On periodic solutions of superquadratic Hamiltonian systems

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1 Electronic Journal of Differential Equations, Vol. 22(22), No. 8, pp ISSN: URL: or ftp ejde.math.swt.edu (login: ftp) On periodic solutions of superquadratic Hamiltonian systems Guihua Fei Abstract We study the existence of periodic solutions for some Hamiltonian systems ż = JH z(t, z) under new superquadratic conditions which cover the case H(t, z) = z 2 (ln(1 + z p )) q with p, q > 1. By using the linking theorem, we obtain some new results. 1 Introduction We consider the superquadratic Hamiltonian system ż = JH z (t, z) (1.1) ( ) where H C 1 ([, 1] R 2N IN, R) is a 1-periodic function in t, J = is I N the standard 2N 2N symplectic matrix, and H(t, z) z 2 + as z + uniformly in t. (1.2) We assume H satisfies the following conditions. (H1) H(t, z), for all (t, z) [, 1] R 2N. (H2) H(t, z) = o( z 2 ) as z uniformly in t. In [12], Rabinowitz established the existence of periodic solutions for (1.1) under the following superquadratic condition: there exist µ > and r 1 > such that for all z r 1 and t [, 1] < µh(t, z) z H z (t, z). (1.3) Since then, the condition (1.3) has been used extensively in the literature; see [1-14] and the references therein. Mathematics Subject Classifications: 58E5, 58F5, 34C25. Key words: periodic solution, Hamiltonian system, linking theorem. c 22 Southwest Texas State University. Submitted September 2, 21. Published Janaury 15, 22. 1

2 2 Periodic solutions of superquadratic systems EJDE 22/8 It is easy to see that (1.3) does not include some superquadratic nonlinearity like H(t, z) = z 2 (ln(1 + z p )) q, p, q > 1. (1.4) In this paper, we shall study the periodic solutions of (1.1) under some superquadratic conditons which cover the cases like (1.4). We assume H satisfies the following condition. (H3) There exist constants β > 1, 1 < λ < 1 + β 1 β, c 1, c 2 > and L > such that z H z (t, z) 2H(t, z) c 1 z β, z L, t [, 1]; H z (t, z) c 2 z λ, z L, t [, 1]. Theorem 1.1 Suppose H C 1 ([, 1] R 2N, R) is 1-periodic in t and satisfies (1.2), (H1) (H3). Then (1.1) possesses a nonconstant 1-periodic solution. A straightforward computation shows that if H satisfies (1.4), for any T >, the system (1.1) has a nonconstant T-periodic solution with minimal period T. One can see Remark 2.2 and Corollary 2.3 for more examples. For the second order Hamiltonian system ü(t) + V (t, u(t)) =, u() u(1) = u() u(1) = (1.5) we have a similar result. Theorem 1.2 Suppose V C 1 ([, 1] R N, R) is 1-periodic in t and satisfies (V1) V (t, x), for all (t, x) [, 1] R N (V2) V (t, x) = o( x 2 ) as x uniformly in t (V3) V (t, x)/ x 2 + as x + uniformly in t (V4) There exist constants 1 < λ β, d 1, d 2 > and L > such that x V (t, x) 2V (t, x) d 1 x β, x L, t [, 1]; V (t, x) d 2 x λ, x L, t [, 1]. (1.6) ( or V (t, x) d 2 x λ+1, x L, t [, 1]). (1.7) Then (1.5) possesses a nonconstant 1-periodic solution. We shall use the linking theorem [13, Theorem 5.29] to prove our results. The idea comes from [11, 12, 13]. Theorem 1.1 is proved in Section 2 while the proof of Theorem 1.2 is carried out in Section 3.

