Implementing forward Euler method
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Matthew Kaplan
el 13 de Nov. de 2017
Comentada: Torsten
el 13 de Nov. de 2017

So I'm working on part(b) and I'm unsure how to plot the error versus step size on a log-log scale. Here is the code I have thus far...
function [tgrid, Y] = euler_method(fun, y_0, n, T)
if nargin(fun) ~=2
error('fun must take two inputs, t and y.');
end
if ~all(size(y_0) == size(fun(0, y_0)))
error('You have not passed appropriate fun or y_0.');
end
%Set up the time grid. ***NOTE THE n+1***
tgrid = linspace(0, T, n+1);
%Compute h from the time grid.
h = tgrid(2) - tgrid(1);
%Orient tgrid as a column vector.
tgrid = reshape(tgrid, n+1, 1);
%How many equations?
m = length(y_0);
%Orient y0 as a row vector.
y_0 = reshape(y_0, 1, m);
% Preallocate an array to hold the approximate solution. Each row
% corresponds to a point in the time grid.
Y = zeros(n+1, m);
% Set the initial conditions.
Y(1,:) = y_0;
% Euler loop
for i = 1: n
% Store the point in time as a temporary variable
t_i = tgrid(i);
% Take the Euler step into the temporary variable
y_1 = y_0 + h * fun(t_i, y_0);
% Store the Euler step
Y(i+1,:) = y_1;
% Update the temporary variable
y_0 = y_1;
end
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Respuesta aceptada
Torsten
el 13 de Nov. de 2017
Call "euler_method" in a loop for n = 8*2^k (k=1,...,15) and store Y(n+1,1) for each run.
Then make the plot.
Best wishes
Torsten.
4 comentarios
Torsten
el 13 de Nov. de 2017
y_0 = 0;
tend = 8.0;
fun=@(t,y) sin(t)-y;
for k = 1:15
n = 8*2^k;
[T Y] = euler_method(fun, y_0, n, tend);
Yend(k) = Y(n+1,1);
end
Now add the plot.
Best wishes
Torsten.
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