Approximate solutions to the Dirichlet problem for harmonic maps between
  hyperbolic spaces
APPROXIMATE SOLUTIONS TO THE DIRICHLET PROBLEM
FOR HARMONIC MAPS BETWEEN HYPERBOLIC SPACES
DUONG MINH DUC AND TRUONG TRUNG TUYEN
Abstract. Our main result in this paper is the following: Given Hm, Hn
hyperbolic spaces of dimensional m ≥ 2 and n corresponding, and given a
Holder function f = (f1, ..., fn−1) : ∂Hm → ∂Hn between geometric bound-
aries of Hm and Hn. Then for each ǫ > 0 there exists a harmonic map
u : Hm → Hn which is continuous up to the boundary (in the sense of Eu-
clidean) and u|∂Hm = (f
1, ..., fn−1, ǫ).
1. Introduction
Let Hm and Hn are hyperbolic spaces with dimensions m ≥ 2 and n correspond-
ingly. For convenience, we use the upper-half space models for Hm and Hn. So
Hm = {(x1, ..., xm) ∈ IRm : xm > 0}, Hn = {(y1, ..., yn) ∈ IRn : yn > 0} with
metrics
d2Hm =
(xm)2
((dx1)2 + ...+ (dxm)2),
d2Hn =
(yn)2
((dy1)2 + ...+ (dyn)2).
So the tension fields of u = (y1, ..., yn) is
τα = (xm)2(∆0y
< ∇0y
α,∇0y
n >),
for 1 ≤ α ≤ n− 1 and
τn(u) = (xm)2(∆0y
α|2 − |∇0y
n|2)),
where ∇0 is the Euclidean gradient and ∆0 is the Euclidean Laplacian.
A C2 map u : Hm → Hn is called a harmonic map if τ(u)s = 0 for all s =
1, 2, ..., n. The literature about harmonic maps between Riemannian manifolds are
abundant, we refer the readers to the classical work [4].
One of the interesting problems for harmonic maps is that of the Dirichlet prob-
lem at infinity: Given ∂Hm and ∂Hn geometric boundaries of Hm and Hn, and
given a continuous map f : ∂M → ∂N (here continuity is understood in the sense
Date: October 22, 2018.
2000 Mathematics Subject Classification. 53A35.
Key words and phrases. Dirichlet problems; Harmonic functions; Hyperbolic spaces.
This work has been initiated when the second author was at Department of mathematics,
University of natural sciences, Hochiminh city, Vietnam. He would like to thank Professor Dang
Duc Trong for his many invaluable helps. He also would like to express his thankfulness to
Professor F. Helein, Professor R. Schoen, and Mr. Le Quang Nam for their generous help.
http://arxiv.org/abs/0704.0087v2
2 DUONG MINH DUC AND TRUONG TRUNG TUYEN
of Euclidean), is there a harmonic map u : Hm → Hn such that in Euclidean sense
u is continuous up to the boundary ∂Hm and takes boundary value f?
For this problem with some more requirements for the smoothness of f , there are
many results. In three papers [8], [9] and [7], Li and Tam established the existence
and uniqueness of a harmonic function u which is C1 up to the boundary and has
boundary value f , provided f is C1. But for more general types of f , according to
our knowledge, there is no answer to the existence of a solution u.
In this paper we establish the existence of approximate solutions to the Dirichlet
problem for harmonic maps between two hyperbolic spaces with prescribed bound-
ary value. More explicitly, we prove the following result
Theorem 1. Let f : Hm → Hn be a bounded uniformly continuous. Let functions
g and ϕ be as in Section 2. Assume that
t−1g(t)dt < ∞, in particular, this
condition is satisfied if f is Holder continuous. For each ǫ > 0, there exists a
harmonic map uǫ : H
m → Hn which is continuous up to the boundary ∂Hm and
u|∂Hm = (f
1, ..., fn−1, ǫ).
Our strategy for proving this result is the follows: First, we construct an initial
map, i.e., a C2 map v = (v1, ..., vn−1, vn) : Hm → Hn which has boundary value
f for any continuous map f : ∂Hm → Hn. For this step we follows the ideas in
[9], with some changes: Since the function f needs not to differentiable, we can
not take vn as in [9], and the function vn of ours is a function of one variable xm.
