of a functional which is modified in the quadratic term of $I|_{H_0^1(B(0,n))}(u)$. Thanks to this modification, we can guarantee both boundedness and compactness of the sequence of solutions. This modification is an important idea in this paper.

We also obtain the following two results as by-products of Theorem 1.1.

Theorem 1.3

Assume $N\ge 3$, (V.1)–(V.4), (f.1$'$)–(f.3$'$), and (f.4). Then (1.1) has a positive solution.

Next, we assume that $\Omega \subset \mathbb {R}^N$ satisfies the following condition.

($\Omega$):

$\Omega \subset \mathbb {R}^N$ is a bounded domain with smooth boundary, $0\in \Omega$, and $(x\cdot \nu (x))>0$ for all $x\in \partial \Omega$, where $\nu (x)$ is the outward unit normal vector at $x\in \partial \Omega$.

We set $\Omega _R=\left\{ x\in \mathbb {R}^N \big | R^{-1}x\in \Omega \right\}$ for $R\ge 1$.

Theorem 1.4

Assume $N\ge 3$, (V.1)–(V.4), (f.1)–(f.4), and ($\Omega$). For any $k \in \mathbb {N}$, there exists $R_0=R_0(k)>0$ such that if $R>R_0, then$

$$\begin{aligned} -\Delta u + V(x) u = f(u) \quad \text { in } \Omega _R, \qquad u \in H^1_0(\Omega _R) \end{aligned}$$

(1.6)

has at least k distinct pairs of nontrivial solutions $\pm u_j$ $(j=1,\dots ,k)$.

This paper consists as follows: In Sect. 2, we modify the functional I(u) and define balanced sequences as sequences of critical points of modified functional. We also present propositions which bring the boundedness and compactness to balanced sequences. Those propositions are proved in Sects. 4–6. In Sect. 3, we prove Theorems 1.1 and 1.3. In Sect. 4, we prove a proposition about the boundedness. Through Sects. 5 and 6 , we prove propositions about the compactness. In Sect. 6, we also prove Theorem 1.4.

2 Preliminaries

In this section, through several subsections, we give balanced sequences which satisfy the boundedness and compactness. In Sect. 2.1, we define notations and a modified functional. In Sect. 2.2, we show properties of the modified functional. In Sect. 2.3, we state propositions about $H^1(\mathbb {R}^N)$-boundedness. In Sect. 2.4, we construct balanced sequences as sequences of critical points of the modified functional. We also state about the compactness for the balanced sequences.

2.1 Notation and modified functional

We use the following notations:

$$\begin{aligned} \begin{aligned}&\langle u,v\rangle =\int _{\mathbb {R}^N} \nabla u \cdot \nabla v +V(x) uv \,dx, \quad \Vert u\Vert =\langle u,u\rangle ^{1/2}, \\&\Vert u\Vert _r = \left( \int _{\mathbb {R}^N} |u|^r \,dx \right) ^{1/r} \hbox { for } r\in [1, \infty ), \quad \Vert u\Vert _\infty =\mathop {{\hbox {ess sup}}}\limits _{x\in \mathbb {R}^N} |u(x)|, \\&B(x,R)=\left\{ y\in \mathbb {R}^N\,|\,|y-x|<R\right\} , \quad B_R=B(0,R). \end{aligned} \end{aligned}$$

We remark that $B_\infty =\mathbb {R}^N$ and we regard $u \in H^1_0(B_R)$ as $u \in H^1(\mathbb {R}^N)$ by expanding $u=0$ on $\mathbb {R}^N\setminus B_R$. Then we also regard $\Vert \cdot \Vert$ and $\Vert \cdot \Vert _r$ as norm on $H^1_0(B_R)$ and $L^r(B_R)$, respectively. We set

$$\begin{aligned} V_0=\inf _{x \in \mathbb {R}^N} V(x), \quad V_1=\sup _{x \in \mathbb {R}^N} V(x). \end{aligned}$$

$0< V_0 \le V_\infty \le V_1 < \infty$ follows from the assumptions (V.1) and (V.2). Then we have

$$\begin{aligned} \Vert u\Vert _2^2 \le \frac{1}{V_0}\int _{\mathbb {R}^N}V(x)u^2\, dx \le \frac{1}{V_0}\Vert u\Vert ^2 \quad \hbox { for all } u\in H^1(\mathbb {R}^N). \end{aligned}$$

(2.1)

We take an auxiliary function g(s) which satisfies

• (g.1) $g \in C^1(\mathbb {R},\mathbb {R})$, $g(s) \ge 0$ for all $s>0$, and $g(-s)=-g(s)$ for all $s \in \mathbb {R}$.

• (g.2) $g'(0)=0$.

• (g.3) There exists $s_0>0$ such that $g(s)=0$ for all $s \ge s_0$.

• (g.4) $f(s) + g(s) >0$ for all $s>0$.

Remark 2.1

If $f(s)>0$ for all $s>0$, then $g(s)\equiv 0$ satisfies (g.1)–(g.4). Otherwise, we can construct g(s) as follows. We define ${\widetilde{g}}(s)$ by

$$\begin{aligned} {\widetilde{g}}(s) = \int _0^s \left( - f'(t)\right) _+ \,dt + s^{p} \quad \text { for all } s \ge 0. \end{aligned}$$

Then ${\widetilde{g}}\in C^1([0,\infty ), [0,\infty ))$ and ${\widetilde{g}}(0)={\widetilde{g}}'(0)=0$. Since

$$\begin{aligned} {\widetilde{g}}(s) \ge \int _0^s (-f'(t)) \,dt + s^{p} = -f(s)+s^p \quad \text { for all } s \ge 0, \end{aligned}$$

we have $f(s)+{\widetilde{g}}(s)>0$ for all $s>0$. From (f.4), there exists $s_0>1$ such that $f(s)>0$ for all $s \ge s_0 -1$. Thus we take an odd function $g\in C^1(\mathbb {R},\mathbb {R})$ satisfying

$$\begin{aligned} \begin{aligned} g(s)&={\left\{ \begin{array}{ll} {\widetilde{g}}(s) &{} \text { for } 0\le s \le s_0-1, \\ 0 &{} \text { for } s_0\le s, \\ \end{array}\right. } \\ g(s)&\ge 0 \quad \text { for } s_0-1 \le s \le s_0. \end{aligned} \end{aligned}$$

Then, g(s) satisfies (g.1)–(g.4).

For $q=2^*=2N/(N-2)>2$ and $L \ge 1$, we define

$$\begin{aligned} a_L(s)= s - (s-L)_+^q \quad \text { for all } s \ge 0. \end{aligned}$$

By using $a_L(s)$, we modify the quadratic term of I(u) as follows. For $L \ge 1$ and $u \in H^1(\mathbb {R}^N)$, we define $J_L \in C^2(H^1(\mathbb {R}^N), \mathbb {R})$ by

$$\begin{aligned} J_{L}(u)= a_L\left( \frac{1}{2} \Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u) \,dx\right) - \int _{\mathbb {R}^N} F(u) + G(u)\,dx, \end{aligned}$$

where $G(u)=\int _0^u g(s)\,ds$. We remark that $J_L(u)$ is written as

$$\begin{aligned} J_{L}(u) =\frac{1}{2} \Vert u\Vert ^2 -\left( \frac{1}{2} \Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u) \,dx-L\right) _+^q - \int _{\mathbb {R}^N} F(u)\,dx. \end{aligned}$$

(2.2)

Therefore $J_L(u)$ satisfies

$$\begin{aligned}\begin{aligned}&J_{L}(u)=I(u) \quad \text { if } \,\frac{1}{2} \Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u) \,dx\le L, \\&J_L(u) \le I(u) \quad \text { for all } u \in H^1(\mathbb {R}^N). \end{aligned}\end{aligned}$$

Our required sequence satisfying the $H^1(\mathbb {R}^N)$-boundedness and the compactness will given as a sequence of critical points of $J_{L}: H^1_0(B_{R_n})\rightarrow \mathbb {R}$ ($R_n\rightarrow \infty$).

At the end of this subsection, we state properties of $a_L(s)$ which are used later.

Lemma 2.2

$a_L(s)\in C^2([0,\infty ),\mathbb {R})$ satisfies the following:

• (i) $a_L'(s)\le 1$ for all $s\ge 0$.

• (ii) If $a_L'(s_1)=0$, then $a_L(s_1)> L$.

Proof

(i) follows from $a_L'(s)=1-q(s-L)_+^{q-1}$. We show (ii). Let $s_1>0$ satisfy $a_L'(s_1)=1-q(s_1-L)_+^{q-1}=0$. Then, we see that $s_1>L$ and

$$\begin{aligned} (s_1-L)_+^{q-1}=(s_1-L)^{q-1}=\frac{1}{q}. \end{aligned}$$

Thus

$$\begin{aligned} a_L(s_1)=s_1-(s_1-L)^q =s_1-\frac{1}{q}(s_1-L) =\left( 1-\frac{1}{q}\right) (s_1-L)+L > L. \end{aligned}$$

Hence, (ii) holds. $\square$

2.2 The properties of modified functional

In this section, we state some properties of $J_L(u)$. First, thanks to the modification, we can easily obtain the boundedness of $\{u \in H^1(\mathbb {R}^N) | J_L(u)\ge 0\}$.

Lemma 2.3

Assume $L \ge 1$. There exists a constant $C_0=C_0(L)>0$ such that

$$\begin{aligned} \Vert u\Vert \le C_0 \quad \hbox { for all } u \in H^1(\mathbb {R}^N) \hbox { with } J_{L}(u) \ge 0. \end{aligned}$$

Proof

From (f.1)–(f.3), there exists a constant $c_1>0$ such that $|F(s)|\le \frac{V_0}{2}|s|^2+\frac{c_1}{p+1}|s|^{p+1}$ for all $s\in \mathbb {R}$. Then, from (2.1), for some $c_2>0$, we have

$$\begin{aligned} \left| \int _{\mathbb {R}^N} F(u) \,dx\right| \le \frac{V_0}{2} \Vert u\Vert _2^2 + \frac{c_1}{p+1}\Vert u\Vert _{p+1}^{p+1} \le \frac{1}{2}\Vert u\Vert ^2+ c_2 \Vert u\Vert ^{p+1}. \end{aligned}$$

Thus, from (2.2), we get

$$\begin{aligned} J_L(u)&\le \Vert u\Vert ^2 -\left( \frac{1}{2}\Vert u\Vert ^2 +\int _{\mathbb {R}^N}G(u) \,dx-L\right) _+^{q} + c_2 \Vert u\Vert ^{p+1} \nonumber \\&\le \Vert u\Vert ^2 - \left( \frac{1}{2}\Vert u\Vert ^2-L\right) _+^{q} + c_2\Vert u\Vert ^{p+1} =h(\Vert u\Vert ). \end{aligned}$$

(2.3)

Here we set $h(s)=s^2-\left( \frac{1}{2}s^2 -L\right) _+^{q}+cs^{p+1}$. Since $\lim _{s \rightarrow \infty } h(s) = -\infty$, there exists a constant $C_0=C_0(L)>0$ such that

$$\begin{aligned} h(t) < 0 \quad \text { for all } t\in [C_0, \infty ). \end{aligned}$$

(2.4)

From (2.3) and (2.4), $J_L(u)\ge 0$ implies $\Vert u\Vert \le C_0$. $\square$

For $R\ge 1$, we also consider functional $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$ that is restricted on $H_0^1(B_R)$. (We use same notation $J_L$.) We see that $J_L\in C^2(H_0^1(B_R),\mathbb {R})$ and

$$\begin{aligned} \begin{aligned} J_L'(u)\varphi&=a_L'\left( \frac{1}{2}\Vert u\Vert ^2 +\int _{B_R}G(u) \,dx\right) \left( \langle u,\varphi \rangle +\int _{B_R}g(u)\varphi \, dx\right) \\&\qquad -\int _{B_R}f(u)\varphi +g(u)\varphi \, dx \quad \text{ for } \text{ all } \varphi \in H_0^1(B_R). \end{aligned} \end{aligned}$$

Thus, if u is a critical point of $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$, then u is a solution of

$$\begin{aligned} a_L'(\beta )\left( -\Delta u+V(x)u\right) =f(u)+\left( 1-a_L'(\beta )\right) g(u) \quad \hbox { in } B_R, \qquad u\in H_0^1\left( B_R\right) , \end{aligned}$$

where $\beta =\frac{1}{2}\Vert u\Vert ^2 +\int _{B_R}G(u) \,dx$.

Moreover $J_L(u)$ satisfies the Palais–Smale condition.

Lemma 2.4

Assume $L \ge 1$ and $R\in [1, \infty )$. Then, for any $c\in (0,L]$, $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$ satisfies (PS)$_c$-condition, that is, every (PS)-sequence of $J_L$ at level c has a convergent subsequence.

