Embed Size (px)
Citation preview
Making Waves in Multivariable Calculus
<http://blogs.ams.org/blogonmathblogs/2013/04/22/the-mathematics-of-planet-earth/>
J. B. ThooYuba College
2014 CMC3 Fall Conference, Monterey, Ca
December 10, 2014
This presentation was produced using LATEX with C. Campani’sBeamer LATEX 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 givenfunction u = u(x , t) satisfies the heat equation, ut = Duxx , or thewave equation, utt = c2uxx . While this is a useful exercise in usingthe chain rule, it is not a very exciting one because it ends there.
The mathematical theory of waves is a rich source of partialdifferential equations. This talk is about introducing somemathematics of waves to multivariable calculus (vector calculus)students. We will show you some examples that we have presentedto 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 ruleIntegrating factorPartial fractions
Other examples (time permitting)
References
Roger Knobel, An Introduction to the Mathematical Theory of Waves, StudentMathematics Library, IAS/Park City Mathematical Subseries, Volume 3, Ameri-can 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 wholerange of wave phenomena it seems preferable to be guided by theintuitive view that a wave is any recognizable signal that istransferred from one part of the medium to another with arecognizable 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 wholerange of wave phenomena it seems preferable to be guided by theintuitive view that a wave is any recognizable signal that istransferred from one part of the medium to another with arecognizable velocity of propagation.
[Whitham]
Some wave equations
The wave equation
The wave equation: utt = c2uxx
Models a number of wavephenomena, e.g., vibrations ofa stretched string
Standing wave solution:
un(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: ut + uux + uxxx = 0
Models shallow water gravitywaves
x
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 → ±∞.
The Sine-Gordon equation
The Sine-Gordon equation: utt = uxx − sin u
Models a mechanicaltransmission line such aspendula 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 =⇒ Dhm×n
= Dgm×p
Dfp×n
if f : Rn → Rp and g : Rp → Rm so that h : Rn → Rm
E.g., f : R → R2 : f (t) = (x , y), g : R2 → R2 : g(x , y) = (w , z),and
h = g ◦ f : R → R2 : h(t) = (w , z)
Then
Dh =
[∂w∂x
∂w∂y
∂z∂x
∂z∂y
]Dg
[dxdt
dydt
]Df
=⇒
[dwdt
dzdt
]=
[∂w∂x
dxdt +
∂w∂y
dydt
∂z∂x
dxdt +
∂z∂y
dydt
]
Chain rule
h = g ◦ f =⇒ Dhm×n
= Dgm×p
Dfp×n
if f : Rn → Rp and g : Rp → Rm so that h : Rn → Rm
E.g., f : R → R2 : f (t) = (x , y), g : R2 → R2 : g(x , y) = (w , z),and
h = g ◦ f : R → R2 : h(t) = (w , z)
Then
Dh =
[∂w∂x
∂w∂y
∂z∂x
∂z∂y
]Dg
[dxdt
dydt
]Df
=⇒
[dwdt
dzdt
]=
[∂w∂x
dxdt +
∂w∂y
dydt
∂z∂x
dxdt +
∂z∂y
dydt
]
Example 1
The wave equation:* utt = auxx , 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 withoutchanging 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
utt = c2uxx .
Let z = x − ct. Then, u(x , t) = exp(−z2) = f (z) and, using the chainrule, we find that
ut =df
dz
∂z
∂t= 2cz exp(−z2),
utt =df ′
dz
∂z
∂t= 4c2z2 exp(−z2),
ux =df
dz
∂z
∂x= −2z exp(−z2),
uxx =df ′
dz
∂z
∂x= 4z2 exp(−z2)
utt = c2uxx =⇒ 4c2z2 exp(−z2) = c2 · 4z2 exp(−z2)
Typical exercise: Show that u(x , t) = exp[−(x − ct)2
]satisfies
utt = c2uxx .
