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Suppose I have the expression:

expr = U''[x] + U'[x]

I would like to be able to write an abstract replacement rule so that I can transform the above expression to:

u2 + u1

I was hoping that the replacement rule:

U''[x] + U'[x] /. Derivative[n_][U][x]->StringJoin["u",ToString[n]]

would work. However, this disappointingly uses a literal n, and not a variable n:

U''[x] + U'[x] /. Derivative[n_][U][x]->StringJoin["u",ToString[n]]
(* 2 un *)

I suspect I'm close with my solution and I need to use some kind of # and & method but I still do not fully understand those functions.

Any help would be appreciated.

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    $\begingroup$ (dexpr = U''[x] + U'[x]) // FullForm and try: dexpr /. Derivative[x_][f_][v_] :> ToLowerCase@ToString[f] <> ToString[x] $\endgroup$
    – Syed
    Commented May 4, 2022 at 17:53
  • $\begingroup$ @Syed ah yes that works. I always forget about the delayed replace. Using :> in my example also makes the replacement works as desired. $\endgroup$
    – akozi
    Commented May 4, 2022 at 17:56

1 Answer 1

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It is generally easier to deal with indexed variables and to format them for display in any desired manner.

expr = U''[x] + U'[x] + U[x];

Format[u[n_]] := Row[{u, n}]

expr2 = expr /. {Derivative[n_][U][_] :> u[n], U[_] :> u[0]}

enter image description here

Or,

Format[u[n_]] := Subscript[u, n]

expr2

enter image description here

Reversing the replacements

expr2 /. u[n_] :> Derivative[n][U][x]

enter image description here

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  • $\begingroup$ Whoops, sorry I have no idea why I did not accept this nearly a year ago, but this was the perfect solution to my problem. $\endgroup$
    – akozi
    Commented Sep 13, 2023 at 14:03

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