Hệ phương trình Hàm 6

Luận án thạc sĩ khoa học -ngành giải tích -Chuyên đê :Hệ phương trình Hàm

CHIfONG 4 " ., '" , ,

TDUA T GIAI SO VA CAC AP DUNG

4.1

(4.1)

(4.2)

(4.3)

THU~T GIAI SO TONG QUAT

Trong ph~n nay, ta xet thu~t giiii (2.8) voi I=[-b, b], ct;1th~ la :

Ta xet day le)p {fV)} cho bdi cong thuc sau :

VOiXE 1= [-b,b] va i = 1,2, , n, ta de)t

n m

j=l k=l J;(O)(x)=o.

Dva vao (4.1), ta tinh toan cac gia tri J;(V)(x/J t~i mQt s6 di~m nUt

roi r~c

X~ = -b + ~ L1x, ~=0, 1, , N,

(4.4 )

(4.5)

(4.6)

(4.7)

(4.8)

N ];(V) (x) = I];(V )(Xfl),Wfl(X)

fl=O

trong d6 cac ham wo(x), WI(X), , wr.ix) du'Qc xac dinh nhu' sau:

(x - Xfl-l)! Ax,

W flex) =~ (Xfl+l0 - x)! Ax,

WO(X)=

{

(Xl - x)! Ax,

0,

WN(X)=

{

(x - XN-I)! Ax,

0, f)~t

];~V) =];(V) (Xfl)

X fl-l ~ X ~ X fl '

X fl ~ X ~ X fl+l'

x ~ [Xfl-I,Xfl+d,

,1 ~ Jl ~ N - 1

- b ~ x ~ Xl'

Xl ~ X ~ b,

XN-I ~ X ~ b, -b~X~XN-I'

(4.9)

(4.10)

O~f.i~N, 1 ~ i ~ n,

~(O)

-J;Jl = 0

v ~ 1

Vi du 1:

(4.11)

(4.12)

HAl VI DC) TINH TOAN CC) THE

Ta ap d\lng tinh toan sf)voi hai vi d\l C\lthg san day

Truong hQp phi tuye' n

I

xl

trong d6

2

2

(4.14 )

Loi giai chinh xac cua h~ (4.11), (4.12) Ia

Tinh loan bdi thu~t giai (4.9), (4.10) voi cae bu'oc l~p v =1,2, saD cho

max

I

J;(V) - {'(v-I)

I

l:O:l:O:n Ji J IJl

0:0:Jl:O:N

< 10-8

Sau do, cho N tang d~n l~n lu'Qtvoi N=5, 10, 15,20, 100

Bang 1, 2 cho ke't qua so sanh fi(V)(x~)voi fiex(x~)t(,licae nut x~ voi i=I,2

Bang 3, 4 cho cae sai s6 thay d6i theo s6 nUtN tang d~n

-1

7(v) tex

lJi - J IJl - J 1 X Ji

-0

1

2

3

4

5

-1.000

-0.600 -0.200 0.200 0.600 1.000

-1.000 -0.600 -0.200 0.200 0.600 1.000

0.000061 0.000149 0.000193 0.000060 0.000149 0.000194

-max O:O:~:O:5 E~l - 1 94E-04

Bang 2: N=5 E 2fl -- lf~(V) 2fl - f ~ex2 ( X fl )1

-0

1

2

3

4

5

1.000 0.360 0.040 0.040 0.359 1.000

1.000 0.360 0.040 0.040 0.360 1.000

0.000004 0.000071 0.000239 0.000372 0.000525 0.000251

-max 0::;11::;5 E21l - O 525E-03

N el=max O::;Il::;NElll e2=max O::;Il::;NE21l

Vi du 2: Truong hqp tuye'n Hnh

a Uk(x,y) = aUk.Y Xet h~ san day voi n= m=2, b =1, x E [-1,1]

II (x) = 100II (2"+ 3) + 200 12(2"+ 4) + 200 12("3+ 2) + g I(x)

(4.15 ) J

12 (x) =200 II ("3+ 2) + 200 J; (2"+ 3) + 100 12 ("3+ 4) + g 2(x)

trong d6

g2(x) = 1200x - 1200

Cac s6 aUk'bijk.Cijkvoi i, j, k b~ng 1 ho~c 2 thml (2.10), (2.13), (2.14).

