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Fundamental solutions

As is well known, any solution ${\bf u}({\bf r}, t)$ of the equation (30) in case of point source can be expressed in the form  
 \begin{displaymath}
{\bf u}({\bf r}, t) = f(t) \ast {\bf g}({\bf r},t)\end{displaymath} (39)
where f(t) is the source function (the function that describes the temporal behavior of a source) and ${\bf
g}({\bf r}, t)$ is the fundamental solution of a correspondent boundary value problem for the equation (30).

Fundamental solution is a response of a medium to the $\delta$-function as the source. It is easily understood that (in case of piecewise smooth media with restricted numbers of interfaces) ${\bf
g}({\bf r}, t)$ consists of some singularities (discontinuities) that propagates with correspondence to eiconal's equation (31) and relations (32) and some very smooth field ${\bf \rho} ( {\bf R}, t)$.

Let us consider a very simple example: a fundamental solution that corresponds to a point vertical force in a homogeneous medium (Figure [*]) is  
 \begin{displaymath}
{\bf g} ( {\bf r}, t ) = {\bf U}_{\rm I}( {\bf r}, t ) +
{\bf U}_{\rm II}( {\bf r}, t )\end{displaymath} (40)
where
\begin{eqnarraystar}
{\bf U}_{\rm I}( {\bf r}, t ) & = & C \left[
 {\delta \left...
 ...theta},\quad C ={1\over 4\pi\rho}, \ \ 
R = \vert {\bf r}\vert\end{eqnarraystar}
and
\begin{eqnarraystar}
{\bf U}_{\rm II}( {\bf r}, t ) & = & -C \left[
 {\delta \le...
 ...over V_s} \right )_+ \over R^3}
\right]
\cos \theta {\bf e}_R.\end{eqnarraystar}
As we see, the wave field entirely consists of standard discontinuities! Figure [*] shows radial (a) and angular (b) component for some distance R and direction $\theta$.

Let us perform the convolution in equation (39), bearing in mind that

 
f(t)*Rq(t) = Iq f(t) = fq(t) (41)

At $R\rightarrow\infty$ we shall have
\begin{eqnarraystar}
u({\bf r}, t) & =
 & {C\over V_P^2 R} f \left( t- {R\over V...
 ...\left( t - {R\over V_s} \right)
 \cos \theta {bf e}_\theta \\ \end{eqnarraystar}
The first term of the last equation is the main component of the P-wave. The second one is an additional component of the P-wave. The third and forth terms are main and additional components of the S-wave. It is interesting to note that at $\theta = \pi /2$, the P-wave has only additional (transverse!) component. In addition, at $\theta = 0$ the S-wave has only components in the ray direction. But what is indeed important for convolution of the amplitude of discontinuity of fundamental solution appears to be an amplitude of the signal! This is the real reason why discontinuities describe not only the arrival of a signal, but also the signal itself: discontinuity is an adequate tool to investigate fundamental solutions.

5: KINEMATICALLY EQUIVALENT OPERATORS OF WAVE-FIELD CONTINUATION

Let $u_0({\bf r}_0,t) $ be the observed field (${\bf r}_0 \! = \! x$ or ${\bf r}_0 \! = \! (x,y)$) that is given on the surface of observation $\Sigma \! = \! \{ {\bf r}_0 \}$ and ${\bf P}(v)$ is a linear operator that transforms the function $u_0({\bf r}_0,t) $into the field $u({\bf r},t)$ in the adjacent domain ${\bf D}= \{ ({\bf r}_0,z), \;\; z \geq 0 \}$

\begin{displaymath}
u({\bf r},t)={\bf P}(v)u_0({\bf r}_0,t).\end{displaymath}

The fields $u_0({\bf r}_0,t) $ and $u({\bf r},t)$ can be scalar or vectorial. The coordinates (x,y) can be curvilinear and the surface of observation can also be curved.

