Ro'yxatga olish sanasi 29 Oktyabr 2024
100 SotishChiziqsiz tenglamalar sistemasini yechish usullari. Nyuton usullari
)TH (k)
(Δx(k))TH (k)(f(k+1)− f(k))
(3.33 )
Bro y den us u lini n g a l g o ri t m i q u y id a gi c h a :
1. x (0)
bos h l a n g ‘i c h y aqinlashi s h t a nla nad i.
2. Yakob m atr it s a s i W
( 0 )
ni ng t es k a r isini his o b l a s h bilan H (0)
m atrits an ing bo sh lang ‘ i c h q iy m ati hisob l a n ad i.
3.
f (k) ¿ f
x(k)
, k = 0 ,1,2, … his o bl a n a di.](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_14.png?v=1)
![o m m a bop b o ‘l i b, h i so b l a s h a m ali y otida k e ng qo ‘ ll an ilib kel i n m oqda.
Tez k or tushish usuli (gra di yen t lar u s uli)
Y u qorid ag i (3 . 1) t e n g l a m alar siste m a si ni q aray m iz. Faraz qil a y li k , fi
f u nksi y a-lar o‘zla r i ni n g u m u m i y an i qlan i sh s o h asida ha qi q i y v a uzluksiz
differ e nsi a l-lanuv c h a n. Qu y idagi funksi y ani qaray m iz:
U
( x ) =
∑
i = 1n [
f
i ( x )] 2
= ( f ( x ) , f ( x ) )
Ravch a nki, (3 . 1) sist e m aning har b i r y echi m i U ( x ) f unk si y ani n o l ga
a y lantir a di; va a ks i ncha U ( x ) funksi y a n ol g a t e ng b o ‘ la d ig a n x
1 , x
2 , … , x
n s o n lar
( 3 . 1) sis t e m an-ing i l dizla r i.
Faraz qi l a m i z k i, ( 3 .1) si ste m a i zo l y atsi y ala n g a n y echi m ga g ina e g a
b o ‘lib, b u y echim U ( x ) f u nksi y an in g qat ’ iy m i nim u m n u qtasi bo ‘ls i n. Bu
b il a n m asala n o ‘ lc h ovli fa z oda U ( x ) funks i y ani n g m ini m u m i n i top i shga olib
kelin ad i.
Faraz qila y lik, x - (1 ) siste m aning v ektor - i l dizi, x (
0)
esa u ning
boshl a ng‘ich y aqin l a s hi s hi b o ‘lsin. x (
0 )
n uqta o rq a li U ( x ) funksi y aning sath
si r t in i o ‘tka z a m iz. A g ar x (
0 )
n uqta x il d i z ga y etarl i c ha y aqin b o ‘l sa, u h olda
biz ni ng fa r az i m iz b o‘ y icha s a t h sirti
U (x)=U (x(0)) ellipsoidga o’xshash.
