BO MON TOAN NG DUNG - HBK-------------------------------------------------------------------------------------

TOAN 4

CHUOI VA PHNG TRNH VI PHAN BAI 6: HE PHNG TRNH VI PHAN CAP 1

TS. NGUYEN QUOC LAN (5/2006)

NOI DUNG---------------------------------------------------------------------------------------------------------------------------

1 HE PHNG TRNH VI PHAN

2 PHNG PHAP KH

3 HE PHNG TRNH VI PHAN CAP 1 TUYEN TNH

THUAN NHAT HE SO HANG: PHNG PHAP TR

RIENG (CHEO HOA MA TRAN)

KHAI NIEM (SGK, TRANG 165)---------------------------------------------------------------------------------------------------------------------------

He m phtrnh vi phan (cap n) vi m ham an: Minh

hoa m = 2

: dang chuan hoa( )( )

=

=

0',',,,0',',,,

yxyxtGyxyxtF ( )

( )

=

=

yxtgyyxtfx

,,'

,,'

VD: He cap 1

++=

+=teyxty

tyxtx

310)('sin3)('

VD: He cap 2

+=

+=

txtytytx

sin10)(''cos)(''

Van e: Bien oi he cap 1 tren ve 1 ptrnh vi phan va giai?

PHNG PHAP KH (SGK, TRANG 166)-------------------------------------------------------------------------------------------------------------------------------------

a he n phng trnh vi phan cap 1 ve 1 phng trnh vi

phan cap n: ao ham len, lan lt kh (n 1) an khac

VD: Giai( )

( )

++=

+=

2310)('123)('

teyxty

yxtx

( ) ( )( ) ( ) )()(0

,,

31023

tbtAXdtdX

etb

tytx

tXAt

+=

=

=

=

Chu y: He phng trnh tuyen tnh Cach viet dang ma tran

He 2 phng trnh cap 1 : Xem nh tng ng 1 phng

trnh cap 2 Nghiem cha ung 2 hang so C1, C2

( ) ( ) ( ) tetxtxtx 211'6'' =

HE PHNG TRNH TUYEN TNH (SGK, TRANG 170)--------------------------------------------------------------------------------------------------------------------------------

He n ham an, n phng trnh vi phan cap 1 tuyen tnh:

( )Etbxtaxtaxtatx

tbxtaxtaxtatxtbxtaxtaxtatx

nnnnnnn

nn

nn

++++=

++++=

++++=

)()()()()('............................................................................

)()()()()(')()()()()('

2211

222221212

112121111

L

L

L

Ma tran: )()( tbXtAdtdX

+= : he pt ttnh khong thuan nhat

=

)()()(.........................................

)()()()()()(

21

22221

11211

tatata

tatata

tatata

A

nnnn

n

n

L

L

L

=

)(.......

)()(

2

1

tx

tx

tx

X

n

=

)(.......

)()(

)( 21

tb

tbtb

tb

n

HE PTVP TTNH THUAN NHAT (SGK, TRANG 170)--------------------------------------------------------------------------------------------------------------------------------------

He n ham an x1(t), x2(t) xn(t) & n phng trnh vi phan cap

1 tuyen tnh thuan nhat (khong co ve phai)

( ) ( ) ( )tXtAdtdXE

xtaxtaxtatx

xtaxtaxtatx

xtaxtaxtatx

nnnnnn

nn

nn

=

+++=

+++=

+++=

0

2211

22221212

12121111

)()()()('.................................................................

)()()()(')()()()('

L

L

L

Cau truc nghiem he thuan nhat: Nghiem tong quat cua he

thuan nhat: Xtq.tn(t) = c1X1(t) + c2X2(t) + + cnXn(t), ck vi he nghiem X1(t), X2(t) Xn(t) la he nghiem c s (tc n vect

{X1(t), X2(t) Xn(t)} oc lap tuyen tnh t (a, b))

TTNH THUAN NHAT HE SO HANG (SGK, TRANG 173)-------------------------------------------------------------------------------------------------------------------------------

He p/trnh vi phan cap 1 tuyen tnh thuan nhat he so hang

( ) ( )02211

22221212

12121111

)('..................................................

)(')('

EtAXdtdX

xaxaxatx

xaxaxatx

xaxaxatx

nnnnnn

nn

nn

=

+++=

+++=

+++=

L

L

L

( )( )

=

=

2

1

2

1

2

1

1232

c

c

c

c

ec

ec

tytx

t

t

Vect v = [c1, c2]T: vect rieng ma tran A ng tr rieng !

