Cfrgtkky

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:

$$Delta^{2}u = Delta Delta u = Delta Big(1-frac{2}{r}Big)u = DeltaBig(u -frac{2u}{r}Big) = Big(1-frac{2}{r}Big)u - DeltaBig(frac{2u}{r}Big) .tag{3}$$

With the rightmost term being the confusing part to me; If I proceed with distributing the $Delta$:

$$DeltaBig(frac{u}{r}Big) overset{?}{=} u DeltaBig(frac{1}{r}Big) +2 nabla u cdot nablaBig(frac{1}{r}Big) + frac{1}{r}Delta u.tag{4}$$

As I understand it all of these objects are defined, with: $$nablaBig(frac{1}{r}Big) = -frac{1}{r^{2}}hat{r}qquad text{and}qquadDeltaBig(frac{1}{r}Big) = -4pidelta^3(r).tag{5}$$

Where $delta^3(r)$ is the 3D Dirac delta distribution.

This all seems to suggest that:

$$-2DeltaBig(frac{u}{r}Big) = -2Big(-4pidelta^3(r) u + 2frac{u}{r^{2}} + frac{1}{r}Big(1 - frac{2}{r}Big)u Big)tag{6}$$

$$require{cancel}-2DeltaBig(frac{u}{r}Big) = -2Big(-4pidelta(r) u + cancel{2frac{u}{r^{2}}} + frac{1}{r}Big(1 - cancel{frac{2}{r}}Big)u Big)tag{7}$$

$$-2DeltaBig(frac{u}{r}Big) = 8pidelta^3(r)u - frac{2u}{r}.tag{8}$$

Leaving me with:

$$Delta^{2}u = Big(1 -frac{4}{r} + 8pidelta^3(r)Big)u.tag{9}$$

Now the heart of my question is:

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




2. Why does $DeltaBig(Delta e^{-r}Big)$ contain a discontinuity (the $delta^3(r)$) when all the $r$-derivatives exist?

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




2. 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 $fin C^{infty}_c(mathbb{R}^3)$.




3. In practice, it may be simpler to regularize OP's function
   $$u({bf r})~=~ lim_{varepsilonto 0^+}u_{varepsilon}({bf r}) tag{C}$$
   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
   $$lim_{varepsilonto 0^+}u_{varepsilon}^{-1} Delta u_{varepsilon}~=~1-2r^{-1}, qquad lim_{varepsilonto 0^+}u_{varepsilon}^{-1} Delta^2 u_{varepsilon}~=~1-4r^{-1}+8pidelta^3,tag{E} $$
   in the sense of generalized functions/distributions, which are similar to
   OP's formulas (2) & (9).