Unique solvability of the Dirichlet problemfor weakly harmonic maps
Received: 9 June 2000 / Revised version: 17 April 2001
Abstract. We prove existence and uniqueness of weakly harmonic maps from the unit ballin R
n (with n ≥ 3) to a smooth compact target manifold within the class of maps with smallscaled energy for suitable boundary data.
1. Introduction and statement of results
LetN be a smooth, compact Riemannian manifold of dimensionk without bound-ary. Letn ≥ 3 andB be the unit ball inRn. We may assume – by the theorem ofNash–Moser – thatN is isometrically embedded in a Euclidean spaceR
K . So wecan define the Sobolev space
H 1,2(B,N) := {u ∈ H 1,2(B,RK) : u(x) ∈ N almost everywhere}.We consider critical points of the energy functional
E(u) =∫B
|∇u|2 dx,
i.e. mapsu ∈ H 1,2(B,N) satisfying
d
dt
∣∣∣∣t=0E(πN ◦ (u+ tφ)) = 0
for all φ ∈ C∞0 (B,RK), whereπN is the nearest point projection ontoN . TheEuler-Lagrange equation for this variational problem is
−�u = A(u)(∇u,∇u) :=n∑α=1
A(u)
(∂u
∂xα,∂u
∂xα
)(1)
in the weak sense (whereA is the second fundamental form ofN ⊂ RK ), as can
easily be verified. We call a mapu ∈ H 1,2(B,N) satisfying (1) weakly harmonic.
R. Moser: Max-Planck-Institute for Mathematics in the Sciences, Inselstr. 22–26,04103 Leipzig, Germany. e-mail: [email protected]
Mathematics Subject Classification (2000): 58E20
380 R. Moser
A weakly harmonic mapu ∈ H 1,2(B,N) is called stationary, if it is a criti-cal point ofE also with respect to variations of the independent variable, i. e. ifddt
∣∣t=0E(u(x + tη(x))) = 0 for all η ∈ C∞0 (B,Rn). This is equivalent to
div
(|∇u|2eα − 2
⟨∂u
∂xα,∇u
⟩)= 0, α = 1, . . . , n,
in the distribution sense, whereeα is theαth standard unit vector inRn.
Our main goal in this work is to establish certain uniqueness results concerningweakly harmonic maps with prescribed boundary data, similar to those obtainedby Struwe in [14], where the casen = 3 is considered. As pointed out in [14], onecannot expect the uniqueness of harmonic maps to hold without further restrictions.A suitable condition to require is
supx0∈Br>0
(r2−n
∫B∩Br(x0)
|∇u|2 dx)≤ ε2 (2)
for sufficiently smallε > 0. Note that this condition is invariant under scaling. Itwill therefore enable us to apply certain regularity results for stationary harmonicmaps with small energy (such as Theorem I.4 in [1] or Theorem 2.2 in [9]) not onlyto u, but to any map that comes fromu by scaling. However it is not immediatelyclear that there exist solutions of (1) that satisfy (2), even for “nice” boundaryconditions. Natural candidates are energy minimizing maps. Indeed, it will turn outto be quite easy to estimate their energy on balls that have their center boundedaway from∂B. If the center is close to the boundary, however, things will be moredifficult.
Our main results are the following.
Theorem 1. There exists a number ε > 0, depending only on n and N , such thatif u, v ∈ H 1,2(B,N) are weakly harmonic maps, both satisfying (2), such thatu− v ∈ H 1,2
0 (B,RK), then u = v.
Theorem 2. Let ε > 0 be the constant from Theorem 1. There exists a numberε0 > 0, depending on n andN , such that for all maps u0 ∈ H 1,2(B,RK) satisfying
supx0∈∂Br>0
(r2−n
∫B∩Br(x0)
|∇u0|2 dx)≤ ε2
0 (3)
and u0|∂B(x) ∈ N almost everywhere (with respect to the surface measure on ∂B),there is a unique energy minimizing map u ∈ H 1,2
u0 (B,N) := H 1,2(B,N)∩ (u0+H
1,20 (B,RK)). Moreover, u satisfies (2).
Theorem 3. Let ε be as before. If the boundary data u0 from Theorem 2 satisfy theinequality
supx0∈Br>0
(rp−n
∫B∩Br(x0)
|∇u0|p dx)≤ εp0 (4)
Unique solvability of the Dirichlet problem for weakly harmonic maps 381
for a number p > 2, and if ε0 is chosen appropriately (now depending also on p),then the unique energy minimizer u ∈ H 1,2
u0 (B,N) satisfies
supx0∈Br>0
(rq−n
∫B∩Br(x0)
|∇u|q dx)≤ εq
for some number q > 2. This number also depends on n, N , and p.
So for suitable boundary data, we have existence, uniqueness within a certain class,and higher regularity (if the boundary data are more regular as well) for solutionsof the Dirichlet problem for weakly harmonic maps.
We choose the ballB as the domain of these maps only for simplicity. Themethods we will use apply also to domains that are smooth Riemannian manifoldswith sufficiently smooth boundaries; the proofs just get technically more laborious.
Note that the image of the mapu0 (which provides the boundary data) is notrequired to be inN , except at the boundary. In particular we cannot apply the directmethod of the calculus of variations to find a corresponding energy minimizer. Thefact that there is one all the same is due to (3). Failing this condition, we could noteven be sure that there exists a map inH 1,2
u0 (B,N) at all. As a matter of fact, thefollowing holds.
