aa r X i v : . [ m a t h . S G ] J u l Fiberwise convexity of Hill’s lunar problem
Junyoung Lee [email protected] of Mathematics and Research Institute of MathematicsSeoul National University
Abstract
In this paper, we prove the fiberwise convexity of the regularized Hill’s lunar problembelow the critical energy level. This allows us to see Hill’s lunar problem of any energy levelbelow the critical value as the Legendre transformation of a geodesic problem on S with afamily of Finsler metrics. Therefore the compactified energy hypersurfaces below the criticalenergy level have the unique tight contact structure on R P . Also one can apply the systolicinequality of Finsler geometry to the regularized Hill’s lunar problem. Studying the motion of the moon has been a challenging problem for a long time. If we consideronly the sun, the earth and the moon, this problem is the three body problem. One can agreethat the three body problem is one of the hardest problems in classical mechanics. For thisreason, many researchers have studied this with some restrictions which depend on the situationof the problem. One problem with reasonable and practical restrictions is the (circular planar)restricted three body problem. The restricted three body problem is obtained by assumingthat two primary particles P , P follow Keplerian circular motion and one massless particle S does not influence these primaries. Namely, the masses of the particles have the relation M , M >> m where M , M are the masses of the two primaries P , P respectively and m isthe mass of S . One can study the motion of the moon in this set-up. However the lunar theoryis a limit case of the restricted three body problem since the sun is much heavier than the othersand the distance between the sun and the earth is much longer than the distance between theearth and the moon. One suggestive formulation for this situation was given by Hill in [8]. Hillintroduced this lunar problem in 1878 to study the stability of the orbit of the moon. This canbe obtained by taking the limit for µ := M M + M in the restricted three body problem. If wetake only µ → of µ when one Precisely, M Sun M Earth ∼ M Earth M Moon ∼ . | P Sun − P Earth || P Earth − P Moon | ∼ µ →
0. We will explain this procedure in section 2.2. After Hill gave a new formulationfor the lunar theory, many researchers have used Hill’s lunar problem to get accurate motion ofthe moon.An important feature on studying Hamiltonian systems is the existence of integrals of thesystem. Integrals of given Hamiltonian system make this problem easier. The extreme caseis a completely integrable system. A Hamiltonian system is called completely integrable if itpossesses the maximal number of Poisson commuting independent integrals. If the dimensionof the phase space is 2 n , then the maximal possible number of Poisson commuting independentintegrals is n . In Hill’s lunar problem case the integrability is equivalent to the existence of asecond integral, which is independent of the first integral. The non-integrability of Hill’s lunarproblem has been determined by some authors. Meletlidou, Ichtiaroglou and Winterberg in [13]proved the analytic non-integrability of Hill’s lunar problem. Morales-Ruiz, Sim´o and Simongave an algebraic proof of meromorphic non-integrability in [15]. Recently, Llibre and Robertoin [11] discussed the C integrability. With these results about non-integrability, Sim´ o andStuchi in their numerical research [17] pointed out the importance of ’numerical methods guidedby the geometry of dynamical systems’ by observing the chaotic behavior in the Levi-Civitaregularization of Hill’s lunar problem. We show in this paper that the Hamiltonian flow of Hill’slunar problem can be interpreted as a Finsler flow. This will provide a geometric feature ofHill’s lunar problem.One difficulty in the study of this problem comes from collision. Namely, this problemhas a singularity at the origin. However, two body collision can be regularized. One way toregularize this problem is Moser regularization. Moser introduced this regularization for theKepler problem in [16]. In this paper, he tells us that the Hamiltonian flow of the Keplerproblem can be interpreted as a geodesic flow on the 2-sphere endowed with its standard metricby interchanging the roles of position and momentum. We will discuss this relation in section2.1. If one replaces the standard metric by a Finsler metric, then this idea can be applied toother problems which admit two body collisions. To get a Finsler metric one needs fiberwiseconvexity. One recent result using this is given in [4]. They prove that the rotating Keplerproblem is fiberwise convex and so can be regarded as the Legendre transformation of the 2-sphere endowed with a Finsler metric. As in the rotating Kepler problem, one can ask whetherHill’s lunar problem has also this property or not. The main theorem of this paper is thefollowing. Theorem 1.1.
The bounded components of the regularized Hill’s lunar problem are fiberwiseconvex for the energy level below the critical value.To understand the meaning of fiberwise convexity below the critical value, we need to seethe Hamiltonian of Hill’s lunar problem. H HLP : R × ( R − { (0 , } ) → R ,H HLP ( q, p ) = 12 | p | − | q | − q + 12 q + p q − p q Here q is the position variable and p is the momentum variable. This Hamiltonian has onecritical value. We can introduce the effective potential U ( q , q ) := − p q + q − q
2n order to see the critical points easily. In fact, we can write the Hamiltonian H HLP ( q, p ) = 12 (( p + q ) + ( p − q ) ) + U ( q , q )using the effective potential. Since the other term is of degree 2, the critical points of H HLP correspond to the critical points of U . It means the correspondence π ( Crit ( H HLP )) =
Crit ( U )where π is the projection to the q -coordinate. We can compute the critical points and the criticalvalue of the effective potential UCrit ( U ) = ( ± − , , U ( ± − ,
0) = − − c . Then one can get the critical point of H HLP
Crit ( H HLP ) = { (3 − , , , − ) , ( − − , , , − − ) } ∈ R ( q ) × R ( p )and the critical value H HLP (( ± − , , , ± − )) = − c . We are interested in the energy level below this critical value − c in order to prove the Theorem1.1. Thus we will assume for the energy level that − c < − c ⇐⇒ c > c = 3 c , we define the Hamiltonians K c K c ( q, p ) = | q | ( H HLP ( q, p ) + c )for the regularization of this problem. It will be proven in section 3 that π ( K − c (0)), the pro-jection of the zero level set of K c to the q -coordinate, has one bounded component and twounbounded components. Let us denote by Σ c the component of K − c (0) which projects to thebounded component. Namely Σ c is a connected component of K − c (0) and π (Σ c ) is bounded.By the symplectomorphism ( q, p ) → ( p, − q ), we can think of p as a position variable and of q as a momentum variable. In this situation p can be regarded as a value in C and so Σ c ⊂ T ∗ C .We can regard T ∗ C as a subset of T ∗ S by the one point compactification of C . Then we canthink of Σ c as a subset of T ∗ S using the stereographic projection. In this situation Theorem1.1 can be rephrased as follows. • ( F1 ) The closure Σ c of Σ c in T ∗ S is a submanifold of T ∗ S for all c > c . • ( F2 ) For any fixed p ∈ S and c > c , Σ c ∩ T ∗ p S bounds a convex region which containsthe origin in the cotangent plane T ∗ p S .In short, the connected component Σ c of the energy hypersurface H − HLP ( − c ) with energy − c < − c can be symplectically embedded into T ∗ S as a fiberwise convex hypersurface aftercompactification. By proving the above statements ( F1 ), ( F2 ), we can show that the regularizedHill’s lunar problem can be regarded as Legendre dual to a geodesic problem in S with Finslermetric. With this definition of fiberwise convexity, we have one obvious Corollary of Theorem1.1. 3 orollary 1.2. The bounded component of the regularized Hill’s lunar problem has a contactstructure for the energy level below the critical value. Moreover, this contact structure is theunique tight contact structure on R P up to contact isotopy.It is clear that fiberwise convexity implies fiberwise starshapedness with respect to the originfor all T ∗ p S . This implies that the restriction of the Liouville 1-form on T ∗ S to Σ c gives acontact form. Since the unit sphere bundle S ∗ S of T ∗ S with respect to any Finsler metric isdiffeomorphic to R P and the contact structure given by the Liouville 1-form is strongly fillable,we have the strongly fillable contact structure on Σ c which has diffeomorphism type R P . Fromthe criterion due to Eliashberg and Gromov in [5] and [7], any symplectically fillable contact3-manifold is tight. Moreover R P admits a unique tight contact structure up to isotopy by theresult of Eliashberg [6].We will prove Theorem 1.1 in section 4 and 5 . As one can see in section 5, by the complexityof computation, it seems hard to take further computations about the corresponding geodesicproblem in spite of our knowledge of the existence of a corresponding Finsler metric. However,this correspondence itself gives us information about closed characteristics. Corollary 1.3.
The Conley-Zehnder indices of the closed characteristics of the regularized Hill’slunar problem below the critical energy level is nonnegative.The Conley-Zehnder indices of the closed characteristics of the Hamiltonian flow includingcollision orbits coincide with the Morse indices of the corresponding geodesics. Therefore, weknow that all closed characteristics of the regularized Hill’s lunar problem have nonnegativeConley-Zehnder indices. Of course, it is well-known that the Conley-Zehnder indices of closedcharacteristics of the unregularized Hill’s lunar problem are nonnegative. Indeed, the Hamilto-nian of the unregularized Hill’s lunar problem is a magnetic Hamiltonian and the Conley-Zehnderindices are nonnegative for any magnetic Hamiltonian. However, this result is new for collisionorbits. Moreover, thanks to the result in [1] using the systolic inequality, we can ensure theexistence of a closed characteristic whose action is less than a volume related constant. We referthe following Theorem.
Theorem 1.4. ( ´Alvarez Paiva-Balacheff-Tsanev) There exists a constant k ( S ) > ⊂ T ∗ S bounding a volume V carries a closed character-istic whose action is less than k ( S ) √ V . Here k ( S ) does not depend on Σ.The volume V in here is the Holmes-Thompson volume that is the symplectic volume withthe canonical symplectic form in the cotangent bundle. This coincides with the contact volumeof Σ c with the canonical contact form α := λ | Σ c where λ is the Liouville one form of T ∗ S by Stokes’ Theorem. Moreover, it is known that the constant k ( S ) is less than √ π andthis constant is independent of Σ. In [1], they explained the beautiful relationship betweencontact and systolic geometry which allows to extend the result of Gromov and Croke in systolicinequality on Riemannian manifolds. As an application of this Theorem, we can formulate thefollowing Corollary. Corollary 1.5.
The regularized Hill’s lunar problem has at least one periodic orbit, possibly acollision orbit, whose action is less than k ( S ) q V ol (Σ c ).One interesting question is what we can get from systolic geometry for our practical Hamil-tonian problems which have contact structures. In particular, one can ask how the systolic4apacity changes under a perturbation of the Hamiltonian, because Hill’s lunar problem is alimit case of the restricted three body problem. Hopefully, if one can answer this question, thenone might get insight into the restricted three body problem using this information and methodin the proof. Definition 1.1.
A contact form on a compact 3-manifold is called dynamically convex, if itscontractible periodic Reeb orbits of positive period have Conley-Zehnder index greater than 2with respect to any filling disk.As a goal on the Hill’s lunar problem, I want to mention dynamical convexity. A motivationof this paper is showing the dynamical convexity of the double cover of Hill’s lunar problem. Itis known that the rotating Kepler problem is dynamically convex and the restricted three bodyproblem is also dynamically convex for some mass ratio and energy. It is still unknown if theHill’s lunar problem is dynamically convex. Because we know an energy hypersurface of theregularized Hill’s lunar problem is tight R P , its double cover has the contact structure of theunique tight structure on S . If the double cover of Hill’s lunar problem is dynamically convex,then this double cover allows a disk-like global surface of section for the Hamiltonian vectorfield. This will simplify the problem dramatically. We will see the related result in section 2.2. Acknowledgements : I thank Urs Frauenfelder and Otto van Koert for encouragements anddiscussions. I am also grateful to the colleagues in Augsburg university for their helps thatmake me adapt well to the new surroundings in Augsburg. I wish to express my thanks to thereferee of this article for many helpful comments. This research is supported by DFG-CI 45/6-1:Algebraic Structures on Symplectic Homology and Their Applications.
