This is a preprint of: “Existence of at most two limit cycles for some non-autonomous differential equations”, Armengol Gasull, Yulin Zhao, Commun. Pure Appl. Anal., vol. 22(3), 970–982, 2023. DOI: [10.3934/cpaa.2023016] EXISTENCE OF AT MOST TWO LIMIT CYCLES FOR SOME NON-AUTONOMOUS DIFFERENTIAL EQUATIONS ARMENGOL GASULL AND YULIN ZHAO Abstract. It is know that the non-autonomous differential equations dx/dt = a(t)+ b(t)|x|, where a(t) and b(t) are 1-periodic maps of class C 1 , have no upper bound for their number of limit cycles (isolated solutions satisfying x(0) = x(1)). We prove that if either a(t) or b(t) does not change sign, then their maximum number of limit cycles is two, taking into account their multiplicities, and that this upper bound is sharp. We also study all possible configurations of limit cycles. Our result is similar to other ones known for Abel type periodic differential equations although the proofs are quite different. 1. Introduction The problem of knowing the number of limit cycles of general planar vector fields is extremely complicated and many efforts have been dedicated to face it for concrete families of planar differential equations. Some of these families are: quadratic systems, cubic systems, Kolmorov systems, rigid systems, Liénard type equations, . . . , see for instance [17, 18, 28, 29]. For this reason, and to try to consider simpler questions that capture the main difficulties of the problem, people has addressed similar problems for one-dimensional non-autonomous and periodic differential equations. More concretely, consider C 1 differential equations of the form dx = S(x, t), (1) dt with x, t ∈ R, that are 1-periodic in the variable t. We are interested on solutions x(t), defined for all t ∈ R, and such that x(0) = x(1). We will call them periodic solutions because they are closed when we consider (1) on the cylinder R × [0, 1]. Moreover, a periodic solution which is isolated in the set of all the periodic solutions of (1) is called a limit cycle of the differential equation. These differential equations are interesting by themselves, but also appear from some families of planar autonomous polynomial differential equations ([5, 6, 15, 22]), in pro- blems of control theory, see for instance [4, 10] and their references, or in other models of the real world ([2, 16]). In particular, the study of the maximum number of limit cycles of (1) when S(x, t) is a polynomial of degree n has a long history. We briefly summarize it. This question was proposed by N. G. Lloyd ([20]), V. A. Pliss ([26]), and C. Pugh ([19]). When n = 1 it is a linear equation and it is well known that it has either a continuum of periodic solutions or at most 1 limit cycle. Similarly, when n = 2 it is a periodic Riccati equation and it has at most two limit cycles, see for instance [19, 21]. When n = 3, it is called Abel equation and the situation is much more complicated. In fact, for all n ≥ 3 it is 2000 Mathematics Subject Classification. Primary: 34C25 Secondary: 34A34, 37C07, 37C27. Key words and phrases. Non-autonomous differential equation; Limit cycle; Periodic orbit. 1 2 A. GASULL AND Y. ZHAO known that there is no upper bound for the number of limit cycles for periodic Abel differential equations, see [19, 24]. Hence, for n ≥ 3 to have results that guarantee an upper bound for the number of limit cycles of polynomial differential equation some additional hypotheses need to be added. Next we recall some results where it can be seen that usually these hypotheses fix the sign invariance of one of the coefficients of some of the monomials xk . • The 1-periodic differential equation, dx = an (t)xn + am (t)xm + a1 (t)x, (2) dt with n > m > 1, where either an (t) or am (t) does not change sign, has at most 4 (resp. 5) limit cycles when n is even (rep. odd). Moreover both bounds are optimal if n ≥ 4, see [12]. When n = 3 and a3 (t) does not change sign, the (optimal) upper bound for the number of limit cycles is 3, see [1, 14, 19, 26]. Also, when n = 3, a1 (t) ≡ 0 and a2 (t) does not change sign, it is proved in [14] that again the maximum number of limit cycles is 3. • The 1-periodic differential equation dx = an (t)xn + a2 (t)x2 + a1 (t)x + a0 (t), (3) dt with n ≥ 3, and where an (t) does not change