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Electronic Journal of Differential Equations, Vol. 2004(2004), No. 137, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp) EXISTENCE OF MULTIPLE SOLUTIONS FOR A CLASS OF SECOND-ORDER ORDINARY DIFFERENTIAL EQUATIONS XIAO-BAO SHU, YUAN-TONG XU Abstract. By means of variational structure and Z2 group index theory, we obtain multiple solutions for the second-order differential equation d du (p(t) ) + q(t)u + f (t, u) = 0, 0 < t < 1, dt dt subject to one of the following two sets of boundary conditions: u0 (0) = u(1) + u0 (1) = 0 or u(0) = u(1) = 0 . 1. Introduction Erbe and Mathsen [6] study the boundary-value problem −(ru0 )0 + qu = λf (t, u), 0 < t < 1, 0 αu(0) − βu (0) = 0 = γu(1) + δu0 (1), where λ > 0 is a parameter, α, β, γ, δ ≥ 0 and αδ + αγ + βγ > 0, f ∈ C((0, 1) × R, R), r ∈ C([0, 1], (0, ∞)) and q ∈ C([0, 1], [0, ∞)). In this paper we are interested in the study of second-order ordinary differential equation d du (p(t) ) + q(t)u + f (t, u) = 0, 0 < t < 1, (1.1) dt dt subject to one of the following two sets of boundary conditions u0 (0) = 0 = γu(1) + u0 (1) (1.2) u(0) = u(1) = 0 (1.3) or By means of variational structure and Z2 group index theory, we obtain multiple solutions of boundary-value problems for (1.1) and lower bound estimate of number for the solutions. 2000 Mathematics Subject Classification. 34B15, 34B05, 65K10, 34B24. Key words and phrases. Variational structure; Z2 group index theory; critical points; boundary value problems. c 2004 Texas State University - San Marcos. Submitted October 5, 2004. Published November 25, 2004. Supported by grant 10471155 from NNSF of China, grant 031608 from the Foundation of the Guangdong province Natural Science Committee, and grant 20020558092 from Foundation for PhD Specialities of Educational Department of China. 1 2 X. B. SHU & Y. T. XU EJDE-2004/137 A critical point of f is a point x0 where f 0 (x0 ) = θ and a critical value is a number c such that f (x0 ) = c for some critical point x0 . Next, we recall the definition of the Palais-Smale condition. Definition Let E be real Banach space and f ∈ C 1 (E, R). We say that f satisfies the Palais-Smale condition if every sequence {xn } ⊂ E such that {f (xn )} is bounded and f 0 (xn ) → θ as n → ∞ has a converging subsequence. Let K = {x ∈ E : f 0 (x) = θ}, Kc = {x ∈ E : f 0 (x) = θ, f (x) = c} and fc = {x ∈ E : f (x) ≤ c}. The class of subsets ofP X \ {θ} ⊂ E closed and symmetric with respect to the origin will be denoted by . Next, we recall the concept of genus. Definition Let E be a real Banach space, and Σ = {A : A ⊂ E \ {θ} is a closed, symmetric set}. Define γ : Σ → Z + ∪ {+∞} as follows n min{n ∈ Z : there exists an odd continuous map ϕ : A → R \ {θ}}; γ(A) = 0 If A = ∅; +∞ If there is no odd continuous map ϕ : A → Rn \ {θ} for n ∈ Z. P Then we say γ is the genus of . Denote i1 (f ) = lima→−0 γ(fa ) and i2 (f ) = lima→−∞ γ(fa ). P We know that if A ∈ and if there exists an odd homeomorphism of n−sphere onto A then γ(A) = n +P1; If X is a Hilbert