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Newsgroups: sci.math
From: Ray Vickson <RGVick...@shaw.ca>
Date: Fri, 23 Oct 2009 22:20:21 -0700 (PDT)
Local: Sat, Oct 24 2009 5:20 am
Subject: Re: ? estimating coef of DE
On Oct 23, 8:09 pm, Cheng Cosine <asec...@gmail.com> wrote:
> Hi: If y(0) = y0, the solution of the DE is y(t) = -(f/a) + (f+y0*a)*exp > Given dy/dt = a*y+f, and we want to estimate the value of a by > the value of y for a given f. The question is how many different f > we set as the input so that we have enough output(s) of y to estimate (a*t)/a, so if you know y0 it is enough to measure y at one non-zero value of t. If you don't know y0, you need two measurements, y1 = y (t1) and y2 = y(t2). This gives two nonlinear equations to be solved for y0 and a. These can be tackled numerically using your favorite equation-solving routine. R.G. Vickson > Suppose we are dealing with a PDE like pdiff(u, t, 1) = a*pdiff(u, x, > where pdiff(u, x, 2) means twice partial derivate of u to x. Again we > to estimate a for given f and measuremnt u. What are the extra > considerations we have, compared with the case when we deal with > only ode? > Thanks, You must Sign in before you can post messages.
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