second order Hamiltonian systems

Existence of solutions for subquadratic convex or B-concave operator equations and applications to

second order Hamiltonian systems

Mingliang Song

School of Mathematical Sciences, Nanjing Normal University, Nanjing, 210097, P. R. China Mathematics and Information Technology School, Jiangsu Second Normal University,

Nanjing, 210013, P. R. China.

Received 12 August 2019, appeared 23 July 2020 Communicated by Gabriele Bonanno

Abstract. This paper investigates solutions for subquadratic convex orB-concave oper- ator equations. First, some existence results are obtained by the index theory and the critical point theory. Then, some applications to second order Hamiltonian systems sat- isfying generalized periodic boundary value conditions and Sturm–Liouville boundary value conditions are pointed out. In particular, some well known theorems about peri- odic solutions for second order Hamiltonian systems are special cases of these results.

Keywords: subquadratic, operator equations, index theory, critical point, second order Hamiltonian systems.

2010 Mathematics Subject Classification: 34B15, 34C25, 58E05, 70H05.

1 Introduction and main results

Mawhin and Willem [9] investigated the second order Hamiltonian system (−x¨(t)−m²ω²x(t) =∇_xV(t,x(t)), a.e.t ∈[0,T],

x(0)−x(T) =x˙(0)−x˙(T) =0, (1.1) where T >_0,ω = ^2π_T,m ∈ {0, 1, 2, . . .}_,V ∈C([_0,T]×_Rⁿ_,R)_,∇_xV denotes the gradient ofV with respect to x,∇_xV ∈ C([0,T]×_Rⁿ,Rⁿ), and for each x ∈_Rⁿ,V(t,x)is periodic in twith period T. Using the dual least action principle and the perturbation technique, the Authors, in theirs excellent book [9], proved some existence theorems of solutions for problem (1.1) with subquadratic convex or concave potential. Recently, using the reduction method, the perturbation argument and the least action principle, Tang and Wu [12] proved an abstract critical point theorem without the compactness assumptions which generalizes the results in [7]. As a main application, they successively obtained some existence theorems of problem

BEmail: mlsong2004@163.com

(1.1) withm=0 and subquadratic convex potential ork(t)-concave potential, which unify and generalize some earlier results in [9,13,14,16,17]. Later on, applying the abstract critical point theory established in [12], Ye [15] proved some existence theorems of problem (1.1), where m ≥ 1 and the potential is convex and satisfies conditions which are more general than the subquadratic conditions in [9]. In this paper we reconsider in the framework of the operator equations some theorems proved in [9,12,15].

LetXbe a real infinite-dimensional separable Hilbert space with inner product(·,·)_Xand the corresponding normk · k_X. Let A: D(A)⊂ X → X be an unbounded linear self-adjoint operator with σ(A) = σ_d(A) bounded from below. Hence, there is an orthonormal basis {e_j}^∞_j=1 of X andλ₁ ≤ λ₂ ≤ · · · such that Ae_j = λ_je_j,D(A) =_∑^∞_j=1c_je_j|_∑^∞_j=1λ²_jc²_j < _∞ . In addition, let Z ≡ D(|A|¹²) = _∑^∞_j=1c_je_j|_∑^∞_j=1|λ_j|c²_j < _∞ equipped with the norm kxk²_Z = kxk²=_∑^∞_j=1(1+|λ_j|)c²_j. For anyx= _∑^∞_j=1c_je_j ∈ Z,y=_∑^∞_j=1d_je_j ∈ Z, we can define a bilinear form

a(x,y) =

∑

λ_jc_jd_j.

Note that (Ax,y)_X = a(x,y) if x ∈ D(A),y ∈ Z, this shows that a(x,y) is the extension of (Ax,y)_X on Z. Moreover, let L_s(X)be the usual space consisting of bounded symmetric operators inX. For givenB∈ L_s(X), we define

ν_A(B) =dim ker(A−B), i_A(B) =

∑

λ<₀

ν_A(B+λId),

as introduced by Dong, see Definition 7.1.1 in [5] or Definition 3.1.1 and Proposition 3.1.4 in [4]. We consider the following operator equation

Ax−B₁x =∇_Φ(x), (1.2)

whereB₁∈ L_s(X),ν_A(B₁)6=0, andΦsatisfies

(_Φ0)_Φ∈C¹(Z,R)is weakly continuous with weakly continuous derivative, that is,x_n * x0 in Z implies that Φ(xn) → _Φ(x0)and Φ⁰(xn) → _Φ⁰(x0). Moreover, for every x ∈ Z there exists∇_Φ(x)∈ Xsuch thatΦ⁰(x)y= (∇_Φ(x),y)_X for ally∈ Z.

