7 Special Functions In this chapter we will look at some additional functions which arise often in physical applications and are eigenfunctions for some Sturm-Liouville bound- ary value problem. We begin with a collection of special functions, called the classical orthogonal polynomials. These include such polynomial functions as the Legendre polynomials, the Hermite polynomials, the Tchebychef and the Gegenbauer polynomials. Also, Bessel functions occur quite often. We will spend more time exploring the Legendre and Bessel functions. These func- tions are typically found as solutions of differential equations using power series methods in a first course in differential equations. 7.1 Classical Orthogonal Polynomials We begin by noting that the sequence of functions 1, x, x2,... is a basis of linearly independent functions. In fact, by the Stone-Weie{ rstrass} Approxima- 2 tion Theorem this set is a basis of Lσ(a,b), the space of square integrable functions over the interval [a,b] relative to weight σ(x). We are familiar with being able to expand functions over this basis, since the expansions are just power series representations of the functions, ∞ f(x) c xn. ∼ n n=0 X However, this basis is not an orthogonal set of basis functions. One can eas- ily see this by integrating the product of two even, or two odd, basis functions with σ(x)=1and (a,b)=( 1, 1). For example, − 1 2 < 1, x2 >= x0x2 dx = . 1 3 Z− Since we have found that orthogonal bases have been useful in determining the coefficients for expansions of given functions, we might ask if it is possible 206 7 Special Functions to obtain an orthogonal basis involving these powers of x. Of course, finite combinations of these basis element are just polynomials! OK, we will ask.“Given a set of linearly independent basis vectors, can one find an orthogonal basis of the given space?” The answer is yes. We recall from introductory linear algebra, which mostly covers finite dimensional vector spaces, that there is a method for carrying this out called the Gram- Schmidt Orthogonalization Process. We will recall this process for finite dimensional vectors and then generalize to function spaces. 3 Fig. 7.1. The basis a1, a2, and a3, of R considered in the text. 3 Let’s assume that we have three vectors that span R , given by a1, a2, and a3 and shown in Figure 7.1. We seek an orthogonal basis e1, e2, and e3, beginning one vector at a time. First we take one of the original basis vectors, say a1, and define e1 = a1. Of course, we might want to normalize our new basis vectors, so we would denote such a normalized vector with a “hat”: e1 eˆ1 = , e1 where e = √e e . 1 1 · 1 Next, we want to determine an e2 that is orthogonal to e1. We take another element of the original basis, a2. In Figure 7.2 we see the orientation of the vectors. Note that the desired orthogonal vector is e2. Note that a2 can be written as a sum of e2 and the projection of a2 on e1. Denoting this projection by pr1a2, we then have e = a pr a . (7.1) 2 2 − 1 2 We recall the projection of one vector onto another from our vector calculus class. a2 e1 pr1a2 = ·2 e1. (7.2) e1 7.1 Classical Orthogonal Polynomials 207 Fig. 7.2. A plot of the vectors e1, a2, and e2 needed to find the projection of a2, on e1. Note that this is easily proven by writing the projection as a vector of length a2 cos θ in direction eˆ1, where θ is the angle between e1 and a2. Using the definition of the dot product, a b = ab cos θ, the projection formula follows. Combining Equations (7.1)-(7.2),· we find that a2 e1 e2 = a2 ·2 e1. (7.3) − e1 It is a simple matter to verify that e2 is orthogonal to e1: a2 e1 e2 e1 = a2 e1 ·2 e1 e1 · · − e1 · = a e a e =0. (7.4) 2 · 1 − 2 · 1 Now, we seek a third vector e3 that is orthogonal to both e1 and e2. Picto- rially, we can write the given vector a3 as a combination of vector projections along e1 and e2 and the new vector. This is shown in Figure 7.3. Then we have, a3 e1 a3 e2 e3 = a3 ·2 e1 ·2 e2. (7.5) − e1 − e2 Again, it is a simple matter to compute the scalar products with e1 and e2 to verify