Inverse problem, Overdetermined, Nonlinear

consider we have this five equations

A(i) = SA/((Kf(i) - L) * phi + (Ka - N));

B(i) = SB/((Kf(i) - L) * phi + (Ka - N));

C(i) = SC/((Kf(i) - L) * phi + (Ka - N));

F(i) = SF/((Kf(i) - L) * phi + (Ka - N));

D(i) = SD/((Kf(i) - L) * phi + (Ka - N));

phi,Ka are known, single value and constant.

SA, SB, SC, SF,SD, L, N are our 7 unknowns, they are scalar and single value, across the equations they all are constant and the do not change with varing Kf

if we vary Kf from 1:3,3 we have

3 values for each A, B, C, F, D

15 equations and 7 unkowns:

A(Kf = 1) = S_A/((Kf(1) - L) * phi + (Ka - N));.

A(Kf = 2) = S_A/((Kf(2) - L) * phi + (Ka - N));

A(Kf = 3) = S_A/((Kf(3) - L) * phi + (Ka - N));

.

.

.

D(Kf = 3) = S_D/((Kf(3) - L) * phi + (Ka - N));

if we vary Kf from 1:5,5 we have

5 values for each A, B, C, F, D

25 equations and 7 unkowns

...

if we vary Kf from 1:20,20 we have

20 values for each A, B, C, F, D

100 equations and 7 unkowns

...

and we have A_measured, B_measured, C_measured, F_measured, D_measured

which they are inputs

I want to calculate these 7 unkowns with many equations that I have, which works for all equations

I want in final

coditions:

measured - predicted = 10^-12

[John D'Errico]

Do you KNOW a solution exists ith the desired accuracty? Or do you just want that? Given that you have measured data, asking for a tight tolerance is likely to be a waste of effort. Is your data measured that accurately?

Answer 1

[Matt J]

You would use fsolve or, if you want specific constraints on the unknowns, e.g., bounds, you would use lsqnonlin.

You could also set the problem up using eqnproblem, and let the problem-based optimization engine choose the best solver for you.

SA = optimvar('SA');
SB = optimvar('SB'); 
SC = optimvar('SC');
SD = optimvar('SD');
SF = optimvar('SF');
L = optimvar('L');
N = optimvar('N');

prob = eqnproblem;
for i=1:nnumEquations
    prob.Equations.eqA(i) = A(i) == SA/((Kf(i) - L) * phi + (Ka - N));
    prob.Equations.eqB(i) = B(i) == SB/((Kf(i) - L) * phi + (Ka - N));
    prob.Equations.eqC(i) = C(i) == SC/((Kf(i) - L) * phi + (Ka - N));
    prob.Equations.eqF(i) = F(i) == SF/((Kf(i) - L) * phi + (Ka - N));
    prob.Equations.eqD(i) = D(i) == SD/((Kf(i) - L) * phi + (Ka - N));
end


sol=solve(prob, initialGuess);

I want in final coditions: measured - predicted = 10^-12

That cannot be guaranteed. Because it is an overdetermined system, the solution will have to be a least squares solution. The minimized least squared error will be whatever the minimization problem permits. You have no control over it.

[Torsten]

@Matt J

You would recommend "fsolve" for overdetermined nonlinear problems without constraints ? It might be applicable, but it's not the first choice I would think of.

[Matt J]

@Torsten yes, if there were constraints, fsolve would not be applicable. You would have to use lsqnonlin.