On Rectifiable Oscillation of Euler Type Second Order Linear Differential Equations

We study the oscillatory behavior of solutions of the second order linear differential equation of Euler type: (E) y + λx −α y = 0, x ∈ (0, 1], where λ > 0 and α > 2. Theorem (a) For 2 ≤ α < 4, all solution curves of (E) have finite arc length; (b) For α ≥ 4, all solution curves of (E) have infinite arc length. This answers an open problem posed by M. Pasic [8] §1. In a recent paper [8], M. Pasic introduced the concept of " rectifiable oscillation " in the study of oscillatory behavior of solutions of the second order linear differential equation of Euler type in the finite interval (0, 1]: (1) y + λx −α y = 0, x ∈ (0, 1]. In case of the Euler's equations, i.e. α = 2, it is well known that equation (1) is oscillatory when λ > 1 4 and nonoscillatory if λ ≤ 1 4. For α > 2, it follows from the Sturm's Comparison Theorem that all solutions of (1) are oscillatory as x → 0. For a real function y(x) defined on the closed interval I = [0, 1], we denote its graph by G(y) = t, y(t) : 0 ≤ t ≤ 1 as a subset in R 2. G(y) is said to be a rectifiable curve in R 2 if its arc length L G (y) is finite where L G (y) is defined by L G (y) = sup m i=1 t i , y(t i) − t i−1 , y(t i−1) 2 ,

In case of the Euler's equations, i.e. α = 2, it is well known that equation (1)  Theorem that all solutions of (1) are oscillatory as x → 0.
For a real function y(x) defined on the closed interval I = [0, 1], we denote its graph by G(y) = t, y(t) : 0 ≤ t ≤ 1 as a subset in R 2 .G(y) is said to be a rectifiable curve in R 2 if its arc length L G (y) is finite where L G (y) is defined by EJQTDE, 2007 No. 20, p. 1 where supremum is taken over all partitions: 0 = t 0 < t 1 < • • • < t m = 1 of the unit interval [0, 1] (See Apostol [1], p. 175).Here • 2 denotes the Euclidean norm in R 2 .
An oscillatory function y(x) on I = (0, 1] is said to be rectifiable (resp.unrectifiable) if its graph G(y) is rectifiable (resp.unrectifiable).In other words, it is rectifiable if it has finite arc length on I and unrectifiable otherwise.Equation ( 1) is said to be rectifiable (resp. unrectifiable) oscillatory on I if all its nontrivial solutions are rectifiable (resp.unrectifiable).
In the familiar case of the Euler's equation, i.e. equation ( 1) when α = 2 and λ > 1 4 , we can write down the explicit form of its general solutions (2) where Consider a typical section of the solution curve y(x) of (1) between its two zeros a k+1 and a k where a k+1 < a k and a k → 0 as k → ∞.It is clear from the geometry that its arc length is bounded above by 2|y(s k )| + a k − a k+1 where y (s k ) = 0 and a k+1 < s k < a k .In the special case (2) of the Euler equation, we can simply estimate sin(ρ log x) as the case of cos(ρ log x) is similar.For x ∈ (0, 1], log x is negative, so zeros of sin(ρ log x) are given by a k = exp(−kπ/ρ) and maxima of sin(ρ log x) occurs at s k = exp(−kπ/2ρ).The curve 1) can also be solved in explicit form, namely, solutions are of the form The typical section of the solution curve y(x) between two zeros a k+1 and a k with its extrema at s k , a k+1 < s k < a k , has arc length exceeding 2|y(s k )|.Let us again consider only the case  This is interesting because the solution curves given in (3) exhibits a phenomenon whereby a one dimensional curve can have infinite length within a bounded region in R 2 , like the Koch's snowflake in fractal geometry, see Devancy [5] p. 182.
Pasic [8] posed as an open problem whether the additional assumptions of boundary layer conditions (I) and (II) can be removed for other values of α, α = 2 and 4. The purpose of this paper is to show that both of these boundary layer conditions are superfluorous in the case of equation (1).We can therefore improve Theorem P to read Theorem 1.
(a) Equation ( 1) is rectifiable oscillatory for 2 < α < 4, and (b) Equation ( 1) is unrectifiable oscillatory for α ≥ 4. §2.In this section, we shall give the proof of our main theorem.In view of Theorem P, it suffices to prove that solutions of (1) satisfy either one of the boundary layer conditions (I) and (II).This can be accomplished by the use of Louiville transformation and an application EJQTDE, 2007 No. 20, p. 3 of an asymptotic integration theorem in the elliptic case.
EJQTDE, 2007 No. 20, p. 4 We can now apply Theorem P to complete the proof of our main Theorem.On the other hand, by using the transformed equation ( 4), we can also give a much simpler proof somewhat different from that gives by Pasic [8].
We first consider the unperturbed equation ( 8) which has its general solution given by Let {x n } be the sequence of consecutive zeros of u(x) = sin √ λ/σx σ , i.e. (12) , so by Sturm Comparison Theorem between two consecutive zeros a k 0 +1 and a k 0 of y(x), {a k } denotes the decreasing sequence of consecutive zeros of y(x) there exists at least one zero Repeating this procedure to all pairs of consecutive zeros a a 0 +j and a k 0 +j−1 , we obtain another zero x n 0 +n j −1 , n j ≥ j such that a k 0 +j < x n 0 +n j −1 < a k 0 +j−1 .