3 EJDE 22/8 Guihua Fei 3 2 First order Hamiltonian system Let S 1 = R/(2πZ) and E = W 1/2,2 (S 1, R 2N ). Then E is a Hilbert space with norm and inner product,. We define Ax, y = ( Jẋ, y) dt, x, y E; (2.1) f(z) = 1 2 Az, z H(t, z) dt, z E. (2.2) Then A is a bounded selfadjoint operator and ker A = R 2N. (H1) (H3) imply that H(t, z) a 1 + a 2 z λ+1, z R 2N. This implies that f C 1 (E, R) and looking for the solutions of (1.1) is equivalent to looking for the critical points of f [12, 13]. Let E = ker(a), E + = positive definite subspace of A, and E = negative definite subspace of A. Then E = E E E +. Lemma 2.1 Under the conditions of Theorem 1.1, f satisfies the (PS) condition. Proof. Let {z m } be a (PS)-sequence, i.e., f(z m ) M; f (z m ) as m. We want to show that {z m } is bounded. Then by a standard argument, {z m } has a convergent subsequence [13]. Suppose {z m } is not bounded, then passing to a subsequence if necessary, + as m +. By (H3), there exists C 3 > such that for all z R 2N, t [, 1] Therefore, we have 2f(z m ) f (z m ), z m = This implies z H z (t, z) 2H(t, z) C 1 z β C 3. [z m H z (t, z m ) 2H(t, z m )]dt [C 1 z m β C 3 ]dt = C 1 z m β dt C 3. z m β dt as m. (2.3) z m Note that from (H3), 1 < λ < 1 + β 1 β 1 β. Let α = β(λ 1). Then α > 1, αλ 1 = α 1 β. (2.4)

4 4 Periodic solutions of superquadratic systems EJDE 22/8 By (H3), there exists C 4 > such that H z (t, z) α C α 2 z λα + C 4, (t, z) [, 1] R 2N. (2.5) Denote z m = z + m + z m + z m E + E E. We have f (z m ), z + m = Az + m, z + m Az + m, z + m Az + m, z + m ( [H z (t, z m ) z + m]dt H z (t, z m ) z + m dt H z (t, z m ) α ) 1 α Cα z + m, (2.6) where C α > is a constant independent of m. By (2.5), H z (t, z m ) α dt C 5 ( C 6 ( (C α 2 z m λα + C 4 )dt z m β dt) 1/β ( z m β ) 1/β (αλ 1) + C 4. Combining this inequality with (2.3) and (2.4) yields that ( H z(t, z m ) α dt) 1 α as m. By (2.6) we have z m (αλ 1) β β 1 dt) 1 1 β + C4 [ C 6( z m β dt) 1/β z m (αλ 1) + C 4 1/β α 1 β α ] 1 α Az m, + z m + z m + f (z m ) z m + z m + + ( H z(t, z m ) α dt) 1 α as m. This implies Cα z + m z + m z m + as m. (2.7) Similary, we have zm as m. (2.8) By (H3) there exist C 7, C 8 > such that z H z (t, z m ) 2H(t, z) C 7 z C 8, (t, z) [, 1] R 2N.

5 EJDE 22/8 Guihua Fei 5 This implies 2f(z m ) f (z m ), z m = Therefore, by (2.7) and (2.8) [z m H z (t, z m ) 2H(t, z m )]dt [C 7 z m C 7 z + m C 7 z m C 8 ]dt C 9 z m C 1 ( z + m + z m + 1). zm as m. Combine this with (2.7) and (2.8), we get 1 = z m z+ m + zm + zm as m, a contradiction. Therefore, {z m } must be bounded. [C 7 z m C 8 ]dt Proof of Theorem 1.1 We prove that f satisfies the conditions of Theorem 5.29 in [13]. Step 1: By (H1) (H3), we have H(t, z) a 1 + a 2 z λ+1, (t, z) [, 1] R 2N. By (H2), for any ε >, there exists δ > such that H(t, z) ε z 2, t [, 1], z δ. Therefore, there exists M = M(ε) > such that H(t, z) ε z 2 + M z λ+1, (t, z) [, 1] R 2N. Note that λ + 1 > 2. By the same arguements as in [13, Lemma 6.16], there exist ρ > and ã >, such that for z E 1 = E + f(z) ã if z = ρ, i.e., f satisfies (I 7 )(i) in [13, Theorem 5.29] with S = B ρ E 1. Step 2: Let e E + with e = 1 and Ẽ = E E span{e}. We denote K = { z Ẽ : z = 1}, λ = inf z E, z =1 Az, z, For z K, we write z = z + z + z + Ẽ. i) If z > γ z + + z, by (H1) we have, for any r >, f(rz) = 1 2 < Arz, rz Arz+, rz λ r 2 z A r2 z + 2. γ = ( A λ )1/2. H(t, z)dt