Proposition 1. Suppose that the first homotopy group π1(N) of N is non-trivial.Then for any integer n ≥ 2, there exists a map u0 ∈ H 1,2(Bn,RK), where Bn
is the unit ball in Rn, that has boundary values g0 = u0|∂Bn contained almost
everywhere in N , but such that H 1,2u0 (B
n,N) is empty.
Proof. Consider first the casen = 2. Chooseg0 to be a smooth representation of anon-trivial member ofπ1(N). Obviously one can find a mapu0 ∈ H 1,2(B2,RK)
with boundary valuesg0, even a smooth one. Suppose there exists a mapv ∈H
1,2u0 (B
2, N). Then an energy minimizeru ∈ H 1,2u0 (B
2, N) can be found by thedirect method. (I. e. choose a minimizing sequence inH
1,2u0 (B,N), then a certain
subsequence converges weakly inH 1,2(B,RK), and its limit is an energy mini-mizer, since the energy is weakly lower semi-continuous.)According to Theorem IIin [10], u is continuous in the interior ofB2, whereas Theorem 2.7 in [11] assertsthat it is continuous in a neighbourhood of the boundary. But this is a contradictionto the choice ofg0.
We proceed by induction. Assume that the claim of this proposition is true forthe dimensionn− 1≥ 2. Instead ofBn, consider the “double cone”
A = {(x′, xn) ∈ Rn−1× R : |x′| < min{1+ xn,1− xn}},
which can be mapped uponBn by means of a bi-Lipschitz map. Choose a mapu0 ∈ H 1,2(Bn−1,RK) that satisfies the assertion of this proposition in dimensionn− 1. Defineu0 ∈ H 1,2(A,RK) by
This construction makes sure thatu0 ∈ H 1,2(A,RK)and that it has boundary valuesin N almost everywhere. Suppose now that there exists a mapu ∈ H 1,2
u0 (A,N).Then for almost allxn ∈ (−1,1), the trace map
uxn = u(·, xn)belongs – up to scaling – toH 1,2
u0(Bn−1, N), in contradiction to the choice ofu0.
The proposition is therefore proved.��On the other hand, quite a large number of boundary datau0 ∈ H 1,2(B,RK) do
have extensions withinH 1,2(B,N). This is for instance true – besides those mapsalready mentioned in Theorem 2 – if thetraceu0|∂B is either of classH 1,2(∂B,N),or if it is continuous. In the former case, the homogeneous extension of degree 0provides a map inH 1,2
u0 (B,N) (we consider again only the casen ≥ 3). The lattercase is considered in the next proposition.
Proposition 2. Suppose that u0 ∈ H 1,2(B,RK) has boundary values g = u0|∂B ∈C0(∂B,N). Then H 1,2
u0 (B,N) is not empty, and hence there exists an energy min-imizer with boundary values g.
Proof. Let v satisfy�v ≡ 0 onB andv = g on ∂B. It is easy to see thatv is aminimizer of the energy among all mapsu ∈ H 1,2
u0 (B,RK). In particular,v belongs
toH 1,2u0 (B,R
K) itself. Sinceg is continuous,v also belongs toC0(B,RK). So forany δ > 0 there exists a radiusr ∈ (0,1), such that dist(v(x),N) ≤ δ for allx ∈ B\Br(0). The mapw, defined by
w(x) =πN ◦ v(x), x ∈ B\Br(0),πN ◦ v
(rx
|x|), x ∈ Br(0),
is clearly inH 1,2u0 (B,N) (sincev|∂Br (0) is smooth), provided thatδ is sufficiently
small. An energy minimizer can now be found by the direct method.��Remark 1. In [15], White gives a condition that is equivalent to the existence ofan extension of a map∂M → N in H 1,2(M,N) for a general smooth compactRiemannian manifoldM – he requires Lipschitz continuous boundary data however.
Notation. Open balls inRj are denotedBjr (x), B
jr = B
jr (0), or Bj = B
j1(0).
In dimensionn, we writeBr(x) = Bnr (x), etc. For any setA ⊂ Rn, let A+ =
A ∩ (Rn−1 × (0,∞)) be the part ofA contained in the upper half-space. So inparticular,B+r is a half-ball of radiusr.
Subsequently, we will often have estimates involving certain constants, de-pending only on the geometry of the manifoldN and on the dimensionn. We willindiscriminately denote them asC. So even within one single chain of estimates,the same letterC may denote different constants.
We denote the standard scalar product inRn by ·, and inR
K by 〈· , ·〉.Finally, letδ > 0 denote a number, such that the nearest point projectionπN is
a Lipschitz map on theδ-neighbourhoodUδ(N) of N . Thus ifu ∈ H 1,2(B,RK)
takes values inUδ(N) almost everywhere, thenπN ◦ u ∈ H 1,2(B,N) and∫B
|∇(πN ◦ u)|2 dx ≤ C∫B
|∇u|2 dx.
Unique solvability of the Dirichlet problem for weakly harmonic maps 383
2. Energy estimates for energy minimizing maps
To prove Theorem 2, we need to estimate the scaled energy
r2−n∫B∩Br(x0)
|∇u|2 dx
on any ballBr(x0) with x0 ∈ B for energy minimizing mapsu ∈ H 1,2(B,N)
with suitable boundary data. We will do that by proving that if the scaled energyis small for some ballBr(x0), then this is still true withr replaced byθr, whereθ ∈ (0,1) is a constant not depending onu. Such (and even better) results arealready known for balls entirely contained inB. For instance, there is the followingmonotonicity formula, which is due to Price [8] (or to Schoen–Uhlenbeck [10] forenergy minimizing maps).