sign, has at most 3 limit cycles when n is odd ([23]) and it has no upper bound for its number of limit cycles when n is even ([12, 23]). See also [8] for some refinements when a2 (t) ≡ 0. It is important to notice for instance that when it is assumed in (2) that a1 (t) does not change sign, there is no upper bound for its number of limit cycles. Hence the role of the coefficients corresponding to different xk can be very different. Moreover, for (3) its role clearly also depends on the parity of n. In this paper we are interested on the number of limit cycles of the differential equation dx = a(t) + b(t)|x|, (4) dt where a(t) and b(t) are real, 1-periodic and C 1 functions, t ∈ R. Although, at a first look, it might look simple, it is proved in [3, 9] that there is no upper bound for its number of limit cycles. More concretely, for instance in [9] it is proved that given any natural number k ≥ 2, for ε small enough, the differential equation dx = 2π sin(2πt) + ε cos(2kπt)|x|, (5) dt has at least k − 2 limit cycles. Therefore it is quite natural to wonder if (4) will have an upper bound for its number of limit cycles when either a(t) or b(t) does not change sign. As we will show, the answer is yes and to prove this fact is the main goal of this work. Notice that the line L = {x = 0} is the locus where the differential equation (4) is not smooth, but this differential equation is Lipschitz continuous with respect to x and as a consequence the existence and uniqueness of solutions of its associated Cauchy initial value problem is guaranteed. We will denote by u(t) = u(t, 0, x) the solution of (4) with the initial condition u(0) = x. Let u+ (t) be the solution of (4) such that u(t) ≥ 0 for every t ∈ R. Then u(t) = u+ (t) satisfies x0 = a(t) + b(t)x and is called a positive solution. If u+ (t) is 1-periodic then it is called a positive periodic orbit. Similarly, let u− (t) be the solution of (4) such that EXISTENCE OF AT MOST TWO LIMIT CYCLES 3 u(t) ≤ 0 for every t ∈ R. Then u− (t) satisfies x0 = a(t) − b(t)x and is called a negative solution. A periodic solution u− (t) is called a negative periodic orbit. The limit cycles of (4) that take positive and negative values will be called crossing limit cycles. This is so because they cross the line L. For instance, the k − 2 limit cycles of (5) are of crossing type. As usual, define the Poincaré return map as P (ρ) = u(1, 0, ρ) and the displacement map H(ρ) = P (ρ) − ρ. Clearly, the periodic orbits of (4) are the fixed points of P (ρ), or equivalently, the zeros of H(ρ). Moreover, if some of these values of ρ is a isolated fixed point (or zero) then the corresponding periodic solution u(t, 0, ρ) is a limit cycle. It is proved in [3] that P (ρ) and H(ρ) are of class C 1 , see Proposition 2.4. If the limit cycle x = u(t, 0, ρ) is such that P (ρ) = ρ and P 0 (ρ) 6= 1 then it is said that it as a hyperbolic limit cycle. A limit cycle that is an atractor from below and a repeller from above (or vice versa) is called semistable. If P (ρ) is twice derivable, P (ρ) = ρ, P 0 (ρ) = 1 and P 00 (ρ) 6= 0, then the corresponding limit cycle is called double limit cycle and it is semistable. For short we will say that the limit cycles configuration of a differential equation (4) with a(t) 6≡ 0 is [k, `, m] if it has respectively, k, ` and m negative, crossing and positive limit cycles, taking into account their multiplicities. In Section 4 we will see examples with all possible configurations when either a(t) or b(t) does not change sign. Notice that x = 0 can be a limit cycle only if a(t) ≡ 0 and in this paper we disregard this trivial case. Theorem 1.1. Consider the differential equation (4), dx = a(t) + b(t)|x|, dt where a(t) and b(t) are real, 1-periodic C 1 functions. Then the following statements hold: (a) If a(t) 6≡ 0 and does not change sign then it has at most two limit cycles. R1 Moreover they exist if and only if sign(a) 0 b(t) dt < 0, and then they have configuration [1, 0, 1] and are hyperbolic. (b) If b(t) does not change sign and a(t) has finitely many zeros in [0, 1], then it has at most two limit cycles, taking into R 1 account their multiplicities. The limit cycles can only exist when sign(b) 0 a(t) dt < 0, and if they exist, the total number, taking into account their multiplicities, is two. Moreover, the only possible configurations are [1, 0, 1], [1, 1, 0], [0, 1, 1] and [0, 2, 0] and all are realizable, where when a 2 appears it means that there are or two hyperbolic crossing limit cycles or a double (semistable) crossing limit cycle, and both cases do happen. The hypothesis that a(t) has finitely many zeros is technical and we have inherited it from [3], where it is also introduced for other reasons. 