ES240
Solid Mechanics
Z. Suo
Waves
Yet another play of two actors: inertia and elasticity
References.
 H. Kolsky, Stress Waves in Solids, Dover Publications, New York.
 K.F. Graff, Wave Motion in Elastic Solids, Dover Publications, New York.
 J.D. Achenbach, Wave propagation in elastic solids. North-Holland, Amsterdam. Also see
Achenbach’s
acceptance
speech
of
the
Timoshenko
Medal
(http://imechanica.org/node/185).
Light and sound. A comparison between light and sound is instructive.
 What’s vibrating? Vibrating electric field and magnetic field transport light. Vibrating
stress field and displacement field transport sound.
 Media. Light can propagate in vacuum, as well as in certain materials. Sound wave must
propagate in elastic media. Spring of air, liquid, solid. When a noisy watch is suspended
in a glass jar with a thread. In the beginning, you can see the watch and hear its noise.
When the air is sucked out of the jar, you can still see the watch, but cannot hear it.
 Speeds. The light wave speed is 3  108 m/s in vacuum. The sound speed is about 340
m/s in air, 1500 m/s in water, and 5000 m/s in steel.
 Frequencies and wavelengths. f  c/  . Visible electromagnetic waves. Red:  0.7
m, f = 4.3 1014 Hz. Violet:  0.4 m, f = 7.5 1014 Hz. Audible sound waves. 20
Hz to 20 kHz. In air, they correspond to 17 m and 17 mm. In water, they correspond to
75 m and 7.5 cm.
Ultrasound. Ultrasound has frequencies too high to be detected by the human ear.
Ultrasound can be generated using piezoelectric materials, which convert an electric field to a
stress field. High frequencies correspond to short wavelengths.
 Ultrasound imaging: see baby inside mother.
 Non-destructive evaluation (NDE): detect flaws inside materials.
 Surface Acoustic Wave (SAW) devices for wireless applications.
t=0

5mm
10mm
x
t = 0.5 s

5mm
10mm

15mm
t = 1 s
5mm
10mm

15mm
x
x
t = 2 s
5mm
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Z. Suo
Longitudinal wave in a rod. The speed of sound in steel is ~ 5 km/s. A frame-by-frame
“movie” shows the stress profiles at several times.
 Before time zero, these is no stress in a steel rod.
 At t = 0, a hammer hits the end of the rod for 1 s.
 At t = 0.5 s, 2.5 mm of rod is under compression, but the rest of the rod is stress-free.
The delay is caused by the inertia of the matter.
 At t = 1 s, 5 mm of rod is under compression, but the rest of the rod is stress-free.
At t = 2 s, the same 5 mm wave packet travels for 5 mm. The rod behind and ahead of the
packet is stress-free.
The D'Alambert solution. So far we have appealed to our daily experience about
waves. What do the equations say? The equation of motion of a rod is
 2u
 2u
E 2  2 .
x
t
Let c be the wave speed. Define a composite variable,
  x  ct .
Let f   be an arbitrary function. The function
ux, t   f  
represents a fixed displacement profile traveling at speed c to the right. At t = 0, the
displacement profile is ux, 0  f x  . At time t1 , the displacement profile is ux, t1   f x  ct1  .
The profile has the same shape as at time t = 0, but moves toward the right by a distance ct1 .
We suspect that any function f will satisfy the wave solution. To ascertain this, we need
to insert ux, t   f   into the equation of motion. Using the chain rule in the differential
calculus, we obtain that
u df  df


