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For a function $u$ on a compact Riemannian manifold, its Dirichlet energy is $\frac{1}{2}{\int }_M|\nabla u{|}_g^{2}d{\mathrm{vol}}_g$. Adding a linear source term $-{\int }_{M}fud{\mathrm{vol}}_g$ gives Euler-Lagrange equation $-{\Delta }_{g}u=f$.

For a scalar function $u$ on a Riemannian manifold, the Dirichlet energy is $E(u)=\frac{1}{2}{\int }_M|\nabla u{|}_g^{2}d{\mathrm{vol}}_g$.

The nonnegative Dirichlet Laplacian is the self-adjoint realization of $-\mathrm{div}\mathrm{grad}$ with zero boundary values. On a bounded domain, its weak form is the Dirichlet energy on $H_0^1$. For polygonal domains the weak operator domain is preferable to a requirement of smoothness at corners.

The nonnegative Neumann Laplacian is the self-adjoint realization of $-\mathrm{div}\mathrm{grad}$ with vanishing outward normal derivative. Its weak quadratic form is the Dirichlet energy on $H^1$. On a connected compact domain the constants give a one-dimensional zero eigenvalue space.

Unbounded-domain Dirichlet problems can need a condition at infinity. On all of ${\mathbb{R}}^n$, constant functions have identical empty boundary data and are distinct finite-energy stationary points of $\frac{1}{2}\int |Du{|}^2$. With nonempty boundary, on the exterior of the unit ball in dimension at least three, $1-|x{|}^{2-n}$ is a nonzero harmonic function vanishing on the sphere, with finite Dirichlet energy and a nonzero limit at infinity. It is stationary for compactly supported variations. The permitted class of variations and the behaviour at infinity must be specified for a global variational uniqueness claim.

Suppose an invertible real matrix $C$ intertwines symmetric orthogonal tile-reflection matrices: $C{M}_s=N_{s}C$. Then $C^{T}C$ commutes with every $M_s$, so its positive inverse square root does too. The polar decomposition of an invertible real matrix therefore gives an orthogonal matrix $O=C(C^{T}C{)}^{-1/2}$ satisfying $O{M}_s=N_{s}O$. On a disjoint union of congruent tiles, $O$ preserves both the $L^2$ inner product and Dirichlet energy. The intertwining identities preserve the face trace conditions defining the assembled Sobolev space, so $O$ gives a unitary equivalence of the assembled Laplacians.

Assume sufficient smoothness and a regular boundary for the integrations below, with $n$ outward. Put $\delta =w-\varphi$, so $\delta =0$ on $S$. Expanding the Dirichlet energy givesBy Green's first identity, the middle integral isThus the Dirichlet principle follows with an explicit remainder:Equality means $w-\varphi$ is constant on each connected component; its zero boundary value forces that constant to be zero.

Now let $w$ solve the heat equation and set $E(t)={\int }_V|\nabla w{|}^{2}dV$. Differentiation followed by Green's first identity givesUnder time-independent Dirichlet boundary conditions, $w_t=0$ on $S$, while $w_t=\Delta w$ in $V$. ThereforeIf equality holds at a particular time, continuity makes $\Delta w=0$ throughout $V$ at that time. Its boundary values are still $f$, so uniqueness of the Dirichlet problem implies $w=\varphi$. Conversely $w=\varphi$ has $\Delta w=0$ and gives equality. This proves the full equality criterion for Dirichlet energy dissipation for the heat equation.

For the dynamic boundary condition for the heat equation, substitute $w_t=-\alpha {\partial }_{n}w$ directly into the boundary term. The interior equation is unchanged, soThis is Dirichlet energy dissipation with a dynamic heat boundary condition. It remains valid when $\alpha$ vanishes, because no division by $\alpha$ is used. The last part preserves energy monotonicity; its stationary states need not have the earlier prescribed Dirichlet data. For example, with strictly positive $\alpha$ on a connected boundary, a stationary harmonic state has zero outward normal derivative and is constant, as follows by applying Green's first identity to that state itself.

