Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

Within this analytic quartic Landau theory, equilibrium of a nonconserved scalar order parameter requires $f^{\prime }=0$. At a continuous phase transition, its restoring curvature vanishes, so $f^{\prime \prime }=0$. A nonzero $f^{\prime \prime \prime }$ would leave a cubic leading term in the Taylor expansion about that state and prevent it from being a local minimum. Thus a stable critical endpoint of a quartic Landau free energy requires all three derivatives to vanish, with $f^{\prime \prime \prime \prime }>0$. The condition $f^{\prime \prime }=0$ alone can instead identify a spinodal point.

For $f=\bar{b}{\lambda }^4+\bar{c}{\lambda }^3+\bar{a}{\lambda }^2+\bar{d}\lambda$, $\bar{b}>0$, the successive equations areThey giveAt these values, the free energy is exactly $f({\lambda }_c)+\bar{b}(\lambda -{\lambda }_c{)}^4$, proving stability. With no electric field, $\bar{d}=0$, so necessarily $\bar{c}=0$, followed by ${\lambda }_c={\bar{a}}_c=0$.

For $\bar{c}<0$ and $\chi >0$, the electric-field coupling to nematic order supplies the required negative ${\bar{d}}_c$ atTuning the quadratic coefficient to ${\bar{a}}_c$ simultaneously reaches the field-induced critical endpoint of the isotropic-nematic transition. The PDF's $\tilde{c}$ in the last clause is undefined; it is interpreted here as the previously defined $\bar{c}$. The endpoint is a prediction of the stable quartic, small-field mean-field approximation; higher-order terms or a field beyond the expansion's validity can shift it.

A scalar coupling linear in the nematic order parameter requires a second-rank anisotropy. A single electric field supplies no such tensor, while $E_i{E}_j$ does. Its isotropic part gives $E^2\mathrm{Tr}Q=0$. Hence the lowest joint order in $E$ and $Q$ of the electric-field coupling to nematic order isTerms depending only on $E$ do not affect minimization over $Q$. With $\vartheta$ the angle between the nematic director and the electric field,For $\chi \lambda >0$, the lowest free energy occurs at ${\cos }^2\vartheta =1$, so the nematic director aligns along either $+\mathbf{E}$ or $-\mathbf{E}$. Then $E_i{Q}_{ij}{E}_j=\lambda E^2$, givingUnlike the linear composition term in a conserved binary fluid mixture, this term changes the equilibrium nematic order parameter: its spatial integral is not fixed.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.

The nematic order parameter is a symmetric second-rank tensor and a traceless second-rank tensor describing orientational anisotropy. For molecular unit axes $\mathbf{s}$, $Q_{ij}$ is proportional to $⟨s_i{s}_j-{\delta }_{ij}/3⟩$. It vanishes in an isotropic phase and is unchanged by head-tail reversal $\mathbf{s}\mapsto -\mathbf{s}$, unlike a polar order parameter. For uniaxial nematic order, the nematic director is thus an unoriented axis.

One convention for the Landau-de Gennes free energy through fourth order, respecting rotational symmetry, isFor a traceless second-rank tensor in three dimensions, the Cayley-Hamilton theorem gives $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$, so there is only one independent quartic rotational invariant of a symmetric traceless tensor. The only possible linear scalar is $\mathrm{Tr}Q=0$; without an external anisotropy no linear term survives.

The cubic rotational invariant of a symmetric traceless tensor distinguishes prolate and oblate forms of uniaxial nematic order. Head-tail reversal leaves $Q$ unchanged and does not impose $Q\mapsto -Q$. Thus a cubic term is allowed in three dimensions and generically produces a first-order phase transition, rather than a symmetry-enforced continuous onset. In two dimensions a symmetric traceless second-rank tensor has eigenvalues $s,-s$, hence $\mathrm{Tr}{Q}^3=0$: the cubic invariant vanishes identically in two dimensions. This is a statement about the Landau free energy; fluctuations in two dimensions require a separate treatment.

