Harmonic function for which the second dilatation is α-spiral

Abstract

Let $f=h+\overline{g}$ be a harmonic function in the unit disc $\mathbb{D}$. We will give some properties of f under the condition the second dilatation is α-spiral.

MSC:30C45, 30C55.

1 Introduction

A planar harmonic mapping in the unit disc $\mathbb{D}=\{z\in \mathbb{C}||z|<1\}$ is a complex-valued harmonic function f which maps $\mathbb{D}$ onto some planar domain $f(\mathbb{D})$. Since $\mathbb{D}$ is simply connected, the mapping f has a canonical decomposition $f=h+\overline{g}$, where h and g are analytic in $\mathbb{D}$. As usual, we call h the analytic part of f and g the co-analytic part of f. An elegant and complete account of the theory of planar harmonic mapping is given in Duren’s monograph [1].

Lewy [2] proved in 1936 that the harmonic function f is locally univalent in a simply connected domain ${\mathbb{D}}_1$ if and only if its Jacobian

$$J_f(z)=|h^{\prime }(z){|}^2-|g^{\prime }(z){|}^2>0$$

is different from zero in ${\mathbb{D}}_1$. In view of this result, locally univalent harmonic mappings in the unit disc are either sense-reversing if

$$|g^{\prime }(z)|>|h^{\prime }(z)|$$

in ${\mathbb{D}}_1$ or sense-preserving if

$$|g^{\prime }(z)|<|h^{\prime }(z)|$$

in ${\mathbb{D}}_1$. Throughout this paper, we will restrict ourselves to the study of sense-preserving harmonic mappings. However, since f is sense-preserving if and only if $\overline{f}$ is sense-reserving, all the results obtained in this article regarding sense-preserving harmonic mappings can be adapted to sense-reversing ones. Note that $f=h+\overline{g}$ is sense-preserving in $\mathbb{D}$ if and only if $h^{\prime }(z)$ does not vanish in the unit disc and the second-complex dilatation $w(z)=\frac{g^{\prime }(z)}{h^{\prime }(z)}$ has the property $|w(z)|<1$ in $\mathbb{D}$; therefore, we can take $h(z)=z+a_2{z}^2+\cdots$ , $g(z)=b_{1}z+b_2{z}^2+\cdots$ . Thus, the class of all harmonic mappings being sense-preserving in the unit disc can be defined by

$$\begin{array}{rcl}{S}_H& =& \{f=h(z)+\overline{g(z)}|h(z)=z+a_2{z}^2+\cdots ,\\ g(z)=b_{1}z+b_2{z}^2+\cdots ,f\text{ sense-preserving}\}.\end{array}$$

Let Ω be the family of functions $\varphi (z)$ which are regular in $\mathbb{D}$ and satisfy the conditions $\varphi (0)=0$, $|\varphi (z)|<1$ for all $z\in \mathbb{D}$. Denote by P the family of functions $p(z)=1+p_{1}z+p_2{z}^2+\cdots$ which are regular in $\mathbb{D}$ such that

$$p(z)=\frac{1+\varphi (z)}{1-\varphi (z)}$$

(1.1)

for some function $\varphi (z)\in \mathrm{\Omega }$ for all $z\in \mathbb{D}$.

Next, let $S^{\ast }$ denote the family of functions $s(z)=z+c_2{z}^2+c_3{z}^3+\cdots$ which are regular in $\mathbb{D}$ such that

$$z\frac{s^{\prime }(z)}{s(z)}=p(z)$$

(1.2)

for some $p(z)\in P$ for all $z\in \mathbb{D}$.

Let $s_1(z)=z+{\alpha }_2{z}^2+{\alpha }_3{z}_3+\cdots$ and $s_2(z)=z+{\beta }_2{z}^2+{\beta }_3{z}^3+\cdots$ be analytic functions in $\mathbb{D}$. If there exists $\varphi (z)\in \mathrm{\Omega }$ such that $s_1(z)=s_2(\varphi (z))$ for all $z\in \mathbb{D}$, then we say that $s_1(z)$ is subordinate to $s_2(z)$ and we write $s_1(z)\prec s_2(z)$, then $s_1(\mathbb{D})\subset s_2(\mathbb{D})$.

Now, we consider the following class of harmonic mappings in the plane:

$$\begin{array}{rcl}{S}_{\mathrm{HPST}}^{\ast }(\alpha )& =& \{f=h(z)+\overline{g(z)}|f\in S_H,h(z)\in S^{\ast },\\ Re(e^{i\alpha }w(z))=Re\left(e^{i\alpha }\frac{g^{\prime }(z)}{h^{\prime }(z)}\right)>0,|\alpha |<\frac{\pi }{2}\}.\end{array}$$

(1.3)

In the present paper, we will investigate the class $S_{\mathrm{HPST}}^{\ast }(\alpha )$.

