Elliptic equations with decaying cylindrical potentials and power-type nonlinearities Marino Badiale - Michela Guida - Sergio Rolando
Abstract We obtain existence, nonexistence and asymptotic results for solutions to cylindrical equations of the form: − u+
A u = f (u) in RN , x = (y, z) ∈ Rk × RN −k , N > k ≥ 2, |y|α
where A > 0, 0 < α ≤ 2 and f is continuous and satisfies power-type growth conditions.
1
Introduction and main results
In the recent mathematical literature a great deal of work has been devoted to the study of nonlinear elliptic equations of the form − u + V (x) u = f (x, u)
in RN
(1)
with lim inf |x|→∞ V (x) = 0 (see e.g. [1]-[13], [17], [20]-[24], [29], [33], [36], [37]). The main motivation of this study is the fact that such equations arise in the search for solitary waves of nonlinear evolution equations of the Schrödinger or Klein-Gordon type (cf. [6], [11], [32], [35]). Roughly speaking, a solitary wave is a nonsingular solution which travels as a localized packet in such a way that the physical quantities corresponding to the invariances of the equation are finite and conserved in time. Accordingly, a solitary wave preserves intrinsic properties of particles such as the energy, the angular momentum and the charge, whose finiteness is strictly related to the finiteness of the L2 norm (cf. Remark 2 below). −2 In particular, equations of type (1) with V (x) = A |y| , x = (y, z) ∈ R2 × R, A > 0, give rise to solitary waves with nonvanishing angular momentum (see [7], [8]). Owing to their particle-like behaviour, solitary waves can be regarded as a model for extended particles and they arise in many problems of mathematical physics, such as classical and quantum field theory, nonlinear optics, fluid mechanics and plasma physics (see, for instance, [31], [40]). In this paper we consider semilinear elliptic scalar equations in RN featuring autonomous power-type nonlinearities and singular cylindrical potentials decaying to zero at infinity. 1
Letting N > k ≥ 2, p ≥ 2 and A, α > 0, the model problem for our results is A − u + α u = |u|p−2 u |y| (2) −α N p N u ∈ X(R , |y| dx) ∩ L (R )
with x = (y, z) ∈ Rk × RN −k and
X(RN , |y|−α dx) := D1,2 (RN ) ∩ L2 (RN , |y|−α dx) ,
(3)
where ∗
2∗ :=
D1,2 (RN ) = u ∈ L2 (RN ) : ∇u ∈ L2 (RN ) ,
2N , N −2
is the usual completion of Cc∞ (RN ) with respect to the L2 norm of the gradient and L2 (RN , |y|−α dx) denotes the Lebesgue space of square integrable functions −α with respect to the measure |y| dx. Actually, we also consider more general nonlinearities f ∈ C (R, R) satisfying (fp ) ∃M > 0 such that |f (s)| ≤ M |s|p−1 for all s ∈ R and refer to the notion of solution given by the following definition: we say that u ∈ X(RN , |y|−α dx) is a weak solution to the equation − u+
A u = f (u) |y|α
if and only if for all h ∈ X(RN , |y| RN
∇u · ∇h dx +
−α
RN
in Rk × RN −k
(4)
dx) one has A uh dx = |y|α
f (u) h dx .
(5)
RN
Our results mainly concern existence, nonexistence and asymptotic behaviour of weak solutions to equations (2) and (4), and they rely on compatibility conditions between the exponents α and p. The radial case of problem (2), namely A − u + α u = |u|p−2 u |x| (6) −α N p N u ∈ X(R , |x| dx) ∩ L (R )
(with N ≥ 3, p ≥ 2 and A, α > 0), has been studied by several authors (see [10], [18], [36], [37], [38]). In particular Terracini proves in [38] that no positive solution is allowed either for p = 2∗ and α = 2, or for p = 2∗ and α = 2. Indeed, as it is shown in [10], problem (6) turns out to have no solution at all whenever the pair (α, p) belongs to the light gray region of the picture below (including
2
boundaries, except for the pair (α, p) = (2, 2∗ ) and the line α = 0), defined as A := A1 ∪ A2 ∪ A3 where / (2α , 2∗ ) , p ≥ 2 \ {(2, 2∗ )} A1 := (α, p) ∈ R2 : α ∈ (0, 2] , p ∈ ∗
2
/ (2 , 2α ) , p ≥ 2 A2 := (α, p) ∈ R : α ∈ (2, N ) , p ∈ ∗
2
A3 := (α, p) ∈ R : α ∈ [N, +∞) , p ∈ [2, 2 ]
(7) (8) (9)
and
2N for every α ∈ (0, N ) . (10) N −α On the other hand, all the positive radial solutions of (6) are explicitely known (see [38] again) for (α, p) = (2, 2∗ ), while existence and multiplicity of radial solutions follow from the results of [36] under the compatibility condition corresponding to the dark gray regions between the lines p = 2∗ and 2α :=
p = 2∗α := 2
2N − 2 + α 4α =2+ 2N − 2 − α 2N − 2 − α
(11)
(boundaries excluded), that is, α ∈ (0, 2) and p ∈ (2∗α , 2∗ ), or α ∈ (2, 2N − 2) and p ∈ (2∗ , 2∗α ), or α ≥ 2N − 2 and p > 2∗ .
Existence for (α, p) in the white region of the picture above is still an open problem. The strictly cylindrical problem (2) has been very much less studied and no existence result is avaliable in the literature, at least at our knowledge. The only result we know is due to Musina [29], who shows that, for α = 2 and p = 2∗ , problem (2) does not admit ground state solutions, namely, the infimum 2
S (N, k, A) :=
RN
inf
|∇u| dx + A
u∈(RN ,|y|−2 dx)\{0}
2∗
RN
3
|u|
dx
u2 dx RN |y|2 2/2∗
is not achieved (for A > 0 and N > k ≥ 2). In the following theorem we complete such a result by establishing that an higher level solution actually exists (see also Remark 3). Theorem 1 Let N > k ≥ 2 and A > 0. Then the equation − u+
A
2u
|y|
∗
= |u|2
−2
u
in Rk × RN−k
(12)
has at least a nonzero nonnegative weak solution, satisfying u (y, z) = u (|y| , z). Writing u (y, z) = u (|y| , z) we naturally mean u (y, z) = u (gy, z) for all g ∈ O (k) and almost every (y, z) ∈ Rk × RN −k . The next theorem concerns the (strictly cylindrical) case of equation (4) with α < 2. For N > k ≥ 2, define pα := 2 and
2k − 2 + α 4α =2+ 2k − 2 − α 2k − 2 − α 2 2+ N p¯α := pα
for every α ∈ (0, 2k − 2)
if 0 < α ≤ 2
k−1 2N + 1
k−1 k−1 if 2 ≤α 0 and 0 < α < 2 (k − 1) / (N − 1). Assume that f ∈ C 1 (R, R) satisfies f (0) = 0 together with the following conditions: p−2
(fp ) ∃M > 0 and ∃p ∈ [¯ pα , 2∗ ) such that |f (s)| ≤ M |s|
for all s ∈ R
(f ) ∃ϑ > 2 such that ϑF (s) ≤ f (s) s < f (s) s2 for all s ∈ R \ {0} (F) F ≥ 0 on (0, +∞) and ∃¯ s > 0 such that F (s) > 0 for all s ≥ s¯ , s
where F (s) := 0 f (t) dt. Then equation (4) has at least a nonzero nonnegative weak solution, which belongs to Lq (RN ) for all q ∈ [¯ pα , 2∗ ], satisfies u (y, z) = u (|y| , |z|) and is nonincreasing with respect to z. Notice that assumption (fp ) together with f (0) = 0 implies condition (fp ) (with the same exponent p). On the other hand, Theorem 2 applies to problem (2) and provides solutions for the pairs (α, p) belonging to the dark gray region of the picture above. As before, u (y, z) = u (|y| , |z|) means u (y, z) = u (g1 y, g2 z) for all (g1 , g2 ) ∈ O (k) × O (N − k) and almost every (y, z) ∈ Rk × RN−k . Then the request of nonincreasingness with respect to z asks for the following condition: |z1 | ≤ |z2 | ⇒ u (y, z1 ) ≥ u (y, z2 ) ≥ 0 for a.e. (y, (z1 , z2 )) ∈ Rk × R2(N −k) . Note that this definition requires nonnegativity. Remark 1 All the hypotheses of Theorem 2 still hold true if one replaces f (s) with f (|s|) s/ |s|. Thus, as we are interested in nonnegative solutions, Theorem 2 also works if assumptions (fp ) and (f ) only hold for s > 0. The existence of solutions to problem (2) for pairs (α, p) in the white region of the picture above is an open issue, but the following result shows that, at least under an additional assumption on the summability of the gradient, no solution is allowed if (α, p) is in the same region of nonexistence for the radial problem (6), namely A = A1 ∪ A2 ∪ A3 defined by (7)-(10). For N > k ≥ 2 and α > 0 we set qk :=
2k , k−1
qkα :=
2k , k−2+α
qk,N :=
2kN . (k − 2) N + 2k
Theorem 3 Let N > k ≥ 2 and A > 0. If (α, p) ∈ A then the equation − u+
A p−2 u α u = |u| |y|
in (Rk \ {0}) × RN−k −α
has no nonzero classical solution u ∈ C 2 ((Rk \{0})×RN−k )∩X(RN , |y| dx)∩ Lp (RN ) satisfying ∇u ∈ Lqloc RN for q = max {qk , qkα } or q = max {qk , qk,N }.
