A Global Compact Result for a Fractional Elliptic Problem with Hardy term and critical non-linearity on the whole space

TL;DR

This paper establishes the existence of positive solutions for a fractional elliptic equation with Hardy potential and critical Sobolev nonlinearity on the entire space, using compactness analysis of the associated functional.

Contribution

It provides a new existence result for fractional elliptic equations with Hardy terms and critical nonlinearity, under specific conditions on the coefficients.

Findings

Existence of positive solutions under certain assumptions.

Compactness analysis of the associated functional.

Extension of results to fractional elliptic equations with Hardy potential.

Abstract

In this paper, we deal with a fractional elliptic equation with critical Sobolev nonlinearity and Hardy term $$ (-\Delta)^{\alpha} u-\mu\frac{u}{|x|^{2\alpha}}+a(x) u=|u|^{2^*-2}u+k(x)|u|^{q-2}u$$ $$ u\,\in\,H^\alpha({\mathbb R}^N),$$ where $2<q< 2^*$, $0<\alpha<1$, $N>4\alpha$, $2^*=2N/(N-2\alpha)$ is the critical Sobolev exponent, $a(x),k(x)\in C({\mathbb R}^N)$. Through a compactness analysis of the functional associated to $(*)$, we obtain the existence of positive solutions for $(*)$ under certain assumptions on $a(x),k(x)$.

Equations

$$\{\begin{array}{l}(-\Delta {)}^{\alpha }u-\mu \frac{u}{|x{|}^{2\alpha }}+a(x)u=|u{|}^{2^{\ast }-2}u+k(x)|u{|}^{q-2}u,\\ u\in H^{\alpha }({\mathbb{R}}^N),\end{array}$$

$$\{\begin{array}{l}(-\Delta {)}^{\alpha }u-\mu \frac{u}{|x{|}^{2\alpha }}+a(x)u=|u{|}^{2^{\ast }-2}u+k(x)|u{|}^{q-2}u,x\in {\mathbb{R}}^N,\\ u\in H^{\alpha }({\mathbb{R}}^N),\end{array}$$

$$\{\begin{array}{ll}(-\Delta {)}^{\alpha }u-\mu \frac{u}{|x{|}^{2\alpha }}=|u{|}^{2^{\ast }-2}u,& x\in {\mathbb{R}}^N,\\ u\in {\dot{H}}^{\alpha }({\mathbb{R}}^N).& \end{array}$$

$$\{\begin{array}{ll}(-\Delta {)}^{\alpha }u-\mu \frac{u}{|x{|}^{2\alpha }}=\frac{|u{|}^{2^{\ast }-2}u}{|x{|}^s}+|u{|}^{2^{\ast }-2}u,& x\in {\mathbb{R}}^N,\\ u>0,u\in {\dot{H}}^{\alpha }({\mathbb{R}}^N),& \end{array}$$

$$\Gamma_{N,\alpha}\big{(}\int_{{\mathbb{R}}^{N}}\frac{|u(x)|^{2}}{|x|^{2\alpha}}dx\bigl{)}\leq c_{N,\alpha}\int_{{\mathbb{R}}^{N}}\frac{|u(x)-u(y)|^{2}}{|x-y|^{N+2\alpha}}dy,\forall u\in C_{0}^{\infty}({\mathbb{R}}^{N}),$$

$${\Gamma }_{N,\alpha }=2^{2\alpha }\frac{{\Gamma }^2(\frac{N+2\alpha }{4})}{{\Gamma }^2(\frac{N-2\alpha }{4})},c_{N,\alpha }=2^{2\alpha -1}{\pi }^{-\frac{N}{2}}\frac{\Gamma (\frac{N+2\alpha }{2})}{|\Gamma (-\alpha )|}.$$

$$\hat{u}(\xi )=\frac{1}{(2\pi {)}^{\frac{N}{2}}}{\int }_{{\mathbb{R}}^N}{e}^{-i\xi \cdot x}u(x)dx.$$

$$\widehat{(-\Delta {)}^{\alpha }u}(\xi )=|\xi {|}^{2\alpha }\hat{u}(\xi ),\forall u\in C_0^{\infty }({\mathbb{R}}^N).$$

$$\Vert u{\Vert }_{{\dot{H}}^{\alpha }({\mathbb{R}}^N)}=\Vert |\xi {|}^{\alpha }\hat{u}{\Vert }_{L^2({\mathbb{R}}^N)},$$

$$\Vert u{\Vert }_{H^{\alpha }({\mathbb{R}}^N)}=\Vert u{\Vert }_{L^2({\mathbb{R}}^N)}+\Vert |\xi {|}^{\alpha }\hat{u}{\Vert }_{L^2({\mathbb{R}}^N)}.$$

$$c_{N,\alpha }{\int }_{{\mathbb{R}}^N}{\int }_{{\mathbb{R}}^N}\frac{|u(x)-u(y){|}^2}{|x-y{|}^{N+2\alpha }}dxdy={\int }_{{\mathbb{R}}^N}|(-\Delta {)}^{\alpha /2}u(x){|}^{2}dx=\Vert u{\Vert }_{{\dot{H}}^{\alpha }({\mathbb{R}}^N)}^2,$$

$$\Vert u^{+}{\Vert }_{{\dot{H}}^{\alpha }}\le \Vert u{\Vert }_{{\dot{H}}^{\alpha }}.$$

$$\Vert u{\Vert }_{L^{p,\nu }({\mathbb{R}}^N)}^p=\underset{r>0,\bar{x}\in {\mathbb{R}}^N}{\sup }{r}^{\nu -N}{\int }_{B(\bar{x},r)}|u(x){|}^{p}dx<\infty .$$

$$L^{2^{\ast }}({\mathbb{R}}^N)\hookrightarrow L^{p,\nu }({\mathbb{R}}^N),\text{ for }1\le p<2^{\ast },$$

$$L^{p,\frac{(N-2\alpha )p}{2}}({\mathbb{R}}^N)\hookrightarrow L^{p_1,\frac{(N-2\alpha )p_1}{2}}({\mathbb{R}}^N),\text{ for }1<p_1<p<2^{\ast }.$$

