Wave propagation in isotropic linear elastic solids

As linear elasticity is a theory of small displacements, it is natural to consider the equation of motion in

isotropic linear elastic solids,

Dv

Dt= (+ )( u) + 2u+ b,

at this limit. In the right-hand side

Dv

Dt= 0(detF )

1D2u

Dt2,

0(1 + u)1h t

+Du

Dt ih t

+Du

Dt iu(x, t) 0

2u(x, t)

t2

where we have neglected terms of higher than linear order in u. Thus, in the absence of body forces, we

obtain the equation

02u

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

This equation is very useful in studying various vibrations and wave propagation in linear solids. We shall

start by considering wave propagation in an infinite isotropic material.

Typeset by FoilTEX 1

Shear waves

Assume, first, that the deformation in the solid is isovoluminous, u = 0. Thus,

2u

t2=

02u.

This is a standard wave equation describing the propagation of isovoluminous or shear waves (S waves).

The velocity of propagation is

Ct =

r

0.

The same result can be obtained by considering the infinitesimal rotation vector, = 12 u, andtaking the curl of the linearized equation of motion,

02u

t2= (+ )( u) + 2u

12

02

t2= 12(+ ) [( u)]| {z }

0+2 = 2

indicating that satisfies the shear-wave equation for an arbitrary (small) displacement field.

Typeset by FoilTEX 2

Dilatational waves

Consider, then, a displacement field not accompanied with rotation, i.e., = 0. Using the vectoridentity,

2u ( u) ( u) =0= ( u)

we get 02u

t2= (+ )( u) + 2u = (+ 2)2u,

which is the wave equation for irrotational or dilatational waves (P waves). The propagation velocity is

Cl =

s+ 2

0.

The same result can be obtained by taking the divergence of the linearized equation of motion,

02u

t2= (+ )( u) + 2u

02

t2= (+ ) | {z }

2+2 = (+ 2)2,

so the dilatation u fulfills the equation of dilatational waves for a general deformation field.

Typeset by FoilTEX 3

Properties of dilatational and shear waves

Let us study the general wave equation

02u

t2= (+ )( u) + 2u

in an infinite medium by using the Fourier transform pair

u(x, t) =1

2pi

Zd

ZZZd3k u(k, ) exp{i(t k x)}

u(k, ) =1

(2pi)3

Zdt

ZZZd3xu(x, t) exp{i(t k x)}

Thus, the transformed wave equation becomes

02u = (+ )k(k u) + k2u.

Write u = ut + ul, where k ut = 0 and k ul = 0. Thus, k(k u) = k2ul and

02ut = k

2ut

02ul = (+ 2)k

2ul.

Typeset by FoilTEX 4

Transforming these equations back to the real space gives

02ut

t2= 2ut

02ul

t2= (+ 2)2ul,

where nowut = 0 andul = 0, showing that the propagation of shear waves (with amplitudeut) is decoupled from the propagation of dilatational waves (with amplitude ul) in a infinite medium.

Sufficiently far from the source, waves can be treated as plane waves. Thus, we can take

u(x, t) = ue exp{i(t k x)},

where the physical displacement corresponds to the real part of the complex quantity. Here e is a unit

vector in the direction of displacement and k is the wave vector giving the propagation direction of the

wave. The density perturbation related to the wave is = 0(1+ u)1 0 ik u0. Wehave

shear waves: phase speed

k= Ct =

r

0, e k, = 0

dilatat. waves: phase speed

k= Cl =

s+ 2

0; e k, = iku0 ei(tkx).

Note that Cl/Ct = [(+ 2)/]1/2 = [(2 2)/(1 2)]1/2.

Typeset by FoilTEX 5

Wave propagation in a linear anisotropic elastic medium

Consistent with the assumed material isotropy, we obtained wave propagation speeds that are independent

of the propagation direction of the waves for both wave modes. In an anisotropic medium, however, the

situation is more complex: waves may have different propagation velocities in different directions as well

as mixed states of polarization (i.e., neither u k nor u k).Example. Consider wave propagation a simple transversely isotropic medium that has the following

constitutive equation:

Tij = (Ekk + E33)ij + 2Eij.

Thus,

T = (+ ) u+ u3

x3

+ 2u

and the wave equation in Fourier space becomes

02u = (+ )k(k u+ k3u3) + k2u.

