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For real test functions $f,g\in C_c^{\infty }(D)$, the Dirichlet inner product in the normalization compatible with the logarithmic Green kernel isIts norm measures gradient energy rather than the values of the functions. A change of variables shows that this energy is invariant under planar conformal maps: the squared derivative in the transformation of the gradient cancels the area Jacobian determinant.

Set $F(w)=(\varphi (w)-\varphi (0))/w$ for $w\ne 0$. The singularity is removable, with $F(0)={\varphi }^{\prime }(0)$. Injectivity of the conformal map makes $F(w)\ne 0$ away from zero, and conformality makes $F(0)\ne 0$ as well. Since the disc is simply connected, $F$ has a holomorphic logarithm. Therefore $\psi =\log |F|$ is a harmonic function, including at zero, and $\psi (0)=\log |{\varphi }^{\prime }(0)|$.

The mean value property for harmonic functions gives, for every $0<r<1$,For general $D$, boundary values mean radial limits, rather than a continuous extension of $\varphi$ to every point of the circle. The boundary logarithmic mean of a univalent function justifies taking $r\uparrow 1$: the Koebe distortion theorem bounds $|F|$ below by $|{\varphi }^{\prime }(0)|/4$, while the standard integral-mean bound for a univalent function, ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, gives uniform integrability of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,The expression on the left is the logarithm of the conformal radius of $D$ at $z=\varphi (0)$.

Interpret the printed $A_0=0$ as the empty initial hull. Write $a(t)=\mathrm{hcap}({\tilde{A}}_t)$ and $F_t={\psi }_t$. Conformal maps preserve inclusion and simple connectedness of the complementary domains, so the image sets form an increasing hull family. Boundedness of $\psi$ on bounded sets ensures that these image hulls are bounded. The Loewner local growth property is preserved under conformal transport: after mapping out the hull at time $t$, the small new hull is transported by $F_t$ near its single boundary growth point. The Schwarz reflection principle extends $F_t$ analytically across that point, with real positive derivative. The image diameters therefore tend to zero as the original ones do. These statements use the usual hull closures and prime ends; the initial image hull is empty.

For completeness, the infinitesimal capacity rule underlying this argument is that a shrinking hull $J$ attached near $u$, transported by a map $F$ analytic there, hasOne obtains this by rescaling at $u$: the transported map tends to its linear part, and half-plane capacity scales by the square of the dilation. Uniform analytic distortion near the growth point controls the error. Applying it to the mapped-out increments gives local absolute continuity of $a$ and $a^{\prime }(t)=2F_t^{\prime }(U_t{)}^2$.

The coefficient can also be read directly from the Chordal Loewner equation. The image driver is ${\tilde{U}}_t=F_t(U_t)$. Differentiate $F_t={\tilde{g}}_t\circ \psi \circ g_t^{-1}$, at a fixed point $w$, to obtainThe left side is regular at $w=U_t$. On the right, the coefficient of $(w-U_t{)}^{-1}$ is $a^{\prime }(t)/F_t^{\prime }(U_t)-2F_t^{\prime }(U_t)$, so it must vanish. This proves the conformal change of half-plane capacity rule and its integrated version:If the derivative at the initial boundary point is singular, the formula is interpreted by integrating from a positive time and taking the lower limit to zero; finite image capacity and kernel continuity give this limit. The duplicated incomplete normalization sentence in the TeX is absent from the PDF.

Let $v_1,v_2\ne 0$ be tangent vectors to the two smooth curves at $p$. complex differentiability at a point givesWhen $f^{\prime }(p)\ne 0$, its real derivative is multiplication by a nonzero complex number, which is a rotation followed by a positive scaling. It therefore preserves the angle between $v_1$ and $v_2$, so $f$ is conformal at $p$. The condition is essential: $f(z)=z^2$ has $f^{\prime }(0)=0$ and sends rays making angle $\pi /4$ at zero to rays making angle $\pi /2$.