$$\Phi (u_n)\to c,{\Phi }^{\prime }(u_n)\to 0\text{ as }n\to \infty .$$

$$I(u)$$

$$-\frac{1}{2^{\ast }}{\int }_{{\mathbb{R}}^N}(u^{+}(x){)}^{2^{\ast }}dx-\frac{1}{q}{\int }_{{\mathbb{R}}^N}k(x)(u^{+}(x){)}^{q}dx.$$

$$\{\begin{array}{l}(-\Delta {)}^{\alpha }u+\bar{a}u=\bar{k}|u{|}^{q-2}u+|u{|}^{2^{\ast }-2}u,\\ u\in H^{\alpha }({\mathbb{R}}^N),\end{array}$$

$$I^{\infty }(u)$$

$$-\frac{1}{2^{\ast }}{\int }_{{\mathbb{R}}^N}(u^{+}(x){)}^{2^{\ast }}dx,u\in H^{\alpha }({\mathbb{R}}^N).$$

$$\{\begin{array}{l}(-\Delta {)}^{\alpha }u-\mu \frac{u}{|x{|}^{2\alpha }}=|u{|}^{2^{\ast }-2}u,\\ u\in {\dot{H}}^{\alpha }({\mathbb{R}}^N),\end{array}$$

$$I_{\mu }(u)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}(|(-\Delta {)}^{\alpha /2}u(x){|}^2-\mu \frac{u^2}{|x{|}^{2\alpha }})dx-\frac{1}{2^{\ast }}{\int }_{{\mathbb{R}}^N}(u^{+}(x){)}^{2^{\ast }}dx,u\in {\dot{H}}^{\alpha }({\mathbb{R}}^N).$$

$$\{\begin{array}{l}(-\Delta {)}^{\alpha }u=|u{|}^{2^{\ast }-2}u,\\ u\in {\dot{H}}^{\alpha }({\mathbb{R}}^N),\end{array}$$

$$I_0(u)=\frac{1}{2}{\int }_{{\mathbb{R}}^N}|(-\Delta {)}^{\alpha /2}u(x){|}^{2}dx-\frac{1}{2^{\ast }}{\int }_{{\mathbb{R}}^N}(u^{+}(x){)}^{2^{\ast }}dx,u\in {\dot{H}}^{\alpha }({\mathbb{R}}^N).$$

$$S_{\alpha ,\mu }=\underset{u\in H^{\alpha }({\mathbb{R}}^N)\backslash \{0\}}{\inf }\frac{c_{N,\alpha }{\int }_{{\mathbb{R}}^{2N}}\frac{|u(x)-u(y){|}^2}{|x-y{|}^{N+2\alpha }}dxdy-\mu {\int }_{{\mathbb{R}}^N}\frac{|u(x){|}^2}{|x{|}^{2\alpha }}dx}{({\int }_{{\mathbb{R}}^N}|u(x){|}^{2^{\ast }}dx{)}^{2/2^{\ast }}},$$

$$\frac{c}{\Bigl{(}|x|^{1-\eta_{\mu}}(1+|x|^{2\eta_{\mu}})\Bigl{)}^{\frac{N-2\alpha}{2}}}\leq U_{\mu}(x)\leq\frac{C}{\Bigl{(}|x|^{1-\eta_{\mu}}(1+|x|^{2\eta_{\mu}})\Bigl{)}^{\frac{N-2\alpha}{2}}},\text{ in }{\mathbb{R}}^{N}\backslash\{0\}$$

$${\eta }_{\mu }=1-\frac{2{\alpha }_{\mu }}{N-2\alpha },$$

$${\phi }_{s,N}({\alpha }_{\mu })=2^{2\alpha }\frac{\Gamma (\frac{{\alpha }_{\mu }+2\alpha }{2})\Gamma (\frac{N-{\alpha }_{\mu }}{2})}{\Gamma (\frac{N-{\alpha }_{\mu }-2\alpha }{2})\Gamma (\frac{{\alpha }_{\mu }}{2})}=\mu ,$$

$${\alpha }_{\mu }={\phi }_{\alpha ,N}^{-1}(\mu ).$$

$$U_{\mu }^{\epsilon }(x):={\epsilon }^{\frac{2\alpha -N}{2}}{U}_{\mu }(x/\epsilon ).$$

Full text

[TABLE]

where $2<q<2^{*}$, $0<\alpha<1$, $N>4\alpha$, $2^{*}=2N/(N-2\alpha)$ is the critical Sobolev exponent, $a(x),k(x)\in C({\mathbb{R}}^{N})$. Through a compactness analysis of the functional associated to $(*)$, we obtain the existence of positive solutions for $(*)$ under certain assumptions on $a(x),k(x)$.

Key words.   Fractional Laplacian, compactness, positive solution, unbounded domain, Hardy term, critical Sobolev nonlinearity.

AMS Classification: 35J10 35J20 35J60

1 Introduction

We consider the following nonlinear elliptic equation:

[TABLE]

where $2<q<2^{*}$, $0<\alpha<1$, $N>4\alpha$, $2^{*}=2N/(N-2\alpha)$ is the critical Sobolev exponent, $a(x),k(x)\in C({\mathbb{R}}^{N})$.

Recently the fractional Laplacian and more general nonlocal operators of elliptic type have been widely studied, both for their interesting theoretical structure and concrete applications in many fields such as optimization, finance, phase transitions, stratified materials, anomalous diffusion and so on (see [2, 5, 7, 8, 14, 21, 23, 24]). In particular, a lot of results have been accumulated for elliptic equations with critical nonlinearity related to (1.1). In [5], Dipierro etc. considere’d the critical problem with Hardy-Leray potential

[TABLE]

where $\dot{H}^{\alpha}({\mathbb{R}}^{N})$ is defined in (1.6). They proved the existence, certain qualitative properties and asymptotic behavior of positive solutions to (1.2). Ghoussoub and Shakerian in [9] investigated the following double critical problem in ${\mathbb{R}}^{N}$

[TABLE]

with $\mu>0,0<s<2$. Through the non-compactness analysis of the Palais-Smale sequence of (1.3), the existence of the solutions were obtained. The authors in [11] established a concentration-compactness result for a fractional Schrödinger equation with the subcritical nonlinearity $f(x,u)$. Motivated by [5, 9, 11, 12, 27] we consider the existence of positive solutions for problem (1.1) in ${\mathbb{R}}^{N}$. The main interest for this type of problems, in addition to the nonlocal fractional Laplacian is the presence of the singular potential $\frac{1}{|x|^{2\alpha}}$ related to the fractional Hardy’s inequality. We recall the Hardy inequality([5]),

