# Dirichlet Boundary Value Problem for the Second Order Asymptotically Linear System

Authors: A. Gritsans; F. Sadyrbaev; I. Yermachenko
Journal: International Journal of Differential Equations
 (2016)

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## Abstract

We consider the second order systemx′′=f(x) with the Dirichlet boundary conditions x(0)=0=x(1), where the vector field f∈C1(Rn,Rn) is asymptotically linear and f(0)=0. We provide the existence and multiplicity results using the vector field rotation theory.


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## Body

## 1. Introduction

The theory of nonlinear boundary value problems (BVPs in short) is intensively developed since the first works on calculus of variations where BVPs naturally appear in a classical problem of minimizing the integral functional considered on curves with fixed end points. The Euler equation for the problems of the calculus of variations can be written in the form(1)x′′=ft,x,x′and the boundary conditions are(2)xa=A,xb=Bif the problem of fixed end points is considered. The methods for investigation of this problem are diverse. For the existence of a solution, a lot of papers use topological approaches. The main scheme is the following. Imagine f in (1) is continuous and one is looking for classical (x∈C2([a,b])) solution of the problem. If f is bounded, then problem (1) and (2) is solvable. This is true for scalar and vectorial cases. If f is not bounded, then a priori estimates for a possible solution should be proved first in order to reduce given problem to that with bounded nonlinearity. The interested reader may consult books [1, Ch. 12] and [2–4] for details. We would like to mention also articles [5–8]. The diverse approaches to the subject were used in relatively recent contributions to the theory [9–16].In all the above-mentioned references, the main question is about the existence of a solution. The problem of the uniqueness of a solution is the next important one, especially for purposes of numerical investigation. It is to be mentioned that both problems (existence and uniqueness) are closely related for linear problems. Indeed, the linear problemx′′+k2x=0,x(0)=A,x(1)=B has at most one solution for any A,B∈R if k is not multiple of π. The condition k≠0(modπ) is also sufficient for solvability of the problem for any A,B.This is not the case for nonlinear problems. The solvability and multiplicity of solutions may be observed simultaneously. The problemx′′=-x3,x(0)=0,x(1)=0 is solvable and has a countable number of solutions. Another phenomenon was observed. Consider the problem x′′+f(x)=0 together with Sturm-Liouville boundary conditions a1x0+a2x′0=0,b1x(1)-b2x′(1)=0. It is convenient to look at this problem in a phase plane (x,x′). Suppose that f(x)≈k2x at zero and f(x)≈l2x at infinity, where k and l are essentially different constants. Then, the problem generally has multiple solutions due to the fact that trajectories of solutions of the equation have essentially different rotation speed near the origin and at infinity. This is evident geometrically and one of the first works employing this type of arguments is in the book [17, Ch. 15].When passing to systems of the second-order differential equations, the analogous approach can be applied. The geometrical interpretation fails however. One should think of a substitute for the rotation (angular) speed. It appears that apparatus of vector fields is good enough. It is possible to construct special vector fields (based on the form of boundary conditions and on the behaviour of nonlinearities of a system) in the vicinity of the origin and “at infinity.” This approach was applied to study BVPs for a system of the two second-order nonlinear differential equations in the work [16]. The considered system was supposed to be asymptotically linear (of one kind) at zero and quasi-linear (linear plus bounded nonlinearity) of another kind at infinity. Special vector fields were considered and the appropriate rotation numbers were invented.The current article considers the case ofn second-order differential equations. The approach is the same. However, there is need for employing the respective results concerning rotation of n-dimensional vector fields. The main object is a system of the second-order ordinary differential equations given together with the Dirichlet type boundary conditions. The main difference compared with paper [16] is that the computation of rotation numbers at zero and “at infinity” is more complicated and uses an advanced technique.The structure of the work is the following. In Section2, the general idea is discussed and useful references and needed definitions are given. In Section 3, the analysis of the vector field at zero (i.e., for solutions with small initial values) is carried out. The similar work is done in Section 4 for the infinity. Section 5 contains the main result. The example and the conclusions complete the article.


