law, Eq. (\[Growth-Rate-Henry\]). Dashed curve corresponds to Sievert’s law, Eq. (\[Growth-Rate\]). For both solubility laws $P_0/\Pi = 10^3$.[]{data-label="Fig:Rate"}](Rate.eps){width="1.0\linewidth"}

Differentiating Eq. (\[Growth-Rate\]) by $R$ we obtain $$\label{d-Growth-Rate}
\frac{d\dot{R}}{dR} = \frac{D s}{(R + R_\sigma)^2} \left[ \frac{3}{4} \frac{R_\sigma (R + R_\sigma)}{\sqrt{1 + \frac{3}{2} \frac{R_\sigma}{R}} R^2} - \left( \zeta + 1 - \sqrt{1 + \frac{3}{2} \frac{R_\sigma}{R}} \right) \right]\; .$$ It can be easily seen, that, as it is for Henry’s law, here bubble radius growth rate as a function of variable $R$ also has the only maximum. Denote the corresponding radius as $R_m$, we can obtain its value from Eq. (\[Extremum\]). Using Eq. (\[d-Growth-Rate\]) we rewrite this equation as $$\label{Extremum-Eq}
RR_\sigma + R_\sigma^2 + \frac{4}{3} R^2 \left( 1 + \frac{3}{2} \frac{R_\sigma}{R} \right) - \frac{4}{3} R^2 \sqrt{1 + \frac{3}{2} \frac{R_\sigma}{R}} (\zeta + 1) = 0\; .$$

Assuming that the sought quantity $R_m$ is considerably less than $R_\sigma$, we will use for $R$ in Eq. (\[Extremum-Eq\]) the strong inequality $3 R_\sigma/ 2 R \gg 1$. Below we will need this inequality to be even more stronger $$\label{Strong-For-Extr}
R_\sigma/ 3 R \gg 1\; .$$ It will allow us to omit the second term in the brackets and the second addend in the square root in Eq. (\[Extremum-Eq\]). 