For $J(z)=z+z^{-1}$,If $z,w$ both satisfy $|z|>1$, or both satisfy $0<|z|<1$, the alternative $zw=1$ is impossible unless the points lie on the omitted unit circle. Hence $J$ is one-to-one on each region. Also $J^{\prime }(z)=1-z^{-2}$ has no zero there, so the restrictions are conformal. Solving $z^2-wz+1=0$ shows that $w\in [-2,2]$ exactly when both roots lie on the unit circle; otherwise one root is inside and the other outside. Thus each region has image

Taking the reciprocal of the interior restriction and filling its removable value at zero givesThe reciprocal sends $\mathbb{C}\setminus [-2,2]$ ontoand $g(0)=0$. Therefore $g$ is the required one-to-one conformal map from the unit disc.

A compact H-hull is a bounded, relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe half-plane capacity is

Let $D$ be simply connected domain with distinct marked boundary points $a,b$, and choose a conformal map $f:D\to \mathbb{H}$ with $f(a)=0$ and $f(b)=\infty$. Chordal ${\mathrm{SLE}}_{\kappa }$ from $a$ to $b$ is the unparameterized curve $f^{-1}(\gamma )$, where $\gamma$ is chordal SLE in $(\mathbb{H},0,\infty )$.

Any other such map is $\tilde{f}=rf$ for some $r>0$. The Scaling invariance of SLE says that $r^{-1}\gamma$ has the same unparameterized law as $\gamma$; only its capacity clock changes. Hence the pullback law is independent of $f$. This proves the Conformal invariance of SLE definition is well-defined.

The Wirtinger derivative $u_z=(u_x-i{u}_y)/2$ is holomorphic, since ${\partial }_{\bar{z}}{u}_z=\Delta u/4=0$ by the Laplace equation. The conformal map $w=z^2$ takes the quadrant to the complex upper half-plane. Integrating the prescribed tangential boundary derivatives, choose a common corner value $c$ and define the real boundary functionThe decaying data are bounded under the usual regularity at the corner, so $f(s)=O(1+\sqrt{|s|})$. Thus its Poisson integral converges. A holomorphic function whose real part is that Poisson integral is supplied by the regularized Schwarz integral formula:For $u_{\ast }(z)=\mathrm{Re}W(z^2)$, the chain rule gives $(u_{\ast }{)}_z=z{W}^{\prime }(z^2)$. Apply integration by parts to obtainSubstituting $s=r^2$ and $s=-r^2$ gives the requested particular integral representation:Both integrals converge absolutely for an interior point of the quadrant. The Sokhotski–Plemelj formula gives $\mathrm{Re}(u_{\ast }{)}_z(x)=g_2(x)/2$ on the horizontal edge and $\mathrm{Im}(u_{\ast }{)}_z(iy)=-g_1(y)/2$ on the vertical edge, verifying the prescribed derivatives.

The printed conditions alone do not determine $u_z$ uniquely. The general answer adds a holomorphic function $H$ satisfyingEquivalently, write $H(z)=ih(z^2)/z$, where $h$ is holomorphic in the complex upper half-plane with real boundary values on the nonzero real axis. An antiderivative of $2H$ has a harmonic real part with zero tangential boundary derivatives. For example, $u=xy$ contributes $H=-iz/2$ without changing either datum. If the two edge constants differ by $c_2-c_1$, the angular harmonic function contributes $i(c_2-c_1)/(\pi z)$. The boxed expression selects the Poisson integral representative with equal edge constants; regularity and suitable growth conditions can be used to select that representative.

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityIts half-plane capacity is $\mathrm{hcap}(A)=a$.

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ such that $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe nonnegative coefficient $a$ is the half-plane capacity $\mathrm{hcap}(A)$.

