Dual exponential polynomials and a problem of Ozawa
Janne Heittokangas, Katsuya Ishizaki, Kazuya Tohge, Zhi-Tao Wen
aa r X i v : . [ m a t h . C V ] J a n Dual exponential polynomials anda problem of Ozawa
J. Heittokangas, K. Ishizaki, K. Tohge and Z.-T. Wen
Abstract
Complex linear differential equations with entire coefficients are studied in thesituation where one of the coefficients is an exponential polynomial and dominatesthe growth of all the other coefficients. If such an equation has an exponentialpolynomial solution f , then the order of f and of the dominant coefficient are equal,and the two functions possess a certain duality property. The results presented inthis paper improve earlier results by some of the present authors, and the paperadjoins with two open problems. Key words:
Dual exponential polynomials, exponential sum, finite order, lineardifferential equation, Ozawa’s problem, value distribution.
Primary 30D15; Secondary 30D35.
Frei [2] has proved that the differential equation f ′′ + e − z f ′ + αf = 0 , α ∈ C \ { } , (1.1)has a subnormal (that is, non-trivial and finite-order) solution if and only if α = − m fora positive integer m . The subnormal solution f is a polynomial in e z of degree m , that is,an exponential sum of the form f ( z ) = 1 + C e z + · · · + C m e mz , C j ∈ C . (1.2)It was discovered in [18, Lemma 1] that in this representation one has C j = 0 for 1 ≤ j ≤ m .Substituting the subnormal solution f into (1.1), we get m X j =1 C j j e jz + m X j =1 C j je ( j − z − m m X j =1 C j e jz = m . By the Borel-Nevanlinna theorem [3, pp. 70, 108], or simply by an elementary observationon three polynomials in e z , this gives rise to the recursive formula C = m , ( m − j ) C j = ( j + 1) C j +1 , ≤ j ≤ m, from which C j = j ! Q j − k =0 ( m − k ) for 1 ≤ j ≤ m . Due to the presence of the transcendentalcoefficient e − z , any solution of (1.1) linearly independent with f in (1.2) must be of infiniteorder [7]. For example, when α = −
1, the function g ( z ) = exp( e − z + z ) is an infinite ordersolution of (1.1) and linearly independent with f ( z ) = 1 + e z . ual exponential polynomials and a problem of Ozawa a = 0, then the non-trivial solutions of f ′′ + e − z f ′ + ( az + b ) f = 0are of infinite order of growth. If P ( z ) is a non-constant polynomial, the question whetherall non-trivial solutions of f ′′ + e − z f ′ + P ( z ) f = 0are of infinite order of growth has been known as the Ozawa problem. This problem hasbeen answered affirmatively for particular polynomials P ( z ) by Amemiya-Ozawa [1] andby Gundersen [4], while the complete solution is by Langley [12].We proceed to state three new examples of Frei-Ozawa type. Example 1.1 If H is an arbitrary entire function, then f ( z ) = e z + 1 solves f ′′ + ( H − He − z ) f ′ − Hf = 0 . Of particular interest is the case when H is a polynomial. Example 1.2
The function f ( z ) = 1 + (1 − c ) (cid:0) e z + (cid:0) − c (cid:1)(cid:1) e z , c ∈ C \ { , } , withtwo exponential terms solves f ′′ + (cid:18) − − c + 23 e − z (cid:19) f ′ + (cid:18) c − (cid:19) f = 0 . Example 1.3
The function f ( z ) = 1 + 3 e z + √ ie z with two exponential terms solves f ′′ + (cid:16) − √ ie − z + 2 e − z (cid:17) f ′ − f = 0 , where the transcendental coefficient has two exponential terms also. By making a changeof variable z → wz , where w ∈ C \ { } , we see that g ′′ + w (cid:16) − √ ie − wz + 2 e − wz (cid:17) g ′ − w g = 0has a solution g ( z ) = 1 + 3 e wz + √ ie wz = f ( wz ).One might wonder about possible examples of solutions f with a single exponentialterm and of transcendental coefficients A ( z ) having at least two exponential terms. Thenon-existence of such examples will be confirmed in Theorem 3.2 below. For example, itwill be shown that a function f ( z ) = 1 + be wz for b, w ∈ C is a solution of f ′′ + (cid:8) P ( z ) + P ( z ) e − wz (cid:9) f ′ − P ( z ) f = 0for P ( z ) , P ( z ) , P ( z ) ∈ C [ z ] if and only if P ( z ) = w P ( z ) and P ( z ) = bw P ( z ).In contrast to Ozawa’s problem and complementing the three examples above, ourprimary focus is on exponential polynomial solutions of linear differential equations, inparticular of second order equations f ′′ + A ( z ) f ′ + B ( z ) f = 0 , (1.3)where A ( z ) and B ( z ) are entire. An exponential polynomial is a function of the form f ( z ) = P ( z ) e Q ( z ) + · · · + P k ( z ) e Q k ( z ) , (1.4) ual exponential polynomials and a problem of Ozawa P j , Q j are polynomials for 1 ≤ j ≤ k . Observe that a polynomial is a special caseof an exponential polynomial. A transcendental exponential polynomial f can be writtenin the normalized form f ( z ) = F ( z ) + F ( z ) e w z q + · · · + F m ( z ) e w m z q , (1.5)where q = max { deg( Q j ) } ≥ f , the frequencies w j are non-zero andpairwise distinct, the multipliers F j are exponential polynomials of order ≤ q − F j ( z ) ≤ j ≤ m , and m ≤ k [8, 16]. Definition 1.4 ([18]) Let f be given in the normalized form (1.5). If the non-zero fre-quencies w , . . . , w m of f all lie on a fixed ray arg( w ) = θ , then f is called a simpleexponential polynomial . If g is another simple exponential polynomial of the same order q as f such that the non-zero frequencies of g all lie on the opposite ray arg( w ) = θ + π ,then f and g are called dual exponential polynomials .For example, the functions f ( z ) = z e − iz + ze z + e z +(1 − i ) z and g ( z ) = 2 e − z +(1+ i ) z + z e − z + iz are dual exponential polynomials of order 2.In studying the differential equation (1.3) with entire coefficients A ( z ) and B ( z ), it isfundamental that each of its solutions f is an entire function also. In this paper we studycases when f can be an exponential polynomial assuming that A ( z ) is an exponentialpolynomial and that B ( z ) grows slowly compared to A ( z ). Naturally, the set E of entirefunctions is a ring closed under differentiation, and the set Exp q of exponential polynomialsof order ≤ q ∈ N together with constants in C and ordinary polynomials in C [ z ] =: Exp becomes a differential subring of E . On the other hand, Exp q is not closed under integrationin general except for the set Exp of exponential polynomials of order ≤