Choose the coordinate system so that k = k(e1 sin + e3 cos ). Thus,

02u1 = (+ )k

2sin [u1 sin + (1 + )u3 cos ] + k

2u1

02u2 = k

2u2

02u3 = (+ )k

2cos [u1 sin + (1 + )u3 cos ] + k

2u3 + k

2cos

2u3

Typeset by FoilTEX 6

Clearly, one solution is obtained by taking u2 6= 0 = u2 = u3 and /k = p/0. This is a

shear wave (k u) with isotropic velocity.The other two solutions are obtained by taking u2 = 0 and solving for non-trivial solutions of the

remaining pair of equations:

[02 (+ )k2 sin2 k2]u1 (+ )(1 + )k2 sin cos u3 = 0

(+ )k2 cos sin u1 + [02 (+ )(1 + )k2 cos2 k2]u3 = 0.

These are obtained from

0 =

k

2 +

0sin

2

0+

0(1 + ) sin cos

+ 0

sin cos

k

2 +

0(1 + ) cos

2

0

=

"

k

2 +

0sin

2

0

#"

k

2 +

0(1 + ) cos

2

0

#

+

0

2(1 + ) sin

2 cos

2

=

k

4 2

+

20(1 + cos

2) +

0

k

2+

0

2+

0

+

0(1 + cos

2).

Typeset by FoilTEX 7

Thus,

k

2=+

20(1 + cos

2) +

0

s

+

20(1 + cos2 ) +

0

2

0

2 0

+

0(1 + cos2 )

=+

20(1 + cos

2) +

0 +

20(1 + cos

2)

i.e.,

k

2=

0and

k

2+

=+

0(1 + cos

2) +

0.

Substituting the first solution, 02 = k2, with isotropic propagation velocity back to the wave

equation gives

u1 sin = (1 + )u3 cos yielding k u = k(u1 sin + u3 cos ) = ku3 cos . Thus, the wave is in general not a pureshear wave nor a pure dilatational wave, but reduces to a shear wave at 0 or 0 or pi/2.The other solution, 0

2 = k2[+ (+ )(1 + cos2 )], gives

u1 cos = u3 sin

implying that k u = e2(k3u1 k1u3) = e2k(u1 cos u3 sin ) = 0 so this is a purelydilatational wave. However, it has a propagation velocity that depends on the propagation direction.

Typeset by FoilTEX 8

Elastic waves at material boundaries

When an elastic bulk wave meets a boundary surface of two

media it is reflected and refracted. In general, a longitudinal

wave is reflected (or refracted) not only as a longitudinal but

also a transverse wave. This is known as mode conversion.

Consider, first, the reflection of longitudinal waves from the in-

terface between a linear elastic isotropic medium and vacuum.

Choose the coordinate system so that the planar interface be-

tween the medium and the vacuum is the plane x1 = 0 and

that the incident wave is propagating in the x1x2 plane at an

angle wrt. to the interface normal. Thus, x1 = 0

inciden

t lwav

e

reflected swave

reflected lwave

uincl = eA1 exp{i(t kl x)} = eA1 exp{i[t kl(x1 cos+ x2 sin)]} = e1

where e = e1 cos + e2 sin and 1 = A1 exp{i[t kl(x1 cos + x2 sin)]} andkl = Cl/ =

p(2+ )/02. The boundary condition is

e1 T (x1 = 0) = e1 T (x1 = 0+) = 0 T11 = T12 = T13 = 0 at x1 = 0

They can be only satisfied if, in general, both longitudinal and shear waves are reflected.

For the reflected waves,

urefl = e

2; u

reft = e

3

Typeset by FoilTEX 9

with

2 = A2 exp{i[t kl(x1 cos + x2 sin)]}, e = e1 cos + e2 sin

3 = A3 exp{i[t kt(x1 cos + x2 sin )]}, e = e1 sin + e2 cos

Here, kt = Ct/ =p/02. Note that vibrations of the shear wave tranverse to the x1x2 plane

are absent, as the incident wave has no such fluctuations.

The total displacement is

u = uincl + u

refl + u

reft = e1 + e

2 + e

3

and the strain components are

E11 =u1

x1= e1

1

x1+ e

1

2

x1+ e

1

3

x1

= i(kl1 cos2 + kl2 cos2 kt3 sin cos )

E22 =u2

x2= i(kl1 sin2 + kl2 sin2 kt3 sin cos )

E12 =1

2

u1

x2+u2

x1

= 12i(kl1 sin 2 kl2 sin 2 kt3 cos 2)

and all others zero.

