Extremal functions and best constants to an inequality involving Hardy potential and critical Sobolev exponent

Nonlinear Analysis 71 (2009) 845–859

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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

Extremal functions and best constants to an inequality involving Hardy potential and critical Sobolev exponent Benjin Xuan ∗ , Jiangchao Wang Department of Mathematics, University of Science and Technology of China, China

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a b s t r a c t

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Article history: Received 5 September 2008 Accepted 31 October 2008

In this paper, we study the asymptotic behavior of radial extremal functions to an inequality involving Hardy potential and critical Sobolev exponent. Based on the asymptotic behavior at the origin and the infinity, we shall deduce a strict inequality between two best constants. Finally, as an application of this strict inequality, we consider the existence of a nontrivial solution of a quasilinear Brezis–Nirenberg type problem with Hardy potential and critical Sobolev exponent. © 2008 Elsevier Ltd. All rights reserved.

MSC: 35J60 Keywords: Asymptotic behavior Extremal functions Hardy potential Critical Sobolev exponent Brezis–Nirenberg type problem

1. Introduction In this paper, we first study the asymptotic behavior of extremal functions to the following inequality involving Hardy potential and critical Sobolev exponent:

Z C RN

|u|p∗ dx |x|bp∗

p/p∗

RN

where 1 < p < N , 0 6 a < following Hardy inequality:

µ

Z RN

|u|p dx 6 |x|(a+1)p



Z 6 N −p p

Z RN

|u|p |Du|p − µ (a+1)p ap |x| |x|



dx,

, a 6 b < (a + 1), p∗ =

Np N −(a+1−b)p

|Du|p dx. |x|ap

(1.1)

, µ < µ. µ = ( N −(ap+1)p )p is the best constant in the

(1.2)

To do this, let us consider the following extremal problem:

n

1 ,p

p

o

S0,µ = inf Qµ (u) : u ∈ Da,b (RN ), ku; Lb∗ (RN )k = 1 , where Qµ (u) =

∗

Z RN

|Du|p dx − µ |x|ap

Z RN

|u|p dx, |x|(a+1)p

Corresponding author. Tel.: +86 551 3601201. E-mail address: [email protected] (B. Xuan).

0362-546X/$ – see front matter © 2008 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2008.10.114

(1.3)

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B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

and 1,p

p

Da,b (RN ) = {u ∈ Lb∗ (RN ) : |Du| ∈ Lpa (RN )} p

is the closure of C0∞ (RN ) under the norm kukD1,p (RN ) = k|Du|; La (RN )k. For any parameters α > 0 and q > 1, the norm of a,b

weighted space Lqα (RN ) is defined as

ku; Lα (R )k = q

N

Z RN

|u|q dx |x|αq

 1q

. 1,p

Let Ω be a smooth bounded open domain in RN with 0 ∈ Ω , define Da,b (Ω ) as the closure of C0∞ (Ω ) under the norm p La

kukD1,p (Ω ) = k|Du|; (Ω )k and a,b

1,p

n

o

p

Sλ,µ (Ω ) = inf Qλ,µ (u) : u ∈ Da,b (Ω ), ku; Lb∗ (Ω )k = 1 ,

(1.4)

where Qλ,µ (u) =

Z Ω

|Du|p dx − µ |x|ap

Z Ω

|u|p dx − λ |x|(a+1)p

|u|p

Z Ω

|x|(a+1)p−c

dx.

If λ = 0, by the rescaling argument, it is easy to show that S0,µ (Ω ) = S0,µ . But for λ > 0, we shall have a strict inequality: Sλ,µ (p, a, b; Ω ) < S0,µ .

(1.5)

In their famous paper [6], Brezis and Nirenberg studied the existence and nonexistence of positive solutions to the problem:

 ∗ −∆u = λu + u2 −1 , u > 0, in Ω ,  u = 0, on ∂ Ω ,

in Ω , (1.6) ∗

where Ω is a bounded open domain in RN . Since the embedding H01 (Ω ) ,→ L2 (Ω ) is not compact where 2∗ = 2N /(N − 2), the associated energy functional does not satisfy the (PS) condition globally, which caused a serious difficulty when applying standard variational methods to study problem (1.6). Inspired by the works [27] of Trudinger and [3] of Aubin concerning the Yamabe problem, Brezis and Nirenberg successfully reduced the existence of solutions of problem (1.6) into the verification of the following strict inequality: Sλ :=

inf {kDu|k2L2 − λkuk2L2 } < S :=

u∈H 1 0 ∗ kuk2L =1

inf {kDu|k2L2 },

(1.7)

u∈H 1 0 ∗ kuk2L =1

∗

for some suitable λ’s, where the constant S corresponds to the best constant for the Sobolev embedding H01 (Ω ) ,→ L2 (Ω ). Clearly, inequality (1.7) is a special version of inequality (1.5) with p = 2, a = b = µ = 0. To verify (1.7) in their case, they applied the explicit expression of the extremal functions to the classical Sobolev inequality, especially the asymptotic behavior of the extremal functions at the origin and the infinity. Brezis–Nirenberg type problems have been generalized to many other situations (see [10–12,16,19,21,22,29,31,32] and references therein). Recently, Jannelli [18] introduced the term µ |xu|2 in the equation, that is,

 u ∗  −∆u − µ 2 = λu + u2 −1 , | x|  u > 0, in Ω , u = 0, on ∂ Ω ,

in Ω , (1.8)

and studied the relation between critical dimensions for λ ∈ (0, λ1 ) and L2loc integrability of the associated Green function, where λ1 is the first eigenvalue of operator −∆ −µ |x1|2 on Ω with zero-Dirichlet condition. Ruiz and Willem [23] also studied problem (1.8) under various assumption on the domain Ω , and even for µ 6 0. Those proofs in [18] and [23] were reduced to verify the strict inequality of (1.5) with p = 2, a = b = 0. In 2001, Ferrero and Gazzola [13] considered the existence of signchanged solution to problem (1.8) for larger λ. They distinguished two distinct cases: resonant and nonresonant cases of the Brezis–Nirenberg type problem (1.8). For the resonant case, they only studied a special case: Ω is the unit ball and λ = λ1 . The general case was left as an open problem. In 2004, Cao and Han [9] accomplished the general case. In [31], the authors considered the more general case with singular coefficients. In all the references cited above, the asymptotic behavior of the extremal functions at the origin and the infinity was applied to derive the local (PS) condition for the associated energy functional.

