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As an example, consider the function
f.x/ D A C B sin.!x/ :
Since f lies between A C B and A B (the absolute value of the sine function is
at most one), a characteristic reference value f r is the average A. The characteristic
magnitude of f A is B so we can introduce the scaling
f.x/ D f.x/ A
B
D sin !x :
We clearly see that max x j f.x/ jD 1, as desired. In f.x/there is only one param-
eter left, !, i.e., A and B are in a sense “parameterized away”. No information is
lost, since we can easily construct f on the basis of f : f.x/ D A C B f.x/.Inthis
example the reference value and the characteristic magnitude were obvious. Finding
the characteristic magnitude is occasionally quite demanding in PDE problems, and
it requires quite a bit of physical insight and approximate mathematical analysis of
the problem.
Let us apply a scaling procedure to ( 7.52 )-( 7.55 ). Our aim to is scale x, t , u ,and
I such that these variables have a magnitude between zero and order unity. The x
parameter varies between a and b,so
N x D x a
b a
is a scaling where the scaled parameter N x varies between 0 and 1, as desired. Since
the initial time is 0, the reference value is 0. The characteristic magnitude of time,
call it t c , is more problematic. This t c can be the time it takes to experience signif-
icant changes in u . More analysis is needed to find candidate values for t c ; thus we
just scale t as
N t D t
t c
;
and remember that t c must be determined later in the procedure. The function I.x/
is easy to scale since it has only two values. We can take U a
as a reference value
and U b U a
as a characteristic magnitude:
I. N x/ D I. N xL/ U a
U b U a
:
Finally we need to scale u . Initially we could use the same scaling as for I ,but,
as time increases, the PDE governs the magnitude of u . Without any idea of how
the solution of the PDE behaves, it is difficult to determine an appropriate scaling.
Choosing a scaling based on the initial data,
N u D u U a
U b U a
;
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