CC BY
77
УДК 517.9
Limit Cycles for a Class of Polynomial Differential Systems Via Averaging Theory
Ahmed Bendjeddou*
Department of Mathematics University of Setif, 19 000 Algeria
Aziza Berbache^
Department of Mathematics University of Bordj Bou Arreridj, 34265
Algeria
Abdelkrim Kina*
Department of Mathematics University of Setif, 19 000 Algeria
Received 02.10.2018, received in revised form 13.12.2018, accepted 26.01.2019 In this paper, we consider the limit cycles of a class of polynomial differential systems of the form
{ x = y - e(gu (x) y2a+1 + fn (x) y2a) - e2 (g12 (x) y2a+1 + f12 (x) y2a), \y = -x - e(g2i (x) y2a+1 + f2i (x) y2a) - e2 (g22 (x) y2a+1 + f22 (x) y2a),
where m,n,k,l and a are positive integers, д1к, g2K,f1K and f2K have degree n,m,l and k, respectively for each к = 1, 2, and e is a small parameter. We obtain the maximum number of limit cycles that bifurcate from the periodic orbits of the linear center x = y, y = -x using the averaging theory of first and second order.
Keywords: limit cycles, averaging theory, Lienard differential systems. DOI: 10.17516/1997-1397-2019-12-2-145-159.
Introduction
One of the main problems in the qualitative theory of real planar differential equations is to determinate the number of limit cycles for a given planar differential system. As we all know, this is a very difficult problem for a general polynomial system. Therefore, many mathematicians study some systems with special conditions. To obtain the number of limit cycles as many as possible for a planar differential system, we usually take in consideration of the bifurcation theory. In recent decades, many new results have been obtained (see [9,10]).
The number of medium amplitude limit cycles bifurcating from the linear center x = y, y = —x for the following three kind of generalized polynomial Lienard differential systems
{i
У
x = y,
y = -x - g2 (x) + /2 (x) y
{X
y
= y - gi (x), 1 x
= -x - g2 (x) + /2 (x) y I y
{x
y
= y - fi (x) У,
= -x - g2 (x) + /2 (x) y
where studied in the papers [1,2,4-6] and [13,14], respectively.
* bendjeddou@uiv-setif.dz 1 azizaberbache@hotmail.fr ^ abdelkrimkina@gmail.com © Siberian Federal University. All rights reserved
In [12] the authors use the averaging theory of first and second order to study the system
fx = y - e(gn (x) + fn (x) y) - £2(gi2 (x) + /12 (x) y), \ y = -x - £(g21 (x) + /21 (x) y) - £2(g22 (x) + /22 (x) y),
where g1i, /1i, g2i, /2i have degree l, k, m, n respectively for each i = 1,2 and £ is a small parameter. They provided an accurate upper bound of the maximum number of limit cycles that the above system can have bifurcating from the periodic orbits of the linear center x = y,y = -x.
In [7], using the averaging theory of first and second order, the authors studied the number of medium amplitude limit cycles bifurcating from the linear center x = y,y = -x of the more generalized polynomial Lienard differential systems.
fx = y,
\ y = -x - £(p1 (x) y + q1 (x) y2) - £2(p2 (x) y + q2 (x) y2),
where p1,q1,p2 and q2 have degree n.
By using the averaging theory we shall study in this work the maximum number of limit cycles which can bifurcate from the periodic orbits of a linear center perturbed inside the class of generalized polynomial differential equations
x = y - £(g11 (x) y2a+1 + /11 (x) y2a) - £2(g12 (x) y2a+1 + /12 (x) y2a),
1 y = -x - £(g21 (x) y2a+1 + /21 (x) y2a) - £2(g22 (x) y2a+1 + /22 (x) y2a),
where m,n,k,l and a are positive integers g1K, g2K,/1k and /2k have degree n,m,l and k, respectively for each k = 1,2, and £ is a small parameter.
Let [x] denotes the integer part function of x G R. Our main result is the following one.
Theorem 1. For sufficiently small, the maximum number of limit cycles of the polynomial differential systems (1) bifurcating from the periodic orbits of the linear center x = y,y = -x using the averaging theory
(a) of first order is
X1 =m&x{ , [f]} ;
(b) of second order is
A = m*x{[f ] ] + [^] + a, [f] + [a, [^]+ M + a,
[¥] , [^n] + №] + 1 + a, [2] + [f] - 1 + a, [f] + M + ^ ,
where m = min { , [-f] } ■
The proof of the above theorem is given in Section 2.
