Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 79 3 v Solution Created 2026-10-03 Updated 2026-10-14
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.