# Radial Green's function

I would like to solve an equation of the form $$\bigg(\frac{d}{dr^2} + m^2 \bigg)f(r) = g(r),$$ for $f(r)$. Normally I would just find the Green's function $G(r,r')$, which is defined by $$\bigg(\frac{d}{dr^2} + m^2 \bigg)G(r,r') = \delta(r-r').$$ This I would do by Fourier transforming over the real line. The solution is then $$f(r) = \int_{-\infty}^{\infty} dr'G(r,r')g(r').$$ However, the coordinate $r$ is radial and thus restricted to $(0,\infty)$. I thought a natural way to solve the problem is to use Hankel transforms. The Hankel transform of a function $f(r)$ of order $\nu$ is given by $$F_{\nu}(k) = \int_0^{\infty}f(r)J_{\nu}(kr)r dr,$$ where $J_{\nu}(kr)$ is a Bessel function of order $\nu$. I think we should then define the Green's function by the property $$\bigg(\frac{d}{dr^2} + m^2 \bigg)G(r,r') = \frac{\delta(r-r')}{r},$$ such that the solution to the problem would be given by $$f(r) = \int_0^{\infty}dr'r'G(r,r')g(r').$$ However, finding the Green's function using Hankel transforms turns out to be a problem. I tried using the Hankel transform of order 0, for which the identity $$\frac{\delta(r)}{r} = \int_0^{\infty}dkJ_0(kr)J_0(kr)k$$ holds. But then the equation for the Greens function becomes $$\int_0^{\infty} d k k\bigg(G(k)\frac{d}{dr^2}J_0(kr) + G(k)m^2J_0(kr) - J_0(kr)J_0(kr) \bigg) = 0.$$ Now I would like to invoke some kind of orthonormality, but due to the $\frac{d}{dr^2}J_0(kr)$ term I have no idea how to do this?

Any ideas?