Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 171, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS FOR BOUNDARY-VALUE PROBLEMS OF IMPULSIVE DIFFERENTIAL EQUATIONS JING WANG, BAOQIANG YAN Abstract. This article presents global properties and existence of multiple solutions for a class of boundary value problems of impulsive differential equations. We first show that the spectral properties of the linearization of these problems are similar to the well-know properties of the standard SturmLiouville problems. These spectral properties are then used to prove two Rabinowitz-type global bifurcation theorems. Finally, we use the global bifurcation theorems to obtain multiple solutions for the above problems having specified nodal properties. 1. Introduction In this article, we study the global properties for the boundary-value problem −x00 (t) = λf (t, x), t ∈ (0, 1), t 6= 1 , 2 1 ∆x|t= 21 = λβx( ), 2 0 0 1 ∆x |t= 21 = −λβx ( − 0), 2 x(0) = x(1) = 0, (1.1) where λ 6= 0, β 6= 0, ∆x|t= 21 = x( 12 + 0) − x( 12 ), ∆x0 |t= 12 = x0 ( 12 + 0) − x0 ( 12 − 0), and f : [0, 1] × R → R is continuous. The theory of impulsive differential equations has been a significant development in recent years and played a very important role in modern applied mathematical models of real processes rising in phenomena studied in physics, population dynamics, chemical technology, biotechnology and economics (see [4, 6, 7, 11, 26, 27]). There have appeared numerous papers on impulsive differential equations and many of them are committed to study the existence of solutions for boundary value problems of second order impulsive differential equations. The general methods to solve these problems include topological degree theory (see [12, 23, 30]), the 2000 Mathematics Subject Classification. 34B09, 34B15, 34B37. Key words and phrases. Eigenvalues; bifurcation technique; global behavior; multiple solutions. c 2013 Texas State University - San Marcos. Submitted June 4, 2013. Published July 26, 2013. 1 2 J. WANG, B. YAN EJDE-2013/171 method of upper and lower solutions (see [9, 10, 13]) and variational method (see [3, 20, 21, 23, 27, 28, 31]). In natural sciences, there are various concrete problems involving bifurcation phenomena, for example, Taylor vortices [2] and catastrophic shifts in ecosystems [25]. Rabinowitz [21] established the global bifurcation theorem from the trivial solution and in [22] the bifurcation from infinity was studied. In recent years, this theory has been successfully applied to Sturm-Liouville problems for ordinary differential equations, integral equations and partial differential equations (see [15, 16, 17, 20]). In [14], using bifurcation techniques, Liu and O’Regan studied the existence of multiple solutions for the second order impulsive differential equation x00 (t) + ra(t)f (t, x(t)) = 0, ∆x|t=ti = αi x(ti − 0), t ∈ (0, 1), t 6= ti , i = 1, 2, . . . , k, (1.2) x(0) = x(1) = 0, in which they convert (1.2) into Y r a(t)f (t, (1 + αi )y(t)) = 0, y 00 (t) + Q 0<ti <t (1 + αi ) 0<t <t i t ∈ (0, 1), (1.3) y(0) = y(1) = 0. The aim of this conversion is to prove the existence of multiple solutions for above problems by using the properties of the eigenvalues and eigenfunctions of the linear equations corresponding to (1.3). In this article, without using this conversion, in section 