Making Waves in Multivariable Calculus <http://blogs.ams.org/blogonmathblogs/2013/04/22/the-mathematics-of-planet-earth/> J. B. Thoo Yuba College 2014 CMC3 Fall Conference, Monterey, Ca
This presentation was produced using L A TEX with C. Campani s Beamer L A TEX class and saved as a PDF file: <http://bitbucket.org/rivanvx/beamer>. See Norm Matloff s web page <http://heather.cs.ucdavis.edu/~matloff/beamer.html> for a quick tutorial. Disclaimer: Our slides here won t show off what Beamer can do. Sorry. :-)
Are you sitting in the right room? A common exercise in calculus textbooks is to verify that a given function u = u(x, t) satisfies the heat equation, u t = Du xx, or the wave equation, u tt = c 2 u xx. While this is a useful exercise in using the chain rule, it is not a very exciting one because it ends there. The mathematical theory of waves is a rich source of partial differential equations. This talk is about introducing some mathematics of waves to multivariable calculus (vector calculus) students. We will show you some examples that we have presented to our students that have given a context for what they are learning.
Outline of the talk Some examples of waves Mathematical definition of a wave Some wave equations Using what we have learnt Chain rule Integrating factor Partial fractions Other examples (time permitting)
References Roger Knobel, An Introduction to the Mathematical Theory of Waves, Student Mathematics Library, IAS/Park City Mathematical Subseries, Volume 3, American Mathematical Society, Providence (2000)
Some examples of waves
Typical Pond Guitar Strings (L) <http://astrobob.areavoices.com/2008/10/12/the-silence-of-crashing-waves/> (R) <http://rekkerd.org/cinematique-instruments-releases-guitar-harmonics-for-kontakt/>
Internal waves Internal wave trains around Trinidad from space Model of an estuary in a lab (T) <http://en.wikipedia.org/wiki/internal_wave> (B) <http://www.ocean.washington.edu/research/gfd/hydraulics.html>
Internal waves Kelvin-Helmholtz instability Clouds In a tank (L) <http://www.documentingreality.com/forum/f241/amazing-clouds-89929/> (R) <http://www.nwra.com/products/labservices/#tiltingtank>
Water gravity waves Deep-water waves Bow waves or ship waves (L) <http://wanderinweeta.blogspot.com/2011/12/bow-wave.html> (R) <http://www.fluids.eng.vt.edu/msc/gallery/waves/jfkkub.jpg>
Water gravity waves Deep-water waves Bow waves or ship waves (L) <http://wanderinweeta.blogspot.com/2011/12/bow-wave.html> (R) <http://www.fluids.eng.vt.edu/msc/gallery/waves/jfkkub.jpg>
Water gravity waves Shallow-water waves Tsunami (2011 Tohoku, Japan, earthquake) Iwanuma, Japan Crescent City, Ca Santa Cruz, Ca (L) <http://www.telegraph.co.uk/news/picturegalleries/worldnews/8385237/ Japan-disaster-30-powerful-images-of-the-earthquake-and-tsunami.html> (C) <http://www.katu.com/news/local/117824673.html?tab=gallery&c=y&img=3> (R) <http://www.conservation.ca.gov/cgs/geologic_hazards/tsunami/ Inundation_Maps/Pages/2011_tohoku.aspx>
Solitary waves Morning glory cloud Ocean wave (L) <http://www.dropbears.com/m/morning_glory/rollclouds.htm> (R) <http://www.math.upatras.gr/~weele/weelerecentresearch_solitarywaterwaves.htm>
Solitary waves Recreation of John Scott Russell s soliton, Hariot-Watt University (1995) <http://www.ma.hw.ac.uk/solitons/soliton1b.html>
Shock waves F-18 fighter jet Schlieren photograph (L) <http://www.personal.psu.edu/pmd5102/blogs/its_only_rocket_science/about/> (R) <http://www.neptunuslex.com/wiki/2007/11/20/more-education/>
Mathematical definition of a wave
Definition No single precise definition of what exactly constitutes a wave. Various restrictive definitions can be given, but to cover the whole range of wave phenomena it seems preferable to be guided by the intuitive view that a wave is any recognizable signal that is transferred from one part of the medium to another with a recognizable velocity of propagation. [Whitham]
Definition No single precise definition of what exactly constitutes a wave. Various restrictive definitions can be given, but to cover the whole range of wave phenomena it seems preferable to be guided by the intuitive view that a wave is any recognizable signal that is transferred from one part of the medium to another with a recognizable velocity of propagation. [Whitham]
Some wave equations
The wave equation The wave equation: u tt = c 2 u xx Models a number of wave phenomena, e.g., vibrations of a stretched string Standing wave solution: u n (x, t) = [A cos(nπct/l) + B sin(nπct/l)] sin(nπx/l) 0 L n = 3, A = B = 0.1, c = L = 1, t = 0 : 0.1 : 1, 0 x 1
The Korteweg-de Vries (KdV) equation The Korteweg-de Vries (KdV) equation: u t + uu x + u xxx = 0 Models shallow water gravity waves u speed c Look for traveling wave solution u(x, t) = f (x ct), c > 0, f (z), f (z), f (z) 0 as z ±. x
The Sine-Gordon equation The Sine-Gordon equation: u tt = u xx sin u Models a mechanical transmission line such as pendula connected by a spring u Look for traveling wave solution: u(x, t) = f (x ct), c > 0, f (z), f (z) 0 as z.
