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

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 112, No. 2, May 2002, pp. 361–365.
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Stability of a bubble expanding and translating through
an inviscid liquid
DINESH KHATTAR and B B CHAKRABORTY ∗
Department of Mathematics, Kirori Mal College, Delhi University, Delhi 110 007, India
∗ Department of Mathematics, Delhi University, Delhi 110 007, India
MS received 27 August 2001; revised 2 February 2002
Abstract. A bubble expands adiabatically and translates in an incompressible and
inviscid liquid. We investigate the number of equilibrium points of the bubble and the
nature of stability of the bubble at these points. We find that there is only one equililbrium
point and the bubble is stable there.
Keywords.
function.
Bubble; Rayleigh’s equation; stability; Hamiltonian; Liapounov’s
1. Introduction
The basic equation, describing the mathematical model for an expanding bubble in an
incompressible liquid has been given by Rayleigh [8], Plesset [6] and others (Noltingk
and Neppiras [5] and Poritsky [7]). This mathematical model, and the equation describing
it, have been generalized (Chakraborty [1], Chakraborty and Tuteja [3]) when the bubble
expands as well as translates in a liquid.
The equation, governing the variation of R with time t, is a highly non-linear equation.
The progress in the study of this equation is, therefore, generally expected numerically.
However, when liquid outside the bubble is inviscid, after transforming the equation suitably,
we prove analytically, that the expanding and translating bubble has only one equilibrium
point and the bubble is stable at this equilibrium point. In our discussion, we use
Liapounov’s first stability theorem [4]. This stability problem in the absence of translation
was eariler studied by Chakraborty and Khattar [2].
2. Mathematical formulation and analysis
As the bubble expands adiabatically and translates with velocity U, its radius R satisfies
the equations (Chakraborty [1], Chakraborty and Tuteja [3])
R d2 R
dt 2 + 3 ( ) {
dR
2
− U 2
2 dt 4 + 1 ( ) }
3γ R0
p e − p g0 + 2σ = 0 (1)
ρ
R R
and
UR 3 = U 0 R0 3 = k, (2)
361

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

Stability of a bubble 363
Equation (9) defines a Hamiltonian system, with Hamiltonian H(x,y) so that
and
dx
dt
dy
dt
= ∂H
∂Y
=− ∂H
∂x . (10)
In view of (8), eqs (9) and (10) give
H = y2
2 + 25
+ 5p g 0
2pU 2 0
48 x−6/5
{ 5
6 p ex 6/5 + 5 2 x−2/5 + 5σ
2p g0 R 0
x 4/5 }
+ C (11)
where C is an arbitrary constant.
The equilibrium point (x, y) for the basic dynamical system, defined by (8), is given by
∂H
∂x = ∂H
∂y
= 0. (12)
Equation (12), in view of (11), shows that an equilibrium point (x, y) satisfies the equations.
(9)
y = 0,
p e x 12/5 + 2σx2
p g0 R 0
−x 4/5 − ρU2 0
4p g0
= 0. (13)
In view of (7) and the fact that r = R 5/2 , we find from (13) that at an equilibrium, R
satisfies the equation
p e R 6 +
2σ
p g0 R 0
R 5 − R 2 − ρU2 0
4p g0
= 0.
There is only one change in sign in the coefficients of powers of R in this algebraic equation.
Therefore, by Decartes’ rule of sign [9], there is atmost one positive real root for R.
When R → 0,
f(R)=p e R 6 +
which is negative and when R →∞,
f(R)→+∞.
2σ
p g0 R 0
R 5 −R 2 − ρU2 0
4p g0
→ −ρU2 0
4p g0
This confirms that f(R) = 0 has only one positive root R and the expanding and
translating bubble has only one equilibrium point given by this root.
Writing (8) as
dx
dt
= F(x), (14)

364 Dinesh Khattar and B B Chakraborty
where
⎡
x = ⎣ x ⎤
⎦
y
and
F(x) =
⎡
⎢
⎣
y
{
5
8 x−11/5 1 − 4p (
g 0
ρU0
2 p e x 12/5 − x 4/5 +
⎤
2σ ) } ⎥
x 2 ⎦ . (15)
p g0 R 0
We find that F(x) vanishes at the equilibrium point. Also, H and its partial derivatives are
continuous at all points except when x = 0. If α is a positive real root of (13), then eq. (15)
shows that x = α and y = 0 is the equilibrium point of (14). Let us choose C in (11) in
such a way that at this equilibrium point, the value of H vanishes. Hence,
H (α, 0) = 25
48 α−6/5
+ 5p g 0
2ρU 2 0
H is minimum at the equilibrium point x = α, y = 0if
∂ 2 H
∂x 2 · ∂2 H
∂y 2
{ 5
6 p eα 6/5 + 5 2 α−2/5 + 5σ
2p g0 R 0
α 4/5 }
+ C = 0. (16)
( ∂ 2 ) 2 − H
> 0. (17)
∂x∂y
The condition (17), in view of (11), gives us the condition
p e x 12/5 −
2σ x 2 + 7x 4/5 + 11ρU2 0
> 0 (18)
p g0 R 0 4p g0
for the existence of a minimum at the equilibrium point (α, 0). Adding the left hand side
of (13) to the left hand side of (18), we find that this condition (18) becomes
p e x 12/5 + 3x 4/5 + 5 ρU0
2 > 0 (19)
4 p g0
which is always satisfied as x is real and positive. Therefore, H is minimum at the equilibrium
point x = α, y = 0, but H vanishes at (α, 0). Thus H is always positive near (α, 0)
and is therefore positive definite in the neighborhood of the equilibrium point (α, 0). Also
−F · grad H vanishes in view of (9), (10) and (14). Thus, −F · grad H is positive semidefinite.
We can therefore choose H as a Liapounov function and by Liapounov’s theorem
[4], we have the result that a bubble expanding and translating through an inviscid liquid
is stable at its equilibrium point.
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Stability of a bubble 365
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