Making Waves in Vector Calculus

<http://blogs.ams.org/blogonmathblogs/2013/04/22/the-mathematics-of-planet-earth/>

J. B. ThooYuba College

2014 MAA MathFest, Portland, OR

July 18, 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 vector calculus students. We will showyou some examples that we have presented to our vector calculusstudents that have given a context for what they are learning.

Outline of the talk

Some examples of waves

Mathematical definition of a wave

Some equations of waves

Using what we have learnt

Chain ruleIntegrating factorPartial fractions

Other examples

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

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]

Some equations of waves

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)

Using what we have learnt

Chain rule

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

(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 (small k = long wave, largek = short wave).

*KdV = Korteweg-de Vries; the KdV equation models shallow-water gravitywaves

Chain rule

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

(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 (small k = long wave, largek = short wave).*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, A 6= 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 adispersive wave. Here, smaller k or longer waves (λ = 2π/k) speedahead, while larger 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, A 6= 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 adispersive wave. Here, smaller k or longer waves (λ = 2π/k) speedahead, while larger 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, A 6= 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 adispersive wave. Here, smaller k or longer waves (λ = 2π/k) speedahead, while larger 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

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)

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

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,

−12cf 2 + 1

6 f 3 + 12 f ′ 2 = 0 =⇒

√3

f√3c − f

f ′ = 1,

where we chose 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,

−12cf 2 + 1

6 f 3 + 12 f ′ 2 = 0 =⇒

√3

f√3c − f

f ′ = 1,

where we chose 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,

−12cf 2 + 1

6 f 3 + 12 f ′ 2 = 0 =⇒

√3

f√3c − f

f ′ = 1,

where we chose 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

But that would have to wait for another day.

Thank you.

Other examples

Water gravity waves

Ship waves

Tsunamis

Shock waves

But that would have to wait for another day.

Thank you.

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)

More slides: http://ms.yccd.edu/~jthoo