Proof

Let $(u_n)_{n=1}^\infty \subset H_0^1(B_R)$ satisfy

$$\begin{aligned} J_L'\left( u_n\right) \rightarrow 0 \quad \hbox { in } \left( H_0^1(B_R)\right) ^*, \quad J_L(u_n)\rightarrow c\in (0,L]. \end{aligned}$$

From Lemma 2.3, since $\Vert u_n\Vert$ is bounded, there exist a subsequence $(u_n)_{n=1}^\infty$ (we use same notation), $\alpha \ge 0$, and $u_0\in H_0^1(B_R)$ such that

$$\begin{aligned}&\Vert u_n\Vert \rightarrow \alpha , \end{aligned}$$

(2.5)

$$\begin{aligned}&u_n \rightarrow u_0 \quad \hbox { weakly in } H_0^1(B_R) \hbox { and strongly in } L^{p+1}(B_R). \end{aligned}$$

(2.6)

From $J_L'(u_n)u_0\rightarrow 0$, we have

$$\begin{aligned} J_L'\left( u_n\right) u_0&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2 +\int _{B_R}G(u_n) \,dx\right) \left( \langle u_n,u_0 \rangle +\int _{B_R}g(u_n)u_0\, dx\right) \nonumber \\&\qquad -\int _{B_R}f(u_n)u_0+g(u_n)u_0\, dx \nonumber \\&\rightarrow a_L'(\beta )\left( \Vert u_0\Vert ^2+\int _{B_R}g(u_0)u_0\, dx\right) -\int _{B_R} f(u_0)u_0+g(u_0)u_0\, dx=0. \end{aligned}$$

(2.7)

Here we set

$$\begin{aligned} \beta =\frac{1}{2}\alpha ^2 +\int _{B_R}G(u_0) \,dx. \end{aligned}$$

(2.8)

Also, from $J_L'(u_n)u_n\rightarrow 0$, we have

$$\begin{aligned} J_L'(u_n)u_n&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2 +\int _{B_R}G(u_n) \,dx\right) \left( \Vert u_n\Vert ^2+\int _{B_R}g(u_n)u_n\, dx\right) \nonumber \\&\qquad -\int _{B_R}f(u_n)u_n+g(u_n)u_n\, dx \nonumber \\&\rightarrow a_L'(\beta )\left( \alpha ^2+\int _{B_R}g(u_0)u_0\, dx\right) -\int _{B_R} f(u_0)u_0+g(u_0)u_0\, dx=0. \end{aligned}$$

(2.9)

Thus, subtracting (2.9) from (2.7), we get

$$\begin{aligned} a_L'(\beta )\left( \Vert u_0\Vert ^2-\alpha ^2\right) =0. \end{aligned}$$

If $a_L'(\beta )\not =0$, then we have $\Vert u_0\Vert ^2=\alpha ^2$. Thus, from (2.5) and (2.6), we see that $u_n\rightarrow u_0$ strongly in $H_0^1(B_R)$ and the proof is completed. Therefore we show $a_L'(\beta )\not =0$ by contradiction, and suppose $a_L'(\beta )=0$. From (2.7), we have

$$\begin{aligned} \int _{B_R} f(u_0)u_0+g(u_0)u_0\, dx=0. \end{aligned}$$

Since $f(s)s+g(s)s>0$ $(s\not =0)$ from (g.1) and (g.4), we see that $u_0=0$. From $J_L(u_n)\rightarrow c\in (0,L]$ and (2.6), we have

$$\begin{aligned} J_L(u_n)&=a_L\left( \frac{1}{2}\Vert u_n\Vert ^2 +\int _{B_R}G(u_n) \,dx\right) -\int _{B_R}F(u_n)+G(u_n)\, dx \rightarrow a_L\left( \frac{1}{2}\alpha ^2 \right) =c\le L. \end{aligned}$$

(2.10)

On the other hand, (2.8) implies $a_L'(\beta )=a_L'\left( \frac{1}{2}\alpha ^2 \right) =0$. Thus, by (ii) of Lemma 2.2, $a_L\left( \frac{1}{2}\alpha ^2 \right) >L$. This contradicts (2.10). Thus, $a_L'(\beta )\not =0$, and the proof was finished. $\square$

Moreover $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$ has a mountain pass geometry which does not depend on $L\ge 1$ and $R\in [1,\infty ]$.

Lemma 2.5

For $L\ge 1$ and $R\in [1,\infty ]$, $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$ satisfies the following:

• (i) $J_L(0)=0$.

• (ii) There exist $\delta >0$ and $\rho >0$ which are independent of $L\ge 1$ and $R\in [1,\infty ]$ such that

  $$\begin{aligned} J_L(u)\ge \delta \quad \hbox { for all } u\in H_0^1(B_R) \hbox { with } \Vert u\Vert =\rho . \end{aligned}$$

• (iii) For any $k\in \mathbb {N}$, there exist subspace $E_k\subset H_0^1(B_1)\subset H_0^1(B_R)$ and $r_k>0$ which are independent of $L\ge 1$ and $R\in [1,\infty ]$ such that $E_k\subset E_{k+1}$ and

  $$\begin{aligned} J_L(u)\le 0 \quad \hbox { for all } u\in E_k \hbox { with } \Vert u\Vert \ge r_k. \end{aligned}$$

Proof

Since the oddness of f and g implies $f(0)=g(0)=0$, (i) is trivial. From (g.1)–(g.3), for some $c_1>0$, we have $|G(s)|\le c_1|s|^2$ for all $s\in \mathbb {R}$. Thus, there is a constant $c_2>0$ such that

$$\begin{aligned} \left( \frac{1}{2} \Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u) \,dx-L\right) _+^q \le \left( \frac{1}{2} \Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u) \,dx\right) ^q\le c_2\Vert u\Vert ^{2q}. \end{aligned}$$

(2.11)

From (f.1)–(f.3), for some $c_3>0$, we have $|F(s)|\le \frac{V_0}{4}|s|^2+c_3|s|^{p+1}$ for all $s\in \mathbb {R}$. Thus, from (2.1), there is a constant $c_4>0$ such that

$$\begin{aligned} \int _{\mathbb {R}^N}F(u) \,dx\le \frac{1}{4}\Vert u\Vert ^2+c_4\Vert u\Vert ^{p+1}. \end{aligned}$$

(2.12)

From (2.2) and (2.11)–(2.12), we have

$$\begin{aligned} J_L(u)\ge \frac{1}{4}\Vert u\Vert ^2-c_2\Vert u\Vert ^{2q}-c_4\Vert u\Vert ^{p+1}. \end{aligned}$$

Since $2<p+1<2q$, by Young’s inequality, for some $c_5>0$,

$$\begin{aligned} J_L(u)\ge \frac{1}{8}\Vert u\Vert ^2-c_5\Vert u\Vert ^{2q} =\frac{1}{8}\Vert u\Vert ^2\left( 1-8c_5\Vert u\Vert ^{2(q-1)}\right) . \end{aligned}$$

Thus, setting $\rho =\left( \frac{1}{16c_5}\right) ^{\frac{1}{2(q-1)}}>0$ and $\delta =\frac{1}{16}\left( \frac{1}{16c_5}\right) ^{\frac{1}{q-1}}>0$, we get (ii). Next, we show (iii). We choose $w_1, \dots , w_k \in C^\infty _0(B_1)\setminus \{0\}$ such that ${\text {supp}}w_i \cap {\text {supp}}w_j =\emptyset$ for $i \not = j$, and set

$$\begin{aligned} E_k= {\text {span}}\left\{ w_1, \dots , w_k\right\} . \end{aligned}$$

From (2.2), for any $w\in E_k$ with $\Vert w\Vert =1$, we have

$$\begin{aligned} J_L(tw)\le \frac{1}{2}t^2\Vert w\Vert ^2-\int _{B_1}F(tw)\, dx \le t^2\left( \frac{1}{2}-\int _{B_1}\frac{F(tw)}{(tw)^2}w^2\, dx\right) . \end{aligned}$$

From (f.1) and (f.4), we have

$$\begin{aligned} \lim _{|t| \rightarrow \infty } \int _{B_1} \frac{F(t w)}{(t w)^2} w^2 \,dx = \infty , \end{aligned}$$

where the above limit is uniformly with respect to $w\in E_k$ with $\Vert w\Vert =1$. Thus we see that (iii) holds. $\square$

For the critical points of $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$, we have the following.

Proposition 2.6

Assume $L\ge 1$ and $R\in [1, \infty ]$. If $u\in H_0^1(B_R)$ satisfies $J_L'(u)=0$ in $(H_0^1(B_R))^*$, then it holds that

$$\begin{aligned} a_L'\left( \frac{1}{2}\Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u)\,dx \right) \in (0,1]. \end{aligned}$$

Moreover, for $\lambda =a_L'\left( \frac{1}{2}\Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u)\,dx \right) ^{-1}\ge 1$, u is a solution of

$$\begin{aligned} -\Delta u+V(x)u=\lambda f(u)+(\lambda -1)g(u) \quad \text { in } B_R, \qquad u\in H_0^1(B_R). \end{aligned}$$

(2.13)

Proof

We set $\beta =\frac{1}{2}\Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u)\, dx\ge 0$. Then (i) of Lemma 2.2 asserts $a_L'(\beta )\le 1$. Since $u\in H_0^1(B_R)$ satisfies $J_L'(u)u=0$, we have

$$\begin{aligned} a_L'\left( \beta \right) \int _{B_R} |\nabla u|^2+V(x)u^2 +g(u)u \,dx =\int _{B_R} f(u)u +g(u)u \,dx. \end{aligned}$$

(2.14)

Since $f(s)s+g(s)s\ge 0$ and $g(s)s\ge 0$ from (g.1) and (g.4), we get $a_L'(\beta )\ge 0$. We suppose $a_L'(\beta )=0$ by contradiction to show $a_L'(\beta )>0$. Then, from (2.14), we have

$$\begin{aligned} \int _{B_R} f(u)u +g(u)u \,dx=0. \end{aligned}$$

Since $f(s)s+g(s)s>0$ $(s\not =0)$, we get $u=0$. This implies $\beta =\frac{1}{2}\Vert u\Vert ^2+\int _{B_R}G(u)\,dx=0$. Thus, $a_L(\beta )=0$ holds. On the other hand, by (ii) of Lemma 2.2, $a_L'(\beta )=0$ implies $a_L(\beta )>L$. This contradicts $a_L(\beta )=0$. Therefore $a_L'(\beta )\in (0,1]$. Set $\lambda =a_L'(\beta )^{-1}\ge 1$. For any $\varphi \in H_0^1(B_R)$, it holds that

$$\begin{aligned} J_L'(u)\varphi =\frac{1}{\lambda }\int _{B_R} \nabla u \cdot \nabla \varphi +V(x)u\varphi +g(u)\varphi \,dx -\int _{B_R} f(u)\varphi +g(u)\varphi \,dx=0. \end{aligned}$$

This means that u is a solution of (2.13). $\square$

2.3 $H^1(\mathbb {R}^N)$-boundedness for modified functional on $\mathbb {R}^{N}$

In our approach, we have to invalidate the modification finally. Therefore we require apriori estimate for critical points of $J_L:H^1(\mathbb {R}^N) \rightarrow \mathbb {R}$ and this estimate must be independent of large $L>0$. For the original functional $I:H^1(\mathbb {R}^N) \rightarrow \mathbb {R}$, Jeanjean-Tanaka [11] had gotten apriori estimate as follows.

Proposition 2.7

(cf. [11, Proposition 4.2]) Assume (f.1)–(f.4) and (V.1)–(V.3). If $(u_n)_{n=1}^\infty \subset H^1(\mathbb {R}^N)$ satisfies $\varlimsup _{n\rightarrow \infty }I(u_n)<\infty$ and $I'(u_n)=0$ in $(H^1(\mathbb {R}^N))^*$, then $(u_n)_{n=1}^\infty$ is bounded in $H^1(\mathbb {R}^N)$.

Proof

This follows from [11, Proposition 4.2] and its proofs. We can apply the almost same proofs of [11, Proposition 4.2] to $(u_n)_{n=1}^\infty \subset H^1(\mathbb {R}^N)$ satisfying $\varlimsup _{n\rightarrow \infty }I(u_n)<\infty$ and $I'(u_n)=0$ in $(H^1(\mathbb {R}^N))^*$. We remark that [11, Proposition 4.2] required that V satisfies $\sup _{x\in \mathbb {R}^N}V(x)\le V_\infty$. However, this assumption is not essential and we can easily remove it. $\square$

We discuss similar argument of Proposition 2.7 to modified functional $J_L:H^1(\mathbb {R}^N) \rightarrow \mathbb {R}$ in next proposition. For $J_L$, we also obtain the following apriori estimate which is independent of large $L>0$.

Proposition 2.8

Assume (f.1)–(f.4) and (V.1)–(V.3). For any $b>0$, there exist constants $C_1=C_1(b)>0$ and ${\mathcal {L}}_1={\mathcal {L}}_1(b)>b$ such that, for any $L\ge {\mathcal {L}}_1$, we have

$$\begin{aligned} \frac{1}{2}\Vert u\Vert ^2+ \int _{\mathbb {R}^N} G(u) \,dx \le C_1 \quad \hbox { for all } u\in K_{L,b}, \end{aligned}$$

where $K_{L,b}=\left\{ u\in H^1(\mathbb {R}^N) \big | 0\le J_L(u) \le b, \ J_L'(u)=0 \hbox { in } (H^1(\mathbb {R}^N))^* \right\} .$

Remark 2.9

In the proofs of Propositions 2.7 and 2.8 , we don’t use $f, g\in C^1(\mathbb {R},\mathbb {R})$ but use only $f,g\in C(\mathbb {R},\mathbb {R})$ (see Sect. 4). Thus, under the assumptions (f.1$'$)–(f.3$'$), (f.4), and (V.1)–(V.3), Propositions 2.7 and 2.8 still hold.

We prove Proposition 2.8 in Sect. 4. As a corollary of Proposition 2.8, we obtain the following.

Corollary 2.10

For any $b>0$, there exists a constant ${\mathcal {L}}_2={\mathcal {L}}_2(b)>b$ such that, for any $L\ge {\mathcal {L}}_2$, if $u\in K_{L,b}$, then we have

$$\begin{aligned} 0 \le I(u)=J_L(u) \le b, \quad I'(u) =0 \quad \text { in } \left( H^1(\mathbb {R}^N)\right) ^*. \end{aligned}$$

Proof

We set ${\mathcal {L}}_2 = \max \left\{ {\mathcal {L}}_1, \,C_1+1\right\}$ where ${\mathcal {L}}_1$ and $C_1$ are constants which were given in Proposition 2.8. For any $L \ge {\mathcal {L}}_2$, if $u \in K_{L,b}$, then we have

$$\begin{aligned} \frac{1}{2}\Vert u\Vert ^2 +\int _{\mathbb {R}^N} G(u) \,dx\le C_1 \le {\mathcal {L}}_2-1 \le L-1. \end{aligned}$$

Therefore, we see that $J_L \equiv I$ in a neighborhood of u and we get the conclusion. $\square$

2.4 Balanced sequence and the compactness

In this section, we consider about balanced sequences and the compactness.

Definition 2.11

Suppose ${\mathcal {I}}\in C^1(H^1(\mathbb {R}^N),\mathbb {R})$. If $u_n \in H_0^1(B_{R_n})$ $(R_n\rightarrow \infty )$ satisfies $\sup _{n\in \mathbb {N}}\left| {\mathcal {I}}(u_n)\right| < \infty$ and ${\mathcal {I}}'(u_n)=0$ in $(H_0^1(B_{R_n}))^*$, then we say that $(u_n)_{n=1}^\infty$ is a balanced sequence for ${\mathcal {I}}$.

