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In the local short-wavelength approximation, derivatives of the slowly varying coefficients are smaller than derivatives of the phase. For the perturbation $v{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, replace ${\partial }_t$ by $-i\omega$ and ${\mathbf{B}}_p\cdot \nabla$ by $i\mathbf{k}\cdot {\mathbf{B}}_p$ in the torsional Alfvén wave equation. Cancel the common nonzero amplitude and $R^2$ to obtainThus the local dispersion relation is an Alfvén wave relation with Alfvén velocity ${\mathbf{v}}_{A,p}={\mathbf{B}}_p/\sqrt{{\mu }_0\rho }$. The approximation requires wavelength small compared with the background variation scales. A wave vector exactly perpendicular to ${\mathbf{B}}_p$ gives zero leading frequency, so cannot simultaneously obey the assumed large-frequency limit.

Without gas drag, the dispersion relation factorizes as $s(s^2+{\omega }^2)=0$. There is a neutral branch and two density-wave branches. If ${\omega }^2>0$, the waves have $s=\pm i\omega$; if ${\omega }^2<0$, one root is $\boxed{s=\sqrt{-{\omega }^2}>0}$ and the dust layer is gravitationally unstable.

For $c>0$, complete the square in the wavenumber magnitude:The minimum occurs at $|k|=\pi G{\sigma }_0/c^2$. Thus the Toomre stability criterion isAt $Q=1$ the minimizing mode is marginal. For $Q>1$ there is no growing axisymmetric wave in this razor-thin, pressure-supported, drag-free model. The unstable band for $Q<1$ isIn a finite layer, this band must contain an admissible mode; the continuum statement assumes an adequately large domain.

For fixed ${\omega }^2>0$ and weak gas drag, expand a wave root as $s=s_0+ϵs_1+O({ϵ}^2)$ with $s_0=\pm i\omega$. At order $ϵ$, the dispersion relation yieldsSince $s_0^2=-{\omega }^2$, this reduces to $-2{\omega }^2{s}_1-\Omega ({\omega }^2+{\Omega }^2)=0$. The weak-drag dust density waves therefore haveThe first-order real part is negative: the two density waves are weakly damped, with their frequencies unchanged to first order. Gas drag removes their perturbation energy relative to the fixed gas flow. The expansion is an asymptotic result at fixed positive ${\omega }^2$; it is not uniform close to ${\omega }^2=0$, where the nominal damping can become comparable to the oscillation frequency and a different scaling is needed.

The branch originating at $s=0$ has $s=ϵs_1+O({ϵ}^2)$. Keeping first-order terms in the cubic dispersion relation gives ${\omega }^2{s}_1+\Omega ({\omega }^2-{\Omega }^2)=0$. Hence the weak-drag secular gravitational growth rate isCarrying forward the positive-${\omega }^2$ regime of the preceding part, this mode grows when $\boxed{0<{\omega }^2<{\Omega }^2}$. This specifies the scope of the source's abbreviated condition. If ${\omega }^2<0$, the small root written here is instead damped, while the separate dynamical branch already grows; the dust is still unstable, but growth is not identified with this particular small-root continuation. The expansion also fails at ${\omega }^2=0$.

At sufficiently small nonzero $|k|$, ${\omega }^2$ is positive and just below ${\Omega }^2$, sinceTherefore an infinite dust layer with nonzero self-gravity and nonzero drag has a long-wavelength secular gravitational instability of an astrophysical disk, even when $Q>1$. Its small-$k$ growth rate tends to $2\pi ϵG{\sigma }_0|k|/\Omega$, and vanishes at $k=0$; the uniform-density mode itself is neutral. Drag transfers perturbation angular momentum to the gas, weakening rotational support while self-gravity exceeds pressure on long wavelengths.

The instability is not limited to weak drag as an existence statement. For any $ϵ>0$, the cubic has $P(0)=ϵ\Omega ({\omega }^2-{\Omega }^2)<0$ and $P(s)\to +\infty$ on the positive real axis. Thus it has a positive real root whenever ${\omega }^2<{\Omega }^2$. This exact positive-root criterion for dust self-gravity with drag distinguishes the growth criterion from the approximation used to compute its slow rate.

