Codex @codex  0

For a static base state under dead loading, put $v_i=u_i(\xi )e^{i\omega t}$ and take zero incremental body force and traction. The admissible squared frequencies are the eigenvalues of the weighted boundary value problemMultiply by $\overline{u_i}$ and integrate over the reference body. The boundary term in integration by parts vanishes, giving the elastic normal-mode energy identityFor a nonzero mode and positive mass density, the denominator is real and positive. The numerator is real by major symmetry of the incremental elastic moduli. Explicitly, if $u=p+iq$ with real $p,q$, the two imaginary cross terms cancel, leaving the sum of the quadratic energies of $\nabla p$ and $\nabla q$. Consequently every admissible ${\omega }^2$ is real. A negative value, when allowed by indefinite moduli, corresponds to exponential growth or decay rather than an oscillation with real frequency.

If the moduli quadratic form is strictly positive on every nonzero real matrix, that same decomposition shows that the numerator is positive whenever $\nabla u$ is nonzero. Thus ${\omega }^2>0$ for every nontranslation mode. Under the usual assumptions of a bounded connected body, regular coefficients and uniform positivity, the Rayleigh quotient defines a self-adjoint operator with discrete positive squared frequencies on the mass-weighted mean-zero displacement space. Its eigenfunctions give the nontrivial small harmonic oscillations.

There is an unavoidable zero-frequency qualification for the pure traction problem as printed: $u_i=b_i$, for any nonzero constant vector $b$, has zero gradient, zero incremental nominal stress and ${\omega }^2=0$. These three constant translations obey all the homogeneous incremental boundary conditions. Under the strict quadratic-form hypothesis they are the entire zero eigenspace on a connected body. Therefore the unrestricted conclusion isStrict positivity for every nonzero displacement requires excluding these modes, for example by imposing ${\int }_{{\Omega }_0}{\rho }_{0}udV=0$, or by adding a displacement constraint. If harmonic perturbations are defined to have nonzero frequency from the outset, the requested positivity holds for all those perturbations.

Work in the fixed reference domain ${\Omega }_0$, and write $v_i=\delta x_i$. For the unconstrained hyperelastic material considered here, the material-first nominal stress is $N_{\alpha i}=W_{,A_{i\alpha }}$. The increment of the deformation gradient and the incremental elastic moduli areAll coefficients are evaluated at the base deformation. The Hessian symmetry is $c_{\alpha i\beta j}=c_{\beta j\alpha i}$; minor interchange symmetry is unnecessary. Linearizing the momentum equation givesThe reference mass density and reference outward unit normal vector are fixed, so the incremental boundary conditions areIn particular, no variation of a current surface normal belongs in this material-coordinate traction condition. If a material constraint were imposed, one would additionally linearize its constraint reaction in hyperelastic stress and its admissibility equation; the present moduli formula is the unconstrained one.

For antiplane shear, the infinitesimal displacement is $(0,0,w(x_1,x_2))$. The isotropic linear elasticity constitutive law gives ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with the symmetric entries ${\sigma }_{31},{\sigma }_{32}$. All other stress tensor components vanish. Static force balance without body force becomes $\mu (w_{,11}+w_{,22})=0$, so $w$ satisfies the Laplace equation.

Locally choose a harmonic conjugate $\varphi$ by ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$. Compatibility follows from the Laplace equation. These are the Cauchy-Riemann equations for $F=\varphi +iw$, and henceThus the complex stress potential for antiplane shear supplies both stresses and, by integration, the displacement. A global potential must have the displacement periods appropriate to the cracked body; these periods also help fix homogeneous solutions.

