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Research article

Solutions to a discrete resonance problem with eigenparameter-dependent boundary conditions


  • Received: 11 September 2023 Revised: 03 February 2024 Accepted: 19 February 2024 Published: 26 February 2024
  • In this paper, we considered the existence of solutions to a discrete second-order resonance problem with eigenparameter-dependent boundary conditions. We first transformed the resonance problem into its corresponding equivalent system using the Lyapunov-Schmidt method. In addition, using Schauder's fixed-point theorem and the connectivity theories of the solution set of compact vector fields, we obtained the existence and multiplicity of solutions to the second-order discrete resonance problem with eigenparameter-dependent boundary conditions.

    Citation: Chenghua Gao, Enming Yang, Huijuan Li. Solutions to a discrete resonance problem with eigenparameter-dependent boundary conditions[J]. Electronic Research Archive, 2024, 32(3): 1692-1707. doi: 10.3934/era.2024077

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  • In this paper, we considered the existence of solutions to a discrete second-order resonance problem with eigenparameter-dependent boundary conditions. We first transformed the resonance problem into its corresponding equivalent system using the Lyapunov-Schmidt method. In addition, using Schauder's fixed-point theorem and the connectivity theories of the solution set of compact vector fields, we obtained the existence and multiplicity of solutions to the second-order discrete resonance problem with eigenparameter-dependent boundary conditions.



    In 2005, Rodríguez [1] used the Lyapunov-Schmidt method and Brower fixed-point theorem to discuss the following discrete Sturm-Liouville boundary value problem

    {Δ[p(t1)Δy(t1)]+q(t)y(t)+λy(t)=f(y(t)), t[a+1,b+1]Z,a11y(a)+a12Δy(a)=0, a21y(b+1)+a22Δy(b+1)=0,

    where λ is the eigenvalue of the corresponding linear problem and the nonlinearity f is bounded.

    Furthermore, in 2007, Ma [2] studied the following discrete boundary value problem

    {Δ[p(t1)Δy(t1)]+q(t)y(t)+λy(t)=f(t,y(t))+h(t), t[a+1,b+1]Z,a11y(a)+a12Δy(a)=0, a21y(b+1)+a22Δy(b+1)=0,

    where f is subject to the sublinear growth condition

    |f(t,s)|A|s|α+B,sR

    for some 0α<1 and A,B(0,). Additional results to the existence of solutions to the related continuous and discrete problems on the nonresonance and the resonance can be found in [3,4,5,6,7,8,9,10,11,12,13] and the references therein. For example, Li and Shu [14] considered the existence of solutions to the continuous Sturm-Liouville problem with random impulses and boundary value problems using the Dhage's fixed-point theorem and considered the existence of upper and lower solutions to a second-order random impulsive differential equation in [15] using the monotonic iterative method.

    Inspired by the above literature, we use the solution set connectivity theory of compact vector field [16] to consider the existence of solutions to discrete resonance problems

    {Δ[p(t1)Δy(t1)]+q(t)y(t)=λkr(t)y(t)+f(t,y(t))+γψk(t)+¯g(t),   t[1,T]Z,(a0λk+b0)y(0)=(c0λk+d0)Δy(0),(a1λk+b1)y(T+1)=(c1λk+d1)y(T+1), (1.1)

    where p:[0,T]Z(0,), q:[1,T]ZR, ¯g:[1,T]ZR, r(t)>0, t[1,T]Z, (λk,ψk) is the eigenpair of the corresponding linear problem

    {Δ[p(t1)Δy(t1)]+q(t)y(t)=λr(t)y(t), t[1,T]Z,(a0λ+b0)y(0)=(c0λ+d0)Δy(0),(a1λ+b1)y(T+1)=(c1λ+d1)y(T+1). (1.2)

    It is worth noting that the difference between the problem (1.1) and the above questions is the eigenvalue that not only appears in the equation but also in the boundary conditions, which causes us considerable difficulties. Furthermore, it should be noted that these problems also apply to a number of physical problems, including those involving heat conduction, vibrating strings, and so on. For instance, Fulton and Pruess [17] discussed a kind of heat conduction problem, which has the eigenparameter-dependent boundary conditions. However, to discuss this kind of problem, we should know the spectrum of the problem (1.2). Fortunately, in 2016, Gao and Ma [18] obtained the eigenvalue theory of problem (1.2) under the conditions listed as follows:

    (A1) δ0:=a0d0b0c0<0,c00, d1b10,

    (A2) δ1:=a1d1b1c1>0,c10, b0+d00,

    which laid a theoretical foundation for this paper.

