# Characterization for Rectifiable and Nonrectifiable Attractivity of Nonautonomous Systems of Linear Differential Equations

Authors: Yūki Naito; Mervan Pašić
Journal: International Journal of Differential Equations
 (2013)

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## Abstract

We study a new kind of asymptotic behaviour neart=0 for the nonautonomous system of two linear differential equations: x'(t)=A(t)x(t),  t∈(0,t0], where the matrix-valued function A=A(t) has a kind of singularity at t=0. It is called rectifiable (resp., nonrectifiable) attractivity of the zero solution, which means that ∥x(t)∥2→0 as t→0 and the length of the solution curve of x is finite (resp., infinite) for every x≠0. It is characterized in terms of certain asymptotic behaviour of the eigenvalues of A(t) near t=0. Consequently, the main results are applied to a system of two linear differential equations with polynomial coefficients which are singular at t=0.


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## Body

## 1. Introduction

We are concerned with the two-dimensional nonautonomous linear differential systems:(1)dxdt=A(t)xfort∈(0,t0],x(t0)=x0∈ℝ2,
where t0>0, and A=A(t), A:(0,t0]→ℳ2 is a matrix-valued function such that A∈C1((0,t0];ℳ2), where ℳ2 denotes the space of all 2×2 real matrices. A solution of linear system (1) is a function x=x(t), x:(0,t0]→ℝ2 such that x∈C1((0,t0],ℝ2).In appendix of the paper, we show, there exists a unique solutionx of linear system (1). By the uniqueness of solution x of (1), it is clear that if x0=(0,0), then x(t)=(0,0) for all t∈(0,t0]. It is called the zero solution of linear system (1). The zero solution of linear system (1) is said to be attractive ast→0 if for every x0∈ℝ2 the corresponding solution x of (1) satisfies ∥x(t)∥2→0 as t→0, where ∥·∥2 denotes the standard Euclidean norm on ℝ2. Since we are dealing with the attractive zero solution of (1), from x∈C1((0,t0],ℝ2) we conclude that x∈C([0,t0],ℝ2)∩C1((0,t0],ℝ2). For a continuous function x:[0,t0]→ℝ2, let Γx denote a curve in ℝ2 corresponding to x determined by
(2)Γx=x([0,t0])={(x,y)∈ℝ2:x=x(t),y=y(t)},
where (x(t),y(t))≡x(t). It is often said that Γx is parametrized by x.Letx∈C([0,t0];ℝ2). The curve Γx⊆ℝ2 is said to be a Jordan curve if x is injective (one-to-one) on [0,t0]. Hence, Γx is nonself-intersecting; it has two ends and it is a compact connected set in ℝ2. In appendix of the paper, see Theorem A.3; we give some simple sufficient conditions on the matrix-valued function A=A(t) such that the solution curve Γx of every nontrivial solution of (1) is a Jordan curve in ℝ2. The assumption that Γx is a Jordan curve is important for the process of measuring the length of Γx (the rectification of Γx) as follows. If Γx is a Jordan curve, then the length of Γx is denoted by length (Γx) and it is defined as usual by (see [1–3])
(3)length(Γx)=sup{∑k=1m‍∥x(tk)-x(tk-1)∥2:m∈ℕ,0=tm<⋯<t1<t0∑k=1m‍},
where the supremum is taken over all finite dissections 0=tm<⋯<t1<t0 of the interval [0,t0].Definition 1.
The zero solution of linear system (1) is said to be  rectifiable (resp.,  nonrectifiable)  attractive ast→0; if for every x0∈ℝ2∖{0} such that the Γx of corresponding solution x of (1) is a Jordan curve in ℝ2, one has ∥x(t)∥2→0 as t→0 and length (Γx)<∞ (resp., length (Γx)=∞).Letλ(t)=Reλ(t)±iImλ(t), t∈(0,t0], be two complex conjugate eigenvalues of A(t). The main purpose of the paper is to characterize attractivity, rectifiable and nonrectifiable attractivity, as t→0 of the zero solution of system (1) in the terms of regular and singular asymptotic behaviour of A(t) near t=0; see Theorems 2 and 3. It is illustrated on a simple two-dimensional linear differential systems with the polynomial coefficients; see Theorem 4. The key point of proofs of the main results is an asymptotic solution formula for system (1) obtained by its asymptotic integration near t=0; see Section 3. The proofs of the main results are given in Sections 4, 5, and 6. Such a kind of topics has been considered for the first time in [4] but only in the case of the so-called integrable systems (systems that allow all solutions in explicit forms), where the asymptotic integration near t=0 is not required. About the asymptotic integration near t=∞ of differential equations and systems, we refer reader to [5–12]. On several asymptotic properties of two-dimensional differential systems near t=∞, let us see, for instance, [13–18]. Recently interest for studying differential equations and systems on compact (finite time) intervals is rapidly increasing because of its application in atmosphere, fluid, and ocean dynamics and in biological science; see, for instance, [19–21] and references therein.