5
We observe that qk = max {qk , qkα } if α ≥ 1 and qk = max {qk , qk,N } if 2k/N ≥ 1, while qkα ≥ qk,N > qk if α ≤ 2k/N < 1 and qk,N ≥ qkα > qk if 2k/N ≤ α < 1. In Theorem 32 of Section 5 we also give a nonexistence result for the case of equation (4) with nonlinearities different from a pure power. From the results of [7, Propositions 5 and 6] we know that all nonnegative solutions of equation (12), N > k ≥ 2, A > 0, are bounded on RN (which is not obvious owing to the singularity of the potential) and satisfy lim sup |x|ν u (x) < ∞
for every ν
k ≥ 2, A > 0 and 0 < α < 2 (k − 1) / (N − 1). Assume that f ∈ C (R, R) satisfies (fp ) for some p ∈ (2 + 2pα /N, 2∗ ). Then any znonincreasing weak solution u (y, z) = u (|y| , |z|) of equation (4) is bounded on RN and satisfies √ 2 A β|x|1−α/2 lim sup e u (x) < ∞ for every β < . (16) 2−α |x|→∞ Observe that, with respect to Theorem 2, Theorem 4 also works in lower dimensions (for which existence is still an open question) but shrinks the range of the admissible exponents p (see picture below).
6
Remark 2 Thanks to Theorem 4 and the already mentioned results of [7] leading to (15), the solutions we find in Theorem 1 (provided that A > 3/4 if N = 3) and Theorem 2 actually belong to H 1 (RN ).
2
Preliminaries and notations
Let N, k ∈ N be such that N > k ≥ 2 and let A, α > 0. As in (3), we define the weighted Sobolev space −α
X := X(RN , |y|
dx) :=
u ∈ D1,2 RN :
RN
u2 dx < +∞ , |y|α
which is a Hilbert space with respect to the norm defined by u
2
:= RN
|∇u|2 dx + A
RN
u2 dx |y|α
for all u ∈ X ,
(17)
whose inner product we denote by (u | v) :=
RN
∇u · ∇v dx + A
RN
uv α dx |y|
for all u, v ∈ X .
Clearly X → D1,2 (RN ), whence, by well known embeddings of D1,2 (RN ), one ∗ derives X → L2 (RN ) and X → Lploc (RN ) for 1 ≤ p ≤ 2∗ . In particular, the latter embedding is compact if p < 2∗ and thus it assures that weak convergence in X implies (up to a subsequence) almost everywhere convergence in RN . 7
Of main interest in the following are the subspaces of symmetric functions Xr := Xr (RN , |y|−α dx) := {u ∈ X : u (y, z) = u (|y| , z)} −α
N
Xs := Xs (R , |y|
dx) := {u ∈ X : u (y, z) = u (|y| , |z|)} .
(18) (19)
Thanks to almost everywhere convergence (up to a subsequence) of X-converging sequences, Xr and Xs are closed in X and thus they are Hilbert spaces with respect to the same norm (17) of X. For any f ∈ C (R, R) satisfying (fp ) for some p > 2, we set I (u) :=
1 u 2
2
−
F (u) dx
(20)
RN
s
where F (s) := 0 f (t) dt. Thanks to condition (fp ), this defines a real functional I : X ∩ Lp (RN ) → R on the Banach space X ∩ Lp (RN ) equipped with the norm · + · Lp (RN ) . Moreover, by standard computations, I is of class C 1 and has Fréchet derivative I (u) at any u ∈ X ∩ Lp (RN ) given by I (u) h := (u | h) −
f (u) h dx RN
for all h ∈ X ∩ Lp RN .
(21)
Notice that I is the Euler functional associated to equation (4) but its critical points are not weak solutions in general, since they do not necessarily satisfy (5) for all h ∈ X. We conclude this preliminary section by summarizing the notations of most frequent use in the following. Notations • Given N, k ∈ N, we shall always write x = (y, z) ∈ Rk × RN−k . s • Given f ∈ C(R, R), we always denote F (s) := 0 f (t) dt for all s ∈ R. • N is the set of natural numbers, including 0. • For any r ∈ R we set r+ := (|r| + r) /2 and r− := (|r| − r) /2, so that r = r+ − r− with r+ , r− ≥ 0. • The open ball Br (ξ 0 ) := ξ ∈ Rd : |ξ − ξ 0 | < r shall be simply denoted by ¯r stands for the closure of Br and σ d denotes the (d − 1)Br when ξ 0 = 0. B dimensional measure of the unit sphere ∂Br of Rd . • |A| and χA respectively denote the d-dimensional Lebesgue measure and the characteristic function of any measurable set A ⊆ Rd , d ≥ 1. We set Ac := Rd \ A. • O (d) is the orthogonal group of Rd . • By → and we respectively mean strong and weak convergence in a Banach space E, whose dual space is denoted by E . • → denotes continuous embeddings. • Cc∞ (Ω) is the space of the infinitely differentiable real functions with compact support in the open set Ω ⊆ Rd , d ≥ 1. • For any open set Ω ⊆ RN , N ≥ 3, D01,2 (Ω) denotes the usual Sobolev space ∗ defined as the closure of Cc∞ (Ω) in D1,2 (Ω) = {u ∈ L2 (Ω) : ∇u ∈ L2 (Ω)} with respect to the norm u L2∗ (Ω) + ∇u L2 (Ω) . Recall that ∇u L2 (Ω) is an equivalent norm on D01,2 (Ω). 8
3
Existence for the critical problem
In this section we focus on equation (2) in the critical case considered in Theorem 1, namely A 2∗ −2 − u + 2 u = |u| u in Rk × RN −k . (22) |y| As we are interested in nonnegative solutions, we set 2∗ −1
f (t) := χ(0,+∞) (t) |t|
for all t ∈ R
(23)
and look at (22) as the Euler-Lagrange equation of the corresponding functional (20), that is, I (u) =
1 2
RN
where F (t) =
|∇u|2 dx +
A 2
u2 RN
1 2∗ χ(0,+∞) (t) |t| ∗ 2
|y|2
dx −
F (u) dx RN
for all t ∈ R .
(24)
∗
By the continuous embedding X = X(RN , |y|−2 dx) → L2 (RN ), this functional belongs to C 1 (X, R) and the variational approach to equation (22) is quite standard (see Lemma 8 below). The only difficulty is that we cannot guarantee the fulfilment of the Palais-Smale condition, owing to the lack of compactness due to z-translation invariance. Such a difficulty will be faced in Subsection 3.1 with the aid of a version of the Concentration-Compactness Principle, due to S. Solimini [34]. In order to state and use the result of Solimini, we have to preliminarly introduce a group of rescaling operators, of which we also remark some basic properties. Definition 5 Fix λ > 0 and x ∈ RN . For any u ∈ Lp RN with 1 < p < ∞ we define Tλ,x u := λ−(N −2)/2 u λ−1 · + x . Clearly Tλ,x u ∈ Lp (RN ) for all u ∈ Lp (RN ) and in particular Tλ,x u ∈ D (RN ) if u ∈ D1,2 (RN ). Moreover, by direct computations, it is easy to ∗ ∗ see that the linear operators u ∈ L2 (RN ) → Tλ,x u ∈ L2 (RN ) and u ∈ D1,2 (RN ) → Tλ,x u ∈ D1,2 (RN ) are isometric. Notice that 1,2
−1 Tλ,x = T1/λ,−λx
and Tλ1 ,x1 Tλ2 ,x2 = Tλ1 λ2 ,x1 /λ2 +x2 .
(25)
Similarly, for any z˜ = (0, z) ∈ RN and λ > 0, one plainly deduces that the linear operator u → Tλ,˜z u is an isometry of both X and the symmetric subspace Xr = Xr (RN , |y|−2 dx) introduced in (18). The following proposition is proved in [7].
9
Proposition 6 Let 1 < p < ∞ and assume that {λn } ⊂ (0, +∞) and {xn } ⊂ RN are such that λn → λ = 0 and xn → x. Then un u in Lp (RN ) implies p N Tλn ,xn un Tλ,x u in L (R ). We now recall the above mentioned result of Solimini [34]. Theorem 7 If {un } ⊂ D1,2 RN is bounded, then, up to a subsequence, either ∗ un → 0 in L2 (RN ) or there exist {λn } ⊂ (0, +∞) and {xn } ⊂ RN such that ∗ Tλn ,xn un u in L2 (RN ) and u = 0.
3.1
Proof of Theorem 1
Here we assume N > k ≥ 2 and A > 0, and we give the proof of Theorem 1, which will be achieved through several lemmas. In the first one we collect some standard but useful properties of I on Xr , which derive from the Principle of Symmetric Criticality [30], the Sobolev embedding and other well known arguments (a detailed proof of iv can be found for instance in [7]). Lemma 8 i) Every critical point of I|Xr is nonnegative and weakly solves equation (22). ii) I|Xr has a mountain-pass geometry, namely, there exist ρ > 0 and u ¯ ∈ Xr with u ¯ > ρ such that min
u∈Xr , u ≤ρ
I (u) = I (0) = 0,
inf
u∈Xr , u =ρ
I (u) > 0
and
I (¯ u) < 0 .
iii) Every Palais-Smale sequence for I|Xr is bounded in Xr . iv) For any h ∈ Xr the mapping I (·) h : Xr → R is sequentially weakly continuous. In order to prove Theorem 1 and according to Lemma 8.i, we look for nonzero critical points of I|Xr . The starting point is the bounded Palais-Smale sequence {vn } ⊂ Xr provided by Lemma 8.ii-iii (see [39, Theorem 2.9]), which is such that I (vn ) → c > 0 and I (vn ) → 0 in Xr , where c is the mountain-pass level of I|Xr defined by c := inf max I (γ (t)) , Γ := {γ ∈ C ([0, 1] , Xr ) : γ (0) = 0, I (γ (1)) < 0} . γ∈Γ t∈[0,1]
(26) As {vn } is bounded, it must satisfy one of the alternatives allowed by Theorem 7. The next lemma shows that the first one cannot occur. ∗
Lemma 9 The sequence {vn } does not tend to zero in L2 (RN ).
