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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.

Take the equilibrium mass density to be constant and absorb any conservative equilibrium force into the pressure. The linearized ideal magnetohydrodynamic induction equation, using the solenoidal vector field conditions, isFor $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$, differentiation of the cylindrical basis is important. Acting on a vector whose cylindrical components have azimuthal dependence $e^{im\varphi }$ gives $(\mathbf{B}\cdot \nabla )\mathbf{u}=imJ\mathbf{u}+J{\mathbf{e}}_z\times \mathbf{u}$. Also $(\mathbf{u}\cdot \nabla )\mathbf{B}=J{\mathbf{e}}_z\times \mathbf{u}$. These basis contributions cancel in induction, giving $i\omega \mathbf{b}=imJ\mathbf{u}$.

The perturbation of the Lorentz force isBoth of the first two terms now contribute a basis-rotation term. With $\tilde{p}=p_1/\rho +\mathbf{B}\cdot \mathbf{b}/({\mu }_0\rho )$, the linearized ideal magnetohydrodynamic momentum equation is thereforeThe magnetic part of $\tilde{p}$ is the perturbation of magnetic pressure, not an extra restoring force to retain after projecting out pressure.

Put $D={\omega }^2-m^2{\tilde{J}}^2$, where $\tilde{J}=J/\sqrt{{\mu }_0\rho }$. For $mJ\ne 0$, substitute $\mathbf{u}=\omega \mathbf{b}/(mJ)$ into the momentum equation and multiply by $mJ$. Taking a curl removes the modified pressure. Sincewe obtain the uniform-current toroidal-field curl reductionA second curl, using $\nabla \times \nabla \times \mathbf{b}=-{\nabla }^2\mathbf{b}$, givesThe divided expression assumes $D\ne 0$. If $mkJ\ne 0$, the undivided curl equation shows that $D=0$ would force $\mathbf{b}=0$, so this restriction loses no such magnetic mode. The $k=0$ degenerate limit can instead be taken in the undivided equations; it has no growing branch of the kind considered below. Axisymmetric $m=0$ disturbances are likewise excluded by the velocity elimination, and induction gives no nonzero-frequency magnetic disturbance in this particular equilibrium.

For a vector Laplacian eigenmode with ${\alpha }^2+k^2>0$, elimination givesThus both squared frequencies are real. The lower branch satisfies ${\omega }_{-}^2\ge {\tilde{J}}^2(|m{|}^2-2|m|)$, so it is positive for $|m|>2$. For $|m|=2$ it is positive when $\alpha \ne 0$, but the printed strict-positivity statement needs an endpoint qualification: $|m|=2$, $\alpha =0$ permits a neutral mode unless boundary conditions exclude it.

This is a genuine counterexample to the unqualified endpoint, not merely a loose algebraic bound. The neutral quadrupolar perturbation of a uniform-current toroidal fieldis regular at the axis, is solenoidal, and has $\nabla \times \mathbf{b}=k\mathbf{b}$, hence ${\nabla }^2\mathbf{b}=-k^2\mathbf{b}$. Also ${\mathbf{e}}_z\times \mathbf{b}=-i\mathbf{b}$, so the non-gradient part of the linearized Lorentz force vanishes for $m=2$. The gas-pressure perturbation $p_1=-\mathbf{B}\cdot \mathbf{b}/{\mu }_0$ cancels its magnetic pressure gradient, as required by $\tilde{p}=0$. Taking its real part gives a real disturbance. Decay at radial infinity or an appropriate homogeneous radial boundary condition would exclude it, restoring strict positivity for the allowed eigenmodes; no such condition is printed.

For $|m|=1$ the lower branch is unstable precisely whenAt equality it is marginal. With the time convention $e^{i\omega t}$, the growing member has $\omega =-is$ andThe other branch remains oscillatory. Actual admissibility of a chosen $\alpha$ depends on the radial domain and its boundary conditions.

Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

First find the positive spatially homogeneous equilibrium. Setting the reactions to zero gives $v=u^2$ and $u/v=b$, and hence the unique positive equilibrium isThe zero pair is not an admissible positive state and is singular in the $u^2/v$ term. For homogeneous deviations, the reaction Jacobian matrix at this equilibrium isThus $\mathrm{tr}J=b-1$, $\det J=b$, and its eigenvalues areFor $0<b<1$, both real parts are negative: a real pair has positive product and negative sum, while a complex pair has real part $(b-1)/2$. The reaction equilibrium is therefore asymptotically stable, with no growing oscillatory mode. This does not exclude damped oscillations. Indeed, for $3-2\sqrt{2}<b<1$ the eigenvalues are complex; for example, $b=1/2$ gives $-1/4\pm i\sqrt{7}/4$, a stable focus with oscillatory transients. Thus a literal exclusion of every oscillatory solution is false; the intended exclusion of sustained or growing oscillations is the meaningful statement.

One can strengthen this to exclude nonlinear periodic oscillations in the positive quadrant without relying only on the local linear stability analysis. For the reactions $f=u^2/v-bu$, $g=u^2-v$, use the Dulac function $H=u^{-2}$. ThenWere there a periodic orbit wholly in the positive quadrant, the flux of $H(f,g)$ through it would vanish, since this vector is tangent to the orbit; Green's theorem would equate that flux to a strictly negative interior integral. This contradiction proves there is no sustained positive periodic oscillation when $0<b<1$, the positive-quadrant Dulac multiplier for quadratic activation argument. It leaves the damped transients entirely consistent with the conclusion.

