Deformations of Annuli on Riemann surfaces with Smallest Mean

arXiv:1005.5269v3 [math.CV] 17 Aug 2010

DEFORMATIONS OF ANNULI ON RIEMANN SURFACES WITH SMALLEST MEAN DISTORTION DAVID KALAJ

A BSTRACT. Let A and A′ be two circular annuli and let ρ be a radial metric defined in the annulus A′ . Consider the class Hρ of ρ−harmonic mappings between A and A′ . It is proved recently by Iwaniec, Kovalev and Onninen that, if ρ = 1 (i.e. if ρ is Euclidean metric) then Hρ is not empty if and only if there holds the Nitsche condition (and thus is proved the J. C. C. Nitsche conjecture). In this paper we formulate a condition (which we call ρ−Nitsche conjecture) with corresponds to Hρ and define ρ−Nitsche harmonic maps. We determine the extremal mappings with smallest mean distortion for mappings of annuli w.r. to the metric ρ. As a corollary, we find that ρ−Nitsche harmonic maps are Dirichlet minimizers among all homeomorphisms h : A → A′ . However, outside the ρ-Nitsche condition of the modulus of the annuli, within the class of homeomorphisms, no such energy minimizers exist. This extends some recent results of Astala, Iwaniec and Martin (ARMA, 2010) where it is considered the case ρ = 1 and ρ = 1/|z|.

1. I NTRODUCTION 1.1. Mappings of finite distortion. A homeomorphism w = f (z) between planar domains Ω and D has finite distortion if 1,1 a) f lies in the Sobolev space Wloc (Ω, D) of functions whose first derivatives are locally integrable, and b) f satisfies the distortion inequality |fz¯| ≤ µ(z)|fz |, 0 ≤ µ(z) < 1 almost everywhere in Ω. Such mappings are generalizations of quasiconformal homeomorphisms where one works with the stronger assumption µ(z) ≤ k < 1. Mappings of finite distortion have found considerable interest in geometric function theory and the mathematical theory of elasticity. A comprehensive overview of the theory of mappings of finite distortion in two-dimensions can be found in [3]. The Jacobian determinant of a mapping f of finite distortion is non-negative almost everywhere, since Jf (z) = |fz |2 − |fz¯|2 = (1 − µ(z)2 )|fz |2 ≥ 0. 2000 Mathematics Subject Classification. 58E20,30F45. Key words and phrases. Harmonic maps, Riemann surfaces, Modulus of annuli, Finite distortion, Radial metric. 1

DEFORMATIONS OF ANNULI WITH SMALLEST MEAN DISTORTION

2

The distortion function of particular interest to us in this article is defined by the rule (1.1)

K(z, f ) =

kDf (z)k2 |fz |2 + |fz¯|2 | = 2 2 |fz | − |fz¯| Jf (z)

if Jf (z) > 0. Here 1 Tr(AT A) 2 is the square of mean Hilbert-Schmidt norm. We conveniently set K(z, f ) = 1 if fz = fz¯ = 0. Notice that then K(z, f ) = 1 and we have the equality K(z, f ) = 1 if and only if f is conformal, by the Looman Menchoff theorem. kAk2 =

1.2. Radial metrics. By U we denote the unit disk {z : |z| < 1}, by C is denoted the extended complex plane. Let Σ be a Riemannian surface over a domain C of the complex plane or over C and let p : C 7→ Σ be a universal covering. Let ρΣ be a conformal metric defined in the universal covering domain C or in some chart D of Σ. It is well-known that C can be one of three sets: U, C and C. Then the distance function is defined by d(a, b) = inf

a,b∈γ

Z

1

ρΣ (˜ γ (t))|˜ γ ′ (t)|dt,

0

where γ˜ , γ˜ (0) = 0, is a lift of γ, i.e. p(˜ γ (t)) = γ(t), γ(0) = a, γ(1) = b. The Gauss curvature of the surface (and of the metric ρΣ ) is given by K=−

∆ log ρΣ . ρ2Σ

In this paper we will consider those surfaces Σ, whose metric have the form ρΣ (z) = h(|z|2 ), defined in some chart A′ = {z : τ < |z| < σ} of Σ (not necessarily in the whole universal covering surface). Here h is an positive twice differentiable function. We call these metrics radial symmetric. Definition 1.1. The radial metric ρ is called a regular metric if inf sρ(s) = lim sρ(s)

τ r).