6 6 Periodic solutions of superquadratic systems EJDE 22/8 ii) If z γ z + + z, we have i.e., 1 = z 2 = z 2 + z + + z 2 (1 + γ 2 ) z + + z 2, Denote K = {z K : z γ z + + z }. Claim: There exists ε 1 > such that, u K, For otherwise, k >, u k K such that z + + z 2 1 >. (2.9) 1 + γ2 meas{t [, 1] : u(t) ε 1 } ε 1. (2.1) meas{t [, 1] : u k (t) 1 k } < 1 k. (2.11) Write u k = u k + u k + u+ k Ẽ. Notice that dim(e span{e}) < + and u k + u+ k 1. In the sense of subsequence, we have Then (2.9) implies that u k + u + k u + u + E span{e} as k +. u + u γ 2 >. (2.12) + 1 Note that u k 1, in the sense of subsequence u k u E as k +. Thus in the sense of subsequences, u k u = u + u + u + as k +. This means that u k u in L 2, i.e., u k u 2 dt as k +. (2.13) By (2.12) we know that u >. Therefore, u 2 dt >. Then there exist δ 1 >, δ 2 > such that Otherwise, for all n >, we must have meas{t [, 1] : u (t) δ 1 } δ 2. (2.14) meas{t [, 1] : u (t) 1 n } =, i.e., meas{t [, 1] : u (t) < 1 n } = 1; < u 2 dt < 1 1 as n +. n2

7 EJDE 22/8 Guihua Fei 7 We get a contradiction. Thus (2.14) holds. Let Ω = {t [, 1] : u (t) δ 1 }, Ω k = {t [, 1] : u k (t) < 1/k}, and Ω k = [, 1]\Ω k. By (2.11), we have meas(ω k Ω ) = meas(ω Ω Ω k ) meas(ω ) meas(ω Ω k ) δ 2 1 k. (2.15) Let k be large enough such that δ 2 1 k 1 2 δ 2 and δ 1 1 k 1 2 δ 1. Then we have This implies that u k (t) u (t) 2 (δ 1 1 k )2 ( 1 2 δ 1) 2, t Ω k Ω. u k u 2 dt u u u 2 dt ( 1 Ω k Ω 2 δ 1) 2 meas(ω k Ω ) ( 1 2 δ 1) 2 (δ 2 1 k ) (1 2 δ 1) 2 ( 1 2 δ 2) >. This is a contradiction to (2.13). Therefore the claim is true and (2.1) holds. For z = z + z + z + K, let Ω z = {t [, 1] : z(t) ε 1 }. By (1.2), for M = A >, there exists L ε 3 1 > such that 1 Choose r 1 L 1 /ε 1. For r r 1, H(t, x) M x 2, x L 1, uniformly in t. H(t, rz(t)) M rz(t) 2 Mr 2 ε 2 1, t Ω z. By (H1), for r r 1 f(rz) = 1 2 r2 Az +, z r2 Az, z H(t, rz)dt 1 2 A r2 H(t, rz)dt 1 Ω z 2 A r2 Mr 2 ε 2 1 meas(ω z ) 1 2 A r2 Mε 3 1r 2 = 1 2 A r2 <. Therefore, we have proved that f(rz), for any z K and r r 1. (2.16) Let E 2 = E E, Q = {re : r 2r 1 } {z E 2 : z 2r 1 }. By (H1) and (2.16) we have f Q, i.e., f satisfies (I 7 )(ii) in [13, Theorem 5.29]. Step 3: By Lemma 2.1, f satisfies the (PS) condition. Similar to the proof of [13, Theorem 6.1], by the linking theorem [13, Theorem 5.29], there exists a critical point z E of f such that f(z ) ã >. Moreover, z is a classical solution of (1.1) and z is nonconstant by (H1).