[TABLE]

where

[TABLE]

The Sobolev embedding $\dot{H}^{\alpha}({\mathbb{R}}^{N})\hookrightarrow L^{2}(|x|^{-2\alpha},{\mathbb{R}}^{N})$ is not compact, even locally, in any neighborhood of zero. As it is well known, the loss of the compactness of the embeddings is one of the main difficulties for elliptic problems with critical nonlinearities. Problem (1.1) has three factors, critical Sobolev term, Hardy term and unbounded domain which lead to the non-compactness of the embeddings. In [5] and [9], the authors can consider the solutions of critical problems in the homogeneous fractional Sobolev space $\dot{H}^{\alpha}({\mathbb{R}}^{N})$, while we must deal with (1.1) in the nonhomogeneous fractional Sobolev space $H^{\alpha}({\mathbb{R}}^{N})$ given the presence of low sub-critical terms in (1.1). This is why the methods in [5] and [9] can not be used directly to (1.1). As far as we know, the existence results for the fractional elliptic problems with a mixture of critical Sobolev terms, Hardy term and subcritical terms are relatively new. To overcome the difficulties caused by the lack of compactness, we carry out a non-compactness analysis which can distinctly express all the parts which cause non-compactness. As a result, we are able to obtain the existence of nontrival solutions of the elliptic problem with the critical nonlinear term on an unbounded domain by getting rid of these noncompact factors. To be more specific, for the Palais-Smale sequences of the variational functional corresponding to (1.1) we first establish a complete noncompact expression which includes all the blowing up bubbles caused by the critical Sobolev nonlinearity, the Hardy term and by the unbounded domain. Then we derive the existence of positive solutions for (1.1). Our methods are based on some techniques of [4, 11, 13, 16, 19, 20, 25, 26].

Before introducing our main results, we give some notations and assumptions.

Notations and assumptions:

Denote $c$ and $C$ as arbitrary constants which may change from line to line. Let $B(x,r)$ denote a ball centered at $x$ with radius $r$ and $B(x,r)^{C}={\mathbb{R}}^{N}\setminus B(x,r)$.

Let $N\geq 1$, $u\in L^{2}({\mathbb{R}}^{N})$, let the Fourier transform of $u$ be

[TABLE]

We define the operator $(-\Delta)^{\alpha}u$ by the Fourier transform

[TABLE]

Let $\dot{H}^{\alpha}({\mathbb{R}}^{N})$ be the homogeneous fractional Sobolev space as the completion of $C^{\infty}_{0}({\mathbb{R}}^{N})$ under the norm

[TABLE]

and denote by $H^{\alpha}({\mathbb{R}}^{N})$ the usual nonhomogeneous fractional Sobolev space with the norm

[TABLE]

For $0<\alpha<1$, a direct calculation (see e.g. [[14], proposition 4.4] or [[5], Proposition 1.2]) gives

[TABLE]

where $c_{N,\alpha}=2^{2\alpha-1}\pi^{-\frac{N}{2}}\frac{\Gamma(\frac{N+2\alpha}{2})}{|\Gamma(-\alpha)|}$.

Let $u^{+}=\max\{u,0\},u^{-}=u^{+}-u$. From the proof of (2.15) in [15], it follows

[TABLE]

We call $u\not\equiv 0$ in ${\mathbb{R}}^{N}$ if the measure of the set $\{x\in{\mathbb{R}}^{N}|u(x)\not=0\}$ is positive.

Recall the definition of Morrey space. A measurable function $u:{\mathbb{R}}^{N}\rightarrow{\mathbb{R}}$ belongs to the Morrey space with $p\in[1,\infty)$ and $\nu\in(0,N]$, if and only if

[TABLE]

By Hölder inequality, we can verify (refer to [14])

[TABLE]

and

[TABLE]

Moreover, we have $L^{p,\nu}({\mathbb{R}}^{N})\hookrightarrow L^{1,\frac{\nu}{p}}({\mathbb{R}}^{N})$.

Next we give the definition of the Palais-Smale sequence. Let $X$ be a Banach space, $\Phi\in C^{1}(X,{\mathbb{R}})$, $c\in{\mathbb{R}}$, we call $\{u_{n}\}\subset X$ is a Palais-Smale sequence of $\Phi$ if

[TABLE]

In this paper we assume that:

(a) $a(x)\in C({\mathbb{R}}^{N})$, $k(x)\in C({\mathbb{R}}^{N})$;

(b) $\displaystyle\lim_{|x|\rightarrow\infty}a(x)=\bar{a}>0,\lim_{|x|\rightarrow\infty}k(x)=\bar{k}>0,\,\,\,\inf_{x\in{\mathbb{R}}^{N}}a(x)={\hat{a}}>0,\,\inf_{x\in{\mathbb{R}}^{N}}k(x)={\hat{k}}>0.$

In the following, we assume that $a(x),k(x)$ always satisfy (a) and (b). The energy functional associated with (1.1) is for all $u\in H^{\alpha}({\mathbb{R}}^{N})$,

[TABLE]

Finally we present some problems associated to (1.1) as follows.

The limit equation of (1.1) involving subcritical and critical terms is

[TABLE]

and its corresponding variational functional is

[TABLE]

The limit equation of (1.1) involving the Hardy term and critical Sobolev nonlinearity is

[TABLE]

and the corresponding variational functional is

[TABLE]

The limit equation of (1.1) involving critical Sobolev nonlinearity is

[TABLE]

and the corresponding variational functional is

[TABLE]

Define

[TABLE]

the Euler equation associated to (1.15) is (1.13). In particular it has been showed in Theorem 1.2 of [5] that for any positive solution $U_{\mu}(x)\in H^{\alpha}({\mathbb{R}}^{N})$ of (1.13), there exist two positive constants $c,C$ such that

[TABLE]

where

[TABLE]

and $\alpha_{\mu}\in(0,\frac{N-2\alpha}{2})$ is a suitable parameter whose explicit value will be determined as the unique solution to the following equation

[TABLE]

and $\varphi_{\alpha,N}$ is strictly increasing. That is

[TABLE]

All the positive solutions of (1.13) are of the form

[TABLE]

In particular, for $\mu=0$, it follows that (refer to [6] )

[TABLE]

where $C>0$ is a constant. These solutions $U_{0}^{\varepsilon,y}:=\varepsilon^{\frac{2\alpha-N}{2}}U_{0}(\frac{x-y}{\varepsilon})$ are also minimizers for the quotient

[TABLE]

Define

[TABLE]

[TABLE]

[TABLE]

[TABLE]

It is known that ${\mathcal{N}}\neq\emptyset$ since problem (1.12) has at least one positive solution if $N>2\alpha$ (see Theorem 1.3 in [28]) for $2<q<2^{*}$ and $\bar{k}>\lambda^{*}(\lambda^{*}>0$ is a positive constant definded in [28]).

