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MathGroup Archive 2005

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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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