Multiple solutions of impulsive Sturm–Liouville boundary value problem via lower and upper solutions and variational methods

J. Math. Anal. Appl. 387 (2012) 475–489

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Multiple solutions of impulsive Sturm–Liouville boundary value problem via lower and upper solutions and variational methods ✩ Yu Tian a,∗ , Weigao Ge b a b

School of Science, Beijing University of Posts and Telecommunications, Beijing 100876, PR China Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081, PR China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 24 March 2011 Available online 25 August 2011 Submitted by J.J. Nieto

In this paper, we prove the existence of multiple solutions for second order Sturm–Liouville boundary value problem



Keywords: Multiple solutions Impulsive Sturm–Liouville boundary value problem Critical point Lower and upper solutions Variational methods

− Lu = f (x, u ),  x ∈ [0, 1]\{  x1 , x2 , . . . , xl }, − p (xi )u  (xi ) = I i u (xi ) , i = 1, 2, . . . , l, R 1 ( u ) = 0, R 2 ( u ) = 0,

where Lu = ( p (x)u  ) − q(x)u is a Sturm–Liouville operator, R 1 (u ) = α u  (0) − β u (0), R 2 (u ) = γ u  (1)+ σ u (1). The technical approach is fully based on lower and upper solutions and variational methods. The interesting point is that the property that the critical points of the energy functional are exactly the fixed points of an operator that involves the Green’s function. Besides, the existence of four solutions is given. © 2011 Elsevier Inc. All rights reserved.

1. Introduction In the real world the phenomena of sudden or discontinuous jumps occur in the dynamics of processes. These processes are often seen in chemotherapy, population dynamics, optimal control, ecology, engineering, etc. The mathematical model that describes the phenomena is impulsive differential equations. Due to their significance, it is important to study the solvability of impulsive differential equations. For relevant and recent references on impulsive differential equations, we refer the reader to [4,15,26–30,41–43]. For the background and applications of the theory of impulsive differential equations to different areas, we refer the reader to the monographs and some recent contributions as [2,10,13,16,22,23,34,45,46,48,49]. Recently, there are many papers studying the existence of solutions for impulsive boundary value problems. In papers [1,17,18,44], the fixed point theorems in cone were applied into impulsive boundary value problems. In papers [3,6,14,19], the method of lower and upper solutions with monotone iterative technique was applied into impulsive boundary value problems. In papers [24,25,31–33,37,39,47], variational methods were applied into impulsive boundary value problems. In paper [35], variational methods and iterative methods were applied into impulsive boundary value problem. By using topological degree theory and variational methods, Sturm–Liouville boundary value problems have been studied in the last few 1, 2 years [5,12,36–39]. In paper [5], the authors proved a theorem on the existence of an unbounded (in W 0 [0; 1]) sequence of   weak solutions for the following Sturm–Liouville boundary value problem ( p (t )u ) + q(t )u = f (u ); u (0) = u (1) = 0; where ✩ Project 11001028 supported by National Science Foundation for Young Scholars, Project BUPT2009RC0704 supported by Chinese Universities Scientific Fund, Project 11071014 supported by National Science Foundation of PR China. Corresponding author. E-mail address: [email protected] (Y. Tian).

*

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2011.08.042

©

2011 Elsevier Inc. All rights reserved.

476

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

p , q ∈ L ∞ [0, 1], λ > 0 is a parameter and f : R → R is almost everywhere continuous (i.e., the Lebesgue measure of its set of discontinuities is zero) and locally essentially bounded. However, to the best of our knowledge, there are few papers concerned with the existence of solutions for impulsive Sturm–Liouville boundary value problems by using variational methods and lower and upper solutions method. Motivated by [9,11,20,21], in this paper we will fill the gap in this area. We study the existence of multiple solutions for second order Sturm–Liouville boundary value problem



− Lu = f (x, u ),  x ∈ [0, 1]\{  x1 , x2 , . . . , xl }, − p (xi )u  (xi ) = I i u (xi ) , i = 1, 2, . . . , l, 5R 1 (u ) = 0, R 2 (u ) = 0,

(1.1)

where Lu = ( p (x)u  ) − q(x)u is a Sturm–Liouville operator, R 1 (u ) = α u  (0) − β u (0), R 2 (u ) = γ u  (1) + σ u (1), α , β, σ , γ  0, α 2 + β 2 > 0, γ 2 + σ 2 > 0, p , q ∈ C 1 [0, 1], p , q > 0, 0 = x0 < x1 < · · · < xl < xl+1 = 1, ( p (xi )u  (xi )) = p (x+ )u  (x+ )− i i −  − + −    p (xi )u (xi ), here u (xi ) (respectively u (xi )) denotes the right limit (respectively left limit) of u (x) at x = xi , (x, u ) → f (x, u ), u → I j (u ) are both continuous in [0, 1] × R, and Lipschitz continuous for u uniformly in x ∈ [0, 1]\{x1 , x2 , . . . , xl }. Our aim of this paper is to apply variational methods and lower and upper solutions method to problem (1.1) and prove the existence of at least four solutions. Besides, we give more accurate characterization of the specific scope of solutions. The results are different from those in the literature. With the impulse effects and the Sturm–Liouville boundary conditions taken into consideration, difficulties, such as how to construct suitable energy functional Φ and operator A such that the critical point of Φ is just the fixed point of operator A, and how to verify that Φ satisfies (PS) condition and Φ(h R (s)) → −∞ as R → +∞. Besides, this paper is a generalization of [12,36–39]. Contrast with paper [36], the impulse effect makes the difference of construction of Φ . So the proof of Lemma 3.2, Lemma 3.4 and Theorem 1.2 are very different. The following lemmas will be needed in our argument, which can be found in [21]. Lemma 1.1. (See [21, Theorem 3.3].) Let H be a Hilbert space and X a Banach space such that X is embedded in H . Let Φ be a C 2−0 functional defined on H , that is, the differential Φ  of Φ is Lipschitz continuous from H to H . Assume that Φ  (u ) = u − Au and Φ  is also Lipschitz continuous as an operator from X to X . Assume also that K = {u ∈ H: Φ  (u ) = 0} ⊂ X . Assume that Φ satisfies the (PS)-condition on H and there are two open convex subsets D 1 and D 2 of X with the properties that D 1 ∩ D 2 = ∅, A (∂ X D 1 ) ⊂ D 1 , and A (∂ X D 2 ) ⊂ D 2 . If there exists a path h : [0, 1] → X such that