## 2. The Vector Fieldϕ Associated with the Dirichlet Boundary Value Problem

Consider the system(3)x′′=fx,given with the boundary conditions(4)x0=0=x1and the initial conditions(5)x0=0,x′0=β,where 0=(0,0,…,0︸n)T∈Rn.We suppose that the following conditions are fulfilled.(A1)
f∈C1(Rn,Rn).(A2)
f(0)=0, and hence system (3) has the trivial solution x=0.(A3)
The vector fieldf isasymptotically linear; that is, there exists n×n matrix f′(∞) with real entries such that(6)limx→∞fx-f′∞xx=0.The norms are standard everywhere. The matrix f′(∞) is calledthe derivative of the vector fieldfat infinity [18].It follows from the above conditions that(7)fx=f′∞x+hx,∀x∈Rn,where h∈C1(Rn,Rn), h(0)=0, and(8)limx→∞hxx=0.It follows from (7) and (8) that the vector field f is asymptotically linear if and only if for any ε>0 there exists M(ε)>0 such that(9)hx≤Mε+εx,∀x∈Rn.The asymptotically linear vector field f islinearly bounded. Indeed, fix ε0>0 and consider the corresponding M0=M(ε0)>0. Then, it follows from (7) and (9) that(10)fx≤f′∞x+M0+ε0x=a1+b1x,∀x∈Rn,where f′(∞)=maxβ=1f′∞β≥0, a1=M0>0, and b1=f′(∞)+ε0>0.Rewrite system (3) in the equivalent form(11)z′=Fz,where F(z)=(y,f(x))T, z=(x,y)T∈RN, y=x′, and N=2n.Proposition 1.
Suppose that conditions(A1),(A2), and(A3) are fulfilled. Then, the vector field F has the following properties. (1)
F∈C1(RN,RN).(2)
F(o)=o∈RN, where o=(0,0)T.(3)
The vector fieldF isasymptotically linear since there exists N×N matrix (12)F′∞=OnInf′∞On,whereIn and On are n×n unity and zero matrices, respectively, such that (13)limzN→∞Fz-F′∞zNzN=0.(4)
The vector fieldF is linearly bounded.Proof.
(1) and (2) follow from(A1) and(A2).
(3) For every z=(x,y)T∈RN, one has that F(z)=F′(∞)z+H(z), where H(z)=(0,h(x))T. Then, for any ε>0 there exists M(ε)>0 such that for every z=(x,y)T∈RN(14)HzN=hx≤Mε+εx≤Mε+εzN.(4) It follows that asymptotic linearity of the vector field F implies its linear boundedness.Since the vector fieldF∈C1(RN,RN) is linearly bounded, then [19, 20] its flow Φt(γ)=z(t;γ) is complete and Φt∈C1(RN,RN) for any t∈R, where z(t;γ) is the solution to the Cauchy problem(15)z′=Fz,z0=γ.Let γ=(α,β)T∈RN. We consider for our purposes the restriction of time one flow Φ1α=0(γ)=(x(1;β),x′(1;β)), where x(1;β) is the solution to Cauchy problem (3) and (5). Denote the first component of Φ1α=0 by ϕ; that is, (16)ϕβ=x1;β,∀β∈Rn.Then, ϕ∈C1(Rn,Rn). Thesingular points of the vector field ϕ are β∈Rn such that ϕ(β)=0 and they are in one-to-one correspondence with the solutions to Dirichlet boundary value problem (3) and (4). It follows from condition(A2) that ϕ(0)=0 and hence the singular point β=0 of the vector field ϕ corresponds to the trivial solution to problem (3) and (4). Any singular point β≠0 of the vector field ϕ generates a nontrivial solution to problem (3) and (4). In what follows, we investigate singular points of the vector field ϕ in terms of rotation numbers and provide the conditions which guarantee the existence of a solution (nontrivial) for the boundary value problem under consideration.Consider a bounded open setΩ⊂Rn. Suppose that the vector field ϕ isnonsingular on the boundary∂Ω; that is, (17)ϕβ≠0,∀β∈∂Ω.Then [21, 22], there is an integer γ(ϕ,Ω), which is associated with the vector field and calledthe rotation of the vector fieldϕon the boundary∂Ω.A singular pointβ0∈Rn of the vector field ϕ is calledisolated [21, 22], if there is neighbourhood Br(β0)={β-β0<r,β∈Rn} containing no other singular points. In this case, the rotation γ(ϕ,Br(β0)) is the same for any sufficiently small radius r. This common value ind(β0,ϕ) is calledthe index of the isolated singular pointβ0∈Rn.If the vector fieldϕ is nonsingular for all β∈Rn of sufficiently large norm, then by definition the point ∞ is an isolated singular point of ϕ. In this case, the rotation γ(ϕ,BR(0)) is the same for sufficiently large radius R. This common value ind(∞,ϕ) is calledthe index of the isolated singular point∞ [21, 22].