The Fourier coefficients satisfy $a_0=a_n=0$ andThus the square wave has Fourier seriesIt equals $f$ away from its jump discontinuities, and converges to $0$ at the jumps. The assigned endpoint values do not affect its Fourier coefficients.

For the Poisson equation, use separation of variables $\varphi ={\sum }_{n\text{ odd}}{b}_n{R}_n(r)\sin (n\theta )$. The radial equation is $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. Its bounded solution with zero Dirichlet boundary condition at $r=1$ isIndeed the negative power is excluded at the centre, and the boundary fixes the coefficient of $r^n$. No resonance at $n=2$ occurs because only odd modes are forced. ThereforeThis is a Poisson equation on a disk with angular forcing. The series converges uniformly on the closed disk: for $n\ge 3$ each term is bounded by a constant times $n^{-3}$. Its first spatial derivatives also converge uniformly. For instance the radial derivative and $r^{-1}$ times the angular derivative of each term are bounded by constants times $n^{-2}$; the $n=1$ harmonic term is smooth in Cartesian coordinates. It defines a function with zero boundary data and the required weak formulation. Applying the Laplacian to finite partial sums gives the Fourier partial sums of $f$, which converge in $L^2$; passing to the limit against smooth compactly supported test functions verifies the Poisson equation. It holds classically off the jump rays and the centre, where the forcing is locally constant. It cannot be interpreted as a global classical equation with continuous second derivatives for discontinuous forcing. Uniqueness follows from the zero-boundary Dirichlet energy identity for the difference of two solutions.

Finally orthogonality gives ${\int }_0^{2\pi }f(\theta )\sin (n\theta )d\theta =\pi b_n$. The radial integral isUniform convergence of $\varphi$ and boundedness of $f$ justify integration term by term, soThe negative sign also agrees with $\int \varphi \Delta \varphi =-\int |\nabla \varphi {|}^2$. The original PDF's radial denominator is $n^2-4$; the converted TeX's $n^2-1$ is a transcription error.

Let $p:N\to M=U\backslash N$ be a finite normal Riemannian covering, so $U$ acts freely by deck isometries. With $x,y$ any lifts of $\bar{x},\bar{y}$, the heat kernel on a finite isometric quotient satisfiesThere is no factor $1/|U|$ in this kernel formula. Simultaneous invariance of $H_N$ under isometries proves independence of the lifts, by reindexing $U$. Each term solves the lifted heat equation. If $F_U$ is a fundamental domain and $f$ a function on $M$, the integral of the right side against $f(\bar{y})$ over $F_U$ equalsbecause the translates $u{F}_U$ tile $N$ up to sets of measure zero. This converges to $f(p(x))$ as $t\downarrow 0$, proving the initial condition. Uniqueness of the heat equation solution on the closed manifold gives the formula. For a noncompact finite cover, use the minimal heat kernels, or compatible Friedrichs heat realizations. The normalized pullback $f\mapsto |U{|}^{-1/2}f\circ p$ is unitary onto the deck-invariant $L^2$ subspace and preserves the Dirichlet energy, so it intertwines the heat semigroups and yields the same kernel formula. No uniqueness of unrestricted noncompact heat-equation solutions is needed.

Set $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$ for an isometry $g$. Diagonal integration and the covering degree giveUnlike the kernel formula, the trace formula has an averaging factor: integration of a quotient function over $N$ is $|U|$ times integration over $U\backslash N$.

Suppose now $U\le T$, with $T$ any larger group of isometries of the closed manifold $N$. Change variables $x=hy$ and use simultaneous heat-kernel invariance to obtainThus $I_t$ is a class function, and the Sunada orbital heat-trace formula can be grouped by conjugacy classes $C$ of $T$:Only classes meeting the finite subgroup $U$ contribute, so this class sum is finite even if the larger isometry group $T$ is infinite.

For finite $T$, Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every conjugacy class. Summing these counts gives $|U_1|=|U_2|$, so the displayed formula gives identical heat traces. Their spectra, with multiplicities, agree: as $t\to \infty$, the constant limit identifies the multiplicity of zero; after subtracting it, the slowest-decaying exponential identifies the least positive eigenvalue and its multiplicity. Repeating determines every subsequent eigenvalue. This proves the Sunada theorem in the stated covering situation.

