A study of the apsidal angle and a proof of monotonicity in the logarithmic potential case
AA study of the apsidal angleand a proof of monotonicity in the logarithmic potential case
Roberto Castelli ∗ Abstract
This paper concerns the behaviour of the apsidal angle for orbits of central force system withhomogenous potential of degree − ≤ α ≤ (cid:96) . The monotonicity of the apsidal angle as function of (cid:96) is discussed and it is proved in the logarithmic potential case. Keywords
Two-body problem · Logarithmic potential · Apsidal angle
Mathematics Subject Classification (2010) · In this paper we investigate the apsidal precession for the orbits of central force systems of the form¨ u = ∇ V α ( | u | ) , u ∈ R (1)where V α ( x ) = 1 α x α , α ∈ [ − , \ { } V ( x ) = − log( x ) α = 0 . (2)The solutions of system (1) preserve the mechanical energy E = | ˙ u | − V and the angular momentum (cid:126)(cid:96) = u ∧ ˙ u and, while rotating, all the bounded non-collision orbits oscillates between the apses, thatis, the points of minimal ( pericenter) and maximal (apocenter ) distance from the origin. Theapsidal angle is defined as the angle at the origin swept by the orbit between two consecutive apses.For an admissible choice of E and (cid:96) ( being (cid:96) the scalar angular momentum, positive for coun-terclockwise orbits and negative for clockwise orbits), the apsidal angle of an orbit u ( t ) is givenby ∆ α θ ( u ) = (cid:90) r + r − (cid:96)r (cid:113) (cid:0) E + V ( r ) (cid:1) − (cid:96) r dr (3)where r ± are the points where the denominator vanishes.The extremal cases α = − α = 1 consist in the harmonic oscillator and the Kepler prob-lem respectively. Systems with − < α < ∗ BCAM - Basque Center for Applied Mathematics, Alameda de Mazarredo, 14 48009 Bilbao, Basque Country,Spain. Email: [email protected] . a r X i v : . [ m a t h . D S ] S e p strophysics, for instance in describing the galactic dynamics in presence of power law densitiesor massive black-holes and in modelling the gravitational lensing, see [1, 2]. Also, the logarithmicpotential appears in particle physics [3, 4] and in a model for the dynamics of vortex filaments inideal fluid, see i.e. [5].Since Newton, the behaviour of the apsidal angle has been extensively studied, in particular forits implications to celestial mechanics. In the Book I of the Principia
Newton derived a formulathat relates the apsidal angle to the magnitude of the attracting force: for “close to circular” orbits,a force fields of the form κr n − leads to an apsidal angle equal to π/ √ n . Hence, the experimentalmeasurement of the precession of an orbit may give the exponent of the force law. Newton itself inBook III, looking at the orbit of the Earth and of the Moon, concludes that the attracting force ofthe Sun and of the Earth must be inverse square of the distance. “Close to circular” means that thisformula has been proved for orbits with small eccentricity, or equivalently, with angular momentumclose to (cid:96) max .In the singular case ( α ≥ π as (cid:96) →
0, seefor instance [6, 7, 8, 9]. Also a variational approach to system (1) may lead to collision avoidanceprovided ∆ α θ ( u ) > π as | u | →
0, see [10]. Moreover, as pointed out in [11], the derivative of theapsidal angle with respect to the angular momentum has common feature with the derivative of thepotential scattering phase shift and applications in a variety of fields, such as nuclear physics andastrophysics.Two are the main technical hurdles that make both the analytical treatment and the numericalinvestigation of the integral (3) a difficult problem: the integrand is singular at the end-points andthe end-points themselves are only implicitly defined as function of E and (cid:96) . Nevetherless, partialand important results are very well known. In 1873, Bertrand proved that there are only two centralforce laws for which all bounded orbits are closed, namely the linear and the inverse square [12]. Itmeans that the apsidal angle of any bounded solution of the elastic and Kepler problem is rationallyproportional to π for any value of E and (cid:96) . In the first case it values π/ π in the second. Inaddition, for α = − , − , , , the apsidal angle can be expressed in terms of elliptic functions, see[13], but in general ∆ α θ ( u ) can not be written in closed form.Valluri et al. [11] propose an expansion of ∆ α θ ( u ) and ∂ (cid:96) ∆ α θ ( u ) in terms of the eccentricity ofthe orbit for α close 1. A study for the logarithmic case using the p − ellipse approximation andthe Lambert W function is presented in [2]. Touma and Tremaine in [1], by means of the Mellintransform, provide an asymptotic series on (cid:96) for any α ∈ [ − , α and E the apsidal angle is a monotonic function of the angular momentum. However, at the bestof the author’s knowledge, no proof of this statement is available in the literature. The purpose ofthis paper is to review the notion of the apsidal angle and to derive a formula for its derivative withrespect to the angular momentum. From where, the monotonicity will be evident for any α ∈ ( − , q plays the role of the angular momentum. Then, section 3 is dedicated to theanalytical study of the integrand, in particular of the function E α ( s, q ). In section 4 the results sofar obtained are used to compute the limit values of the apsidal angle for radial and close to circularorbits. In section 5 we compute the derivative of the apsidal angle and we provide insight into themonotonicity in the α -homogneous case. Finally, in section 6 the apsidal angle for α = 0 is provedto be monotonically increasing as function of the angular momentum. In Appendix some technical2emmas are collected. In polar coordinates u = ( r, θ ) the energy associated to a solution u ( t ) of the problem (1) is E ( u ) = 12 ˙ r + 12 (cid:96) r − V α ( r )hence the motion is allowed only for those r where E > V effα = (cid:96) r − V α ( r ). From the analysis ofthe V effα , see also Fig. 1, we realise that • for α ∈ (0 , E <
0, the modulusof the angular momentum (cid:96) ranges between 0 and (cid:96) max = (cid:16) αα − E (cid:17) − − α α . Values E ≥ • for α = 0: since log( r ) → ∞ as r → ∞ , all the possible orbits are bounded. For any value of E , the motion is possible provided 0 ≤ | (cid:96) | ≤ (cid:96) max = e E − ; • for α ∈ [ − , V α is negative therefore the motion exists only forpositive values of E . In particular all the orbits are bounded and the | (cid:96) | ranges within theinterval [0 , (cid:96) max ] with (cid:96) max = (cid:16) αα − E (cid:17) − − α α .Figure 1: Plot of the function V effα ( r ) for a ) α ∈ (0 , b ) α = 0, c ) α ∈ [ − , r c is theradius where the minimum of V effα ( r ) is achieved and it corresponds to the radius of the circular orbit. Itholds : in a ) and b ) r c = (cid:96) − α , V effα ( r c ) = − − α α (cid:96) − α − α , b) r c = (cid:96) , V eft ( r c ) = + log( (cid:96) ) . The radial coordinate r ( t ) of the orbit u ( t ) solves the differential equation¨ r − (cid:96) r + 1 r α +1 = 03nd oscillates periodically between the extremal values r + and r − given as positive solutions of theapses equation (cid:96) r − V α ( r ) − E = 0. While oscillating, the solution rotates around the centre ofattraction, in direction according to the sign of (cid:96) .From the relation (cid:96) = r ˙ θ it follows that the rate of rotation between two consecutive apses isgiven by (cid:82) r + r − (cid:96)r ˙ r dr , from where formula (3) comes. Remark 2.1.
