# Existence of Three Solutions for Nonlinear Operator Equations and Applications to Second-Order Differential Equations

Authors: Mingliang Song; Shuyuan Mei
Journal: Journal of Function Spaces
 (2021)

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## Abstract

The existence of three solutions for nonlinear operator equations is established via index theory for linear self-adjoint operator equations, critical point reduction method, and three critical points theorems obtained by Brezis-Nirenberg, Ricceri, and Averna-Bonanno. Applying the results to second-order Hamiltonian systems satisfying generalized periodic boundary conditions or Sturm-Liouville boundary conditions and elliptic partial differential equations satisfying Dirichlet boundary value conditions, we obtain some new theorems concerning the existence of three solutions.


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## Body

## 1. Introduction and Main Results

In their excellent paper [1] in 1991, Brezis-Nirenberg proved the existence of two nonzero critical points under suitable assumptions by a negative gradient flow and linking. The obtained result was applied to second-order Hamiltonian systems and yielded the existence of three periodic solutions for these systems. After that, in [2], Tang obtained one existence of at least three periodic solutions for second-order Hamiltonian systems by using two nonzero critical points theorem in [1]. This result generalizes the corresponding result in [1]. In addition, in [3], one new existence of at least three periodic solutions for second-order Hamiltonian systems was obtained by using the critical point reduction method and two nonzero critical points theorem in [1, 4].Moreover, after the three critical points theorem (see [5, 6]) of Ricceri appearing, it has proved to be one of the most often widely used to solve differential equations, such as elliptic partial differential equations satisfying Dirichlet boundary value conditions (see [7–12]) and second-order Hamiltonian systems satisfying periodic boundary value conditions (see [3, 13]). Recently, a general three critical points theorem is given by Averna-Bonanno [14]. After that, in [15], the existence of at least three periodic solutions was established for a class of p-Hamiltonian systems by two general three critical points theorems in [6, 14].It is well known that the above-mentioned problems can be displayed as a self-adjoint operator equation. For the multiplicity of self-adjoint operator equation, Chang proved the existence of three distinct solutions under suitable assumptions via Morse theory in his excellent paper [16] in 1981. The obtained results were applied to second-order Hamiltonian systems and wave equations and yielded the existence of three periodic solutions for these systems. Similar results can be found in [17–21].Inspired by the ideas of [16, 18, 19], in this paper, we reconsider in the framework of the operator equations some theorems proved in [1–3, 9, 10, 15, 22] and generalize these theorems to nonlinear self-adjoint operator equations. Our technical approach is based on some three critical points theorems obtained by Brezis-Nirenberg [1], Averna-Bonanno [14], and Ricceri [6] or the critical point reduction method in [4]. As applications, we consider the existence of three solutions for second-order Hamiltonian systems satisfying generalized periodic boundary conditions or Sturm-Liouville boundary conditions and elliptic partial differential equations satisfying Dirichlet boundary value conditions. Meanwhile, we give some examples (see Examples 1–4 in Section 4.1) to illustrate the effectiveness of our results and point out that our results improve and generalize the corresponding results in [1–3, 9, 10, 15, 22] via several remarks (see Remarks 18, 20, 22, 24, 29, and 33 in Section 4).Before expressing our main results, for the convenience of the readers, we give a brief introduction to the index theory and the setting of the space; for some details, we refer to [23, 24] (or see Section 2).LetX be a real infinite-dimensional separable Hilbert space with inner product ·,·X and the corresponding norm ∥·∥X. Let A:DA⊂X⟶X be an unbounded linear self-adjoint operator with σA=σdA bounded from below. This implies that there is an orthonormal basis ejj=1∞ of X and λ1≤λ2≤⋯ such that Aej=λjej,DA=∑j=1∞cjej∣∑j=1∞λj2cj2<∞. Meanwhile, let Z≡DA1/2=∑j=1∞cjej∣∑j=1∞λjcj2<∞ with the norm xZ2=∑j=1∞1+λjcj2. For any x=∑j=1∞cjej∈Z,y=∑j=1∞djej∈Z, we define a bilinear form
(1)ax,y=∑j=1∞λjcjdj.ByAx,yX=ax,y if x∈DA,y∈Z, we can see that ax,y is the extension of Ax,yX on Z. Moreover, let B∈LsX be the usual space consisting of bounded symmetric operators in X. For any B∈LsX, we define
(2)νAB=dimkerA−B,iAB=∑λ<0νAB+λId,which are introduced by Dong (see Definition 3.1.1 and Proposition 3.1.4 in [23] or Definition 7.1.1 in [24]).Let us consider the following two nonlinear operator equations:(3)Ax−B0x=∇Φx,(4)Ax−B0x+B1x=η∇Φx+μ∇Ψx,where η,μ∈0,+∞,B0,B1∈LsX with iAB0=0,νAB0≠0, and B1 is positive definite for all x∈X, that is, there exists a positive constant λ¯ such that B1x,xX≥λ¯∥x∥X2 for all x∈X. Meanwhile, Φ and Ψ satisfy, respectively.Φ0Φ∈C1Z,R is weakly continuous with compact derivative, that is, xn⇀x0 in Z implies that Φxn⟶Φx0 and Φ′xn⟶Φ′x0. Moreover, for every x∈Z, there exists ∇Φx∈X such that Φ′xy=∇Φx,yX for all y∈Z.Ψ0Ψ∈C1Z,R is weakly continuous with compact derivative. Moreover, for every x∈Z, there exists ∇Ψx∈X such that Ψ′xy=∇Ψx,yX for all y∈Z.Noticing thatiAB0=0,νAB0≠0, we have minσA−B0=0. Then, we can define another inner product
(5)x,y=ax,y−B0x,yX+x,yX,∀x,y∈Z,with the corresponding norm
(6)x=ax,x−B0x,xX+xX21/2,∀x∈Z.Clearly,∥·∥ is equivalent to ∥·∥Z.Let(7)xB1=ax,x−B0x,xX+1/2,∀x∈Z.SinceB1 is positive definite, there exist λ¯≥λ¯>0 such that λ¯∥x∥X2≥B1x,xX≥λ¯∥x∥X2 for all x∈X. So we have
(8)min1,λ¯x2≤xB12≤max1,λ¯x2.Observe that the embeddingZ↪X is compact. Further, assume that there exists a Banach space X0 satisfying Z⊂X0⊂X and the embeddings Z↪X0 and X0↪X are compact and continuous, respectively. Thus, there exists k1>0 such that ∥x∥X0≤k1∥x∥ for all x∈Z. Let
(9)k2=supx∈Z\θxX0xB1.From (8), we have
(10)0<k2≤k~=k1min1,λ¯−1/2.LetX1 be a nontrivial subspace of X. For B1,B2∈LsX, we write B1≤B2 with respect to X1 if and only if B1x,xX≤B2x,xX for all x∈X1; we write B1<B2 w.r.t. X1 if and only if B1x,xX<B2x,xX for all x∈X1\θ. If X1=X, then we just write B1≤B2 or B1<B2.Next, we use the indexiAB,νAB∈N×N defined in [23, 24] (see Section 2) for all B∈LsX to reach our main results.Theorem 1.
Assume thatΦ satisfies Φ0 and
Φ1 there exist c0>0 and B2∈LsX with B2≥0,νAB0+B2=0 and iAB0+B2=νAB0, such that
(11)∇Φx,xX≤B2x,xX+c0∥x∥,for all x∈X
Φ2 there exists a positive definite B3∈LsX such that Φx−1/2B3x,xX is convex
Then, problem (3) has at least one solution with saddle character in Z (i.e., the solution is a saddle point).
Assume in addition that
Φ3 there exist δ>0,B01,B02∈LsX with B02>B01>B0, νAB0i≠0i=1,2 and iAB02=iAB01+νAB01, such that
(12)12B01−B0x,xX≤Φx≤12B02−B0x,xX,for all ∥x∥≤δ. Then, problem (3) has at least three distinct solutions in Z.Theorem 2.
Assume thatΦ satisfies Φ0 and
Φ4 there exist c1,c2>0 and γ0∈0,1 such that
(13)∣∇Φx,yX∣≤c1∥x∥X0γ0·∥y∥X0+c2∥y∥X0,for all x,y∈X
Φ5 there exists c3>c12k12/2ε22>0 such that
(14)limsup∥x∥→+∞Φx∥x∥2γ0<−c3,for all x∈kerA−B0, where the constants k1 and ε2 are given by (9) and (49), respectively.
Then, problem (3) has at least one solution in Z. Further, assume that Φ satisfies Φ3. Then, problem (3) has at least three distinct solutions in Z.Theorem 3.
Assume thatΦ satisfies Φ0 and
Φ6Φθ=0
Φ7 there exist c3,c4∈R+ and γ∈0,2 such that Φx≤c3xX0γ+c4 for all x∈Z
Φ8 there exist a positive constant c~ and a x0∈kerA−B0 with c~<k~x0Xλ¯, such that
(15)supx∈EΦx<c~2λ¯k~2x0X2Φx0,where E=x∈ZxX0≤c~.
Then, there exist an open intervalΛ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every Ψ:X⟶R satisfies Ψ0, there exists ρ>0 such that, for each μ∈0,ρ, problem (4) has at least three distinct solutions whose norms in Z are less than r0.Theorem 4.
Assume thatx0,c3,c4,c~, γ, and E are the same as in Theorem 3, Φ satisfies Φ0, Φ6, Φ7, and Φ9(16)supx∈EΦx<c~2c~2+λ¯k~2x0X2Φx0.
Then, for each(17)η∈λ¯x0X22Φx0−supx∈EΦx,c~22k~2supx∈EΦx,problem (4) admits at least three distinct solutions provided μ=0.Remark 5.
For some examples of the scalar equations corresponding to Theorems1–4, we will give in Section 4, respectively.The paper will be organized as follows. In Section2, we recall first a saddle point reduction theorem under rather general assumptions in [4], a general two nonzero critical points theorem in [1], and two general three critical points theorems in [6, 14] as well as one proposition in [25]. Then, we recall some useful conclusions of index theory for linear self-adjoint operator equations from [23, 24]. Finally, we quote a lemma in [26], which verifies that (3) and (4) possess a variational construction in Z. In Section 3, we prove Theorems 1–4. In Section 4, we will investigate its applications to second-order Hamiltonian systems satisfying generalized periodic boundary conditions or Sturm-Liouville boundary conditions and second-order elliptic partial differential equations satisfying Dirichlet boundary value conditions. Meanwhile, we give some examples and some remarks to illustrate that the corresponding results of [1–3, 9, 10, 15, 22] are special cases of these results in a sense.