1, which plays arole in our discussions.To identify a primitive of each element in Exp , it is convenient to use the formula Z z n e wz dz = w z n + n − X ν =0 ( − n − ν n ! w n − ν +1 ν ! z ν ! e wz + constantfor n ∈ N ∪ { } and w ∈ C \ { } . Of course, an analogous formula is not in generalavailable for z n e wz q when q ≥
2. Indeed, recall the error function erf( z ) defined also forcomplex argument z by erf( z ) = 2 √ π Z z e − ζ dζ . It is the primitive of √ π e − z ∈ Exp , but the function itself is not an exponential poly-nomial. For this reason one needs the special expression erf( z ) for this function as inthe real argument case. This is also the case when q ≥
3. The value distribution of thefunctions R z e − ζ q dζ , q ∈ N , as described in Nevanlinna’s monograph [14, pp. 168–170], isquite different from that of exponential polynomials [8, 10, 16].Along with Exp q − , the set S q ( θ ) of simple exponential polynomials of order q withrespect to a fixed angle θ ∈ [0 , π ) forms a differential subring of Exp q . A unit elementin S q ( θ ) is a single exponential term e wz q + p ( z ) with arg( w ) = θ , p ( z ) ∈ C [ z ] and deg( p ) ≤ q −
1, whose multiplicative inverse belongs to the set S q ( θ + π ) as its dual exponentialpolynomial. It should be observed that if f ∈ S q ( θ ) and g ∈ S q ( θ + π ) are dual exponentialpolynomials, then f g ∈ Exp q − might not hold, but even so, the growth of f g in terms ofthe characteristic function could be somewhat reduced from that of f or g . ual exponential polynomials and a problem of Ozawa Example 1.5 If f ( z ) = e z + e z and g ( z ) = e − z , then T ( r, f ) = π r + O (log r ) and T ( r, g ) = π r + O (log r ), while T ( r, f g ) = π r + O (log r ), see [10]. Alternatively, the choice g ( z ) = e − z gives T ( r, g ) = π r and T ( r, f g ) = π r + O (1).In our setting, the duality of two exponential polynomials is an interdependence amongthem in order to reduce the growth under multiplication, especially when combined withdifferentiation. For example, if A and f are dual exponential polynomials of order q , then A and f ′ are also dual, and at times the growth of Af ′ is reduced to ρ ( Af ′ ) < q .The motivation for studying exponential polynomial solutions of (1.3) arises from thefollowing previous result. Theorem 1.6 ([18])
Suppose that f is a transcendental exponential polynomial solutionof (1.3) , where A ( z ) and B ( z ) are exponential polynomials satisfying ρ ( B ) < ρ ( A ) . Thenthe following assertions hold. (a) f and A ( z ) are dual exponential polynomials of order q ∈ N , and f has the normalizedrepresentation f ( z ) = c + F ( z ) e w z q + · · · + F m ( z ) e w m z q , (1.6) where m ∈ N and c ∈ C \ { } . (b) If ρ ( Af ′ ) < q , then q = 1 and A ( z ) = ae − wz , B ( z ) = − w and f ( z ) = c (cid:16) wa e wz (cid:17) , (1.7) where w = w and a ∈ C \ { } . If a = c = w = 1, then (1.7) reduces to Frei’s equation (1.1) and Frei’s solution (1.2) inthe case m = 1. The following example illustrates that it is not always the case that thedifferential equation (1.3) possesses a non-trivial exponential polynomial solution when A ( z ) and B ( z ) are exponential polynomials satisfying ρ ( B ) < ρ ( A ). Example 1.7
For a fixed n ∈ Z , let A ( z ) = − + n + e − z and B ( z ) = − + n . Then(1.3) has a zero-free solution f ( z ) = exp (cid:26) e − z + (cid:18) − n (cid:19) z (cid:27) . (1.8)Note that f is an exponential of an exponential polynomial. Another solution of (1.3),linearly independent with f , is g ( z ) = f ( z ) Z z e − ζ f ( ζ ) dζ = exp (cid:26) e − z + (cid:18) − n (cid:19) z (cid:27) Z z exp (cid:26) e − ζ + (cid:18) − n (cid:19) ζ (cid:27) dζ , where the integral represents an arbitrary primitive function. We may re-write this as g ( z ) exp (cid:26) − e − z − (cid:18) − n (cid:19) z (cid:27) = Z z exp (cid:26) e − ζ + (cid:18) − n (cid:19) ζ (cid:27) dζ ual exponential polynomials and a problem of Ozawa g solves a first order equation g ′ ( z ) + (cid:26) − e − z + (cid:18) − n (cid:19)(cid:27) g ( z ) = exp (cid:26) e − z + (cid:18) − n (cid:19) z (cid:27) . This shows that g cannot be any exponential polynomial as a function of infinite order.Hence it is necessary in Theorem 1.6 to assume that (1.3) has a nontrivial exceptionalpolynomial solution f .One may also observe that a small perturbation in the above coefficients A ( z ) and B ( z )brings our desired case. In fact, by choosing A ( z ) = − − n + e − z and B ( z ) = − + 2 n for any n ∈ Z , the equation (1.3) permits the exponential polynomial solution f ( z ) = 1 + (1 − n ) e z + (1 − n ) (cid:18) − n (cid:19) e z . (1.9)A difference between these two cases can also be observed in the logarithmic derivatives:If f is the function in (1.8), then f ′ ( z ) f ( z ) = − e − z + − n , while if f is the function in (1.9),then f ′ ( z ) f ( z ) is not an exponential polynomial but an irreducible rational function in e z .After discussing some properties of exponential polynomials in Section 2, we will showin Section 3 that the conclusions in Theorem 1.6(a) can be made stronger under weakerassumptions. Complementing the condition ρ ( Af ′ ) < q in Theorem 1.6(b), some newconditions implying the conclusion q = 1 will be discovered. Examples on higher orderduality as well as on the cases where a solution is dual to more than one coefficient willbe discussed in Sections 4 and 5, respectively. Two open problems are formulated in thehope that these findings would give raise to further discussions in the future. We need to introduce several concepts some of which are new.
