Hodograph plane ($q=u-iv$)

The hodograph plane records velocity rather than position. For a two-dimensional potential flow, the complex coordinate $q=u-iv$ is the complex velocity. A straight impermeable wall fixes the velocity direction and hence maps to a radial line in this plane. A constant-speed free streamline maps to a circle centered at zero. Thus a free-boundary problem in physical space can become a fixed geometric domain in velocity space. Recovering the physical map then uses $dz/dw=1/q(w)$, wherever $q$ is nonzero.

A unit-speed free layer turning downward around a reentrant wall corner has complex velocity $q=u-iv$ in the first-quadrant exterior of the unit disc. Upstream $q=1$, downstream $q=i$, and the corner has infinite speed because its fluid angle is $3\pi /2$. Squaring $q$, applying $(s-1)/(s+1)$, and squaring again maps this domain to the upper half-plane with upstream, corner and downstream at $0,1,\infty$. The potential strip map $e^{\pi w}$ has the same normalization, proving the formula. The sign of the turn and the fluid corner angle determine whether the relevant hodograph is an exterior or interior quarter disc.

A semi-infinite horizontal plate at unit height and a floor at zero give a velocity potential strip of height $a$ and an upper unit semicircle in the complex velocity $q$. Their matching half-plane maps give $(1+e^{\pi w/a}{)}^{-1}=((q-1)/(q+1){)}^2$. Along the fluid free boundary put $q=e^{-i\theta }$, $-\pi <\theta <0$. Then $dz/d\theta =-(a/\pi )e^{i\theta }\cot (\theta /2)$ and $y=1-(a/\pi )(\theta +\pi +\sin \theta )$. Its downstream limit is $1-a$, which must equal the downstream depth $a$. Thus the contraction is exactly one half.

For a planar reservoir draining below a vertical lip of height one into a unit-speed stream, the potential strip $0<\mathrm{Im}w<a$ and first-quadrant unit velocity disk satisfyOn the fluid free boundary put $q=e^{-i\theta }$, $-\pi /2\le \theta <0$. Then $dz/d\theta =-2a{e}^{i\theta }\cot \theta /\pi$, so the downstream height is $1-2a/\pi$. Equating it to $a$ gives $a=\pi /(\pi +2)$.

For an inviscid irrotational unit-speed layer turning through a right angle, the complex potential of a two-dimensional potential flow occupies the strip $0<\mathrm{Im}w<1$ and its complex velocity $q$ occupies a fourth-quadrant quarter disk. The maps $r=q^2$, $s=(r-1)/(r+1)$ and $s^2$ take this disk to the upper half-plane, sending upstream velocity one, corner stagnation zero, and downstream velocity $-i$ to $0,1,\infty$. The map $e^{\pi w}$ has the same normalization, proving the displayed relation.