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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco

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Volume 8, Issue 1

Quasilinear equations with indefinite nonlinearity

Junfang Zhao
• School of Science, China University of Geosciences, Beijing 100083, P. R. China; and Department of Mathematics, University of Texas-Rio Grande Valley, Edinburg, TX 78539, USA
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• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
/ Xiangqing Liu
• Department of Mathematics, Yunnan Normal University, Kunming, Yunnan 650500, P. R. China; and Department of Mathematics, University of Texas-Rio Grande Valley, Edinburg, TX 78539, USA
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• Other articles by this author:
• De Gruyter OnlineGoogle Scholar
/ Zhaosheng Feng
Published Online: 2018-06-20 | DOI: https://doi.org/10.1515/anona-2018-0010

Abstract

In this paper, we are concerned with quasilinear equations with indefinite nonlinearity and explore the existence of infinitely many solutions.

MSC 2010: 35J62; 35A15; 35B20; 35B38

1 Introduction

Consider the quasilinear equation with the indefinite nonlinearity

(1.1)

where $\mathrm{\Omega }\subset {ℝ}^{N}$, $N\ge 3$, is a bounded smooth domain, $r>4$, $4, and ${a}_{±}$ are nonnegative continuous functions in $\overline{\mathrm{\Omega }}$. A great number of theoretical issues concerning nonlinear elliptic equations with indefinite nonlinearity have received considerable attention in the past few decades. In particular, the existence of solutions has been studied extensively. For example, the existence of positive solutions and their multiplicity was studied by variational techniques [2], and the existence of nontrivial solutions was investigated by two different approaches (one involving the Morse theory and the other using the min-max method) [1]. It was shown that the existence of positive solutions, negative solutions and sign-changing solutions could be established by means of the Morse theory [7, 8]. For the results on a priori estimates and more comparable relations among various solutions etc., we refer the reader to [3, 5, 6, 9, 10, 12] and the references therein. However, as far as one can see from the literature, not much has been known about the existence of solutions to quasilinear equations with indefinite nonlinearity. From the variational point of view, there are two main difficulties that arise in the study. One lies in the fact that there is no suitable space in which the corresponding functional enjoys both smoothness and compactness. Compared with quasilinear equations with the definite nonlinearity, the other one is to prove the boundedness of the associated Palais–Smale sequences. In this work, we will get over these obstacles by means of the variational techniques and the perturbation method to study the existence of infinitely many solutions of system (1.1).

Set

${\mathrm{\Omega }}_{±}=\left\{x\in \overline{\mathrm{\Omega }}\mid {a}_{±}\left(x\right)>0\right\}\mathit{ }\text{and}\mathit{ }{\mathrm{\Omega }}_{0}=\mathrm{\Omega }\setminus \left({\overline{\mathrm{\Omega }}}_{+}\cup {\overline{\mathrm{\Omega }}}_{-}\right).$

Assume that

• (a)

${\mathrm{\Omega }}_{+}\ne \mathrm{\varnothing }$ and ${\overline{\mathrm{\Omega }}}_{+}\cap {\overline{\mathrm{\Omega }}}_{-}=\mathrm{\varnothing }$.

We are looking for $u\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ satisfying equation (1.1) in the weak form

${\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right)\nabla u\nabla \phi \mathrm{d}x+{\int }_{\mathrm{\Omega }}u{|\nabla u|}^{2}\phi dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\phi dx={\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\phi dx$(1.2)

for $\phi \in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$, which is formally the variational formulation of the following functional:

$I\left(u\right)=\frac{1}{2}{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx+\frac{1}{r}{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx-\frac{1}{s}{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx.$

In view of the perturbation (regularization) approach [11], due to the lack of a suitable working space, we introduce the corresponding perturbed functionals, which are smooth functionals in the given space and satisfy the necessary compactness property. For $\mu \in \left(0,1\right]$, we define the perturbed functional ${I}_{\mu }$ on the Sobolev space ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$ with $p>N$ by

${I}_{\mu }\left(u\right)=\frac{\mu }{2}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+I\left(u\right)$$=\frac{\mu }{2}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+\frac{1}{2}{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx+\frac{1}{r}{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx-\frac{1}{s}{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx.$

Note that ${I}_{\mu }$ is a ${C}^{1}$- functional. We shall show that ${I}_{\mu }$ satisfies the Palais–Smale condition. The critical points of ${I}_{\mu }$ will be used as the approximate solution of problem (1.1).

Now let us briefly summarize our main results of this paper.

Theorem 1.1.

Assume that condition (a) holds, $r\mathrm{>}\mathrm{4}$, $\mathrm{4}\mathrm{<}s\mathrm{<}\frac{\mathrm{4}\mathit{}N}{N\mathrm{-}\mathrm{2}}$, ${\mu }_{n}\mathrm{>}\mathrm{0}$, ${\mu }_{n}\mathrm{\to }\mathrm{0}$, ${u}_{n}\mathrm{\in }{W}_{\mathrm{0}}^{\mathrm{1}\mathrm{,}p}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$, $D\mathit{}{I}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{=}\mathrm{0}$ and ${I}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{\le }C$. Then the following assertions hold:

• (i)

There exists a function $u\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ satisfying equation ( 1.1 ).

• (ii)

Up to a subsequence, there holds ${\parallel {u}_{n}\parallel }_{{L}^{\mathrm{\infty }}}\le C$, ${u}_{n}\to u$ for a.e. $x\in \mathrm{\Omega }$, ${u}_{n}\to u$ in ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$ , and

Theorem 1.2.

Assume that condition (a) holds, $r\mathrm{>}\mathrm{4}$ and $\mathrm{4}\mathrm{<}s\mathrm{<}\frac{\mathrm{4}\mathit{}N}{N\mathrm{-}\mathrm{2}}$. Then problem (1.1) has infinitely many solutions.

Consider the more general quasilinear equation

(1.3)

In the weak form, we look for $u\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ such that

${\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}\phi \mathrm{d}x+\frac{1}{2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{D}_{s}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}u\phi \mathrm{d}x+{\int }_{\mathrm{\Omega }}{a}_{-}\left(x\right){|u|}^{r-2}u\phi dx$$={\int }_{\mathrm{\Omega }}{a}_{+}\left(x\right){|u|}^{s-2}u\phi dx$(1.4)

for $\phi \in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$, where ${D}_{i}=\frac{\partial }{\partial {x}_{i}}$ and ${D}_{s}{a}_{ij}\left(x,s\right)=\frac{\partial }{\partial s}{a}_{ij}\left(x,s\right)$. For the coefficients ${a}_{ij}$, $i,j=1,\mathrm{\dots },N$, we make the following assumptions:

• (a0)

${a}_{ij}$, ${D}_{s}{a}_{ij}\in {C}^{1}\left(\overline{\mathrm{\Omega }}×ℝ\right)$, ${a}_{ij}={a}_{ji}$, $i,j=1,\mathrm{\dots },N$.

• (a1)

There exist constants ${c}_{1},{c}_{2}>0$ such that

${c}_{1}\left(1+{s}^{2}\right){|\xi |}^{2}\le \sum _{i,j=1}^{N}{a}_{ij}\left(x,s\right){\xi }_{i}{\xi }_{j}\le {c}_{2}\left(1+{s}^{2}\right){|\xi |}^{2}$

for $x\in \overline{\mathrm{\Omega }}$, $s\in ℝ$ and $\xi =\left({\xi }_{i}\right)\in {ℝ}^{N}$.

• (a2)

There exist $\delta >0$ and $0 such that

$\delta \sum _{i,j=1}^{N}{a}_{ij}\left(x,s\right){\xi }_{i}{\xi }_{j}\le \sum _{i,j=1}^{N}\left[{a}_{ij}\left(x,s\right)+\frac{1}{2}s{D}_{s}{a}_{ij}\left(x,s\right)\right]{\xi }_{i}{\xi }_{j}\le q\left(\frac{1}{2}-\delta \right)\sum _{i,j=1}^{N}{a}_{ij}\left(x,s\right){\xi }_{i}{\xi }_{j}$

for $x\in \overline{\mathrm{\Omega }}$, $s\in ℝ$ and $\xi \in {ℝ}^{N}$.

