Re: How to express ODE?
- To: mathgroup at smc.vnet.net
- Subject: [mg52831] Re: How to express ODE?
- From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
- Date: Tue, 14 Dec 2004 05:59:27 -0500 (EST)
- Organization: Uni Leipzig
- References: <cpjr25$nvl$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
a) if V[phi] generate a function phi[t] you has to build the derivative
with respect to phi[t] and not with respect to phi only
b) the parameters are extrem bad scaled for numerical work and you should
rescale
it.
c) Your input say
> f = \[Phi] /. First[NDSolve[
> {Inflaton == 0,
> Derivative[1][\[Phi]][
> t] == 0}, \[Phi],
> {t, 0, 100*m}]]
that phi'[t] (for all t) is zero, and so phi[t] can be only a constant
d) you have missed the initial conditions for phi[0] and phi'[0] and
NDSolve[] can't work with out that.
e) until you don't proper scale the parameters and the variables
NDSolve[] will need a very long time to compute the solution
with the given accuracy
Regards
Jens
"Adam Getchell" <agetchell at physics.ucdavis.edu> schrieb im Newsbeitrag
news:cpjr25$nvl$1 at smc.vnet.net...
> Still (unfortunately) not getting another ODE to work, even using Dr.
> Bob's example. I want to express this equation:
>
> In[143]:=
> m = 10^16;
> mp = 1.2211*10^19;
> G = mp^(-2);
> V[\[Phi]_] := (1/2)*m^2*\[Phi][t]
> \[Rho][\[Phi]_] :=
> Derivative[1][\[Phi]][t]^2/2 +
> V[\[Phi]]
> H := Sqrt[((8*Pi*G)/3)*
> \[Rho][\[Phi]]]
>
> In[149]:=
> Inflaton :=
> Derivative[2][\[Phi]][t] +
> 3*H*Derivative[1][\[Phi]][
> t] + D[V[\[Phi]], \[Phi]]
>
> That is, phi double-dot of t + 3 H phi dot of t + V' of phi, where phi
> is a function of t; V is a function of phi; H is a function of rho; rho
> is a function of phi dot and V.
>
> However, solving Inflaton for phi as a function of t:
>
> In[163]:=
> f = \[Phi] /. First[NDSolve[
> {Inflaton == 0,
> Derivative[1][\[Phi]][
> t] == 0}, \[Phi],
> {t, 0, 100*m}]]
>
> Generates NDSolve::overdet and ReplaceAll::reps.
>
> I *think* part of my problem is that dV/d(phi) is not getting properly
> evaluated (it returns 0 when I look at it explicitly).
>
> I thought that Derivative[1][\[Phi]][x] would give the above, but not
> the way I'm doing it.
>
> --Adam
>