Generalized harmonic functions and Schwarz lemma for biharmonic mappings

Abstract

In this paper, we establish some Schwarz type lemmas for mappings $\Phi$ satisfying the inhomogeneous biharmonic Dirichlet problem $\Delta (\Delta (\Phi )) = g$ in ${\mathbb D}$, $\Phi =f$ on ${\mathbb T}$ and $\partial _n \Phi =h$ on ${\mathbb T}$, where g is a continuous function on $\overline{{\mathbb D}}$, f, h are continuous functions on ${\mathbb T}$, where ${\mathbb D}$ is the unit disc of the complex plane ${\mathbb C}$ and ${\mathbb T}=\partial {\mathbb D}$ is the unit circle. To reach our aim, we start by investigating some properties of generalized harmonic functions called $T_\alpha$-harmonic functions. Finally, we prove a Landau-type theorem for this class of functions, when $\alpha >0$.

1 Preliminaries and main results

Let $\mathbb {C}$ denote the complex plane and $\mathbb {D}$ the open unit disk in $\mathbb {C}$. Let $\mathbb {T}=\partial \mathbb {D}$ be the boundary of $\mathbb {D}$, and $\overline{\mathbb {D}} = {\mathbb D}\cup \mathbb {T}$, the closure of $\mathbb {D}$. Furthermore, we denote by $\mathcal {C}^m(\Omega )$ the set of all complex-valued $m-$times continuously differentiable functions from $\Omega$ into $\mathbb {C}$, where $\Omega$ stands for a domain of $\mathbb {C}$ and $m \in \mathbb {N}$. In particular, $\mathcal {C}(\Omega ):= \mathcal {C}^0(\Omega )$ denotes the set of all continuous functions in $\Omega$.

For a real $2 \times 2$ matrix A, we use the matrix norm

$$\begin{aligned} \Vert A\Vert =\sup \lbrace |Az| : |z|=1\rbrace , \end{aligned}$$

and the matrix function

$$\begin{aligned} \lambda (A)=\inf \lbrace |Az| : |z|=1 \rbrace . \end{aligned}$$

For $z = x+iy \in {\mathbb C}$, the formal derivative of a complex-valued function $\Phi = u+iv$ is given by

$$\begin{aligned} D_\Phi =\begin{pmatrix} u_x &{}&{} u_y\\ v_x &{}&{} v_y \end{pmatrix}, \end{aligned}$$

so that

$$\begin{aligned} \Vert D_\Phi \Vert =|\Phi _z|+|\Phi _{\overline{z}}|\ \ \ \text {and}\ \ \ \lambda (D_\Phi )=\big ||\Phi _z|-|\Phi _{\overline{z}}|\big |, \end{aligned}$$

where

$$\begin{aligned} \Phi _z=\frac{1}{2}(\Phi _x-i\Phi _y)\ \ \ \text {and}\ \ \ \ \Phi _{\overline{z}}=\frac{1}{2}(\Phi _x+i\Phi _y). \end{aligned}$$

We use

$$\begin{aligned} J_\Phi := \det D_\Phi = |\Phi _z|^2-|\Phi _{\overline{z}}|^2. \end{aligned}$$

The main objective of this paper is to establish a Schwarz-type lemma for the solutions to the following inhomogeneous biharmonic Dirichlet problem (briefly, IBDP):

$$\begin{aligned} \left\{ \begin{array}{cccr} \Delta ^2 \Phi &{}=&{} g &{}\text{ in } {\mathbb D},\\ \Phi \ &{}=&{} f &{}\text{ on } \mathbb {T},\\ \partial _n \Phi &{}=&{} h &{}\text{ on } \mathbb {T}. \end{array} \right. \end{aligned}$$

(1.1)

where

$$\begin{aligned} \displaystyle \Delta =\frac{\partial ^2}{\partial x^2}+\frac{\partial ^2}{\partial y^2}, \end{aligned}$$

denotes the standard Laplacian and $\partial _n$ denotes the differentiation in the inward normal direction, $g \in \mathcal {C}(\overline{{\mathbb D}})$ and the boundary data f and h $\in \mathcal {C}(\mathbb {T})$.

We would like to mention that in [13, 14] the authors have considered similar inhomogeneous biharmonic equations but with different boundaries conditions.

In order to state our main results, we introduce some necessary terminologies. For $z,w \in {\mathbb D}$, let

$$\begin{aligned} G(z,w) = |z -w|^2\log \bigg |\frac{1 - z\overline{w}}{z - w}\bigg |^2- (1 - |z|^2)(1 - |w|^2), \end{aligned}$$

and

$$\begin{aligned} P(z) = \frac{1-|z|^2}{|1 - z|^2}, \end{aligned}$$

denote the biharmonic Green function and the harmonic Poisson kernel, respectively.

For $\varphi \in L^1(\mathbb {T})$, we denote by $P[\varphi ]$ the Poisson extension of $\varphi$, defined on ${\mathbb D}$ by

$$\begin{aligned} P[\varphi ](z) =\frac{1}{2\pi }\int ^{2\pi }_0 P(ze^{-i\theta })\varphi (e^{i\theta }) d\theta . \end{aligned}$$

Riesz representation of (super-)biharmonic functions started with Abkar and Hedenmalm [2]. By [26, Theorem 1.1], we see that all solutions of IBDP (1.1) are given by

$$\begin{aligned} \Phi (z) = F_0[f](z) + H_0[h](z)-G[g](z), \end{aligned}$$

where

$$\begin{aligned} F_0[f](z)= & {} \frac{1}{2\pi }\int ^{2\pi }_0 F_0(ze^{-i\theta })f(e^{i\theta }) d\theta ,\quad H_0[h](z) = \frac{1}{2\pi }\int ^{2\pi }_0 H_0(ze^{-i\theta })h(e^{i\theta }) d\theta ,\\ \text {and}\ \ \ G[g](z)= & {} \frac{1}{16}\int _{{\mathbb D}}G(z,\omega )g(\omega )dA(\omega ), \end{aligned}$$

where $dA(\omega )$ denotes the Lebesgue area measure in ${\mathbb D}$. Here the kernels $H_0$ and $F_0$ are given by

$$\begin{aligned} F_0(z)= & {} H_0(z) + K_2(z),\\ H_0(z)= & {} \displaystyle \frac{1}{2}(1-|z|^2)P(z),\\ K_2(z)= & {} \frac{1}{2}\frac{(1-|z|^2)^3}{|1 - z|^4}. \end{aligned}$$

Thus, the solutions of the equation (1.1) are given by

$$\begin{aligned} \Phi (z)=\frac{1}{2} (1-|z|^2)P[f+h](z)+K_2[f](z)-G[g](z). \end{aligned}$$

Obviously $P[f+h]$ is a bounded harmonic function, and Heinz [19] proved the Schwarz lemma for planar harmonic functions: if $\Phi$ is a harmonic mapping from ${\mathbb D}$ into itself with $\Phi (0) = 0$, then for $z \in {\mathbb D}$,

$$\begin{aligned} |\Phi (z)|\le \frac{4}{\pi } \arctan |z|. \end{aligned}$$

Hethcote [20] and Pavlović [34, Theorem 3.6.1] improved Heinz’s result, by removing the assumption $\Phi (0)=0$, and proved the following.

Theorem A

Let $\Phi :{\mathbb D}\rightarrow {\mathbb D}$ be a harmonic function from the unit disc to itself, then

$$\begin{aligned} \bigg |\Phi (z)-\frac{1-|z|^2}{1+|z|^2} \Phi (0)\bigg |\le \frac{4}{\pi } \arctan |z|,\ \ \ \ \ \ \ \ z\in {\mathbb D}. \end{aligned}$$

(1.2)

A higher dimensional version for harmonic functions is proved in [21].

We remark that $K_2[f]$ is a bounded $T_2$-harmonic which is a special type of biharmonic functions. So naturally our first aim is to study the class of $T_\alpha$-harmonic functions [31]. These functions can be seen as generalized harmonic functions as $T_0$-harmonic functions coincide with classical harmonic functions. Other variants of generalized (or weighted) harmonic functions and their properties can be found in [32, 33].

First, let us recall the definition of $T_\alpha$-harmonic functions.

Definition 1

[31] Let $\alpha \in {\mathbb R}$, and let $f\in \mathcal {C}^2({\mathbb D})$. We say that f is $T_\alpha$-harmonic if f satisfies

$$\begin{aligned} T_\alpha (f)=0\quad \text {in} \ {\mathbb D}, \end{aligned}$$

where the $T_\alpha$-Laplacian operator is defined by

$$\begin{aligned} T_\alpha =-\frac{\alpha ^2}{4}(1-|z|^2)^{-(\alpha +1)}+\frac{1}{2}L_\alpha +\frac{1}{2}\overline{L_\alpha }, \end{aligned}$$

with the weighted Laplacian operator $L_\alpha$ is defined by

$$\begin{aligned} L_\alpha =\frac{\partial }{\partial \overline{z}}(1 - |z|^2)^{-\alpha }\frac{\partial }{\partial z}. \end{aligned}$$

Remark 1.1

Let f be a $T_\alpha$-harmonic function.

• (1) If $\alpha =0$, then f is harmonic.

