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I have to solve this linear differential equation $$ rf''(r) + f'(r) + k^2rf(r) = \delta(r) $$ on $\mathbb{R}^+$.

I know the solutions to the homogeneous problem are $cJ_0(kr)+dY_0(kr)$, where $J_0,Y_0$ are the zeroth order Bessel functions are the first and second kind. But how do I go about solving the problem with this right hand side ?

For background, I'm trying to solve the partial differential equation $$\left(\Delta-\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}\right)A(r,t) = -\dfrac{\mu_0 i(t)}{2\pi r}\delta(r)$$ which reduces to what I have after Fourier transform w.r.t. time and fixing $\omega$, and setting $k=\omega/c$.

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    It's actually the electric field for a situation where there's a time-dependant current flowing through a wire on the line of equation $r=0$ (in cylindrical coordinates). – 2017-01-01
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    Already tried that. I get a differential equation that I can solve, but I'm not able to really Fourier transform back the solution. – 2017-01-01
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    Indeed, it leads the linear ODE, where $x$ is the Fourier transform variable w.r.t. r $$ \frac{d}{dx}\left((k^2-x^2)F(x)\right) - xF(x) = 1, $$ which I change to $$G'(x) - \frac{x}{k^2-x^2}G(x) = 1 $$ with $G(x) = (k^2-x^2)F(x)$. The solution of the homogeneous problem is $\sqrt{|k^2-x^2|}$, yes, but the particular solution is $\ln(\sqrt{x^2-k^2}+x)$ for $x>k$ and $\arctan(x/\sqrt{k^2-x^2})$ for $x – 2017-01-01

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For $r\neq 0$, the Dirac delta vanishes, and therefore $$ f(r)=\begin{cases}c_1J_0(kr)+d_1Y_0(kr) & r<0\\ c_2J_0(kr)+d_2Y_0(kr) & r>0 \end{cases} $$

The ODE, being of second order, should have solutions with two arbitrary coefficients; but the expression above apparently has four arbitrary constants. In fact, there are two equations that constraint these four constants, reducing the number down to two, as one would expect.

The two relations are $$ f(0^-)=f(0^+) $$ (why?) and $$ \big[rf'(r)\big]^{0^+}_{0^-} = 1 $$ (why?)

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    My solution function isn't defined on the negative reals, though... – 2017-01-01
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    then you cannot have a delta at $r=0$. Maybe you want to solve $...=\delta(r-r_0)$? – 2017-01-01
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    The function is defined on $\mathbb R^+$: why can't I have a delta ? – 2017-01-01