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- Bu sahifa navigatsiya:
- I bobning qisqacha xulosasi
- II bob. Xususiy hosilali differensial tenglamalarni Mathcad muhitida taqribiy yechish 2.1. Elliptik
- 2.2. Chebishev parametrlar majmuasi bilan oshkor iteratsion metodi M athcad
Birinchi chegaraviy masalani yechish. Biz endi tebranish tenglamasi uchun G = {0 < x < 1,0 < t < T} sohada ushbu birinchi chegaraviy masalani ko’rib chiqamiz. Yani G sohada ikki marta uzluksiz differensiallanuvchi u(x, t)
funksiyanitopishkerakki , bu sohadau = (1.5.10) tenglamani qanoatlantirib, t = 0 to’g’ri chiziqda u(x,0) = Q(x), (x,0) =Ψ (x) dastlabki shartlarniva u(0, t) = μ1 (t), u(1, t) = μ2 (t), 0 < t < T (1.5.11) (1.5.12) Chegaraviy shartlarni qanoatlantirsin. Bu maslani to’r metodi bilan yechish uchun ushbu Ghτ = {xi = ih, i = 0, M, hM = 1;tk = kτ, k = 0, N, Nτ = T} To’rni kiritamiz va 1.3-chizmadagidek uch qatlamli andaza bo’yicha (1.5.1) differensial tenglamani (1.5.3) dagi ayirmali sxema bilan almashtiramiz, bu yerda i va k quyidagi qiymatlarni qabul qiladi: i = 1,2,......, M - 1; k = 1,2,....., N - 1 Dastlabki shartlar uchun (1.5.7) formuladan foydalanamiz. Chegaraviy shartlar quyidagichayoziladi: y0+1 = μ1 (tk+1), yM1 = μ2 (tk+1), k = 0,1,...., N - 1 Bularning hammasini birlashtirib, ayirmali sxemaning quyidagi hisoblash algoritmiga egabo’lamiz: yi = Q(xi ), yi = Q(xi )+τΨ(xi )+ Δ 2Qi , yi+1 = 2yi +τ2 Δ 2 yi - yi -1 , i = 1,2,....., M - 1, y0+1 = μ1 (tk+1), yM1 = μ2 (tk+1), k = 0,1,......, N - 1 (1.5.13) (1.5.14) (1.5.15) Yuqorida ko’rdikki, bu sxema (1.5.1), (1.5.3) chegaraviy masalani 0(τ2 + h2 ) aniqlikda approksimatsiya qiladi. Ko’rsatish mumkinki, agar ixtiyoriy ε > 0 uchun τ va h qadamlar quyidagi < (1.5.16) Shartni qanoatlantirsa , (1.5.7), (1.5.10), (1.5.11) sxema turg’un bo’ladi. Biz buning isbotiga to’xtalib o’tirmaymiz. Misol. To’r metodi bilan G = {0 < x <1, 0 < t < T} sohada δ 2 u δ 2 u = δt 2 δx2 To’rtenglamasining u(0, t) = u(1, t) = 0(0 < t < 1), u(x,0) = sinπx, (x,0) = 0 Chegaraviy va dastlabki shartlarni qanoatlantiradigantaqribiy yechimi topilsin. Yechish. Bu yerda h = 0,1 va Q(x) = sinτx, Q" (x) = 一τ2 sinπx hamda Ψ(x) = 0 quyidagichayozamiz: τ = 0.08 deb olamiz. Keyin liginihisobga olib, (6.5) formulani yi = yi + Q" (xi ) = (1 一 0.0032π2 )sinπxi Endi Δ 2 operatorning ko’rinishini e’tiborga olsak, hisoblash uchun quyidagi algoritm hosilbo’ladi: yi = sinπxi , yi = (1 一 0.0032π2 )sinπxi , i = 1,2,......, M 一 1, y 0+1 = yM1 = 0, k = 0,1,........, N 一 1, yi+1 = 0.64y 1 + 0.72yi + 