Phương pháp bậc Tôpô cho bài toán biên6_2_2

Luận văn thạc sĩ toán học: Phương pháp bậc Tôpô cho bài toán biên

Chu'dng 4

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Ung d\lng

Chung t6i trlnh bay mQt s6 ti'ngdl,lngcua cac b~t qua chuang 2 va chuang 3

GQiC 1£1t~p tiit ca cac phi@mham { : X -7 IR lien tl,lCtang, kh6ng nhiit thi@t tuy@n tfnh, saD cho {(a) = o Cac ph~n tv cua t~p C cling

co mQt tinh chiit nhu cac ph~n tv cua A da dU<;5cnoi tai d chuang 1:

n@u { E C thl {(x) = 0 vai mQt x E X naG do se cho ta x(~) = 0 vai mQt ~E J.

B5 d€J 4.1 Cia SU:v E X, a, {3E C Hai bili loan sau

x" (t) = v (t ) , {3(x) = 0, x' (0) = 0 (4.2)

co nghi~m Xl, X2 tttdng itng H dn nila

IIxiii :::; M2 ' Ilx~ II :::; Ilv II, i = 1,2 (4.3)

Chung minh Phuong trlnh x"(t) = v(t) co nghi~m t6ng quat 1£1

x(t) = Cj +c,t+l' 1" v(T)dnk

x vai bi~u thti'c nhu tIeD 1£1 mQt ph~n tv cua X D~ co khiing dinh

nay ta chi c~n chti'ng minh x lien tl,lC tIeD J Liiy {tn} 1£1day hQi tl,l

v§ t E J, ta danh gia x(tn) - x(t) va ki~m tra x(tn) - x(t) -7 0 khi

n -7 +00 Tli bi~u thti'c cua x(t) ta co

tn {s t (s

x(tn) - x(t) = C2(tn- t) + Jo Jo v(T)dTds - Jo Jo v(T)dTds,

Itn (s

x(tn) - x(t) = C2(tn- t) + t Jo v(T)dTds,

[X(tn) - X(t)[ ~ [C2[.ltn ~ t[ + [" l' v(T)dT[ ds,

IxU.)- x(t)1:( hl.lt. - tl + [n Ilvllds,

Ix(tn)- x(t)1 ~ IC21.ltn- tl + IIVII.ltn - tl.

R6 rang khi tn -+ t thi x(tn) -+ x(t).

a) Xet (4.1).

V6i di~u ki~n bien a(x) = 0,f3(x') = 0 ta Urndl1Qc~,c E J sao cho

x(~) = O,x'(c) = 0:

x'(c) = Cz+ l' v(r)dr = O.

Vdi C2 + l'v(T)dT = 0 ta co 1~ (C2+ l' V(T)dT)ds = O.Mil

Cl+C2~ + l' l' v(T)dTds = 0 nen Cl+ l' l' v(T)dTds = o Nghiem

x(t) tren day dl1Qc vi~t l<;1i:

x(t) = c] + c,t + l' l' v(r)drds,

x(t) = (C1 + t l' V(T)dTdS) + C2t+ l' 1'V(T)dTdS,

xli) = c2t+ ll' v(T)dTds.

Lily t = ~ta c6 "(0 = C2+ l' l' v(T)dTds, hay lit C2 = O Suy fa

nghi~rncua (4.1)lit x(t) = it l' v(T)dTds.

x(t) = l' l' v(r)drds Iiinghi~mcluynMt ciia billtOM (4.1).

Ix(t)1= 11' [ v(r)drds .;: l' [ Iv(r)ldrds .;: l' [ IIvlldrds,

Ix(t)1 :;;; IIVII;;' l' dTds = IIvll;;' (8 - e)d8,

]

Ix(t)I~IIvll" ( ct)-( c~) 2 2 =llvll c(t-'!jJ) .2 2

Dieu nay clan den Ix(t)1 ~ -llvll, tac la Ilxll ~ -IIvll.2 2

Tit x( t) = it l' v( T) dTds ta liiy d"", ham hai v~: x' (t) ~ it v(T)dT.

guy fa Ix'(t)1 ~ [IV(T)!dT = IIvllU - 0) ~ IIvll

Vf}.yta da chang minh (4.1) co nghi~m cluy nh§,t Xl thoa (4.3)

4.1 Biti toan thil nhiit

Xet bai tmin sau

(4) - ( ", ",)

X - P t,x,x,x,x ,

cy(x) = 0, (3(x') = 0, X"(0) = 0, X"(1) = 0

(4.4) (4.5)

vdi p E Car(J x JR4)va cy,(3 E C.

