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Suppose we have the asymptotic

$$ R \sim 1 - \sqrt{\frac{2t}{\beta}} + \frac{1}{\beta}\left[ \frac{V}{\sqrt{\pi}} \sqrt{kt} - \frac{2}{3}(1+kV)t + \mathcal{O}(t^{3/2}e^{-1/kt}) \right] +\mathcal{O}(\beta^{-3/2}) \tag 1$$

as $t \to 0^+$, and $\beta \to \infty$. I would like to know how this equation can be asymptotically solved to find

$$ \frac{t}{\beta} \sim \left[\frac{1}{2} + \frac{V}{\beta^{1/2}}\sqrt{\frac{k}{2\pi}} + \mathcal{O}(\beta^{-1}) \right](1-R)^2 + \left[-\frac{1}{3}(1+Vk) + \mathcal{O}(\beta^{-1/2}) \right](1-R)^3 + \mathcal{O}((1-R)^4) \tag 2 $$

as $R \to 1^-$, $\beta \to \infty$. Note that $V$, $k$ and $\beta$ are constants, and $R$ is a function depending on $t$ that approches $1$ as $t \to 0^+$.

I have tried solving the quadratic $1-R \sim At^{1/2} - Bt + \mathcal{O}(t^{3/2})$ with the appropriate $A$ and $B$ terms, this however results in terms as high as $(1-R)^1$, which is confusing to me since such terms to not appear in (2).

This expansion can be found in the following paper (pages 7-8): https://www.researchgate.net/publication/27473447_Classical_two-phase_Stefan_problem_for_spheres#:~:text=The%20classical%20Stefan%20problem%20for,flows%20in%20one%20phase%20only.&text=In%20this%20study%2C%20the%20full,the%20large%20Stefan%20number%20limit.

I'm new to this kind of analysis, so I'd appreciate all the help I can get.

Cornelius
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    If $t\to 0$ in the first expression, $R \to 1+\mathcal{O}(\beta ^{ - 3/2} )$, whereas the inverted asymptotics suggests that $t\to 0$ corresponds to $R\to 1$. Something is not correct there. If the term $\mathcal{O}(\beta ^{ - 3/2} )$ was not there in the first formula, I would be able tell you how to do the inversion. – Gary May 31 '21 at 18:53
  • I forgot to mention that this asymptotic is for $\beta \to \infty$, so I believe that resolves the issue. – Cornelius May 31 '21 at 23:20
  • If that extra error term was not there, you could basically use series inversion. Write the RHS as a truncated series in powers of $\sqrt{t/\beta}$ and revert for $1-R$. Alternatively, you can use the method of successive approximation. The authors are definitely not very precise with this problem. – Gary Jun 01 '21 at 14:17

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