International Journal of Analysis and Applications
ISSN 2291-8639
Volume 9, Number 1 (2015), 1-8
http://www.etamaths.com

UNIVALENT BIHARMONIC MAPPINGS AND LINEARLY

CONNECTED DOMAINS

Z. ABDULHADI1 AND L. EL HAJJ2,∗

Abstract. A four times continuously differentiable complex-valued function

F = u + iv in a simply connected domain Ω is biharmonic if the laplacian
of F is harmonic. Every biharmonic mapping F in Ω has the representation

F = |z|2G + K, where G and K are harmonic in Ω. This paper investigates
the relationship between the univalence of F and of K using the concept of
linearly connected domains.

1. Introduction

A planar harmonic mapping in a simply connected domain Ω ⊂ C is a complex-
valued harmonic function f(z) defined on Ω, where z = x+iy. The mapping f has
a canonical decomposition f = h + g, where h and g are analytic (holomorphic) in
Ω (see [13, 14]). We say that f is locally univalent and sense preserving if and only
if its Jacobian Jf (z) is positive, where Jf (z) is given by

Jf (z) = |fz(z)|2 −|fz(z)|2 = |h′(z)|2 −|g′(z)|2,
(See Lewy [11]).
Clunie and Sheil-Small made the following important observation : f is locally

univalent and orientation -preserving in D if and only if |g′(z)| < |h′(z)| in Ω; or
equivalently if h′(z) 6= 0 and the dilatation ω(z) = g

′(z)
h′(z)

has the property |ω(z)| < 1.

A four times continuously differentiable complex-valued function F = u + iv in
a simply connected domain Ω is biharmonic if the laplacian of F is harmonic. Note
that 4F is harmonic in Ω, if 4F satisfies Laplace’s equation 4(4F) = 0, where

4 = 4
∂2

∂z∂z
:=

∂2

∂x2
+

∂2

∂y2
.

Every harmonic function is biharmonic but not necessarily the converse. More-
over, it is easy to see that every biharmonic mapping F in Ω has the representation

(1.1) F = |z|2G + K,
where G and K are harmonic in Ω and they can be expressed as,

G = g1 + g2,(1.2)

K = k1 + k2,

2010 Mathematics Subject Classification. 30C35, 30C45, 35Q30.

Key words and phrases. Biharmonic mappings; univalent; linearly connected domains.

c©2015 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

1



2 ABDULHADI AND HAJJ

where g1, g2, k1 and k2 are analytic in Ω (for details see [2]). Note that the
composition F ◦φ of a harmonic function F with analytic function φ is harmonic,
while this is not true when F is biharmonic.

Biharmonic mappings arise in a lot of physical situations, particularly in fluid
dynamics and elasticity problems, and have many important applications in engi-
neering and biology. Most important applications of the theory of functions of a
complex variable were obtained in the plane theory of elasticity and in the approxi-
mate theory of plates subject to normal loading. That is, in cases when the solutions
are biharmonic functions or functions associated with them ( see [15, 16 ]). More-
over, biharmonic Mapping are closely related to the theory of Laguerre Minimal
Surfaces (for details see [5, 7, 8, 9, 17, 18]). Investigation of biharmonic mappings
in the context of geometric function theory started only recently (for details see[ 1,
2, 3, 4, 10 ]). For example, in [2], Abdulhadi, AbuMuhanna and Khuri analyze the
univalence of the solutions of the biharmonic equations. Throughout we consider
harmonic and biharmonic functions defined on the unit disk D = {z : |z| < 1}.

Definition 1. A domain Ω ⊂ C is linearly connected if there exists a constant
M < ∞ such that any two points w1,w2 ∈ Ω are joined by a path γ, γ ⊂ Ω, of
length `(γ) ≤ M|w1 −w2|.

Such a domain is necessarily a Jordan domain, and for piecewise smoothly bound-
ed domains, linear connectivity is equivalent to the boundary having no inward-
pointing cusps.