x d x d
and
u df 
df

 c
t d t
d
Insert into the equation of motion, and we have
d2 f
d2 f
E 2  c 2 2
d
d
The equation is satisfied by any function f provided that
1/ 2
c  E /   .
This relates the speed of the longitudinal elastic wave, c, to Young’s modulus and density.
Similarly, gx  ct  represents a fixed displacement profile traveling at speed c to the left.
Let   x  ct and   x  ct . Any functions f   and g  satisfy the equation of motion. The
general solution for the displacement is
ux,t   f x  ct   gx  ct  .
Denote F  df / d and G  dg/ d . Velocity of a material particle is v  u / t .
The general expression for the velocity is
vx, t   c F x  ct   Gx  ct  .
The stress field is given by   Eu / x . The general expression for the stress is
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 x, t   EF x  ct   Gx  ct  .
Initial value problems. The time-dependent field ux, t  is governed by
 A PDE (i.e., the equation of motion)
 Boundary conditions (i.e., the displacement and the stress values at the two ends of the
rod)
 Initial conditions (i.e., the displacement field and velocity field in the rod at time zero.)
In reaching the D'Alambert solution, we have used the PDE, but not the boundary
conditions and the initial conditions. Here is an example to illustrate how the initial conditions
come into play.
Imagine a long steel rod. Pull a rod with grips at two points in the rod. Held the forces
constant before time zero. Release the grips at time zero. We’d like to find out the waves
generated in the rod afterward.

x

x
Before the grips are released. The rod has a static stress field, sx  . When the grips are
released, at time zero, the stress field is still the same as the static field:
EF x  Gx  sx .
At time zero, there is no velocity, so that
c F x   Gx  0 .
A combination of the two equations gives that
1
F x   G x  
sx  .
2E
The stress field afterward is
 x, t   EF x  ct   Gx  ct 
1
 sx  ct   sx  ct 
2
After the grips are released, the static stress profile splits into two waves, one traveling to the
right, and the other to the left.
Reflection from a free end. We now give an illustration of the boundary condition. An
incident wave packet in the rod, hits the free end of the rod, and reflects back into the rod. Let’s
say the rod lies on the axis, x < 0, and the free end is x  0 . The incident wave travels from the
left to the right. The boundary condition: Stress vanishes at all time at x  0 . An incident
compressive wave, after reflection, becomes a tensile wave. Dynamic fracture.
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How do we obtain the solution by doing algebra? We know the shape of the incident
wave,  x, t   EF x  ct . That is, we know the function F   . The reflected wave travels
from the right to the left. It must be of the form  x, t   EGx  ct  . Its shape G  is to be
determined. The stress in the rod is the sum of the incident and the reflected wave:
 x, t   EF x  ct   EGx  ct  .
How to determine the function G  ? The boundary condition: at the free end x  0 the
stress is zero at all time, namely,  0,t   0 . Put this boundary condition to the general
expression for the stress, and we have
0  EF  ct   EGct  .
Denote ct by Q. We have
GQ  F  Q .
That is, we have expressed the function G in terms of the known function F. The independent
variable can be anything. The stress in the rod is given by
 x, t   EF x  ct   EF  x  ct  .
The first term is the incident wave, running towards the positive x-direction. The second term is
the reflected wave, running towards the negative x-direction. At the free end, x = 0, the stress
indeed vanishes at all time. If an incident wave induces a compressive stress in the rod, the
reflected wave induces a tnesile stress in the rod.
t1
c
Free end
c
t2
t3
Standing waves. A normal mode is a standing wave. For example, consider a rod fixed
at one end and free to move on the other end. From the two boundary conditions, the standing
wave of the longest wavelength has   4L . Only a quarter of this wavelength is inside the rod.
The frequency of the fundamental mode is given by f  c /  , which recovers the result we
obtained before. You can understand other normal modes in the similar way.
Acoustic impedance. An alternative form of the D’Alambert solution is
x 
x 
u  x, t   f   t   g   t  .
c 
c 
The particle velocity v  u / t is
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The axial force P  AEu / x is
Z. Suo
x 
x 
v  x, t    F   t   G   t  .
c 
c 
x 
x 
P x, t   RF   t   RG   t  ,
c 
c 
where
R  A E .
Note that P  Rv for a wave propagates in the positive x direction, and P   Rv for a
wave propagates in the negative x direction. These relations may remind you of Hooke’s law,
except that force is proportional to the velocity, rather than elongation. The quantity R is called
the acoustic impedance, and has the unit of force per unit velocity.
Reflection and transmission at the interface of two materials. Now consider two rods
1/ 2
joined at x = 0. The rod on the left-hand side has properties A1 , E1 , 1 , and c1  E1 / 1  . The
rod on the right-hand side has properties A2 , E2 ,  2 , and c2  E2 /  2  . An incident wave
comes from rod 1 towards the interface. Upon hitting the interface, the wave is partly reflected
back to rod 1, and partly transmitted into rod 2.
1/ 2
Let the incident wave be
x