Lemma 1. Let u ∈ H 1,2(B,N) be a stationary weakly harmonic map and Bs(x0)
⊂ Br(x0) ⊂ B. Then the monotonicity formula
r2−n∫Br(x0)
|∇u|2 dx − s2−n∫Bs(x0)
|∇u|2 dx
= 2∫Br(x0)\Bs(x0)
|x − x0|−n|(x − x0) · ∇u|2 dx
holds.
For balls intersecting the boundary ofB, the situation is more difficult. Thereare some results by Schoen–Uhlenbeck [11] generalizing the monotonicity formulato that situation, but they require boundary data that are at least continuous, whereaswe want to use only (3). We will modify instead an energy decay result of Hardt–Lin[5]. Their method is to compare the energy of an energy minimizer with the one ofa suitably constructed map with the same boundary values.
An important tool for this will be the following lemma due to Luckhaus (see [7]),which allows us to modify a given map in order to obtain certain given boundarydata. The version given here is from [12] (Lemma 2.6.1).
Lemma 2. Let u, v ∈ H 1,2(∂B,RK) and 0 < λ ≤ 12 . There exists a map w ∈
H 1,2(B\B1−λ,RK), coinciding with u on ∂B (in the sense of traces) and withv((1− λ)−1x) on ∂B1−λ, such that∫
B\B1−λ|∇w|2 dx ≤ Cλ
∫∂B
(|∇u|2+ |∇v|2
)do+ Cλ−1
∫∂B
|u− v|2 do
and
dist2(w(x),u(∂B) ∪ v(∂B))
≤ Cλ1−n(∫
∂B
(|∇u|2+ |∇v|2
)do
) 12(∫
∂B
|u− v|2 do) 1
2
+ Cλ−n∫∂B
|u− v|2 doalmost everywhere, where C is a constant depending only on n and K .
384 R. Moser
Suppose now that we have given boundary datau0 ∈ H 1,2(B,N), satisfying (3),for which we want to find an energy minimizeru ∈ H 1,2
u0 (B,N) such that (2) holds.(Note that this situation is not the same as the one of Theorem 2. However, we needto establish a result for this case first, before we proceed to the general situation.)It is easy to find an energy minimizing map taking the boundary values ofu0 bythe direct method, so it suffices to estimate the scaled energy for a given energyminimizer.
Lemma 3. Let u0 ∈ H 1,2(B,N) and suppose that u ∈ H 1,2u0 (B,N) is an energy
minimizing map. For any µ ∈ (0,1), there exists a constant ε0, depending onlyon µ, n, and N , and a constant C > 0, depending on n and N , such that for anyx0 ∈ B and for any r ∈ (0, 1
2], the estimate∫B∩Br/2(x0)
|∇u|2 dx ≤ µ∫B∩Br(x0)
|∇u|2 dx + Cµ
∫B∩Br(x0)
|∇u0|2 dx
+ C
µr2
∫B∩Br(x0)
|u− u0|2 dx(5)
holds, provided that the conditions
r2−n∫B∩Br(x0)
|∇u|2 dx ≤ ε2 ≤ 1 (6)
and
r2−n∫B∩Br(x0)
|∇u0|2 dx ≤ ε20 (7)
are satisfied.
Proof. Let δ0, λ ∈ (0, 12] be two constants, with values not yet determined (but
depending only onn andN ). Let x0 ∈ B andr ∈ (0, 12]. We may assume that
r−n∫B∩Br(x0)
|u− u0|2 dx ≤ δ20, (8)
for the assertion is clear otherwise. For anyρ ∈ ( r2, r), we choose a pointyρ ∈Rn\B that satisfies
∂B ∩ ∂B2ρ(yρ) = ∂B ∩ ∂Bρ(x0)
(see Fig. 1). Note thatyρ is uniquely determined unless the set on the right-handside is empty. Setu0 = 1
|B∩Br(x0)|∫B∩Br(x0)
u0 dx. We can chooseρ ∈ ( r2, r) suchthat
ρ
∫B∩∂Bρ(x0)
|∇u|2 do ≤ C∫B∩Br(x0)
|∇u|2 dx,
ρ
∫B∩∂Bρ(x0)
|u− u0|2 do ≤ C∫B∩Br(x0)
|u− u0|2 dx,
Unique solvability of the Dirichlet problem for weakly harmonic maps 385
For the first two formulas, this is proved by a standard argument. A short com-putation shows that the same method can be applied for the third and the forthformula.
The setBρ(x0)\B2ρ(yρ) can be mapped onto the ballBρ(0) by a bi-Lipschitzmap. having Lipschitz constants that are bounded by a number only depending onn. Therefore, according to Lemma 2, we can find a mapw ∈ H 1,2(B∩Bρ(x0),R
almost everywhere. (The derivation of the latter inequality uses the same steps asthat of the former one – which is why we do not describe it in detail – in additionto the inequalities (6), (7), and (8).) Choosingε0 andδ0 small (depending onλ, n,andN ), we see thatw can be projected ontoN byπN . Asu is an energy minimizer,this implies that∫
B∩Br/2(x0)
|∇u|2 dx ≤∫B∩Bρ(x0)
|∇(πN ◦ w)|2 dx
≤ Cλ∫B∩Br(x0)
|∇u|2 dx + Cλ
∫B∩Br(x0)
|∇u0|2 dx
+ C
λr2
∫B∩Br(x0)
|u− u0|2 dx.
