ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM
D2 TO S 2
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
Abstract. We consider rotationally symmetric 1-harmonic maps from
D2 to S 2 subject to Dirichlet boundary condition. We prove that the
corresponding energy – a degenerate non-convex functional with linear
growth – admits a unique minimizer, and that the minimizer is smooth
in the bulk and continuously differentiable up to the boundary. We also
show that, in contrast with 2-harmonic maps, a range of boundary data
exists such that the energy admits more than one smooth critical point:
more precisely, we prove the existence of a unique (up to scaling and
symmetries) global solution to the corresponding ode, which turns out
to be oscillating, and characterize the minimizer and the smooth critical
points of the energy as monotone, respectively non-monotone, branches
of such solution.
Contents
1. Introduction and main results
2. The relaxed functional
3. Properties of minimizers
4. Existence of a smooth minimizer
5. Local behavior at r = 0
6. Uniqueness of the minimizer
7. Smooth critical points
References
1
5
7
10
13
16
18
22
1. Introduction and main results
Given a domain Ω in RN , a real constant p ≥ 1, a smoothly embedded compact
submanifold (without boundary) M of RN +1 and a mapping u : Ω → M , let
Z
1
Ep (u) :=
|∇u|p dx
p Ω
(if p = 1, E1 is the total variation of u). The so-called p-harmonic flow for Ep
(i.e. the L2 -gradient flow for Ep with constraint of values in M ) is given by
(1)
ut = −πu (−div (|∇u|p−2 ∇u)) ,
1
2
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
where πv denotes the orthogonal projection of RN +1 onto the tangent space
Tv M of M at v ∈ M . Because of πv the values of a solution of (1) are constrained to remain in M if they are in M initially.
Here we shall consider the case in which N = 2, Ω is the unit disk (Ω =
D2 = {(x1 , x2 ) ∈ R2 : x21 + x22 ≤ 1}) and M is the unit sphere (M = S 2 =
{(x1 , x2 , x3 ) ∈ R3 : x21 + x22 + x23 = 1}). In this case (1) may be explicitly
written as
(2)
ut = div (|∇u|p−2 ∇u) + u|∇u|p .
Equation (2) arises in various contexts, such as multi-grain problems [14], theory of liquid crystals [12], ferromagnetism [9] and image processing [16]. In
particular, for p = 1, (2) may be seen as a constrained gradient system of total
variation:
µ
¶
Du
ut = div
(3)
+ u|Du|.
|Du|
Equation (3) was proposed in [17, 18] as a tool to denoise color images by
smoothing the chromaticity data while the contrast is being preserved. The
unconstrained version of (3) – the so-called total variation flow – corresponds
to the gray image denoising. This problem is by now well understood and there
is a vast literature about it (see the monograph [2] and the references therein).
However, much less is known about the constrained problem (3).
The Dirichlet problem for (2) has been widely studied over the last decades;
referenced discussions of the cases p = 2, p ∈ (1, ∞) and p = 1 may be found
e.g. in [3, 4], [13, 15] and [10], respectively. In particular, it is shown in [8]
that for p = 2 and suitable boundary data, classical solutions cease to exist in
finite time. This result was recently extended to p = 1 in [11]. In both papers,
the prototypical case of rotational symmetry and constant (in time) Dirichlet
boundary conditions is considered. A rotationally symmetric solution of (3)
has the form
³x
´
x2
1
u(t, x) =
sin h(t, r),
sin h(t, r), cos h(t, r) ,
r
r
(4)
r = |x| , x = (x1 , x2 ) ∈ D2
while a constant Dirichlet boundary condition turns into the constraint h(t, 1) =
h(0, 1) = `. Upon (4), the energy functional E1 takes the form
Z
(5)
E(h) :=
0
1
q
r2 h2r (r) + sin2 h(r) dr,
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
3
and after an routine computation, equation (3) becomes
µ
¶− 23
sin2 h
2
ht =
hr +
·
r2
µ 2 µ
¶
µ
¶¶
sin h
2hr
sin 2h
h2r
sin 2h
(6)
hrr +
+
hr −
.
·
−
r2
r
2r2
r
r
By comparison with suitable sub-solutions, in [11] it is shown that a solution
of (6) ceases to be smooth in finite time if the initial datum h0 is such that
h0 (0) = 0, h0 (r) ∈ (0, π) for r ∈ (0, 1), and h0 (1) = π. Noticeably, for the
same initial datum the solution of (2) with p = 2 will keep smooth for all
times [7]. Such striking contrast enforces the intuition that the strong singular
character of (3) should give raise to more complex phenomena with respect
to the heat flow of harmonic maps, especially concerning the formation and
the profile of singularities. It also naturally raises the question whether the
constant π = h0 (1) plays any specific role in this respect.
As a step towards a better understanding of these issues, in this note we will
investigate the minimizers of E subject to h(1) = `, as well as its smooth critical
points, i.e. the steady states of (6). In fact, by the periodicity of sin2 h and the
symmetry with respect to π2 (i.e. sin(π − h) = sin(h)), we may assume without
loss of generality that
π
(7)
` ∈ (0, ]
2
(we also rule out the trivial case ` = 0). We incorporate the boundary condition
by letting
½
E(h) if h(1) = `
G(h) =
+∞
otherwise.
In Section 2 we shall prove the following representation formula for the relaxation of G in BV ((0, 1)):
Proposition 1. The relaxation of G in BV ((0, 1)) is given by
G(h) = E(h) + |` − h(1)|,
where
Z
E(h) =
0
1
Z
q
r2 (h0a )2
2
+ sin (h) dr +
1
r |dhc | +
0
X
r|h+ (r) − h− (r)|.
r∈Jh
Here and throughout, primes denote differentiation with respect to the independent variable; as usual, the measure h0 is decomposed into its absolutely
continuous part h0a dr, its Cantor part dhc and its jump part dhj supported in
Jh , and the relaxation J of a functional J in BV (I) is defined as
n
o
J (h) := inf lim inf J (hn ) : hn ∈ W 1,1 (I), hn → h in L1 (I) .
n→∞
The first main result is the following:
4
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
Theorem 1. Upon (7), and up to a reflection with respect to h =
there exists a unique solution h to
h ∈ argmin {G(h) : h ∈ BV ((0, 1))}.
