Conformal metrics on <span class="mathjax-formula">$ {\mathbb{R}}^{2m}$</span> with constant Q-curvature and large volume

Ann. I. H. Poincaré – AN 30 (2013) 969–982

www.elsevier.com/locate/anihpc

Conformal metrics on R ^2m with constant Q-curvature and large volume

Luca Martinazzi

Rutgers University, United States

Received 23 January 2012; accepted 21 December 2012 Available online 8 January 2013

Abstract

We study conformal metricsg_u=e^2u|dx|²onR^2mwith constantQ-curvatureQ_gu≡(2m−1)!(notice that(2m−1)!is the Q-curvature ofS^2m) and finite volume. Whenm=3 we show that there existsV^∗such that for any V ∈ [V^∗,∞)there is a conformal metricg_u=e^2u|dx|²onR⁶withQ_gu≡5!and vol(g_u)=V. This is in sharp contrast with the four-dimensional case, treated by C.-S. Lin. We also prove that whenmis odd and greater than 1, there is a constantV_m>vol(S^2m)such that for every V ∈(0, Vm] there is a conformal metricgu=e^2u|dx|² onR^2m withQg_u≡(2m−1)!, vol(g)=V. This extends a result of A. Chang and W.-X. Chen. Whenmis even we prove a similar result for conformal metrics ofnegativeQ-curvature.

©2013 Elsevier Masson SAS. All rights reserved.

Keywords:Q-curvature; Paneitz operators; GMJS operators; Conformal geometry

1. Introduction and statement of the main theorems

We consider solutions to the equation

(−)^mu=(2m−1)!e^2mu inR^2m, (1)

satisfying V :=

R^2m

e^2mu(x)dx <+∞, (2)

with particular emphasis on the role played byV.

Geometrically, ifusolves (1) and (2), then the conformal metricg_u:=e^2u|dx|²hasQ-curvatureQ_gu≡(2m−1)! and volumeV (by|dx|²we denote the Euclidean metric). For the definition ofQ-curvature and related remarks, we refer to Chapter 4 in[3]or to[8]and[9]. Notice that given a solutionuto (1) andλ >0, the functionv:=u−_2m¹ logλ solves

E-mail address:luca.martinazzi@math.rutgers.edu.

0294-1449/$ – see front matter ©2013 Elsevier Masson SAS. All rights reserved.

http://dx.doi.org/10.1016/j.anihpc.2012.12.007

(−)^mv=λ(2m−1)!e^2mv inR^2m,

R^2m

e^2mv(x)dx=V λ,

hence there is no loss of generality in the particular choice of the constant (2m−1)! in (1). On the other hand this constant has the advantage of being the Q-curvature of the round sphere S^2m. This implies that the function u1(x)=log_1+|²_x|2, which satisfiese^2u1|dx|²=(π⁻¹)^∗g_S2m (hereπ:S^2m→R^2mis the stereographic projection) is a solution to (1)–(2) withV =vol(S^2m). Translations and dilations (i.e. Möbius transformations) actually give us a large family of solutions to (1)–(2) withV =vol(S^2m), namely

u_x0,λ(x):=u₁

λ(x−x₀)

+logλ=log 2λ

1+λ²|x−x₀|², x₀∈R^2m, λ >0. (3) We shall call the functionsu_x0,λstandard orsphericalsolutions to (1)–(2).

The question whether the family of spherical solutions in (3) exhausts the set of solutions to (1)–(2) has raised a lot of interest and is by now well understood. W. Chen and C. Li[6]proved that onR²(m=1) every solution to (1)–(2) is spherical, while for everym >1, i.e. in dimension 4 and higher, it was proven by A. Chang and W.-X. Chen[4]that problems (1)–(2) admit solutions which are non-spherical. In fact they proved

Theorem A.(See A. Chang–W.-X. Chen, 2001,[4].) For everym >1andV ∈(0,vol(S^2m))there exists a solution to (1)–(2).

Several authors have tried to classify spherical solutions or, in other words, to give analytical and geometric condi- tions under which a solution to (1)–(2) is spherical (see[5,23,25]), and to understand some properties of non-spherical solutions, such as their asymptotic behavior, their volume and their symmetry (see[11,14,24]). In particular C.-S. Lin proved:

Theorem B.(See C.-S. Lin, 1998,[11].) Letusolve(1)–(2)withm=2. Then eitheruis spherical(i.e. as in(3))or V <vol(S⁴).

