This section deals with \(\Re (\Theta (\xi ) ):= \Re ((1-\alpha ) ( \frac{\xi }{\sigma (\xi )} )^{\beta }+\alpha ( \frac{\xi }{\sigma (\xi )} )^{\beta +1} \sigma '(\xi )-1 )\).
Theorem 3.1
Consider the class \(\mathcal{M}^{\alpha ,\beta }(\lambda )\). Define a functional
$$\begin{aligned} \Theta (\xi ) &= (1-\alpha ) \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{ \beta }+ \alpha \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{\beta +1} \sigma '(\xi )-1 \\ &=\xi +\sum_{n=2}^{\infty }\phi _{n} \xi ^{n}, \quad \xi \in \cup . \end{aligned}$$
If \(\Re (\Theta (\xi ) )>0\), then
$$ \vert \phi _{n} \vert \leq 2 \int _{0}^{2\pi } \bigl\vert e^{-in\theta } \bigr\vert \,d\mu (\theta ), \quad n \geq 2, $$
where dμ is a probability measure. Moreover, if \(\Re (e^{i \varpi }\Theta (\xi ) )>0\), then
$$ \lambda (\xi )= \frac{(A-B)\xi }{1+B\xi } \in \mathcal{C}, \quad \vert A \vert =1, \vert B \vert =1, $$
where \(\mathcal{C}\) is the class of analytic convex in ∪.
Proof
For the first part of the theorem, we suppose that
$$ \Re \bigl(\Theta (\xi )+1\bigr)= \Re \Biggl(1+\sum_{n=1}^{\infty } \phi _{n} \xi ^{n} \Biggr)>0, \quad \phi _{1}=1. $$
Then, by the Carathéodory positivist method for analytic functions, we have
$$ \vert \phi _{n} \vert \leq 2 \int _{0}^{2\pi } \bigl\vert e^{-in\theta } \bigr\vert d\mu (\theta ), \quad n \geq 2, $$
where dμ is a probability measure. For the second part, we have the assumption
$$ \Re \bigl(e^{i \varpi }\bigl(1+\Theta (\xi )\bigr) \bigr)>0, \quad \xi \in \cup , \varpi \in \mathbb{R}. $$
Then, according to [15]-Theorem 1.6(P22) and for some real numbers ϖ, we obtain
$$ \bigl[\Theta (\xi )\bigr] +1 \approx \frac{A \xi +1}{B\xi +1}, \quad \xi \in \cup , $$
or \(\lambda (\xi )= \frac{(A-B)\xi }{1+B\xi }\). But \(\frac{A\xi +1}{B \xi +1}\) is convex in ∪, then we obtain that \(\lambda (\xi ) \in \mathcal{C}\). □
The next theorem is the converse of Theorem 3.1.
Theorem 3.2
Consider the functional \([\Theta (\xi )]=(1-\alpha ) (\frac{\xi }{\sigma (\xi )} )^{ \beta }+\alpha (\frac{\xi }{\sigma (\xi )} )^{\beta +1} \sigma '(\xi )-1\), where \(\sigma \in \wedge \). Then the subordination
$$\begin{aligned}& \frac{\beta [\Theta (\xi )] ([\Theta (\xi )] +1)+\xi [\Theta (\xi )]' }{\beta ([\Theta (\xi )] +1)} \prec \frac{A\xi +1}{B\xi +1} \\& \quad \bigl(\alpha \in [0,1], \beta \in [1,\infty ), B \in [-1,0], A \in (0,1], \xi \in \cup \bigr) \end{aligned}$$
implies \(\sigma \in \mathcal{M}^{\alpha ,\beta }(\frac{(A-B)\xi }{1+B\xi }) \). Moreover, the subordination
$$ \sigma (\xi )+\xi \sigma ' (\xi ) \bigl(\beta \sigma (\xi )+ \alpha \bigr)\prec \frac{(A-B)\xi }{1+B\xi } $$
yields \(\sigma (\xi )\prec \lambda (\xi )=\frac{(A-B)\xi }{1+B\xi }\).
