Open Access

Existence of positive solutions for a critical nonlinear Schrödinger equation with vanishing or coercive potentials

Boundary Value Problems20132013:201

DOI: 10.1186/1687-2770-2013-201

Received: 2 May 2013

Accepted: 22 August 2013

Published: 8 September 2013

Abstract

In this paper we investigate the existence of positive solutions for the following nonlinear Schrödinger equation:

u + V ( x ) u = K ( x ) | u | p 2 u in  R N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equa_HTML.gif

where V ( x ) a | x | b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq1_HTML.gif and K ( x ) μ | x | s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq2_HTML.gif as | x | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq3_HTML.gif with 0 < a , μ < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq4_HTML.gif, b < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq5_HTML.gif, b 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq6_HTML.gif, 0 < s b < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq7_HTML.gif and p = 2 ( N 2 s / b ) / ( N 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq8_HTML.gif.

MSC:35J20, 35J60.

Keywords

semilinear Schrödinger equation vanishing or coercive potentials

1 Introduction and statement of results

In this paper, we consider the following semilinear elliptic equation:
u + V ( x ) u = K ( x ) | u | p 2 u in  R N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ1_HTML.gif
(1.1)
where N 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq9_HTML.gif. The exponent
p = 2 ( N 2 s b ) / ( N 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ2_HTML.gif
(1.2)
with the real numbers b and s satisfying
b < 2 , b 0 , 0 < s b < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ3_HTML.gif
(1.3)

By this definition, 2 < p < 2 : = 2 N / ( N 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq10_HTML.gif.

With respect to the functions V and K, we assume that

(A1) V , K C ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq11_HTML.gif for every x R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq12_HTML.gif, V ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq13_HTML.gif and K ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq14_HTML.gif.

(A2) There exist 0 < a < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq15_HTML.gif and 0 < μ < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq16_HTML.gif such that
lim | x | | x | b V ( x ) = a and lim | x | | x | s K ( x ) = μ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ4_HTML.gif
(1.4)
A typical example for Eq. (1.1) with V and K satisfying (A1) and (A2) is the equation
u + a ( 1 + | x | ) b u = μ ( 1 + | x | ) s | u | p 2 u in  R N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ5_HTML.gif
(1.5)

When 0 < b < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq17_HTML.gif, the potentials are vanishing at infinity and when b < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq18_HTML.gif, the potentials are coercive.

Equation (1.1) arises in various applications, such as chemotaxis, population genetics, chemical reactor theory and the study of standing wave solutions of certain nonlinear Schrödinger equations. Therefore, they have received growing attention in recent years (one can see, e.g., [16] and [710] for reference).

Under the above assumptions, Eq. (1.1) has a natural variational structure. For an open subset Ω in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif, let C 0 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq20_HTML.gif be the collection of smooth functions with a compact support set in Ω. Let E be the completion of C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq21_HTML.gif with respect to the inner product
( u , v ) E = R N u v d x + R N V ( x ) u v d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equb_HTML.gif
From assumptions (A1) and (A2), we deduce that
( R N | u | 2 ( 1 + | x | ) b d x ) 1 / 2 and ( R N V ( x ) | u | 2 d x ) 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equc_HTML.gif
are two equivalent norms in the space
L V 2 ( R N ) = { u  is measurable in  R N | R N V ( x ) | u | 2 d x < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equd_HTML.gif
Therefore, there exists B 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq22_HTML.gif such that
( R N | u | 2 ( 1 + | x | ) b d x ) 1 / 2 B 1 ( R N V ( x ) | u | 2 d x ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Eque_HTML.gif
Moreover, assumptions (A1) and (A2) imply that there exists B 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq23_HTML.gif such that
K ( x ) B 2 ( 1 + | x | ) s , x R N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equf_HTML.gif
Then, by the Hölder and Sobolev inequalities (see, e.g., [[11], Theorem 1.8]), we have, for every u C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq24_HTML.gif,
( R N K ( x ) | u | p d x ) 1 p C ( R N | u | p ( 1 + | x | ) s d x ) 1 p = C ( R N | u | 2 s b ( 1 + | x | ) s | u | p 2 s b d x ) 1 p C ( R N | u | 2 ( 1 + | x | ) b d x ) s p b ( R N | u | 2 d x ) 1 p ( 1 s b ) C ( R N | u | 2 ( 1 + | x | ) b d x ) s p b ( R N | u | 2 d x ) 2 2 p ( 1 s b ) = C ( R N | u | 2 ( 1 + | x | ) b d x ) 1 2 2 s p b ( R N | u | 2 d x ) 1 2 ( 1 2 s p b ) C ( R N V ( x ) | u | 2 d x ) 1 2 2 s p b ( R N | u | 2 d x ) 1 2 ( 1 2 s p b ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equg_HTML.gif
where C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif is a constant independent of u. It follows that there exists a constant C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq26_HTML.gif such that
( R N K ( x ) | u | p d x ) 1 / p C ( R N | u | 2 d x ) 1 / 2 + C ( R N V ( x ) | u | 2 d x ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equh_HTML.gif
This implies that E can be embedded continuously into the weighted L p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq27_HTML.gif-space
L K p ( R N ) = { u  is measurable in  R N | R N K ( x ) | u | p d x < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equi_HTML.gif
Then the functional
Φ ( u ) = 1 2 u E 2 1 p R N K ( x ) | u | p d x , u E , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equj_HTML.gif

is well defined in E. And it is easy to check that Φ is a C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq28_HTML.gif functional and the critical points of Φ are solutions of (1.1) in E.

In a recent paper [12], Alves and Souto proved that the space E can be embedded compactly into L K p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq29_HTML.gif if 0 < b < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq17_HTML.gif and 2 ( N 2 s / b ) / ( N 2 ) < p < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq30_HTML.gif and Φ satisfies the Palais-Smale condition consequently. Then, by using the mountain pass theorem, they obtained a nontrivial solution for Eq. (1.1). Unfortunately, when p = 2 ( N 2 s / b ) / ( N 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq8_HTML.gif, the embedding of E into L K p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq29_HTML.gif is not compact and Φ no longer satisfies the Palais-Smale condition. Therefore, the ‘standard’ variational methods fail in this case. From this point of view, p = 2 ( N 2 s / b ) / ( N 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq8_HTML.gif should be seen as a kind of critical exponent for Eq. (1.1). If the potentials V and K are restricted to the class of radially symmetric functions, ‘compactness’ of such a kind is regained and ‘standard’ variational approaches work (see [5] and [6]). However, this method does not seem to apply to the more general equation (1.1) where K and V are non-radially symmetric functions.

