5_460_zhichun.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 2, No. 4, 2009, (532-543)

ISSN 1307-5543 – www.ejpam.com

On the Solutions for Grötzsch Annulus Extremal

Problem

Xie Zhichun∗ and Huang Xinzhong

School of Mathematical Science, Huaqiao University, Quanzhou, Fujian, 362021, P. R. C.

Abstract. Using the properties of planar quasiconformal mappings, we obtain the solutions

for the Grötzsch extremal problem on the annulus. We also point out the shortcoming used

in [3] for solving this extremal problem. Moreover, by the hyperbolic area distortion and the

property of domain module, one criterion for the solution of the Grötzsch annulus extremal

problem is given under some conditions.

2000 Mathematics Subject Classifications: Primary 30C62, Secondary 30C55

Key Words and Phrases: Grötzsch extremal problem, quasiconformal mapping, hyperbolic

area distortion

1. Introduction

Let Ω,Ω′ ⊂ C be planar domains, a sense-preserving homeomorphism f : Ω→ Ω′

is said to be K−quasiconformal mapping in Ω, if it satisfies: (1) f is ACL in Ω; (2)

∗Corresponding author.

Email addresses: xiezhi
hun�hqu.edu.
n (X. Zhichun), huangxz�hqu.edu.
n (H. Xinzhong)

http://www.ejpam.com 532 c© 2009 EJPAM All rights reserved.



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 533

fz̄(z) = µ(z) fz(z) a.e., z ∈ Ω, where ess sup
z∈Ω
|µ(z)|= k < 1, K = 1+k

1−k
. Suppose that f (z)

is a quasiconformal mapping in Ω, let D f (z) =
| fz|+| fz̄ |

| fz|−| fz̄ |
. The Grötzsch extremal problem

is to find a quasiconformal mapping f0(z) among all quasiconformal mappings f (z)

in Ω, such that

ess sup
z∈Ω

D f0
(z) = inf

f
sup

z

D f (z). (1)

It is known by [1] that the solution of the Grötzsch extremal problem from square

to rectangle is an affine mapping, and the Grötzsch extremal problem from annulus

{z|r ≤ |z| ≤ 1} onto {z|R ≤ |z| ≤ 1} was investigated in [2] and [3]. The following

result was proved in [2].

Theorem 1. If f (z) be a quasiconformal homeomorphism in the unit disk onto itself

such that

f (0) = 0, lim
z→0

| f (z)|

|z|
1

K

= 1.

Then the solution for the Grötzsch annulus extremal problem is f (z) = eiθz|z|
1

K
−1, where

θ is a real number.

Obviously, the normalized condition f (0) = 0 is unnecessary for solving the Grötzsch

annulus extremal problem. By the method of extremal length, the following result

was proved in [3].

Lemma 1. If f (z) is a K−quasiconformal homeomorphism from {z|r ≤ |z| ≤ 1} to

{z|r
1

K ≤ |z| ≤ 1}, then

f (z) = λz|z|
1

K
−1,

where λ is a constant and |λ|= 1.

We point out that Lemma 1 is not true. For example, let g(z) = r
1

K ei(θ+ksinθ ), where

z = reiθ , k = K−1

K+1
, K ≥ 1, then g(z) is a 2K2

K+1
−quasiconformal homeomorphism from

{z|r ≤ |z| ≤ 1} onto {z|r
1

K ≤ |z| ≤ 1}. Therefore, Lemma 1 is meaningful only under



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 534

considering the Grötzsch extremal problem on the annulus. On the other hand, [3]

tried to prove the solution of the Grö tzsch extremal problem on the annulus by using

the method of extremal length, the property of the module of ring domain and used

the following equality
1

Kmod(RR)
=

1

mod(R
R

1
K
)
.

However, above equality is also not true, since we find that 1

Kmod(RR)
= 1

K 1

2π
log 1

R

and

1

mod(R
R

1
K
)
= 1

1

K

1

2π
log 1

R

.

In this paper, the solution for the Grötzsch extremal problem on the annulus is

solved by using analytic method and the extremal condition. On the other hand, [3]

also tried to find the solution of the Grötzsch extremal problem for area distortion

problem. We also point out that the proved result is not correct. At last, we obtain

one criterion for the solution of the Grötzsch annulus extremal problem considering

hyperbolic area distortion.

2. Main Results and their Proofs

The notations in this paper are adopted as in [3]. Let A(r1, r2) = {z|r1 ≤ |z| ≤

r2}, A(r1, r2;θ1,θ2) = {z|r1 ≤ |z| ≤ r2,θ1 ≤ argz ≤ θ2}. We will prove the following

result.