10
∗
Proof. For sake of contradiction, assume that vn → 0 in L2 (RN ), so that (23) and (24) imply RN
|f (vn ) vn | dx ≤ vn
and 0≤
RN
F (vn ) dx ≤
2∗ L2∗ (RN )
1 vn 2∗
→0
2∗ L2∗ (RN )
→ 0.
Therefore, since I (vn ) → 0 in Xr and {vn } is bounded in Xr , we conclude vn
2
= I (vn ) vn +
and thus I (vn ) =
1 vn 2
2
−
RN
f (vn ) vn dx → 0
RN
F (vn ) dx → 0 ,
which contradicts J (vn ) → c > 0. Thanks to Lemma 9, Theorem 7 implies that there exist {λn } ⊂ (0, +∞), ∗ {xn } ⊂ RN and v ∈ L2 (RN ) such that (up to a subsequence) Tλn ,xn vn
∗
v in L2
RN
and v = 0 .
Letting xn =: (yn , zn ), y˜n := (yn , 0) and z˜n := (0, zn ), so that xn = y˜n + z˜n , we now define un := Tλn ,˜zn vn and exploit the invariances of equation (22) to get the following lemma. Lemma 10 The sequence {un } is bounded in Xr and satisfies I (un ) → c,
I (un ) → 0 in Xr
and
un ( · + λn y˜n )
∗
v in L2
RN .
Proof. Since the operators Tλn ,˜zn are isometries of Xr , {un } ⊂ Xr is bounded because {vn } ⊂ Xr is bounded. Moreover, recalling (25), we have T1,λn y˜n un = T1,λn y˜n Tλn ,˜zn vn = Tλn ,xn vn
∗
v in L2
∗
RN .
Finally, since f (λt) = λ2 −1 f (t) for all t ∈ R and λ > 0, by easy computation we obtain I (un ) = I (vn ) and I (un ) h = (un | h) − − N 2−2
= λn
f (un ) h dx RN
˜n | h + vn λ−1 n · +z
∗ − N−2 2 (2 −1)
−λn
− N 2−2
= λn
RN
f vn λ−1 ˜n n · +z
−2 λN (vn | h (λn · − λn z˜n )) + n
11
h dx
∗ − N−2 2 (2 −1) N λn
−λn
RN
f (vn ) h (λn · − λn z˜n ) dx
N−2
= =
N−2
vn | λn 2 h (λn · − λn z˜n ) − vn | Tλ−1 ˜n h − n ,−λn z
RN
RN
f (vn ) λn 2 h (λn · − λn z˜n ) dx
f (vn ) Tλ−1 ˜n h dx n ,−λn z
= I (vn ) Tλ−1 ˜n h n ,−λn z for all h ∈ Xr , which implies I (un ) Tλ−1 ˜n is an isometry of Xr . n ,−λn z
Xr
= I (vn )
Xr
again by the fact that
The key step in the proof of Theorem 1 is the removal of translations from the sequence un ( · + λn y˜n ). This is the topic of Lemma 12, where we will take advantage of the following elementary proposition (see [7] for a proof). Proposition 11 Let {η n } ⊂ Rk be such that limn→∞ |η n | = +∞ and fix R > 0. Then for any m ∈ N \ {0, 1} there exists nm ∈ N such that for any n > nm one can find g1 , ..., gm ∈ O (k) satisfying the condition: i = j ⇒ BR (gi ηn ) ∩ BR (gj ηn ) = ∅. Lemma 12 There exists u ∈ Xr , u = 0, such that (up to a subsequence) un in Xr .
u
Proof. As {un } ⊂ Xr is bounded (Lemma 10), we can assume that (up to a subsequence) un u in Xr . If u = 0 the proof is complete. We are now going to show by contradiction that u = 0 is impossible. So, assume un
0 in Xr .
(27)
Letting T˜n := T1,λn y˜n for brevity, we recall from Lemma 10 that T˜n un ∗ in L2 (RN ). First, we deduce that lim |λn y˜n | = +∞ .
v=0 (28)
n→∞
Otherwise, up to a subsequence, λn y˜n → y˜0 ∈ Rk × {0} and T1,−λn y˜n T˜n un ∗ T1,−˜y0 v in L2 (RN ) by Proposition 6, but, since T1,−λn y˜n T˜n = T1,0 , this means ∗ un T1,−˜y0 v = 0 in L2 (RN ) and thus it contradicts (27). Since v = 0, there exist δ > 0 and D ⊆ RN with |D| = 0 such that either v > δ or v < −δ almost everywhere in D. Fixing R > 0 such that |BR ∩ D| > 0, by weak convergence we obtain RN
T˜n un χBR ∩D dx →
vχBR ∩D dx ≥ δ |BR ∩ D| > 0 .
RN
(29)
On the other hand RN
T˜n un χBR ∩D dx
≤
T˜n un dx = BR
BR
|un ( · + λn y˜n )| dx
(30) 1/2∗
= BR (λn y ˜n )
|un | dx ≤ C
12
∗
2
BR (λn y ˜n )
|un |
dx
where C > 0 only depends on R and N . From (29) and (30) we now deduce that 2∗ lim inf |un | dx > 0 n
BR (λn y˜n )
and hence, up to a subsequence, we can assume inf n
BR (λn y˜n )
|un |
2∗
dx > ε for some ε > 0 .
(31)
This will yield a contradiction. Indeed, using (28), from Proposition 11 it readily follows that for every m ∈ N \ {0, 1} there exists nm ∈ N such that for any n > nm one can find g1 , ..., gm ∈ O (k) satisfying the condition i = j ⇒ BR (λn (gi yn , 0)) ∩ BR (λn (gj yn , 0)) = ∅. As a consequence, using (31) and the fact that un ∈ Xr , we get 2∗
RN
|un |
m
dx ≥
i=1
BR (λn (gi yn ,0))
|un |
2∗
m
2∗
dx = i=1
BR (λn y ˜n )
|un |
dx > mε
for every m ∈ N \ {0, 1} and n > nm . This finally implies un L2∗ (RN ) → +∞ ∗ as n → ∞, which is a contradiction, since {un } is bounded in L2 (RN ). We are now able to easily conclude the proof of Theorem 1. Proof of Theorem 1. Since un u = 0 in Xr (Lemma 12) and I (un ) → 0 in Xr (Lemma 10), from Lemma 8.iv we deduce that u is a nonzero critical point of I|Xr . The proof is then completed by Lemma 8.i. Remark 3 Since the sequence of approximating solutions {un } is a PalaisSmale sequence at the mountain-pass level c defined in (26) (Lemma 10), by standard arguments exploiting the properties of the projection on the Nehari manifold 2 2∗ Nr := w ∈ Xr \ {0} : w = w L2∗ (RN ) (see [39, Lemma 4.1, Theorem 4.2]) it is not difficult to see that the nonzero nonnegative solution found in Theorem 1 is a ground state solution for equation (22) restricted to Xr , namely, u achieves the infimum 2
Sr (N, k, A) :=
4
inf
RN
|∇w| dx + A
w∈Xr \{0}
2∗
RN
|w|
dx
w2 dx RN |y|2 ∗ 2/2
.
Existence for the sub-critical problem
In this section we will deduce Theorem 2 by studying the non-critical case of equation (4), namely − u+
A α u = f (u) |y| 13
in Rk × RN −k
(32)
where f satisfies (fp ) for some p > 2, p = 2∗ . In this case the presence of a potential vanishing at infinity creates some difficulties in the variational approach to the equation. For example, it prevents the use of well known H 1 (RN ) em−α beddings, so that we cannot assure that the energy space X = X(RN , |y| dx) p N ∗ is contained in any L (R ) with p = 2 . Hence, under the growth condition (fp ) alone (which is the case we are interested in), the Euler functional (20) associated to (32), namely I (u) =
1 2
RN
|∇u|2 dx +
A 2
RN
u2 dx − |y|α
F (u) dx ,
(33)
RN
does not necessarily make sense on X. On the other hand, owing to evident difficulties in bounding criticizing sequences, X ∩ Lp (RN ) is not a good space where to look for critical points of I. Therefore a different approach is nedeed. Dealing with the radial case of equation (32), such obstacles have been overcome in [18] by considering the nonlinear ODE associated to the problem of radial solutions, whose Euler fuctional turns out to be well defined on a suitable subspace of H 1 (R) provided that a suitable compatibility condition between α and p holds. Here, instead, we follow the technique used in [10] and [36]. In Subsection 4.1 we deepen the study of the symmetric subspace Xs = Xs (RN , |y|−α dx) introduced in (19) and give a pointwise estimate (Theorem 13) which allows us to deduce some embedding and compactness properties of Xs in connection with Lebesgue spaces (Proposition 18 and Theorem 20). Then, in Subsection 4.2, we derive a variational principle for equation (32) by proving that Xs ∩ Lp (RN ) is, in some sense, a natural constraint for finding weak solutions (Proposition 26). In Subsection 4.3 we finally give the proof of Theorem 2, by showing, via a Steiner symmetrization argument, that I achieves its infimum over the set of solutions which belongs to Xs ∩ Lp (RN ). Unfortunately, maybe for technical reasons, our arguments need high dimensions and rely on a compatibility condition between α and p which implies α < 2 and p < 2∗ .
