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 >