Then, we use this function to produce harmonic maps uǫ : H
m → Hn which takes
boundary value (f1, ..., fn−1, vn + ǫ) for every ǫ > 0.
2. Initial maps
In this part, we use the techniques in [9] to construct good initial maps v having
the map f : ∂Hm → ∂Hn as the boundary value.
Let f : IRm−1 → IRn−1 be a uniformly continuous bounded function. Let
g : Hm → (0.∞) be C2, bounded and
g(x′, xm) = 0,
uniformly in x′.
We denote by v = {f, g} : Hm → Hn the extension of f defined as follows
vα(x′, xm) =
xmfα(y′)
(|x′ − y′|2 + (xm)2)m/2
for 1 ≤ α ≤ n− 1 and
vn(x′, xm) = g(x′, xm).
By results in [9] (pp. 628-630) we have
(i) v is C2 and up to the boundary given by xm = 0 it is continuous.
(ii) If 1 ≤ α ≤ n− 1 then
xm|∇0v
α| = 0,
uniformly in x′.
Moreover, by estimates of elliptic PDEs (see Theorem 2.10 in [5]), noting that
vα is bounded, there exists constants C > 0 such that
(2.1) max{(xm)3|D3vα|, (xm)2|D2vα|, (xm)|∇0v
α|} ≤ C.
APPROXIMATE SOLUTIONS TO THE DIRICHLET PROBLEM FOR HARMONIC MAPS BETWEEN HYPERBOLIC SPACES3
We put
g(r) = sup
x′,y′∈IRm−1, |x′−y′|≤r
|f(y′)− f(x′)|,
ϕ(r) =
s2 + r2
g(s)ds.
Since g is monotone it follows that g is Lebesgue measurable. Moreover, since g is
bounded, we see that ϕ is well-defined.
Using polar coordinates with center at x′ we see that there exists a constant
C > 0 such that ∫
IRm−1
xm|f(y′)− f(x′)|
(|x′ − y′|2 + (xm)2)m/2
≤ Cϕ(xm),
for all x′ ∈ IRm−1.
Since f is uniformly continuous we see that
g(r) = 0.
Now we show that
ϕ(xm) = 0.
Indeed, for any ǫ > 0, we find δ > 0 such that
g(s) ≤ ǫ,
if 0 < s ≤ δ. So, if K = sup
s∈IR g(s) we have
ϕ(r) =
s2 + r2
g(s)ds+
s2 + r2
g(s)ds
s2 + r2
s2 + r2
= ǫ arctan(δ/r) +K(π/2− arctan(δ/r)).
Letting r → 0 we see that
lim sup
ϕ(r) ≤ ǫπ/2.
Since ǫ > 0 is arbitrary, we see that
ϕ(r) = 0.
Thus, if we put v = {f, ϕ(xm)} we see that v is an extension of f . Moreover we
have the following result
Lemma 1. Let f : ∂Hm → ∂Hn be nonconstant, uniformly continous and
bounded. Put v = {f, ϕ(xm)} as above. Then v is smooth, up to the boundary
it is continuous, v|IRm−1 = f and there exists C > 0 such that for x
m near 0 we
||τ(v)||2 ≤ C.
4 DUONG MINH DUC AND TRUONG TRUNG TUYEN
Proof. By Section 6 in [9] we have
|(xm)∇0v
α| ≤ C3|ϕ(x
where 1 ≤ α ≤ n− 1 and C3 is a positive constant.
Directly computation gives
ϕ′(r) =
s2 − r2
(s2 + r2)2
g(s)ds,
ϕ”(r) =
(s2 + r2)2
g(s)ds+
−4r(s2 − r2)
(s2 + r2)3
g(s)ds.
max{|rϕ′(r)|, |r2ϕ”(r)|} ≤ C4ϕ(r),
where C4 is a constant.
Since g is increasing, g′ exists almost everywhere and g′ ≥ 0. Using integration
by parts, noting that d
s2+r2
) = s
(s2+r2)2
, we have
ϕ′(r) =
s2 − r2
(s2 + r2)2
g(s)ds
s2 + r2
g(s)|∞0 +
s2 + r2
g′(s)ds
s2 + r2
g′(s)ds.