space, and E is an n−dimensional subspace of X, and A ∈ is such that A ∩ E ⊥ = ∅ then γ(A) ≤ n. The following Lemmas play an important role in proving our main results. Lemma 1.1 ([5]). Let f ∈ C 1 (E, R) be an even functional which satisfies the Palais-Smale condition and f (θ) = 0. Then (P1) If there exists an m-dimensional subspace X of E and ρ > 0 such that sup f (x) < 0, x∈X∩Sρ then we have i1 (f ) ≥ m e of E such that (P2) If there exists a j-dimensional subspace X inf f (x) > −∞, e⊥ x∈X we have i2 (f ) ≤ j If m ≥ j, (P1) and (P2) hold, then f has at least 2(m − j) distinct critical points. Lemma 1.2 ([9]). Let f ∈ C 1 (X, R) be an even functional which satisfies the Palais-Smale condition and f (θ) = 0. If (F1) There exists ρ > 0, α > 0 and a finite dimensional subspace E of X, such that f |E ⊥ ∩Sρ ≥ α e of X, there is a r = r(E) e > 0, such (F2) For all finite dimensional subspace E e r that f (x) ≤ 0 for x ∈ E\B Then f possesses an unbounded sequence of critical values. 2. Main Results In this paper, we use Lemma 1.1 and 1.2 to study the boundary-value problems (1.1)-(1.2) and (1.1)-(1.3) Theorem 2.1. Let f , p(t) and q(t) satisfy the following conditions: EJDE-2004/137 (i) (ii) (iii) (iv) (v) (vi) MULTIPLE SOLUTIONS FOR ODE’S 3 p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1] f ∈ C([0, 1] × R, R) limu→0 f (t,u) = ξ(t) > 0 uniformly for t ∈ [0, 1], λ = min0≤t≤1 ξ(t) u There exists α > 0 such that f (t, α) ≤ 0 f (t, u) is odd in u − λ2 < q(t) + p(t) ≤ 0, for all 0 ≤ t ≤ 1. Then (1.1)-(1.2), has at least 2n nontrivial solutions in C 2 [0, 1] whenever 2n2 (M + p(1)|γ|)(1 + π 2 ) < λ ≤ 2(n + 1)2 (M + p(1)|γ|)(1 + π 2 ) m and γ > − 2p(1) Proof. Set h : [0, 1] × R → R, if u > α, f (t, α) h(t, u) = f (t, u) if |u| ≤ α, f (t, −α) if u < −α Let us consider the functional defined on H01 (0, 1) by Z 1 1 1 p(1) 2 [ p(t)|u0 (t)|2 − q(t)|u(t)|2 − G(t, u(t))]dt + γu (1), (2.1) I(u) = 2 2 2 0 Ru Where G(t, u) = 0 h(t, v)dv. The norm k · k and inner product (·, ·) can be defined respectively by Z 1 Z 1 kuk = ( (|u0 (t)|2 + |u(t)|2 )dt)1/2 ; (u, v) = (u0 (t)v 0 (t) + u(t)v(t))dt . 0 0 Thus H01 (0, 1) = W01,2 (0, 1) will be a Hilbert space. Let E = H01 (0, 1), since h(t, u) is an odd continuous map in u, we know that I ∈ C 1 (E, R) is even in u and I(θ) = 0. First, we will show that the critical points of the I(u) are the solutions of (1.1)(1.2) in C 2 [0, 1]. Since Z 1 I(u + sv) = I(u) + s{ [p(t)u0 (t)v 0 (t) − q(t)u(t)v(t) − h(t, u + θ(t)sv)v(t)]dt 0 Z 1 s2 { (p(t)|v 0 (t)|2 2 0 − q(t)|v(t)|2 )dt + p(1)γv 2 (1)} ∀u, v ∈ E, 0 < θ(t) < 1 + p(1)γu(1)v(1)} + (2.2) We have, for all u, v ∈ E, Z 1 (I 0 (u), v) = [p(t)u0 (t)v 0 (t) − q(t)u(t)v(t) − h(t, u(t))v]dt + p(1)γu(1)v(1) . (2.3) 0 0 By I (u) = 0, one gets Z 1 [p(t)u0 (t)v 0 (t) − q(t)u(t)v(t) − h(t, u(t))v]dt + p(1)γu(1)v(1) = 0 0 (2.4) 4 X. B. SHU & Y. T. XU EJDE-2004/137 for all v ∈ E. On the other hand, Z 1 Z 1 du d 0 0 p(t)u (t)v (t)dt + (p(t) )vdt dt 0 0 dt Z 1 Z 1 Z = p(t)u0 (t)v 0 (t)dt + p(t)u00 (t)v(t)dt + 0 0 1 p0 (t)u0 (t)v(t)dt (2.5) 0 = p(1)v(1)u0 (1) − p(0)u0 (0)v(0) = 0 So, it is easy to see that Z 1 du d v[ (p(t) ) + q(t)u(t) + h(t, u(t))]dt dt dt 0 0 = p(1)v(1)(u (1) + γu(1)) − p(0)u0 (0)v(0) = 0 Hence we obtain d du (p(t) ) + q(t)u(t) + h(t, u(t)) = 0 dt dt Thus the critical points of I(u) are the solutions of (1.1)-(1.2) in C 2 [0, 1]. For convenience, we transform (2.1) into Z 1 1 I(u) = [ p(t)(|u0 (t)|2 + |u(t)|2 ) 0 2 1 p(1) 2 − (q(t) + p(t))|u(t)|2 − G(t, u(t))]dt + γu (1) . 