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In this paper, we prove existence and multiplicities of solutions for asymptotically linear ordinary differential equations satisfying Sturm-Liouville boundary value conditions with resonance. Adding assumption H
_{3}
that is similar to (LL) in Theorem 1.1, by index theory and Morse theory, we obtain more nontrivial solutions.

In this paper, we investigate the nontrivial solutions of asymptotically linear ordinary differential equations satisfying Sturm-Liouville BVPs with resonance. Various boundary value problems of asymptotically linear ordinary differential equations have been studied before. Most of them are gotten by the topological degree theory. There are also some papers about resonant problem. But the asymptotically linear ordinary differential equations with resonance aren’t concerned ago. Here, we concern asymptotically linear ordinary differential equations satisfying both Sturm-Liouville boundary value and resonance. We solve the problem to get Theorem 1.1 in the following section.

Now, we consider solutions of the following Sturm-Liouville boundary value problem:

x ″ ( t ) + f ( t , x ( t ) ) = 0, (1.1)

x ( 0 ) cos α − x ′ ( 0 ) sin α = 0, (1.2)

x ( 1 ) cos β − x ′ ( 1 ) sin β = 0, (1.3)

where f ∈ C 1 ( [ 0 , 1 ] × R , R ) , 0 ≤ α < π , 0 < β ≤ π . In this paper, f ′ denotes the derivative with respect to x. Our main result is the following theorem.

Theorem 1.1 Assume that a 0 ∈ H 0 ( 1, ( 0,1 ) , π 2 , π 2 ) in [

and (1.2)-(1.3) has one nontrivial solution x = x 0 ( t ) > 0 , ∀ t ∈ [ 0,1 ] , λ ∈ L ∞ [ 0,1 ] . f satisfies the following two conditions:

(H_{1}) a 0 ( t ) ≤ f ( t , x ) x < λ ( t ) as | x | ≥ r > 0 , r is a constant, i ( λ ) = 1 , v ( λ ) = 0 , i ( a 0 ) = 0 , v ( a 0 ) = 1 ;

(H_{2}) f ( t ,0 ) ≡ 0 , 1 ∈ ¯ [ i ( λ 0 ) , i ( λ 0 ) + v ( λ 0 ) ] , where λ 0 ( t ) : = f ′ ( t , 0 ) ;

For the sake of convenience, we denote f 1 ( t , x ) = f ( t , x ) − a 0 ( t ) x ,

(H_{3})

∫ 0 1 f 1 ( t , + ∞ ) x 0 d t > 0 > ∫ 0 1 f 1 ( t , − ∞ ) x 0 d t ,

where f 1 ( t , + ∞ ) = lim inf x → + ∞ f 1 ( t , x ) , f 1 ( t , − ∞ ) = lim sup x → + ∞ f 1 ( t , x ) , and x 0 ( t ) ( > 0 ) is a nontrivial solution of x ″ + a 0 ( t ) x = 0 and (1.2)-(1.3), ∀ t ∈ [ 0,1 ] . Then (1.1-1.3) has at least one nontrivial solution. Moreover, if we assume

(H_{4}) v ( λ 0 ) = 0 , i ( λ 0 ) ≥ 2 .

Then (1.1-1.3) has at least two nontrivial solutions.

In this paper, for any a ∈ L ∞ [ 0,1 ] , i ( a ) and v ( a ) denote its index and nullity of the associated linear ordinary differential equation (see [

i ( λ ) = { 0 as λ ≤ 0, k + 1 as λ ∈ ( k 2 π 2 , ( k + 1 ) 2 π 2 ] ,

and

v ( λ ) = { 0 as λ ≠ k 2 π 2 , 1 as λ = k 2 π 2 , k ∈ N + ∪ { 0 } .

In [

It is well known [

( 2 k π ) 2 + δ ≤ f ( t , x ) x ≤ ( 2 ( k + 1 ) π ) 2 − δ , as | x | > r > 0 , k ∈ N ,

the existence of solutions of a second order nonlinear ordinary differential equation. Such conditions are called nonresonant. Resonant conditions in [

( 2 k π ) 2 ≤ f ( t , x ) x ≤ ( 2 ( k + 1 ) π ) 2 , as | x | > r > 0 , k ∈ N ,

are not enough for existence solutions of (1.1-1.2). An additional condition called the (LL) condition like (H_{3}) is usually needed. For resonant conditions, we refer to [

In [

In this paper, we study the existence of equations with resonance conditions. In order to prove our theorem, we construct the corresponding functional:

φ ( x ) = 1 2 [ ∫ 0 1 | x ′ ( t ) | 2 d t − ( x ( 1 ) , x ( 1 ) ) k ( β ) + ( x ( 1 ) , x ( 1 ) ) k ( α ) ] − ∫ 0 1 F ( t , x ( t ) ) d t , ∀ x ∈ E . (1.4)

where F ( t , u ) = ∫ 0 u f ( t , s ) d s , u ∈ R , k ( t ) = cot t = cos t sin t , k ( t ) = 0 as t = 0 or

π, and E will be described in Section 2. This functional φ ( x ) is continuous differentiable on E, and any critical point of φ corresponds to a solution of (1.1)-(1.3).

