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Zero-approximation.

Let us consider the equation (32) for n = 0:

\begin{displaymath}
{\bf N} {\bf U}_{0} = 0.\end{displaymath}

With accordance to equation (23), this is a linear algebraic equation with degenerate matrices. It may be solved only in the case when ${\bf U}_{0}$ is an eigenvector. We restrict ourselves with the situation when the eiconal equation for P-waves is valid. In this case, the eigenvector is ${\bf P} = \nabla \tau_{P}$ so vector ${\bf U}_{0}$ can be expressed in the form ${\bf U}_{0} = A_{0} \nabla \tau_{P}$. In order to specify the amplitude A0 let us consider the equation (32) for n = 1:  
 \begin{displaymath}
{\bf N U}_{1} = {\bf M} (A_{0} \nabla \tau_{P})\end{displaymath} (34)

But any vector $\bf U_{1}$ can be decomposed into the sum ${\bf U_{1}} = A_{1} \nabla \tau_{P} + {\bf Q}$ where ${\bf
Q} \perp \nabla \tau_{P}$. Both vectors $\nabla \tau_{P}$and $\bf Q$ are eigenvectors connected with eigenvalues and $\mu \mid \nabla \tau_{P} \mid^{2} - \rho$correspondently, therefore,

\begin{displaymath}
{\bf N U}_{1} \equiv {\bf N} (A_{1} \nabla \tau_{P} + {\bf
Q}) \equiv (\mu \mid {\nabla \tau_P} \mid^{2} - \rho) {\bf
Q}\end{displaymath}

We see that the left side of the equation (34) is orthogonal to the vector $\nabla \tau_{P}$. It means that equation (33) can be resolved if and only if  
 \begin{displaymath}
\nabla \tau_{P} \cdot M\ ( A_{0} \nabla \tau_{P} ) = 0\end{displaymath} (35)

Inserting expression in equation (33) into the last equation, we use the property of the operator $\nabla \tau_{P}$, if s is the natural parameter of a ray ${\bf r} = {\bf r} (s)$, then for any scalar function $\psi ({\bf r})$

\begin{displaymath}
\nabla \tau_{P} \cdot \nabla \psi =\frac{1}{v_P} \frac
{d \psi}{ds}\end{displaymath}

where $d[\cdot]/ds$ means differentiation along the ray that intersects the given point. After simple manipulation we derive from equation (35) the transient equation for the amplitude A = A0/v:  
 \begin{displaymath}
\frac{dA}{ds} + \frac{1}{2}A[v \triangle \tau_{P} + \frac{
d (\ln \rho {v_P}^2)}{ds}] = 0\end{displaymath} (36)
which have solution

\begin{displaymath}
A(s) = \frac{A(0)}{v_P \sqrt{\rho}}
e^{-{1 \over 2} \int_0^s v_P \triangle \tau_{P} ds }\end{displaymath}

The well-known notion of geometrical spreading $L = \sqrt{
J}$ is connected with $\triangle \tau$ by relation (S.V. Goldin, 1986)

\begin{displaymath}
v \triangle \tau = d ( \ln {J} /v ) / ds\end{displaymath}

consequently,  
 \begin{displaymath}
A(s) = A(0)/ \sqrt{\rho v_{P} {J} }\end{displaymath} (37)
Analogous formula is true for S-waves.

The equation (37) describes how the amplitude of the senior discontinuity of a wave is changing along a ray path.

If the value of J is positive in all points of a given ray ${\bf r} = {\bf r} (s)$, then along the ray  
 \begin{displaymath}
{\bf U} \sim \frac{A(0)}{\sqrt{\rho v_{P} {J}
}}R_{q_{0}} [t - t(s)]{\bf t}\end{displaymath} (38)
where $t(s) = \tau_{P}[{\bf r}(s)]$ and ${\bf t} =
v_{P}\nabla\tau_{P}$ is the tangent-vector of the ray. And if J < 0, then

\begin{displaymath}
{\bf U} \sim \frac{A(0)}{\sqrt{\rho v_{P}{J}}}
R_{q_{0}, 1}[t - t(s)]{\bf t}\end{displaymath}

The situation when J = 0 means that $u \sim
A'R_{q', v'}$ at q'<q0, $0 \leq v' \leq
1$.


next up previous print clean
Next: Fundamental solutions Up: 4: CONNECTION WITH RAY Previous: Ray method
Stanford Exploration Project
1/13/1998