Electronic Journal of Differential Equations, Vol. 2006(2006), No. 15, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)
A LIOUVILLE THEOREM FOR F-HARMONIC MAPS WITH FINITE F-ENERGY
M’HAMED KASSI
Abstract. Let (M, g) be a m-dimensional complete Riemannian manifold with a pole, and (N, h) a Riemannian manifold. LetF:R+→R+be a strictly increasingC2 function such thatF(0) = 0 anddF := sup(tF0(t)(F(t))−1)<
∞. We show that ifdF< m/2, then everyF-harmonic mapu:M→Nwith finiteF-energy (i.e a local extremal ofEF(u) :=R
MF(|du|2/2)dVgandEF(u) is finite) is a constant map provided that the radial curvature ofMsatisfies a pinching condition depending todF.
1. Introduction and statement of result
Let (M, g) and (N, h) be two Riemannian manifolds andFbe a givenC2function F :R+→R+. Then, a mapu:M →N of classC2is said to beF-harmonic if for every compactK ofM, the mapuis extremal ofF-energy:
EF(u) :=
Z
K
F(|du|2 2 )dVg.
In a normal coordinate system, the tension field associated with EF(u) by the Euler-Lagrange equations is
τF(u) :=
m
X
i=1
(∇ei(F0(|du|2
2 )du))ei=F0(|du|2
2 )τ(u) +du.n
grad F0(|du|2 2 )o whereτ(u) is the usual tension field ofudefined by
τ(u)k = ∆Muk+
n;m
X
β,γ;i,j
NΓkαγ(u)gij∂uβ
∂xi
∂uγ
∂xj
, k= 1, . . . , n .
Then, the mapuisF-harmonic ifτF(u) = 0. For further properties ofF-harmonic maps, we refer the reader to [1, 2]. For the particular case ofF(t) =t, the Liouville problem for harmonic maps with finite energy have been studied in [4, 6, 7, 8, 9].
While for F(t) = 2ptp/2, with p ≥ 2, this is the problem of p-harmonic maps with finite p-energy (corollary 1.2. If F(t) = √
1 + 2t−1 corresponding to the minimal graph (corollary 1.3). In this paper, we study the same problem for F- harmonic maps with finite F-energy without condition on the curvature for the
2000Mathematics Subject Classification. 58E20, 53C21, 58J05.
Key words and phrases. F-harmonic maps; Liouville propriety; Stokes formula;
comparison theorem.
c
2006 Texas State University - San Marcos.
Submitted March 24, 2005. Published January 31, 2006.
1
target manifold. We assume that F is strictly increasing, F(0) = 0, and dF = suptFF(t)0(t)<∞, “the degree ofF”. Forxin M, we setr(x) =dg(x, x0).
Theorem 1.1. Let (M, g) be a m-dimensional complete Riemannian manifold, m >2, with a pole x0, and let (N, h) be a Riemannian manifold. If dF < m/2, then everyF-harmonic map ofM intoN with finite F-energy is constant provided that the radial curvatureKr ofM satisfies one of the following two conditions:
(i) −α2≤Kr≤ −β2 withα >0, β >0 and1 + (m−1)β−2dFα >0
(ii) −1+rα2 ≤Kr ≤ 1+rβ2 with α≥0 and β ∈[0,14]such that 2 + (m−1)(1 +
√1−4β)−2dF(1 +√
1 + 4α)>0.
Furthermore, we have the following corollaries.
Corollary 1.2. Let (M, g) and (N, h) be as in the theorem. Then, every C2 p- harmonic map ofM intoN with finitep-energy, forp < m, is constant.
Corollary 1.3. Let (M, g) and (N, h) be as in the theorem. Then, for m > 2, every C2 map uof M intoN, with finite energy, solution of
τ(u)
p1 +|du|2 +du.n
grad 1
p1 +|du|2 o
= 0 is constant.
For m = 2, the statement of the theorem is false in general. In fact, for the case (i), there exist holomorphic maps of the hyperbolic disc with finite energy [9].
While for the case (ii) there exist holomorphic maps ofCintoP1with finite energy [8].
2. Proof of Theorem 1.1
LetX andY be two vector fields on M. It is well-known [3, 6], that the stress- energy for harmonic maps is
Su:= |du|2
2 hX, Yig− hdu(X), du(Y)ih
and satisfies
(divSu)(X) =−hτ(u), du(X)ih.
Following [2], we define the stress-energy ofF-harmonic maps by SF,u(X, Y) :=F(|du|2
2 )hX, Yig−F0(|du|2
2 )hdu(X), du(Y)ih.
When F(t) := t we have SF,u := Su. Also (divSF,u)(X) = −hτF(u), du(X)ih
thanks to the following lemma.
Lemma 2.1. For every vector field X onM, we have
(divSF,u)(X) =−hτF(u), du(X)ih, (2.1) div(F(|du|2
2 )X)
= div(F0(|du|2
2 )hdu(X), du(ei)ihei)− hτF(u), du(X)ih+ [SF,u, X],
(2.2)
where
[SF,u, X](x) =
m
X
i,j=1
F(|du|2
2 )δij−F0(|du|2
2 )hdu(ei), du(ej)ih
h∇eiX, ejig.
In particular, ifuisF-harmonic andD⊂⊂M is a C1 boundary domain, then we
have Z
∂D
SF,u(X, ν)dσg= Z
D
[SF,u, X]dVg whereν is the normal to∂D.
Proof. Letx∈M. Chose a normal coordinate system such that atx. gij(x) =δij
dg(x) = 0, where (e1, . . . , em) being a normal basis, we have∇ejek = 0 for allj, k and
(divSF,u)(X)
=
m
X
i=1
n∇eiSF,u(ei, X)−SF,u(ei,∇eiX)−SF,u(∇eiei, X)o
=
m
X
i=1
n∇ei F(|du|2
2 )hei, Xi − hF0(|du|2
2 )du(ei), du(X)i
−F(|du|2
2 )hei,∇eiXi +F0(|du|2
2 )hdu(ei), du(∇eiX)i −SF,u(∇eiei, X)o
=
m
X
i=1
n∇ei F(|du|2
2 )hei, Xi
− ∇ei hF0(|du|2
2 )du(ei), du(X)i
−F(|du|2
2 )hei,∇eiXi +F0(|du|2
2 )hdu(ei), du(∇eiX)i −SF,u(∇eiei, X)o
=
m
X
i=1
n Xm
j=1
F0(|du|2
2 )h∇ei(du(ej)), du(ej)i hei, Xi
+F(|du|2
2 )∇eihei, Xi − h∇ei(F0(|du|2
2 )du(ei)), du(X)i
−F0(|du|2
2 )hdu(ei),∇ei(du(X))i
−F(|du|2
2 )hei,∇eiXi+F0(|du|2
2 )hdu(ei), du(∇eiX)i
−SF,u(∇eiei, X)o . Thus
(divSF,u)(X) =
m
X
i,j=1
n
F0(|du|2
2 )h∇ei(du(ej)), du(ej)iXi
o
+
m
X
i=1
n F(|du|2
2 )h∇eiei, Xi+F(|du|2
2 )hei,∇eiXi
− h∇ei(F0(|du|2
2 )du(ei)), du(X)i
−F0(|du|2
2 )hdu(ei),∇ei(du(X))i −F(|du|2
2 )hei,∇eiXi +F0(|du|2
2 )hdu(ei), du(∇eiX)i −SF,u(∇eiei, X)o
=
m
X
i,j=1
nF0(|du|2
2 )hXi∇ei(du(ej)), du(ej)io
−
m
X
i=1
n
F0(|du|2
2 )hdu(ei),∇ei(du(X))i +F0(|du|2
2 )hdu(ei), du(∇eiX)i+F(|du|2
2 )h∇eiei, Xi +F(|du|2
2 )hei,∇eiXi −F(|du|2
2 )hei,∇eiXi
− h∇ei(F0(|du|2
2 )du(ei)), du(X)i −SF,u(∇eiei, X)o .
Since ∇eiei = 0, with (∇eidu)(X) =∇ei(du(X))−du(∇eiX) and by symmetry (∇eidu)(X) = (∇Xdu)(ei), we have
div(SF,u)(X) =
m
X
j=1
n
F0(|du|2
2 )h∇X(du(ej), du(ej)io
−
m
X
i=1
n
F0(|du|2
2 )hdu(ei),∇ei(du(X))−du(∇eiX)i
− h∇ei(F0(|du|2
2 )du(ei)), du(X)io . Finally,
div(SF,u)(X) =−hτF(u), du(X)i.
Also
div(F(|du|2 2 )X) =
m
X
i=1
h∇ei(F(|du|2
2 )X), eii
=
m
X
i=1
nh∇ei(F(|du|2
2 ))X, eii+F(|du|2
2 )h∇eiX, eiio
=∇XF(|du|2 2 ) +
m
X
i=1
F(|du|2
2 )h∇eiX, eii.
