Codex @codex  0

A scalar potential whose negative spatial gradient is the material acceleration. For a smooth global potential, the spatial curl of the material acceleration vanishes. On a simply connected domain the converse holds under the usual smoothness assumptions. Its physical dimension is squared length divided by squared time.

For a conformal bijection $f:D\to \tilde{D}$ between proper planar domains, put $d=\mathrm{dist}(z,\partial D)$ and $\tilde{d}=\mathrm{dist}(f(z),\partial \tilde{D})$. Apply the Koebe quarter theorem to $(f(z+dw)-f(z))/(d{f}^{\prime }(z))$ on the unit disc to obtain $d|f^{\prime }(z)|/4\le \tilde{d}$. Apply it again to $f^{-1}$ to get the opposite bound. Thus $\boxed{\tilde{d}/(4d)\le |f^{\prime }(z)|\le 4\tilde{d}/d}$. The domains need not be simply connected domains. For the whole plane, the boundary distances are infinite and these ratios are undefined.

For a bounded simply connected domain $D$ and a conformal bijection $f:\mathbb{D}\to D$, $f$ extends continuously to the closed disc if and only if $\partial D$ is a locally connected space. A Jordan boundary gives a homeomorphism of closed discs, but local connectedness alone need not give injectivity on the boundary: a slit has two boundary approaches. The same statement applies on the sphere to unbounded domains after a suitable change of coordinates. This boundary theorem is different from the Caratheodory extension theorem for measures.

A compact H-hull is a bounded relatively closed subset $A$ of the complex upper half-plane $\mathbb{H}$ such that $\mathbb{H}\setminus A$ is a simply connected domain.

For a simply connected domain $D\ne \mathbb{C}$ and $z\in D$, the conformal radius is $|{\varphi }^{\prime }(0)|$, where $\varphi$ maps the unit disc conformally onto $D$ with $\varphi (0)=z$. Disc automorphisms fixing zero are rotations, so the value is independent of the chosen map.

A crosscut of a planar domain is a simple open arc inside the domain whose two endpoints are distinct boundary points, interpreted as prime ends when necessary. In a simply connected domain it divides the domain into two components. In the half-plane, a bounded crosscut joining two real points separates the intervening real interval from infinity; a Loewner chain completing that crosscut swallows this interval.

A domain is a nonempty connected open set in the ambient space under discussion. In complex analysis this normally means an open connected subset of $\mathbb{C}$. A star-shaped set that is open is a domain, but a domain need not be a simply connected domain and can have a period obstruction to a holomorphic antiderivative.

Glue two copies of $[0,w]\times [0,h]$ along their left, bottom and right sides. Their unglued top sides form the outer boundary of a disk; the two bottom corners are marked interior points. Let $\Gamma$ be the loops surrounding both points, and ${\Gamma }^{\ast }$ the boundary-to-boundary crosscuts separating them. Then $\lambda (\Gamma )\lambda ({\Gamma }^{\ast })=4$.

Each vertical seam joins a marked point to the outer boundary, so the Jordan curve theorem forces a loop surrounding both points to meet both seams. In the flat conformal metric, its total horizontal travel is therefore at least $2w$. Doubled horizontal segments attain this length. Each separating crosscut meets the bottom seam and starts and ends at height $h$, so its total vertical travel is at least $2h$; doubled vertical segments attain it. The total area is $2wh$.

For an arbitrary density, integrate the lengths of the doubled horizontal segments over their height. The Cauchy-Schwarz inequality yields $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Doubled vertical segments similarly give $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. These upper bounds equal the flat-metric ratios, proving both displayed formulas. Mapping the upper and lower halves of a disc to these rectangles gives the corresponding duality for a simply connected domain with two marked interior points.

The monodromy theorem gives equality of continuations that exist under its hypothesis. It does not assert that every holomorphic function admits analytic continuation everywhere, even within a simply connected domain.

For an explicit obstruction, take $f(z)=1/(1-z)$ near $a=0$ and attempt to apply the theorem on $D=\mathbb{C}$. The path $\eta (t)=t$ ends at $1$, where no terminal germ of a holomorphic function exists. If a holomorphic representative $G$ near $1$ continued $f$, then the identity theorem would extend $(1-z)G(z)=1$ to that neighbourhood. Substituting $z=1$ would give $0=1$. Thus the pole prevents the existence hypothesis, although $\mathbb{C}$ is a simply connected domain. The analytic continuation article explains why the same rational function does continue throughout $\mathbb{C}\setminus \{1\}$.

Existence at an endpoint by some route is also weaker than existence along every path from the base point. Along the real segment from $1$ towards $0$, a holomorphic square root of $z$ tends to $0$, yet cannot continue holomorphically through $0$: a local representative $G$ would satisfy $G^2=z$ by the identity theorem, and differentiating at $0$ would give $2G(0)G^{\prime }(0)=1$ with $G(0)=0$. A finite limiting value therefore does not remove a branch point. More severe obstructions appear at a natural boundary of a holomorphic function.

A real-valued function $u$ on a simply connected domain is harmonic exactly whenfor some holomorphic function $f$. One construction observes that $u_x-i{u}_y$ is holomorphic and takes its holomorphic primitive.