u x, t   f   t  .
 c1 
The arbitrary function f represents the form of the incident wave, and is known. Because the
displacement and force are continuous across the junction at all time, the reflected wave in rod 1
should take the same wave form, but run in the opposite direction, and have a different
amplitude:
 x

u x, t   af    t  .
 c1 
Everything about the reflected wave is known, except for the amplitude a. The net field in rod 1
is the superposition of the incident and reflected waves:
x

 x

u1 x, t   f   t   af    t  .
 c1 
 c1 
The axial force in rod 1 takes the form
x

 x

P1 x, t   R1 F   t   R1aF    t  ,
 c1 
 c1 
where F    df / d . The transmitted wave in rod 2 takes the form
x

u2 x, t   bf   t  .
 c2

 x

P2 x, t   R2 bF   t 
 c2

Everything about the reflected wave is known, except for the amplitude b.
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To determine the amplitudes of the reflected and transmitted waves, a and b, we invoke
the boundary conditions: at x = 0 the force in rod 1 equals that in rod 2, and the particle velocity
in rod 1 equals that in rod 2. Thus,
1 a  b
R1 1  a   R2b
Solving for the two amplitudes, we find that
R  R2
2R1
.
a 1
, b
R1  R2
R1  R2
We note several special cases:
 When the two rods have the same impedance, R1  R2 , the two amplitudes become a  0
and b  0 . Upon hitting the joint, the wave does not reflect, but fully transmits into rod
2. The two rods are said to have matched impedance.
 When rod 2 has much lower impedance than rod 2, R2 / R1  1 , the two amplitudes
become a  1 and b  2 . The force transmitted into rod 2 is vanishingly small, so that all
the energy of the incident wave in rod 1 will be reflected back into rod 1. The reflected
wave changes the sign of the axial force.
 When rod 2 has much higher impedance than rod 2, R2 / R1  1 , the two amplitudes
become a  1 and b  0 , all the energy of the incident wave in rod 1 will be reflected
back into rod 1. The reflected wave keeps the sign of the axial force as that of the
incident wave.
Bending wave in a beam. The equation of motion of a beam is
4w
2w
EI 4   A 2 .
x
t
This equation cannot be satisfied by an arbitrary function
wx, t   f x  ct 
Instead, we look for solution of a more special form:
wx, t   a sin kx  t  ,
where k is the wave number, and  the frequency. Inserting the above form into the equation of
motion, we obtain
EI 2

k .
A
Phase velocity. The velocity of a pure sinusoidal wave is known as the phase velocity,
written as
cp   / k .
The phase velocity of the pure sinusoidal wave in the beam is
EI
cp 
k.
A
The phase velocity of this wave depends on the wave number. Consequently, sinusoidal waves
of different wave numbers propagate in a beam at different phase velocities. Such a wave is
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known as a dispersive wave. By contrast, the longitudinal wave in a rod is nondispersive, so
that a wave of an arbitrary shape can propagate without change in the rod.
Note that the velocity increases with the wave number, and becomes infinite for very
short wavelengths. This is clearly an artifact of our model. The equation of motion is based on
the classical theory of beams, a theory that breaks down when the wavelength becomes too small.
Group velocity. Another useful description of a dispersive wave is the group velocity,
defined by
cg  d / dk .
For the wave in the beam, the group velocity is
EI
cg  2
k.
A
The definition of the group velocity is motivated as follows. Consider two sinusoidal
waves with the same amplitude but slightly different wave numbers:
a sin k1x  1t , a sin k2 x  2t  .
The overall response is the superposition of the two waves:
w x, t   a sin k1 x  1t   a sin k2 x  2t 
  2   k1  k2
  2 
k k
 2a cos 1 2 x  1
t  sin 
x 1
t
2
2
 2
  2