0.64yik一1 一 yi 一1 Hisoblashni faqat 0 < x < uchun bajarsa yetarli bo’ladi, chunki u = u(x, t) yechimning grafigi x = tekislikkanisbatan simmetrik ravishda joylashgan. I bobning qisqacha xulosasi I bobdaxususiy hosilali differensial tenglamalarnitaqribiy yechish metodlari haqida gapirilgan. II bob. Xususiy hosilali differensial tenglamalarni Mathcad muhitida taqribiy yechish 2.1. Elliptik tenglamalarni Mathcad dasturiyordamida taqribiy yechish. Faraz qilamiz: chegaraviy shartlar u(A, y) = u(y)1 u(x, C) = u(x)3 bilan ux ,x + uy ,y = f (x,y) u(B,y) = u(y)2 u(x,D) = u(x)4 A < x < B C < y < D tenglamani sonli yechish talab qilingan bo’lsin. ui-1,j - 2ui , j + ui+1, j ui ,j-1 - 2ui , j + ui , j+1 h k , 2 + 2 = yi j i = 1,..N - 1 ayirmali sxema bilan almashtiramiz. j = 1,..M - 1 Bu sistemani yechib: yi ,j i = 1,..N taqribiy yechiminitopamiz . Konkret bir masalani qaraymiz. j := 0,1..M u(x1 , x2 ) + u(x1 , x2 ) = 2x2 f (x1 , x2 ) := -2(x1 + x2 ) u(0, x2 ) = 0 μ chap- μleft (x2 ) := 0 u(x1 ,0) = 0 μ quyi- μbottom (x1 ) := 0 u(1, x2 ) = x2 + (x1 )2 u(x1 ,1) = (x ) + x11 μright (x2 ) := x2 + x2 μtop (x1 ) := x12 + x1 tenglamaning aniq yechimi: u(x1 , x2 ) := x12x2 + x1x22 G = {(x1 , x2 ) : 0 < x1 < 1,0 < x2 < 1} ekanligima’lum. N := 5 M := 5 a:= 0 b:= 1 c := 0 d := 1 h := l := h yj ,0 := μleft (x2,j ) = 0.2 l = 0.2 i := 0...N j := 0..M x1,,i := ih x2,j := jl yj ,N := μright (x2,j ) y0,i := μbottom (x1,i ) yM ,i := μtop (x1,i ) j := 0..M - 2 i := 0...N ( | | | | | | | | 「- 0.4 - 0.8 - 1.2 - 1.6 - 2 - | 2.4] | 2.8 | 3.2| | 3.6」 - 0.8) - F = |- 0.8 - 1.2 - 1.6 - 2 - 2.4 - |- 1.2 - 1.6 - 2 - 2.4 - 2.8 - | 1.2 - Fj ,i := f (x1,i , x2,j+1) F<1> = 1.6 ( L- 1.6 - 2 - 2.4 - 2.8 - 3.2 - - 2 ) F0,2 := F0,i + y0,i l2 「- 0.4 - 0.8 - 1.2 - 1.6 - 2 - F | M -2,i := FM -2,i + F = - 1.2 - 1.6 - 126 --224 - 2.8 - | L- 1.6 - 2 - 2.4 - 2.8 - 3.2 - 2.4] | 2.8 | 3.2| | 3.6」 A := identity(M - 1) B:= A m := 0,...M - 2 n := 0,..N - 2 h Λm,n := if (m = n - 1) + (m - 1 = n),1,0 C := identity(M - 1) - Λ a1 := C- 1B β1 := C-1F<0> i := 1,..N ai := (C - Aai )-1 B C-1A + C -1B = 0.01 C-1B = 0.005 ai+1 := (C - Aai )-1B βi+1 := (C - Aai )-1 (Aβi + F<i>) i := 0,..N j := 0,..M - 2 Wj ,i := yj+1,i i := N - 1,...0 W<i> := ai+1W<i+1> + βi+1 i := 0,..N j := 0,...M - 2 yj+1,i := Wj ,i 2.1 chima. Taqribiy yechim grafigi 2.2. chizma. To’rni ichkinuqtalaridagi yechim grafigi 2.2. Chebishev parametrlar majmuasi bilan oshkor iteratsion metodi Mathcad Download 0.78 Mb. Do'stlaringiz bilan baham: |
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