π
2
if ` =
π
2,
(P)
Furthermore h ∈ C ∞ ((0, 1]) ∩ C 1 ([0, 1]), h is positive and nondecreasing in
(0, 1], h(0) = 0, h(1) = `, and h solves
sin 2h 2h0
rh02
h =
−
+
2r2
r
sin2 h
00
µ
¶
sin 2h
0
−h .
r
(ODE)
Here and throughout, by a solution to an ode we mean a twice-differentiable
function which satisfies the ode point-wise.
Of course, Theorem 1 implies the existence of a unique (up to symmetries)
minimizer of G. However, it turns out that G is genuinely non-convex at least
for ` sufficiently close to π2 . Indeed, in our second main result we provide the
existence of smooth critical points of G which are different from the absolute
minimizer.
Theorem 2. There exists a global solution h ∈ C ∞ ((0, ∞)) ∩ C 1 ([0, ∞)) to
(ODE) which satisfies the following properties:
a) h(0) = 0, h(r) → π2 as r ↑ ∞;
b) the counter-image of the critical points of h consists of increasing sequence rn ↑ ∞ such that h(r2n ) ∈ ( π2 , π) are local maxima, h(r2n+1 ) ∈
(0, π2 ) are local minima, and |h(rn ) − π2 | is decreasing.
Furthermore:
c) for any ` ∈ (0, π2 ], the minimizer h` of G is given by h` (r) = h(αr) (or
h` (r) = π − h(αr) if ` = π2 ) for a suitable α > 0;
d) any non-constant global solution ĥ to (ODE) has the form
ĥ(r) = kπ ± h(αr) for some α > 0, k ∈ Z.
Note that (d) characterizes all global non-constant solutions to (ODE) as suitable rescalings of h: its qualitative properties are summarized in Figure 1.
Due to the scaling invariance of (ODE) with respect to r, Theorem 2 shows
that, in addition to h ≡ π2 and to the minimizer given by Theorem 1, there
exist infinitely many other critical points for E subject to h(1) = π2 . This
marks another substantial difference with respect to 2-harmonic maps. Also
in contrast with 2-harmonic maps, Theorem 2 excludes the existence of global
monotone solutions connecting 0 to π2 .
Coming back to the question of the role played by π, Theorem 2 suggests that
a classical solution of (6) subject to h = h0 on the parabolic boundary, with
h0 (0) = 0 and h0 (r) ∈ (0, π) for 0 < r ≤ 1, should cease to exist in finite time
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
5
π
H
π
2
r
Figure 1. The qualitative profile of the global solutions to
(ODE).
at least provided h0 (1) > H = max h, where h is the profile given by Theorem
2 (see Figure 1). Whether or not this expected lower bound may be improved
to h0 (1) > π2 is intuitively less clear and relates to the stability of the smooth
critical points.
The paper is organized as follows: in section 2 we prove Proposition 1. In
section 3 we derive basic properties of any minimizer: positivity, monotonicity,
continuity and attainment of the boundary value. In section 4 we use these
properties, combined with regularity theory, to construct a smooth minimizer.
In Section 5 we characterize the behavior at r = 0 of solutions to (ODE).
The proof of Thoerem 1 is completed in section 6 by showing uniqueness of
monotone solutions to (ODE). Finally, in section 7 we prove Theorem 2.
2. The relaxed functional
In this section we prove Proposition 1. The relaxation G of G is characterized
by:
(a) G is lower semi-continuous;
(b) for any h ∈ BV ((0, 1)) there exist {hn } ⊂ W 1,1 ((0, 1)) with hn (1) = `
such that hn → h in L1 ((0, 1)) and
G(h) = lim G(hn ).
n→∞
1
To see (a), let us introduce EA : L ((0, A)) → R defined by
Z Ap
r2 h02 + sin2 h dr.
EA (h) :=
0
By [1, Theorem 6.1], its relaxation is represented by
Z Aq
Z A
X
r|h+ (r) − h− (r)|.
E A (h) =
r2 (h0a )2 + sin2 (h) dx +
r |dh0c | +
0
0
r∈Jh
6
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
Given h ∈ BV ((0, 1)), we consider h∗ ∈ BV ((0, 2)) given by
½
h(r) if 0 < r ≤ 1
∗
h (r) =
`
1<r<2
Then
E 2 (h∗ ) = E(h) + | sin `| + |h(1) − `| = G(h) + | sin `|,
and (a) is implied by the lower semi-continuity of E 2 .
In order to prove (b), let h ∈ BV ((0, 1)); we first claim that a sequence {fn } ⊂
W 1,1 ((0, 1)) exists such that
(8)
fn (1) = h(1), fn → h in L1 ((0, 1)) and E(h) = lim E(fn ).
n→∞
Indeed, as E 1 = E is the relaxed functional of E1 = E, a sequence f˜n ∈
W 1,1 ((0, 1)) exists such that f˜n → h in L1 ((0, 1)) and
E(h) = lim E(f˜n )
n→∞
A version of the Slicing Lemma of De Giorgi given in [6, Lemma 2.7] then
permits find a sequence fn ∈ W 1,1 ((0, 1)) such that fn (1) = h(1), fn → h in
L1 ((0, 1)) and
lim sup E(fn ) ≤ lim inf E(f˜n ) = E(h).
n→∞
n→∞
Together with the lower semi-continuity of E, this proves (8).
Let now
µ
¶
1
gn (x) = n(` − h(1)) x − 1 +
and hn = fn + gn .
n +
Then
G(hn ) =
=
E(hn )
Z 1− n1 q
Z
(rfn0 )2
0
Z
≤
1
1− n
+ sin fn +
≤ E(fn ) +
(rfn0 )2
Z
1
1
1− n
2
+ sin fn +
|gn0 | +
1
1
1− n
Z
q
0
=
2
1
1
1− n
q
(r(fn0 + gn0 ))2 + sin2 (fn + gn )
(r|fn0 | + r|gn0 | + 1)
1
n
E(fn ) + |` − h(1)| + on (1)
which proves (b) completes the proof of Proposition 1.
