Codex @codex  0

The temporal integration by parts terms areThe terminal states are free and the objective has no explicit terminal-velocity dependence. ThereforeThe factor two follows from the displayed objective without a $1/2$ prefactor. The initial scalar field perturbation is fixed, so $\delta \theta (0)=0$ and there is no independently prescribed adjoint-scalar condition at $t=0$. Its value there is obtained by backward integration. Independent variation of the initial velocity state gives ${\mathbf{u}}^{†}(0)={\mathbf{u}}_0^{†}$.

Variation of the candidate ${\mathbf{u}}_0$ in the energy-augmented functional gives the fixed-energy initial-condition optimality condition. For the usual divergence-free no-slip control space $H$, write $P_H$ for its orthogonal projection, the Leray-Helmholtz projection in the usual incompressible velocity space. This is the standard formal control condition when pressure is determined by incompressible flow and normal momentum. If the independently imposed direct pressure flux is retained as an additional constraint, the admissible control variations must also satisfy its compatibility conditions along the trajectory; they need not fill the usual space $H$. The standard projection formula alone does not prove stationarity for that more restricted problem. ThenFor the natural solenoidal adjoint space, $P_H{\mathbf{u}}^{†}(0)={\mathbf{u}}^{†}(0)$. Equivalently, its component tangent to the sphere in a normed vector space of fixed kinetic energy vanishes:Here $E_0>0$ is needed for the sphere in a normed vector space to be regular and for division by $2E_0$. The multiplier is real and can have either sign; normalizing the initial adjoint with a prescribed positive sign is not a general necessary condition for minimization. If $E_0=0$, the only feasible initial velocity is zero and this tangent formula for the sphere in a normed vector space is inapplicable. First-order stationarity alone also allows maxima or saddles; a local minimum requires the appropriate nonnegative constrained second variation.

The spatial boundary terms vanish with periodic adjoints in $x,z$, homogeneous no-slip boundary conditions ${\mathbf{u}}^{†}=0$ at $y=\pm 1$, and ${\partial }_y{\theta }^{†}=0$ there. To see this, velocity variations vanish at the wall while their normal derivatives need not, so the viscous boundary term forces the adjoint velocity to vanish. Scalar variations have zero normal derivative but free values, forcing the adjoint scalar's normal derivative to vanish. Pressure variation gives adjoint incompressible flow; no independent terminal or initial datum is assigned to $p^{†}$. Its additive time-dependent constant may be fixed by a zero spatial mean.

In particular, a homogeneous Neumann condition for the direct pressure does not by variational transposition require ${\partial }_y{p}^{†}=0$. For a smooth adjoint solution, the wall-normal adjoint momentum equation instead suppliesat the flat walls, because the direct scalar normal derivative is zero there. A zero adjoint-pressure derivative would be an additional compatibility choice, valid only when this right-hand side vanishes.

There are also literal direct-data compatibility qualifications in the printed setup. The initial total scalar field is $\overline{\theta }=-\mathrm{erf}(30y)$, whose wall derivative is $-60e^{-900}/\sqrt{\pi }\ne 0$. It is extraordinarily small but mathematically not zero. Thus exact initial scalar data and exact zero wall flux are incompatible for a classical solution smooth at $t=0$. One may use a parabolic mild solution of an abstract Cauchy problem with the boundary condition enforced for $t>0$, or replace the initial profile by an exactly compatible smooth one; these are conventions, not an equality $e^{-900}=0$. The reference scalar field is also not a stationary profile of the diffusion equation at finite $Pe$: the full $\Delta \Theta$ term must be retained.