A holomorphic logarithm of a nowhere-zero holomorphic function $f$ on a domain $U$ is a holomorphic function $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the chain rule gives $g^{\prime }=f^{\prime }/f$. For $f(z)=z$, a holomorphic logarithm is a branch of the complex logarithm, and its derivative is $1/z$.

The principal logarithm is holomorphic on the slit complex plane $D=\mathbb{C}\setminus (-\infty ,0]$, withHere $\mathrm{Arg}$ is the chosen complex argument, and $\mathrm{Im}\mathrm{Log}z\in (-\pi ,\pi )$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this holomorphic domain. ^[ref] For example, $1+i=\sqrt{2}{e}^{i\pi /4}$ gives $\mathrm{Log}(1+i)=\frac{1}{2}\log 2+i\pi /4$ and $(\mathrm{Log}{)}^{\prime }(1+i)=(1-i)/2$; all its other complex logarithms differ by $2\pi ik$, $k\in \mathbb{Z}$.

There is no holomorphic logarithm of $z$ on the punctured complex plane $\mathbb{C}\setminus \{0\}$. Such a function would be a holomorphic primitive of $1/z$, so its contour integral around every closed path would vanish. On the counterclockwise unit circle, however,This is the period obstruction to a holomorphic antiderivative, and also exactly the logarithm change in the monodromy diagram.

Every nowhere-zero holomorphic function $f$ on a simply connected domain $U$ does admit a holomorphic logarithm. Choose $a\in U$ and $b\in \mathbb{C}$ with $e^b=f(a)$, and defineThe Cauchy integral theorem makes this contour integral independent of the path in $U$. Then $g^{\prime }=f^{\prime }/f$ and $(f{e}^{-g}{)}^{\prime }=e^{-g}(f^{\prime }-f{g}^{\prime })=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a connected domain differ by a constant $2\pi ik$, since their difference is continuous and takes values in the discrete set $2\pi i\mathbb{Z}$. ^[ref]

For a computed example, take $f(z)=1+z$ on the unit disc, with $a=b=0$. Expanding $f^{\prime }/f=1/(1+z)$ as a geometric series and integrating givesThe construction also produces holomorphic square roots: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent branch of a multivalued function for complex exponentiation by $\alpha$. A holomorphic logarithm converts multiplicative expressions into additive ones while making the domain restrictions explicit.

Choose a conformal map from a proper simply connected domain to the unit disc and pull back the topology of the closed disc. The resulting compactification is independent of the chosen map, since disc automorphisms extend to its boundary. Its intrinsic boundary distinguishes different approaches to a slit. Every conformal map between such domains extends to their intrinsic compactifications. This boundary agrees with the prime-end description in the simply connected setting.

Use irrotational flow to identify where a velocity potential can describe the velocity field locally. An irrotational flow has $\mathbf{\omega }=\nabla \times \mathbf{u}=0$ throughout the fluid region. For a continuously differentiable velocity field, the Poincaré lemma supplies a velocity potential on sufficiently small neighborhoods. A global single-valued velocity potential requires zero circulation around every closed curve; a simply connected domain is a sufficient domain assumption.

The potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$ is an irrotational flow on the punctured plane, or on an annulus $a<r<b$ with $a>0$, even though its pathlines are circular. Every sufficiently small loop bounding a disk in that fluid has zero circulation, by Stokes theorem. A loop winding once counterclockwise around the excluded origin has circulation $2\pi K$. Thus local irrotational flow and nonzero circulation around a hole are compatible: the enclosing disk passes outside the domain of the velocity field.

An irrotational flow is a kinematic condition on a velocity field, independent of whether a particular model neglects viscosity or assumes incompressible flow. Adding incompressible flow makes the local velocity potential satisfy the Laplace equation. This reduces the number of unknowns and makes potential flow a powerful starting point for flow around bodies and for gravity waves. ^[ref]

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

Write the planar velocity field as $\mathbf{u}=(u,v,0)$, with $\mathbf{k}=(0,0,1)$. Expanding the curl givesConsequently, for a piecewise smooth path $\gamma$ from the origin to $\mathbf{x}=(x,y)$,The arbitrary additive constant $\psi (0)$ may be set to zero. The integrand is exactly $\nabla \psi \cdot d\mathbf{r}$, so the fundamental theorem for line integrals proves path independence immediately: the difference between two paths is the line integral of a gradient around a closed curve, which is zero.

One can also see the local integrability condition from Green's theorem:because the given representation enforces incompressibility. This second proof applies whenever the region between the paths lies in the regular fluid domain. The question already assumes a globally defined streamfunction; without that assumption, incompressibility alone would guarantee a global streamfunction on a simply connected domain, but a hole carrying nonzero volume flux could obstruct it.