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
7
h
π
h
h∗
π
2
h∗
(a)
1 r
(b)
1 r
Figure 2. Cutting h at levels 0 and π decreases the energy
(a), and reflecting it below π2 does not increase it (b).
3. Properties of minimizers
In this section we prove the following result.
Proposition 2. Assume (7). Then, up to a reflection with respect to h = π2
if ` = π2 , any solution h to problem (P) is nondecreasing, continuous in (0, 1]
with h(1) = `, and positive in (0, 1].
We split the proof into Lemmas. We begin with two straightforward observations based on periodicity and symmetry of sin2 h (see Figure 2).
Lemma 1. Assume (7). Then for any h ∈ BV ((0, 1)), the function
h∗ (r) = min{π, max{0, h(r)}}
is such that G(h∗ ) < G(h) unless h = h∗ .
Lemma 2. Assume (7). Then for any h ∈ BV ((0, 1); [0, π]), the function
¯
π ¯¯ π
¯
h∗ (r) = − ¯ − h¯
2
2
is such that G(h∗ ) ≤ G(h).
Next, we show that the energy may be strictly decreased if h is not monotone
(see Figure 3 (a)).
Lemma 3. Assume (7). If h ∈ BV ((0, 1); [0, π2 ]) is not nondecreasing, there
exists h∗ ∈ BV ((0, 1); [0, π2 ]) such that h∗ is nondecreasing and G(h∗ ) < G(h).
Proof. Since h is not nondecreasing, r1 ∈ (0, 1) exists such that
(9)
h(r1 ) > h∗ = inf h.
(r1 ,1)
We are going to show that the energy may be decreased by “cutting the hill”
around r1 with a straight line (see Figure 3 (a)). Take a minimizing sequence
(10)
(r1 , 1) 3 rn → r2 ∈ [r1 , 1] such that h(rn ) ↓ h∗ as n ↑ ∞.
8
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
π
2
π
2
h
∗
h
h∗
h
r0
r1 r2
(a)
1 r
r∗
(b)
1 r
Figure 3. Cutting a hill (a), as well as healing a discontinuity
(b), decreases the energy.
Let also
(11)
r0 = inf{r ≤ r1 : h(ρ) > h∗ for all ρ ∈ [r, r1 ]}.
Note that
(11)
(12)
lim− h(r) ≤ h∗ .
r→r0
If r2 = r0 (= r1 ), we would have by (10) that limr→r+ h(r) = h∗ , which together
1
with (11) would contradict (9). Therefore r2 − r0 > 0. Let then
½
h∗
r ∈ (r0 , r2 )
h∗ (r) =
h(r) elsewhere.
If r2 < 1, we have
G(h) − G(h∗ )
Z
=
r2
r0
+
Z
q
r2 (h0a )2
X
2
r2
+ sin (h) dx +
r0
r d|h0c | − (r2 − r0 ) sin h∗
r|h+ (r) − h− (r)| − r0 |h∗ − h− (r0 )| − r2 |h+ (r2 ) − h∗ |
r∈Jh ∩[r0 ,r2 ]
=
¡
¢
r0 |h+ (r0 ) − h− (r0 )| − |h∗ − h− (r0 )|
¡
¢
+r2 |h+ (r2 ) − h− (r2 )| − |h+ (r2 ) − h∗ |
¡
¢
¡
¢
r0 h+ (r0 ) − h∗ + r2 |h+ (r2 ) − h− (r2 )| − (h+ (r2 ) − h∗ )
≥
0.
9),(11)
>
(11),(12)
In the last inequality, we have used the fact that, by (10), either h+ (r2 ) = h∗
or h− (r2 ) = h∗ . This proves Lemma 3 if r2 < 1. Drawing the conclusion if
r2 = 1 is simpler and we omit it.
¤
¤
We continue by ruling out jumps in the bulk:
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
9
Lemma 4. Any solution h to problem (P) belongs to C((0, 1]) and satisfies
h(1) = `.
Proof. By Lemmas 1-3, we may assume without loss of generality that h is
nondecreasing and h ≤ π2 . We first observe that
(13)
h(1) ≤ `.
Indeed, if by contradiction h(1) > `, one may argue as in the proof of Lemma
3: letting
r0 = inf{r ≤ 1 : h(ρ) > ` for all ρ ∈ (r, 1]} < 1,
the function
½
h(r) r ∈ (0, r0 )
∗
h (r) =
`
elsewhere
would have less energy. We now show that
(14)
h ∈ C((0, 1]) and h(1) = `.
Suppose by contradiction that (14) is false. Then, by (13), either
• h has a jump at r∗ ∈ (0, 1); then, let A = h+ (r∗ ) − h− (r∗ )
or
• h(1) < `; then, let r∗ = 1 and A = ` − h(1).
We are going to show that the energy may be decreased by moving the discontinuity down to r = 0. Namely, for r ≤ r∗ we raise h (see Figure 3 (b)) by
A:
½
h(r) + A if 0 < r < r∗
θ(r) :=
h(r)
r∗ < r < 1
π
Since A > 0, h ≥ 0 and A + h ≤ 2 , we see that for r ∈ (0, r∗ )
q
q
r2 (h0 )2 + sin2 θ − r2 (h0 )2 + sin2 h
=
sin2 (h + A) − sin2 (h)
q
r2 (h0 )2 + sin2 (h + A) + r2 (h0 )2 + sin2 h
q
sin2 (h + A) − sin2 h
sin(h + A) + sin h
= sin(h + A) − sin h
< sin A
< A.
≤
Hence
Z
G(θ) =
r∗
G(h) +
µq
r2 (h0 )2 + sin2 θ −
¶
q
r2 (h0 )2 (r) + sin2 h(r) − r∗ A
0
<
G(h),
and we have found a contradiction. Hence (14) holds true and the proof is
complete.
¤
¤
10
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
In view of Lemmas 3 and 4, the proof of Proposition 2 is complete once we
have shown that:
Lemma 5. Assume (7). Up to a reflection with respect to
solution h to problem (P) is positive in (0, 1].