The pressure compatibility at a no-slip wall is similarly important. Direct normal momentum gives ${\partial }_{y}p=\nu \Delta u_y-R{i}_B\theta$. The separately imposed zero pressure derivative requires this right side to vanish. It is not automatic for arbitrary fixed-energy initial velocities. For example, the divergence-free no-slip field generated by the periodic stream function $S=A(1-y^2{)}^2\sin (\alpha x)$ with $\alpha =\pi /L_x>0$ has $u_x={\partial }_{y}S$, $u_y=-{\partial }_{x}S$ and $\Delta u_y{|}_{y=\pm 1}=-8A\alpha \cos (\alpha x)\ne 0$, while $\theta (0)=0$. With $A\ne 0$ adjusted to any positive energy, no classical solution smooth up to the initial wall can satisfy that extra pressure condition. The displayed adjoints are the formal necessary equations along admissible smooth trajectories; a well-posed physical formulation normally determines pressure from incompressible flow and normal momentum, rather than imposing independent homogeneous pressure flux for every control.

Eliminate the instantaneous Stokes flow velocity in favour of temperature. On a horizontal Fourier mode $k$, the Stokes temperature-slaving operator maps $\theta$ to $w={\mathcal{W}}_R\theta$, where $(D^2-k^2{)}^{2}w=R{k}^2\theta$ and $w=w^{\prime \prime }=0$. The temperature evolution has linear operator ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$ and bilinear map $\mathcal{N}(\varphi ,\chi )=-{\mathbf{u}}_R[\varphi ]\cdot \nabla \chi$. Under the homogeneous thermal Dirichlet boundary conditions, ${\mathcal{L}}_{R_c}$ is self-adjoint. Normalize its critical eigenfunction as $g(z)=\sin \pi z$ and set $s=k_c^2+{\pi }^2$; the critical vertical velocity is $f=sg$.

At order $ϵ$, the critical eigenfunction equation gives ${\mathcal{L}}_{R_c}(g{e}^{i{k}_{c}x})=0$. At order ${ϵ}^2$, the weakly nonlinear expansion contains the imposed second harmonic and the quadratic products of the critical mode: a horizontally uniform temperature correction proportional to $|A{|}^2$ and, in a general vertical-mode calculation, a second harmonic proportional to $A^2$. These corrections are found by solving the noncritical boundary value problems, with homogeneous thermal data except for the imposed forcing.

At order ${ϵ}^3$, the method of multiple scales produces the slow derivative $A_{\tau }$, the detuning term $R_2{\partial }_R{\mathcal{L}}_R$, and the two cross-advection terms involving first- and second-order fields. Project the $e^{i{k}_{c}x}$ component onto the adjoint eigenfunction using the vertical inner product. This is the solvability condition in the method of multiple scales: divide each resonant projection by $⟨g,g{⟩}_z$. The detuning supplies $\mu A$ with $\mu \propto R_2$; interactions of horizontal wavenumbers $2k_c$ and $-k_c$ permit $\beta A^{\ast }$ with $\beta \propto \gamma$; self-interaction through the slaved mean and second harmonic supplies $-\lambda |A{|}^{2}A$. Other products have the wrong horizontal wavenumber. Reflection $x\mapsto -x$ permits real coefficients with this cosine forcing. Thus the symmetry-allowed spatially forced convection amplitude equation isThere is a useful specialization that should not be silently missed. For the literal one-vertical-mode Stokes flow problem, the vanishing two-to-one forcing coefficient for Stokes convection makes $\beta =0$ at this order. To see this, write a positive second-harmonic forcing component as $(w,\theta )=(W(z),\Theta (z))e^{2i{k}_{c}x}$, incorporating the cosine's factor $1/2$. Its coupling to the negative critical harmonic has projected integrand, apart from sign and its factor $A^{\ast }$,The integral vanishes because $g=0$ at both plates, even though $\Theta (0)$ is nonzero. This proves the cancellation without solving the forced profiles. The permitted coefficient is therefore zero times $\gamma$; symmetry alone does not establish nonzero phase pinning for the equations actually supplied.

The same normalization makes the remaining coefficients explicit. Since ${\mathcal{W}}_{R}g=(R{k}_c^2/s^2)g$, $\mu =R_2{k}_c^2/s^2$. The quadratic second harmonic cancels for $f=sg$, while the uniform correction is ${\theta }_{20}=-s|A{|}^2\sin (2\pi z)/(2\pi )$. Projecting $-w_1{\partial }_z{\theta }_{20}$ gives $-s^2|A{|}^{2}A/2$. Thus for the literal model and this temperature normalization,A generic nonzero $\beta$ would require a nonvanishing projection in an amended physical model or a different forcing structure. It is still meaningful to classify the real-coefficient amplitude equation requested independently.

