Prescribing scalar curvature on <span class="mathjax-formula">${S}^{3}$</span>

Prescribing scalar curvature on S ³

Matthias Schneider

Ruprecht-Karls-Universität, Mathematisches Institut, Im Neuenheimer Feld 288, 69120 Heidelberg, Germany Received 20 May 2005; accepted 30 March 2006

Available online 20 September 2006

Abstract

We give existence results for solutions of the prescribed scalar curvature equation onS³, when the curvature function is a positive Morse function and satisfies an index-count condition.

©2006 Elsevier Masson SAS. All rights reserved.

Résumé

Nous démontrons l’existence de solutions de l’équation courbure scalaire surS³, quand la courbure scalaire est une fonction de Morse positive satisfaisant à une condition d’indice.

©2006 Elsevier Masson SAS. All rights reserved.

MSC:35J60; 35J20; 55C21

Keywords:Prescribed scalar curvature; Leray–Schauder degree; Finite dimensional reduction

1. Introduction

LetS³be the standard sphere with round metricg0induced byS³=∂B1(0)⊂R⁴. We study the problem: Which functionsKonS³occur as scalar curvature of metricsgconformally equivalent tog0? Writingg=ϕ⁴g0andk(θ ):=

1

6(K(θ )−6)this is equivalent to solving fort=1 (see [3])

−8S³ϕ+6ϕ=6

1+t k(θ )

ϕ⁵, ϕ >0 inS³. (1.1)

In stereographic coordinatesSθ(·)centered at some pointθ∈S³, i.e.Sθ(0)=θ, Eq. (1.1) is equivalent to

−u=

1+t kθ(x)

u⁵ inR³, u >0, (1.2)

wherek_θ(x):=k◦Sθ(x)and u(x)=Rθ(ϕ)(x):=3¹⁴

1+ |x|²₋¹₂

ϕ◦Sθ(x). (1.3)

An obvious necessary condition for the existence of solutions to (1.1) is that the function K has to be positive somewhere. Moreover, there are theKazdan–Warnerobstructions [14,7], which imply in particular, that a monotone function of the coordinate functionx₁cannot be realized as the scalar curvature of a metric conformal tog₀.

E-mail address:mschneid@mathi.uni-heidelberg.de (M. Schneider).

0294-1449/$ – see front matter ©2006 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2006.03.004

Numerous studies have been made on Eq. (1.1) and its higher dimensional analogue and various sufficient con- ditions for its solvability have been found (see [2,16,15,11,12,6,4] and the reference therein), usually under a non- degeneracy assumption onK. OnS³a positive functionKis non-degenerate, if

K(θ )=0 if∇K(θ )=0. (nd)

For positive Morse functionsKonS³it is shown in [18,5,10] that (1.1) is solvable ifKsatisfies (nd) and d:= −

1+

∇K(θ )=0, K(θ )<0

(−1)^{ind(K,θ )}

=0, (1.4)

where ind(K, θ )is the Morse index ofKatθ. We are interested in the case when the non-degeneracy assumption (nd) is not satisfied.

As in [10] we use a continuity method and join the curvature functionKto the constant functionK₀≡6 by a one parameter familyK_t(θ ):=6(1+t k(θ )).

The positive solutions of (1.2) fort=0 are completely known (see [9,13]) and given by a non-compact manifold Z:=

z_μ,y(x):=μ⁻¹²3¹⁴

1+ x−y

μ ²₋¹

2

: y∈R³, μ >0 , wherez_μ,y(y)→ ∞asμ→0.

Thus, in general, there are no a prioriL^∞-estimates for (1.1). Fort=0 this lack of compactness stems from the fact that the noncompact group of conformal transformation ofS³acts on solutions. In particular the dilations allow solutions to concentrate in a single point with largeL^∞-norm. One expects that a non-constantkbreaks this symmetry leading to a priori estimates for solutions. Indeed, in [10,11] it is shown that ifK∈C²(S³)is a positive function and satisfies (nd) then forδ >0 there isC=C(K, δ) >0 such that for allt∈ [δ,1]and solutionsϕ_tof (1.1) we have

C⁻¹ϕt(θ )C and ϕtC^2,α(S³)C.

Furthermore, Chang et al. [10] compute the Leray–Schauder degree for (1.1) fort >0 small, and show that it equals

−deg(G, B1(0),0), which is given byd in (1.4), whenK is a Morse function. The mapGis associated toK and defined onB₁(0)⊂R⁴. The a priori estimate implies the invariance of the degree as the parametert moves to 1 and gives a solution to (1.1) ifd=0.