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

847

We shall show that for µ < µ the best constant of inequality (1.1) is achievable. Furthermore, any extremal function of inequality (1.1) is radial symmetric. Then we study the asymptotic behavior of the radial extremal functions of inequality (1.1) at the origin and the infinity. At last, for any smooth bounded open domain Ω ⊂ RN containing 0 in its interior, we shall deduce the strict inequality (1.5). As many simple cases of (1.5), e.g. a = b = µ = 0, have played important roles in the study of the Yamabe problem (cf. [24,27]) and Brezis–Nirenberg problem (cf. [6,30,31,14]), we believe that the asymptotic behavior of the radial extremal functions and the general strict inequality (1.5) will be useful to study the existence of the quasilinear elliptic problem involving Hardy potential and critical Sobolev exponent. As an application of this strict inequality, we consider the existence of a nontrivial solution of a quasilinear Brezis–Nirenberg problem with Hardy potential and critical Sobolev exponent. The rest of this paper is organized as follows. In Section 2, we shall show that the best constant of (1.1) is achieved by some radial extremal functions. Section 3 is concerned with the asymptotic behavior of the radial extremal functions at the origin and the infinity. In Section 4, we first derive various estimates on the approximation extremal functions, and then establish the strict inequality (1.5). In Section 5, based on this strict inequality, we obtain the existence results of a nontrivial solution of a quasilinear Brezis–Nirenberg problem.

2. Radial extremal functions In this section, we shall show that the best constant of (1.1) is achieved by some radial extremal functions. First, similarly 1,p to Lemma 2.1 in [15], one can easily obtain the Hardy inequality (1.2), and the best constant µ is not achievable in Da,b (RN ). 1 ,p

Thus, for µ < µ, Qµ (u) > 0 for all u ∈ Da,b (RN ), and the equality holds if and only if u ≡ 0. From the so-called Caffarelli–Kohn–Nirenberg inequality [8], S0,µ < ∞. The next lemma shows that contrast S0,µ is achievable. 1,p

Lemma 2.1. If µ ∈ (0, µ), b ∈ [a, a + 1), then S0,µ is achieved at some nonnegative function u0 ∈ Da,b (RN ). In particular, there exists a solution to the following ‘‘limited equation’’:

 − div

|Du|p−2 Du |x|ap



−µ

|u|p−2 u |u|p∗ −2 u = . |x|(a+1)p |x|bp∗

(2.1)

1,p

p

Proof. The achievability of S0,µ at some u0 ∈ Da,b (RN ) with ku0 ; Lb∗ (RN )k = 1 is due to [28] for p = 2 and [26] for general p. Without loss of generality, suppose that u0 > 0, otherwise, replace it by |u0 |. It is easy to see that u0 satisfies the following Euler–Lagrange equation:

 −div

|Du|p−2 Du |x|ap



−µ

|u|p−2 u |u|p∗ −2 u =δ , ( a + 1 ) p |x| |x|bp∗ 1 p −p

p

∗ where δ = Qµ (u0 )/ku0 ; Lb∗ (RN )kp∗ = Qµ (u0 ) = S0,µ > 0 is the Lagrange multiplier. Set u = c0 u0 , c0 = S0,µ , then u is a solution to Eq. (2.1). 

− N −(a+1)p

p In fact, all the dilations of u0 of the form σ u0 ( σ· ) are also minimizers of S0,µ . In order to obtain further properties of the minimizers of S0,µ , let us recall the definition of the Schwarz symmetrization (see [17,7]). Suppose that Ω ⊂ RN , and f ∈ C0 (Ω ) is a nonnegative continuous function with compact support, the Schwarz symmetrization S (f ) of f is defined as

S (f )(x) = sup t : µ(t ) > ωN |x|N ,





µ(t ) = | {x : f (x) > t }| , 1,p

where ωN denotes the volume of the standard N-sphere. For v ∈ Da,b (RN ) \ {0}, define

R R(v) =

S N −1

R +∞ 0

p

{k1−p− p∗ (|∂ρ v|2 + R R +∞ S N −1

0

p

|Λv|2 p/2 N −1 ) ρ ρ2

− k1− p∗ µ|v|p ρ N −1−p } dρ dS , p/p∗ (N −p)p∗ |v|p∗ ρ p −1 dρ dS

where ∂ρ is the directional differential operator along direction ρ and Λ is the tangential differential operator on S N −1 . Applying those properties of Schwarz symmetrization in [17], we have the following lemma: Lemma 2.2. If k > 0, µ ∈ (0, µ), then 1 ,p

1,p

inf{R(v) : v ∈ Da,b (RN ) is radial} = inf{R(v) : v ∈ Da,b (RN )}.

848

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

Proof. By the density argument, it suffices to prove the lemma for v ∈ C0∞ (RN ). Let v ∗ be the Schwarz symmetrization of

v . Noting that Λv ∗ = 0, p∗ 6

functions ρ Schwarz symmetrization in [17], we have

Z

+∞

Z S N −1

Z

Np , N −p

(N −p)p∗

|v | ρ ∗ p∗

p

−1

dρ dS >

+∞

S N −1



|∂ρ v ∗ |2 +

0

|Λv ∗ |2 ρ2

p/2

p

Z

and ρ −p are decreasing in (0, +∞). Applying those properties of

−N

+∞

Z S N −1

0

Z

(N −p)p∗

|v|p∗ ρ

(N −p)p∗

−1

p

dρ dS = 1,

0

Z

ρ N −1 dρ dS 6

+∞

Z S N −1

0

 p/2 |Λv|2 |∂ρ v|2 + ρ N −1 dρ dS ρ2

and

Z

+∞

Z S N −1

Z

|v ∗ |p ρ N −1−p dρ dS >

S N −1

0

+∞

Z

|v|p ρ N −1−p dρ dS . 0

Thus, we have R(v ∗ ) 6 R(v), thus, 1,p

1 ,p

inf{R(v) : v ∈ Da,b (RN ) is radial} 6 inf{R(v) : v ∈ Da,b (RN )}. On the other hand, it is trivial that 1,p

1 ,p

inf{R(v) : v ∈ Da,b (RN ) is radial} > inf{R(v) : v ∈ Da,b (RN )}, which implies the desired result.



Lemma 2.3. If µ ∈ (0, µ), b ∈ [a, a + 1), then any minimizer of S0,µ is radial. Proof. We rewrite those integrals in the definition of S0,µ in polar coordinates. Noting that |Du|2 = |∂r u|2 + have Qµ (u) =

Z

Z S N −1

+∞



|∂r u|2 +

0

1 r

|Λu|2 2

Making the change of variables r = ρ k , k = Qµ (u) = k1−p

Z

+∞

Z S N −1



p/2

r N −1−ap dr dS − µ

N −p N −(a+1)p

|Λu|2 ρ2

|∂ρ u|2 + k2

0

Z

+∞

Z

S N −1

|u|p r N −1−(a+1)p dr dS .

1 r2

|Λu|2 , we

(2.2)

0

> 1, from (2.2), we have

p/2

ρ N −1 dρ dS − kµ

Z

+∞

Z S N −1

|u|p ρ N −p−1 dρ dS .

(2.3)

0

p

On the other hand, the restriction condition ku; Lb∗ (RN )k = 1 becomes

Z

+∞

Z

|u|p∗ ρ

k S N −1

(N −p)p∗

−1

p

dρ dS = 1.

(2.4)

0 1

p

To cancel the coefficient k in (2.4), let v = k p∗ u, then for ku; Lb∗ (RN )k = 1, have Qµ (u) =

Z

Z S N −1

+∞

(

p

1−p− p ∗ k



0

|Λv|2 |∂ρ v| + k ρ2 2

2

p/2

ρ

N −1

p

−

1− k p∗

R

S N −1

R +∞ 0

|v|p∗ ρ

(N −p)p∗ p

−1

dρ dS = 1, and we

) µ|v| ρ p

N −1−p

dρ dS ,

(2.5)

that is,

(Z S0,µ = inf

:v∈

1 ,p Da,b

+∞

Z S N −1

(R ), N

(

p

1−p− p ∗ k 0

Z

+∞

Z S N −1

)   2 p/2 p 1− p 2 2 |Λv| N −1 p N − 1 − p |∂ρ v| + k dρ dS ρ − k ∗ µ|v| ρ ρ2 )

|v|p∗ ρ

(N −p)p∗ p

−1

dρ dS = 1 .

(2.6)

0

Since k > 1, we have 1,p

S0,µ > inf{R(v) : v ∈ Da,b (RN )}.

(2.7)

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

849

Let us define another constant 1 ,p

n

o

p

S0r ,µ = inf Qµ (u) : u ∈ Da,b (RN ) is radial, ku; Lb∗ (RN )k = 1 . It is from the definition of infimum that S0r ,µ > S0,µ .

(2.8)

Obviously, a function u defined in RN is radial if and only if Λu = 0. Thus it follows from (2.5) that 1 ,p

S0r ,µ = inf{R(v) : v ∈ Da,b (RN ) is radial}.

(2.9)

From Lemma 2.1, we suppose that u0 is a minimizer of S0,µ with u0 ≥ 0, and without loss of generality assume that 1

ku0 ; Lpb∗ (RN )k = 1. That is, S0,µ = Qµ (u0 ). Assume by contradiction that u0 is not radial. Let v0 = k p∗ u0 . For k > 1, from (2.5) and the definition of R(v) we know that Qµ (u0 ) > R(v0 ),

(2.10)

and 1 ,p

R(v0 ) > inf{R(v) : v ∈ Da,b (RN )}.

(2.11)

Applying Lemma 2.2 and (2.8)–(2.11), we have 1 ,p

inf{R(v) : v ∈ Da,b (RN ) is radial} = S0r ,µ > S0,µ = Qµ (u0 )

> R(v0 ) > inf{R(v) : v ∈ D1a,,bp (RN )} = inf{R(v) : v ∈ D1a,,bp (RN ) is radial}, which is a contradiction. Thus, for the case k > 1, any minimizer of S0,µ is radial. As to the case k = 1, i.e., a = 0, using standard properties of the Schwartz symmetrization, one can obtain a contradiction too, e.g., [7].  Remark 2.4. In Appendix 1, [14] gave a similar result for the extremal functions of S0,µ in the special case where a = b = 0, i.e., any nonnegative minimizer of S0,µ in their case is positive, radially symmetric, radially decreasing with respect to 0 and approaches to zero as |x| → ∞. 3. Asymptotic behavior of extremal functions In this section, we describe the asymptotic behavior of radial extremal functions of S0,µ . Our argument here is borrowed N −(a+1)p

from Section 3.2 of [1]. Note that our choice of δ = p solution to (2.1). Rewriting in polar coordinates, we have

(r

N −1−ap

u) +r

0 p−2 0 0

|u |

N −1



µ

|u|p−2 u r (a+1)p

+

|u|p∗ −2 u r bp∗



is different from δ =

N −p p

= 0.

in [1]. Let u(r ) be a nonnegative radial

(3.1)

Set y(t ) = r δ u(r ),

t = log r , where δ =

N −(a+1)p . p

z (t ) = r (1+δ)(p−1) |u0 (r )|p−2 u0 (r ),

(3.2)

A simple calculation shows that

 2−p dy   = δ y + |z | p−1 z ; dt

(3.3)

  dz = −δ z − |y|p∗ −2 y − µ|y|p−2 y.

dt It follows from (3.3) that y satisfies the following equation:

(p − 1)|δ y − y0 |p−2 (δ y0 − y00 ) + δ|δ y − y0 |p−2 (δ y − y0 ) − µyp−1 − yp∗ −1 = 0.

(3.4)

It is easy to see that the complete integral of the autonomous system (3.3) is V (y, z ) =

1 p∗

|y|p∗ +

µ p

|y|p +

p−1 p

p

|z | p−1 + δ yz .

(3.5)

Applying the phase analysis methods on autonomous system (3.3), similar to Lemmas 3.6–3.9 in [1], we have the following four lemmas. We will omit proofs of the first three lemmas, which are a slight modification of those of Lemmas 3.6–3.8 in [1]. The interested reader can refer to [1]. The idea of the fourth Lemma is also similar to that of Lemma 3.9 in [1], with a different choice of the function ξ (s) and roots of the function equation ξ (s) = 0. We shall write down its complete proof for completeness.

850

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

Lemma 3.1. y > 0 and z < 0 are bounded. Lemma 3.2. For any t ∈ RN , (y(t ), z (t )) ∈ {(y, z ) ∈ R2 : V (y, z ) = 0}. Lemma 3.3. There exists t0 ∈ R, such that y(t ) is strictly increasing for t < t0 ; and strictly decreasing for t > t0 . Furthermore, we have max y(t ) = y(t0 ) =



N N − ( a + 1 − b) p

t ∈R

 p 1−p ∗ (δ − µ) . p

(3.6)

Since system (3.3) is autonomous, by translation transform, we may assume that t0 = 0 without loss of generality. Lemma 3.4. Suppose that y is a positive solution to (3.4) such that y is increasing in (−∞, 0) and decreasing in (0, +∞), then there exist c1 , c2 > 0, such that lim e(l1 −δ)t y(t ) = y(0)c1 > 0;

(3.7)

lim e(l2 −δ)t y(t ) = y(0)c2 > 0,

(3.8)

t →−∞

t →+∞

where l1 , l2 are zeros of function ξ (s) = (p − 1)sp − (N − (a + 1)p)sp−1 + µ such that 0 < l1 < l2 . Proof. First, it is easy to see that ξ (δ) < 0 and l1 < δ < l2 . Next, we prove (3.7) step by step and omit the proof of (3.8). 1. It follows from (3.3) that d dt

1

(e−(δ−l1 )t y(t )) = −(δ − l1 )e−(δ−l1 )t y(t ) + e−(δ−l1 )t (δ y(t ) + |z | p−1 ) 1

=e

−(δ−l1 )t

|z (t )| p−1 y(t ) l1 − y(t )

! .

