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and consequently
ar
n
t
1
0;
r
R
and thus
D
r
n
C
ar
n
t
1
r
n
;
r
R
r
nC1
(2.44)
which resembles the monotonicity property of the analytical solution.
In order to show that
r
0
6
r
n
6
R
for all values of n; we will study the function
x
R
/
g.x/
D
x
C
ax
t.1
for x in the interval from 0 to R: Observe that
g
0
.x/
D
1
C
at
2at
R
x;
so, for x in Œ0; R; we have
2at
R
g
0
.x/
>
1
C
at
R
D
1
at
>0;
where we used the assumption that t < 1=a (see (2.43)). Note that
r
nC1
D
g.r
n
/;
R: Then, since g
0
.x/ > 0 for 0
and assume that 0
6
r
n
6
6
x
6
R; we have
r
nC1
D
g.r
n
/
6
g.R/
D
R;
and
r
nC1
D
g.r
n
/
>
g.0/
D
0:
Hence, if 0
6
r
n
6
R; then also 0
6
r
nC1
6
R; and then it follows by induction
on n that 0
6
r
n
6
R holds for all n
>
0, provided that 0<r
0
<R. Moreover,
since we have already seen that r
nC1
>
r
n
; it follows that
r
0
6
r
n
6
R;
(2.45)
for all n
>
0:
