The asymptotic behavier of the solution of $ f''(r)+\frac{1}{r}f(r)+a^2 f(r)=0 $.

Question

$\;$ Consider the Laplace equation with potential term in $ \mathbb{R}^2 $ as $ \Delta u+a^2u=0 $ where $ a>0 $ is a constant. I am considering the fundamental solution of it, i.e., the solution that is radial symmetric. I denote the fundamental solution as $u(x_1,x_2)=f(r) $ where $ r=\sqrt{x_1^2+x_2^2} $ and I get an ODE, $$f''(r)+\frac{1}{r} f'(r)+a^2f(r)=0 .$$ $\;$ I do not know how to solve this ODE and I even do not know the asymptotic behavier on the $ \infty $ and $ 0 $. As the fundamental soultion for Laplace equation in $ \mathbb{R}^2 $ is exactly $ -\frac{1}{2\pi}\ln|x| $, I guess that the $ f $ here satisfies $ f\sim \ln|x| $, which means that $ \lim_{x\to 0}|f|/(\ln|x|) $ and $ \lim_{x\to \infty}|f|/(\ln|x|) $ exsit and do not equal to $ 0 $. However, I cannot prove it, can you give me some hints or references?

Answer 1

Rescale the equation with $\rho = ar$

$$\rho^2f'' + \rho f' + \rho^2 f = 0$$

This is the Bessel differential equation with $n=0$. The homogenous solution is

$$f(\rho) = C_1J_0(\rho)+C_2Y_0(\rho)$$

For a fundamental solution we require that

$$Df = \frac{1}{2\pi r}\delta(r) = \frac{a^2}{2\pi\rho}\delta(\rho)$$

Can you prove from here which $C$ would give you the desired quantity?

Hint: near $0$, $Y_0(z) \sim \frac{2}{\pi}\left(\gamma+\log\frac{z}{2}\right)$ and $J_0(z) \sim 1 - \frac{z^2}{4}$ whereas the homogeneous solutions for the Poisson equation radially were $f = C_1 \log r + C_2$