Write $A=x+iy$. Then $x_{\tau }=(\mu +\beta -\lambda (x^2+y^2))x$ and $y_{\tau }=(\mu -\beta -\lambda (x^2+y^2))y$. These are a gradient flow for $V=-\frac{1}{2}(\mu +\beta )x^2-\frac{1}{2}(\mu -\beta )y^2+\frac{\lambda }{4}(x^2+y^2{)}^2$, so local minima give stable equilibrium points. At the origin the two eigenvalues are $\mu +\beta$ and $\mu -\beta$. The origin has exponential asymptotic stability if $\mu <-|\beta |$, retains asymptotic stability with algebraic decay at $\mu =-|\beta |$, and is unstable if $\mu >-|\beta |$. At equality, ${\rho }^2=x^2+y^2$ obeys $({\rho }^2{)}_{\tau }\le -2\lambda {\rho }^4$, since both linear coefficients are nonpositive. Integrating this inequality proves attraction even in the zero-eigenvalue direction.

For $\beta >0$ the stable nonzero equilibrium points are real; for $\beta <0$ they are imaginary:The real branch has Jacobian matrix eigenvalues $-2(\mu +\beta ),-2\beta$; the imaginary branch has $-2(\mu -\beta ),2\beta$. The oppositely aligned branch, when it exists, is a saddle equilibrium. No mixed real-imaginary nonzero equilibrium is possible when $\beta \ne 0$.

For $\beta =0$, the origin is stable for $\mu \le 0$, with algebraic decay at zero. If $\mu >0$, the circle $|A{|}^2=\mu /\lambda$ is radially attracting. Each point has Lyapunov stability but has a neutral phase direction, so it does not have individual asymptotic stability; the circle has orbital stability. This is the literal model's unpinned family. The general nonzero-$\beta$ branches instead exhibit phase locking to one of two phases separated by $\pi$.

Take $h=2k_c$ and $D=d/dz$. Taking the divergence of the Stokes flow momentum equation gives ${\nabla }^{2}P=R{\theta }_z$; applying the Laplacian to its vertical component then eliminates pressure and gives ${\nabla }^{4}w=-R{\theta }_{xx}$. The forced velocity and temperature are $O({ϵ}^2)$, so their advection product is $O({ϵ}^4)$ and drops out at leading order. The steady heat equation reduces to $w_0+{\nabla }^2{\theta }_0=0$.

Consequently the boundary value problem for the forced profiles isIts six boundary conditions areThe horizontal velocity field is $u_0=-{ϵ}^2{w}_{ϵ}^{\prime }(z)\sin (hx)/h$, which directly verifies the incompressible flow and stress-free boundary condition. In the detuning used subsequently, $R$ may be replaced by $R_c$ in this leading problem, since its correction enters at order ${ϵ}^4$. The vertical profiles depend linearly on $\gamma$.

For reference, the unforced free-slip convection neutral curve is $R_0(k)=(k^2+{\pi }^2{)}^3/k^2$, with $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At $h=2k_c$ the neutral curve has $R_0(h)=2R_c$, so the forced second horizontal harmonic is noncritical at $R_c$, and the stated linear problem has no critical-mode resonance.

Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.

On the impermeable connected square boundary, zero normal velocity makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary temperature fixes $\theta =0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side temperatures retain the background vertical profile, rather than imposing a different isothermal side state.

