Algorithmic derivation of isochronicity conditions J.M. Hill,N.G. Lloyd,J.M. Pearson? Institute of Mathematical and Physical Sciences, University of Wales Aberystwyth UK, SY23 3BZ Abstract This paper is concerned with obtaining the conditions under which the origin is an isochronous centre for certain planar polynomial systems. The necessity of these conditions is proved using a computer implementation of an algorithm first proposed by I.I.Pleshkan. Their sufficiency is then proved using various methods. Key words: differential equations, isochronous centres 1991 MSC: 34C07, 37G15 1 Introduction We consider the planar differential system ?x = P(x,y), ?y = Q(x,y), (1) where P, Q are polynomials with real coefficients. We also use the complex form of (1) i?z = Z(z,?z), where z = x+iy and ?z is the complex conjugate of z. Critical points of the system occur where P = Q = 0. A critical point is a centre if, within some neighbourhood, all orbits surrounding it are closed; a centre is non-degenerate if the linearised vector field has two non-zero eigenvalues. A centre is said to be isochronous if all the orbits surrounding it have the same period and it was shown in [20] that an isochronous centre is non-degenerate. ? Corresponding author. Tel: +44 1970 622002; fax: +44 1970 611446. Email address: jmp@aber.ac.uk (J.M. Pearson). Preprint submitted to Elsevier 27 February 2007 The question of isochronicity was first discussed by Huygens in Horologium Oscillatorium sive de motu pondulorum (1673). Huygens showed that the cycloidal pendulum has a motion whose period (and hence frequency) is independent of the amplitude. Thus the cycloidal pendulum has isochronous oscillations, in contrast to the simple pendulum in which the period of oscillation increases with amplitude. The simplest form of the equation of the cycloidal pendulum is given in terms of arc length, s, as the independent variable ?s+w2s = 0, (2) where w is a real constant. This is a special case of a class of second order differential equations ?s+g(s) = 0, (3) which arise in the study of conservative mechanical systems. Urabe [29] proved that if g(s) is an odd function then (3) can only have an isochronous centre when it is system (2). Many examples of isochronous systems exist in applications. Needham [21] discussed one such system that arises in the telecommunications industry. Atmospheric distortion can affect the transmission of high-speed digital signals and is removed by introducing tapped delay devices, which act as adaptive equalisers. The differential system modelling this situation has the property that small changes in the initial conditions can alter the amplitude, but not the frequency, of an oscillation. The existence of isochronous centres for several classes of differential systems has been studied - see the survey paper [4] and references therein, for example. In particular, Loud [19] gave the results for quadratic systems, cubic systems with homogeneous nonlinearities were considered by Pleshkan [24] and results for other cubic systems can be found in [26], [3] , [18] for example. Systems of higher degree have received less attention; there are results for quartic and quintic systems with homogeneous nonlinearities [1], [2] and for Li?enard systems [7]. It was shown in [10] that some systems can have coexisting isochronous and non-isochronous centres. We shall present results for particular classes of cubic and quartic systems. Any isochronous family of periodic orbits surrounds a unique non-degenerate critical point of centre type. So, without loss of generality, we can assume that the origin is the critical point of interest and transform system (1) to ?x = y+p(x,y), ?y = ?x+q(x,y), (4) where p, q are polynomials without constant or linear terms. Every system can be transformed to a linear system at a non-singular point, but this is not true in general at singular (critical) points. A key property of isochronous systems, which was proved in [20], is that they can be transformed to a 2 linear system even at critical points. Centres of linear systems are clearly isochronous, which leads to the following result. Theorem 1 A centre of system (4) is isochronous if and only if there exists an analytic change of coordinates of the form u(x,y) = x+o(|(x,y)|), v(x,y) = y+o(|(x,y)|), which reduces the system to ?u = v, ?v = ?u. Proof. See for example [20]. The equivalent linear system in complex notation is i?z = z. Linearisation of differential systems forms the basis of several methods for the determination of the conditions under which a critical point is an isochronous centre. Pleshkan presented an algorithm in [24] which uses the complex form of (4). We describe this algorithm in the next section. Others have used the polar form of (4), see for example [9], or the real form as in [6]. Commuting systems have also been used [28] and in some instances this is a simpler approach than searching for a linearisation. However, not every polynomial system with an isochronous centre has a polynomial commutator. This would require that every centre in the plane was isochronous and Devlin [10] showed that isochronous and non-isochronous centres can coexist. Another approach is to consider the period function; if T(x) is the period of the solution through the point (x,0) then T can be written as T(x) = 2pi+?1x+?2x2 +??? . The origin is an isochronous centre if the period constants ?i = 0, for all i. See [11] for an example of this approach, or an alternative method for calculating