Abstract
In this paper, we establish three sharp versions of the Landau-type theorems for bounded biharmonic mappings \(F(z)=|z|^2G(z)+H(z)\), where G(z) and H(z) are harmonic in the unit disk U with \(G(0)=H(0)=0\) and \(\lambda _{F}(0)=||F_z(0)|-|F_{{\overline{z}}}(0)||=1\). Our results generalize (or improve) the corresponding results given in Liu et al. (Math Methods Appl Sci 40:2582–2595, 2017). Three conjectures for the sharp version of the Landau-type theorem for certain bounded biharmonic mappings are given in third section.
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1 Introduction
A function \(f(z)=u(z)+ i v(z), \, z=x+iy\) is a harmonic mapping on the unit disk \( U=\{z\in {\mathbb {C}}:|z|<1\} \) if and only if f is twice continuously differentiable and satisfies the Laplacian equation
for \(z\in U\), where we use the common notations for its formal derivatives:
A function \(f(z)=u(z)+ i v(z)\) is a biharmonic mapping on U if and only if f is four times continuously differentiable and satisfies the biharmonic equation \(\Delta (\Delta f)=0\) for \(z\in U\).
For such function f, let
and
Biharmonic mappings arise in many physical situations, particularly in fluid dynamics and elasticity problems, and have many important applications in engineering (see [1] and the references therein for more details).
It is known that a harmonic mapping is locally univalent if and only if its Jacobian \(J_f(z)=|f_z(z)|^2-|f_{{\overline{z}}}(z)|^2\not =0\) for \(z\in U \) (cf. [19]). Since U is simply connected, f(z) can be written as \(f=h+{\overline{g}}\) with \(f(0)=h(0)\), where g and h are analytic on U (for details see [11]). Thus,
It is well-known (cf. [1]) that a mapping F(z) is biharmonic in a simply connected domain D if and only if F(z) has the following representation:
where G(z) and H(z) are complex-valued harmonic functions in D.
The classical Landau’s theorem states that if f is an analytic function on the unit disk U with \(f(0)=f^{\prime }(0)-1=0\) and \(|f(z)|<M\) for \(z\in U \), then f is univalent in the disk \( U _{r_0}=\{z\in {\mathbb {C}}:|z|<r_0\}\) with
and \(f( U _{r_0})\) contains a disk \(|w|<R_0\) with \(R_0=M r_0^2\). This result is sharp, with the extremal function \(f_0(z)=M z \frac{1-M z}{M-z}\). The Bloch theorem asserts the existence of a positive constant number b such that if f is an analytic function on the unit disk \(U =\{z\in {\mathbb {C}}:|z|<1\}\) with \(f^{\prime }(0)=1\), then f(U) contains a Schlicht disk of radius b, that is, a disk of radius b which is the univalent image of some region in U. The supremum of all such constants b is called the Bloch constant (see [3, 13]).
In 2000, under a suitable restriction, Chen et al. [3] first established the Bloch and Landau theorems for harmonic mappings. Their results were not sharp. Better estimates were given in [12] and later in [4, 7,8,9,10, 14,15,16, 21, 22, 24, 27, 28]. In 2008, Abdulhadi and Muhanna established two versions of Landau-type theorems of certain bounded biharmonic mappings in [2]. From that on, many authors also considered the Landau-type theorems for certain bounded biharmonic mappings (see [6, 7, 20, 23, 25, 29]) and logharmonic mappings (see [26]). However, few sharp results were found. In 2017, Liu et al. established the following Landau-type theorems for certain biharmonic mappings.
Theorem A
([25]). Let \(F(z)=|z|^2g(z)+h(z)\) be a biharmonic mapping of the unit disk U with \(g(z),\, h(z)\) are harmonic mappings in U, and \(g(0)=h(0)=0, \lambda _{F}(0)=\lambda _{g}(0)=\Lambda _{g}(0)=1\), \(\Lambda _{g}(z)\le \Lambda _{1}\) and \(\Lambda _{h}(z)\le \Lambda _{2}\) for \(z\in U\). Then \(\Lambda _{1}\ge 1\), \(\Lambda _{2}\ge 1\), and F is univalent in the disk \(U_{r_1}\), where \(r_1\) is the minimum positive root in (0, 1) of the equation:
and \(F(U_{r_1})\) contains a Schlicht disk \(U_{R_1}\), with
When \(\Lambda _{1}=\Lambda _{2}=1\), \(r_1=\frac{\sqrt{3}}{3}\) and \(R_1=\frac{2\sqrt{3}}{9}\) are sharp.