It is not easy to deal with Eq. (1.1) directly because there are no known approaches that can be used directly to overcome the difficulty brought by the loss of compactness. However, in this paper, through an interesting transformation, we find an equivalent equation for Eq. (1.1) (see Eq. (2.9) in Section 2). This equation has the advantages that its Palais-Smale sequence can be characterized precisely through the concentration-compactness principle (see Theorem 5.1), and it possesses partial compactness (see Corollary 5.8). By means of these advantages, a positive solution for this equivalent equation and then a corresponding positive solution for Eq. (1.1) are obtained.

Before stating our main result, we need to give some definitions.

Let
V ( x ) = | x | 2 b 2 b V ( | x | b 2 b x ) + C b | x | 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ6_HTML.gif
(1.6)
where
C b = b 4 ( 1 b 4 ) ( N 2 ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ7_HTML.gif
(1.7)
and
K ( x ) = | x | 2 s 2 b K ( | x | b 2 b x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ8_HTML.gif
(1.8)
Let H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif be the Sobolev space endowed with the norm and the inner product
u = ( R N | u | 2 d x + R N u 2 d x ) 1 / 2 and ( u , v ) = R N ( u v + u v ) d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equk_HTML.gif
respectively, and let L p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq32_HTML.gif be the function space consisting of the functions on R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif that are p-integrable. Since 2 < p < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq33_HTML.gif, H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif can be embedded continuously into L p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq32_HTML.gif. Therefore, the infimum
inf v H 1 ( R N ) { 0 } R N | v | 2 d x + a R N v 2 d x ( R N | v | p d x ) 2 / p > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ9_HTML.gif
(1.9)

We denote this infimum by S p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq34_HTML.gif.

Our main result reads as follows.

Theorem 1.1 Under assumptions (A1) and (A2), if b, s and p satisfy (1.3) and (1.2) and
inf u H 1 ( R N ) { 0 } R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x + R N V ( x ) | u | 2 d x ( R N K ( x ) | u | p d x ) 2 / p < ( 1 b / 2 ) p 2 p μ 2 p S p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ10_HTML.gif
(1.10)

then Eq. (1.1) has a positive solution u E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq35_HTML.gif.

Remark 1.2 We should emphasize that condition (1.10) can be satisfied in many situations. For r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq36_HTML.gif, let R r = { x R N r / 2 < | x | < r } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq37_HTML.gif and H 0 1 ( R r ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq38_HTML.gif be the closure of C 0 ( R r ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq39_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif. Under assumptions (A1) and (A2), we have
inf u H 0 1 ( R r ) { 0 } R r | u | 2 d x ( R r K ( x ) | u | p d x ) 2 / p 0 , as  r + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equl_HTML.gif
Then, for any ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq40_HTML.gif, there exist r ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq41_HTML.gif and u ϵ H 0 1 ( R r ) { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq42_HTML.gif such that
R r | u ϵ | 2 d x ( R r K ( x ) | u ϵ | p d x ) 2 / p < ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equm_HTML.gif
It follows from this inequality and R r | x u ϵ | 2 | x | 2 d x R r | u ϵ | 2 d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq43_HTML.gif that if sup R r V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq44_HTML.gif is small enough such that
R r V ( x ) | u ϵ | 2 d x ( R r K ( x ) | u ϵ | p d x ) 2 / p < ϵ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equn_HTML.gif
then
R r | u ϵ | 2 d x + ( b 2 4 b ) R r | x u ϵ | 2 | x | 2 d x + R r V ( x ) | u ϵ | 2 d x ( R r K ( x ) | u ϵ | p d x ) 2 / p < ( 2 + | b 2 4 b | ) ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equo_HTML.gif

This implies that (1.10) is satisfied if ϵ is chosen such that ( 2 + | b 2 4 b | ) ϵ < ( 1 b / 2 ) p 2 p μ 2 p S p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq45_HTML.gif.

Notations Let X be a Banach space and φ C 1 ( X , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq46_HTML.gif. We denote the Fréchet derivative of φ at u by φ ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq47_HTML.gif. The Gateaux derivative of φ is denoted by φ ( u ) , v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq48_HTML.gif, u , v X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq49_HTML.gif. By → we denote the strong and by the weak convergence. For a function u, u + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq50_HTML.gif denotes the functions max { u ( x ) , 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq51_HTML.gif. The symbol δ i j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq52_HTML.gif denotes the Kronecker symbol:
δ i j = { 1 , i = j , 0 , i j . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equp_HTML.gif

We use o ( h ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq53_HTML.gif to mean o ( h ) / | h | 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq54_HTML.gif as | h | 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq55_HTML.gif.

