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I generated a very simple rule that is of the type

{a[0] -> 0,  a[1] -> 0,  a[2] -> 8/3*pi}

And I want to impose this set of rules on an equation. The problem is that in this equation there are some derived terms like a'[0]. How can I use the rule directly by imposing its derivative where needed as well?

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    – bmf
    Feb 2 at 1:14
  • $\begingroup$ You can do this also in the same way i.e. {a[0] -> 0, a[1] -> 0, a[2] -> 8/3*pi, a'[0]->value} $\endgroup$
    – Dunlop
    Feb 2 at 5:06

2 Answers 2

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Clear["Global`*"]

Given rules like {a[0] -> 0, a[1] -> 0, a[2] -> 8/3*Pi} you cannot take their derivative since everything is constant.

An example for which derivatives can be taken.

eqn = {a''[x] + 4 a[x] == 0, a[0] == 1, a'[0] == 4};

Rule:

sol = DSolve[eqn, a[x], x][[1]]

(* {a[x] -> Cos[2 x] + 2 Sin[2 x]} *)

Derivatives:

der = NestList[D[#, x] &, sol, 2] // Flatten

(* {a[x] -> Cos[2 x] + 2 Sin[2 x], 
 a'[x] -> 4 Cos[2 x] - 2 Sin[2 x], a''[x] -> -4 Cos[2 x] - 
   8 Sin[2 x]} *)

Initial conditions:

ic = Most@der /. x -> 0

(* {a[0] -> 1, a'[0] -> 4} *)

Verification,

eqn /. der /. ic // Simplify

(* {True, True, True} *)
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If you want to set derivatives of constants to zero you can write e.g.:

2 a[0] + 3 a[1]' + 4 a[2]'' /. {Derivative[_][_] -> 0}
(* 2 a[0] *)

However, if the derivatives have some values that you want to replace, you can write:

drule = {a[0]' -> 11, a[1]' -> 22, a[2]' -> 33};
2 a[0] + 3 a[1]' + 4 a[2]' /. drule
(* 198 + 2 a[0] *)
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