LetX₁ be a nontrivial subspace ofX. ForB₁,B₂ ∈ L_s(X)we writeB₁≤ B₂with respect to X₁if and only if(B₁x,x)_X≤ (B2x,x)_Xfor all x∈ X₁; we writeB₁ < B2 w.r.t. X₁ if and only if (B₁x,x)_X< (B₂x,x)_Xfor all x∈ X₁\{θ}. IfX₁ =X, then we just writeB₁≤ B₂ or B₁ <B₂. In addition, we writeB₁ < B₂ properly if and only ifB₁ ≤ B₂ andB₁ < B₂ w.r.t. ker(A−B)for allB∈ L_s(X).

Our main results can be stated as follows.

Theorem 1.1. Assume thatΦsatisfies(_Φ0)and (_Φ1) _Φis convex in X;

(_Φ2) _ΦandΦ⁰ are bounded in Z;

(_Φ3) _Φ(x)→+_∞askxk →_∞with x∈ker(A−B₁);

(_Φ4) there exist c>0and B₂ ∈ L_s(X)with B₂≥ B₁ and B₂ > B₁w.r.t.ker(A−B₁),ν_A(B₂)6= 0 and i_A(B₂) =i_A(B₁) +ν_A(B₁), such that

Φ(x)≤ ¹

2((B₂−B₁)x,x)_X+c (1.3)

for all x ∈X, and

Φ(x)−¹

2((B2−B₁)x,x)_X→ −_∞ (1.4) askxk →_{∞, where x}=x_e+x with x∈ker(A−B2)andkx˜kis bounded.

Then problem(1.2)has a solution in Z.

Theorem 1.2. The conclusion of Theorem1.1still holds if we replace(_Φ4)with

(_Φ⁰₄) there exist c> 0and B₂ ∈ L_s(X)with B₂ ≥ B₁ and B₂ > B₁w.r.t.ker(A−B₁),ν_A(B₂) =0 and i_A(B₂) =i_A(B₁) +ν_A(B₁), such that

Φ(x)≤ ¹

2((B₂−B₁)x,x)_X+c (1.5) for all x ∈X.

Theorem 1.3. The conclusion of Theorem1.1still holds if we replace(_Φ1)and(_Φ4)with (_Φ⁰₁) _Φis(B₂−B₁)-concave, that is,−_Φ(x) + ¹₂((B₂−B₁)x,x)_Xis convex in X.

(_Φ⁰⁰₄) there exists B₂ ∈ L_s(X)with B₂≥ B₁and B₂ >B₁w.r.t.ker(A−B₁),i_A(B₁) =0,ν_A(B₂)6=0 and i_A(B₂) =i_A(B₁) +ν_A(B₁), such that

−_Φ(x) + ¹

2((B₂−B₁)x,x)_X→+_∞ (1.6) askxk →_∞with x∈ker(A−B₂), respectively.

Theorem 1.4. The conclusion of Theorem1.1still holds if we replace(_Φ1)and(_Φ4)with(_Φ⁰₁), (_Φ⁰⁰⁰₄ ) there exists B₂∈ L_s(X)with B₂ ≥B₁ and B₂ > B₁ w.r.t.ker(A−B₁)_,i_A(B₁) =_0,ν_A(B₂) =

0, such that

i_A(B₂) =i_A(B₁) +ν_A(B₁), respectively.

The paper is organized as follows. In Section 2, we first recall a critical point theorem as given in [12]. Then, following [4,5], we recall some useful conclusions of index theory for linear self-adjoint operator equations. Finally, we quote a lemma in [3], which shows that (1.2) possesses a variational structure. In Section 3, we prove Theorems 1.1–1.4. In Section 4, we investigate their applications to second order Hamiltonian systems with generalized periodic boundary conditions and Sturm–Liouville boundary conditions. The corresponding results in [9,12,15] are special cases of these results.

2 Preliminaries

In order to prove our main results, we recall first two lemmas due to Tang and Wu [12].

Lemma 2.1([12, Theorem 1.1]). Suppose that X₁ and X₂ are reflexive Banach spaces, I ∈ C¹(X₁× X₂,R). I(x₁,·)is weakly upper semi-continuous for all x₁ ∈ X₁and I(·,x₂): X₁ →_Ris convex for all x2 ∈X2, and I⁰ is weakly continuous. Assume that

I(_θ,x₂)→ −_{∞ (2.1)}

askx₂k →+_∞and, for every M >0

I(x₁,x₂)→+_∞ (2.2)

askx₁k →+_∞uniformly forkx₂k ≤M. Then I has at least one critical point.

Lemma 2.2([12, Lemma 5.1]). Suppose that H is a real Hilbert space, f : H×H→_Ris a bilinear functional. Then g:H →_Rgiven by

g(x) = f(x,x), ∀x ∈ H is convex if and only if

g(x)≥0, ∀x ∈ H.

Now we also recall some definitions and propositions in [4,5].

Definition 2.3([5, Page 108]). For anyB∈ L_s(X), we define

ψ_a,B(x,y) =a(x,y)−(Bx,y)_X, ∀x,y ∈Z.

For any x,y ∈ Z if ψ_a,B(x,y) = 0 we say that x and y are ψ_a,B-orthogonal. For any two subspacesZ₁ and Z₂ of Zif ψ_a,B(x,y) = 0 for any x ∈ Z₁,y ∈ Z₂ we say that Z₁ and Z₂ are ψ_a,B-orthogonal.