It is based on Moser regularization in [16] to understand why the fiberwise convexity is helpfulto study Hill’s lunar problem in Hamiltonian dynamics. Moser regularization tells us the planarKepler problem can be compactified to the geodesic problem on the standard 2-sphere. Thisargument can be improved for the case of a fiberwise convex hypersurface which corresponds tothe geodesic problem on a 2-sphere with Finsler metric. On the other hand, we need to know howHill’s lunar problem can be derived from the restricted three body problem. Since Hill’s lunarproblem is a limit of the restricted three body problem, they have relationships with each other.For example, Meyer and Schmidt in [14] show that any non-degenerate periodic solution of Hill’slunar problem whose period is not a multiple of 2 π can be lifted to the three body problem.This could be proven by observing the derivation of Hill’s lunar problem from the three bodyproblem. Thus, understanding the relation between Hill’s lunar problem and the restricted threebody problem will be helpful to get some ideas for the restricted three body problem from theresult of Hill’s lunar problem. For example, fiberwise convexity of the restricted three bodyproblem is still open. We will review Moser regularization of the Kepler problem in section 2.1and the restricted three body problem with its relation to Hill’s lunar problem in section 2.2.5 .1 Kepler problem and Moser regularization The differential equation of the Kepler problem is given by d qdt = − q | q | using some normalization. Therefore the potential function V : ( R ) ∗ → R is V ( q ) = − | q | andthis induces the Hamiltonian of the Kepler problem as the total energy. H : R × ( R ) ∗ → R , H ( p, q ) = 12 | p | − | q | However, this is not so practical to analyze by geometric methods because this Hamiltonian hasa singularity at q = 0. One preferred way to remove this singularity is Moser regularization. Forconstant c ∈ R , we define the Hamiltonian K c ( p, q ) := | q | ( H ( p, q ) + c )Then we can easily see that K c has no singularity and H − ( − c ) = K − c (0). However, these twoHamiltonian dynamics on this level set arising from H and K c are not exactly same, but havethe same Hamiltonian flow up to time reparametrization. We introduce a new time parameter s = R dt | q | for K c to make these equivalent problems.We briefly explain Moser’s paper [16] which shows that this regularized Kepler problem isequivalent to the geodesic problem on standard 2-sphere. We consider the energy level − c = − . K ( p, q ) = 12 | q | ( | p | + 1) − ⇒ K − (0) = { ( p, q ) ∈ R × ( R ) ∗ |
12 ( | p | + 1) | q | = 1 } Other energy levels can be proven analogously by simple rescaling. Note that ( p, q ) ( q, − p )is symplectic and in our case this seems like interchanging the role of p and q . We can see that ( | p | + 1) | q | = 1 comes from energy hypersurface F ( x, y ) = 1 of T ∗ S where F ( x, y ) = | y | round the Hamiltonian for free particle via the stereographic projection. The flow of the Hamiltonianfor a free particle is the geodesic flow in general. Therefore the Hamiltonian flow of the Keplerproblem corresponds to the geodesic problem on S with the round metric. Precisely, Moserproved the following Theorem for the Kepler problem in n -dimensional space. Theorem 2.1 (Moser) . For a negative energy − c <
0, the energy hypersurface H − ( − c ) canbe mapped bijectively into the unit tangent bundle of S n − { north pole } . Furthermore, theflow defined by the Kepler problem is mapped into the geodesic flow on the punctured sphere S n − { north pole } .The above argument can be extended to the fiberwise convex case. In the case of Keplerproblem case, amazingly, the trajectory of q for fixed position p ∈ S is exactly unit circle in thecotangent space T ∗ p S with the round metric. Thus, if an energy hypersurface of a Hamiltoniansystem is the unit cotangent bundle S ∗ g S of a metric g , then the Hamiltonian system on thishypersurface corresponds to the problem of geodesic on S with the metric g . Moreover, if aproblem has a level set trajectory of q which encircles the convex region containing the originfor any p ∈ S , then this will be the geodesic problem on S with Finsler metric by defining theposition of q in T ∗ p S to be the unit length. Therefore, we set up ( F1 ) to determine whetherthe energy hypersurface of Hill’s lunar problem can be seen as a submanifold in T ∗ S afterregularization and changing the role of q and p . Moreover, we set up ( F2 ) to determine if thehypersurface can define a Finsler metric on T ∗ S .6 .2 The restricted three body problem, the rotating Kepler problem andHill’s lunar problem We can derive the time-independent Hamiltonian of the restricted three body problem by in-troducing rotating coordinates with unit angular velocity. It is important to understand howone can derive Hill’s lunar problem from the restricted three body problem not only to de-cide which problem can be effective with Hill’s setup, but also to get intuitions to know closedcharacteristics of the restricted three body problem from Hill’s lunar problem.First, we explain the derivation of the Hamiltonian for the restricted three body problembriefly. We denote the masses M , M of two primaries P , P . We define µ = M M + M andassume that two primaries have the following motion. P ( t ) = ( − µ cos t, − µ sin t ) , P ( t ) = ((1 − µ ) cos t, (1 − µ ) sin t )We are interested in the motion of a massless particle S ( t ) ∈ R − { P ( t ) , P ( t ) } and we caneasily derive the Hamiltonian H i ( t, q i , p i ) = 12 | p i | − µ | q i − P ( t ) | − − µ | q i − P ( t ) | for the restricted three body problem in inertial system. We put index i to emphasize thatthis Hamiltonian is taken in the inertial system. Note that H i is time-dependent. Now weconsider the rotating system to make this Hamiltonian become time-independent. We expressthe positions of P and P A := ( − µ, , A := (1 − µ,
0) = ⇒ P ( t ) = R t A , P ( t ) = R t A by the rotation R t = (cid:18) cos t − sin t sin t cos t (cid:19) of two fixed points A and A , respectively. We define therotation Ψ t := R t ⊕ R t = cos t − sin t t cos t t − sin t t cos t on T ∗ R = R × R . We can find the following Theorem in many books, for example, see [10]. Theorem 2.2.
Let H r be the Hamiltonian in a rotating system which rotate by Ψ t . Then H r = H i ◦ φ tK − K where K = q p − q p and φ tK are Hamiltonian diffeomorphisms generatedby K . In particular H r is autonomous.We have a time-independent Hamiltonian H µ : R × ( R − { A , A } ) → R ,H µ ( p, q ) = 12 | p | − µ | q − A | − − µ | q − A | + p q − p q for the restricted three body problem of mass ratio µ in the rotating coordinates. Also, we canget this equivalent Hamiltonian H µ : R × ( R − { (0 , , (1 , } ) → R , µ ( p, q ) = 12 | p | − µ | q − (1 , | − − µ | q | + p q − p q − µp by translating in q -coordinates.Many important studies of global properties of the restricted three body problem have beendone in the study of this time-independent Hamiltonian using symplectic geometry. Recentlythere was a remarkable result [2] which tells us the existence of a disk-like global surfaces ofsection for Hamiltonian vector field in the restricted three body problem for µ ∈ ( µ ( c ) ,
1) where − c is the energy below the first Lagrange value. We mention the definition of the disk-like globalsurface of section. Definition 2.1.
Let Σ be a smooth 3-manifold with a nowhere vanishing vector field X . Aglobal disk-like surface of section for X consists of a embedded closed disk D ∈
Σ having thefollowing properties:1. The boundary ∂ D is a periodic orbit, called the spanning orbit.2. The interior int ( D ) of the disk is a smooth submanifold of Σ and is transversal to the flow.3. Every orbit except the spanning orbit intersects int ( D ) in forward and backward time.One can easily recognize from this definition that a global disk-like surface of section reducesthe study of the dynamics on a 3-manifold to the study of the return map on the disk. The resultabout the existence of a global disk-like surface of section for the restricted three body problemin [2] based on the result of Hofer, Wysocki and Zehnder [9] which uses a pseudoholomorphiccurve theory for an energy hypersurface in R . In [9], they prove that strict convexity of anenergy hypersurface implies dynamical convexity and dynamical convexity implies the existenceof global disk-like surfaces of section. As an application of this theory, in [2], they found pairsof ( µ, c ) where the energy hypersurfaces K − µ,c (0) of the regularized Hamiltonian for such pairsbound a strictly convex region. We define the regularized Hamiltonian K µ,c for the precisestatement in [2]. We introduce the Levi-Civita coordinates ( u, v ) using a 2:1 symplectic map,up to a constant factor, q = 2 v , p = uv and apply to H µ . K µ,c ( u, v ) := | v | ( H µ ( u, v ) + c ) = 12 | u | + 2 | v | < u, iv > − µIm ( uv ) − − µ − µ | v | | v − | + c | v | The energy hypersurface K − µ,c (0) coincides with the energy hypersurface ( H µ ) − ( − c ). Thisimplies the Hamiltonian flows φ tK µ,c , φ tH µ are same up to time reparametrization. We state theresult in [2]. Theorem 2.3 (Albers-Fish-Frauenfelder-Hofer-van Koert) . Given c > , there exists µ = µ ( c ) ∈ [0 ,
1) such that for all µ < µ < K − µ,c (0) with its Reeb vector field.One can ask the same question for the limit problems of the restricted three body problem.In [3], they give the answer for the rotating Kepler problem. The rotating Kepler problem isdynamically convex after Levi-Civita regularization for each energy below the critical value ofthe Jacobi energy. Thus, the energy hypersurfaces will have global surfaces of section for theHamiltonian vector field. Since they also proved the failure of strict convexity in [3], the proofis entirely different from the proof in [2]. Instead of using theory in [9], they observed every8eriodic orbit and computed the Conley-Zehnder indices. One main idea in the computation ofthe indices comes from the fiberwise convexity of the rotating Kepler problem. They regardedperiodic orbits of the rotating Kepler problem as periodic Finsler geodesics and used localstability of Morse homology. On the other hand, we do not know the existence of a globalsurface of section for Hill’s lunar problem.Now let us explain briefly the derivation of Hill’s lunar problem. We will borrow the simplederivation from [12]. Apply a coordinate transformation on the Hamiltonian H µ by translating p, q -coordinates in the following way. q → q + 1 − µ, q → q , p → p , p → p + 1 − µ. We have the Hamiltonian H µ ( p, q ) = 12 | p | − µ | q | − − µ | q + (1 , | + p q − p q − (1 − µ ) q up to constant. By Newton’s binomial series (1+ x ) − = 1 − x + x + · · · , we get the expansion − − µ p ( q + 1) + q = − (1 − µ )(1 − q + q − q + · · · )and we apply this on H µ H µ ( p, q ) = 12 | p | − µ | q | + p q − p q − (1 − µ )( q − q + · · · ) . Consider the scaling q → µ q, p → µ p which is symplectic with conformal coefficient µ − . Wemultiply this factor µ − H µ ( µ p, µ q ) = H HLP ( p, q ) + O ( µ ) , then we obtain the Hamiltonian H HLP ( p, q ) = 12 | p | − | q | + p q − p q − q + 12 q for Hill’s lunar problem by taking µ → From now on, we will concentrate on Hill’s lunar problem. Hence we will denote simply by H the Hamiltonian H HLP of Hill’s lunar problem. We showed that H has unique critical value − c := − in section 1. We want to show fiberwise convexity for all − c < − c . We define theHamiltonian K c ( q, p ) := | q | ( H ( q, p ) + c )for the regularization of this problem. The Hamiltonian flow of K c on K − c (0) coincides withthe Hamiltonian flow of H on H − ( − c ). Moreover, K c has no singularity. Observe the structure9f K − c (0).( q, p ) ∈ K − c (0) ⇐⇒
12 (( p + q ) + ( p − q ) ) = 1 p q + q + 32 q − c ⇐⇒ √ q + q + q = b ≥ c ( p + q ) + ( p − q ) = 2( b − c )We introduce polar coordinates q = r cos θ, q = r sin θ , then √ q + q + q = b ⇐⇒ cos θr +1 = br . We can see the structure of the set { ( q , q ) ∈ R | √ q + q + q = b } from the followingLemma. Lemma 3.1.
For b > c = , the polar equation (cos θ ) r + 1 = br consists of one boundedclosed curve and two unbounded curves. Moreover, if we denote the bounded component of (cos θ ) r + 1 = br by σ b , then σ b is contained in the inside of σ c for any b > c > c . Proof.
