2 Vector field φ associated with the boundary value problem

A two-point boundary value problem for third order asymptotically linear systems

Armands Gritsans

and Felix Sadyrbaev

Institute of Life Sciences and Technology, Daugavpils University, Parades street 1, Daugavpils, LV–5400, Latvia

Received 17 December 2018, appeared 27 April 2019 Communicated by Nickolai Kosmatov

Abstract. We consider a third order systemx⁰⁰⁰ = f(x) with the two-point boundary conditions x(₀) = _0, x⁰(₀) = _0, x(₁) = _{0, where} f(₀) = ₀ and the vector field f ∈ C¹(_Rⁿ,Rⁿ)is asymptotically linear with the derivative at infinity f⁰(_∞). We introduce an asymptotically linear vector fieldφsuch that its singular points (zeros) are in a one- to-one correspondence with the solutions of the boundary value problem. Using the vector field rotation theory, we prove that under the non-resonance conditions for the linearized problems at zero and infinity the indices of φ at zero and infinity can be expressed in the terms of the eigenvalues of the matrices f⁰(₀)_and f⁰(_∞), respectively.

This proof constitutes an essential part of our article. If these indices are different, then standard arguments of the vector field rotation theory ensure the existence of at least one nontrivial solution to the boundary value problem. At the end of the article we consider the consequences for the scalar case.

Keywords:boundary value problem, spectrum, asymptotically linear vector field, index of isolated singular point.

2010 Mathematics Subject Classification: 34B15.

1 Introduction

Third order ordinary differential equations play an important role in modeling various pro- cesses in physics, engineering and technology, for example, in modeling compressible flows, thin viscous films, three-layer beams, electric circuits, and many others. The reader may con- sult, for instance, [4,8,9,13,25] and references therein, for more information about applications of third order equations.

The present article regards the existence results for a system of n third order ordinary differential equations

x⁰⁰⁰ = f(x), (1.1)

BCorresponding author. Email: arminge@inbox.lv

satisfying the two-point boundary conditions

x(0) =0, x⁰(0) =0, x(1) =0. (1.2) The vector field f: Rⁿ → _Rⁿ is supposed to present a linear behavior near zero and infinity.

We prove the existence of at least one nontrivial solution to the problem (1.1), (1.2) when appropriate indices associated with the linearized problems at zero and infinity are different.

The same approach was used in the authors papers [10,11] considering the Dirichlet boundary value problem for second order systems with asymptotically linear and asymptotically asym- metric behavior at infinity, respectively. The idea of comparison the behavior of solutions near zero and infinity was developed in [1,24].

The literature on boundary value problems for third order nonlinear systems is not ex- tensive, see, for instance, [5,15,16] and references therein. In the works [6,8,17,19,22,23,27]

boundary values problems form-th order systems are considered.

First, in Section 2, we consider some properties of asymptotically linear vector fields. We assume that a vector field f: Rⁿ→_Rⁿsatisfies the following conditions.

(A1) f ∈C¹(_Rⁿ,Rⁿ).

(A2) f(0) = (0), where0= (0, 0, . . . , 0

| {z }

n

)^T ∈_Rⁿ. (A3) f is asymptotically linear.

Then we introduce a vector field φ: Rⁿ → _Rⁿ associated with the problem (1.1), (1.2). The singular points (zeros) of φ are in a one-to-one correspondence with the solutions of the boundary value problem (1.1), (1.2).

In Section 3, we will prove a number of auxiliary propositions which will be used in the next sections.

In Section 4, we explore a linear vector field ψ : Rⁿ → _Rⁿ associated with the linear boundary value problem

p⁰⁰⁰= Ap, (1.3)

p(0) =0, p⁰(0) =0, p(1) =0, (1.4) where A is a n×n matrix with real entries. If this problem is non-resonant, that is, the problem (1.3), (1.4) has only the trivial solution, then Theorem 4.2 states that the index of ψ at zero can be expressed in the terms of the eigenvalues of A. The proof of Theorem 4.2 constitutes an essential part of our article.

In Section5, we explore the vector field φnear zero. We consider a vector fieldφ₀: Rⁿ → Rⁿassociated with the linearized at zero problem (5.4). The assumptions (A1) to (A3) coupled with the condition that (5.4) is non-resonant ensure that ind(0,φ) =ind(0,φ0).

In Section 6, we prove that the vector field φ is asymptotically linear. We consider the vector field φ_∞: Rⁿ → _Rⁿ associated with the linearized at infinity problem (6.3). The as- sumptions (A1) and (A3) combined with the condition that (6.3) is non-resonant ensure that ind(_∞,φ) =ind(0,φ_∞).

In Section 7, we prove the main theorem7.1 of this paper. The assumptions (A1) to (A3) coupled with the condition that the problems (5.4) and (6.3) are non-resonant provide that zero and infinity are isolated singular points ofφ. The standard arguments of the vector field

rotation theory (the Brouwer degree theory) ensure the existence of at least one nontrivial solution to the problem (1.1), (1.2) whenever ind(0,φ)6=ind(_∞,φ).

In Section8, we consider the example illustrating the main theorem and providing calcu- lations of indices ofφat zero and infinity based on Theorem4.2.

In Section9, we analyze the main theorem in the scalar case.

The concluding remarks in Section10finish our paper.

