Phuong Phap Phan Tu Huu Han

Bai giang

PHNG PHAP PHAN T HU HAN

c Trnh Anh Ngoc

1/9/2009

iMuc ch

. Sinh vien nam c noi dung cua phng phap phan t hu han(PTHH).

. Sinh vien biet ap dung phng phap PTHH e giai so cac bai toan xuathien trong khoa hoc va ky thuat, ac biet, trong c hoc va vat ly.

. Sinh vien co the t oc cac sach, bai bao ve PTHH.

Noi dung

1. Gii thieu

2. Ly thuyet c ban

3. PTHH trong c hoc

Tai lieu oc them

T.A. Ngoc, Hoc Matlab bang th du, 2009.Rao S.S., The finite element method in engineering, Pergamon Press, 1989.Strang G., Fix G.J., An analysis of the finite element method, Prentice-Hall, Inc.,Englewood Cliffs, N.J., 1973.

Cap nhat: 9/2011

ii

Toan hoc khong phai la Ba hoang,Toan hoc la Nang hau, Ngi phuc vu cua cac khoa hoc

(Mathematics is a servant, but not a queen).Nh tat ca chung ta eu biet, hau het cac ly thuyet toan eu co nguon

goc t c hoc va vat ly, ong lc phat trien cua no (cac ly thuyet toan) la apdung vao (phuc vu cho) cac nganh khoa hoc co the "rat xa" toan hoc. Viechoc toan, lam toan, v the, khong the tach ri cac nganh khoa hoc ma toanhoc hng ti. Trong tai lieu nay hau het cac th du c trnh bay lay tnguon cac bai toan xuat hien trong c hoc va vat ly. ieu nay ban au cothe lam "nan long" cac ban sinh vien chon cac hng giai tch, ai so, . . . ;tuy nhien, vi viec tm hieu them cac ly thuyet c hoc, vat ly tng ng, viechoc toan se tr nen thuan li hn, va nhat la kien thc toan cua ngi hocse tr nen toan dien hn.

Trnh Anh Ngoc

Bang ky hieu

nn so nutne so phan tx(e)i nut th i (th t a phng) cua phan t exi nut th i (th t toan cuc)LA ma tran lap ghepCOORD ma tran toa o[x

(e)1 , x

(e)2 ] phan t e

{ue} = {ue1, ue2}T vect chuyen dch nut phan t e{u} = {u1, u2, . . . , unn}T vect chuyen dch nut toan cuc[N e] ma tran ham dang cua phan t e[N ] ma tran ham dang toan cuc[Ke] ma tran o cng cua phan t e[K] ma tran o cng toan cuc[Me] ma tran khoi lng cua phan t e[M ] ma tran khoi lng toan cuc{pe} vect tai phan t e{p} vect tai toan cuc

iii

Chng 1

Gii thieu

1.1 Dan nhap

. Phng phap PTHH la mot phng phap giai so tm nghiem xap xcua cac phng trnh ao ham rieng xuat hien trong khoa hoc va kythuat. Khac vi phng phap sai phan hu han, xap x trc tiep phngtrnh ao ham rieng (xap x ao ham), phng phap phan t hu handung cong thc bien phan cua bai toan bien (dang tch phan cua baitoan). Trong phng phap PTHH Mien cua bai toan (mien xac nhcua phng trnh ao ham rieng) c phan hoach thanh cac mien congoi la phan t hu han, va nghiem cua phng trnh ao ham riengc xap x bang cac a thc n gian tren moi phan t.

. Kien thc can biet: Toan hoc cao cap (giai tch, ai so tuyen tnh),phng trnh vat ly toan, c hoc ket cau1).

. Phng tien tnh toan: phan mem Matlab2).

. Lch s cua PTHH+ Cong thc PTHH au tien c phat trien nh la phng phap matran cua c hoc ket cau.

+ Courant (1943) a ra he gom cac phan t tam giac (triangular ele-ments) lap ghep vi nhau va cc tieu the nang (potential energy) e

1)Hai mon hoc cuoi khong bat buoc phai hoc trc, sinh vien co the tham khao them khican.

2)Sinh vien khong yeu cau phai biet ve Matlab, cach s dung Matlab se c gii thieudan trong giao trnh.

1

2 CHNG 1. GII THIEU

giai xap x bai toan xoan (cf. R. Courant, Variational methods for thesolution of problems of equilibrium and vibrations, Bulletin of AmericanMathematical Society, 49, 1-23,(1943)).

+ Clough e ngh ten goi "phan t hu han" (cf. M.J. Turner, R.W.Clough, H.C. Martin and L.J. Topp, Stiffness and deflection analysis ofcomplex stuctures, Journal of Aeronautical Sciences, 23, 805-824 (1956)).

+ Phat trien ban au cua phng phap c thc hien bi cac ky strong lanh vc c hoc ket cau; ve sau c s toan hoc c thiet lap choPTHH, t o m rong ap dung PTHH cho cac lanh vc khac.

1.2 Phep tnh bien phan

e n gian ta xet trng hp ham mot bien.Cho phiem ham (functional) F (x, u, u) vi u = u(x) va u = du/dx.Bien phan (variation) cua u, ky hieu u, la mot ham bieu dien "bien

thien cua u"u+ u

tng t nh so gia x bieu dien "bien thien cua x"

x+x.

Neu u c cho trc tai mot iem (thng la ieu kien bien) th bienphan cua u phai bang khong tai o, th du

u(x0) = u0 u(x0) = 0.V bien phan u triet tieu tai cac iem c cho trc, nhng lay gia tr tuyy tai cac iem khac nen u con c goi la bien thien ao.

Tng t nh phep tnh vi phan cua ham nhieu bien, bien phan capmot cua phiem ham F tai u c nh ngha la

F = lim0

F (x, u+ u, u + u) F (x, u, u)

. (1.1)

Dung cong thc vi phan toan phan cho F (x, u+v, u+v) tai (x, u, u) vix co nh,

F (x, u+ u, u + u) F (x, u, u) Fu

u+F

uu,

1.2. PHEP TNH BIEN PHAN 3

ta co

F =F

uu+

F

uu. (1.2)

Chu y s tng t gia bien phan cap mot vi vi phan toan phan cua ham F

dF =F

xdx+

F

uu+

F

uu.

Vi phep tnh bien phan x khong thay oi khi u bien thien thanh u + unen dx = 0 va s tng t gia F va dF tr nen ro rang. Ngha la, toan tbien phan tac ong nh toan t vi phan oi vi cac bien phu thuoc.

Ta co cac cong thc sau:

1. (F G) = F G.2. (FG) = FG+ FG.

3. (F

G

)=FG FG

G2.

4. (F )n = nF n1F .

5.d

dx(u) =

(du

dx

).

6. ba

u(x)dx =

ba

u(x)dx.

Th du 1.1. Phiem ham nang lng cua dam chu uon:

I(w) =

l0

[b

2

(d2w

dx2

)2+ wf

]dx dw

dx(l)M0

4 CHNG 1. GII THIEU

S dung cac cong thc tren ta co

I(w) =

l0

(bd2w

dx2d2w

dx2+ wf

)dx dw

dx(l)M0.

Nhan xet 1.1. Trong c hoc moi trng lien tuc, ta co hai khai niem quantrong: chuyen dch kha d va chuyen dch ao. Bi nh ngha, chuyen dch kha dla chuyen dch bat ky thoa ieu kien bien ong hoc, chuyen dch ao la hieucua hai chuyen dch kha d. Nh vay chuyen dch ao thoa ieu kien bienong hoc thuan nhat. Neu goi w la mot chuyen dch kha d, u la chuyendch thc (mot trng hp cua chuyen dch kha d) th chuyen dch ao lau = wu. Noi khac i, chuyen dch ao chnh la bien phan cua chuyen dchthc u, hay chuyen dch kha d la tong cua chuyen dch thc vi chuyen dchao, w = u+ u.Nhan xet 1.2. S tng t gia phep tnh bien phan vi phep tnh vi phan:

Phep tnh vi phan Phep tnh bien phanHam so f(x) Phiem ham I(u)oi so - x (so thc) oi so - u (ham so)So gia vo cung be dx Bien phan uVi phan cua ham so, df Bien phan cua phiem ham, Iieu kien at cc tr cua ham so ieu kien at cc tr cua phiem hamdf = 0 vi moi dx I = 0 vi moi u

1.3 Cac phng phap so

PTHH c xay dng tren c s phng phap bien phan (variational method).Muc nay gii thieu mot so the hien cua no.

1.3.1 Phng phap Ritz

Giai bai toan cc tieu hoa phiem ham I(v), v V (trong cac bai toan c hocth V la tap hp tat ca cac "chuyen dch" kha d).

1.3. CAC PHNG PHAP SO 5

Phng phap Ritz tm nghiem xap x di dang tong hu han

uN =

Nj=1

cjj + 0, (1.3)

trong o i, i = 0, 1, . . . , N , la cac ham chon trc; cac he so cj , goi la he soRitz, c chon sao cho

I(c1, . . . , cN) := I(uN) min (1.4)ieu kien can e I(c1, c2, . . . , cN) at cc tr la cac ao ham rieng cua

no oi vi moi he so Ritz triet tieu:

I

c1=

I

c2= = I

cN= 0. (1.5)

Trong cac bai toan tuyen tnh, he (1.5) gom N phng trnh ai so tuyen tnhvi N an c1, c2, . . . , cN .

Tnh chat cua cac ham i (i = 0, 1, . . . , N). Ham 0 c chon thoa ieu kien bien cot yeu (ieu kien bien Dirichlet)

cua bai toan.. Cac ham i, i = 1, 2, . . . , N , thoa ieu kien bien cot yeu thuan nhat.. Hn na, cac i can thoa them cac ieu kien sau:

(i) I(uN) c xac nh tot,(ii) Vi N bat ky, tap hp {i} la oc lap tuyen tnh,(iii) {i} la ay u.

Cac yeu cau (i) -- (iii) bao am, vi cac bai toan tuyen tnh, s hoi tu cuanghiem Ritz ti nghiem chnh xac khi N tang. S hoi tu c hieu theongha sau

I(uN) I(uM) khi N M. (1.6)

1.3.2 Phng phap d so co trong

Phng phap d so co trong la mot tong quat hoa cua phng phap Ritz.Cho L la toan t ao ham, xet phng trnh toan t

Lu = f trong , (1.7)

6 CHNG 1. GII THIEU

ngoai ra ham u (nghiem) con thoa cac ieu kien bien.Trong phng phap d so co trong, nghiem u c xap x theo cach

giong nh phng phap Ritz

uN =

Nj=1

cjj + 0, (1.8)

trong o 0 phai thoa moi ieu kien bien ch nh cua bai toan, va i phaithoa dang thuan nhat cua cac ieu kien ch nh. Cung nh phng phapRitz, vi muc ch xap x, cac ham i cung phai co cac tnh chat (i) -- (iii).

Thay xap x (1.8) vao phng trnh toan t (1.7) ket qua la d so

E = LuN f 6= 0. (1.9)Ngay khi 0 va cac i a c chon, E la ham cua cac bien oc lap va cactham so cj . Trong phng phap d so co trong, cac tham so c xac nhbang cach at tch phan cua tch d so co trong vi cac ham trong lng bang0,

i(x, y)E(x, y, cj)dxdy = 0, (1.10)

trong o i la cac ham trong lng. Hien nhien {i} phai la tap oc laptuyen tnh.

Noi chung, cac ham trong lng i khong giong cac ham i. Tuy theocach chon i ta co cac phng phap khac nhau:

Phng phap Galerkin i = i. PTHH c thiet lap tren c sphng phap nay. Khi toan t L la toan t ao ham tuyen tnh cap chan, thphng phap Galerkin dan ve phng phap Ritz.

Phng phap Petrov - Galerkin i 6= i.

1.3.3 Phng phap bnh phng toi thieu

Phng phap bnh phng toi thieu (least-squares method) tm nghiem dang(1.8) va xac nh cac tham so bang cach cc tieu hoa tch phan cua bnhphng d so. ieu kien cc tieu:

ci

E2(x, y, cj)dxdy = 0


hay

E

ciEdxdy = 0. (1.11)

1.3.4 Phng phap ong v

Phng phap ong v (collocation) hay phng phap chon iem tm nghiemxap x uN cua phng trnh (1.7) dang (1.8) bang cach oi hoi d so trongphng trnh triet tieu tai N iem chon trc xi = (xi, yi) (i = 1, 2, . . . , N )trong mien :

E(xi, yi, cj) = 0 i = 1, 2, . . . , N (1.12)

S chon la cac iem xi la cot yeu e nhan c mot he phng trnh chnh(well-conditional) va cho nghiem xap x tot. Phng phap ong v la mottrng hp ac biet cua (1.10) vi i(x) = (x xi), trong o la ham deltaDirac

f(x)(x x)dx = f(x).

Th du 1.2. Cho bai toan bien: tm ham u(x), 0 < x < 1, thoa phng trnhvi phan

u + f = 0, (1.13)

va cac ieu kien bien

u(1) = g, (1.14)u(0) = h, (1.15)

trong o f(x) la ham va g, h la cac hang so cho trc.ieu kien bien (1.14) la ieu kien bien cot yeu, ieu kien bien (1.15) la

ieu kien bien t nhien.e ap dung phng phap PTHH giai bai toan nay ta can phat bieu bai

toan di dang bien phan.? Cong thc bien phan (yeu)Ky hieu:

8 CHNG 1. GII THIEU

S = {s|s H1(0, 1), s(1) = g} la tap hp cac ham th (tng t nh taphp cac chuyen dch kha d);

V = {v|v H1(0, 1), v(1) = 0} la tap hp cac ham trong lng (tng tnh tap hp cac chuyen dch ao)3). Ky hieu H1(0, 1) ch khong gian Sobolevcac ham cung vi ao ham cap 1 (ao ham suy rong) bnh phng kha tch(thuoc L2(0, 1)).

