Hardy–Sobolev inequalities in half-space and some semilinear elliptic equations with singular coefficients

Nonlinear Analysis 66 (2007) 324–348 www.elsevier.com/locate/na

Hardy–Sobolev inequalities in half-space and some semilinear elliptic equations with singular coefficients Shaowei Chen a,b,∗ , Shujie Li b a Department of Mathematics, Fujian Normal University, Fuzhou 350007, PR China b Institute of Mathematics, Academy of Mathematics and System Science, Academia Sinica, Beijing 100080, PR China

Received 1 April 2005; accepted 23 February 2006

Abstract We study some semilinear elliptic equations with singular coefficients which relate to some Hardy–Sobolev inequalities. We obtain some existence results for these equations and give a theorem for prescribing the Palais–Smale sequence for these equations. Moreover, we find some interesting connections between these equations and some semilinear elliptic equations in hyperbolic space. Using these connections, we obtain many new results for these equations. c 2006 Published by Elsevier Ltd  MSC: 35J20; 35J70 Keywords: Hardy–Sobolev inequalities; Semilinear equations; Palais–Smale sequence; Hyperbolic space

1. Introduction Recently, there has been much work devoted to studying some semilinear elliptic equations with singular coefficients. These equations often relate to some variational inequalities, such as the Hardy–Sobolev inequality or the Caffarelli–Kohn–Nirenberg inequality (see [6]). See [10,12, 16,26,7,2,11] and [24] for details. In the papers mentioned, the authors investigated the following semilinear elliptic equations in R N : −u =

λ ∗ u + |u|2 −2 u, |x|2

∗

−u =

|u|2 (s)−2u , |x  |s

−div(|x|−2a ∇u) = |x|−bp u p−1 ,

∗ Corresponding author at: Department of Mathematics, Fujian Normal University, Fuzhou 350007, PR China.

E-mail addresses: [email protected] (S. Chen), [email protected] (S. Li). c 2006 Published by Elsevier Ltd 0362-546X/$ - see front matter  doi:10.1016/j.na.2006.02.032

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

325

where x ∈ R N , x  ∈ Rk , 2 ≤ k ≤ N, 2∗ (s) = 2(N − s)/(N − 2) and s ∈ (0, 2), −∞ < a < (N − 2)/2, a ≤ b ≤ a + 1 and p = 2N/(N − 2 + 2(b − a)). And they obtained many results concerning the existence, multiplicity, symmetry and uniqueness of solutions for these equations. In this paper, we study the following semilinear elliptic equations with singular coefficients N: in half-space R+ −u =

|u|2

∗ (s)−2

u

x Ns

N u ∈ D01,2 (R+ ),

N in R+ ,

(1.1)

and ∗

−u =

|u|2 (s)−2u in Ω , d s (x)

u ∈ H01(Ω ),

(1.2)

N ) is the space given by the completion of C ∞ (R N ) under the norm u = where D01,2 (R+ + 0  N = {x = (x , x , . . . , x ) ∈ R N | x 2 ( R N |∇u| dx)1/2, R+ 1 2 N N > 0}, N ≥ 3, is the half-space +

and Ω is a bounded domain with smooth boundary in R N , d(x) = dist(x, ∂Ω ). As far as we are aware, Eqs. (1.1) and (1.2) have never been studied before. Eqs. (1.1) and (1.2) relate to the following Hardy–Sobolev inequalities respectively:  2/2∗ (s)  ∗ |u|2 (s) N S dx ≤ |∇u|2 dx, u ∈ D01,2 (R+ ), (1.3) N N x Ns R+ R+ and

 SΩ

∗

|u|2 (s) dx s Ω d (x)

2/2∗ (s)

 ≤

Ω

|∇u|2 dx,

u ∈ H01(Ω ).

(1.4)

The difficulty of these equations arises from the so-called losses of compactness, i.e., the functionals which these equations correspond to do not satisfy the Palais–Smale condition. Breaking down of compactness arises in many problems naturally appearing in many fields of mathematic and physics, for example, the elliptic equations with critical Sobolev exponent, harmonic mapping equations and Yang–Mills equations. Many authors have developed various methods to overcome these difficulties. In the celebrated paper [5], Brezis and Nirenberg showed the compactness for minimizing sequences with energy below a certain critical level. And in the remarkable paper [3], Brezis and Coron give a precise prescription of the Palais–Smale sequence for the H -system. The idea of the proof of the main result in Sections 2 and 4 was inspired by the work of Brezis, Nirenberg and Coron. In Section 2 of this paper, we show that the best constant S in (1.3) can be achieved with a cylindrically symmetric function and it gives a positive solution for (1.1) (Theorem 2.4). We show that for any bounded domain Ω with smooth boundary in R N , the best constant SΩ in (1.4) is less than or equal to S (Theorem 2.5). In Section 3, we prove that the solutions of (1.1) and (1.2) are bounded (Theorems 3.1 and 0,α 3.3) and under some additional conditions, they are Cloc regular (Theorems 3.4 and 3.6). In Section 4, we give a precise prescription of the Palais–Smale sequence for Eq. (1.2) which indicates that Eq. (1.1) is the “limiting” equation of (1.2) (Theorem 4.1). As a consequence, we deduce that if SΩ < S, then SΩ can be achieved (Corollary 4.5).

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In Section 5, by means of some conformal transformations, we show that Eq. (1.1) is equivalent to one of the following three equations: 2s ∗ |u|2 (s)−2u in B(0, 1), u ∈ H01(B(0, 1)), 2 s (1 − |x| ) 2s ∗ −u = |u|2 (s)−2u in R N \ B(0, 1), u ∈ D01,2 (R N \ B(0, 1)), (|x|2 − 1)s

−u =

(1.5) (1.6)

and − H N u =

N(N − 2) ∗ u + |u|2 (s)−2u, 4

u ∈ H 1(H N ),

(1.7)

where H N is the N-dimensional hyperbolic space. Using these equivalent relations, we get many interesting results. For example, we prove that Eq. (1.5) has radially symmetric and non-radially symmetric positive solutions (Remark 5.17) and that Eq. (1.1) ((1.5)–(1.7) as well) has infinitely many solutions whose “energy” converges to ∞ (Remark 5.12). We show that if Ω = B(0, 1), then the best constant S B(0,1) in inequality (1.4) cannot be achieved. On the other hand, we illustrate that for some domain Ω , the inequality SΩ < S holds (Theorem 5.4). We prove that under condition (A) (see Section 5), every positive solution of (1.5) must be radially symmetric after a transformation (Corollary 5.16) and it belongs to C 2 (B(0, 1))∩C 1 (B(0, 1)) (Remark 5.8). Furthermore, when N = 3, 1 ≤ s < 3/2 or N = 4, s = 1, we prove that after a transformation, the positive solution for (1.5) and (1.1) is unique (Theorem 5.18 and Corollary 5.19). Finally, we give the explicit formula for the radially symmetric positive solution for Eqs. (1.5) and (1.1) in the case N = 3, s = 4/3 and N = 4, s = 1 (see (5.7) and (5.8)). 2. Two extremal problems In this section, we use a concentration–compactness lemma to prove that the following extremal problems which are related to Hardy–Sobolev inequalities (1.3) and (1.4) attaining their infima:    ∗ |u|2 (s) 1,2 2 N S = inf |∇u| dx | u ∈ D0 (R+ ), dx = 1 , (2.1) N N x Ns R+ R+ and

 SΩ = inf

 |∇u| dx | u ∈ 2

Ω

H01(Ω ),

 ∗ |u|2 (s) dx = 1 . s Ω d (x)

(2.2)

Firstly, we give a simple proof of inequality (1.3). N ), the following Lemma 2.1. There exists a constant C > 0, such that for any u ∈ D01,2 (R+ Hardy inequality holds:   |u|2 C dx ≤ |∇u|2 dx. (2.3) N x2 N R+ R+ N N Proof. We only give the proof for u ∈ C0∞ (R+ ).