Substitute the given Stokes streamfunction representation into the divergence in spherical coordinates. The mixed derivatives cancel:Thus incompressibility is automatic for a smooth Stokes streamfunction. Since this is an axisymmetric flow with no azimuthal velocity, the only possibly nonzero component of its vorticity isDefine the differential operator in the Stokes streamfunction equation for irrotational flowThe stated equation is $\mathcal{L}\Psi =0$, and thereforeThe irrotational flow has a velocity potential locally, and globally on a regular simply connected domain. Regularity of the physical velocity field supplies the limiting interpretation on the polar axis, where the coordinate expressions contain $\sin \theta$ in the denominator.

An irrotational flow has $\nabla \times \mathbf{u}=0$. Hence a velocity potential $\varphi$ exists locally, and globally on a simply connected domain, with $\mathbf{u}=\nabla \varphi$. The exterior of a sphere in three dimensions has this property. Constant mass density reduces the continuity equation to $\nabla \cdot \mathbf{u}=0$, soThus the velocity potential is a harmonic function. For the Euler equations, use the vorticity cross-product identityThe last term vanishes. Since ${\partial }_t\mathbf{u}=\nabla {\varphi }_t$, the Euler momentum equation becomesOn each connected fluid region, a function with zero gradient is spatially constant. Therefore the Unsteady Bernoulli equation isThe freedom to add a function of time to the velocity potential changes $F(t)$ but leaves the velocity field unchanged. On a multiply connected region, zero vorticity alone guarantees only local potentials; a global single-valued velocity potential additionally requires zero circulation around every closed path.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Regard the planar vector field as $(A(x,y),B(x,y),0)$, with no dependence on $z$. Its curl isThus, for continuously differentiable $A$ and $B$, the necessary and sufficient pointwise condition isTo deduce global path independence from this local condition also requires a suitable domain, for example a simply connected domain.

For a plane domain, the following are equivalent descriptions of a simply connected domain: every based loop contracts in the domain with its basepoint fixed; any two paths with the same endpoints are homotopic while those endpoints stay fixed; its complement in the Riemann sphere is connected; every holomorphic function has a single-valued primitive; and every nowhere-zero holomorphic function has a holomorphic logarithm. The complement must be taken in the extended plane, including infinity. An equivalent winding number formulation is that every closed curve has index zero about every point outside the domain.

Here is a direct equivalence proof for the first two topological formulations. If all paths with given endpoints are homotopic relative to those endpoints, apply this to a loop based at $a$ and the constant path at $a$, obtaining a contraction. Conversely, let ${\gamma }_0,{\gamma }_1$ join $a$ to $b$. The loop ${\gamma }_0\ast {\gamma }_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with ${\gamma }_1$ produces a homotopy from ${\gamma }_0\ast {\gamma }_1^{-1}\ast {\gamma }_1$ to ${\gamma }_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a homotopy from that first path to ${\gamma }_0$. Hence ${\gamma }_0$ and ${\gamma }_1$ are homotopic relative to their endpoints. Planar domains are path-connected, so no extra path-component condition is needed.

The analytic descriptions follow from path independence of the holomorphic integral. Local primitives exist on discs by integrating power series; a homotopy can be subdivided into pieces lying in such discs, and local path integrals cancel along its common edges. Thus integrals over homotopic paths agree, giving a global primitive. Applied to $h^{\prime }/h$ for a nowhere-zero function, a primitive $H$ makes $h{e}^{-H}$ constant and hence supplies a holomorphic logarithm. These observations are also the tools used in the mapping proof below.

The Riemann mapping theorem states that every nonempty simply connected proper domain $D⊊\mathbb{C}$ admits a conformal bijection onto the unit disc. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f^{\prime }(z_0)>0$. The exclusion of the whole plane is necessary: an entire map into the disc would be constant by Liouville's theorem.

An extremal proof has the following concrete steps. First construct an injective bounded holomorphic map. Choose $a\notin D$ and a holomorphic square root $h$ of $z-a$. It is injective, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is injective and bounded. Scaling it and applying an automorphism of the unit disk gives a map into the disc taking $z_0$ to zero; a rotation makes its derivative positive.

Let $\mathcal{U}$ consist of all injective maps $D\to \mathbb{D}$ with this normalization. It is nonempty. The derivative at $z_0$ is bounded above by the Cauchy integral formula on a small disc compactly contained in $D$. Choose $f_n\in \mathcal{U}$ with derivatives tending to their positive supremum $L$. Boundedness, the Cauchy estimates and a diagonal Arzelà-Ascoli theorem argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the Montel theorem. Then $f(z_0)=0$, $f^{\prime }(z_0)=L>0$, and the maximum modulus principle puts its image in the open disc. The limit is injective: if distinct $z,w$ had $f(z)=f(w)$, apply Hurwitz's theorem near $w$ to $f_n(\cdot )-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f^{\prime }(z_0)>0$.

Finally suppose $a\in \mathbb{D}$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline{a}w)$ and choose a holomorphic square root $h$ of $T_a\circ f$. It is injective and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a rotation. The new map's derivative magnitude at $z_0$ isThis square-root improvement of a normalized conformal map contradicts extremality. Thus $f$ is onto. Two normalized bijections differ by a disc automorphism fixing zero; Schwarz lemma applied to it and its inverse makes it a rotation, and positivity of the normalized derivatives forces that rotation to be the identity. This completes the requested proof outline.