• (2) If $\alpha =2n$, then f is $(n+1)$-harmonic, where $n\in {\mathbb N}$, see [1, 5, 31, 32].

The homogeneous expansion of $T_\alpha$-harmonic functions is giving by

Theorem B

[31] Let $\alpha \in {\mathbb R}$ and $f\in \mathcal {C}^2({\mathbb D})$. Then f is $T_\alpha$-harmonic if and only if it has a series expansion of the form

$$\begin{aligned} f(z)\!=\!\sum _{k=0}^\infty c_k F(-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;|z|^2)z^k\!+\! \sum _{k=1}^\infty c_{-k} F(-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;|z|^2)\overline{z}^k,\nonumber \!\!\!\!\!\!\\ \end{aligned}$$

(1.3)

for some sequence $\{c_k\}$ of complex numbers satisfying $\limsup _{|k|\rightarrow \infty } |c_k|^{\frac{1}{|k|}}\le 1,$ where F is the Gauss hypergeometric function.

For $\alpha >-1$, a Poisson type integral representation for $T_\alpha$-harmonic mappings is provided by the following theorem.

Theorem C

([31] Theorem 3.3) Let $\alpha >-1$ and u be a $T_\alpha$-harmonic in ${\mathbb D}$. Assume that $\displaystyle \lim _{r\rightarrow 1}u_r=u^*$ in $\mathcal {D}'({\mathbb T})$. Then u has a form of a Poisson type integral

$$\begin{aligned} u(z)=K_\alpha [u^*](z)=\frac{1}{2\pi } \int _{0}^{2\pi } K_\alpha (z,e^{i \theta }) u^*(e^{i\theta })\, d\theta . \end{aligned}$$

The integral is understood in the sense of distribution theory and

$$\begin{aligned} K_\alpha (z,e^{i\theta })=c_\alpha \frac{(1-|z|^2)^{\alpha +1}}{|z-e^{i\theta }|^{\alpha +2}},\quad c_\alpha =\displaystyle \frac{\Gamma (\alpha /2+1)^2}{\Gamma (\alpha +1)}. \end{aligned}$$

The factor of normalization $c_\alpha$ is chosen in order to ensure that the integral means

$$\begin{aligned} M_\alpha (r)= \frac{1}{2\pi } \int _{\mathbb {T}} K_\alpha (r,e^{i\theta })d\theta , \quad r\in [0,1) \end{aligned}$$

satisfies

$$\begin{aligned} \lim _{r\rightarrow 1} M_\alpha (r)=1. \end{aligned}$$

Moreover, the function $M_\alpha$ is increasing on [0, 1), see [31, Theorem 3.1].

It is well known that the Schwarz lemma is one of the most influential results in many branches of mathematical research for more than a hundred years. We refer the reader to [6, 13, 22, 29, 30] for generalizations and applications of this lemma.

Define

$$\begin{aligned} U_\alpha (z)=K_\alpha [\chi _{{\mathbb T}^r}- \chi _{{\mathbb T}^l}](z), \end{aligned}$$

(1.4)

where

$$\begin{aligned} {\mathbb T}^r=\{ z \in {\mathbb T}: {{\text {Re}}\,}{z}>0 \}, \text{ and } {\mathbb T}^l=\{ z \in {\mathbb T}: {{\text {Re}}\,}{z}<0 \}. \end{aligned}$$

$U_\alpha$ is a $T_\alpha$-harmonic function on ${\mathbb D}$ with values in $(-1,1)$ such that $U_\alpha (0)=0$.

First, we establish a Heniz-Hethcote theorem for $T_\alpha$-harmonic functions.

Theorem 1

Let $\alpha >-1$ and $u:{\mathbb D}\longrightarrow {\mathbb D}$ be a $T_\alpha$-harmonic function, then

$$\begin{aligned} \bigg |u(z)-\frac{(1-|z|^2)^{\alpha +1}}{(1+|z|^2)^{\frac{\alpha }{2}+1}}u(0)\bigg |\le U_\alpha (|z|), \end{aligned}$$

for all $z\in {\mathbb D}$, where $U_\alpha$ is the function defined in (1.4).

In particular, for $T_2$-harmonic functions, we obtain

Corollary 1.1

Let $u:{\mathbb D}\longrightarrow {\mathbb D}$ be a $T_2$-harmonic function, then

$$\begin{aligned} \bigg |u(z)-\frac{(1-|z|^2)^3}{(1+|z|^2)^2}u(0)\bigg |\le \frac{2}{\pi }\bigg [\frac{|z|(1-|z|^2)}{1+|z|^2}+(1+|z|^2)\arctan |z|\bigg ]. \end{aligned}$$

Next, we prove a sharp estimate of $D_u(0)$, where u is a $T_\alpha$-harmonic function.

Theorem 2

Let $\alpha >-1$ and $u:{\mathbb D}\longrightarrow {\mathbb D}$ be a $T_\alpha$-harmonic function, then

$$\begin{aligned} \Vert D_u(0)\Vert \le \frac{2c_\alpha }{\pi }(\alpha +2). \end{aligned}$$

(1.5)

The inequality (1.5) is sharp and $U_\alpha$ is an extremal function, see (1.4).

Let $\mathcal {A}({\mathbb D})$ the set of all holomorphic functions $\Phi$ in ${\mathbb D}$ satisfying the standard normalization: $\Phi (0) = \Phi '(0)-1 = 0$. Landau [23] showed that there is a constant $r > 0$, independent of elements in $\mathcal {A}({\mathbb D})$, such that $\Phi ({\mathbb D})$ contains a disk of radius r. Later, Landau’s theorem has become an important tool in geometric function theory. Indeed, many authors considered Landau type theorems for harmonic functions i.e., $\alpha =0$ (cf. [7,8,9,10, 12, 28]), for biharmonic functions, $\alpha =2$ (cf. [1, 27]) and for polyharmonic functions $\alpha =2(n-1)$ (see [4, 11]), and in [12], the authors considered the case $\alpha \in (-1,0)$.

Naturally, our next aim is to establish a Landau type theorem for $T_\alpha$-harmonic functions, for $\alpha >0$.

Theorem 3

Let $\alpha >0$, and $u\in \mathcal {C}^2({\mathbb D})$ be a $T_\alpha$-harmonic function satisfying $u(0)=J_u(0)-1=0$ and $\sup _{z\in {\mathbb D}}|u(z)|\le M$, where $M>0$ and $J_u$ is the Jacobian of u. Let $n\ge 1$ be an integer such that $n-1<\frac{\alpha }{2}\le n.$ Then u is univalent on $D_{r_\alpha }$, where $r_\alpha$ satisfies the following equation

$$\begin{aligned} \frac{2c_\alpha (\alpha +2)}{\pi }M\sigma _\alpha (r_\alpha )=1. \end{aligned}$$

(1.6)

Moreover, $u({\mathbb D}_{r_\alpha })$ contains an univalent disk $D_{R_\alpha }$ with

$$R_\alpha \ge \frac{\sigma _\alpha (r_\alpha )r_\alpha }{2},$$

where

$$\begin{aligned} \sigma _\alpha (r):= & {} \dfrac{4Mr}{\pi }\bigg [\frac{6a_\alpha }{(1-r)^3}+\frac{r}{(1-r)^2}+(\frac{2-\alpha }{4})r\Bigg ],\ \text { if } n=1.\\ \sigma _\alpha (r):= & {} \dfrac{4Mr}{\pi }\Bigg [\frac{24a_\alpha }{(1-r)^{4}} +3\alpha a_\alpha \left( 1+ \frac{4r}{3(1-r)^{3}} \right) r+3a_\alpha (\alpha -2)r\Bigg ], \text { if } n=2.\\ \sigma _\alpha (r):= & {} \dfrac{4Mr}{\pi }\Bigg [\frac{(n+1)(n-2)}{2}(1+r)+\frac{a_\alpha (2n)!r^{n-2}}{(1-r)^{2n}} \\&\,+&\frac{\alpha a_\alpha (2n-1)!}{n!}\left( 1+ \frac{2nr}{(n+1)(1-r)^{n+1}} \right) r^{n-1} +(\frac{\alpha -2}{4})r\Bigg ], \text { if } n\ge 3. \end{aligned}$$

with $\displaystyle a_\alpha =\frac{\Gamma (\frac{\alpha }{2}+1)}{\Gamma (\alpha +1)}$.

Remark 1.2

In particular, for $\alpha =2$, we obtain

$$\begin{aligned} \sigma _2(r) =\dfrac{4Mr}{\pi (1-r)^2}\bigg [\frac{3}{1-r}+r\Bigg ]. \end{aligned}$$

(1.7)

Now we are in the position to prove some results related to the Dirichlet problem (1.1).

