I'm trying to set the derivative of a function to zero at a specified point. This will act as an initial condition for an algorithm I'm writing: so I'm specifying the value of the function at x=0, and I'm specifying the value of the slope of the function at x=0. The function is u[n, x] and is unknown. The algorithm, based on Adomian decomposition method, will solve a nonlinear differential equation for this u-function.

u /: D[u[n_, x_], x_] := 0
D[u[n, x], x] == 0

It gives the right output -- says it's True.

The problem I'm having is other queries say they are True when they shouldn't be:

D[u[n, x], 0] == 0 \\True
D[u[n, 0], 0] == 0 \\True

This is a somewhat duplicate of a question here, which is asked in a broader sense: How to set the derivative of a function to zero?

Ideally, what should happen is based on modifying this line of code:

u /: (D[u[n_, x_], x_] := 0)/.x->0

The result would be:

(D[u[n, x], x] == 0)/.x->0 \\True
(D[u[n, x], x] == 0)/.x->5 \\False or indeterminate

Thank you for any tips.

  • $\begingroup$ Why not attach rules with Derivative[] instead? u /: Derivative[0, 1][u][n_, x_] := 0. $\endgroup$ Jul 7, 2016 at 23:36
  • 1
    $\begingroup$ @J.M. In that case, I think u is too deeply nested in the expression, so it won't work. Every time I think UpValues is the solution to one of my problems, I get this error. $\endgroup$
    – march
    Jul 7, 2016 at 23:47
  • $\begingroup$ @march, thanks for trying it out; I currently do not have Mathematica on hand. I guess a simple Derivative[0, 1][u][n_, x_] := 0 should have to suffice. $\endgroup$ Jul 7, 2016 at 23:48
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    $\begingroup$ @J.M. Or, as I read it, Derivative[0, 1][u][n_, 0] := 0. $\endgroup$
    – Michael E2
    Jul 8, 2016 at 0:33
  • $\begingroup$ Thank you guys! Michael's advice worked. I did not know we could set derivatives equal to a number using := notation, but it makes sense. $\endgroup$
    – Buddhapus
    Jul 8, 2016 at 17:08


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