3 Ph÷ìng ph¡p l°p gi£i b i to¡n stick slip têng qu¡t
3.3 Mët sè k¸t qu£ thüc nghi»m
º kiºm tra ë ch½nh x¡c cõa sì ç l°p ¢ · xu§t gi£i b i to¡n Stick - Slip, chóng ta sû döng ph÷ìng ph¡p l÷îi chuyºn c¡c b i to¡n vi ph¥n v· c¡c b i to¡n sai ph¥n tr¶n khæng gian l÷îi, ta thu ÷ñc c¡c h» ph÷ìng tr¼nh vectì 3 iºm sau â sû döng thuªt to¡n thu gån khèi l÷ñng t½nh to¡n [7], [8] x¡c ành nghi»m g¦n óng t¤i méi b÷îc l°p. Trong c¡c sì ç l°p, c¡c ¤o h m ÷ñc t½nh vîi ë ch½nh x¡c c§p 2. Trong thû nghi»m, l÷îi chia luæn luæn ÷ñc chån l M × N = 64 × 64, mi·n ÷ñc x²t
Ω = [−1; 1]×[0; 1].
+ Trong tr÷íng hñp 1: So s¡nh nghi»m x§p x¿ vîi nghi»m ch½nh x¡c. Xu§t ph¡t tø nghi»m ch½nh x¡c u∗(x, y) , chóng ta x¡c ành c¡c h m v¸ ph£i v c¡c i·u ki»n bi¶n cõa b i to¡n, tø â sû döng thuªt to¡n x¡c ành nghi»m x§p x¿ cõa b i to¡n sai sè ε1 = Max
u(i,j∗) = u(i,jk)
.
+ Trong tr÷íng hñp 2: K¸t qu£ khi khæng bi¸t tr÷îc nghi»m óng (b i to¡n gèc Stick _ Slip). Sû döng sì ç l°p º x¡c ành nghi»m trong tr÷íng hñp têng qu¡t khi khæng bi¸t tr÷îc nghi»m óng, chóng ta x¡c ành c¡c h» i·u ki»n bi¶n theo h¼nh 3.3, tø â sû döng thuªt to¡n x¡c ành nghi»m x§p x¿ cõa b i to¡n sai sè ε2 = Max
u(i,jk+1) = u(i,jk)
.
Trong qu¡ tr¼nh t½nh to¡n, ngæn ngú sû döng l ngæn ngú Matlab version 7.0, c¡c h m m¨u gi£i c¡c b i to¡n c§p hai ÷ñc sû döng trong
th÷ vi»n ch÷ìng tr¼nh RC2009 [2].
B£ng 3.1: Tr÷íng hñp têng qu¡t ( bi¸t tr÷îc nghi»m óng)
τ n ε1 0,1 135 9.10−5 0,3 75 1,9.10−4 0,4 76 1,9.10−4 0,5 92 1,9.10−4 0,6 111 1,9.10−4 0,7 134 1,9.10−4 0,8 165 1,9.10−4 0,9 218 1,9.10−4 B£ng 3.1: H¼nh 3.4: ç thà nghi»m óng
τ n ε2 0,1 92 6.10−5 0,2 36 9.10−5 0,6 35 9.10−5 0,7 35 9.10−5 0,8 34 9.10−5 0,9 35 9.10−5 B£ng 3.2:
K¸t luªn chung
Luªn v«n ¢ · cªp ¸n c¡c ph÷ìng ph¡p x§p x¿ x¡c ành nghi»m g¦n óng cõa b i to¡n Stick - Slip. C¡c k¸t qu£ ch½nh cõa luªn v«n gçm câ:
1. Tr¶n cì sð t i li»u [5], luªn v«n ÷a ra mæ h¼nh to¡n håc cõa b i to¡n Stick - Slip vîi i·u ki»n bi¶n ký dà, tr¼nh b y cì sð to¡n håc cõa ph÷ìng ph¡p ti»m cªn, c¡c k¸t qu£ thüc nghi»m èi vîi b i to¡n.
