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Use the ratio definition of cutoff from part (a), and write $T_n=t_{\mathrm{mix}}^{(n)}(1/4)$. In the reversible Markov chain setting of the preceding parts let ${\gamma }_n=1-{\lambda }_{2,n}$ and ${\gamma }_{\ast ,n}=1-{\max }_{j\ge 2}|{\lambda }_{j,n}|$. In particular ${\gamma }_{\ast ,n}\le {\gamma }_n$.

If ${\gamma }_n{T}_n$ did not tend to infinity, there would be a subsequence and $B>0$ for which ${\gamma }_n{T}_n\le B$. Along it part (c) implies, for every fixed $0<\epsilon <1/2$,Since $T_n\to \infty$, choose fixed $\epsilon$ so small that $\log (1/(2\epsilon ))>B$. The lower limit of this ratio is then greater than one, contradicting cutoff for Markov chains. Thus the Peres product condition holds:In fact the argument with ${\gamma }_{\ast ,n}$ proves the stronger absolute-gap product condition. Part (d) does not repeat the reversible Markov chain hypothesis; the ordinary real ordering of eigenvalues above uses that hypothesis. If arbitrary chains are intended, define the spectral gap explicitly: the eigenvalue lower bound for total variation mixing works for complex eigenfunctions too, and proves the same assertion for ${\gamma }_{\ast }$ or for $1-{\max }_{j\ge 2}\mathrm{Re}{\lambda }_j$, which is at least ${\gamma }_{\ast }$. No real ordering of complex eigenvalues is implicit.

For the requested counterexample take nonlazy simple random walk on the complete graph $K_n$, $n\ge 3$. Let $\Pi$ have every row equal to the uniform distribution on a finite set, and let $a=-1/(n-1)$. ThenThus $d_n(0)\to 1$ and $d_n(1)=1/n\to 0$: every fixed-level mixing time is eventually one. This is a bounded-time jump satisfying the ratio definition of cutoff. The ordinary spectral gap is $n/(n-1)$ and the absolute one is $(n-2)/(n-1)$, so both products with $T_n=1$ tend to one, not infinity. The divergence assumption on $T_n$ is therefore essential under this convention.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.

There is a qualification to the printed “little group”. The little group with mixed string boundary conditions must preserve the endpoints as well as momentum. These boundary conditions break the full 26-dimensional Lorentz group, so the massive states above cannot be classified as representations of the unbroken 26-dimensional massive little group $SO(25)$. In a D1–D25 realization the common worldvolume has Lorentz group $SO(1,1)$ and trivial connected massive little group; $SO(24)$ acts on the ND coordinates as an internal rotation group. The scalar, vector and symmetric trace-free decomposition is under the surviving transverse SO(24), with this boundary-background qualification.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

Set $y=\tanh z$ and $a=1-k$, so $dy/dz=1-y^2$ and $\hat{w}=(1-y^2{)}^{k/2}{y}^a$. Away from $y=0$, the logarithmic derivative givesFor $c=0$, ${\overline{U}}^{\prime \prime }/\overline{U}=-2(1-y^2)$ and $N^2/{\overline{U}}^2=J(1-y^2)/y^2$. The Taylor–Goldstein equation residual divided by $\hat{w}$ consequently simplifies toThus the neutral mode of the equal-width Hazel model requiresThe qualifier “solution” needs care at the critical level of an internal gravity wave $z=0$. For $0<k<1$, $\hat{w}\sim z^{1-k}$ and ${\hat{w}}^{\prime }\sim (1-k)z^{-k}$: it is a solution separately on either side, not a classical smooth eigenfunction across the critical level of an internal gravity wave. A branch for negative $\tanh z$ is also needed. For example, a limit from $c_i>0$ assigns $(\tanh z{)}^{1-k}=|\tanh z{|}^{1-k}{e}^{-i\pi (1-k)}$ on $z<0$.

The critical-layer regularity of a neutral Hazel mode further gives local finite horizontal kinetic energy only for $0<k<1/2$, since incompressible flow gives $\hat{u}=i{\hat{w}}^{\prime }/k$ and ${\int }_0^{ϵ}{z}^{-2k}dz$ then converges. At $k=1$, $J=0$ and $\hat{w}=\mathrm{sech}z$ is smooth after the removable ${\overline{U}}^{\prime \prime }/\overline{U}$ quotient is continued. At $k=0$, $J=0$ and the formal $\tanh z$ profile does not decay at infinity, so it fails the remote endpoint condition. These qualifications prevent interpreting the whole closed parameter interval as a family of classical decaying modes.

