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

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)