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AN INDUCTION PRINCIPLE FOR THE WEIGHTED p-ENERGY MINIMALITY OF x/llxll.
Jean-Christophe Bourgoin
To cite this version:
Jean-Christophe Bourgoin. AN INDUCTION PRINCIPLE FOR THE WEIGHTED p-ENERGY MINIMALITY OF x/llxll.. 2006. �hal-00018444�
ccsd-00018444, version 1 - 2 Feb 2006
p-ENERGY MINIMALITY OF x/kxk
JEAN-CHRISTOPHE BOURGOIN
Abstract. In this paper, we investigate minimizing properties of the mapx/kxkfrom the Euclidean unit ballBnto its boundarySn−1, for the weighted energy functionalsEp,αn (u) =R
Bnkxkαk∇ukpdx. We establish the following induction principle: if the map kxkx :Bn+1→Snminimizes En+1p,α among the maps u : Bn+1 → Sn satisfying u(x) = x on Sn, then the map kyky : Bn → Sn−1 minimizes Enp,α+1 among the maps v:Bn→Sn−1satisfyingv(y) =yonSn−1.
This result enables us to enlarge the range of values ofp andαfor whichx/kxkminimizesEnp,α.
1. Introduction and statement of main results
Let Bn be the unit Euclidean ball of Rn and Sn−1 its bourdary. For any couple (α, p) of real numbers, with α ≥ 0 and p ≥ 1, we define the rα-weighted p-energy functional of a mapu :Bn→Sn−1 by
Ep,αn (u) = Z
Bnkxkαk∇ukpdx.
This functional is nothing but the p-energy functional associated with the Riemannian metric rα(n−p)geuc on the ballBn, where geuc is the Euclidean metric, the sphere Sn−1 being endowed with its standard metric. The func- tional Ep,αn is to be considered on the Sobolev space
Wα1,p(Bn,Sn−1) ={u∈W1,p((Bnrα(n−p)geuc),Rn);kuk= 1a.e}. The question is to know which map minimizes Ep,α among the maps in Wα1,p(Bn,Sn−1) satisfyingu(x) =x on Sn−1.
Following the arguments of Hildebrant, Kaul and Widman [11], the map x/kxk is a natural candidate to be the minimizer of Ep,αn , forp∈[1, n+α) (notice that x/kxk ∈Wα1,p(Bn,Sn−1) if and only ifp < n+α).
The minimality ofx/kxk was first established for the 2-energy functional En2 := E2,0n by J¨ager and Kaul [14] in dimension n ≥ 7, then by Brezis, Coron and Lieb [2] in dimension 3. Coron and Gulliver [5] actually proved the minimality of x/kxk for the p-energy functional Epn := Ep,0n for any integer p ∈ {1,· · · , n−1} and any dimension n≥3. An alternative proof using the null Lagrangian method (or calibration method) of this last result was obtained by Lin [15] and Avellaneda and Lin [1].
Hardt and Lin [7, 8] and Hardt, Lin and Wang [9, 10] developed the study of the singularities ofp-harmonic andp-minimizing maps and obtained results extending those of Schoen and Uhlenbelck [16, 17]. A consequence of their results is the minimality of x/kxk for Epn for any p ∈ (n−1, n).
1
2 JEAN-CHRISTOPHE BOURGOIN
Finally, Hong [12] and Wang [18] have proved independently the minimality of x/kxk forEpn in dimensionn≥7 for anyp≤n−2√
n−1.
In order to close the question of the Epn-minimality of x/kxk for the remaining values of p, Hong [13] suggested a new idea. Indeed, he observed that, for any p ∈ (2, n), the minimality of x/kxk for Enp follows from the minimality of x/kxk for the r2−p-weighted 2-energy E2,2−pn . Unfortunately, we have proved in a previous paper [3], the existence of an interval of values of p∈(1, n−1) for which the mapx/kxk fails to be a minimizer of E2,2−pn . Nevertheless, we proved thatx/kxkminimizes therα-weightedp-energyEnp,α for any α ≥0 and any integer p ≤n−1. Actually, we obtained in [4] the minimality of x/kxk for more general weighted p-energy functionals of the form Ep,fn (u) = R
Bnf(kxk)k∇ukpdx, where p ≤ n−1 is an integer and f : [0,1]→R+ is a continuous non-decreasing function.
