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A stability analysis asks whether perturbations in an evolution remain bounded on a fixed physical time interval by a constant independent of the mesh. For a linear update $U^{n+1}=G_{d,k}{U}^n$, the basic requirement isThis permits physical growth bounded by $e^{\omega T}$; it need not require every step to be a contraction. Consistency of a numerical method measures how well its stencil reproduces the differential equation, whereas stability of a numerical method controls the accumulated defects. For a well-posed linear problem, the Lax equivalence theorem connects stability and convergence for a consistent approximation in the specified norm.

On the whole line or a periodic grid, constant-coefficient stencils commute with translations. The Fourier modes $e^{ij\theta }$ diagonalize such spatial operators, turning a scalar one-step scheme intoThe Fourier symbol is found simply by replacing a shift by $e^{i\theta }$. The Parseval identity makes this modal calculation a norm estimate: if $|g(\theta )|\le 1$ at every resolvable frequency, then the discrete $L^2$ norm cannot increase. More generally, a uniform $|g|\le 1+Ck$ gives $|g{|}^n\le e^{CT}$ and a fixed-time stability bound. All frequencies matter, including those near the grid scale; a small-wavenumber expansion establishes consistency but cannot test the entire stability range.

For the heat equation, put $r=k/d^2$ and $s={\sin }^2(\theta /2)$. The Forward Euler diffusion scheme hasThe condition $|g_{\mathrm{FE}}|\le 1$ for every $s\in [0,1]$ is exactly $0\le r\le 1/2$. This illustrates a parabolic time-step restriction $k\le d^2/2$. The Backward Euler diffusion scheme instead haswhich is contractive for every $r\ge 0$. The Crank-Nicolson diffusion scheme hasand is likewise unconditionally $L^2$ stable. However, its high-frequency amplification approaches $-1$ for large $r$, so it need not damp rapidly decaying modes effectively and may produce alternating transients from nonsmooth data. An A-stable time integrator can stabilize a diffusion semidiscretization with left-half-plane spectrum, but damping and accuracy remain separate questions.

For transport $u_t+a{u}_x=0$, define the hyperbolic Courant number $\nu =ak/d$. Forward Euler with a centered spatial difference gives $g=1-i\nu \sin \theta$ and $|g{|}^2=1+{\nu }^2{\sin }^2\theta$. It is unstable under a fixed nonzero CFL ratio. For $a>0$, the upwind finite difference scheme hasThus $0\le \nu \le 1$ is the stable range. Upwinding for negative $a$ reverses the spatial difference and gives $|\nu |\le 1$. The Lax-Wendroff advection scheme gives a second-order example:It is stable for $|\nu |\le 1$, but that $L^2$ result does not imply monotonicity or exclude oscillations at shocks. The Courant–Friedrichs–Lewy condition has a domain-of-dependence interpretation for explicit hyperbolic schemes; the Fourier calculation supplies the sharper algebraic stability criterion for the particular stencil.

For multilevel methods, each frequency gives an amplification polynomial and a companion matrix. Every amplification root must lie in the closed unit disk, with unit-modulus roots simple. Uniformity across frequencies and meshes is essential: one also needs control of the associated matrix powers, for example through a uniformly bounded diagonalization or the uniform power bound from separated amplification roots. A repeated unit root generates a Jordan block and polynomial growth in the number of steps. The same issue occurs for systems: checking only eigenvalue moduli of a nonnormal Fourier amplification matrix is insufficient. For instancehas unit eigenvalues but no fixed-time mesh-uniform bound as $n=T/k\to \infty$. For a hyperbolic system with a constant coefficient matrix admitting a uniformly bounded characteristic diagonalization, scalar modal estimates for each characteristic speed do give a system estimate. A positive energy symmetrizer provides another route. Frozen-coefficient Fourier tests for variable coefficients are useful local diagnostics, but a global estimate must also account for coefficient variation and boundary terms.

