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Put $D=d-1$ for the membrane's internal dimension and assume $\sigma >0$. The small-slope expansion isThe zeroth-order term is the projected area and is independent of the height field. Discarding it changes only an overall normalization of the partition function. Thus the leading fluctuation energy and its statistical weight areOne obtains this Gaussian field theory by discretizing the height field on a microscopic mesh, integrating every height with its Boltzmann weight, and then writing the continuum measure. A microscopic cutoff is retained for quantities dominated by short wavelengths. Quadratic order here really means small slopes: a large rigid height shift is still free because the energy has no dependence on $h$ itself.

The uniform height mode has zero energy. Fix its mean, impose a boundary condition, or divide out the integral over global translations to normalize the partition function. Height differences are independent of that choice. The approximation is the tension-only part of small-slope elastic membrane energy; there is no bending rigidity or confining potential in this part.

Spontaneous symmetry breaking occurs when the statistical Hamiltonian is invariant under a symmetry but a chosen equilibrium pure state is not. A small selecting field or a boundary condition chooses one of several states related by the symmetry. For a finite unconstrained system the symmetric ensemble generally has zero order parameter; spontaneous order is defined by taking the thermodynamic limit before removing the selecting field. A ferromagnet with a selected magnetization direction is the standard example.

If the broken symmetry is continuous, nearby symmetry-related states cost no energy for a uniform displacement along the manifold of states. Locality and a nonzero stiffness then make a slowly varying displacement cost gradient energy rather than a mass term. These gapless long-wavelength fluctuations are Goldstone modes. In the conventional relativistic internal-symmetry setting their count is the number of broken generators. Nonrelativistic dynamics can pair broken generators, as in a ferromagnet; gauge symmetries and broken space-time symmetries require additional qualifications. A discrete broken symmetry alone does not produce a Goldstone mode.

For an interface with a selected mean height, the continuous displacement is a rigid translation normal to the mean plane. The tension gives a stiffness for spatially varying translations. The following calculation concerns this static Goldstone sector. It determines the excitation energy and equilibrium fluctuations, but a frequency dispersion requires a separate inertial or dissipative dynamics. In low dimensions thermal roughness also limits the existence of a strictly localized flat interface in an infinite system.

Work first on the infinite chain, or with a periodized exponential kernel. Put $\rho =e^{-\kappa }$ and use lattice Fourier momentum $q\in [-\pi ,\pi ]$. Summing the geometric series gives the exponentially coupled Ising inverse kernel:ThusAlthough the original kernel is long-ranged, its inverse contains only a constant and a cosine and therefore is nearest-neighbor. Fourier orthogonality givesSubstitution into part (a) yieldsThis exact bulk identity does not require replacing $\sinh \kappa$ by $\kappa$; small $\kappa$ is used later to motivate a smooth continuum field.

For a finite open chain the inverse has off-diagonal entries $-\rho /[J(1-{\rho }^2)]$, interior diagonal entries $(1+{\rho }^2)/[J(1-{\rho }^2)]$ and endpoint diagonal entries $1/[J(1-{\rho }^2)]$. Accordingly the displayed homogeneous potential acquires the endpoint correction $\rho (m_1^2+m_N^2)/[J(1+\rho )]$. These boundary terms do not change the bulk thermodynamic conclusion. Stating infinite-chain or periodized boundary conditions is necessary if the translation-invariant expression is asserted exactly.

With lattice spacing set to one and a smoothly varying auxiliary field, the discrete action becomesThis is precisely the Euclidean path integral of a quantum particle with $\hslash =1$, mass $M$ and potential $U$. With fixed endpoints it gives the transition amplitude $⟨m_f|e^{-\mathcal{L}{\hat{H}}_Q}|m_i⟩$, while periodic classical boundary conditions give its trace. The corresponding operator isThis is classical Ising chain as a quantum particle. It is the continuum approximation to the exact positive transfer kernel, so its numerical spectrum is accurate when the relevant auxiliary configurations vary slowly. The transfer-kernel reasoning and the domain-wall argument below also support the qualitative conclusion without treating a continuum approximation as an exact finite-lattice identity.

At zero field write $a=\tanh (\kappa /2)/J>0$. Expansion givesA double well exists when $a<2$; the stationary points satisfy $am=\tanh (2m)$. For $a\ge 2$ there is a single minimum at zero, with the quadratic curvature marginal at equality. Thus a double well is the low-effective-temperature regime of the model, not an unconditional property for every $J$. The exact potential remains confining because $U(m)=a{m}^2-2|m|+O(1)$ at large $|m|$.