## 3. The Vector Fieldϕ Near Zero

Suppose that conditions(A1) and(A2) hold. Then, there exists the derivative f′(0) (the Jacobian matrix) of the vector field f at zero x=0 and we can consider the linearized system at zero(18)u′′=f′0u,the Dirichlet boundary conditions(19)u0=0=u1,and the initial conditions(20)u0=0,u′0=β.Ifu(t;β) is a solution to Cauchy problem (18) and (20) and P(t) is the solution to the n×n matrix Cauchy problem(21)P′′t=f′0Pt,P0=On,P′0=In,then u(t;β)=P(t)β for every t∈R and β∈Rn. Let us define the linear vector field ϕ0:Rn→Rn:(22)ϕ0β=u1;β=P1β,∀β∈Rn.Hence, ϕ0′(β)=ϕ0′(0)=P(1) for every β∈Rn.Let us consider the following condition.(A4)
The linearized system at zero (18) isnonresonant with respect to boundary conditions (19); that is, linear homogeneous problem (18) and (19) has only the trivial solution.The spectrumσD={-(jπ)2:j∈N} of the scalar Dirichlet boundary value problem(23)x′′=λx,x0=0=x1consists of all λ such that boundary value problem (23) has a nontrivial solution.Proposition 2.
The following statements are equivalent.(1)
Condition(A4) holds.(2)
detϕ0′(0)=detP(1)≠0.(3)
β=0 is the unique singular point of the vector field ϕ0.(4)
No eigenvalue of matrixf′(0) belongs to the spectrum σD of scalar Dirichlet boundary value problem (23).Proof.
The nonzero singular points of the vector fieldϕ0 are in one-to-one correspondence with the nontrivial solutions to Dirichlet boundary value problem (18) and (19). Hence, the equivalence (1)⇔(2)⇔(3) follows from (22).
Let us prove that(2)⇔(4).
IfJ is the real Jordan form [23] of matrix f′(0), then there exists a real nonsingular matrix M such that J=M-1f′(0)M. Cauchy problem (18) and (20) transforms to the Cauchy problem(24)v′′=Jv,v0=0,v′0=η,where v=M-1u and η=M-1β.
Ifv(t;η) is the solution to Cauchy problem (24) and Q(t) is the solution to the n×n matrix Cauchy problem (25)Q′′t=JQt,Q0=On,Q′0=In,then v(t;η)=Q(t)η for every t∈R and η∈Rn. Let us consider the linear vector field ψ0:Rn→Rn such that (26)ψ0η=v1;η=Q1η,∀η∈Rn.Hence, ψ0′(η)=ψ0′(0)=Q(1) for every η∈Rn.
The Jacobian matricesϕ0′(β) and ψ0′(η) are similar. Indeed, since v(t;η)=M-1u(t;β) and η=M-1β, one has that (27)ϕ0′β=Mψ0ηβ′=Mψ0′ηηβ′=Mψ0′ηM-1,and hence P(1)=MQ(1)M-1 and detϕ0′(0)=detP(1)=detQ(1). Next we shall analyze detQ(1).
The blocks of the real Jordan formJ of matrix f′(0) are of two types [23]: a real eigenvalue λ of matrix f′(0) generates blocks(28)Jkλ=λ10⋯000λ1⋯00⋮⋮⋮⋯⋮⋮000⋯λ1000⋯0λ,where k is the size of the block, but a pair λ=a+ib and λ¯=a-ib (b≠0) of complex conjugate eigenvalues of matrix f′(0) is associated with blocks(29)Jkλ=C2mλ=C2λI2O2⋯O2O2O2C2λI2⋯O2O2⋮⋮⋮⋯⋮⋮O2O2O2⋯C2λI2O2O2O2⋯O2C2λ,where k=2m is the size of the block and(30)C2λ=a-bba,I2=1001,O2=0000.