Theorem 2. Let Q be K-quasiconformal mappings from {z|r1 ≤ |z| ≤ 1} to {z|r
1

K

1 ≤

|z| ≤ 1}, and if f ∈Q satisfies

ess sup
z

D f (z) = inf
g∈Q

sup
z

Dg(z),

then

f (z) = r
1

K ei(θ+α), z = reiθ , where α is a constant.



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 535

Proof. For any R1, r1 < R1 < 1, let A= {z|r1 ≤ |z| ≤ R1}, B = {z|R1 ≤ |z| ≤ 1}, by

the property of the module of ring domain [4], we have

1

K

1

2π
log

1

r1

=
1

K
(mod(A) +mod(B))

≤ mod( f (A)) +mod( f (B)) ≤mod(A(r
1

K

1 , 1)) =
1

K

1

2π
log

1

r1

,

thus,

mod( f (A)) +mod( f (B)) =mod(A(r
1

K

1 , 1)) =
1

K

1

2π
log

1

r1

.

By Lemma 1.3 in [3], we see that f (A) and f (B) are concentric annulus, we have

| f (R1eiθ )|= R′, 0≤ θ < 2π. By the quasi-invariant property of module, we have







mod( f (A)) ≥ 1

K
mod(A),

mod( f (B)) ≥ 1

K
mod(B),

that is






R′ ≥ R
1

K

1 ,

R′ ≤ R
1

K

1 .

Thus, R′ = R
1

K

1 . Therefore, we obtain that f (z) = r
1

K eiϕ(r,θ ), z = reiθ . Next, we will

prove that ϕ(r,θ ) depends only on θ . By calculation, we have

fz̄ =
1

2
r

1

K
−1ei(ϕ+θ )(

1

K
+ irϕr −ϕθ ),

fz =
1

2
r

1

K
−1ei(ϕ−θ )(

1

K
+ irϕr +ϕθ ).

Obviously, w = λz|z|
1

K
−1 is a K − q.c. from {z|r1 ≤ |z| ≤ 1} onto {z|r

1

K

1 ≤ |z| ≤ 1}. If

f (z) is an extremal K − q.c., which satisfies the conditions of Theorem 2, we have

|
fz̄

fz

|= |
1

K
+ irϕr −ϕθ

1

K
+ irϕr +ϕθ

| ≤
K − 1

K + 1
,



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 536

thus,

K2ϕ2
θ
− (K2 + 1)ϕθ + 1+ K2r2ϕ2

r
≤ 0, (2)

If ϕr 6= 0, by the inequality (2), we have

K2ϕ2
θ
− (K2 + 1)ϕθ + 1 ≤−K2r2ϕ2

r
< 0,

that is

(K2ϕθ − 1)(ϕθ − 1) < 0,

so we obtain that
1

K2
< ϕθ < 1.

For any r ∈ (0, 1), we have

2π = ϕ(r, 2π)−ϕ(r, 0) =

∫ 2π

0

ϕθ (r,θ ) dθ <

∫ 2π

0

dθ = 2π.

This is impossible. So we have proved that ϕr = 0. Therefore, we may assume that

f (z) = r
1

K eiψ(θ ).

At last, we will prove that ψ(θ ) = θ + α, where α is a constant. For any θ1,θ2, 0 ≤

θ1 < θ2 ≤ 2π, let

A1 = A(r1, 1;θ1,θ2), A2 = A(r1, 1;θ2,θ1+ 2π),

then

f (A1) = A(r
1

K

1 , 1;ψ(θ1),ψ(θ2)),

f (A2) = A(r
1

K

1 , 1;ψ(θ2),ψ(θ1) + 2π).

By the quasi-invariant property of module, we have






1

K
mod(A1)≤mod( f (A1)),

1

K
mod(A2)≤mod( f (A2)),



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 537

that is






ψ(θ2)−ψ(θ1)≤ θ2− θ1,

ψ(θ2)−ψ(θ1)≥ θ2− θ1,

therefore,

ψ(θ2)−ψ(θ1) = θ2 − θ1.

For any θ , 0≤ θ ≤ 2π, since

ψ′(θ ) = lim
h→0

ψ(θ + h)−ψ(θ )

h
= lim

h→0

(θ + h)− θ

h
= 1,

then

ψ(θ ) = θ +α,α is a constant,

thus, we obtain that

f (z) = r
1

K ei(θ+α).

The theorem is proved.

By Theorem 2, the following result can be proved.