Theorem 4

Let $g\in \mathcal {C}(\overline{{\mathbb D}})$, $f, h \in \mathcal {C}(\mathbb {T})$ and suppose that $\Phi \in \mathcal {C}^4({\mathbb D}) \cap \mathcal {C}(\overline{{\mathbb D}})$ satisfies (1.1). Then for $z \in {\mathbb D}$,

$$\begin{aligned} \bigg | \Phi (z)- & {} \frac{1}{2}\frac{(1-|z|^2)^3}{(1+|z|^2)^2}P[f](0)-\frac{1}{2}\frac{(1-|z|^2)^2}{1+|z|^2}P[f+h](0)\bigg |\nonumber \\\le & {} \Big [\frac{2}{\pi }(1-|z|^2)\arctan |z|\Big ]\Vert f+h\Vert _\infty \nonumber \\+ & {} \frac{2}{\pi }\bigg [(1+|z|^2)\arctan |z|+|z|\frac{1-|z|^2}{1+|z|^2}\bigg ]\Vert f\Vert _\infty +\frac{(1-|z|^2)^2}{64}\,\Vert g\Vert _\infty ,\qquad \end{aligned}$$

(1.8)

where $\Vert f \Vert _\infty = \sup _{\zeta \in \mathbb {T}}|f(\zeta )|$, $\Vert f+h \Vert _\infty = \sup _{\zeta \in \mathbb {T}}|f(\zeta )+h(\zeta )|$ and $\Vert g \Vert _\infty = \sup _{\zeta \in \mathbb {D}}|g(\zeta )|$.

Theorem 5

Let $g\in \mathcal {C}(\overline{{\mathbb D}})$, f and $h \in \mathcal {C} (\mathbb {T})$. Suppose that $\Phi \in \mathcal {C}^4({\mathbb D})$ is satisfying (1.1). Then for all $z\in {\mathbb D}$,

$$\begin{aligned} \Vert D_\Phi (z)\Vert \le \frac{2+5|z|}{1-|z|^2}(1+|z|^2)\Vert f\Vert _\infty +\left( \frac{2}{\pi } +|z| \right) \Vert f+h\Vert _\infty +\frac{23}{48}\Vert g\Vert _\infty . \end{aligned}$$

(1.9)

Moreover at $z=0$, we have

$$\begin{aligned} \Vert D_\Phi (0)\Vert \le \frac{4}{\pi }\Vert f\Vert _\infty +\frac{2}{\pi }\Vert f+h\Vert _\infty +\frac{23}{48}\Vert g\Vert _\infty . \end{aligned}$$

(1.10)

The classical Schwarz lemma at the boundary is as follows.

Theorem D

Suppose $f:{\mathbb D}\longrightarrow {\mathbb D}$ is a holomorphic function with $f(0)=0$, and further, f is analytic at $z=1$ with $f(1)=1$. Then, the following two conditions hold:

• (a) $f'(1)\ge 1;$

• (b) $f'(1)=1$ if and only if $f(z)=z.$

The previous theorem is known as the Schwarz lemma on the boundary, and its generalizations have important applications in geometric theory of functions (see, [18, 24, 35]). Among the recent papers devoted to this subject, for example, Burns and Krantz [6], Krantz [22], Liu and Tang [29] explored many versions of the Schwarz lemma at the boundary point of holomorphic functions, Dubinin also applied this latter for algebraic polynomials and rational functions (see [16, 17]). In the present paper, we refine the Schwarz type lemma at the boundary for $\Phi$ satisfies (1.1) as an application of Theorem 4.

Theorem 6

Suppose that $\Phi \in \mathcal {C}^4({\mathbb D})\cap \mathcal {C}(\overline{{\mathbb D}})$ satisfies (1.1), where $g\in \mathcal {C}(\overline{{\mathbb D}})$ and f, $h\in \mathcal {C}(\mathbb {T})$ such that $\Vert f\Vert _\infty \le 1$, and $\Vert f+h\Vert _\infty \le 1$. If $\displaystyle \lim _{r\rightarrow 1} |\Phi (r\eta )|=1$ for $\eta \in \mathbb {T}$, then

$$\begin{aligned} \liminf _{r\rightarrow 1}\frac{|\Phi (\eta )-\Phi (r\eta )|}{1-r}\ge 1-\Vert f+h\Vert _\infty . \end{aligned}$$

In particular if $\Vert f+h\Vert _\infty =0$, then $\liminf _{r\rightarrow 1}\frac{|\Phi (\eta )-\Phi (r\eta )|}{1-r}\ge 1,$ and this estimate is sharp.

For $g\in \mathcal {C}(\overline{{\mathbb D}})$ and $h \in \mathcal {C}(\mathbb {T})$, let $\mathcal {BF}_{g,h}(\overline{{\mathbb D}})$ denote the class of all complex-valued functions $\Phi \in \mathcal {C}^4({\mathbb D})\cap \mathcal {C}(\overline{{\mathbb D}})$ satisfying (1.1) with the normalization $\Phi (0) = J_\Phi (0) - 1 = 0$.

We establish the following Landau-type theorem for $\Phi \in \mathcal {BF}_{g,h}(\overline{{\mathbb D}})$. In particular, if $g \equiv 0$, then $\Phi \in \mathcal {BF}_{g,h}(\overline{{\mathbb D}})$ is biharmonic. In this sense, the following result is a generalization of [1, Theorem 1 and 2].

Theorem 7

Suppose that $M_1>0$, $M_2>0$ and $M_3>0$ are constants, and suppose that $\Phi \in \mathcal {BF}_{g,h}(\overline{{\mathbb D}})$ satisfies the following conditions:

$$\begin{aligned} \sup _{z\in \mathbb {T}} |f(z)|\le M_1,\ \ \ \sup _{z\in \mathbb {T}} |f(z)+h(z)|\le M_2,\ \ \text { and }\sup _{z\in {\mathbb D}} |g(z)|\le M_3. \end{aligned}$$

Then $\Phi$ is univalent in ${\mathbb D}_{r_0}$ and $\Phi ({\mathbb D}_{r_0})$ contains a univalent disk ${\mathbb D}_{R_0}$,where $r_0$ satisfies the following equation:

$$\begin{aligned} \left( \frac{4}{\pi } M_1+\frac{2}{\pi }M_2+\frac{23}{48}M_3\right) \mu (r_0)=1, \end{aligned}$$

with

$$\begin{aligned} \mu (|z|):= & {} (M_1+M_2+\frac{101}{120} M_3)|z|+\frac{2M_2|z|}{\pi }\bigg [\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg ]\\&+\,\frac{4M_1|z|}{ \pi (1-|z|)^3}(|z|^2(1-|z|)+3), \end{aligned}$$

and

$$\begin{aligned} R_0\ge \frac{r_0}{\frac{8}{\pi } M_1+\frac{4}{\pi }M_2+\frac{23}{24}M_3}. \end{aligned}$$

2 Preliminaries

Here we collect some preliminary facts used in the sequel. The Gauss hypergeometric function is defined by the power series

$$\begin{aligned} F(a,b;c;x)=\sum _{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \frac{x^n}{n!}, \quad |x|<1, \end{aligned}$$

for $a,b,c\in {\mathbb R}$, with $c\not = 0,-1,-2,\ldots .,$ where $(a)_0=1$ and $(a)_n=a(a+1)\ldots (a+n-1)$ for $n=1,2,\ldots$ are the Pochhammer symbols.

We list few properties, see for instance [3, Chapter 2]

$$\begin{aligned} F(a,b;c;1)= & {} \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}\, \text{ if } c-a-b>0. \end{aligned}$$

(2.1)

$$\begin{aligned} F(a+1,b+1;c+1;1)= & {} \dfrac{c}{c-a-b-1}F(a,b;c;1), \text{ if } c-a-b>1.\qquad \end{aligned}$$

(2.2)

$$\begin{aligned} F(a,b;c;x)= & {} (1-x)^{c-a-b}F(c-a,c-b;c;x). \end{aligned}$$

(2.3)

$$\begin{aligned} \frac{d}{dx}F(a,b;c;x)= & {} \frac{ab}{c}F(a+1,b+1;c+1,x). \end{aligned}$$

(2.4)

The following lemma about the monotonicity of hypergeometric functions follows immediately from the properties (2.3) and (2.4).

Lemma 1

[31] Let $c > 0$, $a \le c$, $b \le c$ and $ab \le 0$ ($ab \ge 0$). Then the function is decreasing (increasing) on (0, 1).

The following results are useful to establish a Landau theorem for $T_\alpha$-harmonic functions, when $\alpha >0$.

Lemma 2

[25, Formula 5.2.2 (9) p. 697] for $n\ge 1$ and $|x|<1$

$$\begin{aligned} \gamma _n(x):=\sum _{k=0}^\infty (k+1)(k+2)\ldots (k+n)x^k=\frac{n!}{(1-x)^{n+1}}. \end{aligned}$$

(2.5)

As a direct application of Lemma 2, it yields

Lemma 3

For $r\in (0,1)$, and $n\ge 1$, define the sequence

$$\begin{aligned} S_n(r)=\sum _{k\ge 1}(k+1)(k+2)\ldots (k+n)r^k. \end{aligned}$$

Then,

$$\begin{aligned} S_n(r)=\frac{n!r}{(1-r)^{n+1}}\sum _{k=0}^n (1-r)^k. \end{aligned}$$

In particular,

$$\begin{aligned} S_n(r)\le \frac{n!r}{(1-r)^{n+1}}\Big [(n+1)-nr\Big ] \le \frac{(n+1)! r}{(1-r)^{n+1}}. \end{aligned}$$

(2.6)

Proposition 2.1

Let $n\ge 1$ and $n-1<\frac{\alpha }{2}\le n$. Then, we have the following two estimates

$$\begin{aligned}&\displaystyle (a) \sum _{k=n}^\infty \dfrac{\Gamma (k+\frac{\alpha }{2}+1)}{(k-1)!}\,r^{k-1}\le r^{n-2}S_{2n-1}(r) \le \frac{(2n)! r^{n-1}}{(1-r)^{2n}}, \end{aligned}$$

(2.7)

$$\begin{aligned}&\displaystyle (b) \sum _{k=n}^{\infty }\frac{\Gamma (k+\frac{\alpha }{2}+1)(k-\frac{\alpha }{2})}{(k+1)!}\,r^{k+1}\le \frac{(2n)!}{(n+1)!}\frac{r^{n+1}}{(1-r)^{n+1}}, \end{aligned}$$

(2.8)

for $r\in [0,1)$.