Both spherical solutions and the solutions given by Theorem A are radially symmetric (i.e. of the formu(|x−x₀|) for somex₀∈R^2m). On the other hand there also exist plenty of non-radial solutions to (1)–(2) whenm=2.

Theorem C.(See J. Wei and D. Ye, 2006,[24].) For everyV ∈(0,vol(S⁴))there exist(several)non-radial solutions to(1)–(2)form=2.

Remark D. Probably the proof of Theorem C can be extended to higher dimension 2m2, yielding several non- symmetric solutions to (1)–(2) for everyV ∈(0,vol(S^2m)), but failing to produce non-symmetric solutions forV vol(S^2m). As in the proof of Theorem A, the conditionV <vol(S^2m)plays a crucial role.

Theorems A, B, C and Remark D strongly suggest that also in dimension 6 and higher all non-spherical solutions to (1)–(2) satisfyV <vol(S^2m), i.e. (1)–(2) has no solution forV >vol(S^2m)and the only solutions withV =vol(S^2m) are the spherical ones. Quite surprisingly we found out that this is not at all the case. In fact in dimension 6 we found solutions to (1)–(2) with arbitrarily largeV:

Theorem 1.Form=3there existsV^∗>0such that for everyV V^∗there is a solutionuto(1)–(2), i.e. there exists a metric onR⁶of the formg_u=e^2u|dx|²satisfyingQ_gu≡5!andvol(gu)=V.

In order to prove Theorem1we will consider only rotationally symmetric solutions to (1)–(2), so that (1) reduces to and ODE. Precisely, givena, b∈Rletu=u_a,b(r)be the solution of

⎧⎪

⎪⎨

⎪⎪

⎩

³u= −e^6u inR⁶,

u(0)=u(0)=u(0)=u(0)=0, u(0)=^u(0)₆ =a,

u(0)=²₁₆^u(0) =b.

(4)

Here and in the following we will always (by a little abuse of notation) see a rotationally symmetric functionf both as a function of one variabler∈ [0,∞)(when writingf,f, etc.) and as a function ofx∈R⁶(when writingf, ²f, etc.). We also used that

f (0)=6f(0), ²f (0)=16f(0),

see e.g.[14, Lemma 17]. Also notice that in (4) we replaced 5!by 1 to make the computations lighter. As we already noticed, this is not a problem.

Theorem 2.Letu=u_a,3solve(4)for a givena <0andb=3.¹Then

R⁶

e^6ua,3dx <∞ for|a|sufficiently large; lim

a→−∞

R⁶

e^6ua,3dx= ∞. (5)

In particular the conformal metricg_ua,3=e^2ua,3|dx|²of constantQ-curvatureQ_gua,3≡1satisfies

a→−∞lim vol(gu_a,3)= ∞.

Theorem1will follow from Theorem2and a continuity argument (Lemma8below).

Going through the proof of Theorem A it is clear that it does not extend to the caseV >vol(S^2m). With a different approach, we are able to prove that, at least whenm3 is odd, one can extend Theorem A as follows.

Theorem 3.For everym3odd there existsV_m>vol(S^2m)such that for everyV ∈(0, Vm]there is a non-spherical solutionu to(1)–(2), i.e. there exists a metric on R^2m of the form g_u=e^2u|dx|² satisfyingQ_gu≡(2m−1)!and vol(gu)=V.

The conditionm3 odd is (at least in part) necessary in view of Theorem B and[6], but the casem4 even is open. Notice also that whenm=3, Theorems1and3guarantee the existence of solutions to (1)–(2) for

V ∈(0, Vm] ∪ V^∗,∞ ,

but we cannot rule out thatV_m< V^∗(the explicit value ofV_mis given in (37) below) and the existence of solutions to (1)–(2) is unknown forV ∈(V_m, V^∗). Could there be a gap phenomenon?

We now briefly investigate how large the volume of a metricg_u=e^2u|dx|²onR^2mcan be whenQ_gu≡const<0.