The dispersion symmetry elimination of a boundary trace uses the symmetry of the dispersion relationFor $k\in \overline{D^{+}}$, $\mathrm{Im}\nu (k)\le 0$, so the global relation is valid at $\nu (k)$. Since $i\nu +\alpha =-ik$, it givesSubstitution into the contour integral representation produces an unwanted integral ${\int }_{\partial D^{+}}{e}^{ikx}\hat{u}(\nu (k),t)dk$. Its integrand is analytic in $D^{+}$ and decays on closing the contour upwards; Jordan lemma makes this integral zero for $x>0$. Hence the Fokas method eliminates the unknown normal derivative:All quantities here are determined by the prescribed initial and Dirichlet boundary data up to time $t$.

For verification and for numerical evaluation it is useful to evaluate the spectral contour integrals, giving a half-line drift reflection kernel. Putand define the half-line drift boundary kernelFubini's theorem, the Gaussian Fourier transform and contour deformation give the equivalent causal formulaFor the reflected initial term, ${\hat{u}}_0(i\alpha -k)={\int }_0^{\infty }{e}^{\alpha y+iky}{u}_0(y)dy$ on the contour; the evaluated Gaussian supplies the necessary large-$y$ decay. If $e^{\alpha y}{u}_0(y)$ is not integrable, truncate the initial conditions first, evaluate, and pass to the limit using Gaussian bounds. No extra exponential-decay assumption on the original data is needed for this kernel formula.

The boundary kernel follows particularly simply fromThis calculation independently checks both the sign and the coefficient of the boundary forcing.

Apply Fourier inversion to the global relation on the real axis, extending $u$ by zero to negative $x$. At an interior point $x>0$ this givesThe finite-time spectral boundary transforms are entire functions of $k$. Define the upper decay domainOrient $\partial D^{+}$ from its left infinite end through $i\alpha$ to its right infinite end, so that $D^{+}$ lies to the left. Its finite point is $i\alpha$, not zero. On the contour, $\mathrm{Re}\omega =0$; the factor $e^{ikx}$ decays because $x>0$.

For the boundary term, write $e^{-\omega t}{G}_j={\int }_0^t{e}^{-\omega (t-s)}{g}_j(s)ds$. Between the real axis and $\partial D^{+}$, $\mathrm{Re}\omega \ge 0$, so this factor has no exponential growth. Contour deformation and Jordan lemma, with the usual cutoffs for oscillatory integrals, therefore giveThis is the requested complex-plane representation with the unknown finite-time spectral boundary transform still present. The decay domain must agree with the sign of the drift in the dispersion relation.

Use the spectral parameter for a linear boundary value problem $k\in \mathbb{C}$ and define the dispersion relation $\omega (k)=k^2-i\alpha k$. Direct differentiation gives the divergence formIndeed the coefficient of $u$ left over after expansion is $\omega -k^2+i\alpha k=0$, and the remaining factor is $u_t-u_{xx}-\alpha u_x$. Thus this is equivalent to the advection-diffusion equation for every $k$.

Introduce the Half-range Fourier transforms and finite-time spectral boundary transformswhere $g_1(t)=u_x(0,t)$ is the unknown normal derivative with the positive-$x$ convention. The outward normal at zero instead gives $-g_1$. The spatial transforms are analytic for $\mathrm{Im}k<0$ and continuous on the real axis under the stated decay assumptions. Integrating the divergence form on $(0,\infty )\times (0,t)$ gives the global relationThe sign follows from the lower spatial endpoint: the integrated spatial derivative is minus its value at zero. This sign will determine the boundary-forcing term in the solution.

Let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation and $p$ the pressure perturbation divided by ${\rho }_0$. For stable stratification, $N>0$. The nonrotating Linearized Boussinesq equations areEliminating $u,p,b$ gives $({\partial }_x^2+{\partial }_z^2)w_{tt}+N^2{w}_{xx}=0$. Since $w={\zeta }_t$, a nonzero-frequency plane wave obeys the same equation for its displacement. Substituting its phase yields the dispersion relation for a plane internal gravity wave:Thus the frequency depends on the wavevector direction rather than its magnitude.

Advection of the background density gives ${\rho }^{\prime }=-\zeta {\bar{\rho }}_z$ to first order. With constant ${\bar{\rho }}_z<0$, the instantaneous density gradient is ${\rho }_z={\bar{\rho }}_z(1-{\zeta }_z)$. A region has unstable density stratification when this becomes positive, namely when ${\zeta }_z>1$. The maximum of ${\zeta }_z$ is $|m{A}_{\zeta }|$, so the monochromatic internal-wave overturning criterion isEquality gives a locally vanishing gradient. This is the prediction of the displacement field extrapolated to overturning; the small-amplitude approximation itself ceases to be reliable there.

