A NECESSARY AND SUFFICIENT CONDITION FOR THE GLOBAL ASYMPTOTIC STABILITY OF DAMPED HALF-LINEAR OSCILLATORS

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Acta Math. Hungar., 138 (1-2 (213, 156 172. DOI: 1.17/s1474-12-259-7 First published online September 5, 212 A NECESSARY AND SUFFICIEN CONDIION FOR HE GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS S. HAA 1 and J. SUGIE 2 1 Department of Mathematics and Computer Science, Shimane University, Matsue 69-854, Japan e-mail: hata@math.shimane-u.ac.jp 2 Department of Mathematics and Computer Science, Shimane University, Matsue 69-854, Japan e-mail: jsugie@riko.shimane-u.ac.jp Abstract. his paper is concerned with the global asymptotic stability of the equilibrium of the half-linear differential equation with a damped term, ( ϕp (x + h(tϕp (x + (ϕ p (x =. A necessary and sufficient growth condition on h(t is obtained by using the generalized Prüfer transformation and the generalized Riccati transformation. Equivalent planar systems to the above-mentioned equation are also considered in the proof of our main result. 1. Introduction We consider the second-order differential equation (HL (ϕ p (x + h(tϕ p (x + (ϕ p (x =, where the prime denotes d/dt, the function ϕ p (z is defined by ϕ p (z = z p 2 z, z R Corresponding author. Supported in part by Grant-in-Aid for Scientific Research, No. 225419 from the Japan Society for the Promotion of Science. Keywords and phrases: Global asymptotic stability, half-linear differential equations, damped oscillator, growth condition, Prüfer transformation, Riccati transformation. 21 Mathematics Subject Classification: Primary 34D23, 34D45; Secondary 34C15, 37C7.

2 S. HAA and J. SUGIE with p > 1, and the damping coefficient h(t is continuous and nonnegative for t. he only equilibrium of (HL is the origin (x, x = (,. Equation (HL includes the damped linear oscillator (L x + h(tx + x = as a special case in which p = 2. By the demand from a theoretical aspect and a practical aspect, there are a lot of reports concerning equation (L or its general type. For example, see [1, 3, 6, 11, 12, 15, 16, 18, 2, 22]. As is well known, the solution space of (L is additive and homogeneous; namely, (i the sum of two solutions is a solution and (ii the product of a solution and any constant is also a solution. Since the solution space of (HL has only the characteristic (ii mentioned above, equation (HL is often called half-linear. In particular, we will call equation (HL the damped half-linear oscillator because it is a generalization of the damped linear oscillator (L. Let x(t = (x(t, x (t and x R 2, and let be any suitable norm. We denote the solution of (HL through (t, x by x(t; t, x. he global existence and uniqueness of solutions of (HL are guaranteed for the initial value problem. For details, see Došlý [4, p. 17] or Došlý and Řehák [5, pp. 8 1]. he equilibrium of (HL is said to be stable if, for any ε > and any t, there exists a δ(ε, t > such that x < δ implies x(t; t, x < ε for all t t. he equilibrium is uniformly stable if it is stable and δ can be chosen independent of t. he equilibrium is said to be attractive if, for any t, there exists a δ (t > such that x < δ implies x(t; t, x as t. he equilibrium is uniformly attractive if δ in the definition of attractivity can be chosen independent of t, and for every η > there is a (η > such that t and x < δ imply x(t; t, x < η for t t + (η. he equilibrium is said to be globally attractive if, for any t, any η > and any x R 2, there is a (t, η, x > such that x(t; t, x < η for all t t + (t, η, x. he equilibrium is said to be uniformly globally attractive if, for any ρ > and any η >, there is a (ρ, η > such that t and x < ρ imply x(t; t, x < η for all t t + (ρ, η. he equilibrium is globally asymptotically stable if it is stable and globally attractive. he equilibrium is uniformly globally asymptotically stable if it is uniformly stable and is uniformly globally attractive. With respect to the various definitions of stability and boundedness, the reader may refer to the books [2, 8, 9, 17, 21, 27] for example. he purpose of this paper is to establish a criterion for judging whether the equilibrium of (HL is globally asymptotically stable or not. Very recently, Onitsuka and Sugie [19, heorem 3.1] have considered a system of differential equations of the form x = a(tx + b(tϕ p (y, y = (c(tϕ p (x (d(ty, where p is the conjugate number of p; namely, 1 p + 1 p = 1,

GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS 3 and presented some sufficient conditions for the uniform global asymptotic stability. As a new variable, let y be ϕ p (x. Note that ϕ p is the inverse function of ϕ p. hen, equation (HL becomes the above system with a(t =, b(t = c(t = 1 and d(t = h(t/(. By means of their result, we see that if h(t is bounded for t and (1 lim inf t +γ t h(sds > holds for every γ >, then the equilibrium of (HL is uniformly globally asymptotically stable. Condition (1 is a restriction that is considerably stronger than H(t def = h(sds as t. If h(t satisfies condition (1, then it is said to be integrally positive. For example, the function sin 2 t is integrally positive. About the integral positivity, see [1, 23 26]. Needless to say, the concept of the global asymptotic stability is stronger than that of the uniform global asymptotic stability. herefore, to seek a good criterion for judging whether the equilibrium of (HL is globally asymptotically stable, we have to discuss under the assumption that is weaker than condition (1. So far as the special case in which p = 2 is concerned, Hatvani and otik [14, heorem 3.1] have already proved that if there exists a γ with < γ < π such that (2 lim inf t +γ t h(sds >, then the equilibrium of (L is asymptotically stable if and only if (3 eh(s ds dt =. e H(t Besides, they pointed out that < γ < π was the best requirement. Condition (2 is weaker than condition (1. For example, let h(t = max {, a + sin t }. hen, condition (1 never holds unless a 1. On the other hand, condition (2 is satisfied as long as a >. A strong point of condition (2 is that the set {t : h(t = } is permitted to be the union of infinitely many disjoint intervals whose length are less than π. he criterion (3 is the so-called growth condition on h(t. It has first been given by Smith [22, heorems 1 and 2]. He showed that if there exists an h > such that h(t h for t, then condition (3 is necessary and sufficient for the equilibrium of (L to be asymptotically stable. he above-mentioned result of Hatvani and otik is its natural generalization. It is known that (i if h(t is bounded for t or h(t = t, then condition (3 is satisfied;

4 S. HAA and J. SUGIE (ii if h(t = t 2, then condition (3 is not satisfied. (for details, see [12]. In this paper, we intend to extend their result to be able to apply to the damped half-linear oscillator (HL. o this end, we define ω p by sup{(tan p θ p 2 (π p 2θ: < θ < π p /2} if p 2, ω p = sup{(tan p (π p /2 θ p 2 (π p 2θ: < θ < π p /2} if 1 < p < 2 (the constant π p and the function tan p θ are explained in Section 2. hen, we can state our main theorem as follows: heorem 1. Suppose that there exists a γ with < γ < ω p such that (4 lim inf t +γ t h(sds >. hen the equilibrium of (HL is globally asymptotically stable if and only if ( (5 ϕ eh(s ds p dt =. e H(t Remark 1. In the special case in which p = 2, it is clear that π p = ω p = π and ϕ p (z = ϕ 2 (z = z. Hence, assumption (4 becomes assumption (2 and our criterion (5 coincides with the well-known criterion (3. he function tan p θ is continuous and increasing for θ ( π/2, π/2. Hence, there exists the function arctan p that is defined as the inverse function of tan p θ. It turns out that ω p is more than at least π p 2 arctan p 1 for any p > 1. In the proof of Hatvani and otik s result, it is important that every solution x(t of (L is either oscillatory or monotone for t sufficiently large. his important property is easily led from only the form of (L. Although equation (HL has the same property as equation (L, we have to use the so-called Riccati technique to confirm this property (see Lemma 2. Riccati s technique is required also in order to show that condition (5 is necessary for the equilibrium of (HL to be globally asymptotically stable. It is comparatively easy to transform the damped linear oscillator (L into its polar coordinates system and to examine the asymptotic behavior of solutions of (L. In the damped half-linear oscillator (HL, we adopt the generalized Prüfer transformation mentioned in Section 2. At that time, we need to note whether an angle of each solution of (HL becomes nπ p with n Z. heorem 1 tells us that Hatvani and otik s ideas is also available for nonlinear differential equations such as (HL by combining with other techniques and that the Smith-type growth condition (5 becomes a criterion for the equilibrium of (HL to be globally asymptotically stable.

GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS 5 From heorem 1, we can conclude that the growth condition (3 in Smith s result and Hatvani and otik s result is not caused by the special characteristics of the linear solution space, namely, addition and homogeneity. For this reason, there may be a possibility that a criterion such as the growth condition (5 can be found for more general nonlinear differential equations than (HL. 2. Preliminaries We introduce the generalized sine function sin p θ, the generalized cosine function cos p θ and the generalized circular constant π p according to the book of Došlý and Řehák[5, pp. 4 5]. Let S(t be the solution of an elementary half-linear differential equation without the damping term, (ϕ p (x + (ϕ p (x = satisfying the initial condition (S(, S ( = (, 1. hen, it is easy to verify that the generalized Pythagorean identity S(t p + S (t p 1 holds and the function S(t is positive and increasing on [, π p /2] with S(π p /2 = 1 and S (π p /2 =, where 1 2 π p = (1 s p ds = 2π 1/p p sin(π/p. Combining the arc of S(t from to π p /2 as a basic pattern, we define the generalized sine function sin p θ as follows: { S(θ if θ πp /2 sin p θ = S(π p θ if π p /2 < θ π p and { sinp (θ π p if π p θ < 2π p sin p θ = sin p (θ 2nπ p if 2nπ p θ < 2(n + 1π p for n Z. he generalized cosine function cos p θ is defined as cos p θ = d dθ sin p θ. hen, from the above Pythagorean identity it follows that sin p θ p + cos p θ p = 1 for θ R. We define the generalized tangent function tan p θ by tan p θ = sin p θ cos p θ.

6 S. HAA and J. SUGIE About generalized trigonometric functions, see also [4, 7, 13]. Before we advance to the proof of heorem 1, it is very useful to change equation (HL into planar equivalent systems. Because ϕ p is the inverse function of ϕ p as mentioned in Section 1, by putting y = ϕ p (x as a new variable, equation (HL becomes the half-linear system (HS Let x = ϕ p (y, y = (ϕ p (x h(ty. x = r cos p θ and y = ϕ p (r sin p θ (this is slightly different from the generalized Prüfer transformation given in [4, pp. 169 17] and [5, p. 7]. hen, by a straightforward calculation, we can transform system (HS into the system (6 r = h(t r sin p θ p, sin p θ p 2 θ = cos p θ p 2 h(t ϕ p(sin p θϕ p (cos p θ. In particular, when θ nπ p with n Z, system (6 becomes the generalized polar coordinates system (HP r = h(t r sin p θ p, θ = cos p θ p 2 h(t sin p θ p 2 sin p θϕ p (cos p θ. 3. Fundamental characteristic of solutions As is customary, we say that a nontrivial solution x(t of (HL is oscillatory, if there exists a sequence {t n } tending to such that x(t n =. Otherwise, we say that it is nonoscillatory. Sturm s separation theorem holds for equation (HL as well as for equation (L. Equation (HL can be classified as oscillatory or nonoscillatory according to whether all nontrivial solutions of (HL have or do not have a sequence of zeros tending to. Lemma 2. Every nonoscillatory solution x(t of (HL is monotone for t sufficiently large. Proof. Since x(t is nonoscillatory, there exists a such that x(t for t. We may assume without loss of generality that x(t > for t. Let w(t = ϕ p (x (t/x(t for t. Since d dz ϕ p(z = ( z p 2, z R,

GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS 7 we have w (t = ( = ( x (t x(t x (t x(t p 2 x (tx(t (x (t 2 x 2 (t p 2 x (t ( x(t x (t x(t p. aking into account that p = p ( and ( x (t p 2 x (t + h(t x (t p 2 x (t + ( x(t p 2 x(t = (this is an alternative expression of (HL, we obtain the generalized Riccati equation (7 w (t + + h(tw(t + ( w(t p = for t. Suppose that there exists a 1 such that w(t > for t 1. hen, w (t < ( for t 1. Integrating this inequality from 1 to t, we obtain w(t < w( 1 ((t 1, which tends to as t. his is a contradiction. hus, we see that w(t = for some t. From (7 it turns out that w (t = ( <, and hence, we can choose an ε > so that w(t < for t < t t + ε. Suppose that there exists a t > t + ε such that w(t >. Let ˆt be the supremum of all t [t + ε, t ] for which w(t. hen, w(ˆt = and (8 w(t > for ˆt < t t. Using (7 again, we see that w (ˆt = ( <. Hence, w(t < on the rightneighborhood of ˆt. his contradicts (8. hus, we get ( x (t ϕ p = w(t for t t. x(t From the positivity of x(t it follows that x (t for t t ; namely, x(t is monotone for a sufficiently large t. As defined in Section 1, the concept of the global asymptotic stability is the union of the concept of the stability and the concept of the global attractivity. We can show the stability of the equilibrium of (HL only by assuming that h(t for t. Lemma 3. he equilibrium of (HL is uniformly stable. Proof. Let x(t be any solution of (HL with the initial time t. Define v(t = x(t p + ϕ p (x (t p.