## 2. Preliminaries

In order to prove our main results, we recall first four lemmas and one proposition due to [1, 4, 6, 14, 25], respectively.Lemma 6 (see [4], Theorem 2.3).
LetZ1,Z2 be Hilbert spaces, and let ψ∈C1Z1×Z2,R. Suppose that ψ satisfies −D1ψ·,x2:Z1⟶Z1 is α-monotone for all x2∈Z2, that is, there exists α>0 such that
(18)−D1ψx1,1,x2+D1ψx1,2,x2,x1,1−x1,2≥αx1,1−x1,22∀x1,1,x1,2∈Z1.
Then, there exists a mapϕ∈CZ2,Z1 such that ϕx2=x1x2 is the unique saddle point of ψ·,x2 for every x2∈Z2. Moreover, the map φ:Z2⟶R defined by
(19)φx2=ψx1x2,x2=supx1∈Z1ψx1,x2is continuously differentiable, and its derivative is given by
(20)φ′x2=D2ψx1x2,x2foreveryx2∈Z2.Lemma 7 (see [1], Theorem 4).
LetZ be a reflexive real Banach space with a direct sum decomposition Z=Z1⊕Z2 and dimZ1<+∞. Let I∈C1Z,R with Iθ=0, satisfying (PS) condition and assume that I is bounded below and infZI<0. Assume also that there is δ>0 such that
(21)Ix≤0,forx∈Z1,x≤δ,Ix≥0,forx∈Z2,x≤δ.
Then,I has at least two nonzero critical points.Lemma 8 (see [6], Theorem 1).
LetZ be a reflexive real Banach space, I~⊆R an interval, J:Z⟶R a sequentially weakly lower semicontinuous C1 functional whose derivative admits a continuous inverse on Z∗, and J bounded on each bounded subset of Z,Φ:Z⟶R a C1 functional with compact derivative. Assume that
(22)lim∥x∥→+∞Jx−ηΦx=+∞,for all η∈I~ and that there exists β∈R such that
(23)supη∈I~infx∈ZJx+ηβ−Φx<infx∈Zsupη∈I~Jx+ηβ−Φx.
Then, there exist a nonempty open setΛ∈I~ and a positive number r0 with the following property: for every η∈Λ and every C1 functional Ψ:Z⟶R with compact derivative, there exists ρ>0 such that, for each μ∈0,ρ, the equation
(24)J′x−ηΦ′x−μΨ′x=0has at least three solutions whose norms in Z are less than r0.Proposition 9 (see [25], Proposition 3.1).
LetZ be a nonempty set and J,Φ two real functions on Z. Assume that there are r~>0 and x0,x1∈Z such that
(25)Jx1=Φx1=0,Jx0>r~,supx∈J−1−∞,r~Φx<r~Φx0Jx0.
Then, for eachβ∈R satisfying
(26)supx∈J−1−∞,r~Φx<β<r~Φx0Jx0,one has
(27)supη≥0infx∈ZJx+ηβ−Φx<infx∈Zsupη≥0Jx+ηβ−Φx.Lemma 10 (see [14], Theorem B).
LetZ be a reflexive Banach space, J:Z⟶R a continuously Ga^teaux differentiable, coercive and sequentially weakly lower semicontinuous functional whose Ga^teaux derivative admits a continuous inverse on Z∗,Φ:Z⟶R a continuously Ga^teaux differentiable functional whose Ga^teaux derivative is compact. Put, for each r~>infZJ,
(28)φ1r~=infx∈J−1−∞,r~supJ−1−∞,r~¯ωΦ−Φxr~−Jx,φ2r~=infx∈J−1−∞,r~supy∈J−1r~,+∞Φy−ΦxJy−Jx,where J−1−∞,r~¯ω is the closure of J−1−∞,r~ in the weak topology, and assume that
(i)
there isr~∈R such that r~>infZJ and φ1r~<φ2r~(ii)
lim∥x∥→+∞Jx−ηΦx=+∞ for all η∈1/φ2r~,1/φ1r~
Then, for eachη∈1/φ2r~,1/φ1r~, the equation J′x−ηΦ′x=0 has at least three solutions in Z.
Next, we recall some definitions and propositions about index theory in [23, 24].Definition 11 (see [24], page 108).
For anyB∈LsX, we define
(29)ψa,Bx,y=ax,y−Bx,yX,∀x,y∈Z.
For anyx,y∈Z if ψa,Bx,y=0, we say that x and y are ψa,B-orthogonal. For any two subspaces Z1 and Z2 of Z if ψa,Bx,y=0 for any x∈Z1,y∈Z2, we say that Z1 and Z2 are ψa,B-orthogonal.Proposition 12 (see [23], Proposition 2.1.1; [24], Proposition 7.2.1).
For anyB∈LsX, the space Z has a ψa,B-orthogonal decomposition
(30)Z=Za+B⊕Za0B⊕Za−B,such that ψa,B is positive definite, null, and negative definite on Za+B,Za0B, and Za−B, respectively. Moreover, Za0B and Za−B are finitely dimensional.Definition 13 (see [23], Definition 2.1.2; [24], Definition 7.2.1).
For anyB∈LsX, we define νaB=dimZa0B,iaB=dimZa−B.Proposition 14 (1) (see [24], Proposition 7.2.2 (i)).
For anyB∈LsX, we have
(31)νAB=νaB,iAB=iaB,kerA−B=Za0B.
(2) ([27], Proposition 2.6 (5)). For any B1,B2∈LsX, if B1≤B2 and B1<B2 w.r.t. kerA−B1 and iAB2=iAB1+νAB1, then Z=Za−B1⊕Za0B1⊕Za0B2⊕Za+B2, and −ψa,B1x1,x11/2+ψa,B2x2,x21/2 is an equivalent norm on Z for x=x1+x2 with x1∈Za−B1,x2∈Za+B2. In particular, for any B1∈LsX, then Z=Za−B1⊕Za0B1⊕Za+B1 and −ψa,B1x1,x11/2+ψa,B1x2,x21/2 is also an equivalent norm on Z for x=x1+x2 with x1∈Za−B1,x2∈Za+B1.
Finally, let us consider two functionalsI and I1 defined by
(32)Ix=12ax,x−12B0x,xX−Φx,(33)I1x=12ax,x−12B0x,xX+12B1x,xX−ηΦx−μΨx,for every x∈Z and η,μ∈0,+∞, respectively. Under assumption Φ0 and Ψ0, from Theorem 2 in [28], we can verify that I,I1∈C1Z,R are weakly lower semicontinuous on Z and I′,I′1 are weakly continuous with
(34)I′xy=ax,y−B0x,yX−Φ′xy,(35)I1′xy=ax,y−B0x,yX+B1x,yX−ηΦ′xy−μΨ′xy,for every x,y∈Z and η,μ∈0,+∞.
The following important lemma is an immediate conclusion of Lemma6 in [26].Lemma 15.
Assume thatΦ0 and Ψ0 hold. Then, a critical point of Ix (or I1x) is a solution for problem (3) (or (4)).


