Boundary conditions at a horizontal interface

At a horizontal interface, we assume a displacement-stress vector whose variables are continuous across the interface $( \vec{\bf u} , \mbox{\boldmath$\tau_N$} )$, where $\vec{\bf u} = (u_x,u_y,u_z)^T$ is the velocity, and $\mbox{\boldmath$\tau_N$} = -1/(i\omega)\,(\tau_{xz},\tau_{yz},\tau_{zz})^T$ represents the vertical component of the stress tensor. This vector can be divided into

$$\pmatrix{ \vec{\bf u}\cr \mbox{\boldmath$\tau_N$} \cr} = F \cdot \vec{\bf w}$$ (10)

where the elements of F are

$$F_{ij} = \left( \begin{array} {cccccc} v_x^1 & v_x^2 & v_x^3... ...zz}^3 &\tau_{zz}^4 &\tau_{zz}^5 &\tau_{zz}^6 \end{array}\right)$$ (11)

and the elements of the vector $\vec{\bf w}$ are a function of the wave amplitudes, as follows:

$$w_j(z) = S_j e^{ i\omega(t-\vec{\bf x} \cdot \vec{\bf p}^j)}$$ (12)

To calculate the amplitude partitioning at an interface between two layers we equate the displacement-stress vector across the interface, thus:  

$$F^{top} \cdot \vec{\bf w}^{top} = F^{bottom} \cdot \vec{\bf w}^{bottom}$$ (13)

Translating the coordinate frame so that the interface is at z=0, the exponential terms in w are the same in both layers, and we can write equation (13) as

$$F^{top} \cdot \vec{\bf S}^{top} = F^{bottom} \cdot \vec{\bf S}^{bottom}$$ (14)

giving a general relation between the up-going and down-going wave systems in the two media. If we partition $\vec{\bf S}$ so that $\vec{\bf S}_D = ( S_1,S_2,S_3 )$ is a vector of the amplitudes of down-going waves and $\vec{\bf S}_U = ( S_4,S_5,S_6 )$ of up-going waves, we can write the block-matrix equation as

$$\pmatrix{F^{top}_{11} & F^{top}_{12}\cr F^{top}_{21} & E^{t... ...\pmatrix{\vec{\bf S}^{bottom}_D \cr \vec{\bf S}^{bottom}_U\cr}$$ (15)

In order to calculate the up-going reflected wavefield and the down-going transmitted wavefield for a downward propagating wavefield incident on the boundary from above, we need to solve the system

$$\pmatrix{F^{top}_{11} & F^{top}_{12}\cr F^{top}_{21} & E^{t... ...22}\cr } \pmatrix{\vec{\bf S}^{bottom}_D \cr \vec{\bf 0} \cr}$$ (16)

After some manipulation, we obtain Nichols (1991)

$$\begin{eqnarray} \vec{\bf S}^{top}_U & = & R_D \cdot \vec{\bf S}^{top}_D \\ \vec{\bf S}^{bottom}_D & = & T_D \cdot \vec{\bf S}^{top}_D \nonumber\end{eqnarray}$$ (17)

where

$$\begin{eqnarray} R_D & = & ( F^{bottom}_{21}(F^{bottom}_{11})^{-1}F^{top}_{12} ... ...top}_{11} - F^{top}_{11}(F^{top}_{22})^{-1}F^{top}_{21}) \nonumber\end{eqnarray}$$ (18)

The $3\times 3$ matrices R_D and T_D convert the vector of down-going wave amplitudes in layer 1 into a vector of up-going reflected wave amplitudes in layer 1 and a vector of down-going transmitted amplitudes in layer 2. The next section studies the PP wave reflection amplitudes given by the first column, first row element in matrix R_D.