Typeset by FoilTEX 10

Returning to the boundary conditions, we use the constitutive equation

Tij = Ekkij + 2Eij

and notice that T13 = 0 is automatically satisfied. The other T1i at x1 = 0

0 = T12 = 2E12

0 = T11 = (+ 2)E11 + E22 = 0C2l E11 + 0(C

2l 2C2t )E22

From the first relation,

kl1 sin 2 kl2 sin 2 kt3 cos 2 = 0 at x1 = 0 t, x2 (1)

which can be satisfied if the exponents in 1, 2, 3 are all equal at x1 = 0. Thus,

kl sin = kl sin= kt sin or

= and

sin

Ct=

sin

Cl

Substituting this back to (1) gives

kl(A1 A2) sin 2 ktA3 cos 2 = 0. (2)

The second boundary condition gives

kl(A1 + A2)(C2l 2C2t sin2 ) ktA3C2t sin 2 = 0, (3)

and (2) and (3) together give the amplitudes A2 and A3.

Typeset by FoilTEX 11

The amplitude relations,

kl(A1 A2) sin 2 ktA3 cos 2 = 0kl(A1 + A2)(C

2l 2C2t sin2 ) ktA3C2t sin 2 = 0

can be immediately used to see that no shear waves are reflected (A3 = 0) if

(A1 A2) sin 2 = 0(A1 + A2)(C

2l 2C2t sin2 ) = 0

These equations can only be satisfied iff = 0 or = pi/2, whence A2 = A1. In all other cases,the reflected waves consist of both transverse and longitudinal waves. The amplitudes can be solved as

A2

A1=

sin 2 sin 2 (Cl/Ct)2 cos2 2sin 2 sin 2 + (Cl/Ct)2 cos2 2

(may vanish!)

A3

A1=

2(Cl/Ct) sin 2 sin 2

sin 2 sin 2 + (Cl/Ct)2 cos2 2,

A 2/A

2

1

+1

90

[o]

where Cl/Ct = [(2 2)/(1 2)]1/2. The relations can also be used to show that conservationof energy applies, i.e.,

A21Cl cos = A

22Cl cos+ A

23Ct cos .

Typeset by FoilTEX 12

If the interface is between two elastic media, re-

fracted waves have to be considered. The boundary

conditions at the interface (x1 = 0) are

e1 T (1) = e1 T (2) and u(1) = u(2).

For the angles, a law analogous to the first case is

obtained by setting the complex exponents equal at

x1 = 0:

sin

C(1)l

=sin1

C(1)l

=sin 1

C(1)t

=sin2

C(2)l

=sin 2

C(2)t

x1 = 0

112

2

inciden

t lwav

e

reflected swave

reflected lwave

(1) (2)

refracted swave

refracted

lwave

The amplitude relations can also be formed in similar manner as before, but of course now they are more

complicated.

Finally, we note that the any material impurities in form of grains of different density or elastic constants

embedded in the medium causes scattering of waves at the surfaces of the impurities. This scattering, of

course, leads to mode conversion. Scattering and reflection of waves can be used to probe the properties

of materials.

Typeset by FoilTEX 13

Elastic waves in a finite medium. Longitudinal wave

Waves in infinite media are different from those in bounded media. As an illustration, let us consider

longitudinal waves propagating in a thin rod made of linear elastic material.

Consider a thin rod along the x1-axis with the radius of cross-section much shorter than the length,

r0 l. Let one of its ends be subjected to a periodic longitudinal stress T11. Since the rod is verythin, we may assume that other components of stress are zero.

Thus, T11 = EE11 = E u1/x1 and

02u1

t2=T11

x1= E

2u1

x21

i.e., the perturbation propagates at speed

k=

sE

0=

s

0

3+ 2

+

along the rod. The speed is obviously different from both Ct and

Cl. The material in the rod moves in the axial direction and (for

6= 0) in the radial direction.

x1

x1

r

r

For the limit of negligible thickness of the rod to be valid, one needs to assume that the wavelength

2pi/k of the wave is much larger than the thickness of the rod (i.e., kr0 pi).

Typeset by FoilTEX 14

Torsional and flexural waves

The propagation speed of a longitudinal wave is given by

Youngs modulus. If, instead, the applied stress is a shearing

stress, the speed is determined by the shear modulus

k=

sG

0=

r

0.

In this case, the material in the rod moves in a torsional manner,

i.e., only u 6= 0. The mode is called the torsional wave.Longitudinal and torsional waves are of symmetric type.

x1

x1

x2

x2

The most complicated case is that of a flexural wave, where all components of the displacement are

non-zero and one takes u sin and ur, u1 cos. The phase speed of the flexural wavebecomes proportional to k, i.e., the wave mode is dispersive. In a flexural wave, elements of the cylinder

axis are in a lateral motion, and this type of wave is called anti-symmetric.