π
2
if ` =
π
2,
any
Proof. By Lemmas 1-3 and 4, we assume without loss of generality that h ∈
[0, π2 ] is nondecreasing and continuous. Assume by contradiction that a > 0
exists such that h(a) = 0 and h(r) > 0 for a < r ≤ 1. By Lemma 4, a < 1. Let
then ε > 0 such that a − ε ≥ 0 and a + ε ≤ 1. We first notice that
¶
Z a+ε µq
Z a+ε
2
2
0
2
r
r (|h0a | dr + |dhc |)
r (ha ) + sin hdr + |dhc |
≥
0
a
≥ ah(a + ε).
On the other hand, if we take
Ã
θ(r) = 2 arctan
we have
Z
a+ε
)r
tan( h(a+ε)
2
(a + ε)
!
,
q
0
√
ln( cos(h(a+ε))+1
)
2
r2 (θ0 )2 + sin2 θ dr = − 2(a + ε)
.
tan(h(a + ε)/2)
Therefore, letting
½
θ(r) 0 < r < a + ²
h(r) elsewehere
and recalling that h is continuous, we obtain that
∗
h (r) =
G(h∗ ) − G(h)
√
ln( cos(h(a+ε))+1
)
2
≤ − 2(a + ε)
− ah(a + ε)
tan(h(a + ε)/2)
µ
¶
√
cos h(a + ²) − 1
= −a
+ h(a + ²)
2(1 + oε (1))
h(a + ²)
à √
!
2
= −ah(a + ²) −
as ² ↓ 0.
+ 1 + o² (1)
2
Hence h cannot be a minimizer.
¤
¤
4. Existence of a smooth minimizer
In this section we prove the following result:
Proposition 3. Assume (7). There exists a solution h ∈ C ∞ ((0, 1]) to problem
(P), and h is also a solution of (ODE).
We split the proof into Lemmas. Existence of a minimizer follows immediately
from Lemmas 1-3:
Lemma 6. There exists a solution h to problem (P).
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
11
Proof. Let hj ∈ BV ((0, 1)) be a minimizing sequence. By Lemmas 1-3, we may
assume that each hj is nondecreasing and hj ∈ [0, π2 ]. Then,
khj kBV ((0,1)) = khj kL1 ((0,1)) + |Dh|((0, 1)) ≤ 2hj (1) ≤ π.
Thus, we can extract a subsequence, still denoted by {hj } such that hj → h∗ ∈
BV ((0, 1)) in L1 ((0, 1)). Having in mind the lower semi-continuity of G, we
obtain that h∗ is a solution to problem (P).
¤
¤
In order to improve continuity of h (as given by Proposition 2) to Lipschitz
continuity, we shall apply a regularity result proved in [5]. This is made possible
by the fact that any minimizer is positive in (0, 1], i.e. G is non-degenerate in
the bulk. The following holds:
Lemma 7. For any solution h of (P) and any 0 < a < b ≤ 1, there exists a
solution h∗a,b ∈ W 1,∞ ((a, b)) of
Rbq
h ∈ argmin{Fa,b (h) = a r2 (h0 )2 + sin2 h dr :
(15)
h ∈ W 1,1 ((a, b)), h(a) = h(a), h(b) = h(b), h0 ≥ 0}.
Proof. Via the smooth transformation
1
h(be−Lx )
log(tan(
)),
L
2
we see that (15) is equivalent to
Z 1
2Le−Lx eLg p
g ∈ argmin{Ga,b (g) =
1 + |gx |2 dx :
1 + e2Lg
0
g ∈ W 1,1 ((0, 1)), g(0) = sb , g(1) = sa , g 0 ≥ 0}
g(x) =
where
1
h(b)
h(a)
log(tan(
)), sa = log(tan(
)).
L
2
2
Applying [5, Theorem 5.1], we obtain that there exists a solution g ∗ ∈ W 1,∞ ((0, 1))
to
g ∈ argmin{Ga,b (g) + | arctan(eg(0) ) − arctan(esb )|
L = log b − log a, sb =
+ ab | arctan(eg(1) ) − arctan(esa )| : g ∈ W 1,1 ((a, b)), gx ≥ 0}
which implies the existence of a solution h∗a,b ∈ W 1,∞ (a, b)) to
(16)
h ∈ argmin{Fa,b (h) + a|h(a) − h(a)| + b|h(b) − h(b)| :
h ∈ W 1,1 ((0, 1)), h0 ≥ 0}.
To complete the proof, it remains to show that
h∗a,b (a) = h(a) and h∗a,b (b) = h(b).
To this aim, let
hn ∈ W 1,1 ((a, b)), hn (a) = h(a), hn (b) = h(b)
such that hn → h in BV ((a, b))
12
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
and
½
h̃(r) :=
Note that
Z bq
h(r)
r ∈ (0, 1) \ (a, b)
h∗a,b (r)
r ∈ (a, b)
Z
0
r2 (hn )2 + sin2 hn dr →
a
Therefore
Z
G(h̃) =
G(h) −
b
q
G(h) −
a
a
q
Z
0
r2 (ha )2 + sin2 h dr +
0
Z
r2 (ha )2 + sin2 h dr −
a
+Fa,b (h∗a,b ) +
Z bq
=
b
b
r|dhc |.
a
b
r|dhc |
a
a|h(a) − h∗a,b (a)| + b|h(b) − h∗a,b (b)|
0
r2 (hn )2 + sin2 hn dr + on (1)
+Fa,b (h∗a,b ) + a|h(a) − h∗a,b (a)| + b|h(b) − h∗a,b (b)|
≤ G(h) + on (1),
where the last inequality follows from the fact that h∗a,b is a solution to (16).
Then h̃ is also a minimizer of G, hence it is continuous by Lemma 4.
¤ ¤
We may use Lemma 7 to construct a Lipschitz continuous minimizer from a
given one. It will turn out useful in the next section to also fix a value in the
interior.
Lemma 8. For any solution h to problem (P) and any r0 ∈ (0, 1), there exists
1,∞
a solution h̃ to problem (P) such that h ∈ Wloc
((0, 1]) and h̃(r0 ) = h(r0 ).
