Information Technology Reference
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quantities, but the context should imply the right distinction), we can find that
u
j
xDb
U
a
C
Q
k
.b
a/;
resulting in a characteristic magnitude
u
j
xDb
U
a
and a scaling
N
u
D
k.
u
U
a
/
Q
b
.b
a/
:
(7.79)
The new characteristic magnitude cancels out in the PDE and does not influence
our arguments for the choice of t
c
. The boundary condition at x
D
b gets a simpler
form with this new scaling of
u
:
@
@
N
x
N
u
.1;
N
t/
D
1:
(7.80)
The initial condition is also affected:
N
u
.
N
x; 0/
D
0; 0
x
c;
ˇ
1
;
N
c<
N
x
1:
(7.81)
We see that for both scalings, there is one physical parameter in the problem, ˇ,
and it enters the PDE problem in different ways (in the boundary condition or in the
initial condition), depending on the choice of scale for
u
.
If ˇ is of order unity, there is no significant difference between the two scalings.
For ˇ
1, i.e., Q
b
.b
a/ is much larger than k.U
b
U
a
/, the scaling (7.79)
is advantageous, whereas for ˇ
1,(7.73) is the best choice. Project
7.7.4
on
page
356
asks you to perform detailed calculations to understand the difference
between the two suggested scalings.
This discussion shows that the
scales depend on the regimes of the physical
parameters
in the problem. In this example there is no unified scaling that can ensure
values of order unity. This is also the limitation of scaling. In complicated problems,
there can be a need for different scalings for different physical regimes.
When performing numerical experiments in the scaled problem with a deriva-
tive boundary condition, we need to vary one physical parameter, i.e., we compute
N
u
.
N
x;
N
t
I
ˇ/ for various values of the ˇ parameter. The physical solution is obtained by
the “inverse scaling”, i.e., expressing
u
.x; t / in terms of
N
u
.
N
x;
N
t
I
ˇ/ using the various
physical input data in the problem.
Remark
Before writing a computer program, one has to decide whether to use the scaled
form or use the variables with physical dimensions. One solution is to implement
