Reports of the Academy of Sciences of the USSR

1. Vol. 163, No. 6

GEOPHYSICS

M. L. GERVER, V. M. MARKUSHEVICH

INVESTIGATION OF THE NONUNIQUENESS IN DETERMINING, FROM THE HODOGRAPH, THE PROPAGATION VELOCITY OF A SEISMIC WAVE

(Presented by Academician E. K. Fedorov on 6 February 1965)

I. Let a positive, piecewise twice differentiable function \(u(y)\) be defined on the half-axis \(y \ge 0\); let \(u(0)=1\), let \(u(y)\) be bounded on every finite interval, and unbounded on the whole half-axis.

1. Consider the following problem (see Fig. 1). A material point is at the initial moment at the point \(O\) of the plane \(x,y\). It begins to move in a direction forming an angle \(\alpha\) with the \(y\)-axis, \(0<\alpha<\pi/2\). The trajectory of motion \(L\) is determined by the following conditions: 1) denote \(\sin \alpha\) by \(p\); at each point \((x,y)\) of its descending branch the trajectory \(L\) forms with the \(y\)-axis an angle \(\alpha(y) \le \pi/2\) such that \(\sin \alpha(y)=p u(y)\); 2) the deepest point of \(L\) has ordinate \(Y(p)=\inf \{y,\; pu(y)\ge 1\}\); 3) if the abscissa of the deepest point \(X(p)<\infty\), then \(L\) also has an ascending branch, symmetric to the descending one with respect to the straight line \(x=X(p)\). Being at the point \((x,y)\in L\), the material point has velocity \(u(y)\). Let \(T(p)\) be the time of motion along \(L\) from \(0\) to \((X(p),Y(p))\). The trajectory \(L\) will be called a ray, and the number \(p\) the ray parameter. A ray with parameter \(p\) that has an ascending branch ends at the point \(x=2X(p),\ y=0\); the time of motion along it is equal to \(2T(p)\).

Fig. 1

The problem is posed as follows: the curve \(\Gamma\{x=2X(p),\ t=2T(p),\ p\in(0,1)\}\) (we shall call it the hodograph) is known; find \(u(y)\).

1. The geophysical problem of determining the velocity section from the hodograph, more precisely an idealized version of this problem, can be reduced to this problem: the Earth is regarded as a sphere; it is assumed that seismic impulses propagate along seismic rays according to the laws of geometrical optics, and that their propagation velocity depends only on depth.

Let us carry out this reduction. It is clearly sufficient to consider seismic rays not in a sphere, but in a disk. Let \(K\) be a disk of radius \(R\) with center \(C\), and let \(\gamma\) be its circumference (see Fig. 2). Suppose that an impulse arises at a point \(A\in\gamma\). Take polar coordinates \(r,\theta\), and let us measure angles from the radius \(CA\). Let the propagation velocity of the impulse be \(v(r)\), \(r\in[0,R]\). The travel time to a point \(B\in\gamma\) is equal to \(\displaystyle \int_{AB}\frac{ds}{v}\), where

Fig. 2

\[ ds=\sqrt{dr^2+r^2d\theta^2}. \]

If one makes the transformation

\[ x=\frac{R}{v(R)}\theta,\qquad y=\frac{R}{v(R)}\ln\frac{R}{r}, \]

\[ u(y)=\frac{v\left(Re^{-(v(R)/R)y}\right)} {v(R)e^{-(v(R)/R)y}}, \]

then we arrive at the problem on the plane described above.

(We assume that the angular distance between \(A\) and \(B\) along any ray \(AB\)

known exactly, and not modulo \(2\pi\) *. The choice of the function \(u(y)\) is determined by the following requirements: 1) let \(L\) denote the image of the ray \(AB\); then
\[ \int_{AB}\frac{ds}{v}=\int_L\frac{dl}{u}, \qquad dl=\sqrt{dx^2+dy^2}; \]
2) \(u(0)=1\).) The rays \(L\) considered in item I, 1 are, to be sure, only part of the images of the seismic rays \(AB\): where the velocity has a discontinuity, the ray may split into a reflected and a refracted ray; in this case we consider only the refracted rays.