Even when two wells exist, a quantum particle with finite mass and a finite barrier tunnels between them. Its one-dimensional confining Schrödinger operator has a unique, strictly positive ground-state wavefunction. At $h=0$ reflection symmetry makes that state even, so every odd observable has zero expectation. An antisymmetric first excited state is split from it by a positive gap. A spectral expansion of an odd-observable correlation has factors $e^{-x(E_n-E_0)}$; the smallest contributing gap therefore gives a finite classical correlation length. The original spin observable, after decoupling, is $\tanh (2m+h)$, which is odd at zero field. HenceA deep well can give an exponentially small quantum tunnelling gap and an extremely long correlation length, which can mimic order in a finite sample, but it does not produce true long-range order.

The classical counterpart is the finite cost of a domain wall. Because the pair sum is doubled, a sharp wall has dimensionless energyAt any finite temperature such walls have nonzero fugacity, divide a long chain into domains and destroy spontaneous magnetization. The range is large for $\kappa \ll 1$ but finite. The zero-temperature limit and an infinite-range limit are singular and must not be interchanged casually with the thermodynamic limit; an infinite-range interaction also requires an extensive normalization before comparing models.

A nonzero $h$ tilts the potential: to first order $U_h(m)=U_0(m)-h\tanh (2m)+O(h^2)$. A positive field favors the positive well and produces positive magnetization. For fixed finite parameters the ground state and classical free energy vary smoothly, so there is no zero-field discontinuity due merely to a double-well bare potential. A discrete broken-symmetry state appears only in an appropriate singular limit, not by ignoring quantum tunnelling.

More generally a local $d$-dimensional classical statistical model can be sliced along one coordinate and represented by a transfer operator. That coordinate becomes imaginary time; the other $d-1$ coordinates become the spatial coordinates of a quantum system. Infinite extent in the transfer direction corresponds to its ground-state, or zero-quantum-temperature, limit. The classical Ising model is thus related to a quantum Ising system in $d-1$ spatial dimensions, with transverse dynamics generated by the transfer operator. For $d=1$ this is quantum mechanics of finitely many local variables, where quantum tunnelling restores symmetry. For $d\ge 2$ the quantum system has infinitely many spatial degrees of freedom and can have distinct symmetry-broken ground states, allowing a classical finite-temperature ordering transition. Locality and anisotropy of the mapping determine the quantum couplings; the dimension correspondence alone does not fix them.

Irrotational flow means $\nabla \times \mathbf{u}=0$; incompressible flow means $\nabla \cdot \mathbf{u}=0$. The given streamfunction automatically satisfies the second condition, since ${\partial }_x{\psi }_y-{\partial }_y{\psi }_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\mathbf{n}=(dy/ds,-dx/ds)$. ThenThe opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The volume flux between two streamlines is thus their streamfunction difference. A closed contour has zero net flux, expressing mass conservation for constant-density flow without a source inside.

For the channel, a rigid boundary has zero normal velocity and hence constant streamfunction. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi =-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta \psi$, so irrotationality gives $\Delta \psi =0$.

Write $\psi =-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. Separation of variables gives the harmonic sine mode in a half-strip $e^{-n\pi x}\sin n\pi y$. The Fourier sine series coefficients of the left boundary areTherefore the corner sink in a semi-infinite channel isFor $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, ${\sum }_{n\ge 1}\sin (n\pi y)/n=\pi (1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.

The plus sign printed in the final series is inconsistent with the printed boundary conditions. It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi \sim -2m\theta /\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.