We will need the following lemma and theorem in the sequel.

Theorem 1.1 ([3, 4])

Let $h(z)$ be an element of $S^{\ast }$, then

$$\frac{r}{{(1+r)}^2}\le |h(z)|\le \frac{r}{{(1-r)}^2},$$

for all $|z|=r<1$.

$$\frac{1-r}{{(1+r)}^3}\le |h^{\prime }(z)|\le \frac{1+r}{{(1-r)}^3}.$$

These inequalities are sharp because the extremal function is $h(z)=\frac{z}{{(1-z)}^2}$.

Lemma 1.2 ([2, 5])

Let $h(z)$ and $g(z)$ be regular in $\mathbb{D}$, $h(z)$ map $|z|<1$ onto a many-sheeted starlike region, $Re(e^{i\alpha }\frac{g^{\prime }(z)}{h^{\prime }(z)})>0$, $|\alpha |<\frac{\pi }{2}$ for $|z|<1$. $h(0)=g(0)=0$. Then $Re(e^{i\alpha }\frac{g(z)}{h(z)})>0$ for $|z|<1$.

2 Main results

Lemma 2.1 Let $f=h(z)+\overline{g(z)}$ be an element of $S_{\mathrm{HPST}}^{\ast }(\alpha )$ then

$$\frac{|b_1|-r}{1-|b_1|r}\le \left|\frac{g^{\prime }(z)}{h^{\prime }(z)}\right|\le \frac{|b_1|+r}{1+|b_1|r}$$

(2.1)

for all $|z|=r<1$. This inequality is sharp because the extremal function is

$$e^{i\alpha }\frac{g^{\prime }(z)}{h^{\prime }(z)}=\frac{z+b}{1+\overline{b}z},$$

where $b=e^{i\alpha }{b}_1$.

Proof

Since

then the function

$$\varphi (z)=\frac{W(z)-W(0)}{1-\overline{W(0)}W(z)}=\frac{W(0)-b}{1-\overline{b}W(0)}=\frac{b-b}{1-b^2}=0$$

satisfies the condition of the Schwarz lemma. Using the definition of subordination, we have

$$W(z)=e^{i\alpha }w(z)=e^{i\alpha }\frac{g^{\prime }(z)}{h^{\prime }(z)}=\frac{b+\varphi (z)}{1+\overline{b}\varphi (z)}\iff e^{i\alpha }\frac{g^{\prime }(z)}{h^{\prime }(z)}\prec \frac{b+z}{1+\overline{b}z}.$$

On the other hand, the transformation $(\frac{b+z}{1+\overline{b}z})$ maps $|z|<1$ onto the disc with the center

$$C(r)=(\frac{{\alpha }_1(1-r^2)}{1-{|b_1|}^2{r}^2},\frac{{\alpha }_2(1-r^2)}{1-{|b_1|}^2{r}^2}),b={\alpha }_1+i{\alpha }_2$$

and the radius

$$\rho (r)=\frac{(1-{|b_1|}^2)r}{1-{|b_1|}^2{r}^2}.$$

Therefore, we can write

$$|e^{i\alpha }\frac{g^{\prime }(z)}{h^{\prime }(z)}-\frac{b_1(1-r^2)}{1-{|b_1|}^2{r}^2}|\le \frac{(1-{|b_1|}^2)r}{1-{|b_1|}^2{r}^2}$$

(2.2)

which gives (2.1). □

Corollary 2.2 Let $f\in S_{\mathrm{HPST}}^{\ast }(\alpha )$, then

(2.3)

(2.4)

for all $|z|=r<1$.

Proof

Using Lemma 1.2 and Lemma 2.1, then we can write

(2.5)

(2.6)

If we use Theorem 1.1 in the inequalities (2.5) and (2.6), we get (2.3) and (2.4). □

Corollary 2.3 Let $f=h(z)+\overline{g(z)}$ be an element of $S_{HPTS}^{\ast }(\alpha )$, then

$$\frac{(1-|b_1{|}^2){(1-r)}^3}{{(1+r)}^5{(1+|b_1|r)}^2}\le J_{f(z)}\le \frac{(1-|b_1{|}^2){(1+r)}^3}{{(1-r)}^5{(1+|b_1|r)}^2}$$

(2.7)

for all $|z|=r<1$.

Proof

Since

$$J_{f(z)}=|h^{\prime }(z){|}^2-|g^{\prime }(z){|}^2=|h^{\prime }(z){|}^2(1-|w(z){|}^2),$$

(2.8)

using Lemma 2.1 and Theorem 1.1 in the equality (2.8) and after simple calculations, we get (2.7). □

Corollary 2.4 If $f=h(z)+\overline{g(z)}$ is an element of $S_{HPTS}^{\ast }(\alpha )$, then

(2.9)

where $a=|b_1|$ for all $|z|=r<1$.