In Section 3, we will give proofs by the Morse theory following [

For any a ∈ L ∞ [ 0,1 ] , consider the following equation:

{ x ″ ( t ) + a ( t ) x = 0 , x ( 0 ) cos α − x ′ ( 0 ) sin α = 0 , x ( 1 ) cos β − x ′ ( 1 ) sin β = 0 , (2.1)

where 0 ≤ α < π , 0 < β ≤ π . Define a Hilbert space E : = E α , β . Here E α , β = H 1 [ 0 , 1 ] as α , β ∈ ( 0, π ) ; E 0 , β = { x ∈ H 1 [ 0 , 1 ] | x ( 1 ) = 0 } as β ∈ ( 0, π ) ; E α , 0 = { x ∈ H 1 [ 0 , 1 ] | x ( 0 ) = 0 } as α ∈ ( 0, π ) and

E 0 , π = { x ∈ H 1 [ 0 , 1 ] | x ( 1 ) = 0 = x ( 0 ) } . With norm ‖ x ‖ E : = { ∫ 0 1 | x ( t ) | 2 + | x ′ ( t ) | 2 d t } 1 2 and a bilinear form as q a ( ⋅ , ⋅ ) as follows

q a ( x , y ) = ∫ 0 1 [ | x ′ ( t ) | 2 d t − ( x ( 1 ) , x ( 1 ) ) k ( β ) + ( x ( 0 ) , x ( 0 ) ) k ( α ) ] − ∫ 0 1 a ( t ) x ( t ) y ( t ) d t , ∀ x , y ∈ E . (2.2)

From proposition 2.1.1 and 2.3.3 in [

Proposition 2.1 For any a ∈ L ∞ [ 0,1 ] ,

1) The E can be divided into three parts:

E = E + ( a ) ⊕ E 0 ( a ) ⊕ E − (a)

such that q a is positive definite, null and negative definite on E + ( a ) , E 0 ( a ) and E − ( a ) respectively. Furthermore, E 0 ( A ) and E − ( A ) are finitely dimensional. We call ν ( a ) : = dim E 0 ( a ) and i ( a ) : = dim E − ( a ) the nullity and index respectively.

2) ( i ( a ) , ν ( a ) ) ∈ N × { 0,1 } .

3) ν ( a ) is the dimension of the solution subspace of (2.1-2.3), and i ( a ) = ∑ s < 0 ν ( a + s ) .

4) If a 1 ≤ a 2 , then i ( a 1 ) ≤ i ( a 2 ) and i ( a 1 ) + ν ( a 1 ) ≤ i ( a 2 ) + ν ( a 2 ) ; if a 1 < a 2 , then i ( a 1 ) + ν ( a 1 ) ≤ i ( a 2 ) .

5) There exists δ > 0 such that

q a ( u , u ) ≥ δ ‖ u ‖ E 2 , ∀ u ∈ E + ( a ) .

Remarks: 1) The notation ⊕ means that the space E is the direct sum of some subspaces.

2) By 4), we can see the index has monotonicity.

1) Let a ( t ) = 0 , α = π 2 , and β = π 2 . Then (2.1) has a nontrivial solution x = c ( ∈ R ) ≠ 0 . So ν ( 0 ) = 1 , i ( 0 ) = 0 . If a ( t ) = π 2 , α = π 2 , and β = π 2 , then (2.1) has a nontrivial solution c 1 sin π t , c 1 ≠ 0 . So ν ( π 2 ) = 1 , i ( π 2 ) = 1 . If a ( t ) = k 2 π 2 , α = π 2 , and β = π 2 , then (2.1) has a nontrivial solution c k 1 sin k π t , c k 1 ≠ 0 . So by Proposition 2.1 (3), i ( k 2 π 2 ) = ∑ s < 0 ν ( k 2 π 2 + s ) = ∑ n = 1 k − 1 ν ( n 2 π 2 ) = 1 + ( k − 1 ) = k , ν ( k 2 π 2 ) = 1 .

The following lemmas are useful for us to prove the results.

Lemma 2.2 The norm ‖ x ‖ C : = sup 0 ≤ t ≤ 1 | x ( t ) | ≤ C ∗ ‖ x ‖ E , for any x ∈ E , where C * ( ∈ R ) is a positive constant.

Lemma 2.3 If (H_{1}) holds, then we have that E = span { x 0 } ⊕ E + ( λ ) , where x 0 is given in Theorem 1.1.

Proof By Proposition 2.1 (1) and conditions ν ( λ ) = 0 , i ( λ ) = 1 , we have that

E 0 ( λ ) = { θ } , dim E − ( λ ) = 1. (2.3)

By Proposition 2.1 (1) and (3), we know that with respect to λ ∈ L ∞ [ 0,1 ] , the following decomposition holds,

E = E − ( λ ) ⊕ E 0 ( λ ) ⊕ E + ( λ ) = E − ( λ ) ⊕ E + ( λ ) . (2.4)

Since E − ( λ ) is one dimensional space, we can assume that { e − } is a base of E − ( λ ) , i.e. E − ( λ ) = span { e − } . So for any x ∈ E , we have

x = x − + x + = c 0 e − + x + ,

where x − ∈ E − ( λ ) , x + ∈ E + ( λ ) , and c 0 is a constant. For x 0 ∈ E , we have the decomposition x 0 = c 1 e − + e + , where e + ∈ E + ( λ ) and c 1 is a constant. It is obvious that c 1 ≠ 0 . Indeed, if c 1 = 0 , then x 0 = e + . By definition of q λ ( ⋅ , ⋅ ) , we will have a contradiction that

q λ ( x 0 , x 0 ) = ∫ 0 1 ( x ′ 0 , x ′ 0 ) d t − ( x 0 ( 1 ) , x 0 ( 1 ) ) k ( β ) + ( x 0 ( 1 ) , x 0 ( 1 ) ) k ( α ) − ∫ 0 1 λ ( t ) x 0 2 d t = ∫ 0 1 ( a ( t ) − λ ( t ) ) x 0 2 d t < 0

and q λ ( e + , e + ) = ∫ 0 1 ( e + ′ , e + ′ ) d t − ( e + ( 1 ) , e + ( 1 ) ) k ( β ) + ( e + ( 0 ) , e + ( 0 ) ) k ( α ) − ∫ 0 1 λ ( t ) ( e + , e + ) d t ≥ 0 . So we obtain

x = c 0 e − + x + = c 0 c 1 ( c 1 e − + e + ) − c 0 c 1 e + + x + = c 0 c 1 x 0 + ( x + − c 0 c 1 e + ) .

