Hardy-Sobolev-Maz’ya inequalities: symmetry and breaking symmetry of extremal functions
∗
Marita Gazzini† and Roberta Musina‡
Abstract. Denote points in Rk × RN −k as pairs ξ = (x, y), and assume 2 ≤ k < N . In this paper we study problem −∆v = λ|x|−2 v + |x|−b v p−1 in RN , x 6= 0 v>0, “ ”2 N −2 , the Hardy constant. Our results are the following: where p > 2, b = N − p 2 and λ ≤ k−2 2 • Let p < 2∗N −k+1 . Then there exists at least an entire cylindrically symmetric solution. • Let p ≤ 2∗N and λ ≥ 0. Then any solution v ∈ Lp (RN ; |x|−b dξ) is cylindrically symmetric. “ ”2 − k−1 . Then ground state solutions are not cylindrically symmetric, and • Let p < 2∗N and λ ≤ k−2 2 p−2 therefore there exist at least two distinct entire solutions. We prove also similar results for the degenerate problem −div(|x|−2α ∇u) = |x|−βp up−1 u>0,
, x 6= 0
namely, for the Euler-Lagrange equations of the Maz’ya inequality with cylindrical weights.
Key Words: Variational methods, critical growth, weighted Hardy-Sobolev inequalities. 2000 Mathematics Subject Classification: 35J20, 35J70, 35B33
Ref. SISSA 06/2008/M
∗ Work supported by Cofin. M.I.U.R. Progetto di Ricerca “Metodi Variazionali ed Equazioni Differenziali Nonlineari” † SISSA, Via Beirut 4 – 34010 Trieste, Italy, e-mail:
[email protected] ‡ Dipartimento di Matematica ed Informatica, Universit` a di Udine, via delle Scienze, 206 – 33100 Udine, Italy, e-mail:
[email protected] 1
Introduction Let k, N be positive integers with k < N . We put RN = Rk × RN −k , and we denote points ξ in RN as pairs (x, y) ∈ Rk × RN −k . In this paper we study the elliptic problem −∆v = λ|x|−2 v + |x|−b v p−1 in RN , x 6= 0 (0.1) v>0, where p > 2, λ ≤
k−2 2 2
and
N −2 . (0.2) 2 A large number of bibliographical references for (0.1) is available in case k = N : we quote for example [1], [3], [6], [8], [18] and references there-in. In particular, in [6] and [8] one can find a careful analysis on breaking symmetry of ground state solutions. Concerning existence in case k < N we cite [2], [5], [13], [15] and [17]. Existence and multiplicity results can be found in [9]. Symmetry properties of weak entire solutions to (0.1) were proved in [11], under the assumptions k ≥ 2, λ = 0 and p ∈ (2, 2∗N ). As noticed in [5] and in [13], solutions that are radially symmetric in the x-variable receive importance with regard to certain elliptic equations on the n = N −k +1-dimensional hyperbolic space Hn . More precisely, if v(x, y) = v(|x|, y) solves (0.1), then the transform N −2 u(r, y) := r 2 v(r, y) gives a solution of b=N −p
−∆Hn u = µu + up−1 .
(0.3)
Here, ∆Hn is the Laplace-Beltrami operator on Hn and the parameter µ is given by µ=λ+
(N − k)2 − (k − 2)2 . 4
We refer to [5] and to [13] for a discussion on the relevances between equation (0.3) and some significant problems in hyperbolic geometry, Yamabe-type equations of Heisenberg type, Grushing-type equations. Motivated by these considerations, in the present paper we address our attention towards cylindrically symmetric solutions (see Definition 1.1 and [12]). Our first result concerns the existence of entire solutions (see Definition 1.2 and [13]), in case that p is smaller than the critical Sobolev exponent in RN −k+1 . 2 and 2 < p < 2∗N −k+1 . Then problem (0.1) Theorem 0.1 Assume 2 ≤ k < N , λ ≤ k−2 2 has a cylindrically symmetric entire solution. By [13], Theorem 1.1 and Section 6, the bound on λ is a necessary condition for existence. Notice that we can allow p to be supercritical, since 2∗N < 2∗N −k+1 . Theorem 0.1 will be
2
proved in Section 2, where we state also a non-existence/existence result in the limiting case p = 2∗N −k+1 . In Section 3 we investigate the symmetry properties of solutions to (0.1). In case k ≥ 2 and p ∈ (2, 2∗N ), Mancini, Fabbri and Sandeep adopted in [11] the moving plane method to show that nonnegative entire solutions to −∆v = |x|−b v p−1
in RN
(0.4)
are cylindrically symmetric. As a matter of fact, their arguments work as well in case k = 1. We omit the proof of the next result. −1) ∗ Theorem 0.2 Assume k = 1, N ≥ 3, and 2(N N −2 < p < 2N . Then every weak nonnegative solution v ∈ D01 (RN ) to (0.4) is cylindrically symmetric.