Now introduce a perturbation $(\delta u,\delta v)=\mathbf{a}{e}^{\sigma t+ikx}$ in the quadratic activator-inhibitor model. The Fourier mode of squared wavenumber $q=k^2$ evolves underIts matrix trace and determinant areA Turing instability must start with reaction stability, requiring $0<b<1$. Then $\tau (q)<0$ for every $q\ge 0$, so a mode is unstable exactly when $D(q)<0$: its two real eigenvalues have opposite signs. This also proves the diffusion-driven onset is stationary rather than oscillatory.

The upward-opening quadratic $D(q)$ has its minimum at $q_m=(db-1)/(2d)$. To have a negative value at a physically allowed positive squared wavenumber, one needsWriting $r=db$, the second inequality is $r^2-6r+1>0$, with roots $3\pm 2\sqrt{2}$. The first inequality discards the lower branch, givingThese are the Turing threshold of the quadratic activator-inhibitor model conditions. Equivalently, in the $(b,d)$ plane the instability region lies strictly above $d=(3+2\sqrt{2})/b$ and strictly to the left of $b=1$, with $b>0$. The boundary diverges as $b\to 0^{+}$ and approaches $3+2\sqrt{2}$ as $b\to 1^{-}$.

At equality, the determinant has a double root and one growth rate is zero, while the other remains negative. Substituting $db=3+2\sqrt{2}$ into its minimum location givesBeyond onset, the unstable band isThe stated parameter region assumes continuously available wavenumbers, as on an infinite domain. With specified finite-domain boundary conditions, an allowed nonzero spatial eigenmode must additionally lie in this band; homogeneous stability alone does not ensure that a finite box admits the critical wavelength.

Seek an outer asymptotic expansion $y=y_0+ϵy_1+\cdots$ for fixed $x>0$. The leading equation is $x{y}_0^{\prime }=0$, and the endpoint condition gives $y_0=1$. At first order, $x{y}_1^{\prime }=y_0=1$ and $y_1(1)=0$, so $y_1=\log x$. This outer correction cannot be evaluated at $x=0$.

The coefficient $x+ϵ$ selects the boundary layer coordinate $X=x/ϵ$. With $Y(X)=y(ϵX)$, the equation becomes $(1+X)Y_X=ϵY$. Its inner leading term is constant; matching it to the outer solution gives $Y_0=1$. At first order, $Y_{1X}=1/(1+X)$, hence $Y_1=\log (1+X)+C$. In the overlap $1\ll X\ll {ϵ}^{-1}$, the outer solution is $1+ϵ\log ϵ+ϵ\log X$. Matching the constants therefore givesfor fixed $X$. The logarithmic matching in a small-exponent boundary layer must retain $ϵ\log ϵ$, which is larger in magnitude than a plain order-$ϵ$ term. At the endpoint,The coefficient of a separate plain $ϵ$ term is zero.

Direct integration provides the exact check:At zero its logarithm is $ϵ\log ϵ-ϵ\log (1+ϵ)=ϵ\log ϵ-{ϵ}^2+O({ϵ}^3)$. Exponentiating gives $y(0)=1+ϵ\log ϵ+O({ϵ}^2{\log }^2ϵ)$, whose remainder is indeed $o(ϵ)$. The first-order uniform composite $1+ϵ\log [(x+ϵ)/(1+ϵ)]$ also satisfies the boundary condition exactly and reproduces both expansions to the requested order.

For the WKB approximation, put $Q(x)=|x|-E$. Away from the turning points $x=\pm E$, the forbidden-region solutions have the form $Q^{-1/4}\exp (\pm \int \sqrt{Q}dx)$; in the allowed region put $k(x)=\sqrt{E-|x|}$ and use $k^{-1/2}\exp (\pm i\int kdx)$. The decaying tails selected by the boundary conditions areThe simple-turning-point Airy connection formula matches these toCompatibility requires the Bohr-Sommerfeld quantization conditionwith levels indexed from $n=0$. ThereforeThe cusp at zero does not alter the leading high-energy count. An exact matching check is available: on $x\ge 0$ the decaying solution is $\mathrm{Ai}(x-E)$, and parity at zero imposes ${\mathrm{Ai}}^{\prime }(-E)=0$ for even states or $\mathrm{Ai}(-E)=0$ for odd states. Their interlacing large negative zeros give the same combined leading quantization. Reindexing from one changes only the constant phase shift, not the requested growth law.

The proposed stream function gives $u=Cx{H}^{-2}{\varphi }^{\prime }$ and $v=-C{H}^{-1}\varphi$. Since $\eta =y/H$, ${\partial }_t\eta =-\eta H^{\prime }/H$, the vorticity equation terms areFor $H=\sqrt{2Ct}$, $H{H}^{\prime }=C$. Divide by $C^{2}x/H^5$ (or compare coefficients at $x=0$ by continuity) to obtainThe planes move normally with velocities $v(\pm H)=\pm H^{\prime }=\pm C/H$ and have zero tangential velocity. No-slip boundary conditions therefore requireThese ensure both normal impermeability relative to the moving planes and tangential no slip.