4.1
The weighted Sobolev space Xs (RN , |y|−α dx)
Let A, α > 0. The first result of this subsection concerns a pointwise estimate which plays a preminent role in the whole section. However, since quite technical, we displace its proof in the Appendix at the end of the paper. Here we denote m := N − k. Theorem 13 Let N > k ≥ 2. If u ∈ Xs is nonincreasing with respect to z then 2(m−1)/2 (m + 1) 1 u u (y, z) ≤ √ 1/4 (2k−2−α)/4 (m+1)/2 σk σm 2 − 1 A |y| |z|m/2 for almost every (y, z) ∈ Rk × Rm . 14
For sake of brevity, hereafter we denote 2(m−1)/2 (m + 1) 1 CN,k,A := √ . σ k σ m 2(m+1)/2 − 1 A1/4
(34)
Recall from (13) that we define pα := 2
2k − 2 + α 4α =2+ 2k − 2 − α 2k − 2 − α
for k ≥ 2 and α ∈ (0, 2k − 2)
and notice that pα < 2∗ if and only if α < 2 (k − 1) / (N − 1). Proposition 14 Let N > k ≥ 2 and 0 < α < 2k−2. If u ∈ Xs is nonincreasing c with respect to z then u ∈ Lpα (Rk × BR ) for all R > 0 and u
c ) Lpα (Rk ×BR
(p −2)/pα
α ≤ CN,k,A
A−1/pα u R−mα/(2k−2+α) .
Proof. From Theorem 13 we get u (y, z)
pα
pα −2
= u (y, z) ≤
2
u (y, z)
pα −2
pα −2 CN,k,A u (pα −2)(2k−2−α)/4
|y|
pα −2
pα −2 = CN,k,A u
(pα −2)m/2
|z|
u (y, z)2
u (y, z)2 α 2αm/(2k−2−α) |y| |z| 1
for almost every (y, z) ∈ Rk × Rm . Hence c Rk ×BR
pα −2 upα dx ≤ CN,k,A u
pα −2
1 R2αm/(2k−2−α)
c Rk ×BR
u2 α dx |y|
and the result ensues. Corollary 15 Let N > k ≥ 2 and 0 < α < 2 (k − 1) / (N − 1). If u ∈ Xs c is nonincreasing with respect to z then u ∈ Lp (Rk × BR ) for all R > 0 and ∗ p ∈ [pα , 2 ]. Moreover there exists a constant c1 = c1 (N, k, A, α, p) > 0 such that ∗ u Lp (Rk ×B c ) ≤ c1 u R−mN(1/p−1/2 )α/(2k−2+α−αN) . R
2∗
Proof. Since u ∈ L (RN ), the claim follows from Proposition 14 together with Hölder and Sobolev inequalities. We now look for estimates on the strips of RN with |z| bounded. To this end we have to assume m ≥ 3, so that we can use a well known radial lemma [15] in dimension m, that is, there exists a constant Cm > 0 such that all the radial mappings w ∈ D1,2 (RN ) satisfy |w (z)| ≤ Cm ∇w
L2 (Rm )
− m−2 2
|z|
for almost every z ∈ Rm
(35)
(let us observe that in [15] the authors prove (35) for |z| ≥ 1, but their argument actually works for z = 0). 15
Proposition 16 Let N ≥ k + 3 ≥ 4. Then every u ∈ Xs satisfies 2
Rk
2 u (y, z) dy ≤ Cm ∇u
2 L2 (RN )
1 m−2
|z|
for almost every z ∈ Rm .
(36)
In particular, Xs → L2 (Rk × BR ) for all R > 0 and u
L2 (Rk ×BR )
≤ Cm
σ m /2 u R .
(37)
Proof. Note that m = N − k ≥ 3 and let u ∈ Xs . As one can easily check, for almost every y ∈ Rk the radial mapping u (y, ·) has weak derivatives given by ∇ (u (y, ·)) = ∇z u (y, ·) ∈ L2 (Rm ). On the other hand, u ∈ X implies u (y, ·) ∈ L2 (Rm ). Hence u (y, ·) ∈ H 1 (Rm ) ⊂ D1,2 (Rm ) and by (35) we get 1/2
|u (y, z)| ≤ Cm
Rm
|∇z u (y, ζ)|2 dζ
Rm
|∇u (y, ζ)|2 dζ
|z|−(m−2)/2 1/2
≤ Cm
2
2
|z|−(m−2)/2 2
2
for almost every z ∈ Rm , since |∇z u| ≤ |∇y u| + |∇z u| = |∇u| almost everywhere on RN . Thus (36) ensues. Then u ∈ L2 (Rk × BR ) for all R > 0 and (37) readily follows from integrating on BR . Corollary 17 Let N ≥ k + 3 ≥ 4. Then Xs → Lp (Rk × BR ) for all R > 0 and p ∈ [2, 2∗ ]. Proof. As for Corollary 15, it follows from Proposition 16 by interpolation and Sobolev inequality. Proposition 18 Let N ≥ k + 3 ≥ 5 and 0 < α < 2 (k − 1) / (N − 1). If u ∈ Xs is nonincreasing with respect to z, then u ∈ Lp (RN ) for all p ∈ [pα , 2∗ ]. Moreover there exists c0 = c0 (N, k, A, α, p) > 0 such that u Lp (RN ) ≤ c0 u . Proof. It readily ensues from Corollaries 15 and 17. The above results lead to the compactness property contained in Theorem 20, which also relies on the following well known lemma. Lemma 19 Let N ≥ k + 1 ≥ 3. Then, for any R > 0 and p ∈ (2, 2∗ ), the space Hr1 (Rk × BR ) := u ∈ H 1 (Rk × BR ) : u (y, z) = u (|y| , z) is compactly embedded into Lp (Rk × BR ). Proof. See Lemma III.2 of [28] (in a different context, the argument is developed also in Lemma 6.8 of [9], and in [32] with full detail). Theorem 20 Let N ≥ k + 3 ≥ 5 and 0 < α < 2 (k − 1) / (N − 1). Then any bounded sequence {un } ⊂ Xs whose functions are nonincreasing in z is relatively compact in Lp (RN ) for p ∈ [pα , 2∗ ). 16
Proof. Let {un } ⊂ Xs be as in the statement. Then un u in Xs (up to a subsequence), where also u is nonincreasing in z by almost everywhere convergence. Moreover there exists C > 0 such that un , u ≤ C. Now mN (1−p/2∗ )α/(2k−2+α−αN) let ε > 0 and fix Rε > 0 such that Rε > 2p+2 cp1 C p /ε, where c1 is the constant of Corollary 15. Thus, by Corollary 15 itself, for all n one obtains
c Rk ×BR ε
|un − u|p dx ≤ 2p
c Rk ×BR ε
upn dx +
2p cp1
≤
up dx c Rk ×BR ε
mN(1−p/2∗ )α/(2k−2+α−αN )
Rε
2p+1 cp1 C p
≤
mN(1−p/2∗ )α/(2k−2+α−αN ) Rε
( un
k ≥ 2 and 0 < α < 2k − 2. There exists L1 = L1 (N, k, A, α) > 0 such that if u ∈ Xs is nonincreasing with respect to z then Rk ×B1c
upα −1 |v| dx ≤ L1 u
pα −1
v
for all v ∈ X .
Proof. By Hölder inequality we get
Rk ×B1c
upα −1 |v| dx = ≤ ≤
α/2
Rk ×B1c
|y|
α
Rk ×B1c
1 √ A
upα −1
|v|
|y|α/2
2(pα −1)
|y| u
dx 1/2
dx
1 A
RN
Av 2 α dx |y|
1/2 Rk ×B1c
17
|y|α u2(pα −1) dx
v .
1/2
Then we write 2 (pα − 1) = 2 + 8α/ (2k − 2 − α) and use Theorem 13 to obtain α
Rk ×B1c
|y| u2(pα −1) dx =
Rk ×B1c
u2 2α |y| u8α/(2k−2−α) dx |y|α
8α/(2k−2−α)
≤ CN,k,A ·
u
8α/(2k−2−α) 2α
2
Rk ×B1c
·
u |y| dx α 2α 4mα/(2k−2−α) |y| |y| |z|
8α/(2k−2−α)
≤ CN,k,A
8α/(2k−2−α)
≤ CN,k,A
A−1 u A−1 u
8α/(2k−2−α)
2(pα −1)
Rk ×B1c
Au2 α dx |y|
where CN,k,A is given by (34). Lemma 22 Let N ≥ k + 3 ≥ 4. If p ∈ [2 + 2/N, 2∗ ] then there exists L2 = L2 (N, k, A, α, p) > 0 such that p−1
Rk ×B1
|u|
p−1
|v| dx ≤ L2 u
v
for all u ∈ Xs and v ∈ X .
Proof. By Hölder and Sobolev inequalities there exists S = S (N ) > 0 such that p−1
Rk ×B1
|u|
|v| dx ≤
2∗ (p−1)/(2∗ −1)
Rk ×B1 p−1
≤ S u
|u|
2∗ (p−1)/(2∗ −1)
L
dx
(Rk ×B1 )
(2∗ −1)/2∗
v
∗
L2 (RN )
v .