Proof
Let \(p(\xi ):=[\Theta (\xi )] +1 \). Then a computation gives
$$\begin{aligned} p(\xi )+ \frac{\xi p' (\xi )}{\beta (p(\xi )+1)}&= \bigl(\bigl[\Theta ( \xi )\bigr] +1 \bigr) \biggl( \frac{\xi [\Theta (\xi )]' }{\beta ( [\Theta (\xi )] +1 )} \biggr) \\ &= \frac{\beta [ \Theta (\xi )] ([\Theta (\xi )] +1 ) +\xi [ \Theta (\xi )] ' }{\beta ([\Theta (\xi )] +1 )} \\ &\prec \frac{A\xi +1}{B\xi +1}. \end{aligned}$$
According to [2]-Theorem 3.3d., we have
$$ \Theta (\xi )+1 \prec \frac{A\xi +1}{B\xi +1}. $$
That is, \(\sigma \in \mathcal{M}^{\alpha ,\beta }(\frac{(A-B)\xi }{1+B\xi })\).
Now, since \(\Re ( \frac{(A-B)\xi }{1+B\xi } )>0\) for \(0<\Re (z)<1\) and \(B \in [-1,0]\), \(A\in (0,1]\), where \((\frac{(A-B)\xi }{1+B\xi })\in \mathcal{C}\) in ∪, then this together with [2]-Corollary 3.4a.2 implies that \(\sigma (\xi ) \prec \lambda (\xi )\). □
The next result shows that every univalent solution of Bernoulli’s equation is the best dominant for all other solutions.
Theorem 3.3
Let \(\lambda \in \mathcal{C}\). Assume that \(\sigma _{1}\in \wedge \) is a univalent solution in ∪ of Bernoulli’s equation
$$ (1-\alpha ) \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{\beta }+\alpha \biggl( \frac{\xi }{\sigma (\xi )} \biggr)^{\beta +1} \sigma '(\xi )-1= \lambda (\xi ). $$
If σ and \(\sigma _{1} \in \mathcal{M}^{\alpha ,\beta }(\lambda )\), then \(\sigma (\xi ) \prec \sigma _{1}(\xi )\).
Proof
Suppose that
$$ \Theta \bigl[\sigma (\xi )\bigr]=(1-\alpha ) \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{\beta }+\alpha \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{ \beta +1} \sigma '(\xi )-1. $$
Clearly, \(\Theta [\sigma (0)]=\lambda (0)=0\); and since \(\sigma , \sigma _{1} \in \wedge \), then \(\sigma (0)= \sigma _{1}(0)=0\). Moreover, we have
$$ \Theta \bigl[\sigma (\xi )\bigr]=(1-\alpha ) \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{\beta }+\alpha \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{ \beta +1} \sigma '(\xi )-1\prec \lambda (\xi ) $$
and
$$ \Theta \bigl[\sigma _{1}(\xi )\bigr]=(1-\alpha ) \biggl( \frac{\xi }{\sigma _{1}(\xi )} \biggr)^{\beta }+\alpha \biggl( \frac{\xi }{\sigma _{1}(\xi )} \biggr)^{\beta +1} \sigma '_{1}( \xi )-1 \prec \lambda (\xi ). $$
Thus, in view of [2]-Theorem 3.4.c, \(\sigma (\xi ) \prec \sigma _{1}(\xi )\) such that \(\sigma _{1}\) is the best dominant of the last subordination. □
We proceed to present more information about solutions of Bernoulli’s equation. Next two results indicate that a solution of Bernoulli’s equation can be considered as a solution of the Briot–Bouquet equation. A more interesting outcome is that the equation has a positive real and univalent solution.
Theorem 3.4
Let λ be analytic and g be an analytic starlike function in ∪. Assume that \(\sigma \in \wedge \) is a solution of Bernoulli’s equation
$$ \Theta \bigl[\sigma (\xi )\bigr]= \lambda (\xi ), $$
where
$$ \Re \bigl( \Theta \bigl[\sigma (\xi )\bigr] \bigr)=\Re \biggl((1-\alpha ) \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{\beta }+\alpha \biggl( \frac{\xi }{\sigma (\xi )} \biggr)^{\beta +1} \sigma '(\xi )-1 \biggr) >0. $$
Then σ is a solution of the Briot–Bouquet equation
$$ \sigma (\xi )+ \frac{\sigma ' (\xi )g(\xi )}{\sigma (\xi )g'(\xi )}= \lambda (\xi ) $$
such that \(\Re (\sigma (\xi ))>0\).
Moreover, if \(\Theta [\sigma (\xi )]\in \mathcal{S}^{*}(\alpha )\) (starlike of order α), then \(\sigma \in \mathcal{M}^{\alpha ,\beta } ( \frac{\xi }{(1-\xi )^{2-2\alpha }} )\), \(\alpha \in [0,1]\), \(| \xi | \in (0.21,0.3)\), and
$$ \bigl(\Theta \bigl[\sigma (\xi )\bigr] \bigr)'\prec \biggl( \frac{\xi }{(1-\xi )^{2-2\alpha }} \biggr)'. $$
Proof
Since g is a starlike analytic function in ∪, then
$$ \Re \biggl(\frac{\xi g' (\xi )}{g(\xi )} \biggr)>0, \quad \xi \in \cup . $$
Define a function \(Q: \cup \rightarrow \cup \) as follows:
$$ Q(\xi ):= \biggl(\frac{\xi g' (\xi )}{g(\xi )} \biggr) \Theta \bigl[ \sigma (\xi ) \bigr]. $$
Thus, \(\Re (Q(\xi ) )>0 \). According to [2]-Theorem 3.4j, the Briot–Bouquet equation
$$ \sigma (\xi )+ \frac{\sigma ' (\xi )g(\xi )}{\sigma (\xi )g'(\xi )}= \lambda (\xi ) $$
such that \(\Re (\sigma (\xi ))>0\).
Since \(\Theta [\sigma (\xi )]\in \mathcal{S}^{*}(\alpha )\), then in view of [15]-Corollary 2.2, there is a probability measure \(\nu \in \partial \cup \) such that
$$ \Theta \bigl[\sigma (\xi )\bigr]= \int _{\partial \cup } \frac{\xi }{(1-t\xi )^{2-2\alpha }}\,d\nu (\xi ). $$
That is, \(\Theta [\sigma (\xi )]\) satisfies the majority inequality
$$ \Theta \bigl[\sigma (\xi )\bigr]\ll \frac{\xi }{(1-\xi )^{2-2\alpha }}. $$
But \(\frac{\xi }{(1-\xi )^{2-2\alpha }}\) is starlike in ∪, then in virtue of [16]-Corollary 2, we have
$$ \Theta \bigl[\sigma (\xi )\bigr]\prec \frac{\xi }{(1-\xi )^{2-2\alpha }}, \quad \vert \xi \vert \in (0.21,0.3), $$
which leads to \(\sigma \in \mathcal{M}^{\alpha ,\beta } ( \frac{\xi }{(1-\xi )^{2-2\alpha }} )\), \(\alpha \in [0,1]\), \(| \xi | \in (0.21,0.3)\). The last part comes immediately from [16]-Theorem 3. □
Corollary 3.5
Consider Bernoulli’s equation
$$\begin{aligned} \Theta \bigl[\sigma (\xi )\bigr]= \frac{\xi }{(1-\xi )^{2-2\alpha }}, \quad \vert \xi \vert \in (0.21,0.3). \end{aligned}$$
(3.1)
Then the solution is defined by the hypergeometric function as follows:
$$ c= \frac{\alpha ^{2} \xi ^{\frac{(\alpha - 1) \beta }{\alpha }} (\frac{\xi }{\sigma (z)} )^{-\beta } \sigma (\xi )^{-\beta } (\beta \xi \ _{2}F_{1}(2 - 2 \alpha , 1 - \frac{\beta }{\alpha }, 2 -\frac{\beta }{\alpha }; \xi ) + (\alpha -\beta ) ( (\frac{\xi }{\sigma (\xi )} )^{\beta }- 1 ) )}{(\alpha - 1) \beta ^{2} (\alpha - \beta ) }, $$
where c is a nonzero constant.
