# Asymptotic Solutions ofnth Order Dynamic Equation and Oscillations

Authors: Gro Hovhannisyan
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

We establish a new asymptotic theorem for thenth order nonautonomous dynamic equation by its transformation to the almost diagonal system and applying Levinson's asymptotic theorem. Our transformation is given in the terms of unknown phase functions and is chosen in such a way that the entries of the perturbation matrix are the weighted characteristic functions. The characteristic function is defined in the terms of the phase functions and their choice is  exible. Further applying this asymptotic theorem we prove the new oscillation and nonoscillation theorems for the solutions of the nth order linear nonautonomous differential equation with complex-valued coefficients. We show that the existence of the oscillatory solutions is connected with the existence of the special pairs of phase functions.


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## Body

## 1. Introduction

Consider an ordinary nonautonomous differential equation of thenth order
(1)L(u(t))=∑j=0n‍aj(t)u(j)(t)=0,t>t0,an=1,
with complex-valued continuous variable coefficients aj(t),  j=0,…n-1.A solution of (1) is said to be oscillatory if it has an infinite sequence of zeros in (t0,∞) and nonoscillatory otherwise. Equation (1) is said to be nonoscillatory if all nontrivial solutions are nonoscillatory.Oscillation theorems for ordinary differential equation of thenth order in the case of real variable coefficients have been studied in many papers (see [1, 2] and references therein). To the best of the author’s knowledge the oscillations of the solutions of nonautonomous nth order equations with complex coefficients have not been studied yet (except [3]).LetCk(t0,∞) be the set of k times differentiable functions on (t0,∞) and L1(t0,∞) the set of Lebesgue absolutely integrable functions on (t0,∞).To consider the case of complex coefficients we are using asymptotic solutions of (1) in Euler form u(t)=e∫t0t‍f(s)ds.  Define the characteristic function of (1) depending on a phase function fj(t)∈Cn-1(t0,∞)(2)Chj(t)=e-∫t0t‍fj(s)dsL(e∫t0t‍fj(s)ds),j=1,2,…,n.By direct calculations forj=1,2,…,n we get
(3)Chj(t)=∑k=1n+1‍fkj(t)ak-1(t)=fn+1,j(t)+fnjan-1(t)+⋯+f2ja1(t)+f1ja0(t),
where
(4)f1j(t)=1,fk+1,j(t)=(ddt+fj(t))fkj(t),k=1,2,3,…,n.
For example,
(5)f2j(t)=fj(t),f3j(t)=fj2(t)+fj′(t),f4j(t)=fj3(t)+3fj′fj(t)+fj′′(t).
Define the auxiliary square matrix F(t) with the entries fkj(t) (see (4))
(6)F(t)=∥fkj(t)∥k,j=1n,
and denote by Fj(t),  j=1,…,n the (n,j) minors of the matrix F(t).


## 2. Main Theorems

The basic method of this paper is a new version of Levinson’s asymptotic theorem (see [4, 5]).Theorem 1.
Assume there exist complex-valued functionsfj(t)∈Cn-1(t0,∞),  j=1,2,…,n such that for all t>t0 expressions ℜ[fj(t)-fk(t)], 1≤k<j≤n do not change a sign; that is,
(7)ℜ[fj(t)-fk(t)]≤0orℜ[fj(t)-fk(t)]≥0,1≤k<j≤n,t>t0,(8)∫t0∞‍|Fj(t)Chk(t)F0(t)|dt<∞,j=1,2,…,n-1,k=1,2,…,n,(9)F0′(t)F0(t)+∑j=1n‍fj(t)+an-1(t)=0,t>t0,
where characteristic functions Chk(t),  k=1,2,…,n are defined in (2), (3), and
(10)F0(t)≡det(F(t)).
Then solutions of (1) may be represented in the form
(11)u(t)=∑j=1n‍φj(t)(1+ɛj(t))Cj,limt→∞ɛj(t)=0,j=1,2,…,n,
where
(12)φj(t)=e∫t0t‍fj(s)ds,j=1,2,…,n.Note that condition (9) is the well-known Abel’s identity that is satisfied if (12) are exact solutions of (1).Theorem1 means that the error functions ɛj(t)→0 as t→∞ if the weighted characteristic functions Fj(t)Chk(t)/F0(t) are absolutely integrable in (t0,∞).In Theorem1 the weighted characteristic functions are the entries of the perturbation matrix from Levinson’s Theorem (see Remark 18 in Section 3).Note that the error functionsɛj(t) in representation (11) may be estimated via the characteristic function (see, e.g., Theorem 2.2 in [6]).Theorem1 may be used also for the applications mentioned in [5], the stability theory, and the Dirac equation with complex coefficients (see, e.g., [6–8]).We will say that (1) has the asymptotic solutions e∫t0t‍fj(s)ds corresponding to the phase functions fj(t)∈Cn-1(t0,∞), j=1,2,…,n if (7)–(9) are satisfied.Theorem 2.