The apsidal angle ∆ α θ ( u ) is odd as function of (cid:96) . Remark 2.2.
From (3) it descends that ∆ β θ ( u ) = − α ∆ α θ ( u ) , where β = − α − α , see also [15].Thus apsidal angle for α ∈ [ − , is determined by the value for α ∈ (0 , . Inlight of the two remarks and since the case α = 1 is completely solved, henceforth we restrictour analysis to α ∈ [0 , , (cid:96) ∈ (0 , . A standard approach when studying the apsidal angle is to consider the differential equation d dθ z + z = φ (cid:18) z (cid:19) z (cid:96) where z = r and φ ( r ) is the central force. From here, following Bertrand [12] and Griffin [16], itholds ∆ α θ ( u ) = (cid:90) ba (cid:112) w ( b ) − w ( a ) (cid:112) a w ( b ) − b w ( a ) + ( b − a ) w ( z ) − z ( w ( b ) − w ( a )) dz (4)where the function w ( z ) = 2 (cid:82) ψ ( z ) dz and ψ ( z ) = r φ ( r ). In practice, for the force field we areinterest in, it holds φ ( r ) = − r α +1 , ψ ( z ) = z α − , w ( z ) = 2 α z α , α (cid:54) = 0 φ ( r ) = − r , ψ ( z ) = z − , w ( z ) = 2 log( z ) , α = 0 . The dependence on (cid:96) and on the energy of the integral (4) is through the values of the a = 1 /r + and b = 1 /r − . Indeed, in the logarithmic potential case, such values are given as solutions of theequation log( z ) − (cid:96) z = − E , whose roots can only be expressed as z = (cid:114) − W j ( − (cid:96) e − E ) (cid:96) where W j represent the two branches of the Lambert W function, [17]. Similarly, in the homogenouscase, a and b are the solutions of − (cid:96) z + α z α = − E .From (4), by easy manipulation it follows∆ α θ ( u ) = (cid:90) ba dz (cid:113) b − z − ( b − a ) w ( b ) − w ( z ) w ( b ) − w ( a ) = (cid:90) ba dz (cid:115) ( b − z ) (cid:18) ( b + z ) − ( b + a ) w ( b ) − w ( z ) b − zw ( b ) − w ( a ) b − a (cid:19) = (cid:90) ba √ z − a √ b − z (cid:112) ε ( z ) dz where ε ( z ) = b + az − a (cid:34) − w ( b ) − w ( z ) b − zw ( b ) − w ( a ) b − a (cid:35) . (5)4et L := b − a and perform the change of variable s = z − aL ; it results∆ α θ ( u ) = (cid:90) (cid:112) s (1 − s ) 1 (cid:112) (cid:15) ( s ) ds (6)where (cid:15) ( s ) = (2 b − L ) Ls (cid:34) − w ( b ) − w (cid:0) b − L (1 − s ) (cid:1) (1 − s ) (cid:0) w ( b ) − w ( b − L ) (cid:1) (cid:35) . Then, taking q = Lb , we obtain∆ α θ ( u ) = (cid:90) (cid:112) s (1 − s ) 1 (cid:112) E α ( s, q ) ds (7)with E α ( s, q ) = 2 − qqs (cid:34) − − (cid:0) − q (1 − s ) (cid:1) α (1 − s )(1 − (1 − q ) α ) (cid:35) , α ∈ (0 , E ( s, q ) = 2 − qqs (cid:20) − log(1 − q (1 − s ))(1 − s ) log(1 − q ) (cid:21) , α = 0 (8)For any fixed value of the mechanical energy, the parameter q = 1 − ab is a function of the angularmomentum (cid:96) . In particular, as (cid:96) increases from zero to (cid:96) max , the ratio a/b increases from 0 to 1.Therefore q = q ( (cid:96) ) ∈ (0 , , dqd(cid:96) < , ∀ (cid:96) ∈ (0 , . The integral in (7) is the main object of investigation of the present paper. Two are the reasons whyit is advantageous to analyse (7) instead of the classical one (3): in the first integral the end-pointsare fixed and the dependence on the angular momentum is confined and well separated from thesingularities of the integrand function.The behaviour of the apsidal angle as the angular momentum (cid:96) varies is intimately related to thebehaviour of the function E α ( q, s ) when s , q vary in their domain. Hence, we first present a detailedanalysis of the function E α ( s, q ) and its derivative ∂ q E α ( s, q ). E α ( s, q ) It is convenient to introduce the weights ω αn := − (cid:0) αn (cid:1) ( − n α ∈ (0 , n α = 0 ∀ n ≥ n ≥ A n ( s ) := 1 − (1 − s ) n − s , s ∈ (0 , n ≥ α ∈ [0 ,
1) let K αn ( s ) := 2 ω αn +1 ω αn A n +1 ( s ) − A n ( s ) = 2 n − αn + 1 A n +1 ( s ) − A n ( s ) , s ∈ (0 , . (10)5ote that ω αn > n ≥ α ∈ [0 , A n ( s ) and K αn ( s ) used throughout the paper.Recalling the power series expansions:(1 − x ) α = 1 + (cid:88) n ≥ (cid:18) αn (cid:19) ( − n x n , log(1 − x ) = − (cid:88) n ≥ x n n , | x | < α ∈ (0 , E α ( s, q ) = 2 − qqs − (1 − s ) (cid:80) n ≥ (cid:0) αn (cid:1) ( − n q n + (cid:80) n ≥ (cid:0) αn (cid:1) ( − n q n (1 − s ) n (1 − s ) (cid:16) − (cid:80) n ≥ (cid:0) αn (cid:1) ( − n q n (cid:17) = 2 − qqs (cid:16) − (cid:80) n ≥ (cid:0) αn (cid:1) ( − n q n (cid:17) − (cid:88) n ≥ (cid:18) αn (cid:19) ( − n q n + (cid:88) n ≥ (cid:18) αn (cid:19) ( − n q n (1 − s ) n − = 1 (cid:80) n ≥ (cid:0) αn (cid:1) ( − n q n − (cid:88) n ≥ (cid:18) αn (cid:19) ( − n q n − (2 − q ) A n ( s ) (11)and similarly, for α = 0, E ( s, q ) = 1 (cid:80) n ≥ n q n − (cid:88) n ≥ n (2 − q ) q n − A n ( s ) . By means of the weights ω αn , for any α ∈ [0 ,
1) we can write in compact form E α ( s, q ) = 1 (cid:80) n ≥ ω αn q n − (cid:88) n ≥ ω αn (2 − q ) q n − A n ( s ) . (12) Proposition 3.1.
For any α ∈ [0 , and q ∈ (0 , the function E α ( q, s ) is monotonically decreasingand convex in s for any s ∈ (0 , .Proof. Since A (cid:48) ( s ) = 0, for any α ∈ [0 , dds E α ( s, q ) = 1 (cid:80) n ≥ ω αn q n − (cid:88) n ≥ ω αn q n − (2 − q ) A (cid:48) n ( s )and d ds E α ( s, q ) = 1 (cid:80) n ≥ ω αn q n − (cid:88) n ≥ ω αn q n − (2 − q ) A (cid:48)(cid:48) n ( s ) . Since ω αn > iii ) of lemma 7.1, A (cid:48) n < A (cid:48)(cid:48) n >
0, the thesis follows. (cid:4)
Corollary 3.2.
