homework and exercises - Is this problem from Griffiths' "Classical electromagnetism" correct?

I am talking about pbl 2.46 pag 107.

The potential

$$\frac{A}{r} \, \exp{(-\lambda r)}$$ is given I have to find the charge density and the electric field.

The book says the solution for the charge density is:

$$\left[\delta^3(r)-\dfrac{\lambda^2}{r} \, \exp{(-\lambda r)}\right]\, \epsilon_0\, A$$

The first part comes from $\nabla^2 \, \dfrac{1}{r}$ , the second comes from $\nabla^2\,\exp{(-\lambda r)}$.

I think this part is missing: $2\,\nabla\, \dfrac{1}{r} \, \cdot\, \nabla\,\exp{(-\lambda r)}$.

Or am I misunderstanding something?

Answer

This is but the screened Poisson equation. Since the functions are radial, only the radial part of the Laplacian survives,

$$\nabla^2 (e^{-\lambda r}/r) = \frac{1}{r^2}\partial_r ( r^2 \partial_r (e^{-\lambda r}/r))= \frac{1}{r^2}\partial_r \left ( -\lambda r e^{-\lambda r} +e^{-\lambda r} r^2\partial_r \frac{1}{r}\right )\\ = \frac{1}{r^2}\partial_r \left( r^2\partial_r \frac{1}{r}\right ) ~ e^{-\lambda r} -\lambda e^{-\lambda r} \left(\frac{1}{r^2}+\partial_r \frac{1}{r} \right ) + \lambda^2 \frac{ e^{-\lambda r}}{r}.$$ Note the cross $O(\lambda)$ term vanishes.

Since

$$\nabla^2 \frac{1}{r}=-4\pi \delta^{(3)} (\vec{r})= -\frac{\delta(r)}{r^2},$$ one has $$(\nabla^2 -\lambda^2) \left (\frac{e^{-\lambda r}}{r}\right)=-\frac{\delta(r)}{r^2}.$$

So your text is just fine. You probably misunderstand, that, in fact, $\nabla^2 e^{-\lambda r}= (\lambda^2 -2\lambda/r )e^{-\lambda r}$, as evident from above. This additional $O(\lambda)$ term here exactly cancels the cross term you wrote down. (Try evaluating $\nabla^2 r$ to see the point.)