Landau-type theorems and bi-Lipschitz theorems for bounded biharmonic mappings

Abstract

In this paper, we first establish five versions of Landau-type theorems for five classes of bounded biharmonic mappings $F(z)=|z|^2G(z)+H(z)$ on the unit disk ${\mathbb {D}}$ with $G(0)=H(0)=J_F(0)-1=0$, which improve the related results of earlier authors. In particular, two versions of those Landau-type theorems are sharp. Then we derive five bi-Lipschitz theorems for these classes of bounded and normalized biharmonic mappings.

1 Introduction

Let ${\mathbb {D}}$={$z\in {\mathbb {C}}: |z|<1$} denote the unit disk with center at the origin and radius 1. For $r>0$, let ${\mathbb {D}}_r =\{z\in {\mathbb {C}}: |z|<r$}. A function $f(z)=u(z)+i v(z), z=x+i y$ is a harmonic mapping on the unit disk ${\mathbb {D}}$ if and only if F is twice continuously differentiable and satisfies the Laplacian equation

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

for $z \in {\mathbb {D}}$, where the formal derivatives of f are defined by

$$\begin{aligned} f_{z}=\frac{1}{2}\left( f_{x}-i f_{y}\right) , \quad f_{\bar{z}}=\frac{1}{2}\left( f_{x}+i f_{y}\right) . \end{aligned}$$

A function $F(z)=U(z)+i V(z)$ is a biharmonic mapping on ${\mathbb {D}}$ if and only if F is four times continuously differentiable and satisfies the biharmonic equation $\Delta (\Delta F)=0$ for $z \in {\mathbb {D}}$. In other words, F(z) is biharmonic on ${\mathbb {D}}$ if and only if $\Delta F$ is harmonic on ${\mathbb {D}}.$

It is known [1] that a mapping F is biharmonic on ${\mathbb {D}}$ if and only if F can be represented as follow:

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

(1.1)

where G(z) and H(z) are complex-valued harmonic mappings on ${\mathbb {D}}$.

In [15], it’s known that a harmonic mapping f(z) is locally univalent on ${\mathbb {D}}$ if and only if its Jacobian $J_f(z)=|f_z|^2-|f_{\overline{z}}|^2\not =0$ for any $z\in {\mathbb {D}}$. Since ${\mathbb {D}}$ is simply connected, f(z) can be written as $f=h+\overline{g}$ with $f(0)=h(0)$, h and g are analytic on ${\mathbb {D}}$. Thus, we have

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

For such function f, we define

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

and

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

Recall that a mapping $\omega :{\mathbb {D}}\rightarrow \Omega$ is said to be $L_1$-Lipschitz ($L_1>0$) ($l_1$-co-Lipschitz ($l_1>0$)) if

$$\begin{aligned} |\omega (z_1)-\omega (z_2)|\le L_1|z_1-z_2|,\quad z_1,z_2\in {\mathbb {D}}, \end{aligned}$$

(1.2)

$$\begin{aligned} (|\omega (z_1)-\omega (z_2)|\ge l_1|z_1-z_2|,\quad z_1,z_2\in {\mathbb {D}}). \end{aligned}$$

(1.3)

A mapping $\omega$ is bi-Lipschitz if it is Lipschitz and co-Lipschitz (see [14]). In [13], the Lipschitz character of q.c. harmonic self-mappings of the unit disk was established with respect to the hyperbolic metric and this was generalized to an arbitrary domain in [25].

Harmonic mappings techniques have been used to study and solve fluid flow problems (see [4, 11]). For example, in 2012, Aleman and Constantin [4] developed ingenious technique to solve the incompressible two dimensional Euler equations in terms of univalent harmonic mappings. More precisely, the problem of finding all solutions which in Lagrangian variables describing the particle paths of the flow present a labelling by harmonic mappings is reduced to solve an explicit nonlinear differential system in ${\mathbb {C}}^n$ (please refer to [11]).

The classical Landau’s theorem states that if f is an analytic function on the unit disk ${\mathbb {D}}$ with $f(0)=f^{\prime }(0)-1=0$ and $|f(z)|<M$ for $z\in {\mathbb {D}}$, then f is univalent in the disk ${\mathbb {D}}_{r_0}$ with $r_0=\frac{1}{M+\sqrt{M^2-1}}$ and $f( {\mathbb {D}} _{r_0})$ 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 ${\mathbb {D}}$ with $f^{\prime }(0)=1$, then $f({\mathbb {D}} )$ contains a schlicht disk of radius b, that is, a disk of radius b which is the univalent image of some region on ${\mathbb {D}}$. The supremum of all such constants b is called the Bloch constant (see [6, 12]).

For harmonic mappings on ${\mathbb {D}}$, under suitable restriction, Chen, Gauthier and Hengartner [6] obtained two versions of Landau’s theorems. In 2008, Abdulhadi and Muhanna proved the following Landau-type theorem of certain bounded biharmonic mappings in [2].

Theorem A

(Abdulhadi and Muhanna [2])  Let $f(z)=|z|^2 g(z)+h(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}}$, as in (1.1), with $f(0)=h(0)=J_f(0)-1=0$ and $|g(z)|\le M, \, |h(z)|\le M$ for $z\in {\mathbb {D}}$. Then there is a constant $0<r_1<1$ so that f is univalent in the disk ${\mathbb {D}}_{r_1}$. In specific $r_1$ satisfies the following equation

$$\begin{aligned} \frac{\pi }{4M} -2r_1 M -\frac{2M r_1^2}{(1-r_1)^2}-2M\cdot \frac{2r_1-r_1^2}{(1-r_1)^2}=0, \end{aligned}$$

(1.4)

and $f({\mathbb {D}}_{r_1})$ contains a schlicht disk ${\mathbb {D}}_{R_1}$ with

$$\begin{aligned} R_1=\frac{\pi }{4M}r_1- 2M\frac{r_1^3+r_1^2}{1-r_1} \, . \end{aligned}$$

(1.5)

From that on, many authors considered the Landau-type theorems for certain bounded biharmonic mappings (see [5, 7,8,9, 16, 18,19,20,21,22,23, 26]). Liu et al. improved Theorem A by establishing the following theorem.

Theorem B

(Liu [16])  Let $F(z)=|z|^2 g(z)+h(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}}$, as in (1.1), with $F(0)=h(0)=J_F(0)-1=0$ and $|g(z)|\le M_1, \, |h(z)|\le M_2$ for $z\in {\mathbb {D}}$. Then, F is univalent in the disk ${\mathbb {D}}_{r_2}$, and $F({\mathbb {D}}_{r_2})$ contains a schlicht disk ${\mathbb {D}}_{R_2}$, where $r_2$ is the minimum positive root of the following equation

$$\begin{aligned} \lambda _0(M_2) -2r M_1 -\frac{2M_1 r^2}{(1-r)^2}-\sqrt{2M_2^2-2}\cdot \frac{2r-r^2}{(1-r)^2}=0, \end{aligned}$$

(1.6)

and

$$\begin{aligned} R_2=\lambda _0(M_2) r_2-2M_1\cdot \frac{ r_2^3}{1-r_2}-\sqrt{2M_2^2-2}\cdot \frac{ r_2^2}{1-r_2} \, , \end{aligned}$$

(1.7)

where $\lambda _0(M)$ is defined by

$$\begin{aligned} \lambda _0(M)=\left\{ \begin{array}{lll} \frac{\sqrt{2}}{\sqrt{M^2-1}+\sqrt{M^2+1}}&{}&{} \text{ if } 1\le M\le M_0= \frac{\pi }{2\root 4 \of {2\pi ^2-16}},\\ \\ \frac{\pi }{4M} &{}&{} \text{ if } M>M_0= \frac{\pi }{2\root 4 \of {2\pi ^2-16}}\approx 1.1296 \, . \end{array} \right. \end{aligned}$$

(1.8)

Chen et al. established the following theorem, which improved Theorems A and B for the case $M_1=M_2=M$.

Theorem C

(Chen et al. [8])  Let $F(z)=|z|^2 g(z)+h(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}}$, as in (1.1), with $F(0)=h(0)=J_F(0)-1=0$ and $|g(z)|\le M_1, \, |h(z)|\le M_2$ for $z\in {\mathbb {D}}$. Then, F is univalent in the disk ${\mathbb {D}}_{r_3}$, and $F({\mathbb {D}}_{r_3})$ contains a schlicht disk ${\mathbb {D}}_{R_3}$, where $r_3$ is the minimum positive root of the following equation

$$\begin{aligned} \frac{\pi }{4 M_2} -2r M_1 -\frac{4 M_1 r^2}{\pi (1-r)^2}-\sqrt{2M_2^2-2}\cdot \frac{2r-r^2}{(1-r)^2}=0, \end{aligned}$$

(1.9)

and

$$\begin{aligned} R_3=\frac{\pi }{4 M_2} r_3-\frac{ r_3^2(4M_1 r_3 + \pi \sqrt{2M_2^2-2})}{\pi (1-r_3)} \, , \end{aligned}$$

(1.10)

Zhu et al. improved Theorems A, B and C by establishing the following theorem:

Theorem D

(Zhu and Liu [26])  Suppose that $F(z)=|z|^{2}g(z)+h(z)$ is a biharmonic mapping in the unit disk ${\mathbb {D}}$ such that $|g(z)|\le M_{1}$ and $|h(z)|\le M_{2}$ for $z\in {\mathbb {D}}$ with $|J_{F}(0)|=1$.