Now choose a suitableλ, then the claim follows. ��The third term on the right-hand side of (5) can be further estimated as follows.
Lemma 4. Let u0 ∈ H 1,2(B,N), and let u ∈ H 1,2u0 (B,N) be weakly harmonic.
Then for any λ > 0, there exist constants C0, ε > 0 and θ ∈ (0,1), all dependingonly on λ, n, and N , such that for any x0 ∈ ∂B and any r > 0, the estimate
(θr)−n∫B∩Bθr (x0)
|u− u0|2 dx ≤ λr2−n∫B∩Br(x0)
|∇u|2 dx
+ C0r2−n
∫B∩Br(x0)
|∇u0|2 dx
holds, provided that (6) is satisfied.
Proof. Suppose the statement were not true for a particular numberλ > 0. Thenfor anyθ ∈ (0, 1
2), there would be weakly harmonic mapsui ∈ H 1,2(B,N) with
boundary data given byvi ∈ H 1,2(B,N) (meaning thatui − vi ∈ H 1,20 (B,RK)),
such that for certain pointsxi ∈ ∂B and radiiri > 0, we have
r2−ni
∫B∩Bri (xi )
|∇ui |2 dx =: ε2i → 0,
and yet
(θri)−n∫B∩Bθri (xi )
|ui − vi |2 dx > λε2i + Cir2−n
i
∫B∩Bri (xi )
|∇vi |2 dx (9)
for constantsCi that converge to∞ asi →∞. Since on the other hand
(θri)−n∫B∩Bθri (xi )
|ui − vi |2 dx ≤ Cθ2−n(ε2i + r2−n
i
∫B∩Bri (xi )
|∇vi |2 dx)
Unique solvability of the Dirichlet problem for weakly harmonic maps 387
by the Poincaré inequality, we have
r2−ni
ε2i
∫B∩Bri (xi )
|∇vi |2 dx → 0
as i → ∞. We may assume that allxi are the same, namelyxi = x0 :=(0, . . . ,0,−1), and that theri converge to a numberr0 ≥ 0. Otherwise we applyappropriate rotations toB and pick a subsequence. Note that (9) would contradictthe Poincaré inequality if the sequence(ri)i∈N was unbounded. Define
wi(x) = 1
εi(ui − vi)(rix + x0)
and extend everywi by 0 outside ofB1/ri (0, . . . ,0,1ri). These maps satisfy
limi→∞
∫B
|∇wi |2 dx = 1
and∣∣∣∣∫B∩0∇φ · ∇wi dx
∣∣∣∣ ≤ r2−ni
εi
∣∣∣∣∣∫B∩Bri (x0)
φ
(x − x0
ri
)A(ui)(∇ui,∇ui)(x)dx
∣∣∣∣∣+ riεi
∣∣∣∣∫B∩0∇φ(x) · ∇vi(rix + x0)dx
∣∣∣∣≤ C‖φ‖L∞ r
2−ni
εi
∫B∩Bri (x0)
|∇ui |2 dx
+ ‖∇φ‖L2
(r2−ni
ε2i
∫B∩Bri (x0)
|∇vi |2 dx) 1
2
→ 0
for all φ ∈ C∞0 (B ∩0) and for largei, where0 = B1/r0(0, . . . ,0,1r0) in the case
r0 > 0 and0 = (Rn)+ otherwise. We can assume that thewi converge weakly inH 1,2(B,RK) and strongly inL2(B,RK) to a mapw ∈ H 1,2(B,RK)with�w ≡ 0onB ∩0 andw ≡ 0 onB\0. In particular, the mapw|B∩0 extends to a solutionof a linear, homogeneous elliptic system onB. We may therefore apply standardelliptic estimates, along with the Poincaré inequality, to obtain
θ−n∫Bθ (0)
|w|2 dx ≤ Cθ2−n∫Bθ (0)
|∇w|2 dx ≤ Cθ2∫B
|∇w|2 dx ≤ Cθ2.
The constantC does not depend onr0, since we need only consider the caser0 ≤ 2.If r0 > 2, then0 ⊂ B and hencew ≡ 0. Therefore, the estimate
θ−n∫Bθ (0)
|wi |2 dx ≤ Cθ2
holds for largei. This means
(θri)−n∫B∩Bθri (x0)
|ui − vi |2 dx ≤ Cθ2ε2i ,
a contradiction to (9) ifθ is small. This completes the proof.��
388 R. Moser
Now we are able to prove estimate (2) for certain energy minimizers.
Lemma 5. For any ε > 0, there is a number ε0 > 0, such that the following holds. Ifu ∈ H 1,2
u0 (B,N) is an energy minimizing map for boundary data u0 ∈ H 1,2(B,N)
satisfying (3), then (2) holds.
Proof. Let x0 ∈ B andr > 0 be given. According to the monotonicity formula(Lemma 1), we need only show that (6) holds in the case wherer ≥ dist(x0, ∂B).Since such a ball is always contained in another ball of at most twice its radius andcentred on∂B, we may even assume thatx0 ∈ ∂B.
