Citation: Mónica Clapp, Filomena Pacella. Existence of nonradial positive and nodal solutions to a critical Neumann problem in a cone[J]. Mathematics in Engineering, 2021, 3(3): 1-15. doi: 10.3934/mine.2021022
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We consider the Neumann problem
{−Δu=|u|2∗−2uin Σω,∂u∂ν=0on ∂Σω, | (1.1) |
in the unbounded cone Σω:={tx:x∈ω and t>0}, where ω is an open connected subset of the unit sphere SN−1 in RN with smooth boundary, N≥3, and 2∗:=2NN−2 is the critical Sobolev exponent.
It is well known that, if ω=SN−1, i.e., if Σω is the whole space RN, then the only positive solutions to the critical problem
−Δw=|w|2∗−2w,w∈D1,2(RN), | (1.2) |
are the rescalings and translations of the standard bubble U defined in (2.3). Moreover, they are the only nontrivial radial solutions to (1.2), up to sign. It is immediately deduced that, up to sign, the restriction of the bubbles (3.1) to Σω are the only nontrivial radial solutions of (1.1) in any cone; see Proposition 3.4. In addition, if the cone Σω is convex, it was shown in [8,Theorem 2.4] that these are the only positive solutions to (1.1). The convexity property of the cone is crucial in the proof of this result, and it is strongly related to a relative isoperimetric inequality obtained in [7].
The aim of this paper is to establish the existence of nonradial solutions to (1.1), both positive and sign-changing. As mentioned above, the positive ones can only exist in nonconvex cones. On the other hand, nodal radial solutions to (1.1) do not exist, as this would imply the existence of a nontrivial solution to problem (2.5) in the bounded cone Λω:={tx:x∈ω and t∈(0,1)}, which is impossible because of the Pohozhaev identity (2.6) and the unique continuation principle.
For the problem (1.2) in RN various types of sign-changing solutions are known to exist; see [2,3,4,5]. In particular, a family of entire nodal solutions, which are invariant under certain groups of linear isometries of RN, were exhibited in [2]. These solutions arise as blow-up profiles of symmetric minimizing sequences for the critical equation in a ball, and are obtained through a fine analysis of the concentration behavior of such sequences.
Here we use some ideas from [2] to produce sign-changing solutions to (1.1), but we exploit a different kind of symmetry. Our main result shows that, if ω is symmetric with respect to the north pole of SN−1 and if the cone Σω has a point of convexity in the sense of Definition 2.6, then the problem (1.1) has an axially antisymmetric least energy solution, which is nonradial and changes sign; see Theorem 2.8. As far as we know, this is the first existence result of a nodal solution to (1.1).
Next, we investigate the existence of positive nonradial solutions. In this case we do not require the cone to have any particular symmetry. We establish the existence of a positive nonradial solution to (1.1) under some conditions involving the local convexity of Σω at a boundary point and the measure of the bounded cone Λω; see Corollary 3.5 and Theorem 3.6. We refer to Section 3 for the precise statements and further remarks.
If Ω is a domain in RN we consider the Sobolev space
D1,2(Ω):={u∈L2∗(Ω):∇u∈L2(Ω,RN)} |
with the norm
‖u‖2Ω:=∫Ω|∇u|2. |
We denote by JΩ:D1,2(Ω)→R the functional given by
JΩ(u):=12∫Ω|∇u|2−12∗∫Ω|u|2∗, |
and its Nehari manifold by
N(Ω):={u∈D1,2(Ω):u≠0 and ∫Ω|∇u|2=∫Ω|u|2∗}. |
For u∈D1,2(Ω)∖{0} let tu∈(0,∞) be such that tuu∈N(Ω). Then,
JΩ(tuu)=1N[QΩ(u)]N2,where QΩ(u):=∫Ω|∇u|2(∫Ω|u|2∗)2/2∗. | (2.1) |
Hence,
cΩ:=infu∈N(Ω)JΩ(u)=infu∈D1,2(Ω)∖{0}1N[QΩ(u)]N2. | (2.2) |
We set c∞:=cRN. It is well known that this infimum is attained at the function
U(x)=aN(11+|x|2)N−22,aN:=(N(N−2))N−24, | (2.3) |
which is called the standard bubble, and at every rescaling and translation of it, and that
c∞=JRN(U)=1NSN2, |
where S is the best constant for the Sobolev embedding D1,2(RN)↪L2∗(RN).
