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For constant mass density and incompressible flow, use $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ in the Euler equations for an inviscid fluid. Taking the curl removes the pressure and kinetic energy gradients, giving ${\mathbf{\omega }}_t=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. The supplied vector identity, with $\nabla \cdot \mathbf{u}=0$ and $\nabla \cdot \mathbf{\omega }=0$, yields the vorticity equationIn a two-dimensional flow $\mathbf{u}=(u,v,0)$ is independent of $z$, and $\mathbf{\omega }=\omega {\mathbf{e}}_z$. The right side vanishes, so $D\omega /Dt=0$: each material element retains its vorticity. The vortex lines are parallel to $z$ and are carried by the planar velocity. Also planar incompressibility preserves the area of a material cross-section; integrating its conserved vorticity over that area shows that its circulation, or vortex strength, stays constant. Thus the vortex lines move with the fluid without changing strength.

Place a line vortex of circulation $\Gamma$ at $(0,b)$ above the wall $y=0$. Its velocity field is $\Gamma {\mathbf{e}}_z\times (\mathbf{r}-{\mathbf{r}}_0)/(2\pi |\mathbf{r}-{\mathbf{r}}_0{|}^2)$. The method of images puts a vortex of circulation $-\Gamma$ at $(0,-b)$. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$, the resulting stream function isIt is identically zero on the wall, so $u_y=-{\psi }_x=0$ there. In the fluid half-plane it has just the required vortex singularity, with no other singularities, and its velocity decays at infinity. This is the no-background-flow solution with the wall: the difference from another such solution is a regular irrotational, divergence-free decaying field with zero normal velocity. Reflect its tangential component evenly and its normal component oddly across the wall. The reflected components are bounded harmonic functions on the whole plane, so they are constant by harmonic Liouville theorem, and decay forces both constants to be zero. Thus replacing the wall by the image gives the same flow. The real vortex moves under the image velocity, excluding its own singular velocity, soIts signed velocity parallel to the wall is $\Gamma /(4\pi b)$, and its speed is $|\Gamma |/(4\pi b)$.

For the quarter-plane, place images of circulation $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. The paired stream functions make both walls streamlines. Summing their velocities at the real vortex givesIn polar coordinates,It follows directly thatThus the line-vortex trajectory in a quarter-plane is $r\sin 2\theta =\text{constant}$.