On Moment Condition and Center Condition for Abel Equation
aa r X i v : . [ m a t h . C A ] A ug On Moment Condition and Center Condition for Abel Equation
Anderson L. A. de Araujo, Ab´ılio Lemos and Alexandre Miranda Alves
Universidade Federal de Vi¸cosa, CCE, Departamento de Matem´aticaAvenida PH Rolfs, s/nCEP 36570-900, Vi¸cosa, MG, BrasilE-mails: [email protected], [email protected], [email protected]
Abstract . In this paper we consider Abel equation x ′ = g ( t ) x + f ( t ) x , where f and g areanalytical functions. We proved that if the equation has a center at x = 0, then the MomentConditions, i. e., m k = R − f ( t )( G ( t )) k dt = 0 , k = 0 , ,
2, is satisfied where G ( t ) = R t − g ( s ) ds .Besides, we give partial a positive answer to a conjecture proposed by Y. Lijun and T. Yun in2001. AMS Subject Classification 2010 . 30E05, 34A34, 34C07.
Keywords . Abel equation; moment condition; center condition; recurrence relation.1
Introduction
Let the planar system(1) ˙ x = − y + P ( x, y )˙ y = x + Q ( x, y ) , where P ( x, y ) and Q ( x, y ) are polynomials, without constant term, of maximum degree n. Thesingular point (0 ,
0) is a center, if surrounded by closed trajectories; or a focus, if surroundedby spirals. The classical center-focus problem consists in distinguishing when a singular pointis either a center or a focus. The problem started with Poincar´e [29] and Dulac [20], and, inthe present days, many questions remain unanswered. The basic results were obtained by A. M.Lyapunov [26]. He proved that if P ( x, y ) and Q ( x, y ) satisfy an infinite sequence of recursiveconditions, then (1) has a center to the origin. He also presented conditions for the origin of thesystem (1) to be a focus.If we write P ( x, y ) = P li =1 P m i ( x, y ) and Q ( x, y ) = P li =1 Q m i ( x, y ), where P m i ( x, y ) and Q m i ( x, y ) are homogeneous polynomials of degree m i ≥
1, then, from Hilbert’s theorem onthe finiteness of basis of polynomial ideals ([23], Theorem 87, p. 58), it follows that, in thementioned infinite sequence of recursive conditions, only a finite number of conditions for centerare essential. The others result from them.Let us consider a particular case of (1). Namely,(2) ˙ x = − y + P n ( x, y )˙ y = x + Q n ( x, y ) , where P n ( x, y ) and Q n ( x, y ) are homogeneous polynomials of degree n.When n = 2, systems (2) are quadratic polynomial differential systems (or simply quadraticsystems in what follows). Quadratic systems have been intensively studied over the last 30years, and more than a thousand papers on this issue have been published (see, for example,the bibliographical survey of Reyn [30]).A method for investigating if (2) has a center at the origin consists in transforming the planarsystem into an Abel equation. In polar coordinates ( r, θ ) defined by x = r cos θ, y = r sin θ , thesystem (2) becomes(3) ˙ r = A ( θ ) r n ˙ θ = 1 + B ( θ ) r n − , where(4) A ( θ ) = cos θP n (cos θ, sin θ ) + sin θQ n (cos θ, sin θ ) ,B ( θ ) = cos θQ n (cos θ, sin θ ) − sin θP n (cos θ, sin θ ) . We remark that A and B are homogeneous polynomials of degree n + 1 in the variables cos θ and sin θ . In the region R = { ( r, θ ) : 1 + B ( θ ) r n − l > } , drdθ = A ( θ ) r n B ( θ ) r n − . It is known that the periodic orbits surrounding the origin of the system (3) do not intersectthe curve θ = 0 (see the Appendix of [15]). Therefore, these periodic