Let SN−1 be the unit sphere in RN and let ω be a smooth domain in SN−1 with nonempty boundary, i.e., ω is connected and open in SN−1 and its boundary ∂ω is a smooth (N−2)-dimensional submanifold of SN−1. The nontrivial solutions to the Neumann problem (1.1) in the unbounded cone
Σω:={tx:x∈ω and t>0} |
are the critical points of JΣω on N(Σω).
To produce a nonradial sign-changing solution for (1.1) we introduce some symmetries. We write a point in RN as x=(x′,xN)∈RN−1×R, and consider the reflection ϱ(x′,xN):=(−x′,xN). Then, a subset X of RN will be called ϱ-invariant if ϱx∈X for every x∈X, and a function u:X→R will be called ϱ-equivariant if
u(ϱx)=−u(x)∀x∈X. |
Note that every nontrivial ϱ-equivariant function is nonradial and changes sign.
Throughout this section we will assume that ω is ϱ-invariant. Note that (0,±1)∉∂ω because ∂ω is smooth. Hence, ϱx≠x for every x∈∂Σω∖{0}. Our aim is to show that (1.1) has a ϱ-equivariant solution. We set
D1,2ϱ(Σω):={u∈D1,2(Σω):u is ϱ-equivariant}, |
Nϱ(Σω):={u∈N(Σω):u is ϱ-equivariant} |
and
cϱΣω:=infu∈Nϱ(Σω)JΣω(u)=infu∈D1,2ϱ(Σω)∖{0}1N[QΣω(u)]N2. | (2.4) |
Define
Λω:={tx:x∈ω and 0<t<1} |
and set Γ1:=∂Λω∖¯ω. In Λω we consider the mixed boundary value problem
{−Δu=|u|2∗−2uin Λω,u=0on ω,∂u∂ν=0on Γ1. | (2.5) |
We point out that (2.5) does not have a nontrivial solution. Indeed, by the well known Pohozhaev identity, a solution to (2.5) must satisfy
0=∫Γ1∪ω((∇u⋅s)∂u∂ν−|∇u|22s⋅ν+F(u)s⋅ν)ds=∫ω((∇u⋅s)∂u∂ν−|∇u|22s⋅ν)ds=12∫ω|∂u∂ν|2ds | (2.6) |
because s⋅ν=0 for every s∈Γ1 and s⋅ν=1 for every s∈ω. Therefore ∂u∂ν vanishes on ω. But then the trivial extension of u to the infinite cone Σω solves (1.1), contradicting the unique continuation principle.
Let V(Λω) be the space of functions in D1,2(Λω) whose trace vanishes on ω. Note that V(Λω)⊂D1,2(Σω) via trivial extension. Let JΛω:V(Λω)→R be the restriction of JΣω to V(Λω) and set
Nϱ(Λω):=Nϱ(Σω)∩V(Λω)andcϱΛω:=infu∈Nϱ(Λω)JΛω(u). |
To produce a sign-changing solution for the problem (1.1) we will study the concentration behavior of ϱ-equivariant minimizing sequences for (2.5). We start with the following lemmas.
Lemma 2.1. 0<cϱΛω=cϱΣω≤c∞.
Proof. It is shown in [8,Theorem 2.1] that cϱΛω>0.
Since Nϱ(Λω)⊂Nϱ(Σω), we have that cϱΛω≥cϱΣω. To prove the opposite inequality, let φk∈Nϱ(Σω)∩C∞(¯Σω) be such that φk has compact support and J(φk)→cϱΣω as k→∞. Then, we may choose εk>0 such that the support of ˜φk(x):=ε−(N−2)/2kφk(ε−1kx) is contained in ¯Λω∖¯ω. Thus, ˜φk∈Nϱ(Λω) and, hence,
cϱΛω≤J(˜φk)=J(φk)for all k. |
Letting k→∞ we conclude that cϱΛω≤cϱΣω.