For any α ∈ [0 , the function E α ( s, q ) > for any s ∈ (0 , , q ∈ (0 , .Proof. From the previous proposition E α ( s, q ) ≥ E α (1 , q ) = − qq (cid:16) − (1 − q ) α − αq − (1 − q ) α (cid:17) , < α < − qq (cid:16) q log(1 − q ) (cid:17) , α = 0 . Since (1 − q ) α < − αq and − log(1 − q ) > q , it follows E α (1 , q ) > q ∈ (0 , Consider now the derivative of E α ( s, q ) with respect to the variable q . For α ∈ [0 , ∂ q E α ( s, q ) = 1 (cid:16)(cid:80) n ≥ ω αn q n − (cid:17) · (cid:88) n ≥ ω αn (cid:16) ( n − − q ) − q (cid:17) q n − A n ( s ) (cid:88) n ≥ ω αn q n − − (cid:88) n ≥ ω αn (2 − q ) q n − A n ( s ) (cid:88) n ≥ ω αn ( n − q n − = 1 (cid:16)(cid:80) n ≥ ω αn q n − (cid:17) · (cid:88) n ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) A n ( s ) q ( n − m − . (13)Henceforth, let us denote C α ( q ) := (cid:88) n ≥ ω αn q n − − . In the sequel we prove that the function ∂ q E α ( s, q ) is monotonic and convex in s . For that let usintroduce the quantity H α ( n, m ) := ω αn +2 ω αm ( n − m )and we prove the following. Lemma 3.3.
Let n, m ≥ and n > m . Then, for any α ∈ [0 , , H α ( n, m ) + H α ( m, n ) > . Proof.
Case α = 0. H ( n, m ) + H ( m, n ) = (cid:18) n + 2 1 m − m + 2 1 n (cid:19) ( n − m ) = 2 ( n − m ) nm ( n + 2)( m + 2) > . Let be now α ∈ (0 , t , we remind the relations (cid:18) αt + 1 (cid:19) = α − tt + 1 (cid:18) αt (cid:19) , (cid:18) αt + 2 (cid:19) = α − ( t + 1) t + 2 α − tt + 1 (cid:18) αt (cid:19) . Then H α ( n, m ) + H α ( m, n ) = (cid:18) α − ( n + 1) n + 2 α − nn + 1 − α − ( m + 1) m + 2 α − mm + 1 (cid:19) ω αn ω αm ( n − m )= (cid:18) ( n + 1) − αn + 2 n − αn + 1 − ( m + 1) − αm + 2 m − αm + 1 (cid:19) ω αn ω αm ( n − m ) . (14)Since K − αK +1 > P − αP +1 > K > P ≥
1, it follows H α ( n, m )+ H α ( m, n ) > n > m . (cid:4) Proposition 3.4.
For any α ∈ [0 , and q ∈ (0 , i) dds (cid:16) ∂ q E α ( s, q ) (cid:17) < , ∀ s ∈ (0 , i) d ds (cid:16) ∂ q E α ( s, q ) (cid:17) > , ∀ s ∈ (0 , . Proof. i ) From (13), since A (cid:48) ( s ) = 0, dds ( ∂ q E α ( s, q )) = C α ( q ) (cid:88) n ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) A (cid:48) n ( s ) q ( n − m − = C α ( q ) (cid:88) p ≥ (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) A (cid:48) n ( s ) q p − . Denote by
Coef ( s, q, α, p ) := (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) A (cid:48) n ( s ) . We prove that
Coef ( s, q, α, p ) < s ∈ (0 , q ∈ (0 ,
1) and α ∈ [0 ,
1) and p ≥
4. Computefirst the derivative in q : ddq Coef ( s, q, α, p ) = − (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm ( n − m ) A (cid:48) n ( s ) . For point iii ) in lemma 7.1 A (cid:48) n ( s ) <
0, then if p ≤ n ≥ m in any contribution. For p ≥
7, we write ddq Coef ( s, q, α, p ) = − ω αp − ω α ( p − A (cid:48) p − ( s ) − ω αp − ω α ( p − A (cid:48) p − ( s ) − (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm ( n − m ) A (cid:48) n ( s ) . The first and the second terms in the right hand side are positive then, collecting the contributionsto the sum due to the couples ( n, m ) and ( m, n ), and for point iv ) in lemma 7.1, it holds ddq Coef ( s, q, α, p ) > − (cid:88) n + m = pn>m ≥ ω αn ω αm ( n − m )( A (cid:48) n ( s ) − A (cid:48) m ( s )) > . Therefore, ddq
Coef ( s, q, α, p ) > p ≥ s ∈ (0 , q ∈ (0 , α ∈ [0 , Coef ( s, q, α, p ) < Coef ( s, , α, p ) . Now,
Coef ( s, , α, p ) = (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) n − m − (cid:17) A (cid:48) n ( s ) = (cid:88) n + m = p − n ≥ ,m ≥ ω αn +2 ω αm (cid:16) n − m (cid:17) A (cid:48) n +2 ( s )= (cid:88) n + m = p − n>m ≥ (cid:104) ω αn +2 ω αm (cid:16) n − m (cid:17) A (cid:48) n +2 ( s ) + ω αm +2 ω αn (cid:16) m − n (cid:17) A (cid:48) m +2 ( s ) (cid:105) . Since ω αn +2 ω αm > iii ) of lemma 7.1, A (cid:48) n +2 ( s ) < A (cid:48) m +2 ( s ), it follows8 oef ( s, , α, p ) < (cid:88) n + m = pn>m ≥ (cid:104) ω αn +2 ω αm (cid:16) n − m (cid:17) + ω αm +2 ω αn (cid:16) m − n (cid:17)(cid:105) A (cid:48) m +2 ( s )= (cid:88) n + m = pn>m ≥ ( H α ( n, m ) + H α ( m, n )) A (cid:48) m +2 ( s ) < Coef ( s, q, α, p ) < dds ( ∂ q E α ( s, q )) < ii ) Consider now the second derivative d ds (cid:16) ∂ q E ( s, q ) (cid:17) = C α ( q ) (cid:88) p ≥ (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) A (cid:48)(cid:48) n ( s ) q p − . Arguing as before, we realise that q = 1 represents the worst case and, since A (cid:48)(cid:48) n ( s ) >
0, it follows (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) ( n − m ) − (cid:17) A (cid:48)(cid:48) n ( s ) > , ∀ s ∈ (0 , , q ∈ (0 , , α ∈ (0 , , p ≥ . (cid:4) Up to now we investigated the monotonicity and the convexity of ∂ q E α ( q, s ) without mentioningabout the value of ∂ q E α ( q, s ). To the purpose of proving the monotonicity of the apsidal angle, itwould be enough that ∂ q E α ( q, s ) is positive for any s . Unfortunately that is not the case, indeed forany α there is a region of q where ∂ q E α ( q, s ) is not everywhere positive. Nevertheless, we prove nowthat ∂ q E α ( q, s ) > s = 1 / α and q . For the previous result it follows that ∂ q E α ( q, s ) > s ∈ (0 , / q and α . Proposition 3.5.
