Re: Replace in operators once again
- To: mathgroup at smc.vnet.net
- Subject: [mg103363] Re: [mg103341] Replace in operators once again
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Thu, 17 Sep 2009 06:21:33 -0400 (EDT)
- References: <200909160947.FAA13041@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
Like this, I think:
Clear[f, g, a, b, c, F]
f[t_, x_] = a g[t/b, x/c];
eqn1=D[f[t,x],t]==D[f[t,x],{x,2}]+k F[f[t,x]]
(a (g^(1,0))[t/b,x/c])/b==k F[a g[t/b,x/c]]+(a (g^(0,2))[t/b,x/c])/c^2
eqn2=eqn1/.{t->b u,x->c v}
(a (g^(1,0))[u,v])/b==k F[a g[u,v]]+(a (g^(0,2))[u,v])/c^2
There is no F' term, but I think you didn't really mean a derivative of F.
Bobby
On Wed, 16 Sep 2009 04:47:27 -0500, Alexei Boulbitch
<Alexei.Boulbitch at iee.lu> wrote:
> Dear Community members,
>
> I would like to put a question closely related to "[mg102869] Replace in
> operators", the one recently discussed here. See
> http://forums.wolfram.com/mathgroup/archive/2009/Sep/msg00006.html
> and the thread.
> It concerns analytical transformations in differential equations. I need
> to make a replacement, not just f[x]->g[x] as discussed in [mg102869],
> but
> a simple rescaling of both the function and the coordinate. To be more
> concrete, consider a PDE over two variables time (t) and coordinate (x):
>
> df/dt=d^2f/dx^2 + k df/dx + F(f)
>
> Here f is a function f=f(t,x), F=F(f) is another function for instance,
> a polynomial in terms of f, and depending upon some parameters.
> Finally, k is a constant. In order to reduce the number of parameters in
> this equation to the minimum one may wish to rescale both the function
> and the both coordinates as follows:
>
> f[t, x]->a*g[u,v]; t->b*u; x->c*v
>
> where a, b and c are some constants. What one finds after the rescaling
> looks like the following:
>
> (a/b)dg/du=(a/c^2)d^2g/dv^2 + k(a/c) dg/dv + F'(g)
>
> where F' is the transformed polynomial. Then this should be manipulated
> further, and it is important to be able to hold all these manipulations
> on-screen, rather than to go to the paper for intermediate steps.
> It is not difficult to make the first substitution f[t, x]->a*g[u,v].
> One does not even need to use Replace:
>
> In[40]:= SetAttributes[{a, b, c}, Constant];
>
> f[t_, x_] := a*g[t, x]
> D[f[t, x], x]
> D[f[t, x], {x, 2}]
>
> Out[42]= a
> \!\(\*SuperscriptBox["g",
> TagBox[
> RowBox[{"(",
> RowBox[{"0", ",", "1"}], ")"}],
> Derivative],
> MultilineFunction->None]\)[t, x]
>
> Out[43]= a
> \!\(\*SuperscriptBox["g",
> TagBox[
> RowBox[{"(",
> RowBox[{"0", ",", "2"}], ")"}],
> Derivative],
> MultilineFunction->None]\)[t, x]
>
> However, I cannot see how to cope with the second and the third
> substitutions. Evidently, the simple
>
> In[45]:= \!\(
> \*SubscriptBox[\(\[PartialD]\), \(x\)]\ \(f[t, x]\)\) /. x -> c*v
>
> Out[45]= a
> \!\(\*SuperscriptBox["g",
> TagBox[
> RowBox[{"(",
> RowBox[{"0", ",", "1"}], ")"}],
> Derivative],
> MultilineFunction->None]\)[t, c v]
>
> does not make the job.
>
> A closely related question: assume we need to make a substitution of the
> type x->g[v] into derivative. Say, x->Log[v]. I would like to have the
> result in a form
> v D[g[v],v]. Instead I get of coarse,
>
> In[48]:= D[q[x], x] /. x -> Log[v]
>
> Out[48]=
> \!\(\*SuperscriptBox["q", "\[Prime]",
> MultilineFunction->None]\)[Log[v]]
>
> rather than what I need. Could you think of simple solutions for these
> cases?
>
> Thank you, Alexei
>
--
DrMajorBob at yahoo.com
- References:
- Replace in operators once again
- From: Alexei Boulbitch <Alexei.Boulbitch@iee.lu>
- Replace in operators once again