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March 1, 2016 13:53
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%% Метод Эйлера %% | |
% Условие задачи | |
U = dsolve('Du = (t - u) / (t - 2*u)', 'u(0) = 1'); | |
tau0 = 0.03; | |
b = 0.3; | |
% Функция для нахождения аналитического решения | |
u = @(t)eval(U); | |
% Функция для нахождения приближенного решения | |
f = @(t,u) (t - u) / (t - 2*u); | |
T = b - 0; | |
%Метод Эйлера для случаев tau, tau/2, tau/4, tau/8, tau/16 | |
% Порядок аппроксимации метода Эйлера | |
p = 1; | |
for k = 0:4 | |
% Начальное условие | |
y = 1; | |
tau = tau0/2^k; | |
N = T/tau; | |
t = 0; | |
for m = 2:N+1 | |
y = y + tau*f(t,y); | |
t = t + tau; | |
end | |
% Погрешность относительно точного решения u(b) | |
D = abs(u(b)-y); | |
if k > 0 | |
% Апостериорная оценка | |
d = 1/(2^p-1)*abs(y - yprev); | |
end | |
yprev = y; | |
end | |
%Метод Эйлера-Коши для случаев tau, tau/2, tau/4, tau/8, tau/16 | |
p = 2; | |
for k = 0:4 | |
y = 1; | |
tau = tau0/2^k; | |
N = T/tau; | |
t = 0; | |
for m = 2:N+1 | |
% Приближение во вспомогательной точке | |
y_ = y + tau*f(t,y); | |
y = y + tau/2*(f(t,y)+f(t+tau,y_)); | |
t = t + tau; | |
end | |
D = abs(u(b)-y); | |
if k > 0 | |
d = 1/(2^p-1)*abs(y - yprev); | |
end | |
yprev = y; | |
end | |
%Метод Рунге-Кутты для случаев tau, tau/2, tau/4, tau/8, tau/16 | |
p = 4; | |
for k = 0:4 | |
y = 1; | |
tau = tau0/2^k; | |
N = T/tau; | |
t = 0; | |
for m = 2:N+1 | |
k1 = tau*f(t,y); | |
k2 = tau*f(t+0.5*tau, y+0.5*k1); | |
k3 = tau*f(t+0.5*tau, y+0.5*k2); | |
k4 = tau*f(t+tau, y+k3); | |
y = y + 1/6*(k1+2*k2+2*k3+k4); | |
t = t + tau; | |
end | |
D = abs(u(b)-y); | |
if k > 0 | |
d = 1/(2^p-1)*abs(y - yprev); | |
end | |
yprev = y; | |
end |
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