For any q ∈ (0 , and α ∈ [0 , ∂ q E α (cid:18) , q (cid:19) > . Proof. ∂ q E α (cid:18) , q (cid:19) = C α ( q ) (cid:88) p ≥ (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) A n (1 / q p − = C α ( q ) (cid:88) p ≥ (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) n − m − (cid:17) A n (1 / q p − − (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm (cid:16) n − m (cid:17) A n (1 / q p − . When p = 3 the first sum does not contribute, then we can write ∂ q E (cid:18) , q (cid:19) = C α ( q ) (cid:88) p ≥ (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm n − m − A n (1 / − (cid:88) n + m = p − n ≥ ,m ≥ ω αn ω αm (cid:16) n − m (cid:17) A n (1 / q p − . Q p ( q ) := (cid:88) n + m = pn ≥ ,m ≥ ω αn ω αm n − m − A n (1 / − (cid:88) n + m = p − n ≥ ,m ≥ ω αn ω αm ( n − m ) A n (1 / q p − . For p ≥ Q p ( q ) = (cid:88) n + m = p − n ≥ ,m ≥ ω αn +1 ω αm n − m ) A n +1 (1 / − (cid:88) n + m = p − n ≥ ,m ≥ ω αn ω αm ( n − m ) A n (1 / ω α ω αp − − p ) A (1 /
2) + (cid:88) n + m = p − n ≥ ,m ≥ ω αn ω αm ( n − m ) (cid:16) ω αn +1 ω αn A n +1 (1 / − A n (1 / (cid:17) = ω α ω αp − − p ) A (1 /
2) + ω αp − ω α ( p − (cid:32) ω αp − ω αp − A p − (1 / − A p − (1 / (cid:33) + (cid:88) n + m = p − n ≥ ,m ≥ ω αn ω αm ( n − m ) (cid:18) ω αn +1 ω αn A n +1 (1 / − A n (1 / (cid:19) . (15)Reminding the definition (10) of K αn ( s ) and collecting the contributions due to the couple ( n, m )and ( m, n ), it results Q p ( q ) = ω αp − ( p − (cid:34) ω α ω αp − ω αp − A p − (1 / − ω α A p − (1 / − ω α A (1 / (cid:35) + (cid:88) n + m = p − n>m ≥ ω αn ω αm ( n − m ) (cid:16) K αn (1 / − K αm (1 / (cid:17) . (16)For point ii ) in lemma 7.2, the last sum is positive. Moreover, for α ∈ (0 , η α ( p ) := 2 ω α ω αp − ω αp − A p − (1 / − ω α A p − (1 / − ω α A (1 / α ( p − − α ) p − A p − (1 / − αA p − (1 / − α (1 − α ) (17)is such that η α (4) = 0 , ˙ η α ( p ) = 2 α (2 + 2 α )( p − p − (cid:16) p − − − ( p −
1) log(2) (cid:17) > ∀ p ≥ . A similar results holds for η ( p ). Thus Q p ( q ) > p ≥ α ∈ [0 ,
1) and the thesis follows. (cid:4) ∆ α θ ( u ) as (cid:96) → and (cid:96) → (cid:96) max This section concerns the limits of the apsidal angle for near-circular and near-radial orbits, that iswhen (cid:96) → (cid:96) max and (cid:96) → heorem 4.1. For any α ∈ [0 , (cid:96) → ∆ α θ ( u ) = π − α , lim (cid:96) → (cid:96) max ∆ α θ ( u ) = π √ − α . Proof.
According to the discussion in section 2, the limit values for the apsidal angle are given asthe limit for q → q → q → (cid:96) → (cid:96) max and q → (cid:96) →
0. Since E α ( s, q ) > α ∈ [0 , , q ∈ (0 , , s ∈ (0 , q → ∆ α θ = (cid:90) (cid:112) s (1 − s ) 1 (cid:112) q → E α ( s, q ) ds = (cid:90) (cid:112) s (1 − s ) 1 (cid:112) − α ) ds = π √ − α andlim q → ∆ α θ = (cid:90) (cid:112) s (1 − s ) 1 (cid:112) q → E α ( s, q ) ds = (cid:90) (cid:112) s (1 − s ) 1 (cid:113) s s α − s − s ds = π − α . (cid:4) ∆ α θ ( u ) We now examine the derivative of the apsidal angle as function of the angular momentum (cid:96) . Thatis equivalent to compute the derivative of ∆ α θ ( u ) given in (7) with respect to q . From proposition3.4 and 3.5 | ∂ q E α ( s, q ) | < ∂ q E α (0 , q ) = M α ( q )where M α ( q ) = a (1 − q ) a q + (cid:16) (2 − a ) (1 − q ) a + (cid:0) − a + a − (cid:1) (1 − q ) a + 2 (cid:17) q + (cid:16) − − q ) a + 8 (1 − q ) a − (cid:17) q + 2((1 − q ) α − ((1 − q ) a − ( q − q , α ∈ (0 , q − ln (1 − q ) q − q + 2 (ln (1 − q )) q − − q )) q + 2 (ln (1 − q )) (ln (1 − q )) q (1 − q + q ) α = 0is C ((0 , α . Therefore, for any q ∈ (0 ,
1) we can pass the derivative under the integralsign and write ddq ∆ α θ ( u ) = (cid:90) (cid:112) s (1 − s ) ∂ q (cid:32) (cid:112) E α ( s, q ) (cid:33) ds = − (cid:90) (cid:112) s (1 − s ) ∂ q E α ( s, q )(1 + E α ( s, q )) ds. (18)The following analysis aims at proving that the integral in (18) is positive for any q ∈ (0 ,
1) and α ∈ [0 , (cid:96) for any α ∈ [0 , I α ( q ) := (cid:90) (cid:112) s (1 − s ) ∂ q E α ( s, q )(1 + E α ( s, q )) ds. Taking advantage from the symmetry in s , it holds I α ( q ) = (cid:90) / (cid:112) s (1 − s ) (cid:32) ∂ q E α ( s, q )(1 + E α ( s, q )) + ∂ q E α (1 − s, q )(1 + E α (1 − s, q )) (cid:33) ds. ∂ q E α (1 − s, q )could be negative. A sufficient condition for I α ( q ) > ∂ q E α ( s, q )(1 + E α ( s, q )) > − ∂ q E α (1 − s, q )(1 + E α (1 − s, q )) , ∀ s ∈ (0 , / . (19)The convexity of ∂ q E α ( s, q ) in the variable s implies that ∂ q E α ( s, q ) > | ∂ q E α (1 − s, q ) | for any s ∈ (0 , / E α ( s, q ) > E α (1 − s, q ) hence the validity of (19)can not be achieved from the global behaviour of E α and ∂ q E α , rather it deserves a carefully analysis.Consider the function I α ( s, q ) := ∂ q E α ( s, q ) (1 + E α (1 − s, q )) + ∂ q E α (1 − s, q ) (1 + E α ( s, q )) . Clearly, the statement I α ( s, q ) > s ∈ (0 , /
2) is equivalent to (19), then we aim at provingthat I α ( s, q ) is positive. The next lemma allows to replace the power by 2. Lemma 5.1.