h(0) ∈ D 1 \ D 2 ,

h(1) ∈ D 2 \ D 1 ,

and

inf

u ∈ D 1X ∩ D 2X

  Φ(u ) > sup Φ h(t ) , t ∈[0,1]

then Φ has at least four critical points, one in D 1 ∩ D 2 , one in D 1 \ D 2X , one in D 2 \ D 1X , and one in X \ ( D 1X ∪ D 2X ). Here ∂ X D and D X mean respectively the boundary and the closure of D relative to X . Now we shall introduce Sobolev spaces, definitions of lower and upper solutions and eigenvalue problem in order to state our main result. Let the space H = H 1 [0, 1] be the Sobolev space. Let the space

















= u  (xi ), ∃u  x+ X = u ∈ C [0, 1]: pu  (·) ∈ C [0, 1]\{x1 , x2 , . . . , xl } , u  x− i i



with the norm

    

u X = max max u (x), sup u  (x) . x∈[0,1]

x∈[0,1]

Clearly X is a Banach space and densely embedded in H . Definition 1.1. A function φ : [0, 1] → R is a lower solution of (1.1) if φ ∈ X such that



− L φ  f (x, φ),  x ∈[0, 1] \ {x1 , x2 , . . . , xl }, − p (xi )φ  (xi )  I i φ(xi ) , i = 1, 2, . . . , l, α φ  (0) − βφ(0)  0, γ φ  (1) + σ φ(1)  0.

Definition 1.2. A function ψ : [0, 1] → R is an upper solution of (1.1) if ψ ∈ X such that



− L ψ  f (x, ψ), x ∈[0, 1] \ {x1 , x2 , . . . , xl }, − p (xi )ψ  (xi )  I i ψ(xi ) , i = 1, 2, . . . , l, α ψ  (0) − βψ(0)  0, γ ψ  (1) + σ ψ(1)  0.

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

477

Following we introduce linear eigenvalue problem



− Lu = λu , R 1 ( u ) = 0,

(1.2)

R 2 ( u ) = 0.

As is well known, (1.2) possesses a sequence of eigenvalues (λi ) with

0 < λ1  λ 2  · · ·  λ j  · · · . Corresponding eigenfunctions are ϕ1 , ϕ2 , . . . and ϕi H = 1, i = 1, 2, . . . . It is well known that {ϕi }∞ 1 forms an orthonormal base of H . By the Sturm–Liouville theory of ordinary differential equations (see, for example [8,40]), the first eigenvalue λ1 is simple and eigenfunction ϕ1 with respect to λ1 > 0 satisfies ϕ1 > 0 in [0, 1]. Now we state the main result Theorem 1.2. Assume that (H1) there are numbers a, b with λk < a < b < λk+1 for some k  2 such that

a  f t (x, t )  b,

∀x ∈ [0, 1], |t |  R 1 ,

∂ f (x,t ) where f t (x, t ) = ∂ t ;

a

−1

k , (H2) | I i (u )|  M |u |, where M < min{ l( ϕ +···+ ϕk C 0 )2 1 C0 function of

λ



− Lu = 0, R 1 ( u ) = 0,

min{λk+1 −b,a−λk } }, lλk+1 G

G = max(x,s)∈[0,1]×[0,1] |G (x, s)|, G (x, s) is the Green’s

R 2 ( u ) = 0.

From [9], G (x, s) can be written by



G (x, s) =

m(x)n(s)

ω

m(s)n(x)

ω

, 0  x  s  1, , 0  s  x  1.

(1.3)

(H3) (1.1) has a lower solution φ ∈ X and an upper solution ψ ∈ X with φ  ψ ; (H4) f (x, u )  f (x, v ) for u  v; I i (u )  I i ( v ) for u  v, i = 1, 2, . . . , l. Then problem (1.1) has at least four solutions, φ < u 1 < ψ , u 2 > ψ , u 3 < φ , u 4 ∈ X \( D 1X ∪ D 2X ), where D 1 = {u ∈ X: u > φ in [0, 1]} and D 2 = {u ∈ X: u < ψ in [0, 1]}. Remark 1.1. If φ  0 and ψ > 0, then the positive, negative and sign-changing solutions can be obtained. However, signchanging solution of impulsive equations are less studied now. Remark 1.2. If problem (1.1) is studied by using the topological degree theory, the restriction on nonlinearity is compared with the first eigenvalue [12]. However, the assumption (H1) in Theorem 1.2 is between kth to (k + 1)th eigenvalues (k  2). The paper is organized as follows: In Sections 2 and 3 we first discuss the case α , γ > 0 in Sturm–Liouville boundary condition R 1 (u ) = 0, R 2 (u ) = 0 for the convenience of expressing the functional Φ(u ). In Section 2, we give some basic knowledge and related lemmas. In Sections 3, we prove the existence of four solutions by applying descending flow and lower and upper solutions method. In Sections 4, we discuss the case α , γ  0 in Sturm–Liouville boundary condition R 1 (u ) = 0, R 2 (u ) = 0 and give examples to illustrate the main results. At the end Appendix A is given. 2. Basic knowledge with α , γ = 0 In what follows Y 1 and Y 2 are two Banach spaces. Definition 2.1. (See [7, Definition 4.1.1].) A map f : Y 1 → Y 2 is said to be Gâteaux differentiable at x ∈ Y 1 , if and only if there exists A (x) ∈ L(Y 1 ; Y 2 ) such that

lim

λ→0

f (x + λh) − f (x)

λ

= A (x)h,

∀h ∈ Y 1 .