Nhan hai ve phng trnh (2.118) vi ham trong lng v V bat ky, laytch phan t 0 en 1, ta c

10

(u + f)vdx = 0. (1.16)

Dung cong thc tch phan tng phan, sau mot so bien oi, ta c

10

u(x)v(x)dx =

10

f(x)v(x)dx+ hv(0). (1.17)

Goi 0 la ham thuoc S thoa ieu kien bien cot yeu 0(1) = g th u = 0+w,w V , va phng trnh tren thanh

10

w(x)v(x)dx =

10

f(x)v(x)dx+ hv(0) 10

0(x)v(x)dx. (1.18)

Tren V a vao dang song tuyen tnh (oi xng) a : V V R,

a(w, v) =

10

w(x)v(x)dx (1.19)

va dang tuyen tnh l : V R

l(v) =

10

f(x)v(x)dx+ hv(0) 10

0(x)v(x)dx, (1.20)

3)Trong trng hp g 6= 0, V la khong gian vect con S th khong.


bai toan bien (2.118)-(1.15) dan ve bai toan bien phan tuyen tnh: Tm w Vthoa phng trnh bien phan

a(w, v) = l(v) (1.21)

vi moi v V .? Phng phap GalerkinTm nghiem xap x di dang tong hu han (1.8). Ham 0 thoa ieu

kien (1.14), cac ham i thoa ieu kien thuan nhat, i V .Neu dung phng phap Galerkin, i = i (i = 1, . . . , N ), th ta co bai

toan xap x

Ni=1

ci

10

i(x)j(x)dx =

10

f(x)j(x)dx+ hj(0) 10

0(x)j(x)dx (1.22)

vi j = 1, . . . , N .? Phng phap Ritze y rang, ham v trong phng trnh (1.17) la bien phan cua u, nen co

the bien oi nh sau.

10

u(x)(u)(x)dx 10

f(x)u(x)dx hu(0) = 0

[1

2

10

(u(x))2dx 10

f(x)u(x)dx hu(0)]

= 0

I(u) = 0, (1.23)

trong o

I(u) =1

2

10

(u(x))2dx 10

f(x)u(x)dx hu(0) (1.24)

la phiem ham nang lng (energy functional) lien ket vi bai toan.T phng trnh (1.23) ta co the ket luan: nghiem u cua bai toan bien

(2.118)-(1.15) la iem cc tr cua phiem ham I . Phng phap Ritz c ap dung tay.

10 CHNG 1. GII THIEU

Nhan xet 1.3. Cach trnh bay cong thc bien phan trong th du 1.2 la theoly thuyet bien phan trong giai tch ham. Theo ly thuyet nay, bai toan bienphan tuyen tnh tru tng c nh ngha nh sau.

Cho V la khong gian Hilber, cho a : V V R la dang song tuyentnh tren V , ` : V R la dang tuyen tnh tren V . Bai toan: Tm u V thoa

a(u, v) = `(v) (1.25)

vi moi v V c goi la bai toan bien phan tuyen tnh.ieu kien ton tai nghiem cua bai toan bien phan c cho trong nh

ly Lax - Milgram:

nh ly 1.1 (Lax - Milgram). Cho V la khong gian Hilbert vi tch vo hng (, )va chuan tng ng . Neu

(i) a : V V R la dang song tuyen tnh lien tuc tren V , khang t, ngha laton tai hang so > 0 sao cho

a(v, v) v2

vi moi v V .(ii) ` : V R la dang tuyen tnh lien tuc tren V .

Th bai toan bien phan (1.25) ton tai va duy nhat nghiem.

Chng minh nh ly Lax - Milgram da tren nh ly bieu dien Riesz.

nh ly 1.2 (Riesz). Cho V la khong gian Hilbert vi tch vo hng (, ). Neu` : V R la dang tuyen tnh lien tuc tren V , th ton tai duy nhat vect w V saocho:

`(v) = (w, v), v V.Chng minh nh ly Lax - Milgram. Do a(, ) la dang song tuyen tnh lien tucva khang t tren V nen co the xem no la tch vo hng (mi) xac nh trenV vi chuan tng ng:

v =a(v, v).

V dang song tuyen tnh a(, ) lien tuc nen ton tai M > 0 sao cho vi moiv V ,

v2 a(v, v) Mv2 v v Mv,

ngha la chuan tng ng vi chuan tren V .T gia thiet `() lien tuc tren V vi chuan ta suy ra no cung lien

tuc tren V vi chuan mi. Ap dung nh ly Riesz, ton tai u V sao cho`(v) = a(u, v), v V ;

1.4. AP DUNG CUA PHNG PHAP PTHH 11

ngha la bai toan bien phan (1.25) co nghiem.Neu bai toan (1.25) co hai nghiem u1, u2 th

a(u1 u2, v) = 0 u1 u2 = 0 u1 u2 = 0;ngha la u1 = u2. Nh vay, nghiem cua bai toan (1.25) la duy nhat.

Trong trng hp a oi xng th bai toan bien phan tng ng vibai toan toi u, oi tng cua phng phap Ritz: Tm u V cc tieu hoaphiem ham

I(v) =1

2a(v, v) l(v). (1.26)

1.4 Ap dung cua phng phap PTHH

1.4.1 Tong quan

Qua trnh giai quyet mot bai toan trong khoa hoc va ky thuat bang phngphap PTHH c the hien bang s o mo ta tren hnh 1.1 gom hai giai oan.

Hnh 1.1: S o giai mot bai toan khoa hoc va ky thuat.

? Giai oan 1 -- Giai quyet -- Bai toan vat ly c toan hoc hoa bangmo hnh toan hoc. Mo hnh toan hoc, sau o, c ri rac hoa va giai bangphng phap PTHH.

? Giai oan 2 -- anh gia -- Kiem tra o tin cay cua ket qua PTHH.Cac loai sai so: (1) sai so mo hnh (mo hnh toan hoc, mo hnh PTHH); (2) sai


so thuat toan; (3) sai so lam tron (bieu dien so trong may tnh). Hieu chnh(neu can).

Trong c hoc moi trng lien tuc phng phap PTHH co the c dungcho nhieu loai phan tch:

-- Phan tch tuyen tnh, tnh.-- Phan tch tuyen tnh, ong.-- Phan tch tr rieng ve cong hng.-- Phan tch bien dang ln.-- Mo phong phuc vu thiet ke.oi tng cua mon hoc: Ap dung phng phap PTHH cho cac bai toan

bien (phng trnh ao ham rieng), ac biet, cac bai toan xuat hien trong choc, tap trung vao giai oan 1 -- mo hnh PTHH va giai so.

1.4.2 Mot so bai toan c hoc va vat ly

Mot so bai toan c hoc va vat ly se c dung en trong giao trnh.1. Truyen nhiet (1-chieu)Phng trnh truyen nhiet trong thanh chieu dai L, tiet dien A = A(x)

x

(kA

T

x

)+ qA = c

T

t, 0 < x < L, (1.27)

trong o T = T (x, t) la nhiet o (tuyet oi), q toc o phat sinh nhiet trenn v the tch (do nguon nhiet), c la nhiet dung, la mat o khoi.

Cac trng hp ac biet:a) Neu q = 0 ta co phng trnh Fourier

x

(kA

T

x

)= c

T

t. (1.28)

b) Neu qua trnh dng ta co phng trnh Poisson

x

(kA

T

x

)+ qA = 0. (1.29)

1.4. AP DUNG CUA PHNG PHAP PTHH 13

c) Neu qua trnh dng va q = 0 ta co phng trnh Laplace.

x

(kA

T

x

)= 0. (1.30)

2. Dong chay 1-chieu

d

dx(Au) = 0, (1.31)

trong o la mat o khoi, A la dien tch tiet dien ngang, u la van toc dongchay.

Neu dong chay khong nht th ton tai ham (x), goi la the van toc, saocho

u =d

dx(1.32)

va phng trnh (1.31) thanh

d

dx

(A

d

dx

)= 0. (1.33)

3. Thanh chieu dai L chu tai doc truc

AEu

x= lc tac dung, 0 < x < L, (1.34)

trong o E la moun Young, A la dien tch tiet dien ngang, u la chuyen dchdoc truc.

Neu lc tac dung la hang th phng trnh thanh

x

(AE

u

x

)= 0. (1.35)


ieu kien bien cua cac bai toan tren thuoc mot trong cac loai: Dirichlet(cho gia tr ham can tm), Neumann (cho gia tr ao ham cua ham can tm),va Robin (to hp ca hai ieu kien Dirichlet va Neumann). Ngoai ra, vi cacbai toan phu thuoc thi gian can phai cho ca ieu kien au.

1.5 Cac van e toan hoc lien quan

Trc khi i vao ly thuyet cua phng phap PTHH ta xet mot th du tongket cac van e a trnh bay trong cac muc tren, ong thi neu bat cac vane ly thuyet can xet khi ap dung phng phap PTHH cho bai toan khoa hocky thuat. Cac van e c trnh bay bang ngon ng giai tch ham vi mucch gi y cac nghien cu sau hn ve ly thuyet PTHH.

. Bai toan bien

d2u

dx2= f trong (0, 1) (1.36)

u(0) = 0, u(1) = 0. (1.37)

. Bai toan bien phan va s ton tai nghiem yeuNeu u la nghiem va v la ham bat ky (u trn) sao cho v(0) = 0, th tch

phan tng phan cho

a(u, v) = (f, v), (1.38)

trong o

a(u, v) :=

10

u(x)v(x)dx, (1.39)

(f, v) :=

10

f(x)v(x)dx. (1.40)

a vao khong gian ham

V = {v H1(0, 1) : v(0) = 0}, (1.41)trong o H1(0, 1) la khong gian Sobolev cac ham cung vi ao ham cap 1(ao ham suy rong) bnh phng kha tch (thuoc L2(0, 1)). Th ta co the noi

1.5. CAC VAN E TOAN HOC LIEN QUAN 15

nghiem u cua (1.36)-(1.37) c ac trng bi

u V sao cho a(u, v) = (f, v) v V, (1.42)

phat bieu nay c goi cong thc bien phan hay cong thc yeu cua bai toan(1.36)-(1.37). Nghiem cua no goi la nghiem yeu (weak solution) cua bai toan cho.S ton tai va duy nhat nghiem yeu cua bai toan ang xet c suy ra ngaynh nh ly Lax-Milgram.

Mot van e quan trong thng c xet: nghiem cua (1.42) co la nghiemcua bai toan (1.36)-(1.37) hay khong? Van e nay lien quan en tnh chnh quycua nghiem yeu. Vi bai toan ang xet ta co ket qua sau.

nh ly 1.3. Gia s f C0([0, 1]) va u C2([0, 1]) thoa (1.42). Th u thoa(1.36)-(1.37).

Chng minh. Cho v V C1([0, 1]). Th cong thc tch phan tng phan cho

(f, v) = a(u, v) =

10

(u)vdx+ u(1)v(1). (1.43)

Vay, (f (u), v) = 0 vi moi v V C1([0, 1]) sao cho v(1) = 0. atw = f + u C0([0, 1]). Neu w 6 0 th w co mot dau xac nh trong khoang[x0, x1] [0, 1] nao o. Chon v = (x x0)2(x x1)2 trong [x0, x1] va v 0ben ngoai khoang [x0, x1]. Th (w, v) 6= 0 mau thuan. Vay u = f . Bay giap dung (1.43) vi v = x ta co ngay u(1) = 0. Tat nhien, u V nen u(0) = 0,vay u thoa (1.36)-(1.37).

. Bai toan xap x Galerkin va s ton tai nghiem xap xCho S V la khong gian con hu han chieu bat ky. Ta xet bai toan

(1.42) vi V c thay bi S , cu the la

uS S sao cho a(uS, v) = (f, v) v S. (1.44)

nh ly 1.4. Cho trc f L2(0, 1), bai toan (1.5) co nghiem duy nhat.

Chng minh. Viet (1.5) theo cac ham c s {i : 1 i n} cua S . ChouS =

nj=1 Ujj , Kij = a(j, i), Fi = (f, i) vi i, j = 1, 2, . . . , n. at

U = [Uj], K = [Kij] va F = [Fi]. Th (1.5) tng ng vi phng trnh matran

KU = F. (1.45)


V cac ham c s i la oc lap tuyen tnh nen (1.45) duy nhat nghiem, suy raton tai nghiem va nh ly c chng minh.

. anh gia sai soTrc het ta xet he thc trc giao c ban gia u va uS . Tr (1.5) cho

(1.42) ta c

a(u uS, w) = 0 w S. (1.46)

Bay gi nh ngha chuan nang lng cua v V la

vE =a(v, v).

ay la chuan tng ng vi chuan trong H1(0, 1). nh ly di ay ch rarang vi chuan nang lng sai so la toi u.

nh ly 1.5.

u uSE = min{u vE : v S}. (1.47)

Chng minh. Vi v S bat ky

u uS2E = a(u uS, u uS)= a(u uS, u v) + a(u uS, v uS)= a(u uS, u v) (do w = v uS S va (1.46)) u uSEu vE.

Neu u uSE 6= 0 th u uSE u vE . Neu u usE = 0 bat angthc la tam thng. Lay inf tren v S ta co

u uSE infSu vE.

V uS S nen ta cung co

u uSE infSu vE u uSE.

Vay nh ly c chng minh.Ve sau, thng ta phai anh gia sai so tren c s xap x PTHH c

dung (cach chon S). e minh hoa, ta xet sai so theo chuan trong L2(0, 1).

1.5. CAC VAN E TOAN HOC LIEN QUAN 17

Muon anh gia u uS ta dung cach chng minh "oi ngau". Cho w langhiem cua

w = u uS tren [0, 1] vi w(0) = w(1) = 0.Nh tch phan tng phan, ta c

u uS2 = (u uS, u uS)= (u uS,w)= a(u uS, w) (v (u uS)(0) = w(1) = 0)= a(u uS, w v) (do (1.46)

vi moi v S . T bat ang thc Schwarz cho chuan nang lng ta suy rau uS u uSEw vE/u uS

u uSEw vE/w.Bay gi lay inf tren v S ta co

u uS u uSE infvV

w vE/w.

