How to fit an integral with data points in Matlab, where the integration limit can vary

How to fit an integral with data points in Matlab, where the integration limit can vary

By varying the integration limit how do I fit some data points in Matlab?

[Ameer Hamza]

Can you show the mathematical formulation of your model?

[Suman Karmakar]

P = P0 + 4RC(T/C)^5 integration (0, (C/T)) (x^5/(exp(x)-1)*(1-exp(-x))) dx

P0, R and C are constants

[Ameer Hamza]

You are given the values of P and T, and you want to find these constants?

[Suman Karmakar]

Yes, I want to find those constants.

[Shobha Gondh]

My problem is also similar to this fitting so please help.

I want to fit experimental data with equation:

y = C(1) * 74.826 * (x/T(1))^3 * integration (t^4*exp(t)/(exp(t)-1)^2), 0, T(1)/x) + C(2) * 24.942 * (T(2)/x)^2 * exp(T(2)/x)/(exp(T(2)/x)-1)^2 + C(3) * 24.942 * (T(3)/x)^2 * *exp(T(3)/x)/(exp(T(3)/x)-1)^2

with six unknown parameters: C(1), T(1), C(2), T(2), C(3), T(3).

I have data in x and y and want to find these unknown parameters by fitting.

Thank you

Answer 1

[Suman Karmakar]

If I want to add an extra term in the above expression (like -AT^3, where A constant) so the code would be like this or different. can u confirm me?

Tv = linspace(200, 300, 10).'; % vector of T values

Pv = Tv.^2; % vector of P values

P_mdl = @(P0, R, C, A, TT) arrayfun(@(T) P0 - A.*T.^3 + 4*R.*C.*(T/C).^5 * integral(@(x) x.^5./(exp(x)-1).*(1-exp(-x)), 0, (C/T)), TT);

sol = lsqcurvefit(@(param, T) P_mdl(param(1), param(2), param(3), param(4), T), rand(4,1), Tv, Pv);

P0_sol = sol(1);

R_sol = sol(2);

C_sol = sol(3);

A_sol = sol(4);

[Ameer Hamza]

Yes, this is correct.

[Suman Karmakar]

One additional information I want to know, in case of my experimantal data I can not fit the integration properly. So how do I deal with the rand(m,n) things. I mean is there any way to choose the initial value to fit the equation with the experimental data ?

[Ameer Hamza]

Do you have a general guess about the value of each parameter. If yes, use that as the initial point. Something like this

x0 = [P0_guess, R_guess, C_guess, A_guess]
sol = lsqcurvefit(@(param, T) P_mdl(param(1), param(2), param(3), param(4), T), x0, Tv, Pv);

Answer 2

[Ameer Hamza]

One method is to use lsqcurvefit() from Optimization toolbox

Tv = rand(10, 1); % vector of T values
Pv = rand(10, 1); % vector of P values

P_mdl = @(P0, R, C, TT) arrayfun(@(T) P0 + 4*R.*C.*(T/C).^5 * integral(@(x) x.^5./(exp(x)-1).*(1-exp(-x)), 0, (C/T)), TT);

sol = lsqcurvefit(@(param, T) P_mdl(param(1), param(2), param(3), T), rand(3,1), Tv, Pv);
P0_sol = sol(1)
R_sol = sol(2)
C_sol = sol(3)

[Suman Karmakar]

Is it possible to plot the original data with the fitted one in a same graph?

[Ameer Hamza]

Yes, it is possible

Tv = linspace(200, 300, 10).'; % vector of T values
Pv = Tv.^2; % vector of P values

P_mdl = @(P0, R, C, TT) arrayfun(@(T) P0 + 4*R.*C.*(T/C).^5 * integral(@(x) x.^5./(exp(x)-1).*(1-exp(-x)), 0, (C/T)), TT);

sol = lsqcurvefit(@(param, T) P_mdl(param(1), param(2), param(3), T), rand(3,1), Tv, Pv);
P0_sol = sol(1);
R_sol = sol(2);
C_sol = sol(3);

figure()
axes();
hold on;
plot(Tv, Pv, 'b+', 'DisplayName', 'Experimental Data');
plot(Tv, P_mdl(P0_sol, R_sol, C_sol, Tv), 'r', 'DisplayName', 'Estimated Model');
legend('FontSize', 16, 'Location', 'best')