    Under the conditions (A1) and (A2), we assume the following conditions hold:

    (H1) (Sublinear growth condition) f:[1,T]Z×RR is continuous and there exist α[0,1) and A,B(0,), such that

    |f(t,y)|A|y|α+B,

    (H2) (Symbol condition) There exists ω>0, such that

    yf(t,y)>0,t[1,T]Zfor|y|>ω, (1.3)

    or

    yf(t,y)<0,t[1,T]Zfor|y|>ω, (1.4)

    (H3) ¯g:[1,T]ZR satisfies

    Ts=1¯g(s)ψk(s)=0, (1.5)

    (H4) f:[1,T]Z×RR is continuous and

    lim|y|f(t,y)=0

    uniformly for t[1,T]Z.

    The organization of this paper is as follows. In the second section, we construct a completely new inner product space. In the new inner product space, we discuss the basic self-adjointness of the corresponding linear operator and the properties of the eigenpair of (1.2). Finally, under the above properties, the Lyapunov-Schmidit method is used to decompose the inner product space and transform our problem to an equivalent system, that is to say, finding the solutions of (1.1) is equivalent to finding the solutions of this system. Under the sublinear condition and sign conditions on nonlinear terms, an existence result of solutions to the problem (1.1) is obtained using Schauder's fixed-point theorem and the connectivity theories of the solution set of compact vector fields. Based on the first result, the existence of two solutions to the problem (1.1) is also obtained in this section.

    Definition 2.1. ([19]) A linear operator P from the linear space X to itself is called the projection operator, if P2=P.

    Lemma 2.2. ([16]) Let C be a bounded closed convex set in Banach space E, T:[α,β]×CC(α<β) be a continuous compact mapping, then the set

    Sα,β={(ρ,x)[α,β]×C|T(ρ,x)=x}

    contains a connected branch connecting {α}×C and {β}×C.

    Lemma 2.3. ([20])(Schauder) Let D be a bounded convex closed set in E, A:DD is completely continuous, then A has a fixed point in D.

    First, we construct the inner product space needed in this paper.

    Let

    Y:={u|u:[1,T]ZR},

    then Y is a Hilbert space under the following inner product

    y,zY=Tt=1y(t)z(t)

    and its norm is yY:=y,yY.

    Furthermore, consider the space H:=YR2. Define the inner product as follows:

    [y,α,β],[z,ζ,ρ]=y,zY+p(0)|δ0|αζ+p(T)|δ1|βρ,

    which norm is defined as

    y=[y,α,β],[y,α,β]12,

    where is transposition to a matrix.

    Let

    y0,0=b0y(0)d0Δy(0), y0,1=a0y(0)c0Δy(0)

    and

    yT+1,0=b1y(T+1)d1y(T+1), yT+1,1=a1y(T+1)c1y(T+1).

    For y=[y,α,β], define an operator L:DH as follows:

    Ly=[Δ[p(t1)Δy(t1)]+q(t)y(t)y0,0yT+1,0]:=[Lyy0,0yT+1,0],

    where D={[y,α,β]:yY, y0,1=α, yT+1,1=β}. Define S:DH as follows:

    Sy=S[yαβ]=[ryαβ].

    Then, the problem (1.2) is equivalent to the eigenvalue problem as follows:

    Ly=λSy, (2.1)

    that is, if (λk,y) is the eigenpair of the problem (1.2), then (λk,y) is the eigenpair of the opertor L. Conversely, if (λk,y) is the eigenpair of the operator L, then (λk,y) is the eigenpair of the problem (1.2).

