Beispiel - SAM - ETH Zürich

3 % midpoint rule of Sect. 1.4.3 to mathematical pendulum equation in
4 % minimal coordinates and Hamiltonian form for Ex. 1.4.17
5 % Arguments: y0 ˆ= initial position, T ˆ= final time N ˆ=
6 % number of equidistant timesteps
7
8 g = 9.8; % constant of gravity
9 l = 1; % length of pendulum
0
1 % Compute ’exact’ solution by means of high-order single step method with tight
2 % error control
3 odefun = @(t,y) [y(2);-g/l*sin(y(1))];
4 [t,s] =
ode45(odefun,[0,T],y0,odeset(’abstol’,1E-10,’reltol’,1E-10));
5
6 h = T/N; % timestep
7
8 % Explicit Euler (1.4.2)
9 y_expl = y0; y = y0;
0 f o r k=1:N
1 y = y + h*[y(2);-g/l*sin(y(1))];
2 y_expl = [y_expl,y];
3 end
4
Numerische
Mathemtik
R. Hiptmair
rev 35327,
25. April
2011
1.4
p. 90