D~nh ly 4.1 Cia 811LI ~ 0 ~ L2, L3 ~ 0 ~ L4 la cae hling 88,

p(t,x,u,v,Ld ~ 0 ~ p(t,x,u,v,L2),

p(t,x,u,v,L4) ~ 0 ~ p(t,x,u,v,L3) thoa vdi hdu htt t E J va vdi m9i (x, u, v) E [-L/2, L/2] x [-L, L] x

[-L, L], L = max{ -LI, -L3, L2, L4}.

B ai loan (4.4) (4.5) co it nhdt m{jt nghi~m x: (t E J)

Ilxll ~ L/2, IIx'II ~ L,Ilx"ll ~ L, min{LI,L3} ~ x"'(t) ~ max{L2, L4}.

(4.6)

Chung minh.

a) L§,yv E X b§,t ky Thea b6 de 4.1, phuong trlnh x"(t) = v(t) co

nghi~m duy nh§,t Xl thoa a(x) = (3(x') = O D~t Fv = Xl thl ta xac

dinh du<;Jctoan tit F : X ~ X.

Cling vdi v E X b§,t ky, theo b6 d@4.1, x"(t) = v(t) co nghi~m duy

nh§,t X2 thoa j3(x) = x'(O) = 0, va ta d~t Hv = X2, di@unay xac dinh

toan tit H : X ~ X.

b) Ta ki@mtra F, H ED.

+ F va H bi ch~n bai theo (4.3) ta co IIFvl1:::; 2"llvll, IIHvl1 :::; 2"llvll.

N§u IIvll :::; L thlllFvll :::; L/2, IIHvll:::; L/2, do do p(F; [0,L]x) :::; L/2

va p(H; [0,L]x) :::;L/2.

+ Ta se chang minh F lien tl)C b~ng each l§,y {vn} C X la day hQi tl)

v@v E X va chi ra Fvn ~ Fv.

Vi (Fvn)"(t) = vn(t), (Fv)"(t) = v(t) nen ta tlm du<;Jchai day bi

ch~n {an}, {bn} trong JRva a, b E JRsaG cho, vdi t E J, n E N

Fvn(t) = an + bnt + 1'1' vn(T)dTds,

va Fv(t) = a + bt + l' l' v(T)dTds Dieu d6 dU<)cviet khac di Ii;

xn(t) = an+bnt+ i' i' vn(T)dTds, va x(t) = a+bt+ i' i' v(T)dTds.

Vdi m, n E N tuy y:

(Fvm - Fvn)(t) = am-an+(bm -bn)t+ l' l' (Vm-vn)(T)dTds.

Ta co am- an = (Fvm - Fvn)(O).

Suy ra lam - ani = I (Fvm - Fvn)(O)1 :::; IIFvm - Fvnll.

1 Theo (4.3), IIFvm- Fvnll = IIF(vm- vn)11:::;-llvm - vnll.

2

{vn} hQi tl) v@v nen Cauchy trong X Tl1 do {an} cling Cauchy.

{bn} la day Cauchy n§u d@y

bm-bn = (Fvm - Fvn)(l) -(am -an) -[1" (Vm-Vn) (T)dTds,

Ibm- bnl ~ I(Fvm - Fvn)(l) I+lam - ani + [ l' Ilvm - vnllds,

Ibm - bnl :::;2"llvm - vnll + lam- ani+ Jo Jo Ilvm- vnllds.

Hai day {bn},{an} la Cauchy nen phlti hQitl)

Ta chang minh hai day nay hQi tl) l§,n 111Qtv@b, a.

Gia s11ngl1Qc l?-i, lim bn n ->00 i= b, {bn} se c6 mOt day con {bnk} hQitw

lim bnk = b*, b* i= b.

k ->00

Vi j3(x~J = 0, X~k(t) = bnk+ (t vnk(s)ds, nen khi cho k -+ +00 ta

thu dli(!e {3(y') = 0, y'(t) = b' + l v(s)ds.

