On Bank-Laine functions

Transcription

1 Computational Methods and Function Theory Volume ), No. 0, XXYYYZZ On Bank-Laine functions Alastair Fletcher Keywords. Bank-Laine functions, zeros MSC. 30D35, 34M05. Abstract. In this article, we improve results of Langley and Whitehead on Bank-Laine functions. If E is a Bank-Laine function and F is an entire function of finite order which has the same zeros as E in a certain sector S of the complex plane and the set of Bank-Laine points of F in S has positive lower density in the set of zeros of F in S, then F can only be of a certain form. 1. Introduction An entire function E is called a Bank-Laine function if whenever Ez) = 0, we have E z) = ±1. Bank-Laine functions are closely linked to differential equations in the following way. Let Az) be an entire function and consider the differential equation 1) w + Az)w = 0 with linearly independent solutions f 1 and f 2 normalized so that the Wronskian W f 1, f 2 ) satisfies W f 1, f 2 ) := f 1 f 2 f 1f 2 = 1. Then the product of these solutions E = f 1 f 2 will be Bank-Laine. The converse of this is also true; if E is Bank-Laine, then A = E ) 2 2E E 1 4E 2 will be an entire function, since the numerator will have a double zero at any zero of E. Then E is the product of linearly independent normalized solutions of 1) see [2]). In [8], Langley proved the following theorem. Theorem 1.1. Suppose that A is a non-constant polynomial and that E is the product of normalized linearly independent solutions of 2) w + Az)w = 0. Version March 12, Author supported by EPSRC grant RA22AP. The author wishes to thank Professor Jim Langley for many extremely helpful comments and suggestions, and the referee for comments which helped improve the readability of the paper. ISSN /$ 2.50 c 20XX Heldermann Verlag

2 2 A. Fletcher CMFT Suppose further that F is a Bank-Laine function of finite order such that all but finitely many zeros of E are zeros of F, and all but finitely many zeros of F are zeros of E. Then F = ±E. Here, the order ρ of an entire function f is defined by ρf) = lim sup r log + log + Mr, f), log r where Mr, f) = sup{ fz) : z r}. It is necessary that Az) in 2) should be non-constant. For example, sin πz Ez) = π is the product of normalized linearly independent solutions 2 πz 2 f 1 z) = cos π 2, f πz 2z) = sin π 2 of 3) w + π2 4 w = 0. However sin πz F z) = π e2πiz has the same zeros as E, is a Bank-Laine function and has finite order, but it is not true that F = ±E. A method observed by Shen [9] uses the Mittag-Leffler Theorem to show that if a n ) is a complex sequence without repetition which tends to infinity, then there exists a Bank-Laine function with zero sequence a n ). The constructed Bank-Laine function may however have infinite order, even if the sequence a n ) has finite exponent of convergence. Theorem 1.1 implies that if you make finitely many changes to the zero sequence of a Bank-Laine function E, then you do not have the zero sequence of a Bank-Laine function of finite order. Now assume that Az) in 2) is a non-constant polynomial and write 4) Az) = Cz n 1 + o1)) as z, where C 0 and n N. The critical rays are defined to be the rays argz) = θ 0 R for which 5) argc) + n + 2)θ 0 = 0 mod 2π. This yields n + 2 critical rays with angles given by 2πij argc) θ j = n + 2 for j = 0, 1,..., n + 1. If argz) = θ 0 is a critical ray and ɛ > 0 is a positive constant, then the asymptotic integration method due to Hille [6, 7]) can be applied to show that 2) has linearly independent solutions [ z ] 6) u j z) = Az) 1/4 exp 1) j i At) 1/2 dt + o1) t 0