Hence, if (nd) fails, we face two problems: Is the a priori bound still valid and how do critical points ofK with K=0 occur in the index count condition (1.4)? A priori bounds, when (nd) fails, are given in [17]. Here, we will mainly deal with the second question and give a generalized version of (1.4).

In the following, unless otherwise stated, we will assume thatK=6(1+k)∈C⁶(S³)is positive. To give our main results we need the following notation. We writek_θ=k◦Sθ and for a critical pointθofkwe let

a₀(θ ):=

C

R³

k_θ(x)−T_k²

θ,0(x)

|x|⁻⁶,

a₁(θ ):=²k_θ(0)+ ∇ k_θ(0)

·

D²k_θ(0)₋1

∇

k_θ(0) , a2(θ ):=kθ(0)a1(θ )− 15

8π

∂B1(0)

D²kθ(0)(x)²², a₃(θ ):= 12

π²

D²k_θ(0)₋1

∇

k_θ(0)

·

C

R³

∇k_θ(x)−T_∇²_k

θ,0(x)

|x|⁻⁶

+48 π²

D²k_θ(0)₋1

∇ k_θ(0)

·

C

R³

k_θ(x)−T_k³

θ,0(x) x_i

|x|⁸+120 π²

C

R³

k_θ(x)−T_k⁴

θ,0(x) 1

|x|⁸,

where all differentiations are done inR³, themth Taylor polynomial ofkiny is abbreviated by

T_k,y^m(x):=

m

=0

1

!D k(y)(x−y) , and

Cis the Cauchy principal value of the integral,

C

R³

f (x):=lim

r→0

R³\Br(0)

f (x).

Denote byM,M₊, andM₀the sets, M:=

θ∈S³: ∇k(θ )=0, kθ(0)=a0(θ )=0, anda2(θ )=0 , M₊:=

θ∈M: 0<−a1(θ )/a2(θ )1 , M0:=

θ∈M: a1(θ )=0 .

We fixθ∈S³and define for 0< μ <∞andy∈R³the Melnikov functionΓ_θby Γ_θ(μ, y):=1

6

R³

k_θ(x)(z_μ,y)⁶dx=1 6

R³

k_θ(μx+y)(z_1,0)⁶dx.

Theorem 1.1.Suppose1+k∈C⁶(S³)is positive and satisfies D²kθ(0)is invertible, ifθ∈A:=

θ∈S³: ∇k(θ )=0and_S3k(θ )=0 ,

the setM₊is empty, anda₃(θ )=0ifθ∈M₀. Then there isR₀>0such that for anyRR₀andθ∈S³the degree deg(Γ_θ, Ω_R,0)is well defined and independent ofθandR, where

Ω_R:=

(μ, y)∈R³: R⁻¹< μ < Rand|y|< R ,

and(1.1)is solvable, if for some(and hence for any)θ∈S³andRR₀ 0=d:= −deg(Γ_θ, Ω_R,0)+

θ∈M₀:a₃(θ )>0

(−1)^{ind(k,θ )}. (1.5)

The numberdis the Leray–Schauder degree of the problem(1.1). If, moreover,kis a Morse function, then d= −

1+

θ∈Crit₋(k)

(−1)^{ind(k,θ )}

, (1.6)

where

Crit₋(k):=

θ∈S³: ∇k(θ )=0, k(θ )²+a₀(θ )²+a₁(θ )²=0, and lim

μ→0⁺sgn

k(θ )+a₀(θ )μ−a₁(θ )μ²

= −1 .

Under the assumptions of Theorem 1.1 the setMis finite. Thus we need only consider and sum over a finite number of pointsθ.

In [17] it is shown for Morse functionsK that the conditionM₊= ∅is equivalent to the compactness inC²(S³) of the set of solutions to (1.1), whent∈ [δ,1+δ]for smallδ >0. In the general case this condition is only sufficient.

Hence, the Leray–Schauder degree of (1.1) is invariant with respect tot∈(0,1]. By simply replacingk byskfor somes >0 we obtain from Theorem 1.1, if we abandon the conditionM₊= ∅, that the degree of (1.1) is given byd for anytsuch that

0< t <min

−a₁(θ )

a₂(θ ): θ∈Mand −a₁(θ )

a₂(θ )>0 . (1.7)

The first addend in (1.5) gives the degree of the solutions that remain uniformly bounded as t→0⁺, whereas the second addend is the degree of the solutions that blow up ast→0⁺. It is part of the proof to show that the family

of solutions splits in this way. Here, the assumptiona3(θ )=0 at pointsθ∈M, wherea1(θ )=0, assures that the blowing-up solutions lie onC¹-curves(t, ϕ_t)emanating fromt=0 (see [17]).