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327

By the one-dimensional Hardy inequality (see [14]), there exists a constant C > 0 such that N for all u ∈ C0∞ (R+ ),   +∞ 2  +∞   ∂u 2 |u|   C dx ≤ N  ∂ x  dx N , x N2 N 0 0 therefore  C

   +∞ |u|2 |u|2 dx = C dx N dx  2 2 N N−1 x x R+ N R 0 N       +∞  ∂u 2  2   ≤  ∂ x  dx N dx ≤ N |∇u| dx.  N 0 R N−1 R+

Remark 2.2. Inequality (1.3) follows from (2.3) and the Sobolev inequality in the following way:   ∗2  2/2∗ (s) ∗ 2 (s) |u|2 (s) |u|s 2∗ (s)−s S dx = S · |u| dx N N xs x Ns R+ R+ N   ∗s   2−s 2 (s) 2∗ (s) |u|2 ∗ 2 (0) ≤ S dx |u| dx N x2 N R+ R+ N  ≤ |∇u|2 dx. N R+

 The inequality (1.4) can be derived from the Hardy inequality C Ω [21] for details) and the Sobolev inequality in the same way.

|u|2 d 2 (x)

 dx ≤ Ω |∇u|2 dx (see

The proof of the following concentration–compactness lemma is standard, and one can see [27, Lemma 1.40] for details. N ) be a sequence such that Lemma 2.3. Let {u n } ⊂ D01,2 (R+ N in D01,2 (R+ ), N in M(R+ ),

un  u |∇(u n − u)|2  μ 1 ∗ |u n − u|2 (s)  ν s xN un → u

N in M(R+ ), N a.e. on R+

N N where M(R+ ) is the space consisting of finite measures in R+ . Define   |∇u n |2 dx, ν∞ = lim lim sup μ∞ = lim lim sup R→∞ n→∞

R→∞ n→∞

|x|≥R

|x|≥R

1 ∗ |u n |2 (s) dx. (2.4) x Ns

Then it follows that

ν 2/2

∗ (s)

2/2∗ (s) ν∞

≤ S −1 μ ,

≤ S −1 μ∞ ,   |∇u n |2 dx = lim sup n→∞

N R+

(2.5) (2.6) |∇u| dx + μ + μ∞ , 2

N R+

(2.7)

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 1 1 2∗ (s) 2∗ (s) lim sup dx = dx + ν + ν∞ . s |u n | s |u| N N x x n→∞ R+ N R+ N 

Moreover, if u = 0 and ν 2/2

∗ (s)

(2.8)

= S −1 μ , then ν and μ are concentrated at a single point.

Theorem 2.4. For every s ∈ (0, 2) the extremal problem (2.1) attains its infimum at a function u which is cylindrically symmetric, that is, there exists x 0 ∈ R N−1 such that u(x  , x N ) = u(|x  − x 0 |, x N ). N ) be such that Proof. Let {u n } ⊂ D01,2 (R+   1 2∗ (s) |u n | dx = 1, |∇u n |2 → S. N xs N R+ R+ N

Define



Q n (λ) := sup

y∈R N

B(y,λ)

1 ∗ |u n |2 (s) dx. x Ns

Since, for every n, lim Q n (λ) = 0,

λ→0+

lim Q n (λ) = 1,

λ→∞

there exists λn > 0 such that Q n (λn ) = 1/2. Moreover, there exists yn ∈ R N such that  1 2∗ (s) dx = Q n (λn ) = 1/2. s |u n | B(yn ,λn ) x N N−2

Let yn = (yn , yn(N) ), where yn ∈ R N−1 . Set vn (x) = λn 2 u n (λn x + z n ), where z n = (yn , 0). We have   1 1 1 2∗ (s) 2∗ (s) = dx = sup dx, (2.9) s |vn | s |vn | 2 x x B(x n ,1) N y∈R N B(y,1) N where x n = (0, yn(N) /λn ). We distinguish the following two cases. Case 1: {x n } is bounded. Without losing generality, we may assume {vn } satisfies that vn  v |∇(vn − v)|2  μ 1 ∗ |vn − v|2 (s)  ν x Ns vn → v

N in D01,2 (R+ ), N in M(R+ ), N in M(R+ ), N a.e. on R+ .

By Lemma 2.3, we have 2/2∗ (s)

∗ (s)

≤ S −1 μ , ν∞ ≤ S −1 μ∞ ,   |∇vn |2 dx = |∇v|2 dx + μ + μ∞ , S = lim sup

ν 2/2

n→∞

1 = lim sup n→∞



N R+

N R+

 1 1 2∗ (s) 2∗ (s) |v | = dx + ν + ν∞ . n s s |v| N N x x R+ N R+ N

(2.10) (2.11) (2.12)

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

By (1.3), (2.10) and (2.11), we have ⎛ ⎞ 2/2∗ (s)  ∗ 1 ∗ ∗ 2/2 (s) ⎠ |v|2 (s) dx + ν 2/2 (s) + ν∞ S ≥ S⎝ . N xs R+ N

329

(2.13)

 ∗ It follows from (2.12) and (2.13) that R N x1s |v|2 (s) , ν and ν∞ are equal either to 0 or to + N 1. Since {x n } is bounded, by (2.9), ν∞ ≤ 1/2 so that ν∞ = 0. If ν = 1, then v = 0 and ∗

ν 2/2 (s) = S −1 μ . Lemma 2.3 implies that ν concentrated at a single point z. We deduce from (2.12) the contradiction   1 1 1 2∗ (s) 2∗ (s) = sup |v | dx ≥ dx → ν = 1. n s s |vn | 2 B(z,1) x N y∈R N B(y,1) x N  ∗ Thus R N x1s |v|2 (s) dx = 1 and so + N   |∇v|2 dx = S = lim |∇vn |2 dx. n→∞ R N +

N R+

Case 2: x n → ∞, as n → ∞. Define  (N−2)/2 vn (λx)) = G(λ

|x| (λ(N−2)/2 vn (λx))2 N 1 + |x| x Ns R+

From lim G(λ(N−2)/2 vn (λx)) =



λ→0+

N R+

1 ∗ |vn |2 (s) = 1, x Ns

N−2 2

dx.

lim G(λ(N−2)/2 vn (λx)) = 0,

λ→∞

(N−2)/2

we can get that there exists λn > 0 such that G(λn

∗ (s)

vn (λn x)) = 1/2, n = 1, 2, . . .. Set

wn (x) = λn vn (λn x), then  |x| 1 1 2∗ (s) = dx s |wn | N 2 1 + |x| x R+ N and



1 2∗ (s) dx = 1, s |wn | N R+ x N

(2.14)

 N R+

 |∇wn |2 dx =

N R+

|∇vn |2 dx → S.

Without losing generality, we may assume {wn } satisfies that wn  w |∇(wn − w)|2  μ 1 ∗ |wn − w|2 (s)  ν x Ns wn → w

N in D01,2 (R+ ), N in M(R+ ), N in M(R+ ), N a.e. on R+ .

By Lemma 2.3, we have   S = lim sup |∇wn |2 dx = n→∞

N R+

N R+

|∇w|2 dx +

μ + μ∞ ,

(2.15)

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S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

 1 1 2∗ (s) 2∗ (s) 1 = lim sup = dx +

ν + ν∞ , s |wn | s |w| N N x x n→∞ R+ N R+ N 

ν 2/2

∗ (s)

≤ S −1

μ ,

2/2∗ (s)

ν∞

≤ S −1 μ∞ .

