Solving a matrix equation with fixed point iteration method
4 views (last 30 days)
Show older comments
I want to solve the following equation for 


where

I believe Equ.1 can be solved using fixed iteration method. The twist here is that the term
itself have a self-consistent equation:


So, to solve Equ.1, I have to solve Equ.3 for
and then put the value of
in Equ.2 and calculate
and finally put
in Equ.1 to solve it.




In Equ.1,
And H and
are 2-by-2 matrices given in below code.
is identity matrix and E is a scalar parameter.



I'm working on this code. The loop for
is converging well, but the loop for
is very slow for certain u values, and it never converges for others. Any tips to speed up convergence or alternative solution methods?


clear; clc;
% parameters of equations:
E = 1;
n = 0.1;
u = 0.2;
% parameters of this script:
Nk = 1000; % number of points for integrating over kx and ky
max_iter = 2000; % # of maximum iterations
convergence_threshold = 1e-6;
% k-points and limits
xmin = -2*pi/(3*sqrt(3));
xmax = 4*pi/(3*sqrt(3));
ymin = -2*pi/3;
ymax = 2*pi/3;
kxs = linspace(xmin,xmax,Nk);
dkx = kxs(2) - kxs(1);
kys = linspace(ymin,ymax,Nk);
dky = kys(2) - kys(1);
%%%%%%%%%%%%%%%%%%%%%%%% Calculation of Sigma_0 %%%%%%%%%%%%%%%%%%%%%%%%
Sigma_0 = (0.1 + 0.1i)*eye(2); % initial guess
for iter = 1:max_iter
% Calculation of integration in Sigma_0 (Equ.3) via sum:
G_0 = @(kx,ky) inv( E*eye(2) - H(kx,ky) - Sigma_0 );
integral_term_Equ3 = zeros(2);
parfor ikx = 1:Nk
qx = kxs(ikx);
for iky = 1:Nk
qy = kys(iky);
integral_term_Equ3 = integral_term_Equ3 + G_0(qx,qy) * dkx * dky;
end
end
%new value of Sigma_0 (Equ.3):
new_Sigma_0 = n * u * inv( eye(2) - 1/(4*pi^2) * integral_term_Equ3 * u);
diff = norm(new_Sigma_0 - Sigma_0); %difference
fprintf('G_0 Iteration: %d, Difference: %0.9f\n', iter, diff);
if diff < convergence_threshold
fprintf('G_0 converged after %d iterations\n', iter);
break;
end
Sigma_0 = new_Sigma_0; % update Solution
end
G_0 = @(kx,ky) inv( E*eye(2) - H(kx,ky) - Sigma_0 ); %the chosen G_0
%%%%%%%%%%%%%%%%%%%%%%%% Calculation of Sigma_x %%%%%%%%%%%%%%%%%%%%%%%%
Sigma_x = Sigma_0; %taking Sigma_0 as initial guess
for iter = 1:max_iter
% Calculating the integration in Sigma (Equ.1) via sum:
integral_term_Equ1 = zeros(2);
parfor ikx = 1:Nk
qx = kxs(ikx);
for iky = 1:Nk
qy = kys(iky);
integrant = G_0(qx,qy) * (Sigma_x - 1i*E*v_x(qx,qy)) * G_0(qx,qy)' + 1i*E/2 * (G_0(qx,qy)*v_x(qx,qy)*G_0(qx,qy) + G_0(qx,qy)'*v_x(qx,qy)*G_0(qx,qy)');
integral_term_Equ1 = integral_term_Equ1 + integrant * dkx * dky;
end
end
%new value of Sigma_x (Equ.1):
