How to solve 4 equations with 4 unknowns with bounds?

Question

KY 2019-4-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/457367-how-to-solve-4-equations-with-4-unknowns-with-bounds

回答： Alex Sha 2019-5-16

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 1

Alan Weiss 2019-4-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/457367-how-to-solve-4-equations-with-4-unknowns-with-bounds#answer_371679

You set options for fsolve, but then call lsqnonlin. This is a mistake.
You do not pass options to the solver. This might be a mistake.
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

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KY 2019-4-23

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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