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differential difference equation
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rancid moth
Oct 10, 2012, 2:31:50 PM10/10/12
to
hello,
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
cheers
moth
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
cheers
moth
Ken Pledger
Oct 10, 2012, 10:37:20 PM10/10/12
to
In article <k54ev4$qpa$1...@dont-email.me>,
rancid moth <ranci...@yahoo.com> wrote:
> .... how does one go about solving the following finite system of
> 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.
rancid moth <ranci...@yahoo.com> wrote:
> .... how does one go about solving the following finite system of
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.
rancid moth
Oct 14, 2012, 5:25:08 AM10/14/12
to
need to check the conditions.
cheers
moth
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