• (a3)

There exists an $M>0$ such that

$\sum _{i,j=1}^{N}\left[{a}_{ij}\left(x,s\right)+\frac{1}{2}{D}_{s}{a}_{ij}\left(x,s\right)\right]{\xi }_{i}{\xi }_{j}\ge \frac{2{s}^{2}}{M+{s}^{2}}\sum _{i,j=1}^{N}{a}_{ij}\left(x,s\right){\xi }_{i}{\xi }_{j}$

for $x\in \overline{\mathrm{\Omega }}$, $s\in ℝ$ and $\xi \in {ℝ}^{N}$.

• (a4)

There holds

$\underset{|s|\to \mathrm{\infty }}{lim}\frac{{a}_{ij}\left(x,s\right)}{{s}^{2}}={A}_{ij}\left(x\right)$

uniformly in $x\in \overline{\mathrm{\Omega }}$.

Theorem 1.3.

Assume that $r\mathrm{>}\mathrm{4}$, $\mathrm{4}\mathrm{<}s\mathrm{<}\frac{\mathrm{4}\mathit{}N}{N\mathrm{-}\mathrm{2}}$ and conditions (a0)(a4) hold. Then equation (1.3) has infinitely many solutions.

The rest of the paper is organized as follows: The proof of Theorem 1.1 is presented in Section 2, and the proof of Theorem 1.2 is shown in Section 3. Section 4 is dedicated to the existence of infinitely many solutions to the more general quasilinear equation (1.3).

2 Convergence theorem

To prove Theorem 1.1, we need the following two technical lemmas.

Lemma 2.1.

There holds that

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\le C\left\{{I}_{\mu }\left(u\right)+\parallel D{I}_{\mu }\left(u\right)\parallel \cdot \parallel u\parallel +\mu {\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){u}^{2}dx\right\}.$

Proof.

For $\phi \in {W}_{0}^{1.p}\left(\mathrm{\Omega }\right)$, we know that

$〈D{I}_{\mu }\left(u\right),\phi 〉=\mu {\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right]}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}\left[{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p-2}\nabla u\nabla \phi +{\left(1+{u}^{2}\right)}^{\frac{p}{2}-1}u{|\nabla u|}^{p}\phi \right]dx$$+{\int }_{\mathrm{\Omega }}\left[\left(1+{u}^{2}\right)\nabla u\nabla \phi +u{|\nabla u|}^{2}\phi \right]dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\phi dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\phi dx.$(2.1)

Since ${\overline{\mathrm{\Omega }}}_{-}\cap {\overline{\mathrm{\Omega }}}_{+}=\mathrm{\varnothing }$, we can choose $\psi \in {C}_{0}^{\mathrm{\infty }}\left({ℝ}^{N}\right)$ such that $\psi \ge 0$, $\psi \left(x\right)=1$ for $x\in {\overline{\mathrm{\Omega }}}_{-}$, and $\psi \left(x\right)=0$ for $x\in {\overline{\mathrm{\Omega }}}_{+}$. Then

${\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}{\psi }^{p}dx={\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx\mathit{ }\text{and}\mathit{ }{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}{\psi }^{p}dx=0.$

Taking $\psi =u{\psi }^{p}$ as the test function in (2.1), we get

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}-1}\left(1+2{u}^{2}\right){|\nabla u|}^{p}{\psi }^{p}dx+{\int }_{\mathrm{\Omega }}\left(1+2{u}^{2}\right){|\nabla u|}^{2}{\psi }^{p}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx$$=〈D{I}_{\mu }\left(u\right),u{\psi }^{p}〉-p{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right)\nabla uu{\psi }^{p-1}\nabla \psi \mathrm{d}x$$-p\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p-2}\nabla uu{\psi }^{p-1}\nabla \psi \mathrm{d}x$$\le C\parallel D{I}_{\lambda }\left(u\right)\parallel \cdot \parallel u\parallel +\epsilon {\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}{\psi }^{p}dx+C{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){u}^{2}{|\nabla \psi |}^{2}dx$$+\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}\left\{\epsilon {\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}{\psi }^{p}dx+C{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}{|\nabla \psi |}^{p}dx\right\}.$(2.2)

In view of the Sobolev inequality

${\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}dx\le C{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx,$(2.3)

it follows from (2.2) and (2.3) that

${\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx\le C\left\{\parallel D{I}_{\mu }\left(u\right)\parallel \parallel u\parallel +\mu {\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){u}^{2}dx\right\}.$

By choosing $q\in \left(4,s\right)$, we deduce that

${I}_{\mu }\left(u\right)-\frac{1}{q}〈D{I}_{\mu }\left(u\right),u〉=\frac{1}{2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+\frac{1}{2}{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx$$-\frac{1}{q}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}-1}\left(1+2{u}^{2}\right){|\nabla u|}^{p}dx$$-\frac{1}{q}{\int }_{\mathrm{\Omega }}\left(1+2{u}^{2}\right){|\nabla u|}^{2}dx-\left(\frac{1}{q}-\frac{1}{r}\right){\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx+\left(\frac{1}{q}-\frac{1}{s}\right){\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\ge \left(\frac{1}{2}-\frac{2}{q}\right)\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+\left(\frac{1}{2}-\frac{2}{q}\right){\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{2}dx\right)}^{\frac{2}{p}}$$-\left(\frac{1}{q}-\frac{1}{r}\right){\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx+\left(\frac{1}{q}-\frac{1}{s}\right){\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx,$

and thus

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\le C\left\{{I}_{\mu }\left(u\right)+\parallel D{I}_{\mu }\left(u\right)\parallel \cdot \parallel u\parallel +{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx\right\}$$\le C\left\{{I}_{\mu }\left(u\right)+\parallel D{I}_{\mu }\left(u\right)\parallel \cdot \parallel u\parallel +\mu {\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|u|}^{2}dx\right\}.\mathit{∎}$

Lemma 2.2.

Assume that ${\mu }_{n}\mathrm{>}\mathrm{0}$, ${\mu }_{n}\mathrm{\to }\mathrm{0}$, ${u}_{n}\mathrm{\in }{W}_{\mathrm{0}}^{\mathrm{1}\mathrm{,}p}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$, $D\mathit{}{I}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{=}\mathrm{0}$ and ${I}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{\le }{C}_{\mathrm{0}}$. Then there exists a constant $C\mathrm{>}\mathrm{0}$ independent of n such that

${\mu }_{n}{\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|{u}_{n}|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){u}_{n}^{2}dx\le C.$

Moreover, there holds

${\mu }_{n}{\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\le C.$

Proof.

Assume that

(2.4)

By Lemma 2.1, we have

${\mu }_{n}{\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\le C{\rho }_{n}^{4}.$

Let ${v}_{n}={\rho }_{n}^{-1}{u}_{n}$. Then

${\int }_{\mathrm{\Omega }}{v}_{n}^{2}{|\nabla {v}_{n}|}^{2}dx={\rho }_{n}^{-4}{\int }_{\mathrm{\Omega }}{u}_{n}^{2}{|\nabla {u}_{n}|}^{2}dx\le C,$

We have ${v}_{n}\to v$ in ${L}^{q}\left(\mathrm{\Omega }\right)$, $1\le q<\frac{4N}{N-2}$, ${v}_{n}^{2}⇀{v}^{2}$ in ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$, ${\int }_{\mathrm{\Omega }}{a}_{-}{|v|}^{r}dx=0$ and ${\int }_{\mathrm{\Omega }}{a}_{+}{|v|}^{s}dx=0$. Thus, $v\left(x\right)=0$ for $x\in {\overline{\mathrm{\Omega }}}_{+}\cap {\overline{\mathrm{\Omega }}}_{-}$ and ${v}^{2}\in {H}_{0}^{1}\left({\mathrm{\Omega }}_{0}\right)\subset {H}_{0}^{1}\left(\mathrm{\Omega }\right)$.