The operator A (x) is said to be the Gâteaux derivative of f at x. It is usually denoted by f G (x). We say that f is Gâteaux differentiable, if it is Gâteaux differentiable at every x ∈ Y 1 .

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Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

Particularly, for the functional f : D( f ) ⊂ Y 1 → R, we say that f G (x0 ) is the gradient of f at x0 . It is denoted by grad f (x0 ), that is grad f (x0 ) = f G (x0 ). Definition 2.2. (See [7, Definition 4.1.4].) A map f : Y 1 → Y 2 is said to be Fréchet differentiable at x ∈ Y 1 , if and only if there exists A (x) ∈ L(Y 1 ; Y 2 ) such that

f (x + h) − f (x) = A (x)h + u (x, h), where

u (x, h) Y 2 → 0 as h Y 1 → 0.

h Y 1 The operator A (x) is said to be the Fréchet derivative of f at x ∈ Y 1 and it is usually denoted by f  (x). We say that f is Fréchet differentiable, if it is Fréchet differentiable at every x ∈ Y 1 . Remark 2.1. It is clear that if f is Fréchet differentiable at x, it is also Gâteaux differentiable at x. Remark 2.2. (grad f (x), h(x)) = ( f  (x), h(x)), ∀h ∈ Y 1 . We define new inner product of H as follows

1

β p (0) σ p (1) u (1) v (1) + u (0) v (0). γ α

p (x)u  (x) v  (x) + q(x)u (x) v (x) dx +

(u , v ) = 0

The inner product induces the norm

1

u H =

 2  2 2 β p (0)   σ p (1)  u (0)2 p (x)u  (x) + q(x)u (x) dx + u (1) +

γ

12

α

,

0

which is equivalent to the usual one. The norm u H is a part of the functional Φ (2.1), which makes it convenient to estimate Φ . Define the functional Φ : H → R by

Φ(u ) =

1

1



2



2

p (x)u  (x) + q(x)u (x) dx +

2

2 β p (0)   σ p (1)  u (0)2 u (1) + 2γ 2α

0

1 −

l   F x, u (x) dx −



u(xi )

I i (s) ds,

(2.1)

i =1 0

0

u

where F (x, u ) = 0 f (x, s) ds. By [39], the solution u of (1.1) is equivalent to the critical point of Φ , that is Φ  (u ), v  = 0 for v ∈ H , where

   Φ (u ), v =

1

p (x)u  (x) v  (x) + q(x)u (x) v (x) dx +

σ p (1) u (1) v (1) γ

0

+

β p (0)

α

1 u (0) v (0) −





f x, u (x) v (x) dx − 0

l  



I i u (xi ) v (xi ).

i =1

Lemma 2.1. (See [9].) The Green’s function G (x, s) defined by (1.3), possesses the following properties: (i) (ii) (iii) (iv) (v) (vi)

m(x) ∈ C 2 ([0, 1]; R ) is increasing and m(x) > 0, x ∈ (0, 1]; n(x) ∈ C 2 ([0, 1]; R ) is decreasing and n(x) > 0, x ∈ [0, 1); ( Lm)(x) ≡ 0, m(0) = α , m (0) = β ; ( Ln)(x) ≡ 0, n(1) = γ , n (1) = −σ ; p (x)(m (x)n(x) − m(x)n (x)) ≡ ω is a positive constant; G (x, s) is continuous and symmetrical over {(x, s): 0  x  s  1} × {(x, s): 0  s  x  1};

(2.2)

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

479

(vii) G (x, s) has continuously partial derivative over {(x, s): 0  x  s  1}, {(x, s): 0  s  x  1}; (viii) For each fixed s ∈ [0, 1], G (x, s) satisfies LG (x, s) = 0 for x = s, x ∈ [0, 1]. Moreover, R 1 (G ) = R 2 (G ) = 0; (ix) G x has discontinuous point of the first kind at x = s and G x (s + 0; s) − G x (s − 0, s) = − p (1s) , s ∈ (0, 1). 3. Four solutions of (1.1) with α , γ = 0 Lemma 3.1. The function u ∈ H is a critical point of the functional Φ if and only if u ∈ X is a fixed point of the operator A, where 1 l G (x, s) f (s, u (s)) ds + j =1 G (x, x j ) I j (u (x j )). 0

( Au )(x) =

Proof. On one hand, by [39], the function u ∈ H is a critical point of the functional Φ if and only if u is the solution of (1.1). On the other hand, by [12] u is the solution of (1.1) if and only if u ∈ H is a fixed point of the operator A. Following we shall show that the fixed point of the operator A is in X . By (x, u ) → f (x, u ) is continuous in [0, 1] × R, and Lipschitz continuous for u uniformly in x ∈ [0, 1]\{x1 , x2 , . . . , xl }, and Lemma 2.1(vii), one has u (x) = ( Au )(x) ∈ X . The proof is complete. 2 By Lemma 3.1, the critical point set K = {u ∈ H: Φ  (u ) = 0} ⊂ X . Suppose that Φ X is the restriction of Φ on X . By (x, u ) → f (x, u ), u → I j (u ) are both continuous in [0, 1] × R, and Lipschitz continuous for u uniformly in x ∈ [0, 1]\{x1 , x2 , . . . , xl }, Φ  : H → H is a C 2−0 functional defined on H , that is, the differential Φ  of Φ is locally Lipschitz continuous from H to H . Moreover, we obtain that Φ  is locally Lipschitz continuous from X to X . Lemma 3.2. The gradient of Φ is

grad Φ(u ) = u − Au , where A is defined in Lemma 3.1 and G (x, s) is defined in (1.3). The proof is in Appendix A. Similar to Lemma 2.3 in [39], we have Lemma 3.3 (Maximal value principle). Suppose that g ∈ C [0, 1], I i ∈ R + , i = 1, 2, . . . , l, A , B ∈ R. If