Let f b,θ ( r ) = (cos θ ) r − br +1 be polynomial for fixed b and θ . For θ = π , π , we observethe values f b,θ ( −∞ ) = −∞ , f b,θ (0) = 1 > f b,θ ( q b θ ) = 1 − b q b θ < − c q c = 0and f b,θ (+ ∞ ) = + ∞ . This implies f b,θ has one negative zero and two different positive zerosfor any fixed b and for any θ = π , π . Moreover, one can see that the larger positive zero goesto infinity as the term cos θ goes to 0. Since the zeros vary continuously with θ , the smallerpositive zero goes to b as the term cos θ goes to 0. This proves that cos θr + 1 = br consistsof one closed curve and two unbounded curves.We define the positive smaller zero r b,θ of the degree 3 polynomial f b,θ for θ = π , π and r b, π = r b, π = b . Then we have f b,θ ( r b,θ ) = 0 and r b,θ < q b θ by above computation. Wedifferentiate f b,θ ( r b,θ ) = 0 with respect to b
92 (cos θ ) r b,θ dr b,θ db = r b + b dr b,θ db = ⇒ dr b,θ db = r b,θ (cos θ ) r b,θ − b . Since r b,θ < q b θ , dr b,θ db <
0. This implies the bounded component is getting smaller as b increases. This proves the Lemma.We can see that the projection to q -coordinate π ( K − c (0)) of K − c (0) consists of one boundedcomponent and two unbounded components for c > c and the bounded component of π ( K − c (0))is enclosed by the closed curve σ c . We will focus on the case where q is in this bounded componentand so denote the bounded component of π ( K − c (0)) by R c . We define the subsetΣ c = { ( q, p ) ∈ K − c (0) | q ∈ R c } of K − c (0). As in Moser regularization, we regard p as a position variable and q as a momentumvariable by using the symplectomorphism ( q, p ) ( p, − q ). Then we can regard Σ c as a subsetof T ∗ C where p ∈ C is a position variable and q is a momentum variable. We will prove thatthere exist ( p, q ) ∈ Σ c for any p ∈ C and such q ’s form a closed curve in T ∗ p C in the followingLemma. 10 emma 3.2. For any fixed c > c , the projection pr : Σ c → C , pr ( p, q ) = p is surjective.Moreover, the fiber pr − ( p ) at p is a closed curve that encloses the origin for any p ∈ C . Proof.
We can give an easy geometric interpretation for the subsetΣ c = { ( q, p ) ∈ K − c (0) | q ∈ R c } = { ( q, p ) ∈ T ∗ C | q − q + 1 | q | = p q − p q + 12 | p | + c, q ∈ R c } of T ∗ C . If we fix the variable p , then the set { q ∈ T ∗ p C | q − q + | q | = p q − p q + | p | + c } canbe seen as the intersection of the graphs of functions f ( q , q ) = q − q + | q | and g p,c ( q , q ) = p q − p q + | p | + c . Note that the function g p,c of q is a linear function for any fixed p, c andso its graph is a plane. We have f ( q ) > g p,c ( q ) for q with sufficiently small | q | . For fixed q , wealso get f ( q ) > g p,c ( q ) when q → ±∞ . On the other hand, we have the inequality g p,c ( ± − , q ) − f ( ± − , q )= 12 ( p + q ) + 12 p ∓ − p − − − q + 3 − ) + c>
12 ( p + q ) + 12 ( p ∓ − ) + 3 − − − − − = 12 ( p + q ) + 12 ( p ∓ − ) ≥ g p,c > f along the lines q = 3 − for any p, c . Thus the intersection consists of twounbounded components lying in the regions of q > − and q < − − , respectively, and onebounded component lying in − − < q < − . Since the plane does not pass the critical points,that component is a one dimensional submanifold and the topology is same for any p, c . Thus weknow this q -bounded component is a closed curve by observing the case where c is sufficientlylarge. Also, we know this closed curve encloses the origin because f > g p,c near the origin forany p, c . This proves Lemma 3.2.Using Lemma 3.2, we can interpret pr : Σ c → C as a fiber subbundle of T ∗ C with fiber acircle. By one point compactification, we can think of C ⊂ S and also Σ c ⊂ T ∗ C ⊂ T ∗ S usingstereographic projection as in Moser regularization. If every fiber in the cotangent plane boundsa convex region, which contains the origin, then we can think of Σ c as a unit cotangent bundleof some Finsler metric. As a result, its Hamiltonian flow can be interpreted as the geodesic flowon S for a Finsler metric. We formulated two statements ( F1 ), ( F2 ) which are equivalent toTheorem 1.1. For ( F1 ), we have to show that the closure Σ c is a submanifold of T ∗ S . Theproblem of being a submanifold can occur only at the north pole. That is, we have to check ifit has a unique limit in T ∗ S when | p | goes to infinity. This can be verified by looking at thefiber when | p | → ∞ . Let us use the notations in Lemma 3.2. Since q lies on the bounded set, g p,c ( q ) goes to infinity when | p | → ∞ for any c . Thus, if q ν is a sequence in the bounded regionsatisfying f ( q ν ) = g p ν ,c ( q ν ) for p ν → ∞ , then q ν →
0. Therefore the equation f ( q ) = g p,c ( q )converges to the equation | q | = | p | + c which is the equation of the Kepler problem and sothe limit at the north pole in any direction will correspond to the circle with radius √− c of theround metric. Therefore the closure Σ c in T ∗ S is a subbundle over S and this proves ( F1 ).11 ! ! ! ! ! Figure 1: Hill’s region: The bounded partMoreover, we have convex fiber at the north pole. Thus, from now on we can regard p ∈ S asan element of C ∼ = R when we discuss ( F2 ), because we have already proved ( F2 ) at the northpole.We investigate the region that q can lie on. We will call this region Hill’s region and willdenote it by R . By Lemma 3.1 and 3.2, we get the region R := [ c>c R c = [ c>c π ( H − ( − c )) b = { ( q , q ) ∈ R | p q + q + 32 q > c , | q | < − , | q | < · − } where X b means the bounded component of X . It is illustrated as the bounded region enclosedby two curves in Figure 1.Since the coordinate change of a cotangent bundle induced from a coordinate change onits base manifold is linear on each cotangent space and a linear map preserves the convexity,Σ c ∩ T ∗ p ( S ) bounds a strictly convex region if and only if Σ c ∩ T ∗ p C bounds a strictly convexregion under the stereographic projection for every p ∈ S \{ north pole } . Because we havealready proved the strict convexity of the fiber at the north pole, we can reduce the problem toΣ c and we can regard Σ c as a fiber bundle over C for a fixed energy level c > c . For any p ∈ C ,the fiber F c,p = { q ∈ R | ( p, q ) ∈ Σ c } of this bundle is a closed curve. Then we want to show thatthis fiber bounds a strictly convex region which contains the origin. The fact that this enclosesthe origin is already proved in Lemma 3.2.If we define K c,p : R → R by K c,p ( q ) := K c ( q, p ), then we need to prove that the boundedcomponent of K − c,p (0) bounds a strictly convex region for every fixed p ∈ R and c > c . Since12 c and H c have the same 0 energy hypersurface, this is equivalent to prove that the boundedcomponent of H − c,p (0) bounds a strictly convex region for every p ∈ R and c > c where H c,p ( q ) = H c ( q, p ). If the Hessian HessH c,p ( q ) of H c,p is positive definite, then its level curvebounds strictly convex region. However, this is not true and in fact we will see HessH c,p ( q ) hasone positive eigenvalue and one negative eigenvalue. Thus we have to consider the tangentialHessian in order to check the strict convexity of the level curves. The tangential Hessian meanssimply the restriction of Hessian to the tangent space of the energy level set. Because level setsof H c,p are curves, it suffices to show the positivity only for one nonzero tangent vector of H − c,p and this tangent vector can be obtained simply by rotating the gradient vector 90 degrees. Wecan state ( F2 ) numerically by the following Theorem. Theorem 3.3.
Suppose q lies on a fiber at p in Σ c , equivalently q ∈ R ∩ H − c,p (0) for c > c and p ∈ R . Then the inequality( J ▽ H c,p ( q )) t HessH c,p ( q )( J ▽ H c,p ( q )) > J = (cid:18) − (cid:19) is π rotation.Therefore we can reduce our problem into an inequality problem with some constraints.Moreover, we do not need to care about the fiber bundle structure. Namely, it suffices to showthat the inequality (( J ▽ H c,p ( q ))) t HessH c,p ( q )( J ▽ H c,p ( q )) > q, p ) instead ofseeing the bounded component of H − c,p (0) for fixed p . We devote the remaining part of thispaper to the proof of Theorem 3.3. Remark 3.1.
There are symmetries of Hill’s lunar problem. The reflections R : ( q , q , p , p ) ( − q , q , p , − p ) , R : ( q , q , p , p ) ( q , − q , − p , p )are anti-symplectic. H HLP is invariant under these maps, namely H HLP ( R i ( q, p )) = H HLP ( q, p )for all ( q, p ) ∈ R and i ∈ { , } . These will allow us to concentrate only on the first quadrantof q -coordinate. The proof of Theorem 3.3 consists of many complicated computations and notations. In section4.1, we will introduce the necessary notations and derive Theorem 4.3 which is stronger thanTheorem 3.3. We apply many elementary methods to prove Theorem 4.3. The list of Proposi-tions and Lemmas will be given and we will explain their relations and meanings in section 4.2.As one can immediately see in Theorem 3.3, the dimension of parameters for this problem is4. Through computations, we try to reduce this dimension by building lower bounds or findingthe subset where the minimum is attained. When we achieve the reduction to 1 dimensionalproblem, we will provide computer plots of the graph to determine if the final term is positive ornegative. The computer plots will be rigorously verified with a computer program in AppendixA.1. The hardest part of this proof occurs near the critical point because the tangential Hessiangoes to 0 near the critical point. We will use the blow up coordinates to overcome this problem.13 .1 Preparation for the proof of Theorem 3.3.
We do not need to prove p = 0 case separately, because this case will be covered by the generalcase, we will prove this case in order to introduce notations and to help understanding.We compute the gradient and Hessian ▽ H c, ( q ) = − q + q | q | q + q | q | ! , HessH c, ( q ) = 1 | q | (cid:18) − | q | + | q | − q − q q − q q | q | + | q | − q (cid:19) of H c, ( q ) = − q + q − | q | + c when p = 0. As we discussed before, we will see the tangentialHessian and so we need a tangent vector of the level curve. For the notational convenience, wedefine v ( q ) ∈ T q H − c, (0) and H ( q ) for q ∈ H − c, (0) as follows. v ( q ) := J ▽ H c, ( q ) = q + q | q | q − q | q | ! where J = (cid:18) − (cid:19) H ( q ) := HessH c, ( q ) = 1 | q | (cid:18) − | q | + | q | − q − q q − q q | q | + | q | − q (cid:19) Then we can express the tangential Hessian ( J ▽ H c,p ( q )) t HessH c,p ( q )( J ▽ H c,p ( q )) as a function v ( q ) t H ( q ) v ( q ) = 1 | q | h q ( − | q | + | q | − q )(1 + | q | ) − q q (1 + | q | )(2 | q | − q ( | q | + | q | − q )(2 | q | − i of variable q . As we discussed in section 3, the curves K − c, (0) bound strictly convex domainsif and only if the inequalities v ( q ) t H ( q ) v ( q ) > q ∈ H − c,p (0). Therefore, we have toshow the following ’Warm-up Lemma’ in order to prove the case p = 0. Lemma 4.1 (Warm-up Lemma) . Suppose q ∈ R ∩ H − c, (0) for c > c . Then the tangentialHessian of H c, at q is positive definite, namely the inequality v ( q ) t H ( q ) v ( q ) > q ∈ R ∩ H − c, (0) and for every c > c . Proof.