Let z = x − ct. Then, u(x , t) = exp(−z2) = f (z) and, using the chainrule, we find that
ut =df
dz
∂z
∂t= 2cz exp(−z2),
utt =df ′
dz
∂z
∂t= 4c2z2 exp(−z2),
ux =df
dz
∂z
∂x= −2z exp(−z2),
uxx =df ′
dz
∂z
∂x= 4z2 exp(−z2)
utt = c2uxx =⇒ 4c2z2 exp(−z2) = c2 · 4z2 exp(−z2)
Let z = x − ct. Then u(x , t) = f (x − ct) = f (z) and, using thechain rule,
ut =df
dz
∂z
∂t= f ′(z)(−c) = −cf ′(z),
utt =df ′
dz
∂z
∂t= −cf ′′(z)(−c) = c2f ′′(z),
ux =df
dz
∂z
∂x= f ′(z)(1) = f ′(z),
uxx =df ′
dz
∂z
∂x= f ′′(z)(1) = f ′′(z)
utt = auxx =⇒ c2f ′′(z) = af ′′(z)
Let z = x − ct. Then u(x , t) = f (x − ct) = f (z) and, using thechain rule,
ut =df
dz
∂z
∂t= f ′(z)(−c) = −cf ′(z),
utt =df ′
dz
∂z
∂t= −cf ′′(z)(−c) = c2f ′′(z),
ux =df
dz
∂z
∂x= f ′(z)(1) = f ′(z),
uxx =df ′
dz
∂z
∂x= f ′′(z)(1) = f ′′(z)
utt = auxx =⇒ c2f ′′(z) = af ′′(z)
c2f ′′(z) = af ′′(z) =⇒ (c2 − a)f ′′(z) = 0
c2 − 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
c2f ′′(z) = af ′′(z) =⇒ (c2 − a)f ′′(z) = 0
c2 − 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
c2f ′′(z) = af ′′(z) =⇒ (c2 − a)f ′′(z) = 0
c2 − 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: ut + ux + uxxx = 0
Look for wave train solution: u(x , t) = A cos(kx − ωt) ,
where A 6= 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︸ ︷︷ ︸
x−ct
)advects at wave speed
c = ω/k
The number ω is the angular frequency and k is called thewavenumber. The wavelength is 2π/k .
*KdV = Korteweg-de Vries; the KdV equation models shallow-water gravitywaves
Example 2
The linearized KdV* equation: ut + ux + uxxx = 0
Look for wave train solution: u(x , t) = A cos(kx − ωt) ,
where A 6= 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︸ ︷︷ ︸
x−ct
)advects at wave speed
c = ω/k
The number ω is the angular frequency and k is called thewavenumber. The wavelength is 2π/k .
*KdV = Korteweg-de Vries; the KdV equation models shallow-water gravitywaves
Let z = kx − ωt and f (z) = A cos(z). Then
u(x , t) = A cos(kx − ωt) = f (z)
and, using the chain rule,
ut =df
dz
∂z
∂t= f ′(z)(−ω) = ωA sin(z),
ux =df
dz
∂z
∂x= f ′(z)(k) = −kA sin(z),
uxx =df ′
dz
∂z
∂x= f ′′(z)(k) = −k2A cos(z),
uxxx =df ′′
dz
∂z
∂x= f ′′′(z)(k) = k3A sin(z)
ut + ux + uxxx = 0 =⇒ (ω − k + k3)A sin(z) = 0
Let z = kx − ωt and f (z) = A cos(z). Then
u(x , t) = A cos(kx − ωt) = f (z)
and, using the chain rule,
ut =df
dz
∂z
∂t= f ′(z)(−ω) = ωA sin(z),
ux =df
dz
∂z
∂x= f ′(z)(k) = −kA sin(z),
uxx =df ′
dz
∂z
∂x= f ′′(z)(k) = −k2A cos(z),
uxxx =df ′′
dz
∂z
∂x= f ′′′(z)(k) = k3A sin(z)
ut + ux + uxxx = 0 =⇒ (ω − k + k3)A sin(z) = 0
(ω − k + k3)A sin(z) = 0 =⇒ ω − k + k3 = 0
Dispersion relation: ω = k − k3
Wave speed: c =ω
k= 1− k2
Note: That c depends on k means that wave trains of differentfrequencies travel at different speeds. Such a wave is called a dispersivewave. Here, smaller k or longer waves (λ = 2π/k) speed ahead, whilelarger k or shorter waves trail behind.
Group velocity: C = dωdk = 1− 3k2
The group velocity C is the velocity of the energy in the wave andis generally different from the wave speed c
(ω − k + k3)A sin(z) = 0 =⇒ ω − k + k3 = 0
Dispersion relation: ω = k − k3
Wave speed: c =ω
k= 1− k2
Note: That c depends on k means that wave trains of differentfrequencies travel at different speeds. Such a wave is called a dispersivewave. Here, smaller k or longer waves (λ = 2π/k) speed ahead, whilelarger k or shorter waves trail behind.
Group velocity: C = dωdk = 1− 3k2
The group velocity C is the velocity of the energy in the wave andis generally different from the wave speed c
(ω − k + k3)A sin(z) = 0 =⇒ ω − k + k3 = 0
Dispersion relation: ω = k − k3
Wave speed: c =ω
k= 1− k2
Note: That c depends on k means that wave trains of differentfrequencies travel at different speeds. Such a wave is called a dispersivewave. Here, smaller k or longer waves (λ = 2π/k) speed ahead, whilelarger k or shorter waves trail behind.
Group velocity: C = dωdk = 1− 3k2
The group velocity C is the velocity of the energy in the wave andis generally different from the wave speed c
(ω − k + k3)A sin(z) = 0 =⇒ ω − k + k3 = 0
Dispersion relation: ω = k − k3
Wave speed: c =ω
k= 1− k2
Note: That c depends on k means that wave trains of differentfrequencies travel at different speeds. Such a wave is called a dispersivewave. Here, smaller k or longer waves (λ = 2π/k) speed ahead, whilelarger k or shorter waves trail behind.
Group velocity: C = dωdk = 1− 3k2
The group velocity C is the velocity of the energy in the wave andis 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)
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
]
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
]
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: utt = uxx − sin u
Models a mechanicaltransmission line such aspendula 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
utt = uxx − sin u =⇒ c2f ′′(z) = f ′′(z)− sin f
To solve the equation in f , we multiply through by f ′(z), anintegrating factor
c2f ′f ′′ = f ′f ′′ − f ′ sin f =⇒ c2(12 f ′ 2
)′=(1
2 f ′ 2)′+ (cos f )′
Now integrate w.r.t. z
12c2f ′ 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
Let z = x − ct. Then u(x , t) = f (x − ct) = f (z) and
utt = uxx − sin u =⇒ c2f ′′(z) = f ′′(z)− sin f
To solve the equation in f , we multiply through by f ′(z), anintegrating factor
c2f ′f ′′ = f ′f ′′ − f ′ sin f =⇒ c2(12 f ′ 2
)′=(1
2 f ′ 2)′+ (cos f )′
Now integrate w.r.t. z
12c2f ′ 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
Let z = x − ct. Then u(x , t) = f (x − ct) = f (z) and
utt = uxx − sin u =⇒ c2f ′′(z) = f ′′(z)− sin f
To solve the equation in f , we multiply through by f ′(z), anintegrating factor
c2f ′f ′′ = f ′f ′′ − f ′ sin f =⇒ c2(12 f ′ 2
)′=(1
2 f ′ 2)′+ (cos f )′
Now integrate w.r.t. z
12c2f ′ 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
Let z = x − ct. Then u(x , t) = f (x − ct) = f (z) and
utt = uxx − sin u =⇒ c2f ′′(z) = f ′′(z)− sin f
To solve the equation in f , we multiply through by f ′(z), anintegrating factor
c2f ′f ′′ = f ′f ′′ − f ′ sin f =⇒ c2(12 f ′ 2
)′=(1
2 f ′ 2)′+ (cos f )′
Now integrate w.r.t. z
12c2f ′ 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 →∞,
12c2f ′ 2 = 1
2 f ′ 2 + cos f + a → 0 = 0+ cos 0+ a
so that a = −1,
i.e.