2. Preliminary results In this paper when a continuous function w(t) is non-negative, but not identically zero, that is w(t) ≥ 0, w(t) 6≡ 0, we will say that its sign is positive. Similarly we define negative sign. Moreover, for short, when we say that the sign of some function gives the stability of a periodic orbit we will mean that when it is positive (resp. negative) then the orbit is an attractor (resp. a repeller). 4 A. GASULL AND Y. ZHAO Next result characterizes the positive or negative periodic orbits. For simplicity we introduce the following quantities: Z 1  Z 1 Z 1  A = exp a(t) dt > 0, C= a(t) exp b(s)ds dt 0 0 t Z 1  Z 1  Z 1  B = exp b(t) dt > 0, D= a(t) exp − b(s)ds dt (6) 0 0 t Proposition 2.1. (i) The maximum number of positive limit cycles of (4) is one and a positive limit cycle exists if and only if B 6= 1 and x = u+ (t, 0, C/(1 − B)) ≥ 0, where u+ is the solution of x0 = a(t) + b(t)x, with the given initial condition, and B and C are given in (6). Moreover, it is hyperbolic and its stability is given by the sign of B − 1. (ii) The maximum number of negative limit cycles of (4) is one and a negative limit cycle exists if and only if B 6= 1 and x = u− (t, 0, BD/(B − 1)) ≤ 0, where u− is the solution of x0 = a(t) − b(t)x, with the given initial condition, and B and D are given in (6). Moreover, it is hyperbolic and its stability is given by the sign of 1 − B. Proof. (i) Linear differential equations can be solved analytically and its associated Poincaré map can obtained explicitly in terms of the values introduced in (6). The Poincaré map for x0 = a(t) + b(t)x is P (ρ) = Bρ + C. Hence the necessary and sufficient conditions for x = u+ (t, 0, ρ), solution of this linear differential equation, to be a positive limit cycle for (4) are: B 6= 1, Bρ + C = ρ and u(t, 0, C/(1 − B)) ≥ 0 for all t ∈ R. When u+ (t, 0, ρ) P 0 (ρ) is strictly positive the = B and the hyperbolicity and stability of the limit cycle follows because B 6= 1. When x = u+ (t, 0, ρ) ≥ 0 and the solution tangentially touches x = 0 by the results of [3] we know that the Poincaré map for (4) is of class C 1 . As a consequence, its derivative at the initial condition corresponding to this periodic orbit coincides with the derivative from the right that is again B, and the same results follows. (ii) The proof of this item is similar and we skip the details. In this case the Poincaré map is P (ρ) = ρ/B + D.  Remark 2.2. Notice that the simple case a(t) ≡ 0 can be trivially integrated and has always the solution x = 0 as a periodic orbit and it is a (hyperbolic) limit cycles if and only if B = 1. Similarly, the case b(t) ≡ 0 has either no periodic solutions when A 6= 1, or otherwise, it has a continuum of periodic solutions. Next proposition adapts the ideas of the theory of rotated vector fields ([7, 25]) to our type of non-autonomous differential equations. Proposition 2.3. Consider the 1-parameter family of C 1 differential equations of the form dx = S(x, t) + k, (7) dt with k, x, t ∈ R, that are 1-periodic in the variable t. Let γ be a periodic orbit of (7) when k = K. Then the following statements hold: (i) Let Γ be a periodic orbit of of (7) when k 6= K, then Γ ∩ γ = ∅. (ii) If γ is an attractive limit cycle then, when k & K (resp. k . K), it moves up (resp. down) to another attractive limit cycle. (iii) If γ is a repulsive limit cycle then, when k & K (resp. k . K), it moves down (resp. up) to another repulsive limit cycle. EXISTENCE OF AT MOST TWO LIMIT CYCLES 5 (iv) If γ is semistable then, when k ≈ K, k 6= K, it disappears when k increases and breaks into an attractive and a repulsive limit cycle when k decreases, or viceversa. Proof. This differential equation can be seen as the autonomous differential equation dx dt = S(x, t) + k, = 1, ds ds on the cylinder. Then the angle of its associated vector field Xk (x, t) = (S(x, t) + k, 1) varies monotonically with k and all the properties of the families of rotated vector fields on the plane ([7, 25]) can be adapted to this setting. In particular the properties given in the statement hold.  