Existence was proved in [9], Theorem 3.5, under the same bounds on p as in Theorem 0.2. In case k ≥ 2 we are able to prove a stronger result. 2 Theorem 0.3 Assume 2 ≤ k < N , p ∈ (2, 2∗N ], 0 ≤ λ ≤ k−2 and let v ∈ Lp (RN ; |x|−b dξ) 2 ∗ be a a classical solution to (0.1). If λ = 0 and p = 2N assume in addition that v has a nonremovable singularity on {x = 0}. Then v is cylindrically symmetric. Notice that we only require a summability assumption on v. No assumption on ∇v is needed, so that v might be not entire (compare with Theorem 3.1 in [18] for the case k = N ). The existence of singular, non-entire solutions v ∈ Lp (RN ; |x|−b dξ) to (0.1) was proved in [9], Section 4.2, under suitable assumptions on the parameters involved. Next we focus our attention on ground state solutions (see Section 1 for the definition and for the existence results already available in the literature). It is known that a ground 2 state solution v¯ exists for any λ ≤ k−2 , provided that p ∈ (2, 2∗ ). If in addition λ ≥ 0 2 then v¯ is cylindrically symmetric, by Theorem 0.3. A natural question is to ask whether v¯ preserves symmetry as λ decreases. The answer is negative, as it is shown by the next result. Theorem 0.4 Let 2 ≤ k < N . Then ground states solutions to (0.1) are not radially symmetric in x if 2 k−2 k−1 λ≤ − . 2 p−2 Theorem 0.4 will be proved in Section 4. A similar phenomenon was already pointed out by Catrina and Wang [6] (see also [8]) for a related problem involving spherical weights. In the last Section we deal with classical solutions to −div(|x|−2α ∇u) = |x|−βp up−1 , x 6= 0 u>0. 3
As corollaries of our theorems for (0.1) we prove an existence result of symmetric solutions, symmetry properties of solutions and breaking symmetry. Acknowledgments The authors wish to thank Prof. Gianni Mancini for having proposed them the study of the breaking symmetry phenomenon for problem (0.1).
j Notation For any integer j ≥ 1, we denote by BR (z) the j-dimensional ball of radius R centered j at z ∈ R . We denote by ωj the surface measure of the unit sphere Sj−1 in Rj . If p ∈ [1, +∞), α ∈ R, and if Ω is a domain in RN , then Lp (Ω; |x|α dξ) is the space of measurable R maps u on Ω with Ω |x|α |u|p dξ < +∞. Therefore Lp (Ω; |x|0 dξ) ≡ Lp (Ω) is the standard Lebesgue space.
For N ≥ 3 the critical Sobolev exponent is 2∗N = constant. For N = 2 we set
1
2∗N
2N . N −2
As usual, we denote by S the Sobolev
:= +∞.