The result then follows by Corollary 17, because p ∈ [2 + 2/N, 2∗ ] amounts to 2 ≤ 2∗ (p − 1) / (2∗ − 1) ≤ 2∗ . Recall from (14) that, for N > k ≥ 2, we denote 2 2+ N p¯α := pα
if 0 < α ≤ 2 if 2
k−1 2N + 1
k−1 k−1 ≤α 0 such that ∗
RN
|u|2
−1
|v| dx ≤ L3 u
2∗ −1
v
for all u, v ∈ X ,
we finally obtain the following proposition by interpolation.
18
(38)
Proposition 23 Let N ≥ k + 3 ≥ 5 and 0 < α < 2 (k − 1) / (N − 1). For every p ∈ [¯ pα , 2∗ ] there exists L = L (N, k, A, α, p) > 0 such that if u ∈ Xs is nonincreasing with respect to z then
RN
p−1
up−1 |v| dx ≤ L u
v
for all v ∈ X .
Proof. Taking into account that p ∈ [¯ pα , 2∗ ] implies p ∈ [pα , 2∗ ], fix λ, µ ∈ (0, 1) ∗ such that p = λpα + µ2 and λ + µ = 1. Then, by (38) and Lemma 21, we have
Rk ×B1c
∗
uλ(pα −1)+µ(2
up−1 |v| dx = =
Rk ×B1c
upα −1 |v| u
Rk ×B1c
Lµ3 Lλ1
λ
u2
∗
λ
µ
|v| |v| dx −1
µ
|v|
dx
λ pα −1
≤ ≤
−1)
Rk ×B1c
p−1
u
|v| dx
∗
u2 RN
µ −1
|v| dx
v
for all v ∈ X. Therefore, as p ∈ [¯ pα , 2∗ ] also implies p ∈ [2 + 2/N, 2∗ ], the claim follows by Lemma 22. We now let f ∈ C (R, R) satisfy (fp ) and recall from Section 2 that any critical point u ∈ X ∩ Lp (RN ) of the functional I ∈ C 1 X ∩ Lp (RN ), R defined in (33) satisfies A f (u) h dx (39) ∇u · ∇h + α uh dx = |y| RN RN
for all h ∈ X ∩ Lp (RN ). Hence u is a weak solution to equation (32) provided that we can take any h ∈ X as a test function. The next results show that this holds actually true if u also belongs to Xs and is nonincreasing in z.
Lemma 24 Let N > k ≥ 2 and 0 < α < 2. If 1 ≤ p ≤ 2∗ then Xs ∩ Lp (RN ) is dense in Xs . Proof. Let u ∈ Xs and fix φ ∈ C ∞ (R) such that 0 ≤ φ ≤ 1 on R, φ (t) = 1 for t ≤ 1 and φ (t) = 0 for t ≥ 2. Then for every n ∈ N \ {0} the mapping ϕn (x) := φ(|y| /n)φ(|z| /n) satisfies • |ϕn (x)| ≤ 1 for all x ∈ RN • ϕn (x) = 1 for all x ∈ Bn ¯ √ . • supp ϕn ⊆ B 2 2n Moreover, ∇ϕn (x) = φ
1 n
|y| φ
1 n
|z|
y n|y|
• ∇ϕn (x) = 0 for |y| ≥ 2n 19
z + φ( n1 |y|)φ ( n1 |z|) n|z| is such that
• |∇ϕn (x)| ≤
2 n
φ
L∞ (R)
for all x ∈ RN .
−α
Hence, ϕn u → u in L2 (RN , |y| convergence. Moreover
RN
|∇ϕn |2 u2 dx = ≤
|y| 0 such that x ∈ R : |u (x)| > s¯/t¯ > 0, which does exist because u = 0. So one deduces ϑ
0< {t¯|u|>¯ s}
|u| dx < +∞
and we conclude φu (t) ≤
t2 u 2
≤
t2 u 2
2
−
{t|u|>¯ s}
F (¯ s) 2 − tϑ ϑ s¯
F (tu) dx ≤ {t¯|u|>¯ s}
t2 u 2
2
− tϑ
F (¯ s) s¯ϑ
{t|u|>¯ s}
|u|ϑ dx
|u|ϑ dx → −∞
as t → +∞. The first part of the lemma then follows from claims (iii) and (ii), since one infers the existence of t0 > 0 such that φu (t0 ) = 0 = φu (0) and thus of a unique critical point µ > 0 for φu . By uniqueness and claim (i), µ satisfies 2 φu (µ) = maxt≥0 φu (t). Finally, since φu (0) = 0 and φu (0) = u > 0 imply φu (t) > 0 for t > 0 sufficiently small, if φu (1) = I (u) u < 0 then it must be µ < 1. As N is nonempty, it makes sense to consider the following minimization problem: ν := inf I (u) . (40) u∈N
Lemma 28 There exists a bounded minimizing sequence for problem (40) whose functions are nonincreasing with respect to z. Moreover one has ν > 0. Proof. Let {un } ⊂ N be any minimizing sequence. As f is odd, I is even and |un | ∈ N , so that we can assume un ≥ 0. Let u∗n be the (N − k, N )-Steiner symmetrization of un . By well known properties of Steiner symmetrizations (see [16] for instance), u∗n belongs to Xs ∩ Lp , is nonincreasing with respect to z and has the following properties: u∗n
Lp
= un
I
(tu∗n )
I
(u∗n ) u∗n
(41)
Lp
≤ I (tun )
for every t ∈ R
≤ I (un ) un = 0 .
(42) (43)
In particular, (41) implies u∗n = 0 and therefore, by Lemma 27 and (43), for every n there exists µn ∈ (0, 1] such that µn u∗n ∈ N . Hence for all n one has ν ≤ I (µn u∗n ) ≤ I (µn un ) ≤ max I (tun ) = I (un ) , t≥0
22
where we have used (42) for the second inequality, while the last equality follows by Lemma 27 from the fact that un ∈ N . As a consequence, setting vn := µn u∗n , {vn } ⊂ N is a minimizing sequence whose functions are nonincreasing with respect to z. Now we use assumption (f ) to obtain 1 1 − 2 ϑ
vn
2
= ≤
1 vn 2 1 vn 2
2
2
−
1 ϑ
−
f (vn ) vn dx RN
F (vn ) dx = I (vn ) ,
(44)
RN
whence, since ϑ > 2 and I (vn ) → ν, one infers ν ≥ 0 and { vn } bounded. Hence {vn } is bounded in Xs ∩ Lp by Proposition 18. Finally, suppose by contradiction that I (vn ) → 0, so that (44) gives vn → 0. Then from condition (fp ) and Proposition 18 we get vn
2
=
f (vn ) vn dx ≤ M
RN
RN
which yields the contradiction 1 ≤ M cp0 vn
|vn |p dx ≤ M cp0 vn
p−2
p
,
→ 0 since p > 2.
Theorem 29 The minimization problem (40) has a z-nonincreasing solution. Proof. Let {un } ⊂ N be any bounded minimizing sequence such that un ≥ 0 is nonincreasing in z, which exists by Lemma 28. By Theorem 20, one deduces that (up to a subsequence) un u in Xs un → u in Lp and almost everywhere on RN . Then, passing to the limit as n → ∞ in I (un ) =
1 un 2
2
−
F (un ) dx = RN
RN
we obtain ν= RN
1 f (un ) un − F (un ) dx , 2
1 f (u) u − F (u) dx 2
(45)
which gives u = 0 since ν > 0 (Lemma 28). We now show that u ∈ N , which completes the proof since (45) and u 2 = RN f (u) u dx imply I (u) = ν. Note that u is nonincreasing with respect to z by almost everywhere convergence. With a view to deducing a contradiction, assume u ∈ / N . As un ∈ N and un → u in Lp , one has lim
n→∞
un
2
= lim
n→∞
f (un ) un dx = RN
f (u) u dx . RN
Hence, by weak lower semi-continuity of the norm, the weak convergence un in Xs gives I (u) u = u
2
−
RN
f (u) u dx ≤ lim
n→∞
23
un
2
−
f (u) u dx = 0 . RN
u
Since u ∈ / N and u = 0, this implies I (u) u < 0, so that, by Lemma 27, there exists µ ∈ (0, 1) such that µu ∈ N . Thus we have ν ≤ I (µu) =
1 f (µu) µu − F (µu) dx . 2
RN
On the other hand, the mapping Φ (t) := increasing for t > 0, since one has Φ (t) = =
1 2 1 2t
RN
RN
RN
1 2f
(46)
(tu) tu − F (tu) dx is strictly
tf (tu) u2 − f (tu) u dx f (tu) (tu)2 − f (tu) tu dx > 0
by assumption (f ). Hence, by (46) and (45), we get the contradiction ν ≤ Φ (µ) < Φ (1) = ν. Proof of Theorem 2. First we observe that N is a C 1 manifold. Indeed, letting J (u) := I (u) u, one has that J is of class C 1 on Xs ∩ Lp and J (u) h = 2 (u | h) −
(f (u) uh + f (u) h) dx RN
for all u, h ∈ Xs ∩ Lp ,
whence, by assumption (f ), for every u ∈ N we get J (u) u = 2 u = RN
2
−
f (u) u2 + f (u) u dx RN
f (u) u − f (u) u2 dx < 0
(47)
Now we apply Theorem 29 to deduce the existence of u ∈ N satisfying I (u) = min I (N ) and such that u (y, z) = u (|y| , |z|) ≥ 0 is nonincreasing with respect to z. Then there exists a Lagrange multiplier λ ∈ R such that I (u) = λJ (u) in (Xs ∩ Lp ) . As u ∈ N , we have in particular I (u) u = 0, i.e., λJ (u) u = 0, which gives λ = 0 by (47) and hence I (u) = 0 in (Xs ∩ Lp ) . Thanks to Propositions 26 and 18, the proof is thus complete.
