Re: (in)dependent variables in DSolve: need explanation.

*To*: mathgroup at smc.vnet.net*Subject*: [mg73488] Re: [mg73472] (in)dependent variables in DSolve: need explanation.*From*: "Chris Chiasson" <chris at chiasson.name>*Date*: Mon, 19 Feb 2007 06:35:35 -0500 (EST)*References*: <200702190624.BAA13773@smc.vnet.net>

I can't answer your question, but I wanted to mention the function LaplaceTransform. On 2/19/07, barmau.maurice <barmau.maurice at orange.fr> wrote: > Hello, > > I have a partial differential equation which I can solve by hand, but not > yet in Mathematica. I need an explanation: > In queueing theory I have: z dP(z,t)/dt = (1-z)((a - b z)* P(z,t) - bP0(t)) > with P(z,0)=z^i (i given). I can take the Laplace transform of the DE: > (z^(i+1) - b (1-z)Laplace(P0(t)) > --------------------------------- > (sz - (1-z)(b - az) > > The numerator must be zero, so it follows that Laplace(P(0,t)) is known now. > The rest is substitution in the first formula and after that taking the > inverse. > > In fact I have the DE and search for solutions such that P0(t) exists. How > do I do that in Mathematica with the DE and without applying the Laplace > transform? Is that possible? > > The DE comes from the well known Markov equation for a single channel > Poisson input, and exponential holding time. Its equation starts with dPo/dt > = -a Po(t) + b P1(t) and the rest: dPn/dt=aPn-1(t)-(a+b)Pn(t)+bPn+1(t). > The above used P(z,t) is the generating function of Pn: > P(z,t) = Sum from zero to infinity (Pn(t) z^n). > Well one can find it in Saaty's Elements of Queueing Theory, page 88. It is > a standard procedure. I now would like to do it in Mathematica, and my basic > question is: can it be done with DSolve only. > > > > -- http://chris.chiasson.name/

**Follow-Ups**:**netCDF in Mathematica***From:*"M.I. Mead" <mim25@cam.ac.uk>

**References**:**(in)dependent variables in DSolve: need explanation.***From:*"barmau.maurice" <barmau.maurice@orange.fr>