Multi-bump solutions for a class of quasilinear problems involving variable exponents

Abstract

We establish the existence of multi-bump solutions for the following class of quasilinear problems

$$\begin{aligned} - \Delta _{ p(x) } u + \big ( \lambda V(x) + Z(x) \big ) u ^{ p(x)-1 } = f(x,u) \text { in } \mathbb R^N, \, u \ge 0 \text { in } \mathbb R^N, \end{aligned}$$

where the nonlinearity $f :\mathbb R^N \times \mathbb R \rightarrow \mathbb R$ is a continuous function having a subcritical growth and potentials $V, Z :\mathbb R^N \rightarrow \mathbb R$ are continuous functions verifying some hypotheses. The main tool used is the variational method.

1 Introduction

In this paper, we consider the existence and multiplicity of solutions for the following class of problems

$$\begin{aligned} \big ( P_\lambda \big ) \; {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + \big ( \lambda V(x) + Z(x) \big ) u^{ p(x)-1 }= f(x,u),&{} \text { in } \mathbb R^N, \\ u \ge 0,&{} \text { in } \mathbb R^N, \\ u \in W^{ 1,p(x) } \big ( \mathbb R^N \big ),&{} \end{array}\right. } \end{aligned}$$

where $\Delta _{ p(x) }$ is the $p(x)$-Laplacian operator given by

$$\begin{aligned} \Delta _{ p(x) } u = \text {div} \left( \big | \nabla u \big |^{p(x)-2} \nabla u \right) . \end{aligned}$$

Here, $\lambda > 0$ is a parameter, $p :\mathbb R^N \rightarrow \mathbb R$ is a Lipschitz function, $V,Z :\mathbb R^N \rightarrow \mathbb R$ are continuous functions with $V \ge 0$, and $f :\mathbb R^N \times \mathbb R \rightarrow \mathbb R$ is continuous having a subcritical growth. Furthermore, we take into account the following set of hypotheses:

($H_1$):

$1 < p_- = \displaystyle \inf _{ \mathbb R^N} p \le p_+ = \sup _{\mathbb R^N} p< N$.

($H_2$):

$\Omega = \text {int } V^{ -1 } (0) \ne \emptyset$ and bounded, $\overline{\Omega } = V^{ -1 }(0)$ and $\Omega$ can be decomposed in $k$ connected components $\Omega _1, \ldots , \Omega _k$ with $\text {dist} \big ( \Omega _i, \Omega _j \big ) > 0, \, i \ne j$.

($H_3$):

There exists $M > 0$ such that

$$\begin{aligned} \lambda V(x) + Z(x) \ge M, \, \forall x \in \mathbb R^N, \lambda \ge 1. \end{aligned}$$

($H_4$):

There exists $K > 0$ such that

$$\begin{aligned} \big | Z(x) \big | \le K, \, \forall x \in \mathbb R^N. \end{aligned}$$

($f_1$):

$$\begin{aligned} \limsup _{ |t| \rightarrow \infty } \frac{|f(x,t)|}{|t|^{ q(x)-1 }} < \infty , \text { uniformly in } x \in \mathbb R^N, \end{aligned}$$

where $q :\mathbb R^N \rightarrow \mathbb R$ is continuous with $p_+ < q_-$ and $q \ll p^*=\frac{Np}{N-p}$. Here, the notation $q \ll p^*$ means that $\displaystyle \inf _{ \mathbb R^N} (p^*-q) > 0$.

($f_2$):

$f(x,t) = o \big ( |t|^{ p_+ - 1} \big ), \, t \rightarrow 0, \text { uniformly in } x \in \mathbb R^N$.

($f_3$):

There exists $\theta > p_+$ such that

$$\begin{aligned} 0 < \theta F(x,t) \le f(x,t) t, \, \forall x \in \mathbb R^N, t > 0, \end{aligned}$$

where $F(x,t) = \int _0^t f(x,s) \, \mathrm{{d}}s$.

($f_4$):

$\dfrac{f(x,t)}{t^{p_+ - 1}}$ is strictly increasing in $t \in (0,\infty )$, for each $x \in \mathbb R^N$.

($f_5$):

$\forall a, b \in \mathbb R, \, a < b, \, \mathop {\mathop {\sup }\limits _{x \in \mathbb {R}^N}}\limits _{t \in [a,b]} |f(x,t)| < \infty$.

A typical example of nonlinearity verifying $(f_1)-(f_5)$ is

$$\begin{aligned} f(x,t) = |t|^{ q(x)-2 }t, \, \forall \, x \in \mathbb R^N \, \text{ and } \, \forall t \in \mathbb R, \end{aligned}$$

where $p_+ < q_-$ and $q \ll p^*$.

Partial differential equations involving the $p(x)$-Laplacian arise, for instance, as a mathematical model for problems involving electrorheological fluids and image restorations, see [1, 2, 11–13, 29]. This explains the intense research on this subject in the last decades. A lot of works, mainly treating nonlinearities with subcritical growth, are available (see [4–9, 16–18, 20–24, 28] for interesting works). Nevertheless, to the best of the author’s knowledge, this is the first work dealing with multi-bump solutions for this class of problems.

The motivation to investigate problem $\big ( P_\lambda \big )$ in the setting of variable exponents has been the papers [3] and [15]. In [15], inspired by del Pino and Felmer [14] and Séré [30], the authors considered $\big ( P_\lambda \big )$ for $p = 2$ and $f(u) = u^q , q \in \big (1, \frac{N+2}{N-2} \big )$ if $N \ge 3$; $q \in (1, \infty )$ if $N = 1, 2$. The authors showed that $\big ( P_\lambda \big )$ has at least $2^k-1$ solutions $u_\lambda$ for large values of $\lambda$. More precisely, one solution for each non-empty subset $\Upsilon$ of $\{ 1,\ldots ,k \}$. Moreover, fixed $\Upsilon \subset \{ 1,\ldots ,k \}$, it was proved that, for any sequence $\lambda _n \rightarrow \infty$, we can extract a subsequence $( \lambda _{n_i})$ such that $( u_{ \lambda _{n_i} } )$ converges strongly in $H^1 \big ( \mathbb R^N \big )$ to a function $u$, which satisfies $u = 0$ outside $\Omega _\Upsilon = \bigcup _{ j \in \Upsilon } \Omega _j$ and $u_{|_{\Omega _j}}, \, j \in \Upsilon$ is a least energy solution for

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta u + Z(x) u = u^q, \quad \text { in } \Omega _j, \\ u \in H^1_0 \big ( \Omega _j \big ), \, u > 0, \quad \text { in } \Omega _j. \end{array}\right. } \end{aligned}$$

In [3], employing some different arguments than those used in [15], Alves extended the results described above to the $p$-Laplacian operator, assuming that in $\big ( P_\lambda \big )$ the nonlinearity $f$ possesses a subcritical growth and $2 \le p < N$. In particular, fixed $\Upsilon \subset \{ 1,\ldots ,k \}$, for any sequence $\lambda _n \rightarrow \infty$, we can extract a subsequence $(\lambda _{n_i})$ such that $(u_{ \lambda _{n_i} })$ converges strongly in $W^{ 1,p } \big ( \mathbb R^N \big )$ to a function $u$, which satisfies $u = 0$ outside $\Omega _\Upsilon$ and $u_{|_{\Omega _j}}, \, j \in \Upsilon$, is a least energy solution for

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta _p u + Z(x) u = f(u), \quad \text { in } \Omega _j, \\ u \in W^{ 1,p }_0 \big ( \Omega _j \big ), \, u > 0, \quad \quad \text { in } \Omega _j. \end{array}\right. } \end{aligned}$$

In the present paper, we extend the results found in [3] to the $p(x)$-Laplacian operator. However, we would like to emphasize that in a lot of estimates, we have used different arguments from that found in [3]. The main difference is related to the fact that for equations involving the $p(x)$-Laplacian operator it is not clear that Moser’s iteration method is a good tool to get the estimates for the $L^{\infty }$-norm. Here, we adapt some ideas explored in [18] and [25] to get these estimates. For more details see Sect. 5.

Since we intend to find nonnegative solutions, throughout this paper, we replace $f$ by $f^+ :\mathbb {R}^N \times \mathbb {R} \rightarrow \mathbb {R}$ given by

$$\begin{aligned} f^+(x,t) = {\left\{ \begin{array}{ll} f(x,t), &{} \text{ if } t > 0 \\ 0,&{} \text { if } t \le 0. \end{array}\right. } \end{aligned}$$

Nevertheless, for the sake of simplicity, we still write $f$ instead of $f^+$.

The main theorem in this paper is the following:

Theorem 1.1

Assume that $(H_1)-(H_4)$ and $(f_1)-(f_5)$ hold. Then, there exist $\lambda _0 > 0$ with the following property: for any non-empty subset $\Upsilon$ of $\{1, 2, . . . , k \}$ and $\lambda \ge \lambda _0$, problem $\big ( P_\lambda \big )$ has a solution $u_\lambda$. Moreover, if we fix the subset $\Upsilon$, then for any sequence $\lambda _n \rightarrow \infty$, we can extract a subsequence $(\lambda _{n_i})$ such that $(u_{ \lambda _{n_i} })$ converges strongly in $W^{ 1,p(x) } \big ( \mathbb R^N \big )$ to a function $u$, which satisfies $u = 0$ outside $\Omega _\Upsilon = \bigcup _{ j \in \Upsilon } \Omega _j$ and $u_{|_{\Omega _j}}, \, j \in \Upsilon$, is a least energy solution for

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + Z(x) u = f(x,u),&{} \text { in } \Omega _j, \\ u \in W^{ 1,p(x) }_0 \big ( \Omega _j \big ), \, u \ge 0, &{} \text { in } \Omega _j. \end{array}\right. } \end{aligned}$$

Notations: The following notations will be used in the present work:

• $C$ and $C_i$ will denote generic positive constant, which may vary from line to line;

• In all the integrals, we omit the symbol $dx$.

• If $u$ is a measurable function, we denote $u^+$ and $u^-$ its positive and negative part, i.e., $u^+(x) = \max \{ u(x), 0 \}$ and $u^-(x) = \min \{ u(x), 0 \}$.

• If $u,v$ are measurable functions, $u_- = \text {ess} \displaystyle \inf _{ \! \! \! \! \! \mathbb R^N} u , u_+ = \text {ess} \displaystyle \sup _{ \! \! \! \! \! \mathbb R^N} u$ and the notation $u \ll v$ means that $\text {ess} \displaystyle \inf _{ \! \! \! \! \! \mathbb R^N} \left( v-u \right) > 0$ . Moreover, we will denote by $u^*$ the function

  $$\begin{aligned} u^*(x) = {\left\{ \begin{array}{ll} \frac{Nu(x)}{N-u(x)},&{} \text { if } u(x) < N, \\ \infty , &{}\text { if } u(x) \ge N. \end{array}\right. } \end{aligned}$$

2 Preliminaries on variable exponents Lebesgue and Sobolev spaces

In this section, we recall some results on variable exponents Lebesgue and Sobolev spaces found in [8, 19, 21] and their references.

Let $h \in L^\infty \big ( \mathbb R^N \big )$ with $h_- = \text {ess} \displaystyle \inf _{ \! \! \! \! \! \mathbb R^N} h \ge 1$. The variable exponent Lebesgue space $L^{ h(x) } \big ( \mathbb R^N \big )$ is defined by

$$\begin{aligned} L^{ h(x) } \big ( \mathbb R^N \big ) = \left\{ u :\mathbb R^N \rightarrow \mathbb R \, ; \, u \text { is measurable and } \int \limits _{ \mathbb R^N} \left| u \right| ^{ h(x) } < \infty \right\} , \end{aligned}$$

endowed with the norm

$$\begin{aligned} \left| u \right| _{ h(x) } = \inf \left\{ \lambda > 0 \, ; \, \int \limits _{ \mathbb R^N } \left| \frac{u}{\lambda } \right| ^{ h(x) } \le 1 \right\} . \end{aligned}$$

The variable exponent Sobolev space is defined by

$$\begin{aligned} W^{ 1,h(x) } \big ( \mathbb R^N \big ) = \left\{ u \in L^{ h(x) } \big ( \mathbb R^N \big ) \, ; \, \big \vert \nabla u \big \vert \in L^{ h(x) } \big ( \mathbb R^N \big ) \right\} , \end{aligned}$$

with the norm

$$\begin{aligned} \left\| u \right\| _{ 1,h(x) } = \inf \left\{ \lambda > 0 \, ; \, \int \limits _{ \mathbb R^N } \left( \left| \frac{\nabla u}{\lambda } \right| ^{ h(x) } + \left| \frac{u}{\lambda } \right| ^{ h(x) } \right) \le 1 \right\} . \end{aligned}$$

If $h_- > 1$, the spaces $L^{ h(x) } \big ( \mathbb R^N \big )$ and $W^{ 1,h(x) } \big ( \mathbb R^N \big )$ are separable and reflexive with these norms.

We are mainly interested in subspaces of $W^{ 1,h(x) } \big ( \mathbb R^N \big )$ given by

$$\begin{aligned} E_W = \left\{ u \in W^{ 1,h(x) } \big ( \mathbb R^N \big ) \, ; \, \int \limits _{ \mathbb R^N } W(x) |u|^{ h(x) } < \infty \right\} , \end{aligned}$$

where $W \in C \big ( \mathbb R^N \big )$ is such that $W_- > 0$. Endowing $E_W$ with the norm

$$\begin{aligned} \left\| u \right\| _W = \inf \left\{ \lambda > 0 \, ; \, \int \limits _{ \mathbb R^N } \left( \left| \frac{\nabla u}{\lambda } \right| ^{ h(x) } + W(x)\left| \frac{u}{\lambda } \right| ^{ h(x) } \right) \le 1 \right\} , \end{aligned}$$

$E_W$ is a Banach space. Moreover, it is easy to see that $E_W \hookrightarrow W^{ 1,h(x) } \big ( \mathbb R^N \big )$ continuously. In addition, we can show that $E_W$ is reflexive. For the reader’s convenience, we recall some basic results.

Proposition 2.1

The functional $\varrho :E_W \rightarrow \mathbb R$ defined by

$$\begin{aligned} \varrho (u) = \int \limits _{ \mathbb R^N } \left( \big \vert \nabla u \big \vert ^{ h(x) } + W(x) \left| u \right| ^{ h(x) } \right) , \end{aligned}$$

(2.1)

has the following properties:

• (i) If $\left\| u \right\| _W \ge 1$, then $\left\| u \right\| _W^{ h_- } \le \varrho (u) \le \left\| u\right\| _W^{ h_+ }$.

• (ii) If $\left\| u \right\| _W \le 1$, then $\left\| u \right\| _W^{ h_+ }\le \varrho (u) \le \left\| u \right\| _W^{h_- }$.

In particular, for a sequence $(u_n)$ in $E_W$,

$$\begin{aligned}&\left\| u_n \right\| _W \rightarrow 0 \iff \varrho (u_n) \rightarrow 0, \ \text {and}, \\&(u_n) \ \text {is bounded in} \ E_W \iff \varrho (u_n) \ \text {is bounded in} \ \mathbb R. \end{aligned}$$

Remark 2.2

For the functional $\varrho _{ h(x) } :L^{ h(x) } \big ( \mathbb R^N \big ) \rightarrow \mathbb R$ given by

$$\begin{aligned} \varrho _{ h(x) }(u) = \int \limits _{ \mathbb R^N } \left| u \right| ^{ h(x) }, \end{aligned}$$

an analogous conclusion to that of Proposition 2.1 also holds.