Using what we have learnt
Chain rule h = g f = Dh m n = Dg Df m p p n if f : R n R p and g : R p R m so that h : R n R m
Chain rule h = g f = Dh m n = Dg Df m p p n if f : R n R p and g : R p R m so that h : R n R m E.g., f : R R 2 : f (t) = (x, y), g : R 2 R 2 : g(x, y) = (w, z), and h = g f : R R 2 : h(t) = (w, z) Then Dh = [ w x z x w y z y ] [ dx dt dy dt ] = [ dw dt dz dt ] = [ w dx x dt + w y dy dt z dx x dt + z dy y dt ] Dg Df
Example 1 The wave equation:* u tt = au xx, a > 0 Look for traveling wave solution: u(x, t) = f (x ct) i.e., look for a solution that advects at wave speed c without changing its profile E.g., u(x, t) = sin(x ct), u(x, t) = (x ct) 4, u(x, t) = exp [ (x ct) 2] *Models a number of wave phenomena, e.g., vibrations of a stretched string
Typical exercise: Show that u(x, t) = exp [ (x ct) 2] satisfies u tt = c 2 u xx.
Typical exercise: Show that u(x, t) = exp [ (x ct) 2] satisfies u tt = c 2 u xx. Let z = x ct. Then, u(x, t) = exp( z 2 ) = f (z) and, using the chain rule, we find that u t = df z dz t u tt = df dz = 2cz exp( z 2 ), z t = 4c2 z 2 exp( z 2 ), u x = df z dz x = 2z exp( z2 ), u xx = df dz z x = 4z2 exp( z 2 ) u tt = c 2 u xx = 4c 2 z 2 exp( z 2 ) = c 2 4z 2 exp( z 2 )
Let z = x ct. Then u(x, t) = f (x ct) = f (z) and, using the chain rule, u t = df z dz t u tt = df dz = f (z)( c) = cf (z), z t = cf (z)( c) = c 2 f (z), u x = df z dz x = f (z)(1) = f (z), u xx = df dz z x = f (z)(1) = f (z)
Let z = x ct. Then u(x, t) = f (x ct) = f (z) and, using the chain rule, u t = df z dz t u tt = df dz = f (z)( c) = cf (z), z t = cf (z)( c) = c 2 f (z), u x = df z dz x = f (z)(1) = f (z), u xx = df dz z x = f (z)(1) = f (z) u tt = au xx = c 2 f (z) = af (z)
c 2 f (z) = af (z) = (c 2 a)f (z) = 0
c 2 f (z) = af (z) = (c 2 a)f (z) = 0 c 2 a = 0 = c = ± a: u(x, t) = f (x ± at) provided f exists, otherwise f arbitrary e.g., u(x, t) = sin(x + at) or u(x, t) = exp [ (x at) 2]
c 2 f (z) = af (z) = (c 2 a)f (z) = 0 c 2 a = 0 = c = ± a: u(x, t) = f (x ± at) provided f exists, otherwise f arbitrary e.g., u(x, t) = sin(x + at) or u(x, t) = exp [ (x at) 2] f (z) = 0 = f (z) = A + Bz: u(x, t) = A + B(x ct) provided solution is not constant
Example 2 The linearized KdV* equation: u t + u x + u xxx = 0 Look for wave train solution: u(x, t) = A cos(kx ωt), where A 0, k > 0, ω > 0 (a particular type of traveling wave solution, i.e., u(x, t) = f (x ct)) *KdV = Korteweg-de Vries; the KdV equation models shallow-water gravity waves
Example 2 The linearized KdV* equation: u t + u x + u xxx = 0 Look for wave train solution: u(x, t) = A cos(kx ωt), where A 0, k > 0, ω > 0 (a particular type of traveling wave solution, i.e., u(x, t) = f (x ct)) Note: u(x, t) = A cos ( ) k( x (ω/k)t advects at wave speed }{{} x ct c = ω/k The number ω is the angular frequency and k is called the wavenumber. The wavelength is 2π/k. *KdV = Korteweg-de Vries; the KdV equation models shallow-water gravity waves