We also will need the following result proved in [3]. Proposition 2.4. ([3]) Consider the scalar piecewise differential equation ( dx f (x, t), x ≥ 0, = dt g(x, t), x ≤ 0, where f, g : R2 → R are T -periodic with respect to t, f, g are C 1 -functions, f (t, 0) = g(t, 0) = c(t). Suppose that c has finitely many zeros in [0, T ]. Then its Poincaré map is a C 1 -function and Z T  0 ∂h P (ρ) = exp (u(t, 0, ρ), t) dt , 0 ∂x where   ∂f ∂h  (x, t), x ≥ 0, (x, t) = ∂x ∂x  ∂g (x, t), x ≤ 0.  ∂x We will use the following corollary: Corollary 2.5. Let P (x) be the Poincaré map associated to (4). If a(t) has finitely many zeros, then it is of class C 1 and Z 1  0 P (ρ) = exp χ(ρ, t) dt , (8) 0 where ( b(t), if u(t, 0, ρ) ≥ 0, χ(ρ, t) = (9) −b(t), if u(t, 0, ρ) ≤ 0. We end this section by a simple, but useful result on non-existence of periodic orbits. Proposition 2.6. Consider the differential equation (4) and assume that b(t) 6≡ 0 and R1 does not change sign. Then if a(t) 6≡ 0 and sign(b) 0 a(t) dt ≥ 0 it has not periodic orbits. Proof. We prove the result by contradiction. Assume R1 that the differential equation has a periodic orbit x = u(t). Then it holds that 0 u0 (t) dt = u(1) − u(0) = 0. Moreover, since a(t) 6≡ 0, we know that u(t) 6≡ 0. Hence Z 1 Z 1 Z 1 0 0= u (t) dt = a(t) dt + b(t)|u(t)| dt, 0 0 0 6 A. GASULL AND Y. ZHAO R1 R1 R1 or, equivalently, 0 6= 0 b(t)|u(t)| dt = − 0 a(t) dt. In consequence, sign(b) 0 a(t) dt < 6 A. GASULL AND Y. ZHAO 0 in contradiction with our hypothesis.  R1 R1 R1 or, equivalently, 0 6= 0 b(t)|u(t)| dt = − 0 a(t) dt. In consequence, sign(b) 0 a(t) dt < 0 in contradiction with our hypothesis.  3. Proof of Theorem 1.1 Proof. (a) First we observe 3. Proof of Theorem that there 1.1 are no crossing periodic orbits. This is so, simply because the line L = {x = 0} Proof. (a) First we observe that there are no crossing periodic is without contact by the flow This orbits. of (4)isbecause so, dx/dt| simply because = a(t) that does not change sign. Hence all x=0 the line L = {x = 0} is without contact by the flow of (4) periodic orbits are either positive because dx/dt|and negative and by Proposition 2.1 the upper bound for the number of limit cycles in x=0 = a(t) that does not change sign. Hence all periodic orbits are either positive each of the and negative andregions {x ≥ 0} or2.1 by Proposition {xthe ≤ 0} is onebound upper as we for wanted to prove.ofTheir the number limithyperbolicity cycles in is also proved in that proposition. Moreover, the positive (resp. negative) limit cycle each of the regions {x ≥ 0} or {x ≤ 0} is one as we wanted to prove. Their hyperbolicity exists if B 6= 1 and C/(1 − B) > 0 (resp. BD/(B − 1) < 0), where B, C and D are is also proved in that proposition. Moreover, the positive (resp. negative) limit cycle given in (6). Since sign(a) = sign(C) = sign(D) and B > 0, it follows that either both exists if B 6= 1 and C/(1 − B) > 0 (resp. BD/(B − 1) < 0), where B, C and D are limit cycles exist simultaneously when (1 − B) sign(a) > 0 and otherwise, none of them given in (6). Since sign(a) = sign(C) = sign(D) and B > 0, Rit1follows that either both exists. exist limit cycles The result follows because simultaneously when (1 − B)−sign(a) sign(1 B) = −>sign( b(t) dt). none of them 0 and0 otherwise, R1 exists. The(b) If b(t) ≤ result 0, then follows by thesign(1 because change− yB)==−x, equation − sign( (4) can transform to dy/dt = 0 b(t) dt). −a(t)+ b̄(t)|y|, where b̄(t) = −b(t) ≥ (b) If b(t) ≤ 0, then by the change y = −x, equation (4) canloss 0. Therefore, without of generality, transform we assume to dy/dt = b(t) ≥ 0 in the rest of this proof. −a(t)+ b̄(t)|y|, where b̄(t) = −b(t) ≥ 0. Therefore, without loss of generality, we assume b(t) ≥ 0 Let u(t,rest in the t0 , xof 0 ) this be the proof.solution of (4) with the initial condition u(t0 , t0 , x0 ) = x0 . Let u(t, t0 , x0 ) be the solution ofin (4) Assume that a(t) has n zeros [0, 1]. withInspired by the the initial proof of u(t condition Proposition 2.4 of [3], 0 , t0 , x0 ) = x0 . denote by τ1 < τ2 < · · · < τn the zeros of a(t) in [0, 1]. For each τi , i = 1, ...n, set Assume that a(t) has n zeros in [0, 1]. Inspired by the proof of Proposition 2.4 given xi = u(0, τi , 0), xn+1 = 0, xn+2 = u(0, 1, 0). One can reorder the initial conditions xi so in [3, Prop. 3&4], denote by τ1 < τ2 < · · · < τn the zeros of a(t) in [0, 1]. For each that τi , i = 1, ...n, set xi = u(0, τi , 0), xn+1 = 0, xn+2 = u(0, 1, 0). One can reorder the initial conditions xi so that−∞ = x0 < x1 < ... < xr < xr+1 = +∞, 1 ≤ r ≤ n + 2. −∞ = accordingly We also rename, x0 < x1 < ... < values the xr < xτr+1 = +∞, 1 ≤ r ≤ n + 2. j associated to these new xj . As an illustration seerename, We also Figure 1. accordingly the values τj associated to these new xj . As an illustration see Figure 1. a(t) < 0 a(t) > 0 a(t) < 0 J4 −→ P (x) = u(1, 0, x) x4 J3 −→ x 0 = x3 J2 −→ x2 J1 −→x1 J0 −→ 0 t1 (x) τ4 t2 (x) τ1 1 Figure 1. For a given differential equation (4), example of the values xi and Figure τi for a1.givenFor asolution given differential u(t, 0, x), equation with n =(4), example 2 and of Notice r = 4. the values xi and that for x inτeach i . Also, for aintervals of the particularJi solution u(t, 0, the number ofx), withofn u(t, zeroes = 2 0, and x) ris= 4, we show constant. the valuesfor In particular, t1 (x) x ∈ and t2 (x). J3 , this Noticeis that number k3 =for2. x in each of the intervals Ji the number of zeroes of u(t, 0, x) is constant. In particular, By definition for xof∈xJ3 ,andthisthe number is k3 = of uniqueness 2. solution of equation (4), we know that i u(t, 0, xi ) = u(t, τi , 0). Hence u(t, 0, xi ) has a zero at t = τi . It follows from (4) that ∂ u(τi , 0, xi ) = a(τi ) + b(τi )|u(τi , 0, xi )| = a(τi ) = 0. ∂t EXISTENCE OF AT MOST TWO LIMIT CYCLES 7 By definition of xi and the uniqueness of solution of equation (4), we know that u(t, 0, xi ) = u(t, τi , 0). Hence u(t, 0, xi ) has a zero at t = τi . It follows from (4) that ∂ u(τi , 0, xi ) = a(τi ) + b(τi )|u(τi , 0, xi )| = a(τi ) = 0. ∂t Let us study the functions P (x) and H(x) on each of the intervals Ji := (xi , xi+1 ), i = 0, 1, . . . , r. Because of the above construction, the number of zeros of each solution t → u(t, 0, x) for each x ∈ Ji is 0 ≤ ki ≤ n + 1 and only depends on i, see again [3]. Fix one of these intervals, say Ji with ki ≥ 1. When ki ≥ 1, for shortness set ` = ki and denote all these zeroes by t1 (x), t2 (x), · · · , t` (x), and tj (x) < tj+1 (x), j = 1, 2, · · · , ` − 1. In fact all them are simple zeros and all the functions tm (x) are of class C 1 for x ∈ Ji . This is so, because each one of them is defined implicitly by the equation u(tm (x), 0, x) = 0 and since ∂u(tm (x), 0, x)/∂t = a(tm (x)) 6= 0, because x ∈ Ji , it holds that ∂ ∂ u(tm (x), 0, x)t0m (x) + u(tm (x), 0, x) = 0, ∂t ∂x which implies ∂ ∂ u(tm (x), 0, x) u(tm (x), 0, x) t0m (x) = − ∂x ∂ = − ∂x < 0, ∂t u(tm (x), 0, x) a(tm (x)) where notice that we have used that ∂u(tm (x), 0, x)/∂x > 0, fact proved also in [3] and consequence that x → u(t, 0, x) is a flow and preserves orientation. In short, for all m, sign(t0m (x)) = − sign(a(tm (x))). By using Corollary 2.5, let us compute the derivative of the Poincaré map at x ∈ Ji . First notice that since ∂u(t, 0, x)/∂t|u=0 = a(t) and that since tm (x) and tm+1 (x) are two consecutive simple zeros of u(t, 0, x), the monotonicity of u(t, 0, x) at t = tm (x) and t = tm+1 (x) are different and ut (tm (x), 0, x)ut (tm+1 , 0, x) = a(tm (x))a(tm+1 (x)) < 0, which implies t0m (x)t0m+1 (x) < 0. Set σm = − sign(a(tm (x))). Then σm = (−1)m−1 σ1 . Before R approaching  to the general situation, for the sake of clarity, let us compute d 1 dx 0 χ(s) ds for the particular value x ∈ J3 and the particular situation given in Figure 1, where σ1 = +1, ` = 2, t01 (x) > 0 and t02 (x) < 0. Notice that, by Corollary 2.5, Z  Z Z Z ! 