Loi giai chinh xac cua (4.15), (4.16) la

Tinh loan bdi thu~t giai (4.9), (4.10) voi cac btioc l~p v =1,2, sao cho

I /,(V) - .r(v-I)

I

O';'I',;,N

K€t qua cho bdi bang 5 va 6 san day chI cac gia tri tinh loan l~v) so sanh

vOi cac gia tri chinh xac /,~x (x 1') t~i cac diem nut Xo,Xl ., Xs, voi N=5

Bang 5: N=5 E

-1

(

)1

l,u - J l,u - J! X ,u

-J.l J!,u 'l(V) hex (X ,u ) E!,u

-maxo::;J.!::;5EJ.!1 = 3.035150 E-11

-lf

~(V)

f~ex (

)1

2,u- 2,u- 2 X,u

-J.l f~(V)

-0

1

2

3

4

5

-6.000 -3.600 -1.200 0.120 0.360 6.000

-6.000 -3.600 -1.200 0.120 0.360 6.000

0.000000 0.000000 0.000000 O.OOOOOD 0.000000 0.000000

-max O::;J.!::;5 E 2J.! = 3 13 4 2 0 6 E - 11

N el=max o::;J.!::;NE!J.! e2=max O::;J.!::;NE2J.!

(4.18)

(GI)

(G2)

(G3)

(G4)

(G5)

CHU Y VE MOT HI; TUYEN TINH TUONG T(j

Trong ph~n nay chung t6i ehu yd€n h~ phudng trinh ham tuy€n tinh thuQe d~ng san day:

n

Laijf/Sij (x)) =gi(X), j=1

I:::;;i:::;;n

VxElcR,

trong d6 I la khoang bi eh~n hay kh6ng bi eh~n eua R.

ajj, 1 s i,j s n la' cae h~ng s6 eho truDe, fj ,Is i s n Ia cae fin ham

f)~u lien, gi6ng nhu cae ky hi~u d ehudng 2, ta d~t

x = C(I; Rn)

X = Cb(l; Rn)

n€u I = [a, b] ,

n€u I Ia khoang kh6ng bi eh~n eua R.

Ta thanh l~p cae gia thi€t san :

g=(gl, ,gn) EX,

Sjj t6n t~i ham nguQe Sjj-l lien We, V i =1, ,n ,

ajj "* 0, Vi=I, ,n,

n /I

l aH

a = ~ ~ a~I < 1 j=1

Khi do ta co kSt qua sau:

Dinh If 4.1 :

(4.19)

(4.20)

(4.21)

(01)

(02)

Gid sa (G1) - (G5) la dung Khi do t8n tf;li duy nh[{t mQt ham

cung tIn dtnh deli vai g trong X.

Tir gla thiSt (G3), (G4) ta viSt h<$(4.18) v€ d';lng tu'dng du'dng

hay

j=1

vOi

aii =0

a

aij= ,j*l

au

Sij(X) =Sij(Su (x))

gJx) = ~gi(Si~1(X))

au

Thea cae gia thi€t (GI)-(G5) ta co:

g-\g1, ,gn)EX,

S :I~I 1J lien We ,

(4.22)

(4.23)

(4.24)

(4.25)

a = I iJiij I < 1.

i=1 j=1

Ap dl;lng k€t qua t6ng quat trong chudng 2 yoi m=1 Khi d6 dinh ly 4.1 du<;1cchung minh

.

chlnh thu? t giai cho h~ (4.18) nhusau:

ff/") (x) = tiiiff}"-l) (Sif(X))+ gJx),

trong d6

aH =

t

oa.

aii

Sy-Cx)= Sij(Si~lex))

au

j = i

j 7=i

Til thu?t giai (4.22) ta xac dinh cac gia tri roi r(;lc cua Wi giai t(;li cac di€m nut Xjnhusau:

(4.27)

(4.28)

(4.30)

(4.31)

(4.32)

x~ =-1 + ~ ~x

0:::; ~:::; N,

Sau d6 nQi suy cae gia tri t<;ticae di€m nUt (4.26) bdi cae ham lien tl;te

tren [-1,1] va tren m6i do<;tn[X~-l,x~] Ia ham b~e nhc1ttheo x, 0:::;~:::;N,

tae Ia

N

f;(v>cx) = l:f;(V)(x").w,, (x)

,,=0

f)~1

J;~V) = J;(v>c x" )

!