Proof

Using Corollary 2.2 and Theorem 1.1, we obtain

$$(|h^{\prime }(z)|-|g^{\prime }(z)|)\ge \frac{(1-r^4)(1+|b_1|r)-(1+r^4)(|b_1|+r)}{{(1-r)}^3{(1+r)}^3(1+|b_1|r)},$$

and

$$(|h^{\prime }(z)|+|g^{\prime }(z)|)\le \frac{{(1+r)}^2(1+|b_1|)}{{(1-r)}^3(1+|b_1|r)}.$$

Therefore, we have

(2.10)

Integrating the last inequality (2.10), we get (2.9). □

Theorem 2.5 Let $f=h(z)+\overline{g(z)}$ be an element of $S_{HPTS}^{\ast }(\alpha )$, then

$$\sum _{k=1}^n|A_k{|}^2\le {|t+1|}^2+\sum _{k=1}^n|B_k{|}^2$$

(2.11)

where $A_k=(k+1)(\frac{b_{k+1}}{b_1}-a_{k+1})$; $B_k=(k+1)(\frac{b_{k+1}}{b_1}+t{a}_{k+1})$; $a_k$ and $b_k$ are the coefficients of the functions $h(z)$ and $g(z)$; $k=1,2,3,\dots ,n$; $t=2s-1$; $s=e^{-i\alpha }cos\alpha$.

Proof

Since

$$g(z)=b_{1}z+b_2{z}^2+b_3{z}^3+\cdots \Rightarrow g^{\prime }(z)=b_1+2b_{2}z+3b_3{z}^2+\cdots .$$

We denote by $G(z)=\frac{1}{b_1}g(z)$

then we have

$$\{\begin{array}{c}\frac{1}{cos\alpha }(e^{i\alpha }\frac{\frac{1}{b_1}{g}^{\prime }(z)}{h^{\prime }(z)}-isin\alpha )=p(z)\iff e^{i\alpha }\frac{\frac{1}{b_1}{g}^{\prime }(z)}{h^{\prime }(z)}=cos\alpha p(z)+isin\alpha ,\hfill \\ \iff \frac{\frac{1}{b_1}{g}^{\prime }(z)}{h^{\prime }(z)}=1+e^{-i\alpha }cos\alpha (p(z)-1).\hfill \end{array}$$

(2.12)

Since $p(z)$ is in P, there is a function $\varphi (z)$ satisfying the conditions of the Schwarz lemma such that

$$p(z)=\frac{1+\varphi (z)}{1-\varphi (z)}\iff p(z)-1=\frac{2\varphi (z)}{1-\varphi (z)}.$$

(2.13)

Using this equation in (2.12) and after the following calculations given above

$$\frac{\frac{1}{b_1}{g}^{\prime }(z)}{h^{\prime }(z)}=1+e^{-i\alpha }cos\alpha (p(z)-1)=1+s\left(\frac{2\varphi (z)}{1-\varphi (z)}\right)\Rightarrow ,$$

we get the following equality:

$$\frac{1}{b_1}{g}^{\prime }(z)-h^{\prime }(z)=(t{h}^{\prime }(z)+\frac{1}{b_1}{g}^{\prime }(z)).$$

(2.14)

If $\varphi (z)=c_{1}z+c_2{z}^2+c_3{z}^3+\cdots$ , we have

$$\sum _{k=1}^n{A}_k{z}^k+\sum _{k=n+1}^{\mathrm{\infty }}{D}_k{z}^k=[(1+t)+\sum _{k=1}^n{B}_k{z}^k]\varphi (z),$$

(2.15)

where

$$\sum _{k=n+1}^{\mathrm{\infty }}{D}_k{z}^k=\sum _{k=n+1}^{\mathrm{\infty }}{A}_k{z}^k-(c_1{B}_n{z}^{n+1}+c_1{B}_{n+1}{z}^{n+2}+\cdots ).$$

Therefore, the equality (2.15) can be considered in the following form:

$$F(z)=G(z)\varphi (z).$$

(2.16)

Using the Clunie method [6], then we can write

$$\frac{1}{2\pi }{\int }_0^{2\pi }{\left|F\left(r{e}^{i\theta }\right)\right|}^{2}d\theta \le \frac{1}{2\pi }{\int }_0^{2\pi }{\left|G\left(r{e}^{i\theta }\right)\right|}^{2}d\theta ,$$

which gives

$$\sum _{k=1}^n{|A_k|}^2{r}^{2k}+\sum _{k=n+1}^{\mathrm{\infty }}{|D_k|}^2{r}^{2k}\le ({|t+1|}^2+\sum _{k=1}^n{|B_k|}^2{r}^{2k}).$$

(2.17)

Eventually, we will let $r\to 1^{-}$, then we have

$$\sum _{k=1}^n{|A_k|}^2\le {|t+1|}^2+\sum _{k=1}^n{|B_k|}^2.$$

 □