Preliminaries
We start by recalling the following definition from [12] and [11]. Definition 1.1 A smooth map v on (Rk \ {0}) × RN −k is cylindrically symmetric if i) for any choice of y ∈ RN −k , v(·, y) is symmetric decreasing in Rk ; ii) there exists y0 ∈ RN −k such that, for any choice of x ∈ Rk \ {0}, v(x, ·) is symmetric decreasing about y0 in RN −k . We list here a few integrals inequalities (for details we refer to [14], Section 2.1.6 and to [7], [15], [9]). The first inequality we need is the Hardy inequality: 2 Z Z k−2 |x|−2 |v|2 dξ ≤ |∇v|2 dξ , ∀v ∈ D01 (RN ) ∩ L2 (RN ; |x|−2 dξ) . (1.1) 2 N N R R Next, assume N ≥ 3, p ∈ (2, 2∗N ] and define b as in (0.2). The starting point for studying (0.1) is the Hardy-Sobolev-Maz’ya inequality, that is peculiar to the cylindrical case k < N and that was proved by Maz’ya in [14]. It states that there exists a positive constant cp such that # Z 2/p Z " 2 k − 2 cp |x|−b |v|p dξ ≤ |∇v|2 − |x|−2 |v|2 dξ (1.2) 2 RN RN for any v ∈ Cc∞ ((Rk \ {0}) × RN −k ). Thanks to (1.2), for λ ≤ Hilbert space Hλ (Rk × RN −k ) 4
k−2 2 2
we can define the
as the closure of maps v ∈ Cc∞ ((Rk \ {0}) × RN −k ) with respect to the norm Z
|∇v|2 − λ|x|−2 v 2 dξ
kvkλ =
1/2 .
RN
In general, Hλ (Rk × RN −k ) contains the space D01 (RN ) ∩ L2 (RN ; |x|−2 dξ). More precisely, Hλ (Rk × RN −k ) = D01 (RN ) ∩ L2 (RN ; |x|−2 dξ) 2 2 for λ < k−2 , by the Hardy inequality (1.1). Moreover, for λ = k−2 it turns out that 2 2 H0 (R2 × RN −2 ) = D01 (RN ), while D01 (RN ) = Hλ (Rk × RN −k ) ∩ L2 (RN ; |x|−2 dξ) if k ≥ 3. For future convenience we define also the Hilbert space Hλ,cyl (Rk × RN −k ) as the closure in Hλ (Rk × RN −k ) of maps v ∈ Cc∞ ((Rk \ {0}) × RN −k ), such that v(x, y) = v(|x|, y). Now we are in position to give the following definitions (compare with [13]). Definition 1.2 A classical solution v to (0.1) is said to be entire if v ∈ Hλ (Rk × RN −k ). Definition 1.3 An entire solution v to (0.1) is a ground state solution if v achieves the best constant R |∇v|2 − λ|x|−2 |v|2 dξ λ λ RN . (1.3) Sp = Sp (k, N ) := inf 2/p R v∈Hλ (Rk ×RN −k )) |x|−b |v|p dξ RN Notice that Spλ is positive, by (1.2). The existence of ground state solutions were proved 2 , and in [17], [9] in case λ coincides with the Hardy in [15], Theorem 1, in case λ < k−2 2 ∗ constant. If p = 2N and λ 6= 0, one needs the additional assumption 0 0 small enough. More precisely, we choose ε0 such that p−2
2(p − 1)ε0 p ≤ Spλ ,
(3.2)
where the Spλ is the infimum in (1.3). Since |ξ s | < |ξ| in Ωs , it turns out that −∆ws − λ|x|−2 ws ≥ |x|−b A(ξ)ws ,
(3.3)
pointwise on Ωs \ Σ2s , where A(ξ) :=
vsp−1 − v p−1 vs − v
satisfies 0 ≤ A(ξ) ≤ (p − 1)v p−2
on {ws ≤ 0} .
(3.4)
As in [11], the idea is to use ws− := min{ws , 0} ≤ 0 as test function for (3.3), but, differently from [11], the maps ws and ws− are not smooth enough. Thus we have to use suitable cut-off functions. For ε > 0 small set if |x| ≤ ε2 0 log |x|/ε2 ϕε (x) = if ε2 < |x| < ε , ϕ˜ε (x) = ϕε (xs ) . | log ε| 1 if |x| ≥ ε .