Ma (3(x') = 0, x'(t) = b + l' v(s)ds, va (3la ham tang non plll,j c6

y* = x' hay la b*= b, vo ly Do d6 lim bn= b.

n ->+oo

N@ugia S11lim an i= a, se c6 mOt day con hQi tlJ {anJ cua {an}: n ->00

*-i-1m ank = a , a I a.

k ->00

ta tll1l dl1<;!CI>(Z') = 0, z'(t) = a' + bt + l' L'V(T)dT.

Ma a(x) = 0, x(t) = a + bt + l' 1<' V(T)(lT, va a 1a ham tang non phai c6 z* = x, nghla la a* = a, do d6 lim an = a.

n ->+oo

T6m l?-i, xn(t) -+ x(t) v6i moi t E J.

Suy ra Xn -+ X trong X, tac la Fxn -+ Fx trong X.

Ta da chang minh F lien tlJC tren X.

+ H lien tlJC tren X la t110ng tv:.

+ F, H la cac ham lien tlJC va bi ch~n tren X, nen F, H E V.

c) N@uvi@t u = x", v6i a(x) = j3(x') = 0 ta c6 x = Fu V6i x"(O) =

x"(l) = 0 ta dl1Qcu(O) = u(l) = O.

u = x" cho ta x", = u' N@uvi@tx", = u' thanh (x')" = u', do j3(x') = 0 va (x')'(O) = x"(O) = 0, se c6 x' = Hu'.

Nhl1 vi;iy, (4.4)(4.5) dl1Qc dua v@(4.7)(4.8):

u"(t) = p(t, Fu(t), Hu'(t), u(t), u'(t)), (4.7)

LiLyf(t,x,u,v,w) = p(t,u,w,x,v) v8i (t,x,u,v,w) E J X IR4.

Ky hiE;juL' = min{L1, L3}, M' = max{L2, L4} thi M' ~ £, con

-£' = max{ -L1, -L3} ~ L hay L' ~ -L Ta kH:;mtra f thoa di§u

kiE;jncua dinh 1y 2.3:

f(t,x,u,Ll,W) ~ 0 ~ f(t,x,u,L2,w), f(t,x,u,L4'W) ~ 0 ~ f(t,x,u,L3,w) v8i t E J, (x,u,w) E (L',M';F,H)]R.

Cho t E J va (x,u,w) E (L',M';F,H)]R, ta co

L' ~ X ~ M', lul ~ p(F, (L', M')x), Iwl ~ p(H, (L', M')x).

Do L' ~ - L, M' ~ L, n@uL' ~ x ~ M' thi IxJ ~ L.

Vi p(F, [0,L]x) ~ L/2, p(H, [0,L]x) ~ L/2 ,n@u lul ~ p(F, (£', M')x)

va Iwl ~ p(H, (L', M')x) thi lul ~ L/2, Iwl ~ L.

Nhu th@,(u,w,x) E [-L/2,L/2] x [-£,L] x [-L,L], va ro rang

f(t, x, u, L1, w) = p(t, u, w, x, L1) ~ 0 ~ p(t, u, w, x, L2) = f(t, x, u, £2, w), f(t, x, u, L4, w) = p(t, u, w, x, £4) ~ 0 ~ p(t, u, w, x, L3) = f(t, x, u, £3, w).

Ta da ki@mtra di§u kiE;jncho dinh 1y2.3 Bai tmin (4.7) co nghiE;jm

u sao cho, v8i t E J,

Ilull ~ L, min{L1, L3} ~ u'(t) ~ max{L2, L4}.

Khi u 1a nghiE;jmcua (4.7)(4.8), theo b6 d§ 4.1, t6n t~i duy nhiLt

x E AC3(J) thoa a(x) = 0, !J(x') = 0 va x" = u.

x nay chfnh 1a nghiE;jmcua bai tmin (4.4)(4.5) thoa (4.9).

Tren day ta da sU'd\lng dinh 1y 2.3 cho bai tmin thti' nhiLt Bay gid

ta sU'd\lng dinh 1y 3.3 datrinh bay trong chuang 3 L1, L2, L3, L4 la cac ham lien t\lC sao cho L3 < L1 < 0 < L2 < L4 V8i t E J, x, u, v, wEIR d$,t f(t,x,u,v,w) = p(t,u,w,x,v).

