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If $f$ is holomorphic on a neighborhood of a closed disc and $C$ is its positively oriented boundary circle, then every interior point $z$ satisfiesThus boundary values determine interior values. Differentiation under the integral gives $f^{(n)}(z)=n!{\oint }_{C}f(\zeta )(\zeta -z{)}^{-n-1}d\zeta /(2\pi i)$, showing that one complex derivative forces derivatives of every order. For $f\equiv 1$, parametrizing a small circle about $z$ gives the normalization ${\oint }_{C}d\zeta /(\zeta -z)=2\pi i$; the general formula follows by separating this singular term from the removable remainder.

The Dirichlet problem prescribes the value of an unknown function on the boundary of a domain while imposing a partial differential equation in its interior. Its classical form seeks $u$ with $\Delta u=0$ in $\Omega$ and $u=g$ on $\partial \Omega$, where $\Delta$ is the Laplace operator.

On the unit disc, constant boundary data $g=c$ give $u=c$; the data $g(e^{i\theta })=\cos \theta$ give $u(x,y)=x$. The maximum principle makes these classical solutions unique among functions continuous on the closed disc. Existence is a separate issue and depends on the domain and the regularity of the data. Physically, fixing a conductor's electric potential or the boundary temperature gives a Dirichlet boundary condition.

A planar Euclidean disk consists of the points within a fixed radius of a centre, with either the open or closed boundary convention. The closed disk centered at the origin is $D_R=\{x\in {\mathbb{R}}^2:|x|\le R\}$ and has area $\pi R^2$. It is the two-dimensional closed Euclidean ball; the general closed disc may have any dimension.

Take four congruent octagonal regular polygons with four alternate sides labelled $a,b,c,d$. The intervening sides have Neumann boundary conditions. Pair equal face parameters in the positive reference boundary direction. The complete gluing data areAll unpaired labelled faces not in a Dirichlet column also have Neumann boundary conditions. Encode these rules by signed permutation matrices, using a positive swap for a paired interface and diagonal $-1$ or $+1$ for an unpaired Dirichlet boundary condition or Neumann boundary condition. The four-by-four Hadamard matrix $C$ above satisfies $C{M}_s=N_{s}C$ for every label. Explicitly, if its columns are $h_0,h_1,h_2,h_3$, applying $N_a,N_b,N_c,N_d$ gives respectively $(h_0,h_1,h_3,h_2)$, $(h_0,h_2,h_1,h_3)$, $(h_1,h_0,-h_2,-h_3)$, and $(h_0,-h_1,h_2,-h_3)$, which are exactly the columns of the corresponding $C{M}_s$. On every intervening face the face matrix is the identity.

Since $C^{T}C=4I$, the orthogonal matrix $C/2$ preserves the $L^2$ inner product and the sum of Dirichlet energies. Its intertwining relations map matching Sobolev traces and zero Dirichlet traces bijectively to those of the other assembly. It therefore preserves the quadratic form domains and the forms themselves. The associated self-adjoint operators are unitarily equivalent, proving equality of all eigenvalues with their multiplicities for the specified mixed boundary conditions.

The topology of disk tiles glued along disjoint boundary arcs shows that $D$, whose gluing graph is $0,1,2,3$ in a path, is a closed disc and is simply connected and orientable. The gluing graph of $B$ has edges $13,12,02,01,23$; it is connected with two independent cycles, so $\chi (B)=-1$ and ${\pi }_1(B)\cong F_2$. Its odd cycle $1,2,3,1$ makes it a nonorientable surface. Each assembled surface has nonempty polygonal boundary. This explicitly realizes isospectrality between a simply-connected orientable bordered surface and a nonorientable, non-simply-connected one. The boundary corners and mixed boundary conditions are essential qualifications; it does not contradict the Euler characteristic from the heat trace of a bordered surface for smooth boundaries and pure boundary conditions.