1. Averaging theory
In this section we present the basic results from the averaging theory that we shall need for proving the main results of this paper. The averaging theory up to second order for studying specifically periodic orbits was developed in [3,11]. It is summarized as follows. Consider the differential system
x(t) = £F1(t,x) + £2F2(t,x) + £3R(t,x,£),
where F1,F2 : R x D ^ R, R : R x D x (-£f ,£f) ^ R are continuous functions, T-periodic in the first variable, and D is an open subset of R. Assume that the following hypotheses hold.
(i) Fi(t,-) G C2(D), F2(t,-) G C i(D) for all t G R, Fi; F2,R are locally Lipschitz with respect to x, and R is twice differentiable with respect to e. We define Fk0 : D ^ R for k = 1, 2 as
1 rT
Fio(x) = t J Fi(s,x)ds,
T-, / N 1 fT dFi (s, x) F2o(x) = tJ -dx-yi(s,x)+ F2(s,x)ds,
where
yi(s,x)= Fi(t,x)dt.
o
(ii) For V c D an open and bounded set and for each e G ( —ef,ef )\{0}, there exists ae G V such that Fw(ae) + eF20(ae) = 0 and dB(Fi0 + eF20, V, a£) = 0.
Then, for |e| > 0 sufficiently small there exists a T-periodic solution x(t, e) of the system such that x(0, e) ^ ae as e ^ 0.
The expression dB (F10 + eF20,V,aE) = 0 means that the Brouwer degree of the function F10 + eF20 : V ^ 1" at the fixed point ae is not zero. A sufficient condition in order that this inequality holds is that the Jacobian of the function F10 + eF20 at ae is not zero.
If F10 is not identically zero, then the zeros of F10 + eF20 are mainly the zeros of F10 for e sufficiently small. In this case the previous result provides the averaging theory of first order.
If F10 is identically zero and F20 is not identically zero, then the zeros of F10 + eF20 are mainly the zeros of F20 for e sufficiently small. In this case the previous result provides the averaging theory of second order.
2. Proof of Theorem 1
2.1. Proof of statement (a) of Theorem 1
For the proof we shall use the first order averaging theory as it was stated in Section 1. We write
n m l k
gii (x) = ^ a.ix\ g2i (x)=Y; cixl,fii (x) = ^ hxl, f2i (x) = ^ dix1,
i=0 i=0 i=0 i=0 n m l k
gi2 (x) = ^ Aixi, g22 (x) = ^2 Cixi, fi2 (x) = ^2 Bixi, f22 (x) = ^2 Di,jxi■
i=0 i=0 i=0 i=0
Then in polar coordinates (r,0) given by x = r cos 0 and y = r sin 0, the differential system (1) becomes
r = —eGl (r, 0) — e2H1 (r, 0), 2
e e2 0 = —1 — -G2 (r, 0) — —H2 (r, 0),
where
£ '
i=0
Gi (r,0) = J2biri+2a cosi+i 0 sin2a 0 + X^ airi+2a+i cosi+i 0 sin2a+i 0 +
+ J2 Ciri+2a+i cosi 0 sin2a+2
0 + diri+2a cosi 0 sin2a+i 0
s
(l n
Y^ Biri+2a cosi+1 0 sin2a 0 + Y AiTi+2a+1 cosi+1 0 sin2a+1 0 +
i=0 i=0
+ CiTi+2a+1 cosi 0 sin2a+2 0 + Y DiTi+2a cosi 0 sin2a+1 0
i=0 i=0
(m I
^'2ciri+2a+1 cosi+1 0 sin2a+1 0 -Y^ biri+2a cosi 0 sin2a+1 0 -
i=0 i=0
n f \
- Y airi+2a+1 cosi 0 sin2a+2 0 + Y diri+2a cosi+1 0 sin2a 0} ,
i=0 i=0 J
(m I
Y Ciri+2a+1 cosi+1 0 sin2a+1 0 -Y Biri+2a cosi 0 sin2a+1 0 -i=0 i=0
n f \
-Y Airi+2a+1 cosi 0 sin2a+2 0 + Y Diri+2a cosi+1 0 sin2a 0 I ■
i=0 i=0
Taking 0 as the new independent variable, system (1) becomes
dr = £F1 (r, 0) + £2F2 (r, 0) + O (£3) , (2)
where
F1 (r, 0) = G1 (r,0), (3)
F2 (r, 0) = H1 (r, 0) - 1G1 (r, 0) G2 (r, 0) ■
r
First we shall study the limit cycles of the differential equation (2) using the averaging theory of first order. Therefore, by Section 1 we must study the simple positive zeros of the function
1 r 2n
1 f2n
Fio (t) = — Fi (t,0) d0.