orbits are contained in theregion R . Consequently, they are also periodic orbits of equation (5).The transformation ( r, θ ) → ( γ, θ ) with(6) γ = r n − B ( θ ) r n − is a diffeomorphism from the region R into its image. As far as we know, Cherkas was the firstto use this transformation (see [16]). If we write equation (5) in the variable γ , we obtain(7) dγdθ = − ( n − A ( θ ) B ( θ ) γ + [( n − A ( θ ) − B ′ ( θ )] γ , which is a particular case of an Abel differential equation. We notice that f ( θ ) = − ( n − A ( θ ) B ( θ ) and g ( θ ) = ( n − A ( θ ) − B ′ ( θ ) are homogeneous trigonometric polynomials ofdegree 2( n + 1) and n + 1, respectively.Now the Center-Focus problem of equation (1) has a translation in equation (7). That is,given γ small enough, we look for necessary and sufficient conditions on f ( θ ) and g ( θ ) in orderto assure that the solution of equation (7) with the initial condition γ (0) = γ has the propertythat γ (0) = γ (2 π ). We observe that this condition implies the periodicity of this solution.In the present paper, we consider a certain variant of the Center-Focus problem related tothe original one - the Center-Focus problem for the Abel differential equation(8) dxdt = g ( t ) x + f ( t ) x This problem is to provide necessary and sufficient conditions on f and g and on a, b ∈ R forall the solutions x ( t ) of (8) to satisfy x ( a ) = x ( b ). When f and g are polynomial functions,the equation (8) is called polynomial Abel equation. Notice that, now, such condition does notimply the periodicity of x ( t ). In recent years, the Center-Focus problem for the Polynomial Abelequation has advanced substantially, as observed in [8], [9], [12], [13], [14], [18], [32]. Definition 1.1
An Abel differential equation (8) is said to have a center at a ≤ t ≤ b if x ( a ) = x ( b ) for any solution x ( t ) (with the initial value x ( a ) small enough), or equivalently, theequation (8) has a center at x = 0 if all the solutions close to x = 0 are closed. The Center-Focus problem is to give necessary and sufficient conditions on f , g for the Abelequation above to have a center. The only, known to us, sufficient condition for the Center isthe following ”Polynomial Composition Condition” (PCC):3 efinition 1.2 The polynomials f = F ′ , g = G ′ are said to satisfy the Polynomial CompositionCondition (PCC) on [ a, b ] , if there exist polynomials e F , e G and W , such that F ( t ) = e F ( W ( t )) , G ( t ) = e G ( W ( t )) , W ( a ) = W ( b ) . (PCC) is also known to be necessary for the Center for small degrees of f , g and in someother very special situations. Notice that (PCC) is described by a finite number of algebraicequations on the coefficients of f , g . Proposition 1.3 If f , g satisfy the Polynomial Composition Condition (PCC) on A = [ a, b ] ,then the Abel equation (8) has a center on A . Proof:
Indeed, after a change of variables w = W ( t ), we obtain a new polynomial Abel equation(9) d e ydw = e f ( w ) e y + e g ( w ) e y with e f = e F ′ , e g = e G ′ . All the solutions of (8) are obtained from the solutions of (9) by the samesubstitution y ( t ) = e y ( W ( t )). Fulfilling W ( a ) = W ( b ), one has y ( a ) = y ( b ).All existing results in the literature so far supports the following conjecture: COMPOSITION CONJECTURE . The polynomial Abel equation (8) has a center onthe set of points A = [ a, b ] if and only if the Composition Condition (PCC) holds for f and g on A . This conjecture has been verified for small degrees of f and g and in many special cases in[4], [5], [6], [7], [8], [9], [10], [11], [18], [31], [32].The equation (8) was studied in [1], where necessary and sufficient conditions were obtainedfor this equation to have a center at the origin, where f ( t ) and g ( t ) are particular continuousfunctions. More results that ensure the existence of a center at the origin for some subclassesof Abel equations were obtained in [24, 25].When f and g are odd polynomials, then f ( t ) = t ˆ f ( t ) and g ( t ) = t ˆ g ( t ). Results onsufficient conditions presented below were obtained by Alwash and Lloyd [2]. Proposition 1.4 ([2])
Assume f, g ∈ C ([ a, b ]) to be expressed by (10) f ( t ) = ˆ f ( σ ( t )) σ ′ ( t ) , g ( t ) = ˆ g ( σ ( t )) σ ′ ( t ) for some continuous functions ˆ f , ˆ g and a continuously differentiable function σ , which is closed,i.e., σ ( a ) = σ ( b ) . Then, the Abel equation x ′ ( t ) = f ( t ) x ( t ) + g ( t ) x ( t ) , t ∈ [ a, b ] has a center in [ a, b ] . If true, the Composition Conjecture tells us a lot about the nature of the return map for Abelequations and its relationship with the coefficients of the system. However, it also highlightsa significant difference between the polynomial and trigonometric cases. In the latter case, itis known that the class corresponding to the Composition Conjecture (that is, those systemswith P and Q polynomials of a trigonometric polynomial), although a significant class, does notexhaust all possible center conditions, see [3]. 4 .3 Moment conditions and the Parametric composition conjecture In this section, we consider the polynomial Abel differential equation(11) x ′ ( t ) = g ( t ) x ( t ) + ǫf ( t ) x ( t ) , t ∈ [ a, b ] , where x is real, ǫ ∈ R and g ( t ) and f ( t ) are real polynomials. Let us assume that R ba g ( s ) ds = 0.One of the issues that can be tackled is characterizing when (11) has a center in [ a, b ] forall ǫ with | ǫ | small enough. This type of centers are called infinitesimal centers or persistentcenters , see [3], [17]. In [3], it was proved that a necessary and sufficient conditions for (11) tohave a center in [ a, b ] are(12) Z ba f ( t )( G ( t )) k dt = 0 , for all natural numbers k ∈ N ∪ { } . Conditions (12) are called the moment conditions . Thecomposition conjecture for moments is that the moments conditions imply the compositioncondition . Moreover, in [12] it is proved that “at infinity” the center conditions are reduced tothe moment conditions.A counterexample to the composition conjecture for moments in the polynomial case wasgiven in [28], see too [21].In the trigonometric case, that is, if one considers a trigonometric Abel differential equationof the form(13) x ′ ( θ ) = g ( θ ) x ( θ ) + ǫf ( θ ) x ( θ ) , θ ∈ [0 , π ] , where x is real, and ǫ is a real value close to 0. One can define the composition conjecture formoments analogously to the polynomial case. The moment conditions in this case are writtenas(14) Z π f ( θ )( G ( θ )) k dθ = 0 , for all natural numbers k ∈ N ∪ { } . It is also possible to construct a counterexample of thecomposition conjecture for moments, see [21]In Lijun and Yun [27, Theorem 5.5], the authors proved the following result. Proposition 1.5
Let f ( · ) be a polynomial. Then, for ǫ small enough the Abel equation x ′ = 2 tx + ǫf ( t ) x , t ∈ [ − , has a center x = 0 if and only if f ( t ) is an odd polynomial. Also, the authors conjecture that the conclusion of Theorem 1.5 holds without ǫ .5 .4 Main results This paper aims to study the following Abel’s equation(15) x ′ ( t ) = f ( t ) x + g ( t ) x , t ∈ [ − , , where f and g are analytic functions. Now, we state our main results. Theorem 1.6
Consider the Abel’s equation x ′ = f ( t ) x + g ( t ) x , t ∈ [ − , , where f and g are analytic functions. If this equation has a center at x = 0 , then m k = Z − f ( t )( G ( t )) k dt = 0 , k = 0 , , , where G ( t ) = t R − g ( s ) ds . Remark 1.7
The following equation, dxdθ = (sin θ − sin 2 θ + sin 3 θ ) x + (cos θ + 2 cos 2 θ ) x , θ ∈ [0 , π ] , studied by Gin´e, Grau and Santallusia [22], has a center at x = 0 . Some computations showthat m = 0 , m = 0 , m = 0 and m = π Z (sin θ − sin 2 θ + sin 3 θ ) θ Z (cos s + 2 cos 2 s ) ds dθ = π = 0 , which demonstrates that the result of Theorem 1.6 is the best possible. Now, considering f and g real polynomial functions in Abel’s equation (15), we obtain thefollowing result. Theorem 1.8
Consider the Abel’s equation (16) x ′ ( t ) = f ( t ) x + g ( t ) x , t ∈ [ − , , where f ( t ) = P dj =0 a j t j = p ( t ) + tq ( t ) is such that p ( t ) changes sign at most two times in [ − , and g ( t ) = t n − , where n is a positive even integer. If the equation (16) has a center at x = 0 , then m k = Z − f ( t )( G ( t )) k dt = 0 , k = 0 , , , , . . . , where G ( t ) = R t − g ( s ) ds = n ( t n − . g ( t ) = t n − is as in Theorem 1.8, we have the following result. Corollary 1.9
Let f ( · ) be a polynomial of degree d ≤ . Then, the Abel equation (17) x ′ = t n − x + f ( t ) x , t ∈ [ − , , has a center x = 0 if and only if f ( t ) is an odd polynomial. Remark 1.10
The results of Theorem 1.8 and Corollary 1.9 hold if g ( t ) = nt n − , where n is apositive even integer. Hence, G ( t ) = R t − g ( s ) ds = t n − . According to Corollary 1.9 and Remark 1.10, we give a positive answer to the conjecture proposedby Lijun and Yun [27, Remark 5.6] when f ( · ) is a polynomial of degree d ≤ n = 2.The following diagram gives us some relations and implications about the Center problem , Moment conditions and the
Composition Conjecture . (cid:127)~}|xyz{ Composition Condition (1) (cid:4) (cid:4) (2) (cid:28) (cid:28) (cid:127)~}|xyz{
M oment Condition (5) (3) + + (cid:127)~}|xyz{ y ′ = f ( t ) y + g ( t ) y has a center at origin (4) s s (6) i i Implication (1) was proved in [4, p. 13].Implication (2) was proved in [12, p. 442].Implication (3) is not generally true. For example, see [28] and [21].Implication (4) is generally an open problem, namely
Composition Conjecture . Several particularcases were proved. See, for example, [7] and the reference therein. We proved some particularcases, see Corollaries 1.9.Implication (5) is generally an open problem.Implication (6) holds in several particular cases. One of these cases is the main result of thispaper, see Theorem 1.8.
Remark 1.11
Notice that, if (5) is true, then the composition conjecture (see (4) ) is not true. Preliminaries results
Following Yang Lijun and Tang Yun [27], we write the expression below, for a solution x ( t, ρ )of the Abel equation (15) satisfying x ( − , ρ ) = ρ .(18) x ( t, ρ ) = ρ + ∞ X k =2 r k ( t ) ρ k . To prove Theorems 1.6 and 1.8, we apply the following result of Yang Lijun and Tang Yun[27, Lemma 5.2, p 108].
