On Oct 23, 8:09 pm, Cheng Cosine <asec...@gmail.com> wrote:
> Hi:
> Given dy/dt = a*y+f, and we want to estimate the value of a by > measuring
> the value of y for a given f. The question is how many different f > should
> we set as the input so that we have enough output(s) of y to estimate > a?
If y(0) = y0, the solution of the DE is y(t) = -(f/a) + (f+y0*a)*exp (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.
> On Oct 23, 8:09 pm, Cheng Cosine <asec...@gmail.com> wrote:
> > Hi:
> > Given dy/dt = a*y+f, and we want to estimate the value of a by > > measuring
> > the value of y for a given f. The question is how many different f > > should
> > we set as the input so that we have enough output(s) of y to estimate > > a?
> If y(0) = y0, the solution of the DE is y(t) = -(f/a) + (f+y0*a)*exp > (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, > > 2)+f
> > where pdiff(u, x, 2) means twice partial derivate of u to x. Again we > > want
> > 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,- Hide quoted text -
> - Show quoted text -
Thanks, but in practice any measurement come with noise, which is kind of
random. Is this why in practice we need to take many measurements? But why
taking more than the number of measurements required when we have exact
> On Oct 23, 8:09 pm, Cheng Cosine <asec...@gmail.com> wrote:
> > Hi:
> > Given dy/dt = a*y+f, and we want to estimate the value of a by > > measuring
> > the value of y for a given f. The question is how many different f > > should
> > we set as the input so that we have enough output(s) of y to estimate > > a?
> If y(0) = y0, the solution of the DE is y(t) = -(f/a) + (f+y0*a)*exp > (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, > > 2)+f
> > where pdiff(u, x, 2) means twice partial derivate of u to x. Again we > > want
> > 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,- Hide quoted text -
> - Show quoted text -
What if the eqn to be estimated is:
1. dy/dt = a*b*y+f -> estimate a and b
does it mean that one needs to have 2 measurements?
2. dy/dt = ( a+b )*y+f -> estimate a and b
does it mean that one needs to have 2 measurements?
Some say that one cannot estimate a product of a*b because the value of the product
is the same for a smaller a multiplied with large b. But this argument applies as
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