Nonlinear Integral Equation Formulation of Orthogonal Polynomials

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Nonlinear Integral Equation Formulation of Orthogonal Polynomials Eli Ben-Naim Theory Division, Los Alamos National Laboratory with: Carl Bender (Washington University, St. Louis) C.M. Bender and E. Ben-Naim, J. Phys. A: Math. Theor. 40, F9 (2007) Talk, paper available from: http://cnls.lanl.gov/~ebn BenderFest, St. Louis, MO, March 27, 2009

Working with Carl Data: 50 email messages received during collaboration AM: 25%, PM: 75% Midnight singularity 12 # of email messages 10 8 6 4 2 0 Hong Kong 1 11 6 4 3 3 3 3 3 2 2 2 2 2 1 1 1 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 AM time of day PM

Orthogonal Polynomials 101 Definition: A set of polynomials is orthogonal with respect to the measure g(x) on the interval [ : β] if dx g(x)p n (x)p m (x) = 0 P n (x) for all m n Properties: Orthogonal polynomials have many fascinating and useful properties: All roots are real and are inside the interval [ : β] The orthogonal polynomials form a complete basis: any polynomial is a linear combination of orthogonal polynomials of lesser or equal order

Orthogonal Polynomials 101 Specification: Orthogonal polynomials can be specified in multiple ways (typically linear) Example: Legendre Polynomials orthogonal on [ 1 : 1] w.r.t g(x) = 1 1. Differential Equation: 2. Generating Function: 3. Rodriguez Formula: 4. Recursion Formula: (1 x 2 )P n (x) 2xP n(x) + n(n + 1)P n (x) = 0 1 = 1 2tx + t 2 P n (x) = 1 2 n n! n=0 P n (x)t n d n dx n (x2 1) n (n + 1)P n+1 (x) (2n + 1)xP n (x) + np n 1 (x) = 0

Nonlinear Integral Equation Consider the following nonlinear integral equation P (x) = No restriction on integration limits Weight function restricted: non-vanishing integral dy w(y) P (y) P (x + y) dx w(x) 0 Motivation: stochastic process involving subtraction x 1, x 2 x 1 x 2 e x = 2 0 dy e y e (x+y) Reduces to the Wigner function equation when [ : β] = [ : ] w(y) = e iy

Constant Solution (n=0) Consider the constant polynomial A constant solves the nonlinear integral equation When Since a 0 P (x) = a = P 0 (x) = a a 0 dy w(y) P (y) P (x + y) dy w(y)a 2 we can divide by 1 = a dy w(y)a. A constant solution exists

Linear Solution (n=1) Consider the linear polynomial P 1 (x) = a + bx b 0 Let us introduce the shorthand notation f(x) The nonlinear integral equation a + bx = P 1 (y) [a + by + bx] 1. Equate the coefficients of and divide by x 1 b 0 b = b P 1 P 1 (x) = 1 2. Equate the coefficients of and divide by dx w(x)f(x) x 0 P (x) = P (y)p (x + y) a = a P 1 + b yp 1 (y) xp 1 (x) = 0 b 0 reads Linear solution generally exists Nonlinear equation reduces to 2 linear inhomogeneous equations for a,b

Quadratic Solution (n=2) Consider the quadratic polynomial P 2 (x) = a + bx + cx 2 c 0 The nonlinear integral equation becomes a + bx + cx 2 = P 2 (y) [ a + by + cy 2 + bx + 2cxy + cx 2] Successively equating coefficients c = c P 2 (y) P 2 (x) = 1 b = b P 2 (y) + 2c yp 2 (y) xp 2 (x) = 0 a = a P 2 (y) + b yp 2 (y) + c y 2 P 2 (y) x 2 P 2 (x) = 0 Quadratic solution generally exists Miraculous cancelation of terms Nonlinear equation reduces to 3 linear inhomogeneous equations for a,b,c

General Properties The nonlinear integral equation has two remarkable properties: 1. This equation preserves the order of a polynomial 2. For polynomial solutions, the nonlinear equation reduces to a linear set of equations for the coefficients of the polynomials In general, a polynomial of degree n n P n (x) = a n,k x k k=0 Is a solution of the integral equation if and only if its n+1 coefficients satisfy the following set of n+1 linear inhomogeneous equations x k P n (x) = δ k,0 (k = 0, 1,..., n) Infinite number of polynomial solutions

The set of polynomial solutions is orthogonal! The polynomial solutions are orthogonal w.r.t. g(x) = x w(x) Because for x P n P m = m m < n a m,k x k+1 P n (x) = m+1 a m,k 1 x k P n (x) = 0 k=0 As follows immediately from k=1 x k P n (x) = δ k,0 1.The nonlinear integral equation admits an infinite set of polynomial solutions 2.The polynomial solutions form an orthogonal set

The equations for the coefficients The equations for the coefficients x k P n (x) = δ k,0 (k = 0, 1,..., n) Can be compactly written as n j=0 a n,j m k+j = δ k,0 (k = 0, 1,..., n) Or in matrix form m 0 m 1 m n m 1 m 2 m n+1....... m n m n+1 m 2n In terms of the moments of the weight function m n = x n a n,0 a n,1.. a n,n = 1 0. 0

Formulas for the polynomials Using Cramer s rule, the polynomials can be expressed as a ratio of determinants A n = 1 x x n m 1 m 2 m n+1........ m n m n+1 m 2n, B n = m 0 m 1 m n m 1 m 2 m n+1....... m n m n+1 m 2n Explicit expressions P 0 (x) = 1, P 1 (x) = m 2 xm 1 m 2 m 2 1 P n (x) = det A n det B n, P 2 (x) = (m 2m 4 m 2 3) + (m 2 m 3 m 1 m 4 )x + (m 1 m 3 m 2 2)x 2 m 4 (m 2 m 2 1 ) m2 3 + 2m 1m 2 m 3 m 3 2.

Examples Interval [0:1], weight function w(x)=1 P 0 (x) = 1 P 1 (x) = 4 6x P 2 (x) = 9 36x + 30x 2 P 3 (x) = 16 120x + 240x 2 140x 3. Jacobi polynomials P n (x) G n (2, 2, x) orthogonal w.r.t the measure g(x)=x 1. Generalized Laguerre polynomials L (γ) n (x) = 0, β =, w(x) = x γ 1 e x 2. Jacoby polynomials G n (p, q, x) = 0, β = 1, w(x) = x q 2 (1 x) p q 3. Shifted Chebyshev of the second kind polynomials U n(x) = 0, β = 1, w(x) = (1 x) 1/2 x 1/2

Integration in the complex domain To specify the Legendre Polynomials P (x) = 1 1 dy 1 y Perform integration in the complex domain 1 P (y) P (x + y) 1 This integration path gives the Legendre Polynomials P 0 (x) = 1, P 1 (x) = iπ 2 x, dx x = iπ w(x) = 1 iπ x P 2 (x) = 1 3x 2, P 3 (x) = 3iπ 8 (3x 5x3 ), Nonlinear integral equation extends to complex domain

Generalization I: multiplicative arguments Nonlinear equation with multiplicative argument P (x) = dy w(y) P (y) P (x y) Infinite set of polynomial solutions when x k P n (x) = 1 These polynomials are orthogonal (1 x)p n P m = m k=0 a m,k ( x k P n x k+1 P n ) = 0 Now, the orthogonality measure is m < n g(x) = (1 x)w(x) A series of nonlinear integral formulations

Generalization II: iterated integrals Iterate the nonlinear integral equation P (x) = The double integral equation dy g(y) y P (y) P (x + y) P (x) = dy g(y) y dz g(z) z P (y) P (z) P (x + y + z) Similarly specifies orthogonal polynomials

Summary A set of orthogonal polynomials w.r.t the measure g(x) can be specified through the nonlinear integral equation P (x) = dy g(y) y P (y) P (x + y) For polynomial solutions, this nonlinear equation reduces to a linear set of equations Simple, compact, and completely general way to specify orthogonal polynomials Natural way to extend theory to complex domain

Outlook Non-polynomial solutions Multi-dimensional polynomials Matrix polynomials Polynomials defined on disconnected domains Higher-order nonlinear integral equations Use nonlinear formulation to derive integral identities Asymptotic properties of polynomials

Nonlinear Integral Identities Let P n (x) be the set of orthogonal polynomials specified by the nonlinear integral equation P n (x) = dy g(y) y P n(y) P n (x + y) Then, any polynomial Q m (x) of degree m n satisfies the nonlinear integral identity Q m (x) = dy g(y) y P n(y) Q m (x + y)

Asymptotic Properties & Integral Identities Combining the asymptotics of the Laguerre Polynomials And the nonlinear integral equation with scaled variables 1 n γ Lγ n ( x n ( ) lim n n γ L γ x n n ) = 1 Γ(γ) 0 = x γ/2 J γ (2 x) dy y γ 1 e y/n 1 ( y ) 1 n γ Lγ n n n γ Lγ n Gives a standard identity for the Bessel functions 2 γ 1 J γ(z) 2z γ = 1 Γ(γ) 0 ( x + y dw w γ 1 J γ (w) J γ( w 2 + z 2 ) (w 2 + z 2 ) γ/2 n )