Proof. Let a0 = 1, a1 = r0 , and an such that an ↓ 0. Let h0 = h, h∗an ,an−1 as
given by Lemma 7, and
½
hn−1 (r)
r ∈ (0, 1) \ (an , an−1 )
hn (r) =
h∗an ,an−1 (r) r ∈ (an , an−1 ).
By Lemmas 4 and 7, hn (r) ∈ W 1,∞ ([an , 1]) ∩ C((0, 1]), and by construction,
1,∞
hn → h̃ in Wloc
((0, 1]) with h̃(r0 ) = h(r0 ). To see that h̃ is a minimizer of G,
it suffices to show that
(17)
G(hn ) ≤ G(hn−1 ) ≤ G(h) for all n
and that
(18)
hn → h̃
in L1 ((0, 1)).
For (17), let fj ∈ W 1,1 ((an , an−1 )) such that fj (an ) = hn−1 (an ), fj (an−1 ) =
hn−1 (an−1 ) and fj → h0 in BV ((an , an−1 )). Then
G(hn ) − G(hn−1 )
= Fan ,an−1 (h∗an ,an−1 ) − F an ,an−1 ((h0 ))
= Fan ,an−1 (h∗an ,an−1 ) − lim Fan ,an−1 (fj ),
j→∞
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
13
and since h∗an ,an−1 is a minimizer, the right-hand side can not be positive. For
(18), we just note that hn is a Cauchy sequence, since for n ≥ m
Z 1
Z am−1
|hn − hm | ≤
2` = om (1).
0
0
Therefore h̃ is a solution to problem (P).
¤
¤
We are now ready to complete the proof of Proposition 3.
Proposition 3. Existence of a minimizer is guaranteed by Lemma 6. Let then
h̃ as given by Lemma 8. It is straightforward to check that for all a ∈ (0, 1) h̃
is also a solution to
Rbq
h ∈ argmin{Fa,1 (h) = a r2 (h0 )2 + sin2 h dr :
h ∈ W 1,1 ((a, 1)), h(a) = h̃(a), h(1) = `}.
Then, since h̃ ∈ W 1,∞ ((a, 1)), h̃ is a point-wise solution to (ODE), hence it is
smooth.
¤
¤
5. Local behavior at r = 0
In this section we prove the following result:
Proposition 4. Let h be a non-constant solution to (ODE) in (0, ε) for some
ε > 0, such that h ∈ (0, π). Then h ∈ C 1 ([0, ε)) and
(19)
lim h(r) = 0 or lim+ h(r) = π.
r→0+
r→0
A few smooth coordinate transforms will turn out to be useful here as well as
int the next section. With the first one, we pass to logarithmic coordinates by
letting
f (t) := h(e−t ).
(20)
In terms of f , (ODE) reads as
(f 0 )2 0
sin(2f )
.
(f + sin(2f )) +
2
sin2 f
For f ∈ (0, π), we may define
(
(tan f (t))−1 if f (t) 6= π2
(22)
w(t) :=
0
if f (t) = π2 .
(21)
f 00 = f 0 +
Note that
(23)
w0 (t) = −
f0
1
1
et h0 (et )
f0 = − 2 = −
.
2
2
sin2 (h(et ))
tan f cos f
sin f
In terms of w, (21) reads as
(24)
w00 = w0 +
(w0 )3
− w.
1 + w2
14
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
We first prove (19), which is implied by the following:
Lemma 9. Let w(t) be a non-constant solution of (24) in (t² , +∞) for some
t² ∈ R. Then
lim w2 (t) + (w0 )2 (t) = +∞.
t→∞
Proof. We have
µ
¶
d 2
(w0 )2
0 2
0 2
(w + (w ) )) = 2(w ) 1 +
≥ 2(w0 )2 .
dt
1 + w2
(25)
In particular,
∃ lim w2 (t) + (w0 )2 (t) = C − ∈ (0, ∞]
t→∞
and
Z
w2 (t) + w2 (t) ≥ w2 (t0 ) + w02 (t0 ) +
(26)
t
(w0 )2
t0
for all t0 > tε . Assume by contradiction that C < ∞. Then, we claim that
lim (w0 )2 (t) = 0.
(27)
t→∞
Indeed, if not we would have (w0 )2 (tn ) ≥ C1 > 0 for a subsequence; choosing
t0 = tn in (26) with n sufficiently large, we would have
Z t
(w0 )2 (t) ≥ on (1) +
(w0 )2 dt, (w0 )2 (tn ) ≥ C1 ,
tn
0 2
yielding (w ) → ∞ in contradiction with (w0 )2 ≤ C. Therefore (27) holds,
implying that
lim w2 (t) = C 6= 0.
t→∞
Thus
lim w00 (t) = −C
t→∞
in contradiction with (27). Hence C = ∞.
¤
¤
In view of Lemma 9, w is strictly monotone for t sufficiently large. Hence we
may define its inverse
(28)
t(w) : w(t(w)) = w
and the function
(29)
g(w) :=
In terms of g, (24) reads as
(30)
g0 =
1
w
w
.
w0 (t(w))
µ
g3 − g2 + g −
w2
1 + w2
¶
.
The large-w behavior of g is at the core of our argument for the regularity of
h:
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
15
Lemma 10. Let g be a positive solution of (40) in (wε , ∞) for some wε ∈ R.
Then
lim g(w) = 1− .
(31)
w→+∞
Proof. We claim that
(32)
g(w) ≤ 1 for all w ∈ (wε , ∞).
Assume on the contrary that g(w0 ) = 1 + ε for some ε > 0 and w0 > wε . Since,
by (40),
1
1
g 0 > (g 3 − g 2 + g − 1) = (1 + g 2 )(g − 1),
w
w
we would then have that g 0 (w0 ) > 0. Hence g(w) > 1 + ε for all w > w0 , and
therefore
g2
g 0 (w) > ε.
w
Integrating this inequality, we see that
g(w) >
1
g(w0 )
1
− ε log w
which shows that g(w) would blow up at a finite w, a contradiction. This
proves (32).
