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!

Answer by Star Strider
on 18 Oct 2012

Edited by Star Strider
on 19 Oct 2012

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

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

I also calculated ` u` using

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Answer by Qiqi Wang
on 11 May 2017

Edited by Qiqi Wang
on 11 May 2017

The original post was right that the boundary condition was not set correct. Here's a right way to set the boundary condition:

% compute the chebyshev differentiation matrix and x-grid [D,x] = cheb(N); % compute the derivatives at x (i.e. at the chebyshev grid points) f = 2*x.*exp(x.^2 - 1); % boundary condition (correct way) D(end,:) = 0; D(end,end) = 1; f(end) = 1; % solve u = D\f;

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