x(0)
nuqtadan boshlab U ( x ) = U ( x ( 0)
)
sirt normali bo‘ylab harakatlanib, bu
harakatni shu normal qaysidir boshqa bir
U (x)=U (x(1)) sath sirtiga biror
x ( 1)
nuqtada urinmaguncha davom ettiramiz. Keyin yana
x
( 1)
nuqtadan harakatni
U (x)=U (x(1))
sath sirti bo‘ylab davom ettirib, bu harakatni shu normal boshqa
bir yangi U
( x ) = U ( x ( 2)
)
sath sirtiga biror x(2) nuqtada urinmaguncha davom
ettiramiz va hokazo. Vaholanki,
U
( x ( 0))
> U ( x ( 1))
> U ( x( 2))
> … bo‘lsa, u holda biz shu
yo‘l bilan harakatlanib, U(x) ning (3.1) sistemaning izlanayotgan x ildiziga
mos keluvchi eng kichik qiymatli nuqtasiga tezkor yaqinlashib boramiz. Ushbu](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_16.png?v=1)
![∇ U( x ) =
[ ∂ U
∂ x
1
…
∂ U
∂ x
n ] tuzilgan U(x) funksiyaning gradiyenti belgilashini kiritib,
O M 0M 1,O M 1M 2,…
vektorli uchburchaklardan quyidagini yozamiz (xususiy
holda 3.10-rasmga qarang):
x(p+1)= x(p)− λp∇U (x(p)),(p=0,1,2 ,… )
Endi p ko‘paytuvchin aniqlash qoldi. Buning uchun quyidagi skalyar
funksiyani qaraymiz:
Φ '
(
λ ) = d
dλ U [ x ( p)
− λ
p ∇ U ( x ( p))]
= 0 ( 3.40 )
( 3 .40) t e ngla m aning e n g ki c hik m u s bat i l di z i bizga
λp n ing q i y m atini
b er a d i . E n di
λp s o n n i taq ri biy t opish u sulini q a r ab c h iqa y l i k. Faraz
q il a m i zk i ,
λ−¿ ki c h ik para m etr b o ‘ li b , u ni n g kv a dr a ti va u n dan y uq o r i
da r a j ala r ini e ’ tiborga ol m aslik m u m -kin. U ho l da qu y idagiga k ela m iz:
Φ
( λ ) =
∑
i = 1n {
f
i [ x( p)
− λ
p ∇ U ( x( p))]} 2
fi
f u nksi y ani λ n ing c h izi ql i
had l ar i g acha a n iq l ikd ag i d ara j ala r iga
y o y i b , qu y id a giga ega bo ‘ la m iz:
Φ (λ)=∑i=1
n
{fi[x(p)− λ∂ fi(x(p))
∂x ∇ U (x(p))]}
2
Bu yerda
∂ fi
∂x=[
∂ fi
∂x1
,∂ fi
∂x2
,… ,∂ fi
∂xn]
Bulardan
3.10-rasm. Ikki o’zgaruvchili
funksiya holida gradiyent
usulining geometrik talqini](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_17.png?v=1)
![Φ '(
λ ) = − 2
∑
i = 1n [
f
i ( x( p))
− λ ∂ f
i ( x( p))
∂ x ∇ U ( x( p))]
× ∂ f
i
( x( p))
∂ x ∇ U ( x( p))
= 0
Natijada
λ
p = ∑
i = 1n
f
i
( x( p)) ∂ f
i ( x( p))
∂ x ∇ U ( x ( p))
∑
i = 1n
[
∂ f
i
( x( p))
∂ x ∇ U ( x( p))] 2 =
( f
i ( x( p)))
, W ( x( p))
∇ U ( x ( p))
(
W ( x ( p))
∇ U ( x( p)))
,( W ( x( p))
∇ U ( x ( p)))
Bu yerda W
( x ) = ∂ f
i / ∂ x
– vektor funksiya f ning Yakob matritsasi.
Shularga ko’ra quyidagi natijaga kelamiz:
μ
p = 2 λ
p =
( f p
, W
p W
p'
f ( p))
(
W
p W
p'
f ( p)
, W
p W
p'
f ( p)) ( 3.41 )
bu yerda soddalik uchun quyidagicha yozuv qabul qilingan:
f p
= f
( x ( p))
; W
p = W ( x ( p )
)
bularga ko’ra
x
( p + 1 )
= x ( p)
− μ
p W
p'
f ( p)
p = 0,1,2 , … ¿ ( 3.42 )
Gradiyent tushish algoritmining blok-
sxemasi 3.11-rasmda tasvirlangan.
Misol. Tezkor tushish usuli (gradiyent
usuli) yordamida ushu
x + x 2
− 2 yz = 0.1
y − y 2
+ 3 xz = − 0.2
z + z 2
+ 2 xy = 0.3
}
Tenglamalar sistemasini koordinata
boshi atrofida yotgan ildizlarini
taqribiy hisoblang.
Yechish. Berilganlarga ko’ra
f =
[ x + x 2
− 2 yz − 0.1
y − y 2
+ 3 xz + 0.2
z + z 2
+ 2 xy − 0.3 ] , W = [ 1 + 2 x − 2 z − 2 y
3 z 1 − 2 y 3 x
2 y 2 x 1 + 2 z ] , x
( 0)
=
[ 0
0
0 ]
(3.41) va (3.42) formulalar bo‘yicha quyidagi birinchi yaqinlashishni olamiz:](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_18.png?v=1)
![μ0= (f(0),f(0))
(f(0),f(0))
=1,x(1)= x(0)−1∙E f(0)=
[
0.1
− 0.2
0.3 ]Xuddi shunday
x
( 2)
- ikkinchi yaqinlashishni aniqlaymiz. Bu yerdan:
f
( 0)
=
[ 0.13
0.05
0.05 ] , W
1 = [ 1.2 − 0.6 0.4
0.9 1.4 0.3
− 0.4 0.2 1.6 ] , W
1'
f
( 1)
=
[ 0.181
0.002
0.147 ]
W 1W 1'f(1)=
[
0.181
0.002
0.147 ]
,μ= 0.13 ∙0.2748 +0.05 ∙0.2098 +0.05 ∙0.1632
0.2748 2+0.2098 2+0.1632 2 =0.3719
x(2)=
[
0.1
−0.2
0.3 ]
−0.37119 ∙
[
0.181
0.002
0.147 ]
=
[
0.0327
−0.2007
0.2453 ]
Natijaning qanchalik to’g’ri va aniq ekanligini tekshirish uchun tafovut hisoblanadi.
N yuton u suli
( 3 .1 ` ) t e n gla m alar siste m asini y echish u c hun ket m a - k et y aqi n la sh i s h usulid a n
f o y da l a na m i z. Faraz q ila y lik, ( 3 .1 ` ) vektor tengla m aning iz o l y atsi y al a ng a n x = ( x
1 ,
x
2 , …, x
n ) i l di z la r idan bi t ta s i bo ‘lgan us h b u k -inchi y aq i nl a sh i sh
x
( k)
= ( x
1 ( k)
, x
2 ( k)
, … , x
n ( k)
)
top il g a n b o ‘lsin. U h olda ( 3 . 1 ` ) vektor t en gla m aning a n iq i ld iz i ni ushbu
x= x(k)+ε(k)(3.2 )
ko ‘r i ni s hda ifod a l a sh m u m kin, bu y erda ε ( k )
= ( ε
1
( k)
, ε
2 ( k)
, … , ε
n ( k)
)
x a t o likni tuz a tuv c hi
had ( ildi z ni n g x a toli g i).
( 3 . 2) i f oda n i ( 3 . 1 ` ) ga qo ‘ y i b , q u y i d ag i t en g l a m a n i hosil qila m iz:
f
( x( k)
+ ε ( k))
= 0 ( 3.3 )
Faraz q i la y l i k , f ( x ) - bu x va x ( k )
la r ni o ‘ z i c hi ga o lg a n bi r or qovariq D
soha d a uzluksiz differensiallanuvchan funksiya bo’lsin.
(3 . 3) t engla m aning o ‘ng
taraf in i ε
(k) -kic h ik v e ktor dara j ala r i b o ‘ y icha q a to rg a y o y a m iz v a bu qatorn i ng
c hi zi qli hadl a ri bil a ng i na c he k l a na m iz:
f
( x( k)
+ ε ( k))
= f ( x ( k))
+ f ' (
x ( k))
ε( k)
( 3.4 )](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_19.png?v=1)
![( 3 . 4) f or m ul a d a n kelib chiqadiki, f’(x) hosila deb x1,x2,… ,xn−¿ o‘zga r uvc h ilar g a
n isbat a n
f1,f2,… ,fn−¿ funks i y alar siste m asini n g qu y i d a gi Yakob m atritsasi
tu s hu n i l ad i:
y ok i u ni qi s qa c ha v e k t o r sha k lida y oz s ak,
f'(x)=W (x)=[
∂ f1
∂xj],i,j=1,n
(3.4) sistema bu xatolikni tuzatuvchi had ε
i
( k)(
i = 1 , n )
larga nisbatan W(x) matritsali
chiziqli sistema. Bundan (3.4) formulani quyidagicha yozish mumkin:
f(x(k))+W (x(k))ε(k)= 0
Bu yerdan,
W (x(k)) maxsus bo‘lmagan matritsa deb faraz qilib, quyidagiga ega
bo‘lamiz:
ε(k)=−W −1(x(k))∙f(x(k))
Natijada ushbu
x
( k + 1 )
= x ( k)
− W − 1 (
x ( k))
∙ f ( x( k))
( 3.5 )
Nyut o n usuli f o r m ula s iga kela m i z, b unda x ( 0 )
n olin c hi y aqinl as hi s h si f a t ida
i z l a -na y otgan il d izning qo‘pol q i y m atini oli s h m u m ki n.