5
Nonexistence results
In this section we derive a Pohoˇzaev-type identity for the (classical) equation − u+
A u = f (u) |y|α
in (Rk \ {0}) × RN−k
(48)
with N > k ≥ 2, A, α > 0 and f ∈ C (R, R) (Proposition 30). After checking that classical solutions can be chosen as test functions for equality (48) in the distributional sense on RN (Lemma 31), this identity allows us to prove Theorem 24
3, yielding compatibility conditions on the parameters α and p in order that nontrivial solutions are permitted to equation (48) with power nonlinearities p−2 f (u) = |u| u. A nonexistence result for more general nonlinearities also follows (Theorem 32). Throughout the section we assume N > k ≥ 2 and A > 0. Recall that, for α > 0, we define qk :=
2k , k−1
qkα :=
2k , k−2+α
qk,N :=
2kN . (k − 2) N + 2k
Proposition 30 Let α > 0 and f ∈ C (R, R). Assume that u ∈ C 2 ((Rk \{0})× RN −k ) is a classical solution to equation (48). If
RN
|∇u|2 +
u2 + |F (u)| dx < +∞ |y|α
(49)
and ∇u ∈ Lqloc RN then
N −2 2
RN
|∇u|2 dx +
(50)
for some q ≥ qk
N −α 2
RN
Au2 dx = N |y|α
F (u) dx .
(51)
RN
Proof. The proof relies on a standard argument [26], suited to the case under discussion. The starting point are the following identities, which hold true on Rk \ {0} × RN −k : (x · ∇u) (x · ∇u)
u = div (x · ∇u) ∇u −
1 N −2 |∇u|2 x + |∇u|2 , 2 2
1 Au2 N − α Au2 Au , α = div αx − 2 |y| 2 |y|α |y|
(x · ∇u) f (u) = div [F (u) x] − N F (u) . For R2 > R1 > 0, set Ω := ΩR1 ,R2 := {x ∈ BR2 : |y| > R1 }. Upon multiplying equation (48) by (x · ∇u), the divergence theorem yields −
∂Ω
(x · ∇u) (∇u · ν) dσ + =
N −2 2
Ω
1 2
∂Ω
|∇u|2 dx +
|∇u|2 + N −α 2
Ω
Au2 α |y|
x · ν dσ −
Au2 α dx − N |y|
∂Ω
F (u) x · ν dσ
F (u) dx
(52)
Ω
where ν (x) is the outward normal of ∂Ω at x and dσ is the (N − 1)-dimensional measure of ∂Ω. We have ∂Ω = {x ∈ ∂BR2 : |y| ≥ R1 } ∪ {x ∈ BR2 : |y| = R1 } =: ΣR1 ,R2 ∪ ΓR1 ,R2 25
with obvious definitions of ΣR1 ,R2 and ΓR1 ,R2 . Notice that ΣR1 ,R2 ∩ΓR1 ,R2 = ∅. Note also that ΓR1 ,R2 = {x ∈ RN : |y| = R1 , |z| < R22 − R12 } and ν (x) = − (y/R1 , 0) on ΓR1 ,R2 . Since the integral R2
dR1
{|y|=R1 , |z|
0 such that R1,n ΓR1,n ,R2
|∇u|2 + |∇u|q +
u2 + |F (u)| dσ → 0 as n → ∞ . |y|α
Hence |∇u|2 +
ΓR1,n ,R2
Au2 α + |F (u)| x · ν dσ = |y|
= −R1,n
ΓR1,n ,R2
|∇u|2 +
Au2 + |F (u)| dσ → 0 . |y|α
Moreover, since q ≥ qk implies (k − 1) (q − 2) /2 ≥ 1, by Hölder inequality we get
ΓR1,n ,R2
(x · ∇u) (∇u · ν) dσ ≤ R2
2
ΓR1,n ,R2
|∇u| dσ ≤
1−2/q
≤ R2
(53)
2/q q
dσ ΓR1,n ,R2
ΓR1,n ,R2
|∇u| dσ
1−2/q
≤ R2 = R2
2/q q
dσ {|y|=R1,n , |z| 0 and f ∈ C (R, R). Assume that u ∈ X(RN , |y| dx) ∩ C 2 ((Rk \{0})×RN−k ) is a classical solution to equation (48) satisfying f (u) u ∈ L1 (RN ) and ∇u ∈ Lqloc RN for q = max {qk , qkα } or q = qk,N . Then 2
|∇u| +
RN
Au2 α |y|
f (u) u dx .
dx =
(55)
RN
Proof. The argument is analogous to the one of the proof of Proposition 30, so we will use the same notations and will be sketchy in some of the details. 2 Multiplying equation (48) by u, using the identity u u = div [u ∇u] − |∇u| k N−k in (R \ {0}) × R and applying the divergence theorem on Ω = ΩR1 ,R2 , we obtain |∇u|2 + A
f (u) u dx = Ω
Ω
−
ΓR1 ,R2
u2 α |y|
dx +
u (∇u · ν) dσ −
(56)
ΣR1 ,R2
u (∇u · ν) dσ .
Suppose q = max {qk , qkα }. By the same computation of (53), there exists a nonincreasing sequence R1,n → 0, R1,n > 0, such that −[(k−1)(q−2)−2]/q
R1,n
ΓR1,n ,R2
|∇u|2 dσ → 0 as n → ∞ .
Hence 1/2
ΓR1 ,R2
u (∇u · ν) dσ
≤ =
1/2 2
2
u dσ ΓR1,n ,R2
ΓR1,n ,R2
[(k−1)(q−2)−2]/q+α R1,n
ΓR1,n ,R2
|∇u| dσ
u2 α dσ |y|
(57)
1/2
o (1)n→∞
and thus, as q ≥ qkα implies α + [(k − 1) (q − 2) − 2] /q ≥ 1, (56) yields 2
f (u) u dx = BR2
BR2
|∇u| + A
u2 |y|α
27
dx + ∂BR2
u (∇u · ν) dσ .
(58)
Now assume q = qk,N and take R1,n → 0 such that ∗
|u|2 + |∇u|q dσ → 0 as n → ∞ .
R1,n ΓR1,n ,R2
Since 2∗ /(2∗ − 1) = 2N/(N + 2) < qk,N and N +2 1 − 2N q
(k − 1)
N −2 1 + , 2N q
=
we infer that there exists a constant C > 0 such that
ΓR1,n ,R2
u (∇u · ν) dσ ≤ 1/2∗ 2∗
≤
ΓR1,n ,R2
|u|
(N+2)/2N 2N
dσ ΓR1,n ,R2
|∇u| N+2 dσ
1/2∗
≤ C
ΓR1,n ,R2
|u|
2∗
1/q (k−1)[(N +2)/2N −1/q]
q
dσ ΓR1,n ,R2
|∇u| dσ
R1,n
1/2∗
= C
2∗
R1,n ΓR1,n ,R2
|u|
1/q
dσ
q
R1,n ΓR1,n ,R2
|∇u| dσ
,
so that one gets (58) again by passing to the limit as n → ∞. Finally, in a similar way, since 2∗ /(2∗ − 1) < 2 and (N − 1)/N = 1/2∗ + 1/2 we obtain 1/2∗ ∂BR2
u (∇u · ν) dσ
2∗
≤
∂BR2
|u|
1/2 (N −1)/N
2
dσ ∂BR2
|∇u| dσ
R2
1/2∗
=
1/N σN
2∗
R2 ∂BR2
|u|
dσ
1/2 2
R2 ∂BR2
|∇u| dσ
so that, as there exists R2,n → +∞ such that ∗
R2,n ∂BR2,n
|u|2 + |∇u|2 dσ → 0 as n → ∞ ,
the conclusion follows from (58). Proof of Theorem 3. Assuming that the assertion of the theorem is false, p−2 we can apply both Proposition 30 and Lemma 31, with f (u) = |u| u and p F (u) = |u| /p. Thus plugging (55) into (51) one gets N −2 N − 2 p
2
RN
|∇u| dx = 28
N N −α − p 2
1/N
σN
RN
Au2 dx |y|α
,
which is not possible for u = 0 if N −2 N − 2 p
N N −α − p 2
< 0.
As this inequality is actually equivalent to (α, p) ∈ A, we have a contradiction. The same argument proving Theorem 3 also yields the following nonexistence result, which applies, for instance, to nonlinearities of the form f (s) =
|s|p−2 s
p−ϑ
1 + |s|
for all s ∈ R
provided that p > ϑ > 0 are large enough. Theorem 32 Let 0 < α < N and f ∈ C (R, R). Assume that there exists ϑ ≥ max{2∗ , 2α } such that ϑF (s) ≤ f (s) s for all s ∈ R. If α = 2, assume in particular ϑ > 2∗ (= 2α ). Then equation (48) has no nonzero classical solution u ∈ C 2 ((Rk \ {0}) × RN −k ) ∩ X(RN , |y|−α dx) satisfying f (u) u ∈ L1 (RN ) and ∇u ∈ Lqloc RN for q = max {qk , qkα } or q = max {qk , qk,N }. Proof. One proceeds as in the proof of Theorem 3. Indeed, from (55) we get 1 ϑ
RN
|∇u|2 +
Au2 |y|α
dx ≥
F (u) dx RN
so that (51) gives (N − 2) (ϑ − 2∗ )
2
RN
|∇u| dx + (N − α) (ϑ − 2α )
∗
RN
∗
Au2 dx ≤ 0 |y|α
with 2 = 2α if α = 2 and ϑ > 2 = 2α if α = 2.
