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

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Continuum
Continuum 2024-1-27
评论： Star Strider 2024-6-26
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.
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Torsten
Torsten 2024-1-27
编辑：Torsten 2024-1-27
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 2024-1-27
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.
.
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Yeuran
Yeuran 2024-6-26
@Star Strider Hello, I have another question about your code. What should we do if the experimental data(epsilon) is shorter than time and Temp? Do you have any suggestions? For instance, if 'x' contains only 3 data points (like x = dataTable1{[1,5,7],3}), and 'tT' and 't' are of equal legth(where tT = dataTable1{:,[1 2]} and t = tT(:,1)), how should we proceed to use 'lsqcurvefit' in such a scenario?
Star Strider
Star Strider 2024-6-26
@Yeuran — All data have to have the same number of elements (in this instance, rows). I would only use the data with corresponding with the rows in ‘x’, deleting everything else in the longer data.
I would not interpolate or extrapolate to equalise them, since in both instances that creates data where there were no data previously, and the parameter estimates are then useless.
So, if you only have three data points, you can only estimate at most two parameters. If you are estimating more parameters than that, you will need to repeat the experiment and get more data. There are simply no other optioons.

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William Rose
William Rose 2024-1-27

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