In [12], Chuaqui and Hermandez, considered the relationship between the har-
monic mapping f = h + g and its analytic factor h on linearly connected domains.
They show that if h is an analytic univalent function, then every harmonic map-
ping f = h + g with dilatation |ω| < c is univalent if and only if h(D) is linearly
connected.

In this paper, we scrutinize the relationship between the univalence of the bi-
harmonic function F = |z|2G + K and the univalence of the harmonic function
K.

2. Main results

In our first results, we deduce the univalence of F(z) from the univalence of K(z).
We first consider subclasses, where G,K are assumed to be analytic or antianalytic.

Theorem 1. Let F(z) = |z|2G(z) +K(z) be a biharmonic function in the unit disk
D, where G,K are analytic. If K is univalent and K(D) is a linearly connected
domain with constant M, and if

2|G| + |G′|
|K′|

<
1

M
,

then F(z) is univalent.

Proof. Let H(z) = |z|2G(z). We define ϕ = H ◦ K−1. Given w � K(D), we claim
w+ϕ(w) is univalent. Assume w+ϕ(w) is not univalent, then there exists w1 6= w2
such that

ϕ(w2) −ϕ(w1) = w1 −w2.
Let γ be a path in K(D) joining w1,w2 such that l(γ) ≤ M|w2 −w1|.



BIHARMONIC MAPPINGS 3

Then

|ϕ(w2) −ϕ(w1)| ≤
∣∣∣∣
∫
γ

ϕwdw + ϕwdw

∣∣∣∣ .
But

ϕw = Hz(K
−1)w + Hz(K−1)w =

Hz
K′

=
zG + |z|2G′

K′
,

ϕw = Hz(K
−1)w + Hz(K−1)w =

Hz

K
′ =

zG

K
′ ,

where z = K−1(w) ∈ D.
Therefore,

|ϕ(w2) −ϕ(w1)| ≤
∫
γ

sup
D

|2G| + |G′|
|K′|

|dw| <
1

M
l(γ) < |w2 −w1|

which is a contradiction. Hence F(z) is univalent. �

Remark 1. In the above proof , if K(D) is convex we may take M = 1, and thus
F will be univalent as long as

2|G|+|G′|
|K′| < 1.

The special case M = 1 when K is convex, is an important special case and we
will state it separately as a corollary.

Corollary 1. Let F(z) = |z|2G(z) + K(z) be a biharmonic function in the unit
disk D, where G,K are analytic. If K is univalent and convex with

2|G| + |G′|
|K′|

< 1,

then F(z) is univalent.

As a consequence of Theorem 1, we have the following corollary :

Corollary 2. Let F(z) = |z|2G(z) +K(z) be a biharmonic function in the unit disk
D, where G,K are antianalytic. If K is univalent and K(D) is a linearly connected
domain with constant M, and

2|G| + |Gz|
|Kz|

<
1

M
,

then F(z) is univalent.

Our next result is the general case, where G ,K are harmonic in the unit disk
D :

Theorem 2. Let F(z) = |z|2G(z) +K(z) be a biharmonic function in the unit disk
D, where G,K are harmonic. If K is univalent and K(D) is a linearly connected
domain with constant M, and if

2|G| + |g′1|(1 + |ωG|)
|k′1|(1 −|ωK|)

<
1

M
,

then F(z) is univalent. In the above ωK,ωG denotes the dilations ωK =
k′2
k′1

,

ωG =
g′2
g′1
.



4 ABDULHADI AND HAJJ

Proof. Let H(z) = |z|2G(z). We define
ϕ = H ◦K−1.

Given w � K(D), we claim w + ϕ(w) is univalent.
Assume w + ϕ(w) is not univalent, then there exists w1 6= w2 such that

ϕ(w2) −ϕ(w1) = w1 −w2.
Let γ be a path in K(D) joining w1,w2 such that l(γ) ≤ M|w2 −w1|.Then

|ϕ(w2) −ϕ(w1)| ≤
∣∣∣∣
∫
γ

ϕwdw + ϕwdw

∣∣∣∣ ≤
∫
γ

(|ϕw| + |ϕw|) |dw|.

