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1 Centrum voor Wiskunde en Informatica Modelling, Analysis and Simulation Modelling, Analysis and Simulation On the zeros of the Scorer functions A. Gil, J. Segura, N.M. Temme REPORT MAS-R222 SEPTEMBER 3, 22

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3 On the Zeros of the Scorer Functions Amparo Gil Departamento de Matemáticas, Universidad Autónoma de Madrid, 2849, Madrid, Spain Javier Segura Departamento de Matemáticas, Universidad Carlos III de Madrid, Leganés, Madrid, Spain Nico M. Temme CWI, Postbus 9479, 19 GB Amsterdam, The Netherlands ABSTRACT Asymptotic approximations are developed for zeros of the solutions Giz) and Hiz) of the inhomogeneous Airy differential equation w zw = ± 1. The solutions are also called Scorer functions. Tables are given π with numerical values of the zeros. 2 Mathematics Subject Classification: 33C1, 41A6, 3E1, 3C15. Keywords and Phrases: Scorer functions, inhomogeneous Airy functions, asymptotic expansion of zeros. Note: Work carried out under project MAS1.2 Analysis, Asymptotics and Computing. This report has been accepted for publication in Journal of Approximation Theory. 1. Introduction Scorer functions are particular solutions of the non-homogeneous Airy differential equation. Detailed information on these functions can be found in [1] and in references given in [2]. We summarize the properties that are needed in this paper. We have for z IR w zw= 1/π, with solution Giz)= 1 sin zt + 13 ) π t3 dt, 1.1) and for z C w zw=1/π, with solution Hiz)= 1 π e zt 1 3 t3 dt. 1.2) The solutions of the homogeneous Airy equation w zw = are denoted by Aiz) andbiz). They have the integral representations Aiz) Biz) = 1 π = 1 π cos zt t3) dt, sin zt t3) dt + 1 π e zt 1 3 t3 dt, 1.3) where we assume that z is real.

4 2 Initial values are Gi) = 1 2 Hi) = 1 3 Bi) = 1 3 Ai) = 1 3 7/6 Γ 2 3 ), Gi ) = 1 2 Hi ) = 1 3 Bi ) = 1 3 Ai 1 ) = 3 5/6 Γ 1 3 ). 1.4) From 1.1), 1.2) and 1.3) it follows that Giz)+Hiz)=Biz). 1.5) Other relations that we need are see [2] and [3]) Hiz)=e ±2πi/3 Hi ze ±2πi/3) +2e πi/6 Ai ze 2πi/3). 1.6) and Giz)= e ±2πi/3 Hi ze ±2πi/3) ± iaiz). 1.7) Proofs follow easily by verifying that the right-hand sides satisfy the differential equations, and from the initial values. We use the asymptotic expansions [ Giz) πz z 3 s= [ Hiz) πz z 3 s= ] 3s +2)! s!3z 3 ) s, z, ph z 1 π δ, 1.8) 3 ] 3s +2)! s!3z 3 ) s, z, ph z) 2 π δ, 1.9) 3 δ being an arbitrary positive constant. These expansions follow from 1.1) and 1.2) and by using standard methods from asymptotics Watson s lemma; see [4], page 112 and page 431). 2. Qualitative properties of the real zeros of Giz) and Gi z) From 1.2) we see that Hiz) > andhi z) > for real finite z. However, Giz) andgi z) have real zeros. First we show that Giz) does not have positive zeros. Later, we study properties of the negative real zeros and we discuss the properties of the zeros of the derivative. For studying qualitative properties of the zeros, the relation in [1]. W[Gi, Bi]x) Giz)Bi z) Gi z)biz)= 1 π z Bit) dt 2.1) will be useful, together with well-known properties on the interlacing of zeros of functions: Lemma 1 Let fx) and gx) be two continuously differentiable functions in an interval I. Let W[f,g]x)=fx)g x) f x)gx) be such that W[f,g]x) x I. Then, the zeros of fx) and gx) in I are simple. Furthermore, between two consecutive zeros of fx) there is exactly one zero of gx) and vice-versa). As a consequence, if gx) has no zeros in I then fx) has at most one simple) zero in I and vice-versa).

