ODE with Bessel decay

Question

This is related to my previous question, but it is probably less scary and an expert in using Mathematica could figure out an answer easily.

I would like to estimate the asymptotic behaviour of the solution of the following ODE $$ w''(r)+\frac{1}{r}w'(r)-k^{2}w(r)=-g(r)$$ where $g$ is a smooth positive function satisfying $$ g(r) = O\left(\frac{e^{-kr}}{\sqrt{r}}\right) \ \ \ \ \ r\to +\infty . $$

Assuming that $w(r)$ decays exponentially fast, is it true that $w(r)$ has exactly the same asymptotic decay as $g(r)$?

Answer 1

Maple 2018 solves the ODE under consideration:

dsolve(((D@@2)(w))(r)+(D(w))(r)/r-k^2*w(r) = -g(r))

$$w \left( r \right) ={{\ I}_{0}\left(kr\right)}{\it \_C2}+{{ K}_{0 }\left(kr\right)}{\it \_C1}-\int \!{{ K}_{0}\left(kr\right)}g \left( r \right) r\,{\rm d}r{{ I}_{0}\left(kr\right)}+$$ $$\int \!{ { I}_{0}\left(kr\right)}g \left( r \right) r\,{\rm d}r{{ K}_{0} \left(kr\right)} $$

The assumption $w(r)$ decays exponentially fast implies the constant ${\it \_C2} $ equals zero because of

asympt(BesselI(0, k*r), r, 2)

$$\frac 1 2\,{\frac {\sqrt {2}\sqrt {{k}^{-1}}{{\rm e}^{kr}}\sqrt {{r}^{-1}}}{ \sqrt {\pi}}}+O \left( \left( {r}^{-1} \right) ^{3/2} \right) $$

and

asympt(BesselK(0, k*r), r, 1)

$$\frac 1 2\,{\frac {\sqrt {2}\sqrt {\pi}{{\rm e}^{-kr}}\sqrt {{r}^{-1}}}{ \sqrt {k}}}+O \left( \left( {r}^{-1} \right) ^{3/2} \right) $$

Hope that would be useful.

Addition. The Mathematica 11.3 command

AsymptoticDSolveValue[ D[w[r], {r, 2}] +D[w[r], r]/r - k^2*w[r] == -Exp[-k*r]/Sqrt[r],w[r], 
{r, Infinity, 1}, Assumptions -> k > 0]

disproves the suggestion of the question. See the long output here https://www.dropbox.com/s/on0jz2kpb7bin7l/AS.pdf?dl=0 and pay your attention to the term $\frac{\sqrt{r}}{2 k} $.

Answer 2

I don't think this statement is correct. As a counter example, take $$g(r)=\frac{e^{-k r} (8 k r-1)}{4 r^{3/2}}$$ which has the desired $r^{-1/2}e^{-kr}$ decay for $r\rightarrow\infty$. It changes sign for small $r$, but I would think that is irrelevant for the large-$r$ decay of $w(r)$. I then note that the function $$w(r)=e^{-kr}\sqrt{r}$$ solves the differential equation, but decays more slowly than $g(r)$.

Answer 3

This answer provides a slight modification of the excellent answer by Carlo Beenakker, to get $g>0$ on the entire interval $(0,\infty)$: if \begin{equation} w(r)=e^{-k r} \left(32 k \sqrt{r}-\frac{2}{\sqrt{r}}\right) \end{equation} and \begin{equation} g(r)=\frac{e^{-k r} \left(128 k^2 r^2-16 k r+1\right)}{2 r^{5/2}} \end{equation} for some $k\ne0$ and all real $r>0$, then $g>0$ on $(0,\infty)$, the equation $w''(r)+\frac1r\,w'(r)-k^2w(r)=-g(r)$ is satisfied for all real $r>0$, and $g(r) = O(e^{-kr}/\sqrt r)$ as $r\to\infty$, whereas $ w(r)\sim 32k e^{-k r} \sqrt r\ne O(e^{-kr}/\sqrt r)$ as $r\to\infty$.