new_Sigma_x = -1/(4*pi^2*n) * Sigma_0 * integral_term_Equ1 * Sigma_0';
diff = norm(new_Sigma_x - Sigma_x); %difference
fprintf('G Iteration: %d, Difference: %0.9f\n', iter, diff);
if diff < convergence_threshold
fprintf('G converged after %d iterations\n', iter);
break;
end
Sigma_x = new_Sigma_x; % update Solution
end
%%%%%%%%%%%%%%%%%%%%%%%% H and v_x functions %%%%%%%%%%%%%%%%%%%%%%%%
function H = H(kx,ky)
J = 1;
D = 0.5;
S = 1;
a1 = [0,-1]';
a2 = [sqrt(3)/2,1/2]';
a3 = [-sqrt(3)/2,1/2]';
b1 = [sqrt(3)/2,-3/2]';
b2 = [sqrt(3)/2,3/2]';
b3 = [-sqrt(3),0]';
s0 = eye(2,2);
sx = [0,1; 1,0];
sy = [0, -1i; 1i, 0];
sz = [1, 0; 0, -1];
k = [kx,ky];
h0 = 3*J*S;
hx = -J*S*( cos(k*a1) + cos(k*a2) + cos(k*a3) );
hy = -J*S*( sin(k*a1) + sin(k*a2) + sin(k*a3) );
hz = 2*D*S*( sin(k*b1) + sin(k*b2) + sin(k*b3) );
H = s0*h0 + sx*hx + sy*hy + sz*hz;
end
function v_x = v_x(kx,ky)
J = 1;
D = 0.5;
S = 1;
sx = [0,1; 1,0];
sy = [0, -1i; 1i, 0];
sz = [1, 0; 0, -1];
dkx_hx = -J*S*((3^(1/2)*sin(ky/2 - (3^(1/2)*kx)/2))/2 - (3^(1/2)*sin(ky/2 + (3^(1/2)*kx)/2))/2);
dkx_hy = J*S*((3^(1/2)*cos(ky/2 - (3^(1/2)*kx)/2))/2 - (3^(1/2)*cos(ky/2 + (3^(1/2)*kx)/2))/2);
dkx_hz = 2*D*S*((3^(1/2)*cos((3*ky)/2 - (3^(1/2)*kx)/2))/2 - 3^(1/2)*cos(3^(1/2)*kx) + (3^(1/2)*cos((3*ky)/2 + (3^(1/2)*kx)/2))/2);
v_x = sx*dkx_hx + sy*dkx_hy + sz*dkx_hz;
end
The result of above code are:
G_0 Iteration: 1, Difference: 0.127690191
G_0 Iteration: 2, Difference: 0.001871435
G_0 Iteration: 3, Difference: 0.000038569
G_0 Iteration: 4, Difference: 0.000000739
G_0 converged after 4 iterations
G Iteration: 1, Difference: 0.046150821
G Iteration: 2, Difference: 0.050105083
G Iteration: 3, Difference: 0.050493792
G Iteration: 4, Difference: 0.050542706
G Iteration: 5, Difference: 0.050547627
G Iteration: 6, Difference: 0.050543314
G Iteration: 7, Difference: 0.050536343
G Iteration: 8, Difference: 0.050528515
G Iteration: 9, Difference: 0.050520402
G Iteration: 10, Difference: 0.050512195
G Iteration: 11, Difference: 0.050503958
G Iteration: 12, Difference: 0.050495711
G Iteration: 13, Difference: 0.050487461
G Iteration: 14, Difference: 0.050479212
G Iteration: 15, Difference: 0.050470964
G Iteration: 16, Difference: 0.050462717
G Iteration: 17, Difference: 0.050454471
G Iteration: 18, Difference: 0.050446226
G Iteration: 19, Difference: 0.050437983
G Iteration: 20, Difference: 0.050429741
G Iteration: 21, Difference: 0.050421501
G Iteration: 22, Difference: 0.050413262
G Iteration: 23, Difference: 0.050405024
G Iteration: 24, Difference: 0.050396788
G Iteration: 25, Difference: 0.050388553