7
For any large integer h choose a cut off function ψh ∈ Cc∞ (RN ), with ψh (ξ) = 1 if |ξ| ≤ h 4 , 0 ≤ ψ ≤ 1 , ||∇ψh k∞ ≤ . h ψh (ξ) = 0 if |ξ| ≥ 2h We are allowed to use ϕ˜2ε ψh2 ws− as test function for (3.3) on Ωs . Set ωhε := ϕ˜ε ψh ws− . After integration by parts and simple computations one gets Z Z Z ε 2 −2 ε 2 −b ε 2 |∇ωh | − λ|x| |ωh | dξ ≤ |x| A(ξ)|ωh | dξ + Ωs
RN
Ωs
|∇(ϕ˜ε ψh )|2 |ws− |2 dξ . (3.5)
The left hand side in (3.5) is bounded from below by Spλ
Z
|x|−b |ωhε |p dξ
2/p .
Ωs
To estimate the right hand side we notice that Z Z |x|−b A(ξ)|ωhε |2 dξ ≤ (p − 1) Ωs
|x|−b A(ξ)|v|p−2 |ωhε |2 dξ
Ωs
≤
p−2 p
Z |x|
(p − 1)ε0
−b
|ωhε |p
2/p dξ
Ωs
by (3.4), H¨ older inequality and (3.1). Comparing with (3.5) and with (3.2) we infer 1 λ S 2 p
Z
−b
|x| Ωε,h s
|ws− |p
2/p
Z ≤
dξ
RN
|∇(ϕ˜ε ψh )|2 |ws− |2 dξ ,
(3.6)
where s Ωε,h s = {(x, y) ∈ Ωs | |x | > ε , |ξ| < h } .
In order to handle the right hand side in (3.6) we compute Z |∇(ϕ˜ε ψh )|2 |ws− |2 dξ ≤ 2(I1ε,h + I2ε,h ), RN
where I1ε,h =
Z RN
I2ε,h =
|∇ϕ˜ε |2 ψh2 |ws− |2 dξ ,
Z RN
|∇ψh |2 ϕ˜2ε |ws− |2 dξ .
To estimate the first integral we notice that v is smooth on {x 6= 0} and that |ws− | ≤ v on Ωs . Therefore, since k ≥ 2, Z Z ε ch ch ε,h 2 I 1 ≤ ch |∇ϕ˜ε | dξ ≤ rk−3 dr ≤ , 2 | log ε| | log ε| Rk ε2 8
N with constants ch that depend only on the measure of B2h and on the L∞ -norm of v on ε,h N B2h ∩ Ωs . Thus, I1 → 0 for h fixed, as ε → 0. Concerning the second integral, we use H¨ older inequality, b ≥ 0 and |ws− | ≤ v on Ωs to get ! p2 Z p−2 Z Z
I2ε,h ≤
|∇ψh |2 |v|2 dξ
2b
|x|−b |v|p dξ
≤ N \B N B2h h
RN
Z ≤
c
2p
p
|ξ| p−2 |∇ψh | p−2 dξ RN
! p2 |x|−b |v|p dξ
.
N \B N B2h h
Thus I2ε,h → 0 as h → +∞ uniformly in ε, since v ∈ Lp (RN ; |x|−b dξ). In conclusion, we have proved that ! p2 Z 2/p Z c 1 λ h |x|−b |ws− |p dξ S ≤ +c . |x|−b |v|p dξ N \B N 2 p | log ε| Ωε,h B2h s h Since |ws− | ≤ v ∈ Lp (RN ; |x|−b dξ), then passing to the limit in (3.6), first as ε → 0 and then as h → +∞, we get that ws ≥ 0 a.e. on Ωs . More precisely, ws > 0 on Ωs \ Σ2s by (3.3) and by the maximum principle. The proof now can be carried out as in [11].