2, we study the properties of the eigenvalues and eigenfunctions of the linear equations corresponding to (1.1), in section 3, we investigate the global bifurcation results and the existence of multiple solutions for BVP (1.1) and in section 4, we give an example as the applications. Let F : E −→ E1 where E and E1 are real Banach spaces and F is continuous. Suppose the equation F(U ) = 0 possesses a simple curve of solutions Ψ given by {U (t)|t ∈ [a, b]}. If for some τ ∈ (a, b), F possesses zeros not lying on Ψ in every neighborhood of U (τ ), then U (τ ) is said to be a bifurcation point for F with respect to the curve Ψ (see [21]). A special family of such equations has the form u = G(λ, u) (1.4) where λ ∈ R, u ∈ E, a real Banach space with the norm k·k and G : E ≡ R×E → E is compact and continuous. In addition, G(λ, u) = λLu + H(λ, u), where H(λ, u) is o(kuk) for u near 0 uniformly on bounded λ intervals and L is a compact linear map on E. A solution of (1.4) is a pair (λ, u) ∈ E. The known curve of solutions Θ = {(λ, 0)|λ ∈ R} will henceforth be referred to as the trivial solutions. The closure of the set of nontrivial solutions of (1.4) will be denoted by Σ. A component of Σ is a maximal closed connected subset. If there exists µ ∈ R and 0 6= v ∈ E such that v = µLv, µ is said to be a real characteristic value of L. The set of real characteristic values of L will be denoted j by σ(L). The multiplicity of µ ∈ σ(L) is the dimension of ∪∞ j=1 N ((µL−I) ) where I is the identity map on E and N (P ) denotes the null space of P . Since L is compact, µ is of finite multiplicity. It is well known that if µ ∈ R, a necessary condition for (µ, 0) to be a bifurcation point of (1.4) with respect to Θ is that µ ∈ σ(L). EJDE-2013/171 GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS 3 Lemma 1.1 ([21]). If µ ∈ σ(L) is simple, then Σ contains a component Cµ which can be decomposed into two subcontinua Cµ+ , Cµ− such that for some neighborhood B of (µ, 0), (λ, u) ∈ Cµ+ (Cµ− ) ∩ B, and (λ, u) 6= (µ, 0) implies (λ, u) = (λ, αv + w) where α > 0(α < 0) and |λ − µ| = o(1), kwk = o(|α|), at α = 0. Moreover, each of Cµ+ , Cµ− either (1) meets infinity in Σ , or (2) meets (µ̂, 0) where µ 6= µ̂ ∈ σ(L), or (3) contains a pair of points (λ, u), (λ, −µ), u 6= 0. Lemma 1.2 ([22]). Assume that L is compact and linear, H(λ, u) is continuous on R × E, H(λ, u) = o(kuk) at u = ∞ uniformly on bounded λ intervals, and u kuk2 H(λ, kuk 2 ) is compact. If µ ∈ σ(L) is of odd multiplicity, then Σ possesses an unbounded component Dµ which meets (µ, ∞). Moreover if Λ ⊂ R is an interval such that Λ ∩ σ(L) = µ and M is a neighborhood of (µ, ∞) whose projection on R lies in Λ and whose projection on E is bounded away from 0, then either (1) Dµ \M is bounded in R×E in which case Dµ \M meets Θ = {(λ, 0)|λ ∈ R} or (2) Dµ \ M is unbounded. If (2) occurs and Dµ \ M has a bounded projection on R, then Dµ \ M meets (µ, ∞) where µ 6= µ̂ ∈ σ(L). Lemma 1.3 ([22]). Suppose the assumptions of Lemma 1.2 hold. If µ ∈ σ(L) is simple, then Dµ can be decomposed into two subcontinua Dµ+ , Dµ− and there exists a neighborhood O ⊂ M of (µ, ∞) such that (λ, u) ∈ Dµ+ (Dµ− )∩O, and (λ, u) 6= (µ, ∞) implies (λ, u) = (λ, αv + w) where α > 0(α < 0) and |λ − µ| = o(1), kwk = o(|α|), at |α| = ∞. 