The asymptotic solution of (1) (corresponding to the phase function fk(t)) generates the oscillatory solution of (1) if there exists another asymptotic solution with the phase fj(t) such that
(13)ℜ[fk(t)-fj(t)]≡0,
and one of the following conditions is satisfied:
(14)ℑ[fk(t)+fj(t)]≡0,∫t0∞‍ℑ[fk(t)]dt=∞or∫t0∞‍ℑ[fk(t)]dt=-∞,(15)∫t0∞‍ℑ[fk(t)-fj(t)]dt=∞or∫t0∞‍ℑ[fk(t)-fj(t)]dt=-∞.The interesting question is to obtain the converse of Theorem2.Conditions (13) and (14) mean that the phase functions fk(t),fj(t) are complex congugate. For the equations with real coefficients the complex phase functions appear in complex conjugate pairs. But for the equations with complex coefficients this is not true.Since every solution of (1) with the constant complex coefficients (without repeated characteristic roots) is the linear combination of the exponents, in view of Theorem 2, the following questions arise.Question 1.
Describe the complex numbersz1,z2,…,zn such that
(16)ez1+ez2+⋯+ezn=0.Question 2.
For which complex constant coefficientsaj,  j=0,1,…,n a polynomial ∑j=0n‍ajzj,  n≥2 has at least one pair of complex conjugate zeros with nonzero imaginary parts?Question1 is complicated in the multidimensional case (see [9]), but by the substitution eyn-1=ezn-1+ezn one can reduce n dimensional equation to the two equations: (n-1)-dimensional ez1+ez2+⋯+ezn-2+eyn-1=0 and 3-dimensional eyn-1=ezn-1+ezn.For the two dimensional case there is a simple answer to Question1:
(17)ez1+ez2=0
if and only if
(18)(A)ℜ[z1-z1]=0,ℑ[z1-z2]=(1+2k)π,k=0,±1,±2,….One can answer Question2 for quadratic, cubic, and quartic equations with the complex coefficients since the simple formulas for solutions are available for these cases (see Remark 14, Theorem 15 below).Theorem2 shows that the oscillations of the solutions could be produced not only by the complex conjugate phase functions, which one can see from the following example.Example 3.
The equation(19)u′′′′(t)+3+4i2t2u′′(t)-3+4it3u′(t)+45+40i16t4u(t)=0
is oscillatory, but there is no complex conjugate pair among its phase functions with nonzero imaginary parts:
(20)f1(t)=1+2i2t,f2(t)=5-2i2t,f3(t)=12t,f4(t)=52t.
Note that uj(t)=e∫t0t‍fj(s)ds, j=1,2,3,4 are exact solutions of (19), u1(t)-u3(t) is an oscillatory solution of (19), and (13), (15) are satisfied.Example 4.
Ifr(t),  φ(t)∈[0,π) are a real-valued C2(t0,∞) functions such that
(21)(r-1/2(t)e-φ(t)/2)′′r-1/2(t)e-φ(t)/2∈L1(t0,∞),
then the equation
(22)u′′(t)+r2(t)e2iφ(t)u(t)=0
is oscillatory if and only if
(23)φ(t)≡0,∫t0∞‍r(s)ds=∞,t>t0.The following theorem is inspired by the asymptotic theorems for two-term differential equation from [10, 11]. We deduce it from Theorem 1 by choosing the phase function fk(t) as an approximate solution of the characteristic equation Chk(t)=0 (see [10]):
(24)fk(t)=bka(t)-(n-1)a′(t)2a(t),k=1,2,…,n.
Note that Theorem 1 may be used to obtain other oscillation theorems by taking different approximate solutions of the characteristic equation (see [10, 12]).Theorem 5.
