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For nonzero real wavenumber $k$, seek a Fourier mode of the Newtonian gravitational potential in the form ${\Phi }^{\prime }=f(z)e^{ikx}$. Away from the razor-thin sheet, the Poisson equation for Newtonian gravity becomes $f^{\prime \prime }-k^{2}f=0$. Requiring the perturbation to decay on both sides and remain continuous gives $f=A{e}^{-|k||z|}$.

Integrating the Poisson equation for Newtonian gravity through $z=0$ fixes the derivative jump:Thus the perturbing gravitational potential isAt the midplane this reduces to the razor-thin disk Poisson kernel. The observable real disturbance is the real part. Writing the perturbation amplitude as ${\Sigma }_a$ instead of ${\Sigma }_0$ gives the same formula with ${\Sigma }_a$; linearity makes it independent of the background density. The $k=0$ disturbance is a uniformly changed sheet and has potential proportional to $|z|$, rather than a decaying nonzero-wavenumber solution.

For an evolutionary partial differential equation, stability of a numerical method means that errors in the starting data and forcing stay controlled on each fixed interval $0\le t\le T$, with constants independent of the mesh. Fourier stability analysis is especially effective for a constant-coefficient finite difference method on a uniform infinite or periodic grid: translation invariance makes different Fourier modes evolve independently.

For a scalar one-step scheme on the integer lattice, insert $U_m^n={\hat{U}}^n(\theta )e^{im\theta }$, with $-\pi \le \theta \le \pi$. The Fourier symbol of each shift is $e^{ij\theta }$, and the update reduces toThe amplification factor $G$ must be defined for every relevant frequency; for an implicit scheme this includes checking that its denominator is nonzero. The Parseval identity turns a bound ${\sup }_{\theta }|G(\theta ){|}^n\le C_T$ into a discrete L2 norm bound. Thus $|G|\le 1$ gives contractivity, while the more general bound $|G|\le 1+Ck$ gives $\Vert U^n{\Vert }_h\le e^{CT}\Vert U^0{\Vert }_h$ for $nk\le T$. Conversely, frequencies with amplification uniformly greater than one produce unstable wave packets or periodic Fourier modes under refinement.

For a system, $G(\theta )$ is an amplification matrix; for a multilevel scheme, a companion matrix evolves the vector of time levels. The actual requirement is a uniform bound on matrix powers, not just their spectral radius. A nontrivial Jordan block at a unit-modulus eigenvalue creates polynomial growth in the time index. Even simple eigenvalues inside the unit disk can fail to give a uniform bound if the eigenvector matrices become ill-conditioned as the mesh changes. Uniformly controlled diagonalization of a matrix, or a suitable quadratic energy estimate, resolves this issue. The power boundedness of a two-level Fourier scheme illustrates why root multiplicities and conditioning matter.

For the heat equation, put $\mu =k/h_x^2$. Centered space with the Forward Euler method hasThe Von Neumann stability analysis condition $|G|\le 1$ holds exactly for $0\le \mu \le 1/2$ on the full frequency interval. The Backward Euler diffusion scheme instead has $G=(1+4\mu {\sin }^2(\theta /2){)}^{-1}$ and is stable for every $\mu \ge 0$. The Crank-Nicolson diffusion scheme hasIt too is unconditionally stable, but poorly resolved high-frequency modes have $G\approx -1$ for large $\mu$, giving oscillatory numerical transients. A-stability therefore does not guarantee strong damping; L-stability distinguishes the damping of the Backward Euler method.

For the advection equation $u_t+c{u}_x=0$, let $\nu =ck/h_x$. Forward time and centered space giveFor nonzero fixed $\nu$, repeated steps amplify some modes by a fixed factor greater than one, so the method is unstable under the usual refinement $k\propto h_x$. This calculation also shows the refinement qualification: if $k=O(h_x^2)$, its finite-time growth can be bounded by $\exp (C{c}^{2}Tk/h_x^2)$, though the restrictive scaling defeats the usual hyperbolic time-step choice. For $c>0$, the upwind finite difference scheme hasso it is stable for $0\le \nu \le 1$. The Lax-Wendroff advection scheme instead hasgiving $|\nu |\le 1$. These examples separate the effects of the spatial stencil, temporal approximation and Courant number; a higher order of a numerical method by itself does not establish stability.