2. Preliminaries Let P C[0, 1] = {x : [0, 1] → R : x(t) is continuous at t 6= 21 , and left continuous at t = 12 , and x( 12 + 0) = limt→ 1 + x(t) exists } with the norm 2 kxk = sup |x(t)|. t∈[0,1] 0 0 Let P C [0, 1] = {x ∈ P C[0, 1] : x (t) is continuous at t 6= 12 , and x0 ( 21 − 0) = limt→ 1 − x0 (t), and x0 ( 12 + 0) = limt→ 1 + x0 (t) exist } with the norm 2 2 kxk1 = max{ sup |x(t)|, sup |x0 (t)|}. t∈[0,1] 0 t∈[0,1] Let E = {x ∈ P C [0, 1] : x(0) = x(1) = 0}. It is well known that E is a Banach space with the norm k · k1 . Let Sk+ denote the set of functions in E which have exactly k − 1 simple nodal zeros in (0, 1) and are positive near t = 0. (By a nodal zero we mean the function changes sign at the zeros and at at a simple nodal zero, the derivative of the function is nonzero.) And set Sk− = −Sk+ , Sk = Sk+ ∪ Sk− . They are disjoint in E. Finally, ± let Φ± k = R × Sk and Φk = R × Sk . For the rest of the paper, we always assume the initial-value problem 1 −x00 (t) = λf (t, x), t ∈ (0, 1), t 6= , 2 4 J. WANG, B. YAN EJDE-2013/171 1 ∆x|t= 21 = λβx( ), 2 0 1 0 ∆x |t= 12 = −λβx ( − 0), 2 x(t0 ) = x0 (t0 ) = 0, has the unique trivial solution x ≡ 0 on [0, 1] for any t0 ∈ [0, 1]. Lemma 2.1 ([8]). x(t) ∈ P C[J, R] ∩ C 2 [J 0 , R] is the solution of (1.1) equivalent to x(t) ∈ P C 0 [J, R] is the solution of the integral equation ( R1 λ 0 G(t, s)f (s, x(s))ds − λβt[x( 12 ) − 12 x0 ( 21 − 0)], t ∈ [0, 12 ], R1 x(t) = λ 0 G(t, s)f (s, x(s))ds + λβ(1 − t)[x( 12 ) + 12 x0 ( 12 − 0)], t ∈ ( 12 , 1], where J = [0, 1], J 0 = J \ { 21 }, ( G(t, s) = s(1 − t), t(1 − s), 0 ≤ s ≤ t ≤ 1, 0 ≤ t ≤ s ≤ 1. Lemma 2.2. If a > 0, then the linear boundary-value problem 1 −u00 (t) = λau(t), t ∈ (0, 1), t 6= , 2 1 ∆u|t= 12 = λβu( ), 2 0 0 1 ∆u |t= 21 = −λβu ( − 0), 2 u(0) = u(1) = 0, (2.1) possess an increasing sequence of eigenvalues 0 < λ1 < λ2 < · · · < λk < . . . , lim λk = +∞. k→+∞ And eigenfunction uk corresponding to λk has exactly k − 1 nodal zeros on (0, 1). Proof. Let u(t) be the solution of (2.1). We consider the three following cases: Case (i) λ = 0. Then u(t) can be written as ( c1 t + c2 , t ∈ [0, 21 ], u(t) = c3 t + c4 , t ∈ ( 12 , 1]. From (2.1), we have c2 = 0, 1 1 − c1 − c2 + c3 + c4 = 0, 2 2 −c1 + c3 = 0, c3 + c4 = 0, which implies c1 = c2 = c3 = c4 = 0. Then u(t) ≡ 0. Thus λ = 0 is not the eigenvalue of (2.1). Case (ii) λ < 0. Then u(t) can be written as ( √ √ c1 e −λat + c2 e− −λat , t ∈ [0, 12 ], √ √ u(t) = c3 e −λat + c4 e− −λat , t ∈ ( 12 , 1]. EJDE-2013/171 GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS 5 From (2.1), we have c1 + c2 = 0, √ (1 + λβ)e √ (1 − λβ)e −λa 2 −λa 2 c1 + (1 + λβ)e− c1 − (1 − λβ)e √ e −λa √ −λa 2 c2 − e √ − −λa 2 c3 + √ −λa 2 √ −λa 2 c2 − e √ e− −λa c4 = √ c3 − e− √ c3 + e− −λa 2 −λa 2 c4 = 0, c4 = 0, 0, which implies c1 = c2 = c3 = c4 = 0. Then u(t) ≡ 0. Thus λ < 0 is not the eigenvalue of (2.1). Case (iii) λ > 0. Then u(t) can be written as ( √ √ c1 cos( λat) + c2 sin( λat), t ∈ [0, 12 ], √ √ u(t) = c3 cos( λat) + c4 sin( λat), t ∈ ( 12 , 1]. From (2.1) we know c1 = 0 and √ √ √ λa λa λa cos( )c3 + sin( )c4 − (1 + λβ) sin( )c2 = 0, 2 2 2 √ √ √ λa λa λa − sin( )c3 + cos( )c4 − (1 − λβ) cos( )c2 = 0, 2 2 √ 2 √ cos( λa)c3 + sin( λa)c4 = 0. √ The determinant of the coefficient matrix det = sin λa. Letting det = 0, we have 2 2 λk = k aπ , k = 1, 2, . . . . Then c3 = 0, which implies λk = k2 π 2 a c4 = (1 + β(−1)k+1 k2 π2 )c2 , a is the eigenvalues of (2.1), and 0 < λ1 < λ2 < · · · < λk < . . . , lim λk = +∞. k→+∞ The eigenfunction corresponding to λk is ( sin(kπt), uk (t) = 2 2 (1 + β(−1)k+1 k aπ ) sin(kπt), t ∈ [0, 21 ], t ∈ ( 21 , 1]. Now we prove that uk has exactly k − 1 nodal zeros on (0, 1). Actually, if k is odd, 1 k−1 1 then uk (t) has k−1 2 zeros on (0, 2 ], uk (t) has 2 zeros on ( 2 , 1), and if k is even, k 1 k−2 1 then uk (t) has 2 zeros on (0, 2 ], uk (t) has 2 zeros on ( 2 , 1). Thus uk (t) = 0 has k − 1 zeros on (0, 1). Using the fact ( kπ cos(kπt), t ∈ [0, 21 ), 0 uk (t) = 2 2 (1 + β(−1)k+1 k aπ )kπ cos(kπt), t ∈ ( 21 , 1], we see that uk (t) has exactly k − 1 nodal zeros on (0, 1). The proof is complete. Lemma 2.3. For each k ≥ 1 the algebraic multiplicity of eigenvalue λk is equal to 1. Proof. Define the operator K : P C 0 [0, 1] → P C 0 [0, 1] as follows: (R 1 G(t, s)ax(s)ds − βt[x( 21 ) − 21 x0 ( 12 − 0)], (Kx)(t) = R01 G(t, s)ax(s)ds + β(1 − t)[x( 21 ) + 12 x0 ( 12 − 0)], 0 t ∈ [0, 12 ], t ∈ ( 12 , 1]. 6 J. WANG, B. YAN EJDE-2013/171 We need to prove only that ker(I − λk K)2 ⊂ ker(I − λk K). For any y ∈ ker(I − λk K)2 , we have (I − λk K)2 y = 0, so (I − λk K)y ∈ ker(I − λk K). Let β̄k = 1 + β(−1)k+1 λk . From Lemma 2.2, there exists a γ satisfying p 1 (I − λk K)y = γ sin( λk at), t ∈ [0, ], 2 p 1 (I − λk K)y = γ β̄k sin( λk at), t ∈ ( , 1]; 2 that is, y satisfies p 1 y 00 + λk ay = −γλk a sin( λk at), t ∈ [0, ), 2 p 1 00 y + λk ay = −γλk aβ̄k sin( λk at), t ∈ ( , 1], 2 and y(0) = y(1) = 0. Now we prove γ = 0. Actually, the general solution of the above differential equation is of the form p p p γ y(t) = c1 cos( λk at) + c2 sin( λk at) − sin( λk at) 2 √ p γ λk at 1 + cos( λk at), t ∈ [0, ], 2 2 p p p p γ β̄k γ β̄k sin( λk at) + sin( λk a y(t) = c1 cos( λk at) + c2 sin( λk at) − 4 4 √ √ p p p γ λk aβ̄k t γ λk aβ̄k 1 − λk at) + cos( λk at) − cos( λk at), t ∈ ( , 1]. 2 4 2 From y(0) = y(1) = 0, we have c1 = 0 and √ √ p p p p γ β̄k γ λk aβ̄k γ λk aβ̄k sin( λk a) + cos( λk a) − cos( λk a) = 0. c2 sin( λk a) − 4 2 4 √ √ By Lemma 2.2, λk = k 2 π 2 /a satisfies sin( λk a) = 0, cos( λk a) 6= 0 which implies γ = 0. Thus y ∈ ker(I − λk K). Therefore ker(I − λk K)2 ⊂ ker(I − λk K). The proof is complete. Lemma 2.4. For any positive integer k, Sk , Sk+ and Sk− are open in E. Proof. We prove only that Sk is open in E. Let t1 , t2 , . . . , tk−1 ∈ (0, 1) are k − 1 simple nodal zeros on (0, 1) of u(t) ∈ Sk . Suppose u0 (tj ) = cj > 0. Then there exists δj > 0 which satisfies for any c t ∈ [tj − δj , tj + δj ], u0 (t) is continuous, u0 (t) > 2j , u(tj − δj ) < 0 and u(tj + δj ) > 0. Letting ϕ ∈ E and kϕ − uk1 ≤ σ, where n c u(t + δ ) u(t − δ ) 1 j j j j j σ = min , ,− ,− max u(t), 2 2 2 2 t∈[tj−1 +δj−1 ,tj −δj ] o 1 max u(t) , 2 t∈[tj +δj ,tj+1 −δj+1 ] we have cj > 0, t ∈ [tj − δj , tj + δj ], 2 ϕ(tj + δj ) > u(tj + δj ) − σ > 0, ϕ0 (t) = ϕ0 (t) − u0 (t) + u0 (t) ≥ −σ + ϕ(tj − δj ) < u(tj − δj ) + σ < 0. EJDE-2013/171 GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS 7 Since ϕ(t) is continuous on [tj − δj , tj + δj ], there exists unique t∗j ∈ [tj − δj , tj + δj ] which satisfies ϕ(t∗j ) = 0, ϕ0 (t∗j ) > 0. Since for any t ∈ [tj−1 + δj−1 , tj − δj ], u(t) < 0, and for any t ∈ [tj + δj , tj+1 − δj+1 ], u(t) > 0, one has ϕ(t) < u(t) + σ < 0 for any t ∈ [tj−1 + δj−1 , tj − δj ] and ϕ(t) > u(t) + σ > 0 for any t ∈ [tj + δj , tj+1 − δj+1 ]. Therefore for any t ∈ [tj−1 + δj−1 , tj+1 − δj+1 ], ϕ(t) has the unique simple zero t∗j . If u0 (tj ) = cj < 0, we can get for any t ∈ [tj−1 + δj−1 , tj+1 − δj+1 ], ϕ(t) has the unique simple zero t∗j by the same method as above also. Because u(t) has exactly k − 1 simple zeros on (0, 1), ϕ(t) has exactly k − 1 simple zeros on (0, 1). Therefore ϕ(t) ∈ Sk , which implies Sk are open in E. By the same argument, we can prove Sk+ and Sk− are open in E. The proof is complete. 3. Main results In this section, we assume that (H1) There exist two positive numbers f0 and f∞ such that f0 = lim|x|→0 both uniformly with respect to t ∈ [0, 1]. and f∞ = lim|x|→∞ f (t,x) x f (t,x) x Let ζ, ξ ∈ C([0, 1] × R, R) be such that f (t, x) = f0 x + ζ(t, x), f (t, x) = f∞ x + ξ(t, x). (3.1) ξ(t, x) =0 x (3.2) Clearly, if (H1) holds, we have lim |x|→0 ζ(t, x) = 0, x lim |x|→∞ both uniformly with respect to t ∈ [0, 1]. Now we define three operators L0 , L∞ , and A as follows: (R 1 t ∈ [0, 21 ], G(t, s)f0 x(s)ds − βt[x( 21 ) − 12 x0 ( 12 − 0)], (3.3) (L0 x)(t) = R01 1 0 1 1 G(t, s)f0 x(s)ds + β(1 − t)[x( 2 ) + 2 x ( 2 − 0)], t ∈ ( 12 , 1], 0 (R 1 (L∞ x)(t) = R01 0 G(t, s)f∞ x(s)ds − βt[x( 12 ) − 21 x0 ( 12 − 0)], t ∈ [0, 12 ], 1 1 0 1 G(t, s)f∞ x(s)ds + β(1 − t)[x( 2 ) + 2 x ( 2 − 0)], t ∈ ( 21 , 1], (3.4) and (R 1 G(t, s)f (s, x(s))ds − βt[x( 21 ) − 12 x0 ( 12 − 0)], (Ax)(t) = R01 G(t, s)f (s, x(s))ds + β(1 − t)[x( 12 ) + 12 x0 ( 12 − 0)], 0 t ∈ [0, 12 ], t ∈ ( 12 , 1]. (3.5) Obviously, L0 and L∞ are compact and linear. From Lemma 2.1, we know that x(t) is the solution of (1.1) if and only if x(t) is the solution of x = λAx. (3.6) Let Γ = R × E. A solution of (3.6) is a pair (λ, x) ∈ Γ. The known curve of solutions {(λ, 0)|λ ∈ R} will henceforth be referred to as the trivial solutions. The closure of the set on nontrivial solutions of (3.6) will be denoted by Σ as in Lemma 1.1. Now we are ready to give our main results. 8 J. WANG, B. YAN EJDE-2013/171 Theorem 3.1. Let (H1) hold. In addition assume that for some k ∈ N, either k2 π2 k2 π2 <λ< , f0 f∞ k2 π2 k2 π2 <λ< . f∞ f0 or − + Then problem (1.1) has at least two solutions x+ k and xk , xk has exactly k −1 zeros − in (0,1) and is positive near t = 0, and xk has exactly k − 1 zeros in (0, 1) and is negative near t = 0. Theorem 3.2. Let (H1) hold. Suppose that there exist two integer k > 0 and j ≥ 0 such that either 2 2 2 2 π π < λ < kf∞ ; (i) (k+j) f0 (ii) (k+j)2 π 2 f∞ k2 π 2 f0 <λ< − + Then problem (1.1) has at least 2(j + 1) solutions x+ k+i , xk+i , i = 0, 1, . . . , j, xk+i − has exactly k + i − 1 zeros in (0, 1) and is positive near t = 0, xk+i has exactly k + i − 1 zeros in (0, 1) and are negative near t = 0. To use the terminology of Rabinowitz [20, 21], we first give the following lemmas. Lemma 3.3. Suppose that (H1) is satisfied. Then the operator A given in (3.5) is Fréchet differentiable at x = θ, and A0 (θ) = L0 . Proof. From (3.1), we obtain kAx − L0 xk1 Z n = max sup | t∈[0,1] Z sup | t∈[0,1] = max ≤C Z G(t, s)f (s, x(s))ds − 0 Z Gt (t, s)f (s, x(s))ds − 0 n 1 G(t, s)f0 x(s)ds|, 0 1 1 Gt (t, s)f0 x(s)ds| o 0 Z sup | t∈[0,1] Z 1 1 Z G(t, s)ζ(s, x(s))ds|, sup | t∈[0,1] 0 1 o Gt (t, s)ζ(s, x(s))ds| 0 1 |ζ(s, x(s))|ds, 0 where C depends on bounds for G and Gt . Consequently, from (3.2), R1 |ζ(s, x(s))|ds kAx − L0 xk1 ≤ lim 0 = 0, ∀x ∈ E. lim kxk1 kxk1 kxk1 →0 kxk1 →0 This means that the operator A given in (3.5) is Fréchet differentiable at x = θ, and A0 (θ) = L0 . The proof is complete. Lemma 3.4. Suppose that (H1) is satisfied. Then the operator A given in (3.5) is Fréchet differentiable at x = ∞, and A0 (∞) = L∞ . Proof. For each ε > 0, by (3.2), there exists R > 0 such that |ξ(t, x)| ≤ ε|x|, for |x| > R, t ∈ [0, 1]. Let M = max|x|≤R |ξ(t, x)|. Then we have |ξ(t, x)| ≤ ε|x| + M, ∀x ∈ R, t ∈ [0, 1]. EJDE-2013/171 GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS 9 Therefore, for any x ∈ E, t ∈ [0, 1], kAx − L∞ xk1 Z n = max sup | t∈[0,1] Z ≤ C1 1 Z 1 G(t, s)ξ(s, x(s))ds|, sup | t∈[0,1] 0 o Gt (t, s)ξ(s, x(s))ds| 0 1 |ξ(s, x(s))|ds 0 Z ≤ C1 (M + ε 1 |x(s)|ds) 0 ≤ C1 (M + εkxk1 ), which implies kAx − L∞ xk1 = 0, kxk1 where C1 depends on bounds for G and Gt . This means that the operator A given in (3.5) is Fréchet differentiable at x = ∞, and A0 (∞) = L∞ . The proof is complete. lim kxk1 →∞ Under the condition (H1), (3.6) can be rewritten as x = λL0 x + H(λ, x), (3.7) here H(λ, x) = λAx − λL0 x, L0 is defined as (3.3). Clearly by Lemma 2.1, x(t) is the solution of (2.1) if and only if x(t) is the solution of the equation x = λL0 x. (3.8) Therefore, the results of Lemma 2.2 and Lemma 2.3 satisfy (3.8). From Lemma 3.3, it can be seen that H(λ, x) is o(kxk1 ) for x near 0 uniformly on bounded λ. Lemma 3.5. For each integer k > 0 and each υ = +, or −, there exists a continua 2 2 2 2 Ckυ of solutions of (3.6) in Φυk ∪ {( k fπ0 , 0)}, which meets {( k fπ0 , 0)} and ∞ in Σ. 2 2 Proof. By Lemma 2.2 and Lemma 2.3, we know that for each integer k > 0, k fπ0 is a simple characteristic value of operator L0 . So with Lemma 3.3, (3.7) can be considered as a bifurcation problem from the trivial solution. From Lemma 1.1 and condition (H1) it follows that Σ contains a component Ck which can be decomposed 2 2 into two subcontinua Ck+ , Ck− such that for some neighborhood B of ( k fπ0 , 0), (λ, x) ∈ Ck+ (Ck− ) ∩ B, and (λ, x) 6= ( k2 π2 , 0), f0 2 2 implies (λ, x) = (λ, αuk + w), where α > 0(α < 0) and |λ − k fπ0 | = o(1), kwk1 = o(|α|) at α = 0. Since Sk is open and uk ∈ Sk , we know x w = uk + ∈ Sk , α α for α 6= 0 sufficiently small. Then there exists δ0 > 0 such that for δ ∈ (0, δ0 ), we have k2 π2 (Ck \ {( , 0)}) ∩ Bδ ⊂ Φk , f0 10 J. WANG, B. YAN EJDE-2013/171 2 2 where Bδ is an open ball in Γ of radius δ centered at ( k fπ0 , 0). From the assumption 2 2 in section 2 and (3.8) we know Ck \ {( k fπ0 , 0)} ∩ ∂Φk = ∅. Consequently, Ck lies 2 2 in Φk ∪ {( k fπ0 , 0)}. 2 2 From the the same reasoning it can be seen that Ckϑ lies in Φϑk ∪{( k fπ0 , 0)}(ϑ = +, or −). Next we show that the alternative (2) of Lemma 1.1 is impossible. Suppose, on 2 2 the contrary and without loss of generality that Ck+ meets ( j fπ0 , 0) with k 6= j and j 2 π2 f0 zm ∈ σ(L0 ). Then there exists a sequence (γm , zm ) ∈ Ck+ with γm → → 0 as m → +∞. Notice that zm = γm L0 zm + H(γm , zm ). j 2 π2 f0 and (3.9) Dividing this equation by kzm k1 and noticing that L0 is compact on E and that H(γm , zm ) = o(kzm k1 ) as m → ∞, we may assume without loss of generality that zm kzm k1 → z as m → ∞. Thus by (3.9) it follows that z= j 2 π2 L0 z. f0 Since z 6= 0, by Lemma 2.2, z belongs to Sj+ or Sj− . Notice