= Solution
For the <reflection of a P-wave from a rigid plane>, use the <displacement> convention $u_x=\phi_x+\psi_y$, $u_y=\phi_y-\psi_x$, where $\phi$ is the P potential and $\psi$ the SV potential. Put $\omega=kc_p$, $\kappa=k\sin\theta$, $a=k\cos\theta$, and $b=\sqrt{\omega^2/c_s^2-\kappa^2}$. The reflected waves must share the tangential <wavenumber> and <frequency> and propagate towards negative $y$, so write
$$
\phi_R=Ae^{i(\kappa x-ay-\omega t)},\qquad\psi_R=Be^{i(\kappa x-by-\omega t)}.
$$
The rigid barrier requires both <displacement> components to vanish. On $y=0$ this gives $\kappa(1+A)-bB=0$ and $a(1-A)-\kappa B=0$. Solving,
$$
\boxed{A=\frac{ab-\kappa^2}{ab+\kappa^2},\qquad B=\frac{2a\kappa}{ab+\kappa^2}.}
$$
The total P potential is the given incident wave plus $\phi_R$, and the S potential is $\psi_R$. Reversing the sign convention for $\psi$ reverses $B$ but leaves the <displacement> unchanged. At normal incidence $A=1,B=0$.
An evanescent reflected <S wave> would require $b^2<0$, or $\sin\theta>c_p/c_s$. In a stable isotropic elastic solid $c_p>c_s$, so this is impossible for real incidence angles: $b^2=k^2[(c_p/c_s)^2-\sin^2\theta]>0$. \b[The reflected <S wave> is never evanescent in the physical isotropic case.] The inequality identifies the hypothetical critical-angle case if the speed ordering were reversed; it is S-to-P conversion, not this P-to-S conversion, that can become evanescent in an ordinary solid.
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