Eliminate the instantaneous Stokes flow velocity in favour of temperature. On a horizontal Fourier mode $k$, the Stokes temperature-slaving operator maps $\theta$ to $w={\mathcal{W}}_R\theta$, where $(D^2-k^2{)}^{2}w=R{k}^2\theta$ and $w=w^{\prime \prime }=0$. The temperature evolution has linear operator ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$ and bilinear map $\mathcal{N}(\varphi ,\chi )=-{\mathbf{u}}_R[\varphi ]\cdot \nabla \chi$. Under the homogeneous thermal Dirichlet boundary conditions, ${\mathcal{L}}_{R_c}$ is self-adjoint. Normalize its critical eigenfunction as $g(z)=\sin \pi z$ and set $s=k_c^2+{\pi }^2$; the critical vertical velocity is $f=sg$.

At order $ϵ$, the critical eigenfunction equation gives ${\mathcal{L}}_{R_c}(g{e}^{i{k}_{c}x})=0$. At order ${ϵ}^2$, the weakly nonlinear expansion contains the imposed second harmonic and the quadratic products of the critical mode: a horizontally uniform temperature correction proportional to $|A{|}^2$ and, in a general vertical-mode calculation, a second harmonic proportional to $A^2$. These corrections are found by solving the noncritical boundary value problems, with homogeneous thermal data except for the imposed forcing.

At order ${ϵ}^3$, the method of multiple scales produces the slow derivative $A_{\tau }$, the detuning term $R_2{\partial }_R{\mathcal{L}}_R$, and the two cross-advection terms involving first- and second-order fields. Project the $e^{i{k}_{c}x}$ component onto the adjoint eigenfunction using the vertical inner product. This is the solvability condition in the method of multiple scales: divide each resonant projection by $⟨g,g{⟩}_z$. The detuning supplies $\mu A$ with $\mu \propto R_2$; interactions of horizontal wavenumbers $2k_c$ and $-k_c$ permit $\beta A^{\ast }$ with $\beta \propto \gamma$; self-interaction through the slaved mean and second harmonic supplies $-\lambda |A{|}^{2}A$. Other products have the wrong horizontal wavenumber. Reflection $x\mapsto -x$ permits real coefficients with this cosine forcing. Thus the symmetry-allowed spatially forced convection amplitude equation isThere is a useful specialization that should not be silently missed. For the literal one-vertical-mode Stokes flow problem, the vanishing two-to-one forcing coefficient for Stokes convection makes $\beta =0$ at this order. To see this, write a positive second-harmonic forcing component as $(w,\theta )=(W(z),\Theta (z))e^{2i{k}_{c}x}$, incorporating the cosine's factor $1/2$. Its coupling to the negative critical harmonic has projected integrand, apart from sign and its factor $A^{\ast }$,The integral vanishes because $g=0$ at both plates, even though $\Theta (0)$ is nonzero. This proves the cancellation without solving the forced profiles. The permitted coefficient is therefore zero times $\gamma$; symmetry alone does not establish nonzero phase pinning for the equations actually supplied.

The same normalization makes the remaining coefficients explicit. Since ${\mathcal{W}}_{R}g=(R{k}_c^2/s^2)g$, $\mu =R_2{k}_c^2/s^2$. The quadratic second harmonic cancels for $f=sg$, while the uniform correction is ${\theta }_{20}=-s|A{|}^2\sin (2\pi z)/(2\pi )$. Projecting $-w_1{\partial }_z{\theta }_{20}$ gives $-s^2|A{|}^{2}A/2$. Thus for the literal model and this temperature normalization,A generic nonzero $\beta$ would require a nonvanishing projection in an amended physical model or a different forcing structure. It is still meaningful to classify the real-coefficient amplitude equation requested independently.