Proposition 2.4 ([5, Proposition 7.2.1]). For any B ∈ L_s(X), the space Z has a ψ_a,B-orthogonal decomposition

Z= Z⁺_a (B)⊕Z⁰_a(B)⊕Z_a⁻(B)

such thatψ_a,B is positive definite, null and negative definite on Z⁺_a (B),Z⁰_a(B)and Z_a⁻(B)respectively.

Moreover, Z⁰_a(B)and Z⁻_a (B)are finitely dimensional.

Definition 2.5([5, Definition 7.2.1]). For anyB∈ L_s(X), we defineν_a(B) =dimZ⁰_a(B),ia(B) = dimZ⁻_a (B).

Proposition 2.6.

(1) For any B∈ L_s(X), we have

ν_A(B) =ν_a(B), i_A(B) =i_a(B), ker(A−B) =Z_a⁰(B). ([5], Proposition 7.2.2 (i))

(2) For any B₁,B₂ ∈ L_s(X), if B₁ ≤ B₂ with respect to Z⁻_a (B₁), then i_a(B₁) ≤ i_a(B₂); if B₁ ≤ B₂ with respect to Z⁻_a (B₁)⊕Z⁰_a(B₁), then ia(B₁) +νa(B₁) ≤ ia(B₂) +νa(B₂); if B₁ < B₂ with respect to Z⁰_a(B₁)and B₁ ≤ B₂ with respect to Z_a⁻(B₁), then i_a(B₁) +ν_a(B₁) ≤ i_a(B₂). ([5], Proposition 7.2.2 (ii))

(3) For any B₁,B₂ ∈ L_s(X), if B₁(t)≤ B₂(t)and B₁(t)<B₂(t)properly, then ia(B₂)−ia(B₁) =

∑

λ∈[0,1)

νa(B₁+λ(B₂−B₁)). ([5], Proposition 7.2.2 (iii))

(4) (Poincaré inequality.) For any B∈ L_s(X), if i_a(B) =0, then ψ_a,B(x,x)≥0, ∀x∈ Z.

And the equality holds if and only if x∈Z_a⁰(B). ([5], Proposition 7.2.2 (v))

(5) For any B₁,B₂ ∈ L_s(X), if B₁ ≤ B₂ and B₁ < B₂ w.r.t. ker(A−B₁) and i_A(B₂) = i_A(B₁) +ν_A(B₁), then Z = Z⁻_a (B₁)⊕Z⁰_a(B₁)⊕Z⁰_a(B₂)⊕Z⁺_a (B₂), and (−ψ_a,B1(x₁,x₁))¹²+ (ψ_a,B2(x₂,x₂))¹² is an equivalent norm on Z for x = x₁+x₂ with x₁ ∈ Z⁻_a (B₁), x₂ ∈ Z⁺_a (B₂). In particular, for any B₁ ∈ L_s(X), then Z = Z_a⁻(B₁)⊕Z⁰_a(B₁)⊕Z⁺_a (B₁) and (−ψ_a,B1(x₁,x₁))¹² + (ψ_a,B1(x₂,x₂))¹² is also an equivalent norm on Z for x = x₁+x₂ with x₁ ∈Z_a⁻(B₁),x₂∈ Z⁺_a (B₁).

Proof. We only prove (5). Let Z₁ = Z⁻_a (B₁)⊕Z⁰_a(B₁),Z2 = Z_a⁰(B2)⊕Z⁺_a (B2). Noticing that ψ_a,B1(x,x) ≥ ψ_a,B2(x,x)for all x ∈ Z, ψ_a,B1(x,x)≤ 0 for all x ∈ Z₁ andψ_a,B2(x,x) ≥ 0 for all x∈ Z₂, ifx∈ Z₁^TZ₂thenψ_a,B2(x,x) =0= ψ_a,B1(x,x), which shows thatx∈ Z⁰_a(B₂)^TZ_a⁰(B₁). By B₁≤ B2 andB₁< B2w.r.t. ker(A−B₁), we have 0=ψa,B₁(x,x)> ψa,B₂(x,x) =0 provided x ∈ Z_a⁰(B₂)^TZ⁰_a(B₁)\{θ}. This is a contradiction, which implies that Z₁^TZ₂ = {θ}. It remains to prove thatZ=Z₁+Z₂. By Proposition2.4, we haveZ= Z₂⊕Z_a⁻(B₂)and for any x ∈ Z there exists a unique pair(x₁,x2) ∈ Z2×Z_a⁻(B2)such thatx = x₁+x2. Let{e_j}^k_j=1 be a basis of Z₁,e_j = e²_j +e⁻_j withe²_j ∈ Z₂,e⁻_j ∈ Z⁻_a (B₂)forj= 1, 2,· · · ,k =i_A(B₁) +ν_A(B₁). By i_A(B₂) =i_A(B₁) +ν_A(B₁) = k, in order to prove{e⁻_j }^k_j=1 is a basis ofZ_a⁻(B₂)we only need to show that {e⁻_j }^k_j=1 is linear independent. In fact, otherwise there exist not all zero constants c₁, . . . ,c_k such that∑^kj=1c_je⁻_j =0. This leads to∑^kj=1c_je_j ∈Z₁^TZ2, a contradiction. The linear independent shows that there exist constants {α_j}^k_{j=1 such that} x₂ = _∑^k_j=1α_je⁻_j . And hence x= x₁+x₂= x=x₁+_∑^k_j=1α_je⁻_j =_∑^k_j=1α_je_j+ x₁−_∑^k_j=1α_je²_j

.

