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For the exterior of the unit disc in two dimensions, with the sign convention ${\Delta }_{x}G=+\delta (x-y)$, this is the Dirichlet Green function that vanishes on the circle and is bounded at infinity. The method of images places an opposite logarithmic source at the inverse point $y/|y{|}^2$. The constant $-(2\pi {)}^{-1}\log |y|$ enforces the zero boundary value. The inversion in a circle identity $|x-y|=|y||x-y/|y{|}^2|$ for $|x|=1$ verifies that condition directly.

The outward normal of the exterior domain is $-x$ on the unit circle. Its normal derivative gives the positive exterior Poisson kernelFor bounded harmonic functions this kernel integrates boundary data to the interior value. Boundedness at infinity is essential to uniqueness: without it, a multiple of $\log |y|$ can be added without changing the circle's Dirichlet boundary condition.

Normalize three distinct points on a generalized circle by a Möbius transformation $f$ taking them to $1,0,\infty$. The inverse image of the extended real line is their generalized circle, and $J=f^{-1}\circ c\circ f$, with $c(z)=\overline{z}$, fixes that circle pointwise and squares to the identity. Changing the three normalizing points replaces $f$ by $M\circ f$ with a real-coefficient Möbius map $M$. Since $M$ commutes with conjugation, $J$ is unchanged. Thus this construction depends only on the circle. It is inversion in a circle for an ordinary circle and Euclidean reflection for a straight line, each extended to the Riemann sphere.

For bounded piecewise continuous Dirichlet boundary data $f$ on a circle of radius $R$, write its Fourier series as $a_0+{\sum }_{n\ge 1}[a_n\cos (n\theta )+b_n\sin (n\theta )]$. The displayed series gives the bounded exterior harmonic extension for $r>R$. The radial inversion in a circle $r\mapsto R^2/r$ converts it into the ordinary interior Poisson kernel solution. Values at discontinuities are understood through the usual boundary trace, with midpoint radial limits at jumps; changing the assigned values at finitely many boundary points does not change the extension.

The limit at infinity is forced to equal the angular mean $a_0=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(\theta )d\theta$. Indeed the angular mean of any exterior harmonic function solves $(r{\overline{u}}^{\prime }{)}^{\prime }=0$, and boundedness excludes its logarithmic term. Equivalently, inversion makes infinity an isolated removable singularity of a bounded planar harmonic function. Uniqueness follows from interior uniqueness after inversion, including the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions when needed.

Inversion in a circle of centre $a$ and radius $R$ is an orientation-reversing involution of the Riemann sphere fixing that circle pointwise and exchanging its interior and exterior. It is an anti-Möbius transformation. Reflection in a line is the corresponding inversion for a generalized circle through infinity. A composition of two such inversions is a Möbius transformation.

For a Euclidean circle with centre $c$ and radius $r$, two distinct finite points are inverse when they lie on the same ray from $c$ and their distances from $c$ have product $r^2$. Points on the circle are fixed, and $c$ and infinity are inverse to each other. In complex coordinates this uniquely defines inversion in a circle byUniqueness follows from the ray and product conditions, including the specified centre–infinity pair; the formula also gives $I_{\Gamma }^2=\mathrm{id}$. A circle through infinity is a straight line in the complex plane; its inversion is reflection in that line, fixing infinity. For a line through $c$ with direction angle $\theta$, its formula is $c+e^{2i\theta }\overline{z-c}$.

Thus every inversion is an anti-Möbius transformation, of the form $(a\overline{z}+b)/(c\overline{z}+d)$ with nonzero determinant. If $J(z)=\overline{z}$, write such an inversion as $M\circ J$ with $M$ a Möbius transformation. Since $J\circ M\circ J$ is the Möbius transformation with conjugated coefficients, the product of two inversions is $M_1\circ (J\circ M_2\circ J)$, hence Möbius. Pairing consecutive inversions proves every composition of an even number of inversions is a Möbius transformation.

inversion in a circle with center $c$ and radius $r>0$ is $I(z)=c+r^2/(\overline{z}-\overline{c})$, with $c$ and infinity interchanged. It fixes the circle pointwise and reverses complex orientation. Each inversion is fractional-linear in $\overline{z}$, so composing two eliminates the conjugation and gives $(az+b)/(cz+d)$ with nonzero determinant: a Möbius transformation.

