Write , with , and a chosen square root . Squaring givesOne cannot have : the second relation would force , and then the first would require the square of a real number to be negative. ConsequentlyThe bracket has the sign of , proving the causal vertical wavenumber branch equivalenceThis selects the square root analytically in the upper frequency half-plane. At it is ; continuation supplies the outgoing branch for other frequencies. The expression must use this inherited branch rather than an independently chosen square root. For positive real frequency its limit is positive real for a propagating wave and positive imaginary for an evanescent wave.
Let the shear-horizontal wave displacement field be . Its velocity is , and its traction is . The scalar shear-wave equation givesWith and , the first-order velocity–traction system isThese equations remain valid for depth-dependent ; differentiating is essential and one must not incorrectly use when varies.
Where , define the SH input impedance . Differentiating 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: describes the actual superposition of both wave directions. At a zero of , may have a pole while the velocity–traction system stays regular.
Use the quadratic spectral flux . For positive real material mass density and shear modulus, differentiation of the velocity–traction system givesBecause is real, . If , then and . This proves complex-frequency SH flux monotonicity:Strictness holds for a nontrivial wavefield: if 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 , and hence , are continuous; integrating the negative derivative gives strict decrease across any positive-depth interval as well.
Where the SH input impedance is finite, . If , then and all shallower fluxes are larger and positive. They cannot have , since that would make . Thus for every . Likewise, negative real SH input impedance at gives negative flux throughout , excludes velocity zeros there and implies for every .
In a uniform region let on the causal vertical wavenumber branch. The constant Riccati solutions are , and integrating the one-way equation givesSince , the plus solution carries positive downward flux and the minus solution negative upward flux. Writing with , their moduli are proportional to and 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 be either endpoint at which is prescribed, and put , and . Integrating the constant-coefficient velocity–traction system givesTaking the ratio yields the uniform-layer SH impedance update and its accompanying velocity:The initial value sets an arbitrary normalization; specifying impedance alone cannot specify the overall wave wave amplitude. Positive propagates data from , while negative propagates a specified load from , 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 give the pure directional solutions. If a denominator in the impedance formula vanishes, propagate the finite pair through that point; this is an impedance-chart pole, not a failure of the physical solution. The upper-half-plane assumptions ensure .
In the lower uniform half-space the causal vertical wavenumber branch has . The downward factor decays as , 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 of thickness , work upwards from its known bottom load . With and , 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 to obtain . Poles can be handled by carrying 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 . At , an incident velocity and reflected velocity giveTaking their ratio and solving establishes reflection from a layered SH impedance:Thus no individual deeper reflection coefficients are needed after is known: it contains all underlying interfaces and their coherent multiples. A matched load gives ; the real-frequency limits of zero traction and zero velocity give and . 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.
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