Re: ? approximates in different subspaces
- From: spellucci@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx (Peter Spellucci)
- Date: Tue, 1 Aug 2006 10:13:17 +0000 (UTC)
In article <Okvzg.12913$4c7.12485@xxxxxxxxxxxxxxxxxxxxxxxx>,
"Cheng Cosine" <acosine@xxxxxxxxxxxx> writes:
u(0)=0???
Let's go one step back.
I saw someone solved du/dt = A*u+b*q where A is N-by-N, u and b are N-by-1,
and q is a scalar.
Taking Laplace transformation:
s*u(s) = A*u(s)+b*q(s) => u(s) = inv( s*I-A )*b*q(s) = inv(
I-s*inv(A) )*( -inv(A)*b )*q(s)
= sum( inv(A)^n*s^n, n =
0..Inf )*( -inv(A)*b )*q(s)
~ sum( inv(A)^n*s^n, n =
0..k-1 )*( -inv(A)*b )*q(s)
That is, instead of determine u from full system, get kth order
approximation.
That person said that this approximation is spanned by {c, inv(A)*c, ...,
inv(A)^(k-1)*c}, where c = -inv(A)*b.
which certainly is true
So use Gram-Schmidt orthogonal process to get a k-dim orthogonal subspace
from {c, inv(A)*c, ..., inv(A)^(k-1)*c}.
Put it iin a matrix V taht is N-by-k. Then approximated solution is given by
the following.
dw/dt = V'*A*V*w+V'*b*q, the kth order approximation uk = V*w.
Why does this work? In principle, whatever orthnormal matrix V serves, but
you replace u by V*w, V is fixed (since A is fixed)
hence (d/dt)(V*w) = V*(d/dt w)
and multiply from the left by V', using V'*V=I_k
are there some
kind of optimal V?
Thanks,
by Cheng Cosine
Jul/31/2k6 NC
first of all: i hardly can see how it makes sense to solve such a simple
ode _numerically_ using this dangerous numerical process. this is indeed
absolutely contraproductive, stone aged math.
(if it had to deal with an analytical discussion, this is quite different)
with an orthogonal V it makes sense if A is symmetric (and, for stability
reasons, I assume negative definite.)
then taking V as the span of the k eigenvectors to the k algebraically largest,
i.e. absolutely smallest) eigenvalues would make sense, or., say the
first k Lanczos vectors from Lanczos combined with inverse iteration
(in order to approximate the dominant exponential part (not the stiff one)
of the solution.) but, as said, I never would solve an ode this way,
even in the large scale case
hth
peter
.
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