ODE solution using Fourier series
5 views (last 30 days)
Show older comments
Hello,
I am trying to solve first order ode () using Fourier Series. I wrote the code, but I was doing something, please idetify my mistake.
Fourier series:
by putting into original ODE. Here's the code:
clear all;clc;close all;
syms x
N = 3;
for j=1:(2*N+1)
xp(j) = simplify(-sym(pi) + (2*sym(pi)*(j-1))./(2.*N+1)); % Equidistance points
end
for k = 1:2*N+1
for l =1:N
a(k,l) = cos(l*xp(k));
b(k,l) = sin(l*xp(k));
end
end
one_col = ones(1,2*N+1);
sumab = double([a';b']);
A = (1+cos(x)).*double([one_col;sumab])'
x = (A\ones(7,1))
a0 = x(1);
an = x(2:N+1);
bn = x(N+2:2*N+1);
dA = diff(A);
% ODE
A_ODE = dA + A;
yxp = A_ODE*x
plot(xp,yxp);
hold on;
%% ODE45
[a,b] = ode45(@(a,b) -(1+cos(a))*b+1, [0 30],0); %to distinguish variable I choose x = a and y = b
plot(a,b)
Thanks
1 Comment
Jan
on 16 May 2021
Please mention, why you assume, that there is a mistake. You d have this important information already. Then it is useful to share it with the ones, who want to help you.
Answers (1)
Jan
on 16 May 2021
syms x
N = 3;
for j=1:(2*N+1)
xp(j) = simplify(-sym(pi) + (2*sym(pi)*(j-1))./(2.*N+1)); % Equidistance points
end
for k = 1:2*N+1
for l =1:N
a(k,l) = cos(l*xp(k));
b(k,l) = sin(l*xp(k));
end
end
one_col = ones(1,2*N+1);
sumab = double([a';b']);
A = (1+cos(x)).*double([one_col;sumab])'
x = (A\ones(7,1))
Are you sure that this is wanted?
a0 = x(1);
an = x(2:N+1);
bn = x(N+2:2*N+1);
dA = diff(A);
% ODE
A_ODE = dA + A;
yxp = A_ODE*x
plot(xp,yxp);
hold on;
%% ODE45
[a,b] = ode45(@(a,b) -(1+cos(a))*b+1, [0 30],0); %to distinguish variable I choose x = a and y = b
plot(a,b)
Your code does not contain meaningful comments. Guessing its intention is not reliable. Therefore it is hard to estimate, if something is wanted or a mistake.
0 Comments
See Also
Categories
Find more on Ordinary Differential Equations in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!