Translate and rotate two distinct centers to $0,d$, where $d>0$, and let the radii be $r,s$. The composition of inversion in the first circle followed by the second isIts fixed points solve $d{z}^2-(d^2+r^2-s^2)z+d{r}^2=0$. The discriminant isDisjoint circle boundaries require either $d>r+s$ or $d<|r-s|$, so both factors have the same nonzero sign and $\Delta >0$. There are two distinct fixed points on the line of centers. If the circles are concentric, their radii differ and the composition is $z\mapsto (s^2/r^2)z$, with exactly the two fixed points $0,\infty$. Thus the composition has two distinct fixed points, also for nested disjoint circles. Reversing the order inverts the Möbius transformation and leaves its fixed points unchanged.

Use the existing cross-ratio conventionValues involving infinity are defined by limits. More uniformly, represent each point by a nonzero homogeneous vector $u=(u_0,u_1)$ and put $D(u,v)=u_0{v}_1-u_1{v}_0$. Then the same value is $D(z,a_1)D(a_0,a_2)/[D(z,a_2)D(a_0,a_1)]$, interpreted in the Riemann sphere. Rescaling any representative cancels out. The first three points are distinct, so numerator and denominator cannot simultaneously vanish. The fourth point may coincide with one of them, giving the values $1,0,\infty$ at $a_0,a_1,a_2$, respectively.

With $a_0,a_1,a_2$ fixed, this is a nonconstant Möbius transformation $f$. For finite first three points, write $C=(a_0-a_2)/(a_0-a_1)\ne 0$; thenThe determinant of this fractional-linear map is nonzero because $a_1\ne a_2$. Its inverse is defined at every point of the sphere, including poles and infinity, so each $w$ has exactly one preimage. The same conclusion follows directly from the homogeneous formula when an $a_i$ is infinity.

Möbius transformations preserve generalized circles. To verify the fact used here, affine complex changes preserve ordinary circles and lines, while a generalized circle has an equation $A|z{|}^2+Bz+\overline{B}\overline{z}+D=0$ with $A,D$ real. Substitution $z=1/w$ and multiplication by $|w{|}^2$ gives another equation of the same form. Every non-affine Möbius map is a composition of affine maps and $z\mapsto 1/z$, so the claim follows. Consequentlyis a generalized circle. It contains the three distinct $a_i$, so it is their unique ordinary circle, or their line together with infinity when they are collinear. This is the real cross-ratio criterion for a generalized circle.

Let $c(w)=\overline{w}$, with $c(\infty )=\infty$. The defining equation for $J$ now gives the cross-ratio construction of generalized-circle reflection:This is well-defined by bijectivity of $f$, and $J^2=f^{-1}{c}^{2}f$ is the identity. Moreover $J(z)=z$ precisely for $z\in S$.

If the first three points lie on the extended real line, the normalization map has real coefficients, so $f(\overline{z})=\overline{f(z)}$. Hence $J(z)=\overline{z}$. For a second triple $b_0,b_1,b_2$ on the same generalized circle, write its normalization as $g=Mf$. Each $f(b_i)$ is real or infinity, and the unique Möbius map taking this real triple to $1,0,\infty$ has real coefficients. Thus $M$ commutes with $c$, givingTherefore $J$ depends only on the generalized circle, not on the three selected points. For an ordinary circle with centre $q$ and radius $R$, it is inversion in a circle, $J(z)=q+R^2/(\overline{z}-\overline{q})$; for a straight line it is the corresponding Euclidean reflection, extended to infinity.