The combined wave is a wave of the average wave number k1  k2  / 2 , modulated by a wave of
a smaller wave number k1  k2  / 2 (i.e., a longer wave length). The former wave is called the
carrier, and the latter the group. The group propagates at the velocity
1  2 d

.
k1  k2
dk
Sketch a carrier wave modulated by a group wave.
Plane waves in a 3D, isotropic elastic medium. When the length scale of a disturbance
is small compared to size of the body to all three directions, a material particle is undisturbed
before the disturbance arrives. We may as well regarded the body as an infinite body.
A plane wave is a wave that propagates in a given direction, and the displacement field is
fixed in another direction, and the amplitude of the displacement in any plane normal to the
direction of propagation is invariant. Without any calculation, we may make the following
remarks:
 Because the problem has no length scale, a plane wave in a 3D medium is nondispersive.
 Because the medium is isotropic, plane waves in all directions are identical.
 Because the medium is linearly elastic, we can superimpose plane waves to obtain any
complex waves.
Two kinds of plane waves exist in an isotropic elastic solid. A longitudinal wave has the
displacement in the direction of the wave propagation. A transverse wave has the displacement
normal to the direction of wave propagation. We consider the longitudinal wave in this lecture,
and leave the transverse wave as a homework problem.
Deformation geometry. Because all directions are equivalent, we only need to consider
the wave propagate in one direction, denoted by x. For a longitudinal wave, the only nonzero
displacement component is ux,t  . Consequently, the only nonzero strain component is
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u
.
x
That is, the solid is in a uniaxial strain state.
Material law. Because of Poisson’s effect, this strain induces normal stresses in all three
directions. Symmetry dictates that  y   z . Writing Hooke’s law in the y-direction, we obtain
that
1
 y  0   y   z   x  ,
E
so that

y z 
x .
1 
Writing Hooke’s law in the x-direction, we obtain that
1
 x   x   y   z  ,
E
or
E 1   
x 
 .
1   1  2  x
The coefficient is the stiffness under the uniaxial strain conditions.
Newton’s second law reduces to
 x
 2u
 2 .
x
t
Putting the three ingredients together, we obtain the equation of motion:
E 1    2u
 2u
.


1   1  2  x2
t 2
Except for the coefficient on the left-hand side, this equation of motion is identical to that for the
rod. Thus, the same solution procedure applies. The longitudinal wave speed is
E 1   
cl 
.
 1   1  2 
By comparison, for a transverse wave, the only nonzero displacement is normal to the
direction of propagation. We may write vx, t  . Following the same procedure, we find that the
transverse wave speed is
E
ct 
.
2  1   
Take a representative value of Poisson’s ratio,   0.3 , and we obtain that
cl  1.16 E /  , ct  0.62 E /  .
Recall that the wave speed in a rod is E /  .
The general form of plane waves in an isotropic elastic medium. Let s be a unit
vector in the direction along which a plane wave propagates. For a longitudinal wave, the
displacement field takes the form
sx 
ux, t   sf 
 t  ,
 cl

