数值分析上机实验

数值分析实验一数值求解正方形域上的Poisson方程边值问题班级：数值分析1班学号：2012214082姓名：郭璟专业：安全技术及工程算法一：Jacobi点迭代法迭代格式：u—u—u—u—00,j N+1,j i,0 i,N+1forj-1:Nfori—1:Nu(k+1)—(u(Ik)+u(k)+u(k)+u(k)+h2f)/4i,j i-1,j i+1,j i,j-1 i,j+1 iendendfunction[v,k]=xsjacobi(n)%xsjacobi：用点Jacobi迭代法求解线性方程组A*u=f%v：方程组的解；k：迭代次数；n：非边界点数%e：允许误差界；er：迭代误差；f(2:n+1,2:n+1)=(n+1)人(-2)*2;v=zeros(n+2,n+2);u=zeros(n+2,n+2);e=0.000000001;fork=1:1000 %迭代求解er=0;forj=2:n+1fori=2:n+1u(i,j)=(v(i-1,j)+v(i+1,j)+v(i,j-1)+v(i,j+1)+f(i,j))/4;er=er+abs(v(i,j)-u(i,j)); %估计误差endendifer/nA2<e,break;end %判断是否达到计算精度，如果达到则退出循环v=u;end>>n=9;>>[v,k]=xsjacobi(n),surf(v)v=0000000000000.02560.04130.05080.05600.05770.05600.05080.04130.0256000.04130.06860.08590.09550.09860.09550.08590.06860.0413000.05080.08590.10880.12160.12580.12160.10880.08590.0508000.05600.09550.12160.13640.14120.13640.12160.09550.05600

0 0.05770.09860.12580.14120.14620.14120.12580.09860.057700 0.05600.09550.12160.13640.14120.13640.12160.09550.056000 0.05080.08590.10880.12160.12580.12160.10880.08590.050800 0.04130.06860.08590.09550.09860.09550.08590.06860.041300 0.02560.04130.05080.05600.05770.05600.05080.04130.0256000000000000k=3040.20.150.10.0515101000.20.150.10.05151010015>>tic>>[v,k]=xsjacobi(n)；>>ti=tocti=51.1720算法二：块Jacobi迭代迭代格式：v=v=0,0N+1forj=1:NAV(k+1)=v(k)+v(k)+b (可用追赶法求解)jjj j-i j+i jendfunction[v,k]=xsbjacobi(n)%xsbjacobi：用块Jacobi迭代法求解线性方程组A*u=f%u：方程组的解；k：迭代次数；n：非边界点数%a：方程组系数矩阵的下对角线元素；b：方程组系数矩阵的主对角线元素；%c：方程组系数矩阵的上对角线元素；d：追赶法所求方程的右端向量；%e：允许误差界；er：迭代误差；f=2*1/(n+1)人2*ones(n+2,n+2);a=-1*ones(1,n);b=4*ones(1,n);c=-1*ones(1,n);u=zeros(n+2,n+2);v=zeros(n+2,n+2);e=0.00001;fork=1:2000er=0;forj=2:n+1%%Ub=u(:,j);d(1:n)=f(2:n+1,j)+v(2:n+1,j-1)+v(2:n+1,j+1);x=zg(a,b,c,d); %用追赶法求解u(2:n+1,j)=x';er=er+norm(v(:,j)-u(:,j),1);endifer/nA2<e,break;end %判断是否达到计算精度，如果达到则退出循环v=u;end%追赶法求解三对角方程组functionx=zg(a,b,c,d)n=length(b);%LU分解。u(1)=b(1);fork=2:n讦u(k-1)==0,D=0,return;endl(k)=a(k)/u(k-1);u(k)=b(k)-l(k)*c(k-1);end%追赶法求解之追过程，求解Ly=d。y(i)=d(i);fork=2:ny(k)=d(k)-l(k)*y(k-1);end%追赶法求解之赶过程，求解Ux=y讦u(n)==0,D=0,return;endx(n)=y(n)/u(n);fork=n-1:-1:1x(k)=(y(k)-c(k)*x(k+1))/u(k);end>>n=9;>>[v,k]=xsbjacobi(n)

v=0000000000000.02560.04120.05070.05590.05760.05590.05070.04120.0256000.04120.06850.08580.09540.09850.09540.08580.06850.0412000.05070.08580.10870.12150.12560.12150.10870.08580.0507000.05590.09540.12150.13620.14100.13620.12150.09540.0559000.05760.09850.12560.14100.14600.14100.12560.09850.0576000.05590.09540.12150.13620.14100.13620.12150.09540.0559000.05070.08580.10870.12150.12560.12150.10870.08580.0507000.04120.06850.08580.09540.09850.09540.08580.06850.0412000.02560.04120.05070.05590.05760.05590.05070.04120.0256000000000000k=69>>tic>>[u,k]=xsbjacobi(n)；>>ti=tocti=17.2500四、 实验结果分析实验用ti・・toc来实现计算所花费时间的计算，通过比较可知Jacobi点迭代比Jacobi块迭代更费时