Landau-Type Theorems for Certain Bounded Biharmonic Mappings

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.

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

$$\begin{aligned} \Delta f=4f_{z{\overline{z}}}=\frac{\partial ^2 f}{\partial x^2}+\frac{\partial ^2 f}{\partial y^2}=0 \end{aligned}$$

for $z\in U$, where we use the common notations for its formal derivatives:

$$\begin{aligned} f_z=\frac{1}{2}(f_x - i f_y),\quad f_{{\overline{z}}}=\frac{1}{2}(f_x + i f_y). \end{aligned}$$

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

$$\begin{aligned} \Lambda _{f}(z)=\max _{0\le \theta \le 2\pi }|{e}^{i\theta }f_{z}(z)+{e}^{-i\theta }f_{{\overline{z}}}(z)|=|f_{z}(z)| +|f_{{\overline{z}}}(z)| \end{aligned}$$

and

$$\begin{aligned} \lambda _{f}(z)=\min _{0\le \theta \le 2\pi }|{e}^{i\theta }f_{z}(z)+{e}^{-i\theta }f_{{\overline{z}}}(z)|= ||f_{z}(z)|-|f_{{\overline{z}}}(z)||. \end{aligned}$$

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,

$$\begin{aligned} J_f(z)=|h^{\prime }(z)|^2-|g^{\prime }(z)|^2. \end{aligned}$$

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:

$$\begin{aligned} F(z)=|z|^2G(z)+H(z), \end{aligned}$$

(1.1)

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

$$\begin{aligned} r_0=\frac{1}{M+\sqrt{M^2-1}}, \end{aligned}$$

(1.2)

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:

$$\begin{aligned} 1-3r^2-\frac{\Lambda _{2}^2-1}{\Lambda _{2}}\cdot \frac{r}{1-r} +\frac{\Lambda _{1}^2-1}{\Lambda _{1}}\cdot \left[ 2r^2+2r\ln (1-r) -\frac{r^3}{1-r}\right] =0,\nonumber \\ \end{aligned}$$

(1.3)

and $F(U_{r_1})$ contains a Schlicht disk $U_{R_1}$, with

$$\begin{aligned} R_1=r_1-r_1^3+\left( r_1^2\cdot \frac{\Lambda _{1}^2-1}{\Lambda _{1}} +\frac{\Lambda _{2}^2-1}{\Lambda _{2}}\right) \Big [r_1+\ln (1-r_1)\Big ]. \end{aligned}$$

(1.4)

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:

$$\begin{aligned} 1-3r^2-K(M)\cdot \frac{2r-r^2}{(1-r)^2}+\frac{\Lambda ^2-1}{\Lambda }\cdot \bigg [2r^2+2r\ln (1-r) -\frac{r^3}{1-r}\bigg ]=0,\nonumber \\ \end{aligned}$$

(1.5)

and $F(U_{r_2})$ contains a Schlicht disk $U_{R_2}$ with

$$\begin{aligned} R_2=r_2-r_2^3-K(M)\frac{r_2^2}{1-r_2}+\frac{\Lambda ^2-1}{\Lambda }[r_2^3+r_2^2\ln (1-r_2)]. \end{aligned}$$

(1.6)

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

$$\begin{aligned} U_p(z,r) = \left\{ \zeta \in U:\bigg |\frac{\zeta -z}{1-{\bar{z}}\zeta }\bigg |<r\right\} , \end{aligned}$$

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

$$\begin{aligned} a=e^{i\theta '}\cdot \frac{|z|-r}{1-r|z|},\ \ b=e^{i\theta '}\cdot \frac{|z|+r}{1+r|z|},\ \ \theta '=\arg {z}. \end{aligned}$$

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

$$\begin{aligned} \bigg |\int _{\overline{z_1z_2}}H_z(z) dz+H_{\bar{z}}(z) d\bar{z} \bigg |\ge \Lambda \, \frac{1-\Lambda r}{\Lambda -r}\, |z_1-z_2|, \end{aligned}$$

(2.1)

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

$$\begin{aligned}&\Delta _{0\le \theta \le 2\pi } \arg \left\{ {H_1'(0)e^{i(\theta _0+\theta )}+H_2'(0) e^{i(\theta _0-\theta )}}\right\} \\&\quad = \Delta _{0\le \theta \le 2\pi }\arg \left\{ {H_1'(0)e^{i(\theta _0+\theta )}}\right\} =2\pi , \end{aligned}$$

