3Transient growth
III Hydrodynamic Stability
3.4 Orr-Sommerfeld and Squire equations
Let’s now see how this is relevant to our fluid dynamics problems. For this
chapter, we will use the “engineering” coordinate system, so that the
y
direction
is the vertical direction. The
x, y, z
directions are known as the streamwise
direction, wall-normal direction and spanwise direction respectively.
y
x
z
Again suppose we have a base shear flow
U
(
y
) subject to some small perturbations
(u, v, w). We can write down our equations as
∂u
∂x
+
∂v
∂y
+
∂w
∂z
= 0
∂u
∂t
+ U
∂u
∂x
+ vU
0
= −
∂p
∂x
+
1
Re
∇
2
u
∂v
∂t
+ U
∂v
∂x
= −
∂p
∂y
+
1
Re
∇
2
v
∂w
∂t
+ U
∂w
∂x
= −
∂p
∂z
+
1
Re
∇
2
w.
Again our strategy is to reduce these to a single, higher order equation in
v
.
To get rid of the pressure term, we differentiate the second, third and fourth
equation with respect to
x, y
and
z
respectively and apply incompressibility to
obtain
∇
2
p = −2U
0
∂v
∂x
.
By applying
∇
2
to the third equation, we get an equation for the wall-normal
velocity:
∂
∂t
+ U
∂
∂x
∇
2
− U
00
∂
∂x
−
1
Re
∇
4
v = 0.
We would like to impose the boundary conditions
v
=
∂v
∂y
= 0, but together with
initial conditions, this is not enough to specify the solution uniquely as we have
a fourth order equation. Thus, we shall require the vorticity to vanish at the
boundary as well.
The wall-normal vorticity is defined by
η = ω
y
=
∂u
∂z
−
∂w
∂x
.
By taking
∂
∂z
of the second equation and then subtracting the
∂
∂x
of the last, we
then get the equation
∂
∂t
+ U
∂
∂x
−
1
Re
∇
2
η = −U
0
∂v
∂z
.
As before, we decompose our perturbations into Fourier modes:
v(x, y, z, t) = ˆv(y)e
i(αx+βz−ωt)
η(x, y, z, t) = ˆη(y)e
i(αx+βz−ωt)
.
For convenience, we set
k
2
=
α
2
+
β
2
and write
D
for the
y
derivative. We
then obtain the Orr-Sommerfeld equation and Squire equation with boundary
conditions ˆv = Dˆv = η = 0:
(−iω + iαU)(D
2
− k
2
) −iαU
00
−
1
Re
(D
2
− k
2
)
2
ˆv = 0
(−iω + iαU) −
1
Re
(D
2
− k
2
)
ˆη = −iβU
0
ˆv.
Note that if we set Re = ∞, then this reduces to the Rayleigh equation.
Let’s think a bit more about the Squire equation. Notice that there is an
explicit
ˆv
term on the right. Thus, the equation is forced by the wall-normal
velocity. In general, we can distinguish between two classes of modes:
(i)
Squire modes, which are solutions to the homogeneous problem with
ˆv
= 0;
(ii) Orr-Sommerfeld modes, which are particular integrals for the actual ˆv;
The Squire modes are always damped. Indeed, set
ˆv
= 0, write
ω
=
αc
, multiply
the Squire equation by η
∗
and integrate across the domain:
c
Z
1
−1
ˆη
∗
ˆη dy =
Z
1
−1
U ˆη
∗
η dy −
i
αRe
Z
1
−1
ˆη
∗
(k
2
− D
2
)ˆη dy.
Taking the imaginary part and integrating by parts, we obtain
c
i
Z
1
−1
|ˆη|
2
dy = −
1
αRe
Z
1
−1
|Dˆη|
2
+ k
2
|ˆη|
2
dy < 0.
So we see that the Squire modes are always stable, and instability in vorticity
comes from the forcing due to the velocity.
It is convenient to express the various operators in compact vector form.
Define
ˆ
q =
ˆ
v
ˆη
, M =
k
2
− D
2
0
0 1
, L =
L
0S
0
iβU
0
L
SQ
,
where
L
OS
= iαU(k
2
− D
2
) + iαU
0
+
1
Re
(k
2
− D
2
)
2
L
SQ
= iαU +
1
Re
(k
2
− D)
2
.
We can then write our equations as
L
ˆ
q = iωM
ˆ
q.
This form of equation reminds us of what we saw in Sturm–Liouville theory,
where we had an eigenvalue equation of the form
Lu
=
λwu
for some weight
function
w
. Here
M
is not just a weight function, but a differential operator.
However, the principle is the say. First of all, this tells us the correct inner
product to use on our space of functions is
hp, qi =
Z
p
†
Mq dy., (∗)
We can make the above equation look like an actual eigenvalue problem by
writing it as
M
−1
L
ˆ
q = iω
ˆ
q.
We want to figure out if the operator
M
−1
L
is self-adjoint under (
∗
), because
this tells us something about its eigenvalues and eigenfunctions. So in particular,
we should be able to figure out what the adjoint should be. By definition, we
have
hp, M
−1
Lqi =
Z
p
†
MM
−1
Lq dy
=
Z
p
†
Lq dy
=
Z
(q
†
(L
†
p))
∗
dy
=
Z
(q
†
M(M
−1
L
†
p))
∗
dy,
where
L
†
is the adjoint of
L
under the
L
2
norm. So the adjoint eigenvalue
equation is
L
†
q = −iωMq.
Here we introduced a negative sign in the right hand side, which morally comes
from the fact we “took the conjugate” of
i
. Practically, adopting this sign
convention makes, for example, the statement of biorthogonality cleaner.
So mathematically, we know we should take the inner product as
R
p
†
Mq
.