(2.17)



Using the same proof as that in case 1, we can deduce that R N x1s |w| + N either to 0 or to 1. Since   ∗ ∗ |x| |wn |2 (s) |x| |wn |2 (s) 1 = dx ≥ dx N 1 + |x| 2 x Ns x Ns R+ |x|≥R 1 + |x|  ∗ |wn |2 (s) R dx, ≥ 1 + R |x|≥R x Ns we have ν∞ ≤ 1/2, so ν∞ = 0. If

ν = 1, then

ν 2/2

ν is concentrated at a point z. From (2.14), we have  1 |x| 1 |z| 2∗ (s) = , dx → s |wn | N 2 1 + |x| x 1 + |z| R+ N and hence |z| = 1. Moreover, by By (2.9), we know that



|wn | s/2∗ (s) xN

B( λxnn , λ1n )

∗ (s)

(2.16)

2∗ (s)

,

ν and ν∞ are equal

= S −1

μ , Lemma 2.3 implies that

∗

N ), we deduce that z = 0. → 0 in L 2loc(s) (R+ N ∗ (s)

|wn |2 s xN

dx = 1/2. If {λn } is bounded, then

   xn 1 |x n − 1| , → ∞, inf |x| | x ∈ B ≥ λn λn λn ν∞ = 0. On the other hand, if λn → ∞, since x n → ∞ as n → ∞. This contradicts the fact that recalling that x n = (0, yn(N) /λn ), |z| = 1 and x n → z as n → ∞, then we know that the ∗ (s)

|wn |2 s xN

concentrates must satisfy z N = 0. This is also a contradiction. These two  ∗ contradictions imply that ν = 0. Therefore R N x1s |w|2 (s) dx = 1 and + N   2 |∇w| dx = S = lim |∇wn |2 dx. point z where

N R+

n→∞ R N +

We now prove that the minimizers for the extremal problem (2.1) are cylindrically symmetric. The following proof was inspired by [2]. N ) denote by u ∗ (r, x ) the non-increasing rearrangement of the function For u ∈ D01,2 (R+ N N−1 , as given by its Schwarz symmetrization. The non-increasing rearrangement u(·, x N ) in R given by its Schwarz symmetrization has the following properties for the non-negative functions u, v with compact support in Rk (see [17] or [2] for details):   (i) Rk u s v p ≤ Rk (u ∗ )s (v ∗ ) p (s, p ≥ 0).   (ii) Rk |u ∗ − v ∗ |q ≤ Rk |u − v|q , for all q > 1.   (iii) Rk |∇u|2 ≥ Rk |∇u ∗ |2 . From (i), we get that   ∗ ∗ u 2 (s)(x  , x N )  (u ∗ (x  , x N ))2 (s)  dx ≤ dx , x Ns x Ns R N−1 R N−1

N for any u ∈ D01,2 (R+ ),

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331

and then 

 1 2∗ (s) 1 ∗ 2∗ (s) u dx ≤ (u ) dx s N x N xs R+ R+ N N

(2.18)

follows upon a further integration with respect to x Ns . From (iii), we have   N |∇x  u|2 dx  ≥ |∇x  u ∗ |2 dx  , for any u ∈ D01,2 (R+ ). R N−1

R N−1

(2.19)

Considering the gradient with respect to the x N -variable, note by (ii) we have    ∗      u (x , x N + t) − u ∗ (x  , x N ) 2   u(x  , x N + t) − u(x  , x N ) 2   dx ≤  dx ,       t t R N−1 R N−1 it follows that  ∗ 2      ∂u   ∂u 2    dx  ≤   dx ,     R N−1 ∂ x N R N−1 ∂ x N and combining with (2.19) we get   2 |∇u| dx ≥ |∇u ∗ |2 dx, N R+

N R+

N for any u ∈ D01,2 (R+ ).

Then the desired result follows from (2.18) and (2.20) immediately.

(2.20) 

Theorem 2.5. If Ω is a bounded domain with smooth boundary in R N , then SΩ ≤ S. Proof. Without losing generality we may assume that 0 ∈ ∂Ω and the tangent plane of ∂Ω on 0 is x N = 0. Therefore, there exists a neighborhood U of 0 such that ∂Ω ∩ U can be represented as {(x  , x N ) | x N = f (x  ), x  ∈ R N−1 , |x  | < δ0 }, where f is a smooth function and f (x  ) = O(|x  |2 ) as |x  | → 0. Moreover, U ∩ Ω = {x ∈ U | x N > f (x  )}. Without losing generality, we can choose U = Uδ0 ,0 := {x ∈ R N | |x  | < δ0 , |x N | < 0 + δ0 }. For 0 < δ < δ0 , let

δ, := {x ∈ R N | |x  | < δ, δ 3/2 < x N < 0 + δ 3/2 }. U 0

δ, ⊂ Ω ∩ Uδ , . Since f (x  ) = O(|x  |2 ), we have | f (x  )| < δ 3/2 if δ small enough and then U 0 0 0

δ, }. It is easy to verify that Set δ := {x/δ 5/4 | x ∈ U 0 δ = {x ∈ R N | |x  | < δ −1/4 , δ 1/4 < x N < δ −5/4 0 + δ 1/4 } N and that δ  ⊂ δ when δ < δ  . Furthermore, for any compact set K ⊂⊂ R+ , there exists 1,2 N N δ > 0 such that K ⊂⊂ δ . Let {u n } ⊂ D(R+ ) be such that u n → u in D0 (R+ ), where u is a minimizer for (2.1). Choose a positive number sequence {δn } such that δn → 0 as n → ∞ N−2 5/4 5/4

δn , , and that the support of u n is contained in δn . Set vn (x) = (δn )− 2 u n (x/δn ), x ∈ U 0

332

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

we have vn ∈ H01(Ω ) and   2 2 N |∇u n | R+ Ω |∇vn | =  2/2∗ (s)  2/2∗ (s) → S. ∗ ∗ |vn |2 (s) |u n |2 (s) Ω

N R+

s xN

(2.21)

s xN

δ, , where C is a positive constant independent of δ, we deduce that From | f (x  )| ≤ Cδ 2 , x ∈ U 0

δ, . x ∈U 0

d(x) ≤ x N + Cδ 2 ,

δ, , we can get Noticing that x N ≥ δ 3/2 for x ∈ U 0

δ, . x ∈U 0

d(x)/x N ≤ 1 + Cδ 1/2 , Thus



  ∗ ∗ ∗ |vn |2 (s) |vn |2 (s) d s (x) |vn |2 (s) 1/2 · s ≤ · (1 + Cδn )s . = s s (x) s (x) x d x d Ω Ω Ω N N

It follows that



SΩ 1/2

(1 + Cδn )2s/2

∗ (s)

≤  Ω

Ω

|∇vn |2

 ∗ ∗ 1/2 s 2/2 (s) |vn |2 (s) d s (x) (1 + Cδn )

Then by (2.21) we deduce that SΩ ≤ S.

 ≤  Ω

Ω

|∇vn |2  ∗ . 2∗ (s) 2/2 (s)

|vn | s xN



Remark 2.6. There exists a bounded domain with smooth boundary in R N (N ≥ 3) such that SΩ < S. In fact, for any ε > 0 there exists a domain Ω such that SΩ < ε. See [8] or [21] for details. 3. Regularity of solutions In this section, we give some results about the regularity of the solutions of Eqs. (1.1) and (1.2). N ). Theorem 3.1. Suppose that N ≥ 3 and s ∈ (0, 2). If v is a solution of (1.1), then v ∈ L ∞ (R+ N Proof. Obviously, by the standard bootstrap argument, we have v ∈ L ∞ loc (R+ ) since the s N singularity of coefficient 1/x N only occurs in the boundary ∂R+ . So, to prove this theorem, N , there exists δ > 0 such that v ∈ L ∞ (B(x , δ)) it is sufficient to prove that for any x 0 ∈ ∂R+ 0 and v(x) = O(|x|2−N ) as |x| → ∞. For t > 2, k > 0, define ⎧ t /2 0≤r ≤k ⎨r ,   t t h(r ) = t t −1 k2, r ≥ k ⎩ k2 r + 1− 2 2 r and φ(r ) = 0 |h  (s)|2 ds. It is easy to verify that

|r φ(r )| ≤

t2 |h(r )|2 , 4(t − 1)

(3.1)

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

333

and |φ(r ) − h(r )h  (r )| ≤ Ct |h(r )h  (r )|,

(3.2)

t −2 2(t −1)

N where Ct = < 1. Let x 0 ∈ ∂R+ , 0 < r1 < r0 and η ∈ C0∞ (R N ) satisfying 0 ≤ η ≤ 1 N in R , η ≡ 1 in B(x 0 , r1 ), η ≡ 0 in R N \ B(x 0 , r0 ) and |∇η| ≤ 2/(r0 − r1 ). Noticing that N η2 φ(u) ∈ D01,2 (R+ ), we can get    2 2  2 2 ∇u∇(η φ(u)) = η (h (u)) |∇u| + 2 ηφ(u)∇u∇η N R+

 =

N R+

N R+

N R+

 η2 |∇(h(u))|2 + 2

ηφ(u)∇u∇η.