## 2. Main Results and Consequences

Let for eacht∈(0,t0] the matrix elements of A(t) be denoted by (ai,j(t))i,j≤2. In the paper, we suppose the following kind of singular asymptotic behaviour of A(t) near t=0: (4)limε→0∫εt0‍|ddt(t2ai,j(t))|dt<∞,i,j∈{1,2},limt→0[t2A(t)]=A0,
where A0 is a constant matrix having two complex conjugate eigenvalues.Some essential consequences of (4) are presented in Section 3. Since the eigenvalues of A(t) are continuous on t∈(0,t0], by (4) we may assume that A(t) has two complex conjugate eigenvalues λ(t)=Reλ(t)±iImλ(t) for t∈(0,t0].The first two main results of the paper are the following theorems.Theorem 2.
Let (4) be satisfied, and let λ(t)=Reλ(t)±iImλ(t), t∈(0,t0], be two complex conjugate eigenvalues of A(t). Assume that Reλ(t)≥0 near t=0. Then the zero solution of linear system (1) is attractive as t→0 if and only if
(5)limt→0∫tt0‍Reλ(s)ds=∞.Theorem 3.
Let (4) be satisfied, and let λ(t)=Reλ(t)±iImλ(t), t∈(0,t0], be two complex conjugate eigenvalues of A(t). Assume that Reλ(t)≥0 near t=0, and that (5) holds. Then the zero solution of linear system (1) is rectifiable attractive as t→0 if and only if
(6)Imλ(t)e-∫tt0‍Reλ(s)ds∈L1(0,t0).Previous theorem allows us to study the rectifiable and nonrectifiable attractivity for the following model system with polynomial singular coefficients neart=0:
(7)x′(t)=e(t)t-px(t)+f(t)t-2y(t),y′(t)=-g(t)t-2x(t)+h(t)t-qy(t),t∈(0,t0],
where the parameters p,q∈ℝ and coefficients e,f,g,h∈C([0,t0])∩C1((0,t0]) satisfy, respectively,
(8)1≤max{p,q}<2,(9)f(0)g(0)>0,e(t)>0,h(t)>0,t∈[0,t0],(10){∫0t0‍|f′(t)|dt<∞,∫0t0‍|g′(t)|dt<∞,∫0t0‍|ddt(t2-pe(t))|dt<∞,∫0t0‍|ddt(t2-qh(t))|dt<∞.
According to (8), (9), and (10), we will show that the model system (7) satisfies all assumptions of Theorem 3. Hence, we are able to derive and prove the following conditions for the rectifiable and nonrectifiable attractivity of the zero solution of model system (7).Theorem 4.
Let (8), (9), and (10) be fulfilled. (i)
Ifmax{p,q}>1, then the zero solution of (7) is rectifiable attractive as t→0.(ii)
Ifmax{p,q}=1, then the zero solution of (7) is rectifiable attractive as t→0 provided that one of the following conditions is satisfied:
(11)p=1,q<1,e(t)>2;(12)p<1,q=1,h(t)>2;(13)p=1,q=1,e(t)+h(t)>2.(iii)
Ifmax{p,q}=1, then the zero solution of (7) is nonrectifiable attractive as t→0 provided that one of the following conditions is satisfied:
(14)p=1,q<1,e(t)≤2;(15)p<1,q=1,h(t)≤2;(16)p=1,q=1,e(t)+h(t)≤2.As a consequence of previous theorem, we realize that the rectifiable and nonrectifiable attractivity of the zero solution of model system (7) depends on the order of growth of singular behaviour of coefficients near t=0 determined by parameters p and q. Let us remark that if, besides  (8), (9), and (10), it is supposed that e(t)≡h(t), f(t)≡g(t), and p=q, then system  (7) is integrable, that is, all solutions of system (7) are explicitly determined, see appendix of the paper. In such a particular case of (7), Theorem 4 can be explicitly verified.Remark 5.
We suggest reader to consider the rectifiable and nonrectifiable attractivity of the zero solution of system (7) in the case of max{p,q}=2. It is required to assume that 4f(0)g(0)-e2(0)>0 if p=2>q as well as 4f(0)g(0)-h2(0)>0 if q=2>p and 4f(0)g(0)-(e(0)-h(0))2>0 if p=q=2.The most difficult point of this paper is to derive and prove the nonrectifiable attractivity of the zero solution of systems  (1) and  (7); see Sections 3 and 4. On the (non)rectifiability of the solution curve of scalar second-order differential equations, we refer reader to [22–27].


