 Reflection of a P-wave from a rigid plane

For a stable isotropic solid in $y<0$, use $u_x={\varphi }_x+{\psi }_y$, $u_y={\varphi }_y-{\psi }_x$. An incident P wave with frequency $\omega$, tangential wavenumber $\kappa$ and upward normal wavenumber $a$ reflects into a P potential $A{e}^{i(\kappa x-ay-\omega t)}$ and an SV potential $B{e}^{i(\kappa x-by-\omega t)}$, where $b^2={\omega }^2/c_s^2-{\kappa }^2$. Vanishing of both displacements at the rigid plane gives $\kappa (1+A)-bB=0$, $a(1-A)-\kappa B=0$. Hence $A=(ab-{\kappa }^2)/(ab+{\kappa }^2)$ and $B=2a\kappa /(ab+{\kappa }^2)$. Because $c_p>c_s$ and ${\kappa }^2\le {\omega }^2/c_p^2$, $b^2>0$ and the converted S wave propagates rather than being evanescent.