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

$$\begin{aligned} H_1'(0)e^{i(\theta _0+\theta _1)}+H_2'(0) e^{i(\theta _0-\theta _1)}>0. \end{aligned}$$

For $z\in U$, let

$$\begin{aligned} \omega (z)=\frac{H_1'(z)e^{i(\theta _0+\theta _1)}+H_2'(z) e^{i(\theta _0-\theta _1)}}{\Lambda }. \end{aligned}$$

Then $\omega (z)$ is analytic with $|\omega (z)|\le \Lambda _H(z)/\Lambda <1$ for $z\in U$ and

$$\begin{aligned} \alpha :=\omega (0)=\frac{H_1'(0)e^{i(\theta _0+\theta _1)}+H_2'(0) e^{i(\theta _0-\theta _1)}}{\Lambda }\ge \frac{\lambda _H(0)}{\Lambda }=\frac{1}{\Lambda }. \end{aligned}$$

Using Schwarz–Pick Lemma, we have

$$\begin{aligned} \mathrm{Re}\, \omega (z)\ge \frac{\alpha -r}{1-\alpha r}\ge \frac{\frac{1}{\Lambda }-r}{1-\frac{r}{\Lambda }}, \quad z\in U_r. \end{aligned}$$

That is

$$\begin{aligned} \mathrm{Re}\, \left\{ H_1'(z)e^{i(\theta _0+\theta _1)}+H_2'(z) e^{i(\theta _0-\theta _1)}\right\} \ge \Lambda \, \frac{\frac{1}{\Lambda }-r}{1-\frac{r}{\Lambda }}, \quad z\in U_r. \end{aligned}$$

(2.2)

Then

$$\begin{aligned} \bigg |\int _{\overline{z_1z_2}}H_z(z) dz+H_{{\bar{z}}}(z) d\bar{z} \bigg |= & {} \left| \int _{\overline{z_1z_2}}\bigg (H_1'(z)e^{i(\theta _0+\theta _1)} +\overline{H_2'(z)}e^{-i(\theta _0-\theta _1)}\bigg )|dz|\right| \\\ge & {} \int _{\overline{z_1z_2}}\mathrm{Re}\left\{ H_1'(z)e^{i(\theta _0+\theta _1)} +\overline{H_2'(z)}e^{-i(\theta _0-\theta _1)}\right\} |dz|\\= & {} \int _{\overline{z_1z_2}}\mathrm{Re}\left\{ H_1'(z)e^{i(\theta _0+\theta _1)} +H_2'(z)e^{i(\theta _0-\theta _1)}\right\} |dz|\\\ge & {} \int _{\overline{z_1z_2}}\Lambda \, \frac{\frac{1}{\Lambda }-r}{1-\frac{r}{\Lambda }}|dz|=\Lambda \, \frac{1-\Lambda r}{\Lambda -r}\, |z_1-z_2|. \end{aligned}$$

$\square$

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

$$\begin{aligned} \bigg |\int _{\gamma }H_\zeta (\zeta ) d\zeta +H_{\bar{\zeta }}(\zeta ) d\bar{\zeta } \bigg |\ge \Lambda \int _{0}^{r}\frac{\frac{1}{\Lambda }-t}{1-\frac{t}{\Lambda }}dt. \end{aligned}$$

(2.3)

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

$$\begin{aligned} H_1'(0)e^{i(\theta _\zeta +\theta _\zeta ')}+H_2'(0) e^{i(\theta _\zeta -\theta _\zeta ')}>0. \end{aligned}$$

For $z\in U$, define

$$\begin{aligned} \omega _\zeta (z)=\frac{H_1'(z)e^{i(\theta _\zeta +\theta _\zeta ')}+H_2'(z) e^{i(\theta _\zeta -\theta _\zeta ')}}{\Lambda }. \end{aligned}$$

Then $\omega _\zeta (z)$ is analytic in U, with $|\omega _\zeta (z)|\le \Lambda _H(z)/\Lambda <1$ and

$$\begin{aligned} \alpha _\zeta :=\omega _\zeta (0)=\frac{H_1'(0)e^{i(\theta _\zeta +\theta _\zeta ')}+H_2'(0) e^{i(\theta _\zeta -\theta _\zeta ')}}{\Lambda }\ge \frac{1}{\Lambda }. \end{aligned}$$