We have proved that E = span { x 0 } ∪ E + ( λ ) . It is also obvious that span { x 0 } ∩ E + ( λ ) = { θ } . In fact that if x ( ≠ θ ) ∈ span { x 0 } ∩ E + ( λ ) , we have that on the one hand for x ∈ span { x 0 } . Then there exists a c ( ∈ R ) ≠ 0 such that x = c x 0 and

q λ ( x , x ) = ∫ 0 1 c 2 | x ′ 0 | 2 − c 2 ( x 0 ( 1 ) , x 0 ( 1 ) ) k ( β ) + c 2 ( x 0 ( 1 ) , x 0 ( 1 ) ) k ( α ) − ∫ 0 1 c 2 λ ( t ) x 0 2 d t = ∫ 0 1 c 2 ( a 0 ( t ) − λ ( t ) ) x 0 2 d t < 0 ;

on the other hand for x ∈ E + ( λ ) ,

q λ ( x , x ) = ∫ 0 1 | x ′ ( t ) | 2 d t − ( x ( 1 ) , x ( 1 ) ) k ( β ) + ( x ( 1 ) , x ( 1 ) ) k ( α ) − ∫ 0 1 λ ( t ) x 2 ( t ) d t ≥ 0.

By Proposition 2.1 (1) and (2.2), we have x = θ . This is a contradiction. So the proof is completed.

Remark: For x = c x 0 ∈ span { x 0 } , we can define ‖ x ‖ E = | c | .

In order to prove Theorem 1.1, we need some lemmas. Let X be a Hilbert space and ψ ∈ C 1 ( X , R ) . As in [

When ψ ∈ C 2 ( X , R ) and p ∈ K , we have ψ ″ ( p ) is a self-adjoint operator. We call the dimension of negative space corresponding to the spectral decomposing the Morse index of p and denote it by m − ( ψ ″ ( p ) ) , and denote by m 0 ( ψ ″ ( p ) ) = dim ( ker ψ ″ ( p ) ) . If ψ ″ ( p ) has a bounded inverse we say that p is nondegenerate.

From Theorem 3.1 in Chapter 3, Theorem 5.1, 5.2, Corollary 5.2 in Chapter 5 in [

Lemma 2.4. Assume ψ ∈ C 2 ( X , R ) satisfies the (PS) condition, ψ ′ ( θ ) = θ 1 , where θ is the zero vector in X and θ 1 is the zero vector in X * which is the dual space of X, and there is a positive integer γ such that

γ ∈ ¯ [ m − ( ψ ″ ( θ ) ) , m − ( ψ ″ ( θ ) ) + m 0 ( ψ ″ ( θ ) ) ] and H q ( X , ψ m ; R ) = δ q γ R for

some regular m < ψ ( θ ) , here δ q γ = { 1 q = γ 0 q ≠ γ . Then ψ has a critical point

p 0 ≠ θ with C γ ( ψ , p 0 ) ≠ 0 . Moreover, if θ is a nondegenerate critical point, and m 0 ( ψ ″ ( p 0 ) ) ≤ | γ − m − ( ψ ″ ( θ ) ) | , then ψ has another critical point p 1 ≠ p 0 , θ .

The following lemma is also useful for us to prove the main result.

Lemma 2.5 (Fatou’s lemma). Given a measure space ( Ω , Σ , μ ) and a set X ∈ Σ , let f n be a sequence of ( Σ , B R ≥ 0 ) -measurable non-negative functions f n : X → [ 0, + ∞ ] , where B R ≥ 0 denotes the σ-algebra of Borel sets on [ 0, + ∞ ] . Define the function f : X → [ 0, + ∞ ] by setting

f ( x ) = lim inf n → ∞ f n ( x ) , f ( x ) = lim inf n → ∞ f n ( x ) ,

for every x ∈ X . Then f is ( Σ , B R ≥ 0 ) -measurable, and

∫ X f d μ ≤ lim inf n → ∞ ∫ X f n d μ .

Remark The integrals may be finite or infinite.

The proof of Theorem 1.1 will depend on the following lemma.

Lemma 3.1 Under (H_{1}),(H_{2}), and (H_{3}), the functional φ satisfies the (PS) condition.

Proof For { x n } ⊂ E , φ ′ ( x n ) → θ , and φ ( x n ) is bounded, we shall find a convergent subsequence in E. By (1.3), for u ∈ E , we have

〈 φ ′ ( x n ) , u 〉 = ∫ 0 1 x ′ n ( t ) u ′ ( t ) d t − ( x n ( 1 ) , u ( 1 ) ) k ( β ) + ( x n ( 0 ) , u ( 0 ) ) k ( α ) − ∫ 0 1 f ( t , x n ( t ) ) u ( t ) d t . (3.1)

Next, we will prove { ‖ x n ‖ E } 1 ∞ is bounded. Indeed, it suffices to prove that ‖ x n ‖ C is bounded. By a contradiction, we assume that ‖ x n ‖ C → + ∞ , as n → ∞ .