To prove that cϱΣω≤c∞ we fix a point ξ∈∂Σω∖{0} and a sequence of positive numbers εk→0, and we set Σk:=ε−1k(Σω−ξ). Since ∂Σω∖{0} is smooth, the limit of the sequence of sets (Σk) is the half-space
Hν:={z∈RN:z⋅ν<0}, | (2.7) |
where ν is the exterior unit normal to Σω at ξ. Let uk(x):=ε(2−N)/2kU(x−ξεk), where U is the standard bubble (2.3). Then,
limk→∞∫Σω|∇uk|2=limk→∞∫Σk|∇U|2=∫Hν|∇U|2=12NSN2, | (2.8) |
limk→∞∫Σω|uk|2∗=limk→∞∫Σk|U|2∗=∫Hν|U|2∗=12NSN2. | (2.9) |
The function
ˆuk(x)=uk(x)−uk(ϱx)=ε2−N2kU(x−ξεk)−ε2−N2kU(x−ϱξεk) |
is ϱ-equivariant, and from (2.4), (2.8) and (2.9) we obtain
cϱΣω≤limk→∞1N[QΣω(ˆuk)]N2=1NSN2=c∞. |
This concludes the proof.
Lemma 2.2. Given a domain Ω in RN and ε>0, we set Ωε:={ε−1x:x∈Ω}. If ∂Ω is Lipschitz continuous, then there exist linear extension operators Pε:W1,2(Ωε)→D1,2(RN) and a positive constant C, independent of ε, such that
(i)(Pεu)(x)=u(x) for every x∈Ωε.
(ii)∫RN|∇(Pεu)|2≤C∫Ωε|∇u|2.
(iii)∫RN|Pεu|2∗≤C∫Ωε|u|2∗.
(iv) If Ω is ϱ-invariant, then Pεu is ϱ-equivariant if u is ϱ-equivariant.
Proof. The existence of an extension operator Pε:W1,2(Ωε)→D1,2(RN) satisfying (i)−(iii) is well known, and the fact that the constant C does not depend on ε was proved in [6,Lemma 2.1]. To obtain (iv) we replace Pεu by the function x↦12[(Pεu)(x)−(Pεu)(ϱx)].
The following proposition describes the behavior of minimizing sequences for JΛω on Nϱ(Λω).
Proposition 2.3. Let uk∈Nϱ(Λω) be such that
JΛω(uk)→cϱΛωandJ′Λω(uk)→0in(V(Λω))′. |
Then, after passing to a subsequence, one of the following statements holds true:
(i) There exist a sequence of positive numbers (εk), a sequence of points (ξk) in Γ1 and a function w∈D1,2(RN) such that ε−1kdist(ξk,ˉω∪{0})→∞, w|H solves the Neumann problem
−Δw=|w|2∗−2w,w∈D1,2(H), | (2.10) |
in some half-space H, JH(w)=12c∞,
limk→∞‖uk−ε2−N2kw(⋅−ξkεk)+ε2−N2k(w∘ϱ)(⋅−ϱξkεk)‖Σω=0, |
and cϱΣω=cϱΛω=c∞.
(ii) There exist a sequence of positive numbers (εk) with εk→0, and a ϱ-equivariant solution w∈D1,2(Σω) to the problem (1.1) such that
limk→∞‖uk−ε2−N2kw(⋅εk)‖Σω=0, |
and JΣω(w)=cϱΣω=cϱΛω≤c∞.
Proof. Since
1N‖uk‖2Λω=JΛω(uk)−12∗J′Λω(uk)uk≤C+o(1)‖uk‖Λω, | (2.11) |
the sequence (uk) is bounded and, after passing to a subsequence, uk⇀u weakly in V(Λω). Then, J′Λω(u)=0. Since the problem (2.5) does not have a nontrivial solution, we conclude that u=0.
Fix δ∈(0,N2cϱΛω). As
∫Λω|uk|2∗=N(JΛω(uk)−12J′Λω(uk)uk)→NcϱΛω, |
there are bounded sequences (εk) in (0,∞) and (xk) in RN such that, after passing to a subsequence,
δ=supx∈RN∫Λω∩Bεk(x)|uk|2∗=∫Λω∩Bεk(xk)|uk|2∗, |
where Br(x):={y∈RN:|y−x|<r}. Note that, as δ>0, we have that dist(xk,Λω)<εk. We claim that, after passing to a subsequence, there exist ξk∈ˉΛω and C0>0 such that
ε−1k|xk−ξk|<C0∀k∈N, | (2.12) |
and one of the following statements holds true:
(a)ξk=0 for all k∈N.
(b)ξk∈∂ω=¯ω∩¯Γ1 for all k∈N.
(c)ξk∈Γ1 for all k∈N and ε−1kdist(ξk,ˉω∪{0})→∞.
(d)ξk∈ω for all k∈N and ε−1kdist(ξk,Γ1)→∞.
