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A Dirichlet eigenfunction is an eigenfunction of a differential operator satisfying a homogeneous Dirichlet boundary condition. The term applies to more general boundary-value operators as well as the Dirichlet Laplacian.

For the positive Dirichlet Laplacian $A$ on a bounded smooth domain, with normalized eigenfunctions $e_j$ and positive eigenvalues ${\mu }_j$, define $A^{\alpha }u={\sum }_j{\mu }_j^{\alpha }(u,e_j)e_j$ on the vectors for which ${\sum }_j{\mu }_j^{2\alpha }|(u,e_j){|}^2<\infty$. The resulting Hilbert space embeds in the Sobolev space $H^{2\alpha }$, with the boundary conditions appropriate to the exponent. In particular $H_{\alpha }$ embeds in $C^{0,\beta }$ whenever $0<\beta <\min (1,2\alpha -n/2)$. High powers impose further boundary compatibility; ordinary smoothness alone does not imply membership in every fractional domain.

Identical oriented Dirichlet boundary data imply a rotation-invariant solution whenever the Dirichlet problem is unique. For $\lambda \ge 0$, uniqueness follows directly: the difference $h$ of two solutions has zero boundary trace, and integration by parts gives ${\int }_D|\nabla h{|}^2+4\lambda {\int }_D|h{|}^2=0$. Thus $h=0$. The same conclusion holds for a nonresonant negative parameter by invertibility of the Dirichlet Laplacian shift. Then $q(az)=q(z)$, all three normal traces equal a common $q_N(s)$, and ${\Psi }_j=\Psi$.

SetDivide the global relation by $E(-iak)$. Using $1+a+\bar{a}=0$ yields $E(-ik)/E(-iak)=e(\bar{a}k)$ and $E(-i\bar{a}k)/E(-iak)=e(-k)$. Hencewhere the requested known function isFor an explicitly derivative-free form, put $H(k)={\int }_{-\ell /2}^{\ell /2}{e}^{(k+\lambda /k)s}f(s)ds$ and $f_{\ast }=f(-\ell /2)=f(\ell /2)$. An integration by parts givesThe uniqueness qualification cannot be removed for every real $\lambda$: at a Dirichlet Laplacian eigenvalue a homogeneous addition need not preserve rotation invariance. The scalar reduced relation then applies to the rotation-averaged solution, but not necessarily to the originally chosen solution. Part (d) treats this genuine exceptional case.

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.

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

A useful precise form of the transplantation theorem is as follows. Take $N$ congruent copies of a Euclidean tile with labelled flat faces. For each face colour $s$, encode the gluings by a symmetric signed permutation matrix $M_s$: a glued pair gives off-diagonal entries $+1$ in the two corresponding positions; an unglued face gives a diagonal entry $-1$ for a Dirichlet boundary condition, or $+1$ for a Neumann boundary condition. Corresponding face identifications must use the same maps in the tile coordinates. Suppose two assemblies have matrices $M_s$ and $M_s^{\prime }$ and there is an invertible matrix $C$ withThen, for every eigenvalue $\lambda$, $f=(f_1,\dots ,f_N)\mapsto Cf$ is an isomorphism between their Laplacian eigenfunction spaces. The two assemblies are isospectral, including multiplicities.

To see why, pull every tile restriction back to the same Euclidean tile. A boundary value vector $u$ and an outward normal derivative vector $v$ satisfyAt an internal face these are continuity of the function and equality of oppositely directed derivatives. At a diagonal $-1$ they impose $u_i=0$; at a diagonal $+1$ they impose $v_i=0$. The intertwining identities preserve both equations, and the Euclidean Laplacian commutes with constant linear combinations of tile functions. The resulting piecewise function is a weak eigenfunction, with the prescribed boundary conditions, and interior elliptic regularity makes it smooth wherever the assembled manifold is smooth. At polygonal corners one uses the usual weak domain of the Dirichlet Laplacian or Neumann Laplacian, rather than assuming smoothness through a corner. The inverse matrix gives the inverse transplantation.

Here is a completely specified propeller pair. Label the seven triangles $0,\dots ,6$, with central triangle $0$, and colour the three sides $a,b,c$. In the left and right assemblies the internal-edge pairings areEvery pair means reflection of one triangle across the indicated side; the other faces are external boundary. Each assembly has a central triangle and three two-triangle arms. These are the propeller domains. The drawing uses one acute scalene triangle, and each colour always denotes the side opposite the same marked vertex of the starting triangle.

