Non-homogeneous BVPs for second-order symmetric Hamiltonian systems

Yingying Chen; Yujun Dong; Baiqian Wang

doi:10.1515/math-2024-0112

Article Open Access

Non-homogeneous BVPs for second-order symmetric Hamiltonian systems

Yingying Chen , Yujun Dong and Baiqian Wang

Published/Copyright: December 31, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

By making use of Bolle’s method, we show that the following problem has infinitely many solutions:

x ¨ + V ′ ( x ) = 0 , x ( 0 ) cos α − x ˙ ( 0 ) sin α = x 0 , x ( 1 ) cos β − x ˙ ( 1 ) sin β = x 1 ,

where α , β ∈ ( 0 , π ) x 0 , x 1 ∈ R n are given and V ∈ C 2 ( R n , R ) is even and is super-quadratic at infinity.

Keywords: Sturm-Liouville; Hamiltonian system; non-symmetric

MSC 2010: 34B15; 34B24; 34L11

1 Introduction

In 1996, Ekeland et al. [1] investigated the problem

(1) x ¨ ( t ) + V ′ ( x ) = 0 , x ( 0 ) = x 0 , x ( 1 ) = x 1 ,

where x 0 and x 1 ∈ R n are fixed points and V : R n → R and V ′ ( x ) denotes the gradient of V with regard to x . They have proved the following result.

Theorem 1

Suppose that V ∈ C 1 ( R n , R ) is even and that there exists some p ∈ ( 2 , 4 ) such that

0 < p V ( x ) ≤ V ′ ( x ) x for large ∣ x ∣ ,
∣ V ′ ( x ) ∣ ≤ c 1 ∣ x ∣ p − 1 + c 2 , for all x ∈ R n and for some c 1 , c 2 > 0 .

Then, (1) has infinitely many solutions.

In 1999, Bolle [2] investigated a continuous path { I θ } θ ∈ [ 0 , 1 ] of functionals with I 0 being even and having infinitely many critical levels and established an abstract theorem concerning the preservation of a min-max critical level along the path { I θ } under some suitable assumptions on ∂ ∂ θ I θ ( x ) mainly. By this abstract theorem, P. Bolle investigated (1) only under assumption ( i ) with p > 2 . In this study, we discuss the problem

(2) x ¨ ( t ) + V ′ ( x ) = 0 ,

(3) x ( 0 ) cos α − x ˙ ( 0 ) sin α = x 0 ,

(4) x ( 1 ) cos β − x ˙ ( 1 ) sin β = x 1 ,

where α , β ∈ ( 0 , π ) , x 0 , x 1 ∈ R n are given and V ∈ C 2 ( R n , R ) . We have the following theorem.

Theorem 2

Suppose that V ∈ C 2 ( R n , R ) is even and that there exists some p > 2 and r > 0 such that 0 < p V ( x ) ≤ V ′ ( x ) x for ∣ x ∣ > r . Then, systems (2)–(4) have infinitely many solutions.

We will use Bolle’s method to show Theorem 2. Precisely, in Section 2, we will prove the existence of infinitely many solutions of a class of operator equations by making use of [3, Theorem 2.2], and in Section 3, as an application of the existence result, we will prove Theorem 2.

2 Operator equations

In this section, we investigate an operator equation and obtain a multiple solution theorem, which will be used to show Theorem 2 in Section 3. As in [4], let A be an unbounded self-adjoint operator in a Hilbert space X and σ ( A ) = σ d ( A ) bounded from below. Let Z = D ( ∣ A ∣ 1 2 ) with the usual norm satisfying ‖ x ‖ Z 2 = ∣ A ∣ 1 2 x X 2 + ‖ x ‖ X 2 and Φ ∈ C 1 ( Z , R ) , and there exists ∇ Φ : Z → X , which is weakly continuous, i.e., for any sequence { x k } ∈ Z such that x k ⇀ x in Z , we have ∇ Φ ( x k ) → ∇ Φ ( x 0 ) in X such that Φ ′ ( x ) y = ( ∇ Φ ( x ) , y ) X for all x , y ∈ Z . We consider

(5) A x − ∇ Φ ( x + y 0 ) = 0 ,

where y 0 ∈ Z is a given element. We will use the assumptions:

( Φ 1 ) ∃ μ > 2 , c 1 ≥ 0 such that μ Φ ( y ) ≤ Φ ′ ( y ) y + c 1 , ∀ y ∈ Z ,

( Φ 2 ) ∃ c 2 > 0 , c 3 > 0 , μ > 2 such that Φ ( y ) ≥ c 2 ‖ y ‖ X μ − c 3 , ∀ y ∈ Z ,

( Φ 3 ) Φ ( − x ) = Φ ( x ) , ∀ x ∈ Z .