Proof

The inequality (2.7) follows immediately from (2.6).

Now we prove the inequality (2.8). By assumption, we have

$$\begin{aligned} \sum _{k=n}^{\infty }\frac{\Gamma (k+\frac{\alpha }{2}+1)(k-\frac{\alpha }{2})}{(k+1)!}r^{k+1}\le & {} \sum _{k=n}^{\infty }\frac{\Gamma (k+n+1)(k-n+1)}{(k+1)!}r^{k+1}\\= & {} r^{n+1} \sum _{k=0}^\infty (k+1).(k+n+2)(k+n+3)\ldots (k+2n)r^k. \end{aligned}$$

Clearly, for $k\ge 0$ and all $2\le j \le n$, we have

$$\begin{aligned} k+j+n\le \frac{j+n}{j} (k+j). \end{aligned}$$

Thus

$$\begin{aligned} \sum _{k=0}^\infty (k+1).(k+n+2)(k+n+3)\ldots (k+2n)r^k\le \frac{\gamma _n(r)}{n!} \frac{(2n)!}{(n+1)!}. \end{aligned}$$

Therefore, by Lemma 2, it yields

$$\begin{aligned} \sum _{k=n}^{\infty }\frac{\Gamma (k+\frac{\alpha }{2}+1)(k-\frac{\alpha }{2})}{(k+1)!}r^{k+1}\le \frac{(2n)!}{(n+1)!}\frac{r^{n+1}}{(1-r)^{n+1}}. \end{aligned}$$

$\square$

3 Schwarz and Landau type lemmas for $T_\alpha$-harmonic functions

3.1 Schwarz type lemma for $T_\alpha$-harmonic functions

The main purpose of this section is to prove a Schwarz type lemma for $T_\alpha$-harmonic functions.

Proof of Theorem 1

Let $0\le r=|z|<1$. As u is a $T_\alpha$-harmonic function, then

$$\begin{aligned} u(z)= K_\alpha [u^*](z)=\frac{1}{2\pi }\int _0^{2\pi }c_\alpha \frac{(1-r^2)^{\alpha +1}}{|1-ze^{-i\theta }|^{\alpha +2}}u^*(e^{i\theta })d\theta , \end{aligned}$$

where $u^*\in L^\infty (\mathbb {T})$. Thus

$$\begin{aligned} \bigg |u(z)-\frac{(1-r^2)^{\alpha +1}}{(1+r^2)^{\frac{\alpha }{2}+1}}u(0)\bigg |\le & {} \frac{c_\alpha }{2\pi }\int _{{\mathbb T}}\bigg |\frac{(1-r^2)^{\alpha +1}}{(1+r^2-2r\cos \theta )^{\frac{\alpha }{2}+1}}-\frac{(1-r^2)^{\alpha +1}}{(1+r^2)^{\frac{\alpha }{2}+1}}\bigg |d\theta \\= & {} \frac{c_\alpha }{2\pi }\bigg [\int _{-\pi /2}^{\pi /2}\frac{(1-r^2)^{\alpha +1}}{(1+r^2-2r\cos \theta )^{\frac{\alpha }{2}+1}}-\frac{(1-r^2)^{\alpha +1}}{(1+r^2)^{\frac{\alpha }{2}+1}}\,d\theta \\- & {} \int _{\pi /2}^{3\pi /2}\frac{(1-r^2)^{\alpha +1}}{(1+r^2-2r\cos \theta )^{\frac{\alpha }{2}+1}}-\frac{(1-r^2)^{\alpha +1}}{(1+r^2){\frac{\alpha }{2}+1}}\,d\theta \bigg ]\\= & {} K_\alpha [\chi _{{\mathbb T}^r}-\chi _{{\mathbb T}^l}](|z|). \end{aligned}$$

$\square$

To compute $U_2$, we need to evaluate the following integral.

$$\begin{aligned} J(\theta ):=\int _0^{\theta }\frac{(1-r^2)^3}{(1+r^2-2r\cos \varphi )^2}d\varphi . \end{aligned}$$

Easy but tedious computations show that

Lemma 4

For $0\le \theta <\pi$, and $r\in [0,1)$, we have

$$\begin{aligned} J(\theta ):=\int _0^{\theta }\frac{(1-r^2)^3}{(1+r^2-2r\cos \varphi )^2}\,d\varphi =\frac{2r(1-r^2)\sin \theta }{1+r^2-2r\cos \theta }+2(1+r^2)\arctan \bigg (\frac{(1+r)\tan \theta /2}{1-r}\bigg ), \end{aligned}$$

and $\displaystyle J(\pi )= \lim _{\theta \rightarrow \pi } J(\theta )= \pi (1+r^2).$

Proof of Corollary 1.1

By Lemma 4, and using the fact that $\displaystyle \arctan (\frac{1+r}{1-r})-\frac{\pi }{4}=\arctan r$, we have

$$\begin{aligned} U_2(r)= & {} \frac{1}{2\pi }\bigg [2J(\pi /2)-J(\pi )\bigg ]\\= & {} \frac{1}{2\pi }\bigg [4\frac{r(1-r^2)}{1+r^2}+4(1+r^2)\arctan (\frac{1+r}{1-r})-\pi (1+r^2)\bigg ]\\= & {} \frac{2}{\pi }\bigg [\frac{r(1-r^2)}{1+r^2}+(1+r^2)\arctan r\bigg ]. \end{aligned}$$

$\square$

Proof of Theorem 2

Near 0, we have

$$\begin{aligned}&\frac{(1-r^2)^{\alpha +1}}{(1+r^2-2r\cos \theta ){\frac{\alpha }{2}+1}} =1+ (\alpha +2)\cos \theta r +O(r^2),\\&U_\alpha (r)=\frac{2c_\alpha }{\pi }(\alpha +2)r+O(r^2) \text { and } \frac{(1-r^2)^3}{(1+r^2)^2}= 1+O(r^2). \end{aligned}$$

Hence from Theorem 1 and (3.1), we get

$$\begin{aligned} |u(z)-u(0)| \le \frac{2c_\alpha }{\pi }(\alpha +2)|z|+O(|z|^2). \end{aligned}$$

(3.1)

Thus

$$\begin{aligned} \Vert D_u(0)\Vert \le \frac{2c_\alpha }{\pi }(\alpha +2). \end{aligned}$$

To show that the last estimate is sharp. Let us consider the $T_\alpha$-harmonic mapping defined by

$$\begin{aligned} U_\alpha (z)=K_\alpha [\chi _{{\mathbb T}^r}- \chi _{{\mathbb T}^l}](z). \end{aligned}$$

By [31, Theorem 1.1], we have

$$\begin{aligned} \frac{\partial }{\partial z}K_{\alpha }(ze^{-it})= c_\alpha \frac{(1-|z|^2)^\alpha }{|1-ze^{-it}|^{2+\alpha }}\Bigg (-\frac{\alpha }{2}\overline{z}e^{it}+\frac{2+\alpha }{2}\frac{1-\overline{ze^{-it}}}{1-ze^{-it}}\Bigg )e^{-it}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{\partial }{\partial z}K_{\alpha }(ze^{-it})_{\vert {z=0}} =\frac{c_\alpha }{2}(2+\alpha ) e^{-it}. \end{aligned}$$

As

$$\begin{aligned} \frac{1}{2\pi } \int _0^{2\pi } e^{-i\theta } (\chi _{{\mathbb T}^r}- \chi _{{\mathbb T}^l})(\theta ) d\theta =\frac{1}{2\pi } \int _{-\pi /2}^{\pi /2} e^{-i\theta }d\theta -\frac{1}{2\pi } \int _{\pi /2}^{3\pi /2} e^{-i\theta }d\theta =\frac{2}{\pi }, \end{aligned}$$

we conclude that

$$\begin{aligned} |\nabla U_\alpha (0)|=2|\frac{\partial U_\alpha }{\partial z}(0)|=\frac{2c_\alpha }{\pi }(\alpha +2). \end{aligned}$$

$\square$

3.2 Proof of Theorem 3

First, we need the following theorem which provides some estimates on the coefficients of $T_\alpha$-harmonic mappings.