12c2f ′ 2 = 1
2 f ′ 2 + cos f − 1 =⇒ f ′ 2 =2
1− c2 (1− cos f )
Exercise:
1 Show that f (z) = 4 arctan[exp(− z√
1− c2
)]is a solution
2 Solve the equation to obtain the solution above(hint: 1− cos f = 2 sin2(f /2))
Then, as z →∞,
12c2f ′ 2 = 1
2 f ′ 2 + cos f + a → 0 = 0+ cos 0+ a
so that a = −1, i.e.
12c2f ′ 2 = 1
2 f ′ 2 + cos f − 1 =⇒ f ′ 2 =2
1− c2 (1− cos f )
Exercise:
1 Show that f (z) = 4 arctan[exp(− z√
1− c2
)]is a solution
2 Solve the equation to obtain the solution above(hint: 1− cos f = 2 sin2(f /2))
Then, as z →∞,
12c2f ′ 2 = 1
2 f ′ 2 + cos f + a → 0 = 0+ cos 0+ a
so that a = −1, i.e.
12c2f ′ 2 = 1
2 f ′ 2 + cos f − 1 =⇒ f ′ 2 =2
1− c2 (1− cos f )
Exercise:
1 Show that f (z) = 4 arctan[exp(− z√
1− c2
)]is a solution
2 Solve the equation to obtain the solution above(hint: 1− cos f = 2 sin2(f /2))
Then, as z →∞,
12c2f ′ 2 = 1
2 f ′ 2 + cos f + a → 0 = 0+ cos 0+ a
so that a = −1, i.e.
12c2f ′ 2 = 1
2 f ′ 2 + cos f − 1 =⇒ f ′ 2 =2
1− c2 (1− cos f )
Exercise:
1 Show that f (z) = 4 arctan[exp(− z√
1− c2
)]is a solution
2 Solve the equation to obtain the solution above(hint: 1− cos f = 2 sin2(f /2))
Wave front solution:
u(x , t) = 4 arctan[exp(− x − ct√
1− c2
)]
x
u
2π
speed cu
A wave front is a solution u(x , t) for which
limx→−∞
u(x , t) = k1 and limx→∞
u(x , t) = k2
Partial fractions
Given a rational function p(x)/q(x)
p(x)
q(x)=
r1(x)
q1(x)+
r2(x)
q2(x)+ · · ·+ rn(x)
qn(x)
where qi (x) is a linear or an irreducible quadratic factor of q(x) and
ri (x) =
Bi (constant) if qi is linear,
Aix + Bi if qi is quadratic
Example
The KdV equation: ut + uux + uxxx = 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
x
u
speed c
Then
ut + uux + uxxx = 0 =⇒ −cf ′ + ff ′ + f ′′′ = 0
Rewrite,
then integrate
−cf ′ +(1
2 f 2)′ + (f ′′)′ = 0
=⇒ −cf + 12 f 2 + f ′′ = a
To determine a, impose f (z), f ′′(z)→ 0 as z →∞. Then
−cf + 12 f 2 + f ′′ = a → 0+ 0+ 0 = a
so that−cf + 1
2 f 2 + f ′′ = 0
Then
ut + uux + uxxx = 0 =⇒ −cf ′ + ff ′ + f ′′′ = 0
Rewrite, then integrate
−cf ′ +(1
2 f 2)′ + (f ′′)′ = 0 =⇒ −cf + 12 f 2 + f ′′ = a
To determine a, impose f (z), f ′′(z)→ 0 as z →∞. Then
−cf + 12 f 2 + f ′′ = a → 0+ 0+ 0 = a
so that−cf + 1
2 f 2 + f ′′ = 0
Then
ut + uux + uxxx = 0 =⇒ −cf ′ + ff ′ + f ′′′ = 0
Rewrite, then integrate
−cf ′ +(1
2 f 2)′ + (f ′′)′ = 0 =⇒ −cf + 12 f 2 + f ′′ = a
To determine a, impose f (z), f ′′(z)→ 0 as z →∞.