Let $P_s,Q_s$ be the ordinary permutation matrices obtained by adding fixed points to the pairings. The following matrix supplies the Neumann boundary condition transplantation:There is a short way to construct and check it. Start with the support $\{1,2,4\}$ in row $0$, and require the support in row $Q_s(i)$ to be $P_s$ applied to the support in row $i$. Traversing the six edges of the right-hand tree gives exactly the displayed rows; checking the remaining fixed-point relations verifies $N{P}_s=Q_{s}N$ for all three colours. Every row has three ones and every two different rows meet in exactly one column, soThe right-hand side has eigenvalues $9$ once and $2$ six times, proving $N$ is invertible, with $|\det N|=24$. Its incidence pattern is that of the Fano plane, which explains the uniform intersections.

For the Dirichlet boundary condition, put $D=\mathrm{diag}(1,-1,-1,1,-1,1,1)$. In both assemblies this gives opposite signs to adjacent triangles. Thus their signed face matrices are $M_s=-D{P}_{s}D$ and $M_s^{\prime }=-D{Q}_{s}D$: the off-diagonal gluing entries are positive and the unpaired diagonal entries are negative. The matrixthen satisfies $C{M}_s=M_s^{\prime }C$ and is invertible. The transplantation theorem proves equality of the Dirichlet Laplacian spectra; $N$ proves equality of the Neumann Laplacian spectra as well.

Finally vary the three labelled side lengths $(a,b,c)$ of the initial triangle in an open region of strict triangle inequalities, with an acute scalene shape sufficiently close to an equilateral triangle. Reflection constructs both domains continuously from these three independent real parameters. At the equilateral configuration the tiles have disjoint interiors. Choose a sufficiently small neighbourhood of that configuration with every triangle angle ${\alpha }_i$ in $(\pi /4,\pi /2)$. At each central vertex exactly four tile angles meet, forming a sector of angle $4{\alpha }_i<2\pi$, while nonincident pieces stay separated. Thus there is an open three-dimensional parameter region in which both assemblies are embedded planar domains.

For a scalene triangle in this region, the three central vertices are exactly the three reflex boundary vertices: their interior angles are $4{\alpha }_i$, while the other boundary vertices have angles ${\alpha }_i$ or $2{\alpha }_i<\pi$. Hence a planar isometry of the domains must map the central triangle onto the central triangle. Put those two central triangles in the same position. Their unequal side lengths force any such isometry to fix the central triangle pointwise. Yet the first arm across, for example, side $a$ attaches its outer triangle across colour $c$ in $L$, and across colour $b$ in $R$. These occupy different positions, so the identity does not identify the domains. An intrinsic Riemannian isometry between connected planar domains also restricts locally to a rigid Euclidean motion and hence is one globally. The two propellers are nonisometric throughout this open three-parameter family. Scaling is included as one of the parameters; modulo scaling the family has two parameters.

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.

A propeller domain consists of seven congruent reflected triangles: one central triangle and three arms of two triangles each. There are different coloured gluing patterns whose Dirichlet Laplacian and Neumann Laplacian spectra agree by the transplantation theorem. Varying the three side lengths of a scalene triangle in a suitable open region gives three parameters; the central triangle is identifiable from the three reflex corners, allowing a direct nonisometry proof.

Take seven copies of a scalene triangle, with sides labelled $a,b,c$ opposite its three vertices, and reflect adjacent copies across their glued side. For the first planar domain, the glued pairs are $a:(2,3),(6,7)$, $b:(4,6),(5,7)$, $c:(1,5),(3,7)$; for the second they are $a:(1,3),(5,7)$, $b:(2,6),(3,7)$, $c:(4,5),(6,7)$. Both coloured trees have central tile $7$. For triangles sufficiently close to equilateral the interiors do not overlap. If its three angles are distinct and between $\pi /4$ and $\pi /2$, the only reentrant vertices are the original central vertices, with angles four times the corresponding triangle angles. They force any proposed isometry to preserve the labelled central triangle, which is incompatible with the different attachments on its arms.

For the spectral proof index the nonzero vectors and covectors of ${\mathbb{F}}_2^3$ by $1,\dots ,7$, using binary coordinates. The Fano plane incidence matrix is $B_{y,x}=1_{y\cdot x=0}$. It satisfies $B^{T}B=2I+J$, so is invertible, and intertwines the point and line permutations of each of the three elementary transvections. Put $D=\mathrm{diag}(1,1,-1,1,-1,-1,1)$. The signed reflection matrices for the Dirichlet boundary condition are $-D{P}_{s}D$ and $-D{Q}_{s}D$, hence $C=DBD$ intertwines them. Its polar orthogonal factor gives a unitary equivalence of the Dirichlet Laplacians, preserving both the $L^2$ inner product and the Dirichlet energy of the piecewise functions. This produces a three-parameter family of noncongruent isospectral planar domains.