2. Düa tr¶n k¸t qu£ cõa thuªt to¡n chia mi·n èi vîi b i to¡n song i·u háa vîi i·u ki»n bi¶n gi¡n o¤n m¤nh, luªn v«n ¢ ÷a ra sì ç l°p x¡c ành nghi»m x§p x¿ cõa b i to¡n Stick - Slip trong tr÷íng hñp têng qu¡t, ti¸n h nh lªp tr¼nh x¡c ành nghi»m sè cõa b i to¡n, ¡nh gi¡ v· tèc ë hëi tö v ë ch½nh x¡c cõa sì ç l°p.
H÷îng ph¡t triºn cõa luªn v«n l mð rëng c¡c k¸t qu£ èi vîi c¡c b i to¡n song i·u háa vîi c¡c h» i·u ki»n bi¶n phùc t¤p hìn.
Do v§n · ÷ñc · cªp trong luªn v«n l t÷ìng èi phùc t¤p, hìn núa do thíi gian v kh£ n«ng cán h¤n ch¸ n¶n m°c dò ¢ câ nhi·u cè gng nh÷ng luªn v«n khâ tr¡nh khäi nhúng thi¸u sât. T¡c gi£ mong nhªn ÷ñc nhúng þ ki¸n âng gâp quþ b¡u cõa th¦y cæ gi¡o v nhúng ng÷íi quan t¥m º luªn v«n ÷ñc ho n thi»n hìn.
T i li»u tham kh£o
T i li»u Ti¸ng Vi»t
[1] °ng Quang , Tr÷ìng H H£i, Vô Vinh Quang, (4/2011), Ph÷ìng ph¡p l°p gi£i mët b i to¡n bi¶n hén hñp èi vîi ph÷ìng tr¼nh song i·u háa, hëi th£o Tèi ÷u v T½nh to¡n khoa håc l¦n thù 9, Ba v¼. [2] Vô Vinh Quang, Tr÷ìng H H£i, Nguy¹n Thà Tuyºn (2010), X¥y düng bë ch÷ìng tr¼nh RC2009 gi£i mët sè b i to¡n bi¶n elliptic vîi h» sè h¬ng, T¤p ch½ Khoa håc v Cæng ngh» ¤i håc Th¡i Nguy¶n, T.69(07), tr.56 - 63.
[2] Vô Vinh Quang, Tr÷ìng H H£i, Nguy¹n Thà Tuyºn, X¥y düng bë ch÷ìng tr¼nh RC2009 gi£i mët sè b i to¡n bi¶n elliptic vîi h» sè h¬ng, T¤p ch½ Khoa håc v Cæng ngh» ¤i håc Th¡i Nguy¶n, T.69(07):56 - 63. 2010.
T i li»u Ti¸ng Anh
[3] Dang Qang A, Vu Vinh Quang, A domain decomposition method for strongly mixed boundary value problems for the Poisson equa- tion, In book: Modeling, Simulation and Optimization of Complex Processes (Proc. 4th Inter. Conf. on HPSC, 2009, Hanoi, Vietnam), Springer, 2012.
[4] Dang Quang A, Truong Ha Hai, Vu Vinh Quang, Interative Method for a Biharmonic Problem with Crack Singgularities, Applied Math- ematical Sciences, Vol. 6, 2012, no. 62, 3095-3018.
[5] Funaro D., Quarteroni A., Zanolli P. (1998), " An iterative proce- dure with interface relaxation for domain decomposition method", SIAM J. Number. Anal. 25(6), pp. 1213 - 1236.
[6] M. Elliotis, G. Georgiou and C. Xenophontos, Solution of the planar Newtonian stick-slip problem with the singular function boundary integral method, int. J. Numer. Meth. Fluids 2005; 48:1001-1021. [7] Marchuk G.I. (1982), Methods of Numerical Mathematics, Springer,
New York.
[8] Samarskij A. and Nikolaev E., Numerical methods for Grid Equa- tions, vol. 2, Birkhauser, basel, 1989.