(X-X,,_l)/ &, X,,-l ~ X ~ x"'

wo(X)=

{

(Xl - x)/ &,

0,

-l~X~Xl'

Xl ~ X ~ 1,

WN(X) =

{

(x - XN-l)/ &,

0,

XN-l ~ X ~ 1, -l~X~XN-l

.{'(v)- o 1<°< 0 < < N _12

Ta ap dvng thu~t giai (4.33) tren hai vi dVev the:

Vi do 1:

X6t h~

(4.34)

3

(

t+1

2t+1 J

If; 1)+ 2001, (Vt)+ 1,( 2t; 1) = g,(t),

f (CDSt) + /, (sint) + 300/, ((1+21)' -I) = g, (I),

-1::::;t:~1

trong d6

g2(t)=(t-1) /2, g3(t)=cost

Loi giai ehinh xae eua (4.34), (4.35) Ia

Cae s6 thlfe aij , cae ham Siit), gi(t) thoa cae gia thie"t (G 1) - (G5)

(4.38)

Vi6t l,!i h~ (4.34) vi: d,!ng (4.20)

x ~

(

Vx+l

J

-~

(

2Vx+l

1

(

X3 -l\

j 1 C (

2X3 -I

J

1; (x) = -~300 1;(cOS(-1+ ~2(x + 1)))- ~300 12(sin(-1 + ~2(x + 1)))

Tinh toan bdi thu~t giai (4.33) voi cae bu'oc l~p v =1,2, sao cho

maxI J:(V) - f(v-l)

I

O';p,;N

Sau do, cho N tang dc1nlc1nlu'Qtvoi N=5, 10, 15,20, 100

Bang 9, 10, 11 cho k6t qua so sanh fj(V)(x/l)voi fjex(x/l)t,!i cae nut x/l voi i=1,2,3 va I.l=0, , N.

Bang 12, 13, 14 cho cae sai s6 thay d6i theo s6 nut N tang dc1n

Elp = I};~)- };ex(xp)1

===================================================================

====================~.~=============================================

0

1

2

3

4

5

I -0.999999999753

I -0.5999999997i1

I -0.199999999824

I 0.199999999982

I 0.599999999984

I 0.999999999985

-1.01)0000000000 -0.600000000000 -0.200000L}',QOOO 0.200000000000 0.600000000000 1.000000000000

0.000000000247 0.000000000229 0.000000000176 0.000000000018 0.000000000016 0.000000000015

===================================================================

maXO::;;/l::;;5 E 1/l = 2 4 7 E - 1 0

Bang 10: N=5 E - lf~(V)

==================================================================

J.l I f ~(V) 2j.1 f2ex(Xj.1)

==================================================================

E2j.1

0

1

2

3

4

5

I 0.000000000038

I -0.000000000033

I -0.000000000035

I -0.000000000035

I -0.000000000037

I -0.00000000n033

0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000

0.000000000038 0.000000000033 0.000000000035 0.000000000035 0.000000000037 0.000000000033

==================================================================

maxO:::Jl:::5 E2Jl = 3 82E-11

Bang 11: N=5 E 3j.1 - - lf~(V) 3j.1 - f ~ex3 ( Xj.1)1

-J.l I f ~(V) 3j.1 ft(xj.1)

- E3j.1

0

1

2

3

4

5

I 0.000000000038

I -0.000000000033

I -0.000000000035

I -0.000000000037

I -0.000000000036

I -0.000000000029

0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000

0.000000000038 0.000000000033 0.000000000035 0.000000000037 0.000000000036 0.000000000029

===================================================================

maXO:::Jl:::5 E3Jl = 3.82E-11

N EI=maxO:::Jl:::N EIJl e2=max O:::Jl:::NE2Jl e3=max O:::Jl:::NE3Jl

Vi do 2:

(4.39)

(4.40)

(4.41 )

Viet I~i h<$

(4.42)

VOl

100/1(13)+ 12 (sin2 ~J=gl(l),

(

t+I

m

)

-1:=0;t:=o;1 ,

gl(t)=IOOe , g2(t)=(t+ 1) / 2

Loi giai chinh xac cua h<$(4.39) la

f1e\t) = t f2eX(t) = 0

1 l

2 [

77:~

J}

77:

Il

f (x) =-" f

I

.

/

-I:=o;x:=o;1.

Bang 15: N=5 E

-I

'l(v) 'lex

l,u - J l,u - J I X ,u

-EI,u

0

1

2

3

4

5

I -1.000000000000

I -0.600000000000

I -0.200000000000

I 0.200000000000

I 0.600000000000

I 1.000000000000

-1.000000000000 -0.600000000000 -0.200000000000 0.200000000000 0.600000000000 1.000000000000

0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 -

-maXOS;/lS;5 E 1/l = 1, 14 E-13

Bang 16: N=5 E 2,u - - lf~(V) 2,u - f ~ex2 ( X,u)1

-E2,u

0

1

2

3

4

5

0.000000000011

0.000000000008

0.000000000004

0.000000000000

0.000000000000

0.000000000000

0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000 0.000000000000

0.000000000011 0.000000000008 0.000000000004 0.000000000000 0.000000000000 0.000000000000

-maXOS;/lS;5 E2/l = 1 06E-ll

N el=max OS;/lS;NEI/l e2=max OS;/lS;NE2/l