(i) If $M_2>1$ or $M_2=1$ and $M_1>0$, then F is univalent in the disk ${\mathbb {D}}_{r_4}$, and $F(\mathbb {D}_{r_4})$ contains a schlicht disk ${\mathbb {D}}_{R_4}(F(0))$, where $r_4=r_4(M_1, M_2)$ is the minimum positive root of the following equation:

$$\begin{aligned} \lambda _0(M_2)-2M_1r-\frac{4M_1r^2}{\pi (1-r^2)}- \lambda _0(M_2)\sqrt{M_2^4-1}\cdot \frac{r\sqrt{4-3r^2+r^4}}{(1-r^2)^{3/2}}=0,\quad \qquad \end{aligned}$$

(1.11)

and

$$\begin{aligned} R_4=\lambda _0(M_2)r_4-M_1r^2_4-\lambda _0(M_2)\sqrt{M_2^4-1}\cdot \frac{r_4^2}{(1-r^2_4)^{1/2}}, \end{aligned}$$

(1.12)

where $\lambda _0(M)$ is given by (1.8).

(ii) If $M_2=1$ and $M_1=0$, then F is univalent in ${\mathbb {D}}$ and $F({\mathbb {D}})={\mathbb {D}}$.

For the biharmonic mappings with $\lambda _F(0)=1$, many versions of Landau-type theorems, even sharp results have been found. In 2019, Liu and Luo proved the following sharp results.

Theorem E

(Liu and Luo [20])  Suppose that $\Lambda _{1} \ge 0$ and $\Lambda _{2}>1 .$ Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}},$ where G(z) and H(z) are harmonic in ${\mathbb {D}},$ satisfying $G(0)=H(0)=0, \lambda _{F}(0)=1, \Lambda _{G}(z) \le \Lambda _{1}$ and $\Lambda _{H}(z)<\Lambda _{2}$ for all $z \in {\mathbb {D}} .$ Then F(z) is univalent on the disk ${\mathbb {D}}_{r_5}$ and $F\left( {\mathbb {D}}_{r_5}\right)$ contains a Schlicht disk ${\mathbb {D}}_{R_5},$ where $r_5$ is the unique root in (0, 1) of the equation

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

(1.13)

and

$$\begin{aligned} R_5=\Lambda _{2}^{2} r_5+\left( \Lambda _{2}^{3}-\Lambda _{2}\right) \ln \left( 1-\frac{r_5}{\Lambda _{2}}\right) -\Lambda _{1} r_5^{3}. \end{aligned}$$

(1.14)

This result is sharp, with an extremal function given by

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

(1.15)

Theorem F

(Liu and Luo [20])  Suppose that $\Lambda \ge 0 .$ Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of ${\mathbb {D}}$, where G(z), H(z) are harmonic in ${\mathbb {D}},$ satisfying $G(0)=H(0)=0,$ $\lambda _F(0)=1, \Lambda _{G}(z) \le \Lambda$, and $\Lambda _{H}(z) \le 1$ or $|H(z)|<1$ for all $z \in {\mathbb {D}}$. Then F is univalent on the disk ${\mathbb {D}}_{\rho _{1}}$, and $F\left( {\mathbb {D}}_{\rho _{1}}\right)$ contains a schlicht disk ${\mathbb {D}}_{\sigma _{1}}$, where

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

(1.16)

and

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

(1.17)

This result is sharp.

It is natural raise the following.

Problem 1

If $\lambda _F(0)=1$ is replaced by $J_F(0)=1$ in Theorems E and F, can we obtain sharp versions of Landau-type theorems for such bounded and normalized biharmonic mappings?

Problem 2

Can we improve Theorem D?

In this paper, we first establish several new lemmas (see Lemmas 2.1, 2.2, 2.4, 2.5 and 2.9). Then, using these estimates, we prove several new versions of Landau-type theorems of bounded biharmonic mappings F(z) with $J_F(0)=1$. In particular, the results of Theorems 3.1 and 3.2 are sharp, which gives part of affirmative answer to the first question, Theorem 3.5 improves Theorems A, B, C and D, which gives an affirmative answer to the second question. Moreover, we can verify that these biharmonic mappings F(z) are bi-Lipschitz on the univalent disks without changing the hypothesis of the theorems in Sect. 3.

2 Preliminaries

In this section, we establish some lemmas needed in the proof of the main results.

Lemma 2.1

Suppose $\Lambda >1$. Let H(z) be a harmonic mapping of the unit disk ${\mathbb {D}}$ with $J_{H}(0)=1$ and $\Lambda _{H}(z)<\Lambda$ for all $z \in {\mathbb {D}} .$ Then for all $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}\left( 0<r<1, z_{1} \ne z_{2}\right) ,$ we have

$$\begin{aligned} |H(z_2)-H(z_1)|=\left| \int _{\overline{z_{1} z_{2}}} H_{z}(z) d z+H_{\bar{z}}(z) d \bar{z}\right| \ge \Lambda \frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}\left| z_{1}-z_{2}\right| .\quad \end{aligned}$$

(2.1)

Proof

Following the idea from [17] (see also [20, Proof of Lemma 2.2]), let $\theta _{0}=\arg \left( z_{2}-z_{1}\right)$. Since H(z) is a harmonic mapping in the unit disk ${\mathbb {D}},$ H(z) can be written as $H(z)=H_{1}(z)+\overline{H_{2}(z)}$ for $z \in {\mathbb {D}},$ where $H_{1}$ and $H_{2}$ are analytic in ${\mathbb {D}}$. Since $J_{H}(0)=|H_{1}^{\prime }(0)|^2-| H_{2}^{\prime }(0)|^2=1$, we have $|H_{1}^{\prime }(0)|>| H_{2}^{\prime }(0)|$, and

$$\begin{aligned}&\Delta _{0 \le \theta \le 2 \pi } \arg \left\{ H_{1}^{\prime }(0) e^{i\left( \theta _{0}+\theta \right) }+H_{2}^{\prime }(0) e^{i\left( \theta _{0}-\theta \right) }\right\} \\&\quad =\Delta _{0 \le \theta \le 2 \pi } \arg \left\{ H_{1}^{\prime }(0) e^{i\left( \theta _{0}+\theta \right) }\right\} =2 \pi , \end{aligned}$$

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

$$\begin{aligned} H_{1}^{\prime }(0) e^{i\left( \theta _{0}+\theta _{1}\right) }+H_{2}^{\prime }(0) e^{i\left( \theta _{0}-\theta _{1}\right) }>0. \end{aligned}$$

Since $\Lambda _{H}(0)<\Lambda$, we have

$$\begin{aligned} \lambda _{H}(0)=\frac{J_{H}(0)}{\Lambda _{H}(0)}>\frac{1}{\Lambda }>0. \end{aligned}$$

For $z \in {\mathbb {D}},$ let

$$\begin{aligned} \omega (z)=\frac{H_{1}^{\prime }(z) e^{i\left( \theta _{0}+\theta _{1}\right) }+H_{2}^{\prime }(z) e^{i\left( \theta _{0}-\theta _{1}\right) }}{\Lambda }. \end{aligned}$$

Then $\omega (z)$ is analytic with $|\omega (z)| \le \Lambda _{H}(z) / \Lambda <1$ for $z \in {\mathbb {D}}$ and

$$\begin{aligned} \alpha :=\omega (0)=\frac{H_{1}^{\prime }(0) e^{i\left( \theta _{0}+\theta _{1}\right) }+H_{2}^{\prime }(0) e^{i\left( \theta _{0}-\theta _{1}\right) }}{\Lambda } \ge \frac{\lambda _{H}(0)}{\Lambda }. \end{aligned}$$

Using Schwarz–Pick Lemma, we have

$$\begin{aligned} {\text {Re}} \omega (z) \ge \frac{\alpha -r}{1-\alpha r} \ge \frac{\frac{\lambda _{H}(0)}{\Lambda }-r}{1-\frac{\lambda _{H}(0)r}{\Lambda }}=\frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}, \quad z \in {\mathbb {D}}_{r}. \end{aligned}$$

Then

$$\begin{aligned} \left| \int _{\overline{z_{1} z_{2}}} H_{z}(z) d z+H_{z}(z) d \bar{z}\right|&=\left| \int _{\overline{z_{1} z_{2}}}\left( H_{1}^{\prime }(z) e^{i\left( \theta _{0}+\theta _{1}\right) }+\overline{H_{2}^{\prime }(z)} e^{-i\left( \theta _{0}-\theta _{1}\right) }\right) |d z|\right| \\&\ge \int _{\overline{z_{1} z_{2}}} {\text {Re}}\left\{ H_{1}^{\prime }(z) e^{i\left( \theta _{0}+\theta _{1}\right) }+\overline{H_{2}^{\prime }(z)} e^{-i\left( \theta _{0}-\theta _{1}\right) }\right\} |d z| \\&=\int _{\overline{z_{1} z_{2}}} {\text {Re}}\left\{ H_{1}^{\prime }(z) e^{i\left( \theta _{0}+\theta _{1}\right) }+H_{2}^{\prime }(z) e^{i\left( \theta _{0}-\theta _{1}\right) }\right\} |d z| \\&\ge \int _{\overline{z_{1} z_{2}}} \Lambda \frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}|d z|=\Lambda \frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}\left| z_{1}-z_{2}\right| . \end{aligned}$$

$\square$

Applying the analogous proof of Lemma 2.3 in [20], we have the following lemma.

