Classification of solutions to the higher order Liouville's equation on R^{2m}

0.28 MB·English

by  Luca Martinazzi

Classification of solutions to the higher order
Liouville’s equation on R2m
Luca Martinazzi
8
February 2, 2008
0
0
2
Abstract
n
a We classify the solutions to the equation (−∆)mu = (2m−1)!e2mu
J on R2m giving rise to a metric g =e2ugR2m with finite total Q-curvature
7
in terms of analytic and geometric properties. The analytic conditions
1
involve the growth rate of u and the asymptotic behaviour of ∆u(x) as

|x|→∞. Asaconsequencewegiveageometriccharacterization interms
P] of the scalar curvature of the metric e2ugR2m at infinity, and we observe
A that the pull-back of this metric to S2m via the stereographic projection
canbeextendedtoasmoothRiemannianmetricifandonlyifitisround.
.
h
t
a
1 Introduction and statement of the main the-
m
orems
[

1
The study of the Paneitz operators has moved into the center of conformal
v
9 geometry in the last decades, in part with regardto the problem of prescribing
2 the Q-curvature. Given a 4-dimensional Riemannian manifold (M,g), the Q-
7 curvature Q4 and the Paneitz operator P4 have been introduced by Branson-
g g
2
Oersted [BO] and Paneitz [Pan]:
.
1
0 1
Q4 := − ∆ R −R2+3|Ric |2
8 g 6 g g g g
0 (cid:0) 2 (cid:1)
v: Pg4(f) := ∆2gf +div 3Rgg−2Ricg df, ∀f ∈C∞(M),
(cid:16) (cid:17)
i
X where R and Ric denote the scalar and Ricci curvatures of g. Higher order
g g
r Q-curvatures Qn and Paneitz operators Pn have been introduced in [Bra] and
a
[GJMS]. Their interest lies in their covariant nature: considering in dimension
2m the conformal metric g :=e2ug, we have
u
P2m =e−2muP2m, P2mu+Q2m =Q2me2mu, (1)
gu g g g gu
see for instance [Cha] Chapter 4. The last identity is a generalized version of
Gauß’s identity: in dimension 2
−∆ u+K =K e2u,
g g gu
where K is the Gaussian curvature, and ∆ is the Laplace-Beltrami operator
g g
withtheanalysts’sign. Indeed,indimension2wehaveP2 =−∆ andQ2 =K .
g g g g
1
Moreover∆ =e−2u∆ . AnotherinterestingfactisthatthetotalQ-curvature
gu g
is a global conformal invariant: if M is closed and 2m-dimensional,
Q2mdvol = Q2mdvol .
Z gu gu Z g g
M M
Further evidence of the geometric relevance of the Q-curvatures is givenby the
Gauss-Bonnet-Chern’stheorem[Che]: onalocallyconformallyflatclosedmani-
fold of dimension 2m, since Q2m is a multiple of the Pfaffian plus a divergence
g
term (see [BGP]), we have
Q2mdvol =[(2m−2)!!]2vol(S2m−1)χ(M),
Z g g
M
where χ(M) is the Euler-Poincar´echaracteristic of M.
HereweareinterestedinthespecialcasewhenM isR2m withtheEuclidean
metric gR2m. In this case we simply have Pg2Rm2m = (−∆)m and Q2gRm2m ≡ 0. We
consider solutions to the equation
(−∆)mu=(2m−1)!e2mu on R2m, (2)
satisfying e2mudx < ∞. From the above remarks and (1) in particular, it
R2m
follows thaRt (2) has the following geometric meaning: if u solves (2), then the
conformal metric g := e2ugR2m has Q-curvature Q2gm ≡ (2m−1)!. As we shall
see, every solution to (2) with e2mu ∈L1 (R2m) is smooth (Corollary 8).
