A contractible domain changes the story

Circulation has to
live somewhere.

Prescribe a tangential velocity around the boundary of a smooth disk. Because that boundary encloses the disk, Stokes says its circulation must equal the total interior vorticity.

FULL DISK

The loop bounds a patch.

\Gamma(r)=\int_{D_r}\omega

A smooth nonzero circulation at the outer boundary requires nonzero total vorticity inside.

FLAT TORUS

A generator bounds no patch.

d\eta=0,\qquad\oint_\gamma\eta\ne0

The missing circulation can be a smooth harmonic degree of freedom because the loop is noncontractible.

ANNULUS

Two boundary circulations reveal two pieces.

\eta=(Ar^2+B)d\theta,\qquad d\eta=2A\,dx\wedge dy

The difference Γ_out − Γ_in measures total vorticity. A common shift of both circulations changes only the harmonic part.

SMOOTH FULL DISKsolid rotation · globally smooth Clebsch labels

a grid b grid probe loop

PARAMETER GRID

SMOOTH CLEBSCH RECONSTRUCTION

u^\flat(0)=d(-U_bab)+(2U_ba)d b,\qquad(a,b)=(x,y)

At t = 0 the global Cartesian parameter labels give a seam-free Clebsch triple. As a and b are carried, φ must evolve to keep the same Euler velocity.

time / horizon0.00 / 8.00

probe radius—

loop circulation—

enclosed vorticity—

Stokes defect—

Choose the domain

Boundary speed U_bsigned tangential velocity at r = 10.70

Probe-loop radius ρmeasurement loop only · it does not change the flow0.68 Maximum forward timepause at this horizon · exact resampling avoids time-step drift8.0

BOUNDARY AUDIT

u\cdot n=0,\qquad u\cdot t=U_b

The normal component vanishes pointwise. The tangential value is imposed, so the outer circulation is always $2\pi U_b$ .

MOVABLE STOKES AUDIT

The probe measures the field; it does not steer it.

\Gamma(\rho)=2\pi U_b\rho^2=\int_{D_\rho}2U_b\,dA

Slide ρ to choose a different integration loop. Every probe circle bounds a smaller disk, so its circulation must equal the vorticity inside that disk.

A complete smooth test case

The Clebsch labels encode the same rotating one-form.

For solid rotation, the parameter grid can be interpreted directly as two globally smooth label families. The exact term then completes the velocity without changing its vorticity.

01 · BOUNDARY CONDITION

Choose solid rotation.

u=U_b(-y,x)=U_b r\,e_\theta

At the unit boundary, the tangential speed is exactly U_b. The center is completely regular.

02 · GLOBAL LABELS

Use Cartesian scalar fields.

\alpha=2U_ba,\qquad\beta=b,\qquad\phi=-U_bab

The displayed parameter grid carries a = x₀ and b = y₀. In the smooth test, rescaling a gives the Clebsch label α. In the annulus mode the same grid remains a material-coordinate aid; the live decomposition is written directly as a one-form.

03 · ASSEMBLE VELOCITY

The pieces cancel correctly.

d\phi_0+\alpha_0d\beta_0=-U_by\,dx+U_bx\,dy

At t = 0, the extra dy term in dφ cancels half of αdβ. During animation the two labels stay material while φ changes.

04 · TAKE d

The crossing becomes vorticity.

d\alpha\wedge d\beta=2U_b\,dx\wedge dy

Integrating this two-form over the disk gives the prescribed outer circulation.

The Jacobi-sn puzzle

A slit chooses a branch.
A hole changes the domain.

The attached Mathematica experiment maps a rectangle into a disk with two radial slits using $z=c\,\operatorname{sn}(u\mid c^4)$ . Its vertical rectangle lines become curved coordinate lines in the slit disk.

OPEN CONFORMAL FLOWthe notebook's coordinate curves used as streamlines

Re u constant Im u constant tracers

Turn the map into a velocity

\psi(z)=\operatorname{Re}u(z),\qquad (u^\flat)_{\rm flow}=*d\psi

Constant- $\operatorname{Re}u$ curves are streamlines. The flow is divergence-free in the interior and tangent to the slit shores. Tracers enter and leave through the outer circle, so this is an open-boundary experiment—not the impermeable circulating-wall problem above.

Slit-tip position cslits run from ±c to the outer circle0.60 Tracer speedcoordinate-time speed through the conformal sheet0.34

Want closed walls instead?

Choose one streamfunction value on every physical boundary piece. A smooth example for these horizontal slits is $\psi_{\rm wall}(z)=(1-|z|^2)\operatorname{Im}z$ . It vanishes on both slit shores and the outer circle, so $u^\flat=*d\psi_{\rm wall}$ creates vortical recirculation with no boundary flux.

01 · FUNDAMENTAL SHEET

The slits make the inverse coordinate single-valued.

u(z)=\operatorname{sn}^{-1}(z/c\mid c^4)

On the chosen slit domain, one branch of u can be followed continuously. The red curves are level sets of Re u.

02 · MIRRORED SHEETS

Crossing a slit continues to another branch.

u\longmapsto 2jK+(-1)^j u

The stacked surfaces in the notebook visualize analytic continuation, not additional layers of physical fluid.

03 · THE ANSWER

A branch cut is bookkeeping—not an annular hole.

H^1(D)=0,\qquad H^1(D\setminus\{0\})\cong\mathbb R

Glue the slit shores and u becomes multivalued on the original disk. Treat the slits as physical walls and the boundary-attached slit domain is still simply connected. Only removing a compact center creates the annulus and its genuine circulation degree of freedom.