The Seismic Wave Equation

Rick Aster

February 15, 2011

Waves in one dimension. The wave equation is a partial differential equationthat relates second time and spatial derivatives of propagating wave disturbancesin a simple way. For a nondispersive system (where all frequencies of excitationpropagate at the same velocity), the formula for sinusoidal or harmonic wavesof displacement with amplitude A as a function of space and time is just

uy(x, t) = A sin(kx t) = A sin(kx kct) (1)

where k = 2pi/ = /c is the wavenumber for a disturbance of wavelength , = 2pif is the radian frequency, and c is the phase velocity. We can constructthe 1-dimensional wave equation by noting that

2uyx2

= k2A sin(kx kct) (2)

and2uyt2

= k2c2A sin(kx kct) (3)Thus, the proportionality constant between the two second partial derivativesis just c2, so that

2uyx2

=1

c22uyt2

(4)

which is the 1-dimensional scalar wave equation.An instructional 1-dimensional wave system that we will examing before

considering (the considerably more complicated) 3-d seismic wave system istransverse waves on a string aligned in the x direction, with a linear density ,and under a tension, (e.g., a guitar string). To derive the properties of wavesin this system (Figure 1) we apply the equation of motion, F = ma. For a stringelement displaced in the y direction, the net vertical force is

fy = sin 2 sin 1 (5)

If the angles are small (this is the string equivalent of small strain), using

sin = 3

3!+5

5!+ (6)

1

Figure 1: A Tensioned String Element in Disequilibrium

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and

tan = +3

3+

25

15+ (7)

we have

sin tan uyx

(8)

and thus

fy (uyx

x+x

uyx

x

) (9)

where

=uyx

x+x

uyx

x

(10)

is the change in slope over the element. The equation of motion can now bewritten as

fy = ma = x2uyt2

= (11)

or

2uyt2

=

x(12)

Taking the limit as the segment becomes small we have

limx0

x=2uyx2

(13)

so the equation of motion becomes

2uyt2

=

2uyx2

. (14)

(14) is just a 1-dimensional scalar wave equation for waves with a phase propa-gation velocity in the x direction of

c =

. (15)

Note that the simplicity of this result hinges on the small angle approxi-mation (8). This provides the necessary linear relationship between stress(the y component of tension) and strain (y displacement). A similar situa-tion will occur with seismic waves, which commonly have small or infinitesimalstrains in earthquake or exploration seismology, so that first order terms inthe stress-strain relationship are adequate to characterize the constitutive rela-tionship. Because of the linear stress-strain relationship, transverse waves ona string obey the principals of superposition, i.e., if y1(x, t) is a solution, andy2(x, t) is a solution, then so is y1 + y2. On the other hand, if the angles ofthe string perturbation become large, the first-order stress-strain theory will notbe an adequate representation of the balance of forces and displacements, and

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Figure 2: An Inhomogeneous String

the system will become appreciably nonlinear. Nonlinear systems do not obeysuperposition, and, also unlike linear systems, their behavior also depends onthe amplitude of the disturbance.

Superposition makes the analysis of string, seismic, or other linear wavesystems much easier. In particular, it enables us to apply Fourier theory. For alinear system, can decompose any propagating wave, g(x, t) into its constituentharmonic components

g(x, t) =

(f, k)e2pifte2pikxdf dk (16)

(16) is an inverse Fourier transform and is the fk, or frequency-wavenumberspectrum of the wave g(t kx). Note that k in this context is just 1/ ratherthan the customary 2pi/ to make it analagous to f = 1/T , where T is theperiod. For a particular frequency, f0, and wavenumber constituent of g, k0,the complex gives the amplitude and phase. To find , we apply a forwardFourier transform

(f, k) =

g(x, t)e2pifte2pikxdt dx . (17)

For harmonic waves, which have only a single frequency, we can readily see whatthe f-k spectrum is by noting that

g(x, t) = A cos(2pif0t k0x) = A2

(e(2pif0tk0x) + e(2pif0tk0x)

). (18)