    Eventually, we define A:DH as follows:

    Ay=F(t,y)+[γψk+¯g,0,0],

    where F(t,y)=F(t,[y,α,β])=[f(t,y),0,0]. Obviously, the solution of the problem (1.1) is equivalent to the fixed point of the following operator

    Ly=λkSy+Ay. (2.2)

    It can be seen that there is a homomorphism mapping (λk,y)(λk,y) between the problem (1.1) and the operator Eq (2.2).

    Next, we are committed to obtaining the orthogonality of the eigenfunction.

    Lemma 2.4. Assume that (λ,y) and (μ,z) are eigenpairs of L, then

    y,LzLy,z=(μλ)y,Sz.

    Proof Let y=[y,α,β]D, z=[z,ζ,ρ]D, then

    y,Lz=[y,α,β],[Lz,z0,0,zT+1,0]=y,LzY+p(0)|δ0|α(z0,0)+p(T)|δ1|β(zT+1,0)=μy,rzY+p(0)|δ0|α(μζ)+p(T)|δ1|β(μρ)=μy,Sz. (2.3)

    Similarly, we have

    Ly,z=[Ly,y0,0,yT+1,0],[z,ζ,ρ]=Ly,zY+p(0)|δ0|(y0,0)ζ+p(T)|δ1|(yT+1,0)ρ=λry,zY+p(0)|δ0|λαζ+p(T)|δ1|λβρ=λy,Sz. (2.4)

    It can be seen from (2.3) and (2.4)

    y,LzLy,z=(μλ)y,Sz.

    Lemma 2.5. The operator L is the self-adjoint operator in H.

    Proof For y=[y,α,β]D,z=[z,ζ,ρ]D, we just need to prove that y,Lz=Ly,z. By the definition of inner product in H. we obtain

    y,Lz=y,LzY+p(0)|δ0|α(z0,0)+p(T)|δ1|β(zT+1,0),

    and

    Ly,z=Ly,zY+p(0)|δ0|(y0,0)ζ+p(T)|δ1|(yT+1,0)ρ.

    Therefore,

    y,LzLy,z=y,LzYLy,zY+p(0)|δ0|[α(z0,0)(y0,0)ζ]+p(T)|δ1|[β(zT+1,0)(yT+1,0)ρ],

    where

    y,LzY=Tt=1y(t)(Δ[p(t1)Δz(t1)]+q(t)z(t))=Tt=1y(t)p(t1)Δz(t1)Tt=1y(t)p(t)Δz(t)+Tt=1q(t)y(t)z(t)=T1t=0y(t+1)p(t)Δz(t)Tt=1y(t)p(t)Δz(t)+Tt=1q(t)y(t)z(t)=T1t=0p(t)Δy(t)Δz(t)+p(0)y(0)Δz(0)p(T)y(T)Δz(T)+Tt=1q(t)y(t)z(t)

    and

    Ly,zY=T1t=0p(t)Δy(t)Δz(t)+p(0)Δy(0)z(0)p(T)Δy(T)z(T)+Tt=1q(t)y(t)z(t).

    Moreover, from

    α(z0,0)(y0,0)ζ=[a0y(0)c0Δy(0)][d0Δz(0)b0z(0)][d0Δy(0)b0y(0)][a0z(0)c0Δz(0)]=(a0d0b0c0)[y(0)Δz(0)Δy(0)z(0)]

    and

    β(zT+1,0)(yT+1,0)ρ=[a1y(T+1)c1y(T+1)][b1z(T+1)+d1z(T+1)][b1y(T+1)+d1y(T+1)][a1z(T+1)c1z(T+1)]=(a1d1b1c1)[y(T+1)z(T+1)y(T+1)z(T+1)],

    we have

    y,LzLy,z=p(0)|y(0)Δy(0)z(0)Δz(0)|p(T)|y(T)Δy(T)z(T)Δz(T)|p(0)|y(0)Δy(0)z(0)Δz(0)|+p(T)|y(T+1)y(T+1)z(T+1)z(T+1)|=0.

