Curve fitting using custom equation

I got a costum equation in the form of
OPCrtr = @(r) 1.5186e-08.*integral(5.9122e-6.*A.*exp(-5e-7.*a.^2).*(1-exp(-alpha.*10.*a.^2))./(4.*alpha.*a.^2).*besselj(0,r.*a).*a, 0,Inf);
and the experimental data of
y = [5.63519982407452, 4.38965221332586, 3.08480267517907, 2.53580063087177, 1.84559777892743, 1.23011321773770, 0.662503695933817, 0.300962951613869];
r = [0.861841642228739, 1.50647116324536, 2.08103225806452, 2.65559335288368, 3.25818181818182, 3.90281133919844, 4.61750928641251, 5.61248093841642];
with the boundary conditions of min(r) = 0, max(r) = 14e-6.*512, and the initial conditions of A = 0.129, alpha = 5.17e-7.
When I was trying to fit using the Curve ftting toolbox, it returns "The expression is not a valid MATLAB expression, has non-scalar coefficients, or cannot be evaluated:".
How can I fit the data using the above custom equation please?

7 Comments

So you want to fit A and alpha and your vector y should be modelled as y(r) = OPCrtr(r) ? Why is min(r) = 0, max(r) = 14e-6.*512 while your r-vector doesn't satisfy these constraints ?
- So you want to fit A and alpha and your vector y should be modelled as y(r) = OPCrtr(r) ?
- YES
- Why is min(r) = 0, max(r) = 14e-6.*512 while your r-vector doesn't satisfy these constraints ?
- r in the theoretical equation is within the range of [0, 14e-6.*512], but I only got discrete data points in the experiment.
But the range for r of the theoretical model equation and of your data points should be identical. Otherwise, the model equation could not be used for your data.
Think about a function like y(x) = A.*exp(-B.*x)+C and x=linspace(0,10,101). If I extract 9 discrete values of y at x = 0.1:0.1:0.9 as the experimental data, is it impossible to fit the experimental data using the equation y(x) = A.*exp(-B.*x)+C in the range of x=linspace(0,10,101)?
You have experimental data for a range of r between 0.8 and 5.6. So you can fit coefficients A and alpha for the model function for a range of r between 0.8 and 5.6. It doesn't make sense trying to evaluate the model function in the range 0 <= r <= 14e-6 * 512 because you didn't have experimental data for this range.
Make sense. So how can I do it?
Make sense. So how can I do it?
By getting experimental data for the range of interest (0 <= r <= 14e-6.*512).

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 Accepted Answer

t seems that there might be some issues with your custom equation. Let's first rewrite the custom equation using a proper anonymous function definition and make sure it is properly defined. Then, we can use the lsqcurvefit function in MATLAB to fit the data. Here's a step-by-step guide:
1- Save this function as customEquation.m in your MATLAB working directory.
% customEquation.m
function OPCrtr = customEquation(params, r)
A = params(1);
alpha = params(2);
integrand = @(a, r) 1.5186e-08 .* 5.9122e-6 .* A .* exp(-5e-7 .* a .^ 2) .* (1 - exp(-alpha .* 10 .* a .^ 2)) ./ (4 .* alpha .* a .^ 2) .* besselj(0, r .* a) .* a;
OPCrtr = integral(@(a) integrand(a, r), 0, Inf, 'ArrayValued', true);
end
2-Save this script as main_script.m in your MATLAB working directory. Run the main_script.m script in MATLAB to perform curve fitting and plot the data with the fitted curve.
% main_script.m
% Define the data and initial conditions
y = [5.63519982407452, 4.38965221332586, 3.08480267517907, 2.53580063087177, 1.84559777892743, 1.23011321773770, 0.662503695933817, 0.300962951613869];
r = [0.861841642228739, 1.50647116324536, 2.08103225806452, 2.65559335288368, 3.25818181818182, 3.90281133919844, 4.61750928641251, 5.61248093841642];
% Set the initial conditions and bounds
initial_conditions = [0.129, 5.17e-7];
lb = [0, 0];
ub = [14e-6 * 512, Inf];
% Perform the curve fitting
fitted_params = lsqcurvefit(@customEquation, initial_conditions, r, y, lb, ub);
% Plot the data and fitted curve
r_fit = linspace(min(r), max(r), 100);
y_fit = customEquation(fitted_params, r_fit);
figure;
plot(r, y, 'ro', 'MarkerSize', 8, 'LineWidth', 2);
hold on;
plot(r_fit, y_fit, 'b-', 'LineWidth', 2);
xlabel('r');
ylabel('y');
legend('Data', 'Fit');
This should give you the fitted parameters and the plot of the custom equation fit to your data.
Important: There may be some mistakes in this answer Experts can tell if there are any mistakes

2 Comments

Hi Albara, Thanks for your answer. But below is what I got. :-(
Problem solved. Nice! Thanks a lot!

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Asked:

on 27 Apr 2023

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on 28 Apr 2023

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