Existence and Asymptotic Profile of Nodal Solutions to Supercritical Problems

Let N=4 or N≥6. Then, for every p∈(2,2N-1∗), the problem

has infinitely many solutions which satisfy

for all ei⁢ϑ∈S1, ϱ∈O⁢(N-4), z=(z1,z2)∈C2 and y∈RN-4.

Fix R>0. For p∈(2N∗,2N-1∗), let uδ,p be a solution to (1.3) which satisfies (1.4) and has minimal energy among all solutions to (1.3) with these symmetry properties. Then there exist sequences (pk) in (2N∗,2N-1∗), and (δk) and (λk) in (0,∞), and a nontrivial solution ω to problem (1.2) such that the following hold:

1. p k → 2 N ∗ , δk→0 and λk→0.

2. ω ⁢ ( z , y ) = ω ⁢ ( e i ⁢ ϑ ⁢ z , ϱ ⁢ y ) and ω ⁢ ( z 1 , z 2 , y ) = - ω ⁢ ( - z ¯ 2 , z ¯ 1 , y ) for all e i ⁢ ϑ ∈ 𝕊 1 , ϱ∈O⁢(N-4), z=(z1,z2)∈ℂ2 and y∈ℝN-4. Also, ω has minimal energy among all solutions to (1.2) which have these symmetry properties.

3. u δ k , p k has the following asymptotic profile:

   u δ k , p k = λ k 2 - N 2 ⁢ ω ⁢ ( ⋅ λ k ) + o ⁢ ( 1 )   in ⁢ D 1 , 2 ⁢ ( ℝ N ) .

If N-m>2, then D01,2⁢(Ω)G and D01,2⁢(Ω)ϕ are embedded in Lp⁢(Ω) for every p∈[1,2N-m∗] and the embedding is compact for p∈[1,2N-m∗).

Proof.

Note that dim⁡(G⁢x)=dim⁡(Γ⁢x) for every x∈Ω. Hebey and Vaugon showed that the space D01,2⁢(Ω)G is embedded in Lp⁢(Ω) for every p∈[1,2N-m∗] and that the embedding is compact for p∈[1,2N-m∗), see [15, Corollary 2]. Since D01,2⁢(Ω)ϕ is a subspace of D01,2⁢(Ω)G, this is also true for D01,2⁢(Ω)ϕ. ∎

If p∈(2,2N-m∗] and u is a critical point of Jp:D01,2⁢(Ω)ϕ→R, then u is a solution to problem (1.1) which satisfies

Proof.

For a function v:Ω→ℝ and γ∈Γ, we set vγ⁢(x):=ϕ⁢(γ)⁢v⁢(γ-1⁢x). Since ϕ⁢(γ-1⁢g⁢γ)=1 for every g∈G, we have that γ-1⁢g⁢γ∈G. So, if v∈D01,2⁢(Ω)G, then, as

for every g∈G, we have that vγ∈D01,2⁢(Ω)G. This shows that D01,2⁢(Ω)G is invariant under the action of Γ given by (γ,v)↦vγ. The set of fixed points of this action is the space D01,2⁢(Ω)ϕ. Hence, by the principle of symmetric criticality, if u is a critical point of Jp:D01,2⁢(Ω)ϕ→ℝ, then u is also a critical point of Jp:D01,2⁢(Ω)G→ℝ, see [18].

For φ∈𝒞c∞⁢(Ω), set

where μ is the Haar measure. Then, φ~∈𝒞c∞⁢(Ω)G:=𝒞c∞⁢(Ω)∩D01,2⁢(Ω)G. Since u is G-invariant, an easy computation shows that

As u is a critical point of Jp:D01,2⁢(Ω)G→ℝ, this implies that

as claimed. ∎

If p∈(2,2N-m∗), then problem (1.1) has a solution up, which satisfies (2.1) such that Jp⁢(up)=cpϕ. Moreover, (1.1) has an unbounded sequence of solutions which satisfy (2.1).