    In order to obtain the orthogonality of the eigenfunction, we define a weighted inner product related to the weighted function r(t) in H. First, we define the inner product in Y as y,zr=Tt=1r(t)y(t)z(t).

    Similarly, the inner product associated with the weight function r(t) in the space H is defined as follows:

    [y,α,β],[z,ζ,ρ]r=y,zr+p(0)|δ0|αζ+p(T)|δ1|βρ.

    Lemma 2.6. (Orthogonality theorem) Assume that (A1) and (A2) hold. If (λ,y) and (μ,z) are two different eigenpairs corresponding to L, then y and z are orthogonal under the weight inner product related to the weight function r(t).

    Proof Assume that (λ,y) and (μ,z) is the eigenpair of L, then it can be obtained from Lemmas 2.4 and 2.5

    0=(μλ)y,Sz=(μλ)y,zr.

    Therefore, if λμ, then y,zr=0, which implies that y and z are orthogonal to the inner product defined by the weighted function r(t).

    Lemma 2.7. ([18]) Suppose that (A1) and (A2) hold. Then (1.2) has at least T or at most T+2 simple eigenvalues.

    In this paper, we consider that λk is a simple eigenvalue, that is, the eigenspace corresponding to each eigenvalue is one-dimensional. Let ψk=[ψk,α,β]D be the eigenfunction corresponding to λk, and assume that it satisfies

    ψk,ψk=1. (2.5)

    Denote by L:=LλkS, then the operator (2.2) is transformed into

    Ly=Ay. (2.6)

    Define P:DD by

    (Px)(t)=ψk(t)ψk(t),x(t).

    Lemma 2.8. P is a projection operator and Im(P)=Ker(L).

    Proof Obviously, P is a linear operator, next, we need to prove P2=P.

    (P2x)(t)=P(Px)(t)=ψk(t)ψk(t),Px(t)=ψk(t)ψk(t),ψk(t)ψk(t),x(t)=ψk(t)ψk(t),x(t)ψk(t),ψk(t)=ψk(t)ψk(t),x(t)=(Px)(t).

    It can be obtained from the Definition 2.1, P is a projection operator. In addition, Im(P)=span{ψk}=Ker(L).

    Define H:HH by

    H([yαβ])=[yαβ][yαβ],ψkψk.

    Lemma 2.9. H is a projection operator and Im(H)=Im(L).

    Proof Obviously, H is a linear operator, next, we need to prove that H2=H.

    H2([yαβ])=H(H[yαβ])=H[yαβ]H[yαβ],ψkψk=[yαβ][yαβ],ψkψk[yαβ][yαβ],ψkψk,ψkψk=[yαβ]2[yαβ],ψkψk+[yαβ],ψkψk,ψkψk=[yαβ]2[yαβ],ψkψk+[yαβ],ψkψk,ψkψk=H([yαβ]).

    It can be obtained from Definition 2.1 that H is a projection operator. On the one hand, for any [y,α,β]H, we have

    H[yαβ],ψk=[yαβ][yαβ],ψkψk,ψk=[yαβ],ψk[yαβ],ψkψk,ψk=0,

    thus, Im(H)Im(L). On the other hand, for any yIm(L), we have

    y,ψk=0.

    In summary, Im(H)=Im(L).

    Denote that I is a identical operator, then

    D=Im(P)Im(IP),H=Im(H)Im(IH).

    The restriction of the operator L on L|Im(IP) is a bijection from Im(IP) to Im(H). Define M:Im(H)Im(IP) by

    M:=(L|Im(IP))1.

    It can be seen from KerL=span{ψk} that there is a unique decomposition for any y=[y,α,β]D

    y=ρψk+x,

    where ρR,x=[x,α,β]Im(IP).

    Lemma 2.10. The operator Eq (2.6) is equivalent to the following system

    x=MHA(ρψk+x), (2.7)
    Tt=1ψk(t)f(t,ρψk(t)+x(t))=γ(p(0)|δ0|α2+p(T)|δ1|β21):=θ, (2.8)

    where α=a0ψk(0)c0Δψk(0),β=a1ψk(T+1)c1ψk(T+1).