2 An equivalent equation for Eq. (1.1)

For x R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq12_HTML.gif, let y = | x | b / 2 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq56_HTML.gif. To u, a C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq28_HTML.gif function in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif, we associate a function v, a C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq28_HTML.gif function in R N { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq57_HTML.gif, by the transformation
u ( x ) = | x | b 4 ( N 2 ) v ( | x | b 2 x 1 , , | x | b 2 x N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ11_HTML.gif
(2.1)
Lemma 2.1 Under the above assumptions,
x u ( x ) = | y | b ( N + 2 ) 2 ( 2 b ) ( i , j = 1 N y j ( A i j ( y ) v y i ) C b | y | 2 v ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ12_HTML.gif
(2.2)
where
A i j ( y ) = δ i j + ( b 2 4 b ) y i y j | y | 2 , i , j = 1 , , N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ13_HTML.gif
(2.3)
Proof Let r = | x | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq58_HTML.gif. By direct computations,
u x i = r b ( N 2 ) 4 b 2 v y i b 2 r b ( N 2 ) 4 b 2 2 x i j = 1 N x j v y j b 4 ( N 2 ) r b ( N 2 ) 4 2 x i v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ14_HTML.gif
(2.4)
and
2 u x i 2 = b N 2 r b ( N 2 ) 4 b 2 2 x i v y i + r b ( N 2 ) 4 b 2 v y i 2 b r b ( N 2 ) 4 b 2 j = 1 N x j x i 2 v y j y i + ( b 2 4 ( N 1 ) + b ) r b ( N 2 ) 4 b 2 4 x i 2 j = 1 N x j v y j b 2 r b ( N 2 ) 4 b 2 2 j = 1 N x j v y j + b 2 4 r b ( N 2 ) 4 b 4 x i 2 j , k = 1 N x j x k 2 v y j y k + b 4 ( N 2 ) ( b 4 ( N 2 ) + 2 ) r b 4 ( N 2 ) 4 x i 2 v b 4 ( N 2 ) r b 4 ( N 2 ) 2 v . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equq_HTML.gif
Then
x u = i = 1 N 2 u x i 2 = r b ( N 2 ) 4 b { y v + ( b 2 4 b ) r 2 i , j = 1 N x i x j 2 v y i y j + ( b 2 4 b ) ( N 1 ) r b 2 2 i = 1 N x i v y i b 4 ( 1 b 4 ) ( N 2 ) 2 r b 2 v } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ15_HTML.gif
(2.5)
Since y = | x | b / 2 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq56_HTML.gif, we have r = | y | 2 2 b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq59_HTML.gif and x i = | y | b 2 b y i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq60_HTML.gif, 1 i N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq61_HTML.gif. Then
r 2 i , j = 1 N x i x j 2 v y i y j + ( N 1 ) r b 2 2 i = 1 N x i v y i = | y | 2 i , j = 1 N y i y j 2 v y i y j + ( N 1 ) | y | 2 i = 1 N y i v y i = i , j = 1 N y j ( y i y j | y | 2 v y i ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ16_HTML.gif
(2.6)
Substituting (2.6) and r = | y | 2 2 b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq59_HTML.gif into (2.5) results in
x u ( x ) = | y | b ( N + 2 ) 2 ( 2 b ) ( y v + ( b 2 4 b ) i , j = 1 N y j ( y i y j | y | 2 v y i ) C b | y | 2 v ) = | y | b ( N + 2 ) 2 ( 2 b ) ( i , j = 1 N y j ( A i j ( y ) v y i ) C b | y | 2 v ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equr_HTML.gif

 □

Let
H loc 1 ( R N ) = { u |  for every bounded domain  Ω R N , Ω | u | 2 d x + Ω u 2 d x < + } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ17_HTML.gif
(2.7)
From the classical Hardy inequality (see, e.g., [[13], Lemma 2.1]), we deduce that for every bounded C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq62_HTML.gif domain Ω R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq63_HTML.gif, there exists C Ω > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq64_HTML.gif such that, for every u H loc 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq65_HTML.gif,
Ω u 2 | x | 2 d x C Ω ( Ω | u | 2 d x + Ω u 2 d x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ18_HTML.gif
(2.8)
Theorem 2.2 If v H loc 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq66_HTML.gif is a weak solution of the equation
i , j = 1 N y j ( A i j ( y ) v y i ) + V v = K | v | p 2 v in  R N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ19_HTML.gif
(2.9)
i.e., for every ψ C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq67_HTML.gif,
R N i , j = 1 N A i j ( y ) v y i ψ y j d y + R N V ( y ) v ψ d y = R N K ( y ) | v | p 2 v ψ d y , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ20_HTML.gif
(2.10)
and u is defined by (2.1), then u H loc 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq65_HTML.gif and it is a weak solution of (1.1), i.e., for every φ C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq68_HTML.gif,
R N u φ d x + R N V ( x ) u φ d x = R N K ( x ) | u | p 2 u φ d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ21_HTML.gif
(2.11)
Proof Using the spherical coordinates
x 1 = r cos σ 1 , x 2 = r sin σ 1 cos σ 2 , x j = r sin σ 1 sin σ 2 sin σ j 1 cos σ j , 2 j N 1 , x N = r sin σ 1 sin σ 2 sin σ N 2 sin σ N 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equs_HTML.gif
where 0 σ j < π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq69_HTML.gif, j = 1 , 2 , , N 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq70_HTML.gif, 0 σ N 1 < 2 π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq71_HTML.gif, we have
d x = r N 1 f ( σ ) d r d σ 1 d σ N 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equt_HTML.gif
where f ( σ ) = sin N 2 σ 1 sin N 3 σ 2 sin σ N 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq72_HTML.gif. Recall that y = | x | b 2 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq73_HTML.gif. Let R = | y | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq74_HTML.gif. Then r = R 2 2 b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq75_HTML.gif and
d x = r N 1 f ( σ ) d r d σ 1 d σ N 1 = R 2 ( N 1 ) 2 b f ( σ ) d ( R 2 2 b ) d σ 1 d σ N 1 = 2 2 b R 2 N 2 b 1 f ( σ ) d R d σ 1 d σ N 1 = 2 2 b | y | b N 2 b d y . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ22_HTML.gif
(2.12)
Here, we used d y = R N 1 f ( σ ) d R d σ 1 d σ N 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq76_HTML.gif in the last inequality above. From (2.4), (2.12) and (2.8), we deduce that there exists C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif such that for every bounded domain Ω R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq63_HTML.gif,
Ω | u x i | 2 d x C Ω r b ( N 2 ) 2 b ( v y i ( | x | b / 2 x ) ) 2 d x + C Ω r b ( N 2 ) 2 b 4 ( x i j = 1 N x j v y j ( | x | b / 2 x ) ) 2 d x + C Ω r b ( N 2 ) 2 4 x i 2 v 2 ( | x | b / 2 x ) d x = 2 C 2 b Ω ( v ( y ) y i ) 2 d y + 2 C 2 b Ω ( y i | y | j = 1 N y j | y | v ( y ) y j ) 2 d y + 2 C 2 b Ω | y | 4 y i 2 v 2 ( y ) d y C ( Ω | v | 2 d y + Ω v 2 | y | 2 d y ) < + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equu_HTML.gif
Moreover,
Ω u 2 d x = Ω | x | b 2 ( N 2 ) v 2 ( | x | b 2 x ) d x = 2 2 b Ω | y | 2 b 2 b v 2 ( y ) d y < + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equv_HTML.gif
Therefore, u H loc 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq65_HTML.gif. Then, to prove that u satisfies (2.11) for every φ C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq68_HTML.gif, it suffices to prove that (2.11) holds for every φ C 0 ( R N { 0 } ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq77_HTML.gif. For φ C 0 ( R N { 0 } ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq77_HTML.gif, let ψ C 0 ( R N { 0 } ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq78_HTML.gif be such that
φ ( x ) = | x | b 4 ( N 2 ) ψ ( | x | b 2 x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equw_HTML.gif
By using the divergence theorem and Lemma 2.1, we get that
R N u φ d x = R N u φ d x = R N u | y | b ( N + 2 ) 2 ( 2 b ) ( i , j = 1 N y j ( A i j ( y ) ψ y i ) C b | y | 2 ψ ) d x = R N | x | b 4 ( N 2 ) v ( | x | b 2 x ) | y | b ( N + 2 ) 2 ( 2 b ) ( i , j = 1 N y j ( A i j ( y ) ψ y i ) C b | y | 2 ψ ) d x = R N | y | b ( N 2 ) 2 ( 2 b ) v ( y ) | y | b ( N + 2 ) 2 ( 2 b ) ( i , j = 1 N y j ( A i j ( y ) ψ y i ) C b | y | 2 ψ ) 2 2 b | y | b N 2 b d y = 2 2 b R N v ( i , j = 1 N y j ( A i j ( y ) ψ y i ) C b | y | 2 ψ ) d y = 2 2 b R N i , j = 1 N A i j ( y ) v y i ψ y j d y 2 C b 2 b R N v ψ | y | 2 d y . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equx_HTML.gif
Moreover,
R N V ( x ) u φ d x = 2 2 b R N V ( | y | b 2 b y ) u ( | y | b 2 b y ) φ ( | y | b 2 b y ) | y | b N 2 b d y = 2 2 b R N | y | 2 b 2 b V ( | y | b 2 b y ) | y | b ( N 2 ) 2 ( 2 b ) u ( | y | b 2 b y ) | y | b ( N 2 ) 2 ( 2 b ) φ ( | y | b 2 b y ) d y = 2 2 b R N | y | 2 b 2 b V ( | y | b 2 b y ) v ( y ) ψ ( y ) d y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equy_HTML.gif
and
R N K ( x ) | u | p 2 u φ d x = R N K ( | y | b 2 b y ) | u ( | y | b 2 b y ) | p 2 u ( | y | b 2 b y ) φ ( | y | b 2 b y ) 2 2 b | y | b N 2 b d y = 2 2 b R N | y | 2 s 2 b K ( | y | b 2 b y ) | v ( y ) | p 2 v ( y ) ψ ( y ) d y . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equz_HTML.gif
Therefore,
R N u φ d x + R N V ( x ) u φ d x R N K ( x ) | u | p 2 u φ d x = 2 2 b ( R N i , j = 1 N A i j ( y ) v y i ψ y j d y C b R N v ψ | y | 2 d y + R N | y | 2 b 2 b V ( | y | b 2 b y ) v ( y ) ψ ( y ) d y R N | y | 2 s 2 b K ( | y | b 2 b y ) | v ( y ) | p 2 v ( y ) ψ ( y ) d y ) = 2 2 b ( R N i , j = 1 N A i j ( y ) v y i ψ y j d y + R N V ( y ) v ψ d y R N K ( y ) | v | p 2 v ψ d y ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equaa_HTML.gif

This completes the proof. □

This theorem implies that the problem of looking for solutions of (1.1) can be reduced to a problem of looking for solutions of (2.9).

3 The variational functional for Eq. (2.9)

The following inequality is a variant Hardy inequality.

Lemma 3.1 If v H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq79_HTML.gif, then
R N | x v | 2 | x | 2 d x ( N 2 ) 2 4 R N | v | 2 | x | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ23_HTML.gif
(3.1)
Proof We only give the proof of (3.1) for v C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq80_HTML.gif since C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq21_HTML.gif is dense in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif. For v C 0 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq80_HTML.gif, we have the following identity:
| v ( x ) | 2 = 1 d d λ | v ( λ x ) | 2 d λ = 2 1 v ( λ x ) ( x v ( λ x ) ) d λ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equab_HTML.gif
By using the Hölder inequality, it follows that
R N | v ( x ) | 2 | x | 2 d x = 2 1 R N v ( λ x ) | x | 2 ( x v ( λ x ) ) d x d λ = 2 1 d λ λ N 1 R N v ( x ) | x | 2 ( x v ( x ) ) d x = 2 N 2 R N v ( x ) | x | 2 ( x v ( x ) ) d x 2 N 2 ( R N v 2 ( x ) | x | 2 d x ) 1 / 2 ( R N | x v | 2 | x | 2 d x ) 1 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equac_HTML.gif
Then we conclude that
R N | x v | 2 | x | 2 d x ( N 2 ) 2 4 R N | v | 2 | x | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equad_HTML.gif