Then
−cf + 12 f 2 + f ′′ = a → 0+ 0+ 0 = a
so that−cf + 1
2 f 2 + f ′′ = 0
Then
ut + uux + uxxx = 0 =⇒ −cf ′ + ff ′ + f ′′′ = 0
Rewrite, then integrate
−cf ′ +(1
2 f 2)′ + (f ′′)′ = 0 =⇒ −cf + 12 f 2 + f ′′ = a
To determine a, impose f (z), f ′′(z)→ 0 as z →∞. Then
−cf + 12 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 ′ + 12 f 2f ′ + f ′f ′′ = 0
=⇒ −c(1
2 f 2)′ + 12
(13 f 3)′ + (1
2 f ′ 2)′= 0
=⇒ −12cf 2 + 1
6 f 3 + 12 f ′ 2 = b
To determine b, impose f (z), f ′(z)→ 0 as z →∞.
Then
−12cf 2 + 1
6 f 3 + 12 f ′ 2 = b → 0+ 0+ 0 = b
so that−1
2cf 2 + 16 f 3 + 1
2 f ′ 2 = 0
Now multiply through by integrating factor f ′, then integrate
− cff ′ + 12 f 2f ′ + f ′f ′′ = 0
=⇒ −c(1
2 f 2)′ + 12
(13 f 3)′ + (1
2 f ′ 2)′= 0
=⇒ −12cf 2 + 1
6 f 3 + 12 f ′ 2 = b
To determine b, impose f (z), f ′(z)→ 0 as z →∞. Then
−12cf 2 + 1
6 f 3 + 12 f ′ 2 = b → 0+ 0+ 0 = b
so that−1
2cf 2 + 16 f 3 + 1
2 f ′ 2 = 0
Rewrite,
12 f ′ 2 = 1
2cf 2 − 16 f 3 =⇒
√3
f√3c − f
f ′ = 1
where we choose the positive√
and assume that 3c − f > 0.
Now let 3c − f = g2
√3
(3c − g2)g(−2gg ′) = 1 =⇒ 2
√3
3c − g2 g ′ = −1
To integrate, use partial fractions
13c − g2 =
A√3c − g
+B√
3c + g
Rewrite,
12 f ′ 2 = 1
2cf 2 − 16 f 3 =⇒
√3
f√3c − f
f ′ = 1
where we choose the positive√
and assume that 3c − f > 0.
Now let 3c − f = g2
√3
(3c − g2)g(−2gg ′) = 1 =⇒ 2
√3
3c − g2 g ′ = −1
To integrate, use partial fractions
13c − g2 =
A√3c − g
+B√
3c + g
Rewrite,
12 f ′ 2 = 1
2cf 2 − 16 f 3 =⇒
√3
f√3c − f
f ′ = 1
where we choose the positive√
and assume that 3c − f > 0.