Again with no loss of generality we assumeQg_u≡ −(2m−1)!. In other words consider the problem

(−)^mu= −(2m−1)!e^2mu onR^2m. (6)

Although form=1 it is easy to see that problems (6)–(2) admit no solutions for anyV >0, whenm2 problems (6)–(2) have solutions for someV >0, as shown in[15]. Then with the same proof of Theorem3we get:

Theorem 4. For every m2 even there exists V_m>vol(S^2m)such that for V ∈(0, Vm]there is a solution u to (6)–(2), i.e. there exists a metric onR^2mof the formg_u=e^2u|dx|²satisfying

Q_gu≡ −(2m−1)!, vol(gu)=V .

The cases of solutions to (1)–(2) withmeven, or (6)–(2) andmodd seem more difficult to treat since the ODE corresponding to (1) or (6), in analogy with (4) becomes

^mu(r)=(2m−1)!e^2mu(r),

whose solutions can blow up in finite time (i.e. for finiter) if the initial data are not chosen carefully (contrary to Lemma5below).

1 The choiceb=3 is convenient in the computations, but any otherb >0 would work.

2. Proof of Theorem2

Set ω2m−1:=vol(S^2m−1)and let Br denote the unit ball inR^2m centered at the origin. Given a smooth radial functionf =f (r)inR^2mwe will often use the divergence theorem in the form

Br

f dx=

∂Br

∂f

∂νdσ=ω_2m−1r^2m−1f(r). (7)

Dividing byω_2m−1r^2m−1into (7) and integrating we also obtain f (t )−f (s)=

t s

1 ω2m−1ρ^2m−1

Bρ

f dx dρ, 0st. (8)

When no confusion can arise we will simply writeuinstead ofua,3orua,bto denote the solution to (4). In what follows, also other quantities (e.g. R,r0,r1,r2,r3,φ,ξ1,ξ2) will depend ona andb, but this dependence will be omitted from the notation.

Lemma 5.Given anya, b∈R, the solutionuto the ODE(4)exists for all times.

Proof. Applying (8) tof=²u, and observing that(²u)= −e^6u0 we get

²u(t )²u(s)²u(0)=16b, 0st, (9)

i.e.²u(r)is monotone decreasing. This and (8) applied touyield u(r)u(0)+

r 0

1 ω₅ρ⁵

Bρ

16b dx dρ=6a+ r 0

8

3bρ dρ=6a+4 3br². A further application of (8) toufinally gives

u(r) r 0

1 ω₅ρ⁵

B_ρ

6a+4

3b|x|²

dx dρ= r 0

aρ+ρ³b 6

dρ=a

2r²+ b

24r⁴=:φ(r). (10)

Similar lower bounds can be obtained by observing that−e^6u−1 foru0. This proves thatu(r)cannot blow-up in finite time and, by standard ODE theory,u(r)exists for everyr0. 2

Proof of (5) (completed). Fixb=3 and takea <0. The functionφ(r)=^a₂r²+¹₈r⁴vanishes forr=R=R(a):=

2√

−a. In order to prove (5) we shall investigate the behavior ofuin a neighborhood ofR. The heuristic idea is that u^{(j )}(0)=φ^{(j )}(0), for 0j5, ³φ≡0,

and for every ε >0 on [ε, R−ε] we haveφC_εa→ −∞ and|³u|e^Cεa→0 asa→ −∞, hence forr∈ [0, R−ε]we expectu(r)to be very close toφ(r). On the other hand,ucannot stay close toφforrmuch larger than Rbecause eventually−³u(r)will be large enough to make²u,uandunegative according to (8) (see Fig. 1).

Then it is crucial to show thatustays close toφfor somer > R(hence in a region whereφis positive and³uis not necessarily small) and long enough to make the second integral in (5) blow up asa→ −∞.

Step1:Estimates of u(R),u(R)and²u(R). From (10) we infer ³u= −e^6u−e^6φ,

which, together with (8), gives ²u(r)=²u(0)+

r 0

1 ω₅ρ⁵

B_ρ

³u dx dρ48− r 0

1 ω₅ρ⁵

B_ρ

e^{6φ (|x|)}dx dρ. (11)

Fig. 1. The functionsφ(r)=^a₂r²+¹₈r⁴(above) andu_a,3(r)φ(r).