that k kzzmmk1 − zk1 → 0, Sj+ and Sj− are open, so zm ∈ Sj+ or Sj− for m sufficiently large. This is a contradiction with zm ∈ Sk+ (m ≥ 1) and j 6= k. Hence alternative (2) is impossible. Finally since Sk+ and Sk− are disjoint, thus Ckυ does not contain a pair of points (λ, x), (λ, −x), x 6= 0, which means alternative (3) of Lemma 1.1 is impossible. Therefore alternative (1) of Lemma 1.1 holds. This implies that for each integer k > 0, and each υ = + or −, there exists a continua Ckυ of solutions of (3.6) in 2 2 2 2 Φυk ∪ {( k fπ0 , 0)}, which meets {( k fπ0 , 0)} and ∞ in Σ. The proof is complete. Lemma 3.6. For each integer k > 0 and each υ = +, or −, there exists a continua 2 2 2 2 2 2 π π Ĉkυ of Σ in Φυk ∪ {( kf∞ , ∞)} coming from {( kf∞ , ∞)}, which meets {( k fπ0 , 0)} or has an unbounded projection on R. Proof. Let T (λ, x) = λAx − λL∞ x, where L∞ is defined as in (3.4). Then (3.6) can be rewritten as x = λL∞ x + T (λ, x). (3.10) By Lemma 3.4, we know that T (λ, x) is o(kxk1 ) for x ∈ E near ∞ uniformly on bounded λ intervals. Notice that L∞ is a compact linear map on E. By Lemma 2.2 2 2 π and Lemma 2.3, we know that for each k > 0, kf∞ is a simple characteristic value of operator L∞ . So with Lemma 3.4, (3.10) can be considered as a bifurcation problem from infinity. A similar reasoning as in the proof of [22, Theorem 2.4] x shows that kxk21 T (λ, kxk 2 ) is compact. By Lemma 1.2 and Lemma 1.3, Σ contains 1 a component Dk which can be decomposed into two subcontinua Dk+ , Dk− which 2 2 π meets ( kf∞ , ∞). 2 2 π Next we show that for a smaller neighborhood O ⊂ M of ( kf∞ , ∞), (λ, x) ∈ 2 2 π Dk ∩ O and (λ, x) 6= ( kf∞ , ∞) implies that x ∈ Sk . In fact, by Lemma 1.3 we 2 2 π already know that there exists a neighborhood O ⊂ M of ( kf∞ , ∞) satisfying 2 2 π , ∞) implies (λ, x) = (λ, αuk + w) where α > (λ, x) ∈ Dk ∩ O and (λ, x) 6= ( kf∞ EJDE-2013/171 GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS 2 11 2 π 0(α < 0) and |λ − kf∞ | = o(1), kwk1 = o(|α|) at |α| = ∞. Since Sk is open and w α is small compared to uk ∈ Sk near α = +∞, and therefore x = αuk + w ∈ Sk for α 2 2 π near +∞. Therefore, Dk ∩ O ⊂ (R × Sk ) ∪ ( kf∞ , ∞). 2 2 π From the same reasoning it can be seen that Dkυ ∩ O ⊂ (R × Skυ ) ∪ ( kf∞ , ∞), where υ = +, or −. Let Ĉk+ denote the maximal subcontinuum of Dk+ lying in R × Sk+ . First suppose + Ĉk \ O is bounded. Then there exists (λ, x) ∈ ∂ Ĉk+ with x ∈ ∂Sk+ . Hence x has a double zero. By the assumption in section 2 we know x ≡ 0. Thus there exists a sequence (γm , zm ) ∈ Ĉk+ satisfying (3.10) with zm → x ≡ 0 as m → +∞. By 2 2 Lemma 1.2, like in the proof of Lemma 3.5, one can see that Ĉk+ meets ( k fπ0 , 0). Finally suppose Ĉk+ \ O is unbounded. In this case we show Ĉk+ \ O has an unbounded projection on R. Suppose, on the contrary, that there exists a sequence (µm , xm ) ∈ Ĉk+ \ O with µm → µ and kxm k1 → +∞ as m → +∞. Let ym := xm kxm k1 , m ≥ 1. From the fact that xm = µm L∞ xm + T (µm , xm ), it follows that T (µm , xm ) . (3.11) kxm k1 Notice that L∞ is compact. We may assume that there exists y ∈ E with kyk1 = 1 such that kym − yk1 → 0 as m → +∞. m ,xm ) → 0 as m → +∞ one obtains Letting m → +∞ in (3.11) and noticing T (µ kxm k1 ym = µm L∞ ym + y = µL∞ y. (3.12) 2 2 π Since µ 6= 0 is an eigenvalue of operator L∞ and y 6= 0; that is, µ = if∞ for + − some i 6= k. Then by Lemma 2.2 y belongs to Si or Si . Notice the fact that kym − yk1 → 0 as m → +∞. Thus xm ∈ Si+ or Si− for m sufficiently large since Si+ or Si− are open. This is a contradiction with xm ∈ Sk+ (m ≥ 1). Thus Ĉk+ \ O has an unbounded projection on R. Similarly, the same argument works if + is replaced by − in the above cases. The proof is complete. 