The main result of our paper is as follows:

Theorem 1.1**.**

Suppose $a(x),\,k(x)$ satisfy (a) and (b), $\bar{k}>\lambda^{*}$, $2<q<2^{*}$, $0<\alpha<1$, $N>4\alpha$, $0<\mu<\phi_{\alpha,N}(\frac{N-4\alpha}{2})$. Assume that $\{u_{n}\}$ is a positive Palais-Smale sequence of I at level $d\geq 0$, then there exist sequences $\{y^{k}_{n}\}\subset{\mathbb{R}}^{N}\,(1\leq k\leq l_{1})$, $\{R^{j}_{n}\}\subset{\mathbb{R}}^{+}\,(1\leq j\leq l_{2}$), $\{\bar{R}^{i}_{n}\}\subset{\mathbb{R}}^{+},\,\,\,x_{n}^{i}\subset{\mathbb{R}}^{N}(1\leq i\leq l_{3}$) and $u_{k}\in H^{\alpha}({\mathbb{R}}^{N})\,(1\leq k\leq l_{1})(l_{1},l_{2},l_{3}\in{\mathbb{N}})$, $u\in H^{\alpha}({\mathbb{R}}^{N})$, such that up to a subsequence:

$\,\,\displaystyle d=I(u)+\sum^{l_{1}}_{j=1}I^{\infty}(u_{k})+l_{2}D_{\mu}+l_{3}D_{0};$**

[TABLE]

where $u$ and $u_{k}\,(1\leq k\leq l_{1})$ satisfy

[TABLE]

and

[TABLE]

In particular, if $u\not\equiv 0$, then $u$ is a weakly solution of (1.1). Note that the corresponding sum in (1.25) will be treated as zero if $l_{i}=0\,(i=1,2,3).$

Remarks:

1. Similar as Corollary 3.3 in [19], one can show that any Palais-Smale sequence for $I$ at a level which is not of the form $m_{1}D_{0}+m_{2}D_{\mu}+m_{3}J^{\infty}$, $m_{1},m_{2}\in{\mathbb{N}}\bigcup\{0\}$, gives rise to a non-trivial weak solution of equation (1.1).

2. In our non-compactness analysis, we prove that the blowing up positive Palais-Smale sequences can bear exactly three kinds of bubbles. Up to harmless constants, they are either of the form

[TABLE]

or

[TABLE]

or

[TABLE]

where $u$ is the solution of (1.12). For any Palais-Smale sequence ${u_{n}}$ for $I$, ruling out the above two bubbles yields the existence of a non-trivial weak solution of equation (1.1).

Using the compactness results and the Mountain Pass Theorem [1] we prove the following existence result.

Theorem 1.2**.**

Assume that $2<q<2^{*}$, $0<\alpha<1$, $N>4\alpha$, $0<\mu<\phi_{\alpha,N}(\frac{N-4\alpha}{2})$. If $a(x),k(x)$ satisfy (a), (b) and

[TABLE]

Then (1.1) has a nontrivial solution $u\in H^{\alpha}({\mathbb{R}}^{N})$ which satisfies

[TABLE]

This paper is organized as follows. In Section 2, we prove Theorem 1.1 by carefully analyzing the features of a positive Palais-Smale sequence for $I$. Theorem 1.2 is proved in Section 3 by applying Theorem 1.1 and the Mountain Pass Theorem. Finally we put some preliminaries in the last section as an appendix.

2 Non-compactness analysis

In this section, we prove Theorem 1.1 by using the Concentration-Compactness Principle and a delicate analysis of the Palais-Smale sequences of $I$. Firstly we give the following Lemmas.

Lemma 2.1**.**

Let $0<\alpha<N/2$, $\{u_{n}\}\subset{\dot{H}}^{\alpha}({\mathbb{R}}^{N})$ be a bounded sequence such that

[TABLE]

Then, up to subsequence, there exist two sequences $\{r_{n}\}\subset{\mathbb{R}}^{+}$ and $\{x_{n}\}\subset{\mathbb{R}}^{N}$ such that

[TABLE]

where

[TABLE]

Proof.

By Theorem 1 in [16],

[TABLE]

where $\frac{N-2\alpha}{N}\leq\theta<1$.

Then there exists a constant $c>0$ such that

[TABLE]

From (2.5), we may find $r_{n}>0$ and $x_{n}\in{\mathbb{R}}^{N}$ such that for $n$ large enough,

[TABLE]

Denote

[TABLE]

Since $\{u_{n}\}$ is bounded in $\dot{H}^{\alpha}({\mathbb{R}}^{N})$, from the scaling and translation invariance of $\dot{H}^{\alpha}({\mathbb{R}}^{N})$, then $\{\bar{u}_{n}\}$ is bounded in $\dot{H}^{\alpha}({\mathbb{R}}^{N})$, therefore, up to a subsequence (still denoted by $\bar{u}_{n}$),

[TABLE]

If $\frac{x_{n}}{r_{n}}$ is bounded, there exists a $\tilde{R}>1$ such that $B(\frac{x_{n}}{r_{n}},1)\subset B(0,\tilde{R})$, then

[TABLE]

If $|\frac{x_{n}}{r_{n}}|\rightarrow\infty$, then

[TABLE]

where $\tilde{R}>1$. Obviously we have $w\not\equiv 0$. From (2.8) and (2.9), Lemma 2.1 is complete. ∎