Definition 2.1 ([11, p. 214]) Let f in (1.5) be a simple exponential polynomial. If thereexists a constant w ∈ C \ { } such that w j /w is a positive integer for every j = 1 , , . . . , m ,then the (non-zero) frequencies of f are said to be commensurable , and w is called a common factor .For example, f ( z ) = e πz + 3 e πz + ze πz and g ( z ) = e iz + e iz are simple exponentialpolynomials, both of their frequencies are commensurable, and examples for commonfactors are π, π/ f and i, i for g . In particular, a common factor is not unique.Note that it is usual to say that non-zero real numbers a and b are commensurable iftheir ratio a/b is a rational number. Equivalently, there exist a real number c and integers m and n such that a = mc and b = nc . In Definition 2.1 we are concerned with a simpleexponential polynomial and a fixed θ ∈ [0 , π ), and thus all the non-zero frequencies areof the form w j = r j e iθ for r j >
0, and the ratio of w j and w i is w j w i = r j r i = w j /ww i /w . This is a positive rational number for a common factor w . If we consider the dual expo-nential polynomial as well, all those frequencies are commensurable in the usual sense. ual exponential polynomials and a problem of Ozawa f is a simple exponential polynomial of order one with constant multipliers, and ifits frequencies are commensurable as in Frei’s case (1.2), then by the fundamental theoremof algebra, f can be written as f ( z ) = A m Y j =1 ( e wz − α j ) , where A = 0, α j ’s are complex constants and m is a positive integer. In particular, all thezeros of f lie on at most m lines.We note that if the non-zero frequencies of f are commensurable, then they are clearlylinearly dependent over rationals (see [13] for results in this direction), but not the otherway around. For example, the points w = 1 , w = √ , w = √ − Definition 2.2
Suppose that f and g are dual exponential polynomials with commensu-rable frequencies { w j } ( j >
0) and { λ i } ( i > w but with opposite signs. If the points w j + λ i are on one ray including the originfor all i, j >
0, then f and g are called strongly dual exponential polynomials .For example, the functions f ( z ) = 1 + ze z + 2 e z and g ( z ) = 1 − e − z are stronglydual exponential polynomials, while f ( z ) and h ( z ) = g ( z ) + 2 z e − z are not. Note that ifarg( w j ) = θ , then arg( λ j ) = θ + π by duality, and moreover, if w j + λ i = 0, then preciselyone of arg( w j + λ i ) = θ or arg( w j + λ i ) = θ + π holds for all i, j >
0. Alternatively, strongduality of f and g of order q can be expressed as follows: There exists a non-zero constant w such that f ( z ) = m X j =0 F j ( z )( e wz q ) j and g ( z ) = m X i =0 G i ( z )( e − wz q ) i , (2.1)where F j , G i are exponential polynomials of order ≤ q −
1. Hence f is a polynomial in e wz q and g is a polynomial in e − wz q , with smaller exponential polynomials as multipliers.Using the notation above, f ∈ Exp q − [ e wz q ] and g ∈ Exp q − [ e − wz q ]. Differing from thesituation in (1.5), some of the multipliers F j , G i ( i, j >
0) in (2.1) must suitably vanishidentically so that only one of j − i ≥ j − i ≤ F j , G i ( i, j > f ( z ) − F ( z )and g ( z ) − G ( z ) becomes again a commensurable exponential polynomial with either w or − w as a common factor. In the case when both F ( z ) and G ( z ) are constant, eachproduct of the derivatives f ( k ) ( z ) and g ( ℓ ) ( z ), k, ℓ ∈ N , is a commensurable exponentialpolynomial with the same common factor as the product of f ( z ) − F ( z ) and g ( z ) − G ( z ). Definition 2.3 ([18]) Denote the set of complex conjugate frequencies of the function f in (1.5) by W f = { w , w , . . . , w m } , where w = 0 is related to the multiplier F ( z ) W f = { w , . . . , w m } when F ( z ) ≡
0. Denote the convex hull of the set W f by co( W f ),and let C (co( W f )) denote the circumference of co( W f ).The set co( W f ) is defined as the intersection of all closed convex sets containing W f ,and as such it is either a convex polygon or a line segment. The latter occurs when f issimple, and, in particular, when w , . . . , w m are commensurable. The vertices of co( W f ) areformed by some (possibly all) of the points w , w , . . . , w m . The circumference C (co( W f ))of co( W f ) plays an important role in describing the value distribution of f , see [8, 10, 16]. ual exponential polynomials and a problem of Ozawa h be a quotient of two transcendental exponential polynomials, say h ( z ) = f ( z ) /g ( z ) , where f is of the form (1.5) and g is an exponential polynomial of the normalized form g ( z ) = G ( z ) + G ( z ) e w z q + · · · + G m ( z ) e w m z q . In these representations of f and g for the quotient h , we allow that some of the multipliers F j or G j may vanish identically, but we suppose that the matching multipliers F j and G j do not both vanish identically for any j .For the quotient h = f /g , define the set W h = { w , w , . . . , w m } . The proximityfunction of h is m ( r, h ) = (cid:0) C (co( W h )) − C (co( W g )) (cid:1) r q π + o ( r q ) , (2.2)see [17, Satz 1]. In particular, if g ≡
