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Expand $a,u,f$ in the two-dimensional Fourier basis. Applying the derivative and product rules from part (a), the $(m,n)$ coefficient of the equation isThe normalization and compatibility conditions areThus, on ${\mathbb{Z}}^2\setminus \{(0,0)\}$, the infinite Fourier spectral method system isFor a square spectral truncation, chooseretain equations and unknowns indexed by ${\Lambda }_N$, and set omitted coefficients to zero. This is the Fourier–Galerkin method: it projects the residual onto the retained Fourier modes. Other finite mode sets may be used in the same way.

At leading order in $|k\tau |$, the inflationary Fourier mode and its derivative areThus the leading parts of $\hat{f}$ and $\hat{\pi }$ contain the same quadrature of the creation and annihilation operators and satisfyTheir commutator consequently vanishes in this leading approximation. The exact nonzero result in part b is carried by the subleading decaying mode. Relative to the rapidly growing anticommutator, its effect is suppressed by powers of $|k\tau |$, which is the squeezing captured by the classicality parameter of a cosmological perturbation. The perturbation can therefore be treated as an effectively classical stochastic field on superhorizon scales even though its exact quantum commutator is unchanged.

For the Cauchy problem, insert the Fourier mode $u_m=e^{im\vartheta }$. The spatial symbol isIts real part is nonpositive for every $\vartheta$, so each mode has modulus $e^{t\mathrm{Re}{\lambda }_h}\le 1$. By the discrete Fourier transform and Parseval identity, the scheme is stable in the discrete $L^2$ norm.

For the discrete Fourier mode $u_m^n=G^n{e}^{im\theta }$, the centred second difference has multiplier $-4{\sin }^2(\theta /2)$. Hence the amplification factor isFor every $\mu >0$, $|G(\theta )|\le 1$. The discrete Parseval identity then shows directly that the discrete $L^2$ norm cannot increase at any time step, proving unconditional stability for the Cauchy problem.

In the razor-thin disk approximation, the three-dimensional Poisson equation isFor a horizontal Fourier mode with wavevector $\mathbf{k}$ and $k=|\mathbf{k}|>0$, the Fourier transform of this equation isThe solution that decays away from the disk isConsequently the required razor-thin disk Poisson kernel in the midplane isThe spatially uniform $k=0$ background is excluded from this local perturbation formula.

Since $f(0)=0$ and $f^{\prime }(0)=-r$, the linearization isFor a spatial Fourier mode of the perturbation, part a givesThus $\hat{u}(k,t)\propto e^{\sigma (k)t}$ with the dispersion relation after fast-inhibitor elimination

The linearized equations are homogeneous and have constant coefficients in $x$ and $t$. Consequently, spatial Fourier modes evolve independently, and the superposition principle reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the normal modesHere the real $k$ is the wavenumber, $\sigma$ is the possibly complex temporal exponent, and taking the real part produces a physical displacement.

The axisymmetric razor-thin astrophysical disk equations areFor the stationary background, mass and azimuthal momentum conservation are automatic, while radial force balance and the Poisson equation giveHere $d{p}_0/dR=0$ under the stated constant-pressure assumption.

Retaining first-order perturbations and using $c_s^2=dp/d\Sigma$ giveswhereis the squared radial epicyclic frequency. In the local $|k|R\gg 1$ limit, a Fourier mode $e^{i(kR-\omega t)}$ obeysEliminating ${\Sigma }^{\prime }$, $u_{\varphi }^{\prime }$, and ${\Phi }^{\prime }$ yields the local dispersion relation

For a Keplerian orbit, $\Omega \propto R^{-3/2}$, soInstability requires ${\omega }^2<0$, equivalently a mode with positive imaginary frequency. Treating the right-hand side as a quadratic in $x=|k|$, its two roots areReal distinct roots, and hence an unstable interval $x_{-}<|k|<x_{+}$, exist exactly when the Toomre stability criterion has $Q<1$.

With $h=c_s/\Omega$,The constant positive term is epicyclic restoration by rotation and stabilizes long wavelengths. The negative term is the razor-thin disk's self-gravity and drives collapse. The positive $k^2$ term is gas-pressure restoration and stabilizes short wavelengths. Gravitational instability can therefore survive only on an intermediate band of scales.

Write the reaction terms asA nonzero homogeneous equilibrium satisfies $v_{\ast }=\beta u_{\ast }$ and $1+u_{\ast }-\gamma v_{\ast }=0$, henceFor physically positive populations it exists exactly when

The reaction Jacobian matrix at this equilibrium isIts trace and determinant areThe equilibrium is therefore stable to spatially uniform perturbations when

For a spatial Fourier mode of wavenumber $k$, put $q=k^2$. The linearized reaction-diffusion system has matrixIts trace is smaller than $\mathrm{tr}J$, whileThe two-species diffusion-driven instability criterion says that this upward-opening quadratic becomes negative for some $q>0$ precisely whenCombining all conditions, a Turing instability may occur in the region

At onset the discriminant vanishes, and the double root isThe threshold relation givesso the critical wavenumber isIf $d=1$, uniform stability requires $\beta >1$ whereas diffusion-driven instability requires $d>\beta$. These inequalities are incompatible, so the Turing region vanishes.

