How to solve 4 equations with 4 unknowns with bounds?

KY (view profile)

on 19 Apr 2019
Latest activity Answered by Alex Sha

Alex Sha (view profile)

on 16 May 2019
Accepted Answer by Alan Weiss

Alan Weiss (view profile)

Hi everyone, I'm trying to solve this but the message displayed local minimum found and also the values of the x generated did not match with the boundary which was set ie. x(1)+x(2) = Ym. It also says lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.Could anyone help with this? Many thanks!
clear; clc;
x0 = [50; 50; 50; 50;];
options = optimoptions('fsolve','Display','iter');
%[x,fval] = fsolve(@myfun,x0,options)
lb = [0; 0; 0; 0;];
ub = [100; 100; 100; 100;];
x = lsqnonlin(@myfun,x0,lb,ub)
myfun(x)
function F = myfun(x)
A = 87.3145;
B = -0.289762;
C = 0.0000199677;
alpha = 0.4705;
Xm = 20;
Ym = 27.5;
F = [x(1)+x(2)-Ym;
x(3)+x(4)-Xm;
A*exp(B*x(3)^0.5 - C*x(3)^3) - x(1);
A*exp(B*x(4)^0.5 - C*x(4)^3) - x(2);
Ym/alpha - ((1-alpha)/alpha)*x(2) - x(1);
Xm/alpha - ((1-alpha)/alpha)*x(4) - x(3);
]
end

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Answer by Alan Weiss

Alan Weiss (view profile)

on 22 Apr 2019

1. You set options for fsolve, but then call lsqnonlin. This is a mistake.
2. You do not pass options to the solver. This might be a mistake.
3. You have six equations in four unknowns. Generally, you should not expect a solution to such a system, only a point that is a local minimum of the sum of squares.
Alan Weiss
MATLAB mathematical toolbox documentation

KY

KY (view profile)

on 23 Apr 2019
Thank you Alan. I've edited it, add a tighter constraint based on my situation, and it works.
x0 = [10; 10; 10; 10;];
%options = optimoptions('fsolve','Display','iter');
%[x,fval] = fsolve(@myfun,x0,options)
lb = [5.31; 5.31; 0.68; 0.68;];
ub = [68.80; 68.80; 37.24; 37.24;];
x = lsqnonlin(@myfun,x0,lb,ub)
myfun(x)
function F = myfun(x)
A = 87.3145;
B = -0.289762;
C = 0.0000199677;
alpha = 0.3935;
Xm = 20;
Ym = 22.5;
F = [ A*exp(B*x(3)^0.5 - C*x(3)^3) - x(1);
A*exp(B*x(4)^0.5 - C*x(4)^3) - x(2);
Ym/alpha - ((1-alpha)/alpha)*x(2) - x(1);
Xm/alpha - ((1-alpha)/alpha)*x(4) - x(3);
]
end

Answer by Alex Sha

Alex Sha (view profile)

on 16 May 2019

Refer the results below:
x1: 7.73410323214524
x2: 32.0801819920043
x3: 33.4573119565274
x4: 11.2688338748665