Our aim in this paper is to show that the minimizing properties ofx/kxk in dimension n and n+ 1 are not independent. Indeed, we will prove that, for anyp∈[1, n+α+ 1), ifx/kxkminimizes therα-weightedp-energyEp,αn+1 in dimensionn+1, then it also minimizes therα+1-weightedp-energyEp,α+1n in dimension n.
Theorem 1. Let n ≥ 2 be an integer, α ≥ 0 be a real number and p ∈ [1, n+α+ 1). If the map kxkx : Bn+1 7→ Sn minimizes the rα-weighted p- energy Ep,αn+1 among the maps u ∈ Wα1,p(Bn+1,Sn) satisfying u(x) = x on Sn, then the map kyky : Bn 7→ Sn−1 minimizes the rα+1-weighted p-energy Enp,α+1 among the maps v∈Wα+11,p (Bn,Sn−1) satisfyingv(y) =y on Sn−1.
The proof of this theorem relies on a construction which associates to each mapu:Bn→Sn−1such thatu(y) =yonSn−1, a mapu :Bn+1 →Sn such that u(x) =u(x) in Bn× {0} and u(x) = x on the unit sphere Sn, in such a way that, if u0 is the map defined inBn by u0(y) = kyky , then u0 is exactly the map defined on Bn+1 by u0(x) = kxkx . The energy Ep,αn+1(u) of u is estimated above in terms of the energy Ep,α+1n (u) ofu and the equality holds in the estimate for the map u0 = kyky .
Thanks to Theorem 1 and the minimality results mentioned above, we deduce the following :
Corollary 1. The mapy/kykminimizesEp,αn among the maps inWα1,p(Bn,Sn−1) which coincide with y/kyk on Sn−1 in the following cases :
i) α∈Nand p∈(n+α−1, n+α),
ii) α ≥β and p is an integer in {1,· · · , n+β−1}, for any real number β ≥0.
iii) α∈N, n+α≥7 and p≤n+α−2√
n+α−1,
1.1. Construction ofuand proof of Theorem 1.1. LetBn+1andSnbe the unit open ball and the unit sphere of Rn+1 and letBn andSn−1 be the unit open ball and the unit sphere of Rn that we identify with the subspace Rn=Rn× {0}inRn+1. Moreover, we writex= (x1,· · · , xn, xn+1) a vector of Rn+1,y = (y1,· · · , yn) a vector ofRn, h., .i the standard metric of Rn+1 and (e1,· · ·, en, en+1) the standard basis ofRn+1.
Let Πn be the projection defined by : Πn:Bn+1 −→ Bn
(x1,· · · , xn+1) −→ (x1,· · ·, xn,0) = (x1,· · · , xn).
Consider the map ϕn defined onBn\Ren+1 byϕn(x) = kΠΠn(x)
n(x)k. We define the map udefined by
u:Bn+1 −→ Bn x −→ hen+1, x
kxkien+1+hϕn(x), x
kxkiu(kxkϕn(x)).
Lemma 1. For any x∈Bn+1,k∇u(x)k2 = kxk12 +k∇u(kxkϕn(x))k2. Proof. For anyi∈ {1,· · · , n}, we have,
du(x).ei = hΠn(x), eii kxk
1
kΠn(x)k −kΠn(x)k kxk2
u(kxkϕn(x))
−hen+1, xihx, eii kxk3 en+1
+du(kxkϕn(x)). hΠn(x), eiiΠn(x)
kxk2 +ei−hΠn(x), eiiΠn(x) kΠn(x)k2
!
and,
du(x).en+1 = en+1
kxk − hen+1, xi2
kxk3 en+1−hen+1, xi
kxk3 kΠn(x)ku(kxkϕn(x)) +hen+1, xi
kxk2 du(kxkϕn(x)).Πn(x).
Since ku(x)k2 = 1, one has hdu(x).h, u(x)i = 0 for any h∈ Rn. Hence, for any i∈ {1,· · · , n}, we have,
kdu(x).eik2 = hΠn(x), eii2 kxk2
1
kΠn(x)k −kΠn(x)k kxk2
2
+ hen+1, xi2hx, eii2 kxk6
+
du(kxkϕn(x)).
hΠn(x), eii 1
kxk2 − 1 kΠn(x)k2
Πn(x) +ei
2.
kdu(x).en+1k2 = 1 kxk2
1−hen+1, xi2 kxk2
2
+hen+1, xi2
kxk6 kΠn(x)k2 +hen+1, xi2
kxk4 kdu(kxkϕn(x)).Πn(x)k2. Finally, we have,
k∇u(x)k2 = 1
kxk2 +k∇u(kxkϕn(x))k2.