Assume there exists a complex-valued functiona(t)∈Cn-1(t0,∞),  a(t)≠0, and complex numbers bk,  k=1,2,…,n such that
(25)b1+b2+⋯+bn=0(26)ℜ[(bk-bj)a(t)]donotchangeasignon(t0,∞),1≤k<j≤n,(27)∫t0∞‍|a(1-n)/2(t)e-bk∫t0t‍a(s)ds×L(a(1-n)/2(t)ebk∫t0t‍a(s)ds)|dt<∞,k=1,…,n,
the set {bka(t)}k=1n contains a set S={bka(t)}k=12r of r pairs of complex conjugate functions, and
(28)∫t0∞‍ℑ[bka(t)]dt=±∞,k=1,…,2m,m≤r,(29)|∫t0∞‍ℑ[bka(t)]dt|<∞,k=2m+1,m+2,…,2r.
Then (1) with an-1(t)≡0 has 2m oscillatory and (n-2m) nonoscillatory linearly independent solutions.By takinga(t)=1/t from Theorem 5 we get the following corollary.Corollary 6.
Assume that for some complex numbersb1,b2,…,bn, the set {bj}j=1n contains a set S={bj}k=12r of r pairs of complex conjugate numbers, and conditions (25), (26),
(30)ℑ[bj]≠0,j=1,…,2m,ℑ[bj]≡0,j=2m+1,…,2r,(31)a0(t)tn-1+∑j=1n‍aj(t)tn-1-j∏p=1j‍(bk+n+1-2p2)∈L1(t0,∞)
are true for all k=1,2,…,n. Then (1) with an-1(t)≡0 has 2m oscillatory and (n-2m) nonoscillatory linearly independent solutions.From Corollary6 follows the well-known result.Corollary 7 (see [1]).
Assume that conditions(32)∫t0∞‍tn-1-j|aj(t)|dt<∞,j=0,1,2,…,n-2
are satisfied. Then (1) with an-1(t)≡0 is nonoscillatory.For the fourth order equation we deduce the following result.Corollary 8.
Assume that for some real numberɛ>0 and for the fixed number j from the set {0,1,2}(33)t3-j|aj(t)+9j2-15j+18-2j+1ɛ232t4-j|∈L1(t0,∞),t3-kak∈L1(t0,∞),k=0,1,2,k≠j.
Then equation
(34)u′′′′(t)+a2(t)u′′(t)+a1(t)u′(t)+a0(t)u(t)=0
is nonoscillatory.Remark 9.
Ifj=0 conditions (33) turn to
(35)∫t0∞‍t3|a0(t)+9-ɛ216t4|dt<∞,∫t0∞‍|t3-kak(t)|dt<∞,k=1,2.Remark 10.
To compare condition (35) with the well-known real coefficient case, note that (34) with a1(t)=a2(t)≡0 and real valued a0(t) is nonoscillatory if (see [1])
(36)∫t∞‍s2(a0(s)+916s4)ds=∫t∞‍s2a0(s)ds+916t>0.By takingɛ=iξ, j=0 one can get another result.Corollary 11.
Assume that for some real numberξ conditions
(37)∫t0∞‍t3-k|ak(t)|dt<∞,∫t0∞‍t3|a0(t)+9+ξ216t4|dt<∞,k=1,2,
are satisfied. Then (34) has 2 oscillatory and 2 nonoscillatory (linearly independent) solutions.For two terms equation:(38)u(n)(t)+a0(t)u(t)=0,a0(t)=|a0(t)|eiφ,n≥2,φ=0orφ=π,
with real valued coefficient a0(t) by choosing
(39)a(t)=|a0|1/n,bj=eiφ/neiπ(2j+1)/n,j=1,…,n,
from Theorem 5 one can deduce the following theorem.Theorem 12.
Assumea(t)=|a0|1/n∈Cn-1(t0,∞), and for all k=1,…,n(40)∫t0∞‍|a(1-n)/2(t)e-bk∫t0t‍a(s)ds(dndtn+a0(t))×(a(1-n)/2(t)ebk∫t0t‍a(s)ds)|dt<∞,(41)∫t0∞‍|a0(t)|1/ndt=∞.
Then (38) has 2m=2[(n+φ/π)/2]-(2φ/π) oscillatory and n-2m nonoscillatory linearly independent solutions. Here [·] is the integral part of a real number.Remark 13.
Ifa0(t)=tγ, then conditions (40), (41) are simplified to
(42)γ>-2n-1.Remark 14.
For the quadratic equations with complex coefficientsz2+a1z+a0=0, it is easy to see that it has one pair of complex conjugate zeros with nonzero imaginary parts if and only if the coefficients a0,a1 are real, and a12-4a0<0.Theorem 15.
