Faster method for numerical integration in 3D
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I have a very complex expression which includes three integrations:

Please bear with me. This looks very complicated but indeed it is very simple. Allow me to explain.
here ℏ is a constant, Tr means trace of product of matrices
which depend upon
. Also,
and
. And β is also a constant.
which depend upon I want to evaluate these three integration numerically, but the method I am using is very slow and also it is not giving me expected results. I will be very thankful to you if you could have a look at my method and suggest any faster method for this 3D integration. My codes are given below:
main file:
clear; clc;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Some Parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a = 1;
JN = 1;
Dp = 0.75*JN;
T = 600;
eta = 0.01*JN;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Limits
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
xmin = -2*pi/(3*a);
xmax = 4*pi/(3*a);
ymin = -2*pi/(a*sqrt(3));
ymax = 2*pi/(a*sqrt(3));
Emin = -10000; %The function goes to approximately zero beyond this range
Emax = 10000;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Numerical Integration
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
tic
EBZsum = integral(@(E) integral(@(x) integral(@(y) (FUN_Exy(E,x,y,JN,Dp,T,eta)), ymin, ymax, 'ArrayValued', 1) , xmin, xmax, 'ArrayValued', 1), Emin, Emax, 'ArrayValued', 1);
tim = toc;
EBZsum = EBZsum*hbar;
fprintf('time = %f mins, sigma = %f q2/hbar\n',tim/60,EBZsum)
The helping function:
% FUN_Exy.m
function out = FUN_Exy(E,x,y,JN,Dp,T,eta)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Constants & Helping Expressions
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a = 1;
s = 1;
t = pi;
kB = 0.0863;
hbar = 6.582e-13;
beta = 1/(kB*T);
ny = cos(t);
H = [ 4*JN*s, -(s*cos((a*x)/4 - (3^(1/2)*a*y)/4)*(4*JN - Dp*ny*4i))/2, -(s*cos((a*x)/2)*(4*JN + Dp*ny*4i))/2
-(s*cos((a*x)/4 - (3^(1/2)*a*y)/4)*(4*JN + Dp*ny*4i))/2, 4*JN*s, -(s*cos((a*x)/4 + (3^(1/2)*a*y)/4)*(4*JN - Dp*ny*4i))/2
-(s*cos((a*x)/2)*(4*JN - Dp*ny*4i))/2, -(s*cos((a*x)/4 + (3^(1/2)*a*y)/4)*(4*JN + Dp*ny*4i))/2, 4*JN*s];
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Expressions of vx, vy, GA, GR, fE, dfdE
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
vx = 1/hbar * [ 0, (a*s*sin((a*x)/4 - (3^(1/2)*a*y)/4)*(4*JN - Dp*ny*4i))/8, (a*s*sin((a*x)/2)*(4*JN + Dp*ny*4i))/4
(a*s*sin((a*x)/4 - (3^(1/2)*a*y)/4)*(4*JN + Dp*ny*4i))/8, 0, (a*s*sin((a*x)/4 + (3^(1/2)*a*y)/4)*(4*JN - Dp*ny*4i))/8
(a*s*sin((a*x)/2)*(4*JN - Dp*ny*4i))/4, (a*s*sin((a*x)/4 + (3^(1/2)*a*y)/4)*(4*JN + Dp*ny*4i))/8, 0];
vy = 1/hbar * [ 0, -(3^(1/2)*a*s*sin((a*x)/4 - (3^(1/2)*a*y)/4)*(4*JN - Dp*ny*4i))/8, 0
-(3^(1/2)*a*s*sin((a*x)/4 - (3^(1/2)*a*y)/4)*(4*JN + Dp*ny*4i))/8, 0, (3^(1/2)*a*s*sin((a*x)/4 + (3^(1/2)*a*y)/4)*(4*JN - Dp*ny*4i))/8
0, (3^(1/2)*a*s*sin((a*x)/4 + (3^(1/2)*a*y)/4)*(4*JN + Dp*ny*4i))/8, 0];
GR = inv((E+1i*eta)*eye(size(H))-H);
GA = GR';
dGR = -GR*GR; %this is derivative of GR w.r.t. E
dGA = -GA*GA; %this is derivative of GA w.r.t. E
fE = (exp(beta*E) - 1)^(-1);
dfdE = -beta*exp(beta*E)*(exp(beta*E) - 1)^(-2);
TR1 = trace( vx*(GR*vy*GR + GA*vy*GA - 2*GR*vy*GA) )*dfdE; %first term in the given expression
TR2 = trace( vx*(GA*vy*dGA - dGA*vy*GA - GR*vy*dGR + dGR*vy*GR) )*fE; %second term in the given expression
out = real(hbar/(2*(2*pi)^3) * (TR1-TR2) );
end
6 Comments
John D'Errico
on 24 Mar 2023
Why not just use integral3?
Luqman Saleem
on 24 Mar 2023
Torsten
on 24 Mar 2023
I don't know if it's faster, but you might want to try
EBZsum = integral3(@(E,x,y)arrayfun(@(E,x,y)FUN_Exy(E,x,y,JN,Dp,T,eta),E,x,y),Emin, Emax,xmin, xmax,ymin, ymax)
instead of
EBZsum = integral(@(E) integral(@(x) integral(@(y) (FUN_Exy(E,x,y,JN,Dp,T,eta)), ymin, ymax, 'ArrayValued', 1) , xmin, xmax, 'ArrayValued', 1), Emin, Emax, 'ArrayValued', 1);
John D'Errico
on 24 Mar 2023
I suspect integral3 should be faster. But I've never truly tested it to verify that prediction. That it was giving you an error just means you either should tell us what the error was, or do as I suspect might work, to use the code Torsten suggests.
Luqman Saleem
on 24 Mar 2023
Edited: Luqman Saleem
on 24 Mar 2023
Torsten
on 24 Mar 2023
thank you, it worked. I have run the code with integral3 and it indeed is faster.
But the unsatisfactory results shouldn't have changed, have they ?
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