PHN PHÖ LÖC
(C¡c ch÷ìng tr¼nh nguçn tr¶n mæi tr÷íng Matlab) 1. Ch÷ìng tr¼nh kiºm tra b i to¡n Stick-lip
% Chuong trinh giai bai to¡n stick-slip
% Truong hop khong biet truoc nghiem dung % phi la ham ve phai
% b1,b2,b3,b4: la cac gia tri tren bien trai,phai,duoi,tren % Ngay lap 25/11/2015
% Da kiem tra ch½nh xac clear all
clc
teta=0.5;%tham so lap chia mien to=0.95;%tham so lap song dieu hoa a=1; e1=0.5; b=1; k1=1;k2=1; cc=0; count=-1; epxilon=10 (-4);saiso=10; n=6; N=2n;
M=N;
p1=1;p2=M+1;p3=2*M+1;p4=3*M+1;p5=4*M+1;p6=5*M+1;q1=1;q2=N+1;q3=2*N+1; for j=0:N;
csi1(j+1)=0;%khoi tao gia tri lap chia mien eta1(j+1)=0;
end;
for i=0:M;
phi1(i+1)=0; %Khoi tao cho day lap song dieu hoa phi3(i+1)=0; end; h11=e1/M;h21=(a-e1)/M; h12=b/N;h22=b/N; for i=0:2*M; for j=0:N; uluu(i+1,j+1)=0; end; end; thoigian=cputime; while and(count<400,saiso>epxilon); count=count+1; %==================================Giai mien 2=============
% Giai bai toan voi v2
l1=a-e1;l2=b;M2=M;N2=N;n2=n; % Gia tri ve phai v nghiem dung for i=0:M2;
for j=0:N2;
phi(i+1,j+1)=0;% Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N2;
b1(j+1)=csi1(j+1); b2(j+1)=0;
end;
% Dieu kien tren canh duoi va tren for i=0:M2;
b3(i+1)=0; b4(i+1)=0; end;
v2=u1100(phi,b1,b2,b3,b4,l1,l2,k1,k2,cc,M2,N2,n2,p2,p3,q1,q2); % Giai bai toan voi u2
% Gia tri ve phai for i=0:M2;
for j=0:N2;
phi(i+1,j+1)=-v2(p2+i,q1+j); % Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N2;
b1(j+1)=eta1(j+1); b2(j+1)=0;
end;
% Dieu kien tren canh duoi va tren for i=0:M2;
b3(i+1)=-1; b4(i+1)=0; end;
u2=u1100(phi,b1,b2,b3,b4,l1,l2,1,1,0,M2,N2,n2,p2,p3,q1,q2); % Giai bai toan voi v1
l1=e1;l2=b;M1=M;N1=N;n1=n; h11=l1/M1;
h12=l2/N1;
% Gia tri ve phai for i=0:M1;
for j=0:N1;
phi(i+1,j+1)=0; % Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N1;
b1(j+1)=phi1(j+1); b2(j+1)=v2(p2,q1+j); end;
for i=0:M1; b3(i+1)=0;
b4(i+1)=phi3(i+1); end;
v1=u0000(phi,b1,b2,b3,b4,l1,l2,k1,k2,cc,M1,N1,n1,p1,p2,q1,q2); % Giai bai toan voi u1
% Gia tri ve phai for i=0:M1;
for j=0:N1;
phi(i+1,j+1)=-v1(p1+i,q1+j); % Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N1;
x2=0+j*h12;
b1(j+1)=1/2*x2*(3-x22)-1; b2(j+1)=u2(p2,q1+j); end;
% Dieu kien tren canh duoi va tren for i=0:M1; b3(i+1)=-1; b4(i+1)=0; end; u1=u0000(phi,b1,b2,b3,b4,l1,l2,1,1,0,M1,N1,n1,p1,p2,q1,q2); %==============