(e)ξk∈Λω for all k∈N, ε−1kdist(ξk,∂Λω)→∞ and, either ε−1k|ξk−ϱξk|→∞, or ξk=ϱξk for all k∈N.
This can be seen as follows: If the sequence (ε−1k|xk|) is bounded, we set ξk:=0. Then, (2.12) and (a) hold true. If (ε−1kdist(xk,∂ω)) is bounded, we take ξk∈∂ω such that |xk−ξk|=dist(xk,∂ω). Then, (2.12) and (b) hold true. If both (ε−1k|xk|) and (ε−1kdist(xk,∂ω)) are unbounded and (ε−1kdist(xk,Γ1)) is bounded, we take ξk∈Γ1 with |xk−ξk|=dist(xk,Γ1). Then, (2.12) and (c) hold true. If (ε−1kdist(xk,Γ1)) is unbounded and (ε−1kdist(xk,ω)) is bounded, we take ξk∈ω with |xk−ξk|=dist(xk,ω). Then, (2.12) and (d) hold true. Finally, if (ε−1kdist(xk,∂Λω)) is unbounded, we set ξk:=xk+ϱxk2 if (ε−1k|xk−ϱxk|) is bounded and ξk:=xk if (ε−1k|xk−ϱxk|) is unbounded. Then, (2.12) and (e) hold true.
Set C1:=C0+1. Inequality (2.12) yields
δ=∫Λω∩Bεk(xk)|uk|2∗≤∫Λω∩BC1εk(ξk)|uk|2∗. | (2.13) |
We consider uk as a function in D1,2(Σω) via trivial extension, and we define ˆuk∈D1,2(Σω) as ˆuk(z):=ε(N−2)/2kuk(εkz). Since ˆuk is ϱ-equivariant, so is its extension Pεkˆuk∈D1,2(RN) given by Lemma 2.2. Let
wk(z):=(Pεkˆuk)(z+ε−1kξk)∈D1,2(RN). |
Then,
wk(z)=εN−22kuk(εkz+ξk)if z∈Λk:=ε−1k(Λω−ξk), | (2.14) |
wk(z−ε−1kξk)=−wk(ϱz−ε−1kξk)for every z∈RN, | (2.15) |
δ=supz∈RN∫Λk∩B1(z)|wk|2∗≤∫Λk∩BC1(0)|wk|2∗, | (2.16) |
and (wk) is bounded in D1,2(RN). Hence, a subsequence satisfies that wk⇀w weakly in D1,2(RN), wk→w a.e. in RN and wk→w strongly in L2loc(RN). Choosing δ sufficiently small and using (2.16), a standard argument shows that w≠0; see, e.g., [10,Section 8.3]. Moreover, we have that ξk→ξ and εk→0, because uk⇀0 weakly in V(Λω) and w≠0.
Let E be the limit of the domains Λk. Since (wk) is bounded in D1,2(RN), using Hölder's inequality we obtain
|∫E∖Λk∇wk⋅∇φ|≤C(∫E∖Λk|∇φ|2)12=o(1),|∫E∖Λk|wk|2∗−2wkφ|≤C(∫E∖Λk|φ|2∗)12∗=o(1), |
for every φ∈C∞c(RN), and similarly for the integrals over Λk∖E. Therefore, as wk⇀w weakly in D1,2(E), rescaling and using (2.14) we conclude that
∫E∇w⋅∇φ−∫E|w|2∗−2wφ=∫E∇wk⋅∇φ−∫E|wk|2∗−2wkφ+o(1)=∫Λk∇wk⋅∇φ−∫Λk|wk|2∗−2wkφ+o(1)=∫Λω∇uk⋅∇φk−∫Λω|uk|2∗−2ukφk+o(1), | (2.17) |
where φk(x):=ε(2−N)/2kφ(x−ξkεk). Next, we analyze all possibilities, according to the location of ξk.
(a) If ξk=0 for all k∈N, then E=Σω and wk is ϱ-equivariant. Hence, w is ϱ-equivariant. Let φ∈C∞c(RN). Then, φk|Λω∈V(Λω) for large enough k, and from (2.17) we obtain
J′Σω(w)[φ|Σω]=∫Σω∇w⋅∇φ−∫Σω|w|2∗−2wφ=J′Λω(uk)[φk|Λω]=o(1). |
This shows that w|Σω solves (1.1). Therefore,
cϱΣω≤1N‖w‖2Σω≤lim infk→∞1N‖wk‖2Σω=limk→∞1N‖uk‖2Λω=cϱΛω. |
Together with Lemma 2.1, this implies that JΣω(w)=cϱΣω=cϱΛω≤c∞ and
o(1)=‖wk−w‖Σω=‖uk−ε2−N2kw(⋅εk)‖Σω. |
So, in this case, we obtain statement (ii).