We note that for the segment of the solution curve Γ k = x, y(x) : where y(x) attains its extrema s k between a k+1 and a k , i.e. y (s k ) = 0. Piecing together the segments Γ k , we note that the arc length of the Graph of solution curve satisfies is rectifiable or unrectifiable depending on whether the series Writing k = k 0 + j, we obtain the estimate ( 16) Now consider α/4σ as a function of α which for α > 2 is decreasing in α and equals to 1 at α = 4. Combining ( 16)and (17), we obtain for < ∞, since α/4σ > 1 for 2 < α < 4.This proves (a).
Returning to the case when α ≥ 4, we have q(x) ≥ λx −α/2 so we can apply Sturm's Comparison theorem to equations ( 4) and (11) and conclude that between two zeros (12) of solutions of (11), x k 0 +1 < x k 0 , there exists at least one zero a i 0 , i.e. x k 0 +1 < a i 0 < x k 0 .
Repeating this process to all pairs of consecutive zeros, x k 0 +k < x k 0 +k−1 , we obtain a i 0 +i k satisfying x k 0 +k < a i 0 +i k < x k 0 +k−1 where i k ≥ k.Hence by (12) Let w(x) be a linearly independent solution of y(x) chosen such that the Wronskian W (y, w)(x) = y(x)w (x) − w(x)y (x) is a constant, say W (y, w) ≡ 1. Evaluating W (y, w)(x) at x = s k where y (s k ) = 0, we obtain by boundary layer condition (II) when applied to w(x) the following estimate : = ∞, since α/4σ ≤ 1 for α ≥ 4. The divergence of the series in (20) shows that solution g(x) is unrectifiable.This completes the proof of (b).§3.In this section, we show that our Theorem can be further extended to give a somewhat more general result for the harmonic oscillator equation: where f (x) > 0 and f (x) ∼ λx −α , λ > 0, α > 2, as x → 0, i.e. lim x→∞ then conclusion of the Theorem remains valid.
So if α ≥ 4 we have α/4σ ≤ 1.Thus the infinite series appeared as the last term in (29) diverges, so does x → 0 and f (x) satisfies (22), thus (a) for α < 4, all solution of (21) are rectifiable oscillatory; and (b) for α ≥ 4, all solutions of (21) are unrectifiable oscillatory.§4.In this last section, we give another example of unrectifiable oscillation and several remarks concerning results discussed in this paper.
Example.Consider a special case of equation ( 21) where the coefficient f (x) is highly singular at x = 0, but is not of Euler type, i.e. f (x) is not asymptotic to a negative power of x.Furthermore, f (x) = x −4 exp (2/x) satisfies (i) Once again let w(x) be the solution of (30) linearly independent of y(x) such that the Wronkian W (y, w)(x) = (yw − wy )(x) ≡ 1. Denote {x n } the sequence of consecutive zeros of y(x), x n → 0 as n → ∞ and {s n } the corresponding consecutive extrema {s n }, i.e.
y (s n ) = 0, such that x n+1 < s n < x n (the uniqueness of s n between the two zeros x n+1 and x n is guaranteed by the concavity of y(x), i.e. y (x) ≤ 0).
Using the asymptotic formula (23), (24) of any solution y(x) and w(x) of (30), we can determine its extrema s n by the formula (31) e We can now estimate |y(s n )| by using (31) to obtain EJQTDE, 2007 No. 20, p. 9 since log x ≤ √ x for large value of x and e + 1 2 π ≤ 2π.
Summing up the terms |y(s n )| in (32), we find which proves that y(x) is unrectifiable oscillatory.Suppose that f (x) ∼ x −α , α > 2, as x → 0 and f (x) is sufficiently smooth such that its second derivative satisfies f (x) ∼ x −α−2 .Then it is easy to see that lim Remark 2. The proof can be modified if we assume instead of f (x) ∼ λx −α , λ > 0, α > 2, as x → 0 the inequality.
where m 1 , m 2 , and b are positive constants and 0 < b < 1. Remark 4. Oscillation of the harmonic oscillator (21) was traditionally discussed for the semi-infinite interval where the independent variable x relates to time and the dependent variable y relates to the travelling waves.However for studies in some areas such as nuclear physics, the independent variable x relates to radial distance from the centre of the nuclei and the dependent variable y relates to nuclear charge of the atom(s).How does one interpret the physical meaning of unrectifiable oscillation in physical problems is certainly a most interesting question.
Remark 5.The characteristic exhibited by solution y(x) in ( 3) is interesting because it arises from a simple linear differential equations whilst its solution curve over a finite interval has infinite arclength.Such curves are also known as fractals which normally associated with chaos in nonlinear dynamical systems.See e.g., Le Mehaute [7].Addison [2] for further discussions on fractal geometry and chaos.
so the length of of the solution curve x cos( √ λ/x) is infinite hence the oscillatory solution given in (3) is unrectifiable.
so condition (22) of Wintner's asymptotic formula is satisfied.Hence all solutions of (30) satisfy the boundary layer conditions (I) and (II) introduced in Pasic[8].