2 Vector field φ associated with the boundary value problem

Definition 2.1([20]). A vector field f ∈ C(_Rⁿ,Rⁿ)is called asymptotically linear if there exists a realn×nmatrix f⁰(_∞)such that

kxlimk→_∞

f(x)− f⁰(_∞)x

kxk =0, (2.1)

where k · k is the Euclidean norm on Rⁿ. The matrix f⁰(_∞) is called the derivative of f at infinity.

If a vector field f ∈ C(_Rⁿ,Rⁿ)is asymptotically linear, then, see [2], its derivative f⁰(_∞)at infinity is uniquely determined by f.

Definition 2.2 ([31]). A vector field f ∈ C(_Rⁿ,Rⁿ) is called linearly bounded if there exist non-negative constants aandbsuch that

f(x)≤ a+bkxk, ∀x∈_Rⁿ. (2.2) Proposition 2.3. Consider a vector field f ∈C(_Rⁿ,Rⁿ).

(a) If f is asymptotically linear and

g(x) = f(x)− f⁰(_∞)x, ∀x∈_Rⁿ, (2.3) then for everyε>₀there exists M(ε)>_{0such that}

g(x)≤ M(ε) +εkxk, ∀x∈_Rⁿ. (2.4) (b) Suppose that there exist a real n×n matrix B and a vector field g: Rⁿ →_Rⁿ such that f has

the form

f(x) =Bx+g(x), ∀x∈ _Rⁿ. (2.5) If for every ε > 0 there exists M(ε) > 0 such that (2.4) is fulfilled, then f is asymptotically linear and B = f⁰(_∞).

(c) If f is asymptotically linear, then f is linearly bounded.

Proof. (a) The statement is valid due to the reference [20].

(b) Considerε>0. Then there exists M(ε)>0 such that (2.4) fulfills. It follows from (2.4) and (2.5) that

kxlimk→_∞

f(x)−Bx kxk ≤ε.

Sinceε > 0 can be arbitrary, then (2.1) is fulfilled with f⁰(_∞) = B, by the uniqueness of the derivative of f at infinity.

(c) If f is asymptotically linear with the derivative at infinity f⁰(_∞), then it follows from (2.3) that

f(x)≤ f⁰(_∞)x

+g(x)≤f⁰(_∞)kxk+g(x), ∀x∈ _Rⁿ, (2.6) where

kf⁰(_∞)k= max

kxk=1kf⁰(_∞)xk

is the induced matrix norm on the Euclidean spaceRⁿ. Consider ε >0. Then, in accordance with (a), there exists M(ε)>0 such that (2.4) is valid. It follows from (2.4) and (2.6) that (2.2) is fulfilled witha= M(ε)>0 andb=f⁰(_∞)+ε>0.

Corollary 2.4. Consider a vector field f ∈C(_Rⁿ,Rⁿ).

(a) If f is bounded onRⁿ, then f is asymptotically linear with f⁰(_∞) =On, where Onis the n×n zero matrix.

(b) If f is quasi-linear, that is, f has the form(2.5), where gis bounded onRⁿ, then f is asymptoti- cally linear with f⁰(_∞) =B.

From now on, we assume that the conditions (A1) to (A3) are fulfilled.

Let us rewrite the system (1.1) in the equivalent formw⁰ = F(w), where

F(w) = y,z, f(x)^T, w= (x, y, z)^T ∈_R^N, y= x⁰, z=y⁰ =x⁰⁰, N=3n.

Denote byk · k_N the Euclidean norm onR^N.

Proposition 2.5. Suppose that the conditions (A1) to (A3) hold. The vector fieldF:R^N → _R^N has the following properties.

(a) F ∈C¹ R^N,R^N .

(b) F(o) =o, whereo= (0,0,0)^T ∈_R^N, besides

F⁰(o) =





On In On

O_n O_n I_n f⁰(₀) O_n O_n



, (2.7)

where I_nis the n×n unit matrix.

(c) F is asymptotically linear and its derivative at infinity is

F⁰(_∞) =





O_n I_n O_n On On In

f⁰(_∞) On On



. (2.8)

(d) F is linearly bounded.

Proof. The statements (a) and (b) are direct consequences of (A1) and (A2).

(c) Since f is asymptotically linear, then it follows from (a) of Proposition2.3that for every ε>0 there existsM(ε)>0 such that (2.4) fulfills. For eachw= (x,y,z)^T ∈_R^N we have

kF(w)−F⁰(_∞)w

N =g(x)≤ M(ε) +εkxk ≤M(ε) +εkwk_N,

where F⁰(_∞)andg are given by (2.8) and (2.3), respectively. In accordance with (b) of Propo- sition2.3, the vector fieldF is asymptotically linear with the derivative at infinity F⁰(_∞)_.

(d) It follows from (c) of Proposition2.3that f is linearly bounded, that is, there exist non- negative constants aandbsuch that (2.2) is valid. Consider an arbitraryw= (x,y,z)^T ∈_R^N. Since

kyk − kzk² ≥0, kxk −bkyk²≥0, kxk −bkzk²≥0, then

2kyk kzk ≤ kyk²+kzk²_, 2bkxk kyk ≤ kxk²+b²kyk²_, 2bkxk kzk ≤ kxk²+b²kzk²_. Taking into account the last inequalities, we obtain

kyk+kzk+bkxk² ≤(2+b²) kxk²+kyk²+kzk² = (2+b²)kwk²_N. (2.9) In view of (2.2) and (2.9), we have

F(w)

N =

q

kyk²+kzk²+f(x)

2≤ kyk+kzk+f(x)

≤a+ kyk+kzk+bkxk ≤a+^p2+b²kwk_N. Consequently,F is linearly bounded.