Bai toan (4.7)(4.8)

u"(t) = p(t, Fu(t), Hu'(t), u(t), u'(t))

u(O)= 0,u(l) = 0

trd thanh

u"(t) = j(t,x(t), Fu(t), x'(t), Hx'(t))

u(O) = 0, u(l) = O.

Thea dinh 1y 3.3, n~u di@u ki~n sau thai man thl (4.7)(4.8) co nghi~m:

p(t, x, u, W,L1(t)) ~ 0 ~ p(t, x, u, w, L2(t)), p(t, x, u, W,L4(t)) ~ 0 ~ p(t, x, u, W,L3(t)) y6i h§,u h~t t E J, y6i mQi x, u, w E ffi.ma

IxJ ~ p(F, ((L3, L4)),

lul ~ p(H, /L(L3, L4)),

Iwl ~ max{IIL311,IIL411}= L.

Ta co nh~n xet sau

+ N~u y E ((L3, L4) thl 0 ~ Ilyll ~ max{IIL311,IIL411}.Luc do IIFyll ~ ~lIyll~ ~L;

+ N~u z E /L(L3,L4) thl 0 ~ Ilzll~ L Luc do IIHzll ~ IIzll ~ L.

Nghi~m cua phudng trlnh u"(t) = j(t, u(t), Fu(t), u'(t), Hu'(t)), u(O) = 0, u(l) = 0 1uc fLyse thai lu(t)1 ~ L ya L3(t) ~ u'(t) ~ L4(t) y6i mQit E J.

Ta co dinh 1y 4.2.

D~nh ly 4.2 Cia SV:L3 < L1 < 0 < L2 < L4 la cae ham lien t7,lC,

L1,L4 tang, L2, L3 giam tren J thoa

p(t, x, u, v, L1(t)) ~ 0 ~ p(t, x, u, v, L2(t)), p(t, x, u, V,L4(t)) ~ 0 ~ p(t, x, u, v, L3(t))

vdi hh t E J, vdi m9i (x, u, v) E [-L/2, L/2] x [-L, L] x [-L, L],

L = max{IIL311,IIL411}.

The thi (4.4)(4.5) eo nghi~m x,

Ilxll ~ L/2, IIx'll ~ L, Ilx"lI ~ L, L3(t) ~ Xlll(t) ~ L4(t), t E 1. (4.9)

Bai toan thu hai

Tllong tlj nhll tren ta xet bai toan sail

III

(t ' If)

a(x) = 0, x'(O)= 0, x' (1) = 0 (4.11)

trong do q E Car(J x ]R3),a E C.

N~u thay x' bdi u, thi u' = x" Vdi a(x) = 0, x'(O) = 0, theo b6

d~ 4.1 ta vi~t x = Hu'; va tli di~u ki~n bien x'(O) = x'(l) = 0 ta co

u(O) = u(l) = o.

Bai toan (4.10)(4.11) trd thanh

u"(t) = q(t, Hu'(t), u(t), u'(t)), (4.12)

u(O) = u(l) = O.

DM f(t,x,v,w) = q(t,w,x,v), (4.12)(4.13) trd thanh

(4.13)

u"(t) = f(t, u(t), u'(t), Hu'(t)), (4.14)

D!nh 1:5'4.3 Cia SU:L1 ~ 0 ~ L2, L3 ~ 0 ~ L4 la cae hling 58)

q(t,x,u,Ld ~ 0 ~ q(t,x,u,L2), q(t, x, u, L4) ~ 0 ~ q(t, x, u, L3) thoa man v(;i hau htt t E J va v(;i moi (x, u) E [-L, L] x [-L, L], L =

max{ -L1, -L3, L2, L4} Bai loan (4.10)(4.11) co nghi~m x sao cho

IIxii ~ L, IIx'll ~ L, min{L1, L3} ~ x"(t) ~ max{L2, L4}, t E J.

Chung minh. Ta Slt d\mg dinh ly 2.3 ChI c§,n ki~m tra di~u sail

xay ra vdi t E J, L' ~ u ~ M' va Ixl ~ p(H, (L', M')), L' =

min{L1, L3}, M' = max{L2, L4}:

f(t,u,L1,x) ~ 0 ~ f(t,u,L2'X),

f(t,u,L4'X) ~ 0 ~ f(t,u,L3,x).