## 3. Some Asymptotic Properties Neart=0

As a consequence of the main assumption (4), we derive in this section some important asymptotic properties of A(t) as well as of its eigenvalues and solutions of system (1), which will be used in the next sections in the proof of the main results. Since the eigenvalues of A(t) are continuous on (0,t0], by (4) we may assume that A(t) has two complex conjugate eigenvalues λ(t)=Reλ(t)±iImλ(t), t∈(0,t0].Lemma 6.
Let (4) be satisfied. Then the eigenvalues of A(t) satisfy
(17)∥A(t)∥~|Imλ(t)|~t-2ast⟶0,(18)Imλ(t)≠0neart=0,
where f(t)~g(t) means that f(t)/g(t)→1 as t→0.Proof.
Letai,j=ai,j(t), ai,j∈C1((0,t0]), be matrix elements of A(t). Also, let ai,j0 and λ0=Reλ0±iImλ0 be matrix elements and eigenvalues of A0, respectively, where A0 appears in (4). Rewriting (4) in terms of matrix elements and eigenvalues, we get
(19)limt→0[t2ai,j(t)]=ai,j0,Imλ0≠0.
Hence, we can conclude the following:
(20)limt→0[t2|Imλ(t)|]=12limt→0[t2-(a1,1(t)-a2,2(t))2-4a1,2(t)a2,1(t)]=12limt→0-(t2a1,1(t)-t2a2,2(t))2-4t2a1,2(t)t2a2,1(t)=12-(a1,10-a2,20)2-4a1,20a2,10=|Imλ0|≠0.
It proves that |Imλ(t)|~t-2 as t→0. Hence (18) and the second claim in (17) hold. In order to complete (17), it rests to show that ∥A(t)∥~t-2 as t→0. In fact, from the assumption Imλ0≠0, we observe that for at least one matrix element ai,j0 we have ai,j0≠0 and, therefore, ∥A0∥≠0. Moreover, since every matrix norm is a continuous function, from (4) we especially obtain limt→0[t2∥A(t)∥]=∥A0∥≠0. It completes the proof of (17).As the main consequence of (4) we have the following asymptotic formula near t=0 for all solutions x of linear system (1):
(21)x(t)=e-∫tt0‍Reλ(s)ds[h1(t)cos(∫tt0‍Imλ(s)ds)+h2(t)sin(∫tt0‍Imλ(s)ds)],
where h1=h1(t), h2=h2(t), hi:(0,t0]→ℝ2 and there are c1,c2∈ℝ2 such that
(22)limt→0hi(t)=ci≠0,i∈{1,2}.
It is the subject of the next lemma.Lemma 7.
Letλ(t)=Reλ(t)±iImλ(t) be two eigenvalues of A(t). If A(t) satisfies (4), then for every solution x of linear system (1) with x0≠{0} there exist two functions hi(t) and two constants c1,c2∈ℝ2 such that  (21) and  (22) are fulfilled.Proof.
LetA(t) be a matrix satisfying all assumptions of this lemma. Let B(τ)=(bi,j(τ)i,j≤2) be a new matrix depending on a new variable τ:=1/t defined by
(23)B(τ)=-τ-2A(τ-1)∀τ≥τ0:=1t0.
Then bi,j∈C1([τ0,∞)). Let A0 be a constant matrix determined in (4). If B0 is a new matrix defined by B0=-A0, then from (4), (23), and t=1/τ, we conclude
(24)limM→∞∫τ0M‍|ddτbi,j(τ)|dτ<∞,i,j∈{1,2},limτ→∞B(τ)=B0,
where  B0   is  a  constant  matrix  having  two complex  conjugate  eigenvalues.
Also, the eigenvaluesμ1(τ),μ2(τ) of B(τ) and λ1(t)=Reλ(t)+iImλ(t), λ2(t)=Reλ(t)-iImλ(t) of A(t) are obviously related by the equality
(25)μi(τ)=-τ-2λi(τ-1)=-τ-2[Re(λ(τ-1))±iIm(λ(τ-1))],i=1,2,
and hence,
(26)Re(μ1(τ)-μ2(τ))=0∀τ≥τ0.
Putting t=1/τ into system (1) and denoting by
(27)w(τ):=x(1τ)forτ≥τ0,
we obtain the equivalent system of (1)
(28)w′(τ)=B(τ)w(τ)forτ≥τ0,
where matrix B(τ) is defined in (23). For system (28), we will use the next generalized Matell’s theorem appearing as Theorem 11 in Coppel’s book [5, Chapter 4] and Theorem 6.5 in Kiguradze’s monograph [7] (see also [12, Theorem 7] and [8]).
Theorem8 (see [7]). Letμ1=μ1(τ)andμ2=μ2(τ),μ1,μ2:[τ0,∞)→ℂbe two eigenvalues of an arbitrary2×2matrixB(τ),τ∈[τ0,∞), with the matrix elementsbi,j(τ)which are absolutely continuous functions on every compact setI⊂[τ0,∞)and(29)∫τ0∞‍|ddτbi,j(τ)|dτ<∞,i,j∈{1,2}.Letc>0be a constant such that(30)∫σ*τ*‍Re(μ1(τ)-μ2(τ))dτ≤c,τ0≤σ*<τ*,or∫τ0∞‍Re(μ1(τ)-μ2(τ))dτ=∞,∫σ*τ*‍Re(μ1(τ)-μ2(τ))dτ≥-c,τ0≤σ*<τ*.LetB(τ)→B0asτ→∞, whereB0is a constant matrix having two different eigenvalues. Then linear differential system  (28) has a fundamental system of solutionsw1(τ)∈ℂ2andw2(τ)∈ℂ2such that(31)limτ→∞[w1(τ)e-∫τ0τ‍μ1(σ)dσ]=b1,limτ→∞[w2(τ)e-∫τ0τ‍μ2(σ)dσ]=b2,whereb1,b2∈ℂ2are two eigenvectors of constant matrixB0.
Now, with the help of (24) and (26), it is easy to check that the matrix B(τ) defined in (23) and its eigenvalues μ1(τ) and μ2(τ) defined in (25) satisfy all required assumptions of Theorem 8. Hence, by Theorem 8, we observe that linear system (1) has a fundamental system of solutions x1(t) and x2(t) satisfying
(32)limt→0[x1(t)e∫tt0‍λ1(s)ds]=b1,limt→0[x2(t)e∫tt0‍λ2(s)ds]=b2.
Indeed, from (31) and using τ=1/t, τ0=1/t0, (25), (27), and σ=1/s, we obtain
(33)b1=limτ→∞[w1(τ)e-∫τ0τ‍μ1(σ)dσ]=limt→0[w1(1t)e-∫1/t01/t‍μ1(σ)dσ]=limt→0[w1(1t)e∫1/t01/t‍σ-2λ1(σ-1)dσ]=limt→0[x1(t)e∫tt0‍λ1(s)ds].
Analogously, the second equality in (32) can be proved. Now, from (32) we particularly obtain
(34)limt→0[x1(t)e∫tt0‍λ1(s)ds-b1]=0,limt→0[x2(t)e∫tt0‍λ2(s)ds-b2]=0.
Denoting(35)g1(t)=x1(t)e∫tt0‍λ1(s)ds-b1,g2(t)=x2(t)e∫tt0‍λ2(s)ds-b2,
from (34), we get limt→0gi(t)=0,i∈{1,2} and near t=0,
(36)x1(t)=(b1+g1(t))e-∫tt0‍λ1(s)ds,x2(t)=(b2+g2(t))e-∫tt0‍λ2(s)ds.
Since λ1(t)=Reλ(t)+iImλ(t), λ2(t)=Reλ(t)-iImλ(t), bi∈ℂ2 and gi:(0,t0]→ℂ2, previous statements verify the desired asymptotic formula (21). The statement (22) holds because b1 and b2 are two eigenvectors and hence, bi≠0.Now, according to the asymptotic formula (21), we are able to prove Theorem 2.Proof of Theorem2.
Because of (4) and Lemmas 6 and 7, we can use (17) and (18) and the asymptotic solution’s formula (21)-(22).
At first, we suppose that (5) holds. Taking the norm on both sides in (21) and using (22), we obtain the upper estimate
(37)∥x(t)∥2≤ce-∫tt0‍Reλ(s)dsneart=0
with some constant c>0. Now putting the limit as t→0 on both sides in (37), and using (5), we get ∥x(t)∥2→0 as t→0. Thus, it is shown that the zero solution of system (1) is attractive as t→0 provided that (5) holds.
The proof of the reverse claim is slightly complicated. Suppose that(38)∥x(t)∥2⟶0ast⟶0.
Next, from (17), and (18), we especially obtain
(39)limt→0|∫tt0‍Imλ(s)ds|=∞.
Remarking that in (21) and (22), we use the following notations:
(40)x(t)=(x(t),y(t)),hi(t)=(hi,1(t),hi,2(t)),ci=(ci,1,ci,2),i∈{1,2}.
By (22), at least one of ci,1 and ci,2 is nonzero. So, for instance, c1,1≠0, which together by (22) gives
(41)limt→0|h1,1(t)|=|c1,1|>0.
Now, from (21) and (40), we derive the following inequality:
(42)∥x(t)∥≥∥x(t)∥=e-∫tt0‍Reλ(s)ds|h1,1(t)cos(∫tt0‍Imλ(s)ds)+h1,2(t)sin(∫tt0‍Imλ(s)ds)|.
Next, because of (18) and (39), there is a sequence tn∈(0,t0], tn→0 as n→∞, such that
(43)|∫tnt0‍Imλ(s)ds|=nπ,n∈ℕ.
Now, putting t=tn into (42) and using (41), (43), and sin(±nπ)=0, we get the inequality
(44)∥x(tn)∥2≥|c1,1|2e-∫tnt0‍Reλ(s)ds∀largen∈ℕ.
Taking the limit on both sides in previous inequality, with the help of (38), we conclude that
(45)limn→∞e-∫tnt0‍Reλ(s)ds=0,
that is,
(46)limn→∞∫tnt0‍Reλ(s)ds=∞.
Since Reλ(t)≥0 near t=0, the function ∫tt0‍Reλ(s)ds is continuous (since matrix elements of A(t) are continuous functions on (0,t0]) and monotone. Hence, statement (46) proves (5). Thus, the statement (5) holds provided that the zero solution of system (1) is attractive.