 □

From the definition of A i j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq81_HTML.gif (see (2.3)), it is easy to verify that for u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq82_HTML.gif,
R N i , j = 1 N A i j ( x ) u x i u x j d x = R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ24_HTML.gif
(3.2)
Lemma 3.2 There exist constants C 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq83_HTML.gif and C 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq84_HTML.gif such that for every u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq82_HTML.gif,
C 1 u 2 R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x + R N V ( x ) | u | 2 d x C 2 u 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equae_HTML.gif
Proof From conditions (A1) and (A2), we deduce that there exists a constant C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif such that
| x | 2 b 2 b V ( | x | b 2 b x ) C ( 1 + | x | 2 ) , x R N { 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ25_HTML.gif
(3.3)
Since
R N V ( x ) | u | 2 d x = R N | x | 2 b 2 b V ( | x | b 2 b x ) | u | 2 d x + C b R N | u | 2 | x | 2 d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equaf_HTML.gif
by (3.3) and the classical Hardy inequality (see, e.g., [13])
( N 2 ) 2 4 R N | u | 2 | x | 2 d x R N | u | 2 d x , u H 1 ( R N ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equag_HTML.gif
we deduce that there exists a constant C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif such that
R N V ( x ) | u | 2 d x C u 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equah_HTML.gif
This together with the fact that R N | x u | 2 | x | 2 d x R N | u | 2 d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq85_HTML.gif yields that there exists a constant C 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq84_HTML.gif such that
R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x + R N V ( x ) | u | 2 d x C 2 u 2 , u H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ26_HTML.gif
(3.4)
If 0 < b < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq17_HTML.gif, then b 2 4 b < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq86_HTML.gif and
R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x R N | u | 2 d x + ( b 2 4 b ) R N | u | 2 d x = ( 1 b / 2 ) 2 R N | u | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ27_HTML.gif
(3.5)
In this case, C b = b 4 ( 1 b 4 ) ( N 2 ) 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq87_HTML.gif and
R N V ( x ) | u | 2 d x = R N | x | 2 b 2 b V ( | x | b 2 b x ) | u | 2 d x + C b R N | u | 2 | x | 2 d x R N | x | 2 b 2 b V ( | x | b 2 b x ) | u | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ28_HTML.gif
(3.6)
Conditions (A1) and (A2) imply that there exists a constant C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif such that
R N | u | 2 d x + R N | x | 2 b 2 b V ( | x | b 2 b x ) u 2 d x C R N u 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ29_HTML.gif
(3.7)
Combining (3.5)-(3.7) yields that there exists a constant C 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq83_HTML.gif such that
R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x + R N V ( x ) | u | 2 d x C 1 u 2 , u H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ30_HTML.gif
(3.8)
If b < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq18_HTML.gif, (3.7) still holds. From Lemma 3.1 and (3.7), we deduce that there exists a constant C 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq83_HTML.gif such that for every u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq82_HTML.gif,
R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x + R N V ( x ) | u | 2 d x = R N | u | 2 d x + ( b 2 4 b ) ( R N | x u | 2 | x | 2 d x ( N 2 ) 2 4 R N | u | 2 | x | 2 d x ) + R N | x | 2 b 2 b V ( | x | b 2 b x ) | u | 2 d x R N | u | 2 d x + R N | x | 2 b 2 b V ( | x | b 2 b x ) | u | 2 d x C 1 u 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ31_HTML.gif
(3.9)

Then the desired result of this lemma follows from (3.4), (3.8) and (3.9) immediately. □

This lemma implies that
u A = ( R N | u | 2 d x + ( b 2 4 b ) R N | x u | 2 | x | 2 d x + R N V ( x ) | u | 2 d x ) 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ32_HTML.gif
(3.10)
is equivalent to the standard norm https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq88_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif. We denote the inner product associated with A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq89_HTML.gif by ( , ) A https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq90_HTML.gif, i.e.,
( u , v ) A = R N u v d x + R N V ( x ) u v d x + ( b 2 4 b ) R N ( x u ) ( x v ) | x | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ33_HTML.gif
(3.11)
By the Sobolev inequality, we have
S A : = inf u H 1 ( R N ) { 0 } u A 2 ( R N | u | p d x ) 2 / p > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ34_HTML.gif
(3.12)
and
u A S A 1 2 ( R N | u | p d x ) 1 / p , u H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ35_HTML.gif
(3.13)
By conditions (A1) and (A2), if 0 < b < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq17_HTML.gif, then K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq91_HTML.gif is bounded in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif. Therefore, by (3.13), there exists C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif such that
( R N K ( x ) ( u + ) p d x ) 1 / p C u A , u H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ36_HTML.gif
(3.14)
However, if b < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq18_HTML.gif, K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq91_HTML.gif has a singularity at x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq92_HTML.gif, i.e.,
K ( x ) | x | 2 s 2 b K ( 0 ) , as  | x | 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ37_HTML.gif
(3.15)
Recall that p = 2 ( N 2 s / b ) / ( N 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq8_HTML.gif and 2 s / ( 2 b ) > 2 s / b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq93_HTML.gif if b < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq18_HTML.gif. Then, by the Hardy-Sobolev inequality (see, for example, [[14], Lemma 3.2]), we deduce that there exists C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq25_HTML.gif such that (3.14) still holds. Therefore, the functional
J ( u ) = 1 2 u A 2 1 p R N K ( x ) ( u + ) p d x , u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ38_HTML.gif
(3.16)
is a C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq28_HTML.gif functional defined in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif. Moreover, it is easy to check that the Gateaux derivative of J is
J ( u ) , h = ( u , h ) A R N K ( x ) ( u + ) p 1 h d x , u , h H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equai_HTML.gif

and the critical points of J are nonnegative solutions of (2.9).

4 Some minimizing problems

For θ = ( θ 1 , , θ N ) R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq94_HTML.gif with | θ | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq95_HTML.gif, let
B i j ( θ ) = δ i j + ( b 2 4 b ) θ i θ j , i , j = 1 , , N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ39_HTML.gif
(4.1)
By this definition, we have, for u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq82_HTML.gif,
R N i , j = 1 N B i j ( θ ) u x i u x j d x = R N | u | 2 d x + ( b 2 4 b ) R N | θ u | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ40_HTML.gif
(4.2)
From
( 1 + | b 2 4 b | ) R N | u | 2 d x R N | u | 2 d x + ( b 2 4 b ) R N | θ u | 2 d x { ( 1 b / 2 ) 2 R N | u | 2 d x , 0 < b < 2 , R N | u | 2 d x , b < 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equaj_HTML.gif
we deduce that the norm defined by
u θ : = ( R N | u | 2 d x + ( b 2 4 b ) R N | θ u | 2 d x + a R N | u | 2 d x ) 1 / 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ41_HTML.gif
(4.3)
is equivalent to the standard norm https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq88_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif. The inner product corresponding to θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq96_HTML.gif is
( u , v ) θ = R N u v d x + a R N u v d x + ( b 2 4 b ) R N ( θ u ) ( θ v ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equak_HTML.gif
Lemma 4.1 The infimum
inf u H 1 ( R N ) { 0 } u θ 2 ( R N | u | p d x ) 2 / p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ42_HTML.gif
(4.4)

is independent of θ R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq97_HTML.gif with | θ | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq95_HTML.gif.

