Loop over ode45 to find minimum of a parameter

3 views (last 30 days)
I'm trying to loop over an ode45 for different b and k, to find the couple of the two that minimize the error from the analytical solution. But when I run this code it enters in an infinite loop. What am I doing wrong?
T = readtable('samples.csv'); % three column [time,analytical_sol1,analytical_sol2]
test = @(t,y,b,k) [0;0;k/J1 * y(2) - k/J1 * y(1); T0/J2 - b*y(4)/J2 - (k/J2)*(y(2)-y(1))]; %my ode
err = []; % initialize error vector
for b = 0:0.01:10 %loop over different b
for k = 0:0.1:100 %loop over different k
[t,y] = ode45(@(t,y) test(t,y,b,k) , [0:0.01:10] , [0,0,0,0]); %solve my ode
errbk = abs( norm( T{:,3} - y(:,4) ) ); % compute error from the analytical solution 1
err = [err;b,k,errbk];
end
end
% then i would find b and k with the minimum errb and errk
  2 Comments
Walter Roberson
Walter Roberson on 28 Oct 2021
What should happen if the entry with minimum errb is not the entry with the minimum errk ?
Nader Mohamed
Nader Mohamed on 28 Oct 2021
You're right. I edited the code to compute the error only compared to one analytical solution
My idea is then after having the vector err - find the min(errbk) and output the corrisponding b and k

Sign in to comment.

Answers (1)

Star Strider
Star Strider on 28 Oct 2021
There are several examples on fitting differential equations to data, one being Coefficient estimation for a system of coupled ODEs — not trivial, however also not difficult.
.
  2 Comments
Nader Mohamed
Nader Mohamed on 28 Oct 2021
Thanks! I understand that with lsqcurvefit I can fit the data with the results, but I still don't understand how can I retrieve b and k from that
Star Strider
Star Strider on 28 Oct 2021
It would be relatively straightforward to adapt my code to calculate ‘b’ and ‘k’. They become parameters, so the ‘kinetics’ function becomes —
function C=kinetics(theta,t,T0,J2)
% c0=[1;0;0;0];
c0 = theta(3:6);
[T,Cv]=ode45(@DifEq,t,c0);
%
function dC=DifEq(t,c) % k = theta(1), b = theta(2)
dcdt=zeros(4,1);
dcdt(1)= 0;
dcdt(2)= 0;
dcdt(3)= theta(1)/J1 * y(2) - theta(1)/J1 * y(1);
dcdt(4)= T0/J2 - theta(2)*y(4)/J2 - (theta(1)/J2)*(y(2)-y(1));
dC=dcdt;
end
C=Cv;
end
This is my best guess on how to implement your system of differential equations with my existing code. Here, the parameter vector ‘theta’ has ‘k’ and ‘b’ as the first two elements, and the initial conditions for the system of differential equations as the last four elements. All will be estimated by the optimisation funciton (lsqcurvefit, ga, or others). It may be necessary to edit this, because I do not understand what the objective is.
The ‘C’ output will be the result that matches the data to be regressed against. All will be matrices of column vectors.
.

Sign in to comment.

Products


Release

R2018b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!