Laplacian $\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{\partial \phi }{\partial r})= \frac{1}{r} \frac{\partial ^2 }{\partial r^2}(r \phi )$

Question

Does anyone have any intuition on remembering or very quickly deriving that

$$\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{\partial \phi }{\partial r}) = \frac{1}{r} \frac{\partial ^2 }{\partial r^2}(r \phi )$$

holds for the Laplacian in spherical coordinates? Doing the IBP is too long and slow, maybe there is some obvious obvious physical reason for this simplification making it obvious and easy to remember?

Edit: In other words, is there a nice natural way to start with

$$\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{\partial \phi }{\partial r})$$

and derive the other side of the equality intuitively, without remembering it beforehand?

Answer 1

Maybe you want to take a look at how the Laplacian is derived for any system of curvilinear coordinates:

http://en.wikipedia.org/wiki/Curvilinear_coordinates

Answer 2

based on integration by parts with compactly supported functions, assuming you know that $\nabla f = \partial_r f u_r$: $$ \int \phi {\rm div} g dV = -\int \nabla \phi \cdot g dV = -\int \partial_r \phi g_r dV \\ = -\int \partial_r \phi g_r 4\pi r^2 dr = -\int \phi \partial_r(r^2 g_r) 4\pi dr = -\int \phi \frac{\partial_r(r^2 g_r)}{r^2} 4\pi r^2 dr $$and then $$ \Delta f = {\rm div} \nabla f = {\rm div} (\partial_r f_r e_r)= \frac{\partial_r(r^2 \partial_r f_r )}{r^2}$$