(16) tells us that the proper to recover this function must be

(f, k) =A

2((f f0, k k0) + (f + f0, k + k0)) (19)

where (f, k) is the 2-dimensional delta function.We next address the question as to how waves behave when there is a spatial

change in properties (either density or elasticity). Consider an infinite-length

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spliced string where the density takes an abrupt change at x = 0 as x increases(Figure 2). The system is excited by a harmonic wave traveling to the right atx = . To investigate the wave-propagation physics of this simple inhomoge-neous system, we construct a composite solution to the wave equation and applyphysical matching conditions at the splice. Where = 1, we will have a generalwave consisting of right-going (incident) and left-going (reflected) componentswill be

u1(x, t) = Ae(tk1x) +Be(t+k1x) (20)

where

k1 =

c1=

1. (21)

and where = 2, we will have only a right-going (transmitted) component

u2(x, t) = Ce(tk2x) (22)

where

k2 =

c2=

2. (23)

Our first matching condition has already been applied; as we are exciting thesplice by a harmonic disturbance of frequency , all other wave components inthe system will also only consist of harmonic disturbances of the same frequency.This is a consequence of linear systems; it would not generally be true in anonlinear system. The second matching condition is that the string does notdisassociate, so that the y displacement will be continuous at the splice

u1(0, t) = u2(0, t) (24)

which implies thatA+B = C . (25)

The final matching condition is that the y forces are continuous at the splice.This is analogous to the continuity of stress that we discussed earlier for elasticmedia (a force discontinuity within a connected string would result in a stringelement having an infinite acceleration). The force balance is

f0 = f0+ = u1x

x=0

= u2x

x=0+

(26)

which implies thatk1A+ k1B = k2C (27)

ork1(AB) = k2C . (28)

Converting the wavenumbers to frequency, densities, and tension gives

(AB)1

= C

2

= (AB)1

1= (AB)1c1 = C2c2 (29)

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where the density-velocity products are called acoustic impedances.To solve for the reflection and transmission coefficients, R12 = B/A and

T12 = C/A, respectively we need to solve the linear system of equations(1 11 2c21c1

)(R12T12

)=

( 11

). (30)

(30) is easily solved to obtain the coefficients

R12 =1c1 2c21c1 + 2c2

=

1 21 +

2

=B

A(31)

and

T12 =21c1

1c1 + 2c2=

21

1 +2

=C

A. (32)

Note that the coefficients depend only on the density change, which should beexpected as the only other physical property, the tension, is constant in thissystem. Note that if 1 < 2, then T < 1, and the amplitude of the transmittedwave will be less than that of the incident wave, while if 2 < 1, then T > 1,and the amplitude of the transmitted wave is greater than that of the incidentwave.

Reversing the indices, gives the equivalent expressions for waves incidentfrom the right

R21 =2c2 1c11c1 + 2c2

= R12 (33)

and

T21 =22c2

1c1 + 2c2= 2 T12 . (34)

We can evaluate the kinetic and potential energy in waves on a string bynoting that the kinetic energy of an element of length dx is

EK =1

2mv2 =

1

2

(uyt

)2 dx (35)

and its potential energy stored as strain is

EP = dx (36)where is the strain, which in this case is just the normalized lengthening ofthe string element

=

(dx2) + (duy)2 dx

dx=

1 +

(uyx

)2 1 . (37)

For small slopes, this gives

= 1 +1

2

(uyx

)2 1 = 1

2

(uyx

)2(38)

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so that for infinitesimal strains we have

EP =1

2

(uyx

)2dx . (39)

Now consider a harmonic wave

u(x, t) = A cos(t kx) . (40)

The potential energy density (energy/length) averaged over one wavelength is

EP =A2

2

0

k2 sin2(t kx) dx = A2k2

2

0

sin2(t kx) dx . (41)

Substituting = 2pi/k and = t kx gives

EP =A2k3

4pi

2pi/k0

sin2(t kx) dx = A2k2

4pi

t2pit

sin2 d (42)

=A2k2

4pi

2pi0

sin2 d =A2k2

4=A2c2k2

4=A22

4.

Following a similar development for the energy density over one wavelength,we have

EK =A2

2

0

2 sin2(t kx) dx (43)

=A2k

4pi

2pi/k0

2 sin2(t kx) dx = A22

4= EP .