Proof.

Since the embedding D01,2⁢(Ω)ϕ↪Lp⁢(Ω) is compact for p∈(2,2N-m∗), a standard argument shows that Jp satisfies the Palais–Smale condition on D01,2⁢(Ω)ϕ. Therefore, cpϕ is attained on 𝒩pϕ and, as Jp:D01,2⁢(Ω)ϕ→ℝ is even and has the mountain pass geometry, the symmetric mountain pass theorem (see [1]) guarantees the existence of an unbounded sequence of critical values of Jp on 𝒩pϕ. By Lemma 2.2, the corresponding critical points are solutions to (1.1). ∎

Proof of Theorem 1.1.

For N≥4, let Γ be the subgroup of O⁢(N) generated by 𝕊1∪O⁢(N-4)∪{τ}, where ei⁢ϑ∈𝕊1, ϱ∈O⁢(N-4) and τ act on a point (z1,z2,y)∈ℂ×ℂ×ℝN-4≡ℝN by

Let ϕ:Γ→ℤ/2 be the homomorphism given by ϕ(ei⁢ϑ):=1=:ϕ(ϱ), ϕ⁢(τ):=-1.

If N=4, then dim⁡(Γ⁢x)=1 for every x=(z1,z2,y)∈Ar,R, whereas for N≥5, we have that

Hence, m:=min⁡{dim⁡(Γ⁢x):x∈Ar,R}=1 if N=4 and N≥6, and Theorem 1.1 follows from Theorem 2.3. ∎

If pk,q∈(2,2N-m∗), pk→q, and (uk) is a bounded sequence in D01,2⁢(Ω)ϕ, then

Proof.

By the mean value theorem, for each x∈Ω, there exists qk⁢(x) between pk and q such that

Fix η>0 so that [q-η,q+η]⊂(2,2N-m∗). Then, for some positive constant C and k large enough,

Therefore, using Proposition 2.1, we obtain

for some positive constant C¯. As (uk) is bounded in D01,2⁢(Ω), we conclude that limk→∞⁡∫Ω(|uk|pk-|uk|q)=0, as claimed. ∎

For q∈(2,2N-m∗), we have that

Proof.

Set

From Hölder’s inequality, we obtain that

So, as p approaches q from the right, we get

Assume that lim infp→q+⁡Spϕ<Sqϕ. Then there exist ε>0 and sequences (pk) in (q,2N-m∗) and (uk) in D01,2⁢(Ω)ϕ, with |uk|pk=1, such that ∥uk∥2<Sqϕ-ε. Lemma 2.4 implies that |uk|q→1. Hence, ∥uk∥2/|uk|q2<Sqϕ for k large enough, contradicting the definition of Sqϕ. This proves that

The corresponding statement when p approaches q from the left is proved in a similar way. Therefore, limp→q⁡Spϕ=Sqϕ. An easy computation shows that cpϕ=p-22⁢p⁢(Spϕ)pp-2. It follows that

Since Jp⁢(up)=p-22⁢p⁢∥up∥2=cpϕ, we have that (up) is bounded in D01,2⁢(Ω)ϕ for p close to q. The expression (2.2), together with Lemma 2.4, yields limp→q⁡tq,p=1 which, in turn yields

as claimed. ∎

If ΩΓ≠∅, then c∗ϕ=c∞ϕ and c∗ϕ is not attained by J∗ on N∗ϕ.

Proof.

See [6, Theorem 2.3]. ∎

Given sequences (λk) in (0,∞) and (ξk) in RN, there exist a sequence (ζk) in RN and a closed subgroup K of Γ such that, after passing to a subsequence, the following statements hold true:

1. The sequence ( λ k - 1 ⁢ dist ⁡ ( Γ ⁢ ξ k , ζ k ) ) is bounded.

2. Γ ζ k = K for all k ∈ ℕ .