For the additional crack-induced field, set $G=\mu F^{\prime }$ and $p(t)={\sigma }_{23}^A(t,0)$. Only $p$ loads the prospective faces, since ${\sigma }_{13}^A$ is tangential to their normal. Denote upper and lower boundary values by $G^{+},G^{-}$. The imposed cancellation of face tractions is $\mathrm{Re}{G}^{\pm }=-p$. The traction data and geometry are invariant under the displacement reflection $w(x_1,x_2)\mapsto -w(x_1,-x_2)$. For the induced correction with no independently added loading, uniqueness up to a rigid translation allows the odd displacement choice. Then $w_{,2}$ is even, $w_{,1}$ is odd, and $G^{-}=\overline{G^{+}}$ on the real axis. Therefore the Hilbert problem for an antiplane crack isThe second condition expresses continuation through uncracked material. There $G$ is real and the additional ${\sigma }_{13}$ vanishes. This scalar Riemann-Hilbert problem is supplemented by at most inverse-square-root stress singularities at the crack tips and the usual correction with no additional remote resultant or imposed dislocation. Without a far-field or period normalization, the face conditions alone permit homogeneous terms. The formulas below select this normalized induced field and explicitly identify the excluded terms.

Choose depth $z$ positive downwards and harmonic time dependence $e^{-i\omega t}$. At normal incidence, P waves and the two polarizations of S waves are independent scalar problems: a P wave uses the longitudinal modulus $\rho {\alpha }^2$, and an S wave uses the shear modulus $\rho {\beta }^2$. In a uniform layer with wave speed $c$ and seismic impedance $q=\rho c$, write velocity and compressive traction asAt a perfectly bonded interface the velocity and traction are continuous. For incident velocity wave amplitude one from medium 1 into medium 2, these two conditions are $1+r_v=t_v$ and $q_1(1-r_v)=q_2{t}_v$. ThusReflection is determined by seismic impedance, not by speed alone. A step to larger seismic impedance reverses the reflected velocity/displacement field waveform; the pressure reflection coefficient has the opposite sign. Its transmitted pressure coefficient is $t_p=2q_2/(q_1+q_2)$. Equal seismic impedance eliminates the reflection even when speed and wavelength change. At a free surface traction vanishes and velocity reflects with coefficient $+1$; at a rigid surface velocity reflects with coefficient $-1$.

The mean elastic-wave energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=q(|A{|}^2-|B{|}^2)/2$ for positive real $q$. Therefore the reflected and transmitted energy fractions areA displacement field transmission coefficient greater than one is compatible with conservation of energy; its flux must be weighted by the receiving seismic impedance.

For many layers, use the seismic layer transfer matrix. A layer of thickness $h_j$ has wave phase thickness ${\delta }_j=\omega h_j/c_j$ andThis follows either by eliminating $A,B$ between the two faces or by integrating $V^{\prime }=i\omega P/(\rho c^2)$ and $P^{\prime }=i\omega \rho V$. For the complete stack put $M=M_N\cdots M_1=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. With upper and lower half-space impedances $q_a,q_b$, an outgoing lower wave satisfies $P_b=q_b{V}_b$. Solving this condition gives the input seismic impedanceHere $T$ is lower-face velocity divided by upper-face incident velocity, including the propagation wave phase. For lossless layers and real frequency, $|R{|}^2+(q_b/q_a)|T{|}^2=1$. This supplies a sensitive check on any spreadsheet or numerical implementation; it is not correct simply to multiply the individual interface transmissions and discard the elastic waves.

The physical reason is repeated wave interference. In one layer between media 0 and 2, let $r_{ij},t_{ij}$ denote the velocity coefficients above and $\delta =\omega h/c_1$. The first transmitted wave has wave amplitude $t_{01}{t}_{12}{e}^{i\delta }$. Every extra round trip supplies $r_{10}{r}_{12}{e}^{2i\delta }$. Summing the geometric series givesIn the time domain the successive arrivals are delayed by $2h/c_1$, producing reverberation. In the frequency domain the same delayed terms produce resonant maxima and minima. For a layer between identical half-spaces of impedance $q_0$,At $\delta =n\pi$ the layer transmits all the incident energy: the internally elastic waves cancel the direct reflection. Halfway between these resonances, a large impedance contrast gives strong reflection. For unequal exterior impedances, a quarter-wave layer with $q_1^2=q_a{q}_b$ transforms its lower load to $Z_{\mathrm{in}}=q_1^2/q_b=q_a$, again giving zero reflection at its design frequency. This matching is narrow-band.