Vay, ta thay chuan L2(0, 1) cua sai so co the nho hn chuan nang lng ratnhieu, mien la w co the xap x tot bi mot ham nao o trong S . Co ly egia s rang ta co the lay v S gan w, ngha la gia s

infvV

w vE w (1.48)

vi be. Ap dung (1.48) ta c

u uS u uSE .

Ap dung (1.48) mot lan na vi w thay bang u, va dung nh ly 1.5 ta c

u uSE u.To hp cac anh gia tren ta c

nh ly 1.6.

u uS u uSE 2u = 2f. (1.49)

Ta thay u uSE cap trong khi u uS cap 2.


1.6 Khong gian Sobolev

e trnh bay phng phap PTHH ta can en cac khong gian ham Sobolev.

1.6.1 Khong gian Lebesgue

Cho tap m Rd co bien trn tng manh. Theo ly thuyet tch phanLebesgue mot menh e phu thuoc x c goi la "ung hau het" (h.h.)tren neu no ung vi moi x ngoai tr cac x thuoc mot tap co o o khong(khong ang ke). Ta ong nhat hai ham bang nhau h.h.

Vi 1 q , ta nh ngha khong gian Lebesgue, Lq(), la tap hpcac ham thc v xac nh tren sao cho vLq()

1.6. KHONG GIAN SOBOLEV 19

Ky hieu D() hay C0 () la tap hp cac ham co gia compact trong .e nh ngha ao ham yeu, ta a vao khong gian ham:

L1loc() = {v : v L1(K) vi moi compact K trong }.Chu y rang L1loc() cha tat ca cac ham thuoc lp C

0().Ky hieu di ay c dung e ch mot ao ham rieng cua ham v

Dv = ||v

x11 x22 xdd

,

trong o = (1, 2, . . . , d) la mot a ch so, vi 1, 2, . . . , d la cac songuyen khong am, va || = 1 + 2 + + d la o dai cua .

Ham v L1loc() c goi la co ao ham suy rong, Dwv, neu ton taiham u L1loc() sao cho

u()(x)dx = (1)||

v(x)D(x)dx D().

Neu u ton tai ta viet Dwv = u.Neu v C ||(), th ao ham suy rong Dwv ton tai va bang Dv.

1.6.3 Khong gian Sobolev

Vi r = 1, 2, . . . va v L1loc(), gia s cac ao ham suy rong Dv ton tai vimoi || r. Ta nh ngha chuan Sobolev

vW r,q() =||r

DvqLq()

1/q

khi 1 q


W r,q() thoa nh ngha chuan. Hn na, W r,q() la khong gian Banachvi chuan nay.

Ky hieu W r,q0 () la ay u hoa cua D() oi vi chuan W r,q().Vi co bien trn va v W 1,q(), han che tren bien , v|, co the

c bieu dien nh la ham trong Lq(), 1 q . ieu nay khong khangnh gia tr tng iem cua v tren co ngha. T tnh chat nay ta co

W r,q0 () = {v W r,q() : Dv| = 0 trong L2(), || < r}.

Trong cac phat bieu ve sau ta con dung cac na chuan:

|v|W r,q() =||=r

DvqLq()

1/q , 1 q 0,phu thuoc ch vao , sao cho

vL2() C|v|H1(). (1.54)

Neu b chan, bat ang thc nay am ch na chuan | |H1() tng ngvi chuan H1(). Tong quat, Hr() tng ng vi chuan H1().

Bai tap chng 1 21

Xet khong gian Lq(), 1 q

22 Bai tap chng 1

vi cac ieu kien bienu(0) = u(1) = 0

a) Tm phiem ham nang lng lien ket vi bai toan.b) Giai bai toan bang phng phap Galerkin hoac Ritz vi

0(x) = 0, 1(x) = sin(pix), 2(x) = sin(2pix), 3(x) = sin(3pix).

HD: S = V = {v H1(0, 1), v(0) = v(1) = 0}. Bai toan bien phan:

10

[uv + uv + xv]dx = 0

vi moi v V .a) e tm phiem ham nang lng lien ket vi bai toan, thay v bang ky

hieu bien phan u cua u.b) Nghiem xap x dang:

u3 = c11 + c22 + c33

= c1 sin(pix) + c2 sin(2pix) + c3 sin(3pix).

Thay vao phng trnh bien phan ta c he:

c1(pi3 pi) 2 = 0c2(4pi3 pi) + 1 = 0c3(27pi3 3pi) 2 = 0

Giai:

c1 =2

pi(pi2 1) , c2 = 1

pi(4pi2 1) , c3 =2

3pi(9pi2 1) .

Vay,

u3 =2 sin(pix)

pi(pi2 1) sin(2pix)

pi(4pi2 1) +2 sin(3pix)

3pi(9pi2 1) .

So sanh ket qua vi nghiem chnh xac ucx = sinxsin 1 x, hnh 1.2.

Bai tap chng 1 23

Hnh 1.2: Nghiem chnh xac va nghiem xap x.

1.3. Cho phiem ham I(u) nh bi

I(u) =

ba

F (x, u, ux, uxx)dx,

trong o u = u(x), x [a, b], cac ky hieu ux, uxx e ch ao ham cap mot vahai cua u theo bien x.

a) Tnh I .b) Tm ieu kien can e I at cc tr.

S:

a) I = ba

(F

uu+

F

uxux +

F

uxxuxx

)dx.

b) I = 0. Bien oi I ,

I =

ba

[F

u ddx

(F

ux

)+

d2

dx2

(F

uxx

)]udx

+

[F

uxu d

dx

(F

uxx

)u

]ba

+

[(F

uxx

)ux

]ba

.

24 Bai tap chng 1

Do bien phan u tuy y, t ieu kien I = 0 suy ra

F

u ddx

(F

ux

)+

d2

dx2

(F

uxx

)= 0, (1.55)

[F

uxu d

dx

(F

uxx

)u

]ba

= 0, (1.56)

[(F

uxx

)ux

]ba

= 0. (1.57)

Phng trnh (1.55) c goi la phng trnh Euler hay phng trnh Euler-Lagrange; cac phng trnh (1.56), (1.57) cho ieu kien bien.

1.4. Xet bai toan bien

L[u] := ddx

(p(x)

du

dx

)+ q(x)u = f(x), 0 < x < 1, (1.58)

u(0) = u(1) = 0. (1.59)

a) Dung phng phap d so co trong tm nghiem xap x vi

j(x) = (x) = sin jpix, j = 1, 2, . . . , N.

b) Chng to rang: neu dung (1.10) vi

j(x) = (x xj), j = 1, 2, . . . , N,

trong o (x) la ham delta Dirac thoa

(x) = 0, x 6= 0;

(x)dx = 1,

va0 < x1 < x2 < . . . < xN < 1.

th ta c ket qua cua phng phap ong v; ngha la, nghiem thoa phngtrnh vi phan (1.58) tai cac iem xj .

Bai tap chng 1 25

Hnh 1.3: Bai tap 1.5.

1.5. o vong cua dam chieu dai L, chu uon di tac dung cua tai trong phanbo eu p, thoa phng trnh

EId4w

dx4 p = 0, (1.60)

trong o E la moun Young, I la momen quan tnh cua dam.Bang phng phap Galerkin, tm o vong cua dam vi cac ieu kien

bien (ngam):

w(0) = w(L) = 0, (1.61)

EIdw

dx(0) = EI

dw

dx(L) = 0. (1.62)

26 Bai tap chng 1

Chng 2

Ly thuyet c ban

Luon luon co cach bieu dien tot hn!Bat chc roi sang tao.

ay la chng dai nhat trnh bay: cac y tng c ban cua phng phapPTHH, cach ap dung PTHH giai bai toan khoa hoc ky thuat, con goi la mohnh hoa PTHH. Vi nh hng ng dung PTHH giai cac bai toan khoa hocky thuat, cach thc mo hnh hoa va ca lap trnh tnh toan PTHH c trnhbay cho mot bai toan c hoc ket cau n gian - bai toan keo thanh doc truc.Do cach trnh bay van e co ong, dung ngon ng ma tran, nen cong vieccua ngi oc kha nang ne. Hay oc ky cho thau ao va kiem tra can thancac bc bang giay but.

2.1 PTHH nh la phng phap xap x ham

Y tng c ban cua PTHH la xap x ham. ac iem cua phep xap x nay lagia tr xap x cua ham tai mot iem ch phu thuoc vao gia tr (cho trc) cuaham tai vai iem "kha gan" no.

Cho ham f(x) xac nh tren mien Rd (d = 1, 2, 3).

2.1.1 Phep phan hoach PTHH

Xap x ham bat au bang s phan hoach thanh cac mien con (e), goi laphan t hu han (finite element), co hnh hoc n gian. Cac phan t hu han

27

28 CHNG 2. LY THUYET C BAN

c xac nh nh cac iem, goi la nut (node) hnh hoc. Noi chung, phep phanhoach phai thoa hai ieu kien:

1) Hoi cac phan t tao thanh mien h xap x mien ,

h =e

(e);

Hnh 2.1: Xap x PTHH mien .

2) Cac phan t hu han khac nhau khong "dam len" nhau. Gia haiphan t co chung nhau hn hai nut hnh hoc, th phan bien cua chung iqua cac iem chung nay phai trung nhau. Noi khac i, phep phan hoachPTHH khong lam xuat hien cac "lo hong".

Hnh 2.2: Tnh chat cua phep phan hoach.

2.1.2 Cac loai phan t hu han

Tuy theo th nguyen d cua mien va so iem nut (cua phan t) ta co cacloai phan t hu han khac nhau. Viec chon loai phan t gop phan ch nhcach xap x ham trong phan t ay.

1) Phan t 1-chieu la cac oan thang hay cong (hnh 2.3)

2.1. PTHH NH LA PHNG PHAP XAP X HAM 29

Hnh 2.3: Phan t hu han 1-chieu.


2) Phan t 2-chieu la cac tam giac hoac t giac vi cac canh la ngcong bac nhat, bac hai hay bac ba (hnh 2.4).

3) Phan t 3-chieu la cac khoi t dien, luc dien (hnh 2.5).

2.1.3 Phan t tham chieu

Mot ac iem quan trong cua xap x phan t hu han la tnh a phng cuaphep xap x. Xap x cua ham tren moi phan t ch phu thuoc vao cac gia trcua ham tai cac nut, goi la nut noi suy, nam trong phan t ay. V vay ta co thexay dng phep xap x tren mot phan t ac thu (cung loai, n gian) r goila phan t tham chieu. Xap x cua ham tren mot phan t thc (e) bat ky co



the nhan c nh phep bien oi (song anh) (e) bien phan t tham chieuthanh phan t thc.

Th du 2.1. Phan t tham chieu 1-chieu la oan r = [0, 1] tren truc toa o vi cac nut 1 = 0 va 2 = 1. Vi phan t thc (e) tren truc toa o x vicac nut tng ng xi, xj , phep bien oi (e) : 7 x, bien r thanh (e), laanh xa affine

(e)() = a + b,

trong o cac hang so a, b c xac nh bi cac tng ng:

{ (e)(1) = xi, (e)(2) = xj

{a = xj xi,b = xi

Nh vay, phep tng ng gia iem r vi iem x (e):

x = (1 )xi + xj (0 1). (2.1)

Nhan xet 2.1. Trong hnh hoc affine, iem x nam tren oan thang noi haiiem xi, xj c bieu dien di dang mot to hp tuyen tnh cua hai iem nay

x = 1xi + 2xj


Hnh 2.6: Phep bien oi (e).

vi 1, 2 0 va 1 + 2 = 1. Khi o, (1, 2) goi la toa o trong tam(barycentric coordinates) cua x oi vi hai iem xi, xj .

So sanh vi cong thc bien oi (2.1), ta co:

1 = 1 , 2 = .

Hai ham nay xac nh phep bien oi (e) : [0, 1] [xi, xj].Nhan xet 2.2. Cac ham 1, 2 la cac a thc bac nhat theo , trong o cache so c xac nh t ieu kien Kronecker i(k) = ik.Th du 2.2. Phan t tham chieu 2-chieu tam giac la tam giac r vi cac nh1= (0, 0),

2= (1, 0),

3= (0, 1) trong mat phang toa o . Phan t tam

giac bat ky (e) vi cac nut xi = (xi, yi), xj = (xj, yj), xk = (xk, yk) trong matphang toa o xy c bien oi t phan t tham chieu nh phep bien oi

x = A +B, (2.2)

trong o

A =

[a11 a12a21 a22

], B =

[b1b2

](2.3)


la ma tran va vect hang c xac nh bi cac he thc:

xi = A1 +B,

xj = A2 +B,

xk = A3 +B,

Giai ra, ta c

A =

[xj xi xk xiyj yi yk yi

], B =

[xiyi

]. (2.4)

Nh vay, s tng ng gia iem va iem x:

[xy

]= (1 )

[xiyi

]+

[xjyj

]+

[xkyk

](2.5)

hay

x = (1 )xi + xj + xk. (2.6)

Cac ham

1 = 1 , 2 = , 3 = . (2.7)

la cac toa o trong tam cua x oi vi ba iem x i, xj , xk.

Hnh 2.7: Phan t tham chieu tam giac 3 nut.


Tong quat, trong Rd PTHH n gian nhat la d-n hnh - bao loi cuad + 1 iem xi (i=1,2,. . . ,d+1) oc lap tuyen tnh trong Rd . d-n hnh thamchieu co cac nh:

1

= (0, 0, . . . , 0),

2

= (1, 0, . . . , 0),

3

= (0, 1, . . . , 0),

d+1

= (0, 0, . . . , 1).

Ky hieu:x = (x1, x2, . . . , xd) la iem thuoc phan t thc; = [1, 2, . . . , d]

T la iem thuoc phan t tham chieu.

xi = [xi,1, xi,2, . . . , xi,d]T la cac nh cua phan t thc.Phep bien oi Affine gia phan t tham chieu va phan t thc:

x = A +B. (2.8)

T ieu kien Ai+B = xi ta co ngay

B = x1 =

x1,1x1,2...

x1,d

, A(:, i) = xi+1 x1 =

xi+1,1 x1,1xi+1,2 x1,2

...xi+1,d x1,d

, (i = 1, . . . , d).