4
Breaking symmetry
The reason for the phenomenon described in Theorem 0.4 is that cylindrically symmetric solutions become highly unstable as λ → −∞, that is, their Morse index becomes too large (see Remark 4.2). This is a consequence of the next crucial Theorem. Its proof was inspired by the papers by Kawohl [10] and by Smets, Su and Willem [16]. Theorem 4.1 Assume k ≥ 2 and let v 6= 0 be a local minimum for R R |∇v|2 dξ − λ RN |x|−2 v 2 dξ RN Rλ (v) = 2/p R |x|−b |v|p dξ RN on D01 (RN ) ∩ L2 (R2 ; |x|−2 dξ), such that v(x, y) = v(|x|, y) for a.e. y ∈ RN −k . Then Z Z Z k−1 |∇v|2 dξ − λ |x|−2 v 2 dξ ≤ |x|−2 v 2 dξ . p − 2 RN RN RN Proof. Take any h ∈ D01 (RN ) ∩ L2 (RN ; |x|−2 dξ), and set Z Z 2 z(t) = |∇(v + th)| dξ − λ |x|−2 (v + th)2 dξ , RN
Z n(t) =
RN
|x|−b |v + th|p dξ
RN
g(t) =
z(t) = Rλ (v + th) . n(t) 9
2/p ,
Since 0 is a local minima for g, then g 0 (0) = 0 and g 00 (0) ≥ 0. To simplify notations we 00 assume n(0) = 1. Thus we get z(0) ≤ nz 00(0) (0) , that is, Z
−2 2
|∇v| − λ RN
|∇h|2 RN |x|−b |v|p−2 h2 R
Z
2
|x|
≤
v
(p − 1)
RN
R RN
|x|−2 h2 R 2 . (4.1) − (p − 2)( RN |x|−b |v|p−2 vh)
−λ
R
RN
Now, let f1 ∈ H 1 (Sk−1 ) be an eigenfunction of the Laplace operator on Sk−1 with respect to the eigenvalue k − 1. Thus, f1 solves the minimization problem R |∇σ f |2 dσ SRk−1 = k − 1. inf fR∈H 1 (Sk−1 ) |f |2 dσ Sk−1 f =0 Sk−1
To simplify computations it is convenient to take Z Z |f1 |2 dσ = 1 , |∇σ f1 |2 dσ = k − 1 . Sk−1
Sk−1
Notice that we are allowed to use h(x, y) = v(|x|, y)f1 (x/|x|) as test function in (4.1). Indeed, h ∈ D01 (RN ) ∩ L2 (RN ; |x|−2 dξ) since v ∈ D01 (RN ) ∩ L2 (RN ; |x|−2 dξ). It turns out that Z Z Z |∇h|2 = |∇v|2 + (k − 1) |x|−2 v 2 , RN
Z
Z
−2 2
|x|
|x|
h =
RN
RN −2 2
−b
|x|
v ,
RN
|v|
Z
p−2 2
h =1,
RN
Thus from (4.1) we infer that Z Z 2 |∇v| − λ |x|−2 v 2 ≤ RN
RN
Z
RN
1 p−1
|x|−b |v|p−2 vh = 0 .
RN
Z |∇v| RN
2
Z −λ RN
Z v + (k − 1)
−2 2
|x|
−2 2
|x|
v
.
RN
The conclusion easily follows. Proof of Theorem 0.4. Assume that v ∈ Hλ,cyl (Rk × RN −k ) achieves Spλ . Then, by Theorem 4.1 and by Hardy inequality one has !Z 2 Z Z k−2 −λ |x|−2 v 2 dξ ≤ |∇v|2 dξ − λ |x|−2 v 2 dξ 2 RN RN RN Z k−1 ≤ |x|−2 v 2 dξ . p − 2 RN Thus λ ≥
k−2 2 2
−
k−1 p−2 .
Equality can not hold, since the Hardy constant is not achieved.