The action of the operator ${\bf P}(v)$ depends on the choice of the function $v=v({\bf r})$ which in turn is a parameter in a fixed eikonal equation

 
 \begin{displaymath}
\Psi(\nabla\tau({\bf r});v({\bf r}))=1, 
\mbox{\hspace{0.7cm}} {\bf r} \in {\bf D}.\end{displaymath} (42)
The simplest example of eikonal equation is the one for an isotropic medium

\begin{displaymath}
v^2({\bf r})\vert\nabla\tau\vert^2=1.\end{displaymath}

In an anisotropic medium $v({\bf r})$ is a vector-function. For instance, in a medium with elliptic anisotropy, we have

\begin{displaymath}
\nabla \tau \cdot {\bf G} \nabla \tau = 1\end{displaymath}

where ``" stands for the scalar product and ${\bf G}$ is a positive matrix.

We don't demand that the eikonal equation (42) is a characteristic equation of some hyperbolic differential operator. It is given, that's all!

Let us suppose that the field $u_0({\bf r}_0,t) $ contains a discontinuity $\tau_{0}$(of some order q) with a travel time curve $t=\tau_{0}({\bf r}_0)$.It means that in a neighborhood of the surface $t=\tau_{0}({\bf r}_0)$

\begin{displaymath}
u_0({\bf r}_0,t) \sim A_0({\bf r}_0) R_q[t-\tau_{0}({\bf r}_0)].\end{displaymath}

Generally speaking then, the wave field $u({\bf r},t)$ will contain one or several discontinuities (with orders not necessarily equal to q) and at least one of them will coincide with $\tau_{0}$ at ${\bf r}=({\bf r}_0,0)$.If the position of this discontinuity is described by a function $t=\tau({\bf r})$, then  
 \begin{displaymath}
\tau({\bf r}) \vert _{{\bf r} \in \Sigma} = \tau_{0}({\bf r}_0).\end{displaymath} (43)

We shall call the operator ${\bf P}(v)$ a kinematically-equivalent operator of wave field continuation (on short K-operator, KO) if the field $u({\bf r},t)$ necessarily contains at least one discontinuity with eikonal $t=\tau({\bf r})$, that satisfies equations (42) and (43) simultaneously.

The notion of K-operator may be expanded to the case when both fields $u_0({\bf r}_0,t) $ and $u({\bf r},t)$ are vectorial fields. In that case we have two equations (isotropic case) or three equations (anisotropic case):  
 \begin{displaymath}
\Psi_i(\nabla \tau,{\bf v})=0\end{displaymath} (44)
depending on the polarization of the vector-amplitude of the discontinuity. And if the field $u_0({\bf r}_0,t) $ has a discontinuity with i-th polarization at $t=\tau_{0}({\bf r}_0)$, then application of a K-operator ${\bf P}(v)$ will result in the presence of discontinuity of the field $u({\bf r},t)={\bf P}(v)u_0({\bf r}_0,t)$ which:

The question of existence of a K-operator corresponding to a given equation (42) can be simply answered (in a positive sense) if equation (42) is a characteristic equation for some equation in partial derivatives:  
 \begin{displaymath}
{\bf L} u = \rho {\partial^2 u\over\partial t^2}.\end{displaymath} (45)

Let us consider the boundary value problem for equation (45) with condition  
 \begin{displaymath}
u \vert _{{\bf r} \in \Sigma} =u_0\end{displaymath} (46)
and any additional condition  
 \begin{displaymath}
{\bf A} u \vert _{{\bf r} \in \Sigma} =0\end{displaymath} (47)
that supplies a unique solution for the problem (45) - (47).

It is obvious that the solution contains at least one discontinuity which at ${\bf r}=({\bf r}_0,0)$ coincides with $\tau_{0}$ and propagates with the eikonal satisfying equation (42).

Later we shall show that KO exists for any equation (42) if there is a solution for the Cauchy problem for equation (42) with the initial condition (43). As a matter of fact there is a whole family ${\bf P}_{\Psi}$ of K-operators that corresponds to the particular eikonal equation (42).

K-equivalence is a notion which is much wider than q-equivalence or (k)-equivalence. For instance, if ${\bf P}_1(v) , {\bf P}_2(v)$belong to the same family ${\bf P}_{\Psi}$, then it is not necessary that

\begin{displaymath}
{\bf P}_1(v) u_0({\bf r}_0,t) \sim {\bf P}_2(v) u_0({\bf r}_0,t).\end{displaymath}

Fields $u_1={\bf P}_1(v)u_0$ and $u_2={\bf P}_2(v)u_0$ can have different orders and amplitudes of discontinuities (and even their number) which satisfy the equations (42) and (43).

