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Re: variable substitution in differential eqns
*To*: mathgroup at smc.vnet.net
*Subject*: [mg61960] Re: [mg61950] variable substitution in differential eqns
*From*: "Carl K. Woll" <carl at woll2woll.com>
*Date*: Sun, 6 Nov 2005 05:11:35 -0500 (EST)
*References*: <200511050653.BAA02074@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Joseph Fagan wrote:
> For Calculus of Variations, I need to make some hairy variable
> substitutions.
>
> A simple substitution example is shown here in the first few lines.
>
> See
> http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html
>
> If I make the substitution x=1/z how can I get Mathematica to give
> me eqn (2)
> and eqn (5)
> and, being greedy, eqn (6)?
> or point me to where to begin.
>
> Thanks
> Joe
>
The question is how to convert y''[x] to the appropriate expression when
x -> 1/z. One idea is to use Composition:
In[23]:=
y''[x] /. y->Composition[y,1/#&] /. x->1/z
Out[23]=
3 4
2 z y'[z] + z y''[z]
The equation to be transformed has the left hand side:
lhs = y''[x] + P[x]y'[x] + Q[x]y[x]
Using the Composition rule, we find:
In[25]:=
lhs /. y->Composition[y,1/#&] /. x->1/z
Out[25]=
1 3 2 1 4
Q[-] y[z] + 2 z y'[z] - z P[-] y'[z] + z y''[z]
z z
If we want to have Q and P transformed as well, we should include
Composition rules for them too:
In[26]:=
lhs /. h_ /; MemberQ[{P,Q,y},h] -> Composition[h,1/#&] /. x->1/z
Out[26]=
3 2 4
Q[z] y[z] + 2 z y'[z] - z P[z] y'[z] + z y''[z]
This agrees with equation (6) from the MathWorld link.
Carl Woll
Wolfram Research
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