According to Schwarz–Pick Lemma, we get that

$$\begin{aligned} \mathrm{Re}\left\{ \omega _\zeta (\zeta )\right\} \ge \frac{\alpha _\zeta -|\zeta |}{1-\alpha _\zeta |\zeta |}\ge \frac{\frac{1}{\Lambda }-|\zeta |}{1-\frac{|\zeta |}{\Lambda }}, \quad \zeta \in \gamma . \end{aligned}$$

(2.4)

Thus

$$\begin{aligned} \bigg |\int _{\gamma }H_\zeta (\zeta ) d\zeta +H_{{\bar{\zeta }}}(\zeta ) d\bar{\zeta } \bigg |= & {} \left| \int _{\gamma }\bigg (H_1'(\zeta )e^{i(\theta _\zeta +\theta _\zeta ')} +\overline{H_2'(\zeta )}e^{-i(\theta _\zeta -\theta _\zeta ')}\bigg ) |d\zeta |\right| \\\ge & {} \int _{\gamma }\mathrm{Re}\left\{ H_1'(\zeta )e^{i(\theta _\zeta +\theta _\zeta ')} +H_2'(\zeta )e^{-i(\theta _\zeta -\theta _\zeta ')} \right\} |d\zeta |\\= & {} \int _{\gamma }\mathrm{Re}\left\{ H_1'(\zeta )e^{i(\theta _\zeta +\theta _\zeta ')} +H_2'(\zeta )e^{i(\theta _\zeta -\theta _\zeta ')} \right\} |d\zeta |\\= & {} \Lambda \int _{\gamma }\mathrm{Re}\left\{ \omega _\zeta (\zeta )\right\} |d\zeta |\ge \Lambda \int _{0}^{r} \frac{\frac{1}{\Lambda }-t}{1-\frac{t}{\Lambda }}dt. \end{aligned}$$

$\square$

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

$$\begin{aligned} |a_{n}|+|b_{n}|\le \frac{\Lambda ^{2}-1}{n\Lambda },\quad n=2,3,\ldots . \end{aligned}$$

(2.5)

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

$$\begin{aligned} f_{n}(z)=\Lambda ^{2}z-(\Lambda ^{3}-\Lambda )\int _{0}^{z}\frac{dz}{\Lambda +z^{n-1}}. \end{aligned}$$

(2.6)

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

$$\begin{aligned} \begin{array}{lll} \sigma _1=\rho _1-\Lambda \rho _1^3 =\left\{ \begin{array}{lll} 1-\Lambda , &{}&{} \text{ if } 0\le \Lambda \le \frac{1}{3},\\ \frac{2}{3\sqrt{3\Lambda }}, &{}&{} \text{ if } \Lambda >\frac{1}{3}. \end{array} \right. \end{array} \end{aligned}$$

(2.7)

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

$$\begin{aligned} |F(z_{1})-F(z_{2})|= & {} \left| a\Lambda (|z_{1}|^2z_{1}-|z_{2}|^2z_{2}) +b(\overline{z_{1}-z_2})\right| \\\ge & {} |b||z_{1}-z_2|-|a|\Lambda |z_1^2\overline{z_{1}}-z_{2}^2\overline{z_{2}}|\\\ge & {} |z_{1}-z_2|-\Lambda |z_1^2-z_2^2||\overline{z_{1}}|-\Lambda |z_{2}^2|| \overline{z_1-z_{2}}|\\\ge & {} |z_{1}-z_2|(1-\Lambda |z_{1}|(|z_1|+|z_2|)-\Lambda |z_{2}|^2)>0. \end{aligned}$$

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

$$\begin{aligned} |F(z')-F(0)|=|F(z')|\ge |b||z'|-|a|\Lambda |z'|^3=1-\Lambda , \end{aligned}$$

and

$$\begin{aligned} |F(z_0')-F(0)|= |ae^{i\frac{\pi +\theta _0}{2}}(\Lambda -1)|=1-\Lambda . \end{aligned}$$