Proposition 2.3

Let $m \in L^\infty \big ( \mathbb R^N \big )$ with $0 < m_- \le m(x) \le h(x) \text { for a.e. } x \in \mathbb R^N$. If $u \in L^{ h(x) } \big ( \mathbb R^N \big )$, then $|u|^{ m(x) } \in L^{ \frac{h(x)}{m(x)} } \big ( \mathbb R^N \big )$ and

$$\begin{aligned} \left| |u|^{ m(x) } \right| _{ \frac{h(x)}{m(x)} } \le \max \left\{ |u|_{ h(x) }^{ m_- }, |u|_{ h(x) }^{ m_+ } \right\} \le |u|_{ h(x) }^{ m_- } + |u|_{ h(x) }^{ m_+ }. \end{aligned}$$

Related to the Lebesgue space $L^{ h(x) } \big ( \mathbb R^N \big )$, we have the following generalized Hölder’s inequality.

Proposition 2.4 (Hölder’s inequality)

[Hölder’s inequality] If $h_- > 1$, let $h' :\mathbb R^N \rightarrow \mathbb R$ such that

$$\begin{aligned} \frac{1}{h( x)} + \frac{1}{h'(x)} = 1\quad \text { for a.e. } x \in \mathbb R^N. \end{aligned}$$

Then, for any $u\in L^{ h( x) } \big ( \mathbb R^N \big )$ and $v \in L^{ h'(x) } \big ( \mathbb R^N \big )$,

$$\begin{aligned} \int \limits _{\mathbb R^N} | uv | \, dx \le \left( \frac{1}{h_-} + \frac{1}{h'_-} \right) |u| _{ h( x) } |v|_{ h'(x) }. \end{aligned}$$

We can define variable exponent Lebesgue spaces with vector values. We say $u = ( u_1, \ldots , u_L ) :\mathbb R^N \rightarrow \mathbb R^L \in L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big )$ if, and only if, $u_i \in L^{ h(x) } \big ( \mathbb R^N \big )$, for $i = 1, \ldots , L$. On $L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big )$, we consider the norm $| u |_{ L^{ h(x) }( \mathbb R^N, \mathbb R^L) } = \sum _{i=1}^L | u_i |_{ h(x) }$.

We state below lemmas of Brezis–Lieb type. The proof of the two first results follows the same arguments explored at [26], while the proof of the latter can be found at [8].

Proposition 2.5 (Brezis–Lieb lemma, first version)

[Brezis–Lieb lemma, first version] Let $\left( u_n \right)$ be a bounded sequence in $L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big )$ such that $u_n(x) \rightarrow u(x) \text { for a.e. } x \in \mathbb R^N$. Then, $u \in L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big )$ and

$$\begin{aligned} \int \limits _{ \mathbb R^N } \left| \left| u_n \right| ^{ h(x) } - \left| u_n - u \right| ^{ h(x) } - \left| u \right| ^{ h(x) } \right| \, dx = o_n(1). \end{aligned}$$

(2.2)

Proposition 2.6 (Brezis–Lieb lemma, second version)

[Brezis–Lieb lemma, second version] Let $\left( u_n \right)$ be a bounded sequence in $L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big )$ with $h_- > 1$ and $u_n(x) \rightarrow u(x) \text { for a.e. } x \in \mathbb R^N$. Then

$$\begin{aligned} u_n \rightharpoonup u \quad \text { in } L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big ). \end{aligned}$$

Proposition 2.7 (Brezis–Lieb lemma, third version)

[Brezis–Lieb lemma, third version] Let $(u_n)$ be a bounded sequence in $L^{ h(x) } \big ( \mathbb R^N, \mathbb R^L \big )$ with $h_- > 1$ and $u_n(x) \rightarrow u(x)$ for a.e. $x \in \mathbb R^N$. Then

$$\begin{aligned} \int \limits _{ \mathbb R^N } \left| \left| u_n \right| ^{ h(x)-2 } u_n - \left| u_n - u \right| ^{ h(x)-2 } \left( u_n - u \right) - | u |^{ h(x)-2 } u \right| ^{ h'(x) } \, dx = o_n(1), \end{aligned}$$

(2.3)

To finish this section, we notice that for any open subset $\Omega \subset \mathbb R^N$, we can define in the same way the spaces $L^{ h(x) } \big ( \Omega \big )$ and $W^{ 1,h(x) } \big ( \Omega \big )$. Moreover, all the above propositions have analogous versions for these spaces and, besides, we have the following embedding Theorem of Sobolev’s type.

Proposition 2.8

([21, Theorems 1.1, 1.3]) Let $\Omega \subset \mathbb R^N$ an open domain with the cone property, $h :\overline{\Omega } \rightarrow \mathbb R$ satisfying $1 < h_- \le h_+ < N$ and $m \in L^{\infty }_+ \big ( \Omega \big )$.

• (i) If $h$ is Lipschitz continuous and $h \le m \le h^{*}$, the embedding $W^{ 1,h(x) } \big ( \Omega \big ) \hookrightarrow L^{ m(x) } \big ( \Omega \big )$ is continuous;

• (ii) If $\Omega$ is bounded, $h$ is continuous and $m \ll h^{*}$, the embedding $W^{ 1,h(x) } \big ( \Omega \big ) \hookrightarrow L^{ m(x) } \big ( \Omega \big )$ is compact.

3 An auxiliary problem

In this section, we work with an auxiliary problem adapting the ideas explored in del Pino and Felmer [14] (see also [3]).

We start noting that the energy functional $I_\lambda :E_\lambda \rightarrow \mathbb R$ associated with $\big ( P_\lambda \big )$ is given by

$$\begin{aligned} I_\lambda (u) = \int \limits _{ \mathbb R^N } \frac{1}{p(x)} \left( \big | \nabla u \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x) } \right) - \int \limits _{ \mathbb R^N } F(x,u), \end{aligned}$$

where $E_\lambda = \big ( E, \Vert \cdot \Vert _\lambda \big )$ with

$$\begin{aligned} E = \left\{ u \in W^{ 1,p(x) } \big ( \mathbb R^N \big ) \, ; \, \int \limits _{ \mathbb R^N } V(x) |u|^{ p(x) } < \infty \right\} , \end{aligned}$$

and

$$\begin{aligned} \Vert u \Vert _\lambda = \inf \left\{ \sigma > 0 \, ; \, \varrho _{ \lambda } \left( \frac{u}{\sigma } \right) \le 1 \right\} , \end{aligned}$$

being

$$\begin{aligned} \varrho _{ \lambda }(u) = \int \limits _{ \mathbb R^N } \left( \big | \nabla u \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x) } \right) . \end{aligned}$$

Thus, $E_\lambda \hookrightarrow W^{ 1,p(x) } \big ( \mathbb R^N \big )$ continuously for $\lambda \ge 1$ and $E_\lambda$ is compactly embedded in $L_{ loc }^{ h(x) } \big ( \mathbb R^N \big )$, for all $1 \le h \ll p^*$. In addition, we can show that $E_\lambda$ is a reflexive space. Also, being $\mathcal{O} \subset \mathbb R^N$ an open set, from the relation

$$\begin{aligned} \varrho _{ \lambda , \mathcal{O} }(u) = \int \limits _{ \mathcal O } \left( \big | \nabla u \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x) } \right) \ge M \int \limits _{ \mathcal O } |u|^{ p(x) } = M \varrho _{ p(x), \mathcal {O} }(u), \end{aligned}$$

(3.1)

for all $u \in E_\lambda$ with $\lambda \ge 1$, writing $M = ( 1-\delta )^{ -1 } \nu$, for some $0 < \delta < 1$ and $\nu > 0$, we derive

$$\begin{aligned} \varrho _{ \lambda , \mathcal {O} }(u) - \nu \varrho _{ p(x), \mathcal {O} }(u) \ge \delta \varrho _{ \lambda ,\mathcal {O} }(u), \quad \forall u \in E_\lambda , \, \lambda \ge 1. \end{aligned}$$

(3.2)

Remark 3.1

From the above commentaries, in this work the parameter $\lambda$ will be always bigger than or equal to 1.

We recall that for any $\epsilon > 0$, the hypotheses $(f_1), (f_2)$ and $(f_5)$ yield

$$\begin{aligned} f(x,t) \le \epsilon | t |^{ p(x)-1 } + C_\epsilon | t |^{ q(x)-1 }, \quad \forall x \in \mathbb R^N, \, t \in \mathbb R, \end{aligned}$$

(3.3)

and, consequently,

$$\begin{aligned} F(x,t) \le \epsilon | t |^{ p(x) } + C_\epsilon | t |^{ q(x) }, \quad \forall x \in \mathbb R^N, \, t \in \mathbb R, \end{aligned}$$

(3.4)

where $C_\epsilon$ depends on $\epsilon$. Moreover, for each $\nu >0$ fixed, the assumptions $(f_2)$ and $(f_3)$ allow us considering the function $a :\mathbb R^N \rightarrow \mathbb R$ given by

$$\begin{aligned} a(x) = \min \left\{ a > 0 \, ; \, \frac{f(x,a)}{a^{ p(x)-1 }} = \nu \right\} . \end{aligned}$$

(3.5)

From $(f_2)$, it follows that

$$\begin{aligned} 0 < a_- = \inf _{ x \in \mathbb R^N } a(x). \end{aligned}$$

(3.6)

Using the function $a(x)$, we set the function $\tilde{f} :\mathbb R^N \times \mathbb R \rightarrow \mathbb R$ given by

$$\begin{aligned} \tilde{f}(x,t) = {\left\{ \begin{array}{ll} \ \, f(x,t), \ t \le a(x) \\ \nu t^{ p(x)-1 }, \ t \ge a(x) \end{array}\right. }, \end{aligned}$$

which fulfills the inequality

$$\begin{aligned} \tilde{f}(x,t) \le \nu | t |^{ p(x)-1 }, \quad \forall x \in \mathbb R^N, t \in \mathbb R. \end{aligned}$$

(3.7)

Thus

$$\begin{aligned} \tilde{f}(x,t) t \le \nu | t |^{ p(x) }, \quad \forall x \in \mathbb R^N, t \in \mathbb R, \end{aligned}$$

(3.8)

and

$$\begin{aligned} \tilde{F}(x,t) \le \frac{\nu }{p(x)} | t |^{ p(x) }, \quad \forall x \in \mathbb R^N, t \in \mathbb R, \end{aligned}$$

(3.9)

where $\tilde{F}(x,t) = \int _0^t \tilde{f}(x,s) \, \mathrm{{d}}s$.

Now, once that $\Omega =\text {int } V^{ -1 } (0)$ is formed by $k$ connected components $\Omega _1, \ldots , \Omega _k$ with $\text {dist} \big ( \Omega _i, \Omega _j \big ) > 0, \, i \ne j$, then for each $j \in \{ 1, \ldots , k \}$, we are able to fix a smooth bounded domain $\Omega '_j$ such that

$$\begin{aligned} \overline{\Omega _j} \subset \Omega '_j \, \quad \text {and} \quad \overline{\Omega '_i} \cap \overline{\Omega '_j} = \emptyset ,\quad \text {for } i \ne j. \end{aligned}$$

(3.10)

From now on, we fix a non-empty subset $\Upsilon \subset \left\{ 1, \ldots , k \right\}$ and

$$\begin{aligned} \Omega _\Upsilon = \bigcup _{ j \in \Upsilon } \Omega _j, \, \Omega '_\Upsilon = \bigcup _{ j \in \Upsilon } \Omega '_j, \, \chi _\Upsilon = {\left\{ \begin{array}{ll} 1,&{} \text { if } x \in \Omega '_\Upsilon \\ 0,&{} \text { if } x \notin \Omega '_\Upsilon . \end{array}\right. } \end{aligned}$$

Using the above notations, we set the functions

$$\begin{aligned} g(x,t) = \chi _\Upsilon (x) f(x,t) + \big ( 1-\chi _\Upsilon (x) \big ) \tilde{f}(x,t), \, (x,t) \in \mathbb R^N \times \mathbb R \end{aligned}$$

and

$$\begin{aligned} G(x,t) = \int \limits _0^t g(x,s) \, \mathrm{{d}}s, \, (x,t) \in \mathbb R^N \times \mathbb R, \end{aligned}$$

and the auxiliary problem

$$\begin{aligned} \big ( A_\lambda \big ) \, {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x)-2 } u = g(x,u), \text { in } \mathbb R^N, \\ u \in W^{ 1,p(x) } \big ( \mathbb R^N \big ). \end{array}\right. } \end{aligned}$$

The problem $\big ( A_\lambda \big )$ is related to $\big ( P_\lambda \big )$ in the sense that, if $u_\lambda$ is a solution for $\big ( A_\lambda \big )$ verifying

$$\begin{aligned} u_\lambda (x) \le a(x), \, \forall x \in \mathbb R^N \setminus \Omega '_\Upsilon , \end{aligned}$$

then it is a solution for $\big ( P_\lambda \big )$.

In comparison with $\big ( P_\lambda \big )$, problem $\big ( A_\lambda \big )$ has the advantage that the energy functional associated with $\big ( A_\lambda \big )$, namely, $\phi _\lambda :E_\lambda \rightarrow \mathbb R$ given by

$$\begin{aligned} \phi _\lambda (u) = \int \limits _{ \mathbb R^N } \frac{1}{p(x)} \left( \left| \nabla u \right| ^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x) } \right) - \int \limits _{ \mathbb R^N } G(x,u), \end{aligned}$$

satisfies the $(PS)$ condition, whereas $I_\lambda$ does not necessarily satisfy this condition. In this way, the mountain pass level (see Theorem 3.6) is a critical value for $\phi _\lambda$.

Proposition 3.2

$\phi _\lambda$ satisfies the mountain pass geometry.

Proof

From (3.4) and (3.9),

$$\begin{aligned} \phi _\lambda (u) \ge \frac{1}{p_+} \varrho _\lambda (u) - \epsilon \int \limits _{ \mathbb R^N } |u|^{ p(x) } - C_\epsilon \int \limits _{ \mathbb R^N } |u|^{ q(x) } - \frac{\nu }{p_-} \int \limits _{\mathbb R^N} |u|^{ p(x) }, \end{aligned}$$

for $\epsilon > 0$ and $C_\epsilon > 0$ be a constant depending on $\epsilon$. By (3.1), fixing $\epsilon < \frac{M}{p_+}$ and $\nu < p_- M \left( \frac{1}{p_+}-\frac{\epsilon }{M} \right)$ and assuming $\Vert u \Vert _\lambda < \min \left\{ 1, 1/C_q \right\}$, where $| v |_{ q(x) } \le C_q \Vert v \Vert _\lambda , \, \forall v \in E_\lambda$, we derive from Proposition 2.1

$$\begin{aligned} \phi _\lambda (u) \ge \alpha \Vert u \Vert ^{ p_+ }_\lambda - C \Vert u \Vert ^{q_-}_\lambda , \end{aligned}$$

where $\alpha = \left( \frac{1}{p_+} - \frac{\epsilon }{M} \right) - \frac{\nu }{p_-M} > 0$. Once $p_+ < q_-$, the first part of the mountain pass geometry is satisfied. Now, fixing $v \in C^{\infty }_{0}(\Omega _\Upsilon )$, we have for $t \ge 0$

$$\begin{aligned} \phi _{\lambda }(tv)= \int \limits _{ \mathbb R^N } \frac{t^{p(x)}}{p(x)} \left( \left| \nabla v \right| ^{ p(x) } + Z(x) \big ) | v |^{ p(x) } \right) - \int \limits _{ \mathbb R^N } F(x,tv). \end{aligned}$$

If $t>1$, by $(f_3)$,

$$\begin{aligned} \phi _{\lambda }(tv) \le \frac{t^{p^{+}}}{p_{-}} \int \limits _{ \mathbb R^N }\left( \left| \nabla v \right| ^{ p(x) } + Z(x) \big ) | v |^{ p(x) } \right) -C_1t^{\theta }\int \limits _{\mathbb {R}^{N}}|v|^{\theta }-C_2, \end{aligned}$$

and so,

$$\begin{aligned} \phi _\lambda (tv) \rightarrow -\infty \quad \text{ as } \quad t \rightarrow +\infty . \end{aligned}$$

The last limit implies that $\phi _{\lambda }$ verifies the second geometry of the mountain pass. $\square$

Proposition 3.3

All $(PS)_d$ sequences for $\phi _\lambda$ are bounded in $E_\lambda$.