Fig. 3

Taking into account the remarks made, the mechanical problem I, 1 and the geophysical problem I, 2 are equivalent.

1. Natural from the physical point of view is the following additional restriction on \(u(y)\): the set of points \(y^0>0\) where
   \[ u(y^0)<s(y^0)=\sup\{u(y),\ y\in[0,y^0]\} \]
   consists of a finite number of intervals. We shall assume that \(\Gamma\) is not arbitrary, but serves as the hodograph for some function \(u(y)\) satisfying all the requirements of item I and the last restriction, and, moreover, that the function \(X(p)\) (unknown to us in advance) is not constant on any segment.

II. Let us clarify some properties of the functions \(Y(p)\), \(X(p)\), and \(T(p)\).

1a. Put \(f(y)=1/s(y)\). The function \(f(y)\), obviously, does not increase, \(0<f(y)\le 1\). Consider the graph of \(f(y)\) in the plane \(y,p\) (see Fig. 3). Let \(y^0\) be a point of discontinuity of \(f(y)\); join the points \(\{y^0,f(y^0-0)\}\) and \(\{y^0,f(y^0+0)\}\) by a straight-line segment and adjoin it to the graph of \(f(y)\). We proceed similarly with the graph of \(Y(p)\). Then, by virtue of the relation
\[ Y(p)=\inf\{y,\ pu(y)\ge 1\}=\inf\{y,\ f(y)\le p\}, \]
the graphs of \(f(y)\) and \(Y(p)\) (as curves in the plane \(y,p\)) coincide (on an interval of monotonicity these are mutually inverse functions).

b. The function \(Y(p)\) does not increase. If for some \(p\), \(Y(p-0)>Y(p+0)\), then on the interval \((Y(p+0),Y(p-0))\) \(f(y)\) is constant and equal to \(p\). The latter condition is necessary and sufficient for the point \(p\in(0,1)\) to be a point of discontinuity of \(Y(p)\).

From the assumption on the structure of the set of points \(y\) where \(u(y)<s(y)\), it follows that there exist numbers \(p_1>p_2>\cdots>p_n>0\) possessing the following properties: put
\[ y_k=Y(p_k+0), \qquad \bar y_k=Y(p_k-0), \qquad 1\le k\le n; \]
then each of the intervals \(j_k=(y_k,\bar y_k)\) contains points where \(f(y)<1/u(y)\), and outside the intervals \(j_k\), \(f(y)=1/u(y)\). For definiteness we shall assume that \(p_1<1\), so that for \(y\in[0,y_1]\), \(f(y)=1/u(y)\). Put
\[ p_0=1,\qquad p_{n+1}=0,\qquad y_0=\bar y_0=0. \]
We shall soon see that the numbers \(p_1,\ldots,p_n\) play a very important role.

1. Mark on the interval \((0,1)\) of the \(p\)-axis a finite number of points: the points \(f(y-0)\) and \(f(y+0)\), if \(f(y-0)>f(y+0)\), and the points \(f(y^0)\), if \(f(y)\) is continuous but \(f'(y)\) or \(f''(y)\) either does not exist or is discontinuous at \(y=y^0\). Divide the remaining points \(p\in(0,1)\) into two sets \(\Pi\) and \(\Pi'\), assigning to \(\Pi'\) those of them for which \(f'(Y(p))=0\). Using the obvious relations
   \[ X(p)=\int_0^{Y(p)}\tan\alpha(y)\,dy =\int_0^{Y(p)}\frac{p\,dy}{\sqrt{u^{-2}(y)-p^2}}, \]

