Faster method for numerical integration in 3D

4 ビュー (過去 30 日間)
Luqman Saleem
Luqman Saleem 2023 年 3 月 24 日
コメント済み: Torsten 2023 年 3 月 24 日
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
  6 件のコメント
Luqman Saleem
Luqman Saleem 2023 年 3 月 24 日
編集済み: Luqman Saleem 2023 年 3 月 24 日
@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 2023 年 3 月 24 日
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 ?

サインインしてコメントする。

回答 (0 件)

カテゴリ

Help Center および File ExchangeMATLAB についてさらに検索

製品


リリース

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by