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Faster method for numerical integration in 3D

조회 수: 9 (최근 30일)
Luqman Saleem
Luqman Saleem 2023년 3월 24일
댓글: Torsten 2023년 3월 24일
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 is a constant, Tr means trace of product of matrices which depend upon . Also, and . And β is also a constant.
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
  댓글 수: 6
Luqman Saleem
Luqman Saleem 2023년 3월 24일
편집: Luqman Saleem 2023년 3월 24일
@Torsten thank you, it worked. I have run the code with integral3 and it indeed is faster.
@John D'Errico yes, integral3 is faster.
Torsten
Torsten 2023년 3월 24일
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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