N R+

Since |∇(ηh(u))|2 = η2 |∇(h(u))|2 + h 2 (u)|∇η|2 + 2ηh(u)∇(h(u))∇η, by (3.2), we have  ∇u∇(η2 φ(u)) N R+



 2

=

N R+

|∇(ηh(u))| −



−2  ≥ 

N R+

h 2 (u)|∇η|2

ηh(u)h  (u)∇u∇η + 2 

|∇(ηh(u))|2 − 

N R+

2

≥ Since

N R+

N R+

N R+

N R+

N R+

ηφ(u)∇u∇η 

h 2 (u)|∇η|2 − 2

N R+

η|φ(u) − h(u)h  (u)||∇u∇η|



2

|∇(ηh(u))| −



2

h (u)|∇η| − 2Ct

N R+

|ηh(u)∇(h(u))∇η|.

 N R+

|ηh(u)∇(h(u))∇η| 

=  ≤

N R+

N R+

|(∇(ηh(u)) − h(u)∇η)∇η||h(u)|  |h(u)∇(ηh(u))∇η| +

N R+

|h(u)|2 |∇η|2

   1 1 2 2 2 ≤ h (u)|∇η| + |∇(ηh(u))| + |h(u)|2 |∇η|2 , N 2 R+N 2 R+N R+ by (3.3) and Hardy–Sobolev inequality (1.3), we deduce that  ∇u∇(η2 φ(u)) N R+

≥



 N R+

|∇(ηh(u))|2 −

N R+

h 2 (u)|∇η|2

     1 1 2 2 2 2 2 − 2Ct h (u)|∇η| + |∇(ηh(u))| + |h(u)| |∇η| N 2 R+N 2 R+N R+

(3.3)

334

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

 t 2 = |∇(ηh(u))| − (1 + 3Ct ) h 2 (u)|∇η|2 N 2(t − 1) R+N R+   ∗2  ∗ |ηh(u)|2 (s) 2 (s) St ≥ − (1 + 3Ct ) h 2 (u)|∇η|2 . N N 2(t − 1) |x N |s R+ R+ 

By (3.1), we have  ∗ |u|2 (s)−2u 2 η φ(u) N |x N |s R+  ∗ |u|2 (s)−2 t2 |ηh(u)|2 ≤ 4(t − 1) R+N |x N |s  2∗ (s)−2   ∗2  ∗ (s) ∗ 2∗ (s) 2 2 |u| |ηh(u)|2 (s) 2 (s) t ≤ . s N 4(t − 1) |x N |s η=0 |x N | R+

(3.4)

(3.5)

Noticing that u is a solution of (1.1), by (3.4) and (3.5) we have   ∗2 ∗ |ηh(u)|2 (s) 2 (s) N |x N |s R+ t ≤ 2S +



∗

η=0

|u|2 (s) |x N |s

 2∗ (s)−2  2∗ (s)

|ηh(u)|2 N |x N |s R+

∗ (s)



2 2∗ (s)

 2(1 + Ct )(t − 1) h 2 (u)|∇η|2 . N St R+

(3.6)

Let t0 > 2 be a fixed number. Choose r0 small enough such that t0 2S Since



∗

η=0

|u|2 (s) |x N |s

2(1+3C t0 )(t0 −1) t0



 2∗ (s)−2 ∗ 2 (s)

< 8 (note that 0 < Ct0 < 1) and |∇η| < 2/(r0 − r1 ), from (3.6) we have ∗

B(x 0 ,r1 )

< 1/2.

|h(u)|2 (s) |x N |s



2 2∗ (s)

≤

64 S(r0 − r1 )2

Let k → ∞, from (3.7) we deduce that  ∗ |u|2 (s)t0/2 < +∞. |x N |s B(x 0,r1 )

 B(x 0,r0 )

h 2 (u).

(3.7)

(3.8)

Let 0 < r2 < r1 and η1 ∈ C0∞ (R N ) satisfying 0 ≤ η1 ≤ 1 in R N , η1 ≡ 1 in B(x 0 , r2 ), η1 ≡ 0 in R N \ B(x 0 , r1 ) and |∇η1 | ≤ 2/(r1 − r2 ). By (3.8) and (3.1) we have  ∗ |u|2 (s)−2u 2 η1 φ(u) N |x N |s R+

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

 ∗ t2 |u|2 (s)−2 |η1 h(u)|2 4(t − 1) R+N |x N |s  2(2∗∗ (s)−2)  2  ∗ |u|2 (s)t0/2 2 (s)t0 |η1 h(u)|q q t2 ≤ , N 4(t − 1) η1 =0 |x N |s |x N |s R+

335

≤

(3.9)

∗

2 ∗ where 2(22∗ (s)−2) (s)t0 + q = 1. It follows that 2 < q < 2 (s). Since u is a solution of (1.1), by (3.4) and (3.9) we can get

 ∗2 ∗ |η1 h(u)|2 (s) 2 (s) N |x N |s R+  2(2∗∗ (s)−2)  2  ∗ (s)t /2 2 (s)t0 2 0 |u| |η1 h(u)|q q t ≤ N 2S |x N |s |x N |s η1 =0 R+  2(1 + 3Ct )(t − 1) + h 2 (u)|∇η1 |2 N St R+ 2    |η1 h(u)|q q 1 ≤C t+ , N |x N |s (r2 − r1 )2 R+



(3.10)

where C is a positive constant. Define 1/ l  |u|l Φ(l, r ) = . s B(x 0,r) |x N | Let k → ∞ in (3.10). We have  ∗   2   l 2 (s) 1 q Φ . l, r2 ≤ C 2/ l l + l, r Φ 1 2 2 (r1 − r2 )2 Let l = 2t/q, we get  ∗   q t 2 (s) 1 q/t 2t Φ t, r2 ≤ C + Φ(t, r1 ). q q (r1 − r2 )2 Now let tm = 2∗ (s)(2∗ (s)/2)m−1 , rm = r0 (2 + 2−m )/4, m = 1, 2, . . .. Iterating the above procedure, we obtain  q m  ti 1 q/ti 2ti + C Φ(t1 , r1 ). Φ(tm , rm ) ≤ q (r1 − r2 )2 i=2 Since lim

m→∞

m  i=2

 C

q/ti

1 2ti + q (r1 − r2 )2

we deduce that u ∈ L ∞ (B(x 0 , r0 /2)).

q ti

< ∞,

336

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Noticing that Eq. (1.1) is invariant under the Kelvin transform u(x) → |x|2−N u(x/|x|2), we deduce that |x|2−N u(x/|x|2) ∈ L ∞ (B(0, δ)). It follows that u(x) = O(|x|2−N ) as |x| → ∞. So N ).  we get u ∈ L ∞ (R+ Corollary 3.2. If v is a solution of (1.1), then v(x) = O(|x|2−N ) as |x| → ∞. Likewise, we have Theorem 3.3. If u is a solution of Eq. (1.2), then u ∈ L ∞ (Ω ). Theorem 3.4. Suppose that N ≥ 3 and s ∈ (0, 1). If u is a solution of Eq. (1.1), then 0,α N ) for any α ∈ (0, 1). u ∈ Cloc (R+ Proof. Set  u (x) =

 u(x), −u(−x),

xN ≥ 0 x N ≤ 0,

it is easy to verify that  u is a solution of the following equation: ∗

−u =

|u|2 (s)−2u , |x N |s

N u ∈ D1,2 (R N ) and u = 0 on ∂R+ .

