105刘晓燕.doc

文档介绍：105刘晓燕数值线性代数论文1140501105刘晓燕题目:2dydy考虑两点边值问题:,,,,a,0,a,12dxdxy(0),0,y(1),11,a,x/,y,(1,e),ax容易知道它的精确解为,1/,1,e1x,ih为了把微分方程离散,把区间n等分,令,,,[0,1]i,1,2,...,n,1h,iny,2y,yy,yi,1ii,1i,1i,,,a得到差分方程2hh2(,,h)y,(2,,h)y,,y,ah简化为i,ii,11从而离散后得到的线性方程组的系数矩阵为,,,(2,h),h,,,,,,,,(2,h),h,,,,,,A,,(2,h)?,,,,h??,,,,,,,(2,h),,1对,分别用Jacobi,GS和超松弛迭代方法求线性方程组的解,要求有4,,1,a,,n,202位有效数字,然后比较与精确解的误差。对,考虑同样的问题。,,,,,,,,(,bA,D,L,U对于非奇异线性方程组,令,其中D,diag(a11,a22,a33,...,ann)12131n00,,,aa?a,,,,,,,,21232n,a00,,a?a,,,,,,,,L,,U3132,a,a00??,,,,?????an,1,n,,,,,,,,,,an1,an2...,an,n,100,,,,则原方程组可改写成:x,Bx,g,其中JJ,,11B,D(L,U),g,Db,JJ(0)(0)(0)Tx,(x,x,?,x)给定初始向量:,012n构造迭代公式x,Bx,g,k,1,2,3?kJk,1Jn1()(,1)kk其分量形式为x,(ax,b),11jj1a,2j11i,1n1k(k,1)(k,1),x,(ax,ax,b)i,2,?,n,1,,iijjijjiaj,,11j,iiin,11(k)(k,1)x,(ax,b),nnjjnaj,1nnGauss―(0)(0)(0)Tx,(x,x,?,x)给定初始向量011n,1,1B,(D,L)Ug,(D,L)b令,GSGSx,Bx,g,k,1,2,3?则迭代公式为kGSk,1GSn1()(,1)kkx,(ax,b)其分量形式为,11jj1a,2j11i,1n1k(k,1)(k,1)x,(ax,ax,b),i,2,?,n,1,,iijjijjiaj,,11j,iiin,11(k)(k,1)x,(ax,b),nnjjnaj,(SOR迭代法)(k)(k,1)(k)(k)x,(1,,)x,,xSOR迭代法是对GS迭代法的加权平均改造,即,为GS迭x(k)(k,1),1(k),1(k,1),1x,(1,,)x,,(DLx,DUx,Db)代解,即i,1n1(k)(k,1)(k)(k)它的分量形式为x,(1,,)x,,(ax,ax,b),,iiijjijjiaj,,11j,iii,,1,,1,,1其中为松弛因子,当时称为超松弛,时称为低松弛,时就是GS迭代.,二(实验方法,,1,,1n,201,,,时,,,,1n,202,,,时,,,,1n,203,,,时,,1n,20,,,,,时三、,迭代次数k与精确解的误差||x,x||,k*,,1222