Defining

q n ( t ) = { f ( t , x n ( t ) ) x n ( t ) | x n ( t ) | ≥ r λ ( t ) | x n ( t ) | < r , and h n ( t ) = f ( t , x n ( t ) ) − q n ( t ) x n ( t ) , (3.2)

from (H_{1}), (H_{2}) and f : [ 0,1 ] × R → R is continuous, we have

a 0 ≤ q n ( t ) ≤ λ ( t ) and | h n ( t ) | < r | λ ( t ) | + C 0 , (3.3)

where C 0 is a constant. Then we get

f ( t , x ( t ) ) = q n ( t ) x n ( t ) + h n ( t ) . (3.4)

By (3.1), it follows that

∫ 0 1 ( x ′ n ( t ) , u ′ ( t ) ) d t = ∫ 0 1 ( f ( t , x n ( t ) ) , u ( t ) ) d t + ( x n ( 1 ) , u ( 1 ) ) k ( β ) − ( x n ( 0 ) , u ( 0 ) ) k ( α ) + 〈 φ ′ ( x n ) , u 〉 . (3.5)

Assuming y n = x n ‖ x n ‖ C , by (3.4), and multiplying ‖ x n ‖ C − 1 on both sides of (3.5), we can get that

∫ 0 1 ( y ′ n ( t ) , u ′ ( t ) ) d t = ∫ 0 1 q n ( t ) y n u d t + ( y n ( 1 ) , u ( 1 ) ) k ( β ) − ( y n ( 0 ) , u ( 0 ) ) k ( α ) + ‖ x n ‖ C − 1 ( ∫ 0 1 h n ( t ) u ( t ) d t + 〈 φ ( x n ) , u 〉 ) . (3.6)

Furthermore, we add ∫ 0 1 y n ( t ) u ( t ) d t on two sides of (3.6) to obtain that

( y n , u ) E = ∫ 0 1 y ′ u ′ d t + ∫ 0 1 y n u d t = ∫ 0 1 q n ( t ) y n u d t + ∫ 0 1 y n u d t + ( y n ( 1 ) , u ( 1 ) ) k ( β ) − ( y n ( 0 ) , u ( 0 ) ) k ( α ) + ‖ x n ‖ C − 1 ( ∫ 0 1 h n ( t ) u ( t ) d t + 〈 φ ( x n ) , u 〉 ) .

So, by ‖ y n ‖ L 2 = ( ∫ 0 1 y n 2 ( t ) d t ) 1 2 ≤ ‖ y n ‖ C = 1 and (3.3) we have

‖ y n ‖ E = sup ‖ u ‖ E ≤ 1 ( y n , u ) E ≤ ∫ 0 1 ( q n ( t ) y n ( t ) u ) d t + ∫ 0 1 ( y n ( t ) u ) d t + | k ( β ) | + | k ( α ) | + C 2 ≤ C 3 ‖ y n ‖ L 2 ‖ u ‖ L 2 + | k ( β ) | + | k ( α ) | + C 2 ≤ C * ,

where C 2 , C 3 and C * are constants. So { ‖ y n ‖ E } 1 ∞ is bounded. Then { y n } has a convergent subsequence. Without loss of generality, we also denoted by { y n } . Then y n ⇀ y 0 in E and y n → y 0 in C [ 0,1 ] . By inequality a 0 ( t ) ≤ q n ( t ) ≤ λ ( t ) , we have q n ⇀ q 0 in L 2 [ 0,1 ] . Then taking the limits on both sides of (3.6), we have, for any u ∈ E ,

∫ 0 1 y ′ 0 u ′ d t − ( y 0 ( 1 ) , u ( 1 ) ) k ( β ) + ( y 0 ( 0 ) , u ( 0 ) ) k ( α ) − ∫ 0 1 q 0 ( t ) y 0 u d t = 0, ∀ u ∈ E . (3.7)

From (3.7) and [

{ y ″ ( t ) + q 0 ( t ) y = 0, a . e . t ∈ [ 0,1 ] y ( 0 ) cos α − y ′ ( 0 ) sin α = 0, y ( 1 ) cos β − y ′ ( 1 ) sin β = 0, (3.8)

What’s more, since a 0 ( t ) ≤ q n ( t ) ≤ λ ( t ) , we have q 0 = 0 . In fact, by the meaning of the notation “<” and “ ≤ ”, on the one hand, if q 0 ( t ) = λ ( t ) , then ν ( q 0 ) = ν ( λ ) = 0 . Therefore, by the definition ν ( ⋅ ) , this means that (3.8) only has a trivial solution. In fact, by ‖ y n ‖ C = 1 , we obtain ‖ y 0 ‖ C = 1 . So (3.8) has a nontrivial solution. This is a contradiction. On the other hand, if a 0 ( t ) < q 0 ( t ) < λ ( t ) , then 1 = i ( a 0 ) + ν ( a 0 ) ≤ i ( q 0 ) ≤ i ( λ ) = 1 holds. While y 0 is a nontrivial solution of (3.8), this leads ν ( q 0 ) = 1 . So by Proposition 2.1 (4), we get 2 = i ( q 0 ) + ν ( q 0 ) ≤ i ( λ ) = 1 . This is also a contradiction. From discussion above, we obtain the conclusion that q 0 = a 0 . So we immediately get y 0 ∈ span { x 0 } .

Since y 0 ∈ span { x 0 } , there are two cases about y 0 . One is that y 0 > 0 , the other is that y 0 < 0 . Without loss of generality, if y 0 > 0 , we assume y 0 = x 0 , and if y 0 < 0 , we assume y 0 = − x 0 . Firstly, we discuss the situation that y 0 > 0 . If y 0 > 0 , i.e. y 0 = x 0 , then for ∀ ε > 0 , ∃ N = N ( ε ) such that for n > N , | y n − x 0 | < ε holds. Here, we take the ε such that x 0 − ε > 0 , i.e. when n > N , { y n ( t ) } belong to the neighborhood of y 0 , ( x 0 − ε , x 0 + ε ) , for all t ∈ [ 0,1 ] . This means y n ( t ) > x 0 − ε , as n > N , for all t ∈ [ 0,1 ] .