SupposeQk(t)=(qij(t)) solves the k×k matrix Cauchy problem(31)Qk′′t=JkλQkt,Qk0=Ok,Qk′0=Ik.
Letλ be a real eigenvalue of matrix f′(0) and λ=r2sgnλ, where r=λ. Then [15, 24],(32)Qkt=Ikt+13!Jkλt3+15!Jkλ2t5+17!Jkλ3t7+⋯+12j+1!Jkλjt2j+1+⋯=∑j=0∞12j+1!Jkλjt2j+1.Since(33)Jkλj=r2jsgnλj∗∗⋯∗∗0r2jsgnλj∗⋯∗∗⋮⋮⋮⋯⋮⋮000⋯r2jsgnλj∗000⋯0r2jsgnλjare upper triangular matrices, then matrix Qk(t) is upper triangular also with the diagonal elements (34)qt=∑j=0∞12j+1!r2jsgnλjt2j+1.It follows from (31) that function q(t)=qkk(t) solves the Cauchy problem(35)q′′t=r2sgn⁡λqt,q0=0,q′0=1,detQk1=q1k.
(a) Ifλ=0, then the solution to the Cauchy problem(36)q′′t=0,q0=0,q′0=1is q(t)=t. Hence, (37)det⁡Qk1=q1k=1k=1>0.
(b) Ifλ=r2>0(r>0), then the solution to the Cauchy problem(38)q′′t=r2qt,q0=0,q′0=1is q(t)=sinh⁡rt/r. Hence, (39)det⁡Qk1=q1k=sinh⁡rrk=sinhk⁡rrk>0.
(c) Ifλ=-r2<0(r>0), then the solution to the Cauchy problem(40)q′′t=-r2qt,q0=0,q′0=1is q(t)=sin⁡rt/r. Hence, (41)det⁡Qk1=q1k=sin⁡rrk=sink⁡rrk≠0⟺λ∉σD.
(d) Supposeλ=a+ib and λ¯=a-ib (b≠0) are complex conjugate eigenvalues of matrix f′(0) and Jk(λ)=C2m(λ), k=2m. Then [15, 24],(42)Q2mt=I2mt+13!C2mλt3+15!C2mλ2t5+17!C2mλ3t7+⋯+12j+1!C2mλjt2j+1+⋯=∑j=0∞12j+1!C2mλjt2j+1.The matrices (43)Jkλj=C2mλj=C2λj∗∗⋯∗∗O2C2λj∗⋯∗∗⋮⋮⋮⋯⋮⋮O2O2O2⋯C2λj∗O2O2O2⋯O2C2λjare k×k upper triangular block matrices of 2×2 blocks, (44)C2λj=a-bbaj=ρjcos⁡jφ-ρjsin⁡jφρjsin⁡jφρjcos⁡jφ,where a=ρcos⁡φ and b=ρsin⁡φ. Then, Q2m(t) is k×k upper triangular block matrix of 2×2 blocks also with diagonal blocks(45)D2t=ut-vtvtut=I2t+13!C2λt3+15!C2λ2t5+17!C2λ3t7+⋯+12j+1!C2λjt2j+1+⋯=∑j=0∞12j+1!C2λjt2j+1,where (46)ut=∑j=0∞ρjcos⁡jφ2j+1!t2j+1,vt=∑j=0∞ρjsin⁡jφ2j+1!t2j+1.It follows from (31) that the matrix (47)D2t=ut-vtvtut=qk-1,k-1tqk-1,ktqk,k-1tqk,ktsolves the matrix Cauchy problem (48)D2′′t=C2λD2t,D20=O2,D2′0=I2or (49)u′′t-v′′tv′′tu′′t=a-bbaut-vtvtut,u0-v0v0u0=0000,u′0-v′0v′0u′0=1001.Suppose that λ=a+ib=μ2=(α+iβ)2, where a=α2-β2 and b=2αβ≠0(α≠0,β≠0). Then, functions u(t) and v(t) solve the Cauchy problem (50)u′′t=α2-β2ut-2αβvt,v′′t=2αβut+α2-β2vt,u0=0,u′0=1,v0=0,v′0=0,and hence(51)ut=1α2+β2αsinh⁡αtcos⁡βt+βcosh⁡αtsin⁡βt,vt=1α2+β2αcosh⁡αtsin⁡βt-βsinh⁡αtcos⁡βt.Therefore,(52)detD21=u1-v1v1u1=u21+v21=sinh2αcos2⁡β+cosh2αsin2⁡βα2+β2>0,detQ2m1=detD21m=sinh2αcos2⁡β+cosh2αsin2⁡βmα2+β2m>0.
The determinant ofQ(1) is equal to the product of the determinants of the blocks Qk(1) corresponding to the eigenvalues λ of matrix f′(0). It follows from the above-mentioned considerations that detϕ0′(0)=detP(1)=detQ(1)≠0 if and only if the eigenvalues of matrix f′(0) do not belong to the spectrum σD={-(jπ)2:j∈N} of scalar Dirichlet boundary value problem (23). Hence, (2)⇔(4).Proposition 3.