Let λ,µ ∈ (0,1) be constants, the values of which remain to be determined.Suppose thatε andε0 are so small, that Lemma 3 and Lemma 4 can be applied (fortheseλ andµ). Clearly (6) is valid forr ≥ 1
4, if ε20 ≤ 42−nε2. So forr ∈ [18, 1
4],we can apply Lemma 3 and Lemma 4 and obtain
r2−n∫B∩Br(x0)
|∇u|2 dx
≤ µr2−n∫B∩B2r (x0)
|∇u|2 dx + Cµr2−n
∫B∩B2r (x0)
|∇u0|2 dx
+ Cµr−n
∫B∩B2r (x0)
|u− u0|2 dx
≤ Cµε2+ C(1+ C0)
µε2
0 +Cλ
µε2.
For appropriateλ, µ, andε0, we may conclude that (6) holds forr ≥ 18. Namely,
we may first chooseµ, such that the first term on the right-hand side above is atmost1
3ε2. Then for that choice ofµ, we can findλ such that the same is true for the
third term. Finally, for a sufficiently smallε0, the second term can also be estimatedby 1
3ε2. We can now proceed by induction to prove (6) for allr > 0. ��
3. More general boundary data
The results in section 2 require boundary data that extend to a function havingvalues inN almost everywhere. We now want to examine the situation of Theorem2, where we have boundary data given by a mapu0 that is only contained inH 1,2(B,RK), but with boundary values belonging toN almost everywhere. Underthe assumption (3) for those boundary data, we want to find an energy minimizingmap satisfying (2).
First note that we can always replaceu0 by a mapv ∈ u0+H 1,20 (B,RK) having
vanishing Laplacian. Lemma 5 shows that
supx0∈Br>0
(r2−n
∫B∩Br(x0)
|∇v|2 dx)≤ ε2
1 (10)
for anyε1 > 0, provided thatε0 is small enough (depending onε1, n, andN ), forv is an energy minimizer inH 1,2(B,RK). The fact thatRK is not compact causes
Unique solvability of the Dirichlet problem for weakly harmonic maps 389
no trouble, because we only have to deal with a compact subset, according tothe maximum principle. Also assume for the moment that we have already found aweakly harmonic mapu ∈ H 1,2(B,N)with small scaled energy and with boundarydatav.
Lemma 6. Let u ∈ H 1,2(B,N) be a weakly harmonic map, and let v ∈H 1,2(B,RK) satisfy �v = 0. If u − v ∈ H 1,2
0 (B,RK) and if (2) and (10) hold,then
‖u− v‖L∞ ≤ Cmax{ε, ε1} (11)
for a constant C depending only on n andN , provided that ε is small enough (alsodepending only on n and N ).
Proof. We use a method as in [14]. By the results of Bethuel [1],u ∈ C∞(B,N).According to Theorem 2.2 in [9], we even have
|∇u(x)| ≤ Cε
1− |x| , x ∈ B. (12)
We also know thatv ∈ C∞(B,RK).Set ε = max{ε, ε1} andw = u − v. Choose a pointx0 ∈ B and setR =
13 dist(x0, ∂B). Choose furthermore a cut-off functionζ ∈ C∞0 (B2R(x0)) with theproperties 0≤ ζ ≤ 1, ζ ≡ 1 onBR(x0), |∇ζ | ≤ CR−1, and|∇2ζ | ≤ CR−2. LetGx0 be the fundamental solution of Lapace’s equation with singularity inx0, i. e.
Gx0(x) =1
n(n− 2)ωn|x − x0|2−n,
whereωn = |B|. We have the identity
�
( |w|2ζ 2
2
)= 〈w,�w〉ζ 2+ |∇w|2ζ 2
+ 4〈w,∇w · ∇ζ 〉ζ + |w|2ζ�ζ + |w|2|∇ζ |2,so we can conclude
using the facts that�v = 0 and thatu is a weakly harmonic map, as well asYoung’sinequality. We obtain therefore
−�(|w|2ζ 2) ≤ C(|w||∇u|2+ |w|2(|∇ζ |2+ |�ζ |)).Multiply this withGx0 and integrate overB2R(x0). Integration by parts then yields
|w(x0)|2 ≤ CR−n∫B2R(x0)\BR(x0)
|w|2 dx + Cε2R−2∫B2R(x0)
|w|Gx0 dx, (14)
390 R. Moser
because of (12) and the estimates for|∇ζ | and|∇2ζ |. We will now estimate the twointegrals on the right-hand side. For this purpose choose the pointx1 ∈ ∂B havingthe distance 3R to x0. Then we get
R−n∫B2R(x0)\BR(x0)
|w|2 dx ≤ R−n∫B5R(x1)
|w|2 dx
≤ CR2−n∫B5R(x1)
|∇w|2 dx ≤ Cε2,
using the Poincaré inequality, and
R−2∫B2R(x0)
|w|Gx0 dx ≤ C‖w‖L∞ .
Therefore
‖w‖2L∞ ≤ Cε2+ Cε2‖w‖L∞ ≤ Cε2+ Cε2+ Cε2‖w‖2L∞ ≤ Cε2+ Cε2‖w‖2L∞by Young’s inequality. For a smallε this proves (11). ��
This is the key lemma of this section, for it is the one that will reduce the moregeneral case to the results of Sect. 2. The method we will use to achieve this is toapproximate the given boundary values by smooth maps in a certain way. The nextlemma will explain how this is done.
Lemma 7. Let u0 ∈ H 1,2(B,RK) be a map with boundary values g0 = u0|∂B ,such that g0(x) is in N for almost all x ∈ ∂B (with respect to the surface measureon ∂B). Suppose that u0 satisfies (3). Then there exist maps ht ∈ H 1,2(B,RK)
with smooth boundary values gt = ht |∂B for t ∈ (0,1], such that
(i) gt (x) ∈ N for all x ∈ ∂B,(ii) if t0 ∈ (0,1], then gt → gt0 in C∞ as t → t0,(iii) gt → g0 in L2 as t ↘ 0,(iv) sup|∇g1| ≤ Cε0, and(v) the maps ht satisfy
supx0∈∂Br>0
(r2−n
∫B∩Br(x0)
|∇ht |2 dx)≤ Cε2
0,
for a constantC that depends only on n andN , provided that ε0 is sufficiently small(depending on n and N ).
Proof. Extendg0 homogeneously of degree 0 toRn. Denote this extension byg0.
With the help ofu0 we can easily construct a maph0 ∈ H 1,2loc (R
n+1,RK) such thatg0 is the trace ofh0 on the hyperplaneRn × {0}, and with the property
supy0∈B3(0)×{0}
r>0
(r1−n
∫Bn+1r (y0)
|∇h0|2 dy)≤ Cε2
0.
Unique solvability of the Dirichlet problem for weakly harmonic maps 391
This can e. g. be achieved by extending
h(x, s) : = u0((1− |s|)x), x ∈ ∂B, |s| < 12,
homogeneously of degree 0 toRn× (−1
2,12) with respect to the firstn coordinates
and extending the resulting map periodically with respect tos.Choose a functionφ ∈ C∞0 (B) with
∫Bφ dx = 1, and setφt (x) = t−nφ(x/t)
for 0< t ≤ 1. Define now
gt = g0 ∗ φt ,ht (x, s) = (h0(·, s) ∗ φt )(x), x ∈ R
n, s ∈ R.
Of course the mapsgt are the traces ofht for all t ∈ (0,1] with respect toRn ×{0}. Observe that by the continuity of the trace operatorH 1,2(Bn+1
t (x,0)) →L2(Bt (x)), the estimate
t−n∫Bt (x)
|g0(y)− gt (x)|2 dy
= infc∈RK
(t−n
∫Bt (x)
|(g0(y)− c)− (φt ∗ (g0 − c))(x)|2 dy)
≤ C infc∈RK
(t−n
∫Bt (x)
|g0(y)− c|2 dy)
≤ Ct1−n∫Bn+1t (x,0)
|∇h0|2 dz ≤ Cε20
holds for allx ∈ B2(0). Thus, there is always a pointx ∈ Bt(x) such that
|g0(x)− gt (x)| ≤ Cε0.In particular we may assume that dist(gt (x),N) ≤ δ/2 for all x ∈ B2(0) and allt ∈ (0,1], if we just chooseε0 sufficiently small. Furthermore we have
|∇gt (x)| ≤ t−n−1‖∇φ‖L∞∫Bt (x)
∣∣∣g0 − (g0)Bt (x)
∣∣∣ dy ≤ Ct−1ε0
for all x ∈ B2(0), where(g0)Bt (x) = 1|Bt |
∫Bt (x)
g0 dy. Let t0 > 0. Then the mapsgtconverge inC∞ to gt0 for t → t0. Fort → 0, gt still converges tog0 in L2(B3(0)).Finally, if we writez = (x, s) ∈ R
n × R, then∫Bn+1r (x0,0)
|∇ht |2 dz =∫Bn+1r (x0,0)
∣∣∣∣∫Bt (0)
φt (y)∇h0(x − y, s) dy∣∣∣∣2
dz
≤∫Bt (0)|φt |2 dy
∫Bt (0)
∫Bn+1r (x0,0)
|∇h0(x − y, s)|2 dz dy
≤ Crn−1ε20
(15)
for x0 ∈ B2(0) andr > 0.
392 R. Moser
Now set
gt (x) = πN(
1
2λ
∫ 1+λ
1−λgt (σx) dσ
), x ∈ ∂B,
for an appropriateλ ∈ (0,1). Note that sinceg0 is homogeneous of degree 0, wehave
In particular if|1−σ | is small (independently oft), then|gt (σx)−gt (x)| < δ/2,and we can indeed applyπN to 1
2λ
∫ 1+λ1−λ gt (σx) dσ . The properties (ii)–(iv) follow
directly from the corresponding properties ofgt . We are now going to construct themapsht . Set
ht (z) = 1
2λ
∫ 1+λ
1−λht (σz) dσ, z ∈ (∂Bn+1)+.
For any ballBn+1r (z0) with z0 ∈ ∂B × {0} and r ≤ λ, we can cover the set
A = {σz : z ∈ Bn+1r (z0), 1−λ < σ < 1+λ} (see Fig. 2) with at most a constant
timesλ/r balls of radius 2r with center onB2(0)× {0}. Hence∫(∂Bn+1)+∩Bn+1
r (z0)
|∇ht |2 dHn ≤ Cλ
∫A
|∇ht |2 dz ≤ Cε20rn−2,
because of (15). All we have to do now is identify(∂Bn+1)+ with B by employinga bi-Lipschitz map and compose the map thus obtained with a smooth extension ofπN . The condition (v) is then satisfied.��
Unique solvability of the Dirichlet problem for weakly harmonic maps 393
This lemma together with Lemma 6 now finally enables us to prove an existenceresult that allows more general boundary data than Lemma 5.
Proposition 3. Let ε > 0 be fixed. There is a number ε0 > 0, depending on ε, n,and N , such that for all u0 ∈ H 1,2(B,RK) that satisfy (3), there exists an energyminimizing map u ∈ H 1,2(B,N) having the same boundary values as u0, providedthat g0(x) := u0|∂B(x) ∈ N almost everywhere. Furthermore, u can be chosensuch that (2) holds.
Proof. Chooseε0 so small that Lemma 7 can be applied. We use the mapsht andtheir boundary valuesgt constructed in that lemma, with the properties (i)–(v).From (v) we conclude by applying Lemma 5 that the mapsvt ∈ H 1,2(B,RK)
satisfying�vt = 0 andvt |∂B = gt all have the property
supx0∈Br>0
(r2−n
∫B∩Br(x0)
|∇vt |2 dx)≤ ε2
1 (16)
for a numberε1 that we are free to choose, provided thatε0 is small enough.AlthoughR
K is not compact, Lemma 5 may be applied here, since the image ofvtis contained in a compact subset ofR
K by the maximum principle.Let nowI be the set of all numberst ∈ [0,1] that admit an energy minimizing
maput ∈ H 1,2(B,N) with boundary valuesgt and satisfying
supx0∈Br>0
(r2−n
∫B∩Br(x0)
|∇ut |2 dx)≤ ε2. (17)
The estimate (iv) for∇g1 together with Lemma 5 implies that 1∈ I , for thehomogeneous extension of degree 0 ofg1 certainly satisfies the premises of thatlemma, ifε0 is small. Definet0 = inf I .
Let us first show thatt0 ∈ I . There is a sequencetk ↘ t0 with associated en-ergy minimizersutk satisfying (17). Lemma 2.9.1 in [12] tells us that a subsequenceconverges weakly inH 1,2(B,RK) and strongly inH 1,2
loc (B,RK) to a weakly har-
monic maput0. As gtk → gt0 in L2, the limit ut0 can only have boundary valuesgt0. The weak convergence implies (17) forut0. By Lemma 6, we may assume thatdist(vt0(x),N) ≤ δ almost everywhere. Ifε1 is sufficiently small, then Lemma5 can be applied withπN ◦ vt instead ofu0, proving the existence of an energyminimizing map with boundary datagt0 and satisfying (17). Hencet0 ∈ I .
Suppose now thatt0 > 0. According to (16) and Lemma 6, we know thatdist(vt0, N) ≤ δ/2, provided thatε1 andε are sufficiently small. Becausegt → gt0in C∞ for t → t0, there is at1 < t0 such that still dist(vt1, N) ≤ δ. An argumentas before shows thatt1 ∈ I , contradicting the definition oft0. This completes theproof. ��
The proof of Theorem 2 is now just an application of Lemma 5, Proposition 3,and Theorem 1. The former two are already proved, and Theorem 1, i.e. the unique-ness result, will be established in the next section.
394 R. Moser
4. Uniqueness
In this section we will prove Theorem 1. To this end, letu, v ∈ H 1,2(B,N) be twoweakly harmonic maps which satisfyu− v ∈ H 1,2
0 (B,RK) and (2) for a numberε > 0 that remains to be determined. Ifε is sufficiently small, then we can applyTheorem I-4 in [1]. Although the statement of that theorem requiresu resp.v tobe stationary, it applies as well to this case, for in its proof only the consequence(2) of stationarity is used. Thusu, v ∈ C∞(B,N). As before we can then applyTheorem 2.2 in [9] to obtain the bounds
|∇u(x)|, |∇v(x)| ≤ Cε
1− |x| , x ∈ B, (18)
for the gradients ofu andv, provided thatε is small enough. Furthermore, thefollowing holds for the mapw = u− v.Lemma 8. There exists a constant C such that
‖w‖L∞ ≤ C[w]∗,where
[w]∗ = supx0∈∂Br>0
(r2−n
∫B∩Br(x0)
|∇w|2 dx) 1
2
,
provided that ε is sufficiently small (depending onn andN ). The constantC dependsonly on n and N .
Proof. This inequality can be proved like Lemma 6, except that instead of thequantity|〈w,A(u)(∇u,∇u)〉| in (13) we have now
if ε is small. Moreover, sinceh = 0 on∂B∩Br(x0), we may use standard estimatesto find that
s−n∫B∩Bs(x0)
|∇h|2 dx ≤ Cr−n∫B∩Br(x0)
|∇h|2 dx
≤ Cr−n∫B∩Br(x0)
|∇w|2 dx
for s ≤ r, the latter inequality following from the energy minimizing property ofh. Note that this inequality is trivial ifr ≥ 2, for in that caseh ≡ 0. Now compute
r2−n∫B∩Br(x0)
|∇w − ∇h|2 dx
= r2−n∫B∩Br(x0)
〈w − h, (A(u)− A(v))(∇u,∇u)+ A(v)(∇w,∇u+ ∇v)〉 dx
≤ Cr2−n∫B∩Br(x0)
|w − h| (|w| |∇u|2+ |∇w| (|∇u| + |∇v|)) dx
≤ Cε2‖w‖2L∞ + Cε‖w‖L∞[w]∗ ≤ Cε[w]2∗by integration by parts, assuming thatε ≤ 1 and using (19) and Lemma 8. For anys ≤ r, these estimates imply
396 R. Moser
s2−n∫B∩Bs(x0)
|∇w|2 dx ≤ 2s2−n∫B∩Br(x0)
|∇w − ∇h|2 dx
+ 2s2−n∫B∩Bs(x0)
|∇h|2 dx
≤ C(( sr
)2−nε +
( sr
)2)[w]2∗.