Boundary conditions change the allowable modes and can change the conclusion. A periodic calculation is exact only for a periodic closure. On a finite interval with zero Dirichlet boundary conditions, the centered diffusion matrix is diagonalized instead by the discrete sine transform. On $M$ interior points its eigenvalues are $-4{\sin }^2(j\pi /[2(M+1)])$ before scaling by $d^{-2}$. Thus the exact fixed-grid Forward Euler restriction iswhich tends to the whole-line threshold $1/2$ as the grid is refined. Backward Euler and Crank-Nicolson remain stable for all nonnegative $r$. An appropriate Neumann boundary condition uses cosine-type modes and includes a constant mode with eigenvalue zero. The constant mode must be preserved, rather than forced to decay; depending on the boundary stencil, the natural discrete inner product may have endpoint weights.

For hyperbolic transport, the continuous and discrete boundary closures must respect the direction of propagation. If $a>0$ on $[0,L]$, prescribe the left inflow value; prescribing an independent outflow value can overdetermine the problem. The energy method exposes this directly:Homogeneous inflow is dissipative, while forced inflow contributes data to the estimate. A compatible upwind boundary update has the same one-sided information flow. For diffusion,The last term vanishes for homogeneous Dirichlet or Neumann conditions. With Robin boundary conditions of dissipative sign it contributes a further nonpositive term. A non-dissipative boundary can produce genuine PDE growth, which a valid stability estimate should represent rather than incorrectly suppress.

Even a perfectly stable interior symbol cannot detect every defective boundary closure of a difference scheme. As a concrete counterexample, leave a stable explicit heat stencil on all grid points away from the left endpoint, but replace its first interior row by $U_1^{n+1}=2U_1^n$, still keeping $U_0=0$. The interior Fourier symbol is unchanged, yet $U_1^n=2^n{U}_1^0$ makes the finite-interval update unstable. The faulty boundary-adjacent row creates a mode absent from the periodic interior calculation. Thus the entire finite update matrix, or an energy estimate including its boundary rows, must be examined.

A more systematic boundary stability of a finite-difference method analysis transforms time and any tangential spatial variables, then looks for modes $U_j^n=z^n{\kappa }^j$ with $|z|>1$ and $|\kappa |<1$ on a half-line. The interior dispersion relation selects decaying spatial modes; the boundary equations must determine their coefficients without admitting a nonzero growing homogeneous mode. A uniform inverse estimate for the boundary system, expressed by the Uniform Kreiss--Lopatinskii condition, also rules out arbitrarily weak boundary control near limiting frequencies. Merely excluding one obvious unstable mode is weaker than proving a mesh-uniform estimate. Discrete summation by parts and suitable boundary penalties offer an energy-based alternative.

Finally, nonlinear scalar conservation laws can develop discontinuities, so linearized Fourier stability is only one part of the analysis. A monotone conservative scheme, such as the Engquist-Osher method under its CFL restriction, has an invariant state range and L1 contraction of a monotone conservative scheme, with a discrete entropy inequality selecting admissible shocks. These complement the Fourier and boundary estimates. For an evolution problem, the meaningful outcome is a stability bound in a specified norm, uniform over the chosen mesh family and compatible with the actual boundary and initial data.

Use the homogeneous equilibrium point condition $k({\rho }_0){\rho }_0=a_{0}f({\rho }_0)$. Evaluate $D_1,D_2$ at $(a_0,{\rho }_0)$ and abbreviateThe quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence $e^{st}$, put $z=|\mathbf{q}{|}^2$ and writeConsequentlyAt $z=0$, the eigenvalues are $0$ and $-c$. The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.

Write $a=D_2>0$, $b=D_{\rho }>0$, $C=D_1{f}_0$, and interpret production physically as $f_0\ge 0$. If $c\ge 0$, the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when $C>ac$. The unstable band isIf $c<0$, the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since $C\ge 0$, this also implies $C>ac$. Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold isFor $\kappa >0$, dividing by $D_2\kappa$ gives the requested formAt equality there is no strictly growing mode when $c\ge 0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using $C\ge 0$ must be revisited.

The printed hypotheses do not ensure $\kappa >0$. Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho )=e^{-2\rho }$, $f(\rho )=e^{-2}$, $a_0={\rho }_0=1$, and $D_1=D_2=D_{\rho }=1$. The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but $\kappa =c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa =0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.