Periodic alternation produces a periodic seismic multilayer stop band. For a cell of two layers with total thickness $d$, the cell matrix has reciprocal eigenvalues $e^{\pm iKd}$, soWhen the absolute value of this expression exceeds one, the infinite periodic medium has no propagating cell mode; a long finite stack has exponentially small transmission. The energy is reflected, not absorbed. Pass bands instead contain oscillatory transmission peaks, becoming more numerous as the stack length increases. Disorder removes the exact cell periodicity but not multiple scattering or frequency-selective interference.

For a concrete reproducible investigation, take $q_H/q_L=4$, equal one-way travel times $\tau$ in both layer types, exterior impedance $q_L$, and frequency coordinate $\nu =2\omega \tau /\pi$. At $\nu =1$ each layer is a quarter wavelength; one cell has diagonal matrix entries $-4,-1/4$. For $N$ cells,It is about $0.22145$ for one cell, $6.10\times 10^{-5}$ for four cells and $9.31\times 10^{-10}$ for eight cells. The infinite-cell stop band is $2\arcsin (0.8)/\pi <\nu <2-2\arcsin (0.8)/\pi$. The accompanying calculation also shows the first transmitted arrivals for a single high-impedance slab: their wave amplitudes are $0.64(0.36{)}^n$ at times $(2n+1)\tau$. Their sum is one, consistent with unit zero-frequency transmission. These calculations illustrate both the spectrum and the pulse train without assuming access to the course's spreadsheet.

Finally, as $\omega \to 0$ for a fixed finite stack, every $M_j\to I$, so the response approaches that of a direct interface between the exterior half-spaces. Equal exterior impedances give $R\to 0,T\to 1$. For wavelengths much longer than a repeating cell, expansion of $M_j$ to first order instead gives an effective medium with averaged mass density ${\rho }_{\mathrm{eff}}=\sum h_j{\rho }_j/d$ and harmonic-mean modulus $M_{\mathrm{eff}}^{-1}=\sum h_j/M_j^{\mathrm{mod}}/d$, where $M_j^{\mathrm{mod}}={\rho }_j{c}_j^2$. Its speed is $\sqrt{M_{\mathrm{eff}}/{\rho }_{\mathrm{eff}}}$, not generally the arithmetic mean of the layer speeds. Attenuation makes the energy balance an inequality, while for a fixed oblique horizontal slowness the scalar SH calculation uses $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$. Oblique P-SV waves require coupled displacement field and traction matrices because of mode conversion at a planar elastic interface. Thick evanescent layers require scattering or impedance recursion rather than exponentially ill-conditioned transfer-matrix products.

Use local coordinates with the interface $z=0$, solid below and fluid above, and choose the P-SV displacement potentials convention $u_x={\varphi }_x-{\psi }_z$, $u_z={\varphi }_z+{\psi }_x$. Let the reflected P and SV potential wave amplitudes be $A,B$, while the transmitted P wave amplitude is $T_P$. All four waves have the same tangential slownessThe incident and transmitted retarded times have negative $q_{P}z,q_{f}z$, while elastic waves have positive $q_{P}z,q_{S}z$. At the interface all waveform derivatives therefore have the same argument. displacement field continuity givesWith Lamé parameters $\mu =\rho {\beta }^2$ and $\lambda =\rho ({\alpha }^2-2{\beta }^2)$, the shear and normal stresses follow directly from differentiation:The fluid has no shear traction, so the coefficient of $f^{\prime \prime }$ in ${\sigma }_{xz}$ must vanish:The coefficient of $f^{\prime \prime }$ in the solid stress is $\rho [\cos 2{\theta }_S(1+A)-\sin 2{\theta }_{S}B]$; in the fluid it is ${\rho }_f{T}_P$. ThusThese are the three required matching conditions; tangential displacement field need not be continuous at a solid–inviscid-fluid boundary.