% Hieu chinh gia tri csi1 tren bien chia 2 mien 1-2 for j=0:N1; ph01(j+1)=0; end; dv1=dx(q1,q2,p2,ph01,v1,h11,h12,k1,k2,M1,-1); csi1=teta*csi1-(1-teta)*dv1;
% Hieu chinh gia tri eta1 tren bien chia 2 mien 1-2 for j=0:N1;
ph01(j+1)=-v1(p2,q1+j); end;
du1=dx(q1,q2,p2,ph01,u1,h11,h12,1,1,M1,-1); eta1=teta*eta1-(1-teta)*du1;
%Hieu chinh gia tri phi1 for j=0:N2; x2=0+j*h12; dh22(j+1)=0; ph003(j+1)=-v1(p1,q1+j); end; du1=dx(q1,q2,p1,ph003,u1,h11,h12,1,1,N,1); phi1=phi1+to*(du1-dh22);
%Hieu chinh gia tri phi3 for i=0:M1; ph0043(i+1)=-v1(p1+i,q2); end; du3=dy(p1,p2,q2,ph0043,u1,h11,h12,1,1,M,-1); phi3=phi3+to*du3; for i=0:M;
for j=0:N;
uxx(p1+i,q1+j)=u1(p1+i,q1+j);%Nghiem xap xi tren mien 1
uxx(p2+i,q1+j)=u2(p2+i,q1+j);%Nghiem xap xi tren mien 2 end; end; saiso=chuan(uluu-uxx) uluu=uxx; end; %======================= thoigian=cputime-thoigian count
%================ Ve do thi tren toan mien for i=0:M; x1=0+i*h11; xx(p1+i)=x1; x1=e1+i*h21; xx(p2+i)=x1; end; for j=0:N; x2=0+j*h12; yy(q1+j)=x2; end; [X, Y]=meshgrid(xx,yy); mesh(X,Y,uxx');
title('Deflection surfaces') xlabel('x')
ylabel('y')
2. Ch÷ìng tr¼nh Stick_lip têng qu¡t % Chuong trinh giai bai to¡n stick_slip_cx
% Truong hop biet truoc nghiem dung % phi la ham ve phai
% b1,b2,b3,b4: la cac gia tri tren bien trai,phai,duoi,tren % udd la nghiem bai toan:
% Ngay lap 24/11/2015 % Da kiem tra ch½nh xac clear all
clc
teta=0.5;%tham so lap chia mien to=0.95;%tham so lap song dieu hoa a=1; e1=0.5; b=1; k1=1;k2=1; cc=0; count=-1; epxilon=10(-4);saiso=10; n=6; N=2n; M=N; p1=1;p2=M+1;p3=2*M+1;p4=3*M+1;p5=4*M+1;p6=5*M+1;q1=1;q2=N+1;q3=2*N+1; for j=0:N;
eta1(j+1)=0; end;
for i=0:M;
phi1(i+1)=0; %Khoi tao cho day lap song dieu hoa phi3(i+1)=0; end; h11=e1/M;h21=(a-e1)/M; h12=b/N;h22=b/N; for i=0:M; for j=0:N; x1=0+i*h11; x2=0+j*h12;
ud(p1+i,q1+j)=u(x1,x2);%Nghiem dung tren Mien 1 x1=e1+i*h21;
x2=0+j*h22;
ud(p2+i,q1+j)=u(x1,x2);%Nghiem dung tren Mien 2 end; end; thoigian=cputime; while and(count<400,saiso>epxilon); count=count+1; %==================================Giai mien 2============= % Giai bai toan voi v2
l1=a-e1;l2=b;M2=M;N2=N;n2=n; % Gia tri ve phai v nghiem dung for i=0:M2;
for j=0:N2; x1=e1+i*h21; x2=0+j*h22;
phi(i+1,j+1)=vp1(x1,x2,cc,k1,k2);% Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N2;
x2=0+j*h22;
b1(j+1)=csi1(j+1); b2(j+1)=dh3x(a,x2); end;