Take $b>0$ for a stable quartic Landau free energy. The cubic-term removal in a conserved quartic Landau free energy uses $\varphi =\psi -c/(3b)$ and gives, apart from a constant,For a closed binary fluid mixture, the compositional order parameter has fixed integral. Adding $d\int \varphi$ therefore adds the same constant to every allowed state. Equivalently, in the common-tangent construction for phase coexistence, a linear term changes the tangent slope, hence the chemical potential, by $d$, but leaves the coexistence compositions unchanged. This is why linear composition bias leaves coexistence compositions unchanged; it would not mean an unchanged composition at a fixed externally imposed chemical potential.

In the shifted variable, the binodal compositions are ${\psi }_{\pm }=\pm \sqrt{-a^{\prime }/b}$ for $a^{\prime }<0$. They merge atThe continuous phase transition therefore has the same scalar quartic Landau theory as the symmetric binary fluid mixture, with a shifted critical composition and control parameter. In the mean-field approximation, the composition difference vanishes with order-parameter critical exponent $1/2$. This argument presumes the critical composition lies in the physically accessible composition range; conservation affects the dynamics, not this static phase coexistence construction.

At $\xi =0$, the polar order parameter obeysFor $a<0$, $b>0$, the magnitude relaxes to $\sqrt{-a/b}$ while the direction rotates continuously. Thus the likely ordered behaviour is tumbling rather than a fixed alignment.

More generally, for $|\xi |<1$ and $g\ne 0$, $\xi \cos 2\theta -1$ never vanishes, so the tumbling of a polar order parameter persists. The polar order parameter completes a full $2\pi$ rotation inobtained by integrating $dt=2|d\theta |/(|g|(1-\xi \cos 2\theta ))$ over a full rotation. An unoriented nematic director repeats after half this period, whereas a polar order parameter distinguishes opposite directions. For $\xi \ne 0$ its magnitude can also oscillate. If the polar order parameter decays to zero, its orientation is no longer a physical observable; the tumbling conclusion concerns a persistent ordered state.

The polar molecular field isFor a spatially uniform polar order parameter, $\mathbf{h}=(a+b{p}^2)\mathbf{p}$, so it is colinear with $\mathbf{p}$ and its magnitude is $|\mathbf{h}|=|a+b{p}^2|p$. The scalar factor can be negative; colinearity does not necessarily mean parallel orientation.

For the simple shear flow, the paper's velocity gradient convention givesSubstitute these matrices into $\dot{\mathbf{p}}=-\Omega \mathbf{p}+\xi D\mathbf{p}-\Gamma \mathbf{h}$. Projection along and perpendicular to the polar order parameter yieldsAssume $g\ne 0$, $\Gamma >0$, $b>0$ and $p>0$. The flow-alignment angle in planar shear must satisfyFor $\xi =-1$, use the cosine equation: $\theta =\pi /2$ modulo $\pi$, and the displayed tangent is infinite. The radial equation giveswhere the sign distinguishes the angular branches, and only positive right-hand sides are ordered solutions. Linearizing the angular equation gives $\delta \dot{\theta }=-g\xi \sin 2\theta \delta \theta$. For $|\xi |>1$, the stable angular branch therefore hasIts radial relaxation eigenvalue is $-2\Gamma b{p}^2<0$. At $|\xi |=1$ the angular linearization is marginal and $p^2=-a/b$ requires $a<0$. If $g=0$, the shear restriction disappears and any orientation is allowed for $p^2=-a/b>0$. The printed request for dependence only on $a,b,\Gamma ,g$ omits $\xi$: in general the magnitude necessarily depends on it.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

Use the paper's velocity gradient convention $A_{ij}={\partial }_i{v}_j$, so $\Omega =(A-A^T)/2$ and $D=(A+A^T)/2$. In a rigid motion there is rotation but no rate-of-strain tensor: $D=0$. A polar order parameter attached to the material must rotate exactly with it. In this convention its material derivative is $-\Omega \mathbf{p}$, and its corotational derivative of a polar vector must vanish.

