Landau’s theorem for certain biharmonic mappings

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Abstract

In this paper, we show the existence of Landau and Bloch constants for biharmonic mappings of the form L(F). Here L represents the linear complex operator L=zz-z¯z¯ defined on the class of complex-valued C1 functions in the plane, and F belongs to the class of biharmonic mappings of the form F(z)=|z|2G(z)+K(z)(|z|<1), where G and K are harmonic.

Introduction

A four times continuously differentiable complex-valued function F=u+iv in a domain DC is biharmonic if and only if the Laplacian of F is harmonic. Note that ΔF is harmonic in D if F satisfies the biharmonic equation Δ(ΔF)=0, where Δ represents the Laplacian operatorΔ=42zz¯.

Biharmonic functions arise in a lot of physical situations, particularly in fluid dynamics and elasticity problems, and have many important applications in engineering and biology (see [10], [12], [13] for the details).

It has been shown that a mapping F is biharmonic in a simply connected domain D if and only if F has the following representation:F=|z|2G+K,where G and K are complex-valued harmonic functions in D (see [2], [3]). Also, it is known that (see [5], [7]) G and K can be expressed asG=g1+g2¯andK=k1+k2¯,where g1, g2, k1 and k2 are analytic in D. The Jacobian Jf of f is defined byJf=|fz|2-|fz¯|2.

We now introduceλf=||fz|-|fz¯||andΛf=|fz|+|fz¯|.

Then Jf=λfΛf if Jf0.

In [3], the authors considered the following differential operator L defined on the class of complex-valued C1 functions:L=zz-z¯z¯.

Clearly, L is a complex linear operator and satisfies the usual product rule:L(af+bg)=aL(f)+bL(g)andL(fg)=fL(g)+gL(f),where a,b are complex constants, f and g are C1 functions. In addition, the operator L possesses a number of interesting properties. For instance. it is easy to see that the operator L preserves both harmonicity and biharmonicity. Many other basic properties are stated in [3].

Throughout the paper, we use the notation Dr={zC:|z|<r}, D=D1, and we now recall one of the main results from [3].

Theorem A [3, Corollary 1(3)]

Let F be a univalent biharmonic function in D. If F is convex and L(F) is univalent, then L(F) is starlike.

At this place, it is also important to recall that (see [3, Corollary 1(3)]) the operator L(F) for biharmonic functions behaves much like zf(z) for analytic functions, for example in the sense that for F univalent and biharmonic, F is starlike if and only if Re(L(F)(z)/F(z))0. A similar characterization has also been obtained by the authors for convex (biharmonic) functions.

The classical theorem of Landau shows that there exists a ρ=ρ(M)>0 such that every function f, analytic in the unit disk |z|<1 with f(0)=f(0)-1=0 and |f(z)|<M in |z|<1, is univalent in the disk Dρ and in addition, the range f(Dρ) contains a disk of radius Mρ2. Recently, many authors considered Landau’s theorem for planar harmonic mappings (see, for example, [4], [6], [8], [14]) and biharmonic mappings (see [1]). From Theorem A, we see that it is significant to consider Landau’s theorem for biharmonic mappings of the form L(F), where F belongs to the class of biharmonic mappings. The main aim of this paper is to consider this problem. Our main results are the following.

Theorem 1.1

Let F=|z|2G+K be a biharmonic mapping in D such that F(0)=K(0)=0, JK(0)=1, where G and K are harmonic in D. Assume that both |G| and |K| are bounded by M. Then there is a constant ρ (0<ρ<1) such that L(F) is univalent in Dρ, where ρ satisfies the following equation:π4M-6Mρ2(1-ρ)2-4Mρ3(1-ρ)3-16Mπ2m1arctanρ-4Mρ(1-ρ)3=0,where m16.059 is the minimum value of the function2-x2+4πarctanxx(1-x2)for 0<x<1. The minimum is attained at x0.588. Moreover, the range L(F)(Dρ) contains a schlicht disk DR1, whereR1=ρπ4M-2Mρ2(1-ρ)2-16Mπ2m1arctanρ.

It is important to remark that the bounds of G and K are not preserved under the differential operator L. In our next result we deal with the case K=0. Then the result differs from Theorem 1.1, because for K=0, the Jacobian JK(0)=0. Therefore, we need to assume that JG(0)=1 instead.

Theorem 1.2

Let F=|z|2G be a biharmonic mapping in D such that G(0)=0, JG(0)=1 and |G(z)|<M, where G is harmonic in D. Then there is a constant ρ (0<ρ<1) such that L(F) is univalent in Dρ, where ρ satisfies the following equation:π4M-48Mπ2m1arctanρ-2Mρ(1-ρ)3=0,where m1 is defined as in Theorem 1.1. Moreover, L(F)(Dρ) contains a schlicht disk DR2 withR2=ρ3π4M-16Mπ2m1arctanρ.