Lemma 2.2

Suppose $\Lambda >1 .$ Let H(z) be a harmonic mapping of the unit disk ${\mathbb {D}}$ with $J_{H}(0)=1$ and $\Lambda _{H}(z)<\Lambda$ for all $z \in {\mathbb {D}} .$ Set $\gamma =H^{-1}(\overline{o w^{\prime }})$ with $w^{\prime } \in H\left( \partial {\mathbb {D}}_{r}\right) (0<r \le 1)$ and $\overline{o w^{\prime }}$ denotes the closed line segment joining the origin and $w^{\prime },$ then

$$\begin{aligned} \left| \int _{\gamma } H_{\zeta }(\zeta ) d \zeta +H_{\bar{\zeta }}(\zeta ) d \bar{\zeta }\right| \ge \Lambda \int _{0}^{r} \frac{\lambda _{H}(0)-\Lambda t}{\Lambda -\lambda _{H}(0)t} d t. \end{aligned}$$

(2.2)

Lemma 2.3

[16]  Suppose that $f(z)=h(z)+\overline{g(z)}$ is a harmonic mapping of the unit disk ${\mathbb {D}}$ with $|f(z)|\le 1$. If $J_{f}(0)=1$, then $f(z)=\alpha z,$ where $|\alpha |=1$.

Lemma 2.4

Suppose that $f(z)=h(z)+\overline{g(z)}$ is a harmonic mapping of the unit disk ${\mathbb {D}}$ with $J_f(0)=1$. Then $|f(z)|\le 1$ for all $z\in {\mathbb {D}}$ if and only if $\Lambda _f(z)\le 1$ for all $z\in {\mathbb {D}}$.

Proof

If $|f(z)|\le 1$ for all $z\in {\mathbb {D}}$, it follows from Lemma 2.3 that $f(z)=\alpha z$, where $|\alpha |=1$. Hence

$$\begin{aligned} \Lambda _f(z)=|f_z(z)|+|f_{\overline{z}}(z)|=|\alpha |=1\le 1 \end{aligned}$$

for all $z\in {\mathbb {D}}$. Conversely, if $\Lambda _f(z)\le 1$ for all $z\in {\mathbb {D}}$, then for each $z\in {\mathbb {D}}$, we have

$$\begin{aligned} |f(z)|=\left| \int _{[0, z]} f_{z}(z) d z+f_{\bar{z}}(z) d \bar{z}\right| \le \int _{[0, z]}\left| \Lambda _{f}(z)\right| |d z| \le |z|\le 1. \end{aligned}$$

$\square$

Because of its independent interest, we establish the following estimates of coefficients of harmonic mapping f with $f(0)=J_{f}(0)-1=0$ and $\Lambda _{f}(z)\le \Lambda$ for all $z\in {\mathbb {D}}$.

Lemma 2.5

Suppose that $f(z)=h(z)+\overline{g(z)}$ is a harmonic mapping on ${\mathbb {D}}$ with $h(z)=\sum _{n=1}^\infty a_{n}z^{n}$ and $g(z)=\sum _{n=1}^\infty b_{n}z^{n}$ are analytic on ${\mathbb {D}}$, and $f(0)=J_{f}(0)-1=0$, $\Lambda _{f}(z)\le \Lambda$ for all $z\in {\mathbb {D}}$, then $\Lambda \ge 1$, $|a_{1}|+|b_{1}|\le \Lambda$, and

$$\begin{aligned} |a_{n}|+|b_{n}|\le \frac{\Lambda ^{4}-1}{n\Lambda ^3},\,\,\,n=2,3,\ldots . \end{aligned}$$

(2.3)

and

$$\begin{aligned} \frac{1}{\Lambda }\le \lambda _f(0)\le 1. \end{aligned}$$

(2.4)

When $\Lambda =1$, then $f(z)=a_{1}z$ with $|a_1|=1$.

Proof

Since $J_{f}(0)=(|a_{1}|+|b_1|)(|a_{1}|-|b_{1}|)=1$ and $\Lambda _{f}(z)\le \Lambda$ for all $z\in {\mathbb {D}}$, we have

$$\begin{aligned} 0< \frac{1}{|a_{1}|+|b_1|}=|a_{1}|-|b_{1}|\le |a_{1}|+|b_1|=\Lambda _{f}(0)\le \Lambda . \end{aligned}$$

which implies that $\Lambda \ge 1$,

$$\begin{aligned} \lambda _f(0)=||a_{1}|-|b_{1}||=\frac{1}{|a_{1}|+|b_1|}\ge \frac{1}{\Lambda }. \end{aligned}$$

(2.5)

and

$$\begin{aligned} \lambda _f(0)=||a_{1}|-|b_{1}||\le |a_{1}|+|b_1|=\frac{1}{||a_{1}|-|b_1||}\Longrightarrow \lambda _f(0)=||a_{1}|-|b_{1}||\le 1.\nonumber \\ \end{aligned}$$

(2.6)

Fixed $n\in {\mathbb {N}}-\{1\}=\{2,3,\ldots \}$, we choose a real number $\alpha$ such that $|a_n+e^{i\alpha }b_n|=|a_n|+|b_n|$, and set

$$\begin{aligned} F(z)=\frac{1}{\Lambda }[h'(z)+e^{i\alpha }g'(z)]=\frac{a_1+e^{i\alpha }b_1}{\Lambda }+\sum _{n=2}^{\infty }\frac{k(a_k+e^{i\alpha }b_k)}{\Lambda }z^{k-1}. \end{aligned}$$

Since g(z) and h(z) are analytic and $\Lambda _{f}(z)=|h'(z)|+|g'(z)|\le \Lambda$ on ${\mathbb {D}}$, we get F(z) is analytic and $|F(z)|\le \frac{|h'(z)|+|g'(z)|}{\Lambda }\le 1$ on ${\mathbb {D}}$. By Lemma 1.3 in [16] and (2.5), we have

$$\begin{aligned} \Big |\frac{k(a_k+e^{i\alpha }b_k)}{\Lambda }\Big |\le 1-\Big |\frac{a_1+e^{i\alpha }b_1}{\Lambda }\Big |^2\le 1-\frac{||a_1|-|b_1||^2}{\Lambda ^2}\le 1-\frac{1}{\Lambda ^4} \end{aligned}$$

for $k=2,3,\ldots .$ In particular, we have

$$\begin{aligned} n(|a_n|+|b_n|)=n|a_n+e^{i\alpha }b_n|\le \Lambda (1-\frac{1}{\Lambda ^4})=\frac{\Lambda ^4-1}{\Lambda ^3}, \end{aligned}$$

which implies that

$$\begin{aligned} |a_{n}|+|b_{n}|\le \frac{\Lambda ^{4}-1}{n\Lambda ^3},\,\,\,n=2,3,\ldots . \end{aligned}$$

When $\Lambda =1$, we have $|a_{n}|+|b_{n}|\le \frac{\Lambda ^{4}-1}{n\Lambda ^3}=0$ for $n=2,3,\ldots ,$ which implies $a_{n}=b_{n}=0$ for $n=2,3,\ldots$.

Since $0\le |a_{1}|-|b_{1}|\le |a_{1}|+|b_{1}|\le 1$ and $J_f(0)=(|a_{1}|-|b_{1}|)(|a_{1}|+|b_{1}|)=1$, we have $|a_{1}|-|b_{1}|=|a_{1}|+|b_{1}|=1$. Hence $|a_1|=1,\, b_1=0$, and $f(z)=a_{1}z$ with $|a_1|=1$. $\square$

Lemma 2.6

[3, 16, 26]  Let $f(z)=h(z)+\overline{g(z)}=\sum _{n=1}^{\infty }a_{n} z^{n}+\overline{\sum _{n=1}^{\infty }b_{n}z^{n}}$ be a harmonic mapping on the unit disk ${\mathbb {D}}.$

• (i) If $|f(z)|<M$, then

  $$\begin{aligned} \sum _{n=1}^{\infty }\left( \left| a_{n}\right| +\left| b_{n}\right| \right) ^{2} \le 2 M^{2}. \end{aligned}$$

• (ii) If $J_{f}(0)=1$ and $|f(z)|<M$, then

  $$\begin{aligned} \sqrt{\sum _{n=2}^{\infty }\left( \left| a_{n}\right| +\left| b_{n}\right| \right) ^{2}} \le \sqrt{M^{4}-1} \cdot \lambda _{f}(0), \end{aligned}$$

  and $\lambda _f(0)\ge \lambda _0(M)$, where $\lambda _0(M)$ is given by (1.8).

Applying the analogous proof of Lemma 2.5 in [20], we have the following lemma.

Lemma 2.7

Suppose $\Lambda \ge 0$. Let $F(z)=a \Lambda |z|^{2} z+b z$ be a biharmonic mapping of the unit disk ${\mathbb {D}}$ with $|a|=|b|=1 .$ Then F is univalent on the disk ${\mathbb {D}}_{\rho _{1}}$, and $F\left( {\mathbb {D}}_{\rho _{1}}\right)$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{1}}$, where $\rho _1$ and $\sigma _1$ are given by (1.16) and (1.17) respectively. This result is sharp.