To find the fastest-growing Keller--Segel mode, use the larger eigenvalueFor $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation givesA positive maximizing wavenumber exists in either of two growing cases: $c\ge 0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max (ac,-bc)$. In this range $C+dc>0$, andThus the stationary point is the unique maximum. The derivative condition yieldsSolving it, with the root that satisfies the unsquared derivative equation, givesFor $c=0$, the unequal-diffusivity expression simplifies to $z_{\ast }=C/[\sqrt{ab}(\sqrt{a}+\sqrt{b}{)}^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf{q}$ with this length is equivalent by rotational symmetry.

If $c<0$ but $0\le C\le -bc$, the maximum instead occurs at $z=0$: the fastest mode is homogeneous, with $s_{\ast }=-c$ and infinite wavelength. For $C>0$, its initial derivative is $-b+C/(-c)\le 0$. If $C+dc\ge 0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two eigenvalues are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_{+}$ over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.

The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell diffusion and removal of chemical perturbations. Its numerator $D_1{f}_0$ measures chemotaxis together with attractant production; its denominator $D_2\kappa$ measures dispersal together with incremental degradation. This competition between directed chemotaxis and cell diffusion produces spatial aggregation. The second ratio compares the concentration dependence of chemical production, $a_0{f}_0^{\prime }$, with incremental degradation $\kappa$. Positive $f_0^{\prime }$ amplifies chemical fluctuations, whereas negative $f_0^{\prime }$ suppresses them. It changes the chemical relaxation available to the chemotaxis feedback loop and can itself destabilize the homogeneous chemical field. Chemical diffusion $D_{\rho }$ suppresses short scales and sets the selected wavelength, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume $\kappa >0$.

For a nonzero-frequency Fourier mode proportional to $e^{i(kx+mz-\omega t)}$, the time-independent source has no oscillatory part. The pressure equation givesTake $k,m\ne 0$ and ${\omega }^2\ne f^2,N^2$ for the following pressure-amplitude formulas. With $P_0$ the amplitude of $p/{\rho }_0$, the momentum and buoyancy equations implyConsequentlybecause the numerator is $(k^2+m^2){\omega }^2-N^2{k}^2-f^2{m}^2$. Propagating linear inertia-gravity waves carry zero perturbation potential vorticity. In degenerate directions the division formulas must be replaced by the original linear equations, or interpreted by a regular limit; the zero-PV property of the nonzero-frequency wave component still follows from $q_t=0$. A steady balanced mode need not have zero potential vorticity.

Use $x^{\pm }=(x^0\pm x^{D-1})/\sqrt{2}$, transverse coordinates $x^i$, $i=1,\dots ,d=D-2$, and the Minkowski metricThe relativistic particle phase-space action becomesIn the light-cone gauge $x^{+}=t$, solve the mass-shell condition for $p_{+}$, assuming $p_{-}\ne 0$. The reduced phase-space action is $\int dt({\dot{x}}^{-}{p}_{-}+{\dot{x}}^i{p}_i-H)$, withThe last equality selects the future-directed momentum sector and makes positivity transparent. With $\hslash =1$ and $p_a=-i{\partial }_a$, the Schrodinger equation isThe inverse acts only on Fourier modes with nonzero $p_{-}$. Multiplication by $2p_{-}$ gives $2{\partial }_{-}{\partial }_{+}\Psi =({\Delta }_{\perp }-{\mu }^2)\Psi$. ThereforeThe light-cone Hamiltonian thus gives the same Klein-Gordon equation as covariant quantization.