(b) If ξk∈∂ω for all k∈N, then E=Hξ∩Hν, where ξ=limk→∞ξk, \; ν is the exterior unit normal to Σω at ξ, and Hξ and Hν are half-spaces defined as in (2.7). If φ∈C∞c(Hξ), then φk|Λω∈V(Λω) for large enough k, and using (2.17) we conclude that w|E solves the mixed boundary value problem
−Δw=|w|2∗−2w,w=0 on ∂E∩∂Hξ,∂w∂ν=0 on ∂E∩∂Hν. |
Since ξ and ν are orthogonal, extending w|E by reflection on ∂E∩∂Hν, yields a nontrivial solution to the Dirichlet problem
−Δw=|w|2∗−2w,w∈D1,20(Hξ). | (2.18) |
It is well known that this problem does not have a nontrivial solution, so (b) cannot occur.
(c) If ξk∈Γ1 for all k∈N and ε−1kdist(ξk,ˉω∪{0})→∞, then E=Hν, where ν is the exterior unit normal to Σω at ξ=limk→∞ξk. Using (2.17) we conclude that w|Hν solves the Neumann problem (2.10) in Hν. Since ε−1k|ξk|→∞, we have that ε−1k|ξk−ϱξk|→∞. Therefore,
wk−(w∘ϱ)(⋅+ε−1k(ξk−ϱξk))⇀wweakly in D1,2(RN). |
Note also that wk∘ϱ⇀w∘ϱ weakly in D1,2(RN). Using these facts and performing suitable rescalings and translations we obtain
‖uk−ε2−N2kw(⋅−ξkεk)+ε2−N2k(w∘ϱ)(⋅−ϱξkεk)‖2Σω=‖ˆuk−w(⋅−ε−1kξk)+(w∘ϱ)(⋅−ε−1kϱξk)‖2Σω=‖wk−w+(w∘ϱ)(⋅+ε−1k(ξk−ϱξk))‖2Σω−ε−1kξk=‖wk+(w∘ϱ)(⋅+ε−1k(ξk−ϱξk))‖2Σω−ε−1kξk−‖w‖2Hν+o(1)=‖−wk∘ϱ+w∘ϱ‖2Σω−ε−1kϱξk−‖w‖2Hν+o(1)=‖ˆuk‖2Σω−2‖w‖2Hν+o(1)=‖uk‖2Λω−2‖w‖2Hν+o(1)=NcϱΛω−2‖w‖2Hν+o(1). |
Since JHν(w)=1N‖w‖2Hν≥12c∞, applying Lemma 2.1 we conclude that JHν(w)=12c∞, cϱΣω=cϱΛω=c∞, and
limk→∞‖uk−ε2−N2kw(⋅−ξkεk)+ε2−N2k(w∘ϱ)(⋅−ϱξkεk)‖2Σω=0. |
So, in this case we obtain statement (i).
(d) If ξk∈ω for all k∈N and ε−1kdist(ξk,Γ1)→∞, then E=Hξ and using (2.17) we conclude that w|Hξ solves the Dirichlet problem (2.18). So this case does not occur.
(e) If ξk∈Λω for all k∈N and ε−1kdist(ξk,∂Λω)→∞, then E=RN and w solves the problem (1.2). If ρξk=ξk for every k, then wk is ϱ-equivariant, and so is w. Since w is a sign-changing solution to (1.2) we have that
2c∞<1N‖w‖2RN≤limk→∞1N‖wk‖2RN=limk→∞1N‖uk‖2Λω=cϱΛω, |
contradicting Lemma 2.1. On the other hand, if ε−1k|ϱξk−ξk|→∞, then, arguing as in case (c), we conclude that
2c∞≤2N‖w‖2RN≤limk→∞1N‖wk‖2RN=limk→∞1N‖uk‖2Λω=cϱΛω, |
contradicting Lemma 2.1 again. So (e) cannot occur.
We are left with (a) and (c). This concludes the proof.
Proposition 2.3 immediately yields the following result.
