I am looking for a function of variable $r > 0$ (spherical radius), $f(r) \geq 1$, that obeys to:
$$ (\Delta_r + \alpha \delta(\boldsymbol{r}) ) f(r) = 0$$
with $\alpha > 0$ the only condition $f(r \rightarrow \infty) = 1$. The domain of resolution of this equation should be when $r/\alpha >> 1$.
Which is a Helmholtz equation with $\Delta_r = 1/r^2 \partial_r r^2 \partial_r $ the radial part of the Laplacian and:
$$ \delta(\boldsymbol{r}) = \frac{ \delta(r) \delta(\theta) \delta(\phi)}{r^2 \sin \theta} $$
the 3-dimensionnal Dirac delta in spherical coordinates. Do you have an idea ?
Some tracks:
We can simplify the equation by integrating on the solid angle $\sin \theta d \theta d \phi$ to get:
$$ \left( \partial_r r^2 \partial_r + \frac{\alpha}{4 \pi} \delta(r) \right) f(r) = 0 $$
Otherwise we can also use a rectangular distribution function, $h(r)$, instead of the Dirac delta:
$$ \left( \partial_r r^2 \partial_r + \frac{\alpha}{4 \pi} h(r) \right) f(r) = 0 $$
with $ h(r) = \frac{1}{R}$, $r \leq R $ and $ h(r) = 0$, $r > R$. Which is equivalent to a Dirac delta when $R \rightarrow 0$ and we solve $f(r)$ for $r >R$.