Lemma 2.8

[20]  Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}}$ with $G(0)=H(0)=0$ and $\Lambda _{G}(z) \le \Lambda$ for all $z \in {\mathbb {D}},$ where $G(z)=G_{1}(z)+\overline{G_{2}(z)}=\sum _{n=1}^{\infty } a_{n} z^{n}+\sum _{n=1}^{\infty } \overline{b_{n}} \bar{z}^{n}, H(z)=H_{1}(z)+\overline{H_{2}(z)}=$ $\sum _{n=1}^{\infty } c_{n} z^{n}+\sum _{n=1}^{\infty } \overline{d_{n}} \bar{z}^{n}$ are harmonic mappings on ${\mathbb {D}} .$ Then for all $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}(0<r<1)$ with $z_{1} \ne z_{2},$ we have

$$\begin{aligned} \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \ge \left| z_{1}-z_{2}\right| \left[ \left\| c_{1}|-| d_{1}\right\| -\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) n r^{n-1}-3 \Lambda r^{2}\right] . \nonumber \\ \end{aligned}$$

(2.7)

Lemma 2.9

Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}}$ with $G(0)=H(0)=0$ and $\Lambda _{G}(z) \le \Lambda$ for all $z \in {\mathbb {D}},$ where $G(z)=G_{1}(z)+\overline{G_{2}(z)}=\sum _{n=1}^{\infty } a_{n} z^{n}+\sum _{n=1}^{\infty } \overline{b_{n}} \bar{z}^{n}, H(z)=H_{1}(z)+\overline{H_{2}(z)}=$ $\sum _{n=1}^{\infty } c_{n} z^{n}+\sum _{n=1}^{\infty } \overline{d_{n}} \bar{z}^{n}$ are harmonic mappings on ${\mathbb {D}} .$ Then for all $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}(0<r<1)$ with $z_{1} \ne z_{2},$ we have

$$\begin{aligned} \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \le \left| z_{1}-z_{2}\right| \left( \Lambda _{H}(z)+3 \Lambda r^{2}\right) . \end{aligned}$$

(2.8)

Proof

For any $z_{1}, z_{2} \in {\mathbb {D}}_{r}\left( 0<r<1, z_{1} \ne z_{2}\right) ,$ we have

$$\begin{aligned} |G(z)|=\left| \int _{[0, z]} G_{z}(z) d z+G_{\bar{z}}(z) d \bar{z}\right| \le \int _{[0, z]}\left| \Lambda _{G}(z) \Vert d z\right| \le \Lambda \, |z|, \end{aligned}$$

(2.9)

and

$$\begin{aligned} \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right|= & {} \left| \int _{\overline{z_{1}, z_{2}}} F_{z}(z) d z+F_{\bar{z}}(z) d \bar{z}\right| \\= & {} \left| \int _{\overline{z_{1}, z_{2}}}(\bar{z} G(z)+|z|^{2} G_{1}^{\prime }(z)\right. \\&\left. +H_{z}(z)) d z+(z G(z)+|z|^{2} \overline{G_{2}^{\prime }(z)}+H_{\bar{z}}(z)) d \bar{z}\right| \\\le & {} \left| \int _{\overline{z_{1}, z_{2}}} H_{z}(z) d z+H_{z}(z) d \bar{z}\right| \\&+\left| \int _{\overline{z_{1}, z_{2}}}\left( \bar{z} G(z)+|z|^{2} G_{1}^{\prime }(z)\right) d z+(z G(z)+| z|^{2} \overline{G_{2}^{\prime }(z)}) d \bar{z}\right| \\\le & {} \int _{\overline{z_{1}, z_{2}}}(|H_{z}(z)|+|H_{\bar{z}}(z)|)|dz|\\&+\int _{\overline{z_{1}, z_{2}}}(2|z||G(z)|+|z|^2|G_{1}^{\prime }|+|z|^2|G_{2}^{\prime }|)|dz|\\\le & {} |z_1-z_2|\left( \Lambda _H(z)+3\Lambda r^2\right) . \end{aligned}$$

This completes the proof of the lemma. $\square$

Lemma 2.10

[24]  Let G(z) be a harmonic mapping of the unit disk ${\mathbb {D}}$ with $G(0)=0$ and $|G(z)|\le M$. Then for all $z_{1}, z_{2} \in {\mathbb {D}}_{r}(0<r<1)$ with $z_{1} \ne z_{2}$, we have

$$\begin{aligned} ||z_2|^2G(z_2)-|z_1|^2G(z_1)|\le \frac{4M(3r^2-2r^4)}{\pi (1-r^2)}|z_1-z_2|. \end{aligned}$$

Lemma 2.11

[6]  Let G be a harmonic mapping of the unit disk ${\mathbb {D}}$ with $G(0)=0$ and $G({\mathbb {D}})\subset {\mathbb {D}}$. Then

$$\begin{aligned} |G(z)|\le \frac{4}{\pi }\arctan |z|\le \frac{4}{\pi } |z|, \, \text{ for } z\in {\mathbb {D}}. \end{aligned}$$

Lemma 2.12

[10]  Suppose that $f(z)=f_1(z)+\overline{f_2(z)}$ is a harmonic mapping with $f_1(z)=\sum _{n=0}^{\infty }a_nz^n$ and $f_2(z)=\sum _{n=1}^{\infty }b_nz^n$ being analytic in ${\mathbb {D}}$. If $|f(z)|\le M$ for all $z\in {\mathbb {D}}$, then

$$\begin{aligned} \Lambda _f(z)\le \frac{4M}{\pi (1-|z|^2)}. \end{aligned}$$

(2.10)

3 Landau-type theorems of biharmonic mappings

We first prove a new version of Landau-type theorem for biharmonic mappings F(z) under the assumptions $G(0)=H(0)=J_{F}(0)-1=0, \Lambda _{G}(z) \le \Lambda _{1}$ and $\Lambda _{H}(z)<\Lambda _{2}$ for all $z \in {\mathbb {D}}$, which is one of the main results in this paper.

Theorem 3.1

Suppose that $\Lambda _{1} \ge 0$ and $\Lambda _{2}>1 .$ Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of the unit disk ${\mathbb {D}},$ where G(z) and H(z) are harmonic on ${\mathbb {D}},$ satisfying $G(0)=H(0)=J_{F}(0)-1=0, \Lambda _{G}(z) \le \Lambda _{1}$ and $\Lambda _{H}(z)<\Lambda _{2}$ for all $z \in {\mathbb {D}}$. Then $\frac{1}{\Lambda _2}<\lambda _{F}(0)\le 1$, F(z) is univalent on the disk ${\mathbb {D}}_{\rho _{0}}$ and $F\left( {\mathbb {D}}_{\rho _{0}}\right)$ contains a schlicht disk ${\mathbb {D}}_{\sigma _{0}}$, where $\rho _{0}$ is the unique root in (0, 1) of the equation

$$\begin{aligned} \Lambda _2 \frac{\lambda _{H}(0)-\Lambda _2 r}{\Lambda _2-\lambda _{H}(0)r} -3 \Lambda _{1} r^{2}=0, \end{aligned}$$

(3.1)

and

$$\begin{aligned} \sigma _{0}=\frac{\Lambda _{2}^{2}}{\lambda _{H}(0)} \rho _{0}+\left( \frac{\Lambda _{2}^{3}}{\lambda _{H}^2(0)}-\Lambda _{2}\right) \ln \left( 1-\frac{\lambda _{H}(0)\rho _{0}}{\Lambda _{2}}\right) -\Lambda _{1} \rho _{0}^{3}. \end{aligned}$$

(3.2)

This result is sharp for the biharmonic mapping given by (1.15).

Proof

We first prove that F is univalent in the disk ${\mathbb {D}}_{\rho _{0}}$. Indeed, for all $z_{1}, z_{2} \in {\mathbb {D}}_{r}\, (0<r<\rho _{0})$ with $z_{1} \ne z_{2}$, note that $J_{H}(0)=J_{F}(0)=1$ and $\Lambda _{H}(z)<\Lambda _{2}$ for all $z \in {\mathbb {D}}$, we obtain from (2.4), (2.9) and Lemma 2.1 that

$$\begin{aligned} 0<\frac{1}{\Lambda _2}<\lambda _{F}(0)=\lambda _{H}(0)\le 1, \end{aligned}$$

(3.3)

and

$$\begin{aligned} \left| F\left( z_{2}\right) -F\left( z_{1}\right) \right|= & {} \left| \int _{\overline{z_{1} z_{2}}} F_{z}(z) d z+F_{\bar{z}}(z) d \bar{z}\right| \nonumber \\= & {} \left| \int _{\overline{z_{1} z_{2}}}\left( \bar{z} G(z)+|z|^{2} G_{z}(z)+H_{z}(z)\right) d z\right. \nonumber \\&\left. +\left( z G(z)+|z|^{2} G_{\bar{z}}(z)+H_{\bar{z}}(z)\right) d \bar{z}\right| \nonumber \\\ge & {} \left| \int _{\overline{z_{1} z_{2}}} H_{z}(z) d z+H_{\bar{z}}(z) d \bar{z}\right| -\int _{\overline{z_{1} z_{2}}} 3 \Lambda _{1} r^{2}|d z| \nonumber \\\ge & {} \left| z_{1}-z_{2}\right| \left( \Lambda _2 \frac{\lambda _{H}(0)-\Lambda _2 r}{\Lambda _2-\lambda _{H}(0)r}-3 \Lambda _{1} r^{2}\right) . \end{aligned}$$

(3.4)

It is easy to verify that the function

$$\begin{aligned} g_{0}(r):=\Lambda _2 \frac{\lambda _{H}(0)-\Lambda _2 r}{\Lambda _2-\lambda _{H}(0)r}-3 \Lambda _{1} r^{2} \end{aligned}$$

is continuous and strictly decreasing on $[0,1], g_{0}(0)=\lambda _{H}(0)>\frac{1}{\Lambda _2}>0,$ and

$$\begin{aligned} g_{0}(1)=-\left( \Lambda _{2}+3 \Lambda _{1}\right) <0. \end{aligned}$$

Therefore, by the mean value theorem, there is a unique real $\rho _{0} \in (0,1)$ such that $g_{0}\left( \rho _{0}\right) =0 .$ Then, for any $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}\left( 0<r<\rho _{0}\right)$ with $z_{1} \ne z_{2}$, we obtain that

$$\begin{aligned} \left| F\left( z_{2}\right) -F\left( z_{1}\right) \right| \ge \left| z_{1}-z_{2}\right| \left( \Lambda _2 \frac{\lambda _{H}(0)-\Lambda _2 r}{\Lambda _2-\lambda _{H}(0)r}-3 \Lambda _{1} r^{2}\right) >\left| z_{1}-z_{2}\right| g_{0}\left( \rho _{0}\right) =0. \end{aligned}$$

This implies $F\left( z_{1}\right) \ne F\left( z_{2}\right) ,$ which proves the univalence of F in the disk ${\mathbb {D}}_{\rho _{0}}.$

Next, we prove that $\sigma _{0}>0$ and $F\left( {\mathbb {D}}_{\rho _{0}}\right) \supseteq {\mathbb {D}}_{\sigma _{0}}$.