By the assumption on A , there exist an orthonormal basis { e j } j = 1 ∞ of X and a nondecreasing sequence { λ j } j = 1 ∞ such that A e j = λ j e j , j ∈ N * . Then, Z = D ( ∣ A ∣ 1 2 ) = { ∑ j = 1 ∞ ξ j e j ∣ ∑ j = 1 ∞ ∣ λ j ∣ ξ j 2 < ∞ } . And for any x = ∑ j = 1 ∞ ξ j e j ∈ Z , ‖ x ‖ Z = ( ∑ j = 1 ∞ ( 1 + ∣ λ j ∣ ) ξ j 2 ) 1 2 . Define a ( x , y ) = ∑ j = 1 ∞ λ j ξ j ζ j for x = ∑ j = 1 ∞ ξ j e j , y = ∑ j = 1 ∞ ζ j e j . Choose λ 0 = 1 + 2 ∣ λ 1 ∣ . Then, ‖ x ‖ = ( a ( x , x ) + λ 0 ‖ x ‖ X 2 ) 1 2 = ( ∑ j = 1 ∞ ( λ 0 + λ j ) ξ j 2 ) 1 2 is an equivalent norm on Z . Set Z k = { ∑ j = 1 k ξ j e j ∣ ξ j ∈ R } , let G k = { g ∈ C ( Z k , Z ) ∣ g is odd, and there exists R k > 0 such that g ( x ) = x for x ∈ Z k and ‖ x ‖ ≥ R k } and define

I ( x ) = 1 2 a ( x , x ) − Φ ( x ) , ∀ x ∈ Z .

c k = inf g ∈ G k sup x ∈ Z k I ( g ( x ) ) .

Our last assumption is the following:

( A Φ ) c k ⁄ k 2 → + ∞ as k → + ∞ .

The following theorem is the main result of this section.

Theorem 3

Assume ( Φ 1 – Φ 3 ) and ( A Φ ) hold. Then, (5) has infinitely many solutions.

To prove Theorem 3, we will use Bolle’s method. Let E be a real Hilbert space, and let I : [ 0 , 1 ] × E → R be a C 2 functional. I ′ ( θ , x ) denotes the F-derivative of I with respect to x . Assume

( H 1 ) I satisfies ( P S ) , i.e., for any ( θ k , x k ) ∈ [ 0 , 1 ] × E such that ‖ I ′ ( θ k , x k ) ‖ → 0 as k → ∞ and I ( θ k , x k ) is bounded, there is a convergent subsequence.

( H 2 ) For all b > 0 , there is a constant c 1 ( b ) > 0 such that ∣ I ( θ , x ) ∣ ≤ b implies ∣ ∂ ∂ θ I ( θ , x ) ∣ ≤ c 1 ( b ) ( ‖ I ′ ( θ , x ) ‖ + 1 ) ( ‖ x ‖ + 1 ) .

( H 3 ) There exists c > 0 such that ∣ ∂ ∂ θ I ( θ , x ) ∣ ≤ c ( ∣ I ( θ , x ) ∣ 2 + 1 ) 1 4 .

( H 4 ) I ( 0 , ⋅ ) is even, and for any finite dimensional subspace W of E , we have sup θ ∈ [ 0 , 1 ] I ( θ , y ) → − ∞ as y ∈ W and ‖ y ‖ → ∞ .

Assume E 1 ⊂ E 2 ⊂ … ⊂ E k ⊂ … are the subspaces of E and dim E k = k . Set G k = { g ∈ C ( E k , E ) ∣ g is odd and g ( x ) = x for x ∈ E k and ‖ x ‖ large } , and

c k ′ = inf g ∈ G k sup x ∈ E k I ( 0 , g ( x ) ) .

Lemma 1

[3] Assume that I satisfies ( H 1 – H 4 ). Then, there exists c > 0 such that for every k ∈ N * :

either I ( 1 , x ) has a critical level c ¯ k with c ¯ k > c k ′ ;
or c k + 1 ′ − c k ′ ≤ c ( ∣ c k + 1 ′ ∣ 1 2 + ∣ c k ′ ∣ 1 2 + 1 ) .

Proof of Theorem 3

Let I : [ 0 , 1 ] × Z → R be defined by

(6) I ( θ , x ) = 1 2 a ( x , x ) − Φ ( x + θ y 0 ) , ∀ θ ∈ [ 0 , 1 ] , x ∈ Z .

Note that for ν ∈ ( 2 , μ ) ,

I ( θ , x ) − 1 ν I ′ ( θ , x ) x = 1 2 − 1 ν a ( x , x ) + 1 ν Φ ′ ( x + θ y 0 ) ( x + θ y 0 ) − Φ ( x + θ y 0 ) + θ ν I ′ ( θ , x ) y 0 − θ ν a ( x , y 0 ) ≥ 1 2 − 1 ν a ( x , x ) + μ ν − 1 Φ ( x + θ y 0 ) − M 1 ( ‖ I ′ ( θ , x ) ‖ + ‖ x ‖ + 1 ) ≥ M 2 ‖ x ‖ 2 − M 3 ( ‖ I ′ ( θ , x ) ‖ + 1 ) ,

where M 1 , M 2 , and M 3 are positive constants. Here, the first inequality holds because