For the rising packet, distinguish its conserved absolute frequency $\omega$ from its actual intrinsic frequency $\sigma =\omega -kU(z)$. The printed terminology calls $\omega$ intrinsic while also assigning it to a stationary observer; the stationary-observer interpretation is the one consistent with the displayed Doppler shift. On the positive-frequency branch, the ray Hamiltonian isThe Hamiltonian ray-tracing equations giveThe last identity follows also by differentiating the Hamiltonian along its canonical trajectory: the spatial and wavevector terms cancel in pairs. Thus absolute-frequency conservation in steady shear gives constant $k$, constant $\omega$, and constant stationary-observer horizontal phase speed $c_x=\omega /k$. In contrast, $\sigma =\omega -ksz$ decreases as the packet rises. At its initial height,This is the critical level of an internal gravity wave. In fact $m(t)=m_0-kst$ and $z(t)=[\omega -\sigma (k,m(t))]/(ks)$, so the inviscid ray approaches $z_c$ as $t\to \infty$, rather than reaching it at a finite time.

Write $\theta =|\Theta |=\arctan (|m|/k)$, so $\sigma =N\cos \theta$ with $0<\theta <\pi /2$. The intrinsic internal-wave phase and group velocity calculation givesThe observer-frame horizontal ray velocity is $U+c_{gx}$. Dividing it by $c_{gz}$ proves the internal-wave ray in uniform vertical shear:The angle increases toward $\pi /2$ and the vertical group speed tends to zero near the critical level.

The wave-action conservation law fixes the prescribed upward flux. For a nonzero packet, $B>0$, and the given flux relation impliesApply the monochromatic internal-wave overturning criterion, using $|m|=k\tan \theta$. After multiplying by the positive trigonometric factors, the exact instability condition isAt marginal overturning near a critical level, $\sin \theta \simeq 1$, so the wave-action criterion for critical-level overturning givesWith fixed $B,k,N$, this is the requested quarter-power order estimate; the prefactor supplies the dimensions suppressed in that notation. It is an onset balance, not a replacement for $\cos \theta =\sigma /N$. Combining the two relations instead gives ${\cos }^3\theta =(B{k}^3/N^2)\sin \theta$ at onset. Since $m^2{A}_{\zeta }^2$ diverges as ${\sigma }^{-3}$ toward $z_c$, any nonzero packet flux eventually violates the linear overturning criterion before reaching that level, within this nondissipative ray model.

For the second-harmonic feedback in a long-wave convection amplitude equation, at $\mu =2$ the linear steady operator is $L_0=-(1+{\partial }_x^2{)}^2$. Its eigenvalue on $e^{inx}$ is $-(1-n^2{)}^2$, so the critical Fourier modes are $n=\pm 1$. More generally the linear dispersion relation is $\lambda (k)=-1+\mu k^2-k^4$, giving first onset at $\mu =2,k=1$.

At order ${ϵ}^2$, the weakly nonlinear expansion givesWith ${\Theta }_1=A{e}^{ix}+\overline{A}{e}^{-ix}$,The constant part disappears under differentiation. Inverting $L_0$ on the second harmonic, whose eigenvalue is $-9$, givesup to a critical-harmonic correction absorbed into the definition of $A$. At order ${ϵ}^3$,The coefficient of $e^{ix}$ in the last three terms is respectivelyThe Fredholm solvability condition requires their sum to vanish, because $L_0$ annihilates the critical Fourier mode. For a nonzero amplitude,This requires ${\mu }_2/(3-8s^2/9)>0$. The cubic amplitude coefficient is positive for $s^2<27/8$, yielding a small-amplitude branch in a supercritical bifurcation; it is negative for $s^2>27/8$, yielding a subcritical bifurcation branch in this leading approximation. At $s^2=27/8$, the cubic coefficient vanishes and the quoted relation is singular: higher-order nonlinear terms and a different detuning balance are needed. One must not assert a finite $p$ there.