* In the opposite case we arrive at the following generalization of problem I, 1: instead of the hodograph a curve \(\widetilde{\Gamma}\) is given \(\{x=2X(p),\ t=2T(p),\ p\in(0,1)\}\), where \(0\le X(p)<\pi R/v(R)\), \(X(p)\equiv X(p)\pmod{\pi R/v(R)}\). Denote by \(\Gamma^*\) the spatial curve \(\{x=2X(q),\ t=2T(q),\ p=q;\ q\in(0,1)\}\), for which \(\widetilde{\Gamma}\) serves as the projection onto the plane \(x,t\). Determining the hodograph from \(\widetilde{\Gamma}\) is done nonuniquely. But in the important case for applications, when \(\Gamma^*\) contains a finite number of components (the points \((x,t,p)\) and \((x+2\pi R/v(R),t,p)\) are identified), there exists only a finite number of hodographs \(\Gamma\) corresponding to \(\widetilde{\Gamma}\).

\[ T(p)=\int_0^{Y(p)} \frac{dy}{u(y)\cos\alpha(y)} =\int_0^{Y(p)} \frac{dy}{u^2(y)\sqrt{u^{-2}(y)-p^2}}, \]

one can show that for \(p\in\Pi\), \(X'(p)\) and \(T'(p)\) exist and are continuous, and \(T'(p)=pX'(p)\); for \(p\in\Pi'\), \(X(p)=X(p\pm0)=T(p\pm0)=T(p)=+\infty\). Let \(\Pi_0\) be the subset of \(\Pi\) where \(X'(p)=0\); \(\Pi_0\) is closed and, according to I, 3, is nowhere dense in \(\Pi\).

One can prove the following property of the hodograph \(\Gamma\). Let \((x_0,t_0)\in\Gamma\), and let \(\Gamma\) contain a smooth arc \(\gamma\) passing through \((x_0,t_0)\) at an angle \(\varphi\) to the \(x\)-axis; if \(p=\tg\varphi\in\Pi\cup\Pi'\), then \(2X(p)=x_0\), \(2T(p)=t_0\). Thus, from the given hodograph one can determine \(X(p)\) and \(T(p)\) for all \(p\in\Pi\setminus\Pi_0\), and then (by continuity) also for \(p\in\Pi_0\cup\Pi'\). Hence, \(X(p)\) and \(T(p)\) are uniquely determined for all \(p\in(0,1)\) (except, perhaps, for a finite number of points).

III, 1. Introduce the following notation:

\[ \Phi(q)=\frac{2}{\pi}\int_q^1 \frac{X(p)\,dp}{\sqrt{p^2-q^2}}, \qquad \tau(q)=\int_0^{Y(q)} \sqrt{u^{-2}(y)-q^2}\,dy, \qquad q\in(0,1); \]

\[ \Psi(q)=\sum_{i=1}^k \frac{2}{\pi}\int_{y_i}^{\bar y_i} \arctg \sqrt{\frac{u^{-2}(y)-p_i^2}{p_i^2-q^2}}\,dy, \]

\[ \sigma_k=\int_{y_k}^{\bar y_k}\sqrt{u^{-2}(y)-p_k^2}\,dy, \qquad q\in(p_{k+1},p_k),\quad 1\le k\le n. \]

Below, along with \(u(y)\), \(Y(q)\), \(y_k\), and \(\bar y_k\), the functions \(u^*(y)\), \(Y^*(q)\) and the numbers \(y_k^*\) and \(\bar y_k^*\) are considered. The expressions obtained from \(\tau(q)\), \(\Psi(q)\), and \(\sigma_k\) by replacing \(u(y)\), \(Y(q)\), \(y_k\), and \(\bar y_k\) with \(u^*(y)\), \(Y^*(q)\), \(y_k^*\), and \(\bar y_k^*\) are denoted by \(\tau^*(q)\), \(\Psi^*(q)\), and \(\sigma_k^*\).