In fact, considering the real differentiable function

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

(3.5)

Since the continuous function

$$\begin{aligned} h'(x)=\frac{\Lambda _{2}^{2}}{\lambda _{H}(0)}-3\Lambda _{1} x^{2}+\frac{\Lambda _{2}\lambda _{H}(0)^2-\Lambda _{2}^{3}}{\lambda _{H}(0)(\Lambda _{2}-\lambda _{H}^2(0)x)} \end{aligned}$$

(3.6)

is strictly decreasing on [0, 1] and $h'(\rho _0)=g_0(\rho _0)=0$, we see that $h'(x)=0$ for $x\in [0,1]$ if and only if $x=\rho _0$. Thus h(x) is strictly increasing on $[0,\rho _0]$ and strictly decreasing on $[\rho _0, 1]$. Since $h(0)=0$, we have

$$\begin{aligned} \sigma _{0}= & {} \frac{\Lambda _{2}^{2}}{\lambda _{H}(0)} \rho _{0}+\left( \frac{\Lambda _{2}^{3}}{\lambda _{H}^2(0)}-\Lambda _{2}\right) \nonumber \\&\ln \left( 1-\frac{\lambda _{H}(0)\rho _{0}}{\Lambda _{2}}\right) -\Lambda _{1} \rho _{0}^{3}=h(\rho _0)>h(0)=0. \end{aligned}$$

(3.7)

In addition, note that $F(0)=0$, for any $z\in \partial {\mathbb {D}}_{\rho _{0}}$, taking $z_0=\rho _0e^{i\theta } \in \partial {\mathbb {D}}_{\rho _{0}}$ with $w_0=F\left( z_0\right) \in F\left( \partial {\mathbb {D}}_{\rho _{0}}\right)$ and $\left| w_0\right| =\min \left\{ |w|: w \in F\left( \partial {\mathbb {D}}_{\rho _{0}}\right) \right\}$. Let $\gamma =F^{-1}(\overline{o w})$, by Lemma 2.2 and (2.9), we have

$$\begin{aligned} |F(z)-F(0)|\ge & {} |w_0|= ||\rho _0e^{i\theta }|^{2} G(\rho _0e^{i\theta })+H(\rho _0e^{i\theta })|\ge |H(\rho _0e^{i\theta })|-\Lambda _{1} \rho _{0}^{3} \\= & {} \left| \int _{\gamma } H_{\zeta }(\zeta ) d \zeta +H_{\bar{\zeta }}(\zeta ) d \bar{\zeta }\right| -\Lambda _{1} \rho _{0}^{3} \\\ge & {} \Lambda _{2} \int _{0}^{\rho _{0}} \frac{\lambda _{H}(0)-\Lambda _{2}t}{\Lambda _{2}-\lambda _{H}(0) t} d t-\Lambda _{1} \rho _{0}^{3} \\= & {} \frac{\Lambda _{2}^{2}}{\lambda _{H}(0)} \rho _{0}+\left( \frac{\Lambda _{2}^{3}}{\lambda _{H}^2(0)}-\Lambda _{2}\right) \ln \left( 1-\frac{\lambda _{H}(0)\rho _{0}}{\Lambda _{2}}\right) -\Lambda _{1} \rho _{0}^{3}=\sigma _{0}. \end{aligned}$$

which implies that $F\left( {\mathbb {D}}_{\rho _{0}}\right) \supseteq {\mathbb {D}}_{\sigma _{0}}$.

Now, we prove the sharpness of $\rho _0$ and $\sigma _0$ for the biharmonic mapping $F_0(z)$ given by (1.15). In fact, 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 $\mathbb {D}_{\rho _0}$, and $F_0(\mathbb {D}_{\rho _0}) \supseteq \mathbb {D}_{\sigma _0}$.

Note that for the biharmonic mapping $F_0(z)$, $\lambda _{H}(0)=\lambda _{F}(0)=J_F(0)=1$, the Eqs. (3.1), (3.2) reduce to (1.13) and (1.14) respectively. Thus we obtain $\rho _0=r_5$ and $\sigma _0=R_5$. By Theorem E, we conclude that $\rho _0$ and $\sigma _0$ are sharp. This completes the proof. $\square$

Now we give an example to show that for each a value $\alpha \in (1/\Lambda _2,\, 1)$, there exits a biharmonic mapping F satisfying the hypothesis of Theorem 3.1 such that $\lambda _{F}(0)=\alpha$.

Example 3.1

Suppose that $\Lambda _{1} \ge 0$, $\Lambda _{2}>1$ and $\alpha \in (1/\Lambda _2,\, 1)$. Let $|a|=1, |b|=\frac{1}{2}(1/\alpha -\alpha )$, and

$$\begin{aligned} |c|=\frac{\sqrt{|b|^2+1}}{|b|}=\frac{1+\alpha ^2}{1-\alpha ^2}. \end{aligned}$$

Consider the biharmonic mapping

$$\begin{aligned} F(z)=a\Lambda _{1} |z|^2z+b(cz+{\overline{z}}),\quad z\in {\mathbb {D}}. \end{aligned}$$

Then F(z) satisfies the hypothesis of Theorem 3.1, $\lambda _{F}(0)=\alpha$, F(z) is univalent on ${\mathbb {D}}_{\rho _{0}''}$, and $F({\mathbb {D}}_{\rho _{0}''})$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{0}''}$, where

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

(3.8)

and

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

(3.9)

and when $\arg c =\pi - \arg \frac{b}{a}$, this result is sharp.

Proof

Set $G(z)=a\Lambda _{1} |z|^2z, H(z)=b(cz+{\overline{z}})$. Direct computation yields

$$\begin{aligned}&G(0)=H(0)=0, \Lambda _{G}(z)=|a\Lambda _{1}|=\Lambda _{1}\le \Lambda _{1},\\&\Lambda _{H}(z)=|bc|+|c|=|b|(|c|+1)=1/\alpha <\Lambda _{2}, \end{aligned}$$

and $J_{F}(0)=|b|^2(|c|^2-1)=1$, thus F(z) satisfies the hypothesis of Theorem 3.1, and

$$\begin{aligned} \lambda _{F}(0)=|b|(|c|-1)=\alpha . \end{aligned}$$

Applying the analogous proof of Lemma 2.5 in [20] (please also refer to example 2.1 in [3]), we may verify that if $\Lambda _{1}\le \frac{\alpha }{3}$, then F(z) is univalent on ${\mathbb {D}},$ and $F({\mathbb {D}})$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{0}''}$, where

$$\begin{aligned} \sigma _{0}''= \alpha -\Lambda _1. \end{aligned}$$

If $\Lambda _{1}>\frac{\alpha }{3},$ then F(z) is univalent on ${\mathbb {D}}_{\rho _{0}''}$, and $F({\mathbb {D}}_{\rho _{0}''})$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{0}''},$, where

$$\begin{aligned} \rho _{0}''=\sqrt{\frac{\alpha }{3\Lambda _{1}}},\quad \sigma _{0}'' =\frac{2\alpha }{3}\sqrt{\frac{\alpha }{3\Lambda _{1}}}, \end{aligned}$$

and when $\arg c =\pi - \arg \frac{b}{a}$, the radii $\rho _{0}''$ and $\sigma _{0}''$ are sharp. $\square$

Remark 3.1

For the biharmonic mapping F(z) of the unit disk ${\mathbb {D}}$ with $J_{F}(0)=1$ and $\Lambda _{H}(z) \le \Lambda _{2}$, it follows from Lemma 2.5 that $\Lambda _{2} \ge 1$. Theorem 3.1 provides a sharp version of Landau-type theorem of biharmonic mappings for the case $J_F(0)=1, \Lambda _{1} \ge 0$ and $\Lambda _{2}>1$. If $J_F(0)=1, \Lambda _{1} \ge 0$ and $\Lambda _{2}=1$, then we prove Theorem 3.2 using Lemmas 2.3, 2.4 and 2.7, which is the sharp version of Landau-type theorem of biharmonic mappings and is also one of the main results in this paper.

Theorem 3.2

Suppose that $\Lambda \ge 0$. Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of ${\mathbb {D}},$ where G(z), H(z) are harmonic on ${\mathbb {D}},$ satisfying $G(0)=H(0)=J_F(0)-1=0$, $\Lambda _{G}(z) \le \Lambda$ and $\Lambda _H(z)\le 1$ or $|H(z)|\le 1$ for all $z \in {\mathbb {D}}$. Then F is univalent on the disk ${\mathbb {D}}_{\rho _{1}}$, and $F\left( {\mathbb {D}}_{\rho _{1}}\right)$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{1}}$, where $\rho _1$ and $\sigma _1$ are given by (1.16) and (1.17) respectively. This result is sharp.

