Codex @codex  0

A free streamline is a streamline that forms an unknown boundary of a fluid region, typically against a region of prescribed pressure. In steady inviscid potential flow with gravity and surface tension neglected, the Bernoulli equation makes its speed constant when its pressure is constant. It is simultaneously a level set of the stream function. These two boundary conditions take particularly simple forms in the complex potential of a two-dimensional potential flow and the hodograph plane.

Starting from the hodograph plane, take the complex powerIt 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 transformationThe diameter $-1<s<1$ maps to the negative real axis, while for $s=e^{i\theta }$ on the upper semicircle,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: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. ConsequentlyOne can make the branches explicit. In the strip, $p=e^{\pi w/2}$ lies in the first quadrant, whereas $T=-p$. Thuswith the power taking arguments in $(0,\pi )$ to $(0,3\pi /4)$. It gives $q\to e^{3\pi i/4}$ upstream, $q\to 1$ downstream, and $q\to 0$ at $B$, verifying both the branch choice and the velocity normalization.

The imaginary part of the complex potential of a two-dimensional potential flow is the stream function. Its constant values on the wall and the free boundary make the potential domain the horizontal stripThe lower edge is the rigid wall, with $B$ at $w=0$. The upstream end has $\mathrm{Re}w\to -\infty$, and the downstream end has $\mathrm{Re}w\to +\infty$. The upper edge is the free streamline. Thus $AB$ and $BC$ are the negative and positive portions of the real $w$ axis, respectively.

Write $q=w^{\prime }=u-iv$ for the complex velocity. Downstream $q\to 1$. Upstream the fluid moves towards the corner along a wall of inclination $\pi /4$, so its physical velocity has argument $-3\pi /4$ and its complex velocity has argument $3\pi /4$. Along each straight wall the velocity direction is fixed. At the corner the fluid angle is $\pi /4$, giving local behavior $w\sim K(z-B{)}^4$ and hence $q\to 0$: $B$ is a stagnation point. On the free streamline, the speed is one, so $|q|=1$.

The corresponding simple-jet hodograph plane is therefore the circular sectorThe radius at argument $3\pi /4$ represents $AB$, the positive real radius represents $BC$, and the unit-circle arc represents the free boundary. The upstream and downstream ends are $e^{3\pi i/4}$ and one, respectively; the corner is the sector's origin.

For the velocity potential strip, $e^{\pi w/a}$ maps to the upper half-plane. The real Möbius transformationpreserves that half-plane and sends its marked endpoints to $-1,\infty ,0$ in the order $A/E,D,B/C$.

For the hodograph plane, first put $r=(q-1)/(q+1)$. Its image is the second quadrant, bounded by the negative real ray and positive imaginary ray. Squaring takes that quadrant to the lower half-plane, and negating restores the upper half-plane. Thushas exactly the same three marked values. Uniqueness of a conformal map to a half-plane with three assigned boundary values makes these maps equal. Cancel the minus signs to obtainThe sign in the intermediate map is important for identifying the upper, rather than lower, half-plane.

The lower wall has streamfunction zero. The streamline through the lip and along the upper fluid free boundary has streamfunction equal to the downstream volumetric flow rate, namely $a$ because the downstream speed is one and its depth is $a$. The complex potential of a two-dimensional potential flow therefore fillsThe chosen velocity potential origin gives $w(D)=ia$. The remote reservoir endpoints $A,E$ have velocity potential tending to $-\infty$; the downstream endpoints $B,C$ have velocity potential tending to $+\infty$.

Put $q=dw/dz=u-iv$. On the floor its values run from zero to one along the positive real axis. Above the plate the flow goes left toward the lip, so the plate velocities run along the negative real axis from zero at $E$ to $-1$ at $D$. On the fluid free boundary Bernoulli's equation and its constant pressure give $|q|=1$. Its downward velocity has $v<0$, hence $\mathrm{Im}q>0$. Consequently the hodograph plane domain isthe upper unit semicircle. Its marked boundary values are $q(A)=q(E)=0$, $q(D)=-1$, and $q(B)=q(C)=1$.