Theorem E

[12] For $\alpha >-1$, let $u\in \mathcal {C}^2({\mathbb D})$ be a $T_\alpha$-harmonic function with the series expansion of the form (1.3) and $\sup _{z\in {\mathbb D}}|u(z)| \le M$, where $M>0$. Then, for $k\in \{1,2,\ldots \}$,

$$\begin{aligned} \left( |c_k|+|c_{-k}|\right) F(-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;1) \le \frac{4M}{\pi }, \end{aligned}$$

(3.2)

and

$$\begin{aligned} |c_0|F(-\frac{\alpha }{2},-\frac{\alpha }{2};1;1)\le M. \end{aligned}$$

(3.3)

Therefore for $k\ge 1$ and $\alpha >-1$, using (2.1), we have

$$\begin{aligned} F\big (-\frac{\alpha }{2}, k-\frac{\alpha }{2};k+1;1\big )=\dfrac{k!\Gamma (\alpha +1)}{\Gamma (\frac{\alpha }{2}+1)\Gamma (k+\frac{\alpha }{2}+1)}. \end{aligned}$$

Thus if u is $T_\alpha$-harmonic such that $|u(z)|\le M$, then by (3.2) it yields

$$\begin{aligned} |c_k|+|c_{-k}|\le \dfrac{4M a_\alpha }{\pi }\dfrac{ \Gamma (k+\frac{\alpha }{2}+1)}{k!} \ \ \text{ for } k\ge 1. \end{aligned}$$

(3.4)

Proof of Theorem 3

Let us compute $u_z$ and $u_{\overline{z}}$, for u is a $T_\alpha$-harmonic with $\alpha >0$ and $u(0)=c_0=0$. The power series expansion is provided by

$$\begin{aligned} u(z)=\sum _{k=1}^\infty c_k F(-\dfrac{\alpha }{2}, k-\dfrac{\alpha }{2};k+1;|z|^2)z^k+\sum _{k=1}^\infty c_{-k} F(-\dfrac{\alpha }{2}, k-\dfrac{\alpha }{2};k+1;|z|^2)\overline{z}^k, \end{aligned}$$

and the series converges for $\mathcal {C}^{\infty }$-topology. Hence

$$\begin{aligned} u_z(z)= & {} \sum _{k=2}^\infty k c_k F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )z^{k-1}+\sum _{k=1}^\infty c_k \frac{d}{d\omega }F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )\overline{z}z^k\nonumber \\+ & {} \sum _{k=1}^\infty c_{-k} \frac{d}{d\omega } F\big (-\frac{\alpha }{2}, k-\frac{\alpha }{2};k+1;\omega \big )\overline{z}^{k+1}+c_1F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;\omega \big ), \end{aligned}$$

(3.5)

and

$$\begin{aligned} u_{\overline{z}}(z)= & {} \sum _{k=1}^\infty c_k\frac{d}{d\omega }F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )z^{k+1}+\sum _{k=2}^\infty c_{-k} k F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )\overline{z}^{k-1}\nonumber \\+ & {} \sum _{k=1}^\infty c_{-k} \frac{d}{d\omega }F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )z\overline{z}^k+c_{-1}F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;\omega \big ), \end{aligned}$$

(3.6)

where $\omega =|z|^2$.

We have $u_z(0)=c_1F\big (-\frac{\alpha }{2}, k-\frac{\alpha }{2};k+1;0\big )=c_1$, and similarly $u_{\overline{z}}(0)=c_{-1}$. Thus combining (3.5) and (3.6), we obtain

$$\begin{aligned} |u_z(z)-u_z(0)|+|u_{\overline{z}}(z)-u_{\overline{z}}(0)|\le & {} \sum _{k=2}^\infty k(|c_k|+|c_{-k}|) F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )|z|^{k-1}\\+ & {} 2\sum _{k=1}^\infty (|c_k|+|c_{-k}|)\, \Big |\frac{d}{d\omega }F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )\Big ||z|^{k+1}\\+ & {} (|c_1|+|c_{-1}|)\bigg |F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;\omega \big )-1\bigg |. \end{aligned}$$

By (2.4) and (2.3), we see that

$$\begin{aligned} \Big |\dfrac{d}{d\omega }F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )\Big |=\dfrac{\alpha \big |\frac{\alpha }{2}-k\big |}{2(k+1)}F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;\omega \big ), \end{aligned}$$

as the mapping is positive. We denote

$$\begin{aligned} E_\alpha (r):= & {} \sum _{k=2}^\infty k (|c_k|+|c_{-k}|) F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;r^2\big )r^{k-1}, \end{aligned}$$

(3.7)

$$\begin{aligned} F_\alpha (r):= & {} \sum _{k=1}^\infty (|c_k|+|c_{-k}|)\, \frac{\alpha \big |\frac{\alpha }{2}-k\big |}{(k+1)}F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;r^2\big )r^{k+1}, \end{aligned}$$

(3.8)

$$\begin{aligned} G_\alpha (r):= & {} (|c_1|+|c_{-1}|)\bigg |F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;r^2\big )-1\bigg |. \end{aligned}$$

(3.9)

In the sequel, we will estimate each of these expressions. $\square$

3.3 Estimate of $E_\alpha (r)$

Lemma 5

Let $n\in {\mathbb N}$, $n\ge 1$ and $\frac{\alpha }{2}\in (n-1,n]$.

If $n=1$, then

$$\begin{aligned} E_\alpha (r)\le \frac{4M a_\alpha }{\pi }S_2(r)\le \frac{24M a_\alpha r}{\pi (1-r)^3}. \end{aligned}$$

(3.10)

If $n\ge 2$, then

$$\begin{aligned} E_\alpha (r)\le \frac{4M}{\pi }\bigg [ \frac{(n+1)(n-2)}{2}r+\frac{a_\alpha (2n)!r^{n-1}}{(1-r)^{2n}}\bigg ]. \end{aligned}$$

(3.11)

Proof

A straightforward application of Lemma 1 implies that the monotonicity properties of depends on $\alpha \big (\dfrac{\alpha }{2}-k\big )$. Therefore the function is decreasing on [0, 1) when $\alpha \in (0,4]$ and $k\ge 2$. Thus for $\omega \in [0,1)$,

$$\begin{aligned} F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;\omega \big )\le F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;0\big )= 1. \end{aligned}$$

First, we estimate $E_\alpha (r)$ for $\alpha \in (0,4]$, then we will consider the case $\alpha >4$.

Case 1. $0<\alpha \le 4$

The decreasing property of and (3.4) imply that

$$\begin{aligned} E_\alpha (r)= & {} \sum _{k=2}^\infty k (|c_k|+|c_{-k}|) F(-\frac{\alpha }{2}, k-\frac{\alpha }{2};k+1;r^2)r^{k-1}\\\le & {} \sum _{k=2}^\infty k (|c_k|+|c_{-k}|)r^{k-1}\\\le & {} \dfrac{4Ma_\alpha }{\pi } \sum _{k=2}^\infty \dfrac{\Gamma (k+\frac{\alpha }{2}+1)}{(k-1)!}r^{k-1}. \\ \end{aligned}$$

Subcase 1. $0<\alpha \le 2$

Remark that $\Gamma (k+\frac{\alpha }{2}+2)\le \Gamma (k+3)$. Thus

$$\begin{aligned} \sum _{k=2}^\infty \dfrac{\Gamma (k+\frac{\alpha }{2}+1)}{(k-1)!}r^{k-1} = \sum _{k=1}^\infty \dfrac{\Gamma (k+\frac{\alpha }{2}+2)}{k!}r^k \le \sum _{k=1}^\infty \dfrac{\Gamma (k+3)}{k!}r^k =S_2(r). \end{aligned}$$

Hence

$$\begin{aligned} \displaystyle E_\alpha (r) \le \frac{4Ma_\alpha }{\pi }S_2(r) \le \frac{24Mr a_\alpha }{\pi (1-r)^3}. \end{aligned}$$

Subcase 2. $2<\alpha \le 4$

By Proposition 2.1, for $n=2$, it yields $\displaystyle \sum _{k=2}^\infty \dfrac{\Gamma (k+\frac{\alpha }{2}+1)}{(k-1)!}r^k \le S_3(r)$. Therefore,

$$\begin{aligned} E_\alpha (r)\le \frac{4Ma_\alpha }{\pi } S_3(r)\le \frac{96Ma_\alpha r}{(1-r)^4}. \end{aligned}$$

Case 2. $2(n-1)< \alpha \le 2n$, $n\ge 3$, that is, $n-1=\lceil \frac{\alpha }{2}\rceil$.

According to the discussion at the beginning of the proof, we see that the function is increasing for $2\le k\le n-1$, and, decreasing, for $k\ge n$ on [0, 1).

Consequently, we split $E_\alpha$ in two sums according to the monotonicity of . On one hand, we have

$$\begin{aligned} \sum _{k=2}^{n-1} k (|c_k|+|c_{-k}|) F(-\frac{\alpha }{2}, k-\frac{\alpha }{2};k+1;r^2)r^{k-1}\le & {} \dfrac{4M}{\pi }\sum _{k=2}^{n-1} k r^{k-1} \\\le & {} \dfrac{2Mr}{\pi }(n+1)(n-2). \end{aligned}$$

On the other hand, using the estimate of the coefficients (3.4), we have

$$\begin{aligned} \sum _{k=n}^\infty k (|c_k|+|c_{-k}|) F(-\frac{\alpha }{2}, k-\frac{\alpha }{2};k+1;r^2)r^{k-1}\le & {} \sum _{k=n}^\infty k (|c_k|+|c_{-k}|) r^{k-1}\\\le & {} \dfrac{4M a_\alpha }{\pi }\sum _{k=n}^\infty \dfrac{\Gamma (k+\frac{\alpha }{2}+1)}{(k-1)!}r^{k-1}. \end{aligned}$$

Finally, by Proposition 2.1 (a), we conclude

$$\begin{aligned} E_\alpha (r)\le & {} \frac{4M}{\pi } \Big [ \frac{(n+1)(n-2)}{2}r +a_\alpha \frac{(2n)! r^{n-1}}{(1-r)^{2n}} \Big ]. \end{aligned}$$

We remark that this formula is still valid for $n=2$. $\square$

3.4 Estimate of $F_\alpha (r)$

Lemma 6

Let $n\in {\mathbb N}$, $n\ge 1$ and $\frac{\alpha }{2}\in (n-1,n]$.