Corollary 1. Let Q be K−quasiconformal mappings from {z|r1 ≤ |z| ≤ 1} to {z||rK
1
≤

|z| ≤ 1}, and if f ∈Q satisfies

ess sup
z

D f (z) = inf
g∈Q

sup
z

Dg(z),

then

f (z) = rK ei(θ+α), z = reiθ ,

where α is a constant.

On the other hand, [3] also considered the Grötzsch extremal problem on annulus

under some restriction of area distortion, and proved the following result.



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 538

Theorem 3. Let f :∆= {z||z|< 1} →∆ be a K−quasiconformal mapping and f (∆) =

∆, f (0) = 0. For any ∆r = {z||z|< r} ⊂∆, if

Area( f (∆r))

Area(∆r)
1

K

≥ π1− 1

K ,

then, f (z) = λz|z|
1

K
−1 for any z ∈ Rr = {z|r < |z| < 1}, where λ is a constant with

|λ|= 1.

We will point out that Theorem 3 is not correct. For example, if we again take

g(z) = r
1

K ei(θ+ksinθ ), z = reiθ , it is easy to see that
Area(g(∆r ))

Area(∆r )
1
K

= π1− 1

K , but g(z) is not

the required form, thus Theorem 3 is fault.

Next, by considering the relationship between the hyperbolic area of ring domain

and its module, we can obtain one criterion for extremal mapping under the hyper-

bolic area distortion condition.

Suppose that E ⊂∆ is a measurable subset, let

|E|hyp =

∫∫

E

1

(1− |z|2)2
|dz|2,

be the hyperbolic area of E, and Area(E) be its Euclidean area. We need the following

result made in [5].

Lemma 2. Let R be a ring domain bounded by two mutually disjoint closed curves B0

and B1. For any arbitrary concentric ring R∗ bounded by concentric circles B∗
0

and B∗
1
, if

Area(R) = Area(R∗), then mod(R)≤mod(R∗).

Lemma 3. If f (x) is a positive continuous function in [0, 2π], then

∫ 2π

0

1

f (x)
d x ≥

4π2

∫ 2π

0
f (x) d x

,

with equality if and only if f (x)≡ c > 0.



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 539

Proof. If f (x) ∈ C[0, 2π] and f (x)> 0, we have

[

∫ 2π

0

p

f (x) ·
1
p

f (x)
d x]2 ≤

∫ 2π

0

f (x) d x ·

∫ 2π

0

1

f (x)
d x ,

thus

4π2 ≤

2π
∫

0

f (x)d x ·

2π
∫

0

1

f (x)
d x ,

that is,
∫ 2π

0

1

f (x)
d x ≥

4π2

∫ 2π

0
f (x) d x

.

Obviously, with equality if and only if f (x)≡ c > 0. The proof of Lemma 3 is finished.

Next we will prove the following

Theorem 4. Suppose that S1 = A(r1, r2), 0 < r1 < r2 < 1, and S2 ⊂∆ is a ring domain

bounded by two mutually disjoint closed curves C1 and C2, where C1 = {z||z| = r1}

and C2 = {z|z = f (θ )eiθ }, where f (θ ) is a continuous function in [0, 2π] and f (0) =

f (2π), are the inner and outer boundaries of S2. If

|S1|hyp ≥ |S2|hyp,

then

mod(S1) ≥mod(S2).

With equality if and only if S2 = S1 = A(r1, r2).

Proof. Since

|S1|hyp =

∫∫

S1

1

(1− |z|2)2
|dz|2 =

∫ 2π

0

∫ r2

r1

r

(1− r2)2
drdθ

= π(
1

1− r2
2

−
1

1− r2
1

),



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 540

|S2|hyp =

∫∫

S2

1

(1− |z|2)2
|dz|2 =

∫ 2π

0

∫ f (θ )

r1

r

(1− r2)2
drdθ

=
1

2

∫ 2π

0

(
1

1− f 2(θ )
−

1

1− r2
1

) dθ ,

by the condition |S1|hyp ≥ |S2|hyp, we have

π

1− r2
2

≥
1

2

∫ 2π

0

1

1− f 2(θ )
dθ .

By Lemma 3, we obtain

π
1

1− r2
2

≥
1

2

∫ 2π

0

1

1− f 2(θ )
dθ ≥

1

2

4π2

∫ 2π

0
1− f 2(θ ) dθ

,

thus,
1

2π

1

1− r2
2

≥
1

∫ 2π

0
1− f 2(θ ) dθ

,

by calculation, we get

Area(S1 ∪∆r1
)≥

1

2

∫ 2π

0

f (θ )2 dθ = Area(S2 ∪∆r1
),

thus

Area(S1) ≥ Area(S2),

owing to Lemma 2, we obtain

mod(S1) ≥mod(S2).

Next, we will discuss the equality.

