Electronic Journal of Differential Equations, Vol. 2007(2007), No. ??, pp. 1???.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
CENTRES AND LIMIT CYCLES FOR AN EXTENDED KUKLES
SYSTEM
JOE M. HILL, NOEL G. LLOYD, JANE M. PEARSON
Abstract. We present conditions for the origin to be a centre for a class
of cubic systems. Some of the centre conditions are determined by finding
complicated invariant functions. We also investigate the coexistence of fine
foci and the simultaneous bifurcation of limit cycles from them.
1. Introduction
In this paper we establish some properties of the cubic differential system
?x = P(x,y) = ?x+y +kxy,
?y = Q(x,y) = ?x+?y +a1x2 +a2xy +a3y2 +a4x3 +a5x2y +a6xy2 +a7y3,
(1.1) kukplus
where the ai and k are real. We first became interested in this class of systems when
considering transformations to generalised Li?enard form [1]. It was also brought to
our attention that a system used to model predator-prey interactions with intrat-
rophic predation could be transformed so that it is an example of a system of type
(1.1). We investigated this particular case of system (1.1) in [7]. In [6] we found
conditions for the origin to be an isochronous centre for system (1.1).
When ? = 0 the origin is said to be a fine focus; then system (1.1) is derived from
a second order scalar equation and it has an invariant line kx = ?1. When k = 0,
(1.1) is often referred to as the Kukles system; this system has been extensively
studied, see [2], [12] and [14] for example.
Here we derive conditions for the origin to be a centre for system (1.1) and
consider the simultaneous bifurcation of limit cycles from several fine foci. We shall
see, for example, that at most two fine foci of (1.1) can coexist; when one fine focus
is of order one, the other is of maximum order six and when one fine focus is of
order two, the other is of maximum order two. We show that in the latter case a
large amplitude limit cycle can surround the two fine foci and conjecture that this
is also true in the former.
We obtain necessary conditions for a critical point to be a centre for (1.1) by
calculating the focal values, which are polynomials in the coefficients k,ai. There is
a function V, analytic in a neighbourhood of the origin, such that its rate of change
along orbits, ?V, is of the form ?2r2+?4r4+???, where r2 = x2+y2. The ?2j are the
2000 Mathematics Subject Classification. 34C07, 37G15.
Key words and phrases. Nonlinear differential equations; invariant curves; limit cycles.
c?2007 Texas State University - San Marcos.
Submitted August 10, 2007. Published ??.
1
2 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
focal values and the origin is a centre if, and only if, they are all zero. The relations
?2 = ?4 = ??? = ?2j = 0 are used to eliminate some of the variables from ?2j+2.
This reduced focal value ?2j+2, with strictly positive factors removed, is known as
the Liapunov quantity L(j). We note that L(0) = ?. The circumstances under
which the calculated L(j) are zero yield possible centre conditions. The origin is a
fine focus of order j if L(i) = 0 for i = 0,1,...,j?1 and L(j) negationslash= 0; at most j small
amplitude limit cycles can bifurcate from a fine focus of order j.
Various methods are used to prove the sufficiency of centre conditions; in this
paper we require three of them. The simplest is that the origin is a centre if
the system is symmetric in either axis, that is, it remains invariant under the
transformation (x,y,t) mapsto? (x,?y,?t) or (x,y,t) mapsto? (?x,y,?t). Another technique
which we employ involves a transformation of the system to Li?enard form
?x = y, ?y = ?f(x)y?g(x). (1.2) lienard
The relevant results are as follows; proofs can be found in [3].
lemma1 Lemma 1.1. Consider system (1.2) where f,g are analytic, g(0) = 0, xg(x) > 0
for x negationslash= 0 and gprime(0) > 0. Let F(x) = integraltextx0 f(?)d? and G(x) = integraltextx0 g(?)d?.
(i) The origin is a centre for system (1.2) if and only if there is an analytic
function ? with ?(0) = 0 such that G(x) = ?(F(x)) in a neighbourhood of
x = 0.
(ii) The origin is a centre for system (1.2) if and only if there is a function z(x)
satisfying z(0) = 0, zprime(0) < 0 such that F(z) = F(x) and G(z) = G(x).
The third approach, and the one which is of particular interest to us here, is the
possibility of finding an integrating factor. If the origin is a critical point of focus
type then it is a centre if there is a function D negationslash= 0 such that
?
?x (DP)+
?
?y (DQ) = 0 (1.3) dulac
in a neighbourhood of the origin. Such a function is called an integrating factor
or Dulac function. The existence of the function D means the system is integrable
and the origin is a centre.
We make a systematic search for an integrating factor of the form D = ?ni=1C?ii ,
where each Ci is an invariant algebraic function. In this context an invariant func-
tion is such that ?Ci = CiLi, where Li, known as the cofactor of Ci, is of degree one
less than that of the system. We require
?
?x (DP)+
?
?y (DQ) = D(Px +Qy +?1L1 +???+?nLn) = 0. (1.4) dulacn
The ?i and the coefficients in the Ci,Li are functions of the coefficients k,ai. We
note that the Ci,Li,?i may be complex; a real Dulac function is then constructed
from these together with their conjugates. In any given situation there may well
be no invariant functions and even though there is an upper bound for the possible
degree of an invariant curve it is not known how to determine this bound. Darboux
[5] showed that if n ? 12m(m+1)+2 invariant functions exist, where m is the degree
of the system, then the n functions can be combined to form a first integral. In
practice we find that fewer such functions are required. As will be apparent later,
finding such functions is non-trivial. However it is a relatively straightforward
matter to confirm that the functions found actually satisfy the relation (1.3).
EJDE-2007/?? EXTENDED KUKLES SYSTEM 3
These techniques for finding centre conditions are well established but the com-
putational problems encountered are often formidable. We are constantly pushing
the available software to its limits. The reduction of the focal values to obtain the
Liapunov quantities is one area which causes difficulties and here we demonstrate
the usefulness of our suite of programs INVAR [11] in the search for invariant func-
tions. We are unable to complete the reduction the focal values for (1.1) to obtain
the conditions that are necessary for the origin to be a centre, but we can find
sufficient conditions by searching for invariant functions. We then try to determine
whether or not we have a complete set of conditions for the origin to be a centre.
The necessary and sufficient conditions for the origin to a centre for the Kukles
system are known; we summarise them in Theorem 1.2. We note that the condition
given in [8, Theorem 3.3], with a2 = 0, is covered by condition (v) of Theorem 1.2.
In [9] it was conjectured that, when a7 negationslash= 0, the origin is a centre for the Kukles
system if and only if one of the conditions given in Theorems 2.1 or 2.2 therein is
satisfied. This was verified in [10].
kuklestheorem Theorem 1.2. Let ? = k = 0. The origin is a centre for system (1.1) if and only
if one of the following conditions holds:
(i) a2 = a5 = a7 = 0;
(ii) a1 = a3 = a5 = a7 = 0;
(iii) a4 = a3(a1+a3), a5 = ?a2(a1+a3), (a1+2a3)a6+a23(a1+a3) = 0, a7 = 0;
(iv) a5 +3a7 +a2(a1 +a3) = 0, 9a6a22 +2a42 +27a7?+9?2 = 0, a4a22 +a5? = 0,
(3a7?+?2 +a6a22)a5 ?3a7?2 ?a6a22? = 0, where ? = 3a7 +a2a3;
(v) a5 + 3a7 + a2(a1 + a3) = 0, 18a4a5 ? 27a4a7 + 9a5a21 + 9a5a6 + 2a5a22 =
0, 27a4a1 + 4a5a2 + 9a31 + 2a1a22 = 0, 18a24 + 9a4a21 + 2a4a22 + 2a25 = 0,
18a4a2 +9a5a1 +9a5a3 +9a21a2 ?27a1a7 +9a6a2 +2a32 = 0.
For a proof of the above theorem, see [2, 8, 9, 10].
This is one particular sub-class of system (1.1). In the next section we shall
present conditions that are necessary and sufficient for the origin to be a centre
for two other sub-classes of system (1.1); one with a7 = 0 and one with a2 = 0.
Presenting the results in this way allows for a clearer description of the general case
and gives us insight into the types of invariant functions we should seek for system
(1.1) in general. In section 3 we derive sufficient conditions for the origin to be a
centre for system (1.1) and in section 4 we investigate whether there are any other
conditions. The coexistence of fine foci and the bifurcation of small amplitude limit
cycles is considered in section 5, and in section 6 we investigate the possibility of
the existence of large amplitude limit cycles.
2. Sub-classes a7 = 0 and a2 = 0
In this section we consider the sub-classes of system (1.1) with a7 = 0 or a2 = 0.
We find that the origin is a fine focus of maximum order six when a7 = 0 and
maximum order seven when a2 = 0.
a7=0 Theorem 2.1. Let ? = a7 = 0. The origin is a centre for system (1.1) if and only
if one of the following conditions holds:
(i) a2 = a5 = a7 = 0;
(ii) k = a1 = a3 = a5 = a7 = 0;
(iii) k = ?12a1, a3 = ?23a1, a4 = ?14a21, a5 = ?13a1a2, a6 = a7 = 0;
4 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
(iv) k = ?a1, a3 = ?23a1, a4 = 0, a5 = ?13a1a2, a6 = a7 = 0;
(v) k = ?14a1, a3 = ?34a1, a4 = ?14a21, a5 = ?14a1a2, a6 = a7 = 0;
(vi) k = ?12a1, a3 = ?a1, a5 = a6 = a7 = 0;
(vii) a4 = a3(a1+a3), a5 = ?a2(a1+a3), (a1+2a3)a6?a3(k?a3)(a1+a3) = 0,
a7 = 0;
(viii) k = ?(a1 + a3), a6 = a1(a1 + a3), a5 = ?a2(a1 + a3), (3a1 + 2a3)a4 +
a21(a1 +a3) = 0, a7 = 0.
Proof. Calculation of the focal values for system (1.1) with a7 = 0, up to ?14, and
their reduction to give the corresponding Liapunov quantities is routine. We do
not present the details here. We find that L(0) = L(1) = ??? = L(6) = 0 only if
one of the conditions of Theorem 2.1 holds. The sufficiency of these conditions is
confirmed as follows.
When (i) holds the system is invariant under the transformation (x,y,t) mapsto?
(x,?y,?t); the system is symmetric in the x-axis, hence the origin is a centre.
Similarly, when condition (ii) holds the system is invariant under the transformation
(x,y,t) mapsto? (?x,y,?t); the system is symmetric in the y-axis, so the origin is a
centre.
Conditions (iii), (iv), (v) and (vi) have a6 = a7 = 0, in which case system (1.1)
is of the form
?x = (1 +kx)y, ?y = x(?1+a1x+a4x2)+x(a2 +a5x)y +a3y2. (2.1) a6=0
If k = 0 in these cases then condition (ii) is satisfied. When k negationslash= 0, we are able
to transform (2.1) to a Li?enard system. The required transformation (see [3]) is
(x,y,t) mapsto? (x,(1+kx)y?(x),?), where
?(x) = dtd? = (1 +kx)?1 exp
parenleftBig
?
integraldisplay x
0
a3(1+ks)?1ds
parenrightBig
= (1 +kx)?1?a3k .
Then system (2.1) becomes a system of the form (1.2) with
f(x) = ?x(a2 +a5x)(1+kx)?1?a3k , g(x) = x(1?a1x?a4x2)(1+kx)?1?2a3k .
We compute the integrals of f, g and denote these by F, G respectively. For
condition (iii) we have
F(x) = a2a2
1
parenleftBig
9?parenleftbig 2a
1x?2
parenrightbig4/3(a
1x?3)2
parenrightBig
,
G(x) = ? 6a2
1
parenleftBig
3+parenleftbig 2a
1x?2
parenrightbig2/3parenleftbiga
1x?3
parenrightbigparenrightBig.
Let u3 = a1x?2 and v3 = a1z?2 then
F(x)?F(z) = 2
4/3a2
a21u4v4(v?u)(u
3v2 +u2v3 ?u2 ?v2)?,
G(x)?G(z) = 3 2
5
3a2
a21u2v2(v?u)?,
where
? = u2v2 +u+v = ((a1x?2)(a1z?2))2/3 +(a1x?2)1/3 +(a1z?2)1/3 .
When x = z = 0,?x = ?z = ?2?23a1. By the Implicit Function Theorem there is
z(x) with zprime(x) < 0 such that F(x) = F(z(x)), G(x) = G(z(x)). The origin is a
centre by Lemma 1.1 (ii).
EJDE-2007/?? EXTENDED KUKLES SYSTEM 5
Similarly for condition (iv) we find
F(x) = a24a2
1
parenleftBig
9? (a1x?3)
2
(1?a1x)2/3
parenrightBig
, G(x) = ? 32a2
1
parenleftBig
9+ (a1x?3)
(1?a1x)1/3
parenrightBig
and
? = ((1?a1x)(1?a1z))1/3
parenleftBig
(1?a1x)1/3 +(1?a1z)13
parenrightBig
?2.
When condition (v) holds
F(x) = ?8 a2x
2
(a1x?4)2,
G(x) = 8 x
2
(a1x?4)6
parenleftbiga4
1x
4 ?24a3
1x
3 +240a2
1x
2 ?768a1x+768parenrightbig
and ? = a1xz?2(x+z). For condition (vi)
F(x) = ?2 a2x
2
(a1x?2)2,
G(x) = ?4(a
21a4x4 +4a21x2 ?8a1x+4)
(a1x?2)4
and ? = a1xz ? x ? z. In each case ? is a common factor of F(x) ? F(z) and
G(x)?G(z); the origin is a centre by Lemma 1.1 (ii).
To prove the sufficiency of the remaining conditions we use INVAR to help us
find appropriate invariant functions and to build Dulac functions. Confirmation
that the functions obtained are indeed Dulac functions is routine. When condition
(vii) holds we find the Dulac function
D = (1 +kx)?1e?2xC?3,
where
C = 1 +a3x??y, ?1 = (a2 ?2?)(a3k?a6)?k
2?
k2? ,
?2 = a6(a2 ?2?)k? , ?3 = ?a2? ,
and ? satisfies ?2 ?a2? ?a23 + a3k ?a6 = 0. Hence, when k? negationslash= 0, the origin is
a centre. When k = 0 condition (iii) of Theorem 1.2 holds. When ? = 0, then
a6 = a3(k ? a3) = 0 and the system can be transformed to Li?enard form with
f(x) = ?a2g(x); the origin is a centre by Lemma 1.1 (i).
For condition (viii) we find the Dulac function
D = (1 +kx)?1e?2xC?3,
where
C = 1?a1x+ a2? y?a4x2 + a5? xy,
?1 = 1, ?2 = a1(? +2), ?3 = ?,
and ? is a root of a31?2 ?(3a1 +2a3)a22(? +1) = 0. If ? negationslash= 0, the origin is a centre.
When ? = 0, then one of conditions (i), (ii) or (vii) is satisfied. This completes the
proof. square
6 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
When none of the conditions of Theorem 2.1 holds and L(i) = 0, for i =
0,1,2,...,5, then L(6) negationslash= 0; the origin is then a fine focus of maximum order
six and at most six small amplitude limit cycles can be bifurcated from the origin.
We now consider the sub-class of system (1.1) with a2 = 0 and a3a7 negationslash= 0. We
exclude the possibility that a3 = 0 because, when a2 = a3 = 0, the origin is a centre
for system (1.1) only if a7 = 0.
a2=0 Theorem 2.2. Let ? = a2 = 0, with a3a7 negationslash= 0. The origin is a centre for system
(1.1) if and only if one of the following conditions holds:
(i) a2 = 0, k = ?(2a1 + a3), (a1 + 2a3)a4 + a21(a1 + a3) = 0, a5 = ?3a7,
(a1+2a3)a6?2a1(a1+a3)(2a1+a3) = 0, 2(a1+2a3)2a27+a31(a1+a3)2(3a1+
2a3) = 0;
(ii) a2 = 0, k = ?(a1 + a3), 2a3a4 + a1(a1 + a3)(a1 + 3a3) = 0, a5 = ?3a7,
2a3a6 ?a1(a1 +a3)(3a1 +5a3) = 0, 4a23a27 +a1(a1 +a3)4(a1 +2a3) = 0.
Proof. When a2 = 0 and a3a7 negationslash= 0 we find that L(0) = L(1) = ??? = L(7) = 0 only
if one of the conditions of Theorem 2.2 holds. The sufficiency of these conditions is
confirmed by constructing integrating factors from invariant functions. Again we
use INVAR to find these functions. When condition (i) holds there exists a Dulac
function
D = (1 +kx)?1e?2xC?3,
where
C = 1?a1x+ a
21(a1 +a3)
(2a1 +a3) x
2 +a7xy,
?1 = (a1 +a3)
2
(2a1 +a3)2,?2 = ?
a1(a1 +a3)
(2a1 +a3) ,
and hence the origin is a centre. We note that when 2a1 +a3 = 0, then k = 0 and
condition (v) of Theorem 1.2 is satisfied.
The Dulac function for condition (ii) is somewhat more complicated. It consists
of an invariant line, an invariant conic, an invariant degree three curve and an
invariant exponential. We have
D = (1 +kx)?1e?2xC?31 C?42 , (2.2) dulaca2=0
with
C1 = 1?a1x+ 2a3a7k2 y + k?(2a3 ?k)2a
3
x2 ? 2a1a7k xy + k
2?
2a3y
2,
C2 = 1+ ?112a2
3?2w?
x??y + ?272a5
3?2w2?
x2 + ??38a3
3?2w?
xy + ?412a2
3?2?
y2
+ ?(2a3 ?k)?5144a5
3?2w2?
x3 + ?648a5
3?2w2?
x2y + ?736a3
3?2w?
xy2 +wy3,
?1 = 9a3k
2??0 +3k2??1? ?a3?2?2 ?6a3k2?3?3 ?4a23k2??4(?4 ?a3a7?2)
?48a43?2w2(3a3a7 +k2?) ,
?2 = 9a3k
3?F0 +3k3?F1? +a3?F2?2 ?6a3k2?F3?3 ?4a23k3??4(F4 ?a3a7?2)
48a43?2w2(3a3a7 +k2?) ,
?3 = ? 6a3a7?3k2?+4a
3a7?
,?4 = ? 3k
2?
3k2?+4a3a7?,
EJDE-2007/?? EXTENDED KUKLES SYSTEM 7
where ? = a23 ?k2, ? = a3 +k and ? = 4a23a7? ?3k3?. Here ?,w are roots of
4a23?4 ?36a1a3(a21 +a1a3 ?a23)?2 +81a21k4 = 0,
64a63w4 ?16a31a33parenleftbiga21 +a1a3 ?a23parenrightbig
?parenleftbiga41 ?4a31a3 ?22a21a23 ?20a1a33 +a43parenrightbigw2 +a61k12 = 0,
respectively and the ?i,?i and Fi are as given in the Appendix.
To complete the proof we consider what happens when any of the denominators
in the above are zero. When k?w = 0, then a7 = 0. When ? = 0 then either a7 = 0
or a21 + 7a1a3 + 8a23 = 0. Let a1 = 12(?17?7)a3. We find a Dulac function that
consists of an invariant exponential function and three invariant lines. We have
D = (1 +kx)e?1xC?21 C?32 ,
with
C1 = 1+ (4a
23 +?2)(5?4 +329a23?2 ?4a43)
2?2a3? x??y,
C2 = 1+ ?
4?1
16a33??x?ny, ?1 =
?2?2
4a3?3,
?2 = ?12a
23?
pi1 , ?3 =
12?a23?
npi1
where ? = 812a23 +33?2, ? = 4a43 +39a23?2 ?73?4, pi1 = 16a63 ?301a23?4 +5?6,
n = 4a
23?
?(156a43 +9a23?2 ?5?4),
? is a root of (16a43 ?32a23?2 ??4)(4a43 ?103a23?2 ??4) = 0 and ?1, ?2, ?3 are
polynomials of degree six in a3,?.
When (3a3a7 + k2?)(3k2? + 4a3a7?) = 0 then either a7 = 0 or 2a1 + 3a3 = 0.
Let a1 = ?32a3. Then there exists a Dulac function
D = (1 +kx)C?21 C?12
where
C1 = 1+ 34a3x+ 4a7a
3
y,
C2 = 1 + 32a3x? 8a7a
3
y + 916a23x2 ?3a7xy.
This completes the proof. square
When none of the conditions of Theorem 2.2 holds and L(i) = 0, for i =
0,1,2,...,6, then L(7) negationslash= 0; the origin is a fine focus of maximum order seven,
at most seven small amplitude limit cycles can be bifurcated from the origin.
3. Sufficient centre conditions
Now we return to the full system and derive some sufficient conditions for the
origin to be a centre. We have obtained the necessary and sufficient conditions
for the origin to be a centre for three sub-classes of system (1.1); with k = 0,
with a2 = 0 or with a7 = 0. In these sub-classes we determined possible centre
conditions by considering the focal values and then proved that the conditions
we had found were sufficient. As the reduction of the focal values in the general
8 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
case requires the calculation of some resultants that cannot be obtained with the
currently available hardware and software we adopt a different approach. We use
the knowledge gained from consideration of the sub-classes to give us an insight into
the probable centre conditions in the general case. We search for invariant functions
and corresponding integrating factors for the general system without introducing
a condition for which the origin may be a centre. The relationships between the
coefficients in system (1.1) that must be satisfied to ensure that ?Ci = CiLi and
(1.4) holds, for D = ?ni=1C?ii , are sufficient conditions for the origin to be a centre.
We find three sufficient conditions for the origin to be a centre for system (1.1),
with ka2a7 negationslash= 0, using this approach.
Knowledge gained from the sub-classes suggests the type of invariant functions
we should seek in order to determine integrating factors. In particular, for the
Kukles system and the sub-class with a7 = 0 combinations of invariant exponential
functions, invariant lines and invariant conics are required. The Dulac functions for
the class with a2 = 0 are more complicated and include invariant lines, conics and
cubic functions. The line kx = ?1 is invariant with respect to system (1.1), with
? = 0, and is included in each Dulac function we seek in the general case. Where
the degrees of the equations in the system are not equal it is often found that an
exponential function is also required and this is so in all cases here.
We search for functions that are invariant with respect to system (1.1). Both
f(x) = ex and g(x) = kx+1 are invariant without any constraints on the coefficients
ai,k. Next we look for functions that are invariant only when some relationships
between the coefficients are satisfied. For the three sub-classes we knew from the
reduction of the focal values what these relationships were. Here we aim to find
the relationships by satisfying ?Ci = CiLi and equation (1.4).
We start with the simplest invariant curve, namely a line. Let C = 1+c10x+c01y,
with cofactor L = m10x + m01y + m20x2 + m11xy + m02y2. We have c10 = m01
and c01 = ?m10; then seven equations must be satisfied for C = 0 to be invariant
with respect to (1.1). We assume that m10 negationslash= 0, otherwise we recover the line
kx = ?1. We determine m01,m20,m11,m02 in terms of m10 and the ai,k. There
are three remaining equations which must be satisfied. At this stage we try to build
a Dulac function using this line together with f and g. Five additional equations
must hold if D = g?1f?2C?3 is a Dulac function that satisfies (1.4). If a7 negationslash= 0, we
must have ?3 = ?3. Then m10 = 13a2 and the other ?i are given by two of these
equations. We have determined all the coefficients of C and L, and the ?i. The
four relationships between the coefficients that must hold to satisfy the remaining
equations are those of condition (i) of Theorem 3.1 below.
In a similar manner we search for invariant conics. Let C = 1 + c10x + c01y +
c20x2 +c11xy+c02y2, with cofactor L as above. Again c10 = m01 and c01 = ?m10.
Twelve equations in the remaining eight coefficients of the conic and its cofactor
must be satisfied if C = 0 is invariant with respect to system (1.1). Five additional
equations must hold if D = g?1f?2C?3 is a Dulac function that satisfies (1.4).
Consideration of all possible situations in which the conic does not reduce to a line,
or become the product of two lines, leads us to conclude that either c02 = 0 or
m02 = 2a7. Let c02 = 0. If a7 negationslash= 0, then ?3 = ?3, the coefficients of the cofactor
are m10 = 13a2,m01 = ?a1,m20 = 13a5,m11 = ?a21 ? a1k ? 2a7 ? 29a22 and C,
?1,?2 are as given in the proof of condition (ii) of Theorem 3.1 below. As two
of the equations are linearly dependent in this situation this leaves five equations
EJDE-2007/?? EXTENDED KUKLES SYSTEM 9
in the coefficients k,ai that must be satisfied if ?C = CL and (1.4) holds. These
equations lead to precisely the relationships of condition (ii) of Theorem 3.1. When
c02 negationslash= 0 and m02 = 2a7 these same five equations must be satisfied together with
an additional equation; this is a specific instance of condition (ii).
We know that, when a2 = 0, there is a Dulac function which consists of powers
of f,g, an invariant conic and an invariant cubic curve. We search for this type of
Dulac function in the general case. Again the linear coefficients in each invariant
curve can be given in terms of the linear coefficients in the corresponding cofactor.
We have thirty-five equations in the twenty-four unknowns. In this instance the
invariant conic in which c02 negationslash= 0 and m02 = 2a7 is used. We determine all the coef-
ficients of the invariant conic and its cofactor, in terms of the coefficients of system
(1.1), from the twelve equations that must be satisfied for the conic to be invariant
with respect to (1.1). This leaves four relationships between the coefficients in (1.1)
that must hold.
We then proceed to determine the coefficients of the cubic function and its co-
factor. Here there are eighteen equations in twelve unknowns. We eliminate all but
two of the unknowns, namely the coefficient of x in the cofactor (say ?) and the
coefficient of y3 in the invariant cubic (say w). One of the remaining equations is
quadratic in ? and independent of w. Attempts to eliminate ? from all remaining
equations using this equation lead to expressions being generated that result in
stack overflow. We turn our attention to the five equations that must be satisfied
if (1.4) holds. We find that if a7 negationslash= 0, then ?3 = ?32(?4 + 1) and with ?4 in terms
of ?,w and the coefficients k,ai we must have
(a1a2 +a2a3 +a5 +4a7)(a1a2 +a2a3 +a5 +3a7)(a1a3 +a23 ?a4) = 0.
The two remaining equations that must hold to satisfy (1.4) give ?1,?2. We cal-
culate that ?4 = a1a2 + a2a3 + a5 + 3a7; ?4 = 0 is necessary for the origin to be
a centre. This, with the four relationships from the requirement for the conic to
be invariant, yield condition (iii) of Theorem 3.1 below. We use these relation-
ships to replace k,a4,a5,a6,a7 in the remaining equations. We note that we have
introduced another unknown, r, where r2 = a22 +4a23 ?4k2.
We use the quadratic in ? mentioned above to eliminate r and, for consistency,
we equate this expression for r with radicalbiga22 +4a23 ?4k2. This consistency condition
has
V =(a22 +4a23)?4 ?3a2(a22 +4a23)?3
?9(4a31a3 ?2a21a22 +4a21a23 ?4a1a22a3 ?4a1a33 ?a22a23)?2
?54a1a2(a1 +a3)3? +81a21(a1 +a3)4
as a factor. We know from consideration of a specific example that this factor
will ultimately lead to an appropriate Dulac function. This is the only remaining
equation that is independent of w.
We factorise each of the equations and remove any factors that involve only the
remaining coefficients a1,a2,a3; we are able to show that such factors being zero
lead to specific instances of conditions that are already known to us. Other than
V = 0, the simplest of the remaining equations has over 7000 terms. We use a
polynomial remainder sequence to eliminate ? (see Section 4 for more details on
polynomialremaindersequences). The laterstagescanonlybe completedbyfurther
simplifying the expressions by replacing a1 by ?(k+a3), a22 +4a23 by t and scaling
10 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
such that k = 1. For example, at the second stage of the polynomial remainder
sequence, where a quadratic in ? is produced with approximately 30000 terms, the
size of the expressions can be almost halved by these changes of variable. We note
however that in order to check for factors that can be removed we need to replace
t by a22 + 4a23 before attempting the factorisation. The calculations are repetitive,
but formidable. In some cases, in order to multiply two expressions together, we
have to split each expression into smaller units and multiply each unit then sum the
results. Near the final stage we produce an expression with 191690 terms, which
we need to factorise. Fortunately we can predict that one of the factors will be the
coefficient of ?2 at the quadratic stage of the polynomial remainder sequence, an
expression with 9411 terms. There are four other factors, one of which is
W =(a22 +4a23)3w4 +a2(a22 +4a23)2(a42 +7a22a23 ?6a22k2 +12a43 ?24a23k2
?6a3k3 +6k4)w3 +(?a62a63 +6a62a43k2 +6a62a33k3 ?3a62a23k4 ?6a62a3k5
?a62k6 ?12a42a83 +72a42a63k2 +60a42a53k3 ?69a42a43k4 ?90a42a33k5
+9a42a23k6 +36a42a3k7 +6a42k8 ?48a22a103 +288a22a83k2 +192a22a73k3
?408a22a63k4 ?432a22a53k5 +120a22a43k6 +276a22a33k7 +72a22a23k8
?12a22a3k9 ?64a123 +384a103 k2 +192a93k3 ?720a83k4 ?672a73k5
+272a63k6 +528a53k7 +192a43k8 +16a33k9)w2
= a2k7(a3 +k)3(6a22a23 +3a22a3k?a22k2 +24a43 +12a33k?36a23k2
?24a3k3)w +k12(a3 +k)6.
We can show that when V = W = 0 all remaining equations are satisfied. We have
found an appropriate Dulac function and condition (iii) of Theorem 3.1 is sufficient
for the origin to be a centre.
sufficiency Theorem 3.1. Let ? = 0. The origin is a centre for system (1.1) if one of the
following conditions holds:
(i) a5 = ?a2(a1 +a3)?3a7,
a6 = ?2a
42 ?9a22a23 +9a22a3k?81a2a3a7 +27a2a7k?162a27
9a22 ,
a4 = (?2a
42 ?9a22a23 +9a22a3k?54a2a3a7 +27a2a7k?81a27)?2
2a62 ,
a1 = (?4a
42 ?9a22a23 +9a22a3k?54a2a3a7 +27a2a7k?81a27)?
2a52 ,
where ? = a2a3 +3a7 and a2 negationslash= 0;
(ii) a5 = a2k?3a7, k = ?(a1 +a3),
a7 = k
2(a2k?a3r)
a22 +4a23 ,
a6 = k(a
22(a1 +3a3)?4a1a3(3a1 +5a3))?3a2k2r
2(a22 +4a23) ,
a4 = k(a
22(a1 ?a3)+4a1a3(a1 +3a3))+a2k2r
2(a22 +4a23) ,
EJDE-2007/?? EXTENDED KUKLES SYSTEM 11
where r2 = a22 +4a23 ?4k2 and a22 +4a23 negationslash= 0;
(iii) a5 = ?a2(a1 +a3)?3a7,
a7 = 9a
21(a1 +k)?2a22(a1 +2a3)+9a4(3a1 +2k)
12a2 ,
a6 = 9(18a4kepsilon1?2a1a
22?+9a1(a21(?+2k)+a4(3?+4k)+a1kepsilon1))?8a22?
36a22 ,
9(16a22 +(9a1 +6k)2)a24 +2?(27a21 +18a1k +4a22)a4 +a21?2 = 0,
9(3a1 +2k)(4a22 +9?(3a1 +2k))a24 +2
parenleftBig
81a21?(a1 +k)(3a1 +2k)
+36a1a22(2a1 +k)(a1 +k)?4a42(a1 +2a3)
parenrightBig
a4
+a21?parenleftbig2a22(k?a3)+9a1?(a1 +k)parenrightbig = 0,
where ? = 9a21 + 9a1k + 2a22, ? = 2a1 + a3 + k, epsilon1 = a3 + k, ? = a22 + 9a4
and a2 negationslash= 0.
Proof. When either condition (i) or (ii) holds we find a Dulac function which con-
sists of the line kx = ?1, an exponential function and either another line or a
conic. The Dulac function then takes the form D = (1 + kx)?1e?2xC?3, where C
and the ?i are given below for each condition. For condition (iii) an invariant line,
an invariant conic and an invariant cubic together with an exponential function are
needed.
When condition (i) holds, and k negationslash= 0, we find
C = 1+ (a2a3 +3a7)a
2
x? a23 y,
?1 = (2a42 +54a2a7k?81a27 +9(a23 ?k2)a22)/9a22k2,
?2 = (?2a42 +27a2a7k +81a27 +9(k?a3)a22a3)/9a22k.
When k = 0, condition (iv) of Theorem 1.2 holds.
For condition (ii) we find
C = 1?a1x? a23 y?a4x2 ? a53 xy,
?1 = (9a21 +2a22 ?6a3k +18a4 +6a6 ?3k2)/3k2,
?2 = ?(9a21 +9a1k +2a22 +18a4 +6a6)/3k,
with k negationslash= 0. When k = 0, condition (v) of Theorem 1.2 holds.
The Dulac function required when condition (iii) holds is by some way the most
complicated we have encountered. Here, in addition to the invariant exponential
function and the line kx = ?1, we require an invariant conic, C1 = 0, and an
invariant cubic, C2 = 0. The Dulac function is
D = (1 +kx)?1e?2xC?31 C?42 .
The invariant curves are not of high degree but have thousands of terms, and the
powers ?i are non-trivial. The expressions are too lengthy to be given here. We
note that when a2 = 0, condition (iii) becomes condition (ii) of Theorem 2.2 and
the Dulac function reduces to that of equation (2.2). square
12 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
4. Focal values
Having established a set of sufficient conditions for the origin to be a centre for
system (1.1) with ka2a7 negationslash= 0 we endeavour to ascertain if we have found the nec-
essary conditions. If we could find a basis for the focal values for system (1.1) we
would be able to determine the necessary and sufficient conditions for the origin
to be a centre. However the computations soon become too large for the currently
available software and hardware systems. We reduce the focal values as far as is
possible and, by using examples, determine whether or not the sufficient conditions
we have found are indeed the only conditions for the origin to be a centre. We con-
jecture that the conditions given in Theorem 3.1 are both necessary and sufficient
for the origin to be a centre for system (1.1).
We calculate the focal values up to ?16 and in order to simplify them we set
a1 = m ? a3. We assume throughout this section that ka2a7 negationslash= 0. We aim to
establish under what conditions the L(i) are zero simultaneously. We have
L(1) = a2m+a5 +3a7.
Let a5 = ?a2m?3a7. Then L(1) = 0 and
L(2) = Aa6 +B,
where
A = a2(a3 +m)?3a7
and
B =?2a22a7 +2a2a23m?a2a3a4 ?3a2a3km?5a2a3m2 +2a2a4k
+5a2a4m+6a3a7m?9a4a7 ?9a7km?15a7m2.
Assume that A negationslash= 0 and let a6 = ?B/A. Then
L(3) = M0 +M1a4 +M2a24,
L(4) = N0 +N1a4 +N2a24 +N3a34,
L(5) = P0 +P1a4 +P2a24 +P3a34 +P4a44,
L(6) = Q0 +Q1a4 +Q2a24 +Q3a34 +Q4a44 +Q5a54,
where the Mi,Ni,Pi,Qi are polynomials in k,a1,a2,a3,a7.
In this instance calculating resultants to eliminate a4 is not feasible because of
the degrees to which the variables occur and the number of terms in the polyno-
mials involved. We employ a polynomial remainder sequence approach, the main
advantage being that we can work with the individual coefficients of the variable
being eliminated rather than the entire polynomial. Also factors of the reduced
polynomials can be removed at each stage in the process and some such factors can
be predicted. We use the following result to establish what these factors are.
Lemma 4.1. Suppose we have two univariate polynomials ?1,?2. We can deter-
mine a sequence of polynomials ?3,...,?j, of decreasing degree, such that
remainder(epsilon1?i?1+1i ?i?1,?i) = ?i+1?i+1; i greaterorequalslant 2,
where ?i is the difference in degrees between ?i and ?i+1; epsilon1i is the leading coefficient
of ?i; ?3 = 1,?i+1 = epsilon1?i?2+1i?1 . Hence we have that ?i+1 divides the pseudo remainder
of ?i?1 and ?i.
EJDE-2007/?? EXTENDED KUKLES SYSTEM 13
The Proof of the above lemma can be found in [4].
Assume that M2 negationslash= 0. Let a24 = ?(M0 +M1a4)/M2 such that L(3) = 0. Then
L(4) = A2(?0 +?1a4),
L(5) = A3(?0 +?1a4),
L(6) = A4(?0 +?1a4),
where the ?i,?i,?i are polynomials in m,a2,a3,a7,k. In particular ?0,?1 are poly-
nomials with 936,654 terms respectively.
Assume that ?1 negationslash= 0 and let a4 = ??0/?1. Then
L(3) = M22A2a7Digamma?,
L(5) = M22Aa7Digamma?,
L(6) = M22Aa7Digamma?,
where
Digamma =?81a2a27k?54a22a3a7k?9a32a23k +243a37 +12a42a7
+2a52a1 +243a2a3a27 +4a52a3 +81a22a23a7 +9a32a33
and ?,?,? are polynomials in m,a2,a3,a7.
When Digamma = 0, the focal values ?8,...,?14 have a common factor
? =2a62a23 +2a62a4 +12a52a3a7 +9a42a43 ?9a42a33k +18a42a27 +108a32a33a7
?81a32a23a7k +486a22a23a27 ?243a22a3a27k +972a2a3a37 ?243a2a37k +729a47.
Let
a1 = (?4a
42 ?9a22a23 +9a22a3k?54a2a3a7 +27a2a7k?81a27)?
2a52 ,
a4 = (?2a
42 ?9a22a23 +9a22a3k?54a2a3a7 +27a2a7k?81a27)?2
2a62 ,
where ? = a2a3 +3a7. Then Digamma = ? = 0 and
a6 = ?2a
42 ?9a22a23 +9a22a3k?81a2a3a7 +27a2a7k?162a27
9a22 .
These, together with a5 = ?a2m?3a7, are condition (i) of Theorem 3.1; the origin
is a centre for system (1.1). We note that in the special case when A = B = 0 this
condition is still satisfied and there are no other conditions with ka2a7 negationslash= 0.
The polynomials ?,?,? have 2294,2895 and 7674 terms respectively. The de-
grees to which each of the remaining variables occur in ?,?,? are as shown in the
following table:
a2 a3 m a7 k
? 12 13 19 11 11
? 13 14 20 12 12
? 18 19 25 15 16
Clearly any further progress in the reduction of the focal values is going to be
difficult, if not impossible, but we note that ? = ? = ? = 0 if either of the
conditions (ii) or (iii) of Theorem 3.1 holds.
Suppose that we could calculate the resultants of ?,? and ?,? with respect to
a3. Any common factor of the leading coefficients of a3 in ?, ?, ? will be a factor
14 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
of the resultants, but this common factor being zero may not be sufficient for the
vanishing of the polynomials. We have that a2a7m(k + m) is such a factor. In
particular, when k = ?m, we find
?1 = (a22 +4a23)a27 +2a2m3a7 +a1m4(a1 +2a3)
is a common factor of ?, ? and ?. Let
? = a27 = ?2a2m
3a7 ?a1m4(a1 +2a3)
a22 +4a23 .
Then L(3) = q0 + q1a7. Assume q1 negationslash= 0 and let a7 = ?q0/q1. For consistency we
must have ? = q21??q20 = 0. We find
?2 = (a22+4a23)a24+(4a1a3(a1+3a3)+a22(a1?a3))ma4+a1m2(a1(a1+3a3)2?a22a3)
is a common factor of L(4) and ?. Now k = ?m, a5 = ?a2m?3a7, a6 = ?B/A,
?1 = ?2 = 0 is precisely condition (iii) of Theorem 3.1; the origin is a centre.
We have seen how conditions (i) and (iii) of Theorem 3.1 emerge from the reduc-
tion of the focal values. However we are unable to locate condition (ii) by a similar
argument. It is possible that conditions (ii) and (iii) are specific instances of more
general conditions. We show that this is not the case by considering a particular
example. Each of these conditions has five relationships between the eight coeffi-
cients ai,k. We can choose values for three of the variables without imposing new
relationships.
Let a1 = 1, a2 = 1, a3 = ?2. Now ?,?,? are polynomials in a7,k, and a4, a5,
a6 are given in terms of a7,k also. Let R(f,g,x) denote the resultant of f and g
with respect to x and # represent a (large) integer. We calculate resultants with
respect to a7 and find
R(?,?,a7) = #(k?1)3(2k?3)6(k2 +6k +10)4?K1K2,
R(?,?,a7) = #(k?1)3(2k?3)4(k2 +6k +10)6?K1K3,
where ? = 162k3 + 152k2 ?28k ?84 and K1,K2,K3 are irreducible polynomials
of degrees 52,64,95 in k, respectively. When K1 = 0, then ?0 = ?1 = 0 and when
2k ?3 = 0, then a7 = 0; both situations are excluded under current assumptions.
The leading coefficients of a7 in ?,?,? have k2+6k+10, which is positive definite,
as a common factor. For the general case this factor is a22 + (a1 ?a3 + k)2, which
is non-zero when a2 negationslash= 0. Clearly K2,K3 cannot be zero simultaneously. So we
must have k = 1(= ?m) or ? = 0; the former is true when condition (iii) holds, the
latter when condition (ii) holds. Furthermore there are no other centre conditions
with five or fewer relationships that satisfy ? = ? = ? = 0.
After extensive consideration of all the possible situations in which the Liapunov
quantities up to L(6) are zero simultaneously we have not found any centre condi-
tions other than those given in Theorems 1.2, 2.1, 2.2 and 3.1. However there are
several instances where we are unable to calculate resultants or eliminate variables
using a polynomial remainder sequence, so we cannot be certain that we have a
complete set of necessary conditions. We have shown, by considering an example,
that when ? = ? = ? = 0 there are no other such conditions which contain five or
fewer relationships between the coefficients. We can also demonstrate that when
one of k,a2 or a7 is zero in ?,?,? then all the centre conditions found, by consid-
ering ? = ? = ? = 0, can be obtained from one of the conditions in Theorem 3.1
EJDE-2007/?? EXTENDED KUKLES SYSTEM 15
with the appropriate variable set to zero. This wealth of evidence leads us to the
following claim.
conj7 Conjecture 4.2. The origin is a centre for system (1.1) when ka2a7 negationslash= 0 if and
only if one of the conditions of Theorem 3.1 holds.
5. Coexisting fine foci
One of the significant features of a planar dynamical system is the possible
configuration of limit cycles. Information is sought on the number of critical points
that can be encircled by closed orbits and on the number of closed orbits that can
encircle one such critical point. This may be phrased as asking how many nests of
limit cycles can there be and how many limit cycles make up each nest.
Suppose that ? = 0 in system (1.1), so that the origin is a fine focus. From
the equation for ?x, critical points can occur only on the x-axis and on kx = ?1.
However kx = ?1 is invariant, so any critical point on it cannot be of focus type.
Thus any fine focus must have y = 0 and x(a4x2 +a1x?1) = 0. The condition for
a fine point is x(a5x + a2) = 0. Thus only one critical point other than the origin
can be a fine focus.
lem8 Lemma 5.1. System (1.1) with ? = 0 can have at most two fine foci.
Suppose that there are two fine foci. We can scale coordinates so that they are
(0,0) and (1,0). Then a4 = 1?a1,a5 = ?a2 and the system is
?x = y(1+kx),
?y = ?x+a1x2 +a2xy +a3y2 +(1?a1)x3 ?a2x2y +a6xy2 +a7y3. (5.1) coexist2
The point (1,0) is a fine focus, as opposed to a fine col, if (k +1)(a1 ?2) > 0. We
denote the Liapunov quantities associated with the origin by L(i) and those for the
point (1,0) by M(i).
sixone Theorem 5.2. Suppose that the origin and (1,0) coexist as fine foci in system
(5.1). If (1,0) is of order one then the origin is of order at most six.
Proof. We calculate that
L(1) = a2(a1 +a3 ?1)+3a7, (5.2) L(1)
M(1) = 3(a1 ?2)2a7 +a2((2?a1)a6 +(a1 ?1)(k?a3 +1)+a3). (5.3) M(1)
For (1,0) to be a fine focus of order one we must ensure that M(1) negationslash= 0. Let
a7 = 13a2(1?a1 ?a3), then L(1) = 0 and
L(2) = a2(3a6(2a1 +3a3 ?1)+?),
where
? =15a31 +24a21a3 +9a21k?39a21 +2a1a22 +9a1a23 +9a1a3k?45a1a3
?15a1k +33a1 +2a22a3 ?2a22 ?9a23 ?9a3k +21a3 +6k?9.
If a2 = 0, then a5 = a7 = 0 and the origin is a centre by Theorem 2.1 (i). We require
a2 negationslash= 0 for the origin to be a fine focus of order greater than two. If 2a1 + 3a3 = 1
then ? = a2?, where
? = (a1 ?2)(2a22 +9(a1 ?1)2)+9(k +1)(a1 ?1)2, (5.4) phi
which is non-zero if a2 negationslash= 0 and (1,0) is a fine focus.
16 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
So, for the origin to be a fine focus of order greater than two, we also need
2a1 +3a3 negationslash= 1. Let a6 = ?3(1?2a1?3a3). Now we require M(1) = a2(a1 +a3?1)? to
be non-zero and we have
L(3) = M(1)(??2a22(a1 +a3 ?1)),
where
? =3(25a31 +97a21a3 +7a21k?32a21 +117a1a23 +22a1a3k?88a1a3
?4a1k +14a1 +45a33 +15a23k?54a23 ?8a3k +22a3 ?2).
Let a22 = ?2(a1+a3?1). Then
L(4) = a2(Ak2 +Bk +C)?,
where
A = 15(a1 +a3)2 ?20(a1 +a3)+8 > 0,
B = 15(a1 +a3)2(11a1 +13a3)?5(a1 +a3)(59a1 +67a3)+4(45a1 +49a3 ?8),
C =90(a1 +a3)3(4a1 +5a3)?45(a1 +a3)2(19a1 +23a3)
+30(a1 +a3)(25a1 +29a3)?8(35a1 +38a3 ?5),
? =15(a1 ?2)a23 +(a1 ?2)(29a1 +5k?13)a3 +14a31 +5a21k?40a21
?11a1k +28a1 +3k?7.
When a3 is either root of ? = 0, we have a22 < 0. We require a2 to be real, so
? negationslash= 0. As A is non-zero let k = (?B+r)2A , where
r =
radicalbig
B2 ?4AC. (5.5) disc
Then
L(5) = a2?(?r +?),
where ?,? are polynomials of degrees nine and twelve in a1,a3. Assume for the
time being that ? negationslash= 0. Let r = ???. Then
L(6) = a2?(3a1 +3a3 ?2)2?,
where ? is a polynomial of degree sixteen in a1 and a3. For consistency, r as given
by (5.5) must also be equal to ???. This is true if (3a1 + 3a3 ?2)? = 0, where ?
is a polynomial of degree twelve in a1,a3. When 3a1 + 3a3 = 2 then k = ?1, and
(1,0) is not a fine focus. The origin can be a fine focus of order greater than six
only if ? = ? = 0. We calculate the resultants of ? and ? with respect to a1 and
a3. We find
R(?,?,a3) =#(a1 ?2)3(a1 ?1)3(a1 ?5)2(2a1 ?1)(6a31 ?58a21 +183a1 ?191)
?(8a21 ?41a1 +53)(27a41 ?180a31 +478a21 ?620a1 +343)2Digamma?,
where Digamma,? are polynomials of degrees eighteen and forty in a1. The quadratic
and degree four factors are positive definite. The one real root of the cubic factor,
together with the corresponding value for a3, are such that a1 + a3 = 1, which is
excluded if (1,0) is a fine focus of order one. When a1 = 12 (or a1 = 5) then a3 = 0
(or a3 = ?3) and 2a1 +3a3 = 1, which is also excluded. Similarly, when a1 = 1 or
a1 = 2 then (k +1)(a1 ?2) ? 0.
EJDE-2007/?? EXTENDED KUKLES SYSTEM 17
Using Sturm sequences [13] we find that ? = 0 has six distinct real roots. We
also locate the corresponding roots a3, and find that all the root pairs are such
that (k + 1)(a1 ?2) ? 0. When Digamma = 0, we have ? = ? = 0. A similar analysis to
that for the case ? negationslash= 0 leads us to conclude that, when ? = 0, the origin can be of
maximum order five when (1,0) is of order one.
We conclude that L(6) negationslash= 0 under current assumptions; the origin is of maxi-
mum order six and at most seven small amplitude limit cycles can be bifurcated
simultaneously from the two fine foci. square
thm10 Theorem 5.3. Suppose that the origin and (1,0) coexist as fine foci for system
(5.1). If (1,0) is of order greater than one then both the fine foci are of maximum
order two, or both are centres.
Proof. We have L(1),M(1) given by (5.2), (5.3) respectively. Again we make a
substitution for a7 from L(1) = 0. Then
M(1) = a2(a6(2?a1)+?),
where
? = (1?a1)(a21 +a1a3 ?4a1 ?2a3 ?k +3).
If (1,0) is a fine focus then a1 negationslash= 2. Let a6 = ?a1?2. Then
L(2) = M(2) = a2(a1 ?2)(a1 +a3 ?1)?,
where ? is given by (5.4). If a2 = 0, then a5 = a7 = 0 and both critical points
are centres by Theorem 2.1 (i). Similarly, when a1 + a3 = 1 then a4 = a3,a6 =
a3(k?a3)
a3+1 ,a7 = 0 and both critical points are centres by Theorem 2.1 (iv). If a2 negationslash= 0
and (1,0) is a fine focus then (a1?2)? negationslash= 0. If L(2) is non-zero, so is M(2) . Hence
both points are fine foci of maximum order two or they are both centres. square
We can demonstrate that four small amplitude limit cycles can be bifurcated
from the two fine foci of order two of system (1.1). We begin with system (1.1)
with
? = 0, a4 = 1?a1, a5 = ?a2, a7 = ?13a2(a1 +a3 ?1), a6 = ?a
1 ?2
.
Hence the origin and (1,0) are fine foci of order two, and L(0) = M(0) = L(1) =
M(1) = 0,M(2) = L(2) negationslash= 0. First we perturb a6 such that M(1) becomes non-zero
and of opposite sign to M(2). If (2?a1)(a1+a3?1) > 0 we decrease a6, otherwise
we increase a6. The stability of (1,0) is reversed and a limit cycle bifurcates. Next
we perturb a7 such that L(1) becomes non-zero and of opposite sign to L(2), so
reversing the stability of the origin. If a2(a1 + a3 ? 1) > 0 then we decrease a7,
otherwise we increase a7. A second limit cycle bifurcates, but this time from the
origin. Another limit cycle can be bifurcated from the origin, by increasing ? if
a2(a1 + a3 ? 1) > 0 or decreasing ? otherwise. To bifurcate a fourth limit cycle
from (1,0) we require the stability of (1,0) to be reversed, so then it has the same
stability as the origin. Hence we require ?M(1) < 0, which is the case when
M(1)M(2) < 0.
Similarly seven small amplitude limit cycles can be bifurcated from the two fine
foci when one is of order one and the other is of order six.
18 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
6. Large amplitude limit cycles
We have proved that the origin can be a fine focus of order seven and we have
investigated the possibility of small amplitude limit cycles bifurcating from two
coexisting fine foci. In this case we found that the maximum number of small
amplitude limit cycles that can exist simultaneously is seven. By considering the
global phase portrait within a particular parameter range, we shall demonstrate
that a large amplitude limit cycle can surround two fine foci in system (1.1). It is
known that the Kukles system, with two fine foci of order two, can have a large
amplitude limit cycle surrounding both critical points [14].
thm11 Theorem 6.1. If the fine foci at the origin and (1,0) are both of order two for
system (1.1) then at least five limit cycles exist under certain conditions.
Proof. We begin with two fine foci, each of maximal order two. Therefore system
(1.1) is of the form
?x = y(1+kx),
?y = ?x+a1x2 +a2xy +(1?a1 +?)y2 +(1?a1)x3 ?a2x2y +Axy2 ? a2?3 y3,
(6.1) 22dist
where
? = a1 +a3 ?1, A = (1?a1)
parenleftBig
?? (a1 +k?1)(a
1 ?2)
parenrightBig
, (k +1)(a1 ?2) > 0.
One can consider system (6.1) as a perturbation of the system
?x = y(1+kx),
?y = ?x+a1x2 +(1?a1 +?)y2 +(1?a1)x3 +Axy2, (6.2) a2perturb
with the introduction of the term
? = ?a2y
parenleftBig
(x? 12)2 + ?3y2 ? 14
parenrightBig
,
when a2 is perturbed from zero. System (6.2) has centres at the origin and (1,0),
and a col at ( 1a1?1,0). We consider a particular global phase portrait of system
(6.2). We can arrange for there to be no critical points on the line kx = ?1 by
choosing values of the parameters such that
k(a1 ?2)(??a1)?k +a21(??1)+a1(2?3?)+2??1 ? 0.
For example, we can take k = 1,a1 = 3,? = 3. We then use polar coordinates to
consider the critical points at infinity. Provided that A ? 0, that is ? ? a1+k?1a1?2 ,
the only critical points at infinity lie on ? = ?pi2.
At infinity ?? ? 0, so the motion is clockwise in the region kx > ?1 and the
outward separatrices of the col cannot tend to a critical point at infinity. The
system is symmetric in the x-axis, so the separatrices form homoclinic loops and
the orbits outside the ?figure of eight? so formed are closed. Take one of these closed
orbits; ?, say. Increase a2 so that system (6.2) becomes system (6.1). For system
(6.1) the flow is inwards across ? because the vector product of the two fields is
?y2(1+kx)a2
parenleftBig
(x? 12)2 + ?3y2 ? 14
parenrightBig
.
If a2 > 0, the two fine foci are both unstable and hence there is a large amplitude
limit cycle inside ?.
EJDE-2007/?? EXTENDED KUKLES SYSTEM 19
In the previous section we demonstrated that a total of four small amplitude
limit cycles can be bifurcated from the origin and (1,0) simultaneously. Our current
assumptions are consistent with the argument therein. Therefore system (1.1) can
have five limit cycles. square
This leads us to the following conjecture.
conj12 Conjecture 6.2. When system (1.1) has two fine foci of orders six and one than
at least eight limit cycles can exist.
Concluding remarks. We have presented various properties of system (1.1). In
particular we have found sufficient conditions for the origin to be a centre by finding
complicated invariant functions that can be combined to form a Dulac function.
We conjecture that we have found the necessary and sufficient conditions for the
origin to be a centre for system (1.1) even though we were unable to complete the
reduction of the focal values because of the size of the expressions generated. We
also proved some results on the possible configurations of limit cycles.
7. Appendix
The polynomials required in the proof of condition (ii) of Theorem 2.2 are as
follows:
?0 =a43a7k4 +2a33a7k5 +2a33k4w +4a23a7w2 +2a23k5w
?2a3a7k7 ?2a3k6w?a7k8 ?2k7w,
?1 =2a63k4 +20a53a7w +2a53k5 ?5a43k6 +12a43w2 ?24a33a7k2w?6a33k7
?12a33kw2 ?4a23a7k3w +2a23k8 +4a3k9 +k10,
?2 =?4a83a7k2 ?12a73k2w +16a63a7k4 +12a63k3w +4a53a7k5 +30a53k4w
?21a43a7k6 ?72a43a7w2 ?24a43k5w?10a33a7k7 +216a33a7kw2
?24a33k6w +8a23a7k8 +12a23k7w +6a3a7k9 +6a3k8w +a7k10,
?3 =a53k4 ?8a43a7w +a43k5 ?2a33k6 +8a33w2 +8a23a7k2w
?2a23k7 +a3k8 +k9,
?4 =2a33a7 +3a23w?3a3a7k2 ?3a3kw?a7k3,
F0 =a43a7k4 +2a33a7k5 +2a33k4w +4a23a7w2 +2a23k5w?2a3a7k7 ?2a3k6w
?a7k8 ?2k7w,
F1 =2a63k4 +20a53a7w +2a53k5 +12a43a7kw?5a43k6 +12a43w2 ?12a33a7k2w
?6a33k7 ?4a23a7k3w +2a23k8 +4a3k9 +k10,
F2 =4a73a7k3 ?4a63a7k4 +12a63k3w?12a53a7k5 ?12a53k4w +8a43a7k6
+144a43a7w2 ?18a43k5w +13a33a7k7 ?216a33a7kw2 +12a33k6w
?3a23a7k8 +6a23k7w?5a3a7k9 ?a7k10,
F3 =a43k5 ?8a33a7kw +4a33w2 +8a23a7k2w?2a23k7 +k9,
F4 =2a33a7 +3a23w?3a3a7k2 ?a7k3,
?1 =96a63a7k3?2 ?32a63k4?3 +12a53a7k4?2 +144a53a7?3w?9a53k7?
20 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
?108a53k3?2w?243a43a7k7 ?282a43a7k5?2 ?36a43a7k3?w +72a43k8?
+98a43k6?3 +12a43k4?2w?486a33a7k8 ?114a33a7k6?2 +180a33k9?
+34a33k7?3 +108a33k5?2w +186a23a7k7?2 +36a23a7k5?w +18a23k10?
?66a23k8?3 ?12a23k6?2w +486a3a7k10 +102a3a7k8?2 ?171a3k11?
?34a3k9?3 +243a7k11 ?90k12?,
?2 =512a113 a7k4?3 ?384a113 k7?2 ?36a103 a7k7? +2016a103 a7k3?2w
?96a103 k8?2 ?672a103 k4?3w?1152a93a7k8? ?2624a93a7k6?3
+228a93a7k4?2w +768a93a7?3w2 +837a93k11 +2256a93k9?2
?225a93k7?w +96a93k5?3w?576a93k3?2w2 ?2106a83a7k9?
?1052a83a7k7?3 ?5211a83a7k7w?5958a83a7k5?2w?540a83a7k3?w2
+1917a83k12 +1206a83k10?2 +1575a83k8?w +2058a83k6?3w
+180a83k4?2w2 +2556a73a7k10? +4164a73a7k8?3 ?10071a73a7k8w
?2316a73a7k6?2w +1368a73a7k4?w2 +792a73a7k2?3w2 ?288a73a7?2w3
?1917a73k13 ?4572a73k11?2 +3726a73k9?w +402a73k7?3w
?783a73k7w2 ?126a73k5?2w2 +216a73k3?w3 +8226a63a7k11?
+2936a63a7k9?3 +702a63a7k9w +4008a63a7k7?2w +2484a63a7k5?w2
?168a63a7k3?3w2 ?7695a63k14 ?3462a63k12?2 ?216a63k10?w
?1488a63k8?3w?1458a63k8w2 ?54a63k6?2w2 +3528a53a7k12?
?1672a53a7k10?3 +10422a53a7k10w +2088a53a7k8?2w +576a53a7k6?w2
?2187a53k15 +3600a53k13?2 ?3861a53k11?w?480a53k9?3w
+108a53k9w2 +702a53k7?2w2 ?216a53k5?w3 ?4806a43a7k13?
?1884a43a7k11?3 +4509a43a7k11w?66a43a7k9?2w +8667a43k16
+3690a43k14?2 ?1377a43k12?w +102a43k10?3w +1458a43k10w2
?126a43k8?2w2 ?4932a33a7k14? ?380a33a7k12?3 ?351a33a7k12w
+6777a33k17 ?588a33k15?2 +360a33k13?w?18a33k11?3w +675a33k11w2
?1278a23a7k15? ?1917a23k18 ?1338a23k16?2 +18a23k14?w?3510a3k19
?312a3k17?2 ?972k20,
?3 =192a83k4?2 ?576a73a7?2w?192a73k5?2 ?64a63a7k3?3 +576a63a7k?2w
?432a63k8 ?780a63k6?2 +56a53a7k4?3 +1296a53a7k4w +576a53a7k2?2w
?432a53k9 +390a53k7?2 +48a53k3?3w +162a43a7k7? +132a43a7k5?3
+1296a43a7k5w?48a43a7k3?2w +1323a43k10 +1134a43k8?2
?32a43k4?3w +162a33a7k8? ?56a33a7k6?3 +48a33a7k4?2w +1782a33k11
?54a33k9?2 ?16a33k5?3w?162a23a7k9? ?68a23a7k7?3 ?432a23k12
?546a23k10?2 ?162a3a7k10? ?1350a3k13 ?144a3k11?2 ?459k14,
EJDE-2007/?? EXTENDED KUKLES SYSTEM 21
?4 =128a63a7?3 ?96a63k3?2 +36a53a7k3? ?12a53k4?2 ?288a43a7k4?
?264a43a7k2?3 ?48a43a7?2w +243a43k7 +282a43k5?2 +36a43k3?w
?684a33a7k5? ?136a33a7k3?3 +486a33k8 +114a33k6?2 ?360a23a7k6?
?186a23k7?2 ?36a23k5?w?486a3k10 ?102a3k8?2 ?243k11,
?5 =512a103 a7k4?3 ?96a93k8?2 ?480a93k4?3w?1116a83a7k8?
?2624a83a7k6?3 +1020a83a7k4?2w +192a83a7?3w2 ?96a83k9?2
+96a83k5?3w?2232a73a7k9? ?1088a73a7k7?3 +144a73a7k5?2w
+243a73k12 +390a73k10?2 +1197a73k8?w +1470a73k6?3w
?324a73k4?2w2 +2412a63a7k10? +4128a63a7k8?3 ?1944a63a7k8w
?2748a63a7k6?2w +828a63a7k4?w2 +648a63a7k2?3w2 ?288a63a7?2w3
+729a63k13 +492a63k11?2 +2169a63k9?w +216a63k7?3w
+108a63k5?2w2 +8352a53a7k11? +3008a53a7k9?3 ?3888a53a7k9w
?996a53a7k7?2w +1944a53a7k5?w2 ?24a53a7k3?3w2 +243a53k14
?390a53k12?2 ?450a53k10?w?1092a53k8?3w +324a53k6?2w2
+3708a43a7k12? ?1600a43a7k10?3 +1728a43a7k8?2w +1116a43a7k6?w2
?1215a43k15 ?696a43k13?2 ?2394a43k11?w?312a43k9?3w
?108a43k7?2w2 ?4824a33a7k13? ?1920a33a7k11?3 +3888a33a7k11w
+852a33a7k9?2w?1215a33k16 ?6a33k14?2 ?747a33k12?w
+102a33k10?3w?5004a23a7k14? ?416a23a7k12?3 +1944a23a7k12w
+243a23k17 +300a23k15?2 +225a23k13?w?1296a3a7k15?
+729a3k18 +102a3k16?2 +243k19,
?6 =?3072a133 a7k4?2 +2880a123 k4?2w +6912a113 a7k8 +21888a113 a7k6?2
?1152a113 a7?2w2 +256a113 k7?3 ?576a113 k5?2w +13824a103 a7k9
+9408a103 a7k7?2 ?960a103 a7k3?3w +64a103 k8?3 ?6480a103 k8w
?14580a103 k6?2w?28512a93a7k10 ?55704a93a7k8?2 ?488a93a7k4?3w
+2592a93a7k4w2 ?1584a93a7k2?2w2 ?558a93k11? ?1504a93k9?3
?11664a93k9w?2514a93k7?2w +96a93k3?3w2 ?85050a83a7k11
?44580a83a7k9?2 +2394a83a7k7?w +2844a83a7k5?3w +5184a83a7k5w2
+648a83a7k3?2w2 ?1278a83k12? ?804a83k10?3 +15957a83k10w
+24042a83k8?2w +216a83k4?3w2 ?7614a73a7k12 +51984a73a7k10?2
+5634a73a7k8?w +2264a73a7k6?3w +11340a73a7k6w2
+7440a73a7k4?2w2 +192a73a7?3w3 +1278a73k13? +3048a73k11?3
+42930a73k11w +9954a73k9?2w +414a73k7?w2 +156a73k5?3w2
?144a73k3?2w3 +137862a63a7k13 +66036a63a7k11?2 +1692a63a7k9?w
22 J. M. HILL, N. G. LLOYD, J. M. PEARSON EJDE-2007/??
?1520a63a7k7?3w +17172a63a7k7w2 +2952a63a7k5?2w2 +5130a63k14?
+2308a63k12?3 +8991a63k12w?12690a63k10?2w +972a63k8?w2
?228a63k6?3w2 +112914a53a7k14 ?2520a53a7k12?2 ?4788a53a7k10?w
?1776a53a7k8?3w +8100a53a7k8w2 ?1104a53a7k6?2w2 +1458a53k15?
?2400a53k13?3 ?29808a53k13w?6462a53k11?2w +144a53k9?w2
?252a53k7?3w2 +144a53k5?2w3 ?30942a43a7k15 ?29436a43a7k13?2
?4086a43a7k11?w?364a43a7k9?3w?324a43a7k9w2 ?5778a43k16?
?2460a43k14?3 ?17901a43k14w +774a43k12?2w?972a43k10?w2
+12a43k8?3w2 ?78570a33a7k16 ?12576a33a7k14?2 ?846a33a7k12?w
?4518a33k17? +392a33k15?3 ?486a33k15w?402a33k13?2w
?558a33k11?w2 ?35694a23a7k17 ?1428a23a7k15?2 +1278a23k18?
+892a23k16?3 ?567a23k16w?426a23k14?2w?5130a3a7k18
+2340a3k19? +208a3k17?3 ?972a3k17w +648k20?,
?7 =768a93a7k3?2 ?256a93k4?3 +192a83a7k4?2 +384a83a7?3w +18a83k7?
?288a83k3?2w?1674a73a7k7 ?3744a73a7k5?2 +180a73a7k3?w
+576a73k8? +1312a73k6?3 ?492a73k4?2w?3834a63a7k8
?2220a63a7k6?2 ?936a63a7k4?w?792a63a7k2?3w?144a63a7?2w2
+1053a63k9? +526a63k7?3 +783a63k7w +846a63k5?2w +108a63k3?w2
+2160a53a7k9 +5400a53a7k7?2 ?2304a53a7k5?w?264a53a7k3?3w
+108a53a7k3w2 ?1278a53k10? ?2082a53k8?3 +2484a53k8w
+1698a53k6?2w?36a53k4?w2 +11556a43a7k10 +4704a43a7k8?2
?1044a43a7k6?w?4113a43k11? ?1468a43k9?3 +1836a43k9w
?114a43k7?2w?108a43k5?w2 +6534a33a7k11 ?1800a33a7k9?2
+180a33a7k7?w?144a33a7k5?3w?108a33a7k5w2 ?1764a33k12?
+836a33k10?3 ?1566a33k10w?1206a33k8?2w +36a33k6?w2
?5778a23a7k12 ?2676a23a7k10?2 +36a23a7k8?w +2403a23k13?
+942a23k11?3 ?2619a23k11w?444a23k9?2w?7020a3a7k13
?624a3a7k11?2 +2466a3k14? +190a3k12?3 ?918a3k12w?1944a7k14
+639k15?.
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Journal of Differential Equation, 2000 (2000), No. 76 , pp. 1-11.
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Fran?coise and R Roussarie, Springer LNM 1455 (1990), pp. 230-242.
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Joe M. Hill
Institute of Mathematical and Physical Sciences, Aberystwyth University UK SY23
3BZ, U.K.
E-mail address: jyh@aber.ac.uk
Noel G. Lloyd
Institute of Mathematical and Physical Sciences, Aberystwyth University UK SY23
3BZ, U.K.
E-mail address: ngl@aber.ac.uk
Jane M. Pearson
Institute of Mathematical and Physical Sciences, Aberystwyth University UK SY23
3BZ, U.K.
E-mail address: jmp@aber.ac.uk