Ifk satisfies (nd) the solutions are uniformly bounded with respect tot ∈(0,1], the set M is empty, and one recovers (1.4). The derivative of the Melnikov functionΓ_θ is closely related to the above mentioned mapGin [10]

(see Remark 5.1 below).

It is interesting to note that ifKis a Morse function, then the formula for the degree in (1.6) is independent of the coefficientsa₃(θ ). This gives the perspective that for Morse functions andtsatisfying (1.7) the degree is always given by (1.6).

We sketch the strategy of the proof of Theorem 1.1 and outline the remaining part of the paper. The transformations in (1.3) gives rise to a Hilbert space isomorphism between H^1,2(S³)and D^1,2(R³), where D^1,2(R³) denotes the closure ofC_c^∞(R³)with respect to

u²=

R³

|∇u|².

Due to elliptic regularity (see [8]) and Harnack’s inequality it is enough to find a weak nonnegative solution of (1.1) inH^1,2(S³), or of the equivalent equation. We take advantage of both formulations: We fixθ∈S³, consider (1.2), and find solutions as critical points off_t^θ:D^1,2(R³)→R, where

f_t^θ(u):=1 2

|∇u|²−1

6 1+t k_θ(x)

|u|⁶.

To avoid cumbrous indexing we will suppress the dependenceθand writeftinstead off_t^θ. Fort=0 the functionalf0

possesses, as seen above, a(3+1)-dimensional manifold of critical pointsZ. We recall some facts about the spectrum off₀(z)in Section 2. We use a finite dimensional reduction of Melnikov type developed in [1,2]. In Section 3 we recall without proof that a sequence of solutions to (1.1) can only blow-up in a single point (see [18,15]) and fit this result into our framework. Section 4 contains the finite dimensional reduction of our problem, where we sum up the results in [17] and obtain a one-dimensional function that describes the blow-up behavior of solutions. The computation of the Leray–Schauder degree is done in Section 5. Appendix A contains the proof of differentiability of the curve of blowing-up solutions, which is done briefly by using the computations and estimates in [17].

2. The unperturbed problem

We define forμ >0 andy∈R³the mapsUμ,Ty:D^1,2(R³)→D^1,2(R³)by Uμ(u):=μ⁻¹²u

· μ

and Ty(u):=u(· −y).

With this notation the critical manifoldZis given by Z=

z_μ,y=Ty◦Uμ(z_1,0): y∈R³, μ >0 .

It is easy to check that the dilationUμand the translationTyconserve the norms · and · L⁶. Thus for everyμ >0 andy∈R³

(Uμ)⁻¹=(Uμ)^t=U_μ−1, (Ty)⁻¹=(Ty)^t=T−y, and f0=f0◦Uμ=f0◦Ty (2.1) where(·)^t denotes the adjoint. Twice differentiating the identities forf0in (2.1) yields

f₀(v)=(Ty◦Uμ)⁻¹◦f₀

Ty◦Uμ(v)

◦(Ty◦Uμ) ∀v∈D^1,2 R³

. (2.2)

Moreover, we see thatU (μ, y, z):=Ty◦Uμ(z)maps(0,∞)×R³×ZintoZ, hence

∂U

∂z(μ, y, z)=Ty◦Uμ:T_zZ→T_Ty◦Uμ(z)Z and Ty◦Uμ:(T_zZ)^⊥→(T_Ty◦Uμ(z)Z)^⊥. (2.3) The tangent spaceT_zμ,yZat a pointz_μ,y∈Zis spanned by 4 orthonormal functions,

T_zμ,yZ=(˙ξ_μ,y)_i: i=0. . .3 ,

where(ξ˙μ,y)idenotes fori=0 the normalized tangent vector_dμ^d zμ,y and for 1i3 the normalized tangent vector

d

dyiz_μ,y= −_∂x^∂_iz_μ,y. By (2.3) we obtain (˙ξ_μ,y)_i =Ty◦Uμ

(˙ξ_1,0)_i .

An explicit calculation gives for 1i3 (˙ξ_1,0)_i(x)= 8

π√ 15

1+ |x|²₋¹

2x_i. Fori=0 we find

(˙ξ1,0)0(x)= 4 π√

15

1+ |x|²₋¹

2 1−2

1+ |x|²₋1 .

Using the canonical identification of the Hilbert space D^1,2(R³)with its dual induced by the scalar-product and denoted byK:(D^1,2(R³))→D^1,2(R³),

K(ϕ), u

=ϕ(u) ∀(ϕ, u)∈ D^1,2

R³

×D^1,2 R³

,

we shall considerf_t(u)as an element ofD^1,2(R³)andf_t(u)as one ofL(D^1,2(R³)). With this identificationf_t(u) is of the formidentity–compact(see [2]) and hence a Fredholm operator of index zero.