The cubic equation with complex constant coefficients(43)P(z)=z3+a2z2+a1z+a0=0
has a pair of complex conjugate solutions with nonzero imaginary parts if and only if one of the following conditions is true:
(44)a0¯=a0,a1¯=a1,a2¯=a2,D3(t)>0
or
(45)a2¯≠a2,D2(t)<0,P(z1)=0,
where
(46)D3=-108(a13-a229)3-108(a02-a1a26+a2327)2=a12a22-4a13-4a23a0-27a02+18a0a1a2,(47)D2=(a1-a1¯2(a2-a2¯))2-a0¯-a0a2¯-a2,
are the discriminants of (43) and P(z1)-P¯(z1)=0 correspondingly,
(48)z1=D2-a1-a1¯2(a2-a2¯).Example 16.
For the equation(49)z3-(5+4i)z2+(11+8i)z-15-20i=0
we have
(50)a0-a0¯=-40i,a1-a1¯=16i,a2-a2¯=-8i≠0,D2=(a1-a1¯2(a2-a2¯))2-a0¯-a0a2¯-a2=1-5<0,z1=1-5+1=2i+1,P(z1)=P(z1¯)=0
and condition (45) is satisfied.


## 3. Proofs

We are going to use Levinson’s asymptotic theorem as it appears in [5].Theorem 17 (see [5]).
LetΛ(t)=diag(λ1(t),…,λn(t)) be an n×n diagonal matrix-function which satisfies dichotomy condition.
For each pair of integersi and j in [1,n] (i≠j) exist constants K1,K2 such that for all x and t, t0≤t≤x<∞(51)∫tx‍ℜ[λi(s)-λj(s)]ds≤K1,or∫tx‍ℜ[λi(s)-λj(s)]ds≥K2.
Let the n×n matrix P(t) satisfy P(t)∈L1(t0,∞) or
(52)∫tx‍|P(t)|ds<∞,
by which one means that each entry in P(t) has an absolutely convergent infinite integral. Then the system
(53)Y′(t)=(Λ(t)+P(t))Y(t)
has a vector solution Y(t) with the asymptotic form
(54)Y(t)=(E+ɛ(t))e∫t0t‍Λ(s)dsC,limt→∞ɛ(t)=0,
where E is the identity matrix, ɛ(t) is the n×n error matrix-function, and C=(C1,…Cn)tr is a constant column-vector.Proof of Theorem1.
Rewrite (1) as a system
(55)y′(t)=A(t)y(t),(56)A(t)=(0100⋯00010⋯0⋯⋯⋯0⋯1-a0(t)-a1(t)-a2(t)-a3⋯-an-1(t)),y(t)=(u(t)u′(t)⋮u(n-1)).
By transformation
(57)y(t)=Φ(t)z(t),
where matrix-function Φ(t) is defined via phase functions fj(t):
(58)Φ(t)=F(t)e∫t0t‍f1(s)ds,
and the entries of the matrix F(t) are defined in (4), (6), we get
(59)z′(t)=Φ-1(t)(A(t)Φ(t)-Φ′(t))z(t),
or, in view of identity,
(60)A(t)Φ(t)-Φ′(t)=Φ(t)(Λ(t)+B(t)F0(t)),
we have
(61)z′(t)=(Λ(t)+B(t)F0(t))z(t),
where
(62)Λ(t)=(00000f2(t)-f1(t)0000f3(t)-f1(t)00⋯⋯00⋯0fn(t)-f1(t)),(63)B(t)=(-F1Ch1-F1Ch2⋯-F1ChnF2Ch1F2Ch2⋯F2Chn-F3Ch1-F3Ch2⋯-F3Chn⋯⋯⋯⋯(-1)nFnCh1(-1)nFnCh2⋯(-1)nFnChn).
Here and further we often suppress the time variable t for the simplicity.Remark 18.