To complete the proof, it suffices to show that lim inf w→+∞ g(w) ≥ 1. Assume
on the contrary that a sequence wn ↑ ∞ exists such that g(wn ) ≤ 1 − ε for
2
wn
ε
some ε > 0, and choose n so large that 1+w
2 ≥ 1 − 2 . Note that
n
1 h
εi
g0 ≤
(1 + g 2 )(g − 1) +
w
2
1 h
εi
w2
ε
<
g−1+
for all w :
≥ 1 − and g(w) < 1.
2
w
2
1+w
2
Hence g 0 (wn ) < 0, g(w) ≤ 1 − ε for all w > wn , and moreover
ε
g 0 (w) < − .
2w
After integration, this yields
ε
g(w) < g(wn ) − log(w)
2
in contradiction with the positivity of g.
¤
¤
We are now ready to prove Proposition 4.
Proposition 4. The limit (19) follows immediately from Lemma 9. Up to exchanging h with π − h, we may assume that h ↓ 0, so that (9) reads as
(33)
lim w(t) = ∞,
t→∞
lim w0 (t) = ∞.
t→∞
16
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
In particular, for t sufficiently large w is increasing, so that (28) and (29) make
sense, and g is positive. To complete the proof, it suffices to show that
∃ lim h0 (r) ∈ R.
(34)
r→0+
We write
(35)
h0 (r) ∼
rh0 (r)
tan h(r)
h0 (r)
=
·
cos2 h(r)
sin h(r) cos h(r)
r
as r ↓ 0.
On one hand, we have
lim+
r→0
rh0 (r)
sin h(r) cos h(r)
(20)
=
f 0 (t)
t→∞
sin f (t) cos f (t)
0
w (t)
lim
t→∞ w(t)
w
lim 0
w→∞ w (t(w))
lim −
(23)
=
(33),(28)
=
(31)
= 1− .
(36)
On the other hand, we have
(31)
¡ −t
¢0
e w(t) = e−t (w0 (t) − w(t)) ≥ 0 for t À 1,
and therefore
1
≤C
w(t)e−t
which in terms of h means that
(37)
tan h(r)
≤C
r
for t À 1,
for r ¿ 1.
Plugging (36) and (37) into (35) and passing to the limit yields (34) and completes the proof of Proposition 4.
¤
6. Uniqueness of the minimizer
Uniqueness of the smooth minimizer is the central result of this section:
Proposition 5. Assume (7). Then, there exists a unique positive, non-constant
and non-decreasing solution to

µ
¶
sin 2h 2h0
r(h0 )2 sin 2h
 00
0
h =
−
+
−h
r ∈ (0, 1)
(38)
2r2
r
r
sin2 h

h(1) = `.
Furthermore h is increasing.
As a consequence, we obtain Theorem 1:
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
17
Theorem 1. By Propositions 2, 3 and 4, there exists a solution h to problem
(P), and furthermore h ∈ C ∞ ((0, 1])∩C 1 ([0, 1]) with h(0) = 0 and h(1) = `, h is
positive and nondecreasing in (0, 1], and h solves (ODE). In order to show that
h is unique, let h be any solution to problem (P): by Proposition 3, h ∈ C((0, 1])
with h(1) = `, hence we may assume by contradiction that r0 ∈ (0, 1) exists
such that h(r0 ) 6= h(r0 ).
∞
By Lemma 8, there exists a minimizer h̃ ∈ Cloc
((0, 1]) of G such that h̃(r0 ) =
h(r0 ) and h̃ is a solution of (ODE). By Propositions 2 and 4, h̃(1) = `, h̃ is
positive and non-decreasing in (0, 1], and h̃ is non-constant since h̃(0) = 0.
Hence, by Proposition 5, h̃ ≡ h, in contradiction with h̃(r0 ) = h(r0 ) 6= h(r0 ).
As a consequence h is increasing, and the proof is complete.
¤
¤
The rest of the section is concerned with the proof of Proposition 5. We pass
to logarithmic coordinates by (20). Then (38) reads as

0 2

 f 00 = f 0 + (f ) (f 0 + sin(2f )) + sin(2f )
t ∈ (0, ∞)
2
sin2 f
(39)

 lim f (t) = 0, f (0) = ` ≤ π
t→∞
2
with f positive, non-constant and non-increasing.
Lemma 11. Assume (7). Then any positive and non-increasing solution of
(39) is decreasing.
Proof. Suppose by contradiction that there exists t0 ∈ (0, +∞) such that
f 0 (t0 ) = 0. We have 0 < f (t0 ) < π2 (otherwise f ≡ π2 ). Then f 00 (t0 ) =
sin(2f (t0 ))
> 0 and therefore f will be increasing, a contradiction. Hence
2
f 0 (t) < 0 for all t ∈ (0, +∞).
¤
¤
Since f ∈ (0, π2 ) in (0, ∞), we may perform the transformations (22) and (29).
In terms of g, (39) turns into
µ
¶
1
w2
0
3
2
ˆ ∞)
(40)
g =
g −g +g−
w ∈ (`,
w
1 + w2
with g positive, where
( 1
if ` ∈ (0, π2 )
tan `
ˆ
(41)
`=
0
if ` = π2 .
This reformulation is particularly useful since its solutions satisfy a comparison
principle:
Lemma 12. Assume (7) and (41). If g1 and g2 are two positive solutions of
ˆ > g2 (`),
ˆ then
(40) and g1 (`)
1
ˆ
(42)
(g1 − g2 )0 >
(g1 − g2 ) for all w > `.
2w
ˆ ∞).
In particular, g1 > g2 in (`,
18
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
Proof. We have
(g1 − g2 )0
=
=
=
1 3
(g − g23 − (g12 − g22 ) + g1 − g2 )
w 1
g1 − g2 2
(g + g1 g2 + g22 + g1 + g2 + 1)
w "1
#
µ
¶2 µ
¶2
g1 − g2
1
1
1
g1 −
+ g2 −
+ g1 g2 +
w
2
2
2
ˆ > 0, and (42) follows.
Hence (g1 − g2 )0 (`)
¤
¤
We are now ready to prove Proposition 5.