## 3. Proofs of the Theorems

In this section, we give the proofs of Theorems1–4, respectively. First, with the aid of the critical point reduction method in [4] and the three critical points theorem in [1], we prove Theorem 1. For Theorem 2, we use Theorem 1 in [28] and the three critical points theorem in [1] to give its proof. In the Proof of Theorem 3, our technical approach is based on the three critical points theorem of Ricceri [6]. Finally, we use Proposition 3.1 in [25] and Theorem B in [14] to prove Theorem 4.Proof of Theorem 1.
By assumptioniAB0=0,νAB0≠0, Definition 13, and Proposition 12, we know that Z=Za0B0⊕Za+B0. Let Z1=Za0B0,Z2=Za+B0. The functional φ is defined by
(36)φx2=supx1∈Z1ψx1+x2=supx1∈Z1Ix1+x2,x2∈Z2.
From the convexity ofΦx−1/2B3x,xX and assumption Φ0, we can obtain
(37)∇Φx−∇Φy,x−yX≥B3x−y,x−yX,x,y∈Z.
SinceB3∈LsX is positive definite, there exists ε>0 such that B3x,xX≥εxX2 for all x∈X, which shows that
(38)∇Φx−∇Φy,x−yX≥εx−yX2,x,y∈Z.
Thus, for each fixedx2∈Z2 and any x1,1,x1,2∈Z1, we obtain
(39)∇Φx1,1+x2−∇Φx1,2+x2,x1,1−x1,2X≥εx1,1−x1,2X2.
SinceZ=Z1⊕Z2=Za0B0⊕Za+B0, from (39), Definition 11, and Proposition 12, we have
(40)−ψ′x1,1+x2−−ψ′x1,2+x2,x1,1−x1,2=−I′x1,1+x2−−I′x1,2+x2,x1,1−x1,2=−ax1,1−x1,2,x1,1−x1,2+B0x1,1−x1,2,x1,1−x1,2X+∇Φx1,1+x2−∇Φx1,2+x2,x1,1−x1,2X=−ψa,B0x1,1−x1,2,x1,1−x1,2+∇Φx1,1+x2−∇Φx1,2+x2,x1,1−x1,2X≥εx1,1−x1,2X2,via ψa,B0x,y=0 for all x,y∈Za0B0. Noticing that Z1=Za0B0 is finite-dimensional, we know that there exists ε0>0 such that
(41)−ψ′x1,1+x2−−ψ′x1,2+x2,x1,1−x1,2≥εε0x1,1−x1,22,which implies that ψx=Ix satisfies the εε0-monotonicity of Lemma 6. Thus, by Lemma 6, there exists a continuous mapping ϕ:Z2⟶Z1 such that φx2=ψϕx2+x2 for all x2∈Z2, where φ:Z2⟶R defined by (19) is continuous differentiable and φ′x2=ψ′ϕx2+x2Z2 for all x2∈Z2. This means that if x2∈Z2 is a critical point of φ, ϕx2+x2 is a critical point of ψ and I.
Again by assumptionΦ1, for every x2∈Z2, we have
(42)Φx2−Φθ=∫01∇Φsx2,x2Xds≤12B2x2,x2X+c0x2.
Thus, from (19), we have
(43)φx2≥ψx2=Ix2≥12ax2,x2−B0+B2x2,x2X−c0x2−Φθ=12ψa,B0+B2x2,x2−c0x2−Φθ,as x1=θ. Since νAB0+B2=0 and iAB0+B2=νAB0, by (2) of Proposition 14, we can see that ψa,B0+B2x2,x21/2 is an equivalent norm on Z for x2∈Z2=Za+B0=Za+B0+B2. Hence, there exists ε1>0 such that
(44)φx2≥ε12x22−c0x2−Φθ≥ε12x22−c0x2−Φθ⟶+∞,as x2⟶+∞. Noticing that I is weakly lower semicontinuous on Z, then there exists x2,0∈Z2 such that φx2,0=minx2∈Z2φx2, and hence, x2,0+ϕx2,0 is a solution with saddle point character of problem (3) in Z.
Moreover, if conditionΦ3 holds, then Z=Za−B01⊕Za0B01⊕Za0B02⊕Za+B02 via (2) of Proposition 14. Since B01>B0, from Definition 11, we have ψa,B0x,x>ψa,B01x,x for all x∈Z\θ and ψa,B0x,x>0 for all x∈Za0B01\θ with Za0B01⊂Za+B0. Let Z2,1=Za−B01⊕Za0B01∩Za+B0 and Z2,2 be the orthogonal complement of Z2,1 in Z2. We claim that ϕθ=θ. Indeed, (Φ3) means that Φθ=0 and ∇Φθ=θ. By (Φ2), (39), and ϕθ∈Z1=Za0B0, we have
(45)0=ψ′ϕθ,−ϕθ=I′ϕθ,−ϕθ=−∇Φϕθ,−ϕθX=∇Φθ−∇Φϕθ,−ϕθX≥εϕθX2≥0,which implies that ϕθ=θ. By the continuity of ϕ, we know that there exists 0<r<δ/2 such that ϕx2<δ/2 as x2≤r. Thus, from (12) of (Φ3) and Definition 11, we obtain
(46)φx2,1=ψx2,1+ϕx2,1=Ix2,1+ϕx2,1≤12ax2,1+ϕx2,1,x2,1+ϕx2,1−B01x2,1+ϕx2,1,x2,1+ϕx2,1X=12ψa,B01x2,1+ϕx2,1,x2,1+ϕx2,1≤0,for all x2,1∈Z2,1 with x2,1≤r via B01>B0 and x2,1+ϕx2,1∈Za−B01⊕Za0B01∪Z0B0, and
(47)φx2,2≥ψx2,2=Ix2,2≥12ax2,2,x2,2−B02x2,2,x2,2X=12ψa,B02x2,2,x2,2≥0,for all x2,2∈Z2,2 with x2,2≤r via Z2,2=Za+B01∩Za+B0 and Za+B01=Za+B02⊕Za0B02.
SinceI is weakly lower semicontinuous on Z, so is φ on Z2. From the weakly lower semicontinuity and coerciveness of φ, we can see that φ satisfies (PS) condition and is bounded below on Z2.
Ifinfφx2:x2∈Z2=0, then all x2,1∈Z2,1 with x2,1≤r are minima of φ, which shows that φ has infinite critical points. If infφx2:x2∈z2<0, then φ has at least two nonzero critical points via Lemma 7. Consequently, problem (3) has at least two nontrivial solutions in Z. Further, since Φθ=0, we know that problem (3) has trivial solution θ. Hence, problem (3) has at least three distinct solutions in Z. The proof is complete.Proof of Theorem 2.
By assumptioniAB0=0,νAB0≠0, Definition 13, and Proposition 12, we have Z=Za0B0⊕Za+B0. Let Z1=Za0B0,Z2=Za+B0. For x∈Z, we have x=x1+x2 with x1∈Z1 and x2∈Z2. It follows from Φ4 that
(48)Φx−Φx1=∫01∇Φx1+sx2,x2Xds≤c1x2X0∫01x1+sx2X0γ0ds+c2x2X0≤c1x2X0x1X0γ0+c1γ0+1x2X0γ0+1+c2x2X0≤c1k1ε2x22+c1k12εx12γ0+c1k1γ0+1x2γ0+1+c2k1x2,where ε is any positive real number. Since x2∈Z2=Za+B0, by (2) of Proposition 14, we know that ψa,B0x2,x21/2 is an equivalent norm on Z for x2∈Z2. Hence, there exists ε2>0 such that
(49)ψa,B0x2,x21/2≥ε2x2,for all x2∈Z2, and
(50)Ix=12ψa,B0x2,x2−Φx−Φx1−Φx1≥ε222x22−Φx−Φx1−Φx1≥ε22−c1k1ε2x22−x12γ0c1k12ε+Φx1x12γ0−c1k1γ0+1x2γ0+1−c2k1x2.
Byc3>c12k12/ε22>0, we can choose an ε0∈0,1 such that c3>1+ε0c12k12/ε22. Taking ε=ε22/c1k11+ε0, by Φ5, we have ε22−c1k1ε=1−1/1+ε0ε22>0 and
(51)c1k12ε+Φx1x12γ0<c1k12ε−c3<−1+ε02·c12k12ε22<0,for x1∈Z1=kerA−B0. As x⟶+∞ if and only if x1+x2⟶+∞, from the above equality and Φ5, we obtain that
(52)Ix⟶+∞,as x⟶+∞. Noticing that I is weakly lower semicontinuous on Z, by Theorem 1 in [28], we know that problem (3) has one solution in Z.
Moreover, if condition (Φ3) holds, then Z=Za−B01⊕Za0B01⊕Za0B02⊕Za+B02 via (2) of Proposition 14. Let Z¯1=Za−B01⊕Za0B01 and Z¯2=Za0B02⊕Za+B02. Then, we have Z=Z¯1⊕Z¯2 and dimZ¯1<+∞. From (12) of (Φ3) and Definition 11, we obtain
(53)Ix1≤12ax1,x1−B01x1,x1X=12ψa,B01x1,x1≤0,for all x1∈Z¯1 with x1≤δ, and
(54)Ix2≥12ax2,x2−B02x2,x2X=12ψa,B02x2,x2≥0,for all x2∈Z¯2 with x2≤δ.
By the weakly lower semicontinuity and coerciveness ofI, we can see that I satisfies (PS) condition and is bounded below on Z. Noticing that Ix=−Φx for all x∈kerA−B0, from (12) of (Φ3), we have Ix<0 for all x∈kerA−B0 with 0<x<δ and Iθ=0, which implies that infZI<0. Thus, I has at least two nonzero critical points via Lemma 7. Thanks to θ is also a trivial solution of problem (3), we know that problem (3) has at least three distinct solutions in Z. The proof is complete.Proof of Theorem 3.
Let(55)Jx=12ax,x−12B0x,xX+12B1x,xX=xB122,for each x∈Z. From (33) and (35), we can see that a solution of J′x−ηΦ′x−μΨ′x=0 must be a critical point of I1x, which means that problem (4) also has a solution via Lemma 15.
Obviously,J is a continuously Ga^teaux differentiable and sequentially weakly lower semicontinuous functional. From the conditions (Φ0) and (Ψ0), we know that Φ and Ψ are continuously Ga^teaux differentiable functional whose Ga^teaux derivative is compact. Moreover, J is bounded on each bounded subset of Z via (8).
Next, we prove thatJ admits a continuous inverse on Z. In fact, from (8) and (35), we have
(56)J′x−J′y,x−y=x−yB12≥min1,λ¯x−y2,for all x,y∈Z, which shows that J′ is uniformly monotone on Z. Moreover, using standard arguments ensure that J′ also turns out to be coercive and hemicontinuous in Z. Applying Theorem 26.A of [29], we know that J admits a continuous inverse on Z.
Thanks to (Φ7) and (8), for each η>0, we have
(57)lim∥x∥→+∞Jx−ηΦx≥lim∥x∥→+∞xB122−ηc3xX0γ+c4=+∞,via γ∈0,2 and xX0≤k1x. This implies that the assumption (22) of Lemma 8 holds.
Forx0∈kerA−B0 in (Φ8), by the definition of ∥·∥B1, we have
(58)λ¯x0X≥x0B1=B1x0,x0X1/2≥λ¯x0X.
Fromc~<k~x0Xλ¯, we see that
(59)x0B1>c~k~.
Hence, thanks to (15) of (Φ8) and (58), we have
(60)supx∈EΦx<c~2λ¯k~2∥x0∥X2Φx0≤c~2k~2∥x0∥B12Φx0.
Now, takingr~=1/2c~/k~2. By (59), we have
(61)Jx0=x0B122>r~>0=Jθ.
From (9) and (10), we obtain ∥x∥X0≤k~∥x∥B1 for all x∈Z. Hence, for each x∈Z such that
(62)Jx=xB122≤r~,we have 1/2xX0/k2≤r~=1/2c~/k2, which implies that
(63)xX0≤c~.
It follows from (60) and (63) that
(64)supx∈J−1−∞,r~Φx≤supx∈EΦx<c~2k~2x0B12Φx0=r~Φx0Jx0.
Putβ satisfying
(65)supx∈J−1−∞,r~Φx<β<r~Φx0Jx0,and x1=θ, by (61), (64), and Proposition 9, we have
(66)supη≥0infx∈ZJx+ηβ−Φx<infx∈Zsupη≥0Jx+ηβ−Φx,which implies that the assumption (23) of Lemma 8 holds as I~=0,+∞.
Finally, setI~=0,+∞, we can see that all the assumptions of Lemma 8 are satisfied. Hence, our conclusion follows from Lemma 8.Proof of Theorem 4.
In order to use Lemma10 to prove Theorem 4, we consider the functional I1 defined by (33) as μ=0. From the Proof of Theorem 3, it is not difficult to see that we only need to verify the validity of (i) in Lemma 10.
From the Proof of Theorem3, we know that Jθ=Φθ=0, r~=1/2c~/k2, and
(67)Jx0=x0B122>r~>0=infZJ.
This shows thatθ∈J−1−∞,r~ and x0∈J−1r~,+∞. From (63) and (Φ9), we have
(68)supx∈J−1−∞,r~Φx≤supx∈EΦx<c~2c~2+λ¯k~2∥x0∥X2Φx0<Φx0,which implies that
(69)Φx0−Φx>0,for each x∈J−1−∞,r~⊂Z. Consequently, it follows from (Φ9), (58) and (69) that
(70)φ1r~=infx∈J−1−∞,r~supJ−1−∞,r~Φ−Φxr~−Jx≤supJ−1−∞,r~Φ−Φθr~−Jθ=supJ−1−∞,r~Φr~≤supx∈EΦxr~=2k~2supx∈EΦxc~2<2k~2c~2+λ¯k~2∥x0∥X2Φx0,(71)φ2r~≥infx∈J−1−∞,r~Φx0−ΦxJx0−Jx≥infx∈J−1−∞,r~Φx0−ΦxJx0≥2∥x0∥B12Φx0−supx∈J−1−∞,r~Φx≥2λ¯∥x0∥X2Φx0−supx∈EΦx>2λ¯∥x0∥X2Φx0−c~2c~2+λ¯k~2∥x0∥X2Φx0≥2k~2c~2+λ¯k~2∥x0∥X2Φx0.
From (70) and (71), we obtain
(72)φ2r~>φ1r~,η∈λ¯x0X22Φx0−supx∈EΦx,c~22k~2supx∈EΦx⊂1φ2r~,1φ1r~.
Therefore, (i) of Lemma10 is satisfied. The proof is complete.


## 4. Applications to Second-Order Hamiltonian Systems and Elliptic Partial Differential Equations

In this section, as three examples of applications of the main results, we investigate second-order Hamiltonian systems satisfying generalized periodic boundary value conditions, second-order Hamiltonian systems satisfying Sturm-Liouville boundary value conditions, and elliptic partial differential equations satisfying Dirichlet boundary value conditions. For systematic researches of Hamiltonian systems, we refer to the excellent books [28, 30, 31]. Meanwhile, for researches of elliptic partial differential equations, we refer to [32–34].