If we take q ∈ H − c, (0), then q satisfies the equation q − q + 1 | q | = c > c . This implies the inequality | q | + 1 | q | > c ⇐⇒ | q | − c | q | + 1 > | q | . We have that | q | is less than the smallest positive zero, say α , of the polynomial x − x + 1 and so | q | < α < .
54. Thus it suffices to prove the function v ( q ) t H ( q ) v ( q ) ispositive for every | q | < .
54. We have the expression of v ( q ) t H ( q ) v ( q ) v ( q ) t H ( q ) v ( q ) = 1 | q | − q | q | − q q | q | − q | q | + 4 q − q .45 0.50 0.55 0.60100200300400 Figure 2: Graph of f ( x ) = x − x −
274 1 x − x shows that it is positive on [0, 0.54].in terms of q . Since the following inequalities3 q | q | + 3 q | q | ≤ q + 3 q | q | = 3 | q | , q q | q | ≤
274 1 | q | hold for all | q | < .
54. We get the following estimate. v ( q ) t H ( q ) v ( q ) ≥ | q | − | q | −
274 1 | q | − | q | As we can see in the graph of y = x − x −
274 1 x − x in Figure 2, we have x − x −
274 1 x − x > x ∈ (0 , . v ( q ) t H ( q ) v ( q ) > | q | < .
54 and this proves Lemma4.1.Now we consider general p ∈ R . We recall the function H c,p : R → R H c,p ( q ) = 12 | p | + p t J q − q + 12 q − | q | + c defined for each p ∈ R and c > c . We calculate the gradient, tangent vector and Hessian of H c,p . ▽ H c,p ( q ) = − q + q | q | − p q + q | q | + p ! , J ▽ H c,p ( q ) = q + q | q | + p q − q | q | + p , ! = v ( q ) + pHessH c,p ( q ) = HessH c, ( q ) = 1 | q | (cid:18) − | q | + | q | − q − q q − q q | q | + | q | − q (cid:19) = H ( q )for every q ∈ H − c,p (0). We can express the tangential Hessian( J ▽ H c,p ( q )) t HessH c,p ( q )( J ▽ H c,p ( q )) = ( v ( q ) + p ) t H ( q )( v ( q ) + p )using v ( q ) and H ( q ). We can rewrite Theorem 3.3 with this notations.15 heorem 4.2. Suppose that q ∈ R ∩ H − c,p (0) for p ∈ R and c > c . Then the inequality( v ( q ) + p ) t H ( q )( v ( q ) + p ) > p, q and c in Theorem 4.2. In particular,it is difficult to describe the trajectory of q for a fixed p and for some c > c . However,the corresponding p to a fixed q ∈ R form a disk with center ( − q , q ). From the followingequivalences q ∈ R ∩ H − c,p (0) for some c > c ⇐⇒ ( q, p ) ∈ H − c (0) for some c > c ⇐⇒ √ q + q + q = b > c and( p + q ) + ( p − q ) < b − c ) , we have the set { p ∈ R | q ∈ R ∩ H − c,p (0) for some c > c } = { p ∈ R | ( p + q ) +( p − q ) <
2( 1 p q + q + 32 q − c ) } of p for a fixed q with a simple inequality. We introduce new variables w ( q ) , s obtained bytranslations. If we set s := p + J q , then we can simplify the equation as follows.12 | p | + p t J q − q + 12 q − | q | + c = 0 ⇐⇒ | s | = 3 q + 2 | q | − c. This induces the following equivalent condition | s | < q + 2 | q | − c ⇐⇒ q ∈ H − c, − Jq + s (0) for some c > c for being a point of trajectory. With this substitution, we define the vector w ( q ) = v ( q ) − J q and we have that v ( q ) + p = v ( q ) − J q + s = q | q | q − q | q | ! + s =: w ( q ) + s, ( v ( q ) + p ) t H ( q )( v ( q ) + p ) = ( w ( q ) + s ) t H ( w ( q ) + s )where w ( q ) = q | q | q − q | q | ! . Theorem 4.2 has the following stronger statement. Here ’stronger’means that | s | < q + | q | − c is replaced by | s | ≤ q + | q | − c and it will be helpful forour argument. Theorem 4.3.
We define w ( q ) = q | q | q − q | q | ! , H ( q ) = | q | (cid:18) − | q | + | q | − q − q q − q q | q | + | q | − q (cid:19) for q ∈ R = { ( q , q ) ∈ R | √ q + q + q > c , | q | < − , | q | < · − } . Then the inequality( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R and | s | ≤ q + | q | − c .It is suffices to prove Theorem 4.3 for the proof of our main Theorem. This is nothing butan inequality problem with constraints and restrictions of parameters. At this moment we have4 dimensional parameters. We will explain the strategy to reduce the dimension in the nextsection. 16 tep1Step2Step3 ! ! ! ! ! ! Figure 3: Partition of R by radius In this section, we discuss the strategy for the proof of Theorem 4.3. First, we divide Theorem4.3 into the following three steps. • Step 1 : ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R ∩ B . (0) and | s | ≤ q + | q | − c . • Step 2 : ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R ∩ ( B . (0) \ B . (0)) and | s | ≤ q + | q | − c . • Step 3 : ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R \ B . (0) and | s | ≤ q + | q | − c .Here, B r (0) is the open disk with center the origin and radius r . This division of steps is visu-alized in Figure 3. Note that the radii 0 . , .
63 are taken only for computational convenience.Obviously, these three steps imply Theorem 4.3.
Step 1 can be proven directly by using simpleestimates. In fact, this can be proven by the following simple argument.
Proposition 4.4.
We have the following estimate.( w ( q ) + s ) t H ( q )( w ( q ) + s ) ≥ r −
103 1 r −
296 1 r + 4 r − r −
2( 1 r − r ) r r + 2 r − − (2 + 2 r )(3 r + 2 r − )for every q ∈ R ∩ B . (0) where r = | q | . Moreover, the inequality1112 1 r −
103 1 r −
296 1 r + 4 r − r −
2( 1 r − r ) r r + 2 r − − (2 + 2 r )(3 r + 2 r − ) > r ∈ (0 , . Step 2 and
Step 3 because of the behaviors of w ( q ), q q + | q | − c and H ( q ) near the critical point.One can see that w ( q ) and q q + | q | − c go to zero and H ( q ) goes to (cid:18) − (cid:19) as q goesto the critical point (3 − , w ( q ) and q q + | q | − c but also the direction of w ( q ) to see that the tangential Hessians( w ( q ) + s ) t H ( q )( w ( q ) + s ) are positive for all | s | < q q + | q | − c . We interpret Theorem 4.3.as a minimum value problem for the parameter s . Namely, we have the following equivalentform. ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R and | s | ≤ q + 2 | q | − c ⇐⇒ min | s | ≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R . We can concentrate only on the first quadrant of R including axes by the symmetry argumentin Remark 3.1. We define R + := { ( q , q ) ∈ R | √ q + q + q > c , | q | < − , | q | < · − , q ≥ , q ≥ } the first quadrant, including q , q -axis, of R . Moreover, we can reduce the domainof s which needs to be examined by one dimension by proving the following Proposition. Wewill use Proposition 4.5 for the proofs of Step 2 and
Step 3 . Proposition 4.5.
The following equality holds for every q ∈ R + \ B . (0).min | s | ≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) = min α ∈ [ θ,θ + π ] ( w ( q ) + s q,α ) t H ( q )( w ( q ) + s q,α )where s q,α = q q + | q | − / (cid:18) cos α sin α (cid:19) is a point of ∂B r q + | q | − (0) and θ is the angle of q in polar coordinates.Proposition 4.5 means that the minimum occurs at a specific part of the boundary of the disk.The proof of Proposition 4.5 consists of Lemma 4.6, 4.7 and 4.8. To establish these Lemmas,we define F q : D q → R , F q ( s ) = ( w ( q ) + s ) t H ( q )( w ( q ) + s )for a fixed q . Here, D q := B r q + | q | − (0) denotes the closed disk of radius q q + | q | − with center at the origin. Lemma 4.6.
The function F q : D q → R has no local minimum in int ( D q ) for all q ∈ R + \ B . (0).Lemma 4.6 can be easily shown by examining the Hessian of F q in the variable s . If we proveLemma 4.6, then we only need to see F q on the boundary of D q . We define F q | ∂D q : S → R by restricting F q to ∂D q , that is, F q | ∂D q ( α ) = F q ( s q,α ) = F q ( q q + | q | − (cid:18) cos α sin α (cid:19) . Here weuse an abuse of notation that ignores the reparametrization of angle. With this notation, weintroduce the following Lemmas. 18 emma 4.7. There exists a unique local minimum and a unique local maximum of the restrictedfunction F q | ∂D q : S → R for each q ∈ R + \ B . (0). Lemma 4.8.
The unique minimum of F q | ∂D q : S → R of Lemma 4.7 is attained in [ θ, θ + π ]where q = r cos θ, q = r sin θ .Above three Lemmas will prove Proposition 4.5. Once we prove Proposition 4.5, it is enoughto show the following inequalitymin α ∈ [0 , π ] ( w ( q ) + s q,θ + α ) t H ( q )( w ( q ) + s q,θ + α ) > q ∈ R + \ B . (0)in order to prove Step 2 and
Step 3 where θ is the angle of q in polar coordinate. Thus wedefine the translated function f q ( α ) := F q | D q ( θ + α ) = ( w ( q ) + s q,θ + α ) t H ( q )( w ( q ) + s q,θ + α )of α for each q . It suffices to prove that min ≤ α ≤ π f q ( α ) > q ∈ R + \ B . (0). Proposition 4.5 will be applied to both of Step 2 and
Step 3 . Butwe will use different techniques for the proof of
Step 2 and
Step 3 from this point. We willuse ’supporting tangent line inequality’ for the proof of
Step 2 and ’blowing up the corner’ forthe proof of
Step 3 .We will use a sharp estimate in the proof of
Step 2 . In general, it is hard to know where theminimum is attained for the problem min ≤ α ≤ π f q ( α ). Thus we need the following geometricobservations to give another sufficient condition which allows us to forget α . Lemma 4.9.
The function f q is convex on [0 , π ] for each q ∈ R + \ B . (0).Using Lemma 4.9, we know that the tangent line at any point in this interval will lie belowthe graph of f q . Let l q be the function for the tangent line at π , that is, l q is linear, l q ( π ) = f q ( π )and dl q dα ( π ) = df q dα ( π ). Then the inequality f q ( α ) ≥ l q ( α ) holds for every α ∈ [0 , π ]. Moreover,we have the following obvious inequalitymin α ∈ [0 , π ] l q ( α ) ≥ min α ∈ [ π − , π +1] l q ( α ) = min { l q ( π − , l q ( π } using π − < π < π + 1 for the line l q . We summarize the above arguments to get thefollowing lower boundmin α ∈ [0 , π ] f q ( α ) ≥ min α ∈ [0 , π ] l q ( α ) ≥ min { l q ( π − , l q ( π } . Thus it is enough to prove l q ( π ± > α ∈ [0 , π ] f q ( α ) > q . Wewill prove l q ( π ± > q . Proposition 4.10.
The inequality l q ( π + 1) > q ∈ R + \ B . (0). Proposition 4.11.