∣ a ( x , y ) ∣ ≤ ∑ j = 1 ∞ ∣ λ j ∣ ξ j 2 1 2 ∑ j = 1 ∞ ∣ λ j ∣ ζ j 2 1 2 ≤ ‖ x ‖ ‖ y ‖ ,

for all x , y ∈ Z ; the second inequality holds because ∃ M 4 > 0 , M 5 > 0 such that Φ ( x + θ y 0 ) ≥ M 4 ‖ x ‖ X 2 − M 5 for all x ∈ Z and

1 2 − 1 ν a ( x , x ) + μ ν − 1 Φ ( x + θ y 0 ) ≥ 1 2 − 1 ν a ( x , x ) + 2 ( μ − ν ) ν − 2 M 4 ‖ x ‖ X 2 − 2 ( μ − ν ) ν − 2 M 5 ≥ 1 2 − 1 ν ‖ x ‖ 2 − μ ν − 1 M 5 ,

if ν − 2 > 0 is small enough such that 2 ( μ − ν ) ν − 2 c 3 ≥ λ 0 . Therefore,

(7) ‖ x ‖ 2 ≤ M 4 ( 1 + ∣ I ( θ , x ) ∣ + ∣ I ′ ( θ , x ) x ∣ + ‖ I ′ ( θ , x ) ‖ ) .

Assume that ( θ k , x k ) ∈ [ 0 , 1 ] × Z satisfies ‖ I ′ ( θ k , x k ) ‖ → 0 and I ( θ k , x k ) is bounded. Then, ‖ x k ‖ is bounded. We may assume x k ⇀ x 0 in Z and x k → x 0 in X along a subsequence. We also assume θ k → θ 0 in [ 0 , 1 ] . Because

(8) I ′ ( θ , x ) ϕ = a ( x , ϕ ) − ( ∇ Φ ( x + θ y 0 ) , ϕ ) X , ∀ x , ϕ ∈ Z ,

(9) a ( x k , ϕ ) = ( ∇ Φ ( x k + θ k y 0 ) , ϕ ) X + I ′ ( θ k , x k ) ϕ , ∀ x , ϕ ∈ Z .

It follows that

a ( x 0 , ϕ ) = ( ∇ Φ ( x 0 + θ 0 y 0 ) , ϕ ) X , ∀ ϕ ∈ Z ,

and

a ( x k − x 0 , ϕ ) + λ 0 ( x k − x 0 , ϕ ) X = I ′ ( θ k , x k ) ϕ + ( ∇ Φ ( x k + θ k y 0 ) − ∇ Φ ( x 0 + θ 0 y 0 ) , ϕ ) X + λ 0 ( x k − x 0 , ϕ ) X .

By definition, the inner product in Z associated with the norm ‖ ⋅ ‖ is ( x , y ) = a ( x , y ) + λ 0 ( x , y ) X for all x , y ∈ Z , and

‖ x k − x 0 ‖ = sup ‖ ϕ ‖ ≤ 1 ∣ ( x k − x 0 , ϕ ) ∣ ≤ ‖ I ′ ( θ k , x k ) ‖ + ‖ ∇ Φ ( x k + θ k y 0 ) − ∇ Φ ( x 0 + θ 0 y 0 ) ‖ X + λ 0 ‖ x k − x 0 ‖ X .

Therefore, I satisfies ( P S ) . Since

J θ ( x ) = ∂ ∂ θ I ( θ , x ) = − ( ∇ Φ ( x + θ y 0 ) , y 0 ) X = I ′ ( θ , x ) y 0 − a ( x , y 0 ) ,

(10) ∣ J θ ( x ) ∣ ≤ ( ‖ I ′ ( θ , x ) ‖ + ‖ x ‖ ) ‖ y 0 ‖ .

By (7),

∣ J θ ( x ) ∣ ≤ M 5 ( 1 + ∣ I ( θ , x ) ∣ + ‖ I ′ ( θ , x ) ‖ + ∣ I ′ ( θ , x ) x ∣ ) ≤ M 5 ( 1 + ∣ I ( θ , x ) ∣ ) ( 1 + ‖ I ′ ( θ , x ) ‖ ) ( 1 + ‖ x ‖ ) .