5 3 From Lemma 1 we can check that Lemma 2 Gix) is positive for x. Proof. Bix) > forx see [1], for example the series expansion in 1.4.3) and then Eq. 2.1)) W[Gi, Bi]x) > forx. From Lemma 1, Gix) can have at most one simple) positive real zero, but Gi) > andgix) > for large positive x Eq. 1.9)); therefore Gix) > forx. To consider negative values of z,we first remark that Biz) has an infinite number of negative zeros which we denote by {b n }. Lemma 3 Gib n ) < n. Proof. From 1.5), and the fact that Hix) > wehavebix) > Gix) for all real x. Lemma 4 Gix) has exactly one simple zero in b 1, ). Proof. Gib 1 ) < whereasgi)> see 1.4)); then Gix) has at least one zero in b 1, ). Furthermore Bix) > andthenw[gi, Bi]x) < inb 1, ). Then, Lemma 1 implies that there is only one zero in this interval and that it is simple. Between b 1 = and b 2 = , the function Bix) is negative, and so Gix) is negative in that interval Bix) > Gix)). More generally, we have that: Lemma 5 Gix) has no zeros in the intervals [b 2n+2,b 2n+1 ], n =, 1,... We are only left with the possibility of having zeros in intervals b 2n+1,b 2n ), n =1, 2,..,where Bix) >. Numerical experiments show that these zeros are simple. The proof that all real or complex zeros of Giz) andhiz) are simple does not follow from the inhomogeneous differential equations 1.1) and 1.2). Recall that for functions defined by homogeneous linear differential equations of second order, as the Airy functions, such a proof is trivial. The essential difference between these two cases is that the existence and uniqueness theorem for solutions of a linear second order homogeneous ODE guarantees that the only solution having a double zero at a point x = x is the trivial solution; contrary, for 1.1) and 1.2), there is always one solution with a double zero at x = x and it is not a trivial solution. We can see this explicitly: Lemma 6 The function yz)=αz )Aiz)+βz )Biz)+Giz), with αz )= z Bit)dt, andβz )= z Ait)dt 1 3 is the solution of ω zω = 1/π with a double zero at z Proof. Solve the system yz )=,y z )=forα and β and use the Wronskian relations 1.4.1, and of [1]. In fact, x Ait)dt is numerically seen to be negative for negative x, which indicates that the negative real zeros of Gix) are simple because βz ) < z <. If there were any real double zero which is not the case), it necessarily would be an extremum: Lemma 7 The double real zeros of a real solution of ω zω = ±1/π are necessarily local extrema of the function. Proof. Let x be a double) zero of a solution ωx). Then, using the differential equation, ω x )= ±1/π.

6 4 Lemma 8 The number of simple zeros of Gix) in each interval b 2n+1,b 2n ), n =1, 2,..., is, at most, two. Proof. Given that Gix) is negative at the zeros of Bix) and that the double zeros, if any, are extrema of the function, we see that the number of simple zeros if any) must be even. Let us show there can be no more than two simple zeros. d The fact that Bix) > inb 2n+2,b 2n+1 ) implies that W[Gi, Bi]x) >, which means that dx W[Gi, Bi]x) has at most one zero in b 2n+2,b 2n+1 ). Then, if Gix) had2n zeros, n>1, at least two of these zeros would lie in an interval where W[Gi, Bi]x) does not change sign; this would imply Lemma 1) that there would be a zero of Bix) between these two zeros of Gix), but Bix) > in b 2n+2,b 2n+1 ). In fact, numerical calculations show that in the intervals b 2n+1,b 2n ), n =1, 2,..., exactly two zeros of Gix) occur, which means that there are no double zeros of Gix)andthatW[Gi, Bi]x)= x Bit)dt has exactly one zero in the intervals b 2n+1,b 2n ). This, together with the monotony of x Bit)dt in the intervals b n+1,b n ), indicates that: Conjecture 1 The real zeros of Bix) and x Bit)dt are interlaced. and we also propose that Conjecture 2 There are exactly two zeros of Gix) in the intervals b 2n+1,b 2n ). We can prove that this holds for large negative zeros: Lemma 9 For large n each interval b 2n+1,b 2n ) has two zeros of Gix). Proof. This follows from known asymptotic estimates. Bi x) has negative zeros, denoted by b n. Then see [1], page 45), Bib n)= 1) n O n 1/6), n, 2.2) From 1.9) we see that Hib n)=o1/b n)asn. Hence, Gib 2n)=Bib 2n) Hib 2n) is positive for large values of n,andgix) has at least two zeros in the interval b 2n+1,b 2n ). The fact that the real zeros of Gix) are simple is also supported by the fact that the zeros of Gix) and Gi seem to be interlaced, which can be easily proved for large x using the forthcoming asymptotic expansions. Conjecture 3 The negative zeros of Gix) and Gi x) are interlaced. Assuming this conjecture to be true, together with the fact that, numerically, we observe that g 1 >g 1 where g 1 and g 1 are, respectively, the first negative zeros of Gix)andGi x), we see that: Lemma 1 The negative zeros of Gi x) are simple Proof. Differentiating the differential equation it is easy to see that the double zeros of Gi x) can not be extrema of Gix): if x is such that Gi x )=Gi x )=thengi x )=x Gi x )+Gix )= Gix ) because we assume that the zeros of Gix) andgi x) are interlaced). However, by this same assumption, between two zeros of Gix) there must be only one zero of Gi x) which, clearly, must be a local extrema and therefore can not be a double zero of Gi x). We use an additional numerical fact to prove Lemma 12; also, the following result is used: Lemma 11 W[Hi, Gi ]x) has, at most, one positive real zero.