1, then W g = { } and C (co( W g )) = 0. This yields[16, Satz 1] as a special case, namely T ( r, f ) = m ( r, f ) = C (co( W f )) r q π + o ( r q ) , (2.3)where W f = W f ∪ { } . The estimates (2.2) and (2.3) are consistent with the estimate m (cid:18) r, f ′ f (cid:19) = o ( T ( r, f )) , known as the lemma on the logarithmic derivative, since W f ′ /f = W f holds for any givenexponential polynomial f of the form (1.5). We also point out that W f/f ′ = W f ′ . Thisfact will be used in proving our main results in Section 3. Motivated by Example 1.3, we improve Theorem 1.6(a) under weaker assumptions on B ( z ). Theorem 3.1
Suppose that f and A ( z ) in (1.3) are transcendental exponential polynomi-als, and that B ( z ) is an entire function satisfying T ( r, B ) = o ( T ( r, A )) . Then the followingassertions hold. (a) f and A ( z ) are dual exponential polynomials of order q ∈ N , f has the normalizedrepresentation (1.6) , and B ( z ) is an exponential polynomial of order ρ ( B ) ≤ q − . (b) The frequencies of f are commensurable if and only if the frequencies of A ( z ) arecommensurable. In both cases, f and A ( z ) are strongly dual exponential polynomials.Proof. (a) Suppose that 0 ≤ ρ ( f ) ≤ ρ ( A ) −
1. The case ρ ( f ) = 0 is not possible because f is transcendental. Hence ρ ( f ) ≥
1. But now | A | ≤ | f ′′ /f ′ | + | B || f /f ′ | and the assumption T ( r, B ) = o ( T ( r, A )) imply T ( r, A ) = m ( r, A ) ≤ m ( r, B ) + O (cid:0) r ρ ( A ) − (cid:1) = o ( T ( r, A )) , which is a contradiction. Here we have used (2.2) for h = f /f ′ and g = f ′ , as well as(2.3) for A in place of f . The following two cases are also impossible by the proof of [9,Theorem 3.6]: ual exponential polynomials and a problem of Ozawa ρ ( f ) = ρ ( A ) and either F ( z ) ≡ F ′ ( z ) ρ ( f ) ≥ ρ ( A ) + 1.Thus ρ ( f ) = ρ ( A ) = q ≥ f has the representation (1.6).We proceed to prove that f and A ( z ) are dual exponential polynomials. Using (1.3),we find that m (cid:18) r, Af ′ f (cid:19) = O (log r ) + m ( r, B ) = o ( T ( r, A )) = o ( r q ) . The formula (7.3) in [18] should be replaced by this. Thus the formula (7.7) in [18] holds,and the reasoning in [18] shows that f and A ( z ) are dual exponential polynomials.To complete the proof of (a), it suffices to prove that B ( z ) is an exponential polynomialof order ρ ( B ) ≤ q −
1. Since the frequencies w j of f are all on one ray, we may appealto a rotation, and suppose that w , . . . , w m ∈ R + . By renaming the frequencies w j , ifnecessary, we may further suppose that 0 < w < · · · < w m . Thus the dual coefficientmust be of the form A ( z ) = A ( z ) + k X j =1 A j ( z ) e − λ j z q , (3.1)where A j ( z ) j ∈ { , . . . , k } and λ , . . . , λ k ∈ R + . Renaming the frequencies λ j ,if necessary, we may suppose that 0 < λ < · · · < λ k . Write f ′ ( z ) = m X j =1 G j ( z ) e w j z q and f ′′ ( z ) = m X j =1 H j ( z ) e w j z q , where G j ( z ) = F ′ j ( z ) + qw j z q − F j ( z ) H j ( z ) = G ′ j ( z ) + qw j z q − G j ( z )
0. Next,write − Af ′ = Bf + f ′′ in the form − k X j =1 A j e − λ j z q ! m X j =1 G j e w j z q ! = cB + m X j =1 ( A G j + BF j + H j ) e w j z q . (3.2)From (3.2) we find that B is an exponential polynomial of order ρ ( B ) ≤ q . In fact, from(2.3) and the assumption T ( r, B ) = o ( T ( r, A )), it follows that ρ ( B ) ≤ q − m r, k X j =1 A j e − λ j z q ! = T r, k X j =1 A j e − λ j z q ! = λ k π r q + o ( r q ) , and m r, (cid:26) cB + m X j =1 ( A G j + BF j + H j ) e w j z q (cid:27). m X j =1 G j e w j z q ! = 2 w m − w m − w )2 π r q + o ( r q ) = w π r q + o ( r q ) . Therefore, we deduce that 0 < λ < · · · < λ k = w < · · · < w m . (3.3) ual exponential polynomials and a problem of Ozawa − A k G = cB. (3.4)If A G m + BF m + H m
0, then from [10, Theorem 2.2] and (3.3), we get N ( r, , L ) = 2( w m − w ) + 2( λ k − λ )2 π r q + O ( r q − + log r )= w m − λ π r q + O ( r q − + log r ) ,N ( r, , R ) = w m π r q + O ( r q − + log r ) , where N ( r, , L ) and N ( r, , R ) are the counting functions of zeros of the exponentialpolynomials on the left-hand side and on the right-hand side of (3.2), respectively. Thisimplies w m = w m − λ , which is impossible. Thus we have A G m + BF m + H m ≡ . (3.5)Now (3.2) reduces to − k X j =1 A j e − λ j z q ! m X j =1 G j e w j z q ! = cB + m − X j =1 ( A G j + BF j + H j ) e w j z q . (3.6)From the Borel-Nevanlinna theorem, and from A i G j j ∈ { , , . . . , m } and i ∈{ , , . . . , k } , it follows that there are only two possibilities:(I) For some pairs ( j, i ), where j ∈ { , , . . . , m } and i ∈ { , , . . . , k } , there exists ℓ ∈ { , , . . . , m − } such that w j − λ i = w ℓ . (3.7)(II) For some pairs ( j, i ), where j ∈ { , , . . . , m } and i ∈ { , , . . . , k } , there exist s ∈ { , , . . . , m } \ { j } and t ∈ { , , . . . , k } \ { i } such that w j − λ i = w s − λ t . (3.8)After these preparations we proceed to prove that the frequencies of f are commensu-rable if and only if the frequencies of A ( z ) are commensurable. By appealing to (3.3) andto a change of variable as in Example 1.3, we may suppose that w = λ k ∈ N . Thus weprove that w j ∈ N for j ∈ { , . . . , m } if and only if λ i ∈ N for i ∈ { , . . . , k } .