4
$\begingroup$

Ultimately, I wish to get the higher generalized product rule for Laplacian. That is, we have $\Delta(fg)=f\Delta g+2\nabla f\cdot\nabla g+g\Delta f$ but what is $\Delta^n(fg)$?

I am tying to guess the general formula using Mathematica.

We get

Laplacian[f[x, y, z] g[x, y, z], {x, y, z}]

(*
===> 

2 (f^(0,0,1))[x,y,z] (g^(0,0,1))[x,y,z]+g[x,y,z] (f^(0,0,2))[x,y,z]+f[x,y,z] (g^(0,0,2))[x,y,z]+2 (f^(0,1,0))[x,y,z] (g^(0,1,0))[x,y,z]+g[x,y,z] (f^(0,2,0))[x,y,z]+f[x,y,z] (g^(0,2,0))[x,y,z]+2 (f^(1,0,0))[x,y,z] (g^(1,0,0))[x,y,z]+g[x,y,z] (f^(2,0,0))[x,y,z]+f[x,y,z] (g^(2,0,0))[x,y,z]

*)

Is there a way to automatically rewrite this as more readable "$f\Delta g+2\nabla f\cdot\nabla g+g\Delta f$"?

Further, 

Laplacian[Laplacian[f[x, y, z] g[x, y, z], {x, y, z}],{x,y,z}]

(*
===>

6 (f^(0,0,2))[x,y,z] (g^(0,0,2))[x,y,z]+4 (g^(0,0,1))[x,y,z] (f^(0,0,3))[x,y,z]+4 (f^(0,0,1))[x,y,z] (g^(0,0,3))[x,y,z]+g[x,y,z] (f^(0,0,4))[x,y,z]+f[x,y,z] (g^(0,0,4))[x,y,z]+8 (f^(0,1,1))[x,y,z] (g^(0,1,1))[x,y,z]+4 (g^(0,1,0))[x,y,z] (f^(0,1,2))[x,y,z]+4 (f^(0,1,0))[x,y,z] (g^(0,1,2))[x,y,z]+2 (g^(0,0,2))[x,y,z] (f^(0,2,0))[x,y,z]+2 (f^(0,0,2))[x,y,z] (g^(0,2,0))[x,y,z]+6 (f^(0,2,0))[x,y,z] (g^(0,2,0))[x,y,z]+4 (g^(0,0,1))[x,y,z] (f^(0,2,1))[x,y,z]+4 (f^(0,0,1))[x,y,z] (g^(0,2,1))[x,y,z]+2 g[x,y,z] (f^(0,2,2))[x,y,z]+2 f[x,y,z] (g^(0,2,2))[x,y,z]+4 (g^(0,1,0))[x,y,z] (f^(0,3,0))[x,y,z]+4 (f^(0,1,0))[x,y,z] (g^(0,3,0))[x,y,z]+g[x,y,z] (f^(0,4,0))[x,y,z]+f[x,y,z] (g^(0,4,0))[x,y,z]+8 (f^(1,0,1))[x,y,z] (g^(1,0,1))[x,y,z]+4 (g^(1,0,0))[x,y,z] (f^(1,0,2))[x,y,z]+4 (f^(1,0,0))[x,y,z] (g^(1,0,2))[x,y,z]+8 (f^(1,1,0))[x,y,z] (g^(1,1,0))[x,y,z]+4 (g^(1,0,0))[x,y,z] (f^(1,2,0))[x,y,z]+4 (f^(1,0,0))[x,y,z] (g^(1,2,0))[x,y,z]+2 (g^(0,0,2))[x,y,z] (f^(2,0,0))[x,y,z]+2 (g^(0,2,0))[x,y,z] (f^(2,0,0))[x,y,z]+2 (f^(0,0,2))[x,y,z] (g^(2,0,0))[x,y,z]+2 (f^(0,2,0))[x,y,z] (g^(2,0,0))[x,y,z]+6 (f^(2,0,0))[x,y,z] (g^(2,0,0))[x,y,z]+4 (g^(0,0,1))[x,y,z] (f^(2,0,1))[x,y,z]+4 (f^(0,0,1))[x,y,z] (g^(2,0,1))[x,y,z]+2 g[x,y,z] (f^(2,0,2))[x,y,z]+2 f[x,y,z] (g^(2,0,2))[x,y,z]+4 (g^(0,1,0))[x,y,z] (f^(2,1,0))[x,y,z]+4 (f^(0,1,0))[x,y,z] (g^(2,1,0))[x,y,z]+2 g[x,y,z] (f^(2,2,0))[x,y,z]+2 f[x,y,z] (g^(2,2,0))[x,y,z]+4 (g^(1,0,0))[x,y,z] (f^(3,0,0))[x,y,z]+4 (f^(1,0,0))[x,y,z] (g^(3,0,0))[x,y,z]+g[x,y,z] (f^(4,0,0))[x,y,z]+f[x,y,z] (g^(4,0,0))[x,y,z]

*)

Is there a way to rewrite this with $\Delta^2$, $\Delta$, or $\nabla$ etc. if applicable?

I bet somewhere around this How to collect terms with z-derivative? but I have no clue...

(I am assuming I can apply the possible answer to Laplacian[Laplacian[Laplacian...etc.)

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Please take time to format your code: mathematica.stackexchange.com/editing-help $\endgroup$
    – Yves Klett
    Commented Jul 5, 2016 at 14:21
  • $\begingroup$ Instead of guessing, you could read about the generalized Leibniz rule. $\endgroup$
    – QuantumDot
    Commented Jul 5, 2016 at 17:20
  • $\begingroup$ @QuantumDot I don't really see how far that is from guessing, as my intention of using mathematica is to guess the assumption (using $\Delta$ and $\nabla$) for the inductive proof, which I believe is how the generalized Leibniz rule is shown. $\endgroup$
    – user41467
    Commented Jul 5, 2016 at 22:34

1 Answer 1

Reset to default
1
$\begingroup$

Try this:

Nest[Laplacian[#, {x, y}] &, f[x, y], n]

where instead of f[x,y] you will substitute whatever you need. I used 2D for shortness, but even then it is long enough. Then, for instance, with n=2 and f=u*v you get

Nest[Laplacian[#, {x, y}] &, u[x, y]*v[x, y], 2] // Simplify

with the effect

enter image description here

Have fun!

Improve this answer
$\endgroup$
1
  • $\begingroup$ Sorry but this isn't quite what I intended to ask...but it seems useful indeed in its own right, thanks. $\endgroup$
    – user41467
    Commented Jul 5, 2016 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.