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My problem can be easily solved without Mathematica, however, I am curious how to do it in Mathematica. Moreover, I can be stuck with a similar, but more cumbersome problems in the future.

Following this topic, I defined the two complex differential operators of interest:

b1[r_, ph_] = Exp[I ph] (D[#, r] + I/r D[#, ph]) &
b2[r_, ph_] = Exp[-I ph] (D[#, r] - I/r D[#, ph]) &

Now I can easily apply those to any functions of interest: b1[r, ph][r^3].

Now I want to define an operator that is a product (more precisely, superposition) of the two operators above. Note, that in my case the superposition is Laplacian operator in polar coordinates.

I've tried to do:

b3[r_, ph_] = b1[r, ph][b2[r, ph] [&]]

And some similar things. Can anyone help me to figure out how to define operators' superposition (so that b3[r, ph][r^3] would return me the expression of interest)?

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    $\begingroup$ How about using Composition: b3[r_, ph_] = b1[r,ph]@*b2[r,ph]? This is pretty close to what you almost had: b3[r_, ph_] = (b1[r, ph][b2[r, ph] [#]]&). $\endgroup$
    – jjc385
    Commented Nov 10, 2017 at 23:50
  • $\begingroup$ Thanks, that works for me. You can post this as an answer if you'd like $\endgroup$
    – Mikhail Genkin
    Commented Nov 14, 2017 at 2:07

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For the sake of an answer:

How about using Composition: b3[r_, ph_] = b1[r,ph]@*b2[r,ph]? This is pretty close to what you almost had: b3[r_, ph_] = (b1[r, ph][b2[r, ph] [#]]&). – jjc385 Nov 10 '17 at 23:50

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  • $\begingroup$ Hey folks: An upvote on this CW answer will prevent the Community bot from bumping it to the top of the stack from time to time. Alternatively, you could close the question, but that seems harsher... $\endgroup$
    – Michael E2
    Commented Mar 6, 2021 at 19:56

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