Theorem B
([25]). Let \(F(z)=|z|^2g(z)+h(z)\) be a biharmonic mapping of the unit disk U with \(g(z),\, h(z)\) are harmonic mappings in U, and \(g(0)=h(0)=0, \lambda _{F}(0)=\lambda _{g}(0)=\Lambda _{g}(0)=1\), \(\Lambda _{g}(z)\le \Lambda \) and \(|h(z)|\le M\) for \(z\in U\). Then \(\Lambda \ge 1\), \(M\ge 1\), and F is univalent in the disk \(U_{r_2}\), where \(K(M)=\min \{\frac{4M}{\pi },\sqrt{2M^2-2}\}\), and \(r_2\) is the minimum positive root in (0, 1) of the equation:
and \(F(U_{r_2})\) contains a Schlicht disk \(U_{R_2}\) with
When \(\Lambda =M=1\), \(r_2=\frac{\sqrt{3}}{3}\) and \(R_2=\frac{2\sqrt{3}}{9}\) are sharp.
However, Theorems A and B are not sharp if \(\Lambda>1\, (\Lambda _1>1)\) or \(M>1\, (\Lambda _2>1)\), and they both have the strong hypothesis \(\lambda _{g}(0)=\Lambda _{g}(0)=1\). In this paper, by extending the method and technique in [24], we will establish several sharp versions of the Landau-type theorems for bounded biharmonic mappings.
This paper is organized as follows. In Sect. 2, we should recall several notions and lemmas, and establish four new lemmas, which play a key role in the proofs of our main results. In Sect. 3, by establishing Theorems 3.1 and 3.3, we first establish the sharp versions of Landau-type theorems for Theorem A without the hypothesis \(\lambda _{g}(0)=\Lambda _{g}(0)=1\). Next, by establishing Theorem 3.4, we establish the sharp version of Landau-type theorem for the case \(\Lambda \ge 1, M=1\) of Theorem B without the hypothesis \(\lambda _{g}(0)=\Lambda _{g}(0)=1\). Then, by establishing Theorem 3.7 and Corollary 3.8, we provide two sharp versions of Landau-type theorems of biharmonic mappings for the case \(\Lambda \ge 1, M>1\). Finally, we also provide three conjectures for the sharp versions of Landau-type theorems of bounded harmonic mappings or biharmonic mappings.
2 Preliminaries
In order to establish our main results, we need the following notions and lemmas.
We first introduce the notion of pseudo-disk [5, 17, 18]. For \(z\in U\) and \(0<r<1\), the pseudo-disk of pseudo-center z and pseudo-radius r is defined by
and \({\overline{U}}_p(z, r):=\left\{ \zeta \in U:\big |\frac{\zeta -z}{1-{\bar{z}}\zeta }\big |\le r\right\} \), \(\partial U_p(z, r):=\left\{ \zeta \in U:\big |\frac{\zeta -z}{1-{\bar{z}}\zeta }\big |=r\right\} \).
It is obvious that \(U_p(0,r)=U_r\), and that if \(z\ne 0\), it is easy to verify that \(U_p(z,r)\) is the Euclidian disk of diameter (a, b), where
Next, we recall the classical Schwarz–Pick Lemma as follows.
Lemma 2.1
(Schwarz–Pick Lemma). Suppose f(z) is an analytic function in U and \(f(U)\subset U\).
- (1)
For \(z\in U\) and \(0<r<1\), we have
$$\begin{aligned} f\left( U_p(z, r)\right) \subseteq U_p\left( f(z), r\right) ,\ \ f\left( {\overline{U}}_p(z, r)\right) \subseteq {\overline{U}}_p\left( f(z), r\right) ; \end{aligned}$$ - (2)
For \(z'\in \partial U_p(z, r), \ f(z')\in \partial U_p\left( f(z), r\right) \) if and only if f is a Möbius transformation of U onto itself.
- (3)
\(\left| f'(z)\right| /\left( 1-|f(z)|^2\right) \le 1/(1-|z|^2)\) holds for \(z\in U\), and the equality holds for some \(z\in U\) if and only if f is a Möbius transformation.
Now we establish two new lemmas, which play a key role in our proofs of the main results in this paper.
Lemma 2.2
Suppose \(\Lambda >1\). Let H(z) be a harmonic mapping of the unit disk U with \(\lambda _H(0)=1\) and \(\Lambda _H(z)<\Lambda \) for all \(z\in U\). Then for all \(z_1,z_2\in U_r\, (0<r<1\), \(z_1\ne z_2)\), we have
where \(\overline{z_1z_2}\) is the line segment joining \(z_1\) and \(z_2\).
