Bessel Functions. A Touch of Magic. Fayez Karoji 1 Casey Tsai 1 Rachel Weyrens 2. SMILE REU Summer Louisiana State University

Transcription

1 Bessel s Function A Touch of Magic Fayez Karoji 1 Casey Tsai 1 Rachel Weyrens 2 1 Department of Mathematics Louisiana State University 2 Department of Mathematics University of Arkansas SMILE REU Summer 2010

2 Outline Introduction Terminology Drum Example

3 General Form Introduction Terminology Bessel s differential equation is x 2 y + xy + (x 2 n 2 )y = 0

4 General Form Introduction Terminology Bessel s differential equation is x 2 y + xy + (x 2 n 2 )y = 0 The linearly independent solutions are J n and Y n.

5 General Form Introduction Terminology Bessel s differential equation is x 2 y + xy + (x 2 n 2 )y = 0 The linearly independent solutions are J n and Y n. The zeros are j n and y n.

6 of Order Zero Terminology /Bessel/j0.pdf Figure: Bessel Function of the First Kind, J 0

7 of Order Zero Terminology /Bessel/y0.pdf Figure: Bessel Function of the Second Kind, Y 0

8 Key Terms Introduction Terminology Separation of variables

9 Terminology Key Terms Separation of variables Regular singular

10 Terminology Key Terms Separation of variables Regular singular Superposition

11 Physical Description Introduction Drum Example Radially symmetric

12 Drum Example Physical Description Radially symmetric Radius r = 1

13 Drum Example Physical Description Radially symmetric Radius r = 1 Beginning at rest

14 Drum Example Physical Description Radially symmetric Radius r = 1 Beginning at rest Edges fixed

15 Boundary Valued Problem Drum Example This physical problem can be represented by the following boundary valued problem:

16 Boundary Valued Problem Drum Example This physical problem can be represented by the following boundary valued problem: u tt = u rr + 1 r u r

17 Boundary Valued Problem Drum Example This physical problem can be represented by the following boundary valued problem: u tt = u rr + 1 r u r u(r, 0) = f (r)

18 Drum Example Boundary Valued Problem This physical problem can be represented by the following boundary valued problem: u tt = u rr + 1 r u r u(r, 0) = f (r) u t (r, 0) = 0

19 Drum Example Boundary Valued Problem This physical problem can be represented by the following boundary valued problem: u tt = u rr + 1 r u r u(r, 0) = f (r) u t (r, 0) = 0 u(1, t) = 0

20 Drum Example Separation of Variables We have u(r, t) = R(r)T (t) T + μt = 0 R + 1 r R + μr = 0

21 Drum Example Separation of Variables We have u(r, t) = R(r)T (t) T + μt = 0 R + 1 r R + μr = 0 μ = α 2 > 0

22 Solutions Introduction Drum Example The solutions of the given ODE s are

23 Solutions Introduction Drum Example The solutions of the given ODE s are T (t) = c 1 cos(αt) + c 2 sin(αt)

24 Drum Example Solutions The solutions of the given ODE s are T (t) = c 1 cos(αt) + c 2 sin(αt) R(r) = c 3 J 0 (αr) + c 4 Y 0 (αr)

25 Evaluation Introduction Drum Example Using initial and boundary conditions, we have u n (r, t) = A n J 0 (j n r) cos(j n t)

26 Evaluation Introduction Drum Example Using initial and boundary conditions, we have General solution u n (r, t) = A n J 0 (j n r) cos(j n t)

27 Evaluation Introduction Drum Example Using initial and boundary conditions, we have u n (r, t) = A n J 0 (j n r) cos(j n t) General solution u(r, t) = A n J 0 (j n r) cos(j n t) n=1

28 Amplitude Introduction Drum Example The amplitude of displacement, from u(r, 0) = f (r) is:

29 Amplitude Introduction Drum Example The amplitude of displacement, from u(r, 0) = f (r) is: A n = 1 0 rj 0(j n r)f (r)dr 1 0 rj 0(j n r)j 0 (j n r)dr

30 Frequencies Introduction Drum Example Fundamental pitch j 1 2π First overtone j 2 2π Second overtone j 3 2π

31 Property of /Bessel/jnspdf.pdf Figure: of the First Kind

32 Problems in Mathematical Physics PDE s model physical phenomena. Example: Steady Temperatures in Circular Cylinder (Laplacian in Cylindrical Coordinates). Example: The Vibrating Drumhead (Wave Equation in Polar Coordinates).

33 Methods of Solution PDE s are difficult to solve. Fourier s Method: Linear and homogeneous PDE s with homogeneous boundary conditions. Also known as Separation of Variables.