(3.9)

Rewriting the above equation into the integral form, we have e−(δ−l1 )t y(t ) = y(0)e−

R0 t

(l1 −y(s)−1 |z (s)|1/p−1 )ds

.

(3.10)

1

|z (s)| p−1 y(s)

2. Claim: H (s) =

is a increasing function from (−∞, 0] into (l1 , δ].

In fact, we shall prove that H 0 (s) > 0 for s < 0. A direct computation shows that 2−p

H (s) = 0

1

− p−1 1 y(s)z 0 (s)|z (s)| p−1 − |z (s)| p−1 y0 (s) y2 (s)

.

Replacing formulas of y0 (s) and z 0 (s) from (3.3), and noting (3.5) and Lemma 3.2, it follows that H 0 (s) =



1 p

−

1 p∗



yp∗ (s) > 0.

Thus, H is strictly increasing on (−∞, 0]. On the other hand, from (3.3) and y0 (0) = 0, we have H (0) = δ . From (3.3) and (3.5), a direct computation shows that H 0 (s) = −

(a + 1 − b)p H (s)2−p ξ (H (s)), (p − 1)(N − (a + 1 − b)p)

(3.11)

where

ξ (s) = (p − 1)sp − (N − (a + 1)p)sp−1 + µ. From the definitions of l1 , l2 and (3.11), it follows that l1 and l2 are two trivial solutions of (3.11) on (0, +∞), and limt →−∞ H (s) = l1 , which proves our claim. 3. (3.7) holds. From the above claim and (3.10), it follows that e−(δ−l1 )t y(t ) > 0 is decreasing on (−∞, 0], and hence the limit limt →−∞ e−(δ−l1 )t y(t ) exists. Set

α ≡ lim e−(δ−l1 )t y(t ) = y(0)e t →−∞

R0

−∞ (H (s)−l1 )ds

.

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

851

To prove (3.7), it suffices to show that α < +∞. From the definitions of l1 , l2 , we may suppose that H 0 (s) = (H (s) − l1 )(H (s) − l2 )g (H (s)), where g is a continuous negative function on the interval [l1 , δ] and thus satisfies |g (H (s))| > c1 > 0. From (3.10), it follows that

α = lim e−(δ−l1 )t y(t ) = y(0)e

R0

−∞ (H (s)−l1 )ds

t →−∞

= y(0)e

Rδ

l1 [(H (s)−l2 )g (H (s))]

−1 dH (s)

.

Since l2 > δ and |g (H (s))| > c1 on [l1 , δ], we know that

Z

δ

[(H (s) − l2 )g (H (s))]−1 dH (s) < +∞,

l1

that is, α < +∞, thus (3.7) follows.



In the following corollary, we rewrite these conclusions on y into those on the positive solution u ∈ D1,p (RN ) of Eq. (3.1). Corollary 3.5. Let u ∈ D1,p (RN ) be a positive solution of Eq. (3.1). Then there exist two positive constants C1 , C2 > 0 such that lim r l1 u(r ) = C1 > 0,

lim r l2 u(r ) = C2 > 0

(3.12)

r →+∞

r →0

and lim r l1 +1 |u0 (r )| = C1 l1 > 0,

lim r l2 +1 |u0 (r )| = C2 l2 > 0.

r →+∞

r →0

(3.13)

Proof. From (3.2), we know u(r ) = r −δ y(t ). Applying Lemma 3.4 directly, we have lim r l1 u(r ) = lim e(l1 −δ)t y(t ) = y(0)c1 = C1 > 0, t →−∞

r →0

lim r l2 u(r ) = lim e(l2 −δ)t y(t ) = y(0)c2 = C2 > 0.

r →+∞

t →+∞

Noting that limt →−∞ H (t ) = l1 and limt →+∞ H (t ) = l2 , it follows that 1

lim r l1 u(r ) · H (t ) = lim r l1 u(r ) ·

r →0

r →0

r 1+δ |u0 (r )| |z (t )| p−1 = lim r l1 u(r ) · r →0 y(t ) r δ u(r )

= lim r l1 +1 |u0 (r )| = C1 l1 > 0

(3.14)

r →0

and 1

|z (t )| p−1 r 1+δ |u0 (r )| lim r u(r ) · H (t ) = lim r u(r ) · = lim r l2 u(r ) · r →+∞ r →+∞ r →+∞ y(t ) r δ u(r ) l2

l2

= lim r l2 +1 |u0 (r )| = C2 l2 > 0. 

(3.15)

r →+∞

Next, we shall give a uniqueness result of the positive solution of Eq. (3.1). Theorem 3.6. Suppose that u1 (r ) and u2 (r ) are two positive solutions of Eq. (3.1). Let (y1 (t ), z1 (t )) and (y2 (t ), z2 (t )) be two solutions to ODE system (3.5) corresponding to u1 (r ) and u2 (r ) respectively. If max t ∈(∞,+∞)

y1 (t ) = y1 (0) =



N N − (a + 1 − b)p

(δ p − µ)

 p 1−p ∗

,

(3.16)

and y2 (0) = y1 (0), then (y1 (t ), z1 (t )) = (y2 (t ), z2 (t )). Hence u1 = u2 . Proof. The proof is similar to that of Theorem 3.11 in [1].



Similarly to Theorem 3.13 in [1], we resum the above results together and obtain the following theorem which describes the asymptotic behavior of all the radial solutions to Eq. (3.1) at the origin and the infinity.