Lemma 2.2**.**

Assume $N>4\alpha,2<q<2^{*},0<\alpha<1$. Let $\{v_{n}\}\subset H^{\alpha}({\mathbb{R}}^{N})$ be a Palais-Smale sequence of $I$ at level $d_{1}$ and $v_{n}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$. If there exists two sequence $\{r_{n}\}\subset{\mathbb{R}}^{+}$ and $\{x_{n}\}\in{\mathbb{R}}^{N}$ $\text{ with }r_{n}\rightarrow 0,\,\frac{|x_{n}|}{r_{n}}\rightarrow\infty$ as $n\rightarrow\infty$ such that $\bar{v}_{n}(x):=r_{n}^{\frac{N-2\alpha}{2}}v_{n}(r_{n}x+x_{n})$ converges weakly in $\dot{H}^{\alpha}({\mathbb{R}}^{N})$ and almost everywhere to some $v_{0}\in\dot{H}^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$ with $v_{0}\not\equiv 0$, then $v_{0}$ solves (1.14), the sequence $z_{n}(x):=v_{n}(x)-v_{0}(\frac{x-x_{n}}{r_{n}})r_{n}^{\frac{2\alpha-N}{2}}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})$ and $z_{n}(x)$ is a Palais-Smale sequence of $I$ at level $d_{1}-I_{0}(v_{0})$.

Proof.

First, we prove that $v_{0}$ solves (1.14) and $I(z_{n})=I(v_{n})-I_{0}(v_{0})$. Fix a ball $B(0,r)$ and a test function $\phi\in C^{\infty}_{0}(B(0,r))$. Since

[TABLE]

it implies

[TABLE]

where $\phi_{n}=r_{n}^{-\frac{N-2\alpha}{2}}\phi(\frac{x-x_{n}}{r_{n}})$. The last equality in (2.10) holds since

[TABLE]

[TABLE]

Thus $v_{0}$ is a nontrival critical point of $I_{0}$. By Lemma 4.5, (1.20) and the fact $N>4\alpha$, it follows

[TABLE]

which implies that $v_{0}\in L^{2}({\mathbb{R}}^{N})$. Let

[TABLE]

From (2.11), $v_{0}\in L^{p}({\mathbb{R}}^{N})$ for all $p\in[2,2^{*})$. Then it follows

[TABLE]

Thus $z_{n}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$. Now we prove that $\{z_{n}\}$ is a Palais-Smale sequence of $I$ at level $d_{1}-I_{0}(v_{0})$. By the Brézis-Lieb Lemma and the weak convergence, similar to Lemma 4.6 in the Appendix, we can prove that

[TABLE]

[TABLE]

as $n\rightarrow\infty$. It completes the proof. ∎

Lemma 2.3**.**

Assume $N>4\alpha,2<q<2^{*},0<\alpha<1$, $0<\mu<\phi_{\alpha,N}(\frac{N-4\alpha}{2})$. Let $\{v_{n}\}\subset H^{\alpha}({\mathbb{R}}^{N})$ be a Palais-Smale sequence of $I$ at level $d_{1}$ and $v_{n}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$. If there exists a sequence $\{r_{n}\}\subset{\mathbb{R}}^{+},\text{ with }r_{n}\rightarrow 0$ as $n\rightarrow\infty$ such that $\bar{v}_{n}(x):=r_{n}^{\frac{N-2\alpha}{2}}v_{n}(r_{n}x)$ converges weakly in $\dot{H}^{\alpha}({\mathbb{R}}^{N})$ and almost everywhere to some $v_{0}\in\dot{H}^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$ with $v_{0}\not\equiv 0$, then $v_{0}$ solves (1.13), the sequence $z_{n}(x):=v_{n}(x)-v_{0}(\frac{x}{r_{n}})r_{n}^{\frac{2\alpha-N}{2}}\rightharpoonup 0\text{ in }\dot{H}^{\alpha}({\mathbb{R}}^{N})$ and $z_{n}(x)$ is a Palais-Smale sequence of $I$ at level $d_{1}-I_{\mu}(v_{0})$.

Proof.

First, we prove that $v_{0}$ solves (1.13) and $I(z_{n})=I(v_{n})-I_{\mu}(v_{0})$. Fix a ball $B(0,r)$ and a test function $\phi\in C^{\infty}_{0}(B(0,r))$. Since

[TABLE]

it implies

[TABLE]

where $\phi_{n}=r_{n}^{-\frac{N-2\alpha}{2}}\phi(\frac{x}{r_{n}})$. The last equality in (2.14) holds since

[TABLE]

[TABLE]

Thus $v_{0}$ is a nontrival critical point of $I_{\mu}$. Noting the fact $N>4\alpha$, $\mu<\phi_{\alpha,N}(\frac{N-4\alpha}{2})$ and $\phi_{\alpha,N}$ is a strictly increasing, it follows

[TABLE]

then by Lemma 4.5 and (1.16), it follows

[TABLE]

which implies that $v_{0}\in L^{2}({\mathbb{R}}^{N})$. Let

[TABLE]

From (2.15) and $v_{0}\in L^{p}({\mathbb{R}}^{N})$ for all $p\in[2,2^{*})$, it follows

[TABLE]

Thus $z_{n}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$. Now we prove that $\{z_{n}\}$ is a Palais-Smale sequence of $I$ at level $d_{1}-I_{\mu}(v_{0})$. By the Brézis-Lieb Lemma and the weak convergence, similar to Lemma 4.6 in the Appendix, we can prove that

[TABLE]

[TABLE]

as $n\rightarrow\infty$. It completes the proof. ∎

Proof of Theorem 1.1. By Lemma 4.3 in the appendix, we can assume that $\{u_{n}\}$ is bounded. Up to a subsequence, $\text{ let }n\rightarrow\infty$, we assume that

[TABLE]

Denote $v_{n}(x)=u_{n}(x)-u(x)$, then $\{v_{n}\}$ is a Palais-Smale sequence of $I$ and $v_{n}\rightharpoonup 0$ in $H^{\alpha}({\mathbb{R}}^{N})$ and

[TABLE]

Then by Lemma 4.6 we know that

[TABLE]

Without loss of generality, we may assume that

[TABLE]

In fact if $l=0$, Theorem 1.1 is proved for $l_{1}=0,l_{2}=0,l_{3}=0$.