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

$$\begin{aligned} |F(z_{1})-F(z_{2})|= & {} \left| a\Lambda (|z_{1}|^2z_{1}-|z_{2}|^2z_{2}) +b(\overline{z_{1}-z_2})\right| \\\ge & {} |b||z_{1}-z_2|-|a|\Lambda |z_1^2\overline{z_{1}}-z_{2}^2\overline{z_{2}}|\\\ge & {} |z_{1}-z_2|-\Lambda |z_1^2-z_2^2||\overline{z_{1}}| -\Lambda |z_{2}^2||\overline{z_1-z_{2}}|\\\ge & {} |z_{1}-z_2|(1-\Lambda |z_{1}|(|z_1|+|z_2|)-\Lambda |z_{2}|^2)\\> & {} |z_{1}-z_2|\bigg (1-3\Lambda \bigg (\frac{1}{\sqrt{3\Lambda }}\bigg )^2\bigg )= 0. \end{aligned}$$

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

$$\begin{aligned} \delta =\frac{\frac{3}{\sqrt{3\Lambda }}+\varepsilon -\sqrt{\left( \frac{3}{\sqrt{3\Lambda }} -3\varepsilon \right) \left( \frac{3}{\sqrt{3\Lambda }}+\varepsilon \right) }}{2}\in (0, 2\varepsilon )\subseteq \left( 0,\frac{1}{\sqrt{3\Lambda }}\right) . \end{aligned}$$

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

$$\begin{aligned} F(z_{1})= & {} a\Lambda \left( r_1^3e^{i\theta _1}+e^{i\arg \frac{b}{a}}\frac{1}{\Lambda }r_1 e^{-i\theta _1}\right) =a\Lambda e^{i\theta _1}\left( r_1^3-\frac{r_1}{\Lambda }\right) \\= & {} a\Lambda e^{i\theta _1}\left( r_2^3-\frac{r_2}{\Lambda }\right) =F(z_2), \end{aligned}$$

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

$$\begin{aligned} |F(z'')-F(0)|=|F(z'')|\ge |b||z''|-|a|\Lambda |z''|^3=\frac{1}{\sqrt{3\Lambda }}-\Lambda \bigg (\frac{1}{\sqrt{3\Lambda }}\bigg )^3 =\frac{2}{3\sqrt{3\Lambda }}, \end{aligned}$$

and

$$\begin{aligned} |F(z_0'')-F(0)|= \bigg |ae^{i\frac{\pi +\theta _0}{2}}\bigg (\Lambda \bigg (\frac{1}{\sqrt{3\Lambda }}\bigg )^3-\frac{1}{\sqrt{3\Lambda }}\bigg )\bigg |=\frac{2}{3\sqrt{3\Lambda }}. \end{aligned}$$

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

$$\begin{aligned} |a_n|+|b_n|\le \sqrt{2M^2-2},\quad n=2, 3,\ldots . \end{aligned}$$

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

$$\begin{aligned} |F(z_1)-F(z_2)|\ge |z_1-z_2|\bigg [||c_1|-|d_1||-\sum _{n=2}^{\infty }(|c_n|+|d_n|)nr^{n-1}-3\Lambda r^2\bigg ]. \end{aligned}$$

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

$$\begin{aligned} |G(z)|=\bigg |\int _{[0,z]}G_z(z)dz+G_{{\bar{z}}}(z)d{\bar{z}}\bigg |\le \int _{[0,z]}|\Lambda _G(z)||dz|\le \Lambda |z|, \end{aligned}$$

(2.8)

and

$$\begin{aligned}&|F(z_1)-F(z_2)|=\bigg |\int _{[z_1,z_2]}F_z(z)dz+F_{{\bar{z}}}(z)d{\bar{z}}\bigg |\\&\quad =\bigg |\int _{[z_1,z_2]}({\bar{z}}G(z)+|z|^2G_1'(z)\!+\!H_z(z))dz\!+\!(zG(z) \!+\!|z|^2\overline{G_2'(z)}+H_{{\bar{z}}}(z))d{\bar{z}}\bigg |\\&\quad =\bigg |\int _{[z_1,z_2]}(H_{z}(0)dz+H_{{\overline{z}}}(0)d{\overline{z}}) +\int _{[z_{1},z_{2}]} (H_{z}(z)-H_{z}(0))dz\\&\qquad +(H_{{\overline{z}}}(z)-H_{{\overline{z}}}(0))d{\overline{z}}+\int _{[z_1,z_2]}({\bar{z}}G(z)+|z|^2G_1'(z))dz+(zG(z) +|z|^2\overline{G_2'(z)})d{\bar{z}}\bigg |\\&\quad \ge \bigg |\int _{[z_{1},z_{2}]}H_{z}(0)dz+H_{{\overline{z}}}(0) d{\overline{z}}\bigg |-\bigg |\int _{[z_{1},z_{2}]} (H_{z}(z)-H_{z}(0))dz\\&\qquad +(H_{{\overline{z}}}(z)-H_{{\overline{z}}}(0)) d{\overline{z}}\bigg |-\int _{[z_{1},z_{2}]}(2|z||G(z)|+|z|^2|G_1'(z)|+|z|^2 |G_2'(z)|)|dz|\\&\quad \ge |z_1-z_2|\bigg [||c_1|-|d_1||-\sum _{n=2}^{\infty }(|c_n|+|d_n|)nr^{n-1} -3\Lambda r^2\bigg ]. \end{aligned}$$