1 t1 (x) t2 (x) 1 d d χ(s) ds = χ(s) ds + χ(s) ds + χ(s) ds dx 0 dx 0 t1 (x) t2 (x) Z Z Z ! t1 (x) t2 (x) 1 d = b(s) ds − b(s) ds + b(s) ds dx 0 t1 (x) t2 (x) = 2b(t1 (x))t01 (x) − 2b(t2 (x))t02 (x) > 0, because both terms, t01 (x) and −t02 (x) > 0 are positive. In general, it holds that 8 A. GASULL AND Y. ZHAO   Z 1  d P 00 (x) d 0  d = ln(P (x)) = χ(s) ds dx P 0 (x) dx dx 0  Z t1 (x) `−1 X Z tm+1 (x) Z 1  d = σ1 b(s) ds + σm+1 b(s) ds + σ`+1 b(s) ds dx 0 tm (x) t` (x) m=1  `−1 X  = σ1 b(t1 (x))t01 (x) + (−1)m b(tm+1 (x))t0m+1 (x) − b(tm (x))t0m (x) m=1  + (−1) `−1 b(t` (x))t0` (x) ! X̀ = 2σ1 (−1)m−1 b(tm (x))t0m (x) > 0. m=1 Notice that the above quantity is positive because all terms have the same sign and σ1 b(t1 (x))t01 (x) > 0. This is so, because b(t) ≥ 0, for each m, t0m (x)t0m+1 (x) < 0, and also sign(t0m (x)) = − sign(a(tm (x))) = σm . In particular σ1 t01 (x) > 0. The above equality implies Z 1  Z 1  00 00 d H (x) = P (x) = exp χ(s)ds χ(s)ds > 0, x ∈ Ji , 0 dx 0 provided that u(t, 0, x) has at least one zero in [0, 1]. Notice that in particular this implies that H 0 is increasing on Ji . On the other hand, by (8) and (9), we have that P 0 > 0 and H 0 are constant if x ∈ Ji , where Ji is any of the intervals such that when x ∈ Ji then u(t, 0, x) has not zeros in [0, 1]. Even more, by the proof of Proposition 2.1, H 0 (x) ≡ (1 − B)/B, when u(t, 0, x) < 0 and H 0 (x) ≡ B − 1, when u(t, 0, x) > 0, where B is given in (6), and since we are assuming that b(t) ≥ 0, B > 1. Furthermore, clearly H 00 (x) = P 00 (x) = 0 for x in these intervals Ji . We recall that H(x) and P (x) are of class C 1 , see Proposition 2.4. Their second derivatives can be discontinuous at the points xi , i = 0, 1, . . . , r, but they are well defined on the intervals Ji . As a simple example, see the map P for the differential equation dx/dt = −1 + |x|, in Section 4.1. In short, from all the above properties we get that in general H 0 (x) is negative and constant in J0 , positive and constant in Jr , strictly increasing in R \ (J1 ∪ Jr ), and continuous. As a consequence H 0 (x) has at most one zero, and by Rolle’s theorem H(x) has at most two zeros, as we wanted to prove. This fact is equivalent to say that (4) has at most two limit cycles. Because of our proof we also know that two is the maximum number of limit cycles, taking into account their multiplicities. R1 That the limit cycle can exist only when sign(b) 0 a(t) dt < 0 is a consequence of Proposition 2.6. Let us prove that the total number of limit cycles is either 0 or 2. Consider the straight lines x = ±R, R > 0. Since b(t) ≥ 0 (and not identically zero), for R big enough both lines are without contact by the flow the differential equation. In particular, this implies that the number of zeros of H must be even, as this result follows. To end the proof we only need to show that examples with two limit cycles do exist and study all their possible configurations. Recall that in Proposition 2.1 it is proved that there are never two positive (or two negative) limit cycles. Hence we know that EXISTENCE OF AT MOST TWO LIMIT CYCLES 9 the only possible configurations are [1, 0, 1], [1, 1, 0], [0, 1, 1] and [0, 2, 0]. In next section we prove that all them exist. Hence, the theorem follows.  4. Examples of all possible configurations In this section we collect several examples to illustrate the different configurations of limit cycles of the differential equation considered in Theorem 1.1. 4.1. Examples with configuration [1, 0, 1]. The differential equation dx = −1 + |x| (10) dt has two limit cycles x = 1 and x = −1 and both a(t) ≡ −1 and b(t) ≡ 1 do not change their signs. In fact (10) is a very simple example for which is not difficult to get the Poincaré map explicitly. It is    ρ+1   − 1, ρ ≤ 0,  e P (ρ) = 1  − 1, 0 ≤ ρ ≤ 1 − 1/e,   e(1 − ρ)  e(ρ − 1) + 1, ρ ≥ 1 − 1/e. From the above expressions it is easy to see that ρ = ±1 are the only two fixed point of P, that correspond to the two limit cycles x = ±1. Moreover, as it is proved in [3] for the general differential equation (4), P 0 is continuous, P 0 (−1) = 1/e < 1 and P 0 (1) = e > 1. Hence, both limit cycles are hyperbolic and with different stabilities. Finally, if H = P − Id is the displacement map, H 00 = 0 outside (−1, 1) and H 00 > 0 on the interval (−1, 1), as it is proved in Section 3 when b does not change sign. Notice that P 00 and H 00 are discontinuous. In fact, imbedding the differential equation (10) into the 1-parameter family of dif- ferential equations, dx = −1 + k + |x|, dt we have a nice illustration of the results proved in Proposition 2.3. For k < 1 it has two hyperbolic limit cycles, x = ±(1 − k), that collapse into