Let us consider K-operators for the classical eikonal equation  
 \begin{displaymath}
\vert\nabla \tau \vert^2 = {1\over v^2}.\end{displaymath} (48)
Let us pose for that equation the Cauchy problem with initial condition  
 \begin{displaymath}
\tau({\bf r}) \vert _{{\bf r} \in \Sigma} = \tau_{0}({\bf r}_0).\end{displaymath} (49)
It is easy to show that if for all ${\bf r}_0$

\begin{displaymath}
\vert\nabla_{\Sigma} \tau_0({\bf r}_0)\vert < {1 \over v_0({\bf r}_0)} \end{displaymath}

(here $v_0({\bf r}_0) \equiv v({\bf r}_0,0) $), then in a neighbourhood of the surface $\Sigma$ there are two solutions for this problem: $\tau^{+}({\bf r}) $ and $\tau^{-}({\bf r})$. In order to prove this we shall search for solution of the problem (48) - (49)) using characteristics of equation (48). Characteristics of the equation (48) (that is rays, or characteristics of the proper wave equation (45)) satisfy to the first order differential equation  
 \begin{displaymath}
\left\{ \begin{array}
{l}{d{{\bf r}} \over ds}=v
{{\bf p}} 
...
 ...bla({1 \over v}),\end{array}\right. \ \ \ {\bf p}= \nabla \tau.\end{displaymath} (50)

In order to get a unique solution of the equation (50) it is necessary (and sufficient) to have a definite value ${\bf p}={\bf p}_0$in a starting point ${\bf r}_0$.We can express the vector ${\bf p}_0$ in the form of a sum

\begin{displaymath}
{\bf p}_0 = p r_{\Sigma} \nabla \tau \vert _{{\bf r}={\bf r}_0} + a {\bf n} \end{displaymath}

where ${\bf n}$ is a normal to $\Sigma$ at the point ${\bf r}_0$, and $a={\bf p}_0 \cdot {\bf n}$ (instead of ${\bf r} \in \Sigma$ we shall write ${\bf r}={\bf r}_0$ ). The value $p r_{\Sigma} \nabla \tau({\bf r}_0)$ can be expressed through the derivatives of the function $ \tau_0(x,y) $. If $\Sigma$ is a horizontal plane and x and y are rectangular coordinates, then

\begin{displaymath}
p r_{\Sigma} \nabla \tau({\bf r}_0) = {\partial \tau_0 \over...
 ...tau_0 \over \partial y} {\bf e}_y \equiv \nabla_{\Sigma} \tau_0\end{displaymath}

where $\nabla_{\Sigma}$ means gradient in the plane $\Sigma$ and ${\bf e}_x$ and ${\bf e}_y$ are unit vectors.

Let us consider the case when $\Sigma$ is a curved surface and x and y are a curvilinear system of coordinates. We propose that the axis z coincides with the direction of a vector ${\bf n}$. Let us determine a tangent plane P to $\Sigma$ in the point ${\bf r}_0$ and let's introduce the rectangular coordinates $\xi$ and $\eta$in P, then

\begin{displaymath}
p r_{\Sigma} \nabla \tau({\bf r}_0) = {\partial \tau_0 
\ove...
 ...}_{\xi} + {\partial \tau_0 \over \partial \eta} {\bf e}_{\eta} \end{displaymath}

where

\begin{displaymath}
{\partial \tau_0 \over \partial \xi}=\left( {\partial \tau_0...
 ...au_0 \over \partial y}\right) 
{\partial y \over \partial \xi} \end{displaymath}

and

\begin{displaymath}
{\partial \tau_0 \over \partial \eta}=\left( {\partial \tau_...
 ..._0 \over \partial y}\right) 
{\partial y \over \partial \eta}. \end{displaymath}

This construction we will also denote as $ \nabla_{\Sigma} \tau_0 $. So,

\begin{displaymath}
p r_{\Sigma} \nabla \tau({\bf r}_0) = \nabla_{\Sigma} 
\tau_0({\bf r}_0).\end{displaymath}