A periodic boundary condition is ideal for this analysis. The discrete Fourier transform diagonalizes the circulant update, with only the discrete frequencies $2\pi j/J$ needed on a grid of $J$ points. A bound on all $\theta$ is a convenient guarantee over every such grid. On the whole line, the Fourier transform uses a continuous frequency interval and the Parseval identity proves the corresponding square-summable-data estimate.

Homogeneous Dirichlet boundary conditions do not admit arbitrary complex exponential Fourier modes. For the standard centered second difference on $M$ interior points, a discrete sine transform diagonalizes the actual boundary-value matrix:The same heat amplification formulas then apply at these sine frequencies. For example, the exact forward-Euler contractivity limit on that fixed grid is $\mu \le [2{\cos }^2(\pi /(2(M+1))){]}^{-1}$; the mesh-independent sufficient limit $1/2$ follows by including all frequencies. Homogeneous Neumann boundary conditions can similarly permit a discrete cosine transform, provided the endpoint discretization and its weighted inner product are chosen consistently. The constant mode is then present, reflecting conservation of the heat equation's spatial mean. For inhomogeneous boundary conditions, subtract a suitable lifting of the prescribed boundary data and estimate the induced source using the homogeneous evolution's bound and a Duhamel principle estimate. Bounds for the lifting and source must themselves be uniform in the mesh.

General boundary conditions require separate boundary stability of a finite-difference method. For an advection equation, incoming data are prescribed at the inflow boundary, while an outflow closure must respect the outgoing characteristics. An interior periodic Fourier symbol cannot detect a growing mode confined near a boundary. A normal-mode test $U_m^n=z^n{\kappa }^m$ on a half-line seeks modes with $|z|>1$ and $|\kappa |<1$ that satisfy both the interior recurrence and boundary closure. Their existence proves instability, and uniform control requires more than merely excluding isolated growing roots; the Uniform Kreiss--Lopatinskii condition addresses boundary resolvent bounds. Alternatively, a direct energy method using summation by parts can include the boundary terms and establish the required estimate.

Variable coefficients and nonuniform meshes usually destroy exact Fourier transform diagonalization. Frozen-coefficient Fourier stability analysis is then a useful diagnostic, but it is not automatically a proof for the full variable-coefficient boundary problem. The mesh-uniform energy method in question 5 is an example of a direct proof. Also, an L2 norm proof is not automatically a mesh-uniform maximum-norm proof: finite-dimensional norm-equivalence constants may grow with the number of grid points.

Finally, stability measures error propagation; consistency of a numerical method measures the defect of inserting the exact solution. For a well-posed linear initial-value problem in the chosen norm, the Lax equivalence theorem says that a consistent approximation is convergent exactly when it is stable. Boundary discretization and starting data must be included in that assertion. A Fourier calculation proves convergence only after consistency, well-posedness and the actual boundary treatment have also been checked.

Use the Fourier coefficientsThey are continuous for $0\le y\le 1$ and bounded by $2\Vert u{\Vert }_{\infty }$. On compact subintervals of $0<y<1$, the equation is uniformly elliptic with smooth coefficients. Local elliptic regularity at the flat vertical sides with zero boundary values permits differentiation and integration by parts there. ThusFor completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\epsilon ,\pi -\epsilon )$ against $\sin (nx)$ and a test function of $y$ with support away from zero and one. The interior gradient estimate near a zero Dirichlet side gives $|u_x|\le C{\epsilon }^{-1}\sup |u|$ on a nearby ball of radius comparable to $\epsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin (nx)$ and $u\cos (nx)$ tend to zero. Move the $y$ derivatives to the test function before taking $\epsilon \downarrow 0$. This proves the ordinary differential equation, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.