34 Fourier s Method: PDE ODE s PDE: Wave Equation in Polar Coordinates Apply Fourier s Method Two second order ODE s Simple Harmonic Motion T + μt = 0 Bessel s Equation R + 1 r R + R = 0

35 Orthogonal Functions Introduction Analysis of solutions to ODE s Underlying Theme: Orthogonal Functions Examples: Sine and Cosine Functions Legendre Polynomials (Special Function) (A "Very" Special Function)

36 What is? Introduction Dot Product or Inner Product in R n

37 What is? Introduction Dot Product or Inner Product in R n Given x, y R n

38 What is? Dot Product or Inner Product in R n Given x, y R n Define x y = n i=1 x iy i

39 What is? Dot Product or Inner Product in R n Given x, y R n Define x y = n i=1 x iy i x and y are orthogonal when n i=1 x iy i = 0

40 Generalize Inner Product in R[a, b]

41 Generalize Inner Product in R[a, b] Given f, g R[a, b]

42 Generalize Inner Product in R[a, b] Given f, g R[a, b] Define f, g = b a f (x)g(x) dx

43 Generalize Inner Product in R[a, b] Given f, g R[a, b] Define f, g = b a f (x)g(x) dx f and g are orthogonal when b a f (x)g(x) dx = 0

44 Example: Simple Harmonic Motion Consider T + n 2 T = 0, (μ = n 2 )

45 Example: Simple Harmonic Motion Consider T + n 2 T = 0, (μ = n 2 ) Solutions are sin(nx) and cos(nx)

46 Example: Simple Harmonic Motion Consider T + n 2 T = 0, (μ = n 2 ) Solutions are sin(nx) and cos(nx) Easy to show that π π sin(mx) cos(nx) dx = 0 for any n, m Z

47 Example: Legendre Polynomials It was shown that the Legendre Polynomials satisfy 1 1 P n(x)p m (x) dx = 0 for n, m Z, n = m

48 Example: property of J n (λx) and J n (μx)

49 Example: property of J n (λx) and J n (μx) of the First Kind of Order n

50 Example: property of J n (λx) and J n (μx) of the First Kind of Order n λ and μ are distinct positive roots of J n (x) = 0

51 Example: property of J n (λx) and J n (μx) of the First Kind of Order n λ and μ are distinct positive roots of J n (x) = 0 Will show: 1 0 xj n(λx)j n (μx) dx = 0

52 Theorem Introduction Theorem If λ and μ are distinct positive roots of J n (x) = 0 then { 1 0, if λ = μ xj n (λx)j n (μx) dx = J2 n+1 (λ), if λ = μ

53 Proof Introduction Proof. Suppose λ = μ, then λ and μ are distinct positive roots of J n (x) = 0. Since J n (λx) and J n (μx) are solutions of the Bessel equation in parametric form, we can write and x 2 J n (λx) + xj n(λx) + (λ 2 x 2 n 2 )J n (λx) = 0 (1) x 2 J n (μx) + xj n(μx) + (μ 2 x 2 n 2 )J n (μx) = 0 (2) Equations (1) and (2) may be written in the form

54 Proof Introduction x d dx [ x d ] dx J n(λx) + (λ 2 x 2 n 2 )J n (λx) = 0 (3) and x d dx [ x d ] dx J n(μx) + (μ 2 x 2 n 2 )J n (μx) = 0 (4) Multiplying (3) by Jn(μx) x and (4) by Jn(λx) x we get

55 Proof Introduction J n (μx) d dx [ x d ] dx J n(λx) + 1 x (λ2 x 2 n 2 )J n (λx)j n (μx) = 0 (5) and J n (λx) d dx [ x d ] dx J n(μx) + 1 x (μ2 x 2 n 2 )J n (μx)j n (λx) = 0 (6) Then subtracting, (5) - (6) we get

56 Proof Introduction J n (μx) d [ x d ] dx dx J n(λx) J n (λx) d [ x d ] dx dx J n(μx) + (λ 2 μ 2 )xj n (λx)j n (μx) = 0 (7) With some more manipulation, equation (7) may be written as

57 Proof Introduction [ d dx d dx J n (μx)x d ] dx J n(λx) [ J n (λx)x d ] dx J n(μx) + (λ 2 μ 2 )xj n (λx)j n (μx) = 0 (8) Finally integrating (8) from 0 to 1 noting that J n (λ) = J n (μ) = 0, we get

58 Proof Introduction (λ 2 μ 2 ) 1 0 xj n (λx)j n (μx) dx = 0 And since λ = μ, then we may divide to get the desired result 1 0 xj n (λx)j n (μx) dx = 0 (9)

59 Coefficients Introduction Theorem If λ and μ are distinct positive roots of J n (x) = 0 then { 1 0, if λ = μ xj n (λx)j n (μx) dx = J2 n+1 (λ), if λ = μ 1 0 xj n(λx)j n (λx) dx = 1 2 J2 n+1 (λ) = 0

60 Coefficients Introduction Theorem If λ and μ are distinct positive roots of J n (x) = 0 then { 1 0, if λ = μ xj n (λx)j n (μx) dx = J2 n+1 (λ), if λ = μ 1 0 xj n(λx)j n (λx) dx = 1 2 J2 n+1 (λ) = rj 0(j n r)j 0 (j n r) dr = 1 2 J2 1 (j n) = 0

61 Coefficients Introduction Theorem If λ and μ are distinct positive roots of J n (x) = 0 then { 1 0, if λ = μ xj n (λx)j n (μx) dx = J2 n+1 (λ), if λ = μ 1 0 xj n(λx)j n (λx) dx = 1 2 J2 n+1 (λ) = rj 0(j n r)j 0 (j n r) dr = 1 2 J2 1 (j n) = 0 A n = 1 0 rj 0(j nr)f (r) dr 1 0 rj 0(j nr)j 0 (j nr) dr

62 Thank You!