There is an important convention when the universal cover $N$ has an infinite deck group $T$. For compact quotient manifolds, the subgroups must have finite index; the appropriate generalized Gassmann condition is equality of the characters of the finite coset permutation representations, not equality of possibly infinite cardinalities of conjugacy-class intersections. Let $K$ be the intersection of the two subgroup cores. Each core is the kernel of the corresponding finite coset action, so $K$ is normal and of finite index in $T$, and $K\subseteq U_1\cap U_2$. The manifold $X=K\backslash N$ is then a compact finite normal cover of $M_0$. The finite group $G=T/K$ acts on $X$, and $H_i=U_i/K$ have equal coset permutation characters in $G$. For a finite group, the fixed-coset count isEquality at the identity gives equal subgroup orders, and then equality for all $g$ gives almost-conjugacy. Applying the proved finite theorem to $H_i\backslash X=U_i\backslash N$ establishes isospectrality also in this finite-index interpretation. It avoids taking the divergent heat trace of a noncompact universal cover.

To realize a finitely presented group $T=⟨a_1,\dots ,a_r\mid R_1,\dots ,R_s⟩$, begin with the closed smooth four-manifold $B={\#}_{j=1}^r(S^1\times S^3)$, whose fundamental group is free on the $a_j$. Represent the relator words by disjoint smoothly embedded circles; general position permits this in dimension four, including separating self-intersections of representatives. Their oriented normal rank-three bundles are trivial, so perform framed surgeries replacing $S^1\times D^3$ by $D^2\times S^2$. Removing the circles' tubular neighborhoods leaves the fundamental group unchanged, by codimension-three general position. The Seifert-van Kampen theorem shows that each replacement kills exactly the normal closure of its relator. The resulting closed smooth $M_0$ has ${\pi }_1(M_0)\cong T$. This gives a concrete closed-manifold realization of a finitely presented fundamental group. Equip it with any smooth Riemannian metric and lift that metric to the universal and intermediate covers; all covering maps become local isometries.

To ensure nonisometry, one must choose nonconjugate subgroups. If $U_1=U_2$, or they are conjugate, the quotients are isometric for every lifted metric, so Gassmann equivalence alone cannot imply nonisometry. For nonconjugate subgroups, choose the base metric to have no nonidentity local isometry between open sets, using the Sunada local isometry lemma in dimension at least two. If $F:M_1\to M_2$ were an isometry, a local inverse of the first covering followed by $F$ and the second covering would be a local isometry of $M_0$, hence the identity. Thus the covering maps satisfy $p_{2}F=p_1$ everywhere. Classification of connected covers then forces $U_1,U_2$ to be conjugate in ${\pi }_1(M_0)=T$, a contradiction. Therefore nonconjugate almost-conjugate subgroups and a locally rigid base metric give isospectral nonisometric quotients.

We construct the two domains explicitly from seven congruent triangles. Let a scalene reference triangle have sides $a,b,c$ opposite its three labelled vertices. Choose its three side lengths in a small open neighbourhood of $(1,1,1)$, with a fixed strict ordering. Thus the three angles are distinct and lie between $\pi /4$ and $\pi /2$. Each copy keeps these side labels.

For the first assembly glue the following tile pairs, reflecting across the indicated side:Each coloured adjacency graph is a tree with central tile $7$ and three two-tile arms. Place tile $7$ in the plane and obtain every other tile by the prescribed reflections. At the equilateral reference shape the copies have disjoint interiors and form a polygonal disk; the same holds throughout a sufficiently small neighbourhood, since the only contacts between tiles are the prescribed edges and their incident vertices. In particular each assembly is a bounded connected planar domain.

To prove equality of spectra, index tiles by the seven nonzero vectors $x\in {\mathbb{F}}_2^3$, using $x=(x_1,x_2,x_3)$ with label $x_1+2x_2+4x_3$. LetThese involutions act on points by $x\mapsto s_{s}x$ and on nonzero covectors by $y\mapsto s_s^{-T}y=s_s^{T}y$. Their swaps are exactly the two columns of the table. Write $P_s,Q_s$ for the corresponding permutation matrices. The Fano plane incidence matrixsatisfies $B{P}_s=Q_{s}B$, since incidence is preserved by a linear transformation and its inverse transpose. Each point lies on three lines, while two different points have one common line. Thereforewhere $J$ is the all-ones matrix, proving that $B$ is invertible.

For the Dirichlet boundary condition, a glued side has a positive swap and an exposed side has diagonal entry $-1$. Both trees have the same bipartition, with sign matrixTheir signed reflection matrices are $M_s=-D{P}_{s}D$ and $N_s=-D{Q}_{s}D$. On a glued edge its endpoint signs are opposite, so the off-diagonal entries are $+1$; on an exposed edge the entry is $-1$. Consequently the invertible matrix $C=DBD$ satisfies

We spell out why this gives isospectrality. Pull each function back from the seven tiles to the reference triangle. Its seven boundary values along side $s$ form a vector $u_s$ satisfying $M_s{u}_s=u_s$: this imposes equal values at glued edges and zero values at exposed edges. Outward normal derivatives of a smooth eigenfunction satisfy $M_s{v}_s=-v_s$, giving opposite normal derivatives on glued edges. The identities above preserve both conditions under $u\mapsto Cu$. Since $C$ is constant, it also commutes with the scalar Laplacian on each triangle.

One can include all weak eigenfunctions without any corner regularity issue by orthogonalization of a transplantation matrix. The matrix $C^{T}C$ commutes with the symmetric $M_s$, so $O=C(C^{T}C{)}^{-1/2}$ is orthogonal and still intertwines $M_s,N_s$. On the disjoint tiles, $O$ preserves the $L^2$ norm and the sum of the Dirichlet energies; it maps the trace-matching form domain $H_0^1({\Omega }_P)$ bijectively to $H_0^1({\Omega }_L)$. Thus it is a unitary equivalence of the Dirichlet Laplacians, proving equality of every eigenvalue with its multiplicity.

Finally, the only reentrant vertices of either polygon are the three vertices of tile $7$, with interior angles four times the corresponding reference-triangle angles. All other boundary angles are less than $\pi$. These three distinct angles force any hypothetical isometry to map each labelled central vertex to its counterpart. An isometry between connected open planar domains is a restriction of a rigid motion: its derivative is orthogonal, and preservation of the flat Levi-Civita connection makes that derivative constant. After aligning the two central triangles, this rigid motion fixes three noncollinear points and is the identity. But the two domains do not coincide: the tile beyond side $a$ of the central triangle has its next attachment along $b$ in the point assembly and along $c$ in the line assembly. These occupy different open regions. This contradiction proves nonisometry.

Therefore the three freely varying side lengths give a three-parameter family of nonisometric Dirichlet-isospectral bounded planar domains. The parameters include scale; no congruence normalization removes it.

Put $\delta =w-u$, which vanishes on the boundary. Expanding the Dirichlet energy givesThe cross term vanishes by integration by parts: it is ${\int }_{\partial S}\delta {\partial }_{n}uds-{\int }_S\delta \Delta udA=0$, using both the zero boundary values of $\delta$ and Laplace's equation for $u$. Thus the Dirichlet principle gives the stronger identityWith the usual regular boundary, equality forces $w=u$.

In the unit disc, take boundary data $\cos \theta$. The harmonic function $u=r\cos \theta =x$ realizes these data and has constant gradient of length one, so its energy is $\pi$. For $w=g(r)\cos \theta$, the polar gradient givesThe energy principle therefore yields the sharp radial Dirichlet energy inequality

There is a small regularity point at the centre: finite energy need not make the polar expression classically smooth there. The bound still holds under the stated finite-energy condition. Indeed for $0<s<t$,The tails tend to zero, so $g^2$ has a limit at zero. That limit must be zero, since a positive limit would make $\int g^2/r$ diverge. Completing the square on $[\epsilon ,1]$ and letting $\epsilon \to 0$ givesThis proves the finite-energy extension directly, and also shows that equality holds exactly for $g(r)=r$.

