Geoscience Reference
In-Depth Information
In case that the Eq. (
2.49
) has the form:
Z
x
y
0
y
ð
m
l
Þ
dx
L½y
¼f
ð
x
;
y
Þþ
F
ð
x
;
y
;
; ...;
;
x
2
½a
;
b
;
a
then the Eq. (
2.62
) will read
p
ð
1
Þ
h
n
p
ð
0
Þ
k
k
E
k
ð
1
þ
hp
ð
0
Þ
Þ
½
ð
1
þ
hp
ð
0
Þ
Þ
e
0
þ
1
;
where
"
#
p
ð
0
Þ
¼
X
n
1
b
þ
X
n
i¼1
ð
p
i
þð
m
l
þ
1
Þc
i
b
n
i
Þþð
b
a
Þ
c
ð
m
l
þ
1
Þ
h
s
1
s
!
h
n
1
n
!
þ
;
s¼1
!
a
þ
b
b
1
þ
2B
þ
X
n
þ
ð
b
a
Þ
q
n
1
ð
n
þ
1
Þ!
p
ð
1
Þ
¼
p
0i
;
!
i¼0
;
;
;
n
o
;
y
ð
i
Þ
@
f
@
@
f
@
@
P
i
@
y
ð
s
Þ
b
¼
max
½a
;
b
y
ð
i
Þ
a
¼
max
½a
;
b
c
i
¼
max
s
;
½a
;
b
b
i
¼
max
½a
;
b
;
x
y
!
!
;
5 A
þ
C
X
m
l
s¼0
b
s
þ
1
l
i
þ c
i
X
m
l
s¼0
b
s
þ
1
@
F
c
¼
max
s
;
½a
;
b
q
¼
0
:
;
p
oi
¼
b
n
i
þ
2
a
i
b
n
þ
1
i
;
@
y
ð
s
Þ
;
@
F
j ;
F
A
¼
max
½a
B
¼
max
½a
@
x
;
b
;
b
2.12.5.2 Solution of Equation y
ðÞ
¼ f
ð
x
y
0
; ...;
y
n
1
ð
Þ
Þ
;
y
;
Let us apply the approximate method of the solution presented in Sect.
2.12.5.1
to
integro-differential equations to solve the initial value problem:
y
ðÞ
¼ f
ð
x
y
0
y
ð
n
1
Þ
Þ;
;
y
;
; ...;
ðÞ2
G
x
;
;
ð
2
:
63
Þ
y
ð
j
Þ
ð
x
0
Þ
¼y
ð
j
Þ
yx
ðÞ
¼y
0
;
;
j ¼ 1
; ...;
n
1
;
ð
x
0
;
y
0
Þ
G
;
ð
2
:
64
Þ
0
where the function f satis
es the Lipschitz condition
K
X
n
1
i¼0
djj ð
2
y
ð
n
1
Þ
þ d
n
1
Þ
f
ð
x
y
0
y
ð
n
1
Þ
Þ
f
ð
x
;
y
þ d
0
; ...;
;
y
;
; ...;
:
65
Þ
Let us divide the interval [a,b] by a sequence of points x
0
= a, x
1
,
…
, x
s
= b into
elementary intervals. Let E ={x
0
,
…
, x
s
}. On each interval [x
ʽ
,x
ʽ
+1
], let us solve the
initial value problem:
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