A m ali y otda ( 3 . 1 ` ) noc h izi qli t e ngla m alar
si ste m a s ini bu u s u l b i lan y echi s h u c hun hi s obla s hlar
( 3.5) for m ula bo ‘ y icha qu y ida g i sha r t b a j a ril g unga q a d ar
d avo m et t i r iladi:
|
x ( k + 1 )
− x ( k)|
∞ < ε
(3 . 6 ) Y uqoridagila r d a n k e lib c hiqib, N y u t on usul i n -
ing a lgo r i t m i ni q u y idagic h a y oz a m i z:](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_20.png?v=1)
![{
x1= 4.6 − x12
10 − ln x2
3
x2=3.1 −e−x1−√x2
;
{
φ1(x1,x2)= 4.6 − x12
10 − ln x2
3
φ2(x1,x2)=3.1 −e−x1−√x2Endi qidirilayotgan o‘zgaruvchilar bo‘yicha hususiy hosilalar olinadi:
∂φ1
∂x1
= − x1
5 ,∂φ1
∂x2
= −1
3x2
,∂φ2
∂x1
= e−x1,∂φ2
∂x2
= −1
2√x2
Aytaylik, boshlang’ich yaqinlashish
x1 va x2 lar bo‘yicha [1,4] k е smada bo‘lsin. U
holda hosilalar uchun quyidagi t е ngsizliklar o‘rinli bo‘ladi:
|
∂ φ
1
∂ x
1 | ≤ 4
5 = 0.8 = p
1 | ∂ φ
1
∂ x
2 | ≤ 1
12 = 0.083 = q
1
|
∂φ2
∂x1|≤ 1
e4≈0.0069 = p2|
∂φ2
∂x2|≤1
4= 0.25 =q2
Dеmak, qaralayotgan kvadratda:
p1+p2=0.8 +0.0 069 =0.8069 <1
q
1 + q
2 = 0.083 + 0.25 = 0.333 < 1
yaqinlashish shartlari bajariladi.
U holda dastlabki yaqinlashish sifatida x
1
( 0)
= 3.5 x
2 ( 0)
= 1.7
ni olib, k е yingi
yaqinlashishlarni oddiy it е ratsiya usuliga mos dastur ta`minoti yordamida
aniqlanadi.
Buning uchun MathCAD dasturining ishchi oynasiga quyidagi buyruqlar
kiritiladi.
1 x1 x2( ) 4.6 x1 2
10 ln x2( )
3
2 x1 x2 ( ) 3.1 e x1 x2 iter x1 y1 ( ) k 0
x0 x1
y0 y1
x1 1 x0 y0( )
y1 2 x0 y0( )
x x1 x0
y y1 y0
k k 1
break max x y( )( ) if1while
x1
y1
iter 3.5 2.2 0.001 ( )
3.315
1.743
](https://docx.uz/documents/984002b5-23df-44d1-b151-5f35ec98136e/page_25.png?v=1)
Mavzu: Chiziqsiz tenglamalar sistemasini yechish usullari. Nyuton usullari
Mundarija:
- Kirish
- Iteratsiyalar usuli (ketma-ket yaqinlashish usuli)
- Oddiy iteratsiya usuli
- Chiziqsiz tenglamalar sistemasini Mathcad dasturi yordamida taqribiy yechish
Foydalanilgan adabiyotlar ro’yxati