3Transient growth
III Hydrodynamic Stability
3.2 A toy model
Let’s try to understand transient dynamics in a finite-dimensional setting. Ul-
timately, the existence of transient growth is due to the non-normality of the
operator.
Recall that a matrix
A
is normal iff
A
†
A
=
AA
†
. Of course, self-adjoint
matrices are examples of normal operators. The spectral theorem says a normal
operator has a complete basis of orthonormal eigenvectors, which is the situation
we understand well. However, if our operator is not normal, then we don’t
necessarily have a basis of eigenvectors, and even if we do, they need not be
orthonormal.
So suppose we are in a 2-dimensional world, and we have two eigenvectors
that are very close to each other:
Φ
2
Φ
1
Now suppose we have a small perturbation given by
ε
= (
ε,
0)
T
. While this
perturbation is very small in magnitude, if we want to expand this in the basis
Φ
1
and Φ
2
, we must use coefficients that are themselves quite large. In this case,
we might have ε = Φ
2
− Φ
1
, as indicated in red in the diagram above.
Let’s let this evolve in time. Suppose both Φ
1
and Φ
2
are stable modes, but
Φ
1
decays much more quickly than Φ
2
. Then after some time, the perturbation
will grow like
Note that the perturbation grows to reach a finite, large size, until it vanishes
again as the Φ
1
component goes away.
Let’s try to put this down more concretely in terms of equations. We shall
consider a linear ODE of the form
˙
x = Ax.
We first begin by considering the matrix
A =
0 1
0 0
,
which does not exhibit transient growth. We first check that
A
is not normal.
Indeed,
AA
T
1 0
0 0
6=
0 0
0 1
= A
T
A.
Note that the matrix
A
has a repeated eigenvalue of 0 with a single eigenvector
of (1 0).
To solve this system, write the equations more explicitly as
˙x
1
= x
2
˙x
2
= 0.
If we impose the initial condition
(x
1
, x
2
)(0) = (x
10
, x
20
),
then the solution is
x
1
(t) = x
20
t + x
10
x
2
(t) = x
20
.
This exhibits linear, algebraic growth instead of the familiar exponential growth.
Let’s imagine we perturb this system slightly. We can think of our previous
system as the
Re
=
∞
approximation, and this small perturbation as the effect
of a large but finite Reynolds number. For ε > 0, set
A
ε
=
−ε 1
0 −2ε
.
We then have eigenvalues
λ
1
= −ε
λ
2
= −2ε,
corresponding to eigenvectors
e
1
=
1
0
, e
2
=
1
−ε
.
Notice that the system is now stable. However, the eigenvectors are very close
to being parallel.
As we previously discussed, if we have an initial perturbation (
ε,
0), then it
is expressed in this basis as
e
1
−e
2
. As we evolve this in time, the
e
2
component
decays more quickly than
e
1
. So after some time, the
e
2
term is mostly gone,
and what is left is a finite multiple of
e
1
, and generically, we expect this to have
magnitude larger than ε!
We try to actually solve this. The second row in
˙
x = Ax gives us
˙x
2
= −2εx
2
.
This is easy to solve to get
x
2
= x
20
e
−2εt
.
Plugging this into the first equation, we have
˙x
1
= −εx
1
+ x
20
e
−2εt
.
We would expect a solution of the form
x
1
=
Ae
−εt
−Be
−2εt
, where the first term
is the homogeneous solution and the second comes from a particular solution.
Plugging this into the equation and applying our initial conditions, we need
x
1
=
x
10
+
x
20
ε
e
−εt
−
x
20
ε
e
−2εt
.
Let us set
y
10
= x
10
+
x
20
ε
y
20
= −
x
20
ε
.
Then the full solution is
x = y
10
e
−εt
e
1
+ y
20
e
−2εt
e
2
.
For
εt
1, we know that
x ∼ y
10
e
−εt
e
2
. So our solution is an exponentially
decaying solution.