### 4.1. Second-Order Hamiltonian Systems Satisfying Generalized Periodic Boundary Value Conditions

As a first example, we discuss the solvability of the following two generalized periodic boundary value problems(73)−x¨−B¯0tx=∇xVt,x,a.e.t∈0,1,x1=Mx0,ẋ1=Nẋ0,(74)−x¨−B¯0tx+B¯1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x1=Mx0,ẋ1=Nẋ0,where η,μ∈0,+∞,B¯0t,B¯1t∈L∞0,1,LsRn=Bt=bjkn×nbjkt=bkjt,t∈0,1,bjkt∈L∞0,1, M,N∈GLn=A=ajkn×najk∈RanddetA≠0, and MNT=In, where In is the unit matrix of order n, and ∇xVt,x and ∇xFt,x denote the gradient of Vt,x and Ft,x w.r.t. x∈Rn, respectively.LetX=L20,1,Rn. We define A1:DA1⟶X by A1xt=−x¨t where DA1=x∈H20,1,Rnxsatisfiesx1=Mx0,ẋ1=Nẋ0. Set Bixt=B¯itxti=0,1. Then, DBi=X. From Corollary 1.21 in [26], we see that A1 is self-adjoint and σA1=σdA1⊂0,+∞. Define iMB¯0=iA1B0,νMB¯0=νA1B0, that is, νMB¯0 is the dimension of the solution subspace of (73) as Vt,x≡0 and iMB¯0=∑λ<0νMB¯0+λIn.Remark 16 (see [23], Example 2.4.3; [24], Remark 7.1.3).
Letα1≤α2≤⋯≤αn be the eigenvalues of a constant n×n symmetric matrix B. For ζ∈R\0 with ξ0=arccos2/ζ−1+ζ, we have
(75)iζInB=∑k=1n#j∈N:2jπ+ξ02<αk+∑k=1n#j∈N:2π−ξ0+2jπ2<αk,(76)νζInB=∑k=1n#j∈N:2jπ+ξ02=αk+∑k=1n#j∈N:2π−ξ0+2jπ2=αk,where #S denotes the number of elements in a set S. In particular, formulae (75) and (76) when ζ=1 were given first by Mawhin and Willem in [28].
Now, we assume thatB¯1t is positive definite for all x∈Rn, that is, there exists a positive constant λ¯ such that B¯1tx,x≥λ¯x2 for all x∈Rn, and V,F:0,1×Rn⟶R satisfy the following condition:
H0Vt,x and Ft,x are measurable in t for every x∈Rn and continuously differentiable in x for a.e. t∈0,1.
Further, assume thatiMB¯0=0,νMB¯0≠0. Meanwhile, set Z1=x∈H10,1,Rn∣x1=Mx0. By Corollary 1.21 in [26] again, we know that Z1=DA11/2 and the norm
(77)xZ1=∫01ẋt2+xt2dt1/2,for each x∈Z1. Noticing that iMB¯0=0,νMB¯0≠0, we can define another equivalent norm
(78)x=∫01ẋt2−B¯0txt,xt+xt2dt1/2,for each x∈Z1.
Let(79)xB¯1=∫01ẋt2−B¯0txt,xt+B¯1txt,xtdt1/2,∀x∈Z1.
SinceB¯1t∈L∞0,1,LsRn is positive definite, there exist λ¯>λ¯>0 such that λ¯xL22≥∫01B¯1txt,xtdt≥λ¯xL22 for all x∈X. So, we have
(80)min1,λ¯x2≤xB¯12≤max1,λ¯x2.
Observe that the embeddingZ1°L2=X is compact. Let ∥·∥∞ be the norm of L∞0,1,Rn. Then, there is a constant δ0>0 such that
(81)x≤x∞≤δ0x,for any x∈Z1. Let
(82)k2=supx∈Z1\θ∥x∥∞∥x∥B¯1.
From (80) and (81), we have
(83)0<k2≤k~=δ0min1,λ¯−1/2.
For anyB¯1t,B¯2t∈L∞0,1,LsRn, we write B¯1≤B¯2 if B¯1t≤B¯2t for a.e. t∈0,1 and define B¯1<B¯2 if B¯1t<B¯2t a.e. t∈0,1. From Theorems 1–4, we have the following four results hold.Theorem 17.
Assume thatVt,x satisfies H0 and
H1 5there exist g∈L10,1,R+ and B¯2∈L∞0,1,LsRn with B¯2≥0,νMB¯0+B¯2=0 and iMB¯0+B¯2=νMB¯0 such that
(84)∇xVt,x,x≤B¯2tx,x+gtx,for all x∈Rn and a.e. t∈0,1
H2 there exists a positive definite B¯3∈L∞0,1,LsRn such that Vt,x−1/2B¯3tx,x is convex in x for a.e. t∈0,1
Then, problem (73) has at least one solution with saddle character in Z1.
Assume in addition that
H3 there exist δ1>0,B¯01,B¯02∈L∞0,1,LsRn such that B¯02>B¯01>B¯0 and νMB¯0i≠0i=1,2, iMB¯02=iMB¯01+νMB¯01, and for all ∣x∣≤δ1(85)12B¯01t−B¯0tx,x≤Vt,x≤12B¯02t−B¯0tx,x,for a.e. t∈0,1. Then, problem (73) has at least three distinct solutions in Z1.Proof.
Set(86)Φx=∫01Vt,xdt,∀x∈Z1.
Then,Φ∈C1Z1,R is weakly continuous with compact derivative, and for every x∈Z1,
(87)Φ′xy=∫01∇xVt,x,ydt,∀y∈Z1,because of H0 and (84). Hence, Φ0 holds.
Thanks to (81) and (84) of H1, we have
(88)∇Φx,xL2≤∫01B¯2txt,xtdt+∫01gtxtdt≤B2x,xL2+x∞∫01gtdt≤B2x,xL2+c0x,for all x∈L2, where c0=δ0∫01gtdt. This shows that Φ1 holds.
SinceB¯3t is positive definite, that is, there exists a positive constant ε0 such that B¯3tx,x≥ε0x2 for all x∈Rn, which implies that B3x,xL2≥ε0∥x∥L22 for all x∈L2. Thus, Φ2 follows from H2.
Clearly,B¯02>B¯01>B¯0⟹B02>B01>B0. Noticing that ∣x∣≤∥x∥∞≤δ0∥x∥ for any x∈Z1, we have ∣x∣≤δ1 as ∥x∥≤δ1/δ0=δ. So (12) follows from (85), which shows that Φ3 holds. Now, we can apply Theorem 1 to obtain Theorem 17.Remark 18.
In particular, setB¯0t≡0,B¯2t=ftIn,B¯3t=htIn,B¯01t=m22π2In,B¯02t=m+122π2In,m∈1,2,⋯ and M=In, where ft,ht∈L10,1,R+ with 0<ft<4π2 and inft∈0,1ht>0. Then, Z1=x∈H10,1,Rn∣x1=x0, σA1=2mπ2∣m∈N, and kerA1−B¯0=Rn. Hence, the following problem
(89)−x¨t=∇xVt,xt,x0−x1=ẋ0−ẋ1=0has a similar conclusion of Theorem 17. In addition, for the interval 0,T considered in second-order Hamiltonian systems satisfying periodic boundary value conditions. As T=1, the conclusion of Theorem 2.2 in [3] holds if we replace 0<ft<4π2 and inft∈0,1ht>0 with 0<∫01ftdt<12 and ∫01htdt>0, respectively. Clearly, inft∈0,1ht>0⟹∫01htdt>0. But for f∈L10,1,R+, we can see that ∫01ftdt<12⟹0<ft<4π2 and 0<ft<4π2⟹∫01ftdt<12. Indeed, if
(90)ft=8π2,x∈0,14π2,1−14π2,x∈14π2,1,then ∫01ftdt=3 and ft>4π2 for x∈0,1/4π2; if 12<ft<4π2, then ∫01ftdt>12. So Theorem 17 is a new result and in a sense a development of Theorem 2.2 in [3].
Next, we give some examples about the problem (89).Example 1.
Letft=2π21+κ0sin2t,gt=4π2,ht=2π2,B¯01t=2π2In,B¯02t=42π2In. For each κ0∈0,1/sin21, by the convexity of
(91)Ys=2π2ks−π2,s≥1,π2ks2,0≤s≤1,where k>0, we can verify that
(92)Vt,x=π21+κ0sin2tx2+2π21+κ0sin2t∣x∣−π21+k0sin2t,∣x∣≥1,2π21+κ0sin2tx2,∣x∣≤1satisfies the conditions of Theorem 17. So problem (89) has at least three distinct solutions. Noticing that ∫01ftdt>2π2>12, we know that Example 1 does not satisfy the conditions of Theorem 2.2 in [3].Theorem 19.
Assume thatVt,x satisfies H0 and
H4 there exist f,g∈L10,1,R+ and γ0∈0,1 such that
(93)∣∇xVt,x∣≤ftxγ0+gt,for all x∈Rn and a.e. t∈0,1
H5 there exists c3>δ02/2ε22∫01ftdt2>0 such that
(94)limsup∥x∥→+∞x−2γ0∫01Vt,xdt<−c3,for all x∈kerA1−B¯0, where the constants δ0 and ε2 are given by (81) and (49), respectively.
Then, problem (73) has at least one solution in Z1. Further, assume that Vt,x satisfies H3. Then, problem (73) has at least three distinct solutions in Z1.Proof.
Similar to the proof of Theorem17, we need only to show that Φ4 follows from H4. By H4, we have
(95)∣∇Φx,yL2∣≤∫01ftxtγ0·ytdt+∫01gtyt∣dt≤c1x∞γ0·y∞+c2x∞,for all x,y∈L2, where c1=∫01ftdt,c2=∫01gtdt. This shows that Φ4 holds. Hence, we can apply Theorem 2 to obtain Theorem 19.Remark 20.
In particular, setB¯0t≡0,B¯01t=m22π2In,B¯02t=m+122π2In,m∈1,2,⋯ and M=In. Then, the following problem (89) has a similar conclusion of Theorem 19. For the interval 0,T considered in second-order Hamiltonian systems satisfying periodic boundary value conditions, as T=1, the conclusion of Theorem 4 in [2] holds if we replace H5 with
(96)lim∣x∣→+∞x−2γ0∫01Vt,xdt=−∞.
Clearly, our result generalizes Theorem 4 in [2].
In addition, asT=1, in Theorem 7 of [1], assume that the smooth periodic function Vt,x satisfies H3,
(97)H5′Vt,x⟶−∞asx⟶+∞uniformlyint,and for some x0∈Rn,
(98)∫01Vt,x0dt>0.
Then, problem (89) has at least three distinct solutions. Clearly, condition H4 is stronger than condition of Theorem 7 in [1], and condition H5 is weaker than condition H′5. Moreover, we can see that the below Example 2 satisfies the conditions of Theorem 19 but does not satisfy Theorem 7 in [1]. So Theorem 19 is also a generalization of Theorem 7 in [1] in a sense. Next, we give an example about problem (89).Example 2.
Let(99)Vt,x=sin2πtx74+15−tx3/2+4π2sin2πt−74sin2πt+8π2,+32t−15∣x∣−2π2sin2πt+34sin2πt−4π2−12t−15,∣x∣≥1,2π22+sin2πtx2,∣x∣≤1,for all x∈Rn and t∈0,1. It is easy to see that
(100)∇xVt,x=74sin2πtxx1/4+3215−txx1/2+4π2sin2πt−74sin2πt,+8π2+32t−15x∣x∣,∣x∣≥1,4π22+sin2πtx,∣x∣≤1,∇xVt,x≤74sin2πt·x∣34+65x12+12π2+65,for all x∈Rn and a.e. t∈0,1. Note that for ε>0, we have
(101)65x1/2≤74εx3/4,as x∈Rn and ∣x∣≥1/ε4, which implies that 6/5x1/2≤7/4εx3/4+6/5ε2 and
(102)∇xVt,x≤74sin2πt+εx3/4+65ε2+13π2,for all x∈Rn and a.e. t∈0,1. The above shows H4 holds with γ0=3/4,ft=7/4sin2πt+ε,gt=6/5ε2+13π2. By Wirtinger’s inequality and Sobolev inequality of Proposition 1.3 in [28], we have kerA1−B¯0=Rn, x=x∞=x for all x∈Rn, and δ02=1/12,ε22=4π2/1+4π2, ∫01ftdt2=7/2π+7ε/42 via a simple calculation. Noticing that
(103)limsupx→+∞x−2γ0∫01Vt,xdt=−310<−1+4π296π272π+7ε42=−δ022ε22∫01ftdt2as ε>0 sufficiently small, we know that H5 holds. Set B¯01t=2π2In,B¯02t=42π2In. By B¯0t≡0 and Vt,x=2π22+sin2πtx2 as ∣x∣≤1, we can verify that H3 holds. By Theorem 19, problem (89) has at least three distinct solutions. Obviously, Example 19 does not satisfy the conditions of Theorem 4 in [2] and Theorem 7 in [1].Theorem 21.
Assume thatVt,x satisfies H0 and
H6Vt,0=0 for all t∈0,1
H7 there exist qt∈L10,1,R+ and γ∈0,2 such that Vt,x≤qt1+∣xγ for all x∈Rn
H8 there exist a positive constant c~ and a x0∈kerA1−B¯0 with c~<k~x0L2λ¯, such that
(104)supt,x∈0,1×E1Vt,x<c~2λ¯k~2∥x0∥L22∫01Vt,x0tdt,where E1=x∈Rn∣x≤c~.
Then, there exist an open intervalΛ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every Ft,x:0,1×Rn⟶R satisfies H0, there exists ρ>0 such that, for each μ∈0,ρ, problem (74) has at least three distinct solutions whose norms in Z1 are less than r0.Proof.
Clearly,H6 implies that Φθ=0. Similar to the proof of Theorem 17, set
(105)Ψx=∫01Ft,xdt,∀x∈Z1.
Then,Ψ0 holds. Again set X0=L∞0,1,Rn, we have Z1⊂X0⊂X=L2 and the embeddings Z1°X0 and X0°X=L2 are compact and continuous, respectively. By H7, we obtain Φx≤∫01qtdt+x∞γ∫01qtdt=c2+c1xX0γ for all x∈Z1, which shows that Φ7 holds. Thanks to (81), we know that E=x∈Z∥x∥∞≤c~⊂E1=x∈Rn∣x∣≤c~. So (15) follows from (104), which shows that Φ8 holds. Now, we can apply Theorem 3 to obtain Theorem 21.Remark 22.
In particular, setB¯0t≡0 and M=In. Then, the following problem
(106)−x¨+B¯1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x1−x0=ẋ1−ẋ0=0,has a similar conclusion of Theorem 21, where k~=2. In addition, our result slightly generalizes Theorem 1 in [15] as T=1,p=2 in two aspects. At first, Theorem 21 requires Vt,x being measurable in t for every x∈Rn instead of continuous in t for every x∈Rn; secondly Theorem 21 requires B¯1t∈L∞0,1,LsRn instead of B¯1t∈C0,1,LsRn.
An example of problem (106) corresponding to Theorem 21 is given below.Example 3.
Letn=2,
(107)B¯1t=f1t00f1t,and f1t∈L∞0,1,R with supf1t=2,inff1t=1. Then, λ¯=2,λ¯=1. Let
(108)Vx1,x2=x12+x2212e1−x12+x22+x12+x2234,∀x1,x2∈R2.
Then, for everyFt,x:0,1×Rn⟶R satisfies H0, all the hypotheses of Theorem 21 are satisfied with k~=2,c~=1,x0=3,6,γ=3/2 and qt is a large enough constant. So, problem (106) has at least three distinct solutions. Noting that f1t∈C0,1,R is not necessarily true, we can see that the above example does not satisfy the conditions of Theorem 1 in [15].Theorem 23.
Assume thatx0,c~, γ, qt, and E1 are the same as in Theorem 21, Vt,x satisfies (H0), (H6), (H7), and H9(109)supt,x∈0,1×E1Vt,x<c~2c~2+λ¯k~2∥x0∥L22∫01Vt,x0tdt.
Then, for each(110)η∈λ¯∥x0∥L222∫01Vt,x0tdt−supt,x∈0,1×E1Vt,x,c~22k~2supt,x∈0,1×E1Vt,x=E2,problem (74) admits at least three distinct solutions provided μ=0.Proof.
Similar to the proof of Theorem21, we need only to show that Φ9 follows from (H9). Noticing that E⊂E1, we have
(111)supx∈EΦx=supx∈E∫01Vt,xdt≤supt,x∈0,1×E1Vt,x,which means that Φ9 holds and
(112)η∈E2⊂λ¯∥x0∥L222Φx0−supx∈EΦx,c~22k~2supx∈EΦx.
Therefore, we can apply Theorem4 to obtain Theorem 23.Remark 24.
In particular, setB¯0t≡0 and M=In. Then, the following problem (106) has a similar conclusion of Theorem 23. By the above Remark 5, we can see that our result slightly generalizes Theorem 2 of [15] as T=1,p=2 in two aspects.
Finally, we give an example of problem (106) corresponding to Theorem 23.Example 4.
Letn=2,
(113)B¯1t=f1t00f1t,and f1t∈L∞0,1,R with supf1t=2,inff1t=1. Then, λ¯=2,λ¯=1. Let
(114)Vx1,x2=19x12+x2212e1−x12+x22,∀x1,x2∈R2.
Then, all the hypotheses of Theorem23 are satisfied with k~=2,c~=1,x0=3,6,γ=3/2 and qt is a large enough constant. So, for each η∈81e8/912−e8,9/4, problem (106) has at least three distinct solutions provided μ=0. Noting that f1t∈C0,1,R is not necessarily true, we can see that the above example does not satisfy the conditions of Theorem 2 in [15].


