Following the question in here, now I would like to make the terms with derivatives to vanish. To shortly explain, given a variable decomposed as follows:

$a(t,r) = a(r) + \delta a(t,r)$

in my long calculations several derivatives of the term $\delta a(t,r)$ appears like $\partial_t \delta a(t,r)$ or $\partial_r \delta a(t,r)$. But I need that multiplied $\delta$-terms with another to vanish, like

$\partial_r \delta a(t,r) \partial_r \delta b(t,r)=0$ or $(\partial_r \delta a(t,r))^2 =0$ or $\partial_r \delta a(t,r) \partial_t \delta a(t,r)$.

In my previous question, the method showed was able to answer only when there are not derivative terms and I would like to generalize it. If I follow the method showed, Mathematica takes the derivative of the variable $\delta a(t,r)$ as:

\partial_r \delta[a[t,r]] = \delta'[a[t,r]] \partial_r a[t,r].

following the chain rule. And, obviously, it does not results in zero when two terms of this type are multiplied.

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    $\begingroup$ Multiply all your deltas by some ϵ. Set ϵ /: ϵ^n_ /; n >= 2 = 0;. Voilà. $\endgroup$ Commented Sep 21, 2019 at 13:40
  • $\begingroup$ @AccidentalFourierTransform That is what I have suggested in the comment to the original post. I have seen the original accepted solution. I am sure it works, but I have a feeling that simple algebraic approach is more appropriate here. Not reinventing the wheel. $\endgroup$
    – yarchik
    Commented Sep 21, 2019 at 15:40
  • $\begingroup$ Well, it worked. Thanks a lot. Do you know any way to set $\epsilon = 1$ by the end of the whole set of calculations? My equations are kinda messy and huge, any simplifications, or anything to let me look at them easier, would be greatly appreciated. $\endgroup$ Commented Sep 23, 2019 at 19:24


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