Study of limit cycles for some non-smooth Liénard systems

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1 ( ) Journal of East China Normal University (Natural Science) No. 3 May 011 Article ID: (011) Study of limit cycles for some non-smooth Liénard systems YANG Lu, LIU Xia, XING Ye-peng (Department of Mathematics, Shanghai Normal University, Shanghai 0034, China) Abstract: Algebraic method was used to study the Hopf cyclicity of non-smooth Liénard systems on the plain. Some new formulas for computing focus values were presented. Based on the formulas, the number of limit cycles bifurcating from some non-smooth Liénard systems was obtained. The results improve the known results. Key words: Liénard systems; non-smooth; limit cycle; Hopf cyclicity CLC number: O19 Document code: A DOI: /j.issn Liénard,, (, 0034) : Liénard Hopf,, Liénard Hopf,. : Liénard ; ; ; Hopf 1 Introduction and preliminary Several papers have been concerned with the bifurcation problems of non-smooth systems, see [1-7] for example. In this paper, we consider a non-smooth Liénard system of the form ẋ = p(y) F(x, a), ẏ = g(x), (1) where a R n, F(x, a) = F (x, a), x 0, F (x, a), x 0, g(x) = g (x), x 0, g (x), x 0. : : ( ), (10YZ7) :,,,. :,,,. :,,,,. ypxing@shnu.edu.cn.

2 3, : Liénard 45 Here, F ± and g ± are all C functions and satisfy F ± (0, a) = 0, p(0) = 0, g ± (0) = 0, (g ± ) (0) = g ± 1 > 0, p (0) = p 0 > 0, (F ± x (0, a 0 )) 4p 0 g ± 1 < 0, a 0 R n. () For s > 0 small, the Poincaré return maps h ± (s) were defined in a neighborhood of the origin of general non-smooth systems in [7] and a succession function d was introduced as follows Definition 1.1 [7] If there is a k 1 such that d(s) = h (h (s)) s = V k s k O(s k1 ), V k 0 for s > 0 close to the zero, then V k is called the kth Liapunov constant or focus value. The origin is called fine or weak focus of order k if k > 1. as follows In [1] the definition of succession function of non-smooth systems is presented for s 0 Definition 1. [1] Define h (h (s)) s, for s > 0, d(s) = 0, for s = 0, h (h (s)) s, for s < 0. The function d(s) is called a succession function or a displacement function. Let ( g ) x x 1 g (x), x > 0, G(x) = g(u)du = ( 0 g ) x 1 g (x), x 0, where g ± (0) = 0. Using the above definitions, for system (1), the authors obtained the following lemmas and theorems in [1] Lemma 1.1 [1] and only if Let (1) satisfy (). Then the origin is a fine or weak focus for a = a 0 if g1 F x (0, a 0) g 1 F x (0, a 0) = 0. (3) Lemma 1. [1] Let () and (3) hold. Suppose formally for 0 < x 1 F(α(x), a) F(x, a) = F (α(x), a) F (x, a) = i 1 B i (a)x i, (4) where α(x) = g 1 xo(x ) satisfies G(α(x)) G(x) for 0 < x 1. Then we have formally g 1 d(r 0, a) = i 1 d i (a)r i 0 for a a 0 small, where d 1 (a) = B 1 N 1 (a), d i(a) = B i N i (a) O( B 1, B 1,, B i 1 ),

3 46 ( ) 011 with N i C and N i (a 0) > 0 for i 1. Theorem 1.1 [1] Suppose () and (3) hold. If there exists k 1 such that B j (a 0 ) = 0, j = 1,, k, B k1 (a 0 ) < 0(> 0), then the origin is a stable (unstable) focus of order k 1 of system (1) for a = a 0 and there are at most k limit cycles near the origin for all a a 0 small and k limit cycles can appear if rank (B 1,, B k ) (a 1,, a n ) a=a 0 = k. Corollary 1.1 [1] Let () and (3) hold. If there exists k 1 such that F (α(x), a) F (x, a) when B j1 = 0, j = 0,, k (5) for all a R m, then the origin is a focus of order at most k 1 of system (1) unless it is a center. further Theorem 1. [1] Suppose () and (3) are satisfied. Let (5) hold for some k 1. If B j1 (a 0 ) = 0, j = 0,, k, rank (B 1,, B k1 ) a=a0 = k 1, (6) (a 1,, a n ) for some a 0 R m, then system (1) has Hopf cyclicity k at the origin for a a 0 small. Theorem 1.3 [1] Let () and (3) hold. Suppose there exists k 1 such that (5) holds for all a R m and (6) holds for some a 0 R m. If F is linear in a then for any constant N > a 0, system (1) has Hopf cyclicity k for all a N. For general system (1), let G G(x) = x G 3 x3 G 4 x4 G 5 x5, x > 0 G x G 3 x3 G 4 x4 G 5 x5, x 0, F(x, a) = F 1 x F x F 3 x3 F 4 x4 F 5 x5, x > 0 F 1 x F x F 3 x3 F 4 x4 F 5 x5, x 0. Suppose α(x) = α 1 x α x α 3 x 3 α 4 x 4 α 5 x 5 for 0 < x 1. By inserting α(x) into the equation G(α(x)) = G(x), it was obtained in [1] G α 1 =, α = G 3 α3 1G 3 G α 1 G, α 3 = G 4 G 4 α4 1 3G 3 α 1α G α α 1 G, α 4 = G 5 G 5 α5 1 4G 4 α3 1 α 3G 3 (α 1 α 3 α 1 α ) G α α 3 α 1 G, α 5 = G 6 G 6 α6 1 5G 5 α4 1 α G 4 (4α3 1 α 3 6α 1 α ) G 3 (3α 1 α 4 α 3 6α 1α α 3 ) α 1 G G (α α 4 α 3) α 1 G, (7)