Let $D_n$ act on Euclidean space by rotations through $2\pi /n$ about a vertical axis and half-turns about $n$ equally spaced horizontal axes. A round solid torus about the vertical axis, drilled through radially along the $2n$ places where the horizontal axes cross its tube, has a smooth boundary of genus $2n+1$ after equivariant rounding of its rims. All nonidentity rotation axes lie in the complement of this boundary, so the induced group action is a free action of a group. Each bore replaces two closed discs by an annulus, lowering Euler characteristic by two and increasing genus by one. The boundary orientation is preserved; its induced Riemann surface has an unramified quotient space of genus two by the Riemann-Hurwitz formula.

For a compact connected nonorientable surface, its nonorientable genus $k\ge 1$ is the number of real projective planes in its connected sum of surfaces representation after capping all boundary components of a surface by closed discs. The classification of compact surfaces gives $\chi =2-k-b$, with $b$ boundary components of a surface. The real projective plane has $k=1$ and the Klein bottle has $k=2$.

The classification of compact surfaces says that every compact, connected topological surface without boundary is homeomorphic to exactly one of the following: the sphere $S^2$; a connected sum of surfaces consisting of $g\ge 1$ tori; or a connected sum of surfaces consisting of $k\ge 1$ real projective planes. The first two cases are orientable; the last case is a nonorientable surface. Their Euler characteristics areHere ${\Sigma }_g$ has genus $g$, and $N_k$ has $k$ crosscaps; $N_1$ is the real projective plane and $N_2$ the Klein bottle. Thus orientability together with the Euler characteristic determines the closed surface up to homeomorphism.

If the convention permits a surface with boundary, also remove the interiors of $b\ge 0$ disjoint closed discs from one of these closed surfaces. The Euler characteristic decreases by $b$, giving $2-2g-b$ in the orientable case and $2-k-b$ in the nonorientable surface case. The number of boundary components is then an additional invariant.

A strict Lyapunov function is a continuously differentiable function $V$ with $V(0)=0$, $V(x)>0$ off the origin, and $\nabla V\cdot f<0$ off the origin in the domain. Its compact sublevel sets inside the domain are forward invariant and force convergence to the equilibrium. The domain of stability, or basin of attraction, consists of initial points whose forward solutions exist for all positive time and converge to the fixed point.

For $V=(x^2+y^2)/2$, direct differentiation and factorization giveIf $R^2=x^2+y^2<1/2$, then $x^2+4y^2\le 4R^2<2$, so $\dot{V}<0$ off zero. For any initial point in that disc, choose a slightly larger closed disc still inside it. The vector field points inward on its boundary; the trajectory is bounded, exists globally, and the strict Lyapunov decrease forces its limit to be zero. Thus the origin is asymptotically stable and the whole open disc lies in its basin.

If $R^2>2$, then $x^2+4y^2\ge R^2>2$, so $\dot{V}>0$. A solution starting there cannot cross inward through $R^2=2$ and cannot approach zero. Therefore

Fix a bounded tile in Euclidean space $\Omega$ with labelled flat boundary faces. For each face label $s$, encode an assembly of $n$ copies by an involutive real symmetric signed permutation matrix $M_s$: a paired interface contributes a two-by-two swap, an unglued face with a Neumann boundary condition contributes $+1$ on the diagonal, and an unglued face with a Dirichlet boundary condition contributes $-1$. Use the same face-identification convention on each copy. If a second assembly has matrices $N_s$ and there is an invertible constant matrix $C$ satisfyingthen the map $u=(u_1,\dots ,u_n)\mapsto Cu$ is an isomorphism between corresponding Laplace-Beltrami operator eigenspaces. This is the transplantation theorem.

