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is such that
u
.x; T I f/D g.x/
for all
x 2 .0; 1/:
In short, we want to use
g.x/
and the diffusion equation to compute
f.x/
. Conse-
quently,
g.x/
is our observation data, and the output least squares formulation of
the problem becomes
Z
1
.
u
.x; T I f/ g.x//
2
dx
min
f
(9.24)
0
subject to
u
D
u
.x; t I f/
satisfying (
9.21
)-(
9.23
).
At first glance one can get the impression that (9.24) is extremely hard to solve.
However, we have put ourselves in the fortunate position such that the problem can
be studied with Fourier analysis. (In most practical situations this is not the case and
some sort of minimization algorithm must be employed.)
Recall that the solution
u
.x; t I f/
of (
9.21
)-(
9.23
) can be written in the form
X
c
k
e
k
2
2
t
sin
.kx/;
u
.x;t;f/ D
kD1
where
X
f.x/ D
c
k
sin
.kx/
for
x 2 .0; 1/;
kD1
and (9.24) can therefore be expressed in terms of the Fourier coefficients,
2
3
X
!
2
Z
1
c
k
e
k
2
2
T
sin
.kx/ g.x/
4
5
:
min
c
1
;c
2
;:::
dx
(9.25)
0
kD1
Next, we insert the Fourier sine expansion
X
g.x/ D
d
k
sin
.kx/
for
x 2 .0; 1/
kD1
of
g
into (9.25) and obtain the following form of our problem:
2
3
X
!
2
Z
1
X
4
c
k
e
k
2
2
T
sin
.kx/
5
:
min
c
1
;c
2
;:::
d
k
sin
.kx/
dx
(9.26)
0
kD1
kD1
The Fourier coefficients of
g
can be computed by invoking formula (8.74) - keep in
mind that
g
is the given observation data.