2 2 By condition (iv) of Theorem 2.1, we have uh(u) ≤ 0 when |u(t)| ≥ α. So Z 1 Z 1Z u Z 1Z α G(t, u) = h(t, v)dvdt ≤ |h(t, v)|dv dt . 0 0 0 0 −α R1Rα Denote by c the value of 0 −α |h(t, v)|dvdt. On the other hand, q(t) + p(t) ≤ 0, R1 then − 0 (q(t) + p(t))|u(t)|2 dt ≥ 0. So, we have p(1) 2 m kuk2 − c + γu (1) ∀u ∈ E. 2 2 Next, we show that I(u) has lower bound. We divide our proof into two parts (I) When γ ≥ 0, it is easy to see m I(u) ≥ kuk2 − c ∀u ∈ E (2.6) 2 m (II) When − 2p(1) < γ < 0, we divide again our proof into two parts in order to show I(u) has lower bound: (a) If there exist t0 ∈ [0, 1] such that u(t0 ) = 0, then Z 1 Z 1 √ |u(1)| = | u0 (s)ds| ≤ |u0 (s)|ds ≤ 2kuk . I(u) ≥ t0 0 So, we get 1 (m + 2γp(1))kuk2 − c ∀u ∈ E, (2.7) 2 (b) If does not exist t1 ∈ [0, 1] such that u(t1 ) = 0, then u(t) > 0 or u(t) < 0, for all t ∈ [0, 1]. We might as well let u(t) > 0 for all t ∈ [0, 1]. When max0≤t≤1 u(t) ≤ 1,we have u(1) ≤ 1 and I(u) ≥ I(u) ≥ m 1 kuk2 − c + γp(1), 2 2 EJDE-2004/137 MULTIPLE SOLUTIONS FOR ODE’S 5 i.e., I(u) has lower bound. When max0≤t≤1 |u(t)| > 1, since u ∈ C 2 [0, 1], there exist t2 ∈ [0, 1] such that u(t2 ) = min0≤t≤1 u(t). So Z 1 Z 1 0 u(1) − u(t2 ) = | u (s)ds| ≤ |u0 (s)|ds t2 0 i.e. Z u(1) ≤ u(t2 ) + 1 Z |u (s)|ds ≤ ( 0 Z u (t)dt) + ( 2 1 2 1 u2 (t)dt + 0 Z 1 1 |u0 (t)|2 ) 2 0 0 0 √ Z ≤ 2( 1 1 |u0 (t)|2 )1/2 = √ 2kuk 0 As in the proof of (I), we have 1 I(u) ≥ (m + 2γp(1))kuk2 − c. ∀u ∈ E. (2.8) 2 m < γ < 0. From By (a) and (b), it is easy to see I(u) has lower bound when − 2p(1) 1 (I) and (II), we get that I(u) has lower bound for all u ∈ H0 (0, 1), i.e., i2 (I) = 0. Next, we verify that I(u) satisfies the Palais-Smale condition. Suppose that {un } ⊂ E with and c1 ≤ I(un ) ≤ c2 , 0 I (un ) → 0 as n → ∞. (2.9) (2.10) Then Z sup{ 1 [p(t)u0n v 0 − q(t)un v − h(t, un (t))v]dt + γp(1)un (1)v(1)} → 0, 0 as n → ∞, (2.11) ∀u, v ∈ E, kvk = 1 with kzn k = kI 0 (xn )k. Let us denote εn = kzn k, then εn → 0 as n → ∞. Replace v by un in above equality. By (2.3) and (2.11), we have Z 1 Z 1 [p(t)|u0n (t)|2 − q(t)|un (t)|2 ]dt = h(t, un )un (t)dt + hzn , un i. 0 0 The above equality is equivalent to Z 1 p(t)[|u0n (t)|2 + |un (t)|2 ]dt 0 Z = 1 [(q(t) + p(t))|un (t)|2 + h(t, un )un (t)]dt + hzn , un i 0 So, there exist ξ ∈ [0, 1] such that Z 1 2 p(ξ)kun k = [(q(t) + p(t))|un (t)|2 + h(t, un )un (t)]dt + hzn , un i . (2.12) 0 Next, we show that {un } satisfying condition (2.9) and (2.10) is bounded. We divide again our proof into two parts. (c) When γ ≥ 0, by (2.6), one gets 2 2 kun k2 ≤ (I(un ) + c) ≤ (c2 + c), m m q 2 (c2 + c). i.e., kun k ≤ m 6 X. B. SHU & Y. T. XU EJDE-2004/137 m (d) When − 2p(1) < γ < 0, by the above proof and (2.7) and (2.8), we have kun k2 ≤ 2 2 (I(un ) + c) ≤ (c2 + c) m + 2γp(1) 1 + 2γp(1) or kuk2 ≤ 2 1 [c2 + c − γp(1)] m 2 i.e., s kun k ≤ 2 (c2 + c) m + 2γp(1) r or kuk ≤ 2 1 (c2 + c − γp(1)). m 2 By (c) and (d), it is easy to see {un } is bounded in the space H01 (0, 1). Reflexivity of H01 (0, 1) implies that there exists a subsequence of {un } which is weak convergent in H01 (0, 1). We still denote it by {un } and suppose that un * u0 in H01 (0, 1) as n → ∞. On the one hand, by boundedness of {un } and (2.12), we have Z 1 p(ξ)kun k2 − [(q(t) + p(t))|un (t)|2 + h(t, un )un (t)]dt → 0 as n → ∞ 0 Note that the weak convergent of {un } in H01 (0, 1) implies the uniform convergence of {un } in C([0, 1], R) [8, Proposition 1.2 ]. Hence Z 1 p(ξ)kun k2 → [(q(t) + p(t))|u0 (t)|2 + h(t, u0 )u0 (t)]dt as n → ∞ 0 This means that {un } converges in H01 (0, 1). So the P.S. condition holds. Thirdly, we show that Theorem 2.1 holds by Lemma 1.1. Denote βk (t) = √ 2 kπ cos kπt, k = 1, 2, 3, . . . , n, . . . , then Z 1 Z 1 1 2 |βk (t)| dt = 2 2 , |βk0 (t)|2 dt = 1 k π 0 0 Definite n-dimensional linear space En = span{β1 (t), β2 (t), . . . βn (t)}. It is obvious that En is a symmetric set. Suppose ρ > 0, then E n ∩ Sρ = n X bk βk : k=0 n X k=0 b2k (1 + 1 ) = ρ2 k2 π2 1 Let g(t, u) = ξ(t) h(t, u) − u, by condition (iii) of Theorem 2.1, limu→0 uniformly for t ∈ [0, 1]. We choose ε such that g(u) u = 0, 1 2(M + p(1)|γ|)(1 + π 2 ) − . n2 λ By condition (iii) of Theorem 2.1, there exist δ > 0 such that |g(t, u)| ≤ ε|u| whenever |u| ≤ δ. We can choose ρ such that 0 < ρ < min{α, δ}, and have n n √ X X 2 max u(t) ≤ |bk | ≤ kuk = k bk βk k 0≤t≤1 kπ 0<ε< k=0 n X =( k=0 k=0 b2k (1 + 1 ))1/2 = ρ < min{α, δ} k2 π2 EJDE-2004/137 MULTIPLE SOLUTIONS FOR ODE’S 7 when u ∈ En ∩ Sρ . So Z u G(t, u) = ξ(t) [v + g(t, v))]dv Z u 1 = ξ(t)|u(t)|2 + ξ(t) g(t, v)dv 2 0 Z u 1 ≥ ξ(t)|u(t)|2 − ξ(t) εvdv 2 0 1 = ξ(t)(1 − ε)|u(t)|2 2 ≥ λ(1 − ε)|u(t)|2 0 From q(t) + p(t) ≥ − λ2 , we get that n 1 Z (q(t) + p(t))|u(t)|2 dt ≤ − 0 λ X b2k 2 k2 π2 k=0 So 1 Z 1 1 [ p(t)(|u0 (t)|2 + |u(t)|2 ) − (q(t) + p(t))|u(t)|2 ]dt 2 2 I(u) = 0 1 Z − G(t, u)dt + 0 Z p(1) 2 γu (1) 2 1 1 1 [ (|u0 (t)|2 + |u(t)|2 ) − λ(1 − ε)|u(t)|2 ]dt 2 0 2 n 2 X λ bk p(1) + + |γ|kuk2C 4 k2 π2 2 ≤M k=0 n n n X X M X 2 1 1 b2k p(1) ) + |γ|( bk (1 + 2 2 ) − λ(1 − 2ε)( ≤ 2 k π 4 k2 π2 2 k=0 k=0 k=0 n √ 2|bk | 2 ) kπ n X b2 1 1 M + p(1)|γ| X 2 k bk (1 + 2 2 ) − λ(1 − 2ε) ≤ 2 k π 4 k2 π2 < M + p(1)|γ| 2 k=0 n X k=0 k=0 n X 1 1 b2k 2 bk (1 + 2 ) − λ(1 − ε) π 4 n2 π 2 k=0 n X λ M + p(1)|γ| π 2 + 1 1 ≤ ( b2k − 2 2 + ε) 2 2 λ π 2n π k=0 n 2 ≤ X λ (M + p(1)|γ|)(1 + π ) 1 ( − 2 + ε) b2k 2 2π λ 2n 2 ≤ λ 2(M + p(1)|γ|)(1 + π ) 1 ( − 2 + ε) 4π 2 λ n k=0 n X b2k < 0 k=0 By Lemma 1.1 and the above result, we have i1 (I) ≥ n and I has 2n distinct critical points, i.e., boundary-value problem (1.1)-(1.2) has at least 2n nontrivial solutions in C 2 [0, 1]. 