If $p>N$, then ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)↪{C}^{\alpha }\left(\mathrm{\Omega }\right)$ for some $\alpha \in \left(0,1\right)$. Given $\psi \ge 0$ and $\psi \in {C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{0}\right)$, we take

${\phi }_{n}=\frac{\psi {u}_{n}}{1+{u}_{n}^{2}}\in {W}_{0}^{1,p}\left({\mathrm{\Omega }}_{0}\right)\subset {W}_{0}^{1,p}\left(\mathrm{\Omega }\right),$

and have

$0=〈D{I}_{{\mu }_{n}}\left({u}_{n}\right),{\phi }_{n}〉$$={\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{1+{u}_{n}^{2}}\mathrm{d}x$$+{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}-2}{|\nabla {u}_{n}|}^{p}\psi dx$$+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right)\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{1+{u}_{n}^{2}}\mathrm{d}x+{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{-1}{|\nabla {u}_{n}|}^{2}\psi dx$$\ge {\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p-1}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{{\left(1+{u}_{n}^{2}\right)}^{\frac{1}{2}}}\mathrm{d}x+{\int }_{\mathrm{\Omega }}{u}_{n}\nabla {u}_{n}\nabla \psi \mathrm{d}x.$

We further obtain the estimates as

${\rho }_{n}^{-2}{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}|{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p-1}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{{\left(1+{u}_{n}^{2}\right)}^{\frac{1}{2}}}\mathrm{d}x|$$\le {\rho }_{n}^{-2}{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{1}{p}}{\left({\int }_{\mathrm{\Omega }}{|\nabla \psi |}^{p}dx\right)}^{\frac{1}{p}}$

and

Hence, it gives

Since ${v}^{2}\in {H}_{0}^{1}\left({\mathrm{\Omega }}_{0}\right)$, we have ${v}^{2}\equiv 0$ in Ω. It follows from (2.4) and Lemma 2.1 that

${\rho }_{n}^{-4}\left[{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx\right]\le C,$

and

${\rho }_{n}^{-4}\left[{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|{u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){u}_{n}^{2}dx\right]=1.$(2.5)

In view of $1\le q<\frac{4N}{N-2}$, by taking ${v}_{n}\to v=0$ in ${L}^{q}\left(\mathrm{\Omega }\right)$, we get

${\rho }_{n}^{-4}{\int }_{\mathrm{\Omega }}{u}_{n}^{4}dx={\int }_{\mathrm{\Omega }}{v}_{n}^{4}dx\to {\int }_{\mathrm{\Omega }}{v}^{4}dx=0,$

and

${\rho }_{n}^{-4}{\int }_{\mathrm{\Omega }}{u}_{n}^{2}dx\le \frac{1}{2}{\rho }_{n}^{-4}{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{4}\right)dx\to 0.$

Let ${z}_{n}\in {W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$ and

$D{z}_{n}={\rho }_{n}^{-2}{\mu }_{n}^{\frac{1}{2}}{\left(1+{u}_{n}^{2}\right)}^{\frac{1}{2}}D{u}_{n}.$

It is easy to see that

${\left({\int }_{\mathrm{\Omega }}{|D{z}_{n}|}^{p}dx\right)}^{\frac{2}{p}}\le {\rho }_{n}^{-4}{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}\le C.$(2.6)

So we further deduce that

${C}_{1}{\rho }_{n}^{-2}{\mu }_{n}^{\frac{1}{2}}{\left(1+{u}_{n}^{2}\right)}^{\frac{1}{2}}|{u}_{n}|\le |{z}_{n}|\le {C}_{2}{\rho }_{n}^{-2}{\mu }_{n}^{\frac{1}{2}}{\left(1+{u}_{n}^{2}\right)}^{\frac{1}{2}}|{u}_{n}|,$

Consequently, the left-hand side of (2.5) converges to zero, which yields a contradiction. ∎

Proof of Theorem 1.1.

By Lemmas 2.1 and 2.2, we have

${\mu }_{n}{\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right]}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx\le C.$

Assume that ${u}_{n}\to u$ in ${L}^{q}\left(\mathrm{\Omega }\right)$, $1\le q<\frac{4N}{N-2}$ and ${u}_{n}^{2}\to {u}^{2}$ in ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$. We separate the proof into three steps.

Step 1. Moser’s iteration shows that the sequence $\left\{{u}_{n}\right\}$ is uniformly bounded.

Note that for $p>N$, the convergence ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)↪{C}^{\alpha }\left(\overline{\mathrm{\Omega }}\right)$ holds for some $\alpha \in \left(0,1\right)$. The function ${u}_{n}$ belongs to ${L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$. Here we prove that ${\parallel {u}_{n}\parallel }_{{L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)}$ is uniformly bounded. The term ${a}_{+}{|u|}^{s-2}u$ in equation (1.2) is subcritical ($s<\frac{4N}{N-2}$), while the term of ${a}_{-}{|u|}^{r-2}u$ in the equation does not cause any trouble to us at this step. The sequence $\left\{{u}_{n}\right\}$ is bounded in ${L}^{4N/\left(N-2\right)}\left(\mathrm{\Omega }\right)$. Starting from this ${L}^{4N/\left(N-2\right)}\left(\mathrm{\Omega }\right)$-bound, by Moser’s iteration we see the ${L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$-bound.

Step 2. Choose a suitable test function and show that the limit function u satisfies equation (1.2).

Let $\psi \ge 0$, $\psi \in {C}_{0}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ and $\phi =\psi {e}^{-{u}_{n}}\in {W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$. Take φ as the test function. Then we have

$0=〈D{I}_{{\mu }_{n}}\left({u}_{n}\right),\phi 〉$$={\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi {e}^{-{u}_{n}}\mathrm{d}x$$-{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}-1}\left(1+{u}_{n}^{2}-{u}_{n}\right){|\nabla {u}_{n}|}^{p}\psi {e}^{-{u}_{n}}dx$$+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right)\nabla {u}_{n}\nabla \psi {e}^{-{u}_{n}}\mathrm{d}x-{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}-{u}_{n}\right){|\nabla {u}_{n}|}^{2}\psi dx$$+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r-2}{u}_{n}\psi {e}^{-{u}_{n}}dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s-2}{u}_{n}\psi {e}^{-{u}_{n}}dx.$(2.7)

We estimate each term in (2.7). From (2.6) we get

$|{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi {e}^{-{u}_{n}}\mathrm{d}x|$$\le {\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{1}{p}}\left({\int }_{\mathrm{\Omega }}{|\nabla \psi |}^{\frac{1}{p}}dx\right)$

By the weak convergence, it gives

${\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right)\nabla {u}_{n}\nabla \psi {e}^{-{u}_{n}}\mathrm{d}x\to {\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right)\nabla u\nabla \psi {e}^{-u}\mathrm{d}x.$

By the lower semi-continuity, we have

$\underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}-{u}_{n}\right){|\nabla {u}_{n}|}^{2}\psi {e}^{-{u}_{n}}dx\ge {\int }_{\mathrm{\Omega }}\left(1+{u}^{2}-u\right){|\nabla u|}^{2}\psi {e}^{-u}dx.$

Using Lebesgue’s dominated convergence theorem leads to

${\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r-2}{u}_{n}\psi {e}^{-{u}_{n}}dx\to {\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\psi {e}^{-u}dx$

and

${\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s-2}{u}_{n}\psi {e}^{-{u}_{n}}dx\to {\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\psi {e}^{-u}dx.$

It follows from (2.7) and the above estimates that

${\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right)\nabla u\nabla \psi {e}^{-u}\mathrm{d}x-{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}-u\right){|\nabla u|}^{2}\psi {e}^{-u}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\psi {e}^{-u}dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\psi {e}^{-u}dx\ge 0$

for $\psi \ge 0$ and $\psi \in {C}_{0}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$. Moreover, we get

${\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right)\nabla u\nabla \left(\psi {e}^{-u}\right)\mathrm{d}x+{\int }_{\mathrm{\Omega }}u{|\nabla u|}^{2}\psi {e}^{-u}dx3+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\psi {e}^{-u}dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\psi {e}^{-u}dx\ge 0.$(2.8)

Given $\phi \ge 0$ and $\phi \in {C}_{0}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$, we choose ${\psi }_{n}\in {C}_{0}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ such that ${\psi }_{n}\to \phi {e}^{u}$ in ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$, $\parallel {\psi }_{n}\parallel \le C$ and ${\psi }_{n}\to \phi {e}^{-u}$ for a.e. $x\in \mathrm{\Omega }$. Taking ${\psi }_{n}$ as the test function in (2.8) and letting n tend to infinity, we obtain

${\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right)\nabla u\nabla \phi \mathrm{d}x+{\int }_{\mathrm{\Omega }}u{|\nabla u|}^{2}\phi dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\phi dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\phi dx\ge 0$

for $\phi \ge 0$ and $\phi \in {C}_{0}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$.