⎧   ⎨ − p (x)u  (x) +q(x)u (x) = g (x)  0, t ∈ [0, 1] \ {x1 , x2 , . . . , xl },    ⎩ − p (xi )u (xi ) = I i > 0, i = 1, 2, . . . , l, α u (0) − β u (0) = A  0, γ u (1) + σ u (1) = B  0,

(3.1)

then u (x)  0, x ∈ [0, 1]. Lemma 3.4. Under the conditions (H1), (H2) given to Theorem 1.2, Φ satisfies (PS)-condition. Proof. First we shall show that (un ) is bounded in L 2 [0, 1]. Let λ = 

1

Φ (u ) = u −





G (·, s) f s, u (s) ds −



2

. For u ∈ H , by appendix Theorem A.1,



G (·, x j ) I j u (x j )

j =1

0

= u − (− L )−1 f (·, u ) −

l 

λk +λk+1

l 





G (·, x j ) I j u (x j )

j =1 l      G (·, x j ) I j u (x j ) = (− L − λ) (− L − λ)−1 u − (− L )−1 (− L − λ)−1 f (·, u ) − j =1 l      = (− L − λ)(− L )−1 (− L )(− L − λ)−1 u − (− L − λ)−1 f (·, u ) − G (·, x j ) I j u (x j ) j =1 l      G (·, x j ) I j u (x j ) = (− L − λ)(− L )−1 u + λ(− L − λ)−1 u − (− L − λ)−1 f (·, u ) − j =1 l       = (− L − λ)(− L )−1 u − (− L − λ)−1 f (·, u ) − λu − G (·, x j ) I j u (x j ) . j =1

480

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

Then for u ∈ H ,

−1

u = (− L )(− L − λ)

Φ (u ) +



= Φ (u ) +

l 

   G (·, x j ) I j u (x j ) + (− L − λ)−1 f (·, u ) − λu 

l     −1  G (·, x j ) I j u (x j ) + λ(− L − λ) G (·, x j ) I j u (x j ) Φ (u ) +

j =1

−1

+ (− L − λ)





j =1



f (·, u ) − λu .

H satisfies Φ  (un ) H

u n C 0

l  j =1



If (un ) ⊂



(3.2)

→ 0 as n → ∞, then

 l             Φ (u n ) C 0 +  G (·, x j ) I j un (x j )    0 j =1 C      Φ (un ) + λ(− L − λ)−1  0 0

 l       + G (·, x j ) I j un (x j )  L(C [0,1],C [0,1]) C0   0 j =1 C       −1        f (·, un ) − λun C 0  Φ (un ) C 0 + lMG un C 0 + (− L − λ) L(C 0 [0,1],C 0 [0,1])       − 1 + λ(− L − λ) L(C 0 [0,1],C 0 [0,1]) Φ (un )C 0 + lMG un C 0     + (− L − λ)−1 L(C 0 [0,1],C 0 [0,1])  f (·, un ) − λun C 0 . (3.3)          −1       Φ (un ) L 2 [0,1] + lMG un C 0

un L 2 [0,1]  Φ (un ) L 2 [0,1] + lMG un C 0 + λ (− L − λ) L( L 2 [0,1], L 2 [0,1])     − 1 + (− L − λ) L( L 2 [0,1], L 2 [0,1])  f (·, un ) − λun L 2 [0,1] . (3.4) We compute

  (− L − λ)−1 

L(C 0 [0,1],C 0 [0,1])

= =

  (− L − λ)−1 

L( L 2 [0,1], L 2 [0,1])

=

sup u ∈C 0 [0,1], u

C 0 [0,1]

=1

  (− L − λ)−1 u (x) 0 C [0,1]

   u (x)    λ − λ

sup

C0

u ∈ L 2 [0,1], u L 2 [0,1] =1

1

=

2

=

λk+1 − λk =1 C 0 [0,1]   (− L − λ)−1 u  2 sup L [0,1]

u ∈C 0 [0,1], u

(3.5)

1/2 −1

|(− L − λ)

sup u ∈ L 2 [0,1], u L 2 [0,1] =1

.

2

u | dx

0

   1  2 = = sup  .  λ − λk 2 λ − λ k + 1 u ∈ L [0,1]

(3.6)

By (H1),

  λk+1 − λk − − (a − λk ) u − C 1 2



= (a − λ)u − C 1  f (x, u ) − λu  (b − λ)u + C 1 =   λk+1 − λk − (λk+1 − b) u − C 1

λk+1 − λk 2

 − (λk+1 − b) u + C 1 for u > 0,

2

  λk+1 − λk = (b − λ)u − C 1  f (x, u ) − λu  (a − λ)u + C 1 = − − (a − λk ) u + C 1 for u < 0. 2

So

   f (x, u ) − λu   And then



λk+1 − λk 2

 − min{λk+1 − b, a − λk } |u | + C 1 .