The Cauchy-Euler differential equation has indicial rootsFor $n\ge 1$ the negative root is strictly negative. Boundedness as $y\downarrow 0$ forces $B_n=0$. Continuity at $y=1$ identifiesBecause $p_n^{+}>n$, the series ${\sum }_{n\ge 1}{A}_n{y}^{p_n^{+}}\sin (nx)$ converges absolutely and uniformly for $0\le y\le \rho <1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its Fourier coefficients equal those of $u(\cdot ,y)$; completeness of the Fourier sine basis in $L^2(0,\pi )$ and continuity in $x$ make the two functions equal. ThereforeThis is also the requested sum over integers: take all coefficients with $n\le 0$ to be zero. Negative indices give redundant Fourier modes, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary Fourier series on the top boundary is needed.

An important consequence is a forced zero trace at a quadratically degenerate boundary. Indeed the absolute sum is bounded by $2\Vert u{\Vert }_{\infty }{\sum }_{n\ge 1}{y}^n=2\Vert u{\Vert }_{\infty }y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow 0$, uniformly in $x$. Thus $u(x,0)=0$ on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.

When $u_0=0$, the field theory is a Gaussian field theory, so integrating out short-wavelength modes can be done exactly rather than merely at a saddle. Introduce an ultraviolet momentum cutoff $\Lambda$ and a scale factor $b>1$, then split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into $|q|<\Lambda /b$ and $\Lambda /b<|q|<\Lambda$. Orthogonality of Fourier modes eliminates cross terms in the quadratic Hamiltonian. A constant source couples only to the zero-momentum mode and therefore does not couple to ${\varphi }_{>}$.

Before rescaling, the Gaussian functional integral factorizes:where $H_{<}$ has the original $r_0$ and $h$, restricted to low momenta. Up to a field-independent normalization, the eliminated modes contributeThe determinant affects the additive free energy, but generates neither a mass correction nor a source correction, since no interaction mixes low and high modes. This remains true beyond the Landau approximation: it is an exact integration of a Gaussian field theory, not neglect of the high-mode fluctuations.

To restore the original cutoff, use $x=b{x}^{\prime }$, $q^{\prime }=bq$, and choose the field rescalingSubstitution into the low-mode Hamiltonian gives the Gaussian momentum-shell scalingThe gradient coefficient is unchanged: the volume factor is $b^D$, the two derivatives contribute $b^{-2}$, and the two fields contribute $b^{2-D}$. The mass term gains $b^2$, while the source term gains $b^{(D+2)/2}$. ThusFor a nonconstant source the corresponding formula is $h^{\prime }(x^{\prime })=b^{(D+2)/2}h(b{x}^{\prime })$, after restricting its coupling to retained modes. The constant-source case avoids that extra source filtering.

Writing $b=e^{\ell }$, the exact Gaussian renormalization-group flow hasBoth perturbations are relevant at the massless zero-source Gaussian fixed point, with $y_t=2$ and $y_h=(D+2)/2$. There is no anomalous field rescaling here, so $\eta =0$ and the correlation-length critical exponent is again $\nu =1/y_t=1/2$. The coefficient changes arise from rescaling, not from an interaction-induced shift in $r_0$ during shell integration.

A stable real Gaussian functional integral requires $r_0>0$, or an infrared prescription that treats the zero mode at the massless point. With $u_0=0$ and $r_0<0$, the field energy is unbounded below; the Gaussian field theory alone cannot describe a stable ordered phase. The scaling laws are therefore interpreted about the Gaussian fixed point from the stable side, with finite-volume zero-mode regularization when needed.

Under the first-order smoothing approximation, neglect the fluctuating product in the resistive induction equation while retaining its contribution to the mean electromotive force. Since the test field ${\mathbf{B}}_0$ is constant and the velocity is a solenoidal vector field, the fluctuation equation isLet $a=\eta k^2>0$, $d=a^2+{\Omega }^2$, and ${\mathbf{C}}_j=i({\mathbf{B}}_0\cdot {\mathbf{k}}_j){\hat{\mathbf{u}}}_j$. For each Fourier mode, the diffusion equation becomes a scalar linear relaxation equation applied to each vector component. Take its long-time periodic response, with $\Omega \ne 0$. For the first mode,while for the second,Thus the complex response coefficients areThey giveAn arbitrary initial fluctuation also contains a homogeneous diffusive transient. In the forced modes this is $\mathrm{Re}{\sum }_j{\mathbf{r}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}-at}$; more general initial modes also diffuse. The displayed harmonic expression is the periodic particular solution after those transients, not the most general solution at finite time. The constants $a,d$ and the factors of $i$ are unchanged by taking the final real part.

