This is a preprint of: “The limit cycles of the Higgins-Selkov systems”, Hebai Chen, Jaume Llibre, Yilei Tang, J. Nonlinear Sci., vol. 31, 85 (25 pages), 2021. DOI: [10.1007/s00332-021-09742-0] THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS HEBAI CHENA JAUME LLIBREB AND YILEI TANGC Abstract. In this paper we investigate the problem of limit cycles for general Higgins-Selkov systems with degree n + 1. In particular, we first prove the unique- ness of limit cycles for a general Liénard system which allows for discontinuity. Then, by changing the Higgins-Selkov systems into Liénard systems, theorems and some techniques for Liénard systems can be applied. After we prove the nonex- istence of limit cycles if the bifurcation parameter is outside an open interval. Finally we complete the analysis of limit cycles for the Higgins-Selkov systems showing its uniqueness. 1. Introduction and main results In the qualitative theory of planar polynomial differential systems it is well known how difficult is to study the famous Hilbert’s 16th problem, see [8, 10, 16]. Up to now there are seldom works having solved the problem of exact number of limit cycles for polynomial differential systems. The most important physiological function of carbohydrates is to provide energy for organisms’ life activities. Glucose catabolism is the main way for organisms to obtain energy. There are three main pathways for the oxidative decomposition of glucose in organisms. Among them, the anaerobic oxidation of glucose is called glycolysis. We consider the following polynomial differential system of arbitrary degree ẋ = 1 − xy n , (1) ẏ = ay(−1 + xy n−1 ) which was proposed first by Higgins [7] and modified further by Selkov [13] for studying the biological nonlinear glycolytic oscillations, and was called the Higgins- Selkov system. Here n is a positive integer and a is a real parameter. Artés, Llibre and Valls in [1] characterized the global dynamics described in the Poincaré disc for system (1) as n = 2 and a ∈ R\(1, 3). Moreover there are two conjectures stated in [1] on the the number of limit cycles of systems (1) when a ∈ (1, 3). After Chen and Tang in [5] proved these conjectures which complete the global phase portraits of system (1) when n = 2. Recently Brechmann and Rendall in [3] researched the uniqueness of limit cycles for system (1) and additionally proved that no limit cycles exist when a ∈ (0, 1/(1 − n)). Llibre and Mousavi [11] classified the phase portraits of system (1) for n = 3, 4, 5, 6 in the Poincaré disc for all the values of the parameter a and determined in function of the parameter a the regions of the phase space with biological meaning. 2010 Mathematics Subject Classification. Primary 34C07, 34C23, 49J52. Key words and phrases. Higgins-Selkov system, Liénard system of arbitrary degree, uniqueness of limit cycles, nonexistence of Limit cycles. 1 2 H. CHEN, J. LLIBRE AND Y. TANG The aim of this paper is to give a clearer study and answer for the existence and the exact number of limit cycles of system (1). We have the following main results. Theorem 1. For every positive integer n ≥ 3 and some bounded constant a∗ < 1, system (1) has no periodic orbits when a ∈ (−∞, 1/(n − 1)] ∪ [a∗ , +∞) and has a unique limit cycle when a ∈ (1/(n − 1), a∗ ), which is stable and hyperbolic. An outline of this paper is as follows. A theorem on the uniqueness of limit cycles for general Liénard systems are presented in section 2, which we need in our study of the limit cycles of the Higgins-Selkov system. In section 3 we obtain the existence and the exact number of limit cycles of the Higgins-Selkov system and then prove our main theorem. 2. Preliminaries In order to study the number of limit cycles for system (1), we need the following preliminary results. We first recall the uniqueness theorem of Zhifen Zhang in [14] or in [15] on the number of limit cycles of the following generalized Liénard systems ẋ = −φ(y) − F̂ (x), (2) ẏ = ĝ(x). Let Z x Ĝ(x) := ĝ(s)ds. 0 Theorem 2. Consider the generalized Liénard system (2) for x ∈ (−∞, +∞), when φ(y), F̂ (x) and ĝ(x) satisfy the following conditions: (i): ĝ(x) is Lipschitz in any finite interval, xĝ(x) > 0 for all x 6= 0, and Ĝ(−∞) = Ĝ(+∞) = +∞. (ii): fˆ(x) = F̂ 0 (x) is C 0 , F̂ (0) = 0, fˆ(x)/ĝ(x) is nondecreasing in (−∞, 0) ∪ (0, +∞) and fˆ(x)/ĝ(x) is not a constant when |x| is small. (iii): φ(y) is Lipschitz in any finite interval, yφ(y) > 0 for all y 6= 0, φ(y) is nondecreasing, φ(−∞) = −∞, φ(+∞) = +∞, φ(y) has right-derivative φ0+ (0) and left-derivative φ0− (0) at y = 0, φ0− (0)φ0+ (0) 6= 0 when fˆ(0) = 0. Then system (2) has at most one limit cycle. Moreover the limit cycle is stable when it exists. In fact we can find many differential systems of the form (2) but many of them do not satisfy the conditions of Theorem 2. Thus we propose