(3.7)

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489



   f (x, un ) − λun 

C0

   f (x, un ) − λun 

 L 2 [0,1]



λk+1 − λk 2

and



481

 − min{λk+1 − b, a − λk } un C 0 + C 1 ,

λk+1 − λk 2

(3.8)

 − min{λk+1 − b, a − λk } un L 2 [0,1] + C 1 .

(3.9)

By (3.3), (3.5), (3.8)

 

un C 0  Φ  (un )C 0 + lMG un C 0 +

    Φ (un ) 0 + lMG un 0 + C C

2λ

λk+1 − λk

   2 min{λk+1 − b, a − λk } 2λ

u n C 0 .  C2 + 1 − + lMG 1 + λk+1 − λk λk+1 − λk

2

λk+1 − λk

   f (·, un ) − λun 

C0

By (H2), (un ) is bounded in C 0 [0, 1]. By (3.4), (3.6), (3.9)

      2λ Φ (un ) 2

un L 2 [0,1]  Φ  (un ) L 2 [0,1] + lMG un C 0 + + lMG un C 0 L [0,1] λk+1 − λk    λk+1 − λk 2 + − min{λk+1 − b, a − λk } un L 2 + C 1 , λk+1 − λk 2 which together with (un ) bounded in C 0 [0, 1], implies (un ) is bounded in L 2 [0, 1]. Next we will show that (un ) is bounded in H . By Φ(un ) is bounded, (H1), (H2), one has



un 2H

1

= 2 Φ(un ) +

l   F x, un (x) dx +





u (x j )

I j (s) ds

j =1 0

0

   2 M 1 + M 2 un 2L 2 + 1/2lM un 2C 0 ,

for M 1 , M 2 > 0. Since (un ) is bounded in C 0 [0, 1] and L 2 [0, 1] , one has (un ) is bounded in H . By standard argument as [39], (un ) strongly converges to u in H . Therefore, Φ satisfies the (PS)-condition.

2

Proof of Theorem 1.2. It is clear that D 1 and D 2 are open convex sets in X and D 1 ∩ D 2 = ∅. If u ∈ ∂ X D 1 and v = Au, then by (H4)

1 v (x) =





G (x, s) f s, u (s) ds +

l 





j =1

0

1

G (x, x j ) I j u (x j ) 





G (x, s) f s, φ(s) ds + 0

l 





G (x, x j ) I j φ(x j ) = A φ(x).

j =1

So

Au  A φ.

(3.10)

Moreover, let w = A φ . Following we shall show that A φ  φ . In fact,

      − L A φ + L φ = − p (x)( A φ) (x) + q(x) A φ(x) + − p (x)φ  (x) + q(x)φ(x) 1 

 l        G x (x, y ) f y , φ( y ) dy + G x (x, x j ) I j u (x j ) = − p (x) 0

1 + q(x) 

j =1

l     G (x, y ) f y , φ( y ) dy + G (x, x j ) I j u (x j ) + L φ 

j =1

0

 = f x, φ(x) + L φ  0,



l        − p (xi )( A φ) (xi ) − p (xi )φ (xi ) = − p (xi ) G x (xi , x j ) I j φ(x j ) +  p (xi )φ  (xi ) 







j =1

   = I i φ(xi ) +  p (xi )φ  (xi )  0. By Lemma 2.1

482

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

1

    G x (0, s) f s, φ(s) ds − φ (0) − β G (0, s) f s, φ(s) ds − φ(0)

1





R 1 ( w ) − R 1 (φ) = α 0

0

= −α φ  (0) + βφ(0)  0. So R 1 ( w )  R 1 (φ). Similarly R 2 ( w )  R 2 (φ). By Lemma 3.3 w  φ . So

A φ  φ.

(3.11)

By (3.10), (3.11) Au  φ and A (∂ X D 1 ) ⊂ D 1 . Similarly A (∂ X D 2 ) ⊂ D√ 2. + R cos π s + R sin π sϕ2 (x). We take R sufficiently large Assume (H1) with k = 2,√we define h R (s) = R cos π sϕ1 (x)√ such that h R (0) = R ϕ1 (x) + R ∈ D 1 \ D 2 , h R (1) = − R ϕ1 (x) − R ∈ D 2 \ D 1 . Following we shall show Φ(h R (s)) → −∞ as R → +∞, s ∈ [0, 1]. First we compute

 1 h R (s)2 = 1 H 2 2

1

2





2

p (x)h R (s)  + q(x)h R (s) dx +

2 β p (0)   σ p (1)  h R (s)(0)2 h R (s)(1) + 2γ 2α

0

=

1

1

2



2



p (x) R cos π sϕ1 (x) + R sin π sϕ2 (x) + q(x) R cos π sϕ1 (x) +

√

2

R cos π s + R sin π sϕ2 (x) dx

0

+ + =

1

√ 2 σ p (1)  R cos π sϕ1 (1) + R cos π s + R sin π sϕ2 (1) 2γ

β p (0)  2α



2

R cos π sϕ1 (0) +

1

2

2

πs

R cos

2

R cos π s + R sin π sϕ2 (0)

   2  2 σ p (1) 2 β p (0) 2     ϕ1 (1) + ϕ1 (0) p (x) ϕ1 (x) + q(x) ϕ1 (x) dx +

γ

0

1 2

2

√

+ R sin π s

α

   2  2 β p (0) 2 σ p (1) 2     p (x) ϕ2 (x) + q(x) ϕ2 (x) dx + ϕ2 (1) + ϕ2 (0)

γ

0

1

2

+ 2R cos π s sin π s



α

1



p (x)ϕ1 (x)ϕ2 (x) dx + 0

q(x)ϕ1 (x)ϕ2 (x) dx 0



β p (0) σ p (1) + ϕ1 (1)ϕ2 (1) + ϕ1 (0)ϕ2 (0) + 2R 3/2 cos2 π s γ α

1 q(x)ϕ1 (x) dx 0

1 + 2R

3/2

1 2

q(x)ϕ2 (x) dx + R cos

cos π s sin π s 0

+ +

2σ p (1)

γ 2β p (0)

α

q(x) dx +

πs

2σ p (1)

γ

R 3/2 cos2 π sϕ1 (1)

0

σ p (1) 2β p (0) 3/2 R 3/2 cos π s sin π sϕ2 (1) + R cos2 π s + R cos2 π sϕ1 (0) γ α R

3/2

cos π s sin π sϕ2 (0) +

β p (0)

α



2

R cos

πs .

