1Linear stability analysis
III Hydrodynamic Stability
1.4 Finite depth shear flow
We now consider a finite depth shear flow. This means we have fluid moving
between two layers, with a z-dependent velocity profile:
z = L
z = −L z = −L
U(z)
We first show that it suffices to work in 2 dimensions. We will assume that the
mean flow points in the
ˆ
x
direction, but the perturbations can point at an angle.
The inviscid homogeneous incompressible Navier–Stokes equations are again
∂u
∂t
+ u · ∇u
=
Du
Dt
= −∇
p
0
ρ
, ∇ · u = 0.
We linearize about a shear flow, and consider some 3D normal modes
u =
¯
U(z)
ˆ
x + u
0
(x, y, z, t),
where
(u
0
, p
0
/ρ) = [
ˆ
u(z), ˆp(z)]e
i(kx+`y−kct)
.
The phase speed is then
c
p
=
ω
κ
=
ω
(k
2
+ `
2
)
1/2
=
kc
r
(k
2
+ `
2
)
1/2
and the growth rate of the perturbation is simply σ
3d
= kc
i
.
We substitute our expression of
u
and
p
0
/ρ
into the Navier–Stokes equations
to obtain, in components,
ik(
¯
U − c)ˆu + ˆw
d
¯
U
dz
= −ikˆp
ik(
¯
U − c)ˆv = −ilˆp
ik(
¯
U − c) ˆw = −
dˆp
dz
ikˆu + i`ˆv +
d ˆw
dz
= 0
Our strategy is to rewrite everything to express things in terms of new variables
κ =
p
k
2
+ `
2
, κ˜u = kˆu + `ˆv, ˜p =
κˆp
k
and if we are successful in expressing everything in terms of
˜u
, then we could
see how we can reduce this to the two-dimensional problem.
To this end, we can slightly rearrange our first two equations to say
ik(
¯
U − c)ˆu + ˆw
d
¯
U
dz
= −ik
2
ˆp
k
i`(
¯
U − c)ˆv = −i`
2
ˆp
k
which combine to give
i(
¯
U − c)κ˜u + ˆw
d
¯
U
dz
= −iκ˜p.
We can rewrite the remaining two equations in terms of ˜p as well:
iκ(
¯
U − c) ˆw = −
d
dz
˜p
κ
˜
U +
d ˆw
dz
= 0.
This looks just like a 2d system, but with an instability growth rate of
σ
2d
=
κc
i
> kc
i
=
σ
3d
. Thus, our 3d problem is “equivalent” to a 2d problem with
greater growth rate. However, whether or not instability occurs is unaffected.
One way to think about the difference in growth rate is that the
y
component
of the perturbation sees less of the original velocity
¯
U
, and so it is more stable.
This result is known as Squire’s Theorem.
We now restrict to two-dimensional flows, and have equations
ik(
¯
U − c)ˆu + ˆw
d
¯
U
dz
= ikˆp
ik(
¯
U − c) ˆw = −
dˆp
dz
ikˆu +
d ˆw
dz
= 0.
We can use the last incompressibility equation to eliminate
ˆu
from the first
equation, and be left with
−(
¯
U − c)
d ˆw
dz
+ ˆw
d
¯
U
dz
= −ikˆp.
We wish to get rid of the
ˆp
, and to do so, we differentiate this equation with
respect to z and use the second equation to get
−(
¯
U − c)
d
2
ˆw
dz
2
−
d ˆw
dz
d
¯
U
dz
+
d ˆw
dz
d
¯
U
dz
+ ˆw
d
2
¯
U
dz
2
= −k
2
(
¯
U − c) ˆw.
The terms in the middle cancel, and we can rearrange to obtain the Rayleigh
equation
(
¯
U − c)
d
2
dz
2
− k
2
−
d
2
¯
U
dz
2
ˆw = 0.
We see that when
¯
U
=
c
, we have a regular singular point. The natural boundary
conditions are that ˆw → 0 at the edge of the domain.
Note that this differential operator is not self-adjoint! This is since
¯
U
has non-
trivial
z
-dependence. This means we do not have a complete basis of orthogonal
eigenfunctions. This is manifested by the fact that it can have transient growth.
To analyze this scenario further, we rewrite Rayleigh equation in conventional
form
d
2
ˆw
dz
2
− k
2
ˆw −
d
2
¯
U/dz
2
¯
U − c
ˆw = 0.
The trick is to multiply by
w
∗
, integrate across the domain, and apply boundary
conditions to obtain
Z
L
−L
¯
U
00
¯
U − c
|ˆw|
2
dz = −
Z
L
−L
(|ˆw
0
|
2
+ k
2
|ˆw|
2
) dz
We can split the LHS into real and imaginary parts:
Z
L
−L
¯
U
00
(
¯
U − c
r
)
|
¯
U − c|
2
|ˆw|
2
dz + ic
i
Z
L
−L
¯
U
00
|
¯
U − c|
2
|ˆw|
2
dz.