Similar to the proof of Proposition 7.2.2 (iv) in [5], we can prove that (−ψ_a,B1(x₁,x₁))¹² + (ψa,B₂(x2,x2))¹² is an equivalent norm on Z for x = x₁+x2 with x₁ ∈ Z_a⁻(B₁),x2 ∈ Z_a⁺(B2), and (−ψ_a,B1(x₁,x₁))¹² + (ψ_a,B1(x₂,x₂))¹² is also an equivalent norm on Zfor x = x₁+x₂ with x₁ ∈Z_a⁻(B₁),x₂∈ Z⁺_a (B₁).

Finally, let us consider the functional I defined by I(x) =−¹

2a(x,x) + ¹

2(B₁x,x)_X+_Φ(x), (2.3) for every x ∈ Z. Under assumption (_Φ0), from Theorem 1.2 in [9] it is easy to verify that I ∈ C¹(Z,R)is weakly upper semi-continuous on ZandI⁰ is weakly continuous with

I⁰(x)y=−a(x,y) + (B₁x,y)_X+_Φ⁰(x)y, (2.4) for every x,y∈Z.

The following important lemma is an immediate conclusion of Lemma 2.1 in [3].

Lemma 2.7. Assume that(_Φ0)holds. Then a critical point of I(x)is a solution for problem(1.2).

3 Proofs of the Theorems

In this section, we present the proof of Theorems1.1–1.4.

Proof of Theorem1.1. By ν_A(B₁)6=0, B₁ ≤ B₂ andB₁ < B₂ w.r.t. ker(A−B₁)andi_A(B₂) = iA(B₁) +ν_A(B₁), we have Z= Z_a⁻(B₁)⊕Z⁰_a(B₁)⊕Z⁰_a(B2)⊕Z⁺_a (B2)via (5) of Proposition2.6.

SetX₁ =Z_a⁻(B₁)⊕Z⁰_a(B₁),X₂ = Z⁰_a(B₂)⊕Z⁺_a (B₂) =Z_a⁺(B₁),x∈ Z,x= x₁+x₂ withx₁ ∈X₁ andx₂∈ X₂. Next, we divide the proof into three steps.

Step 1. We show that I(·,x₂) : X₁ → _R is convex for all x₂ ∈ X₂. By (_Φ1), it is obvious that Φ(x₁+x₂)is convex in x₁ ∈ X₁. From Definition2.3 and Proposition2.4 we can see that for everyx₁ ∈X₁,

−¹

2ψ_a,B1(x₁,x₁) =−¹

2a(x₁,x₁) + ¹

2(B₁x₁,x₁)_X ≥0,

which implies that −¹₂ψa,B₁(x₁,x₁) is convex in x₁ ∈ X₁ via Lemma 2.2. Hence, for every x₂ ∈ X₂,

I(x₁+x₂) =−¹

2a(x₁+x₂,x₁+x₂) +¹

2(B₁(x₁+x₂),x₁+x₂)_X+_Φ(x₁+x₂)

=−¹

2ψ_a,B1(x₁,x₁) +_Φ(x₁+x₂)−¹

2ψ_a,B1(x₂,x₂) is convex inx₁ ∈X₁.

Step 2. By contradiction, we prove that (2.2) of Lemma2.1holds. Assume that (2.2) of Lemma 2.1 does not hold. Then there exist M > 0,c₀ > 0 and two sequences {x_1,n} ⊂ X₁ and {x_2,n} ⊂X₂ withkx_1,nk →+_∞asn→_∞andkx_2,nk ≤Mfor all nsuch that

I(x_1,n+x_2,n)≤c₀, ∀n∈N. (3.1) For x₁ ∈ X₁, write x₁ = x⁻₁ +x⁰₁, where x₁⁻ ∈ Z_a⁻(B₁) and x⁰₁ ∈ Z⁰_a(B₁). We consider the functionalΦ|_Z0

a(B₁). By (_Φ0), we easily see that Φ|_Z0

a(B₁) is weakly lower semi-continuous on Z⁰_a(B₁). Using (_Φ3), by the least action principle (see Theorem 1.1 in [9]), Φ|_Z0

a(B1) has a minimum at somex⁰_1,0∈ Z⁰_a(B₁)for which

0= _Φ⁰(x⁰_1,0)x⁰₁= (∇_Φ(x⁰_1,0),x₁⁰)_X, ∀x⁰₁∈ Z⁰_a(B₁). By assumption(_Φ0)and the convexity ofΦ, we have

Φ(x₁+x₂)−_Φ(x⁰_1,0)≥(∇_Φ(x⁰_1,0),x⁻₁ +x₂+x₁⁰−x⁰_1,0)_X

= (∇_Φ(x⁰_1,0),x⁻₁ +x2)_X, and then, fromkxk_X ≤ kxkfor all x∈ Z,

Φ(x₁+x₂)≥_Φ(x⁰_1,0)− k∇_Φ(x_1,0⁰ )k_X· kx⁻₁ +x₂k_X

≥_Φ(x⁰_1,0)− k∇_Φ(x_1,0⁰ )k_X·(kx⁻₁k+kx₂k)

=c₁−c₂·(kx⁻₁k+kx₂k)

where c₁ = _Φ(x_1,0⁰ ),c₂ = k∇_Φ(x_1,0⁰ )k_X ≥ 0. Rewrite x_1,n = x⁻_1,n+x⁰_1,n, where x⁻_1,n ∈ Z⁻_a (B₁) andx⁰_1,n∈ Z⁰_a(B₁). By (3.1), we have

c₀ ≥ I(x_1,n+x_2,n)

=−¹

2ψ_a,B1(x⁻_1,n,x⁻_1,n)− ¹

2ψ_a,B1(x_2,n,x_2,n) +_Φ(x_1,n+x_2,n)

≥ −¹

2ψ_a,B1(x⁻_1,n,x⁻_1,n)− ¹

2ψ_a,B1(x_2,n,x_2,n) +c₁−c₂·(kx_1,n⁻ k+kx_2,nk).

From(_Φ4)and (5) of Proposition2.6, we know that(−ψa,B₁(x₁⁻,x⁻₁))¹² is an equivalent norm on Z forx⁻₁ ∈ Z⁻_a (B₁)_and(ψ_a,B1(x₂,x₂))¹² is an equivalent norm on Zfor x₂ ∈ Z⁺_a (B₁)_{. This} means that there existc₃>0 andc₄>0 such that

c₀≥ I(x_1,n+x2,n)

≥ ^c

23

2kx_1,n⁻ k²− ^c

24

2kx2,nk²+c₁−c₂·(kx_1,n⁻ k+kx2,nk)

≥ ^c

23

2kx_1,n⁻ k²− ^c

2 4M²

2 +c₁−c2·(kx_1,n⁻ k+M)

viakx_2,nk ≤ M, which shows that{kx⁻_1,nk}is bounded. Combining this with assumption(_Φ2) and the convexity ofΦ, we see that there existc₅>0 andc₆=sup

n Φ(−x⁻_1,n−x_2,n)such that c0≥ I(x1,n+x2,n)

=−¹

2ψ_a,B1(x_1,n⁻ ,x⁻_1,n)−¹

2ψ_a,B1(x_2,n,x_2,n) +_Φ(x_1,n+x_2,n)

≥ (c₃c₅)²

2 − ^c

24M² 2 +2Φ

1 2x_1,n⁰

−_Φ(−x_1,n⁻ −x_2,n)

≥ (c₃c₅)²

2 − ^c

24M² 2 +2Φ

1 2x_1,n⁰

−c₆.

By (_Φ3), we know that {kx_1,n⁰ k} is also bounded. This contradicts the fact that kx_1,n⁻ k+ kx⁰_1,nk ≥ kx_1,nk →+_∞asn→_∞. Therefore (2.2) of Lemma2.1 holds.

Step 3. We check that (2.1) of Lemma2.1holds. If not, there exist a constantc₇and a sequence {x_2,n}in X₂such thatkx_2,nk →+_∞asn→_∞and

I(x_2,n)≥ c₇ (3.2)

for all n. For x₂ ∈ X₂, write x₂ = x⁰₂+x⁺₂, wherex⁰₂ ∈ Z⁰_a(B₂)andx⁺₂ ∈ Z⁺_a (B₂). Notice that ν^s_M(B₂) 6= 0 and X₂ = Z⁰_a(B₂)⊕Z⁺_a (B₂). Let x_2,n = x⁰_2,n+x⁺_2,n,x⁰_2,n ∈ Z_a⁰(B₂),x⁺_2,n ∈ Z_a⁺(B₂). Then by (1.3) of(_Φ4), (3.2), Definition2.3and Proposition2.4, we have

c₇≤ I(x_2,n)