Remarkable formulas (61)–(63) show that any nth order (1) may be transformed to the first order system z′(t)=A1(t)z(t), where the matrix A1(t) is a sum of diagonal matrix Λ(t) and perturbation matrix P(t); that is, A1(t)=Λ(t)+P(t) (Levinson form). Moreover, if the diagonal matrix is chosen in terms of phase functions as in (62), then the perturbation matrix P(t)=B(t)/F0(t) is the weighted characteristic function (i.e., the entries of matrix B are proportional to the characteristic functions Chj(t)). Our conjecture is that it is true not only for (1), but also for any first order system. For planar systems it was proved in [7].To prove basic formula (60) note that by using Laplace expansion by the minors of the determinants:
(64)F0=det(f11f12⋯f1nf21f22⋯f2n⋮⋮⋮⋮fn1fn2⋯fnn),0=det(f11f12⋯f1nf21f22⋯f2n⋮⋮⋮⋮fk1fk2⋯fkn),k=1,…,n-1,
we get identities:
(65)F1fn1-F2fn2+⋯+(-1)n-1Fnfnn=F0(t),F1fk1-F2fk2+⋯+(-1)n-1Fnfkn=0,k=1,…,n-1,
or (f1j≡1,f2j=fj)(66)F1-F2+F3+⋯+(-1)n-1Fn=0,f1F1-f2F2+⋯+(-1)n-1fnFn=0,(f12+f1′)F1-(f22+f2′)F2+(f32+f3′)F3+(-1)n-1(fn2+fn′)Fn=0.
By direct calculations we get
(67)(AΦ-Φ′)11=e∫tf1ds(f21-f1)=0,(Φ(Λ+BF0))11=(-F1+F2-F3+⋯+(-1)nFn)Ch1F0=0,(AΦ-Φ′)21=(f2-f1)e∫tf1ds(Φ(Λ+BF0))21=(f2-f1)e∫tf1ds+e∫tf1ds(-F1+F2+⋯+(-1)nFn)Ch2F0=(f2-f1)e∫tf1ds,(AΦ-Φ′)n1=(fn1′-a0-a1f1-a2f31-⋯-an-1fn1-fn1-fn1′f1)e∫tf1ds=-e∫tf1dsCh1,(Φ(Λ+BF0))n1=e∫tf1ds(-fn1F1+fn1F2+⋯+(-1)nfnnFn)Ch1F0=-e∫tf1dsCh1,
and the same way we get
(68)(AΦ-Φ′)jk=(Φ(Λ+BF0))jk,j,k=1,2,…,n.To apply Theorem17 to system (61) note that from (7) follows dichotomy condition (51) of Theorem 17. Condition (52) of Theorem 17 turns to B(t)/F0(t)∈L1(t0,∞):
(69)FjChk(t)F0(t)∈L1(t0,∞),k,j=1,2,3,…,n,
or (8) in view of  -F1+F2-⋯+(-1)nFn=0.From Theorem17 applied to system (61) we get
(70)z(t)=z0(t)(E+ɛ(t))C,limt→∞ɛ(t)=0,z0(t)=e∫t0t‍Λ(s)ds=(10000e∫t0t‍(f2-f1)ds0000e∫t0t‍(f3-f1)ds0000e∫t0t‍(f4-f1)ds).
From (57):
(71)y(t)=Φ(t)z(t)=Φ(t)z0(t)(E+ɛ(t))C,
we get representation (11). If functions {fj(t)}j=1n are exact solutions of Ch(fj)=0, j=1,2,…n, then F(t)e∫t0t‍f1dse∫t0t‍Λds is the fundamental solution of (55). Abel’s identity
(72)C=det(F(t))e∫t0t‍f1dse∫t0t‍Λdse-∫t0t‍Tr(A(s))ds=F0(t)e∫t0t‍(f1+f2+⋯+fn+an-1)(s)ds
may be written in the form (9). We always choose (approximate) phase functions fj(t), j=1,2,…,n such that (9) is satisfied.Proof of Theorem2.
From the assumptions (13), (14) of Theorem 2 one can generate an asymptotic solution as the difference of two asymptotic solutions:
(73)u0(t)=e∫t0t‍fk(s)ds-e∫t0t‍fj(s)ds=2ie∫t0t‍ℜ[fk(s)]dssin(∫t0t‍ℑ[fk(s)]ds),
which is oscillatory. Indeed if ∫t0t‍ℑ[fk(s)]ds=∞, or ∫t0t‍ℑ[fk(s)]ds=-∞, then the infinite sequence of zeros tp of u0(t) is given by
(74)∫t0tp‍ℑ[fk(s)]ds=2pπ,p=1,2,3,…orp=-1,-2,-3,….
From the assumptions (13), (15) for the solution u0(t) one can generate another infinite sequence of zeros tm that is given by
(75)∫t0tm‍ℑ[fk(s)-fj(s)]ds=2mπ,m=1,2,3,…,orm=-1,-2,-3,….
Indeed we have the following:
(76)u0(tm)=e∫t0tm‍fj(s)ds(e∫t0t‍ℜ[fk(s)-fj(s)]dsei∫t0t‍ℑ[fk(s)-fj(s)]ds-1)=0.
Theorem 2 is followed from representation (11).Proof of Example4.
Asymptotic solutions of (22) we choose in the form
(77)uj(t)=e∫t0t‍fj(s)ds,fj=xj-xj′(t)2xj(t),x1(t)=ir(t)eiφ(t),x2(t)=-x1(t)
or
(78)uj(t)=Cxj-1/2(t)e∫t0t‍xj(s)ds,j=1,2.
From (4) we have the following:
(79)F(t)=(11f1(t)f2(t)),F0(t)=det(F)=f2(t)-f1(t)=-2x1(t),F1=F2=1,Chj=fj′(t)+f2(t)-x12(t)=(x1-1/2)′′(t)x11/2(t).
So if (8) is satisfied, that is,
(80)(x1-1/2)′′(t)x11/2(t)∈L1(t0,∞),
then solutions of u′′(t)+r2(t)e2iφ(t)u(t) may be written in the form following:
(81)u(t)=C1(1+ɛ1(t))u1(t)+C2(1+ɛ2(t))u2(t),limt→∞ɛj(t)=0,j=1,2.
If u(t) has a zero for some complex numbers C1,2,|C1|+|C2|≠0, then
(82)x1-1/2(t)(C1e∫t0t‍x1(s)ds+C2e∫t0t‍x2(s)ds)=0,
or
(83)C1eiy(t)+C2e-iy(t)=0,y(t)=∫t0t‍r(s)[cos(φ(s))+isin(φ(s))]ds,
or
(84)-C2C1=e2iy(t).
By taking the derivative by t of both sides we get
(85)0=2iy′(t)e2iy(t)
and since
(86)y′(t)=r(t)[cos(φ(t))+isin(φ(t))]≠0,
we get
(87)0=e2iy(t)=e2i∫t0t‍r(s)cos(φ(s))dse-2∫t0t‍rsin(φ(s))ds,
or
(88)0=e-2∫t0t‍rsin(φ(s))ds.
By taking the derivative by t of the both sides we get
(89)0=-2r(t)sin(φ(t))e-2∫t0t‍rsin(φ(s))ds
or
(90)φ(t)≡0.
So condition (8) turns to (21).Proof of Theorem5.
We deduce Theorem5 from Theorem 1 by choosing fj(t) as in (24), (25). By calculations we get
(91)F0(t)=∏1≤k<m≤n‍(bk-bm)a1+2+⋯+n-1=∏1≤k<m≤n‍(bk-bm)a(n-1)n/2.
Note that (9) is satisfied in view of b1+⋯+bn=0:
(92)f1+f2+⋯+fn+F0′F0=(b1+⋯+bn-n(n-1)a′2a2)a(t)+n(n-1)a′2a=0.
Further we have the following:
(93)Fj=∏k<m<n;k,m≠j‍(bk-bm)a1+2+⋯+n-2=∏k<m<n;k,m≠j‍(bk-bm)a(n-1)(n-2)/2,
and condition (8), in view of Definition (2), turns to (27):
(94)Fj(t)Chk(t)F0(t)=CjChk(t)a1-n(t)=Cja1-na(n-1)/2e-bk∫t0t‍a(s)ds×L(a(1-n)/2ebk∫t0t‍a(s)ds),Cj=∏k<m<n;k,m≠j(bk-bm)∏1≤k<m≤n(bk-bm).
Further (7) turns to (26). So under conditions of Theorem 5 conditions (7)–(9) are satisfied and Theorem 1 is applicable. Further from (28) we get (13), (14) is satisfied for k=1,2,…,2m, and Theorem 5 is followed from Theorem 1 and Theorem 2.Proof of Corollary6.
Corollary6 is followed from Theorem 5 by taking a(t)=1/t. From (27) we have
(95)t(n-1)/2-bk∑j=0naj(t)(tbk+(n-1)/2)(j)∈L1(t0,∞),k=1,…,n,
or, since (tβ)(j)=β(β-1)⋯(β-j+1)tβ-j,  j=1,2,…,n, we get
(96)∑j=1n‍aj(t)tn-1-j(bk+n-12)×(bk+n-32)⋯(bk+n+1-2j2)+a0(t)tn-1∈L1(t0,∞),
or condition (27) turns to (31). Further conditions (28),(29) turn to (30).Proof of Corollary7.