loc
Given such a solution u, define the auxiliary function
(2m−1)! |y|
v(x):= log e2mu(y)dy, (3)
γm ZR2m (cid:18)|x−y|(cid:19)
whereγ isdefined bythe followingproperty: (−∆)m 1 log 1 =δ inR2m,
m γm |x| 0
see Proposition 22 below. Then (−∆)mv =(2m−1)!e(cid:0)2mu. We p(cid:1)rove
Theorem 1 Let u be a solution of (2) with
1
α:= e2mu(x)dx<+∞. (4)
|S2m|ZR2m
Then
u(x)=v(x)+p(x), (5)
where p is a polynomial of even degree at most 2m−2, v is as in (3) and
sup p(x) < +∞,
x∈R2m
lim ∆jv(x) = 0, j =1,...,m−1,
|x|→∞
v(x) = −2αlog|x|+o(log|x|), as |x|→+∞.
It is well known that the function
2λ
u(x):=log (6)
1+λ2|x−x |2
0
2
solves (2) and (4) with α = 1 for any λ > 0, x ∈ R2m. We call the func-
0
tions of the form (6) standard solutions. They all arise as pull-back under the
stereographic projection of metrics on S2m which are round, i.e. conformally
diffeomorphic to the standard metric. A. Chang and P. Yang [CY] proved that
the round metrics are the only metrics on S2m having Q-curvature identically
equal to (2m−1)!.
In the next theorem we give conditions under which an entire solution of
Liouville’s equation satisfying (4) is necessarily a standard solution.
Theorem 2 Let u be a solution of (2) satisfying (4). Then the following are
equivalent:
(i) u is a standard solution,
(ii) lim ∆u(x)=0
|x|→∞
(ii’) lim ∆ju(x)=0 for j =1,...,m−1,
|x|→∞
(iii) u(x)=o(|x|2) as |x|→∞,
(iv) degp=0, where p is the polynomial in (5).
(v) liminf|x|→+∞Rgu >−∞, where gu =e2ugR2m.
(vi) π∗g can be extended to a Riemannian metric on S2m, where π :S2m →
u
R2m is the stereographic projection.
Moreover, if u is not a standard solution, there exist 1 ≤ j ≤ m −1 and a
constant a<0 such that
∆ju(x)→a as |x|→+∞. (7)
The 2-dimensionalcase (m=1) of Theorem 2 was treated by W. Chen and
C.Lin[CL],whoprovedthatevery solutionwithfinitetotalGaussiancurvature
is a standardone. The 4-dimensionalcasewastreatedby C-S.Lin[Lin], witha
classification of u in terms of its growth, or of the behaviour of ∆u at ∞. The
classification of C-S. Lin in terms of ∆u was used by F. Robert and M. Struwe
[RS] to study the blow-up behaviour of sequences of solutions u to
k
∆2uk =λuke32π2u2k in Ω⊂R4
(cid:26) u = ∂uk =0 on ∂Ω,
k ∂n
and by A. Malchiodi [Mal] to show a compactness criterion for sequences of
solutions u to the equation
k
P4u +Q4 =h e4uk, h constant
g k k k k
on a closed 4-manifold. The same criterion could be used in higher dimension
in the proof of an analogous compactness result. This was observed by C. B.
Ndiaye [Ndi], who then used a different technique to show compactness. We
will discuss this in a forthcoming paper.
In higher dimension (m>2), J. Wei and X. Xu [WX] treated a special case
ofTheorem2: if u(x)=o(|x|2)atinfinity, thenu isalwaysastandardsolution.
This result is not sufficient to prove compactness. Moreover,the proof appears
3
to be overly simplified. For instance, in their Lemma 2.2 the argument for
showingthatu≤C isnotconclusive,andinthecrucialLemma2.4theysimply
refer to [Lin] for details. This latter lemma corresponds to Lemma 13 here and
it is the main regularity result, as it implies that u≤C, hence that the volume
of the metric e2ugR2m cannot concentrate in small balls. Its generalization is a
major issue, because Lin’s analysis is focused on the function ∆u, and it makes
useoftheHarnack’sinequalityandofthefactthat∆(u−v)≡C. Inthegeneral
case, Harnack’s inequality does not work and there are no uniform bounds for
∆(m−2)(u−v) (while it is still true that ∆(m−1)(u−v)≡C). To overcomethis
difficulties,wespendafewpagesinthefollowingsectiontostudypolyharmonic
functions. As a reward we obtain a Liouville-type theorem for polyharmonic
functions (Theorem 6) which allows us to make the proof of [Lin] more direct
and transparent.
