I am looking for the Green's function of the problem in two dimensions $r =(x,z)$, \begin{equation} \nabla^2g + \frac{v}{D}\frac{\partial g}{\partial z} = -\delta (r-r_0) \end{equation}
Which corresponds to finding temperature distribution in a moving frame (moving at constant velocity $v$ in $z$ direction and diffusion coefficient of the medium is $D$) because of the constant source and $|r-r_0| = \sqrt{(x-x_0)^2+(z-z_0)^2}$
research paper gives solution as,
\begin{eqnarray} g(r,r_0) = \frac{1}{2\pi}\exp\left(-\frac{z-z_1}{l} \right)K_0\left(\frac{|r-r_0|}{l} \right) \end{eqnarray}
Where, $K_0$ is a modified Bessel function of second kind and $l = \frac{2D}{v}$. How can one arrive at this solution? I tried to varify the solution but I could not go anywhere close.
\begin{eqnarray} \frac{\partial g(r,r_0)}{\partial z} &=& \exp\left(-\frac{z-z_1}{l} \right)\left[- \frac{1}{l}K_0 + \frac{\partial K_0}{\partial z} \right]\\ \frac{\partial^2 g(r,r_0)}{\partial z^2} &=& \exp\left(-\frac{z-z_1}{l} \right)\left[ \frac{1}{l^2}K_0 -\frac{2}{l} \frac{\partial K_0}{\partial z} + \frac{\partial^2 K_0}{\partial z^2}\right] \\ \frac{\partial^2 g(r,r_0)}{\partial x^2} &=& \exp\left(-\frac{z-z_1}{l} \right)\left[ \frac{\partial^2 K_0}{\partial x^2}\right] \end{eqnarray}
then putting the terms together, \begin{eqnarray} \nabla^2g + \frac{2}{l}\frac{\partial g}{\partial z} &=& \exp\left(-\frac{z-z_1}{l} \right)\left[\frac{\partial^2 K_0}{\partial x^2}+ \frac{\partial^2 K_0}{\partial z^2} -\frac{1}{l^2}K_0 \right] \end{eqnarray}
rhs of the equation should be zero unless $r=r_0$.
possible definition of $K_0$ is \begin{eqnarray} K_0([r-r_0]/l) &=& \int^{\infty}_{0}\frac{\cos([r-r_0] \frac{u}{l})}{\sqrt{1+u^2}} du \\ &=& \int^{\infty}_{0}\frac{\cos(\sqrt{(x-x_0)^2+(z-z_0)^2} \frac{u}{l})}{\sqrt{1+u^2}} du \end{eqnarray}
rhs is then, \begin{eqnarray} rhs &=& -\exp\left(-\frac{z-z_1}{l} \right)\left[ \int^{\infty}_{0}\frac{\frac{1+u^2}{l^2}\cos([r-r_0] \frac{u}{l}) + \frac{u}{l(r-r_0)}\sin([r-r_0] \frac{u}{l})}{\sqrt{1+u^2}} du \right] \end{eqnarray}
Which I do not see how is equal to $\delta(r-r_0)$. What am I missing here?