It follows from Proposition2.5thatF ∈C¹ R^N,R^N

andF is linearly bounded. Therefore, see [3,31], the flow Φ^t(ξ) = w(t;ξ)of F is complete and Φ^t ∈ C¹ R^N,R^N

for everyt ∈ _R, wherew(t;ξ) = x(t;ξ),x⁰(t;ξ),x⁰⁰(t;ξ)^T, is the solution of the Cauchy problem

w⁰ = F(w), w(0) =ξ. Considerξ = (α,β,γ)^T ∈_R^N. Ifα= β=0, then

x(t;γ)_:= x(t;ξ) _(2.10)

solves the Cauchy problem (1.1),

x(0) =0, x⁰(0) =0, x⁰⁰(0) =γ. (2.11) Consider a vector field φ: Rⁿ→_Rⁿ,

φ(γ) =x(1;γ), ∀γ∈_Rⁿ. Sinceφis the first component of the restrictionΦ¹

_α=0

β=0

, thenφ∈C¹(_Rⁿ,Rⁿ).

Definition 2.6. A pointγ∈_Rⁿis called a singular point of the vector fieldφifφ(γ) =0.

The singular points of φ are in a one-to-one correspondence with the solutions of the boundary value problem (1.1), (1.2), since φ(_γ) = ₀if and only if x(t;γ)solves (1.1), (1.2). It follows from (A2) thatγ =0is a singular point ofφand it corresponds to the trivial solution of (1.1), (1.2). Each singular pointγ 6=0ofφgenerates a nontrivial solution of (1.1), (1.2).

In this paper, we will prove that under the conditions, formulated in the main theorem 7.1, the vector fieldφhas a singular pointγ6=0. For this we will use the vector field rotation theory. The reader may consult, for instance, [21,30], for definitions of isolated singular points of vector fields and their indices.

3 Auxiliary results

Consider the function

h(z) =e^z+ε₁e^ε1z+ε₂e^ε2z =e^z−2e^−z2 sin

√3 2 z+ ^π

6

!

, ∀z∈_C, (3.1)

whereε_1,2 =−¹₂±i

√3

2 are cube roots of unity ande^z is the complex exponential function. The functionhis analytic onCand

h(ε₁z) =ε2h(z), h(ε2z) =ε₁h(z), ∀z ∈_C. (3.2) Ifτis a zero ofh, then it follows from (3.2) thatε₁τandε₂τare zeros ofh, too.

The next lemma plays an important role in our considerations.

Lemma 3.1. The function h has no complex zeros outside of the linesImz=0andImz= ±√ 3 Rez.

For the proof of this lemma we will prove a number of auxiliary propositions, some of which are of independent interest and will be used in the next sections.

3.1 The change of the argument along a path

Definition 3.2. Letαandβbe real numbers withα<β. A continuous mappingw: [α,β]→_C is called a path from w(α) _to w(β)and its image [w] = w

α,β] is called the trace of w. A pathw: [α,β]→_Cis called a loop ifw(α) =w(β). A subset L⊂_Cis called a Jordan curve if there exists a loopw: [α,β]→_Csuch that L = [w]and the mappingw, restricted to[α,β), is injective.

For each pathw: [α,β]→_C\ {0}there exist, see [29], continuous functionsρ: [α,β]→_R+

andϕ: [α,β]→_Rsuch that

w(t) =ρ(t)e^iϕ(t), ∀t∈ [α,β]. (3.3) Definition 3.3([29]). Every continuous functionϕsatisfying (3.3) is called a continuous branch of the argument along the pathw.

If ϕandψare two continuous branches of the argument along the path w, then, see [29], there exists k ∈ _Z such that ψ(t) = ϕ(t) +2πk for every t ∈ [α,β]. Therefore, if ϕ is a continuous branch of the argument along the path w: [_α,β] → _C\ {0}, then the difference

∆warg= ϕ(β)−ϕ(α)does not depend on the choice of the branch ϕalongw.

Definition 3.4. The difference∆warg is called the change of the argument along the pathw.

If w: [_α,β] → _C\ {0}is a loop, then, see [29], the change of the argument alongw is an integer multiple of 2π, and, as already mentioned above, it does not depend on the choice of the branch ϕalongw.

Definition 3.5. The integer windw= ^∆w_2π^arg is called the winding number ofw.

The reader may consult, for instance, [29], for more information about winding numbers.

Proposition 3.6. If p,q: [α,β]→_Care paths such that

p(t)−q(t) <q(t), ∀t∈[α,β], (3.4) p(α)q(β) = p(β)q(α), (3.5) then

[p],[q]⊂_C\ {0}, (3.6)

∆parg=_∆qarg . (3.7)

Proof. It follows from (3.4) that (3.6) fulfills. Taking into account (3.5), the pathω= ^p_q: [_α,β]→ C\ {0}is a loop. The inequality (3.4) yields

ω(t)−1

<1, ∀t∈[α,β]. (3.8)

Consider the constant loopκ(t)≡1 and a mappingH: [0, 1]×[α,β]→_C, H(s,t) = (1−s)κ(t) +sω(t), ∀s ∈[0, 1], ∀t ∈[α,β].