If a term $ϵ{\mu }_1$ were included in $\mu$, order ${ϵ}^2$ would also contain $-{\mu }_1{\Theta }_{1xx}={\mu }_1{\Theta }_1$. The quadratic nonlinearity produces only the zeroth and second harmonics, with the zeroth differentiated away, so it cannot balance a first-harmonic contribution at that order. Solvability would give ${\mu }_{1}A=0$. Hence a nontrivial critical-mode expansion forces ${\mu }_1=0$ and first balances detuning against cubic amplitude effects at order ${ϵ}^3$.

For an axisymmetric vertical mode of a shearing sheet with real $k\ne 0$, the two divergence constraints give $ik{v}_z=ik{b}_z=0$. The vertical momentum equation then yields $ikp=0$. Thus all three amplitudes vanish:Define the magnetic amplitudes $a_i=b_i/\sqrt{4\pi {\rho }_0}$ and the signed vertical Alfvén velocity $v_A=B_0/\sqrt{4\pi {\rho }_0}$. Write $A=k^2{v}_A^2$. The remaining equations for the normal mode areFor a growing or oscillatory mode with $\sigma \ne 0$, eliminating $a_x,a_y,\vartheta$ givesA nonzero velocity requires the determinant to vanish. After removing its factor $\sigma$, the dynamical dispersion relation isThis is the radially stratified magnetorotational dispersion relation, with $A=k^2{B}_0^2/(4\pi {\rho }_0)$. Keeping the original five-amplitude system instead gives characteristic polynomial $\sigma$ times this quartic. There is also a stationary balanced normal mode; division by $\sigma$ excludes it but loses no exponentially growing mode. At $k=0$ the divergence argument for vanishing vertical components does not apply, so that spatially uniform case must be treated separately.

When the imposed magnetic field vanishes, the dynamical dispersion relation becomesThe nonstationary hydrodynamic branch therefore has ${\sigma }^2=-(N^2+{\Omega }^2)$. The radial Solberg–Høiland instability criterion in this local model isThe radial buoyancy frequency supplies either restoration or a driving force, while the radial epicyclic frequency $\kappa =\Omega$ supplies rotational restoration. Zero sum is marginal, and positive sum gives stable oscillations. The neutral roots do not themselves signify exponential growth.

Because both equilibrium pressure and dimensionless specific entropy decrease outward, their radial gradients have the same sign. The leading minus sign in the expression for $N^2$ therefore makes $N^2<0$: the radial stratification is adverse. For smooth profiles varying over radius $r$, $|{\partial }_{r}P|\sim P/r$ and $|{\partial }_{r}S|\sim 1/r$, soFor a thin disk, the magnitude of this negative radial buoyancy frequency squared is much less than ${\Omega }^2$. Rotation stabilizes the hydrodynamic mode despite the adverse entropy gradient. This estimate assumes gradients on the global radial scale; a sharp thermal feature may be different. It addresses the ideal nondiffusive equations here. Processes such as convective overstability require additional thermal relaxation and are not ruled out by this particular criterion.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.

Use the inviscid, nonrotating Boussinesq approximation, with a stable background mass density $\bar{\rho }(z)$ and reference density ${\rho }_{\ast }$. Write $B=-g{\rho }^{\prime }/{\rho }_{\ast }$ for the perturbation buoyancy and $\pi =p^{\prime }/{\rho }_{\ast }$ for the kinematic pressure. The buoyancy frequency is $N_0^2=-(g/{\rho }_{\ast }){\bar{\rho }}_z>0$. Dropping products of perturbations in the Boussinesq equations gives the Linearized Boussinesq equationsThe last equation expresses incompressible flow; the second follows by advecting the background mass density gradient. Neglecting rotation and viscosity is part of this internal gravity wave model.

For a plane internal gravity wave proportional to $\exp [i({\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t)]$, put $\kappa =|{\mathbf{k}}_h|>0$ and $K^2={\kappa }^2+m^2$. Eliminating the horizontal velocity, pressure and buoyancy from the linear equations givesFor example, the horizontal momentum and continuity equations give $\pi =-\omega mw/{\kappa }^2$; substituting $B=-i{N}_0^{2}w/\omega$ into vertical momentum yields the displayed dispersion relation. Thus an IGW has $0<|\omega |\le N_0$ in this model.

On the positive-frequency branch, the phase velocity normal to a constant-phase plane and the group velocity areConsequently $({\mathbf{k}}_h,m)\cdot {\mathbf{c}}_g=0$, and phase and group velocity are perpendicular. Equivalently, the dispersion relation is homogeneous of degree zero in the wave vector, so differentiating with respect to its scale proves the same orthogonality. Energy travels with the group velocity, along the phase planes. At the degenerate limit $m=0$, $\omega =N_0$ and the group velocity vanishes; the orthogonality statement then has this limiting interpretation.

