PNAS-1955-Pekeris-469-80

GEOPHYSICS: C. L. PEKERIS

data available to us in advance of publication. Fay Jones, Emily Hermann,Elizabeth S. Skinner, and Annette Trefzer assisted in reducing the data and pre-paring the manuscript.

This work was carried out under Contract N6-onr-271 Task Order 8 with theOffice of Naval Research, Department of the Navy.

* This is Contribution No. 132 of the Lamont Geological Observatory of Columbia University.1 F. A. Vening Meinesz, Theory and Practice of Pendulum Observations at Sea (Delft: Tech-

nische Boekhandel en Drukkerij, J. Waltman, Jr., 1929).2 F. A. Vening Meinesz, Theory and Practice of Pendulum Observations at Sea, Part II: Second

Order Corrections, Terms of Browne and Miscellaneous Subjects (Delft: Drukkerij, A. J. Waltmans1941).

3J. Lamar Worzel and Maurice Ewing, Trans. Am. Geophys. Union, 31, 917, 1950.4J. Lamar Worzel and G. Lynn Shurbet, Crust of the Earth (Geological Society of America

[in press]).6 Charles L. Drake, J. Lamar Worzel, and Walter C. Beckmann, Bull. Geol. Soc. Amer., 65,

957, 1954.6 C. B. Officer and Maurice Ewing, Bull. Geol. Soc. Amer., 65, 653, 1954.7 Maurice Ewing, J. L. Worzel, N. C. Steenland, and Frank Press, Bull. Geol. Soc. Amer., 61,

877, 1950.8 D. B. Ericson, Maurice Ewing, and Bruce C. Heezen, Bull. Geol. Soc. Amer., 62, 961, 1951.9 Maurice Ewing, George P. Woollard, and A. C. Vine, Bull. Geol. Soc. Amer., 51, 1821, 1940.

10 Maurice Ewing, A. P. Crary, and H. M. Rutherford, Bull. Geol. Soc. Amer., 48, 753, 1937.D. C. Skeels, Geophysics, 15, 413, 1950.

THE SEISMIC SURFACE PULSE

BY C. L. PEKERIS

DEPARTMENT OF APPLIED MATHEMATICS, WEIZMANN INSTITUTE, REHOVOT, ISRAEL

1. Introduction.-The problem under investigation is to determine the motionof the surface of a uniform elastic half-space produced by the application at thesurface of a point pressure pulse varying with time like the Heaviside unit function.The original formulation of the problem is due to Lamb,I who synthesized thesolution for the pulse from the periodic solution. Lamb's method is, however,very intricate. In a previous publication2 the author gave an exact and closedexpression for the vertical component of displacement for the case when the pressurepulse varies like the Heaviside unit function H(t). The derivation of this result,as well as the solution for the horizontal displacement, are given in this paper.The seismic pulse problem was treated nearly simultaneously by Cagniard,3 andmore recently by Pinney4 and Dix.5 Because of the complexity of the analysis,it was thought worth while to reproduce in this and a subsequent publication theoriginal solution for the surface source and the buried source.

2. Formal Solution.-In this section we derive a formal solution for the problemof the motion produced by a seismic source buried below the surface in a uniformelastic half-space, when the time variation of the pulse is H(t). The solution forthe surface source will then be obtained by letting the depth of source H approachzero. Referring to Figure 1, we choose a cylindrical system of co-ordinates withorigin at the level of the source and the surface situated at z = -H. Quantities

VOL'. 41, 1955 469

GEOPHYSICS: C. L. PEKERJS

referring to the portions of the medium below and above the source will be desig-nated by the subscripts 1 and 2, respectively.The space variation of the source will be specified by the condition that at the

level of the source (z = 0) the surface integral of the applied stress (Pzz - Pzz2)shall be equal to Z (negative):

27r- (pzzl - pzz2)r dr = Z.