It follows from (3.6) and (3.8) that H(s,t)6= 0 for everys ∈[0, 1]andt∈ [α,β]. The mapping H(s,·): [α,β] → _C\ {0} is a loop for every s ∈ [0, 1]. Therefore, the loop ω is homotopic through loops to the constant loop κ in the region C\ {0}. In view of the reference [26], windw=0. Hence, for every continuous branch ϕof the argument along the loopωwe have

∆ωarg= ϕ(β)−ϕ(α) =0. Taking into account (3.6), there exist continuous branchesϕp and ϕ_q of the argument along the paths pandq, respectively. Sinceω = ^p_q, then ϕ= ϕ_p−ϕ_q is a continuous branch of the argument along the loopω. We have

0=_∆ωarg= ϕ(β)−ϕ(α) = ϕp(β)−ϕq(β)− ϕp(α)−ϕq(α)

= ϕp(β)−ϕp(α)− ϕq(β)−ϕq(α) =_∆parg−_∆qarg . Consequently, (3.7) fulfills.

Remark 3.7. If p,q: [α,β] → _Care loops, then Lemma 3.6 actually is the Rouché’s Theorem for loops, since (3.5) fulfills and (3.7) yields windp=windq.

3.2 The number of zeros ofh in the interior of the triangle T_k Consider the function

g(z) =ε₁e^ε1z+ε₂e^ε2z =−2e^−2zsin

√3 2 z+^π

6

!

, ∀z∈ _C. (3.9)

If

ξ_k = −√^2π 3

2+3k

3 (k∈_N0), (3.10)

then sin

√3

2 ξ_k+^π₆= (−1)^k+1(_k∈_N0). For everyk∈_N0consider a mappingw_k: [−1, 5]→_C,

w_k(t) =















w_1,k(t) =ξ_k+i√

3ξ_kt, if t∈[−1, 1],

w_2,k(t) =ε₁w_1,k(t−2) =−^ξ₂^k(−5+3t) +i

√ 3ξk

2 (3−t), if t∈[1, 3], w_3,k(t) =ε₁w_2,k(t−2) =ε₂w_1,k(t−4)

= ^ξ₂^k(−13+3t)−i

√3ξk

2 (−3+t), if t∈[3, 5].

(3.11)

The mappingw_k (k ∈ _N0)is a loop and its trace [w_k] =T_k ⊂_C\ {0}is a positively oriented Jordan curve. The setT_k is a triangle with the vertices

z_1,k =w_1,k(−1) =ξ_k−i√

3ξ_k, (3.12)

z_2,k =w_2,k(1) =ε₁z_1,k =ξ_k+i√

3ξ_k, (3.13)

z_3,k =w_3,k(3) =ε₁z_2,k =ε₂z_1,k = −2ξ_k, and it consists of the line segments

L[z_1,k;z_2,k], L[z_2,k;z_3,k] =ε₁L[z_1,k;z_2,k], L[z_3,k;z_1,k] =ε2L[z_1,k;z_2,k] (3.14) with the parametrizationsw_1,k,w_2,k,w_3,k, respectively.

Proposition 3.8. For every k ∈ _N0 the number of zeros of h in the interior of the Jordan curve T_k, counted with multiplicity, is equal to2+3k.

Proof. Suppose thatk∈_N0. Consider the loop p_k = h◦w_k: [−1, 5]→_C,

p_k(t) =











lcl p_1,k(t) = (h◦w_1,k)(t), ift∈ [−1, 1], p_2,k(t) = (h◦w_2,k)(t) =ε₂p_1,k(t−2), ift∈ [1, 3], p_3,k(t) = (h◦w_3,k)(t) =ε₁p_1,k(t−4), ift∈ [3, 5],

(3.15)

and the pathq_1,k = g◦w_1,k: [−1, 1]→_C. Let us prove that

p_1,k(t)−q_1,k(t)<q_1,k(t), ∀t ∈[−1, 1]. (3.16) For everyt∈[−1, 1]we have

p_1,k(t)−q_1,k(t) = e^w1,k(t)

=e^ξk <1. (3.17)

Since

q_1,k(t) =g w_1,k(t)= −2e⁻

w1,k(t)

2 sin

√3

2 w_1,k(t) +^π 6

!

=2(−1)^kcosh 3

2ξ_kt

e^−ξ2ke⁻ⁱ

√ 3ξk t

2 ,

(3.18) then for everyt∈ [−1, 1]_{we obtain}

q_1,k(t)=2e^−ξ2k cosh 3

2ξ_kt

>2. (3.19)

In view of (3.17) and (3.19) we deduce that (3.16) fulfills.

From (3.1), (3.9), (3.10) and (3.11) it follows that p_1,k(−1)

q_1,k(−1) =₁+ ¹

1+e^−3ξk = ^p1,k(1) q_1,k(1)^. Therefore,

p_1,k(−1)q_1,k(1) =p_1,k(1)q_1,k(−1). (3.20) On account of (3.16) and (3.20), we infer from Proposition3.6that

[p_1,k],[q_1,k]⊂_C\ {0}, (3.21)

∆p1,karg=_∆q1,karg . (3.22)

It follows from (3.18) that

q_1,k(t) =ν(t)e^iθ(t), ∀t ∈[−1, 1], where

ν(t) =2 cosh 3

2ξ_kt

e^−ξ2k >0, θ(t) =







−

√ 3ξ_kt

2 , if k=0, 2, 4, . . . , π−

√3ξkt

2 , if k=1, 3, 5, . . . ,

are continuous functions on the interval [−1, 1]. Therefore, θ is a continuous branch of the argument along the pathq_1,k and

∆q1,karg=θ(1)−θ(−1) = ²+3k 3 2π.