We construct balanced sequences for $J_L:H^1(\mathbb {R}^N)\rightarrow \mathbb {R}$. For any $k\in \mathbb {N}$, let subspace $E_k\subset H_0^1(B_1)$ and $r_k>0$ be as in (iii) of Lemma 2.5. For $L\ge 1$ and $R\in [1,\infty ]$, we define minimax values as follows:

$$\begin{aligned} D^k&=\{u \in E_k | \Vert u\Vert \le r_k\}, \nonumber \\ \Gamma _R^k&=\left\{ \gamma \in C\left( D^k, H_0^1(B_R)\right) \Bigg | \begin{aligned}&\gamma (-u)=-\gamma (u) \text { for all } u \in D^k, \\&\gamma (u)=u \text { for all } u \in \partial D^k \end{aligned}\right\} , \nonumber \\ b_{L,R}^k&=\inf _{\gamma \in \Gamma _R^k} \max _{u \in D^k}J_L(\gamma (u)), \end{aligned}$$

(2.15)

$$\begin{aligned} b_R^k&=\inf _{\gamma \in \Gamma _R^k} \max _{u \in D^k}I(\gamma (u)). \end{aligned}$$

(2.16)

Lemma 2.12

For $k\in \mathbb {N}$, $L\ge 1$, and $R\in [1,\infty )$, we have the following:

• (i) $b_{L,R}^k \le b_R^k \le b_1^k$.

• (ii) $b_{L,R}^k$ is a critical value of $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$ if $b_{L,R}^k \le L$.

Proof

Since $\Gamma _1^k\subset \Gamma _R^k$, we have $b_R^k\le b_1^k$. Also, since $J_L(u) \le I(u)$, we have $b_{L,R}^k\le b_R^k$. Thus, we get (i). From Lemmas 2.4 and 2.5 , by a standard method, we see that $b_{L,R}^k$ is a critical value of $J_L:H_0^1(B_R)\rightarrow \mathbb {R}$. $\square$

Then there exist critical points having the estimates from the below of the Morse indexes.

Lemma 2.13

For $k\in \mathbb {N}$, $L\ge b_1^k$, and $R\in [1,\infty )$, there exists $w_{L,R}^k\in H_0^1(B_R)$ such that

$$\begin{aligned} \begin{aligned}&0\le J_L\left( w_{L,R}^k\right) \le b_{L,R}^k\le b_1^k, \\&J_L'\left( w_{L,R}^k\right) =0 \quad \hbox { in } \left( H_0^1(B_R)\right) ^*, \\&{\text {index}}_0\, J_L''\left( w_{L,R}^k\right) \ge k, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} {\text {index}}_0 J_L''(w_{L,R}^k)=\max \left\{ \dim H \left| \begin{aligned}&H \subset H_0^1(B_R) \text { is a subspace such that } \\&J_L''(w_{L,R}^k)(h,h) \le 0 \text { for } h \in H_0^1(B_R) \end{aligned}\right. \right\} . \end{aligned}$$

Proof

This follows from [18, Theorem B]. We remark that [18, Theorem B] is true, if we replace the assumption $\mathrm (I_4)$ in [18] to the following $\mathrm (I_4)'$.

$\mathrm (I_4)'$ :

For any u with $I'(u)=0$, $I''(u)$ is represented as $I''(u)=a_uid+K_u$, where $a_u>0$ and $K_u$ is a compact operator.

In fact, in the proof of [18, Theorem B], we use only $\mathrm (I_4)'$. $J_L(u)$ satisfies $\mathrm (I_4)'$ because $J_L''(u)$ is written as $\langle J_L''(u)\varphi , \psi \rangle =\langle a_L'(P(u))\varphi , \psi \rangle +\langle K_u\varphi , \psi \rangle ,$ where $P(u)=\frac{1}{2}\Vert u\Vert ^2+\int _{B_R}G(u)\, dx$ and

$$\begin{aligned} \begin{aligned} \langle K_u\varphi , \psi \rangle&= a_L''\left( P(u)\right) P'(u)\varphi P'(u)\psi \\&\quad -\int _{B_R}f'(u)\varphi \psi \, dx -\left( 1-a_L'\left( P(u)\right) \right) \int _{B_R}g'(u)\varphi \psi \, dx. \end{aligned} \end{aligned}$$

From Proposition 2.6, $J_L'(u)=0$ implies $a_L'(P(u))>0$. We can find that $K_u:H_0^1(B_R)\rightarrow H_0^1(B_R)$ is a compact operator. Thus we get Lemma 2.13. $\square$

The following proposition guarantees the compactness of $(w_{L,R}^k)_{R\ge 1}$.

Proposition 2.14

We assume (f.1)–(f.4), (V.1), (V.2), and (V.4). Let $u_n\in H_0^1(B_{R_n})$ $(R_n\rightarrow \infty )$ satisfy

$$\begin{aligned} 0 \le J_L(u_n)\le L, \quad J_L'\left( u_n\right) =0 \quad \hbox { in } \left( H_0^1(B_{R_n})\right) ^*. \end{aligned}$$

(2.17)

Then there exist a subsequence $(u_n)_{n=1}^\infty$ (we use same notation) and $u_0\in H^1(\mathbb {R}^N)$ such that

$$\begin{aligned} \Vert u_n-u_0\Vert \rightarrow 0 \quad \hbox { and } \quad J_L'(u_0)=0 \quad \hbox { in } \left( H^1(\mathbb {R}^N)\right) ^*. \end{aligned}$$

For the original functional I(u), the following similar compactness holds.

Proposition 2.15

We assume (f.1)–(f.4), (V.1), (V.2), and (V.4). Let $u_n\in H_0^1(B_{R_n})$ $(R_n\rightarrow \infty )$ be bounded in $H^1(\mathbb {R}^N)$ and satisfy

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }I(u_n)<\infty \quad \hbox { and } \quad I'(u_n)=0 \quad \hbox {in } \left( H_0^1(B_{R_n})\right) ^*. \end{aligned}$$

Then there exist a subsequence $(u_n)_{n=1}^\infty$ (we use same notation) and $u_0\in H^1(\mathbb {R}^N)$ such that

$$\begin{aligned} \Vert u_n-u_0\Vert \rightarrow 0 \quad \hbox { and } \quad I'(u_0)=0 \quad \hbox { in } \left( H^1(\mathbb {R}^N)\right) ^*. \end{aligned}$$

Proof

The proof is almost same as the proof of Proposition 2.14. Thus we omit it. $\square$

Remark 2.16

In the proofs of Propositions 2.14 and 2.15 , we don’t use differentiability of f or g (see Sects. 5 and 6). Thus, under the assumptions (f.1$'$)–(f.3$'$), (f.4), (V.1)–(V.2), and (V.4), Propositions 2.14 and 2.15 still hold.

In order to prove Proposition 2.14, we use the concentration compactness arguments. The key of getting the compactness is the assumption (V.4). We argue with the concentration compactness in Sect. 5 and prove Proposition 2.14 in Sect. 6.

3 Proof of main theorems

First, we give a proof of Theorem 1.1. We need the following lemma which is similar with [3, Lemma A.1].

Lemma 3.1

If $w \in H^1(\mathbb {R}^N)$ satisfies $I'(w)=0$, then there exists a finite dimensional subspace $M\subset H^1(\mathbb {R}^N)$ such that

$$\begin{aligned} I''(w)(h,h) \ge \frac{1}{2}\Vert h\Vert ^2 \quad \text { for all } h \in M^\perp . \end{aligned}$$

Proof

We suppose, by contradiction, that there exists a sequence $(h_n)_{n=1}^\infty \subset H^1(\mathbb {R}^N)$ such that

$$\begin{aligned} \langle h_m, h_n\rangle =\delta _{mn}, \quad I''(w)(h_n,h_n) < \frac{1}{2} \Vert h_n\Vert ^2 \quad \text { for all } m,n \in \mathbb {N}. \end{aligned}$$

(3.1)

Here we have

$$\begin{aligned} I''(w)(h_n,h_n)&= \Vert h_n\Vert ^2 - \int _{\mathbb {R}^N} f'(w) h_n^2 \,dx \nonumber \\&\ge \frac{3}{4} \Vert h_n\Vert ^2 +\int _{\mathbb {R}^N}\left( \frac{1}{4}V(x)-f'(w)\right) h_n^2 \,dx. \end{aligned}$$

(3.2)

From (3.1)–(3.2), we have

$$\begin{aligned} \int _{\mathbb {R}^N}\left( \frac{1}{4}V(x)-f'(w)\right) h_n^2 \,dx <-\frac{1}{4}\Vert h_n\Vert ^2=-\frac{1}{4}. \end{aligned}$$

(3.3)

Since |w| satisfies $-\Delta |w|+V(x)|w|\le f(|w|)$ in $\mathbb {R}^N$, by a subsolution estimate (cf. Lemma 6.1), $\Vert w\Vert _{L^{p+1}(\mathbb {R}^N\setminus B_R)}\rightarrow 0$ ($R\rightarrow \infty$) implies $\Vert w\Vert _{L^{\infty }(\mathbb {R}^N\setminus B_R)}\rightarrow 0$ ($R\rightarrow \infty$). Thus, there exists $r>0$ such that

$$\begin{aligned} \frac{1}{4}V(x)-f'(w)>0 \quad \text { if } |x|\ge r. \end{aligned}$$

Therefore, there exists $c(x) \in C^\infty _0(B_r)$ such that

$$\begin{aligned} \frac{1}{4}V(x)-f'(w) \ge c(x) \quad \text { in } \mathbb {R}^N. \end{aligned}$$

(3.4)

From (3.3)–(3.4), we have

$$\begin{aligned} \int _{B_r} c(x) h_n^2 \,dx \le \int _{\mathbb {R}^N} \left( \frac{1}{4}V(x)-f'(w)\right) h_n^2 \,dx <-\frac{1}{4}. \end{aligned}$$

(3.5)

On the other hand, since $h_n \rightarrow 0$ weakly in $H^1(\mathbb {R}^N)$, we have $h_n \rightarrow 0$ in $L^2(B_r)$ and

$$\begin{aligned} \int _{B_r} c(x) h_n^2 \,dx \rightarrow 0 \quad \text { as } n \rightarrow \infty . \end{aligned}$$

This contradicts (3.5). Thus we get the conclusion. $\square$

Now we prove Theorem 1.1.

Proof of Theorem 1.1

For any $k\in \mathbb {N}$, we define a minimax value $b_1^k$ as (2.16). We choose and fix $L^k={\mathcal {L}}_2(b_1^k)>0$ in Corollary 2.10. We consider the modified functional $J_{L^k}(u)$ and define minimax values $b_{L^k,R}^k$ as (2.15). From Lemma 2.13, for $R\in [1,\infty )$, there exists $w_{L^k,R}^k\in H_0^1(B_R)$ such that

$$\begin{aligned} \begin{aligned}&0 \le J_{L^k}\left( w_{L^k,R}^k\right) \le b_{L^k,R}^k \le b_1^k,\\&J_{L^k}'\left( w_{L^k,R}^k\right) =0 \quad \hbox { in } \left( H_0^1(B_R)\right) ^*,\\&{\text {index}}_0 J_{L^k}''\left( w_{L^k,R}^k\right) \ge k. \end{aligned} \end{aligned}$$

Here we set

$$\begin{aligned} b^k=\varliminf _{R\rightarrow \infty }J_{L^k}\left( w_{L^k,R}^k\right) . \end{aligned}$$

From Proposition 2.14, there exist a subsequence $(w_{L^k,R_n}^k)_{n=1}^{\infty }$ ($R_n\rightarrow \infty$) and $w_{L^k}^k\in H^1(\mathbb {R}^N)$ such that

$$\begin{aligned}&\left\| w_{L^k,R_n}^k-w_{L^k}^k\right\| \rightarrow 0, \nonumber \\&J_{L^k}\left( w_{L^k}^k\right) =b^k, \nonumber \\&J_{L^k}'\left( w_{L^k}^k\right) =0 \quad \hbox { in } \left( H^1(\mathbb {R}^N)\right) ^*. \end{aligned}$$

(3.6)

From the choice of $L^k={\mathcal {L}}_2(b_1^k)$ in Corollary 2.10, $w_{L^k}^k$ satisfies

$$\begin{aligned}&\frac{1}{2}\left\| \omega ^k_{L^k}\right\| ^2+\int _{\mathbb {R}^N}G\left( \omega ^k_{L^k}\right) \, dx <L^k-1 \nonumber \\&b^k=I\left( w_{L^k}^k\right) =J_{L_k}\left( w_{L^k}^k\right) \le b_1^k, \nonumber \\&I'\left( w_{L^k}^k\right) =J_{L_k}'\left( w_{L^k}^k\right) =0 \quad \hbox { in } \left( H^1(\mathbb {R}^N)\right) ^*. \end{aligned}$$

(3.7)

Thus $w_{L^k}^k$ is a critical point of I(u). If we get $b^k\rightarrow \infty$ as $k\rightarrow \infty$, then the proof of Theorem 1.1 is finished. To show $b^k\rightarrow \infty$ by contradiction, suppose that there exists ${\overline{b}}>0$ such that

$$\begin{aligned} b^k \le {\overline{b}} \quad \hbox { for all } k\in \mathbb {N}. \end{aligned}$$

Then, from Proposition 2.7, $(w_{L^k}^k)_{k=1}^\infty$ is bounded in $H^1(\mathbb {R}^N)$. Furthermore, from Proposition 2.15, there exist a subsequence $(w_{L^k}^k)_{k=1}^\infty$ (we use same notation) and ${\overline{w}}\in H^1(\mathbb {R}^N)$ such that

$$\begin{aligned} \left\| w_{L^k}^k-{\overline{w}}\right\| \rightarrow 0 \hbox { as } k\rightarrow \infty , \quad I'({\overline{w}})=0 \hbox { in } \left( H^1(\mathbb {R}^N)\right) ^*. \end{aligned}$$

From Lemma 3.1, there exists a finite dimensional subspace $M\subset H^1(\mathbb {R}^M)$ such that

$$\begin{aligned} I''({\overline{w}})(h,h) \ge \frac{1}{2} \Vert h\Vert ^2 \quad \text { for all } h \in M^\perp . \end{aligned}$$

(3.8)