One representation of this point source is

(Pzzl - Pzz2) = lim f(e, r),eo

(1)

f(E, r) =- - Jo( r)e e dt = (2)Z2 (W2 + r2)-3/2

Z=O

I )

SURFACE

t SOURCE

FIG. 1.-A point source is situated at a depth H below thesurface of a uniform elastic half-space.

V2,- h2 = 0,

The vertical displacementw and the horizontal displace-ment q can be derived (Lamb,p. 29) from

q = okT + Xrz,w= SUZ + XZZ-k2X, (3)

where the subscripts denotepartial differentiation, and thepotentials d and x satisfy thewave equations for dilatationaland equivoluminal motion, re-spectively:

V2X- = 0, (4)

C2c2 = +Cp2 - = 3c2.

p p(5)

Here c, denotes the velocity of compressional waves, c the velocity of shear waves,and p denotes the operator 6b/t. X and j& are the elastic constants of the medium,which in the sequel we shall assume to be equal.The normal stress pzz and the shear stress prz are given by

(6)Pzz = Xh2o + 2A(6bzz + xzzz - kxz)

PTz = A (2z + 2xzz -k2X)

It is understood that after we have obtained the operational representation ofw(r, z, p), say, the actual w(r, z, t) will be obtained by performing the integrationover the Bromwich contour

1 (a+ic (ePt\

w(r, z, t) = w(r, z, p)dp.(8)

470 PRoc. N. A. S.

Z =-H

2

h2 =21

p

(7)


Appropriate solutions of equations (4) in regions 1 and 2 are

& = Ae'a"Jo(tr), xI = Bek"zJo(ur), (9)

2 = (Cek + De)Jo(ar) X x = (Ee kIz + Fek3z)Jo(4r)j (10)

where

ka = (Q2 + h2)'12, key = (Q2 + k2)'/A (11)

and the functions are to be integrated over a suitable path in the t-plane. Thispath, as well as the functions (of t and p) A, B, C, D, E, and F, are determinedfrom the following six boundary conditions. At the level z = 0 the required con-tinuity of q, w, and Prz yields

(db + Xz)1 = (b + X.)2, (12)

('0Z + Xzz - k2X)I = (4z + Xzz - k2X)2, (13)

(24oz + 2Xzz - k2X)i = (24z + 2Xzz - k2X)2, (14)

while the source condition (1) is met, in the limit of vanishing E, by putting

(Pzz'l-Pzz2) = (2 )Jo(tr)t (15)

and integrating with respect to t from zero to infinity. At the surface z = -H,Pzz and prz, as given by equations (6) and (7), must vanish. Solving these sixboundary equations for A, B, C, D, E, and F, we get

F -1 D =-kF, B = E + F A = C-D, (16)F127jk~h2)kOFekaH = - { [(2t2 + k2)2 + 4k2t2aO]e-kaH - 4t2(2t2 + k2)e-kH}l (17)M

ek HE = M {4k(2t2 + k2) aoge kaH - [(2t2 + k2)2 + 4k2t2calIe- klH} (18)M

M-[(242 + k2)2 - 4k2t2a3]. (19)

Substituting these values into formulas (9) and (10), we can derive from equations(3) the expressions for the operational representation of the displacements every-where in the medium. The values of the latter at the surface are

w(p) w(r, 0, p) = f Jo(tr)t[-(242 + k2)ekaH + 22eklHI (j) dt(20)

q(p) q(r, 0, p) = - f J1(tr)42[-2k2at8e-kaH + (2t2 + k2)e-kPHI X

d. (21)

VOL. 41, 1955 471


3. Vertical Displacement.-The expression for the operational representation -ofthe vertical displacement in the case of a surface source can be obtained fromformula (20) by letting the depth of source H approach zero:

Zk-1t2 2 22Xaw(p) = -K Jo(r)ta[(242 + k2)2- 4k2d2a]t. (22)

By writing t = kx, this expression simplifies to

(Z()~ )N(kr), (23)