In view of (3.22),

∆p_1,karg= ²+3k

3 2π. (3.23)

Taking into account (3.21), there exist continuous functions r: [−1, 1]→_R+, a: [−1, 1]→_R such that

p_1,k(t) =h w_1,k(t)=r(t)e^ia(t), ∀t ∈[−_{1, 1}]_{. (3.24)} Since (3.23) is valid for every continuous branch of the argument along the path p_1,k, then

a(1)−a(−1) = ²+3k

3 2π. (3.25)

It follows from (3.2), (3.12), (3.13) and (3.24) that

r(1)e^ia(1) = p_1,k(1) =h w_1,k(1) =h ε₁w_1,k(−1)

=ε₂h w_1,k(−1) =ε₂p_1,k(−1) =ε₂r(−1)e^ia(−1). Hence,

r(1) =r(−1). (3.26)

Consider functionsρ_k,ϕ_k: [−1, 5]→_R,

ρ_k(t) =







ρ_1,k(t) =r(t)_{, if} t∈ [−_{1, 1}]_, ρ_2,k(t) =r(t−2), if t∈ [1, 3], ρ_3,k(t) =r(t−4), if t∈ [3, 5],

(3.27)

ϕ_k(t) =







ϕ_1,k(t) =a(t), if t ∈[−1, 1], ϕ_2,k(t) = ^4π₃ +a(t−2) +2πk, if t∈ [1, 3], ϕ_3,k(t) = ^2π₃ +a(t−4) +2π(1+2k), if t∈ [3, 5].

(3.28) Since r(_t) >0 for every t ∈ [−1, 1]_{, then}ρ_k(_t)> 0 for everyt ∈ [−1, 5]. It follows from (3.2), (3.11), (3.15), (3.27) and (3.28), that

ρ_k(t)e^iϕk(t)= p_k(t), ∀t∈ [1, 5].

On account of (3.25) and (3.26), the functionsρ_kand ϕ_k are continuous on the interval[−1, 5]. Consequently, ϕ_k is a continuous branch of the argument along the loopp_k and the change of the argument along the loop p_k is

∆p_karg= ϕ_k(₅)−ϕ_k(−₁) = ϕ_3,k(₅)−ϕ_1,k(−₁) = (₂+3k)_2π.

The winding number of the functionhalong the Jordan curve T_k is windT_kh:=_windp_k = ^∆pkarg

2π =₂+3k.

It follows from the argument principle for analytic functions, see [29], that the number of zeros ofhin the interior ofT_k, counted with multiplicity, is equal to 2+3k.

3.3 The scalar eigenvalue problem Proposition 3.9.

(a) The number r0=0is a double zero of h. Every zeroτ∈_C\ {0}of h, if any, is simple.

(b) The function h is positive on the interval(0,+_∞).

(c) The function h on the real axis has a countable number of zeros r_k (k∈_N0)which can be ordered as

· · · <ξ_k+1<r_k+1 <ξ_k <r_k <· · · <ξ₂< r2 <ξ₁<r₁< ξ₀<r0=0, (3.29) whereξ_k (k ∈_N0)are given by(3.10).

(d) The function h is positive on(r_k+1,r_k)if k ∈_N0is even and negative if k∈_N0 is odd.

Proof. (a) The numberr₀=0 is a double zero ofh, since

h(0) =1+ε₁+ε₂=0, h⁰(0) =1+ε²₁+ε²₂ =0, h⁰⁰(0) =1+ε³₁+ε³₂ =36=0. (3.30) Consider τ ∈ _C\ {0}such that h(τ) = 0. Let us prove that h⁰(τ) 6= 0. Suppose, on the contrary, that h⁰(τ) = 0. Since h(z) = e^z +g(z) for every z ∈ _{C, then} e^τ+g(τ) = 0 and e^τ+g⁰(τ) =0, where the functiongis given by (3.9). Therefore,

ε₂e^ε1τ+ε₁e^ε2τ =g⁰(τ) =g(τ) =ε₁e^ε1τ+ε₂e^ε2τ. We deduce thate^{(ε1−ε2)τ} =1 and thus there existsm∈_Zsuch thatτ= ^2πm√

3 . If follows from 0= h(τ) =e^−πm√3

e

√3πm−(−1)^m

thate

√3πm =_{1. Hence,m}=_{0 and}τ=0. The contradiction obtained proves the statement.

(b) Let us consider the functionhon the real axis. Taking into account (3.1), we can write h(t) =2e^−2t q(t), ∀t ∈_R, (3.31) where

q(t) =q₁(t)−q₂(t)_, q₁(t) = ¹

2e^3t2, q₂(t) =_sin

√ 3 2 t+ ^π

6

!

, ∀t∈_R.

It follows from (3.31) that the functionsh andqhave the same zeros.

1) Sinceq₂(ξ_k) = (−1)^k+1 (k ∈ _Z), thenξ_k (k =0,±2,±4, . . .)are local minima of q₂ and ξ_k (k = ±1,±3,±5, . . .) are local maxima ofq₂. The pointsη_k = −^√2π

3 1+6k

6 (k ∈ _Z)_{are zeros} ofq2and

· · ·< ξ_k+1 <η_k+1 <ξ_k <η_k <· · · <η₂ <ξ₁ <η₁ <ξ₀ <η₀ <₀<ξ₋₁ <η₋₁ <· · ·

The functionq₁is positive and strictly increasing onR, besides, lim

t→−_∞q₁(t) =0 andq₁(0) = ¹₂. Therefore, for each k∈_N0 the pointsξ_k andη_k are not zeros of the functionq, since

q(η_k) =q₁(η_k)>0 (k ∈_N0), q(ξ_k) =q₁(ξ_k) + (−1)^k

(<0, ifk=1, 3, 5, . . . ,

>0, ifk=0, 2, 4, . . .