Proof
Let \(\theta _0=\arg (z_2-z_1)\). Since H(z) is a harmonic mapping in the unit disk U, we have H(z) can be written as \(H(z)=H_1(z)+\overline{H_2(z)}\) for \(z\in U\), where \(H_1\) and \(H_2\) are analytic in U. Without lost of the generality, we may assume that \(|H_1'(0)|>|H_2'(0)|\), since \(\lambda _H(0)=||H_1'(0)|-|H_2'(0)||=1\) (if \(|H_1'(0)|<|H_2'(0)|\), we can consider the harmonic mapping \({\overline{H}}=\overline{H_1}+H_2\) instead of H). Then
where \(\Delta _{0\le \theta \le 2\pi }\) denotes the increment of the succeeding function as \(\theta \) increasing from 0 to \(2\pi \). Thus there exists a \(\theta _1\in [0, 2\pi ]\) such that
For \(z\in U\), let
Then \(\omega (z)\) is analytic with \(|\omega (z)|\le \Lambda _H(z)/\Lambda <1\) for \(z\in U\) and
Using Schwarz–Pick Lemma, we have
That is
Then
\(\square \)
Lemma 2.3
Suppose \(\Lambda >1\). Let H(z) be a harmonic mapping of the unit disk U with \(\lambda _H(0)=1\) and \(\Lambda _H(z)<\Lambda \) for all \(z\in U\). Set \(\gamma =H^{-1}(\overline{ow'})\) with \(w'\in H(\partial U_r)\ (0<r\le 1)\) and \(\overline{ow'}\) denotes the closed line segment joining the origin and \(w'\), then
Proof
Let \(d\zeta =|d\zeta |e^{i\theta _\zeta }, \ \zeta \in \gamma \). Since H(z) is a harmonic mapping in the unit disk U, we see that H(z) can be written as \(H(z)=H_1(z)+\overline{H_2(z)}\) for \(z\in U\), where \(H_1\) and \(H_2\) are analytic in U. Similar to the proof of Lemma 2.2, we may assume that \(|H_1'(0)|>|H_2'(0)|\) since \(\lambda _H(0)=||H_1'(0)|-|H_2'(0)||=1\). Then for a fixed \(\zeta \in \gamma \), there is a \(\theta _\zeta '\in [0, 2\pi ]\) such that
For \(z\in U\), define
Then \(\omega _\zeta (z)\) is analytic in U, with \(|\omega _\zeta (z)|\le \Lambda _H(z)/\Lambda <1\) and
According to Schwarz–Pick Lemma, we get that
Thus
\(\square \)
Lemma 2.4
([20]). Suppose that \(f(z)=g(z)+\overline{h(z)}\) is a harmonic mapping in U with \(g(z)=\sum _{n=1}^\infty a_{n}z^{n}\) and \(h(z)=\sum _{n=1}^\infty b_{n}z^{n}\) are analytic in U, and \(\lambda _{f}(0)=1\). If \(\Lambda _{f}(z)\le \Lambda \) for \(z\in U\), then \(\Lambda \ge 1\) and
When \(\Lambda >1\), the above estimates are sharp for all \(n=2,3,\ldots \), with the extremal functions \(f_{n}(z)\) and \(\overline{f_{n}(z)}\), where
When \(\Lambda =1\), then \(f(z)=a_{1}z+\overline{b_{1}}{\overline{z}}\) with \(||a_{1}|-|b_{1}||=1\).
Lemma 2.5
Suppose \(\Lambda \ge 0\). Let \(F(z)=a\Lambda |z|^2 z +b{\overline{z}}\) be a biharmonic mapping of the unit disk U with \(|a|=|b|=1\). Then F is univalent in the disk \(U_{\rho _1}\), and \(F(U_{\rho _1})\) contains a Schlicht disk \(U_{\sigma _1}\), where \(\rho _1=1\) when \(0\le \Lambda \le \frac{1}{3}\), \(\rho _1=\frac{1}{\sqrt{3\Lambda }}\) when \(\Lambda >\frac{1}{3}\), and
This result is sharp.
Proof
We first prove the case of \(0\le \Lambda \le \frac{1}{3}\).
To this end, for every \(z_1, z_2\in U\) with \(z_1\ne z_2\), because \(|a|=|b|=1\), we have
This implies \(F(z_{1})\ne F(z_{2})\), which proves the univalence of F(z) in the disk U.
On the other hand, set \(\theta _0=\arg \frac{b}{a}\), for each \(z'\in \partial U\) and \(z_0'=e^{i\frac{\pi +\theta _0}{2}}\in \partial U\), because \(F(0)=0\), we have
and
Hence F(U) contains a Schlicht disk \(U_{1-\Lambda }\), and the radius \(1-\Lambda \) is sharp.
Next, we prove the case of \(\Lambda >\frac{1}{3}\).
To this end, for every \(z_1, z_2\in U_{\frac{1}{\sqrt{3\Lambda }}}\) with \(z_1\ne z_2\), because \(|a|=|b|=1\), we have
This implies \(F(z_{1})\ne F(z_{2})\), which proves the univalence of F in the disk \(U_{\frac{1}{\sqrt{3\Lambda }}}\).
Now we prove that F is not univalent in the disk \(U_r\) for each \(r\in (\frac{1}{\sqrt{3\Lambda }}, 1]\).