Let A ≥ C > , B, D > . If AB − CD > ⇒ AB − CD > . Proof. AB > CD ⇒ A B > C D ⇒ AB > C A D > CD . (cid:4) For those s where ∂ q E α ( q, − s ) ≥ I α ( s, q ) > s where ∂ q E α ( q, − s ) < ∂ q E α ( q, s ) > − ∂ q E α ( q, − s ), then for the lemma, we can replacethe above function I α ( s, q ) by I α ( s, q ) := ∂ q E α ( q, s ) (1 + E α ( q, − s )) + ∂ q E α ( q, − s ) (1 + E α ( q, s )) (20)and we investigate whether the new defined I α ( s, q ) is positive.From (13) it descends, for α ∈ (0 , I α ( q, s ) = C α ( q ) (cid:88) n ≥ ,m ≥ ω αn ω αm (cid:16) (2 − q )( n − m ) − (cid:17) (cid:20) A n ( s ) (1 + E α (1 − s, q )) + A n (1 − s ) (1 + E α ( s, q )) (cid:21) q ( n − m − . (21)Algebraic manipulations lead to the expansion I α ( s, q ) = C α ( q ) (cid:88) p ≥ Q α ( p, s ) q p − (22)where Q α ( p, s ) = (cid:88) n ≥ ,m ≥ n + m = p ω αn ω αm n − m − (cid:20) A n ( s ) (cid:16) E α (1 − s, q ) (cid:17) + A n (1 − s ) (cid:16) E α ( s, q ) (cid:17) (cid:21) − (cid:88) n ≥ ,m ≥ n + m = p − ω αn ω αm ( n − m ) (cid:20) A n ( s ) (cid:16) E α (1 − s, q ) (cid:17) + A n (1 − s ) (cid:16) E α ( s, q ) (cid:17) (cid:21) . (23)Note that the quantities in square brackets are positive and increasing in n and it can be proved thatboth the sums appearing in Q α ( p, s ) are positive for any p ≥
4. However, Q α ( p, s ) has not a definite12ign as q ranges in (0 ,
1) then we can not conclude about the sign of I α ( s, q ) looking at the signof each of the Q α ( p, s )’s. Therefore we have to extrapolate the dependence on q from each of the E α ( s, q ) and E α (1 − s, q ). By substituting the expansion (12) in (21), straightforward computationsgive I α ( q, s ) = C α ( q ) (cid:88) n ≥ ,m ≥ k ,k ≥ ω αn ω αm ω αk ω αk (cid:16) (2 − q )( n − m ) − (cid:17) q ( n − m − k − k − ·· A n ( s ) (cid:16) ω αk − ω αk + (2 − q ) A k (1 − s ) (cid:17) (cid:16) ω αk − ω αk + (2 − q ) A k (1 − s ) (cid:17) + A n (1 − s ) (cid:16) ω αk − ω αk + (2 − q ) A k ( s ) (cid:17) (cid:16) ω αk − ω αk + (2 − q ) A k ( s ) (cid:17) . (24)Note that in the series there is no the term q − . Indeed the only possibility for this term to appearis for the choice ( n, m, k , k ) = (2 , , ,
2) but the first bracket results − q . Also the coefficient of q is equal to zero. Therefore, collecting on the powers of q , we can write I α ( q, s ) = C α ( q ) (cid:88) p ≥ Coef p ( α, s ) q p where, for any p , only a finite number of quart-uple ( n, m, k , k ) contributes to Coef p ( α, s ). Herewe list the first few of these coefficients, for the case α ∈ (0 , Coef ( α, s ) = 19 (2 − α ) (cid:0) (5 + α ) s α − (5 + α ) s + α + 2 (cid:1) Coef ( α, s ) = 118 α (1 − α )(7 − α )(2 − α ) (cid:0) (5 + α ) s α − (5 + α ) s + α + 2 (cid:1) Coef ( α, s ) = 190 a (2 − a ) (1 − a ) (cid:0) a − a − a + 69 (cid:1) s + (cid:0) a − − a + 34 a (cid:1) s + (cid:0) − a + 15 a + 23 a (cid:1) s + (cid:0) − a − a −
413 + 329 a (cid:1) s +21 a − a − a + 156 Coef ( α, s ) = 1540 α (1 − α )(2 − α ) (9 − a ) (cid:0) a − a − a + 207 (cid:1) s + (cid:0) − a + 30 a + 102 a − (cid:1) s + (cid:0) a − a − a + 746 (cid:1) s + (cid:0) − a − a + 377 a − (cid:1) s +23 a − a − a + 188 . The first two coefficients are clearly positive for any s ∈ (0 , /
2) and α ∈ (0 ,
1) and the same itholds for
Coef ( α, s ), Coef ( α, s ). By symbolic computation and numerical visualisation it appearsevident that even for larger p the coefficients Coef p ( α, s ) are positive and, therefore, that I α ( q, s ) >
0. On the other side, although rigorous, the check of the positivity of any large but finite numberof coefficients does not provide a proof for I α ( q, s ) >
0, unless a proof that
Coef p ( α, s ) > p > p (cid:63) is provided.That is exactly what we are presenting in the next section, where the monotonicity of the apsidalangle is proved in the case α = 0. We aim at proving that, for any fixed value of the energy, the apsidal angle ∆ θ ( u ) is monotonicallyincreasing as function of the angular momentum (cid:96) . According to the discussion of the previous13ections, it is enough to prove that the function I ( s, q ) = ∂ q E ( s, q ) (1 + E (1 − s, q )) + ∂ q E (1 − s, q ) (1 + E ( s, q )) (25)is positive for any s ∈ (0 , /
2) and q ∈ (0 , Matlab with the interval arithmetic package
Intlab . [19]. That assures thatthe results we obtain are reliable in the strict mathematical sense.First we show that ∂ q E (1 − s, q ) > s ∈ (0 , /
2) for any q ∈ [0 . , I ( s, q ) > s ∈ (0 , / q ∈ [0 . , Proposition 6.1.
For any q ∈ [0 . , , s ∈ (0 , / , ∂ q E (1 − s, q ) > .Proof. From proposition 3.4, it is enough to show that lim s → ∂ q E ( s, q ) > q ∈ [0 . , F ( q ) := lim s → ∂ q E ( s, q ) = − q − − q ) + 2 − q (1 − q ) log (1 − q ) . Compute F (cid:48) ( q ) = 4(1 − q ) log (1 − q ) + q log(1 − q ) + 2 q (2 − q ) q (1 − q ) log (1 − q ) =: N ( q ) D ( q ) . The denominator D ( q ) is negative for any q ∈ (0 , N ( q ) < − − q ) q + q log(1 − q ) + 2 q (2 − q ) = q (cid:16) log(1 − q ) − q + 6 (cid:17) =: q B ( q ) . The function B ( q ) is decreasing in q and, by rigorous computation it holds B (0 . < − . q ∈ [0 . , .
91] it holds N ( q ) < − .
2. Therefore N ( q ) < F ( q ) is increasingfor any q ∈ [0 . , F (0 . ∈ [0 . , . F ( q ) > q ∈ [0 . , (cid:4) From now on, we restrict our analysis to the case q ∈ (0 , . q the function ∂ q E (1 − s, q ) is not everywhere positive for s ∈ (0 , /
2) then the argument above adopted is notanymore valid to prove that I ( s, q ) > I ( s, q ) = C ( q ) (cid:16) I f ( s, q ) + I ∞ ( s, q ) (cid:17) = C ( q ) (cid:88) p =4 Q ( s, p ) q p − + (cid:88) p ≥ Q ( s, p ) q p (26)and we prove separately, in section 6.1 and 6.2, that the series I ∞ ( s, q ) and the finite sum I f ( s, q )are positive for any s ∈ (0 , /
2) and q ∈ (0 , . Q ( p, s ) = (cid:88) n ≥ ,m ≥ n + m = p n m n − m − (cid:18) A n ( s ) (cid:16) E (1 − s, q ) (cid:17) + A n (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) (cid:18) A n ( s ) (cid:16) E (1 − s, q ) (cid:17) + A n (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) . (27)14t is convenient to express the coefficients Q ( p, s ) in a different way: by a change of index n → n + 1in the first sum, it follows Q ( p, s ) = (cid:88) n ≥ ,m ≥ n + m = p − n + 1 1 m n − m ) (cid:18) A n +1 ( s ) (cid:16) E (1 − s, q ) (cid:17) + A n +1 (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) (cid:18) A n ( s ) (cid:16) E (1 − s, q ) (cid:17) + A n (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) = 12 1 p − − ( p − (cid:18) A ( s ) (cid:16) E (1 − s, q ) (cid:17) + A (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) + (cid:88) n ≥ ,m ≥ n + m = p − n + 1 1 m n − m ) (cid:18) A n +1 ( s ) (cid:16) E (1 − s, q ) (cid:17) + A n +1 (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) (cid:18) A n ( s ) (cid:16) E (1 − s, q ) (cid:17) + A n (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:19) . Collecting all the contributions and reminding the definition (10) it results Q ( p, s ) = (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) (cid:20) K n ( s ) (cid:16) E (1 − s, q ) (cid:17) + K n (1 − s ) (cid:16) E ( s, q ) (cid:17) (cid:21) − p − p − (cid:32)(cid:16) E (1 − s, q ) (cid:17) + (cid:16) E ( s, q ) (cid:17) (cid:33) . (28) The goal of this section is to prove that Q ( s, p ) ≥ s ∈ (0 , / q ∈ (0 ,
1) and p ≥ I ∞ ( s, q ) ≥ s ∈ (0 , /
2) and q ∈ (0 , (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n ( s ) > p − p − , ∀ p ≥ , s ∈ (0 , . A first estimates is the following:
Lemma 6.2.
For any p ≥ (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n ( s ) ≥ (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) n − n + 1 , ∀ s ∈ (0 , . Proof.
Set g ( s ) := (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n ( s ) . p is odd, p = 2 l + 1. Then g ( s ) = l − (cid:88) n = l +1( m =2 l − n ) n m ( n − m ) (cid:16) K n ( s ) − K m ( s ) (cid:17) + 2 l − l − K l − ( s )= l − (cid:88) n = l +1( m =2 l − n ) n m ( n − m ) (cid:16) K n ( s ) − K m ( s ) (cid:17) + 2 l − l − K l − ( s ) + l − l − (cid:16) K l − ( s ) − K ( s ) (cid:17) . Any term ( K n ( s ) − K m ( s )) inside the series is such that m ≥ n ≥ m + 2. By adding andsubtracting equal terms, we obtain K n ( s ) − K m ( s ) = (cid:16) K n ( s ) − K n − ( s ) (cid:17) + (cid:16) K n − ( s ) − K n − ( s ) (cid:17) · · · + (cid:16) K m +1 ( s ) − K m ( s ) (cid:17) , if m > K n ( s ) − K m ( s ) = (cid:16) K n ( s ) − K n − ( s ) (cid:17) + (cid:16) K n − ( s ) − K n − ( s ) (cid:17) · · · + (cid:16) K ( s ) − K ( s ) (cid:17) , if m = 3 . In both the cases, for point iii ) of proposition 7.2 and for remark 7.3, we infer K n ( s ) − K m ( s ) ≥ K n (1) − K m (1) . Again, for point i ) of proposition 7.2, K l − ( s ) ≥ K l − (1)and, for lemma 7.4,2 l − l − K l − ( s ) − l − l − K ( s ) ≥ l − l − K l − (1) − l − l − K (1)for any l ≥ g ( s ) ≥ g (1) = (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n (1) = (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) (cid:20) nn + 1 − (cid:21) . The case p even is completely equivalent. Indeed setting p = 2 l + 2 we obtain g ( s ) = l − (cid:88) n = l +1( m =2 l +1 − n ) n m ( n − m ) (cid:16) K n ( s ) − K m s ) (cid:17) + 2 l − l − K l − ( s ) + 2 l − l K l ( s ) − l − l − K ( s ) . Note that when m = 3 then n = 2 l − p − ≥
6, therefore the term K ( s ) − K ( s ) is not present.Arguing as before, we conclude that g ( s ) ≥ g (1). (cid:4) Define S ( p ) := (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) 1 n + 1 . emma 6.3. S ( p ) is decreasing for any p > and S ( p ) < for any p ≥ .Proof. S ( p ) = p − (cid:88) n =2 (cid:18) p − − n − n (cid:19) n + 1 . Then S ( p ) − S ( p + 1) = p − (cid:88) n =2 (cid:18) p − − n − n (cid:19) n + 1 − p − (cid:88) n =2 (cid:18) p − n − n (cid:19) n + 1= p − (cid:88) n =2 (cid:18) p − − n − p − n (cid:19) n + 1 − (cid:18) − p − (cid:19) p . Since (cid:16) p − − n − p − n (cid:17) > n = 2 . . . , p −
2, it holds S ( p ) − S ( p + 1) ≥ p − p − (cid:88) n =2 (cid:18) p − − n − p − n (cid:19) − p − p ( p − p − (cid:34) p − (cid:88) n =3 p − n − p − (cid:88) n =2 p − n (cid:35) − p − p ( p − p − (cid:18) − p − (cid:19) − p − p ( p −
1) = p − p − p − p > ∀ p > . Then S ( p ) is decreasing for any p >
4. In particular S (11) = − . (cid:4) We are now in the position to prove
Theorem 6.4.
For any s ∈ (0 , and any p ≥ (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n ( s ) > p − p − . Proof.
For lemma 6.2, for any p ≥ (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n ( s ) ≥ (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) n − n + 1 = (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) (cid:18) − n + 1 (cid:19) = (cid:88) n ≥ ,m ≥ n + m = p − (cid:18) m − n (cid:19) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) 1 n + 1 . The first sum gives (cid:88) n ≥ ,m ≥ n + m = p − (cid:18) m − n (cid:19) = p − (cid:88) n =2 (cid:18) p − − n − n (cid:19) = 1 − p − p − (cid:88) n =2 p − − n − p − (cid:88) n =2 n = p − p − (cid:4) orollary 6.5. For any s ∈ (0 , ] and p ≥ Q ( p, s ) ≥ . Proof.