Proposition 5. Let h1 and h2 be two different positive, non-constant and nondecreasing solutions of (38). Then h01 (1) 6= h02 (1), so that without loss of
ˆ >
generality the corresponding solutions g1 and g2 to (40) are such that g1 (`)
ˆ
g2 (`). By Lemma 12
g1 − g2
ˆ for all w > `.
ˆ
(g1 − g2 )0 ≥
for all w > `.
2w
After integrating, we get for a fixed w0 > `ˆ
r
g1 (w) − g2 (w)
w
≥
.
g1 (w0 ) − g2 (w0 )
w0
Hence
lim (g1 (w) − g2 (w)) = ∞,
w→∞
in contradiction with Lemma 10.
¤
¤
7. Smooth critical points
In this section we prove Theorem 2. The existence of the profile is based on
the analysis of the continuation of solutions to problem (P) beyond r = 1.
This amounts to studying the following problem, which follows from (24) upon
exchanging the t-axis direction:

0 3
 w00 = −w0 − (w ) − w
t>0
1 + w2
(43)

w(0) = 0, w0 (0) = −α.
The following holds:
Lemma 13. For any α > 0 there exists a unique global solution to (43).
Furthermore
(44)
lim w(t) = 0
t→∞
and the counter-image of the critical points of w consists of an increasing sequence tn ↑ ∞ such that w(t2n ) < 0 are local minima, w(t2n+1 ) > 0 are local
maxima, and |w(tn )| is decreasing.
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
19
Proof. Let
W(w) = w2 + (w0 )2 .
Since
d
2w04
W(w(t)) = 2ww0 + 2w0 w00 = −2w02 −
< 0,
dt
1 + w2
we have that kwkC 1 < α, and therefore the solution to (43) is global. The
oscillatory properties of w (43) are based on the following observations:
(A) Given a point s∗ such that w(s∗ ) = 0 and w0 (s∗ ) 6= 0, there exists
t∗ ∈ (s∗ , ∞) such that w0 (t∗ ) = 0 and w(t∗ ) < 0 if w0 (s∗ ) < 0 [resp.
w(t∗ ) > 0 if w0 (s∗ ) > 0], and t∗ is a local minimum [resp. maximum]
point for w.
(B) Given a point t∗ such that w0 (t∗ ) = 0 and w(t∗ ) 6= 0, there exists
s∗ ∈ (t∗ , ∞) such that w(s∗ ) = 0 and w0 (s∗ ) > 0 if w(t∗ ) < 0 [resp.
w0 (s∗ ) < 0 if w(t∗ ) > 0].
To see (A), let
w(t)
g(t) := 0 .
w (t)
Then (43) reads as
(45)
g0
=
=
ww00
w
w03
0
=
1
+
(w
+
+ w)
w02
w02
1 + w2
ww0
1 + g + g2 +
,
(1 + w2 )
1−
with g(s∗ ) = 0 and g 0 (s∗ ) = 1. Therefore g is locally increasing, hence positive,
which means that ww0 > 0 and that g 0 ≥ 1 + g 2 . Consequently, g remains
positive as long as it is defined and blows up to ∞ at a finite t. This means
that t∗ ∈ (s∗ , ∞) exists such that w0 (t∗ ) = 0. The sign of w(t∗ ) follows from the
positivity of g. Moreover, from (43), we get that w00 (t∗ ) > 0 [resp. w00 (t∗ ) < 0]
and therefore t∗ is a local minimum [resp. maximum] for w.
To see (B), let
w0 (t)
j(t) :=
.
w(t)
Then (43) reads as
w00
w02
j 3 w2
− 2 = −j −
− 1 − j2.
w
w
1 + w2
We have j(t∗ ) = 0 and so j 0 (t∗ ) = −1. Therefore j is locally decreasing, hence
negative, with j 0 ≤ −1 − j 2 . Therefore j is negative as long as it is defined, and
blows down to −∞ at a finite t, that is, s∗ ∈ (t∗ , ∞) exists such that w(s∗ ) = 0.
The sign of w0 (s∗ ) is determined by that of j.
j0 =
It follows from an iterated application of (A) and (B), starting from s∗ = s0 = 0,
that an increasing sequence tn exists such that w0 (tn ) = 0, with w(t2n ) < 0 local
20
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
minima, w(t2n+1 ) local maxima, and w monotone in between, with w(sn ) = 0,
sn ∈ (tn , tn+1 ). We have that
lim tn = ∞.
n→∞
Indeed, else we would have tn ↑ T , sn ↑ T , and therefore w(T ) = w0 (T ) = 0,
which is impossible since w is not constant. Furthermore, it holds that
|w(tn+1 )| < |w(tn )|.
Else, we would have W(tn ) ≥ W(tn+1 ), which recalling (45) would imply W 0 =
0, i.e. w0 = 0, in (tn , tn+1 ), a contradiction. Finally, we observe that by (45)
there exists a constant C ≥ 0 such that limt→+∞ W(t) = C. If by contradiction
C > 0, we would have that W(tn ) = w2 (tn ) ≥ C 2 . Thus,
Z xn+1
2C 2 ≤ |w(xn )| + |w(xn + 1)| = |w(xn+1 ) − w(xn )| =
|w0 |,
xn
which implies that
Z
∞
|w0 | = +∞.
0
On the other hand, we see from (45) that W 0 (t) ≤ −2(w0 )2 , and therefore
µ Z t ¶ 12
Z t
√
02
W(t) ≤ W(0) −
w ≤ W(0) − t
w0
≤ W(0)(1 − t),
0
0
in contradiction with W ≥ 0. Hence C = 0 and (44) follows.
¤
¤
In order to charactrize global solutions to (ODE), we shall also need to rule
out that solutions may smoothly cross h = kπ, the values where the equation
becomes singular (which is conceivable for a generic ode, think of h00 = h02 /2h,
which is solved by h = r2 ).
Lemma 14. There exists no nontrivial solution h of (ODE) such that h(r0 ) = 0
for some r0 > 0.
Proof. Since h is non-trivial, r1 ∈ (0, ∞) exists such that h(r1 ) 6= 0. We assume
without loss of gnerality that h(r1 ) ∈ (0, π/2], pass to logarithmic coordinates
by letting f (t) = h(et ), and define
F (t) = log(tan(f /2)).
