Making Waves in Vector Calculus

Transcription

1 Making Waves in Vector Calculus < J. B. Thoo Yuba College 2014 MAA MathFest, Portland, OR

2 This presentation was produced using L A TEX with C. Campani s Beamer L A TEX class and saved as a PDF file: < See Norm Matloff s web page < for a quick tutorial. Disclaimer: Our slides here won t show off what Beamer can do. Sorry. :-)

3 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 vector calculus students. We will show you some examples that we have presented to our vector calculus students that have given a context for what they are learning.

4 Outline of the talk Some examples of waves Mathematical definition of a wave Some equations of waves Using what we have learnt Chain rule Integrating factor Partial fractions Other examples

5 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)

6 Some examples of waves

7 Typical Pond Guitar Strings (L) < (R) <

8 Internal waves Internal wave trains around Trinidad from space Model of an estuary in a lab (T) < (B) <

9 Internal waves Kelvin-Helmholtz instability Clouds In a tank (L) < (R) <

10 Water gravity waves Deep-water waves Bow waves or ship waves (L) < (R) <

11 Water gravity waves Shallow-water waves Tsunami (2011 Tohoku, Japan, earthquake) Iwanuma, Japan Crescent City, Ca Santa Cruz, Ca (L) < Japan-disaster-30-powerful-images-of-the-earthquake-and-tsunami.html> (C) < (R) < Inundation_Maps/Pages/2011_tohoku.aspx>

12 Solitary waves Morning glory cloud Ocean wave (L) < (R) <

13 Solitary waves Recreation of John Scott Russell s soliton, Hariot-Watt University (1995) <

14 Shock waves F-18 fighter jet Schlieren photograph (L) < (R) <

15 Mathematical definition of a wave

16 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]

17 Some equations of waves

18 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

19 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

20 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)

21 Using what we have learnt

22 Chain rule 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 (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

23 Chain rule 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 (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 (small k = long wave, large k = short wave). *KdV = Korteweg-de Vries; the KdV equation models shallow-water gravity waves

24 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)

25 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

26 (ω k + k 3 )A sin(z) = 0, A 0 = ω k + k 3 = 0 Dispersion relation: ω = k k 3

27 (ω k + k 3 )A sin(z) = 0, A 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.

28 (ω k + k 3 )A sin(z) = 0, A 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

29 Integrating factor To solve: y (x) + p(x)y(x) = q(x) for y = y(x)

30 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

31 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)

32 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

33 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 )

34 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

35 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

36 Then, as z, 1 2 c2 f 2 = 1 2 f 2 + cos f + a 0 = 0 + cos 0 + a so that a = 1,

37 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

38 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

39 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))

40 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

41 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

42 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

43 Then u t + uu x + u xxx = 0 = cf + ff + f = 0 Rewrite, cf + ( 1 2 f 2) + (f ) = 0

44 Then u t + uu x + u xxx = 0 = cf + ff + f = 0 Rewrite, then integrate cf + ( 1 2 f 2) + (f ) = 0 = cf f 2 + f = a To determine a, impose f (z), f (z) 0 as z. Then cf f 2 + f = a = a so that cf f 2 + f = 0

45 Now multiply through by integrating factor f, then integrate cff 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 f f 2 = b To determine b, impose f (z), f (z) 0 as z. Then 1 2 cf f f 2 = b = b so that 1 2 cf f f 2 = 0

46 Rewrite, 1 2 cf f f 2 = 0 = 3 f 3c f f = 1, where we chose the positive and assume that 3c f > 0.

47 Rewrite, 1 2 cf f f 2 = 0 = 3 f 3c f f = 1, where we chose 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

48 Rewrite, 1 2 cf f f 2 = 0 = 3 f 3c f f = 1, where we chose 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

49 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)

50 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

51 Solve for g: g(z) = 3c exp( cz + d) 1 exp( cz + d) + 1 Recall: f = 3c g 2

52 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) ]

53 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

54 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.

55 Other examples Water gravity waves Ship waves Tsunamis Shock waves

56 Other examples Water gravity waves Ship waves Tsunamis Shock waves But that would have to wait for another day. Thank you.

57 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) More slides: