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April 5, 2013, 15:18 |
Boundary condition in Spectral method
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#1 |
Member
Join Date: Feb 2011
Posts: 41
Rep Power: 15 |
Hi All
I am playing with the boundary conditions in spectral method using Chebyshev grids. Say, I want to solve the eigenfunction of . Analytically, the eigenfunction will be sin or cosine. Indeed, in Matlab, the following small code will generate the eigenfunction with three different boundary conditions. I choose to plot from each boundary condition case an eigenfunction and attach here. clear; nrmod=357; [y,DM]=chebdif(nrmod,4); D2=DM(:,:,2); % q is the eigenfunction. % bc: q(-1)=0,q(1)=0 % [ef,ev]=eig(D2(2:end-1,2:end-1)); % bc: q(-1)=0,q(1)=1. Just change the last 1 to 2 for q(-1)=0,q(1)=2 % [ef,ev]=eig(D2(2:end-1,2:end-1)+diag(D2(2:end-1,1))*1); % bc: q(-1)=1,q(1)=1 [ef,ev]=eig(D2(2:end-1,2:end-1)+diag(D2(2:end-1,1))*1+diag(D2(2:end-1,end))*1); for i=nrmod-2:-1:1 plot(y(2:end-1),real(ef(:,i)),'-*b') title([num2str(i) ' eigenvalue is ' num2str(ev(i,i))]) pause end My question is: Does it make sense to require the eigenfunction to have boundary condition q(-1)=0,q(1)=2? I tried in Matlab, but strange things happens, it returns the results of q(-1)=0,q(1)=0. Why is this? Thanks a lot in advance! Jo |
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April 5, 2013, 16:58 |
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#2 |
Member
Join Date: Feb 2011
Posts: 41
Rep Power: 15 |
I am sorry.
I just realized that the way imposing the boundary condition as implemented here is not right. Could anyone tell me the right way?......Thanks. Jo |
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