S. Ghorai 1. Lecture XV Bessel s equation, Bessel s function. e t t p 1 dt, p > 0. (1)

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1 S Ghorai 1 1 Gamma function Gamma function is defined by Lecture XV Bessel s equation, Bessel s function Γp) = e t t p 1 dt, p > 1) The integral in 1) is convergent that can be proved easily Some special properties of the gamma function are the following: i It is readily seen that Γp + 1) = pγp), since ii Γ1) = 1 trivial proof) Γp + 1) = lim T T [ = lim = p T e t t p dt e t t p ] T T + p e t t p 1 dt e t t p 1 dt = pγp) iii If p = m, a positive integer, then Γm + 1) = m! use i repeatedly) iv Γ1/2) = π This can be proved as follows: I = Γ1/2) = e t t 1/2 dt = 2 Hence I 2 = 4 e u2 du e v2 dv = 4 Using polar coordinates ρ, θ, the above becomes I 2 = 4 π/2 e u2 du e x2 +y 2) dx dy e ρ2 ρ dρ dθ I 2 = π I = π v Using relation in i, we can extend the definition of Γp) for p < Suppose N is a positive integer and N < p < N + 1 Now using relation of i, we find Γp) = Γp + 1) p = Γp + 2) pp + 1) = = Γp + N) pp + 1) p + N 1) Since p + N >, the above relation is well defined vi Γp) is not defined when p is zero or a negative integer For small positive ɛ, Γ±ɛ) = Γ1 ± ɛ) ±ɛ 1 ±ɛ ± as ɛ Since Γ) is undefined, Γp) is also undefined when p is a negative integer

2 S Ghorai 2 2 Bessel s equation Bessel s equation of order ν ν ) is given by x 2 y + xy + x 2 ν 2 )y = 2) Obviously, x = is regular singular point Since p) = 1, q) = ν 2, the indicial equation is given by r 2 ν 2 = Hence, r 1 = ν, r 2 = ν and r 1 r 2 = 2ν A Frobenius series solution exists for the larger root r = r 1 = ν To find this series, we substitute y = x r into 2) and after some manipulation) find a n x n, x > ρn + r)a n x n + a n 2 x n = n=2 where ρr) = r 2 ν 2 This equation is rearranged as ρr)a + ρr + 1)a 1 x + Hence, we find since a ) ) ρn + r)an + a n 2 x n = n=2 ρr) =, ρr + 1)a 1 =, ρr + n)a n = a n 2, n 2 From the first relation, we get r 1 = ν, r 2 = ν Now with the larger root r = r 1 we find a 1 =, a n = a n 2 nn + 2ν), n 2 Iterating we find by induction), Hence a 2n+1 =, a 2n = 1) n 1 2 2n n!ν + 1)ν + 2) ν + n) a, n 1 y 1 x) = a x ν 1 + n=1 1) n x 2n ) 3) 2 2n n!ν + 1)ν + 2) ν + n) Here it is usual to choose instead of a = 1 as was done in lecture 14 ) a = 1 2 ν Γν + 1) Then the Frobenius series solution 3) is called the Bessel function of order ν of the first kind and is denoted by J ν x): J ν x) = 1) n ) x 2n+ν 4) n!γn + ν + 1) 2 To find the second independent solution, we consider the following three cases:

3 S Ghorai 3 A r 1 r 2 = 2ν is not a nonnegative integer: We know that a second Frobenius series solution for r 2 = ν exist We do similar calculation as in the case of r 1 and it turns out that the resulting series is given by 4) with ν replaced by ν Hence, the second solution is given by J ν x) = 1) n ) x 2n ν 5) n!γn ν + 1) 2 B r 1 = r 2 : Obviously this corresponds to ν = and a second Frobenius series solution does not exist C r 1 r 2 = 2ν is a positive integer: Now there are two cases We discuss them separately Ci ν is not a positive integer: Clearly ν = 2k+1)/2, where k {, 1, 2, } Now we have found earlier that since a ) ρr) =, ρr + 1)a 1 =, ρr + n)a n = a n 2, n 2 With r = r 2 = ν, we get ρr) = ; 1 1 2k + 1) ) a 1 = ; n n 2k + 1) ) a n = a n 2, n 2 It is clear that the even terms a 2n can be determined uniquely For odd terms, a 1 = a 3 = = a 2k 1 = but for a 2k+1 we must have n a 2k+1 = a 2k 1 a 2k+1 = This can be satisfied by taking any value of a 2k+1 and for simplicity, we can take a 2k+1 = Rest of the odd terms thus also vanish Hence, the second solution in this case is also given by 5), ie J ν x) = 1) n ) x 2n ν 6) n!γn ν + 1) 2 Cii ν is a positive integer: Clearly ν = k, where k {1, 2, 3, } Now we find since a ) ρr) =, ρr + 1)a 1 =, ρr + n)a n = a n 2, n 2 With r = r 2 = ν, we get ρr) = ; 1 1 2k ) a 1 = ; n n 2k ) a n = a n 2, n 2 It is clear that all the odd terms a 2n+1 vanish For even terms, a 2, a 4,, a 2k 2 each is nonzero For a 2k we must have n a 2k = a 2k 2 a 2k, which is a contradiction Thus, a second Frobenius series solution does not exist in this case