In fact, fixed \(r\in (\frac{1}{\sqrt{3\Lambda }}, 1]\), set \(\theta _1=(\pi +\arg \frac{b}{a})/2, \varepsilon =\min \bigg \{(r-\frac{1}{\sqrt{3\Lambda }})/2,\frac{1}{2\sqrt{3\Lambda }}\bigg \}>0\) and \(r_1=\frac{1}{\sqrt{3\Lambda }}+\varepsilon , r_2=\frac{1}{\sqrt{3\Lambda }}-\delta \) with
Direct computation yields \(r_1^3-\frac{1}{\Lambda }r_1=r_2^3-\frac{1}{\Lambda }r_2\). Thus there exist two points \(z_1=r_1e^{i\theta _1},\, z_2=r_2e^{i\theta _1}\) in \(U_r\) with \(z_1\ne z_2\) such that
which implies that F(z) is not univalent in the disk \(U_r\) for each \(r\in (\frac{1}{\sqrt{3\Lambda }}, 1]\). Hence, the univalent radius \(\frac{1}{\sqrt{3\Lambda }}\) is sharp.
On the other hand, for each \(z''\in \partial U_{\frac{1}{\sqrt{3\Lambda }}}\) and \(z_0''=\frac{1}{\sqrt{3\Lambda }}e^{i\frac{\pi +\theta _0}{2}}\in \partial U_{\frac{1}{\sqrt{3\Lambda }}}\), where \(\theta _0=\arg \frac{b}{a}\), because \(F(0)=0\), we have
and
Hence \(F(U_{\frac{1}{\sqrt{3\Lambda }}})\) contains a Schlicht disk \(U_{\frac{2}{3\sqrt{3\Lambda }}}\), and the radius \(\frac{2}{3\sqrt{3\Lambda }}\) is sharp. This completes the proof. \(\square \)
Lemma 2.6
([20]). Suppose that \(f(z)=h(z)+\overline{g(z)}\) is a harmonic mapping of the unit disk U with \(h(z)=\sum _{n=1}^{\infty }a_nz^n\) and \(g(z)=\sum _{n=1}^{\infty }b_nz^n\). If \(\lambda _f(0)=1\) and \(|f(z)|<M\) for all \(z\in U\), then \(M\ge 1\), and
Lemma 2.7
Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of the unit disk U with \(G(0)=H(0)=0\) and \(\Lambda _G(z)\le \Lambda \) for all \(z\in U\), where \(G(z)=G_1(z)+\overline{G_2(z)} =\sum _{n=1}^\infty a_nz^n+\sum _{n=1}^\infty \overline{b_n}{\bar{z}}^n\), \(H(z)=H_1(z)+\overline{H_2(z)} =\sum _{n=1}^\infty c_nz^n+\sum _{n=1}^\infty \overline{d_n}{\bar{z}}^n\) are harmonic mappings in U. Then for all \(z_1,z_2\in U_r(0<r<1)\) with \(z_1\ne z_2\), we have
Proof
According to Lemma 2.4, for any \(z_1,z_2\in U_r(0<r<1\), \(z_1\ne z_2)\), we have
and
This completes the proof of Lemma 2.7. \(\square \)
3 The Landau-Type Theorems of Biharmonic Mappings
We first prove the sharp version of the Landau-type theorem for biharmonic mappings F(z) under the assumptions \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G}(z)\le \Lambda _1\) and \(\Lambda _{H}(z)<\Lambda _2\) for all \(z\in U\), which is one of the main results in this paper.
Theorem 3.1
Suppose that \(\Lambda _1\ge 0\) and \(\Lambda _2>1\). Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of the unit disk U, where G(z) and H(z) are harmonic in U, satisfying \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1,~\Lambda _{G}(z)\le \Lambda _1\) and \(\Lambda _{H}(z)<\Lambda _2\) for all \(z\in U\). Then F(z) is univalent in the disk \(U_{\rho _0}\) and \(F(U_{\rho _0})\) contains a Schlicht disk \(U_{\sigma _0}\), where \(\rho _0\) is the unique root in \((0,\,1)\) of the equation
and
This result is sharp, with an extremal function given by
Proof
By the hypothesis of Theorem 3.1, we have
We first prove that F is univalent in the disk \(U_{\rho _0}\). Indeed, for all \(z_1,z_2\in U_r(0<r<\rho _0\), \(z_1\ne z_2)\), note that \(\lambda _F(0)=\lambda _H(0)=1\) and \(\Lambda _{H}(z)<\Lambda _2\) for all \(z\in U\), we obtain from Lemma 2.2 that
It is easy to verify that the function
is continuous and strictly decreasing on [0, 1], \(g_0(0)=1>0\), and
Therefore, by the mean value theorem, there is a unique real \(\rho _0\in (0, 1)\) such that \(g_0(\rho _0)=0\). We obtain that
This implies \(F(z_1)\ne F(z_2)\), which proves the univalence of F in the disk \(U_{\rho _0}\).
Next, we prove that \(F(U_{\rho _0}) \supseteq U_{\sigma _0}\).
Indeed, note that \(F(0)=0\), for \(z'\in \partial U_{\rho _0}\) with \(w'=F(z')\in F(\partial U_{\rho _0})\) and \(|w'|=\min \left\{ |w|:w\in F\left( \partial U_{\rho _0}\right) \right\} \). Let \(\gamma =F^{-1}\left( \overline{ow'}\right) \), by Lemma 2.3, we have
which implies that \(F(U_{\rho _0})\supseteq U_{\sigma _0}\).
