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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.

Because $s$ is arc length, $|d\mathbf{x}/ds|=1$. Thus $\mathbf{y}=\mathbf{n}/\alpha$, where $\mathbf{n}$ is the unit tangent to the elastic ray. In an isotropic medium this is also the wavefront normal. Consequently $\mathbf{y}$ is the vector slowness, of magnitude $1/\alpha$; locally $\mathbf{y}=\nabla T=\mathbf{k}/\omega$, with $T$ the travel time and $\mathbf{k}$ the wavevector.

For depth-dependent wave speed $\alpha (z)$, the ray equation gives $d{y}_x/ds=d{y}_y/ds=0$. The horizontal component of $d\mathbf{x}/ds$ therefore has a fixed direction, and the elastic ray lies in the vertical plane determined by that direction and its initial point. A vertical elastic ray is the limiting case. If $i$ is the angle to the vertical, its conserved horizontal slowness givesAt a horizontal interface, matching the tangential wavevector at fixed frequency gives the same invariant on both sides: this is Snell's law.

For radial wave speed $\alpha (r)$, take the cross product of position with slowness. The ray equations givesince the second vector is radial. A nonzero constant $\mathbf{L}=\mathbf{x}\times \mathbf{y}$ is perpendicular to every position on the elastic ray, so the ray lies in a fixed plane through the centre. Radial rays obey the same conclusion with a nonunique plane. The spherical elastic ray invariant is thereforewhere $i$ is measured from the radial direction. Across a spherical interface the radius is common and tangential slowness is continuous, so this form of Snell's law survives an interface discontinuity.

In a homogeneous sphere a P wave follows the straight chord between the two surface points. If their principal angular separation is $0\le \Delta \le \pi$, its length is $2R\sin (\Delta /2)$. For constant P wave speed $c_P$,Now put $a=R^{\prime }$ and denote the mantle and core P wave speeds by $c_m>c_c$. Equal incidence and reflection angles put the core-boundary reflection point on the angular bisector. Each mantle leg has length $L$ given by $L^2=R^2+a^2-2Ra\cos (\Delta /2)$. Hence the core-reflected travel time isThe stated domain is essential. At the reflection point the incoming line approaches from the mantle only if $R\cos (\Delta /2)\ge a$. Otherwise it has already crossed the core boundary. The mantle-only direct chord also exists only for $\Delta \le {\Delta }_g$, with $T_D=2R\sin (\Delta /2)/c_m$. In this interval,so the reflected arrival is later. Its intercept is $2(R-a)/c_m$, whereas the direct intercept is zero. At grazing incidence the branches meet at $T=2\sqrt{R^2-a^2}/c_m$, and both have slope $dT/d\Delta =a/c_m$. A continuation of the homogeneous-sphere direct formula beyond this endpoint is not a mantle-only arrival in the modified model.

The core-transmitted elastic rays consist of three straight legs. Their impact distances in mantle and core are $b_m=p{c}_m$ and $b_c=p{c}_c$, respectively. At core entry,Since $c_c<c_m$, the elastic ray bends towards the normal on entry and away from it on exit. The mantle leg subtends $i_m-i_R$ at the centre, and the core chord subtends $\pi -2i_c$. Thus the total angular sweep along the transmitted path is $2(i_m-i_R)+\pi -2i_c$. This sweep need not equal the principal separation if it exceeds $\pi$; the diagram uses a path with sweep below $\pi$.

Use a linear, lossless arterial pressure wave model with an incompressible fluid of mass density $\rho$, small wall displacement and negligible mean-flow advection compared with the wave speed. Let the parent vessel have equilibrium area $A_1$ and speed $c_1$. Linearizing mass conservation and axial momentum gives $a_t+A_1{u}_x=0$ and $\rho u_t=-p_x$. With wall compliance $p=(dp/dA)a$, these imply $c_1^2=(A_1/\rho )dp/dA$. A stiffer wall therefore has larger wave speed. A downstream travelling pressure wave carries flux $Q_{+}=Y_1{p}_{+}$, where $Y_1=A_1/(\rho c_1)$ is the characteristic admittance of an arterial wave; a returning wave has $Q_{-}=-Y_1{p}_{-}$.