    Proof (ⅰ) For any y=ρψk+x, we have

    Ly=Ay  H(L(ρψk+x)A(ρψk+x))=0LxHA(ρψk+x)=0x=MHA(ρψk+x). 

    (ⅱ) Since Ly,ψk=0, we have Ay,ψk=0. Therefore,

    f(t,y)+γψk+¯g,ψkY=Tt=1f(t,ρψk(t)+x(t))ψk(t)+Tt=1γψk(t)ψk(t)+Tt=1¯g(t)ψk(t)=0.

    Combining (H3) with (2.5), we have

    Tt=1ψk(t)f(t,ρψk(t)+x(t))=γ(p(0)|δ0|α2+p(T)|δ1|β21)=θ,

    where α=a0ψk(0)c0Δψk(0),β=a1ψk(T+1)c1ψk(T+1).

    Let

    A+={t{1,2,,T} s.t. ψk(t)>0},
    A={t{1,2,,T} s.t. ψk(t)<0}.

    Obviously,

    A+A, min{|ψk(t)||tA+A}>0.

    Lemma 3.1. Supposed that (H1) holds, then there exist constants M0 and M1, such that

    xM1(|ρ|ψkY)α,

    where (ρ,x) is the solution of (2.7) and satisfies |ρ|M0.

    Proof Since

    A(ρψk+x)=F(t,ρψk+x)+[γψk+¯g,0,0]=[f(t,ρψk+x)+γψk+¯g,0,0],

    we have

    xMIm(H)Im(IP)HHIm(H)[¯gY+γψkY+A(|ρ|ψkY+xY)α+B]=MIm(H)Im(IP)HHIm(H)[¯gY+A(|ρ|ψkY)α(1+xY|ρ|ψkY)α+Bθ]MIm(H)Im(IP)HHIm(H)[¯gY+A(|ρ|ψkY)α(1+αxY|ρ|ψkY)+Bθ]=MIm(H)Im(IP)HHIm(H)[¯gY+A(|ρ|ψkY)α(1+α(|ρ|ψkY)1αxY(|ρ|ψkY)α)+Bθ]. 

    Denote that

    D0=MIm(H)Im(IP)HHIm(H)(¯gY+Bθ),D1=AMIm(H)Im(IP)HHIm(H).

    Furthermore, we have

    x(|ρ|ψkY)αD0(|ρ|ψkY)α+D1+αD1(|ρ|ψkY)1αxY(|ρ|ψkY)αD0(|ρ|ψkY)α+D1+αD1(|ρ|ψkY)1αx(|ρ|ψkY)α.

    So, if we let

    αD1(|ρ|ψkY)1α12,

    we have

    |ρ|(2αD1)11αψkY:=M0.

    Thus,

    x(|ρ|ψkY)α2D0(M0ψkY)α+2D1:=M1.

    This implies that

    xM1(|ρ|ψkY)α.

    Lemma 3.2. Suppose that (H1) holds, then there exist constants M0 and Γ, such that

    xΓ(|ρ|min{|ψk(t)||tA+A})α,

    where (ρ,x) is the solution of (2.7) and satisfies |ρ|M0.

    According to Lemma 3.2, choose constant ρ0, such that

    ρ0>max{M0,Γ(|ρ0|min{|ψk(t)||tA+A})α}. (3.1)

    Let

    K:={xIm(IP)|x=MHA(ρψk+x),|ρ|ρ0}.

    Then, for sufficiently large ρρ0, there is

    ρψk(t)+x(t)ω, tA+,xK, (3.2)
    ρψk(t)+x(t)ω, tA,xK, (3.3)

    and for sufficiently small ρρ0, there is

    ρψk(t)+x(t)ω, tA+,xK, (3.4)
    ρψk(t)+x(t)ω, tA,xK. (3.5)

    Theorem 3.3. Suppose that (A1), (A2) and (H1)(H3) hold, then there exists a non-empty bounded set Ω¯gR, such that the problem (1.1) has a solution if and only if θΩ¯g. Furthermore, Ω¯g contains θ=0 and has a non-empty interior.

