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For a homogeneous isotropic solid with constant mass density $\rho >0$ and Lamé parameters $\lambda ,\mu$, the Navier-Cauchy equation isTaking the divergence, $d=\nabla \cdot u$ obeys $d_{tt}=c_P^2\Delta d$ with $c_P^2=(\lambda +2\mu )/\rho$. Taking the curl, $w=\nabla \times u$ obeys $w_{tt}=c_S^2\Delta w$ with $c_S^2=\mu /\rho$, because the curl of a gradient vanishes. These are the longitudinal P wave and transverse S wave speeds.

A suitable Helmholtz decomposition into a scalar-potential gradient and a divergence-free part separates the corresponding motions. In two-dimensional in-plane propagation, the P-SV displacement potentials implement this separation with two scalar wave equations. The constant-coefficient and isotropy assumptions matter: interfaces can couple P and SV modes, and arbitrary heterogeneous coefficients cannot be passed through the divergence and curl as if they were constants.

Use local coordinates with the interface $z=0$, solid below and fluid above, and choose the P-SV displacement potentials convention $u_x={\varphi }_x-{\psi }_z$, $u_z={\varphi }_z+{\psi }_x$. Let the reflected P and SV potential wave amplitudes be $A,B$, while the transmitted P wave amplitude is $T_P$. All four waves have the same tangential slownessThe incident and transmitted retarded times have negative $q_{P}z,q_{f}z$, while elastic waves have positive $q_{P}z,q_{S}z$. At the interface all waveform derivatives therefore have the same argument. displacement field continuity givesWith Lamé parameters $\mu =\rho {\beta }^2$ and $\lambda =\rho ({\alpha }^2-2{\beta }^2)$, the shear and normal stresses follow directly from differentiation:The fluid has no shear traction, so the coefficient of $f^{\prime \prime }$ in ${\sigma }_{xz}$ must vanish:The coefficient of $f^{\prime \prime }$ in the solid stress is $\rho [\cos 2{\theta }_S(1+A)-\sin 2{\theta }_{S}B]$; in the fluid it is ${\rho }_f{T}_P$. ThusThese are the three required matching conditions; tangential displacement field need not be continuous at a solid–inviscid-fluid boundary.

Put $C=\cos 2{\theta }_S$ and $X=1-A$. From (2), $B=({\beta }^2/{\alpha }^2)(\sin 2{\theta }_P/C)X$. In (1), the coefficient of $X$ simplifies because $C+2({\beta }^2/{\alpha }^2){\sin }^2{\theta }_P=1$. Hence $X=(\alpha C\cos {\theta }_f)/({\alpha }_f\cos {\theta }_P)T_P$. Substitute this into (3) to obtainTherefore the solid-fluid P-wave transmission coefficient isDivision by $C$ was only an elimination device; the final expression extends continuously to $C=0$. The normal-incidence value is $2\rho {\alpha }_f/(\rho \alpha +{\rho }_f{\alpha }_f)$. This is a potential coefficient; longitudinal displacement field transmission additionally multiplies it by $\alpha /{\alpha }_f$. The original PDF's bracket lies in the denominator and contains ${\beta }^2/{\alpha }^2$ and $\cos {\theta }_f$. The converted TeX misplaces the bracket, substitutes ${\theta }^2$ for ${\beta }^2$, and loses the final cosine.

For downward-phase reversal, the P-SV displacement potentials instead produceEliminating the potentials as before givesThe Hermitian part is negative definite, and consequentlyfor a nonzero backward displacement field. The P potential must use vertical wavenumber $k_{\alpha }$ and the S potential $k_{\beta }$, as in the original PDF; the TeX incorrectly gives the shear wavenumber for both.

In a uniform isotropic material, these scalar P-SV displacement potentials separate longitudinal P waves from vertically polarized SV-waves in an $x,z$ plane. They satisfy ${\varphi }_{tt}={\alpha }^2\Delta \varphi$ and ${\psi }_{tt}={\beta }^2\Delta \psi$, with compressional and shear speeds $\alpha ,\beta$. Reversing the sign convention for $\psi$ changes its polarization column but not the physical P-SV directional impedance matrix or the separate modal energy fluxes.