852

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

Theorem 3.7. All positive radial solutions to Eq. (2.1) have the form: u(·) = ε

− N −(ap+1)p

u0

· ε

,

(3.17) 1

where u0 is a solution to Eq. (2.1) satisfying u0 (1) = y(0) = [ N −(a+N1−b)p (δ p − µ)] p∗ −p . Furthermore, there exist constants C1 , C2 > 0 such that 0 < C1 6

u0 (x) 6 C2 , (|x|l1 /δ + |x|l2 /δ )−δ

(3.18)

where l1 , l2 are the two zeros of the function ξ (s) = (p − 1)sp − (N − (a + 1)p)sp−1 + µ satisfying 0 < l1 < l2 . 4. Strict inequality In this section, applying the asymptotic behavior of the solutions to Eq. (2.1) obtained in the previous section, we give p some estimates on the extremal function of S0,µ . Let u0 be an extremal function of S0,µ with ku0 ; Lb∗ (RN )k = 1. From the discussion in Sections 2 and 3, we know that u0 is radial, and for all ε > 0, Uε (r ) = ε

− N −(ap+1)p

u0

r  ε

is also an extremal function of S0,µ , and there exists a positive constant Cε such that Cε Uε is a solution to Eq. (2.1). In fact, 1 p −p

∗ from the proof of Lemma 2.1, we know that Cε = S0,µ , which is independent of ε , denoted by C0 . Set u∗ε = C0 Uε , then from Eq. (2.1) we have p∗ p −p

p

∗ Qµ (u∗ε ) = ku∗ε ; Lb∗ kp∗ = S0,µ

N

a+1−b)p = S0(,µ .

(4.1)

For any ε > 0, and m ∈ N large enough such that B 1 ⊆ Ω , define m

um ε (x) =

   u∗ (x) − u∗ 1 , ε

x ∈ B 1 \ {0};

ε

m x ∈ Ω \B1.

0,

m

(4.2)

m

Lemma 4.1. Set ε = m−h , h > 1. Then as m → ∞, we have N

+ O (m−(h−1)[(a+1+l2 )p−N ] ),

(a+1−b)p

Qµ (um ε ) 6 S0,µ

(4.3)

and N

a+1−b)p ku∗ε ; Lpb∗ kp∗ > S0(,µ − O (m−(h−1)[(b+l2 )p∗ −N ] ),

(4.4)

where and afterward O (m−α ) denotes a positive quality which is O(m−α ), but is not o(m−α ), as m → ∞. Proof. We shall only prove (4.3), and omit the proof of (4.4). Since Qµ (um ε)=

Z Ω

R

RN

p |Dum ε| dx = |x|ap

p |Dum ε| |x|ap

dx − µ

p |um ε| RN |x|(a+1)p

R

dx, we estimate each term in Qµ (um ε ) as follows:

|Du∗ε |p dx |x|ap

Z B1

m

Z = RN

|Du∗ε |p dx − |x|ap

6

R N \B 1

|Du∗ε |p dx |x|ap

m

|Du∗ε |p

Z

Z

|x|ap

RN

dx

and

Z Ω

p |um ε| dx = |x|(a+1)p

Z B1

m


(u∗ε (x) − u∗ε m1 )p dx |x|(a+1)p

(4.5)

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

u∗ε (x)p − pu∗ε

Z >


1 m

u∗ε (x)p−1

|x|(a+1)p

B1

853

dx

m

u∗ε (x)p

Z RN

u∗ε (x)p

R N \B 1

R N \B 1

dx − pu∗ε |x|(a+1)p

p

dx = C0 ωN |x|(a+1)p

1

m

B1

(4.6)

r p

ε

r (a+1)p

+∞

dx.

we have

1 m

Z

|x|(a+1)p

m

ε −[N −(a+1)p] u0

+∞

Z

m

= C0p ωN

u∗ε ,

u∗ε (x)p−1

 Z

m

On the other hand, from the definition of

Z

u∗ε (x)p

Z

dx − |x|(a+1)p

=




r N −1 dr

u0 (t )p t N −1−(a+1)p dt

mh−1

= O (m−(h−1)[(a+1+l2 )p−N ] ),

(4.7)

where in the second equality, we make the change of variable t = εr , and in the last equality, we use the asymptotic behavior p−1 p−2 of u0 at infinity. Since h > 1, mh−1 → ∞ as m → ∞. Note that ξ 0 (l2 ) = p(p − 1)l2 − (p − 1)(N − (a + 1)p)l2 > 0, that is (a + 1 + l2 )p − N > 0. Similarly, we can estimate the last integration in (4.6) as follows: u∗ε

u∗ε (x)p−1

 Z 1

m

B1

dx =

|x|(a+1)p

p C0

ωN u0



1 m

Z

1 mε

ε −[N −(a+1)p] u0 r (a+1)p

0

r p−1

ε




r N −1 dr

m

= C0p ωN u0 (mh−1 )

mh−1

Z

u0 (t )p−1 t N −1−(a+1)p dt

0

6 C0 ωN C2 m−(h−1)l2 p [C + m(h−1)[N −(a+1)p−(p−1)l2 ] ] p

= O (m−(h−1)[(a+1+l2 )p−N ] ),

(4.8) p−1

where the last equality is from ξ (l2 ) = 0 and so N −(a+1)p−(p−1)l2 = µ/l2

> 0. Thus, (4.3) follows from (4.5)–(4.8).



Lemma 4.2. Set ε = m−h , h > 1. If c < (a + 1 + l2 )p − N, then

Z RN

p |um ε (x)| dx > O (m−ch ). ( a + 1 ) p − c |x|

(4.9)

Proof. A direct computation shows that

Z RN

p |um ε (x)| dx = ( a + 1 ) p − c |x|

u∗ε (x) − u∗ε

Z

1 m (a+1)p−c



p dx

|x|

B1

m

u∗ε (x)p − pu∗ε

Z >

|x|

B1

1 u∗ε m (a+1)p−c




(x)p−1

dx

m

u∗ε (x)p

Z = RN

u∗ε (x)p

Z

dx − |x|(a+1)p−c

R N \B 1

dx − |x|(a+1)p−c

pu∗ε

m

u∗ε (x)p−1

 Z 1

m

B1

|x|(a+1)p−c

dx.

m

We estimate each of the above integrations as follows: u∗ε (x)p

Z RN

p

dx = C0 ωN ε c |x|(a+1)p−c u∗ε (x)p

Z R N \B 1

m

p

∞

Z

u0 (t )p t N −1−(a+1)p+c dx = O (m−ch ),

(4.10)

0

dx = C0 ωN ε c |x|(a+1)p−c

Z

∞

mh−1

u0 (t )p t N −1−(a+1)p+c dx

= O (m−(h−1)[(a+1+l2 )p−N ]−c )

(4.11)

854

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

and u∗ε

u∗ε (x)p−1

 Z 1

m

B1

|x|(a+1)p−c

dx =

p C0

ω N u0



1 mε

ε −[N −(a+1)p] u0

1 m

Z

r (a+1)p−c

0

r p−1

ε




r N −1 dr

m

= C0p ωN u0 (mh−1 )εc

Z

mh−1

0 −(h−1)l2 p−ch

p

6 C0 ω N C2 m

u0 (t )p−1 t N −1−(a+1)p+c dt

[C + m(h−1)[N −(a+1)p−(p−1)l2 ] ]

= O (m−(h−1)[(a+1+l2 )p−N ]−c ).