4 S Ghorai 4 Summary of solutions for Bessel s equation: The Bessel s equation of order ν ν ) x 2 y + xy + x 2 ν ) y =, has two independent Frobenius series solutions J ν and J ν when ν is not an nonnegative) integer: J ν x) = 1) n ) x 2n+ν, J νx) = n!γn + ν + 1) 2 Thus the general solution, when ν is not an nonnegative) integer, is yx) = C 1 J ν x) + C 2 J ν x) 1) n ) x 2n ν n!γn ν + 1) 2 When ν is a nonnegative) integer, a second solution, which is independent of J ν, can be found This solution is called Bessel function of second kind and is denoted by Y ν Hence, the general solution, when ν is an nonnegative) integer, is yx) = C 1 J ν x) + C 2 Y ν x) 3 Linear dependence of J m and J m, m is a +ve integer When ν = m is a positive integer, then J m x) = since Γn + m + 1) = n + m)! 1) n ) x 2n+m 1) n ) x 2n+m =, n!γn + m + 1) 2 n!n + m)! 2 Since Γ±) = ±, we define 1/Γk) to be zero when k is nonpositive integer Now J m x) = 1) n ) x 2n m n!γn m + 1) 2 Now each term in the sum corresponding to n = to n = m 1 is zero since 1/Γk) is zero when k is nonpositive integer Hence, we write the sum starting from n = m: J m x) = Substituting n m = k, we find J m x) = n=m k= 1) n ) x 2n m n!γn m + 1) 2 1) k+m x m + k)!γk + 1) 2 = 1) m k= = 1) m J m x) 1) k k!m + k)! x 2 ) 2m+k) m ) 2k+m Hence J m and J m becomes linearly dependent when m is a positive integer

5 S Ghorai 5 4 Properties of Bessel function Few important relationships are very useful in application These are described here A From the expression for J ν given in 4), we find x ν J ν x) = Taking derivative with respect to x we find x ν J ν x) ) = 1) n ) x 2n+2ν n!γn + ν + 1) 2 1) n ) n + ν) x 2n+2ν 1 1) n ) x 2n+2ν 1 =, n!γn + ν + 1) 2 n!γn + ν) 2 where we have used Γn + ν + 1) = n + ν)γn + ν) We can write the above relation as x ν J ν x) ) = x ν 1) n n!γ n + ν 1) + 1 ) x 2 ) 2n+ν 1 Hence, x ν J ν x) ) = x ν J ν 1 x) 7) B From 4), we find x ν J ν x) = Taking derivative with respect to x we find x ν J ν x) ) = n=1 1) n 2 2n+ν n!γn + ν + 1) x2n 1) n 2 2n+ν 1 n 1)!Γn + ν + 1) x2n 1 Note that the sum runs from n = 1 in contrast to that in A) Let k = n 1, then we obtain x ν J ν x) ) ) 2k+ν+1 = x ν 1) k+1 x k= k!γ k + ν + 1) + 1 ) 2 = x ν 1) k ) x 2k+ν+1 k!γ k + ν + 1) + 1 ) 2 k= Hence, x ν J ν x) ) = x ν J ν+1 x) 8) Note: In the first relation A, while taking derivative, we keep the sum running from n = This is true only when ν > In the second relation B, we only need ν Taking ν = in B, we find J = J 1 If we put ν = in A, then we find J = J 1 But J 1 = J 1 and hence we find the same relation as that in B Hence, the first relation is also valid for ν

6 S Ghorai 6 C From A and B, we get J ν + ν x J ν = J ν 1 J ν ν x J ν = J ν+1 Adding and subtracting we find J ν 1 J ν+1 = 2J ν 9) and J ν 1 + J ν+1 = 2ν x J ν 1)