≤ −¹

2a(x⁰_2,n+x⁺_2,n,x⁰_2,n+x⁺_2,n) +¹

2(B₂(x⁰_2,n+x⁺_2,n),x⁰_2,n+x⁺_2,n)_X+c

=−¹

2ψ_a,B2(x_2,n⁺ ,x⁺_2,n) +c

which implies that{x_2,n⁺ }is bounded since(−ψ_a,B1(x₁,x₁))¹² + (ψ_a,B2(x₂,x₂))¹² is an equivalent norm on Z for x = x₁+x₂ with x₁ ∈ Z⁻_a (B₁) and x₂ ∈ Z⁺_a (B₂), where x₁ = θ. Since kx_2,nk ≤ kx⁰_2,nk+kx_2,n⁺ k, we havekx⁰_2,nk → _∞as n →+_{∞. By}x_2,n ∈ X₂ = Z⁰_a(B₂)⊕Z_a⁺(B₂), we have ψ_a,B2(x_2,n,x_2,n) ≥ _{0 for all} n via Proposition 2.4. From kx⁰_2,nk → _{∞ as} n → +_{∞ we} have

I(x_2,n)≤_Φ(x_2,n)−¹

2((B₂−B₁)x_2,n,x_2,n)_X→ −_∞ via (1.4) of(_Φ4), which contradicts (3.2). Hence (2.1) of Lemma2.1holds.

By Lemma 2.1, I has at least one critical point. Hence problem (1.2) has at least one solution in Zvia Lemma2.7. The proof is complete.

Proof of Theorem1.2. By ν_A(B₁)6=0, B₁ ≤ B₂ andB₁ < B₂ w.r.t. ker(A−B₁)andi_A(B₂) = iA(B₁) +ν_A(B₁), we have Z= Z_a⁻(B₁)⊕Z⁰_a(B₁)⊕Z⁰_a(B2)⊕Z⁺_a (B2)via (5) of Proposition2.6.

Note that ν_A(B₂) = 0, we have Z⁰_a(B₂) = {θ}, which implies that Z = Z⁻_a (B₁)⊕Z⁰_a(B₁)⊕ Z⁺_a (B₂) and Z_a⁺(B₂) = Z⁺_a (B₁). Set X₁ = Z_a⁻(B₁)⊕Z_a⁰(B₁),X₂ = Z⁺_a (B₂) = Z_a⁺(B₁),x ∈ Z,x=x₁+x2 withx₁∈ X₁andx2∈ X2.

Let us follow the proof of Theorem1.1 until (3.2). Forx_2,n ∈Z_a⁺(B₂) =Z⁺_a (B₁), by (1.5) of (_Φ⁰₄), (3.2), Definition2.3and Proposition2.4, we have

c7≤ I(x2,n)

≤ −¹

2a(x2,n,x2,n) + ¹

2(B2x2,n,x2,n)_X+c

= −¹

2ψ_a,B2(x_2,n,x_2,n) +c.

Since (−ψ_a,B1(x₁,x₁))¹² + (ψ_a,B2(x₂,x₂))¹² is an equivalent norm on Z for x = x₁+x₂ with x₁ ∈ Z⁻_a (B₁)andx2 ∈ Z⁺_a (B2), wherex₁ =θ, we have ψ_a,B2(x2,n,x2,n) →+_∞via kx2,nk → _∞ asn→+∞. Thus, we have

I(x2,n)≤ −¹

2ψ_a,B2(x2,n,x2,n) +c→ −_∞ asn→+∞, which contradicts (3.2). Hence (2.1) of Lemma2.1holds.

By Lemma 2.1, I has at least one critical point. Hence problem (1.2) has at least one solution inZvia Lemma2.7. The proof is complete.

Proof of Theorem1.3. We apply Lemma2.1. Consider the functional I₁ defined by I₁(x) =−I(x) = ¹

2a(x,x)− ¹

2(B₁x,x)_X−_Φ(x)_{, (3.3)} for every x ∈ Z. Under assumption (_Φ0), it is easy to verify that I₁ ∈ C¹(Z,R) and I₁⁰ is weakly continuous.

Note that i_A(B₁) =0, we have Z_a⁻(B₁) ={θ}. Byν_A(B₁)6= 0, B₁ ≤ B₂ and B₁ < B₂ w.r.t.

ker(A−B₁)andi_A(B2) =i_A(B₁) +ν_A(B₁), we have Z=Z_a⁰(B₁)⊕Z⁰_a(B2)⊕Z⁺_a (B2)via (5) of Proposition2.6. SetX₁ = Z⁰_a(B₂)⊕Z⁺_a (B₂) = Z⁺_a (B₁),X₂ = Z⁰_a(B₁),x ∈ Z,x = x₁+x₂ with x₁ ∈ X₁ andx₂ ∈X₂. From Definition2.3 and Proposition2.4, we have

I₁(x) = I₁(x₁+x₂) = ¹

2a(x₁,x₁)− ¹

2(B₁x₁,x₁)_X−_Φ(x₁+x₂),

for every x ∈ Z. Thus, I₁(x₁,·) is weakly upper semi-continuous for all x₁ ∈ X₁ via Φ ∈ C¹(Z,R)is weakly continuous.

Next, we still divide the proof into three steps.