So by y n = x n ( t ) ‖ x n ‖ C , we can get that for any t ∈ [ 0,1 ] , x n ( t ) = y n ( t ) ‖ x n ‖ C > 0

for n > N . Then x n ( t ) → + ∞ for all t ∈ [ 0,1 ] , as ‖ x n ‖ C → ∞ . By the assumption that ‖ x n ‖ C → ∞ , as n → ∞ , taking the limits on both sides of (3.1) and letting u = x 0 , we can obtain

〈 φ ′ ( x n ) , x 0 〉 = ∫ 0 1 x ′ n x ′ 0 d t − ( x n ( 1 ) , x 0 ( 1 ) ) k ( β ) + ( x n ( 0 ) , x 0 ( 0 ) ) k ( α ) − ∫ 0 1 f ( t , x n ( t ) ) x 0 d t = ∫ 0 1 ( q 0 ( t ) x n − f ( t , x n ) ) x 0 d t = − ∫ 0 1 f 1 ( t , x n ) x 0 d t . (3.9)

So, by (H_{3}), and (3.9), the following holds

∫ 0 1 f 1 ( t , x n ( t ) ) x 0 d t → 0, as n → ∞ . (3.10)

Furthermore, by the Fatou’s Lemma and (3.10), we have

0 = lim _ n → ∞ ∫ 0 1 f 1 ( t , x n ( t ) ) x 0 d t ≥ ∫ 0 1 lim _ n → ∞ f 1 ( t , x n ( t ) ) x 0 d t = ∫ 0 1 f 1 ( t , + ∞ ) x 0 d t ,

a contradiction to assumption (H_{3}). Hence, if y 0 > 0 , this leads to a contradiction. Secondly, in a similar way, we can show that if y 0 < 0 , there also be a contradiction. Therefore, the sequence { ‖ x n ‖ C } 1 ∞ is a bounded sequence. By the equality x n = y n ‖ x n ‖ C and the fact that ‖ y n ‖ E is bounded, we can get that { ‖ x n ‖ E } 1 ∞ is bounded in E. Furthermore, { x n } 1 ∞ has a weak convergent subsequence in E, without loss of generality, still denoted by { x n } 1 ∞ . So we have x n ⇀ x * in E and x n → x * in C [ 0,1 ] . In addition, by (3.1), we also have

∫ 0 1 x ′ * ( t ) u ′ ( t ) d t − ( x * ( 1 ) , u ( 1 ) ) k ( β ) + ( x * ( 0 ) , u ( 0 ) ) k ( α ) − ∫ 0 1 ( f ( t , x * ( t ) ) , u ( t ) ) d t = 0. (3.11)

At last, we only need to finish the mission that x n → x * in E. Indeed, by (3.5), (3.11) and x n ⇀ x * , we obtain the fact that

‖ x n − x * ‖ E = sup ‖ u ‖ E ≤ 1 ( x n − x * , u ) E = sup ‖ u ‖ E ≤ 1 [ ∫ 0 1 ( x ′ n − x ′ * , u ′ ) d t + ∫ 0 1 ( x n − x * , u ) d t ] = sup ‖ u ‖ E ≤ 1 { ∫ 0 1 ( f ( t , x n ) − f ( t , x * ) , u ( t ) ) d t + 〈 φ ′ ( x n ) − φ ′ ( x * ) , u 〉 + ( x n ( 1 ) − x * ( 1 ) , u ) k ( β ) − ( x * ( 0 ) − x n ( 0 ) , u ) k ( α ) + ∫ 0 1 ( x n − x * , u ) d t } → 0, as n → ∞ .

The (PS) condition is verified.

After the preliminary work, we can prove Theorem 1.1.

Proof of Theorem 1.1. Since ν ( a ) ≤ 1 for any a ∈ L ∞ [ 0,1 ] , by Lemma 2.4, we only need to prove

H q ( E , φ z ; R ) ≅ δ q γ R (3.12)

for − z > − φ ( θ ) large enough, where γ = ν ( 0 ) = i ( λ ) = 1 . By Lemma 2.3, we know that E can be split into two subspaces span { x 0 } and E + ( λ ) , i.e.

E = span { x 0 } ⊕ E + ( λ ) .

Next, we will take two steps to obtain the proof of (3.12).

First step: For − z > − φ ( θ ) large enough, we have

H q ( E , φ z ; R ) ≅ H q ( M , M ∩ φ z ; R ) , q = 0,1,2, ⋯ , (3.13)

where M ⊂ H will be defined later. By assumption, for any y ∈ E , we have

〈 φ ′ ( x ) , y 〉 = ∫ 0 1 x ′ y ′ d t − ( x ( 1 ) , y ( 1 ) ) k ( β ) + ( x ( 0 ) , y ( 0 ) ) k ( α ) − ∫ 0 1 f ( t , x ) y d t . (3.14)

We will consider the behavior of f in two subintervals of [ 0,1 ] . One is { t | x ( t ) ≥ r } ⊂ [ 0,1 ] , the other is { t | | x ( t ) | < r } ⊂ [ 0 , 1 ] . Since f is continuous on [ 0,1 ] × ( − ∞ , + ∞ ) , it is obvious that f ( t , x ( t ) ) is bounded on { t | | x ( t ) | < r } . So there exists a constant M 1 ∈ R such that | f ( t , x ( t ) ) | < M 1 when | x ( t ) | < r .