3 ), No. 0 On Bank-Laine functions 3 for j = 1, 2, where t 0 = R 0 e iθ 0 with R 0 a large positive constant, in a sectorial region { 7) S = z : z > R 1, argz) θ 0 < 2π } n + 2 ɛ, where R 1 is chosen large enough so that 8) ψz) := z t 0 At) 1/2 dt is univalent in S. In [10], Whitehead proves the following improvement to Theorem 1.1. Theorem 1.2. Let Az) be a non-constant polynomial and let E be the product of linearly independent solutions f 1, f 2 of 2), normalized to have Wronskian equal to 1. Let A satisfy 4) and let argz) = θ 0 be a critical ray for 2) so that θ 0 satisfies 5). Assume that E has infinitely many zeros in the sectorial region S defined in 7), where ɛ is a small positive constant. Further, R 1 is chosen large enough so that u 1 and u 2, as defined in 6), have no zeros in S. Let F be a Bank-Laine function of finite order, having the same zeros in S as E. Then either F = ±E, or there exist non-zero constants A 1, A 2, B 1, B 2 such that 9) E = A 1 U 1 f 2, F = A 2 U 2 f 2, f 2 = B 1 U 1 + B 2 U 2, where U 1 and U 2 are linearly independent solutions of 2) having finitely many zeros in S. The solutions U 1, U 2 are each constant multiples of u 1 or u 2. Let X =: z n ) n N be a sequence of points in C ordered so that z 1 z 2..., with no finite limit point. We will say that a subset X I X, indexed by a subset I N, has lower density at least D in X if I {1,..., n} D o1))n as n. For example, if f is an entire function, we can consider the lower density of subsets of the zero set of f since the zeros of an entire function can have no finite limit point by the identity theorem. Remark 1.1. If f is a linear combination of u 1, u 2 defined by 6) with infinitely many zeros in S, then the zeros z j can be calculated asymptotically from 6) and z j is eventually strictly increasing. Therefore the zeros of f 1 can be ordered in a natural way, as can the zeros of f 2, recalling that f 1 and f 2 are the normalized solutions of 2). We will say that z is a Bank-Laine point of an entire function F if F z) = 0 and F z) = ±1. The improvement to Theorem 1.2 is the following. Theorem 1.3. Let A, E, θ 0, S, ɛ, R 1, u 1 and u 2 be as in the hypotheses of Theorem 1.2 recalling 4), 5), 6) and 7)), and let X be the zero set of E in S. Let F be an entire function of finite order with the following properties: F has the same zeros as E in S, the zeros of F in S are simple and there exists a

4 4 A. Fletcher CMFT subset X I of X, indexed by a subset I N of positive lower density, such that for z X I, F z) = ±1 i.e. the set of Bank-Laine points of F in S has positive lower density in the zeros of F in S). Then either: i) F = ±E; ii) there exist constants α, β C with α β and q N such that 10) E = ± U ) q 2 U 1 )αu 1 βu 2 ) U2, F = E α β)w U 1, U 2 ) where U 1 and U 2 are linearly independent solutions of 2) having finitely many zeros in S and W U 1, U 2 ) is the Wronskian of U 1 and U 2. The solutions U 1, U 2 are each constant multiples of u 1 or u 2. Further, if U 1 has a zero in C, then 11) E = ± U 1U 2 U 1 ) W U 1, U 2 ), F = ±U 2U 2 U 1 ) W U 1, U 2 ), where U 1 and U 2 are as above. Corollary 1.4. Theorem 1.3 implies Theorem 1.2. Remark 1.2. If A satisfies 4) with n an odd integer, then the order of the linearly independent solutions U 1, U 2 to 2) is n + 2)/2 see [6]) and hence not an integer. This means that both U 1 and U 2 must have infinitely many zeros see for example [3]) and so, except in the case where q = 1, β = 0, F given in 10) must have poles, recalling that U 1 and U 2 cannot share zeros), which is a contradiction since F is entire. Therefore conclusion ii) of Theorem 1.3 cannot happen when n is odd, unless q = 1, β = 0, in which case 11) implies E and F are both Bank-Laine. Assuming that n is even, case ii) can always occur. Remark 1.3. The assumption in Theorems 1.2 and 1.3 that Az) is non-constant cannot be deleted, as the following example shows. Let sin πz Ez) = π, corresponding to Az) = π 2 /4 see 3)), and F z) = e 2πiz2 Ez). Both E and F are Bank-Laine functions and F has the same zeros as E. The functions u 1, u 2 given in 6) are constant multiples of e iπz/2 and e iπz/2 respectively. Hence U 2 /U 1 is a constant multiple of e iπz recalling that U 1, U 2 are constant mupltiples of u 1, u 2 respectively). However, there is no q N such that U 2 /U 1 ) q = e 2πiz2. We conclude that we have neither F = ±E nor F satisfying 10). The proof of Theorem 1.3 relies on the following result, a special case of a result proved in [4]. Theorem 1.5. Let I N be a subset of positive lower density and let f be analytic of zero exponential type in the right-half plane Ω. Suppose that fn) Z for every n I. Then f is a polynomial. U 1