Let z = kx ωt and f (z) = A cos(z). Then and, using the chain rule, u(x, t) = A cos(kx ωt) = f (z) u t = df z dz t u x = df z dz x u xx = df dz u xxx = df dz = f (z)( ω) = ωa sin(z), = f (z)(k) = ka sin(z), z x = f (z)(k) = k 2 A cos(z), z x = f (z)(k) = k 3 A sin(z)
Let z = kx ωt and f (z) = A cos(z). Then and, using the chain rule, u(x, t) = A cos(kx ωt) = f (z) u t = df z dz t u x = df z dz x u xx = df dz u xxx = df dz = f (z)( ω) = ωa sin(z), = f (z)(k) = ka sin(z), z x = f (z)(k) = k 2 A cos(z), z x = f (z)(k) = k 3 A sin(z) u t + u x + u xxx = 0 = (ω k + k 3 )A sin(z) = 0
(ω k + k 3 )A sin(z) = 0 = ω k + k 3 = 0 Dispersion relation: ω = k k 3
(ω k + k 3 )A sin(z) = 0 = ω k + k 3 = 0 Dispersion relation: ω = k k 3 Wave speed: c = ω k = 1 k2
(ω k + k 3 )A sin(z) = 0 = ω k + k 3 = 0 Dispersion relation: ω = k k 3 Wave speed: c = ω k = 1 k2 Note: That c depends on k means that wave trains of different frequencies travel at different speeds. Such a wave is called a dispersive wave. Here, smaller k or longer waves (λ = 2π/k) speed ahead, while larger k or shorter waves trail behind.
(ω k + k 3 )A sin(z) = 0 = ω k + k 3 = 0 Dispersion relation: ω = k k 3 Wave speed: c = ω k = 1 k2 Note: That c depends on k means that wave trains of different frequencies travel at different speeds. Such a wave is called a dispersive wave. Here, smaller k or longer waves (λ = 2π/k) speed ahead, while larger k or shorter waves trail behind. Group velocity: C = dω dk = 1 3k2 The group velocity C is the velocity of the energy in the wave and is generally different from the wave speed c
In general, a wave train solution is u(x, t) = f (kx ωt), where k > 0, ω > 0, and f is periodic (a particular type of traveling wave solution, i.e., u(x, t) = f (x ct)) In general, not a solution for every possible k or ω Note: u(x, t) = f ( k(x (ω/k)t ) advects at wave speed c = ω/k
Integrating factor To solve: y (x) + p(x)y(x) = q(x) for y = y(x)
Integrating factor To solve: y (x) + p(x)y(x) = q(x) for y = y(x) Multiply through by integrating factor µ = µ(x) µy + µpy = µq
Integrating factor To solve: y (x) + p(x)y(x) = q(x) for y = y(x) Multiply through by integrating factor µ = µ(x) µy + µpy = µq If µ = µp, then µy + µpy = µy + µ y, so that (µy) = µq = µy = µq dx and hence y(x) = 1 µ(x) [ µ(x)q(x) dx where µ(x) = exp ] p(x) dx
Example The Sine-Gordon equation: u tt = u xx sin u Models a mechanical transmission line such as pendula connected by a spring u Look for traveling wave solution: u(x, t) = f (x ct), c > 0, f (z), f (z) 0 as z.