Phep bien oi:

x =

(1

dk=1

k

)x1 + 1x2 + 2x3 + + dxd+1

= 1x1 + 2x2 + 3x3 + + d+1xd,trong o i la toa o trong tam cua x:

1 = 1d

k=1

k, 2 = 1, 3 = 2, . . . , d+1 = d.


Th du 2.3. Phan t tham chieu 2-chieu t giac. Xet phan t t giac bat ky(e) vi cac nut xi = (xi, yi), xj = (xj, yj), xk = (xk, yk), xl = (xl, yl) trongmat phang toa o xy, a vao he toa a phng co cac ng toa o , lahai ho ng thang phan eu bon canh cua t giac (hnh 2.8a)). Lay iemtrung tam cua hai ho nay lam goc bang cach at = = 0, ve cac truc , theo hng tang cua ,, ong thi lay gia tr tren bon canh la 1, th ta thuc hnh vuong r vi cac nh:

1= (1,1),

2= (1,1),

3= (1, 1),

4= (1, 1)

trong he toa o (hnh 2.8b)). Hnh vuong (e) la phan t tham chieu2-chieu t giac.

Hnh 2.8: Phan t hu han t giac.

e xay dng phep bien oi t phan t tham chieu r sang phan t thc(e) ta can en mot so khai niem cua ly thuyet noi suy sieu han.

nh ngha 2.1. Cho L la khong gian vect va L la khong gian con ong cuano. Mot anh xa tuyen tnh P : L L co tnh chat luy ang (idempotent),ngha la

P P = P,c goi la phep chieu t L len L.

Cho L la khong gian cac ham (hai bien) lien tuc xac nh tren r , vaL la khong gian cac ham lien tuc tren r , sao cho ao ham theo ton tai vala hang so. Xet phep chieu P : L L xac nh bi

P(f) =1 2

f(1, ) + + 12

f(1, ). (2.9)


Ta thay, P(f) gi nguyen gia tr cua ham f doc theo cac canh thang ng = 1 cua phan t tham chieu, va noi suy tuyen tnh gia tr cua no gia = 1 va = 1 doc theo cac canh nam ngang = const.

Tng t cho cac ng nam ngang = 1

P(f) =1 2

f(,1) + + 12

f(, 1). (2.10)

Tch cua hai phep chieu P va P ,

P P(f) = (1 )(1 )4

f(1,1) + (1 + )(1 )4

f(1,1)

+(1 + )(1 + )

4f(1, 1) +

(1 )(1 + )4

f(1, 1)

=(1 )(1 )

4f(

1) +

(1 + )(1 )4

f(2)

+(1 + )(1 + )

4f(

3) +

(1 )(1 + )4

f(4), (2.11)

cung la mot phep chieu. Bang cach kiem trc tiep, dung (2.9) - (2.11), co thech ra rang

P(f) + P(f) P P(f) =

=1 2

f(1, ) + + 12

f(1, )

+1 2

f(,1) + + 12

f(, 1)

(1 )(1 )4

f(1) (1 + )(1 )

4f(

2)

(1 + )(1 + )4

f(3) (1 )(1 + )

4f(

4) (2.12)

noi suy ham f chnh xac tren ca bon canh cua hnh vuong r .


Bay gi, ta xay dng phep bien oi phan t tham chieu r thanh phant thc (e) , {

x = x(, ),y = y(, ).

e y rang cac iem (1, ), 1 1 (tren canh 23), tng ng vi cac

iem (x, y) tren canh x2x3; do o,

x(1, ) =x2 + x3

2+

x3 x22

. (2.13)

Tng t vi cac canh con lai, ta co:

x(1, ) = x1 + x42

+ x4 x1

2, (2.14)

x(, 1) =x4 + x3

2+

x3 x42

, (2.15)

x(,1) = x2 + x12

+ x2 x1

2. (2.16)

Dung cong thc (2.12) vi f thay bang x, sau mot so bien oi, ta thu c:

x = x1(1 )(1 )

4+ x2

(1 + )(1 )4

+x3(1 + )(1 + )

4+ x4

(1 )(1 + )4

. (2.17)

Do tnh oi xng, bang cach thay x, x1, x2, x3, x4, tng ng, bang y, y1, y2, y3, y4trong cong thc tren ta thu c cong thc bien oi cho thanh phan tung


o. at:

N1(, ) =(1 )(1 )

4, (2.18)

N2(, ) =(1 + )(1 )

4, (2.19)

N3(, ) =(1 + )(1 + )

4, (2.20)

N4(, ) =(1 )(1 + )

4. (2.21)

Khi o cong thc bien oi co the viet di dang vect:

x =4

i=1

xiNi(). (2.22)

Nhan xet 2.3. Phep chieu (2.9) co the de dang c tong quat hoa thanh noisuy ham f chnh xac doc theo m+ 1 ng thang ng = i,

1 = 0 < 1 < . . . < m = 1,

Pv(f) =mi=0

f(i, )vi (), (2.23)

trong o

vi () =j 6=i

ji j ,

i = 0, 1, . . . , m la nhng ham noi suy Lagrange c s. Trong khuon kho lythuyet noi suy sieu han (transfinite interpolation), cac ham vi c goi la cacham uon (blending function). V ca hai ham so P(f) va Pv(f) trung vi f tai


mot so khong em c cac iem nen chung co ten goi la noi suy sieu han,toan t tng ng c goi la toan t noi suy sieu han.

Cong thc tng t (2.23) cho cac ng nam ngang = j ,

1 = 0 < 1 < . . . < n = 1,

Ph(f) =n

j=0

f(, j)hj (), (2.24)

trong o

hj () =j 6=i

ij i , = 0, 1, . . . , n.

Tch cua hai phep chieu Pv va Ph:

Ph Pv(f) =mi=0

nj=0

f(i, j)vi ()

hj (). (2.25)

Toan t tch Ph Pv cung la mot phep chieu, va Ph Pv(f) noi suy ham fchnh xac tai (m+1)(n+1) iem (i, j), i = 0, 1, . . . , m, j = 0, 1, . . . , n. HamPh Pv(f) c goi la noi suy Lagrange lng a thc cua ham f tai cac iem(i, j).

Phep chieu Ph Pv khong la noi suy sieu han. Ta co mot cach to hphai toan t Pv va Ph e co c mot toan t noi suy sieu han ma tap chnhxac cha toan bo cac ng = i, = j . o la tong Boolean

Pv Ph = Pv + Pv Pv Ph. (2.26)nh ly 2.1. Cho cac toan t Pv va Ph. Th (Pv Ph)(f) noi suy ham f chnhxac doc theo cac ng = i, = j , i = 0, 1, . . . , m, j = 0, 1, . . . , n.

Ham (Pv Ph)(f) c goi la noi suy Lagrange hai chieu sieu han.Nhan xet 2.4. Ta co the tm cong thc bien oi di dang mot a thc bacnhat theo tng bien:

x = a0 + a1 + a2 + a3.


Vi cach chon nay th khi = const (hay = const) a thc tr thanh bacnhat theo y (hay x), tng ng vi ng thang.

2.1.4 Xap x ham tren phan t hu han

Phep xap x ham bang phan t hu han s dung cac ham noi suy Lagrange.Bac toi a cua a thc xap x qui nh loai phan t hu han va so nut noi suy 1)can thiet.

Th du 2.4. Trng hp 1-chieu, a thc bac nhat. Cac a thc p1(x) = 1,p2(x) = x lap thanh c s cua khong gian cac a thc bac 1. Ham xap xcua u = u(x) co the bieu dien di dang

uh(x) = [p1(x) p2(x)]

{a1a2

}= [p(x)]{a}. (2.27)

ay ta s dung cac ky hieu thu gon:

[p(x)] = [p1(x) p2(x)], {a} = {a1 a2}T .

T yeu cau ham xap x va ham chnh xac co gia tr bang nhau tai cac nutnoi suy xi, xj , ta co:

[p(xi)]{a} = ui, [p(xj)]{a} = ujhay di dang ma tran

[p1(xi) p2(xi)p1(xj) p2(xj)

]{a1a2

}=

{uiuj

}[1 xi1 xj

]{a1a2

}=

{uiuj

}.

1)Phan biet vi nut hnh hoc.


Giai ra ta c

a1 =uixj ujxixj xi ,

a2 =uj uixj xi .

Suy ra

uh(x) = uixj xxj xi + uj

x xixj xi .

a vao cac ky hieu:

Ni(x) =xj xxj xi , Nj(x) =

x xixj xi ,

ta co the viet lai

uh(x) = uiNi(x) + ujNj(x) = [N ]{u}.Cac ham Ni(x), Nj(x) c goi la cac ham dang (shape function), [N ] la matran ham dang.

Nhan xet 2.5. Cac ket qua tren co the tm c khi i t phan t thamchieu. Ham xap x:

uh() = [1 ]

{a1a2

}.

Chu y, dau ganh tren au ham xap x e ch no la ham xac nh tren phant tham chieu. Cac ham u(), uh() lien he vi u(x), uh(x) qua phep oi bien(th du 2.1),

u(x()) = u(), uh(x()) = uh().

Thc hien cac bc tnh toan tng t, ta c:

uh() = u1N1() + u2N2() = [N ]{u},trong o

N1() = 1 , N2() = .


Cac ham dang trung vi cac ham toa o trong tam cua x. Nhng phant hu han vi phep xap x co tnh chat nay c goi la phan t hu hanang tham so (isoparametric).

Tong quat -- xap x bang a thc bac N . Cac a thc c s cua khonggian cac a thc bac N : p1(x) = 1, p2(x) = x, . . . , pN+1(x) = xN . exay dng xap x ta can N + 1 nut noi suy. Trong trng hp nay ngoai hainut bien (nut hnh hoc) ta can a vao N 1 nut nam ben trong phant. e n gian, ta ky hieu x1, x2, . . . , xN+1 la cac nut noi suy cua phan t,u1, u2, . . . , uN+1 la gia tr cua ham u(x) tai cac nut tng ng. Da vao th du2.4 ta thiet lap cac ham dang nh sau.

Ham xap x

uh(x) = [p1(x) p2(x) . . . pN+1(x)]

a1a2...

aN+1

= [p(x)]{a}, (2.28)

trong o

[p(x)] = [p1(x) p2(x) . . . pN+1(x)], {a} = {a1 a2 . . . aN+1}T . (2.29)

T ieu kien ham xap x va ham chnh xac co gia tr bang nhau tai cacnut noi suy, ta co he phng trnh ai so tuyen tnh xac nh a1, . . . , aN+1

p1(x1) p2(x1) . . . pN+1(x1)p1(x2) p2(x2) . . . pN+1(x2)...

......

p1(xN+1) p2(xN+1) . . . pN+1(xN+1)

a1a2...

aN+1

=

u1u2...

uN+1

1 x1 . . . xN1

1 x2 . . . xN2

......

...1 xN . . . x

NN

a1a2...

aN+1

=

u1u2...

uN+1

P{a} = {u}, (2.30)


trong o

P =

1 x1 . . . xN1

1 x2 . . . xN2

......

...1 xN . . . x

NN

, {u} =

u1u2...

uN+1

. (2.31)

Giai phng trnh (2.30) ta c

{a} = P1{u},roi thay vao (2.28) ta thu c cac ham dang

uh = ([p(x)]P1){u} = [N(x)]{u} (2.32)

vi

[N(x)] = [N1(x) N2(x) . . . NN+1(x)] = [p(x)]P1 (2.33)

la ma tran cac ham dang.

Nhan xet 2.6. Do cac a thc p1(x), p2(x), . . . , pN+1(x) oc lap tuyen tnhnen cac ham dang N1(x), N2(x), . . . , NN+1(x) cung oc lap tuyen tnh. Theongon ng ai so tuyen tnh, cac gia tr nut u1, u2, . . . , uN+1 co the xem la toao cua vect uh(x) trong c s gom cac ham dang; con cac tham so a1, a2, . . . ,aN+1 la toa o cua uh(x) trong c s gom cac a thc p1(x), p2(x), . . . , pN+1.Theo o, ta goi u1, u2, . . . , uN+1 la cac tham so nut, con a1, a2, . . . , aN+1 la cactham so suy rong, xap x theo cong thc (2.28) goi la xap x phi nut.Th du 2.5. Trng hp 2-chieu, a thc bac nhat. Xap x ham u(x, y)bang a thc bac nhat th phan t thch hp la tam giac 3 nut. Cac a thcp1(x, y) = 1, p2(x, y) = x, p3(x, y) = y lap thanh c s cua khong gian cac athc hai bien co bac 1. Ky hieu:

[p(x, y)] = [p1(x, y) p2(x, y) p3(x, y)], {a} = {a1 a2 a3}T .Ham xap x uh(x, y) c bieu dien di dang:

uh(x, y) = [p(x, y)]{a}. (2.34)He phng trnh xac nh {a}

P{a} = {u}, (2.35)


trong o

P =

1 x1 y11 x2 y2

1 x3 y3

, {a} =

a1a2a3

, {u} =

u1u2u3

.

Giai he phng trnh (2.35), roi thay vao (2.34) ta thu c

uh(x, y) = [N(x, y)]{u}

trong o

N1(x, y) =x2y3 y2x3 + x(y3 + y2) y(x3 + x2)x2y3 y2x3 x1y3 + y1x3 + x1y2 y1x2 ,

N2(x, y) =x1y3 + y1x3 + x(y3 + y1) + y(x3 + x1)x2y3 y2x3 x1y3 + y1x3 + x1y2 y1x2 ,

N3(x, y) =x1y2 y1x2 + x(y2 + y1) y(x2 + x1)x2y3 y2x3 x1y3 + y1x3 + x1y2 y1x2 .