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

$$\begin{aligned} \Lambda _2\ \frac{1-\Lambda _2 r}{\Lambda _2-r}-3\Lambda _1 r^2=0, \end{aligned}$$

(3.1)

and

$$\begin{aligned} \sigma _0=\Lambda _2^2\rho _0+\left( \Lambda _2^3-\Lambda _2\right) \ln \left( 1-\frac{\rho _0}{\Lambda _2}\right) -\Lambda _1 \rho _0^3. \end{aligned}$$

(3.2)

This result is sharp, with an extremal function given by

$$\begin{aligned} F_0(z)= & {} \Lambda _2\int _{[0,z]}\frac{1-\Lambda _2 z}{\Lambda _2-z}\ dz -\Lambda _1|z|^2z \nonumber \\= & {} \Lambda _2^2 z-\Lambda _1|z|^2z+\left( \Lambda _2^3-\Lambda _2\right) \ln \bigg (1-\frac{z}{\Lambda _2}\bigg ),\quad z\in U. \end{aligned}$$

(3.3)

Proof

By the hypothesis of Theorem 3.1, we have

$$\begin{aligned} |G(z)|=\bigg |\int _{[0,z]}G_z(z)dz+G_{{\bar{z}}}(z)d{\bar{z}}\bigg |\le \int _{[0,z]}|\Lambda _G(z)||dz|\le \Lambda _1 |z|,\ \ z\in U. \end{aligned}$$

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

$$\begin{aligned}&|F(z_2)-F(z_1)|=\left| \int _{\overline{z_1z_2}}F_z(z)dz+F_{{\bar{z}}} (z)d{\bar{z}}\right| \nonumber \\&\quad =\left| \int _{\overline{z_1z_2}}\left( {\bar{z}}G(z)+|z|^2G_z(z) \!+\!H_z(z)\right) dz\!+\!\left( zG(z)+|z|^2G_{{\bar{z}}}(z) +H_{{\bar{z}}}(z)\right) d{\bar{z}}\right| \nonumber \\&\quad \ge \bigg |\int _{\overline{z_1z_2}}H_z(z)dz+H_{{\bar{z}}}(z)d{\bar{z}}\bigg | -\int _{\overline{z_1z_2}}3\Lambda _1r^2 |dz|\nonumber \\&\quad \ge |z_1-z_2|\left( \Lambda _2 \frac{1-\Lambda _2 r}{\Lambda _2-r}-3\Lambda _1r^2\right) . \end{aligned}$$

(3.4)

It is easy to verify that the function

$$\begin{aligned} g_0(r):=\Lambda _2 \frac{1-\Lambda _2 r}{\Lambda _2-r}-3\Lambda _1r^2 \end{aligned}$$

is continuous and strictly decreasing on [0, 1], $g_0(0)=1>0$, and

$$\begin{aligned} g_0(1)=-(\Lambda _2+3\Lambda _1)<0. \end{aligned}$$

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

$$\begin{aligned} |F(z_2)-F(z_1)|>|z_1-z_2|\left( \Lambda _2 \frac{1-\Lambda _2\rho _0}{\Lambda _2-\rho _0}-3\Lambda _1\rho _0^2\right) =0. \end{aligned}$$

