Bessel Functions of the First Kind

Jim Lambers MAT 415/515 Fall Semester 2013-14 Lecture 12 and 13 Notes

These notes correspond to Section 14.1 in the text.

Recall the Bessel equation x2y00 + xy0 + (x2 − n2)y = 0. For a ﬁxed value of n, this equation has two linearly independent solutions. One of these solutions, that can be obtained using Frobenius’ method, is called a Bessel function of the ﬁrst kind, and is denoted by Jn(x). This solution is regular at x = 0. The second solution, that is singular at x = 0, is called a Bessel function of the second kind, and is denoted by Yn(x).

Generating Function for Integral Order

A generating function for a sequence {an} is a power series

X n g(t) = ant , n of which the terms of the sequence are the coeﬃcients. Similarly, a generating function for a sequence of functions {fn(x)} is a power series

X n g(x, t) = fn(x)t , n whose coeﬃcients are now functions of x. The generating function for the sequence of Bessel functions of the ﬁrst kind, of integer order, is ∞ (x/2)(t−1/t) X n g(x, t) = e = Jn(x)t . n=−∞ x To obtain an expression for Jn(x), we use the Maclaurin series for e to obtain

g(x, t) = ext/2e−x/(2t) ∞ ∞ X 1 xr X (−1)s xs = tr t−s r! 2 s! 2 r=0 s=0 ∞ ∞ X X (−1)s xr+s = tr−s r!s! 2 r=0 s=0 ∞  ∞  X X (−1)s xn+2s = tn  (n + s)!s! 2  n=−∞ s=max{0,−n} It follows that for n ≥ 0, ∞ X (−1)s xn+2s J (x) = , n (n + s)!s! 2 s=0

1 while for n ≥ 0, we have ∞ X (−1)s x−n+2s J (x) = . −n (s − n)!s! 2 s=n Using an index shift, we obtain

∞ X (−1)s+n xn+2s J (x) = = (−1)nJ (x). −n (n + s)!s! 2 n s=0 For noninteger n, rather than using the generating function, we can use Frobenius’ method to obtain ∞ X (−1)s xν+2s J (x) = , ν s!Γ(ν + s + 1) 2 s=0 where Γ is the Gamma function. As Γ(s) = (s−1)! for s a positive integer, this formula is consistent with the one for integer order.

Recurrence Relations Using the generating function g(x, t), we can obtain some useful recurrence relations involving Bessel functions of the ﬁrst kind. Diﬀerentiating g(x, t) with respect to x and t yields

∞ ∂g(x, t) 1 1 X = t − e(x/2)(t−1/t) = J 0 (x)tn, ∂x 2 t n n=−∞ ∞ ∂g(x, t) x 1 X = 1 + e(x/2)(t−1/t) = nJ (x)tn−1. ∂t 2 t2 n n=−∞ Replacing the formula for g(x, t) with its power series expansion in the above equations yields

∞ ∞ ∞ X 1 1 X 1 X J 0 (x)tn = t − J (x)tn = J (x)(tn+1 − tn−1), n 2 t n 2 n n=−∞ n=−∞ n=−∞ ∞ ∞ ∞ X x 1 X x X nJ (x)tn−1 = 1 + J (x)tn = J (x)(tn + tn−2). n 2 t2 n 2 n n=−∞ n=−∞ n=−∞ Using index shifts, we obtain

∞ ∞ X 1 X J 0 (x)tn = J (x)(tn+1 − tn−1) n 2 n n=−∞ n=−∞ ∞ ∞ 1 X 1 X = J (x)tn+1 − J (x)tn−1 2 n 2 n n=−∞ n=−∞ ∞ ∞ 1 X 1 X = J (x)tn − J (x)tn 2 n−1 2 n+1 n=−∞ n=−∞ ∞ 1 X = [J (x) − J (x)]tn, 2 n−1 n+1 n=−∞ ∞ ∞ X x X nJ (x)tn−1 = J (x)(tn + tn−2) n 2 n n=−∞ n=−∞

2 ∞ ∞ x X x X = J (x)tn + J (x)tn−2 2 n 2 n n=−∞ n=−∞ ∞ ∞ x X x X = J (x)tn−1 + J (x)tn−1 2 n−1 2 n+1 n=−∞ n=−∞ ∞ x X = [J (x) + J (x)]tn−1. 2 n−1 n+1 n=−∞ Matching power series coeﬃcients, we obtain the recurrence relations

0 2Jn(x) = Jn−1(x) − Jn+1(x), 2n J (x) = J (x) + J (x). x n n−1 n+1 From these recurrence relations, we can obtain the formulas d [xnJ (x)] = xnJ (x), dx n n−1 d [x−nJ (x)] = x−nJ (x), dx n n+1 n ± 1 J (x) = ±J 0 (x) + J (x). n n±1 x n±1