(3.12)

By ϕi H = 1, (ϕi , ϕ j ) = 0, i = j,

the above equality 

1 2

where M 3 , M 4 > 0. By (H1), there exist M 5 , M 6 > 0 satisfying

1 0

 F x, h R (s) dx = 

1



R 2 cos2 π s + R 2 sin2 π s + 2M 3 R 3/2 + 2M 4 R =



a 2 t 2

2

R 2 + M 3 R 3/2 + M 4 R ,

− M 5 t − M 6  F (x, t )  b2 t 2 + M 5 t + M 6 . So

F x, R cos π sϕ1 (x) + 0

1

√



R cos π s + R sin π sϕ2 (x) dx

(3.13)

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489



1

a 2

483

√    R cos π sϕ1 (x) + R cos π s + R sin π sϕ2 (x)2 dx

0

1 − M5

√    R cos π sϕ1 (x) + R cos π s + R sin π sϕ2 (x) dx − M 6

0

=

1

a



2

R 2 cos2 π sϕ12 (x) + R 2 sin2 π sϕ22 (x) + R cos2 π s + 2R 3/2 cos2 π sϕ1 (x)

0

 + 2R 3/2 cos π s sin π sϕ2 (x) + 2R 2 cos π s sin π sϕ1 (x)ϕ2 (x) dx 1 − M5

√    R cos π sϕ1 (x) + R cos π s + R sin π sϕ2 (x) dx − M 6 .

(3.14)

0

It is clear that

1

ϕ1 (x)ϕ2 (x) dx = 0

(3.15)

0

and

1

ϕi2 (x) dx = 0

1

λi

ϕi 2H =

1

λi

(3.16)

.

Substituting (3.15), (3.16) into (3.14), one has

1





F x, h R (s) dx 0



a



R 2 cos2 π s

2

λ1 1

− M5



a



+ R cos π s + 2R

λ1

− M5 a R

2

cos

− M5

cos π s sin π s

ϕ2 (x) dx 0

+

λ2

1 2

+ R cos π s − 2R

3/2

2

cos

1/2 2 1 (x) dx

πs

ϕ

1 −R

3/2

0

1/2  2 2 (x) dx

ϕ 0

√    R cos π sϕ1 (x) + R cos π s + R sin π sϕ2 (x) dx − M 6

+ 1

ϕ1 (x) dx + 2R

πs



1 3/2

0

R 2 sin2 π s

0 2

2 λ2

R 2 cos2 π s



λ2

  λ2 cos2 π s R 3/2 − 1 + R cos2 π s − 2R 3/2 √ −√ λ1 λ1 λ2

√    R cos π sϕ1 (x) + R cos π s + R sin π sϕ2 (x) dx − M 6 .

0

 l hR (s)   l       I i (θ) dθ  =      i =1 0

3/2

√    R cos π sϕ1 (x) + R cos π s + R sin π sϕ2 (x) dx − M 6

R 2 cos2 π s

1

=

λ2

1 2

0

2



+

R 2 sin2 π s

R cos π sϕ1 (xi )+ R sin  π sϕ2 (xi )+

i =1

 1/2lM

0

max

√

   I i (θ) dθ  

R cos π s

√    R cos π sϕ1 (xi ) + R sin π sϕ2 (xi ) + R cos π s2

s∈[0,1],xi ∈[0,1]

(3.17)

484

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

√ 2  = 1/2lM R ϕ1 C 0 + R ϕ2 C 0 + R  2   = 1/2lM R 2 ϕ1 C 0 + ϕ2 C 0 + lM R 3/2 ϕ1 C 0 + ϕ2 C 0 + 1/2RlM .

(3.18)

By (3.13), (3.17), (3.18), (H2)

2  1 Φ h R (s) = h R (s) H − 

1

2

l   F x, h R (s) dx −



0



1 2

2

R + M3 R

1 + M5

3/2



+ M4 R −

a R2 2 λ2

hR (s)

I i (θ) dθ

i =1 0

+

R 2 cos2 π s



λ2

  λ2 2R 3/2 cos2 π s R 3/2 2 − 1 + R cos π s − −√ √ λ1 λ1 λ2

√    R cos π sϕ (x) + R cos π s + R sin π sϕ2 (x) dx + M 6

0

 2    + 1/2lM R 2 ϕ1 C 0 + ϕ2 C 0 + lM R 3/2 ϕ1 C 0 + ϕ2 C 0 + 1/2RlM      2 R2 a a cos2 π s λ2 = 1− − − 1 + lM ϕ1 C 0 + ϕ2 C 0 + M 7 R 3/2 + M 8 R 2 λ2 λ2 λ1     R2 a a cos2 π s λ2 a  1− − −1 + − 1 + M 7 R 3/2 + M 8 R 2 λ2 λ2 λ1 λ2   R2 1 1  − a cos2 π s − + M 7 R 3/2 + M 8 R → −∞ 2 λ1 λ2 

as R → +∞, s ∈ [0, 1]. So for R sufficiently large, h R meets infu ∈ D X ∩ D X Φ(u ) > supt ∈[0,1] Φ(h R (t )). Besides, Φ satisfies (PS)1

2

condition on H by Lemma 3.4. Applying Lemma 1.1, problem (1.1) has at least four solutions, u 1 ∈ D 1 ∩ D 2 , u 2 ∈ D 1 \ D 2X , u 3 ∈ D 2 \ D 1X , u 4 ∈ X \( D 1X ∪ D 2X ). 2 4. Four solutions of (1.1) with α , γ  0 For

α = 0, β = 1, γ = 0, σ = 1, the boundary condition in (1.1) is reduced to u (0) = 0,

For

u  (1) = 0.