Step 1: getting rid of the blowing up bubbles caused by unbounded domains.

Suppose there exists a constant $0<\delta<\infty$ such that

[TABLE]

By interpolation inequality, it follows

[TABLE]

where $0<\lambda<1$. Thus there exists a $\tilde{\delta}>0$ such that

[TABLE]

By Lemma 4.1, there exists a subsequence still denoted by $\{v_{n}\}$, such that one of the following two cases occurs.

i) Vanish occurs.

[TABLE]

By Lemma 4.2, (4.7) and Sobolev inequality, it follows

[TABLE]

which contradicts (2.26).

ii) Nonvanish occurs.

There exist $\beta>0,\,0<\bar{R}<\infty,\,\{y_{n}\}\subset{\mathbb{R}}^{N}\text{ such that }$

[TABLE]

We claim that $|y_{n}|\rightarrow\infty$ as $n\rightarrow\infty$. Otherwise, if there exists a constant $M>0$ such that $|y_{n}|\leq M$, then we can choose a $R_{2}>0$ large enough such that

[TABLE]

which contradicts (2.27).

To proceed, we first construct the Palais-Smale sequences of $I^{\infty}$.

Denote $\bar{v}_{n}(x)=v_{n}(x+y_{n})$. Since $\|\bar{v}_{n}\|_{H^{\alpha}({\mathbb{R}}^{N})}=\|v_{n}\|_{H^{\alpha}({\mathbb{R}}^{N})}\leq c$, without loss of generality, we assume that $\text{ as }n\rightarrow\infty$,

[TABLE]

Then $\forall\phi\in C^{\infty}_{0}(B(0,r))\text{ as }n\rightarrow\infty$,

[TABLE]

Similarly we have

[TABLE]

Since $v_{n}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})$ and $\displaystyle\lim_{n\rightarrow\infty}a(x+y_{n})=\bar{a}$, we have as $n\rightarrow\infty$,

[TABLE]

and

[TABLE]

that is,

[TABLE]

Similarly we have

[TABLE]

Recall that $v_{n}$ is a Palais-Smale sequence of $I$, by (2.29) and (2.31)-(2.33) we have

[TABLE]

This shows that $v_{0}$ is a weak solution of (1.12).

We claim that $v_{0}\not\equiv 0$. From (2.26), we may assume that there exists a sequence $\{y_{n}\}$ satisfying (2.27) and

[TABLE]

where $b>0$ is a constant. If $v_{0}\equiv 0$, we have

[TABLE]

which contradicts (2.35).

Denote $z_{n}(x)=v_{n}(x)-v_{0}(x-y_{n})$. Since

[TABLE]

where the last equality but one is a result of (2.31), therefore, as $n\rightarrow\infty$,

[TABLE]

Hence $z_{n}\rightharpoonup 0\text{ in }H^{\alpha}({\mathbb{R}}^{N})\text{ as }n\rightarrow\infty$, and $z_{n}$ is a Palais-Smale sequence of $I$. From (4.7) in Lemma 4.4, it follows $\|v_{0}^{-}\|_{H^{\alpha}}=0$, that is $v_{0}\geq 0$ a.e. in ${\mathbb{R}}^{N}$. Then by Brezis-Lieb Lemma and (4.7), there exists a constant $c>0$ such that

[TABLE]

where the last inequality follows from the fact $v_{0}\not\equiv 0$. If $\|z_{n}\|_{L^{q}({\mathbb{R}}^{N})}\rightarrow\delta_{2}>0\text{ as }n\rightarrow\infty$, from (2.38) and the boundedness of $\|v_{n}\|_{L^{q}}$, then one can repeat Step 1 for finite times ($l_{1}$ times). Thus we obtain a new Palais-Smale sequence of $I$, without loss of generality still denoted by $v_{n}$, such that

[TABLE]

[TABLE]

[TABLE]

as $n\rightarrow\infty$.

Step 2:  Getting rid of the blowing up bubbles caused by the critical terms.

Suppose there exists $0<\delta<\infty$ such that

[TABLE]

It follows from Lemma 2.1 that there exist two sequences $\{r_{n}\}\subset{\mathbb{R}}^{+}$ and $\{x_{n}\}\subset{\mathbb{R}}^{N}$, such that

[TABLE]

where

[TABLE]

Now we claim that $r_{n}\rightarrow 0\text{ as }n\rightarrow\infty.$ In fact there exists a $R_{1}>0$ such that

[TABLE]

From the Sobolev compact embedding, (2.18), (2.43) and (2.45), we have that for all $r>0$,

[TABLE]

[TABLE]

[TABLE]

If $|\frac{x_{n}}{r_{n}}|\rightarrow\infty$, then there exists a constant $\bar{c}$ such that

[TABLE]

Then from (2.41) (2.47) and the fact $q<2^{*}$, it follows that $r_{n}\rightarrow 0$. Similarly, if $\frac{x_{n}}{r_{n}}$ is bounded, we also have that $r_{n}\rightarrow 0$.

For the case that $\frac{x_{n}}{r_{n}}$ is bounded and $\bar{v}_{n}(x)=r_{n}^{\frac{N-2\alpha}{2}}v_{n}(r_{n}x)$, define $z_{n}(x)=v_{n}(x)-v_{0}(\frac{x}{r_{n}})r_{n}^{\frac{2\alpha-N}{2}}$. It follows from Lemma 2.2 that $\{z_{n}\}$ is a Palais-Smale sequence of $I$ satisfying

[TABLE]

and $z_{n}\rightharpoonup 0$ in $H^{\alpha}({\mathbb{R}}^{N})$. Since $v_{0}$ satisfies (1.13), from Lemma 4.5, (1.19) and (1.21) there exists $\varepsilon_{1}>0$ such that

[TABLE]

Let $R_{n}^{1}=r_{n}\varepsilon_{1}$, from (2.49), it follows

[TABLE]

with $R^{1}_{n}\rightarrow 0$. Then from (2.23) it follows

[TABLE]

with $R^{1}_{n}\rightarrow 0$. From (4.7), we have that $z_{n}\geq 0,\text{ a.e. in }{\mathbb{R}}^{N}$. From Lemma 4.7, let $a=v_{n},b=U_{\mu}^{R^{1}_{n}}$, it follows

[TABLE]

where $\Omega=\{x|z_{n}(x)\geq 0\}\bigcap{\mathbb{R}}^{N}$.