By Lemma 2.3, we have a decomposition with respect to x ( t ) ∈ E , i.e. there exist x + ( t ) ∈ E + ( λ ) , and c ∈ R such that x = x + + c x 0 . When | x + + c x 0 | < r , we have | c x 0 | < r + | x + | < r + ‖ x + ‖ C . Furthermore, we get

| c x 0 ∫ | x + + c x 0 | < r f ( t , x + + c x 0 ) d t | < M 1 ( r + ‖ x + ‖ C ) . (3.15)

So by (15), we have

∫ | x + + c x 0 | < r f ( t , x + + c x 0 ) ( x + − c x 0 ) d t = ∫ | x + + c x 0 | < r f ( t , x + + c x 0 ) x + d t − c x 0 ∫ | x + + c x 0 | < r f ( t , x ) d t ≤ M 1 ‖ x + ‖ C + M 1 ( r + ‖ x + ‖ C ) . (3.16)

By (H_{1}), we have

∫ | x + + c x 0 | ≥ r f ( t , x + + c x 0 ) ( x + − c x 0 ) d t = ∫ | x + + c x 0 | ≥ r f ( t , x + + c x 0 ) x + + c x 0 ( x + + c x 0 ) ( x + − c x 0 ) d t = ∫ | x + + c x 0 | ≥ r f ( t , x + + c x 0 ) x + + c x 0 ( x + 2 − c 2 x 0 2 ) d t = ∫ | x + + c x 0 | ≥ r f ( t , x + + c x 0 ) x + + c x 0 x + 2 d t − ∫ | x + + c x 0 | ≥ r c 2 x 0 2 f ( t , x ) x d t ≤ ∫ | x + + c x 0 | ≥ r λ ( t ) | x + | 2 d t − ∫ | x + + c x 0 | ≥ r c 2 a 0 ( t ) x 0 2 d t ≤ ∫ 0 1 λ ( t ) | x + | 2 d t + M ′ , (3.17)

where M ′ is a constant. By (3.14), (3.16), (3.17), Proposition 2.1 (5) and Lemma 2.3, we obtain

〈 φ ′ ( x ) , x + − c x 0 〉 = ∫ 0 1 ( x ′ + + c x ′ 0 , x ′ + − c x ′ 0 ) d t − ( x + ( 1 ) + c x 0 ( 1 ) , x + ( 1 ) − c x 0 ( 1 ) ) k ( β ) + ( x + ( 0 ) + c x 0 ( 0 ) , x + ( 0 ) − c x 0 ( 0 ) ) k ( α ) − ∫ 0 1 f ( t , x + + c x 0 ) ( x + − c x 0 ) d t = ∫ 0 1 | x ′ + | 2 d t − ( x + ( 1 ) , x + ( 1 ) ) k ( β ) + ( x + ( 0 ) , x + ( 0 ) ) k ( α ) − c 2 [ ∫ 0 1 | x ′ 0 | 2 d t − ( x 0 ( 1 ) , x 0 ( 1 ) ) k ( β ) + ( x 0 ( 0 ) , x 0 ( 0 ) ) k ( α ) ]

− { ∫ | x + + c x 0 | ≥ r f ( t , x + + c x 0 ) ( x + − c x 0 ) d t + ∫ | x + + c x 0 | < r f ( t , x + + c x 0 ) ( x + − c x 0 ) d t } ≥ ∫ 0 1 | x ′ + | 2 d t − ( x + ( 1 ) , x + ( 1 ) ) k ( β ) + ( x + ( 0 ) , x + ( 0 ) ) k ( α ) − ∫ 0 1 c 2 a 0 ( t ) x 0 2 d t − ∫ 0 1 λ ( t ) x + 2 d t − M ′ − M 1 ‖ x + ‖ C − M 1 ( r + ‖ x + ‖ C ) = q λ ( x + , x + ) − 2 M 1 ‖ x + ‖ C − r M 1 − ∫ 0 1 c 2 a 0 ( t ) x 0 2 d t − M ′ ≥ C 1 ‖ x + ‖ E 2 − C 4 ‖ x + ‖ E − r M 1 + M 0 ,

where C 1 ( > 0 ) , C 4 ( > 0 ) and M 0 are constants. And hence, there exists R 0 > 0 such that

〈 φ ′ ( x ) , x + − c x 0 〉 > 1, ∀ x ∈ E with ‖ x + ‖ E > R 0 .

Set M = ( E + ( λ ) ∩ B R 0 ) ⊕ span { x 0 } , where B R 0 = { x ∈ E | ‖ x ‖ E ≤ R 0 } . We want to define a deformation from ( E , φ z ) to ( M , M ∩ φ z ) . Since for every x = x + + c x 0 ∈ M , f is decreasing along vector field V ( x ) = − x + + c x 0 , we can define the flow σ = σ ( t , x ) = e − t x + + e t c x 0 and T x = ln ( ‖ x + ‖ E ) − ln R 0 , which is the first time that σ ( t , x ) arrives at M . Then the deformation is

η ( t , x + + c x 0 ) = { x + + c x 0 , ‖ x + ‖ E ≤ R 0 , σ ( T x t , x ) , ‖ x + ‖ E > R 0 .

One can verify that η : [ 0,1 ] × E → E is continuous and satisfy

η ( 0, ⋅ ) = i d E , η ( 1, E ) ⊂ M , η ( 1, φ z ) ⊂ M ∩ φ z , η ( t , φ z ) ⊂ φ z , η ( t , ⋅ ) | M = i d M ∀ t ∈ [ 0,1 ] .

Then, ( M , M ∩ φ z ) is a deformation retract of ( E , φ z ) . So (3.13) is verified.

Second step: we will prove the following

H q ( M , M ∩ φ z ; R ) ≅ δ q γ R , (3.18)

for any − z > − φ ( θ ) large enough. In fact, assuming that | x ( t ) | = | x + + c x 0 | ≥ r , by (H_{1}), we will have two cases: one is f 1 ( t , x ) ≥ 0 as x ≥ r , another is f 1 ( t , x ) ≤ 0 as x ≤ − r . Firstly, we analyze the situation that f 1 ( t , x ) ≥ 0 as x = x + + c x 0 ≥ r . Since f 1 ( t , + ∞ ) x 0 = lim inf x → + ∞ f 1 ( t , x ) x 0 = lim inf x → + ∞ y ≥ x f 1 ( t , y ) x 0 and inf y ≥ x f 1 ( t , y ) x 0 is a monotonically increasing nonnegative function with respect to x ≥ r > 0 , by (H_{3}), we have

∫ 0 1 f 1 ( t , ∞ ) x 0 d t = ∫ 0 1 lim inf x → + ∞ y ≥ x f 1 ( t , y ) x 0 d t = lim x → + ∞ ∫ 0 1 inf y ≥ x f 1 ( t , y ) x 0 d t > 0.