## 4. Rectifiable Attractivity of the Zero Solution

Unlike the nonrectifiable attractivity which will be studied in the Section5, the rectifiable attractivity of the zero solution is proved without any essential difficulties. It is because we can use here the following known result.Lemma 9.
LetΓx be a Jordan curve in ℝ2. Then one has length(Γx)<∞ if and only if x′∈L1((0,t0),ℝ2).For the proof of previous lemma, we suggest reader to consult the appropriate scalar case shown in [28].The following theorem derives some properties ofA(t) and its eigenvalues λ(t) which ensure that the zero solution of system (1) is rectifiable attractive.Theorem 10.
Letλ(t)=Reλ(t)±iImλ(t) be two complex conjugate eigenvalues of A(t) for t∈[0,t0]. Assume that (5) holds. Let every solution x of system (1) satisfies (37). If
(47)∥A(t)∥e-∫tt0‍Reλ(s)ds∈L1(0,t0),
then the zero solution of  (1) is rectifiable attractive as t→0.Proof.
It is clear that the attractivity of the zero solution of system (1) follows from (5), (21), and (37). Next, by taking the matrix norm in (1) and using a priori estimate (37), we obtain,
(48)∥x′(t)∥2=∥A(t)x(t)∥2≤∥A(t)∥∥x(t)∥2≤c∥A(t)∥e-∫tt0‍Reλ(s)ds.
Now, from the assumption  (47) and the inequality (48), we observe that
(49)limε→0∫εt0‍∥x′(t)∥2dt≤limε→0c∫εt0‍∥A(t)∥e-∫tt0‍Reλ(s)dsdt<∞.
Hence, Lemma 9 proves this theorem.IfA(t) admits the asymptotic behaviour given in (4), then a priori estimate  (37) is fulfilled because of the asymptotic formula (21) and thus, the rectifiable attractivity of the zero solution of system (1), in such a case, holds without supposing  (37), as follows.Theorem 11.
Let (4) be satisfied, and let λ(t)=Reλ(t)±iImλ(t) be two complex conjugate eigenvalues of A(t) for t∈[0,t0]. Assume that (5) holds. If the condition (47) is fulfilled, then the zero solution of linear system (1) is rectifiable attractive as t→0.Proof.
It is clear that all assumptions of Theorem11 ensure that Lemma 7 can be applied. Hence, by Lemma 7, that is, by asymptotic formula (21), one can easily show that a priori estimate  (37) holds for all solutions x of system (1). Thus, all assumptions of Theorem 10 are fulfilled. Now, Theorem 10 proves Theorem 11.