6
Behaviour of weak solutions
This section is devoted to the proof of Theorem 4, which relies on a comparison argument in the spirit of the maximum principle. Accordingly, we assume all the −α hypothesis of the theorem and let u ∈ Xs (RN , |y| dx) be any z-nonincreasing weak solution of equation (4). We need the following result from [19, Theorem 2 and Lemma 9]. Proposition 33 Assume that v ∈ D1,2 (RN ) is nonnegative and satisfies RN
∇v · ∇ϕ dx ≤
φ (x, v) ϕ dx RN
for all ϕ ∈ Cc∞ RN , ϕ ≥ 0,
where φ : RN × R → R is a measurable function such that 0 ≤ φ (x, s) ≤ b (x) sa−1 2∗
for all s ≥ 0 and almost every x ∈ RN
with a ∈ (2, 2∗ ) and b ∈ L 2∗ −a (RN ). Then v is bounded in a neighbourhood of the origin and satisfies lim sup|x|→∞ |x|N −2 v (x) < ∞. 29
Since α < 2 ≤ k implies Cc∞ (RN ) ⊆ X, every weak solution actually solves equation (4) in the distributional sense on RN . The boundedness of u and a first decay estimate then follow from Proposition 33, as we show in the following lemma. Lemma 34 u ∈ L∞ (RN ) and lim sup|x|→∞ |x|N −2 u (x) < ∞. Proof. First recall from Proposition 18 that u ∈ Lq (RN ) for all q ∈ [pα , 2∗ ] and observe that p ∈ (2 + 2pα /N, 2∗ ) is equivalent to pα < 2∗ (p − 2) / (2∗ − 2) < 2∗ , so that we can fix a ∈ (2, 2∗ ) such that u∈L
2∗ (p−a) 2∗ −a
RN .
Then, with a view to applying Proposition 33, define p−a
a−1
for all s ∈ R and almost every x ∈ RN .
φ (x, s) := f u (x) p−1 |s| p−1
Since φ (x, u (x)) = |f (u (x))| for almost every x ∈ RN , by definition of weak solution one has A ∇u · ∇ϕ dx ≤ ∇u · ∇ϕ dx + f (u) ϕ dx α uϕ dx = |y| N N N R R R RN ≤
φ (x, u) ϕ dx RN
for every ϕ ∈ Cc∞ (RN ), ϕ ≥ 0. Moreover (fp ) yields p−a
a−1
φ (x, s) ≤ M u (x) p−1 s p−1
p−1
= M u (x)p−a sa−1
for all s ≥ 0 and almost every x ∈ RN . Therefore, by Proposition 33, we conclude that u is bounded in a neighbourhood of the origin and satisfies lim sup|x|→∞ |x|N−2 u (x) < ∞. Thus u is bounded in a neighbourhood of {0} × RN −k by z-nonincreasingness and, as u ∈ C 2 (RN \ ({0} × RN −k )) by standard elliptic regularity theory, the proof of the lemma is complete. In order to improve the asymptotic estimate √ of Lemma 34 and deduce (16), we need some preliminaries. First, let β < 2 A/ (2 − α) and fix ε ∈ (0, A) such that √ 2 A−ε β ε := > β. (59) 2−α 2
Notice that β 2ε (1 − α/2) = A − ε. Second, by (fp ) and Lemma 34, let C1 > 0 and R1 > 0 be such that f (u) ≤ M up−2 u ≤
C1 (N −2)(p−2)−α
|x|
u α |x|
for almost every |x| ≥ R1 . (60)
Finally, thanks to the fact that p ∈ (2 + 2pα /N, 2∗ ) implies (N − 2) (p − 2) > α, α−(N−2)(p−2) ≤ ε and set fix Rε ≥ R1 such that |x| ≥ Rε implies C1 |x| ¯Rε . Ωε := RN \ B 30
Lemma 35 For all nonnegative h ∈ X with supp h ⊂ Ωε one has Ωε
∇u · ∇h dx ≤ − (A − ε)
Ωε
uh α dx . |x|
(61)
Proof. By (60) and the definition of Ωε we have f (u) ≤
C1 (N −2)(p−2)−α
|x|
u u α ≤ε α |x| |x|
A A α ≥ α |y| |x|
and
for almost every x ∈ Ωε . Hence, by definition of weak solution, one obtains ∇u · ∇h dx =
Ωε
Ωε
f (u) h dx − A
= − (A − ε)
Ωε
uh dx |x|α
Ωε
uh dx ≤ ε |y|α
Ωε
uh dx − A |x|α
Ωε
uh dx |x|α
for every h as in the statement of the lemma. In order to develope our comparison argument, define 1−α/2
vε (x) := e−β ε |x|
for all x ∈ RN .
Lemma 36 vε ∈ D1,2 (RN ) and for all nonnegative h ∈ D01,2 (Ωε ) one has Ωε
∇vε · ∇h dx ≥ − (A − ε)
Ωε
vε h dx . |x|α
(62)
Proof. For every x ∈ RN \ {0} straightforward computations give |∇vε (x)| = β ε 1 −
√ α |x|−α/2 vε (x) = A − ε |x|−α/2 vε (x) 2
and vε (x) =
√ √ α A−ε A−ε− N −1− |x|α/2−1 |x|−α vε (x) . 2
Hence |∇vε |2 = (A − ε) |x|−α vε2 ∈ L1 (RN ) and for all h ∈ Cc∞ (Ωε ) one has
Ωε
√ ∇vε · ∇h dx = − A − ε
−α
−1−α/2
∗
Ωε
√ N − 1 − α/2 A−ε− |x|1−α/2
∗
vε h dx . |x|α
(63)
vε ∈ L2 /(2 −1) (Ωε ), an obvious density argument shows As |x| vε , |x| that (63) also holds for all h ∈ D01,2 (Ωε ) and the conclusion then follows since vε is positive and 1 + α/2 < 2 < N . We can now conclude the proof of Theorem 4.
31
Proof of Theorem 4. Define Ω0 := x ∈ RN : Rε < |x| < Rε + 1 ⊂ Ωε and introduce the mappings w := u/vε L∞ (Ω0 ) vε − u ∈ D1,2 (RN ) and w ¯ := χΩε w− , defined almost everywhere in RN . Clearly w ¯ is nonnegative and supp w ¯ ⊂ Ωε . Moreover from 0 ≤ w− = −χ{w 0 for i = 1, 2 and set Λ1 := y ∈ Rk : |y| ∈ Λ1 , (m) Λ2 := {z ∈ Rm : |z| ∈ Λ2 }. Since u ∈ L1loc (RN ), by Fubini’s theorem we readily have
Λ1 ×Λ2
|˜ u (r, t)| drdt ≤
1 λm−1 2
dr Λ1
Λ2
32
tm−1 |˜ u (r, t)| dt
= ≤
1 dr σ m λm−1 Λ1 2 1 σ k σ m λk−1 λm−1 1 2
(m)
Λ2
|˜ u (r, |z|)| dz
(k)
(m)
Λ1 ×Λ2
|˜ u (|y| , |z|)| dx < ∞ .
Now we exploit the density of Cc∞ (RN ) ∩ Ds1,2 (RN ) in Ds1,2 (RN ) (which follows from standard convolution and regularization arguments) in order to infer that, by the symmetries of u, there exist ϕ, ψ : R2 → R such that ∇y u (y, z) · y = ϕ (|y| , |z|) and ∇z u (y, z) · z = ψ (|y| , |z|) for almost every (y, z) ∈ Rk × Rm . Then we have
Λ1 ×Λ2
|ϕ (r, t)| drdt = r ≤
1 σk σm
(k)
(m)
|∇y u (y, z) · y|
Λ1 ×Λ2
1 σ k σ m λk−1 λm−1 1 2
k
m−1
|y| |z| (k)
(m)
Λ1 ×Λ2
dx
|∇y u (y, z)| dx < ∞ . (k)
Finally, letting φ ∈ Cc∞ (Λ1 × Λ2 ) and setting Φ (x) := φ (|y| , |z|) ∈ Cc∞ (Λ1 × (m) Λ2 ), we get u ˜ (r, t) Λ1 ×Λ2
∂φ (r, t) drdt = ∂r
1 σk σm
=
1 σk σm
∂u ˜ ∂t
u ˜ (|y| , |z|) ∂φ (|y| , |z|) dx |y|k−1 |z|m−1 ∂r k
u (x) (k) (m) Λ1 ×Λ2
= −
1 σk σm
= −
1 σk σm
= −
1 σk σm
= − Similarly one obtains complete.
(k) (m) Λ1 ×Λ2
k−1
|y|
k
j=1
(k) (m) Λ1 ×Λ2
(k) (m) Λ1 ×Λ2
(k)
(m)
Λ1 ×Λ2
Λ1 ×Λ2
m−1
|z|
j=1
yj ∂Φ (x) dx ∂yj |y|
yj ∂u (x) k m−1 Φ (x) dx ∂yj |y| |z|
∇y u (x) · y k
m−1
k
m−1
|y| |z| ϕ (|y| , |z|)
|y| |z|
Φ (x) dx φ (|y| , |z|) dx
ϕ (r, t) φ (r, t) drdt . r
(r, t) = t−1 ψ (r, t) ∈ L1 (Λ1 × Λ2 ) and the proof is thus
For any z ∈ Rm , z = 0, define u ¯ (r) :=
|z|
u ˜ (r, t) t
m−1 2
|z|/2
Let a, b ∈ R be such that 0 < a < b.