But
ϕw = Hz(K

−1)w + Hz(K−1)w

ϕw = Hz(K
−1)w + Hz(K−1)w.

Differentiating K−1(K(z)) = z, we show that

(K−1)w =
k′1

|k′1|2 −|k′2|2
, (K−1)w =

−k′2
|k′1|2 −|k′2|2

.

It follows

ϕw = Hz(K
−1)w + Hz(K−1)w

= (zG + |z|2g′1)
k′1

|k′1|2 −|k′2|2
+ (zG + |z|2g′2)

−k′2
|k′1|2 −|k′2|2

,

ϕw = Hz(K
−1)w + Hz(K−1)w

= (zG + |z|2g′1)
−k′2

|k′1|2 −|k′2|2
+ (zG + |z|2g′2)

k′1
|k′1|2 −|k′2|2

.

Therefore,

|ϕw| + |ϕw| ≤
2|z||G|(|k′1| + |k′2|)
|k′1|2 −|k′2|2

+
|z|2(|g′1||k′2| + |g′2||k′1|)

|k′1|2 −|k′2|2

=
2|z||G| + |z|2 (|g′1| + |g′2|)

|k′1|− |k′2|

=
2|z||G| + |z|2|g′1|(1 + |ωG|)

|k′1|(1 −|ωK|)
,

where z = K−1(w) ∈ D. Then we have

|ϕ(w2) −ϕ(w1)| ≤
∫
γ

sup
D

2|G| + |g′1|(1 + |ωG|)
|k′1|(1 −|ωK|)

|dw| <
1

M
l(γ) < |w2 −w1|

which is a contradiction. Hence F(z) is univalent. �

Corollary 3. Let F(z) = |z|2G(z) + K(z) be a biharmonic function in the unit
disk D, where G,K are harmonic. If K is univalent and convex with

2|G| + |g′1|(1 + |ωG|)
|k′1|(1 −|ωK|)

< 1,

then F(z) is univalent.



BIHARMONIC MAPPINGS 5

The following Corollary follows immediately from Theorem 2, for the case when
G is analytic.

Corollary 4. Let F(z) = |z|2G(z) + K(z) be a biharmonic function in the unit
disk D, where G is analytic, K is harmonic. If K is univalent and K(D)is a
linearly connected domain with constant M, and if

2|G| + |G′|
|k′1|(1 −|ωK|)

<
1

M
,

then F(z) is univalent. In the above ωk denotes the dilation ωK =
k′2
k′1
.

In each of the previous results, it follows under the same conditions that F(D)
will also be linearly connected. We will prove it for the general case and the proof
is along the same lines for the special cases.

Proposition 1. Let F(z) = |z|2G(z) + K(z) be a biharmonic function in the unit
disk D, where G ,K are harmonic. If K is univalent and K(D) is a linearly
connected domain with constant M, and if

2|G| + |g′1|(1 + |ωg|)
|k′1|(1 −|ωk|)

≤ C,

where C < 1
M
, then F(D) is linearly connected.

Proof. Given w ∈ Ω = K(D), we let Ψ(w) = w + ϕ(w), where ϕ = H ◦ K−1,
and H = |z|2G. Since K is univalent, we may look at R = F(D), as the image of
Ω = K(D) under the mapping Ψ, and we show Ψ(Ω) is linearly connected. Let
ς1 = Ψ(w1), ς2 = Ψ(w2), w1,w2 ∈ Ω. Since K(D) is a linearly connected domain,
then there exists a curve γ ⊂ Ω satisfying l(γ) ≤ M|w2 −w1|.

Let Γ = Ψ(γ). In the proof of Theorem 2, we have showed that

|ϕw| + |ϕw| ≤
2|G| + |g′1|(1 + |ωG|)
|k′1|(1 −|ωK|)

< C,

it follows

|Ψw| + |Ψw| ≤ 1 + |ϕw| + |ϕw| < 1 + C.
Hence we have,

l(Γ) =

∫
Γ

dς ≤
∫
γ

(|Ψw| + |Ψw|) dw < (1 + C)l(γ) ≤ (1 + C)M|w2 −w1|.