orthogonality. We can easily generalize the procedure to the N-dimensional case. Gram-Schmidt Orthogonalization in N-Dimensions N Let an,n =1, ..., N be a set of linearly independent vectors in R . Then, an orthogonal basis can be found by setting e1 = a1 and for n> 1, n 1 − a e e = a n · j e . (7.6) n n − e2 j j=1 j X 208 7 Special Functions Fig. 7.3. A plot of the vectors and their projections for determining e3. Now, we can generalize this idea to (real) function spaces. Gram-Schmidt Orthogonalization for Function Spaces Let fn(x), n N0 = 0, 1, 2,... , be a linearly independent se- quence of continuous∈ functions{ defined} for x [a,b]. Then, an ∈ orthogonal basis of functions, φn(x), n N0 can be found and is given by ∈ φ0(x)= f0(x) and n 1 − <f , φ > φ (x)= f (x) n j φ (x), n =1, 2,.... (7.7) n n − φ 2 j j=0 j X k k Here we are using inner products relative to weight σ(x), b <f,g>= f(x)g(x)σ(x) dx. (7.8) Za Note the similarity between the orthogonal basis in (7.7) and the expression for the finite dimensional case in Equation (7.6). Example 7.1. Apply the Gram-Schmidt Orthogonalization process to the set f (x)= xn, n N , when x ( 1, 1) and σ(x)=1. n ∈ 0 ∈ − First, we have φ0(x)= f0(x)=1. Note that 1 2 1 φ0(x) dx = . 1 2 Z− We could use this result to fix the normalization of our new basis, but we will hold off on doing that for now. Now, we compute the second basis element: 7.2 Legendre Polynomials 209 <f , φ > φ (x)= f (x) 1 0 φ (x) 1 1 − φ 2 0 k 0k < x, 1 > = x 1= x, (7.9) − 1 2 k k since < x, 1 > is the integral of an odd function over a symmetric interval. For φ2(x), we have <f , φ > <f , φ > φ (x)= f (x) 2 0 φ (x) 2 1 φ (x) 2 2 − φ 2 0 − φ 2 1 k 0k k 1k < x2, 1 > < x2,x> = x2 1 x − 1 2 − x 2 k k k k 1 2 2 1 x dx = x − − 1 R 1 dx − 1 = x2 R. (7.10) − 3 So far, we have the orthogonal set 1, x, x2 1 . If one chooses to nor- { − 3 } malize these by forcing φn(1) = 1, then one obtains the classical Legendre polynomials, Pn(x)= φ1(x). Thus, 1 P (x)= (3x2 1). 2 2 − Note that this normalization is different than the usual one. In fact, we see that P2(x) does not have a unit norm, 1 2 2 2 P2 = P2 (x) dx = . k k 1 5 Z− The set of Legendre polynomials is just one set of classical orthogonal polynomials that can be obtained in this way. Many had originally appeared as solutions of important boundary value problems in physics. They all have similar properties and we will just elaborate some of these for the Legendre functions in the next section. Other orthogonal polynomials in this group are shown in Table 7.1. For reference, we also note the differential equations satisfied by these functions. 7.2 Legendre Polynomials In the last section we saw the Legendre polynomials in the context of or- thogonal bases for a set of square integrable functions in L2( 1, 1). In your first course in differential equations, you saw these polynomials− as one of the solutions of the differential equation 210 7 Special Functions Polynomial Symbol Interval σ(x) −x2 Hermite Hn(x) (−∞, ∞) e α −x Laguerre Ln (x) [0, ∞) e Legendre Pn(x) (-1,1) 1 λ 2 λ−1/2 Gegenbauer Cn (x) (-1,1) (1 − x ) 2 −1/2 Tchebychef of the 1st kind Tn(x) (-1,1) (1 − x ) 2 −1/2 Tchebychef of the 2nd kind Un(x) (-1,1) (1 − x ) (ν,µ) ν µ Jacobi Pn (x) (-1,1) (1 − x) (1 − x) Table 7.1. Common classical orthogonal polynomials with the interval and weight function used to define them. Polynomial Differential Equation Hermite y′′ − 2xy′ + 2ny = 0 ′′ ′ Laguerre xy +(α + 1 − x)y + ny = 0 ′′ ′ Legendre (1 − x2)y − 2xy + n(n + 1)y = 0 Gegenbauer (1 − x2)y′′ − (2n + 3)xy′ + λy = 0 ′′ ′ Tchebychef of the 1st kind (1 − x2)y − xy + n2y = 0 ′′ ′ Jacobi (1 − x2)y +(ν − µ +(µ + ν + 2)x)y + n(n +1+ µ + ν)y = 0 Table 7.2. Differential equations satisfied by some of the common classical orthog- onal polynomials. 2 (1 x )y′′ 2xy′ + n(n + 1)y =0, n N . (7.11) − − ∈ 0 Recall that these were obtained by using power series expansion methods. In this section we will explore a few of the properties of these functions. For completeness, we recall the solution of Equation (7.11) using the power series method.