7 5 Proof. We use the differential equations and 1.5); with this: Wx) W[Hi, Gi ]x)= 1 π Bi x) x π x Bit)dt Then, given that W) > andw x)= x Bit)dt < forx>, Wx) has at most one positive zero. Indeed, we observe that such zero exists x Lemma 12 Gi x) has exactly one positive real zero. Proof. Indeed, Gi ) > 1.4) while 1.9) shows that Gi x) < for large positive x, which implies that there must be at least one positive zero of Gi x). This, together with the fact that the double zeros of Gi are not extrema, implies that there must be an odd number of positive zeros. Let us assume for the moment that all the positive zeros of Gi x) are simple; in this case, we show that there is only one positive zero. Gi x) can not have three or more simple positive zeros because, W[Hi, Gi ] has at most one positive zero Lemma 11) and Hi x) >. The possible zero of W[Hi, Gi ]x) cannotcoincidewithanyzero of Gi x), because we are assuming by now that the zeros of Gi x) are simple; thus, if Gi x) had at least three zeros, at least two of them would lie in an interval where W[Hi, Gi ]x) does not change sign. This is in contradiction with the fact that Hi x) > see Lemma 1). On the other hand, the only possible double zero of Gi x) should coincide with the positive zero of W[Hi, Gi ]x). However, it is numerically observed that Gi x ) < beingx the positive zero of W[Hi, Gi ]x). The numerical value of this isolated zero of Gi x), is g = Asymptotics of the negative zeros of Giz) We write Gi z) =Bi z) Hi z), and use the asymptotic expansion of Bi z) as given in [[1], page 449] and of Hi z) that follows from 1.9). We write Hi z)= 1 πz Haz), Haz) 1 h s z, h 3s+1) s = 1) s3s +2)! s!3 s, 3.1) s= Bi z)= ] 1 [cosζ + 14 π)pζ)+ 1ζ sinζ + 14 π) Qζ), 3.2) πz 1/4 Pζ) s= 1) s c 2s ζ 2s, Qζ) s= 1) s c 2s+1 ζ 2s, 3.3) ζ = 2 3 z 3 2, c =1, c s = Γ3s ) 54 s s!γs ) 2 ). We explain the method by taking Haζ) =Pζ) =1andQζ) =. This gives for the equation Gi z) = Bi z) Hi z) = a first equation cosζ π)= 1 πz 3/4. 3.5)

8 6 Using z 3/4 = 3ζ/2, we obtain cosζ π)= 2 3πζ. 3.6) For large ζ solutions occur when the cosine function is small. We put ζ = ζ n + ε, ζ n =n 3 )π, n =1, 2, 3, ) 4 The equation for ε reads sin ε = c ζn + ε = ct, t=1/ ζ n, c= 1) n 2/3π). 3.8) 1+εt 2 For small values of t this equation can be solved by substituting a power series ε = ε 1 t + ε 2 t , and the coefficients can be obtained by standard methods. For example, ε 1 = c. By using the asymptotic expansions for Haz), Pζ)andQζ) a few extra technicalities are introduced. With the help of a computer algebra package the general coefficients ε s are easy to calculate. Finally we find for z =3ζ/2) 2/3,andforg n, the zeros of Giz), the expansion or ) 2 g n 3 2 ζ 3 [ n 1+ε1 t 3 + ε 2 t ] 2/3, n=1, 2, 3,..., 3.9) [ g n [3π4n 3)/8] γ3 t 3 + γ 4 t ] 1, t=, 3.1) n 3/4)π where γ 3 = 2c 3, γ 4 = 5 18, γ 5 = c3 9, γ 6 = 4c2 9, γ 7 = c81c4 16) 162, γ 8 = 189c The expansion in 3.1) reduces to the expansion of the zeros b n of Biz) ifwetakec =. 3.11) 3.1 The real zeros of Gi z) For the real zeros of Gi z) we can use the same procedure. For this case we need Hi z)= 1 πz 2 Haz), Haz) hs 1 z, 3s+1) hs =3s +4)h s 3.12) s= [ ] Bi z)= z1/4 π sinζ π)rζ) 1 ζ cosζ π) Sζ), 3.13) Rζ) s= 1) s d 2s ζ 2s, Sζ) 1) s d 2s+1 6s +1 ζ 2s, d s = 6s 1 c s 3.14) s= where ζ and c s are defined in 3.4). Using Gi z) =Bi z) Hi z) we obtain the equation for determining the zeros: sinζ π)=1 ζ Sζ) Rζ) cosζ π)+ 1 πz 9/4 Haz) Rζ). 3.15)