Processing in a similar manner, one can also obtain an inequality with an opposite direction. Equation (1.2) holds for all $\phi \ge 0$ and $\phi \in {C}_{0}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$. By the density argument, equation (1.2) also holds for all functions $\phi \in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$.

Step 3. Since $u\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ satisfies equation (1.2), we have

${\int }_{\mathrm{\Omega }}\left(1+2{u}^{2}\right){|\nabla u|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx={\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx.$

By $D{I}_{{\mu }_{n}}\left({u}_{n}\right)=0$, we have $〈D{I}_{{\mu }_{n}}\left({u}_{n}\right),{u}_{n}〉=0$. That is,

${\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}-1}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{p}dx+{\int }_{\mathrm{\Omega }}\left(1+2{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx$$={\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx.$

In view of $s<\frac{4N}{N-2}$ and ${\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\to {\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$, as n tends to infinity, by the lower semi-continuity we deduce that

${\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}\to 0,\mathrm{ }{\int }_{\mathrm{\Omega }}{|\nabla {u}_{n}|}^{2}dx$$\to {\int }_{\mathrm{\Omega }}{|\nabla u|}^{2}dx,$${\int }_{\mathrm{\Omega }}{u}_{n}^{2}{|\nabla {u}_{n}|}^{2}dx\to {\int }_{\mathrm{\Omega }}{u}^{2}{|\nabla u|}^{2}dx,\mathrm{ }{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx$$\to {\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx,$

and

${I}_{{\mu }_{n}}\left({u}_{n}\right)\to I\left(u\right).$

Hence, we obtain ${u}_{n}\to u$ in ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$ and ${u}_{n}\nabla {u}_{n}\to u\nabla u$ in ${L}^{2}\left(\mathrm{\Omega }\right)$ as $n\to \mathrm{\infty }$. ∎

In the following context, we call $c\in ℝ$ a critical value of the functional I, provided there exists a function $u\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ satisfying equation (1.2) and $I\left(u\right)=c$. Theorem 1.1 implies that if ${\mu }_{n}\to 0$, ${c}_{n}$ is a critical value of ${I}_{{\mu }_{n}}$ and $c={lim}_{n\to \mathrm{\infty }}{c}_{n}$, then c is a critical value of I.

3 Proof of Theorem 1.2

In this section, we prove Theorem 1.2. We construct a sequence of critical values of the functional ${I}_{\mu }$ with $\mu >0$. The corresponding critical points will be used as the approximate solutions of equation (1.2).

Lemma 3.1.

Suppose that $\mathrm{\left\{}{u}_{n}\mathrm{\right\}}$ is a Palais–Smale sequence of the functional ${I}_{\mu }$ with $\mu \mathrm{>}\mathrm{0}$. Then ${u}_{n}$ is bounded in ${W}_{\mathrm{0}}^{\mathrm{1}\mathrm{,}p}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$.

Proof.

The proof is similar to the one of Lemma 2.2. Since $\mu >0$ is fixed and $\parallel D{I}_{\mu }\left({u}_{n}\right)\parallel \to 0$, by Lemma 2.1 we have

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx$$\le C\left[1+{I}_{\mu }\left({u}_{n}\right)+\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|{u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){u}_{n}^{2}dx\right].$

As in the proof of Lemma 2.2, we prove it by way of contradiction. Assume that

(3.1)

Then it is easy to see that

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\le C{\rho }_{n}^{4}.$

Let ${v}_{n}={\rho }_{n}^{-1}{u}_{n}$. As in the proof of Lemma 2.1, we have ${v}_{n}\to v$ in ${L}^{q}\left(\mathrm{\Omega }\right)$, $1\le q<\frac{4N}{N-2}$, ${v}_{n}^{2}⇀{v}^{2}$ in ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$, and $v=0$ in ${\overline{\mathrm{\Omega }}}_{+}\cup {\overline{\mathrm{\Omega }}}_{-}$. Since $p>N$ and μ is fixed, we find

$\mu {\left({\int }_{\mathrm{\Omega }}{|{v}_{n}|}^{p}{|\nabla {v}_{n}|}^{p}dx\right)}^{\frac{2}{p}}\le {\rho }_{n}^{-4}\mu {\left({\int }_{\mathrm{\Omega }}{|{u}_{n}|}^{p}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}\le C.$

So we see ${v}_{n}^{2}\to {v}^{2}$ in ${C}^{\alpha }\left(\overline{\mathrm{\Omega }}\right)$ for some $\alpha \in \left(0,1\right)$. Given $\psi \ge 0$ and $\psi \in {C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{0}\right)$, we take ${\phi }_{n}=\psi {u}_{n}/\left(1+{u}_{n}^{2}\right)$ as the test function, and thus have

$\nabla {\phi }_{n}=\nabla \psi \frac{{u}_{n}}{1+{u}_{n}^{2}}-\psi \frac{1-{u}_{n}^{2}}{{\left(1+{u}_{n}^{2}\right)}^{2}}\nabla {u}_{n},$$\parallel {\phi }_{n}\parallel ={\left({\int }_{\mathrm{\Omega }}{|\nabla {\phi }_{n}|}^{p}dx\right)}^{\frac{1}{p}}\le C{\left({\int }_{\mathrm{\Omega }}{|\nabla \psi |}^{p}dx\right)}^{\frac{1}{p}}+C{\parallel \psi \parallel }_{{L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)}{\left({\int }_{\mathrm{\Omega }}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{1}{p}}\le C{\rho }_{n}^{2}\parallel \psi \parallel ,$

and

$o\left(1\right)\parallel \psi \parallel ={\rho }_{n}^{-2}〈D{I}_{\mu }\left({u}_{n}\right),{\phi }_{n}〉$$\ge {\rho }_{n}^{-2}\left[\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}-1}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}{u}_{n}\nabla \psi \mathrm{d}x+{\int }_{\mathrm{\Omega }}{u}_{n}\nabla {u}_{n}\nabla \psi \mathrm{d}x\right]$$=\mu {\left({\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {v}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}-1}{|\nabla {v}_{n}|}^{p-2}\nabla {v}_{n}{v}_{n}\nabla \psi \mathrm{d}x+{\int }_{\mathrm{\Omega }}{v}_{n}\nabla {v}_{n}\nabla \psi \mathrm{d}x$(3.2)

for $\psi \ge 0$ and $\psi \in {C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{0}\right)$.

If

$\underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {v}_{n}|}^{p}dx={C}_{0}=0,$

then

${\int }_{\mathrm{\Omega }}{|v|}^{p}{|\nabla v|}^{p}dx\le {\underset{¯}{\mathrm{lim}}}_{n\to \mathrm{\infty }}{\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {v}_{n}|}^{p}dx=0.$

So one can see ${v}^{2}\equiv 0$. Otherwise, we assume that

${\underset{¯}{\mathrm{lim}}}_{n\to \mathrm{\infty }}{\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {v}_{n}|}^{p}dx={C}_{0}>0.$

By a density argument, inequality (3.2) also holds for $\psi \ge 0$ and $\psi \in {W}_{0}^{1,p}\left({\mathrm{\Omega }}_{0}\right)$.