Typeset by FoilTEX 15

Surface waves

Other types of waves in bounded media include, e.g.,

surface waves. Examples of these are seismic waves

called Rayleigh waves and Love waves. In these

waves, the amplitude of the disturbance decays ex-

ponentially with depth from the surface.

Figure source: Wikipedia

Typeset by FoilTEX 16

Wave propagation in fluids

We will start the study of wave propagation in fluids by considering Newtonian fluids. As for the elastic

waves, we start by introducing the equation of motion. This is, of course, the NavierStokes equation:

Dv

Dt= bp+ (+ )( v) + 2v,

where and are now the viscosity coefficients (i.e., not the Lames constants).

The NS equation alone does not form a closed system of equations. In addition, we need the continuity

equation,

t+ (v) = 0

and an equation of state. In case of wave propagation, we will use the equation of state,

p = P ()

In addition, in wave propagation problems we will set the body force to zero, b = 0.

The next step is to derive the dispersion relation for linear waves in Newtonian fluids.

Typeset by FoilTEX 17

Linearized wave equation for Newtonian fluids

Start by linearizing the NS equation, term by term. For this purpose, write

= 0 + 1(x, t); v = v1(x, t); p = P ()

and consider the terms of the NS equation:

Dv

Dt= (0 + 1)

v1

t+ v1 v1

0

v1

t

p = P1

P

(0)1

(+ )( v) = (+ )( v1); 2v = 2v1

Thus,

0v1

t= P

(0)1 + (+ )( v1) + 2v1.

Together with the linearized continuity equation,

1

t+ 0 v1 = 0,

we now have a closed set of equations.

Typeset by FoilTEX 18

Taking a time derivative of the linearized NS eq. and using the continuity equation we get

02v1

t2=

P

0

1t

+ (+ )( v1t

) + 2v1t

2v1

t2=

P

0

( v1) ++

0( v1

t) +

02v1

t

This is the linearized equation of wave-propagation for Newtonian fluids. For inviscid fluids ( = = 0)

we get2v1

t2=

P

0

( v1)

By taking a divergence of this equation and denoting G = v1 = 10 (1/t) we get

2G

t2=

P

0

2G

This is the wave equation for dilatational waves, which have a propagation velocity

C0 =

sP

0

=

sbulk modulus

0sound speed.

By taking the curl of the wave equation, we see immediately that shear waves do not propagate in inviscid

fluids.

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Properties of sound waves in inviscid fluids

Wave equation for inviscid fluids:2v1

t2= C

20( v1).

In infinite media,

2v1 = C

20k(k v1).

Thus, it is immediately clear that sound waves are longitudinal (v1 k) waves. Fourier transform ofthe continuity equation gives

i1 = i0k v1Thus, the condition of linearity can be obtained by requiring that 1/0 1, i.e.,

v1

k= C0 or v1/C0 1

The intensity (= flux of energy F = 120v21C0) of plane sound waves produced in laboratory variesbetween 0.1 and 0.3 W cm2. In air (0 = 1.2 103 g cm3; C0 = 340 m s1), this gives (forF = 0.1 W cm2)

v1/C0 7 103and in water (0 = 1.0 g cm

3; C0 = 1500m s1)

v1/C0 2.4 105,

both clearly in the linear regime.

Typeset by FoilTEX 20

Effect of viscosity

The full equation for wave propagation in viscous fluids reads

2v1

t2= C

20( v1) +

+

0( v1

t) +

02v1

t

We are analyzing the propagation of a longitudinal wave, so assume v1 = v1(x1, t) e1. Thus,

2v1

t2= C

20

2v1

x21++ 2

0

2

x21

v1

t= C

20

2v1

x21+C20v

t

2v1

x21,

where v 0C20/( + 2) is called the viscosity relaxation frequency. For a plane wave, v1 =v1 exp{i(t kx1)}, we get

2= C

20k

2(1 + i/v) or k

2=

2

C20(1 + i/v),

which gives the dispersion relation of waves in a viscous fluid. Writing k = kr + iki we get

v1 = v1 exp(kix1) exp{i(t krx1)} and

k2r k2i =

2

C20(1 2/2v); 2krki =

3/v

C20(1 2/2v)= (k2r k2i )

v

which may be solved for the phase speed C = /kr and the attenuation length k1i .

Typeset by FoilTEX 21

At low frequencies, v and ki kr, we find

C C0 and ki 2

2vC0.