4 3, : Liénard 47 α 6 = G 7 G 7 α7 1 6G 6 α5 1 α 5G 5 (α3 1 α α4 1 α 3) 4G 4 (3α 1 α α 3 α 1 α 3 α3 1 α 4) α 1 G 3G 3 (α 1α 5 α 1 α α 4 α 1 α 3 α α 3 ) G (α 3α 4 α α 5 ) α 1 G. Then by using (4), the following formulas were obtained in [1] B 1 (a) = α 1 F1 F 1, B (a) = α F1 α 1F F, B 3 (a) = α 3 F1 α 1α F α3 1F3 F 3, B 4 (a) = α 4 F1 (α 1α 3 α )F 3α 1 α F3 α4 1 F 4 F 4, B 5 (a) = α 5 F1 (α 1α 4 α α 3 )F 3(α 1 α 3 α 1 α )F 3 4α3 1 α F4 α5 1 F 5 F 5, B 6 (a) = α 6 F1 (α 1α 5 α α 4 α 3 )F (3α 1 α 4 6α 1 α α 3 α 3 )F 3 (6α 1α 4α 3 1α 3 )F4 5α4 1α F5 α6 1F6 F 6. (8) Applying Theorem 1.3 to the following system, ẋ = y F n (x, a), ẏ = g(x), (9) where F n (x, a) = n a i xi, x > 0, n a i xi, x 0, i=1 i=1 g(x) = x g x, x > 0, x g x, x 0, the authors in [1] proved that for n = 1, or 3 system (9) respectively has Hopf cyclicity 1, 3, 5 at the origin. In this paper we first give a formula for computing B i, i = 7, 8, 9, 10, and then apply Theorems to study the Hopf bifurcation of some non-smooth Liénard systems. In particular, we found the Hopf cyclicity for system (9) in the case of n = 3 can be 4, which improve the result in [1], see Theorem. below. Main results and proofs For general system (1) we have Theorem.1 Let (7) and (8) hold. We can further obtain α 7 = [G 8 G 8 α8 1 7G 7 α6 1α G 6 (6α5 1α 3 15α 4 1α ) 5G 5 (α4 1α 4 4α 3 1α α 3 α 1α 3 ) G 4 (1α 1α α 4 6α 1α 3 4α 3 1α 5 1α 1 α α 3 α 4 ) 3G 3 (α 1α α 5 α 1 α 3 α 4 α 1α 6 α α 3 α α 4 ) G (α α 6 α 3 α 5 α 4 )]/(α 1G ),

5 48 ( ) 011 α 8 = [G 9 G 9 α9 1 8G 8 α7 1α 7G 7 (α6 1α 3 3α 5 1α ) G 6 (6α5 1α 4 30α 4 1α α 3 0α 3 1α 3 ) 5G 5 (α4 1α 5 4α 3 1α α 4 α 3 1α 3 6α 1α α 3 α 1 α 4 ) 4G 4 (3α 1 α 3α 4 3α 1 α α 5 3α α 1 α 3 α3 1 α 6 α 3 α 3 3α α 1α 4 ) G 3 (3α 1 α 7 3 α 1 α 4 6 α 1α 3 α 5 6 α 1 α α 6 α α α 3 α 4 3 α α 5) G (α α 7 α 3 α 6 α 4 α 5 )]/(α 1 G ), α 9 = [ G 10 α10 1 G 10 9α8 1α G 9 4(7α6 1α α 7 1α 3 )G 8 7(α6 1α 4 5α 4 1α 3 6α 5 1α 3 α )G 7 (6α 5 1α 5 15α 4 1α 3 15α 4 α 1 30α 4 1α 4 α 60α 3 1α α 3 )G 6 (5α 6α α 1α α 3 0α 1 α 3 α 3 0α 3 1α α 5 0α 3 1α 3 α 4 30α 1α α 4 α 5 )G 5 (4α 3 α 4 4α 3 3α 1 4α 7 α 3 1 6α α 3 6α 1α 4 1α 1 α α 5 1α 1α 3 α 5 1α 6 α 1 α 4α 1 α α 4 α 3 )G 4 3(α 3 α 4 α α 6 α 1 α 8 α α 4 α 1α α 7 α 1 α 4 α 5 α 1 α 3 α 6 α α 3 α 5 )G 3 (α 3 α 7 α 4 α 6 α α 8 α 5 )G ]/( α 1G ), α 10 = [ G 11 α11 1 G 11 10α9 1α G 10 9(α8 1α 3 4α 7 1α )G 9 8(α7 1α 4 7α 5 1α 3 7α 6 1α α 3 )G 8 7(α 6 1α 5 5α 3 1α 4 3α 5 1α 3 15α 4 1α 3 α 6α 5 1α 4 α )G 7 6(α5 α 1 α 6 α 5 1 5α 4 1α 3 α 4 10α 1α 3 α 3 10α 3 1α 3α 10α 3 1α α 4 5α 4 1α 5 α )G 6 5(α4 α 3 α 7 α 4 1 α 3 3α 1 α 3 1 α 4 6α 1α α 3 6α 1 α α 5 4α 3 1 α 3α 5 4α 1 α 3 α 4 4α 6 α 3 1 α 1α 1 α α 4 α 3 )G 5 4(α3 3 α α 3 α 5 α 8 α 3 1 3α 7α 1 α 3α 6 α 1 α 3 3α 6 α 1 α 3α α 4α 3 3α 1 α α 4 3α 1α 3 α 4 3α 1 α 4α 5 6α 1 α α 3 α 5 )G 4 3(α 1α 5 α 3α 4 and α α 7 α 3α 5 α 1α 9 α α 4 α 5 α 1 α α 8 α 1 α 3 α 7 α 1 α 4 α 6 α α 3 α 6 )G 3 (α 4 α 7 α 5 α 6 α 3 α 8 α α 9 )G ]/( α 1G ) B 7 (a) = α 7 F 1 (α 1α 6 α α 5 α 3 α 4 )F 3(α 1 α 5 α 1 α α 4 α 1 α 3 α α 3)F 3 4(α 3 1α 4 3α 1α α 3 α 1 α 3 )F 4 5(α4 1α 3 α 3 1α )F 5 6α5 1α F 6 α7 1F 7 F 7, B 8 (a) = α 8 F 1 (α 1α 7 α α 6 α 3 α 5 α 4)F 3(α 1α 6 α 1 α α 5 α 1 α 3 α 4 α α 4 α α 3)F 3 (4α3 1α 5 1α 1α α 4 6α 1α 3 1α α 1 α 3 α 4 )F 4 5(α4 1α 4 4α 3 1 α α 3 α 1 α3 )F 5 3(α5 1 α 3 5α 4 1 α )F 6 7α6 1 α F 7 α8 1 F 8 F 8

6 3, : Liénard 49 B 9 (a) = α 9 F 1 (α α 7 α 4 α 5 α 1 α 8 α 3 α 6 )F (3α α 5 3α 1 α 4 3α 1α 7 6α 1 α α 6 6α 1 α 3 α 5 6α α 3 α 4 α 3 3)F 3 4(α3 α 3 α 6 α 3 1 3α 1α α 5 3α 1 α α 3 3α 1 α α 4 3α 1 α 3α 4 )F 4 5(α4 α 1 α 4 1 α 5 α 3 1 α 3 4α3 1 α α 4 6α 1 α α 3)F 5 (0α 3 1 α3 6α5 1 α 4 30α 4 1 α 3α )F 6 7(α6 1 α 3 3α 5 1 α )F 7 8α7 1 α F 8 α 9 1 F 9 F 9, B 10 (a) = α 10 F 1 (α 3α 7 α 1 α 9 α 4 α 6 α α 8 α 5 )F 3(α 1 α 8 α 3 α 4 α α 4 α α 6 α 1 α 4 α 5 α α 3 α 5 α 1 α α 7 α 1 α 3 α 6 )F 3 (4α 7α 3 1 4α3 α 4 6α α 3 4α 1α 3 3 6α 1 α 4 1α 1 α 3α 5 1α 6 α 1 α 1α 1 α α 5 4α 1 α α 3 α 4 )F 4 Then (5α 6 α 4 1 0α 3 1α 3 α 4 0α 1 α 3 α 3 30α 1α α 4 0α 3 1α α 5 30α 1α α 3 α 5 )F 5 3(5α 4 α 1 5α 4 1α 3 α 5 1α 5 10α 4 1α 4 α 0α 3 1α α 3 )F 6 [mm] 7(5α 4 1α 3 α 6 1α 4 6α 5 1α 3 α )F 7 (8α7 1α 3 8α 6 1α )F 8 9α8 1α F 9 α 10 1 F 10 F 10. Proof Let α(x) = α 1 x α x α 3 x 3 α 4 x 4 α 5 x 5, xα(x) < 0. G(α(x)) = G (α(x)) G 3 (α(x))3 G 4 (α(x))4 G 5 (α(x))5 G 6 (α(x))6 G 7 (α(x))7 G 8 (α(x))8, G(x) = G x G 3 x3 G 4 x4 G 5 x5 G 6 x6 G 7 x7 G 8 x8. Note that G(α(x)) = G(x). Comparing the coefficients of x 8 in the both sides, we obtain G 8 = G (α 1α 7 α α 6 α 3 α 5 α 4 ) G 3 (6α 1α α 5 6α 1 α 3 α 4 3α 1 α 6 3α α 3 3α α 4) G 4 (1α 1 α α 4 6α 1 α 3 4α3 1 α 5 1α 1 α α 3 α 4 ) G 5 (5α4 1 α 4 0α 3 1 α α 3 10α 1 α3 ) G 6 (6α5 1 α 3 15α 4 1 α ) 7G 7 α6 1 α G 8 α8 1. Then α 7 can be solved from the above. The coefficients α 8, α 9, α 10 can be obtained in the same way. By F (α(x), a) F (x, a) = F1 (α(x)) F (α(x)) F3 (α(x))3 F4 (α(x))4 F5 (α(x))5 F6 (α(x))6 F7 (α(x))7 F8 (α(x))8 (F 1 x F x F 3 x3 F 4 x4 F 5 x5 F 6 x6 F 7 x7 F 8 x8 ),

7 50 ( ) 011 and using (4), by combining the coefficients of x i, i = 7, 8, 9, 10 respectively, we can obtain the expressions of B i (a), i = 7, 8, 9, 10. The proof is completed. where Let Theorem. Consider the following system F n (x, a) = ẋ = y F n (x, a), ẏ = g(x), (10) n a i xi, x > 0, n a i xi, x 0, i=1 i=1 g(x) = x g x, x > 0, x g x, x 0. M 3 = 7 97 (g ) (g )5 g (g )4 (g ) (g )3 (g ) (g ) (g ) (g )(g )5, M 4 = (g )5 (g ) (g )3 (g ) (g )9 (g ) (g )10 g (g )7 (g ) (g ) (g ) (g ) (g )4 (g ) (g )(g ) (g )8 (g )3, M 5 = (g )9 (g ) (g )14 (g 3935 ) (g )4 (g ) (g )5 (g ) (g ) g (g ) (g )17 g (g )6 (g ) (g )7 (g ) (g )16 ( ) (g )11 (g 995 ) (g )16 (g ) (g )3 (g ) (g )1 (g 3935 ) (g )15 (g ) (g )13 (g ) (g )10 (g ) (g )8 (g )10. Then system (10) has Hopf cyclicity n at the origin if M n 0 for n = 3, 4 and 5 respectively. Proof In fact, for (10) we have G = G = 1, G 3 = g 3, G 3 = g 3, G± j By (7), (8) and Theorem.1, we can obtain = 0, j 4. α 1 = 1, α = g g, 3 α 3 = 9 g g 5 18 (g ) 1 18 (g ), α 4 = 5 18 (g ) g 8 7 (g ) (g )3, α 5 = (g ) (g )3 g (g ) (g ) (g )4,

8 3, : Liénard 51 α 6 = (g ) (g ) (g )4 g 3 43 (g )3 (g ) 5 97 (g ) (g )3, α 7 = (g ) (g ) (g )5 g (g )4 (g ) (g ) (g )4, α 8 = (g )5 (g ) (g )6 (g ) (g ) (g ) (g )4 (g ) (g ) (g )5, α 9 = (g )4 (g ) (g )6 (g ) (g ) (g ) (g )7 g (g )5 (g ) (g ) (g )8, and α 10 = (g )6 (g ) (g ) (g ) (g )4 (g ) (g )7 (g ) (g )8 g (g ) (g )9, B 1 (a) = a 1 a 1, B (a) = α a 1 a a, B 3 (a) = α 3 a 1 α a a 3 a 3, B 4 (a) = α 4 a 1 ( α 3 α )a 3α a 3 a 4 a 4, B 5 (a) = α 5 a 1 ( α 4 α α 3 )a 3(α 3 α )a 3 4α a 4 a 5 a 5, B 6 (a) = α 6 a 1 ( α 5 α α 4 α 3)a (3α 4 6α α 3 α 3 )a 3 (6α 4α 3 )a 4 5α a 5, B 7 (a) = α 7 a 1 ( α 6 α α 5 α 3 α 4 )a 3(α 5 α α 4 α 3 α α 3 )a 3 4( α 4 3α α 3 α 3 )a 4 5(α 3 α )a 5, B 8 (a) = α 8 a 1 ( α 7 α α 6 α 3 α 5 α 4)a 3(α 6 α α 5 α 3 α 4 α α 4 α α 3)a 3 ( 4α 5 1α α 4 6α 3 1α α 3 α 4 )a 4 5(α3 α 4 4α α 3 )a 5, B 9 (a) = α 9 a 1 (α 3α 6 α 4 α 5 α α 7 α 8 )a (α3 3 6 α α 3 α 4 3 α α 5 3 α 7 3 α 4 6 α α 6 6 α 3 α 5 )a 3 4(3 α α 5 α 3 α 3 3 α 3 α 4 3 α α 4 3 α α 3 α 6 )a 4 5( α 3 α 5 α 4 4 α α 4 6 α α 3 )a 5, B 10 (a) =α 10 a 1 ( α 4α 6 α 5 α α 8 α 3 α 7 α 9 )a 3( α α 7 α 4 α 5 α α 4 α α 3 α 5 α 3 α 6 α 3 α 4 α α 6 α 8 )a 3 (α3 α 4 α 7 1α α 3 α 4 3α α 3 3α 4 6α α 6 6α 3 α 5 α 3 3 6α α 5)a 4 (5α 6 0α α 5 30α α 4 0α 3 α 3 0α 3 α 4 30α α 3 α5 )a 5. We take g, g as constants. For n = 3, take a = (a 1, a 1, a, a, a 3, a 3 ) R6, we have det (B 1, B, B 3, B 4, B 5, B 6 ) (a 1, a 1, a, a, a 3, a 3 ) = 0, and det (B 1, B, B 3, B 4, B 5 ) (a 1, a 1, a, a, a 3 ) = M 3.

9 5 ( ) 011 Then by Solving the equations B i (a) = 0, i = 1,, 3, 4, 5, we have and (11), we have a 1 = 0, a 1 = 0, a = 3 a 3, a = 3 a 3, a 3 = g a 3. (11) G (α(x)) = G (x) 1 α (x) g 3 α3 (x) = 1 x g 3 x3, F (α(x)) F (x) = a 1 α(x) a α (x) a 3 α3 (x) a 1 x a x a 3 x3 = a 3 3 g ( 1 α (x) g 3 α3 (x) 1 x g 3 x3 ) = 0. Hence, for i = 1,, 3, 4, 5, B i (a) = 0 implies F(α(x)) = F(x). Taking a 0 = ( 0, 0, 3 g the conclusion for n = 3 follows from Theorem 1.3. For n = 4, we have ) a 3, 3 a 3, g a 3, a 3, det (B 1, B, B 3, B 4, B 5, B 6, B 7, B 8 ) (a 1, a 1, a, a, a 3, a 3, a 4, a 4 ) = 0, Solve B i (a) = 0, i = 1,,, 7, we obtain det (B 1, B, B 3, B 4, B 5, B 6, B 7 ) (a 1, a 1, a, a, a 3, a 4, a 4 ) = M 4. a 1 = a 1 = a 4 = a 4 = 0, a = a = 3 a 3, a 3 = g a 3, which follows F (α(x)) = F (x) from G (α(x)) = G (x) using the discussion for the case n = 3. Thus, when B i (a) = 0, i = 1,,, 7, one has F(α(x)) = F(x). The conclusion follows for n = 4 by taking For n = 5, we have a 0 = ( 0, 0, 3 g ) a 3, 3 a 3, g a 3, a 3, 0, 0. det (B 1, B, B 3, B 4, B 5, B 6, B 7, B 8, B 9, B 10 ) (a 1, a 1, a, a, a 3, a 3, a 4, a 4, a 5, a 5 ) = 0, det (B 1, B, B 3, B 4, B 5, B 6, B 7, B 8, B 9 ) (a 1, a 1, a, a, a 3, a 4, a 4, a 5, a 5 ) = M 5. Solve B i (a) = 0, i = 1,,, 9, we obtain a 1 = a 1 = a 4 = a 4 = a 5 = a 5 = 0, a = a = 3 a 3, a 3 = g a 3.

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