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

$$\begin{aligned} |w'|=\big ||z'|^2G(z')+H(z')\big |\ge & {} |H(z')|-\Lambda _1 \rho _0^3\\= & {} \bigg |\int _{\gamma }H_\zeta (\zeta ) d\zeta +H_{{\bar{\zeta }}}(\zeta ) d\bar{\zeta } \bigg |-\Lambda _1\rho _0^3\\\ge & {} \Lambda _2\int _{0}^{\rho _0}\frac{\frac{1}{\Lambda _2}-t}{1-\frac{t}{\Lambda _2}}dt-\Lambda _1\rho _0^3\\= & {} \Lambda _2^2\rho _0+\left( \Lambda _2^3-\Lambda _2\right) \ln \left( 1-\frac{\rho _0}{\Lambda _2}\right) -\Lambda _1\rho _0^3=\sigma _0, \end{aligned}$$

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

$$\begin{aligned} h_0(x)=\Lambda _2^2x-\Lambda _1x^3+\left( \Lambda _2^3-\Lambda _2\right) \ln \left( 1-\frac{x}{\Lambda _2}\right) , \quad x\in [0, 1]. \end{aligned}$$

(3.5)

Because the continuous function

$$\begin{aligned} h_0'(x)=\Lambda _2^2-3\Lambda _1x^2+\frac{\Lambda _2-\Lambda _2^3}{\Lambda _2-x}=g_0(x) \end{aligned}$$

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

$$\begin{aligned} \sigma _0=\Lambda _2^2\rho _0+\left( \Lambda _2^3-\Lambda _2\right) \ln \left( 1-\frac{\rho _0}{\Lambda _2}\right) -\Lambda _1\rho _0^3=h_0(\rho _0)>h_0(0)=0. \end{aligned}$$

(3.6)

For every fixed $r\in (\rho _0, 1]$, set $x_1=\rho _0+\varepsilon$, where

$$\begin{aligned} \varepsilon =\left\{ \begin{array}{lll} \min \left\{ \frac{r-\rho _0}{2}, \frac{r_3-\rho _0}{2}\right\} , &{}&{} \text{ if } f_0(1)\le 0,\\ \frac{r-\rho _0}{2}, &{}&{} \text{ if } f_0(1)>0, \end{array} \right. \end{aligned}$$

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

$$\begin{aligned} F_0(z_1)=F_0(x_1)=h_0(x_1)=h_0(x_2)=F_0(z_2). \end{aligned}$$

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

$$\begin{aligned} |F_0(z')-F_0(0)|=|F_0(\rho _0)|=|h_0(\rho _0)|=h_0(\rho _0)=\sigma _0. \end{aligned}$$

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

$$\begin{aligned} \lambda _{F}(0)=\lambda _{H}(0)=||c_1|-|d_1||=1. \end{aligned}$$

Since $\Lambda _H(z)\le 1$, using Lemma 2.4, we have

$$\begin{aligned} c_n=d_n=0,\quad n=2,3,\ldots . \end{aligned}$$

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

$$\begin{aligned} |F(z_1)-F(z_2)|\ge & {} \bigg |\int _{\overline{z_1z_2}}H_z(z)dz +H_{{\bar{z}}}(z)d{\bar{z}}\bigg |-\int _{\overline{z_1z_2}}3\Lambda _1r^2 |dz|\\\ge & {} |z_1-z_2|(||c_1|-|d_1||-3\Lambda r^2)\\= & {} |z_1-z_2|(1-3\Lambda r^2)>0. \end{aligned}$$

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

$$\begin{aligned} |F(z)|= & {} ||z|^2G(z)+H(z)|\ge |H(z)|-\rho _1^{2}|G(z)|\\\ge & {} \rho _1||c_1|-|d_1||-\Lambda \rho _1^{3}=\rho _1-\Lambda \rho _1^{3}=\sigma _1. \end{aligned}$$

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

$$\begin{aligned} G(0)=H(0)=0, \, \lambda _{F}(0)=|-\overline{d_1}|=1, \, \Lambda _{G}(z)=|a_{1}\Lambda |\le \Lambda \end{aligned}$$

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

$$\begin{aligned} c_n=d_n=0,\quad n=2,3,\ldots . \end{aligned}$$

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

$$\begin{aligned} \rho _1=\left\{ \begin{array}{lll} 1&{}&{} \text{ if } 0\le \Lambda \le \frac{1}{3},\\ \frac{1}{\sqrt{3\Lambda }} &{}&{} \text{ if } \Lambda >\frac{1}{3}. \end{array} \right. \end{aligned}$$

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

$$\begin{aligned} |F(z_1)-F(z_2)|\ge |z_1-z_2|(1-3\Lambda r^2)> |z_1-z_2|(1-3\Lambda \rho _0^2)\ge 0. \end{aligned}$$