To justify the helicity relation, write ${\hat{\mathbf{u}}}_j={\mathbf{a}}_j+i{\mathbf{c}}_j$ with real vectors. Solenoidality gives ${\mathbf{k}}_j\cdot {\mathbf{a}}_j={\mathbf{k}}_j\cdot {\mathbf{c}}_j=0$. Their cross product is parallel to ${\mathbf{k}}_j$, andThis is the helicity vector of a solenoidal Fourier mode. In these conventions the spatially averaged kinetic helicity density of $\mathrm{Re}({\hat{\mathbf{u}}}_j{e}^{i{\mathbf{k}}_j\cdot \mathbf{x}})$ is $-H_j{k}^2$; the sign is set by the specified cross-product order.

For equal wavevectors, spatial averaging of two real harmonic fields givesTime averaging gives another factor $1/2$, because $⟨{\cos }^2\Omega t{⟩}_t=⟨{\sin }^2\Omega t{⟩}_t=1/2$ and the mixed temporal average is zero. Consequently each diagonal mode contributes to the mean-field electromotive forceDistinct modes with ${\mathbf{k}}_1\ne \pm {\mathbf{k}}_2$ have no spatially averaged cross term. The printed condition also permits the opposite-wavevector case. The possible cross terms still cancel after time averaging: the out-of-phase response coefficients are ${\mathbf{p}}_2=-\Omega {\mathbf{C}}_2/d$ and ${\mathbf{q}}_1=\Omega {\mathbf{C}}_1/d$, while integration by parts gives $⟨{\mathbf{u}}_1\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_2{⟩}_x=⟨{\mathbf{u}}_2\times ({\mathbf{B}}_0\cdot \nabla ){\mathbf{u}}_1{⟩}_x$ for the real spatial fields. This handles all the distinct wavevectors allowed by the question, assuming the usual periodic-cell or whole-space spatial average.

Thus the alpha tensor isBoth averaging factors matter. The alpha tensor is real and symmetric, and this contribution vanishes when the mode kinetic helicity densities vanish.

For the two perpendicular wavevectors and common $H_1=H_2=H$, put ${\alpha }_0=\eta H{k}^4/[2({\Omega }^2+{\eta }^2{k}^4)]$. ThenMean-field dynamo action is possible when ${\alpha }_0\ne 0$ and sufficiently long mean-field wavelengths are allowed. The missing $zz$ entry does not prevent an alpha-squared dynamo. Indeed a slowly varying mean magnetic field transverse to $z$, with dependence $e^{iKz+\lambda t}$, obeysThe two growth rates are $-\eta K^2\pm {\alpha }_{0}K$, so the larger is positive if $0<|K|<|{\alpha }_0|/\eta$. This supplies a brief constructive reason; scale separation also requires $|K|\ll k$. A spatially uniform test field itself has zero curl of its mean-field electromotive force and does not grow. If $H=0$, or if boundary conditions exclude all unstable long wavelengths, this particular alpha effect does not yield growth. The conclusion concerns the stated first-order smoothing approximation; no uncomputed turbulent-diffusion correction is assumed.

Write $W(\Sigma )=\bar{\nu }\Sigma$. In a small radial patch $r=r_c+\chi$ with disturbance scale much smaller than $r_c$, the leading disk diffusion equation is ${\partial }_t\Sigma \simeq 3{\partial }_{\chi }^{2}W$. Taking the linearization around a uniform surface density ${\Sigma }_c$ givesFor a local Fourier mode $\delta \Sigma \propto e^{st+ik\chi }$,because $W\propto {\Sigma }^{-1}$ has $W^{\prime }<0$. Thus the effective density diffusion coefficient is negative and the disk has a viscous instability of an accretion disk.