(3.11)

For any x 0 ∈ R N and R > r > 0, let v ∈ H 1,2(B(x 0 , r )) satisfy v =  u on ∂ B(x 0 , r ) and  ∇v · ∇η = 0 (3.12) B(x 0,r)

for any η ∈ C0∞ (B(x 0 , r )). Then by the standard regularity theory (cf. [13, Corollary 3.11]) we know that there exists C = C(N) > 0 such that for any ρ ∈ (0, r ],       ρ N 2 2 2 |∇ u| ≤ C |∇ u| + |∇( u − v)| . (3.13) r B(x 0,ρ) B(x 0 ,r) B(x 0,r) Since  u is a solution of the Eq. (3.11), by (3.12) we have    |∇( u − v)|2 = ∇ u · ∇( u − v) − B(x 0,r)

 =

B(x 0 ,r)

| u |2 B(x 0 ,r)

∗ (s)−2

B(x 0,r)

∇v · ∇( u − v)

· u · ( u − v) . s |x N |

(3.14)

∗

Let M := sup{| u |2 (s)−1| u − v|; x ∈ B(x 0 , r )}. By s ∈ (0, 1), we have   ∗ (s)−2 2 | u| · | u | · | u − v| 1 ≤ M ≤ Cr N−s s s |x N | B(x 0,r) B(x 0 ,r) |x N |

(3.15)

for some constant C > 0. Thus by (3.14), (3.13) and (3.15), we know that there exists B > 0 such that for any ρ ∈ (0, r ],   ρ N  |∇ u |2 ≤ C |∇ u |2 + Br N−s . r B(x 0,ρ) B(x 0 ,r) Then by Lemma 3.4 of [13], we know that for any γ ∈ (N − s, N), there exists constant C > 0 such that for any ρ ∈ (0, r ],  γ    ρ 2 2 N−s |∇ u| ≤ C |∇ u | + Bρ . R γ B(x0,r) B(x 0,ρ)

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

337

So, by the Morrey’s Theorem (cf. [13, Corollary 3.2]) we have 0,1− 2s

 u ∈ Cloc

(R N ). 0,1− s

Noticing that when  u ∈ Cloc 2 (R N ), s  ∗  ∗ s | u |2 (s)−2 · | u | · | u − v| |x N |(1− 2 )(2 (s)−1) ∗ ≤ M ≤ Cr N−s+(2 (s)−1)(1− 2 ) , s s |x N | |x N | B(x 0 ,r) B(x 0 ,r) 0,α (R N ) for any α ∈ (0, 1). iterating the above procedure, we can get  u ∈ Cloc



Remark 3.5. When N = 3 and N = 4, the result of this theorem can be improved. In fact, in 1 (R N ) (see Remark 5.8 and Section 5, we will show that if N = 4 and s = 1, then u ∈ Cloc + Theorem 5.15). The following theorem gives an improvement of Theorem 3.4 for the case N = 3. Theorem 3.6. Suppose that N = 3, s ∈ (0, 3/2). If u is a solution of Eq. (1.1), then u ∈ 0,α N Cloc (R+ ) for any α ∈ (0, 1). Proof. The proof is the same as the proof of Theorem 3.4 essentially. For N = 3, s ∈ (1, 3/2), let ε = s − 1 and choose δ ∈ (0, 1 − ε). By the H¨older inequality we have  ∗ | u |2 (s)−2 · | u | · | u − v| s |x 3 | B(x 0 ,r)   1   ε  1 1−δ q | u |2 2 1 q(2∗ (s)−2) q ≤ | u| | u − v| 2 1−δ B(x 0,r) |x 3 | B(x 0,r) |x 3 | B(x 0 ,r)   ε 1−δ N 1 ≤ Cr q 1−δ |x B(x 0 ,r) 3 |   ε 1−δ x 0,1 +r  x 0,2 +r  x 0,3 +r N 1 ≤ Cr q dx dx dx 1 2 3 1−δ x 0,1 −r x 0,2 −r x 0,3 −r |x 3 | 3

5

≤ Cr 2 −ε = Cr 2 −s , ε 1−δ

(3.16) 0, 3 − s

= 1. As in the proof of Theorem 3.4, we can get that  u ∈ Cloc4 2 (R3 ). 1  3 s ∗ ∗ q u |q(2 (s)−2)| u − v|q ≤ Cr (2 (s)−2)( 4 − 2 ) for some positive constant C Noticing that B(x0,r) | and iterating the above procedure, we can get the desired result.  where

1 2

+

+

1 q

In the same way, we have the following theorems: Theorem 3.7. Suppose N ≥ 3 and s ∈ (0, 1). If u is a solution of (1.2), then for any α ∈ (0, 1), u ∈ C 0,α (Ω). Theorem 3.8. Suppose N = 3, s ∈ (0, 3/2). If u is a solution of (1.2), then for any α ∈ (0, 1), u ∈ C 0,α (Ω).

338

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

4. Prescribing the Palais–Smale sequence of Eq. (1.2) In this section we will prove the following theorem: Theorem 4.1. Suppose Ω is a bounded domain with smooth boundary in R N and ϕλ (u) =   |u|2∗ (s) 1 λ 2 1 1 2 Ω |∇u| dx − 2∗ (s) Ω d s (x) , u ∈ H0 (Ω ), λ > 0. If {u n } ⊂ H0 (Ω ) satisfies: ϕλ (u n ) → 0 (H −1(Ω )),

ϕλ (u n ) → c > 0,

then there exist a solution u of Eq. (1.2), k solutions u 1 , . . . , u k of the equation −u = λ

|u|2

∗ (s)−2

x Ns

u

N in R+ ,

N u ∈ D01,2 (R+ ),

(4.1) j

k orthogonal transformations A1 , . . . , Ak of R N , k sequences {yn } ⊂ Ω , j = 1, . . . , k, and k j positive number sequences {λn }, j = 1, . . . , k, such that j

j

j

j

(i) For every 1 ≤ j ≤ k, yn → z j ∈ ∂Ω , λn → 0 as n → ∞, and {dist(yn , ∂Ω )/λn } is bounded. j  N−2 j (ii) u n − u − kj =1 (λn )− 2 v j ( ·−yj n ) D1,2 (R N ) → 0 as n → ∞, where v j = u j ◦ A j , j = λn

1, . . . , k.  (iii) ϕλ (u) + kj =1 ψλ (u j ) = c, where ψλ (v) =



1 N 2 R+

N D01,2 (R+ ).

|∇v|2 dx −

 |v|2 (s) λ ,v N s 2∗ (s) R+ xN ∗

∈

To prove the theorem, we need the following lemmas. Lemma 4.2 (Brezis–Lieb Lemma, See [4] or [27]). Let Ω be an open subset of R N and let {u n } ⊂ L p (Ω ), 1 ≤ p < ∞. If (a) {u n } is bounded in L p (Ω ), (b) u n → u almost everywhere on Ω , then p

p

p

lim ( u n L p (Ω ) − u n − u L p (Ω )) = u L p (Ω ) .

n→∞

Lemma 4.3. Suppose that {u n } is a bounded sequence of H01(Ω ) and u n  u in H01(Ω ). If u ∈ L ∞ (Ω ), then ∗

∗

∗

|u n |2 (s)−2 u n |u n − u|2 (s)−2(u n − u) |u|2 (s)−2u − − → 0 in H −1(Ω ). d s (x) d s (x) d s (x) Proof. It is sufficient to prove that    ∗ ∗ ∗ |u n − u|2 (s)−2(u n − u) |u|2 (s)−2u |u n |2 (s)−2 u n − − lim ·h =0 n→∞ Ω d s (x) d s (x) d s (x) uniformly for h ∈ H01(Ω ), h ≤ 1. By the Vitali convergence theorem, it is equivalent to prove the following claim: (A) for any  > 0, there exists δ > 0 such that for any Lebesgue measurable set e ⊂ Ω , if |e| ≤ δ, then for any h ∈ H01(Ω ), h ≤ 1 and any n,    ∗ ∗ 2∗ (s)−2 u |u n − u|2 (s)−2(u n − u) |u|2 (s)−2u  n  |u n | − − (4.2)  ·h ≤ d s (x) d s (x) d s (x)  e

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

Let us define f (v) :=

∗

|v|2 (s)−2 v d s (x) .

339

By the mean value theorem, we have, a.e. on R N ,

| f (u n ) − f (u n − u)| ≤ (2∗ (s) − 1)(|u n | + |u|)2

∗ (s)−2

|u|/d s (x).