To prove it, restrict an eigenfunction to the tile copies and identify each copy with $\Omega$. Along a face, let $a$ be its vector of boundary values and $b$ its vector of outward normal derivatives. At a glued interface the values agree and the outward derivatives are opposite. At an unglued face with a Neumann boundary condition the derivative vanishes, and at a face with a Dirichlet boundary condition the value vanishes. All three cases are encoded byThe intertwining identities give $N_s(Ca)=Ca$ and $N_s(Cb)=-Cb$, so the transplanted function satisfies the second assembly's interface and boundary conditions. On each tile its Laplace-Beltrami operator is the same constant linear combination of the original Laplacians, hence it has the same eigenvalue. Across an interface, matching values and opposite normal derivatives make it a weak solution; elliptic regularity, or the allowed reflection principle, makes it a genuine smooth solution there. At corners the operator is understood by its finite-energy, Friedrichs extension; the same matching argument applies to its quadratic form domain. Applying $C^{-1}$ reverses the construction. Thus dimensions of all eigenspaces agree, including multiplicities, and the two specified boundary-value spectra coincide.

The requested pair exists with explicitly specified mixed boundary conditions and polygonal boundary. Take four copies $Q_0,Q_1,Q_2,Q_3$ of an octagonal regular polygon in Euclidean space. Label four alternate sides $a,b,c,d$ in cyclic order; the four intervening sides always have a Neumann boundary condition. The labelled sides are disjoint even at their endpoints. Parameterize every side in the positive boundary direction of the reference octagon. When pairing same-labelled sides, identify equal parameters. This produces a locally flat surface with boundary: a neighbourhood of an interior point of an interface consists of two half-disks glued along their diameters. All tile vertices remain boundary points, with boundary angles $3\pi /4$ or $3\pi /2$; there are no interior cone points.

Here are the complete gluing and boundary conditions. An entry $ij$ pairs the corresponding sides of $Q_i,Q_j$. The indicated unpaired sides have a Dirichlet boundary condition; every other unpaired side has a Neumann boundary condition.Both assemblies are compact and connected. The Laplace-Beltrami operator is the nonnegative operator $P=-\Delta$ associated with the finite-energy quadratic form $\int |\nabla u{|}^2$, with zero Sobolev trace on the specified Dirichlet arcs. Thus the mixed boundary condition is defined without requiring classical differentiability at a corner or at a Dirichlet–Neumann junction.

The four signed permutation matrices for $D$ areFor $B$ they areUse the Hadamard matrixThe intertwining relations can be checked on all columns at once. Writing the columns of $C$ as $h_0,h_1,h_2,h_3$, direct application of the displayed $N_s$ givesFor the four intervening faces both assemblies have the identity face matrix, so these intertwining relations hold as well. Also $C^{T}C=C^2=4I$. This four-tile mixed-boundary transplantation between a disk and a nonorientable surface gives a bijection of every pair of eigenspaces, preserving eigenvalues and multiplicities.

There is also a direct argument on the quadratic form domains that handles all boundary corners. The orthogonal matrix $U=C/2$ acts on tile restrictions, preserves their L2 inner product and the sum of their Dirichlet energies, and commutes with taking a Sobolev trace. The condition $M_{s}u=u$ on each face encodes equality of the two traces at a paired interface and vanishing trace at an unpaired Dirichlet face. Intertwining takes these conditions exactly to $N_s(Uu)=Uu$. Consequently $U$ maps the whole form domain onto the other form domain and preserves its quadratic form. The associated self-adjoint operators are unitarily equivalent, proving isospectrality for the stated mixed boundary conditions without any assumption of smoothness at a Dirichlet–Neumann junction.

We now verify the actual topology using the topology of disk tiles glued along disjoint boundary arcs. Because the gluing arcs in each tile are disjoint, the resulting surface with boundary is a thickening of its gluing graph, with one vertex for each closed disc tile and one edge for each paired arc. One can construct this homotopy equivalence by taking a small central closed disc in each tile and narrow strips to its paired arcs. What remains is a collar neighbourhood of the boundary, which retracts onto this connected system of closed discs and strips. Shrinking each central closed disc and each strip then gives the gluing graph. Equivalently, begin with a spanning tree of the gluing graph: its closed disc tiles joined along boundary arcs form one closed disc, and every remaining pair of boundary arcs adds one band. Thus for $n$ tiles and $e$ paired arcs the fundamental group is a free group of rank $e-n+1$, and the Euler characteristic is $n-e$.

For $D$, the gluing graph is the pathGluing a new closed disc to a closed disc along one boundary arc again gives a closed disc; applying this three times proves that $D$ is homeomorphic to a closed disc. In particularFor $B$, the five paired arcs give the graph edges $13,12,02,01,23$. It is connected, and henceHere $F_2$ is the free group on two generators, so $B$ is not simply connected. To decide orientation of a surface, give every reference tile its counterclockwise orientation. Our same-parameter gluing preserves boundary directions, so a global orientation requires opposite orientation signs on paired tiles: the two induced boundary orientations at an interior interface must be opposite. Such signs exist exactly when the gluing graph is bipartite. In $B$ the cyclehas length three. Following it forces the sign to reverse three times and return to its starting tile with the opposite sign. This proves that $B$ is a nonorientable surface. It has nonempty boundary because every intervening octagon side remains unpaired. Thus $D$ is a simply-connected orientable surface with boundary, whereas its isospectral partner $B$ is neither simply connected nor orientable, exactly as required, for the specified mixed boundary problem.

The boundary qualification is essential. With smooth boundary and the same pure Dirichlet boundary condition or pure Neumann boundary condition on both surfaces, the following existing obstruction applies. Polygonal corners and the mixed boundary conditions in the construction above place it outside that obstruction.

With the interpretation of compact smooth Riemannian surfaces with the same pure Dirichlet boundary condition or pure Neumann boundary condition, the printed simply-connected/nonorientable pair cannot exist. The two-dimensional heat trace has the following expansion (the smooth-boundary coefficients are given in the heat-coefficient formulas, Theorems 5.1–5.2):where the boundary sign is minus for Dirichlet and plus for Neumann. The Gauss-Bonnet theorem identifies the constant term with $\chi (M)/6$. Isospectrality therefore forces equal Euler characteristics. A compact connected simply-connected surface with boundary is a disk and has $\chi =1$. A connected nonorientable surface with $c\ge 1$ crosscaps and $b\ge 1$ boundary components has $\chi =2-c-b\le 0$. Thus no such smooth pure-Dirichlet or pure-Neumann pair exists. This is the Euler characteristic from the heat trace of a bordered surface.

Piecewise-smooth boundaries introduce corner terms, and mixed boundary conditions change the heat coefficients, as the explicit construction above demonstrates. Moreover, in the usual pure-Neumann polygonal reflection-transplantation setup, the pure Neumann reflection transplantation preserves Euler characteristic argument applies even with corners. The numbers of tiles agree. For each face label the number of edge classes is $(n+\mathrm{tr}{M}_s)/2$, so these agree too. For each tile-vertex label, vertex classes are orbits of the subgroup generated by incident-face involutions. Their number is the dimension of invariant vectors in that permutation representation, which is unchanged by the intertwiner. Hence the numbers of vertices agree, proving equal $V-E+F$. Thus this standard tiled construction also cannot supply a disk on one side and a nonorientable surface on the other.

For comparison, the following existing construction gives an orientable and a nonorientable pair with the same spectrum for the pure Neumann boundary condition. Both have nontrivial fundamental groups; this is a separate example, not the construction used to meet the simply-connected requirement above. This eight-tile Neumann transplantation across orientability makes the comparison explicit. The finite seed is the affine modulo-eight example in the original orientability construction; the intertwiner and the topology checks are derived below. Take eight copies of a regular Euclidean hexagon and label three pairwise nonadjacent sides $\sigma ,t,u$. Every remaining side is a side with a Neumann boundary condition. Index the copies by $j\in \mathbb{Z}/8\mathbb{Z}$. For the first assembly pair same-labelled faces according tofor the second, usewith all arithmetic modulo eight. A fixed point means an unglued side with a Neumann boundary condition. Pair faces by the reflection convention, matching the same endpoint parameter in the reference tiles. The distinguished sides are disjoint, so no interior vertex singularities are created; the resulting surfaces have polygonal boundary corners.