Place the reflecting junction at $x=L$. Treat the daughters as outgoing-wave loads, with combined admittance $Y_d={\sum }_j{A}_j/(\rho c_j)$ and no further returning waves. At the junction, pressure continuity and flux conservation give $p_I+p_R=p_T$ and $Y_1(p_I-p_R)=Y_d{p}_T$. Thus the bifurcation reflection of an arterial pressure wave hasA decrease in total area gives $0<r<1$ if the parent and daughter wave speeds are equal. More generally this requires $Y_d<Y_1$, an assumption about both area and stiffness. To isolate the effect of ageing, assume all relevant wave speeds increase by the same factor, so their ratios and hence $r$ stay fixed. Unchanged geometry alone does not ensure unchanged $r$ if the vessels stiffen by different factors.

At a station $x<L$, put $\varphi =\omega (t-x/c_1)$ and $\vartheta =2\omega (L-x)/c_1$, the round-trip phase to the reflector. The incident pressure harmonic and its reflection giveFor time convention $e^{i\varphi }$, the two phasors are $P=1+r{e}^{-i\vartheta }$ and $V=1-r{e}^{-i\vartheta }$. Their maximum times are shifted by $-\arg P/\omega$ and $-\arg V/\omega$, respectively. Hence the pressure-flow phase lag from one arterial reflection isIndeed $V/P$ has the argument of $1-r^2+2ir\sin \vartheta$. On the branch $0<\vartheta <\pi$, this phase lag is positive: the pressure maximum follows the flow maximum. There is no cycle-wrap ambiguity on this branch because $1-r^2>0$.

The age trend need not have the same sign at every station. Differentiating the phase givesIncreasing $c_1$ decreases $\vartheta$. At a distal station with $0<\vartheta <\pi /2$, the lag therefore decreases. At a more proximal, ascending-aortic station with $\pi /2<\vartheta <\pi$, it increases. Thus one sufficient set of assumptions reproducing both trends isAs long as ageing keeps the stations in these ranges, the distal lag falls while the proximal lag rises. The effect is interference between pressure waves that add and flow waves that subtract, not merely the shortening of one travel time. The lag is largest at a round-trip phase of $\pi /2$.

This is a conditional explanation, with fixed representative harmonic frequency, fixed reflector location, negligible attenuation, negligible additional reflections and common stiffness scaling. The given stiffness and geometry facts alone do not specify the phase ranges, so they do not force either age trend without these extra assumptions. Here the two stations lie on the incident-plus-reflected segment. A station beyond the sole junction in an ideal outgoing daughter has only a transmitted wave and zero pressure-flow lag in this model. Actual peripheral measurements beyond that junction require returning waves from further junctions or terminal loads; a single-reflector calculation cannot establish their lag. Non-sinusoidal waveforms, distributed viscosity, mean flow and differently evolving reflection coefficients can also shift peak times. These limitations matter when applying the harmonic explanation to a real arterial network.

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.

Let the positive real material parameters depend on $x$ but not on time. Define the flux-normalized elastic characteristic amplitudes and logarithmic seismic impedance gradientWrite $P={\varphi }_{+}+{\varphi }_{-}$ and $Q={\varphi }_{+}-{\varphi }_{-}$. Then $v=Z^{-1/2}P$, $\sigma =-Z^{1/2}Q$, and direct substitution givesFor example, the first relation uses $(Z^{1/2}{)}^{\prime }=g{Z}^{1/2}$ and $Z/\rho =\beta$; the second uses $\mu /Z=\beta$. Adding and subtracting now provesThe original PDF has these two first-order equations. The TeX mistakenly combines them into ratios, which must not be used. A seismic impedance gradient couples forward and backward wave amplitudes even when the local wave speed is allowed to vary.