Suppose that condition(A4) holds. If matrix f′(0) does not have negative eigenvalues with odd algebraic multiplicities, then ind(0,ϕ0)=1. If matrix f′(0) has k(1≤k≤n) different negative eigenvalues λj(1≤j≤k) with odd algebraic multiplicities, then(53)ind⁡0,ϕ0=sgndetϕ0′0=sgndetP1=sgn∏j=1ksin⁡λj.Proof.
Suppose that condition(A4) holds. It follows from Proposition 2 that det⁡ϕ0′(0)=det⁡P(1)≠0 and β=0 is the unique singular point of the vector field ϕ0. Hence [21, 22], (54)ind0,ϕ0=sgndetϕ0′0=sgndetQ1.The sign of det⁡Q(1) is equal to the product of the signs of detQk(1) for the blocks Qk(1) corresponding to the eigenvalues λ of matrix f′(0). It follows from the proof of Proposition 2 that sgndetQk(1)=1 for the blocks Qk(1) corresponding to nonnegative and complex eigenvalues λ of matrix f′(0). Let λ=-r2<0r=|λ| be a negative eigenvalue of matrix f′(0) with algebraic multiplicity μ and geometric multiplicity γ, 1≤γ≤μ≤n. Then, matrix Q(1) has γ blocks Qk1(1),…,Qkγ(1) corresponding to the eigenvalue λ and (55)detQk11·…·detQkγ1=sink1rrk1·…·sinkγrrkγ=sink1+⋯+kγrrk1+⋯+kγ=sinμrrμ≠0.Therefore, (56)sgndetQk11·…·sgndetQkγ1=+1,ifμiseven,sgnsin⁡λ,ifμisodd.If matrix f′(0) does not have negative eigenvalues with odd algebraic multiplicities, then ind(0,ϕ0)=1. If matrix f′(0) has k(1≤k≤n) different negative eigenvalues λj(1≤j≤k) with odd algebraic multiplicities, then formula (53) is valid.Theorem 4.
Suppose that conditions(A1),(A2), and(A4) hold. Then, β=0 is an isolated singular point of the vector field ϕ and ind(0,ϕ)=ind(0,ϕ0).Proof.
We already mentioned that the flowΦt(γ)=z(t;γ) of the vector field F is of class C1 for every t∈R, where z(t;γ) is the solution to Cauchy problem (15). Then, there exist continuous partial derivatives (∂zi/∂γj)(t;γ)(i,j=1,2,…,N) for every t∈R and γ∈RN. Matrix Z(t;γ)=((∂zi/∂γj)(t;γ)) solves [4, 19] the N×N matrix Cauchy problem(57)Z′t;γ=F′zt;γZt;γ,Z0;γ=IN,where F′(z(t;γ)) is the Jacobian matrix of the vector field F along the solution z(t;γ). One has for γ=(0,β), taking into account that z=(x,x′), that matrix X(t;β)=((∂xi/∂βj)(t;β)) solves the n×n matrix Cauchy problem (58)X′′t;β=f′xt;βXt;β,X0;β=On,X′0;β=In,where f′(x(t;β)) is the Jacobian matrix of the vector field f along the solution x(t;β) to Cauchy problem (3) and (5). If β=0, then it follows from condition(A2) that x(t;0)=0 and the matrix X(t;0)=((∂xi/∂βj)(t;0)) solves the n×n matrix Cauchy problem(59)X′′t;0=f′0Xt;0,X0;0=On,X′0;0=In.Uniqueness of solutions to n×n matrix Cauchy problems (21) and (59) implies that X(t;0)=P(t) for every t∈R. Hence, X(1;0)=P(1). Notice that X(1;0)=((∂xi/∂βj)(1;0))=ϕ′(0). Therefore, ϕ′(0)=P(1). Since P(1)=ϕ0′(0), one has that ϕ′(0)=ϕ0′(0). It follows from Proposition 2 that detϕ′(0)=detϕ0′(0)≠0. Hence, [21, 22] β=0 is an isolated singular point of the vector field ϕ and (60)ind⁡0,ϕ=sgndetϕ′0=sgndetϕ0′0=ind0,ϕ0.