For the massive two-form field, take $\mu \ne 0$. Apply ${\partial }^n$ to its field equation. Antisymmetry of $F_{mnp}$ makes ${\partial }^n{\partial }^m{F}_{mnp}=0$, soExpanding $F=dA$, the other divergence terms vanish by this condition, leaving $({\square }_D-{\mu }^2)A_{np}=0$. The light-cone decomposition of a massive two-form makes its dependent components explicit. The divergence equation isTaking $n=-$ and $n=i$, respectively, givesThe $n=+$ equation follows from these expressions: the two terms containing ${\partial }_{+}{\partial }_i{A}_{-i}$ cancel and ${\partial }_i{\partial }_j{A}_{ji}=0$. Consequently $A_{-i}$ and $A_{ij}$ are independent, each satisfying the Klein-Gordon equation with mass $\mu$. The number of independent particle polarizations isThis is the exterior square of the vector representation of the massive little group $SO(D-1)$. In the analogous Proca equation, ${\partial }^m{A}_m=0$ determines $A_{+}$ from $A_{-}$ and $A_i$, leaving $D-1$ components. A massive field has no gauge freedom that would justify setting these longitudinal components to zero. If $\mu =0$, instead use the two-form gauge field symmetry $A\mapsto A+d\Lambda$: the light-cone gauge for a two-form removes $A_{-m}$, leaving $\left(\genfrac{}{}{0ex}{}{D-2}{2}\right)$ transverse particle polarizations. The massive and massless counts are different.

In the closed-string mode expansion, $x^m,p_m$ are center-of-mass canonical variables, while ${\alpha }_k^i,{\tilde{\alpha }}_k^i$ are independent left- and right-moving transverse string oscillators. Their complex conjugates are ${\alpha }_{-k}^i,{\tilde{\alpha }}_{-k}^i$. The two zero-mode Lagrange multipliers impose the remaining mass-shell condition and closed-string level matching. The string level operators areTheir quantum definitions use normal ordering. The symplectic terms in the phase-space action givewith all brackets between distinct sectors zero. The nonzero-index string oscillators obey ${\alpha }_n^{i†}={\alpha }_{-n}^i$ and similarly for the right-moving sector. Define the momentum-labelled oscillator vacuum byFor $k>0$, $a_k^i={\alpha }_k^i/\sqrt{k}$ has $[a_k^i,a_l^{j†}]={\delta }_{kl}{\delta }^{ij}$. HenceStarting with $N|0;p⟩=0$, a finite product with $r_{ki}$ creation operators of mode $k$ has eigenvalue ${\sum }_{k,i}k{r}_{ki}$. The Fock space is generated by these products; both level operators have nonnegative integer eigenvalues. This establishes the integer string oscillator level property. Subtracting their physical zero-mode constraints enforces $N=\tilde{N}$.

There is a distinction between the displayed classical zero modes and their quantum constraints. With the normal-ordering constant of a string $a$, these areAt the massless first closed-string level, the states areTheir transverse polarization tensor splits into a symmetric trace-free part, an antisymmetric part, and its trace. These are the graviton, Kalb–Ramond field, and dilaton, with respective particle polarization counts $d(d+1)/2-1$, $d(d-1)/2$, and one. They have the transverse little group representations of massless particles. In a Lorentz-consistent bosonic string theory, the first chiral level is a massless vector, not a massive vector with one missing physical polarization; the closed-string products are therefore massless. This fixes $a=1$. Equivalently, regularized transverse zero-point energy gives $a=(D-2)/24$, and Lorentz consistency fixes the critical dimension of the bosonic string $D=26$.

It follows that the bosonic string mass spectrum isThe ground state has $M_0^2=-8\pi T$ and is a tachyon; level one is massless; for $N\ge 2$ the mass is $M_N=\sqrt{8\pi T(N-1)}$. The masslessness claim uses the consistent quantum theory, rather than an unshifted reading of the classical $L_0$.