From (28) Q ( p, s ) = (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n ( s ) − p − p − (cid:16) E (1 − s, q ) (cid:17) + (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) K n (1 − s ) − p − p − (cid:16) E ( s, q ) (cid:17) . (29)For theorem 6.4 both the terms are non negative whenever p ≥ (cid:4) We are now concerning the finite part in the splitting (26). We wish to prove that I f ( s, q ) = (cid:88) p =4 Q ( p, s ) q p − ≥ , ∀ s ∈ (0 , / , q ∈ (0 , . . By inserting Q ( p, s ) in the form (27), it results I f ( s, q ) = (cid:88) p =4 (cid:88) n ≥ ,m ≥ n + m = p n m n − m − A n ( s ) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) A n ( s ) q p − (cid:16) E (1 − s, q ) (cid:17) + (cid:88) p =4 (cid:88) n ≥ ,m ≥ n + m = p n m n − m − A n (1 − s ) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) A n (1 − s ) q p − (cid:16) E ( s, q ) (cid:17) = R ( s, q ) (cid:16) E (1 − s, q ) (cid:17) + R (1 − s, q ) (cid:16) E ( s, q ) (cid:17) (30)where R ( s, q ) := (cid:88) p =4 (cid:88) n ≥ ,m ≥ n + m = p n m n − m − A n ( s ) − (cid:88) n ≥ ,m ≥ n + m = p − n m ( n − m ) A n ( s ) q p − .
18f we perform all the computations, we obtain R ( s, q ) = − s + 13 + (cid:18) − s + s (cid:19) q + (cid:18) − s − s + 4310 s (cid:19) q + (cid:18) − s + 3110 + 43 s + 68960 s − s (cid:19) q + (cid:18) − s − s + 4093168 s − s + 17321 s (cid:19) q + (cid:18) s + 3739126 s − s + 3749630 − s − s + 28313630 s (cid:19) q + (cid:18) − s + 4748896300 s − s + 512563 s − s + 43936 s − s (cid:19) q . Note that the coefficient of q is odd in s = 1 / Theorem 6.6.
For any s ∈ (0 , / and q ∈ (0 , . I f ( s, q ) > . Proof.
For any q ∈ (0 ,
1) the function E ( s, q ) is decreasing in s , that is E (1 , q ) ≤ E ( s, q ) ≤ E (0 , q ).We know that R ( s, q ) ≥ q ∈ [0 ,
1] and s ∈ [0 , ] while it could be negative for larger valuesof s . Therefore, I f ( s, q ) > s where R (1 − s ) is positive, otherwise I f ( s, q ) ≥ B ( s, q ) := R ( s, q ) (cid:16) E (1 , q ) (cid:17) + R (1 − s, q ) (cid:16) E (0 , q ) (cid:17) . (31)Hence, it is enough to prove that B ( s, q ) >
0. Since B ( s, q ) is made up a finite number of termsand s, q are within bounded intervals, the problem B ( s, q ) > B ( s, q ) > s ∈ (0 , /
2) and q ∈ (0 , .
1] then we address the case q ∈ [0 . , . R ( s, q ) + R (1 − s, q ) = O ( q ), meaning B ( s, q ) → s when q →
0. This behaviour underlie anobstacle for a rigorous computational scheme to be completely successful: indeed any numerical toolbased on interval arithmetics will definitely result a non-positive output when computing B ( s, q ) atsmall values of q . Therefore, the case when q is small deserves a particular treatment. First computation: q ∈ [0 , / B ( s, q ) ≥ s ∈ [0 , q ∈ [0 , ]. We compute E (1 , q ) := lim s → E ( s, q ) = 2 q − − q ln(1 − q ) E (0 , q ) := lim s → E ( s, q ) = 1 − q − − q (1 − q ) ln(1 − q ) . Using the relation ln(1 − x ) − > − (cid:16) q + q + q + q + q (cid:17) − it follows E (1 , q ) > − q + 190 q − q , ∀ q ∈ (0 , . Similarly, E (0 , q ) < q + 13 q + 13 (cid:0)
58 + 87 q + 8 q + 15 q + 12 q (cid:1) (1 − q ) (60 + 30 q + 20 q + 15 q + 12 q ) q < q + 13 q + 0 . q , ∀ q ∈ [0 , . .
19y inserting the above estimates into (31) B ( s, q ) ≥ R ( s, q ) (cid:16) − q + 190 q − q (cid:17) + R (1 − s, q ) (cid:16) q + 13 q + 0 . q (cid:17) , ∀ q ∈ [0 , . . As expected, the 0 th order term in q vanishes. Let be introduced ˜ R ( s, q ), ˜ E (1 , q ), ˜ E (0 , q ) so that R ( s, q ) = (cid:18) − s (cid:19) + q ˜ R ( s, q ) (cid:16) − q + 190 q − q (cid:17) = 4 + q ˜ E (1 , q ) , (cid:16) q + 13 q + 0 . q (cid:17) = 4 + q ˜ E (0 , q ) . It results B ( s, q ) ≥ q (cid:16) ˜ R ( s, q ) + ˜ R (1 − s, q ) (cid:17) + (cid:0) − s (cid:1) (cid:16) ˜ E (1 , q ) − ˜ E (0 , q ) (cid:17) + q (cid:16) ˜ R ( s, q ) ˜ E (1 , q ) + ˜ R (1 − s, q ) ˜ E (0 , q ) (cid:17) . For the choice ∆( s ) = 0 .
02, ∆( q ) = 0 .
01, define the intervals s j := [( j − s ) , j ∆ s ] , j = 1 , . . . , , q k := [( k − q , k ∆ q ] , k = 1 , . . . , (cid:83) j s j ⊃ [0 , /
2] and (cid:83) k q k ⊃ [0 , / j, k in their range, we compute usinginterval arithmetics the quantities M ( j, k ) =4 (cid:16) ˜ R ( s j , q k ) + ˜ R (1 − s j , q k ) (cid:17) + (cid:18) − s j (cid:19) (cid:16) ˜ E (1 , q k ) − ˜ E (0 , q k ) (cid:17) + q k (cid:16) ˜ R ( s j , q k ) ˜ E (1 , q k ) + ˜ R (1 − s j , q k ) ˜ E (0 , q k ) (cid:17) . (32)It results min j,k (min( M ( j, k ))) > . B ( s, q ) > s ∈ (cid:16) , (cid:17) , q ∈ (cid:16) , (cid:105) . Second computation, q ∈ [0 . , . q ∈ [ , ]. We rigorously compute thelower bound for B ( s, q ), as given in (31), for s ∈ (0 , q ∈ [0 . , . s = 2 · − and ∆ q = 2 · − it results that B ( s, q ) ≥ . (cid:4) We can now state the theorem:
Theorem 6.7.