Then ∃ x * ( ≥ r ) ∈ R , for all l > x * , ∫ 0 1 inf y ≥ l f 1 ( t , y ) x 0 d t > 0 holds. What’s more, since x 0 ( t ) > 0 for all t ∈ [ 0,1 ] . So there exists a M : = sup t ∈ [ 0 , 1 ] | x 0 ( t ) | > 0 , such that M ∫ 0 1 f 1 ( t , y ) d t ≥ ∫ 0 1 inf y ≥ l f 1 ( t , y ) x 0 d t > 0 , i.e. ∫ 0 1 inf y ≥ l f 1 ( t , y ) d t > 0 .

So letting x = x + + c x 0 > l , where c ∈ R , l is fixed and l > x * ≥ r , we have

F ( t , x ) = ∫ 0 x f ( t , s ) d s = ∫ 0 x + + c x 0 [ f ( t , s ) − a 0 ( t ) s ] d s + 1 2 a 0 ( t ) ( x + + c x 0 ) 2 = ∫ 0 l [ f ( t , s ) − a 0 ( t ) s ] d s + ∫ l x + + c x 0 [ f ( t , s ) − a 0 ( t ) s ] d t + 1 2 a 0 ( t ) x 2 ≥ M 4 + ( x + + c x 0 − l ) inf x + + c x 0 ≥ y ≥ l f 1 ( t , y ) + 1 2 a 0 ( t ) x 2 ≥ M 4 + ( x + + c x 0 − l ) inf y ≥ l f 1 ( t , y ) + 1 2 a 0 ( t ) x 2 . (3.19)

Furthermore, by (3.19), we obtain

φ ( x ) = 1 2 [ ∫ 0 1 | x ′ | 2 d t − ( x ( 1 ) , x ( 1 ) ) k ( β ) + ( x ( 0 ) , x ( 0 ) ) k ( α ) ] − ∫ 0 1 F ( t , x ) d t = 1 2 [ ∫ 0 1 ( x ′ + 2 + 2 c x ′ + x ′ 0 + c 2 x ′ 0 2 ) d t − ( x + ( 1 ) + c x 0 ( 1 ) ) 2 k ( β ) + ( x + ( 0 ) + c x 0 ( 0 ) ) 2 k ( α ) ] − ∫ 0 1 F ( t , x ) d t

≤ 1 2 ∫ 0 1 ( x ′ + 2 + 2 c x ′ + x ′ 0 + c 2 x ′ 0 2 ) d t − x + ( 1 ) 2 k ( β ) + x + ( 0 ) 2 k ( α ) − 1 2 ∫ 0 1 a 0 ( t ) ( x + 2 + 2 c x + x 0 + c 2 x 0 2 ) d t − 1 2 c 2 x 0 ( 1 ) 2 k ( β ) + 1 2 c 2 x 0 ( 0 ) 2 k ( α ) − 1 2 c x + ( 1 ) x 0 ( 1 ) k ( β ) + 1 2 c x + ( 0 ) x 0 ( 0 ) k ( α ) − M 4 − ( x + + c x 0 − l ) ∫ 0 1 inf y ≥ l f 1 ( t , y ) d t

= 1 2 [ ∫ 0 1 x ′ + 2 − a 0 ( t ) x + 2 d t − x + ( 1 ) 2 k ( β ) + x + ( 0 ) 2 k ( α ) ] + c ∫ 0 1 x ′ + x ′ 0 − a 0 ( t ) x + x 0 d t − c x + ( 1 ) x 0 ( 1 ) k ( β ) + c x + ( 0 ) x 0 ( 0 ) k ( α ) + c 2 2 [ ∫ 0 1 ( x ′ 0 2 − a 0 ( t ) x 0 2 ) d t − x 0 ( 1 ) 2 k ( β ) + x 0 ( 0 ) 2 k ( α ) ] − M 4 − ( x + + c x 0 − l ) ∫ 0 1 inf y ≥ l f 1 ( t , y ) d t

= 1 2 ∫ 0 1 ( x ′ + 2 − a 0 ( t ) x + 2 ) d t − x + ( 1 ) 2 k ( β ) + x + ( 0 ) 2 k ( α ) − M 4 − ( x + + c x 0 − l ) ∫ 0 1 inf y ≥ l f 1 ( t , y ) d t .

So we get φ ( x ) → − ∞ , as c → + ∞ , uniformly in x + ∈ E + ( λ ) ∩ B R 0 . Secondly, we analyze the situation that f 1 ( t , x ) ≤ 0 as x = x + + c x 0 ≤ − r . In a similar way, we also get φ ( x ) → − ∞ , as c → − ∞ , uniformly in x + ∈ E + ( λ ) ∩ B R 0 . So we obtain that

φ ( x ) → − ∞ ⇔ | c | → + ∞ uniformly in x + ∈ E + ( λ ) ∩ B R 0 .

Thus, there exist T > 0 , z 1 < z 2 < − T , R 1 > R 2 > R 0 such that

( E + ( λ ) ∩ B R 0 ) ⊕ ( span { x 0 } \ B ′ R 1 ) ⊂ φ z 1 ∩ M ⊂ ( E + ( λ ) ∩ B R 0 ) ⊕ ( span { x 0 } \ B ′ R 2 ) ⊂ φ z 2 ∩ M , (3.20)

where B ′ R = { c x 0 | c ∈ R and ‖ c x 0 ‖ E = | c | ≤ R } by remark. For the sake of convenience, we set N R = ( E + ( λ ) ∪ B R 0 ) ⊕ ( span { x 0 } \ B ′ R ) . Then (3.20) can also be denoted as

N R 1 ⊂ φ z 1 ∩ M ⊂ N R 2 ⊂ φ z 2 ∩ M .

