In case of pulsating bubble arising from underwater explosion, bubble radius satisfies the following equation.
$x^3\dot{x}^{2} + x^3 + \frac{k}{x^{3(\gamma-1)}} = 1$
The minimum and maximum bubble radii can be obtained by substituting $\dot{x}=0$ in the above equation.
$x^3 + \frac{k}{x^{3(\gamma-1)}} = 1$
As per [1], for small $k$, the roots of the above equation are
$x_{0} \sim k^{\frac{1}{3(\gamma-1)}}\bigg[1+k^{\frac{1}{\gamma-1}}/3(\gamma-1)\bigg]$; minimum
$x_{m} \sim 1-\frac{k}{3}\bigg[1+\big(\gamma-\frac{2}{3}\big)k\bigg]$; maximum
I would like to know the derivation of these approximate roots.
Screenshot from the original source [1] has been included for more details.
[1] Hicks, A.N. 1972. The Theory of Explosion Induced Ship Whipping Motions. NCRE Report R579.
Case $x$ maximum and $k$ small :
$x^3$ is large compared to $\frac{k}{x^{3(\gamma-1)}}$
$x^3=1-\frac{k}{x^{3(\gamma-1)}} \quad\to\quad x=\left(1-\frac{k}{x^{3(\gamma-1)}}\right)^{1/3} $
First approximate: $x\simeq 1 \quad\to\quad x\simeq \left(1-\frac{k}{1}\right)^{1/3} \simeq 1-\frac{k}{3}$
Second approximate: $x\simeq 1-\frac{k}{3} \quad\to\quad x\simeq\left(1-\frac{k}{(1-\frac{k}{3} )^{3(\gamma-1)}}\right)^{1/3}$
Expending to series of powers of $k$ leads to : $$x\simeq 1-\frac{k}{3}-\frac{\gamma-\frac{2}{3}}{3}k^2$$
Case $x$ small :
$x^3$ is small compared to $\frac{k}{x^{3(\gamma-1)}}$
$\frac{k}{x^{3(\gamma-1)}}=1-x^3 \quad\to\quad x=\left(\frac{k}{1-x^3}\right)^{\frac{1}{3(\gamma-1)}}$
First approximate: $x\simeq 0 \quad\to\quad x\simeq\left(k\right)^{\frac{1}{3(\gamma-1)}}$
Second approximate: $x\simeq (k)^{\frac{1}{3(\gamma-1)}} \quad\to\quad x\simeq \left(\frac{k}{1-\left(k^{\frac{1}{3(\gamma-1)}} \right)^3}\right)^{\frac{1}{3(\gamma-1)}} \simeq (k)^{\frac{1}{3(\gamma-1)}} \left(1+(k)^{\frac{1}{(\gamma-1)}} \right)^{\frac{1}{3(\gamma-1)}} $
$$x\simeq (k)^{\frac{1}{3(\gamma-1)}} \left(1+ {\frac{1}{3(\gamma-1)}} (k)^{\frac{1}{(\gamma-1)}} \right) $$