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For two Fourier elastic wave fields with the same real elastic stiffness tensor, its major symmetry cancels the cross-strain terms in the divergence displayed above. The force-balance equations supply the remaining volume terms, and the divergence theorem gives their boundary integral of reciprocal tractions. Taking one field to be a complex conjugate gives energy-work identities and weighted modal orthogonality. The bilinear identity uses no complex conjugation until that field is chosen explicitly.

The bulk modulus relates hydrostatic compressive pressure to fractional volume compression. In isotropic three-dimensional linear elasticity, the spherical part of the stress tensor is $\kappa \mathrm{tr}(\epsilon )I$, so uniform dilation has strain energy density $\kappa (\mathrm{tr}\epsilon {)}^2/2$. Positive bulk modulus and positive shear modulus make the isotropic elastic stiffness tensor positive on nonzero symmetric strains.

An elastic laminate has planar perfectly bonded layers with a common normal. Its piecewise constant strains differ by $\mathrm{sym}(a\otimes n)$, while its normal traction is common to every layer. For positive elastic stiffness tensors, these conditions and the mean strain uniquely determine the layer strains: the difference of two solutions has zero mean and zero total elastic work, and positivity forces it to vanish.

The elastic polarization is the excess stress relative to a homogeneous comparison elastic stiffness tensor. In heterogeneous linear elasticity, $\tau =(C-C^0)\epsilon$. The comparison static elastic equilibrium equation gives the integral equation $\epsilon ={\epsilon }^0-\Gamma \tau$, where ${\epsilon }^0$ is the reference strain under the same external loading.

An elastic stiffness tensor maps a small displacement gradient tensor to the stress tensor. Minor symmetries in each index pair ensure symmetric stress and dependence only on symmetric strain; major symmetry $c_{ijpq}=c_{pqij}$ follows from an elastic energy and underlies reciprocity. Stability requires positive elastic energy on nonzero strains. In an isotropic material, $c_{ijpq}=\lambda {\delta }_{ij}{\delta }_{pq}+\mu ({\delta }_{ip}{\delta }_{jq}+{\delta }_{iq}{\delta }_{jp})$, with Lamé parameters $\lambda ,\mu$.

For a homogeneous comparison elastic stiffness tensor $C^0$, the elastic strain Green operator maps an elastic polarization to minus the induced infinitesimal strain tensor. In a bounded domain the induced displacement field has zero boundary data; in an infinite medium it is the decaying perturbation. The defining static elastic equilibrium equation is $\nabla \cdot (C^0\mathrm{sym}\nabla u+\tau )=0$. The bounded-domain operator is self-adjoint, annihilates constant elastic polarizations, has zero mean output, and satisfies $\Gamma C^0\Gamma =\Gamma$.

For unconstrained hyperelastic material, the Hessian of the strain energy density gives $\delta N_{\alpha i}=c_{\alpha i\beta j}\delta A_{j\beta }$. Commuting the second derivatives gives major symmetry $c_{\alpha i\beta j}=c_{\beta j\alpha i}$. These material-coordinate moduli linearize about a finite base deformation; they need not have the minor symmetries of a small-strain elastic stiffness tensor. Material constraints require their additional reaction increments and admissibility equations.

The two minor symmetries of the elastic stiffness tensor make it a map between six-dimensional spaces of symmetric strain and stress. An elastic energy gives major symmetry, so the matrix of this map is symmetric and has 21 independent entries. For a symmetry operation acting on strains with parity spaces of dimensions $m$ and $6-m$, invariance eliminates the mixed block and leaves $m(m+1)/2+(6-m)(7-m)/2$ independent coefficients.

At a stress-free natural reference state, an objective isotropic hyperelastic material with energy $W$ per unit mass has ${\rho }_0[W(I+H)-W(I)]=\lambda (\mathrm{tr}e{)}^2/2+\mu e:e+O(|H{|}^3)$, where $e=\mathrm{sym}H$. Twice differentiating gives the displayed Lamé parameters and the usual elastic stiffness tensor. A prestressed reference requires initial-stress corrections.

