FSOLVE GIVES SAME VALUE

clc;clearvars;close all; format short g;format compact;
tfinal=30;
pars.D=0.00005611;
pars.x2f=20;
pars.Y=0.4;
pars.beta=0.000055;
pars.k1=0.04545;
pars.alpha=2.2;
pars.mumax=0.000133;
pars.km=1.2;
pars.x3max=50;
csol2=fsolve(@(c) chemofun(c,pars),[5 4.5 15]);
function f=chemofun(c,pars)
f=zeros(3);
x1=c(1);
x2=c(2);
x3=c(3);
mumax=pars.mumax;
x3max=pars.x3max;
km=pars.km;
k1=pars.k1;
mu=((mumax)*x2*(1-(x3/(x3max))))/(km+x2+(x2^2)*(k1));
D=pars.D;
x2f=pars.x2f;
Y=pars.Y;
A=pars.alpha;
B=pars.beta;
f(1)=(mu-(D));
f(2)=(D)*(x2f-x2)-(mu*x1)/(Y);
f(3)=(-1)*(D)*x3+((A)*mu+B)*x1;
end

4 Comments

need convergence to, x1=5.99, x2=5.02 and x3=19.05
You are trying to solve for x1, x2 and x3, but they only appear in two equations. There can be an infinite number of solutions.
Mu have x1
Actually, it seems to have x2 and x3; but you are right, three equations are involved. The results seem very sensitive to the initial guesses though.

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Answers (1)

clc;clearvars;close all; format short g;format compact;
tfinal=30;
pars.D=0.00005611;
pars.x2f=20;
pars.Y=0.4;
pars.beta=0.000055;
pars.k1=0.04545;
pars.alpha=2.2;
pars.mumax=0.000133;
pars.km=1.2;
pars.x3max=50;
fun=@(c) chemofun(c,pars);
opts=optimoptions('fsolve','StepTolerance',1e-12,'FunctionTolerance',1e-12,'OptimalityTolerance',1e-12);
[csol2,fsol2]=fsolve(fun,[5 4.5 15],opts)
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
csol2 = 1×3
5.9905 5.0238 19.051
fsol2 = 1×3
-8.9331e-12 1.1261e-10 -9.9097e-11
function f=chemofun(c,pars)
f=zeros(1,3); %<-------
x1=c(1);
x2=c(2);
x3=c(3);
mumax=pars.mumax;
x3max=pars.x3max;
km=pars.km;
k1=pars.k1;
mu=((mumax)*x2*(1-(x3/(x3max))))/(km+x2+(x2^2)*(k1));
D=pars.D;
x2f=pars.x2f;
Y=pars.Y;
A=pars.alpha;
B=pars.beta;
f(1)=(mu-(D));
f(2)=(D)*(x2f-x2)-(mu*x1)/(Y);
f(3)=(-1)*(D)*x3+((A)*mu+B)*x1;
end

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