(4.12)

Note that since c < (a + 1 + l2 )p − N, we have −ch > −(h − 1)[(a + 1 + l2 )p − N ] − c, that is, we prove the lemma.



Theorem 4.3. If µ ∈ (0, µ), λ > 0, b ∈ [a, a + 1), c ∈ (0, (a + 1 + l2 )p − N ), then the strict inequality (1.5) holds for all λ > 0. Proof. We shall study Qλ,µ (um ε)

p∗ p k um ε ; Lb (Ω )k

.

It follows from Lemmas 4.1 and 4.2 that Qλ,µ (uε ) = Qµ (uε ) − λ m

m

N (a+1−b)p

6 S0,µ

Z Ω

p |um ε (x)| dx |x|(a+1)p−c

+ O (m−(h−1)[(a+1+l2 )p−N ] ) − O (m−ch )

(4.13)

and N

p∗ (a+1−b)p∗ p k um − O (m−(h−1)[(b+l2 )p∗ −N ]p/p∗ ) ε ; Lb (Ω )k > S0,µ N

a+1−b)p∗ = S0(,µ − O (m−(h−1)[(a+1+l2 )p−N ] ).

(4.14)

Thus, we have N

(a+1−b)p

Qλ,µ (um ε)

k

p∗ um ε Lb

;

(Ω )kp

6

S0,µ

+ O (m−(h−1)[(a+1+l2 )p−N ] ) − O (m−ch ) N

(a+1−b)p∗

S0,µ

− O (m−(h−1)[(b+l2 )p∗ −N ]p/p∗ )

= S0,µ + O (m−(h−1)[(a+1+l2 )p−N ] ) − O (m−ch ).

(4.15)

If c ∈ (0, (a + 1 + l2 )p − N ), we can choose h large enough such that c < ((h − 1)(a + 1 + l2 )p − N )/h and so −ch > −(h − 1)[(a + 1 + l2 )p − N ], thus as m large enough, (1.5) holds.  5. Application In this section, as an application of the strict inequality of (1.5), we consider the existence of nontrivial solutions to the following quasilinear Brezis–Nirenberg type problem involving Hardy potential and Sobolev critical exponent:

 

|Du|p−2 Du |x|ap  u = 0, on ∂ Ω , 

−div



−µ

|u|p−2 u |u|p∗ −2 u |u|p−2 u + λ = , |x|(a+1)p |x|bp∗ |x|(a+1)p−c

in Ω ,

(5.1)

where Ω ⊂ RN is an open bounded domain with C 1 boundary and 0 ∈ Ω , 1 < p < N , p∗ = N −(a+1−b)p , 0 ≤ a < p , a 6 b < (a + 1), c > 0; λ, µ are two positive real parameters. Many authors had considered the semilinear case of problem (5.1), see, e.g., [19,23,13,9,31] and references therein. Note that in the semilinear case, the Hardy potential can be canceled by a simple change of variables. But in our quasilinear case, the Hardy potential cannot be canceled. In [1], the authors obtained the existence and multiplicity results for a quasilinear elliptic equation involving the p-Laplacian with a Hardy type singular potential and a critical nonlinearity, which is similar and related to (5.1), but on the whole space RN and a = b = 0. 1,p To obtain the existence result, let us define the energy functional Eλ,µ on Da,b (Ω ) as Np

Eλ,µ (u) =

1

Z 

p

Ω

|Du|p |u|p |u|p − µ − λ |x|ap |x|(a+1)p |x|(a+1)p−c

 dx −

1 p∗

Z Ω

|u|p∗ dx. |x|bp∗

N −p

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859 1,p

855

1 ,p

It is easy to see that Eλ,µ is well-defined in Da,b (Ω ), and Eλ,µ ∈ C 1 (Da,b (Ω ), R). Furthermore, all the critical points of Eλ,µ are weak solutions to (5.1). We shall apply the Mountain Pass Lemma without the (PS) condition due to Ambrosetti and Rabinowitz [2] to ensure the existence of the (PS)β sequence of Eλ,µ at some Mountain Pass type minimax value level β . Then N

(a+1−b)p

the strict inequality (1.5) implies that β < a+N1−b S0,µ . Finally, combining the generalized concentration compactness principle and a compactness property called the singular Palais–Smale condition due to Boccardo and Murat [4] (cf. also [15]), we shall obtain the existence of nontrivial solutions to (5.1). 1,p Let us define two more functionals on Da,b (Ω ) as follows: 1

Iµ (u) =

Z

p

Ω

|Du|p µ dx − ap |x| p

Z Ω

|u|p dx, |x|(a+1)p

J ( u) =

|u|p

Z Ω

|x|(a+1)p−c

dx,

1,p

|Du|p

µ

|u|p

and denote M = {u ∈ Da,b (Ω ) : J (u) = 1}. For µ ∈ (0, µ), the Hardy inequality shows that 1p |x|ap dx − p (a+1)p dx is a |x| nonnegative measure on Ω . The classical results in the Calculus of Variations (cf. [25]) show that Iµ is lower semicontinuous on M . On the other hand the compact embedding theorem in [30] implies that M is weakly closed. Thus the direct method ensures that Iµ attains its minimum on M . Denote λ1 = min{Iµ (u) : u ∈ M } > 0. From the homogeneity of Iµ and J, λ1 is the first nonlinear eigenvalue of the problem:

 

|Du|p−2 Du |x|ap  u = 0, on ∂ Ω . 



−div

−µ

|u|p−2 u |u|p−2 u = λ , |x|(a+1)p |x|(a+1)p−c

in Ω ,

(5.2)

The following lemma indicates that Eλ,µ satisfies the geometric condition of the Mountain Pass Lemma without the (PS) condition due to Ambrosetti and Rabinowitz [2]. The proof is direct and is omitted. Lemma 5.1. If µ ∈ (0, µ), λ ∈ (0, λ1 ), then: (i) Eλ,µ (0) = 0; (ii) ∃ α, r > 0, s.t. Eλ,µ (u) > α , if kuk = r; 1 ,p

(iii) For any v ∈ Da,b (Ω ), v 6= 0, there exists T > 0 such that Eλ,µ (t v) 6 0 if t > T . 1,p

For v ∈ Da,b (Ω ) with kvk > r and Eλ,µ (v) 6 0, set

β := inf max Eλ,µ (γ (t )), γ ∈Γ t ∈[0,1]

where 1,p

Γ := {γ ∈ C ([0, 1], Da,b (Ω )) | γ (0) = 0, γ (1) = v}. It is easy to see that β is independent of the choice of v such that Eλ,µ (v) 6 0, and furthermore β > α . If β is finite, from Lemma 5.1 and the Mountain Pass Lemma, there exists a (PS)β sequence {um }∞ m=1 of Eλ,µ at level β , that is, Eλ,µ (um ) → β 1 ,p

1,p

0 and Eλ,µ (um ) → 0 in the dual space (Da,b (Ω ))0 of Da,b (Ω ) as m → ∞.