We set $k_0=\dim M$. Since I is $C^2$, there exists $k_1>k_0$ such that

$$\begin{aligned} \left\| I''({\overline{w}}) - I''\left( w^{k_1}_{L^{k_1}}\right) \right\| \le \frac{1}{8}. \end{aligned}$$

From (3.6), there exists $R_n>0$ such that

$$\begin{aligned} \left\| I''\left( w^{k_1}_{L^{k_1}}\right) -I''\left( w_{L^{k_1},R_n}^{k_1}\right) \right\| \le \frac{1}{8}, \quad {\text {index}}_0 I''\left( w_{L^{k_1},R_n}^{k_1}\right) \ge k_1. \end{aligned}$$

From the definition of ${\text {index}}_0$, there exists a finite dimensional subspace ${\widehat{M}} \subset H^1_0(B_R)$ with $\dim {\widehat{M}} =k_1$ such that

$$\begin{aligned} I''\left( w_{L^{k_1},R_n}^{k_1}\right) (h, h) \le 0 \quad \text { for all } h \in {\widehat{M}}. \end{aligned}$$

Since $\dim M=k_0 < k_1 = \dim {\widehat{M}}$, there exists $h \in M^\perp \cap {\widehat{M}}$ with $\Vert h\Vert =1$. From (3.8), we have

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \Vert h\Vert ^2 \le I''({\overline{w}})(h,h) \\&\quad = \left[ \left( I''({\overline{w}})-I''\left( w^{k_1}_{L^{k_1}}\right) \right) +\left( I''\left( w^{k_1}_{L^{k_1}}\right) -I''\left( w_{L^{k_1},R_n}^{k_1}\right) \right) +I''\left( w_{L^{k_1},R_n}^{k_1}\right) \right] (h,h) \\&\quad \le \left\| I''({\overline{w}}) - I''\left( w^{k_1}_{L^{k_1}}\right) \right\| \Vert h\Vert ^2 +\left\| I''\left( w^{k_1}_{L^{k_1}}\right) - I''\left( w_{L^{k_1},R_n}^{k_1}\right) \right\| \Vert h\Vert ^2 \le \frac{1}{4} \Vert h\Vert ^2. \end{aligned} \end{aligned}$$

This is a contradiction. Thus we get $b^k \rightarrow \infty$ and the proof was finished. $\square$

Remark 3.2

In the above proof, from (3.6)–(3.7), there exists $R^k>0$ such that, for any $R>R^k$, $b^k-1\le J_{L^k}(\omega ^k_{L^k,R})$ and

$$\begin{aligned} \frac{1}{2}\left\| \omega ^k_{L^k,R}\right\| ^2+\int _{\mathbb {R}^N}G\left( \omega ^k_{L^k,R}\right) \, dx <L_k- \frac{1}{2}. \end{aligned}$$

(3.9)

Indeed, if (3.9) does not hold, there exists $(R_n)_{n=1}^\infty$ with $R_n \rightarrow \infty$ such that

$$\begin{aligned} \frac{1}{2}\left\| \omega ^k_{L^k,R_n}\right\| ^2+\int _{\mathbb {R}^N}G\left( \omega ^k_{L^k,R_n}\right) \, dx \ge L_k - \frac{1}{2}. \end{aligned}$$

(3.10)

On the other hand, similarly as in the proof of Theorem 1.1, taking a subsequence if necessary, there exists $w_{L^k}^k \in H^1(\mathbb {R}^N)$ such that (3.6) and (3.7) hold. Taking $n \rightarrow \infty$ in (3.10), we have

$$\begin{aligned} \frac{1}{2}\left\| \omega ^k_{L^k}\right\| ^2+\int _{\mathbb {R}^N}G\left( \omega ^k_{L^k}\right) \, dx \ge L_k - \frac{1}{2}. \end{aligned}$$

This contradicts (3.7). From (3.9), $w_{L^k,R}^k$ satisfies

$$\begin{aligned}\begin{aligned}&b^k-1 \le I\left( w_{L^k,R}^k\right) =J_{L^k}\left( w_{L^k,R}^k\right) \le b_{L^k,R}^k \le b^k_1, \\&I'\left( w_{L^k,R}^k\right) =J_{L^k}'\left( w_{L^k,R}^k\right) =0 \quad \hbox { in } (H_0^1(B_R))^*. \end{aligned}\end{aligned}$$

Thus $w_{L^k,R}^k$ $(R\ge R^k)$ is a solution of (1.6) with $\Omega _R=B_R$. Since $\lim _{k\rightarrow \infty }b^k=\infty$, we obtain Theorem 1.4 for the case $\Omega =B_{R}$.

As a by-product of Theorem 1.1, we can obtain Theorem 1.3. We state only outline of the proof.

Outline of proof of Theorem 1.3

We can show Theorem 1.3 as a similar way to the proof of Theorem 1.1. To obtain positive solutions, we put $f \equiv 0$ on $(-\infty , 0)$. Under (f.1$'$)–(f.3$'$), we take an auxiliary function $g(s) = (-f(s))_+ + |s|^{p-1} s$ near $s=0$. We also define $J_L(u)$ as in (2.2). We note that Lemmas 2.2, 2.4, 2.5, Propositions 2.6, 2.8, and 2.14 still hold. Indeed, in this setting, (g.4) does not hold. However, we have $f(s) s + g(s) s \ge 0$ for any $s \in \mathbb {R}$ and $s=0$ is an isolated solution of $f(s)s + g(s) s=0$. Thus $\int _{B_R} f(u) u + g(u) u \,dx =0$ and $u \in H^1_0(B_R)$ imply $u = 0$ in $B_R$. Hence these lemmas and propositions hold. We define the following minimax value

$$\begin{aligned}\begin{aligned} b_{L,R}&=\inf _{\gamma \in \Gamma _R} \max _{u \in [0,1]} J_L(\gamma (u)), \\ \Gamma _R&=\left\{ \gamma \in C\left( [0,1], H_0^1(B_R)\right) \big | \gamma (0)=0, \ \gamma (1)=e\right\} . \end{aligned} \end{aligned}$$

Here we choose $e\in H_0^1(B_1)$ satisfying $J_L(e)<0$. Let $w_{L,R}\in H_0^1(B_R)$ be a critical point for $b_{L,R}$. By the similar way to the proof of Theorem 1.1, for $R_n\rightarrow \infty$, $(w_{L,R_n})_{n=1}^\infty$ is a balanced sequence and, after extracting a subsequence (we use same notation), we can show that $(w_{L,R_n})_{n=1}^\infty$ converges to a nontrivial critical point $w_{L}\in H^1(\mathbb {R}^N)$ of I(u). Moreover, by the maximum principle, we see that $w_{L,R_n}>0$. Thus we got a positive solution of (1.1). $\square$

4 Apriori estimate for critical points of $J_L$

In this section, we prove Proposition 2.8. The fundamental idea of the proof comes from [11, Proposition 4.2]. First, we show the following.

Lemma 4.1

Let $u \in H^1(\mathbb {R}^N)$ satisfy $J_L'(u)=0$ in $(H^1(\mathbb {R}^N))^*$. Then

$$\begin{aligned} NL \left( \frac{1}{2}\Vert u\Vert ^2 + \int _{\mathbb {R}^N}G(u)\,dx-L\right) _+^{q-1} +\Vert \nabla u\Vert _2^2 \le NJ_L(u)+\frac{1}{2}\int _{\mathbb {R}^N}u^2\eta ^2\, dx. \end{aligned}$$

(4.1)

Here $\eta \in L^2(\mathbb {R}^N)\cap W^{1,\infty }(\mathbb {R}^N)$ is the function in (V.3).

Proof

From Proposition 2.6, u is a solution of

$$\begin{aligned} -\Delta u+V(x)u=\lambda f(u)+(\lambda -1) g(u) \quad \text { in } \mathbb {R}^N, \qquad u \in H^1(\mathbb {R}^N), \end{aligned}$$

where $\lambda =a_L'\left( \frac{1}{2}\Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u)\,dx\right) ^{-1}\ge 1$. Thus u satisfies the following Pohozaev identity;

$$\begin{aligned}&\left( \frac{1}{2}-\frac{1}{N}\right) \Vert \nabla u\Vert _2^2 +\frac{1}{2} \int _{\mathbb {R}^N} V(x) u^2 \,dx + \frac{1}{2N} \int _{\mathbb {R}^N} (x \cdot \nabla V(x)) u^2 \,dx \nonumber \\&\quad = \int _{\mathbb {R}^N} \lambda F(u) +(\lambda -1)G(u)\,dx. \end{aligned}$$

(4.2)

Set $\beta =\frac{1}{2}\Vert u\Vert ^2 + \int _{\mathbb {R}^N}G(u) \,dx$. (4.2) is written as

$$\begin{aligned} a_L'(\beta )\left( \beta - \frac{1}{N} \Vert \nabla u\Vert _2^2 +\frac{1}{2N} \int _{\mathbb {R}^N} (x \cdot \nabla V(x)) u^2 \,dx\right) =\int _{\mathbb {R}^N} F(u) +G(u)\,dx. \end{aligned}$$

(4.3)

From (4.3), we have

$$\begin{aligned} \begin{aligned} J_L(u)&= a_L(\beta )-\int _{\mathbb {R}^N} F(u) + G(u) \,dx \\&=a_L(\beta ) -a_L'(\beta )\left( \beta -\frac{1}{N} \Vert \nabla u\Vert _2^2 +\frac{1}{2N} \int _{\mathbb {R}^N} \left( x \cdot \nabla V(x)\right) u^2 \,dx\right) . \end{aligned} \end{aligned}$$

From $0 <a'(\beta )\le 1$ and the definition of $\eta$, we have

$$\begin{aligned} a_L(\beta ) -a_L'(\beta )\left( \beta -\frac{1}{N}\Vert \nabla u\Vert _2^2\right)&=J_L(u)+\frac{a_L'(\beta )}{2N} \int _{\mathbb {R}^N} (x \cdot \nabla V(x)) u^2 \,dx \nonumber \\&\le J_L(u)+\frac{1}{2N} \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx. \end{aligned}$$

(4.4)

Now we calculate the left hand side of (4.4).

$$\begin{aligned}\begin{aligned} a_L(\beta ) - a_L'(\beta )\left( \beta -\frac{1}{N}\Vert \nabla u\Vert _2^2\right)&=\beta -(\beta -L)_+^q-\left( 1-q(\beta -L)_+^{q-1}\right) \left( \beta -\frac{1}{N} \Vert \nabla u\Vert _2^2\right) \nonumber \\&= \frac{1}{N} \Vert \nabla u\Vert _2^2+(\beta -L)_+^{q-1} \left( -\beta +L+q\beta -\frac{q}{N} \Vert \nabla u\Vert _2^2\right) . \end{aligned} \end{aligned}$$

Recalling $\beta =\frac{1}{2}\Vert u\Vert ^2 + \int _{\mathbb {R}^N}G(u) \,dx$, we have

$$\begin{aligned}&-\beta + L+q\beta -\frac{q}{N} \Vert \nabla u\Vert _2^2 \nonumber \\&\quad = \left( \frac{N-2}{2N}q-\frac{1}{2}\right) \Vert \nabla u\Vert _2^2+L+(q-1) \left( \frac{1}{2} \int _{\mathbb {R}^N}V(x)u^2 \,dx +\int _{\mathbb {R}^N} G(u) \,dx\right) \nonumber \\&\quad \ge L. \end{aligned}$$

(4.5)

Combining (4.4) and (4.5), we get

$$\begin{aligned} J_L(u)+\frac{1}{2N} \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx \ge \frac{1}{N} \Vert \nabla u\Vert _2^2+L (\beta -L)_+^{q-1}. \end{aligned}$$

Thus we got (4.1). $\square$

Next we show the following lemma by the argument in [11, Proposition 4.2].

Lemma 4.2

For any $b>0$, there exists a constant $C_2=C_2(b)>0$ such that

$$\begin{aligned} \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx \le C_2 \quad \hbox { for all } u\in K_{L,b}, \end{aligned}$$

where $K_{L,b}$ is defined in Proposition 2.8.

Proof

Let $u\in K_{L,b}$. We set $\lambda =a_L'\left( \frac{1}{2}\Vert u\Vert ^2+\int _{\mathbb {R}^N}G(u)\,dx\right) ^{-1}\ge 1$. From $J_L'(u)(u\eta ^2)=0$, it holds that

$$\begin{aligned} \int _{\mathbb {R}^N} \nabla u \cdot \nabla (u \eta ^2)+V(x)u^2 \eta ^2 \,dx&= \lambda \int _{\mathbb {R}^N} (f(u)+g(u))u\eta ^2 \,dx -\int _{\mathbb {R}^N} g(u)u\eta ^2 \,dx \nonumber \\&\ge \int _{\mathbb {R}^N} (f(u)+g(u))u\eta ^2 \,dx - \int _{\mathbb {R}^N} g(u)u\eta ^2 \,dx \nonumber \\&=\int _{\mathbb {R}^N} f(u) u\eta ^2 \,dx. \end{aligned}$$

(4.6)

Here we used $(f(s)+g(s))s \ge 0$ and $\lambda \ge 1$. From (f.1) and (f.4), for any $M>0$, there exists $c_M>0$ such that

$$\begin{aligned} f(s)s \ge M s^2 - c_M \quad \text { for all } s \in \mathbb {R}. \end{aligned}$$

Thus we have

$$\begin{aligned} \int _{\mathbb {R}^N} f(u) u\eta ^2 \,dx \ge M \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx- c_M \Vert \eta \Vert _2^2. \end{aligned}$$

(4.7)

On the other hand, we have

$$\begin{aligned} \int _{\mathbb {R}^N} V(x) u^2 \eta ^2 \,dx \le V_1 \int _{\mathbb {R}^N}u^2 \eta ^2 \,dx \end{aligned}$$

(4.8)

and

$$\begin{aligned} \int _{\mathbb {R}^N} \nabla u \cdot \nabla (u \eta ^2) \,dx&\le \int _{\mathbb {R}^N} |\nabla u|^2 \eta ^2 \,dx +\int _{\mathbb {R}^N} 2 |\nabla u|\, |\nabla \eta |\, |u\eta | \,dx \nonumber \\&\le \int _{\mathbb {R}^N} |\nabla u|^2 \left( \eta ^2 + |\nabla \eta |^2 \right) \,dx + \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx \nonumber \\&\le \left( \Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2\right) \Vert \nabla u\Vert _2^2 +\int _{\mathbb {R}^N} u^2 \eta ^2 \,dx. \end{aligned}$$

(4.9)

From (4.1) and $J_L(u)\le b$, we have

$$\begin{aligned} \Vert \nabla u\Vert _2^2 \le Nb+\frac{1}{2}\int _{\mathbb {R}^N} u^2 \eta ^2 \,dx. \end{aligned}$$

(4.10)

Combining (4.6), (4.7), (4.8), (4.9), and (4.10), we get

$$\begin{aligned}\begin{aligned} M\int _{\mathbb {R}^N} u^2 \eta ^2 \,dx- c_M \Vert \eta \Vert _2^2 \le&V_1 \int _{\mathbb {R}^N}u^2 \eta ^2 \,dx +\left( \Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2\right) Nb \\&+\frac{\left( \Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2\right) }{2} \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx +\int _{\mathbb {R}^N} u^2 \eta ^2 \,dx. \end{aligned}\end{aligned}$$

Thus we have

$$\begin{aligned}\begin{aligned}&\left( M-V_1-1-\frac{\Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2}{2}\right) \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx \le \left( \Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2\right) Nb +c_M \Vert \eta \Vert _2^2. \end{aligned} \end{aligned}$$

Since $M>0$ is arbitrary, we set $M=V_1 + 2 + \left( \Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2\right) /2$. Then

$$\begin{aligned} \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx \le \left( \Vert \eta \Vert _\infty ^2+\Vert \nabla \eta \Vert _\infty ^2\right) Nb +c_M \Vert \eta \Vert _2^2. \end{aligned}$$

Thus we get the conclusion. $\square$

Here we observe that, when L is large, $\lambda$ is restricted.

Lemma 4.3

For any $b>0$, there exists a constant ${\mathcal {L}}_1={\mathcal {L}}_1(b)>b$ such that, for any $L \ge {\mathcal {L}}_1$,

$$\begin{aligned} \lambda =a_L'\left( \beta \right) ^{-1}\in [1,2] \quad \hbox { for all } u\in K_{L,b}, \end{aligned}$$

where $\beta =\frac{1}{2}\Vert u\Vert ^2 + \int _{\mathbb {R}^N} G(u) \,dx$.

Proof

We set ${\mathcal {L}}_1=2q\left( b+\frac{C_2}{2N}\right)$. From (4.1) and Lemma 4.2, we have

$$\begin{aligned} L(\beta -L)_+^{q-1} \le b + \frac{1}{2N} \int _{\mathbb {R}^N} u^2 \eta ^2 \,dx \le b+ \frac{C_2}{2N} = \frac{{\mathcal {L}}_1}{2q} \le \frac{L}{2q}. \end{aligned}$$

Thus $a_L'(\beta ) = 1-q(\beta -L)_+^{q-1} \ge 1-\frac{q}{2q} = \frac{1}{2}$ holds. $\square$

Next we show the boundedness for $\Vert u\Vert$.

Lemma 4.4

For any $b>0$, there exists a constant $C_3=C_3(b)>0$ such that, for any $L \ge {\mathcal {L}}_1$,

$$\begin{aligned} \Vert u\Vert \le C_3 \quad \hbox { for all } u\in K_{L,b}. \end{aligned}$$

Proof

From (f.1)–(f.3) and (g.1)–(g.3), for any $\epsilon >0$, there exists $c_\epsilon >0$ such that

$$\begin{aligned} |f(s)|, \ |g(s)| \le \epsilon |s|+c_\epsilon |s|^{\frac{N+2}{N-2}} \quad \hbox { for all } s \in \mathbb {R}. \end{aligned}$$

Let $u\in K_{L,b}$ and $L \ge {\mathcal {L}}_1$. Then $\lambda =a_L'\left( \frac{1}{2}\Vert u\Vert ^2 + \int _{\mathbb {R}^N} G(u) \,dx\right) ^{-1}\in [1,2]$ by Lemma 4.3. From $J_L'(u)u=0$, we have

$$\begin{aligned} \begin{aligned} \Vert u\Vert ^2&=\int _{\mathbb {R}^N}\lambda f(u)u+(\lambda -1)g(u)u\, dx \\&\le 3\left( \epsilon \Vert u\Vert _2^2+c_\epsilon \Vert u\Vert _{\frac{2N}{N-2}}^{\frac{2N}{N-2}}\right) \\&\le 3\left( \epsilon \Vert u\Vert _2^2+c_\epsilon C\Vert \nabla u\Vert _2^{\frac{2N}{N-2}}\right) . \end{aligned} \end{aligned}$$

From Lemmas 4.1 and 4.2 ,

$$\begin{aligned} \Vert \nabla u\Vert _2^2 \le Nb+\frac{C_2}{2}. \end{aligned}$$

Thus we see that $\Vert u\Vert$ is bounded. $\square$

Finally, we prove Proposition 2.8.

Proof of Proposition 2.8

From (g.1)–(g.3), there exists $c>0$ such that $G(s) \le c s^2$ for all $s \in \mathbb {R}$. Thus we have

$$\begin{aligned} \begin{aligned} \frac{1}{2}\Vert u\Vert ^2 +\int _{\mathbb {R}^N} G(u) \,dx \le \frac{1}{2}\Vert u\Vert ^2 + c \Vert u\Vert _2^2. \end{aligned} \end{aligned}$$

From Lemma 4.4, we obtain Proposition 2.8. $\square$

5 Concentration compactness

In this section, in order to prove Proposition 2.14, we argue about the concentration compactness for balanced sequences. We assume that $(u_n)_{n=1}^\infty$ with $u_n \in H_0^1(B_{R_n})$ is a balanced sequence which satisfies

$$\begin{aligned} 0 \le J_L(u_n)\le L, \quad J_L'(u_n)=0 \hbox { in } \left( H_0^1(B_{R_n})\right) ^*. \end{aligned}$$

(5.1)

Then, from Lemma 2.3 and (5.1), $(u_n)_{n=1}^\infty$ is bounded in $H^1(\mathbb {R}^N)$ and from Proposition 2.6, $u_n$ satisfies

• (a) $J_L(u_n)\le L$.

• (b) $J_L'(u_n)\varphi =0$ for all $\varphi \in H^1_0(B_{R_n})$.

• (c) $\displaystyle 0<a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) \le 1$.

The main theorem of this section is the following.

Theorem 5.1

We assume (f.1)–(f.4), (V.1), and (V.2). Let $(u_n)_{n=1}^\infty$ with $u_n \in H_0^1(B_{R_n})$ be a bounded sequence in $H^1(\mathbb {R}^N)$ and satisfy (a)–(c). Then, there exist a subsequence $(u_n)_{n=1}^\infty$ (we use same notation), $\ell \in \mathbb {N}\cup \{0\}$, $u_0 \in H^1(\mathbb {R}^N)$, $\omega ^1,\dots , \omega ^\ell \in H^1(\mathbb {R}^N)\setminus \{0\}$, $(z_n^k)_{n=1}^\infty \subset \mathbb {R}^N$ with $z_n^k \in B_{R_n}$, $|z_n^k|\rightarrow \infty$ $(k=1,\dots ,\ell )$ and $|z_n^k-z_n^{k'}|\rightarrow \infty$ $(k\not =k')$, and $\lambda _0\ge 1$ such that

• (i) $\displaystyle \Big \Vert u_n-u_0-\textstyle {\sum }_{k=1}^\ell \omega ^k(\cdot +z_n^k)\Big \Vert \rightarrow 0$.

• (ii) $u_0$ is a solution of

  $$\begin{aligned} -\Delta u+V(x)u=\lambda _0f(u)+(\lambda _0-1)g(u) \hbox { in } \mathbb {R}^N, \quad u\in H^1(\mathbb {R}^N). \end{aligned}$$

  (5.2)

• (iii) $\omega ^k$ $(k=1,\dots ,\ell )$ are solutions of

  $$\begin{aligned} -\Delta u+V_\infty u=\lambda _0f(u)+(\lambda _0-1)g(u) \hbox { in } \mathbb {R}^N, \quad u\in H^1(\mathbb {R}^N). \end{aligned}$$

  (5.3)

In particular, when $\ell =0$, we have $\Vert u_n-u_0\Vert \rightarrow 0$ and $J_L'(u_0)=0$.

We state some lemmas which are repeatedly used.

Lemma 5.2

Let $(u_n)_{n=1}^\infty$ and $(v_n)_{n=1}^\infty$ be bounded in $H^1(\mathbb {R}^N)$ and

$$\begin{aligned} v_n \rightarrow 0 \quad \text { strongly in } L^{p+1}(\mathbb {R}^N). \end{aligned}$$

(5.4)

Then we have

$$\begin{aligned} \int _{\mathbb {R}^N}f(u_n)v_n\, dx \rightarrow 0, \quad \int _{\mathbb {R}^N}g(u_n)v_n\, dx \rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

(5.5)

Proof

Since $(u_n)_{n=1}^\infty$ and $(v_n)_{n=1}^\infty$ are bounded in $H^1(\mathbb {R}^N)$, there exists $M>0$ such that

$$\begin{aligned} \Vert u_n\Vert \le M, \qquad \Vert u_n\Vert _{p+1}\le M, \qquad \Vert v_n\Vert \le M. \end{aligned}$$

From (f.1)–(f.3) and (g.1)–(g.3), for any $\epsilon >0$, there exists $c_\epsilon >0$ such that

$$\begin{aligned} |f(s)|, \ |g(s)| \le \epsilon |s|+c_\epsilon |s|^p \quad \hbox { for all } s \in \mathbb {R}. \end{aligned}$$

Thus we have

$$\begin{aligned} \begin{aligned} \left| \int _{\mathbb {R}^N}f(u_n)v_n\, dx\right| , \ \left| \int _{\mathbb {R}^N}g(u_n)v_n\, dx\right|&\le \epsilon \int _{\mathbb {R}^N}|u_n\Vert v_n|\, dx +c_\epsilon \int _{\mathbb {R}^N}|u_n|^p|v_n|\, dx \\&\le \epsilon \Vert u_n\Vert _2\Vert v_n\Vert _2 +c_\epsilon \Vert u_n\Vert _{p+1}^p\Vert v_n\Vert _{p+1} \\&\le \epsilon M^2+c_\epsilon M^p\Vert v_n\Vert _{p+1}. \end{aligned} \end{aligned}$$

From (5.4), we see that

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }\left| \int _{\mathbb {R}^N}f(u_n)v_n\, dx\right| , \ \varlimsup _{n\rightarrow \infty }\left| \int _{\mathbb {R}^N}g(u_n)v_n\, dx\right| \le \epsilon M^2. \end{aligned}$$

Since $\epsilon >0$ is arbitrary, we get (5.5). $\square$

Lemma 5.3

([13, Lemma 1.1]) Assume $p \in (1, (N+2)/(N-2))$ if $N\ge 3$, $p\in (1,\infty )$ if $N=1, 2$. Let $(u_n)_{n=1}^\infty \subset H^1(\mathbb {R}^N)$ be bounded in $H^1(\mathbb {R}^N)$ and satisfy

$$\begin{aligned} \sup _{z\in \mathbb {R}^N} \int _{B(z,1)}|u_n|^{p+1}\, dx \rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

Then we have $\Vert u_n\Vert _{p+1}\rightarrow 0$ as $n\rightarrow \infty$.

Proof

This follows from [13, Lemma 1.1]. $\square$

We prove Theorem 5.1 through several steps.

Proof of Theorem 5.1

Let $(u_n)_{n=1}^\infty$ be a sequence satisfying the assumption of Theorem 5.1. In this proof, we repeatedly choose subsequence of $(u_n)_{n=1}^\infty$. Thus, for simplicity, we use same notation for subsequence. Since $(u_n)_{n=1}^\infty$ is bounded in $H^1(\mathbb {R}^N)$, there exist a subsequence $(u_n)_{n=1}^\infty$, $u_0\in H^1(\mathbb {R}^N)$, and $\beta \ge 0$ such that

$$\begin{aligned}&u_n \rightarrow u_0 \quad \hbox { weakly in } H^1(\mathbb {R}^N), \nonumber \\&\frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\rightarrow \beta . \end{aligned}$$

(5.6)

Here we define functionals $\Phi (u)$ and $\Psi (u)$ as follows:

$$\begin{aligned}\begin{aligned} \Phi (u)&=a_L'\left( \beta \right) \left( \frac{1}{2}\Vert u\Vert ^2-\int _{\mathbb {R}^N}G(u)\, dx\right) -\int _{\mathbb {R}^N}F(u)+G(u)\, dx, \\ \Psi (u)&=a_L'\left( \beta \right) \left( \frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2 +V_\infty u^2\, dx -\int _{\mathbb {R}^N}G(u)\, dx\right) -\int _{\mathbb {R}^N}F(u)+G(u)\, dx. \end{aligned} \end{aligned}$$

Step 1

Then we have

$$\begin{aligned}&0\le a_L'(\beta ) \le 1, \end{aligned}$$

(5.7)

$$\begin{aligned}&\Phi '(u_0)\varphi =0 \quad \hbox { for all } \varphi \in H^1(\mathbb {R}^N). \end{aligned}$$

(5.8)

Proof of Step 1

Since $a_L'$ is continuous, (5.7) follows from (c). From (b), for any $\varphi \in C_0^\infty (\mathbb {R}^N)$, we have

$$\begin{aligned}\begin{aligned} J_L'(u_n)\varphi&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) \left( \left\langle u_n, \varphi \right\rangle +\int _{\mathbb {R}^N}g(u_n)\varphi \, dx \right) \\&\quad -\int _{\mathbb {R}^N}f(u_n)\varphi +g(u_n)\varphi \, dx \\&\rightarrow a_L'\left( \beta \right) \left( \left\langle u_0, \varphi \right\rangle +\int _{\mathbb {R}^N}g(u_0)\varphi \, dx \right) -\int _{\mathbb {R}^N}f(u_0)\varphi +g(u_0)\varphi \, dx \\&=\Phi '(u_0)\varphi =0. \end{aligned} \end{aligned}$$

Thus we get (5.8). $\square$

Step 2

We set $v_n^1=u_n-u_0$. Then, either (A) or (B) holds.

(A) $(v_n^1)_{n=1}^\infty$ satisfies

$$\begin{aligned} \sup _{z\in \mathbb {R}^N} \int _{B(z,1)}|v_n^1|^{p+1}\, dx \rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

Then, we have

$$\begin{aligned}&0<a_L'(\beta )\le 1, \end{aligned}$$

(5.9)

$$\begin{aligned}&u_n \rightarrow u_0 \quad \text { strongly in } H^1(\mathbb {R}^N), \nonumber \\&J_L'(u_0)\varphi =0 \quad \hbox { for all } \varphi \in H^1(\mathbb {R}^N). \end{aligned}$$

(5.10)

In particular, for $\lambda _0=a_L'(\beta )^{-1}\ge 1$, $u_0$ is a solution of (5.2).

(B) There exist a subsequence $(v_n^1)_{n=1}^\infty$ and a sequence $(z_n^1)_{n=1}^\infty \subset \mathbb {R}^N$ with $z_n^1 \in B_{R_n}$ for each n such that

$$\begin{aligned} \int _{B(z_n^1,1)}|v_n^1|^{p+1}\, dx \rightarrow d_1>0 \qquad (n\rightarrow \infty ). \end{aligned}$$

(5.11)

Then, after extracting a subsequence, there exists $\omega ^1\in H^1(\mathbb {R}^N)\setminus \{0\}$ such that

$$\begin{aligned}&|z_n^1|\rightarrow \infty , \end{aligned}$$

(5.12)

$$\begin{aligned}&u_n(\cdot +z_n^1)\rightarrow \omega ^1 \ \text { weakly in } H^1(\mathbb {R}^N) \hbox { and } \Vert \omega ^1\Vert _{p+1}^{p+1}\ge d_1, \end{aligned}$$

(5.13)

$$\begin{aligned}&\Psi '(\omega ^1)\varphi = 0 \quad \hbox { for all } \varphi \in H^1(\mathbb {R}^N), \end{aligned}$$

(5.14)

$$\begin{aligned}&0<a_L'(\beta )\le 1. \end{aligned}$$

(5.15)

In particular, for $\lambda _0=a_L'(\beta )^{-1}\ge 1$, $u_0$ is a solution of (5.2) and $\omega ^1$ is a solution of (5.3).

Proof of Case (A) of Step 2. From Lemma 5.3, we have

$$\begin{aligned} \Vert v_n^1\Vert _{p+1}\rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

(5.16)

We suppose by contradiction that $a_L'(\beta )=0$. From (5.8), we have

$$\begin{aligned} \begin{aligned} 0&=\Phi '(u_0)u_0 \\&=a_L'(\beta )\left( \Vert u_0\Vert ^2+\int _{\mathbb {R}^N}g(u_0)u_0\, dx \right) -\int _{\mathbb {R}^N}f(u_0)u_0+g(u_0)u_0\, dx \\&=-\int _{\mathbb {R}^N}f(u_0)u_0+g(u_0)u_0\, dx. \end{aligned} \end{aligned}$$

Since $(f(s)+g(s))s>0$ $(s\not =0)$ from (g.1) and (g.4), we see that $u_0=0$ in $\mathbb {R}^N$. Since $v_n^1=u_n-u_0=u_n$, (5.16) implies

$$\begin{aligned} \Vert u_n\Vert _{p+1}\rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

(5.17)

From (a), we have

$$\begin{aligned} J_L(u_n)=a_L\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) -\int _{\mathbb {R}^N}F(u_n)+G(u_n)\, dx\le L. \end{aligned}$$

Since $\displaystyle \lim _{n\rightarrow \infty }\int _{\mathbb {R}^N}F(u_n)+G(u_n)\, dx=0$ by (5.17), we get $a_L\left( \beta \right) \le L$. However $a_L'(\beta )=0$ implies $a_L\left( \beta \right) > L$ by (ii) of Lemma 2.2. This is a contradiction. Thus (5.9) holds. Next, we show (5.10). By Lemma 5.2, we have

$$\begin{aligned} \begin{aligned} J_L'(u_n)v_n^1&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) \left\langle u_n, v_n^1\right\rangle +o(1),\\ \Phi '(u_0)v_n^1&=a_L'(\beta )\left\langle u_0, v_n^1\right\rangle +o(1), \end{aligned} \end{aligned}$$

where $o(1)\rightarrow 0$ as $n\rightarrow \infty$. Thus

$$\begin{aligned} \begin{aligned}&J_L'(u_n)v_n^1-\Phi '(u_0)v_n^1 \\&\quad =a_L'(\beta )\Vert v_n^1\Vert ^2 +\left\{ a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) -a_L'(\beta )\right\} \left\langle u_n, v_n^1\right\rangle +o(1). \end{aligned} \end{aligned}$$

From (b) and (5.8), we see that $J_L'(u_n)v_n^1-\Phi '(u_0)v_n^1\rightarrow 0$. Consequently, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty } a_L'(\beta )\Vert v_n^1\Vert ^2=0. \end{aligned}$$

Since $a_L'(\beta )>0$, we get (5.10) which implies $a_L'(\beta )=a_L'\left( \frac{1}{2}\Vert u_0\Vert +\int _{\mathbb {R}^N}G(u_0)\, dx\right)$. Thus, we have $J_L'(u_0)=\Phi '(u_0)$. From (5.8), we get $J_L'(u_0)=0$. From (5.9), $u_0$ is a solution of (5.2) with $\lambda _0=a_L(\beta )^{-1}\ge 1$. $\square$

Proof of Case (B) of Step 2. First, we show (5.12). From (5.6), $v_n^1 =u_n-u_0 \rightarrow 0$ weakly in $H^1(\mathbb {R}^N)$. Thus

$$\begin{aligned} v_n^1 = u_n-u_0 \rightarrow 0 \quad \text{ strongly } \text{ in } L^{p+1}_{\mathrm {loc}}(\mathbb {R}^N). \end{aligned}$$

(5.18)

If $(z_n^1)_{n=1}^\infty$ is bounded, then (5.11) and (5.18) contradict each other. Thus (5.12) holds. Since $(v_n^1(\cdot +z_n^1))_{n=1}^\infty$ is bounded in $H^1(\mathbb {R}^N)$, after extracting a subsequence, there exists $\omega ^1\in H^1(\mathbb {R}^N)\setminus \{0\}$ such that

$$\begin{aligned} v_n^1\left( \cdot +z_n^1\right) \rightarrow \omega ^1 \hbox { weakly in } H^1(\mathbb {R}^N) \hbox { and } \Vert \omega ^1\Vert _{p+1}^{p+1}\ge d_1. \end{aligned}$$

Thus we have

$$\begin{aligned} u_n\left( \cdot +z_n^1\right) =v_n^1\left( \cdot +z_n^1\right) +u_0\left( \cdot +z_n^1\right) \rightarrow \omega ^1 \quad \text{ weakly } \text{ in } H^1(\mathbb {R}^N), \end{aligned}$$

and (5.13) holds. Here we show that $(z_n^1)_{n=1}^\infty$ satisfies

$$\begin{aligned} \hbox {dist}\left( z_n^1, \partial B_{R_n}\right) \rightarrow \infty . \end{aligned}$$

(5.19)

By a contrary, we assume $\varliminf _{n\rightarrow \infty }\hbox {dist}(z_n^1, \partial B_{R_n})=:r^1<\infty$. We may also assume $\lim _{n\rightarrow \infty }{\frac{z_n^1}{|z_n^1|}}=:e^1\in \mathbb {R}^N$. Set $H_1=\{x\in R^N | (x\cdot e^1) < r^1\}$. Then $H_1$ is a half space in $\mathbb {R}^N$. For any $\varphi \in C_0^\infty (H_1)$, $\varphi (\cdot -z_n^1)\in C_0^\infty (B_{R_n})$ for large n. From (b), we have

$$\begin{aligned}&J_L'(u_n)\varphi \left( \cdot -z_n^1\right) \nonumber \\&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{B_{R_n}}G(u_n)\, dx\right) \left( \left\langle u_n, \varphi \left( \cdot -z_n^1\right) \right\rangle +\int _{B_{R_n}}g(u_n)\varphi \left( \cdot -z_n^1\right) \, dx \right) \nonumber \\&\quad -\int _{B_{R_n}}f(u_n)\varphi \left( \cdot -z_n^1\right) +g(u_n)\varphi \left( \cdot -z_n^1\right) \, dx \nonumber \\&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{B_{R_n}}G(u_n)\, dx\right) \nonumber \\&\left( \int _{H_1}\nabla u_n\left( \cdot +z_n^1\right) \cdot \nabla \varphi + V\left( x+z_n^1\right) u_n\left( \cdot +z_n^1\right) \varphi \, dx +\int _{H_1}g\left( u_n\left( \cdot +z_n^1\right) \right) \varphi \, dx \right) \nonumber \\&\quad -\int _{H_1}f\left( u_n\left( \cdot +z_n^1\right) \right) \varphi +g\left( u_n\left( \cdot +z_n^1\right) \right) \varphi \, dx \nonumber \\&\rightarrow a_L'\left( \beta \right) \left( \int _{H_1}\nabla \omega ^1 \cdot \nabla \varphi +V_\infty \omega ^1 \varphi \, dx+\int _{H_1}g(\omega ^1)\varphi \, dx \right) \nonumber \\&\quad -\int _{H_1}f(\omega ^1)\varphi +g(\omega ^1)\varphi \, dx \nonumber \\&= 0. \end{aligned}$$

(5.20)

If $a_L'\left( \beta \right) =0$, then $\int _{H_1}f(\omega ^1)\omega ^1+g(\omega ^1)\omega ^1\, dx=0$ that implies $\omega ^1=0$. This is a contradiction. Thus $a_L'\left( \beta \right) >0$ and $\omega ^1$ is a non-trivial solution of

$$\begin{aligned} -\Delta u+V_\infty u=\lambda _0 f(u)+(\lambda _0-1)g(u) \quad \hbox { in } H_1 \quad u\in H_0^1(H_1) \end{aligned}$$

(5.21)

where $\lambda _0=a_L'(\beta )^{-1}\ge 1$. However, since (5.21) has only a trivial solution by [7], this is a contradiction. Thus (5.19) holds. From (5.19), for any $\varphi \in C_0^\infty (\mathbb {R}^N)$, $\varphi (\cdot +z_n^1)\in C_0^\infty (B_{R_n})$ for large n. By similar calculations to (5.20), we find that

$$\begin{aligned} J_L'(u_n)\varphi \left( \cdot -z_n^1\right) \rightarrow \Psi (\omega ^1)\varphi =0 \end{aligned}$$

and $a_L'(\beta )>0$. Thus (5.14) and (5.15) hold. From (5.8), (5.14), and (5.15), $u_0$ and $\omega ^1$ are solutions of (5.2) and (5.3) respectively. $\square$

In Step 2, if the case (A) occurs, Theorem 5.1 holds as $\ell =0$. If the case (B) occurs, we proceed next step.

Step 3

We set

$$\begin{aligned} \begin{aligned} v_n^2 = v_n^1-\omega ^1\left( \cdot -z_n^1\right) =u_n-u_0-\omega ^1\left( \cdot -z_n^1\right) . \end{aligned} \end{aligned}$$

Then, either (A) or (B) holds.

(A) $(v_n^2)_{n=1}^\infty$ satisfies

$$\begin{aligned} \sup _{z\in \mathbb {R}^N} \int _{B(z,1)}|v_n^2|^{p+1}\, dx \rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

(5.22)

Then, we have $\Vert v_n^2\Vert \rightarrow 0$.

(B) There exist a subsequence $(v_n^2)_{n=1}^\infty$ and a sequence $(z_n^2)_{n=1}^\infty \subset \mathbb {R}^N$ with $z_n^2 \in B_{R_n}$ for each n such that

$$\begin{aligned} \int _{B(z_n^2,1)}|v_n^2|^{p+1}\, dx \rightarrow d_2>0 \qquad (n\rightarrow \infty ). \end{aligned}$$

(5.23)

Then, after extracting a subsequence, there exists $\omega ^2\in H^1(\mathbb {R}^N)\setminus \{0\}$ such that

$$\begin{aligned}&|z_n^2|\rightarrow \infty , \quad |z_n^2-z_n^1|\rightarrow \infty , \end{aligned}$$

(5.24)

$$\begin{aligned}&u_n(\cdot +z_n^2) \rightarrow \omega ^2 \ \text{ weakly } \text{ in } H^1(\mathbb {R}^N) \hbox { and } \Vert \omega ^2\Vert _{p+1}^{p+1}\ge d_2, \end{aligned}$$

(5.25)

$$\begin{aligned}&\Psi '(\omega ^2)\varphi = 0 \quad \hbox { for all } \varphi \in H^1(\mathbb {R}^N). \end{aligned}$$

(5.26)

In particular, $\omega ^2$ is a solution of (5.3) with $\lambda _0=a_L'(\beta )^{-1}\ge 1$.