And I satisfies ( H 2 ) . Moreover, if I ′ ( θ , x ) = 0 , (7) and (10) imply that

(11) ∣ J θ ( x ) ∣ ≤ M 6 ( 1 + ∣ I ( θ , x ) ∣ 2 ) 1 4 ,

so ( H 3 ) holds. And ( Φ 2 ) – ( Φ 3 ) show that I satisfies ( H 4 ) . By Lemma 1 and ( A Φ ) , if the set of critical values of I 1 is bounded, then for large k , we have

(12) c k + 1 − c k ≤ M ( c k + 1 + c k + 1 ) ,

and c k ≤ M 7 k 2 for some M 7 > 0 and k ∈ N * . This contradicts ( A Φ ) . And the proof is complete.□

Remark 1

Assume Ψ ∈ C 1 ( Z , R ) , and there exists M > 0 such that ∣ Ψ ′ ( x ) ϕ ∣ ≤ M ‖ ϕ ‖ X , ∀ ϕ ∈ Z . Then, there exists ∇ Ψ ( x ) ∈ X such that Ψ ′ ( x ) ϕ = ( ∇ Ψ ( x ) , ϕ ) X ∀ ϕ ∈ Z . Then, if we replace (5) with the equation

A x = ∇ Φ ( x + y 0 ) + ∇ Ψ ( x ) ,

the conclusion of Theorem 3 is also valid.

In fact, set

I ( θ , x ) = 1 2 a ( x , x ) − Φ ( x + θ y 0 ) − θ Ψ ( x ) , x ∈ Z .

Then, all the assumptions of Lemma 1 can be checked, and the set of critical values of I ( 1 , ⋅ ) is unbounded as in the proof of Theorem 3.

3 Proof of Theorem 2

In this section, we give the proof of Theorem 2 by making use of Theorem 3. To this end, we consider the more general problem

(13) x ¨ ( t ) + V ′ ( t , x ) = F ′ ( t , x ) ,

(14) x ( 0 ) cos α − x ˙ ( 0 ) sin α = x 0 ,

(15) x ( 1 ) cos β − x ˙ ( 1 ) sin β = x 1 ,

where α , β ∈ ( 0 , π ) , x 0 , x 1 ∈ R n are given and V , F : [ 0 , 1 ] × R n → R , V ′ , F ′ : [ 0 , 1 ] × R n → R n , V ′ ′ , F ′ ′ : [ 0 , 1 ] × R n → R n × n are continuous.

Theorem 4

Assume V and F satisfy

( V 1 ) there exist p > 2 , r > 0 such that 0 < p V ( t , x ) ≤ V ′ ( t , x ) ⋅ x , ∀ x ∈ R n , ∣ x ∣ ≥ r , ∀ t ∈ [ 0 , 1 ] , and

( V 2 ) V ( t , − x ) = V ( t , x ) , ∀ ( t , x ) ∈ [ 0 , 1 ] × R n ,

( F 1 ) there exists M > 0 such that ∣ F ′ ( t , x ) ∣ ≤ M , ∀ ( t , x ) ∈ [ 0 , 1 ] × R n .

Then, systems (13)–(15) have infinitely many solutions.

Remark 2

As F ≡ 0 and V ( t , x ) = V ( x ) , Theorem 4 reduces to Theorem 2.

Proof of Theorem 4

Let x = u + z , where z = z ( t ) = t ( a 2 t 2 + a 1 t + a 0 ) , for some a 0 , a 1 , a 2 ∈ R n satisfies (14)–(15). Then, systems (13)–(15) are equivalent to

(16) u ¨ ( t ) + V ′ ( t , u + z ) = F ′ ( t , u + z ) − z ¨ ≡ F 1 ′ ( t , u ) ,

(17) u ( 0 ) cos α − u ˙ ( 0 ) sin α = 0 ,

(18) u ( 1 ) cos β − u ˙ ( 1 ) sin β = 0 ,

where F 1 ( t , u ) = F ( t , u + z ) − z ¨ ( t ) u , ∀ ( t , u ) ∈ [ 0 , 1 ] × R n .

Set ( A u ) ( t ) = − u ¨ ( t ) with D ( A ) = { u ∈ H 2 ( [ 0 , 1 ] , R n ) : u satisfies (17)–(18) } and X = L 2 ( [ 0 , 1 ] , R n ) .

By Proposition 7.1.1 of [5] A is self-adjoint, σ ( A ) = σ d ( A ) is bounded from below, and D ( ∣ A ∣ 1 2 ) = H 1 ( [ 0 , 1 ] , R n ) ≡ Z via Proposition 7.3.1 of [5].

By definition, for u ∈ D ( A ) and v ∈ Z ,

a ( u , v ) = ( A u , v ) X = ∫ 0 1 ( − u ¨ ⋅ v ) d t = ∫ 0 1 u ˙ ⋅ v ˙ d t − u ( 1 ) ⋅ v ( 1 ) cot β + u ( 0 ) ⋅ v ( 0 ) cot α .

In particular, for u ∈ Z ,

a ( u , u ) = ∫ 0 1 ∣ u ˙ ∣ 2 d t − ∣ u ( 1 ) ∣ 2 cot β + ∣ u ( 0 ) ∣ 2 cot α .

For u ∈ Z , define Φ ( u ) = ∫ 0 1 V ( t , u ( t ) ) d t , Ψ ( u ) = − ∫ 0 1 F 1 ( t , u ( t ) ) d t ,

I ( θ , u ) = 1 2 a ( u , u ) − ∫ 0 1 V ( t , u ( t ) + θ z ( t ) ) d t + θ ∫ 0 1 F 1 ( t , u ( t ) ) d t .