Chooses = ε1/nr, then
s2−n∫B∩Bs(x0)
|∇w|2 dx ≤ Cε2/n[w]2∗.
Since this holds for anyx0 ∈ ∂B and for anyr > 0, we have in fact
[w]∗ ≤ Cε1/n[w]∗,
which meansw ≡ 0 if ε is small. The proof of Theorem 1 is thus complete.
Remark 2. As mentioned in the proof of Lemma 6, the method we use to obtaintheL∞-estimate of Lemma 8 is from [14]. It is used there to get an estimate ofthe seminorm ofu in the space BMO of functions of bounded mean oscillationas introduced by John–Nirenberg in [6]. Similar estimates (also forL∞) can beproved for the heat flow of harmonic maps (i. e. the evolution problem associatedto (1)), provided that a regularity result corresponding to the one of Bethuel [1]holds. This is true e. g. if the target manifoldN is a sphereSK−1 (see Feldman [3]or Chen–Li–Lin [2]). Then uniqueness for solutions of the heat flow of harmonicmaps with small scaled energy and given initial values can be proved, as done byStruwe in [13].
5. Proof of Theorem 3
We use a method due to Giaquinta–Modica to prove the higher regularity result.Let u0 satisfy the conditions of Theorem 2 and
supx0∈Br>0
(rp−n
∫B∩Br(x0)
|∇u0|p dx)≤ εp0
for an exponentp > 2, whereε0 is chosen such that all the results of Sects. 2 and3 can be applied. Letv ∈ u0 +H 1,2
0 (B,RK) satisfy�v = 0. We show first thatvsatisfies a higher integrability condition like the one we want to prove foru. Oncethis is done, we may work withv instead ofu0, which has the advantage that wewill be able to apply the results of Sect. 2.
According to Lemma 2.8.1 in [12], Lemma 3, and the Sobolev inequality,v
satisfies the estimate
Unique solvability of the Dirichlet problem for weakly harmonic maps 397
∫Br/2(x0)
|∇v|2 dx ≤ C
r2
∫Br(x0)
|v − vBr (x0)|2 dx
≤ C
r2
(∫Br(x0)
|∇v| 2nn+2 dx
) n+2n
if Br(x0) ⊂ B, and
∫B∩Br/2(x0)
|∇v|2 dx ≤ µ∫B∩Br(x0)
|∇v|2 dx + Cµ
∫B∩Br(x0)
|∇u0|2 dx
+ C
µr2
∫B∩Br(x0)
|v − u0|2 dx
≤ µ∫B∩Br(x0)
|∇v|2 dx + Cµ
∫B∩Br(x0)
|∇u0|2 dx
+ C
µr2
(∫B∩Br(x0)
|∇v| 2nn+2 dx
) n+2n
for x0 ∈ ∂B andr ∈ (0, 12], and for an arbitrarily small constantµ > 0, provided
that ε0 is small enough. (As a matter of fact,v would satisfy this even ifε0 wasnot small, because it is the solution of a linear elliptic equation. However, we aregoing to repeat this argument withu instead ofv, and then it would be no longertrue without that condition. For the same reason, the inequalities above are derivedin a rather unusual way.) Define
f = |∇v| 2nn+2 , g = |∇u0| 2n
n+2 .
Then the following inequality holds forp = n+2n
and for anyx0 ∈ B and anyr ∈ (0, 1
2]:( r
8
)−n ∫B∩Br/8(x0)
f p dx ≤ Cµr−n∫B∩Br(x0)
f p dx + Cµr−n
∫B∩Br(x0)
gpdx
+ Cµ
(r−n
∫B∩Br(x0)
f dx
)p.
According to a lemma due to Giaquinta–Modica, based upon an idea of Gehring(see e. g. Proposition V.1.1 in [4]), this already guarantees that there is a numberq > p, such that
(r−n
∫B∩Br/8(x0)
f q dx
) 1q
≤ C(r−n
∫B∩Br(x0)
f p dx
) 1p
+ C(r−n
∫B∩Br(x0)
gq dx
) 1q
398 R. Moser
for all x0 ∈ B andr > 0, if justµ is chosen small enough. In terms ofv andu0,this means
rq−n∫B∩Br/8(x0)
|∇v|q dx ≤ C(r2−n
∫B∩Br(x0)
|∇v|2 dx) q
2
+ Crq−n∫B∩Br(x0)
|∇u0|q dx
for q = 2nn+2 q. Note thatq > 2. If we chooseq no bigger thanp, then we can
achieve, by makingε0 smaller, that
supx0∈Br>0
(rq−n
∫B∩Br(x0)
|∇v|q dx)≤ εq0
for any ε0 > 0. Because of Lemma 6, we may assume thatv ∈ H 1,2(B,N)
(compose it withπN ). In particular, we can apply Lemma 3 tou andv. So we canrepeat the whole argument above withu instead ofv andv instead ofu0. Eventuallythis proves the claim of Theorem 3.
Acknowledgements. This work covers parts of my Ph.D. thesis at the ETH Zürich. I wouldlike to thank M. Struwe for drawing my attention to this topic and for his support. I am alsomost grateful to T. Ilmanen and P. Harpes for their numerous remarks on my work and forvarious fruitful discussions.
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