Write the integrand, without its common overall factor, as $\mathcal{L}={\partial }_{+}X\cdot {\partial }_{-}X-{\psi }_1\cdot {\partial }_{-}{\psi }_1-{\psi }_2\cdot {\partial }_{+}{\psi }_2$. Keep each constant Grassmann variable on the left. For the first supersymmetry component, varying the bosonic term givesVarying its fermion term gives $-{\epsilon }_1{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1+{\epsilon }_1{\psi }_1\cdot {\partial }_{-}{\partial }_{+}X$. The plus sign in the second term comes from moving the odd parameter past ${\psi }_1$. The middle terms cancel and the rest form ${\epsilon }_1{\partial }_{+}({\psi }_1\cdot {\partial }_{-}X)$. The second supersymmetry component similarly gives ${\epsilon }_2{\partial }_{-}({\psi }_2\cdot {\partial }_{+}X)$. Hence, off shell,This proves the local total-derivative identity for rigid worldsheet supersymmetry; the action is invariant when the corresponding boundary flux vanishes and the transformation preserves the allowed fields. A constant parameter preserves a periodic boson and a periodic fermion. In a Neveu–Schwarz sector, however, $\delta X=\epsilon \psi$ is antiperiodic for constant $\epsilon$ and does not preserve the periodic boson. Thus the bulk identity does not supply a global constant supercharge in every spin sector. This is the spin-structure obstruction to constant worldsheet supersymmetry; the local superconformal formulation handles the different sectors consistently.

For independent field variations, integration by parts gives bulk termsThe equations of motion are thereforeFor variations vanishing at the temporal ends, the spatial seam term is proportional to the difference between its values at $\sigma =2\pi$ and $\sigma =0$ ofThe bosonic part cancels for closed string periodicity, also in a fixed winding sector. For a real fermion impose ${\psi }_a(2\pi )=s_a{\psi }_a(0)$ and the same relation on its variation. Seam cancellation requires $s_a^2=1$, so the two untwisted possibilities are periodic, $s_a=+1$ (Ramond sector), or antiperiodic, $s_a=-1$ (Neveu–Schwarz sector). The two chiralities of a closed string can choose these signs independently, giving R–R, NS–NS, R–NS and NS–R sectors. This discussion assumes the ordinary Lorentz-invariant untwisted boundary conditions of the flat theory.

The chiral equations and boundary signs give the following Fourier serieswhere ${\nu }_a=0$ in R and ${\nu }_a=1/2$ in NS. Thus integer or half-integer modes follow directly from the Fourier series monodromy. With the conventional string oscillator normalization their anticommutation relations and reality conditions areFor an open string the endpoint gluing identifies the two sets of coefficients, leaving one R or NS tower.

In ten-dimensional light-cone gauge in string theory, the eight transverse bosonic modes have positive integer level and each NS fermionic creator has a positive half-integer level. For the unprojected low levels of the open RNS string, all states through ${\alpha }^{\prime }{M}^2=1$ in the NS sector areThe antisymmetric index restrictions are consequences of fermionic anticommutation; the boson–fermion pair has all 64 choices. The displayed partitions exhaust levels at most $3/2$. Here the NS normal-ordering constant of a string is $a=1/2$.

In the R sector $a=0$, and the zero modes satisfy $\{{\psi }_0^i,{\psi }_0^j\}={\delta }^{ij}$. Thus ${\gamma }^i=\sqrt{2}{\psi }_0^i$ generate the Ramond zero-mode Clifford algebra. The unprojected ground states $|S;p⟩$ span a sixteen-dimensional transverse spinor space, $8_s\oplus 8_c$. All states in the requested range areThere is no intervening half-level, since all nonzero R modes have integer level. Acting with the zero modes only moves within the ground spinor space and does not generate further independent copies.

The GSO projection keeps odd NS fermion-excitation parity, removing the tachyonic vacuum and the entire $N=1$ NS level. In R it keeps one eigenvalue of total fermion parity, whose zero-mode part is spinor chirality. It leaves eight massless spinor states. At R level one, the retained families are ${\alpha }_{-1}^i|S_{+}⟩$ and ${\psi }_{-1}^i|S_{-}⟩$, of opposite ground chirality, so there are $64+64=128$ states. The retained NS states are spacetime bosons and the R states spacetime fermions. Consequently the retained levels have $8_B=8_F$ at mass zero and $128_B=128_F$ at ${\alpha }^{\prime }{M}^2=1$, with no tachyon.

Equality at every higher mass level follows from the standard Jacobi abstruse identity, or equivalently the equality of the transverse generating functionsThe left side counts the projected NS bosons, the right side the projected R fermions. Their common coefficients start $8+128q+\cdots$. This consistent truncation yields spacetime supersymmetry; the bulk worldsheet symmetry alone would not remove the tachyon or impose the projection.