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where f is the profile of the displacement field.
For a transverse wave, the displacement can be in any direction normal to the direction of
propagation, and takes the form
sx 
ux, t   ag 
 t  ,
 ct

where a is a unit vector normal to s.
The general form of a plane wave propagates in direction s in an isotropic material is the
superposition of three waves:
sx 
sx 
sx 
ux, t   sf 
 t   ag 
 t   bh
 t  ,
 cl

 ct

 ct

where a and b are unit vectors normal to s. Vectors a and b are also normal to each other.
Plane waves in 3D, anisotropic elastic medium. A combination of the momentum
balance equation
 ij
 2u
  2i
x j
t
and the stress-strain relation
u
 ij  Cijkl k
xl
leads to the the equation of motion
 2u k
 2u
Cijkl
  2i .
xl x j
t
Because these is no length scale in the problem, the wave is nondispersive. Let the plane
wave propagate in a direction specified by a unit vector normal to the plane, s. The displacement
field takes the form
sx 
ui x, t   ai f 
t .
 c

where a is a unit vector in the direction of the displacement, and f is the profile of the
displacement.
Insert this expression into the equation of motion, and we obtain that
Cijkl sl s j ak  c 2ai .
Consequently, the wave velocity is determined by an eigenvalue of the matrix Cijkl nl n j . Because
this matrix is symmetric and positive-definite, each direction of propagation will have three
distinct plane waves. Associated with each eigenvalue is an eigenvector, ai , which is determined
up to a scalar. This eigenvector gives the direction of the displacement of the plane wave. The
three eigenvectors associated with the three velocities are orthogonal to one another. However,
none of them need be longitudinal or transverse to the direction of propagation.
Let the three wave velocities be c1 , c2 , c3 , and the associated eigenvectors be a1 , a 2 , a3 .
These quantities depend on the direction of the propagation, s. The general form of a plane wave
takes the form
sx 
sx 
sx 
ux, t   a1 f1 
 t   a2 f 2 
 t   a3 f3 
 t  .
 c1

 c2

 c3

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Direction of reflection or refraction. Snell’s law. Two materials are bonded on a
planar interface. An incident plane wave propagates in direction s and velocity c. We would like
to determine the directions of reflected and refracted waves. Let one of such waves propagate in
sx
t .
direction s at velocity c . The field of the incident wave is a function of the argument
c
s  x
 t . The continuity at the interface
The field of the other wave is a function of the argument
c
must be maintained at all times. Consequently, the two arguments must be equal for all time and
for all x on the interface, namely,
s  x s  x

,
c
c
or
 s s 
  x  0
 c c 
s s
for all x on the interface. That is, the vector  is normal to the interface. Thus, the vector
c c
s is in the plane spanned by s and the normal vector of the interface, known as the plane of
incidence.
Let  be the angle from the normal vector of the interface to s, and   be the angle from
s s
 is normal to the interface, in the direction
the normal vector to s . Because the vector
c c
s
s
parallel to the interface, the two vectors
and must have
c
c
an equal magnitude and be in opposite directions.
n
Consequently,
  s
sin   sin 

.
c
c
s
This is known as Snell’s law. These equations determine the
directions of the reflected and refracted waves. The

arguments we have used are general, so that Snell’s law is
applicable to any waves, e.g., elastic waves and
electromagnetic waves.
In general, an incident plane wave will generate three
refracted
reflected waves and three refracted waves. These waves
propagate in directions in the plane of incidence, with the
angles determined by Snell’s law. For an isotropic material,
the wave speed is known, so that Snell’s law determines the incident
direction of the reflected or refracted wave. For an
anisotropic material, c itself is a function of   , and the
function c (   ) is determined by the eigenvalue problem.
reflected
The combination of Snell’s law and the function c (   )
determines both the wave speed c and the angle   .
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We can also give explicit equations for the unit vector in the direction of a reelected or
refracted wave. Let n be the unit vector normal to the interface, pointing away from the half
space that contains the incident wave. Thus, for the reflected wave,
s s
 cos  cos 
  

n
c c
c 
 c
For a refracted wave
s s
 cos  cos 
  

n .
c c
c 
 c
Amplitude of reflection. Consider a half space, and a known wave
sx 
ux, t   af 
t.
 c

