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I have a differential equation of the form

$$f'(x)=g(x,f(x))$$

(where $g$ is has a known explicit form, e.g. $x f(x)$) and I have an expression which contains several derivatives of $f$. I want to use the differential equation to substitute all the derivatives in order to leave the expression as a function of $f(x)$ and $x$ only. For this purpose I have written the following rule:

ruleDerF = Derivative[n_][f][x_] -> D[x f[x], {x, n - 1}];

This works ok, e.g.,

f''[x] /. ruleDerF

(*f[x] + x f'[x]*)

and

f''[x] //. ruleDerF

(*f[x] + x^2 f[x]*)

The problem appears when I want to apply this rule when the function is evaluated for a value of $x$. If I try naively to apply this rule on $f(1)$ it gives an obvious error:

f''[1] /. ruleDerF

General::ivar: 1 is not a valid variable. >>

(* \!\(\*SubscriptBox[\(\[PartialD]\), \({1, 1}\)]\(f[1]\)\) *)

This is obvious because you cannot derive wrt 1. So I tried to redefine the rule in the following way, 

ruleDerF = Derivative[n_][f][x_] -> (D[y f[y], {y, n - 1}]/.y:>x);

believing that the delayed rule would cause to first perform the derivation wrt the symbolic variable y and then evaluate it to the introduced value, but it doesn't work this way (I get the same result as before).

Any idea that could help me?

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    $\begingroup$ Perhaps this should be marked as a duplicate of (22917) or one of the many questions it is linked to? $\endgroup$
    – Mr.Wizard
    Jul 21, 2016 at 11:17
  • $\begingroup$ You might be interested in DifferentialRoot[]: f = DifferentialRoot[Function[{y, x}, {y'[x] == x y[x], y[0] == 0}]] and then look at f'' for example. This will only work if you have given initial conditions and your DEs are linear. $\endgroup$
    – J. M.'s eventual burnout ♦
    Jul 21, 2016 at 11:58
  • $\begingroup$ Are you trying to prevent the derivative D[y f[y], {y,n-1}] from being carried out every time the rule is applied? One can imagine applying this rule thousands of times for a fairly complicated f[x], and wishing for the derivative to only be carried out once, and stored thereafter. This is how I interpreted in intent behind -> rather than :>, but perhaps it's simply a mistake that I'm overthinking. $\endgroup$
    – jjc385
    Aug 6, 2016 at 1:56

1 Answer 1

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Define your rule as a function:

ruleDerF = (# /. Module[{x}, 
   Derivative[n_][f][var_] :> (D[x f[x], {x, n - 1}] /. x -> var)
])&

f''[x] // ruleDerF

f[x] + x Derivative[1][f][x]

 FixedPoint[ruleDerF, f''[x]]

f[x] + x^2 f[x]

FixedPoint[ruleDerF, f''[1]]

2 f[1]

(havent tested heavily)

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  • $\begingroup$ Shouldn't you localize x? And wouldn't mentioning RuleDelayed be useful to the OP? $\endgroup$
    – Mr.Wizard
    Jul 21, 2016 at 11:04
  • $\begingroup$ Maybe I'm crazy but at first blush it looks like he is using :> and -> backward, or at least haphazardly. Certainly you changed -> to :> following Derivative[n_][f][x_]. $\endgroup$
    – Mr.Wizard
    Jul 21, 2016 at 11:08
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    $\begingroup$ Trying to understand why this solution works I realized that defining the rule as Derivative[n_][f][x_] :> (D[y f[y], {y, n - 1}]/.y->x) does the job. The only thing left, as mentioned, is to localize the dummy variable y. $\endgroup$
    – dpravos
    Jul 21, 2016 at 11:11
  • $\begingroup$ Finally I have opted for defining the rule as Derivative[n_][f][var_] :> Module[{x}, (D[x f[x], {x, n - 1}] /. x -> var)]. I will mark the answer as right because it's equivalent. Thank you for your insights. $\endgroup$
    – dpravos
    Jul 21, 2016 at 11:15
  • $\begingroup$ @DavidPravos no need to accept so quickly, I will add more explanations later. Some things are needed and some may be confusing, like using FixedPoint, which isn't necessary. $\endgroup$
    – Kuba ♦
    Jul 21, 2016 at 11:17

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