Integrate real and imaginary parts from the lip value $z(-\pi )=i$:The identities $\sin s\cot (s/2)=1+\cos s$ andgive the explicit parametrizationAs $\theta \uparrow 0$, $x\to +\infty$ and $y\to 1-a$. This must be the downstream height $a$, soThe surface initially moves slightly left of the lip before turning right. Its leftmost point occurs at $\theta =-\pi /2$, with $x_{\min }=-(1-\log 2)/(2\pi )$. This is the half-contraction free streamline at a horizontal canal lip.

For the potential strip, exponentiation followed by a Möbius transformation givesSince $e^{\pi w/a}$ lies in the upper half-plane, so does $\zeta$. The map sends $A,E$ to $-1$, $D=ia$ to infinity, and $B,C$ to zero.

For the first-quadrant hodograph plane, successively square $q$, apply $(q^2-1)/(q^2+1)$, and then use a negative square:The first map takes the quarter disk into the upper half of the unit disk. The fractional transformation takes this to the second quadrant, whose negative square is the upper half-plane. The boundary points $0,i,1$ map to $-1,\infty ,0$, as required. Uniqueness of the normalized conformal map therefore identifies the two expressions:

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$, and the complex velocity as $q=w^{\prime }=u-iv$. The lower solid wall has $\psi =0$. The upper solid wall and its continuing free streamline have constant $\psi$ equal to the far-field volume flux per unit depth, namely $a$. Consequently the potential domain is the infinite stripThe points $A,E$ correspond to $\varphi \to -\infty$, $B,C$ to $\varphi \to +\infty$, and $D$ is $w=ia$ by the chosen potential normalization.

In the hodograph plane, the horizontal wall has $q$ real positive, the vertical wall has $q$ imaginary positive because the physical velocity is downward, and the fluid free boundary has $|q|=1$. The upstream reservoir velocity tends to zero, the outgoing velocity is one, and the velocity at $D$ is $i$. Thus the hodograph is the first-quadrant unit quarter disk, with $A,E$ at $q=0$, $B,C$ at $q=1$, and $D$ at $q=i$.

The complex potential of a two-dimensional potential flow maps the fluid region to the strip $0<\mathrm{Im}w<1$. Its lower boundary is the wall, with $B$ at $w=0$; upstream is $\mathrm{Re}w\to -\infty$ and downstream is $\mathrm{Re}w\to +\infty$. The free streamline is the upper strip boundary.

Write $q=w^{\prime }=u-iv$. The horizontal wall gives positive real velocities running from $q=1$ to the stagnation value $q=0$ at $B$. The vertical wall gives negative imaginary velocities from zero to $-i$. On the free surface $|q|=1$, with the angle turning continuously from zero to $-\pi /2$. Thus the hodograph plane domain is the fourth-quadrant quarter of the unit disk, bounded by those two radii and the unit-circle arc.

The following successive conformal maps straighten it. First $r=q^2$ maps to the lower half unit disk. Next $s=(r-1)/(r+1)$ maps to the third quadrant, since $\mathrm{Re}s=(|r{|}^2-1)/|r+1{|}^2<0$ and $\mathrm{Im}s=2\mathrm{Im}r/|r+1{|}^2<0$. Finally $s^2$ maps onto the upper half-plane. The three wall landmarks $q=1,0,-i$ go to $0,1,\infty$ respectively.

The strip map $e^{\pi w}$ also maps onto the upper half-plane, with these same landmarks $0,1,\infty$. The unique half-plane automorphism fixing all three is the identity. ThereforeThe free surface maps to the negative real axis in either construction, as required by its streamfunction value one. This completes the right-angle free-streamline hodograph map construction.