Image vortex at a plane wall ($U=\frac{\Gamma }{4\pi b}$)

The method of images enforces impermeability at a plane wall by reflecting a line vortex across it and reversing its circulation. The two logarithmic stream functions cancel on the wall, which is consequently a streamline. For a vortex of circulation $\Gamma$ at $(0,b)$ above $y=0$, the image of circulation $-\Gamma$ at $(0,-b)$ induces velocity $(\Gamma /(4\pi b),0)$ at the real vortex. The signed drift parallel to the wall is therefore $\Gamma /(4\pi b)$.

In the half-plane $y>0$ with an impermeable wall, place a point vortex of circulation $\Gamma$ in the background velocity $(\kappa x,-\kappa y)$, where $\kappa >0$. The image vortex at a plane wall has circulation $-\Gamma$ at the reflected location and contributes horizontal velocity $\Gamma /(4\pi y)$. Excluding the vortex's own singular velocity gives the displayed motion equations.

Put $c=\Gamma /(4\pi )$. Their first integral is $\kappa xy+c\log y$, since its time derivative is $\kappa [(\kappa x+c/y)y-\kappa xy]+c(-\kappa )=0$. The explicit trajectory is $y=y_0{e}^{-\kappa t}$ and $x=e^{\kappa t}(x_0+ct/y_0)$. These formulas show how image-induced drift changes the hyperbolic trajectories of the background flow while preserving a modified stream-function-type invariant.

In the quadrant $x,y>0$, a line vortex of circulation $\Gamma$ at $(x,y)$ has three images: circulations $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. Their induced velocity isIn polar coordinates, these equations give $\dot{r}=\Gamma \cot (2\theta )/(2\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation yields $\frac{d}{dt}(r\sin 2\theta )=0$, so the trajectory obeys $r\sin 2\theta =\text{constant}$.