5 ), No. 0 On Bank-Laine functions 5 2. Proof of Theorem 1.3 We first note some lemmas. Lemma 2.1 is well-known and is proved in, for example, [5]. Lemma 2.2 is proved in [10], but is in any case elementary. Lemmas 2.3 and 2.4 follow from the asymptotics, see [1] or [10]. An R-set is a countable union of disks Dz j, r j ) = {z C : z z j < r j } such that z j as j and r j <. Lemma 2.1. Let f be transcendental entire of order ρ < L < M <. Let z j be the zeros of f in z > 2, repeated according to multiplicity. Then the union U of the disks Dz j, z j M ) is an R-set and f z) fz) for all z with z large and z / U. = o z L+M ) Lemma 2.2. Let ψz) be defined by 8), and recall C from 4) and S from 7) for the critical ray argz) = θ 0. Then ψz) = 2C1/2 n + 2 zn+2)/2 1 + o1)) as z in S, where the square root C 1/2 is chosen so that C 1/2 z n+2)/2 is real and positive on argz) = θ 0. The function ψ is univalent on S, for R 1 sufficiently large. Lemma 2.3. Let h 1 and h 2 be linearly independent solutions of 2), where Az) is a non-constant polynomial. Then h 1 h 2 has infinitely many zeros. Lemma 2.4. Let Az) be a non-constant polynomial and argz) = θ 0 a critical ray. Then a solution of 2) has, for each δ > 0, only finitely many zeros in S with argz) θ 0 > δ. That is, most zeros of a solution of 2) in S lie near the critical ray argz) = θ 0. Now, since E and F have the same zeros in S and the zeros of E and F are simple in S, we are able to define an analytic branch of log F/E in S. Define and G = F E 12) V = log G = log F E. Lemma 2.5. There exists M > 0 such that V z) = O z M ) as z in a sectorial region 13) S = { z : z > R 2 >> R 1, argz) θ 0 < 2π } n + 2 2ɛ.