No: the conclusion need not survive without an additional restriction at infinity. A counterexample that keeps the functional finite is $\Omega ={\mathbb{R}}^n$ and $F(p)=\frac{1}{2}|p{|}^2$. The distinct functions $u\equiv 0$ and $v\equiv 1$ both lie in $C^2(\overline{\Omega })$ and have energy zero. Their boundary data agree because $\partial \Omega$ is empty. For any variation with finite gradient energy,so both are genuine critical points, not merely formal solutions with infinite functional value.

With nonempty boundary, the classical Euler-Lagrange equation also illustrates the missing condition: in dimension $n\ge 3$, on $\Omega =\{|x|>1\}$, both $0$ and $1-|x{|}^{2-n}$ are harmonic functions with zero boundary values. The latter has finite Dirichlet energy, because its gradient-energy integral is proportional to ${\int }_1^{\infty }{r}^{1-n}dr$, and is stationary under compactly supported variations. It approaches $1$ at infinity while the zero solution approaches $0$.

This is nonuniqueness of Dirichlet data without control at infinity. For a global variational formulation one must specify the admissible variations: the exterior example is not critical against every finite-energy zero-boundary variation, since varying it in its own direction changes its energy to first order. Prescribing an appropriate limit at infinity, or a variational space in which the difference is an admissible zero-boundary test function and constants are excluded, can restore uniqueness.

We construct a family by seven-triangle Dirichlet transplantation. Start with a scalene triangle $\tau$ sufficiently close to an equilateral triangle. Its three side lengths are the three independent parameters; restrict to an open set where its angles $\alpha ,\beta ,\gamma$ are distinct and lie between $\pi /4$ and $\pi /2$. Label the sides $a,b,c$ opposite the corresponding vertices $A,B,C$.

For each of two planar domains, begin with tile $7$ and attach the remaining congruent copies by Euclidean reflection across the indicated sides. The attachment tables areFor example, $(6,7)$ in the first $a$ row means to reflect tile $7$ across its $a$ side to obtain tile $6$. Both attachment graphs are trees: there is a central tile with three arms, each consisting of two tiles. Every side not used in the table is an exterior boundary side.

At the equilateral triangle, these seven reflected tile interiors are disjoint; at each central vertex four sectors of angle $\pi /3$ meet, leaving an exterior gap $2\pi /3$. Other boundary vertices have angles $\pi /3$ or $2\pi /3$, and distinct nonincident pieces are separated. The positive angular gaps and separations persist under sufficiently small perturbations of the three side lengths. Thus this construction really gives connected polygonal planar domains on an open three-dimensional parameter set. Their only reentrant vertices are the central vertices $A,B,C$, with interior angles $4\alpha ,4\beta ,4\gamma$; all other boundary angles are one or two triangle angles and are less than $\pi$.

Because these three reentrant angles are distinct, an isometry of the two planar domains would send each labelled central vertex to its counterpart. A planar isometry is the restriction of a Euclidean rigid motion: its differential preserves the Euclidean connection, hence is constant on the connected domain. After aligning the central triangles, such a motion fixes three noncollinear points and is the identity. But the domains are not identical. On the arm attached to side $a$, the point construction attaches the last tile across side $b$, whereas the line construction attaches it across side $c$. The former side is therefore interior in one domain and exterior in the other. This proves the two domains are nonisometric throughout this three-parameter family.

We now prove equality of the Dirichlet Laplacian spectra, including multiplicities. Index $1,\dots ,7$ by the nonzero binary vectors $x=(x_1,x_2,x_3)$, with integer label $x_1+2x_2+4x_3$. The transformationsare involutions over ${\mathbb{F}}_2$. Their point permutations $P_s$ are precisely the transpositions in the left attachment table, with the remaining points fixed. On covectors the inverse-transpose transformations give permutations $Q_s$ with exactly the right table. Let $B$ be the point-line incidence matrix of a set system of the Fano plane:The incidence relation is invariant under transforming $x$ and its covector $y$ contragrediently, so $B{P}_s=Q_{s}B$ for all three sides. Each point lies on three lines, and two distinct points lie on exactly one common line. Consequentlywhere $J$ is the all-ones matrix. Its eigenvalues are $9$ once and $2$ six times, so $B$ is invertible.