is incident upon the surface of the half space. Here, a is the unit vector in the direction of the
displacement, s is the unit vector pointing in the direction of the propagation, c is the wave speed,
and f  is the wave form.
Three waves reflected back. The net displacement field in the body is the superposition
of the four waves:
sx 
 s  x 
 s  x 
 s  x 
ux, t   af 
 t   Aaf 
 t   Aaf 
 t   Aaf 
t
 c

 c

 c

 c

Except for the amplitudes A, A, A , everything else about the reflected waves has been
determined. To determine the amplitudes, we use the traction free boundary conditions. In
particular, when x is on the surface of the half space, all the arguments of the functions are
equal:
sx
s  x
s  x
s  x
t 
t 
t 
t 
c
c
c
c
Let n be the unit vector normal to the surface of the half space. The traction vector of the surface
is
u
Aap sq Aap sq Aapsq  df  
a s

ti  n j Cijpq p  n j Cijpq  p q 


.
xq
c
c
c  d
 c
The traction free condition ti  0 gives 3 linear algebraic equations for the three amplitudes:
Aap sq Aap sq Aapsq 
a s
0.
n j Cijpq  p q 


c
c
c 
 c


Reflection from a free surface of an isotropic material.
If the incident wave is a transverse wave with the displacement vector perpendicular to
the plane of incidence, the wave is entirely reflected as a wave of the same kind, and the
amplitude is 1.
If the incident wave is a transverse wave with the displacement vector in the plane of
incidence, the reflected consists of a transverse wave of the same kind and a longitudinal
wave. The amplitudes of the two reflected waves are
2
2cl / ct sin 2 cos 2
sin 2l sin 2  cl / ct  cos 2 2
,
.
At 
A

l
2
2
sin 2l sin 2  cl / ct  cos 2 2
sin 2l tan 2  cl / ct  cos 2 2
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If the incident wave is a longitudinal wave, the reflected wave consists of a longitudinal
wave and a transverse wave with the displacement vector in the plane of incidence. The
amplitudes of the two reflected waves are
2
sin 2 t sin 2  cl / ct  cos 2 2 t
2cl / ct sin 2 cos 2t
, At  
.
Al 
2
2
2
sin 2t sin 2  cl / ct  cos 2 2t
sin 2 t sin 2  cl / ct  cos 2 t
Exercise. A wave incident upon a bimaterial interface generates three reflected waves
and three refracted waves. The amplitudes of the six waves are determined by the continuity of
the displacement vector and the traction vector. Establish a set of six equations that determine
the six amplitudes.
Stroh representation. (Stroh, A.N., 1962. Steady state problems in anisotropic
elasticity. Math. Phys. 41, 77-103.) A body is in a state of stress independent of the coordinate
x3, and the source of stress is moving through the body at a constant speed v in the x1 direction.
Let ( x, y ) be a coordinate system moving at the source speed, relating to the fixed coordinates as
x  x1  vt ,
y  x2 .
To an observer moving at speed v, the steady state field is time-independent. That is, the
displacements are functions of x, y  .
Consider a specific steady state displacement field,
ui  Ai f ( x  py ) ,
where A1 , A2 , A3 and p are constants, and f (  ) is an arbitrary, one-variable function. The
displacement field will satisfy the equation of motion provided
Ci1k1  v 2 ik  pCi1k 2  pCi 2k1  p 2Ci 2k 2 Ak  0 .
This is a set of linear algebraic equations for Ak . To represent a nontrivial displacement field,
Ak cannot be all zero. Consequently, the determinant of the above equations must vanish, which
leads to a polynomial equation of degree six in p. Denote the six roots by p , labeling
  1, 2,  3 so that, if complex roots occur, p  , p  are complex conjugates, and giving the
positive  to the root with positive imaginary part. The labeling for real roots will be specified
later. Following Stroh, we will only consider the case that the p are all distinct; equal roots
may be regarded as the limiting case of distinct roots. For each p , we can determine a column
Ak up to a scaling factor. Make Ak real when p is real, and Ak , Ak ,  complex conjugates
when p is complex.
For any six arbitrary functions f  (  ) , the linear combination
ui 
3
 A  f   z 
  1
i
satisfies the equation of motion. Here z  x  p y . Summation over a Greek suffix will
always be indicated explicitly. Any steady state solution can be represented in this form; the six
Stroh functions f  are to be determined by boundary conditions. Make f  real when p is real,
and f  , f  complex conjugate when p is complex, so that the displacements will always be
real-valued.
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One can expresses the stresses in terms of the Stroh functions:
 i2 
3
 L  f z  ,
i
  1
 i1    Li p  v 2 Ai  f   z  ,
3
  1
with
Li  Ci 2 k1  p Ci 2 k 2 Ak .
We use (  ) to indicate the differentiation of any one-variable function.
The above equationshold for any steady source speed whether greater or less than the
sonic speeds of the solid. When v  0 , all the p are complex. When v is sufficiently large, all
the p are real. There are three critical speeds, V3  V2  V1 . When v passes V , a pair of roots
p  change from complex to real. If then v  V , the roots p  are complex, and the functions
f  are complex analytic functions. If v  V , the two roots p  are real, and the equations,
x  p y  constant , are the characteristic lines, along which the real functions f  have
constant values.
For a half space, the anti-plane and the in-plane deformation decouple. We will only
consider the in-plane deformation, for which the determinant has four roots:
p 21  v 2 cl2  1 ,
p2 2  v 2 ct2  1 .
The longitudinal and shear wave speeds are given by
12
12
 2(1   )  

