Stability of a bubble expanding and translating through an inviscid ...

362 Dinesh Khattar and B B Chakraborty
where k is a constant, σ and ρ are the surface tension coefficient and density of the liquid
outside the bubble respectively, p g0 , R 0 and U 0 are the gas pressure, radius of the bubble
and its speed initially, respectively, p e is the pressure in the liquid at a large distance from
the bubble and γ is the ratio of the two specific heats of the gas.
We find that eq. (1) can be written, in view of (2), as
{
d 2 (
dt 2 R 5/2) − 5 8 k2 R −11/2 + 5
( ) }
3γ R0
2ρ R1/2 p e − p g0 + 2σ = 0. (3)
R R
Defining r as
r = R 5/2
and taking γ = 4/3 for simplicity, we finally get from (3) the equation
d 2 r
dt 2 − 5 8 k2 r −11/5 + 5 { ( r0
) 8/5
2ρ r1/5 p e − p g0 + 2σr
−2/5}
= 0 (4)
r
where
r 0 = R 5/2 .
We use p g0 and U 0 as characteristic pressure and speed respectively and r 0 as the characteristic
value of r and T 0 as that of time, where T 0 = R 0 /U 0 . We define the dimensionless
quantities r ′ ,t ′ and p ′ e as
r ′ = r/r 0 , t ′ = t/T 0 , p ′ e = p e/p g0 .
Using the relation k = U 0 R0 3 [cf. (2)], we can write eq. (4) as
{
d 2 r ′
dt ′2 = 5 8 r′−11/5 1 + 4p (
g 0
ρU0
2 p e ′ r′12/5 − r ′4/5 +
2σ ) }
r ′2 . (5)
p g0 R 0
Omitting the dashes from r ′ ,t ′ and P e ′ and using from now on these undashed symbols for
the corresponding quantities, we find that eq. (5) can be written in dimensionless form as
{
d 2 r
dt 2 = 5 8 r−11/5 1 − 4p (
g 0
ρU0
2 p e r 12/5 − r 4/5 +
2σ ) }
r 2 . (6)
p g0 R 0
Finally, defining x and y by the equations
x = r, y = dr
dt
we can write eq. (6) as
⎡
⎢
⎣
dx
dt
dy
dt
⎤ ⎡
⎥
⎦ = ⎢
⎣
y
{
5
8 x−11/5 1 − 4p (
g 0
ρU0
2 p e x 12/5 − x 4/5 +
(7)
⎤
2σ ) } ⎥
x 2 ⎦ . (8)
p g0 R 0