the following three questions: (a): When Ĝ(−∞) = Ĝ(+∞) 6= +∞ and the other conditions of Theorem 2 hold, does the conclusion of Theorem 2 still hold? (b): When either fˆ(x) or ĝ(x) has a discontinuity point x0 of the second kind (i.e., limx→x0 + ĝ(x) or limx→x0 − ĝ(x) does not exist) and the other conditions of Theorem 2 hold, does the conclusion of Theorem 2 still hold? (c): When ĝ(x) has a discontinuity point at x = 0 of the first kind (i.e., limx→0+ ĝ(x) 6= limx→0− ĝ(x)) and the other conditions of Theorem 2 hold, does the conclusion of Theorem 2 still hold? THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 3 For example we have that G(−∞) = G(+∞) 6= +∞ when ĝ(x) = x/(1 + x2 )2 . Either fˆ(x) or ĝ(x) has a discontinuity point at x = −1 of the second kind when fˆ(x) = 1/a − (n − 1)/(x + 1)n or ĝ(x) = x/(x + 1)n . Here we will show why the condition G(−∞) = G(+∞) = +∞ is necessary in the proof of Theorem 2 of [15]. Zhang in [15] only need to research the following special Liénard system u̇ = −φ(y) − F̂ (x(u)), (3) ẏ = u, because p system (2) can be changed p into system (3) through the transformation u = 2G(x)Sgn(x) and dt → 2G(x)sgn(x)/ĝ(x) dt. However, the transformation is not an 1-1 transformation in (−∞, +∞) when G(−∞) = G(+∞) 6= +∞. For these reasons we give the following theorem without the aforementioned con- ditions. Theorem 3. Consider system (2) in the interval (α, β), where α and β eventually can be −∞ and +∞, respectively. Assume that φ(y), F̂ (x) and ĝ(x) satisfy the following conditions: (i): ĝ(x) := g0 (x) + c sgn(x), xĝ(x) > 0 for all x 6= 0, where c ≥ 0 and g0 (x) is Lipschitz in any finite interval and g0 (0) = 0. (ii): fˆ(x) = F̂ 0 (x) is C 0 (α, β), F̂ (0) = 0, fˆ(0) 6= 0, fˆ(x)/ĝ(x) is nondecreasing in (α, 0) ∪ (0, β) and fˆ(x)/ĝ(x) is not a constant when |x| is small. (iii): φ(y) is Lipschitz in any finite interval, yφ(y) > 0 for all y 6= 0, φ(y) is increasing, φ(−∞) = −∞, φ(+∞) = +∞, φ(y) has right-derivative φ0+ (0) and left-derivative φ0− (0) at y = 0, φ0− (0)φ0+ (0) 6= 0 when fˆ(0) = 0. Then system (2) has at most one limit cycle in (α, β). Moreover the limit cycle is stable when it exists. Proof. Since the vector field of system (2) is Lipschitz for c = 0, its solutions exist and are unique except at x = 0. Since the vector field of system (2) is discontinuous at the line Σ := {(x, y) : x = 0} for c > 0, we need to study the dynamics on Σ and we will adapt the Filippov method, see [2, 9]. Let δ := h(1, 0), (−φ(y), −c)ih(1, 0), (−φ(y), c)i = φ2 (y), where h·, ·i denotes the inner product. As defined in [2, 9], the crossing set is Σc = {(x, y) ∈ Σ : δ > 0} = {(x, y) ∈ Σ : y 6= 0}. The sliding set Σs is the complement to Σc , which is given by Σs = {(x, y) ∈ Σ : δ ≤ 0} = {(x, y) ∈ Σ : y = 0}. Therefore except at the origin, all orbits crossing any point are unique. In other words all periodic orbits are crossing. HAssume that γ is a periodic orbit of system (2). Then we have that γ is hyperbolic if γ div(−φ(y) − F̂ (x), ĝ(x))dt 6= 0, see for instance Theorem 1.23 of [6]. Moreover 4 H. CHEN, J. LLIBRE AND Y. TANG γ is stable (resp. unstable) if I div(−φ(y) − F̂ (x), ĝ(x))dt < 0 (resp. > 0). γ In order to prove the uniqueness of limit cycles of system (2), assume that system (2) has at least two limit cycles, where γ1 , γ2 are the innermost limit cycles and γ1 lies in the bounded region surrounded by γ2 . Actually we must have fˆ(0) < 0. Otherwise, if fˆ(0) > 0 we can obtain fˆ(x) > 0 since fˆ(x)/ĝ(x) is non-decreasing and xĝ(x) > 0. For x > 0 near the origin, we can get fˆ(x) > 0 by the continuity of fˆ(x) at x = 0 and then fˆ(x)/ĝ(x) > 0, implying fˆ(x)/ĝ(x) > 0 for all x > 0 by the monotonicity of this function. Thus we have fˆ(x) > 0 for all x > 0. Similarly for all x < 0 we can also get fˆ(x) > 0. Then, by Green formula we have I I 0= (−φ(y) − F̂ (x))dy + ĝ(x)dx = − fˆ(x)dxdy, γi Di which contradicts the fact that fˆ(x) > 0, where Di is the bounded region surrounded by γi for i = 1, 2. Here, note that the Dulac criterion cannot be applied because the vector field of system (2) is not C 1 . Thus we get fˆ(0) < 0 if a periodic orbit exists. Moreover we claim that the equation fˆ(x) = 0 has at most one positive root and one negative root, where a connect set of roots is viewed as one root. Otherwise, assume that fˆ(x) has two positive zeros x1 and x2 such that 0 < x1 < x2 . Then there exists a real x0 ∈ (x1 , x2 ) satisfying fˆ(x0 )/ĝ(x0 ) > 0 = fˆ(x2 )/ĝ(x2 ), which contradicts the non-decreasing of fˆ(x)/ĝ(x). Thus the claim is proved. Applying Green formula again we have that system (2) has no periodic orbits when fˆ(x) ≤ 0 for all x ∈ (α, β). Therefore if system (2) exhibits periodic orbits, there is an x3 ∈ (α, β) such that fˆ(x3 ) > 0. In the following we divide the proof of the uniqueness of limit cycles of system (2) in three