How to find fitting optimized parameters to fit a system of non-linear ODEs to experiment.

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Hi
I have a set of ODEs (attached), I have been able to solve them using ode45, however, my issue now is my experimental results don't match the integrated values of the equations. So, I am looking to fit only the solution for epsilon with it's experimental results to find the best parameters A, B, (A0/alpha), k0, Q, and QG. Attached is my code based on an answer from another thread, but it just runs continuously but I couln't figure out what the problem is. Could it be that the there are too many parameters to fit? Any help is greatly appreciated. Thank you.
  7 Commenti
Continuum
Continuum il 27 Gen 2024
If you observe the data, there is a period where the temp stayed almost constant for some time.
Torsten
Torsten il 27 Gen 2024
Modificato: Torsten il 27 Gen 2024
But it's one long experiment with a dynamic development in time. So your measurements have a history. It's usually necessary that you start at t = 0 with a fixed temperature which is kept constant over time until the experiment has finished.
But we are in a MATLAB forum here ...

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Star Strider
Star Strider il 27 Gen 2024
You will need to post the ‘modified_data.xlsx’ file as well to run this here.
My edit of your code —
Model_Exp_Fit
function Model_Exp_Fit
% 2016 12 03
% NOTES:
%
% 1. The ‘theta’ (parameter) argument has to be first in your
% ‘kinetics’ funciton,
% 2. You need to return ALL the values from ‘DifEq’ since you are fitting
% all the values
dataTable1=readtable('modified_data.xlsx','Sheet','Sheet1', 'VariableNamigRule','preserve');
dataTable1 = dataTable1(:,4:end);
tempFunc = griddedInterpolant(dataTable1.time1_min_,dataTable1.T1_K_);
% Temp = tempFunc(t);
% tT = [t Temp];
tT = dataTable1{:,[1 2]}; % Time And Temperature
t = tT(:,1);
x = dataTable1{:,3}; % Epsilon
% return
function X = evolution(theta,tT)
T = tT(:,2);
tv = tT(:,1);
x0=[20;0.4363;0.0000001];
[T,Xv] = ode15s(@DifEq,t,x0);
k0 = theta(1);
QG = theta(2);
A0_alpha = theta(3);
Q = theta(4);
A = theta(5);
B = theta(6);
function dX = DifEq(t,x)
%thet_c = 0.508;
rho_c = 0.948;
R = 8.314;
Temp = interp1(tv, T, t); % Interpolates 'T' For Each Value Of 't'
% dataTable1=readtable('modified_data.xlsx','Sheet','Sheet1')
% tempFunc = griddedInterpolant(dataTable1.time1_min_,dataTable1.T1_K_);
% Temp = tempFunc(t);
% tT = [t Temp];
dxdt = zeros(3,1);
phi1 = (theta(1)/3)*(1-rho_c)^1.5;
phi2 = 9/(2*theta(3));
phi3 = 3/(2*theta(3));
dxdt(1) = (exp(-theta(2)./(R*Temp))).*(phi1./x(1).^2).*1./((1-rho_c+x(2)).^1.5);
dxdt(2) = (-phi2./(x(1).*Temp)).*(1./exp(theta(4)./(R*Temp))).*x(2).^theta(6).*(((1-x(2)).^3)./(1-x(2)).^theta(5));
dxdt(3) = (-phi3./(x(1).*Temp)).*(1./exp(theta(4)./(R*Temp))).*x(2).^theta(6).*(((1-x(2)).^2)./(1-x(2)).^theta(5));
dX = dxdt;
end
X=Xv(:,3);
end
% return
% eps_exp = load("exp_epsilon.txt");
% t = eps_exp(:,1);
% t = [0;t];
% % x = eps_exp(:,2)/100;
% x = [0; x];
theta0 = [(29.65e-5)/(1/60);164.8e3;2.03*((1/60)/(10^6));217.2e3;11.35;0.49];
[theta,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat]=lsqcurvefit(@evolution,theta0,tT,x);
fprintf(1,'\tRate Constants:\n')
for k1 = 1:length(theta)
fprintf(1, '\t\tTheta(%d) = %8.5f\n', k1, theta(k1))
end
% tv = linspace(min(t), max(t));
Xfit = evolution(theta, tT);
figure(1)
plot(t, x, 'p')
hold on
hlp = plot(t, Xfit);
hold off
grid
xlabel('Time')
ylabel('Concentration')
% legend(hlp, 'G(t)', 'theta(t)', 'epsilon(t)', 'Location','N')
end
I left in the sections that I commented-out so that you can understand my edits. I will give this a shot with some standardized code that I use to run genetic algorithm parameter estimations to see if I can improve on these results. If I get good results with it, I will post the results and the code here. For the time being, this code runs, and you can experiment with it.
.

Più risposte (1)

William Rose
William Rose il 27 Gen 2024
  2 Commenti
William Rose
William Rose il 27 Gen 2024
@Buhari Ibrahim, You are fitting six parameters. That is not too many. I have fitted ODE models with more parameters. You say you have attached code, but none is attached.

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