need convergence to, x1=5.99, x2=5.02 and x3=19.05
FSOLVE GIVES SAME VALUE
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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 Commenti
Alan Stevens
il 27 Nov 2020
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.
Risposte (1)
Matt J
il 27 Nov 2020
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)
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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