Corollary 2.4. If cϱΣω<c∞, then the problem (1.1) has a ϱ-equivariant least energy solution in D1,2(Σω).
Equality is not enough, as the following example shows. Set
SN−1+:={(x1,…,xN)∈SN−1:xN>0}. |
Example 2.5. If ω=SN−1+, then problem (1.1) does not have a ϱ-equivariant least energy solution in D1,2(Σω).
Proof. Σω is the upper half-space RN+:={(x1,…,xN)∈RN:xN>0}. If u were a ϱ-equivariant least energy solution to (1.1) in RN+ then, extending u by reflection on ∂(RN+), would yield a sign-changing solution ˜u to the problem (1.2) in RN with JRN(˜u)≤2c∞. But the energy of any sign-changing solution to (1.2) is >2c∞; see [9].
The following local geometric condition guarantees the existence of a minimizer. It was introduced by Adimurthi and Mancini in [1].
Definition 2.6. A point ξ∈∂ω is a point of convexity of Σω (of radius r>0) if Br(ξ)∩Σω⊂Hν and the mean curvature of ∂Σω at ξ with respect to the exterior unit normal ν at ξ is positive.
As in [1] we make the convention that the curvature of a geodesic in ∂Σω is positive at ξ if it curves away from the exterior unit normal ν. The half-space Hν is defined as in (2.7). Examples of cones having a point of convexity are given as follows.
Proposition 2.7. If ¯ω⊂SN−1+, then Σω has a point of convexity.
Proof. Let β be the smallest geodesic ball in SN−1, centered at the north pole (0,…,0,1), which contains ω. Then, ∂ω∩∂β≠∅ and ¯β⊂SN−1+. Hence, every point on ∂β is a point of convexity of Σβ. As ω⊂β, we have that any point ξ∈∂ω∩∂β is a point of convexity of Σω.
Theorem 2.8. If Σω has a point of convexity, then cϱΣω<c∞. Consequently, the problem (1.1) has a ϱ-equivariant least energy solution in D1,2(Σω). This solution is nonradial and changes sign.
Proof. Let ξ∈∂ω be a point of convexity of Σω of radius r>0. It is shown in [1,Lemma 2.2] that, after fixing r small enough and a radial cut-off function ψ∈C∞c(RN) with ψ(x)=1 if |x|≤r4 and ψ(x)=0 if |x|≥r2, the function uε,ξ(x):=ψ(x−ξ)ε(2−N)/2U(ε−1(x−ξ)), with U as in (2.3), satisfies
QΣω(uε,ξ)={S22/N−dNHω(ξ)Sεln(ε−2)+O(ε)if N=3,S22/N−dNHω(ξ)Sε+O(ε2ln(ε−2))if N≥4, | (2.19) |
where dN is a positive constant depending only on N and Hω(ξ) is the mean curvature of ∂Σω at ξ. Hence, for ε small enough,
JΣω(tε,ξuε,ξ)=1N[QΣω(uε,ξ)]N2<12NSN2=12c∞, |
where tε,ξ>0 is such that tε,ξuε,ξ∈N(Σω); see (2.1). Choosing r so that Br(ξ)∩Br(ϱξ)=∅ we conclude that tε,ξ(uε,ξ−uε,ξ∘ϱ)∈Nϱ(Σω) and
cϱΣω≤JΣω(tε,ξ(uε,ξ−uε,ξ∘ϱ))<c∞. |
The existence of a ϱ-equivariant least energy solution to (1.1) follows from Corollary 2.4.
In this section ω is not assumed to have any symmetries.
We are interested in positive solutions to the problem (1.1). Note that this problem has always a positive radial solution given by the restriction to Σω of the standard bubble U defined in (2.3). The question we wish to address in this section is whether problem (1.1) has a positive nonradial solution.
Recall the notation introduced in Section 2 and set
cΣω:=infu∈N(Σω)JΣω(u)=infu∈D1,2(Σω)∖{0}1N[QΣω(u)]N2, |
N(Λω):=N(Σω)∩V(Λω)andcΛω:=infu∈N(Λω)JΛω(u). |
It is shown in [8,Theorem 2.1] that cΛω>0. As in Lemma 2.1 one shows that cΣω=cΛω≤12c∞. We start by describing the behavior of minimizing sequences for JΛω on N(Λω).