If mod(S1) = mod(S2), we conclude that Area(S1) = Area(S2). Otherwise, if

Area(S1) > Area(S2), then there exists an A(r1, r ′) ⊂ S1 = A(r1, r2), r ′ < r2, such

that Area(A(r1, r ′)) = Area(S2). By Lemma 2, we get

mod(S1)>mod(A(r1, r ′))≥mod(S2),



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 541

this is a contradiction with mod(S1) = mod(S2). Therefore, if mod(S1) =mod(S2), by

Lemma 3 and the proof above, we have







|S1|hyp = |S2|hyp,

f (θ ) ≡ c = r2.

⇔ S1 = S2.

The proof is complete.

We have pointed out that the area distortion used in [3, Theorem 2.1] could not

characterize extremal quasiconformal mapping. We will prove the following

Theorem 5. Let w = f (z) be a quasiconformal mapping from S = {z|R1 ≤ |z| ≤ R2, 0<

R1 < R2 < 1} onto S′, where S′ ⊂ D = {w||w| < 1} is a ring domain bounded by two

mutually disjoint closed curves Γ1 = {w||w| = R
1

K

1 } and Γ2 = {w|w = f (R2eiθ ), 0 ≤ θ ≤

2π}. If f (z) satisfies

| f (S)|hyp

|S|hyp

=
(R

2

K

2 − R
2

K

1 )(1− R2
2
)(1− R2

1
)

(R2
2− R2

1)(1− R
2

K

2 )(1− R
2

K

1 )

,

and K[ f ] = K, then f (z) is an extremal quasiconformal mapping from S onto S′, and

f (z) = r
1

K ei(θ+α), z = reiθ ,

where α is a constant.

Proof. Let S′′ = A(R
1

K

1 , R
1

K

2 ), by calculation, we have

|S|hyp = π
R2

2
− R2

1

(1− R2
2)(1− R2

1)
, |S′′|hyp = π

R
2

K

2 − R
2

K

1

(1− R
2

K

2 )(1− R
2

K

1 )

,

by the hypothesis that

| f (S)|hyp

|S|hyp

=
(R

2

K

2 − R
2

K

1 )(1− R2
2
)(1− R2

1
)

(R2
2− R2

1)(1− R
2

K

2 )(1− R
2

K

1 )

,



X. Zhichun and H. Xinzhong / Eur. J. Pure Appl. Math, 2 (2009), (532-543) 542

we have

| f (S)|hyp = |S|hyp

(R
2

K

2 − R
2

K

1 )(1− R2
2
)(1− R2

1
)

(R2
2 − R2

1)(1− R
2

K

2 )(1− R
2

K

1 )

= π
R2

2
− R2

1

(1− R2
2)(1− R2

1)
·
(R

2

K

2 − R
2

K

1 )(1− R2
2
)(1− R2

1
)

(R2
2− R2

1)(1− R
2

K

2 )(1− R
2

K

1 )

= π
R

2

K

2 − R
2

K

1

(1− R
2

K

2 )(1− R
2

K

1 )

= |S′′|hyp.

Using Theorem 4, we get

mod(S′′) ≥mod( f (S)),

and by the quasiconformallity of f (z), we have

mod( f (S)) ≥
1

K
mod(S),

thus, mod( f (S)) = 1

K
mod(S) =mod(S′′), again by Theorem 4, we obtain

f (S) = S′′ = A(R
1

K

1 , R
1

K

2 ).

Therefore, as it is proved in Theorem 2, we have

f (z) = r
1

K ei(θ+α), z = reiθ ,

where α is a constant.

Using Theorem 5, we have the following

Corollary 2. Let w = f (z) be a quasiconformal mapping from S onto S′ = {w|RK
1
≤

|w| ≤ RK
2
, 0 < R1 < R2 < 1}, where S ⊂ D = {z||z| < 1} is a ring domain bounded by

two mutually disjoint closed curves Γ1 = {z||z| = R1} and Γ2 = {z|| f (z)| = RK
2
}. If f (z)

satisfies
|S|hyp

| f (S)|hyp

=
(R2

2
− R2

1
)(1− R2K

2
)(1− R2K

1
)

(R2K
2 − R2K

1 )(1− R2
2)(1− R2

1)
,



REFERENCES 543

and K[ f ] = K, then f (z) is an extremal quasiconformal mapping from S onto S′, and

f (z) = rK ei(θ+α), z = reiθ ,

where α is a constant.

ACKNOWLEDGEMENTS The authors would like to thank the referee very much for

his many valuable suggestions.

This Project was supported by the Natural Science Foundation of Fujian Province, PR

China (2008J0195).

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