33
dt for all r > 0 .
(64)
Lemma 38 u ¯ ∈ W 1,1 (a, b) and u ¯ (r) =
|z| |z|/2
m−1 ∂u ˜ (r, t) t 2 dt ∂r
for almost every r ∈ (a, b) .
(65)
Proof. Set Λ1 := (a, b) and Λ2 := (|z| /2, |z|). From Lemma 37 one readily infers that u ¯, u ¯ ∈ L1 (Λ1 ), where u ¯ is given by (65). Now fix any ϕ ∈ Cc∞ (Λ1 ). ∞ Then, for every ψ ∈ Cc (Λ2 ), the product ϕ (r) ψ (t) belongs to Cc∞ (Λ1 × Λ2 ) and we have ψ (t) Λ2
u ˜ (r, t) ϕ (r) dr dt =
u ˜ (r, t) ϕ (r) ψ (t) drdt Λ1 ×Λ2
Λ1
∂u ˜ (r, t) ϕ (r) ψ (t) drdt ∂r ∂u ˜ ψ (t) = − (r, t) ϕ (r) dr dt ∂r Λ2 Λ1
= −
Λ1 ×Λ2
˜ (r, t) ϕ (r) dr = − Λ1 ∂∂ru˜ (r, t) ϕ (r) dr for alagain by Lemma 37. Hence Λ1 u most every r ∈ Λ2 , whence, upon multiplying by t(m−1)/2 and integrating over Λ2 , one readily deduces that u ¯ is the weak derivative of u ¯ on Λ1 . 2
For any α > 0, we now define v (r) := rk−1−α/2 u ¯ (r) for all r > 0. Then, by Lemma 38, v ∈ W 1,1 (a, b) and v (r) = k − 1 −
α α k−2− α 2 2 u ¯ (r) + 2rk−1− 2 u ¯ (r) u ¯ (r) r 2
(66)
for almost every r ∈ (a, b). Moreover, v ∈ C ([a, b]) and b
v (b) − v (a) = Lemma 39 Assume
(Bb \Ba )×Rm
∇u
v (b) − v (a) ≤
v (r) dr .
(67)
a
|y|−α u2 dx < ∞. Then
L2 (RN )
σk σm
(Bb \Ba )×Rm
u2 α dx |y|
1/2
or v (b) − v (a) ≥ −
∇u
L2 (RN )
σk σm
(Bb \Ba )×Rm
u2 dx |y|α
|z|
(68)
1/2
|z|
(69)
according as α ≥ 2k − 2 or α < 2k − 2. Proof. First, according as α ≥ 2k − 2 or α < 2k − 2, i.e., k − 1 − α/2 ≤ 0 or k − 1 − α/2 > 0, by (66) and (65) we have α
α
¯ (r) u ¯ (r) ≤ 2rk−1− 2 |¯ u (r)| |¯ u (r)| v (r) ≤ 2rk−1− 2 u α
≤ 2rk−1− 2
|z|
|z|/2
|˜ u (r, t)| t 34
m−1 2
dt
|z|
|z|/2
m−1 ∂u ˜ (r, t) t 2 dt ∂r
or α
α
¯ (r) u ¯ (r) ≥ −2rk−1− 2 |¯ u (r)| |¯ u (r)| v (r) ≥ 2rk−1− 2 u |z|
α
≥ −2rk−1− 2
|z|/2
|˜ u (r, t)| t
m−1 2
|z|
dt
|z|/2
m−1 ∂u ˜ (r, t) t 2 dt ∂r
for almost every r ∈ (a, b). Then, by Hölder inequalities, we compute b
|z|
α
rk−1− 2
|z|/2
a b
r
=
|˜ u (r, t)| t |z|
k−1 α 2 −2
|z|/2
a
rk−1−α
· ≤
rk−1
|z|/2
a b
r
2
a
r
· |z| 2
2
m−1 2
dt r
|˜ u (r, t)| t
|z|
k−1 2
|z|/2
1/2
m−1 2
dr
dt
2 m−1
u ˜ (r, t) t
|z| |z|/2
(a,b)×(0,+∞)
dr
=
≤ ≤
(a,b)×(0,+∞)
|z| 2σk σ m |z| 2σk σ m ∇u
1/2
2
u ˜ (r, t)2 k−1 m−1 r t drdt rα
1/2
· 1/2
∂u ˜ (r, t) rk−1 tm−1 drdt ∂r
(Bb \Ba )×Rm
(Bb \Ba )×Rm
u2 α dx |y|
L2 (RN )
2σ k σ m
·
∂u ˜ (r, t) tm−1 dt dr ∂r
u2 dx |y|α
(Bb \Ba )×Rm
·
1/2
dt dr
2
·
1/2
2
(Bb \Ba )×Rm 1/2 RN
u2 α dx |y|
m−1 ∂u ˜ (r, t) t 2 dt dr ∂r
1/2
2
m−1 ∂u ˜ (r, t) t 2 dt ∂r
|z|/2
k−1 |z|
a
|z|/2
|z|
k−1−α |z|
b
≤
|z|
b
m−1 ∂u ˜ (r, t) t 2 dt dr = ∂r
2
|z|/2
a
|z|
dt
|˜ u (r, t)| t
|z|
b
≤
m−1 2
∂u ˜ (|y| , |z|) dx ∂r
y ∇y u (x) · |y|
1/2
1/2
2
dx
1/2
|z|
and the conclusion finally follows from (67). In order to give the proof of Theorem 13, we now assume N > k ≥ 2 and let A, α > 0. 35
Proof of Theorem 13. Let u ∈ Xs (RN , |y|−α dx) be nonincreasing with respect to z. First we observe that +∞
rk−1−α u ¯ (r)2 dr
0
|z|
+∞
≤
|z| 2
≤
|z| 2σ k σm
dr
rk−1
|z|/2
0
RN
u ˜ (r, t)2 m−1 t dt rα
u2 α dx < ∞ . |y|
Hence, according as α ≥ 2 or α < 2, there exists r1,n → 0+ or r2,n → +∞ 2 such that v (ri,n ) → 0 as n → ∞. Indeed, λ := lim inf r→0+ rk−1−α/2 u ¯ (r) > 0 2 implies rk−1−α u ¯ (r) ≥ λr−α/2 /2 for r close to 0, and a contradiction then ensues if α ≥ 2; similarly for α < 2 and r near +∞. Moreover, since u ¯ (r)
2∗
≤
|z| 2
=
|z| 2
≤
|z|2
∗
2∗ −1
2∗ −1
|z|
u ˜ (r, t)
2∗
t
m−1 ∗ 2 2
dt
|z|/2 |z|
∗
u ˜ (r, t)2 tm−1+
(m−1)(2∗ −2) 2
dt
|z|/2
(m+1)/2−1
22∗ −1
|z|
∗
u ˜ (r, t)2 tm−1 dt
|z|/2
implies +∞
r
k−1
2∗
u ¯ (r)
2∗ (m+1)/2−1
dr
0
≤
|z|
+∞
dr
2∗ −1
2
≤
|z| 22∗ −1 σ k σm
∗
rk−1 u ˜ (r, t)2 tm−1 dt
|z|/2
0
2∗ (m+1)/2−1
|z|
∗
RN
u2 dx < ∞ ,
the same argument yields that a sequence r2,n → +∞ such that v (r2,n ) → 0 as n → ∞ also exists for α ≥ 4k/N − 2. Thus it exists for every α > 0, because k < N implies 4k/N − 2 < 2. Now suppose α ≥ 2k − 2. Then one has α ≥ 2 (recall that k ≥ 2) and we can apply Lemma 39 with b = r > 0 arbitrary and a = r1,n (with n large enough to assure a < b). Thus (68) gives v (r) − v (r1,n ) ≤ ≤
∇u
L2 (RN )
σk σm
∇u
(Br \Br1,n )×Rm
L2 (RN )
√ σk σm A
RN
Au2 dx |y|α
u2 dx |y|α
1/2
|z| ≤
1/2
|z|
u 2 √ |z| σk σm A
and letting n → ∞ we get 2
2
rk−1−α/2 u ¯ (r) ≤
u √ |z| σk σm A
36
for all r > 0 .
(70)
Similarly, if α < 2k − 2, for a = r > 0 arbitrary and b = r2,n with n large, (69) yields 2 u √ |z| , v (r2,n ) − v (r) ≥ − σk σm A so that, as n → ∞, we get (70) again. On the other hand, by nonincreasingness, for all r > 0 one has u ¯ (r) ≥ u ˜ (r, |z|)
|z| |z|/2
t
m−1 2
dt =
m+1 2(m+1)/2 − 1 u ˜ (r, |z|) |z| 2 . (m−1)/2 2 (m + 1)
Hence we conclude rk−1−α/2
m+1 2(m+1)/2 − 1 u ˜ (r, |z|) |z| 2 (m−1)/2 2 (m + 1)
2
2
≤
u √ |z| σk σm A
and the result ensues.
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Marino Badiale, Michela Guida, Sergio Rolando Dipartimento di Matematica Università degli Studi di Torino Via Carlo Alberto 10 10123 Torino, Italia
[email protected] [email protected] [email protected] Research of M. Badiale supported by MIUR, project “Metodi variazionali e equazioni differenziali non lineari”
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