Take the wave speed $c>0$ and introduce characteristic coordinates $\xi =x+ct$, $\eta =x-ct$. The chain rule givesThe wave equation therefore says $y_{\xi \eta }=0$. Integrating first with respect to $\eta$ gives $y_{\xi }=F(\xi )$, and integrating with respect to $\xi$ gives $y=f(x+ct)+g(x-ct)$ with twice differentiable $f,g$. Conversely substitution shows that every such sum solves the wave equation. The two terms are travelling waves of unchanged shape, moving respectively to the left and right at speed $c$.

The zero initial displacement gives $g(x)=-f(x)$, and the initial velocity gives $2c{f}^{\prime }(x)=\psi (x)$. Hence the D'Alembert formula with initial velocity isFor a classical twice differentiable solution one may take $\psi \in C^1$. Differentiation gives $y_t(x,0)=\psi (x)$ and $y(x,0)=0$ directly. The coordinate calculation also works for any nonzero signed $c$, with physical speed $|c|$. If the printed constant is allowed to be $c=0$, the equation instead gives $y=A(x)+tB(x)$ and these initial data give $y=t\psi (x)$; the displayed travelling-wave representation requires $c\ne 0$.

For the wave equation with d'Alembert operator $\square =-{\partial }_t^2+\Delta$, the wave speed is one. The domain of dependence of $(t,x)$ is the initial ball $\overline{B}(x,t)$. More precisely, two solutions whose Cauchy data agree on a neighbourhood of that ball have the same value at $(t,x)$. Thus, if both Cauchy data have support in $K$, finite propagation speed givesThe Strong Huygens principle is sharper in three spatial dimensions: only the initial data on the sphere $|y-x|=t$ contribute, including the first spatial derivatives of the initial displacement there. The Kirchhoff formula makes this precise:In particular, if the Cauchy data have compact support in $\overline{B}(0,R)$, thenFor $t>R$, the solution therefore vanishes behind the inward edge $|x|=t-R$ as well as outside $|x|=t+R$. The absence of an interior tail is the additional content of the Strong Huygens principle.

For $\mathbf{r}=R\hat{\mathbf{r}}$ with the observation distance much larger than the scatterer, expand $|\mathbf{r}-{\mathbf{r}}^{\prime }|=R-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }+O(R^{-1})$. The far-field pattern in the Born approximation for scalar wave scattering is consequentlyThis convention includes the factor $1/(4\pi )$ in $f_{\infty }$; moving that factor into the outgoing-wave definition would change the pattern's normalization.

The absolute energy flux depends on what physical amplitude $\psi$ represents. For acoustic pressure in a homogeneous fluid of mass density ${\rho }_0$ and wave speed $c_0$, the momentum equation gives the particle-velocity amplitude $\mathbf{v}=\nabla \psi /(i{\rho }_0\omega )$. The time average of harmonic power and acoustic intensity then giveThe radial derivative of the outgoing factor is $(ik-1/R)e^{ikR}/R$. Its imaginary current is therefore $k|f_{\infty }{|}^2/R^2$ to leading order. The incident unit-amplitude plane wave has imaginary current $k{\hat{\mathbf{r}}}_0$. Hence, with $k=\omega /c_0$,The scattered energy flux points along $\hat{\mathbf{r}}$ at leading order, and the incident energy flux along ${\hat{\mathbf{r}}}_0$. If $\psi$ is a velocity-potential amplitude instead, both leading expressions use the common prefactor ${\rho }_0\omega k/2$ in place of $k/(2{\rho }_0\omega )$. A normalized scalar current uses $k$ as the common prefactor. The paper does not specify this amplitude normalization; the ratio in part (c) is independent of that choice.