Then, by straightforward computation, we obtain
∇XF(|du|2 2 ) =
m
X
i=1
1
2F0(|du|2
2 )∇Xhdu(ei), du(ei)i
=
m
X
i=1
F0(|du|2
2 )h∇X(du(ei)), du(ei)i
=
m
X
i=1
F0(|du|2
2 )h(∇Xdu)(ei) +du(∇Xei), du(ei)i
=
m
X
i=1
F0(|du|2
2 )h(∇Xdu)(ei), du(ei)i
=
m
X
i=1
F0(|du|2
2 )h(∇eidu)(X), du(ei)i (by symmetry)
=
m
X
i=1
nh∇ei(du(X)), F0(|du|2
2 )du(ei)i
−F0(|du|2
2 )hdu(∇eiX), du(ei)io Thus
∇XF(|du|2 2 ) =
m
X
i=1
n∇eihdu(X), F0(|du|2
2 )du(ei)i
− hdu(X),∇ei(F0(|du|2
2 )du(ei))i
−F0(|du|2
2 )idu(∇eiX), du(ei)io
=
m
X
i=1
ndiv F0(|du|2
2 )idu(X), du(ei)iei
+hdu(X),−∇ei(F0(|du|2
2 )du(ei))i
−F0(|du|2
2 )hdu(∇eiX), du(ei)io
=
m
X
i=1
n
div F0(|du|2
2 )hdu(X), du(ei)ieio
− hdu(X), τF(u)i −
m
X
i=1
F0(|du|2
2 )hdu(∇eiX), du(ei)i Thus
div(F(|du|2 2 )X) =
m
X
i=1
n
div F0(|du|2
2 )hdu(X), du(ei)ieio
− hdu(X), τF(u)i+ [SF,u, X]
with
[SF,u, X] =
m
X
i,j=1
F(|du|2
2 )δij−F0(|du|2
2 )hdu(ei), du(ej)ih
h∇eiX, ejig
because ∇eiX =h∇eiX, ejiej. If D ⊂⊂M is a C1 boundary domain, we get by the use of Stokes formula
Z
D
(divSF,u)(X) + Z
D
[SF,u, X]
= Z
D
div(F(|du|2 2 )X)−
Z
D m
X
i=1
div F0(|du|2
2 )< du(X), du(ei)> ei
= Z
∂D
F(|du|2
2 )hX, νi − Z
∂D
F0(|du|2
2 )hdu(X), du(ν)i. Thus, ifuisF-harmonic:
Z
∂D
F(|du|2
2 )hX, νi −F0(|du|2
2 )hdu(X), du(ν)i
= Z
D
[SF,u, X].
This completes the proof.
Lemma 2.2. Letu:M →N be aF-harmonic with finiteF-energy andX a vector field onM such that|X| ≤φ(r)forφ:R+→R+ satisfying
Z +∞
1
dt
φ(t) = +∞.
Then there exists an increasing strictly sequence(Rn)such that
n→∞lim Z
B(x0,Rn)
[SF,u, X]dVg= 0.
Proof. SincetF0(t)≤dFF(t) we have
Z
B(x0,R)
[SF,u, X]
≤ Z
∂B(x0,R)
F(|du|2
2 )hX, νi +
Z
∂B(x0,R)
F0(|du|2
2 )idu(X), du(ν)i
≤ Z
∂B(x0,R)
F(|du|2
2 )|hX, νi|+ Z
∂B(x0,R)
F0(|du|2
2 )|idu(X), du(ν)i|
≤(1 + 2dF) Z
∂B(x0,R)
F(|du|2 2 )|X|. By the Co-area formula and|X| ≤φ(r(x)),
Z ∞
0
1 φ(t)
Z
∂B(x0,t)
F(|du|2 2 )|X|
dt= Z
M
|X||∇r|
φ(r) F(|du|2 2 )
≤ Z
M
F(|du|2 2 )<∞ Since R∞
1 dt
φ(t) = ∞, there exists a increasing strictly sequence (Rn) such that limn→∞R
∂B(x0,Rn)F(|du|22)|X|= 0. Hence
n→∞lim Z
B(x0,Rn)
[SF,u, X]dVg= 0.
This completes the proof of Lemma 2.2.
For the theorem, it suffices to chooseX satisfying Lemma 2.2 and the condition [SF,u, X]≥cF(|du|2/2) wherec >0 is a constant. For that we takeX =r∇rand using the comparison theorem of the Hessian [5].
Theorem 2.3(Comparison theorem). Let(M, g)be a complete Riemannian man- ifold with a pole x0 and k1, k2 be two continuous functions on R+ such that
k2(r)≤Kr≤k1(r), whereKr is the radial curvature of M, i.e., the sectional cur- vature of the tangent planes containing the radial vector∇r. Also, let Ji (i= 1,2) be the solution of classical Jacobi equation
Ji00+kiJi= 0; Ji(0) = 0 and Ji0(0) = 1.
