Nonlinear equation solver issue

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?

[John D'Errico]

Fsolve is happy with the starting values. It told you that fact. They meet the convergence criteria, so why look beyond that point?

Answer 1

[Matt J]

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.

[Rick]

But why are there 3 elements in D? I only have two unknowns

[Alan Weiss]

I only see two elements in D. The first is 1e-4*0.1013, the second is 1e-4*0.1391.

Alan Weiss

MATLAB mathematical toolbox documentation

[John D'Errico]

Rick - I think you miss that the first "number" shown is not an element of D, but an exponent applied to those numbers. Note the little * after that number.