By ( V 1 ) , there exist M 1 , M 2 > 0 such that

V ′ ( t , x ) ⋅ x ≥ p V ( t , x ) − M 2 , V ( t , x ) ≥ M 1 ∣ x ∣ p − M 2 .

So for u ∈ Z ,

Φ ′ ( u ) u = ∫ 0 1 V ′ ( t , u ( t ) ) ⋅ u ( t ) d t ≥ ∫ 0 1 ( p V ( t , u ( t ) ) − M 2 ) d t = p Φ ( u ) − M 2

and

Φ ( u ) = ∫ 0 1 V ( t , u ( t ) ) d t ≥ M 1 ∫ 0 1 ∣ u ∣ p d t − M 2 ≥ M 1 ∫ 0 1 ∣ u ∣ 2 d t p 2 − M 2 = M 1 ‖ u ‖ X p − M 2 .

By Theorem 3, it suffices to show ( A Φ ) holds.

From Lemma 5.1 of [6], there exists G ∈ C 2 ( R , R ) such that G ′ = g is odd, G ( 0 ) = g ( 0 ) = g ′ ( 0 ) = 0 , g is increasing on [ 0 , + ∞ ) and 0 < G ( s ) ≤ 1 3 g ( s ) s , s g ′ ( s ) ≥ 2 g ( s ) , ∀ 0 < s ∈ R , V ( t , x ) ≤ ∑ i = 1 n G ( x i ) + c , ∀ x = ( x 1 , … , x n ) ∈ R n . Set

a 1 ( u , v ) = ∫ 0 1 u ˙ v ˙ d t − u ( 1 ) v ( 1 ) cot β + u ( 0 ) v ( 0 ) cot α ,

K ( u ) = 1 2 a 1 ( u , u ) − ∫ 0 1 G ( u ) d t ,

for u , v ∈ H 1 ( [ 0 , 1 ] , R ) , I ¯ ( u ) = ∑ j = 1 n K ( u j ) for u = ( u 1 , … , u n ) ∈ Z and d k = inf g ∈ G k sup u ∈ Z k I ¯ ( g ( u ) ) . Then, c k ≥ d k − c for all k ∈ N * . We have the following proposition, and its proof will be given in the end of this section.

Proposition 1

There exists { u k } ⊂ Z such that d k ≥ I ¯ ( u k ) , I ¯ ′ ( u k ) = 0 and i 0 ( u k ) ≥ k , where as in [1], i 0 ( u k ) denotes the number of non-positive eigenvalues of I ¯ ″ ( u k ) , i.e.,

i 0 ( u k ) = max { dim H ; H ⊂ Z is a s u b s p a c e s u c h t h a t I ¯ ″ ( u k ) ( h , h ) ≤ 0 for h ∈ H } .

Write u k = ( u k 1 , u k 2 , … , u k n ) , and define d ¯ k = I ¯ ( u k ) , d ¯ k j = K ( u k j ) . Then, d ¯ k = d ¯ k 1 + … + d ¯ k n , i 0 ( u k ) = i 0 ( u k 1 ) + … + i 0 ( u k n ) , and there exists some j 0 such that i 0 ( u k j 0 ) ≥ [ k n ] .

Considering the form of the functional I ¯ , we obtain that K ′ ( u k j ) = 0 , i.e.,

(19) a 1 ( x , y ) − ∫ 0 1 g ( x ) y d t = 0 ,

for all y ∈ H 1 [ 0 , 1 ] with x = u k j ( t ) . We have the following.

Lemma 2

[5, Lemma 2.6.4] Assume u ∈ H 1 [ 0 , 1 ] and f ∈ L 2 [ 0 , 1 ] satisfy (19) with x and g ( x ) replaced with u and f. Then, u ∈ D ( A ) with n = 1 and

(20) u ¨ ( t ) + f ( t ) = 0 , a.e. t ∈ [ 0 , 1 ] .

By this lemma and (19), x = u k j ( t ) is a solution of the following problem:

(21) x ¨ + g ( x ) = 0 ,

(22) x ( 0 ) cos α − x ˙ ( 0 ) sin α = 0 ,

(23) x ( 1 ) cos β − x ˙ ( 1 ) sin β = 0 .