Then, ifJ1>0 onR+, we have on M\ {x0} J10(r)
J1(r)(g−dr⊗dr)≤Hess(r)≤ J20(r)
J2(r)(g−dr⊗dr). Case (i) of Theorem 2.3: Withk1(r) =−β2 andk2(r) =−α2, we have
βcoth(βr)(g−dr⊗dr)≤Hess(r)≤αcoth(αr)(g−dr⊗dr).
Case (ii) of Theorem 2.3: Withk1(r) = rβ2 and k2(r) =−rα2, and the fact that on M\ {x0},
−α
r2 ≤ − α
1 +r2 ≤Kr≤ β 1 +r2 ≤ β
r2 we have
1 +√ 1−4β 2r
(g−dr⊗dr)≤Hess(r)≤1 +√ 1 + 4α 2r
(g−dr⊗dr). Lemma 2.4. Under hypothesis of Theorem 2.3, in case (1), we have
[SF,u, X]≥(1 + (m−1)β−2dFα)F(|du|2 2 ) and in case (ii),
[SF,u, X]≥ 1
2(2 + (m−1)(1 +p
1−4β)−2dF(1 +√
1 + 4α)F(|du|2 2 ).
Proof. First note that [SF,u, X] =
m
X
i,j=1
F(|du|2
2 )δij−F0(|du|2
2 )hdu(ei), du(ej)ih
<∇eiX, ejig, where (e1, . . . , em−1,∂r∂) with em = ∂r∂ , being a normal basis on B(x0, R). Then, sinceX=r∂r∂ , it follows that∇∂
∂rX = ∂r∂ and so we get h∇∂
∂rX, ∂
∂rig= 1,
h∇eiX, eiig=rHess(r)(ei, ei), fori= 1, . . . , m−1,
∇eiX =
m−1
X
j=1
rHess(r)(ei, ej)ej, fori= 1, . . . , m−1.
Therefore,
[SF,u, X] =F(|du|2 2 )(1 +
m−1
X
i=1
rHess(r)(ei, ei))
−
m−1
X
i,j=1
F0(|du|2
2 )hdu(ei), du(ej)ihh∇eiX, ejig
−F0(|du|2 2 )hdu(∂
∂r), du(∂
∂r)ihh∇∂
∂rX, ∂
∂rig
−
m−1
X
j=1
F0(|du|2 2 )hdu(∂
∂r), du(ej)ihh∇∂
∂rX, ejig
−
m−1
X
i=1
F0(|du|2
2 )hdu(ei), du( ∂
∂r)ihh∇eiX, ∂
∂rig
=F(|du|2 2 )(1 +
m−1
X
i=1
rHess(r)(ei, ei))
−
m−1
X
i,j=1
F0(|du|2
2 )hdu(ei), du(ej)irHess(r)(ei, ej)
−F0(|du|2 2 )hdu(∂
∂r), du(∂
∂r)i For the case (i), we have
[SF,u, X]≥F(|du|2
2 ) + (m−1)(βr) coth(βr)F(|du|2 2 )
−F0(|du|2
2 )|du|2(αr) coth(αr) +F0(|du|2
2 )((αr) coth(αr)−1)hdu(∂
∂r), du(∂
∂r)i
≥F(|du|2
2 ) +F(|du|2
2 )((m−1)(βr) coth(βr)−2dF(αr) coth(αr))
≥F(|du|2
2 ) +F(|du|2
2 )rcoth(βr)((m−1)β−2dFαcoth(αr) coth(βr)).
Since the function coth(x) is decreasing and, xcoth(x) is bounded below by a positive constant inR+, we have
[SF,u, X]≥(1 + (m−1)β−2dFα)F(|du|2 2 ) For the case (ii), we have
[SF,u, X]≥F(|du|2
2 ) + (m−1)aF(|du|2
2 )−bF0(|du|2 2 )|du|2 + (b−1)F0(|du|2
2 )hdu(∂
∂r), du( ∂
∂r)i
≥(1 + (m−1)a−2dFb)F(|du|2 2 ), where we have set
a=1 +√ 1−4β
2 and b= 1 +√
1 + 4α
2 ≥1.
Acknowledgements. I would like to express my gratitude to Professor S. Asserda for his valuable suggestions and warm encouragement. Also I thank the anonymous referee for many valuable comments.
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M’hamed Kassi
Equipe d’Analyse Complexe, Laboratoire d’Analyse Fonctionnelle, Harmonique et Com- plexe, D´epartement de Math´ematiques, Facult´e des Sciences, Universit´e Ibn Tofail, K´enitra, Maroc
E-mail address:[email protected]