The stress-energy tensor generates translations. One may improve it by adding a divergence without changing the translation charges under suitable boundary conditions, and choose it symmetric in a relativistic theory. If ${\Theta }^{\mu \nu }$ is symmetric and conserved, a spacetime vector field $\omega$ defines the current $j^{\mu }={\Theta }^{\mu \nu }{\omega }_{\nu }$. For a Conformal Killing vector field,Thus a traceless conserved stress-energy tensor produces all the conformal charges. The dilation current is $x_{\nu }{\Theta }^{\mu \nu }$, and the special conformal currents are $(2x_{\lambda }{x}_{\nu }-x^2{\eta }_{\lambda \nu }){\Theta }^{\mu \nu }$. A trace that is merely the divergence of a virial current gives scale invariance; conformal invariance requires the additional improvement that removes the trace. These statements distinguish an actual traceless tensor from a trace discarded without justification.

For a free massless scalar field, the canonical stress-energy tensor has trace $(1-d/2)(\partial \varphi {)}^2$. For $d>1$, the improved stress-energy tensor of a free massless scalar isIts improvement is identically conserved. On $\square \varphi =0$, the identity $\square {\varphi }^2=2(\partial \varphi {)}^2$ shows that its trace vanishes. In $d=2$, no improvement is needed: the canonical stress-energy tensor is already traceless. In Lorentzian null coordinates, conservation and tracelessness imply that its two surviving components depend separately on $x^{+}$ and $x^{-}$. After Euclidean continuation, these become the holomorphic stress-energy tensor $T(z)$ and its antiholomorphic counterpart $\bar{T}(\bar{z})$, withLocal conformal changes are arbitrary holomorphic and antiholomorphic coordinate changes, not just the three global Möbius transformation generators in either sector.

To make the free-field quantum normalization explicit, rescale the field to $\phi =\sqrt{4\pi }\varphi$, so that its plane contraction is $\phi (z,\bar{z})\phi (w,\bar{w})\sim -\log |z-w{|}^2$. Define $J=i\partial \phi$, so $J(z)J(w)\sim (z-w{)}^{-2}$. Normal ordering givesA single Wick contraction shows that $J$ is a weight-one primary operator. The two double contractions in $T(z)T(w)$ produce $\frac{1}{2}(z-w{)}^{-4}$; single contractions, followed by Taylor expansion at $w$, produce $2T(w)/(z-w{)}^2+\partial T(w)/(z-w)$. Thus the operator product expansion isThe same holds independently for $\bar{T}$, with $\bar{c}=1$. The undifferentiated noncompact scalar has a logarithmic contraction and a zero-mode subtlety; its derivative current has the ordinary primary transformation law.

The holomorphic stress-energy tensor implements infinitesimal conformal transformations through contour charges $Q_v=\frac{1}{2\pi i}\oint dzv(z)T(z)$. A primary field $\mathcal{O}$ of weight $h$ obeysFor example the free-field vertex operator $:e^{i\alpha \phi }:$ has holomorphic weight $h={\alpha }^2/2$, as follows by contracting $\partial \phi$ with the exponential twice. In correlators the contour may encircle each insertion, giving the conformal Ward identity $⟨T(z){\prod }_i{\mathcal{O}}_i(z_i)⟩={\sum }_i[h_i/(z-z_i{)}^2+{\partial }_{z_i}/(z-z_i)]⟨{\prod }_i{\mathcal{O}}_i(z_i)⟩$. This also yields the global conformal Ward identities for chiral correlators.

Writing $T(z)={\sum }_n{L}_n{z}^{-n-2}$ and integrating the stress-tensor operator product expansion gives the Virasoro algebraIts central charge is the quantum correction to the classical conformal algebra; the correction vanishes on $L_{-1},L_0,L_1$. Equivalently, in the pullback convention a finite holomorphic change $z\mapsto f(z)$ sends $T$ to $(f^{\prime }{)}^{2}T(f)+\frac{c}{12}\{f,z\}$, where $\{f,z\}=f^{\prime \prime \prime }/f^{\prime }-\frac{3}{2}(f^{\prime \prime }/f^{\prime }{)}^2$ is the Schwarzian derivative. This anomalous term vanishes for Möbius transformations. On a curved background the related trace anomaly is proportional to its curvature; it does not destroy the flat-space conformal symmetry of this massless theory.