N = frO Jo(krx)xm(x) dx, (24)

where

a = VX2 + 1/3, t =\Vx2 +1, k =P (25)c

and k appears only in the external factor and in Jo(krx).The interpretation of the external factor is simply /abt. In order to interpret

the integral, we transform it into the form f exp [- kf(r, x) Jg(x) dx, with f(r, x)real. The interpretation of the latter is fg(x)H [t - (1/c)f(r, x) I dx, H denotingthe Heaviside unit function. This can be accomplished by transforming the pathof integration in equation (24) into one along the imaginary axis v in the complexx-plane, when Jo(krx) goes over into Ko(krv), the interpretation of which followsfrom the representation

Ko(krv) = exp (-krv cosh 0) d@. (26)

Now m(x) has branch points at x = i (i/V/3), Pi, and poles at

x8=/23 + /3m. (27)

We shall cut up the x-plane in such a manner as to leave m(x) regular to the rightof the imaginary axis, in order to be able to apply to the integral in ecuation (24)the theorem of Bateman and Pekeris,6 leading to

N = f Jo(krx)xm(x) dx = - I()1 f Ko(krv)vm(iv) dv, (28)

I denoting the imaginary part. This is shown in Figure 2, where the values ofa and 0 in the various portions of the imaginary axis are also indicated. Nearthe pole iy the path is indented by a semicircle passed in the positive direction.It is easy to see that the integral over the semicircle is real and therefore does notcontribute to equation (28). We must, however, take the principal value of theintegral in equation (28) once we pass the pole. Since, moreover, Im(iv) = 0 forv < 1/V3, we obtain

PRoc. N. A. S.472


IP J Ko(krv)vm(iv) dv,Tv/a/3

where P denotes the principal value.

(29)

'm~~~viVIi X, +vI i-1.0

I/V3_

X-* PLANE, X U+iV-o

5 U - IO

VI--VI TV--10

is _VTjs )

FIG. 2.-The complex x-plane is cut up by branch lines (thick) atthe branch points v =_(1/V\3) and v = 41.0. The poles are situ-ated at v = '1/2A3 + \3. a and ia denote the values of VI/'3 + X2and of VY2 on the various portions of the v-axis.

Now the interpretation of Ko(krv) is

o (rp cosh 0) O V> T,

cosh-' ( -) v < a, (30)w(

where r = (ct/r), and ---denotes operational interpretation. Hence

VOL. 41, 1955 473


kKo(krv) -- ! (s) cosh-' (-) = (re2 - v2)-1/2,= 0,

kN -- 0,

kN --o(- )-IP twrr 1/-,-

vm(iv) (T2 V2) -1/2 dv,

Using the values of a, 8 shown in Figure 2 to evaluate Im(iv), and substituting inequation (23), we get

r < -1/3

w(r) = 0,

-T < 1,V=3<W(T) = ( 2 (r)

I3Z

w(r) = ( 2r) [Gi(r) + G2(r) ],7r2,ur

(33)

r > 1, (34)

where

Gi(r) = P V/V'2 -'/3(1 - 2V2)2 dv

Jiv V/T2 - V2(3 - 24V2 + 56v4 - 32v6)'

G2(T) Pv3V"2 1[4V2 - 4/3]dv

1 -IT2 V2(3 - 24V2 + 56v4 - 32v6)

(35)

(36)

The integrals Gi(T) and G2(r) can be evaluated in closed form by partial fractiondecomposition. Writing, in equation (35),

V2 = 1/3 + (c2 sin2 0; w2 = T2 1/3, (37)

the integral is transformed into

1 r/2Gl(r) = -Pf dO

{ 12+ [1/12 +C2 sin2 0] (-b + W2 sin2 6) (c + w2 sin20)}' (38)

B = 3 +

3b = 5/12 + -\ C= 3- 5

½'/iC = 4 - 5/12. (39)

Using the result

fJ/2 dO(a2 +a,2 sin2 0)-1 = (a2 + o2)Y1/(420

v < -,

V> T (31)

