The code speed is too slow
이전 댓글 표시
I need plot(z, W_res) but my code is incredibly slow and I can't do it.
How can I make the code faster so that it can calculate everything?
This is the expression I want to calculate:

My code:
D = 100;
e_f = 1.88e-18;
m = 9.11e-31;
k_b = 1.38e-23;
h_bar = 1.05e-34;
ksi = 1e-9;
t = 1;
A = 2*m/h_bar^2;
k_f = sqrt(e_f / (pi * k_b * 1.2));
N_0 = 10e28;
v_f = 1.57e6;
l = 1.5e-10;
tau = l / v_f;
gamma = h_bar / (2*pi * k_b * 1.2 * tau);
E = e_f / (pi * k_b * 1.2);
norm = 2*m / h_bar^2;
k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*k_b.*1.2);
coef = 1;
w_n = @(n) t*(2*n+1);
n_max = 49;
cuA = m / (pi*h_bar*N_0*tau);
coef_K = cuA / (2*pi*ksi);
coef_w = (cuA * ksi * A) / (4*pi);
wt_sum = wt(1.2, 49);
k_pl = @(e) k0*sqrt( e + wt_sum); % k_+ off dim
s_w_func = @(e,z)( sin(k_pl(e).*z) .* sin(k_pl(e).*(z-D)) ) ./ (k_pl(e) .* sin(k_pl(e)*D));
S_w = @(z) imag(integral(@(e)s_w_func(e,z), 0, E,'ArrayValued',1 )); % off dim
K = @(e, z) k0*sqrt(e + 1i*wt_sum - coef_K * S_w(z));
inner_fun = @(e,z) 1/K(e,z) .* ( cos(integral(@(z_)K(e,z_), z-D, z)) ./ sin(integral(@(z_)K(e,z_), 0, D)) - cot(integral(@(z_)K(e,z_), 0, D)));
W_res = @(z)wt_sum - 1i*coef_w .* integral(@(e)inner_fun(e,z), 0, E,'ArrayValued',1); % off-dim, norm on pi*k_b*Tc
% Определение вектора z от 5 до 100 с шагом 0.1
z_values = 5:0.1:100;
% Инициализация вектора для хранения значений W_res
W_res_values = zeros(size(z_values));
% Вычисление значений W_res для каждого значения z
for i = 1:length(z_values)
W_res_values(i) = W_res(z_values(i));
end
% Построение графика
figure;
plot(z_values, real(W_res_values), 'LineWidth', 2);
xlabel('z');
ylabel('Re(W\_res)');
title('График W\_res от z');
grid on;
function result = wt(t, n)
result = 0;
for i = 1:n
result = result + t * (2 * i + 1);
end
end
댓글 수: 11
Alexander
2024년 2월 25일
Could you format your code as code? Mark it and press on code button on the insert of the part of the upper line.
Torsten
2024년 2월 25일
Did you get senseful values for W_res_values with your code and only the speed is too slow ? Or are the results unsatisfactory, too ?
Dmitry
2024년 2월 25일
Dmitry
2024년 2월 25일
Alexander
2024년 2월 25일
Every thing okay now. My former view of you code was text. Now it is code.
Torsten
2024년 2월 25일
If the code does not produce a result, you should work on this first (e.g. by checking your constants and equations). What is striking are your extreme parameter values used in the setup.
Dmitry
2024년 2월 26일
Dmitry
2024년 2월 26일
Torsten
2024년 2월 26일
Even when I try to compute
sin(integral(@(z_)K(e,z_), 0, D))
for different values of e in between 0 and E in order to see whether it becomes 0 somewhere, integral does not finish.
Could you test it on your computer and report the result of
D = 100;
e_f = 1.88e-18;
m = 9.11e-31;
k_b = 1.38e-23;
h_bar = 1.05e-34;
ksi = 1e-9;
t = 1;
A = 2*m/h_bar^2;
k_f = sqrt(e_f / (pi * k_b * 1.2));
N_0 = 10e28;
v_f = 1.57e6;
l = 1.5e-10;
tau = l / v_f;
gamma = h_bar / (2*pi * k_b * 1.2 * tau);
E = e_f / (pi * k_b * 1.2);
norm = 2*m / h_bar^2;
k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*k_b.*1.2);
coef = 1;
w_n = @(n) t*(2*n+1);
n_max = 49;
cuA = m / (pi*h_bar*N_0*tau);
coef_K = cuA / (2*pi*ksi);
coef_w = (cuA * ksi * A) / (4*pi);
wt_sum = wt(1.2, 49);
k_pl = @(e) k0*sqrt( e + wt_sum); % k_+ off dim
s_w_func = @(e,z)( sin(k_pl(e).*z) .* sin(k_pl(e).*(z-D)) ) ./ (k_pl(e) .* sin(k_pl(e)*D));
S_w = @(z) imag(integral(@(e)s_w_func(e,z), 0, E,'ArrayValued',1 )); % off dim
K = @(e, z) k0*sqrt(e + 1i*wt_sum - coef_K * S_w(z));
e = linspace(0,E,20)
result = arrayfun(@(e)sin(integral(@(z_)K(e,z_), 0, D)),e)
function result = wt(t, n)
result = 0;
for i = 1:n
result = result + t * (2 * i + 1);
end
end
Dmitry
2024년 2월 26일
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