For $d\ge 2$, let $\omega$ be a Conformal Killing vector field, put $\sigma =\partial \cdot \omega /d$, and assign the massless scalar field scaling dimension $\Delta =(d-2)/2$. In a finite conformal transformation with metric scale ${\Omega }^2$, the transformation law is ${\varphi }^{\prime }(x^{\prime })=\Omega (x{)}^{-\Delta }\varphi (x)$. Its infinitesimal fixed-coordinate form isFor $\mathcal{L}=(\partial \varphi {)}^2/2$, contract the Conformal Killing equation with ${\partial }_{\mu }\varphi {\partial }_{\nu }\varphi$. Direct differentiation then givesHere $2(1+\Delta )=d$. Flat conformal Killing integrability gives $\square \sigma =0$, with the Lorentzian version obtained by replacing the flat metric consistently. Thus the action variation is a boundary integral, vanishing for suitable boundary conditions. This proves conformal invariance with field dimension $(d-2)/2$, without using the field equation. The connected finite symmetry follows by integrating the infinitesimal generators. Equivalently its conformal currents follow from the improved stress-energy tensor of a free massless scalar.

In one dimension the metric conformality condition is vacuous. The same action-variation calculation contains $-{\varphi }^2{\omega }^{\prime \prime \prime }/4$, so the free action is invariant under the projective subgroup ${\omega }^{\prime \prime \prime }=0$, rather than arbitrary reparametrizations. The usual field-theory conformal-invariance claim above concerns $d\ge 2$.

In $d=2$, $\Delta =0$, so the field transforms simply as a scalar. There is no improvement term and no derivative of the conformal scale in the variation. All local independent reparametrizations of the two null coordinates preserve the action; after Euclidean continuation they are holomorphic and antiholomorphic maps. The conformal algebra is therefore infinite-dimensional, in contrast to $d>2$.

Now use Lorentzian coordinates with spatial period $\ell$. The canonical momentum is $\pi =\dot{\varphi }$, and canonical quantization imposes $[\varphi (t,x),\pi (t,y)]=i{\delta }_{\ell }(x-y)$, with the periodic delta distribution ${\delta }_{\ell }(s)={\ell }^{-1}{\sum }_{n\in \mathbb{Z}}{e}^{2\pi ins/\ell }$. Solving the wave equation and imposing reality gives a convenient oscillator normalization:Strict spatial periodicity excludes a winding term. The canonical commutation relation is equivalent towith ${\alpha }_n^{†}={\alpha }_{-n}$ and the corresponding tilded relation. Both oscillator families commute with $q,p$. To verify the normalization, the zero-mode commutator supplies $i/\ell$, while each oscillator family supplies $i(2\ell {)}^{-1}{\sum }_{n\ne 0}{e}^{2\pi in(x-y)/\ell }$. Their sum is exactly $i{\delta }_{\ell }(x-y)$. The positive-index modes are annihilation operators and the negative-index modes are creation operators, giving two independent bosonic Fock spaces together with the free-particle zero mode.

Define ${\alpha }_0={\tilde{\alpha }}_0=p/\sqrt{4\pi }$ and the normal ordering quadratic generatorsThese are the Fourier modes of the two chiral components of the stress-energy tensor. The oscillator operator commutators immediately give $[L_m,{\alpha }_n]=-n{\alpha }_{m+n}$. Applying this to both factors of $L_n$ gives $(m-n)L_{m+n}$, except for a scalar term when $m+n=0$, produced by reordering creators and annihilators. Its value is independent of the momentum sector. Evaluate it in the formal zero-momentum oscillator vacuum. For $m>0$,The two contractions of each pair explain the factor $1/2$. Hence the free-boson Virasoro central term yieldsThe tilded modes give a second commuting Virasoro algebra with $\bar{c}=1$. The adjoint condition $L_m^{†}=L_{-m}$ and the positive oscillator norms make this a unitary highest-weight Virasoro representation in each momentum sector. In a fixed momentum sector, $L_0=p^2/(8\pi )+{\sum }_{n>0}{\alpha }_{-n}{\alpha }_n$, so the oscillator ground state is a highest-weight state of weight $p^2/(8\pi )$ in either sector. For a noncompact field the zero mode is a free particle: definite momentum ground states are generalized states, rather than a normalizable zero-mode vacuum. The oscillator representation and central term remain well defined. With the standard cylinder vacuum-energy subtraction, the Hamiltonian isThe constant is the Casimir energy of a two-dimensional conformal field theory; omitting it amounts to a different additive Hamiltonian convention, not a change of the plane Virasoro algebra.