## 5. Nonrectifiable Attractivity of the Zero Solution

In this section we study the nonrectifiable attractivity for linear system (1). Unlike the rectifiable attractivity, the nonrectifiable attractivity is a more difficult case. It is because the required property x′∉L1(0,t0) cannot be derived from system (1) and therefore, we are not in an opportunity to use Lemma 9. Hence, instead of Lemma 9, we state and prove the next new lemma, which plays an essential role in the proof of the main results of this section.Lemma 12.
Letu=u(t), u∈C([0,t0]), and let there be a sequence sn∈(0,t0], a number n0∈ℕ, and two functions φ=φ(t), a=a(t), t∈(0,t0], such that(H1)
sequencesn is decreasing, sn→0 as n→∞ and u(sn)u(sn+1)<0 for all n≥n0;(H2)
functionφ(t), φ∈C1((0,t0]), is strictly monotone near t=0, that is, φ′(t)≠0 near t=0 and there is a constant c2>0 such that
(50)|φ(sn+1)-φ(sn)|≤c2∀n≥n0;(H3)
functiona(t), a∈C([0,t0]), is nondecreasing and there is a constant c1>0 such that
(51)|u(sn)|≥c1a(sn)∀n≥n0;(H4)
limε→0∫εt0‍a(t)|φ′(t)|dt=∞.
Then the graph(52)Γu={(x,y)∈ℝ2:x=t,y=u(t)fort∈[0,t0]}
of function u(t) is a nonrectifiable curve in ℝ2.Proof.
Because of(H1), there is a sequence an∈(sn+1,sn) such that u(an)=0 for all n≥n0 and hence we have
(53)0<⋯<sn+1<an<sn<an-1<⋯<sn0+1<an0<sn0<t0.
Let tn∈(0,t0] be a sequence defined by
(54)t2n-1:=sn,t2n:=anforn≥1.
Obviously for all m≥n0, the set {0,t2m-1,…,t2n0-1,t0} makes a finite partition of interval [0,t0] and since length(Γu) is a supremum of all sums ∑‍|u(τn)-u(τn-1)| taken over all finite partitions {τn} of [0,t0]; we conclude that
(55)length(Γu)≥|u(0)-u(t2m-1)|+∑n=2n02m-1‍|u(tn)-u(tn-1)|+|u(t2n0-1)-u(t0)|≥∑n=2n02m-1|u(tn)-u(tn-1)|=∑n=n0m‍|u(sn)|.
Next, let, for instance, φ(t) be strictly decreasing near t=0 (the case when φ(t) is strictly increasing could be analogously considered). Hence, from (H2) and (H3), it follows that
(56)∑n=n0m‍|u(sn)|≥c1∑n=n0m‍a(sn)≥c1c2∑n=n0m‍a(sn)(φ(sn+1)-φ(sn))=c1c2∑n=n0m‍∫φ(sn)φ(sn+1)‍a(sn)dτ=c1c2∑n=n0m‍∫φ(sn)φ(sn+1)‍[maxt∈[sn+1,sn]a(t)]dτ≥c1c2∑n=n0m‍∫φ(sn)φ(sn+1)‍a(φ-1(τ))dτ=c1c2∑n=n0m‍∫sn+1sn‍a(t)|φ′(t)|dt.
Now from (55) and previous inequality, for all m≥n0 we obtain
(57)length(Γu)≥c1c2∑n=n0m‍∫sn+1sn‍a(t)|φ′(t)|dt=c1c2∫sm+1sn0‍a(t)|φ′(t)|dt.
Letting m→∞ in previous inequality, the hypothesis (H4) proves this lemma.Now, we are able to prove the following result.Theorem 13.
Letλ(t)=Reλ(t)±iImλ(t) be two complex conjugate eigenvalues of A(t) for t∈(0,t0]. Assume that Reλ(t)≥0 and Imλ(t)≠0 near t=0 and that (5) holds. Let every solution x(t)=(x(t),y(t)) of system  (1) satisfy  (37) as well as the following assumption: let u(t) be defined by u(t):=x(t) or u(t):=y(t) and let there be two constants c1>0,c2>0, a number n0∈ℕ, and a decreasing sequence sn∈(0,t0], sn→0 as n→∞, which all depend on function u(t), such that for all n≥n0, one has
(58)u(sn)u(sn+1)<0,|u(sn)|≥c1e-∫snt0‍Reλ(t)dt,|∫sn+1sn‍Imλ(t)dt|≤c2.
If
(59)Imλ(t)e-∫tt0‍Reλ(s)dsdt∉L1(0,t0),
then the zero solution of  (1) is nonrectifiable attractive as t→0.Proof.
