fsolve error using trustnleqn (line 28)

Hello everyone,

I'm trying to solve non linear equations system using fsolve. I'm beginner in matlab I just have no idea what's going on with my program when I tried to run it, there's an error message in the command window:

Error using trustnleqn (line 28)
Objective function is returning undefined values at initial point. FSOLVE cannot continue.

Error in fsolve (line 408)
        trustnleqn(funfcn,x,verbosity,gradflag,options,defaultopt,f,JAC,...

Error in singletray (line 10)
x = fsolve(fun,x0);

what does that mean? and what am I supposed to do to get it solved?

here is the function file:

function F = equilibrium(x)
rec_a=0.7;
rec_b=0.9;

HCOin=1.72e-8;
HSin=0.1128;
a=(1-rec_a)*HCOin;
b=(1-rec_b)*HSin;

F(1)=x(1)*a/(x(2)*x(3)) - 32.1421;
F(2)=x(1)*b/(x(2)*x(4)) - 16.7567;
F(3)=x(5)/(a*x(6)) - 3.508e-13;
F(4)=x(1)*x(6)/x(2) - .32134475;
F(5)=x(7)/(x(8)*x(9)) - 2.246e10;
F(6)=x(9)*x(6) - 7.573e-11;
F(7)=x(1)+x(9)+x(7)-b-a-2*x(5)-x(6);
F(8)=2.93-x(8)-x(7);
F(9)=28.27-x(2)-x(1);
end

I run it in :

%Reactive desorption modeling
%assumption: isothermal & only 1 tray

clc;
clear;

%solving 7 unknowns
options = optimoptions('fsolve','Display','iter');
[x,fval] = fsolve(@equilibrium,[0;0;0;0;0;0;0;0;0],options);

Any help would be appreciated. Thank you so much

Answer 1

[Matt J]

The initial point is [0;0;0;0;0;0;0;0;0]. Calling your function at this point gives something with NaNs

>> equilibrium([0;0;0;0;0;0;0;0;0])

ans =

       NaN       NaN       NaN       NaN       NaN   -0.0000   -0.0113    2.9300   28.2700

because of divisions by zero. You must use a different initial point, where your function is actually defined.

[Matt J]

You might also want to consider using lsqnonlin instead of fsolve, so that you can put upper and lower bounds on the x(i) that need to be kept away from zero.

[Bertiningrum Devi]

Thank you for your help

so the initial points influences the result. actually I have no idea of in what range my ans will be, any tips to choose the initial points?

[Matt J]

It does influence the result, but the problem here is that your guess can't even be evaluated. Your objective function has no legitimate value there.

You must have some approximate idea on physical grounds of what the result should be. How else will you know if an answer given to you makes any sense?

[Walter Roberson]

The terms used in the variable names and comments are seldom associated with negative values, so you can probably impose a lower bound of 0 for all of you values.

You divide by several of your x values, so you can probably impose a lower bound of realmin() for those, or more likely sqrt(realmin)

[Bertiningrum Devi]

Thank you Matt, Thank you Walter, your information helps me so much.