(4.3)

Assume |u| L ∞ (Ω ) ≤ M, then by the H¨older inequality and Hardy–Sobolev inequalities we have  ∗ (|u n | + |u|)2 (s)−2 |u| · |h| d s (x)  ∗ (|u n | + |u|)2 (s)−2 ≤M · |h| d s (x) ⎛ ⎞ 2∗ (s)−1 1/2∗ (s)  2∗ (s)−2 ∗ 2∗ (s)  ∗ ·2 (s) ∗ (|u n | + |u|) 2 (s)−1 |h|2 (s) ⎠ ≤ M⎝ · d s (x) d s (x) ⎛ ≤ M SΩ h ⎝



2∗ (s)−2 ∗ ·2 (s) 2∗ (s)−1

(|u n | + |u|) d s (x)

⎞ 2∗ (s)−1 ∗ 2 (s)

⎠

(4.4)

Since the embedding H01(Ω ) → L p (Ω , d −s (x)) is compact for 2 ≤ p < 2∗ (s), where  p L p (Ω , d −s (x)) := {u | Ω d|u| s (x) < ∞}, we know that claim (A) follows from (4.3) and (4.4) immediately.  Likewise, we have Lemma 4.4. Suppose that {u n } and {vn } are bounded sequences in H01(Ω ) and u n  0 in H01(Ω ). If vn L ∞ (Ω ) ≤ M, n = 1, 2, . . ., then ∗

∗

∗

|u n − vn |2 (s)−2 (u n − vn ) |vn |2 (s)−2 vn |u n |2 (s)−2u n − − → 0 in H −1(Ω ). s d (x) d s (x) d s (x) We now give the proof of Theorem 4.1. Our proof is inspired by the proof of Theorem 8.13 in the book [27]. Proof of Theorem 4.1. It is easy to deduce that {u n } is bounded. We may assume that u n  u in H01(Ω ), u n → u a.e. on Ω . Then ϕλ (u) = 0. Let u 1n = u n − u, by Lemma 4.2, Theorem 3.3 and Lemma 4.3, we have

u 1n 2 = u n 2 − u 2 + o(1) ϕλ (u 1n ) → c − ϕλ (u)

(4.5)

ϕλ (u 1n ) → 0 in H −1 (Ω ). ∗

If u 1n → 0 in L 2 (s)(Ω , d −s (Ω )), then, since ϕλ (u 1n ) → 0, it follows that u 1n → 0 in H01(Ω ) ∗ and the proof is complete. If u 1n → 0 in L 2 (s)(Ω , d −s (Ω )) we may assume that  ∗ |u 1n |2 (s) > δ, n = 1, 2, . . . , s Ω d (x) where 0 < δ < (SΩ /2λ)2  Q n (r ) := sup y∈ R N

∗ (s)/(2∗(s)−2)

B(y,r)

. Let us define

2∗ (s)

|u 1n | . d s (x)

340

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

Since Q n (0) = 0 and Q n (∞) > δ, there exist sequences {yn1 }, {λ1n } such that yn1 ∈ Ω , λ1n > 0 and   ∗ ∗ |u 1n |2 (s) |u 1n |2 (s) = . δ = sup s s B(yn1 ,λ1n ) d (x) y∈ R N B(y,λ1n ) d (x) N−2 2

Set vn1 (x) = (λ1n )  δ = sup y∈ R N

u 1n (λ1n x + yn1 ), x ∈ Ωn = (Ω − yn1 )/λ1n and dn (x) = dist(x, ∂Ωn ), then  ∗ ∗ |vn1 |2 (s) |vn1 |2 (s) = . (4.6) s s B(y,1) dn (x) B(0,1) dn (x)

Assume that vn1  v in D1,2 (R N ), vn1 → v1 a.e. on R N . Let f n ∈ D01,2 (Ω ) be such that    ∗ |u 1n |2 (s)−2 1 u ∀h ∈ H01(Ω ), ∇u 1n ∇h − λ · h = ∇ fn ∇h. n d s (x) Ω Ω Ω Then gn (x) := (λ1n )(N−2)/2 f n (λ1n x + yn1 ) satisfies    ∗ |vn1 |2 (s)−2 1 1 1 vn · h = ∀h ∈ H0 (Ωn ), ∇vn ∇h − λ ∇gn ∇h dns (x) Ωn Ωn Ωn and   2 |∇ f n | dx = |∇gn |2 dx = o(1). Ω

Ωn

(4.7)

If v1 = 0 then vn1 → 0 in L 2loc (R N ). Let h ∈ C0∞ (R N ) be such that supp h ⊂ B(y, 1) for some y ∈ R N . It follows then from (4.7) and Hardy–Sobolev inequality (1.4) that    |∇(hvn1 )|2 dx = ∇vn1 · ∇(h 2 vn1 ) dx + (vn1 )2 · |∇h|2 dx Ωn Ωn Ωn  = ∇vn1 · ∇(h 2 vn1 ) dx + o(1) Ωn



= = ≤ ≤ ≤

 ∗ h 2 |vn1 |2 (s) λ dx + ∇gn · ∇(h 2 vn1 ) dx + o(1) dns (x) Ωn Ωn  ∗ h 2 |vn1 |2 (s) dx + o(1) λ dns (x) Ωn    2∗ (s)−2  ∗2 ∗ (s) ∗ 2∗ (s) 2 (s) 1 2 |vn | |hvn1 |2 (s) λ dx dx + o(1) s s h=0 dn (x) Ωn dn (x)  2∗ (s)−2 −1 SΩ λδ 2∗ (s) |∇(hvn1 )|2 dx + o(1) Ωn  1 |∇(hvn1 )|2 dx + o(1). 2 Ωn s/2∗ (s)

2∗ (s)

We obtain ∇vn1 → 0 in L 2loc (R N ) and vn1 /dn → 0 in L loc (R N ). This contradicts (4.6). Hence v1 = 0. Using the fact that Ω is bounded, we may assume that λ1n → λ10 ≥ 0, yn1 → z 1 ∈ Ω. If 1 λ0 > 0, the fact that u 1n  0 implies that vn1  0 in D1,2 (R N ). This contradicts v1 = 0.

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

341

If λ1n → 0 and lim sup n→∞

1 dist(yn1 , ∂Ω ) = ∞, λ1n

it is then easy to verify that v1 is a solution of −v = 0 in D1,2 (R N ). So v1 must vanish identically. This is also a contradiction. Hence lim sup n→∞

1 dist(yn1 , ∂Ω ) < ∞, λ1n

and it is then easy to verify that after a orthogonal transformation A−1 1 , v1 is a solution of −v = λ

|v|2

∗ (s)−2

x Ns

v

,

N v ∈ D01,2 (R+ ).

Let us define u 2n (x) = u 1n (x) − (λ1n )(2−N)/2 wn ( u = v1 in Ωn ,

(4.8) x−yn1 ), λ1n

x ∈ Ω , where wn is the solution of

u = 0 on ∂Ωn .

It is easy to deduce that wn → v1 in D1,2 (R N ) and by Theorem 3.1 and Corollary 3.2, there exists M > 0 such that wn L ∞ (R N ) ≤ M, n = 1, 2, . . .. Then by Lemma 4.2 and Lemma 4.4 we have

u 2n 2 = u n 2 − u 2 − v1 2 + o(1) ϕλ (u 2n ) → c − ϕλ (u) − ψλ (v1 ) ϕλ (u 2n ) → 0 in H −1 (Ω ). Any nontrivial critical point of ψλ satisfies  S

∗

|u|2 (s) dx N x Ns R+

2/2∗ (s)

 ≤

 N R+

|∇u|2 dx = λ

∗

|u|2 (s) dx, N x Ns R+

so that ψλ (u) ≥ cλ∗ := λ



1 1 − 2 2∗ (s)

  2∗ (s)/(2∗(s)−2) S . λ j

j

Iterating the above procedure, we construct sequences {v j }, {λn }, {A j } and {yn }. Since for every j , ψλ (v j ) ≥ cλ∗ , the iteration must terminate at some finite index.  By Theorems 2.5 and 4.1, we have Corollary 4.5. If SΩ < S, then the extremal problem (2.2) attains its infimum.