The characterization in terms of the scalar curvature at infinity is new and
quiteinteresting,asitshowsthatnon-standardsolutionshaveageometryessen-
tially different from standard solutions, and it also shows that the Q-curvature
and the scalar curvature are independent of each other in dimension 4 and
higher. On the other hand, since in dimension 2 we have 2Q = R , (v) is
g g
consistent with the result of [CL].
Thecharacterizationin(vi)impliestheresultofA.ChangandP.Yang[CY]
described above, which here follows from the general case.
The paper is organized as follows. In Section 2 we collect some relevant
results about polyharmonic functions which will be needed later. Section 3
containstheproofofTheorems1and2;attheendofthepaperwegiveexamples
to show that the hypothesis of Theorem 2 are sharp in terms of the growth at
infinity and ofthe degreeof p. Recently J. C. Wei and D. Ye [WY] provedthat
already in dimension 4 there is a great abundance of non-radially symmetric
solutions.
In the following, the letter C denotes a generic constant, which may change
from line to line and even within the same line.
Aknowledgments
I wish to thank my advisor, Prof. M. Struwe for stimulating discussions and
for introducing me to this very interesting subject. I also thank my friend D.
Saccavino for referring me to the result of Gorin, which we use in Lemma 11.
2 A few remarks on polyharmonic functions
We briefly recallsome propertiesof polyharmonicfunctions, whichwill be used
in the sequel. For the standard elliptic estimates for the Laplace operator, we
refer to [GT] or [GM]. The next lemma can be considered a generalized mean
value inequality. We give the short prooffor the convenienceof the reader,and
because identity (12) will be used in the next section.
Lemma 3 (Pizzetti [Piz]) Let ∆mh = 0 in B (x ) ⊂ Rn, for some m,n
R 0
4
positive integers. Then
m−1
h(z)dz = c R2i∆ih(x ), (8)
i 0
Z
Xi=0
BR(x0)
where
n (n−2)!!
c =1, c = , i≥1. (9)
0 i
n+2i(2i)!!(2i+n−2)!!
Proof. We can translate and assume that x = 0. We first prove by induction
0
on m that there are constants b(m),...,b(m) such that
0 m−1
m−1
h(z)dS = b(m)r2i∆ih(0), 0<r <R, B :=B (0). (10)
Z i r r
Xi=0
∂Br
For m = 1 this reduces to the mean value theorem for harmonic functions.
Assumenowthattheassertionhasbeenproveduptom−1,andthat∆mh=0.
Let G be the Green function of ∆m in B :
r r
∆mG =δ in B , G =∆G =...=∆m−1G =0 on ∂B . (11)
r 0 r r r r r
For simplicity, let us only consider the case n=2m. Then G (x)=G x ,
r 1 r
(cid:0) (cid:1)
G (x)=α log|x|+α |x|2+...+α |x|2m−2,
1 0 1 m−1
where the constants can be computed inductively starting with α up to α
0 m−1
in order to satisfy (11). Notice that G is radial. Integrating by parts
1
0 = G ∆mhdx
r
Z
Br
m−1 ∂∆m−1−iG
= h(0)− r∆ihdS (12)
Z ∂n
Xi=0 ∂Br
m−1
= h(0)− a r2i∆ihdS,
i
Z
Xi=0
∂Br
where each a depends only on n and m. For each term on the right-hand side
i
with i≥1, we can use the inductive hypothesis
m−i−1
r2i ∆ihdS =r2i b(m−1)r2j∆j+ih(0), 0≤i≤m−1,
Z j
Xj=0
∂Br
and substituting we obtain (10). To conclude the induction it is enough to
multiply (10) by rn−1, integrate with respect to r from 0 to R and divide by
Rn.
n
Tocomputethec ’s,wetestwiththefunctionsh(x)=r2i :=|x|2i,i≥1(for
i
the case i=0 use the function h(x)≡1). Since ∆r2i =2i(2i+n−2)r2i−2, we
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have that ∆kh(0) = 0 for k 6= i and ∆ih(0) = (2i)!!(2i+n−2)!!. Hence Pizzetti’s
(n−2)!!
formula reduces to
(2i)!!(2i+n−2)!! n
c R2i = r2idx= R2i,
i
(n−2)!! Z n+2i
BR
whence (9). (cid:3)
Remark. From (12), moreover,for an arbitrary C2m-function u it follows that
m−1
u(z)dz = c R2i∆iu(x )+c R2m∆mu(ξ), (13)
i 0 m
Z
Xi=0
BR(x0)
for some ξ ∈B (x ). •
R 0
Proposition 4 Let ∆mh = 0 in B ⊂ Rn. For every 0 ≤ α < 1, p ∈ [1,∞)
4
and k ≥0 there are constants C(k,p),C(k,α) independent of h such that
khk ≤ C(k,p)khk
Wk,p(B1) L1(B4)
khk ≤ C(k,α)khk .
Ck,α(B1) L1(B4)
The proof of Proposition 4 is given in the appendix. As a consequence
of Proposition 4 and Pizzetti’s formula we have the following Liouville-type
theorem, compare [ARS].
Theorem 5 Consider h : Rn → R with ∆mh = 0 and h(x) ≤ C(1+|x|ℓ), for
some ℓ≥2m−2. Then h(x) is a polynomial of degree at most ℓ.
Proof. Thanks to Proposition 4, we have for any x∈Rn
C C
|Dℓ+1h(x)|≤ |h(y)|dy =− h(y)dy+O(R−1), as R→∞.
Rℓ+1 Z Rℓ+1 Z
BR(x) BR(x)
(14)
On the other hand, Pizzetti’s formula implies that
m−1
h(y)dy = c R2i∆ih(x)=O(R2m−2),
i
Z
Xi=0
BR(x)
and letting R→∞, we obtain Dℓ+1h=0. (cid:3)
A variant of the above theorem, which will be used later is the following
Theorem 6 Consider h : Rn → R with ∆mh = 0 and h(x) ≤ u−v, where
epu ∈ L1(Rn) for some p > 0, v ∈ L1 (Rn) and −v(x) ≤ C(log(1+|x|)+1).
loc
Then h is a polynomial of degree at most 2m−2.
6
Proof. The only thing to change in the proof of Theorem 5, is the estimate of
the term 2C – h+dy, corresponding to the O(R−1) in (14). We have
R2m−1 BR(x)
R
h+dy ≤ u+dy+C log(1+|y|)dy+C
Z Z Z
BR(x) BR(x) BR(x)
1
≤ epudy+ClogR+C,
p Z
BR(x)
and all terms go to 0 when divided by R2m−1 and for R→∞. (cid:3)
The following estimate has been obtained by Br´ezis and Merle [BM] in di-
mension 2 and by C.S. Lin [Lin] and J. Wei [Wei] in dimension 4. Notice that
the constant γ , defined by the relation
m
1 1
(−∆)m log =δ , in R2m
0
(cid:18)γ |x|(cid:19)
m
(see Proposition 22 in the appendix), plays an important role.
Theorem 7 Let f ∈L1(B (x )) and let v solve
R 0
(−∆)mv =f in B (x )⊂R2m,
R 0
(cid:26) v =∆v =...=∆m−1v =0 on ∂BR(x0).
Then, for any p∈ 0, γm , we have e2mp|v| ∈L1(B (x )) and
(cid:16) kfkL1(BR(x0))(cid:17) R 0
e2mp|v|dx≤C(p)R2m,
Z
BR(x0)
where γ is given by (48).
m
Proof. We can assume x =0 and, up to rescaling, that kfk =1. Define
0 L1(BR)
1 2R
w(x):= log |f(y)|dy, x∈R2m.
