将TFQMR与矩阵或函数手柄输入输入

This example shows how to usetfqmrwith a matrix input and with a function input.

n = 100;on =一个（n，1）;a = spdiags（[-2*on 4*on-on]，-1：1，n，n）;b = sum（a，2）;TOL = 1E-8;最大值= 15;m1 = spdiags（[on/（-2）on]，-1：0，n，n）;m2 = spdiags（[4*on -on]，0：1，n，n）;x = tfqmr（a，b，tol，maxit，m1，m2，[]）;

您还可以将矩阵矢量产品功能作为输入：

函数y = afun（x，n）y = 4 * x;y（2：n）= y（2：n）-2 * x（1：n -1）;y（1：n-1）= y（1：n-1）-x（2：n）;x1 = tfqmr（@（x）afun（x，n），b，tol，maxit，m1，m2）;

如果申请是适合与QMR, it may be used withtfqmrby wrapping it in an anonymous function:

x1 = tfqmr（@（x）applaop（x，'notransp'），b，tol，maxit，m1，m2）;

Using tfqmr with a Preconditioner

此示例演示了使用预处理。

LoadA = west0479, a real 479-by-479 nonsymmetric sparse matrix.

loadwest0479; A = west0479;

Definebso that the true solution is a vector of all ones.

b = full（sum（a，2））;

设置公差和最大迭代次数。

tol = 1e-12; maxit = 20;

采用tfqmrto find a solution at the requested tolerance and number of iterations.

[x0，fl0，rr0，it0，rv0] = tfqmr（a，b，tol，maxit）;

fl0is 1 becausetfqmr不收敛到请求的公差1e-12之内the requested 20 iterations. The seventeenth iterate is the best approximate solution and is the one returned as indicated byit0 = 17。MATLAB®将剩余历史存储在rv0。

Plot the behavior oftfqmr。

半学（0：Maxit，RV0（1：Maxit+1）/Norm（b），'-o'); xlabel('迭代编号'); ylabel(“相对残留”);

Note that likebicgstab,tfqmrkeeps track of half iterations. The plot shows that the solution does not converge. You can use a preconditioner to improve the outcome.

用伊卢, since the matrixAis nonsymmetric.

[L,U] = ilu(A,struct('type','ilutp','droptol',1e-5));

Error using ilu There is a pivot equal to zero. Consider decreasing the drop tolerance or consider using the 'udiag' option.

MATLAB无法构建不完整的LU，因为它会导致一个奇异的因素，这是无用的预处理。

如错误消息所示，您可以使用降低的降低公差重试。

[L,U] = ilu(A,struct('type','ilutp','droptol', 1 e-6));(x1, fl1 rr1、it1 rv1] =tfqmr(A,b,tol,maxit,L,U);

fl1is 0 becausetfqmrdrives the relative residual to4.1410E-014(the value ofRR1). The relative residual is less than the prescribed tolerance of1e-12at the sixth iteration (the value ofit1) when preconditioned by the incomplete LU factorization with a drop tolerance of1e-6。输出RV1（1）isnorm(b), and the outputrv1(7)isnorm(b-A*x2)。

You can follow the progress oftfqmr通过从初始估计开始（迭代号0）开始在每次迭代处绘制相对残差。

半学（0：0.5：IT1，RV1/Norm（b），'-o'); xlabel('迭代编号'); ylabel(“相对残留”);