Abstract
Let \(w=P[F]\) be a harmonic mapping of the unit disk \({\mathbb {D}}\) with the boundary function \(F\). By using Poisson formula, we obtain some better estimates on Bloch constants for planar harmonic mappings.
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1 Introduction
Let \(H({\mathbb {D}})\) denote the class of holomorphic functions in the unit disk \({\mathbb {D}}=\{z:\, |z|<1\}\). Given a function \(f\in H({\mathbb {D}})\) define \(B_f\) to be the least upper bound of all numbers \(b>0\) such that there exists a number \(z_0\in {\mathbb {C}}\) and a region \(\Omega \subseteq {\mathbb {D}}\) which is univalently mapped onto \(\{z\in {\mathbb {C}}: |z-z_0|<b\}\) by \(f\). The Bloch’s constant \(B\) is defined as (cf. [2])
The best known bounds for \(B\) at present are (cf. [1, 3])
where \(\Gamma \) is the Gamma function. However, the exact value of \(B\) is still unknown.
Let \(H_M({\mathbb {D}})\) denote the class of functions \(f\in H({\mathbb {D}})\) with \(|f(z)|<M\) for \(z\in {\mathbb {D}}\). The classical Landau theorem states that if \(f\in H_M({\mathbb {D}})\) with \(f(0)=f'(0)-1=0\), then \(f\) is univalent in the disk \(|z|<\rho _0\) with \(\rho _0=\frac{1}{M+\sqrt{M^2-1}}\) and \(f(|z|<\rho _0)\) contains a disk \(|w|<\sigma _0\) with \(\sigma _0=M\rho _0^2\). This result is sharp (cf. [8]).
Define complex derivatives of \(w(z)\) as follows:
where \(z=x+iy\). A complex-valued function \(w\) on \({\mathbb {D}}\) is harmonic if it is twice continuously differentiable and satisfies Laplace’s equation:
We refer to the book of Duren [5, 6] for good setting of harmonic mappings. Let
and
It is well known that \(w(z)\) is locally univalent and sense-preserving in \({\mathbb {D}}\) if and only if its Jacobian satisfies the following condition.
We know from Poisson formula that every bounded harmonic mapping \(w\) defined in \({\mathbb {D}}\) has the representation
where \(F\) is the boundary function defined on the unit circle \({\mathbb {T}}:=\{z:\, |z|=1\}\) and
denote the Poisson kernel. In what follows we write \(F(t)\) instead of \(F(e^{it})\) for the boundary function (cf. [9, 10]).
For harmonic mappings in \({\mathbb {D}}\), under suitable restriction we can obtain its Bloch and Landau theorems. For \(r>0\), let \({\mathbb {D}}_r=\{z\in {\mathbb {C}}:\ |z|<r\}\). In [4], Chen et al. proved the Landau theorem for harmonic mappings as follows.
Theorem A
Let \(w\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|<M\) for \(z\in {\mathbb {D}}\). Then \(w\) is univalent in the disk \({\mathbb {D}}_{r_1}\), where
and \(w({\mathbb {D}}_{r_1})\) contains a schlicht disk \({\mathbb {D}}_{R_1}\), where
Here \(m\approx 6.85\) is the minimum of the function \((3-r^2)/(r(1-r^2))\) for \(0<r<1\).
By using sharp coefficients estimate, the authors in [7] improved a version of Landau theorem for the class of bounded harmonic mappings.
Theorem B
Let \(w\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|<M\) for \(z\in {\mathbb {D}}\). Then \(w\) is close-to-convex (univalent) in the disk \({\mathbb {D}}_{r_2}\), where
and \(w({\mathbb {D}}_{r_2})\) contains a schlicht disk \({\mathbb {D}}_{R_2}\), where
By using Poisson formula, we obtain better estimates on Bloch constants for planar harmonic mappings.
2 Auxiliary Results
Lemma 2.1
Let \(z=\rho e^{i\theta }\in {\mathbb {D}}\). Then
where \(0\le \rho <1\) and \(0\le \theta \le 2\pi \).
Proof
Let \(\zeta =e^{it}\in {\mathbb {T}}\). According to the residue theorem, we see that
This completes the proof. \(\square \)
Lemma 2.2
Let \(0\le \alpha \le 2\pi \) and \(w=P[F]\) be a harmonic mapping in \({\mathbb {D}}\) with the boundary function \(F\). Then
where \(z=re^{i\theta }\in {\mathbb {D}}\).
Proof
For each \(z=re^{i\theta }\in {\mathbb {D}}\), it follows from (1.2) that
Then \(w_z(z)=\frac{1}{2\pi }\int \limits _0^{2\pi } \frac{e^{it}F(t)}{(e^{it}-z)^2}dt\) and \(w_{\bar{z}}(z)=\frac{1}{2\pi }\int \limits _0^{2\pi } \frac{e^{-it}F(t)}{(e^{-it}-\bar{z})^2}dt\). Let \(\alpha \in [0, 2\pi ]\) be an arbitrary constant. Then
Similarly, we can obtain
Equality (2.2) holds directly from the above two equalities. \(\square \)
3 Main Results
Theorem 3.1
Let \(w=P[F]\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|< M\) for \(z\in {\mathbb {D}}\), where \(M>1\) and \(F\) is the boundary function. Then \(w\) is univalent in the disk \({\mathbb {D}}_{r_0}\), where
is the root of the equation \(1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}=0\) and \(w({\mathbb {D}}_{r_0})\) contains a schlicht disk \({\mathbb {D}}_{R_0}\), where
Proof
For \(0<r<1\), take \(z_1\), \(z_2\in {\mathbb {D}}_r\). Let \(\ell : z(x)=z_1+(z_2-z_1)x=\rho (x)e^{i\theta (x)}\) be the segment line of \(z_1\) and \(z_2\), where \(0\le x\le 1\). Then
where \(\alpha =\arg (z_1-z_2)\in [0, 2\pi ]\). Since \(0\le \rho (x)<r<1\) and the function \(\frac{s\sqrt{2-s^2}}{1-s^2}\) is an increasing function for \(0\le s<1\), we see that
Applying \(w_{\bar{z}}(0)=w_z(0)-1=0\) together with (2.2) and Lemma 2.1 we see that
Hence,
Let \(1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}=0\). Then \(r_0=\sqrt{1-\frac{2M}{\sqrt{1+4M^2}}}\). This shows that \(w(z)\) is univalent in the disk \({\mathbb {D}}_{r_0}\). Let \(z=r_0e^{i\varphi }\in \partial {\mathbb {D}}_{r_0}\). Then according to (2.2) and Lemma 2.1, we have
We see that \(w({\mathbb {D}}_{r_0})\) contains the schlicht disk \({\mathbb {D}}_{R_0}\).
The proof is completed. \(\square \)
Remark 3.2
For \(1<M\le 5.07\), we see that \(r_2<r_0\). Let
Then \(w(z)\) is univalent in the disk \(D_{r_0^*}\), where \(r_0^{*}\ge r_2>r_1\). This shows that our Theorem 3.1 has improved the former results.
Theorem 3.3
Let \(w=P[F]\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|<M\) for \(z\in {\mathbb {D}}\), where \(M>1\) and \(F\) is the boundary function. Then
and
Proof
Let \(z=re^{i\theta }\in {\mathbb {D}}.\) According to (2.2) and Lemma 2.1, we have
This shows that \(\Lambda _w\le 1+\frac{2Mr\sqrt{2-r^2}}{1-r^2}\). Similarly,
This implies that
It follows from (3.3) that \(1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}>0\) holds for \(0\le r<r_0\).
The proof is completed. \(\square \)
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The author expresses his appreciation to the reviewer’s helpful advices.
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Communicated by Ronen Peretz.
The author of this work was supported by the National Natural Science Foundation of China under Grant 11471128 and the Natural Science Foundation of Fujian Province of China (No. 2014J01013).
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Zhu, JF. Landau Theorem For Planar Harmonic Mappings. Complex Anal. Oper. Theory 9, 1819–1826 (2015). https://doi.org/10.1007/s11785-015-0449-8
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DOI: https://doi.org/10.1007/s11785-015-0449-8