Lemma 5.2. If µ ∈ (0, µ), λ ∈ (0, λ1 ), then the strict inequality (1.5) is equivalent to

β<

a+1−b N

N

(a+1−b)p

S0,µ

.

(5.3)

Proof. 1. (1.5) H⇒ (5.3). p Let v1 be a function such that kv1 ; Lb∗ (Ω )k = 1, and Qλ,µ (v1 ) < S0,µ . We have

 p  t t p∗ β 6 sup Eλ,µ (t v1 ) = sup Qλ,µ (v1 ) − p p∗ 0
<

p∗

a+1−b N

N

N

(a+1−b)p

S0,µ

.

2. (5.3) H⇒ (1.5). Since λ < λ1 , for u = g (t ) = t v with t closed to 0, we have (DEλ,µ (u), u) > 0; while for u = g (1) = v , we have

(DEλ,µ (v), v) < pEλ,µ (v) 6 0.

(5.4)

856

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

Considering the function f (t ) = Eλ,µ (t v) ∈ C 1 ([0, 1], R), we have that f 0 (t ) > 0 for t closed to 0, and f−0 (1) 6 0. From the medium value theorem, there exists t0 ∈ (0, 1) such that f 0 (t0 ) = 0, that is, for u = t0 v , we have

(DEλ,µ (u), u) = Qλ,µ (u) − ku; Lpb∗ (Ω )kp∗ = 0. Thus a direct computation shows that Qλ,µ (u)



= Qλ,µ (u)1−p/p∗ =

ku; Lpb∗ (Ω )kp

 (a+1N−b)p

N a+1−b

Eλ,µ (u)

,

that is,

β = inf max Eλ,µ (γ (t )) >

a+1−b

γ ∈Γ t ∈[0,1]

Hence (5.3) H⇒ (1.5).

N

N

Sλ,µ (p, a, b, Ω ) (a+1−b)p .

 1,p

Lemma 5.3. If µ ∈ (0, µ), λ ∈ (0, λ1 ), then any (PS)β sequence of Eλ,µ is bounded in Da,b (Ω ). Proof. Suppose that {um }∞ m=1 is a (PS)β sequence of Eλ,µ . As m → ∞, we have

β + o(1) = Eλ,µ (um )  Z  Z 1 |Dum |p |um |p |um |p 1 |um |p∗ = − µ − λ dx − dx ap ( a + 1 ) p ( a + 1 ) p − c p Ω |x| | x| |x| p∗ Ω |x|bp∗

(5.5)

and o(1)kϕk = (DEλ,µ (um ), ϕ)

Z  = Ω

|Dum |p−2 Dum · Dϕ |um |p−2 um ϕ |um |p−2 um ϕ − µ − λ |x|ap |x|(a+1)p |x|(a+1)p−c



Z dx − Ω

|um |p∗ −2 um ϕ dx, |x|bp∗

(5.6)

1,p

for any ϕ ∈ Da,b (Ω ). From (5.5) and (5.6), as m → ∞, it follows that p∗ β + o(1) − o(1)kum k = p∗ Eλ,µ (um ) − (DEλ,µ (um ), um )



p∗



p∗

=

p

>

p

 >

p∗ p

Z   |Dum |p |um |p |um |p −1 − µ − λ dx |x|ap |x|(a+1)p |x|(a+1)p−c Ω  Z   λ |Dum |p |um |p −1 1− − µ dx λ1 |x|ap |x|(a+1)p Ω    λ µ −1 1− 1− kum kp . λ1 µ

1,p

Thus, {um }∞ m=1 is bounded in Da,b (Ω ) if µ ∈ (0, µ), λ ∈ (0, λ1 ).



1 ,p

From the boundedness of {um }∞ m=1 in Da,b (Ω ), we have the following medium convergence: 1,p

p

p

um * u

in Da,b (Ω ), L1 (Ω ) and Lb∗ (Ω ),

um → u

in Lα (Ω ) if 1 ≤ r <

um → u

a.e. in Ω .

r

Np

,

α

N −p r

< (a + 1) + N



1 r

−

1 p



,

p

∗ In order to obtain the strong convergence of {um }∞ m=1 in Lb (Ω ), we need the following generalized concentration compactness principle (cf. also [26] and [28] and references therein). The proof is similar to that in [20] and we omit it.

Lemma 5.4 (Concentration Compactness Principle). Suppose that M (RN ) is the space of bounded measures on RN , and {um } ⊂ 1,p Da,b (Ω ) is a sequence such that: 1 ,p

um * u

−(a+1)p

ξm := |x| |Dum | − µ|x| νm := |x|−bp∗ |um |p∗ dx * ν um → u −ap

p

p

| um |




dx * ξ

in Da,b (Ω ), in M (RN ), in M (RN ), a.e. on RN .

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

857

Then there are the following statements: (1) There exist some at most countable set J, a family {x(j) : j ∈ J } of distinct points in RN , and a family {ν (j) : j ∈ J } of positive numbers such that

ν = |x|−bp∗ |u|p∗ dx +

X

ν (j) δx(j) ,

(5.7)

j∈J

where δx is the Dirac-mass of mass 1 concentrated at x ∈ RN . (2) The following inequality holds

ξ ≥ (|x|−ap |Du|p − µ|x|−(a+1)p |u|p ) dx +

X

ξ (j) δx(j) ,

(5.8)

j∈J

for some family {ξ (j) > 0 : j ∈ J } satisfying S0,µ ν (j) In particular,


p/p∗

P

j∈J

6 ξ (j) ,

ν

for all j ∈ J .


(j) p/p∗

(5.9)

< ∞.