Lemma 2.1
The origin x = 0 is a center of the Abel equation (15) if and only if (19) Z − g ( t ) dt = 0 and Z − f ( t ) r k ( t ) dt = 0 , k ≥ or, equivalently, if and only if (20) Z − g ( t ) dt = 0 and Z − f ( t ) x ( t, ρ ) dt = 0 , | ρ | < ρ for ρ small enough, where x ( t, ρ ) = ρ − ρ Z t − ( f ( s ) x ( s ) + g ( s )) ds . Now, suppose that x = 0 is a center of the equation (15). Notice that a solution of (15) isequivalent to a solution of the integral equation x ( t, ρ ) = ρ − ρ Z t − ( f ( s ) x ( s, ρ ) + g ( s )) ds , t ∈ [ − , x ( −
1) = ρ , for ρ small enough such that ρ R t − ( f ( s ) x ( s, ρ ) + g ( s )) ds < t ∈ [ − , T ρ : C [ − , → C [ − , T ρ ( x )( t ) = ρ − ρ R t − ( f ( s ) x ( s, ρ ) + g ( s )) ds (23)where f, g ∈ C [ − ,
1] and ρ ∈ R . Of course, T ρ is well defined on arbitrary bounded set of C [ − ,
1] if ρ is small enough. In [27], Lijun and Yun proved that T ρ is a contraction. Accordingto the well known Banach contraction theorem, T ρ has a unique fixed point in B = { f ∈ C [ − , || f || ≤ } , where || f || = sup t ∈ [ − , | f ( t ) | . In addition, Lijun and Yun proved that, for ρ ≤ ρ = ( √ g + || g || + || f || + 1) − , this fixed point is the solution x ( t, ρ ) of the Abel equation(15), with x ( − , ρ ) = ρ and || x || ≤
1. 8ccording to the Abel equation (15), f ( t, x ) = f ( t ) x + g ( t ) x , with t ∈ [ − ,
1] and x ∈ R ,is analytic. Thus, the solution x ( t, ρ ) is also analytic in ( t, ρ ) (see in [19, Th 8.2, p. 35]).Set(24) H ( t, ρ ) = Z t − ( f ( s ) x ( s, ρ ) + g ( s )) ds. Since f , g and x are analytic functions, we have that H is analytic. Furthermore, as || x ( t, ρ ) || ≤ f and g are limited on the interval [ − , ρ , we have − < ρ R t − ( f ( s ) x ( s, ρ ) + g ( s )) ds <
1. Thus, the following identities are well defined x ( t, ρ ) = ρ − ρ Z t − ( f ( s ) x ( s, ρ ) + g ( s )) ds = ρ − ρH ( t, ρ )= ρ (cid:0) H ( t, ρ ) ρ + H ( t, ρ ) ρ + H ( t, ρ ) ρ + . . . (cid:1) = ρ + H ( t, ρ ) ρ + H ( t, ρ ) ρ + H ( t, ρ ) ρ + . . . = ρ ∞ X k =0 H k ( t, ρ ) ρ k . By Lemma 2.1, we obtain Z − f ( t ) x ( t, ρ ) dt = 0 , | ρ | < ρ The last identities lead us to conclude that Z − f ( t ) x ( t, ρ ) dt = ρ ∞ X k =0 Z − f ( t ) H k ( t, ρ ) dtρ k = 0 , | ρ | < ρ . In the proof of Theorem 1.6, we need to show that R − f ( t ) H k ( t, ρ ) dt = 0 in ρ = 0 and k = 0 , ,
2. For such, we need some lemmas, which are presented below.