Proof

Let $(u_n)$ be a $(PS)_d$ sequence for $\phi _\lambda$. So, there is $n_0 \in \mathbb N$ such that

$$\begin{aligned} \phi _\lambda (u_n) - \frac{1}{\theta } \phi _\lambda '(u_n) u_n \le d+1 + \Vert u_n \Vert _\lambda , \text{ for } n \ge n_0. \end{aligned}$$

On the other hand, by (3.8) and (3.9)

$$\begin{aligned} \tilde{F}(x,t) - \frac{1}{\theta } \tilde{f}(x,t)t \le \left( \frac{1}{p(x)} - \frac{1}{\theta } \right) \nu | t |^{ p(x) }, \, \forall x \in \mathbb R^N,\quad t \in \mathbb R, \end{aligned}$$

which together with (3.2) gives

$$\begin{aligned} \phi _\lambda (u_n) - \frac{1}{\theta } \phi _\lambda '(u_n) u_n \ge \left( \frac{1}{p_+} - \frac{1}{\theta } \right) \delta \varrho _\lambda (u_n), \, \forall n \in \mathbb N. \end{aligned}$$

Hence

$$\begin{aligned} d+1 + \max \left\{ {\varrho _\lambda (u_n)}^{1/p_-}, {\varrho _\lambda (u_n)}^{1/p_+} \right\} \ge \left( \frac{1}{p_+} - \frac{1}{\theta } \right) \delta \varrho _\lambda (u_n), \, \forall n \ge n_0, \end{aligned}$$

from where it follows that $(u_n)$ is bounded in $E_\lambda$.$\square$

Proposition 3.4

If $(u_n)$ is a $(PS)_d$ sequence for $\phi _{\lambda }$, then given $\epsilon >0$, there is $R>0$ such that

$$\begin{aligned} \limsup _n \int \limits _{ \mathbb R^N \setminus B_R (0) } \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) < \epsilon . \end{aligned}$$

(3.11)

Hence, once that $g$ has a subcritical growth, if $u \in E_\lambda$ is the weak limit of $(u_n)$, then

$$\begin{aligned} \int \limits _{\mathbb R^N}g(x,u_n)u_n\,dx \rightarrow \int \limits _{\mathbb R^N} g(x,u)u \, dx \, \text { and } \, \int \limits _{\mathbb R^N} g(x,u_n)v \, dx \rightarrow \int \limits _{\mathbb R^N} g(x,u)v \, dx, \, \forall v \in E_\lambda . \end{aligned}$$

Proof

Let $(u_n)$ be a $(PS)_d$ sequence for $\phi _\lambda , R > 0$ large such that $\Omega '_\Upsilon \subset B_{ \frac{R}{2} }(0)$ and $\eta _R \in C^\infty \big ( \mathbb {R}^N \big )$ satisfying

$$\begin{aligned} \eta _R (x) = {\left\{ \begin{array}{ll} 0, &{} x \in B_{ \frac{R}{2} }(0) \\ 1, &{} x \in \mathbb {R}^N {\setminus } B_R (0) \end{array}\right. }, \end{aligned}$$

$0 \le \eta _R \le 1$ and $\big | \nabla \eta _R \big | \le \dfrac{C}{R}$, where $C > 0$ does not depend on $R$. This way,

$$\begin{aligned}&\int \limits _{ \mathbb {R}^N } \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \eta _R \\&\quad = \phi _\lambda '(u_n) \left( u_n \eta _R \right) - \int \limits _{ \mathbb {R}^N } u_n \big | \nabla u_n \big |^{ p(x)-2 } \nabla u_n \cdot \nabla \eta _R + \int \limits _{ \mathbb {R}^N \setminus \Omega '_\Upsilon } \tilde{f}(x,u_n) u_n \eta _R. \end{aligned}$$

Denoting

$$\begin{aligned} I =\int \limits _{ \mathbb {R}^N } \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \eta _R, \end{aligned}$$

it follows from (3.8),

$$\begin{aligned} I \le \phi _\lambda '(u_n) \left( u_n \eta _R \right) + \frac{C}{R} \int \limits _{ \mathbb {R}^N } | u_n | \big | \nabla u_n \big |^{ p(x)-1 } + \nu \int \limits _{ \mathbb {R}^N } | u_n |^{ p(x) } \eta _R. \end{aligned}$$

Using Hölder’s inequality 2.4 and Proposition 2.3, we derive

$$\begin{aligned} I \le \phi _\lambda '(u_n) \left( u_n \eta _R \right) + \frac{C}{R} | u_n |_{ p(x) } \max \left\{ \big | \nabla u_n \big |^{ p_- -1 }_{ p(x) }, \big | \nabla u_n \big |^{ p_+ -1 }_{ p(x) } \right\} + \frac{\nu }{M}I. \end{aligned}$$

Since $(u_n)$ and $\Big ( \big | \nabla u_n \big | \Big )$ are bounded in $L^{ p(x) } \big ( \mathbb {R}^N \big )$ and $\frac{\nu }{M}=1-\delta$, we obtain

$$\begin{aligned} \int \limits _{ \mathbb {R}^N \setminus B_R(0) } \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \le o_n(1) + \frac{C}{R}. \end{aligned}$$

Therefore

$$\begin{aligned} \limsup _n \int \limits _{ \mathbb {R}^N \setminus B_R(0) } \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \le \frac{C}{R}. \end{aligned}$$

So, given $\epsilon > 0$, choosing a $R > 0$ possibly still bigger, we have that $\dfrac{C}{R} < \epsilon$, which proves (3.11). Now, we will show that

$$\begin{aligned} \int \limits _{\mathbb {R}^N}g(x,u_n)u_n \rightarrow \int \limits _{\mathbb {R}^N}g(x,u)u. \end{aligned}$$

Using the fact that $g(x,u)u \in L^{1}(\mathbb {R}^N)$ together with (3.11) and Sobolev embeddings, given $\epsilon >0$, we can choose $R>0$ such that

$$\begin{aligned} \limsup _{n \rightarrow +\infty }\int \limits _{\mathbb {R}^N \setminus B_R(0)}|g(x,u_n)u_n| \le \frac{\epsilon }{4} \quad \text{ and } \quad \int \limits _{\mathbb {R}^N \setminus B_R(0)}|g(x,u)u| \le \frac{\epsilon }{4}. \end{aligned}$$

On the other hand, since $g$ has a subcritical growth, we have by compact embeddings

$$\begin{aligned} \int \limits _{B_{R}(0)}g(x,u_n)u_n \rightarrow \int \limits _{B_{R}(0)}g(x,u)u. \end{aligned}$$

Combining the above information, we conclude that

$$\begin{aligned} \int \limits _{\mathbb {R}^N}g(x,u_n)u_n \rightarrow \int \limits _{\mathbb {R}^N}g(x,u)u. \end{aligned}$$

The same type of arguments works to prove that

$$\begin{aligned} \int \limits _{\mathbb {R}^N}g(x,u_n)v \rightarrow \int \limits _{\mathbb {R}^N}g(x,u)v \quad \forall v \in E_{\lambda }. \end{aligned}$$

$\square$

Proposition 3.5

$\phi _\lambda$ verifies the $(PS)$ condition.

Proof

Let $(u_n)$ be a $(PS)_d$ sequence for $\phi _\lambda$ and $u \in E_\lambda$ such that $u_n \rightharpoonup u$ in $E_{\lambda }$. Thereby, by Proposition 3.4,

$$\begin{aligned} \int \limits _{ \mathbb R^N } g(x,u_n)u_n \rightarrow \int \limits _{ \mathbb R^N } g(x,u)u \quad \text { and } \quad \int \limits _{ \mathbb R^N } g(x,u_n)v \rightarrow \int \limits _{ \mathbb R^N } g(x,u)v, \, \forall v \in E_\lambda . \end{aligned}$$

Moreover, the weak limit also gives

$$\begin{aligned} \int \limits _{ \mathbb R^N } \big | \nabla u \big |^{ p(x)-2 } \nabla u \cdot \nabla ( u_n-u ) \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} \int \limits _{ \mathbb R^N } \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x)-2 }u ( u_n-u ) \rightarrow 0. \end{aligned}$$

Now, if

$$\begin{aligned} P_n^1 (x) = \left( \big | \nabla u_n \big |^{ p(x)-2 } \nabla u_n - \big | \nabla u \big |^{ p(x)-2 } \nabla u \right) \cdot \left( \nabla u_n - \nabla u \right) \end{aligned}$$

and

$$\begin{aligned} P_n^2 (x) = \left( | u_n|^{ p(x)-2 } u_n - | u |^{ p(x)-2 } u \right) ( u_n - u ), \end{aligned}$$

we derive

$$\begin{aligned}&\int \limits _{ \mathbb R^N } \Big ( P_n^1(x) + \big ( \lambda V(x) + Z(x) \big ) P_n^2(x) \Big ) = \phi _\lambda '(u_n)u_n + \int \limits _{ \mathbb R^N } \! g(x,u_n)u_n - \phi _\lambda '(u_n)u - \int \limits _{ \mathbb R^N } \! g(x,u_n)u \\&\quad - \int \limits _{ \mathbb R^N } \left( \big | \nabla u \big |^{ p(x)-2 } \nabla u \cdot \nabla ( u_n-u ) + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x)-2 }u ( u_n-u ) \right) . \end{aligned}$$

Recalling that $\phi _\lambda '(u_n)u_n=o_n(1)$ and $\phi _\lambda '(u_n)u=o_n(1)$, the above limits lead to

$$\begin{aligned} \int \limits _{ \mathbb R^N } \Big ( P_n^1(x) + \big ( \lambda V(x) + Z(x) \big ) P_n^2(x) \Big ) \rightarrow 0. \end{aligned}$$

Now, the conclusion follows as in [8].$\square$

Theorem 3.6

The problem $\big ( A_\lambda \big )$ has a (nonnegative) solution, for all $\lambda \ge 1$.

Proof

The proof is an immediate consequence of the Mountain Pass Theorem due to Ambrosetti and Rabinowitz [10].$\square$

4 The $(PS)_\infty$ condition

A sequence $(u_n) \subset W^{ 1,p(x) } \big ( \mathbb R^N \big )$ is called a $(PS)_\infty$ sequence for the family $\left( \phi _\lambda \right) _{\lambda \ge 1}$, if there is a sequence $( \lambda _n ) \subset [1, \infty )$ with $\lambda _n \rightarrow \infty$, as $n \rightarrow \infty$, verifying

$$\begin{aligned} \phi _{ \lambda _n }(u_n) \rightarrow c\quad \text { and }\quad \left\| \phi '_{ \lambda _n }(u_n) \right\| \rightarrow 0, \text { as } n \rightarrow \infty . \end{aligned}$$

Proposition 4.1

Let $(u_n) \subset W^{ 1,p(x) } \big ( \mathbb R^N \big )$ be a $(PS)_\infty$ sequence for $\left( \phi _\lambda \right) _{\lambda \ge 1}$. Then, up to a subsequence, there exists $u \in W^{ 1,p(x) } \big ( \mathbb R^N \big )$ such that $u_n \rightharpoonup u$ in $W^{ 1,p(x) } \big ( \mathbb R^N \big )$. Furthermore,

• (i) $\varrho _{ \lambda _n } (u_n-u) \rightarrow 0$ and, consequently, $u_n \rightarrow u$ in $W^{ 1,p(x) } \big ( \mathbb R^N \big )$;

• (ii) $u = 0$ in $\mathbb R^N \setminus \Omega _\Upsilon , u \ge 0$ and $u_{|_{\Omega _j}}, \, j \in \Upsilon$, is a solution for

  $$\begin{aligned} (P_j) \; {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + Z(x) | u |^{ p(x)-2 } u = f(x,u), \text { in } \Omega _j, \\ u \in W^{ 1,p(x) }_0 \big ( \Omega _j \big ); \end{array}\right. } \end{aligned}$$

• (iii) $\displaystyle \int _{\mathbb R^N} \lambda _n V(x) | u_n |^{ p(x) } \rightarrow 0$;

• (iv) $\varrho _{ \lambda _n, \Omega '_j } (u_n) \rightarrow \displaystyle \int _{ \Omega _j } \left( \big | \nabla u \big |^{ p(x) } + Z(x) | u |^{ p(x) } \right) , \text { for } j \in \Upsilon$;

• (v) $\varrho _{ \lambda _n, \mathbb R^N \setminus \Omega _\Upsilon } (u_n) \rightarrow 0$;

• (vi) $\phi _{ \lambda _n } (u_n) \rightarrow \displaystyle \int _{ \Omega _\Upsilon } \frac{1}{p(x)} \left( \big | \nabla u \big |^{ p(x) } + Z(x) | u |^{ p(x) } \right) - \int _{ \Omega _\Upsilon } F(x,u)$.

Proof

Using the same reasoning as in the proof of Proposition 3.3, we obtain that $\big ( \varrho _{ \lambda _n }(u_n) \big )$ is bounded in $\mathbb R$. Then $\big ( \Vert u_n \Vert _{ \lambda _n } \big )$ is bounded in $\mathbb R$ and $(u_n)$ is bounded in $W^{ 1,p(x) } \big ( \mathbb R^N \big )$. So, up to a subsequence, there exists $u \in W^{ 1,p(x) } \big ( \mathbb R^N \big )$ such that

$$\begin{aligned} u_n \rightharpoonup u \quad \text { in } W^{ 1,p(x) } \big ( \mathbb R^N \big ) \quad \text {and} \quad u_n(x) \rightarrow u(x) \quad \text {for a.e. } x \in \mathbb R^N. \end{aligned}$$

Now, for each $m \in \mathbb N$, we define $C_m = \left\{ x \in \mathbb R^N \, ; \, V(x) \ge \dfrac{1}{m} \right\}$. Without loss of generality, we can assume $\lambda _n < 2 ( \lambda _n-1 ), \, \forall n \in \mathbb N$. Thus

$$\begin{aligned} \int \limits _{ C_m } | u_n |^{ p(x) } \le \frac{2m}{\lambda _n} \int \limits _{ C_m } \big ( \lambda _n V(x)+Z(x) \big ) | u_n |^{ p(x) } \le \frac{2m}{\lambda _n} \varrho _{ \lambda _n }(u_n) \le \frac{C}{\lambda _n}. \end{aligned}$$

By Fatou’s lemma, we derive

$$\begin{aligned} \int \limits _{ C_m } | u |^{ p(x) } = 0, \end{aligned}$$

which implies that $u = 0$ in $C_m$ and, consequently, $u = 0$ in $\mathbb R^N \setminus \overline{\Omega }$. From this, we are able to prove $(i)-(vi)$.

$(i)$ :

Since $u = 0$ in $\mathbb R^N \setminus \overline{\Omega }$, repeating the argument explored in Proposition 3.5 we get

$$\begin{aligned} \int \limits _{ \mathbb R^N } \Big ( P_n^1(x) + \big ( \lambda _n V(x) + Z(x) \big ) P_n^2(x) \Big ) \rightarrow 0, \end{aligned}$$

where

$$\begin{aligned} P_n^1 (x) = \left( \big | \nabla u_n \big |^{ p(x)-2 } \nabla u_n - \big | \nabla u \big |^{ p(x)-2 } \nabla u \right) \cdot \left( \nabla u_n - \nabla u \right) \end{aligned}$$

and

$$\begin{aligned} P_n^2 (x) = \left( | u_n|^{ p(x)-2 } u_n - | u |^{ p(x)-2 } u \right) ( u_n - u ). \end{aligned}$$

Therefore, $\varrho _{ \lambda _n } ( u_n-u ) \rightarrow 0$, which implies $u_n \rightarrow u$ in $W^{ 1,p(x) } \big ( \mathbb R^N \big )$.