### 4.2. Second-Order Hamiltonian Systems Satisfying Sturm-Liouville Boundary Value Conditions

As a second example, we discuss the solvability of the following two Sturm-Liouville boundary value problems(115)−x¨−B~0tx=∇xVt,x,a.e.t∈0,1,x0cosα−ẋ0sinα=0,x1cosβ−ẋ1sinβ=0,(116)−x¨−B~0tx+B~1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x0cosα−ẋ0sinα=0,x1cosβ−ẋ1sinβ=0,where η,μ∈0,+∞,B~0t,B~1t∈L∞0,1,LsRn, α∈0,π,β∈0,π, and ∇xVt,x and ∇xFt,x denote the gradient of Vt,x and Ft,x for x∈Rn, respectively.LetX=L20,1,Rn. We define A2:DA2⟶X by A2xt=−x¨t where DA2=x∈H20,1,Rn∣xsatisfiesx0cosα−ẋ0sinα=0,x1cosβ−ẋ1sinβ=0. Set Bixt=B~itxti=0,1. Then, DBi=X. A2 is self-adjoint and σA2=σdA2 is bounded from below via Proposition 1.17 in [26]. Define iα,βB~0=iA2B0,να,βB~1=νA2B1, that is, να,βB~0 is the dimension of the solution subspace of (115) with Vt,x≡0.Now, we assume thatiα,βB~0=0,να,βB~0≠0, B~1t is positive definite for all x∈Rn, and V,F:0,1×Rn⟶R satisfies H0. Meanwhile, set
(117)Z2=x∈H10,1,Rn∣x1=0,α=0,β∈0,π,x∈H10,1,Rn∣x0=0,α∈0,π,β=π,x∈H10,1,Rn∣x1=x0=0,α=0,β=π,H10,1,Rn,α,β∈0,π.Thanks to Proposition 1.17 in [26], we have Z2=DA21/2 and the norm
(118)xZ2=∫01ẋt2+xt2dt1/2,for each x∈Z2. Noticing that iα,βB~0=0,να,βB~0≠0, we can define another equivalent norm
(119)x=∫01ẋt2−B~0txt,xt+xt2dt1/2,for each x∈Z2.Let(120)xB~1=∫01ẋt2−B~0txt,xt+B~1txt,xtdt1/2,∀x∈Z2.SinceB~1t∈L∞0,1,LsRn is positive definite, there exist λ¯>λ¯>0 such that λ¯xL22≥B1x,xX≥λ¯xL22 for all x∈X. Similar to Section 4.1, we know that (80)–(83) hold. Hence, the following four results also hold. Because of the similarities with the proofs of Theorems 17, 19, 21, and 23, we omit them here.Theorem 25.
Assume thatVt,x satisfies (H0), (H1), and (H2) with B¯0,B¯2, and B¯3 replaced with B~0,B~2, and B~3, respectively, then problem (115) has at least one solution with saddle character in Z2. In addition, assume that Vt,x satisfies H3 with B¯01 and B¯02 replaced with B~01 and B~02, respectively, then problem (115) has at least three distinct solutions in Z2.Theorem 26.
Assume thatVt,x satisfies H0, H4, and H5 with B¯0 replaced with B~0, then problem (115) has at least one solution in Z2. In addition assume that Vt,x satisfies H3 with B¯01 and B¯02 replaced with B~01 and B~02, respectively, then problem (115) has at least three distinct solutions in Z2.Theorem 27.
Assume thatVt,x satisfies (H0), (H6), (H7), and (H8) with B¯0 and A1 replaced with B~0 and A2, respectively, then there exist an open interval Λ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every Ft,x:0,1×Rn⟶R satisfies H0, there exists ρ>0 such that, for each μ∈0,ρ, problem (116) has at least three distinct solutions whose norms in Z2 are less than r0.Theorem 28.
Assume thatVt,x satisfies (H0), (H6), (H7), and (H9) with B¯0 and A1 replaced with B~0 and A2, respectively, then for each η∈E2 problem (116) admits at least three distinct solutions provided μ=0.Remark 29.
In particular, setα=0,β=π and B~0t≡π2In. Then, we have Z2=H01, σA2=k2π2∣k∈N\0 and kerA2−B~0=asintπ∣a∈Rn. Hence, the following problems
(121)−x¨t−π2x=∇xVt,xt,x0=x1=0,(122)−x¨−π2x+B~1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x0=x1=0,have the conclusion corresponding to Theorems 25–28, respectively, where B~01t≡kπ2In and B~02t=k+1π2In with k≥2. In (122), if B~1t=π2,μ=0, then Theorem 27 corresponding to problem (122) reduces to Theorem 2 in [10] and Theorem 2 in [22], respectively, and our result requires Vt,x being measurable in t for every x∈Rn instead of continuous in t for every x∈Rn.