But since the RHS is purely real, we know the imaginary part must vanish.
One way for the imaginary part to vanish is for
c
i
to vanish, and this
corresponds to stable flow. If we want
c
i
to be non-zero, then the integral must
vanish. So we obtain the Rayleigh inflection point criterion:
d
2
dz
2
¯
U
must change
sign at least once in −L < z < L.
Of course, this condition is not sufficient for instability. If we want to get
more necessary conditions for instability to occur, it might be wise to inspect
the imaginary part, as Fjortoft noticed. If instability occurs, then we know that
we must have
Z
L
−L
¯
U
00
|
¯
U − c|
2
|ˆw|
2
dz = 0.
Let’s assume that there is a unique (non-degenerate) inflection point at
z
=
z
s
,
with
¯
U
s
=
¯
U
(
z
s
). We can then add (
c
r
−
¯
U
s
) times the above equation to the
real part to see that we must have
−
Z
L
−L
(|ˆw
0
|
2
+ k
2
|ˆw|
2
) dz =
Z
L
−L
¯
U
00
(
¯
U −
¯
U
s
)
|
¯
U − c|
2
|ˆw|
2
dz.
Assuming further that
¯
U
is monotonic, we see that both
¯
U −
¯
U
s
and
U
00
change
sign at
z
s
, so the sign of the product is unchanged. So for the equation to be
consistent, we must have
¯
U
00
(
¯
U −
¯
U
s
) ≤ 0 with equality only at z
s
.
We can look at some examples of different flow profiles:
In the leftmost example, Rayleigh’s criterion tells us it is stable, because there is
no inflection point. The second example has an inflection point, but does not
satisfy Fjortoft’s criterion. So this is also stable. The third example satisfies
both criteria, so it may be unstable. Of course, the condition is not sufficient, so
we cannot make a conclusive statement.
Is there more information we can extract from them Rayleigh equation?
Suppose we indeed have an unstable mode. Can we give a bound on the growth
rate c
i
given the phase speed c
r
?
The trick is to perform the substitution
ˆ
W =
ˆw
(
¯
U − c)
.
Note that this substitution is potentially singular when
¯
U
=
c
, which is the
singular point of the equation. By expressing everything in terms of
ˆ
W
, we
are essentially hiding the singularity in the definition of
ˆ
W
instead of in the
equation.
Under this substitution, our Rayleigh equation then takes the self-adjoint
form
d
dz
(
¯
U − c)
2
d
ˆ
W
dz
!
− k
2
(
¯
U − c)
2
ˆ
W = 0.
We can multiply by
ˆ
W
∗
and integrate over the domain to obtain
Z
L
−L
(
¯
U − c)
2
d
ˆ
W
dz
2
+ k
2
|
ˆ
W
2
| {z }
≡Q
dz = 0.
Since Q ≥ 0, we may again take imaginary part to require
2c
i
Z
L
−L
(
¯
U − c
r
)Q dz = 0.
This implies that we must have
U
min
< c
r
< U
max
, and gives a bound on the
phase speed.
Taking the real part implies
Z
L
−L
[(
¯
U − c
r
)
2
− c
2
i
]Q dz = 0.
But we can combine this with the imaginary part condition, which tells us
Z
L
−L
¯
UQ dz = c
r
Z
L
−L
Q dz.
So we can expand the real part to give us
Z
L
−L
¯
U
2
Q dz =
Z
L
−L
(c
2
r
+ c
2
i
)Q dz.
Putting this aside, we note that tautologically, we have
U
min
≤
¯
U ≤ U
max
. So
we always have
Z
L
−L
(
¯
U − U
max
)(
¯
U − U
min
)Q dz ≤ 0.
expanding this, and using our expression for
R
L
−L
¯
U
2
Q dz, we obtain
Z
L
−L
((c
2
r
+ c
2
i
) − (U
max
− U
min
)c
r
+ U
max
U
min
)Q dz ≤ 0.
But we see that we are just multiplying
Q
d
z
by a constant and integrating.
Since we know that
R
Q dz > 0, we must have
(c
2
r
+ c
2
i
) − (U
max
+ U
min
)c
r
+ U
max
U
min
≤ 0.
By completing the square, we can rearrange this to say
c
r
−
U
max
+ U
min
2
2
+ (c
i
− 0)
2
≤
U
max
− U
min
2
2
.
This is just an equation for the circle! We can now concretely plot the region of
possible c
r
and c
i
:
c
r
c
i
U
max
+U
min
2
U
min
U
max
Of course, lying within this region is a necessary condition for instability to
occur, but not sufficient. The actual region of instability is a subset of this
semi-circle, and this subset depends on the actual
¯
U
. But this is already very
helpful, since if we way to search for instabilities, say numerically, then we know
where to look.