MATLAB: Nonlinear equation solver issue

fsolvenonlinear

Hello, I have an equation
N = (DP/RT * ln(P/(P-P*)))/(L/A) + (l/a))
P,R,T,P*,(L+l), and A are known. L+l is always .05
There is some experimental data
l = .004, N = 6.6e-11
l = .008, N = 4.9e-11
l = .012, N = 4.0e-11
I am trying to find D and a. So I make a function and I try to use fsolve, but when I do I just get an erroneous answer. I am saying D(1) = D, and D(2) = a in my function
function F = stefanflow(D)
P = 1e5;
T = 294;
R = 8.314e3;
A = 1e-4;
P_sat = 1.12e4;
O = log(P/(P-P_sat));
I = P/(R*T);
F = zeros(3,1);
F(1) = 6.6e-11 - (D(1)*I*O)/(((.05-.004)/A)+(.004/D(2)));
F(2) = 4.9e-11 - (D(1)*I*O)/(((.05-.008)/A)+(.008/D(2)));
F(3) = 4.0e-11 - (D(1)*I*O)/(((.05-.012)/A)+(.012/D(2)));
I run this in my script file
options = optimoptions('fsolve','Display','iter');
[D,fval] = fsolve(@stefanflow,[1e-5;1e-5],options);
And this is my output
Warning: Trust-region-dogleg algorithm of FSOLVE cannot handle
non-square systems; using Levenberg-Marquardt algorithm instead. 
> In fsolve at 286 
                                          First-Order                    Norm of 
   Iteration  Func-count    Residual       optimality      Lambda           step
       0           3     2.59598e-22        1.19e-16         0.01
       1           6     2.59598e-22        1.19e-16        0.001    1.37935e-14
Equation solved, fsolve stalled.
fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance and the vector of function values
is near zero as measured by the default value of the function tolerance.
<stopping criteria details>
EDU>> D
D =
     1.0e-04 *
      0.1000
      0.1000
What is going on, and how can I solve this system?

Best Answer

The solution fsolve gave you appears to satisfy the equations very well,
>> stefanflow(D)
    ans =
       1.0e-11 *
        0.9493
        0.9167
        0.9243
So, either you have incorrectly coded your equations, or they have multiple solutions that you don't expect.
You could also try making the tolerances more strict,
   options = optimoptions('fsolve','Display','iter','TolFun',1e-30,'TolX',1e-30);
When I try this, it produces a small change in the solution,
    D =
       1.0e-04 *
        0.1013
        0.1391
but maybe that's significant to you.
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