## 4. The Vector Fieldϕ at Infinity

Suppose that conditions(A1) and(A3) hold. Then, there exists the derivative f′(∞) of the vector field f at infinity and we can consider the linearized system at infinity(61)w′′=f′∞w,the Dirichlet boundary conditions(62)w0=0=w1,and the initial conditions(63)w0=0,w′0=β.Ifw(t;β) is the solution to Cauchy problem (61) and (63) and S(t) is the solution to n×n matrix Cauchy problem(64)S′′t=f′∞St,S0=On,S′0=In,then w(t;β)=S(t)β for every t∈R and β∈Rn. Let us define the linear vector field ϕ∞:Rn→Rn,(65)ϕ∞β=w1;β=S1β,∀β∈Rn.Hence, ϕ∞′(β)=ϕ∞′(0)=S(1) for every β∈Rn.Let us consider the following condition.(A5)
The linearized system at infinity (61) isnonresonant with respect to boundary conditions (62); that is, linear homogeneous problem (61) and (62) has only the trivial solution.Proposition 5.
The following statements are equivalent.(1)
Condition(A5) holds.(2)
det⁡ϕ∞′(0)=detS(1)≠0.(3)
β=0 is the unique singular point of the vector field ϕ∞.(4)
No eigenvalue of the matrixf′(∞) belongs to the spectrum σD of scalar Dirichlet boundary value problem (23).Proposition 6.
Suppose that condition(A5) holds. If the matrix f′(∞) does not have negative eigenvalues with odd algebraic multiplicities, then ind(0,ϕ∞)=1. If the matrix f′(∞) has s(1≤s≤n) different negative eigenvalues μi(1≤i≤s) with odd algebraic multiplicities, then(66)ind⁡0,ϕ∞=sgndetϕ∞′0=sgn⁡det⁡S1=sgn∏i=1ssin⁡μi.The proofs of Propositions5 and 6 are analogous to the proofs of Propositions 2 and 3, respectively.Theorem 7.
Suppose that conditions(A1),(A3), and(A5) hold. Then, the point ∞ is an isolated singular point of the vector field ϕ and ind(∞,ϕ)=ind(0,ϕ∞).Proof.
First of all, we shall prove that the vector fieldϕ is asymptotically linear with the derivative at infinity ϕ′(∞)=ϕ∞′(0)=S(1). We proceed in the following steps.
Step 1 (auxiliary linear nonhomogeneous initial value problem). Let us consider the function p(t;β)=(1/‖β‖)x(t;β)-wt;β/‖β‖ for every t∈R and β∈Rn∖{0}, where x(t;β) is the solution to Cauchy problem (3) and (5) and w(t;β) is the solution to Cauchy problem (61) and (63). The function p(t;β) solves the Cauchy problem(67)p′′t;β=f′∞pt;β+gt;β,p0;β=0,p′0;β=0β≠0,where g(t;β)=(1/β)h(x(t;β)) for every t∈R and β∈Rn∖{0}. One can find [24, 25] that(68)pt;β=∑k=0∞f′∞kφ2k+1t;β,φmt;β=∫0tt-τmm!gτ;βdτ,m=0,1,2,…,(69)φ1′t;β=φ0t;β,φ3′t;β=φ2t;β,⋮φ5′t;β=φ4t;β,⋮φ1′′t;β=gt;β,φ3′′t;β=φ1t;β⋮φ5′′t;β=φ3t;β,⋮.
Step 2 (estimates for φ m ( 1 ; β )   ( m = 0,1 , 2 , … ) ). Suppose β≠0 and consider (70)φm1;β=∫01t-τmm!gτ;βdτm=0,1,2,….Taking into account (9) for any ε>0, one concludes that there exists M(ε)>0 such that (71)hxτ;β≤Mε+εxτ;β,∀τ∈0,1.Since 0≤1-τ≤1, one has(72)1-τmm!gτ;β=1-τmm!hxτ;ββ≤1m!βhxτ;β≤1m!βMε+εxτ;β,∀τ∈0,1.In accordance with Proposition 1, the vector field F is asymptotically linear, and hence there exist A,B≥0 such that F(z)≤A+Bz for every z=(x,y)T∈RN, y=x′. Consider the integral equation z(τ;γ)=γ+∫0τF(z(s;γ))ds equivalent to the Cauchy problem z′(τ;γ)=F(z(τ;γ)), z(0;γ)=γ=(0,β)T. Then, (73)zτ;γ≤γ+Aτ+B∫0τzs;γds.Using Grönwall’s inequality [20], one has that (74)zτ;γ≤γ+AτeBτ.Therefore, (75)zτ;γ≤AeB+eBγ,∀τ∈0,1.Since x(τ;β)≤z(τ;γ) and γ=β, we obtain(76)xτ;β≤AeB+eBβ,∀τ∈0,1.It follows from (72) and (76) that (77)1-τmm!gτ;β≤Mε+εAeBm!β+εeBm!,∀τ∈0,1.Therefore, (78)φm1;β=∫011-τmm!gτ;βdτ≤∫011-τmm!gτ;βdτ≤Mε+εAeBm!β+εeBm!,φm1;β≤Mε+εAeBm!β+εeBm!.
Step 3. Let us prove that limβ→∞p(1;β)=0.
(1) Suppose that f′(∞)=On. It follows from (68) and (78) that (79)p1;β=φ11;β≤Mε+εAeBβ+εeB.Hence, (80)limβ→∞p1;β≤εeB.Since ε>0 can be arbitrary, limβ→∞p1;β=0.
(2) Suppose that B=f′(∞)≠On. Then, (81)B=maxβ=1Bβ>0.Let us prove that the series ∑k=0∞Bkφ2k+1(1;β) is absolutely convergent; that is, the number series ∑k=0∞Bkφ2k+1(1;β) is convergent. It follows from (78) that(82)Bkφ2k+11;β≤Bkφ2k+11;β≤Bk2k+1!Mε+εAeBβ+εeB.The series ∑k=0∞(Bk/(2k+1)!)(M(ε)+εAeB)/β+εeB converges and the sum is(83)∑k=0∞Bk2k+1!Mε+εAeBβ+εeB=sinh⁡BBMε+εAeBβ+εeB.One can conclude from (82) by using the comparison test that the number series ∑k=0∞Bkφ2k+1(1;β) is convergent also and the sum is (84)∑k=0∞Bkφ2k+11;β≤sinh⁡BBMε+εAeBβ+εeB.Hence, the series ∑k=0∞Bkφ2k+1(1;β) is absolutely convergent and(85)p1;β=∑k=0∞Bkφ2k+11;β≤∑k=0∞Bkφ2k+11;β≤sinh⁡BBMε+εAeBβ+εeB.Therefore, limβ→∞p(1;β)≤εeB(sinh⁡(B)/B). Since ε>0 can be chosen arbitrary, one has that limβ→∞p(1;β)=0.
Step 4 (asymptotic linearity of the vector field ϕ ). If β≠0, then(86)ϕβ-ϕ∞β=x1;β-w1;β=x1;β-S1β=x1;β-βS1ββ=x1;β-βw1;ββ=β1βx1;β-w1;ββ=βp1;β.Hence, (87)ϕβ-ϕ∞ββ=ϕβ-ϕ∞′0ββ=p1;β.Since limβ→∞p(1;β)=0, (88)limβ→∞ϕβ-ϕ∞′0ββ=0;that is, the vector field ϕ is asymptotically linear with the derivative at infinity ϕ′(∞)=ϕ∞′(0). It follows from Proposition 5 that detϕ′(∞)=detϕ∞′(0)≠0. Hence [21, 22], the point ∞ is an isolated singular point of the vector field ϕ and (89)ind⁡∞,ϕ=sgndetϕ′∞=sgndetϕ∞′0=ind⁡0,ϕ∞.