Proof

Because $F(z)=|z|^{2} G(z)+H(z)$ satisfies the hypothesis of Theorem 3.2, 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 ${\mathbb {D}}$. Then

$$\begin{aligned} J_{H}(0)=J_{F}(0)=|c_{1}|^2-|d_{1}|^2=1. \end{aligned}$$

By the hypothesis of Theorem 3.2 and Lemmas 2.3 and 2.4, we have

$$\begin{aligned} H(z)=c_{1} z,\quad |c_{1}|=1. \end{aligned}$$

Now we prove that F is univalent in the disk ${\mathbb {D}}_{\rho _{1}}$, where

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

To this end, for any $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}\left( 0<r<\rho _{1}\right)$ with $z_{1} \ne z_{2}$, by (3.4), we have

$$\begin{aligned} \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right|&\ge \left| \int _{\overline{z_{1} z_{2}}} H_{z}(z) d z+H_{\bar{z}}(z) d \bar{z}\right| -\int _{\overline{z_{1} z_{2}}} 3 \Lambda _{1} r^{2}|d z| \\&= \left| z_{1}-z_{2}\right| \left( |c_{1}|-3 \Lambda r^{2}\right) \\&=\left| z_{1}-z_{2}\right| \left( 1-3 \Lambda r^{2}\right) >0. \end{aligned}$$

Then, we have $F\left( z_{1}\right) \ne F\left( z_{2}\right) ,$ which proves the univalence of F in the disk ${\mathbb {D}}_{\rho _{1}}$.

Noting that $F(0)=0,$ for any $z=\rho _{1} e^{i \theta } \in \partial {\mathbb {D}}_{\rho _{1}}$, we have

$$\begin{aligned} |F(z)-F(0)|&=\left. || z\right| ^{2} G(z)+H(z)|\ge |H(z)|-\rho _{1}^{2}| G(z) | \\&=\rho _{1}|c_{1}|-\Lambda \rho _{1}^{3}=\rho _{1}-\Lambda \rho _{1}^{3}=\sigma _{1}. \end{aligned}$$

Hence, $F\left( {\mathbb {D}}_{\rho _{1}}\right)$ contains a schlicht disk ${\mathbb {D}}_{\sigma _{1}}.$

Finally, for $F(z)=a_{1} \Lambda |z|^{2} z+c_1 z$ with $\left| a_{1}\right| =\left| c_{1}\right| =1,$ we have $G(z)=$ $a_{1} \Lambda z, H(z)=c_1 z.$ Direct computation yields

$$\begin{aligned} G(0)=H(0)=0, J_{F}(0)=|c_1|=1, \Lambda _{G}(z)=\left| a_{1} \Lambda \right| \le \Lambda . \end{aligned}$$

and $|H(z)|=|c_1 z| \le 1$ for all $z \in {\mathbb {D}}$. Applying Lemma 2.7, we obtain that the radii $\rho _{1}$ and $\sigma _{1}$ are sharp. This completes the proof. $\square$

Next, we establish another new version of Landau-type theorem for biharmonic mappings F(z) under the assumptions $G(0)=H(0)=J_{F}(0)-1=0, \Lambda _{G}(z) \le \Lambda$ and $|H(z)|\le M, (M>1)$ for all $z \in {\mathbb {D}}$.

Theorem 3.3

Suppose that $\Lambda \ge 0,\, M>1$. Let $F(z)=|z|^{2} G(z)+H(z)$ be a biharmonic mapping of ${\mathbb {D}},$ where G(z), H(z) are harmonic on ${\mathbb {D}},$ satisfying $G(0)=H(0)=J_F(0)-1=0$, $\Lambda _{G}(z) \le \Lambda$ and $|H(z)|\le M$ for all $z \in {\mathbb {D}}$. Then F is univalent on the disk ${\mathbb {D}}_{\rho _2}$, where $\rho _{2}$ is the minimum positive root in (0, 1) of the equation

$$\begin{aligned} \lambda _0(M)-\lambda _{0}(M) \sqrt{M^{4}-1} \sqrt{\frac{4 r^{2}-3 r^{4}+r^{6}}{\left( 1-r^{2}\right) ^{3}}}-3 \Lambda r^{2}=0, \end{aligned}$$

(3.10)

and $F\left( {\mathbb {D}}_{\rho _{2}}\right)$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{2}},$ where

$$\begin{aligned} \sigma _{2}=\lambda _{0}(M)\rho _2-\lambda _{0}(M) \sqrt{M^{4}-1}\frac{\rho _2^{2}}{\sqrt{1-\rho _2^{2}}}-\rho _{2}^{3}\Lambda , \end{aligned}$$

(3.11)

where $\lambda _0(M)$ is given by (1.8).

Proof

By the hypothesis of Theorem 3.3, we can assume that

$$\begin{aligned} H(z)=\sum _{n=1}^{\infty }c_{n} z^{n}+\overline{\sum _{n=1}^{\infty }d_{n}z^{n}},z\in {\mathbb {D}}. \end{aligned}$$

Since $J_{H}(0)=J_{F}(0)=1$ and $|H(z)|\le M$, by Lemma 2.6, we have

$$\begin{aligned} \sqrt{\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) ^{2}} \le \sqrt{M^{4}-1} \cdot \lambda _{H}(0), \end{aligned}$$

and $\lambda _H(0)\ge \lambda _0(M),$ where $\lambda _0(M)$ is given by (1.8).

Now we prove that F is univalent in the disk ${\mathbb {D}}_{\rho _{2}}.$ For all $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}\left( 0<r<\rho _{2}, z_{1} \ne z_{2}\right) ,$ we obtain from Lemmas 2.8 and 2.6 that

$$\begin{aligned}&\left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \\&\quad \ge \left| z_{1}-z_{2}\right| \left[ \left\| c_{1}|-| d_{1}\right\| -\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) n r^{n-1}-3 \Lambda r^{2}\right] \\&\quad \ge \left| z_{1}-z_{2}\right| \left[ \lambda _{H}(0)-\left( \sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) ^{2}\right) ^{1 / 2}\left( \sum _{n=2}^{\infty } n^{2} r^{2 n-2}\right) ^{1 / 2}-3 \Lambda r^{2}\right] \\&\quad \ge \left| z_{1}-z_{2}\right| \left( \lambda _{H}(0)-\lambda _{H}(0) \sqrt{M^{4}-1} \sqrt{\frac{4 r^{2}-3 r^{4}+r^{6}}{\left( 1-r^{2}\right) ^{3}}}-3 \Lambda r^{2}\right) \\&\quad \ge \left| z_{1}-z_{2}\right| \left( \lambda _0(M)-\lambda _{0}(M) \sqrt{M^{4}-1} \sqrt{\frac{4 r^{2}-3 r^{4}+r^{6}}{\left( 1-r^{2}\right) ^{3}}}-3 \Lambda r^{2}\right) >0. \end{aligned}$$

Then, we have $F\left( z_{1}\right) \ne F\left( z_{2}\right) ,$ which proves the univalence of F in the disk ${\mathbb {D}}_{\rho _{2}}$.

Noting that $F(0)=0,$ for any $z=\rho _{2} e^{i \theta } \in \partial {\mathbb {D}}_{\rho _{2}}$, by (2.9), we have

$$\begin{aligned} |F(z)|&=\left. || z\right| ^{2} G(z)+H(z)|\ge | H(z)|-\rho _{2}^{2}| G(z) | \\&\ge \left\| c_{1}|-| d_{1}\right\| \rho _2-\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) \rho _2^{n}-\rho _{2}^{3}\Lambda \\&\ge \lambda _{H}(0)\rho _2-\left( \sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) ^{2}\right) ^{\frac{1}{2}}\left( \sum _{n=2}^{\infty } \rho _2^{2 n}\right) ^{\frac{1}{2}}-\rho _{2}^{3}\Lambda \\&\ge \lambda _{H}(0)\rho _2-\lambda _{H}(0) \sqrt{M^{4}-1}\frac{\rho _2^{2}}{\sqrt{1-\rho _2^{2}}}-\rho _{2}^{3}\Lambda \\&\ge \lambda _{0}(M)\rho _2-\lambda _{0}(M) \sqrt{M^{4}-1}\frac{\rho _2^{2}}{\sqrt{1-\rho _2^{2}}}-\rho _{2}^{3}\Lambda =\sigma _{2}. \end{aligned}$$

Hence, $F\left( {\mathbb {D}}_{\rho _{2}}\right)$ contains a schlicht disk ${\mathbb {D}}_{\sigma _{2}}.$ This completes the proof (Table 1). $\square$

Table 1 The values of $\rho _2,\sigma _2$ are in Theorem 3.3

Now, we will consider the Landau-type theorem for the case $|G(z)|\le M,\, \Lambda _H(z)<\Lambda .$

Theorem 3.4

Suppose that $M\ge 0, \Lambda \ge 1$, $F(z)=|z|^2G(z)+H(z)$ is a biharmonic mapping on the unit disk ${\mathbb {D}}$, where G(z), H(z) are harmonic mappings on ${\mathbb {D}}$, satisfying $G(0)=H(0)=J_F(0)-1=0$ and $|G(z)|\le M, \, \Lambda _H(z)<\Lambda$ for $z\in {\mathbb {D}}$.

• (i) If $M\ge 0, \Lambda >1$ or $M>0, \Lambda =1$, then F(z) is univalent on ${\mathbb {D}}_{\rho _3}$, where $\rho _{3}$ is the minimum positive root in (0, 1) of the equation

  $$\begin{aligned} \frac{\Lambda (1-\Lambda ^2 r)}{\Lambda ^2-r}-\frac{4M}{\pi }\frac{3r^2-2r^4}{1-r^2}=0, \end{aligned}$$

  (3.12)

  and $F({\mathbb {D}}_{\rho _3})$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _3}$, where

  $$\begin{aligned} \sigma _{3}=\Lambda ^{3} \rho _{3}+\left( \Lambda ^{5}-\Lambda \right) \ln \left( 1-\frac{\rho _{3}}{\Lambda ^2}\right) - \frac{4M}{\pi }\rho _3^3. \end{aligned}$$

  (3.13)

• (ii) If $M=0, \Lambda =1$, then F(z) is univalent on ${\mathbb {D}}$ and $F({\mathbb {D}})= {\mathbb {D}}$.