A localized overdensity is a maximum of $\Sigma$ but a minimum of $W$. At its center ${\partial }_{\chi }^{2}W>0$, so ${\partial }_t\Sigma >0$: it grows. The local radial mass flux is proportional to $-{\partial }_{\chi }W$, directing matter toward that minimum from both sides. This is the physical feedback: an overdense patch has a smaller viscous torque in an accretion disk and attracts further accumulation. The arbitrarily rapid $k\to \infty$ growth in this ideal negative-diffusion model is not a resolved prediction below the validity scale of the thin-disk and local viscous closure.

Substitute the axisymmetric Fourier mode into the linearized equations. With hats denoting amplitudes,For a nondegenerate nonzero frequency, eliminate the azimuthal velocity and useContinuity then givesEquivalently, taking the determinant of the original four amplitude equations avoids division by zero and gives the full dispersion relationThe two squared frequencies areThe discriminant is $({\Omega }^2-c^2{k}^2{)}^2+4{\Omega }^2{c}^2{k}_x^2\ge 0$. For $c^2>0$, both squared frequencies are nonnegative. The high-frequency branch is predominantly acoustic and the low-frequency branch is an inertial wave. Zero-frequency limiting modes, including $k_z=0$, are retained by the determinant calculation.

Let ${\mathbf{d}}_{jn}=h(j-n){\mathbf{e}}_y$ and $\Delta \mathbf{\xi }={\mathbf{\xi }}_j-{\mathbf{\xi }}_n$, with ${\mathbf{\xi }}_n=(x_n^{\prime },y_n^{\prime })$. Linearizing the Newton's law of universal gravitation usesFor an azimuthal separation, the matrix is $\mathrm{diag}(1,-2)$. After canceling the equilibrium forces, the linearized shearing sheet equations areThe transverse gravitational response tends to restore a radial displacement, while the longitudinal response reinforces an azimuthal density concentration. Both sums converge for the bounded Fourier modes considered next.

Set $\xi =hk$. For the discrete Fourier mode, the displacement difference in the $l$th neighbor pair carries $e^{il\xi }-1$, andThus the infinite system reduces toThe gravitating lattice Fourier kernel $F$ is nonnegative, even and $2\pi$-periodic, with $F(0)=0$. The uniform convergence and absolute convergence of its differentiated series givesTo prove the maximum, it is not sufficient to maximize each cosine term separately: the even terms do not reach their individual maxima at $\pi$. Instead use the convergent integral representation $l^{-2}={\int }_0^{\infty }t{e}^{-lt}dt$. Summing the geometric series for sines givesFor $0<\xi <\pi$, the denominator is positive and $\sin \xi >0$, so $F^{\prime }>0$. For $\pi <\xi <2\pi$, $F^{\prime }<0$. The interchange is justified by ${\sum }_l{\int }_0^{\infty }t{e}^{-lt}|\sin (l\xi )|dt\le {\sum }_l{l}^{-2}<\infty$. HenceTheir value iswhere $\zeta$ is the Riemann zeta function. This maximizing Fourier mode alternates the signs of neighboring particle displacements.

A direct elimination of pressure gives the same result as taking a double curl of the momentum equation. Incompressibility gives ${\nabla }^{2}p=\mathrm{Ra}\mathrm{Pr}{\partial }_z\theta$. Apply ${\nabla }^2$ to the vertical momentum equation and subtract this pressure derivative to obtainHere ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$ is the horizontal Laplacian. At $z=0,1$, impermeability gives $w=0$ and fixed temperature gives $\theta =0$. The stress-free boundary condition gives ${\partial }_{z}u={\partial }_{z}v=0$, since horizontal derivatives of the boundary value $w=0$ vanish. Differentiating incompressibility once in $z$ therefore givesFor a horizontal Fourier mode $X=e^{i(k_{x}x+k_{y}y)}$, ${\nabla }_h^{2}X=-{\lambda }^{2}X$ with real $\lambda =\sqrt{k_x^2+k_y^2}$. More generally the nonnegative spectrum of $-{\nabla }_h^2$ follows by integration by parts for periodic or square-integrable horizontal disturbances. Substituting the separated normal mode gives, with $D=d/dz$,The boundary conditions admit a complete Fourier sine series in $z$. For a nonzero horizontal wavenumber, take $W=W_n\sin (n\pi z)$, $\Theta ={\Theta }_n\sin (n\pi z)$, and set $a_n^2={\lambda }^2+n^2{\pi }^2$. The two amplitude equations areThus the stress-free convection growth-rate polynomial isIts discriminant is $(\mathrm{Pr}-1{)}^2{a}_n^4+4\mathrm{Ra}\mathrm{Pr}{\lambda }^2/a_n^2$, so both growth rates are real for $\mathrm{Ra}>0$. For $\mathrm{Ra}<0$, its two coefficients after the leading term are positive: their sum gives ${\sigma }_1+{\sigma }_2=-(1+\mathrm{Pr})a_n^2<0$, and their product is $\mathrm{Pr}(a_n^4-\mathrm{Ra}{\lambda }^2/a_n^2)>0$. Real roots are both negative; nonreal roots have real part $-(1+\mathrm{Pr})a_n^2/2<0$. Hence stable heating from above damps every mode. Horizontally uniform thermal modes and uncoupled horizontal viscous modes are also purely damped; impermeability and incompressibility exclude a horizontally uniform nonzero $w$.

