= Solution
Starting from the <hodograph plane>, take the <complex power>
$$
s=q^{4/3},\qquad0<\arg q<3\pi/4.
$$
It takes the <velocity> sector bijectively to the upper half of the <unit disk>. The <boundary> points representing the upstream end, the corner and the downstream end become $-1$, zero and one. Next use the <Möbius transformation>
$$
T=\frac{s+1}{s-1}.
$$
The diameter $-1<s<1$ maps to the negative real axis, while for $s=e^{i\theta}$ on the upper semicircle,
$$
T=-i\cot(\theta/2),
$$
which is on the negative imaginary axis. The interior maps to the third quadrant. Squaring therefore gives a <conformal> bijection from that quadrant to the <upper half-plane>:
$$
\eta=T^2=\left(\frac{q^{4/3}+1}{q^{4/3}-1}\right)^2.
$$
The upstream end, corner and downstream end map to zero, one and <infinity>, respectively. The two walls map to the positive real axis, and the free <boundary> maps to the negative real axis.
Independently, the <exponential function> $\eta=e^{\pi w}$ takes the potential strip to the same <upper half-plane> with the same <boundary> correspondence. The strip height fixes the coefficient $\pi$, and the normalization $w(B)=0$ fixes the remaining positive multiplicative constant in $\eta$ to one. Consequently
$$
\boxed{e^{\pi w}=\left(\frac{(w')^{4/3}+1}{(w')^{4/3}-1}\right)^2.}
$$
One can make the branches explicit. In the strip, $p=e^{\pi w/2}$ lies in the first quadrant, whereas $T=-p$. Thus
$$
q^{4/3}=\frac{p-1}{p+1}=\tanh\left(\frac{\pi w}{4}\right),\qquad
q=\left[\tanh\left(\frac{\pi w}{4}\right)\right]^{3/4},
$$
with the power taking <arguments> in $(0,\pi)$ to $(0,3\pi/4)$. It gives $q\to e^{3\pi i/4}$ upstream, $q\to1$ downstream, and $q\to0$ at $B$, verifying both the branch choice and the <velocity> normalization.
Back to article page