A massive two-form at closed-string level two is present. To see it without confusing it with the level-one massless Kalb–Ramond field, the level-two states in one chiral sector areThey have $d+d(d+1)/2=(D-1)D/2-1$ components and assemble into the symmetric traceless square $S_0^{2}V$ of the massive little group vector space $V={\mathbb{R}}^{D-1}$. The full closed-string level is $S_0^{2}V\otimes S_0^{2}V$. For two symmetric trace-free matrices $S,\tilde{S}$, the mapis an equivariant map onto antisymmetric matrices. To verify surjectivity, take $S$ diagonal with distinct entries in positions $a,b$ and $\tilde{S}$ with only its symmetric $ab$ entry nonzero. Their commutator gives the $ab$ antisymmetric basis element. Finite-dimensional representations of the compact little group are completely reducible, so this quotient representation is also a subrepresentation. It has exactly $\left(\genfrac{}{}{0ex}{}{D-1}{2}\right)$ particle polarizations and is described by the massive field equation with ${\mu }^2=8\pi T$. At $D=26$ this gives 300 particle polarizations, consisting in light-cone coordinates of 24 components $A_{-i}$ and 276 components $A_{ij}$.

Let $d=\Delta x$, impose $u_0=u_{M+1}=0$, and write the method of lines system as $U^{\prime }=LU$, whereThus $D_2$ is a negative definite symmetric matrix and $D_1$ is a skew-symmetric matrix. Use the mesh-weighted Euclidean norm $\Vert U{\Vert }_d^2=d{\sum }_{m=1}^M|u_m{|}^2$. Discrete summation by parts yields the centered Dirichlet drift-diffusion energy identityConsequentlyThe same estimate controls perturbations and is uniform in the number of grid points and in the fixed drift coefficient. Finite-dimensional linear ODE theory guarantees existence, so this proves stability of a numerical method for the semidiscretization.

The factor $d^{1/2}$ simply rescales the vector norm and does not change the induced matrix norm. Equivalently the symmetric part is $(L+L^{\ast })/2=D_2$, whose largest eigenvalue is $-4d^{-2}{\sin }^2[\pi /(2(M+1))]<0$. This is the Euclidean logarithmic norm, rather than generally the spectral abscissa of a nonnormal matrix. No periodic Fourier mode assumption has been made: the zero endpoint terms are part of the proof. In particular positivity of both off-diagonal coefficients is not needed for this $L^2$ stability result.

Use the Fourier transform for the spatial Cauchy problem, or its periodic analogue, and regard the two starting levels as independently perturbed data. A Fourier mode with spatial factor $e^{im\theta }$ has amplification roots $G$ satisfyingPut $c=2\mu -1$ and $G=e^{i\theta /2}q$. ThenIf $|c|<1$, both roots have modulus one and their separation is bounded below uniformly in frequency:The uniform power bound from separated amplification roots now controls the two-level companion matrix for every time step. Its entries are uniformly bounded, and its eigenvector conditioning is bounded by the reciprocal root gap. The Parseval identity transfers this frequency-uniform bound to the spatial ${\ell }^2$ norm. This proves stability, rather than merely checking each root's modulus.

If $|c|>1$, the frequency $\theta =\pi$ has a root outside the unit disk, so there is exponential instability. If $|c|=1$, the two roots at $\theta =\pi$ coincide on the unit circle. The companion matrix is not a scalar matrix and has a nontrivial Jordan block; its powers grow linearly in the number of steps. Frequencies arbitrarily near that value produce the same lack of a uniform bound for localized Fourier packets, so this also invalidates Cauchy ${\ell }^2$ stability, not only periodic plane-wave stability. At $\mu =1$ the double amplification root is $-1$, and at $\mu =0$ it is $1$.

Therefore the full two-level stability range for a fixed positive Courant ratio isThe endpoint $\mu =0$ is moreover not a positive time step. Bounds deteriorate as $\mu$ approaches either endpoint; the displayed range is not a uniform claim over ratios arbitrarily close to one. A prescribed starter that removes one special parasitic component can change behavior for selected initial data, but does not establish the requested unrestricted two-level stability.

Take a long periodic section of length $L$, or neglect end effects and choose an integer number of wavelengths. The amplitude $u_q$ here is a single real sine amplitude, fixing the normalization of the thermal membrane undulation spectrum. For a lipid vesicle with an axisymmetric membrane deformation, conservation of the enclosed volume givesThe mean radius therefore decreases at second order. This adjustment is essential: fixing the mean radius instead of the volume would miss the unstable term in the fixed-volume capillary spectrum of a cylindrical membrane.