period constants as given in [16]. Although Pleshkan used his algorithm to find necessary and sufficient conditions for the origin to be an isochronous centre, we prefer to use the algorithm to show necessity only. We then prove sufficiency using various methods, some of which we describe in section 3. Using this approach avoids the need to find a basis for a set of very large multivariate polynomials and also serves as a check on the validity of the computer assisted calculation of the necessary conditions. In section 4 we demonstrate the use of our implementation of Pleshkan?s algorithm by confirming the known results for the homogeneous cubic system i?z = z+A30z3 +A21z2?z+A12z?z+A03?z3, (5) 3 where the Aij are complex coefficients. We have also used the algorithm to confirm the known results for the system ?x = y, (6) ?y = ?x+a1x2 +a2xy+a3y2 +a4x3 +a5x2y+a6xy2 +a7y3, where the ai, 1 ?i? 7 are arbitrary real coefficients. The conditions for the origin to be an isochronous centre for (6), which is often referred to as the Kukles system, were established in [20] and [6]. We present two examples for which the conditions for isochronicity were previously unknown. In section 5, we consider ?x = y(1 +x), (7) ?y = ?x?a1x2 ?a2x3 ?a3x4 ?a4(x+a5x2)y?a6y2, where the ai,1 ?i? 6 are arbitrary real coefficients. System (7) was investigated by Cherkas in [5]. This example belongs to a class of systems of the form ?x = p0(x) +p1(x)y, ?y = q0(x) +q1(x)y+q2(x)y2. These systems can be transformed to Li?enard form using the transformation (x,y,t) ? (x,(p0(x) +p1(x)y)?(x),?), where dt d? = ?(x) = 1 p1(x) exp parenleftBigg ? integraldisplay x 0 q2(s) p1(s)ds parenrightBigg . Such a transformation can be used to show the sufficiency of certain centre conditions, see [8]. However, since it is a spatially-dependent transformation of time, it does not preserve isochronicity and cannot be used in this context. Finally, in section 6, we consider the system ?x = y(1 +kx), (8) ?y = ?x+a1x2 +a2xy+a3y2 +a4x3 +a5x2y+a6xy2 +a7y3. This is an extension of the Kukles system in that the term kxy is added to the ?x equation, whilst the ?y equation is unchanged. A specific example of system (8) can be used to model predator-prey populations with the inclusion of intratrophic predation, see [13]. We note that (7) and (8) have an invariant line parallel to the y-axis. 4 2 Pleshkan?s Algorithm In [24], Pleshkan presented an algorithm for deriving conditions under which a critical point of a polynomial system is an isochronous centre. His approach is to find the conditions under which the system can be linearised. We shall restate the algorithm, and we begin by writing (4) in complex form: i?z = z+summationdisplay r?2 ?r(z,?z) (9) where ?r(z,?z) = rsummationdisplay j=0 cr?j,jzr?j?zj, z = x+iy, the cr?j,j are complex-valued and a bar denotes complex conjugation. We now seek a transformation ? = z+summationdisplay r?2 ?r(z,?z), (10) where ?r(z,?z) = rsummationdisplay j=0 dr?j,jzr?j?zj and the dr?j,j are complex-valued, which reduces (9) to i?? = ?. (11) Substituting (10) into (11), we find summationdisplay k?2 parenleftBigg z??k?z ? ?z??k??z + ?k parenrightBigg +summationdisplay k?2 ? ?summationdisplay r?2 ??k ?z ?r ? ??k ??z ??r ? ? = summationdisplay k?2 ?k. (12) Considering terms of degree k in (12), we have ?k ?z??k?z + ?z??k??z = ?k + k?1summationdisplay r=2 parenleftBigg?? r ?z ?k?r+1 ? ??r ??z ??k?r+1 parenrightBigg . (13) Equating the coefficients of zk?j?zj in (13), we have k?1summationdisplay r=2 rsummationdisplay p=0 (r?p)ck?j+p?r+1,j?pdr?p,p ??ck?p?j,j+p?r+1dp,r?p +ck?j,j = (2j + 1?k)dk?j,j, (14) where terms with negative subscripts are zero. It follows from the above that if (9) is to have an isochronous centre at the origin, it is necessary and 5 sufficient that the equations given by (14) are consistent. The coefficients dj+1,j can be chosen arbitrarily; we take dj+1,j = 0 (j ? 1). Putting k = 2j + 1 in (14), we obtain a set of conditions for the consistency of (14), namely ?j = 0 (j ? 1), where ?j ?cj+1,j + 2jsummationdisplay r=2 rsummationdisplay p=0 (r?p)cj+p?r+2,j?pdr?p,p ??cj?p+1,j+p?r+1dp,r?p. We call the ?j the Pleshkan polynomials. Pleshkan showed that the conditions ?j = 0 (j ? 1) were necessary and sufficient for the origin to be an isochronous centre. There are infinitely many conditions but, by the Hilbert Basis Theorem, there is M such that ?j = 0 for j ?M implies that ?j = 0 for all j. Thus calculating all the conditions for isochronicity is a terminating process, though M is not known a priori, and is equivalent to finding a basis for the Pleshkan polynomials for a given system. However the polynomials grow rapidly as j increases and finding the appropriate basis is non-trivial computationally. We prefer to calculate the first few polynomials, reducing each one modulo the preceding polynomials, until we have ?m negationslash= 0, when ?j = 0 for j = 1,...,m?1 except under certain conditions. The sufficiency of these conditions is then proved separately using a variety of methods. This approach has the added advantage of being an independent check on the computer assisted calculation of the Pleshkan polynomials and the subsequent derivation of the necessary conditions. We have implemented the algorithm in REDUCE. This choice of computer algebra system was primarily due to the fact that the suite of programs developed by the Aberystwyth group is written in REDUCE. A copy of the code is available on request from the authors. It can also be found in [12]. Pleshkan proved that if the origin is a centre then the conditions Re(?j) = 0 (j ? 1), where Re denotes the real part, are necessary and sufficient for the origin to be an isochronous centre. Introducing the centre conditions, if known, can simplify the calculations but often leads to several cases having to be considered. For example, the derivation of the conditions for the origin to be an isochronous centre for system (5) is relatively straightforward without introducing the centre conditions. For some cases it is possible to calculate the conditions for isochronicity where derivation of the centre conditions proves intractable. Next we describe some methods we use to prove the sufficiency of conditions obtained through the use of this algorithm. 6 3 Sufficiency As noted previously, the vanishing of all the Pleshkan polynomials is necessary and sufficient for isochronicity, but in practice only a finite number can be calculated. In our approach, the sufficiency of the conditions we find must be proved separately. This is achieved using different methods, and we summarise some of these in this section. In many cases one of the first two theorems below is used. For some systems the transformation to the linear isochrone, which is known to exist, can be found and for others we make use of Urabe?s result [29]. Explicit solutions of particular systems can sometimes be found, see [19]. Theorem 2 Suppose that system (4) is holomorphic, that is i?z = z+f (z) where f (z) is an analytic function without linear terms (equivalently, that px = qy, py = ?qx). Then the origin is an isochronous centre. Proof. The origin is a centre, as the system is Hamiltonian. The period of a trajectory ? is given by T(?) = vextendsinglevextendsingle vextendsinglevextendsingle integraldisplay ? dt vextendsinglevextendsingle vextendsinglevextendsingle = vextendsinglevextendsingle vextendsinglevextendsingle vextendsingle integraldisplay ? i z+f(z)dz vextendsinglevextendsingle vextendsinglevextendsingle vextendsingle = 2piResidue parenleftBigg 1 z+f(z),0 parenrightBigg = 2pi, and so the origin is an isochronous centre. Theorem 3 Let x = rcos?, y = rsin?. Then system (4) has an isochronous centre at the origin if ?? = ?1. Proof. The result is immediate. When choosing transformations to linearise systems, we must find a first integral of the right form. This is not a straightforward process, and the first integrals are not unique. First integrals can be found in numerous ways; for example using invariant functions and integrating factors. In this context, a function C is invariant with respect to system (4) if ?C = Cx ?x+Cy ?y = LC, where the degree of the polynomial L (known as the cofactor of C) is at most n?1, where n is the degree of the system. The first stage is to search for invariant functions, products of which form an integrating factor, see [22] for examples of this technique. The integrating factor is then used to find a first integral, using the result of Lemma 4 below. A transformation is sought such that the first integral is of the form H(x,y) = X2 +Y2. Then system (4) becomes ?X = Y, ?Y = ?X. 7 Lemma 4 Suppose that B(x,y) is an integrating factor of system (1). Let H(x,y) = B(x,y)P(x,y)dy+f(x), (15) where f(x) is such that Hx(x,y) = ?B(x,y)Q(x,y). Then H(x,y) is a first integral of system (1). Proof. By the definition of H(x,y) in (15), we have ?H = ?BQP + parenleftbiggintegraldisplay BPdy+f(x) parenrightbigg y Q = ?BQP +BPQ = 0, and so H(x,y) is a first integral of system (1). In some cases, one may be able to use more specialised results such as the following, which is attributed to Urabe. Consider the equation ?x+g(x) = 0, which is clearly equivalent to the first order planar system ?x = y, ?y = ?g(x). (16) If xg(x) > 0 for xnegationslash= 0, then 12y2 +integraltextx0 g(u)du = constant is a family of closed orbits surrounding the origin. Hence the origin is a centre of system (16). Theorem 5 (Urabe) Define X by 1 2X 2 = integraldisplay x 0 g(u)du. If the relationship above can be inverted, with x and X having the same sign, and we obtain x = ?X +h(X), where ? is a constant and h(X) is an even function of X, then the origin is an isochronous centre of system (16). Proof. See [29] or [25]. More recently other methods for establishing sufficiency have been considered in [27]. However, these results are not required for the systems investigated in this paper. 4 Homogeneous Cubic Systems We have tested our implementation of Pleshkan?s algorithm on systems whose isochronicity conditions are already known. We have successfully 8 replicated the results for both quadratic and homogeneous cubic systems (see [19] and [24] respectively) amongst others. To illustrate our approach, we shall present the details of our investigation of the homogeneous cubic system (5). As the reduction of the Pleshkan polynomials is straightforward, in this example, we do not introduce the known conditions for the origin to be a centre. In [24], Pleshkan proved that the origin is an isochronous centre of system (5) if and only if one of the following holds: (1) A21 = A12 = A03 = 0; (2) A21 = A03 = 0, A12 = ? ?A30; (3) A21 = 0, 3 ?A30+7A12 = 0, 9A03 ?A03?16A12 ?A12 = 0, 9A203 ?A12?16A312 = 0. We use our algorithm to compute the first six Pleshkan polynomials; we find ?1 = A21. Let A21 = 0, then ?2 = ?3A03 ?A03 ?4A12 ?A12 ?4A30A12. If A12 = 0, then ?2 = A03 ?A03 is a factor of ?j, j = 3,...,6. With A21 = A12 = A03 = 0 the system becomes i?z = z+A30z3 and the origin is an isochronous centre by Theorem 2; this is condition (1) above. Assume that A12 is non-zero and let A30 = ?3A03 ?A03 ?4A12 ?A12 4A12 . Then ?3 = ?A03(8A12(?9A203 ?A12 + 24A312 + 8A212 ?A30) + 27A303 ?A03). We first consider A03 = 0. We find that ?4 = ?A212 ?A212(A12 + ?A30) is a factor of ?6 and that ?5 = 0. When A21 = A03 = 0, A12 = ? ?A30 = ?a+ib, then, in polar coordinates, system (5) becomes ?r = 2r3(asin(2?) +bcos(2?)), ?? = ?1 and the origin is an isochronous centre by Theorem 3; this is condition (2) above. Assume now that A12A03 is non-zero and let ?A03 = 8A12(9A203 ?A12 ?24A312 ?8A212 ?A30) 27A303 . 9 Let R(A, B, c) denote the resultant of A and B with respect to c, and # represent a (large) integer. Then ?4 = ?5 = 0 only if R(?4,?5, ?A12) = #A1212A1203(3A12 + ?A30)4(7A12 + 3 ?A30)?1 = 0, where ?1 is a homogeneous polynomial of degree six in A12 and ?A30. If ?A30 = ?3A12, then ?4 = A12 ?A212A403 negationslash= 0 and similarly if ?1 = 0, then we would require A12 = 0 in order to achieve ?4 = ?5 = ?6 = 0. When ?A30 = ?73A12, we find that ?4 = A12?(243A203 ?A12 ?64A312), ?5 = A12??2 and ?6 = A12??3, where ? = 9A203 ?A12 ?16A312 and ?2, ?3 are homogeneous polynomials of degree six in A03, A12 and ?A12. If 243A203 ?A12 ?64A312 = 0, then ?2 = ?3 = 0 only if A12 = 0. Let A12 = R1ei?1 and w = ze?12?1, then when A21 = 0, ?A30 = ?73A12, 9A03 ?A03 ?16A12 ?A12 = 0, 9A203 ?A12 ?16A312 = 0 the system becomes i ?w = w? 73R1w3 +R1w?w2 + 43R1 ?w3. We note that such a transformation preserves isochronicity. The real form of the transformed system is ?x = y parenleftbigg 1?12R1x2 + 83R1y2 parenrightbigg , ?y = ?x(1 + 4R1y2), which admits C = 1 + 4R1y2 = 0 as an invariant curve, with cofactor L = ?8R1xy. Hence we find B(x,y) = C?4 is an integrating factor and H(x,y) = (x2 +y2(1 + 83R1y2)2)C?3 is a first integral. Let X = xC?32, Y = y(1 + 83R1y2)C?32, such that H(x,y) = X2 +Y2. Then ?X = Y, ?Y = ?X and the system is transformed to the linear isochrone. Since this spatial transformation preserves isochronicity, the origin is an isochronous centre of the original system when condition (3) above holds. The ability to replicate the known results for systems such as (5), (6) and others gives confidence in the integrity of our algorithm and reduction process. In the following two sections we determine conditions for the origin to be an isochronous centre for two systems for which such results were previously unknown. 10 5 The Cherkas System We now derive necessary and sufficient conditions for the origin to be an isochronous centre for system (7) with at least one of the coefficients, ai non-zero. System (7) was originally investigated by Cherkas [5], but it was later found that the centre conditions obtained were incomplete. The complete set of necessary and sufficient centre conditions is stated in [17] and [8]. Theorem 6 The origin is a centre for system (7) if and only if one of the following conditions hold: (i) a1 = a5?a6,a3 = a5(a2+a5a6), a5a6(2a5?2a6?3)+a2(2a5?a6?2) = 0; (ii) a4 = 0; (iii) a5 = 0,a6 = 1,a1 = ?1,a3 = ?a2; (iv) a5 = 1,a6 = 1,a1 = 0,a3 = 23a2; (v) a5 = 13,a6 = ?23,a1 = 1,a3 = 19a2; (vi) a5 = 23,a6 = ?43,a1 = 2,a3 = 89(a2 ?1); (vii) a5 = 0,a6 = ?3,a1 = 3,a3 = 13a2; (viii) a5 = 1,a6 = ?3,a1 = 4,a3 = 2a2 ?8. Proof. See [8]. We shall proceed by analysing each of the eight centre conditions separately, first obtaining necessary conditions and then turning to the question of sufficiency. We note that knowledge of the centre conditions is not essential, but their inclusion eases the computation of the Pleshkan polynomials and their subsequent reduction. Theorem 7 The origin is an isochronous centre for system (7) if and only if one of the following conditions holds: (1) a1 = 12,a2 = a3 = a4 = 0,a6 = ?12; (2) a1 = a2 = a3 = a4 = 0,a6 = ?1; (3) a1 = a2 = a3 = a4 = 0,a6 = ?14; (4) a1 = 12,a2 = a3 = a4 = 0,a6 = ?2; (5) a1 = 2,a2 = 1,a3 = a4 = 0,a6 = ?2; (6) a1 = 1,a2 = 13,a3 = a4 = 0,a6 = ?3; (7) a1 = 32,a2 = 1,a3 = 14,a4 = 0,a6 = ?4; (8) a1 = 32,a2 = 1,a3 = 14,a4 = 0,a6 = ?1; (9) a1 = 1,a2 = 13,a3 = a4 = 0,a6 = ?34; (10) a1 = 2,a2 = 19(a24 + 9),a3 = a5 = 0,a6 = ?2. Proof. The necessity and sufficiency of these conditions are proved in the following two sub-sections. 11 5.1 Necessary conditions for Isochronicity Lemma 8 If any one of the conditions (iii) to (viii) of Theorem 6 holds, then the origin cannot be an isochronous centre. Proof. In each case, if the first Pleshkan polynomial is zero, then the second must be non-zero, hence the origin cannot be an isochronous centre. Lemma 9 Suppose that condition (ii) of Theorem 6 holds. Then the origin is an isochronous centre for system (7) only if one of the conditions (1) to (9) of Theorem 7 holds. Proof. We calculate Pleshkan polynomials, ?i, for system (7) with a4 = 0. We find ?1 = 10a21 + 10a1a6 +a1 + 4a26 + 5a6 + 1?9a2. Let a2 = 19(10a21 + 10a1a6 +a1 + 4a26 + 5a6 + 1) then ?2 = ??27a3(14a1 + 12a6 ?3), where ? = 280a41 + 770a31a6 ?70a31 + 804a21a26 ?15a21a6 + 9a21 + 362a1a36 + 12a1a26 ?108a1a6 ?28a1 + 52a46 ?35a36 ?120a26 ?35a6 ?2. When 14a1 + 12a6 = 3 then ?2 = ?, a polynomial of degree 4 in a6, and when ? = 0, ?3 = ?4 = 0 is not possible. So we assume that (14a1 + 12a6 ?3)?negationslash= 0 and let a3 = ?/(27(14a1 + 12a6 ?3)). Now ?3 is a non-homogeneous polynomial of degree 8 in a1,a6, ?4 is of degree 10 and ?5 is of degree 13. Let T1 = R(?3,?4,a1) and T2 = R(?4,?5,a1). Then T1 = #(a6 + 1)3(a6 + 2)3(a6 + 3)(a6 + 4)(2a6 + 1)6(4a6 + 3)(4a6 + 1)?4?1, (17) T2 = #(a6 + 1)3(a6 + 2)3(a6 + 3)(a6 + 4)(2a6 + 1)6(4a6 + 3)(4a6 + 1)?6?2, where ?1, ?2 are polynomials in a6 of degrees 48 and 64 respectively. Considering each linear factor of the right hand side of (17) we find that the first five Pleshkan polynomials are zero, when a4 = 0, only if one of the conditions (1) to (9) of Theorem 7 holds. Lemma 10 Suppose that condition (i) of Theorem 6 holds and a4 negationslash= 0. Then the origin is an isochronous centre for system (7) only if condition (10) of Theorem 7 holds. Proof. We have a1 = a5 ?a6. Assume for the time being that (2?2a5 +a6) negationslash= 0 and let a2 = a5a6(2a5 ?2a6 ?3)(2?2a5 +a6)?1. Now 12 a3 = a25a6(1 +a6)(2a5 ?a6 ?2)?1 and we compute that ?1 = ??(2a5 ?a6 ?2)a24, where ? = 2 + 9a6 + 18a25 + 12a26 ?20a35 + 12a25a6 + 4a36. Let a24 = ?(2a5 ?a6 ?2)?1 > 0. We calculate that ?2 = ?(2a5 ?1)a25?1 and ?3 = (2a5 ?1)a25 (?2 + 315a4a5?1(2a5 ?a6 ?2)i), where ?1 = 10a25 ?10a5 ?2a26 ?3a6 and ?2 =?26200a55 + 28600a45a6 + 45000a45 + 9620a35a26 ?34910a35a6 ?11260a35 ?5436a25a36 ?8762a25a26 + 24692a25a6 ?7644a25 ?972a5a46 + 1524a5a36 ?5299a5a26 ?18496a5a6 + 68a5 ?1656a46 ?5772a36 ?4872a26 + 114a6 + 36. If a5(2a5 ?1) = 0 then we must have a4 = 0, which is contrary to assumptions. We calculate the resultant R(?1,?2,a6) = #a5(a5 + 1)(a5 ?1)2(2a25 ?2a5 ?1)?3, where ?3 = 21610a45 ?4470a35 ?5640a25 + 2107a5 ?153. When a5 = 1, we have a6 = 0 and if a5 = ?1 then a6 = ?4. In both cases 2a5 ?a6 = 2, which is contrary to assumptions. When 2a25 ?2a5 = 1, then ?1 = ?(2a6 + 5)(a6 ?1). If a6 = 1, then a24 < 0 and if a6 = ?52 then ?2 negationslash= 0. The factor of R(?1,?2,a5) which corresponds to ?3 is ?4 = 17288a46 + 4548a36 ?18282a26 ?34597a6 ?11532. Both ?3 = 0 and ?4 = 0 have two real roots; however when a5 and a6 take any of the four possible pairs of values then a24 < 0. We now return to the one excluded case. Let a6 = 2(a5 ?1). To satisfy condition (i) of Theorem 6, we require ?2a5(a5 ?1)(2a5 ?1) = 0. Let a5 = 0, then ?1 = 9a2 ?a24 ?9 and ?1 is a factor of ?j, j = 2,...,6. Hence a1 = 2,a2 = 19(a24 + 9),a3 = a5 = 0,a6 = ?2, 13 is a necessary condition for the origin to be an isochronous centre for system (7) and this is condition (10) of Theorem 7. When a5 = 1, then ?1 = 9a2 ?a24 ?12. If a2 = 19(a24 + 12), then ?2 = ?(2a24 + 69) negationslash= 0, and so the origin cannot be an isochronous centre. Finally, if a5 = 12, then ?1 = 9a2 ?a24 ?9. Let a2 = 19(a24 + 9), then ?2 = ?4a24, which is non-zero under current assumptions. This completes the proof of the necessity of the conditions in Theorem 7. 5.2 Sufficient Conditions for Isochronicity We now show, using various methods, that each of the necessary conditions listed in Theorem 7 is indeed also sufficient. We note that the sufficiency of all the conditions for the origin to be a centre in Theorem 6 can be proved under transformation to a system of Li?enard form and subsequent symmetry arguments. However, such a transformation cannot be used to prove the sufficiency of the conditions for isochronicity, as it uses a space-dependent transformation of time which destroys isochronicity. Lemma 11 If any one of conditions (1) to (10) stated in Theorem 7 holds, then the origin is an isochronous centre of system (7). Proof. System (7) reduces to a quadratic system when any one of the conditions (1) to (4) of Theorem 7 holds. The conditions for isochronicity in such cases were resolved by Loud in [19]. In particular, system (7) can be transformed to Loud?s systems S1, S2, S3 and S4 respectively when conditions (1) to (4) hold. We prove the sufficiency of the isochronicity conditions (5) to (9) of Theorem 7 by transforming the system to a second order differential equation and then applying Urabe?s Result. We need to transform system (7) to an equation of the form ?x+g(x) = 0. When any one of the conditions (5) to (9) holds, system (7) can be written in the form ?x = y(x+ 1), ?y = f(x+ 1) +my2, where f is a function and m is a constant. Without loss of generality we can restrict our attention to x>?1. We have y = ?x/(x+ 1), hence ?x = (x+ 1)f(x+ 1) + (m+ 1) (?x) 2 x+ 1. 14 Let x+ 1 = (w+ 1)? 1m. We take the positive root for w, so that x = 0 if and only if w = 0. We have ?w+m(w+ 1)f parenleftBig (w+ 1)? 1m parenrightBig = 0, which is a system of the required form. According to Urabe?s result, if we define X2 = 2m integraldisplay w 0 (z+ 1)f((z+ 1)? 1m)dz, where w has the same sign as X, then the origin is an isochronous centre of system (7) if and only if w = ?X +h(X), where ? is a constant and h(X) is an even function of X. We present the details of this transformation for condition (5) of Theorem 7. In this case f (x+ 1) = (1?(x+ 1))(x+ 1)2 and m = 2. Therefore, we define X2 = 4 integraldisplay w 0 (z+ 1) parenleftBig 1?(z+ 1)?12 parenrightBig (z+ 1)?1dz = 4 parenleftBig (w+ 1)12 ?1 parenrightBig2 . We must ensure that w has the sign of X, so w = X + 14X2 and hence the origin is an isochronous centre of system (7) by Urabe?s result. Similarly, for conditions (6) and (7), we find h(X) = 0, and for conditions (8) and (9) we have h(X) = ?1 +?X2 + 1. We find that ? = 1 in all cases. We conclude that the origin is an isochronous centre of system (7) when any one of the conditions (5) to (9) of Theorem 7 holds. We now turn our attention to the one remaining condition in Theorem 7. We note that this occurs when centre condition (i) of Theorem 6 holds, whereas the other nine conditions occur when centre condition (ii) holds. When condition (10) of Theorem 7 holds, system (7) has an invariant curve C = 9(x+ 1)2 +a24x2 + 3a4y = 0, with cofactor L = 2y? 13a4x. The line x = ?1 is also invariant with respect to system (7). We find that these two invariant curves can be used to create an integrating factor, B(x,y) = 729(x+ 1)C?3. We note that C negationslash= 0 in a neighbourhood of the origin. By Lemma 4, a first integral H(x,y) = 12(81x2(x+ 1)2 + 9(a4x2 + 3y)2)C?2, can be constructed from this integrating factor. Let X = 9x(x+ 1)(?2C)?1 and Y = 3(a4x2 + 3y)(?2C)?1, so that H(x,y) = X2 +Y2. Then system (7) becomes ?X = (x+ 1)Y, ?Y = ?(x+ 1)X, 15 or in polar coordinates, ?r = 0, ?? = ?(x+ 1). (18) Simple manipulation gives x+ 1 = 3? ?2a 4Y 3??2a4Y ?3?2X. We note that if 3??2a4Y ?3?2X = 0, then we must also have x = ?1, which is not true in a neighbourhood of the origin. Therefore we may assume that the denominator is non-zero. The orbits of system (18) are given by r = ? (constant), which are the level curves of H(x,y). Thus, on an orbit X2 +Y2 = ?2 and ?? = ? parenleftBigg 3??2a 4?sin? 3??2a4?sin??3?2?cos? parenrightBigg . The period of this orbit is given by T(?) = integraldisplay 2pi 0 3??2a4?sin??3?2?cos? 3??2a4?sin? d? = 2pi. Thus, the origin is an isochronous centre of system (7). This completes the proof of Theorem 7. Remark If the origin is an isochronous centre of system (7), then at most one other critical point of focus type exists, that is (?2,0). If this point is of focus type, it is an isochronous centre and the system is quadratic. 6 The Extended Kukles System We now turn our attention to system (8). We consider here the conditions for the origin to be an isochronous centre without prior knowledge of the centre conditions, which are not known in general. Theorem 12 The origin is an isochronous centre for system (8) if one of the following conditions holds: (1) a1 = a2 = 0, a3 = k, a4 = a5 = a6 = a7 = 0; (2) a1 = a2 = 0, a3 = k4, a4 = a5 = a6 = a7 = 0; (3) a1 = ?k2, a2 = 0, a3 = 2k, a4 = a5 = a6 = a7 = 0; (4) a1 = ?k2, a2 = 0, a3 = k2, a4 = a5 = a6 = a7 = 0; 16 (5) a1 = ?k, a2 = 0, a3 = 3k4 , a4 = ?k23 , a5 = a6 = a7 = 0; (6) a1 = ?k, a2 = 0, a3 = 3k, a4 = ?k23 , a5 = a6 = a7 = 0; (7) a1 = ?2k, a3 = 2k, a4 = ?19(a22 + 9k2), a5 = a6 = a7 = 0. Proof. System (8) with k = 0 is the Kukles system, for which the conditions for the origin to be an isochronous centre are known. Therefore, without loss of generality, we can scale x, y by k negationslash= 0 (noting that such a scaling does not destroy isochronicity). Hence we consider (8) with k = 1. Conditions (1) to (4) give the quadratic systems covered by Loud, so in each case the origin is an isochronous centre. Under condition (5) the scaled system becomes system (7) when condition (9) of Theorem 7 holds, for which sufficiency is proved using Urabe?s result. Similarly, when condition (6) holds, system (8) is equivalent to system (7) when condition (6) of Theorem 7 holds. Again Urabe?s result is used to prove sufficiency. Condition (7) leads to a special case of system (7) when condition (10) of Theorem 7 holds; in this case a first integral was obtained and all the orbits surrounding the origin were shown to have a period of 2pi, hence the origin is an isochronous centre. We now consider whether or not the conditions contained in Theorem 12 are necessary, as well as sufficient, for the origin to be an isochronous centre for system (8). We calculate the first five Pleshkan polynomials for system (8) with k = 1. In this example we do not introduce conditions for the origin to be a centre, so the Pleshkan polynomials do not become real-valued upon reduction. We denote the real and imaginary parts of each polynomial by Re(?) and Im(?) respectively. We find Im(?1) = a2(a1 +a3)+a5 +3a7. Let a7 = ?(a2(a1 +a3)+a5)/3 then Re(?1) = ?9a4 +?, where ? = ?10a21 ?10a1a3 +a1 ?a22 ?4a23 + 5a3 ?3a6 ?1. Let a4 = ?/9. Then Im(?2) = 6a2(a3 ?a1 ?1)a6 + ?, where ? =?35a31a2 ?78a21a2a3 + 15a21a2 + 15a21a5 ?2a1a32 ?75a1a2a23 + 54a1a2a3 ?6a1a2 + 42a1a3a5 + 30a1a5 ?a32a3 ?2a32 + 3a22a5 ?28a2a33 + 27a2a23 + 3a2a3 ?2a2 + 15a23a5 + 42a3a5 ?3a5. Assume for the time being that a2(a3 ?a1 ?1) negationslash= 0 and let a6 = ?/(6a2(a1 ?a3 + 1)). As no variable occurs linearly in the remaining Pleshkan polynomials we use resultant calculations to eliminate further variables. Some of the resultants we need to compute involve multivariate polynomials of high degree and cannot be obtained using the procedures available in Computer Algebra systems such as REDUCE and Maple. We use procedures we have developed for this purpose, see [23]; these exploit the 17 possibility of removing factors of a resultant as they arise during its calculation. We calculate R(Re(?2),Im(?3),a5) = #a62(a1 ?a3 + 1)4S1T R(Re(?2),Re(?3),a5) = #a62(a1 ?a3 + 1)6S2T R(Re(?2),Im(?4),a5) = #a62(a1 ?a3 + 1)6S3. Now T = a42 + 2?a22 +?2, where ? = 7a21+10a1a3+14a1+7a23+10a3+1, ? = 5a21+14a1a3+10a1+5a23+14a3?1. Clearly with a2 negationslash= 0, we cannot have T = 0 unless ? < 0. For a22 to be a positive real root of T = 0 we require a1 +a3 ? 0 or a1 +a3 ? ?2. However if a1,a3 satisfy either of these conditions then ? > 0. We conclude that T negationslash= 0, under current assumptions. We investigate the possibility that S1 = S2 = S3 = 0. These are polynomials in a1,a2,a3 with a2 occurring to even powers only. We calculate R(S1,S2,a22) = #m29(m+ 2)(4m+a1 + 2)?4?1?2, R(S1,S3,a22) = #m35(m+ 2)2(4m+a1 + 2)21?4?3?4, where m = a1 +a3 and ?1 is degree 69 in a1, 93 in m; ?2 is degree 32 in a1, 46 in m; ?3 is degree 94 in a1, 127 in m; ?4 is degree 40 in a1, 58 in m. We cannot calculate the resultants of ?1 (or ?2), ?3 (or ?4) with respect to either m or a1 as the expressions generated exceed the space available on the Compaq Alpha XP1000 workstation, with 1Gb of memory, that we use. We have investigated the possibility of eliminating two variables simultaneously from S1,S2,S3 using an implementation of the Kapur-Saxena-Yang variant of the Dixon resultant due to Lewis [15]. This approach yields some simple factors and one, not necessarily irreducible, factor of degree 2069 in a1. The simple factors confirm the results we have obtained, the other factor remains to be considered further. We conjecture that there are no conditions for the origin to be an isochronous centre, with ?1?2 = ?3?4 = 0, that are not covered elsewhere. When (m+ 2)? = 0, all possible isochronous centres have a2 = 0. Consider m = a1 +a3 = 0; either a2 = 0, which is contrary to current assumptions, or a1 = ?2, a3 = 2, a4 = ?(a 2 2 + 9) 9 , a5 = a6 = a7 = 0 (19) is a necessary condition for the origin to be an isochronous centre. If 4m+a1 +2 = 0 then the origin can only be an isochronous centre when (19) holds. 18 Next we consider the situation with a2 = 0. If a5 = ?3a7 then Im(?1) = 0 and a7 is a factor of all the imaginary parts of the calculated Pleshkan polynomials. Consider first a7 = 0, with a4 = ?/9. We calculate the following resultants to eliminate a6; R1 = R(Re(?2),Re(?3),a6), R2 = R(Re(?2),Re(?4),a6) and R3 = R(Re(?2),Re(?5),a6) where R1,R2,R3 are polynomials of degrees 12,20,25 respectively in a1,a3. Further resultant calculations with respect to a1 give: R(R1,R2,a1) = #(4a3 ?3)(a3 ?3)(4a3 ?1)(a3 ?1)(a3 ?2)(2a3 ?1)?5?6, R(R1,R3,a1) = #(4a3 ?3)(a3 ?3)(4a3 ?1)3(a3 ?1)(a3 ?2)(2a3 ?1)?7?8, where ?5, ?6, ?7, ?8 are irreducible polynomials of degrees 72, 86, 96, 110 respectively in a3 and multiplicity of the linear factors is ignored. Clearly we cannot have ?5?6 = ?7?8 = 0. Considering the linear factors of these resultants we find the following necessary conditions for the origin to be an isochronous centre: (i) a1 = a2 = 0, a3 = 1, a4 = a5 = a6 = a7 = 0; (ii) a1 = a2 = 0, a3 = 1/4, a4 = a5 = a6 = a7 = 0; (iii) a1 = ?1/2, a2 = 0, a3 = 2, a4 = a5 = a6 = a7 = 0; (iv) a1 = ?1/2, a2 = 0, a3 = 1/2, a4 = a5 = a6 = a7 = 0; (v) a1 = ?1, a2 = 0,a3 = 3/4, a4 = ?1/3, a5 = a6 = a7 = 0; (vi) a1 = ?1, a2 = 0,a3 = 3, a4 = ?1/3, a5 = a6 = a7 = 0; (vii) a1 = ?2, a2 = 0,a3 = 2, a4 = ?1, a5 = a6 = a7 = 0. When a2 = 0, but a7 negationslash= 0 we have a5 = ?3a7, a6 = ?5a1 ?8a1a3 ?3a1 ?3a23 ?3a3 ?3a4, Re(?1) = ? and Re(?2), Im(?3), Re(?3), Im(?4) are polynomials in a1, a3, a4, a7. We calculate R(Re(?2),Im(?3),a4) = #a27U1, R(Re(?2),Re(?3),a4) = #U2, R(Re(?2),Im(?4),a4) = #a27U3, where U1, U2, U3 are polynomials in a1, a3, a7. Further calculation gives R(U1,U2,a7) = #(a1 ?a3 + 1)4V21 V22 , R(U1,U3,a7) = #(a1 ?a3 + 1)4V23 V24 . When a1 ?a3 = ?1 we find that U1 = U2 = 0 if and only if a1 = ?12, a3 = 12. Finally R(?,V1,a3) = #(2a1 + 1)4W1, R(?,V2,a3) = #(2a1 + 1)2W2, R(?,V3,a3) = #(2a1 + 1)4W3, R(?,V4,a3) = #(2a1 + 1)3W4, 19 where the Wi are distinct irreducible polynomials of degree 14 in a1 alone. When a1 = ?12 we must have a3 = 12 and Re(?2) = a24 +a27, which is non-zero under current assumptions. We conclude that there are no isochronous centres with a2 = 0, a7 negationslash= 0. When a2 negationslash= 0, but a3 ?a1 = 1 and a4 = ?/9, we have Im(?2) = ?+a5?, where ? = ?4a2(72a31 + 72a21 +a1a22 + 18a1 +a22) and ? = 24a21 + 48a1 + 18 +a22. We find that if ? = ? = 0 there are no isochronous centres for real values of the coefficients ai. Similarly, when ? negationslash= 0 and a5 = ??/? there are no isochronous centres under current hypotheses. In terms of the coefficients before scaling, conditions (i) to (vi) and (19) are precisely conditions (1) to (7) of Theorem 12. We note that when k = 0, the origin is an isochronous centre for system (8) if and only if condition (7) of Theorem 12 holds. This gives us the following result. Theorem 13 If ?1?2 negationslash= 0 or ?3?4 negationslash= 0 then the origin is an isochronous centre for system (8) if and only if one of the conditions of Theorem 12 holds. Proof. We have shown above that the conditions of Theorem 12 are necessary, sufficiency follows from Theorem 12. Theorem 14 If a7 = 0 then the origin is an isochronous centre for system (8) if and only if one of the conditions of Theorem 12 holds. Proof. The proof follows from Theorems 12 and 13 . Remark We have obtained all the necessary and sufficient conditions for the origin to be an isochronous centre for the sub-class with a7 = 0. Moreover when a7 negationslash= 0, for all the known centre conditions which are given in [14], there are no isochronous centres. This leads us to conjecture that the origin is an isochronous centre for system (8) if and only if one of the conditions of Theorem 12 holds. Then, if it is an isochronous centre, at most one other critical point of focus type can exist, that is parenleftBig ?2k,0 parenrightBig . If this point is of focus type, it is an isochronous centre and the system is quadratic. 20 7 Concluding Remarks We are confident in the validity of our computer implementation of Pleshkan?s algorithm. This conviction stems from the agreement with known results and also the independent proofs of sufficiency. We have investigated differential systems of the form ?x = y(1 +x), ?y = ?x+q(x,y), (20) where q is a polynomial without constant or linear terms. The conditions for the origin to be an isochronous centre when q is a quadratic polynomial were established by Loud [19]. For system (8), where q is a polynomial of degree 3, we find that the origin is an isochronous centre if and only of the origin is a centre (not necessarily an isochronous centre) for the quadratic part of (8). Similarly, for (7) (where q is of degree 4) the origin must be a centre for the quadratic part, and for the quadratic plus cubic parts, for the origin to be an isochronous centre. This is certainly not the case for systems in general, see [18] for example. We pose the question - is it true for systems of the form (20) whatever the degree of q? That is, if q is of degree n is it a necessary condition for isochronicity that the origin be a centre for the sub-systems of degree 2 to degree n?1? Assuming this is the case we consider the quartic system with q = a1x2 +a2xy+a3y2 +a4x3 +a5x2y+a6xy2 +a7y3 +a8x4 +a9x3y+a10x2y2 +a11xy3 +a12y4, with at least one of the degree 4 terms present. Now the origin is a centre for the quadratic part of (20) if a2(a1 +a3) = 0. We find that, when a1 = ?a3, and the origin is a centre for the cubic part of (20), the origin cannot be an isochronous centre for (20). Furthermore, when a2 = 0, with the origin as a centre for the cubic part of (20), the origin can only be an isochronous centre if a5 = a7 = 0 also. Finally we suggest that, if the origin is an isochronous centre for systems of the form (20), there are no critical points at infinity other than those on the invariant line x = ?1. References [1] J Chavarriga, J Gin?e and I A Garc??a, Isochronous centers of a linear center perturbed by fourth degree homogeneous polynomial, Bull. Sci. Math. 123 (1999), pp 77-96. 21 [2] J Chavarriga, J Gin?e and I A Garc??a, Isochronous centers of a linear center perturbed by fifth degree homogeneous polynomials, J. Comput. Appl. Math. 126 (2000), pp 351-368. [3] J Chavarriga, I A Garc??a and J Gin?e, Isochronicity into a family of time-reversible cubic vector fields, App. Maths and Comp. 121 (2001), pp 129-145. [4] J Chavarriga and M Sabatini, A Survey of Isochronous Centers, Qualitative Theory of Dynamical Systems 1 (1999), pp 1-70. [5] L A Cherkas, Conditions for a Li?enard equation to have a centre, Differensial?nye Uravneniya 12 (1976), pp 201-206. [6] C J Christopher and J Devlin, Isochronous centers in planar polynomial systems, SIAM J. Math. Anal. 28 (1) (1997), pp 162-177. [7] C Christopher and J Devlin, On the Classification of Li?enard Systems with Amplitude-Independent Periods, J. Diff. Equations 200 (2004), pp 1-17. [8] C J Christopher, N G Lloyd and J M Pearson, On Cherkas?s method for centre conditions, Nonlin. World 2 (1995), pp 459-469. [9] A Cima, A Gasull, V Ma?nosa and F Ma?nosas, Algebraic properties of the Liapunov and period constants, Rocky. Mount. J. Math. 27 (1997), pp 471-501. [10] J Devlin, Coexisting isochronous and nonisochronous centres, Bull. Lond. Math. Soc. 28 (1996), pp 495-500. [11] A Gasull and V Ma?nosa, An explicit expression of the first Liapunov and period constants with applications, J. Math. Anal. Appl. 211 (1997), pp 190-212. [12] J M Hill, Isochronicity and Coexisting Fine Foci, PhD Thesis, The University College of Wales, Aberystwyth (2004). [13] J M Hill, N G Lloyd and J M Pearson, Properties of a predator-prey model with intratrophic predation, preprint. [14] J M Hill, N G Lloyd and J M Pearson, The extended Kukles system, preprint. [15] R H Lewis, Fermat, Computer Algebra System, www.bway.net/ lewis . [16] Y Liu and W Huang, A new method to determine isochronous center conditions for polynomial differential systems, Bull. Sci. Math. 127 (2003), pp 133-148. [17] N G Lloyd and J M Pearson, Limit Cycles and Centres: An Example, Diff. Equations, Dyn. Systems and Control Science, ed. Elworthy, K D, Everitt, W N & Lee, E B (Marcel Dekker, New York 1993), pp 143-158. [18] N G Lloyd, C J Christopher, J Devlin, J M Pearson and N Yasmin, Quadratic-like cubic systems, Diff. Equations and Dyn. Systems 5 (3/4) (1997), pp 329-345. 22 [19] W S Loud, Behaviour of the period of solutions of certain plane autonomous systems near centers, Contributions to Diff. Equations 3 (1) (1964), pp 21-36. [20] P Marde?si?c, C Rousseau and B Toni, Linearization of Isochronous Centers, J. Diff. Equations 121 (1995), pp 67-108. [21] D J Needham, A centre theorem for two-dimensional complex holomorphic systems and its generalizations, Proc. Roy. Soc. London Ser. A 450 (1995), pp 225-232. [22] J M Pearson, N G Lloyd and C J Christopher, Algorithmic derivation of centre conditions, SIAM Review 38 (4) (1996), pp 619-636. [23] J M Pearson and N G Lloyd, Space saving calculation of the resultant of multivariate polynomials, Institute of Mathematical and Physical Sciences, University of Wales, Aberystwyth, internal report. [24] I I Pleshkan, A new method of investigating the isochronicity of a system of two differential equations, Differential Equations 5 (1969), pp 796-802. [25] M Robnik and V G Romanovski, On Urabe?s criteria of isochronicity, J. of Phys. A: Math. Gen. 32 (1999), pp 1279-1283. [26] V G Romanovski, Y H Wu and C Christopher, Linearizability of three families of cubic systems, J. Phys. A: Math. Gen. 36 (2003), pp 145-152. [27] M Sabatini, On the period function of Li?enard systems, J. Diff. Equations 152 (2) (1999), pp 467-487. [28] M Sabatini, Quadratic isochronous centers commute, Applicationes Mathematicae 26, (3) (1999), pp 357-362. [29] M Urabe, Potential forces which yield periodic motions of a fixed period, J. Math. Mech. 10 (1961), pp 569-578. 23