Integral Representation It is quite useful to have an integral representation of Bessel functions. From complex analysis, the residue theorem states that if a function f(z) deﬁned on the complex plane has the Laurent series representation ∞ X n f(z) = an(z − z0) , n=−∞ and if C is any positively oriented simple closed curve in the complex plane containing z0, then I f(z) dz = 2πia−1. C Therefore, if we apply this theorem to the generating function for the Bessel functions of integer order, we obtain

∞ I e(x/2)(t−1/t) I X dt = J (x)tm−n−1 dt = 2πiJ (x), tn+1 m n C m=−∞ where C is any positively oriented simple closed curve in the complex plane containing the origin. Now, suppose that we choose C to be the the unit circle, and we use the substitution t = eiθ, for which dt = ieiθ dθ. Then we have Z 2π e(x/2)(eiθ−e−iθ) Z 2π 2πiJ (x) = i eiθ dθ = i ei(x sin θ−nθ) dθ n i(n+1)θ 0 e 0 which yields Z 2π Z 2π 1 i(x sin θ−nθ) 1 Jn(x) = e = cos(x sin θ − nθ) + i sin(x sin θ − nθ) dθ. 2π 0 2π 0

3 Taking the real and imaginary parts of both sides of the equation, we obtain 1 Z 2π 1 Z π Jn(x) = cos(x sin θ − nθ) dθ = cos(x sin θ − nθ) dθ, 2π 0 π 0 where the second integral is obtained via symmetry, and Z 2π sin(x sin θ − nθ) dθ = 0. 0 A special case of particular interest is 1 Z π J0(x) = cos(x sin θ) dθ. π 0 Also, it is worth noting that although these integral representations were derived only for Bessel functions of integer order, the relation on which they are based also applies to Bessel functions of noninteger order. Speciﬁcally, 1 I e(x/2)(t−1/t) Jν(x) = ν+1 dt, 2πi C t where C is a contour in the complex plane that encircles the origin t = 0.

Bessel Functions of Noninteger Order Bessel functions of noninteger order satisfy the same recurrence relations as those of integer order, as can be proven using the power series representation given earlier. However, one key diﬀerence between Bessel functions of integer and noninteger order is that if ν is not an integer, then Jν and J−ν are linearly independent solutions of the Bessel equation of order ν, which means that the ν relation J−ν = (−1) Jν does not hold. However, if ν is an integer, then we must use some technique other than Frobenius’ method, such as reduction of order, to obtain a second linearly independent solution of the Bessel equation of order ν.

The Bessel Equation We have seen that Bessel functions of the ﬁrst kind, which are solutions of the Bessel equation, satisfy the recurrence relations 2ν J (x) − J (x) = 2J 0 (x),J (x) + J (x) = J (x). ν−1 ν+1 ν ν−1 ν+1 x ν

Now, suppose that a sequence of functions {Zν(x)}, but not necessarily Bessel functions, all satisfy the above recurrence relations. Then, we have Z − Z 0 Z − Z x2Z00 + xZ0 − ν2Z = x2 ν−1 ν+1 + x ν−1 ν+1 − ν2Z ν ν ν 2 2 ν x2 1 1 x = Z0 + Z − Z0 − Z − ν2 [Z + Z ] 2 ν−1 x ν−1 ν+1 x ν+1 2ν ν−1 ν+1 x2 ν − 1 ν + 1 = Z0 − Z − Z0 − Z 2 ν−1 x ν−1 ν+1 x ν+1 x2 = [−2Z ] 2 ν 2 = −x Zν.

4 In other words, 2 00 0 2 2 x Zν + xZν + (x − ν )Zν = 0, meaning that Zν is a solution of the Bessel equation. Now, suppose that Zν(x) is replaced by Zν(kx) for some constant k. Proceeding as before, and using the Chain Rule, we obtain

Z − Z 0 Z − Z x2Z00 + xZ0 − ν2Z = k2x2 ν−1 ν+1 + kx ν−1 ν+1 − ν2Z ν ν ν 2 2 ν k2x2 1 1 kx = Z0 + Z − Z0 − Z − ν2 [Z + Z ] 2 ν−1 kx ν−1 ν+1 kx ν+1 2ν ν−1 ν+1 k2x2 ν − 1 ν + 1 = Z0 − Z − Z0 − Z 2 ν−1 kx ν−1 ν+1 kx ν+1 k2x2 = [−2Z ] 2 ν 2 2 = −k x Zν, where all Zν, Zν−1 and Zν+1, and their derivatives, are evaluated at kx. It follows that

2 00 0 2 2 2 x Zν (kx) + xZν(kx) + (k x − ν )Zν(kx) = 0.

This slightly modiﬁed form of the Bessel equation will arise when solving partial diﬀerential equations (PDE) using separation of variables.