At first, the attractivity of the zero solution of linear system (1) follows from the assumptions (5) and (37).
Next, letφ(t) and a(t) denote two new functions defined by
(60)φ(t)=∫tt0‍Imλ(s)ds,a(t)=e-∫tt0‍Reλ(s)ds.
Let u(t):=x(t) or u(t):=y(t), where x(t)=(x(t),y(t)) is a solution of system  (1). With the help of notation (60) and assumption (58), it is easy to check that the function u(t) satisfies all hypotheses (H1), (H2), and (H3) from Lemma 12. Indeed, the existence of a sequence sn∈(0,t0) satisfying (H1) follows from the first inequality in (58). Also, from  (18), (58), and (60) we obtain
(61)φ′(t)=-Imλ(t)≠0neart=0,|φ(sn+1)-φ(sn)|=|∫snsn+1‍Imλ(s)ds|≤c2∀n≥n0,|u(sn)|≥c1e-∫snt0‍Reλ(s)ds=c1a(sn)∀n≥n0,
which show that (H2) and (H3) from Lemma 12 are fulfilled too. Moreover, the equality
(62)a(t)|φ′(t)|=|Imλ(t)|e-∫tt0‍Reλ(s)ds
together with assumption (59) ensure that the hypothesis (H4) from Lemma 12 is also fulfilled. Now, by Lemma 12 we obtain that the graph of u(t):=x(t) or u(t):=y(t) is a nonrectifiable curve in ℝ2. In particular, it follows that the solution curve Γx is also a nonrectifiable curve in ℝ2, where x(t)=(x(t),y(t)).In the next theorem, the nonrectifiable attractivity is obtained without supposing the a priori estimates  (37) and  (58), since they immediately follow from the asymptotic behaviour of A(t) near t=0 given in (4).Theorem 14.
Let (4) be satisfied, and let λ(t)=Reλ(t)±iImλ(t) be two complex conjugate eigenvalues of the matrix A(t) for t∈(0,t0]. Assume that Reλ(t)≥0 near t=0 and that (5) holds. If the condition (59) holds, then the zero solution of  (1) is nonrectifiable attractive as t→0.Proof.
This proof is a consequence of Theorem13 and the asymptotic formula (21). At first, according to Lemma 7, every solution x(t)=(x(t),y(t)) of system (1) satisfies (21). If we denote by u(t):=x(t) or u(t):=y(t), then (21) ensures that u(t) could be written in the form
(63)u(t)=[c1+ε1(t)]a(t)cosφ(t)+[c2+ε2(t)]a(t)sinφ(t)neart=0,
where c1,c2∈ℝ with c12+c22>0, ε1,ε2∈C(0,t0], limt→0ε1(t)=limt→0ε2(t)=0, and a(t),φ(t) are defined in (60).
Sinceλ(t) is a continuous complex-valued function on (0,t0], from (18), we conclude that
(64)eitherImλ(t)<0orImλ(t)>0neart=0.
Therefore, there are two real constants c3,c4∈ℝ and δ∈C(0,t0] such that c3≠0, limt→0δ(t)=0 and
(65)u(t)=[c3+δ(t)]e-∫tt0‍Reλ(σ)dσ×sin(∫tt0‍|Imλ(σ)|dσ+c4)neart=0.
Because of (59), we have ∫tt0‍|Imλ(σ)|dσ→∞ as t→0 and therefore there is a decreasing sequence sn∈(0,t0] and a number n0∈ℕ such that
(66)∫snt0‍|Imλ(σ)|dσ+c4=π2+nπ∀n≥n0.
Putting (66) into (65) and retaking n0∈ℕ as sufficiently large, we get
(67)u(sn)u(sn+1)<0,|u(sn)|≥|c3|2e-∫snt0‍Reλ(s)ds∀n≥n0,
and because of (64) and (66),
(68)|∫sn+1sn‍Imλ(s)ds|=∫sn+1sn‍|Imλ(s)|ds=∫sn+1t0‍|Imλ(s)|ds-∫snt0‍|Imλ(s)|ds=π.
Now, statements (67) and (68) show that the function u(t) and sequence sn defined, respectively, in (65) and (66) satisfy the required condition (58). Therefore, the main conclusion of this theorem immediately follows from Theorem 13.


## 6. Proof of Theorem4

In this section, we prove the rectifiable and nonrectifiable attractivity for the zero solution of model system (7) as a direct consequence of Theorem 3.LetA(t) be the matrix of the model system (7) given by
(69)A(t)=(e(t)t-pf(t)t-2-g(t)t-2h(t)t-q)fort∈(0,t0],
where the real numbers p, q, and functions e, f, g, and h satisfy (8), (9), and (10), respectively. We will show that such a matrix A(t) together with its eigenvalues λ(t) satisfy the required conditions Reλ(t)≥0 near t=0, (5), and (4).At first, because of (9), we have Reλ(t)=(e(t)t-p+h(t)t-q)/2>0.Next, sincemax{p,q}≥1, let, for instance, max{p,q}=p≥1, and t0∈(0,1]; from (9) we obtain a constant c0>0 such that
(70)∫tt0Reλ(s)ds=12∫tt0(e(s)s-p+h(s)s-q)ds≥12∫tt0‍e(s)s-pds≥c0∫tt0s-1ds=c0lnt0t⟶∞ast⟶0.
Now, it is clear that (70) proves the desired statement (5).Finally, sincep<2 and q<2,
(71)limt→0[t2A(t)]=limt→0(e(t)t2-pf(t)-g(t)h(t)t2-q)=(0f(0)-g(0)0)=:A0,
where A0 is a constant matrix. Since f(0)g(0)>0, matrix A0 has two different eigenvalues. So, the condition  (4) is fulfilled too.Thus, matrixA(t) together with its eigenvalues satisfy all required conditions so that Theorem 3 can be applied to the model system (7) in all cases of p and q given in (11)–(16). Since by Theorem 3 the rectifiable and nonrectifiable attractivity of the zero solution of (1) depends on the integrability of function
(72)Φ(t)=|Imλ(t)|e-∫tt0‍Reλ(s)dsfort∈(0,t0],
we can conclude that the zero solutions of (7) is rectifiable attractive if Φ∈L1(0,t0) and nonrectifiable attractive if Φ∉L1(0,t0). Let us remark that in Lemma 6, it has been proved that
(73)|Imλ(t)|~t-2ast⟶0.(i) Assume thatmax{p,q}>1. For instance, we assume p>1. Since e(t)>0 on [0,t0], we have e(t)t-p+1→∞ as t→0. In particular, there exist t1∈(0,t0] and δ>0 such that
(74)e(t)t-p+1≥2+δfort∈(0,t1].
Since
(75)e-∫tt0‍Reλ(s)ds=e(-1/2)∫tt0‍(e(s)s-p+h(s)s-q)ds,
we have
(76)e-∫tt0‍Reλ(s)ds≤e(-1/2)∫tt0‍e(s)s-pds=e(-1/2)∫tt0‍e(s)s-p+1s-1ds≤e-(1+(δ/2))∫tt0‍s-1ds=ct1+(δ/2),
where the constant c=t0-1-(δ/2). Now from (72), (73), and (76), we observe that
(77)Φ(t)≤c1t-2t1+(δ/2)=c1t-1+(δ/2)∈L1(0,t0).
Thus, it is shown that Φ∈L1(0,t0).(ii) Assume that (11) is fulfilled. Since e(t)>2 on [0,t0], there exists δ>0 such that e(t)≥2+δ on [0,t0]. Since p=1 and h(t)>0 on (0,t0], from (75) we have
(78)e-∫tt0‍Reλ(s)ds≤e(-1/2)∫tt0‍e(s)s-1ds≤e-(1+(δ/2))∫tt1‍s-1ds=ct1+(δ/2),
where the constant c=t0-1-(δ/2). Now from (72), (73), and (78), we obtain Φ∈L1(0,t0). When (12) is fulfilled, we obtain Φ∈L1(0,t0) by the similar argument.Assume that (13) is fulfilled. Since e(t)+h(t)>2 on [0,t0], there exists δ>0 such that e(t)+h(t)≥2+δ on [0,t0]. From (75) we have
(79)e-∫tt0‍Reλ(s)ds=e(-1/2)∫tt0e(s)+h(s)‍s-1ds≤e-(1+(δ/2))∫tt0‍s-1ds=ct1+(δ/2),
where the constant c=t0-1-(δ/2). Thus we obtain Φ∈L1(0,t0) by the argument above.(iii) Let (14) be fulfilled. Since h∈C([0,t0]), there is a constant m0>0 such that h(t)≤m0 on [0,t0], and since q<1 there is a constant c0>0 such that
(80)e(-1/2)∫tt0‍h(s)s-qds≥e-(m0/2)∫tt0‍s-qds≥c0ast⟶0.Hence,(81)e-∫tt0‍Reλ(s)ds=e(-1/2)∫tt0‍(e(s)s-1+h(s)s-q)ds=e-∫tt0‍s-1dse(-1/2)∫tt0‍h(s)s-qds≥c0e-∫tt0‍s-1ds=c0tt0.
Now, by combining (72), (73), and (81), we observe
(82)∫tt0‍Φ(s)ds≥c1t0∫tt0‍s-2sds=c1t0lnt0t⟶∞ast→0.
Thus, Φ∉L1(0,t0). When (15) is fulfilled, we similarly obtain Φ∉L1(0,t0).Let (16) be fulfilled. Then, from (75) we have
(83)e-∫tt0Reλ(s)ds=e(-1/2)∫tt0(e(s)+h(s))s-1ds≥e-∫tt0s-1ds=tt0.
By combining (72), (73), and (83), we obtainΦ∉L1(0,t0). Thus, the proof is completed.



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*Source: 740980-2013-06-16.xml*