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multidimensional numerical integration without large matrices

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I have a function of 4 variables which neeeds to be integrated over a 3D space. The integral does not have an analytical solution so it needs to be done numerically. The function is also very sensitive and easily becomes unstable if the step size is too small. The way I've done it previously is as
f=numerically evaluated function using above mesh (so this is a 4D matrix)
This however, bottlenecks very fast when I need to increase the evaluation points to make it stable (runs out of memory). So my question is if there is another way to do this, without having 4D matrices.
I tried something like
g1=@(x1,x2,x3) integral(@(x4) f(x1,x2,x3,x4,x4(1),x4(end));
g2=@(x1,x2) integral(@(x3) g1(x1,x2,x3,x3(1),x3(end));
g3=@(x1) integral(@(x1) g2(x1,x2,x3,x4, x2(1),x2(end));
which does not work (gets matrix dimension mismatch error despite correctly vectorized function), as well as
g=@(x1) integral3(@(x2,x3,x4) F(x1,x2,x3,x4),x2(1),x2(end),x3(1),x3(end),x4(1),x4(end))
which fails the global error test, returning NaN.
The function in question is
F=@(x1,x2,x3,x4) sqrt(c./(1i.*x4.*a.*b)) .* exp((-1i.*pi./(x4.*b)).* (2.*x3.*x1+b./a.*x2-c./a.*x3.^2-a./c.*(x1+b./a.*x2).^2) ).*heaviside(abs(x3)-d) ;
a,b,c,d are constants
Would highly appriciate any suggestions!
Walter Roberson
Walter Roberson on 19 Mar 2018
"which fails the global error test"
? Could you explain that a bit more?
kristoffer svendsen
kristoffer svendsen on 20 Mar 2018
It states "Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test."
By decreasing the tolerances this is solved but the integration is incorrect as it shows only noise.

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Answers (3)

Walter Roberson
Walter Roberson on 19 Mar 2018
  1 Comment
kristoffer svendsen
kristoffer svendsen on 20 Mar 2018
Tried this but this integrates over all variables, I want to integrate over 3 out of 4 variables to be left with either a vector or another function F @(x1). Maybe I misunderstood this script but I could not get it to work this way.

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Unai San Miguel
Unai San Miguel on 20 Mar 2018
I don't like meshgrid even for 2 variables, so I would use ndgrid instead
[X1, X2, X3, X4] = ndgrid(x1, x2, x3, x4);
This way X1(i, j, k, l) = x1(i), X2(i, j, k, l) = x2(j) ans so on. Your array f would be of size [n1, n2, n3, n4], being n1 = length(x1), .... Then you could use trapz but with the lowercase-single dimensional variables.
g(x1, x2, x3) = int(f, dx4) ~ |trapz(x4, f, 4)|
h(x1, x2) = int(g, dx3) ~ |trapz(x3, trapz(x4, f, 4), 3)|, ...
So your final function would be
g_1 = trapz(x2, trapz(x3, trapz(x4, f, 4), 3), 2);
an array of size n1. I haven't done integrals of more than 2 variables, but if you can handle the 4D f array this looks doable (you are always reducing the size of the arrays).
  1 Comment
kristoffer svendsen
kristoffer svendsen on 20 Mar 2018
Edited: kristoffer svendsen on 20 Mar 2018
My apologies, yes I do indeed use ndgrid and not meshgrid, as meshgrid does not accept more than 3 dimensions. The 4D array however takes up about 60 Gb of memory and can not be stored for computation.

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Torsten on 20 Mar 2018
Edited: Torsten on 20 Mar 2018
x2min = ...;
x2max = ...;
x3min = ...;
x3max = ...;
x4min = ...;
x4max = ...;
a = ...;
b = ...;
c = ...;
d = ...;
F=@(x1,x2,x3,x4) sqrt(c./(1i.*x4.*a.*b)) .* exp((-1i.*pi./(x4.*b)).* (2.*x3.*x1+b./a.*x2-c./a.*x3.^2-a./c.*(x1+b./a.*x2).^2) ).*heaviside(abs(x3)-d) ;
g=@(x1) integral3(@(x2,x3,x4) F(x1,x2,x3,x4),x2min,x2max,x3min,x3max,x4min,x4max);
does not work ?
Which values do you use for the "..." indicated constants ?
Best wishes
kristoffer svendsen
kristoffer svendsen on 20 Mar 2018
Edited: kristoffer svendsen on 20 Mar 2018
This works if the 'AbsTol' and 'RelTol' are decreased (to roughly 1e-3) but at this point the integration does not converge to the correct solution and the result is not as it should be.
The constants varies (hehe) but an example would be
c=0.15; b=0.1; a=0.05; d=25e-6;
Torsten on 20 Mar 2018
What if you split the integral into three integrals
x3min <= x3 <= -d
-d <= x3 <= d
d <= x3 <= x3max
and remove the heaviside term in F ?

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