Step 1. We show that I₁(·,x₂) : X₁ →_R is convex for allx₂ ∈ X₂. By(_Φ⁰₁), it is obvious that

−_Φ(x₁+x₂) +¹₂((B₂−B₁)(x₁+x₂),x₁+x₂)_X is convex inx₁ ∈ X₁. From Definition2.3and Proposition2.4we know that for everyx₁ ∈X₁,

1

2ψ_a,B2(x₁,x₁) = ¹

2a(x₁,x₁)−¹

2(B₂x₁,x₁)_X≥0,

which shows that ¹₂ψ_a,B2(_x1,x1)is convex in x1∈ X1via Lemma2.2. Hence, for everyx2 ∈X2, I₁(x₁+x₂) = ¹

2a(x₁+x₂,x₁+x₂)−¹

2(B₁(x₁+x₂),x₁+x₂)_X−_Φ(x₁+x₂)

= ¹

2ψ_a,B2(x₁,x₁)−_Φ(x₁+x₂) + ¹

2((B₂−B₁)(x₁+x₂),x₁+x₂)_X+¹

2ψ_a,B2(x₂,x₂)

is convex in x₁∈ X₁.

Step 2. By contradiction, we verify that (2.2) of Lemma 2.1holds. If (2.2) of Lemma2.1 does not hold, there exist M > 0,c8 > 0 and two sequences {x1,n} ⊂ X₁ and {x2,n} ⊂ X2 with kx_1,nk →+_∞asn→_∞andkx_2,nk ≤ Mfor alln such that

I₁(x_1,n+x_2,n)≤c₈, ∀n∈N. (3.4) For x₁ ∈ X₁, write x₁ = x⁰₁+x₁⁺, where x⁰₁ ∈ Z⁰_a(B₂) and x⁺₁ ∈ Z_a⁺(B₂). Let us consider the functional

ϕ(x) =−_Φ(x) + ¹

2((B₂−B₁)x,x)_X

for all x ∈ X. By (_Φ0) and (_Φ⁰₁), we easily see that ϕ ∈ C¹(Z,R) and ϕ is weakly lower semi-continuous onZ_a⁰(B₂). Using (1.6) of(_Φ⁰⁰₄), by the least action principle (see Theorem 1.1 in [9]), ϕhas a minimum at somex⁰_1,0∈ Z⁰_a(B₂)for which

0= ϕ⁰(x⁰_1,0)x₁⁰=−(∇_Φ(x⁰_1,0),x⁰₁)_X+ ((B₂−B₁)x⁰_1,0,x₁⁰)_X, ∀x⁰₁∈ Z⁰_a(B₂). By ϕ∈C¹(Z,R)and the(B₂−B₁)-concavity ofΦ, we have

ϕ(x₁+x₂)−ϕ(x_1,0⁰ )

≥ −(∇_Φ(x_1,0⁰ ),x⁺₁ +x₂+x⁰₁−x⁰_1,0)_X+ ((B₂−B₁)x⁰_1,0,x₁⁺+x₂+x⁰₁−x⁰_1,0)_X

=−(∇_Φ(x_1,0⁰ ),x⁺₁ +x2)_X+ ((B2−B₁)x⁰_1,0,x⁺₁ +x2)_X, and then, fromkxk_X ≤ kxkfor all x∈ Z,

ϕ(x₁+x2)≥ ϕ(x⁰_1,0)−(k∇_Φ(x_1,0⁰ )k_X+k(B2−B₁)x⁰_1,0k_X)· kx₁⁺+x2k_X

≥ ϕ(x⁰_1,0)−(k∇_Φ(x_1,0⁰ )k_X+k(B₂−B₁)x⁰_1,0k_X)·(kx₁⁺k+kx₂k)

=c₉−c₁₀·(kx⁺₁k+kx₂k)

where c₉ = ϕ(x⁰_1,0),c₁₀ = k∇_Φ(x⁰_1,0)k_X+k(B₂−B₁)x⁰_1,0k_X ≥ 0. Rewrite x_1,n = x_1,n⁺ +x⁰_1,n, where x⁺_1,n ∈Z_a⁺(_B2)_andx⁰_1,n ∈Z_a⁰(_B2). By (3.4), we have

c₈≥ I₁(x_1,n+x_2,n) = ¹

2ψ_a,B2(x_1,n+x_2,n,x_1,n+x_2,n) + ¹

2((B₂−B₁)(x_1,n+x_2,n),x_1,n+x_2,n)_X−_Φ(x_1,n+x_2,n)

= ¹

2ψ_a,B2(x⁺_1,n,x⁺_1,n) + ¹

2ψ_a,B2(x_2,n,x_2,n) +ϕ(x_1,n+x_2,n)

≥ ¹

2ψ_a,B2(x⁺_1,n,x⁺_1,n) + ¹

2ψ_a,B2(x2,n,x2,n) +c9−c₁₀·(kx_1,n⁺ k+kx2,nk).