Let

${C}_{n}=\mathrm{max}\left\{{v}_{n}^{2}\left(x\right)\mid x\in \partial {\mathrm{\Omega }}_{0}\right\}.$

Since ${v}_{n}^{2}\to {v}^{2}$ in ${C}^{\alpha }\left(\overline{\mathrm{\Omega }}\right)$ and ${v}^{2}=0$ in ${\overline{\mathrm{\Omega }}}_{+}\cup {\overline{\mathrm{\Omega }}}_{-}$, we have ${v}^{2}\equiv 0$ on $\partial {\mathrm{\Omega }}_{0}$ and ${C}_{n}\to 0$ as $n\to \mathrm{\infty }$.

Define

${\psi }_{n}\in {W}_{0}^{1,p}\left({\mathrm{\Omega }}_{0}\right)\mathit{ }\text{by}\mathit{ }{\psi }_{n}\left(x\right)={\left({v}_{n}^{2}\left(x\right)-{C}_{n}\right)}_{+},$

for $x\in {\overline{\mathrm{\Omega }}}_{0}$, $|\nabla {\psi }_{n}|\le 2|{v}_{n}\nabla {v}_{n}|$ and $\parallel {\psi }_{n}\parallel \le C$. Taking ${\psi }_{n}$ as the test function in inequality (3.2), we have

$o\left(1\right)\ge \mu {C}_{0}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{|{v}_{n}|}^{p-2}{v}_{n}{|\nabla {v}_{n}|}^{p-2}\nabla {v}_{n}\nabla {\left({v}_{n}^{2}-{C}_{n}\right)}_{+}\mathrm{d}x+{\int }_{\mathrm{\Omega }}{v}_{n}\nabla {v}_{n}\nabla {\left({v}_{n}^{2}-{C}_{n}\right)}_{+}\mathrm{d}x$$=\mu {C}_{0}^{\frac{2}{p}-1}{2}^{1-p}{\int }_{\mathrm{\Omega }}{|\nabla {\left({v}_{n}^{2}-{C}_{n}\right)}_{+}|}^{p}\mathrm{d}x+\frac{1}{2}{\int }_{\mathrm{\Omega }}{|\nabla {\left({v}_{n}^{2}-{C}_{n}\right)}_{+}|}^{2}\mathrm{d}x.$

Taking $n\to \mathrm{\infty }$, we obtain

$0\ge \mu {C}_{0}^{\frac{2}{p}-1}{2}^{1-p}{\int }_{\mathrm{\Omega }}{|\nabla {v}^{2}|}^{p}dx+\frac{1}{2}{\int }_{\mathrm{\Omega }}{|\nabla {v}^{2}|}^{2}dx.$

Hence, ${v}^{2}=0$ in $\overline{\mathrm{\Omega }}$, which leads to a contradiction by virtue of Lemma 2.2. In fact, it follows from (3.1) that

$\mu {\left({\int }_{\mathrm{\Omega }}{|v|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}{v}^{4}dx=1.\mathit{∎}$

Lemma 3.2.

The functional ${I}_{\mu }$ with $\mu \mathrm{>}\mathrm{0}$ satisfies the Palais–Smale condition.

Proof.

Let $\left\{{u}_{n}\right\}$ be a Palais–Smale sequence of the functional ${I}_{\mu }$ with $\mu >0$. By Lemma 3.1, $\left\{{u}_{n}\right\}$ is bounded in ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$. When $p>N$, we have ${u}_{n}\to u$ in ${C}^{\alpha }\left(\overline{\mathrm{\Omega }}\right)$ for some $\alpha \in \left(0,1\right)$.

Let

$\underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx={C}_{0}.$

If ${C}_{0}=0$, then ${u}_{n}\to 0$ in ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$. Otherwise, there holds

$o\left(1\right)=〈D{I}_{\mu }\left({u}_{n}\right)-D{I}_{\mu }\left({u}_{m}\right),{u}_{n}-{u}_{m}〉$$=\mu {C}_{0}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}\left({|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}-{|\nabla {u}_{m}|}^{p-2}\nabla {u}_{m},\nabla {u}_{n}-\nabla {u}_{m}\right)dx$$+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla {u}_{n}-\nabla {u}_{m}|}^{2}dx+o\left(1\right)$$\ge C{\int }_{\mathrm{\Omega }}{|\nabla {u}_{n}-\nabla {u}_{m}|}^{p}dx+o\left(1\right).$

So, $\left\{{u}_{n}\right\}$ is a Cauchy sequence of ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$ . ∎

Proof of Theorem 1.2.

We define a sequence of critical values of the functional ${I}_{\mu }$ with $\mu \in \left(0,1\right]$ by

${c}_{k}\left(\mu \right)=\underset{\phi \in {\mathrm{\Gamma }}_{k}}{inf}\underset{t\in {B}_{k}}{sup}{I}_{\mu }\left(\phi \left(t\right)\right),k=1,2,\mathrm{\dots },$

where

and ${B}_{k}$ is the unit ball of ${ℝ}^{k}$. Then we have

${I}_{\mu }\left(u\right)=\frac{1}{2}\mu {\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right]}^{\frac{2}{p}}+I\left(u\right)$$\ge \frac{1}{2}{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx+\frac{1}{r}{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx-\frac{1}{s}{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\ge \frac{1}{2}{\int }_{\mathrm{\Omega }}{|\nabla w|}^{2}dx-C{\int }_{\mathrm{\Omega }}{|w|}^{\frac{s}{2}}dx$$:=J\left(w\right),$

where w is defined by $Dw={\left(1+{u}^{2}\right)}^{1/2}Du$ and $w\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)$, and J is a ${C}^{1}$-functional defined on ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$.

Define the critical values of J by

${\alpha }_{k}=\underset{\phi \in {G}_{k}}{inf}\underset{t\in {B}_{k}}{sup}J\left(\phi \left(t\right)\right),$

where

By Lemma 3.2 and the symmetric mountain pass lemma [4], ${\alpha }_{k}$, $k=1,2,\mathrm{\dots }$, are critical values of J and ${\alpha }_{k}\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$. We have the estimate ${c}_{k}\left(\mu \right)\ge {\alpha }_{k}$. On the other hand, let ${\beta }_{k}={c}_{k}\left(1\right)$. Then we get

Let

${c}_{k}=\underset{\mu \to 0}{lim}{c}_{k}\left(\mu \right).$

By virtue of Theorem 1.1, ${c}_{k}$, $k=1,2,\mathrm{\dots }$, are the critical values of I and ${c}_{k}\to +\mathrm{\infty }$ as $k\to \mathrm{\infty }$. ∎

4 More general cases

In this section, we consider the more general quasilinear equation (1.3) and prove Theorem 1.3.

Equation (1.3) has a variational structure, given by the functional

$H\left(u\right)=\frac{1}{2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{i,j}\left(x,u\right){D}_{i}u{D}_{j}u\mathrm{d}x+\frac{1}{r}{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx-\frac{1}{s}{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx.$

Again we apply the perturbation method and introduce the perturbed functional ${H}_{\mu }$ with $\mu \in \left(0,1\right]$:

${H}_{\mu }\left(u\right)=\frac{1}{2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+H\left(u\right)$$=\frac{1}{2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+\frac{1}{2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}u\mathrm{d}x+\frac{1}{r}{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx-\frac{1}{s}{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx.$

Note that ${H}_{\mu }$ is defined on the Sobolev space ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$ with $p>N$. It is a ${C}^{1}$-functional on ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$, and satisfies

$〈D{H}_{\mu }\left(u\right),\phi 〉=\mu {\left[{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right]}^{\frac{2}{p}-1}\cdot {\int }_{\mathrm{\Omega }}\left[{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p-2}\nabla u\nabla \phi +{\left(1+{u}^{2}\right)}^{\frac{p}{2}-1}u{|\nabla u|}^{p}\phi \right]dx$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}\phi \mathrm{d}x+\frac{1}{2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{D}_{s}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}\phi \mathrm{d}x$$+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r-2}u\phi dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s-2}u\phi dx$

for $\phi \in {W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$.

As we have seen in the preceding section, for the quasilinear equations with indefinite nonlinearity, compared with ones with definite nonlinearity, the difficulty is to prove the boundedness of some associated sequences, either the sequence of approximate solutions (Lemmas 2.1 and 2.2) or the Palais–Smale sequence of the perturbed functional (Lemma 3.1). When we have proved the boundedness of these sequences, we can deal with the quasilinear equations as before to obtain the convergence and the existence results.

In the following, we will prove the boundedness of sequences of the approximate solutions, and the boundedness of the Palais–Smale sequences of the functional ${H}_{\mu }$.

Lemma 4.1.

For ${H}_{\mu }\mathit{}\mathrm{\left(}u\mathrm{\right)}$, there holds

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\le C\left[{H}_{\mu }\left(u\right)+\parallel D{H}_{\mu }\left(u\right)\parallel \parallel u\parallel +\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){u}^{2}dx\right].$(4.1)

Proof.

The proof is closely analogous to the one of Lemma 2.1. Choose $\psi \in {C}_{0}^{\mathrm{\infty }}\left({ℝ}^{N}\right)$ such that $\psi \ge 0$, $\psi \left(x\right)=1$ for $x\in {\mathrm{\Omega }}_{-}$, and $\psi \left(x\right)=0$ for $x\in {\mathrm{\Omega }}_{+}$. Taking $\phi =u{\psi }^{p}$ as the test function, we have

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}-1}\left(1+2{u}^{2}\right){|\nabla u|}^{p}{\psi }^{p}dx$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\left({a}_{ij}\left(x,u\right)+\frac{1}{2}u{D}_{s}{a}_{ij}\left(x,u\right)\right){D}_{i}u{D}_{j}u{\psi }^{p}\mathrm{d}x+{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx$$=〈D{H}_{\mu }\left(u\right),u{\psi }^{p}〉-p\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p-2}\nabla uu{\psi }^{p-1}\nabla \psi \mathrm{d}x$$-p{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}uu{\psi }^{p-1}{D}_{j}\psi \mathrm{d}x$$\le \mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}\left[\epsilon {\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}{\psi }^{p}dx+C{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}{|\nabla \psi |}^{p}dx\right]+C\parallel C{H}_{\mu }\left(u\right)\parallel \parallel u\parallel$$+\epsilon {\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}u{\psi }^{p}\mathrm{d}x+C{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}\psi {D}_{j}\psi {u}^{2}{\psi }^{p-2}\mathrm{d}x.$

In view of assumptions (a2) and (a3), it follows from the Sobolev inequality (2.3) that

${\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx\le C\left[\parallel D{H}_{\mu }\left(u\right)\parallel \parallel u\parallel +\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|u|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){u}^{2}dx\right].$(4.2)

Hence, we further have

${H}_{\mu }\left(u\right)-\frac{1}{q}〈D{H}_{\mu }\left(u\right),u〉=-\frac{1}{q}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+2{u}^{2}\right)}^{\frac{2}{p}-1}\left(1+2{u}^{2}\right){|\nabla u|}^{p}dx$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\left(\frac{1}{2}{a}_{ij}\left(x,u\right)-\frac{1}{q}\left({a}_{ij}\left(x,u\right)+\frac{1}{2}u{D}_{s}{a}_{ij}\left(x,u\right)\right)\right){D}_{i}u{D}_{j}u\mathrm{d}x$$+\frac{1}{2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}-\left(\frac{1}{q}-\frac{1}{r}\right){\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx+\left(\frac{1}{q}-\frac{1}{s}\right){\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\ge \left(\frac{1}{2}-\frac{2}{q}\right){\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+\delta {\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}u\mathrm{d}x$$-\left(\frac{1}{q}-\frac{1}{r}\right){\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx+\left(\frac{1}{q}-\frac{1}{s}\right){\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx.$(4.3)

By virtue of (4.2) and (4.3), we arrive at (4.1). ∎

Lemma 4.2.

Assume that ${\mu }_{n}\mathrm{>}\mathrm{0}$, ${\mu }_{n}\mathrm{\to }\mathrm{0}$, ${u}_{n}\mathrm{\in }{W}_{\mathrm{0}}^{\mathrm{1}\mathrm{,}p}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$, $D\mathit{}{H}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{=}\mathrm{0}$ and ${H}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{\le }C$. Then there exists a constant C independent of n such that

${\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|{u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){u}_{n}^{2}dx\le C.$

Moreover, we have

${\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\le C.$

Proof.

As in the proof of Lemma 2.2, we apply the indirect argument by assuming that

By Lemma 4.1, we get

${\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|\nabla {u}_{n}|}^{2}dx+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r}dx+{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\le C{\rho }_{n}^{4}.$(4.4)

Let ${v}_{n}={\rho }_{n}^{-1}{u}_{n}$. Then ${v}_{n}\to u$ in ${L}^{q}\left(\mathrm{\Omega }\right)$, $1\le q<\frac{4N}{N-2}$ and ${v}_{n}D{v}_{n}⇀v\nabla v$ in ${L}^{2}\left(\mathrm{\Omega }\right)$, where ${v}^{2}\in {H}_{0}^{1}\left({\mathrm{\Omega }}_{0}\right)\subset {H}_{0}^{1}\left(\mathrm{\Omega }\right)$. Given $\psi \ge 0$ and $\psi \in {C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{0}\right)$, we set ${\phi }_{n}=\psi {u}_{n}/\left(M+{u}_{n}^{2}\right)$ with M being the constant given in condition $\left({a}_{3}\right)$. From condition (a3) we have

$0=〈D{H}_{{\mu }_{n}}\left({u}_{n}\right),{\phi }_{n}〉$$={\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{M+{u}_{n}^{2}}\mathrm{d}x$$+{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}\psi$$\cdot \left(\frac{M}{{\left(M+{u}^{2}\right)}^{2}}+\frac{\left(M-1\right){u}_{n}^{2}}{\left(1+{u}_{n}^{2}\right){\left(M+{u}_{n}^{2}\right)}^{2}}\right)\mathrm{d}x+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{u}_{n}^{2}}{u}_{n}{D}_{i}{u}_{n}{D}_{j}\psi \mathrm{d}x$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\left({a}_{ij}\left(x,{u}_{n}\right)\frac{M-{u}_{n}^{2}}{{\left(M+{u}_{n}^{2}\right)}^{2}}+\frac{1}{2}{u}_{n}{D}_{s}{a}_{ij}\left(x,{u}_{n}\right)\frac{1}{M+{u}_{n}^{2}}\right){D}_{i}{u}_{n}{D}_{j}{u}_{n}\psi \mathrm{d}x$$\ge {\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{M+{u}_{n}^{2}}\mathrm{d}x$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{u}_{n}^{2}}{u}_{n}{D}_{i}{u}_{n}{D}_{j}{u}_{n}\psi \mathrm{d}x.$

By (4.4), we find

${\rho }_{n}^{-2}{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}|{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{M+{u}_{n}^{2}}\mathrm{d}x|$$\le {\rho }_{n}^{-2}{\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{1}{p}}{\left({\int }_{\mathrm{\Omega }}{|\nabla \psi |}^{p}dx\right)}^{\frac{1}{p}}$

It follows from Lemma 4.3 that

That is,

${\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{A}_{ij}\left(x\right)v{\partial }_{i}v{\partial }_{j}\psi \mathrm{d}x\le 0$

for $\psi \ge 0$ and $\psi \in {C}_{0}^{\mathrm{\infty }}\left({\mathrm{\Omega }}_{0}\right)$.

Since ${v}^{2}\in {H}_{0}^{1}\left({\mathrm{\Omega }}_{0}\right)\subset {H}_{0}^{1}\left(\mathrm{\Omega }\right)$, by the density argument we have

${\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{A}_{ij}\left(x\right)v{\partial }_{i}v{\partial }_{j}{v}^{2}\mathrm{d}x=0$

and ${v}^{2}\equiv 0$ in Ω. The remainder is the same as that shown in Lemma 2.2, so we omit it here. ∎

Lemma 4.3.

For $\psi \mathrm{\in }{H}_{\mathrm{0}}^{\mathrm{1}}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$, there holds

$\underset{n\to \mathrm{\infty }}{lim}{\rho }_{n}^{-2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{u}_{n}^{2}}{u}_{n}{\partial }_{i}{u}_{n}{\partial }_{j}\psi \mathrm{d}x={\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{A}_{ij}\left(x\right)v{\partial }_{i}v{\partial }_{j}\psi \mathrm{d}x.$

Proof.

For $T>0$, let ${u}_{n}^{T}$ be the truncated function of ${u}_{n}$, that is, ${u}_{n}^{T}={u}_{n}$ if $|{u}_{n}|\le T$, and ${u}_{n}^{T}=±T$ if $±{u}_{n}\ge T$. Taking ${u}_{n}^{T}$ as the test function, we have

$0=〈D{H}_{{\mu }_{n}}\left({u}_{n}\right),{u}_{n}^{T}〉$$={\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}\left[{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}^{T}|}^{p}+{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}-1}{|\nabla {u}_{n}|}^{2}{u}_{n}{u}_{n}^{T}\right]dx$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\left({a}_{ij}\left(x,{u}_{n}\right)+\frac{1}{2}{u}_{n}{D}_{s}{a}_{ij}\left(x,{u}_{n}\right)\right){\partial }_{i}{u}_{n}^{T}{\partial }_{j}{u}_{n}^{T}\mathrm{d}x+\frac{1}{2}{\int }_{|{u}_{n}|\ge T}\sum _{i,j=1}^{N}{u}_{n}^{T}{D}_{s}{a}_{ij}\left(x,{u}_{n}\right){D}_{i}{u}_{n}{D}_{j}{u}_{n}\mathrm{d}x$$+{\int }_{\mathrm{\Omega }}{a}_{-}{|{u}_{n}|}^{r-2}{u}_{n}{u}_{n}^{T}dx-{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s-2}{u}_{n}{u}_{n}^{T}dx.$(4.5)

In view of assumption $\left({a}_{3}\right)$, there exists a ${T}_{0}>0$ such that

$\sum _{i,j=1}^{N}s{D}_{s}{a}_{ij}\left(x,s\right){\xi }_{i}{\xi }_{j}\ge 0$

for $|s|\ge T>{T}_{0}$ and $x\in \overline{\mathrm{\Omega }}$, where $\xi \in {ℝ}^{N}$. By condition (a2) and (4.5), we get

${\int }_{\mathrm{\Omega }}{|\nabla {u}_{n}^{T}|}^{2}dx\le C{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\left({a}_{ij}\left(x,{u}_{n}\right)+\frac{1}{2}{u}_{n}{D}_{s}{a}_{ij}\left(x,{u}_{n}\right)\right){\partial }_{i}{u}_{n}^{T}{\partial }_{j}{u}_{n}^{T}\mathrm{d}x$$\le C{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s-2}{u}_{n}{u}_{n}^{T}dx\le CT{\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s-1}dx$$\le CT{\left({\int }_{\mathrm{\Omega }}{a}_{+}{|{u}_{n}|}^{s}dx\right)}^{\frac{s-1}{s}}$$\le CT{\rho }_{n}^{4-\frac{4}{s}}.$

Hence, we obtain the estimate

${\rho }_{n}^{-2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{u}_{n}^{2}}{u}_{n}{\partial }_{i}{u}_{n}{\partial }_{j}\psi \mathrm{d}x$$={\rho }_{n}^{-2}{\int }_{|{u}_{n}|\ge T}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{u}_{n}^{2}}{u}_{n}{\partial }_{i}{u}_{n}{\partial }_{j}\psi \mathrm{d}x+{\rho }_{n}^{-2}{\int }_{|{u}_{n}|$={\rho }_{n}^{-2}\left({\int }_{|{u}_{n}|\ge T}{A}_{ij}\left(x\right){u}_{n}{\partial }_{i}{u}_{n}{\partial }_{j}\psi \mathrm{d}x+{o}_{T}\left(1\right){\parallel {u}_{n}\nabla {u}_{n}\parallel }_{{L}^{2}\left(\mathrm{\Omega }\right)}\right)+{\rho }_{n}^{-2}CT{\parallel \nabla {u}_{n}^{T}\parallel }_{{L}^{2}\left(\mathrm{\Omega }\right)}$$={\int }_{\mathrm{\Omega }}{A}_{ij}\left(x\right){v}_{n}{\partial }_{i}{v}_{n}{\partial }_{j}\psi \mathrm{d}x+{o}_{T}\left(1\right)+{\rho }_{n}^{-2}C{T}^{\frac{3}{2}}{\rho }_{n}^{2-\frac{2}{s}}$

The following proposition can be regarded as a counterpart of Theorem 1.1.

Proposition 4.4.

Assume that ${\mu }_{n}\mathrm{>}\mathrm{0}$, ${\mu }_{n}\mathrm{\to }\mathrm{0}$, ${u}_{n}\mathrm{\in }{C}_{\mathrm{0}}^{\mathrm{1}}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$, $D\mathit{}{H}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{=}\mathrm{0}$ and ${H}_{{\mu }_{n}}\mathit{}\mathrm{\left(}{u}_{n}\mathrm{\right)}\mathrm{\le }C$. Then the following assertions hold:

• (i)

There exists a function $u\in {H}_{0}^{1}\left(\mathrm{\Omega }\right)\cap {L}^{\mathrm{\infty }}\left(\mathrm{\Omega }\right)$ satisfying equation ( 1.4 ).

• (ii)

Up to a subsequence, there holds

and

${H}_{{\mu }_{n}}\left({u}_{n}\right)\to H\left({u}_{n}\right).$

By Lemma 4.2, there holds

${\mu }_{n}{\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}\mathrm{d}x\right)}^{\frac{2}{p}}\right)+\int {}_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right)|\nabla {u}_{n}|{}^{2}\mathrm{d}x\le C.$(4.6)

Proposition 4.4 can be proved similarly to Theorem 1.1, so we omit it and refer to [11]. As we have seen in the proof of Theorem 1.1, with the help of estimate (4.6), the proof of Theorem 1.1 can also be done in the same way as that for quasilinear equations with definite nonlinearity.

Next, we consider the perturbed functional ${H}_{\mu }$ with $\mu \in \left(0,1\right]$.

Lemma 4.5.

Let $\mathrm{\left\{}{u}_{n}\mathrm{\right\}}$ be a Palais–Smale sequence of the functional ${H}_{\mu }$ with $\mu \mathrm{>}\mathrm{0}$. Then ${u}_{n}$ is bounded in ${W}_{\mathrm{0}}^{\mathrm{1}\mathrm{,}p}\mathit{}\mathrm{\left(}\mathrm{\Omega }\mathrm{\right)}$.

Proof.

By Lemma 4.1, we only need to prove

$\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|{u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|{u}_{n}|}^{2}dx\le C.$

Otherwise, we have

${\rho }_{n}^{4}=\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|{u}_{n}|}^{p}dx\right)}^{\frac{2}{p}}+{\int }_{\mathrm{\Omega }}\left(1+{u}_{n}^{2}\right){|{u}_{n}|}^{2}dx\to +\mathrm{\infty }.$

Let

Then ${v}_{n}\to v$ in ${L}^{q}\left(\mathrm{\Omega }\right)$, $1\le q<\frac{4N}{N-2}$, ${v}_{n}^{2}\to {v}^{2}$ in ${C}^{\alpha }\left(\overline{\mathrm{\Omega }}\right)$ for some $\alpha \in \left(0,1\right)$, and ${v}_{n}^{2}⇀{v}^{2}$ in ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$ (and ${H}_{0}^{1}\left(\mathrm{\Omega }\right)$). As in the proof of Lemma 2.2, we know that $v\left(x\right)=0$ for $x\in {\overline{\mathrm{\Omega }}}_{+}\cap {\overline{\mathrm{\Omega }}}_{-}$, and ${v}^{2}\in {W}_{0}^{1,p}\left({\mathrm{\Omega }}_{0}\right)\subset {W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$. For $\psi \ge 0$ and $\psi \in {W}_{0}^{1,p}\left({\mathrm{\Omega }}_{0}\right)$, we let ${\phi }_{n}=\psi {u}_{n}/\left(M+{u}_{n}^{2}\right)$, ${\phi }_{n}\in {W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$ and