Now let 3c − f = g2
√3
(3c − g2)g(−2gg ′) = 1 =⇒ 2
√3
3c − g2 g ′ = −1
To integrate, use partial fractions
13c − g2 =
A√3c − g
+B√
3c + g
13c − g2 =
A√3c − g
+B√
3c + g
=⇒ 1 = A(√3c + g) + B(
√3c − g)
=⇒ A =1
2√3c, B =
12√3c
=⇒ 13c − g2 =
1/2√3c√
3c − g+
1/2√3c√
3c + g
=⇒ 2√3
3c − g2 g ′ =g ′
√c(√3c − g)
+g ′
√c(√3c + g)
2√3
3c − g2 g ′ = −1
=⇒ g ′√
c(√3c − g)
+g ′
√c(√3c + g)
= −1
=⇒ g ′√3c − g
+g ′√
3c + g= −√
c
=⇒ − ln(√3c − g) + ln(
√3c + g) = −
√cz + d
=⇒ ln√3c + g√3c − g
= −√
cz + d
Solve for g : g(z) =√3c
exp(−√
cz + d)− 1exp(−
√cz + d) + 1
Recall: f = 3c − g2
Use: tanh ζ =sinh ζcosh ζ
=12(e
ζ − e−ζ)12(e
ζ + e−ζ)= −exp(−2ζ)− 1
exp(−2ζ) + 1
Substitute −2ζ = −√
cz + d :
g(z) = −√3c tanh
[12(√
cz − d)]
Use f = 3c − g2 and choose d = 0:
f (z) = 3c sech2[12√
cz]
=⇒ u(x , t) = 3c sech2[√
c
2(x − ct)
]
Solve for g : g(z) =√3c
exp(−√
cz + d)− 1exp(−
√cz + d) + 1
Recall: f = 3c − g2
Use: tanh ζ =sinh ζcosh ζ
=12(e
ζ − e−ζ)12(e
ζ + e−ζ)= −exp(−2ζ)− 1
exp(−2ζ) + 1
Substitute −2ζ = −√
cz + d :
g(z) = −√3c tanh
[12(√
cz − d)]
Use f = 3c − g2 and choose d = 0:
f (z) = 3c sech2[12√
cz]
=⇒ u(x , t) = 3c sech2[√
c
2(x − ct)
]
Solve for g : g(z) =√3c
exp(−√
cz + d)− 1exp(−
√cz + d) + 1
Recall: f = 3c − g2
Use: tanh ζ =sinh ζcosh ζ
=12(e
ζ − e−ζ)12(e
ζ + e−ζ)= −exp(−2ζ)− 1
exp(−2ζ) + 1
Substitute −2ζ = −√
cz + d :
g(z) = −√3c tanh
[12(√
cz − d)]
Use f = 3c − g2 and choose d = 0:
f (z) = 3c sech2[12√
cz]
=⇒ u(x , t) = 3c sech2[√
c
2(x − ct)
]
x
u
amplitude 3c
speed c
Soliton solution: u(x , t) = 3c sech2[√
c
2(x − ct)
]
Note: That amplitude is 3c means that taller waves move fasterthan shorter waves.
Other examples
Water gravity waves
Ship waves
Tsunamis
Shock waves
Water gravity waves
Consider water (inviscid incompressible fluid) in a constantgravitational 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: ω2 = gk tanh(kd) , k = |~k | = 2π/λ
Here, ω is the frequency, ~k is the wavenumber vector, λ is thewavelength, g is gravity, and d is the depth of the water
Water gravity waves
Consider water (inviscid incompressible fluid) in a constantgravitational 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: ω2 = gk tanh(kd) , k = |~k | = 2π/λ
Here, ω is the frequency, ~k is the wavenumber vector, λ is thewavelength, g is gravity, and d is the depth of the water
d
λ
“shallowness parameter” δ = dλ
deep water: δ > 0.28 shallow water: δ < 0.07
ω2 = gk tanh(kd)
=⇒ c =ω
k=
√g
ktanh(kd),
C =dω
dk=
12
√g
ktanh(kd) +
d√
gk sech2(kd)
2√
tanh(kd)
d
λ
“shallowness parameter” δ = dλ
deep water: δ > 0.28 shallow water: δ < 0.07
ω2 = gk tanh(kd)
=⇒ c =ω
k=
√g
ktanh(kd),
C =dω
dk=
12
√g
ktanh(kd) +
d√
gk sech2(kd)
2√
tanh(kd)
d
λ
“shallowness parameter” δ = dλ
deep water: δ > 0.28 shallow water: δ < 0.07
ω2 = gk tanh(kd)
=⇒ c =ω
k=
√g
ktanh(kd),
C =dω
dk=
12
√g
ktanh(kd) +
d√
gk sech2(kd)
2√
tanh(kd)
Deep water: δ →∞ at fixed k
Using limθ→∞
tanh(θ) = 1
c =ω
k=
√g
ktanh(2πδ) →
√g
k
C =dω
dk=
12
√g
ktanh(2πδ) +
d√
gk sech2(2πδ)2√
tanh(2πδ)
→ 12
√g
k=
12
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
Ship waves: C = 12c
travels at wave speed c = ω/k
travels at group velocity C = dω/dk
Ship waves: C = 12c
(now) B A (earlier)
C
(now) B A (earlier)
C
(now) B A (earlier)
C
B AD
C
E
OA′
D ′
C ′
O ′
Kelvin wedge
19.5◦ 35◦
B AD
C
E
O
A′D ′
C ′
O ′
Kelvin wedge
19.5◦ 35◦
B AD
C
E
OA′
D ′
C ′
O ′
Kelvin wedge
19.5◦ 35◦