    Proof We prove only the case of (1.3) in (H2), and the case of (1.4) can be similarly proved.

    From (1.3) and (3.2)–(3.5), it is not difficult to see that

    f(t,ρψk(t)+x(t))>0,   tA+, xK,
    f(t,ρψk(t)+x(t))<0,   tA, xK,

    for sufficiently large ρρ0 and for sufficiently small ρρ0,

    f(t,ρψk(t)+x(t))<0,   tA+, xK,
    f(t,ρψk(t)+x(t))>0,   tA, xK.

    Therefore, if ρρ0 is sufficiently large,

    ψk(t)f(t,ρψk(t)+x(t))>0, tA+A, xK, (3.6)

    if ρρ0 is sufficiently small,

    ψk(t)f(t,ρψk(t)+x(t))<0, tA+A, xK. (3.7)

    Let

    C:={xIm(IP)|xρ0}.

    Define Tρ:Im(IP)Im(IP) by

    Tρ:=MHA(ρψk+x).

    Obviously, Tρ is completely continuous. By (3.1), for xC and ρ[ρ0,ρ0],

    TρxΓ(|ρ|min{|ψk(t)||tA+A})αΓ(|ρ0|min{|ψk(t)||tA+A})αρ0,

    i.e.,

    Tρ(C)C.

    According to Schauder's fixed point theorem, Tρ has a fixed point on C, such that Tρx=x. It can be seen from Lemma 2.10 that the problem (1.1) is equivalent to the following system

    Ψ(s,x)=θ,   (s,x)S¯g,

    where

    S¯g:={(ρ,x)R×Im(IP)|x=MHA(ρψk+x)},
    Ψ(ρ,x):=Ts=1ψk(s)f(s,ρψk(s)+x(s)).

    At this time, the Ω¯g in Theorem 3.3 can be given by Ω¯g=Ψ(S¯g). There exists a solution to the problem (1.1) for θΩ¯g.

    From (3.6), (3.7) and A+A, we can deduce that for any xK

    Ts=1ψk(s)f(s,ρ0ψk(s)+x(s))<0, Ts=1ψk(s)f(s,ρ0ψk(s)+x(s))>0.

    Thus,

    Ψ(ρ0,x)<0<Ψ(ρ0,x), xK. (3.8)

    According to Lemma 2.2, S¯gRׯBρ0 contains a connected branch ξρ0,ρ0 connecting {ρ0}×C and {ρ0}×C. Combined with (3.8), Ω¯g contains θ=0 and has a non-empty interior.

    Theorem 3.4. Suppose that (A1), (A2), (H2)(H4) hold. Ω¯g as shown in Theorem 3.3, then there exists a nonempty set Ω¯gΩ¯g{0}, such that problem (1.1) has at least two solutions for θΩ¯g.

    Proof We prove only the case of (1.3), and the case of (1.4) can be similarly proved. Since the condition (H4) implies that (H1), using Theorem 3.3, we know that there exists ρ0>0, such that

    Ψ(ρ0,x)>0, xK.

    Let

    δ:=min{Ψ(ρ0,x)|xK},

    then δ>0.

    Next, we prove that problem (1.1) has at least two solutions for any θ(0,δ).

    Let

    S¯g:={(ρ,x)R×Im(IP)|x=MHA(ρψk+x)},
    ¯K:={xIm(IP)|(ρ,x)S¯g}.

    By (H4), there exists a constant A0 such that

    xA0, xK.

    Similar to the derivation of Theorem 3.3, there exists ρ>ρ0 such that the following results hold:

    (ⅰ) For ρρ, there is

    ψk(t)f(t,ρψk(t)+x(t))>0, tA+A, x¯K, (3.9)

    (ⅱ) For ρρ, there is

    ψk(t)f(t,ρψk(t)+x(t))<0, tA+A, x¯K. (3.10)

    Let

    C:={xIm(IP)|xA0}.

