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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