Proof In this proof, we always view a vector in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif as a 1 × N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq98_HTML.gif matrix, and we use A T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq99_HTML.gif to denote the conjugate matrix of a matrix A.

For any θ , θ R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq100_HTML.gif with | θ | = | θ | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq101_HTML.gif, let G be an N × N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq102_HTML.gif orthogonal matrix such that θ G T = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq103_HTML.gif. For any u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq82_HTML.gif, let v ( x ) = u ( x G ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq104_HTML.gif, x R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq12_HTML.gif. The assumption that G is an N × N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq102_HTML.gif orthogonal matrix implies that G G T = I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq105_HTML.gif, where I is the N × N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq102_HTML.gif identity matrix. Then it is easy to check that
R N | v | 2 d x = R N | u | 2 d x , R N | v | p d x = R N | u | p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ43_HTML.gif
(4.5)
Note that
v ( x ) = ( u ) ( x G ) G . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ44_HTML.gif
(4.6)
By G G T = I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq105_HTML.gif, we have
| v ( x ) | 2 = v ( x ) ( v ( x ) ) T = ( u ) ( x G ) G G T ( ( u ) ( x G ) ) T = | ( u ) ( x G ) | 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equal_HTML.gif
It follows that
R N | v ( x ) | 2 d x = R N | ( u ) ( x G ) | 2 d x = R N | u ( x ) | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ45_HTML.gif
(4.7)
By (4.6) and θ G T = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq103_HTML.gif, we get that
i = 1 N θ i v x i = θ ( ( u ) ( x G ) G ) T = θ G T ( ( u ) ( x G ) ) T = θ ( ( u ) ( x G ) ) T = i = 1 N θ i ( u y i ) ( x G ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equam_HTML.gif
It follows that
R N | θ v | 2 d x = R N | i = 1 N θ i v x i | 2 d x = R N | i = 1 N θ i ( u y i ) ( x G ) | 2 d x = R N | i = 1 N θ i u x i | 2 d x = R N | θ u | 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ46_HTML.gif
(4.8)

By (4.5), (4.7) and (4.8), we get that v θ 2 = u θ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq106_HTML.gif. This together with (4.5) leads to the result of this lemma. □

Since the infimum (4.4) is independent of θ R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq97_HTML.gif with | θ | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq95_HTML.gif, we denote it by S.

Lemma 4.2 Let S p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq34_HTML.gif be the infimum in (1.9). Then S = ( 1 b / 2 ) p 2 p S p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq107_HTML.gif.

Proof Choosing θ = ( 1 , 0 , , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq108_HTML.gif in θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq96_HTML.gif, we have
u θ 2 = ( 1 b 2 ) 2 R N | u x 1 | 2 d x + i = 2 N R N | u x i | 2 d x + a R N u 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equan_HTML.gif
By Lemma 4.1, we have
S = inf u H 1 ( R N ) { 0 } ( 1 b 2 ) 2 R N | u x 1 | 2 d x + i = 2 N R N | u x i | 2 d x + a R N u 2 d x ( R N | u | p d x ) 2 / p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equao_HTML.gif
Let
v ( x ) = u ( ( 1 b / 2 ) x 1 , x 2 , , x N ) , x R N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equap_HTML.gif
Then
( 1 b 2 ) 2 R N | u x 1 | 2 d x + i = 2 N R N | u x i | 2 d x + a R N u 2 d x ( R N | u | p d x ) 2 / p = ( 1 b / 2 ) p 2 p R N | v | 2 d x + a R N v 2 d x ( R N | v | p d x ) 2 / p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equaq_HTML.gif
It follows that
S = ( 1 b / 2 ) p 2 p inf v H 1 ( R N ) { 0 } R N | v | 2 d x + a R N v 2 d x ( R N | v | p d x ) 2 / p = ( 1 b / 2 ) p 2 p S p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equar_HTML.gif

 □

Since the functionals u θ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq109_HTML.gif and R N | u | p d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq110_HTML.gif are invariant by translations, the same argument as the proof of [[11], Theorem 1.34] yields that there exists a positive minimizer U θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq111_HTML.gif for the infimum S. Moreover, from the Lagrange multiplier rule, it is a solution of
i , j = 1 N y j ( B i j ( θ ) u y i ) + a u = S ( u + ) p 1 in  R N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equas_HTML.gif
and ( μ / S ) 1 / ( p 2 ) U θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq112_HTML.gif is a solution of
i , j = 1 N y j ( B i j ( θ ) u y i ) + a u = μ ( u + ) p 1 in  R N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ47_HTML.gif
(4.9)
In the next section, we shall show that Eq. (4.9) is the ‘limit’ equation of
i , j = 1 N y j ( A i j ( x ) u y i ) + V ( x ) u = K ( x ) ( u + ) p 1 in  R N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ48_HTML.gif
(4.10)
It is easy to verify that
J θ ( u ) = 1 2 u θ 2 μ p R N ( u + ) p d x , u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ49_HTML.gif
(4.11)
is a C 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq28_HTML.gif functional defined in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif, the Gateaux derivative of J θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq113_HTML.gif is
J θ ( u ) , h = ( u , h ) θ μ R N ( u + ) p 1 h d x , u , h H 1 ( R N ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equat_HTML.gif

and the critical points of this functional are solutions of (4.9).