2 2 2 2 π Proof of Theorem 3.1. Case 1. k fπ0 < λ < kf∞ . Consider (3.7) as a bifurcation problem from the trivial solution. To obtain the desired results we need only to show that Ckϑ ∩ ({λ} × E) 6= ∅, ϑ = +, −. 2 2 By Lemma 3.5 we know that Ckϑ joins ( k fπ0 , 0) to infinity Φϑk . Therefore, there exists a sequence (µn , xn ) ∈ Ckϑ such that lim (µn + kxn k1 ) = ∞. n→∞ We note that µn > 0 for all n ∈ N since (λ, x) = (0, 0) is the unique solution of (3.6) with λ = 0 in E and Ckϑ ∩ ({0} × E) = ∅. If lim µn = ∞, n→∞ then Ckϑ ∩ ({λ} × E) 6= ∅. 12 J. WANG, B. YAN EJDE-2013/171 Assume that there exists M > 0, such that for all n ∈ N, µn ∈ (0, M ]. In this case it follows that kxn k1 → ∞. We divide the equation xn = µn L∞ xn + T (µn , xn ) by kxn k1 and set yn = kxxnnk1 . Since yn is bounded in E, choosing a subsequence and relabelling, if necessary, we see that yn → y for some y ∈ E with kyk1 = 1, and y = µL∞ y where µ = limn→∞ µn . From the proof of Lemma 3.6 one can see µ= 2 2 2 k2 π2 . f∞ 2 π , ∞) which implies Thus Ckϑ joins ( k fπ0 , 0) to ( kf∞ Ckϑ ∩ ({λ} × E) 6= ∅. 2 2 2 2 π Case 2. kf∞ < λ < k fπ0 . Consider (3.10) as a bifurcation problem from infinity. As above we need only to prove that Ĉkϑ ∩ ({λ} × E) 6= ∅, ϑ = +, −. 2 2 2 2 π From Lemma 3.6, we know that Ĉkϑ comes from ( kf∞ , ∞) meets ( k fπ0 , 0) or has an unbounded projection on R. 2 2 2 2 If it meets ( k fπ0 , 0), then the connectedness of Ĉkϑ and k fπ0 > λ guarantee that Ĉkϑ ∩ ({λ} × E) 6= ∅, ϑ = +, −. If Ĉkϑ has an unbounded projection on R, notice that (λ, x) = (0, 0) is the unique solution of (3.6), so Ĉkϑ ∩ ({λ} × E) 6= ∅, ϑ = +, −. The proof is complete. Proof of Theorem 3.2. Repeating the arguments used in the proof in Theorem 3.1, we see that for each ϑ ∈ {+, −} and each i ∈ {1, 2, . . . , j} ϑ ϑ (Ck+i ∪ Ĉk+i ) ∩ ({λ} × E) 6= ∅. Remark 3.7. From [5], we know if β = 0, we have also the same results as Theorem 3.1 and Theorem 3.2. 4. Applications In this section, we give an example to illustrate the applications of Theorem 3.2. EJDE-2013/171 GLOBAL PROPERTIES AND MULTIPLE SOLUTIONS Example 4.1. Consider the second-order impulsive differential equation 1 x00 (t) + f (x(t)) = 0, t ∈ (0, 1), t 6= , 2 1 ∆x|x= 12 = x( ), 2 1 ∆x0 |x= 21 = −x0 ( − 0), 2 x(0) = x(1) = 0, 13 (4.1) where f (x) = 100 sin x + 5x. It is not difficult to see that f here satisfies the assumption in section 2 and (H1) with f0 = 105, f∞ = 5 and the eigenvalues of the boundary-value problem 1 x00 (t) + λx(t) = 0, t ∈ (0, 1), t 6= , 2 1 ∆x|x= 21 = λx( ), 2 0 0 1 ∆x |x= 21 = −λx ( − 0), 2 x(0) = x(1) = 0, can be written as λk = k 2 π 2 , k = 1, 2, . . . . 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Jing Wang School of Mathematical Sciences, Shandong Normal University, Jinan, 250014, China E-mail address: wnself36@gmail.com Baoqiang Yan (corresponding author) School of Mathematical Sciences, Shandong Normal University, Jinan, 250014, China E-mail address: yanbaoqiang666@gmail.com