Why is Laplacian sometimes defined as $-\operatorname{div}(\nabla f)$ and sometimes as $\operatorname{div}(\nabla f)$?

Question

Why is Laplacian sometimes defined as $-\operatorname{div}(\nabla f)$ and sometimes as $\operatorname{div}(\nabla f)$? What are the respective pros and cons of defining it with a negative and without negative sign?

Answer 1

Both definitions exist for the same reason: we prefer positive numbers to negative ones. They admit real square roots, they are closed under multiplication, they are easier to write ($-3$ vs $3$: two characters vs one).

So, it's nice to have $\Delta f = f_{x_1x_1} + \dots + f_{x_nx_n}$: positive coefficients, the coefficient matrix is positive-definite, the formula is shorter.

But then we start thinking of the Laplacian as a self-adjoint operator on a function space (Sobolev space $H^1_0$, usually) and discover that if it's defined as above, it is a negative operator: $$ \langle \Delta f, f\rangle = \int (\Delta f) f = - \int |\nabla f|^2 $$ So all of the eigenvalues are negative. And the spectrum of Laplacian turns out to be an important feature of its domain (the function space on which it acts, and the underlying point space). For example, the lowest eigenvalue is important... but what do I mean by lowest when they are negative... smallest in absolute value, right? Also, the square root of an operator is an important thing to have, so we want a positive operator.

For these reasons, as soon as we focus on the Laplacian as an operator, the choice $\Delta f = -(f_{x_1x_1} + \dots + f_{x_nx_n})$, or generally $\Delta f = - \operatorname{div}(\nabla f)$ (where the divergence and gradient may be taken with respect to some Riemannian metric) becomes preferable.