fsolve error using trustnleqn (line 28)

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Bertiningrum Devi 2019-5-10

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评论： Bertiningrum Devi 2019-5-19

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

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Answer 1

Matt J 2019-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/461369-fsolve-error-using-trustnleqn-line-28#answer_374465

编辑：Matt J 2019-5-10

在 MATLAB Online 中打开

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.

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Walter Roberson 2019-5-11

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 2019-5-19