Corollary7 is followed from Corollary 6 by taking bk=k-1-((n-1)/2),  k=1,2,…,n, m=0.Proof of Corollary8.
We deduce Corollary8 from Corollary 6.
In the casen=4,  a3≡0 condition (29) turns to
(97)t-1(bk+32)(bk+12)(bk-12)(bk-32)+ta2(bk+32)(bk+12)+t2a1(bk+32)+t3a0∈L1(t0,∞),k=1,2,3,4.
Considering the casej=0 of Corollary 8 assume t3-kak∈L1(t0,∞),  k=1,2. Then condition (97) turns to
(98)t3(a0+(bk+3/2)(bk+1/2)(bk-1/2)(bk-3/2)t4)∈L1(t0,∞).
For any real ɛ by choosing
(99)b1=125-25-ɛ2,b2=-b1,b3=125+25-ɛ2,b4=-b3,
we get (bk2-9/4)(bk2-1/4)=(9-ɛ2)/16 and (97) turns to (33) with j=0.
In the casej=1, assuming t3-kak∈L1(t0,∞), k=0,2 (97) turns to
(100)t2(bk+3/2)(a1+(bk2-1/4)(bk-3/2)t3)∈L1(t0,∞).
By choosing b4=-3/2, and b1, b2, and b3 as real solutions of
(101)(bk2-14)(bk-32)=3-ɛ28,b1+b2+b3=32
with the negative discriminant (see (46)), we get conditions (33) with j=1:
(102)t2(a1+3-ɛ28t3),t3-kak∈L1(t0,∞),k=0,2.
In the case j=2, assume t3-kak∈L1(t0,∞),k=0,1. Condition (97) turns to
(103)t(bk+3/2)(bk+1/2)(a2+(bk-3/2)(bk-1/2)t2)∈L1(t0,∞).
By choosingb3=-3/2, b4=-1/2, and b1,2 as real solutions of
(104)(bk-32)(bk-12)=3-ɛ24,b1+b2+b3+b4=0,
or
(105)b1=1-1-ɛ24,b2=1+1-ɛ24,
we get nonoscillation conditions (33) with j=2:
(106)t(a2+3-ɛ28t2),t3-kak∈L1(t0,∞),k=0,1.Proof of Corollary11.
Proof of Corollary11 is similar to proof of Corollary 8 in case j=0 with the choice b1=(1/2)5-25+ξ2, b2=-b1, b3=(1/2)5+25+ξ2, and b4=-b3. Note since Im[b1]>0,  Im[b2]<0,  and Im[b3]=Im[b4]=0 we get 2m=2.Proof of Theorem12.
We deduce Theorem12 from Theorem 5.
From the choice (39) we get
(107)bj=eiφ/neiπ(2j+1)/n,(-a0(t))1/n=bja(t),j=1,…,n.
It is easy to see that condition (25) of Theorem 5 is satisfied. Condition (27) turns to (40). To check condition (26) consider
(108)ℜ[bk-bj]=cos(φ+πn+2πkn)-cos(φ+πn+2πjn)=2sin(φ+π(1+j+k)n)sin(π(j-k)n).
Assuming 1≤k<j≤n, we have
(109)sin(π(j-k)n)>0,
and since φ≡0 or φ≡π, the sign of ℜ[bk-bj] does not depend on t and (26) is true. Let m be the number of indices j∈{1,…,n} for which
(110)ℑ[bja(t)]=|a0(t)|1/nsin(φ+π(2j+1)n)≠0.
Then (28) is true in view of (41).
If for somej∈{1,…,n}(111)sin(φ+π(2j+1)n)=0,
then a nonoscillatory solution exists. This condition
(112)α+2j+1n∈Z,α=φπ,
that is, α=0,1 since φ=0 or φ=π.
Consider the case(113)φ=0,α=0.
If n is odd, then (2j+1)/n is integer if and only if j=(n-1)/2 which means that the solution with
(114)j=n-12,n≥2
is nonoscillatory, so other 2m=n-1=2[n/2] solutions are oscillatory.
Ifφ=0 and n is even, then and all 2m=n=2[n/2] solutions are oscillatory.
Further in the case(115)φ=π,α=1,
and n is even we have the solutions with
(116)j=n2-1,j=n-1
are nonoscillatory, and the other n-2 solutions are oscillatory.