cases. Case (I). First we consider the case x3 > 0 if fˆ(x3 ) > 0. Then there is a unique value x4 ∈ (0, β) such that F̂ (x4 ) = 0. Moreover if there exist two different points x41 , x42 ∈ (0, β) such that F̂ (x41 ) = 0 and F̂ (x42 ) = 0, we can get an x̃4 ∈ (x41 , x42 ) satisfying fˆ(x̃4 ) = 0, which contradicts the non-decreasing of fˆ(x)/ĝ(x) for x > 0. We claim that any periodic orbit must surround the point (x4 , 0). So no periodic orbits exist if x4 does not exist. Let Z x y2 (4) E(x, y) = + ĝ(s)ds, 2 0 which implies that dE(x, y) = −ĝ(x)F̂ (x). dt It is to note that ĝ(x)F̂ (x) < 0 for all x ∈ (α, x4 ). Assume that system (2) exhibits a periodic orbit γ, which lies in the strip x ∈ (α, x4 ). Then we can find that I I 0 = dE = −ĝ(x)F̂ (x)dt > 0. γ γ Thus the claim is proved. THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 5 Now we will prove that I I (5) fˆ(x)dt < fˆ(x)dt. γ1 γ2 y P2 A2 B2 A1 C2 I2 B1 I1 N2 C1 a O x4 b x J1 H1 M2 D1 Q2 H2 D2 Figure 1. Limit cycles of system (2) in the Case (I). Consider the two limit cycles γ1 = A1 B1 C\ \ 1 D1 H1 I1 A1 and γ2 = A2 B2 C2 D2 H2 I2 A2 of Figure 1. Notice that the limit cycle γi intersects the graphic of the function y = φ−1 (−F̂ (x)) at the points Ci and Ii for i = 1, 2, respectively. Since I I I I ĝ(x)dt = dy = 0 = dy = ĝ(x)dt, γ1 γ1 γ2 γ2 we only need to prove I I (6) f1 (x)dt < f1 (x)dt, γ1 γ2 which is equivalent to (5), where (7) f1 (x) := fˆ(x) + bĝ(x) for any constant b ∈ R. It is clear that f1 (x)/ĝ(x) is still non-decreasing if f (x)/ĝ(x) is non-decreasing. Fixing b = −fˆ(xI1 )/ĝ(xI1 ) < 0, we have f1 (xI1 ) = 0. Moreover we have f1 (x)/ĝ(x) ≥ 0 for xI1 < x < 0 and f1 (x)/ĝ(x) ≤ 0 for x < xI1 , because f1 (xI1 )/ĝ(xI1 ) = 0 and f1 (x)/ĝ(x) is non- decreasing. Thus f1 (x) ≤ 0 if xI1 < x < 0, and f1 (x) ≥ 0 if x < xI1 , because ĝ(x) < 0 for x < 0. Denote by P = (xP , yP ) for an arbitrary point P . We can find a point J1 (xJ1 , yJ1 ) in the curve y = φ−1 (−F̂ (x)) such that f1 (xJ1 ) = 0 and xJ1 ∈ (0, xC1 ). Otherwise, 6 H. CHEN, J. LLIBRE AND Y. TANG f1 (x) < 0 for all x ∈ (0, xC1 ), and if the point J1 does not exist then f1 (x) ≤ 0 for all x ∈ (xI1 , xC1 ). Thus we obtain I (8) f1 (x)dt < 0. γ1 However the origin is a source and H the periodic orbit γ1 is internally stable because f1 (x) < 0 for small x, implying γ1 f1 (x)dt ≥ 0. It induces a contradiction with the inequality (8). Thus the point J1 exists. Moreover we have f1 (x) ≥ 0 for x > xJ1 , and f1 (x) ≤ 0 for all 0 < x < xJ1 , because ĝ(x) > 0 for x > 0 and f1 (x)/ĝ(x) is non-decreasing. Assume that the line x = xJ1 intersects with the graphic of the function y = −1 φ (−F̂ (x)) at the points Bi and Di for i = 1, 2, respectively. Notice that xB1 = xB2 = xD1 = xD2 = xJ1 . Let y = y1 (x) and y = y2 (x) be the orbit segments A [ [ 1 B1 and A2 B2 , respectively. Since y1 < y2 and the function φ(x) is increasing, we have φ(y1 ) < φ(y2 ). Then we have (9)Z Z Z xB Z xB 1 f1 (x) 2 f1 (x) f1 (x)dt − f1 (x)dt = dx − dx B\ 1 A1 B\ 2 A2 0 φ(y1 ) + F̂ (x) 0 φ(y2 ) + F̂ (x) Z xB 1 f1 (x)(φ(y2 ) − φ(y1 )) = < 0. 0 (φ(y1 ) + F̂ (x))(φ(y2 ) + F̂ (x)) It is similar to prove that Z Z f1 (x)dt − f1 (x)dt < 0, \ H 1 D1 \ H 2 D2 Z Z (10) f1 (x)dt − f1 (x)dt < 0, Ad 1 I1 A\2 P2 Z Z f1 (x)dt − f1 (x)dt < 0, I\ 1 H1 \ Q2 H2 where P2 , Q2 ∈ γ2 and xP2 = xQ2 = xI1 . Let x = x1 (y) and x = x2 (y) be the orbit segments D\ \ 1 C1 B1 and D2 C2 B2 , respec- tively. Then we have Z Z Z Z f1 (x)dt − f1 (x)dt < fˆ(x)dt − fˆ(x)dt D\ 1 C1 B1 D\ 2 C2 B2 D\ 1 C1 B1 \ M 2 N2 (11) Z ! y N2 fˆ(x1 ) fˆ(x2 ) = − dy < 0, yM 2 ĝ(x1 ) ĝ(x2 ) where M2 , N2 ∈ γ2 , yM2 = yD1 and yN2 = yB1 . Since f1 (x) ≥ 0 for all x < xI1 , we have Z (12) f1 (x)dt > 0. P\ 2 I2 Q2 Therefore (6) holds from (9)-(12). Notice that H the origin is a source and the periodic orbit γ1 is internally stable. Thus γ1 fˆ(x)dt ≥ 0. It follows from (5) THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 7 H that γ2 fˆ(x)dt > 0. Consequently γ2 is stable and hyperbolic. By the Poincaré- Bendixson Theorem (see for instance Corollary 1.30 of [6]), it is impossible the existence of two consecutive stable limit cycles. Therefore system (2) has at most two limit cycles. Moreover, γ1 is semi-stable and γ2 is stable if they exist. In order to induce a contradiction for the case that γ1 is semi-stable, we construct an auxiliary vector field (−φ(y) − F̃ (x), ĝ(x)), where F̃ (x) = F̂ (x) + R(x) and  0, if x ≤ x4 , R(x) := F̂ (x), if x > x4 , for small ||. We can check that the vector field (−φ(y) − F̃ (x), ĝ(x)) is rotated with respect to the parameter ; see [16, Chapter 4.3] or [12]. Consider the following system ẋ = −φ(y) − F̃ (x), (13) ẏ = ĝ(x). System (13) is exactly system (2) if  = 0. Moreover, we can check that system (13) still satisfies all assumptions of Theorem 3. In other words system (13) has at most two limit