## 5. The Main Theorem

Let us recall that the singular points of the vector fieldϕ are in one-to-one correspondence with solutions to Dirichlet boundary value problem (3) and (4). A solution x(t;β) of problem (3) and (4) is callednondegenerate, if the singular point β of the vector field ϕ is nondegenerate; that is, detϕ′(β)≠0.Theorem 8.
Suppose that conditions(A1) to(A5) hold. Then, the points β=0 and ∞ are isolated singular points of the vector field ϕ.(a)
Ifind(0,ϕ)≠ind(∞,ϕ), then boundary value problem (3) and (4) has a nontrivial solution.(b)
Ifind(0,ϕ)≠ind(∞,ϕ) and boundary value problem (3) and (4) has a nontrivial nondegenerate solution, then there exists yet another nontrivial solution to problem (3) and (4).Proof.
(a) It follows from Theorems4 and 7 that the points β=0 and ∞ are isolated singular points of the vector field ϕ. Hence, one can find positive r,R such that r<R and the sets (90)B¯r0∖0=0<β≤r,β∈Rn,B¯R∞=β≥R,β∈Rncontain no singular points of the vector field ϕ. The vector field ϕ is nonsingular on the spheres Sr(0)=∂Br(0) and SR(0)=∂BR(0) and the rotations on these spheres are different: (91)γϕ,Br0=ind⁡0,ϕ≠ind⁡∞,ϕ=γϕ,BR0.Using [22, Theorem 2], one can conclude that the n-dimensional annulus (92)Ann⁡r,R=r<β<R,β∈Rncontains a singular point β0≠0 of the vector field ϕ, which generates a nontrivial solution to Dirichlet boundary value problem (3) and (4).
(b) Letind(0,ϕ)≠ind(∞,ϕ) and suppose x(t;β0) is a nontrivial nondegenerate solution to boundary value problem (3) and (4), or equivalently β0∈Rn is a nonzero nondegenerate singular point of the vector field ϕ. Then [21, 22], ind(β0,ϕ)=sgndetϕ′(β0)∈{-1,1}. Suppose the contrary that x(t;β0) is the unique nontrivial solution to boundary value problem (3) and (4) or equivalently β0 is the unique singular point of the vector field ϕ in the set Rn∖{0}. Hence [21, 22], (93)ind⁡∞,ϕ=ind⁡0,ϕ+ind⁡β0,ϕ.If ind(∞,ϕ)=1 and ind(0,ϕ)=-1, then ind(β0,ϕ)=ind(∞,ϕ)-ind(0,ϕ)=1-(-1)=2. If ind(∞,ϕ)=-1 and ind(0,ϕ)=1, then ind(β0,ϕ)=ind(∞,ϕ)-ind(0,ϕ)=-1-1=-2. The contradiction proves that there exists a singular point β1∈Rn∖{0} of the vector field ϕ such that β1≠β0 or equivalently that there exists a solution x(t;β1) to boundary value problem (3) and (4), which is different from x(t;β0).Remark 9.
The practical implementation of Theorem8 is based on Propositions 3 and 6 and Theorems 4 and 7. Firstly the eigenvalues of the matrices f′(0) and f′(∞) must be calculated. If the eigenvalues do not belong to spectrum σD={-(jπ)2:j∈N} of scalar Dirichlet boundary value problem (23), then the indices ind(0,ϕ) and ind(∞,ϕ) must be calculated accordingly with Propositions 3 and 6. If these indices are different, then Theorem 8 is applicable and the existence of a nontrivial solution to boundary value problem (3) and (4) can be concluded.Remark 10.
Suppose that conditions(A1) to(A5) hold and ind(0,ϕ)≠ind(∞,ϕ). If boundary value problem (3) and (4) has an odd number of nontrivial nondegenerate solutions x(t;βi)(i=0,1,…,2k), where βi≠0 and detϕ′(βi)≠0(i=0,1,…,2k), then there exists yet another nontrivial solution to problem (3) and (4). Suppose the contrary that the set Rn∖{0} contains only an odd number of singular points βi(i=0,1,…,2k) of the vector field ϕ and these points are nondegenerate. Then [21, 22], ind(∞,ϕ)-ind(0,ϕ)=∑i=02kind(βi,ϕ), where the left hand side is equal to ±2, but the right hand side is odd.


