fsolve and current step

hello. i have a problem with fsolve.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
fsolve stopped because the relative norm of the current step, 1.117892e-17, is less than
max(options.StepTolerance^2,eps) = 2.220446e-16. However, the sum of squared function
values, r = 7.870134e-09, exceeds sqrt(options.FunctionTolerance) = 1.000000e-25.

Optimization Metric                                           Options
relative norm(step) =   1.12e-17              max(StepTolerance^2,eps) =   2e-16 (selected)
r =   7.87e-09                                 sqrt(FunctionTolerance) =  1.0e-25 (selected)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

any one can help me???

Answer 1

[Walter Roberson]

Your function is steeper than can be handled with the fsolve options you are using. fsolve has found two points separated by 1e-17 where there is a zero between the two, but the difference in function values is more than 1e-25.

I would suspect that you passed options to fsolve telling it to use a function tolerance of 1e-50, which is way way too small a tolerance unless you know your function values have absolute magnitude below 1e-35 or so.

[alex alexia]

thank you for your answer.

 i use this code.but i dont know where to change.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
fun = @root3d;
>>options=optimset('Display','iter','MaxFunEvals',1e10,'MaxIter',1e10,'TolFun',1e-50,'TolX',1e-50);
>> x0 = [1e6,11];
>> x = fsolve(fun,x0,options)

function F = root3d(x)
l=1.23e-9;
ntr=1.23e12;
tph=2.58e-12;
vgr=9e9;
gama=0.02;
beta=1e-5;
B=1.33e-10;  %Recombination coefficient
q = 1.602e-19;% e = electron charge
A=9e-6;
tnr=100e-9;
b=5e-5;
c=6.25e-18;

I=1e-11;
 F(1) =(x(2)/tph)-(vgr*gama*x(2)*(l*(x(1)-ntr)))-(beta*B*(x(1)^2));
 F(2) =(I/(q*A))-(x(1)/tnr)-(b*(x(1)^2))-(c*(x(1)^3))-(vgr*gama*(l*(x(1)-ntr))*x(2));

[Walter Roberson]

If you use the symbolic toolbox, you can get closed-form solutions. They are approximately

                x1                    x2
           2980663501499.4         -1649993401.34255
          693575.067371611      9.69500687359391e-16
         -205267520134.293      7.94471142311242e-05
         -7794733173260.32        0.0338721889970218

You only used single location initialization (1e6 and 11 respectively) so you could end up getting any of those when you work numerically.

However, the function is so steep that if you take the double-precision approximations from above and substitute them into F, then you will get results that can be quite different from 0:

             F1 value                  F2 value
          47612.8298891023          30417.6297938751
      2.51664170452454e-20      0.000450530583097153
       2.9726319961552e-10          26.1923721370618
     -2.38857466441952e-07          120080.281965203

If you take eps() of the closed-form solutions for x1 and x2, you will find that for the 4th of the solutions, the adjacent representable numbers near that x1 are separated by 0.0009765625 . But for some reason you are expecting to be able to have fsolve calculate down to a function precision of 1E-50 .

Your TolFun', 1e-50 and 'TolX' 1e-50 simply cannot be justified numerically. If you need your values to be that precise, you need to switch to the symbolic toolbox.

I would, however, remind you that none of your input coefficients have more than 3 significant digits. For example, what does

ntr=1.23e12;

convey? Does it mean that ntr is to be exactly 1230000000000, or does it mean that the true value of ntr is somewhere between 1225000000000 and 1234999999999.99999999999 ? When you only have 3 significant figures for your coefficients, you are going to get out numeric garbage if you ask for solutions beyond (at best) 4 significant figures.