A simple computation shows that F solves
¶
µ
1 − e2F
0
−
F
F 00 = (1 + (F 0 )2 )
1 + e2F
in a neighbourhood of t1 = ln r1 , with F (t1 ) ∈ (−∞, 0]. Note that
µ
¶
d 1
ln(1 + (F 0 )2 ) + G(F ) = −(F 0 )2 ,
dt 2
ROTATIONALLY SYMMETRIC 1-HARMONIC MAPS FROM D 2 TO S 2
21
where G(F ) = log(1 + e2F ) − F is non-negative and such that G(F ) → ∞ as
|F | → ∞. Therefore F is defined for all t > t1 , which in terms of h means that
h(r) > 0 for all r > r1 . Hence, it remains to show that
h(r) > 0 for all r < r1 .
If not, we may assume without loss of generality that F (t) → −∞ as t ↓ 0.
Note that
F 00 < (1 + (F 0 )2 )(1 − F 0 ).
Note also that F 0 → +∞ at least for a sequence. Hence there exists a point t0
such that F 0 (t0 ) ≥ 2. At t0 , F 00 < 0, hence F 0 > 2 and F 00 < 0 for all t < t0 ,
and therefore F 0 ↑ +∞ as t ↓ 0. Integrating
F 00
< −1,
(1 + F 02 )(F 0 − 1)
we see that
1
− 02 + C . −t as t ↓ 0,
F
that is
1
F 02 .
as t ↓ 0.
1+t
Then F would be finite at t = 0, a contradiction.
¤
¤
We are now ready to prove Theorem 2.
Theorem 2. Let h∗ ∈ C ∞ ((0, 1]) be the solution to (P) with h∗ (1) = π2 , as
0
0
given by Theorem 1. Since h∗ is increasing, h∗ (1) = α > 0 (h∗ (1) = 0 would
imply h∗ ≡ π2 , whereas h∗ (0) = 0). Consider now the Cauchy problem

µ
¶
sin 2h 2h0
r(h0 )2 sin 2h
 00
0
h =
−
+
−
h
r>1
(46)
2r2
r
r
sin2 h

h(1) = π2 , h0 (1) = α.
By standard ode theory, (46) admits a maximal solution. We pass to logarithmic coordinates by letting f (t) := h(et ), t ∈ (0, ∞). As long as f ∈ (0, π), we
may define as in section 5
½
(tan f (t))−1 if f (t) 6= π2
w(t) :=
0
if f (t) = π2 .
In terms of w, (46) turns into (43), which by Lemma 13 admits a unique global
solution. Hence the solution of (46) is global, too, and moreover h ∈ (0, π).
Patching this solution with h∗ and translating the properties of w back to h
immediately gives the existence of a global solution h which satisfies (a) and
(b) of Theorem 2.
To prove (c), for ` < π2 we scale r so that ĥ(r̂) = h(αr̂) satisfies ĥ(1) = `
with α < 1. By the scale invariance of (ODE), ĥ is also a solution, and ĥ is
increasing and positive in (0, 1). Hence, by Proposition 5, it coincides with the
minimizer h` of G.
22
ROBERTA DAL PASSO, LORENZO GIACOMELLI AND SALVADOR MOLL
To complete the proof, it remains to characterize global non-constant solutions
ĥ to (ODE). Take one of them: by Lemma 14, we may assume without loss of
generality that ĥ ∈ (0, π). We claim that
∃ r1 : ĥ0 (r1 ) = 0.
(47)
If not, ĥ is monotone. Up to exchanging ĥ with π − ĥ, we may assume that
ĥ is non-decreasing. By Proposition 4 ĥ → 0 as r → 0, hence ĥ(ε) ≤ π2 for
some ε > 0. Then h̃(r) = ĥ(εr) is a positive, non-decreasing and non-constant
solution to (38). On the other hand, we may also scale r so that h(r) = h(αr)
satisfies h(1) = ĥ(ε) with α < 1. Therefore h̃ and h are two positive, nonconstant and non-decreasing solutions of (38) in (0, 1) with the same boundary
value. By Proposition 5 they coincide in (0, 1), hence they coincide everywhere.
But this is absurd since h oscillates, and (47) holds.
Next, we show that
(48)
∃ r0 ≤ r1 : ĥ0 (r0 ) = 0 and ĥ si monotone in (0, r0 ).
Up to exchanging ĥ with π − ĥ, we may assume by Proposition 4 that h(r) → 0,
which means that ε > 0 exists such that 0 < h(r) < π2 for all 0 < r < ε. It
follows from the equation that all critical points in (0, ε) have positive second
derivative, i.e. are local minima, which is impossible. Therefore there is no
critical point in (0, ε), and (48) follows from (47).
To conclude, up to exchanging ĥ with π − ĥ we may assume by (48) that ĥ
is non-decreasing in (0, r0 ), with ĥ0 (r0 ) = 0. If ĥ(r0 ) = π2 then ĥ would be a
constant, and if ĥ(r0 ) < π2 then ĥ00 (r0 ) > 0. Both are impossible, and therefore
ĥ(r0 ) > π2 . Then, again by Proposition 4, r∗ < r0 exists such that ĥ(r∗ ) = π2 .
Scaling r so that r∗ = 1 and applying Proposition 5 with ` = π2 we conclude
that ĥ ≡ h.
¤
¤
Acknowledgement. Supported by EC through the RTN-Programme “FrontsSingularities” (HPRN-CT-2002-00274). The third author acknowledges partial support by the Spanish Ministerio de Educación y Ciencia and FEDER
projects, references MTM2005-00620 and MTM2006-14836. He also would like
to thank the Istituto per le Applicazioni del Calcolo “Mauro Picone” in Rome
and its director M. Bertsch for the kind hospitality during his stay when this
work began.
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R. Dal Passo: Dipartimento di Matematica - Università di Roma Tor Vergata, [email protected]
L. Giacomelli: Dipartimento Me.Mo.Mat. - Università di Roma La Sapienza, [email protected]
S. Moll: Departament de Tecnologia - Universitat Pompeu Fabra, [email protected]