The amplitude attenuation per wavelength, thus, becomes

= ki = 2piki/kr = pi/vNotes:

(i) The observed attenuation of ultrasonic waves can be used to obtain the value of

v =0C

20

+ 2=

0C20

b+ 43,

where b = + 23 is the bulk viscosity. As the shear viscosity is, in general, easy to measure,

the ultrasonic attenuation experiment provides a means to determine b. The ratio of bulk-to-shear

viscosity measured this way for some liquids is given in the table below.

Liquid Glycerol Water Methyl alcohol Toluene Benzene

b/ 1.1 2.5 3.2 13 100

(ii) Attenuation length for plane waves is proportional to 2 (or 2). Thus, to probe the propertiesof matter, one should use as low a frequency as possible. (However, the wavelength needs to be

much smaller than the typical scale sizes under investigation.)

Typeset by FoilTEX 22

(iii) Highly viscous fluids. For large enough values of b+43,

v becomes comparable to . Thus, the condition

w is no longer satisfied and the values of krand ki have to be solved exactly from the dispersion

relation. Thus (exercise),

C2=2

k2r= 2C

20

1 + 2/2w1 + (1 + 2/2w)

1/2

ki =C

2C20

/w

1 + 2/2w

As an example, the phase speed of the wave is plotted in the figure for Glycerol (5% of water).

Clearly, the theory does not well describe the observations. This means that the NavierStokes

equation is not valid for highly viscous systems. This is to be expected, as in the constitutive

equation for Newtonian fluids, all higher-than-first-order derivatives of the velocity were neglected.

By applying the same reasoning in the present case, one expectsC20v t

2v1

x2i

C202v1x2i

,

but this for plane harmonic waves implies /w 1. Thus, this limitation is built-in limitationfor Newtonian fluids.

Typeset by FoilTEX 23

Diffusion of vorticity. Viscous waves

Consider the rotational motion, i.e., vorticity in a fluid. Recall the linearized NS equation (omit the

subscript 1)2v

t2= C

20( v) +

t

+

0( v) +

02v

Taking the curl of both sides gives

2w

t2=

t

02w

w

t=

02w,

which is a diffusion equation for vorticity. Consider a transverse wave v = v2 exp{i(t kx1)}e2.Obviously, v = 0 and, thus,

2v2

t2=

0

t

2v2

x21so

2= ik

2/0 or k

2= i0/

Setting, again, k = kr + iki gives

k2r k2i = 0 and krki = 0/2, i.e., kr = ki = (0/2)1/2

The attenuation per wavelength, = 2pi, so the waves, called viscous waves, are very rapidly damped.

Thus, transverse fluctuations can penetrate a viscous fluid by only about one wave length thickness.

Typeset by FoilTEX 24

Energy equation and heat flow.

So far we have applied an equation of state for the pressure. Next, instead, recall the energy equation

derived in Ch. 5:

De

Dt= TijDij q,

where e is the internal energy per unit mass, Tij = (p + Dkk)ij + 2Dij is the stress tensor,Dij =

12(vi/xj + vj/xi) the rate-of-strain tensor, and q is the heat flux.

The internal energy in a fluid can be obtained from thermodynamics as e = cV , where cV is the

specific heat capacity in constant volume and is the thermodynamic temperature. Using this relation

and the Fouriers law of heat conduction, q = , where is the coefficient of heat conductivity,allows us to write the energy equation as

cV D

Dt= TijDij + 2.

Now, assuming that the pressure is given by p = P (, ) allows one to write

p =P

+P

Thus, the effect of heat flow on sound propagation can be studied by subtituting this into the Navier

Stokes equation and using it together with the energy equation and the equation of continuity after

linearising the equations.

Typeset by FoilTEX 25

Effect of heat flow on wave propagation in inviscid fluids

For simplicity, consider an inviscid fluid, Tij = pij, with a thermodynamic pressure given by theideal gas law, P (, ) = R. Thus,

D

Dt= v

Dv

Dt= R R

cV D

Dt= R v + 2 = RD

Dt+ 2.

Linearize, i.e., substitute = 0 + 1(x, t), = 0 + 1(x, t) and v = v1(x, t) to these

equations:

1

t= 0 v1

0v1

t= R01 R01

cV 01

t= R0

1

t+ 21.

Typeset by FoilTEX 26

Take a divergence of the second equation and use the first to eliminate v1 to get

21

t2= R021 + R021

cV 01

t= R0

1

t+ 21.