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

$$\begin{aligned} |F(z)-F(0)|= & {} ||z|^2G(z)+H(z)|\ge |H(z)|-\rho _1^{2}|G(z)|\\\ge & {} \rho _1||c_1|-|d_1||-\Lambda \rho _1^{3}=\rho _1-\Lambda \rho _1^{3}=\sigma _1. \end{aligned}$$

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

$$\begin{aligned} M^2-\frac{M^2(M^2-1)}{(M-r)^2}-3\Lambda r^2=0, \end{aligned}$$

(3.7)

and $F_2(U_{\rho _2 })$ contains a Schlicht disk $U_{\sigma _2 }$, with

$$\begin{aligned} \sigma _2 =M\rho _2 \frac{1-M\rho _2 }{M-\rho _2 }-\Lambda \rho _2 ^{3}. \end{aligned}$$

(3.8)

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

$$\begin{aligned} |F_2(z_1)-F_2(z_2)|\ge & {} |H(z_1)-H(z_2)|-\big |\Lambda |z_1|^2z_1 -\Lambda |z_2|^2z_2\big |\\= & {} M^2|z_1-z_2|\bigg |\frac{1-M(z_1+z_2)+z_1z_2}{(M-z_1)(M-z_2)}\bigg |\\&-\,\Lambda \Big ||z_1|^2(z_1-z_2)+z_2(|z_1|+|z_2|)(|z_1|-|z_2|)\Big |\\\ge & {} M^2|z_1-z_2|\bigg |1-\frac{M^2-1}{(M-z_1)(M-z_2)}\bigg |-3\Lambda r^2 |z_1-z_2|\\\ge & {} |z_1-z_2|\bigg [M^2\bigg (1-\frac{M^2-1}{(M-r)^2}\bigg )-3\Lambda r^2\bigg ]. \end{aligned}$$

It is easy to verify that the function

$$\begin{aligned} g_1(r):=M^2-\frac{M^2(M^2-1)}{(M-r)^2}-3\Lambda r^2 \end{aligned}$$

is continuous and strictly decreasing on [0, 1], $g_0(0)=1>0$, and

$$\begin{aligned} g_1(1)=-\Big (\frac{2M^2}{M-1}+3\Lambda \Big )<0. \end{aligned}$$

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

$$\begin{aligned} |F_2(z_1)-F_2(z_2)|>|z_1-z_2|\bigg (M^2-\frac{M^2(M^2-1)}{(M-\rho _2 )^2}-3\Lambda \rho _2 ^2\bigg )=0. \end{aligned}$$

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

$$\begin{aligned} h_1(x)=M x\frac{1-M x}{M-x}-\Lambda x^3, \quad x\in [0, 1]. \end{aligned}$$

(3.9)

Because the continuous function

$$\begin{aligned} h_1'(x)=M \frac{1-M x}{M-x}+Mx\frac{1-M^2}{(M-x)^2}-3\Lambda x^2=g_1(x) \end{aligned}$$

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

$$\begin{aligned} \sigma _2 =M\rho _2 \frac{1-M\rho _2 }{M-\rho _2 }-\Lambda \rho _2^{3}=h_1(\rho _2)>h_1(0)=0. \end{aligned}$$

(3.10)

For every fixed $r\in (\rho _1, 1]$, set $x_1=\rho _1 +\varepsilon$, where

$$\begin{aligned} \varepsilon =\left\{ \begin{array}{lll} \min \left\{ \frac{r-\rho _2 }{2}, \frac{r_4-\rho _2 }{2}\right\} , &{}&{} \text{ if } h_1(1)\le 0,\\ \frac{r-\rho _2 }{2}, &{}&{} \text{ if } h_1(1)>0, \end{array} \right. \end{aligned}$$

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

$$\begin{aligned} F_2(z_1)=F_2(x_1)=h_1(x_1)=h_1(x_2)=F_2(z_2), \end{aligned}$$

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

$$\begin{aligned} |F_2(z)-F_2(0)|= & {} \bigg |-\Lambda |z|^2 z+ M z\frac{1-M z}{M-z}\bigg |\\\ge & {} M\rho _2 \bigg |\frac{1-M z}{M-z}\bigg |-\Lambda \rho _2 ^3 =M \rho _2 \bigg |\frac{\frac{1}{M}-z}{1-\frac{1}{M}z}\bigg |-\Lambda \rho _2 ^3\\\ge & {} M \rho _2 \frac{\frac{1}{M}-|z|}{1-\frac{1}{M}|z|}-\Lambda \rho _2 ^3= M\rho _2 \frac{1-M\rho _2 }{M-\rho _2 }-\Lambda \rho _2 ^{3}=\sigma _2, \end{aligned}$$

and

$$\begin{aligned} |F_2(z')-F_2(0)|= & {} |F_2(\rho _2 )|=|h_1(\rho _2 )|=h_1(\rho _2 )=\sigma _2. \end{aligned}$$