We now begin to define a deformation from M ∩ φ z 2 to M ∩ φ z 1 . For every x ∈ M ∩ ( φ z 2 \ φ z 1 ) , since the flow is defined by σ ( t , x ) = e − t x + + e t c x 0 , φ ( σ ( t , x ) ) is continuous with respect to t, φ ( σ ( 0 , x ) ) = φ ( x ) > z 1 and φ ( σ ( t , x ) ) → − ∞ as t → + ∞ , so the time t = T 1 ( x ) arriving at φ z 1 ∩ M exists uniquely and is defined by φ ( σ ( t , x ) ) = z 1 . Since

d φ ( σ ( t , x ) ) d t = 〈 φ ′ ( σ ( t , x ) ) , σ ′ ( t , x ) 〉 = 〈 φ ′ ( e − t x + + e t c x 0 ) , − e − t x + + e t c x 0 〉 ≤ − 1

as t > 0 , the continuity of t = T 1 ( x ) comes from the implicit function theorem.

Define

η 1 ( t , x ) = x , ∀ x ∈ φ z 1 ∩ M = σ ( T 1 ( x ) t , x ) , ∀ x ∈ M ∩ ( φ z 2 \ φ z 1 )

then η 1 : [ 0,1 ] × φ z 2 ∩ M → φ z 2 ∩ M is continuous, and is a deformation from φ z 2 ∩ M to φ z 1 ∩ M and τ 1 = η ( 1, ( ⋅ ) ) : φ z 2 ∩ M → φ z 1 ∩ M is a strong deformation retract. Hence,

H q ( φ z 2 ∩ M , φ z 1 ∩ M ; R ) ≅ 0. (3.21)

Recall that for any topological spaces Z ⊆ Y ⊆ X , we have exact sequences

H q ( Y , Z ; R ) → H q ( X , Z ; R ) → H q ( X , Y ; R ) → H q − 1 ( Y , Z ; R ) .

From (3.20), in order to prove

H q ( M , φ z 2 ∩ M ; R ) ≅ H q ( M , N R 1 ; R ) (3.22)

we only to prove

H q ( φ z 2 ∩ M , N R 1 ; R ) ≅ 0.

And from (3.21), it suffices to verify

H q ( φ z 1 ∩ M , N R 1 ; R ) ≅ 0.

Let τ 2 : [ 0,1 ] × N R 2 → N R 2 satisfy

τ 2 ( t , x + + c x 0 ) = x + + c x 0 , | c | > R 1 , = x + + c | c | ( t R 1 + ( 1 − t ) c ) x 0 , R 2 < | c | ≤ R 1 .

We can verify that τ : = τ 1 ∘ τ 2 : [ 0 , 1 ] × φ z 1 ∩ M → φ z 1 ∩ M is continuous, where τ 1 ∘ τ 2 ( t , x ) = τ 1 ( τ 2 ( t , x ) ) , t ∈ [ 0 , 1 ] , x ∈ φ z 1 ∩ M , and satisfies

τ [ 0 , x ] = η 1 ( 1 , τ 2 ( 0 , x ) ) = η 1 ( 1 , x + + c x 0 ) = x + + c x 0 = x

for any x ∈ φ z 1 ∩ M . So τ [ 0, ⋅ ] = i d φ z 1 ∩ M . And

τ [ t , x + + c x 0 ] = τ 1 ∘ τ 2 [ t , x ] = η 1 ( 1, x ) = x ,

for any x ∈ N R 1 . So τ [ t , ⋅ ] | N R 1 = i d N R 1 . We can also see that τ satisfy τ ( 1, φ z 1 ∩ M ) ⊂ N R 1 , τ ( 1, N R 1 ) ⊂ N R 1 , τ ( t , N R 1 ) ⊂ N R 1 . Then ( N R 1 , N R 1 ) is a deformation retract of ( φ z 1 ∩ M , N R 1 ) . This means (3.22) and hence (3.21) holds. Finally from (3.21) we have

H q ( M , M ∩ φ z 2 ; R ) ≅ H q ( ( E + ( λ ) ∩ B R 0 ) ⊕ span { x 0 } , ( E + ( λ ) ∩ B R 0 ) ⊕ ( span { x 0 } \ B ′ R 1 ) ; R ) ≅ H q ( span { x 0 } ∩ B ′ R 1 ; ∂ ( span { x 0 } ∩ B ′ R 1 ) ; R ) = δ q γ R , q = 0 , 1 , 2 , 3 , ⋯ .

Here in the second ≅ we used the deformation ζ : [ 0,1 ] × M → M defined by ζ ( t , x ) = t x + + c x 0 , and excision property. So (3.18) is proved. And by (3.13) and (3.18), (3.12) is obtained. The proof is completed.

In our theorem, we get one nontrivial solution of Equations (1.1)-(1.3). By adding assumption (H_{4}), we get two nontrivial solutions of Equations (1.1)-(1.3).

By index theories established in this paper, and Morse theory, we study the functional corresponding to the problem to obtain more nontrivial solutions of asymptotically linear ordinary differential equations satisfying Sturm-Liouville BVPs with resonance. It’s better than the results obtained by topological degree method.

The authors would like to express their sincere thanks to the editors and reviewers for their remarkable comments, suggestions, and ideas that helped to improve this paper.

This research work was partially supported by the National Science Foundation of China (11501178).

The authors declare that they have no competing interests.

Li, K.Q. and Tong, R.H. (2019) Existence and Multiplicities of Solutions for Asymptotically Linear Ordinary Differential Equations Satisfying Sturm-Liouville BVPs with Resonance. Journal of Applied Mathematics and Physics, 7, 1197-1211. https://doi.org/10.4236/jamp.2019.75080