An inversion in a circle on the Riemann sphere is the anti-Möbius involution fixing that generalized circle pointwise. For a Euclidean circle with centre $a$ and radius $R>0$, it isIt reverses the argument relative to the centre and sends radius $r$ to $R^2/r$, so exactly the points of the circle are fixed. For a straight line, regarded as a circle through infinity, inversion is the ordinary reflection in that line. These formulas prove existence. For uniqueness, the composition of two anti-Möbius transformations fixing the same circle pointwise is a Möbius transformation fixing at least three distinct points, hence the identity. Therefore the two inversions agree.

Every inversion has the form $(a\overline{z}+b)/(c\overline{z}+d)$ with $ad-bc\ne 0$. Composing two conjugate-linear fractional expressions cancels the complex conjugations and gives $(Az+B)/(Cz+D)$ with nonzero determinant. Induction proves that an even number of inversions gives a Möbius transformation.

Conversely translations are compositions of reflections in parallel lines, rotations are compositions of reflections in two lines through their centre, and positive dilations are compositions of two concentric circle inversions: $J_R{J}_1(z)=R^{2}z$. Thus every nonconstant affine map $z\mapsto az+b$ is an even composition. The map $z\mapsto 1/z$ is $J_{|z|=1}$ composed with reflection in the real axis, also an even composition. If $c\ne 0$, factor a general Möbius transformation asThis is a composition of affine maps and $1/z$. If $c=0$, it is already affine. Every Möbius transformation is therefore a composition of an even number of inversions, with the identity represented by zero inversions or by any inversion repeated twice.

Use separation of variables, $\Phi =F(r)H(\theta )$, in the Laplace equation in polar coordinates. Dividing by $FH/r^2$ givesSingle-valued $2\pi$-periodic angular factors have integer $n\ge 0$. For $n\ge 1$, they are $\cos n\theta ,\sin n\theta$, and the radial equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ has solutions $r^n,r^{-n}$. The zero angular mode gives $F=a_0+b_0\log r$. These separated products give the requested solutions; arbitrary sums may use independent coefficients for each radial-angular combination.

Boundedness at the origin removes the logarithm and all negative radial powers. The boundary function is even, so its sine Fourier coefficients vanish. With the convention $f(\theta )=a_0+{\sum }_{n\ge 1}{a}_n\cos n\theta$, its Fourier coefficients areThus the bounded interior harmonic extension isFor $r<R$ the series and its derivatives converge locally uniformly, so it is harmonic. It is a rotated and scaled instance of the harmonic measure of a semicircle in a disk. For example, summing the power series for the complex arctangent, or differentiating the following expression, gives the equivalent real formThis also proves boundedness directly.

At the two jumps, the boundary interpretation matters. At every continuity point the radial limit equals the prescribed value. At $\theta =\pm \pi /2$, the radial limit is $C/2$, the midpoint of the two one-sided values. Consequently for $C\ne 0$ there is no harmonic solution continuous on the closed disk with exactly the printed boundary values. The standard bounded Dirichlet problem imposes the trace away from the jumps, or almost everywhere; assigning zero separately at those two boundary points does not alter the interior solution.

For uniqueness let $w=\tilde{\Phi }-\Phi$. Formally the divergence theorem applied to $w\nabla w$ givesFor continuous zero boundary data, this immediately forces $w=0$. With the actual discontinuous data, one should justify the two exceptional points rather than assume finite Dirichlet energy there. Let $|w|\le M$ and choose a cutoff ${\eta }_{ϵ}$ which is zero within ${ϵ}^2$ of those points, one beyond $ϵ$, and linear in logarithmic distance between. Its gradient satisfies $\int |\nabla {\eta }_{ϵ}{|}^2=O(1/|\log ϵ|)$. Applying the divergence theorem to ${\eta }_{ϵ}^{2}w\nabla w$ givesThe resulting energy tends to zero. Hence $\nabla w=0$ on every compact subset and $w$ is constant; its zero boundary trace on the open arcs makes it zero. This is the logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions.