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

$$\begin{aligned}&\sum \limits _{n=2}^{\infty }n(|c_n|+|d_n|)r^{n-2}\le \frac{(M^2-1)(2M-r)}{(M-r)^2},\quad 0\le r\le r_0\nonumber \\&\quad =\frac{1}{M+\sqrt{M^2-1}}. \end{aligned}$$

(3.11)

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

$$\begin{aligned} ||c_1|-|d_1||=\lambda _{H}(0)=\lambda _{F}(0)=1. \end{aligned}$$

(3.12)

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

$$\begin{aligned} |F(z_1)-F(z_2)|\ge & {} |z_1-z_2|\bigg [||c_1|-|d_1||-\sum _{n=2}^{\infty } (|c_n|+|d_n|)nr^{n-1}-3\Lambda r^2\bigg ]\\\ge & {} |z_1-z_2|\bigg [1-\frac{(M^2-1)(2Mr-r^2)}{(M-r)^2}-3\Lambda r^2\bigg ]\\= & {} |z_1-z_2|\bigg [M^2-\frac{M^2(M^2-1)}{(M-r)^2} -3\Lambda r^2\bigg ]\\> & {} |z_1-z_2|\bigg [M^2-\frac{M^2(M^2-1)}{(M-\rho _2)^2} -3\Lambda \rho _2 ^2\bigg ]=0. \end{aligned}$$

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

$$\begin{aligned} |F(z)-F(0)|= & {} ||z|^2G(z)+H(z)|\ge \bigg |\sum _{n=1}^{\infty }(c_{n}z^{n} +\overline{d_{n}z^{n}})\bigg |-\rho _2 ^{2}|G(z)|\\\ge & {} \rho _2 ||c_1|-|d_1||-\sum _{n=2}^{\infty }(|c_n|+|d_n|)\rho _2 ^{n} -\Lambda \rho _2 ^{3}\\= & {} \rho _2 -\int _0^{\rho _2 }\sum _{n=2}^{\infty }n(|c_n|+|d_n|)r^{n-1}dr -\Lambda \rho _2 ^{3}\\\ge & {} \rho _2 -\int _0^{\rho _2 }\frac{(M^2-1)(2Mr-r^2)}{(M-r)^2}dr -\Lambda \rho _2 ^{3}\\= & {} \rho _2 -\frac{(M^2-1)\rho _2 ^2}{M-\rho _2 }-\Lambda \rho _2 ^{3} = M\rho _2 \frac{1-M\rho _2 }{M-\rho _2 }-\Lambda \rho _2 ^{3} =\sigma _2. \end{aligned}$$

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

$$\begin{aligned} M z\frac{1-M z}{M-z}= & {} z-\sum _{n=2}^{\infty }\frac{M^2-1}{M^{n-1}}z^{n}, \quad z\in U,\\ \sum _{n=2}^{\infty }\frac{M^2-1}{M^{n-1}}nr^{n-1}= & {} \frac{(M^2-1)(2Mr-r^2)}{(M-r)^2}, \quad 0\le r<1, \end{aligned}$$

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

$$\begin{aligned} |c_n|+|d_n|\le \frac{M^2-1}{M^{n-1}},\quad n=2,3,\ldots . \end{aligned}$$

(3.13)

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

$$\begin{aligned} |H_0(z)|=M|z|\Big |\frac{1-M z}{M-z}\bigg |<M\quad \text{ for } \text{ all } z\in U, \end{aligned}$$

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

$$\begin{aligned} \sum \limits _{n=2}^{\infty }n(|c_n|+|d_n|)r^{n-2}\le \frac{(M^2-1)(2M-r)}{(M-r)^2},\quad 0\le r\le r_0=\frac{1}{M+\sqrt{M^2-1}}. \end{aligned}$$

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.