3. If | Γ / K | < ∞ , then λ k - 1 ⁢ | α ⁢ ζ k - β ⁢ ζ k | → ∞ for any α , β ∈ Γ with α - 1 ⁢ β ∉ K .

4. If | Γ / K | = ∞ , then there is a closed subgroup K ′ of Γ such that K⊂K′, |Γ/K′|=∞ and λk-1⁢|α⁢ζk-β⁢ζk|→∞ for any α,β∈Γ with α-1⁢β∉K′.

Proof.

See [7, Lemma 3.3]. ∎

Assume that Ω∖ΩΓ and ΩΓ are nonempty, and that every Γ-orbit in Ω∖ΩΓ has positive dimension. Let(uk) be sequence in N∗ϕ such that J∗⁢(uk)→c∗ϕ. Then, after passing to a subsequence, there exist a nontrivial solution ω to problem (3.1), a sequence (ζk) in ΩΓ and a sequence (λk) in (0,∞) with the following properties:

1. λ k - 1 ⁢ dist ⁡ ( ζ k , ∂ ⁡ Ω ) → ∞ ,

2. J ∞ ⁢ ( ω ) = c ∞ ϕ ,

3. lim k → ∞ ⁡ ∥ u k - λ k 2 - N 2 ⁢ ω ⁢ ( ⋅ - ζ k λ k ) ∥ = 0 .

Proof.

By Ekeland’s variational principle, we may assume that (uk) is a Palais–Smale sequence. Then, (uk) is bounded in D01,2⁢(Ω) and, after passing to a subsequence, uk⇀u weakly in D01,2⁢(Ω). If u≠0, an easy argument shows that u∈𝒩∗ϕ and J∗⁢(u)=c∗ϕ, contradicting Theorem 3.1. Therefore, u=0.

Fix δ∈(0,N2⁢c∗ϕ). Then there exist bounded sequences (λk) in (0,∞) and (ξk) in ℝN such that, after passing to a subsequence,

For (λk) and (ξk), we choose K and (ζk) as in Lemma 3.2. Then, Γζk=K and dist⁡(Γ⁢ξk,ζk)<C⁢λk for some positive constant C and all k∈ℕ. Therefore, (ζk) is bounded and, as |vk| is Γ-invariant, we have that

Set Ωk:={z∈ℝN:λk⁢z+ζk∈Ω} and, for z∈Ωk, define

The sequence (wk) is bounded in D1,2⁢(ℝN) so, after passing to a subsequence, wk⇀ω weakly in D1,2⁢(ℝN), wk→ω strongly in Lloc2⁢(ℝN), and wk→ω a.e. in ℝN. A standard argument, using inequality (3.2), shows that ω≠0. Moreover, since Γζk=K, we have that wk⁢(γ⁢z)=ϕ⁢(γ)⁢wk⁢(z) for all γ∈K. Hence,

Using the fact that the equation -Δ⁢u=|u|2N∗-2⁢u does not have a nontrivial solution in a half-space, it is also standard to show, after passing to a subsequence, that λk-1⁢dist⁡(ζk,∂⁡Ω)→∞, ζk∈Ω and ω is a solution to -Δ⁢u=|u|2N∗-2⁢u in ℝN.

Now, if ζk∉ΩΓ, then dim⁡(Γ⁢ζk)>0, and hence |Γ/K|=∞. But then, for any given m∈ℕ, statement (d) of Lemma 3.2 allows us to choose m elements α1,α2,…,αm∈Γ such that λk-1⁢|αj⁢ζk-αi⁢ζk|→∞ for i≠j. Arguing as in the proof of [6, inequality (2.7)], we obtain that

This is a contradiction. Therefore, ζk∈ΩΓ and K=Γ. From (3.3), we conclude that ω is a nontrivial solution to (3.1).

Note that

Since wk⇀ω weakly in D1,2⁢(ℝN), we have that

and, performing the change of variable z=y-ζkλk, we obtain