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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/

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