10
R Remark 4.2 Set Σ = {u ∈ Hλ (Rk × RN −k ) | RN |x|−b |v|p dξ = 1 } and Z E(v) = |∇v|2 − λ|x|−2 v 2 dξ , E : Σ −→ R . RN
Then the Morse index of any cylindrically symmetric solution v to (0.1) diverges to +∞ as λ → −∞. More precisely, for j ≥ 1 let Λj be the eigenspace of −∆Sk−1 relative to the eigenvalue µj = j(k + j − 2). Then E 00 (v) is negative definite on Λi for any i = 1, ..., j, 2 µj . − p−2 provided that λ < k−2 2 2 Remark 4.3 Assume 2 ≤ k < N , λ ≤ k−2 and p ∈ (2, 2∗N ]. We compare here the best 2 λ λ k N −k constants Sp and Sp,cyl . Since Hλ,cyl (R × R ) ⊂ Hλ (Rk × RN −k ), then λ Spλ ≤ Sp,cyl . λ The infimum Sp,cyl is always achieved on Hλ,cyl (Rk × RN −k ) by Lemma 2.1. By the results in [9], [15], [17] we have that Spλ is achieved on Hλ (Rk × RN −k ) if p < 2∗N , or if p = 2∗N and 2 k−2 0≤λ≤ . (4.2) 2
Notice that (4.2) is a necessary condition for existence in in the limiting case p = 2∗N , since for λ < 0 it happens that S2λ∗ = S. In particular, if k = 2 then S2λ∗ is never achieved, unless N 2 N λ = 0. Finally, S2λ∗ < S if and only if k ≥ 3 and 0 < λ ≤ k−2 . 2 N Next, by the uniqueness result in [13] it turns out that, up to dilations and translations, problem (0.1) has at most one entire cylindrically symmetric solution. Taking into account also Theorem 0.3, we can state that 2 k−2 λ Spλ = Sp,cyl if 0 ≤ λ ≤ , p ∈ (2, 2∗N ] 2 k−2 2 − k−1 p ∈ (2, 2∗N ) or λ ≤ 2 p−2 , λ λ Sp < Sp,cyl if λ < 0 , p = 2∗N .
Sλcyl
Sλcyl Sλ
S Sλ 0
(k−2 2 )
2
λ
0
Fig.1 k ≥ 3, p < 2∗N .
Fig.2 k ≥ 3, p = 2∗N . 11
(k−2 2 )
2
λ
We conclude by noticing the following multiplicity result. 2 k−1 Corollary 4.4 Assume 2 ≤ k < N , p ∈ (2, 2∗N ) and λ ≤ k−2 − p−2 . Then problem (0.1) 2 has at least two distinct (modulo dilations and translations) entire solutions.
5
On a degenerate problem
In this Section we deal with problem −div(|x|−2α ∇u) = |x|−βp up−1 u>0, where p=
, x 6= 0
(5.1)
2N . N − 2 + 2(β − α)
On the exponents α, β we assume, as in [9], that N −2 N − 2 − 2α α 2. In the spherically symmetric case k = N problem (5.1) is related to the Caffarelli-KohnNirenberg inequalities [4]. Existence, non existence and breaking symmetry of extremals functions were discussed in [6]. For k < N the counterpart of the Caffarelli-Kohn-Nirenberg inequalities are the Maz’ya inequalities ([14], Section 2.1.6). Existence results can be found k−2 in [17], where β = α = k−2 2 , and in [15] for α 6= 2 . Some multiplicity results for (5.1) were proved in [9]. We say that a classical solution u to (5.1) is α-cylindrically symmetric if i) for any choice of y ∈ RN −k , u(·, y) is radially symmetric in Rk , and the map |x| → |x|α u(|x|, y) is decreasing. ii) there exists y0 ∈ RN −k such that, for any choice of x ∈ Rk \ {0}, u(x, ·) is symmetric decreasing about y0 in RN −k . In order to state our results for problem (5.1) we need to introduce a suitable Sobolev −2 we denote by D01 (RN ; |x|−2α dξ) the completion of space (see [9] for details). For α < k N2N Cc∞ (RN ) with respect to the scalar product Z hu, vi = |x|−2α ∇u · ∇v dξ . RN
As a Corollary to Theorem 0.1 we easily get the following result. 12
Corollary 5.1 Assume 2 ≤ k < N and (5.2). If N ≥ k + 2 assume in addition that α−
k−1