8 X. B. SHU & Y. T. XU EJDE-2004/137 Next we consider the boundary-value problem (1.1)-(1.3). Similar to Theorem 2.1, we have the following result. Theorem 2.2. Let f , p(t) and q(t) satisfy the following conditions: (i) p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1] (ii) f ∈ C([0, 1] × R, R) (iii) limu→0 f (t,u) = ξ(t) > 0 uniformly for t ∈ [0, 1], λ = min0≤t≤1 ξ(t) u (iv) There exists α > 0, such that f (t, α) ≤ 0 (v) f (t, u) is odd in u (vi) − λ2 < q(t) + p(t) ≤ 0 for all 0 ≤ t ≤ 1. Then (1.1)-(1.3) has at least 2n nontrivial solutions in C 2 [0, 1] whenever 2n2 M (1 + π 2 ) < λ ≤ 2(n + 1)2 M (1 + π 2 ). Next, we consider the boundary-value problem (1.1)-(1.2). Theorem 2.3. Let f , p(t) and q(t) satisfy the following conditions: (i) p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1] (ii) f ∈ C([0, 1] × R, R) (iii) There exists T such that lim supu→0 f (t,u) ≤T u ≥ 2 and α > 0 such that (iv) There exists θ > 2M m Z u 1 0 < G(t, u) = f (t, v)dv ≤ uf (t, u), ∀|u| ≥ α θ 0 (v) f (t, u) is odd in u (vi) q(t) ∈ C[0, 1], q(t) + p(t) ≤ 0 for all 0 ≤ t ≤ 1. mθ−2M , then (1.1)-(1.2) has infinite nontrivial solutions in C 2 [0, 1]. If γ > − 2(θ−2)p(1) Proof. It is easy to see that for u ∈ H01 (0, 1), the functional Z 1 1 1 p(1) 2 I(u) = [ p(t)|u0 (t)|2 − q(t)|u(t)|2 G(t, u(t))]dt + γu (1) 2 2 2 0 (2.13) is well defined. The solutions of boundary-value problems (1.1)-(1.2) are the critical points of the functional I(u). Note that I(u) is equivalent to Z 1 1 1 p(1) 2 I(u) = [ p(t)(|u0 (t)|2 + |u(t)|2 ) − (q(t) + p(t))|u(t)|2 − λG(t, u)]dt + γu (1) 2 2 2 0 We show that Theorem 2.3 holds by using Lemma 1.2. Since f (t, u) is an odd continuous map in u, we know that I ∈ C 1 (E, R) is even in u and I(θ) = 0. Moreover, As in the proof of Theorem 2.1, one gets Z 1 0 (I (u), v) = [p(t)u0 (t)v 0 (t) − q(t)u(t)v(t)) − f (t, u)v(t)]dt + γp(1)u(1)v(1), 0 for all u, v ∈ E. The above equality is equivalent to Z 1 Z 1 f (t, u)v(t)dt = p(t)(u0 (t)v 0 (t) + u(t)v(t))dt 0 0 Z − 0 1 (p(t) + q(t))u(t)v(t)dt − (I 0 (u), v) + p(1)γu(1)v(1). EJDE-2004/137 MULTIPLE SOLUTIONS FOR ODE’S 9 Next, we verify that I(u) satisfies the Palais-Smale condition. Suppose that un ⊂ E with c1 ≤ I(un ) ≤ c2 , 0 I (un ) → 0 as n → ∞ (2.14) (2.15) Now, we show that {un } of satisfying (2.14) and (2.15) is bounded. Denote E1 = {t ∈ [0, 1]||un (t)| ≥ α}, E2 = [0, 1] \ E1 . We divide our proof into two parts . mθ−2M (A) When − 2(θ−2)p(1) < γ < 0, by (iv), we have 1 p(t) (|un (t)|2 + |u0n (t)|2 )dt − 2 Z 1 Z I(un ) = 0 Z 0 1 1 (q(t) + p(t))|un (t)|2 dt 2 p(1) 2 − G(t, un (t))dt + γun (1) 2 0 Z Z m ≥ kun k2 − G(t, un (t))dt − G(t, un (t))dt 2 E1 E2 Z 1 1 p(1) 2 γun (1) − (q(t) + p(t))|un (t)|2 dt + 2 0 2 Z m 1 ≥ kun k2 − un (t)f (t, un (t))dt − c3 2 E1 θ Z 1 1 p(1) 2 γun (1) − (q(t) + p(t))|un (t)|2 dt + 2 0 2 Z 1 m 1 2 ≥ kun k − un (t)f (t, un (t))dt − c4 2 0 θ Z 1 p(1) 2 1 + γun (1) − (q(t) + p(t))|un (t)|2 dt 2 0 2 Z 1 m 1 = kun k2 − ( p(t)(|u0n (t)|2 + |un (t)|2 )dt − (I 0 (un ), un ) 2 θ 0 Z 1 p(1) 2 1 1 + p(1)γu2n (1)) − c4 + γun (1) − ( − ) (q(t) + p(t))|un (t)|2 dt 2 2 θ 0 m M 1 1 1 ≥( − )kun k2 + (I 0 (un ), un ) − c4 + ( − )p(1)γu2n (1) 2 θ θ 2 θ m M 1 0 1 1 2 ≥( − )kun k + kI (un )kkun k − c4 + ( − )p(1)γu2n (1) 2 θ θ 2 θ Remarks: (1) for the rest of this article, ci > 0. (2) The above equality makes use R1 of −( 12 − θ1 ) 0 (q(t) + p(t))|un (t)|2 dt ≥ 0 To show that {un } is bounded, we divide again our proof into two parts. (I’) When max0≤t≤1 |u(t)| ≤ 1, as in the proof of Theorem 2.1, we have u2 (1) ≤ 1 and m M 1 1 1 − )kun k2 + kI 0 (un )kkun k − c4 + ( − )γ 2 θ θ 2 θ m M 1 0 1 1 2 ( − )kun k ≤ I(un ) − kI (un )kkun k + c4 − ( − )γ 2 θ θ 2 θ I(un ) ≥ ( Using θ > 2M m , (2.14) and (2.15), it is not difficulty to see that {kun k} is bounded. 10 X. B. SHU & Y. T. XU EJDE-2004/137 (II’) When max0≤t≤1 |u(t)| > 1, as in the proof of Theorem 2.1, we have u2 (1) ≤ 2kuk2 and m M 1 1 1 − ) + ( − )2p(1)γ]kun k2 ≤ I(un ) − kI 0 (un )kkun k + c4 ≤ c5 kun k + c5 . 2 θ 2 θ θ 2M mθ−2M Since θ > m ≥ 2 and γ > − 2(θ−2)p(1) , it follows that {kun k} is bounded. From mθ−2M (I’) and (II’), we get that {kun k} is bounded when − 2(θ−2)p(1) < γ < 0. (B) when γ > 0, as in the proof of (A), it is not difficult to see that Z 1 Z 1 p(t) I(un ) = (|un (t)|2 + |u0n (t)|2 dt − G(t, un (t))dt 2 0 0 Z 1 p(1) 2 1 + γun (1) − (q(t) + p(t))|un (t)|2 dt 2 2 0 Z 1 m 1 2 ≥ kun k − un (t)f (t, un (t))dt 2 0 θ Z 1 p(1) 2 1 + γun (1) − (q(t) + p(t))|un (t)|2 dt 2 0 2 m M 1 1 1 ≥( − )kun k2 + kI 0 (un )kkun k − c4 + ( − )p(1)γu2n (1) 2 θ θ 2 θ m M 1 0 2 )kun k + kI (un )kkun k − c4 ≥( − 2 θ θ (We remark that the above equality use ( 12 − θ1 )p(1)γu2 (1) ≥ 0 ) and [( m M 1 − )kun k2 ≤ I(un ) − kI 0 (un )kkun k + c4 ≤ c5 kun k + c5 . 2 θ θ So, we get that {kun k} is bounded when γ > 0. From (A) and (B), we obtain that {un } satisfying (2.14) and (2.15) is bounded. So the P.S. condition holds. Thirdly, we show that Theorem 2.3 holds by using Lemma 1.2. First, we verify condition (F1) of Lemma 1.2. Let βj (t) = cos jt, j = 1, 2, . . . . Consider the ndimensional subspace ( En = span{β1 (t), β2 (t), . . . , βn (t)} ⊥ and let X = V . By (ii), we have δ > 0 such that |f (t, u(t))| ≤ T |u|, whenever |u| ≤ δ. Let ρ with ρ = δ. For any u ∈ Sρ ∩ X, we have kukC ≤ kuk = ρ = δ. From Z 1 Z 1 1 |u(t)|2 dt ≤ 2 |u0 (t)|2 dt n 0 0 it is easy to see Z 0 1 |u(t)|2 dt ≤ ρ2 n2 + 1 So, when γ > 0, we have Z 1 Z 1 p(t) 2 0 2 I(un ) = (|un (t)| + |un (t)| )dt − G(t, un (t))dt 2 0 0 Z 1 p(1) 2 1 + γun (1) − (q(t) + p(t))|un (t)|2 dt 2 0 2 EJDE-2004/137 MULTIPLE SOLUTIONS FOR ODE’S 11 Z 1 m kuk2 − G(t, u(t))dt 2 0 Z 1 Z |u(t)| m 2 ( T vdv)dt ≥ ρ − 2 0 0 T 1 T m ρ2 = (m − 2 )ρ2 > 0 . ≥ ρ2 − 2 2(n2 + 1) 2 n +1 ≥ T T Note that in the above equality, we use n2 > max{ m , m+p(1)γ } and Z 1 1 − (q(t) + p(t))|un (t)|2 dt > 0. 2 0 mθ−2M When − 2(θ−2)p(1) < γ < 0, we get Z 1 Z 1 p(t) (|un (t)|2 + |u0n (t)|2 )dt − G(t, un (t))dt I(un ) = 2 0 0 Z 1 p(1) 2 1 + γun (1) − (q(t) + p(t))|un (t)|2 dt 2 2 0 Z 1 m p(1) 2 ≥ kuk2 − γun (1) G(t, u(t))dt + 2 2 0 Z 1 Z |u(t)| m + p(1)γ 2 ρ − ≥ ( T vdv)dt 2 0 0 m + p(1)γ 2 T ρ2 ≥ ρ − 2 2(n2 + 1) 1 T = (m + p(1)γ − 2 )ρ2 > 0 . 