1. It is clear that \(\varepsilon(p)\) has jumps \(\sigma_k\) at the points \(p_k\), \(1\le k\le n\), and is continuous at the remaining points \(p\in(0,1)\) (see II, 16). Since \(\tau(p)=T(p)-pX(p)\), the numbers \(p_k\) and \(\sigma_k\) are known. Since \(\tau'(p)=-X(p)\), specifying \(\Gamma\) is equivalent to specifying \(\tau(p)\).

2. We show that \(Y(p)=\Phi(p)+\Psi(p)\), \(p\in(0,1)\). Let \(D_q\), \(D_q^0\), and \(D^k\) be the following sets in the \(y,p\)-plane: \(\{0<y<Y(q),\ q<p<u^{-1}(y)\}\), \(\{0<y<Y(q),\ q<p<f(y)\}\), and \(\{y\in\bar j_k,\ p_k<p<u^{-1}(y)\}\) (see Fig. 3); \(F(p,y,q)=\dfrac{2p}{\pi}[(u^{-2}(y)-p^2)(p^2-q^2)]^{-1/2}\). For \(q\in(p_{k+1},p_k)\), \(0\le k\le n\), we have

\[ Y(q)=\int_0^{Y(q)}\left[\int_q^{u^{-1}(y)} F(p,y,q)\,dp\right]dy =\int_{D_q} F\,dS = \]

\[ =\int_{D_q^0} F\,dS+\sum_{i=1}^k \int_{D^i} F\,dS =\Phi(q)+\Psi(q); \]

in particular, for \(q\in(p_1,1)\), \(Y(q)=\Phi(q)\).

1. Suppose it has been possible to choose intervals \(j_k^*=(y_k^*,\bar y_k^*)\), \(1\le k\le n\), and to specify \(u^*(y)\) on them so that \(\sigma_k^*=\sigma_k\), the function \(Y^*(q)=\Phi(q)+\Psi^*(q)\) is nonincreasing, and \(Y^*(p_k+0)=y_k^*\), \(Y^*(p_k-0)=\bar y_k^*\) for any \(k=1,\ldots,n\). From \(Y^*(q)\) we determine \(u^*(y)\) outside \(j_k^*\) so that \(Y^*(p)=\inf\{y,\ pu^*(y)\ge1\}\). It can be proved that \(\tau^*(y)=\tau(y)\), i.e., \(u^*(y)\) is a solution; moreover, it is clear that any solution can be constructed in this way.

IV, 1. If \(y_i^*<y_i^*+h_i<\bar y_i^*\), \(u^*(y)=u_i=\mathrm{const}\) for \(y\in(y_i^*,\,y_i^*+h_i)\) and \(u^*(y)=p_i^{-1}\) for \(y\in(y_i^*+h_i,\,\bar y_i^*)\), \(1\le i\le n\), then \(\Psi^*(q)\) is given—

depends only on \(h_i\) (and not on \(y_i^*, \bar y_i^*\)). If, in addition, \(h_i\sqrt{u_i^{-2}-p_i^2}=\sigma_i\) and the \(h_i\) are small, then \(\Psi^*(q)<\Psi(q)\), and \(Y^*(q)=\Phi(q)+\Psi^*(q)\) does not increase on \((0,1)\). Put \(y_k^*=Y^*(p_k+0)\), \(\bar y_k^*=Y^*(p_k-0)\), and define \(u^*(y)\) from \(Y^*(q)\) outside \(\bar j_k^*\).