Proof. Let z = seit , s = |z|, t ∈ [0, 2π) and A+ = {z : Jf (z) > 0}. We will apply Lemma 4.1 with Z s dy p ϕ(s) = , τ ≤ s ≤ σ. 2 y + c̺2 σ Proof of a). Since ̺ = 1/ρ, we have

1 − s2 ϕ′ (s)2 =

c

. +c In view of this fact, we divide the proof into two cases. • The case c ≥ 0. Observe first that, for almost every z = seit ∈ A′ s ≤ 1 ≤ K(z, f ). (4.8) sϕ′ (s) = p s2 + c̺2 (s) s2 ρ2

According to (4.5) and (4.8), we have Z Z Z 2 ′ 2 sρ (s)ϕ (s) + K(z, f )ρ (s) ≥ A′ \A+

A′

1− 2s2 Jf (z) A+ Z Z 2 ′ sρ (s)ϕ (s) + =

+

(4.9)

Z

ρ2 (s)

sρ2 (s)ϕ′ (s)

A+ 2 ′ 2 s (ϕ (s))

A+

A′

|ft |2

c |fs |2 . 2s2 (s2 + c̺2 (s)) Jf (z)

By Hölder inequality we obtain (4.10) Z 1/2 Z Z Jf (z) 1/2 1 |ft | |ft |2 p ≤ . 2 2 2 2 A+ 2s (s + c̺ (s)) Jf (z) A+ s|f | 2(s2 + c̺2 (s)) A+ |f | By [2, Lemma 1] it follows that Z Z Jf (z) Jf (z) ≤ ≤ Mod (A). (4.11) 2 2 A′ |f | A+ |f |

Since

Z

2π 0

Z |ft (seit )| 2π ft (seit ) ≥ dt = 2π, |f (seit )| f (seit ) 0

from (4.11), we obtain Z 1 |ft |2 1 ≥ 2 2 2 Mod (A) A′ 2s (s + c̺ (s)) Jf (z) ≥

4π 2 Mod (A)

Z

|ft |

!2

p s|f | 2(s2 + c̺2 (s)) !2 Z σ ds p . 2(s2 + c̺2 (s)) τ A′


On the other hand, Z

sρ2 (s)ϕ′ (s)dm(z) = 2π

A′

2

K(z, f )ρ (s)dm(z) ≥ 2π

s2 ρ2 (s) p ds. s2 + c̺2 (s)

Z

σ

s2 ρ2 (s)

ds s2 + c̺2 (s) !2 Z σ ds 4π 2 p . + Mod (A) 2(s2 /c + ̺2 (s)) τ

Since

p

τ

1 Mod (A) = 2π log = 2π r

(4.12)

σ

τ

A′

Therefore Z

Z

13

it follows that Z Z 2 K(z, f )ρ (s)dm(z) ≥ 2π =π +π

Z

=π

Z

p

τ σ

p

τ σ

For f c (seit ) = eϕ(s)+it , where Z ϕ(s) =

τ

s2

+ c̺2 (s) Z

ds + πc

s2 ρ2 (s) ds s2 + c̺2 (s) s2 ρ2 (s) +

c̺2 (s)

ds + πc

s2 ρ2 (s) s2 + c̺2 (s)

s σ

ds p

s2 + c̺2 (s)

s2

p

τ

σ

s2 ρ2 (s) p

τ σ

A′

Z

σ

Z

dy p

y2

+ c̺2

ds + π

Z

Z

,

σ τ

ds p

s2 + c̺2 (s))

σ τ σ

τ

ds p

s2

+ c̺2 (s)) p ρ2 (s) s2 + c̺2 (s)ds.

, τ ≤ s ≤ σ,

by making use of formula (4.4), we have Z Z 2π σ 2 1 c 2 ′ K(z, f )ρ (s)dm(z) = sρ (s) ϕ (s)s + ′ ds 2 τ sϕ (s) A′ ! p Z σ 2 + c̺2 (s) s s + sρ2 (s) p =π ds. s s2 + c̺2 (s) τ

Thus

Z

2

A′

K(z, f )ρ (s)dm(z) ≥

Z

K(z, f c )ρ2 (s)dm(z).