8 S. HAA and J. SUGIE Recall that ϕ p is the inverse function of ϕ p. aking into account that d dz z q = qϕ q (z, z R with q = p or p, we obtain ( v (t = pϕ p (x(tx (t + p ϕ p ϕp (x (t ( ϕ p (x (t for t t. Hence, it follows that = pϕ p (x(tx (t p x (t ( h(tϕ p (x (t + (ϕ p (x(t = p h(tϕ p (x (tx (t = p h(t x (t p (9 v(t v(t for t t. Since ϕ p (x (t p = x (t p (p 1 = x (t p, we can rewrite v(t as v(t = x(t p + x (t p. Let be the p-norm p. hen, from (9 it is easy to confirm that for any ε >, there exists a δ(ε > such that t and x < δ imply x(t; t, x < ε for all t t ; that is, the equilibrium of (HL is uniformly stable. Remark 2. By virtue of Lemma 3, to prove our main theorem, we have only to show that the growth condition (5 is necessary and sufficient for the equilibrium of (HL to be globally attractive under the assumption (4. 4. Proof of heorem 1 We first prove that if the equilibrium of (HL is globally attractive, then the growth condition (5 holds. We then prove the converse. Necessity. Suppose that (5 does not hold. hen, we can choose so large that ( (1 ϕ eh(s ds 1 p dt < e H(t 2(. p 1 Consider the solution x(t of (HL satisfying the initial condition (x(, x ( = (1,. We will show that x(t > 1/2 for all t. By way of contradiction, we suppose that there exists a 1 > such that x( 1 = 1/2 and x(t > 1/2 for t < 1. Let w(t = ϕ p (x (t/x(t for t 1. hen, w( =. As shown in the proof of Lemma 2, we see that w(t satisfies the generalized Riccati equation (7. aking into account that w( = and following the same process as in the proof of Lemma 2, we conclude that x(t is nonincreasing, and hence, (11 1 2 = x( 1 x(t x( = 1 for t 1.

GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS 9 From (11 it turns out that ( ϕp (x (t + h(tϕp (x (t = (ϕ p (x(t ( for t 1. Hence, ( ϕp (x (te H(t (e H(t for t 1. Integrating both sides of this inequality from to t, we get ϕ p (x (te H(t ϕ p (x ( e H( ( for t 1. Since ϕ p x (t ϕ p e H(s ds = ( is the inverse function of ϕ p, we see that ( ( eh(s ds e H(t for t 1. Integrate both sides of this inequality from to 1 to obtain 1 ( x( 1 x( ϕ p ( eh(s ds dt 1 ( p 1 1 ( p 1 ϕ p ϕ p ϕ p e H(t ( eh(s ds dt ( e H(t eh(s ds e H(t dt. e H(s ds Hence, it follows from (1 that x( 1 > 1/2, which contradicts the assumption that x( 1 = 1/2. hus, the solution x(t does not tend to zero as t, and hence, the equilibrium of (HL is not globally attractive. Sufficiency. Let x(t be any solution of (HL with the initial time t and let (r(t, θ(t be the solution of (6 corresponding to x(t. We prove that if (4 and (5 hold, then (x(t, x (t tends to the origin (, as t. For this purpose, it is enough to show that r(t = p x(t p + x (t p as t. Since (12 r (t = h(t r(t sin p θ(t p for t t, r(t is nonincreasing for t t and therefore, it has the limiting value r. Suppose that r is positive. hen, we get a contradiction as shown below.