(i) Suppose that w j ∈ N for j ∈ { , . . . , m } . From (3.3), we see that w m − λ =max j,i { w j − λ i } and w m − λ > w j − λ i for any j = m and i = 1. Hence, from (3.7)and (3.8), there exists p < m such that w m − λ = w p , which implies that λ ∈ N .In addition, from (3.3), we have w m − λ > w j − λ i for any j = m and i >
2. Thus,from (3.7) and (3.8), there are only two possibilities: (1) There exists p < m suchthat w m − λ = w p − λ . (2) There exists p < m such that w m − λ = w p . In bothcases, it follows that λ ∈ N . Repeating this argument for k times gives us λ i ∈ N for i ∈ { , . . . , k } . ual exponential polynomials and a problem of Ozawa λ i ∈ N for i ∈ { , . . . , k } . From (3.3), we have λ k = w , and conse-quently w ∈ N . Moreover, from (3.3), we have w − λ k < w j − λ i for any j > i = k . Thus, from (3.7) and (3.8), there are only two possibilities: There exists p < k such that either w − λ k = w − λ p or w − λ k = w . In both cases, we have w ∈ N . Repeating this argument for m times gives us w j ∈ N for i ∈ { , . . . , m } .If the frequencies are commensurable for one of f, A ( z ), then they are commensurablefor both of f, A ( z ) by the reasoning above. The remaining fact that f and A ( z ) arestrongly dual exponential polynomials now follows by (3.3). ✷ The assumption ρ ( Af ′ ) < ρ ( f ) in Theorem 1.6(b) seems to be the only known sufficientcondition for the conclusion q = 1. However, in the case of Frei’s result (1.1), we have A ( z ) f ′ ( z ) = e − z m X j =1 jC j e jz = m − X j =0 ( j + 1) C j +1 e jz , and so ρ ( Af ′ ) = ρ ( f ) = 1. This shows that q = 1 may happen even if ρ ( Af ′ ) = ρ ( f ).Theorem 3.2 below shows that f having only one large exponential term is also a sufficientcondition for q = 1. In contrast, if A ( z ) has only one large exponential term, then f canhave multiple large exponential terms as in (1.1). Theorem 3.2
Suppose that f ( z ) = F ( z ) + F ( z ) e wz q is a solution of (1.3) , where A ( z ) isan exponential polynomial and B ( z ) is an entire function satisfying T ( r, B ) = o ( T ( r, A )) .Then q = 1 , and there are constants c, b ∈ C \ { } and a non-trivial polynomial P ( z ) suchthat f ( z ) = c + be wz , A ( z ) = bc P ( z ) − w + P ( z ) e − wz and B ( z ) = − wbc P ( z ) . Proof.
We proceed similarly as in the proof of Theorem 3.1 until (3.6), which now reducesto the form − k X j =1 A j e − λ j z q ! G e wz q = cB, (3.9)where F ( z ) ≡ c ∈ C \ { } . Hence k = 1, and consequently A ( z ) reduces to the form A ( z ) = A ( z ) + A ( z ) e − wz q . From (3.9) and (3.5), with k = 1 = m , we find that − A G = cB and − A G = BF + H . In other words, c − A G F = A G + H = A G + G ′ + qwz q − G . (3.10)Dividing both sides of (3.10) by G , we observe that at every zero of G the right-handside has a pole but the left-hand side does not. Thus G has no zeros, and so we maywrite it in the form G = e g , where g ( z ) = a q − z q − + · · · + a is a polynomial of degree ≤ q −
1. Since G = F ′ + qwz q − F = e g , ual exponential polynomials and a problem of Ozawa (cid:0) F ( z ) e wz q (cid:1) ′ = e wz q + g ( z ) , and consequently F ( z ) e wz q = Z z e wζ q + a q − ζ q − + ··· + a dζ . (3.11)Here the right-hand side is an exponential polynomial, which happens only if q = 1.Since q = 1, we see from (3.11) that F ( z ) reduces to a non-zero constant, say F ( z ) ≡ b .Thus f ( z ) = c + be wz , and we have G ( z ) ≡ wb and H ( z ) ≡ w b . A substitution to (3.10)followed by a simplification gives bc A = A + w. There is no restriction for A other than the fact that A is an exponential polynomial.Thus we may suppose that A is any non-trivial polynomial, say A = P . This gives us A = bc P − w , and finally B = − wbc P . ✷ Example 1.1 shows that the coefficient B ( z ) in (1.3) can be a polynomial. Next, weprove that this is equivalent to A ( z ) in (3.1) being a polynomial, and reveal anothersufficient condition for the conclusion q = 1. Proposition 3.3