Proof

(i) Note that $J_{H}(0)=J_{F}(0)=1$, we split into two case to prove.

Case 1. When $M\ge 0, \Lambda >1$. We first prove that F is univalent on the disk ${\mathbb {D}}_{\rho _{3}}$. To this end, for all $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r} \left( 0<r<\rho _{3}, z_{1} \ne z_{2}\right)$, we obtain from Lemmas 2.1 and 2.10 that

$$\begin{aligned} |F(z_2)-F(z_1)|&=|(|z_2|^2G(z_2)+H(z_2))-(|z_1|^2G(z_1)+H(z_1))|\\&\ge |H(z_2)-H(z_1)|-||z_2|^2G(z_2)-|z_1|^2G(z_1)|\\&\ge |z_2-z_1|\left( \Lambda \frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}-\frac{4M}{\pi }\frac{3r^2-2r^4}{1-r^2}\right) . \end{aligned}$$

Since $\Lambda \frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}$ is continuous and increasing about $\lambda _{H}(0)$ and by (3.3), we have

$$\begin{aligned} |F(z_2)-F(z_1)|&\ge |z_2-z_1|\left( \Lambda \frac{\lambda _{H}(0)-\Lambda r}{\Lambda -\lambda _{H}(0)r}-\frac{4M}{\pi }\frac{3r^2-2r^4}{1-r^2}\right) \\&>\frac{\Lambda (1-\Lambda ^2 r)}{\Lambda ^2-r}-\frac{4M}{\pi }\frac{3r^2-2r^4}{1-r^2}>0. \end{aligned}$$

This shows that F is univalent on the disk ${\mathbb {D}}_{\rho _{3}}$.

Next, we prove $F({\mathbb {D}}_{\rho _3})\supset \mathbb {{\mathbb {D}}}_{\sigma _3}$. For $z=\rho _3e^{i\theta }\in \partial {\mathbb {D}}_{\rho _3}$, by Lemmas 2.2 and 2.11, we have

$$\begin{aligned} |F(z)|&\ge \Lambda \int _{0}^{\rho _3} \frac{\lambda _{H}(0)-\Lambda t}{\Lambda -\lambda _{H}(0)t} d t- \frac{4M}{\pi }\rho _3^3\\&\ge \Lambda \int _{0}^{\rho _{3}} \frac{1-\Lambda ^2t}{\Lambda ^2-t} d t-\frac{4M}{\pi }\rho _3^3 \\&=\Lambda ^{3} \rho _{3}+\left( \Lambda ^{5}-\Lambda \right) \ln \left( 1-\frac{\rho _{3}}{\Lambda ^2}\right) -\frac{4M}{\pi }\rho _3^3=\sigma _{3}. \end{aligned}$$

Case 2. When $M>0, \Lambda =1$. Using Lemma 2.5, we have

$$\begin{aligned} H(z)=c_{1} z, |c_{1}|=1. \end{aligned}$$

Similarly, we first prove that F is univalent on the disk ${\mathbb {D}}_{\rho _{3}}$. In fact, for all $z_{1}, z_{2} \in {\mathbb {D}}_{r}\left( 0<r<\rho _{3}, z_{1} \ne z_{2}\right)$, we have

$$\begin{aligned} |F(z_2)-F(z_1)|&\ge |H(z_2)-H(z_1)|-||z_2|^2G(z_2)-|z_1|^2G(z_1)|\\&\ge |z_2-z_1|\left( 1-\frac{4M}{\pi }\frac{3r^2-2r^4}{1-r^2}\right) >0. \end{aligned}$$

This shows that F is univalent on the disk ${\mathbb {D}}_{\rho _{3}}$.

Next, we prove $F({\mathbb {D}}_{\rho _3})\supset \mathbb {{\mathbb {D}}}_{\sigma _3}$. For $z=\rho _3e^{i\theta }\in \partial {\mathbb {D}}_{\rho _3}$, by Lemma 2.11, we have

$$\begin{aligned} |F(z)| \ge |H(z)|-\rho _{3}^{2} |G(z)| =\rho _3- \frac{4M}{\pi }\rho _3^3=\sigma _3. \end{aligned}$$

Hence, $F\left( {\mathbb {D}}_{\rho _{3}}\right)$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{3}}.$

Now we prove (ii). If $M=0, \Lambda =1$, by Lemma 2.5, we have

$$\begin{aligned} F(z)=c_1z ,|c_1|=1. \end{aligned}$$

It’s easy to verify that F(z) is univalent on the unit disk ${\mathbb {D}}$, and $F({\mathbb {D}})={\mathbb {D}}$. This completes the proof (Table 2). $\square$

Table 2 The values of $\rho _3,\sigma _3$ are in Theorem 3.4

Finally, we improve Theorem D as follows.

Theorem 3.5

Suppose that $M_1\ge 0, M_2\ge 1$, $F(z)=|z|^2G(z)+H(z)$ is a biharmonic mapping on the unit disk ${\mathbb {D}}$, where G(z), H(z) are harmonic mappings on ${\mathbb {D}}$, satisfying $G(0)=H(0)=J_F(0)-1=0$ and $|G(z)|\le M_1$, $|H(z)|\le M_2$ for $z\in {\mathbb {D}}$.

• (i) If $M_1\ge 0, M_2>1$ or $M_1>0, M_2=1$, then F(z) is univalent on ${\mathbb {D}}_{\rho _4}$, where $\rho _{4}$ is the minimum positive root in (0, 1) of the equation

  $$\begin{aligned} \lambda _0(M_2)-\lambda _{0}(M_2) \sqrt{M_2^{4}-1} \sqrt{\frac{4 r^{2}-3 r^{4}+r^{6}}{\left( 1-r^{2}\right) ^{3}}}-\frac{4M_1}{\pi }\frac{3r^2-2r^4}{1-r^2}=0,\qquad \end{aligned}$$

  (3.14)

  and $F\left( {\mathbb {D}}_{\rho _{4}}\right)$ contains a Schlicht disk ${\mathbb {D}}_{\sigma _{4}},$ where

  $$\begin{aligned} \sigma _{4}=\lambda _{0}(M_2)\rho _4-\lambda _{0}(M_2) \sqrt{M_2^{4}-1}\frac{\rho _4^{2}}{\sqrt{1-\rho _4^{2}}}-\frac{4M_1}{\pi }\rho _4^3, \end{aligned}$$

  (3.15)

  where $\lambda _0(M)$ is given by (1.8).

• (ii) If $M_1=0, M_2=1$, then F(z) is univalent on ${\mathbb {D}}$ and $F({\mathbb {D}}) = {\mathbb {D}}$.

Proof

By the hypothesis of Theorem 3.5, we can assume that

$$\begin{aligned} H(z)=\sum _{n=1}^{\infty }c_{n} z^{n}+\overline{\sum _{n=1}^{\infty }d_{n}z^{n}},z\in {\mathbb {D}}. \end{aligned}$$

Since $|H(z)|\le M_2$ and $J_{H}(0)=J_{F}(0)=1,$ by Lemma 2.6, we have

$$\begin{aligned} \sqrt{\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) ^{2}} \le \sqrt{M_2^{4}-1} \cdot \lambda _{H}(0), \end{aligned}$$

and $\lambda _H(0)\ge \lambda _0(M_2)$, where $\lambda _0(M_2)$ is given by (1.8).

Now we prove that F is univalent in the disk ${\mathbb {D}}_{\rho _{4}}.$ For all $z_{1}, z_{2} \in$ ${\mathbb {D}}_{r}\left( 0<r<\rho _{4}, z_{1} \ne z_{2}\right)$, by Lemma 2.10, we have

$$\begin{aligned}&\left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \\&\quad \ge |H(z_2)-H(z_1)|-||z_2|^2G(z_2)-|z_1|^2G(z_1)|\\&\quad \ge \left| \int _{\overline{z_{1}, z_{2}}} H_{z}(0) d z+H_{\overline{z}}(0) d \bar{z}\right| -\left| \int _{\overline{z_{1}, z_{2}}}( H_{z}(z)- H_{z}(0))d z+(H_{\overline{z}}(z)-H_{\overline{z}}(0) )d \bar{z}\right| \\&\qquad -\Big ||z_2|^2G(z_2)-|z_1|^2G(z_1)\Big |\\&\quad \ge \left| z_{1}-z_{2}\right| \left[ \left\| c_{1}|-| d_{1}\right\| -\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) n r^{n-1}\right] -\Big ||z_2|^2G(z_2)-|z_1|^2G(z_1)\Big |\\&\quad \ge \left| z_{1}-z_{2}\right| \left[ \lambda _{H}(0)-\left( \sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) ^{2}\right) ^{1 / 2}\left( \sum _{n=2}^{\infty } n^{2} r^{2 n-2}\right) ^{1 / 2}\right] \\&\qquad -\Big ||z_2|^2G(z_2)-|z_1|^2G(z_1)\Big |\\&\quad \ge \left| z_{1}-z_{2}\right| \left( \lambda _{H}(0)-\lambda _{H}(0) \sqrt{M_2^{4}-1} \sqrt{\frac{4 r^{2}-3 r^{4}+r^{6}}{\left( 1-r^{2}\right) ^{3}}}-\frac{4M_1}{\pi }\frac{3r^2-2r^4}{1-r^2}\right) \\&\quad \ge \left| z_{1}-z_{2}\right| \left( \lambda _0(M_2)-\lambda _{0}(M_2) \sqrt{M_2^{4}-1} \sqrt{\frac{4 r^{2}-3 r^{4}+r^{6}}{\left( 1-r^{2}\right) ^{3}}}-\frac{4M_1}{\pi }\frac{3r^2-2r^4}{1-r^2}\right) \\&\quad >0. \end{aligned}$$

Then, we have $F\left( z_{1}\right) \ne F\left( z_{2}\right) ,$ which proves the univalence of F in the disk ${\mathbb {D}}_{\rho _{4}}$.