For the exterior harmonic extension of disk boundary data, boundedness instead excludes $r^n$ and the logarithm. The same boundary coefficients giveThis is ${\Phi }_{\mathrm{in}}(R^2/r,\theta )$, using inversion in a circle. Its limit at infinity is $C/2$. To see why that value is forced independently of this construction, average any bounded exterior harmonic function over a circle. The mean satisfies $(r{\overline{\Phi }}^{\prime }{)}^{\prime }=0$, hence is $A+B\log r$. Boundedness forces $B=0$, and the boundary mean fixes $A=C/2$. A constant limit at infinity must therefore be $C/2$.

Finally suppose two bounded exterior solutions have this limit. Their difference $w$ tends to zero. Under planar inversion in a circle, it is bounded and harmonic near the origin, so the isolated point there is removable. Its local expansion gives $w=O(r^{-1})$, ${\partial }_{r}w=O(r^{-2})$ in the original exterior. Apply the preceding divergence theorem energy argument on $R<r<L$, with logarithmic cutoffs at the two boundary jumps. After removing those cutoffs, the inner boundary term vanishes; the outer one is ${\int }_{r=L}w{\partial }_{r}wds=O(L^{-2})$. Letting $L\to \infty$ forces $\nabla w=0$, and the zero limit gives $w=0$. Thus the bounded exterior solution is unique as well.

Let $s=|{\mathbf{r}}^{\prime }|>0$ and $\mathbf{p}={\mathbf{r}}^{\prime }/s^2$. For $|\mathbf{r}|=1$, expanding both squared Euclidean distances givesTaking nonnegative square roots provesThis is the distance identity behind inversion in a circle and the method of images for a circular boundary.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.

Put $w=1/z$, so $z=1/w$ and $w\ne 0$. Substitute into the equation in part (a) and multiply by $|w{|}^2$:This is the reciprocal transformation of a generalized circle. In the original coefficient convention the new triple is $(C,\overline{B},A)$: the linear coefficient is conjugated, which reflects the distinction between $1/z$ and inversion in a circle $1/\overline{z}$.

A circle through zero. Here $A\ne 0$, $C=0$, and nondegeneracy forces $|B{|}^2>0$. Its image satisfiesThis is a straight line, and $w=0$ does not satisfy it because $A\ne 0$. Conversely every point of this straight line has a nonzero reciprocal on the original circle. Thus the original circle with zero omitted maps onto the full finite straight line.

A circle missing zero. Now $A\ne 0$, $C\ne 0$ and $\Delta =|B{|}^2-AC>0$. Applying part (a) to the transformed coefficients gives centre $-\overline{B}/C$ and radius $\sqrt{\Delta }/|C|$, so the image is again a genuine circle. It misses zero, because the transformed equation has nonzero constant $A$. The inverse $z=1/w$ proves that the whole of this circle is attained.

A straight line through zero. Here $A=C=0$ and $B\ne 0$. The transformed equation is $2\mathrm{Re}(Bw)=0$, another straight line through zero. Precisely, the finite map takes the punctured original straight line onto the punctured image straight line, since neither the domain nor the image of $1/z$ contains zero.

In the Riemann sphere convention $0\leftrightarrow \infty$, all three become exact statements about whole generalized circles:Finite straight lines acquire their point at infinity in this convention. Stating it avoids accidentally adding $w=0$ merely because it satisfies a multiplied equation.

Write a generalized circle equation as $a|z{|}^2+b\bar{z}+\bar{b}z+c=0$, where $a,c$ are real. Under the reciprocal Möbius transformation $z=1/w$, with $w\ne 0$, multiply by $|w{|}^2$ to obtainThus the coefficient triple transforms as $(a,b,c)\mapsto (c,\bar{b},a)$. A nondegenerate circle through zero has $c=0$ and becomes a straight line; a straight line missing zero becomes a circle through zero on the Riemann sphere. When working only with finite nonzero coordinates, retain the excluded point $w=0$ even if it satisfies the multiplied equation.

This reciprocal transformation is different from inversion in a circle, whose unit-circle formula is $z\mapsto 1/\bar{z}$. The latter reverses orientation; the reciprocal transformation is holomorphic away from its pole.

For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.