Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 82 / 1

Put $Z=\rho \beta =\sqrt{\rho \mu }>0$, the seismic impedance. In a uniform medium, define $w_{+}=v-\sigma /Z$ and $w_{-}=v+\sigma /Z$. The field equations giveThus $w_{+}=2F(x-\beta t)$ and $w_{-}=2G(x+\beta t)$, with arbitrary independent profiles $F,G$. Reconstruction givesThe two characteristic components therefore have $v_{+}=F$, ${\sigma }_{+}=-Z{v}_{+}$ and $v_{-}=G$, ${\sigma }_{-}=Z{v}_{-}$.

The kinetic plus elastic energy mass density is $e=\rho v^2/2+{\sigma }^2/(2\mu )$. Multiplying the equations by $v$ and $\sigma /\mu$ yields $e_t=(v\sigma {)}_x$. Hence the elastic-wave energy flux is $S=-v\sigma$. For a pure forward wave $S_{+}=Z{v}_{+}^2\ge 0$; for a pure backward wave $S_{-}=-Z{v}_{-}^2\le 0$. In the mixture the cross terms cancel, giving $S=Z(v_{+}^2-v_{-}^2)$. Strict sign holds for nonzero components.

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.

Use a time-harmonic wave with time dependence $e^{-i\omega t}$ and let the same symbols denote phasors. The wave amplitude vector obeysAt fixed material contrast, the stated low-frequency condition permits neglecting the diagonal frequency term to zeroth order. All remaining coefficient matrices are scalar multiples of the same exchange matrix, so they commute. Withthe transfer relation isSolve the second row for the outgoing left wave amplitude and substitute in the first. Since $C^2-H^2=1$, this gives the zero-frequency scattering through an elastic layer coefficientsThus the zero-frequency limit depends only on the endpoint seismic impedances. Monotonicity supports the given scale comparison but is not needed to integrate the zero-frequency equation. If $Z_a=Z_b$ in a monotone profile, seismic impedance is constant: there is no directional coupling, with only wave phase accumulation at nonzero frequency.

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.