Sincef₀(z_μ,y)is a self-adjoint, compact perturbation of the identity map inD^1,2(R³), its spectrumσ (f₀(z_μ,y)) consists of point-spectrum, possibly accumulating at 1. In [17] the spectrum and the eigenfunctions off₀(z_μ,y)are computed,

σ

f₀(zμ,y)

=

λi,j:=1− 15

4(1+i+j )²−1: i, j∈N0 , (2.4)

and the dimension of the eigenspace,Φ_i,j,l^μ,y, corresponding to the eigenvalueλi,jis given by_i+2

2

−_i

2

, the dimen- sion of the spherical harmonics of degreei. SinceZ is a manifold of critical points off₀, the tangent spaceT_zZ at a pointz∈Zis contained in the kernelN (f₀(z))off₀(z). Asλ_i,j =0 if and only ifi+j =1, the dimension of N (f₀(z))is 4, which implies that

T_zZ=N f₀(z)

for allz∈Z. (2.5)

If (2.5) holds the critical manifoldZis called non-degenerate (see [1]) and the self-adjoint Fredholm operatorf₀(z) maps the spaceD^1,2(R³)intoT_zZ^⊥and is invertible inL(T_zZ^⊥). From (2.2) and (2.3), we obtain in this case

f₀(z_1,0)₋1

L(T_z1,0Z^⊥)=f₀(z)₋1

L(T_zZ^⊥) ∀z∈Z. (2.6)

Moreover,f₀(z)andf₀(z)|TzZ^⊥have precisely one negative eigenvalue−4 with one-dimensional eigenspacez. 3. Blow up analysis

Based on the results in [18,15] we have the following lemma (see [17, Corollary 3.2])

Lemma 3.1.Suppose1+k∈C¹(S³)is positive. If(t_i, ϕ_i)∈ [0,1] ×C²(S³)solve (1.1) witht=t_i, then after passing to a subsequence either(ϕ_i)is uniformly bounded inL^∞(S³)(and hence inC^2,α(S³)by standard elliptic regularity) or there existθ∈S³and sequences(μ_i)∈(0,∞),(y_i)∈R³satisfyinglimi→∞μ_i=0andlimi→∞y_i=0, such that (in stereographic coordinatesSθ(·)aboutθ)

Rθ(ϕ_i)−

1+t_ik_θ(y_i)₋¹

4z_μi,yiis orthogonal toT_{zμi ,yi}Z, Rθ(ϕ_i)−

1+t_ik_θ(y_i)₋¹

4z_μi,yi

D^1,2(R³)=o(1).

4. Finite dimensional reduction

For the rest of the paper, unless otherwise indicated, integration extends overR³and is done with respect to the variablex.

In this section we state without proof results obtained in [17], which yield a finite dimensional reduction of our problem.

Lemma 4.1.[17, Lemma 4.2] Suppose 1+k∈C⁵(S³) is positive andθ ∈S³. Then there existρ0=ρ0(k) >0, t0=t0(k) >0,b=b(k) >0, an upper continuous function μ0:R→R₊∪ {∞}, depending only on k, and two functionsw:Ω→D^1,2(R³)andα:Ω→R⁴depending onkandθ, where

Ω:=

(t, μ, y)∈ [−b,1+b] ×(0,+∞)×R³: 0< μ < μ₀(t ) , μ₀(t )= +∞ if|t|t₀,

such that for any(t, μ, y)∈Ω

w(t, μ, y) is orthogonal to T_zμ,yZ (4.1)

f_t

zμ,y+w(t, μ, y)

= α(t, μ, y)· ˙ξμ,y∈Tz_μ,yZ, (4.2)

w(t, μ, y)−w0(t, μ, y)+α(t, μ, y) < ρ0, (4.3)

where{(˙ξ_μ,y)_i: i=0. . .3}denotes the basis ofT_zμ,yZgiven in(1.6)and w₀(t, μ, y):=

1+t k_θ(y)₋¹

4 −1 z_μ,y.

The functionswandαare of classC²and unique in the sense that if(v,β) satisfies(4.1)–(4.3)for some(t, μ, y)∈Ω then(v,β) is given by(w(t, μ, y),α(t, μ, y)).

Moreover, we have for1j3

α(t, μ, y) ₀− 2 i=1

α_i(t, μ, y)₀ O

t∇k_θ(y)²min 1, μ²

+min

1, μ⁹⁴ ,

α(t, μ, y) _j− 2

i=1

α_i(t, μ, y)_j O

tmin 1, μ⁹⁴

, (4.4)

where

α₁(t, μ, y):= −tmin(1, μ)

1+t k_θ(y)₋⁵

4 π

3¹⁴√ 5

0

∇k_θ(y)

,

α₂(t, μ, y):= −tmin 1, μ²

1+t k_θ(y)₋⁵

4 π

3¹⁴√ 5

k_θ(y) 0

.