In accordance with eikonal equation (48)

\begin{displaymath}
\vert{\bf p}_0\vert^2 = {1 \over v_0^2({\bf r}_0)},\end{displaymath}

($ v_0({\bf r}_0)\equiv v({\bf r}_0,0) 
\equiv v({\bf r})\vert _{{\bf r}={\bf r}_0} $), then

\begin{displaymath}
\vert\nabla_{\Sigma} \tau_0\vert^2 + a^2 = {1 \over v_0^2({\bf r}_0)} \end{displaymath}

and

\begin{displaymath}
a = \pm \sqrt{{1 \over v_0^2({\bf r}_0)}-\vert\nabla_{\Sigma} 
\tau_0({\bf r}_0)\vert^2}. \end{displaymath}

We have obtained that

\begin{displaymath}
\nabla \tau \vert _{{\bf r}={\bf r}_0} = \nabla_{\Sigma}\tau...
 ...({\bf r}_0)} - \vert\nabla_{\Sigma} 
\tau_0({\bf r}_0)\vert^2}.\end{displaymath}

The sign `+' determines the solution $\tau^{+}({\bf r}) $ which will be called forward eikonal continuation . The sign ``-'' determines reverse eikonal continuation $\tau^{-}({\bf r})$, (the relationship of these continuations is shown in Figure [*]).

It is easily understood that $\tau^{\pm}({\bf r})=
\tau_0({\bf r}_0)\pm T({\bf r}_0,{\bf r})$ where $T({\bf r}_0,{\bf r})$ is the time of propagation from ${\bf r}_0$ to ${\bf r}$.We shall call K-operator ${\bf P}(v)$ as the K-operator of forward (reverse) wave field continuation and denote it as ${\bf P}^{(+)}(v)$ (similarly ${\bf P}^{(-)}(v)$ ) if in the neighbourhood of $\Sigma$ the field

\begin{displaymath}
u^{(+)}({\bf r},t) = {\bf P}^{(+)}(v) u_0({\bf r}_0,t)\end{displaymath}

(similarly $ u^{(-)}={\bf P}^{(-)}(v)u_0 $ ) contains a discontinuity with eikonal $\tau^{+}$ ( $\tau^{-}$ ) and does not contain a discontinuity with the opposite eikonal $\tau^{-}$ ($\tau^{+}$) .

Each operator ${\bf P}^{(+)}$ can be used for determination of the operator ${\bf P}^{(-)}$ 
 \begin{displaymath}
{\bf P}^{(-)}={\bf I}^{-1}_{T}{\bf P}^{(+)}{\bf I}_{T}\end{displaymath} (51)
where ${\bf I}_{T}u({\bf r},t)=u({\bf r},T-t)$. And vice versa:

\begin{displaymath}
{\bf P}^{(+)}={\bf I}_{T}^{-1}{\bf P}^{(-)}{\bf I}_{T}.\end{displaymath}

There are, of course, such K-operators that contain (in a neighbourhood of $\Sigma$) discontinuities with both eikonals $\tau^{+}$ and $\tau^{-}$ (mixed type operators). This classification of K-operators may be expanded for many other eikonal equations.

The classes of this classification are still very wide. If operators ${\bf P}_1(v) $ and ${\bf P}_2(v)$ belong to the same class (forward, reverse or mixed) of K-operators and for some k  
 \begin{displaymath}
{\bf P}_2(v) u_0 \sim {\bf D}^k {\bf P}_1(v) u_0\end{displaymath} (52)
then we shall call these operators amplitude-equivalent (KA-operators). The most important case of the amplitude equivalence is the situation when the amplitude $A({\bf r})$ is a solution of a transport equation for the definite wave equation, for instance at ${\bf L} \equiv \triangle $ (classical wave equation). In the latter case discontinuity's amplitudes change with accordance to geometrical spreading (as ray's amplitudes). If in equation (52) k=0 we have dynamic equivalence.


next up previous print clean
Next: 6: WHY SO MANY Up: 4: CONNECTION WITH RAY Previous: Zero-approximation.
Stanford Exploration Project
1/13/1998