For the case that $|\frac{x_{n}}{r_{n}}|\rightarrow\infty$ and $\bar{v}_{n}(x)=r_{n}^{\frac{N-2\alpha}{2}}v_{n}(r_{n}x+x_{n})$, define $z_{n}(x)=v_{n}(x)-v_{0}(\frac{x-x_{n}}{r_{n}})r_{n}^{\frac{2\alpha-N}{2}}$. It follows from Lemma 2.3 that $\{z_{n}\}$ is a Palais-Smale sequence of $I$ satisfying

[TABLE]

and $z_{n}\rightharpoonup 0$ in $H^{\alpha}({\mathbb{R}}^{N})$. Since $v_{0}$ satisfies (1.14), from Lemma 4.5, (1.19) and (1.22) there exists $\varepsilon_{1}>0$ such that

[TABLE]

Let $\bar{R}_{n}^{1}=r_{n}\varepsilon_{1}$ and $x_{n}^{1}=y+\varepsilon_{1}x_{n}$, from (2.54), it follows

[TABLE]

with $\bar{R}^{1}_{n}\rightarrow 0$. Then from (2.23) it follows

[TABLE]

with $\bar{R}^{1}_{n}\rightarrow 0$. Similar to (2.52), it follows

[TABLE]

If still there exists a $\bar{\delta}>0,\text{ such that }$

[TABLE]

then repeat the previous argument. From (2.52) and the fact

[TABLE]

we deduce that the iteration must stop after finite times. That is to see, there exist nonnegative constants $l_{2},l_{3}$ and a new Palais-Smale sequence of $I$, (without loss of generality) denoted by $\{v_{n}\}$, such that as $n\rightarrow\infty$,

[TABLE]

with $R^{j}_{n}\rightarrow 0$, $\bar{R}^{i}_{n}\rightarrow 0$ and $|x_{n}^{i}/\bar{R}_{n}^{i}|\rightarrow\infty$.

[TABLE]

and

[TABLE]

Then from the fact $<I^{\prime}(v_{n}),v_{n}>=o(1)$, it follows

[TABLE]

as $n\rightarrow\infty$. From (2.61), it gives that

[TABLE]

From (2.58)-(2.62), the proof of Theorem 1.1 is complete.

3 Proof of Theorem 1.2

Now we are ready to prove Theorem 1.2 by Mountain Pass Theorem [1] and Theorem 1.1.

Proof of Theorem 1.2: From

[TABLE]

we deduce that for a fixed $u\not\equiv 0$ in $H^{\alpha}({\mathbb{R}}^{N})$, $I(tu)\rightarrow-\infty$ if $t\rightarrow\infty$. Since

[TABLE]

we have

[TABLE]

Hence, there exists $r_{0}>0$ small such that I(u)\Bigl{|}_{\partial B(0,r_{0})}\geq\rho>0 for $q,\,{2^{*}}>2$.

As a consequence, $I(u)$ satisfies the geometry structure of Mountain-Pass Theorem. Now define

[TABLE]

where $\Gamma=\{\gamma\in C([0,1],H^{\alpha}({\mathbb{R}}^{N})):\gamma(0)=0,\gamma(1)=\psi_{0}\in H^{\alpha}({\mathbb{R}}^{N})\}$ with $I(t\psi_{0})\leq 0$ for all $t\geq 1$.

To complete the proof of Theorem 1.2, we need to verify that $I(u)$ satisfies the local Palais-Smale conditions. According to Remarks 1), we only need to verify that

[TABLE]

Set $\displaystyle v_{\varepsilon}(x)=\frac{U_{\mu}^{\varepsilon}(x)}{(\int_{{\mathbb{R}}^{N}}{|U_{\mu}^{\varepsilon}(x)|^{{2^{*}}}}dx)^{1/{2^{*}}}}$. We claim

[TABLE]

In fact, from (1.20) it is easy to calculate the following estimates

[TABLE]

[TABLE]

[TABLE]

Since $2^{*}>q>2$ we have

[TABLE]

Denote $t_{\varepsilon}$ the attaining point of $\displaystyle\max_{t>0}I(tv_{\varepsilon})$, similar to the proof of Lemma 3.5 in [3] we can prove that $t_{\varepsilon}$ is uniformly bounded. In fact, we consider the function

[TABLE]

Since $\displaystyle\lim_{t\rightarrow+\infty}h(t)=-\infty$ and $h(t)>0$ when $t$ is closed to [math], then $\displaystyle\max_{t>0}h(t)$ is attained for $t_{\varepsilon}>0$. From the fact $\displaystyle\int_{{\mathbb{R}}^{N}}{(v_{\varepsilon}(x))^{{2^{*}}}}dx=1$, it follows

[TABLE]

Since $k(x)>0$, from (3.3) and (3.4) for $\varepsilon$ sufficiently small, we have

[TABLE]

Then

[TABLE]

Choosing $\varepsilon>0$ small enough, by (3.3)-(3.5), there exists a constant $\alpha_{1}>0$ such that $t_{\varepsilon}>\alpha_{1}>0$. Combining this with (3.9), it implies that $t_{\varepsilon}$ is bounded for $\varepsilon>0$ small enough.

Hence, for $\varepsilon>0$ small,

[TABLE]

This completes the proof of (3.2). By the definition of $c^{*}$, we have $c^{*}<\frac{\alpha}{N}S_{\alpha,\mu}^{\frac{N}{2\alpha}}$.

Next we verify

[TABLE]

Let $\{u_{0}\}$ be the minimizer of $J^{\infty},\,I^{\infty}(u_{0})=J^{\infty}$ and

[TABLE]

Let

[TABLE]

[TABLE]

Thus $g^{\prime}(t)\geq 0$ if $t\in(0,1)$; $g^{\prime}(t)\leq 0$ if $t\geq 1$. Then

[TABLE]

Since there exists a $t_{0}>0$ such that $\displaystyle\sup_{t\geq 0}I(tu_{0})=I(t_{0}u_{0})$, from (3.12) and the assumptions of $a(x)\text{ and }k(x)$, we have

[TABLE]

It proves (3.11). By (3.2) and (3.11) we have (3.1). Then the proof is completed.