Now, we prove the sharpness of \(\rho _0\) and \(\sigma _0\). To this end, we consider a biharmonic mapping \(F_0(z)\) which is given by (3.3). It is easy to verify that \(F_0(z)\) satisfies the hypothesis of Theorem 3.1, and thus, we have that \(F_0(z)\) is univalent in the disk \(U_{\rho _0}\), and \(F_0(U_{\rho _0}) \supseteq U_{\sigma _0}\).
To show that the univalent radius \(\rho _0\) is sharp, we need to prove that \(F_0(z)\) is not univalent in \(U_r\) for each \(r\in (\rho _0, 1]\). In fact, considering the real differentiable function
Because the continuous function
is strictly decreasing on [0, 1] and \(h_0'(\rho _0)=g_0(\rho _0)=0\), we see that \(h_0'(x)=0\) for \(x\in [0, 1]\) if and only if \(x=\rho _0\). So \(h_0(x)\) is strictly increasing on \([0, \rho _0)\) and strictly decreasing on \([\rho _0, 1]\). Since \(h_0(0)=0\), there is a unique real \(r_3\in (\rho _0, 1]\) such that \(h_0(r_3)=0\) if \(h_0(1)\le 0\), and
For every fixed \(r\in (\rho _0, 1]\), set \(x_1=\rho _0+\varepsilon \), where
by the mean value theorem, there is a unique \(\delta \in (0, \rho _0)\) such that \(x_2:=\rho _0-\delta \in (0, \rho _0)\) and \(h_0(x_1)=h_0(x_2)\).
Let \(z_1=x_1\) and \(z_2=x_2\). Then \(z_1,\ z_2\in U_r\) with \(z_1\ne z_2\). Directly computation leads to
Hence \(F_0\) is not univalent in the disk \(U_r\) for each \(r\in (\rho _0, 1]\), and the univalent radius \(\rho _0\) is sharp.
Finally, note that \(F_0(0)=0\) and picking up \(z'=\rho _0\in \partial U_{\rho _0}\), by (3.3), (3.5) and (3.6), we have
Hence, the covering radius \(\sigma _0\) is also sharp. \(\square \)
Remark 3.2
Corollary 1 in [24] is just a special case of Theorem 3.1 when \(\Lambda _1=0\).
For the harmonic mapping H(z) of the unit disk U with \(\lambda _{H}(0)=1\) and \(\Lambda _H(z)\le \Lambda _2\) for all \(z\in U\), it follows from Lemma 2.4 that \(\Lambda _2\ge 1\). Theorem 3.1 provides the sharp version of Landau-type theorem of biharmonic mappings for the case \(\Lambda _1\ge 0\) and \(\Lambda _2>1\). If \(\Lambda _1\ge 0\) and \(\Lambda _2=1\), then we prove the following sharp version of Landau-type theorem for biharmonic mappings using Lemmas 2.4 and 2.5.
Theorem 3.3
Suppose that \(\Lambda \ge 0\). Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of U, where G(z), H(z) are harmonic in U, satisfying \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G}(z)\le \Lambda \) and \(\Lambda _H(z)\le 1\) for all \(z\in U\). Then F is univalent in the disk \(U_{\rho _1}\) and \(F(U_{\rho _1})\) contains a Schlicht disk \(U_{\sigma _1}\), where \(\rho _1=1\) when \(0\le \Lambda \le \frac{1}{3}\), \(\rho _1=\frac{1}{\sqrt{3\Lambda }}\) when \(\Lambda >\frac{1}{3}\), and \(\sigma _1=\rho _1-\Lambda \rho _1^{3}\) is defined by (2.7). This result is sharp.
Proof
Because \(F(z)=|z|^2G(z)+H(z)\) satisfies the hypothesis of Theorem 3.3, where \(G(z)=G_{1}(z)+\overline{G_{2}(z)}\) and \(H(z)=H_{1}(z)+\overline{H_{2}(z)}\) with \(G_{1}(z)=\sum _{n=1}^{\infty }a_{n}z^{n}\), \(G_{2}(z)=\sum _{n=1}^{\infty }b_{n}z^{n}\) and \(H_{1}(z)=\sum _{n=1}^{\infty }c_{n}z^{n}\), \(H_{2}(z)=\sum _{n=1}^{\infty }d_{n}z^{n}\) are analytic on U. Then
Since \(\Lambda _H(z)\le 1\), using Lemma 2.4, we have
Hence \(H(z)=c_1 z+\overline{d_1 z}\) with \(||c_1|-|d_1||=1\).
Now we prove F is univalent in the disk \(U_{\rho _1}\). To this end, for any \(z_1,z_2\in U_r(0<r<\rho _1)\) with \(z_1\ne z_2\), by (3.4), we have
Then, we have \(F(z_1)\ne F(z_2)\), which proves the univalence of F in the disk \(U_{\rho _1}\).