We can explicitly compute (see Lemma 6 below and simplify (29) using that φ(R) =0 and ₋^√_3a

√3a e^t2dt = 2₋^√_3a

0 e^t2dt) R 0

1 ω5ρ⁵

Bρ

e^{6φ (|x|)}dx dρ= 1

48a +(18a²+1)√ 3 144a² e^−3a2

−√

3a 0

e^t2dt.

Then by (9) and Lemma7below we conclude that ²u(r)²u(R)48

1+O a⁻¹

for 0rR=2√

−a. (12)

Here and in the following|a^kO(a^−k)|C=C(k)asa→ −∞for everyk∈R. Then applying (8) as before we also obtain

u(r)6a+4 1+O

a⁻¹

r² for 0rR and

u(r)a

2r²+1+O(a⁻¹)

8 r⁴=φ(r)+O a⁻¹

r⁴ for 0rR.

Atr=Rthis reduces to u(R)O(a).

Step2:Behavior of u(r),u(r),²u(r)forrR. Definer₀(depending ona <0) as r₀:=inf

r >0:u(r)=0

∈ [R,∞].

We first claim thatr₀<∞. We have by Lemmas6and7

B_R

e^6φdx=ω5

−4a

3 +4(6a²−1)√ 3 9 e^−3a2

−√

3a 0

e^t2dt

=O(a). (13)

Since onB_r0 we haveu0, hence³u−1, using (7)–(8) and (13) we get forr∈ [R, r₀]

²u(r)²u(R)− r R

1 ω₅ρ⁵

B_R

e^6φdx+

B_ρ\B_R

1dx

dρ

48+O(a) 1

R⁴− 1 r⁴

− r R

ρ⁶−R⁶

6ρ⁵ dρ. (14)

Assumingr∈ [R,2R]we can now bound with a Taylor expansion 1

R⁴− 1

r⁴=R⁻⁴O˜ r−R

R

(15) and

ρ⁶−R⁶r⁶−R⁶=R⁶O˜ r−R

R

, forρ∈ [R, r], which together with (15) yields

r R

ρ⁶−R⁶ 6ρ⁵ dρ

r R

r⁶−R⁶

6ρ⁵ dρR²O˜

r−R R

2

, (16)

where for anyk∈Rwe have|t^−kO(t˜ ^k)|C=C(k)uniformly for 0t1. Using (15) and (16) we bound in (14) ²u(r)48+O

a⁻¹O˜ r−R

R

+R²O˜

r−R R

2

, r∈ R,min{r₀,2R} ,

whence

²u(r)48+O a⁻¹

+R²O˜

r−R R

2

χ_(R,∞)(r), r∈ 0,min{r₀,2R} ,

whereχ_(R,∞)(r)=0 forr∈ [0, R]andχ_(R,∞)(r)=1 forr > R. Then with (8) we estimate forr∈ [0,min{r₀,2R}]

u(r)6a+4 1+O

a⁻¹

r²+χ_(R,∞)(r) r R

1 ω₅ρ⁵

Bρ\BR

R²O˜

|x| −R R

2 dx dρ

=6a+4 1+O

a⁻¹

r²+R⁴O˜

r−R R

4

χ(R,∞)(r) (17)

and

u(r)a

2r²+1+O(a⁻¹)

8 r⁴+χ(R,∞)(r) r R

1 ω₅ρ⁵

B_ρ\B_R

R⁴O˜

|x| −R R

4 dx dρ

=φ(r)+O a⁻¹

r⁴+R⁶O˜

r−R R

6

χ_(R,∞)(r), (18)

where the integrals in (17) and (18) are easily estimated bounding|x|withrand applying (16).

Making a Taylor expansion ofφ(r)atr=Rand using thatφ(R)=0, we can further estimate the right-hand side of (18) forr∈ [R,min{r₀,2R}]as

u(r)φ(R)(r−R)+R²O˜

r−R R

2 +O

a⁻¹

r⁴+R⁶O˜

r−R R

6

= −aR(r−R)+O a⁻¹

R⁴+R²O˜

r−R R

2 +R⁶O˜

r−R R

6

=:ψa(r).