For a barotropic equation of state, $\delta P=c_s^2\delta \rho =c_s^2\bar{\rho }\delta$. A spatial Fourier mode of comoving wavenumber $k$ therefore satisfiesThe Jeans wavenumber is defined by equality of the pressure and self-gravity terms:For $k\gg k_J$, pressure dominates and produces acoustic oscillations whose amplitude is affected by Hubble friction. For $k\ll k_J$, self-gravity dominates and a growing mode develops: this is Jeans instability.

For $\mathbf{u}=\nabla \times (\psi \hat{\mathbf{y}})$,Hence $u_x+w_z=-{\psi }_{zx}+{\psi }_{xz}=0$, so mass conservation holds identically. Removing the hydrostatic part by writingand taking the curl of the Stokes flow equation gives the biharmonic equation

For a Fourier mode proportional to $e^{ikx}$ with $k>0$, decay as $z\to -\infty$ selectsThe velocity and pressure fields are thereforeDirect substitution verifies the momentum equation.

The magnitude of the phase velocity of a real-frequency plane Fourier mode is $|\omega |/|\mathbf{k}|$. In an anisotropic medium it depends on the direction of the wave vector and need not equal the group velocity magnitude.

In a weakly nonlinear expansion of the long-wave convection equation with broken Boussinesq symmetry at $\mu =2$, a critical Fourier mode ${\Theta }_1=A{e}^{ix}+c.c.$ generates ${\Theta }_2=(2is/9)A^2{e}^{2ix}+c.c.$. The quadratic interaction of these first and second harmonics feeds back $8s^2|A{|}^{2}A/9$ into the critical Fourier mode, while the cubic gradient nonlinearity contributes $-3|A{|}^{2}A$. The Fredholm solvability condition is consequently ${\mu }_{2}A-g|A{|}^{2}A=0$. The sign of $g$ distinguishes supercritical and subcritical branches; $g=0$ requires a higher-order amplitude equation.

A space-time average over periodic modes integrates over the spatial periodic cell and one temporal period. For real spatial Fourier modes sharing a wavevector, $⟨\mathrm{Re}(\mathbf{v}{e}^{i\mathbf{k}\cdot \mathbf{x}})\times \mathrm{Re}(\mathbf{w}{e}^{i\mathbf{k}\cdot \mathbf{x}}){⟩}_x=\mathrm{Re}({\mathbf{v}}^{\ast }\times \mathbf{w})/2$. A matching sine or cosine temporal factor supplies another $1/2$. Opposite wavevectors can have nonzero spatial cross averages, so their temporal coefficients must be checked rather than discarded merely because the wavevectors are distinct.

A steady spatial forcing at twice the critical wavenumber permits coupling of the negative critical Fourier mode to the positive critical mode. A weakly nonlinear expansion then permits a term $\beta \overline{A}$ in the Landau amplitude equation. Its coefficient is found by projecting the resonant forcing-advection terms onto the adjoint eigenfunction. Reflection-symmetric forcing permits real coefficients. The symmetry permission does not prove a nonzero coefficient: it can vanish for a particular model or mode structure.

A primordial bispectrum is squeezed when one external Fourier mode has much smaller wavenumber than the other two. Momentum conservation then makes the two short wavevectors almost opposite. This configuration measures the response of short-scale correlations to a long perturbation.

For a nonzero Fourier mode satisfying $d{h}_q=-\alpha q^2{h}_{q}dt+\sigma d{W}_q$ with $\alpha >0$, the explicit Ornstein-Uhlenbeck solution and Itô isometry give the stationary second moment ${\sigma }^2/(2\alpha q^2)$. The prefactor depends on the normalization of the mode noise, while the inverse-square dependence follows from diffusive relaxation.

For a nonzero horizontal Fourier mode $k$, solve $(D^2-k^2{)}^{2}w=R{k}^2\theta$, $w=w^{\prime \prime }=0$, to obtain $w={\mathcal{W}}_R\theta$. Under homogeneous temperature Dirichlet boundary conditions, the sine modes diagonalize this self-adjoint operator, with eigenvalues $R{k}^2/(k^2+n^2{\pi }^2{)}^2$. The temperature linear operator is ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$. At a marginal mode $g=\sin (n\pi z)$ it has ${\mathcal{W}}_{R}g=(k^2+n^2{\pi }^2)g$.

With thermal diffusivity $\xi$, a Fourier mode of wavenumber $k$ decays by diffusion at this rate.