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 Incoming links: Rigid boundary condition for an elastic wave

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 40C / a / Solution Created 2026-09-24 Updated 2026-09-25

Let

k_{S} = ω / c_{S}

and

k_{P} = ω / c_{P}

. Snell law for elastic waves fixes the reflected waves' common tangential wavenumber:

k_{S} sin θ = k_{P} sin ϕ, sin ϕ = \frac{c _{P}}{c _{S}} sin θ .

Choose the reflected SV-wave and converted reflected P-wave as

u_{RS} = Re {R (cos θ, 0, sin θ) e^{i k_{S} (x s i n θ - z c o s θ) - iω t}},

u_{RP} = Re {T (sin ϕ, 0, - cos ϕ) e^{i k_{P} (x s i n ϕ - z c o s ϕ) - iω t}} .

The polarizations are respectively perpendicular and parallel to their wavevectors. Phase matching makes all three exponential factors equal at

z = 0

. The rigid boundary condition for an elastic wave

u_{I} + u_{RS} + u_{RP} = 0

there gives

cos θ (1 + R) + T sin ϕ = 0,

- sin θ + R sin θ - T cos ϕ = 0.

Solving this two-by-two linear system yields

R = - \frac{cos ( θ + ϕ )}{cos ( θ - ϕ )}, T = - \frac{sin 2 θ}{cos ( θ - ϕ )} .

These fields give the Reflection of an SV-wave from a rigid plane.

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