Let z = x ct. Then u(x, t) = f (x ct) = f (z) and u tt = u xx sin u = c 2 f (z) = f (z) sin f
Let z = x ct. Then u(x, t) = f (x ct) = f (z) and u tt = u xx sin u = c 2 f (z) = f (z) sin f To solve the equation in f, we multiply through by f (z), an integrating factor c 2 f f = f f f sin f = c 2( 1 2 f 2) = ( 1 2 f 2) + (cos f )
Let z = x ct. Then u(x, t) = f (x ct) = f (z) and u tt = u xx sin u = c 2 f (z) = f (z) sin f To solve the equation in f, we multiply through by f (z), an integrating factor c 2 f f = f f f sin f = c 2( 1 2 f 2) = ( 1 2 f 2) + (cos f ) Now integrate w.r.t. z 1 2 c2 f 2 = 1 2 f 2 + cos f + a
Let z = x ct. Then u(x, t) = f (x ct) = f (z) and u tt = u xx sin u = c 2 f (z) = f (z) sin f To solve the equation in f, we multiply through by f (z), an integrating factor c 2 f f = f f f sin f = c 2( 1 2 f 2) = ( 1 2 f 2) + (cos f ) Now integrate w.r.t. z 1 2 c2 f 2 = 1 2 f 2 + cos f + a To determine a, impose the conditions f (z), f (z) 0 as z i.e., pendula ahead of the wave are undisturbed
Then, as z, 1 2 c2 f 2 = 1 2 f 2 + cos f + a 0 = 0 + cos 0 + a so that a = 1,
Then, as z, 1 2 c2 f 2 = 1 2 f 2 + cos f + a 0 = 0 + cos 0 + a so that a = 1, i.e. 1 2 c2 f 2 = 1 2 f 2 + cos f 1 = f 2 = 2 (1 cos f ) 1 c2
Then, as z, 1 2 c2 f 2 = 1 2 f 2 + cos f + a 0 = 0 + cos 0 + a so that a = 1, i.e. 1 2 c2 f 2 = 1 2 f 2 + cos f 1 = f 2 = 2 (1 cos f ) 1 c2 Exercise: [ ( )] z 1 Show that f (z) = 4 arctan exp is a solution 1 c 2
Then, as z, 1 2 c2 f 2 = 1 2 f 2 + cos f + a 0 = 0 + cos 0 + a so that a = 1, i.e. 1 2 c2 f 2 = 1 2 f 2 + cos f 1 = f 2 = 2 (1 cos f ) 1 c2 Exercise: [ ( )] z 1 Show that f (z) = 4 arctan exp is a solution 1 c 2 2 Solve the equation to obtain the solution above (hint: 1 cos f = 2 sin 2 (f /2))
Wave front solution: [ ( u(x, t) = 4 arctan exp x ct )] 1 c 2 2π u speed c u x A wave front is a solution u(x, t) for which lim x u(x, t) = k 1 and lim x u(x, t) = k 2
Partial fractions Given a rational function p(x)/q(x) p(x) q(x) = r 1(x) q 1 (x) + r 2(x) q 2 (x) + + r n(x) q n (x) where q i (x) is a linear or an irreducible quadratic factor of q(x) and B i (constant) if q i is linear, r i (x) = A i x + B i if q i is quadratic
Example The KdV equation: u t + uu x + u xxx = 0 Look for traveling wave solution that is a pulse: u(x, t) = f (x ct), f (z), f (z), f (z) 0 as z ±, where z = x ct u speed c x
Then u t + uu x + u xxx = 0 = cf + ff + f = 0 Rewrite, cf + ( 1 2 f 2) + (f ) = 0
Then u t + uu x + u xxx = 0 = cf + ff + f = 0 Rewrite, then integrate cf + ( 1 2 f 2) + (f ) = 0 = cf + 1 2 f 2 + f = a
Then u t + uu x + u xxx = 0 = cf + ff + f = 0 Rewrite, then integrate cf + ( 1 2 f 2) + (f ) = 0 = cf + 1 2 f 2 + f = a To determine a, impose f (z), f (z) 0 as z.
Then u t + uu x + u xxx = 0 = cf + ff + f = 0 Rewrite, then integrate cf + ( 1 2 f 2) + (f ) = 0 = cf + 1 2 f 2 + f = a To determine a, impose f (z), f (z) 0 as z. Then cf + 1 2 f 2 + f = a 0 + 0 + 0 = a so that cf + 1 2 f 2 + f = 0
Now multiply through by integrating factor f, then integrate cff + 1 2 f 2 f + f f = 0 = c ( 1 2 f 2) + 1 2( 1 3 f 3) + ( 1 2 f 2) = 0 = 1 2 cf 2 + 1 6 f 3 + 1 2 f 2 = b To determine b, impose f (z), f (z) 0 as z.