Noticing that \(F(0)=0\), for any \(z=\rho _1e^{i\theta }\in \partial U_{\rho _1}\), we have
Hence, \(F(U_{\rho _1})\) contains a Schlicht disk \(U_{\sigma _{1}}\).
Finally, for \(F(z)=a_{1}\Lambda |z|^2z+\overline{d_1}{\bar{z}}\) with \(|a_{1}|=|d_{1}|=1\), we have \(G(z)=a_{1}\Lambda z, H(z)=\overline{d_1}{\bar{z}}\). Direct computation yields
and \(|\Lambda _H(z)|=|\overline{d_1}|\le 1\) for all \(z\in U\). Applying Lemma 2.5, we obtain that \(\rho _1, \sigma _1\) are sharp. This completes the proof. \(\square \)
Next, we prove the sharp version of Landau-type theorems for certain biharmonic mappings \(F(z)=|z|^2G(z)+H(z)\) with \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G}(z)\le \Lambda \) and \(|H(z)|< 1\) for all \(z\in U\), which is also one of the main results in this paper.
Theorem 3.4
Suppose that \(\Lambda \ge 0\). Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of U, where G(z), H(z) are harmonic in U, satisfying \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G}(z)\le \Lambda \) and \(|H(z)|< 1\) for all \(z\in U\). Then F is univalent in the disk \(U_{\rho _1}\) and \(F(U_{\rho _1})\) contains a Schlicht disk \(U_{\sigma _1}\), where \(\rho _1=1\) when \(0\le \Lambda \le \frac{1}{3}\), \(\rho _1=\frac{1}{\sqrt{3\Lambda }}\) when \(\Lambda >\frac{1}{3}\), and \(\sigma _1=\rho _1-\Lambda \rho _1^{3}\) is defined by (2.7). This result is sharp.
Proof
Because \(F(z)=|z|^2G(z)+H(z)\) satisfies the hypothesis of Theorem 3.4, where \(G(z)=g_{1}(z)+\overline{g_{2}(z)}\) and \(H(z)=h_{1}(z)+\overline{h_{2}(z)}\) with \(g_{1}(z)=\sum _{n=1}^{\infty }a_{n}z^{n}\), \(g_{2}(z)=\sum _{n=1}^{\infty }b_{n}z^{n}\) and \(h_{1}(z)=\sum _{n=1}^{\infty }c_{n}z^{n}\), \(h_{2}(z)=\sum _{n=1}^{\infty }d_{n}z^{n}\) are analytic on U. Then
Since \(|H(z)|<1\), using Lemma 2.6, we have
Hence \(H(z)=c_1 z+\overline{d_1 z}\) with \(||c_1|-|d_1||=1\).
Now we prove F is univalent in the disk \(U_{\rho _1}\), where
To this end, for any \(z_1,z_2\in U_r(0<r<\rho _1)\) with \(z_1\ne z_2\), by the means of Lemma 2.7, we have
Then, we have \(F(z_1)\ne F(z_2)\), which proves the univalence of F in the disk \(U_{\rho _1}\).
Noticing that \(F(0)=0\), for any \(z=\rho _0e^{i\theta }\in \partial U_{\rho _1}\), it follows from (2.8) that
Hence, \(F(U_{\rho _1})\) contains a Schlicht disk \(U_{\sigma _1}\).
Finally, for \(F_1(z)=a_{1}\Lambda |z|^2{\overline{z}}+\overline{d_1} z\) with \(|a_{1}|=|d_{1}|=1\), we have \(G_1(z)=a_{1}\Lambda {\overline{z}}, H_1(z)=\overline{d_1} z\). It is easy to verify that \(G_1(0)=H_1(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G_1}(z)=|a_{1}\Lambda |\le \Lambda \) and \(|H_1(z)|=|\overline{d_1}||z|< 1\) for all \(z\in U\). Applying Lemma 2.5, we obtain that both of \(\rho _1\) and \(\sigma _1\) are sharp. This completes the proof. \(\square \)
Remark 3.5
For the harmonic mapping H(z) in the unit disk U with \(\lambda _{H}(0)=1\) and \(|H(z)|< M\) for all \(z\in U\), it follows from Lemma 2.6 that \(M\ge 1\). Theorem 3.4 provides the sharp version of Landau-type theorem of biharmonic mappings for the case \(\Lambda \ge 0\) and \(M=1\). For the case of \(\Lambda \ge 0\) and \(M >1\), we first consider an example as follows.
Example 3.6
Suppose that \(M>1\) and \(\Lambda \ge 0\). Let \(F_2(z)=-\Lambda |z|^2 z+ M z\frac{1-M z}{M-z}\) be a biharmonic mapping of U. Then \(F_2(z)\) is univalent in the disk \(U_{\rho _2 }\), where \(\rho _2\) is the unique positive root in (0, 1) of the equation
and \(F_2(U_{\rho _2 })\) contains a Schlicht disk \(U_{\sigma _2 }\), with
Both of \(\rho _2 \) and \(\sigma _2 \) are sharp.