The inequality l q ( π − > q ∈ R + ∩ ( B . (0) \ B . (0)).19ecause the inequality in Proposition 4.11 holds only for q ∈ R + ∩ ( B . (0) \ B . (0)),Proposition 4.10 and 4.11 cannot cover the whole Hill’s region and can imply only Step 2 .We will use blow-up coordinates for the proof of
Step 3 to resolve the convergence of f q to 0 at the critical point. We will see f q ( α ) as a function of q and α again in order to factorout the zeros at the critical point (3 − , G ( q, α ) := f q ( α ).We introduce a lower bound function D ( q, α ) for G ( q, α ), that is G ( q, α ) ≥ D ( q, α ) by removingsmall terms. Roughly speaking, we remove the high degree terms of (3 − − | q | ) from G ( q, α ).Since we have lim q → (3 − , D ( q, α ) = 0, we want to factor out the factor (3 − − | q | ) as manytimes as possible in order to obtain the positivity of the function D easier. We will see thatthe quotient D ( q,α )(3 − −| q | ) is well-defined on R + \ B . (0). However it does not have a continuousextension to the boundary of R + \ B . (0) because lim q → (3 − , D ( q,α )(3 − −| q | ) does not exist. Thuswe want to enlarge near this critical point. We introduce a coordinate change which blowsup the critical point by taking into account the direction to the critical point. We will seethis coordinate change as a composition of two coordinate changes and we will prove its well-definedness in section 5. We summarize the result here. We define the coordinate changes Φand its inverse Ψ Φ : (0 . , − ) × [0 , → R + \ B . , Φ( r, k ) = ( r cos θ ( r, k ) , r sin θ ( r, k )) where cos θ ( r, k ) = 1 + 3 k (3 r − k (3 r − R + \ B . → (0 . , − ) × [0 , , Ψ( q ) = ( r, sin θ r cos θ − r + 3 − cos θ ) where ( q , q ) = ( r cos θ, r sin θ )between a rectangle (0 . , − ) × [0 ,
1) and R + \ B . .As we can see in Figure 4, the critical point (3 − ,
0) corresponds to one side of the rectangleand also this side keeps the information of direction to the critical point like a ”blow-up” proce-dure. We define the function d ( r, k, α ) := D (Φ( r,k ) ,α )(3 − − r ) on the new coordinate (0 . , − ) × [0 , d ( r, k, α ) > r, k, α ) ∈ (0 . , − ) × [0 , × [0 , π ]. We willprove this in the following procedure.Claim 1. The function d : (0 . , − ) × [0 , × [0 , π ] → R can be extended continuously tothe boundary of its domain.We will denote this extension also by d . Thus we have the function d : [0 . , − ] × [0 , × [0 , π ] → R up to the boundary.Claim 2. The function d is monotone decreasing on [0 . , − ] × [0 , × [0 , π ] with respectto r .We will show that ∂d∂r < . , − ] × [0 , × [0 , π ] to prove Claim 2.Claim 3. The inequality d (3 − , k, α ) ≥ k, α ) ∈ [0 , × [0 , π ].20 .55 0.60 0.650.000.050.100.150.200.25 0.55 0.60 0.65 Figure 4: Coordinate change which blows up the corner (3 − , d ( r, k, α ) > d (3 − , k, α ) ≥ . , − ) × [0 , × [0 , π ]. This will imply the desired inequality G ( q, α ) > R + \ B . (0) for Step 3 .This will complete the proof of
Step 3 by Proposition 4.5.We have introduced a numerical form of fiberwise convexity and the strategy of its proof.We will give the details of computations in section 5, .
We introduced the notations to state and modified the main Theorem in section 4.1. We willuse the notations again. We determined the region where the bounded component of the curve H − c,p (0) can lie on. We denoted the Hill’s region by R := { ( q , q ) ∈ R | √ q + q + q > c , | q | < − , | q | < · − } . We defined a tangent vector v ( q ) of H − c, (0) and the Hessian H ( q ) of H c, at each q ∈ H − c, (0). With these notations, we could get the following expressions J ▽ H c,p ( q ) = q + q | q | + p q − q | q | + p ! = v ( q ) + p ∈ T q H − c,p (0) ,HessH c,p ( q ) = HessH c, ( q ) = 1 | q | (cid:18) − | q | + | q | − q − q q − q q | q | + | q | − q (cid:19) = H ( q )for a tangent vector and Hessian of H c,p at q ∈ H − c,p (0) for every p ∈ R . Then the convexity ofthe closed curve H − c,p (0) ∩ R is equivalent to the positivity of the tangential Hessian. Hence wehave to prove the inequality( J ▽ H c,p ( q )) t HessH c,p ( q )( J ▽ H c,p ( q )) = ( v ( q ) + p ) t H ( v ( q ) + p ) > q ∈ H − c,p (0) ∩ R . It is hard to describe the trajectory of q ∈ H − c,p (0) ∩ R . We will see thisinequality with fixing q rather than p and c . We have seen that the equivalent condition. q ∈ H − c,p (0) ∩ R ⇐⇒ q ∈ R and | p + J q | < q + 2 | q | − c We also defined that w ( q ) := v ( q ) − J q , s := p + J q and so v ( q ) + p = w ( q ) + s for thecomputational convenience. We have established Theorem 4.3. Theorem 4.3.
With above notations, the inequality ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R and | s | ≤ q + | q | − c .We divided Theorem 4.3 into three steps by the position of q . • Step 1 : q ∈ R ∩ B . (0) • Step 2 : q ∈ R ∩ ( B . (0) \ B . (0)) • Step 3 : q ∈ R \ B . (0)As we mentioned in section 4.2, the proof of Step 1 can be done by proving Proposition 4.4.
Proposition 4.4.
We have the following estimate.( w ( q ) + s ) t H ( q )( w ( q ) + s ) ≥ r −
103 1 r −
296 1 r + 4 r − r −
2( 1 x − x ) r r + 2 r − − (2 + 2 r )(3 r + 2 r − )for all q ∈ R ∩ B . (0) where r = | q | . Moreover, the following inequality1112 1 r −
103 1 r −
296 1 r + 4 r − r −
2( 1 x − x ) r r + 2 r − − (2 + 2 r )(3 r + 2 r − ) > r ∈ (0 , . Proof of Proposition 4.4.
We will achieve the estimate. After that, the second inequality can beseen simply by its graph. We will omit q of w ( q ) and H ( q ) for notational convenience.( w + s ) t H ( w + s ) = w t H w + 2 w t H s + s t H s We will make several estimate for each of the terms. We express the first term w t H w = 1 | q | − q + 2 q | q | + 3 q | q | − q q | q | + 9 q in the polar coordinates q = r cos θ, q = r sin θ . Then we have the function f ( r, θ ) w t H w = 1 r − θr − θr + 3 cos θr −
27 cos θ sin θr + 9 r cos θ =: f ( r, θ )22n terms of r, θ . Differentiating f with respect to θ∂f ∂θ = 6 r cos θ sin θ − r cos θ sin θ − r cos θ sin θ (cos θ − sin θ ) − r cos θ sin θ = cos θ sin θ (cid:0) r − r − r (2 cos θ − − r (cid:1) gives us the candidates for the minimum points. Namely, w t H w attains its minimum at one ofthese cases: cos θ = 1 , r + − r for a fixed r . w t H w ≥ min r − r + r + 9 r when cos θ = 1 , r − r when cos θ = 0 , r −
103 1 r −
296 1 r + 4 r − r when cos θ = r + − r . Claim: 2) ≥ ≥ Proof. ≥
3) : 12 r (( 1 r − r + 3 r + 9 r ) − ( 1112 1 r −
103 1 r −
296 1 r + 4 r − r ))= 9 r + 60 r + 94 r − r + 1= (3 r + 10 r − ≥ ≥
1) : r (( 1 r − r ) − ( 1 r − r + 3 r + 9 r ))= 3 − r − r > − × − ×
19 = 1Here we use r < . < .By above Claim, we know that 3) is a lower bound for w t H w . Although 3) makes sense onlywhen 0 ≤ r + − r ≤
1, this is not required to get a lower bound. We have an estimate w t H w ≥ r −
103 1 r −
296 1 r + 4 r − r for w t H w .We make an estimate for the second term. We have |H w | H w = 1 | q | q | q | − q | q | − q q − q | q | + 2 q | q | − q q + 3 q | q | ! = ⇒ |H w | = 1 | q | − | q | + 4 | q | + 81 q q | q | − q | q | + 12 q | q | − q q | q | + 9 q = 1 r − r + 4 r + 81 r cos θ sin θ − r cos θ + 12 r cos θ − r cos θ sin θ + 9 r cos θ
23n terms of r, θ . This has its maximum at one of these cases: cos θ = 0 , − r − r +218 r − by thesimilar computation as before. Therefore we have |H w | ≤ max ′ ) r − r + r when cos θ = 0 , ′ ) r − r − r + r + 9 r when cos θ = 1 , ′ ) r − r + r + r +4 r − r when cos θ = − r − r +218 r − . an upper bound for |H w | . As before, it is easy to see that 1 ′ ) > ′ ) , ′ ). We get an estimate |H w | ≤ r − r + 4 r = ( 1 r − r ) for |H w | .Finally, we will investigate the third term which is related with the eigenvalue of H . Thecharacteristic polynomial p H ( λ ) of H = − | q | − q | q | − q q | q | − q q | q | | q | − q | q | has the followingform. p H ( λ ) = λ + (1 + 1 | q | ) λ + ( − − | q | − | q | + 6 q − q | q | )We claim that det( H ) = − − | q | − | q | + q − q | q | < q in the Hill’s region by the followingcomputation. − − | q | − | q | + 6 q − q | q | = − − | q | + 5 q − q | q | = − q − q − c | q | < − − c | q | < q − q + | q | = c ⇐⇒ | q | = c − q + q | q | , c > c > q < λ + , λ − respectively. Wehave a lower bound λ − = 12 (cid:16) − (1 + 1 | q | ) − s | q | + 9 | q | −
4( 6 q − q | q | ) (cid:17) ≥ − (2 + 2 | q | )for λ − . If we summarize all these results, then we can get an estimate for ( w + s ) t H ( w + s ).( w + s ) t H ( w + s ) = w t H w + 2 w t H s + s t H s ≥ w t H w − |H w || s | + λ − | s | ≥ r −
103 1 r −
296 1 r + 4 r − r −
2( 1 x − x ) r r + 2 r − − (2 + 2 r )(3 r + 2 r − )This proves the first statement. We can see that r −
103 1 r −
296 1 r + 4 r − r − x − x ) q r + r − − (2 + r )(3 r + r − ) > r ∈ (0 , .
54) from its graph in Figure 5.Therefore we have proven Proposition 4.4.We have finished the proof of
Step 1 . Now we have to prove
Step 2 , . By the symmetryargument in Remark 3.1, we will see only the first quadrant of R \ B . (0). The first quadrantof the region for Step 2 , is shown in Figure 6. Therefore, we will assume that q , q ≥ ≤ θ ≤ π in the polar coordinate, in the rest of this paper.24 .48 Figure 5: Graph of f ( x ) = x −
103 1 x −
296 1 x + 4 x − x − x − x ) q x + x − − (2 + x )(3 x + x − ) shows that it is positive on [0, 0.54]. Remaining Part ofHill's Region
Figure 6: The remaining Hill’s region that we have to show on25e consider the inequality0 ≤ | s | = 3 q + 2 | q | − c = 3 r cos θ + 2 r − c < r cos θ + 2 r − c for the Hill’s region. This implies that the boundary of the Hill’s region satisfies the equation3 r cos θ + 2 r = 3 ⇐⇒ cos θ = 3 − r r in the polar coordinates. This gives the parametrization of the boundary. Since polar equations r = 0 .
54 and cos θ = − r r intersect at cos θ = − . . > .
7, we may assume cos θ > . s , if we fix the variable q . All possible s forms a disk for a fixed q . The followingProposition allows us to reduce the domain of s that we have to consider for the minimumvalue. Proposition 4.5.
The following equality holds for every q ∈ R + \ B . (0).min | s | ≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) = min α ∈ [ θ,θ + π ] ( w ( q ) + s q,α ) t H ( q )( w ( q ) + s q,α )where s q,α = q q + | q | − / (cid:18) cos α sin α (cid:19) is a point of ∂B r q + | q | − (0) and θ is the angle of q in polar coordinates. Proof of Proposition 4.5.
As we mentioned before, we will show Lemma 4.6, 4.7 and 4.8. Recallthe function F q : D q → R defined by F q ( s ) = ( w ( q ) + s ) t H ( q )( w ( q ) + s ) for each fixed q where D q = B r q + | q | − (0). Lemma 4.6.
The function F q : D q → R has no local minimum in int ( D q ) for all q ∈ R + \ B . (0). Proof of Lemma 4.6.