G Iteration: 26, Difference: 0.050380319
G Iteration: 27, Difference: 0.050372087
G Iteration: 28, Difference: 0.050363856
G Iteration: 29, Difference: 0.050355626
G Iteration: 30, Difference: 0.050347398
G Iteration: 31, Difference: 0.050339171
G Iteration: 32, Difference: 0.050330945
G Iteration: 33, Difference: 0.050322721
G Iteration: 34, Difference: 0.050314498
G Iteration: 35, Difference: 0.050306276
G Iteration: 36, Difference: 0.050298056
G Iteration: 37, Difference: 0.050289837
G Iteration: 38, Difference: 0.050281619
G Iteration: 39, Difference: 0.050273403
G Iteration: 40, Difference: 0.050265188
G Iteration: 41, Difference: 0.050256975
G Iteration: 42, Difference: 0.050248762
G Iteration: 43, Difference: 0.050240552
G Iteration: 44, Difference: 0.050232342
G Iteration: 45, Difference: 0.050224134
G Iteration: 46, Difference: 0.050215927
G Iteration: 47, Difference: 0.050207721
G Iteration: 48, Difference: 0.050199517
G Iteration: 49, Difference: 0.050191315
G Iteration: 50, Difference: 0.050183113
G Iteration: 51, Difference: 0.050174913
G Iteration: 52, Difference: 0.050166714
G Iteration: 53, Difference: 0.050158517
G Iteration: 54, Difference: 0.050150321
G Iteration: 55, Difference: 0.050142126
G Iteration: 56, Difference: 0.050133932
G Iteration: 57, Difference: 0.050125740
G Iteration: 58, Difference: 0.050117549
G Iteration: 59, Difference: 0.050109360
G Iteration: 60, Difference: 0.050101172
G Iteration: 61, Difference: 0.050092985
G Iteration: 62, Difference: 0.050084800
0 Comments
Accepted Answer
Torsten
on 29 Dec 2023
Edited: Torsten
on 29 Dec 2023
main()
function main
clear; clc;
format long
% parameters of equations:
E = 1;
n = 0.1;
u = 0.2;
% parameters of this script:
Nk = 300; % number of points for integrating over kx and ky
% k-points and limits
xmin = -2*pi/(3*sqrt(3));
xmax = 4*pi/(3*sqrt(3));
ymin = -2*pi/3;
ymax = 2*pi/3;
kxs = linspace(xmin,xmax,Nk);
dkx = kxs(2) - kxs(1);
kys = linspace(ymin,ymax,Nk);
dky = kys(2) - kys(1);
%%%%%%%%%%%%%%%%%%%%%%%% Calculation of Sigma_0 %%%%%%%%%%%%%%%%%%%%%%%%
Sigma_0 = (0.1 + 0.1i)*eye(2); % initial guess
sigma0 = [Sigma_0(:,1);Sigma_0(:,2)];
sigma0 = fsolve(@fun_Sigma0,sigma0,optimset('TolFun',1e-12,'TolX',1e-12));
Sigma_0 = [sigma0(1:2),sigma0(3:4)]
%%%%%%%%%%%%%%%%%%%%%%%% Calculation of Sigma_x %%%%%%%%%%%%%%%%%%%%%%%%
Sigma_x = Sigma_0;
sigmax = [Sigma_x(:,1);Sigma_x(:,2)];
sigmax = fsolve(@fun_Sigmax,sigmax,optimset('TolFun',1e-12,'TolX',1e-12));
Sigma_x = [sigmax(1:2),sigmax(3:4)]
function res = fun_Sigma0(sigma0)
Sigma_0 = [sigma0(1:2),sigma0(3:4)];