9 Using z 9/4 =3ζ/2) 3/2, we see that the main part of this equation is obtained by neglecting the term with the function Haz), but we can proceed in the same manner as before. We put ζ = ζ n + ε, ζ n =n 1 )π, n =1, 2, 3,..., 3.16) 4 and we can obtain for ε an expansion. Finally we obtain for g n, the zeros of Gi z), the expansion 7 or ) 2 g n 3 2 ζ 3 [ n 1+ε3 τ 5 + ε 4 τ ] 2/3, n=1, 2, 3,..., 3.17) g n [ [3π4n 1)/8] γ 4 t 4 + γ 5t ] 1, t=, 3.18) n 1/4)π where γ 4 = 7 18, γ 5 = 2c 3, γ 6 = γ 7 =, γ 8 = , γ 9 = 719c 324, γ 1 = 1c2 9. This expansion reduces to that of b n, the zeros of Bi z) ifwetakec =. 3.19) 4. The complex zeros of Giz) Giz)andGi z) have infinite many complex zeros {χ n } just below the half line phz = 1 3 π,andatthe conjugate values. Asymptotic estimates can be obtained by using the connection formula 1.7) with z replaced with ze πi/3.thatis, Gi ze πi/3) = e ±2πi/3 Hi z)+iai ze πi/3). 4.1) We write Hi z) as in 3.1) and for Aiz) we obtain from the standard asymptotic expansion of this Airy function Ai ze πi/3) 1 = 2 πz 1/4 e πi/12 iη A a η), 4.2) where for c s see 3.4)) η = 2 3 z 3 2, Aa η) s= 1) s c s iη) s. 4.3) The equation for deriving the asymptotic expansion of χ n for large n then reads We write e iη = 1 2 πz 3/4 e πi/4 A aη) Haz). 4.4) η = η n 1 2 i lncη n)+ε, η n =2n 1 4 )π, c = 3 π, 4.5) 8 and obtain for ε the equation e iε = 1 iδt + εt A aη) Haz), t= 1 η n, δ= 1 2 lncη n). 4.6)

10 8 The next step is substituting a power series ε = ε 1 t + ε 2 t , considering δ as a fixed parameter. A few straightforward manipulations give the expansion χ n [3π8n 1)/8] 2/3 e πi/3 1+ γ 1 + γ 2 η n ηn 2 + γ ) 3 ηn ) and the first few coefficients are γ 1 = 2 3 iδ, γ 2 = δ +12δ2 ), γ 3 = i δ 45δ2 +8δ 3 ), γ 4 = δ 1182δ2 + 72δ 3 84δ 4 ). 4.8) 4.1 The complex zeros of Gi z) For the complex zeros χ n of Gi z) we use cf. 1.7)) Gi ze πi/3) = e πi/3 Hi z)+ie πi/3 Ai ze πi/3). 4.9) We need the expansion of Hi z) given in 3.12) and Ai ze πi/3) = z1/4 2 π eπi/12 iη Ãaη), Ãaη) where d s is given in 3.14). We put s= 1) s d s iη) s. 4.1) η = η n iδ + ε, η n =2n )π, c = 3 2 π 4 ) 1 3, 4.11) and the equation for ε reads e iε =1 iδ t + ε 3/2 Ãaη) t) Haη), δ = 3 2 lncη n). 4.12) The expansion for the zeros reads ) χ n [3π8n +1)/8] 2/3 e πi/3 1+ γ 1 η n + γ 2 η n 2 + γ 3 η n ,n=1, 2, 3, ) and the first few coefficients are γ 1 = 2 3 iδ, γ 2 = δ +12δ 2 ), γ 3 = i δ 27δ 2 +16δ 3 ), γ 4 = δ 1698δ δ 3 168δ 4 ). 4.14)