Nhan xet 2.7. Viec tm cac ham dang tuy n gian nhng rat de sai sot khitnh toan bang "tay chan". Ta co the dung Matlab e thc hien cac tnhtoan nay. Script file hd2D3N.m di ay tm cac ham dang trong th du tren

hd2D3N.m

% xay dung ham dang cho PTHH 2-D, 3 nut

clear all

syms x y x1 y1 x2 y2 x3 y3 a1 a2 a3

% da thuc co so

p=[1 x y];

% cac he so

a=[a1; a2; a3];

% da thuc xap xi

uh=p*a

syms u1 u2 u3

% ma tran gia tri da thuc co so tai cac nut

PP(1,:)=subs(p,x,y,x1,y1);




% gia tri nut

u=[u1;u2;u3];

% tinh cac he so

aa=inv(PP)*u;

% xac dinh cac ham dang

kq=simplify(subs(uh,a,aa))

N1=simple(maple(coeff,kq,u1))



Nhan xet 2.8. Phng phap xay dng ham dang trong cac th du tren co thetong quat hoa cho trng hp ham can xap x xac nh trong Rd . Goi M laso chieu cua khong gian cac a thc tren Rd co bac N , PN(Rd). e xaydng cac ham dang "ay u" ta can M nut noi suy. Ky hieu:

x = (x1, x2, . . . , xd) la iem trong Rd;{p1(x), p2(x), . . . , pM(x)} la mot c s cua PN(Rd);xi (i = 1, 2, . . . ,M ) la cac nut noi suy.S o thiet lap cac ham dang1. [p(x)] = [p1(x) p2(x) . . . pM (x)];

2. P =

p1(x1) p2(x1) . . . pM (x1)p1(x2) p2(x2) . . . pM (x2)...

......

p1(xM) p2(xM) . . . pM (xM)

;

3. [N ] = [N1(x) N2(x) . . . NM(x)] = [p(x)]P1.

2.1.5 ao ham va tch phan theo toa o tham chieu

Trong cac bai toan c hoc, vat ly, thng cac ham can tm va cac ao hamcua no co mat trong phng trnh xac nh. Neu dung ham xap x tren cacphan t thc (e) se rat phc tap (khong thuan tien). Bang cach a vao khainiem phan t tham chieu r va anh xa (e), ham va cac ao ham cua noc tnh toan tren phan t tham chieu. Di ay ta xet trng hp 2-chieu.


Goi uh la ham xap x cua u tren phan t tham chieu, ta co

uh(, ) = [N(, )]{u}.

T cong thc bien oi:

x = (e)() x = x(, ), y = y(, ) (2.36)va cong thc ao ham ham hp ta co:

=

x

x

y

y

T

x

y

=

{

}= JT

{

x

}, (2.37)

trong o J la ma tran Jacobi cua phep bien oi (e),

J =

x

x

y

y

.

T cong thc (2.37) ta suy ra

x

y

=

x

x

y

y

hay

{

x

}= JT

{

}. (2.38)

va ta co cong thc tnh ao ham cua ham u trong he toa o thc bang cacao ham cua u trong he toa o tham chieu:

u

xu

y

=

x

x

y

y

u

u

= JT

u

u

. (2.39)

Cong thc oi bien (2.36) cho phep ta chuyen tch phan cua mot hamu tren phan t thc (e) thanh tch phan tren phan t tham chieu r n


gian hn:

(e)

u(x, y)dxdy =

ru(, )|J|dd, (2.40)

trong o |J| ky hieu tr tuyet oi nh thc cua ma tran Jacobi.Th du 2.6. Cho phan t tam giac vi cac nh x1 = (1, 1), x2 = (3, 2),x3 = (2, 3) (trong toa o xy). Cho u = x+ y, v = xy la hai ham xac nh tren.

1) Thiet lap cong thc oi toa o, bien phan t tham chieu r , tam giacvi cac nh

1= (0, 0),

2= (1, 0),

3= (0, 1) (toa o ), thanh tam giac .

Xac nh J va J1.2) Dung u va v kiem tra lai cong thc (2.39).

3) Tnh

u vd.

Giai. 1) Nh trong th du 2.2, c bien oi t r nh phep bien oi

{x = 1 + 2 + ,y = 1 + + 2

Ma tran Jacobi va nh thc cua no:

J =[2 11 2

], det J = 3.

T ay suy ra

J1 =1

3

[2 1

1 2].

2) Bieu dien u va v theo toa o :

u = u(x(, ), y(, )) = 2 + 3( + ),

v = v(x(, ), y(, )) = 1 + 3( + ) + 5 + 2(2 + 2).


ao ham u va v trong toa o xy

u

x=u

y= 1,

v

x= y,

v

y= x.

ao ham u va v trong toa o

u

=u

= 3,

v

= 3 + 5 + 4,

v

= 3 + 5 + 4.

Kiem cong thc (2.39) trng hp ham u:Ve trai: {

11

}.

Ve phai:

1

3

[2 1

1 2]{

33

}=

{11

}.

Kiem cong thc (2.39) trng hp ham v:Ve trai: {

yx

}=

{1 + + 21 + 2 +

}.

Ve phai:

1

3

[2 1

1 2]{

3 + 5 + 43 + 5 + 4

}=

{1 + + 21 + 2 +

}.

3) T ket qua tren ta co

u v = (1, 1)(1 + + 2, 1 + 2 + )T = 2 + 3( + ).


Ap dung cong thc oi bien so trong tch phan (2.40)

u vd =r(2 + 3( + ))|3|dd

= 3

10

d

10

(2 + 3( + ))d

= 3

10

((2 + 3) +

3

2210

)d

= 3

10

(7

2 2 3

22)d =

21

2 32 3

2310

= 6.

2.2 Cac bc thiet lap mo hnh PTHH

Muc nay trnh bay cach thiet lap mo hnh PTHH cho cac bai toan khoa hocky thuat. Ta se tm hieu chi tiet cac bc thiet lap, bao gom c s ly thuyetva cach thc thc hien. Van e c trnh bay thong qua mot bai toan ngian cua c hoc ket cau nham lam ro ly do va y ngha cua cac bc thc hien.Nh se thay, sau o, cach ap dung PTHH cho cac bai toan bien cung tngt.

2.2.1 Phan tch ng suat cua thanh nhieu nac

Th du 2.7. Tm ng suat phan bo trong thanh nhieu nac oi xng truc,b ngam chat mot au, au con lai b keo bi lc doc truc p. Thanh cocac dien tch tiet dien ngang A(1) = 2cm2, A(2) = 1cm2 tren cac oan dail(1) = l(2) = 10cm. Cho moun Young E(1) = E(2) = E = 2 106kg/cm2,p = 1kg.

Cong thc bien phan2)

Theo ly thuyet thanh chu keo nen doc truc, bien dang cua thanh cxac nh bi chuyen dch q = q(x) tai cac iem thuoc no. Ta co cac he thcsau:

2)Ngi oc khong quen vi c hoc ket cau co the chap nhan cac cong thc c giithieu ay.

2.2. CAC BC THIET LAP MO HNH PTHH 49

Hnh 2.9: Thanh nhieu nac.

Bien dang (o gian tng oi)

=dq

dx. (2.41)

ng suat

= E = Edq

dx, (2.42)

trong o E la moun Young (nh luat Hooke).The nang toan phan cua thanh

I = Nang lng bien dang Cong cua lc ngoai= Wp, (2.43)

trong o

=

L0

A(x)

1

2(x)(x)dAdx, (2.44)

Wp = pq(L). (2.45)

Bai toan c giai da tren nguyen ly cong ao: chuyen dch thc la chuyendch lam cc tieu hoa the nang toan phan cua thanh.

Ky hieu:L = l(1) + l(2) la chieu dai thanh;S = {q|q H1(0, L), q(0) = 0} la tap hp cac chuyen dch kha d (ham

th);


V = {s|s H1(0, L), s(0) = 0} la khong gian cac chuyen dch ao (hamtrong lng). Do ieu kien bien cot yeu la thuan nhat nen V = S .

Dung cac cong thc (2.41), (2.42) cho bieu thc the nang (2.43)-(2.45) ngvi chuyen dch kha d q, ta c

I(q) =

L0

A(x)

E

2

(dq

dx

)2dAdx pq(L). (2.46)

Nhan xet 2.9. Ve phng dien toan hoc, bai toan phan tch ng suat cuathanh nhieu nac ay la bai toan toi u: tm ham q(x), 0 x L, lam cctieu phiem ham I(q) xac nh bi (2.46).

Ap dung PTHHBc 1 - Phan hoachThanh c xem nh mot ket cau gom 2 phan t: phan t th nhat la

phan thanh gii han bi cac tiet dien x1 = 0, x2 = 10, phan t th hai laphan thanh gii han bi cac tiet dien x2 = 10, x3 = 20 (ve phng dien toanhoc, oan [0, L] c phan hoach thanh cac oan con [x1, x2], [x2, x3]).

Hnh 2.10: Phan t thanh [xi, xi+1], lc tac dung len phan t.

Bc 2 - Xap x ham tren tng phan tChuyen dch (ham q(x)) tren moi phan t c xap x bi a thc bac

nhat (phan t e)

q(e) = q(e)1 N

(e)1 (x) + q

(e)2 N

(e)2 (x) = [N ]

(e){q}(e). (2.47)

trong o [N ](e) = [N (e)1 (x) N(e)2 (x)] la ma tran ham dang, [q](e) = {q(e)1 q(e)2 }T

goi la chuyen dch nut phan t3).

3)Ky hieu u(e) ch nham muc ch "nhan manh" no la thu hep cua ham u tren phan t e.


Nhan xet 2.10. Do ieu kien ngam cua chuyen dch (ieu kien bien cua bai

toan), q(1)1 = 0, nen ham N(1)1 (x) khong co mat trong bieu thc xap x cua

q. Tuy nhien, e thuan tien, trong thc hanh ta van tnh en q(1)1 trong qua

trnh thiet lap phng trnh xap x. Viec kh q(1)1 khoi phng trnh se cthc hien sau.

Bien dang va ng suat tren moi phan t c xap x bi

(e) =d

dx[N ](e){q}(e) = [B]e{q}(e), (2.48)

(e) = E(e)[B]e{q}(e), (2.49)

trong o [B]e = (d/dx)[N ](e) (ao ham cua mot ma tran la ma tran vi cacphan t la ao ham cua phan t tng ng trong ma tran au).

Bc 3 - Thiet lap phng trnh PTHHNeu xet rieng tng phan t th the nang cua phan t c tnh theo

cong thc (tng t nh the nang toan phan)

I (e)(q(e)1 , q

(e)2 ) =

E(e)A(e)

2

xe+1xe

(dq(e)

dx

)2dx p(e)2 q(e)2 + p(e)1 q(e)1 , (2.50)

trong o p(e)1 , p(e)2 la lc tac dung len cac au mut cua phan t. ay ta quy

c hng dng cua lc tac dung len phan t hng ra ngoai phan t (hnh2.10)

Do tnh chat cong tnh cua nang lng, cong thc the nang toan phanco the viet:

I =2

e=1

I (e) =2

e=1

[E(e)A(e)

2

xe+1xe

(dq(e)

dx

)2dx p(e)2 q(e)2 + p(e)1 q(e)1

].

a vao ky hieu {p}(e) = {pe1 pe2}T , goi la vect tai phan t, va dung cong


thc (2.47) ta co

I =2

e=1

[E(e)A(e)

2

xe+1xe

(d[N ](e)

dx{q}(e)

)2dx {q}(e)T{p}(e)

]. (2.51)

e y rang,

d[N ](e)

dx{q}(e) = [B](e){q}(e)

la vo hng nen ta co the viet

(d[N ](e)

dx{q}(e)

)2= {q}(e)T [B](e)T [B](e){q}(e).

Neu lay bien phan ta se c:

(d[N ](e)

dx{q}(e)

)2= {q}(e)T [B](e)T [B](e){q}(e) + {q}(e)T [B](e)T [B](e){q}(e)

= 2{q}(e)T [B](e)T [B](e){q}(e)

ang thc cuoi nhan c t nhan xet: cac vo hng {q}(e)T [B](e)T [B](e){q}(e),{q}(e)T [B](e)T [B](e){q}(e) la chuyen v cua nhau.

ieu kien cc tr cho phiem ham I :

I = 0

2e=1

{q}(e)T[(E(e)A(e)

xe+1xe

[B](e)T [B](e)dx

){q}(e) {p}(e)

]= 0

2e=1

{q}(e)T ([k](e){q}(e) {p}(e)) = 0, (2.52)trong o ma tran

[k](e) = E(e)A(e) xe+1xe

[B](e)T [B](e)dx, (2.53)


c goi la ma tran o cng phan t.a vao vect chuyen dch toan cuc, bien phan cua no va vect tai toan cuc

co thanh phan la chuyen dch, bien phan cua chuyen dch va noi lc tai cacnut cua toan bo thanh

{q} = {q1 q2 q3}T , {q} = {q1 q2 q3}T , {p} = {p1 p2 p3}T ,

ta co the viet

2e=1

{q}(e)T [k](e){q}(e) = {q}T [K]{q},2

e=1

{q}(e)T{p}(e) = {q}T{p},

va phng trnh (2.52) thanh

{q}T([K]{q}+ {p}) = 0, (2.54)

trong o [K] co dang

[K] =

K11 K12 K13K21 K22 K23K31 K32 K33

,

goi la ma tran o cng toan cuc. Tr nut th nhat do ieu kien rang buocq1 = 0, cac bien phan con lai q2, q3 la tuy y, bang cac lay cac bien phannay lan lt bang 1, ta suy ra:

0K11q1 +K12q2 +K13q3 = 0 p1K21q1 +K22q2 +K23q3 = p2

K31q1 +K32q2 +K33q3 = p3

Phng trnh th nhat tng ng vi 0 = 0. He phng trnh suy bien, canphai loai phng trnh nay khoi he. Ta se tr lai van e nay sau.

Nh vay, noi chung, v bien phan la tuy y, phng trnh (2.54) thanh

[K] {q} = {p}. (2.55)

Phng trnh (2.55) c goi la phng trnh PTHH. Do hai phan t kenhau co chung nhau mot iem nut nen phep lay tong trong cac phng trnh(2.54) phai c hien theo mot cach ac biet, goi la phep lap ghep.


Trc khi tien hanh cac bc tiep theo ta cu the hoa cac bc 1-3 quao gii thieu s o tnh toan PTHH.