1Tr < z-\i (

Tr > I-. (32)-\/3-

474 PROC. N. A. S.

(40))


p fr/2 dO(-32 + 02 Sin2o) -1 = 0,

= - _ (p32 - ,)-12 $ >'o, (41)

G1(T) = 9 X96

( ' V/3 93V6 - +V 2 -1/4 \ '/4(3 +

Gi(T) = - -6 + V/3-

_/4

/3 + 5

N/- -2

3-1/3 - 5

F 2T _ 1/4(3 3)J)

3J -i -5r

I T 2 /4(3 +/3) J

Similarly, using the substitution

I = 1 + 2 Si12 , -2 = 2 1,

in equation (36), we obtain

G2(T) = 1/24fd7/2 0 [ 3 - 3 * (3 + 4CW2 sin2 G) - (1 + V3).

(1 - V3 + 4-2 sin2 )-1 - (1 - -3)(i+ V3 + 4,2 Sin2 0)-I]

= -1 +96 V/T2 - 1/4

\3/3 + 5 +\ 1/4(3 + V/3) - r2

7r< 7,II3-/3 - 5 1

Ir '/4(3 /3)J

=Ir 6 -

96

It follows that

w(r) = 0

V /34 +N/T2 1/4

W(T) = - -321rUr

6 - __H -N33N/3 + 5

/3 V 3I + 4 2v4 4

2 4 - 4Ir44/

1-V:: < T < 1,

we obtain

T < y,

T > -. (42)

(43)

T> 7. (44)

IT< ,/3

(45)

(46)

VOL. 41, 1955 475

'.3N/3 5

. Ir 2 '/.(3 N/3)


w(T) = - 16 rS6 - \ 1, 1 < r < -y = /2\3 + A/3, (47)

W(T) = - 3Z T> (. (48)87r/.r'

4. The Horizontal Displacement.-The operational representation q(p) for thehorizontal displacement at the surface in case of a surface source can be obtainedfrom formula (21) by putting H = 0:

(p) = _Z Em Ji(tr)42[(2t2 + k2) - 2k2C13]dk4q~p) = - 27r2A Jo [(2t2 + k2)2 - 4k2a21](

which simplifies again, by the substitution t = kx, to

q(p) = 6Q(aQ) (50)

Q = fOTJo(krx)n(x)x dx, (51)(2x2 + 1)-2VX2 + 1 \X2 + 1/3

(2X2 + 1)2 - 4X2V/x2 + 1 V/X2 +1/3(By cutting up the complex x-plane in the manner shown in Figure 2, we may

again apply the Bateman-Pekeris theorem to equation (51), leading to

Q = f Jo(krx)n(x)x dx = - (2) I f Ko(krv)n(iv)v dv - '/4Ko(k'y), (53)

where the last term arises from the residues at the poles x = i iy. Now

In(iv) = 0, v <-\3

'A/12- 1/3 '/1 -V2(12V2 - 6) 1 < V < 1 (54)(3-24V2 + 56V4-32V6) ' A/3

0, v > 1,

so that

1Cr1 Ko(krv)-\V2 - 1/3 \/1I-v2(12 - 24v2)v dv Ko(k-y)=o7r Ji,,v'3 (3 - 24V2 + 56V4 - 32v6) 4 (55)

In order to obtain the interpretation of 6Q/br, we differentiate equation (30)with respect to r and get

(a)Ke(krv) _ O. v > (56

- .~7v < a, (56)

r-1T2 - V2

476 PRoc. N. A. S.


which, when substituted into equation (50), yields

q(T) = 0,

q (T) = TR1 (T),

1Tr < -<

3

-< T < 1,

q(r) = - (2 )TR2(T), 1 < T < y, (57)

q(r) = - (2 2 T)7R2(T) + 8i (T2 - y2)/, T> r

where

RI(r) = (r VV'v2 - 1/3 V1 - v2(12 - 24V2) dvJ/l/3V\//2 - v2(3 - 24V2 + 56v4 - 32v6)