For a fixed q ∈ R + \ B . (0), F q is a quadratic function in variable s . Thuswe get the Hessian HessF q ( s ) = H ( q ) of F q and we proved that H ( q ) has one positive eigenvalueand one negative eigenvalue in the proof of Proposition 4.4. This implies that there is no localminimum and no local maximum in the interior of the range. This proves Lemma 4.6.As a result of Lemma 4.6, F q attains its minimum at the boundary of D q . We define s q,α = q r cos θ + r − u α where u α = (cid:18) cos α sin α (cid:19) for a fixed α . Then Lemma 4.6 implies thatmin | s | ≤ q + | q | − / ( w + s ) t H ( w + s ) = min α ∈ [0 , π ) ( w + s q,α ) t H ( w + s q,α ) . For convenience of computation, we will consider the translation of α by θ where ( q , q ) =( r cos θ, r sin θ ). Recall the function f q : S → R , f q ( α ) := F q | D q ( θ + α ) = ( w + s q,θ + α ) t H ( w + s q,θ + α )26efined by restriction and translation. We have to prove the following statementmin α ∈ [0 , π ) f q ( α ) = min α ∈ [0 , π ] f q ( α )in order to prove Proposition 4.5. We need the following Lemma. Lemma 4.7.
There exists a unique local minimum and a unique local maximum of the restrictedfunction F q | ∂D q : S → R for each q ∈ R + \ B . (0). Proof of Lemma 4.7.
We have the following expression for the function f q : S → R . f q ( α ) = ( w + s θ + α ) t H ( w + s θ + α )= w t H w + 2 √ c − c w t H u θ + α + (2 c − c ) u tθ + α H u θ + α = w t H w + 2 √ c − c (cos α (3 r − r ) cos θ sin θ + sin α ( − r + 2 cr ))+(2 c − c )(cos α (1 − cr ) + sin α ( − r + 2 cr −
2) + 2 cos α sin α (3 cos θ sin θ ))We differentiate f q with respect to α . Then we have df q dα ( α ) = ∂∂α (( w + s θ + α ) t H ( w + s θ + α ))= 2 √ c − c ( − sin α (3 r − r ) cos θ sin θ + cos α ( − r + 2 cr ))+(2 c − c )( − α sin α (1 − cr ) + 2 cos α sin α ( − r + 2 cr − α − sin α )(3 cos θ sin θ )= 2 √ c − c (( 9 r − r ) cos θ sin θ ) sin α + ( − r + 2 cr ) cos α )+(2 c − c )(( − r + 4 cr −
3) sin 2 α + (3 cos θ sin θ ) cos 2 α )=: A sin 2 α + A cos 2 α + B sin α + B cos α Claim 1 : | B | ≥ | A | Proof of Claim 1. | B | ≥ | A |⇐⇒ √ c − c ( 9 r − r ) cos θ sin θ ≥ (2 c − c )(6 cos θ sin θ ) ⇐⇒ ( 9 r − r − √ c − c ) ≥ r ∈ (0 . , − ) and 2 c − c ≤ . + . − < . . This proves Claim 1.Claim 2 : | B | > | A | roof of Claim 2. | B | ≥ | A |⇐⇒ ( − r + 2 cr ) − (2 c − c )( − r + 4 cr − > ⇐⇒ ( − r − r − r cos θ ) − (3 r cos θ + 2 r − )(6 cos θ + 3 r − > θ =: y and g ( r, y ) := ( r − r − ry ) − (3 r y + r − )(6 y + r − , then ∂g∂y = 2( r − r − ry )( − r ) − r (6 y + r − − r y + r − )(6 y + r − <
0. We caneasily check three terms are all negative, and therefore it is enough to show that g ( r, > r − r − r ) − (3 r + r − )(3 + r ) >
0. This is clear from a simple calculation.This proves Claim 2.Now we know 2 p A + A < p B + B from Claim 1, 2. We need the following Lemma toget the number of local extrema. I borrow the following geometric proof of Lemma 5.1 from UrsFrauenfelder. This Lemma can be proven with analytic way as well. Lemma 5.1. If A, B ∈ R satisfy 2 | A | < | B | , then the equation for the unknown αA sin(2 α + φ ) + B sin( α + µ ) = 0has exactly 2 solutions on [0 , π ) for any constant φ, µ ∈ R . Proof.
Without loss of generality, we may assume that B = 1 , A = t ∈ [0 , ) and µ = 0.In the case of t = 0, the above equation becomes sin α = 0 and this has 2 solutions.Suppose that there exist t ∈ [0 , ) such that t sin(2 α + φ ) + sin α = 0 does not have 2 solutions.We define a function T : S × [0 , t ] → R , T ( α, t ) = t sin(2 α + φ ) + sin α. Then a critical point ( α, t ) of T satisfies ∂ t T = sin(2 α + φ ) = 0 , ∂ α T = 2 t cos(2 α + φ ) + cos α = 0 . and this implies the equations sin(2 α + φ ) = 0 , cos α = ± t. Since 0 ≤ t < t sin(2 α + φ )+ sin α =0. Thus 0 is the regular value for T . Then we get T − (0) is a smooth manifold with boundary.Because it has a different number of points in S × { } and S × { t } by the assumption of t .There must be an appearance or disappearance of curve, so-called, ’birth and death’ of curve.Let ( α , t ) be one of these points. Then T ( α , t ) = 0 and ∂ α T ( α , t ) = 0, that is, we have ( t sin(2 α + φ ) + sin α = 02 t cos(2 α + φ ) + cos α = 0 = ⇒ ( t sin (2 α + φ ) = sin α t cos (2 α + φ ) = cos α By adding these two equations, we get 1 = t + 3 t cos (2 α + φ ) ≤ t < ∂∂α [( w + s θ + α ) t H ( w + s θ + α )] = A sin 2 α + A cos 2 α + B sin α + B cos α = A sin(2 α + φ )+ B sin( α + µ ) where A = p A + A , B = p B + B . We proved 2 | A | < | B | by Claim 1, 2. Thus we get that ∂∂α [( w + s θ + α ) t H ( w + s θ + α )] = 0 has exactly 2 solutions on α ∈ [0 , π ) by applying Lemma 5.1. This implies f q ( α ) = ( w + s θ + α ) t H ( w + s θ + α ) has exactly two critical points. Since the domain of f q is S ,there exist the unique local maximum and minimum respectively on S . This proves Lemma4.7.Now we need the following Lemma to reduce the region where the minimum is attained. Thefollowing Lemma will finish the proof of Proposition 4.5. Lemma 4.8.
The unique minimum of the function f q : S = R / π Z → R is attained in [0 , π ]. Proof.
Now we know f q has only one local minimum for fixed q and so this will be the globalminimum. We calculate the first derivative of f q ddf q (0) = ∂∂α (cid:12)(cid:12) α =0 ( w t H w + 2 √ c − c w t H u θ + α + (2 c − c ) u tθ + α H u θ + α )= 2 √ c − c ( − r + 2 cr ) + (2 c − c )(6 sin θ cos θ ))at α = 0 , π . We can obtain ddf q (0) < r − cr > √ c − c (6 cos θ + r −
3) from Claim 2 in the proof of Lemma 4.7 and the inequality 6 cos θ + r − > θ cos θ .Next, we compute the derivative of f q at π ddf q ( π ∂∂α (cid:12)(cid:12) α = π ( w t H w + 2 √ c − c w t H u θ + α + (2 c − c ) u tθ + α H u θ + α )= 2 √ c − c ( 9 r − r ) cos θ sin θ + (2 c − c )( − θ cos θ ))= 2 √ c − c cos θ sin θ ( 9 r − r − √ c − c )and similarly we can obtain ddf q ( π ) >
0. Therefore, there exists a unique local minimum on α ∈ (0 , π ] and this is the global minimum because the function f q : S → R has only one localminimum. This proves Lemma 4.8.Now we can prove Proposition 4.5 by combining Lemma 4.6, 4.7 and 4.8. We know that F q attains its minimum on the boundary of D q for any fixed q ∈ R + \ B . (0) by Lemma4.6. Moreover, we know f q , the restriction of F q to ∂D q with the translation of angle by θ ,has only one local minimum and so it is global minimum and this minimum is attained in[0 , π ] by Lemma 4.7 and 4.8. Therefore, we get min | s | ≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) =min α ∈ [0 , π ] ( w ( q ) + s q,θ + α ) t H ( q )( w ( q ) + s q,θ + α ) for all q ∈ R + \ B . (0). This completes the proofof Proposition 4.5.We will use the previous notations again. We recall that f q ( α ) := ( w + s θ + α ) t H ( w + s θ + α )for each fixed q ∈ R + \ B . (0) where s q,α = q r cos θ + r − (cid:18) cos α sin α (cid:19) ∈ ∂D q . Lemma 4.9.
The function f q is convex on [0 , π ] for each q ∈ R + \ B . (0).29 roof. We calculate the second derivative ∂ ∂α f q ( α ) = ∂ ∂α ( w + s θ + α ) t H ( w + s θ + α )= 2 √ c − c [(( 9 r − r ) cos θ sin θ ) cos α + ( 1 r − cr ) sin α ]+(2 c − c )[2( − r + 4 cr −
3) cos 2 α + 2( − θ cos θ ) sin 2 α ]and we want to prove that it is positive on α ∈ [0 , π ].Claim 1: ( r − cr ) sin α + √ c − c ( − r + cr −
3) cos 2 α > α ∈ [0 , π ]. Proof of Claim 1.
We already know that ( r − cr ) > √ c − c ( − r + cr − >
0. On the otherhand, we have the following inequality. If b > a >
0, then a cos 2 α + b sin α > α ∈ [0 , π ]. In fact, we have that a cos 2 α + b sin α = − a sin α + b sin α + a and − at + b + a > t ∈ [0 , b > a >
0. Because 0 ≤ sin α ≤ α ∈ [0 , π ]. This proves Claim1. Claim 2: ( r − r ) cos θ sin θ cos α ≥ √ c − c (6 sin θ cos θ ) sin 2 α for α ∈ [0 , π ]. Proof of Claim 2.
It suffices to prove that ( r − r − √ c − c sin α ) cos α ≥ α ∈ [0 , π ].This is clear, because we have that r − r > · and 2 c − c < . r − r > √ c − c . This proves Claim 2We have shown that ∂ ∂α f q ( α ) > α ∈ [0 , π ] by Claim 1, 2. This completes the proofof Lemma 4.9.We have proven that min | s |≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) = min ≤ α ≤ π f q ( α ) and f q is convex on [0 , π ] for all q ∈ R + \ B . (0). Let l q ( α ) = f ‘ q ( π )( α − π ) + f q ( π ) be the tangent lineof f q at α = π then this tangent line will be below the function. In particular, one of the endpoints of this line will be less than or equal to the minimum value of the function, see Figure 7.Thus we have thatmin | s |≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) = min ≤ α ≤ π f q ( α ) ≥ min { l q ( π , l q ( π − } . Therefore we shall show that min | s |≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > l q ( π +1) > l q ( π − > x := r, y := cos θ which are well-defined coordinates on the first quadrant of ( q , q )-coordinate. The domain R ′ of ( x, y ) corresponding to the domain R + \ B . (0) of ( q , q ) is given by R ′ := { ( x, y ) ∈ R | . < x < − , − x x < y ≤ } . We will define a change of variables in terms of x, y in the following Lemma.30 .5 1.0 1.5 ! ! Figure 7: Tangent lines of a convex function - The strategy is ”One of the end points of a tangentline is below the minimum point of convex function.”
Lemma 5.2 (Blow-up coordinates change) . If we define the map φ : R ′′ := (0 . , − ) × [0 , → R ′ by ( x, k ) → ( x, y ) where y = 1 + 3 k (3 x − k (3 x − , then φ is a diffeomorphism. Proof.
