using lsqnonlin with constraints

clc
clear all

%read in text data
D=readmatrix('Treloar_data.xlsx');
stretch=D(1:end,1); %this is lambda
lambda=stretch;
stress=D(1:end,2); %this is stress
T_0=stress;
u=.5673; %small strain shear modulus, mu

%constrained
fun=@(x)(2*(lambda-lambda.^-2).*(x(1)+(lambda.^-1).*x(2))-T_0); 
x0=[-100000,100000]; %initial guess
%constraints
% A=2*(c(1)+c(2))==u  this is the constraint i want to use
lb=[0,0]
ub=[10,10]
x=lsqnonlin(fun,x0,lb,ub);
c(1)=x(1)
c(2)=x(2)
T_MR=2*(lambda-lambda.^-2).*(c(1)+(lambda.^-1).*c(2));

figure
plot(lambda, T_0,'o'), xlabel('stretch'), ylabel('stress'), title('UT')
hold on
plot(lambda, T_MR)

i want to employ a constraint that involves both constants if possible as labeled

Answer 1

[Torsten]

% A=2*(c(1)+c(2))==u this is the constraint i want to use

Then optimize with one parameter c(1) and insert c(2) = u/2 - c(1) for the second parameter in your model function.

And if you want the solution to be in [0 10], why do you supply an initial guess of -100000 and 100000 ?

[John D'Errico]

There are bounds of 0 and 10 on both variables. However, if you use the "constraint" to eliminate one of the variables, With that value of u though, consider what MAY happen.

u=.5673;

If you use this to eleminate one of the unknowns using this constraint,

2*(c(1) + c(2)) = u

so then we will have

c(1) = u/2 - c(2)

But c(2) can be any number within 0 and 10. And that means if c(2) is greater than u/2, so roughly 0.28, then c(1) will be negative. And that would fail the bound constraints.

So you CANNOT use this trick to eliminate one unknown.

[Torsten]

Because of the constraint 2*(c(1)+c(2))==u, one can set lb = 0 and ub = u/2 for c(1) (note here that the lower bound for c(2) is also 0). Then I think one can proceed as suggested.

[John D'Errico]

Yes. That will suffice.

Answer 2

[Jon]

From my understanding lsqnonlin only allows for bound constraints on the components of x. For more general problems you will have to use fmincon

[Jon]

I didn't look at the details of your constraint. I would follow @Torsten's suggestion to see if you can in fact simplify the problem to eliminate the additional constraint.