Why decompose a function with eigenvectors of Laplace operator? [closed]

On periodic domain, people always use Fourier basis, which eigenvectors of Laplace operator. On sphere, people use spherical harmonics, which also are eigenvectors of Laplace operator. In applied science, people decompose functions on a graph using eigenvectors of graph laplacian.

What makes eigenvectors of Laplace operator widely used compared to other orthogonal basis? Are there any other operators also provide orthogonal basis which are also useful? Are there any example that we are not using Laplace operator?

On non-periodic domain, we have many orthogonal polynomial systems, say, Legendre polynomials, Chebyshev polynomials, Jacobi polynomials. So, we have more than just one set of orthogonal basis, in this case. It motivates me to ask those above questions.

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closed as unclear what you're asking by Qiaochu Yuan, Franz Lemmermeyer, Alexey Ustinov, Wolfgang, Anton DeitmarFeb 24 at 14:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

A choice of basis depends on what problem one is trying to solve. When you ask "does any orthogonal basis do the same job" it is not clear what job you are referring to. – Yemon Choi Feb 24 at 0:56
Is it still unclear after I edited the questions? @qiaochu-yuan – Po C. Feb 29 at 4:34

The exponentials used in Fourier series are eigenvalues of shifts, and thus of any operator commuting with shifts, not just Laplacian. Similarly, spherical harmonics carry irreducible representations of $SO(3)$, and so they are eigenfunctions of any rotationally invariant operator.