Assume u = u k j 0 ( t ) has m = m k zeros exactly in ( 0 , 1 ) . We claim i 0 ( u k j 0 ) = m + 1 . In fact, i 0 ( u k j 0 ) is the number of nonpositive eigenvalues of K ″ ( u k j 0 ) . If λ ∈ σ ( K ″ ( u k j 0 ) ) , then there exists x ∈ H 1 ( [ 0 , 1 ] , R ) \ { 0 } such that K ″ ( u k j 0 ) x = λ x , i.e., K ″ ( u k j 0 ) ( x , y ) = λ ( x , y ) , ∀ y ∈ H 1 ( [ 0 , 1 ] , R ) . Recall the inner product in H 1 ( [ 0 , 1 ] , R ) , we have

a 1 ( x , y ) − ∫ 0 1 q 1 ( t ) x y d t = K ″ ( u k j 0 ) ( x , y ) = λ ( a ( x , y ) + λ 0 ( x , y ) L 2 ) ,

i.e.,

a 1 ( x , y ) − λ λ 0 1 − λ + 1 1 − λ q 1 ( t ) x , y L 2 = 0 , ∀ y ∈ H 1 ( [ 0 , 1 ] , R ) ,

where q 1 ( t ) = g ′ ( u k j 0 ( t ) ) . And by Lemma 2 again, x satisfies (22)–(23) and

(24) x ¨ + ( ( 1 − θ ˆ ) ( − λ 0 ) + θ ˆ q 1 ( t ) ) x = 0 , for t ∈ ( 0 , 1 ) ,

where θ ˆ = 1 1 − λ ∈ [ 0 , 1 ] as λ ≤ 0 . Note that the dimension of the solution space of (22)–(24) is at most 1. So i 0 ( u k j 0 ) = the number of θ ˆ ∈ [ 0 , 1 ) such that problems (22)–(24) have a nonzero solution.

Let ϕ ( t ) = ϕ ( t , q , α ) be the unique solution of the problem

(25) ϕ ′ ( t ) = cos 2 ϕ ( t ) + q ( t ) sin 2 ϕ ( t ) , t ∈ ( 0 , 1 ) ϕ ( 0 ) = γ ,

where q ∈ L ∞ [ 0 , 1 ] , γ ∈ R . Note that ϕ ( t , q , γ ) can be used to discuss the nonzero solutions of problems (22)–(23) and

(26) x ¨ + q ( t ) x = 0 .

In fact, assume x = x ( t ) is a nonzero solution of (22)–(23) and (26). Let ρ ( t ) = ( x ( t ) 2 + x ′ ( t ) 2 ) 1 2 and ϕ ( t ) = arccot x ′ ( t ) x ( t ) be continuous with ϕ ( 0 ) = α . Then, ϕ = ϕ ( t , q , γ ) satisfies (25) with γ = α and ϕ = ϕ ( t , q , α ) . Moreover, x ( t ) = ρ ( t ) ϕ ( t ) or x ( t ) = − ρ ( t ) ϕ ( t ) . So ϕ ( 1 , q , α ) ≡ β ( mod π ) and x ( t ) = 0 if and only if ϕ ( t , q , α ) ≡ 0 ( mod π ) .

Define ψ ( θ ˆ ) = ϕ ( 1 , ( 1 − θ ˆ ) ( − λ 0 ) + θ ˆ q 1 , α ) . Then, problems (22)–(24) have a nonzero solution iff ψ ( θ ˆ ) = j π + β for some j ∈ N via [7, Theorem 11.4.1(i)(ii)] and its proof. Note that ψ ( θ ˆ ) is strictly monotone increasing and continuous and recall that − λ 0 = − ( 1 + 2 ∣ λ 1 ∣ ) < λ 1 and λ 1 is the smallest eigenvalue of A . So ψ ( 0 ) = ϕ ( 1 , − λ 0 , α ) < ϕ ( 1 , λ 1 , α ) = β , and to show i 0 ( u k j 0 ) = m + 1 , it suffices to show ϕ ( 1 , q 1 , α ) = ψ ( 1 ) ∈ ( m π + β , m π + π + β ) .

By Proposition 2.2.1 of [5] and the fact that q 1 ( t ) ≡ g ′ ( u ( t ) ) ≥ 2 g ( u ( t ) ) ⁄ u ( t ) ≡ 2 q 2 ( t ) , for u ( t ) ≠ 0 , we have ϕ ( 1 , q 1 , α ) > ϕ ( 1 , q 2 , α ) = m π + β . We need only to show ϕ ( 1 , q 1 , α ) < ( m + 1 ) π + β .

We have four cases:

Case 1: α ∈ ( 0 , π 2 ] , β ∈ ( 0 , π 2 ) ;

Case 2: α ∈ ( 0 , π 2 ] , β ∈ [ π 2 , π ) ;

Case 3: α ∈ ( π 2 , π ) , β ∈ ( 0 , π 2 ) ;

Case 4: α ∈ ( π 2 , π ) , β ∈ [ π 2 , π ) .

We only give the proof for Case 1.

Assume u has m zeros on ( 0 , 1 ) denoted by 0 < t 1 < … < t m < 1 , u ′ ( t ) has m zeros { τ j } j = 1 m on [ 0 , 1 ] and 0 ≤ τ 1 < t 1 < τ 2 < … < τ m < t m < 1 . Since g ( u ) is odd, we can prove that t 2 − t 1 = t 3 − t 2 = … = t m − t m − 1 and u ( t ) is origin symmetric with regard to t = t i , i = 1 , 2 , … , m . Moreover, q 1 ( t ) = g ′ ( u ( t ) ) > 0 for t ≠ τ i and is periodic with the minimal period τ = τ 2 − τ 1 = t 2 − t 1 . Since τ 2 − t 1 ≥ τ 1 − 0 = τ 1 , ϕ ( τ 1 , q 1 , α ) ≤ π and ϕ ( τ m , q 1 , α ) ≤ m π . Note that 1 − τ m < τ m − τ m − 1 , we have ϕ ( 1 , q 1 , α ) ≤ ( m + 1 ) π .