8 8 Periodic solutions of superquadratic systems EJDE 22/8 Remark 2.2 i) Suppose H(t, z) = 1 2 B(t)z, z + H(t, z) with B(t) being a 2N 2N matrix, continuous and 1-periodic in t and H(t, z) satisfies (1.2) and (H1)-(H3). We have the same conclusion as Theorem 1.1. The proof is similar and we omit it. ii) Suppose H(t, z) = H(z) is independent on t, i.e., (1.1) is an autonomous Hamiltonian system. Then under similar conditions as (1.2) and (H1)-(H3), for any T >, the system (1.1) has a nonconstant T -periodic solution. Moreover, if H(z) C 2 (R 2N, R) and satisfies some strictly convex conditions such as H (x) is positive defininte for x, then for any T >, (1.1) has a nonconstant T -periodic solution with minimal period T. We omit the proof which is similar to the one in [4, 5]. iii) Suppose (1.4) holds, i.e., H(t, z) = H(z) = z 2 (ln(1 + z p )) q, (t, z) [, 1] R 2N, where p > 1 and q > 1. Obviously, (1.2), (H1) and (H2) hold. Note that z H z (z) 2H(z) = z 2 q(ln(1 + z q )) q 1 p z p pq(ln 2)q 1 z 2, z z p 2 H z (z) 2(ln(1 + z p )) q z + p z p 1 + z p q(ln(1 + z p )) q 1 z 2 z 5 4, z L, for L being large enough. This implies (H3). By directly computation, H (z) is positive definite for z. Therefore, for any T >, (1.1) possesses a T -periodic solution with minimal period T. iv) There are many examples which satisfy (H1)-(H3) and (1.2) but do not satisfy (1.3). For example H(t, z) = z 2 ln(1 + z 2 ) ln(1 + 2 z 3 ). Corollary 2.3 Suppose H(t, z) = z 2 h(t, z) with h C 1 ([, 1] R 2N, R) being 1-periodic in t and satisfies (H1 ) h(t, z), for all (t, z) [, 1] R 2N. (H2 ) h(t, z) as z ; h(t, z) + as z +. (H3 ) There exist δ < 1, L >, ε > and M > such that z δ h z (t, z) z ε, z h z (t, z) Mh, z L; h(t, z) z γ as z for any γ >. Then system (1.1) possesses a nonconstant 1-periodic solution. Proof Obviously, (H1 ) (H3 ) imply (1.2), (H1) and (H2). z H z (t, z) 2H(t, z) = z 2 h z (t, z) z ε z 2 δ, z L; H z (t, z) 2h(t, z) z + z 2 h z (t, z) (2 + M) z h(t, z) (2 + M) z 1+γ, z L.

9 EJDE 22/8 Guihua Fei 9 Let β = 2 δ and λ = 1 + γ with < γ < (1 δ)/(2 δ). Then (H3) holds. By Theorem 1.1 we get the conclusion. 3 Second order Hamiltonian System Let E = W 1,2 (S 1, R N ) with the norm and inner product,. E C(S 1, R N ) and u 2 = ( u 2 + u 2 )dt. Define Kx, y = x ydt, x, y E; f(z) = 1 2 (id K)z, z V (t, z)dt, z E. Then Then K is compact, ker(id K) = R N, and the negative definite subspace of id K, M (id K) = {}, i.e., E = E E + where E = ker(id K) and E + is the positive definite subspace of id K. Note that (V1) (V4) imply V (t, x) d 2 x λ+1 + d 3. (3.1) This implies that f C 1 (E, R) and critical points of f are 1-periodic solutions of (1.5) [11]. Lemma 3.1 Suppose (V1) (V4) hold. Then f satisfies the (PS) condition. Proof Let {z m } be a (PS) sequence. Suppose {z m } is not bounded. Passing to a subsequence if necessary, + as m. Then by (V4) 2f(z m ) f (z m ), z m = This implies If (1.6) holds, we have [z m V (t, z m ) 2V (t, z m )]dt d 1 z m β dt d 4. z m β dt f (z m ), z + m = (id K)z + m, z + m Since λ β, we have as m +. V (t, z m ) z + mdt (id K)z m, + z m + z m + V (t, z m ) dt (id K)z + m, z + m d 5 z + m ( z m λ dt + d 6 ). z m + as m +. (3.2)

10 1 Periodic solutions of superquadratic systems EJDE 22/8 If (1.7) holds, we have f(z m ) = 1 2 (id K)z+ m, z + m V (t, z m )dt 1 2 (id K)z+ m, z m + d 5 z m 1+λ dt d 7 (id K)z + m, z + m d 8 z m λ dt d 7. Since λ β, we obtain (3.2). On the other hand, (V1) (V4) imply This implies x V (t, x) 2V (t, x) d 9 x d 1, t S 1 R N. 2f(z m ) f (z m ), z m = [z m V (t, z m ) 2V (t, z m )]dt d 9 z m dt d 1 d 9 zm dt d 9 z m dt + d 1 d 9 z m d 11 z + m d 1. zm as m +. (3.3) By (3.2) and (3.3), we get a contradiction. Therefore {z m } is bounded. By a standard argument, {z m } has a convergent subsequence [11]. Proof of Theorem 1.2 As in Step 1 of the proof of Theorem 1.1, by (V2) and (3.1), there exist ã >, ρ > such that f(z) ã, z E + with z = ρ. Choose e E + with e = 1. Let Ẽ = span{e} E and K = {u Ẽ : u = 1}. Note that dim Ẽ < +. By using similar arguments as in the proof of (2.1), there exists ε 1 > such that meas{t [, 1] : u(t) ε 1 } ε 1, u K. (3.4) By (V1), (V3) and similar arguments as in the proof of Theorem 1.1, there exists r 1 > such that f Q, where Q = {re : r 2r 1 } {z E : z 2r 1 }. Now by Lemma 3.1, [13, Theorem 5.29], and (V1), f has a nonconstant critical point z such that f(z ) ã >. z is 1-periodic solution of (1.5).