A′

• The case −τ 2 ρ2 (τ ) ≤ c ≤ 0. In this case we make use of (4.6). Observe also that, for almost every z ∈ A′ p s2 + c̺2 (s) 1 = ≤ 1 ≤ K(z, f ). (4.13) ′ sϕ (s) s


For z ∈ A+ , we obtain K(z, f )ρ2 (s) ≥

14

ρ2 (s) −cρ2 (s) |fs |2 + . sϕ′ (s) 2s2 ρ2 (s) Jf

Hence by (4.13) Z Z Z Z ρ2 (s) −c |fs |2 ρ2 (s) 2 + + K(z, f )ρ (s) ≥ ′ ′ 2 A+ sϕ (s) A+ 2s Jf A′ \A+ sϕ (s) A′ (4.14) Z Z ρ2 (s) −c |fs |2 = + . ′ 2 A′ sϕ (s) A+ 2s Jf By [2, Lemma 3] and (4.12) we have Z σ Z ds 1 |fs |2 p . ≥ Mod (A) = 2π (4.15) 2 J 2 s + s + c̺2 (s) f τ A On the other hand (4.16)

Z

A′

Let

ρ2 (s) = 2π sϕ′ (s)

Z

σ

τ

p ρ2 (s) s2 + c̺2 (s)ds.

p k(s) := s2 + c̺2 (s). From (4.14), (4.15) and (4.16) we obtain (4.17) Z σ Z σ Z ds ρ2 (s)k(s)ds + 2π K(z, f )ρ2 (s) ≥ −cπ k(s) ′ τ τ A Z σ 2 2 Z σ s ρ (s) 2 ρ (s)k(s)ds + π =π ds k(s) τ τ Z σ Z σ 2 2 Z σ ds s ρ (s) 2 − cπ ρ (s)k(s)ds − π +π ds k(s) k(s) τ τ Z τ K(z, f c )ρ2 (s)dm(z) + X, = A′

where

X=π

Z

σ τ

s2 ρ2 (s) c + ρ2 (s)k(s) − − k(s) k(s)

ds = 0.

This concludes the proof of a). The proof of b). We proceed similarly as in the proof of a). For ϕ = ϕ# , according to (4.6), for z ∈ A+ we have K(z, f )ρ2 (s) ≥

τ 2 ρ2 (τ ) |fs |2 ρ2 (s) + . sϕ′ (s) 2s2 Jf

In this case, for c# = −τ 2 ρ2 (τ ), instead of (4.15) we have Z σ Z ds 1 |fs |2 1 p ≥ Mod (A) = 2π log > 2π . (4.18) 2 2 r s + c# ̺2 (s) A+ s Jf τ


15

On the other hand, the relation (4.16) holds. Hence Z Z 2 K(z, f # )ρ2 (s)dm(z) + Y, K(z, f )ρ (s)dm(z) ≥ (4.19) A′

A′

where

−c# Y = 2

1 1 2π log − 2π log ′ r r

=

τ 2 ρ2 (τ ) Mod A(r, r ′ ). 2

This completes the proof of the lemma.

Proof of Theorem 3.1. In view of Lemma 4.2, a), it remains to prove the equality statement. By [2, Lemma 3], the equality in (4.15) occurs if and only if almost everywhere on A+ := {z ∈ A′ : Jf (z) > 0} we have fs fs (4.20) ℜ = f f and

(4.21)

ℑ

ft f

=k

for some constant k > 0. On other hand, by Lemma 4.1 we have the equality (4.22)

fs = −iϕ′ (s)ft ,

for z ∈ A+ . Combining (4.20), (4.21) and (4.22) we arrive at the following system of PDE’s fs ft = ik and = kϕ′ (s). (4.23) f f Integrating the first equality over the unit circle gives k = 1. Let g(s, t) = log f (seit ). Then g is well defined function from [τ, σ] × [0, 2π] onto [log r, 0] × [0, 2π]. Then from (4.23) we obtain g = it + φ(s) and g = ϕ(s) + ψ(t), for some functions φ and ψ. We obtain that, the general solution of system of PDE’s (4.23) is f (seit ) = Ceϕ(s)+it . Proof of Theorem 3.3. The second part of Theorem 3.3 follows by Lemma 4.2, b). It remains to show that, the inequality is sharp and it is not attained by any homeomorphism. Otherwise, if f ∈ F(A′ , A) would satisfy (3.5), f has to be of the form ! Z s dy it p f (se ) = C exp it + , τ ≤ s ≤ σ. y 2 − τ 2 ρ2 (τ )̺2 (y) σ