Noting that $F(0)=0,$ for any $z=\rho _{4} e^{i \theta } \in \partial {\mathbb {D}}_{\rho _{4}}$, by Lemmas 2.6 and 2.11, we have

$$\begin{aligned} |F(z)|&=\left. || z\right| ^{2} G(z)+H(z)|\ge |H(z)|-\rho _{5}^{2}|G(z)| \\&\ge \left\| c_{1}|-| d_{1}\right\| \rho _4-\sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) \rho _4^{n}- \frac{4M_1}{\pi }\rho _4^3\\&\ge \lambda _{H}(0)\rho _4-\left( \sum _{n=2}^{\infty }\left( \left| c_{n}\right| +\left| d_{n}\right| \right) ^{2}\right) ^{\frac{1}{2}}\left( \sum _{n=2}^{\infty } \rho _4^{2 n}\right) ^{\frac{1}{2}}- \frac{4M_1}{\pi }\rho _4^3\\&\ge \lambda _{H}(0)\rho _4-\lambda _{H}(0) \sqrt{M_2^{4}-1}\frac{\rho _4^{2}}{\sqrt{1-\rho _4^{2}}}- \frac{4M_1}{\pi }\rho _4^3\\&\ge \lambda _{0}(M_2)\rho _4-\lambda _{0}(M_2) \sqrt{M_2^{4}-1}\frac{\rho _4^{2}}{\sqrt{1-\rho _4^{2}}}- \frac{4M_1}{\pi }\rho _4^3 =\sigma _{4}. \end{aligned}$$

Hence, $F\left( {\mathbb {D}}_{\rho _{4}}\right)$ contains a schlicht disk ${\mathbb {D}}_{\sigma _{4}}.$

Finally, if $M_1=0, M_2=1$, then by Lemma 2.3, we have

$$\begin{aligned} F(z)=c_1z ,|c_1|=1. \end{aligned}$$

It is evident that F(z) is univalent on ${\mathbb {D}}$, and $F({\mathbb {D}})={\mathbb {D}}$. This completes the proof.

$\square$

Remark 3.2

Note that for $r=\rho _4$, we have

$$\begin{aligned} \frac{4M_1}{\pi }\frac{3r^2-2r^4}{1-r^2}=\frac{4M_1 r^2}{\pi (1-r^2)}+\frac{8M_1 r^2}{\pi }<\frac{4M_1 r^2}{\pi (1-r^2)}+2M_1 r, \end{aligned}$$

it is easy to verify that $\rho _4>r_4,\, \sigma _4>R_4$, where $r_4, R_4$ are given in Theorem D, please also see Table 3.

The Computer Algebra System Mathematica has calculated the numerical solutions to Eqs. (1.11) and (3.14). From Table 3 as follow, it is easy to see that the result of Theorem 3.5 is better than that of Theorem D. From Table 1, Table 2 and Table 3, it is easy to see that both of the result of Theorem 3.3 and that of Theorem 3.4 are better than that of Theorem 3.5.

Table 3 The values of $r_4, R_4$ and $\rho _4,\sigma _4$ are in Theorems D and 3.5 respectively

4 The bi-Lipschitz theorems of biharmonic mappings

In this section, we will establish the Lipschitz characters of certain biharmonic mappings in their univalent disks.

Theorem 4.1

Suppose F(z) satisfies the hypothesis of Theorem 3.1. Then for each $r_0\in (0, \rho _0)$, the biharmonic mapping F(z) is bi-Lipschitz on $\overline{{\mathbb {D}}}_{r_0}$, where $\rho _{0}$ is given by (3.1).

Proof

Fixed $r_0\in (0, \rho _0)$, set

$$\begin{aligned} l_{0}= \Lambda _2 \frac{\lambda _{H}(0)-\Lambda _2 r_0}{\Lambda _2-\lambda _{H}(0)r_0}-3 \Lambda _{1} r_0^{2}. \end{aligned}$$

Then, for any $z_{1}, z_{2} \in \overline{{\mathbb {D}}}_{r_0}$, it follows from the proof of Theorem 3.1 and Lemma 2.9 that $l_{0}>g_0(\rho _0)=0$, and

$$\begin{aligned} l_{0}\left| z_{1}-z_{2}\right|= & {} \Big (\Lambda _2 \frac{\lambda _{H}(0)-\Lambda _2 r_0}{\Lambda _2-\lambda _{H}(0)r_0} -3 \Lambda _{1} r_0^{2}\Big )|z_1-z_2|\\\le & {} \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \\\le & {} \left( \Lambda _{H}(z)+3 \Lambda _1 r_0^{2}\right) \left| z_{1}-z_{2}\right| \\\le & {} \left( \Lambda _2+3 \Lambda _1 r_0^{2}\right) \left| z_{1}-z_{2}\right| . \end{aligned}$$

Hence f is bi-Lipschitz on $\overline{{\mathbb {D}}}_{r_0}$. $\square$

By means of Theorems 3.2–3.5 and Lemmas 2.9 and 2.12, using the analogous proof of Theorem 4.1, we have the following four theorems.

Theorem 4.2

Suppose F(z) satisfies the hypothesis of Theorem 3.2. Then for each $r_1\in (0, \rho _1)$, the biharmonic mapping F(z) is bi-Lipschitz on $\overline{{\mathbb {D}}}_{r_1}$, i.e. for any $z_{1}, z_{2} \in \overline{{\mathbb {D}}}_{r_1}$, there exists $l_{1}=1-3 \Lambda r_1^{2}>0$ such that

$$\begin{aligned} l_{1}\left| z_{1}-z_{2}\right| \le \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \le \left( \frac{4}{\pi (1-r_1^2)}+3 \Lambda r_1^{2}\right) \left| z_{1}-z_{2}\right| , \end{aligned}$$

where $\rho _1$ is given by (1.16).

Theorem 4.3

Suppose F(z) satisfies the hypothesis of Theorem 3.3. Then for each $r_2\in (0, \rho _2)$, the biharmonic mapping F(z) is bi-Lipschitz on $\overline{{\mathbb {D}}}_{r_2}$, i.e. for any $z_{1}, z_{2} \in \overline{{\mathbb {D}}}_{r_2}$, there exists

$$\begin{aligned} l_2=\lambda _0(M)-\lambda _{0}(M) \sqrt{M^{4}-1} \sqrt{\frac{4 r_2^{2}-3 r_2^{4}+r_2^{6}}{\left( 1-r_2^{2}\right) ^{3}}}-3 \Lambda r_2^{2}>0 \end{aligned}$$

such that

$$\begin{aligned} l_2\left| z_{1}-z_{2}\right| \le \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \le \left( \frac{4M}{\pi (1-r_2^2)}+3 \Lambda r_2^{2}\right) \left| z_{1}-z_{2}\right| , \end{aligned}$$

where $\rho _2$ is given by (3.10) and $\lambda _0(M)$ is given by (1.8).

Theorem 4.4

Suppose F(z) satisfies the hypothesis of Theorem 3.4. Then for each $r_3\in (0, \rho _3)$, the biharmonic mapping F(z) is bi-Lipschitz on $\overline{{\mathbb {D}}}_{r_3}$, i.e. for any $z_{1}, z_{2} \in \overline{{\mathbb {D}}}_{r_3}$, there exists

$$\begin{aligned} l_3=\frac{\Lambda (1-\Lambda ^2 r_3)}{\Lambda ^2-r_3}-\frac{4M}{\pi }\frac{3r_3^2-2r_3^4}{1-r_3^2}>0 \end{aligned}$$

such that

$$\begin{aligned} l_3\left| z_{1}-z_{2}\right| \le \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \le \left( \Lambda +\frac{12M}{\pi (1-r_3^2)} r_3^{2}\right) \left| z_{1}-z_{2}\right| , \end{aligned}$$

where $\rho _3$ is given by (3.12).

Theorem 4.5

Suppose F(z) satisfies the hypothesis of Theorem 3.5. Then for each $r_4\in (0, \rho _4)$, the biharmonic mapping F(z) is bi-Lipschitz on $\overline{{\mathbb {D}}}_{r_4}$, i.e. for any $z_{1}, z_{2} \in \overline{{\mathbb {D}}}_{r_4}$, there exists

$$\begin{aligned} l_4=\lambda _0(M_2)-\lambda _{0}(M_2) \sqrt{M_2^{4}-1} \sqrt{\frac{4 r_4^{2}-3 r_4^{4}+r_4^{6}}{\left( 1-r_4^{2}\right) ^{3}}}-\frac{4M_1}{\pi }\frac{3r_4^2-2r_4^4}{1-r_4^2}>0 \end{aligned}$$

such that

$$\begin{aligned} l_4\left| z_{1}-z_{2}\right| \le \left| F\left( z_{1}\right) -F\left( z_{2}\right) \right| \le \left( \frac{4M_2}{\pi (1-r_4^2)}+\frac{12M_1}{\pi (1-r_4^2)} r_4^{2}\right) \left| z_{1}-z_{2}\right| , \end{aligned}$$

where $\rho _4$ is given by (3.14) and $\lambda _0(M_2)$ is given by (1.8).