System of 2nd order ODE with Euler.

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Hi,
I had a system of 2 2nd order ODE.
I got to this point :
I need to find the approximate solutions of y2(t).
M1, M2, G, L1,L2 are variables given by the user.
These are the initials conditions which are given by the user also(i guess ?)
Im a bit lost in what should i do. I know how euler works but not with this type of system.
thanks !
  4 个评论
Guillaume Theret
Guillaume Theret 2021-5-27
explicit ! Im writing some code, i'll send asap !
Guillaume Theret
Guillaume Theret 2021-5-27
编辑:Guillaume Theret 2021-5-27
after writing some code :
function main
xinit = 0;
xfinal = 3;
h = .5;
y0 = 1;
[x,y] = euler_explicit(@fnc,xinit,xfinal,h,y0);
plot(x,y(:,1));
end
function [x,y_e]=euler_explicit(f,xinit,xfinal,h,y0)
x = xinit : h : xfinal;
n = length(x);
disp(n);
y_e =zeros(1,n);
disp(y_e)
y_e(1) = y0;
% compute y_e : euler explicit
for i = 1:n - 1
y_e(i+1) = y_e(i) + h *f(x(i),y_e(i));
end
end
function dy = fnc(t,Y)
L1 = 1;
L2 = 2;
M1 = 2;
M2 = 3;
g = 1;
K = 1/(L1*L2(M1+M2*sin(Y(1) - Y(2)).^2));
Y4 = K*((M1+M2)*g*L1*sin(Y(1))*cos(Y(1)-Y(2)) - (M1+M2)*g*L1*sin(Y(2)) + (M1+M2)*L1^2*sin(Y(1) - Y(2))*Y(3)^2 + M2*L1*L2*sin(Y(1)-Y(2))*cos(Y(1)-Y(2))*Y(4)^2);
Y3 = K*(-(M1+M2)*g*L2*sin(Y(1)) + M2*g*L2*sin(Y(2))*cos(Y(1)-Y(2)) - M2*L1*L2*sin(Y(1) - Y(2))*cos(Y(1)-Y(2))*Y(3)^2 - M2*L2^2*sin(Y(1)-Y(2))*Y(4)^2);
dy = [Y(3),Y(4),Y3,Y4];
end
Im getting these erros. I check at my array and it looks like it has seven 0 which is normal.
I don't know if im going in the good direction or not to solve my equations :(
Thanks !

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采纳的回答

Jan
Jan 2021-5-27
编辑:Jan 2021-5-27
You got it almost. I've fixed a typo and expanded the Euler method to collect the output as matrix.
% function main
xinit = 0;
xfinal = 3;
h = 0.05;
y0 = [1, 0, 0, 0]; % As many elements as the system has
[x, y] = euler_explicit(@fnc, xinit, xfinal, h, y0);
plot(x, y);
% end
function [x, y] = euler_explicit(f, xinit, xfinal, h, y0)
x = xinit : h : xfinal;
n = length(x);
y = zeros(n, numel(y0));
y(1, :) = y0;
for k = 1:n - 1
y(k + 1, :) = y(k, :) + h * f(x(k), y(k, :));
end
end
function dy = fnc(t,Y)
L1 = 1;
L2 = 2;
M1 = 2;
M2 = 3;
g = 1;
K = 1 / (L1 * L2 * (M1 + M2*sin(Y(1) - Y(2)).^2));
% ^ was missing
Y4 = K*((M1+M2)*g*L1*sin(Y(1))*cos(Y(1)-Y(2)) - (M1+M2)*g*L1*sin(Y(2)) + (M1+M2)*L1^2*sin(Y(1) - Y(2))*Y(3)^2 + M2*L1*L2*sin(Y(1)-Y(2))*cos(Y(1)-Y(2))*Y(4)^2);
Y3 = K*(-(M1+M2)*g*L2*sin(Y(1)) + M2*g*L2*sin(Y(2))*cos(Y(1)-Y(2)) - M2*L1*L2*sin(Y(1) - Y(2))*cos(Y(1)-Y(2))*Y(3)^2 - M2*L2^2*sin(Y(1)-Y(2))*Y(4)^2);
dy = [Y(3), Y(4), Y3, Y4];
end

更多回答(1 个)

Torsten
Torsten 2021-5-27
  1. y0 must be a 4x1 vector, not a scalar.
  2. ye = zeros(4,n) instead of ye=zeros(1,n)
  3. ye(:,1) = y0 instead of ye(1) = y0
  4. ye(:,i+1) = ye(:,i) + h*f(x(i),ye(:,i)) instead of the expression in your loop
  5. dy = [Y(3);Y(4);Y3;Y4] instead of the row vector in your code

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