Spherical Laplacians on an Exponential I looked around a bit and couldn't find a resolution to this.
I was curious about the scalar function $u(r) = e^{-r}$ with $r \in [0,\infty)$ and acting Spherical Laplacians on it.
$$\Delta u(r) = \frac{1}{r^{2}}\frac{\partial}{\partial r}\Big(r^{2}\frac{\partial u}{\partial r}\Big).\tag{1}$$
Laplacian is straight forward to compute:
$$\Delta u = \Big(1 - \frac{2}{r}\Big)u.\tag{2}$$
But computing a second one seems to introduce some ambiguity:
$$\Delta^{2}u = \Delta \Delta u = \Delta \Big(1-\frac{2}{r}\Big)u = \Delta\Big(u -\frac{2u}{r}\Big) = \Big(1-\frac{2}{r}\Big)u - \Delta\Big(\frac{2u}{r}\Big) .\tag{3}$$
With the rightmost term being the confusing part to me; If I proceed with distributing the $\Delta$:
$$\Delta\Big(\frac{u}{r}\Big) \overset{?}{=}\  u\ \Delta\Big(\frac{1}{r}\Big) +2\ \nabla u \cdot \nabla\Big(\frac{1}{r}\Big) + \frac{1}{r}\Delta u.\tag{4}$$
As I understand it all of these objects are defined, with: $$\nabla\Big(\frac{1}{r}\Big) = -\frac{1}{r^{2}}\hat{r}\qquad \text{and}\qquad\Delta\Big(\frac{1}{r}\Big) = -4\pi\delta^3(r).\tag{5}$$
Where $\delta^3(r)$ is the 3D Dirac delta distribution.
This all seems to suggest that:
$$-2\Delta\Big(\frac{u}{r}\Big) = -2\Big(-4\pi\delta^3(r) u + 2\frac{u}{r^{2}} + \frac{1}{r}\Big(1 - \frac{2}{r}\Big)u \Big)\tag{6}$$
$$\require{cancel}-2\Delta\Big(\frac{u}{r}\Big) = -2\Big(-4\pi\delta(r) u + \cancel{2\frac{u}{r^{2}}} + \frac{1}{r}\Big(1 - \cancel{\frac{2}{r}}\Big)u \Big)\tag{7}$$
$$-2\Delta\Big(\frac{u}{r}\Big) = 8\pi\delta^3(r)u - \frac{2u}{r}.\tag{8}$$
Leaving me with:
$$\Delta^{2}u = \Big(1 -\frac{4}{r} + 8\pi\delta^3(r)\Big)u.\tag{9}$$
Now the heart of my question is:


*

*Are these manipulations correct, or have I assumed something I should have not?

*Why does $\Delta\Big(\Delta e^{-r}\Big)$ contain a discontinuity (the $\delta^3(r)$) when all the $r$-derivatives exist?
 A: *

*No, OP's manipulations are a priori ill-defined since OP's function $u:\mathbb{R}^3\to \mathbb{R}$ given by $$ u({\bf r})~:=~\exp(-|{\bf r}|), \qquad {\bf r}~\in~\mathbb{R}^3, \tag{A}$$
is not differentiable in the origin ${\bf r}={\bf 0}$.

*Let us define distributions
$$\delta^3[f]~:=~ f[{\bf 0}], \qquad  1[f]~:=~\int_{\mathbb{R}^3} \!d^3 {\bf r} ~f({\bf r}), $$ $$ r^{-1}[f]~:=~\int_{\mathbb{R}^3} \!d^3 {\bf r} \frac{f({\bf r})}{|{\bf r}|}, \qquad  r^{-2}[f]~:=~\int_{\mathbb{R}^3} \!d^3 {\bf r} \frac{f({\bf r})}{|{\bf r}|^2}, \tag{B}$$ for test functions $f\in C^{\infty}_c(\mathbb{R}^3)$.

*In practice, it may be simpler to regularize OP's function
$$u({\bf r})~=~ \lim_{\varepsilon\to 0^+}u_{\varepsilon}({\bf r}) \tag{C}$$
to a $C^{\infty}$-function 
$$u_{\varepsilon}({\bf r})~:=~\exp(-\sqrt{|{\bf r}|^2+\varepsilon}), \qquad {\bf r}~\in~\mathbb{R}^3,\qquad \varepsilon ~>~0.\tag{D}$$
One may then show the formulas
$$\lim_{\varepsilon\to 0^+}u_{\varepsilon}^{-1} \Delta u_{\varepsilon}~=~1-2r^{-1}, \qquad \lim_{\varepsilon\to 0^+}u_{\varepsilon}^{-1} \Delta^2 u_{\varepsilon}~=~1-4r^{-1}+8\pi\delta^3,\tag{E} $$
in the sense of generalized functions/distributions, which are similar to 
OP's formulas (2) & (9).
