On some properties of bi-univalent functions in the unit disc

Lemma 1. Grunsky coefficient $w_{13}$ for a function $f \in \mathcal{S}$ satisfies $$|w_{13}| \le \begin{cases} \frac{1}{2}(1+|w_{11}|^2) & , |w_{11}| \in [0,b] \\ \sqrt{\frac{1}{3}(1-|w_{11}|^2)} & , |w_{11}| \in [b,1] \end{cases},$$ where $$b = \sqrt{\frac{2\sqrt{7}-5}{3}} = 0.311...$$ The first inequality is a consequence of the inequality $|2w_{13} - w_{11}^2| \le 1$ , obtained from $|a_3 - a_2^2| \le 1$ , valid for $f \in \mathcal{S}$ . The second one follows from the inequality (7). As it has been shown in [7, p. 57], if f is given by (1) then the coefficients $a_2$ , $a_3$ , $a_4$ and $a_5$ are expressed by Grunsky's coefficients $\omega_{2p-1,2q-1}$ of the function $f_*$ given by (4) in the following way: $$a_{2} = 2\omega_{11},$$ $$a_{3} = 2\omega_{13} + 3\omega_{11}^{2},$$ $$a_{4} = 2\omega_{33} + 8\omega_{11}\omega_{13} + \frac{10}{3}\omega_{11}^{3},$$ $$a_{5} = 2\omega_{35} + 8\omega_{11}\omega_{33} + 5\omega_{13}^{2} + 18\omega_{11}^{2}\omega_{13} + \frac{7}{3}\omega_{11}^{4},$$ $$0 = 3\omega_{15} - 3\omega_{11}\omega_{13} + \omega_{11}^{3} - 3\omega_{33},$$ $$0 = \omega_{17} - \omega_{35} - \omega_{11}\omega_{33} - \omega_{13}^{2} + \frac{1}{3}\omega_{11}^{4}.$$ A comprehensive overview of the application of Grunsky coefficients in the general class of univalent functions is given in [15].

Theorem 1. Let $f \in \mathcal{S}_b$ be given by (1). Then - (i) $|a_3| \leq 2.427...$ , - $(ii) |a_4| \leq 3.461 \dots,$ - $(iii) |a_5| \le 4.993 \dots$