Write $A=x+iy$. Then $x_{\tau }=(\mu +\beta -\lambda (x^2+y^2))x$ and $y_{\tau }=(\mu -\beta -\lambda (x^2+y^2))y$. These are a gradient flow for $V=-\frac{1}{2}(\mu +\beta )x^2-\frac{1}{2}(\mu -\beta )y^2+\frac{\lambda }{4}(x^2+y^2{)}^2$, so local minima give stable equilibrium points. At the origin the two eigenvalues are $\mu +\beta$ and $\mu -\beta$. The origin has exponential asymptotic stability if $\mu <-|\beta |$, retains asymptotic stability with algebraic decay at $\mu =-|\beta |$, and is unstable if $\mu >-|\beta |$. At equality, ${\rho }^2=x^2+y^2$ obeys $({\rho }^2{)}_{\tau }\le -2\lambda {\rho }^4$, since both linear coefficients are nonpositive. Integrating this inequality proves attraction even in the zero-eigenvalue direction.

For $\beta >0$ the stable nonzero equilibrium points are real; for $\beta <0$ they are imaginary:The real branch has Jacobian matrix eigenvalues $-2(\mu +\beta ),-2\beta$; the imaginary branch has $-2(\mu -\beta ),2\beta$. The oppositely aligned branch, when it exists, is a saddle equilibrium. No mixed real-imaginary nonzero equilibrium is possible when $\beta \ne 0$.

For $\beta =0$, the origin is stable for $\mu \le 0$, with algebraic decay at zero. If $\mu >0$, the circle $|A{|}^2=\mu /\lambda$ is radially attracting. Each point has Lyapunov stability but has a neutral phase direction, so it does not have individual asymptotic stability; the circle has orbital stability. This is the literal model's unpinned family. The general nonzero-$\beta$ branches instead exhibit phase locking to one of two phases separated by $\pi$.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.

On the impermeable connected square boundary, zero normal velocity makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary temperature fixes $\theta =0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side temperatures retain the background vertical profile, rather than imposing a different isothermal side state.

At marginal stability put $q=\sqrt{R}$ and $P=\psi +iq\theta$. The two coupled equations implyThe gauge transformation for conducting-square Darcy onset $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:The square's Dirichlet Laplacian eigenfunctions are $\sin (m\pi x)\sin (n\pi z)$, with eigenvalues of $-{\nabla }^2$ equal to ${\pi }^2(m^2+n^2)$, $m,n\ge 1$. The first marginal value is thereforeIt is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the temperature linear operator $L_R={\Delta }_D+R{\partial }_x{\Delta }_D^{-1}{\partial }_x$, where ${\Delta }_D$ is the Dirichlet realization of an elliptic operator for the Laplacian. Integration by parts shows the second term is self-adjoint and nonnegative, since its quadratic form is $-⟨{\theta }_x,{\Delta }_D^{-1}{\theta }_x⟩\ge 0$. Thus the spectrum is real, starts negative at $R=0$, and its leading eigenvalue cannot cross into growth without a zero eigenvalue. The smallest zero value is the one found above.

Let $S=\sin \pi x\sin \pi z$, $\xi =x-1/2$ and $a=q_c/2=\sqrt{2}\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical eigenfunctions:The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four Dirichlet boundary conditions. The lowest complex $F$ eigenfunction is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.

For homogeneous corrections at the square boundary, integration by parts gives $⟨{\psi }_0\Delta {\psi }_1⟩=⟨{\psi }_1\Delta {\psi }_0⟩$ and $⟨{\theta }_0\Delta {\theta }_1⟩=⟨{\theta }_1\Delta {\theta }_0⟩$. The first-derivative cross terms give $⟨{\psi }_0{\theta }_{1x}⟩=-⟨{\psi }_{0x}{\theta }_1⟩$ and $⟨{\theta }_0{\psi }_{1x}⟩=-⟨{\theta }_{0x}{\psi }_1⟩$. Using $\Delta {\psi }_0=q_c^2{\theta }_{0x}$ and $\Delta {\theta }_0=-{\psi }_{0x}$, these four contributions cancel in pairs. This proves exactly the supplied solvability condition, including the weight $q_c^2$ on its second row. Equivalently $({\psi }_0,q_c^2{\theta }_0)$ is the relevant adjoint eigenfunction for this coupled system.