Proof of Case (A) of Step 3. From (5.22) and Lemma 5.3, we see that $\Vert v_n^2\Vert _{p+1}\rightarrow 0$ as $n\rightarrow \infty$. By Lemma 5.2, we have

$$\begin{aligned} \begin{aligned} J_L'(u_n)v_n^2&=a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) \left\langle u_n, v_n^2\right\rangle +o(1),\\ \Phi '(u_0)v_n^2&=a_L'(\beta )\left\langle u_0, v_n^2\right\rangle +o(1), \\ \Psi '\left( \omega ^1\left( \cdot -z_n^1\right) \right) v_n^2&=a_L'(\beta )\int _{\mathbb {R}^N}\nabla \omega ^1\left( \cdot -z_n^1\right) \cdot \nabla v_n^2 +V_\infty \omega ^1\left( \cdot -z_n^1\right) v_n^2\, dx+o(1), \end{aligned} \end{aligned}$$

where $o(1)\rightarrow 0$ as $n\rightarrow \infty$. Thus

$$\begin{aligned} \begin{aligned} J_L'(u_n)v_n^2&-\Phi '(u_0)v_n^2-\Psi '\left( \omega ^1\left( \cdot -z_n^1\right) \right) v_n^2 \\&=a_L'(\beta )\Vert v_n^2\Vert ^2 +\left\{ a_L'\left( \frac{1}{2}\Vert u_n\Vert ^2+\int _{\mathbb {R}^N}G(u_n)\, dx\right) -a_L'(\beta )\right\} \left\langle u_n, v_n^2\right\rangle \\&\quad +a_L'(\beta )\int _{\mathbb {R}^N}\left( V(x)-V_\infty \right) \omega ^1(\cdot -z_n^1) v_n^2\, dx +o(1). \end{aligned} \end{aligned}$$

From (b) and (5.13),

$$\begin{aligned} \begin{aligned} J_L'\left( u_n\right) v_n^2&=J_L'(u_n)u_n-J_L'(u_n)u_0-J_L'(u_n)\omega ^1\left( \cdot -z_n^1\right) \\&=-J_L'(u_n)u_0-J_L'\left( u_n\left( \cdot +z_n^1\right) \right) \omega ^1 \\&\rightarrow 0-\Psi '(\omega _1)\omega _1=0 \qquad (n\rightarrow \infty ). \end{aligned} \end{aligned}$$

Since $J_L'(u_n)v_n^2-\Phi '(u_0)v_n^2-\Psi '(\omega ^1(\cdot -z_n^1))v_n^2\rightarrow 0$, we obtain $\lim _{n\rightarrow \infty } a_L'(\beta )\Vert v_n^2\Vert ^2=0$. Thus we get $\Vert v_n^2\Vert \rightarrow 0$. $\square$

Proof of Case (B) of Step 3. First, we show (5.24). From (5.6) and (5.13), we see that

$$\begin{aligned} \begin{aligned} v_n^2&=u_n-u_0-\omega ^1(\cdot -z_n^1) \rightarrow 0 \quad \text{ weakly } \text{ in } H^1(\mathbb {R}^N), \\ v_n^2(\cdot +z_n^1)&=u_n(\cdot +z_n^1)-u_0(\cdot +z_n^1)-\omega ^1 \rightarrow 0 \quad \text{ weakly } \text{ in } H^1(\mathbb {R}^N). \end{aligned} \end{aligned}$$

Thus we have

$$\begin{aligned} v_n^2&\rightarrow 0 \quad \hbox { strongly in } L^{p+1}_{\mathrm {loc}}(\mathbb {R}^N), \end{aligned}$$

(5.27)

$$\begin{aligned} v_n^2(\cdot +z_n^1)&\rightarrow 0 \quad \hbox { strongly in } L^{p+1}_{\mathrm {loc}}(\mathbb {R}^N). \end{aligned}$$

(5.28)

If $(z_n^2)_{n=1}^\infty$ is bounded, then (5.23) and (5.27) contradict each other. If $(z_n^2-z_n^1)_{n=1}^\infty$ is bounded, then (5.23) and (5.28) contradict each other. Thus (5.24) holds. Since $(v_n^2(\cdot +z_n^2))_{n=1}^\infty$ is bounded in $H^1(\mathbb {R}^N)$, after extracting a subsequence, there exists $\omega ^2\in H^1(\mathbb {R}^N)\setminus \{0\}$ such that

$$\begin{aligned} v_n^2\left( \cdot +z_n^2\right) \rightarrow \omega ^2 \quad \text{ weakly } \text{ in } H^1(\mathbb {R}^N) \hbox { and } \Vert \omega ^2\Vert _{p+1}^{p+1}\ge d_2. \end{aligned}$$

Then (5.25) follows from

$$\begin{aligned} u_n\left( \cdot +z_n^2\right) =v_n^2\left( \cdot +z_n^2\right) +u_0\left( \cdot +z_n^2\right) -\omega ^1\left( \cdot +z_n^2-z_n^1\right) \rightarrow \omega ^2 \quad \text{ weakly } \text{ in } H^1(\mathbb {R}^N). \end{aligned}$$

Also, by similar calculations to (5.19), we get $\lim _{n\rightarrow \infty }\hbox {dist}(z_n^2,\partial B_{R_n})=\infty$. Thus, for any $\varphi \in C_0^\infty (\mathbb {R}^N)$, we obtain

$$\begin{aligned} J_L'\left( u_n\right) \varphi \left( \cdot -z_n^2\right) \rightarrow \Psi '\left( \omega ^2\right) \varphi = 0. \end{aligned}$$

Thus (5.26) holds. $\square$

In Step 3, if the case (A) occurs, Theorem 5.1 holds as $\ell =1$. If the case (B) occurs, we repeat similar arguments. That is, the following induction holds.

Step 4

We suppose that there exist a subsequence of $(u_n)_{n=1}^\infty$, $m \in \mathbb {N}\cup \{0\}$, $u_0 \in H^1(\mathbb {R}^N)$, $\omega ^1,\dots , \omega ^m \in H^1(\mathbb {R}^N)\setminus \{0\}$, $(z_n^k)_{n=1}^\infty \subset \mathbb {R}^N$ with $z_n^k \in B_{R_n}$, $|z_n^k|\rightarrow \infty$ $(k=1,\dots ,m)$ and $|z_n^k-z_n^{k'}|\rightarrow \infty$ $(k\not =k')$ such that

$$\begin{aligned} \begin{aligned}&u_n \rightarrow u_0 \hbox { weakly in } H^1(\mathbb {R}^N), \\&u_n(\cdot +z_n^k) \rightarrow \omega ^k \hbox { weakly in } H^1(\mathbb {R}^N) \hbox { and } \Vert \omega ^k\Vert _{p+1}^{p+1}\ge d_k>0 \quad (k=1,\dots ,m),\\&\Phi '(u_0)\varphi = 0 \hbox { and } \Psi '(\omega ^k)\varphi = 0 \hbox { for all } \varphi \in H^1(\mathbb {R}^N) \quad (k=1,\dots ,m). \end{aligned} \end{aligned}$$

We set

$$\begin{aligned} v_n^{m+1}=u_n-u_0-\sum _{k=1}^m\omega ^k\left( \cdot -z_n^k\right) . \end{aligned}$$

Then, either (A) or (B) holds.

(A) $(v_n^{m+1})_{n=1}^\infty$ satisfies

$$\begin{aligned} \sup _{z\in \mathbb {R}^N} \int _{B(z,1)}\left| v_n^{m+1}\right| ^{p+1}\, dx \rightarrow 0 \qquad (n\rightarrow \infty ). \end{aligned}$$

Then, we have $\Vert v_n^{m+1}\Vert \rightarrow 0$.

(B) There exist a subsequence $(v_n^{m+1})_{n=1}^\infty$ and a sequence $(z_n^{m+1})_{n=1}^\infty \subset \mathbb {R}^N$ such that

$$\begin{aligned} \int _{B(z_n^{m+1},1)}\left| v_n^{m+1}\right| ^{p+1}\, dx \rightarrow d_{m+1}>0 \qquad (n\rightarrow \infty ). \end{aligned}$$

Then, after extracting a subsequence, there exists $\omega ^{m+1}\in H^1(\mathbb {R}^N)\setminus \{0\}$ such that

$$\begin{aligned}\begin{aligned}&\left| z_n^{m+1}\right| \rightarrow \infty , \quad \left| z_n^{m+1}-z_n^k\right| \rightarrow \infty \qquad (k=1,\ldots ,m),\\&u_n\left( \cdot +z_n^{m+1}\right) \rightarrow \omega ^{m+1} \ \text{ weakly } \text{ in } H^1(\mathbb {R}^N) \hbox { and } \left\| \omega ^{m+1}\right\| _{p+1}^{p+1}\ge d_{m+1}>0,\\&\Psi '\left( \omega ^{m+1}\right) \varphi = 0 \quad \hbox { for all } \varphi \in H^1(\mathbb {R}^N). \end{aligned} \end{aligned}$$

In particular, $\omega ^{m+1}$ is a solution of (5.3) with $\lambda _0=a_L'(\beta )^{-1}\ge 1$.

Since the proof of Step 4 is almost same as Step 3, we omit it.

As long as the case (B) occurs, we repeat Step 4. If the case (A) occurs, Theorem 5.1 holds as $\ell =m$. Finally, after repeating Step 4 a finite times, we observe that the case (A) always occurs.

Step 5

When Step 4 is repeated a finite times, the case (A) occurs.

Proof of Step 5

We suppose, by contradiction, that the case (B) of Step 4 repeated infinite time. Then, there exist a subsequence $(u_n)_{n=1}^\infty$, $u_0 \in H^1(\mathbb {R}^N)$, $(\omega ^k)_{k=1}^\infty \subset H^1(\mathbb {R}^N)\setminus \{0\}$, $(z_n^k)_{n=1}^\infty \subset \mathbb {R}^N$ with $|z_n^k|\rightarrow \infty$ $(k\in \mathbb {N})$ and $|z_n^k-z_n^{k'}|\rightarrow \infty$ $(k\not =k')$ such that

$$\begin{aligned}&u_n \rightarrow u_0 \hbox { weakly in } H^1(\mathbb {R}^N), \end{aligned}$$

(5.29)

$$\begin{aligned}&u_n\left( \cdot +z_n^k\right) \rightarrow \omega ^k \hbox { weakly in } H^1(\mathbb {R}^N) \hbox { and } \left\| \omega ^k\right\| _{p+1}^{p+1}\ge d_k>0 \quad (k\in \mathbb {N}),\nonumber \\&\Phi '(u_0)\varphi = 0 \hbox { and } \Psi '\left( \omega ^k\right) \varphi = 0 \hbox { for all } \varphi \in H^1(\mathbb {R}^N) \quad (k\in \mathbb {N}). \end{aligned}$$

(5.30)

From (5.29) and (5.30), for any $m\in \mathbb {N}$, we see that

$$\begin{aligned} \begin{aligned} 0&\le \varlimsup _{n\rightarrow \infty }\left\| u_n-u_0 -\sum _{k=1}^{m-1}\omega ^k\left( \cdot -z_n^k\right) \right\| _{H^1(\mathbb {R}^N)}^2 \nonumber \\&=\varlimsup _{n\rightarrow \infty }\Vert u_n\Vert _{H^1(\mathbb {R}^N)}^2 -\Vert u_0\Vert _{H^1(\mathbb {R}^N)}^2 -\sum _{k=1}^{m-1}\Vert \omega ^k\Vert _{H^1(\mathbb {R}^N)}^2, \end{aligned} \end{aligned}$$

where $\Vert u\Vert _{H^1(\mathbb {R}^N)}^2=\Vert \nabla u\Vert _2^2+\Vert u\Vert _2^2$ which is equivalent to $\Vert \cdot \Vert$. Thus we have

$$\begin{aligned} \sum _{k=1}^\infty \Vert \omega ^k\Vert _{H^1(\mathbb {R}^N)}^2 \le \varlimsup _{n\rightarrow \infty }\Vert u_n\Vert _{H^1(\mathbb {R}^N)}^2 -\Vert u_0\Vert _{H^1(\mathbb {R}^N)}^2 <\infty . \end{aligned}$$

On the other hand, from (f.2) and (g.2), 0 is an isolated critical point of $\Psi (u)$. Thus there exists $\delta _{\beta }>0$ such that $\Vert \omega ^k\Vert \ge \delta _\beta$ $(k\in \mathbb {N})$. This is a contradiction. Thus (B) of Step 4 is not repeated infinite time. $\square$

Through Step 1 to Step 5, the proof of Theorem 5.1 was completed. $\square$

Remark 5.4

In Theorem 5.1, we also have $J_L(u_n)\rightarrow \Phi (u_0)+\sum _{k=1}^\ell \Psi (\omega ^k)$.

6 The compactness for balanced sequence

In this section, we prove Proposition 2.14. The fundamental idea of the proof comes from [3, Proposition 4.1].

Proof of Proposition 2.14

Let $u_n\in H_0^1(B_{R_n})$ satisfy (2.17). Then, $(u_n)_{n=1}^\infty$ satisfies the assumptions of Theorem 5.1. It is sufficient to show that, adding assumption (V.4) to Theorem 5.1, then only $\ell =0$ occurs. Suppose, by contradiction, that Theorem 5.1 holds for $\ell \ge 1$. Since $u_n =0$ in $\mathbb {R}^N \setminus B_{R_n}$, choosing a subsequence (we use same notation) and replacing the order $k=1,\dots ,\ell$, we may assume $|z_n^1| \le |z_n^2| \le \cdots \le |z_n^\ell | \le R_n$. Furthermore, choosing a subsequence, we can also assume that there exist $d_k \in [0,\infty ]$ with

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\left| z_n^k-z_n^1\right| }{\left| z_n^1\right| } =d_k \quad (k=1,2,\dots ,\ell ). \end{aligned}$$

We set $r_n$ and $d>0$ such that

$$\begin{aligned} 11r_n=|z_n^1|, \quad d=\min \left\{ 1,\, 11\rho /6,\, d_k \big | d_k>0 \right\} . \end{aligned}$$

Here $\rho$ is a constant defined by (V.4). Then, for large n, it holds the following.

• (i) If $d_k=0$, then $B(z_n^k, dr_n)\subset B(z_n^1, 2dr_n)$.

• (ii) If $d_k>0$, then $B(z_n^k, dr_n)\subset \mathbb {R}^N\setminus B(z_n^1, 9dr_n)$.

From (i) of Theorem 5.1, we see that

$$\begin{aligned} \left\| u_n\right\| _{L^{p+1}(B(z_n^1, 9dr_n)\setminus B(z_n^1, 2dr_n))}\rightarrow 0. \end{aligned}$$

(6.1)

Since $u_n\in H_0^1(B_{R_n})$ is a solution of

$$\begin{aligned} -\Delta u_n+V(x)u_n=\lambda _0 f(u_n)+(\lambda _0-1)g(u_n) \quad \hbox { in } B_{R_n}, \end{aligned}$$

$|u_n|\in H^1(\mathbb {R}^N)$ (expanding 0 on $\mathbb {R}^N\setminus B_{R_n}$) is a subsolution of

$$\begin{aligned} -\Delta u_n+V(x)u_n=\lambda _0 f(u_n)+(\lambda _0-1)g(u_n) \quad \hbox { in } \mathbb {R}^N. \end{aligned}$$

(6.2)

Here we use the subsolution estimate below.

