Limit cycles in generalized Lie´nard systemspublish.uwo.ca/~pyu/pub/preprints/YH_CSF2006.pdf ·...

Chaos, Solitons and Fractals 30 (2006) 1048–1068

www.elsevier.com/locate/chaos

Limit cycles in generalized Lienard systems

P. Yu a,b,*, M. Han a

a Institute of Mathematics, Shanghai Normal University, Shanghai 200234, Chinab Department of Applied Mathematics, The University of Western Ontario London, Ont., Canada N6A 5B7

Accepted 1 September 2005

Abstract

This paper presents some new results which we obtained recently for the study of limit cycles of nonlinear dynamicalsystems. Particular attention is given to small limit cycles of generalized Lienard systems in the vicinity of the origin.New results for a number of cases of the Lienard systems are presented with the Hilbert number, bH ði; jÞ ¼ bH ðj; iÞ, forj = 4, i = 10,11,12,13; j = 5, i = 6,7,8,9; and j = 6, i = 5,6. Detailed proofs for the existence of limit cycles are given infour cases.� 2005 Elsevier Ltd. All rights reserved.

1. Introduction

Since Hilbert [1] presented the well-known 23 mathematical problems to the Second International Congress ofMathematicians in 1990, great advances inspired by the famous address have been achieved in many different mathe-matical research areas. However, two of the 23 problems remain unsolved after more than 100 years. One of them is the16th problem titled as the Problem of the topology of algebraic curves and surfaces. This problem includes two parts. Inthe first part, Hilbert suggested a thorough investigation of the relative positions of separate branches of algebraiccurves in nth-order vector fields, which is in the area of real algebraic geometry. In the second part of the problem, Hil-bert asked to consider the upper bound of the number of limit cycles and their relative locations in polynomial vectorfields of order n. This part of problem is related to ordinary differential equations and dynamical systems. Generally, thesecond part of the problem is what usually meant when talking about Hilbert�s 16th problem. Recently, a modern ver-sion of the second part of Hilbert�s 16th problem was formulated by Smale, chosen as one of the 18 challenging math-ematical problems for the 21st century [2]. To be more specific, consider the following planar system:

0960-0doi:10

* CoE-m

_x ¼ P nðx; yÞ; _y ¼ Qnðx; yÞ; ð1Þ

where the dot denotes differentiation with respect to time t, and Pn and Qn are nth-degree polynomials of x and y. Thesecond part of Hilbert�s 16th problem is to find the upper bound K = H(n) 6 nq on the number of limit cycles that thesystem can have, where q is a universal constant. More precisely, the statement of second part of Hilbert�s 16th problemis given below.

779/$ - see front matter � 2005 Elsevier Ltd. All rights reserved..1016/j.chaos.2005.09.008

rresponding author. Fax: +1 519 661 3523.ail address: [email protected] (P. Yu).

mailto:[email protected]

P. Yu, M. Han / Chaos, Solitons and Fractals 30 (2006) 1048–1068 1049

The second part of Hilbert’s 16th problem estimate a uniform upper bound for the number of limit cycles of system(1), H(n), for any n 2 Z+.

Traditionally, the second part of 16th problem is split into three problems.

Problem 1. Is it true that a planar polynomial vector field has but a finite number of limit cycles?

Problem 2. Is it true that the number of limit cycles of a planar polynomial vector field is bounded by a constantdepending on the degree of polynomials only?

Problem 3. Give an upper bound for H(n).

These three problems are related in chain. Solving Problem 3 implies both Problems 1 and 2 solved; while solvingProblem 2 only means Problem 1 solved. Problem 1 has been solved if it is restricted to a given individual polynomialsystem, and Problem 2 is partially solved; while Problem 3 is still open even for n = 2. In fact, it is only known so farthat H(2) P 4.

Although it has not been possible to obtain the uniform upper bound for H(n), various efforts have been made infinding the maximal number of limit cycles and raising the lower bound of Hilbert number H(n) for general planar poly-nomial systems or for individual degree of systems, through which people hope to get close to the estimation of theupper bound of H(n). Even for estimating the lower bound of H(n) it is, in general, a very difficult problem, in partic-ular, for determining large (global) limit cycles. A recent survey article [3] (and more references therein) has compre-hensively discussed this problem and reported the latest progress.

If Hilbert�s 16th problem is restricted to the neighborhood of isolated fixed points, then the problem is reduced tostudying degenerate Hopf bifurcations, which gives rise to studying fine focus points of nonlinear dynamical systems. Inthe past 50 years, many researchers have considered the local problem and obtained a series of results (e.g., see [4–11]).In the last two decades, much progress on finite cyclicity near a fine focus point or a homoclinic loop has been achieved.For a quadratic system, it is well known that the maximal number of small limit cycles is three [4]. For cubic ordersystems, the best result published so far is twelve limit cycles, found from the following system [12–17]:

_x ¼ axþ by þ a30x3 þ a21x2y þ a12xy2 þ a03y3;

_y ¼ bxþ ay þ b30x3 þ b21x2y þ b12xy2 þ b03y3;ð2Þ

where a, b, aij�s and bij�s are constant coefficients.The key step in finding the number of limit cycles of a dynamical system in the neighborhood of a fixed point (which

is a linear center) is to compute the focus values of the point. This is equivalent to calculating the normal form of thesystem associated with Hopf singularity. Many methods have recently been developed for computing normal formswith the aid of computer algebra systems such as Maple, Mathematica, Reduce, etc. Among various methods, a per-turbation technique [18] has been proved computationally efficient, which combines the computations of normal formand center manifold into a unified approach.

Since this paper does not intend to discuss normal form computation, we assume that the normal form of system (1)has been obtained in polar coordinates (interest readers can find the details of normal form computation in [18]), givenby

_r ¼ rðv0 þ v1r2 þ v2r4 þ � � � þ vkr2kÞ; ð3Þ_h ¼ xþ t1r2 þ t2r4 þ � � � þ tkr2k ; ð4Þ

up to the (2k + 1)th order term, where both vk and tk are expressed in terms of the original system�s coefficients. vk iscalled the kth-order focus value of Hopf-type critical point (the origin). The zeroth order focus value, v0, is obtainedfrom linear perturbation.

The bifurcation solutions and their stabilities of system (1) near the origin are determined by Eq. (3). The basic ideaof finding k small limit cycles around the origin is as follows: First, find the conditions such that v1 = v2 = � � � = vk�1 = 0(v0 = 0 is automatically satisfied at the critical point), but vk 5 0; and then perform appropriate small perturbations toprove the existence of exact k limit cycles. This indicates that the procedure for finding multiple limit cycles involves twosteps: Computing the focus values (i.e., computing the normal form) and solving the coupled nonlinear polynomialequations: v1 = v2 = � � � = vk�1 = 0. In the following, we give two theorems for sufficient conditions of the existenceof small amplitude limit cycles, which will be used in the following sections. (The proofs of the theorems can be foundin [15–17].)

1050 P. Yu, M. Han / Chaos, Solitons and Fractals 30 (2006) 1048–1068

Theorem 1. If the focus values vi given in Eq. (3) satisfy the following conditions:

viviþ1 < 0 and jvij � jviþ1j � 1; for i ¼ 0; 1; 2; . . . ; k � 1;

then the polynomial equation given by _r ¼ 0 in Eq. (3) has k positive real roots for r2, and thus system (1) has exactly k limit

cycles.

In many cases, vj depends on k parameters:

vj ¼ vjðc1; c2; . . . ; ckÞ; j ¼ 0; 1; . . . ; k; ð5Þ

and at the critical point (c1,c2, . . . ,ck) = (c1c,c2c, . . . ,ckc), vj = 0, j = 0,1, . . . ,k. One can then perturb these parameters asci = cic + �i, i = 1,2, . . . ,k to obtain k limit cycles. In this case, the following theorem is more convenient in applications.

Theorem 2. Suppose that the condition (5) holds, and further assume that

vkðc1c; c2c; . . . ; ckcÞ 6¼ 0; vjðc1c; c2c; . . . ; ckcÞ ¼ 0; j ¼ 0; 1; . . . ; k � 1;

and

detoðv0; v1; . . . ; vk�1Þoðc1; c2; . . . ; ckÞ

ðc1c; c2c; . . . ; ckcÞ� �

6¼ 0. ð6Þ

Then for any given �0 > 0, there exist �1, �2, . . . , �k and d > 0 with j�jj < �0, j = 1,2, . . . , k such that equation _r ¼ 0 has exactly

k real positive roots of r2, and thus the corresponding original dynamical system has exactly k limit cycles in a d-ball

centered at the origin.

2. The Lienard system

In this section, we consider a simplified version of Hilbert�s 16th problem—the Lienard equation. Most of the earlyhistory in the theory of limit cycles was stimulated by practical problems displaying periodic behaviour. For example,the differential equation derived by Rayleigh [19] in 1877, related to the oscillation of a violin string, is given by

€xþ � 13_x2 � 1

� �_xþ x ¼ 0; ð7Þ

where � is a small perturbation parameter. Following the invention of the triode vacuum tube, which was able to pro-duce stable self-excited oscillations of constant amplitude, van der Pol [20] obtained the following differential equationto describe this phenomenon:

€xþ �ðx2 � 1Þ _xþ x ¼ 0. ð8Þ

Systems (7) and (8) can both display periodic behaviour.Perhaps the most famous class of differential equations, which generalize Eq. (8), are those first investigated by Lie-

nard [21] in 1928:

€xþ f ðxÞ _xþ gðxÞ ¼ 0; ð9Þ

where f(x) and g(x) are polynomial functions of x. Letting _x ¼ �y, one obtains the above system described in phase plane:

_x ¼ �y; _�y ¼ �gðxÞ � f ðxÞ�y. ð10Þ

Further, let �y ¼ y � F ðxÞ, where F ðxÞ ¼R x

0f ðsÞds. Then we have the following equivalent system:

_x ¼ �y ¼ y � F ðxÞ; _y ¼ _�y þ dFdx

_x ¼ �gðxÞ � f ðxÞ�y þ f ðxÞ�y ¼ �gðxÞ. ð11Þ

Let bH ði; jÞ denote the maximal number of small-amplitude limit cycles bifurcating from the origin of system (11) (or(9)), where i and j are the degrees of f and g, respectively. Then the existing results as well as the new results whichwe obtained recently for the Lienard system (11) are summarized in Table 1, where the numbers marked by * are givenin [22], while those bold-faced numbers are our new results. Note that the numbers given in this table are symmetricwith respect to f and g [23], i.e., bH ði; jÞ ¼ bH ðj; iÞ. Thus, one only needs to prove for cases i P j. It should be also notedthat the notation bH ði; jÞ as used in [23] denotes the possible maximal number of small limit cycles which may exist in thevicinity of the origin. It does not include global (large) limit cycles, nor contain other local (small) limit cycles, whichmay appear in the neighborhood of other non-zero focus points. It will be shown in our proofs given in the next sectionthat the maximal numbers listed in Table 1 can be actually reached since the sufficient conditions given in Theorems 1and 2 can be satisfied by choosing appropriate parameter values.

Table 1The values of bH ði; jÞ for the generalized Lienard systems when f and g are of varying degrees


Recently, there has been a great deal of progress when considering the bifurcation of small amplitude limit cyclesfrom the origin for generalized Lienard systems of the form:

_x ¼ /ðyÞ � F ðxÞ;_y ¼ �gðxÞ;

�ð12Þ

where /(y) is in the polynomial form:

/ðyÞ ¼ y þ c2y2 þ c3y3 þ � � � . ð13Þ

For definite, further let

f ðxÞ ¼ a1 þ 2a2xþ 3a3x2 þ 4a4x3 þ � � � ;gðxÞ ¼ b1xþ b2x2 þ b3x3 þ b4x4 þ � � � ;

ð14Þ

then

F ðxÞ ¼ a1xþ a2x2 þþa3x3 þ � � � . ð15Þ


Without loss of generality, one may assume that b1 = 1. Next, introduce the following transformation [24]:

u2 ¼ 2GðxÞ ¼ 2

Z x

0

gðsÞds; uð0Þ ¼ 0; u0ð0Þ > 0; ð16Þ

so that u2 = x2 + O(x3) as x! 0. Therefore, this transformation defines an analytic change of coordinates in a neigh-borhood of the origin. Let the inverse transformation be given by

x ¼ nðuÞ ¼ uþOðu2Þ as u! 0; ð17Þ

and thus,

gðnðuÞÞu

! ðg0ð0ÞÞ1=2

1. ð18Þ

Finally, in the (y,u) coordinate plane, we obtain the new system:

_y ¼ �gðnðuÞÞ;_u ¼ u�1gðnðuÞÞ½/ðyÞ � F �ðuÞ�.

�ð19Þ

By the condition (18), the local qualitative behaviour of system (19) is identical to the following system:

_y ¼ �gðnðuÞÞ ¼ �u;

_u ¼ /ðyÞ � F �ðuÞ ¼ /ðyÞ � A1uþ A2u2 þ A3u3 þ A4u4 þ A5u5 þ � � �ð Þ;

�ð20Þ

where A1 = a1, A2 = a2, and other Ai�s are given explicitly in terms of ai�s and bi�s. The explicit expressions of Ai�sobtained from a Maple program are listed in Appendix A.

It is easy to see from Eqs. (13) and (20) that the origin (0,0) is a linear center. Further, one can apply the Mapleprogram developed in [18] to system (20) to obtain the normal form (3) with the following focus values:

v0 ¼ A1;

v1 ¼ �3

8A3;

v2 ¼ �5

16A5 �

5

24A2

2A3;

v3 ¼ �35

128A7 �

205

1152A2

4A5 �1885

13824A2

4 þ2

3A2A4 þ

999

8192A2

3

� �A3;

v4 ¼ �63

256A9 �

413

2304A2

2A7 �47

96A2A4 þ

2115

4096A2

3 þ4297

41472A4

2

� �A5

� 141

160A2A6 þ

149

240A2

4 þ1093

1152A3

2A4 þ20599

49152A2

2A23 þ

109483

1244160A6

2

� �A3;

..

.

ð21Þ

It follows from Eq. (21) that

v0 ¼ v1 ¼ v2 ¼ v3 ¼ 0; v4 6¼ 0 () A1 ¼ A3 ¼ A5 ¼ A7 ¼ 0; A9 6¼ 0.

In general, one can show that

v0 ¼ v1 ¼ v2 ¼ � � � ¼ vk�1 ¼ 0; vk 6¼ 0 () A1 ¼ A3 ¼ A5 ¼ � � � ¼ A2k�1 ¼ 0; A2kþ1 6¼ 0.

Therefore, in order for system (20) or (12) to have k small limit cycles, one must have that v0 = v1 = v2 = � � � = vk�1 = 0,but vk 5 0, or A1 = A3 = A5 = � � � = A2k�1 = 0, but A2k+1 5 0. For example, when A1 = 0, but A3 5 0, the system (12)has at most one limit cycle around the origin; when A1 = A3 = 0, A5 5 0, the system (12) has at most two limit cycles;etc. Since the coefficients Ai�s are given in terms of ai�s and bi�s, one needs to determine the values of ai�s and bi�s suchthat the above necessary conditions are satisfied. Then, with the sufficient conditions given in Theorems 1 and 2, we canapply appropriate perturbations to obtain exactly k limit cycles.

The new results we have obtained are shown in Table 1 with bold-faced numbers. There are in total 17 cases: bH ði; jÞfor j = 4, i = 10,11,12,13; j = 5, i = 6,7,8,9; and j = 6, i = 5,6. Due to symmetry, only nine cases need proof. In thenext section, we give proofs for four cases: bH ð13; 4Þ, bH ð6; 5Þ, bH ð9; 5Þ and bH ð6; 6Þ, which are circled in Table 1. Othercases can be proved in a similar manner.


3. Proofs for cases bHð13; 4Þ, bHð6; 5Þ, bHð9; 5Þ and bHð6; 6ÞIn this section, we will prove the cases bH ð13; 4Þ, bH ð6; 5Þ, bH ð9; 5Þ and bH ð6; 6Þ. We start from the case H(6,5), and

then consider in sequence, H(9,5), bH ð13; 4Þ and bH ð6; 6Þ.3.1. bH ð6; 5Þ ¼ 8

To show H(6,5) = 8, we need to solve the equations:

A1 ¼ A3 ¼ A5 ¼ A7 ¼ A9 ¼ A11 ¼ A13 ¼ A15 ¼ 0 ðbut A17 6¼ 0Þ.
Note that A1 = a1 = 0 leads to a1 = 0 (so v0 = 0), and thus in the rest of the paper we assume that a1 = 0. Also note thatfor H(6,5) = 8, we must show that it is not possible to have A17 = 0. Therefore, once we find all solutions from the seven(since A1 = a1 = 0) nonlinear polynomial equations, we need to verify A17 5 0 for all of these solutions. (Alternatively,we may add the equation A17 = 0 to the seven equations, and then show that there are no solutions for the eight poly-nomial equations.) For this case ai = 0 for i P 8, and bj = 0 for j P 6.
It is seen from Appendix A that one can use the coefficients a3, a5 and a7 to solve the equations A3 = 0, A5 = 0 andA7 = 0, respectively, to obtain

a3 ¼2

3a2b2;

a5 ¼2

5a2b4 þ

4

3a4b2 �

2

3a2b2b3;

a7 ¼ 2b2a6 �2

3a2b2b5 þ

4

5a4 �

2

5a2b3

� �b4 �

8

9b3

2 þ4

3b2b3

� �a4 þ

2

3b3 þ

4

9b2

2

� �a2b2b3.

ð22Þ

Then substituting the above results into the equation A9 = 0 and solving for a6 yields

a6 ¼1

9ð81b4 � 135b2b3 � 220b32Þð810a4b2 þ 243a2b4 � 810a2b2b3 � 660a2b3

2Þb5

þð486a4b3 þ 1188a4b2

2 � 594a2b22b3 � 243a2b2

3Þb4 � ð810b23 þ 2520b3b2

2 þ 880b42Þb2a4

þð405b23 þ 1260b2

2b3 þ 440b42Þa2b2b3

. ð23Þ

Having determined a3, a5, a7 and a6, the remaining three equations A11 = A13 = A15 = 0 have four coefficients b2, b3, b4

and b5. This seems to suggest that we may use the four coefficients to find the conditions such thatA11 = A13 = A15 = A17 = 0 but A19 5 0, which implies that H(6,5) = 9. However, a careful examination reveals thatthe three equations A11 = A13 = A15 = 0 are actually homogeneous under the following substitutions:

b3 ¼ B3b22; b4 ¼ B4b3

2; b5 ¼ B5b42. ð24Þ

Under the above transformation, A11, A13 and A15 can be rewritten as

A11 ¼ b102bA11ðB3;B4;B5Þ; A13 ¼ b12

2bA13ðB3;B4;B5Þ; A15 ¼ b14

2bA15ðB3;B4;B5Þ;

where bA11, bA13 and bA15 are given as follows (from computer output):

Since b2 5 0 (otherwise it leads to a center), there exist only three independent coefficients, B3, B4 and B5, and thus the
best possibility is that A11 = A13 = A15 = 0, but A17 5 0, implying bH ð6; 5Þ 6 8. Note that if there is a solution such thatA11 = A13 = A15 = 0 and in addition, A17 = 0, then it is a center.


Now eliminating B5 from the two equations: bA11 ¼ bA13 ¼ 0 results in the solution:

and a resultant:

Then substituting the solution of B5 into bA15 yields another resultant:

Notice that f1 and f2 have common factors (5B3 � 3B4)(135B3 � 81B4 + 220) which generate centers. Thus, the solu-tions must come from the remaining factors. Finally, eliminating B4 from the two remaining factors yields the followingsolution:


and a single variable resultant polynomial for B3:

Then all possible solutions are given by the roots of f3 = 0. However, it is easy to check that for the solution B3 = 0, A17

and higher A�s are zero, indicating that this solution generates a center. For the next four roots obtained from the sec-ond, third and fourth factors: B3 ¼ � 496

63;� 256

189; 3592�56

ffiffiffiffiffiffiffi3729p

945, one can show that they are not solutions of the original

equations A11 = A13 = A15 = 0. Therefore, possible solutions are only given by the last factor of f3, which gives threereal roots:

B3 ¼ �2:775113717338634317465290081021;

� 0:106020068761590624161788916358; 0:338225308918069273869422373474.

However, since b2, a2 and a4 are free, there exist infinite solutions such that A11 = A13 = A15 = 0, but A17 5 0. In otherwords, bH ð6; 5Þ 6 8. For example, for the following critical parameter values:

a2c ¼ a4c ¼ b2c ¼ 1; a3c ¼ 0:4;

a5c ¼ 0:73016022981614408282188677826;

a6c ¼ 0:30571999879402330901453525809;

a7c ¼ 0:07340553967200663656881659409;

b3c ¼ 0:12176111121050493859299205445;

b4c ¼ �0:52838314249134854352290999887;

b5c ¼ �0:04507061696602359682310293153;

ð25Þ

we have A3 = A5 = � � � = A15 = 0, but A17 = 0.0000444424 5 0, which leads to v1 = v2 = � � � = v7 = 0, but v8 5 0. Itshould be noted that all the computations performed in this paper with Maple are symbolic, except for the last stepin which a polynomial equation of single variable is solved numerically.

For the parameter values given in Eq. (25) which are up to 30 decimal places, we applied our Maple program [18] toobtain the focus values as follows (using accuracy up to 30 digits):

v1 ¼ 0:0;

v2 ¼ �0:5� 10�30;

v3 ¼ �0:35� 10�29;

v4 ¼ 0:435� 10�28;

v5 ¼ �0:169045� 10�26;

v6 ¼ 0:8986185� 10�25;

v7 ¼ �0:29858535� 10�23;

v8 ¼ �0:00000824275114436717659599733.

Note that due to numerical accumulating errors, the accuracy in the focus values vi, i = 1,2, . . . , 7 are not all up to 10�30.When the accuracy is increased, say, to 1000 decimal places, the errors in vi, i = 1,2, . . . , 7 are close to 10�1000, while v8 isnot changed up to 30 digits.

The above procedure only shows bH ð6; 5Þ 6 8. In order to prove bH ð6; 5Þ ¼ 8, we need to choose proper perturba-tions. To achieve this, we perturb the critical parameter values given in Eq. (25) to obtain exactly eight small amplitudelimit cycles around the origin. The main idea is as follows: First, it is observed from Eqs. (21)–(23) and the formulasgiven in Appendix A that one may perturb v4, v3, v2, and v1 in backward order by using the parameters a6, a7, a5 and a3

one after another. However, one cannot perturb v7, v6 and v5 one by one since the three equations A11 = A13 = A15 = 0are solved simultaneously using b3, b4 and b5. We may use Theorem 2 to show that such perturbations exist. To achievethis, we calculate the Jacobian of the three equations A11 = A13 = A15 = 0 with respect to b3, b4 and b5 at the criticalpoint to obtain


Jc ¼0:01332022483200854032 0:02194521419830844260 0:04972440391917752951

0:00127180514301147982 0:02099197081467864011 0:03779060967872986360

�0:00576530176067044521 0:01592514867446592125 0:00753349557472660287

264

375

and so det(Jc) = �0.0000038765 5 0. Combing the above two perturbation steps shows that there exist perturbations:�i, i = 1,2, . . . , 8 such that b3 = b3c + �1, b4 = b4c + �2, b5 = b5c + �3, a6 = a6c + �4, a7 = a7c + �5, a5 = a5c + �6, a3 =a3c + �7, a1 = �8, under which

viviþ1 < 0 and jvij � jviþ1j � 1; for i ¼ 0; 1; 2; . . . ; 7.

This indicates that

bH ð6; 5Þ ¼ 8.

Similarly, we can prove that bH ð7; 5Þ ¼ 9 and bH ð8; 5Þ ¼ 10.

3.2. bH ð9; 5Þ ¼ 11

In this subsection, we prove bH ð9; 5Þ ¼ 11. For this case, ai = 0 for i P 10, and bj = 0 for j P 6. Similar to the prooffor the case bH ð6; 5Þ, we need to show that for properly chosen parameter values, A3 = A5 = � � � = A21 = 0, but A23 5 0,and further to show that it is not possible to have A3 = A5 = � � � = A21 = A23 = 0. Thus, bH ð9; 5Þ ¼ 11. We need to solve10 nonlinear polynomial equations.

Similarly, we can solve the equations A3 = A5 = � � � = A15 = 0 one by one by using the coefficients a3, a5, a7, a9, a10,a8 and a6, respectively. The solutions for a3, a5 and a7 are given in Eq. (22), while the lengthy expressions for a9, a10, a8

and a6 are omitted here for simplicity. Then the remaining three equations are A17 = A19 = A21 = 0. Similarly, one canverify that these equations are homogeneous under the following transformation:

b3 ¼ B3b22; b4 ¼ B4b3

2; b5 ¼ B5b42;

which leads to the expressions

A17 ¼ b242bA17ðB3;B4;B5Þ; A19 ¼ b26

2bA19ðB3;B4;B5Þ; A21 ¼ b28

2bA21ðB3;B4;B5Þ.

Since b2 5 0 (otherwise it leads to a center), there are only three free coefficients, B3, B4 and B4. Thus, it is not possibleto have solutions for A17 = A19 = A21 = A23 = 0. The best result will be A17 = A19 = A21 = 0, but A23 5 0, implyingthat bH ð9; 5Þ 6 11.

Next, eliminating B5 from the three equations bA17 ¼ bA19 ¼ bA21 ¼ 0 results in two polynomial equations: ~f i ¼ Ff i,i = 1,2, where F is a common factor, and fi, i = 1,2 are functions of B3 and B4. The computer output of the two func-tions is given in Appendix B. Then eliminating B4 from the two equations: f1 = f2 = 0 yields an equationf3 = B3f31(B3)f32(B3) = 0, where f31(B3) is a 9th-degree polynomial of B3 while f32(B3) is a 108th-degree polynomialof B3. With the built-in program in Maple, fsolve, one can show that f31(B3) has three real solutions, and f32(B3) has24 real solutions. However, verifying the original equations shows that A21 5 0 for all the 24 solutions. Hence, the sec-ond polynomial f32(B3) can only have solutions for bH ð9; 5Þ 6 10. The three real solutions obtained from f31(B3) = 0 dosatisfy A17 = A19 = A21 = 0, but A23 5 0, and hence bH ð9; 5Þ 6 11. Actually, we have infinitely many solutions since a2,a4 and b2 can be chosen arbitrarily.

For an example, we choose the following critical parameter values:

a2c ¼ a4c ¼ b2c ¼ 1; a3c ¼2

3;

a5c ¼ 1:34796265977228916492295656132;

a6c ¼ �4:47654700356991805278499402760;

a7c ¼ �9:79768652019339892191312380939;

a8c ¼ �6:46099608912640900311153589268;

a9c ¼ �1:41527303119400585959220295145;

a10c ¼ �0:0000002973729537349992497518;

b3c ¼ �2:97633366517505075983960102126;

b4c ¼ �4:92398279252769502075861029881;

b5c ¼ �1:94146639271750592836680707408;

ð26Þ


under which A3 = A5 = � � � = A21 = 0, but A23 5 0, and so v1 = v2 = � � � = v10 = 0, but v11 5 0. For the parameter val-ues given in Eq. (26), one can execute the Maple program [18] to obtain the focus values, given below (up to 30 decimalplaces):

v1 ¼ 0:0;

v2 ¼ 0:125� 10�28;

v3 ¼ 0:489335138474086742105756054277� 10�28;

v4 ¼ 0:249526710531812489005475943356� 10�27;

v5 ¼ 0:14935� 10�25;

v6 ¼ �0:157139398481021133435166890563� 10�23;

v7 ¼ 0:190592875� 10�21;

v8 ¼ �0:274376373214617212563568726550� 10�19;

v9 ¼ 0:5402669376055� 10�17;

v10 ¼ �0:114560531574081022421386528274� 10�15;

v11 ¼ 0:000001071677889113200802290758.

As the final step, similar to the proof for the case bH ð6; 5Þ, one can choose proper perturbations to show that there existexactly 11 small limit cycles for this case, i.e.,

bH ð9; 5Þ ¼ 11.

3.3. bH ð13; 4Þ ¼ 13

Now we turn to the case bH ð13; 4Þ. For this case, we need to show that there exist parameter values satisfyingA3 = A5 = � � � = A25 = 0, but A27 5 0, and it is not possible to have A3 = A5 = � � � = A25 = A27 = 0. Thus,bH ð13; 4Þ 6 13. Here, ai = 0 for i P 15, and bj = 0 for j P 5. It requires to solve 12 nonlinear polynomial equations.

Similarly, we can solve the equations A3 = A5 = � � � = A21 = 0 one by one using the coefficients a3, a5, a7, a9, a11, a13,a14, a12, a10 and a8, respectively. The solutions for a3, a5 and a7 are the same as that given in Eq. (22) and other lengthyexpressions are omitted here. The remaining two equations are A23 = A25 = 0. Again, an examination shows that thesetwo equations are homogeneous under the following substitutions:

b3 ¼ B3b22; b4 ¼ B4b3

2;

which results in the following expressions:

A23 ¼ �8b17

2 18a6 þ 9a2B23b4

2 þ 4a2b42B3 � 18a4B3b2

2 � 8a4b22

� �B3

4

11390625F ðB3;B4Þ f1ðB3;B4Þ;

A25 ¼1798b19

2 18a6 þ 9a2B23b4

2 þ 4a2b42B3 � 18a4B3b2

2 � 8a4b22

� �B3

4

854296875F ðB3;B4Þ f2ðB3;B4Þ;

where F(B3,B4) is a polynomial of B3 and B4. Thus, it is not possible to have solutions satisfying A23 = A25 = A27 = 0since there are only two free parameters. Hence, bH ð13; 4Þ 6 13. Note from the above expressions of A23 and A25 thatthe common factor involves the coefficients a2, a4, a6 and b2, implying that these four coefficients are not used and canbe chosen arbitrarily. Thus, we only need to find solutions such that A23 = A25 = 0, but A27 5 0. That is, solvef1 = f2 = 0 for B3 and B4. The computer output of the polynomials f1 and f2 are given by


Eliminating B3 from the two equations: f1 = f2 = 0 yields an equation f3 = B4f31(B4)f32(B4) = 0 where f31(B4) is a 22nd-
degree polynomial of B4 while f32(B4) is a 38th-degree polynomial of B4. With the Maple program, fsolve, one can showthat both f31(B4) and f32(B4) have four real solutions. However, the 4 solutions obtained from f31(B4) = 0 give A23 5 0.Hence, the first polynomial f31(B4) have solutions only for bH ð13; 4Þ 6 12. The four real solutions obtained from


f32(B4) = 0 do satisfy A23 = A25 = 0, but A27 5 0, and hence bH ð13; 4Þ 6 13. Actually, we have infinitely many solutionssince a2, a4, a6 and b2 can be chosen arbitrarily.

For an example, we take the following critical parameter values:

a2c ¼ a4c ¼ a6c ¼ 1; b2c ¼ 0:6; a3c ¼ 0:4;

a5c ¼ 0:92577493875946589222204765642;

a7c ¼ 1:31683316572770041230933563844;

a8c ¼ �0:69104631585744219661617073754;

a9c ¼ �1:42088630512148238929036418631;

a10c ¼ �4:44585161428937711400653059843;

a11c ¼ �7:95270132242953255167828316545;

a12c ¼ �4:38693164849870678874782234217;

a13c ¼ �0:43827437598651465881164929251;

a14c ¼ 0:05627618126491330173708737328;

b3c ¼ �1:92387593712707219576747330285;

b4c ¼ �1:60943859022840746521235416180;

ð27Þ

under which A3 = A5 = � � � = A25 = 0, but A27 5 0, and thus v1 = v2 = � � � = v12 = 0, but v13 5 0. For the parametervalues given in Eq. (27), the computed focus values are given below (noting that the accuracy for this case is takenup to 100 decimal places since this case has more focus values):

v1 ¼ 0:0;

v2 ¼ 0:2� 10�99;

v3 ¼ 0:7� 10�99;

v4 ¼ 0:37� 10�98;

v5 ¼ 0:2165� 10�97;

v6 ¼�0:385� 10�97;

v7 ¼ 0:130275� 10�94;

v8 ¼�0:28609925� 10�92;

v9 ¼ 0:78087682� 10�90;

v10 ¼�0:1786023085� 10�87;

v11 ¼ 0:61468531009315� 10�85;

v12 ¼�0:2114286075965054� 10�82;

v13 ¼�0:00073252213821505282085563581471476925109268994079231794163577001601047895350909233731067569529815.

It should be noted that the focus values given above, vi, i = 1,2, . . . , 12 are supposed to be zero, but they are not zero dueto accumulating errors. If we increase the accuracy to, say, 1000 digit points, then all these focus values are close to10�1000, except for v13 which is not changed up to the 100 digits. Finally, one can employ appropriate perturbationsto show that there exist exactly 13 small amplitude limit cycles. This proves that

bH ð13; 4Þ ¼ 13.

3.4. bH ð6; 6Þ ¼ 8

Finally, we consider the case bH ð6; 6Þ and prove that bH ð6; 6Þ ¼ 8. For this case, ai = 0 for i P 8 and bi = 0 for i P 7.Here we need to find parameter values which satisfy A3 = A5 = � � � = A15 = 0, but A17 5 0, and should also prove thatit is not possible to have A3 = A5 = � � � = A15 = A17 = 0. For this case, we will show that there do not exist solutionssatisfying A3 = A5 = � � � = A15 = A17 = 0. Therefore, we need to solve eight nonlinear polynomial equations.


Similarly, we can solve the equations A3 = A5 = A7 = A9 = 0 one by one using the coefficients a3, a5, a7 and b6,respectively. The solutions for a3, a5 and a7 are given in Eq. (22), and b6 is given below:

b6 ¼7

1215ð2a4 � a2b3Þð1215b2b3 þ 1980b3

2 � 729b4Þa6 þ ð810a4b2 þ 243a2b4 � 810a2b2b3 � 660a2b32Þb5

þð486b3b4 þ 1188b4b2

2 � 810b23b2 � 2520b3b3

2 � 880b52Þa4 � 243b3b4

þð440b42 þ 1260b2

2b3 þ 405b23 � 594b2b4Þa2b2b3

.

Then the remaining four equations are A11 = A13 = A15 = A17 = 0. To solve these equations, we may first eliminate thecoefficient a6 from these four equations to obtain three resultant equations:

~f 1 ¼ ða2b3 � 2a4Þf1ðb2; b3; b4; b5Þ ¼ 0;

~f 2 ¼ ða2b3 � 2a4Þf2ðb2; b3; b4; b5Þ ¼ 0;

~f 3 ¼ ða2b3 � 2a4Þð81b4 � 135b2b3 � 220b32Þf3ðb2; b3; b4; b5Þ ¼ 0.

Ignoring the common factor, we need to solve the equations f1(b2,b3,b4,b5) = f2(b2,b3,b4,b5) = f3(b2,b3,b4,b5) = 0. Notethat now there are no coefficients ai�s involved in these equations. It seems that there are four coefficients in the threeequations. However, an examination shows that these equations are actually homogeneous under the followingsubstitutions:

b3 ¼ B3b22; b4 ¼ B4b3

2; b5 ¼ B5b42.

Then, further ignoring the free parameter b2, we need to solve the three equations: f 1 ¼ f 2 ¼ f 3 ¼ 0, where f i arefunctions of B3, B4 and B5. The computer output of the three functions are given in Appendix C.

Eliminating B5 from the three equations: f 1 ¼ f 2 ¼ f 3 ¼ 0 yields two equations:

f4 ¼ ð5B3 � 3B4Þf41ðB3;B4Þf42ðB3;B4Þ;

f5 ¼ ð5B3 � 3B4Þð81B4 � 135B3 � 220Þf51ðB3;B4Þf52ðB3;B4Þ;

where f41, f42, f51 and f52 are, respectively, 5th-, 11th-, 8th- and 15th-degree polynomials of B3 and B4. To solvef4 = f5 = 0, first note that the factor 81B4 � 135B3 � 220 does not give a solution. So we need to solve four groupsof equations:

first group: (f41, f51) = (0,0),second group: (f41, f52) = (0,0),third group: (f42, f51) = (0,0),fourth group: (f42, f52) = (0,0).

For each of these groups, first eliminate B4 and then solve the resulting equation for B3. The first group yields the poly-nomial equation for B3:

F ¼ 1371807839396311815366800796367125B53 � 12745608963016320639029647390357350B4

3

þ 46489461997527105443258799808067400B33 � 83034261563486218004432710846072368B2

3

þ 72584263326198471403294587088309392B3 � 24848924199546145630434699613010656;

which has three real solutions, satisfying f 1 ¼ f 2 ¼ 0, but f 3 6¼ 0. Substituting these solutions back into the originalequations indeed yields A11 = A13 = A15 = 0, but A17 5 0. Similarly, it can be shown that the second group has 16 realsolutions, and the result is similar to the first group. That is, all the 16 solutions satisfy A11 = A13 = A15 = 0, butA17 5 0. The third and fourth groups do not have real solutions satisfying A11 = A13 = A15 = 0. They, instead, havesolutions only for A11 = A13 = 0, but A15 5 0. Therefore, combining the other parameter values, the best result for thiscase is to choose parameter values such that v1 = v2 = � � � = v7 = 0, but v8 5 0, and thus bH ð6; 6Þ 6 8.


An example for selecting the critical parameter values is given below:

a2c ¼ a4c ¼ b2c ¼ 1; a3c ¼2

3;

a5c ¼ 1:01000106396363001306429064012;

a6c ¼ 0:67235076750720532893297490162;

a7c ¼ 0:30067436107555057220956101983;

b3c ¼ 1:19684216379096679633485259370;

b4c ¼ 1:18640626622735302655214758979;

b5c ¼ 0:87410359260179985583519841682;

b6c ¼ 0:54374053517347504259667364709;

ð28Þ

under which v1 = v2 = � � � = v7 = 0, but v8 5 0, with the computer output (up to 30 digits) given below:

v1 ¼ 0:0;

v2 ¼ 0:5� 10�30;

v3 ¼ 0:85� 10�29;

v4 ¼ �0:285� 10�29;

v5 ¼ 0:10419� 10�26;

v6 ¼ �0:459291715721954342488697462314� 10�25;

v7 ¼ 0:405099498118759898913218704449� 10�23;

v8 ¼ �0:002698708680733446404939431712.

Similarly, one can use Theorems 1 and 2 with proper perturbations to show that there exist exactly eight small ampli-tude limit cycles for this case. That is,

bH ð6; 6Þ ¼ 8.

4. Conclusion

In this paper, we have applied an efficient method with the aid of normal form computation and solving coupledpolynomial equations to obtain new results on the limit cycles of generalized Lienard systems. Particular attention isgiven to the small amplitude limit cycles around the origin. Seventeen cases have been considered, and four typical casesare proved to have the exactly number of small limit cycles. Other cases, which are not proved in this paper, can besimilarly proved following the procedure developed in this paper. The method can be extended to consider other morecomplicated cases. This may help develop an approach for finding the general formula, bH ði; jÞ, of generalized Lienardsystems.

Acknowledgments

This work was supported by the Natural Sciences and Engineering Research Council of Canada (NSERC No.R2686A02) and the National Natural Science Foundation of China (NNSFC No. 10371072).


Appendix A

The computer output of the coefficients Ai�s from Maple program is listed below.


Appendix B

The computer output of the functions f1 and f2 for the case bH ð9; 5Þ are given as follows:


Appendix C

The computer output of the functions fi, i = 1,2,3 for the case bH ð6; 6Þ is given below:


References

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