From(_Φ⁰⁰₄)and (5) of Proposition2.6, we know that(ψ_a,B2(x,x))¹² is an equivalent norm onZ for x∈ Z_a⁺(B2). Noticing that −ψ_a,B2(x,x)> 0 for allx∈ Z_a⁻(B2)\{θ}, so (−ψ_a,B2(x,x))¹² is a norm onZ⁻_a (B₂), which is equivalent tok · k_Z =k · kbecause of the finiteness of the subspace Z⁻_a (B₂). This means that there existc₁₁>0 andc₁₂ >0 such that

c₈≥ I₁(x_1,n+x_2,n)

≥ ^c

211

2 kx_1,n⁺ k²− ^c

212

2 kx_2,nk²+c₉−c₁₀·(kx⁺_1,nk+kx_2,nk)

≥ ^c

211

2 kx_1,n⁺ k²− ^c

212M²

2 +c₉−c₁₀·(kx_1,n⁺ k+M)

viakx_2,nk ≤ M, which shows that{kx_1,n⁺ k}is bounded. Combining this with assumption(_Φ2) and the(B2−B₁)-concavity ofΦ, we see that there existc₁₃>0 andc₁₄=sup_nϕ(−x⁺_1,n−x2,n) such that

c₈≥ I₁(x_1,n+x_2,n)

= ¹

2ψ_a,B2(x⁺_1,n,x_1,n⁺ ) + ¹

2ψ_a,B2(x_2,n,x_2,n) +ϕ(x_1,n+x_2,n)

≥ (c₁₁c₁₃)²

2 −^c

212M² 2 +2ϕ

1 2x⁰_1,n

−ϕ(−x⁺_1,n−x_2,n)

≥ (c₁₁c₁₃)²

2 −^c

212M² 2 +2ϕ

1 2x⁰_1,n

−c₁₄.

By (1.6) of(_Φ⁰⁰₄), we know that{kx⁰_1,nk}is also bounded. This contradicts the fact thatkx⁺_1,nk+ kx_1,n⁰ k ≥ kx1,nk →+_∞as n→∞. Therefore (2.2) of Lemma2.1holds.

Step 3. By X₂ = Z⁰_a(B₁)_{, we have} I₁(x₂) =−_Φ(x₂)_{for all} x₂ ∈ X₂. Thus, (2.1) of Lemma2.1 holds via(_Φ3).

By Lemma 2.1, I₁ has at least one critical point. Hence problem (1.2) has at least one solution inZvia Lemma2.7. The proof is complete.

Proof of Theorem1.4. we still consider the functional I₁ defined by (3.3). Under assumption (_Φ0), it is easy to verify that I₁∈ C¹(Z,R)and I₁⁰ is weakly continuous.

Byν_A(B₁)6=0, B₁ ≤ B2 andB₁ < B2 w.r.t. ker(A−B₁)andi_A(B2) = i_A(B₁) +ν_A(B₁), we haveZ= Z⁻_a (B₁)⊕Z⁰_a(B₁)⊕Z⁰_a(B₂)⊕Z⁺_a (B₂)via (5) of Proposition2.6. Note thati_A(B₁) =0 andν_A(B₂) = 0, we haveZ_a⁻(B₁) = Z⁰_a(B₂) = {θ}, which implies that Z = Z⁰_a(B₁)⊕Z⁺_a (B₂), Z⁻_a (B2) =Z_a⁰(B₁)andZ⁺_a (B2) =Z_a⁺(B₁). SetX₁= Z⁺_a (B2) =Z_a⁺(B₁),X2 =Z_a⁰(B₁),x∈ Z,x= x₁+x₂ withx₁ ∈ X₁ andx₂ ∈X₂.

From the proof of Theorem 1.3, it is not difficult to see that we only need to verify the validity of (2.2) in Lemma2.1. If (2.2) of Lemma2.1does not hold, there exist M >0,c₁₅> 0 and two sequences{x_1,n} ⊂X₁and{x_2,n} ⊂X₂withkx_1,nk →+_∞asn→_∞andkx_2,nk ≤M for allnsuch that

I₁(x_1,n+x2,n)≤c₁₅, ∀n∈N. (3.5) We consider the functional

ϕ(x) =−_Φ(x) +¹

2((B₂−B₁)x,x)_X

for allx∈ X. By(_Φ0)and(_Φ⁰₁), we easily see thatϕ∈C¹(Z,R). From the(B₂−B₁)-concavity ofΦ, we have

ϕ(x₁+x₂)−ϕ(θ)≥ −(∇_Φ(θ),x₁+x₂)_X+ ((B₂−B₁)θ,x₁+x₂)_X

= −(∇_Φ(θ),x₁+x₂)_X, and then, fromkxk_X ≤ kxkfor all x∈ Z,

ϕ(x₁+x₂)≥ ϕ(θ)− k∇_Φ(θ)k_X· kx₁+x₂k_X

≥ ϕ(θ)− k∇_Φ(θ)k_X(kx₁k+kx₂k).

From(_Φ⁰⁰⁰₄ )and (5) of Proposition2.6, we know that(ψ_a,B2(x,x))¹² is an equivalent norm onZ forx ∈Z⁺_a (B₂). Noticing that−ψ_a,B2(x,x)>_{0 for all}x ∈ Z⁻_a (B₂)\{θ}_{, so}(−ψ_a,B2(x,x))¹² _{is a}