= Solution
All wavevectors lie in the $xz$ plane and have the same tangential component. The incident P-wave travels upward toward $z=0$. In the lower solid there are a downward reflected P-wave, polarized parallel to its wavevector, and a downward reflected SV-wave, polarized in the $xz$ plane perpendicular to its wavevector. In the upper solid there are corresponding upward transmitted P- and SV-waves. The SH polarization decouples from this in-plane incident field. Their angles obey the <Snell law for elastic and acoustic waves>.
For a welded interface, the two in-plane displacement components are continuous:
$$
[u_x]=[u_z]=0.
$$
The two in-plane traction components are also continuous:
$$
[\sigma_{xz}]=[\sigma_{zz}]=0.
$$
These four independent scalar conditions determine the four reflected and transmitted amplitudes. This is <mode conversion at a planar elastic interface>.
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