The surface area of an axisymmetric graph is $\mathcal{S}=2\pi \int R\sqrt{1+R_z^2}dz$. Expanding for $|u_q|\ll R_0$ and $q|u_q|\ll 1$ givesMultiplication by the membrane tension yields a quadratic potential energy $\Delta E=K_q{u}_q^2/2$, whereFor $q{R}_0>1$, the equipartition theorem gives the stable-mode fixed-volume capillary spectrum of a cylindrical membraneHere $k_B$ is the Boltzmann constant and $T$ is the temperature. The cosine amplitude has the same variance and is an independent real coordinate at quadratic order. If instead $R-{\rho }_0={\sum }_{q\ne 0}{a}_q{e}^{iqz}$ with $a_{-q}=a_q^{\ast }$, the corresponding positive-$q$ complex coefficient has $⟨|a_q{|}^2⟩=k_{B}T{R}_0/[2\pi \sigma L((q{R}_0{)}^2-1)]$. This is the same thermal membrane undulation spectrum in a different Fourier mode normalization.

For $q{R}_0<1$, the cylinder is unstable rather than having a negative fluctuation variance. The quadratic surface area change is negative: a sufficiently long-wavelength modulation reduces area while retaining volume. This is the Rayleigh–Plateau instability, expressed here as pearling of a tension-dominated lipid vesicle. A cylinder supporting such modes has no unconstrained harmonic thermal equilibrium about the uniform state. The $q=0$ uniform radius change is forbidden by fixed volume, and $q{R}_0=1$ is marginal in the tension-only approximation. The finite length and endpoint constraints determine which nonzero Fourier modes are allowed.

The large membrane tension assumption controls modes with $q{R}_0$ of order one. For large $q$ the omitted membrane bending modulus contribution grows approximately as $\kappa q^4$ and overtakes the tension term when $\kappa q^2\gtrsim \sigma$. Thus the displayed tension-only spectrum requires $q\ll \sqrt{\sigma /\kappa }$ as well as small amplitudes and a positive stiffness. Close enough to the marginal wavelength, bending corrections must also be retained; their relative scale at $q{R}_0$ of order one is $\kappa /(\sigma R_0^2)$.

For the circular phase boundary in a lipid bilayer, the energy is its perimeter times the line tension $\gamma$. Let $\phi$ be the polar angle and write $r(\phi )={\rho }_0+u_m\sin (m\phi )$, where the periodicity requires an integer $m=q{R}_0\ge 1$. Fixed enclosed area impliesThe perimeter, including the mean-radius change, isThus the fixed-area capillary spectrum of a circular boundary, again for a single real sine or cosine amplitude, isThe complex Fourier mode coefficient convention again divides this result by two. The mode $m=0$ is excluded by the fixed-area constraint, so the negative-stiffness interval $0<m<1$ is not an allowed mode of a closed circular boundary.

At $q{R}_0=1$, the zero stiffness is the translation mode of a circular boundary, not a shape instability. Displacing the centre by a small distance $h$ changes the radius to first order by $h\cos \phi$ or $h\sin \phi$ while leaving area and perimeter unchanged. For example, an exactly translated circle hasThe higher harmonics complete a true translation mode of a circular boundary. The $m=1$ denominator therefore represents free centre motion: an unconfined domain's centre has no restoring force or finite equilibrium variance in an infinite membrane. Radius fluctuations about a recentered domain omit this mode; external confinement would give it a separate restoring stiffness.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.

The Fourier coefficient of the velocity-integrated density is its mixed transform at zero velocity frequency. ThusThe constant ${\rho }_{\infty }$ contributes only to the zero Fourier mode, where it equals ${\hat{f}}_0(0,0)$. ThereforeIt is important to remove the zero mode, which is conserved rather than mixed away.

Take the positive integer $n\ge 1$ and put $m=n+1\ge 2$. From the preceding estimate and the absent zero Fourier mode, for $t\ne 0$,Here $\zeta$ is the Riemann zeta function, and its displayed series is finite because $m>1$. One may replace the last derivative norm by the given full mixed Sobolev norm to obtain the requested constant depending only on $n$ and $f_0$.