Let $R{e}_j=e_{j+1}$ on the eight-dimensional tile-index space and setIt commutes with the permutation matrices for $t$ and $\sigma$, because its coefficients are invariant under reversing cyclic order. For $u_1,u_2$, write $C_{ij}=c_{i-j}$, where $c_0=c_4=1$, $c_1=c_7=2$, and all other $c_r$ vanish. The required relation $C{M}_{u_1}=M_{u_2}C$ is exactly $c_r=c_{3r+4}$, verified from these four nonzero entries. Thus all three intertwining relations hold.

The cyclic Fourier vectors diagonalize $C$. Its eigenvalues arenamely $6,2\sqrt{2},2,-2\sqrt{2},-2,-2\sqrt{2},2,2\sqrt{2}$. None is zero, so $C$ is invertible and the two assemblies are isospectral manifolds for the Neumann boundary condition.

We prove their topology as well. Their gluing graphs are connected: the first contains the spanning path $0,1,3,6,2,7,5,4$, and the second contains the spanning cycle $0,1,7,2,6,3,5,4,0$. Each has eight tile vertices and ten paired-face edges. Since tiles are disks and gluing arcs are disjoint, each surface retracts up to homotopy to this multigraph; thus its Euler characteristic is $8-10=-2$ and its fundamental group is a free group of rank three.

An orientation assigns a sign to each reference tile, and a reflection gluing requires opposite signs at the ends of every graph edge. The second graph admits this assignment: take sign $+$ on $\{0,5,6,7\}$ and sign $-$ on $\{1,2,3,4\}$. Every paired face joins opposite signs, proving orientability. In the first graph the cyclehas odd length. Transporting orientation around it reverses the initial sign, proving nonorientability. Hence the first surface is also not simply connected. The second surface is orientable but likewise not simply connected; the disk required in the question is supplied instead by the four-tile mixed-boundary construction above.

The four-tile construction resolves the printed simply-connected requirement under explicit mixed boundary conditions. The eight-tile comparison only concerns orientation of a surface under a pure Neumann boundary condition.

Milne-Thomson's circle theorem inserts an impermeable circular boundary into a two-dimensional irrotational incompressible flow. If $f(z)$ is analytic in a neighborhood of the closed disc and its flow singularities lie outside, the new complex potential outside $|z|=a$ isReflection and coefficient conjugation make the second term analytic outside the circle, with its image singularities inside. On $|z|=a$, $a^2/\bar{z}=z$, so the two terms sum to $2\mathrm{Re}f(z)$: the streamfunction is constant there. Thus no velocity crosses the circle. At infinity the image term tends to the constant $\overline{f(0)}$, which has no effect on velocity, and the far-field flow is retained. This proves the construction, with an additive potential constant immaterial.

Put $A=\mu e^{i\alpha }$, $B=\mu e^{-i\alpha }$. For the dipole the reflected potential is $B/(a^2/z-b)=Bz/(a^2-bz)$, givingIts image pole is $c=a^2/b$, inside the cylinder. The complex velocity isNear $c$, the second term is $C(z-c{)}^{-2}$ with $C=B{a}^2/b^2$, while $v_0(z)=-A/(z-b{)}^2$ is analytic there. The square of the pole term has no residue, nor does $v_0^2$. The cross term has residue $2C{v}_0^{\prime }(c)$, soTake the contour counterclockwise around the cylinder, as in the supplied Blasius force theorem. The residue theorem gives $\frac{1}{2}i\rho \oint (w^{\prime }{)}^{2}dz=-\pi \rho \mathrm{Res}(w^{\prime }{)}^2$. ThereforeThe force points toward the dipole. Its independence of dipole orientation follows from the product $AB={\mu }^2$ in the sole residue contribution.

For the sphere model of the real projective plane, letThen ${\mathbb{RP}}^2=S^2/{\sim }_S$ with the quotient topology: each equivalence class is an antipodal pair.

For the closed disc model, letGive $D^2/{\sim }_D$ the quotient topology. Only antipodal pairs on the boundary circle are identified in the closed disc model; interior points remain distinct.