(4.2)

α = 1, β = 0, γ = 0, σ = 1, the boundary condition in (1.1) is reduced to u  (0) = 0,

For

(4.1)

α = 0, β = 1, γ = 1, σ = 0, the boundary condition in (1.1) is reduced to u (0) = 0,

For

u (1) = 0.

u (1) = 0.

(4.3)

α , β , σ = 0, γ = 0, the boundary condition in (1.1) is reduced to

α u  (0) − β u (0) = 0, For β ,

u (1) = 0.

(4.4)

γ , σ = 0, α = 0, the boundary condition in (1.1) is reduced to

u (0) = 0,

γ u  (1) + σ u (1) = 0.

(4.5)

Results. Theorem 1.2 holds for problems (1.1) with boundary conditions (4.1), (4.2), (4.3), (4.4), (4.5). Example 4.1. Consider the problem

⎧ ⎨ −u  = 10u + 16(1 + u 1.6 )−1 − 16, −u  (1/2) = 0.0003u (1/2), ⎩ u (0) = 0, u  (1) = 0, problem (4.6) has at least four solutions.

x ∈ [0, 1], (4.6)

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

485

Contrast to (1.1), α = 0, β = 1, γ = 1, σ = 0, f (u ) = 10u + 16(1 + u 1.6 )−1 − 16, I 1 (u ) = 0.0003u. The eigenvalue of (2k+1)π eigenvalue problem of (4.6) λk = ( k2π )2 , k = 1, 2, . . . . Corresponding eigenfunction is sin( x). f  (u ) = 10 − 1.62 × 2 0.6

10 (1+uu 1.6 )2 ∈ (λ2 , λ3 ) with |u | sufficiently large. let a = 9.9, b = 10. So (H1) is satisfied for |u | sufficiently large. By computa-

tion M = 0.0003926, | I 1 (u )| = 0.0003|u |  M |u |. So (H2) is satisfied. Clearly f  (u ) > 0 for u ∈ R and I 1 (u ) = 0.0003 > 0. So f and I 1 are increasing about u and (H4) is satisfied. Let φ(x) = 0. φ(x) is a lower solution. Let



ψ(x) =

3x − 3/2x2 , 2x − x2 ,

x ∈ [0, 1/2], x ∈ (1/2, 1].

ψ(x) is an upper solution with ψ(x) > 0 on [0, 1]. Thus (H3) is satisfied. By Theorem 1.2, problem (4.6) has at least four solutions with u 1 < φ , u 2 > ψ , 0 < u 3 < ψ , u 4 is sign-changing. Remark 4.1. The main results in [12,37–39] cannot be used to problem (4.6) because the nonlinearity f and I 1 do not satisfy the assumptions in the literature. Appendix A In this appendix, we give the proof of Lemma 3.2. Proof of Lemma 3.2. For any h ∈ H ,

Φ(u + th) − Φ(u ) t

=

1

1 p (x)

2

|u  (x) + th (x)|2 − |u  (x)|2 t

+ q(x)

|u (x) + th(x)|2 − |u (x)|2 t

dx

0

+

σ p (1) |u (1) + th(1)|2 − |u (1)|2 β p (0) |u (0) + th(0)|2 − |u (0)|2 + 2γ t 2α t 1

−

F (x, u (x) + th(x)) − F (x, u (x)) t

dx −

u ( x j +1 )  l 

I j (s) ds.

(A.1)

j =1 u ( x ) j

0

By mean value theorem for integration

Φ(u + th) − Φ(u ) t

=

1

1

2









p (x) 2u  (x) + th (x) h (x) + q(x) 2u (x) + th(x) h(x) dx

0

+

  σ p (1)  β p (0)  2u (1) + th(1) h(1) + 2u (0) + th(0) h(0) 2γ 2α 1

−

 







h(x) f x, u (x) + θ1 th(x) − f x, u (x)

dx

0

1 −





f x, u (x) h(x) dx −

l  



I j u (x j ) + θ th(x j ) h(x j )

j =1

0

1





p (x)u (x)h (x) + q(x)u (x)h(x) dx +

= 0

1

1



2

p (x)t h (x)

2

+ q(x)th2 (x) dx

0

2 β p (0) β p (0)(h(0))2 t σ p (1) σ p (1)  + u (1)h(1) + t h(1) + u (0)h(0) + γ 2γ α 2α 1 −

 







h(x) f x, u (x) + θ1 th(x) − f x, u (x)

dx

0

1 − 0

for 0 < θ1 < 1.





f x, u (x) h(x) dx −

l  



I j u (x j ) + θ th(x j ) h(x j )

j =1

(A.2)

486

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

For any h ∈ H , by the definition of inner product



1

(u − Au , h) = u −

l     G (x, s) f s, u (s) ds − G (x, x j ) I j u (x j ) , h



j =1

0

1 =

p (x)u  (x)h (x) + q(x)u (x)h(x) dx +

β p (0) σ p (1) u (1)h(1) + u (0)h(0) γ α

0

1

1 −

p (x) 0

1 − 0

−

1

1     G x (x, s) f s, u (s) ds h (x) dx − q(x) G (x, s) f s, u (s) ds h(x) dx 



0 l 

0





p (x)G x (x, x j ) I j u (x j ) h (x) dx −

j =1

1

1





G (1, s) f s, u (s) ds −

l 

β p (0)h(0)