There are two chemical atoms in the primitive unit cell, hence six displacement coordinates and six phonon branches at each wavevector. Three rigid translations cost no restoring potential energy in the long-wavelength limit. ConsequentlyFor propagation along $[100]$, one acoustic phonon has longitudinal polarization along $x$ and two have transverse polarization along $y,z$. The cubic crystal symmetry makes those two transverse directions equivalent. In terms of the elastic stiffness tensor and the mass density,Thus there are two distinct acoustic velocities in general, one longitudinal and a doubly degenerate transverse value. This degeneracy of normal modes concerns the two transverse polarizations; it does not imply equality with the longitudinal speed of sound.

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

Use the usual minor and major symmetries of the elastic stiffness tensor, positive density, and strong ellipticity in elasticity. For the plane elastic wave $u=a{e}^{ik(n\cdot x-vt)}$, substitution into momentum balance gives the acoustic tensor problemThe major symmetry makes $\Gamma$ a real symmetric matrix. Moreover $a^{\mathsf{T}}\Gamma a=c_{ijpq}{a}_i{n}_j{a}_p{n}_q>0$ for nonzero $a$ under strong ellipticity in elasticity. Hence its three eigenvalues are positive and the three positive wave speeds are real. Symmetry alone proves only reality of $v^2$: an unstable isotropic tensor with negative shear modulus has negative transverse $v^2$. The requested reality therefore uses stability as part of the physical elastic-solid assumption.

The elastic slowness surface consists of three sheets, with $s=n/v_{\alpha }(n)$, satisfyingIt is centrosymmetric. Sheets can meet at degenerate polarizations; they need not be spherical or convex in an anisotropic solid. The group velocity is normal to the corresponding smooth sheet: for a sheet tangent $ds$, differentiation of $a^{\mathsf{T}}[c_{ijpq}{s}_j{s}_q-\rho I]a=0$ gives $a_i{c}_{ijpq}{a}_p{s}_{q}d{s}_j=0$. The coefficient vector is proportional to the group velocity, so its scalar product with every tangent vanishes.

Fix horizontal slowness $p=(n_1/c,n_2/c)$ and put $q=s_3$. DefineThe symbol $p$ on a tensor index is an index, whereas $p_{\alpha }$ is a horizontal slowness component. The vertical slowness equation is the vertical slowness sextic of an anisotropic solidThe leading coefficient is $\det T>0$, so it is a degree-six polynomial with real coefficients. Nonreal roots occur in conjugate pairs, leaving between zero and six real roots counted with multiplicity. Generically the counts are $0,2,4,6$; a grazing repeated root can give an odd count of distinct real roots. Geometrically these are the intersections of a vertical line of fixed horizontal slowness with the three sheets.

All four generic counts are actually possible in a stable material. For example, take $\rho =1$ and orthotropic Voigt stiffnessesThe stiffness matrix is positive definite. For horizontal slowness $(p,0)$, the three diagonal acoustic equations areThe choices $p=0.1,0.3,0.8,1.2$ give respectively six, four, two and zero real vertical roots.

For the general depth-dependent amplitude, substitution of the horizontal harmonic factor yieldsFor six simple roots $q_{\ell }$, choose nonzero null vectors $a_{\ell }$ of the corresponding matrix. The complete solution isComplex vertical roots give evanescent waves. At defective repeated roots the usual polynomial-in-depth factors multiplying exponentials must be included, or the solution can be defined by a limiting procedure.