Because the unstable-side roots are real, the onset is stationary, an exchange of stabilities in stress-free convection. Setting $\sigma =0$ gives the free-slip convection neutral curveWith $x={\lambda }^2$, its derivative has the sign of $2x-n^2{\pi }^2$. Its minimum is at $x=n^2{\pi }^2/2$, with value $27n^4{\pi }^4/4$. Minimizing also over $n\ge 1$ gives the critical values

In the linear complex Ginzburg-Landau equation, take $U,c_d,\mu$ real. $U$ gives advection, carrying the envelope downstream at speed $U$. $c_d$ gives wave dispersion: the imaginary part of the second-derivative coefficient changes wave phase without directly dissipating a real-wavenumber Fourier mode. $\mu$ gives local linear production or damping: a spatially uniform perturbation grows for $\mu >0$ and decays for $\mu <0$.

The real part of the diffusion coefficient is the scaled value one. It smooths the envelope and damps a real-wavenumber Fourier mode at rate $k^2$. Thus the linear complex Ginzburg-Landau equation models competing advection, smoothing through a diffusion equation, wave dispersion and local growth, rather than nonlinear saturation.

Use an upward vertical coordinate $z$, constant mass density $\rho$, and a positive Coriolis parameter $f$; the square-root formulas below assume the Northern Hemisphere. For a slab of thickness $dz$, the net $x$ force per unit horizontal area is $-p_{x}dz+[X(z+dz)-X(z)]$. Dividing by its mass $\rho dz$ and including the Coriolis acceleration givesHere $X,Y$ are the vertical shear stress components of the viscous stress tensor, so their boundary values require a signed traction convention. Linearization removes advective acceleration. Decompose the horizontal velocity into an exterior pressure response and an Ekman layer correction. On an f-plane these satisfyandThe exterior pressure response is in geostrophic balance when steady. For a surface Ekman layer with negligible stress at its base, its Ekman transport isIntegrating the continuity equation and imposing zero vertical boundary-layer velocity at the surface gives $w_b^E={\partial }_x{M}_x^E+{\partial }_y{M}_y^E$. Thus the base velocity, positive upward, isThe PDF prints the opposite sign. Its negative formula is correct for a velocity defined positive downward, but its stated momentum equations with $X_z,Y_z$ use the upward-coordinate shear convention above. This sign must be accounted for when interpreting the subsequent pumping equation.

For constant kinematic viscosity, write $W_E=u^E+i{v}^E$ and $\delta =\sqrt{2\nu /f}$. The steady Ekman layer equation is $\nu W_E^{\prime \prime }=if{W}_E$. Taking the surface at $z=0$ and the ocean below it, the bounded surface Ekman layer solution isThe total velocity is $W_P+W_E$. Prescribed wind stress determines the coefficient of this Ekman layer correction; it does not determine a relation between the wind and the independent pressure-driven velocity. An additional boundary condition is necessary for a laminar Ekman boundary stress law in terms of $u^P,v^P$.

The two printed laminar stress formulas and the positive pressure-Laplacian pumping formula are instead the standard Bottom Ekman layer relations. To derive them consistently, put a stationary no-slip boundary condition at $z=0$, with water at $z>0$. ThenConsequentlyThese are shear stress values; the actual bottom traction on the fluid has the opposite sign. Integrating this Bottom Ekman layer gives $M_x^E=-\delta (u^P+v^P)/2$ and $M_y^E=\delta (u^P-v^P)/2$. Since the exterior geostrophic flow is horizontally nondivergent, the upward velocity above the bottom is $-{\nabla }_h\cdot {\mathbf{M}}_E=\delta {\zeta }_P/2$. Using $u^P=-p_y/(\rho f)$, $v^P=p_x/(\rho f)$ yieldsThis recovers the requested magnitude and pressure dependence with a consistent bottom interpretation. The surface version would require its own specified boundary velocity and corresponding signs.

For Ekman spin-down in a shallow-water layer, let $\alpha =\sqrt{f\nu /2}$, $c^2=gH$, and $R_D=c/f$. Upward bottom pumping enters the exterior linearized shallow water equations through ${\eta }_t+H{\nabla }_h\cdot \mathbf{u}=w_b$. Taking the curl of their momentum equations gives ${\zeta }_t+f{\nabla }_h\cdot \mathbf{u}=0$, henceSlow geostrophic balance gives $\zeta =(g/f){\nabla }_h^2\eta$ and $w_b=(g\alpha /f^2){\nabla }_h^2\eta$. Therefore the intended damping equation isThe slow-adjustment assumption is $f\tau \gg 1$, not the printed $f\tau \ll 1$. The printed negative pumping term in continuity can alternatively describe downward extraction at the upper boundary, but it cannot be combined unchanged with the upward bottom pumping just derived. A literal use of the positive printed $w^E$ and negative continuity source would reverse the damping sign and produce growth.

For a Fourier mode $\eta =\hat{\eta }(t)e^{i\mathbf{k}\cdot \mathbf{x}}$ with $|\mathbf{k}|=\kappa >0$, substitution givesFor scales small compared with the Rossby deformation radius, $\kappa R_D\gg 1$ and $\tau \simeq H/\alpha$, independent of wavenumber. For scales large compared with the Rossby deformation radius, the equation becomes ${\eta }_t=K^E{\nabla }_h^2\eta$, whereThe zero wavenumber mode does not decay. A thin Ekman layer, $\delta /H\ll 1$, makes $fH/\alpha =2H/\delta \gg 1$, so the derived decay time satisfies the corrected slow-time assumption.

Use $u={\partial }_y\psi$ and $v=-{\partial }_x\psi$ in the frame translating with the sheet. At each instant, the incompressible Stokes equations areTaking their curl gives the biharmonic stream function for planar Stokes flow:in the fluid above the actual sheet. The no-slip boundary condition is imposed on material points, whose label is $x$ and whose actual horizontal position is $x_s$. With $q=x-t$,The second sign follows from $v={\partial }_t{y}_s=-ϵb\cos q$. Far above the sheet,with bounded velocity and no imposed mean pressure gradient. The positive far-field velocity arises because the sheet moves with laboratory velocity $-U{\mathbf{e}}_x$. All fields are $2\pi$-periodic in $x$; an additive function of time in $\psi$ is a harmless gauge. These conditions formulate the Taylor swimming sheet problem with both types of travelling deformation.

Expand $U=ϵU_1+{ϵ}^2{U}_2+\cdots$ and $\psi =ϵ{\psi }_1+{ϵ}^2{\psi }_2+\cdots$. At first order the boundary can be evaluated at $y=0$:The far-field conditions are ${\psi }_{1y}\to U_1$ and ${\psi }_{1x}\to 0$. For a nonzero Fourier mode of wavenumber $k$, bounded velocity selects $(E+Gy)e^{-ky}{e}^{ikq}$; the exponentially growing modes are excluded. Matching the modes $k=1$ and $k=3$ givesAn additive gauge has been set to zero. The first term is a transverse mode of a Taylor swimming sheet, and the second is a longitudinal mode of a Taylor swimming sheet. Their velocities decay at infinity. The mean first-order boundary velocity is zero, and the bounded zero Fourier mode is a constant plus a multiple of $y$, so $U_1=0$.