Now multiply through by integrating factor f, then integrate cff + 1 2 f 2 f + f f = 0 = c ( 1 2 f 2) + 1 2( 1 3 f 3) + ( 1 2 f 2) = 0 = 1 2 cf 2 + 1 6 f 3 + 1 2 f 2 = b To determine b, impose f (z), f (z) 0 as z. Then 1 2 cf 2 + 1 6 f 3 + 1 2 f 2 = b 0 + 0 + 0 = b so that 1 2 cf 2 + 1 6 f 3 + 1 2 f 2 = 0
Rewrite, 1 2 f 2 = 1 2 cf 2 1 6 f 3 = 3 f 3c f f = 1 where we choose the positive and assume that 3c f > 0.
Rewrite, 1 2 f 2 = 1 2 cf 2 1 6 f 3 = 3 f 3c f f = 1 where we choose the positive and assume that 3c f > 0. Now let 3c f = g 2 3 (3c g 2 )g ( 2gg ) = 1 = 2 3 3c g 2 g = 1
Rewrite, 1 2 f 2 = 1 2 cf 2 1 6 f 3 = 3 f 3c f f = 1 where we choose the positive and assume that 3c f > 0. Now let 3c f = g 2 3 (3c g 2 )g ( 2gg ) = 1 = 2 3 3c g 2 g = 1 To integrate, use partial fractions 1 3c g 2 = A 3c g + B 3c + g
1 3c g 2 = A 3c g + B 3c + g = 1 = A( 3c + g) + B( 3c g) = A = 1 2 3c, B = 1 2 3c = = 1 3c g 2 = 1/2 3c + 1/2 3c 3c g 3c + g 2 3 3c g 2 g = g c( 3c g) + g c( 3c + g)
2 3 3c g 2 g = 1 = g g + = 1 c( 3c g) c( 3c + g) = g g + = c 3c g 3c + g = ln( 3c g) + ln( 3c + g) = cz + d = ln 3c + g 3c g = cz + d
Solve for g: g(z) = 3c exp( cz + d) 1 exp( cz + d) + 1 Recall: f = 3c g 2
Solve for g: g(z) = 3c exp( cz + d) 1 exp( cz + d) + 1 Recall: f = 3c g 2 Use: tanh ζ = sinh ζ 1 cosh ζ = 2 (eζ e ζ ) 1 2 (eζ + e ζ ) = exp( 2ζ) 1 exp( 2ζ) + 1 Substitute 2ζ = cz + d: g(z) = 3c tanh [ 1 2 ( cz d) ]
Solve for g: g(z) = 3c exp( cz + d) 1 exp( cz + d) + 1 Recall: f = 3c g 2 Use: tanh ζ = sinh ζ 1 cosh ζ = 2 (eζ e ζ ) 1 2 (eζ + e ζ ) = exp( 2ζ) 1 exp( 2ζ) + 1 Substitute 2ζ = cz + d: g(z) = 3c tanh [ 1 2 ( cz d) ] Use f = 3c g 2 and choose d = 0: f (z) = 3c sech 2[ ] 1 2 cz [ ] c = u(x, t) = 3c sech 2 (x ct) 2
u amplitude 3c speed c [ ] c Soliton solution: u(x, t) = 3c sech 2 (x ct) 2 x Note: That amplitude is 3c means that taller waves move faster than shorter waves.