cycles. Further, we can find that γ1 will split into at least two limit cycle for  < 0 by [16, Theorem 3.4 of Chapter 4]. Then system (13) can have three limit cycles, a contradiction with the previous result. Therefore we have proven that system (2) has at most one periodic orbit in the case x3 > 0 if fˆ(x3 ) > 0. a x- b x+ Figure 2. Limit cycles of system (2) in the Case (III). Case (II). Second, we consider the case that there must be x3 < 0 if fˆ(x3 ) > 0. Since the proof is similar to the Case (I), we omit it. Case (III). We consider the case that x3 can be negative or positive if fˆ(x3 ) > 0. We claim that the equation F̂ (x) = 0 has either one non-zero root or two non-zero 8 H. CHEN, J. LLIBRE AND Y. TANG roots. Notice that the equation F̂ (x) = 0 can not have two positive roots or two negative roots. Otherwise if there exist two different points x41 , x42 ∈ (0, β) or ∈ (α, 0) such that F̂ (x41 ) = 0 and F̂ (x42 ) = 0, we can get a point x̃4 ∈ (x41 , x42 ) satisfying fˆ(x̃4 ) = 0, which contradicts the non-decreasing of the function fˆ(x)/ĝ(x). On the other hand, if the equation F̂ (x) = 0 has not non-zero roots, we get dE/dt ≥ 0 for x ∈ (α, β) from (4), implying that no periodic orbits exist. If the equation F̂ (x) = 0 has a unique non-zero root, we consider that the non-zero root is x+ ∈ (0, β) for simpcility. If system (2) has a periodic orbit, it must surround (x+ , 0). Otherwise this is a contradiction with the fact that dE/dt ≥ 0. When equation F̂ (x) = 0 has one positive root x+ ∈ (0, β) and a negative root x− ∈ (α, 0), if system (2) has a periodic orbit, it must surround at least one of the points (x+ , 0) and (x− , 0). Otherwise again we have a contradiction with the fact that dE/dt ≥ 0. Without loss of generality we can assume that any limit cycle surrounds (x+ , 0). Assume that system (2) has three limit cycles γ1 , γ2 , γ3 as the ones shown in Figure 2, where γ1 is the innermost one, γ3 surrounds γ1 and γ2 , the points Ai , Bi , Ci , Di , Hi , Ii Ji , Ki ∈ γi , the periodic orbit γi intersects the graphic of the function y = φ−1 (−F̂ (x)) at the points Ci and Ji for i = 1, 2, 3, respectively. Notice that xBi = xDi , xAi = xHi = 0, xK2 = xK3 = xI2 = xI3 = xJ4 , f (xBi ) = 0, f (xJ4 ) = 0, α < xJ3 < xJ2 < xJ4 < xJ1 < 0, for i = 1, 2, 3. In a similar way to the proof of Case (I) we shall obtain that system (2) has at most one periodic orbit in the strip x ∈ (xJ4 , β). Moreover the periodic orbit is stable if it exists. Now we shall prove that I I I (14) fˆ(x)dt < fˆ(x)dt < fˆ(x)dt. γ1 γ2 γ3 Notice that the function y = φ(x) has the same properties as in Case (I) when x > 0, as it is shown in Figures 1 and 2. Thus we can obtain that R R \ ˆ(x)dt < \ fˆ(x)dt, f (15) R 1 1 1 H C A RH2 C2 A2 ˆ f (x)dt < fˆ(x)dt. H\2 C2 A 2 H\ 3 C3 A3 To prove the first inequality of (14) it suffices to prove the inequality (6). Using the auxiliary function f1 (x) in (7) for the Case (I) again, we can prove that Z Z (16) f1 (x)dt < f1 (x)dt. A\ 1 J1 H 1 A\ 2 J2 H 2 From (15) and (16) we get I I (17) fˆ(x)dt < fˆ(x)dt. γ1 γ2 THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 9 Moreover we can calculate that Z Z ˆ f (x)dt − fˆ(x)dt < 0, A\K A\K Z 2 2 Z3 3 (18) fˆ(x)dt − fˆ(x)dt < 0, K\ J I K\J I Z 2 22 Z 3 33 fˆ(x)dt − fˆ(x)dt < 0 I\ 2 H2 I\ 3 H3 doing a similar calculation as in Case (I) for x > 0. Thus, from (15) and (18), we get the second inequality of (14), i.e. I I (19) ˆ f (x)dt < fˆ(x)dt. γ2 γ3 It follows from (17) and (19) that (14) holds. Since the origin is a source, we have I I I ˆ f (x)dt ≥ 0, implying ˆ f (x)dt > 0 and fˆ(x)dt > 0, γ1 γ2 γ3 from the inequality (14). However it is impossible to have two consecutive stable limit cycles. Therefore system (2) cannot have three periodic limit cycles and there are at most two limit cycles. We divide the rest of the proof in two subcases. First, we consider the subcase that γ1 only surrounds one of the points (x+ , 0) and (x− , 0). In Case (I) we have proved that for this kind of periodic orbits, as γ1 , at most one can exist and it is H stable. Thus its consecutive periodic orbit γ2 is internally unstable and then γ2 fˆ(x)dt ≤ 0. Moreover the inequality (17) holds and γ2 is stable, indicating a contradiction. Therefore the periodic orbit γ2 does not exist and system (2) has exact one periodic orbit γ1 if such periodic orbit exists. Now we consider the subcase that system (2) has no such kind of periodic orbits like γ1 . We assume that system (2) has two periodic orbits γ2 and γ3 , which surround both points (x+ , 0) and (x− , 0). Since the origin is a source, we have I I ˆ f (x)dt ≥ 0, implying fˆ(x)dt > 0, γ2 γ3 by the inequality (19). Therefore γ2 is semi-stable and γ3 is stable. Using the auxil- iary vector field (13) again, we can get that system (13) still satisfies all conditions of this theorem and has at most two limit cycles. However by the rotated properties of system (13), the semi-stable γ2 will split into at least two limit cycles for  6= 0 by [16, Theorem 3.4 of Chapter 4]. Then system (13) can have three limit cycles, a contradiction. Thus we have proven that system (2) has at most one periodic orbit in the case (III) and the proof is completed.  