11 5. The complex zeros of Hiz) Hiz)andHi z) have infinite many complex zeros {κ n } just above the half line phz = 1 3 π,andatthe conjugate values. For Hiz) weuse1.6)intheform Hi ze πi/3) = e 2πi/3 Hi z)+2e πi/6 Ai ze πi/3). 5.1) The analysis is analogous to the case for Giz) and gives 4.6) with i replaced by i, alsoinaaη), and c by c = 3 2 π.thisgivesforκ n, the zeros of Hiz), κ n [3π8n 1)/8] 2/3 e πi/3 1+ γ 1 + γ 2 η n ηn 2 + γ ) 3 ηn where η n is given in 4.5), δ = 1 2 lncη n), with c = 3 2 π,andthefirstfewγ k are given in 4.8). For Hi z) we find equation 4.12) with i replaced by i, alsoinãaη), and c = 3 2 π 1 3.Forκ n,the zeros of Hi z), we obtain ) κ n [3π8n +1)/8] 2/3 e πi/3 1+ γ 1 η n + γ 2 η n 2 + γ 3 η n ),n=1, 2, 3, ) where η n is given in 4.11), δ = 3 2 lncη n), with c = 3 2 π 1 3,andthefirstfewγ k are given 4.14). 6. Numerical verifications and tables Now we will illustrate the accuracy of the asymptotic approximations for the real and complex zeros of Gix), Gi x) except the positive zero of Gi x)) and the complex zeros of Hix) andhi x). For the complex zeros, by complex conjugation, we only need to consider Iz>. We use the asymptotic approximations as starting values for a Newton method, obtaining convergence in all cases. The code [3] has been used for the calculations. The accuracy of the code is better than 1 12 and we expect that the zeros can be computed with at least 12 exact digits. Table 1 shows the relative error of the asymptotic estimates. n Error g n Error g n Error χ n Error χ n Error κ n Error κ n < Table 1. Relative error of the modulus of the zeros from the asymptotic estimations, compared with numerical computations. The notation is as in the text. The number of non-zero coefficients of the asymptotic expansions considered is as follows: for g n we take 2 coefficients for n = 1 and 6 coefficients for the rest of values of n; for g n, χ n, χ n, κ n and κ n we take the first 4 non zero coefficients. We now compare the approximate values of the first 1 zeros with the numerical values.

12 1 n g n asymp.) g n numer.) g nasymp.) g nnumer.) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) Table 2. Asymptotic estimations of the first 1 negative real zeros of Gix)andGi x) versus their numerical value 12 digits). Between brackets, the number of the first non zero coefficients taken in the calculation of the asymptotic expansion is given. Additionally, G x) has a positive zero: g = n χ n asymptotic) χ n numerical) i i i i i i i i i i i i i i i i i i i i Table 3. Asymptotic estimations of the first 1 complex zeros of Giz) versus their numerical value 12 digits). The expansion is calculated using the first 4 coefficients. n χ nasymptotic) χ nnumerical) i i i i i i i i i i i i i i i i i i i i Table 4. Asymptotic estimations of the first 1 complex zeros of Gi z) versus their numerical value 12 digits). The expansion is calculated using the first 4 coefficients.

13 11 n κ n asymptotic) κ n numerical) i i i i i i i i i i i i i i i i i i i i Table 5. Asymptotic estimations of the first 1 complex zeros of Hiz) versus their numerical value 12 digits). The expansion is calculated using the first 4 coefficients. n κ nasymptotic) κ nnumerical) i i i i i i i i i i i i i i i i i i i i Table 6. Asymptotic estimations of the first 1 complex zeros of Hi z) versus their numerical value 12 digits). The expansion is calculated using the first 4 coefficients. In all cases, as could be expected, the asymptotic estimations are closer to the true value as larger zeros in modulus) are considered. Furthermore, as commented, the asymptotic estimations can be used as starting values to compute accurately the zeros of Scorer functions. The only exception is the positive real zero of Gi x), which can not be estimated via the asymptotic expansions for the zeros. Acknowledgments A. Gil acknowledges financial support from Ministerio de Ciencia y Tecnología BFM C2-1).

14 12 References 1. Abramowitz M., I. Stegun Eds). Handbook of Mathematical Functions. National Bureau of Standards. Applied Mathematics Series, no. 55. U.S. Government Printing Office, Washington DC 1964). 2. Gil A., J. Segura, N.M. Temme. On non-oscillating integrals for computing inhomogeneous Airy functions. Math. Comput. 7, Gil A., J. Segura, N.M. Temme. GIZ, HIZ: Two Fortran 77 routines for the computation of complex Scorer functions. Accepted for publication ACM Trans. Math. Soft. 4. F.W.J. Olver. Asymptotics and Special Functions. Academic Press, New York. Reprinted in 1997 by A.K. Peters.