Trc het, trong thc hanh, viec phan hoach bc 1 c thc hienbang cach a vao hai ma tran:

* Ma tran toa o cac nut COORD

COORD = [x1, x2, x3] = [0, 10, 20].

* Ma tran lap ghep LA mo ta s lien ket gia cac phan t. Ma tran naygom 2 dong (so phan t) va 2 cot (so nut hnh hoc cua phan t). Moi dongcua LA ng vi mot phan t, LA(e, 1) va LA(e, 2) la th t cua nut 1 va 2trong cach anh so toan cuc.

LA =[1 22 3

].

Bc th 2 va 3 c thc hien nh nhau cho moi phan t (ay la aciem cua PTHH). Thanh gom 3 nut nen chuyen dch xap x c xac nhbi vect chuyen dch toan cuc (toan bo thanh),

{q} =

q1q2q3

.

Trong khi o moi phan t ch co hai nut, nen vect chuyen dch phan t, thdu phan t e, co dang

{q}(e) ={q(e)1

q(e)2

}.

Nhan xet tng t vi vect tai toan cuc va vect tai phan t,

{p} =

p1p2p3

, {p}(e) =

{p(e)1

p(e)2

}.


Ma tran o cng phan t, c tnh nh sau.

[N ](e) =

[xe+1 xxe+1 xe ,

x xexe+1 xe

],

[B](e)

=1

xe+1 xe [1, 1] [B](e)T

[B](e)

=1

(xe+1 xe)2[

1 11 1

],

[k](e)

=E(e)A(e)

le

[1 1

1 1].

Nh a noi tren, e thiet lap phng trnh PTHH ta phai lay "tong"hay lap ghep cac phan t ca biet lai vi nhau. Phep lap ghep ma tran o cngva vect tai phan t c thc hien nh sau. Trc het khi tao ma tran ocng va vect tai toan cuc

[K] =

0 0 00 0 0

0 0 0

, [p] =

00

0

.

Phan t th nhat c ghep vao da tren lien he gia th t a phng vath t toan cuc theo ma tran lap ghep,

th t a phng 1 2th t toan cuc 1 2

k(1)11 7 K11 k(1)12 7 K12 p(1)1 7 p1k(1)21 7 K21 k(1)22 7 K22 p(1)2 7 p2

(mui ten 7 am ch s cong vao). Do o

[K] =

k(1)11 k(1)12 0k(1)21 k(1)22 0

0 0 0

, {p} =

p(1)1p(1)2

0

.

Tng t vi phan t th hai

th t a phng 1 2th t toan cuc 2 3

k(2)11 7 K22 k(2)12 7 K23 p(2)1 7 p2k(2)21 7 K32 k(2)22 7 K33 p(2)2 7 p3


[K] =

k

(1)11 k

(1)12 0

k(1)21 k

(1)22 + k

(2)11 k

(2)12

0 k(2)12 k

(2)22

, [p] =

p(1)1p(1)2 p(2)1p(2)2

.

Cho en bay gi gia tr cua tai at tai cac nut van cha c ban ti.That ra, theo d lieu cua bai toan, ta ch biet gia tr cua tai at tai nut th

ba, p(2)2 = p. Tuy nhien, da vao nguyen ly tac dung tng ho ta co ngay

p(1)2 = p

(2)1 , ngha la

[p] =

p(1)10p

vi p(1)1 la tai (cha biet) tac dung tai nut b ngam (nut cho ieu kien biencot yeu). Nh se thay trong bc ke tiep phng trnh co cha p1 se ckh mat, con p se c gan gia tr lc cho trc. V vay, trong tnh toan thchanh, ta co the at

{p} =

000

,

ngha la khong can lap ghep vect tai phan t (xem s o khoi, hnh 2.11).Bc 4 - Kh ieu kien bien (ieu kien chuyen dch va lc)ieu kien chuyen dch: do nut 1 c cho trc chuyen dch, q1 = 0,

phng trnh th nhat tng ng vi 0 = 0 nen e he phng trnh khongsuy bien ta phai kh an q1 va loai phng trnh tng ng khoi he phngtrnh.

Cach kh an a biet va loai phng trnh tng ng ma khong lam thayoi kch thc cua cac ma tran tham gia trong he phng trnh (??). e lamro phng phap ta trnh bay cho trng hp tong quat, phng trnh PTHH


co dang

K11q1 +K12q2 + . . . +K1dqd + . . .+K1nqn = p1K21q1 +K22q2 + . . . +K2dqd + . . .+K2nqn = p2

...Kd1q1 +Kd2q2 + . . . +Kddqd + . . . +Kdnqn = pd

...Kn1q1 +Kn2q2 + . . .+Kndqd + . . . +Knnqn = pn

trong o qd = g c cho trc. Thay qd bang g vao cac phng trnh trphng trnh th d, roi chuyen so hang a biet sang ve hai th ve hai trthanh

pj Kjdg (j 6= d);rieng phng trnh d c thay bang phng trnh 1qd = g; cuoi cung ta co:

K11q1 +K12q2 + . . .+ 0 qd + . . .+K1nqn = p1 K1dgK21q1 +K22q2 + . . .+ 0 qd + . . .+K2nqn = p2 K2dg

...0 q1 + 0 q2 + . . . + 1 qd + . . .+ 0 qn = g

...Kn1q1 +Kn2q2 + . . . + 0 qd + . . .+Knnqn = pn Kndg

Vi cach lam nay th ma tran o cng [K] chuyen thanh ma tran nhanc t no bang cach cho cac phan t dong d va cot d bang khong, riengphan t dong d, cot d cho bang 1

K11 K12 . . . 0 . . . K1nK21 K22 . . . 0 . . . K2n

...0 0 . . . 1 . . . 0

...Kn1 Kn2 . . . 0 . . . Knn

.

Cac thanh phan cua vect tai c hieu chnh theo cong thc tren, riengphan t dong d c lay bang g.


ieu kien lc: do nut 3 cho trc lc keo p, ta "cong them" p(3) =p(3) + p.

Bc 5 - Giai he phng trnh STT.Bc 6 - Hau x lyTnh ng suat trong cac phan t nh cong thc (2.49).

Cac bc tnh toan PTHH co the tom lc bang s o tren hnh 2.11.

2.2.2 Lap trnh tnh toan PTHH

Nh mot th du ve cach lap trnh tnh toan PTHH, di ay la chng trnhthanhnn.m c viet bang ngon ng Matlab. Tat nhien, nh cau cham ngon au chng, ay khong phai la chng trnh tot nhat e thc hien mucch - phan tch ng suat thanh nhieu nac; chng trnh c viet theo satcac bc chung cua viec mo hnh hoa PTHH e ngi mi bat au de theodoi, "bat chc roi sang tao" cac chng trnh tnh khac.

Su lieu chng trnh thanhnn.m4)

Muc ch: phan tch ng suat thanh nhieu nac.Ten: . . . . . .Ngay cap nhat: . . . . . .Bien - Mo ta bien

nn tong so nutne tong so phan tcoord vect toa o cac nutla ma tran lap gheppara ma tran cac tham so hnh hoc va vat lieudcond ieu kien chuyen dchfcond ieu kien lcke ma tran o cng phan tgk ma tran o cng toan cucgp vect tai toan cucq vect chuyen dch nut toan cucsigma vect ng suat cac phan t

4)Su lieu chng trnh la phan khong the thieu khi lap trnh, no cung cap nhng thongtin can thiet e chuyen giao chng trnh cho ngi s dung. Thng bon muc au ca vao trong phan au cua chng trnh.


Hnh 2.11: S o khoi mo hnh hoa PTHH.


Cau truc d lieu cua vect, ma tran1) Vect toa o cac nut coord, kch thc 1nn,

coord(i) = toa o nut toan cuc th i.

2) Ma tran lap ghep la, kch thc ne2

la(e,1) = th t toan cuc cua nut th nhat cua phan t e,la(e,2) = th t toan cuc cua nut th hai cua phan t e.

3) Ma tran cac tham so hnh hoc va vat lieu para, kch thc ne3

para(e,1) = chieu dai cua phan t e,para(e,2) = dien tch tiet dien cua phan t e,para(e,3) = moun Young cua phan t e.

4) ieu kien chuyen dch dcond, kch thc 12

dcond(1) = th t toan cuc cua nut cho trc chuyen dch,dcond(2) = gia tr cua chuyen dch.

5) ieu kien lc fcond, kch thc 12

fcond(1) = th t toan cuc cua nut cho lc ngoai,fcond(2) = gia tr cua lc.

Chng trnh thanhnn.m

thann.m

% chuong trinh phan tich ung suat cua thanh nhieu nac

% T.A. Ngoc

% update: 25/9/2009

% bien - mo ta

% nn ....... tong so nut

% ne ....... tong so phan tu

% coord .... vecto toa do cac nut

% la ....... ma tran lap ghep

% para ..... ma tran cac tham so hinh hoc va vat lieu

% dcond .... dieu kien chuyen dich

% fcond .... dieu kien luc

% ke ....... ma tran do cung phan tu

% pe ....... vecto tai phan tu


% gk ....... ma tran do cung toan cuc

% gp ....... vecto tai toan cuc

% q ........ vecto chuyen dich nut toan cuc

% sigma .... vecto ung suat phan tu

clear all

% TIEN XU LY

nn=3;

ne=2;

coord=[0 10 20];

la=[1 2; 2 3];

para=[10 2 2*10^6; 10 1 2*10^6];

fcond=[3 1000];

dcond=[1 0];

% XU LY

gk=zeros(nn,nn);

gp=zeros(nn,1);

% tinh ma tran do cung va vecto tai toan cuc

for e=1:ne

i=la(e,1);

j=la(e,2);

x1=coord(i);

x2=coord(j);

le=para(e,1);

ae=para(e,2);

ee=para(e,3);

% ma tran do cung phan tu

ke=ae*ee/le*[1 -1; -1 1];

% lap ghep

gk(i,i)=gk(i,i)+ke(1,1);

gk(i,j)=gk(i,j)+ke(1,2);

gk(j,i)=gk(j,i)+ke(2,1);

gk(j,j)=gk(j,j)+ke(2,2);

% vecto tai phan tu

pe=[0; 0];

% lap ghep

gp(i)=gp(i)+pe(1);

gp(j)=gp(j)+pe(2);

end;

% dua vao dieu kien nut

for i=1:size(fcond,1)

gp(fcond(i,1))=gp(fcond(i,1))+fcond(i,2);


end

for i=1:size(dcond,1)

for j=1:nn

gp(j)=gp(j)-gk(j,dcond(i))*dcond(i,2);

end

for j=1:nn

gk(dcond(i,1),j)=0.0;

gk(j,dcond(i,1))=0.0;

end

gk(dcond(i,1),dcond(i,1))=1.0;

gp(dcond(i,1))=dcond(i,2);

end

q=inv(gk)*gp;

% HAU XU LY

disp(sprintf(\n %s,KET QUA))% xuat chuyen dich

disp(sprintf(\n %s,CHUYEN DICH CUA CAC NUT))disp(sprintf(%s,nut))

for i=1:nn

disp(sprintf(%d\t\t %f,i,q(i)))end

% xuat ung suat cua cac phan tu

disp(ung suat trong cac phan tu (kg/cm^2));

for e=1:ne

sigma(e)=para(e,3)*(q(la(e,2))-q(la(e,1)))/para(e,1);

end

disp(sprintf(\n %s,UNG SUAT CUA CAC PHAN TU))disp(sprintf(%s,phan tu))

for e=1:ne

disp(sprintf(%d\t\t%f,e,sigma(e)))end

Ket qua chay chng trnh:

KET QUA

CHUYEN DICH CUA CAC NUT

nut

1 0.000000

2 0.002500

3 0.007500

ung suat trong cac phan tu (kg/cm2)

UNG SUAT CUA CAC PHAN TU


phan tu

1 500.000000

2 1000.000000

2.2.3 Mot cach tiep can khac

Khi thiet lap mo hnh PTHH cua bai toan phan tch ng suat cua thanhnhieu nac ta a dung cach at bai toan theo nguyen ly nang lng (nguyenly cong ao) cua c hoc ket cau phat bieu cho thanh chu keo nen. Cac phant hu han theo cach at nay la cac oan thanh cau thanh thanh nhieu nac,chung co y ngha vat ly cu the. Bay gi ta xet bai toan theo cach at micua ly thuyet cac bai toan bien.

Ap dung phng trnh vi phan mo ta bien dang cua thanh chu tai doctruc, bo qua lc khoi tac dung len thanh, phng trnh (1.35), ta c cacphng trnh vi phan xac nh chuyen dch cua thanh nhieu nac

d

dx

(A(1)E

dq

dx

)= 0, 0 < x < l(1), (2.56)

d

dx

(A(2)E

dq

dx

)= 0, l(1) < x < L. (2.57)

Theo gia thiet cua bai toan ieu kien bien cho nghiem:

q(0) = 0, (2.58)

Edq

dx(L) =

p

A(2). (2.59)

Ngoai ra, chuyen dch q con phai thoa ieu kien lien tuc va tng tac taiiem phan cach x = l(1)

q(l(1)) = q(l(1)+), (2.60)

A(1)Edq

dx(l(1)) = A(2)E

dq

dx(l(1)+). (2.61)

a vao khong gian ham V = {s|s H1(0, L), s(0) = 0}. Lay s Vtuy y, nhan hai ve phng trnh (2.56) (hay phng trnh (2.57)) vi s, lay tch


Hnh 2.12: Cac ham c s.

phan t 0 en L

A(1)E

l(1)0

d2q

dx2sdx+ A(2)E

Ll(1)

d2q

dx2sdx = 0.

