Metody numeryczne cz. 10 - Instytut Metod Komputerowych w ...

k
Michał Pazdanowski
Instytut Technologii Informacyjnych w Inżynierii Lądowej
Wydział Inżynierii Lądowej
Politechnika Krakowska
{} 3
= f ( t , z )
1
1
( t + ⋅h,
z + ⋅h⋅k
)
{} 3
{} 3
2
k
k
k
k
k
k
=
f
2
2
2
k
2
1
1
2
2
3⋅
=
{} 3
Iteracja trzecia:
3⋅
z2
− 5
= =
t + 1
2
1 {} 3
( z + ⋅h⋅k
) − 5 3⋅( 2,447619+
0,5⋅0,500⋅0,585714)
2
t
2
+
2
1
2
1
⋅h
+ 1
3⋅2,447619−5
3,000+
1
4
=
= 0,585714
3,000+
0,5⋅0,500+
1
−5
= 0,654622
, (28)
z = z + h⋅k
= 2,447619 + 0,500⋅0,654622
2,774930 . (29)
3 2 2
=
{} 1
= f ( t , z )
1
1
( t + ⋅h,
z + ⋅h⋅k
)
{} 1
{} 1
2
1
1
( t + ⋅h,
z + ⋅h⋅k
)
{} 1
{} 1
3
=
=
k
f
f
( t + h,
z + h⋅k
)
{} 1
{} 1
4
0
0
=
f
2
2
0
0
0
1
0
=
k
2
2
1
z = z
0
1
2
3
+ h⋅
Metoda Rungego – Kutty IV rzędu
0
0
3⋅
=
3⋅
=
3⋅
=
2,000000 + 0,500 ⋅
Iteracja pierwsza:
3⋅
z0
− 5
= =
t + 1
0
1 {} 1
( z + ⋅h⋅k
) − 5 3⋅( 2,000000+
0,5⋅0,500⋅0,333333)
0
t
⋅h
+ 1
0
1 {} 1
( z + ⋅h⋅k
) − 5 3⋅( 2,000000+
0,5⋅0,500⋅0,384615)
0
t
⋅h
+ 1
0
{} 1
( z + h⋅k
) − 5 3⋅( 2,000000+
0,500⋅0,396450)
0
t
0
+
+
2
2
1
2
1
2
3
+ h + 1
1
2
3⋅2,000000−5
2,000+
1
1 {} 1 1 { 1} 1 { 1} 1 { 1}
( ⋅k
+ ⋅k
+ ⋅k
+ ⋅k
)
6
1
=
=
=
= 0,333333
2,000+
0,5⋅0,500+
1
2,000+
0,5⋅0,500+
1
2,000+
0,500+
1
−5
1
( ⋅0,333333+
1⋅0,384615+
1⋅0,396450+
1⋅0,455621) = 2, 195924
6
{} 2
= f ( t , z )
1
1
( t + ⋅ h,
z + ⋅ h⋅
k )
{} 2
{} 2
2
1
1
( t + ⋅ h,
z + ⋅ h⋅
k )
{} 2
{} 2
3
=
=
k
( t + h,
z + h⋅
k )
{} 2
{} 2
4
f
f
1
1
=
f
2
2
1
1
1
1
1
=
2
2
k
0
1
z = z
1
2
3
+ h⋅
1
1
3⋅
=
3⋅
=
3⋅
=
2,195924 + 0,500 ⋅
3
2
3
3
Iteracja druga:
3⋅
z1
− 5
= =
t + 1
3
3
6
1
1 {} 2
( z + ⋅ h⋅
k ) − 5 3⋅( 2,195924+
0,5⋅0,500⋅0,453649)
1
⋅ h + 1
1
1 {} 2
( z + ⋅ h⋅
k ) − 5 3⋅( 2,195924+
0,5⋅0,500⋅0,514135)
1
⋅ h + 1
1
{} 2
( z + h⋅k
) − 5 3⋅( 2,195924+
0,500⋅0,526233)
1
t
t
1
2
+
2
+
1
2
1
2
3
t + h + 1
1
2
4
3⋅2,195924−5
2,500+
1
1 {} 1 1 { 1} 1 { 1} 1 { 1}
( ⋅k
+ ⋅k
+ ⋅k
+ ⋅k
)
6
1
=
=
=
6
=
= 0,453649
2,500+
0,5⋅0,500+
1
2,500+
0,5⋅0,500+
1
2,500+
0,500+
1
−5
1
( ⋅0,453649+
1⋅0,514135+
1⋅0,526233+
1⋅0,594280) = 2, 456646
6
{} 3
= f ( t , z )
1
1
( t + ⋅h,
z + ⋅h⋅k
)
{} 3
{} 3
2
1
1
( t + ⋅h,
z + ⋅h⋅k
)
{} 3
{} 3
3
=
=
k
( t + h,
z + h⋅k
)
{} 3
{} 3
4
f
f
2
2
=
f
2
2
2
2
2
2
k
2
2
1
1
2
3
2
2
3⋅
=
3⋅
=
3⋅
=
3
2
3
3
3
Iteracja trzecia:
3⋅
z2
− 5
= =
t + 1
3
6
4
2
1 {} 3
( z + ⋅h⋅k
) − 5 3⋅( 2,456646+
0,5⋅0,500⋅0,592484)
2
⋅h
+ 1
2
1 {} 3
( z + ⋅h⋅k
) − 5 3⋅( 2,456646+
0,5⋅0,500⋅0,662188)
2
⋅h
+ 1
2
{} 3
( z + h⋅k
) − 5 3⋅( 2,456646+
0,500⋅0,674489)
2
t
t
t
2
+
+
2
2
1
2
1
2
3
+ h + 1
1
2
3⋅2,456646−5
3,000+
1
=
=
=
6
=
= 0,592484
3,000+
0,5⋅0,500+
1
3,000+
0,5⋅0,500+
1
3,000+
0,500+
1
−5
−5
−5
= 0,384615
= 0,396450
= 0,455621
−5
−5
, (30)
. (31)
= 0,514135
= 0,526233
= 0,594280
−5
−5
, (32)
. (33)
= 0,662188
= 0,674489
= 0,751483
, (34)