Theorem 2. Let $f \in \mathcal{S}_b$ be given by (1). Then - (i) $|a_4| |a_3| \le 1.174...$ , - $(ii) |a_5| |a_4| \le 1.822 \dots$ Proof. (i) At the beginning we consider critical points of $f_4$ on $\Omega$ . For this reason we look for the solutions of the following system $$\begin{cases} \frac{\partial f_4(x,y)}{\partial x} = \left(\frac{9}{a} - 12\right)x^2 + \left(6 - \frac{2}{a}\right)y - \frac{2}{\sqrt{5}}\frac{x}{\sqrt{1 - x^2 - 3y^2}} = 0\\ \frac{\partial f_4(x,y)}{\partial y} = \left(6 - \frac{2}{a}\right)x - \frac{6}{\sqrt{5}}\frac{y}{\sqrt{1 - x^2 - 3y^2}} = 0 \end{cases}$$ Hence. $$3y \frac{\partial f_4(x,y)}{\partial x} - x \frac{\partial f_4(x,y)}{\partial y} = \frac{1}{a} \left[ -x^2 [9y(4a-3) + 6a - 2] + 6(3a-1)y^2 \right] = 0$$ $$x = \frac{y\sqrt{6}\sqrt{3a-1}}{\sqrt{9y(4a-3)+6a-2}} := h_1(y)$$ Putting it into the second equation of the system we receive an equation of variable y. Numerical calculation shows that this equation has only one real solution in (0,1), that is $y'_6 = 0.358...$ , such that $(h_1(y'_6), y'_6) =$ $(0.634..., 0.358...) \in \Omega$ and $f_4(h_1(y_6), y_6) = 1.174...$ - Finally, we need to find the greatest value of $f_4$ on the edges of $\Omega$ : $f_4(0,y) = \frac{2}{\sqrt{5}} \sqrt{1-3y^2} \le f_4(0,0) = \frac{2}{\sqrt{5}} = 0.894...,$ $f_4(a,y) \le f_4(a,y_7') = 1.139...$ , where $y_7' = 0.327...$ is the unique real solution of the equation $f'_4(a,y) = 0$ on the interval (0,d), - $-f_4(x,0) = \left(\frac{3}{a}-4\right)x^3 + \frac{2}{\sqrt{5}}\sqrt{1-x^2} \le f_4(0,0) = \frac{2}{\sqrt{5}} = 0.894..., \text{ since }$ - $f_4(x,0)$ is a decreasing function on (0,a), $g_5(x) := f_4(x,(1+x^2)/2) = \frac{1}{5}\sqrt{5-50x^2-15x^4}-x\left[\left(1-\frac{2}{a}\right)x^2-3+\frac{1}{a}\right]$ , so $g_5(x) \le g_5(x_8') = 0.709\dots$ , where $x_8' = 0.252\dots$ is the unique real solution of $g'_5(x) = 0$ on the interval (0, b), - $g_6(x) := f_4(x, \sqrt{1-x^2}/\sqrt{3}) = 2\left(1-\frac{1}{3a}\right)\sqrt{3(1-x^2)} + \left(\frac{3}{a}-4\right)x^3$ , so $g_6(x) \le g_6(x_9') = 0.969\ldots$ , where $x_9' = 0.715\ldots$ is the unique real solution of $g_3'(x) = 0$ on the interval (b, a). So, on the domain $\Omega$ , the function $f_4$ achieves the greatest value 1.174.... Consequently, $$|a_4| - |a_3| \le 1.174\dots$$ - (ii) Using Wolfram Mathematica we receive that the system of equations $\frac{\partial f_5(x,y)}{\partial x}=0$ and $\frac{\partial f_5(x,y)}{\partial y}=0$ , has only one solution in the interior of $\Omega$ , that is (0.717..., 0.312...) with value 1.822... On the edges of $\Omega$ we - $f_5(0,y) = 3y^2 + \frac{2}{\sqrt{7}}\sqrt{1-3y^2} \le f_5(0,1/2) = 1.127...$ since $f_5(0,y)$ is increasing on (0, 1/2), - $f_5(a,y) \le f_5(a,y'_{10}) = 1.819...$ , where $y'_{10} = 0.300...$ is the unique real solution of the equation $f_5'(a, y) = 0$ on the interval (0, ), - $f_5(x,0) = \left(\frac{4}{a} 5\right) x^4 + \left[\frac{2}{\sqrt{7}} + \frac{1}{\sqrt{5}} \left(6 \frac{2}{a}\right) x\right] \sqrt{1 x^2} \le f_5(a, y'_{11}) =$ 1.374..., where $y'_{11} = 0.667...$ is the unique real solution of $f'_5(x,0) =$ 0 on (0, a), $$\begin{array}{l} -g_7(x):=f_5(x,(1+x^2)/2)=\left[\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{5}}\left(3-\frac{1}{a}\right)x\right]\sqrt{1-10x^2-3x^4}+\\ \frac{1}{4}(3+30x^2+7x^4)+\frac{1}{a}(x^4-3x^2), \text{ so } g_7(x)\leq g_7(x_{12}')=1.317\ldots, \text{ where } x_{12}'=0.247\ldots \text{ is the unique real solution of } g_7'(x)=0 \text{ on } (0,b),\\ -g_8(x):=f_5(x,\sqrt{1-x^2}/\sqrt{3})=2\left(2-\frac{1}{a}\right)x^2\sqrt{3(1-x^2)}+1-x^2+\left(\frac{4}{a}-5\right)x^4 \text{ and } g_8(x)\leq g_8(a)=1.402\ldots, \text{ since } g_8 \text{ is strictly increasing} \end{array}$$ So, on the domain $\Omega$ , the function $f_5$ takes the greatest value 1.822... and for this reason $$|a_5| - |a_4| \le 1.822 \dots$$

Theorem 3. Let $f \in \mathcal{S}_b$ be given by (1). Then $$|H_{2,2}(f)| \le 1.280\dots$$

Theorem 4. Let $f \in \mathcal{S}_b$ be given by (1). Then $$|\gamma_3| \leq 0.551....$$