% Calculation of integration in Sigma_0 (Equ.3) via sum:
G_0 = @(kx,ky) inv( E*eye(2) - H(kx,ky) - Sigma_0 );
integral_term_Equ3 = zeros(2);
for ikx = 1:Nk
qx = kxs(ikx);
for iky = 1:Nk
qy = kys(iky);
integral_term_Equ3 = integral_term_Equ3 + G_0(qx,qy) * dkx * dky;
end
end
Res = Sigma_0 - n * u * inv( eye(2) - 1/(4*pi^2) * integral_term_Equ3 * u);
res = [Res(:,1);Res(:,2)];
end
function res = fun_Sigmax(sigmax)
Sigma_x = [sigmax(1:2),sigmax(3:4)];
% Calculation of integration in Sigma_0 (Equ.3) via sum:
G_0 = @(kx,ky) inv( E*eye(2) - H(kx,ky) - Sigma_0 );
integral_term_Equ1 = zeros(2);
for ikx = 1:Nk
qx = kxs(ikx);
for iky = 1:Nk
qy = kys(iky);
G_0_num = G_0(qx,qy);
v_x_num = v_x(qx,qy);
integrant = G_0_num * (Sigma_x - 1i*E*v_x_num) * G_0_num' + 1i*E/2 * (G_0_num*v_x_num*G_0_num + G_0_num'*v_x_num*G_0_num');
integral_term_Equ1 = integral_term_Equ1 + integrant * dkx * dky;
end
end
Res = Sigma_x - (-1/(4*pi^2*n) * Sigma_0 * integral_term_Equ1 * Sigma_0');
res = [Res(:,1);Res(:,2)];
end
%%%%%%%%%%%%%%%%%%%%%%%% H and v_x functions %%%%%%%%%%%%%%%%%%%%%%%%
function H = H(kx,ky)
J = 1;
D = 0.5;
S = 1;
a1 = [0,-1]';
a2 = [sqrt(3)/2,1/2]';
a3 = [-sqrt(3)/2,1/2]';
b1 = [sqrt(3)/2,-3/2]';
b2 = [sqrt(3)/2,3/2]';
b3 = [-sqrt(3),0]';
s0 = eye(2,2);
sx = [0,1; 1,0];
sy = [0, -1i; 1i, 0];
sz = [1, 0; 0, -1];
k = [kx,ky];
h0 = 3*J*S;
hx = -J*S*( cos(k*a1) + cos(k*a2) + cos(k*a3) );
hy = -J*S*( sin(k*a1) + sin(k*a2) + sin(k*a3) );
hz = 2*D*S*( sin(k*b1) + sin(k*b2) + sin(k*b3) );
H = s0*h0 + sx*hx + sy*hy + sz*hz;
end
function v_x = v_x(kx,ky)
J = 1;
D = 0.5;
S = 1;
sx = [0,1; 1,0];
sy = [0, -1i; 1i, 0];
sz = [1, 0; 0, -1];
dkx_hx = -J*S*((3^(1/2)*sin(ky/2 - (3^(1/2)*kx)/2))/2 - (3^(1/2)*sin(ky/2 + (3^(1/2)*kx)/2))/2);
dkx_hy = J*S*((3^(1/2)*cos(ky/2 - (3^(1/2)*kx)/2))/2 - (3^(1/2)*cos(ky/2 + (3^(1/2)*kx)/2))/2);
dkx_hz = 2*D*S*((3^(1/2)*cos((3*ky)/2 - (3^(1/2)*kx)/2))/2 - 3^(1/2)*cos(3^(1/2)*kx) + (3^(1/2)*cos((3*ky)/2 + (3^(1/2)*kx)/2))/2);
v_x = sx*dkx_hx + sy*dkx_hy + sz*dkx_hz;
end
end
15 Comments
Torsten
on 16 Jan 2024
Edited: Torsten
on 16 Jan 2024
I believe that if I could write a code that takes a lot of points near these points, we can achieve accurate integration.
That's exactly what an adaptive ODE integrator like ode45 does. If it didn't succeed, I doubt you will find a way to handle this problem with existing MATLAB codes.
Are you sure that the matrix G00 has no singularities in the domain of integration ?
More Answers (0)
See Also
Categories
Find more on Loops and Conditional Statements in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!