How to explain a difference in solutions of integrals between MATLAB and MAPLE?

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Dear community,

after solving an integral via MAPLE 2022 (see the solution in blue),

I tried to reproduce the calculation in MATLAB R2021b as a numerical and symbolic expression:

%% integral: numerical expression
            
            %empirical coefficents
            vrel = 0.000018;        
            alpha = 0.08;           
            beta = -6.75;           
            h =1;                   
            d =1;                   
            %function
            fun = @(theta) (alpha*beta*cos(theta).^4).*(pi*d-8*beta*vrel*cos(theta).^2).^(-2);
            
            %integral
            F = integral(fun,0,pi/2)
F = -0.0322
%% integral: symbolic expression
            %system parameters
            syms theta alpha beta d vrel
            %function
            fun = (alpha*beta*cos(theta).^4).*(pi*d-8*beta*vrel*cos(theta).^2).^(-2);
            %substiution of empirical coefficents
            fsub = subs(fun,{alpha, beta, d, vrel}, {0.0814,-6.75,1,0.000018});
            %integral
            f = int(fsub,theta,0,pi/2);
            %evaluation for complex number
            c = subs(f, 1+i);
            F = double(c)
F = -0.0328

However, the solution between MATLAB and MAPLE differs by 8 magnitudes. My colleauge and me double-checked for typos but couldn't figure out any. I verified the result from MAPLE via comparison to a simplfied solution for the integral.

I am curious to learn about my mistake(s) when adapting the integral into MATLAB!

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Tim Hammer 2022년 11월 10일

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Hi Jiri,

thanks for your suggestion! I checked it, and it seems that both softwares evaluate the function correctly:

MAPLE:

MATLAB:

%empirical coefficents

vrel = 0.000018;

alpha = 0.08;

beta = -6.75;

h =1;

d =1;

%function

fun = @(theta) (alpha*beta*cos(theta).^4).*(pi*d-8*beta*vrel*cos(theta).^2).^(-2);

close all

figure

fplot(@(theta) fun(theta),[0 1.6])

Bjorn Gustavsson 2022년 11월 10일

After following @Jiri Hajek's suggestion one gets a smoothly varying curve starting at approximately -0.054 at theta=0 that aproaches zero at pi/2. This curve is everywhere in this interval 0<theta<=1 under the line y = -0.05 + 0.05*x. Therefore the integral has to have a smaller value than the area of the triangle with corners at [0,0], [0,-0.05] and [1,0] which should be -0.05*1/2 or -0.025.

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Answer 1

Jiri Hajek 2022년 11월 10일

0 개 추천

Nicely done graphs! So, just by looking at the graphs, it is clear that:

The integral of your function is a negative value, with lower limit on the value being the area of the graph, i.e. about 1,6x(-0,05)= - 0,08
Comparing your original results from MATLAB with the above, it seems that both provide good estimates, although it remains to see which one is more precise
The MAPLE result is a complete nonsense, 8 orders of magnitude off the target.You have to search for error in your MAPLE implementation, not in MATLAB.