Lemma 5.5. If µ ∈ (0, µ), λ ∈ (0, λ1 ), let {um }∞ m=1 be a (PS)β sequence of Eλ,µ at level β defined above. (5.3) implies that ν (j) = 0 for all j ∈ J, that is, up to a subsequence, um → u in Lpb∗ (Ω ) as m → 0. 1,p

p−2 Proof. From Lemma 5.3, {um }∞ Dum is bounded in Lp0 (Ω ; |x|−ap ) m=1 is bounded in Da,b (Ω ), then we have that |Dum |

0

where p is the conjugate exponent of p, i.e. such that

1 p

+

1 p0

= 1. Without loss of generality, we suppose that T ∈ L (Ω ; |x| p0

−ap


N

,


N )


N |Dum |p−2 Dum * T in Lp0 (Ω ; |x|−ap ) . Also, |um |p−2 um is bounded in Lp0 (Ω ; |x|−(a+1)p ), |um |p∗ −2 um is bounded in Lp∗ 0 (Ω ; |x|−bp∗ ), and um → u almost everywhere in Ω , thus it follows that

|um |p−2 um * |u|p−2 u in Lp0 (Ω ; |x|−(a+1)p ) and

|um |p∗ −2 um * |u|p∗ −2 u in Lp∗ 0 (Ω ; |x|−bp∗ ). From the compactness embedding theorem in [30], it follows that

|um |p−2 um → |u|p−2 u in Lp0 (Ω ; |x|−(a+1)p+c ). Taking m → ∞ in (5.6), we have T · Dϕ

Z Ω

|x|ap

dx = µ

Z Ω

|u|p−2 uϕ dx + λ |x|(a+1)p

|u|p−2 uϕ dx + |x|(a+1)p−c

Z Ω

Z Ω

|u|p∗ −2 uϕ dx, |x|bp∗

(5.10)

1,p

¯ ), and take m → ∞, it follows that for any ϕ ∈ Da,b (Ω ). Let ϕ = ψ um in (5.6), where ψ ∈ C (Ω Z Ω

ψ dξ +

uT · Dψ

Z Ω

|x|ap

Z dx = Ω

ψ dν + λ

Z Ω

|u|p ψ dx. |x|(a+1)p−c

(5.11)

Let ϕ = ψ u in (5.10), it follows that uT · Dψ

Z Ω

|x|

ap

Z dx + Ω

ψ T · Du dx = µ |x|ap

Z Ω

|u|p ψ dx + λ |x|(a+1)p

Z Ω

|u|p ψ dx + |x|(a+1)p−c

Z Ω

|u|p∗ ψ dx, |x|bp∗

(5.12)

Thus, from (5.7) and Lemma 5.4, (5.11) and (5.12) imply that

Z Ω

ψ dξ =

Z Ω

Z = Ω

ψ T · Du dx − µ |x|ap

Z

ψ T · Du dx − µ |x|ap

Z

Letting ψ → δx(j) , we have

ξ (j) = ν (j) .

Ω

|u|p ψ dx + |x|(a+1)p

Ω

X |u|p ψ dx + ν (j) ψ(x(j) ). ( a + 1 ) p |x| j∈J

Z Ω

ψ dν −

Z Ω

|u|p∗ ψ dx |x|bp∗ (5.13)

858

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

Combining with (5.9), it follows that ν (j) > S0,µ ν (j)


p/q

, which means that

N

a+1−b)p ν (j) > S0(,µ ,

if ν

(j)

(5.14)

6= 0. On the other hand, taking m → ∞ in (5.5), and using (5.13) with ψ ≡ 1, (5.7) and (5.10), it follows that Z Z Z 1 λ |u|p 1 dξ − dν − dx β = p Ω p∗ Ω p Ω |x|(a+1)p−c ! ! Z Z Z Z |u|p |u|p∗ T · Du 1 X (j) |u|p 1 X (j) λ dx − µ dx − ν + dx = ν + dx − ap ( a + 1 ) p bp ( a ∗ p j∈J p∗ p Ω |x| +1)p−c Ω |x| Ω | x| Ω |x| j∈J   Z  |u|p∗ 1 X (j) 1 1 1 − ν + − = dx bp∗ p p∗ p p∗ Ω |x| j∈J   1 a + 1 − b X (j) 1 X (j) > ν = − ν . (5.15) p

p∗

N

j∈J

j∈J

From (5.14) and (5.15), (5.3) implies that ν p∗

Z Ω

|um | dx → |x|bp∗

(j)

= 0 for all j ∈ J. Hence we have

p∗

Z Ω

|u| dx, |x|bp∗ p

as m → ∞. Thus, the Brezis–Lieb Lemma [5] implies that, up to a subsequence, um → u in Lb∗ (Ω ) as m → 0.



In order to deduce the almost everywhere convergence of Dum in Ω and to obtain the existence of a nontrivial solution to (5.1), we shall apply the variational approach supposed in [15] and a convergence theorem due to Boccardo and Murat 1 ,p 1,p (cf. Theorem 2.1 in [4]). So we suppose that a = 0, and Da,b (Ω ) = W0 (Ω ). Theorem 5.6. If a = 0, µ ∈ (0, µ), λ ∈ (0, λ1 ), b ∈ [0, 1), c ∈ (0, (1 + l2 )p − N ), then there exists a nontrivial solution to (5.1). Proof. Apply the variational approach supposed in [15] and a convergence theorem in [4]. There exists a subsequence of ∞ {um }∞ m=1 , still denoted by {um }m=1 , such that um → u

1 ,q

in W0 (Ω ), q < p, 1,p

which implies that u is a solution to (5.1) in the sense of distributions. Since u ∈ W0 (Ω ), by the density argument, u is a weak solution to (5.1). Next, we shall show that u 6≡ 0. In fact, from the homogeneity and Lemma 5.5, we have 0 < α 6 β = lim Eλ,µ (um ) = lim m→∞

m→∞



1

1

Z



Eλ,µ (um ) −

1 p

(DEλ,µ (um ), um )



|um |p∗ dx |x|bp∗

= lim − m→∞ p p∗ Ω  Z 1 1 |u|p∗ = − dx. bp∗ p p∗ Ω |x| Thus, u 6≡ 0.



Remark 5.7. [14] considered the existence and nonexistence of a weak solution to a special case of (5.1) in the whole space, i.e., Ω = Rn . In light of Theorem 5.6, we conjecture that the conclusion is also true for 0 6 a < Conjecture 5.8. If 0 6 a < nontrivial solution to (5.1).

N −p p

N −p . p

, µ ∈ (0, µ), λ ∈ (0, λ1 ), b ∈ [a, a + 1), c ∈ (0, (a + 1 + l2 )p − N ), then there exists a

Acknowledgments The authors would like to thank the anonymous referee for drawing their attention to paper [14]. The first author was supported by Grant 10871187 from the National Natural Science Foundation of China.

B. Xuan, J. Wang / Nonlinear Analysis 71 (2009) 845–859

859

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