Info

Cette question est clôturée. Rouvrir pour modifier ou répondre.

Solving a first order ODE using the Euler backward method (implicit)

3 vues (au cours des 30 derniers jours)
Ibrahim Ali
Ibrahim Ali le 1 Oct 2021
Clôturé : Cris LaPierre le 1 Oct 2021
I'm trying to solve a first order ODE with the initial condition y(5) = 0. To do this, I'm using the Euler backward method, but I'm getting these two errors:
Error using fzero (line 214)
Second argument must be finite.
Error in backwardeuler (line 22)
y(i+1) = fzero(@(Y) y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y, y(i), options);
------------------------------------------------------------------------------------
% y_true = log(t^2 -4*t-4); Initial condition y(5) = 0;
% F_ty = @(t,y) (2*t-4)*exp(-y);
dt = 0.01;
t0 = 0;
tf = 3;
t = t0:dt:tf;
y(5) = 0;
% using the formula for backward euler: y(i+1) = y(i) + dt*f(y(i+1),t(i+1))
% we get
%y(i+1) = y(i) + dt*((2*t(i+1)-4)*exp(-y(i+1)));
% setting the LHS equal to zero so we can use fsolve:
% 0 = y(i) + dt*((2*t(i+1)-4)*exp(-y(i+1))) - y(i+1);
% We define y(i+1) = Y, so that
% 0 = y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y;
options = optimset('TolX',1e-06);
for i = 1:length(t)-1
y(i+1) = fzero(@(Y) y(i) + dt*((2*t(i+1)-4)*exp(-Y)) - Y, y(i), options);
y_exact(i+1) = log((t(i+1))^2-4*t(i+1)-4);
end
figure(1)
hold on
plot(t,y,'bo')
plot(t,y_exact,'r-')
xlabel('time')
ylabel('y(t)')
title('Backward Euler method vs exact solution')
legend('Backward Euler', 'Exact')

Réponses (0)

Cette question est clôturée.

Produits

Community Treasure Hunt

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

Start Hunting!

Translated by