Tch phan tng phan, dung cac ieu kien (2.59)-(2.61), ta c

A(1)E

l(1)0

dq

dx

ds

dxdx+ A(2)E

Ll(1)

dq

dx

ds

dxdx ps(L) = 0. (2.62)

Bai toan bien phan: tm q V thoa (2.62) vi moi s V . Dung phngphap Galerkin, lu y en nhan xet 2.10, vi cac ham c s (hnh 2.12)

1(x) =

{x2xx2x1

neu 0 = x1 x x20 neu x2 < x x3 = L

2.3. AP DUNG PTHH CHO CAC BAI TOAN BIEN 65

2(x) =

{ xx1x2x1

neu x1 x x2x3xx3x2

neu x2 < x x3

3(x) =

{0 neu x1 x x2xx2x3x2

neu x2 < x x3

Thay

q 3

i=1

qii =

2e=1

[N ](e){q}(e)

vao (2.62) vi s lay lan lt la j (j = 1, 2, 3) ta c

2e=1

(E(e)A(e)

xe+1xe

[B](e)T [B](e)dx

){q}(e) {p} = 0

2e=1

[k]e{q}(e) {p} = 0

[K]{q} = {p},trong o {p} = {0 0 p}T .

Sau khi kh ieu kien bien cot yeu ta nhan lai c phng trnh PTHHtrong th du 2.7.

2.3 Ap dung PTHH cho cac bai toan bien

Nh a trnh bay tren, phng phap phan t hu han co the ap dungcho cac bai toan bien. Cac thanh phan chnh cua phng phap PTHH chonghiem cua bai toan bien:

i. Phat bieu bien phan hay phat bieu yeu bai toan; va

ii. Giai xap x phng trnh bien phan nh PTHH.

Muc nay gii thieu cach tm nghiem xap x PTHH cua bai toan bien thongqua mot so th du, qua o gii thieu cach nhn van e theo quan iem giaitch ham.


2.3.1 Bai toan 1-chieu

Th du 2.8. Ky hieu = (0, 1). Giai phng trnh vi phan cho u:

d2u

dx2+ f = 0, x , (2.63)

thoa cac ieu kien bien

u(1) = g, (2.64)

dudx

(0) = h. (2.65)

ay f : R la ham cho trc, con g, h la cac so thc cho trc. ayla bai toan trong th du 1.2, chng 1.

Bai toan nay co nghiem giai tch:

u(x) = g + (1 x)h+ 1x

[ y0

f(z)dz

]dy. (2.66)

Ky hieu:S = {s|s H1(), s(1) = g} la tap hp cac ham th ;V = {v|v H1(), v(1) = 0} la tap hp cac ham trong lng.Bai toan bien phan:Cho f, g, h nh trc. Tm u S sao cho vi moi w V

10

dw

dx

du

dxdx =

10

wfdx + w(0)h. (2.67)

Di ay, e n gian ta goi bai toan bien (2.63) - (2.65) la bai toan (S), conbai toan bien phan (2.67) la bai toan (W)5). Ve phng dien toan hoc (tnhchat che) ta can chng minh:

5)S la ch au cua Strong, am ch phat bieu (hay dang) manh cua bai toan. W la ch aucua Weak, am ch phat bieu (hay dang) yeu cua bai toan.


nh ly 2.2. Di cac gia thiet thch hp ta co:

(a) Neu u la nghiem cua (S) th u cung la nghiem cua (W).

(b) Neu u la nghiem cua (W) th u cung la nghiem cua (S).

Chng minh(a) V u la nghiem cua (S) nen

0 = 10

w

(d2u

dx2+ f

)dx

vi moi w V . Tch phan tng phan ve phai,

0 =

10

dw

dx

du

dxdx

10

wfdx wdudx

10

.

Dung ieu kien bien (2.65) va w(1) = 0, ta c:

10

dw

dx

du

dxdx =

10

wfdx + w(0)h. (2.68)

Hn na, v u la nghiem cua (S) nen no thoa u(1) = g va v vay u S . Cuoicung, v u thoa (2.68) vi moi w V nen u thoa nh ngha nghiem yeu chobi (W).

(b) Bay gi gia s u la nghiem yeu. Vay u S , suy ra u(1) = g, va

10

dw

dx

du

dxdx =

10

wfdx+ w(0)h

vi moi w V . Tch phan t phan ve trai va dung w(1) = 0, ta c

0 =

10

w

(d2u

dx2+ f

)dx + w(0)

[du

dx(0) + h

]. (2.69)

e chng minh u la nghiem cua (S) ch can chng minh rang (2.69) dan en:


(i) d2udx2

+ f = 0 tren ; va

(ii) dudx(0) + h = 0.

au tien ta chng minh (i). nh ngha w trong (2.69) bi

w =

(d2u

dx2+ f

),

trong o trn; > 0 vi moi x ; va (0) = (1) = 0. Chang han, ta cothe lay (x) = x(1 x). Do w(1) = 0 nen w V . Thay w vao (2.69) th c

0 =

10

(d2u

dx2+ f

)2

0

dx+ 0.

Suy ra (i) phai c thoa.Dung (i) trong (2.69) e chng minh (ii), cu the la

0 = w

[du

dx(0) + h

].

Ta co the chon w V sao cho w(0) 6= 0. Vay (ii) cung c chng minhat:

a(w, u) =

10

dw

dx

du

dxdx, (2.70)

(w, f) =

10

wfdx (2.71)

lan lt la dang song tuyen tnh oi xng lien tuc, dang tuyen tnh lien tuctren V . Phng trnh bien phan (W) co the viet lai:

a(w, u) = (w, f) + w(0)h. (2.72)


Ap dung PTHH cho bai toan bien phan la xay dng cac xap x hu hanchieu cua S va V , ky hieu Sh va V h tng ng. Ch so tren h lien he vili hay phep phan hoach mien xac nh cua bai toan thanh cac phan thu han. Thong thng ta muon Sh S va V h V , xap x nh vay goi laxap x trong. T ieu kien bien ta thay neu uh Sh, wh V h th uh(1) = g,wh(1) = 0. Cho trc ham gh Sh. Moi ham u Sh co the bieu dien didang uh = vh + gh, trong o vh V h. T phng trnh (2.72), ta co phngtrnh

a(wh, uh) = (wh, f) + wh(0)h (2.73)

xac nh nghiem yeu xap x uh thoa (2.73) vi moi wh V h; hay xac nhvh V h (uh = vh + gh) thoa

a(wh, vh) = (wh, f) + wh(0)h a(wh, gh) (2.74)

Tom lai, ta co dang Galerkin cua bai toan (xap x), ky hieu (G), cphat bieu nh sau: Cho trc f, g, h nh tren, tm uh = vh + gh, vh V h, saocho vi moi wh V h

a(wh, vh) = (wh, f) + wh(0)h a(wh, gh).

Cho en ay, ta co bai toan xap x cua bai toan bien ban au. Ap dungPTHH cho bai toan, thc chat, la xay dng khong gian xap x V h bang PTHH(xap x ham).Xay dng khong gian xap x V h bang PTHH

Mien c chia thanh ne phan t hu han (oan con) bi nn = ne+1iem nut:

0 = x1 < x2 < . . . < xnn,

phan t th e la oan [xe, xe+1], chieu dai he = xe+1 xe.Ham wh V h (khong gian xap x) la a thc bac nhat tren tng phan

t

wh = c(e)1 N

(e)1 (x) + c

(e)2 N

(e)2 (x) = [N ]

(e){c}(e) xe < x < xe+1. (2.75)Vi cach xap x nay ta co ho ham tuyen tnh t manh: N1, N2, . . . , Nnn, goila ham dang toan cuc, xac nh nh sau:

? N1 lien ket vi nut 1:

N1(x) =

{x2xh1

neu 0 = x1 x x20 neu x > x2


? Ni (2 i nn 1) lien ket vi nut i (nut trong):

Ni(x) =

xxi1hi1

neu xi1 x xixi+1x

hineu xi < x xi+1

0 neu x < xi1 hay x > xi+1

? Nnn lien ket vi nut nn:

Nnn(x) =

{0 neu x < xnn1xxnn1hnn1

neu xnn1 x xnn

Cac ham dang Ni la s "lap ghep" ham dang cua cac phan t co chung nut

Hnh 2.13: Cac ham dang toan cuc.

i. De thay cac ham Ni thoa ieu kien kronecker

Ni(xj) = ij

va v vay ho ham nay oc lap tuyen tnh. Do wh(1) = 0, ta nh ngha:V h =< N1, N2, . . . , Nnn1 >. ieu nay co ngha la neu wh V h th ton taicac hang so thc ci , i = 1, 2, . . . , nn 1 sao cho

wh = c1N1 + c2N2 + . . .+ cnn1Nnn1 =nn1i=1

ciNi = [N ]{c}, (2.76)


trong o [N ] = [N1 N2 . . . Nnn1], {c}T = {c1 c2 . . . cnn1}.e xac nh cac ham uh Sh ta a vao ham gh nh ngha nh sau

gh = gNnn, khi o neu uh Sh th ton tai cac so thc qi, i = 1, 2, . . . , nn 1,sao cho

uh =nn1i=1

qiNi + gNnn = [N ]{q}+ gNnn, (2.77)

trong o {q}T = {q1 q2 . . . qnn1}.Bay gi thay (2.76) va (2.77) vao phng trnh Galerkin (2.74), roi dung

tnh song tuyen tnh cua a, va tnh tuyen tnh cua cac bieu thc con lai, tac:

nn1i,j=1

cia(Ni, Nj)qj =nn1i=1

ci(Ni, f) + hnn1i=1

ciNi(0) gnn1i=1

cia(Ni, Nnn).

V phng trnh tren thoa vi moi wh V h nen no thoa vi cac ci tuy y, suyra:

nn1j=1

a(Ni, Nj)qj = (Ni, f) + hNi(0) ga(Ni, Nnn) (2.78)

vi moi i = 1, 2, . . . , nn.Ky hieu: Kij = a(Ni, Nj), pi = (Ni, f) + hNi(0) ga(Ni, Nnn) th (2.78)

tr thanh:

nn1j=1

Kijqj = pi. (2.79)

hay di dang ma tran[K]{q} = {p},

trong o [K] = [Kij], {q} = {q1 q2 . . . qnn}T , {p} = {p1 p2 . . . pnn}T , oichieu vi muc trc, lan lt la ma tran o cng toan cuc, vect chuyen dch,vect tai toan cuc (sau khi kh ieu kien bien).


Nhan xet 2.11. Loi trnh bay ay, theo quan iem toan cuc, neu bat ytng c ban cua PTHH la xap x khong gian ham V bang khong gian huhan chieu V h. Tuy nhien, trong thc hanh giai cac bai toan bien ngi tatien hanh theo quan iem a phng.Thiet lap phng trnh PTHH (Galerkin) theo quan iem a phng

Sau khi a phan hoach mien (bc 1), chon a thc noi suy cho tngphan t (bc 2). Bc 3, thiet lap phng trnh PTHH c thc hien bangcach xet phng trnh (2.63) tren phan t hu han (e) = (xe, xe+1). Thchien tnh toan tng t nh khi thiet lap cong thc bien phan,

xe+1xe

(d2u

dx2+ f

)wdx.

Tch phan tng phan:

xe+1xe

dw

dx

du

dxdx =

xe+1xe

wfdx+du

dx(xe+1)w(xe+1) du

dx(xe)w(xe).

Dung cac ky hieu

a(e)(w, u) =

xe+1xe

dw

dx

du

dxdx, (w, f)(e) =

xe+1xe

wfdx,

cong thc tren co the viet lai:

a(e)(w, u) = (w, f)(e) +du

dx(xe+1)w(xe+1) du

dx(xe)w(xe).

Theo cach xap x ham tren (e), wh = [N ](e){c}(e) va uh = [N ](e){q}(e), th

2,=1

c(e) a(e)(N (e) , N

(e) )q

(e) =

2=1

c(e) (N(e) , f)

(e) +2

=1

c(e)

[du(e)

dx(xe+1)N

(e) (xe+1)

du(e)

dx(xe)N

(e) (xe)

].


Do cac c(e) la tuy y nen

2=1

a(e)(N (e) , N(e) )q

(e) =

(N (e) , f)(e) +

[du(e)

dx(xe+1)N

(e) (xe+1)

du(e)

dx(xe)N

(e) (xe)

].

Ky hieu:

k(e) = a

(e)(N (e) , N(e) ),

p(e) = (N(e) , f)

(e) +

[du(e)

dx(xe+1)N

(e) (xe+1)

du(e)

dx(xe)N

(e) (xe)

]

so hang lien quan en bien phan t

,

phng trnh thanh:[k](e){q}(e) = {p}(e),

trong o [k](e) = [k(e) ] va {p}(e) = [p(e) ] lan lt la ma tran o cng va vecttai phan t. Cu the, trong tnh toan thc hanh, ma tran o cng phan tc tnh bi cong thc:

[k](e) =1

he

[1 1

1 1];

con vect tai phan t, xap x PTHH ham f cung kieu, c tnh nh sau.

p(e) = (N(e) , f)

(e)) =

xe+1xe

N (e) fdx =2

=1

( xe+1xe

N (e) N(e) dx

)f(x

(e) ),

suy ra

{p}(e) = he6

{2f(xe) + f(xe+1)f(xe) + 2f(xe+1)

}.


Chu y, ay ta khong a vao so hang lien quan en bien cua phan t, v nose b triet tieu (kh lan nhau) khi lap ghep neu bien nam ben trong, hoac sec xac nh t ieu kien bien bai toan neu bien nam tren bien cua mien.Trong bai toan ang xet, tai nut au (phan t th nhat),

dudx

(x1)N(1)1 (x1) = h,

du

dx(x1)N

(1)2 (x1) = 0.

Do o, ta ch can "cong them" h vao thanh phan th nhat (ng vi nut 1)cua vect tai toan cuc (xem chng trnh pthh26.m).

Phan con lai c thc hien tng t nh a trnh bay trong muc trc.