If $n=1$, then

$$\begin{aligned} F_\alpha (r)\le \frac{4Mr^2}{\pi (1-r)}\Big (\frac{1}{1-r}-\frac{\alpha }{2}\Big )\le \frac{4Mr^2}{\pi (1-r)^2}. \end{aligned}$$

(3.12)

If $n \ge 2$, then

$$\begin{aligned} F_\alpha (r)\le \frac{2M}{\pi } (n-2)(n+1)r^2 +\frac{4M\alpha a_\alpha (2n-1)!}{\pi n!}\left( 1+ \frac{2nr}{(n+1)(1-r)^{n+1}} \right) r^n. \end{aligned}$$

(3.13)

Proof

We start by investigating the monotonicity of . By Lemma 1, we infer that its monotonicity depends on

$$\begin{aligned} \big (-\frac{\alpha }{2}+1\big )(k-\frac{\alpha }{2}+1),\ \ \ \ \ \ k\ge 1,\ \ \ \ \alpha >0. \end{aligned}$$

Therefore the function is increasing for $0<\alpha \le 2$, and decreasing for $2\le \alpha < 4$ on [0, 1).

Case 1. $0<\alpha \le 2.$

As the function is increasing, we have

$$\begin{aligned} F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;\omega \big )\le F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;1\big ) \end{aligned}$$

According to (2.2), we obtain

$$\begin{aligned} F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;1\big )=\frac{\alpha }{k+1}F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;1\big ) \end{aligned}$$

(3.14)

Finally, using (3.2), (3.8) and (3.14), we have

$$\begin{aligned} F_\alpha (r)\le & {} \sum _{k=1}^\infty (|c_k|+|c_{-k}|)\, \frac{\alpha \big |\frac{\alpha }{2}-k\big |}{(k+1)}F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;1\big )r^{k+1},\\= & {} \sum _{k=1}^{\infty } (k-\frac{\alpha }{2})(|c_k|+|c_{-k}|)F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;1\big )r^{k+1}\\\le & {} \frac{4M}{\pi }\sum _{k=1}^{\infty }(k-\frac{\alpha }{2}) r^{k+1}=\frac{4Mr^2}{\pi (1-r)}\Big (\frac{1}{1-r}-\frac{\alpha }{2}\Big ). \end{aligned}$$

Case 2. $2<\alpha \le 4.$

As the function is decreasing on [0, 1), and using (3.4), it follows

$$\begin{aligned} F_\alpha (r)\le & {} \sum _{k=1}^{\infty } (|c_k|+|c_{-k}|)\dfrac{\alpha \big |k-\frac{\alpha }{2}\big |}{(k+1)}r^{k+1}\\= & {} (|c_1|+|c_{-1}|)\frac{\alpha (\alpha -2)r^2}{4}+ \sum _{k=2}^{\infty } (|c_k|+|c_{-k}|)\dfrac{\alpha \big (k-\frac{\alpha }{2}\big )}{(k+1)}r^{k+1}\\\le & {} \frac{12M \alpha a_\alpha r^2}{\pi } +\frac{4M\alpha a_\alpha }{\pi } \sum _{k=2}^{\infty }\frac{\Gamma (\frac{\alpha }{2}+k+1)(k-\frac{\alpha }{2})}{(k+1)!}r^{k+1}. \end{aligned}$$

Using Proposition 2.1 (b) for $n=2$, we deduce that $\displaystyle F_\alpha (r)\le \frac{12M\alpha a_\alpha r^2}{\pi }\Bigg (1+ \frac{ 4r}{3(1-r)^3}\Bigg ).$

Case 3. $n-1< \frac{\alpha }{2}\le n$, $n\ge 3$

By Lemma 1, we deduce that the function is increasing for $1\le k\le n-2$ and decreasing for $k\ge n-1$. We split the summation in $F_\alpha$ in two sums according to the monotonicity of .

Let us start the first sum. Using (3.14), we get

$$\begin{aligned} \Sigma _1:= & {} \sum _{k=1}^{n-2} (|c_k|+|c_{-k}|)\dfrac{\alpha \big |k-\frac{\alpha }{2}\big |}{(k+1)}F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;r^2\big )r^{k+1}\\\le & {} \sum _{k=1}^{n-2} (|c_k|+|c_{-k}|)(\frac{\alpha }{2}-k) F\big (-\frac{\alpha }{2},k-\frac{\alpha }{2};k+1;1\big )r^{k+1}\\\le & {} \frac{4M}{\pi }\sum _{k=1}^{n-2} (\frac{\alpha }{2}-k) r^{k+1}\\\le & {} \frac{4Mr^2}{\pi }\sum _{k=1}^{n-2}(n-k) = \frac{2Mr^2}{\pi } (n-2)(n+1). \end{aligned}$$

For the second sum, using Proposition 2.1 (b), we have

$$\begin{aligned} \Sigma _2:= & {} \sum _{k=n-1}^{\infty } (|c_k|+|c_{-k}|)\dfrac{\alpha \big |k-\frac{\alpha }{2}\big |}{(k+1)}F\big (-\frac{\alpha }{2}+1,k-\frac{\alpha }{2}+1;k+2;r^2\big )r^{k+1}\\\le & {} \sum _{k=n-1}^{\infty } (|c_k|+|c_{-k}|)\dfrac{\alpha \big |k-\frac{\alpha }{2}\big |}{(k+1)}r^{k+1}\\\le & {} \frac{4M\alpha a_\alpha }{\pi }\sum _{k=n-1}^{\infty } |k-\frac{\alpha }{2}|\frac{\Gamma (k+\frac{\alpha }{2}+1)}{(k+1)!} r^{k+1}\\\le & {} \frac{4M\alpha a_\alpha }{\pi }\left( \frac{(2n-1)!}{n!}r^n+ \sum _{k=n}^{\infty } (k-\frac{\alpha }{2})\frac{\Gamma (k+\frac{\alpha }{2}+1)}{(k+1)!} r^{k+1} \right) \\\le & {} \frac{4M\alpha a_\alpha }{\pi }\left( \frac{(2n-1)!}{n!}r^n+ \frac{(2n)!}{(n+1)!}\frac{r^{n+1}}{(1-r)^{n+1}} \right) . \end{aligned}$$

Finally

$$\begin{aligned} F_\alpha (r)\le \frac{2M}{\pi } (n-2)(n+1)r^2 +\frac{4M\alpha a_\alpha (2n-1)!}{\pi n!}\left( 1+ \frac{2nr}{(n+1)(1-r)^{n+1}} \right) r^n. \end{aligned}$$

We remark that his inequality remains valid for $n=2$. $\square$

3.5 Estimate of $G_\alpha (r)$

Lemma 7

Let $n\in {\mathbb N}$, $n\ge 1$ and $\frac{\alpha }{2}\in (n-1,n]$.

If $n=1$ or $n\ge 3$ , then

$$\begin{aligned} G_\alpha (r)\le \frac{2Mr^2|1-\frac{\alpha }{2}|}{\pi }. \end{aligned}$$

(3.15)

If $n=2$, then

$$\begin{aligned} G_\alpha (r)\le \frac{24M}{\pi }a_\alpha (\frac{\alpha }{2}-1)r^2. \end{aligned}$$

(3.16)

Proof

By the mean value theorem, there exists $c\in (0,r^2)$ such that

$$\begin{aligned} G_\alpha (r)= & {} r^2(|c_1|+|c_{-1}|)\bigg |\frac{d}{d\omega }F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;c\big )\bigg |\\= & {} r^2(|c_1|+|c_{-1}|)\frac{\alpha |\frac{\alpha }{2}-1|}{4}F\big (1-\frac{\alpha }{2},2-\frac{\alpha }{2};3;c\big ). \end{aligned}$$

Lemma 1 shows that the function is increasing for $0<\alpha \le 2$ or $\alpha >4$, and decreasing for $2<\alpha \le 4$.