J0
For every one of these zeros we will have a limit cycle of the polynomial differential system (1).
If F10(r) is identically zero, applying the theory of averaging of second order (see again Section 2) every simple positive zero of the function
F20 (r) = 2n I (dr.F1 (r, 0)^ j F1 (r, s)d^ + F2 (r, 0)^ d0,
will provide a limit cycle of the polynomial differential System (1).
Taking into account the expression of (3), in order to obtain F10 (r) is necessary to evaluate the integrals of the form
/ cosi 0 sin^' 0d0.
0
In the following lemmas we compute these integrals.
Lemma 2. Let Bi,j (0) = cosi 0 sin' 0 and Ai,j(0) = J Bi,j(s)ds. Then the following equalities
hold:
Aij(2n)
(j - 1)(j - 3)... 1 1
, . .w . . -, i i n if i and j are even,
(j + i)(j + i - 2)... (i + 2) 2i-1 il J J '
(i>
if i is odd or j is odd,
f i\ i! 2i +1 2a + 1 where i = \2 , A2i+2,2a(2n) = ---— A2i}2a(2n), A2i,2a+2(2n) = ---— A2i,2a(2n),
\2J i D 2 2(i+a+1)
then,
^ m
2(i + a+1)
2i + 1
A2i+2,2a(2n) = Ô-TT A2i,2a+2(2n).
2a + 1
Proof. The integral Aitj(2n) can be calculated using the integrals (11), (9), (12) and (10) of the of appendix. □
Using this lemma we shall obtain in the next proposition the integral the function F10(r). Proposition 3. We have
Fio(r)
where &i,2a =
r
2a+1
([ — ]
[ m ]
E2i + ^ , 2i+ , 2a + 1 2i
—,-—r02i+ir &i,2a + > —-; 7C2ir &i,2c
i+a+1 i
i=0
i=0
i+a+1
(4)
(22a - 1)(2a - 3) ... 1 1
(2a + 2i) (2a + 2i - 2) ... (2i + 2) 22i-1 Proof. Using Lemma 2, the function F10(r) is given by / i ™
2nF10 (r) = /
0
]T biri+2aBi+h2a (0) + J2 Ciri+2a+1Bi,2a+2 (0) | d0
i=0 \i odd
i=0 i even
c2n (I—] [?]
J2 b2i+1 r2a+2i+1B2i+2,2a (0) + J2 C2ir2i+2a+1B2i,2*.+2 (0) | d0 i=0 i=0
m m
= b2i+1r2a+2i+1A2i+2,2a(2n) +J2 C2ir2i+2a+1A2i,2a+2(2n) =
r
0
2a+1
([ — ]
2
[ mm ]
E ^b2i+1r2iA2i,2a(2n) + J2 C2ir2iA2i,2a.(2n) | .
i=0
i=0
This completes the proof of Proposition 3. □
From Proposition 3, the polynomial F10(r) has at most X1 = ma^ l1-^], [m] } positive
roots, and we can choose b2i+1 and c2i a way that F10(r) has exactly X1 simple positive roots, hence statement (a) of Theorem 1 is proved.
0
4
2.2. Proof of statement (b) of Theorem 1
Now using the results stated in Section 1 we shall apply the second order averaging theory to the previous differential equation, but for doing that we need that Fi0(r) = 0. Therefore, from (4) in what follows we must take
\C2i = — , i =0 11, (5)
I b2i+i = C2i = 0, i = n + 1,...,Xi J
where ^ = min j , [y] }.
We must study the simple positive zeros of the function
F20 (r) = 2n I (dF (r, 0^0 Fi(r, s)dSj + F2 (r, 0)^ dB.
We split the computation of the function F20(r) in two pieces i.e. we define 2nF20(r) = L(r)+ +J(r), where
r2n r, / ,■ e
L tr) = l iFi(r
0^ j0 Fi(r, s)ds j d0, J (r)= j0 F2 (r, 0) d0.
Lemma 4. The integral J(r) can be expressed by
J (r) = r2a+1 (Pi (r2) + r2aP2 (r2)) , where P1 and P2 are polynomials of degree
Xi =mx{ [—] , [ f ]},
and
X2 = [*] + [*—] , [*—] + M + 1, [2] + , [—] + ,, [?] + ,, [f] + [|] - l},
respectively.
2n
Proof. First we calculate / H1 (r, 0) dd.
0
p2n l p2n k p2n
/ Hi (r, 0) d0 = V Biri+2a Bi+h2a. (0) d0 + V Diri+2a Bi2a.+i (0) d0+ 0 i=0 0 i=0 0
m 2n n 2n
+ Y, Ciri+2a+i Bi,2a+2 (0) d0 + J2 Airi+2a+i Bi+i,2a+i (0) d0 = i=0 0 i=0 0 l 2n m 2n
= Y. Biri+2a Bi+i,2a (0) d0 + J2 Ciri+2a+i Bi,2a+2 (0) d0 = 00
0
l 2n m r 2n
iri+2a Bi+iM (0) d0 + J2 Ciri+2a+i . ^,2*+
=0 0 i=0 0 i odd i even
M [m]
= J2 B2i+iA2i+2,2i (2n) r2i+2a+i + £ C2iA2i,2i+2 (2n) r2i+2a+i = i=0 i=0
= r2a+iPi if-2) ■
where P1 is a polynomial in the variable r2 of degree X1. Finally we shall study the contribution of
In 1
the second part f -G1 (r, 0) G2 (r, 0) d0 of F2(0, r) to F20(r). Taking into account the expression
0r
of (2), then
Gi (t, 0) = £ (ß2i+2,2a (0) - B2i,2a+2 (0^ b2i+lT2i+2a+1 + i=o ^ a +
M [l]
+ J2 C2i+lT2i+2a+2B2+1,2a+2 (0)+J2 b2iT2i+2aB2i+1,2a. (0) + i=0 i=0
^^V1] [V]
?2i+1,2a+1 (0) + + S, a2i+1 T ' ' B2i+2,2a+1 I
+ £ d2i+1T2i+2a+1B2i+1,2a+1 (0)++ J2 a2i+1 T2i+2a+2B2i+2,2a+1 (0) +
0
ikl
+ E d2iT2i+2aB2i,2a+1 (0) + Y, a2iT2i+2a+1B2i+1,2a+1 (0) .
i=0 i=0
[k] m
G2 (t, 0) = J2 d2pT2p+2aB2p+1,2a (0) + £ C2p+1T2P+1+2a+1B2p+2,2a+1 (0) -p=0 p=0
[2] M . -I . .
a + 1 + i
-J2 b2pT2p+2aB2p,2a+1 (0) - £ b2p+1T2(p+a) + 1B2i+1,2a+1 (0) -
p=0 p=0 +
[n] [^
a2prp+2a+1B2p,2a+2 (e)+J2 d2p+ir2p+2a+1B2p+2,2a (6) -
p=0 p=0 [ ^ ]
- £ a2p+1 r2p+2a+2B2p+1,2a+2 (6) . p=0
From the 49 products between the different sums only 12 will not be zero after the integration
2n 1
with respect to 6 between 0 and 2n. So the terms of f -G1 (r, 6) G2 (r, 6) d6 which will contribute
0 r
to F20 (r) are
/2n 1 [ 1 ] [k ]
- G1 (r, 6) G2 (r, 6) d6 = d2pb2iA2i+2p+2,4a (2n) r2p+2i+4a-1 +
r i=0 p=0
[ ^ ] [ k ]
+ E Y,d2pC2i+1Ä2i+2p+2Aa+2 (2n) r2p+2i+4a+1-i=0 p=0
[ 1 ] [ ^ ]
-EX) a2i+1b2p A2i+2p+2 4 4a+2 (2n) r2p+2i+4a+1 -p=0 i=0
[ 2 ] [ k ]
- X X d2ib2pA2i+2p,4a+2 (2n) r2p+2i+4a-1 -p=0 i=0
L 2 J M
- b2i+1a2p (A2i+2p+2,4a+2 (2n) - Ä2i+2p,4a+4 №)) r2p+2i+4a+1-
2a+1J
p=0i=0
i k-l
M
^i +1 + •
E 2 a£T:r d2i+1b2i+1Ä2i+2p+2Aa+2 (2n) r2p+2i+4a+1-p=0 i=0 a +
M [ 2 i
E 2 a20Tr a2ib2i+1Ä2i+2p+2Aa+2 (2n) r2p+2i+4a+1 +
p=0 i=0
[ — i
M L 2 J / 2 ' + 1 \
^ E d2p+1b2i+1( Ä2i+2p+4,4a (2n) - Ä2i+2p+2,4a+2 (2*)\ r2p+2i+4a+1
i=0 p=0 ^ a + '
[V] [2]
^ E Y.b2ia2p+lA2i+2p+2A*+2 (2n) r2p+2i+4a+1 — p=0 i=0
[ ] [ m-1 ]
— y^ y^ C2i+ia2p+lA2i+2p+2,4a+4 (2n) r2p+2i+4a+3 + p=0 i=0 [ m-1 ] [ ^ ]
^ E E a2i+lC2p+lA2i+2p+4Aa+2 (2n) r2p+2i+4a+3 + 2p+1=0 i=0
[ m-1 ] [ k ]
+ E E d2iC2p+lA2i+2p+2Aa+2 (2n) r2p+2i+4a+1 = r4a+1P2 (r2) , 2p+1=0 i=0
where P2 is a polynomial in the variable r2 of degree
A3 = max {[ I ] + [ r— ] , [ ] + [ m-1 ] + 1, [ 2 ] + [ ] , [ — ] + ,, [" ] + ,, [ 2 ] + [ | ] — 1}. Finally, we obtain J (r) is a polynomial in the variable r2
J (r) = r2a+1 (Pi (r2) + r2aP2 (r2)) ,
of degree
\j(r) = max {A1, A3 + a} . This completes the proof of the lemma. □
In order to complete the computation of F20(r) we must determine the function L(r). First we
compute the integrals f Aiyj (0)Bpq(0)d0. In the following lemmas we compute these integrals.
0
2i +1
Lemma 5. Let Si (0) = A2i+2,2a(0) — --7A2it2a+2(0), then
2a + 1
Si (0)
1 ( a-1 \
nf . • . n B2i+3,2a+1(0) + y /j2i+2,2aB2i+3,2a-2l-1(0)\ +
2(a + ' T1) \ = J
1 ( i \
+ ^ , s V2i+2,2a B2i+1A(0)+y Ö2i+2,0B2i-2l + 1,1(0) -
2(' T1) \ = J
2i + 1 ( a \ ja+T+~L)(2a+T) \ B2i+1,2a+3(0) + £ Y2i,2a+2B2i+1,2a-2l+1(0) j -
2i + 1 ( i-1 \
7T~-—Tn2i,2a+2\ B2i-1,1(0) + / y^2i,0B2i-2l-1,1(0)\ ,
1 (2a + 1) \ 1=1 J
where
(2a - 1)(2a - 3) ... (2a - 2l + 1) (2a - 1) (2a - 3) ... 1
Yi,2a = 77;-—-tttjt:-—-rr-T~-—-XJT, Vi,2. —
(2a + i - 2) (2a + i - 4) ... (2a + i - 2l) ' ' (2a + i) (2a + i - 2) ... (i +2)' = (2i - 1)(2i - 3)... (2i - 2l +1)
2i'0 = ¥(—j(—)77:(—) .
Moreover
Si (2n) = 0.
Proof. Using the integrals (11) and (9) of appendix and, taking into account that (2j - 1)(2j - 3)... 1 1 (2i + 2
(2i + 2\
(2j + 2i + 2) (2j + 2i)... (2i + 4) 22(i+1) V i + 1 J1
-O 0,
0
2i +1 (2j + 1)(2j - 1)... 1 2. 2j + 1 (2j + +2i2) (2j + 2i)... (2i + 2) 22i
it follows the expression of Si (0). □
2n
Lemma 6. Let (2n) = f Aij(0)Bp,q(0)d0 Then the following equalities hold:
a) The integral ^P'iq+1 0(2n) is zero if p is odd or q is even, and equal to 1 (A (2n) + - 2l+1i (i - 1)... (i - l)
A2i+p,q+1 (2n) + y, IK--TTT^V---07-7T A2i+p+2l-2,q+1 (2n)
,
2i +11^.«-^ v {=0 (2i - 1) (2i - 3)... (2i - 2l - 1) if p is even and q is odd.
b) The integral ^P'iq+1 2j+1(2n) is zero if p is odd or q is odd, and equal to
- 2j+I+1)( A2i+P+j (2n) +
+ - 2lj (j - 1)... (j -1 + 1) ( )
+ ¿1 (2j + 2i) (2j +2i - 2)... (2j +2i - 2l + 2) A2i+p+2,2j-2l+q (2n)
if p is even and q is even.
c) The integral ^p'iq2j+1(2n) is zero if p is even or q is odd, and equal to
A2i+p+1,2j+q (2n) +
2j + 2i + 1 j-1
+ ^_2lj (j - 1)... (j - l +1)_ ( )
+ = A2i+p+1,2j-2l+q (2n)
if p is odd and q is even.
d) The integral P'iq+i 2j(2n) is zero if p is odd or q is even, and equal to
- 2j + 2i + i( A2i+p+2'2j+q+i (2n) -
— (2j - 1)(2j - 3)... (2j - 2l + 1) ) ) +
- ¿f (2j + 2i - 1)(2j + 2i - 3)... (2j + 2i - 2l + 1) A2i+2+p,2j—2l+q—i (2n) ) +
+_(2j - 1)(2j - 3) ^_^ (2n)
(2j + 2i + 1)(2j + 2i - 1)... (2i + 3^ 2i+i'0
if p is even and q is odd.
2n 2i +1
e) The integral Tf£ (2n) = J Si(0)Bp,q (0)d0 = <f2?+2,2a(2n) - 2a + 1 ¥>2i92a+2(2n) is zero if
p is even or q is even, and equal to
1 ( a—i \
I A2i+p+3,2a+q+i(2n) + ^ A2i+p+3,2a+q—2l — i(2n)\ +
2a + 2i + 2 ,
l=i
2i + 1
+ V2i+2,2a 2(i+ 1) (^A2i+p+i,q+i(2n) + ^ Ö2i+2,0Ap+i — 2l+2i,q+i(2n^ -
^'Ap+2i+i,q+2a+3(2n) + ^^ Y2i,2a+2Ap+2i+i,q+2a—2l+i(2n-
2i + 1 ( i — i \
2i (200+1) n2i,2a+2 ( Ap+2i—i,q+i(2n) + S2i,0APp+2i — 2l — i,q+i (2n)j ,
(2a + 2 + 2i)(2a + 1) ^ ^
where
(2a - 1)(2a - 3) ... (2a - 2l + 1) (2a - 1) (2a - 3) ... 1
Yi,2a = ~T7,-—■-77777,---77,-1--7777, Vi,2c
(2a + i - 2) (2a + i - 4) ... (2a + i - 21)'' (2a + i) (2a + i - 2) ... (i + 2)'
(2i - 1)(2i - 3)... (2i - 2l +1)
hi
0
2l (i — 1)(i — 2)... (i — l) if p is odd and q is odd. f) fpq (2n) = w2p+2,2a (2n) 2p +1 2p,2a+2 (2n) = 2(2p + 5a + 2ia + 6pa + 2 + i) _
f) Tia (2n) = ^2i+1,2a+1 (2n) — V2i+1,2a+1(2n) = 2 ^ n
Proof. Using the integrals of the appendix, the six equalities are easily deduced by direct calculation. □
Lemma 7. The integral L(r) can be expressed by
L(r) = r1+4aP3 (r2),
where P3 is a polynomial of degree
A3 = max {» + [—] , [2] + [] , [m-1 ] + [] + 1, [2] + [|] — 1, [] + [|] ,, + ["]} .
Proof. Using (2.), we get
F1 (T, 0) = Y b2i+1 T2i+2a+1 (B2i+2,2a (0) - 22a+\B2i,2a+2 (0)) + [2]
+ X b2iT2i+2aB2i+1,2a (0)+ X c2i+1T2i+2a+2B2i+1,2a+2 (0) + i=0 i=0
[^ [n]
+ X d2i+1T2i+2a+1B2i+1,2a+1 (0) + X a2iT2i+2a+1B2i+1,2a+1 (0) + i=0 i=0
[ k ] m
+ X) d2iT2i+2aB2i,2a.+ 1 (0)+ J2 a2i+1T2i+2a+2B2i+2,2a+1 (0) . 0 i=0
Next we calculate the terms of this integral. First we have that
dF1r 0 = X) (2i + 2a + 1) b2i+1r2i+2a [B2i+2,2a (0) - 2+1 B2i,2a+2 (0)) + dr i=0
[I]
+ Y,2(i + a) b2ir2i+2a-1B2i+1,2a (0)+ J2 2 (i + a +1) C2i+1r2i+2a+1 B2i+1,2a+2 (0) +
i=0 i=0
[^ n
+ Y,(2i + 2a + 1) d2i+1r2i+2aB2i+1,2a+1 (0) +Y,(2i + 2a + 1) a,2ir2i+2aB2i+1,2a+1 (0) + i=0 i=0
[I] [V]
+ Y,2(i + a) d2ir2i+2a-1B2i,2a+1 (0) + X/ 2 (i + a +1) a2i+1r2i+2a+1 B2i+2,2a+1 (0) i=0 i=0
and
/U ^ L 2 J
F1(T, s)ds = J2 b2i+1T2i+2a+1Si (0) + ^ b2iT2i+2aÄ2i+1,2a (0) + i=0 i=0
m m
+ X C2i+1T2i+2a+2Ä2i+1,2a+2 (0)+J2 d2i+1T2i+2a+1Ä2i+1,2a+1 (0) +
0 k
+ X) a2iT2i+2a+1 Ä2i+1,2a+1 (0) + J2 d2iT2i+2aÄ2i,2a+1 (0) +
i=0 i=0 [n- ]
+ £ a2i+1T2i+2a+2Ä2i+2,2a+1 (0) .
i=0
From the 49 products between the different sums only 12 will not be zero after the integration with respect to 0 between 0 and 2n. So the terms of L (r) which will contribute to F20(r) are
m [
L (r) = YH (2p + 2a +1) d2i+1b2p+1T?aq (2n) r2i+2p+4a+1+ p=0 i=0
+ ^J2(2p + 2a + 1) a2i b2p+1(2n) r2i+2p+4a+1+ i=0 p=0
[ 2 ] [ V ]
+ £ £ 2(p + a) b2pa2i+1v2ip^^^+1(2n)r2i+4a+2p+1 + p=0 i=0
[ 2 ] [ k ]
+ ^J22(p + a) b2pd2i4p+T1(2n)r2i+2p+4a-1+
p=0 i=0
[ m-1 ] [ ^ ]
+ £ T,2(P + a +1) C2p+1a2i+1v2i+2,2a+1 (2n)r2i+2p+4a+3+ p=0 i=0
[ m—1 ] [ k ]
+ £ T,2(p + a + 1) C2p+1d2iV2+a+i+'2(2n)r2i+2p+4a+1 + p=0 i=0
[ k—] M
+ £ J2(2p + 2a +1) d2p+1b2i+1f2pa+1'2a+1 (2n) r2i+2p+4a+1+ p=0 i=0
+ E it (2p + 2a + 1) a2pb2i+1f2p0+1'2a+1 (2n) r2i+2p+4a+1 + p=0 i=0
[ 2 ] [ ^ ]
+ £ E (2P + 2a) d2pC2i+1V22P^+2(2n)r2i+2p+4a+1 + p=0 i=0
[ k ] [ 2 ]
+ E E 2(p + a) d2pb2v2++Ca-(2n)r'2i+2p+4a-1 + p=0 i=0
[ V ] [ 1 ]
+ £ £ 2(p + a +1) 02p+1C2i+1v2ptl2at1(2n)r2i+2p+4a+3+ p=0 i=0
[V] [2]
+ ^Y,2(p + a +1) a2p+ib2^22a+1(22n)r2i+2p+4a+1 = p=0 i=0
= r1+4aP3 (r2) , where P3 is a polynomial in the variable r2 of degree
A3 = max {, + [—] , [2] + [r-] , [] + [] + 1, [2] + [|] — 1,
[m^ ] + [I] + ["]}.
This completes the proof of the lemma. □
Finally, we obtain F20 (r) is a polynomial in the variable r2 of the form
2nF20 (r) = r2a+1(r2aP1 (r2) + P2 (r2) + r2aP3 (r2)).
Then, to find the real positive roots of F20, we must find the zeros of a polynomial in r2 of degree A = max {AJ(r), A3 + a}. This yields that F20 has at most A real positive roots. Moreover, we
can choose the coefficients ai, bi, ci, d^, Ai, Bi: Ci and Di in such a way that F20 has exactly A real positive roots. Hence, the statement (b) of Theorem 1 is proved. In fact, we consider the example with n = k = 1 and m = l = a = 1
We have that F10(r) is identically zero, so to look for the limit cycles, we must solve the equation F20(r) = 0 which is equivalent to
^ , n 1 9 7 7 49 5 3 3
2nF02 (r) = -nr--nr +--nr--nr .
w 960 480 960 80
This equation has exactly the three positive roots r1 = 1,r2 = 2 and r3 = 3.
According with Theorem 1, that system (6) has exactly there limit cycles bifurcating from the periodic orbits of the linear differential system (6) with e = 0, using the averaging theory of second order.
We consider the differential system with n = k = 1 and m = l = 3, a = 2
* = y - e ((1 + x3) y5 + - y^j - e2((1 + x3) y5 +(l + 395*) y4)
(( 1 ) 5 ( I929 3 13 ) 4) 2 ( 5 3 4)
y = -x - e 1+--x \yb + -x3--yM - e2 (xyb + x3y4) .
y \\ 90 )y \50960 2 ) ) V '
(7)
An easy computation shows that F10(r) = 0 and
n r, ^ 1 13 3 n 39 9 41 7 9 5
2nF02 (r) = -nr--nr +--nr--nr +--nr .
w 17 920 1792 2560 896 280
Therefore from the periodic orbits of radius 1, 2, 3 and 4 of the linear center X = y, y = —x, it bifurcates three limit cycles. Consequently for system (7) we have that A =
+a +1 = 4.
n—1
+
m—1
+
3. Appendix
Here we list some important formulas used in this article , for more details see [8]. For i > 0 and j > 0 , we have
f8 i ■ a , cos4-1 e sma+1 e i — 1 f8 2 . a .
cos4 s sin sds =--1--cos4 2 s sm sds =
J0 i + a i + a J 0
cosi+1 e sina—1 e a- 1 rd
+- cos1 s sina—2 sds. (8)
i + a i + a J0
f9 COs2i sds = sne (cos2i—1 e + g (2i - V(2i - 3) ... (2i - ^ ^ COS™ ß) + lo 2i y ^ = 2' (i - 1)(i - 2)... (i - l)
(2i - 1)(2i - 3)... 1 = 1 — (2i\ sin2(i - l) e 1 (2i\
+ 2H\ = ¿0V l) 2 (i - l) + y i) .
6 2i+1 , sin 0 / 2ia , - 2l+1i (i - 1) ... (i - l) 2i-2l-2
cos2i+1 sds = - cos 0 + > ----------2l 2 '
~' M + ^ (2i
l=0
i0 2i + 1 \ = (2i - 1)(2i - 3)... (2i - 2l - 1)
1 i—1{2i + 1\ sin (2i - 2l + 1) 0
f (M+T
l=0 l
22i l (2i - 2l + 1)
l=0
¡6 .
/ cos. s sin2a sds = '0
cosi+1 0 ((2a - 1)(2a - 3)... (2a - 2l + 1) ^-2-1 0 + sin2*.+ 1 ^ +
0 (f
l=1
2a + i (2a + i - 2) (2a + i - 4) ... (2a + i - 2l)
6
¡6 .
/ cos. s sin2a+1 sds
0
(10)
(2a - 1)(2a - 3) ... 1 . . ,
+ 7-——-r-—;-s cos. sds. (11)
^ (2a + i)(2a + i - 2) ... (i + 2) J0 V '
(sin2a 0 + t_^ (a - 1) ..• (a - l +1)_- sin2a-2l ^ .
y (2a + i - 1)(2a + i - 3) ... (2a + i - 2l + 1) J
cosi+1 0 (sn2a 0 , ^1 2a (a - 1) ... (a - l +1) sin2a-2l 0j (12)
2a + i + 1 \ (2a + i - 1) (2a + i - 3) ... (2a + i - 2l + 1)
References
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Предельные циклы для одного класса полиномиальных дифференциальных систем, использующие теорию усреднения
Ахмед Бенджедду
Математический факультет Университет Сетиф, 19 000 Алжир
Азиза Бербаке
Математический факультет Университет Бордж-Бу-Аррэридж, 34265
Алжир
Абделькрим Кина
Математический факультет Университет Сетиф, 19 000 Алжир
В данной 'работе рассматриваются предельные циклы одного класса полиномиальных дифференциальных систем вида
{ х = у - е(дц (х) у2а+1 + /ц (х) у2а) - е2 (д12 (х) у2а+1 + /12 (х) у2а), \у = -х - е(д21 (х) у2а+1 + /21 (х) у2а) - е2 (д22 (х) у2а+1 + /22 (х) у2а),
где д1к, д2к,/1к и /2к имеют степень п,т,1 и к, где т,п,к,1 и являются положительными целыми числами, соответственно, для каждого к = 1,2 и е — малый параметр. Мы получаем максимальное число предельных циклов, которые раздваиваются от периодических орбит линейного центра х = у, у = -х, используя теорию усреднения первого и второго порядка.
Ключевые слова: предельные циклы, теория усреднения, лиенардовы дифференциальные системы.