6 6 A. Fletcher CMFT Proof. Since F is of finite order and E has finite order n + 2)/2 see [6]), it follows that G has finite order. Since G has no zeros or poles in S, we can take R 2 large enough so that S is outside the exceptional set U in Lemma 2.1. Therefore, by Lemma 2.1, in S 2 and integrating gives G z) Gz) = O z M 1 ) log Gz) = V z) = O z M ). Since E is a product of linearly independent solutions of 2), we can therefore write E = f 1 f 2, where f 1 = A 1 u 1 + A 2 u 2, f 2 = B 1 u 1 + B 2 u 2, u 1 and u 2 are the solutions to 2) given by 6) and A 1, A 2, B 1, B 2 C. By hypothesis, E has infinitely many zeros in S. A 1 and A 2 cannot both be zero, since then E 0. If exactly one of A 1 and A 2 is zero, then f 1 has no zeros in S, since u 1 and u 2 do not have any zeros in S. Since either f 1 or f 2 must have infinitely many zeros, it follows that B 1 B 2 0. Hence either A 1 A 2 0 or B 1 B 2 0. Let X be the zero set of f 1 in S. We can assume that the zeros of f 1 which are Bank-Laine points of F in S have positive lower density in X since this must be true at least one of f 1 and f 2, and if it is not true for f 1 then interchange f 1 and f 2 ). We can also assume that A 1 A 2 0. Then we write 14) E = A 1 u 1 + A 2 u 2 )B 1 u 1 + B 2 u 2 ) = C 1 u 1 1 C ) 2u 2 B 1 u 1 + B 2 u 2 ), where C 1, C 2 are non-zero constants. We now define Lz) by 15) explz)) = C 2 u 2 u 1. u 1 The zero set X of f 1 then corresponds to points where Lz) = 2πij for j Z. Using 6), it follows that for some D C, 16) Lz) = D + o1) + 2i z t 0 At) 1/2 dt, recalling that t 0 = R 0 e iθ 0 and that the integral on the right hand side of 16) is univalent in S. Note that if z S and explz)) = 1, then Ez) = 0 and so F z) = 0. Lemma 2.6. We have 17) log F z) Ez) = P where P is a polynomial. log u ) 2 u 1

7 ), No. 0 On Bank-Laine functions 7 Proof. Write ψz) = z t 0 At) 1/2 dt. Then by Lemma 2.2, ψ is univalent on { S = z : z > R 1 : argz) θ 0 < 2π } n + 2 ɛ. Without loss of generality, we can assume that θ 0 = 0. There exists δ > δ > 0 such that 18) {w : w > R 2, argw) < π δ} ψs) {w : w > R 2, argw) > π δ } for some R 2, R 2 > 0. Clearly, ψs) contains a half-plane Using Lemma 2.2 and 16), H 1 = {w : Rew) > R 2 }. 19) Lz) = D + o1) + 2i 2C1/2 n + 2 zn+2)/2 1 + o1)) = 4iC1/2 n + 2 zn+2)/2 1 + o1)) and so Lz)/2πi maps a sub-domain Ω of S univalently to a half-plane H 2 = {w : Rew) > R 3 }. Note also that Lz)/2πi)S) is contained in a set of the form {w : w D > R 2, argw D) > π δ } by 18) for some R 2 > 0 large and δ > 0 small. This implies that 20) Lz)/2πi)S) R [R 4, ) for some finite R 4. For j > R 3, we can find z j Ω such that Lz j )/2πi = j. These points z j are in the zero set X of f 1 as noted after 15). By 20), Z Lz)/2πi)S) \ H 2 ) is finite. Therefore, X\{z j } is finite. Since the Bank-Laine points of F intersected with X have positive lower density in X, we conclude that the Bank-Laine points of F intersected with {z j } also have positive lower density in {z j }. We can therefore find a subsequence z jk ) of z j ) of positive lower density in z j ) such that z jk ) are Bank-Laine points of F. Hence using 14), 15) and 16) we have that F z jk ) = 0 and F z jk ) = ±1. Since F = e V E, we have F z jk ) = e V z j k ) E z jk ) + V z jk )e V z j k ) Ez jk ) which implies that e V z j ) k = ±1. Therefore, V z jk ) Z πi for all j k I. By Lemma 2.5 and 19), we have that V and L are of polynomial growth. In particular, they are of zero exponential type and so we can apply Theorem 1.5 to see that V is a polynomial in L. We therefore see that there is a