Case 1. $0<\alpha \le 2$ or $\alpha >4$

As the function is increasing on [0, 1), and using (3.14), we get

$$\begin{aligned} F\big (1-\frac{\alpha }{2},2-\frac{\alpha }{2};3;c\big )\le F\big (1-\frac{\alpha }{2},2-\frac{\alpha }{2};3;1\big ) = \frac{2}{\alpha }F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;1\big ). \end{aligned}$$

By (3.2), we have

$$\begin{aligned} G_\alpha (r)= & {} r^2(|c_1|+|c_{-1}|)\frac{\alpha |\frac{2-\alpha }{2}|}{4}F\big (1-\frac{\alpha }{2},2-\frac{\alpha }{2};3;c\big )\\\le & {} r^2(|c_1|+|c_{-1}|)\frac{\alpha |1-\frac{\alpha }{2}|}{4}\frac{2}{\alpha }F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;1\big )\\= & {} \frac{r^2|1-\frac{\alpha }{2}|}{2}(|c_1|+|c_{-1}|)F\big (-\frac{\alpha }{2},1-\frac{\alpha }{2};2;1\big )\\\le & {} \frac{2Mr^2|1-\frac{\alpha }{2}|}{\pi }. \end{aligned}$$

Case 2. $2<\alpha \le 4$

As the function is decreasing on [0, 1) and using (3.4), we get

$$\begin{aligned} G_\alpha (r)\le & {} r^2(|c_1|+|c_{-1}|)\frac{\alpha (\frac{\alpha }{2}-1)}{4}\\\le & {} r^2(|c_1|+|c_{-1}|) (\frac{\alpha }{2}-1)\\\le & {} \frac{24Ma_\alpha }{\pi }r^2(\frac{\alpha }{2}-1). \end{aligned}$$

$\square$

Finally, combining (3.10–3.16), we conclude that

$$\begin{aligned} \Vert D_{u}(z)-D_{u}(0)\Vert \le E_\alpha (|z|) + F_\alpha (|z|)+G_\alpha (|z|)\le \sigma _\alpha (|z|), \end{aligned}$$

(3.17)

where $\sigma _\alpha$ is defined in Theorem 3.

It is clear that $\sigma _\alpha$ is strictly increasing on [0, 1) for all $\alpha >0$. Applying Theorem 2 (1.5), we get

$$\begin{aligned} 1=J_u(0)=\Vert D_u(0)\Vert \lambda _u(0)\le \frac{2c_\alpha (\alpha +2)M}{\pi }\lambda _u(0). \end{aligned}$$

Therefore,

$$\begin{aligned} \lambda _u(0)\ge \frac{\pi }{2 M c_\alpha (\alpha +2)} . \end{aligned}$$

(3.18)

We will prove that u is univalent in ${\mathbb D}_{r_\alpha }$, where $r_\alpha$ satisfies the following equation:

$$\begin{aligned} \frac{2c_\alpha (\alpha +2)}{\pi }M\sigma (r_\alpha )=1. \end{aligned}$$

Indeed, let $z_1,z_2\in {\mathbb D}_{r_\alpha }$ such that $z_1\ne z_2$ and $[z_1,z_2]$ denote the line segment from $z_1$ to $z_2$, by using (3.17) and (3.18), we get

$$\begin{aligned} |u(z_1)-u(z_2)|= & {} \bigg |\int _{[z_1,z_2]}u_z(z)\,dz+u_{\overline{z}}(z)\,d\overline{z}\bigg |\\\ge & {} \bigg |\int _{[z_1,z_2]}u_z(0)\,dz+u_{\overline{z}}(0)\,d\overline{z}\bigg |\\- & {} \bigg |\int _{[z_1,z_2]}(u_z(z)-u_z(0))\,dz+(u_{\overline{z}}(z)-u_{\overline{z}}(0))\,d\overline{z}\bigg |\\> & {} |z_2-z_1|\bigg \{ \frac{\pi }{2 M c_\alpha (\alpha +2)} -\sigma (r_\alpha ) \bigg \}\\= & {} 0. \end{aligned}$$

Thus $u(z_1)\ne u(z_2)$. The univalence of u follows from the arbitrariness of $z_1$ and $z_2$. This implies that u is univalent in ${\mathbb D}_{r_\alpha }$. As the mapping $\frac{\sigma _\alpha (|z|)}{|z|}$ is increasing, we deduce

$$\begin{aligned} \int _{[0,\xi ]}\sigma _\alpha (|z|)|dz|\le \frac{\sigma _\alpha (r_\alpha )r_\alpha }{2}. \end{aligned}$$

For any $\xi \in \partial {\mathbb D}_{r_\alpha }$, we have

$$\begin{aligned} |u(\xi )|= & {} \bigg |\int _{[0,\xi ]}u_z(z)dz+u_{\overline{z}}(z)d\overline{z}\bigg |\\\ge & {} \bigg |\int _{[0,\xi ]}u_z(0)dz+u_{\overline{z}}(0)d\overline{z}\bigg |\\- & {} \bigg |\int _{[0,\xi ]}(u_z(z)-u_z(0))dz+(u_{\overline{z}}(z)-u_{\overline{z}}(0))d\overline{z}\bigg |\\\ge & {} \lambda ( D_u (0))r_\alpha -\int _{[0,\xi ]}\sigma _\alpha (|z|)|dz|\\\ge & {} \sigma _\alpha (r_\alpha )r_\alpha -\frac{\sigma _\alpha (r_\alpha )r_\alpha }{2}=\frac{\sigma _\alpha (r_\alpha )r_\alpha }{2}. \end{aligned}$$

Hence $u({\mathbb D}_{r_\alpha })$ contains a univalent disk ${\mathbb D}_{R_\alpha }$ with $R_\alpha \ge \frac{\sigma _\alpha (r_\alpha )r_\alpha }{2}.$

4 Schwarz-type lemmas for solutions to inhomogeneous biharmonic equations

Proof of Theorem 4

The solution of (1.1) can be written in the following form

$$\begin{aligned} \Phi (z)=\frac{1}{2}(1-|z|^2)P[f+h](z)+K_2[f](z)-G[g](z). \end{aligned}$$

As $z\longmapsto K_2[f](z)$ is $T_2$-harmonic function, then by Theorem 1, we have

$$\begin{aligned} \bigg |K_2[f](z)-\frac{(1-|z|^2)^3}{(1+|z|^2)^2}K_2[f](0)\bigg | \le \frac{2}{\pi }\bigg [(1+|z|^2)\arctan |z|+ \frac{|z|(1-|z|^2)}{1+|z|^2}\bigg ]\Vert f\Vert _\infty . \end{aligned}$$

(4.1)

Using the estimate (1.2) for the harmonic mapping $P[f+h]$, we get

$$\begin{aligned} \bigg |P[f+h](z)-\frac{1-|z|^2}{1+|z|^2}P[f+h](0)\bigg |\le \frac{4}{\pi }\arctan |z|\,\Vert f+h\Vert _\infty . \end{aligned}$$

(4.2)

In addition, using [13, inequality 2.3], we obtain

$$\begin{aligned} |G[g](z)| \le \frac{(1-|z|^2)^2}{64}\Vert g\Vert _\infty . \end{aligned}$$

(4.3)

Finally as $\displaystyle K_2[f](0)=\frac{1}{2}P[f](0)$, then the inequality (1.8) follows directly from (4.1–4.3). $\square$

Proof of Theorem 5

The solution of (1.1) can be written in the following form

$$\begin{aligned} \Phi (z)= & {} \frac{1}{2} (1-|z|^2)P[f+h](z)+ K_2[f](z) - G[g](z). \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert D_\Phi (z)\Vert \le \frac{1}{2}(1-|z|^2)\Vert D_{P[f+h]} (z)\Vert +|z||P[f+h](z)| +\Vert D_{K_2[f]}(z)\Vert +\Vert D_{G[g]}(z)\Vert . \end{aligned}$$

By Colonna [15], we have

$$\begin{aligned} \Vert D_{P[f+h]}(z) \Vert \le \frac{4}{\pi } \frac{1}{1-|z|^2}\Vert f+h\Vert _\infty . \end{aligned}$$

(4.4)

It follows from [26, Lemma 2.5], that

$$\begin{aligned} \Vert D_{G[g]}(z)\Vert \le \frac{23}{48}\Vert g\Vert _\infty , \end{aligned}$$

(4.5)

since

$$\begin{aligned} \displaystyle \int _{\mathbb D}|G_z(z,\omega ) g(\omega )|dA(\omega ) \le \frac{23}{6} \Vert g\Vert _\infty \text { and } \int _{\mathbb D}|G_{\overline{z}}(z,\omega ) g(\omega )|dA(\omega ) \le \frac{23}{6} \Vert g\Vert _\infty . \end{aligned}$$

In addition by [12, Theorem 1], we have

$$\begin{aligned} \Vert D_{K_2[f]}(z)\Vert \le \frac{(2+5|z|)(1+|z|^2)}{1-|z|^2}, \text { for all } z\in {\mathbb D}. \end{aligned}$$

(4.6)

Therefore, combining (4.4–4.6), we obtain

$$\begin{aligned} \Vert D_\Phi (z)\Vert\le & {} \frac{1}{2}(1-|z|^2)\Vert D_{P[f+h]} (z)\Vert +|z||P[f+h](z)| +\Vert D_{K_2[f]}(z)\Vert +\Vert D_{G[g]}(z)\Vert \\\le & {} \frac{2}{\pi }\Vert P[f+h]\Vert _\infty +|z| \Vert f+h\Vert _\infty + \frac{(2+5|z|)(1+|z|^2)}{1-|z|^2} \Vert f\Vert _\infty +\frac{23}{48}\Vert g\Vert _\infty \\\le & {} (\frac{2}{\pi } +|z|)\Vert f+h\Vert _\infty +\frac{(2+5|z|)(1+|z|^2)}{1-|z|^2} \Vert f\Vert _\infty +\frac{23}{48}\Vert g\Vert _\infty . \end{aligned}$$