At first order in the detuning, the inhomogeneous equations areInsert these right-hand sides into the verified solvability condition. One obtainsA further integration by parts, using the leading heat equation, givesHence the growth-rate solvability for conducting-square Darcy convection yieldsThe PDF prints the reciprocal of the required integral ratio. With its declared ${\theta }_t=ϵ\lambda \theta$, the displayed reciprocal does not follow and is false for $q_1\ne 0$. Both integrals are positive for these nonzero modes. The square's Poincaré inequality gives their ratio at least $2{\pi }^2$, so it cannot equal its reciprocal. For $q_1=1$, the two explicit parities give approximately $\lambda =11.34770$ for odd $\theta$ and $\lambda =7.30164$ for even $\theta$; the printed expression instead gives about $0.00446439$ and $0.00693825$. These values provide direct mode-based counterexamples.

The corrected growth rate is positive for $q_1>0$ and negative for $q_1<0$, as expected at a convection threshold. Because the critical eigenspace is two-dimensional, an arbitrary superposition generally splits into two different first-order growth rates. The specified parity modes remain independent under the detuning: the temperature linear operator preserves reflection parity, so the cross-parity projection vanishes. This justifies applying the scalar solvability condition to either listed eigenfunction; it should not be applied as a single common eigenvalue to an arbitrary mixture.

Consider a horizontal fluid layer under the Boussinesq approximation heated from below, rotating uniformly about its vertical axis. To make the linear formulas definite, take depth one, stress-free impermeable fixed-temperature plates, a horizontally infinite or sufficiently large periodic domain, and negligible centrifugal-buoyancy modifications. Rigid plates change the vertical eigenfunctions and thresholds and require their own boundary value problem; the explicit formulas below use the free-slip model. Let $P=\nu /{\kappa }_T$ be the Prandtl number, $\mathrm{Ta}=(2\Omega d^2/\nu {)}^2$ the Taylor number, and $R$ the Rayleigh number. Time is again measured on the thermal diffusion time.

The linearized rotating Rayleigh-Bénard convection equations about conduction areThe Coriolis force does no work on the fluid, but it couples vertical motion to vertical vorticity and thereby changes the balance of buoyancy and viscous damping. On the plates require $w=\theta =0$ and ${\partial }_z{u}_x={\partial }_z{u}_y=0$. Write $\zeta =(\nabla \times \mathbf{u}{)}_z$. Taking a curl and a double curl removes pressure and givesA normal mode with horizontal wavenumber $k>0$, vertical index $n\ge 1$ and complex growth rate $s$ has $w=W\sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$, $\theta =\Theta \sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$ and $\zeta =Z\cos (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$. Put $a=k^2+n^2{\pi }^2$. The three equations become $a(s/P+a)W+\sqrt{\mathrm{Ta}}n\pi Z=R{k}^2\Theta$, $(s/P+a)Z=\sqrt{\mathrm{Ta}}n\pi W$, and $(s+a)\Theta =W$. Their determinant gives the rotating-convection growth-rate polynomialThis cubic includes viscous and thermal modes as well as the convective instability. It avoids excluding a root by division by $s/P+a$. Linear stability means every root has negative real part; stationary onset has $s=0$, while overstability has $s=\pm i\omega$ with nonzero angular frequency.

Setting $s=0$ gives the stationary neutral curve of rotating convectionFor oscillatory convection, set $s=i\omega$ and equate real and imaginary parts. Eliminating $R$ yieldsThe oscillatory neutral curve of rotating convection is physically admissible only when ${\omega }^2>0$. Thus $P<1$ and sufficiently rapid rotation are necessary; a formal minimum with ${\omega }^2\le 0$ is not a Hopf bifurcation. Convection begins at the smaller of the stationary minimum and the admissible oscillatory minimum. The angular frequency approaches zero where a fixed-wave-number stationary and oscillatory threshold meet; that is a double-zero limit of the cubic, requiring a different slow-time reduction.

For wavenumber selection in rotating convection, the lowest vertical index is $n=1$. To compare indices, set $k^2=n^2{\pi }^{2}y$ at fixed $y>0$. Both neutral thresholds have a positive term proportional to $n^4$ and a rotation term independent of $n$, while the admissible angular frequency squared decreases with $n$. Their minima therefore cannot improve on $n=1$. Set $t=k^2$, $m={\pi }^2$. Minimizing each neutral curve givesFor the latter equation also check ${\omega }^2>0$, or minimize over the admissible set instead. Without rotation, $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At rapid rotation, $k_s\sim (\mathrm{Ta}{\pi }^2/2{)}^{1/6}$ and $R_{s,c}\sim 3(\mathrm{Ta}{\pi }^2/2{)}^{2/3}$: rotation narrows the horizontal convection rolls and raises their threshold. In a finite box only discrete wavenumbers are permitted, so the minimum must be taken over those allowed modes.

