multidimensional numerical integration without large matrices

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kristoffer svendsen
评论： Torsten 2018-3-20
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
[X1,X2,X3,X4]=meshgrid(x1,x2,x3,x4)
f=numerically evaluated function using above mesh (so this is a 4D matrix)
trapz(X1,f,3)
trapz(X1,f,2)
trapz(X1,f,1)
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));
g3(x1)
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!
  2 个评论
Walter Roberson
Walter Roberson 2018-3-19
"which fails the global error test"
? Could you explain that a bit more?
kristoffer svendsen
kristoffer svendsen 2018-3-20
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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回答（3 个）

Walter Roberson
Walter Roberson 2018-3-19
  1 个评论
kristoffer svendsen
kristoffer svendsen 2018-3-20
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 2018-3-20
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 个评论
kristoffer svendsen
kristoffer svendsen 2018-3-20
编辑：kristoffer svendsen 2018-3-20
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
Torsten 2018-3-20
编辑：Torsten 2018-3-20
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);
g(2.5)
does not work ?
Which values do you use for the "..." indicated constants ?
Best wishes
Torsten.
  2 个评论
kristoffer svendsen
kristoffer svendsen 2018-3-20
编辑：kristoffer svendsen 2018-3-20
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;
x2min=-7.5e-5;
x2max=7.5e-5;
x3min=x2min;
x3max=x2max;
x4min=1.2e-10;
x4max=1.2e-9;
Torsten
Torsten 2018-3-20
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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