Now choosingr=R(1+1/√

−a ), so that(r−R)/R→0 asa→ −∞, we get

a→−∞lim ψ_a

R(1+1/√

−a )

lim

a→−∞

4(−a)³² +O(a)−C

= ∞. In particular

r0∈ R, R(1+1/√

−a ) . We now claim that

a→−∞lim u(r0)= ∞. (19)

Indeed we infer from (17) u(r₀)6a+4

1+O a⁻¹

r₀²−C6a+4 1+O

a⁻¹

R²−C−10a−C, for−alarge enough, whence (19). Set

r₁=r₁(a):=inf

r > r₀: u(r)=0 .

Applying (7) to (17), and recalling that ^r0−_R^R√1_a, similar to (18) we obtain u(r₀)ar₀+1+O(a⁻¹)

2 r₀³−Car₀+1+O(a⁻¹)

2 r₀R²−C−ar₀−C.

In particular for−alarge enough we haveu(r₀) >0, which impliesr₁> r₀. Using (7)–(8) and that³u(r)−1 for r∈ [r₀, r₁], it is not difficult to see thatr₁<∞. Moreover there exists at least a pointr₂=r₂(a)∈(r₀, r₁]such that u(r₂)0, which in turn implies that

u(r₃) <0 for somer₃=r₃(a)∈(r₀, r₂], (20)

since otherwise we would have by (7) u(r₂)= 1

ω₅r₂⁵

Br0

u dx+ 1 ω₅r₂⁵

Br2\Br0

u dxr₀⁵

r₂⁵u(r₀) >0, contradiction.

Step3:Conclusion. We now use the estimates obtained in Steps 1 and 2 to prove (5).

From (8), (19) and (20) we infer

a→−∞lim

r₃

r₀

1 ω5r⁵

Br

²u dx dr= lim

a→−∞

u(r₃)−u(r₀)

= −∞, (21)

hence by the monotonicity of²u(r)(see (9))

a→−∞lim ²u(r₃)

r₃²−r₀²

= −∞. (22)

We now claim that

a→−∞lim

Br3

e^6udx= ∞. (23)

Indeed consider on the contrary an arbitrary sequencea_k with limk→∞a_k= −∞and

k→→∞lim

Br3

e^6udx <∞, (24)

where herer₃andudepend ona_kinstead ofaof course. Sinceu0 inB_r3\B_r0we have

B_r3

e^6udx

B_r3\B_r0

1dx=ω₅ 6

r₃⁶−r₀⁶ .

Now observe that(r₃⁶−r₀⁶)(r₃²−r₀²)r₀⁴to conclude that (24) implies

klim→∞

r₃²−r₀²

lim

k→∞

r₃⁶−r₀⁶

r₀⁴ =0. (25)

Then (8), (12) and (22) yield r₃²−r₀^2r3

R

1 ω5r⁵

Br

e^6udx dr=

r₃²−r₀²

²u(R)−²u(r₃) −²u(r₃)

r₃²−r₀²

→ ∞ ask→ ∞. By (25) we also have

klim→∞

r₃

R

1 ω5r⁵

Br

e^6udx dr= ∞,

which implies at once

klim→∞

Br3

e^6udx lim

k→∞4R⁴ω₅

r3

R

1 ω₅r⁵

Br3

e^6udx dr= ∞, contradicting (24). Then (23) is proven.

It remains to show that

R⁶

e^6udx <∞,

at least for−alarge enough. It follows from (22) and the monotonicity of²uthat for−alarge enough we have

²u(r) < B <0, forrr3, (26)

and, using (7)–(8) as already done several times, we can findrar3such that (u)(r) <B

6r, u(r) < B

12r², u(r) < B

96r³, u(r) < B

384r⁴, forrr_a. (27) Then

R⁶

e^6udx

B_ra

e^6udx+

R⁶\B_ra

e^64B|x|2dx <∞,

as wished. 2

2.1. Two useful lemmas

We now state and prove two lemmas used in the proof of Theorem2. Their proof is based on elementary calculus, but we provide it for completeness and because Lemma6in particular was crucial for the estimates of the previous section.

