Lecture 4.6: Some special orthogonal functions

Transcription

1 Lecture 4.6: Some special orthogonal functions Matthew Macauley Department of Mathematical Sciences Clemson University Math 4340, Advanced Engineering Mathematics M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 1 / 13

2 Motivation Recall that every 2nd order linear homogeneous ODE, y + P(x)y + Q(x)y = 0 can be written as a Sturm-Liouville equation, called its self-adjoint or Sturm-Liouville form: d ( p(x)y ) + q(x)y = λw(x)y, where p(x), w(x) > 0. dx Many of these ODEs require the Frobenius method to solve. Examples from physics and engineering Legendre s equation: (1 x 2 )y 2xy + n(n + 1)y = 0. Used for modeling spherically symmetric potentials in the theory of Newtonian gravitation and in electricity & magnetism (e.g., the wave equation for an electron in a hydrogen atom). Parametric Bessel s equation: x 2 y + xy + (λx 2 α 2 )y = 0. Used for analyzing vibrations of a circular drum. Chebyshev s equation: (1 x 2 )y xy + n 2 y = 0. Arises in numerical analysis techniques. Hermite s equation: y 2xy + 2ny = 0. Used for modeling simple harmonic oscillators in quantum mechanics. Laguerre s equation: xy + (1 x)y + ny = 0. Arises in a number of equations from quantum mechanics. Airy s equation: y k 2 xy = 0. Models the refraction of light. M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 2 / 13

3 Legendre s differential equation Consider the following Sturm-Liouville problem, defined on ( 1, 1): d dx [(1 x 2 ) d dx y ] = λy, [ ] p(x) = 1 x 2, q(x) = 0, w(x) = 1. The eigenvalues are λ n = n(n + 1) for n = 1, 2,..., and the eigenfunctions solve Legendre s equation: (1 x 2 )y 2xy + n(n + 1)y = 0. For each n, one solution is a degree-n Legendre polynomial P n(x) = 1 d n [ (x 2 2 n n! dx n 1) n]. 1 They are orthogonal with respect to the inner product f, g = f (x)g(x) dx. 1 It can be checked that 1 P m, P n = P m(x)p n(x) dx = 2 1 2n + 1 δmn. By orthogonality, every function f, continuous on 1 < x < 1, can be expressed using Legendre polynomials: f (x) = f, Pn c np n(x), where c n = P n, P = (n + 1 ) f, Pn n 2 M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 3 / 13

4 Legendre polynomials P 0 (x) = 1 P 1 (x) = x P 2 (x) = 1 2 (3x2 1) P 3 (x) = 1 2 (5x3 3x) P 4 (x) = 1 8 (35x4 30x 2 + 3) P 5 (x) = 1 8 (63x5 70x x) P 6 (x) = 1 8 (231x6 315x x 2 5) P 7 (x) = 1 16 (429x7 693x x 3 35x) M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 4 / 13

5 Parametric Bessel s differential equation Consider the following Sturm-Liouville problem on [0, a]: d dx ( xy ) ν2 x y = λxy, [ p(x) = x, For a fixed ν, the eigenvalues are λ n = ω 2 n := α 2 n/a 2, for n = 1, 2,.... q(x) = ν2 x, w(x) = x ]. Here, α n is the n th positive root of J ν(x), the Bessel functions of the first kind of order ν. The eigenfunctions solve the parametric Bessel s equation: x 2 y + xy + (λx 2 ν 2 )y = 0. Fixing ν, for each n there is a solution J νn(x) := J ν(ω nx). a They are orthogonal with repect to the inner product f, g = f (x)g(x) x dx. 0 It can be checked that a J νn, J νm = J ν(ω nx)j ν(ω mx) dx = 0, if n m. 0 By orthogonality, every continuous function f (x) on [0, a] can be expressed in a Fourier-Bessel series: f, Jνn f (x) c nj ν(ω nx), where c n = J νn, J. νn M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 5 / 13

6 Bessel functions (of the first kind) J ν(x) = ( 1) m 1 ( x ) 2m+ν. m!(ν + m)! 2 m=0 M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 6 / 13