342

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

5. Some equivalent equations and its applications Let us define mappings H : R N \ {(0, . . . , 0, −1)} → R N \ {(0, . . . , 0, −1)}   2x 1 2x N−1 1 − |x|2 x → ,..., , 1 + 2x N + |x|2 1 + 2x N + |x|2 1 + 2x N + |x|2

(5.1)

and K : R N \ {(0, . . . , 0, 1)} → R N \ {(0, . . . , 0, 1)}   2x 1 2x N−1 |x|2 − 1 x → . , . . . , , 1 − 2x N + |x|2 1 − 2x N + |x|2 1 − 2x N + |x|2

(5.2)

H and K can be decomposed as H = π ◦ T ◦ π −1 and K = π ◦ L ◦ π −1 , where π : S N \ {(0, . . . , 0, 1)} → R N   y1 yN (y1 , . . . , y N , y N+1 ) → ,..., 1 − y N+1 1 − y N+1 is the stereographic projection, and its inverse is π −1 : R N → S N \ {(0, . . . , 0, 1)}   2x |x|2 − 1 , (x 1 , . . . , x N ) → , 1 + |x|2 |x|2 + 1 and  (x 1 , . . . , x N−1 , −x N+1 , −x N ) T : SN → SN , x → N N L : S → S , x → (x 1 , . . . , x N−1 , x N+1 , x N ). Denote the standard Euclidean metric in R N as h¯ . It is easy to verify that H and K have the following properties: (i) H = H −1, K = K −1 ; N ) = B(0, 1), H (∂R N ) = ∂ B(0, 1) and H (∞) = (0, . . . , 0, −1); K (R N ) = (ii) H (R+ + + N N R \ B(0, 1), K (∂R+ ) = ∂ B(0, 1) and K (∞) = (0, . . . , 0, 1); (iii) H and K are conformal transformations and H ∗h¯ = ρ 4/(N−2) (x)h¯ , K ∗ h¯ = φ 4/(N−2) (x)h¯ , where  N−2  N−2   2 2 2 2 , φ(x) = , ρ(x) = 2 2 1 + 2x N + |x| 1 − 2x N + |x| where H ∗h¯ and K ∗ h¯ denote the pull-back metrics of mappings H and K respectively. N ) and Proposition 5.1. For v ∈ H01(B(0, 1)), let u = (v ◦ H )ρ, then u ∈ D01,2 (R+   (i) R N |∇u(x)|2 dx = B(0,1) |∇v(y)|2 dy, so v → u = (v ◦ H )ρ is a Hilbert space +

N isomorphism between H01(B(0, 1)) and D01,2 (R+ ). ∗  |u(x)|2 (s)  ∗ 2s (ii) R N dx = B(0,1) (1−|y|2 )s |v(y)|2 (s) dy. xs +

N

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

343

Proof. (i) Let h = H ∗h¯ = ρ 4/(N−2) (x)h¯ , we have  1 v = √ ∂ j ( |h|h i j ∂i (v ◦ H )) |h|  1 = √ ∂ j ( |h|h i j ∂i (u/ρ)) |h| ∗

= ρ −2 |h¯ |−1/2 ∂ j (ρ 2 |h¯ |1/2 h¯ i j ∂i (u/ρ)) ∗

= ρ −2 (ρu − uρ). Since ρ = 0 in R N \ {(0, . . . , 0, −1)}, we can get ∗

v(x) = ρ −2 (H −1(x))ρ(H −1(x))u(H −1(x)). Hence 

 |∇v|2 dx = B(0,1)

vv dx 

B(0,1) ∗

vρ −2 (H −1(x))ρ(H −1(x))u(H −1(x)) dx B(0,1)   = uu dx = |∇u|2 dx.

=

N R+

N R+

∗

(ii) The result (ii) can be deduced from dVh = ρ 2 dVh¯ by integral transformation.



In the same way, we have N ) and Proposition 5.2. For v ∈ H01(R N \ B(0, 1)), let u = (v ◦ K )φ, then u ∈ H01(R+   (i) R N |∇u(x)|2 dx = R N \B(0,1) |∇v(y)|2 dy, so v → u = (v ◦ K )φ is a Hilbert space +

N ). isomorphism between H01(R N \ B(0, 1)) and H01(R+   |u(x)|2∗(s) ∗ (s) 2s 2 dx = R N \B(0,1) (|y|2 −1)s |v(y)| dy. (ii) R N xs +

N

Let us define    2s 2∗ (s)

S := inf |∇v|2 | v ∈ H01(B(0, 1)), |v| = 1 , 2 s B(0,1) B(0,1) (1 − |x| )    2s 2∗ (s)  |∇v|2 | v ∈ H01(R N \ B(0, 1)), |v| = 1 . S := inf 2 s R N \B(0,1) R N \B(0,1) (|x| − 1) By Propositions 5.1 and 5.2, we have the following corollary: Corollary 5.3. S = S= S. Theorem 5.4. (i) S B(0,1) = S and the extremal problem (2.2) for Ω = B(0, 1) cannot attain its infimum.(ii) SR N \B(0,1) < S. Proof. (i) Since



 2

S B(0,1) := inf

|∇v| | v ∈ B(0,1)

H01(B(0, 1)),

 ∗ |v|2 (s) =1 , s B(0,1) (1 − |x|)

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2 by the fact that 1+|x| > 1 when |x| < 1 and Corollary 5.3, we get S B(0,1) ≥ S = S. And by Theorem 2.5, we deduce that S B(0,1) = S. If the infimum of the extremal problem (2.2) for Ω = B(0, 1) can be attained by some u ∈ H01(B(0, 1)), then   ∗ |u|2 (s) |∇u|2 = S B(0,1), = 1. s B(0,1) B(0,1) (1 − |x|)

It follows that 

S≤

 2

|∇u| dx

B(0,1)

This contradicts S B(0,1) = S. The proof of (ii) is similar.

∗

|u|2 (s) 2 s B(0,1) (1 − |x| )



2 2∗ (s)

< S B(0,1).



Now let us recall some facts coming from conformal geometry. See [9] for more details. Theorem 5.5 (Cf. [9]). Let (M, g) and (N, h) be two Riemannian manifolds of dimension m ≥ 3. Suppose that there exists a conformal diffeomorphism f from M onto N, i.e. f ∗ h = 4/(m−2) g for some positive function  ∈ C ∞ (M). The scalar curvatures of (M, g) and (N, h) are Rg and Rh respectively. Consider an equation on M as follows: g u − β Rg (x)u + F(x, u) = 0,

u ∈ C 2 (M),

(5.3)

where g is the Laplacian on (M, g), β = (m − 2)/4(m − 1), and F : M × R → R is a C 1 function. If u is a solution of (5.3) and u = (v ◦ f ), then v is a solution of the following equation: h v − β Rh (y)v +  (y)1−q F( f −1 (y), (y)v) = 0,

v ∈ C 2 (N),

(5.4)

where q = 2m/(m − 2) and  =  ◦ f −1 . Using Theorem 5.5 and the properties of the mappings H and K , we can easily deduce that: Proposition 5.6. (i) If u is a solution of Eq. (1.1), then v = (u ◦ H )ρ is a solution of Eq. (1.5). (ii) If v is a solution of Eq. (1.5), then u = (v ◦ H )ρ is a solution of Eq. (1.1). Proposition 5.7. (i) If u is a solution of Eq. (1.1), then v = (u ◦ K )φ is a solution of Eq. (1.6). (ii) If v is a solution of Eq. (1.6), then u = (v ◦ K )φ is a solution of Eq. (1.1). Remark 5.8. In [23], an equation which is similar to (1.5) is considered. The authors showed that Eq. (1.5) has a radially symmetric positive solution and every radially symmetric positive solution belongs to C 2 (B(0, 1)) ∩ C 1 (B(0, 1)). In the following theorems, we will prove that Eq. (1.5) has infinitely many radially symmetric solutions. However, for some N and s, the positive solution of Eq. (1.5) is unique after some transformation. By the standard concentration–compactness argument (see [19,20] and [25] for details), we have the following lemma: Lemma 5.9. Suppose {u n } ⊂ H01(B(0, 1)) and u n  u in H01(B(0, 1)). Let |∇u n |2  μ in M(B(0, 1)), 2s ∗ |u n |2 (s)  ν in M(B(0, 1)), 2 s (1 − |x| )

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

345

then we have (i) There exists some at most countable set J , a family {x ( j ); j ∈ J } of distinct points in ∂ B(0, 1), and a family {ν ( j ) : j ∈ J } of positive numbers such that  2s 2∗ (s) ν= |u| dx + ν ( j )δx ( j ) , (1 − |x|2 )s j ∈J where δx is the Dirac-mass 1 concentrated at x ∈ R N . (ii) In addition we have  μ( j ) δ x ( j ) μ ≥ |∇u|2 dx + j ∈J

for some family S(ν ( j ) )2/2

{μ( j ); ∗ (s)

j ∈ J }, μ( j ) > 0 satisfying

≤ μ( j ) ,

for all j ∈ J.

Let us define 1 H0,r (B(0, 1)) := {u ∈ H01(B(0, 1)) | u is radially symmetric},

then by Lemma 5.9 we have: 1 (B(0, 1)) and u  u in H 1 (B(0, 1)), then Lemma 5.10. If {u n } ⊂ H0,r n 0  s 2 ∗ |u − u|2 (s) → 0. 2 s n B(0,1) (1 − |x| )

By Lemma 5.10 and the principle of symmetric criticality (see [22] or [27]), we have Theorem 5.11. Eq. (1.5) has a sequence of radially symmetric solutions {u n } which satisfies that as n → ∞,  |∇u n |2 → ∞. (5.5) B(0,1)

Remark 5.12. By Propositions 5.6 and 5.7, we know that both Eqs. (1.1) and (1.6) have infinitely many solutions satisfying   2 |∇u n | → ∞ and |∇u n |2 → ∞ N R+

R N \B(0,1)

respectively. Let R N,1 = (R N+1 , g) be the Minkowski space, where g is the metric with signature (+ · · · , +, −). The hyperbolic space H N of dimension N is the submanifold {x ∈ R N,1 ; g(x, x) = −1, x N+1 > 0}. The mapping   2x 1 + |x|2 ξ : B(0, 1) → H N , x → , 1 − |x|2 1 − |x|2 is a conformal transformation and ξ ∗ h = σ 4/(N−2) h¯ , 2 (N−2)/2 . Since the scalar curvature of H N where h is the Riemannian metric on H N , σ = ( 1−|x| 2) is −N(N − 1), by Theorem 5.5, we have

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Proposition 5.13. (i) If v is a solution of Eq. (1.7), then u = (v ◦ ξ ) · σ is a solution of Eq. σ is a solution of Eq. (1.7), where (1.5).(ii) If u is a solution of Eq. (1.5), then v = (u ◦ ξ −1 ) ·   σ = 1/(σ ◦ ξ −1 ). Recently, Almeida, Damascelli and Ge obtained some results concerning the symmetry of the positive solution for a semilinear equation on a manifold with “group symmetries”. Let Isom(H N ) be the isometric transform group of H N . They obtained the following result: Theorem 5.14 (Cf. [1]). Let f ∈ C 1 ([0, +∞]) satisfy: ∃δ > 0, C > 0 : f  (s) ≤

N(N − 2) + Cs α , 4

∀s ∈ [0, δ].

If v is a positive solution of − H N u = f (u) in H N , L 2 (H N )

and v ∈ symmetric.

∩

u → 0 when x → ∞,

L α N/2 (H N ),

then there exists χ ∈

(5.6) Isom(H N )

such that v ◦ χ is radially

We describe the following four conditions as conditions (A): (1) (2) (3) (4)

N N N N

= 3 and s = 4 and s = 5 and s ≥ 6 and s

∈ (0, 3/2); ∈ (0, 1]; ∈ (0, 1); ∈ (0, 6N−4 ]. N2

Theorem 5.15. If one of the conditions (A) is satisfied, then for any positive solution v of Eq. (1.7), there exists χ ∈ Isom(H N ) such that v ◦ χ is radially symmetric. ∗

Proof. By Theorem 5.14, it is sufficient to prove that v ∈ L 2 (H N ) ∩ L (2 (s)−2)N/2 (H N ). When s ∈ (0, 4/N], we have 2 ≤ (2∗ (s) − 2)N/2 ≤ 2∗ (0). Then by the Sobolev embedding ∗ theorem on manifolds (see [15]), we have v ∈ L 2 (H N ) ∩ L (2 (s)−2)N/2 (H N ). Thus the theorem is proved under condition (2) of (A). When N = 3 and s ∈ (0, 3/2), by Theorem 3.8 and Proposition 5.13 we get v ∈ C 0,α (B(0, 1)) N−2 (1 − |x|2 ) 2 for any α ∈ (0, 1). It follows that for any α ∈ (0, 1), ∃Cα > 0 such that |v(x)| ≤ Cα (1 − |x|2 )α+

N−2 2

.

So we can choose α close enough to 1 such that   ∗ v (2 (s)−2)N/2 (2∗ (s)−2)N/2 N |v| dVH N = 2 dx < ∞. 2 N HN B(0,1) (1 − |x| ) Thus the theorem is proved under condition (1) of (A). The proof of the theorem under condition (3) or (4) of (A) is similar.  Let G := {ξ −1 ◦ χ ◦ ξ | χ ∈ Isom(H N )}. By Proposition 5.13, we obtain Corollary 5.16. Eq. (1.5) is invariant under the action of the group G. And if one of the conditions (A) is satisfied, then for any positive solution v of Eq. (1.5), there exists ζ ∈ G such that v ◦ ζ is radially symmetric.

S. Chen, S. Li / Nonlinear Analysis 66 (2007) 324–348

347

Remark 5.17. It is easy to verify that the orbit {v ◦ ζ ; ζ ∈ G} contains a non-radially symmetric function. Therefore Eq. (1.5) admits a non-radially symmetric solution. If u is a radially symmetric positive solution of Eq. (1.5), let u(r ) = u(|x|), then u(r ) is a positive solution of the following O.D.E.: ⎧ N −1  2s ∗ ⎨  u (r ) + u 2 (s)−1(r ) = 0, u (r ) − 2 s (∗) r (1 − r ) ⎩u > 0 in [0, 1), u  (0) = u(1) = 0. Let z(t) = tu(t 1/(N−2) ), t ∈ [0, 1), then Eq. (∗) transforms to ⎧ 2s 1 ∗ ⎨  · z 2 (s)−1(t) = 0 in [0, 1), z (t) + (N − 2)2 (1 − t 2/(N−2) )s · t γ ⎩z(0) = 0, z(1) = 0, z  (0) = u(0), z(t) > 0 in [0, 1),

(∗∗)

2 where γ = 2∗ (s) − N−2 . By Theorem 9 of [18] we know that when γ ≤ 2, i.e. s ≥ 1, Eq. (∗∗) has a unique solution. Thus we obtain the following theorem:

Theorem 5.18. If N = 3, s ∈ [1, 3/2) or N = 4, s = 1, then the positive solution of Eq. (1.5) is unique (after a transformation). By Proposition 5.6, we get Corollary 5.19. If N = 3, s ∈ [1, 3/2) or N = 4, s = 1, then the positive solution of Eq. (1.1) is unique (after a transformation). By Theorem 4.1 and Corollary 5.19, we get the following corollary: Corollary 5.20. If N = 3, s ∈ [1, 3/2) or N = 4, s = 1, then the functional corresponding to  2∗ (s)  Eq. (1.2) satisfies the (P S)c condition for any c ∈ (c∗ , 2c∗ ), where c∗ = 12 − 2∗1(s) S 2∗ (s)−2 . α  1−|x|2 4 Let (H N , h) = (B(0, 1), (1−|x| , α > 0, is a 2 )2 dx). Notice that (cf. [1]) u α (x) := 2 1+|x| radially symmetric positive solution of  α+2 − H N u = α(N − 1 − α)u + α(α + 1)u α , u → 0 when x → ∂ B(0, 1), u > 0 in B(0, 1). α+2 α

= 2∗ (s) − 1, we get α = 3/2 and α(N − 1 − α) = 3/4 =   2 3/2 N(N − 2)/4, so we deduce that w(x) = 2−3/2 · 153/4 1−|x| is a positive solution of Eq. 2 1+|x| (1.7). Then by Propositions 5.13 and 5.6 , we have

When N = 3, s = 4/3, let

ω(x) = 2−1 · 153/4

1 − |x|2 (1 + |x|2)3/2

and U (x) = 153/4

x 3 (1 + 2x 3 + |x|2 ) (1 + |x|2)3/2

(5.7)

are positive solutions of Eqs. (1.5) and (1.1) respectively. In the same way, we can deduce that when N = 4, s = 1, ω1 (x) =

12(1 − |x|2) (1 + |x|2 )2

and U1 (x) =

24x 4 (1 + |x|2 )2

are solutions of Eqs. (1.5) and (1.1) respectively.

(5.8)

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