Strain and kinetic energy in elastic waves are thus equal and are each pro-portional to density, and proportional to the square of amplitude and frequency.The total energy per wavelength is

ET = EP + EK =A22

2(44)

Although all motion in string waves is perpendicular to the direction ofpropagation, the traveling wave still carries energy. The energy flux is given by

E = ET c =A22c

2. (45)

The energy fluxes for incident, reflected, and transmitted waves for the in-homogeneous string are thus

Eincident =A21

2c12

(46)

Erefl =A2R2121

2c12

(47)

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Figure 3: Harmonic Elastic Wave Displacement, Velocity, and Energy

Etrans =A2T 2122

2c22

. (48)

The total energy flux for waves generated at the splice is thus

Erefl + Etrans (49)

= A2(

(1c1 2c2)2(1c1 + 2c2)2

12c1

2+

(21c1)2

(1c1 + 2c2)2 2

2c22

)

=A22

2

(421c

212c2 + (1c1 2c2)2 1c1

(1c1 + 2c2)2

)

=A221c1

2

((1c1)

2 + 212c1c2 + (2c2)2

(1c1 + 2c2)2

)=A221c1

2= Eincident

so we have, as expected, energy flux conservation for the system.Seismic waves. So far we have discussed some aspects of the static elastic

behavior of elastic media, but have not dealt with dynamic properties. We willnow return to applying the equation of motion

f =

S

fsdS +

V

fbdV = ma = m2u

t2(50)

for elastic media subject to surface and body forces, fs and fb, respectively.

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Figure 4: An Accelerated Element

Consider the total force acting on a small volume element of a connectedmedium with density and volume dV = dx dy dz, which will be some combi-nation of contact and body forces. The contact force from the x and x faces isthe sum of the tractions times their respective face areas (dS = dy dz in bothcases)

Sx

fsx dSx = (x+ + x) dy dz =(( +

xdx

)x+ (x)

)dy dz =

xxdV (51)

where we have assumed that first derivatives are sufficient to propagate stressacross the volume (this is o.k. because we know that tractions are continuousin a connected medium).

The body force term isV

fbx dV = fbxdx dy dz . (52)

The terms are similar for the other face pairs, giving a total force expression of(ijxj

+ fbi

)dV = m

2uit2

= 2uit2

dV . (53)

Dividing through by dV gives the equation of motion for continuous media

ijxj

+ fbi =m

dV

2uit2

= 2uit2

ij,j + fbi (54)

where the comma notation, ij,j , indicates differentiation with respect to xj , andrepeated indices are summed in accordance with the Einstein summation con-vention. This equation must be satisfied for all points in a continuous medium.

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Recall, for an isotropic medium, we have the constitutive relationship be-tween stress and strain

ij = ij + 2ij . (55)

Because

ij =1

2

(uixj

+ujxi

)(56)

the stresses on the x face are just

xx = + 2uxx

(57)

xy =

(uxy

+uyx

)(58)

xz =

(uxz

+uzx

). (59)

The equation of motion (54) has terms which are the spatial derivatives ofthe stress, which are now seen to be

xxx

=

x+ 2

2uxx2

(60)

xyy

=

(2uxy2

+2uyxy

)(61)

xzz

=

(2uxz2

+2uzxz

). (62)

If there are no body forces, and the medium is homogeneous and isotropic,we thus have, for the x component of the equation of motion,

2uxt2

=

x+ 2

2uxx2

(63)

+

(2uxy2

+2uyxy

)+

(2uxz2

+2uzxz

).

Because

x=

x5 u =

2uxx2

+2uyyx

+2uzzx

(64)

we can group terms to obtain

2uxt2

= (+ )

x+

(2uxx2

+2uxy2

+2uxz2

)(65)

= (+ )

x+ 52 ux .

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The development is identical for the y and z components, so the generalequation of motion in three dimensions for isotropic media is

2uit2

= (+ )

xi+ 52 ui (66)

or

2u

t2= (+ )5 (5 u) + 52 u (67)

Using a vector calculus identity for the vector Laplacian

52u = (52ux, 52 uy, 52 uz) = 5(5 u)5 (5 u) (68)gives

2u

t2= (+ 2)5 (5 u) 5(5 u) . (69)

To make sense of this complicated looking expression, we make use of aclever decomposition (Helmholtz Theorem) whereby we write the displacementfield as a sum of the gradient of a scalar displacement potential and the curl ofa vector displacement potential

u(x, t) = 5(x, t) +5(x, t) (70)The physical reason for applying this decomposition is to separate the displace-ment field into two parts. One part will have zero divergence, as

5 (5) = 0 (71)for any vector field, , and the other part will have zero curl, as

5 (5) = 0 (72)for any scalar field, .

