Global error and local error of euler method

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?

[Torsten]

These questions are answered in every standard book about the numerical treatment of ordinary differential equations.

Answer 1

[Sulaymon Eshkabilov]

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')