Lemma 6.1

Let $\Omega$ be a domain and ${\widetilde{V}}\in L^\frac{p+1}{p-1}_{\mathrm {loc}}(\Omega )$. Suppose that $u\in H^1(\Omega )$ satisfies

$$\begin{aligned} \int _{\Omega }\nabla u\cdot \nabla \varphi \, dx -{\widetilde{V}}(x)u\varphi \, dx \le 0 \quad \hbox { for all } \varphi \in H^1_0(\Omega ) \hbox { with } \varphi \ge 0. \end{aligned}$$

Then, for any $B(x_0, 2r)\subset \Omega$, there exist constants $C=C(p,N,r)>0$ and $\sigma =\sigma (p,N)>0$ such that

$$\begin{aligned} \Vert u_+\Vert _{L^\infty (B(x_0,r))} \le C\left( 1+\Vert {\widetilde{V}}_+\Vert _{L^\frac{p+1}{p-1}(B(x_0,2r))}^\sigma \right) \Vert u_+\Vert _{L^{p+1}(B(x_0, 2r))}. \end{aligned}$$

Proof

This was shown in [15, Theorem 2.26]. (Also see [8, Theorem 8.15].) $\square$

From (6.1) and Lemma 6.1, we see that

$$\begin{aligned} \left\| u_n\right\| _{L^{\infty }(B(z_n^1, 8dr_n) \setminus B(z_n^1, 3dr_n))}\rightarrow 0. \end{aligned}$$

(6.3)

Since $|u_n|$ is a subsolution of (6.2), from (6.3), (f.2), and (g.2), by the comparison theorem, there exist constants $C>0$ and $\mu >0$ such that

$$\begin{aligned} \left\| u_n\right\| _{L^{\infty }(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))}\le C e^{-\mu r_n}. \end{aligned}$$

(6.4)

Furthermore, replacing $C>0$ and $\mu >0$, we also have

$$\begin{aligned}&\left\| u_n\right\| _{L^{2}(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))}\le C e^{-\mu r_n}, \end{aligned}$$

(6.5)

$$\begin{aligned}&\left\| u_n\right\| _{L^{p+1}(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))}\le C e^{-\mu r_n}. \end{aligned}$$

(6.6)

Lemma 6.2

There exist constants $C'>0$ and $\mu '>0$ such that

$$\begin{aligned} \left\| \nabla |u_n|\right\| _{L^2(B(z_n^1, 6dr_n) \setminus B(z_n^1, 5dr_n))}\le C' e^{-\mu ' r_n}. \end{aligned}$$

Proof

From (f.1)–(f.3) and (g.1)–(g.3), there exists $c_1>0$ such that

$$\begin{aligned} |\lambda _0f(u)+(\lambda _0-1)g(u)|\le |u|+c_1|u|^p \quad \hbox { for all } u \in \mathbb {R}. \end{aligned}$$

(6.7)

We take a cut-off function $\psi _n\in C_0^\infty (\mathbb {R}^n, [0,1])$ satisfying

$$\begin{aligned} \begin{aligned}&\psi _n(x)=1 \quad \text { for } x\in B(z_n^1, 6dr_n) \setminus B(z_n^1, 5dr_n), \\&\psi _n(x)=0 \quad \text { for } x\not \in B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n), \\&|\nabla \psi _n(x)|\le 1 \quad \text { for } x\in \mathbb {R}^N. \end{aligned} \end{aligned}$$

Since $|u_n|$ is a subsolution of (6.2), for any $\varphi \in H^1(\mathbb {R}^N)$ with $\varphi \ge 0$, we have

$$\begin{aligned} \int _{\mathbb {R}^N}\nabla |u_n| \cdot \nabla \varphi +V(x)|u_n|\varphi -\lambda _0f(u_n)\varphi -(\lambda _0-1)g(u_n)\varphi \, dx \le 0. \end{aligned}$$

Setting $\varphi =|u_n|\psi _n$ in the above, from (6.7), we have

$$\begin{aligned} \begin{aligned}&\int _{B(z_n^1, 7dr_n)\setminus B(z_n^1, 4dr_n)}|\nabla |u_n||^2\psi _n +\left( \nabla |u_n| \cdot \nabla \psi _n\right) |u_n|-V(x)|u_n|^2\psi _n\, dx \\&\quad \le \int _{B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n)} |u_n|^2\psi _n+c_1|u_n|^{p+1}\psi _n\, dx. \end{aligned} \end{aligned}$$

Thus we get

$$\begin{aligned} \begin{aligned}&\int _{B(z_n^1, 6dr_n) \setminus B(z_n^1, 5dr_n)}|\nabla |u_n||^2\, dx \le \int _{B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n)}|\nabla |u_n||^2\psi _n\, dx \\&\quad \le \Vert \nabla |u_n|\Vert _{L^2(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))} \Vert u_n\Vert _{L^2(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))} \\&\qquad +\left( \Vert V\Vert _{\infty }+1\right) \Vert u_n\Vert _{L^2(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))}^2 +c_1\Vert u_n\Vert _{L^{p+1}(B(z_n^1, 7dr_n) \setminus B(z_n^1, 4dr_n))}^{p+1}. \end{aligned} \end{aligned}$$

Lemma 6.2 follows from (6.5) and (6.6). $\square$

Lemma 6.3

There exist constants $C''>0$, $\mu ''>0$, and $s_n \in (5dr_n, 6dr_n)$ such that

$$\begin{aligned} \left\| \nabla u_n\right\| _{L^2(\partial B(z_n^1, s_n))} =\left\| \nabla |u_n|\right\| _{L^2(\partial B(z_n^1, s_n))} \le C'' e^{-\mu '' r_n}. \end{aligned}$$

(6.8)

Proof

From Lemma 6.2, it holds that

$$\begin{aligned}\begin{aligned} \left( C'\right) ^2e^{-2\mu 'r_n}&\ge \left\| \nabla |u_n|\right\| _{L^2(B(z_n^1, 6dr_n) \setminus B(z_n^1, 5dr_n))}^2 \\&=\int _{5dr_n\le |x-z_n^1|\le 6dr_n}|\nabla |u_n||^2\, dx \\&=\int _{5dr_n}^{6dr_n}\left( \int _{|x-z_n^1|=r}|\nabla |u_n||^2\, d\sigma \right) \, dr \\&=\int _{5dr_n}^{6dr_n}\Vert \nabla |u_n|\Vert _{L^2(\partial B(z_n^1,r))}^2\, dr. \end{aligned} \end{aligned}$$

Since $r \mapsto \Vert \nabla \left| u_n\right| \Vert _{L^2(\partial B(z^1_n,r))}$ is continuous, by the mean value theorem for integration, there exists $s_n\in (5dr_n,6dr_n)$ such that

$$\begin{aligned} \left( 6dr_n-5dr_n\right) \Vert \nabla |u_n|\Vert _{L^2(\partial B(z_n^1,s_n))}^2 \le (C')^2e^{-2\mu 'r_n}. \end{aligned}$$

Thus we see that Lemma 6.3 holds. $\square$

Here, we use the following local Pohozaev identity.

Lemma 6.4

([3]) Let $\Omega \subset \mathbb {R}^N$ be a bounded domain with piecewise smooth boundary and $\nu \in \mathbb {R}^N$ be the outward unit normal vector on $\partial \Omega$. We suppose that $V\in C(\mathbb {R}^N, \mathbb {R})$ and $h\in C(\mathbb {R},\mathbb {R})$. If $u\in C^2(\Omega )\cap C^1({\overline{\Omega }})$ satisfies

$$\begin{aligned} -\Delta u+V(x)u=h(u) \quad \hbox { in } \Omega , \end{aligned}$$

(6.9)

then, for any $\xi \in \mathbb {R}^N$, it holds that

$$\begin{aligned} \frac{1}{2} \int _{\Omega }\left( \xi \cdot \nabla V(x)\right) u^2\,dx&=\frac{1}{2} \int _{\partial \Omega }(\xi \cdot \nu )|\nabla u|^2\,d\sigma - \int _{\partial \Omega }(\nabla u \cdot \nu )(\xi \cdot \nabla u)\,d\sigma \nonumber \\&\quad - \int _{\partial \Omega }(\xi \cdot \nu )\left( V(x)\frac{u^2}{2}- H(u)\right) \,d\sigma , \end{aligned}$$

(6.10)

where $H(u)=\int _0^uh(\tau )\, d\tau$.

Proof

Multiplying $(\xi \cdot \nabla u)$ to the both sides of (6.9), integrating over $\Omega$, we get (6.10). (Also see [3, Lemma 4.1].) $\square$

Applying Lemma 6.4 to $u_n$ as $\Omega =B(z_n^1, s_n)\cap B_{R_n}$, $h(u)=\lambda _0 f(u)+(\lambda _0-1)g(u)$, and $\xi =z_n^1$, we calculate as below.

$$\begin{aligned}&\frac{1}{2} \int _{B(z_n^1, s_n)\cap B_{R_n}}\left( z_n^1 \cdot \nabla V(x)\right) u_n^2\,dx \nonumber \\&\qquad =\frac{1}{2} \int _{\Gamma _1\cup \Gamma _2}\left( z_n^1 \cdot \nu \right) |\nabla u_n|^2\,d\sigma - \int _{\Gamma _1\cup \Gamma _2}\left( \nabla u_n \cdot \nu \right) \left( z_n^1 \cdot \nabla u_n\right) \,d\sigma \nonumber \\&\qquad \qquad - \int _{\Gamma _1\cup \Gamma _2}\left( z_n^1 \cdot \nu \right) \left( V(x)\frac{u_n^2}{2}-\lambda _0 F(u_n)-(\lambda _0-1)G(u_n)\right) \,d\sigma \nonumber \\&\qquad =:\int _{\Gamma _1}\text {(I)}\,d\sigma +\int _{\Gamma _2}\text {(II)}\,d\sigma , \end{aligned}$$

(6.11)

where $\Gamma _1 =(\partial B(z_n^1, s_n))\cap B_{R_n}$, $\Gamma _2 =B(z_n^1, s_n)\cap (\partial B_{R_n})$. We note $s_n \le 6dr_n < 11r_n=|z_n^1| \le R_n$. Since $\nu =\nu (x) \in \mathbb {R}^N$ is the outward unit normal vector at $x\in \partial \Omega$, $\nu (x)=x$ on $\Gamma _2$. Thus we have

$$\begin{aligned} \left( z_n^1 \cdot \nu (x)\right) >0 \quad \hbox { for all } x\in \Gamma _2. \end{aligned}$$

(6.12)

Moreover, since $u_n=0$ and $\nu =-\frac{\nabla u_n}{|\nabla u_n|}$ on $\Gamma _2$, we see that

$$\begin{aligned} \int _{\Gamma _2}\text {(II)}\, d\sigma =-\frac{1}{2} \int _{\Gamma _2}\left( z_n^1 \cdot \nu \right) |\nabla u_n|^2\,d\sigma \le 0. \end{aligned}$$

From (6.4) and (6.8), for $\mu _0\in (0, 2\min \left\{ \mu , \mu '' \right\} )$, we see that

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }e^{\mu _0 r_n}\times \left| \int _{\Gamma _1}\text {(I)}\, d\sigma \right| <\infty . \end{aligned}$$

Thus we have

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }e^{\mu _0 r_n} \int _{B(z_n^1, s_n)\cap B_{R_n}}\left( z_n^1 \cdot \nabla V(x)\right) u_n^2\,dx < \infty . \end{aligned}$$

(6.13)

On the other hand, from (i) of Theorem 5.1, we have

$$\begin{aligned} \varliminf _{n\rightarrow \infty }\int _{B(z_n^1, s_n)\cap B_{R_n}}u_n^2\,dx \ge \Vert \omega ^1\Vert _{L^2(\mathbb {R}^N)}^2>0. \end{aligned}$$

From (V.4) and $s_n<6dr_n\le \rho |z^1_n|$, the left hand side of (6.11) satisfies

$$\begin{aligned}\begin{aligned}&e^{\mu _0 r_n}\frac{1}{2} \int _{B(z_n^1, s_n)\cap B_{R_n}}\left( z_n^1 \cdot \nabla V(x)\right) u_n^2\,dx \\&\qquad \qquad \ge \frac{1}{2} e^{\mu _0 |z_n^1|/11} \left( \inf _{x\in B(z_n^1, \rho |z_n^1|)}\left( z_n^1 \cdot \nabla V(x)\right) \right) \int _{B(z_n^1, s_n)\cap B_{R_n}}u_n^2\,dx \\&\qquad \qquad \rightarrow \infty \qquad (n\rightarrow \infty ). \end{aligned}\end{aligned}$$

This contradicts (6.13). Consequently, we see that $\ell =0$, and Proposition 2.14 was proved. $\square$

At the last, we give outline of the proof of Theorem 1.4.

Outline of proof of Theorem 1.4

In order to prove Theorem 1.1, we used the approximating problem on $B_R$. But, even if we approximate by a problem on $\Omega _R$, the proof of Theorem 1.1 is exactly the same if (6.12) hold. Moreover, (6.12) holds by the assumption $(\Omega )$. Indeed, by $(\Omega )$, there exist $\delta , C>0$ such that

$$\begin{aligned} \Omega \subset B_{C} \quad \hbox { and } \quad y \cdot \nu _{\Omega }(x) >0 \text { for } x \in \partial \Omega , \, |x-y|<\delta , \end{aligned}$$

(6.14)

where $\nu _{\Omega }(x)$ is the outward unit normal vector of $\Omega$ at x. We take d with $Cd < \delta$. Since $z_n^1 \in B_{C R_n}$, we obtain

$$\begin{aligned} s_n \le 6 d r_n < \frac{\delta |z_n^1|}{C} \le \delta R_n. \end{aligned}$$

Thus, for $x \in \Gamma _2 =B(z_n^1,s_n) \cap \partial \Omega _{R_n}$, we have

$$\begin{aligned} \frac{1}{R_n}\left( z_n^1 \cdot \nu _{\Omega _{R_n}}(x) \right) = \frac{z^1_n}{R_n} \cdot \nu _{\Omega } \left( \frac{x}{R_n} \right) , \quad \left| \frac{z^1_n}{R_n} - \frac{x}{R_n} \right| \le \frac{s_n}{R_n} \le \delta . \end{aligned}$$

(6.15)

(6.12) follows from (6.14) and (6.15). Thus, we can prove Theorem 1.4 in the same way as Remark 3.2. $\square$