The asymptotic behavier of the solution of

fsrlaihq55

fsrlaihq55

Answered question

2022-04-02

The asymptotic behavier of the solution of f(r)+1rf(r)+a2f(r)=0
Consider the Laplace equation with potential term in R2 as Δu+a2u=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(x1,x2)=f(r) where r=x12+x22 and I get an ODE,
f(r)+1rf(r)+a2f(r)=0.
I do not know how to solve this ODE and I even do not know the asymptotic behavier on the and 0. As the fundamental soultion for Laplace equation in R2 is exactly 12πln|x|, I guess that the f here satisfies fln|x|, which means that limx0|f|ln|x| and limx|f|ln|x| exsit and do not equal to 0. However, I cannot prove it, can you give me some hints or references?

Answer & Explanation

delai59qk

delai59qk

Beginner2022-04-03Added 8 answers

Step 1Rescale the equation with ρ=ar
ρ2f+ρf+ρ2f=0
This is the Bessel differential equation with n=0. The homogenous solution is
f(ρ)=C1J0(ρ)+C2Y0(ρ)
Step 2
For a fundamental solution we require that
Df=12πrδ(r)=a22πρδ(ρ)
Can you prove from here which C would give you the desired quantity?
Hint: near 0, Y0(z)2π(γ+logz2) and J0(z)1z24 whereas the homogeneous solutions for the Poisson equation radially were f=C1logr+C2.

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