Here $D^2$ must denote the closed disc $\{(x,y):x^2+y^2\le 1\}$, as its boundary points participate in the identification. For $r=\sqrt{x^2+y^2}$, defineThe Pythagorean trigonometric identity gives $\Vert F(x,y){\Vert }^2=1$, so the image lies in the sphere $S^2$. Away from the origin the formula is continuous. At the origin, the first two coordinates have absolute value at most $|\sin (\pi r)|\to 0$ and the third tends to $1$, proving continuity there as well.

Every boundary point maps to $(0,0,-1)$, while the origin maps to $(0,0,1)$. For a point $(u,v,w)\in S^2$ with $-1<w<1$, there is a unique $r\in (0,1)$ with $\cos (\pi r)=w$, and its unique preimage isThis point has radius $r$, because $u^2+v^2=1-w^2={\sin }^2(\pi r)$. The two poles have the preimages already described. Thus $F$ is surjective and its fibers are exactly the prescribed equivalence classes.

The closed disc is compact by the Heine-Borel theorem, and $S^2$ is Hausdorff with the subspace topology from Euclidean space. The preceding result gives the homeomorphism

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

Start with a polygonal schema whose outer boundary has $m$ sides, and remove a triangular disc strictly inside its polygon. Cut the resulting annulus along an arc from an outer vertex to an inner vertex. The cut space is a closed disc with $m+5$ marked sides: the original $m$ outer sides, two banks of the cut, and three inner sides. Parametrize it by a convex polygon with that many sides. A boundary homeomorphism carrying the marked vertices in cyclic order extends over the disc by radial extension after identifying both discs with a standard closed disc.

Transport the original side identifications to this polygon and identify the two cut banks, leaving the three inner sides unpaired. The quotient space is the original surface with an open triangular disc removed. In boundary-word notation, if $w$ is the original word, the new word is $wcdef{c}^{-1}$, with $d,e,f$ unpaired. The torus has $w=ab{a}^{-1}{b}^{-1}$; a square model of the real projective plane has $w=abab$, pairing antipodal points of the square's boundary.

The stability function $R$ of a one-step numerical method is its amplification factor on the Dahlquist test equation $y^{\prime }=\lambda y$: one step gives $y_{n+1}=R(h\lambda )y_n$. Its absolute stability region is $\{z:|R(z)|\le 1\}$, using boundedness of repeated steps as the convention.

For explicit Euler method, $R(z)=1+z$, so the region is the closed disc centered at $-1$ with radius one. For Backward Euler method, $R(z)=1/(1-z)$, and the entire left half-plane is stable. These calculations explain why a step can be accurate locally yet generate a growing long-time solution. Multistep methods instead have several amplification roots and require a root condition.^[ref]

An $n$-dimensional topological disc is a space homeomorphic to a Euclidean ball in ${\mathbb{R}}^n$, with its open or closed convention specified. A closed disc is homeomorphic to $D^n=\{x\in {\mathbb{R}}^n:|x|\le 1\}$. The planar disk is the Euclidean two-dimensional instance, rather than the whole abstract topological class.

Consider a connected assembly of $n$ closed discs formed by identifying $e$ pairs of boundary arcs, each arc used once and all arcs in a given tile having disjoint endpoints. Its gluing graph has one vertex per tile and one edge per paired arc. A spanning tree joins the initial discs into a single closed disc; each of the remaining $e-n+1$ pairings attaches a band. This is a thickening of the gluing graph, so it is homotopy equivalent to that graph. Thus its Euler characteristic is $n-e$ and its fundamental group is the free group $F_{e-n+1}$. If the gluing graph is a tree, the assembled surface is itself a closed disc.

Suppose each pairing preserves the positive boundary parameter of the reference discs. A global orientation must then assign opposite orientation signs to adjacent tiles, because their induced boundary orientations on an interior seam must oppose one another. Such signs exist exactly when the gluing graph is bipartite. An odd gluing cycle therefore proves that the assembly is a nonorientable surface. A Dirichlet boundary condition or a Neumann boundary condition on an unpaired arc has no effect on this topological calculation.