A complex function is holomorphic at a point if it is complex differentiable throughout some open neighbourhood of that point. Writing $\zeta =\xi +i\eta$ and $\psi =P+iQ$, a convenient sufficient and necessary criterion under continuously differentiable real components is the Cauchy-Riemann equationsThey must hold throughout an open neighbourhood, rather than at an isolated point. Merely having partial derivatives that satisfy these equalities at one point is not enough without an appropriate differentiability hypothesis. For the composition $\psi =G\circ f$, holomorphy of $f$ and $G$ on the relevant open sets suffices; the additional condition $f^{\prime }\ne 0$ is needed when using a local inverse conformal map, not for holomorphy of the composition itself.

The three contacts in the strip. Work with the external-contact arrangement drawn in the original diagram: both disks lie between the two parallel lines and touch opposite boundaries. A rigid motion puts those boundaries at $x=0$ and $x=d$, with $d>0$, and makes $P=(0,0)$. Write $R=(d,h)$, and let the positive radii be $r_1,r_2$. Because a tangent to a circle is perpendicular to the corresponding radius, the centres areAt the external tangency, the two radial segments form the segment $O_1{O}_2$, with $|O_{1}Q|=r_1$ and $|Q{O}_2|=r_2$. Hence the internal division formula givesThusThe coefficient is strictly between zero and one, so $Q$ lies on the actual line segment $PR$. This proves collinearity for tangent circles between parallel lines. We will also use the fact that $PQ$ and $QR$ are chords of the respective circles; the entire line segment $PR$ therefore lies in the union of their closed disks.

The four contacts. Label the successive contact points $T_{12},T_{23},T_{34},T_{41}$, with $T_{jk}$ belonging to $C_j$ and $C_k$. The original diagram has four positive-radius circles with disjoint interiors, successive external tangencies, and distinct contact points. Set $T=T_{41}$ and apply the reciprocal Möbius transformationBy part (b), the two circles $C_1,C_4$ through $T$ become straight lines ${\ell }_1,{\ell }_4$. These straight lines cannot meet at a finite point, since their inverse images would give a second intersection of $C_1,C_4$. They are therefore distinct parallel lines. The other two circles avoid $T$, so their images $D_2,D_3$ are ordinary circles missing zero.

The three remaining contacts retain their tangencies. Indeed, away from $T$ the map and its inverse are holomorphic functions with nonzero derivatives, hence are local conformal maps. In particular they preserve a shared tangent line. Equivalently, bijectivity preserves the unique intersection of two generalized circles, which characterizes their tangency here.

The finite closed disks bounded by $D_2,D_3$ lie between ${\ell }_1,{\ell }_4$, have external tangency to each other, and touch opposite boundaries at the images of $T_{12}$ and $T_{34}$. To track the sides as well as the boundaries, a closed disk avoiding $T$ maps to the bounded closed disk of its image circle, while each disk with $T$ on its boundary maps to a closed half-plane. The two latter half-planes have disjoint interiors, so their complement is the strip containing zero. Disjointness of the original interiors puts $D_2,D_3$ inside that strip. Neither image closed disk contains zero, since the reciprocal Möbius transformation sends only infinity to zero.

The first part now makes the three finite transformed contact points collinear, on a straight line $\ell$. It is essential that $\ell$ misses zero. If it passed through zero, its segment between ${\ell }_1$ and ${\ell }_4$ would contain zero, since zero lies strictly within the strip. But the first part shows that this whole line segment is covered by the two image closed disks, neither of which contains zero. This contradiction proves $0\notin \ell$.

The inverse image of a straight line missing zero is a genuine circle through $T$, by the same coefficient transformation in part (b), or by applying the involution $u\mapsto 1/u$ in reverse and translating back. This circle contains $T_{12},T_{23},T_{34}$, while $T$ is the inverse image of the point at infinity on $\ell$. ThereforeThis is the four contact points of a circle chain result; the four contacts form a cyclic quadrilateral.