γ Z |x−y|
m BR
Extend f to be zero outside B (x ); then
R 0
(−∆)mw=|f| in R2m.
We claim that w ≥ |v| in B . Indeed by (49) and from |x − y| ≤ 2R for
R
x,y ∈B , we immediately see that
R
(−∆)jw ≥0, j =0,1,2,...
In particular the function z :=w−v satisfies
(−∆)mz ≥0 in B
R
(cid:26) (−∆)jz ≥0 on ∂BR for 0≤j ≤m−1.
By Proposition 21, (−∆)jz ≥ 0 in B , 0 ≤ j ≤ m −1 and the case j = 0
R
corresponds w ≥v. Working also with −v we complete the proof of our claim.
7
Now it suffices to show that for p ∈ (0,γ ) we have ke2mpwk ≤
m L1(BR)
C(p)R2m. By Jensen’s inequality we have
e2mpwdx = e2γmmpRBRlog|x2−Ry||f(y)|dydx
Z Z
BR BR
≤ |f(y)|e2γmmplog|x2−Ry|dydx
Z Z
BR BR
2mp
2R γm
= |f(y)| dx dy
Z (cid:18)Z (cid:18)|x−y|(cid:19) (cid:19)
BR BR
On the other hand
2mp 2mp
2R γm 2R γm
dx ≤ dx
Z (cid:18)|x−y|(cid:19) Z (cid:18)|x|(cid:19)
BR BR
R
= ω2m r2m−1−2γmmp(2R)2γmmpdr
Z
0
= ω2m γm R2m22γmmp.
2mγ −2mp
m
We then conclude
C(m)
e2mpwdx≤ R2m.
Z γ −p
BR m
(cid:3)
Corollary 8 Every solution u to (2) with e2mu ∈L1 (R2m) is smooth.
loc
Proof. Given B (x )⊂R2m, write (2m−1)!e2mu =f +f with
4 0 B4(x0) 1 2
(cid:12)
kf k <γ , f ∈L∞(cid:12)(B (x )),
1 L1(B4(x0)) m 2 4 0
and u=u +u +u , with
1 2 3
(−∆)mu =f in B (x )
i i 4 0
(cid:26) ui =∆ui =...=∆m−1ui =0 on ∂B4(x0)
for i = 1,2, and ∆mu3 = 0. Then, by Theorem 7, e2mu1 ∈ Lp(B4(x0)) for
some p > 1, while, by standard elliptic estimates u ∈ L∞(B (x )) and u
2 4 0 3
is smooth, hence u ∈ L∞(B (x )). Then e2mu ∈ Lp(B (x )). Write now
3 3 0 3 0
u =v +v , where
B3(x0) 1 2
(cid:12)
(cid:12)
(−∆)mv =(2m−1)!e2mu in B (x )
1 3 0
(cid:26) v1 =∆v1 =...=∆m−1v1 =0 on ∂B3(x0)
and ∆mv = 0. Then, by Lp-estimates and Sobolev’s embedding theorem,
2
v ∈ W2m,p(B (x )) ֒→ C0,α(B (x )) for some 0 < α < 1, while v is smooth.
1 3 0 3 0 2
Thenu∈C0,α(B (x ))andwiththe sameprocedureofwritingu asthe sumof
2 0
a polyharmonic (hence smooth) function plus a function with vanishing Navier
boundary condition, we can bootstrap and use Schauder’s estimate to prove
that u∈C∞(B (x )). (cid:3)
1 0
8
3 Proof of Theorems 1 and 2
The proof of Theorems 1 and 2 will be divided into several lemmas. It consists
of a careful study of the functions v, defined in (3), and u−v. In what follows
the generic constant C may depend also on u.
Remark. In general v 6=u, even if u is a standard solution. To see that, rescale
u by a factor r >0 as follows:
u(x):=u(rx)+logr.
