Geoscience Reference
In-Depth Information
this set has a single solution. First of all we note that Eq. (
11.72
) can be solved for
f under a requirement that f is finite when r
!
0
Z
r
1
2r
3
r
1
@
r
1
g.r
1
;t/dr
1
:
f
D
(11.75)
0
Now we prove that if g satisfies Eq. (
11.71
) then the function f given by
Eq. (
11.75
) must satisfy Eq. (
11.70
). For this purpose we take the operator @
t
C
2LJ=LJ
in order to act on Eq. (
11.75
). Using Eq. (
11.71
) we find
r
1
@
r
1
g
00
C
dr
1
:
Z
r
@
t
f
C
2f
LJ
1
2
0
p
LJ
2
r
3
2g
0
r
1
LJ
D
(11.76)
0
Taking the integral several times by parts and applying Eq. (
11.75
) we can reduce
Eq. (
11.76
) to the form which is identical with Eq. (
11.70
). If the solution of
Eq. (
11.71
) under requirements by Eqs. (
11.73
) and (
11.74
) is found, then substi-
tuting this solution; that is the function g,inEq.(
11.75
) gives the function f . Thus
one can obtain the unique solution of the problem.
We expand this solution into a series
X
1
r
n
.t/ sin
nr
g.r;t/
D
R
0
;
(11.77)
nD1
where the eigenfunctions sin
.
nr=R
0
/
form a complete orthogonal system which
satisfies the boundary conditions given by Eq. (
11.74
). The undetermined functions
n
.
t
/
can be found by substituting Eq. (
11.77
)forg into Eq. (
11.71
). Taking into
account the initial conditions given by Eq. (
11.73
), we obtain
0
1
Z
t
Z
t
.
1/
n
6R
0
nLJ
2
.t/
dLJ
2
dt
0
2
n
2
0
p
R
0
LJ
2
dt
00
@
A
dt
0
:
n
.t/
D
exp
(11.78)
0
t
0
Substituting Eqs. (
11.77
) and (
11.78
)forg and
n
into Eq. (
11.75
), and performing
integration with respect to r, yield
(
3nr
R
0
X
R
0
2
2
r
3
n
.t/
n
2
cos
nr
R
0
f .r;t/
D
nD1
"
nr
R
0
3
#
sin
nr
R
0
)
:
2
C
(11.79)
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