1Linear stability analysis
III Hydrodynamic Stability
1.3 Classical Kelvin–Helmholtz instability
Let’s return to the situation of Rayleigh–Taylor instability, but this time, there
is some horizontal flow in the two layers, and they may be of different velocities.
ρ
2
ρ
1
U
1
U
2
Our analysis here will be not be very detailed, partly because what we do
is largely the same as the analysis of the Rayleigh–Taylor instability, but also
because the analysis makes quite a few assumptions which we wish to examine
more deeply.
In this scenario, we can still use a velocity potential
(u, w) =
∂φ
∂x
,
∂φ
∂z
.
We shall only consider the system in 2D. The far field now has velocity
φ
1
= U
1
x + φ
0
1
φ
2
= U
2
x + φ
0
2
with φ
0
1
→ 0 as z → ∞ and φ
0
2
→ 0 as z → −∞.
The boundary conditions are the same. Continuity of vertical velocity requires
∂φ
1,2
∂z
z=η
=
Dη
Dt
.
The dynamic boundary condition is that we have continuity of pressure at the
interface if there is no surface tension, in which case Bernoulli tells us
ρ
1
∂φ
1
∂t
+
1
2
ρ
1
|∇φ
1
|
2
+ gρ
1
η = ρ
2
∂φ
2
∂t
+
1
2
ρ
2
|∇φ
2
|
2
+ gρ
2
η.
The interface conditions are non-linear, and again we want to linearize. But
since U
1,2
is of order 1, linearization will be different. We have
∂φ
0
1,2
∂z
z=0
=
∂
∂t
+ U
1,2
∂
∂x
η.
So the Bernoulli condition gives us
ρ
1
∂
∂t
+ U
1
∂
∂x
φ
0
1
+ gη
= ρ
2
∂
∂t
+ U
2
∂
∂x
φ
0
2
+ gη
This modifies our previous eigenvalue problem for the phase speed and wavenum-
ber
ω
=
kc
,
k
=
2π
λ
. We go exactly as before, and after some work, we find that
we have
c =
ρ
1
U
1
+ ρ
2
U
2
ρ
1
+ ρ
2
±
1
ρ
1
+ ρ
2
g(ρ
2
2
− ρ
2
1
)
k
− ρ
1
ρ
2
(U
1
− U
2
)
2
1/2
.
So we see that we have instability if c has an imaginary part, i.e.
k >
g(ρ
2
2
− ρ
2
1
)
ρ
1
ρ
2
(U
1
− U
2
)
2
.
So we see that there is instability for sufficiently large wavenumbers, even for
static stability. Similar to Rayleigh–Taylor instability, the growth rate
kc
i
grows monotonically with wavenumber, and as
k → ∞
, the instability becomes
proportional to the difference
|U
1
−U
2
|
(as opposed to Rayleigh–Taylor instability,
where it grows unboundedly).
How can we think about this result? If
U
1
6
=
U
2
and we have a discrete
change in velocity, then this means there is a
δ
-function of vorticity at the
interface. So it should not be surprising that the result is unstable!
Another way to think about it is to change our frame of reference so that
U
1
=
U
=
−U
2
. In the Boussinesq limit, we have
c
r
= 0, and instability arises
whenever
g∆ρλ
ρU
2
< 4π. We can see the numerator as the potential energy cost if
we move a parcel from the bottom layer to the top layer, while the denominator
as some sort of kinetic energy. So this says we are unstable if there is enough
kinetic energy to move a parcel up.
This analysis of the shear flow begs a few questions:
–
How might we regularize this expression? The vortex sheet is obviously
wildly unstable.
– Was it right to assume two-dimensional perturbations?
– What happens if we regularize the depth of the shear?