Then u is again a solution,ewith the same energy. On the other hand the
corresponding v satisfies
e
v(x) e= (2m−1)! log |y| e2mu(ry)r2mdy
γm ZR2m (cid:18)|x−y|(cid:19)
e (2m−1)! |y′|
= log e2mu(y′)dy′ =v(rx). (15)
γm ZR2m (cid:18)|rx−y′|(cid:19)
That shows that after rescaling, u−v changes by a contant. •
Lemma 9 Let u be a solution of (2), (4). Then, for |x|≥4,
v(x)≥−2αlog|x|+C. (16)
Proof. The proof is similar to that in dimension 4, compare [Lin]. Fix x with
|x|≥4, and decompose R2m =A ∪A ∪B , where B =B (0) and
1 2 2 2 2
A :=B (x), A :=R2m\(A ∪B ).
1 |x|/2 2 1 2
For y ∈A we have
1
|x| |y|
|y|≥|x|−|x−y|≥ ≥|x−y|, log ≥0,
2 |x−y|
hence
|y|
log e2mu(y)dy ≥0. (17)
Z |x−y|
A1
For y ∈A , since |x|,|y|≥2, we have
2
|y| 1
|x−y|≤|x|+|y|≤|x||y|, log ≥log ,
|x−y| |x|
hence
|y|
log e2mu(y)dy ≥−log|x| e2mu(y)dy. (18)
Z |x−y| Z
A2 A2
For y ∈B , log|x−y|≤log|x|+C and, since u is smooth, we find
2
|y|
log e2mu(y)dy ≥ log|y|e2mu(y)dy−log|x| e2mudy
Z |x−y| Z Z
B2 B2 B2
−C e2mudy
Z
B2
≥ −log|x| e2mudy+C. (19)
Z
B2
9
Puttingtogether(17),(18)and(19)andobservingthatlog 1 <0,weconclude
|x|
that
(2m−1)! |y|
v(x) ≥ log e2mu(y)dy
γ Z (cid:18)|x−y|(cid:19)
m A2∪B2
(2m−1)!
≥ − log|x| e2mudy+C
γ Z
m A2∪B2
(2m−1)!|S2m|
≥ − αlog|x|+C.
γ
m
Finally, observing that (2m−2)!!=2m−1(m−1)!, we infer
(2m−1)!|S2m| (2m−1)!2(2π)m(2m−2)!!
= =2.
γ (2m−1)!!23m−2[(m−1)!]2πm
m
(cid:3)
Lemma 10 Let u be a solution of (2) and (4), with m ≥ 2. Then u = v+p,
where p is a polynomial of degree at most 2m−2. Moreover
∆ju(x) = ∆jv(x)+p
j
22j(j−1)!(m−1)! e2mu(y)
= (−1)j dy+p ,
(m−j−1)!|S2m| ZR2m |x−y|2j j
where p is a polynomial of degree at most 2(m−1−j).
j
Proof. Let p:=u−v. Then ∆mp=0. By Lemma 9 we have
p(x)≤u(x)+2αlog|x|+C,
and Theorem 6 implies that p is a polynomial of degree at most 2m−2. To
compute ∆jv, one can use (49) and the definition of γ . (cid:3)
m
Lemma 11 Let p be the polynomial of Lemma 10. Then
sup p(x)<+∞.
x∈R2m
In particular degp is even.
Proof. Define
f(r):=supp.
∂Br
If sup p=+∞, there exists s>0 such that
R2m
f(r)
lim =+∞, (20)
r→+∞ rs
see [Gor, Theorem 3.1].1 Moreover |∇p(x)| ≤C|x|2m−3 hence, also taking into
account Lemma 9, there is R > 0 such that for every r ≥ R, we can find x
r
with |x |=r such that
r
1
u(y)=v(y)+p(y)≥rs for |y−x |≤ .
r r2m−3
1The statement of Theorem 3.1 in [Gor] is about µ(r):=inf∂Br|p|, but the proof works
inourcasetoo.
10