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

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.

[Torsten]

Since you want to estimate activation energies, I hope you have measurement data for different temperatures, but under adiabatic conditions.

https://www.tutorchase.com/answers/ib/chemistry/how-can-the-activation-energy-be-determined-experimentally

[Continuum]

In this study, we are taking Q to be a constant at a specific temperature of interest. Same applies to QG. Thanks.

[Torsten]

Your data seem to indicate that you are performing non-adiabatic experiments where temperature increases over time because of the heat of reaction. That's inappropriate to estimate activation energies and preexponential factors.

[Continuum]

Yes, you are right that there are different activation energies at the different temperatures, but for this study, the relevant process is considered only at a single temperature. So, what I am looking for are estimates of these parameters to get a 'feel' of the model vs my current data . I will then compare the obtained model with data from another experiment of the same material (think of it like a different specimen). Thanks.

[Torsten]

but for this study, the relevant process is considered only at a single temperature.

Then you must refer to data different from those that you included as modified_data.xlsx here:

https://uk.mathworks.com/matlabcentral/answers/314965-parameter-estimation-for-a-system-of-differential-equations?s_tid=srchtitle

[Continuum]

If you observe the data, there is a period where the temp stayed almost constant for some time.

[Torsten]

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 ...

Answer 1

[Star Strider]

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.

.

[Continuum]

Thanks very much, it works now, I will play around with it a bit and see if I can get some relatively good results. Thanks very much for your input.

[Star Strider]

As always, my pleasure!

[Yeuran]

@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]

@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.

Answer 2

[William Rose]

@Buhari Ibrahim

Matlab has excellent tools for this.

https://www.mathworks.com/help/optim/ug/optimizing-a-simulation-or-ordinary-differential-equation.html

and

https://www.mathworks.com/help/optim/ug/fit-differential-equation-ode.html

explain ways to do it.

[William Rose]

@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.

[Continuum]

Thanks for the pointers, I reattached the code, sorry about that.