Differentiating the last term in above equality we get
ϕ”(r) = −
(s2 + r2)2
g′(s)ds.
Since f is nonconstant we see easily that g′ 6≡ 0 (in fact, we don’t need this
restriction since we can add g with a non-constant positive function, for example
(xm)1/2 ). So since g′ ≥ 0, it follows from above equalities that
ϕ′(r) > 0,
|rϕ”(r)| ≤ C5ϕ
′(r),
where C5 is a positive constant. Then use the formula for the tension field we are
done. �
3. Proof of Theorem 1
Proof. Fixed ǫ > 0. We define vǫ : H
m → Hn as follows:
vǫ(x) = (v
1(x), v2(x), ..., vn−1(x), ϕ(xm) + ǫ).
For each δ > 0 denote uǫ,δ : H
m k Ωδ = {x
m > δ} → Hn the harmonic map
taking value vǫ on ∂Ωδ.
By inequality (2.1) in [2] and properties of v and vǫ (see Lemma 1) we have
∆HmdHn(uǫ,δ, vǫ) ≥ −|τ(vǫ)| ≥ −C
ϕ(xm)
ϕ(xm) + ǫ
ϕ(xm),
for all x ∈ Ωδ, and here C is one constant from Lemma 1.
APPROXIMATE SOLUTIONS TO THE DIRICHLET PROBLEM FOR HARMONIC MAPS BETWEEN HYPERBOLIC SPACES5
We claim that the function
ψ(r) =
u−2ϕ(u) du ds
is well-defined for r ≥ 0. In fact, using the formula for ϕ we have
ψ(r) =
u−2ϕ du ds =
u−1(u2 + t2)−1g(t)dt du ds.
Since the integrand is non-negative, using Fubini’s theorem we have∫ r
u−1(u2 + t2)−1g(t)dt du ds =
u−1(u2 + t2)−1g(t)du dt ds
log(1 +
)g(t)dt ds
log(1 +
)g(t)ds dt
πt2 − 2 arctan( t
)t2 + r log(1 + t
g(t)dt.
Now since g(t) is bounded we have
πt2 − 2 arctan( t
g(t)dt
is convergent. Fixed r ≥ 0, near t = 0 we have
r log(1 + t
g(t) ≈ t−1g(t),
and when t→ ∞ we have
r log(1 + t
g(t) ≈ t−3g(t),
hence since g(t) is bounded and the assumption that
t−1g(t) converges, our claim
is verified.
We use the same ψ to denote the function ψ : Hm → IR defined by ψ(x) = ψ(xm)
for x = (x1, ..., xm−1, xm) ∈ Hm. Now we have ψ′(r) =
u−2ϕ du > 0 and
ψ”(r) = −r−2ϕ(r), since m ≥ 2 we have
∆Hm(−ψ(x)) = −(x
m)2[ψ”(xm)−
(m− 2)
ψ′(xm)] ≥ −(xm)2ψ”(xm) = ϕ(r).
Hence
∆Hm (dHn(uǫ,δ, vǫ)− C
ψ) ≥ 0,
for x ∈ Ωδ. Hence by maximum principle we have
dHn(uǫ,δ, vǫ) ≤ C
ψ(xm).
This bound for dHn(uǫ,δ, vǫ) is independent of δ, hence by standard arguments (see
the proof of Theorem 6.4 in [9]) we have a harmonic map uǫ : H
m → Hn which is
the subsequent limit of uǫ,δ. Moreover for all x ∈ H
m we have
dHn(uǫ, vǫ) ≤ C
ψ(xm).
6 DUONG MINH DUC AND TRUONG TRUNG TUYEN
Hence
dHn(uǫ, vǫ) = 0,
which shows that uǫ is continuous up to the boundary and takes boundary value
1, ..., xm−1, 0) = (f1, f2, ..., fn−1, ǫ). �
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Department of Mathematics, University of natural sciences, Hochiminh city, Viet-
E-mail address: dmduc@hcmuns.edu.vn
Department of Mathematics, Indiana University, Rawles Hall, Bloomington, IN 47405
E-mail address: truongt@indiana.edu
	1. Introduction
	2. Initial maps
	3. Proof of Theorem ??
	References