Denote the ρ values in Theorem 1.1, Theorem 1.2 by ρ1 and ρ2, respectively. In the left half of Table 1, we indicate the precise values of these ρ1’s and the corresponding values of R1’s for various choices of M. These values are obtained using Mathematica.

Remark 1.3

Theorem 1.1, Theorem 1.2 are not sharp.

Also, it could be of some interest to see some numerical examples to compare the situation for biharmonic F obtained in [1] and for L(F) obtained in the present article. Thus, for a computational comparison with the recent results from [1], we recall their results in Theorem B, Theorem C to present the corresponding numerical values in Table 1, Table 2. The numerical values given in these two tables are self-explanatory.

Theorem B

Let F=|z|2G+K be a biharmonic mapping of the unit disk D, where G and K are harmonic in D such that F(0)=K(0)=0, JK(0)=1, and both |G| and |K| are bounded by M. Then there is a constant 0<ρ3<1 so that F is univalent for |z|<ρ3. In specific ρ3 satisfiesπ4M-2ρ3M-2Mρ32(1-ρ3)2+1(1-ρ3)2-1=0and F(Dρ3) contains a schlicht disk DR3, whereR3=πρ34M-2M(ρ33+ρ32)1-ρ3.This result is not sharp.

Theorem C

Let G be harmonic in D such that G(0)=0, JG(0)=1 and |G(z)|<M. Then there is a constant 0<ρ4<1 so that F=|z|2G is univalent in the disk |z|<ρ4. ρ4 is the solution of the equationπ4M-4ρ4M1-ρ4-2M1(1-ρ4)2-1=0and F(Dρ4) contains a disk DR4 withR4=πρ434M-2Mρ441-ρ4.The result is not sharp.

In Section 2, some elementary results will be recalled. The proofs of the main theorems will be presented in Sections 3 Proof for the case, 4 Proof for the case.

Section snippets

Some lemmas

In [4], [14], the following version of Schwarz lemma was obtained.

Lemma D

Let f be a harmonic mapping of the unit disk D such that f(0)=0 and f(D)D. ThenΛf(z)4π(1+|f(z)|)(1-|z|2)forzDand|f(z)|4πarctan|z|4π|z|forzD.

The inequality (2.1) is due to Heinz [11, Lemma]. By Lemma D and the Heinz Lemma, the following is obvious.

Corollary E

Let f be a harmonic mapping of the unit disk D such that f(0)=0 and f(D)=D. Thenc|fz(0)|+|fz¯(0)|4πfor some absolute constant c>0.

In [9], Hall proved that the sharp value c in

Proof for the case F=|z|2G+K

We begin with the following three preliminary results. As an immediate consequence of Lemma F, we have

Corollary 3.1

Let f be a harmonic function in D which is n times continuously differentiable and f(z)=g(z)+h(z)¯. If h(z)=n=1anzn, g(z)=n=1bnzn and |f(z)|M, then for any n1,nfznn!M(1-|z|)n+1andnfz¯nn!M(1-|z|)n+1,zD.

The conclusion of the following lemmas are obvious.

Lemma 3.2

Let M and q be positive real constants. Then the functionf(x)=6Mx2(1-x)2+4Mx3(1-x)3+16Mπ2qarctanx+4Mx(1-x)3is continuous and

Proof of Theorem 1.2

Assume that G is harmonic in D. Then, we may set G(z)=g1(z)+g2(z)¯, whereg1(z)=n=1anzn,g2(z)=n=1bnzn,zDand JG(0)=|a1|2-|b1|2=1. By the hypothesis on G and Lemma D, we obtain thatΛG(0)4Mπ.

Now, we let H(z)=L(F)=|z|2L(G). ThenHz=2|z|2Gz-z¯2Gz¯+z|z|2GzzandHz¯=-2|z|2Gz¯+z2Gz-z¯|z|2Gz¯z¯.

Fix ρ with 0<ρπ364mM2r0, where m is the same as that in the proof of Theorem 1.1. Choose two distinct points z1 and z2 in the disk Dρ, z:=γ(t)=(z2-z1)t+z1 (0t1) and z2-z1=|z1-z2|eiθ.

As before, we use Lemma D

Acknowledgement

The authors would like to thank the referee for the careful reading of this paper and the useful suggestions.

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The research was partly supported by NSFs of China (No. 10771059) and of Hunan Province (No. 05JJ10001), and NCET (No. 04-0783).

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