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From: rancid moth <rancidm...@yahoo.com>
Newsgroups: sci.math
Subject: Re: differential difference equation
Date: Sun, 14 Oct 2012 20:25:08 +1100
Organization: A noiseless patient Spider
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References: <k54ev4$qpa$1@dont-email.me> <ken.pledger-30A160.15372011102012@news.eternal-september.org>
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On 11/10/2012 1:37 PM, Ken Pledger wrote:
> In article <k54ev4$qp...@dont-email.me>,
>   rancid moth <rancidm...@yahoo.com> wrote:
>
>> .... how does one go about solving the following finite system of
>> differential difference equations (i can solve the infinite system easy
>> enough...i don't see how to bring in the 0<x<n restriction)
>>
>> The unknown function if P(x,t) x = integer such that 0<x<=n.
>> Exp=exponential, L =constant, t=variable
>>
>> d/dt ( Exp(L*t)P(1,t) ) =0
>> d/dt ( Exp(2*L*t)P(2,t) ) = L*P(1,t)*exp(2*L*t)
>> ...
>> d/dt ( Exp(n*L*t)P(n,t) ) = L*(n-1)*P(n-1)*exp(n*L*t)
>>
>> such that sum(k=1,k=n) P(k,t) = 1   and P(n,0)=1 if n=1, 0 otherwise....
>
>
>     Presumably the  P(n-1)  should be  P(n-1, t).
>
>     If you leave out the restriction   sum(k=1,k=n) P(k,t) = 1
> then the system appears to have the unique solution
> P(x,t)  =  exp(-Lt)*(1 - exp(-Lt))^(x-1).
>
>     But then imposing the condition   sum(k=1,k=n) P(k,t) = 1
> leads to   (for every t) (1 - exp(-Lt)  =  0)
> which forces L to be 0, so makes everything trivial.
>
>     Are you sure about all your conditions?
>
>        Ken Pledger.
>
hmm.  yes i get that too.  things are not making sense...you're right i 
need to check the conditions.

cheers
moth