### 4.3. Elliptic Partial Differential Equations Satisfying Dirichlet Boundary Value Conditions

As a third example, we discuss the solvability of the following two Dirichlet boundary value problems(123)−Δx−b0ux=fu,x,u∈Ω,x∂Ω=0,(124)−Δx−b0ux+b1ux=ηfu,x+μgu,x,u∈Ω,x∂Ω=0,where Ω∈Rn is a bounded open domain with smooth boundary, Δx=Σi=1n∂2/∂ui2x,b0u,b1u∈L∞Ω,R and η,μ∈0,+∞. As usual, Ω¯=Ω∪∂Ω.LetX=L2Ω¯,R. We define A3:DA3⟶X by A3xu=−Δxu where DA3=H02Ω¯. Set Bixu=biuxui=0,1. Then, DBi=X. It is well known that A3 is self-adjoint and σA1=σdA1⊂0,+∞. Define iΔb0=iA3B0,νΔb0=νA3B0 (see Definition 2.5.1 in [23]), that is, νΔb0 is the dimension of the solution subspace of (123) as fu,x≡0 and iΔb0=∑λ<0νΔb0+λ.Now, we assume thatb1u is positive definite for all x∈R, that is, there exists a positive constant λ¯ such that b1ux2≥λ¯x2 for all x∈R, and two Carathe′odory functions f,g:Ω×R⟶R satisfy the following condition:P0 there exist ai>0i=1,2 such that
(125)fu,x≤a11+∣xs1,∀u,x∈Ω×R,gu,x≤a21+∣xs2,∀u,x∈Ω×R,where s1,s2∈0,n+2/n−2 if n>2 and s1,s2∈0,+∞ if n=2.Further, assume thatiΔb0=0,νΔb0≠0. Meanwhile, set Z3=H01Ω¯=W01,2Ω. Obviously, Z3=DA31/2 and the norm
(126)xZ3=∫Ω∇xu2+x2udu1/2,for each x∈Z3. Noticing that iΔb0=0,νΔb0≠0, we can define another equivalent norm
(127)x=∫Ω∇xu2−b0ux2u+x2udu1/2,for each x∈Z3.Let(128)xb1=∫Ω∇xu2−b0ux2u+b1ux2udu1/2,∀x∈Z3.Sinceb1u∈L∞Ω,R is positive definite, there exist λ¯>λ¯>0 such that λ¯xL22≥∫Ωb1ux2udu≥λ¯xL22 for all x∈X. So we have
(129)min1,λ¯x2≤xb12≤max1,λ¯x2.Observe that the embeddingZ3↪L2=X is compact. Let ∥·∥∞ be the norm of L∞Ω,R. Then, there is a constant δ0>0 such that
(130)x≤x∞≤δ0x,for any x∈Z3. Let
(131)k2=supx∈Z3\θ∥x∥∞∥x∥b1.From (129) and (130), we have 0<k2≤k~=δ0min1,λ¯−1/2.For anyb1u,b2u∈L∞Ω,R, we write b1≤b2 if b1u≤b2u for a.e. u∈Ω and define b1<b2 if b1u<b2u a.e. u∈Ω. From Theorems 1–4, we have the following four results hold.Theorem 30.
Assume thatfu,x satisfies P0 and
P1 there exist g∈L1Ω,R+ and b2∈L∞Ω,R with b2≥0,νΔb0+b2=0 and iΔb0+b2=νΔb0 such that
(132)fu,x≤b2ux+gu,for all x∈R and a.e. u∈Ω
P2 there exists a positive definite b3∈L∞Ω,R such that ∫0xfu,tdt−1/2b3ux2 is convex in x for a.e. u∈Ω
Then, problem (123) has at least one solution with saddle character in Z3.
Assume in addition that
P3 there exist δ1>0,b01,b02∈L∞Ω,R such that b02>b01>b0 and νΔb0i≠0i=1,2, iΔb02=iΔb01+νΔb01, and for all ∣x∣≤δ1,
(133)12b01u−b0ux2≤∫0xfu,tdt≤12b02u−b0ux2,for a.e. u∈Ω. Then, problem (123) has at least three distinct solutions in Z3.Proof.
Set(134)Φx=∫Ω∫0xfu,tdtdu,∀x∈Z3.
Then,Φ∈C1Z3,R is weakly continuous with compact derivative, and for every x∈Z3,
(135)Φ′xy=∫Ωfu,xydu,∀y∈Z3,because of P0 and (132). Hence, Φ0 holds.
Thanks to (130) and (132) of P1, we have
(136)∇Φx,xL2≤∫Ωb2ux2udu+∫Ωguxudu≤B2x,xL2+x∞∫Ωgudu≤B2x,xL2+c0x,for all x∈L2, where c0=δ0∫Ωgudu. This shows that Φ1 holds.
Next, similar to the proof of Theorem17, we can get the conclusion of Theorem 30 via Theorem 1.Theorem 31.
Assume thatfu,x satisfies P0 and
P4 there exist f1,g∈L1Ω,R+ and γ0∈0,1 such that
(137)fu,x≤f1uxγ0+gu,for all x∈R and a.e. u∈Ω
P5 there exists c3>δ02/2ε22∫Ωfudu2>0 such that
(138)limsup∥x∥→+∞x−2γ0∫Ω∫0xfu,tdtdu<−c3,for all x∈kerA3−b0, where the constants δ0 and ε2 are given by (130) and (49), respectively.
Then, problem (123) has at least one solution in Z3. Further, assume that fu,x satisfies P3. Then, problem (123) has at least three distinct solutions in Z3.Proof.
Similar to the proof of Theorem30, we need only to show that Φ4 follows from P4. By P4, we have
(139)∇Φx,yL2≤∫Ωf1uxuγ0·yudt+∫Ωguyudu≤c1x∞γ0·y∞+c2x∞,for all x,y∈L2, where c1=∫Ωf1udu,c2=∫Ωgudu, which shows that Φ4 holds. So, Theorem 31 follows from Theorem 2.Theorem 32.
Assume thatfu,x satisfies P0 and
P7 there exist qu∈L1Ω,R+ and γ∈0,2 such that ∫0xfu,tdt≤qu1+xγ for all x∈R
P8 there exist a positive constant c~ and a x0∈kerA3−b0 with c~<k~∥x0∥L2λ¯, such that
(140)supu,x∈Ω×E1∫0xfu,tdt<c~2λ¯k~2∥x0∥L22·measΩ∫Ω∫0x0fu,tdtdu,where E1=x∈R∣x≤c~.
Then, there exist an open intervalΛ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every gu,x:Ω×R⟶R satisfies P0, there exists ρ>0 such that, for each μ∈0,ρ, problem (124) has at least three distinct solutions whose norms in Z3 are less than r0.Proof.
Clearly,Φθ=0 via the definition of Φx. Set X0=L∞Ω,R, we have Z3⊂X0⊂X=L2 and the embeddings Z3°X0 and X0°X=L2 are compact and continuous, respectively. Similar to the proofs of Theorem 21 and Theorem 30, we can get the conclusion of Theorem 32 via Theorem 3.Remark 33.
Asp=2 in [9], Theorem 32 reduces to Theorem 2.4 in [9], and the continuity of fu,x in u and the condition (i) of Theorem 2.4 in [9] is discarded.Theorem 34.
Assume thatx0,c~, γ, qu, and E1 are the same as in Theorem 32, fu,x satisfies P0, P7, and P9(141)supu,x∈Ω×E1∫0xfu,tdt<c~2c~2+λ¯k~2∥x0∥L22·measΩ∫Ω∫0x0fu,tdtdu.
Then, for each(142)η∈λ¯∥x0∥L222ρ0,c~22k~2ρ1=E2,problem (124) admits at least three distinct solutions provided μ=0, where
(143)ρ0=∫Ω∫0x0fu,tdtdu−ρ1,ρ1=measΩsupu,x∈Ω×E1∫0xfu,tdt.Proof.
Similar to the proof of Theorem32, we need only to show that Φ9 follows from P9. Noticing that E⊂E1, we have
(144)supx∈EΦx=supx∈E∫Ω∫0xfu,tdtdu≤measΩsupu,x∈Ω×E1∫0xfu,tdt,which means that Φ9 holds and
(145)η∈E2⊂λ¯x0L222Φx0−supx∈EΦx,c~22k~2supx∈EΦx.
Therefore, we can apply Theorem4 to obtain Theorem 34.