Proof
We first prove \(F_2(z)\) is univalent in the disk \(U_{\rho _2 }\). To this end, for any \(z_1,z_2\in U_r(0<r<\rho _2 )\) with \(z_1\ne z_2\), simple computation yields that
It is easy to verify that the function
is continuous and strictly decreasing on [0, 1], \(g_0(0)=1>0\), and
There, by the mean value theorem, there is a unique real \(\rho _2 \in (0, 1)\) such that \(g_1(\rho _2 )=0\). We obtain that
Then, we have \(F_2(z_1)\ne F_2(z_2)\), which proves the univalence of \(F_2(z)\) in the disk \(U_{\rho _2}\).
Next, we prove the sharpness of \(\rho _2\). To this end, we need to prove that \(F_2(z)\) is not univalent in the disk \(U_r\) for each \(r\in (\rho _2, 1]\).
In fact, consider the real differentiable function
Because the continuous function
is strictly decreasing on [0, 1] and \(h_1'(\rho _2 )=g_1(\rho _2 )=0\), we obtain that \(h_1'(x)=0\) for \(x\in [0, 1]\) if and only if \(x=\rho _2 \). So \(h_1(x)\) is strictly increasing on \([0, \rho _2 ]\) and strictly decreasing on \([\rho _2, 1]\). Since \(h_1(0)=0\), there is a unique real \(r_4\in (\rho _2, 1]\) such that \(h_1(r_4)=0\) if \(h_1(1)\le 0\), and
For every fixed \(r\in (\rho _1, 1]\), set \(x_1=\rho _1 +\varepsilon \), where
by the mean value theorem, there is a unique \(\delta \in (0, \rho _2 )\) such that \(x_2:=\rho _2 -\delta \in (0, \rho _2 )\) and \(h_1(x_1)=h_1(x_2)\).
Let \(z_1=x_1\) and \(z_2=x_2\), then \(z_1,\ z_2\in U_r\, (r\in (\rho _1 , 1])\) with \(z_1\ne z_2\). Directly computation leads to
which implies that \(F_2(z)\) is not univalent in the disk \(U_r\) for each \(r\in (\rho _2, 1]\). Hence, the univalent radius \(\rho _2 \) is sharp.
Finally, noticing that \(F_2(0)=0\), for any \(z=\rho _2 e^{i\theta }\in \partial U_{\rho _2 }\) and \(z'=\rho _2 \in \partial U_{\rho _2 }\), we have
and
Hence, \(F_2(U_{\rho _2 })\) contains a Schlicht disk \(U_{\sigma _2 }\), and the radius \(\sigma _2 \) is sharp. This completes the proof. \(\square \)
Next, applying Lemma 2.7 and Example 3.6, we may verify the following sharp form of the Landau-type theorem for certain biharmonic mapping in the unit disk U.
Theorem 3.7
Suppose that \(M>1\) and \(\Lambda \ge 0\). Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of U, where G(z) is harmonic in U, satisfying \(G(0)=\lambda _{F}(0)-1=0, \Lambda _{G}(z)\le \Lambda \) for all \(z\in U\), and \(H(z)=\sum _{n=1}^{\infty }c_{n}z^{n}+\overline{\sum _{n=1}^{\infty }d_{n}z^{n}}\) is harmonic in U, satisfying the following inequality
Then F is univalent in the disk \(U_{\rho _2 }\) and \(F(U_{\rho _2 })\) contains a Schlicht disk \(U_{\sigma _2 }\), where \(\rho _2 \) is the unique positive root in (0, 1) of Eq. (3.7) and \(\sigma _2 \) is given by (3.8). This result is sharp with \(F_2(z)=-\Lambda |z|^2 z+ M z\frac{1-M z}{M-z}\) being an extremal mapping.
Proof
Since \(\lambda _{F}(0)=1\), we have
We first prove F is univalent in the disk \(U_{\rho _2}\). To this end, note that \(\rho _2 \le r_0\), for any \(z_1,z_2\in U_r\, (0<r<\rho _2 )\) with \(z_1\ne z_2\), by all hypotheses of Theorem 3.7 and Lemma 2.7, we obtain that
Then, we have \(F(z_1)\ne F(z_2)\), which proves the univalence of F in the disk \(U_{\rho _1 }\).
Next, noticing that \(F(0)=0\), for any \(z=\rho _2 e^{i\theta }\in \partial U_{\rho _2 }\), it follows from (2.8) that
Hence, \(F(U_{\rho _2 })\) contains a Schlicht disk \(U_{\sigma _2 }\).