Physically, what does this mean? From incompressibility
∂u
∂x
+
∂v
∂y
+
∂w
∂z
= 0,
plugging in our series expansions of v, η etc. gives us
iαˆu + iβ ˆw = −Dˆv, iβ ˆu − iα ˆw = ˆη.
So we find that
ˆu =
i
k
2
(αDˆv − βη), ˆw =
i
k
2
(βDˆv + αη).
Thus, we have
1
2
(|u|
2
+ |w|
2
) =
1
2k
2
|Dˆv|
2
+ |η|
2
,
and the total energy is
E =
Z
1
2k
2
|Dˆv|
2
+ k
2
|ˆv|
2
+ |η|
2
=
1
2k
2
Z
1
−1
(ˆv
∗
ˆη
∗
)
k
2
− D
2
0
0 1
ˆv
ˆη
dy =
1
2k
2
hq, qi.
So the inner product the mathematics told us to use is in fact, up to a constant
scaling, the energy!
We can now try to discretize this, do SVD (numerically), and find out what
the growth rates are like. However, in the next chapter, we shall see that there
is a better way of doing so. Thus, in the remainder of this chapter, we shall look
at ways in which transient growth can physically manifest themselves.
Orr’s mechanisms
Suppose we have a simple flow profile that looks like this:
Recall that the z-direction vorticity is given by
ω
3
=
∂v
∂x
−
∂u
∂y
,
and evolves as
∂ω
3
∂t
+ U
∂ω
3
∂x
= U
0
∂u
∂x
+
∂v
∂y
+ vU
00
+
1
Re
∇
2
ω
3
.
Assuming constant shear and
β
= 0 at high Reynolds number, we have
Dω
3
Dt
'
0.
Suppose we have some striped patch of vorticity:
In this stripe, we always have
ω
3
>
0. Now over time, due to the shear flow, this
evolves to become something that looks like
Now by incompressibility, the area of this new region is the same as the area
of the old. We also argued that
ω
3
does not change. So the total quantity
R
D
ω
3
dA =
R
D
∇ × u · dA does not change. But Stokes’ theorem says
Z
D
∇ × u · dA =
Z
∂D
u · d`.
Since the left-hand-side didn’t change, the same must be true for the right hand
side. But since the boundary
∂D
decreased in length, this implies
u
must have
increased in magnitude! This growth is only transient, since after some further
time, the vorticity patch gets sheared into
But remember that Stokes theorem tells us
Z
D
[∇ × u] dA =
Z
∂D
u · d`.
Thus, if we have vortices that are initially tilted into the shear, then this gets
advected by the mean shear. In this process, the perimeter of each vortex sheet
gets smaller, then grows again. Since
R
∂D
u ·
d
`
is constant, we know
u
grows
transiently, and then vanishes again.
Lift up
Another mechanism is lift-up. This involves doing some mathematics. Suppose
we instead expand our solutions as
v = ˜v(y, t)e
iαx+iβz
, η = ˜η(y, t)e
iαx+iβz
.
Focusing on the
α
= 0 and
Re
=
∞
case, the Orr-Sommerfeld and Squire
equations are
∂
∂t
˜η(y, t) = −iβU
0
˜v
∂
∂t
(D
2
− k
2
)˜v = 0.
Since we have finite depth, except for a few specific values of
k
, the only solution
to the second equation is ˜v(y, t) = ˜v
0
(y), and then
˜η = ˜η
0
− iβU
0
˜v
0
t.
This is an algebraic instability. The constant
˜v
(
y, t
) means fluid is constantly
being lifted up, and we can attribute this to the effects of the
ω
1
of streamwise
rolls.
We should be a bit careful when we consider the case where the disturbance
is localized. In this case, we should consider quantities integrated over all
x
. We
use a bar to denote this integral, so that, for example,
¯v
=
R
∞
−∞
v
d
x
. Of course,
this makes sense only if the disturbance is local, so that the integral converges.
Ultimately, we want to understand the growth in the energy, but it is convenient
to first understand ¯v, ¯u, and the long-forgotten ¯p. We have three equations
∇
2
p = −2U
0
∂v
∂x
∂v
∂t
+ U
∂v
∂x
= −
∂p
∂y
∂u
∂t
+ U
∂u
∂x
+ vU
0
= −
∂p
∂x
.
Note that
U
does not depend on
x
, and all our (small lettered) variables vanish
at infinity since the disturbance is local. Thus, integrating the first equation
over all
x
, we get
∇
2
¯p
= 0. So
¯p
= 0. Integrating the second equation then tells
us
∂¯v
∂t
= 0. Finally, plugging this into the integral of the last equation tells us
¯u = ¯u
0
− ¯v
0
U
0
t.
Thus,
¯u
grows linearly with time. However, this does not immediately imply
that the energy is growing, since the domain is growing as well, and the velocity
may be spread over a larger region of space.
Let’s suppose
u
(
x,
0) = 0 for
|x| > δ
. Then at time
t
, we would expect
u
to
be non-zero only at
U
min
t − δ < x < U
max
t
+
δ
. Recall that Cauchy–Schwarz
says
Z
∞
−∞
f(x)g(x) dx
2
≤
Z
∞
−∞
|f(x)|
2
dx
Z
∞
−∞
|g(x)|
2
dx
.
Here we can take f = u, and
g(x) =
(
1 U
min
t − δ < x < U
max
t + δ
0 otherwise
.
Then applying Cauchy–Schwarz gives us
¯u
2
≤ [∆U + t + 2δ]
Z
∞
−∞
u
2
dx.
So we can bound
E ≥
[¯u]
2
2∆U
t =
[¯v
0
U
0
]
2
t
2∆U
provided
t
2δ
∆U
. Therefore energy grows at least as fast as
t
, but not necessarily
as fast as t
2
.