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I'm trying to find the solution to the 2D stationary diffusion equation

$$-D\nabla^2P(\vec{\rho_2})=\delta(\vec{\rho_1}-\vec{\rho_2})$$

where $\vec{\rho}=(x,y)$ and $D$ is the diffusion coefficient.

Help would be very much appreciated. Also, I'd be happy to get some references as well.

Thanks in advance.

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I assume that your sign is correct: that there is a sink rather than a source at ${\bf r}_2$?

Then this is standard result from electrostatics --- the potential of a line charge: $$ P({\bf r}_1, {\bf r}_2)= \frac 1{2\pi D} \ln (\mu|{\bf r}_1- {\bf r}_2|), $$ where $\mu$ is a constant that it is not determined by your equation. You need to specify the radius $\mu^{-1}$ at which $P$ becomes zero. This has to be bigger than zero for a solution to exist.

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  • $\begingroup$ Hi Mike, and thanks for the answer. Is this the 2D solution? Does $\vec{r}$ have a $z$ component as well? $\endgroup$
    – FlyGuy
    Nov 16, 2020 at 13:59
  • $\begingroup$ I was intending ${\bf r}$ to be two dimensional. In three D with a line sink on the $z$ axis, $|r_1-r_2|$ should be the radial distance from the sink, ie the $r$ in cylindrical polar coordinates. $\endgroup$
    – mike stone
    Nov 16, 2020 at 14:02
  • $\begingroup$ I just noticed that I messed up the sign. Clearly there should be a negative sign before the diffusion coefficient. Is this the solution to the corrected equation (with the negative sign before D)? $\endgroup$
    – FlyGuy
    Nov 16, 2020 at 14:05
  • $\begingroup$ Feels like this solution still holds if, say, I define $W=-P$. $\endgroup$
    – FlyGuy
    Nov 16, 2020 at 16:25
  • $\begingroup$ Yes. That's why I asked if there was a sink rather than a source at the origin. Just change the sign of $P$ and you are OK. $\endgroup$
    – mike stone
    Nov 16, 2020 at 16:41

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