Lemma 6.Forφ(r)=^a₂r²+¹₈r⁴,a0, we have

Br

e^{6φ (|x|)}dx=ω₅ 2

3a+1 3e^{6φ (r)}

−2a+r²

+(12a²−2)√ 3 9 e^−3a2

−√

3a

−√ 3(a+r²/2)

e^t2dt

=:ξ₁(r) (28)

and r 0

1 ω₅ρ⁵

Bρ

e^{6φ (|x|)}dx dρ

=−2a−e^{6φ (r)}(−2a+r²)

12r⁴ +(2−12a²+3r⁴)√ 3 36r⁴ e^−3a2

−√

3a

−√ 3(a+r²/2)

e^t2dt:=ξ₂(r). (29)

Proof. Patiently differentiating, using thate^−3a2_dr^d ₋^√3a

−√

3(a+r²/2)e^t2dt=√

3re^{6φ (r)}, one sees that ξ₁(r)=ω₅r⁵e^{6φ (r)}, ξ₂(r)=ξ1(r)

ω₅r⁵.

Using thatφ(0)=0 it is also easy to see thatξ1(0)=0.

Sinceξ2(0)is not defined, we will compute the limit ofξ2(r)asr→0. We first compute the Taylor expansions e^{6φ (r)}=1+3ar²+3

4

1+6a²

r⁴+r⁴o(1), and

√3e^−3a2

−√

3a

−√ 3(a+r²/2)

e^t2dt=3 2r²+9

4ar⁴+r⁴o(1), with errorso(1)→0 asr→0. Then

−2a−e^{6φ (r)}(−2a+r²)

12r⁴ =(1−6a²)r²+(³₂a−9a³)r⁴

12r⁴ +o(1)

= −(2−12a²+3r⁴)√ 3 36r⁴ e^−3a2

−√

3a

−√ 3(a+r²/2)

e^t2dt+o(1),

witho(1)→0 asr→0. Hence limr→0ξ2(r)=0. 2 Lemma 7.We have

rlim→∞re^−r2 r 0

e^t2dt=1

2. (30)

Proof. This is a simple calculus exercise. For instance one notices that (30) is equivalent to

rlim→∞re^−r2 r 2

e^t2dt=1

2, (31)

which follows integrating by parts twice and noticing thatre^−r2_r

2t⁻⁴e^t2dt→0 asr→ ∞. 2

3. Proof of Theorem1

We start with the following lemma.

Lemma 8.Set V (a)= 1

5!

R⁶

e^6ua,3dx

whereu=u_a,3is the solution to(4)for givena <0andb=3. Then there existsa^∗<0such thatV is continuous on (−∞, a^∗].

Proof. It follows from (21) and the monotonicity of²uthat we can fix−a^∗so large that

rlim→∞²ua,3(r) <0, for everyaa^∗.

Fix nowε >0. Givenaa^∗it is not difficult to findr_a>0 andB=B(a) <0 such that

²u_a,3(r) < B <0, forrr_a (32)

and, possibly choosingralarger, using (7)–(8) as already done in the proof of Theorem2, we get (u_a,3)(r) <B

6r, u_a,3(r) < B

12r², u_a,3(r) < B

96r³, u_a,3(r) < B

384r⁴, forrr_a. (33) By possibly choosingraeven larger we can also assume that

R⁶\B_ra

e^64B|x|4dx <ε

2. (34)

By ODE theory the solutionua,3to (4) is continuous with respect toa inC_loc^k (R⁶)for everyk0, in the sense that for any r>0,u_a,3→u_a,3 inC^k(B_r)as a→a. In particular we can findδ >0 (depending on ε) such that if

|a−a|< δthen (32)–(33) witha replaced bya are still satisfied forr=r_a(notr_a) and (32) holds also for every r > r_asince²u_a,3(r)is decreasing inr(see (9)). Then, with (7)–(8) we can also get the bounds in (33) for every rr_a(andu_a,3instead ofu_a,3). For instance

(u_a,3)(r)= 1 ω₅r⁵

B_r

³u_a,3dx= r_a

r 5

(u_a,3)(ra)+ 1 ω₅r⁵

B_r\B_ra

²u_a,3dx

<

r_a r

5Br_a

6 +B(r⁶−r_a⁶) 6r⁵ =B

6r.

