Information Technology Reference
In-Depth Information
Since
2
3
X
!
2
Z
1
X
c
k
e
k
2
2
T
sin
.kx/
4
5
0
d
k
sin
.kx/
dx
0
kD1
kD1
for all choices of
c
1
;c
2
;:::
, we can solve (9.26) by determining
c
1
;c
2
;:::
such that
c
k
e
k
2
2
T
sin
.kx/ D d
k
sin
.kx/
for
k D 1;2;:::;
which is satisfied if
D e
k
2
2
T
d
k
c
k
for
k D 1;2;::::
The solution
f.x/
of (9.24)is
X
e
k
2
2
T
d
k
sin
.kx/
f.x/ D
for
x 2 .0; 1/;
kD1
where
X
g.x/ D
d
k
sin
.kx/
for
x 2 .0; 1/:
kD1
From a mathematical point of view, one might argue that the backward diffusion
equation is a simple problem, since an analytical solution is obtainable. On the other
hand, the problem itself has an undesirable property. More specifically, the distri-
bution
f.x/
at time
t D 0
is determined by multiplying the Fourier coefficients of
g.x/
by factors of the form
e
k
2
2
T
. These factors are very large, even for moderate
k
, e.g. with
T D 1
,
e
2
1:93 10
4
;
e
2
2
2
1:40 10
17
;
e
3
2
2
3:77 10
38
:
For example, if
T D 1
and
g.x/ D
sin
.3x/;
then the solution of the backward diffusion equation is
f.x/ D e
3
2
2
sin
.3x/ 3:77 10
38
sin
.3x/:
This is quite amazing, or what?