the double (semistable) limit cycle x = 0 when k = 1. For the rest of values of k the equation has not periodic orbits. Although the limit cycles of (10) are explicit and simple, it is not difficult to construct examples under each of the hypotheses of Theorem 1.1, with two limit cycles (one positive and one negative), which in general have much more involved expressions. For instance we can consider dx = −1 + εf (t) + (1 + εg(t))|x|, dt where f and g are arbitrary 1-periodic functions of C 1 and ε is small enough. The two limit cycles exist because for ε = 0 they exist and are both hyperbolic and, as a consequence, for ε small enough and near each of one them, a hyperbolic limit cycle remains. Moreover, for suitable small ε it is clear that both −1 + εf (t) and 1 + εg(t) do not change sign. 10 A. GASULL AND Y. ZHAO 4.2. Examples with configuration [0, 2, 0]. It is clear that if we impose that a 1- periodic differentiable function x = u(t) is solution of a differential equation (4) we get that u0 (t) = a(t) + b(t)|u(t)|. Hence, for any 1-periodic function b it holds that x = u(t) is a 1-periodic solution of dx = u0 (t) − b(t)|u(t)| + b(t)|x|. dt When u(t) > 0 or u(t) < 0 the above differential equation is of class C 1 and so, is of the type of equations considered in this paper. Here we are more interested to construct C 1 differential equations (4) for which x = u(t) is a crossing limit cycle. To force that b(t)|u(t)| is of class C 1 , and that we are under the hypotheses of Theorem 1.1, a natural way is to assume that all zeros of u are simple, and moreover to choose b = u2 . Then we have the C 1 differential equation dx = u0 (t) − u2 (t)|u(t)| + u2 (t)|x|. dt To simplify the computations, take u such that u(0) = 1, positive for t ∈ [0, τ1 ) ∪ (τ2 , 1], negative for t ∈ (τ1 , τ2 ) and with simple zeros at t = τ1 and t = τ2 . Then, ρ = 1 is the initial condition of the periodic orbit, and by Corollary 2.5, the derivative of the Poincaré map is Z τ1 Z τ2 Z 1  0 2 2 2 P (1) = exp u (t) dt − u (t) dt + u (t) dt . 0 τ1 τ2 Therefore it is obvious that we can choose u(t) such that P 0 (1) takes any positive value, obtaining in particular cases where x = u(t) is a hyperbolic and stable, or unstable, crossing limit cycle. Because we have proved in Theorem 1.1 that when there are limit cycles its exact number is two, we know that in these situations the differential equation has another limit cycle. We do not know if it is of crossing type or not. For the case where P 0 (1) = 1 (for instance this happens taking u(t) = cos(2πt), and then τ1 = 1/4, τ2 = 3/4) we know from the proof of Theorem 1.1 that P 00 (1) > 0 and hence x = u(t) is a double (semistable) crossing limit cycle. Therefore, if we consider the 1-parameter family of differential equations, dx = k − 2π sin(2πt) − cos2 (2πt)| cos(2πt)| + cos2 (2πt)|x|, (11) dt it holds that: • It has only a limit cycle, that is double and of crossing type when k = 0. • If k < 0 and |k| small enough, it has exactly two limit cycles, that are hyperbolic, of crossing type and with different stabilities. In fact, in the second item, the existence of both limit cycles is a consequence of Propo- sition 2.3 and all the other facts, consequence of Theorem 1.1. Notice that differential equations (11) prove that the two possibilities with all limit cycles of crossing type are realizable for (4) under the hypotheses of item (b) of Theorem 1.1. 4.3. Examples with configurations [0, 1, 1], [1, 1, 0] and [1, 0, 1]. It is clear that if a differential equation (4) has configuration [0, 1, 1], by changing x by −x we have another one of the same form but with configuration [1, 1, 0]. Hence we focus our attention to find an example with configuration [0, 1, 1]. EXISTENCE OF AT MOST TWO LIMIT CYCLES 11 To construct our example, we start imposing that u(t) ≥ 0 is a periodic orbit of the differential equation (4) with b = 1. Then it writes as dx = u0 (t) − u(t) + |x|. (12) dt We fix u(t) = sin4 (2πt) + p, with p ≥ 0. Then, the solutions of dx = u0 (t) − u(t) + x. dt are x(t, 0, x0 ) = u(t) + (x0 − p) exp(t) = sin4 (2πt) + p + (x0 − p) exp(t), x0 ∈ R, and clearly x = u(t) = x(t, 0, p) is a positive limit cycle. Hence its limit cycle configu- ration is [k, `, 1] with k + ` = 1, because from Theorem 