Global error and local error of euler method

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Erm
Erm 2023-1-28
评论： Torsten 2023-1-29
The local error is proportional to h^2. I did not understand the relationship between the h and the error. Is it as h increase the error increase? Also the global error is proportional to h , how is that?
  1 个评论
Torsten
Torsten 2023-1-29
These questions are answered in every standard book about the numerical treatment of ordinary differential equations.

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回答（1 个）

Sulaymon Eshkabilov
Ran in:
Yes, error increase if h step size increases. See this simple example:
dy = y(t) with y(0) = 3;
y(n+1) =y(n)+h*f(t(n), y(n));
from the givend ICs: f(t(0), y(0)) = 3, and therefore, y(1+1) = y(1)+h*3; Let's see in simulation:
h=0.1; % Case 1
f=@(t,y)y; % from the given exercise dy = y(t) and thus f(t,y) = dy;
t=0:h:5;
y(1)=3; % Initial Condition
for ii=2:numel(t) % calculations
y(ii) = y(ii-1)+h*f(t(ii-1),y(ii-1));
end
t=t.';
Solution = array2table(t);
Solution.Y = y.';
plot(t, y, 'r-', 'DisplayName', 'Case 1: h=0.1'), hold on
% Analytical solution:
syms t y(t)
Sol=dsolve(diff(y(t), t)==y, y(0)==3);
Sol_yt = vectorize(Sol)
Sol_yt = '3.*exp(t)'
t=(0:h:5).';
Sol = (eval(Sol_yt));
Error = abs(Sol)-Solution.Y; % Error = Analytical solution - Numerical Solution
Solution.Error=Error
Solution = 51×3 table
t Y Error ___ ______ ________ 0 3 0 0.1 3.3 0.015513 0.2 3.63 0.034208 0.3 3.993 0.056576 0.4 4.3923 0.083174 0.5 4.8315 0.11463 0.6 5.3147 0.15167 0.7 5.8462 0.19511 0.8 6.4308 0.24586 0.9 7.0738 0.30497 1 7.7812 0.37362 1.1 8.5594 0.45315 1.2 9.4153 0.54507 1.3 10.357 0.65108 1.4 11.392 0.7731 1.5 12.532 0.91332
Case1 = [t, Error];
clearvars y
h = 0.5; % Case 2
f=@(t,y)y; % from the given exercise dy = y(t) and thus f(t,y) = dy;
t=0:h:5;
y(1)=3; % Initial Condition
for ii=2:numel(t) % calculations
y(ii) = y(ii-1)+h*f(t(ii-1),y(ii-1));
end
plot(t, y, 'b-', 'DisplayName', 'Case 2: h=0.5'), grid on
t=t.';
Solution = array2table(t);
Solution.Y = y.';
t=(0:h:5).';
Sol = (eval(Sol_yt));
Error = abs(Sol)-Solution.Y; % Error = Analytical solution - Numerical Solution
Solution.Error=Error
Solution = 11×3 table
t Y Error ___ ______ _______ 0 3 0 0.5 4.5 0.44616 1 6.75 1.4048 1.5 10.125 3.3201 2 15.188 6.9797 2.5 22.781 13.766 3 34.172 26.085 3.5 51.258 48.089 4 76.887 86.908 4.5 115.33 154.72 5 173 272.24
Case2 = [t, Error];
clearvars y Solution
h = 1; % Case 3
f=@(t,y)y; % from the given exercise dy = y(t) and thus f(t,y) = dy;
t=0:h:5;
y(1)=3; % Initial Condition
for ii=2:numel(t) % calculations
y(ii) = y(ii-1)+h*f(t(ii-1),y(ii-1));
end
plot(t, y, 'k-', 'DisplayName', 'Case 3: h=1')
legend('show', 'location', 'NW')
xlabel('time')
ylabel('Solution, y(t)')
t=t.';
Solution = array2table(t);
Solution.Y = y.';
t=(0:h:5).';
Sol = (eval(Sol_yt));
Error = abs(Sol)-Solution.Y; % Error = Analytical solution - Numerical Solution
Solution.Error=Error
Solution = 6×3 table
t Y Error _ __ ______ 0 3 0 1 6 2.1548 2 12 10.167 3 24 36.257 4 48 115.79 5 96 349.24
Case3 = [t, Error];
figure
plot(Case1(:,1), Case1(:,2), 'r'), hold on
plot(Case2(:,1), Case2(:,2), 'b')
plot(Case3(:,1), Case3(:,2), 'k'), grid on
legend('Case1: h=0.1', 'Case2: h=0.5', 'Case3: h=1', 'location', 'NW')
xlabel('time')
ylabel('Error')

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