We compute the Jacobian of φ . First, we note that ∂y∂k is not zero. In fact, ∂y∂k = ∂∂k ( 1 + 3 k (3 x − k (3 x −
1) )= − x + 3 x − k (3 x − < x ∈ (0 . , − )Then the Jacobian of this map is given by ∂ ( x, y ) ∂ ( x, k ) = ∗ − x +3 x − k (3 x − ! Thus we know that the Jacobian is nonsingular for every ( x, k ) ∈ (0 . , − ) × [0 ,
1) by theabove computation. This proves that φ is a local diffeomorphism. We need to show that φ is abijective map. If we assume φ ( x , k ) = φ ( x , k ), then we have x = x and k (3 x − k (3 x − = k (3 x − k (3 x − . We get k = k from the monotonicity of y with respect to k and so ( x , k ) =( x , k ). This proves the injectivity of φ . For the surjectivity, we extend the map φ to the mapon (0 . , − ) × [0 ,
1] in obvious way. Then we have that k = 0 = ⇒ y = 1 , k = 1 = ⇒ y = 3 x − x . This proves the surjectivity from the monotonicity. Therefore the map φ is a diffeomorphism.This proves Lemma 5.2. 31 ' (x, y) R'' (x, k)
Figure 8: The domains R ′ , R ′′ of new variablesThis diffeomorphism φ : R ′′ → R ′ cannot be extended to the boundary as a diffeomorphism.As one can see in Figure 8, the critical point (3 − ,
1) at the boundary of R ′ corresponds tothe one side x = 3 − , ≤ k ≤ R ′′ . If we use this map φ as a coordinatechart, then we can handle our problem on a rectangle domain. This coordinate chart will playan important role in the proof of Step 3 as well as Proposition 4.10 and 4.11.We compute the evaluation at π f q ( π w t H w + 2 √ c − c ( 1 √ r − r ) cos θ sin θ + 1 √ − r + 2 cr ))+(2 c − c )( − − r + 3 sin θ cos θ ) f ′ q ( π √ c − c ( 1 √ r − r ) cos θ sin θ ) + 1 √ − r + 2 cr )) + (2 c − c )( − r + 4 cr − f q and f ′ q to express the tangent line l q at π in terms of q . For the tangent line l q ( t ) = f q ( π f ′ q ( π t − π , of f q at π , we can express the values of l q l q ( π w t H w + 2 √ c − c ( √ − r + 2 cr )) + (2 c − c )( − − r + 4 cr + 3 sin θ cos θ )at π + 1 in terms of r, θ, c explicitly. Proposition 4.10.
The inequality l q ( π + 1) > q ∈ R + \ B . (0). Proof.
First, we note that we can express w t H w in terms of r, c using the equation 3 r cos θ + r − c = 0 ⇐⇒ cos θ = cr − r where ( r, θ ) is the polar coordinate system for q , namely32 = r cos θ, q = r sin θ . w t H w = 1 r − r cos θ − r sin θ + 3 r cos θ − r cos θ sin θ + 9 r cos θ = 1 r − r − r ( 2 cr − r ) + 3 r ( 2 cr − r ) − r ( 2 cr − r )(1 − cr − r ) + 9 r ( 2 cr − r )= 1 r − r − r ( 2 c r − r ) + 3 r ( 2 c r − r ) − r ( 2 c r − r )(1 − c r − r ) + 9 r ( 2 c r − r ) − r ( 2 c − c r ) + 3 r ( 2 c − c r ) − c − c r ) − (( 2 cr − r ) − ( 2 c r − r ) )] + 9 r ( 2 c − c r )= 15 r − √ r + 27 √ r + 14 r − √ r − r + 9 √ c − c )( − r + 18 √ r − r + 3) + (2 c − c ) ( 3 r )We have to see that w t H w + 2 √ c − c ( √ − r + cr )) + (2 c − c )( − − r + cr ) > c = , we get the following estimate. w t H w + 2 √ c − c ( √ − r + 2 cr )) + (2 c − c )( − − r + 4 cr )= 15 r − √ r + 27 √ r + 14 r − √ r − r + 9 √ c − c )( − r + 18 √ r − r + 3) + (2 c − c ) ( 3 r )+2 √ c − c ( √ − r + 3 √ r )) + 2(2 c − c ) ( √ r )+(2 c − c )( − r −
72 + 6 √ r ) + (2 c − c ) ( 2 r )= 15 r − √ r + 27 √ r + 14 r − √ r − r + 9 √ √ c − c ( √ − r + 3 √ r ))+(2 c − c )( − r + 18 √ r − r + 6 √ r −
12 )+2(2 c − c ) ( √ r ) + (2 c − c ) ( 3 r + 2 r ) ≥ r − √ r + 27 √ r + 14 r − √ r − r + 9 √ √ c − c ( √ − r + 3 √ r ))+(2 c − c )( − r + 18 √ r − r + 6 √ r −
12 )Therefore, it suffices to prove the following inequality15 r − √ r + 27 √ r + 14 r − √ r − r + 9 √ √ c − c ( √ − r + 3 √ r ))+ (2 c − c )( − r + 18 √ r − r + 6 √ r −
12 ) > x, k ) in Lemma 5.2 which have the relation of x := r, y :=cos θ, y = k (3 x − k (3 x − . Note that the following identities.2 c − c = 3 x y + 2 x − = 3 x ( 1 + 3 k (3 x − k (3 x −
1) ) + 2 x − = 1 − k k (3 x −
1) (3 x + 2 x − )= 1 − k k (3 x −
1) (3 − − x ) (3 + 2 · x ) (2)Using the above identities, the inequality (1) can be written as follows.15 x − √ x + 27 √ x + 14 x − √ x − x + 9 √ s − k k (3 x −
1) (3 − − x ) (3 + 2 · x )( √ − x + 3 √ x ))+( 1 − k k (3 x −
1) (3 − − x ) (3 + 2 · x ))( − x + 18 √ x − x + 6 √ x −
12 ) > x − √ x + 27 √ x + 14 x − √ x − x +9 √ − − x ) ( 15 · x − x − · x + 5 · x + 12 x + 9 · x ) , (3) − x + 3 √ x = − (3 − − x )( 3 x + 3 x + 3 x + 3 x ) , (4)we can factor out the term (3 − − x ) . Then inequality (1) is equivalent to the followinginequality 15 · x − x − · x + 5 · x + 12 x + 9 · x − √ s − k k (3 x −
1) (3 + 2 · x )( 3 x + 3 x + 3 x + 3 x )+ ( 1 − k k (3 x −
1) (3 + 2 · x ))( − x + 18 √ x − x + 6 √ x −
12 ) > g x ( t ) := ( − x + 18 · x − x + 6 · x −
12 ) t − √
2( 3 x + 3 x + 3 x + 3 x ) t +( 15 · x − x − · x + 5 · x + 12 x + 9 · x ) . in variable t . The coefficients of g x are functions of x . We note that g x ( r − k k (3 x − (3 + · x ))is the left hand side of inequality (5) which we have to show. Thus we want to prove that the34 .60 0.65 ! Figure 9: Graph of f ( x ) = − x + · x − x + · x − shows that it is positive on [0 . , − ). ! ! ! Figure 10: Graph of f ( x ) = D x shows that it is negative on [0 . , − ).inequality g x ( s − k k (3 x −
1) (3 + 2 · x )) > x, k ) ∈ (0 . , − ) × [0 , D x of the polynomial g x . D x x + 3 x + 3 x + 3 x ) − ( − x + 18 · x − x + 6 · x −
12 )( 15 · x − x − · x + 5 · x + 12 x + 9 · x )We can see that the coefficient of t for g x is positive for all x ∈ (0 . , − ), namely − x + · x − x + · x − > x ∈ (0 . , − ), from its graph in Figure 9 and this discriminant D x < x ∈ [0 . , − ) from the graph of D x in Figure 10. This means that degree2 polynomial g has a positive coefficient for t and has no real root for all x ∈ [0 . , − ).35 .54 0.55 0.56 0.57 ! ! ! ! ! Figure 11: Graph of f ( x ) = dg x dt ( q · x ) shows that it is negative on [0.54, 0.56].Therefore, we have proven that g x ( s − k k (3 x −
1) (3 + 2 · x )) > x, k ) ∈ [0 . , − ) × [0 , . The inequality for ( x, k ) ∈ [0 . , . × [0 ,
1) is still left. To complete the proof, we note thatthe possible values of t for the proof satisfy the inequality0 < s − k k (3 x −
1) (3 + 2 · x ) < s · x We compute the derivative dg x dt ( s · x ) = 2( − x + 18 · x − x + 6 · x −
12 )( s · x ) − √
2( 3 x + 3 x + 3 x + 3 x )of g x at q · x . Then we have that dg x dt ( s · x ) < x ∈ [0 . , . dg x dt ( q · x ) in Figure 11.Thus we have the inequality g x ( t ) > g x ( q · x ) for all t ∈ (0 , q · x ), when x ∈ (0 . , . g x ( s − k k (3 x −
1) (3 + 2 · x )) > g x ( s · x )36 .54 0.55 0.56 0.57186188 Figure 12: Graph of f ( x ) = g ( q · x ) shows that it is positive on [0 . , − ).for all ( x, k ) ∈ [0 . , . × [0 , g x ( s · x ) := ( − x + 18 · x − x + 6 · x −
12 )(3 + 2 · x ) − √
2( 3 x + 3 x + 3 x + 3 x )( s · x )+( 15 · x − x − · x + 5 · x + 12 x + 9 · x ) > x ∈ (0 . , . g x ( r − k k (3 x − (3 + · x )) >
0. This inequality g x ( q · x ) > x ∈ (0 . , .
56] can be seen from its graph in Figure 12.Therefore, we have proven g x ( s − k k (3 x −
1) (3 + 2 · x )) > x, k ) ∈ [0 . , − ). This implies inequality (5) and so (1). This completes the proof ofProposition 4.10.As in the computation for l q ( π + 1), we can express l q ( π −
1) = w t H w + 2 √ c − c ( √ r − r ) cos θ sin θ ) + (2 c − c )( 52 + 12 r − cr + 3 sin θ cos θ )in terms of r, θ . Proposition 4.11.
The inequality l q ( π − > q ∈ R + ∩ ( B . (0) \ B . (0)). Proof.
