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

댓글 수: 4

JYOTI PRAKASH BEHERA
JYOTI PRAKASH BEHERA 2020년 11월 27일
need convergence to, x1=5.99, x2=5.02 and x3=19.05
Alan Stevens
Alan Stevens 2020년 11월 27일
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.
JYOTI PRAKASH BEHERA
JYOTI PRAKASH BEHERA 2020년 11월 27일
Mu have x1
Alan Stevens
Alan Stevens 2020년 11월 27일
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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답변 (1개)

Matt J
Matt J 2020년 11월 27일

1 개 추천

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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2020년 11월 27일

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2020년 11월 27일

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