2 n +1 T T Note that the above equality uses n2 > max{ m , m+p(1)γ }. We sum up the conclusions above to obtain that I(u) > 0 for all u ∈ Sρ ∩ X, i.e., condition (F1) of Lemma 1.2 holds. Finally, we verify condition (F2) of Lemma 1.2. By (iv), one gets G(t, u(t)) ≥ c7 |u|θ − c8 . For all finite dimensional subspace E1 of E, there exist c9 such that Z 1 1/θ |u(t)|θ dt ≥ c9 kuk, ∀u ∈ E1 . 0 On the other hand, since p(t) ∈ C 1 [0, 1], q(t) ∈ C[0, 1] and p(t) + q(t) ≤ 0, there exist a positive number Q such that −Q = min0≤t≤1 p(t) + q(t), so Z 1 Z 1 2 − (p(t) + q(t))|u(t)| dt ≤ Q |u(t)|2 dt < Qkuk2 . 0 When u ∈ E1 and 0 mθ−2M − 2(θ−2)p(1) Z I(u) = 1 < γ < 0, from the above result, it is easy to obtain p(t) 0 2 (|u (t)| + |u(t)|2 ) − G(t, u(t))]dt 2 0 Z 1 p(1) 2 1 + γu (1) − (q(t) + p(t))|u(t)|2 dt 2 2 0 [ 12 X. B. SHU & Y. T. XU ≤ M +Q kuk2 − 2 Z EJDE-2004/137 1 G(t, u(t))dt 0 Z 1 M +Q 2 ≤ kuk − c7 |u(t)|θ dt + c8 2 0 M +Q kuk2 − c7 cθ9 kukθ + c8 ≤ 2 M +Q =( − c7 cθ9 kukθ−2 )kuk2 + c8 . 2 When u ∈ E1 and γ ≥ 0, as in the proof of (I’) and (II’), we have the following two results. (1) when max0≤t≤1 |u(t)| ≤ 1, we have Z 1 p(t) 0 2 I(u) = [ (|u (t)| + |u(t)|2 ) − G(t, u(t))]dt 2 0 Z 1 p(1) 2 1 γu (1) − (q(t) + p(t))|u(t)|2 dt + 2 2 0 Z 1 M +Q p(1) 2 ≤ kuk − c7 |u(t)|θ dt + c8 + γ 2 2 0 M +Q p(1) ≤ kuk2 − c7 cθ9 kukθ + c8 + γ 2 2 M +Q p(1) =( − c7 cθ9 kukθ−2 )kuk2 + c8 + γ 2 2 (2) when max0≤t≤1 |u(t)| > 1, we have u2 (1) ≤ 2kuk2 and Z 1 p(t) 0 2 (|u (t)| + |u(t)|2 ) − G(t, u(t))]dt I(u) = [ 2 0 Z 1 p(1) 2 1 + γu (1) − (q(t) + p(t))|u(t)|2 dt 2 0 2 Z 1 M + Q + 2γ ≤ kuk2 − c7 |u(t)|θ dt + c8 2 0 M + Q + 2γ ≤ kuk2 − c7 cθ9 kukθ + c8 2 M + Q + 2γ =( − c7 cθ9 kukθ−2 )kuk2 + c8 2 We sum up the conclusions above to obtain that I(u) ≤ 0, for all u ∈ E1 \BR when R = R(E1 ) is adequately big, i.e., condition (F2) of Lemma 1.2 holds. So I possesses infinite critical point, i.e. the boundary-value problem (1.1)-(1.2) has infinitely many nontrivial solutions in C 2 [0, 1]. Using a technique similar to the one above, we can show that the following theorem. Theorem 2.4. Let f , p(t) and q(t) be the function satisfying the following conditions: (i) p(t) ∈ C[0, 1] and 0 < m ≤ p(t) ≤ M for t ∈ [0, 1] (ii) f ∈ C([0, 1] × R, R) (iii) There exists T such that lim supu→0 f (t,u) ≤T u EJDE-2004/137 MULTIPLE SOLUTIONS FOR ODE’S 2M m ≥ 2 and α > 0 such that Z u 1 0 < G(t, u) = f (t, v)dv ≤ uf (t, u), θ 0 (iv) There exists θ > 13 ∀|u| ≥ α (v) f (t, u) is odd in u (vi) q(t) ∈ C[0, 1], q(t) + p(t) ≤ 0 for 0 ≤ t ≤ 1. Then boundary-value problem (1.1)-(1.3) has infinitely solutions in C 2 [0, 1]. 3. Examples Example 3.1. For 0 < t < 1, consider the boundary-value problem d du ((6 + sin t) ) + (−100 + cos t)u + (1000(1 + t2 ) sin u − 10u3 ) = 0, dt dt u0 (0) = 0, u(1) + u0 (1) = 0 (3.1) Note that f (t, u) = 1000(1 + t2 ) sin u − 10u3 , p(t) = 6 + sin t, q(t) = −100 + cos t So f (t, u) satisfy conditions (ii) and (v) of Theorem 2.1. In addition, 0 < 5 ≤ p(t) = 6 + sin t ≤ 7 ∀t ∈ [0, 1]. then (i) and (vi) of Theorem 2.1 hold. When |u(t)| = 4, we have f (t, 4) = 1000(1 + t2 ) sin 4 − 10 × 43 < 