## 4.1. Second-Order Hamiltonian Systems Satisfying Generalized Periodic Boundary Value Conditions

As a first example, we discuss the solvability of the following two generalized periodic boundary value problems(73)−x¨−B¯0tx=∇xVt,x,a.e.t∈0,1,x1=Mx0,ẋ1=Nẋ0,(74)−x¨−B¯0tx+B¯1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x1=Mx0,ẋ1=Nẋ0,where η,μ∈0,+∞,B¯0t,B¯1t∈L∞0,1,LsRn=Bt=bjkn×nbjkt=bkjt,t∈0,1,bjkt∈L∞0,1, M,N∈GLn=A=ajkn×najk∈RanddetA≠0, and MNT=In, where In is the unit matrix of order n, and ∇xVt,x and ∇xFt,x denote the gradient of Vt,x and Ft,x w.r.t. x∈Rn, respectively.LetX=L20,1,Rn. We define A1:DA1⟶X by A1xt=−x¨t where DA1=x∈H20,1,Rnxsatisfiesx1=Mx0,ẋ1=Nẋ0. Set Bixt=B¯itxti=0,1. Then, DBi=X. From Corollary 1.21 in [26], we see that A1 is self-adjoint and σA1=σdA1⊂0,+∞. Define iMB¯0=iA1B0,νMB¯0=νA1B0, that is, νMB¯0 is the dimension of the solution subspace of (73) as Vt,x≡0 and iMB¯0=∑λ<0νMB¯0+λIn.Remark 16 (see [23], Example 2.4.3; [24], Remark 7.1.3).
Letα1≤α2≤⋯≤αn be the eigenvalues of a constant n×n symmetric matrix B. For ζ∈R\0 with ξ0=arccos2/ζ−1+ζ, we have
(75)iζInB=∑k=1n#j∈N:2jπ+ξ02<αk+∑k=1n#j∈N:2π−ξ0+2jπ2<αk,(76)νζInB=∑k=1n#j∈N:2jπ+ξ02=αk+∑k=1n#j∈N:2π−ξ0+2jπ2=αk,where #S denotes the number of elements in a set S. In particular, formulae (75) and (76) when ζ=1 were given first by Mawhin and Willem in [28].
Now, we assume thatB¯1t is positive definite for all x∈Rn, that is, there exists a positive constant λ¯ such that B¯1tx,x≥λ¯x2 for all x∈Rn, and V,F:0,1×Rn⟶R satisfy the following condition:
H0Vt,x and Ft,x are measurable in t for every x∈Rn and continuously differentiable in x for a.e. t∈0,1.
Further, assume thatiMB¯0=0,νMB¯0≠0. Meanwhile, set Z1=x∈H10,1,Rn∣x1=Mx0. By Corollary 1.21 in [26] again, we know that Z1=DA11/2 and the norm
(77)xZ1=∫01ẋt2+xt2dt1/2,for each x∈Z1. Noticing that iMB¯0=0,νMB¯0≠0, we can define another equivalent norm
(78)x=∫01ẋt2−B¯0txt,xt+xt2dt1/2,for each x∈Z1.
Let(79)xB¯1=∫01ẋt2−B¯0txt,xt+B¯1txt,xtdt1/2,∀x∈Z1.
SinceB¯1t∈L∞0,1,LsRn is positive definite, there exist λ¯>λ¯>0 such that λ¯xL22≥∫01B¯1txt,xtdt≥λ¯xL22 for all x∈X. So, we have
(80)min1,λ¯x2≤xB¯12≤max1,λ¯x2.
Observe that the embeddingZ1°L2=X is compact. Let ∥·∥∞ be the norm of L∞0,1,Rn. Then, there is a constant δ0>0 such that
(81)x≤x∞≤δ0x,for any x∈Z1. Let
(82)k2=supx∈Z1\θ∥x∥∞∥x∥B¯1.
From (80) and (81), we have
(83)0<k2≤k~=δ0min1,λ¯−1/2.
For anyB¯1t,B¯2t∈L∞0,1,LsRn, we write B¯1≤B¯2 if B¯1t≤B¯2t for a.e. t∈0,1 and define B¯1<B¯2 if B¯1t<B¯2t a.e. t∈0,1. From Theorems 1–4, we have the following four results hold.Theorem 17.
Assume thatVt,x satisfies H0 and
H1 5there exist g∈L10,1,R+ and B¯2∈L∞0,1,LsRn with B¯2≥0,νMB¯0+B¯2=0 and iMB¯0+B¯2=νMB¯0 such that
(84)∇xVt,x,x≤B¯2tx,x+gtx,for all x∈Rn and a.e. t∈0,1
H2 there exists a positive definite B¯3∈L∞0,1,LsRn such that Vt,x−1/2B¯3tx,x is convex in x for a.e. t∈0,1
Then, problem (73) has at least one solution with saddle character in Z1.
Assume in addition that
H3 there exist δ1>0,B¯01,B¯02∈L∞0,1,LsRn such that B¯02>B¯01>B¯0 and νMB¯0i≠0i=1,2, iMB¯02=iMB¯01+νMB¯01, and for all ∣x∣≤δ1(85)12B¯01t−B¯0tx,x≤Vt,x≤12B¯02t−B¯0tx,x,for a.e. t∈0,1. Then, problem (73) has at least three distinct solutions in Z1.Proof.
Set(86)Φx=∫01Vt,xdt,∀x∈Z1.
Then,Φ∈C1Z1,R is weakly continuous with compact derivative, and for every x∈Z1,
(87)Φ′xy=∫01∇xVt,x,ydt,∀y∈Z1,because of H0 and (84). Hence, Φ0 holds.
Thanks to (81) and (84) of H1, we have
(88)∇Φx,xL2≤∫01B¯2txt,xtdt+∫01gtxtdt≤B2x,xL2+x∞∫01gtdt≤B2x,xL2+c0x,for all x∈L2, where c0=δ0∫01gtdt. This shows that Φ1 holds.
SinceB¯3t is positive definite, that is, there exists a positive constant ε0 such that B¯3tx,x≥ε0x2 for all x∈Rn, which implies that B3x,xL2≥ε0∥x∥L22 for all x∈L2. Thus, Φ2 follows from H2.
Clearly,B¯02>B¯01>B¯0⟹B02>B01>B0. Noticing that ∣x∣≤∥x∥∞≤δ0∥x∥ for any x∈Z1, we have ∣x∣≤δ1 as ∥x∥≤δ1/δ0=δ. So (12) follows from (85), which shows that Φ3 holds. Now, we can apply Theorem 1 to obtain Theorem 17.Remark 18.
In particular, setB¯0t≡0,B¯2t=ftIn,B¯3t=htIn,B¯01t=m22π2In,B¯02t=m+122π2In,m∈1,2,⋯ and M=In, where ft,ht∈L10,1,R+ with 0<ft<4π2 and inft∈0,1ht>0. Then, Z1=x∈H10,1,Rn∣x1=x0, σA1=2mπ2∣m∈N, and kerA1−B¯0=Rn. Hence, the following problem
(89)−x¨t=∇xVt,xt,x0−x1=ẋ0−ẋ1=0has a similar conclusion of Theorem 17. In addition, for the interval 0,T considered in second-order Hamiltonian systems satisfying periodic boundary value conditions. As T=1, the conclusion of Theorem 2.2 in [3] holds if we replace 0<ft<4π2 and inft∈0,1ht>0 with 0<∫01ftdt<12 and ∫01htdt>0, respectively. Clearly, inft∈0,1ht>0⟹∫01htdt>0. But for f∈L10,1,R+, we can see that ∫01ftdt<12⟹0<ft<4π2 and 0<ft<4π2⟹∫01ftdt<12. Indeed, if
(90)ft=8π2,x∈0,14π2,1−14π2,x∈14π2,1,then ∫01ftdt=3 and ft>4π2 for x∈0,1/4π2; if 12<ft<4π2, then ∫01ftdt>12. So Theorem 17 is a new result and in a sense a development of Theorem 2.2 in [3].
Next, we give some examples about the problem (89).Example 1.
Letft=2π21+κ0sin2t,gt=4π2,ht=2π2,B¯01t=2π2In,B¯02t=42π2In. For each κ0∈0,1/sin21, by the convexity of
(91)Ys=2π2ks−π2,s≥1,π2ks2,0≤s≤1,where k>0, we can verify that
(92)Vt,x=π21+κ0sin2tx2+2π21+κ0sin2t∣x∣−π21+k0sin2t,∣x∣≥1,2π21+κ0sin2tx2,∣x∣≤1satisfies the conditions of Theorem 17. So problem (89) has at least three distinct solutions. Noticing that ∫01ftdt>2π2>12, we know that Example 1 does not satisfy the conditions of Theorem 2.2 in [3].Theorem 19.
Assume thatVt,x satisfies H0 and
H4 there exist f,g∈L10,1,R+ and γ0∈0,1 such that
(93)∣∇xVt,x∣≤ftxγ0+gt,for all x∈Rn and a.e. t∈0,1
H5 there exists c3>δ02/2ε22∫01ftdt2>0 such that
(94)limsup∥x∥→+∞x−2γ0∫01Vt,xdt<−c3,for all x∈kerA1−B¯0, where the constants δ0 and ε2 are given by (81) and (49), respectively.
Then, problem (73) has at least one solution in Z1. Further, assume that Vt,x satisfies H3. Then, problem (73) has at least three distinct solutions in Z1.Proof.
Similar to the proof of Theorem17, we need only to show that Φ4 follows from H4. By H4, we have
(95)∣∇Φx,yL2∣≤∫01ftxtγ0·ytdt+∫01gtyt∣dt≤c1x∞γ0·y∞+c2x∞,for all x,y∈L2, where c1=∫01ftdt,c2=∫01gtdt. This shows that Φ4 holds. Hence, we can apply Theorem 2 to obtain Theorem 19.Remark 20.
In particular, setB¯0t≡0,B¯01t=m22π2In,B¯02t=m+122π2In,m∈1,2,⋯ and M=In. Then, the following problem (89) has a similar conclusion of Theorem 19. For the interval 0,T considered in second-order Hamiltonian systems satisfying periodic boundary value conditions, as T=1, the conclusion of Theorem 4 in [2] holds if we replace H5 with
(96)lim∣x∣→+∞x−2γ0∫01Vt,xdt=−∞.
Clearly, our result generalizes Theorem 4 in [2].
In addition, asT=1, in Theorem 7 of [1], assume that the smooth periodic function Vt,x satisfies H3,
(97)H5′Vt,x⟶−∞asx⟶+∞uniformlyint,and for some x0∈Rn,
(98)∫01Vt,x0dt>0.
Then, problem (89) has at least three distinct solutions. Clearly, condition H4 is stronger than condition of Theorem 7 in [1], and condition H5 is weaker than condition H′5. Moreover, we can see that the below Example 2 satisfies the conditions of Theorem 19 but does not satisfy Theorem 7 in [1]. So Theorem 19 is also a generalization of Theorem 7 in [1] in a sense. Next, we give an example about problem (89).Example 2.
Let(99)Vt,x=sin2πtx74+15−tx3/2+4π2sin2πt−74sin2πt+8π2,+32t−15∣x∣−2π2sin2πt+34sin2πt−4π2−12t−15,∣x∣≥1,2π22+sin2πtx2,∣x∣≤1,for all x∈Rn and t∈0,1. It is easy to see that
(100)∇xVt,x=74sin2πtxx1/4+3215−txx1/2+4π2sin2πt−74sin2πt,+8π2+32t−15x∣x∣,∣x∣≥1,4π22+sin2πtx,∣x∣≤1,∇xVt,x≤74sin2πt·x∣34+65x12+12π2+65,for all x∈Rn and a.e. t∈0,1. Note that for ε>0, we have
(101)65x1/2≤74εx3/4,as x∈Rn and ∣x∣≥1/ε4, which implies that 6/5x1/2≤7/4εx3/4+6/5ε2 and
(102)∇xVt,x≤74sin2πt+εx3/4+65ε2+13π2,for all x∈Rn and a.e. t∈0,1. The above shows H4 holds with γ0=3/4,ft=7/4sin2πt+ε,gt=6/5ε2+13π2. By Wirtinger’s inequality and Sobolev inequality of Proposition 1.3 in [28], we have kerA1−B¯0=Rn, x=x∞=x for all x∈Rn, and δ02=1/12,ε22=4π2/1+4π2, ∫01ftdt2=7/2π+7ε/42 via a simple calculation. Noticing that
(103)limsupx→+∞x−2γ0∫01Vt,xdt=−310<−1+4π296π272π+7ε42=−δ022ε22∫01ftdt2as ε>0 sufficiently small, we know that H5 holds. Set B¯01t=2π2In,B¯02t=42π2In. By B¯0t≡0 and Vt,x=2π22+sin2πtx2 as ∣x∣≤1, we can verify that H3 holds. By Theorem 19, problem (89) has at least three distinct solutions. Obviously, Example 19 does not satisfy the conditions of Theorem 4 in [2] and Theorem 7 in [1].Theorem 21.
Assume thatVt,x satisfies H0 and
H6Vt,0=0 for all t∈0,1
H7 there exist qt∈L10,1,R+ and γ∈0,2 such that Vt,x≤qt1+∣xγ for all x∈Rn
H8 there exist a positive constant c~ and a x0∈kerA1−B¯0 with c~<k~x0L2λ¯, such that
(104)supt,x∈0,1×E1Vt,x<c~2λ¯k~2∥x0∥L22∫01Vt,x0tdt,where E1=x∈Rn∣x≤c~.
Then, there exist an open intervalΛ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every Ft,x:0,1×Rn⟶R satisfies H0, there exists ρ>0 such that, for each μ∈0,ρ, problem (74) has at least three distinct solutions whose norms in Z1 are less than r0.Proof.
Clearly,H6 implies that Φθ=0. Similar to the proof of Theorem 17, set
(105)Ψx=∫01Ft,xdt,∀x∈Z1.
Then,Ψ0 holds. Again set X0=L∞0,1,Rn, we have Z1⊂X0⊂X=L2 and the embeddings Z1°X0 and X0°X=L2 are compact and continuous, respectively. By H7, we obtain Φx≤∫01qtdt+x∞γ∫01qtdt=c2+c1xX0γ for all x∈Z1, which shows that Φ7 holds. Thanks to (81), we know that E=x∈Z∥x∥∞≤c~⊂E1=x∈Rn∣x∣≤c~. So (15) follows from (104), which shows that Φ8 holds. Now, we can apply Theorem 3 to obtain Theorem 21.Remark 22.
In particular, setB¯0t≡0 and M=In. Then, the following problem
(106)−x¨+B¯1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x1−x0=ẋ1−ẋ0=0,has a similar conclusion of Theorem 21, where k~=2. In addition, our result slightly generalizes Theorem 1 in [15] as T=1,p=2 in two aspects. At first, Theorem 21 requires Vt,x being measurable in t for every x∈Rn instead of continuous in t for every x∈Rn; secondly Theorem 21 requires B¯1t∈L∞0,1,LsRn instead of B¯1t∈C0,1,LsRn.
An example of problem (106) corresponding to Theorem 21 is given below.Example 3.
Letn=2,
(107)B¯1t=f1t00f1t,and f1t∈L∞0,1,R with supf1t=2,inff1t=1. Then, λ¯=2,λ¯=1. Let
(108)Vx1,x2=x12+x2212e1−x12+x22+x12+x2234,∀x1,x2∈R2.
Then, for everyFt,x:0,1×Rn⟶R satisfies H0, all the hypotheses of Theorem 21 are satisfied with k~=2,c~=1,x0=3,6,γ=3/2 and qt is a large enough constant. So, problem (106) has at least three distinct solutions. Noting that f1t∈C0,1,R is not necessarily true, we can see that the above example does not satisfy the conditions of Theorem 1 in [15].Theorem 23.
Assume thatx0,c~, γ, qt, and E1 are the same as in Theorem 21, Vt,x satisfies (H0), (H6), (H7), and H9(109)supt,x∈0,1×E1Vt,x<c~2c~2+λ¯k~2∥x0∥L22∫01Vt,x0tdt.
Then, for each(110)η∈λ¯∥x0∥L222∫01Vt,x0tdt−supt,x∈0,1×E1Vt,x,c~22k~2supt,x∈0,1×E1Vt,x=E2,problem (74) admits at least three distinct solutions provided μ=0.Proof.
Similar to the proof of Theorem21, we need only to show that Φ9 follows from (H9). Noticing that E⊂E1, we have
(111)supx∈EΦx=supx∈E∫01Vt,xdt≤supt,x∈0,1×E1Vt,x,which means that Φ9 holds and
(112)η∈E2⊂λ¯∥x0∥L222Φx0−supx∈EΦx,c~22k~2supx∈EΦx.
Therefore, we can apply Theorem4 to obtain Theorem 23.Remark 24.
In particular, setB¯0t≡0 and M=In. Then, the following problem (106) has a similar conclusion of Theorem 23. By the above Remark 5, we can see that our result slightly generalizes Theorem 2 of [15] as T=1,p=2 in two aspects.
Finally, we give an example of problem (106) corresponding to Theorem 23.Example 4.
Letn=2,
(113)B¯1t=f1t00f1t,and f1t∈L∞0,1,R with supf1t=2,inff1t=1. Then, λ¯=2,λ¯=1. Let
(114)Vx1,x2=19x12+x2212e1−x12+x22,∀x1,x2∈R2.
Then, all the hypotheses of Theorem23 are satisfied with k~=2,c~=1,x0=3,6,γ=3/2 and qt is a large enough constant. So, for each η∈81e8/912−e8,9/4, problem (106) has at least three distinct solutions provided μ=0. Noting that f1t∈C0,1,R is not necessarily true, we can see that the above example does not satisfy the conditions of Theorem 2 in [15].


