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derivatives of
N
u
to also be of order unity. Looking at (7.61), we should then have ı of
order unity (since both derivative terms are of order unity). Choosing ı
D
1 implies
t
c
D
1
k
%c
v
.b
a/
2
;
which determines the time scale. Newcomers to scaling may find these arguments a
bit too vague. A more quantitative reasoning is based on a typical solution of (
7.52
).
Inserting
u
.x; t /
D
e
t
sin
x
a
b
a
in (
7.52
) reveals that this is a solution for the special case U
a
D
U
b
D
0 and
I.x/
D
sin
xa
ba
, provided that
k
2
%c
v
.b
a/
2
D
:
The boundary and initial values are not important here; what is important is the
temporal evolution of
u
, described by the factor e
t
. The solution is exponentially
damped. A common choice of the characteristic time t
c
is the time that reduces the
initial (unit) amplitude by a factor of 1=e (the e-folding time
17
):
D
%c
v
.b
a/
2
e
t
c
D
e
1
)
t
c
D
1
:
k
2
This means that significant changes (here reduction of the amplitude by a factor of
e
1
0:37)in
u
take place during a time interval of length t
c
. For esthetic reasons,
many prefer to remove the factor
2
from the expression for t
c
and write
t
c
D
1
k
%c
v
.b
a/
2
:
(7.62)
The consequence is that we just use a different amplitude reduction factor to derive
t
c
. There are many different solutions of (
7.52
), with different (typically shorter)
time scales, so there is no single correct value for t
c
. The main point is that t
c
is not
orders of magnitude away from the dominating time scales of the problem.
You should observe that two different ways of reasoning led to the same time
scale t
c
, thus providing some confidence in the result. Numerical experiments or
available analytical solutions of the problem can be used to test if our choice of t
c
is appropriate. In general, the dominating time scales depend on the PDE, the initial
conditions, and the boundary conditions. Hopefully we have managed to explain
17
The specific choice 1=e as a suitable reduction factor is mostly inspired by esthetics. For a func-
tion e
t
we get
1
as the characteristic time. A reduction factor of 10
5
gives a characteristic
time
1
5 ln 10.
