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Solution: method of image for a 2D source near a wall

We verify if a 2D source and its mirror image with respect to a wall located at $y = 0$ satisfies the no flux boundary condition at the wall. First we need to evaluate the component of the velocity vector in the $y$ direction.


\begin{displaymath}\frac{\partial \phi}{\partial y} = \frac{m}{2\pi}\left(\frac{...
...(y-b)^{2}}}+\frac{(y+b)}{\sqrt{x^{2}+(y+b)^{2}}}\right) \notag
\end{displaymath}  

Next, we set $y = 0$, and


\begin{displaymath}\frac{\partial \phi}{\partial y} = \frac{m}{2\pi}\left(\frac{...
...t{x^{2}+b^{2}}}+\frac{b}{\sqrt{x^{2}+b^{2}}}\right) = 0 \notag
\end{displaymath}  

as expected.