In the caseφ=π and n is odd we have only one nonoscillatory solution with j=n-1 and 2m=n-1.
So(117)2m={2[n2],φ≡0,n-2,φ≡π,niseven,n-1,φ≡π,nisodd,
or
(118)2m=2[n+φ/π2]-2φπ.Proof of Remark14.
Ifz2+a1z+a0=0 has 1 pair of complex conjugate zeros z1, z2=z1¯, then a0, a1 are real numbers since a0=z1z2=|z1|2≥0 and a1=-z1-z2=-z1-z1¯, and quadratic equation z2+a1z+a0=0 has two complex conjugate roots if a12-4a0<0. Assuming that a0, a1 are real numbers and a12-4a0<0 from quadratic formula the equation z2+a1z+a0=0 has two complex conjugate roots.Proof of Theorem15.
Assume there exists a pair of complex conjugate solutionsz1, z1¯ of P(z)=z3+a2z2+a1z+a0=0; that is,
(119)P(z1)=0,P(z1¯)=0,
then denoting
(120)P¯(z)=z3+a2¯z2+a1¯z+a0¯,
we have P¯(z1)=0 as well, and
(121)P(z1)-P¯(z1)=(a2-a2¯)z12+(a1-a1¯)z1+a0-a0¯=0.
Consider 3 cases.
First case:a2=a2¯, a1=a1¯, a0=a0¯, D3>0.
In this caseP(z1)-P¯(z1)=0 is satisfied. The coefficients of P(z)=0 are real and (43) has two complex conjugate solutions since discriminant D3 of P(z)=0 is positive. We drop the case a2=a2¯, a1=a1¯, a0=a0¯, D3<0 since in that case all solutions are real.
Second case:a2≠a2¯.
In this case by solving quadratic equationP(z1)-P¯(z1)=0 we get
(122)z1=-a1-a1¯2(a2-a2¯)+D2,D2=(a1-a1¯2(a2-a2¯))2-a0¯-a0a2¯-a2,
and P(z1)=P¯(z1)=0 implies another solution z2 can be found by division:
(123)P(z)=(z-z1)Q(z),Q(z)=z2+(z1+a2)z+a1+a2z1+z12,Q(z2)=0.
Note that if discriminant D2 of the quadratic equation P(z1)-P¯(z1)=0 is negative, then by solving this equation we get two complex conjugate roots with nonzero imaginary parts:
(124)z2=z1¯=-a1-a1¯2(a2-a2¯)-D2.
If(125)D2≥0,
then z1 is real, and to find another (complex) solution z3 one should solve
(126)Q(z3)=z32+(z1+a2)z3+a1+a2z1+z12=0.
Since z3¯ must be a solution as well, we have
(127)Q(z3)-Q¯(z3)=(a2-a2¯)(z3+z1)+a1-a1¯=0,
and z3 is a real number given by
(128)z3=-z1-a1-a1¯a2-a2¯.
Further since z1, z3 are solutions by division
(129)Q(z)=(z-z3)R(z),R(z)=z+z1+z3+a2,
and one can find the third solution z4:
(130)z4=-z1-z3-a2.
This case does not work since we have two real solutions z1, z3 and one complex solution z4 of P(z)=0.
Third case:a2=a2¯, a1≠a1¯.
In this case assumingz5, z5¯ are complex conjugate solutions of P(z)=0 from P(z5)-P¯(z5)=(a1-a1¯)z5+a0-a0¯=0 we get z5 which is a real number given by
(131)z5=-a0-a0¯a1-a1¯
and P(z5)=0.
Ifz6 is another solution of P(z)=0, by division we get
(132)P(z)=(z-z5)Q(z),Q(z)=z2+(z5+a2)z+a1+a2z5+z52,Q(z6)=0.
Since z6=z5¯ is also solution of Q(z)=0, we get
(133)Q(z5)-Q¯(z5)=a1-a1¯=0,
and we come back to the case 1.
To prove that (45) is sufficient condition in view of P(z1¯)=P(z1), it is enough to show that
(134)z1¯=-a1-a1¯2(a2-a2¯)-(a1-a1¯2(a2-a2¯))2-a0¯-a0a2¯-a2,
but it is true since ((a1-a1¯)/2(a2-a2¯))2-(a0¯-a0)/(a2¯-a2)<0.
It is easy to see that (44) is a sufficient condition as well.



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*Source: 946453-2013-09-15.xml*