Remark 4.2.From the proof of Lemma 4.1 in [17] and (4.4) we see that 1

t μα(t, μ, y)

is a well defined, continuous function for(t, μ, y)∈Ω.

Lemma 4.3.[17, Lemma 5.1]Under the assumptions of Lemma4.1we have for all(t, μ, y)∈Ω with|μ|1and 1i, j3

1 t μ

∂α(t, μ, y)_i

∂y_j = − π

3¹⁴√ 5

1+t k(y)₋⁵

4∂²k_θ(y)

∂x_i∂x_j +O∇k_θ(y)²+μ¹⁺¹⁴ .

Under the assumptions of Lemma 4.1 suppose(ti, ϕi)∈(0,1] ×C²(S³)solve (1.1) witht=ti and blow up at θ∈S³, i.e.t_i→t₀∈ [0,1]and there is a sequence(θ_i)inS³such thatθ_i→θandϕ_i(θ_i)→ ∞asi→ ∞. Setting

u_i:=Rθ(ϕ_i)

we obtain from Lemma 3.1 that for largeithe tuple(u_i−z_μi,yi,0)satisfies (4.1)–(4.3) for(t_i, μ_i, y_i)∈Ω. From the uniqueness part in Lemma 4.1 we get that for largei

1

t_iμ_iα(t i, μi, yi)=0.

Consequently to exclude or to construct blow-up sequences it is enough to exclude or construct zeros ofα(t, μ, y) for smallμ. From the expansion ofαin Lemma 4.1 we immediately get that∇k_θ(0)=0 andk_θ(0)=0 ifθ∈S³is a blow-up point.

Thus the remaining possible blow-up points are the critical pointsθ of kwherek_θ(0)=0=_S3k(θ ). Ifθ is a nondegenerate critical point ofk, then the determination of blow-up points can be reduced to a one dimensional problem.

Lemma 4.4.[17, Lemma 6.1]Under the assumptions of Lemma4.1suppose0is a nondegenerate critical point ofk_θ, i.e.

∇k_θ(0)=0 and D²k_θ(0)is invertible.

Moreover, assumekθ(0)=0. Consider the functionα, defined byˆ ˆ

α(t, μ, y):=3¹⁴√ 5 t μπ

1+t k(θ )⁵₄

α(t, μ, y)₁, . . . ,α(t, μ, y) ₃T

,

which is well defined and continuous inΩ(see Remark4.2). Then there areδ₁=δ₁(k) >0and aC²-functionβ, β:

(t, μ): t∈ [−b,1+b], 0< μ < δ₁

→R³, such thatβ(t, μ)=O(μ²)asμ→0and

ˆ α

t, μ, β(t, μ)

=0 for allt∈ [−b,1+b], 0< μ < δ₁.

Moreover,βis unique in the sense that, ify∈Bδ₁(0)satisfiesα(t, μ, y)ˆ =0for somet∈ [−b,1+b]and0< μ < δ1, theny=β(t, μ).

Hence, to exclude or to construct blow-up sequences, which blow-up at a nondegenerate critical pointθofkwith k_θ(0)=0 it suffices to studyα(t, μ, β(t, μ))0.

Lemma 4.5.[17, Lemma 6.2]Under the assumptions of Lemma4.4andk∈C⁶(S³)we have α

t, μ, β(t, μ)

0= −t μ³

1+t k(θ )₋⁵

4 3³⁴4 π√

5a₀(θ ) +t μ⁴

1+t k(θ )₋⁹

4π3³⁴√ 5 30

a₁(θ )+t a₂(θ )

+t μ⁵

1+t k(θ )₋⁵

4π3³⁴√ 5

30 a3(θ )+O t μ⁵⁺¹²

+O

t²μ⁴⁺¹⁴ .

Consequently, under the assumptions of Lemma 4.4, if(ti, ϕi)blow up atθ∈S³then necessarilykθ(0)=0= a0(θ )anda2(θ )=0. The necessary conditiona2(θ )=0 follows from that fact, that ifa2(θ )=0 then from its defin- itiona₁(θ )=0, which is impossible by Lemma 4.5, ifθ is a blow-up point. Moreover, the expansion in Lemma 4.5 leads to restrictions on(t_i), that is(t_i)has to converge to−a₁(θ )/a₂(θ ).

It remains the question if there exist(t_i, ϕ_i)that blow up atθ∈S³, ifk_θ(0)=0=a₀(θ )anda₂(θ )=0. This is true ifa₁(θ )=0, in the casea₁(θ )=0 one needs to assumea₃(θ )=0.