Proposition 3.1. Let uk∈N(Λω) be such that
JΛω(uk)→cΛωandJ′Λω(uk)→0in(V(Λω))′. |
Then, after passing to a subsequence, one of the following statements holds true:
(i) There exist a sequence of positive numbers (εk), a sequence of points (ξk) in Γ1 and a function w∈D1,2(RN) such that ε−1kdist(ξk,ˉω∪{0})→∞, w|H solves the Neumann problem
−Δw=|w|2∗−2w,w∈D1,2(H), |
in some half-space H, JH(w)=12c∞,
limk→∞‖uk−ε2−N2kw(⋅−ξkεk)‖Σω=0, |
and cΣω=cΛω=12c∞.
(ii) There exist a sequence of positive numbers (εk) with εk→0 and a solution w∈D1,2(Σω) to the problem (1.1) such that
limk→∞‖uk−ε2−N2kw(⋅εk)‖Σω=0, |
and JΣω(w)=cΣω=cΛω≤12c∞.
Proof. The proof is similar, but simpler than that of Proposition 2.3.
The following statement is an immediate consequence of this proposition.
Corollary 3.2. If cΣω<12c∞, then the problem (1.1) has a positive least energy solution in D1,2(Σω).
Theorem 3.3. If Σω has a point of convexity, then cΣω<12c∞. Consequently, the problem (1.1) has a positive least energy solution in D1,2(Σω).
Proof. The proof is similar to that of Theorem 2.8.
Let D1,2rad(Σω) be the subspace of radial functions in D1,2(Σω), and define Nrad(Σω):=N(Σω)∩D1,2rad(Σω) and
cradΣω:=infu∈Nrad(Σω)JΣω(u)=infu∈D1,2rad(Σω)∖{0}1N[QΣω(u)]N2. |
It was shown in [8,Theorem 2.4] that, if Σω is convex, then cradΣω=cΣω and the only positive minimizers are the restrictions of the rescalings
Uε(x)=aN(εε2+|x|2)N−22,ε>0, | (3.1) |
of the standard bubble to Σω. In fact, the proof of [8,Theorem 2.4] shows that these are the only positive solutions of (1.1) in a convex cone. Moreover, the following statement holds true.
Proposition 3.4. For any cone Σω, the restrictions to Σω of the functions Uε defined in (3.1) are minimizers of JΣω on Nrad(Σω). These are the only nontrivial radial solutions to (1.1), up to sign. Moreover,
cradΣω=bN|Λω|,wherebN=c∞|B1(0)| |
and |X| is the Lebesgue measure of X. In particular, cradΣω increases with |Λω|.
Proof. A radial function u solves (1.1) in Σω if and only if the function ˉu given by ˉu(r):=u(x) with r=‖x‖ solves
ddr(rN−1ˉu′(r))=rN−1|ˉu(r)|N−2ˉu(r) in (0,∞),ˉu(0)=u(0),ˉu′(0)=0. |
This last problem does not depend on ω. It is well known that, up to sign, the functions Uε are the only nontrivial radial solutions to the problem (1.2) in RN=ΣSN−1. Hence, their restrictions to Σω are the only nontrivial radial solutions to (1.1).
As in Lemma 2.1 one shows that cradΣω=cradΛω:=infu∈Nrad(Λω)JΛω(u). For u∈Vrad(Λω):=D1,2rad(Λω)∩V(Λω),u≠0, we have that
QΛω(u)=∫Λω|∇u|2(∫Λω|u|2∗)2/2∗=N|Λω|∫10|ˉu′(r)|2rN−1dr(N|Λω|∫10|ˉu(r)|2∗rN−1dr)2/2∗. |
Therefore,
cradΛω=infu∈Vrad(Λω)∖{0}1N[QΛω(u)]N2=infu∈Vrad(Λω)∖{0}∫10|ˉu′(r)|2rN−1dr(∫10|ˉu(r)|2∗rN−1dr)2/2∗|Λω|=:bN|Λω|. |
The same formula holds true when we replace ω by SN−1. In this case, the left-hand side is c∞. Hence, bN=c∞|B1(0)|, as claimed.
Corollary 3.5. If Σω has a point of convexity and |Λω|≥12|B1(0)|, then
(i) the problem (1.1) has a positive least energy solution in D1,2(Σω),
(ii) every least energy solution of (1.1) is nonradial.
Proof. From Theorem 3.3 and Proposition 3.4 we get that cΣω is attained and
cΣω<12c∞=cradRN+=bN2|B1(0)|≤bN|Λω|=cradΣω, |
where RN+:={(x1,…,xN)∈RN:xN>0}. So every least energy solution is nonradial.