## 4.2. Second-Order Hamiltonian Systems Satisfying Sturm-Liouville Boundary Value Conditions

As a second example, we discuss the solvability of the following two Sturm-Liouville boundary value problems(115)−x¨−B~0tx=∇xVt,x,a.e.t∈0,1,x0cosα−ẋ0sinα=0,x1cosβ−ẋ1sinβ=0,(116)−x¨−B~0tx+B~1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x0cosα−ẋ0sinα=0,x1cosβ−ẋ1sinβ=0,where η,μ∈0,+∞,B~0t,B~1t∈L∞0,1,LsRn, α∈0,π,β∈0,π, and ∇xVt,x and ∇xFt,x denote the gradient of Vt,x and Ft,x for x∈Rn, respectively.LetX=L20,1,Rn. We define A2:DA2⟶X by A2xt=−x¨t where DA2=x∈H20,1,Rn∣xsatisfiesx0cosα−ẋ0sinα=0,x1cosβ−ẋ1sinβ=0. Set Bixt=B~itxti=0,1. Then, DBi=X. A2 is self-adjoint and σA2=σdA2 is bounded from below via Proposition 1.17 in [26]. Define iα,βB~0=iA2B0,να,βB~1=νA2B1, that is, να,βB~0 is the dimension of the solution subspace of (115) with Vt,x≡0.Now, we assume thatiα,βB~0=0,να,βB~0≠0, B~1t is positive definite for all x∈Rn, and V,F:0,1×Rn⟶R satisfies H0. Meanwhile, set
(117)Z2=x∈H10,1,Rn∣x1=0,α=0,β∈0,π,x∈H10,1,Rn∣x0=0,α∈0,π,β=π,x∈H10,1,Rn∣x1=x0=0,α=0,β=π,H10,1,Rn,α,β∈0,π.Thanks to Proposition 1.17 in [26], we have Z2=DA21/2 and the norm
(118)xZ2=∫01ẋt2+xt2dt1/2,for each x∈Z2. Noticing that iα,βB~0=0,να,βB~0≠0, we can define another equivalent norm
(119)x=∫01ẋt2−B~0txt,xt+xt2dt1/2,for each x∈Z2.Let(120)xB~1=∫01ẋt2−B~0txt,xt+B~1txt,xtdt1/2,∀x∈Z2.SinceB~1t∈L∞0,1,LsRn is positive definite, there exist λ¯>λ¯>0 such that λ¯xL22≥B1x,xX≥λ¯xL22 for all x∈X. Similar to Section 4.1, we know that (80)–(83) hold. Hence, the following four results also hold. Because of the similarities with the proofs of Theorems 17, 19, 21, and 23, we omit them here.Theorem 25.
Assume thatVt,x satisfies (H0), (H1), and (H2) with B¯0,B¯2, and B¯3 replaced with B~0,B~2, and B~3, respectively, then problem (115) has at least one solution with saddle character in Z2. In addition, assume that Vt,x satisfies H3 with B¯01 and B¯02 replaced with B~01 and B~02, respectively, then problem (115) has at least three distinct solutions in Z2.Theorem 26.
Assume thatVt,x satisfies H0, H4, and H5 with B¯0 replaced with B~0, then problem (115) has at least one solution in Z2. In addition assume that Vt,x satisfies H3 with B¯01 and B¯02 replaced with B~01 and B~02, respectively, then problem (115) has at least three distinct solutions in Z2.Theorem 27.
Assume thatVt,x satisfies (H0), (H6), (H7), and (H8) with B¯0 and A1 replaced with B~0 and A2, respectively, then there exist an open interval Λ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every Ft,x:0,1×Rn⟶R satisfies H0, there exists ρ>0 such that, for each μ∈0,ρ, problem (116) has at least three distinct solutions whose norms in Z2 are less than r0.Theorem 28.
Assume thatVt,x satisfies (H0), (H6), (H7), and (H9) with B¯0 and A1 replaced with B~0 and A2, respectively, then for each η∈E2 problem (116) admits at least three distinct solutions provided μ=0.Remark 29.
In particular, setα=0,β=π and B~0t≡π2In. Then, we have Z2=H01, σA2=k2π2∣k∈N\0 and kerA2−B~0=asintπ∣a∈Rn. Hence, the following problems
(121)−x¨t−π2x=∇xVt,xt,x0=x1=0,(122)−x¨−π2x+B~1tx=η∇xVt,x+μ∇xFt,x,a.e.t∈0,1,x0=x1=0,have the conclusion corresponding to Theorems 25–28, respectively, where B~01t≡kπ2In and B~02t=k+1π2In with k≥2. In (122), if B~1t=π2,μ=0, then Theorem 27 corresponding to problem (122) reduces to Theorem 2 in [10] and Theorem 2 in [22], respectively, and our result requires Vt,x being measurable in t for every x∈Rn instead of continuous in t for every x∈Rn.


## 4.3. Elliptic Partial Differential Equations Satisfying Dirichlet Boundary Value Conditions

As a third example, we discuss the solvability of the following two Dirichlet boundary value problems(123)−Δx−b0ux=fu,x,u∈Ω,x∂Ω=0,(124)−Δx−b0ux+b1ux=ηfu,x+μgu,x,u∈Ω,x∂Ω=0,where Ω∈Rn is a bounded open domain with smooth boundary, Δx=Σi=1n∂2/∂ui2x,b0u,b1u∈L∞Ω,R and η,μ∈0,+∞. As usual, Ω¯=Ω∪∂Ω.LetX=L2Ω¯,R. We define A3:DA3⟶X by A3xu=−Δxu where DA3=H02Ω¯. Set Bixu=biuxui=0,1. Then, DBi=X. It is well known that A3 is self-adjoint and σA1=σdA1⊂0,+∞. Define iΔb0=iA3B0,νΔb0=νA3B0 (see Definition 2.5.1 in [23]), that is, νΔb0 is the dimension of the solution subspace of (123) as fu,x≡0 and iΔb0=∑λ<0νΔb0+λ.Now, we assume thatb1u is positive definite for all x∈R, that is, there exists a positive constant λ¯ such that b1ux2≥λ¯x2 for all x∈R, and two Carathe′odory functions f,g:Ω×R⟶R satisfy the following condition:P0 there exist ai>0i=1,2 such that
(125)fu,x≤a11+∣xs1,∀u,x∈Ω×R,gu,x≤a21+∣xs2,∀u,x∈Ω×R,where s1,s2∈0,n+2/n−2 if n>2 and s1,s2∈0,+∞ if n=2.Further, assume thatiΔb0=0,νΔb0≠0. Meanwhile, set Z3=H01Ω¯=W01,2Ω. Obviously, Z3=DA31/2 and the norm
(126)xZ3=∫Ω∇xu2+x2udu1/2,for each x∈Z3. Noticing that iΔb0=0,νΔb0≠0, we can define another equivalent norm
(127)x=∫Ω∇xu2−b0ux2u+x2udu1/2,for each x∈Z3.Let(128)xb1=∫Ω∇xu2−b0ux2u+b1ux2udu1/2,∀x∈Z3.Sinceb1u∈L∞Ω,R is positive definite, there exist λ¯>λ¯>0 such that λ¯xL22≥∫Ωb1ux2udu≥λ¯xL22 for all x∈X. So we have
(129)min1,λ¯x2≤xb12≤max1,λ¯x2.Observe that the embeddingZ3↪L2=X is compact. Let ∥·∥∞ be the norm of L∞Ω,R. Then, there is a constant δ0>0 such that
(130)x≤x∞≤δ0x,for any x∈Z3. Let
(131)k2=supx∈Z3\θ∥x∥∞∥x∥b1.From (129) and (130), we have 0<k2≤k~=δ0min1,λ¯−1/2.For anyb1u,b2u∈L∞Ω,R, we write b1≤b2 if b1u≤b2u for a.e. u∈Ω and define b1<b2 if b1u<b2u a.e. u∈Ω. From Theorems 1–4, we have the following four results hold.Theorem 30.
Assume thatfu,x satisfies P0 and
P1 there exist g∈L1Ω,R+ and b2∈L∞Ω,R with b2≥0,νΔb0+b2=0 and iΔb0+b2=νΔb0 such that
(132)fu,x≤b2ux+gu,for all x∈R and a.e. u∈Ω
P2 there exists a positive definite b3∈L∞Ω,R such that ∫0xfu,tdt−1/2b3ux2 is convex in x for a.e. u∈Ω
Then, problem (123) has at least one solution with saddle character in Z3.
Assume in addition that
P3 there exist δ1>0,b01,b02∈L∞Ω,R such that b02>b01>b0 and νΔb0i≠0i=1,2, iΔb02=iΔb01+νΔb01, and for all ∣x∣≤δ1,
(133)12b01u−b0ux2≤∫0xfu,tdt≤12b02u−b0ux2,for a.e. u∈Ω. Then, problem (123) has at least three distinct solutions in Z3.Proof.
Set(134)Φx=∫Ω∫0xfu,tdtdu,∀x∈Z3.
Then,Φ∈C1Z3,R is weakly continuous with compact derivative, and for every x∈Z3,
(135)Φ′xy=∫Ωfu,xydu,∀y∈Z3,because of P0 and (132). Hence, Φ0 holds.
Thanks to (130) and (132) of P1, we have
(136)∇Φx,xL2≤∫Ωb2ux2udu+∫Ωguxudu≤B2x,xL2+x∞∫Ωgudu≤B2x,xL2+c0x,for all x∈L2, where c0=δ0∫Ωgudu. This shows that Φ1 holds.
Next, similar to the proof of Theorem17, we can get the conclusion of Theorem 30 via Theorem 1.Theorem 31.
Assume thatfu,x satisfies P0 and
P4 there exist f1,g∈L1Ω,R+ and γ0∈0,1 such that
(137)fu,x≤f1uxγ0+gu,for all x∈R and a.e. u∈Ω
P5 there exists c3>δ02/2ε22∫Ωfudu2>0 such that
(138)limsup∥x∥→+∞x−2γ0∫Ω∫0xfu,tdtdu<−c3,for all x∈kerA3−b0, where the constants δ0 and ε2 are given by (130) and (49), respectively.
Then, problem (123) has at least one solution in Z3. Further, assume that fu,x satisfies P3. Then, problem (123) has at least three distinct solutions in Z3.Proof.
Similar to the proof of Theorem30, we need only to show that Φ4 follows from P4. By P4, we have
(139)∇Φx,yL2≤∫Ωf1uxuγ0·yudt+∫Ωguyudu≤c1x∞γ0·y∞+c2x∞,for all x,y∈L2, where c1=∫Ωf1udu,c2=∫Ωgudu, which shows that Φ4 holds. So, Theorem 31 follows from Theorem 2.Theorem 32.
Assume thatfu,x satisfies P0 and
P7 there exist qu∈L1Ω,R+ and γ∈0,2 such that ∫0xfu,tdt≤qu1+xγ for all x∈R
P8 there exist a positive constant c~ and a x0∈kerA3−b0 with c~<k~∥x0∥L2λ¯, such that
(140)supu,x∈Ω×E1∫0xfu,tdt<c~2λ¯k~2∥x0∥L22·measΩ∫Ω∫0x0fu,tdtdu,where E1=x∈R∣x≤c~.
Then, there exist an open intervalΛ⊆0,+∞ and a positive number r0 with the following property: for each η∈Λ and for every gu,x:Ω×R⟶R satisfies P0, there exists ρ>0 such that, for each μ∈0,ρ, problem (124) has at least three distinct solutions whose norms in Z3 are less than r0.Proof.
Clearly,Φθ=0 via the definition of Φx. Set X0=L∞Ω,R, we have Z3⊂X0⊂X=L2 and the embeddings Z3°X0 and X0°X=L2 are compact and continuous, respectively. Similar to the proofs of Theorem 21 and Theorem 30, we can get the conclusion of Theorem 32 via Theorem 3.Remark 33.
Asp=2 in [9], Theorem 32 reduces to Theorem 2.4 in [9], and the continuity of fu,x in u and the condition (i) of Theorem 2.4 in [9] is discarded.Theorem 34.
Assume thatx0,c~, γ, qu, and E1 are the same as in Theorem 32, fu,x satisfies P0, P7, and P9(141)supu,x∈Ω×E1∫0xfu,tdt<c~2c~2+λ¯k~2∥x0∥L22·measΩ∫Ω∫0x0fu,tdtdu.
Then, for each(142)η∈λ¯∥x0∥L222ρ0,c~22k~2ρ1=E2,problem (124) admits at least three distinct solutions provided μ=0, where
(143)ρ0=∫Ω∫0x0fu,tdtdu−ρ1,ρ1=measΩsupu,x∈Ω×E1∫0xfu,tdt.Proof.
Similar to the proof of Theorem32, we need only to show that Φ9 follows from P9. Noticing that E⊂E1, we have
(144)supx∈EΦx=supx∈E∫Ω∫0xfu,tdtdu≤measΩsupu,x∈Ω×E1∫0xfu,tdt,which means that Φ9 holds and
(145)η∈E2⊂λ¯x0L222Φx0−supx∈EΦx,c~22k~2supx∈EΦx.
Therefore, we can apply Theorem4 to obtain Theorem 34.



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