But then r ′ = r which is a contradiction with (2.11). To show that the inequality is sharp we construct the minimizing sequence.


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4.2. The minimizing sequence. We are now given two round annuli A and A′ with inner and outer radii r, 1 and τ , σ respectively. Moreover we assume that the fatness condition (2.11) is satisfied. To show that the inequality (3.6) is sharp we construct a family fn : A′ → A of mappings of finite distortion. Let ! Z τ ρ(y)dy ′ p . r = exp y 2 ρ2 (y) − τ 2 ρ2 (τ ) σ

Then r < r ′ < 1. Let z = seit and n ∈ N such that σ n 1−r < 0. τ Define  ! Z s  dy  p exp it + if sn ≤ s ≤ σ, 2 − τ 2 ρ2 (τ )̺2 (y) fn (z) = y σ   rz|z|n−1 /τ n if τ ≤ s ≤ sn ,

where sn is a solution of the equation ! Z sn s n dy n p (4.24) exp =r . 2 2 2 2 τ y − τ ρ (τ )̺ (y) σ

To show that sn : τ < sn < σ exists satisfying the condition (4.24) take ! Z s s n dy p −r . p(s) = exp τ y 2 − τ 2 ρ2 (τ )̺2 (y) σ Then

σ n < 0, p(τ )p(σ) = (r ′ − r) 1 − r τ if n is big enough. From (4.24) we easily get the relation

r′ . n→∞ r Further, by a similar analysis as in [2, Section 9] we obtain that  √2 2 2 2 s −τ ρ (τ )̺ (s)  √ s if sn ≤ s ≤ σ, + s s2 −τ 2 ρ2 (τ )̺2 (s) K(z, fn ) =  1 1 if τ ≤ s ≤ sn . 2 n+ n (4.25)

lim n(sn − τ ) = τ log

Finally by (4.25) Z Z Z # 2 2 K(z, fn )ρ2 (s) K(z, f )ρ (s) + lim K(z, fn )ρ (s) = lim n→∞ A′ (τ,s ) n→∞ A′ (τ,σ) A′ (τ,σ) n Z Z n sn 2 # 2 K(z, f )ρ (s) + 2π lim = ρ (s)sds n→∞ 2 τ ′ ZA n K(z, f # )ρ2 (s) + πρ2 (τ ) lim (sn 2 − τ 2 ) = n→∞ 2 A′ Z 2 2 τ ρ (τ ) K(z, f # )ρ2 (s) + = Mod A(r, r ′ ). 2 ′ A


This finishes the proof of Theorem 3.3. 5. P ROOF

OF

17

C OROLLARIES 3.6

AND

3.4

In order to apply Theorems 3.1 and 3.3 to energy minimization problems for the inverse mappings (see (3.7)), proving in this way Corollaries 3.6 and 3.4, we need to establish the following Lemma 5.1 (The correction lemma). Let D be doubly connected domain in C and let A′ = {z : τ < |z| < σ}. Let h : D → A′ , be a homeomorphism of finite Dirichlet energy, Z ρ2 ◦ h(|hz |2 + |hz¯|2 )dm(z) < ∞. Eρ [h] = D

Assume that ρ is a radial metric with bounded Gauss curvature satisfying (5.1)

0 < inf ′ ρ2 (w) ≤ sup ρ2 (w) < ∞. w∈A

w∈A′

˜ : D → A′ such that Then there exists a homeomorphism h ˜ ≤ Eρ [h]. Eρ [h]