Letxn+1be a real number in (0,1), consider the setAxn+1 = (xn+1en+1+ e⊥n+1)∩Bn+1\Ren+1, where e⊥n+1 is the orthogonal subspace to Ren+1 for
4 JEAN-CHRISTOPHE BOURGOIN
h., .i. Let θbe the map :
θ:Axn+1 −→ Cxn+1 ={y∈Rn;kyk>|xn+1|}
x= (x1,· · · , xn+1) −→ kxkϕn(x) =y = (y1,· · ·, yn).
Lemma 2. For any y∈Rn, the Jacobian determinant of θ−1 is : Jac(θ−1)(y) = kyk2−x2n+1n−22
kykn−2 . Proof For any i∈ {1,· · ·, n}and for any x∈Axn+1,
dθ(x).ei = 1
kxk − kxk kΠn(x)k2
hΠn(x), eii
kΠn(x)k Πn(x) + kxk kΠn(x)kei. Let us set, for any i∈ {1,· · · , n},
λi= 1
kxk − kxk kΠn(x)k2
hΠn(x), eii
kΠnk and α= kxk kΠn(x)k. Hence, for any i∈ {1,· · ·, n},
dθ(x).ei=λiΠn(x) +αei. Then, for any x∈Axn+1,
Jac(θ)(x) = det(λ1Πn(x) +αe1, λ2Πn(x) +αe2,· · · , λnΠn(x) +αen)
=
n
X
i=1
det(αe1,· · · , αei−1, λiΠn(x), αei+1,· · · , αen) +αndet(e1,· · · , en)
=
n
X
i=1
αn−1λidet(e1,· · · , ei−1,Πn(x), ei+1,· · ·, en) +αn,
where det is the determinant in the basis (e1,· · ·, en+1).
Jac(θ)(x) =
n
X
i=1
αn−1 1
kxk − kxk kΠn(x)k2
hΠn(x), eii2 kΠn(x)k +αn
= αn−1 1
kxk− kxk kΠn(x)k2
kΠn(x)k2
kΠn(x)k +αn =αn−2. Therefore, we have
Jac(θ)(x) = kxkn−2
kΠn(x)kn−2 = kykn−2 kyk2−x2n+1n−22
.
We deduce that,
Jac(θ−1)(y) = kyk2−x2n+1n−22 kykn−2 .
Lemma 3. For any x∈Bn+1, we have, Z
Bn+1kxkαk∇u(x)kpdx1· · ·dxn+1 ≤ np/2−1 Z
Bn+1
1
kxkp−αdx1· · ·dxn+1 +2(1−1/n)1−p/2Wn−1
Z
Bnkykα+1k∇u(y)kpdy, where Wn−1=Rπ2
0 (cos(γ))n−1dγ.
Proof. Writing kxk12+k∇u(kxkϕn(x))k2 = n1(nkxk12)+(1−1n)( 1
1−1
n
k∇u(kxkϕn(x))k2) and using that x→xp2 is a convex map, for anyp∈[1, n+α+ 1), we have,
k∇u(x)kp = 1
kxk2 +k∇u(kxkϕn(x))k2 p/2
≤ np/2−1 1
kxkp + (1−1/n)1−p/2k∇u(kxkϕn(x))kp. Hence,
Z
Bn+1kxkαk∇u(x)kpdx1· · ·dxn+1 ≤ np/2−1 Z
Bn+1kxkα 1
kxkpdx1· · ·dxn+1 +(1−1/n)1−p/2
Z
Bn+1kxkαk∇u(kxkϕn(x))kpdx1· · ·dxn+1
≤ np/2−1 Z
Bn+1kxkα 1
kxkpdx1· · ·dxn+1 + 2(1−1/n)1−p/2
Z 1 0
dxn+1
Z
Bnkxkαk∇u(kxkϕn(x))kpdx1· · ·dxn. Using the change of variables y=θ(x) and Lemma 1.2 we get,
Z
Bn+1kxkαk∇u(x)kpdx1· · ·dxn+1 ≤ np/2−1 Z
Bn+1
1
kxkp−αdx1· · ·dxn+1
+ 2(1−1/n)1−p/2 Z 1
0
dxn+1 Z
Cn+1
kyk2−x2n+1n−22
kykn−2 kykαk∇u(y)kpdy1· · ·dyn, Z
Bn+1kxkαk∇u(x)kpdx1· · ·dxn+1 ≤ np/2−1 Z
Bn+1
1
kxkp−αdx1· · ·dxn+1
+ 2(1−1/n)1−p/2 Z
Bnkykαk∇u(y)kp
Z kyk
0
kyk2−x2n+1n−22 kykn−2 dxn+1
dy.
