Initial value problem Help Please [Math]

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From: mo on 2 Jun 2010 04:04

Can someone help me with this.
I need to solve an exponential equation for a project of mine. It's
an initial value problem where I know the value of y at t=0 as well as
y at t=60s and 120s, but I don't know the value of y=n at t=infinity.
The equation is y=C exp^(-kt) + n
I need to know if the value of n can be solved given the initial value
and one or two intermediate values of y . At infinity the exponential
term will disappear leaving only n, which is what I need. I just need
to know if it can be done or is there insufficient data to determine
both the exponential term k and either C or n, one will lead to the
other.
Thanks

From: Torsten Hennig on 2 Jun 2010 00:20

> Can someone help me with this.
> I need to solve an exponential equation for a
> project of mine. It's
> an initial value problem where I know the value of y
> at t=0 as well as
> y at t=60s and 120s, but I don't know the value of
> y=n at t=infinity.
> The equation is y=C exp^(-kt) + n
> I need to know if the value of n can be solved given
> the initial value
> and one or two intermediate values of y . At
> infinity the exponential
> term will disappear leaving only n, which is what I
> need. I just need
> to know if it can be done or is there insufficient
> data to determine
> both the exponential term k and either C or n, one
> will lead to the
> other.
> Thanks

If you have y at three different times, you can
in principle solve for the three unknown parameters:
(1) y(t=0) = C+n
(2) y(t=60) = C*exp(-k*60)+n
(3) y(t=120) = C*exp(-k*120)+n.

First solve (1) for n and insert in (2) and (3).
You'll end up in a quadratic equation for exp(-60*k)
which can be solved analytically for k.

But more measurements and a nonlinear regression
to your function y=C*exp(-k*t)+n may give better
results for the parameters C,k and n due to
measurement errors at only three given times.

Best wishes
Torsten.

From: mo on 2 Jun 2010 04:41

On Jun 2, 2:20 am, Torsten Hennig <Torsten.Hen...(a)umsicht.fhg.de>
wrote:
> > Can someone help me with this.
> > I need to solve an exponential equation for a
> > project of mine. It's
> > an initial value problem where I know the value of y
> > at t=0 as well as
> > y at t=60s and 120s, but I don't know the value of
> > y=n at t=infinity.
> > The equation is y=C exp^(-kt) + n
> > I need to know if the value of n can be solved given
> > the initial value
> > and one or two intermediate values of y . At
> > infinity the exponential
> > term will disappear leaving only n, which is what I
> > need. I just need
> > to know if it can be done or is there insufficient
> > data to determine
> > both the exponential term k and either C or n, one
> > will lead to the
> > other.
> > Thanks
>
> If you have y at three different times, you can
> in principle solve for the three unknown parameters:
> (1) y(t=0) = C+n
> (2) y(t=60) = C*exp(-k*60)+n
> (3) y(t=120) = C*exp(-k*120)+n.
>
> First solve (1) for n and insert in (2) and (3).
> You'll end up in a quadratic equation for exp(-60*k)
> which can be solved analytically for k.
>
> But more measurements and a nonlinear regression
> to your function y=C*exp(-k*t)+n may give better
> results for the parameters C,k and n due to
> measurement errors at only three given times.
>
> Best wishes
> Torsten.

Thanks for your help Torsten.
I will try that.

From: mo on 2 Jun 2010 04:53

From: Torsten Hennig on 2 Jun 2010 01:37

> On Jun 2, 2:20 am, Torsten Hennig
> <Torsten.Hen...(a)umsicht.fhg.de>
> wrote:
> > > Can someone help me with this.
> > > I need to solve an exponential equation for a
> > > project of mine. It's
> > > an initial value problem where I know the value
> of y
> > > at t=0 as well as
> > > y at t=60s and 120s, but I don't know the value
> of
> > > y=n at t=infinity.
> > > The equation is y=C exp^(-kt) + n
> > > I need to know if the value of n can be solved
> given
> > > the initial value
> > > and one or two intermediate values of y . At
> > > infinity the exponential
> > > term will disappear leaving only n, which is what
> I
> > > need. I just need
> > > to know if it can be done or is there
> insufficient
> > > data to determine
> > > both the exponential term k and either C or n,
> one
> > > will lead to the
> > > other.
> > > Thanks
> >
> > If you have y at three different times, you can
> > in principle solve for the three unknown
> parameters:
> > (1) y(t=0) = C+n
> > (2) y(t=60) = C*exp(-k*60)+n
> > (3) y(t=120) = C*exp(-k*120)+n.
> >
> > First solve (1) for n and insert in (2) and (3).
> > You'll end up in a quadratic equation for
> exp(-60*k)
> > which can be solved analytically for k.
> >
> > But more measurements and a nonlinear regression
> > to your function y=C*exp(-k*t)+n may give better
> > results for the parameters C,k and n due to
> > measurement errors at only three given times.
> >
> > Best wishes
> > Torsten.
>
> Problem is C and n are both unknown.
>

I get

k = 1/60 * log((y(t=60)-y(t=0))/(y(t=120)-y(t=0)))
C = (y(t=60)-y(t=0))^2 / (y(t=120)-y(t=60))
n = y(t=0) - (y(t=60)-y(t=0))^2 / ((y(t=120)-y(t=60))

ln: natural logarithm

Best wishes
Torsten.

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