Under the assumptions of Theorem 3.1, the term A ( z ) of A ( z ) in (3.1) is a polynomial if and only if B ( z ) is a polynomial. Moreover, if the multipliers of f andof A ( z ) are constants, then q = 1 and B ( z ) is a constant function.Proof. From the proof of Theorem 3.1 we find that (3.5) holds, that is, A G m + BF m + H m = 0 , (3.12)where G m = F ′ m + w m qz q − F m ,H m = F ′′ m + 2 w m qz q − F ′ m + (cid:0) w m q ( q − z q − + w m q z q − (cid:1) F m . Thus F m solves the second order differential equation F ′′ m + P ( z ) F ′ m + Q ( z ) F m = 0 , (3.13)where P ( z ) = 2 w m qz q − + A ,Q ( z ) = w m qz q − A + w m q ( q − z q − + w m q z q − + B. Suppose first that A ( z ) is a polynomial. If B ( z ) is transcendental, then it follows from(3.13) and [5, Corollary 1] that ρ ( F m ) = ∞ , which is a contradiction. Hence B ( z ) mustbe a polynomial. Conversely, suppose that B ( z ) is a polynomial. Suppose on the contraryto the assertion that A ( z ) is a transcendental exponential polynomial. Then there existsan open sector S such that A ( z ) blows up exponentially in S . Using [6, Corollary 1] and ρ ( F m ) ≤ q − S that | w m qz q − || A ( z ) | ≤ O (cid:0) | z | max { q, deg( B ) } (cid:1) + O (cid:0) | z | q − ε | A ( z ) | (cid:1) . However, this is obviously a contradiction, and hence A ( z ) is a polynomial. ual exponential polynomials and a problem of Ozawa f and of A ( z ) are constants. From (3.4), wefind that B ( z ) = Cz q − for some constant C ∈ C \ { } . Since F m ( z ) is a non-zero constantfunction, it follows that the coefficient Q ( z ) in (3.13) vanishes identically. But this is notpossible because A ( z ) is a constant function, unless q = 1. ✷ Remark. (a) The equation (3.13) implies that every possible zero of F m is simple.(b) Assuming that A ( z ) is a polynomial, we give an alternative proof for the fact that B ( z ) is a polynomial. We already know from Theorem 3.1 that B ( z ) is of order ≤ q − A ( z ) are all on one ray by duality, it follows that theplane divides into 2 q sectors of opening π/q such that in every other sector A ( z ) eitherblows up exponentially or is asymptotic to the polynomial A ( z ). In the latter case, if A ( z ) ≡
0, then A ( z ) decays to zero exponentially. Thus, from [5, Theorem 7], we deducethat B ( z ) is a polynomial. Note, in particular, that the constant µ in [5, Theorem 7]satisfies µ = π/q . Open problem 1.
Under the assumptions of Theorem 3.1, is it always true that q = 1 and B ( z ) is a polynomial? This problem is fragile in the sense that the desired conclusion is not valid if a minormodification in the assumptions of Theorem 3.1 is performed. For example, the differentialequation f ′′ − (cid:0) qwz q − + z − e − wz q (cid:1) f ′ − q ( q − wz q − f = 0possesses an exponential polynomial solution f ( z ) = e wz q − / ( q −
1) for any q ≥ f ( z ) = e z + 1 satisfies the differential equations f ′′ + e − z − z − z ! f ′ − f = 0 ,f ′′ − e − z ( z −
1) + 4 z + z + 12 z f ′ + ( z − f = 0 . (3.14)The transcendental coefficients in (3.14) are entire exponential polynomials with rationalmultipliers because z = 0 is a removable singularity for both. Next we construct examples of differential equations of order n ≥ f of order ρ ( f ) = n − Example 4.1 If H is an arbitrary entire function, then f ( z ) = e z + 1 solves f ′′′ + (cid:16) e − z (cid:17) Hf ′′ − (6 + 4 z ) f ′ − (2 + 4 z ) Hf = 0 ,f ′′′ − zf ′′ + ( H − He − z ) f ′ − zHf = 0 . (4.1)A particularly interesting case is when H is either a polynomial or an exponential poly-nomial of order one. Thus either of the two possible coefficients can be dual with f .Examples of second order dual solutions for third order equations can be found in [18] butfor polynomial coefficients only. ual exponential polynomials and a problem of Ozawa zf ′′ ( z ) = (2 z + 1) f ′ to see that, in addition to (4.1), thefunction f ( z ) = e z + 1 satisfies the equations f ′′′ ( z ) − zf ′ ( z ) − f ′ ( z ) = 0 , (1 + e − z ) f ′ ( z ) − zf ( z ) = 0 , (1 + e − z ) f ′′ ( z ) − (2 + 4 z ) f ( z ) = 0 . Example 4.2 If H is an arbitrary entire function, then f ( z ) = e z + 1 solves f (4) + (cid:16) e − z (cid:17) Hf ′′′ − z f ′′ − (cid:0) z (cid:1) f ′ − (cid:0) z + 27 z (cid:1) Hf = 0 ,f (4) − z f ′′′ + (cid:16) H − z + He − z (cid:17) f ′′ − f ′ − H (cid:0) z + 9 z (cid:1) f = 0 ,f (4) − z f ′′′ − zf ′′ + (cid:16) H + 27 z + He − z (cid:17) f ′ − z Hf = 0 . A particularly interesting case is when H is an exponential polynomial of order at mosttwo. Thus all three of the possible coefficients can be dual with f . Previous examples ofthird order dual solutions do not seem to be known.As in the previous example, we can use the relations zf ′′ ( z ) = (3 z + 2) f ′ ( z ) and zf ′′′ ( z ) = (3 z + 1) f ′′ ( z ) + 9 z f ′ ( z ) to see that f ( z ) = e z + 1 satisfies the equations f (4) ( z ) − z f ′′′ ( z ) − zf ′′ ( z ) − f ′ ( z ) = 0 , (1 + e − z ) f ′ ( z ) − z f ( z ) = 0 , (1 + e − z ) f ′′ ( z ) − (6 z + 9 z ) f ( z ) = 0 , (1 + e − z ) f ′′′ ( z ) − (6 + 54 z + 27 z ) f ( z ) = 0 . In light of Open problem 1 and the examples just discussed, it is natural to pose oursecond open problem.
Open problem 2.