But we have, Z kyk
0
kyk2−x2n+1 kyk2
n−22
dxn+1 = Z kyk
0
1− xn+1
kyk
2!n−22 dxn+1
= Z 1
0 kyk(1−t2)n−22 dt=kyk Z π/2
0
(cosγ)n−1dγ.
6 JEAN-CHRISTOPHE BOURGOIN
Let us set Wn−1 =Rπ/2
0 (cosγ)n−1dγ. Then, we have the inequality, Z
Bn+1kxkαk∇u(x)kpdx1· · ·dxn+1 ≤ np/2−1 Z
Bn+1
1
kxkp−αdx1· · ·dxn+1 + 2(1−1/n)1−p/2Wn−1
Z
Bnkykα+1k∇u(y)kpdy.
Lemma 4. Let Γ be the Gamma function defined by, Γ(x) =
Z +∞
0
tx−1e−tdt.
We have,
Wn−1
Γ(n+12 ) Γ(n2) =
√π 2 .
Proof From the equality Γ(x+ 1) =xΓ(x) for anyx∈(0,+∞), we have, ifn= 2l, where l∈N,
Γ(2l+ 1
2 ) = 2l−1 2
2l−3 2 · · ·3
2 1 2Γ(1
2) and
Γ(2l
2) = 2l−2 2
2l−4 2 · · ·2
2Γ(1).
Moreover we have,
W2l−1 = (2l−2)(2l−4)· · ·2 (2l−1)(2l−3)· · ·3.1 then,
W2l−1Γ(2l+12 ) Γ(2l2) =
√π 2 . If n= 2l+ 1 where l∈N, we obtain,
Γ(2l+ 2 2 ) = 2l
2 2l−2
2 · · ·2 2Γ(1), Γ(2l+ 1
2 ) = 2l−1 2 · · ·3
2 1 2Γ(1
2) and
W2l= (2l−1)(2l−3)· · ·3.1 (2l)(2l−2)· · ·2 .π
2. We deduce that
W2lΓ(2l+22 ) Γ(2l+12 ) =
√π
2 .
Proof of Theorem 1.1. Suppose that u0 is a minimizer map of Ep,α. Since the measure of Sn is |Sn| = 2π
n+1 2
Γ(n+12 ) and since k∇u0k(x) = n
p 2
kxk, by
Lemma 1.3, for any map u∈Wα+11,p (Bn,Sn−1) satisfying u(y) =y on Sn−1, we have,
np/2 n+ 1 +α−p
2πn+12 Γ(n+12 ) =
Z
Bn+1kxkαk∇u0(x)kpdx1· · ·dxn+1
≤ Z
Bn+1kxkαk∇u(x)kpdx1· · ·dxn+1
≤ np/2−1 Z
Bn+1
1
kxkp−αdx1· · ·dxn+1
+2(1−1/n)1−p/2Wn−1
Z
Bnkykα+1∇u(y)kpdy
≤ 1 n
np/2 n+1+α−p
2πn+12 Γ(n+12 ) + 2Wn−1(1− 1
n)( n n−1)p/2
Z
Bnkykα+1k∇u(y)kpdy.
Then we have, 2Wn−1
1−1
n
n n−1
p/2R
Bnkykα+1k∇u(y)kpdy R
Bnkykα+1k∇ky)ky kpdy ≥ np/2 n+1+α−p
2πn+12 Γ(n+12 )
×n+1+α−p (n−1)p/2
Γ(n2) 2πn2
1−1
n
By Lemma 1.4 we finally get, Z
Bnkykα+1k∇u(y)kpdy≥ Z
Bnkykα+1k∇ y
kykkpdy
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