Remark 4. The conditions φ(−∞) = −∞, φ(+∞) = +∞ are needed only if (α, β) is unbounded. If (α, β) is a bounded interval, these conditions can be deleted in Theorem 3. Notice that the vector field is Lipschitz if c = 0 in Theorem 3. Thus the results of Theorem 3 also hold when system (2) is Lipschitz or further smooth. 10 H. CHEN, J. LLIBRE AND Y. TANG The modified Liénard system (21) of the Higgins-Selkov (1) is Lipschitz except at the line x = 1, which is a discontinuity point of the second kind for the functions in the system. So we need to apply Theorem 3 for showing the uniqueness of the limit cycles. 3. Proof of Theorem 1 Notice that system (1) cannot have periodic orbits when a ≤ 0, because the unique equilibrium (1, 1) is a saddle as a < 0 or ẏ ≡ 0 as a = 0. Thus in the following we only consider the case a > 0. Moreover, the periodic orbits of system (1) must lie in the region {(x, y) ∈ R2 | x > 0, y > 0}, since the x-axis is invariant and ẋ|x=0 = 1. In order to simplify the computations and the analysis we do the following change of coordinates   y 1 − x1 t1 (x, y, t) → , x1 , , a a changing system (1) into xn1 y1 xn+1 (20) ẋ1 = −x1 + − 1 , a a ẏ1 = 1 − x1 . Obviously the periodic orbit of system (20) only exists in the region x1 > 0, because ẏ1 = 1 > 0 and ẋ1 = 0 on the line x1 = 0. Moreover system (20) can be changed into the following Liénard system ẋ = y − F (x), (21) ẏ = −g(x), where 1 x x F (x) := n−1 + − 1 and g(x) := , (x + 1) a a(x + 1)n doing the transformation (x1 , y1 , t1 ) → (x + 1, ay + (a + 1), t/(x + 1)n ). Here we only need to consider x > −1 for the problem of limit cycles of system (21), because system (21) is equivalent to system (1) as a > 0 and x > −1. From [3] system (21) has no periodic orbits when a ∈ (0, 1/(n − 1)). In the following we prove that system (21) may have periodic orbits only if a > 1/(n − 1). Here we cannot use the methods of [5], because it is difficult to decide when the equations g(z1 ) g(z2 ) = and F (z1 ) = F (z2 ), f (z1 ) f (z2 ) have solutions or not for an arbitrary integer n. We need use a new method and technique. Proposition 5. For a > 0, the amplitude of the stable or unstable limit cycle of sys- tem (21) surrounding the origin varies monotonically with respect to the parameter a. THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 11 Proof. Notice that we can change system (21) into the following equivalent form ẋ = y − F̌ (x), (22) ẏ = −ag(x)  √ √  by the transformation of coordinates (x, y, t) → x, y/ a, at , where √  1  x F (x) := a n−1 −1 + √ . (x + 1) a From the calculation in [11], we have the value of the following determinant   √  1  x x y− a1 n−1 −1 + √ − (x + 1) a1 (x + 1)n   √  1  x x y− a2 n−1 −1 + √ − (x + 1) a2 (x + 1)n     n−1 √ √ x 1 − (x + 1) 1 x 2 = ( a1 − a2 )  2n−1 −√ ≤0 (x + 1) a1 a2 (x + 1)n for a2 < a1 , a2 , a1 ∈ (0, +∞) and x > −1. Thus the vector field of system (22) is a generalized rotated vector field (see [16, Chapter 4.3] or [12]) with respect to the parameter a if x > −1. Moreover from [16, Theorem 3.5, Chapter 4], the amplitude of the stable or unstable limit cycle of system (22) surrounding the origin varies monotonically with respect to the positive parameter a.  Proposition 6. System (21) has no periodic orbits when a ∈ (−∞, 1/(n − 1)]. Proof. By [1, 5], system (21) has no periodic orbits for a ≤ 1 when n = 2. In the rest of this proof, we only consider n ≥ 3. We only need to consider system (21) and its limit cycles in the region x > −1. Assume that (23) F (x1 ) = F (x2 ), G(x1 ) = G(x2 ) for n ≥ 3 and −1 < x1 < 0 < x2 , where Z x 1 nx − x + 1 G(x) := g(s)ds = − . 0 a(n − 1)(n − 2) a(x + 1)n−1 (n − 1)(n − 2) It follows from (23) that 1 x1 1 x2 (24) n−1 + −1= n−1 + − 1, (x1 + 1) a (x2 + 1) a (25) 1 1 1 1 1 1 1 1 n−2 + n−1 = n−2 + . (2 − n) (x1 + 1) (n − 1) (x1 + 1) (2 − n) (x2 + 1) (n − 1) (x2 + 1)n−1 From (24) we have 1 1 x1 − x2 (26) n−1 − n−1 =− (x1 + 1) (x2 + 1) a 1 (x1 + 1)(x1 − x2 ) x1 + 1 (27) ⇔ n−2 =− + . (x1 + 1) a (x2 + 1)n−1 12 H. CHEN, J. LLIBRE AND Y. TANG Furthermore it follows from (25) and (26) that     1 1 1 1 1 1 − = − (2 − n) (x1 + 1)n−2 (x2 + 1)n−2 (n − 1) (x2 + 1)n−1 (x1 + 1)n−1 x1 − x2 (28) = . a(n − 1) Moreover we calculate from (27) and (28) that   1 (x1 + 1)(x1 − x2 ) x1 + 1 1 x1 − x 2 − + n−1 − n−2 = , (2 − n)  a (x2 + 1) (x2 + 1)  a(n − 1) 1 (x1 + 1)(x1 − x2 ) 1 x1 − x2 − + n−1 (x1 − x2 ) = , (29) (2 − n)  a (x2+ 1) a(n − 1) 1 (x1 + 1) 1 1 − + n−1 = , (2 − n) a (x2 + 1) a(n − 1) a 1 x1 = − . (x2 + 1)n−1 n − 1 Substituting (29) into (24) we have 1 1 x2 (30) − − = 0. a n − 2 n−1 a(n − 1) a n−1 + (x2 + 1) n−1 Let 1 x 1 (31) H(x, a) := − − . a n − 2 n−1 a a(n − 1) + (x + 1)n−1 n − 1 Then we have that dH a(n − 1)2 1 (32) (x, a) = n − , dx H1 (x, a) a where  a n − 2 H1 (x, a) = + (x + 1). (x + 1)n−1 n − 1 Now we consider the case a = 1/(n − 1). From (31) and (32) we get 1 H(x, 1/(n − 1)) = − (n − 1)x − 1 1 n − 2 n−1 + (n − 1)(x + 1)n−1 n − 1 and dH n−1 (x, 1/(n − 1)) = n − (n − 1), dx H1 (x, 1/(n − 1)) where  1 n − 2 H1 (x, 1/(n − 1)) = + (x + 1). (n − 1)(x + 1)n−1 n − 1 Then we claim that dH (x, 1/(n − 1)) < 0. dx THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 13 Actually, we have dH1  1 n−2 0 (x, 1/(n − 1)) = + (x + 1) dx (n − 1)(x + 1)n−2 n − 1 n − 2 −1  = + 1 > 0, n − 1 (x + 1)n−1 for n ≥ 3 and x ≥ 0, implying min{H1 (x, 1/(n − 1))}x≥0 = H1 (0, 1/(n − 1)) = 1. Thus we get H 0 (x, 1/(n − 1)) ≤ 0 and then max{H(x, 1/(n − 1))}x≥0 = H(0, 1/(n − 1)) = 0. In other words equation (30) has no solutions for x2 > 0, and then equations (23) have no solutions {x1 , x2 } such that −1 < x1 < 0 < x2 if n ≥ 3 and a = 1/(n − 1). Thus from continuity we have F (x1 ) > F (x2 ), or F (x1 ) < F (x2 ) if G(x1 ) = G(x2 ). Moreover we have that F (0) = 0 and xg(x) > 0. Therefore by Proposition 2.1 of [4], system (21) has no periodic orbits for a = 1/(n − 1). Now consider the case a < 1/(n − 1). When a ≤ 0, either the unique equilibrium (1, 1) of system (1) is a saddle or the system has an invariant line through equilibrium (1, 1), which implies non-existence of periodic orbits. The vector field of equivalent system (22) of system (21) is a generalized rotated vector field with respect to a for x > −1 and a > 0 by the proof of Proposition 5. Moreover the amplitude of the stable or unstable limit cycle surrounding the origin of (21) varies monotonically with respect to a. Assume that system (21) exhibits limit cycles for a = a0 ∈ (0, 1/(n−1)), where γ is the innermost limit cycle. Since the origin of (21) is stable, then γ is internally unstable. Note that the amplitude of an unstable limit cycle decreases as a increases by [16, Theorem 3.5, Chapter 4]. When a increases from a = a0 to a = 1/(n − 1), the origin keeps stability. Therefore, γ does not disappear for a = 1/(n − 1). This is a contradiction to our above analysis as a = 1/(n − 1), and the proof is completed.  When 2n − 1 a = an := , 2n − 2 we give the following lemma for the region where periodic orbits exist. Obviously, an > 1 for n ≥ 3. Lemma 7. When a = an for n ≥ 3, periodic orbits of system (21) only exist in the strip x ∈ (−1, 1.6). Proof. Note that any periodic orbit of system (1) must lie in the first quadrant and the y-axis of system (1) is changed into the line y = x − 1 of system (21). Therefore, the periodic orbits of system (21) cannot intersect the line y = x − 1. Assume that Γ is a periodic orbit of system (21) and Γ intersects with the curve y = F (x) at the point (x∗ , F (x∗ )) in the right half-plane. Then x ≤ x∗ as (x, y) ∈ Γ. If x∗ ≤ 1, we have that Γ lies in the strip x ∈ (−1, 1] and this lemma is proven. In the following, we consider the case x∗ > 1. Let y = ỹ(x) < F (x) denote the orbit segment of Γ as 0 ≤ x ≤ x∗ . For x ≥ 1 and a = an , we have that ỹ(x) > x − 1 and dỹ(x) g(x) g(x) (33) = > ≥ x, dx F (x) − ỹ(x) F (x) − x + 1 14 H. CHEN, J. LLIBRE AND Y. TANG which implies 1 (34) ỹ(x) < (x2 − (x∗ )2 ) + ỹ(x∗ ) 2 for 1 ≤ x < x∗ . Actually, the inequality g(x)/(F (x)−x+1) ≥ x in (33) is equivalent to the inequality ϕ1 (x) (35) a≥ , ϕ2 (x) 1 1 where ϕ1 (x) = x − (x+1)n and ϕ2 (x) = x − (x+1)n−1 . Notice that for x ≥ 1 we get the maximum value of the positive function ϕ1 (x) − ϕ2 (x) at x = 1, implying that the function ϕ1 (x)/ϕ2 (x) has its maximum value an also at x = 1 and the inequality (33) is obtained. We can find that the curve 1 1 Υ : y = x2 − 2 2 is tangent to the line y = x − 1 at the point (1, 0). Moreover, the curves Υ and y = F (x) have a unique intersection point at (x̃∗ , ỹ ∗ ) for x ≥ 1. Actually, from F (x) = y = 21 x2 − 21 and a = an we get that 1 (2n − 2)x x2 1 P (x) := + − − = 0. (x + 1)n−1 2n − 1 2 2 Applying 1 2n − 2 P 0 (x) = (1 − n) + −x<0 (x + 1)n 2n − 1 for x ≥ 1, P (1) = (2n−1 − 1)/(22n−1 − 2n−1 ) > 0 and P (1.6) = 2.61−n − 0.18 − 1.6/(2n − 1) < 0, we get a unique value x̃∗ such that P (x̃∗ ) = 0 and 1 < x̃∗ < 1.6. In the following, we prove that x∗ < x̃∗ . Otherwise, if x∗ ≥ x̃∗ , we have −(x∗ )2 /2+ ỹ(x∗ ) ≤ −1/2, inducing 1 1 1 ỹ(x) < (x2 − (x∗ )2 ) + ỹ(x∗ ) ≤ x2 − 2 2 2 for x = 1 by (34). Hence, we can obtain that the curve y = ỹ(x) has intersection points with the line y = x − 1, indicating a contradiction. Therefore, x∗ < x̃∗ . In other words, the periodic orbits of system (21) must lie in the region x ∈ (−1, 1.6). The lemma is proven.  