1 S. HAA and J. SUGIE If x(t is nonoscillatory, then by means of Lemma 2 we see that it is monotone for t sufficiently large. Hence, we can find a t so large that or Since x(t > and x (t for t, x(t < and x (t for t. tan p θ(t = sin p θ(t cos p θ(t = x (t x(t for t, we see that π p /2 < θ(t (mod 2π p or π p /2 < θ(t π p (mod 2π p for t. If x (t = for t, then x(t is a nonzero constant. his is impossible, because the only equilibrium of (HL is the origin (,. Hence, in either case, there exists a such that θ(t nπ p for t with n Z. Hence, (r(t, θ(t is a solution of (HP for t. Since x(t is nonincreasing for t, there exists a c R with c r such that x(t c as t. Hence, it follows that x (t p r p c p as t. If c < r, then we can choose a 2 so that x (t p > rp c p 2 for t 2. Hence, by (12 we have (r p (t = pr p 1 (tr (t = p h(trp (t sin p θ(t p = p h(t x (t p < p (r p c p h(t 2 for t 2. Integrating this inequality from 2 to t, we obtain r p r p ( 2 r p (t r p ( 2 < p (r p c p t h(sds, 2 2 which tends to as t. his is a contradiction. hus, we see that c = r. We therefore conclude that x(t r and x (t as t. here are two cases that should be considered: (i < r x(t x( and x (t for t and (ii x( x(t r < and x (t for t. Note that (13 ( ϕp (x (te H(t = (ϕp (x(te H(t for t.

GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS 11 In the former case, integrate both sides of (13 from to t to obtain ϕ p (x (te H(t ϕ p (x (te H(t ϕ p (x ( e H( (ϕ p (r for t. Since eh(s ds as t, there exists a 3 > such that e H(s ds = Hence, we can estimate that namely, e H(s ds ϕ p (x (te H(t < 1 p ϕ p(r e H(s ds > 1 p ( p ( 1 1 t x (t < r p ϕ eh(s ds p e H(t e H(s ds for t 3. e H(s ds for t 3 ; for t 3. Integrating this inequality from 3 to t, we get ( p 1 ( u r x( 3 x(t x( 3 < 1r p ϕ eh(s ds p du 3 e H(u e H(s ds for t 3. his estimation and condition (5 lead to a contradiction. In the latter case, we have ( p 1 ( u r x( 3 x(t x( 3 > 1r eh(s ds p du, e H(u 3 ϕ p which tends to as t. his is a contradiction. hus, x(t have to oscillate. Since x(t = r(t cos p θ(t, there exist two sequences {τ n } and {σ n } with t τ n < σ n < τ n+1 and τ n as n such that θ(τ n = π p /2 (mod 2π p and θ(σ n = π p /2 (mod 2π p. Recall that y(t = ϕ p (r(t sin p θ(t and (x(t, y(t is the solution of (HS corresponding to (r(t, θ(t. From the second equation of (HP it is clear that θ = 1 h(t tan p θ p 2 sin p θϕ p (cos p θ < for < θ < π p /2 and π p < θ < π p /2 (mod 2π p ; namely, θ is negative in the first quadrant and the third quadrant. his means that the solution (x(t, y(t moves clockwise in the first quadrant and the third quadrant. We divide our argument into two cases: (i p 2 and (ii 1 < p < 2. Case (i: p 2. From the definition of ω p and the assumption that < γ < ω p, there exists a θ (, π p /2 such that (14 γ < (tan p θ p 2 (π p 2θ < ω p.