Therefore, to show ( A Φ ) , it suffices to show d ¯ k ⁄ m k 2 → ∞ as k → ∞ . From (21)–(23) and the properties of G and g ,

K ( x ) = 1 2 a 1 ( x , x ) − ∫ 0 1 G ( x ) d t ≥ 1 2 a 1 ( x , x ) − ∫ 0 1 x g ( x ) d t = 0 .

Thus, d ¯ k j = K ( u k j ) ≥ 0 . Moreover, since x = u k j 0 ( t ) satisfies (21)–(22) and x ( t 1 ) = 0 ,

∫ 0 t 1 ( ∣ x ˙ ( t ) ∣ 2 − x ( t ) g ( x ( t ) ) ) d t + ∣ x ( 0 ) ∣ 2 cot α = 0

and

1 2 ∫ 0 t 1 ∣ x ˙ ( t ) ∣ 2 d t + ∣ x ( 0 ) ∣ 2 cot α − ∫ 0 t 1 G ( x ( t ) ) d t ≥ 0 .

Similarly,

1 2 ∫ t m 1 ∣ x ˙ ( t ) ∣ 2 d t − ∣ x ( 1 ) ∣ 2 cot β − ∫ t m 1 G ( x ( t ) ) d t ≥ 0 .

Thus,

d ¯ k ≥ d ¯ k j 0 ≥ ∫ t 1 t m 1 2 ∣ u ˙ k j 0 ∣ 2 − G ( u k j 0 ) d t ≡ d k ′ .

We need only to show that

(27) d k ′ ⁄ m k 2 → ∞ , as k → ∞ .

This follows from [6, Proposition 4.2] essentially. In fact, as in its proof, let v k ( t ) = u k j 0 ( t − t 1 m − 1 + t 1 ) , v = v k , and u = u k j 0 . Then, v ( t ) satisfies

− v ¨ = 1 ( m − 1 ) 2 g ( v ) , v ( t 1 ) = v ( t m ) = 0 .

And

(28) ∫ t 1 t m ( v ˙ ( t ) ) 2 d t = 1 ( m − 1 ) 2 ∫ t 1 t m g ( v ) v d t .

Also,

∫ t 1 t m ( v ˙ ( t ) ) 2 d t = 1 ( m − 1 ) 2 ∫ t 1 t m ( u ˙ ( s ) ) 2 d s ,

∫ t 1 t m G ( u ) d t = ∫ t 1 t m G ( v ) d t .

We obtain

d k ′ = ∫ t 1 t m 1 2 ( u ˙ ( t ) ) 2 − G ( u ) d t ≥ ∫ t 1 t m 1 2 ( v ˙ ( t ) ) 2 − 1 3 g ( v ) v d t ≥ 1 6 ( m − 1 ) 2 ∫ t 1 t m ( v ˙ ( t ) ) 2 d t .

If ∫ t 1 t m ( v ˙ ( t ) ) 2 d t is bounded, so is ‖ v ‖ C [ t 1 , t m ] . By the properties of g , there exists c > 0 such that ∣ g ( v ( t ) ) ∣ ≤ c ∣ v ( t ) ∣ for t ∈ [ t 1 , t m ] . And

0 < ∫ t 1 t m ( v ( t ) ) 2 d t ≤ t m − t 1 π 2 ∫ t 1 t m ( v ˙ ( t ) ) 2 d t ≤ c t m − t 1 π ( m − 1 ) 2 ∫ t 1 t m ( v ( t ) ) 2 d t ,

via (28), which is a contradiction. So ∫ t 1 t m ( v ˙ ( t ) ) 2 d t → ∞ , and (27) holds. The proof is complete.□

In order to show Proposition 1, we turn to the work [8] by K. Tanaka. As before, let E be a Hilbert space, and let I ∈ C 2 ( E , R ) . Let E j and G j be as in Section 2. Define

b k = inf g ∈ G k sup x ∈ E k I ( g ( x ) ) .

We will use the following assumptions.

( I 1 ) For each finite dimensional subspace F ⊂ E , there is an R = R ( F ) > 0 such that I ( u ) < 0 for all u ∈ F with ‖ u ‖ E ≥ R ( F ) .

( I 2 ) I ′ ( u ) v = ( u + K ( u ) , v ) E for u , v ∈ E , where K : E → E is a compact operator.

( P S ) Whenever a sequence { u j } j = 1 ∞ in E satisfies for some M > 0 ,

∣ I ( u j ) ∣ ≤ M for all j , I ′ ( u j ) → 0 , in E * , as j → ∞ ,

there is a subsequence of { u j } j = 1 ∞ , which converges in E .