Proposition 8. System (21) has no periodic orbits when a ≥ an . Proof. By [1, 5], there is a a∗ ∈ (1, 3) such that system (21) has no periodic orbits for a > a∗ when n = 2. When n = 3 and a = an , we can calculate numerically that the function H(x, a) in (31) has not zeros as x lies between x̃∗ and the positive zero of F (x), where the curves Υ and y = F (x) intersect at (x̃∗ , ỹ ∗ ) for x ≥ 1, as shown in the proof of Lemma 7. By Proposition 2.1 of [4] system (21) has no periodic orbits when n = 3 and a = an . In the rest of this proof, we only need consider the case n ≥ 4. THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 15 From (32), we have that dH a(n − 1)2 1 (x, a) = n − dx H1 (x, a) a and  < 0, 0 < x < x0 , dH1 (x, a) a(n − 2) n−2 =− + = 0, x = x0 , dx (x + 1)n−1 n − 1  > 0, x > x , 0 where p x0 = n−1 a(n − 1) − 1 > 0 as a ≥ an > 1. Further, we obtain min {H1 (x, a)} = H1 (x0 , a), x>−1 implying dH dH n−1 1 max{ (x, a)} = (x0 , a) = p − > 0. x>−1 dx dx n−1 a(n − 1) a Moreover, we have dH lim (x, a) < 0 x→+∞ dx and dH a(n − 1)2 1 H2 (a) (0, a) =   − =  < 0, dx n−2 n a n − 2 n a+ a a+ n−1 n−1 where  n − 2 n H2 (a) = a2 (n − 1)2 − a + , n−1 because dH2 (a)  n − 2 n−1 dH2 (a) = 2a(n − 1)2 − n a + < |a=1 < 0, da n−1 da d2 H2 (a)  n − 2 n−2 d2 H2 (a) 2 = 2(n − 1) − n(n − 1) a + < |a=1 < 0, da2 n−1 da2 d3 H2 (a)  n − 2 n−3 = −n(n − 1)(n − 2) a + < 0. da3 n−1 Notice that 1 1 H(0, a) = − <0 n − 2 n−1 a(n − 1) a+ n−1 from (31) and a similar discussion as dH(0, a)/dx < 0. If the inequality H(x, a) < 0 always holds as a ≥ 1, we can get that system (21) has no periodic orbits by Proposition 2.1 of [4] and a similar discussion as the proof of Proposition 6. So, in the following, we consider the case that there exists a value x∗ > x0 such that max{H(x, a)} = H(x∗ , a) > 0. x>−1 Without loss of generality we assume that x0 1 H(x0 , a) = 1 − − > 0. a a(n − 1) 澳 澳 澳 澳 澳 澳 澳 澳 澳 濃澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 濬澔 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 濭澱澼澜濬澠濕澝澔 澳 澳 澳 澳 16 澳 H. CHEN, J. LLIBRE AND Y. TANG 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 濭澔 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 濃澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 #澔 !澔 澔 澔 澔 澔 澔 澔 澔 澔 "澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 澔 濬澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 澳 濭澱澼澜濬澠濕澝澳 澳 澳 澳 澳 澳 澳 澳 Figure 3. y = H(x, a) 澳 澳 澳 When H(x0 , a) ≤ 0, we can research by a similar way. Thus, there are values x1 and x2 such that x1 > x2 > 0 and H(x, a) > 0 (resp. < 0) for x ∈ (x2 , x1 ) (resp. x ∈ (0, x2 ) ∪ (x1 , +∞)) when n ≥ 4, as shown in Figure 3. p The function F 0 (x) has a unique zero at x = n a(n − 1) − 1 > 0 and there exists a unique positive x3 such that F (x3 ) = 0 and F (x)p> 0 (resp. < 0) for x ∈ (−1, 0) ∪ (x3 , +∞) (resp. x ∈ (0, x3 )). Letting z0 = n−1 a(n − 1) we can find that z0 > 1 and p 1 n−1 a(n − 1) − 1 F (x0 ) = + −1 a(n − 1) a 1 z0 − 1 = n−1 − 1 + z0 a   1 + z0 + · · · + z0n−2 1 = (1 − z0 ) − z0n−1 a  n−2  1 + z0 + · · · + z0 n−1 = (1 − z0 ) − n−1 < 0, z0n−1 z0 indicating x0 < x3 . Consider the case a = an > 1. Calculating the equation F (x3 ) = 0 from (21), we can let (n − k(n))an (36) x3 = , n−1 where k ∈ (1, 1.5) and k = k(n) is decreasing in n. Specially, x3 = 1 as n → +∞ and x3 ≈ 0.92 as n = 4. Further, by (36) we have that 1 k − n − 1/an H(x3 , an ) = (k−1)an −1 n−1 + (1 + ) n−1 n−1 1 − H3 (n) = (k−1)an −1 n−1 > 0, (1 + n−1 ) THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 17 where k − 1 − 1/an (k − 1)an − 1 n−1 H3 (n) = (1 − )(1 + ) n−1 n−1  1 ((k − 1)an − 1) ((k − 1)an − 1)2  (k − 1)an − 1 n−2 = 1 + (1 − ) − (1 + ) an n−1 an (n − 1)2 n−1 < 1, because an > 1 and (k − 1)an − 1 < 0. Thus, we can get x0 < x3 < x1 . From Lemma 7, any limit cycle of system (21) must lie in the region {(x, y) ∈ R2 |x < 1.6}. Hence, we consider the solution of (23) satisfying x3 < x < 2 for x > 0. In order to prove H(x, an ) > 0 for x3 < x < 2, we only need to show H(2, an ) > 0 by Figure 3. We can compute that 1 1− ∂(aH(2, a)) 1/(n − 1) +3n−1 (n − 2)/(a(n − 1)2 ) =  n−1 >0 ∂a a31−n + (n − 2)/(n − 1) as n > 3 and a ∈ [1, an + ε] for small enough ε > 0. It implies that the function aH(2, a) is increasing in a for a ∈ [1, an + ε]. Thus from H(2, 1) > 0 we can get H(2, an ) > 0. Moreover, we have 1 1 1 − Ĥ(n) H(2, 1) = −2− = , 1−n n − 2 n−1 n−1 1−n n − 2 n−1 (3 + ) (3 + ) n−1 n−1 where 1 n − 2 n−1 Ĥ(n) = (2 + )(31−n + ) . n−1 n−1 In the following, we prove that Ĥ(n) < 1 and then H(2, 1) > 0. First, we get q n−1 H4 (n) = ln Ĥ(n) 1 1 1 = ln(2 + ) + ln(31−n + 1 − ) n−1 n−1 n−1 1 = u ln(2 + u) + ln(3− u + 1 − u), 1 where u = n−1 ∈ (0, 31 ]. Noticing that Ĥ(4) ≤ 0.814 < 1, Ĥ(5) ≤ 0.76 < 1, Ĥ(6) ≤ 0.74 < 1, Ĥ(7) ≤ 0.738 < 1, Ĥ(8) ≤ 0.736 < 1 and Ĥ(n) → 2e as n → +∞. We can only consider the case n ≥ 9 and u ∈ (0, 81 ] for proving the inequality Ĥ(n) < 1. Moreover, −1 3 u ln 3 0 u 2 −1 H4 (n) = ln(2 + u) + + − 1u 2+u 3 u +1−u = ξ1 (u) + ξ2 (u), 18 H. CHEN, J. LLIBRE AND Y. TANG where u ξ1 (u) = ln(2 + u) + − 1, 2+u 1 1 3− u ln 3 3− u − u + u2 ξ3 (u) ξ2 (u) = 1 −u =: 1 −u . 