\D;\label{EQ_PDE1}\\
 &\Phi(z,z)  = f(z), & z\in{\mathbb{D}},\label{EQ_PDE2}\end{aligned}$$ with a coefficient $\varphi$ meromorphic in $\C \times \D$ and satisfying the following two conditions:

-   $\varphi(0,0)=0$;

-   $r\!\left|\varphi(w/r,w)\right| \le k$ for all $w\in\D$ and all $r\in\big(|w|^2,1\big)$.

Suppose also that there exists $\varepsilon\in(0,1)$ and $M>0$ such that $$\label{EQ_Phi-at-infinity}
|\Phi(z,w)|\le M |z|\quad\text{whenever~$|w|\le |z|$ and $|z\,w|\le \varepsilon^2$.}$$ Then $f$ admits a $k$-q.c. Becker extension given by $$F(z) := \Phi(z, 1/\bar{z}), \hspace{15pt} |z|>1.$$ In particular, $f\in{\mathcal{S}}_k^B$.

Since $\Phi(0,0)=f(0)=0$, it is sufficient to check condition  only for $z$ large enough.

Before proving Theorem \[TH\_PDE\_THM\], let us consider a few examples.

Let $f$ be a holomorphic function in ${\mathbb{D}}$ with $f'(0)-1=f(0)=0$. Set $\varphi(z,w):=(z-w)f''(w)/f'(w)$. Then $\Phi(z,w):=f(w)+(z-w)f'(w)$ solves problem , and satisfies . Condition (i) in Theorem \[TH\_PDE\_THM\] holds trivially, while (ii) is equivalent to $(1-|w|^2)|wf''(w)/f'(w)|\le k$, which is a classical sufficient condition for q.c.-extendibility.

Similarly, setting $\varphi(z,w):=f'(w)-1$ and $\Phi(z,w):=f(w)+z-w$, we recover another well-known sufficient condition for q.c.-extendibility $\big|f'(w)-1\big|\le k$, $w\in{\mathbb{D}}$, see [@Brown §3].

The following corollary represents another example.

\[CR\_AW-Becker\] Fix $k\in(0,1)$. Let $f(z)=z+a_2 z^2+\ldots$ be holomorphic in ${\mathbb{D}}$. If $$\label{EQ_AW-Becker}
\tfrac{4\sqrt3}{9}(1-|z|^2)|a_2|\,+\,(1-|z|^2)^2\big|a_2^2+\tfrac12S_f(z)\big|~\le~ k\quad\text{for all~$z\in{\mathbb{D}}$},$$ then $f\in{\mathcal{S}}_k^B$, with its Becker extension given by $F(z)=\Phi(z,1/\bar z)$ for all $z\in{\mathbb{C}}\setminus\overline{\mathbb{D}}$, where $$\label{EQ_Phi}
\Phi(z,w):=f(w)+\frac{f'(w)}{\frac{1}{z-w}+a_2-\frac{1}{2}\frac{f''(w)}{f'(w)}}.$$

Let $\varphi(z,w):=2a_2(z-w)+(z-w)^2\big(a_2^2+\tfrac12S_f(w)\big)$, where $S_f$ stands for the Schwarzian derivative of 