$(ii)$ :

Since $u \in W^{ 1,p(x) } \big ( \mathbb R^N \big )$ and $u = 0$ in $\mathbb R^N \setminus \overline{\Omega }$, we have $u \in W^{ 1,p(x) }_0 \big ( \Omega \big )$ or, equivalently, $u_{ |_{\Omega _j} } \in W^{ 1,p(x) }_0 \big ( \Omega _j \big )$, for $j = 1, \ldots , k$. Moreover, the limit $u_n \rightarrow u$ in $W^{1,p(x)}(\mathbb R^N)$ combined with $\phi '_{\lambda _n}(u_n)\varphi \rightarrow 0$ for $\varphi \in C^{\infty }_0 \big ( \Omega _j \big )$ implies that

$$\begin{aligned} \int \limits _{\Omega _j} \left( \big | \nabla u \big |^{ p(x)-2 } \nabla u \cdot \nabla \varphi + Z(x) | u |^{ p(x)-2 } u \varphi \right) - \int \limits _{\Omega _j} g(x,u) \varphi = 0, \end{aligned}$$

(4.1)

showing that $u_{ |_{\Omega _j} }$ is a solution for

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + Z(x) | u |^{ p(x)-2 } u = g(x,u), \text { in } \Omega _j, \\ u \in W^{ 1,p(x) }_0 \big ( \Omega _j \big ). \end{array}\right. } \end{aligned}$$

This way, if $j \in \Upsilon$, then $u_{ |_{\Omega _j} }$ satisfies $(P_j)$. On the other hand, if $j \notin \Upsilon$, we must have

$$\begin{aligned} \int \limits _{ \Omega _j } \left( \big | \nabla u \big |^{ p(x) } + Z(x) | u |^{ p(x) } \right) - \int \limits _{ \Omega _j } \tilde{f}(x,u)u = 0. \end{aligned}$$

The above equality combined with (3.8) and (3.2) gives

$$\begin{aligned} 0 \ge \varrho _{ \lambda , \Omega _j }(u) - \nu \varrho _{ p(x), \Omega _j }(u) \ge \delta \varrho _{ \lambda , \Omega _j }(u) \ge 0, \end{aligned}$$

from where it follows $u_{|_{\Omega _j}} = 0$. This proves $u = 0$ outside $\Omega _\Upsilon$ and $u \ge 0$ in $\mathbb {R}^N$.

$(iii)$ :

It follows from (i), since

$$\begin{aligned} \int \limits _{ \mathbb R^N } \lambda _n V(x) | u_n |^{ p(x) } = \int \limits _{ \mathbb R^N } \lambda _n V(x) | u_n-u |^{ p(x) } \le 2 \varrho _{ \lambda _n }(u_n-u). \end{aligned}$$

$(iv)$ :

Let $j \in \Upsilon$. From (i),

$$\begin{aligned} \varrho _{ p(x), \Omega '_j }( u_n-u ), \varrho _{ p(x), \Omega '_j } \big ( \nabla u_n - \nabla u \big ) \rightarrow 0. \end{aligned}$$

Then by Proposition 2.5,

$$\begin{aligned} \int \limits _{ \Omega '_j } \big ( \big | \nabla u_n \big |^{ p(x) } - \big | \nabla u \big |^{ p(x) } \big ) \rightarrow 0 \quad \text{ and } \quad \int \limits _{ \Omega '_j } Z(x) \big ( | u_n |^{ p(x) } - | u |^{ p(x) } \big ) \rightarrow 0. \end{aligned}$$

From (iii),

$$\begin{aligned} \int \limits _{ \Omega '_j } \lambda _n V(x) \big ( | u_n |^{ p(x) } - | u |^{ p(x) } \big ) = \int \limits _{ \Omega '_j \setminus \overline{\Omega _j} } \lambda _n V(x) | u_n |^{ p(x) } \rightarrow 0. \end{aligned}$$

This way

$$\begin{aligned} \varrho _{ \lambda _n, \Omega '_j } (u_n) - \varrho _{ \lambda _n, \Omega '_j } (u) \rightarrow 0. \end{aligned}$$

Once $u = 0 \text { in } \Omega '_j \setminus \Omega _j$, we get

$$\begin{aligned} \varrho _{ \lambda _n, \Omega '_j } (u_n) \rightarrow \int \limits _{ \Omega _j } \left( | \nabla u |^{ p(x) } + Z(x) | u |^{ p(x) } \right) . \end{aligned}$$

$(v)$ :

By (i), $\varrho _{ \lambda _n }( u_n-u ) \rightarrow 0$, and so,

$$\begin{aligned} \varrho _{ \lambda _n, \mathbb R^N \setminus \Omega _\Upsilon }(u_n) \rightarrow 0. \end{aligned}$$

$(vi)$ :

We can write the functional $\phi _{\lambda _n}$ in the following way

$$\begin{aligned} \phi _{ \lambda _n } (u_n)&= \sum _{ j \in \Upsilon } \int \limits _{ \Omega '_j } \frac{1}{p(x)} \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda _n V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \\&+ \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \frac{1}{p(x)} \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda _n V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) - \int \limits _{ \mathbb R^N } G(x,u_n). \end{aligned}$$

From $(i)-(v)$,

$$\begin{aligned}&\int \limits _{ \Omega '_j } \frac{1}{p(x)} \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda _n V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \\&\rightarrow \int \limits _{ \Omega _j } \frac{1}{p(x)} \left( \big | \nabla u \big |^{ p(x) } + Z(x) | u |^{ p(x) } \right) ,\\&\int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \frac{1}{p(x)} \left( \big | \nabla u_n \big |^{ p(x) } + \big ( \lambda _n V(x) + Z(x) \big ) | u_n |^{ p(x) } \right) \rightarrow 0. \end{aligned}$$

and

$$\begin{aligned} \int \limits _{ \mathbb R^N } G(x,u_n) \rightarrow \int \limits _{ \Omega _\Upsilon } F(x,u). \end{aligned}$$

Therefore

$$\begin{aligned} \phi _{ \lambda _n } (u_n) \rightarrow \int \limits _{ \Omega _\Upsilon } \frac{1}{p(x)} \left( | \nabla u |^{ p(x) } + Z(x) | u |^{ p(x) } \right) - \int \limits _{ \Omega _\Upsilon } F(x,u). \end{aligned}$$

$\square$

5 The boundedness of the $\big ( A_\lambda \big )$ solutions

In this section, we study the boundedness outside $\Omega '_\Upsilon$ for some solutions of $\big ( A_\lambda \big )$. To this end, we adapt for our problem arguments found in [18] and [25].

Proposition 5.1

Let $\big ( u_\lambda \big )$ be a family of solutions for $\big ( A_\lambda \big )$ such that $u_\lambda \rightarrow 0$ in $W^{ 1,p(x) } \big ( \mathbb R^N \setminus \Omega _\Upsilon \big )$, as $\lambda \rightarrow \infty$. Then, there exists $\lambda ^* > 0$ with the following property:

$$\begin{aligned} \left| u_\lambda \right| _{ \infty , \mathbb R^N \setminus \Omega '_\Upsilon } \le a_-, \, \forall \lambda \ge \lambda ^*. \end{aligned}$$

Hence, $u_{\lambda }$ is a solution for $(P_\lambda )$ for $\lambda \ge \lambda ^*$.

Before to prove the above proposition, we need to show some technical lemmas.

Lemma 5.2

There exist $x_1, \ldots , x_l \in \partial \Omega '_\Upsilon$ and corresponding $\delta _{x_1}, \ldots , \delta _{x_l} > 0$ such that

$$\begin{aligned} \partial \Omega '_\Upsilon \subset \mathcal{N} \left( \partial \Omega '_\Upsilon \right) : = \bigcup _{ i=1 }^l B_{ \frac{\delta _{x_i}}{2} } (x_i). \end{aligned}$$

Moreover,

$$\begin{aligned} q^{x_i}_+ \le \big ( p^{x_i}_- \big )^*, \end{aligned}$$

(5.1)

where

$$\begin{aligned} q^{x_i}_+ = \sup _{ B_{ \delta _{x_i} }({x_i}) } q, \ p^{x_i}_- = \inf _{ B_{\delta _{x_i}}(x_i) } p \, \text {and }\, \big ( p^{x_i}_- \big )^* = \frac{N p^{x_i}_-}{N-p^{x_i}_-}. \end{aligned}$$

Proof

From (3.10), $\overline{\Omega _\Upsilon } \subset \Omega '_\Upsilon$. So, there is $\delta > 0$ such that

$$\begin{aligned} \overline{B_{\delta }(x)} \subset \mathbb R^N \setminus \overline{\Omega _\Upsilon }, \, \forall x \in \partial \Omega '_\Upsilon . \end{aligned}$$

Once $q \ll p^*$, there exists $\epsilon > 0$ such that $\epsilon \le p^*(y) - q(y)$, for all $y \in \mathbb R^N$. Then, by continuity, for each $x \in \partial \Omega '_\Upsilon$, we can choose a sufficiently small $0 < \delta _x \le \delta$ such that

$$\begin{aligned} q^x_+ \le \big ( p^x_- \big )^*, \end{aligned}$$

where

$$\begin{aligned} q^x_+ = \sup _{ B_{ \delta _x }(x) } q, \ p^x_- = \inf _{ B_{ \delta _x }(x) } p \, \text {and }\, \big ( p^x_- \big )^* = \frac{N p^x_-}{N-p^x_-}. \end{aligned}$$

Covering $\partial \Omega '_\Upsilon$ by the balls $B_{ \frac{\delta _x}{2} }(x), \, x \in \partial \Omega '_\Upsilon$, and using its compactness, there are $x_1, \ldots , x_l \in \partial \Omega '_\Upsilon$ such that

$$\begin{aligned} \partial \Omega '_\Upsilon \subset \bigcup _{ i=1 }^l B_{ \frac{\delta _{x_i}}{2} }(x_i). \end{aligned}$$

$\square$

Lemma 5.3

If $u_\lambda$ is a solution for $\big ( A_\lambda \big )$, in each $B_{ \delta _{x_i} }(x_i), \, i = 1, \ldots , l$, given by Lemma 5.2, it is fulfilled

$$\begin{aligned} \int \limits _{ A_{k,\overline{\delta },x_i} } \big | \nabla u_\lambda \big |^{ p^{x_i}_- } \le C \left( \big ( k^{ q_+ } + 2 \big ) \big | {A_{k,\widetilde{\delta },x_i} } \big | +\left( \widetilde{\delta }-\overline{\delta } \right) ^{ -\big (p^{x_i}_- \big )^* } \int \limits _{ A_{k,\widetilde{\delta },x_i} } \left( u_\lambda -k \right) ^{ \big ( p^{x_i}_- \big )^* } \right) , \end{aligned}$$

where $0 < \overline{\delta } < \widetilde{\delta } < \delta _{ x_i } , k \ge \dfrac{a_-}{4} , C = C \big ( p_-, p_+, q_-, q_+, \nu , \delta _{ x_i } \big ) > 0$ is a constant independent of $k$, and for any $R > 0$, we denote by $A_ { k,R,x_i }$ the set

$$\begin{aligned} A_ { k,R,x_i } = B_R(x_i) \cap \left\{ x \in \mathbb R^N \, ; \, u_\lambda (x) > k \right\} . \end{aligned}$$

Proof

We choose arbitrarily $0 < \overline{\delta } < \widetilde{\delta } < \delta _{x_i}$ and $\xi \in C^{ \infty } \big ( \mathbb R^N \big )$ with

$$\begin{aligned} 0 \le \xi \le 1, \, \text { supp } \xi \subset B_{\widetilde{\delta } }(x_i), \, \xi = 1 \text { in } B_{\overline{\delta } }(x_i) \quad \text {and} \quad \big | \nabla \xi \big | \le \frac{2}{\widetilde{\delta }-\overline{\delta }}. \end{aligned}$$

For $k \ge \dfrac{a_-}{4}$, we define $\eta = \xi ^{ p_+ } ( u_\lambda -k )^+$. We notice that

$$\begin{aligned} \nabla \eta = p_+ \xi ^{ p_+-1 } (u_\lambda -k) \nabla \xi + \xi ^{ p_+ } \nabla u_\lambda \end{aligned}$$

on the set $\left\{ u_\lambda > k \right\}$. Then, writing $u_\lambda = u$ and taking $\eta$ as a test function, we obtain

$$\begin{aligned}&p_+ \int \limits _{ A_{k,\widetilde{\delta },x_i} } \xi ^{ p_+-1 } (u-k) \big | \nabla u \big |^{ p(x)-2 } \nabla u \cdot \nabla \xi + \int \limits _{ A_{k,\widetilde{\delta },x_i} } \xi ^{ p_+ } \big | \nabla u \big |^{ p(x) } \\&\quad + \int \limits _{ A_{k,\widetilde{\delta },x_i} } \big ( \lambda V(x) + Z(x) \big ) u^{ p(x)-1 } \xi ^{ p_+ } (u-k) = \int \limits _{ A_{k,\widetilde{\delta },x_i} } g(x,u) \xi ^{ p_+ } (u-k). \end{aligned}$$

If we set

$$\begin{aligned} J = \int \limits _{ A_{k,\widetilde{\delta },x_i} } \xi ^{ p_+ } \big | \nabla u \big |^{ p(x) }, \end{aligned}$$

using that $\nu \le \lambda V(x) + Z(x), \, \forall x \in \mathbb R^N$, we get

$$\begin{aligned}&J \le p_+ \int \limits _{ A_{k,\widetilde{\delta },x_i} } \xi ^{ p_+-1 } (u-k) \big | \nabla u \big |^{ p(x)-1 } \big | \nabla \xi \big |\nonumber \\&\qquad - \int \limits _{ A_{k,\widetilde{\delta },x_i} } \nu u^{ p(x)-1 } \xi ^{ p_+ } (u-k) + \int \limits _{ A_{k,\widetilde{\delta },x_i} } g(x,u) \xi ^{ p_+ } (u-k). \end{aligned}$$

(5.2)

From (5.2), (3.3) and (3.7),

$$\begin{aligned} J&\le p_+ \int \limits _{ A_{k,\widetilde{\delta },x_i} } \xi ^{ p_+-1 } (u-k) \big | \nabla u \big |^{ p(x)-1 } \big | \nabla \xi \big | - \int \limits _{ A_{k,\widetilde{\delta },x_i} } \nu u^{ p(x)-1 } \xi ^{ p_+ } (u-k) \\&\quad + \int \limits _{ A_{k,\widetilde{\delta },x_i} } \big ( \nu u^{ p(x)-1 } + C_\nu u^{ q(x)-1 } \big ) \xi ^{ p_+ } (u-k), \end{aligned}$$

from where it follows

$$\begin{aligned} J \le p_+ \int \limits _{ A_{k,\widetilde{\delta },x_i} } \xi ^{ p_+-1 } (u-k) \big | \nabla u \big |^{ p(x)-1 } \big | \nabla \xi \big | + C_\nu \int \limits _{ A_{k,\widetilde{\delta },x_i} } u^{ q(x)-1 } (u-k). \end{aligned}$$

