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Trying to solve dy/dx = 2xy, y(1) = 1, using Chebyshev differentiation matrix
The exact solution is y = exp(x.^2 -1);
Heres what I have:
% compute the chebyshev differentiation matrix and x-grid [D1,x] = cheb(N);
% boundary condition (I don't know if this is correct?) D1 = D1(1:end-1,1:end-1); x = x(1:end-1);
% compute the derivatives at x (i.e. at the chebyshev grid points) f = 2*x.*ones(size(x));
% solve u = D1\f;
% set the boundary condition u = [u;1];
Where cheb.m is from Trefethen (spectral methods in matlab)
function [D,x] = cheb(N)
% check the base case if N == 0; D = 0; x = 1; return; end
% create the Chebyshev grid x = cos(pi*(0:N)/N)';
c = [2; ones(N-1,1);2].*(-1).^(0:N)'; X = repmat(x,1,N+1); dX = X-X'; D = (c*(1./c)')./(dX+(eye(N+1))); D = D - diag(sum(D'));
This solution (u = D1\f) does not match the exact solution at all.
I think what I have is close ... Any help would be awesome. Thanks in advance!
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It took longer than I thought it would for me to learn something about spectral methods and particularly Chebyshev differentiation matrices. Fortunately, the relevant chapters of Spectral Methods in MATLAB are available online.
The problem you are having seems to be in the way you define your f variable:
f = 2*x.*ones(size(x));
in the context of your original differential equation:
dy/dx = 2*x*y
Specifically, y is not a vector of ones. (I am not sure how you would define it in that situation.) When I defined f as:
f = 2*x.*exp(x.^2 - 1);
and then calculated:
u = lsqr(D1,f);
I got a value for u that differed from the analytic integrated differential equation by a constant of 0.6773 but otherwise matched it with acceptable precision. (I haven't accounted for that particular constant of integration.)
I also calculated u using lsqr because using the mldivide appropriate for dense matrices threw a condition number warning when applied to this problem.