1

σ p (1)h(1) γ







G (0, s) f s, u (s) ds

α

0

−



q(x)G (x, x j ) I j u (x j ) h(x) dx

j =1

0

σ p (1)h(1) γ

0

0

l 



β p (0)h(0)  l



G (1, x j ) I j u (x j ) −

α

j =1





G (0, x j ) I j u (x j ) .

j =1

By integrating by parts, Lemma 2.1, one has

1 (u − Au , h) =

p (x)u  (x)h (x) + q(x)u (x)h(x) dx +

β p (0) σ p (1) u (1)h(1) + u (0)h(0) γ α

0

1 − p (1)h(1)





G x (1, s) f s, u (s) ds + p (0)h(0)

0

+

l 

1

1



  p (x j )G x (x j , s)h(x j ) f s, u (s) ds +

1 1

j =1 0

0

1



−

1 =





G (0, s) f s, u (s) ds 0





p (1)G x (1, x j ) I j u (x j ) h(1) +

l 





p (0)G x (0, x j ) I j u (x j ) h(0)

j =1

l l  



  p (x j )G x (x j , xk ) I k u (xk ) h(x j ) + 

j =1 k =1

−

1

α

j =1

+

  LG (x, s) f s, u (s) ds h(x) dx 

0

 β p (0)h(0) G (1, s) f s, u (s) ds −

0 l 



0



σ p (1)h(1) − γ



G x (0, s) f s, u (s) ds

σ p (1)h(1) γ

1  l k =1

0 l 





G (1, xk ) I k u (xk ) −

k =1

p (x)u  (x)h (x) + q(x)u (x)h(x) dx +

 LG (x, xk ) I k u (xk ) h(x) dx 

β p (0)h(0)  l

α



k =1

σ p (1) β p (0) u (1)h(1) + u (0)h(0) γ α

0 l  

1

+

j =1 0

  p (x j )G x (x j , s)h(x j ) f s, u (s) ds − 





G (0, xk ) I k u (xk )

1





f x, u (x) h(x) dx 0

(A.3)

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

+

l l  



  p (x j )G x (x j , xk ) I k u (xk ) h(x j ) + 

j =1 k =1

1  l 0

487

 LG (x, x j ) I j u (x j ) h(x) dx. 

(A.4)

k =1

By Lemma 2.1(viii)

1  l 0

 LG (x, x j ) I j u (x j ) h(x) dx = 0. 

(A.5)

k =1

By the expression of Green function l  

1

l     p (x j )G x (x j , s)h(x j ) f s, u (s) ds =

j =1 0

1





p (x j )G x (x j , s)h(x j ) f s, u (s) ds = 0.

(A.6)

j =1 0

By Lemma 2.1(ix) l l  

l l  l          p (x j )G x (x j , xk ) I k u (xk ) h(x j ) = p (x j )h(x j )G x (x j , xk ) I k u (xk ) = − h (x j ) I j u (x j ) .

j =1 k =1

j =1 k =1

(A.7)

j =1

Substituting (A.5), (A.6), (A.7) into (A.4), one has

1 (u − Au , h) =

p (x)u  (x)h (x) + q(x)u (x)h(x) dx +

β p (0) σ p (1) u (1)h(1) + u (0)h(0) γ α

0

1 −





f x, u (x) h(x) dx −

l  



I j u (x j ) h(x j ).

j =1

0

By (A.2) (A.8)

    Φ(u + th) − Φ(u )  → 0 as t → ∞.  − ( u − Au , h )   t Therefore, grad Φ(u ) = u − Au. Theorem A.1. Let λ =

λk +λk+1 2

2

. The operators (− L − λ)−1 and (− L )−1 satisfy

(− L − λ)−1 (− L )−1 = (− L )−1 (− L − λ)−1 . Proof. The proof is divided into two steps. Step 1. (− L − λ)(− L ) = (− L )(− L − λ). ∀u ∈ C ([0, 1], R ), there exists g (·, u ) ∈ C ([0, 1] × R , R ) satisfying (− L − λ)(− L )u = g (·, u ). Then

(− L − λ)(− L − λ)u + λ(− L − λ)u = g (·, u ), i.e. (− L )(− L − λ)u = g (·, u ). So (− L − λ)(− L ) = (− L )(− L − λ). Step 2. [(− L − λ)(− L )]−1 = (− L )−1 (− L − λ)−1 ; [(− L )(− L − λ)]−1 = (− L − λ)−1 (− L )−1 . ∀ g ∈ C ([0, 1] × R , R ), [(− L − λ)(− L )]−1 g are all solutions of (− L − λ)(− L )x = g. Let x ∈ [(− L − λ)(− L )]−1 g satisfying (− L − λ)(− L )x = g. So (− L )x ∈ (− L − λ)−1 g, and then x ∈ (− L )−1 (− L − λ)−1 g. On the other hand, ∀x ∈ (− L )−1 (− L − λ)−1 g, by (− L )x ∈ (− L − λ)−1 g, one has

(− L − λ)(− L )x = g . So x ∈ [(− L − λ)(− L )]−1 g. Therefore, [(− L − λ)(− L )]−1 = (− L )−1 (− L − λ)−1 . Similarly, [(− L )(− L − λ)]−1 = (− L − λ)−1 (− L )−1 holds. 2

(A.8)

488

Y. Tian, W. Ge / J. Math. Anal. Appl. 387 (2012) 475–489

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