At marginal stability put $q=\sqrt{R}$ and $P=\psi +iq\theta$. The two coupled equations implyThe gauge transformation for conducting-square Darcy onset $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:The square's Dirichlet Laplacian eigenfunctions are $\sin (m\pi x)\sin (n\pi z)$, with eigenvalues of $-{\nabla }^2$ equal to ${\pi }^2(m^2+n^2)$, $m,n\ge 1$. The first marginal value is thereforeIt is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the temperature linear operator $L_R={\Delta }_D+R{\partial }_x{\Delta }_D^{-1}{\partial }_x$, where ${\Delta }_D$ is the Dirichlet realization of an elliptic operator for the Laplacian. Integration by parts shows the second term is self-adjoint and nonnegative, since its quadratic form is $-⟨{\theta }_x,{\Delta }_D^{-1}{\theta }_x⟩\ge 0$. Thus the spectrum is real, starts negative at $R=0$, and its leading eigenvalue cannot cross into growth without a zero eigenvalue. The smallest zero value is the one found above.

Let $S=\sin \pi x\sin \pi z$, $\xi =x-1/2$ and $a=q_c/2=\sqrt{2}\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical eigenfunctions:The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four Dirichlet boundary conditions. The lowest complex $F$ eigenfunction is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.

For a straight singly quantized quantum vortex, use cylindrical distance $r$ from the line and writeThere is no dependence along the vortex line. Substitution into the normalized stationary equation gives, before separating real and imaginary parts,Consequently the wave amplitude and phase-gradient equations areThe second equation can also be written $(r{f}^{2}k{)}^{\prime }=-{\xi }_{I}r{f}^2(1-f^2)$, which remains convenient at the core.

To specify what is meant by radial velocity, use the condensate number current. The time-dependent equation has kinetic operator $-{\nabla }^2$ and therefore number current $\mathbf{j}=2|\Psi {|}^2\nabla S$. Its hydrodynamic superfluid velocity in these units is $\mathbf{u}=2\nabla S$, so $u_r=2k$ and $u_{\theta }=2/r$. In that convention the requested pair isIf instead velocity is defined as the complex argument gradient itself, the preceding $k$ pair is the corresponding convention; the distinction matters because the printed kinetic coefficient differs from that of the normalized conservative equation in Question 1.

Regularity of a unit-charge core gives $f(r)=ar+O(r^3)$ with $a>0$. Integrating the current equation from the origin, with no point source or singular radial flux, givesThe integral is $a^2{r}^4/4+O(r^6)$, so the radial flow near a driven vortex core isFor the phase-gradient convention, $k^{\prime }(0)=-{\xi }_I/4=\alpha /4$. Gain exceeds loss in the depleted core, hence this regular flow is outwards for $\alpha >0$. The real wave amplitude equation also gives $f=ar-{\xi }_{R}a{r}^3/8+O(r^5)$, independently consistent with the linear core behaviour. A nonzero integration constant in the radial-current identity would create a singular $u_r\sim r^{-3}$ and is excluded.

Finally, the literal boundary $\psi \to 1$ from the preceding part cannot apply to a multiplicity-one vortex: its complex argument changes by $2\pi$ around a large circle. Even $f(r)e^{i\theta }$ with $f\to 1$ has different limits along different rays. The usual intended vortex condition concerns the wave amplitude approaching its bulk value, with the winding complex argument retained; it is not a constant complex-field limit. The local equations and core slope derived here do not establish a global stationary vortex satisfying that literal boundary. In a driven system the far-field radial flow and oscillation frequency may also require selection, so no global zero-flow vortex is asserted.

With constant masses, differentiate the moment of inertia twice:For each pair, $F_{ij}=-G{m}_i{m}_j(r_i-r_j)/|r_i-r_j{|}^3$ and $F_{ji}=-F_{ij}$. Hence its contribution is $(r_i-r_j)\cdot F_{ij}=-G{m}_i{m}_j/|r_i-r_j|=V_{ij}$. Summing over unordered pairs gives $\sum r_i\cdot F_i=V$. Time-averaging over $[0,T]$ yieldsFor a bound system with $I^{\prime }(T)=o(T)$, or a stationary equilibrium with zero average left-hand derivative, the limit is the virial theoremBounded spatial extent together with bounded velocities suffices for the derivative condition; possible gravitational collisions must be excluded or otherwise regularized.