2) Since the functionq₁is positive on(η₁,η₀)and the functionq₂is negative on this interval, then the functionqis positive on(η₁,η₀).

3) The functionq₁ is strictly convex onR andq₁(0) = ¹₂, q⁰₁(0) = ³₄. Therefore, for t ∈ _R the graph of q₁is strictly above the tangent line y= ¹₂ +³₄t to the graph ofq₁ at 0,¹₂

except at the point of tangency. The function q₂ is strictly concave on the interval (η₀,ξ−1) and q₂(0) = ¹₂,q⁰₂(0) = ³₄. Therefore, fort ∈ (η₀,ξ−1)the graph ofq₂ is strictly below the tangent liney = ¹₂+³_4t to the graph ofq2 at 0,¹₂

except at the point of tangency. Consequently, the functionqis positive on(η₀,ξ−1)except at the pointr₀=0, whereq(r₀) =0.

4) Sinceq₁(ξ−1) = ¹₂e^√π3 = 3.0668 and the function q₁ is strictly increasing onR, then for everyt ∈[ξ−1,+_∞)we haveq₁(t)≥q₁(ξ−1)> 1≥q2(t). Therefore, the function qis positive on [ξ−1,+_∞).

It follows from 1) to 4) that h(t) > 0 for every t ∈ [η₁,+_∞) except at the point r₀ = 0, where h(r0) =0. In particular,h(t)>0 for every t>0.

(c) To prove the statement, it is sufficient, taking into account (b), to prove that for every k∈_N0the functionqhas a unique zeror_k+1 in the interval(ξ_k+1,ξ_k).

Since the functionq₁is positive on Rand the function q2is nonnegative onC_k = [η_k+1,η_k] (k=1, 3, 5, . . .), then

q⁰⁰(t) = ⁹

8q₁(t) + ³

4q₂(t)>0, ∀t∈ C_k (k=1, 3, 5, . . .).

Therefore, qis a strictly convex function on the intervalC_k (k =1, 3, 5, . . .), and its restriction to every subinterval ofC_k (k=1, 3, 5, . . .)is a strictly convex function as well.

Suppose thatk=0, 2, 4, . . . The functionqhas no zeros in the subinterval I_k = (η_k+1,ξ_k)⊂ (ξ_k+₁,ξ_k), since the function q₁ is positive on I_k and the function q₂ is negative on I_k. The functionqhas a unique zero in the subintervalI_k+1 = (ξ_k+1,η_k+1)⊂(ξ_k+1,ξ_k), sinceq(ξ_k+1)<

0, q(η_k+1) > 0 and the function qis strictly convex and continuous on I_k+1 = [ξ_k+1,η_k+1] ⊂ C_k+1. We have(ξ_k+1,ξ_k) = I_k+1∪ {η_k+1} ∪I_k andq(η_k+1) > 0. Consequently, the function q has a unique zero r_k+1in the interval (ξ_k+1,ξ_k).

Suppose that k = 1, 3, 5, . . . The function q has a unique zero in the subinterval J_k−1 = (η_k+1,ξ_k) ⊂ (ξ_k+1,ξ_k), since q(η_k+1) > 0, q(ξ_k) < 0 and the function q is strictly convex and continuous on J_k−1 = [η_k+1,ξ_k] ⊂ C_k. The function q has no zeros in the subinterval J_k = (ξ_k+1,η_k+1) ⊂ (ξ_k+1,ξ_k), since the function q₁ is positive on J_k and the function q₂ is negative on J_k. We have (ξ_k+1,ξ_k) = J_k+1∪ {η_k+1} ∪J_k and q(η_k+1) > 0. Consequently, the functionqhas a unique zero r_k+1in the interval (ξ_k+1,ξ_k).

(d) It follows from the proof of (b) and (c) that the statement is valid.

Consider the scalar boundary value problem

x⁰⁰⁰= λx, x(0) =0, x⁰(0) =0, x(1) =0. (3.32) Definition 3.10. A numberλis called an eigenvalue of (3.32) if there exists a nontrivial solu- tion of (3.32). The setσof all eigenvalues of (3.32) is called the spectrum of (3.32).

Proposition 3.11. The setσ={r³_j : j∈_N}is the spectrum of (3.32), where r_j (j∈_N)are negative zeros of the function h.

Proof. Ify(t;λ)solves the Cauchy problem

y⁰⁰⁰=λy, y(0) =0, y⁰(0) =0, y⁰⁰(0) =1, (3.33) thenx(t;λ;γ) =γy(t;λ)is the solution of the Cauchy problem

x⁰⁰⁰= λx, x(0) =0, x⁰(0) =0, x⁰⁰(0) =γ.

The boundary value problem (3.32) has a nontrivial solutionx(t;λ;γ)if and only ifγ6=0 and y(1;λ) =0.

1) Ifλ=0, theny(t;λ) = ^t₂² solves (3.33). Since y(_1;λ) = ¹

2 >_{0, (3.34)}

thenλ=0 does not belong to the spectrumσof (3.32).