Shallow water: δ → 0 at fixed d
Using limθ→0
tanh(θ)θ
= 1
c =ω
k=
√gd
tanh(2πδ)2πδ
→√
gd
C =dω
dk=
12
√gd
tanh(2πδ)2πδ
+d√
g · 2πδ/d sech2(2πδ)2√
tanh(2πδ)
→ 12
√gd +
d
2
√g
d=√
gd
Energy in the wave propagates at the wave speed
Tsunamis
Typical wavelength of several hundred kilometers and deepest pointin the ocean in the Marianas Trench (Western Pacific Ocean) about11 kilometers makes a tsunami a shallow-water wave (long wave)
Wave speed c =√
gd
E.g., ocean depth 4 kilometers, gravity 9.8 m/s2 yields a wavespeed c =
√39 200 m/s ≈ 200 m/s or about 445 mph
Typical amplitude in the open ocean about 1 m, but rises up to10 m to 15 m as approaches shore
d d
λ
a
Energy in the wave proportional to a2c ≈ a2√gd remains constant, so a
increases as d decreases
Hat Ray Leach beach, Thailand, December 2004
<http://geol105naturalhazards.voices.wooster.edu/page/32/>
d d
λ
a
Energy in the wave proportional to a2c ≈ a2√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?
Typically the wave will break.*
*G. B. Witham, Linear and Nonlinear Waves, p. 22.
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 “equalarea 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
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1shock forms
x
u
−1 1
1no longer a function
x
u
−1 1
1
Time series
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1shock forms
x
u
−1 1
1no longer a function
x
u
−1 1
1
Time series
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1shock forms
x
u
−1 1
1no longer a function
x
u
−1 1
1
Time series
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1shock forms
x
u
−1 1
1no longer a function
x
u
−1 1
1
Time series
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1shock forms
x
u
−1 1
1no longer a function
x
u
−1 1
1
Time series
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1
x
u
−1 1
1shock forms
x
u
−1 1
1no longer a function
x
u
−1 1
1
To define solution beyond shock formation: equal area rule
x
u
−1 1
1shock forms
x
u
−1 1
1shock propogates
x
u
−1 1
1
x
u
−1 1
1
Note: The amplitude diminishes as the shock propogates, i.e., thewave collapses after a shock forms
To define solution beyond shock formation: equal area rule
x
u
−1 1
1shock forms
x
u
−1 1
1shock propogates
x
u
−1 1
1
x
u
−1 1
1
Note: The amplitude diminishes as the shock propogates, i.e., thewave collapses after a shock forms
More
Can find time when shock forms (breaking time)
Can find the shock speed
But that would have to wait for another day.
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-CurrentInteractions and Tsunamis, CBMS-NSF Regional Conference Series in AppliedMathematics, Volume 81, Society for Industrial and Applied Mathematics,Philadelphia (2011).
Roger Knobel, An Introduction to the Mathematical Theory of Waves, StudentMathematics Library, IAS/Park City Mathematical Subseries, Volume 3,American Mathematical Society, Providence (2000).
James Lighthill, Waves in Fluids, Cambridge Mathematical Library, CambridgeUniversity 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)
![2011 NEWSLETTERreuther.wayne.edu/pdf/fall11.pdf · Stephen Lighthill Film Collection 1979-1980 [21 linear feet] Stephen Lighthill has been involved in a number of film produc- tions](https://img.pdfslide.us/doc/110x75/5f557daa32e28d45f15a1f87/2011-stephen-lighthill-film-collection-1979-1980-21-linear-feet-stephen-lighthill.jpg)
