fsolve解非线性方程组，只迭代了8次就退出了，得到的结果进行复算不等于0，存在较大的误差。(是不是需要修改算法？)

运行程序得到的结果提示如下：

x =

1.0e+003 *

Columns 1 through 4

0.008229968721187   0.000000000008083   0.000006858621379   1.206234368839563

Column 5

0.000029820218392

fval =

-0.003404951183545

0.109922506812515

0.055200484333056

0.009207767273016

-0.000813641543979

exitflag =

-2

output =

iterations: 8

funcCount: 54

algorithm: 'trust-region dogleg'

firstorderopt: 3.163833174781097e+003

message: [1x775 char]

ans =

No solution found.

fsolve stopped because the relative size of the current step is less than the

selected value of the step size tolerance squared, but the vector of function values

is not near zero as measured by the selected value of the function tolerance.

Stopping criteria details:

fsolve stopped because the relative norm of the current step, 2.142692e-011, is less than

max(options.TolX^2,eps) = 2.500000e-011. However, the sum of squared function values,

r = 1.522709e-002, exceeds sqrt(options.TolFun) = 1.000000e-025.

Optimization Metric                                               Options

relative norm(step) =  2.14e-011                max(TolX^2,eps) =  3e-011 (selected)

r =  1.52e-002                                    sqrt(TolFun) = 1.0e-025 (selected)

程序代码如下：(其中调用了mafun函数，mafun函数中包括五个方程组，Iph，Io，Rs，Rp，Vt是要求的参数，已经赋初值，可以保证初值选择正确，因为所有解的范围是可以估计到的)

options = optimset('Display','iter','TolFun',1e-50,'TolX',0.5e-5);

[x,fval,exitflag,output] = fsolve(@mafun,x0,options)

output.message

format long, x,

如果我将程序中的optimset括号中的“0.5e-5”改为“0.5e-20”，则得到的结果为(好像说是变成奇异方程了)：

No solution found.

fsolve stopped because the problem appears to be locally singular.

Stopping criteria details:

fsolve stopped because the trust-region radius, 1.248209e-016, is less than 2*eps.

Optimization Metric                                             Options

trust-region radius =  1.25e-016                              2*eps = 4.4e-016

上面的问题已经说的很详细了，希望高手能帮忙解决。谢谢了。 只有那么多Q币，全给了。

[Last edited by mataohk on 2012-9-30 at 09:32]