( 1 V\V2 - 1/3 /I - V2(12 - 24V2) dv

J/V/3V/T2 - V2(3 - 24V2 + 56v4 - 32v6)

(58)

(59)

By partial fraction decomposition, Ri(T) and R2(T) can be expressed in terms ofelliptic integrals, as follows: Using the substitution (37) in equation (58), Ri(T)is transformed into

RI?(r) = - 18/3('u2 dO 18

<2 Jo /I -k2 sin2 0 (1 +8k2 sin2 6)

(6 - 4 i/3) (6 -+- 4V/3)[1 - (2V3 - 20)k2 sin2 0] [1 + (12V/3 + 20)k2 sin2 0]

= - i J {6K(k) - 1811(8k2, k) + (6 - 4/3)H[-(12V/3 - 20)k2, k] +

(6 + 4V/3)II[(12Vi/3 + 20)k2, k)},

k12 = 1/2(372 - 1),

r/2 dOK(k) =

V/I - kd sin'

(60)

(61)/2 dO

11(n, k) =2

d'o (1 + n sin2 ) /1- k2 sin2O

(62)

H(n, k) can be expressed in terms of incomplete elliptic integrals,7 for which tablesare available. Similarly, the substitution

21=+ 2sin2 0v2

3(63)

transforms the integral in equation (59) into

where

Voi,. 41, 1955 477


i i2 dO (9R2(7) = 3 - .+

4wJ -VI - K2 sin2 l (8 sin2' + 1)

(3 + \/3) + (3- 3)

(8 sin2 0 5-3V/3) (8 sin2 O 5 + 3V/3))

- - 4- {3K(K) - 911(8, K) - (2V/3 - 3)II [-(12V3 - 20), K] +

(2V/3 + 3)HI[(12V/3 + 20), K] }, (64)

K 3T22 (65)

Substituting these into equations (57), we finally get

q(T) = 0, r<V

q(r) = V/32TZ {6K(k) -181(8k2, k) + (6 - 4V3)HI-(12/3 -20)k2, k] +

(6 + 4V/3)H(12V3 + 20)k2, k]}, < T < <<

q(r) = V/3/2TKZ {16K(K) - 18H1(8, K) + (6 - 4V3)HI[-(12V/3 - 20), K] +l67r2Mr

(6 + 4V3)II[(l2V/3 + 20), K]}, 1 < T < y = '/2 13 + /3,q(r) = V/ /27KZ {6K(K) - 18H(8, K) + (6 - 4V3)H[-(12V3 - 20), K] +

oT2Hr~ ~ ~ ~ ~ ~ ~ Z

(6 ,- 4V/3)1I[(12V3 + 20), K] + ,T rr>y (66)

5. Discussion of Results.-The vertical displacement at the surface w(T) due tothe application of a point surface pressure pulse of the form H(t) was computed fromequations (45)-(48), and is plotted in Figure 3. The applied pressure being down-ward, the steady-state vertical displacement is downward. The initial displace-ment, however, at the time of the arrival of the P-wave, is upward. We notethat the shear wave is marked only by a discontinuity in slope of w(r). The dis-placement becomes infinite at the time of the arrival of the Rayleigh wave, andsubsequently it reverts to the steady-state value.

Similarly, the horizontal displacement q(r) was computed from equations (66),and is plotted in Figure 4. The ultimate displacement is inward, but the initialdisplacement at the P-epoch is outward. The arrival of the shear wave is markedonly by a change of slope of q(r), which is even less marked than in w(r). Nearthe time of the arrival of the Rayleigh wave the infinity in q(r), as well as in w(T),varies with distance like 1/xr/, which is characteristic of surface waves.