By virtue of III, 4, \(u^*(y)\) is a solution. For different sets \((h_1,\ldots,h_n)\), the solutions \(u^*(y)\) are different. Note that for any \(\varepsilon>0\), for sufficiently small \(h_i\),
\[ 0<Y^*(q)-\Phi(q)<\varepsilon . \]

1. Let \(u^0(y)\) and \(u^1(y)\) be any two solutions; \(Y^i(p)=\inf\{y,\;pu^i(y)\ge 1\}\), \(i=0,1\). Fix arbitrarily \(\omega\in(0,1)\). Put
   \[ Y^\omega(p)=(1-\omega)Y^0(p)+\omega Y^1(p), \]
   \(y_k^\omega=Y^\omega(p_k+0)\), \(\bar y_k^\omega=Y^\omega(p_k-0)\),
   \[ s_k^\omega=(1-\omega)Y^0(p_k-0)+\omega Y^1(p_k-0); \]
   \[ u^\omega(y)=u^0\!\left[\frac{y-y_k^\omega}{1-\omega}+y_k^0\right]\quad\text{for }y_k^\omega<y<s_k^\omega, \]
   \[ u^\omega(y)=u^0\!\left[\frac{y-y_k^\omega}{1-\omega}+y_k^0\right]\quad\text{for }y_k^\omega<y<s_k^\omega, \]
   \[ u^\omega(y)=u^1\!\left[\frac{y-s_k^\omega}{\omega}+y_k^1\right]\quad\text{for }s_k^\omega<y<\bar y_k^\omega; \]
   outside the intervals \((y_k^\omega,\bar y_k^\omega)\), define \(u^\omega(y)\) by \(Y^\omega(p)\). According to III, 4, \(u^\omega(y)\) is a solution for any \(\omega\in[0,1]\).

Fig. 4

1. Put
   \[ F_k^i(r)=\operatorname{mes}\{y,\; y\le Y^i(p_k+0),\; u^i(y)\le r\},\quad i=0,1,\quad 1\le k\le n+1. \]
   From III, 3 it follows: if \(F_{k+1}^0(r)=F_{k+1}^1(r)\) for \(r\le p_k^{-1}\), then \(Y^0(p)=Y^1(p)\) for \(p\in(p_{k+1},p_k)\).

Note that the converse is also true: if, for some segment \([a,b]\subset(p_{k+1},p_k)\), for \(p\in[a,b]\) one has \(Y^0(p)=Y^1(p)\), then \(F_{k+1}^0(r)=F_{k+1}^1(r)\) for \(r\le p_k^{-1}\).

1. Let \(\{u_\nu(y)\}\) be the set of all solutions;
   \[ Y_\nu(p)=\inf\{y,\;pu_\nu(y)\ge 1\}. \]
   Put
   \[ V(p)=\sup Y_\nu(p),\qquad N(p)=\inf Y_\nu(p). \]
   By virtue of III, 3 and IV, 1, \(N(p)=\Phi(p)\), \(p\in(0,1)\). Clearly, \(Y_\nu(p)\le T(p)/p\); therefore
   \[ V(p_k-0)\le \inf\{T(p)/p,\;p<p_k\}<\infty \]
   for every \(k=1,\ldots,n\). Put \(H_k=V(p_k-0)-N(p_k-0)\). It is not difficult to verify that
   \[ V(p)\le N(p)+H_k\quad\text{for }p\in(p_{k+1},p_k). \]
   If, in addition, \(N(p)+\Psi^*(p)\) does not increase, then
   \[ V(p)=N(p)+\Psi^*(p). \]

V. Assign the point \((y,u)\) to the set \(G\) (see Fig. 4) if
\[ 1\le u\le p_1^{-1},\qquad Y(u^{-1}+0)\le y\le Y(u^{-1}-0), \]
or if
\[ u>p_1^{-1},\qquad N(u^{-1})<y<V(u^{-1}), \]
or if
\[ 0<u\le p_k^{-1},\qquad N(p_k+0)<y<V(p_k-0). \]
It follows from IV that:

1) for any point \((y^0,u^0)\in G\) there exists \(u_\nu(y)\) such that \(u_\nu(y^0)=u^0\);

2) for any \(u_\nu(y)\) and for any \(y\ge0\), the point \((y,u_\nu(y))\in\bar G\).

Schmidt Institute of Physics of the Earth
Academy of Sciences of the USSR

Received
6 II 1965