Lemma 4.3 Let θ R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq97_HTML.gif satisfy | θ | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq95_HTML.gif. If u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq114_HTML.gif is a critical point of J θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq113_HTML.gif, then
J θ ( u ) ( 1 2 1 p ) μ 2 p 2 S p p 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ50_HTML.gif
(4.12)
Proof Since u is a critical point of J θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq113_HTML.gif, we have
0 = J θ ( u ) , u = u θ 2 μ R N ( u + ) p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ51_HTML.gif
(4.13)
It follows that
J θ ( u ) = ( 1 2 1 p ) μ R N ( u + ) p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ52_HTML.gif
(4.14)
Since u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq114_HTML.gif, by u θ 2 = μ R N ( u + ) p d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq115_HTML.gif and u θ 2 S ( R N ( u + ) p d x ) 2 / p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq116_HTML.gif, we get that
R N ( u + ) p d x ( S / μ ) p / ( p 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equau_HTML.gif

This together with (4.14) yields the result of this lemma. □

5 The Palais-Smale condition for the functional J

Recall that J is the functional defined by (3.16). By a ( P S ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq117_HTML.gif sequence of J, we mean a sequence { u n } H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq118_HTML.gif such that J ( u n ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq119_HTML.gif and J ( u n ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq120_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq121_HTML.gif as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq122_HTML.gif, where H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq121_HTML.gif denotes the dual space of H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif. J is called satisfying the ( P S ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq117_HTML.gif condition if every ( P S ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq117_HTML.gif sequence of J contains a convergent subsequence in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif.

Our main result in this section reads as follows.

Theorem 5.1 Under assumptions (A1) and (A2), let { u n } H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq118_HTML.gif be a ( P S ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq117_HTML.gif sequence of J. Then replacing { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq123_HTML.gif if necessary by a subsequence, there exist a solution u 0 H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq124_HTML.gif of Eq. (4.10), a finite sequence { θ l R N | θ l | = 1 , 1 l k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq125_HTML.gif, k functions { u l 1 i k } H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq126_HTML.gif and k sequences { y n l } R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq127_HTML.gif satisfying:
  1. (i)

    i , j = 1 N y j ( B i j ( θ l ) u l y i ) + a u l = μ ( u l + ) p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq128_HTML.gif in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif,

     
  2. (ii)

    | y n l | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq129_HTML.gif, | y n l y n l | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq130_HTML.gif, l l https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq131_HTML.gif, n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq122_HTML.gif,

     
  3. (iii)

    u n u 0 l = 1 k u l ( y n l ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq132_HTML.gif,

     
  4. (iv)

    J ( u 0 ) + i = 1 l J θ l ( u l ) = c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq133_HTML.gif.

     

This theorem gives a precise representation of the ( P S ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq117_HTML.gif sequence for the functional J. Through it, partial compactness for J can be regained (see Corollary 5.8).

To prove this theorem, we need some lemmas. Our proof of this theorem is inspired by the proof of [[11], Theorem 8.4].

Lemma 5.2 Let u H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq82_HTML.gif. Then, for any sequence { y n } R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq134_HTML.gif,
lim R sup n | x | > R K ( x + y n ) | u | p d x = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equav_HTML.gif
If | y n | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq135_HTML.gif, n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq122_HTML.gif, then
lim n R N | K ( x + y n ) μ | | u | p d x = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equaw_HTML.gif
Proof If 2 > b > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq136_HTML.gif, then K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq91_HTML.gif is bounded in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif. In this case, the result of this lemma is obvious. If b < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq18_HTML.gif, then K ( x ) | x | 2 s 2 b K ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq137_HTML.gif as | x | 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq138_HTML.gif. Since 2 s / ( 2 b ) > 2 s / b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq93_HTML.gif, by Lemma 3.2 of [14], the map v K 1 / p v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq139_HTML.gif from H 1 ( R N ) L loc p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq140_HTML.gif is compact. Therefore, for any ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq40_HTML.gif, there exists δ ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq141_HTML.gif such that
sup n | x | δ ϵ K ( x ) | u ( x y n ) | p d x ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equax_HTML.gif
And there exists D ( ϵ ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq142_HTML.gif depending only on ϵ such that K ( x ) D ( ϵ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq143_HTML.gif, | x | δ ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq144_HTML.gif. Then, for every n,
| x | > R K ( x + y n ) | u | p d x { x | | x + y n | δ ϵ , | x | > R } K ( x + y n ) | u | p d x + { x | x + y n | > δ ϵ , | x | > R } K ( x + y n ) | u | p d x ϵ + C ( ϵ ) | x | > R | u | p d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equay_HTML.gif

It follows that lim sup R sup n | x | > R K ( x + y n ) | u | p d x ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq145_HTML.gif. Now let ϵ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq146_HTML.gif.

Using the same argument as above, for any ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq40_HTML.gif, there exist δ ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq147_HTML.gif and D ( ϵ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq148_HTML.gif such that
sup n | x + y n | δ ϵ | K ( x + y n ) μ | | u | p d x ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equaz_HTML.gif
and
| K ( x + y n ) μ | | u | p d x ( D ( ϵ ) + μ ) | u | p , | x + y n | δ ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equba_HTML.gif
Since y n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq149_HTML.gif, we have lim K ( x + y n ) = μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq150_HTML.gif. Then, using the Lebesgue theorem and the above two inequalities, we get that
lim sup n R N | K ( x + y n ) μ | | u | p d x ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbb_HTML.gif

Let ϵ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq146_HTML.gif. Then we get the desired result of this lemma. □

Lemma 5.3 Let ρ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq151_HTML.gif. If { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq123_HTML.gif is bounded in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif and
sup y R N B ( y , ρ ) | u n | 2 d x 0 , n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ53_HTML.gif
(5.1)

then K 1 / p u n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq152_HTML.gif in L p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq32_HTML.gif.

Proof Since 2 s / ( 2 b ) > 2 s / b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq93_HTML.gif, by Lemma 3.2 of [14], the map v K 1 / p v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq139_HTML.gif from H 1 ( R N ) L loc p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq140_HTML.gif is compact. Therefore, for any ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq40_HTML.gif, there exists δ ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq141_HTML.gif such that
sup n | x | δ ϵ K ( x ) | u n | p d x ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbc_HTML.gif
And there exists D ( ϵ ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq142_HTML.gif depending only on ϵ such that K ( x ) D ( ϵ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq143_HTML.gif, | x | δ ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq144_HTML.gif. By (5.1) and the Lions lemma (see, for example, [[11], Lemma 1.21]), we get that
| x | δ ϵ K ( x ) | u n | p d x D ( ϵ ) R N | u n | p d x 0 , n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbd_HTML.gif

Therefore, lim sup n R N K ( x ) | u n | p d x ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq153_HTML.gif. Now let ϵ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq146_HTML.gif. □

Lemma 5.4 Let { y n } R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq134_HTML.gif. If u n u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq154_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif, then
K ( x + y n ) ( u n + ) p 1 K ( x + y n ) ( ( u n u ) + ) p 1 K ( x + y n ) ( u + ) p 1 0 in  H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Eqube_HTML.gif

One can follow the proof of [[11], Lemma 8.1] step by step and use Lemma 5.2 to give the proof of this lemma.