But how about early times? Let’s consider the magnitude of x. We have
kxk
2
= y
2
10
e
−2εt
e
1
· e
2
+ 2y
10
y
20
e
−3εt
e
1
· e
2
+ y
2
20
e
−4εt
e
2
e
2
= y
2
10
e
−2εt
+ 2y
10
y
20
e
−3εt
+ (1 + ε
2
)y
2
20
e
−4εt
.
If y
10
= 0 or y
20
= 0, then this corresponds to pure exponential decay.
Thus, consider the situation where
y
20
=
−ay
10
for
a 6
= 0. Doing some
manipulations, we find that
kxk
2
= y
2
10
(1 − 2a + a
2
(1 + ε
2
)) + y
2
10
(−2 + 6a − 4a
2
(1 + ε
2
))εt + O(ε
2
t
2
).
Therefore we have initial growth if
4a
2
(1 + ε
2
) − a + 2 < 0.
Equivalently, if
a
−
< a < a
+
with
a
±
=
3 ±
√
1 − 8ε
2
4(1 + ε
2
)
.
Expanding in ε, we have
a
+
= 1 − 2ε
2
+ O(ε
4
),
a
−
=
1 + ε
2
2
+ O(ε
4
).
These correspond to
s
20
'
x
10
2ε
and
x
20
' εx
10
respectively, for
ε
1. This is
interesting, since in the first case, we have
x
20
x
10
, while in the second case,
we have the opposite. So this covers a wide range of possible x
10
, x
20
.
What is the best initial condition to start with if we want the largest possible
growth? Let’s write everything in terms of
a
for convenience, so that the energy
is given by
E =
x · x
2
=
y
2
10
2
(e
−2εt
− 2ae
−3εt
+ a
2
(1 + ε
2
)e
−4εt
).
Take the time derivative of this to get
dE
dt
= −ε
y
2
10
2
e
−2εt
(2 − 6(ae
−εt
) + 4(1 + ε
2
)(ae
−εt
)
2
).
Setting ˆa = ae
−εt
, we have
dE
dt
> 0 iff
2 − 6ˆa + 4ˆa
2
(1 + ε
2
) < 0.
When
t
= 0, then
ˆa
=
a
, and we saw that there is an initial growth if
a
−
< a < a
+
.
We now see that we continue to have growth as long as
ˆa
lies in this region. To
see when E reaches a maximum, we set
dE
dt
= 0, and so we have
(a
−
− ae
−εt
)(ae
−εt
− a
+
) = 0.
So a priori, this may happen when
ˆa
=
a
−
or
a
+
. However, we know it must
occur when
ˆa
hits
a
−
, since
ˆa
is decreasing with time. Call the time when this
happens t
max
, which we compute to be given by
εt
max
= log
a
a
−
.
Now consider the energy gain
G =
E(t
max
)
E(0)
=
a
2
−
a
2
a
2
−
(1 + ε
2
) − 2a
1
+ 1
a
2
(1 + ε
2
) − 2a + 1
.
Setting
dG
da
= 0, we find that we need a = a
+
, and so
max
a
G =
(3a
−
− 1)(1 − a
−
)
(3a
+
− 1)(1 − a
+
)
.
We can try to explicitly compute the value of the maximum energy gain,
using our first-order approximations to a
±
;
max
a
G =
3
1+ε
2
2
1 −
1+ε
2
2
(3(1 + 2ε
2
) − 1)(1 − (1 − 2ε
2
))
=
(1 + 3ε
2
)(1 − ε
2
)
16(1 − 3ε
2
)ε
2
∼
1
16ε
2
.
So we see that we can have very large transient growth for a small ε.
How about the case where there is an unstable mode? We can consider a
perturbation of the form
A =
ε
1
1
0 −ε
2
,
with eigenvectors
e
1
=
1
0
, e
2
=
1
p
1 + (ε
1
+ ε
2
)
2
1
−(ε
1
+ ε
2
)
We then have one growing mode and another decaying mode. Again, we have
two eigenvectors that are very close to being parallel. We can do very similar
computations, and see that what this gives us is the possibility of a large initial
growth despite the fact that
ε
1
is very small. In general, this growth scales as
1
1−e
1
·e
2
.