The rapid-rotation comparison has $R_{o,c}/R_{s,c}\sim 2P^{4/3}/(1+P{)}^{1/3}$. Its equality gives $8P^4-P-1=0$, with positive root about $0.67660$. Below this value, sufficiently rapid rotation can make oscillatory convection the primary instability; low Prandtl number allows inertial motions to interact with the thermal field before viscosity damps them. This estimate describes the selected neutral-curve comparison in the stated free-slip asymptotic model, rather than claiming every $P<1$ has oscillatory primary onset at every rotation rate.

Linear selection does not determine the nonlinear planform or its stability. A weakly nonlinear expansion with a solvability condition gives amplitude equations. A stationary convection roll amplitude has the Landau amplitude equation $A_T=rA-g|A{|}^{2}A$; a supercritical branch requires $g>0$ and has $|A{|}^2=r/g$, whereas $g<0$ requires higher-order saturation and permits subcritical behaviour. The coefficient depends on the physical parameters and boundary conditions, and should not be assumed positive merely because the linear threshold is known.

A nonzero-frequency Hopf bifurcation supplies oppositely travelling convection roll amplitudes $Z_{+},Z_{-}$. At a fixed orientation, a cubic normal form iswith complex $g,h$ and the understood symmetry exchanging propagation directions. For $r>0$ and $g_r=\mathrm{Re}g>0$, a travelling-roll branch has one nonzero amplitude, of squared modulus $r/g_r$; it is stable to the opposite travelling amplitude when $h_r=\mathrm{Re}h>g_r$. A standing-roll branch has equal squared moduli $r/(g_r+h_r)$, and is stable within this pair when $g_r>h_r$ and $g_r+h_r>0$. The imaginary parts shift angular frequencies. These conditions concern amplitude perturbations in that reduced subspace; orientation and modulation modes still have to be tested. This is the competition of travelling and standing convection rolls.

Near simultaneous stationary and oscillatory thresholds, a codimension-two bifurcation requires retaining both types of mode. After selecting an orientation and fixing a steady spatial phase, a schematic nonresonant steady–Hopf mode interaction isHere $a$ is real, $z$ is the chosen complex oscillatory amplitude, $g_s,h_s$ are real, and $g_o,h_o$ may be complex. Assume $r_s>0$, $g_s>0$ for a radially stable pure steady branch, and $r_o>0$, $g_{o,r}=\mathrm{Re}(g_o)>0$ for a radially stable pure oscillatory branch. Steady convection rolls suppress the oscillatory mode if $r_o-h_{o,r}{r}_s/g_s<0$, where $h_{o,r}=\mathrm{Re}(h_o)$; oscillatory convection rolls suppress the steady mode if $r_s-h_s{r}_o/g_{o,r}<0$. These are transverse tests for branches whose existence and radial stability have already been checked. Mixed states have positive intensities $u=a^2$, $v=|z{|}^2$ solving $g_{s}u+h_{s}v=r_s$, $h_{o,r}u+g_{o,r}v=r_o$. If $D=g_s{g}_{o,r}-h_s{h}_{o,r}\ne 0$, then $u=(r_s{g}_{o,r}-h_s{r}_o)/D$, $v=(g_s{r}_o-h_{o,r}{r}_s)/D$. For positive $u,v$ and the positive self-saturation coefficients above, the intensity Jacobian matrix has negative trace and determinant $4uvD$, so the mixed state is attracting in intensities when $D>0$ and is a saddle equilibrium when $D<0$; phase and other-mode perturbations remain separate tests. Depending on the cross-couplings, one obtains coexistence or competition and bistability; degeneracies or resonances require extra terms. A full travelling/standing-wave competition must retain both $Z_{\pm }$, not just this illustrative one-mode $z$. If the Hopf angular frequency tends to zero, averaging over fast oscillations fails and a double-zero normal form must keep the corresponding two slow variables. A crossing of global minima at distinct wavenumbers is instead an interaction of distinct modes, not automatically the same double-zero problem.