The following lemma is a variant Brézis-Lieb lemma (see [15]) and its proof is similar to that of [[11], Lemma 1.32].

Lemma 5.5 Let { u n } H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq118_HTML.gif and { y n } R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq134_HTML.gif. If
  1. (a)

    { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq123_HTML.gif is bounded in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif,

     
  2. (b)

    u n u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq155_HTML.gif a.e. on R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq19_HTML.gif, then

     
lim n R N K ( x + y n ) | ( u n + ) p ( ( u n u ) + ) p ( u + ) p | d x = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbf_HTML.gif
Proof Let
j ( t ) = { t p , t 0 , 0 , t < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbg_HTML.gif
Then j is a convex function. From [[15], Lemma 3], we have that for any ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq40_HTML.gif, there exists C ( ϵ ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq156_HTML.gif such that for all a , b R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq157_HTML.gif,
| j ( a + b ) j ( b ) | ϵ j ( a ) + C ( ϵ ) j ( b ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ54_HTML.gif
(5.2)
Hence
f n ϵ : = ( K ( x + y n ) | ( u n + ) p ( ( u n u ) + ) p ( u + ) p | ϵ K ( x + y n ) ( ( u n u ) + ) p ) + ( 1 + C ( ϵ ) ) K ( x + y n ) ( u + ) p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbh_HTML.gif
By Lemma 3.2 of [14], the map v K 1 / p v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq139_HTML.gif from H 1 ( R N ) L loc p ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq140_HTML.gif is compact. We get that there exists δ ϵ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq141_HTML.gif such that for any n,
| x + y n | < δ ϵ f n ϵ d x < ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ55_HTML.gif
(5.3)
And there exists D ( ϵ ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq142_HTML.gif depending only on ϵ such that K ( x ) D ( ϵ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq143_HTML.gif, | x | δ ϵ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq144_HTML.gif. Then
f n ϵ ( 1 + C ( ϵ ) ) D ( ϵ ) ( u + ) p , | x + y n | δ ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbi_HTML.gif
By the Lebesgue theorem, | x + y n | δ ϵ f n ϵ d x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq158_HTML.gif, n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq122_HTML.gif. This together with (5.3) yields
lim sup n R N f n ϵ d x ϵ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbj_HTML.gif

The left proof is the same as the proof of [[11], Lemma 1.32]. □

Lemma 5.6 If
u n u in  H 1 ( R N ) , u n u a.e. on  R N , J ( u n ) c , J ( u n ) 0 in  H 1 ( R N ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbk_HTML.gif
then J ( u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq159_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq121_HTML.gif and v n : = u n u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq160_HTML.gif is such that
v n A 2 = u n A 2 u A 2 + o ( 1 ) , J ( v n ) c J ( u ) , J ( v n ) 0 in  H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbl_HTML.gif
Proof (1) Since u n u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq154_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif, we get that as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq122_HTML.gif,
v n A 2 u n A 2 = ( u n u , u n u ) A u n A 2 = u A 2 2 ( u n , u ) A u A 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbm_HTML.gif
Therefore,
v n A 2 = u n A 2 u A 2 + o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ56_HTML.gif
(5.4)
  1. (2)
    Lemma 5.5 implies
    R N K ( x ) ( v n + ) p d x = R N K ( x ) ( u n + ) p d x R N K ( x ) ( u + ) p d x + o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ57_HTML.gif
    (5.5)
     
By (5.4), (5.5) and the assumption J ( u n ) c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq119_HTML.gif, we get that
J ( v n ) c J ( u ) , n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbn_HTML.gif
  1. (3)
    Since J ( u n ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq120_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq121_HTML.gif and u n u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq154_HTML.gif, it is easy to verify that J ( u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq159_HTML.gif. For h H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq161_HTML.gif,
    J ( v n ) , h = ( v n , h ) A R N K ( x ) ( v n + ) p 1 h d x = ( u n , h ) A ( u , h ) A R N K ( x ) ( v n + ) p 1 h d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ58_HTML.gif
    (5.6)
     
By Lemma 5.4, we have
sup h 1 | R N K ( x ) ( v n + ) p 1 h d x R N K ( x ) ( u n + ) p 1 h d x + R N K ( x ) ( u + ) p 1 h d x | 0 , n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equ59_HTML.gif
(5.7)

Combining (5.6) and (5.7) leads to J ( v n ) = J ( u n ) J ( u ) + o ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq162_HTML.gif. Then, by J ( u n ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq120_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq121_HTML.gif and J ( u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq159_HTML.gif, we obtain that J ( v n ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq163_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq121_HTML.gif. □

Lemma 5.7 If | y n | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq135_HTML.gif and as n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq122_HTML.gif,
u n ( + y n ) u in  H 1 ( R N ) , u n ( + y n ) u a.e. on  R N , J ( u n ) c , J ( u n ) 0 in  H 1 ( R N ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbo_HTML.gif
then there exists θ R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq97_HTML.gif with | θ | = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq95_HTML.gif such that J θ ( u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq164_HTML.gif and v n = u n u ( y n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq165_HTML.gif is such that
v n 2 = u n 2 u 2 + o ( 1 ) , J ( v n ) c J θ ( u ) , J ( v n ) 0 in  H 1 ( R N ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbp_HTML.gif
Proof We divide the proof into several steps.
  1. (1)
    Since u n ( + y n ) u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq166_HTML.gif in H 1 ( R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_IEq31_HTML.gif, it is clear that
    v n 2 = v n ( + y n ) 2 = u n ( + y n ) 2 + u 2 2 ( u n ( + y n ) , u ) = u n 2 u 2 + o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-201/MediaObjects/13661_2013_Article_454_Equbq_HTML.gif
     
  2. (2)
    For any h H 1