, ct    ,
cl  


 (1  2 )  
It is evident that cl and cs are also the critical speeds. When the crack speed v surpasses cl or cs, a
pair of roots change from complex to real.
The related matrices are all 2 by 2, as given by
 2 p1
1  p 22 
 1  p2 
, L  
A
,
2
1 
2 p2 
 p1
 1  p 2


For a subsonic crack, v  ct , all roots are imaginary numbers:
p1  i 1  v 2 cl2  i l ,
Thus,
p2  i 1  v 2 ct2  it .
z1  x  i l y, z2  x  i t y .
Rayleigh wave. A wave can propagate undiminished along the surface of a half space.
However, the wave is localized near the surface, and the amplitude decays beneath the surface.
There is no intrinsic length scale in this problem, so that the surface wave is nondispersive.
We will only consider isotropic materials, although the method below is applicable to
anisotropic materials. Say we seek a surface wave with the velocity v below the transverse body
wave, so that the eigenvalues are both complex. Let the complex potential be f1 z1 , f 2 z2  . The
displacement field is
2
2
 1
 1
ui   Ai f  z    Ai f  z  .
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The traction vector is
2
2
 1
 1
 i 2   Li f  z    Li f  z  .
On the surface of the half space, y  0 , the traction vanishes, so that
Lf x   Lf x   0 .
The first term is analytic in the lower half plane, and the second term is analytic in the upper half
plane. Consequently,
Lf z   0
for any z. To have nonvanishing field, we must require that
det L  0 .
Consequently, the speed of surface wave is determined by


 

2
16 1  v 2 cl2 1  v 2 ct2  1  v 2 ct2 .
This is an algebraic equation for v. An approximate solution of the Rayleigh wave speed is
0.87  1.12 
vR 
.
1 

Waves in laminates. Let us now look at waves propagating in a direction parallel to the
interfaces of a laminate of different materials. An example is the Love wave, propagating in a
laminate of a layer bonded to a half space. The thickness of each layer provides a length scale, so
that the waves are dispersive. Each material has its own set of complex functions. We look for
solutions of the form
f z   a exp ikz   b exp  ikz  .
Such a function ensures that the x dependence is sinusoidal, with the wave number k. The wave
velocity v and the constants a and b are determined by requiring continuity of traction and
displacements, which leads to an eigenvalue problem. The algebra is messy but straightforward.
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