# Vanishing with the derivative terms in perturbation theory

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.

• Multiply all your deltas by some ϵ. Set ϵ /: ϵ^n_ /; n >= 2 = 0;. Voilà. Commented Sep 21, 2019 at 13:40
• @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. Commented Sep 21, 2019 at 15:40
• 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. Commented Sep 23, 2019 at 19:24