For a two-dimensional internal gravity wave with fixed horizontal wave number $k\ne 0$, the vertical velocity amplitude satisfiesThis follows from the Linearized Boussinesq equations just as in part (a), allowing the background buoyancy frequency to vary. The WKB approximation gives locallyIt requires slowly varying stratification compared with the local vertical wavelength, in particular $|m_z|\ll m^2$, and a small-amplitude wave in the inviscid, nonrotating regime. Propagation requires $N>|\omega |$; where $N<|\omega |$, the local vertical wave number is imaginary and the solution is evanescent. Turning levels and abrupt changes need a separate connection calculation. Finite viscosity, background flow, rotation or nonlinear effects require modifications of this model.

The ray tracing equations use the local dispersion relation as a Hamiltonian:The stationary, horizontally uniform background conserves the frequency and horizontal wave number. In two dimensions the group velocity gives $dx/dz=-m/k$. For the specified buoyancy frequency and $\omega =1$, $m^2=256k^2{z}^2$. Hence for $z>0$ the two geometric internal-wave rays through the specified point areEach curve has both directed propagation senses, giving four local directed rays. Their formal intersections with $z=0$ are $x=-8$ and $x=8$.

At $z=0$, $N=\omega$ and $m=0$: this is a double-zero internal-wave turning level, since $m^2$ touches zero without changing sign. The ray becomes vertical geometrically, but its group velocity tends to zero. Taking $k>0$ without loss of generality, the downward branch hasThe formal travel time diverges logarithmically as $z\to 0$. Meanwhile $|m_z|/m^2=1/(16|k|z^2)$ grows, so the WKB approximation fails before that limit; its predicted $|m{|}^{-1/2}$ amplitude cannot be extrapolated to infinity. The spatial ray reaches a finite limiting position, but finite-wavelength behaviour near the zero must be found beyond WKB. There is no evanescent half-space supplied by this particular $N(z)$, so ordinary simple-turning-point reflection cannot be assumed solely from the local ray construction.

Use the Fourier transform convention $\hat{p}(k,y)={\int }_{\mathbb{R}}{e}^{ikx}p(x,y)dx$. Let $\gamma (k{)}^2=k^2-k_0^2$, choosing $\mathrm{Re}\gamma >0$ on the real contour with $\mathrm{Im}\omega <0$. For time dependence $e^{i\omega t}$, this is the decaying continuation of the outgoing Sommerfeld radiation condition.

The transformed Helmholtz equation has upper and lower solutions $P(k)e^{-\gamma y}$ and $Q(k)e^{\gamma y}$. The two kinematic traces give $-\gamma P={\rho }_0{\omega }^2\hat{\eta }=\gamma Q$, henceThe elastic membrane equation gives $[\hat{p}]=(m{\omega }^2-T{k}^2)\hat{\eta }$. Therefore the acoustic wave on a tensioned massive membrane has dispersion relationEquivalently, $T{k}^2={\omega }^2(m+2{\rho }_0/\gamma )$. The fluid on both sides supplies a positive added mass of an evanescent fluid layer, which is a useful independent check on the sign. In particular, the incident pressure isThe symbol $m$ here denotes elastic membrane mass per area, rather than the fluctuating Mach number used in Question 1.

Linearity permits subtraction of the incident acoustic membrane wave. The scattered pressure solves the homogeneous Helmholtz equation, and the kinematic relation defines its line displacement ${\eta }^{\prime }$ on both halves. On $x<0$, subtracting the incident elastic membrane equation gives the scattered dynamic equation. On $x>0$, there is no elastic membrane, so the total pressure jump must vanish, giving $[p^{\prime }]=-[p_I]$. The pinned endpoint has $\eta (0)=0$, hence ${\eta }^{\prime }(0)=-1$, rather than zero.

Use full and Half-range Fourier transforms with the same $e^{ikx}$ convention as part (a), and putWrite $P^{-}$ for the left transform of the upper scattered pressure, $P^{+}$ for its right transform, and ${\eta }^{\pm }$ for the corresponding transforms of ${\eta }^{\prime }$. Outgoing antisymmetry in $y$ impliesThe second relation comes from the prescribed pressure cancellation on $x>0$.