$\square$

Proof of Theorem 6

Suppose that $|z|=r$, it follows from Theorem 4 that

$$\begin{aligned} |\Phi (\eta )-\Phi (r\eta )|\ge & {} 1-\frac{1}{2}(1-r^2) \Vert f+h\Vert _\infty -\frac{2}{\pi }\bigg [(r^2+1)\arctan r+\frac{r(1-r^2)}{1+r^2}\bigg ]\\- & {} \frac{\Vert g\Vert _\infty (1-r^2)^2}{64}-\frac{1}{2}\frac{(1-r^2)^3}{(1+r^2)^2}|P[f](0)| -\frac{1}{2}\frac{(1-r^2)^2}{1+r^2}|P[f+h](0)|. \end{aligned}$$

Divide by $1-r$ and used the Hospital rule, we obtain

$$\begin{aligned} \liminf _{r\rightarrow 1}\frac{|\Phi (\eta )-\Phi (r\eta )|}{1-r}\ge & {} \lim _{r\rightarrow 1}\frac{1-\frac{2}{\pi }(r^2+1)\arctan r}{1-r}-\lim _{r\rightarrow 1}\frac{2}{\pi }\frac{r(1-r^2)}{(1-r)(1+r^2)}\\- & {} \frac{1}{2}\lim _{r\rightarrow 1}(1+r)\Vert f+h\Vert _\infty \\= & {} \varphi '(1)-\frac{2}{\pi }-\Vert f+h\Vert _\infty , \end{aligned}$$

where $\varphi (r)=\frac{2}{\pi }(r^2+1)\arctan r.$ Hence $\liminf _{r\longrightarrow 1}\frac{|\Phi (\eta )-\Phi (r\eta )|}{1-r}\ge 1-\Vert f+h\Vert _\infty .$ $\square$

5 A Landau-type theorem for solutions to inhomogeneous biharmonic equations

First, let us recall the following result.

Theorem F

([11], Lemma 1) Suppose f is a harmonic mapping of ${\mathbb D}$ into ${\mathbb C}$ such that $|f(z)|\le M$ for all $z\in {\mathbb D}$ and $f(z)=\sum _{n=0}^\infty a_n z^n+\sum _{n=1}^\infty \overline{b}_n \overline{z}^n.$

Then $|a_0|\le M$ and for all $n\ge 1,$ $|a_n|+|b_n|\le \frac{4M}{\pi }.$

Proof of Theorem 7

The solution of (1.1) can be written in the following form

$$\begin{aligned} \Phi (z)=H_0[f+h](z)+K_2[f](z)-G[g](z), \end{aligned}$$

where

$$\begin{aligned} H_0[f+h](z)= \frac{1}{2} (1-|z|^2)P[f+h](z). \end{aligned}$$

(5.1)

Since $P[f+h]$ is harmonic in ${\mathbb D}$, we have $P[f+h](z)=\sum _{n=0}^\infty a_n z^n+\sum _{n=1}^\infty \overline{b}_n \overline{z}^n.$ As $|P[f+h](z)|\le M_2$ for all $z\in {\mathbb D}$, by Theorem F, we have

$$\begin{aligned} |a_n|+|b_n|\le \frac{4M_2}{\pi } \quad \text { for } n\ge 1. \end{aligned}$$

(5.2)

Using the chain rule and by (5.1) and (5.2), we have

$$\begin{aligned}&\Vert D_{H_0[f+h]}(z)-D_{H_0[f+h]}(0)\Vert \nonumber \\&\quad \le |z| |P[f+h](z)|+\frac{1}{2}\Vert D_{P[f+h]}(z)-D_{P[f+h]}(0)\Vert +\frac{1}{2}|z|^2\Vert D_{P[f+h]}(z)\Vert \nonumber \\&\quad \le M_2|z|+\frac{1}{2}\sum _{n\ge 2}n(|a_n|+|b_n|) |z|^{n-1} +\frac{1}{2}|z|^2 \sum _{n\ge 1}n(|a_n|+|b_n|) |z|^{n-1} \nonumber \\&\quad \le M_2|z| +\frac{1}{2}(1+|z|^2) \sum _{n\ge 2}n(|a_n|+|b_n|) |z|^{n-1}+\frac{1}{2}|z|^2(|a_1|+|b_1|) \nonumber \\&\quad \le M_2|z|+\frac{2M_2|z|}{\pi }\bigg [\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg ]. \end{aligned}$$

(5.3)

Since $K_2$ is $T_2$-harmonic, then

$$\begin{aligned} K_2(z)=\sum _{k=0}^\infty c_k F(-1,k-1;k+1;|z|^2)z^k+ \sum _{k=1}^\infty c_{-k} F(-1,k-1;k+1;|z|^2)\overline{z}^k. \end{aligned}$$

Let us denote

$$\begin{aligned} K_2^0[f](z):=K_2(z)-c_0F(-1,-1;1;|z|^2)=K_2[f](z)-c_0(1+|z|^2). \end{aligned}$$

Hence

$$\begin{aligned} K_2[f]=K_2^0[f](z)+c_0(1+|z|^2), \end{aligned}$$

and

$$\begin{aligned} \Vert D_{K_2[f]}(z)-D_{K_2[f]}(0)\Vert \le \Vert D_{K_2^0[f]}(z)-D_{K_2^0[f]}(0)\Vert +2|c_0||z|. \end{aligned}$$

By (3.3), we have $2|c_0| \le M_1.$ On the other hand, as $K_2^0(f)$ is a $T_2$-harmonic function with $K_2^0(0)=0$, it yields

$$\begin{aligned} \Vert D_{K_2^0[f]}(z)-D_{K_2^0[f]}(0)\Vert \le \sigma _2(r), \end{aligned}$$

where $\sigma _2$ is defined by $\sigma _2(r)= \frac{4M_1r}{ \pi (1-r)^3}(r^2(1-r)+3),$ see Remark 1.2. Thus

$$\begin{aligned} \Vert D_{K_2[f]}(z)-D_{K_2[f]}(0)\Vert \le \frac{4M_1|z|}{\pi (1-|z|)^3}\big (|z|^2(1-|z|)+3\big )+M_1|z|. \end{aligned}$$

(5.4)

Let

$$\begin{aligned} \psi _1(z)=\bigg |\frac{1}{16\pi }\int _{\mathbb D}g(\omega )(G_z(z,\omega )-G_z(0,\omega ))dA(\omega )\bigg |, \end{aligned}$$

and

$$\begin{aligned} \psi _2(z)=\bigg |\frac{1}{16\pi }\int _{\mathbb D}g(\omega )(G_{\overline{z}}(z,\omega )-G_{\overline{z}}(0,\omega ))dA(\omega )\bigg |. \end{aligned}$$

Then by [13, Inequality (3.6)], we have

$$\begin{aligned} \max (\psi _1(z),\psi _2(z))\le \bigg (\frac{1-|z|^2}{16}+\frac{43}{120}\bigg )\Vert g\Vert _\infty |z|. \end{aligned}$$

(5.5)

Now, it follows from (5.3)–(5.5) that

$$\begin{aligned} \Vert D_{\Phi }(z)-D_{\Phi } (0)\Vert\le & {} M_2|z|+\frac{2M_2|z|}{\pi }\bigg [\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg ]\\+ & {} \sigma _2(|z|)+M_1|z|+\psi _1(z)+\psi _2(z)\\\le & {} \mu (|z|), \end{aligned}$$

where

$$\begin{aligned} \mu (|z|)= & {} (M_1+M_2+\frac{101}{120} M_3)|z|+\frac{2M_2|z|}{\pi }\bigg [\frac{(2-|z|)(1+|z|^2)}{(1-|z|)^2}+|z| \bigg ] \\+ & {} \frac{4M_1|z|}{ \pi (1-|z|)^3}(|z|^2(1-|z|)+3). \end{aligned}$$

Remark that not only $\mu (|z|)$ is increasing but also $\displaystyle \frac{\mu (|z|)}{|z|}$ is increasing with respect to |z| in [0, 1). By Theorem 5, we obtain that

$$\begin{aligned} 1=J_\Phi (0)=\Vert D_\Phi (0)\Vert \lambda ( D_\Phi (0))\le \lambda ( D_\Phi (0))\bigg (\frac{4}{\pi }M_1+\frac{2}{\pi }M_2+\frac{23}{48}M_3\bigg ) \end{aligned}$$

yields $\lambda ( D_\Phi (0))\ge \frac{1}{\frac{4}{\pi }M_1+\frac{2}{\pi }M_2+\frac{23}{48}M_3}.$ As in Theorem 3, we prove that $\Phi$ is univalent in ${\mathbb D}_{r_0}$, where $r_0$ satisfies $(\frac{4}{\pi } M_1+\frac{2}{\pi }M_2+\frac{23}{48}M_3)\mu (r_0)=1,$ and $\Phi ({\mathbb D}_{r_0})$ contains an univalent disk ${\mathbb D}_{R_0}$ with the radius $R_0$ satisfying $R_0\ge \frac{r_0}{\frac{8}{\pi } M_1+\frac{4}{\pi }M_2+\frac{23}{24}M_3}.$ $\square$