Do L' = min{L1, L3}, M' = max{L2, L4} va L = max{ -L1, -L2, L3, L4} nen -L' = max{ -L1, -L2} ~ L, M' ~ L.

N@uL' ~ u ~ M' thi ta co -L ~ u ~ L.

Bai p(H,(L',M')x) = sup{IIHxll,xE X,L' ~ x ~ M'}, n@uIxl ~ p(H, (L', M')x) thi -L ~ x ~ L.

Tom I~i, n@uL' ~ u ~ M', Ix! ~ p(H, (L', M')) thi

(x, u) E [-L, L] x [-L, L].

L§.yt E J, L' ~ u ~ M', Ixl ~ p(H, (L', M')) Ta co

f(t,u,L1,x) = q(t,x,u,L1) ~ 0 ~ q(t,x,u,L2) = f(t,u,L2'X),

f(t,u,L3,x) = q(t,x,u,L3) ~ 0 ~ q(t,x,u,L4) = f(t,u,L4'X).

Cac di@uki~n trong dinh Iy 2.3 dll<;1C dam baa Bai toan (4.14)(4.15)

co nghi~m u thoa L' ~ u ~ M' va L' ~ u' ~ M' Tit L' ~ u' ~ M'

ta dll<;1CL' ~ x"(t) ~ M' Tit L' ~ u ~ M' ta co -L ~ x'(t) ~ L hay

la Ilx'll ~ L Danh gia -L ~ x'(t) ~ L k@th<;1p vdi a(x) = 0 clan d@n -L ~ x(t) ~ L, tlic la Ilxll ~ L Dinh Iy 4.3 chling minh xong. D

Khi L1, L2, L3, L4 la cac ham lien t\lC tren J ta co dinh Iy 4.4.

D!nh ly 4.4 Cia sa L1, L2, L3, L4 la cac ham lien t7J,C,L1, L4 tang,

L2, L3 giam tren J sao rho L3 < L1 < 0 < L2 < L4,

q(t,x,u,L1(t)) ~ 0 ~ q(t,x,u,L2(t)), q(t, x, u, L4(t)) ~ 0 ~ q(t, x, U,L3(t))

thda man vdi hau h~t t E J va vdi m9i (x, u) E [-L, L] x [-L, L], L =

max{IIL311,IIL411}.Bai toan (4.10)(4.11) co nghi~m x sao rho IIxll ~ L,

Ilx'll ~ L, L3(t) ~ x"(t) ~ L4(t),t E J.

Chung minh Ta ki~m tra ding, vdi h§.u h@t t E J, lul ~ L va Ixl ~ p(H, Jl(L3,L4)) thi co di@usail day,

f(t, u, L1(t), x) ~ 0 ~ f(t, u, L2(t), x), f(t, u, L4(t),x) ~ 0 ~ f(t, u, L3(t), x).

R6 rang, n@u lul ~ L va Ixl ~ p(H, /-L(L3,L4)), se co

(x, u) E [-L, L] x [-L, L].

Th@thl,

j(t,u,L1(t),x) = q(t,x,u,L1(t)) ~ 0 ~ q(t,x,u,L2(t)) = j(t,u,L2(t),x), j(t, u, L3(t), x) = q(t, x, U,L3(t)) ~ 0 ~ q(t, x, U,L4(t)) = j(t, u, L4(t), x).

Cac di§u ki~n trong dinh 1y3.3 du'Qcdam baa Bai toan (4.14)(4.15)

co nghi~m u th6a -IIL311~ u ~ IIL411 va L3(t) ~ u'(t) ~ L4(t), t E 1 (4.10)(4.11) co nghi~m x = Hu'.

Bdi x' = u nen khi -IIL311~ u(t) ~ IIL411, t E J thl

-IIL311~ x'(t) ~ IIL411,t E J.

Tl1L3(t) ~ u'(t) ~ L4(t), t E J ta co Ilu'll~ L va du'QCIIHu'll~ L

hay 1a Ilxll ~ L.

Cling tl1 L3(t) ~ u'(t) ~ L4(t), t E J suy ra

L3(t) ~ x"(t) ~ L4(t), t E J.