Using Young’s inequality, we obtain, for $\chi \in (0,1)$,

$$\begin{aligned}&J \le \frac{p_+ (p_+-1)}{p_-} \chi ^{ \frac{p_-}{p_+-1}} J + \frac{2^{ p_+ } p_+}{p_-} \chi ^{ -p_+ } \int \limits _{ A_{k,\widetilde{\delta },x_i} } \left( \frac{u-k}{\widetilde{\delta }-\overline{\delta }} \right) ^{ p(x) } \\&\quad \quad + \frac{C_\nu (q_+-1)}{q_-} \int \limits _{ A_{k,\widetilde{\delta },x_i} } u^{ q(x) } + \frac{C_\nu \left( 1 + \delta _{ x_i }^{ q_+ } \right) }{q_-} \int \limits _{ A_{k,\widetilde{\delta },x_i} } \left( \frac{u-k}{\widetilde{\delta }-\overline{\delta }}\right) ^{ q(x) }. \end{aligned}$$

Writing

$$\begin{aligned} Q = \int \limits _{ A_{k,\widetilde{\delta },x_i} } \left( \frac{u-k}{\widetilde{\delta }-\overline{\delta }} \right) ^{ \left( p^{x_i}_- \right) ^* }, \end{aligned}$$

for $\chi \approx 0^+$ fixed, due to (5.1), we deduce

$$\begin{aligned} J&\le \frac{1}{2} J + \frac{2^{ p_+ } p_+}{p_-} \chi ^{ -p_+ } \Big ( \big | { A_{k,\widetilde{\delta },x_i} } \big | + Q \Big ) + \frac{C_\nu 2^{ q_+ } (q_+-1) \left( 1 + \delta _{ x_i }^{ q_+ } \right) }{q_-} \Big ( \big | { A_{k,\widetilde{\delta },x_i} } \big | + Q \Big ) \\&\quad + \frac{C_\nu 2^{ q_+ } (q_+-1) \left( 1+k^{ q_+ } \right) }{q_-} \big | { A_{k,\widetilde{\delta },x_i} } \big | + \frac{C_\nu \left( 1 + \delta _{ x_i }^{ q_+ } \right) }{q_-} \Big ( \big | { A_{k,\widetilde{\delta },x_i} } \big | + Q \Big ). \end{aligned}$$

Therefore

$$\begin{aligned} \int \limits _{ A_{k,\overline{\delta },x_i} } \big | \nabla u \big |^{ p(x) } \le J \le C \left[ \big ( k^{ q_+ } + 1 \big ) \big | A_{k,\widetilde{\delta },x_i} \big | + Q \right] , \end{aligned}$$

for a positive constant $C = C \big ( p_-, p_+, q_-, q_+, \nu , \delta _{ x_i } \big )$ which does not depend on $k$. Since

$$\begin{aligned} \big | \nabla u \big |^{ p^{x_i}_- } - 1 \le \big | \nabla u \big |^{ p(x) }, \, \forall x \in B_{\delta _{x_i}}(x_i), \end{aligned}$$

we obtain

$$\begin{aligned} \int \limits _{ A_{k,\overline{\delta },x_i} } \big | \nabla u \big |^{ p^{x_i}_- }&\le C \left[ \big ( k^{ q_+ } + 1 \big ) \big | A_{k,\widetilde{\delta },x_i} \big | + Q \right] + \big | A_{k,\widetilde{\delta },x_i} \big | \\&\le C \left( \big ( k^{ q_+ } + 2 \big ) \big | A_{k,\widetilde{\delta },x_i} \big | + \left( \widetilde{\delta }-\overline{\delta } \right) ^{ -\big ( p^{x_i}_- \big )^* } \int \limits _{ A_{k,\widetilde{\delta },x_i} } \left( u-k \right) ^{ \big ( p^{x_i}_- \big )^* } \right) , \end{aligned}$$

for a positive constant $C = C \big ( p_-, p_+, q_-, q_+, \nu , \delta _{ x_i } \big )$ which does not depend on $k$.$\square$

The next lemma can be found at ([27, Lemma 4.7]).

Lemma 5.4

Let $(J_n)$ be a sequence of nonnegative numbers satisfying

$$\begin{aligned} J_{ n+1 } \le C B^n J_n^{ 1+\eta }, \, n=0,1,2,\ldots , \end{aligned}$$

where $C, \eta > 0$ and $B > 1$. If

$$\begin{aligned} J_0 \le C^{ - \frac{1}{\eta } } B^{ - \frac{1}{{\eta }^2} }, \end{aligned}$$

then $J_n \rightarrow 0$, as $n \rightarrow \infty$.

Lemma 5.5

Let $\big ( u_\lambda \big )$ be a family of solutions for $\big ( A_\lambda \big )$ such that $u_\lambda \rightarrow 0$ in $W^{ 1,p(x) } \big ( \mathbb R^N \setminus \Omega _\Upsilon \big )$, as $\lambda \rightarrow \infty$. Then, there exists $\lambda ^* > 0$ with the following property:

$$\begin{aligned} \left| u_\lambda \right| _{ \infty , \mathcal{N} \left( \partial \Omega '_\Upsilon \right) } \le a_-, \, \forall \lambda \ge \lambda ^*. \end{aligned}$$

Proof

It is enough to prove the inequality in each ball $B_{\frac{\delta _{x_i}}{2}} (x_i), \, i = 1, \ldots , l$, given by Lemma 5.2. We set

$$\begin{aligned} \widetilde{\delta }_n = \frac{\delta _{x_i}}{2} + \frac{\delta _{x_i}}{2^{n+1}}, \ \overline{\delta }_n = \frac{\widetilde{\delta }_n + \widetilde{\delta }_{n+1}}{2}, \ k_n = \frac{a_-}{2} \left( 1-\frac{1}{2^{n+1}} \right) , \, \forall n = 0, 1, 2, \ldots . \end{aligned}$$

Then

$$\begin{aligned} \widetilde{\delta }_n \downarrow \frac{\delta _{x_i}}{2}, \quad \widetilde{\delta }_{ n+1 } < \overline{\delta }_n < \widetilde{\delta }_n, \quad k_n \uparrow \frac{a_-}{2}. \end{aligned}$$

From now on, we fix

$$\begin{aligned} J_n(\lambda ) = J_n = \int \limits _{ A_{k_n,\widetilde{\delta }_n,x_i} } \big ( u_\lambda (x) - k_n \big )^{ \left( p^{x_i}_- \right) ^* }, \, n = 0, 1, 2, \ldots . \end{aligned}$$

and $\xi \in C^1 \big ( \mathbb R \big )$ such that

$$\begin{aligned} 0 \le \xi \le 1,\quad \xi (t) = 1, \text { for } t \le \frac{1}{2},\quad \text {and}\quad \xi (t) = 0, \text { for } t \ge \frac{3}{4}. \end{aligned}$$

Setting

$$\begin{aligned} \xi _n(x) = \xi \Bigg ( \frac{2^{ n+1 }}{\delta _{x_i}} \bigg ( \big | x-x_i \big |-\frac{\delta _{x_i}}{2} \bigg ) \Bigg ), \quad x \in \mathbb R^N, \quad n = 0, 1, 2, \ldots , \end{aligned}$$

we have $\xi _n = 1$ in $B_{ \widetilde{\delta }_{n+1} }(x_i)$ and $\xi _n = 0$ outside $B_{ \overline{\delta }_n }(x_i)$. Writing $u_\lambda = u$, we get

$$\begin{aligned} J_{ n+1 }&\le \int \limits _{ A_{k_{n+1},\overline{\delta }_n,x_i} } \big ( (u(x) - k_{ n+1 } ) \xi _n(x) \big )^{ \left( p^{x_i}_- \right) ^* } \\&= \int \limits _{ B_{ \delta _{ x_i } }(x_i) } \big ( ( u-k_{ n+1 } )^+(x) \xi _n(x) \big )^{ \left( p^{x_i}_- \right) ^* } \\&\le C \big ( N, p^{x_i}_- \big ) \left( \int \limits _{ B_{ \delta _{ x_i } }(x_i) } \big | \nabla \big ( ( u-k_{ n+1 } )^+ \xi _n \big )(x) \big |^{ p^{x_i}_- } \right) ^{ \frac{\left( p^{x_i}_- \right) ^*}{ p^{x_i}_-}} \\&\le C \big ( N, p^{x_i}_- \big ) \left( \int \limits _{ A_{k_{n+1},\overline{\delta }_n,x_i} } \big | \nabla u \big |^{ p^{x_i}_- } + \int \limits _{ A_{k_{n+1},\overline{\delta }_n,x_i} } ( u-k_{ n+1 } )^{ p^{x_i}_- } \big | \nabla \xi _n \big |^{ p^{x_i}_- } \right) ^{ \frac{\left( p^{x_i}_- \right) ^*}{ p^{x_i}_-} }. \end{aligned}$$

Since

$$\begin{aligned} \big | \nabla \xi _n(x) \big | \le C \big ( \delta _{ x_i } \big ) 2^{ n+1 }, \, \forall x \in \mathbb R^N, \end{aligned}$$

writing $J_{ n+1 }^{ \frac{p^{x_i}_-}{\left( p^{x_i}_- \right) ^*} } = \widetilde{J}_{ n+1 }$, we obtain

$$\begin{aligned} \widetilde{J}_{ n+1 } \le C \Big ( N, p^{x_i}_-, \delta _{ x_i } \Big ) \left( \int \limits _{ A_{k_{n+1},\overline{\delta }_n,x_i} } \big | \nabla u \big |^{ p^{x_i}_- } + 2^{ n p^{x_i}_- } \int \limits _{ A_{k_{n+1},\overline{\delta }_n,x_i} } ( u-k_{ n+1 } )^{ p^{x_i}_- } \right) . \end{aligned}$$

Using Lemma 5.3,

$$\begin{aligned} \widetilde{J}_{ n+1 }&\le C \Big ( N, p^{x_i}_-, \delta _{ x_i } \Big ) \bigg ( \left( k_{ n+1 }^{ q_+ }+2 \right) \big | A_{ k_{ n+1},\widetilde{\delta }_n,x_i } \big | \\&\quad + \left( \frac{2^{ n+3 }}{\delta _{ x_i }} \right) ^{ \left( p^{x_i}_- \right) ^* } \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_{ n+1 } )^{ \left( p^{x_i}_- \right) ^* } + 2^{ n p^{x_i}_- } \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_{ n+1 } )^{ p^{x_i}_- } \bigg )\\&\le C \Big ( N, p^{x_i}_-, \delta _{ x_i } \Big ) \bigg ( \left( k_{ n+1 }^{ q_+ }+2 \right) \big | A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } \big | \\&\quad + 2^{ n \left( p^{x_i}_- \right) ^* } \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_{ n+1 } )^{ \left( p^{x_i}_- \right) ^* } + 2^{ n p^{x_i}_- } \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_{ n+1 } )^{ p^{x_i}_- } \bigg ). \end{aligned}$$

From Young’s inequality

$$\begin{aligned} \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_{ n+1 } )^{ p^{x_i}_- } \le C \Big ( p^{x_i}_- \Big ) \left( \big | A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } \big | + \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_{ n+1 } )^{ \left( p^{x_i}_- \right) ^* } \right) . \end{aligned}$$

Thus

$$\begin{aligned} \widetilde{J}_{ n+1 } \le C \Big ( N, p^{x_i}_-, \delta _{ x_i } \Big ) \Bigg ( \bigg ( \left( \frac{a_-}{2} \right) ^{ q_+ }+2+2^{ n p^{x_i}_- } \bigg ) \big | A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } \big | + 2^{ n \left( p^{x_i}_- \right) ^* } J_n + 2^{ n p^{ x_i }_- } J_n \Bigg ). \end{aligned}$$

Now, since

$$\begin{aligned} J_n \ge \int \limits _{ A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } } ( u-k_n)^{ \left( p^{x_i}_- \right) ^* } \ge ( k_{ n+1 }-k_n )^{ \left( p^{x_i}_- \right) ^* } \big | A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } \big | \end{aligned}$$

it follows that

$$\begin{aligned} \big | A_{ k_{ n+1 },\widetilde{\delta }_n,x_i } \big | \le \left( \frac{2^{ n+3 }}{a_-} \right) ^{ \left( p^{x_i}_- \right) ^* } J_n, \end{aligned}$$

and so,

$$\begin{aligned} \widetilde{J}_{ n+1 }&\le C \Big ( N, p^{x_i}_-, \delta _{ x_i }, a_-, q_+ \Big ) \left( 2^{ n \left( p^{x_i}_- \right) ^* } J_n + 2^{ n \big ( p^{ x_i }_- + \left( p^{x_i}_- \right) ^* \big ) } J_n + 2^{ n \left( p^{x_i}_- \right) ^* } J_n + 2^{ n p^{ x_i }_- } J_n \right) . \end{aligned}$$

Fixing $\alpha = \big ( p^{ x_i }_- + \left( p^{x_i}_- \right) ^* \big )$, it follows that

$$\begin{aligned} J_{ n+1 } \le C \Big ( N, p^{x_i}_-, \delta _{ x_i }, a_-, q_+ \Big ) \left( 2^{ \alpha \frac{\left( p^{x_i}_- \right) ^*}{p^{ x_i }_-} } \right) ^n { J_n }^{ \frac{\left( p^{x_i}_- \right) ^*}{p^{ x_i }_-}}, \end{aligned}$$

and consequently

$$\begin{aligned} J_{ n+1 } \le C B^n J_n^{ 1+\eta }, \end{aligned}$$

where $C = C \Big ( N, p^{x_i}_-, \delta _{ x_i }, a_-, q_+ \Big ) , B = 2^{ \alpha \frac{\left( p^{x_i}_- \right) ^*}{p^{ x_i }_-} }$ and $\eta = \frac{\left( p^{x_i}_- \right) ^*}{p^{ x_i }_-} -1$. Now, once that $u_\lambda \rightarrow 0$ in $W^{ 1,p(x) } \big ( \mathbb R^N \setminus \Omega _\Upsilon \big )$, as $\lambda \rightarrow \infty$, there exists $\lambda _i > 0$ such that

$$\begin{aligned} \int \limits _{ A_{ \frac{a_-}{4}, \delta _{ x_i }, x_i } } \left( u_\lambda -\frac{a_-}{4} \right) ^{ \left( p^{x_i}_- \right) ^* } = J_0(\lambda ) \le C^{ - \frac{1}{\eta } } B^{ - \frac{1}{{\eta }^2} }, \quad \lambda \ge \lambda _i. \end{aligned}$$

From Lemma 5.4, $J_n(\lambda ) \rightarrow 0 , n \rightarrow \infty$, for all $\lambda \ge \lambda _i$, and so,

$$\begin{aligned} u_\lambda \le \frac{a_-}{2} < a_-, \text { in } B_{\frac{\delta _{x_i}}{2}}, \text { for all } \lambda \ge \lambda _i. \end{aligned}$$

Now, taking $\lambda ^* = \max \{ \lambda _1, \ldots , \lambda _l \}$, we conclude that

$$\begin{aligned} \left| u_\lambda \right| _{ \infty , \mathcal{N} \left( \partial \Omega '_\Upsilon \right) } < a_-, \, \forall \lambda \ge \lambda ^*. \end{aligned}$$

$\square$

Proof of Proposition 5.1

Fix $\lambda \ge \lambda ^*$, where $\lambda ^*$ is given at Lemma 5.5, and define $\widetilde{u}_\lambda :\mathbb R^N \setminus \Omega '_\Upsilon \rightarrow \mathbb R$ given by

$$\begin{aligned} \widetilde{u}_\lambda (x) = \left( u_\lambda -a_- \right) ^+ (x). \end{aligned}$$

From Lemma 5.5, $\widetilde{u}_\lambda \in W^{ 1,p(x) }_0 \big ( \mathbb R^N \setminus \Omega '_\Upsilon \big )$. Our goal is showing that $\widetilde{u}_\lambda = 0$ in $\mathbb R^N \setminus \Omega '_\Upsilon$. This implies

$$\begin{aligned} \left| u_\lambda \right| _{ \infty , \mathbb R^N \setminus \Omega '_\Upsilon } \le a_-. \end{aligned}$$

In fact, extending $\widetilde{u}_\lambda = 0$ in $\Omega '_\Upsilon$ and taking $\widetilde{u}_\lambda$ as a test function, we obtain

$$\begin{aligned} \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \! \! \big | \nabla u_\lambda \big |^{ p(x)-2 } \nabla u_\lambda \cdot \nabla \widetilde{u}_\lambda + \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \! \! \! \! \big ( \lambda V(x) + Z(x) \big ) u_{\lambda }^{ p(x)-2 } u_\lambda \widetilde{u}_\lambda = \! \! \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } g \left( x, u_\lambda \right) \widetilde{u}_\lambda . \end{aligned}$$

Since

$$\begin{aligned} \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \big | \nabla u_\lambda \big |^{ p(x)-2 } \nabla u_\lambda \cdot \nabla \widetilde{u}_\lambda&= \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \big | \nabla \widetilde{u}_\lambda \big |^{ p(x)}, \\ \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \! \! \! \!\big ( \lambda V(x) + Z(x) \big ) u_{\lambda }^{ p(x)-2 } u_\lambda \widetilde{u}_\lambda&= \int \limits _{ \left( \mathbb R^N \setminus \Omega '_\Upsilon \right) _+ } \! \! \! \! \big ( \lambda V(x) + Z(x) \big ) u_{\lambda }^{ p(x)-2 } \left( \widetilde{u}_\lambda +a_- \right) \widetilde{u}_\lambda \end{aligned}$$

and

$$\begin{aligned} \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } g \left( x, u_\lambda \right) \widetilde{u}_\lambda = \int \limits _{ \left( \mathbb R^N \setminus \Omega '_\Upsilon \right) _+ } \frac{g \left( x, u_\lambda \right) }{u_\lambda } \left( \widetilde{u}_\lambda +a_- \right) \widetilde{u}_\lambda , \end{aligned}$$

where

$$\begin{aligned} \left( \mathbb R^N \setminus \Omega '_\Upsilon \right) _+ = \left\{ x \in \mathbb R^N \setminus \Omega '_\Upsilon \, ; \, u_\lambda (x) > a_- \right\} , \end{aligned}$$

we derive

$$\begin{aligned} \int \limits _{ \mathbb R^N \setminus \Omega '_\Upsilon } \! \! \big | \nabla \widetilde{u}_\lambda \big |^{ p(x) } + \int \limits _{ \left( \mathbb R^N \setminus \Omega '_\Upsilon \right) _+ } \! \! \! \! \left( \big ( \lambda V(x) + Z(x) \big ) u_{\lambda }^{ p(x)-2 } -\frac{g \left( x, u_\lambda \right) }{u_\lambda } \right) \left( \widetilde{u}_\lambda +a_- \right) \widetilde{u}_\lambda = 0, \end{aligned}$$

Now, by (3.7),

$$\begin{aligned} \big ( \lambda V(x) + Z(x) \big ) u_{\lambda }^{ p(x)-2 } - \frac{g \left( x, u_\lambda \right) }{u_\lambda } > \nu u_{\lambda }^{ p(x)-2 } - \frac{\tilde{f} \left( x, u_\lambda \right) }{u_\lambda } \ge 0 \quad \text{ in } \quad \left( \mathbb R^N \setminus \Omega '_\Upsilon \right) _+ . \end{aligned}$$

This form, $\widetilde{u}_\lambda = 0$ in $\left( \mathbb R^N \setminus \Omega '_\Upsilon \right) _+$. Obviously, $\widetilde{u}_\lambda = 0$ at the points where $u_\lambda \le a_-$, consequently, $\widetilde{u}_\lambda = 0$ in $\mathbb R^N \setminus \Omega '_\Upsilon$.

6 A special critical value for $\phi _\lambda$

For each $j = 1, \ldots , k$, consider

$$\begin{aligned} I_j(u) = \int \limits _{ \Omega _j } \frac{1}{p(x)} \left( \big | \nabla u \big |^{ p(x) } + Z(x) | u |^{ p(x) } \right) - \int \limits _{ \Omega _j } F(x,u), \ u \in W^{ 1,p(x) }_0 \big ( \Omega _j \big ), \end{aligned}$$

the energy functional associated to $(P_j)$, and

$$\begin{aligned} \phi _{ \lambda ,j }(u) = \int \limits _{ \Omega '_j } \frac{1}{p(x)} \left( \big | \nabla u \big |^{ p(x) } + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x) } \right) - \int \limits _{ \Omega '_j } F(x,u), \ u \in W^{ 1,p(x) } \big ( \Omega '_j \big ), \end{aligned}$$

the energy functional associated to

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + \big ( \lambda V(x) + Z(x) \big ) | u |^{ p(x)-2 } u = f(x,u),&{} \text { in } \Omega '_j, \\ \frac{\partial u}{\partial \eta } = 0,&{} \text { on } \partial \Omega '_j. \end{array}\right. } \end{aligned}$$

It is fulfilled that $I_j$ and $\phi _{ \lambda ,j }$ satisfy the mountain pass geometry and let

$$\begin{aligned} c_j = \inf _{ \gamma \in \Gamma _j } \max _{ t \in [0,1] } I_j \big ( \gamma (t) \big ) \, \text { and } \, c_{ \lambda ,j } = \inf _{ \gamma \in \Gamma _{ \lambda ,j } } \max _{ t \in [0,1] } \phi _{ \lambda ,j } \big ( \gamma (t) \big ), \end{aligned}$$

their respective mountain pass levels, where

$$\begin{aligned} \Gamma _j = \left\{ \gamma \in C \Big ( [0,1], W^{ 1,p(x) }_0 \big ( \Omega _j \big ) \Big ) \, ; \, \gamma (0) = 0 \text { and } I_j \big ( \gamma (1) \big ) < 0 \right\} \end{aligned}$$

and

$$\begin{aligned} \Gamma _{ \lambda ,j } = \left\{ \gamma \in C \Big ( [0,1], W^{ 1,p(x) } \big ( \Omega '_j \big ) \Big ) \, ; \, \gamma (0) = 0 \text { and } \phi _{ \lambda ,j } \big ( \gamma (1) \big ) < 0 \right\} . \end{aligned}$$

Invoking the $(PS)$ condition on $I_j$ and $\phi _{ \lambda ,j}$, we ensure that there exist $w_j \in W^{ 1,p(x) }_0 \big ( \Omega _j \big )$ and $w_{ \lambda ,j } \in W^{ 1,p(x) } \big ( \Omega '_j \big )$ such that

$$\begin{aligned} I_j \big ( w_j \big ) = c_j \, \text { and } \, I'_j \big ( w_j \big ) = 0 \end{aligned}$$

and

$$\begin{aligned} \phi _{ \lambda ,j } \big ( w_{ \lambda ,j } \big ) = c_{ \lambda ,j } \, \text { and } \, \phi '_{ \lambda ,j } \big ( w_{ \lambda ,j } \big ) = 0. \end{aligned}$$

Lemma 6.1

There holds that

• (i) $0 < c_{ \lambda ,j } \le c_j, \, \forall \lambda \ge 1, \, \forall j \in \left\{ 1, \ldots , k \right\}$;

• (ii) $c_{ \lambda ,j } \rightarrow c_j, \text { as } \lambda \rightarrow \infty , \, \forall j \in \left\{ 1, \ldots , k \right\}$.

Proof

• (i) Once $W^{ 1,p(x) }_0 \big ( \Omega _j \big ) \subset W^{ 1,p(x) } \big ( \Omega '_j \big )$ and $\phi _{ \lambda ,j } \big ( \gamma (1) \big ) = I_j \big ( \gamma (1) \big )$ for $\gamma \in \Gamma _j$, we have $\Gamma _j \subset \Gamma _{ \lambda ,j }$. This way

  $$\begin{aligned} c_{ \lambda ,j } = \inf _{ \gamma \in \Gamma _{ \lambda ,j } } \max _{ t \in [0,1] } \phi _{ \lambda ,j } \big ( \gamma (t) \big ) \le \inf _{ \gamma \in \Gamma _j } \max _{ t \in [0,1] } \phi _{ \lambda ,j } \big ( \gamma (t) \big ) = \inf _{ \gamma \in \Gamma _j } \max _{ t \in [0,1] } I_j \big ( \gamma (t) \big ) = c_j. \end{aligned}$$

• (ii) It suffices to show that $c_{ \lambda _n,j } \rightarrow c_j, \text { as } n \rightarrow \infty$, for all sequences $( \lambda _n )$ in $[1,\infty )$ with $\lambda _n \rightarrow \infty , \text { as } n \rightarrow \infty$. Let $\left( \lambda _n \right)$ be such a sequence and consider an arbitrary subsequence of $\left( c_{ \lambda _n,j } \right)$ (not relabeled) . Let $w_n \in W^{ 1,p(x) } \big ( \Omega '_j \big )$ with

  $$\begin{aligned} \phi _{ \lambda _n,j } \big ( w_n \big ) = c_{ \lambda _n,j } \, \text { and } \, \phi '_{ \lambda _n,j } \big ( w_n \big ) = 0. \end{aligned}$$

  By the previous item, $\big ( c_{ \lambda _n,j } \big )$ is bounded. Then, there exists $\big ( w_{ n_k } \big )$ subsequence of $\big ( w_n \big )$ such that $\phi _{ \lambda _{ n_k },j } \big ( w_{ n_k } \big )$ converges and $\phi '_{ \lambda _{ n_k },j } \big ( w_{ n_k } \big ) = 0$. Now, repeating the same type of arguments explored in the proof of Proposition 4.1, there is $w \in W^{ 1,p(x) }_0 \big ( \Omega _j \big ) \setminus \{0\} \subset W^{ 1,p(x) } \big ( \Omega '_j \big )$ such that

  $$\begin{aligned} w_{ n_k } \rightarrow w \text { in } W^{ 1,p(x) } \big ( \Omega '_j \big ), \text { as } k \rightarrow \infty . \end{aligned}$$

  Furthermore, we also can prove that

  $$\begin{aligned} c_{ \lambda _{ n_k },j } = \phi _{ \lambda _{ n_k },j } \big ( w_{ n_k } \big ) \rightarrow I_j(w) \end{aligned}$$

  and

  $$\begin{aligned} 0 = \phi '_{ \lambda _{ n_k },j } \big ( w_{ n_k } \big ) \rightarrow I'_j(w). \end{aligned}$$

  Then, by $(f_4)$,

  $$\begin{aligned} \lim _k c_{ \lambda _{ n_k },j } \ge c_j. \end{aligned}$$

  The last inequality together with item (i) implies

  $$\begin{aligned} c_{ \lambda _{ n_k },j } \rightarrow c_j, \text { as } k \rightarrow \infty . \end{aligned}$$

  This establishes the asserted result.

$\square$

In the sequel, let $R > 1$ verifying

$$\begin{aligned} 0< I_j \left( \frac{1}{R} w_j \right) , I_j(R w_j)< c_j, \text { for } j = 1, \ldots , k. \end{aligned}$$

(6.1)

There holds that

$$\begin{aligned} c_j = \max _{ t \in [1/R^2,1] } I_j (t R w_j ), \text { for } j = 1, \ldots , k. \end{aligned}$$

Moreover, to simplify the notation, we rename the components $\Omega _j$ of $\Omega$ in way such that $\Upsilon = \{ 1, 2, \ldots , l \}$ for some $1 \le l \le k$. Then, we define:

$$\begin{aligned}&\gamma _0 ( t_1, \ldots , t_l )(x) = \sum _{j=1}^l t_j R w_j(x), \, \forall ( t_1, \ldots , t_l )\in [1/R^2,1]^l, \\&\Gamma _*= \Big \{ \gamma \in C \big ( [1/R^2,1]^l, E_\lambda \setminus \{ 0 \} \big ) \, ; \, \gamma = \gamma _0 \text { on } \partial [1/R^2,1]^l \Big \} \end{aligned}$$

and

$$\begin{aligned} b_{ \lambda , \Upsilon } = \inf _{ \gamma \in \Gamma _*} \max _{ ( t_1, \ldots , t_l )\in [1/R^2,1]^l } \phi _\lambda \big ( \gamma ( t_1, \ldots , t_l ) \big ). \end{aligned}$$

Next, our intention is proving that $b_{ \lambda , \Upsilon }$ is a critical value for $\phi _\lambda$. However, to do this, we need to some technical lemmas. The arguments used are the same found in [3]; however, for reader’s convenience, we will repeat their proofs

Lemma 6.2

For all $\gamma \in \Gamma _*$, there exists $(s_1, \ldots , s_l ) \in [1/R^2,1]^l$ such that

$$\begin{aligned} \phi '_{ \lambda ,j } \big ( \gamma ( s_1, \ldots , s_l ) \big ) \big ( \gamma ( s_1, \ldots , s_l ) \big ) = 0, \, \forall j \in \Upsilon . \end{aligned}$$

Proof

Given $\gamma \in \Gamma _*$, consider $\widetilde{\gamma } :[1/R^2,1]^l \rightarrow \mathbb R^l$ such that

$$\begin{aligned} \widetilde{\gamma } ( \mathbf t ) = \Big ( \phi '_{ \lambda ,1 } \big ( \gamma ( \mathbf t ) \big ) \gamma ( \mathbf t ), \ldots , \phi '_{ \lambda ,l } \big ( \gamma ( \mathbf t ) \big ) \gamma ( \mathbf t ) \Big ), \text { where } \mathbf t = ( t_1, \ldots , t_l ). \end{aligned}$$

For $\mathbf t \in \partial [1/R^2,1]^l$, it holds $\widetilde{\gamma } ( \mathbf t ) = \widetilde{\gamma _0} ( \mathbf t )$. From this, we observe that there is no $\mathbf t \in \partial [1/R^2,1]^l$ with $\widetilde{\gamma } ( \mathbf t ) = 0$. Indeed, for any $j \in \Upsilon$,

$$\begin{aligned} \phi '_{ \lambda ,j } \big ( \gamma _0 ( \mathbf{{t}} ) \big ) \gamma _0 ( \mathbf{{t}} ) = I'_j ( t_j R w_j ) ( t_j R w_j ). \end{aligned}$$

This form, if $\mathbf{{t}} \in \partial [1/R^2,1]^l$, then $t_{j_0} =1$ or $t_{j_0} = \frac{1}{R^2}$, for some $j_0 \in \Upsilon$. Consequently,

$$\begin{aligned} \phi '_{ \lambda ,j_0 } \big ( \gamma _0 ( \mathbf{{t}} ) \big ) \gamma _0 ( \mathbf{{t}} ) = I'_{j_0} ( R w_{j_0} ) ( R w_{j_0} ) \, \text { or } \, \phi '_{ \lambda ,j_0 } \big ( \gamma _0 ( \mathbf{{t}} ) \big ) \gamma _0 ( \mathbf{{t}} ) = I'_{j_0} \left( \frac{1}{R} w_{j_0} \right) \left( \frac{1}{R} w_{j_0} \right) . \end{aligned}$$

Therefore, if $\phi '_{ \lambda ,j_0 } \big ( \gamma _0 ( \mathbf{{t}} ) \big ) \gamma _0 ( \mathbf{{t}} ) = 0$, we get $I_{j_0} ( R w_{j_0} ) \ge c_{j_0}$ or $I_{j_0} \left( \frac{1}{R} w_{j_0} \right) \ge c_{j_0}$, which is a contradiction with (6.1).