Use the Galerkin method: substitute the retained trigonometric modes and project onto them with the $L^2$ inner product over the rectangle. Each mode satisfies the printed boundary conditions. Write $X=\alpha x$, $Z=\pi z$, and abbreviate the four mode prefactors by $C=2\sqrt{2}\beta /\alpha$, $D=2\sqrt{2}/\beta$, $V_1=2\sqrt{2}\sigma \pi R_{\Omega }/(\alpha \beta )$ and $V_2=\sigma R_{\Omega }/\alpha$. The Jacobian determinant in the advection terms is $J(f,g)=f_x{g}_z-f_z{g}_x$.

The stream function is a single Laplacian eigenfunction, so ${\nabla }^2\psi =-{\beta }^2\psi$ and $J(\psi ,{\nabla }^2\psi )=0$ exactly. For the temperature modes, direct multiplication givesFor the transverse-velocity modes, the retained products areFor example, $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$ and $\sin X\cos 2X=(\sin 3X-\sin X)/2$ supply the signs of the two retained $ac$ and $ae$ couplings. The third harmonics are orthogonal to the retained modes and are discarded by this Galerkin method; they are not claimed to vanish in the full partial differential equations.

Before changing time, the five coefficient equations areDividing by ${\beta }^2$ gives the stated five-mode rotating-convection truncation, withThe squared rotation parameter is $q=r_{\Omega }^2\ge 0$. The printed $R_{\Omega }$ is proportional to rotation rate, so it is a square-root-type rotation parameter rather than the more usual Taylor number proportional to squared rotation rate; the algebra uses the coefficient as printed. At zero rotation the physical transverse velocity modes vanish; the normalized $d,e$ equations can still be understood through their continuous zero-rotation limit.

For a stationary finite-wavenumber instability, the relevant Landau–Ginzburg model is the real Ginzburg–Landau equation: its coefficients are real although its pattern amplitude is complex. The complex amplitude records both the strength and the position of the underlying pattern. Assume a simple critical pair of Fourier modes $\pm k_c$, a quadratic nondegenerate maximum of the linear growth rate, and no additional conserved or marginal mode. The last assumption is essential: the conserved field in the previous problem requires an additional evolution equation.

To derive the model, write a real evolution equation schematically as $u_t=L(r)u+N_2(u,u)+N_3(u,u,u)+\cdots$, with symmetric multilinear nonlinearities. Put $r={\epsilon }^2{r}_2$, $X=\epsilon x$, $T={\epsilon }^{2}t$ and use the method of multiple scales expansionHere $L(k_c;0)v=0$. At the next order the noncritical mean and second harmonic are slaved to the critical mode:where $L(0;0)w_0=-2N_2(v,\overline{v})$ and $L(2k_c;0)w_2=-N_2(v,v)$. These inverses exist under the no-other-critical-mode hypothesis. Let $v^{\ast }$ be an adjoint null vector normalized by $⟨v^{\ast },v⟩=1$. At order ${\epsilon }^3$, the Fredholm solvability condition requires the resonant forcing to be orthogonal to $v^{\ast }$. The cubic resonant contribution isThis shows why the cubic coefficient includes feedback from slaved harmonics as well as any original cubic nonlinearity. The resulting amplitude equation, after naming its linear control coefficient $\mu$, isHere $\lambda (k)$ is the critical linear dispersion relation for growth. The condition $g>0$ gives a supercritical bifurcation; $g<0$ calls for higher-order saturation, for example the quintic real Ginzburg-Landau equation.

A concrete check is the cubic Swift–Hohenberg equation $u_t={\epsilon }^2\mu u-(1+{\partial }_x^2{)}^{2}u-g_0{u}^3$. With $u=\epsilon (A{e}^{ix}+\overline{A}{e}^{-ix})+\cdots$, its linear operator contributes $4A_{XX}$ at resonant order, while the cubic resonance is $3|A{|}^{2}A$. Hence $A_T=\mu A+4A_{XX}-3g_0|A{|}^{2}A$. This directly realizes the perturbation-theory coefficients above.

Symmetry independently constrains the form. A fast spatial translation rotates $A$ by a constant phase, so every nondifferential monomial must transform like $A$. The leading allowed nonlinear term is $|A{|}^{2}A$, and a quadratic term is forbidden. A spatial reflection maps $A(X)$ to $\overline{A}(-X)$; it makes the leading coefficients real and rules out a drift term in a reflection-symmetric rest frame. The vanishing first derivative of the linear growth rate at $k_c$ is consistent with the absence of a leading first spatial derivative. These arguments constrain terms but do not calculate $\xi ,g$ or replace the solvability calculation. Breaking reflection or studying an oscillatory instability generally leads to the complex Ginzburg–Landau equation with genuinely complex coefficients.