Substituting the Helmholtz potentials into the equation of motion gives

2(5 +5)

t2= (+2)5(5(5+5))5(5(5+5))

(73)= (+ 2)5 (5 5) 5(55) .

We can simplify this somewhat by noting that

5(55) = 5 (5 (5)) +52(5) = 52(5) (74)and

5 (5) = 52 (75)so that

2(5 +5)

t2= (+ 2)5 (52) + 52 (5) . (76)

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We next group the terms and pull out the gradient and curl operators toobtain

5(

(+ 2)52 2

t2

)= 5

(52

2

t2

). (77)

The left-hand side is the gradient of a function of , while the right-hand side isthe curl of a function of . As these sides are always equal for all t and x, theymust be equal to some constant, which we can take as zero. Thus, the Helmholtzformalization has enabled us (with much malice aforethought) to separate theelastodynamic equation of motion for an isotropic medium into two differentialequations

52(x, t) = + 2

2(x, t)

t2(78)

and

52(x, t) =

2(x, t)

t2. (79)

The first of these two differential equations in 3-space and time is a scalar waveequation, the second is a vector wave equation. They describe the two types ofseismic body waves in isotropic, homogeneous elastic media.

Recall that the basic form of a wave equation in one dimension is

2u(x, t)

2x=

1

c22u(x, t)

2t(80)

where c is the phase velocity. Likewise, we can see that the two types of waveswhich we have derived for solid media have phase velocities of

=

+ 2

(81)

and

=

(82)

respectively. Note that these velocities are proportional to the square root ofthe moduli, and inversely proportional to the square root of the density. Phasevelocities in crustal rocks are typical on the order of 2-6 km/s for P waves. For a

typical Poisson solid rock with = = 31010 Pa, and density = 3000 kg/m2,we have 4.47 km/s and = /3 2.58 km/s. At 1 Hz, the correspondingwavelengths are = 4.47 km and = 2.58 km.

Recall, in the 1-dimensional wave equation, a harmonic plane wave solutionis simply

u(x, t) = Ae(tkx) (83)

where k = 2pi/ is the wavenumber, and = 2pif is the radian frequency.

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For plane waves in three spatial dimensions, we can write displacement har-monic solutions to the wave equation as

u(x, t) = Ae(tkx) (84)

where the wavenumber object is now a wavevector specifying the direction ofpropagation

k = (kx, ky, kz) (85)

(propagation is in the +k = k/|k| direction if the negative sign is used and inthe k direction if the positive sign is used) and a length proportional to thereciprocal of the wavelength

|k| = 2pi, (86)

andx = (x, y, z) . (87)

The phase velocity of wave propagation can be easily seen by noting that pointsof constant phase in the wave advance in space at a rate

c =

|k| (88)

A plane-wave solution to the scalar wave equation for elastic media is easilyseen to be

(x, t) = 0(x, t) = Ae(tkx) . (89)

= Ae(t(kx, ky, kz)(x, y, z)) .

To obtain the P-wave displacement field due to the displacement potential, we apply the gradient

u = 5 = iA(kxx+ kyy + kzz) Be(tkx)k . (90)

where B = iA|k|. This is a harmonic displacement disturbance with all dis-placement in the propagation direction k (Figure 6a) and an amplitude pro-portional to A|k|. The change in sign of u means that the P-wave displacementconsists of once-per-wavelength compressions and rarefactions. The dilatationof the P-wave as a function of space and time is

5 u = A|k|2e(tkx) , (91)

which does not vary perpendicular to the k direction. All volumetric strain isthis caused by shortening in the k direction.