˜ −1 belongs to W 1,1 (A′ , D) and has finite distortion. We have The inverse f˜ = h the identity ˜ Kρ [f˜] = Eρ [h]. Proof. We proceed as in [2, Lemma 7], however due to the different intrinsic geometry, we should provide a different proof. A domain G is said to be convex with respect to the metric ρ, if for z, w ∈ G, there exists a geodesic line l[z, w] ⊂ G, z, w ∈ l[z, w] with respect to the metric ρ. By following an argument of Buser ([5, Pages 116-121], see also [6, The proof of Theorem 1]) there exists a triangulation of the Riemann surface (A′ , ρ) by a locally finite set of triangles ∆j j = 1, . . . , m with pairwise disjoint interiors, such that the edges of triangles are geodesic arcs or part of boundary (such domains are lipschitz and convex domains (if the diameter is small enough)). Also we can assume that diamρ (∆j ) := sup{dρ (z, w) : z, w ∈ ∆j } < ε

where ε is a positive constant to be chosen later. Let us sketch the construction of these geodesic triangles. Let P be a finite set of points in A such that any two points of P lie at ρ-distance ≤ ε and ≥ ε/2 from each other. Such a set P is easily obtained by successively marking points in A′ at pairwise distances ≥ ε/2 until there is no more room for such points. The set P is finite because the ρ-diameter of A′ is finite. We add to the set P a maximal finite set of boundary points at pairwise distance between ε/2 and ε. For arbitrary triple of points a, b, c from P we can construct a (possible degenerated)′ triangle ∆i , i = 1, . . . , n, n = |P| defined as follows. If a, b, c are inside A then the wedges 3 are geodesic. If for example a, b belong to the same connected component of ∂A′ , then the wedge ab is the smaller boundary arc bounded by a and b. We exclude all


18

degenerated triangles, those who contain more than three points inside and those whose diameter is larger than ε. Thus we get the family ∆j , j = 1, . . . , m. Then each ∆j is contained in a geodesic disk Bε (p) with center at p ∈ ∆j and radius ε. Through the homeomorphism h we have a decomposition m [

D=

Dk ,

k=1

h−1 (∆

where Dk = k) Let the Gauss curvature of the metric ρ satisfies K ≤ κ2 . Let C be a positive constant, smaller or equal to the minimal distance of a given point p ∈ A′ from its cut-locus. Because the metric is radial, the cut-locus of a point p ∈ A′ is the arc [−τ p, −σp]. Since the metric satisfies the condition (5.1) it follows that C = inf{dρ (x, −y) : τ ≤ x, y ≤ σ} > 0.

Take nπ o ,C . 2κ Consider the Dirichlet problem of finding a ρ−harmonic map hj : Dj → A′ with the given boundary values: hj |∂Dj = h. Since hj (∂Dj ) is contained in a geodesic disk Bε (p) with: (1) radius ε < π/(2κ); (2) the cut locus of the centre p disjoint from Bε (p), by a result of Hildebrandt, Kaul and Widman [11] this Dirichlet problem has a solution contained in Bε (p). Moreover by a result of Jost [15] we obtain that, since hj : ∂Dj → A′ is a homeomorphism onto a Lipschitz convex curve ∂∆j , then the above solution hj is a homeomorphism. Let m X ˜= hj (z)χD j (z). h ε < min

(5.2)

j=1

˜ is a homeomorphism by construction. By using the well known energy Then h estimates we have Z Z kDhj k2 ρ2 (z)dm(z) ≤ kDhk2 ρ2 (z)dm(z). Dj

Dj

˜ −1 . Proceeding as in [2, Lemma 7], we obtain Let f˜ = h Kρ [f˜] ≤ Eρ [h]. To continue observe that Z Z 2 2 (|hz | + |hz¯| )dm(z) ≤ E[h] = D

D

ρ2 ◦ h (|hz |2 + |hz¯|2 )dm(z) < ∞. inf w∈A′ ρ2 (w)

By using a recent result of Hencl, Koskela and Onninen [9], we obtain that the homeomorphism f : A′ → D of integrable distortion in the Sobolev class W 1,1 (A′ , D)


19

˜ ∈ W 1,2 (D, A′ ). By introducing the change of variables w = h(z), ˜ has its inverse h proceeding as in [9], by making use of formulas −1 kBk = kB T k, AT det A = adj A, kadj Ak = kAk we obtain the identity

˜ Kρ [f˜] = Eρ [h]. This completes the proof of the Lemma 5.1.