For any value of the energy E , the apsidal angle ∆ θ ( u ) is monotonically increasingas function of the angular momentum.Proof. From proposition 6.1, corollary 6.5 and theorem 6.6 it follows that I ( s, q ) > s ∈ (0 ,
1) and q ∈ (0 , q ∈ (0 , q = q ( (cid:96) ) is decreasing, it follows that dd(cid:96) ∆ θ ( u ) > , ∀ (cid:96) ∈ (0 , (cid:96) max ) . (cid:4) This Appendix is intended to collect all the technical results concerning the function A n ( s ) and K αn ( s, q ). 20 .1 Properties of A n ( s ) A n ( s ) = 1 − (1 − s ) n − s , n ≥ , s ∈ (0 , Lemma 7.1. i) A ( s ) = 1 , A n (1) = 1 for any n ≥ .ii) A n +1 ( s ) − A n ( s ) = (1 − s ) n − , A n +1 (1 − s ) − A n (1 − s ) = s ( n − .iii) for any n ≥ and s ∈ (0 , , A n ( s ) is decreasing and convex, i.e. A (cid:48) n ( s ) < and A (cid:48)(cid:48) n ( s ) > .iv) for any n > m A (cid:48) n ( s ) < A (cid:48) m ( s ) for any s ∈ (0 , .Proof. i ) , ii ) obvious. iii ) dds A n ( s ) = − − s ) n − + ( n − − s ) ( n − ss . Looking at the numerator, say N ( s ), we see that N (0) = 0, N (1) = − N (cid:48) ( s ) = − ( n − − s ) n − − ( n − n − − s ) n − s + ( n − − s ) ( n − < A n ( s ) is decreasing. d ds A n ( s ) = 2 s − s (1 − s ) ( n − − n − − s ) n − s − ( n − n − − s ) n − s s = 2 − − s ) ( n − − n − − s ) n − s − ( n − n − − s ) n − s s . Looking at the numerator, N ( s ), we note that N (0) = 0 , N (1) = 2 and N (cid:48) = ( n − n − n − − s ) n − > , s ∈ (0 , A n ( s ) is convex. iv ) ddn A (cid:48) n ( s ) = (1 − s ) n − log(1 − s ) + (1 − s ) n − s + ( n − − s ) n − s log(1 − s ) s = (1 − s ) ( n − s (log(1 − s )(1 − s ) + s + ( n − s log(1 − s ))= (1 − s ) ( n − s (log(1 − s ) + s + ( n − s log(1 − s )) < . (cid:4) K αn ( s ) K αn ( s ) = 2 n − αn + 1 A n +1 ( s ) − A n ( s ) , s ∈ (0 , , α ∈ [0 , , n ≥ . Lemma 7.2.
For any α ∈ [0 , and n ≥ :i) K αn ( s ) is decreasing and convex for any s ∈ (0 , . i) K αn +1 ( s ) − K αn ( s ) > for any s ∈ (0 , .For any α ∈ [0 , and n ≥ iii) K αn +1 ( s ) − K αn ( s ) is decreasing for any s ∈ (0 , .Proof. i ) By direct computation the property is proved for n = 2 ,
3. Then, K αn ( s ) = 2( n − α ) n + 1 A n +1 ( s ) − − (1 − s ) n − + (1 − s ) n − (1 − s ) n s (cid:18) n − α ) n + 1 − (cid:19) A n +1 ( s ) + (1 − s ) n − , (33)hence dds K αn ( s ) = (cid:18) n − α − n + 1 (cid:19) A (cid:48) n +1 ( s ) − ( n − − s ) n − < , ∀ s ∈ (0 , , n ≥ . Also, d ds K αn ( s ) = (cid:18) n − α − n + 1 (cid:19) A (cid:48)(cid:48) n +1 ( s ) + ( n − n − − s ) n − > , ∀ s ∈ (0 , , n ≥ .ii ) The case n = 2 gives K α ( s ) − K α ( s ) = − s + s − a + αs − αs that is positivefor any s ∈ (0 ,
1) and α ∈ [0 , K αn +1 ( s ) − K αn ( s ) = 2( n + 1 − α ) n + 2 A n +2 ( s ) − n + 1 − αn + 1 A n +1 ( s ) + A n ( s )= 1( n + 1)( n + 2) [2( n + 1)( n + 1 − α ) A n +2 ( s ) − (3 n + 1 − α )( n + 2) A n +1 ( s ) + ( n + 1)( n + 2) A n ( s )] . Using twice property ii ) of lemma 7.1 we obtain K αn +1 ( s ) − K αn ( s ) = 1( n + 1)( n + 2) · (cid:2) α + 1) A n ( s ) + 2( n + 1)( n + 1 − α )(1 − s ) n − ( n + 3 n − α )(1 − s ) n − (cid:3) . Denote by η αn ( s ) := 2( n + 1)( n + 1 − α )(1 − s ) n − ( n + 3 n − α )(1 − s ) n − . Its derivative ˙ η αn ( s ) =(1 − s ) n − (( n + 3 n − α )( n − − n ( n + 1)( n + 1 − α )(1 − s )) is zero for (1 − s ) = ( n +3 n − α )( n − n ( n +1)( n +1 − α ) where it corresponds the minimum for η αn ( s ). We infer η αn ( s ) ≥ n + 1)( n + 1 − α ) ( n + 3 n − α ) n ( n − n (2 n ) n ( n + 1 − α ) n ( n + 1) n − ( n n − α ) n ( n − n − (2 n ) n − ( n + 1 − α ) n − ( n + 1) n − = − ( n + 3 n − α ) n ( n − n − n − n n ( n + 1 − α ) n − ( n + 1) n − . (34)For any α ∈ [0 ,
1) we can uniformly bound η αn ( s ) ≥ η ( n ) := − ( n + 3 n ) n ( n − n − n − n n ( n ) n − ( n + 1) n − = − ( n + 3) n ( n − n − (2 n ) n − ( n + 1) n − . n ≥
1, and it equals at n = 3. Therefore η αn ( s ) ≥ − for any s ∈ (0 ,
1) and n ≥ α ∈ [0 , A n ( s ) ≥ n and s ∈ (0 , K αn +1 ( s ) − K αn ( s ) ≥ n + 1)( n + 2) (cid:18) − (cid:19) > .iii ) The case n = 4 , n ≥ dds (cid:16) K αn +1 ( s ) − K αn ( s ) (cid:17) = 1( n + 1)( n + 2) · (cid:2) (2 + α ) A (cid:48) n ( s ) − n ( n + 1)( n + 1 − α )(1 − s ) n − + ( n − n n − α )(1 − s ) n − (cid:3) . Denote by ν αn ( s ) := − n ( n + 1)( n + 1 − α )(1 − s ) n − + ( n − n n − α )(1 − s ) n − . Then, arguingas before, we prove that max s ∈ (0 , ( ν αn ( s )) = ( n − n − ( n + 3 n − α ) n − (2 n ) n − ( n + 1) n − ( n + 1 − α ) n − . Therefore, for any α ∈ [0 , ν αn ( s ) ≤ ν ( n ) := ( n − n − ( n + 3 n ) n − (2 n ) n − ( n + 1) n − ( n ) n − = ( n − n − n ( n + 3) n − (2 n ) n − ( n + 1) n − . The function ν ( n ) is decreasing in n and ν (6) <
2. Since A (cid:48) n ( s ) ≤ − s and n , it followsthat K αn +1 ( s ) − K αn ( s ) is decreasing in s for any n ≥ (cid:4) Remark 7.3. K α ( s ) − K α ( s ) is not decreasing in s ∈ (0 , and the minimum is not achieved at s = 1 . However the function K ( s ) − K ( s ) , for s ∈ [0 , attains the minimum at s = 1 . Indeed K ( s ) − K ( s ) = 116 − s + 676 s − s + 53 s and the function has a local minimum and a local maximum at s , s respectively, where s ∈ I =[0 . , . and s ∈ I = [0 . , . . For any s ∈ I , K ( s ) − K ( s ) ∈ [0 . , . > = K (1) − K (1) . Lemma 7.4.
For any n ≥ , the function g n ( s ) := n − n K n ( s ) − n − n − K ( s ) is monotonically decreasing for any s ∈ [0 , .Proof. g n ( s ) = n − n K n ( s ) − n − n − (cid:18) A ( s ) − (cid:19) then, since A (cid:48) ( s ) = −
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