If a solution and the dominant coefficient are dual exponential poly-nomials of order q , then is the differential equation in question of order at least q + 1 ? For the fragility of this problem, recall the equations (3.14) satisfied by f ( z ) = e z + 1.Moreover, the function f ( z ) = e z + 1 satisfies the third order equation f ′′′ + e − z − z ! f ′′ − z (3 z + 5) f ′ −
32 (3 z + 2) f = 0with entire coefficients.As the first initial step to knowing more about Open problem 2, we make a summaryof the fundamental ideas in constructing Examples 4.1 and 4.2. Lemma 4.3
The function f ( z ) = e z q + 1 , q ∈ N , possesses the following two properties: (i) (1 + e − z q ) f ( j +1) ( z ) = j X k =0 P j,k ( z ) f ( k ) ( z ) , j ∈ N ∪ { } , (ii) f ( q +1) ( z ) = q X ℓ =1 Q ℓ ( z ) f ( ℓ ) ( z ) , ual exponential polynomials and a problem of Ozawa where the P j,k ( z ) and Q ℓ ( z ) are non-zero polynomials satisfying (a) P j +1 ,j +1 ( z ) = P j,j ( z ) − qz q − , P , ( z ) = qz q − ,P j +1 ,k ( z ) = P ′ j,k ( z ) + qz q − P j,k ( z ) + P j,k − ( z ) , P j, − ( z ) ≡ , ≤ k ≤ j,P j +1 , ( z ) = P ′ j, ( z ) + qz q − P j, ( z ) , (b) Q ℓ ( z ) = − (cid:18) qℓ − (cid:19) ( e − z q ) ( q − ℓ +1) e z q . Proof.
First, let us prove (i) by induction on j . Of course, by taking their logarithmicderivatives, we have (1 + e − z q ) f ′ ( z ) = qz q − f ( z ) immediately, that is, the case when j = 0follows with P , ( z ) = qz q − . Assume (i) is true for each j = 0 , , . . . , n . Then(1 + e − z q ) f ( n +2) ( z ) = qz q − e − z q f ( n +1) + n X k =0 (cid:8) P ′ n,k ( z ) f ( k ) ( z ) + P n,k ( z ) f ( k +1) ( z ) (cid:9) = qz q − (1 + e − z q ) f ( n +1) ( z ) + (cid:8) P n,n ( z ) − qz q − (cid:9) f ( n +1) ( z ) ++ n X k =1 (cid:8) P ′ n,k ( z ) f ( k ) ( z ) + P n,k − ( z ) (cid:9) f ( k ) ( z ) + P ′ n, ( z ) f ( z )= qz q − n X k =0 P n,k ( z ) f ( k ) ( z ) + (cid:8) P n,n ( z ) − qz q − (cid:9) f ( n +1) ( z ) ++ n X k =1 (cid:8) P ′ n,k ( z ) f ( k ) ( z ) + P n,k − ( z ) (cid:9) f ( k ) ( z ) + P ′ n, ( z ) f ( z )= (cid:8) P n,n ( z ) − qz q − (cid:9) f ( n +1) ( z ) ++ n X k =1 (cid:8) P ′ n,k ( z ) f ( k ) ( z ) + qz q − P n,k ( z ) + P n,k − ( z ) (cid:9) f ( k ) ( z ) ++ (cid:8) P ′ n, ( z ) + qz q − P n, ( z ) (cid:9) f ( z ) , which is the one to be proved.Second, let us calculate the q -th order derivative of the product f ′ ( z ) e − z q = qz q − . TheLeibniz rule gives q X ℓ =0 (cid:18) qℓ (cid:19) f ( ℓ +1) ( z )( e − z q ) ( q − ℓ ) ≡ . Denoting Q ℓ +1 ( z ) = − (cid:0) qℓ (cid:1) ( e − z q ) ( q − ℓ ) e z q for 0 ≤ ℓ ≤ q −
1, we have f ( q +1) ( z ) = q − X ℓ =0 Q ℓ +1 ( z ) f ( ℓ +1) ( z ) = q X ℓ =1 Q ℓ ( z ) f ( ℓ ) ( z ) , as desired. ✷ Example 4.4
We may apply the two identities in Lemma 4.3 to construct differentialequations of arbitrary order. Given any entire function H , we have the identity f ( q +1) ( z ) − q X ℓ =1 Q ℓ ( z ) f ( ℓ ) ( z ) = H ( z ) (1 + e − z q ) f ( j ) ( z ) − j − X k =0 P j − ,k ( z ) f ( k ) ( z ) ! , ual exponential polynomials and a problem of Ozawa f ( z ) = e z q + 1 solves f ( q +1) ( z ) − q X ℓ = j +1 Q ℓ ( z ) f ( ℓ ) ( z ) − (cid:0) (1 + e − z q ) H ( z ) + Q j ( z ) (cid:1) f ( j ) ( z )+ j − X ℓ =1 (cid:0) P j − ,ℓ ( z ) H ( z ) − Q ℓ ( z ) (cid:1) f ( ℓ ) ( z ) + P j − , ( z ) H ( z ) f ( z ) = 0 , where 1 ≤ j ≤ q , the sum P j − ℓ =1 is empty if j = 1 and the sum P qℓ = j +1 is empty if j = q . The possibility that a solution f would be dual to more than one coefficient has notbeen studied rigorously. In this case there would be at least two equally strong dominantcoefficients, or, in the case of (1.3), both coefficients A ( z ) , B ( z ) would be equally strong.For example, f ( z ) = e − z solves f ′′ + e z f ′ + ( e z − f = 0and is dual to both coefficients. Obviously the coefficients are not dual to each other.More examples can be produced from Example 1.1. Note that f ( z ) = e z solves (1.3) if A ( z ) = − B ( z ) −
1. Hence f is not necessarily dual with either of A ( z ) , B ( z ).If H is any entire function, then f ( z ) = e z q solves f ′′ + (cid:0) H ( z ) − qz q − (cid:1) f ′ − (cid:0) q ( q − z q − + qz q − H ( z ) (cid:1) f = 0 . This example is from [5]. Note that f ( z ) = e z q satisfies both f ′ ( z ) − qz q − f ( z ) = 0 and f ′′ ( z ) − qz q − f ′ ( z ) − q ( q − z q − f ( z ) = 0.Recall [9, Theorem 2.1], according to which there cannot be even one ray on which B ( z ) would be stronger than A ( z ) in the sense of the Phragm´en-Lindel¨of indicator, forotherwise all solutions of (1.3) are of infinite order. This happens, for example, when A ( z )and B ( z ) are dual to each other. Example 5.1