Consider the propagation of plane waves, i.e., t i and ik, to get

(2 R0k2)1 R0k21 = 0

i0R1 + (k2 + cV 0i)1 = 0,

so

(2 R0k2)(ik2 cV 0) + R200k2 = 0

gives the dispersion relation. For = 0 we get

cV2= R(cV + R)0k

2

and using the thermodynamic relation cp = cV +R, where cp is the specific heat capacity in constant

pressure, we get

C20 =

cp

cVR0 = R0

consistent with the earlier result for an adiabatic equation of state, p = const.

Typeset by FoilTEX 27

For > 0, we obtain again a complex wavenumber for a real frequency. Denoting = /0 we

have

(2 R0k2)(ik2 cV ) + R20k2 = 0

and solving for gives

2= R0k

2 R20k

2

ik2 cV= R0k

2 ik2 cV R

ik2 cV= R0k

2 cp ik2cV ik2

Taking, again k2 k2r +2ikrki and approximating that ik2 ik2r i/C200 = icp/with 0C20cp/, we get

k2r + 2ikrki

2

R0

cV icp/cp icp/

=2

R0

1 i/1 i/

and separating the real and imaginary parts gives

C2= C

20

1 + 2/21 + 2/2

= R01 + 2/21 + 2/2

ki =C( 1)2R0

2/

1 + 2/2= kr

12

/

1 + 2/2

Thus, thermal conductivity produces dispersion at frequencies & and at the highest frequenciesC2 R0, i.e., isothermal sound propagation. The attenuation has a similar dependence at lowfrequencies on as attenuation by viscosity.

The value of is 1013 rad s1 for water and 1010 rad s1 for air.

Typeset by FoilTEX 28

The viscous attenuation and the attenuation due to heat conduction are additive, i.e.,

kclassi = kvisi kthiand the obtained attenuation of waves is called classical absorption. For monoatomic fluids, classical

values are found to be in good agreement with experiments. For polyatomic fluids, the observed at-

tenuation is significantly greater than classical values. This occurs, e.g., because sound waves disturb

the equipartition of energy between the internal and external degrees of freedom of the molecules. The

phenomenon is called thermal relaxation.

Typeset by FoilTEX 29

Shallow-water waves

Also in fluids, we can consider surface waves. Consider a layer of incompressible, inviscid fluid of thickness

d on top of a solid horizontal floor. Let h = d+(x, t) be the height of the surface from the bottom.

Consider a shallow layer, so that the velocity component along z can be taken as small. Conservation of

mass implies

t(h(x, t) dx) = [h(x+ dx, t)v(x+ dx, t) h(x, t)v(x, t)] dx

x(hv)

h

t+

x(hv) = 0

Similarly, conservation of momentum implies (exercise)

(hv)

t+

x

hv

2+ 12gh

2= 0

Linearising

t+ d

v

x= 0;

v

t+ g

x= 0

giving2

t2= gd

2

x2

i.e., a wave propagating at speed C =gd is found. The solution is valid for d and d.

Typeset by FoilTEX 30

Solitary waves and solitons

If the non-linearities are not neglected but taken into account to the lowest order, one can derive after

transforming to frame moving with the linear waves (x = xCt) and rescaling of the dependent andindependent variables the Kortewegde Vries equation for the propagation of the wave (see, e.g., G.L.

Lamb, Elements of Soliton Theory, Wiley, 1980). In dimensionless form

t + 6x + xxx = 0,

where the subscript denotes differentiation. Seek for wave-like solutions of the equation, i.e., take

= f(x ct)

and substitute to the equation:

cf + 6ff + f = 0 cf + 3f2 + f = A | f

12cf2 + f3 + 12f 2 = Af + BB = 12f 2 + f3 12cf2 Af = 12f 2 + V (f)

where V (f) = f3 12cf2Af andA andB are integration constants. Thus, f(x) behaves like theposition z(t) of a particle with total energy of B in a cubic potential V (z) [i.e., 12z

2 + V (z) = B].

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As V (f) = 3f2 cf A, there are two extrema of thepotential at

f =cc2 + 12A

6; A > 112c2,

where f is a maximum and f+ is a minimum. [Note: forA = 112c2 only one extremum and for A < 112c2 none.]If B < V (f), solutions oscillating between two roots ofV (f) = B are obtained.