% Dieu kien tren canh duoi va tren for i=0:M2; x1=e1+i*h21; b3(i+1)=delta(x1,0); b4(i+1)=delta(x1,b); end; v2=u1100(phi,b1,b2,b3,b4,l1,l2,k1,k2,cc,M2,N2,n2,p2,p3,q1,q2); % Giai bai toan voi u2
% Gia tri ve phai for i=0:M2;
for j=0:N2;
phi(i+1,j+1)=-v2(p2+i,q1+j); % Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N2;
x2=0+j*h22;
b1(j+1)=eta1(j+1); b2(j+1)=dhx(a,x2); end;
% Dieu kien tren canh duoi va tren for i=0:M2; x1=e1+i*h21; b3(i+1)=u(x1,0); b4(i+1)=u(x1,b); end; u2=u1100(phi,b1,b2,b3,b4,l1,l2,1,1,0,M2,N2,n2,p2,p3,q1,q2); % Giai bai toan voi v1
l1=e1;l2=b;M1=M;N1=N;n1=n; h11=l1/M1;
h12=l2/N1;
% Gia tri ve phai for i=0:M1;
for j=0:N1; x1=0+i*h11; x2=0+j*h12;
phi(i+1,j+1)=vp1(x1,x2,cc,k1,k2); % Ham ve phai end;
% Dieu kien tren canh trai va phai for j=0:N1; x2=0+j*h12; %b1(j+1)=delta(0,x2); b1(j+1)=phi1(j+1); b2(j+1)=v2(p2,q1+j); end;
% Dieu kien tren canh duoi va tren for i=0:M1; x1=0+i*h11; b3(i+1)=delta(x1,0); b4(i+1)=phi3(i+1); end; v1=u0000(phi,b1,b2,b3,b4,l1,l2,k1,k2,cc,M1,N1,n1,p1,p2,q1,q2); % Giai bai toan voi u1
% Gia tri ve phai for i=0:M1;
for j=0:N1;
phi(i+1,j+1)=-v1(p1+i,q1+j); % Ham ve phai end;
end;
% Dieu kien tren canh trai va phai for j=0:N1;
x2=0+j*h12; b1(j+1)=u(0,x2);
b2(j+1)=u2(p2,q1+j); end;
% Dieu kien tren canh duoi va tren for i=0:M1; x1=0+i*h11; b3(i+1)=u(x1,0); b4(i+1)=u(x1,b); end; u1=u0000(phi,b1,b2,b3,b4,l1,l2,1,1,0,M1,N1,n1,p1,p2,q1,q2); %==============
% Hieu chinh gia tri csi1 for j=0:N1; x2=0+j*h12; ph01(j+1)=vp1(e1,x2,cc,k1,k2); end; dv1=dx(q1,q2,p2,ph01,v1,h11,h12,k1,k2,M1,-1); csi1=teta*csi1-(1-teta)*dv1;
% Hieu chinh gia tri eta1 for j=0:N1;
ph01(j+1)=-v1(p2,q1+j); end;
du1=dx(q1,q2,p2,ph01,u1,h11,h12,1,1,M1,-1); eta1=teta*eta1-(1-teta)*du1;
%Hieu chinh gia tri phi1 for j=0:N2;
dh22(j+1)=dhx(0,x2); ph003(j+1)=-v1(p1,q1+j); end;
du1=dx(q1,q2,p1,ph003,u1,h11,h12,1,1,N,1); phi1=phi1+to*(du1-dh22);
%Hieu chinh gia tri phi3 for i=0:M1; x1=0+i*h11; dh33(i+1)=dhy(x1,b); ph0043(i+1)=-v1(p1+i,q2); end; du3=dy(p1,p2,q2,ph0043,u1,h11,h12,1,1,M,-1); phi3=phi3+to*(du3+dh33); for i=0:M; for j=0:N;
uxx(p1+i,q1+j)=u1(p1+i,q1+j);%Nghiem xap xi tren mien 1 uxx(p2+i,q1+j)=u2(p2+i,q1+j);%Nghiem xap xi tren mien 2 end; end; saiso=chuan(ud-uxx) end; %======================= thoigian=cputime-thoigian count %===================================