To classify the modes, take $x_3$ positive downwards and $\omega >0$. The reduced traction for a mode isMechanical energy crosses a plane at rate $-{\sigma }_{i3}{u}_{i,t}$. Averaging the real harmonic fields uses $⟨\mathrm{Re}(f{e}^{-i\omega t})\mathrm{Re}(g{e}^{-i\omega t})⟩=\mathrm{Re}(f{g}^{\ast })/2$. With velocity amplitude $-i\omega a_{\ell }$ and traction amplitude $i\omega b_{\ell }$, the time-averaged vertical acoustic energy flux isFor a real propagating mode, write its physical wavevector as $k=\omega (p_1,p_2,q)$. Differentiate $a^{\mathsf{T}}\Gamma (k)a=\rho \omega (k{)}^2{a}^{\mathsf{T}}a$ in $k_3$. The derivatives of $a$ cancel by the eigenvalue equation, while $a^{\mathsf{T}}({\partial }_{k_3}\Gamma )a=2\omega a^{\mathsf{T}}(Tq+R)a$. Thus its vertical group velocity and flux areThus the elastic mode classification by vertical energy flux is upgoing if $J_3<0$, downgoing if $J_3>0$. In anisotropy, the sign of vertical phase slowness alone need not classify energy propagation. For evanescent modes, select $\mathrm{Im}q>0$ for decay into a downward half-space and $\mathrm{Im}q<0$ for decay into an upward half-space. Away from grazing and degeneracy, continuation from zero horizontal slowness, with a small radiation damping if needed, separates three outgoing modes on each side. Superpose only the appropriate three to construct the most general upgoing or downgoing field; limiting modes handle exceptional cases.

For a free surface at $x_3=0$ with solid occupying $x_3>0$, let $(q_0,a_0)$ be a unit-amplitude incident upgoing mode. The reflected field consists of the three downgoing propagating or depth-decaying modes:Frequency and horizontal slowness are common to all modes, by translation invariance of the surface. Zero surface traction imposes the three equationsHence the free-surface anisotropic reflection amplitudes arewhenever $B$ is invertible. This specifies mode conversion as well as reflection. If $B$ is singular, a nonzero outgoing homogeneous surface mode exists; these are surface-wave poles, including Rayleigh waves in the isotropic case. At such parameters one must use the appropriate limiting radiation solution and check solvability instead of assuming a unique finite inverse.

At a perfectly bonded elastic interface without an interfacial force, displacement and traction are continuous:Tangential derivatives of the common displacement trace are also continuous. Thus the jump of a piecewise constant displacement gradient tensor annihilates every tangential vector, and has the form $\alpha \otimes n$. Taking its symmetric part shows that the strain jump is $B\alpha$, whereThe adjoint formula follows from $\tau :B\alpha =(\tau n)\cdot \alpha$ for symmetric stress tensors. Tangential-tangential strain components are therefore common to all layers. Conversely, such gradient jumps can be realized by continuous piecewise affine displacement in the layered geometry.

Take positive elastic stiffness tensors with their usual minor and major symmetries, and a positive comparison elastic stiffness tensor $C^0$. A fictitious layer can have the same tangential displacement derivatives and the same traction as the physical layers. Hence ${\epsilon }^r={\epsilon }^0+B{\alpha }^r$. The common traction condition giveswhere $K^r=B^T{C}^{r}B$ is the positive acoustic tensor, with components $K_{ik}^r=C_{ijkl}^r{n}_j{n}_l$. Solving for ${\alpha }^r$ givesExpanding the two occurrences of $B$ gives exactly the symmetrization over $(ij)$ and $(kl)$ in the directional elastic strain Green operator specified in the paper.

For any two layers, the inverse identity for their acoustic tensors proves the required resolvent relation without interchanging noncommuting tensors:With $D_r=C^r-C^0$, this identity also gives ${\tilde{\Gamma }}^0{D}_r{\tilde{\Gamma }}^r={\tilde{\Gamma }}^0-{\tilde{\Gamma }}^r$. ThereforeThese finite-dimensional operators are inverses. Define $A_r=[I+{\tilde{\Gamma }}^0(C^r-C^0){]}^{-1}$. Then ${\epsilon }^r=A_r{\epsilon }^0$, and averaging strains and stresses givesEliminating the fictitious strain yieldsAll products have the displayed order; the elastic stiffness tensors need not commute.

It remains to justify the inverse and comparison independence. Consider two compatible equilibrated layer solutions with the same mean strain. Their difference has zero common tangential-tangential strain, so $\delta {\epsilon }^r=B{\beta }^r$, with ${\sum }_r{c}_r{\beta }^r=0$ because $B$ is injective and the mean difference is zero. Their traction difference is one common vector $g$. ThereforePositive elastic energy forces every $\delta {\epsilon }^r$ to vanish. In particular, if $⟨A⟩{\epsilon }^0=0$, all physical layer strains vanish. Their common tangential components then force ${\epsilon }^0=B{\beta }^0$, and their zero traction gives $K^0{\beta }^0=0$. Hence ${\epsilon }^0=0$, proving that $⟨A⟩$ is invertible. Finally, every positive choice of $C^0$ constructs the unique layer solution for the same $E$. Its averaged stress is unique, so the displayed effective stiffness is independent of the fictitious comparison material. This is the exact layered constitutive response, rather than a claim that piecewise constant fields meet arbitrary finite-body boundary data.

Put $\mu ={\mu }^0$, $\lambda ={\kappa }^0-2\mu /3$, and $L=\lambda +2\mu ={\kappa }^0+4\mu /3$. Assume positive bulk modulus and shear modulus. For a unit vector $\xi$, direct contraction of the isotropic elastic stiffness tensor gives the acoustic tensor and its inverse:These are its transverse and longitudinal orthogonal projections. Symmetrizing the definition of the directional elastic strain Green operator consequently givesSubstitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as distributions. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ givesThe first identity holds as a locally integrable distribution. Since the second derivative of $|s|$ is $2\delta (s)$, taking two derivatives in the first identity and four in the second givesThus an explicit complete isotropic elastic strain Green kernel isAll derivatives here are distributional derivatives, which retain the contribution at the origin.

For the requested contraction, ${\tilde{\Gamma }}_{ijkk}={\xi }_i{\xi }_j/L$, or equivalently trace the complete kernel and use $\Delta r=2/r$. ThereforeMore explicitly, its distribution form iswhere the Cauchy principal value uses spherical exclusions about the origin. The coefficient of the Dirac delta can be checked by integrating by parts over a punctured ball: ${\partial }_i{\partial }_j(r^{-1})$ has contact term $-4\pi {\delta }_{ij}\delta /3$. Taking the trace, the principal-value part has zero trace and $\Delta (r^{-1})=-4\pi \delta$, soKeeping only the ordinary kernel away from the origin would incorrectly give zero for this double trace.

Now extend the elastic polarization by zero outside $V$. Since the shear modulus is unchanged, the stiffness contrast acts only on the spherical strain:Hence $\tau =(\kappa -{\kappa }^0)\mathrm{tr}\epsilon$ is its scalar coefficient. Relative to the same external loading, the elastic strain Green operator gives $\epsilon ={\epsilon }^0-\Gamma \ast (\tau I)$. Taking the trace and using the contact identity reduces this nonlocal equation toSolving the scalar equation gives the constant-shear bulk-modulus inclusion formulaUsing the single contraction instead of the double trace givesFinally eliminate the actual trace using the scalar formula:The reference trace ${\epsilon }_{kk}^0$ in this last formula is essential and is what the original PDF prints; the converted TeX loses its superscript. The second derivatives are interpreted distributionally, or with their principal-value and contact terms explicitly included. With ordinary regularity they give the corresponding pointwise formula at regular interior points.

An elastic deformation is reversible: after removal of the deforming loads the material returns to its reference state, and its instantaneous stress is determined by its current deformation rather than a plastic or viscous history. In the infinitesimal approximation the displacement field $u_i$ gives the infinitesimal strain tensor $e_{ij}=(u_{i,j}+u_{j,i})/2$. The Cauchy stress tensor is defined by traction $t_i={\sigma }_{ij}{n}_j$ on a plane of unit normal $\mathbf{n}$; absence of body couples makes ${\sigma }_{ij}={\sigma }_{ji}$.

Let $\mathcal{E}(e)$ be the recoverable elastic energy per unit mass. The local mechanical power per unit volume isbecause the symmetric Cauchy stress tensor annihilates the antisymmetric part of the velocity gradient. Conservation of energy during the deformation gives $\rho \dot{\mathcal{E}}={\sigma }_{ij}{\dot{e}}_{ij}$. Since $\dot{\mathcal{E}}=(\partial \mathcal{E}/\partial e_{ij}){\dot{e}}_{ij}$ for arbitrary symmetric strain rates,Here the derivative is the gradient on the space of symmetric tensors, with the double-contraction inner product. This convention accounts for the two occurrences of each off-diagonal entry; treating six independent coordinates without their corresponding factors would give an erroneous factor of two in shear components.

Take a stress-free reference state of mass density ${\rho }_0$. The Taylor series of the elastic energy starts quadratically:Using $\rho ={\rho }_0+O(|e|)$ gives ${\sigma }_{ij}=c_{ijkl}{e}_{kl}+O(|e{|}^2)$, the relation of linear elasticity. If the reference state has initial stress, the first-order expansion also retains that constant stress. Symmetry of stress and strain gives the minor symmetries $c_{ijkl}=c_{jikl}=c_{ijlk}$, while equality of the mixed second derivatives of elastic energy gives the major symmetry $c_{ijkl}=c_{klij}$. Thus the elastic stiffness tensor is a symmetric map on a six-dimensional strain space: there are $6\cdot 7/2=21$ independent coefficients.

For twofold elastic symmetry, choose the symmetry axis as $x_3$. Rotation through $\pi$ is $Q=\mathrm{diag}(-1,-1,1)$, and transforms the infinitesimal strain tensor into $Qe{Q}^T$. Components $e_{11},e_{22},e_{33},e_{12}$ are even and $e_{23},e_{13}$ are odd. Invariance of elastic energy excludes every coupling between these parity classes. In the engineering strain ordering $(e_{11},e_{22},e_{33},2e_{23},2e_{13},2e_{12})$, with corresponding stress ordering $({\sigma }_{11},{\sigma }_{22},{\sigma }_{33},{\sigma }_{23},{\sigma }_{13},{\sigma }_{12})$, the symmetric stiffness matrix has the formIts symmetric even block has ten independent entries and its symmetric odd block has three. Hence a twofold axis leaves $13$ independent coefficients. No invariance under other rotation angles has been assumed.

Let $\Omega$ be the solid exterior to the cavity, and choose $\mathbf{n}$ outward from the solid, hence into the cavity on its boundary $S$. Subtraction of the incident solution from the total solution gives the scattered displacement field equationsThe traction-free cavity condition for the total field gives ${\sigma }_{ij}^s{n}_j=-{\sigma }_{ij}^0{n}_j$ on $S$. The scattered field has zero initial displacement and initial velocity before the pulse reaches the cavity and satisfies an outgoing radiation condition, excluding an additional incoming elastic wave. Equivalently one seeks the causal response to this prescribed boundary traction. Although the incident plane pulse extends across an infinite plane, the causal scattered field from the bounded cavity has finite propagation distance at each finite time.

To prove energy uniqueness for traction-driven elasticity, take the difference $\mathbf{w}$ of two solutions with the same initial conditions, body force, boundary traction and outgoing condition. It satisfiesFor a large enclosing ball $B_L$, multiply by ${\dot{w}}_i$ and integrate over $\Omega \cap B_L$. Integration by parts givesSymmetry of stress replaces the last derivative by ${\dot{e}}_{ij}(w)$, and major symmetry of the elastic stiffness tensor makes that integral the derivative of the strain energy. ThereforeThe cavity contribution is zero. On any fixed finite time interval choose $L$ beyond the causal support of the scattered difference; the outer contribution is then zero as well. This gives conservation of its nonnegative elastic energy. Positive mass density and positive strain energy imply that zero initial energy forces $\dot{\mathbf{w}}=0$ everywhere; the zero initial displacement field then gives $\mathbf{w}=0$. The same argument works for finite-energy outgoing solutions by taking the limit of enclosing surfaces and excluding incoming energy. Hence the causal scattered field ${\mathbf{u}}^s$ is unique. Specifying the incident field and cavity traction alone without the initial and outgoing conditions would not establish uniqueness.