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*: Sat, 5 Nov 2005 22:41:45 -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

**Follow-Ups**:**Re: Re: variable substitution in differential eqns***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**References**:**variable substitution in differential eqns***From:*"Joseph Fagan" <noemailplease@nowhere.ru>