Finally, we prove the sharpness of \(\rho _2 \) and \(\sigma _2 \). We consider the biharmonic mapping \(F_2(z)=-\Lambda |z|^2 z+ M z\frac{1-M z}{M-z}\). Let \(G(z)=-\Lambda z\), \(H(z)= M z\frac{1-M z}{M-z}\) for \(z\in U\). Then G(z), H(z) are harmonic mappings in U, and \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G}(z)=\Lambda \le \Lambda \) for all \(z\in U\). Note that
we obtain that \(F_2(z)\) satisfies all hypotheses of Theorem 3.7. Applying Example 3.6, we obtain that both of \(\rho _2 \) and \(\sigma _2 \) are sharp. The proof of Theorem 3.7 is complete. \(\square \)
Corollary 3.8
Suppose that \(M>1\) and \(\Lambda \ge 0\). Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of U, where G(z) is harmonic in U, satisfying \(G(0)=\lambda _{F}(0)-1=0, \Lambda _{G}(z)\le \Lambda \) for all \(z\in U\), and \(H(z)=\sum _{n=1}^{\infty }c_{n}z^{n}+\overline{\sum _{n=1}^{\infty }d_{n}z^{n}}\) is harmonic in U, satisfying the following inequality
Then F is univalent in the disk \(U_{\rho _2 }\) and \(F(U_{\rho _2 })\) contains a Schlicht disk \(U_{\sigma _2}\), where \(\rho _2 \) is the unique positive root in (0, 1) of Eq. (3.7) and \(\sigma _2\) is given by (3.8). This result is sharp with \(F_2(z)=-\Lambda |z|^2 z+ M z\frac{1-M z}{M-z}\) being an extremal mapping.
Setting \(\Lambda =0\) in Theorem 3.7, we have the following corollary.
Corollary 3.9
Suppose that \(M>1\). Let H(z) be a harmonic mapping of U with \(\lambda _{H}(0)=1\), and \(H(z)=\sum _{n=1}^{\infty }c_{n}z^{n}+\overline{\sum _{n=1}^{\infty }d_{n}z^{n}}\) satisfying the inequality (3.11). Then H is univalent in the disk \(U_{r_0 }\) and \(H(U_{r_0 })\) contains a Schlicht disk \(U_{R_0}\), where \(R_0=M r_0^2 \) and \(r_0\) is given by (1.2). This result is sharp with \(H_0(z)=M z\frac{1-M z}{M-z}\) being an extremal mapping.
Finally, note that
and \(H_0(z)\) satisfies (3.11) and (3.12). It is natural to pose three conjectures as follows:
Conjecture 3.10
Suppose that \(M>1\) and \(\Lambda \ge 0\). Let \(F(z)=|z|^2G(z)+H(z)\) be a biharmonic mapping of the unit disk U, where G(z), H(z) are harmonic in U, satisfying \(G(0)=H(0)=0\), \(\lambda _{F}(0)=1, \Lambda _{G}(z)\le \Lambda \) and \(|H(z)|< M\) for all \(z\in U\). Then F is univalent in the disk \(U_{\rho _2 }\) and \(F(U_{\rho _2})\) contains a Schlicht disk \(U_{\sigma _2 }\), where \(\rho _2 \) is the unique positive root in (0, 1) of Eq. (3.7) and \(\sigma _2 \) is given by (3.8). The two radiuses \(\rho _2, \sigma _2 \) are sharp, with the extremal mapping \(F_2(z)=-\Lambda |z|^2 z+ M z\frac{1-M z}{M-z}\).
Conjecture 3.11
Suppose that \(M>1\). Let H(z) be a harmonic mapping of the unit disk U with \(H(0)=0, \lambda _{H}(0)=1\), and \(|H(z)|<M\) for all \(z\in U\). Then H is univalent in the disk \(U_{r_0 }\) and \(H(U_{r_0 })\) contains a Schlicht disk \(U_{R_0}\), where \(R_0=M r_0^2 \) and \(r_0\) is given by (1.2). This result is sharp with \(H_0(z)=M z\frac{1-M z}{M-z}\) being an extremal mapping.
Conjecture 3.12
Suppose that \(M>1\). Let \(H(z)=\sum _{n=1}^{\infty }c_{n}z^{n}+\overline{\sum _{n=1}^{\infty }d_{n}z^{n}}\) be a harmonic mapping of the unit disk U with \(\lambda _{H}(0)=1\) and \(|H(z)|< M\) for all \(z\in U\). Then
The inequality is sharp, with the extremal mapping \(H_0(z)=M z\frac{1-M z}{M-z}\).
Remark 3.13
If Conjecture 3.12 holds true, it follows from Theorem 3.7 and Corollary 3.9 that Conjectures 3.10 and 3.11 also hold true.
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Acknowledgements
The research of the first author was supported by Guangdong Natural Science Foundation of China (No. 2018A030313508). The authors of this paper thank the referee very much for his valuable comments and suggestions to this paper.
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Liu, MS., Luo, LF. Landau-Type Theorems for Certain Bounded Biharmonic Mappings. Results Math 74, 170 (2019). https://doi.org/10.1007/s00025-019-1095-7
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DOI: https://doi.org/10.1007/s00025-019-1095-7