12 S. HAA and J. SUGIE We can choose two divergent sequences {t n } and {s n } with τ n < t n < σ n < s n < τ n+1 such that θ(t n = π p θ, θ(s n = θ and θ < θ(t < π p θ for t n < t < s n. Since θ is not necessarily negative in the second quadrant and the fourth quadrant, the behavior of (x(t, y(t may be not so simple. For this reason, the point in the set {t (τ n, σ n : θ(t = π p θ } might not be only one. In such a case, we should select the supremum of all t (τ n, σ n for which θ(t π p θ as the point t n. Suppose that there exists an N N such that s n t n γ for n N. By (12 we have r (t r (sin p θ p h(t for t n t s n. Needless to say, r (t otherwise. Hence, we can estimate that r(s n r(t n r (sin p θ p sn t n h(tdt r (sin p θ p n +γ t n h(tdt for n N and r(t n+1 r(s n for n N. Adding these two evaluations, we obtain his inequality yields that r(t n+1 r(t n r (sin p θ p n +γ r r(t N r(t n+1 r(t N r (sin p θ p t n h(tdt for n N. n i=n i +γ t i h(tdt. Since t n tends to as n, this is a contradiction. hus, there exists a sequence {n k } with n k N and n k as k such that (15 s nk t nk < γ. Since sin p θ 1 and cos p θ 1 for θ R, we see that θ (t = 1 h(t tan p θ(t p 2 sin 1 h(t p θ(tϕ p (cos p θ(t (tan p θ p 2 for t nk t s nk. It follows from (15 that θ (π p θ = θ(s nk θ(t nk s n k t nk (tan p θ 1 p 2 > γ (tan p θ 1 p 2 snk t nk snk t nk h(tdt h(tdt

GLOBAL ASYMPOIC SABILIY OF DAMPED HALF-LINEAR OSCILLAORS 13 for each k N. Hence, from (14 it turns out that 1 snk t nk h(tdt > π p 2θ Using (12 and this estimation, we obtain namely, snk γ > for k N. (tan p θ p 2 log r(s n k snk r(t nk = r (t t nk r(t dt = 1 h(t sin p θ(t p dt t nk (sin p θ p snk ( h(tdt (sin p θ p t nk π p 2θ ( } r(s nk { (sin r(t nk exp p θ p π p 2θ γ < 1 (tan p θ p 2 γ ; (tan p θ p 2 for each k N. his shows that r(t tends to zero as t. Because the limiting value r of r(t is positive, this is a contradiction. Case (ii: 1 < p < 2. From the definition of ω p and the assumption that < γ < ω p, there exists a θ (, π p /2 such that γ < Let ε > be so small that (16 γ < ( ( πp p 2(πp tan p 2 θ 2θ < ω p. ( ( πp p 2(πp tan p 2 θ 2θ 2ε < ω p. Since the solution (x(t, y(t of (HS rotates around the origin (, infinite many times, we can choose four divergent sequences { t n }, {t n }, {s n } and { s n } with t n < t n < s n < s n such that θ( t n = π p ε, θ(t n = π p /2 + θ, θ(s n = π p /2 θ, θ( s n = ε, and π p 2 + θ < θ(t < π p ε for t n < t < t n, ε < θ(t < π p 2 θ for s n < t < s n. Here, we select the supremum of all t (τ n, σ n for which θ(t π p ε as the point t n, and the infimum of all t ( t n, σ n for which θ(t π p /2 + θ as the point t n. Hence, it naturally follows that ε < θ(t < π p ε for t n < t < s n. As in the proof of case (i, we can find a sequence {n k } with n k N and n k as k such that (17 s nk t nk < γ.

14 S. HAA and J. SUGIE Since sin p θ 1 and cos p θ 1 for θ R, we see that θ 1 h(t (t = tan p θ(t p 2 sin p θ(tϕ p (cos p θ(t 1 h(t (tan p (π p /2 θ p 2 for t nk t t nk and s nk t s nk. Integrating this inequality, we obtain and π p 2 + θ (π p ε = θ(t nk θ( t nk t nk t nk (tan p (π p /2 θ 1 p 2 nk t nk h(tdt ε (π p /2 θ = θ( s nk θ(s nk s nk s nk (tan p (π p /2 θ 1 p 2 for each k N. Hence, it follows from (16 and (17 that 1 snk t nk h(tdt > π p 2θ 2ε Using (12 and this estimation, we obtain snk s nk h(tdt γ > for k N. (tan p (π p /2 θ p 2 namely, log r( s n k snk r( t nk = r (t t nk r(t dt = 1 snk t nk (sin p ε p (π p 2θ 2ε snk h(t sin p θ(t p dt (sin p ε p h(tdt t nk γ ; (tan p (π p /2 θ p 2 ( } r( s nk { (sin r( t nk exp p ε p π p 2θ γ 2ε < 1 (tan p (π p /2 θ p 2 for each k N. his shows that r(t tends to zero as t. Because the limiting value r of r(t is positive, this is a contradiction. he proof of heorem 1 is thus complete.

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