8 8 A. Fletcher CMFT polynomial P such that we have 17), which completes the proof of the lemma. The rest of the proof of Theorem 1.3 follows the proof of Theorem 1.2 in [10], which we present for completeness. Lemma 2.7. With P the polynomial in the statement of Lemma 2.6, we have either P z) = qz + c where q Z and c is a constant, or F z) = ±Ez). Proof. The relation 17) holds by analytic continuation by taking any path in the complement of the zero sets of u 1, u 2, E and F. If P is a constant, then 17) implies that log F z) Ez) = C 1, where C 1 is a constant, and so F z) = C 2 Ez) and F z) = C 2 E z), where C 2 = e C 1. Looking at the values of F and E at a point z jk for j k I, we see that C 2 = ±1 and so F z) = ±Ez). This gives conclusion i) in Theorem 1.3 and we now assume that P is nonconstant. By Lemma 2.3, either u 1 or u 2 must have infinitely many zeros in C since they are linearly independent solutions of 2). Further, they cannot both be zero at the same point, since the Wronskian is constant and non-zero. Suppose that u 1 w 0 ) = 0, u 2 w 0 ) 0. Then we have ) u 2 z) 1 u 1 z) = O z w 0 as z w 0. Letting z w 0 along a path γ on which argz w 0 ) is bounded, then log u 2z) u 1 z) = log 1 z w 0 + O1) = log z w 0 + O1). Suppose that P z) = d m z m + d m 1 z m d 0, d m 0. Then as z w 0 along γ, we get P log u ) 2 = d m log z w 0 ) m 1 + o1)). u 1 Therefore, P log u ) 2 as z w 0 on γ. Using 17), we have F z) log Ez) u 1

9 ), No. 0 On Bank-Laine functions 9 as z w 0 on γ. There are three possible cases: i) F w 0 ) = 0 and Ew 0 ) 0 ii) F w 0 ) 0 and Ew 0 ) = 0 or iii) Ew 0 ) = 0 and F has a zero at w 0 of order p > 1. Therefore, we have log z w 0 q Z as z w 0 along γ. Note that in case i) above q is the order of the zero of F at w 0, in case ii) q is 1 and in case iii) q = p 1. We also have 21) d m log z w 0 ) m 1 + o1)) = P log u 2 /u 1 ) d m log z w 0 ) m 1 o1)) 1 as z w 0 on γ. There are two cases. a) Suppose that m > 1. Then d m log z w 0 ) m 1 + o1)) 1 = log z w 0 ) d m log z w 0 ) m o1)) q 0 = 0 as z w 0 along γ, which contradicts 21). b) Suppose that m = 1. Then 21) becomes d 1 log z w 0 )1 + o1)) 1 as z w 0 along γ. However, we also have d 1 log z w 0 )1 + o1)) = log z w 0 1 d o1)) q d 1 as z w 0 along γ, which forces d 1 = q Z. If u 1 has no zeros, then the case where u 2 w 0 ) = 0 and u 1 w 0 ) 0 runs analogously. This proves the lemma. We now assume that F z) ±Ez), so that Lemma 2.7 gives 22) log F z) Ez) = q log u 2z) u 1 z) + c, where q Z and c is a constant. Lemma 2.8. We have 23) F z) Ez) = ) q U2 z), U 1 z) where U 1, U 2 are linearly independent solutions of 2), with no zeros in S as defined in 13) and q Z +. Proof. From 22), we have either F z) Ez) = C 3 ) q u2 z) u 1 z)