Now, we compute the degree $\deg \big ( \widetilde{\gamma }, (1/R^2,1)^l, (0, \ldots , 0 ) \big )$. Since

$$\begin{aligned} \deg \big ( \widetilde{\gamma }, (1/R^2,1)^l, (0, \ldots , 0 ) \big ) = \deg \big ( \widetilde{\gamma _0}, (1/R^2,1)^l, (0, \ldots , 0 ) \big ), \end{aligned}$$

and, for $\mathbf t \in (1/R^2,1)^l$,

$$\begin{aligned} \widetilde{\gamma _0} ( \mathbf t ) = 0 \iff \mathbf{{t}} = \left( \frac{1}{R}, \ldots , \frac{1}{R} \right) , \end{aligned}$$

we derive

$$\begin{aligned} \deg \big ( \widetilde{\gamma }, (1/R^2,1)^l, (0, \ldots , 0 ) \big ) \ne 0. \end{aligned}$$

This shows what was stated.$\square$

Proposition 6.3

If $c_{ \lambda ,\Upsilon } = \displaystyle \sum _{ j=1 }^l c_{ \lambda ,j } \, \text { and } \, c_\Upsilon = \sum _{ j=1 }^l c_j$, then

• (i) $c_{ \lambda ,\Upsilon } \le b_{ \lambda ,\Upsilon } \le c_\Upsilon , \, \forall \lambda \ge 1$;

• (ii) $b_{ \lambda ,\Upsilon } \rightarrow c_\Upsilon , \text { as } \lambda \rightarrow \infty$;

• (iii) $\phi _\lambda \big ( \gamma (\mathbf{{t}}) \big ) < c_\Upsilon , \, \forall \lambda \ge 1, \gamma \in \Gamma _*\text { and } \mathbf{{t}} = (t_1, \ldots , t_l ) \in \partial [1/R^2,1]^l$.

Proof

• (i) Once $\gamma _0 \in \Gamma _*$,

  $$\begin{aligned} b_{ \lambda ,\Upsilon } \le \max _{ ( t_1, \ldots , t_l ) \in [1/R^2,1]^l } \phi _\lambda \big ( \gamma _0 ( t_1, \ldots , t_l ) \big ) = \max _{ ( t_1, \ldots , t_l )\in [1/R^2,1]^l } \sum _{ j=1 }^l I_j ( t_j R w_j ) = c_\Upsilon . \end{aligned}$$

  Now, fixing $\mathbf{s} = (s_1, \ldots , s_l) \in [1/R^2,1]^l$ given in Lemma 6.2 and recalling that

  $$\begin{aligned} c_{ \lambda ,j } = \inf \left\{ \phi _{ \lambda ,j } (u) \, ; \, u \in W^{ 1,p(x) } \big ( \Omega '_j \big ) \setminus \{ 0 \} \text { and } \phi '_{ \lambda ,j }(u)u = 0 \right\} , \end{aligned}$$

  it follows that

  $$\begin{aligned} \phi _{ \lambda ,j } \big ( \gamma ( \mathbf{s } ) \big ) \ge c_{ \lambda ,j }, \, \forall j \in \Upsilon . \end{aligned}$$

  From (3.9),

  $$\begin{aligned} \phi _{ \lambda , \mathbb R^N \setminus \Omega '_\Upsilon } (u) \ge 0, \, \forall u \in W^{ 1,p(x) } \big ( \mathbb R^N \setminus \Omega '_\Upsilon \big ), \end{aligned}$$

  which leads to

  $$\begin{aligned} \phi _\lambda \big ( \gamma ( \mathbf{t} ) \big ) \ge \sum _{ j=1 }^l \phi _{ \lambda ,j } \big ( \gamma ( \mathbf{t} ) \big ), \, \forall \mathbf t = (t_1, \ldots , t_l) \in [1/R^2,1]^l. \end{aligned}$$

  Thus

  $$\begin{aligned} \max _{ ( t_1, \ldots , t_l )\in [1/R^2,1]^l } \phi _\lambda \big ( \gamma ( t_1, \ldots , t_l ) \big ) \ge \phi _\lambda \big ( \gamma ( \mathbf s ) \big ) \ge c_{ \lambda ,\Upsilon }, \end{aligned}$$

  showing that

  $$\begin{aligned} b_{ \lambda ,\Upsilon } \ge c_{ \lambda ,\Upsilon }; \end{aligned}$$

• (ii) This limit is clear by the previous item, since we already know $c_{ \lambda ,j } \rightarrow c_j$, as $\lambda \rightarrow \infty$;

• (iii) For $\mathbf t = ( t_1, \ldots , t_l ) \in \partial [1/R^2,1]^l$, it holds $\gamma ( \mathbf t ) = \gamma _0 ( \mathbf t )$. From this,

  $$\begin{aligned} \phi _\lambda \big ( \gamma ( \mathbf t ) \big ) = \sum _{j=1}^l I_j ( t_j R w_j ). \end{aligned}$$

  Writing

  $$\begin{aligned} \phi _\lambda \big ( \gamma ( \mathbf t ) \big ) = \mathop {\mathop {\sum }\limits _{j=1}}\limits _{j \ne j_0 }^l I_j ( t_j R w_j ) + I_{j_0} ( t_{j_0} R w_{j_0} ), \end{aligned}$$

  where $t_{j_0} \in \left\{ \frac{1}{R^2}, 1 \right\}$, from (6.1) we derive

  $$\begin{aligned} \phi _\lambda \big ( \gamma ( \mathbf t ) \big ) \le c_\Upsilon - \epsilon , \end{aligned}$$

  for some $\epsilon > 0$, so (iii).

$\square$

Corollary 6.4

$b_{ \lambda ,\Upsilon }$ is a critical value of $\phi _\lambda$, for $\lambda$ sufficiently large.

Proof

Assume $b_{ \widetilde{\lambda },\Upsilon }$ is not a critical value of $\phi _{\widetilde{\lambda }}$ for some $\widetilde{\lambda }$. We will prove that exists $\lambda _1$ such that $\widetilde{\lambda } < \lambda _1$. Indeed, by item (iii) of Proposition 6.3, we have seen that

$$\begin{aligned} \phi _\lambda \big ( \gamma _0 ( \mathbf t ) \big ) < c_\Upsilon , \, \forall \lambda \ge 1, \, \mathbf t \in \partial [1/R^2,1]^l. \end{aligned}$$

This way

$$\begin{aligned} \mathcal{M} = \max _{ \mathbf t \in \partial [1/R^2,1]^l } \phi _{ \widetilde{\lambda } } \big ( \gamma _0 ( \mathbf t ) \big ) < c_\Upsilon . \end{aligned}$$

Since $b_{ \lambda ,\Upsilon } \rightarrow c_\Upsilon$ (item (ii) of Proposition 6.3), there exists $\lambda _1 > 1$ such that if $\lambda \ge \lambda _1$, then

$$\begin{aligned} \mathcal{M} < b_{ \lambda ,\Upsilon }. \end{aligned}$$

So, if $\widetilde{\lambda } \ge \lambda _1$, we can find $\tau = \tau ( \widetilde{\lambda } ) > 0$ small enough, with the ensuing property

$$\begin{aligned} \mathcal{M} < b_{ \widetilde{\lambda },\Upsilon } - 2\tau . \end{aligned}$$

(6.2)

From the deformation’s lemma [31, Page 38], there is $\eta :E_\lambda \rightarrow E_\lambda$ such that

$$\begin{aligned} \eta \left( \phi _{\widetilde{\lambda }}^{ b_{ \widetilde{\lambda },\Upsilon } +\tau } \right) \subset \phi _{\widetilde{\lambda }}^{ b_{ \widetilde{\lambda },\Upsilon } -\tau } \, \text { and } \, \eta (u) = u, \text { for } u \notin \phi _{\widetilde{\lambda }}^{-1} \big ( [b_{ \widetilde{\lambda },\Upsilon }-2 \tau , b_{ \widetilde{\lambda },\Upsilon }+2 \tau ] \big ). \end{aligned}$$

Then, by (6.2),

$$\begin{aligned} \eta \big ( \gamma _0 ( \mathbf t )\big ) = \gamma _0 ( \mathbf t ), \, \forall \mathbf t \in \partial [1/R^2,1]^l. \end{aligned}$$

Now, using the definition of $b_{ \widetilde{\lambda },\Upsilon }$, there exists $\gamma _*\in \Gamma _*$ satisfying

$$\begin{aligned} \max _{ \mathbf t \in [1/R^2,1]^l } \phi _{ \widetilde{\lambda } } \big ( \gamma _*( \mathbf t ) \big ) < b_{ \widetilde{\lambda },\Upsilon }+\tau . \end{aligned}$$

(6.3)

Defining

$$\begin{aligned} \widetilde{\gamma } ( \mathbf t ) = \eta \big ( \gamma _*( \mathbf t ) \big ), \, \mathbf t \in [1/R^2,1]^l, \end{aligned}$$

due to (6.3), we obtain

$$\begin{aligned} \phi _{ \widetilde{\lambda } } \big ( \widetilde{\gamma }( \mathbf t ) \big ) \le b_{ \widetilde{\lambda },\Upsilon } - \tau , \, \forall \mathbf t \in [1/R^2,1]^l. \end{aligned}$$

But since $\widetilde{\gamma } \in \Gamma _*$, we deduce

$$\begin{aligned} b_{ \widetilde{\lambda },\Upsilon } \le \max _{ \mathbf t \in [1/R^2,1]^l } \phi _{\widetilde{\lambda }} \big ( \widetilde{\gamma } ( \mathbf t ) \big ) \le b_{ \widetilde{\lambda },\Upsilon }-\tau , \end{aligned}$$

a contradiction. So, $\widetilde{\lambda } < \lambda _1$.$\square$

7 The proof of the main theorem

To prove Theorem 1.1, we need to find nonnegative solutions $u_\lambda$ for large values of $\lambda$, which converges to a least energy solution in each $\Omega _j$ $(j \in \Upsilon )$ and to $0$ in $\Omega _\Upsilon ^{c}$ as $\lambda \rightarrow \infty$. To this end, we will show two propositions which together with the Propositions 4.1 and 5.1 will imply that Theorem 1.1 holds.

Henceforth, we denote by

$$\begin{aligned} r = R^{ p_+ } \sum _{ j=1 }^l \left( \frac{1}{p_+}-\frac{1}{\theta } \right) ^{-1} c_j, \quad \mathcal{B}_r^\lambda = \big \{ u \in E_\lambda \, ; \, \varrho _\lambda (u) \le r \big \} \end{aligned}$$

and

$$\begin{aligned} \phi _\lambda ^{ c_\Upsilon } = \big \{ u \in E_\lambda \, ; \, \phi _\lambda (u) \le c_{ \Upsilon } \big \}. \end{aligned}$$

Moreover, for small values of $\mu$,

$$\begin{aligned} \mathcal{A}_\mu ^\lambda = \left\{ u \in \mathcal{B}_r^{\lambda }\, ; \, \varrho _{ \lambda , \mathbb R^N \setminus \Omega _\Upsilon } (u) \le \mu , \, \left| \phi _{ \lambda ,j }(u)-c_j \right| \le \mu , \, \forall j \in \Upsilon \right\} . \end{aligned}$$

We observe that

$$\begin{aligned} w = \sum _{ j=1 }^l w_j \in \mathcal{A}_\mu ^\lambda \cap \phi _\lambda ^{ c_\Upsilon }, \end{aligned}$$

showing that $\mathcal{A}_\mu ^\lambda \cap \phi _\lambda ^{ c_\Upsilon } \ne \emptyset$. Fixing

$$\begin{aligned} 0 < \mu < \frac{1}{4} \min _{ j \in \Gamma } c_j, \end{aligned}$$

(7.1)

we have the following uniform estimate of $\big \Vert \phi '_{ \lambda }(u) \big \Vert$ on the region $\left( \mathcal{A}_{ 2 \mu }^\lambda \setminus \mathcal{A}_\mu ^\lambda \right) \cap \phi _\lambda ^{ c_\Upsilon }$.

Proposition 7.1

Let $\mu > 0$ satisfying (7.1). Then, there exist $\Lambda _*\ge 1$ and $\sigma _0 >0$ independent of $\lambda$ such that

$$\begin{aligned} \big \Vert \phi '_{ \lambda }(u) \big \Vert \ge \sigma _0, \text { for } \lambda \ge \Lambda _*\quad \text {and all}\quad u \in \left( \mathcal{A}_{ 2 \mu }^\lambda \setminus \mathcal{A}_\mu ^\lambda \right) \cap \phi _\lambda ^{ c_\Upsilon }. \end{aligned}$$

(7.2)

Proof

We assume that there exist $\lambda _n \rightarrow \infty$ and $u_n \in \left( \mathcal{A}_{ 2 \mu }^{\lambda _n} \setminus \mathcal{A}_\mu ^{\lambda _n} \right) \cap \phi _{\lambda _n}^{ c_\Upsilon }$ such that

$$\begin{aligned} \big \Vert \phi '_{ \lambda _n }(u_n) \big \Vert \rightarrow 0. \end{aligned}$$

Since $u_n \in \mathcal{A}_{ 2 \mu }^{ \lambda _n }$, this implies $\big ( \varrho _{ \lambda _n } (u_n) \big )$ is a bounded sequence and, consequently, it follows that $\big ( \phi _{ \lambda _n }(u_n) \big )$ is also bounded. Thus, passing a subsequence if necessary, we can assume $\phi _{ \lambda _n }(u_n)$ converges. Thus, from Proposition 4.1, there exists $0 \le u \in W^{ 1,p(x) }_0 \big ( \Omega _\Upsilon \big )$ such that $u_{ |_{ \Omega _j } }, \, j \in \Upsilon$, is a solution for $(P_j)$,

$$\begin{aligned} \varrho _{ \lambda _n, \mathbb R^N \setminus \Omega _\Upsilon } (u_n) \rightarrow 0 \quad \text {and} \quad \phi _{ \lambda _n,j } (u_n) \rightarrow I_j(u). \end{aligned}$$

We know that $c_j$ is the least energy level for $I_j$. So, if $u_{ |_{ \Omega _j } } \ne 0$, then $I_j(u) \ge c_j$. But since $\phi _{ \lambda _n } (u_n) \le c_\Upsilon$, we must analyze the following possibilities:

• (i) $I_j(u) = c_j, \, \forall j \in \Upsilon$;

• (ii) $I_{ j_0 }(u) = 0$, for some $j_o \in \Upsilon$.

If (i) occurs, then for $n$ large, it holds

$$\begin{aligned} \varrho _{ \lambda _n, \mathbb R^N \setminus \Omega _\Upsilon } (u_n) \le \mu \, \text { and } \, \left| \phi _{ \lambda _n,j }(u_n)-c_j \right| \le \mu , \, \forall j \in \Upsilon . \end{aligned}$$

So $u_n \in \mathcal{A}_\mu ^{\lambda _n}$, a contradiction.

If (ii) occurs, then

$$\begin{aligned} \left| \phi _{ \lambda _n,j_0 }(u_n)-c_{ j_0 } \right| \rightarrow c_{ j_0 } > 4 \mu , \end{aligned}$$

which is a contradiction with the fact that $u_n \in \mathcal{A}_{ 2 \mu }^{\lambda _n}$. Thus, we have completed the proof.$\square$

Proposition 7.2

Let $\mu > 0$ satisfying (7.1) and $\Lambda _*\ge 1$ given in the previous proposition. Then, for $\lambda \ge \Lambda _*$, there exists a solution $u_\lambda$ of $(A_\lambda )$ such that $u_\lambda \in \mathcal{A}_\mu ^\lambda \cap \phi _\lambda ^{ c_\Upsilon }$.

