How come $\nabla^2 f=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial f}{\partial r})$ yet $f=\int \int f'' dr dr$?

Question

In spherical coordinate, suppose $f$ dependents only on $r$. Then $\nabla^2 f=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial f}{\partial r})$. However, to integral $f''$ back the $f$, $\int \int f'' dr dr$.

How could that be?

Answer 1

It's not true that $\nabla^2 f(r) = f''(r)$ in 3D. In 1D we do have $\nabla^2 f(x) = \frac{\partial^2 f(x)}{\partial x^2} = f''(x)$ and even if spherical symmetry is effectively 1D it's not as simple as to assume the 1D Laplacian with $r$ instead of $x$ (something changing radially is fundamentally different than something changing along one of the axes). The definition of the Laplacian as (here in 3D) $$\nabla^2 f \equiv \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}$$ is for Cartesian coordinates $(x,y,z)$. If you start from this and change to spherical coordinates $(r,\theta,\phi)$ and assume $f$ only depends on $r$ then you will get the expression you have in your question $$\nabla^2f(r) = f''(r) + \frac{2}{r}f'(r)$$ Try to derive this using the chain-rule $\frac{\partial f(r)}{\partial x} = \frac{x}{r}\frac{\partial f(r)}{\partial r}$ and so on. If you manage this then you can try to generalize it by doing it in $n$ dimensions (for which the result is the same as above with the $2$ above becoming $n-1$).

Secondly if you have any (two times continuously differentiable) function of a single variable then we always have "$f(z) = \iint f''(z){\rm d}z{\rm d}z$", or more precisely $$f(z) = f(0) + f'(0)z + \int_0^z\left(\int_0^{z'} f''(w){\rm d}w\right){\rm d}z'$$ This expression doesn't care about which kind of coordinates it is, it's simply a consequence of the fundamental theorem of calculus.