11 EJDE 22/8 Guihua Fei 11 Remark 3.2 (i) Suppose V (t, x) = V (x) is independent on t and V (x) satisfies (V1) (V4). Then for any T >, (1.5) possesses a nonconstant T -periodic solution. (ii) There are many examples which satisfy (V1) (V4) but do not satisfy a condition similar to (1.3). For example, V (t, x) = [1 + (sin 2πt) 2 ] x 2 ln(1 + 2 x 2 ); V (t, x) = x 2 ln(1 + x 2 ) ln(1 + 2 x 4 ). or By using similar arguments as in the proof of Theorem 1.2, we can prove the following corollary. Details are omited. Corollary 3.3 Suppose V (t, x) = x 2 h(t, x) with h C 1 (S 1 R N, R) satisfies (V1 ) h(t, x), (t, x) S 1 R N. (V2 ) h(t, x) as x ; h(t, x) + as x +. (V3 ) There exist L >, λ >, C 1, C 2 > such that for t S 1 C 1 x (h (t, x) x) h(t, x), h(t, x) C 2 x λ, x L. Then (1.5) possesses a nonconstant 1-periodic solution. References [1] K. C. Chang, Infinite dimensional Morse theory and multiple solution problems, Progress in nonlinear differential equations and their applications, V.6 (1993). [2] I. Ekeland, Convexity Method in Hamiltonian Mechanics, Springer-Verlag. Berlin. (199). [3] I. Ekeland & H. Hofer, Periodic solutions with prescribed period for convex autonomous Hamiltonian systems, Invent. Math. 81 (1985), [4] G. Fei & Q. Qiu, Minimal periodic solutions of nonlinear Hamiltonian systems, Nonlinear Analysis T. M. A. 27 (1996), [5] G. Fei, S. Kim & T. Wang, Minimal period estimates of periodic solutions for superquadratic Hamiltonian systems, J. Math. Anal. Appl. 238 (1999), [6] G. Fei, S. Kim & T. Wang, Solutions of minimal period for even classical Hamiltonian systems, Nonlinear Analysis T. M. A. 43 (21), [7] M. Giradi & M. Matzeu, Periodic solutions of convex autonomous Hamiltonian systems with a quadratic growth at the origin and superquadratic at infinity, Ann. Mat. pura ed appl. 147 (1987),

12 12 Periodic solutions of superquadratic systems EJDE 22/8 [8] Y. Long, The minimal period problem of periodic solutions for autonomous superquadratic second order Hamiltonian systems, J. Diff. Eq. 111 (1994), [9] Y. Long, The minimal period problem of classical Hamiltonian systems with even potentials, Ann. IHP. Anal. non Lineaire vol.1, no.6 (1993), [1] Y. Long, Multiple solutions of perturbed superquadratic second order Hamiltonian systems, Trans. Amer. Math. Soc. 311 (1989), [11] J. Mawhin & M. Willem, Critical Point Theory and Hamiltonian Systems, Appl. Math. Sci. Springer-Verlag, 74(1989). [12] P. H. Rabinowitz, Periodic solutions of Hamiltonian systems, Comm. Pure Appl. Math. 31 (1978), [13] P. H. Rabinowitz, Minimax methods in critical point theory with applications to differential equations, CBMS Regional Conf. Ser. in Math. no.65 A.M.S. (1986). [14] M. Struwe, Variational methods. Applications to nonlinear partial differential equations and Hamiltonian systems, Springer-Verlag, Berlin, 199. Guihua Fei Department of Mathematics and statistics University of Minnesota Duluth, MN 55812, USA gfei@d.umn.edu