Remark 5.2. We think that some of the conditions of Lemma 5.1 concerning the metric ρ are superfluous. 6. A PPENDIX In this section we will give some examples of regular metrics. 6.1. Hyperbolic metrics. For every hyperbolic Riemann surface, the fundamental group is isomorphic to a Fuchsian group, and thus the surface can be modeled by a Fuchsian model U/Γ, where U is the unit disk and Γ is the Fuchsian group ([1]). If Ω is a hyperbolic region in the Riemann sphere C; i.e., Ω is open and connected with its complement Ωc := C \ Ω possessing at least three points. Each such Ω carries a unique maximal constant curvature −1 conformal metric λ|dz| = λΩ (z)|dz| referred to as the Poincaré hyperbolic metric in Ω. The domain monotonicity property, that larger regions have smaller metrics, is a direct consequence of Schwarz’s Lemma. Except for a short list of special cases, the actual calculation of any given hyperbolic metric is notoriously difficult. By the formula ρΣ (z) = h(|z|2 ), we obtain that the Gauss curvature is given by 4(|z|2 h′ 2 − |z|2 hh′′ − hh′ ) . K= h4 Setting t = |z|2 , we obtain that 1 4th′ (t) ′ (6.1) K=− 2 . h h As K ≤ 0 it follows that

Therefore the function

4th′ (t) h

′

≥ 0.

4th′ (t) h

is increasing, i.e. (6.2)

t≥s⇒

4sh′ (s) 4th′ (t) ≥ , h(t) h(s)


20

and in particular t≥0⇒

(6.3)

4th′ (t) ≥ 0. h(t)

In this case we obtain that h is an increasing function. The examples of hyperbolic surfaces are: a) The Poincaré disk U with the hyperbolic metric λ=

2 . 1 − |z|2

b) The punctured hyperbolic unit disk ∆ = U \ {0}. The linear density of the hyperbolic metric on ∆ is 1 λ∆ = 1 . |z| log |z| c) The hyperbolic annulus A(1/R, R), R > 1. The hyperbolic metric is given by π/2 π log |z| sec . |z| log R 2 log R In all these cases the Gauss curvature is K = −1. hR (|z|2 ) = λR (z) =

6.2. Riemann metrics. In the case of the Riemann sphere, the Gauss-Bonnet theorem implies that a constant-curvature metric must have positive curvature K. It √ follows that the metric must be isometric to the sphere of radius 1/ K in R3 via stereographic projection. d) In the z-chart on the Riemann sphere, the metric with K = 1 is given by ds2 = h2R (|z|2 )|dz|2 =

4|dz|2 . (1 + |z|2 )2

e) Another important case is Hamilton cigar soliton or in physics is known as Wittens’s black hole. It is a Kähler metric defined on C. ds2 = h2 (|z|2 )|dz|2 = The Gauss curvature is given by K=

|dz|2 . 1 + |z|2

2 . 1 + |z|2

In both these cases K > 0. This means that 4th′ (t) ′ ≤ 0. h Therefore the function

4th′ (t) h


21

is decreasing i.e. (6.4)

t≥s⇒

4sh′ (s) 4th′ (t) ≤ . h(t) h(s)

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[23] RUH , E.; V ILMS , J.: The tension field of the Gauss map. Trans. Amer. Math. Soc. 149 (1970) 569–573. [24] S CHOEN , R.; YAU , S. T.: Lectures on harmonic maps. Conference Proceedings and Lecture Notes in Geometry and Topology, II. International Press, Cambridge, MA, 1997. vi+394 pp. [25] WAN , T.: Constant mean curvature surface, harmonic maps, and universal Teichmller space. J. Differential Geom. 35 (1992), no. 3, 643–657. [26] W EITSMAN , A.: Univalent harmonic mappings of annuli and a conjecture of J.C.C. Nitsche. Israel J. Math. 124, 327–331 (2001) U NIVERSITY OF M ONTENEGRO , FACULTY OF NATURAL S CIENCES C ETINJSKI PUT B . B . 8100 P ODGORICA , M ONTENEGRO E-mail address: [email protected]

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