Thus the compensating term is $+\Omega \mathbf{p}$ with coefficient exactly one. This follows from an objective time derivative and is independent of material properties. By contrast, the response to $D\mathbf{p}$ describes deformation and flow alignment of a polar order parameter; its coefficient $\xi$ can depend on the material. The transposed velocity gradient convention is essential to the sign of $\Omega$.

For $q\ne 0$ and $\alpha >0$, each Fourier mode is an Ornstein-Uhlenbeck process with decay rate $r_q=\alpha q^2$. Its explicit Ornstein-Uhlenbeck solution isso the Itô isometry givesTo identify the overdamped Langevin dynamics, choose a friction ${\zeta }_q>0$, a harmonic oscillator potential $V_q={\zeta }_q\alpha q^2{h}_q^2/2$, and force noise $f_q={\zeta }_q{\eta }_q$. The fluctuation-dissipation relation for a Langevin particle then defines $k_B{T}_{\mathrm{eff}}={\zeta }_q{\sigma }^2/2$. Its Boltzmann distribution isThis is the stationary spectrum of a linear fluctuating interface, also following from the equipartition theorem. The numerical prefactor uses exactly the mode-noise normalization given in the paper. The nonzero modes can therefore have a stationary Gaussian distribution even though the unpinned zero mode cannot.

The Neumann boundary conditions give Fourier modes $q=n\pi /L$, including the spatially uniform mode $q=0$. In the frame translating with the mean deposition height, that mode has no restoring force and obeysHere $W_t$ is Brownian motion, independent of the initial height. More generally $⟨h_0(t{)}^2⟩=⟨h_0(0{)}^2⟩+{\sigma }^{2}t$ when these second moments exist. Removing the mean deposition drift does not remove the Gaussian white noise.

The Brownian zero mode of a fluctuating interface has no stationary distribution on the real height axis for ${\sigma }^2>0$. This conclusion does not depend on assuming finite variance: a stationary characteristic function would satisfy $\hat{\mu }(k)=\hat{\mu }(k)e^{-{\sigma }^2{k}^{2}t/2}$, hence vanish for every $k\ne 0$, contradicting its continuity at zero and $\hat{\mu }(0)=1$. Any stationary joint distribution of the whole height would have a stationary zero-mode marginal, which is impossible. Thus the full unpinned height has no Boltzmann equilibrium. Pinning the mean height, or retaining only the nonzero Fourier modes, removes this obstruction. The PDF has ${\dot{h}}_q$; the TeX's ${\hslash }_q$ is a transcription error.

At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

Write $E(t)=\delta A/\delta x(t)$ for the functional derivative of the action. Under time reversal in classical mechanics, $\dot{x}$ changes sign whereas $E$ is even by assumption. Thus the forward and reversed Gaussian white noise histories associated with the same geometric path areThe Onsager--Machlup path probability therefore givesHere the path probabilities are densities conditional on their respective initial states; they are not separately normalized bridges conditioned on both endpoints. The noise-to-path Jacobian determinant must be treated with a consistent discretization: it cancels when it is invariant under time reversal in classical mechanics. This is the usual additive-noise Langevin dynamics convention, including a constant-mass Underdamped Langevin dynamics or the time-reversal invariance of a path Jacobian in midpoint overdamped Langevin dynamics. A completely arbitrary velocity-dependent Lagrangian would require specifying that measure rather than deducing its cancellation solely from the parity of $E$.