3 +1−u 3 +1−u In addition, for u ∈ (0, 81 ] we get 1 2 ξ10 (u) = + > 0, 2 + u (2 + u)2 1 1 1 3− u ln 3 3− u ln2 3 3− u ln 3 ξ30 (u) = −1 + + − 2 u2 u4 u3 1 2 1 −u −u 3 ln 3 3 ln 3 2 = (−1 + 4 )+ 2 (1 − ) < 0. u u u It follows that 1 ξ1 (u) ≤ ξ1 ( ) = ln(17/8) − 16/17 < 0, 8 ξ3 (u) ≤ lim ξ3 (u) = 0, u→0+ indicating H40 (n) < 0 and then Ĥ(n) < 1. Therefore, we get H(2, 1) > 0 and then H(2, an ) > 0, implying that H(x, an ) > 0 for x ∈ [x3 , 2]. In other words, the equations (23) have no roots for a = an . By Proposition 2.1 of [4] system (21) has no periodic orbits when n ≥ 3 and a = an . Since the amplitude of the stable or unstable limit cycle of system (21) varies monotonically in a by Proposition 5, system (21) has no periodic orbits when n ≥ 3 and a ≥ an .  When a ∈ (1/(n − 1), an ) we will study the existence and uniqueness of limit cycle in the following proposition. Proposition 9. There exists a constant a∗ ∈ (1/(n − 1), an ) such that system (21) has a unique limit cycle when a ∈ (1/(n − 1), a∗ ) and no periodic orbits when a ∈ (a∗ , +∞). Moreover the limit cycle is stable and hyperbolic, and the amplitude of the limit cycle increases as a increases. Proof. By [16, Theorem 3.5, Chapter 4] and Proposition 5 the amplitude of the stable limit cycle surrounding the origin of (21) is monotonous with respect to a as x > −1 and a > 0. From [11] the Hopf bifurcation occurs and a stable limit cycle appears when a varies from 1/(n − 1) to 1/(n − 1) + , where  > 0 is small. The amplitude of the stable limit cycle is sufficiently small for small enough  > 0. Thus the amplitude of the stable limit cycle increases as a increases. On the other hand, system (21) has no periodic orbits when a ∈ (−∞, 1/(n − 1)] ∪ [an , +∞) by Propositions 6 and 8, and has a unique finite equilibrium at the origin. Therefore, there exists a∗ ∈ (1/(n − 1), an ) such that the amplitude of the stable limit cycle approaches infinity when a = a∗ −  for sufficiently small  > 0 by the continuity of the vector field and the monotonous properties of amplitude of the stable limit cycle in parameter a. THE LIMIT CYCLES OF THE HIGGINS-SELKOV SYSTEMS 19 In the following we will prove the uniqueness of periodic orbits when a ∈ (1/(n − 1), a∗ ). From system (21) we calculate  d f (x)/g(x) n − 1 ((n − 1)x − 1)(x + 1)n−1 Q(x) (37) =a 2 + 2 = , dx x x x2 where Q(x) := a(n − 1) + ((n − 1)x − 1)(x + 1)n−1 , and (38) Q0 (x) := n(n − 1)x(x + 1)n−2 . Notice that Q(0) = a(n − 1) − 1 > 0 since a > 1/(n − 1). Moreover, it is easy to see that Q(x) > 0 when x ∈ [1/(n − 1), +∞). When x ∈ [0, 1/(n − 1)), we get Q0 (x) ≥ 0 from (38). Thus min{Q(x)}x∈[0,1/(n−1)) = Q(0) > 0, inducing Q(x) > 0 in this case. When x ∈ (−1, 0] we get Q0 (x) ≤ 0 from (38). Thus, min{Q(x)}x∈(−1,0] = Q(0) > 0, also inducing Q(x) > 0 in this case. Therefore we obtain Q(x) > 0 when x ∈ (−1, +∞) and then the function f (x)/g(x) is increasing from (37) when x ∈ (−1, 0)∪ (0, +∞). Moreover we can verify that xg(x) > 0 for all x 6= 0, F (0) = 0, F 0 (0) 6= 0 and f (x)/g(x) is not a constant when |x| is small. Hence all conditions in Theorem 3 hold and we can get that system (21) has at most one limit cycle in (−1, +∞). Moreover the limit cycle is stable if it exists. Notice that for our system (21), the function φ(y) = y in the general system (2). The proposition is proven.  From Propositions 5-9 we can obtain Theorem 1. Acknowledgements The first author is supported by the National Natural Science Foundation of Chi- na (No. 11801079). The second author is partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Re- cerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA- RISE-2017-777911. The third author is supported by the National Natural Science Foundations of China (No. 11931016, 11871041), and the International Cooperation Fund of Ministry of Science and Technology of China. References [1] J.C. Artés, J. Llibre, C. Valls, Dynamics of the Higgins-Selkov and Selkov systems, Chaos Solitons Fract. 114 (2018), 145-150. [2] M. di Bernardo, C.J. Budd, A.R. Champneys, P. Kowalczyk, Piecewise-smooth Dynamical Systems: Theory and Applications, Springer-Verlag, London, 2008. [3] P. Brechmann, A.D, Rendall, Dynamics of the Selkov oscillator, Mathematical Biosciences 306 (2018), 152-159. [4] H. Chen, X. 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China E-mail address: chen hebai@csu.edu.cn b Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain E-mail address: jllibre@mat.uab.cat c School of Mathematical Sciences, MOE-LSC, Shanghai Jiao Tong University, Shanghai, 200240, P. R. China E-mail address: mathtyl@sjtu.edu.cn