## 6. Example

Consider the system(94)x1′′=-k2arctanx1+x2,x2′′=-m2arctanx1-x2,where k and m are nonzero integers, together with the boundary conditions(95)x10=x20=0=x11=x21.Consider the vector field f:R2→R2: (96)fx=-k2arctanx1+x2,-m2arctanx1-x2T,∀x=x1,x2T∈R2.Obviously conditions(A1) and(A2) are fulfilled. Due to the boundedness of arctan function, the vector field f is asymptotically linear and f′(∞)=O2, and hence condition(A3) is fulfilled also.Matrixf′(∞) has the only eigenvalue μ=0∉σD, and hence matrix f′(∞) does not have negative eigenvalues with odd algebraic multiplicities. It follows from Proposition 6 and Theorem 7 that ind(∞,ϕ)=1.The matrixf′(0)=-k2-k2-m2m2 has the characteristic equation (97)pλ=λ2-m2-k2λ-2k2m2=0and the eigenvalues are (98)λ1,2=m2-k22±k2+m22+4k2m22.Since (k2+m2)2+4k2m2>k2+m2>m2-k2, one has that λ1>0 and λ2<0. Obviously λ1∉σD={-(jπ)2:j∈N}. Note that λ2∉σD also since λ2 is an algebraic number (λ2 is the root of the characteristic polynomial p(λ) with rational coefficients), but the spectrum σD consists of the transcendental numbers (the product -(jπ)2 of the algebraic number -j2 with the transcendental number π2 is transcendental number). Theorem 8 is applicable, taking into account Proposition 3 and Theorem 4, if ind(0,ϕ)=sgn⁡sin⁡|λ2|=-1. Hence, Theorem 8 guarantees the existence of a nontrivial solution to boundary value problem (94) and (95) for all nonzero integers k and m such that(99)sin⁡k2+m22+4k2m22-m2-k22<0.The pairs (k,m)(1≤k,m≤10) with integer coordinates which satisfy condition (99) are depicted in Figure 1.Figure 1
The points(k,m)(1≤k,m≤10) with integer coordinates such that k and m satisfy condition (99).


## 7. Conclusions

For an asymptotically linear system ofn the second-order ordinary differential equations that are assumed to have the trivial solution to the conditions for existence of nontrivial solutions of the Dirichlet boundary value problem are given. The technique and concepts of the theory of rotation of n-dimensional vector fields are used. The existence conditions are formulated in terms of eigenvalues of coefficient matrices of linearized systems at zero (at the trivial solution) and at infinity. The proposed approach is applicable to other two-point boundary conditions such as the Neumann problem and mixed problem.



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