Proof

Let $\lambda \ge \Lambda _*$. Assume that there are no critical points of $\phi _\lambda$ in $\mathcal{A}_\mu ^\lambda \cap \phi _\lambda ^{ c_\Upsilon }$. Since $\phi _\lambda$ is a $(PS)$ functional, there exists a constant $d_\lambda > 0$ such that

$$\begin{aligned} \big \Vert \phi '_\lambda (u) \big \Vert \ge d_\lambda , \text { for all } u \in \mathcal{A}_\mu ^\lambda \cap \phi _\lambda ^{ c_\Upsilon }. \end{aligned}$$

From Proposition 7.1, we have

$$\begin{aligned} \big \Vert \phi '_\lambda (u) \big \Vert \ge \sigma _0, \text { for all } u \in \left( \mathcal{A}_{ 2 \mu }^{\lambda } \setminus \mathcal{A}_\mu ^{\lambda } \right) \cap \phi _{\lambda }^{ c_\Upsilon }, \end{aligned}$$

where $\sigma _0 > 0$ does not depend on $\lambda$. In what follows, $\Psi :E_\lambda \rightarrow \mathbb R$ is a continuous functional verifying

$$\begin{aligned} \Psi (u) = 1, \text { for } u \in \mathcal{A}_{\frac{3}{2} \mu }^\lambda , \ \Psi (u) = 0, \text { for } u \notin \mathcal{A}_{2 \mu }^\lambda \, \text { and } \, 0 \le \Psi (u) \le 1, \, \forall u \in E_\lambda . \end{aligned}$$

We also consider $H :\phi _\lambda ^{ c_\Upsilon } \rightarrow E_\lambda$ given by

$$\begin{aligned} H(u) = {\left\{ \begin{array}{ll} - \Psi (u) \big \Vert Y(u) \big \Vert ^{ -1 } Y(u),&{} \text { for } u \in \mathcal{A}_{2 \mu }^\lambda , \\ 0,&{} \text { for } u \notin \mathcal{A}_{2 \mu }^\lambda , \\ \end{array}\right. } \end{aligned}$$

where $Y$ is a pseudo-gradient vector field for $\Phi _\lambda$ on $\mathcal{K} = \left\{ u \in E_\lambda \, ; \, \phi '_\lambda (u) \ne 0 \right\}$. Observe that $H$ is well defined, once $\phi '_\lambda (u) \ne 0$, for $u \in \mathcal{A}_{2 \mu }^\lambda \cap \phi _\lambda ^{ c_\Upsilon }$. The inequality

$$\begin{aligned} \big \Vert H(u) \big \Vert \le 1, \, \forall \lambda \ge \Lambda _* \quad \text {and}\quad u \in \phi _\lambda ^{ c_\Upsilon }, \end{aligned}$$

guarantees that the deformation flow $\eta :[0, \infty ) \times \phi _\lambda ^{ c_\Upsilon } \rightarrow \phi _\lambda ^{ c_\Upsilon }$ defined by

$$\begin{aligned} \frac{d \eta }{dt} = H(\eta ), \ \eta (0,u) = u \in \phi _\lambda ^{ c_\Upsilon } \end{aligned}$$

verifies

$$\begin{aligned}&\frac{d}{dt} \phi _\lambda \big ( \eta (t,u) \big ) \le - \frac{1}{2} \Psi \big ( \eta (t,u) \big ) \big \Vert \phi '_\lambda \big ( \eta (t,u) \big ) \big \Vert \le 0, \end{aligned}$$

(7.3)

$$\begin{aligned}&\left\| \frac{d \eta }{dt} \right\| _\lambda = \big \Vert H(\eta ) \big \Vert _\lambda \le 1 \end{aligned}$$

(7.4)

and

$$\begin{aligned} \eta (t,u) = u \text { for all } t \ge 0 \text { and } u \in \phi _\lambda ^{ c_\Upsilon } \setminus \mathcal{A}_{2 \mu }^\lambda . \end{aligned}$$

(7.5)

We study now two paths, which are relevant for what follows:

$\bullet$ The path $\mathbf{t} \mapsto \eta \big ( t, \gamma _0( \mathbf{t} ) \big ), \text { where } \mathbf t = (t_1,\ldots ,t_l) \in [1/R^2, 1]^l$.

The definition of $\gamma _0$ combined with the condition on $\mu$ gives

$$\begin{aligned} \gamma _0( \mathbf{t } ) \notin \mathcal{A}_{2 \mu }^\lambda , \, \forall \mathbf{t } \in \partial [1/R^2, 1]^l. \end{aligned}$$

Since

$$\begin{aligned} \phi _\lambda \big ( \gamma _0( \mathbf{t } ) \big ) < c_\Upsilon , \, \forall \mathbf{t } \in \partial [1/R^2, 1]^l, \end{aligned}$$

from (7.5), it follows that

$$\begin{aligned} \eta \big ( t, \gamma _0( \mathbf{t} ) \big ) = \gamma _0( \mathbf{t} ), \, \forall \mathbf{t} \in \partial [1/R^2, 1]^l. \end{aligned}$$

So, $\eta \big ( t, \gamma _0( \mathbf{t} ) \big ) \in \Gamma _*$, for each $t \ge 0$.

$\bullet$ The path $\mathbf{t} \mapsto \gamma _0( \mathbf{t} ), \text { where } \mathbf t = (t_1,\ldots ,t_l) \in [1/R^2, 1]^l$.

We observe that

$$\begin{aligned} \text {supp} \big ( \gamma _0 ( \mathbf{t} ) \big )\subset \overline{\Omega _\Upsilon } \end{aligned}$$

and

$$\begin{aligned} \phi _\lambda \big ( \gamma _0 ( \mathbf{t} ) \big ) \text { does not depend on } \lambda \ge 1, \end{aligned}$$

forall $\mathbf{t} \in [1/R^2, 1]^l$. Moreover,

$$\begin{aligned} \phi _\lambda \big ( \gamma _0 ( \mathbf{t} ) \big ) \le c_\Upsilon , \, \forall \mathbf{t} \in [1/R^2, 1]^l \end{aligned}$$

and

$$\begin{aligned} \phi _\lambda \big ( \gamma _0 ( \mathbf{t} ) \big ) = c_\Upsilon \text {if}, \quad \text { and }\quad \text { only if,}\quad t_j = \frac{1}{R}, \, \forall j \in \Upsilon . \end{aligned}$$

Therefore

$$\begin{aligned} m_0 = \sup \left\{ \phi _\lambda (u) \, ; \, u \in \gamma _0 \big ( [1/R^2,1]^l \big ) \setminus A_\mu ^\lambda \right\} \end{aligned}$$

is independent of $\lambda$ and $m_0 < c_\Upsilon$. Now, observing that there exists $K_*> 0$ such that

$$\begin{aligned} \big | \phi _{ \lambda ,j }(u) - \phi _{ \lambda ,j }(v) \big | \le K_* \Vert u-v \Vert _{ \lambda , \Omega '_j }, \, \forall u,v \in \mathcal{B}_r^\lambda \text { and } \forall j \in \Upsilon , \end{aligned}$$

we derive

$$\begin{aligned} \max _{ \mathbf{t } \in [1/R^2,1]^l } \phi _\lambda \Big ( \eta \big ( T, \gamma _0 ( \mathbf{t} ) \big ) \Big ) \le \max \left\{ m_0, c_\Upsilon -\frac{1}{2 K_*} \sigma _0 \mu \right\} , \end{aligned}$$

(7.6)

for $T > 0$ large.

In fact, writing $u = \gamma _0( \mathbf{t} ) , \mathbf{t } \in [1/R^2,1]^l$, if $u \notin A_\mu ^\lambda$, from (7.3),

$$\begin{aligned} \phi _\lambda \big ( \eta ( t, u ) \big ) \le \phi _\lambda (u) \le m_0, \, \forall t \ge 0, \end{aligned}$$

and we have nothing more to do. We assume then $u \in A_\mu ^\lambda$ and set

$$\begin{aligned} \widetilde{\eta }(t) = \eta (t,u), \ \widetilde{d_\lambda } = \min \left\{ d_\lambda , \sigma _0 \right\} \text { and } T = \frac{\sigma _0 \mu }{K_*\widetilde{d_\lambda }}. \end{aligned}$$

Now, we will analyze the ensuing cases:

Case 1: $\widetilde{\eta }(t) \in \mathcal{A}_{\frac{3}{2} \mu }^\lambda , \, \forall t \in [0,T]$.

Case 2: $\widetilde{\eta }(t_0) \in \partial \mathcal{A}_{\frac{3}{2} \mu }^\lambda , \text { for some } t_0 \in [0,T]$.

Analysis of Case 1

In this case, we have $\Psi \big ( \widetilde{\eta }(t) \big ) = 1$ and $\big \Vert \phi '_\lambda \big ( \widetilde{\eta }(t) \big ) \big \Vert \ge \widetilde{d_\lambda }$ for all $t \in [0,T]$. Hence, from (7.3),

$$\begin{aligned} \phi _\lambda \big ( \widetilde{\eta }(T) \big ) = \phi _\lambda (u) + \int \limits _0^T \frac{\mathrm{{d}}}{\mathrm{{d}}s} \phi _\lambda \big ( \widetilde{\eta }(s) \big ) \, \mathrm{{d}}s \le c_\Upsilon - \frac{1}{2} \int \limits _0^T \widetilde{d_\lambda } \, \mathrm{{d}}s, \end{aligned}$$

that is,

$$\begin{aligned} \phi _\lambda \big ( \widetilde{\eta }(T) \big ) \le c_\Upsilon - \frac{1}{2} \widetilde{d_\lambda } T = c_\Upsilon - \frac{1}{2 K_*} \sigma _0 \mu , \end{aligned}$$

showing (7.6).

Analysis of Case 2

In this case, there exist $0 \le t_1 \le t_2 \le T$ satisfying

$$\begin{aligned}&\widetilde{\eta }(t_1) \in \partial \mathcal{A}_\mu ^\lambda , \\&\widetilde{\eta }(t_2) \in \partial \mathcal{A}_{\frac{3}{2} \mu }^\lambda , \end{aligned}$$

and

$$\begin{aligned} \widetilde{\eta }(t) \in \mathcal{A}_{\frac{3}{2} \mu }^\lambda \setminus \mathcal{A}_\mu ^\lambda , \, \forall t \in (t_1,t_2]. \end{aligned}$$

We claim that

$$\begin{aligned} \big \Vert \widetilde{\eta }(t_2)-\widetilde{\eta }(t_1) \big \Vert \ge \frac{1}{2 K_*} \mu . \end{aligned}$$

Setting $w_1 = \widetilde{\eta }(t_1)$ and $w_2 = \widetilde{\eta }(t_2)$, we get

$$\begin{aligned} \varrho _{ \lambda , \mathbb R^N \setminus \Omega _\Upsilon } (w_2) = \frac{3}{2} \mu \ \text { or } \, \big | \phi _{ \lambda , j_0 } (w_2) - c_{j_0} \big | = \frac{3}{2} \mu , \end{aligned}$$

for some $j_0 \in \Upsilon$. We analyze the latter situation, once that the other one follows the same reasoning. From the definition of $\mathcal{A}_\mu ^\lambda$,

$$\begin{aligned} \big | \phi _{ \lambda , j_0 } (w_1) - c_{j_0} \big | \le \mu , \end{aligned}$$

consequently,

$$\begin{aligned} \Vert w_2-w_1 \Vert \ge \frac{1}{K_*} \big | \phi _{ \lambda , j_0 } (w_2) - \phi _{ \lambda , j_0 } (w_1) \big | \ge \frac{1}{2 K_*} \mu . \end{aligned}$$

Then, by mean value theorem, $t_2-t_1 \ge \frac{1}{2 K_*} \mu$ and, this form,

$$\begin{aligned} \phi _\lambda \big ( \widetilde{\eta }(T) \big ) \le \phi _\lambda (u) - \int \limits _0^T \Psi \big ( \widetilde{\eta }(s) \big ) \big \Vert \phi '_\lambda \big ( \widetilde{\eta }(s) \big ) \big \Vert \, \mathrm{{d}}s \end{aligned}$$

implying

$$\begin{aligned} \phi _\lambda \big ( \widetilde{\eta }(T) \big ) \le c_\Upsilon - \int \limits _{t_1}^{t_2} \sigma _0 \, \mathrm{{d}}s = c_\Upsilon - \sigma _0 (t_2-t_1) \le c_\Upsilon - \frac{1}{2 K_*} \sigma _0 \mu , \end{aligned}$$

which proves 7.6. Fixing $\widehat{\eta } (t_1, \ldots , t_l) = \eta \big ( T, \gamma _0 (t_1,\ldots ,t_l) \big )$, we have that $\widehat{\eta } \in \Gamma _*$ and, hence,

$$\begin{aligned} b_{ \lambda , \Gamma } \le \max _{ (t_1,\ldots ,t_l) \in [1/R^2, 1] } \phi _\lambda \big ( \widehat{\eta } (t_1,\ldots ,t_l) \big ) \le \max \left\{ m_0, c_\Upsilon - \frac{1}{2 K_*} \sigma _0 \mu \right\} < c_\Upsilon , \end{aligned}$$

which contradicts the fact that $b_{ \lambda , \Upsilon } \rightarrow c_\Upsilon$.$\square$

Proof of Theorem 1.1

According Proposition 7.2, for $\mu$ satisfying (7.1) and $\Lambda _*\ge 1$, there exists a solution $u_\lambda$ for $(A_\lambda )$ such that $u_\lambda \in \mathcal{A}_\mu ^\lambda \cap \phi _\lambda ^{ c_\Upsilon }$, for all $\lambda \ge \Lambda _*$.

Claim: There are $\lambda _0 \ge \Lambda _*$ and $\mu _0>0$ small enough, such that $u_\lambda$ is a solution for $\big ( P_\lambda \big )$ for $\lambda \ge \Lambda _0$ and $\mu \in (0, \mu _0)$.

Indeed, assume by contradiction that there are $\lambda _n \rightarrow \infty$ and $\mu _n \rightarrow 0$, such that $(u_{\lambda _n})$ is not a solution for $(P_{\lambda _n})$. From Proposition 7.2, the sequence $(u_{\lambda _n})$ verifies:

• (a) $\phi '_{ \lambda _n }(u_{\lambda _n}) = 0, \, \forall n \in \mathbb {N}$;

• (b) $\varrho _{ \lambda _n, \mathbb {R}^N \setminus \Omega _\Upsilon }(u_{\lambda _n}) \rightarrow 0$;

• (c) $\phi _{ \lambda _n,j } (u_{\lambda _n}) \rightarrow c_j, \, \forall j \in \Upsilon .$

The item (b) ensures we can use Proposition 5.1 to deduce $u_{\lambda _n}$ is a solution for $\big ( P_{\lambda _n} \big )$, for large values of $n$, which is a contradiction, showing this way the claim.

Now, our goal is to prove the second part of the theorem. To this end, let $(u_{\lambda _n})$ be a sequence verifying the above limits. Since $\phi _{ \lambda _n }(u_{ \lambda _n } )$ is bounded, passing a subsequence, we obtain that $\phi _{ \lambda _n }(u_{ \lambda _n } ) \rightarrow c$. This way, using Proposition 4.1 combined with item (c), we derive $u_{ \lambda _n }$ converges in $W^{ 1,p(x) } \big ( \mathbb {R}^N \big )$ to a function $u \in W^{ 1,p(x) } \big ( \mathbb {R}^N \big )$, which satisfies $u = 0$ outside $\Omega _\Upsilon$ and $u_{|_{\Omega _j}}, \, j \in \Upsilon$, is a least energy solution for

$$\begin{aligned} {\left\{ \begin{array}{ll} - \Delta _{ p(x) } u + Z(x) u = f(u), &{}\text { in } \Omega _j, \\ u \in W^{ 1,p(x) }_0 \big ( \Omega _j \big ), \, u \ge 0,&{} \text { in } \Omega _j. \end{array}\right. } \end{aligned}$$