7 Fourier-Bessel series from J 0 (x) f (x) c nj 0 (ω nx), J 0 (x) = ( 1) m 1 ( x ) 2m (m!) m=0 2 2 Figure: First 5 solutions to (xy ) = λx 2. M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 7 / 13

8 Fourier-Bessel series from J 3 (x) The Fourier-Bessel series using J 3 (x) of the function f (x) = { x 3 0 < x < 10 0 x > 10 f (x) c nj 3 (ω nx/10), J 3 (x) = ( 1) m 1 ( x ) 2m+3. m!(3 + m)! 2 m=0 is Figure: First 5 partial sums to the Fourier-Bessel series of f (x) using J 3 M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 8 / 13

9 Chebyshev s differential equation Consider the following Sturm-Liouville problem on [ 1, 1]: d [ 1 x 2 d ] dx dx y 1 = λ [p(x) y, = 1 x 2, q(x) = 0, w(x) = 1 x 2 The eigenvalues are λ n = n 2 for n = 1, 2,..., and the eigenfunctions solve Chebyshev s equation: (1 x 2 )y xy + n 2 y = 0. For each n, one solution is a degree-n Chebyshev polynomial, defined recursively by T 0 (x) = 1, T 1 (x) = x, T n+1 (x) = 2xT n(x) T n 1 (x). 1 f (x)g(x) They are orthogonal with repect to the inner product f, g = dx. 1 1 x 2 It can be checked that 1 T m, T n = 1 T m(x)t n(x) 1 x 2 dx = { 1 πδmn m 0, n 0 2 π m = n = x 2 ]. By orthogonality, every function f (x), continuous for 1 < x < 1, can be expressed using Chebyshev polynomials: f (x) f, Tn c nt n(x), where c n = T n, T = 2 f, Tn, if n, m > 0. n π M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 9 / 13

10 Chebyshev polynomials (of the first kind) T 0 (x) = 1 T 4 (x) = 8x 4 8x T 1 (x) = x T 5 (x) = 16x 5 20x 3 + 5x T 2 (x) = 2x 2 1 T 6 (x) = 32x 6 48x x 2 1 T 3 (x) = 4x 3 3x T 7 (x) = 64x 7 112x x 3 7x M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 10 / 13

11 Hermite s differential equation Consider the following Sturm-Liouville problem on (, ): d dx [e x2 d dx y ] = λe x2 y, [ p(x) = e x2, q(x) = 0, w(x) = e x2 ]. The eigenvalues are λ n = 2n for n = 1, 2,..., and the eigenfunctions solve Hermite s equation: y 2xy + 2ny = 0. For each n, one solution is a degree-n Hermite polynomial, defined by H n(x) = ( 1) n e x2 dn ( dx n e x2 = 2x d ) n 1 dx 1 They are orthogonal with repect to the inner product f, g = f (x)g(x)e x2 dx. 1 It can be checked that H m, H n = H m(x)h n(x)e x2 dx = π2 n n!δ mn. By orthogonality, every function f (x) satisfying f 2 e x2 dx < can be expressed using Hermite polynomials: f (x) f, Hn f, Hn c nh n(x), where c n = = H n, H n π2 n n!. M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 11 / 13

12 Hermite polynomials H 0 (x) = 1 H 4 (x) = 16x 4 48x H 1 (x) = 2x H 5 (x) = 32x 5 160x x H 2 (x) = 4x 2 2 H 6 (x) = 64x 6 480x x H 3 (x) = 8x 3 12x H 7 (x) = 128x x x x M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 12 / 13

13 Hermite functions The Hermite functions can be defined from the Hermite polynomials as ψ n(x) = ( 2 n n! π ) 1 2 e x2 2 H n(x) = ( 1) n( 2 n n! π ) 1 2 e x2 d n 2 dx n e x2. They are orthonormal with respect to the inner product f, g = f (x)g(x) dx Every real-valued function f such that f 2 dx < can be expressed uniquely as f (x) c nψ n(x) dx, where c n = f, ψ n = f (x)ψ n(x) dx. These are solutions to the time-independent Schrödinger ODE: y + x 2 y = (2n + 1)y. M. Macauley (Clemson) Lecture 4.6: Some special orthogonal functions Advanced Engineering Mathematics 13 / 13