Theorem 4.6.Suppose1+k∈C⁶(S³)is positive and satisfies D²k_θ(0)is invertible, ifθ∈A:=

θ∈S³: ∇k(θ )=0and_S3k(θ )=0 and

a₁(θ )²+a₃(θ )²=0 for allθ∈M.

Then there isδ >0such that for anyθ∈M₊∪M₀there exists a uniqueC¹-curve {0< μ < δ} μ→

t^θ(μ), ϕ^θ(μ,·)

∈

(−δ,1+δ)\ {0}

×C^2,α S³

, such that asμ→0

t^θ(μ)= − 1 a₂(θ )

a₁(θ )+O(μ¹⁴) ifa₁(θ )=0, a₃(θ )μ+O(μ¹⁺¹⁴) ifa₁(θ )=0 andϕ^θ(μ,·)solves(1.1)fort=t^θ(μ)and blows up like

Rθ

ϕ^θ(μ, x)

−

1+t^θ(μ)k(θ )₋¹₄

zμ,0(x)

D^1,2(R³)∩C²(B1(0))=O μ²

.

The curves are unique, in the sense that, if(t_i, ϕ_i)∈((−δ,1+δ)\ {0})×C^2,α(S³)blow up at someθ∈S³thenθ∈ M₊∪M₀and there is a sequence of positive numbers(μ_i)converging to zero such that(t_i, ϕ_i)=(t^θ(μ_i), ϕ^θ(μ_i,·)) for all but finitely manyi∈N.

Proof. The claim follows directly from [17, Theorems 1.2, 1.3, 7.1]; we only need to note that if(t_i, ϕ_i)blow up at someθ∈S³with limi→∞t_i=0 then by our assumptionsa₃(θ )=0 andθhas to be inM₀. 2

In order to compute the degree of the concentrating solutions, whent→0⁺, we need to compute the derivative of t^θ(·)asμ→0⁺.

Lemma 4.7.Under the assumptions of Theorem4.6supposeθ∈M₀. Then

∂t^θ(μ)

∂μ = −a3(θ ) a2(θ )+O

μ¹⁴ .

The proof is given in Appendix A. The next observation is important for the calculation of the degree of solutions which remain uniformly bounded ast→0⁺.

Lemma 4.8.Under the assumptions of Theorem4.6supposeθ∈Aand letαbe as in Lemma4.1. Then there isδ₂>0 such that for all0< δδ₂existt₂(δ) >0andd₂(δ) >0satisfying

|t|<t2(δ), δinfμδ2,|y|δ2

1

t μα(t, μ, y)

d2(δ) >0.

Proof. The assumptions of Theorem 4.6 implya0(θ )²+a1(θ )²+a3(θ )²=0. In the casea0(θ )²+a1(θ )²=0 we may chooset2(δ)independently ofδ: Ifa0(θ )=0 we sett2(δ):=1 and ifa0(θ )=0 anda1(θ )=0 we lett2(δ):=

1

2max(1,|a2(θ )|)⁻¹|a1(θ )|such that a₁(θ )+t a₂(θ )>1

2a₁(θ ) for|t|t₂. Ifa0(θ )²+a1(θ )²=0 we set

t₂(δ):=min

|a₃(θ )| 4|a₂(θ )|,1

δ.

From the expansion in Lemma 4.5 there exists 0< δ2< δ1such that h₁(t, μ):= 1

t μα

t, μ, β(t, μ)

0=0 forδμδ₂and|t|t₂(δ). (4.5)

To obtain a contradiction assume there are(t_n, μ_n, y_n)∈(−t₁, t₁)×(δ, δ₂)×B_δ2(0)such that 1

tnμnα(t _n, μ_n, y_n)→0 asn→ ∞.

We may assume(t_n, μ_n, y_n)→(¯t ,μ,¯ y)¯ asn→ ∞, whereδμ¯ δ₂. Thus,α(¯ˆ t ,μ,¯ y)¯ =0. The uniqueness part of Lemma 4.4 givesy¯=β(¯t ,μ). Hence,¯ h₁(¯t ,μ)¯ =0 contradicting (4.5). 2

5. Leray–Schauder degree

We recall that forθ∈S³the Melnikov functionΓθ:(0,∞)×R³→Ris defined by Γ_θ(μ, y)=1

6

k_θ(x)(z_μ,y)⁶=1 6

k_θ(μx+y)(z_1,0)⁶. It is known (see [1,2]) thatΓ_θextends via

Γ_θ(−μ, y):=Γ_θ(μ, y) and Γ_θ(0, y):=k_θ(y)1 6

(z_1,0)⁶ to a function inC²(R×R³)and

∂Γ_θ

∂y_i (0, y)= ∂k

∂x_i(y)1 6

(z1,0)⁶=π²√ 3 8

∂k

∂x_i(y), ∂Γ_θ

∂μ(0, y)=0,

∂²Γ_θ

∂y_i∂y_j(0, y)=π²√ 3 8

∂²k

∂x_i∂x_j(y), ∂²Γ_θ

∂μ∂y_j(0, y)=0,

∂²Γ_θ

∂μ²(0, y)=k_θ(y)1 18

|x|²(z_1,0)⁶=π²√ 3

8 k_θ(y). (5.1)

Using the Kelvin transformz_μ,y→ |x|⁻¹z_μ,y(x/|x|²), we see Γ_θ

(μ, y)

μ²+ |y|²₋1

=1 6

k_θ x/|x|²

z⁶_μ,y.

Consequently, we may extendΓ_θ to a function inC²(S⁴)by identifyingR×R³withS⁴\ {(0,−θ )}viaS(0,θ )and setting

Γ_θ

(0,−θ )

:=k(−θ )1 6

(z_1,0)⁶=π²√ 3 8 k(−θ ).

Hence there is a function Γ ∈C²(S⁴,R)such that Γ_θ is the functionΓ in stereographic coordinates centered at (0, θ )∈S⁴.

Remark 5.1.The functionΓ is related to the mapG, which was used in [10] to compute the degree of (1.1), via (up to an unimportant constant)

∂Γ_θ

∂y_i (μ, y)=μ⁻¹G(μ, y)i and ∂Γ_θ

∂μ(μ, y)=μ⁻¹G(μ, y)4,

when the set of conformal transformations ofS³is parametrized byμ∈(0,∞)andy∈R³using the coordinatesSθ. Obviously the factorμ⁻¹does not change the degree. We make use ofΓ as it is convenient to have a potential and because it fits perfectly in our perturbative approach.

By standard elliptic regularity the operatorLt, defined by L_t:ϕ→(−8S³+6)⁻¹

6

1+t k(θ ) ϕ⁵

,

is compact fromC²(S³)intoC²(S³). Under the assumptions of Theorem 1.1 for anyδ >0 there is a positive constant C_k,δsuch that the Leray–Schauder degree deg(Id−L_t,Bk,δ,0)is well-defined and independent oft∈ [δ,1](see [17]), where

Bk,δ:=

ϕ∈C² S³

: ϕ_C²_(S³₎< C_k,δandC_k,δ⁻¹< ϕ .

We will compute deg(Id−Lt,Bk,δ,0)for smalltusing the Melnikov functionΓ. To this end we use the transformation (1.3) and get for anyθ∈S³

deg(Id−L_t,Bk,δ,0)=deg

f_t,Rθ(Bk,δ),0 .

In the sequel we will denote by∂₀ the derivation with respect toμ and by∂_i the derivation with respect toy_i for 1i3. A direct calculation gives∂_iz_μ,y=μ⁻¹c_ξ(˙ξ_μ,y)_i, where

c_ξ=π3¹⁴√ 15

8 .

We first show that there is an open neighborhoodU of the equator{(0, θ ): θ∈S³}inS⁴such that all critical points of Γ inU lie on the equator.

Lemma 5.2.Under the assumptions of Theorem4.6there isδ₃=δ₃(k) >0such thatΓ(s, θ )=0for all(s, θ )∈U_δ3, where

U_δ3:=

(s, θ )∈S⁴: 0<dist (s, θ ),

(0, θ ): θ∈S³

< δ₃ . Proof. Fixθ∈S³. We have by (5.1)

Γ_θ(μ, y)=π²√ 3 8

k_θ(0)μ

∇kθ(0)+D²kθ(0)y

+o

μ²+ |y|² .

Hence, forθ /∈Athere isδ(θ ) >0 such thatΓ_θ(μ, y)=0 for all(μ, y)satisfying 0<|μ|< δand|y|< δ.