( P S ) m If for some M > 0 , { u j } j = 1 ∞ satisfies u j ∈ E m ,

I ( u j ) ≤ M for all j , ∥ ( I ∣ E m ) ′ ( u j ) ∥ ( E m ) * → 0 , as j → ∞ ,

then { u j } j = 1 ∞ is relatively compact in E m .

( P S ) * If for some M > 0 , { u j } j = 1 ∞ satisfies

u j ∈ E j , I ( u j ) ≤ M for all j , ∥ ( I ∣ E j ) ′ ( u j ) ∥ ( E j ) * → 0 , as j → ∞ ,

then { u j } j = 1 ∞ is relatively compact in E .

We have the following result.

Lemma 3

Assume I ∈ C 2 ( E , R ) is even, I ( 0 ) = 0 and satisfies ( I 1 – I 2 ), ( P S ) , ( P S ) m , ( P S ) * . Then, there exists { v k } k = 1 ∞ ⊂ E such that

I ( v k ) ≤ b k , I ′ ( v k ) = 0 , i 0 ( I ″ ( v k ) ) ≥ k .

Remark 3

As in [8], set R k = R ( E k ) , D k = { u ∈ E k ∣ ‖ u ‖ ≤ R k } ,

Γ k = { γ ∈ C ( D k , E ) ; γ ( − u ) = − γ ( u ) for all u ∈ D k , γ ( u ) = u for all u ∈ ∂ D k }

and define

b k ′ = inf γ ∈ Γ k sup u ∈ D k I ( γ ( u ) ) .

In [8, Theorem B], K. Tanaka has showed that there exists { v k ′ } k = 1 ∞ ⊂ E such that I ( v k ′ ) ≤ b k ′ , I ′ ( v k ′ ) = 0 , and i 0 ( I ″ ( v k ′ ) ) ≥ k . By their definitions, b k ′ ≤ b k and Lemma 3 cannot be obtained from [8, Theorem B] directly. However, the proof of [8, Theorem B] is suitable for Lemma 3.

Proof of Proposition 1

For u ∈ Z ≡ H 1 ( [ 0 , 1 ] , R n ) , recall that a ( u , u ) = 1 2 ∫ 0 1 ∣ u ˙ ∣ 2 d t − 1 2 cot ( β ) ∣ u ( 1 ) ∣ 2 + 1 2 cot ( α ) ∣ u ( 0 ) ∣ 2 . Since there exists λ 0 > 0 such that ‖ u ‖ = ( a ( u , u ) + λ 0 ‖ u ‖ L 2 2 ) 1 2 is an equivalent norm on Z , ( u , v ) ≡ a ( u , v ) + λ 0 ( u , v ) L 2 is an inner product in Z . Recall that

I ¯ ( u ) = 1 2 a ( u , u ) − ∫ 0 1 V ( t , u ( t ) ) d t ,

I ¯ ′ ( u ) ν = a ( u , ν ) − ∫ 0 1 V ′ ( t , u ( t ) ) ⋅ ν ( t ) d t = ( u , ν ) + ( N ( u ) , ν ) ,

where ( N ( u ) , ν ) = − ∫ 0 1 ( u + V ′ ( t , u ) ) ⋅ ν d t and N : Z → Z is compact. So ( I 2 ) holds. Assume for M > 0 , { u k } k = 1 ∞ satisfies u k ∈ Z k , I ¯ ( u k ) ≤ M and ∥ ( I ¯ ∣ Z k ) ′ ( u k ) ∥ ( Z k ) * → 0 as k → ∞ . As before, { ‖ u k ‖ } is bounded. Assume u k j ⇀ u 0 in Z and u k j → u 0 in C ( [ 0 , 1 ] , R n ) . To show that ‖ u k j − u 0 ‖ → 0 , it suffices to show that a ( u k j , u k j ) → a ( u 0 , u 0 ) as j → ∞ . This follows from

a ( u k j , u k j ) = I ¯ ′ ( u k j ) u k j + ∫ 0 1 V ′ ( t , u k j ( t ) ) ⋅ u k j d t = ( I ¯ ∣ E k j ) ′ ( u k j ) u k j + ∫ 0 1 V ′ ( t , u k j ( t ) ) ⋅ u k j d t

and

a ( u 0 , ν ) − ∫ 0 1 V ′ ( t , u 0 ) ⋅ ν d t = 0 , for all ν ∈ Z .

Hence, I ¯ satisfies ( P S ) * . It is easy to see that all the other assumptions of Lemma 3 are also verified. The proof is complete.□

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Conflict of interest: The authors state no conflict of interest.

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Received: 2023-01-11

Revised: 2024-11-29

Accepted: 2024-12-01

Published Online: 2024-12-31

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2024-0112

Keywords for this article

Sturm-Liouville; Hamiltonian system; non-symmetric

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