The pictured arrangement matters. Arbitrary mixtures of internal tangency and external tangency can instead give four distinct collinear contacts: take four circles whose diameter endpoints on the real axis are respectively $[0,3]$, $[0,1]$, $[1,2]$, and $[2,3]$. Their successive contacts are $0,1,2,3$, so no ordinary circle contains all four. The proof above establishes the requested genuine-circle conclusion for the external-contact arrangement actually drawn, rather than silently treating every generalized circle as an ordinary circle.

On the first quadrant choose the holomorphic logarithm with $0<\arg w<\pi /2$. It maps $w=\rho e^{i\theta }$ to $\log \rho +i\theta$, so its image is precisely the strip $0<\mathrm{Im}\zeta <\pi /2$. Its inverse there is the exponential function; both derivatives are nonzero. Composing with part (i) givesThis is a bijective conformal map from the lens onto the strip, with inverseThe logarithm's branch is fixed by its argument range on the quadrant. Using a different branch would shift the strip vertically by an integer multiple of $2\pi$. This is the lens-to-strip logarithmic conformal map.

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

A conformal map at a point $z_0$ is a map that preserves oriented angles between regular differentiable curves through $z_0$. In the differential formulation, it is real differentiable at $z_0$ and its derivative is an orientation-preserving similarity: a rotation followed by multiplication of lengths by a positive constant. The map is conformal on $U$ when it has this property at every point:Here $R_{\alpha }$ is a planar rotation. Conformality is a local property and does not itself require global injectivity. We use the oriented convention standard in complex analysis; a reflection reverses oriented angles.

The conformal map $\zeta =z^3$ sends the wedge $0<\arg z<\pi /3$ to the upper half-plane. The physical vortex maps to $\zeta =c^3$; an opposite image vortex at ${\overline{c}}^{,3}$ makes the real axis a streamline. Pulling this potential back givesSince $z^3-c^3=(z-c)(z^2+cz+c^2)$, splitting the first logarithm, up to an additive constant, gives exactly the required form. The image construction enforces zero normal velocity on both rays.

To find the vortex motion, hold $c$ fixed while differentiating with respect to $z$, remove the self-induced singular term $-i\Gamma /(2\pi (z-c))$, and then evaluate at $z=c$. The remaining complex velocity isA complex velocity $u-iv$ equals $d\overline{c}/dt$, rather than $dc/dt$. ThusThe overbar on the source derivative is essential to the trajectory calculation.

Write $c=r{e}^{i\theta }$. Since $c^3-{\overline{c}}^{,3}=2i{r}^3\sin (3\theta )$, multiply the equation by $e^{i\theta }$ to obtainTaking real and imaginary parts givesThereforeproving the trajectory invariantThe calculation also covers $\Gamma =0$, when the vortex position is stationary.

This is the sixty-degree case of a point vortex in a wedge.

On $\zeta =a{e}^{i\theta }$, the Joukowski map gives $z=2a\cos \theta$, so the image is the segment $S=[-2a,2a]$. Solving the resulting quadratic gives the inverse candidatesOn $D=\mathbb{C}\setminus [-2a,2a]$, choose the analytic square root $s(z)$ of $z^2-4a^2$ characterized by $s(z)/z\to 1$ at infinity. One explicit definition iswhere the square root on the right is the principal square root. Its argument never lies on the nonpositive real axis for $z\in D$, so this defines the required holomorphic function. Then $g(z)=(z+s(z))/2\sim z$.

The two quadratic roots have product $a^2$. For $z\notin S$, neither root lies on $|\zeta |=a$, since that circle maps into $S$. Consequently exactly one root lies outside the circle, and it is the branch $g$ chosen at infinity. Also, if two exterior points have the same image, thenwhose second factor cannot vanish. Thus the Joukowski map is a conformal map from $|\zeta |>a$ onto $D$, and $g$ is conformal throughout $D$. The two endpoints are branch points and are excluded.