V

ff+

f

Choosing A = B = 0, i.e.,

V (f) = f2(f 12c)

gives a potential with local maximum V = 0 at f = 0. This admits a solution with f, f 0 asx and with a maximum value of f = 12c somewhere in between.Its a easy exercise to show that this solution is given by

(x, t) = f(x ct) = c2 cosh2{

c2 (x ct a)}

,

where a is an arbitrary constant. This describes a solitary wave, i.e., a non-periodic, localized propagating

wave with a permanent form.

The solitary-wave solution of the KdV equation is an example of a soliton. Solitons are solitary waves

which can interact with other solitons emerging from the collision unchanged, except for a phase shift.

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Burgers equation. Shock waves

The KdV equation t + 6x + xxx = 0 without the nonlinear term represents a dispersive wave,

i.e., the linear dispersion relation becomes

+ k3= 0 or

k= k2

showing that high-frequency fluctuations are lagging behind the low-frequency fluctuations. The soliton

solutions describe a system, which attains a balance between nonlinear and dispersive effects.

Another non-linear equation of considerable importance in fluids is the Burgers equation,

vt + vvx = vxx,

where is viscosity. It represents a system that can attain a balance between non-linearity and dissipation.

A propagating wave solution can be obtained by v = f(x ct)

cf + ff = f 12A cf + 12f2 = f df

f2 2cf + A =dx

2

x x02

=

Zdf

c2 A (f c)2A

Therefore, at x , f c c2 A = v, soc2 A = 12(v+ v) = 12v and

c = 12(v+ + v) = v. Thus,

v(x, t) = v 12v tanhx vt x04/v

This represents a solution propagating from left to right at speed v, where the speed of the fluid increases

from right to left (i.e., a fast fluid is overtaking a slower one). The transition layer from the slower to

the faster speed has a thickness of 4/v. This kind of a wave is a shock wave.

Actually, as Burgers equation contains no pressure gradient term, it is not a completely valid description

of shocks. By using p = P () we get as a one-dimensional momentum equation

vt + vvx = P ()x/+ vxx,where the equation of continuity, t + (v)x = 0, has to be used to eliminate the extra term. Write

P () = P ()/ and seek for a propagating solution in form v = v(x ct), = (x ct).

c + v + v = 0 (c v) =M = const. = Mc v

c v + vv = P () + v

v = P() cv + 12v2 = P Mc v

cv + 12v2

which shows that the speed cannot cross v = c without yielding an infinite density and that the pressure

gradient term keeps the flow speed from crossing the speed of propagation of the shock wave.

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For a more physical description of shock waves in hydro-

dynamics, one should use an energy equation instead of

an equation of state.

For 0, the thickness of the shock becomes verysmall, and the solution can be treated as a discontinuity.

The conservation laws for mass, momentum and energy

then read

v =+v

+

v2+ P =+v

+

2+ P+

12v

3+ (P + U)v

=

12+v

+

3+ (P+ + U+)v

+,

v+

v+

v

4/v

v

x

v

x

v

c

Burgers equation

Real shock wave

~/v

c

where v = c v is the velocity measured in the shock frame and U is the internal energy per unitvolume. For an ideal gas it is given as

U =P

1where = cp/cV . One can solve the conservations laws for the values behind the shock (subscript )for known values in the ambient medium (subscript +).

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Wave-propagation in inhomogeneous media

So far, we have considered wave propagation in homogeneous media, infinite, semi-infinite and finite.

We briefly mentioned scattering of elasic wave in granular material, but otherwise inhomogenieties have

not been considered. Next, focus on wave propagation in an inhomogeneous medium. We will restrict

to the weakly inhomogeneous case, where the scale length of the medium is large compared to the wave

length.

Consider a wave propagating in such a medium. Let f be the quantity that fluctuates, and write it as

f = Aei.

The quantity is called the eikonal. For a monochromatic plane wave in a homogeneous medium, we

may write

= k x t,where k and are constants, but in an inhomoneous medium, a more general treatment is needed.

In a small region, we may expand the eikonal as

= 0 + x + tt,

where the origin of the coordinate system lies within the region. Thus,

k = and = t,

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which corresponds to the fact that in each small region, the wave can be regarded as planar, but now

the frequency and wavenumber may vary from one region to the other.

The wavelength and the frequency are related to each other by a dispersion relation of form

= (k;x, t),

where now the slow dependence of the propagation characteristics (like phase speed) on the properties

of the medium (i.e., x and t) is explicitly shown. Thus, we obtain

t + (,x, t) = 0, ( ? )

which shows a direct analogy to the HamiltonJacobi theory: identifying

eikonal Hamiltons principal function S(q, t;),xi generalized coordinate qik = xi canonical momentum pi = qiS, and(k,x, t) Hamiltonian H(q, p, t)of the system, the equation (?) is just the HamiltonJacobi equation H(q, qS, t) + tS = 0 for a

particle,which now corresponds to a wave packet propagating in the medium.