AsTz_μ,yZ=ker(f₀(zμ,y))by (2.5) we obtain for anyv∈Tz_μ,yZ^⊥ 5

(zμ,y)⁴v∂izμ,y= ∂izμ,y, v −f₀(zμ,y)∂izμ,yv=0 fori=0, . . . ,3. (5.2) Supposeθ∈A. From Lemma 4.1, the fact thatw(t, μ, y)andzμ,y are orthogonal to∂izμ,y, and (5.2) we see for

|μ| + |y|1 in coordinatesSθ

c_ξμ⁻¹α(t, μ, y) _i=f_t

z_μ,y+w(t, μ, y)

∂_iz_μ,y

= − 1+t k_θ(x)

z_μ,y+w(t, μ, y)5

∂_iz_μ,y

= −

(zμ,y)⁴zμ,y∂izμ,y−t

kθ(x)(zμ,y)⁵∂izμ,y−5

(zμ,y)⁴w(t, μ, y)∂izμ,y+O t²

= −t ∂_iΓ_θ(μ, y)+O t²

. (5.3)

We apply Lemma 4.8 and obtain from (5.3) Γ_θ(μ, y)d₂(δ) >0

for allδ < μ < δ₂and|y|δ₂, which gives the claim. 2

Thus, under the assumptions of Theorem 4.6 there isRΓ >0 independent ofθ∈S³such that (Γ_θ)⁻¹(0)∩(0,∞)×R³⊂Ω_RΓ,

whereΩRis defined by Ω_R:=

R⁻¹, R

×B_R(0).

Lemma 5.3.Under the assumptions of Theorem4.6we fix some pointϑ∈S³such that∇k(−ϑ )=0and use coor- dinatesSϑand the transformation in(1.3). Then there ist₁=t₁(k, ϑ ) >0andR₀=R₀(k, ϑ )R_Γ >0such that if 0< tt₁then any solutionu_t of (1.2)is of the formz_μ,y+w(t, μ, y), where(μ, y)∈(0, R0)×B_R0(0). Moreover, either(μ, y)∈Ω_R0 oru_t =Rϑ(ϕ^θ(μ,·))andt=t^θ(μ)for some0< μ < (R₀)⁻¹/2andθ∈M₀, whereϕ^θ andt^θ are given in Theorem4.6.

Proof. From the expansion oft^θ in Theorem 4.6 we may takeR˜₁>0 such that we have for allθ∈M₀and 0< μ (R˜₁)⁻¹

μ2|a₂(θ )|

|a₃(θ )|t^θ(μ). (5.4)

We first show that the solutions of (1.1), which do not belong to L₀:=

ϕ^θ(μ,·): 0< μ < (R˜₁)⁻¹, θ∈M₀ ,

remain bounded ast→0. Let(t_i, ϕ_i)be a sequence such thatϕ_i solves (1.1) witht=t_i, 0< ti→0 and ϕi_∞→ ∞ asi→ ∞.

The uniqueness part of Theorem 4.6 shows that(t_i, ϕ_i)equals(t^θ(μ_i), ϕ^θ(μ_i,·))for someθ∈M₀and(μ_i)tending to 0.

Consequently, there are two types of solutions to (1.1), the solutions which lie inL₀and blow up ast→0 and solutions which remain uniformly bounded above (and by Remark 5.4 below) ast→0. From our analysis above there ist˜₁>0 such that the solutions inL_b, defined by

L_b:=

ϕ: ϕsolves (1.1) for some 0< tt˜₁andϕ /∈L₀ , are uniformly bounded below and above.

Using the transformation in (1.3) with coordinatesSϑ there isR0>0 such that for any solutionut ∈Rϑ(Lb)of (1.2) with 0< t <t˜1

u_t−z_μ,y²_D1,2(R³)1

2z_1,0²_D1,2(R³) if(μ, y) /∈Ω_R0.

Due to the classical result of Caffarelli et al. [9], Gidas et al. [13] all positive solutions of (1.2) are given byZ. Hence, using the uniform bound, there exists 0< t1t˜1such that any solutionu_t∈Rϑ(L_b)of (1.2) with 0< t < t1satisfies dist(ut, Z) < ρ0, whereρ0is given in Lemma 4.1. Shrinkingt1if necessary, we see that for any solutionu_t∈Rϑ(L_b) of (1.2) with 0< t < t1there isz_μ,ysuch that(μ, y)∈Ω_R0 and

u_t−z_μ,yis orthogonal toZ_zμ,yZ, u_t−z_μ,yD^1,2(R³)< ρ₀. Forf_t(z_μ,y+u_t−z_μ,y)=0 the uniqueness in Lemma 4.1 yields for

ut=zμ,y+w(t, μ, y), where(μ, y)∈ΩR₀.

The solutions inRϑ(L₀)given by Theorem 4.6 are by construction of the formz_μ,y+w(t, μ, y), wherey(μ)→ Sϑ(θ )for someθ∈M₀asμ→0. EnlargingR₀depending on dist(−ϑ, M₀)and shrinkingt₁we infer from (5.4) that any solution inu_t∈Rϑ(L₀)of (1.2) with 0< t < t₁satisfies,

u_t=z_μ,y+w(t, μ, y), where 0< μ <1

2(R₀)⁻¹and|y|< R₀. This finishes the proof. 2