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Mine is more a technical curiosity, because the $\text{Laplace operator}$ is one of the many differential operators from which we can start, we also have the $\text{Jacobi Theta}$ $\theta$ operator $\rightarrow$ also that is an elliptical function (also Ramanuajan generalization preserves its properties).

Why start from Laplace (homogeneous Poisson equation)?
What's so special and more than other differential operators for example compared to that of Jacobi?

For example, if I started from the $\text{Whittaker equation}$ or from the $\text{Bessel equations}$ (ordinary linear equations of the second order homogeneous) what changes for a physicist that deals with potentials and harmonic functions?

Super Sonic
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  • Inverse square law forces + forces are proportional to the gradient of the potential + 3 dimensional universe = Laplace/Poisson equations for the potential. The second thing is a relatively natural thing, if you think about it. The first and third are AFAIK just quirks of the universe being the way it is. – Ian Jan 31 '18 at 21:03

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The equations that physicists use encode physical 'stuff' which is required to respect the symmetries of the universe. In simple models, we assume space is just $\mathbb R^n$ and the symmetrices are the Euclidean motions.

It turns out that one can describe explicitly all differential operators that are invariant under the Euclidean group. They have to be translation invariant, which is easy, and then invariant under the orthogonal group, which is a bit more complicated. It turns out that the Laplacian is the simplest invariant differential operator — the only one of order $2$.

Mariano Suárez-Álvarez
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  • mm..interesting, so to have not non-translation-invariant we shoudn't select Euclidean group, can you give me an example of non-invariant differential operator not under Euclidean group ? Is possible to integrate Laplacian equation with a non-invariant differential operator ? – Super Sonic Jan 31 '18 at 22:12
  • I don't know hat a «non-invariant differential operator not under Euclidean group» means. As for your second question, I have no idea what it means. – Mariano Suárez-Álvarez Jan 31 '18 at 22:45
  • sorry, maybe I misunderstanding something, is possible describe a differential operators under Euclidean group like non-translation invariant ? I try to find an operator that is not translation invariant like Laplace operator, I don't know if I need to follow something like this. I try to find a non-invariant differential operator, is possible to build it ? I try to see something like this also – Super Sonic Feb 01 '18 at 00:31
  • There are no invariant differential operators of first order, so that gives you tons of examples of non-invariant operators. – Mariano Suárez-Álvarez Feb 01 '18 at 00:32
  • Non-invariant operators on second order exist? I never heard for 2nd order . Just last thing related about translation invariant under Euclidean Groups - is also possible to generate a non-invariant operator using G-Modules ? There are some connections between translation invariants and trivial action ? – Super Sonic Feb 01 '18 at 00:48
  • I am sorry but I don't understand. – Mariano Suárez-Álvarez Feb 01 '18 at 00:52
  • As I wrote, the laplacian is the only invariant differential operator (up to a scalar factor) so any other differential operator is non-invariant. – Mariano Suárez-Álvarez Feb 01 '18 at 00:53
  • So Theta operator is non-invariant, and this is a first or second order ? Is possible to integrate a non-invariant operator into Laplace operator (only invariant operator) ? Is there a method to do that ? – Super Sonic Feb 01 '18 at 00:54