As the wave-packet can be treated as a particle with a canonical momentum of k and a Hamiltonian ,

the equations of motion of a linear wave become

x =

k; k =

k; =

t

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Sound propagation in a non-isothermal atmosphere

Consider a polytropic atmosphere with absolute temperature = (z). Thus,

= (k, z) C(z)k with C(z) 1/2

gives the dispersion relation of sound waves in the atmosphere. The equations of motion for sound waves

with C(z)/|C (z)| become

x =

k=C(z)k

k; k =

x= C (z)ez; =

t= 0

Thus, the canonical momenta kx and ky, corresponding to the cyclic coordinates x and y, and the fre-

quency corresponding to the energy of the wave packet are constants of motion. Consider propagation

in the xz plane. We have

x =Ckx

k=kx

C

2(z); z =

Ckz

k=kz(z)

C

2(z); k

2z(z) =

2

C2(z) k2x

dx

dz=kx

kz= kxC(z)p

2 k2xC2(z),

dt

dz=

C2kz=

C(z)p2 k2xC2(z)

These equations can be integrated to give the ray path, x(z), and the propagation time, t(z), of the

sound waves from a source in the atmosphere.

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In the simplest case of C = C0z/h we get (for upward propagating waves, kz > 0)

x x0 =Z zz0

C0kxz/h dzq2 k2xC20z2/h2

= z0C(z0)kz(z0) C(z)kz(z)

C(z0)kx

t =

Z zz0

h/C0 dz

zq2 k2xC20z2/h2

=h

C0ln

/kx +

p2/k2x C2(z0)

/kx +p2/k2x C2(z)

z

z0

!

It is interesting to note that non-vertically propagating waves (kx 6= 0) will be reflected from upperlayers of the atmosphere if C(z) is a function increasing without a limit. The reflection height, zref , is

determined by

0 = k2z(zref) =

2

C2(zref) k2x i.e., C2(zref) =

2

k2x,

which in the case of linearly increasing sound speed, C = C0z/h, gives

zref =h

kxC0.

If there is a maximum value of the sound speed in the atmosphere, only waves with k2x > 2/C2max =

[k2x + k2z(z0)]C

2(z0)/C2max k2x/k2z(z0) > C2(z0)/[C2max C2(z0)] will be reflected.

Wave propagation is a powerful tool to analyze sound speed in inhomogeneous fluids. An example of this

is provided by helioseismology, where the internal structure of the Sun can be probed by analysing sound

waves propagating below the surface and observed as fluctuations of different quantities at the surface.

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Evolution of wave intensity

So far we have analyzed only the propagation of individual wave packets, but we still need informa-

tion about the evolution of the intensity of the waves. To obtain this, we will take the wave-particle

analogy one step further. We will consider the wave field as being composed of quanta with energy

~ and momentum ~k. The wave field can be seen as a gas of such quanta. We will, however, useclassical mechanics to describe the motion of these quanta in phase space (x, ~k). This is called thesemi-classical approximation. For simplicity, we will consider a time-independent medium.

The energy density of the wave field isU(x, t) =Rd3kUk

Rd3kA2k, whereAk is the amplitude

of the waves in the wave-number range from k to k + dk. In the semi-classical approximation, the

energy density can be given as

Uk = ~Nk,whereNk is the number density of the wave quanta in the 6-D phase space and

is the wave frequencyin the rest-frame of the medium. According to statistical mechanics, the phase space density of the wave

quanta follows a continuity equation in phase space, i.e.,

Nk

t+ (xNk) +

k (kNk) = 0,

where x = /k and k = /x. Thus,

Nk

t+

k Nk

x Nkk

= 0.

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In a static medium (v0 = 0, = = const.), we can multiply the equation with the constant

wave frequency and writeUk

t+

k Uk

x Ukk

= 0.

However, if the medium is not at rest, we have to use the Doppler-shift formula

= v0 k

to calculate the frequency in the rest frame to relate the number density of the wave quanta to the

observable wave intensity Uk = ~Nk.Weakly non-linear wave propagation in inhomogeneous media can be treated by adding terms describing

collisions of the wave packets to the right-hand side of the continuity equation, i.e., using Boltzmanns

equation for the wave-quantum gas. These wave-wave interactions are the basis of the weak-turbulence

theory of often used, e.g., in plasma physics.

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