2) Ifλ >0, thenλ = r³, wherer = √³

λ> 0. The functiony(t;λ) = ¹

3r² h(rt)solves (3.33).

Taking into account (b) of Proposition3.9, y(1;λ) = ¹

3r²h(r)>0. (3.35)

Therefore, positiveλdo not belong to the spectrumσof (3.32).

3) If λ < 0, then λ = −r³, where r = ^p3 |λ| > 0. The functiony(t;λ) = _3r¹₂ h(−rt)solves (3.33). Taking into account (c) of Proposition3.9, y(1;λ) = _3r¹₂ h(−r) =0 if and only if there exists j∈ _Nsuch that −r = r_j. Therefore, a negative λ belongs to the spectrumσof (3.32) if and only if there exists j∈_{Nsuch that}λ=−r³=r³_j.

If follows from 1) to 3) that the statement is valid.

Remark 3.12. The numerical values ofr_j (j=1, 2, 3, 4, 5)are

r₅ =−18.7426, r₄=−15.115, r₃=−11.4874, r₂ =−7.85979, r₁ =−4.23321.

The numerical values of the first five eigenvalues of the spectrumσ are

r³₅= −6583.99, r³₄ =−3453.22, r³₃= −1515.88, r³₂ =−485.549, r³₁= −75.8593.

3.4 The proof of Lemma3.1

Proof. In accordance with (c) of Proposition3.9, the numbersr_j (j∈ _N)are negative zeros of h. Taking into account (3.2), we deduce thatε₁r_j,ε₂r_j (j∈_N)are zeros ofhas well. It follows from (a) of Proposition3.9thatr₀ =0 is a double zero ofhandr_j,ε₁r_j,ε₂r_j (j∈_N)are simple zeros ofh. All these zeros

r₀,r_j,ε₁r_j, ε₂r_j (j∈_N) (3.36) ofhare located on the lines Imz=0 and Imz= ±√

3 Rez.

SinceL[z_1,k;z_2,k] (k∈_N0)is the vertical line segment with the end pointsz_1,k = ξ_k−i√ 3ξ_k andz_2,k = ξ_k+i√

3ξ_k, then it follows from (3.29) that the interior of the triangle T_k (k ∈ _N0) with the edges (3.14) contains 2+3kzeros (3.36) ofh, counted with multiplicity. More detailed,

• the interior ofT₀ contains the double zeror₀ =0 ofh,

• the interior of T_k (k ∈ _N) contains the double zero r₀ = 0 of h and 3k simple zeros r_j,ε₁r_j,ε₂r_j (₁≤ j≤k)_ofh.

It follows from Proposition 3.8 that for every k ∈ _N0 the number of zeros of h in the interior of T_k, counted with multiplicity, is equal to 2+3k and thus the numbers (3.36) are exactly the zeros ofh. Ifz∈ _Cis not located on the lines Imz =_{0 and Im}z =±√

3 Rez, then zis not a zero ofhand thush(z)6=0.

Remark 3.13. Actually, we have proved more than Lemma3.1claims, namely that the numbers (3.36) form the set of all zeros ofh.

4 Vector field ψ associated with the linear boundary value problem

Definition 4.1. The vectorial boundary value problem (1.3), (1.4) is called non-resonant if the problem (1.3), (1.4) has only the trivial solution.

Suppose thatPsolves then×nmatrix Cauchy problem

P⁰⁰⁰ = A P, P(0) =O_n, P⁰(0) =O_n, P⁰⁰(0) =I_n. (4.1) If p(t;γ)is the solution of the vectorial Cauchy problem (1.3),

p(0) =0, p⁰(0) =0, p⁰⁰(0) =γ, (4.2) then p(t;γ) = P(t)γ for every t ∈ _R and γ ∈ _Rⁿ. Let us introduce a linear vector field ψ: Rⁿ →_Rⁿ,

ψ(γ) =p(1;γ) = P(1)γ, ∀γ∈_Rⁿ. (4.3) Consequently,ψ⁰(γ) =ψ⁰(₀) =P(1)for every γ∈_Rⁿ.

Theorem 4.2.

(A) The following statements are equivalent.

(1) The boundary value problem(1.3),(1.4)is non-resonant.

(2) detP(1)6=0.

(3) γ=0is a unique singular point of the vector fieldψ.

(4) No eigenvalue of the matrix A belongs to the spectrumσof (3.32).

(B) Suppose that one of the statements(₁)–(₄)holds. Then

ind(0,ψ) =sgn detP(1). (4.4)

If the matrix A does not have negative eigenvalues with odd algebraic multiplicities, thenind(0,ψ) =1.

If the matrix A has m(1≤ m≤n)different negative eigenvaluesλ_s(1≤s≤ m)with odd algebraic multiplicities, then

ind(0,ψ) =

∏

sgnh

−³ q

|λ_s|

= (−1)^j1+···+jm, (4.5) where

r_js+1<−^q3 |λ_s|<r_js (1≤ s≤m) (4.6) and r_j (j∈_N0)are real zeros of h ordered as in(3.29).

Proof. (A) Since the nonzero singular points of the linear vector field ψ are in one-to-one correspondence with the nontrivial solutions of the boundary value problem (1.3), (1.4), then equivalencies(1)⇔(2)⇔(3)hold by (4.3).

Let us prove(2)⇔(4).