The Dirichlet boundary condition requires a sign change at an unglued side. Give each tile the sign $(-1{)}^{\text{distance from tile }7}$ and putThe signs alternate across each glued side in both domains. The signed side matrices are thereforeThey have entry $+1$ across a glued pair and entry $-1$ on a fixed exterior tile. The invertible seven-triangle Dirichlet transplantation matrix $C=DBD$ satisfiesThis is an actual Dirichlet boundary condition intertwiner, not merely the unsigned incidence relation.

To justify the spectral conclusion even at the reentrant corners, use the closed Dirichlet energy form. Pull a function on either domain back to its seven reference triangles and write its components as a vector $F$. The form domain consists of $H^1$ components whose side traces agree on glued pairs and vanish on exterior sides; equivalently $F=S_{s}F$ on each side for the point domain, and $G=T_{s}G$ for the line domain. These trace conditions exactly characterize $H_0^1$ of the glued polygon. Both its $L^2$ norm and its Dirichlet energy are the sums of their seven reference-tile integrals.

Since the signed side matrices are symmetric, the intertwining identities also give $S_s{C}^{T}C=C^{T}C{S}_s$. Thus $(C^{T}C{)}^{-1/2}$ commutes with every $S_s$, and the polar factoris an orthogonal matrix with $U{S}_s=T_{s}U$. Multiplication of the component vector by this constant matrix preserves the $L^2$ inner product and the integrated squared gradients, and maps the two form domains bijectively. The self-adjoint operators associated with these forms are consequently unitarily equivalent, so their eigenvalues and multiplicities agree. The eigenfunctions are smooth in the interiors; no unwarranted smoothness at polygonal corners is required. We have proved

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.

Put $h=u-\varphi -C$. Then $h=0$ on $S$, and the constant does not change a gradient, so $\nabla u=\nabla \varphi +\nabla h$. Expand the unnormalized Dirichlet energy:The cross term is zero by Green's first identity:Thus the useful stronger identity isEquality holds exactly when $h$ has zero gradient. It is then constant on every connected component and zero on that component’s boundary, hence $h=0$. Therefore equality holds precisely when $u=\varphi +C$ throughout $V$. This is the Dirichlet principle; an additive boundary constant does not alter the minimizing energy. The energy here omits the optional factor $1/2$, which does not affect minimizers or the inequality.

A connected open subset of the Euclidean plane. For a bounded polygonal example, the Dirichlet Laplacian is the self-adjoint operator associated with the Dirichlet energy on $H_0^1(\Omega )$; its eigenfunctions are smooth in the interior but need not be smooth at reentrant boundary corners.

For the Poisson equation $\Delta u={\sum }_{n\ge 1}{b}_n\sin (n\theta )$ on the unit disk with zero Dirichlet boundary condition, separation of variables in polar coordinates gives $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. For $n\ne 2$, boundedness at $r=0$ and $R_n(1)=0$ select $R_n=(r^n-r^2)/(n^2-4)$. The resonant mode is $R_2=r^2\log r/4$. For discontinuous angular data, the Poisson equation is interpreted in its weak formulation; its Dirichlet energy gives uniqueness. Orthogonality of the angular Fourier modes gives $\int ufdA=-{\sum }_n\pi b_n^2/[4(n+2{)}^2]$, with the resonant case obtained directly or by the limit $n\to 2$.

For $g(1)=1$ and finite displayed energy, the angular mode $g(r)\cos \theta$ has Dirichlet energy equal to $\pi$ times that integral. The harmonic function $r\cos \theta$ minimizes the energy with the same boundary data by the Dirichlet principle, giving the bound. Finite energy forces $g(r)\to 0$ as $r\to 0$: Cauchy–Schwarz bounds changes of $g^2$ by twice the product of the two energy tails, so $g^2$ has a limit, and a nonzero limit would make $\int g^2/r$ divergent. Consequently completing the square gives the exact identity ${\int }_0^1[r{g}^{\prime 2}+g^2/r]=1+{\int }_0^{1}r(g^{\prime }-g/r{)}^2$. Equality occurs exactly for $g(r)=r$. This identity also covers finite-energy functions whose polar expression is not classically smooth at the centre.