0 , = 1 + t2 uniformly i.e., (iv) of Theorem 2.1 holds. On the other that limu→0 f (t,u) u 2 for t ∈ [0, 1], λ = min0≤t≤1 1 + t = 1000, i.e., (iii) holds, and 2 × 12 × (13 + sin 1)(1 + π 2 ) < λ < 2 × 22 × (13 + sin 1)(1 + π 2 ) By Theorem 2.1 we have (3.1) has at least 2 nontrivial solutions in C 2 [0, 1]. Example 3.2. For 0 < t < 1, consider boundary-value problem du d ((6 + sin t) ) + (−100 + cos t)u + (1000(1 + t2 ) sin u − 10u3 ) = 0, dt dt u(0) = u(1) = 0, (3.2) As in Example 3.1, it is easy to verify all conditions of Theorem 2.2 hold and 2 × 22 × 7(1 + π 2 ) < λ < 2 × 32 × 7(1 + π 2 ) By Theorem 2.2, we have (3.2) has at least 4 nontrivial solutions in C 2 [0, 1]. Example 3.3. Consider the boundary-value problem d du ((6 + sin t) ) + (−9 + t2 )u(t) + t(u3 (t) + u(t)) = 0, dt dt u0 (0) = 0, u(1) + u0 (1) = 0 0 < t < 1, It is easy to see that f (t, u) = t(u3 (t) + u(t)), p(t) = 6 + sin t q(t) = −9 + t2 So, f (t, u) satisfies conditions (ii) and (v) of Theorem 2.3. In addition lim sup u→0 f (t, u) t(u3 + u) = lim sup =1 u u u→0 (3.3) 14 X. B. SHU & Y. T. XU EJDE-2004/137 i.e., conditions (iii) of Theorem 2.3 hold. Moreover, Z u Z u √ 1 1 1 f (t, v)dv = t(v 3 + v)dv = t( u4 + u2 ) ≤ ut(u3 + u) ∀|u(t)| > 2 4 2 3 0 0 2M 14 = 0 < 5 ≤ p(t) ≤ 7, θ = 3 > m 5 So conditions (iv) holds. On the other hand, it is easy to see that p(t) + q(t) = 6 + sin t + (−9 + t2 ) = −3 + t2 + sin t ≤ 0 ∀t ∈ [0, 1] So conditions (i) and (vi) of Theorem 2.3 hold. By Theorem 2.3, we obtain that (3.3) has infinitely many nontrivial solutions in C 2 [0, 1]. Example 3.4. Consider boundary-value problem d du ((6 + sin t) ) + (−9 + t2 )u(t) + t(u3 (t) + u(t)) = 0, dt dt u(0) = u(1) = 0, 0 < t < 1, (3.4) As in Example 3.3, it is easy to verify that all conditions of Theorem 2.4 hold. By Theorem 2.4, we obtain that (3.4) has infinitely many nontrivial solutions in C 2 [0, 1]. References [1] I. Addou and Shin-Hwa Wang. Exact multiplicity results for a p-Laplacian positone problem with concave-convex-concave nonlinearities, Electron J. diff. Eqns., Vol. 2004(2004), No. 72, 1-24. [2] R. P. Agarwal and D. O’Regan; Existence theory for single and multiple solutions to singular positone boundary value problems. J. Differential Equations 175 (2001), no. 2, 393-414. [3] G. Anello; A multiplicity theorem for critical points of functionals on reflexive Banach spaces. Arch. Math. 82 (2004), no. 2, 172-179. [4] L. E. Bobisud, J. E. Calvert, and W. D. 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[10] Xu Yuan Tong and Guo Zhi-Ming; Applications of a Zp index theory to periodic solutions for a class of functional differential equations. J. Math. Anal. Appl. 257 (2001), no. 1, 189-205. [11] Xu Yuan Tong; Subharmonic solutions for convex nonautonomous Hamiltonian systems. Nonlinear Anal. 28 (1997), 1359-1371. Xiao-Bao Shu Department of Mathematics, Sun Yat-Sen University, Guangzhou, 510275, China E-mail address: [email protected] Yuan-Tong Xu Department of Mathematics, Sun Yat-Sen University, Guangzhou, 510275, China E-mail address: [email protected]