Ma tran A (cap 2): 2 gia tr rieng thc 1, 2 & 2 vect rieng oc lap tuyen tnh: v1, v2 2211tq.tn 21 vecvecX

tt +=

VD: Giai

+=

+=

yxtyyxtx

2)('32)('

NHAC LAI: TR RIENG, VECT RIENG-------------------------------------------------------------------------------------------------------------

Tr rieng: det(A I) = 0. Vect rieng v: (A I)v = 0

VD:

=

1232

A ( )

=

=

=

=

41

012

32det

2

1

IA

VTR v1 = [, ]T ng 1 = 1: Av1 = 1v1 (A 1I)v1 = 0

( )

==+=

=+

11

002233

0 11 vvIA

2 tr rieng thc, phan biet 2 VTR LTT Cheo hoa1

2131

4001

2131

1232

=

=A

Vect rieng v2 = [, ]T ng vi 2 = 4: v2 = [3, 2]T

KET QUA TONG QUAT--------------------------------------------------------------------------------------------------------------------------------------

nh Ly: He X = AX(t), ma tran A n gia tr rieng thc 1, 2 n (khong bat buoc phan biet), tng ng n vect rieng v1, v2 vn oc lap tuyen tnh Nghiem tong quat thuan nhat:

( ) ( ) ( ) ( )[ ] =

==

n

kk

tk

Tn vectxtxtxtX k

121 ,,,

K

VD: Giai he

+=

+=

yxyyxx

2'32'

=

1232

A : 2 GTP thc, VTR LTT

( )( )

+=

+=

+

=

=

tt

tttt

ececty

ecectxecec

yx

tX4

21

4214

21 23

23

11)(

[ ] [ ]TT vv 23,4;11,1 2211 ==== Tr rieng, vect rieng:

TONG QUAT: PHNG PHAP CHEO HOA MA TRAN--------------------------------------------------------------------------------------------------------------------------------

Ma tran A cua he c cheo hoa bi ma tran P: A = PDP-1

X(t) = AX(t) = (PDP-1)X(t) P-1X(t) = D.P-1X(t). oi bien

( )= tXPY 1 = )()(' tDYtY( )( )

( )

( )( )

( )

=

ty

tyty

ty

tyty

nnn

........

00.....................

0000

'

........

'

'

2

1

2

1

2

1

K

K

K

=

=

=

)()('........................

)()(')()('

222

111

tyty

tytytyty

nnn

({v1, vn}: vect rieng)

( )( )

( )

=

=

=

tnn

t

t

necty

ecty

ecty

....................

2

1

22

11

=

==n

kk

tk vecPYX k

1

Phai tnh cac vect rieng v1, vn; Khong tnh ma tran P1

GIA TR RIENG PHC (THAM KHAO)--------------------------------------------------------------------------------------------------------------------------------

Cap gia tr rieng phc, lien hp = i tng ng cap vect rieng v = a ib (a, b: vect) 2 vect nghiem c s

[ ] [ ]tbtaetbtae tt cossin,sincos +

VD: Giai he

+=

=

yxyyxx25'

6'

=

2516

A

i= 42,1

+=

=i

vi

v21

,

21

21

=

=

10

,

21

ba

( )( )

+

+

=

tteCtteC

tytx tt cos

10

sin21

sin1

0cos

21 4

24

1

HE KHONG THUAN NHAT (THAM KHAO) ------------------------------------------------------------------------------------------------------------------------------------------

Ma tran A cua X = AX + b(t) cheo hoa bi ma tran P: A =

PDP-1. He ban au X(t) = (PDP-1)X + b P-1X(t) = D.P-1X(t)

+ P-1b. oi bien:

( )( )

( )

+=

+=

+=

tbtyty

tbtytytbtyty

nnnn

~)()('..................................

~)()('

~)()('2222

1111

Phai tnh ma tran P

= [v1, vn] va P1:

e tnh vect P1b

( ) ( ) :tPYtX =

= XPY 1 ( ) += bDYtY ~'

+

=

nnnn b

bb

y

yy

y

yy

~

...

~

~

....

00.....................

0000

'

.....

'

'

2

1

2

1

2

1

2

1

K

K

K

Tnh tay: cong kenh! n gian hn nhieu: Phng phap kh!

V DU GIAI HE KHONG THUAN NHAT (THAM KHAO)-------------------------------------------------------------------------------------------------------------------------------------

Giai he khong thuan nhat

++=

++=

tyxyeyxx t

sin52'532'

=

1232

A cheo hoa (2 VTR oc lap tuyen tnh) viMa tran he:

[ ] [ ]TT vv 23,4;11,1 2211 ==== Tr rieng, vect rieng:

11

4001

1232

51

51

53

52

2131

=

=

= PPPP

D43421

: Cheo hoa

ma tran

Y = P1X

He mi:

~

1~

'

sinsin32

sin55 bDYY

te

tebPb

t

ebt

tt

+=

+

==

=

V DU GIAI HE KHONG THUAN NHAT (TIEP THEO)-------------------------------------------------------------------------------------------------------------------------------------

Giai he khong thuan nhat

++=

++=

tyxyeyxx t

sin52'532'

He mi:

+

+

=

+=

=

te

te

v

u

v

ubDYY

v

uY

t

t

sinsin32

4001

'

'

'.

~

He

++=

+=

tevv

teuut

t

sin4'sin32' ( ) ( )

( ) ( )

+=

++=

tteeCtv

tteeCtu

tt

tt

sin4cos171

31

sincos23

42

1

Quay ve bien X: Y = P1X

( )( )

( )( )

+

++

==

=

17sin4cos32sincos3

2131

42

1

tteeCtteeC

PYtytx

Xtt

tt