The piecewise-linear hat functions have disjoint supports unless their nodes coincide or are neighbours. Direct integration on the two adjacent intervals givesSince $x=x_i+(x-x_i)$ and the hat is symmetric about $x_i=ih$, its load is $F_i=x_i\int {\phi }_i=x_{i}h=i{h}^2$. Thus the reaction-diffusion finite element matrix gives the explicit equationsThe coefficient matrix is a tridiagonal matrix and a positive-definite matrix by the energy identity in part (b), hence nonsingular. A separate spectral verification uses the sine basis with ${\theta }_j=j\pi /(M+1)$ and givesBoth terms are positive for the Dirichlet modes. This also identifies precisely how the diffusion stiffness matrix and reaction mass matrix combine.

The trial space must be a subspace of $H_0^1(0,1)$, not merely of $L^2(0,1)$. The printed assumptions on the basis do not ensure square-integrable weak derivatives, and boundary traces are not defined for arbitrary L2 space functions. For example, $\phi (x)=x^{1/4}(1-x)$ is continuous, vanishes at both endpoints and belongs to $L^2$, but $|{\phi }^{\prime }(x){|}^2\sim x^{-3/2}/16$ is not integrable at zero. Assume the basis belongs to the zero-boundary Sobolev space, as it does for the hat functions in part (c).

Write $u_H={\sum }_{j=1}^M{c}_j{\phi }_j$. Differentiating the restricted energy with respect to $c_i$, or imposing the Galerkin method with test function ${\phi }_i$, givesThe derivative term is the usual diffusion stiffness matrix and the second is the mass matrix. The full matrix is real symmetric, and for every nonzero coefficient vector $c$,because a basis is linearly independent. Thus it is a positive-definite matrix, so the restricted Ritz method has a unique solution and minimum.

Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

Arrange the interior values into a vector $U$. Represent the five-point Dirichlet Laplacian as a Kronecker sum $L_h=h^{-2}(D\otimes I+I\otimes D)$, where $D$ is the tridiagonal matrix with diagonal $-2$ and adjacent entries one. This Kronecker sum is real symmetric; the real sampled potential has a real diagonal matrix $D_V$. ThusThe generator $-i{H}_h$ is a skew-Hermitian matrix. ConsequentlyThe matrix exponential is a unitary matrix, either by differentiating its product with its adjoint or by unitary diagonalization of a normal matrix. For the two-dimensional discrete L2 norm $\Vert U{\Vert }_h^2=h^2{\sum }_{m,n}|U_{m,n}{|}^2$, this givesApplying the same equation to a difference proves stability of a numerical method with a mesh-independent constant one. This is norm conservation of a semidiscrete Schrödinger equation. It concerns continuous time after spatial discretization; an arbitrary subsequent time integrator need not preserve this stability.

The printed coordinates do not discretize the stated square. For $[-1,1{]}^2$ with $M$ interior points in each direction, use $h=2/(M+1)$, $x_m=-1+mh$, $y_n=-1+nh$, and sample $V$ there. The printed $h=1/(M+1)$ and unshifted coordinates instead describe a grid on $[0,1{]}^2$. The matrix proof is valid for either geometry with its corresponding boundary values, so this transcription-independent statement flaw does not alter the stability conclusion.

For a sufficiently regular solution with zero Dirichlet boundary conditions, differentiate its squared L2 norm. Since $u_t=-i(\Delta u-Vu)$,Integration by parts and the zero boundary trace give $\int \overline{u}\Delta u=-\int |\nabla u{|}^2$. The potential contribution is real because $V$ is real. The quantity in parentheses is consequently real, and its product with $-i$ has zero real part. HenceThis is the energy method for a skew-adjoint evolution. For general L2 space data, the same result follows from the strongly continuous unitary group generated by a self-adjoint realization of $\Delta -V$. A sufficient assumption is a bounded real potential with the Dirichlet Laplacian; arbitrary real potentials require appropriate domain and self-adjointness assumptions. The differentiated proof presumes the regularity needed for the displayed integrals, rather than asserting it for every pointwise real function $V$.