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@
t
B
C
2
B
LJ
1
0
p
LJ
2
r
2
LJ
D
r
0
B
;
r
r
0
B
D
0;
(11.67)
where LJ denotes the time-derivative and the subscript
r
0
stands for derivatives with
respect to Lagrange's variables. For simplicity, we will omit the subscript 0 having
in mind that now
r
is the Lagrange's variable. We seek for the solution of Eq. (
11.67
)
in the form
B
D
cos B
1
.r;t/
O
r
sin B
2
.r;t/
O
:
(11.68)
Substituting the Eq. (
11.68
) into Eq. (
11.67
) we come to the following equations for
the unknown functions B
1
and B
2
:
0
p
LJ
2
h
B
0
1
C
i
;
@
t
B
1
C
2B
1
2B
1
4.B
1
B
2
/
1
LJ
D
r
r
2
0
p
LJ
2
h
B
0
2
C
i
;
@
t
B
2
C
2B
2
2B
2
(11.69)
2.B
1
B
2
/
1
LJ
D
r
C
r
2
2
r
.B
1
B
2
/
D
0:
B
1
C
Here the primes denote derivatives with respect to r.
Now we turn to new unknown dimensionless functions f
D
.B
1
B
2
/=B
0
and
g
D
.B
1
C
2B
2
/=B
0
3. These functions satisfy the new set of equations
0
p
LJ
2
f
00
C
r
2
@
t
f
C
2f
LJ
2f
0
r
1
6f
LJ
D
I
(11.70)
g
00
C
@
t
g
C
2.g
C
3/
LJ
2g
0
r
1
0
p
LJ
2
LJ
D
I
(11.71)
2f
0
C
g
0
C
6f=r
D
0:
(11.72)
Suppose at the initial time there is a uniform magnetic field
B
0
everywhere. The
initial conditions for the functions f and g are as follows then
f .r;0/
D
g.r;0/
D
0:
(11.73)
The normal and tangential components of
B
have to be continuous at the ball
surface. Considering this requirement and combining Eqs. (
11.66
) and (
11.68
)we
come to the following boundary conditions:
f .R
0
;t/
D
3=LJ
3
;
.R
0
;t/
D
0:
(11.74)
The set of Eqs. (
11.70
)-(
11.72
) with initial and boundary conditions (
11.73
) and
(
11.74
) contains only two unknown function. However, we shall demonstrate that
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