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
2 .t/
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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