:

517.91
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A novel method for numerical solving autonomous ordinary differential equation dx/dt = f () is suggested. The solution is found as a function t(x) for that the incrementation is independent from previous points of solution trajectory. This fact is used for parallel realization of computational method.
The numerical experiments show the applicability of suggested method for area with rapid acceleration of right-hand function.
. . , . . , , . , . , , - [1]- [4], [8]- [11]. , , . , [5]- [7], [12].
. , , .
, .
Copyright 2009 .. , .. .
.
E-mail: korobits@rambler.ru
1.

,
= ;{), (0) = 0, > 0. (1)
() : ^ (1), , -
/ .
()
2 2
(0 + ) = (*) + ^() + ^-() (2)
(2) (,),
2 2
. . I I
(*0) := ( + ^ 2 + + '

( + ) - () = (, ) + 0(+1).

/ = /, /' = ^, / = ^
, = 0,1, 2,...:
(, ) = 0,
1(0,) = 0(0,) + ^-() = ,
2 2 2
2^,) = (,) + () = /' + //,
3 3 2 3
&,) = 2^0,) + ^3 () = /' + // + (// + /2/)^>

( + ),
( + ) - () = .

(, )= . (3)
(,) , (3) . , + .
, (1) 0 0 + dx dx > 0 f () > 0 [0; 0 + dx]. 1 f (1) = 0, , (1) . 0+dx. f (0) < 0, [0; 0 dx] ,
2.
, f () > 0 [0; 0 + dx].
fi = f(0 + dx), f-i = f(0 dx).
f 0

/(4) := / /) = /l~2^ + /-1. (4)
O(dx).
(3) k. k = 1 f0T = dx. = , , O(dx).
k = 2
f0 f0 2 I 1 n
-^-T + for -cix = 0,

D = /2 + 2//. = 4 (-1 ± Jl+2d.,f).
f0f0 f0 V V f0/
f0 > 0,
I-1)- (5)
f0 < 0, 1 + 2dxfg /f0 > 0 dx < f0 /(2f^), f (4), dx < f0dx/(f1 f-1).
f0 / ( f1 f- 1 ) > 1 dx
f1 f-1 > f^. ^^ 0 + dx
(5),
k > 3 k. .

+ dx.
, . 1 = (1), 1 = 0 + , 0 = 1 4,
2 ^ X
,((0) = *( - ((1). + + + (-1)^((1) . + . . .
(0) (^) = ( ) + 0(+1), (3), 1 0 = (0,) 0 1 = (1, ),
2dx = Pk(to, ) - Pk (ti, -). (6)

^ df / ^,/2-/0 ,// ^,/2-2/1 +/
/2 = /( + 24), = ~ , = ~---------- (7)
dx 2dx dx2 dx

k.
k = 1 2dx = /0 fi (), = 2dx/(/0 + /1 ), O(dX).
= 2
(// /i/i) - + (/ + /i)t 2dx = 0. (8)
> 3 , .
(8)
/
, (7),

t t, t _ f\~ /-1 t /2 /0 t _ f-ifo + 2/o/i /1/2 J0J0 - J1J1 - wi J0 wi J1 - wi
2dx 2dx 2dx
(8)
D = (f0 + fi) 4(/0/0 f1 fi) ( 2dx) = (f0 + fi) + 4( f-i/o + 2/0/1 /1/2) , _-(/ + /1)±v/D_J ( + /1) T \/(/ + )2 - 4(/-i/o - 2/0/1 + /1/2)
rl,2 7777---- dx
2(/0/0 /i/i) x (/-i/0 2/0/1 + /1 /2 )

/ + /
/-1 / 2// + //2

-7--4 < 1, (/ + /0^ <

> 0, /0 + /0 ^ > 4, < 0, /0 + /0 ^ < 0,
4? /, /0 > 0, > 0, .
(9), -/
(^). , ^ (^), (^).
3.

2^,
71, 2 , 4, 3 , 2^, ,
7------ ^
(+) ^ (/ + /1)^ > 4,
(9)
( + 2^) - = + 2 + (4)
( + 0 = 3 + (24)
1 + 2 3 = (4 (2? 1).

^ = (4)
~1 + 2 - 2 - 1
4.
.
:
&
= , (0) = , I > 0. (10)
() = 0*, () = 1(/0).
(9), : 0 = 1, = 0.19, (, 1-) (). -1,833 -2,845 (, 1-6), , , .
\gBerr
) )
. 1. ) (10). )

-:
(^
= ( ), (0) = , I > 0. (11)
, :
= I + -) 0 < * < = I _ *-2) 1 ^ > -

1 1
/~1 __ _______ _ ______
' ' 2
: = 3, = 2, ^ = 0.04, : ∞ = 1, ∞ = 5, (, 2-) , ,
X
) )
. 2. ) (11). ) ; 0 < < , > .
(11) = , . , , , . (, 2-6),
. . - , ( + ^=1 {) = + :. , , ,
5.
2:
^ = /(*1.*2),
*%- = ), .-
?(∞) = ?,
2() = ∞.
,
I ~^~~\= I => 2) = [ ^1 \ +
3 / (?,2) 3 ] / (?,2)

2(1,2)= [ 2 + 2.
9 (1, 2)
?(?, 2) 2(?, 2) (12), ? 2 .
, (12) 2
= {(?,2) 2| 1 (?,2) - 2(1,2) = 0}.
= 1(1,2) = 2(1,2), (1,2) ,
(12) , ( ^, 3 Z) 4:
^ = {(1,2) 2| ∞ + ^ < 1 < 1 + ( +1)^,2 + 34 < 2 < 2 + (3 + 1)4}.
^ . , 1; 2.
1 2
. , , , . , , .
, ^ , . , , , 1 2
, , . .
,
..
- . 1 1,

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9. , . . - / . , . . - .: , 1999.
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