478 PROC. N. A. S.


r*21T __I 14 IT__ r v

P S R

0 2 A * *8_r t.o.0 2 L4 .6 1. .0

FIG. 3.-Vertical displacement at the surface w(t) due to the ap-plication of a Heaviside unit pressure pulse H(t) at the surface.w(t) = -(Z/7r/Ar)G(r), T = (c~t/r). Z (negative) is the surfaceintegral of the applied pressure. w(t) was computed from equations(45)-(48). P denotes arrival time of the compressional wave, S of theshear wave, and R of the Rayleigh wave.

u A F T U

FIG. 4.-Horizontal displacement at the surface q(t) due to theapplication of a Heaviside unit pressure pulse H(t) at the surface..q(t) = -(Z/T7r.)E(r); T = (c8t/r) Z (negative) is the surface in-tegral of the applied pressure. q(t) was computed from equation(66). P, S, and R denote times of arrival of compressional, shear,and Rayleigh waves, respectively.

6. Summary.-Exact and closed expressions are derived for both the hori-zontal displacement q(t) and the vertical displacement w(t) of the surface of a

uniform elastic half-space due to the application at the surface of a point pressure

pulse varying with time like the Heaviside unit function H(t). The- applied pres-

.6

.4d P S R

.2IA

_y .' v i i

VOL. 41, 1055 479

MATHEMATICS: A. A. ALBERT

sure pulse is specified in equation (2) and is such that its integral over the surfaceis finite. Both w(t) and q(t) turn out to be, in this case of a surface pulse, of theform (1/r)f('r), where r = (ct/r), c denoting the shear velocity. The solution forw(t) is given in equations (45)-(48) and is plotted in Figure 3. The solution forq(t) is given in equation (66) and is plotted in Figure 4. Both w(t) and q(t) becomeinfinite at the time of arrival of the Rayleigh wave, but the arrival of the shearwave is marked only by a change in slope of the displacements.

1 H. Lamb, Phil. Trans. Roy. Soc., ser. A, 203, 1, 1904.2 C. L. Perkeris, these PROCEEDINGS, 26, 433, 1940.3 L. Cagniard, Rtflexion et refraction des ondes seismiques progressives (Paris: Gauthier-Villars,

1939).4 E. Pinney, Bull. Seis. Soc. Amer., 44, 571, 1954.5 C. H. Dix, Geophysics, 19, 722, 1954.6 H. Bateman and C. L. Perkeris, J. Opt. Soc. Amer., 35, 655, 1945.7 P. F. Byrd and M. D. Friedman, Handbook of Elliptic Integrals (Berlin: J. Springer, 1954),

p. 225.

ON INVOLUTORIAL ALGEBRAS

BY A. A. ALBERT

UNIVERSITY OF CHICAGO*

Communicated May 18, 1955

The main result of this note is the proof of a conjecture of A. Weil.'Let 21 be an n-dimensional associative algebra over a real field 0. It will be con-

venient to regard 21 as consisting of the set of all vectors x = (i,, .... ,7) withco-ordinates at in a and the product xy as being defined by

xy = xRy, (1)

where R, is an n-rowed square matrix whose elements are linear forms in the co-ordinates sj of y = (n,, . . ., t Since 21 is associative, we have

Rxy = RxRy. (2)

A linear transformation x -- x* of 2 over a is said to define an involution overof 21 if

(xy)* = *X (x*)* = x, (3)for every x and y of W. Assume that x x* is an involution of 21 over a and thatthere exists a linear functional 8(x) on 21 to a such that

6(xx*) > 0 (4)

for every nonzero x of 21. Then we have the following result.THEOREM 1. The algebra 21 is semisimple.For the radical 91 of 21 is the set of all properly nilpotent elements z of W. If z

is in 9, we have (ZX*)n = 0 for every x of 21, [(zx*)fI* = (xz*)n = 0, and z* is in91. Then y = z + z* is in % for every z ofN and y* - y. If y $ O for some

480 PROC. N. A. S.

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PNAS-1955-Pekeris-469-80