Note that the hypothesis that |Λω|≥12|B1(0)| implies that Σω is not convex.
A closer look at the estimate (2.19) allows to refine Corollary 3.5 and to produce examples of cones Σω with |Λω|<12|B1(0)| for which the problem (1.1) has a positive nonradial solution.
To this end, we fix a smooth domain ω0 in SN−1 for which Σω0 has a point of convexity ξ∈∂ω0 of radius r>0, and we define
ℓ(ω0,ξ,r):={ω:ω is a smooth domain in SN−1,Br(ξ)∩Σω0⊂Br(ξ)∩Σωand dist(Br(ξ)∩Σω0,Br(ξ)∩(Σω∖Σω0))>0}. |
Then, we have the following result.
Theorem 3.6. There exists αξ∈(0,12|B1(0)|), depending only on Br(ξ)∩Σω0, such that, for every ω∈ℓ(ω0,ξ,r) with |Λω|>αξ, the following statements hold true:
(i) the problem (1.1) has a positive least energy solution in D1,2(Σω),
(ii) every least energy solution of (1.1) is nonradial,
(iii)Σω is not convex.
Proof. Recall that the functions uε,ξ, introduced in the proof of Theorem 2.8, vanish outside the ball Br/2(0). Moreover, the value QΣω0(uε,ξ) and the estimate (2.19) depend only on the value of uε,ξ in Br(ξ)∩Σω0. We fix ε0>0 small enough so that
Qξ:=QΣω0(uε0,ξ)<S22/N, |
and we set αξ:=1NbNQN/2ξ with bN as in Proposition 3.4. Then,
αξ<12NbNSN2=12|B1(0)|. |
Given ω∈ℓ(ω0,ξ,r), we fix a function ˆuε0,ξ∈C∞c(Br(0)) such that ˆuε0,ξ(x)=uε0,ξ(x) if x∈Br(ξ)∩Σω0 and ˆuε0,ξ(x)=0 if x∈Br(ξ)∩(Σω∖Σω0). So, if |Λω|>αξ, we have that
cΣω≤1N[QΣω(ˆuε0,ξ)]N2=1NQN2ξ=bNαξ<bN|Λω|=cradΣω. |
Note that ξ is a point of convexity of ω. Hence, by Theorem 3.3 and the previous inequality, cΣω is attained at a nonradial solution of (1.1). Finally, recall that, if Σω were convex, then cΣω=cradΣω; see [8,Theorem 2.4]. This completes the proof.
Corollary 3.7. There exists a smooth domain ω⊂SN−1+ such that the problem (1.1) has a positive nonradial solution in Σω.
Proof. Let ω0 be the geodesic ball in SN−1 of radius π/4 centered at the north pole and let ξ be any point on ∂ω0. Fix r>0 such that Br(ξ)∩SN−1⊂SN−1+. Clearly, ξ is a point of convexity of Σω0 of radius r, so we may fix αξ>0 as in Theorem 3.6. As αξ<12|B1(0)|, there exists ω∈ℓ(ω0,ξ,r) with ω⊂SN−1+ and |Λω|>αξ. Now, Theorem 3.6 yields a positive nonradial solution to problem (1.1) in Σω.
Remark 3.8. Let ω≠SN−1+ be such that Σω is convex. Then, every point ξ∈∂ω is a point of convexity of radius r for any r>0. Fix r=1, and fix ε>0 such that
Qξ:=QΣω(uε,ξ)<S22/N∀ξ∈∂ω. |
Now, define αξ:=1NbNQN/2ξ, as in Theorem 3.6. Since Σω is convex, we must have that
|Λω|≤αξ=|B1(0)|SN/2QN/2ξ,∀ξ∈∂ω, |
where the equality follows from the definition of bN; see Proposition 3.4. Hence, for any convex cone Σω, we obtain the upper bound
|Λω|≤|B1(0)|SN/2minξ∈∂ωQξ |
for the measure of Λω, which is given in terms of the Sobolev constant and the local energy of the standard bubbles.
M. Clapp was partially supported by UNAM-DGAPA-PAPIIT grant IN100718 (Mexico) and CONACYT grant A1-S-10457 (Mexico). F. Pacella was partially supported by PRIN 2015 (Italy) and INDAM-GNAMPA (Italy).
The authors declare no conflict of interest.
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