Other examples Water gravity waves Ship waves Tsunamis Shock waves
Water gravity waves Consider water (inviscid incompressible fluid) in a constant gravitational field Spatial coordinates (x, y, z), fluid velocity u = (u, v, w) Sinusoidal wave train solution, where the wave oscillates in x = (x, y) and t, but not in z
Water gravity waves Consider water (inviscid incompressible fluid) in a constant gravitational field Spatial coordinates (x, y, z), fluid velocity u = (u, v, w) Sinusoidal wave train solution, where the wave oscillates in x = (x, y) and t, but not in z Dispersion relation: ω 2 = gk tanh(kd), k = k = 2π/λ Here, ω is the frequency, k is the wavenumber vector, λ is the wavelength, g is gravity, and d is the depth of the water
λ d shallowness parameter δ = d λ
λ d shallowness parameter δ = d λ deep water: δ > 0.28 shallow water: δ < 0.07
λ d shallowness parameter δ = d λ deep water: δ > 0.28 shallow water: δ < 0.07 ω 2 = gk tanh(kd) = c = ω k = g k tanh(kd), C = dω dk = 1 2 g k tanh(kd) + d gk sech 2 (kd) 2 tanh(kd)
Deep water: δ at fixed k Using lim θ tanh(θ) = 1 c = ω k = g k tanh(2πδ) g k C = dω dk = 1 g 2 k tanh(2πδ) + d gk sech 2 (2πδ) 2 tanh(2πδ) 1 2 g k = 1 2 c Energy in the wave propagates at half the wave speed
travels at wave speed c = ω/k travels at group velocity C = dω/dk
travels at wave speed c = ω/k travels at group velocity C = dω/dk Ship waves: C = 1 2 c
C (now) B A (earlier)
C (now) B A (earlier)
C (now) B A (earlier)
C E B D A
C E B D O A
C Kelvin wedge C E B 19.5 A 35 D O D O A
Shallow water: δ 0 at fixed d tanh(θ) Using lim = 1 θ 0 θ c = ω k = gd tanh(2πδ) 2πδ gd C = dω dk = 1 gd tanh(2πδ) + d g 2πδ/d sech 2 (2πδ) 2 2πδ 2 tanh(2πδ) 1 2 gd + d 2 g d = gd Energy in the wave propagates at the wave speed
Tsunamis Typical wavelength of several hundred kilometers and deepest point in the ocean in the Marianas Trench (Western Pacific Ocean) about 11 kilometers makes a tsunami a shallow-water wave (long wave) Wave speed c = gd E.g., ocean depth 4 kilometers, gravity 9.8 m/s 2 yields a wave speed c = 39 200 m/s 200 m/s or about 445 mph Typical amplitude in the open ocean about 1 m, but rises up to 10 m to 15 m as approaches shore
λ a d d Energy in the wave proportional to a 2 c a 2 gd remains constant, so a increases as d decreases
λ a d d Energy in the wave proportional to a 2 c a 2 gd remains constant, so a increases as d decreases Hat Ray Leach beach, Thailand, December 2004 <http://geol105naturalhazards.voices.wooster.edu/page/32/>
What happens when a wave approaches the shore?
What happens when a wave approaches the shore? Typically the wave will break.* *G. B. Witham, Linear and Nonlinear Waves, p. 22.
Mathematically, to remove the multivalued part of the wave profile, we introduce a discontinuity or shock. We do this using the equal area rule so that conservation is satisfied, i.e., ρ dx is the same before and after a shock is introduced.* *G. B. Witham, Linear and Nonlinear Waves, p. 42.
Time series u 1 1 1 x
Time series u u 1 1 1 1 x 1 1 x
Time series u u 1 1 1 1 x 1 1 x u 1 1 1 x
Time series u u 1 1 1 1 x 1 1 x u u 1 1 shock forms 1 1 x 1 1 x
Time series u u 1 1 1 1 x 1 1 x u u 1 1 shock forms 1 1 x u 1 no longer a function 1 1 x 1 1 x
Time series u u 1 1 1 1 x 1 1 x u u 1 1 shock forms 1 1 x u 1 no longer a function 1 1 1 u x 1 1 x 1 1 x
To define solution beyond shock formation: equal area rule u u 1 shock forms 1 shock propogates 1 1 x 1 1 x
To define solution beyond shock formation: equal area rule u u 1 shock forms 1 shock propogates 1 1 x 1 1 x u u 1 1 1 1 x 1 1 x Note: The amplitude diminishes as the shock propogates, i.e., the wave collapses after a shock forms
More Can find time when shock forms (breaking time) Can find the shock speed
More Can find time when shock forms (breaking time) Can find the shock speed But that would have to wait for another day.
References Adrian Constantin, Nonlinear Water Waves with Applications to Wave-Current Interactions and Tsunamis, CBMS-NSF Regional Conference Series in Applied Mathematics, Volume 81, Society for Industrial and Applied Mathematics, Philadelphia (2011). Roger Knobel, An Introduction to the Mathematical Theory of Waves, Student Mathematics Library, IAS/Park City Mathematical Subseries, Volume 3, American Mathematical Society, Providence (2000). James Lighthill, Waves in Fluids, Cambridge Mathematical Library, Cambridge University Press, Cambridge (1978). Bruce R. Sutherland, Internal Gravity Waves, Cambridge University Press, Cambridge (2010). G. B. Whitham, Linear and Nonlinear Waves, A Wiley-Interscience Publication, John Wiley & Sons, Inc., New York (1999)