One may observe the necessity of the assumption on the duality of f and A ( z ) as well as that on the dominance of A ( z ) over B ( z ) by the following example: Thefunction f ( z ) = (cid:0) e z + e − z (cid:1) e z q , q ∈ N , satisfies f ′′ + (cid:8) H ( z ) e z + H ( z ) e − z − qz q − (cid:9) f ′ − (cid:8) ( qz q − + 1) H ( z ) e z + ( qz q − − H ( z ) e − z − q z q − + q ( q − z q − + 1 (cid:9) f = 0for any entire function H . When q = 1, this becomes f ′′ + (cid:8) H ( z ) e z + H ( z ) e − z − (cid:9) f ′ − H ( z ) e z f = 0with f ( z ) = e z + 1. Thus we may use it in order to observe the duality of A ( z ) and B ( z )by several choices of H such as H ( z ) = e nz for n ∈ Z or H ( z ) = e iz .Here we note that f ( z ) = F ( z ) e z q satisfies f ′ f = F ′ F + qz q − and f ′′ f = F ′′ F + 2 qz q − f ′ f + (cid:0) q ( q − z q − − q z q − (cid:1) ual exponential polynomials and a problem of Ozawa F . For example, taking an Airyfunction as F , we cannot have our desired equation f ′′ + A ( z ) f ′ + B ( z ) f = 0 with theexponential polynomial coefficients A ( z ) and B ( z ). Acknowledgements.
Ishizaki was supported by JSPS KAKENHI Grant Number 20K03658.Wen was supported by the National Natural Science Foundation of China (No. 11971288and No. 11771090) and Shantou University SRFT (NTF18029).
References [1] Amemiya I. and M. Ozawa,
Non-existence of finite order solutions of w ′′ + e − z w ′ + Q ( z ) w = 0. Hokkaido Math. J. (1981), Special Issue, 1–17.[2] Frei M., ¨Uber die subnormalen L¨osungen der Differentialgleichung w ′′ + e − z w ′ +konst · w = 0. Comment. Math. Helv. (1961), 1–8. (German)[3] Gross F., Factorization of Meromorphic Functions . Mathematics Research Center,Naval Research Laboratory, Washington, D. C., 1972.[4] Gundersen G. G.,
On the question of whether f ′′ + e − z f ′ + B ( z ) f = 0 can admita solution f of finite order . Proc. Roy. Soc. Edinburgh Sect. A (1986),no. 1–2, 9–17.[5] Gundersen G. G., Finite order solutions of second order linear differential equations .Trans. Amer. Math. Soc. (1988), no. 1, 415–429.[6] Gundersen G. G.,
Estimates for the logarithmic derivative of a meromorphic function,plus similar estimates . J. London Math. Soc. (2) (1988), no. 1, 88–104.[7] Gundersen G. G., E. Steinbart and S. Wang, The possible orders of solutions of lineardifferential equations with polynomial coefficients . Trans. Amer. Math. Soc. (1998), no. 3, 1225–1247.[8] Heittokangas J., K. Ishizaki, K. Tohge and Z.-T. Wen,
Zero distribution and divisionresults for exponential polynomials . Israel J. Math. (2018), no. 1, 397–421.[9] Heittokangas J., I. Laine, K. Tohge and Z.-T. Wen,
Completely regular growthsolutions of second order complex linear differential equations . Ann. Acad. Sci. Fenn.Math. (2015), no. 2, 985–1003.[10] Heittokangas J. and Z.-T. Wen, Generalization of P´olya’s zero distribution theory forexponential polynomials, and sharp results for asymptotic growth . Comput. MethodsFunct. Theory, online publ., https://doi.org/10.1007/s40315-020-00336-7, 26 pp.[11] Langer R. E.,
On the zeros of exponential sums and integrals . Bull. Amer. Math.Soc. (1931), no. 4, 213–239.[12] Langley J. K., On complex oscillation and a problem of Ozawa . Kodai Math. J. (1986), no. 3, 430–439.[13] Moreno C. J., The zeros of exponential polynomials . I. Compositio Math. (1973),69–78.[14] Nevanlinna R., Analytic Functions . Translated from the second German editionby Phillip Emig. Die Grundlehren der mathematischen Wissenschaften, Band 162Springer-Verlag, New York-Berlin, 1970. ual exponential polynomials and a problem of Ozawa
On a solution of w ′′ + e − z w ′ + ( az + b ) w = 0. Kodai Math. J. (1980),no. 2, 295–309.[16] Steinmetz N., Zur Wertverteilung von Exponentialpolynomen . Manuscripta Math. (1978/79), no. 1–2, 155–167. (German)[17] Steinmetz N., Zur Wertverteilung der Quotienten von Exponentialpolynomen . Arch.Math. (Basel) (1980), no. 5, 461–470. (German)[18] Wen Z. T., G. G. Gundersen and J. Heittokangas, Dual exponential polynomials andlinear differential equations . J. Differential Equations (2018), no. 1, 98–114.
J. Heittokangas
University of Eastern Finland, Department of Physics and Mathematics,P.O. Box 111, 80101 Joensuu, Finland email:[email protected]
K. Ishizaki
The Open University of Japan, Faculty of Liberal Arts, Mihama-ku, Chiba,Japan email:[email protected]
K. Tohge
Kanazawa University, College of Science and Engineering, Kakuma-machi,Kanazawa 920-1192, Japan email:[email protected]
Z.-T. Wen