Lemma 2.2 ∂ j ∂ρ j ( H (1 , ρ )) (cid:12)(cid:12) ρ =0 = j ! Z − f ( t ) G j − ( t ) dt for each j = 1 , and ∂ ∂ρ ( H (1 , ρ )) (cid:12)(cid:12) ρ =0 = 3! Z − f ( t ) G ( t ) dt + 3! Z − f ( t ) F ( t ) dt, where G ( t ) = R t − g ( s ) ds and F ( t ) = R t − f ( s ) ds . roof: Since x ( t, ρ ) = ρ + H ( t, ρ ) ρ + H ( t, ρ ) ρ + H ( t, ρ ) ρ + . . . (25)we obtain(26) ∂∂ρ x ( t, ρ ) = 1 + ∂∂ρ ( H ( t, ρ ) ρ ) + ∂∂ρ ( H ( t, ρ ) ρ ) + . . . , (27) ∂ ∂ρ x ( t, ρ ) = ∂ ∂ρ ( H ( t, ρ ) ρ ) + ∂ ∂ρ ( H ( t, ρ ) ρ ) + . . . = 2 H ( t, ρ ) + R ( t, ρ ) , where R ( t, ρ ) | ρ =0 = 0 and(28) ∂ ∂ρ x ( t, ρ ) = ∂ ∂ρ ( H ( t, ρ ) ρ ) + ∂ ∂ρ ( H ( t, ρ ) ρ ) + . . . = 3! H ( t, ρ ) + 3! ∂∂ρ H ( t, ρ ) + R ( t, ρ ) , where R ( t, ρ ) | ρ =0 = 0.By definition of H and (26), we obtain ∂∂ρ H ( t, ρ ) = t Z − f ( s ) ∂∂ρ x ( s, ρ ) ds and(29) ∂∂ρ H ( t, ρ ) | ρ =0 = t Z − f ( s ) ds = F ( t ) . Hence,(30) ∂∂ρ H (1 , ρ ) | ρ =0 = Z − f ( t ) dt. By definition of H and (27), we obtain ∂ ∂ρ H ( t, ρ ) = 2 t Z − f ( s ) H ( s, ρ ) ds + t Z − R ( s, ρ ) ds and(31) ∂ ∂ρ H ( t, ρ ) | ρ =0 = 2 t Z − f ( s ) H ( s, ds = 2 t Z − f ( s ) G ( s ) ds. ∂ ∂ρ H (1 , ρ ) | ρ =0 = 2 Z − f ( t ) G ( t ) dt. By definition of H , (28) and (29), we obtain ∂ ∂ρ H ( t, ρ ) = 3! t Z − f ( s ) H ( s, ρ ) ds + 3! t Z − f ( s ) ∂∂ρ H ( s, ρ ) ds + t Z − R ( s, ρ ) ds and(33) ∂ ∂ρ H ( t, ρ ) | ρ =0 = 3! t R − f ( s ) H ( s, ds + 3! t R − f ( s ) ∂∂ρ H ( s, ds = 3! t R − f ( s ) G ( s ) ds + 3! t R − f ( s ) F ( s ) ds. Hence, ∂ ∂ρ H (1 , ρ ) | ρ =0 = 3! Z − f ( t ) G ( t ) dt + 3! Z − f ( t ) F ( t ) dt. Lemma 2.3
Let H be defined by (24), then ∂ j ∂ρ j ( H (1 , ρ )) (cid:12)(cid:12) ρ =0 = j ! Z − f ( t ) r j ( t ) dt for each j = 1 , , . Proof:
If we replace the expression (18) by the formula defining H ( t, ρ ), we obtain H ( t, ρ ) = Z t − [ f ( s ) x ( s, ρ ) + g ( s )] ds = (cid:18)Z t − f ( s ) ds (cid:19) ρ + ∞ X k =2 t Z − f ( s ) r k ( s ) ds ρ k + Z t − g ( s ) ds = G ( t ) + F ( t ) ρ + ∞ X k =2 t Z − f ( s ) r k ( s ) ds ρ k . Hence, ∂ j ∂ρ j ( H ( t, ρ )) (cid:12)(cid:12) ρ =0 = j ! t Z − f ( s ) r j ( s ) ds for each j = 1 , , . Therefore, ∂ j ∂ρ j ( H (1 , ρ )) (cid:12)(cid:12) ρ =0 = j ! Z − f ( t ) r j ( t ) dt for each j = 1 , , . Proof of Theorem 1.6
By Lemma 2.1, we have Z − f ( t ) r k ( t ) dt = 0 , k ≥ . By Lemmas 2.2 and 2.3, we obtain ∂ j ∂ρ j ( H (1 , ρ )) (cid:12)(cid:12) ρ =0 = j ! Z − f ( t ) G j − ( t ) dt = j ! Z − f ( t ) r j ( t ) dt = 0 , for each j = 1 , ∂ ∂ρ ( H (1 , ρ )) (cid:12)(cid:12) ρ =0 = 3! Z − f ( t ) G ( t ) dt + 3! Z − f ( t ) F ( t ) dt = 3! Z − f ( t ) r ( t ) dt = 0 . Since by Lemma 2.1 we obtain F (1) = R − f ( t ) dt = 0, we conclude that Z − f ( t ) F ( t ) dt = Z − F ′ ( t ) F ( t ) dt = 12 Z − ddt F ( t ) dt = 12 [ F (1) − F ( − . Therefore, Z − f ( t ) G ( t ) dt = Z − f ( t ) r ( t ) dt = 0 . Hence, we obtain m k = Z − f ( t )( G ( t )) k dt = 0 , k = 0 , , . In the proof of Theorem 1.6, we will use the result due to Briskin, Francoise and Yomdin [8,Theorem 4.1]. By [8], it is sufficient to analyze the sign changes of the function ψ ( u ) = m X i =1 sgn ( g ( t i ( u ))) f ( t i ( u )) g ( t i ( u ))where t ( u ) , t ( u ) , . . . , t m ( u ) are the solutions in [ − ,
1] of the equation G ( t ) = u where u ∈ [ u , u ], with u = min [ − , G ( t ) and u = max [ − , G ( t ).Since g ( t ) = t n − , we obtain G ( t ) = n ( t n − u = − n and u = 0. In particular,if u ∈ [ − n , ≤ nu ≤
1. The solutions in [ − ,
1] of the equation G ( t ) = 1 n ( t n −
1) = u t ( u ) = (1 + nu ) n and t ( u ) = − (1 + nu ) n . Therefore, since n is even, ψ ( u ) = sgn ( g ( t ( u ))) f ( t ( u )) g ( t ( u )) + sgn ( g ( t ( u ))) f ( t ( u )) g ( t ( u )) = f ( t ( u )) g ( t ( u )) − f ( t ( u )) g ( t ( u )) = nu ) n − n [ f ((1 + nu ) n ) + f ( − (1 + nu ) n )] . Now, we define the polynomial h ( t ) = f ( t ) + f ( − t ) = p ( t ). Since according assumptions, thepolynomial h ( t ) changes sign at most two times in [ − , ψ ( u ) changes signat most two times in [ − , x = 0, then by Theorem 1.6, m k = Z − f ( t )( G ( t )) k dt = 0 , k = 0 , , . Hence, ψ ( u ) changes sign at most two times in [ − ,
0] and m k = 0 for k = 0 , ,
2. According to[8, Theorem 4.1], we obtain that m k = 0 for k = 1 , , . . . , that is m k = Z − f ( t )( G ( t )) k dt = 0 , k = 0 , , , . . . , where G ( t ) = R t − g ( s ) ds = n ( t n − We can write the polynomial f as f ( t ) = a + a t + a t + a t + a t + a t . By Theorem 1.6 m k = Z − f ( t )( G ( t )) k dt = 0 , k = 0 , , G ( t ) = R t − g ( s ) ds = n ( t n − Z − f ( t ) dt = 0 , n Z − f ( t )( t n − dt = 0and 1 n Z − f ( t )( t n − dt = 0 . t n − Z − [ a + a t + a t ] dt = 0 , n Z − [ a + a t + a t ]( t n − dt = 0and 1 n Z − [ a + a t + a t ]( t n − dt = 0 . By solving the integrals, we obtain the following system(34) a + a + a = 0 − n n a − n n ) a − n n ) a = 0 n (1+2 n )(1+ n ) a + n n )(3+ n ) a + n n )(5+ n ) a = 0 , or equivalently(35)
13 1511+ n n ) 15(5+ n )1(1+2 n )(1+ n ) 13(3+2 n )(3+ n ) 15(5+2 n )(5+ n ) a a a = . Notice that the matrix associate M =
13 1511+ n n ) 15(5+ n )1(1+2 n )(1+ n ) 13(3+2 n )(3+ n ) 15(5+2 n )(5+ n ) is nonsingular, with det ( M ) = − n +1)( n +3)( n +5)(1+2 n )(3+2 n )(5+2 n ) . Then, the system (35) hasonly the trivial solution, namely, a = a = a = 0. Hence, f ( · ) has only odd powers of t andthis finishes the proof of Corollary 1.9. References [1]
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