10 10 A. Fletcher CMFT when q > 0 or F z) Ez) = C 3 ) q u1 z), u 2 z) when q < 0 and where C 3 is a non-zero constant. In the first case, let U 2 z) = C 4 u 2 z), U 1 z) = u 1 z), where C q 4 = C 3 and the particular q -th root will be chosen below see 24)). In the second case, let U 2 z) = C 4 u 1 z), U 1 z) = u 2 z), again with C q 4 = C 3. This yields 23). Note that the Wronskian of U 1 and U 2 is a non-zero constant, and f 1, f 2 are linear combinations of U 1 and U 2. Let w 1 be a zero of E and a Bank-Laine point of F in S. Without loss of generality, F w 1 ) = E w 1 ) = ±1. If none of the Bank-Laine points of F has the same derivative as E, consider F instead. Then 24) U2 w 1 ) U 1 w 1 ) ) q = F w 1) Ew 1 ) = 1. We now choose the q-th root of C 3 so that U 2 w 1 ) = U 1 w 1 ). Since w 1 is a zero of E, then either f 1 w 1 ) = 0 or f 2 w 1 ) = 0. By our assumption following Lemma 2.5, we may take f 1 w 1 ) = 0. Therefore f 1 and U 2 U 1 share a zero at w 1. This, however, means that f 1 and U 2 U 1 must be linearly dependent since they are both solutions to 2). Therefore, 25) E = C 5 U 2 U 1 )f 2, F = C 5 U2 U 1 ) q U 2 U 1 )f 2, where C 5 is a non-zero constant. Since f 2 is a linear combination of U 1 and U 2, we can write f 2 = αu 1 βu 2, with α β since f 1 and f 2 are linearly independent. Differentiating 25) when E w 1 ) = ±1 shows that C 5 = ±1 α β)w U 1, U 2 ), which gives 10). Further, if U 1 has a zero in C, at w 2 say, then since F is entire and U 1, U 2 cannot both have a zero at the same point, we must have q = 1 and f 2 w 2 ) = 0 by 25) and the fact that f 2 can only have simple zeros. This means that f 2 and U 1 are linearly dependent and so we have E = C 5 U 1 U 2 U 1 ), F = C 5 U 2 U 2 U 1 ). By considering the Wronskian at w 1, then since E w 1 ) = ±1, we therefore have C 5 = ±1/W U 1, U 2 ) which gives 11).

11 ), No. 0 On Bank-Laine functions Proof of Corollary 1.4 Assume that F ±E. If F is Bank-Laine, then the proof of Lemma 2.7 shows that q must be ±1 and hence F E = U 2 U 1 where U 2 = C 3 u 2 and U 1 = u 1 if q = 1, and U 2 = C 3 u 1 and U 1 = u 2 if q = 1. If U 1 has a zero, then 11) implies 9) so we just have to consider the case where U 1 has no zeros, in which case U 2 must have infinitely many zeros. First, if α = 0 in 10), this would imply that F has a zero of order 2 at a zero of U 2, which contradicts F being Bank-Laine. If β = 0, then 10) implies 9). Now assume α, β are both non-zero. By 10), 26) F = W U 1, U 2 )E + U 2 U 1 E. At a zero w 2 of E, 26) and the fact F is Bank-Laine gives ±1 = F w 2 ) = ± U 2. U 1 It follows that f 2 = CU 2 + U 1 ) and so α = β in 10). This gives 27) E = ± U 2 U 1 )U 1 + U 2 ), F = 2W U 1, U 2 ) U 2 1 U2 U 1 ) E. Let w 3 be a zero of U 2 and hence of F. Then 26) and 27) together imply that F w 3 ) = ±1/2, which contradicts F being Bank-Laine. This completes the proof of Corollary 1.4. References 1. S.Bank and I.Laine, On the oscillation theory of f + Af = 0 where A is entire, Trans. Amer. Math. Soc., 273, ). 2. S.Bank and I.Laine, On the zeros of meromorphic solutions of second-order linear differential equations, Comment. Math. Helv., 58, ). 3. R.P.Boas Jr., Entire functions, Academic Press Inc., New York, A.Fletcher and J.K.Langley, Integer points of analytic functions in a half-plane, submitted. 5. W.K. Hayman, Meromorphic functions, Oxford at the Clarendon Press, E.Hille, Lectures on ordinary differential equations, Mass., E.Hille Ordinary differential equations in the complex domain, New York, J.K.Langley, Quasiconformal modifications and Bank-Laine functions, Archiv der Mathematik, 71, ). 9. L.C.Shen, Construction of a differential equation y + Ay = 0 with solutions having prescribed zeros, Proc. Amer. Math. Soc., 95, ). 10. A. Whitehead, Differential equations and differential polynomials in the complex plane, PhD Thesis, Nottingham, Alastair Fletcher alastair.fletcher@nottingham.ac.uk Address: School of Mathematical Sciences, University of Nottingham, Nottingham, NG7 2RD, UK.