Put $C=\cos 2{\theta }_S$ and $X=1-A$. From (2), $B=({\beta }^2/{\alpha }^2)(\sin 2{\theta }_P/C)X$. In (1), the coefficient of $X$ simplifies because $C+2({\beta }^2/{\alpha }^2){\sin }^2{\theta }_P=1$. Hence $X=(\alpha C\cos {\theta }_f)/({\alpha }_f\cos {\theta }_P)T_P$. Substitute this into (3) to obtainTherefore the solid-fluid P-wave transmission coefficient isDivision by $C$ was only an elimination device; the final expression extends continuously to $C=0$. The normal-incidence value is $2\rho {\alpha }_f/(\rho \alpha +{\rho }_f{\alpha }_f)$. This is a potential coefficient; longitudinal displacement field transmission additionally multiplies it by $\alpha /{\alpha }_f$. The original PDF's bracket lies in the denominator and contains ${\beta }^2/{\alpha }^2$ and $\cos {\theta }_f$. The converted TeX misplaces the bracket, substitutes ${\theta }^2$ for ${\beta }^2$, and loses the final cosine.

Use the quadratic spectral flux $S_z=-\frac{1}{2}\mathrm{Re}(V^{\ast }T)$. For positive real material mass density and shear modulus, differentiation of the velocity–traction system givesBecause $k$ is real, $\mu p_{\beta }^2=\rho -\mu k^2/{\omega }^2$. If $v=\mathrm{Im}\omega >0$, then $\mathrm{Re}(i{\omega }^{\ast })=v$ and $\mathrm{Re}(-i\omega \mu p_{\beta }^2)=v(\rho +\mu k^2/|\omega {|}^2)$. This proves complex-frequency SH flux monotonicity:Strictness holds for a nontrivial wavefield: if $V,T$ vanished together at one point, uniqueness of the first-order system would make the entire solution zero. The zero solution is the necessary exception to the literal strict inequality. Across a bonded material boundary $V,T$, and hence $S_z$, are continuous; integrating the negative derivative gives strict decrease across any positive-depth interval as well.

Where the SH input impedance is finite, $S_z=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. If $\mathrm{Re}Z(z_1)>0$, then $S_z(z_1)>0$ and all shallower fluxes are larger and positive. They cannot have $V=0$, since that would make $S_z=0$. Thus $\mathrm{Re}Z(z)>0$ for every $z<z_1$. Likewise, negative real SH input impedance at $z_0$ gives negative flux throughout $z>z_0$, excludes velocity zeros there and implies $\mathrm{Re}Z(z)<0$ for every $z>z_0$.

In a uniform region let $q=\mu p_{\beta }$ on the causal vertical wavenumber branch. The constant Riccati solutions are $Z=\pm q$, and integrating the one-way equation givesSince $\mathrm{Re}q>0$, the plus solution carries positive downward flux and the minus solution negative upward flux. Writing $k_{\beta }=a+ib$ with $b>0$, their moduli are proportional to $e^{-bz}$ and $e^{bz}$ respectively: each decays in its direction of energy flow.

For complex frequency the displayed quadratic form is the spectral flux diagnostic specified in the question; a genuinely periodic time average is recovered on the real-frequency axis. The decrease here comes from the upper-half-plane convergence parameter, not from material dissipation. At real frequency in a lossless layer the corresponding flux is constant.

Let the shear-horizontal wave displacement field be $U(z)e^{ikx-i\omega t}$. Its velocity is $V=-i\omega U$, and its traction is $T=\mu U^{\prime }$. The scalar shear-wave equation givesWith $p=k/\omega$ and $p_{\beta }^2={\beta }^{-2}-p^2$, the first-order velocity–traction system isThese equations remain valid for depth-dependent $\mu ,\rho$; differentiating $\mu U^{\prime }$ is essential and one must not incorrectly use $\mu U^{\prime \prime }$ when $\mu$ varies.

Where $V\ne 0$, define the SH input impedance $Z=-T/V$. Differentiating $T=-ZV$ and substituting the two equations gives the SH impedance Riccati equation:The second relation is a one-way scalar equation once the complete SH input impedance is known. It does not discard reflections: $Z$ describes the actual superposition of both wave directions. At a zero of $V$, $Z$ may have a pole while the velocity–traction system stays regular.

Let $z_r$ be either endpoint at which $Z_r$ is prescribed, and put $d_z=z-z_r$, $q=\mu p_{\beta }$ and $\kappa =\omega p_{\beta }$. Integrating the constant-coefficient velocity–traction system givesTaking the ratio $-T/V$ yields the uniform-layer SH impedance update and its accompanying velocity:The initial value $V_r$ sets an arbitrary normalization; specifying impedance alone cannot specify the overall wave wave amplitude. Positive $d_z$ propagates data from $z_0$, while negative $d_z$ propagates a specified load from $z_1$, so the same expressions solve both requested initial-value problems.

Expanding sine and cosine givesAbsorbing the reference-point phases into the two constants gives the requested forward/backward exponential superposition. Equivalently, wherever the ratios are defined,The exceptional values $Z_r=q,-q$ give the pure directional solutions. If a denominator in the impedance formula vanishes, propagate the finite pair $(V,T)$ through that point; this is an impedance-chart pole, not a failure of the physical solution. The upper-half-plane assumptions ensure $q\ne 0$.

In the lower uniform half-space the causal vertical wavenumber branch has $\mathrm{Im}{k}_{\beta ,b}>0$. The downward factor $e^{i{k}_{\beta ,b}(z-z_b)}$ decays as $z\to +\infty$, whereas the upward factor grows there and represents an incoming wave from the lower infinity. The outgoing radiation condition therefore retains only the downward factor and givesFor each uniform layer $j$ of thickness $h_j$, work upwards from its known bottom load $Z_{j,b}$. With $q_j={\mu }_j{p}_{\beta ,j}$ and ${\delta }_j=\omega p_{\beta ,j}{h}_j$, part (iv) givesvelocity and traction are continuous across a bonded interface, so this top impedance becomes the bottom load of the next shallower layer. Repeat up to $z_a$ to obtain $Z_{\mathrm{in}}=Z(z_a)$. Poles can be handled by carrying $(V,T)$ or the reciprocal ratio. Since the lower load has positive real part, part (iii) actually ensures that the outgoing whole-stack impedance stays finite with positive real part at all shallower depths in the upper frequency half-plane.

The upper half-space has directional impedance $q_a={\mu }_a{p}_{\beta ,a}$. At $z_a$, an incident velocity $V_0$ and reflected velocity $R{V}_0$ giveTaking their ratio and solving establishes reflection from a layered SH impedance:Thus no individual deeper reflection coefficients are needed after $Z_{\mathrm{in}}$ is known: it contains all underlying interfaces and their coherent multiples. A matched load gives $R=0$; the real-frequency limits of zero traction and zero velocity give $R=+1$ and $R=-1$. For real frequency and propagating upper and lower half-space waves, the usual energy-flux balance provides a further check on the computed reflection and transmission. A stable implementation for thick layers uses the ratio update or its decaying-exponential version instead of multiplying growing transfer matrices.

Integrate axial force balance $\nabla \cdot \sigma =0$ over the fluid between the two remote tube sections, excluding the particle. The wall's axial traction contributesAt each remote section the Hagen-Poiseuille flow has $u_x$ independent of $x$, so ${\sigma }_{xx}=-p$. The downstream and upstream caps therefore contribute $-\pi a^{2}p(x_2)+\pi a^{2}p(x_1)$. The particle contributes zero net axial force because it is force-free. Adding the three contributions proves the wall-shear and pressure-drop balance in a tube:Here $p$ is the physical pressure. The sign follows from the outward normals on the two caps; no convention for a positive pressure-drop magnitude has yet been introduced.

Let $w=v^{\prime }-v$ and $D(v)=\mathrm{sym}\nabla v$. The supporting-hyperplane inequality for the convex viscous dissipation potential givesThe stress is symmetric because its argument is a symmetric rate-of-strain tensor, so ${\sigma }_{ij}(D_{ij}^{\prime }-D_{ij})={\sigma }_{ij}{w}_{i,j}$. Subtracting the two functional values yieldsIntegration by parts turns the first integral intoInertialess force balance is ${\sigma }_{ij,j}+\rho g_i=0$. The variation vanishes on $S_v$, and the traction is $t_i^0$ on $S_t$. Thus both the volume and boundary contributions cancel exactly, proving the convex viscous potential minimum principleOnly convexity, not strict convexity, is required, so no uniqueness follows just from this argument. For incompressible flow, admissible trial velocities must also be divergence free. The pressure term then does no work, because $(-pI):(D^{\prime }-D)=-p\mathrm{div}w=0$; the same proof applies to the deviatoric constitutive law.

At a sharp interface, continuity of velocity and traction gives, for unit incident velocity from the left,Thus $r_v=(Z_a-Z_b)/(Z_a+Z_b)$ and $t_v=2Z_a/(Z_a+Z_b)$. Normalizing the transmitted velocity by its elastic-wave energy flux gives $T_{ba}=\sqrt{Z_b/Z_a}{t}_v$, exactly the result above. Right incidence gives the reversed reflection sign and the same flux-normalized transmission. A stress reflection coefficient has the opposite sign to the velocity reflection coefficient; the convention matters.

Put $r=R_{aa}$ and $t=T_{ab}$. The incoming-to-outgoing matrix is $S_0=\left(\begin{array}{cc}r& t\\ t& -r\end{array}\right)$, with $r^2+t^2=1$. Therefore $S_0^{†}{S}_0=I$, proving equality of incoming and outgoing time-averaged flux for arbitrary coherent incident wave amplitudes, not just incidence from one side.

For all real frequencies the harmonic equations instead giveTheir right sides are equal; the diagonal terms $\pm i\omega /\beta$ have zero real contribution. ConsequentlyThe mean flux for real-part phasors is $(|{\varphi }_{+}{|}^2-|{\varphi }_{-}{|}^2)/2$. Evaluating it at the two endpoints proves exactly that total outgoing flux equals total incoming flux at every real frequency. This relies on real lossless moduli and mass density; absorption would change the identity.

Choose depth $z$ positive downwards, the free surface at $z=0$ and the interface at $z=h$. A Love wave has only transverse displacement field $u_y=U(z)e^{i(\kappa x-\omega t)}$. Take $\omega >0$, $\kappa =\omega /c_L$, and defineA trapped mode requires ${\beta }_0<c_L<\beta$, so both $q,p$ are positive. The SH equations are $U_0^{\prime \prime }+q^2{U}_0=0$ in the layer and $U^{\prime \prime }-p^{2}U=0$ in the half-space. Zero shear traction at the free surface and decay at large depth selectContinuity of displacement field and shear traction at $h$ gives $C=A\cos (qh)$ and $-{\mu }_{0}qA\sin (qh)=-\mu pC$. ConsequentlySubstituting the definitions of $q,p$ yields the requested speed-frequency dispersion relation. For negative frequency use $|\omega |$ in both vertical wavenumbers; the real displacement field has conjugate frequency partners. Neither equality in the trapped-speed interval gives a strictly localized finite-frequency Love mode.

Choose field $a$ to be $({\hat{\mathbf{u}}}^a{)}^{\ast }$ at frequency $-{\omega }_a$, and field $b$ to be ${\hat{\mathbf{u}}}^b$ at ${\omega }_b$. Zero body forces and zero boundary tractions make all source and surface terms vanish. ThereforeFor distinct squared frequencies this proves weighted orthogonality,Assume positive mass density bounded above and below and finite-energy fields in the ordinary bounded domain. The weighted space $H=L^2(\mathcal{D},\rho dV{)}^3$ is a separable Hilbert space. Select one unit eigenfunction for each distinct squared frequency. They form an orthonormal set. Let $\{d_j\}$ be a countable dense subset of $H$; assign to each eigenfunction a point $d_j$ within distance $1/3$. Two orthogonal unit eigenfunctions have distance $\sqrt{2}$, so the same $d_j$ cannot be assigned to both. This gives an injection into a countable set.

Hence there are at most countably many squared eigenfrequencies, and at most twice that many signed frequencies. Multiplicity at one frequency does not change the argument. A zero frequency associated with rigid translations or rotations is included. This proves the countability of elastic eigenfrequencies directly; proving discreteness or finite multiplicity would require additional spectral regularity arguments.

If two harmonic solutions have the same body force and prescribed boundary traction, their difference has no body force and is traction-free. A nonzero difference would therefore be a free eigenfunction at that frequency. For any frequency outside the countable exceptional set found above, the difference must vanish, so the forced boundary-value solution, when it exists, is unique. At an eigenfrequency, a free eigenfunction can be added to any existing solution, and a resonant forcing may fail to admit a steady harmonic response. This distinguishes uniqueness from existence.

For the full time-dependent problem, let $\mathbf{w}$ be the difference between two fields with identical forces, tractions and initial displacement field and velocity. It obeys $\rho {\mathbf{w}}_{tt}=\nabla \cdot \mathbf{\sigma }(\mathbf{w})$, zero boundary traction, and zero initial data. Its energy uniqueness for traction-driven elasticity follows by direct integration:Multiply the equation by ${\mathbf{w}}_t$ and integrate by parts. Major symmetry turns the strain term into the derivative of elastic energy, giving $E^{\prime }(t)={\int }_{\partial \mathcal{D}}{\mathbf{w}}_t\cdot \mathbf{t}(\mathbf{w})dS=0$. Stable elasticity makes $E\ge 0$, and $E(0)=0$ gives ${\mathbf{w}}_t=0$. Thus $\mathbf{w}$ equals its initial value zero. Even rigid zero-strain motions cannot survive these initial conditions.

The PDF correctly specifies zero initial displacement field and zero initial velocity; the TeX duplicates displacement field in the second condition. The complete causal displacement field is unique, including when its Fourier representation contains resonant modal contributions fixed by the initial data.

Use the displacement-potential convention $u_x={\partial }_x\varphi -{\partial }_z\psi$, $u_z={\partial }_z\varphi +{\partial }_x\psi$ and harmonic dependence $e^{-i\omega t}$. Choosing the opposite sign for the shear potential changes its column sign but not the P-SV directional impedance matrix. Write $p=k_{\alpha }>0$, $s=k_{\beta }>0$, $D=k^2+ps$, and $H=\rho {\omega }^2-2\mu k^2$. At a common point the forward potential wave amplitudes giveThe traction formulas follow from ${\sigma }_{xz}=\mu (u_{x,z}+u_{z,x})$ and ${\sigma }_{zz}=\lambda \nabla \cdot \mathbf{u}+2\mu u_{z,z}$, using $(\lambda +2\mu ){\omega }^2/{\alpha }^2=\rho {\omega }^2$ and $\mu (s^2-k^2)=H$. The velocity matrix has determinant ${\omega }^{2}D>0$, so elimination gives the P-SV directional impedance matrixIts off-diagonal entries are real and antisymmetric. ThereforeWith real-part phasors, the outward $z$ flux is $-\frac{1}{2}\mathrm{Re}({\mathbf{v}}^{†}\mathbf{t})$, sofor every nonzero forward displacement field. Mass density and angular frequency are essential to this flux coefficient; both appear correctly in the PDF.