The principal steady solutions are the zero state and the Ginzburg-Landau plane wavesTheir physical carrier wavenumber is $k_c+\epsilon q$. For $q=0$ the constant phase is arbitrary, giving a circle of uniform-amplitude states. The zero state has Fourier growth rate $\mu -\xi p^2$ and is stable for $\mu <0$. For $g>0$, each existing plane wave is stable to spatially homogeneous amplitude perturbations, with radial eigenvalue $-2g{R}^2$, and neutral to a constant phase shift.

Steady modulated states are also possible. Writing $A=a(X)e^{i\varphi (X)}$ gives the conserved spatial phase current $j=a^2{\varphi }_X$ andIts first integral is $\xi a_X^2/2+\mu a^2/2-g{a}^4/4+\xi j^2/(2a^2)=E$. Boundary conditions select nonuniform periodic amplitudes, fronts or defects from this spatial dynamics. In the invariant real subspace, for example,is a steady kink joining opposite real states. It is not stable in the unrestricted complex equation: the imaginary perturbation $v=\mathrm{sech}(\sqrt{\mu /(2\xi )}X)$ has eigenvalue $\mu /2>0$ under $v_T=\xi v_{XX}+(\mu -g{A}^2)v$. This illustrates why a real-amplitude calculation can miss a phase instability. With real coefficients the Lyapunov functionalsatisfies $A_T=-\delta \mathcal{F}/\delta \overline{A}$ and $d\mathcal{F}/dT=-2\int |A_T{|}^{2}dX$ under compatible boundary conditions. It supplies a variational structure absent from a generic complex-coefficient model.

To test full linear stability of a plane wave, set $A=e^{iqX}(R+u+iv)$ with real perturbations $u,v$. Let $b=g{R}^2=\mu -\xi q^2>0$. The linear equations areFor perturbation wavenumber $p$, the sideband spectrum of a real Ginzburg-Landau plane wave isThe phase branch at small $p$ isTherefore the robust stable band on the infinite line is $q^2<\mu /(3\xi )$, narrower than the existence band $q^2<\mu /\xi$. Crossing its boundary gives the longitudinal Eckhaus instability. At the boundary the quadratic phase-diffusion coefficient vanishes and the leading nonzero term is negative of order $p^4$. More exactly, a nonzero sideband grows only when $p^2<6q^2-2\mu /\xi$; a finite periodic domain for the wave envelope may exclude that interval. Constant-amplitude stability is understood modulo its neutral global phase, and the allowed perturbation wavenumbers must always be specified.

In a second spatial dimension, an anisotropic medium that selects one roll orientation permits an equation with ${\xi }_x{A}_{XX}+{\xi }_y{A}_{YY}$. For positive coefficients its plane-wave analysis is similar, with the second-dimensional contribution adding damping. But simply replacing $A_{XX}$ by an isotropic Laplacian is not a faithful extension for rotationally invariant stationary rolls. There the critical wavevectors form a circle. Near $(k_c,0)$,This forces the anisotropic transverse scaling of an isotropic roll envelope: $X=\epsilon x$, $Y={\epsilon }^{1/2}y$, $T={\epsilon }^{2}t$. For the positive carrier convention used here the Newell–Whitehead–Segel equation isThe opposite fast-carrier convention reverses the accompanying imaginary sign and detuning convention. For a transverse phase perturbation of $A=R{e}^{iqX}$ our convention gives $\sigma =-\xi q{p}^2/k_c-\xi p^4/(4k_c^2)$. Thus $q<0$ permits the transverse zigzag instability, in addition to the longitudinal Eckhaus instability. The fourth transverse derivative and mixed derivative in the square encode the critical-circle geometry.

One complex amplitude still describes only a narrow cone of nearby roll orientations. Competing well-separated wavevectors require coupled amplitude equations; resonant triples can admit quadratic interactions and produce the hexagonal convection amplitude equations when the underlying sign symmetry allows them. Strong disorder, defect cores with large gradients, far-from-onset patterns, boundaries, mean flows and conserved fields may all lie outside the single-amplitude approximation. The method of multiple scales gives controlled leading behavior for small amplitudes and slow modulations, rather than a universal model of all two-dimensional pattern selection.