Chng trnh pthh28.mChng trnh c viet cho th du 2.8 vi cac so lieu cu the: f(x) = 6x,

g = 1, h = 2. Nghiem chnh xac: u = x3 + 2x 4. Ngi oc nen oichieu vi chng trnh thanhnn.m e thay cac thanh phan chung cua cacchng trnh PTHH.

pthh28.m

% chuong trinh thi du 2.8, f=-6x, g=-1, h=-2

% T.A. Ngoc

% update: 25/9/2009

% bien - mo ta

% nn ....... tong so nut

% ne ....... tong so phan tu

% coord .... vecto toa do cac nut

% la ....... ma tran lap ghep

% dcond .... dieu kien Dirichlet

% fcond .... dieu kien Neumann

% ke ....... ma tran do cung phan tu

% pe ....... vecto tai phan tu

% gk ....... ma tran do cung toan cuc

% gp ....... vecto tai toan cuc

% q ........ vecto chuyen dich nut toan cuc

% chuong trinh goi function ffunc - ham f

clear all

% TIEN XU LY

ne=10;

nn=ne+1;

coord=0:1/ne:1;


for e=1:ne

la(e,1)=e;

la(e,2)=e+1;

end

fcond=[1 -2];

dcond=[nn -1];

% XU LY

gk=zeros(nn,nn);

gp=zeros(nn,1);

% tinh ma tran do cung va vecto tai toan cuc

for e=1:ne

i=la(e,1);

j=la(e,2);

x1=coord(i);

x2=coord(j);


ke=1/abs(x2-x1)*[1 -1; -1 1];

% lap ghep

gk(i,i)=gk(i,i)+ke(1,1);

gk(i,j)=gk(i,j)+ke(1,2);

gk(j,i)=gk(j,i)+ke(2,1);

gk(j,j)=gk(j,j)+ke(2,2);

% vecto tai phan tu

pe=abs(x2-x1)/6*[2*ffunc(x1)+ffunc(x2); ffunc(x1)+2*ffunc(x2)];

% lap ghep

gp(i)=gp(i)+pe(1);

gp(j)=gp(j)+pe(2);

end;

% dua vao dieu kien nut

for i=1:size(fcond,1)

gp(fcond(i,1))=gp(fcond(i,1))+fcond(i,2);

end


for j=1:nn

gp(j)=gp(j)-gk(j,dcond(i,1))*dcond(i,2);

end

for j=1:nn

gk(dcond(i,1),j)=0.0;

gk(j,dcond(i,1))=0.0;

end




end

q=inv(gk)*gp;

% HAU XU LY

disp(sprintf(\n %s, KET QUA SO VOI NGHIEM CX))% nghiem chinh xac u=x^3+2*x-4

ucx=coord.^3+2*coord-4;

% xuat chuyen dich

disp(sprintf(%s,nut uxx ucx))

for i=1:nn

disp(sprintf(%d\t\t%f\t%f,i,q(i),ucx(i)))end

disp(sprintf(%s%e,sai so cuc dai: , max(abs(q-transpose(ucx)))))

Ham c goi trong th du 2.8:

ffunc.m

function v=ffunc(x)

% ham duoc goi trong thi du 2.6

v=-6*x;

Ket qua chay chng trnh:

KET QUA SO VOI NGHIEM CX

nut uxx ucx

1 -4.000000 -4.000000

2 -3.592000 -3.592000

3 -3.136000 -3.136000

4 -2.584000 -2.584000

5 -1.888000 -1.888000

6 -1.000000 -1.000000

sai so cuc dai: 1.776357e-015

Chu thch ve lap trnh Matlab(1) Mot vect X vi cac thanh phan nguyen dng c Matlab "xem

nh" anh xa t N vao N,

1 7 X(1), 2 7 X(2), . . . , n 7 X(n).Matlab cho phep thc hien phep toan hp noi hai anh xa kieu G(X) miensao G la vect co chieu dai ln hn cac X(i). Th du:

>> X=[2 3]


X =

2 3

>> G=[0.1 0.2 0.3 0.4; 0.2 0.3 0.4 0.5; 0.3 0.4 0.5 0.6]

G =

0.1000 0.2000 0.3000 0.4000

0.2000 0.3000 0.4000 0.5000

0.3000 0.4000 0.5000 0.6000

>> G(1,X)

ans =

0.2000 0.3000

>> G(X,1)

ans =

0.2000

0.3000

>> G(X,X)

ans =

0.3000 0.4000

0.4000 0.5000

T nhan xet nay oan ma trong vong lap theo phan t cua chng trnhtren co the viet:

i=la(e,:);

x=coord(i);


ke=1/abs(x(2)-x(1))*[1 -1; -1 1];

% lap ghep

gk(i,i)=gk(i,i)+ke;

% vecto tai phan tu

pe=abs(x(2)-x(1))/6*[2*ffunc(x(1))+ffunc(x(2));

ffunc(x(1))+2*ffunc(x(2))];

% lap ghep

gp(i)=gp(i)+pe;

(2) Dung ky t gop ':' oan ma kh ieu kien bien Dirichlet co the viet:


gp(:)=gp(:)-gk(:,dcond(i,1))*dcond(i,2);

gk(dcond(i,1),:)=0.0;

gk(:,dcond(i,1))=0.0;



end


2.3.2 Bai toan 2-chieu

Th du 2.9. Cho la tap m b chan trong R2 vi bien , xet bai toanbien elliptic:

4 u = f(x), x , (2.80)u = 0 tren , (2.81)

trong o f = f(x), x = (x, y)), la ham cho trc thuoc lp L2().Phng trnh ao ham rieng trong bai toan tren goi la phng trnh Poisson.

No la mo hnh toan hoc cua nhieu bai toan gap trong vat ly. Th du, goi u la nhieto phan bo trong vat the chiem mien R3. Neu qua trnh dan nhiet la dng taco cac he thc sau:

qi = ki(x)u

xi, x , i = 1, 2, 3 (nh luat Fourier), (2.82)

5 q = f, x (nh luat bao toan nang lng), (2.83)

ay q = (q1, q2, q3), qi ky hieu dong nhiet theo hng xi, ki(x) la o dan nhiettai x theo hng xi, f(x) la mat o nguon nhiet tai x. Neu ki(x) 1, x ,i = 1, 2, 3, kh qi trong (2.82) - (2.83) ta c 4 u = f trong .

. Phat bieu yeua vao khong gian ham H10 () = {v H1()| v = 0 tren }. Lay

w H10 () tuy y, tch vo hng w vi hai ve phng trnh (2.80),

w4 udx =

wfdx. (2.84)

V

w4 u = x

(wu

x

)+

y

(wu

y

) wx

u

x wy

u

y,

[

x

(wu

x

)+

y

(wu

y

)]dx =

wu

nd = 0,

(trong bien oi cuoi ta a dung cong thc Gauss vi n la vect phap tuyen


n v ngoai nen (2.84) tr thanh

5w 5udx =

wfdx +

wu

nd. (2.85)

Dung ieu kien w = 0 tren , cuoi cung, ta c cong thc bien phan cuabai toan bien:

5w 5udx =

wfdx. (2.86)

a vao dang song tuyen tnh (oi xng) a(, ) va dang tuyen tnh `() trenH10 () nh bi:

a(w, u) =

5w 5udx, (2.87)

`(w) =

wfdx; (2.88)

bai toan bien phan tng ng: tm u H10 () sao choa(w, u) = `(w) w H10 (). (2.89)

. S ton tai va duy nhat cua nghiem yeuTrong nhan xet 1.3 ngay sau th du 1.2, chng 1, ta a "thong bao": s

ton tai va duy nhat nghiem (yeu) cua bai toan bien phan c chng minhnh nh ly Lax - Milgram, va trong trng hp dang song tuyen tnh a(, )oi xng th bai toan bien phan tng ng vi bai toan toi u. Bay gi,trc khi ap dung PTHH cho bai toan ang xet, ta se chng minh s ton tainghiem, s tng ng vi bai toan toi u cua bai toan nay.

Bo e 2.1 (Poincare -Fredrichs). Gia s Rn la tap m b chan vi bien u trn va cho u H10 (); th ton tai mot hang so c(), oc lap oi vi u sao cho

|u(x)|2dx c()ni=1

uxi (x)2 dx. (2.90)


Chng minh. V moi ham u H10 () la gii han trong H1() cua mot day{um}m=1 C0 (), nen ch can chng minh bat ang thc nay vi u C0 ().

e n gian, ta trnh bay chng minh cho trng hp = (a, b)(c, d) R2. Chng minh cho tong quat la tng t.

Hien nhien,

u(x, y) = u(a, y) +

xa

u

x(, y)d =

xa

u

x(, y)d, c < y < d.

Roi, bi bat ang thc Cauchy - Schwarz,

|u(x, y)|2dxdy = ba

dc

xa

u

x(, y)d

2 dxdy

ba

dc

(x a)( x

a

ux(, y)2 d

)dxdy

ba

(x a)dx ba

dc

ux(, y)2 ddy

12(b a)2

ux(x, y)2 dxdy.

Tng t,

|u(x, y)|2dxdy 12(d c)2

uy (x, y)2 dxdy.

Cong hai bat ang thc va tm, ta c

|u(x, y)|2dxdy c()

(ux(x, y)2 +

uy (x, y)2)dxdy,

trong o c() = ( 2(ba)2

+ 2(dc)2

)1.


De dang kiem tra bai toan bien phan (2.89) thoa cac ieu kien cua nhly Lax - Milgram nen co nghiem (yeu) duy nhat (dung bat ang thc Poincare-Fredrichs).

. Bai toan cc tieu hoa (toi u)Do tnh oi xng cua dang song tuyen tnh a(, ) nen bai toan bien phan

(2.89) co the phat bieu lai nh la bai toan cc tieu hoa. e chnh xac hn,ta nh ngha phiem ham toan phng (quadratic functional) J : H10 () Rbi

J(v) =1

2a(v, v) `(v), v H10 (). (2.91)

Bo e 2.2. Cho u la nghiem yeu (duy nhat) cua (2.89) trong H10 () va gia s ranga(, ) la dang song tuyen tnh oi xng tren H10 (); th u la cai cc tieu hoa duynhat cua J() tren H10 ().

Chng minh. Cho u la nghiem yeu duy nhat cua (2.89) trong H10 () va viv H10 (), xet J(v) J(u):

J(v) J(u) = 12a(v, v) 1

2a(u, u) + `(v u)

=1

2a(v, v) 1

2a(u, u) a(u, v u)

=1

2[a(v, v) 2a(u, v) + a(u, u)]

=1

2[a(v, v) a(v, u) a(u, v) + a(u, u)] (do tnh oi xng)

=1

2a(v u, v u).

V (2.90) nen ton tai hang so dng c0 sao cho a(vu, vu) c0vuH1().Vay,

J(v) J(u) c0v uH1() 0 J(v) J(u) v H10 ();

ngha la u cc tieu hoa J() tren H10 ().


Hn na, de thay u la cai cc tieu hoa duy nhat.De dang chng minh J() loi, ngha la

J((1 )v + w) (1 )J(v) + J(w) [0, 1], v, w H10 ().

[Dung ong nhat thc

(1 )J(v) + J(w) = J((1 )v + w) + 12(1 )a(v w, v w)

va bat ang thc a(v w, v w) 0.]Hn na, neu u cc tieu hoa J() th J() co mot iem dng (stationary

point) tai u; cu the,

J (u)v = lim0

J(u+ v) J(u)

= 0

vi moi v H10 (). V

J(u+ v) J(u)

= a(u, v) `(v) + 2a(v, v),

ta ket luan rang neu u cc tieu hoa J() th

lim0

[a(u, v) `(v) + 2a(v, v)] = a(u, v) `(v) = 0 v H10 (),

ieu nay chng minh bo e di ay.

Bo e 2.3. Gia s u H10 () cc tieu hoa J() tren H10 (); th u la nghiem (duynhat) cua bai toan (2.89). Bai toan (2.89) c goi la phng trnh Euler - lagrangecho bai toan cc tieu hoa nay.

Bo e nay la ao cua bo e trc. Ca hai bo e bieu dien s tngng cua bai toan (2.89) vi bai toan cc tieu hoa: tm u H 10 () sao cho

J(u) J(v) v H10 (). (2.92)


. Xap x PTHHBay gi ta ap dung PTHH e xap x bai toan bien phan (2.89), dung

phan t tam giac vi noi suy tuyen tnh.Bc 1: phan hoach mien bi li cac tam giac e.Bc 2: xap x ham bang a thc bac nhat tren tng phan t (phan t

tam giac 3 nut ), vi phan t e la tam giac vi cac nut c anh so th ta phng ngc chieu kim ong ho6) xi = (xi, yi), i = 1, 2, 3,

u(e) = [N ](e){q}(e)

trong o N = [N (e)1 (x, y) N(e)2 (x, y) N

(e)3 (x, y)],

N(e)1 (x, y) =

x2y3 y2x3 + x(y3 + y2) y(x3 + x2)x2y3 y2x3 x1y3 + y1x3 + x1y2 y1x2 ,

N(e)2 (x, y) =

x1y3 + y1x3 + x(y3 + y1) + y(x3 + x1)x2y3 y2x3 x1y3 + y1x3 + x1y2 y1x2 ,

N(e)3 (x, y) =

x1y2 y1x2 + x(y2 + y1) y(x2 + x1)x2y3 y2x3 x1y3 + y1x3 + x1y2 y1x2 .

{q}(e) = {q(e)1 q(e)2 q(e)3 }T vi q(e)i = u(e)(xi), i = 1, 2, 3.Bc 3: t phng trnh

4 u = f(x), x e

ta co cong thc bien phan cua bai toan phat bieu cho phan t e vi biene:

5w 5udx =ewfdx +

ewu

nde. (2.93)

Ky hieu nh trong th du trc,

a(e)(w, u) =

5w 5udx, (w, f)(e) =ewfdx,

6)Cach anh so nay nham muc ch phep bien oi t phan t e sang phan t tham chieu(xem th du 2.2) co Jacobian dng.


cong thc tren co the viet lai:

a(e)(w, u) = (w, f)(e) +

ewu

nde.

Tng t nh th du trc, tnh ma tran o cng phan t,

k(e) = a

(e)(N (e) , N(e) ) =

e5N (e) (5N (e) )Tdx, , = 1, 2, 3.

Nhng e tnh tch phan tren e th, khac vi th du trc, ta can oi biene bien oi mien e thanh mien n gian hn giong nhau cho moi phan t.Mot cach t nhien, mien n gian c chon la phan t tham chieu 2-chieu(xem th du 2.2), tam giac r vi cac nh

1= (0, 0),

2= (1, 0),

3= (0,

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