0
$\begingroup$

I was reading up in the help about the Laplacian operator. and tried to understand the one example I saw: 

Laplacian[Sin[r^2],{r,\[Theta]},"Polar"]

results in:

$4 Cos[r^2]-4 r^2 Sin[r^2]$

which is not what I expected. However, 

Laplacian[Sin[r^2],{r,\[Theta]}]

does, ${2 Cos[r^2] - 4 r^2 Sin[r^2]}$

What does adding "Polar" to the Laplacian instruction do? Where does the "4" come from (instead of 2)?

thanks

|improve this question
$\endgroup$
  • $\begingroup$ Compare the result of evaluating Laplacian[f[r, \[Theta]], {r, \[Theta]}, "Polar"] vs. Laplacian[f[r, \[Theta]], {r, \[Theta]}] to see where the difference comes from. It becomes even clearer when you assume only a radially changing function like in your example: Laplacian[f[r], {r, \[Theta]}, "Polar"] vs. Laplacian[f[r], {r, \[Theta]}]. Essentially, the "Polar" version has an additional f'[r]/r term that the cartesian version doesn't have. $\endgroup$ – Thies Heidecke Apr 24 at 15:35
  • $\begingroup$ Thank you @Thies Heidecke....but I still don't get it. Isn't the Laplacian operator just the sum of the 2nd partial derivatives? I see what it does, but don't understand why.....---wait....lemme read up on the operator on Wikipedia.... $\endgroup$ – jrive Apr 24 at 17:30
  • 2
    $\begingroup$ Yes, in cartesian coordinates. But in general, that is in a Curvilinear coordinate system, you have to take into account the distortion of the geometry (local area or volume element) with respect to the coordinate. For example, in polar coordinates, for a bigger radius, a change in theta causes a different (higher) jump in euclidean distance than for small radii. These coordinate dependent differences in length (see Metric tensor) is why the Polar Laplacian looks different. $\endgroup$ – Thies Heidecke Apr 24 at 17:47
  • $\begingroup$ Thank you....copy your answer to "Answer" this question when you can so I can give you the proper credit! Thanks again. $\endgroup$ – jrive Apr 24 at 18:03
2
$\begingroup$

As Thies Heidecke explains, you need to account for the curvilinear coordinate system. In this case it is easy to verify that we get the same answer computing the Laplacian in Cartesian coordinates.

In Polar coordinates:

Laplacian[Sin[r^2], {r, θ}, "Polar"]
(* 4 Cos[r^2] - 4 r^2 Sin[r^2] *)

In Cartesian coordinates:

Laplacian[Sin[x^2 + y^2], {x, y}, "Cartesian"] // Simplify
(* 4 Cos[x^2 + y^2] - 4 (x^2 + y^2) Sin[x^2 + y^2] *)

which gives the same answer

% /. x^2 + y^2 -> r^2
(* 4 Cos[r^2] - 4 r^2 Sin[r^2] *)
|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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