1.1 we know that when b does not vanish and the differential equation (4) has some limit cycle, it has exactly 2 and, moreover, that there are never two positive or two negative limit cycles. We claim that when p ∈ [0, P ), with P ≈ 0.24823, where P is the positive root of a polynomial of degree 4 given below, dx = u0 (t) − u(t) − x, (13) dt has not negative periodic orbits. Assuming this claim, for these values of p, the differential equation (12) has the positive hyperbolic limit cycle u(t), no negative limit cycle, and a crossing limit cycle, that is it has the limit cycle configuration [0, 1, 1], as we wanted to show. This is so, because if (12) would have a negative periodic orbit it also would be a negative periodic orbit of (13). Let us prove the claim. We only need to find the limit cycle of (13) and prove that it is not negative. After some tedious computations we get it and it is x = vp (t), where 1 2  vp (t) = 8π sin(4πt) − (16π − 1) cos(4πt) 2(16π 2 + 1) 1 2  3 = (64π − 1) cos(8πt) − 16π sin(8πt) − − p. 8(64π 2 + 1) 8 To control the sign of vp (t) we will use the usual trick to reduce the problem to a polynomial one. This can be done by introducing a rational parameterization of this function, see for instance [13] for more examples of use of rational parameterizations. Let s be such that 2s 1 − s2 sin(4πt) = , cos(4πt) = . 1 + s2 1 + s2 Since sin(8πt) = 2 sin(4πt) cos(4πt) and cos(8πt) = 2 cos2 (4πt) − 1, it holds that −Vp (s) vp (t(s)) = , (1 + s2 )2 (16π 2 + 1)(64π 2 + 1) where   Vp (s) = 1024π 4 p − 256π 4 + 80π 2 p + 48π 2 + p + 1 s4 − 16π 40π 2 + 1 s3   + 2 16π 2 + 1 64π 2 p + 48π 2 + p s2 − 384π 3 s + 1024π 4 p + 768π 4 + 80π 2 p + p. It is easy to see that the polynomial of degree 4, V0 (s) has two simple real roots. Therefore v0 (t) changes sign as we wanted to see. In fact, this also can be seen directly 12 A. GASULL AND Y. ZHAO because v0 (0)v0 (1/4) < 0. We need the rational parameterization to know until which value of p the same property is satisfied. To prove that the same holds for all p ∈ [0, P ) we use the method detailed in the appendix of [11, App. II]. In a few words, these two real roots remain until a value of p, say p = P, for which they collide into a double zero. This value is one of the values such that Vp has a multiple zero of Vp . By the properties of the resultant (see for instance [27]) this value of p has to be a zero of R(q) = Res(Vp (s), Vp0 (s), s) = 256(16π 2 + 1)2 (64π 2 + 1)4 R1 (q)R4 (q), where Res denotes the resultant of both polynomials, see again [27] for more details of how to compute it. Here,   R1 (q) = 16π 2 + 1 64π 2 + 1 p − 256π 4 + 48π 2 + 1, 2 4 R4 (q) = 64π 2 + 1 16π 2 + 1 p4   2 + 64π 2 + 1 32768π 6 + 2304π 4 + 96π 2 + 1 16π 2 + 1 p3  2 + 18432π 6 64π 2 + 1 16π 2 + 1 p2 − 28311552π 12 . The polynomial R1 has only a simple and explicit real root p1 = (256π 4 − 48π 2 − 1)/((16π 2 + 1)(64π 2 + 1)) ≈ 0.24331. The polynomial R4 has two real roots, a negative one, p2 ≈ −0.74410, and another one positive, say p3 ≈ 0.24823. Clearly, we are only interested on values of p > 0, because otherwise u(t) would change sign. To see that the value p = p1 is not interesting for our purposes it suffices to see that for a given fixed p ∈ (0, p1 ) and another one in pb ∈ (p1 , p3 ) it holds that vp (t) and vpb(t) do change sign in [0, 1]. Hence, the value that gives the desired property is p = p3 = P, which is the value given above. Because R4 has degree 4, the value P can be obtained in closed algebraic form. Finally, notice that the limit cycle configuration for differential equation (12) with u(t) = sin4 (2πt) + p, is [0, 1, 1] when p ∈ [0, P ) and [1, 0, 1] when p ≥ P. 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Dong, Qualitative theory of differential equations, Translations of Mathematical Monographs, Vol 101, American Mathematical Society, Providence, RI, 1992. Dept. de Matemàtiques, Universitat Autònoma de Barcelona, Edifici C 08193 Cerda- nyola del Vallès (Barcelona), Spain; and Centre de Recerca Matemàtica, Edifici Cc, Campus de Bellaterra, 08193 Cerdanyola del Vallès (Barcelona), Spain E-mail address: gasull@mat.uab.cat School of Mathematics (Zhuhai), Sun Yat-sen University, Zhuhai Campus, Zhuhai, 519082, China E-mail address: mcszyl@mail.sysu.edu.cn