Since the matrix A has real elements, then, see [14], there exists nonsingular matrix M with real elements such that J = M⁻¹AM, where J is the real Jordan form of A. If p = Mq andγ= Mη, then the Cauchy problem (1.3), (4.2) transforms to the Cauchy problem

q⁰⁰⁰= Jq, q(0) =0, q⁰(0) =0, q⁰⁰(0) =η. (4.7) Ifq(t;η)is the solution of (4.7) andQsolves the n×nmatrix Cauchy problem

Q⁰⁰⁰= J Q, Q(₀) =O_n, Q⁰(₀) =O_n, Q⁰⁰(₀) =I_n, (4.8) thenq(t;η) = Q(t)ηfor everyt ∈ _R andη∈ _Rⁿ. It follows from q(1;η) = M⁻¹p(1;γ) that Q(1)η = M⁻¹P(1)Mηfor every η ∈ _Rⁿ. Therefore, Q(1) = M⁻¹P(1)M. The matrices Q(1) andP(1)are similar and have the same eigenvalues, counted with multiplicity. Consequently,

detQ(1) =detP(1). (4.9)

Next we will analyze detQ(₁)_.

The blocks of the real Jordan form J of A are of two types. A real eigenvalue λ of A generates blocks

J_k(λ) =















λ 1 0 · · · 0 0 0 λ 1 · · · 0 0 ... ... ... · · · ^... ... 0 0 0 · · · λ 1 0 0 0 · · · ₀ λ















(4.10)

of the sizek. A pair of complex conjugate eigenvaluesλ=a+ibandλ=a−ib, whereb6=0, generates blocks

J_k(λ) =















C₂(λ) I₂ O₂ · · · O₂ O₂ O₂ C₂(λ) I₂ · · · O₂ O₂ ... ... ... · · · ^... ... O2 O2 O2 · · · C2(λ) I2

O₂ O₂ O₂ · · · O₂ C₂(λ)















(4.11)

of the sizek=2m, where C₂(λ) =

a −b

b a

, I₂ =

1 0 0 1

, O₂ =

0 0 0 0

. Suppose thatQ_k solves thek×kmatrix Cauchy problem

Q⁰⁰⁰_k = J_k(λ)Q_k, Q_k(₀) =O_k, Q⁰_k(₀) =O_k, Q⁰⁰_k(₀) = I_k. (4.12) If λ = r³ sgnλ, where r = ^p3 |λ| ≥ 0, is a real eigenvalue of A, then corresponding to λ Jordan blocks have the form (4.10). Taking into account [28], we obtain

Q_k(t) =

∑

t^3j+2 (_3j+₂)_!

J_k(λ)^j, ∀t∈_R, (4.13)

where [J_k(λ)^j (j ∈ _N0) is an upper triangulark×k matrix with the same elementλ^j on the main diagonal. Therefore, the matrix Q_k(t)is an upper triangular matrix also with the same element

y(t;λ) =

∑

t^3j+2

(3j+2)!λ^j, ∀t∈_R,

on the main diagonal. It follows from (4.12) thaty(t;λ)solves the Cauchy problem (3.33) and detQ_k(1) =y(1;λ)^k.

(a) Ifλ=0, then it follows from (3.34) that detQ_k(1) = ¹

2^k >0.

(b) Ifλ>0, then it follows from (3.35) that detQ_k(1)>0.

(c) Ifλ<0, then λ= −r³. It follows from Proposition3.11that detQ_k(1) =

h(−r) 3r²

k

6=0⇔ −r 6=r_j (j∈_N)⇔λ6∈σ.

(d) Ifλ= a+ibandλ= a−ib, whereb6=0, is a pair of complex conjugate eigenvalues of A, then corresponding toλandλJordan blocksJ_k(λ)have the form (4.11). The power

J_k(t)^j (k = 2m; j ∈ _N0) is an m×mupper triangular block matrix of 2×2 blocks with the same block

C₂(λ)^j on the main diagonal. Therefore, the matrixQ_k(t), given by (4.13), is anm×m upper triangular block matrix of 2×2 blocks also with the same block

D₂(t) =

u₂(t) −v₂(t) v₂(t) u₂(t)

=

∑

t^3j+2 (3j+₂)_!

C₂(λ)^j

on the main diagonal, where u2(_t) =

∑

t^3j+2

(3j+2)!|λ|^jcos jargλ

, v2(_t) =

∑

t^3j+2

(3j+2)!|λ|^jsin jargλ , and

detQ_k(1) =detD₂(1)^m =u²₂(1) +u²₂(1)^m ≥0. (4.14) It follows from (4.12) thatD₂(t)solves the 2×2 matrix Cauchy problem

D₂⁰⁰⁰(t) =C2(λ)D2(t), D2(0) =O2, D⁰₂(0) =O2, D₂⁰⁰(0) = I2.

Let us introduce a complex-valued function w₂(t) = u₂(t) +i v₂(t)of a real variable t. Then we can rewrite the last Cauchy problem in the complex form

w₂⁰⁰⁰(t) =λw2(t), w2(0) =0, w⁰₂(0) =0, w⁰⁰₂(0) =1. (4.15) The function

w₂(t;λ) = ¹

3µ² e^µt+ε₁e^ε1µt+ε₂e^ε2µt

= ¹

3µ²h(µt), t ∈_R, (4.16) solves (4.15), where µis a fixed cube root ofλ∈ _C\_R. The solution (4.16) of (4.15) does not depend on the particular choice of the cube rootµ, since, in view of (3.2),

1

3µ²₁h(µ₁t) = ¹

3µ²₂h(µ₂t) =w₂(t;λ),