The displacement field from the displacement potential is

u = 5 = 5 (Ax, Ay, Az)e(tkx) (92)

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Figure 5: A Plane Wave

14

=(zy y

z

)x+

(xz z

x

)y +

(yx x

y

)z

= i ((Azky Aykz)x+ (Axkz Azkz)y + (Aykx Axky)z) e(tkx)The displacement in the S-wave is perpendicular to the propagation direction,k, and the dilatation, as expected, is zero everywhere and at all times

5 u = 5 (5) = 0 (93)so the shear wave propagates no volumetric strain, only shear strain.

As with transverse elastic waves on a string, seismic waves carry energy intwo forms: 1) the strain potential energy in the elastically deformed material;and 2) the kinetic energy of moving material. Consider a shear plane wavepropagating in the z direction with displacement

uy(z, t) = B cos(t kz) . (94)The kinetic energy per unit volume, V is

EK =1

2

V

(uyt

)2dV =

1

2

V

B22 sin2(t kz) dV (95)

Averaging this density over one wavelength gives the average kinetic energy perunit area of wavefront

EK =B22

2

0

sin2(wt kz) dz = B22

4. (96)

The potential (strain) energy in a stressed elastic medium can be found bynoting that the work done by each increment of strain per unit volume is

dW = ij dij . (97)

If we replace the stress tensor by its constitutive expression (in terms of strain),we have

dW = cijklkl dij . (98)

Integrating over volume, we get the strain energy per unit volume

W =1

2cijklklij =

1

2ijij . (99)

For a plain shear wave propagating in the z direction with displacement inthe y direction, the nonzero elements of the strain tensor are

ij = ji =1

2

uyz

=Bk

2sin(t kz) (i, j) = (2, 3) (100)

while the corresponding nonzero stress components for an isotropic medium are

ij = ji = 2ij = Bk sin(t kz) (i, j) = (2, 3) . (101)

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Figure 6: Fundamental Seismic Waves and Their Displacement Fields

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The average potential energy per unit area of wavefront is thus

EP =1

2

0

(yzyz + zyzy) dz =B2k2

2

0

sin2(t kz) dz (102)

=B2k2

4=B22

4= EK .

So, as with the elastic string system, we have an equipartition of energy betweenkinetic and potential types (Figure 3). Also note that, as with the string system,the energy density is proportionate to the density and to the square of amplitudeand frequency. The total energy per unit wavefront is thus

ETS = EK + EP =B22

2. (103)

Performing the equivalent calculation for P waves propagating in the z di-rection, we now have a plane wave

uz(z, t) = A sin(t kz) . (104)The kinetic energy is essentially the same expression as for the S wave, only thedirection of the displacement has changed

EK =1

2

V

(uzt

)2dV =

1

2

V

A22 sin2(t kz) dV = A22

4. (105)

To calculate the strain (potential energy) in an equivalent manner, we notethat the only non-zero stress tensor element is

zz = (+ 2)zz = (+ 2)uzz

(106)

(where in this context is obviously the Lame parameter, not the wavelength,). The strain energy per unit volume is thus

EP =1

2

0

zzzz dz =A2(+ 2)k2

2

0

cos2(t kz) dz (107)

but

2 = + 2 =2

k2(108)

so

EP =A22

4(109)

which is the exact equation as for S-waves (102) and produces the same resultfor the total energy

ETP = EK + EP =A22

2. (110)

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The rate at which energy is carried by seismic waves is given by the energyflux expressions

EP = ETP =A22

2. (111)

and

ES = ETS =B22

2. (112)

so the P-wave propagates energy faster by the ratio /.There are, of course, other solutions to the wave equation in homogeneous

media. The most important of these is spherical waves. We rewrite the scalarwave equation in spherical coordinates (r, , )

52(r, t) = 1r2

r

(r2

r

)=

1

c22

t2. (113)

A solution to (113) can be found by trying a solution of the form

(r, t) =(r, t)

r(114)

which gives

1

r2

r

(r2

(r r r2

))(115)

=1

r2

r

(r

r )

=1

rc22

t2,

or1

r

(r2

r2+

r r

)=2

r2=

1

c22

t2(116)

which is just the 1-dimensional wave equation again, this time in terms of rrather than a Cartesian coordinate such as x! So we have a solution for anyfunction of the form (r, t)/r. Note that this function has a singularity at r = 0.

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