error when using lsqnonlin to function containing fzero function

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Haotian
Haotian 2024-9-13
移动：Torsten 2024-9-13
Ran in:
I want to use lsqnonlin analysis to fit a model to two sets of data and obtain values of 4 variables. The function i created contains fzero function which only allows scalar values and i finally got the error 'operands to the || and && operators ust be convertible to logical scalar values. ' So is there any other method that can replace lsqnonlin to solve my problem? Or any corrections, changes to be made in my code to solve this with lsqnonlin? Thanks! Here's my function file and code.
function Result = myfunc1(a,p,G,m,fraction)
totalconcentration = 0.0000291;
n = 2;
K = exp(-((G + m * fraction)./(8.314*298)));
g_lb = 0;
g_ub = 0.9999;
g =[g_lb g_ub];
monomerconcentration = fzero(@denaturationfun,g);
function y = denaturationfun(x)
y = a.^(-1).*(((a.*x).^(n+1)).*(n.*a.*x-n-1)/((a.*x-1).^2)+a.*x/((a.*x-1).^2))-a.^(n-1).*((x^(n+1)).*(n.*x-n-1)/((x-1).^2))-K.*totalconcentration;
end
Result = p * (1- (monomerconcentration / ( K * totalconcentration )));
end
clc;
clear;
close all;
fractiondata = [0.1505 0.1546 0.1587 0.1628 0.1668 0.1708 0.1748 0.1787 0.1825 0.1864 0.1902 0.194 0.1977 0.2014 0.205 0.2087 0.2123];
DegreeofAggdata = [1 0.9087 0.8658 0.83453 0.79569 0.67979 0.62031 0.53043 0.39722 0.25888 0.15171 0.04759 0.00109 3.20E-04 0.00144 7.77E-04 0];
fun = @(x)DegreeofAggdata - myfunc1(x(1),x(2),x(3),x(4),fractiondata);
x0 = [0.01,0.9,-50000,100000];
lb = [0.00001,0.9,-100000,20000];
ub = [1,1.1,-10000,130000];
x = lsqnonlin(fun,x0,lb,ub);
Error using fzero (line 236)
Function values must be scalar.

Error in solution>myfunc1 (line 8)
monomerconcentration = fzero(@denaturationfun,g);

Error in solution>@(x)DegreeofAggdata-myfunc1(x(1),x(2),x(3),x(4),fractiondata) (line 19)
fun = @(x)DegreeofAggdata - myfunc1(x(1),x(2),x(3),x(4),fractiondata);

Error in lsqnonlin (line 221)
initVals.F = feval(funfcn{3},xCurrent,varargin{:});

Caused by:
Failure in initial objective function evaluation. LSQNONLIN cannot continue.

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采纳的回答

Torsten
Torsten 2024-9-13
移动：Torsten 2024-9-13
"K" is a vector because "fraction" is a vector. Thus the "y" you compute in "denaturationfun" is a vector. But "fzero" cannot solve systems of nonlinear equations.
So either use "fsolve" or call "fzero" in a loop over all K-values.

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