4 Appendix

In this appendix, we give some lemmas and detailed proofs for the convenience of the reader.

Lemma 4.1**.**

(Lemma 2.1, [25]) Let $\{\rho_{n}\}_{n\geq 1}$ be a sequence in $L^{1}({\mathbb{R}}^{N})$ satisfying

[TABLE]

where $\lambda>0$ is fixed. Then there exists a subsequence $\{\rho_{n_{k}}\}$ satisfying one of the following two possibilities:

(1) (Vanishing):

[TABLE]

(ii) (Nonvanishing): $\exists\alpha>0,R<+\infty$ and $\{y_{k}\}\subset{\mathbb{R}}^{N}$ such that

[TABLE]

Lemma 4.2**.**

(Lemma 2.2, [17])If $\{u_{n}\}$ is bounded in $H^{\alpha}({\mathbb{R}}^{N})$ and for some $R>0$, we have

[TABLE]

Then $u_{n}\rightarrow 0$ in $L^{q}({\mathbb{R}}^{N})$, for $2<q<\frac{2N}{N-2\alpha}$.

Lemma 4.3**.**

Let $\{u_{n}\}$ be a Palais-Smale sequence of $I$ at level $d\in\mathbb{R}$. Then $d\geq 0$ and $\{u_{n}\}$ $\subset$ $H^{\alpha}({\mathbb{R}}^{N})$ is bounded. Moreover, every Palais-Smale sequence for $I$ at a level zero converges strongly to zero.

Proof.

Since $a(x)\geq 0$, $\bar{a}>0$ and $\displaystyle\inf_{x\in{\mathbb{R}}^{N}}a(x)=\hat{a}>0$, we have

[TABLE]

and hence for $2<q<2^{*}$

[TABLE]

It follows that $\{u_{n}\}$ is bounded in $H^{\alpha}({\mathbb{R}}^{N})$ for $2<q<2^{*}$. Since

[TABLE]

we have $d\geq 0$. Suppose now that $d=0$, we obtain from the above inequality that

[TABLE]

∎

Lemma 4.4**.**

Let $\{u_{n}\}$ be a Palais-Smale sequence of $I$ at level $d\in\mathbb{R}$ and $u^{+}_{n}=\max\{u_{n},0\}$. Then $\{u_{n}^{+}\}$ is also a Palais-Smale sequence of $I$ at level $d$.

Proof.

By the definition of $I$ we have that as $n\rightarrow\infty$

[TABLE]

and

[TABLE]

Taking $\phi=-u_{n}^{-}=\min\{u_{n},0\}$, from the fact

[TABLE]

we have

[TABLE]

from (4.6) and the fact $u^{+}_{n}(x)\geq 0,\,u^{-}_{n}(x)\geq 0,\,a(x)>0$, then

[TABLE]

and

[TABLE]

Then from (4.5) and (4.7)-(4.8), we have

[TABLE]

That is

[TABLE]

Thus

[TABLE]

and

[TABLE]

as $n\rightarrow\infty$. This complete the proof. ∎

Lemma 4.5**.**

All nontrivial critical points of $I_{\mu}$ are the positive solutions of (1.14).

Proof.

Let $u\not\equiv 0$ and $u\in H^{\alpha}({\mathbb{R}}^{N})$ be a nontrivial critical point of $I_{\mu}$. First, arguing as in the proof of Lemma 4.4 (similar to (4.6) and (4.7)), we can obtain that $\|u^{-}\|_{H^{\alpha}}=0$ which gives that

[TABLE]

Then for any $x_{0}\in{\mathbb{R}}^{N}$,

[TABLE]

from Proposition 2.2.6 in [22], we have $u$ is lower semicontinous in $B(x_{0},1)$. Combining this with (4.11), it follows $u(x_{0})\geq 0$. Then $u(x)\geq 0$ pointwise in ${\mathbb{R}}^{N}$.

Next we claim that $u>0$ in ${\mathbb{R}}^{N}$. Otherwise there exist $x_{1}\in{\mathbb{R}}^{N}$ such that $u(x_{1})=0$. Since $u$ is lower semicontinuous in $\overline{B(x_{1},1/2)}$, from Proposition 2.2.8 in [22], it follows $u\equiv 0\text{ in }{\mathbb{R}}^{N}$. This contradicts the assumption $u$ is nontrivial. ∎

Let $\{u_{n}\}$ be a Palais-Smale sequence at level $d$. Up to a subsequence, we assume that

[TABLE]

Obviously, we have $I^{\prime}(u)=0$. Let $v_{n}(x)=u_{n}(x)-u(x)$, as $n\rightarrow\infty$,

[TABLE]

As a consequence, we have the following Lemma.

Lemma 4.6**.**

$\{v_{n}\}$* is a Palais-Smale sequence for $I$ at level $d_{0}=d-I(u)$.*

Proof.

For $\phi(x)\in C^{\infty}_{0}({\mathbb{R}}^{N})$, there exists a $B(0,r)$ such that $\displaystyle\mathrm{supp}\phi\subset B(0,r)$. Then$\text{ as }n\rightarrow\infty$,

[TABLE]

[TABLE]

where $\frac{N}{N-2\alpha}<\bar{q}<2^{*}$.

By (4.13), (4.16) and (4.17), we have $\langle\phi,I^{\prime}(v_{n})\rangle=o(1)\text{ as }n\rightarrow\infty$. Then similar to (4.7), we have

[TABLE]

By Sobolev inequality, (4.7) and (4.18) it follows

[TABLE]

Then by the Brézis-Lieb Lemma in [1] as $n\rightarrow\infty$, we have

[TABLE]

Similarly

[TABLE]

[TABLE]

Hence from (4.19)-(4.21), it follows $I(v_{n})=I(u_{n})-I(u)+o(1)=d-I(u)+o(1)$. ∎

Lemma 4.7**.**

Assume $t\geq b>0\text{ and }q>1$, then

[TABLE]

Proof.

Let $f(t)=t^{q}-(t-b)^{q}$, it follows

[TABLE]

Then $f(t)=t^{q}-(t-b)^{q}\geq f(b)=b^{q}.$ ∎

Acknowledgement   The research was supported by the Natural Science Foundation of China (11271141) and the China Scholarship Council (201508440330).