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 handlenon-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-14Equation solved, fsolve stalled.fsolve stopped because the relative size of the current step is less than thedefault value of the step size tolerance and the vector of function valuesis near zero as measured by the default value of the function tolerance.<stopping criteria details>EDU>> DD = 1.0e-04 * 0.1000 0.1000`

What is going on, and how can I solve this system?

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