Furthermore, up to takingδ >0 even smaller, we can assume that

B_ra

e^6ua,3dx−

B_ra

e^6ua,3dx <ε

2. (35)

Finally, the last bound in (33) and (34) imply at once

R⁶\B_ra

e^6ua,3dx−

R⁶\B_ra

e^6ua,3dx <ε

2, which together with (35) completes the proof. 2

Proof of Theorem1(completed). SetV^∗=V (a^∗), wherea^∗is given by Lemma8. By Lemma8, Theorem2and the intermediate value theorem, for everyV V^∗there existsaa^∗such that

1 5!

R⁶

e^6ua,3dx=V ,

hence the metricgu_a,3=e^2ua,3|dx|²has constantQ-curvature equal to 1 and vol(gu_a,3)=5!V. Applying the transfor- mation

u=u_a,3−1 6log 5!

it follows at once that the metricg_u=e^2u|dx|²satisfies vol(gu)=V andQ_gu≡5!, henceusolves (1)–(2). 2 4. Proof of Theorems3and4

Whenf:Rⁿ→Ris radially symmetric we havef (x)=f(|x|)+ⁿ_|⁻_x|¹f(|x|). In particular we have

^mr^2m=2^2mm(2m−1)! inR^2m. (36)

Form2 andb0 letu_bsolve

⎧⎨

⎩

^mu_b= −(2m−1)!e^2mub inR^2m,

u^{(j )}_b (0)=0 for 0j2m−1, j=2m−2, u^(2m_b ⁻²⁾=b.

From (7)–(8) it follows thatu₀0, hence^mu₀−(2m−1)!. We claim that u₀(r)ψ (r):= − r^2m

2^2mm. Indeed according to (36)ψsolves

^mψ= −(2m−1)!^mu₀ inR^2m and

ψ^{(j )}(0)=0=u^{(j )}₀ (0) for 0j2m−1, which implies

^jψ (0)=0=^ju₀(0) for 0jm−1,

see[14, Lemma 17]. Then the claim follows from (7)–(8) and a simple induction.

Now integrating we get

R^2m

e^2mu0dx

R^2m

e^2mψdx=ω_2m−1 ∞

0

r^2m−1exp

− r^2m 2^2m−1

dr=2^2m−2ω2m−1

m =:V_m.

Using the formulas ω2m−1=vol

S^2m−1

= 2π^m

(m−1)!, ω2m=vol S^2m

=2^2m(m−1)!π^m

(2m−1)! , m1 we verify

V_m= (2m)!

4(m!)²ω_2m, V₂ ω₄=3

2 >1, V_m+1 ω_2m+2

V_m ω_2m

₋1

=(2m+2)(2m+1)

(m+1)² >1, (37)

hence by induction Vm>vol

S^2m

form2. (38)

With the same argument used to prove Lemma8we can show that the function V (b):=

R^2m

e^6ubdx, b∈(−∞,0]

is finite and continuous. Indeed it is enough to replace (32) with ^m−1u_b(r)B <0 forrr_b,

and (33) with ^m−1−jub

(r) < Cm,jBr^2j−1, ^m−1−jub(r) < Dm,jBr^2j, forrrb, 1jm−1 wherer_bis chosen large enough and

Cm,1= 1

2m, Dm,j=C_m,j

2j , Cm,j+1= D_m,j 2m+2j, whence

C_m,j= (m−1)!

2^2j−1(j−1)!(m+j−1)!, D_m,j= (m−1)! 2^2jj!(m+j−1)!.

Moreover, using that^m−1u_b(0)=C_mbfor some constantC_m>0,^mu_b(r)0 forr0 and (7)–(8) as before, we easily obtain

u_b(r)E_mbr^2m−2, (39)

whereE_m:=C_mC_m,m−1>0, hence

b→−∞lim V (b) lim

b→−∞

R⁶

e^{6Emb|x|2m−2}dx=0.

By continuity we conclude that for everyV ∈(0, Vm]there existsb0 such thatu=u_bsolves (1)–(2) ifmis odd or (6)–(2) ifm is even. Taking (38) into account it only remains to prove that the solutionsu_b corresponding to V =vol(S^2m)is not a spherical one. This follows immediately from (39), which is not compatible with (3). 2 5. Applications and open questions

5.1. Possible gap phenomenon

Theorems1and3guarantee that form=3 there exists a solution to (1)–(2) for everyV ∈(0, V3] ∪ [V^∗,∞), with possiblyV₃< V^∗. Could it be that for someV ∈(V₃, V^∗)problems (1)–(2) admit no solution?

If we restrict to rotationally symmetric solutions, some heuristic arguments show that the volume of a solution to (4), i.e. the function

V (a, b):=

R⁶

e^6ua,b(|x|)dx

need not be continuous for all(a, b)∈R², hence the image of the functionV might not be connected.

5.2. Higher dimensions and negative curvature

It is natural to ask whether Theorems1and2generalize to the casem >3 or whether an analogous statement holds whenm2 and (6) is considered instead of (1). Since the sign on the right-hand side of the ODE (4) plays a crucial role, we would expect that part of the proof of Theorem2can be recycled for (1) whenm5 is odd, or for (6) when mis even.

For instance letu_a=u_a(r)be the solution inR⁴of

⎧⎨

⎩

²ua= −6e^4ua,

ua(0)=u_a(0)=u_a(0)=0, u_a(0)=a.

It should not be difficult to see thatua(r)exists for allr0 and that

R⁴e^4ua(|x|)dx <∞. Do we also have

a→+∞lim

R⁴

e^4ua(|x|)dx= ∞?

5.3. Non-radial solutions

The proof of Theorem C cannot be extended to provide non-radial solutions to (1)–(2) form3 andV vol(S^2m), but it is natural to conjecture that they do exist.

5.4. Concentration phenomena

The classification results of the solutions to (1)–(2),[6,11,25]and[14], have been used to understand the asymp- totic behavior of unbounded sequences of solutions to the prescribed Gaussian curvature problem on 2-dimensional domains (see e.g.[2]and[10]), onS²(see[21]) and to the prescribedQ-curvature equation in dimension 2m(see e.g.

[7,12,13,18–20,16,17]).

For instance consider the following model problem. LetΩ⊂R^2mbe a connected open set and consider a sequence (u_k)of solutions to the equation

(−)^mu_k=Q_ke^2muk inΩ, (40)

where

Q_k→Q₀ inC_loc¹ (Ω), lim sup

k→∞

Ω

e^2mukdx <∞, (41)

with the following interpretation:g_k:=e^2uk|dx|²is a sequence of conformal metrics onΩwithQ-curvaturesQ_gk= Q_kand equibounded volumes.

As shown in[1]unbounded sequences of solutions to (40)–(41) can exhibit pathological behaviors in dimension 4 (and higher), contrary to the elegant results of[2]and[10]in dimension 2. This is partly due to Theorem A. In fact form2 andα∈(0, (2m−1)!vol(S^2m)]one can found a sequence(u_k)of solutions to (40)–(41) withQ₀>0 and

Rlim→0 lim

k→∞

BR(x0)

|Q_k|e^2mukdx=α for somex₀∈Ω. (42)

Form=2 this was made very precise by F. Robert[19]in the radially symmetric case. In higher dimension or when Q0is not necessarily positive, thanks to Theorems1–4we see thatαcan take values larger than(2m−1)!vol(S^2m).

Indeed if u is a solution to (1)–(2) or (6)–(2), then uk :=u(kx)+logk satisfies (40)–(41) with Ω=R^2m, Qk ≡

±(2m−1)!and

|Q_k|e^2mukdx (2m−1)!V δ₀, weakly as measures.

Whenm=2, Q0>0 (say Q0≡6) it is unclear whether one could have concentration points carrying more Q- curvature than 6 vol(S⁴), i.e. whether one can take α >6 vol(S⁴)in (42). Theorem B suggests that if the answer is affirmative, this should be due to the convergence to the same blow-up point of two or more blow-ups. Such a phenomenon is unknown in dimension 4 and higher, but was shown in dimension 2 by Wang[22]with a technique which, based on the abundance of conformal transformations ofCinto itself, does not extend to higher dimensions.