But,

|ς1 − ς2| = |w1 −w2 + ϕ(w1) −ϕ(w2)| ≥ |w1 −w2|− |ϕ(w1) −ϕ(w2)|

≥ |w1 −w2|−
∫
γ

(|ϕw| + |ϕw|) dw

> |w1 −w2|−Cl(γ) ≥ (1 −CM)|w1 −w2|.
Then we get,

l(Γ) ≤
(1 + C)M

1 −CM
|ς1 − ς2|

and so R is linearly connected with constant
(1+C)M
1−CM . �

In our next result, we deduce the univalence of K from the univalence of F.



6 ABDULHADI AND HAJJ

Theorem 3. Let F(z) = |z|2G(z) +K(z) be a biharmonic function in the unit disk
D. Suppose F is univalent and F(D) is a linearly connected domain with constant
M and satisfies

2|G| + |Gz| + |Gz|
||Fz|− |Fz||

<
1

M
,

then K(z) is univalent.

Proof. Let H(z) = |z|2G(z). Aiming for a contradiction assume K(z) is not univa-
lent, then there exists z1 6= z2 , such that K(z1) = K(z2). Hence we get

F(z1) −F(z2) = H(z1) −H(z2).
Given w = F(z)� F(D), the above equation is equivalent to

w1 −w2 = ϕ(w2) −ϕ(w1),
where

ϕ = H ◦F−1.
Let γ be a path in F(D) joining w1,w2 such that l(γ) ≤ M|w2 −w1|.
Then

|ϕ(w2) −ϕ(w1)| ≤
∣∣∣∣
∫
γ

ϕwdw + ϕwdw

∣∣∣∣ ≤
∫
γ

(|ϕw| + |ϕw|) |dw|.

But
ϕw = Hz(F

−1)w + Hz(F−1)w

ϕw = Hz(F
−1)w + Hz(F−1)w.

Differentiating F−1(F(z)) = z, we get the following two equations

(F−1)wFz + (F
−1)wFz = 1

(F−1)wFz + (F
−1)wFz = 0.

Solving the above system we get

(F−1)w =
Fz
JF

, (K−1)w =
−Fz
JF

,

where JF denotes the jacobian Jf = |Fz|2 −|Fz|2.
It follows,

ϕw = Hz(F
−1)w + Hz(F−1)w = Hz

Fz
JF
−Hz

Fz
JF

ϕw = Hz(F
−1)w + Hz(F−1)w = −Hz

Fz
JF

+ Hz
Fz
JF

.

Therefore,

|ϕw| + |ϕw| ≤
|Hz||Fz| + |Hz||Fz| + |Hz||Fz| + |Hz||Fz|

|JF |

=
(|Hz| + |Hz|)(|Fz| + |Fz|)

|JF |

=
|Hz| + |Hz|
||Fz|− |Fz||



BIHARMONIC MAPPINGS 7

≤
2|G| + |Gz| + |Gz|
||Fz|− |Fz||

.

Hence

|ϕ(w2) −ϕ(w1)| ≤
∫
γ

sup
D

2|G| + |Gz| + |Gz|
||Fz|− |Fz||

|dw| <
1

M
l(γ) < |w2 −w1|,

which is a contradiction. Thus K(z) is univalent. �

Corollary 5. Let F(z) = |z|2G(z) + K(z) be a biharmonic function in the unit
disk D, where G ,K are harmonic. If K is univalent and convex with

2|G| + |g′1|(1 + |ωG|)
|k′1|(1 −|ωK|)

< 1,

then K(z) is univalent.

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8 ABDULHADI AND HAJJ

1Department of Mathematics and Statistics, AUS, P.O.Box 26666, Sharjah, UAE

2Mathematics Division , AUD, P.O.Box 28282 , Dubai, UAE

∗Corresponding author