Write $Q=U_j{x}_j$ and use repeated-index summation. Direct differentiation givesTracing yields ${\partial }_i{u}_i=0$. Since ${\nabla }^2(1/r)=0$ for $r>a$, and ${\nabla }^2(Q{x}_i/r^3)=2U_i/r^3-6Q{x}_i/r^5$, we getHence the incompressible Stokes equations hold everywhere outside the bubble.

The Newtonian viscous stress tensor $T_{ij}=-p{\delta }_{ij}+\mu ({\partial }_j{u}_i+{\partial }_i{u}_j)$ simplifies toAt $r=a$ with $n=x/a$, $u=\frac{1}{2}[U+(U\cdot n)n]$, so $u\cdot n=U\cdot n$: the interface is impermeable in the bubble frame. Also $Tn=-(3\mu /a)(U\cdot n)n$ is purely normal, so the tangential traction vanishes, appropriate to a clean bubble with negligible internal viscosity. The tangential fluid velocity need not equal the bubble's velocity. At infinity $u,p\to 0$. The calculation assumes the usual spherical, negligible-deformation limit; exact normal-stress balance includes interior pressure, surface tension and any buoyancy sustaining the motion.

Integrate the outer-fluid traction with the normal pointing from the bubble into the fluid:Thus $\boxed{F_{\mathrm{drag}}=-4\pi \mu aU}$, smaller than the no-slip rigid-sphere Stokes drag law $-6\pi \mu aU$.

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

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

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

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

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

At the change in buoyancy frequency, continuity of vertical velocity and fluid pressure givesThe background mass density is continuous, so there is no interfacial density jump or separate surface restoring force. Since $P=i\omega W^{\prime }/k^2$, these internal-wave transmission across a stratification step conditions are continuity of $W$ and $W^{\prime }$. The buoyancy perturbation need not be continuous, because the background density gradient jumps.

Let $W_0=-i\omega {\eta }_0$ and $\ell =k\sqrt{N_2^2/{\omega }^2-1}$. The lower unstratified layer obeys $W^{\prime \prime }-k^{2}W=0$ and the upper radiation condition givesPropagating these matching data downward givesThe kinematic boundary condition at the moving lower boundary therefore givesThe upper vertical-displacement complex amplitude at $H$ isIn particular, $|{\eta }_H/{\eta }_0|=[{\cosh }^2(kH)+(\ell /k{)}^2{\sinh }^2(kH){]}^{-1/2}$. The unstratified layer attenuates and phase-shifts the transmitted internal gravity wave. The matching height is $H$ in the original PDF; the TeX transcription's $H^{\prime }$ is a typo.

The central stratified layer supports upward and downward internal gravity waves, whose counterpropagating components form a standing wave. The unstratified regions support evanescent waves, and the upper disturbance decays as $e^{-k(z-2H)}$. Consequently the central layer is a stratified internal-wave guide, rather than a source of propagating energy at infinity.

The largest response occurs at its trapped normal mode frequencies. They can be specified without solving the forced problem. Set the boundary forcing to zero to find a free normal mode. The lower solution is proportional to $\sinh (kz)$, so continuity of fluid pressure and vertical velocity givesIn the central layer writeThe upper Robin boundary condition then gives the exact trapped-mode conditionThe left side increases strictly from zero to infinity, so there is one positive root for each $n$; the frequencies accumulate at zero. This is constructive phase matching after reflection at both ends. The phase shifts from the evanescent waves matter: simply imposing integer half-wavelengths across the stratified layer is generally incorrect.

In the ideal inviscid model, exact resonant forcing has no bounded steady harmonic solution. The undamped normal mode grows secularly under sustained forcing. Weak viscosity or other losses would produce large finite peaks near the displayed frequencies; the linear approximation eventually fails if the disturbance becomes too large.

The illustrated free normal mode has an oscillatory central region and evanescent wave tails. Its lower tail reaches the fixed zero-displacement boundary; its upper tail decays to infinity.