$\nabla {\phi }_{n}=\nabla \psi \frac{{u}_{n}}{M+{u}_{n}^{2}}+\psi \frac{M-{u}_{n}^{2}}{{\left(M+{u}_{n}^{2}\right)}^{2}}\nabla {u}_{n},\parallel {\phi }_{n}\parallel \le C{\rho }_{n}^{2}\parallel \psi \parallel .$

Then we deduce that

$o\left(1\right)\parallel \psi \parallel ={\rho }_{n}^{-2}〈D{H}_{\mu }\left({u}_{n}\right),{\phi }_{n}〉$$={\rho }_{n}^{-2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}\nabla \psi \frac{{u}_{n}}{M+{u}_{n}^{2}}\mathrm{d}x$$+{\rho }_{n}^{-2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx\right)}^{\frac{2}{p}-1}$$\cdot {\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{p}{|\nabla {u}_{n}|}^{p}\psi \left(\frac{M}{{\left(M+{u}_{n}^{2}\right)}^{2}}+\frac{\left(M-1\right){u}_{n}^{2}}{\left(1+{u}_{n}^{2}\right){\left(M+{u}_{n}^{2}\right)}^{2}}\right)dx$$+{\rho }_{n}^{-2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{u}_{n}^{2}}{u}_{n}{\partial }_{i}{u}_{n}{\partial }_{j}\psi \mathrm{d}x$$+{\rho }_{n}^{-2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\left({a}_{ij}\left(x,{u}_{n}\right)\frac{M-{u}_{n}^{2}}{{\left(M+{u}_{n}^{2}\right)}^{2}}+\frac{1}{2}{D}_{s}{a}_{ij}\left(x,{u}_{n}\right)\frac{{u}_{n}}{M+{u}_{n}^{2}}\right){D}_{i}{u}_{n}{D}_{j}{u}_{n}\psi \mathrm{d}x$$\ge \mu {\left({\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {v}_{n}|}^{p}\right)}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left({\rho }_{n}^{-2}+{v}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {v}_{n}|}^{p-2}\nabla {v}_{n}\nabla \psi \frac{{v}_{n}}{M+{u}_{n}^{2}}\mathrm{d}x$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}\frac{{a}_{ij}\left(x,{u}_{n}\right)}{M+{v}_{n}^{2}}{v}_{n}{D}_{i}{v}_{n}{D}_{j}\psi \mathrm{d}x.$

Choose $\psi =\frac{1}{2}{\left({v}_{n}^{2}-{C}_{n}\right)}_{+}\in {W}_{0}^{1,p}\left({\mathrm{\Omega }}_{0}\right)$ and ${C}_{n}={\mathrm{max}}_{\partial {\mathrm{\Omega }}_{0}}{v}_{n}^{2}\to 0$. By the lower semi-continuity, we obtain

$0\ge \mu {\left({\int }_{\mathrm{\Omega }}{|v|}^{p}{|\nabla v|}^{p}dx\right)}^{\frac{2}{p}}+c{\int }_{\mathrm{\Omega }}{v}^{2}{|\nabla v|}^{2}dx.$

Hence, ${v}^{2}=0$. This yields a contradiction according to Lemma 3.1. ∎

Proposition 4.6.

The functional ${H}_{\mu }$with $\mu \mathrm{>}\mathrm{0}$ satisfies the Palais–Smale condition.

Proof.

Let $\left\{{u}_{n}\right\}$ be a Palais–Smale sequence of ${H}_{\mu }$. By Lemma 4.5, ${u}_{n}$ is bounded in ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$. Assume that ${u}_{n}\to u$ in ${C}^{\alpha }\left(\overline{\mathrm{\Omega }}\right)$ for some $\alpha \in \left(0,1\right)$. Set

${C}_{0}=\underset{n\to \mathrm{\infty }}{lim}{\int }_{\mathrm{\Omega }}{\left(1+{u}_{n}^{2}\right)}^{\frac{p}{2}}{|\nabla {u}_{n}|}^{p}dx.$

If ${C}_{0}=0$, we are done. Otherwise, there holds

$o\left(1\right)=〈D{H}_{\mu }\left({u}_{n}\right)-D{H}_{\mu }\left({u}_{m}\right),{u}_{n}-{u}_{m}〉$$=\mu {C}_{0}^{\frac{2}{p}-1}{\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}\left({|\nabla {u}_{n}|}^{p-2}\nabla {u}_{n}-{|\nabla {u}_{m}|}^{p-2}\nabla {u}_{m},\nabla {u}_{n}-\nabla {u}_{m}\right)dx$$+{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}\left({u}_{n}-{u}_{m}\right){D}_{j}\left({u}_{n}-{u}_{m}\right)\mathrm{d}x+o\left(1\right)$$\ge c{\int }_{\mathrm{\Omega }}{|\nabla {u}_{n}-\nabla {u}_{m}|}^{p}dx+o\left(1\right).$

So, $\left\{{u}_{n}\right\}$ is a Cauchy sequence of ${W}_{0}^{1,p}\left(\mathrm{\Omega }\right)$. ∎

Proof of Theorem 1.3.

Define

${c}_{k}\left(\mu \right)=\underset{\phi \in {\mathrm{\Gamma }}_{k}}{inf}\underset{t\in {B}_{k}}{sup}{H}_{\mu }\left(\phi \left(t\right)\right),k=1,2,\mathrm{\dots },$

where

A straightforward estimate on ${H}_{k}\left(u\right)$ gives

${H}_{k}\left(u\right)=\frac{1}{2}\mu {\left({\int }_{\mathrm{\Omega }}{\left(1+{u}^{2}\right)}^{\frac{p}{2}}{|\nabla u|}^{p}dx\right)}^{\frac{2}{p}}+H\left(u\right)$$\ge \frac{1}{2}{\int }_{\mathrm{\Omega }}\sum _{i,j=1}^{N}{a}_{ij}\left(x,u\right){D}_{i}u{D}_{j}u\mathrm{d}x+\frac{1}{r}{\int }_{\mathrm{\Omega }}{a}_{-}{|u|}^{r}dx-\frac{1}{s}{a}_{+}{|u|}^{s}\mathrm{d}x$$\ge \frac{{c}_{1}}{2}{\int }_{\mathrm{\Omega }}\left(1+{u}^{2}\right){|\nabla u|}^{2}dx-\frac{1}{s}{\int }_{\mathrm{\Omega }}{a}_{+}{|u|}^{s}dx$$\ge \frac{{c}_{1}}{2}{\int }_{\mathrm{\Omega }}{|\nabla w|}^{2}dx-c{\int }_{\mathrm{\Omega }}{|w|}^{\frac{s}{2}}dx$$:=J\left(w\right).$

One can find ${\alpha }_{k}\le {\beta }_{k}$ such that ${\alpha }_{k}\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$, and ${\alpha }_{k}\le {c}_{k}\left(\mu \right)\le {\beta }_{k}$.

Let

${c}_{k}=\underset{\mu \to 0}{lim}{c}_{k}\left(\mu \right).$

According to Proposition 4.4, ${c}_{k}$, $k=1,2,\mathrm{\dots }$, are critical values of H and ${c}_{k}\to +\mathrm{\infty }$ as $k\to \mathrm{\infty }$. Consequently, the proof is completed. ∎

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About the article

Received: 2018-01-14

Accepted: 2018-04-07

Published Online: 2018-06-20

Funding Source: National Natural Science Foundation of China

Award identifier / Grant number: 11601493

Award identifier / Grant number: 11761082

Junfang Zhao is supported by National Science Foundation of China under No. 11601493 and Beijing Higher Education Young Elite Faculty Project. Xiangqing Liu is supported by National Science Foundation of China under No. 11761082 and Yunnan Young Academic and Technical Leaders’ Program (2015HB028). This work is also partially supported by UTRGV Faculty Research Council Award 110000327.

Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 1235–1251, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496,

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© 2019 Walter de Gruyter GmbH, Berlin/Boston. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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