    According to (H4), (3.9) and (3.10), we have

    lim|ρ|Ts=1ψk(s)f(s,ρψk(s)+x(s))=0

    uniformly for x¯K, i.e.

    lim|ρ|Ψ(ρ,x)=0,  x¯K.

    Therefore, there exists a constant l:l>ρ>ρ0>0 such that S¯g contains a connected branch between {l}×C and {l}×C, and

    max{|Ψ(ρ,x)||ρ=±l, (ρ,x)ξl,l}max{|Ψ(ρ,x)||(ρ,x){l,l}ׯK}θ3.

    It can be seen from the connectivity of ξl,l that there exist (ρ1,x1) and (ρ2,x2) in ξl,l(S¯g), such that

    Ψ(ρ1,x1)=θ,    Ψ(ρ2,x2)=θ,

    where ρ1(l,ρ0),ρ2(ρ0,l). It can be proved that ρ1ψk+x1 and ρ2ψk+x2 are two different solutions of problem (1.1).

    In this section, we give a concrete example of the application of our major results of Theorems 3.3 and 3.4. We choose T=3,a0,d0,b1,c1=0 and a1,d1,b0,c0=1, which implies that the interval becomes [1,3]Z and the conditions (A1),(A2) hold.

    First, we consider the eigenpairs of the corresponding linear problem

    {Δ2y(t1)=λy(t),   t[1,3]Z,y(0)=λΔy(0),   λy(4)=y(4). (4.1)

    Define the equivalent matrix of (4.1) as follows,

    Aλ=(λ2+λ1+λ101λ2101λ2+11λ)

    Consequently, Aλy=0 is equivalent to (4.1). Let |Aλ|=0, we have

    λ1=1.4657,λ2=0.1149,λ3=0.8274,λ4=2.0911,λ5=3.4324,

    which are the eigenvalues of (4.1). Next, we choose λ=λ1=1.4657, then we obtain the corresponding eigenfunction

    ψ1(t)={1,t=1,3.4657,t=2,3.465721,t=3.

    Example 4.1. Consider the following problem

    {Δ2y(t1)=1.4657y(t)+f(t,y(t))+ψ1(t)+¯g(t),   t[1,3]Z,y(0)=1.4657Δy(0),   1.4657y(4)=y(4), (4.2)

    where

    f(t,s)={ts3,s[1,1],t5s,s(,1)(1,+),

    and

    ¯g(t)={0,t=1,3.465721,t=2,3.4657,t=3.

    Then, for f(t,y(t)), we have |f(t,y(t))|3|y(t)|13. If we choose ω=1, yf(t,y)>0 for |y(t)|>1. For ¯g(t), we have 3s=1¯g(s)ψ1(s)=0.

    Therefore, the problem (4.2) satisfies the conditions (A1),(A2), (H1)(H3), which implies that the problem (4.2) has at least one solution by Theorem 3.3.

    Example 4.2. Consider the following problem

    {Δ2y(t1)=1.4657y(t)+f(t,y(t))+ψ1(t)+¯g(t),   t[1,3]Z,y(0)=1.4657Δy(0),   1.4657y(4)=y(4), (4.3)

    where

    f(t,s)=tse|s|,   t[1,3]Z

    and

    ¯g(t)={0,t=1,13.46572,t=2,3.4657,t=3.

    Then, for f(t,y(t)), we always have yf(t,y)>0 for all y(t)>0 or y(t)<0, f is continuous and satisfies

    lim|y|f(t,y)=0.

    For ¯g(t), we have 3s=1¯g(s)ψ1(s)=0.

    Therefore, the problem (4.3) satisfies the conditions (A1),(A2), (H2)(H4), which implies that the problem (4.3) has at least two solutions by Theorem 3.4.

    The authors declare that they have not used Artificial Intelligence (AI) tools in the creation of this article.

    Supported by National Natural Science Foundation of China [Grant No. 11961060] and Natural Science Foundation of Qinghai Province(No.2024-ZJ-931).

    The authors declare that there are no conflicts of interest.



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