Example 3.6
-
Let \(\alpha =0.5\) and \(\beta =1\), the solution of (3.1) is given by the formula (see Fig. 1)
$$ \sigma (\xi ) = \frac{1}{c \xi + \xi \int _{\cup } (\frac{2}{\xi ^{3}(\xi -1)} )\, d \xi }; $$
-
Let \(\alpha =0.5\) and \(\beta =2\), the solutions of (3.1) are formulated by the structures
$$ \sigma (\xi ) =\pm \frac{1}{\sqrt{c \xi ^{2} +2 \xi ^{2} \int _{\cup } (\frac{2}{\xi ^{5}(\xi -1)} )\, d \xi }}; $$
-
Let \(\alpha =0.75\) and \(\beta =2\), the solutions of (3.1) are introduced by the formulas
$$ \sigma (\xi ) =\pm \frac{ (1.73205 \xi )}{\sqrt{3 c \xi ^{8/3} + 16 \sqrt{1 - \xi } \xi ^{8/3} _{2}F_{1}(0.5, 2.66667, 1.5;1 - \xi ) + 3)}}; $$
-
Let \(\alpha =0.99\) and \(\beta =2\), the solutions of (3.1) are presented by the formulas
$$ \sigma (\xi ) =\pm \frac{ (7.141 \xi )}{\sqrt{51 c \xi ^{101/50} + 101 _{2}F_{1}(-1.02, 0.02, -0.02; \xi ) + 51)}}. $$
To present our next result, we need the following notion.
Definition 3.7
Two functions \(f, g \in \wedge \) are called convoluted if and only if they satisfy the following operation:
$$\begin{aligned} f(\xi )*g(\xi )&= \Biggl(\xi + \sum_{n=2}^{\infty }a_{n} \xi ^{n} \Biggr)* \Biggl(\xi + \sum_{n=2}^{\infty }b_{n} \xi ^{n} \Biggr) \\ &= \Biggl(\xi + \sum_{n=2}^{\infty }a_{n}b_{n} \xi ^{n} \Biggr). \end{aligned}$$
And a function \(f \in \mathcal{R}_{\alpha }\) if and only if
$$ f* \biggl( \frac{\xi }{(1-\xi )^{2-2\alpha }} \biggr)\in \mathbb{S}^{*}( \alpha ), \quad 0\leq \alpha < 1; $$
and
$$ \Re \biggl(\frac{f(\xi )}{\xi } \biggr)> \frac{1}{2}. $$
Theorem 3.8
Let \(\Theta [\sigma (\xi )] \in \mathcal{R}_{\alpha }\). Then \(\sigma (\xi ) \in \mathcal{M}^{\alpha ,\beta } ( \frac{\xi }{(1-\xi } )\), \(\alpha \in [0,1]\), \(|\xi | \in (0.28, \sqrt{2}-1)\).
Proof
Since \(\Theta [\sigma (\xi )] \in \mathcal{R}_{\alpha }\), then in view of [15]-Corollary 2.1, there occurs a probability measure μ on ∂∪ such that
$$ \Theta \bigl[\sigma (\xi )\bigr] = \int _{\partial \cup } \frac{\xi }{1-\xi } \,d \mu (\xi ). $$
Consequently, we indicate that
$$ \Theta \bigl[\sigma (\xi )\bigr] \ll \frac{\xi }{1-\xi }. $$
Now, since \(\frac{\xi }{1-\xi } \in \mathcal{C}\), then by using [16]-Corollary 1, we obtain
$$ \Theta \bigl[\sigma (\xi )\bigr] \prec \frac{\xi }{1- \xi }, \quad \vert \xi \vert \in (0.28, \sqrt{2}-1), $$
which means that \(\sigma (\xi ) \in \mathcal{M}^{\alpha ,\beta } ( \frac{\xi }{(1- \xi } )\), \(\alpha \in [0,1]\), \(|\xi | \in (0.28, \sqrt{2}-1)\). □
Example 3.9
-
Let \(\alpha =0.5\) and \(\beta =1\), the solution of
$$\begin{aligned} \Theta \bigl[\sigma (\xi )\bigr] = \frac{\xi }{1-\xi }, \quad \vert \xi \vert \in (0.28, \sqrt{2}-1) \end{aligned}$$
(3.2)
is given by the formula (see Fig. 2)
$$ \sigma (\xi ) = \frac{\xi }{(c \xi ^{2} + 2 \xi ^{2} \log (1 - \xi ) - 2 \xi ^{2} \log (\xi ) + 2 \xi + 1)}; $$
-
Let \(\alpha =0.25\) and \(\beta =1\), the solution becomes (see Fig. 3)
$$ \sigma (\xi )\approx \frac{ \xi }{1-\xi ^{3}}=\xi + \xi ^{4}+\xi ^{7}+ \xi ^{10}+O\bigl(\xi ^{1}3\bigr). $$
Solutions of perturbed Bernoulli’s equation are formulated in the following theorem.