The complex potential in the circle plane is the sum of the uniform flow term $U{e}^{-i\alpha }\zeta$, its circle-image term $U{a}^2{e}^{i\alpha }/\zeta$, and the point vortex term $-i\Gamma \log \zeta /(2\pi )$. On $\zeta =a{e}^{i\theta }$, the first two sum to $2Ua\cos (\theta -\alpha )$, which is real, while the imaginary part of the vortex term is the constant $-\Gamma \log a/(2\pi )$. Hence the circle is a streamline, giving no normal flow. At infinity the first term gives the required uniform flow; the logarithmic term supplies circulation $\Gamma$. The potential may be multivalued by a real constant, but its velocity is single-valued.

Conformal transport therefore givesIn the frame where the plate is horizontal, its length is $4a$ and the stream has angle $\alpha$; rotating the frame gives the equivalent incidence description. The complex velocity isAt the trailing edge $z=2a$, corresponding to $\zeta =a$, the denominator has a simple zero. Finite velocity requiresso the Kutta condition isWith this choice, cancellation of the common zero givesThus the limiting velocity has zero transverse component and is tangent to the plate.

Choose each factor with its argument between zero and $\pi$ in the upper half-plane. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f^{\prime }$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg (Z-X_j)$ by $\pi$, hence increases $\arg f^{\prime }$ by ${\beta }_j\pi$, giving the prescribed exterior turn. Locally,so an upper half-disc maps to angle $(1-{\beta }_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the prevertex positions must also satisfy the side-length conditions: the derivative formula fixes angles but arbitrary choices do not fix arbitrary side lengths.

For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm 1$, leading to $f^{\prime }(Z)$ proportional to $Z(1-Z^2{)}^{-3/2}$. Integrating and imposing the tip value givesThe square root is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The derivative is nonzero in the open half-plane. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive imaginary part, proving bijectivity.

The Möbius map $(Z-1)/(Z+1)$ takes the upper half-plane to itself, so its principal argument lies in $(0,\pi )$. The imaginary part of its analytic logarithm is harmonic, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. ThusIt lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this harmonic function back by the inverse conformal map yieldsThis specifies all branches and remains bounded even as $z$ tends to either end at infinity.

To exhibit nonuniqueness without the growth restriction, putThis is harmonic in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm 1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm 1$ non-tangentially. Therefore $u+cH$ gives distinct solutions with the same finite boundary data when boundedness at infinity is removed.

Near a prevertex of a Schwarz-Christoffel map $X_j$, a conformal map to an interior angle $(1-{\beta }_j)\pi$ has local formIts derivative therefore has the factor $(Z-X_j{)}^{-{\beta }_j}$. On each real interval between prevertex, the argument of the product of these factors is constant, so its integral traces a straight polygon side. Passing a prevertex of a Schwarz-Christoffel map changes the argument by the prescribed exterior angle ${\beta }_j\pi$. Thushas exactly the required straight sides and corner angles, with the branches chosen in the upper half-plane. Equivalently $f^{\prime \prime }/f^{\prime }$ has simple poles $-{\beta }_j/(Z-X_j)$; removing them leaves the constant normalization determined by behavior at infinity. The additive integration constant translates the polygon and $C$ rotates and scales it. This is the Schwarz-Christoffel formula.

The automorphisms of the upper half-plane are real Möbius transformations with positive determinant, modulo a common nonzero scalar. They have three real degrees of freedom, so three prevertex of a Schwarz-Christoffel map positions may be assigned independently. The remaining positions are constrained by the side-length ratios.

If $X_n=\infty$, omit its finite factor:Since ${\sum }_j{\beta }_j=2$, the derivative behaves like $Z^{{\beta }_n-2}$ at infinity; in the coordinate $U=1/Z$, this is exactly the required corner exponent there. Thus the missing angle is encoded by behavior at infinity.

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.

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: