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Science Forum Index » Math - Numerical Analysis Forum » ? deconv and Poisson
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| Cheng Cosine |
 Posted: Wed Feb 28, 2007 6:36 pm |
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hi:
Given a Poisson eqn div( a(x,y,z)*grad(u(x,y,z)) ) = q(x,y,z), its soln
can be expressed in terms of Green's function as u = conv( G, q ).
Since we know Fourier transform of conv is a product of two functions,
we can obtain the Green's function by G(x,y,z) = ifft( fft(u)./fft(q) ).
So we only need to solve Poisson eqn once with some complicate method
like FEM and then we can determine other solns by the following.
u = ifft( fft(G).*fft(q1) ) = ifft( fft(u)./fft(q).*fft(q1) ).
Will the above work? Are there some technical difficulties I missed?
Are there more efficient way to do the deconvolution?
Thanks,
by Cheng Cosine
Feb/28/2k7 NC |
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| William R. Frensley |
| Posted: Thu Mar 01, 2007 12:14 pm |
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Cheng Cosine wrote:
Quote: hi:
Given a Poisson eqn div( a(x,y,z)*grad(u(x,y,z)) ) = q(x,y,z), its soln
can be expressed in terms of Green's function as u = conv( G, q ).
Since we know Fourier transform of conv is a product of two functions,
we can obtain the Green's function by G(x,y,z) = ifft( fft(u)./fft(q) ).
So we only need to solve Poisson eqn once with some complicate method
like FEM and then we can determine other solns by the following.
u = ifft( fft(G).*fft(q1) ) = ifft( fft(u)./fft(q).*fft(q1) ).
Will the above work? Are there some technical difficulties I missed?
Are there more efficient way to do the deconvolution?
It works if your boundary conditions are at infinity. The Fourier basis
functions do not in general satisfy the boundary conditions (whatever they
may be) over finite surfaces. In that case, you have to do your expansion
over the set of orthogonal functions that satisfy your boundary conditions.
See Jackson's Classical Electrodynamics (the first few chapters) to get an
idea of how much complexity this topic can produce.
- Bill Frensley |
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| Cheng Cosine |
| Posted: Fri Mar 02, 2007 9:23 am |
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"William R. Frensley" <frensley@utdallas.edu> wrote in message
news:DZCFh.6$BX.0@newssvr11.news.prodigy.net...
Quote: Cheng Cosine wrote:
hi:
Given a Poisson eqn div( a(x,y,z)*grad(u(x,y,z)) ) = q(x,y,z), its soln
can be expressed in terms of Green's function as u = conv( G, q ).
Since we know Fourier transform of conv is a product of two functions,
we can obtain the Green's function by G(x,y,z) = ifft( fft(u)./fft(q) ).
So we only need to solve Poisson eqn once with some complicate method
like FEM and then we can determine other solns by the following.
u = ifft( fft(G).*fft(q1) ) = ifft( fft(u)./fft(q).*fft(q1) ).
Will the above work? Are there some technical difficulties I missed?
Are there more efficient way to do the deconvolution?
It works if your boundary conditions are at infinity. The Fourier basis
functions do not in general satisfy the boundary conditions (whatever they
may be) over finite surfaces. In that case, you have to do your expansion
over the set of orthogonal functions that satisfy your boundary
conditions.
See Jackson's Classical Electrodynamics (the first few chapters) to get an
idea of how much complexity this topic can produce.
Hmm, did not expect that complicate issues. So maybe start with du/dt =
A*u+q is
simpler. The main issue here is to get A or to estimate A or to estimate
exp( A*t )
since u = exp( -A*t )*u_0+conv( exp( -A*t ), q ). Once one get its kernel
function
one can know soln of u for any different q. The problem is how to get this
from output
vector u and input vector q.
Thanks,
by Cheng Cosine
Mar/02/2k7 NC |
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