Theorem 3.10
Let λ be analytic and g be an analytic starlike function in ∪. Assume that \(\sigma \in \wedge \) is a solution of Bernoulli’s equation
$$ \Theta \bigl[\sigma (\xi )\bigr]= \lambda (\xi )+\epsilon , \quad \epsilon > 0, $$
where \((\Theta [\sigma (\xi ) )^{-1} \in \mathcal{C}\) and
$$ \Re \bigl( \Theta \bigl[\sigma (\xi )\bigr] \bigr)=\Re \biggl((1-\alpha ) \biggl(\frac{\xi }{\sigma (\xi )} \biggr)^{\beta }+\alpha \biggl( \frac{\xi }{\sigma (\xi )} \biggr)^{\beta +1} \sigma '(\xi )-1 \biggr) >0. $$
Then σ is a univalent solution of the Briot–Bouquet equation
$$ \sigma (\xi )+ \frac{\sigma ' (\xi )g(\xi )}{g'(\xi )[\sigma (\xi )+\epsilon ]}= \lambda (\xi ) $$
such that \(\Re (\sigma (\xi )+\epsilon )>0\) admitting the structure
$$\begin{aligned} \sigma (\xi )= \Upsilon ^{\epsilon }(\xi ) \biggl( \int _{0}^{\xi }\frac{g' (\tau )\Upsilon (\tau )}{g(\tau )}\,d\tau \biggr)^{-1}- \epsilon , \end{aligned}$$
(3.3)
where
$$ \Upsilon (\xi )= g(\xi ) \exp \biggl( \frac{\int _{0}^{\xi }\frac{g'(\tau )\lambda (\tau )}{g(\tau )}\,d\tau }{\epsilon } \biggr). $$
Proof
Since g is a starlike analytic function in ∪, then
$$ \Re \biggl(\frac{\xi g' (\xi )}{g(\xi )} \biggr)>0, \quad \xi \in \cup . $$
Define a function \(Q: \cup \rightarrow \cup \) as follows:
$$ Q(\xi ):= \biggl(\frac{\xi g' (\xi )}{g(\xi )} \biggr) \Theta \bigl[ \sigma (\xi ) \bigr]. $$
Thus, \(\Re (Q(\xi ) )>0 \). According to [2]-Theorem 3.4j, the Briot–Bouquet equation
$$ \sigma (\xi )+ \frac{\sigma ' (\xi )g(\xi )}{g'(\xi )[\sigma (\xi )+\epsilon ]}= \lambda (\xi ) $$
such that \(\Re (\sigma (\xi )+\epsilon )>0\). Since \((\Theta [\sigma (\xi ) )^{-1} \in \mathcal{C}\), then we have
$$ \Re \biggl(Q(\xi )+ 1+ \frac{\xi ( (\Theta [\sigma (\xi ) )^{-1} )'' }{ ( (\Theta [\sigma (\xi ) )^{-1} )'} \biggr) >0. $$
According to [2]-Theorem 3.4k, the solution of the Briot–Bouquet equation is univalent in ∪ taking formula (3.3). □