Following the computations in the proof of Proposition 4.10, we can get a lower bound37or l q ( π − l q ( π − ≥ w t H w + 2 √ c − c ( √ r − r ) cos θ sin θ ) + (2 c − c )( 52 + 12 r − cr )= 15 r − · r + 27 · r + 14 r − · r − r + 9 · +2 √ c − c ( √ r − r ) cos θ sin θ )+(2 c − c )( − r + 6 · r − r − · r + 112 + 3 cos θ sin θ )+(2 c − c ) ( 3 r − r ) ≥ r − · r + 27 · r + 14 r − · r − r + 9 · +2 √ c − c ( √ r − r ) sin θ )+(2 c − c )( − r + 6 · r − r − · r + 112 )for all q ∈ R + \ B . (0). Thus it is enough to prove the following inequality15 r − · r + 27 · r + 14 r − · r − r + 9 · + 2 √ c − c ( √ r − r ) sin θ ) + (2 c − c )( − r + 6 · r − r − · r + 112 ) > q ∈ R + ∩ ( B . (0) \ B . (0)) in order to prove Proposition 4.11. Using the variables inLemma 5.2, we have that y = 1 + 3 k (3 x − k (3 x − , sin θ = 1 − y = xk k (3 x −
1) (3 x + 2 x − ) . (7)For notational convenience, we define f ( x ) := x − · x + · x + x − · x − x + 9 · . Usingthe identity (2) with above computations, we can write the left hand side of inequality (6)15 r − · r + 27 · r + 14 r − · r − r + 9 · +2 √ c − c ( √ r − r ) sin θ ) + (2 c − c )( − r + 6 · r − r − · r + 112 )= f ( x ) + 2 √ √ x √ k − k k (3 x −
1) (3 x + 2 x − )(3 x − x )+ 1 − k k (3 x −
1) (3 x + 2 x − )( − x + 6 · x − x − · x + 112 )in terms of x, k . We note that − k k (3 x − decreases as k increases for fixed x . We compute thepartial derivative of √ k − k k (3 x − with respect to k∂∂k ( √ k − k k (3 x −
1) ) = 1 − k (3 x + 1)2 √ k − k (1 + k (3 x − . √ k − k k (3 x − attain its maximum at k = x +1 > . Moreover, √ k − k k (3 x − increases for k < x +1 and decreases for k > x +1 with respect to k when we fix the othervariable x . We recall the decompositions (3) in the proof of Proposition 4.10. Then we canfactor out the term (3 − − x ) from inequality (6). With these notations and discussions, wewill prove the following inequality g ( x ) + 2 √ √ x √ k − k k (3 x −
1) (3 + 2 · x )(3 x − x )+ 1 − k k (3 x −
1) (3 + 2 · x )( − x + 6 · x − x − · x + 112 ) > x, k ) ∈ (0 . , . × [0 ,
1] where g ( x ) = ( · x − x − · x + · x + x + · x ). This isequivalent with inequality (6).The strategy for the proof of (8) can be described as follows. . We divide the region into several cases in terms of k . Case 1 ) 0 ≤ k ≤ , Case 2 ) 13 ≤ k ≤ , Case 3 ) 23 ≤ k ≤ Case 4 ) 34 ≤ k ≤ , Case 5 ) 45 ≤ k ≤ . We make the following estimates √ x √ k − k k (3 x −
1) (3 + 2 · x ) ≤ U i ( x ) , m i ( x ) ≤ − k k (3 x −
1) (3 + 2 · x ) ≤ M i ( x )for each Case i of i = 1 , , , , . We construct lower bounds g ( x ) + 2 √ √ x √ k − k k (3 x −
1) (3 + 2 · x )(3 x − x )+ 1 − k k (3 x −
1) (3 + 2 · x )( − x + 6 · x − x − · x + 112 ) ≥ L i ( x )for each Case i where the function L i of x is defined by L i ( x ) = min { L mi ( x ) := g ( x ) + 2 √ U i ( x )(3 x − x ) + m i ( x )( − x + 6 · x − x − · x + 112 ) ,L Mi ( x ) := g ( x ) + 2 √ U i ( x )(3 x − x ) + M i ( x )( − x + 6 · x − x − · x + 112 ) } . .56 0.58 0.60 0.62 0.64100150200250 0.56 0.58 0.60 0.62 0.64100120140160180200220 Figure 13: Case 1) Graphs of f m ( x ) = L m ( x )(left) and f M ( x ) = L M ( x )(right) show that theyare positive on [0 . , . . We prove the inequality L i ( x ) > x ∈ [0 . , .
63] and i = 1 , , , , L mi ( x ) = g ( x ) + 2 √ U i ( x )(3 x − x ) + m i ( x )( − x + 6 · x − x − · x + 112 ) > L Mi ( x ) = g ( x ) + 2 √ U i ( x )(3 x − x ) + M i ( x )( − x + 6 · x − x − · x + 112 ) > x ∈ [0 . , . L mi , L Mi to show L i ( x ) > i = 1 , , , , Case 1 ) 0 ≤ k ≤ For each fixed x ∈ [0 . , . √ x √ k − k k (3 x − (3 + · x ) attains its maximum at k = for k ∈ [0 , ] and the maximum value is given by the function U ( x ) = √ x x +2 (3 + · x ) of x . Theterm − k k (3 x − (3 + · x ) has the value between its value m ( x ) = x +2 (3 + · x ) at and itsvalue M ( x ) = (3 + · x ) at k = 0 for k ∈ [0 , ]. It suffices to show that the functions L m ( x ) = g ( x ) + 2 √ √ x x + 2 (3 + 2 · x )(3 x − x )+ 23 x + 2 (3 + 2 · x )( − x + 6 · x − x − · x + 112 ) ,L M ( x ) = g ( x ) + 2 √ √ x x + 2 (3 + 2 · x )(3 x − x )+(3 + 2 · x )( − x + 6 · x − x − · x + 112 )of x are positive for all x ∈ [0 . , . L m ( x ) > , L M ( x ) > x ∈ [0 . , .
63] from their graphs in Figure 13.
Case 2 ) ≤ k ≤ .56 0.58 0.60 0.62 0.6450100150200250 0.56 0.58 0.60 0.62 0.64406080100120140160180 Figure 14: Case 2) Graphs of f m ( x ) = L m ( x )(left) and f M ( x ) = L M ( x )(right) show that theyare positive on [0 . , . x ∈ [0 . , . √ x √ k − k k (3 x − (3+ · x ) attains its maximum at k = x +1 among k ∈ [ , ] and the maximum value is the function U ( x ) = √ x (3 + · x ) of x . The term − k k (3 x − (3+ · x ) has the value between m ( x ) = x +1 (3+ · x ) and M ( x ) = x +2 (3+ · x )for k ∈ [ , ]. It suffices to show that the functions L m ( x ) = g ( x ) + √ √ x (3 + 2 · x )(3 x − x )+ 16 x + 1 (3 + 2 · x )( − x + 6 · x − x − · x + 112 ) ,L M ( x ) = g ( x ) + √ √ x (3 + 2 · x )(3 x − x )+ 23 x + 2 (3 + 2 · x )( − x + 6 · x − x − · x + 112 )of x are positive for all x ∈ [0 . , . L m ( x ) > , L M ( x ) > x ∈ [0 . , .
63] from their graphs in Figure 14.
Case 3 ) ≤ k ≤ For each fixed x ∈ [0 . , . √ x √ k − k k (3 x − (3 + · x ) attains its maximum at k = among k ∈ [ , ] and the maximum value is the function U ( x ) = √ x x +1 (3 + · x ) of x . The term − k k (3 x − (3+ · x ) has the value between m ( x ) = x +1 (3+ · x ) and M ( x ) = x +1 (3+ · x )41 .56 0.58 0.60 0.62 0.6450100150200250300 0.56 0.58 0.60 0.62 0.6450100150200250 Figure 15: Case 3) Graphs of f m ( x ) = L m ( x )(left) and f M ( x ) = L M ( x )(right) show that theyare positive on [0 . , . k ∈ [ , ]. It suffices to show that the functions L m ( x ) = g ( x ) + 2 √ √ x x + 1 (3 + 2 · x )(3 x − x )+ 19 x + 1 (3 + 2 · x )( − x + 6 · x − x − · x + 112 ) ,L M ( x ) = g ( x ) + 2 √ √ x x + 1 (3 + 2 · x )(3 x − x )+ 16 x + 1 (3 + 2 · x )( − x + 6 · x − x − · x + 112 )of x are positive for all x ∈ [0 . , . L m ( x ) > , L M ( x ) > x ∈ [0 . , .
63] from their graphs in Figure 15.
Case 4 ) ≤ k ≤ For each fixed x ∈ [0 . , . √ x √ k − k k (3 x − (3 + · x ) attains its maximum at k = among k ∈ [ , ] and the maximum value is the function U ( x ) = √ x x +1 (3 + · x ) of x . The term − k k (3 x − (3+ · x ) has the value between m ( x ) = x +1 (3+ · x ) and M ( x ) = x +1 (3+ · x )for k ∈ [ , ]. It suffices to show that the functions L m ( x ) = g ( x ) + 2 √ √ x x + 1 (3 + 2 · x )(3 x − x )+ 112 x + 1 (3 + 2 · x )( − x + 6 · x − x − · x + 112 ) ,L M ( x ) = g ( x ) + 2 √ √ x x + 1 (3 + 2 · x )(3 x − x )+ 19 x + 1 (3 + 2 · x )( − x + 6 · x − x − · x + 112 )of x are positive for all x ∈ [0 . , . L m ( x ) > , L M ( x ) > x ∈ [0 . , .
63] from their graphs in Figure 16.42 .56 0.58 0.60 0.62 0.6450100150200250300 0.56 0.58 0.60 0.62 0.6450100150200250300
Figure 16: Case 4) Graphs of f m ( x ) = L m ( x )(left) and f M ( x ) = L M ( x )(right) show that theyare positive on [0 . , . Figure 17: Case 5) Graphs of f m ( x ) = L m ( x )(left) and f M = L M ( x )(right) show that they arepositive on [0 . , . Case 5 ) ≤ k ≤ x ∈ [0 . , . √ x √ k − k k (3 x − (3 + · x ) attains its maximum at k = among k ∈ [ ,
1] and the maximum value is the function U ( x ) = √ x x +1 (3 + · x ) of x . Theterm − k k (3 x − (3 + · x ) has the value between m ( x ) = 0 and M ( x ) = x +1 (3 + · x ) for k ∈ [ , L m ( x ) = g ( x ) + 2 √ √ x x + 1 (3 + 2 · x )(3 x − x ) ,L M ( x ) = g ( x ) + 2 √ √ x x + 1 (3 + 2 · x )(3 x − x )+ 112 x + 1 (3 + 2 · x )( − x + 6 · x − x − · x + 112 )of x are positive for all x ∈ [0 . , . L m ( x ) > , L M ( x ) > x ∈ [0 . , .
63] from their graphs in Figure 17.These cases complete the proof of Proposition 4.11.We can prove
Step 2 by combining the above results.43 roof of
Step 2 . By Proposition 4.5, we only need to show that f q ( α ) > q ∈ R ∩ ( B . \ B . (0)) and α ∈ [0 , π ]. The function f q is convex on [0 , π ] by Lemma 4.9. This impliesthat the tangent line l q of f q at π is below f q . Thus we havemin α ∈ [0 , π ] f q ( α ) ≥ min α ∈ [0 , π ] l q ( α ) ≥ min α ∈ [ π − , π +1] l q ( α ) = min { l q ( π , l q ( π − } for any q ∈ R \ B . (0)Proposition 4.10 and 4.11 prove that min { l q ( π +1) , l q ( π − } > q ∈ R ∩ ( B . \ B . (0)).Therefore, min α ∈ [0 , π ] f q ( α ) > q ∈ R ∩ ( B . \ B . (0)). This proves Step 2 .We will prove
Step 3 from now. We can consider only when r > .
63. We recall that f q ( α ) := ( w + s θ + α ) t H ( w + s θ + α )= w t H w + 2 √ c − c (cos α (3 r − r ) cos θ sin θ + sin α ( − r + 2 cr ))+(2 c − c )(cos α (1 − cr ) + sin α ( − r + 2 cr −
2) + 2 cos α sin α (3 cos θ sin θ )) Proof of
Step 3 . It suffices to show that f q ( α ) > q ∈ R \ B . (0) and α ∈ [0 , π ] byProposition 4.5. We define the function G ( q, α ) := f q ( α ) of q and α . We have that G ( q, α ) := f q ( α )= w t H w + 2 √ c − c (cos α (3 r − r ) cos θ sin θ + sin α ( − r + 2 cr ))+(2 c − c )(cos α (1 − cr ) + sin α ( − r + 2 cr −
2) + 2 cos α sin α (3 cos θ sin θ )) ≥ w t H w + 2 √ c − c (cos α (3 r − r ) sin θ + sin α ( − r + 2 cr ))+(2 c − c )(cos α (1 − cr ) + sin α ( − r + 2 cr − E ( q, α )for α ∈ [0 , π ] and q ∈ R \ B . (0). It suffices to prove that E ( q, α ) > q ∈ R \ B . (0)and α ∈ [0 , π ]. We use the variables x := r, y := cos θ in Lemma 5.2 and will denote again E ( x, y, α ) by ignoring the composition of this change of variables. Then we can express the44unction E in terms of x, y and α as follows. E ( x, y, α ) = 15 x − √ x + 27 √ x + 14 x − √ x − x + 9 √ c − c )( − x + 18 √ x − x + 3) + (2 c − c ) ( 3 x )+2 √ c − c (cos α (3 x − x ) p − y + sin α ( − x + 2 cx ))+(2 c − c )(cos α (1 − cx ) + sin α ( − x + 2 cx − x − √ x + 27 √ x + 14 x − √ x − x + 9 √ c − c )( − x + 18 √ x − x + 3) + (2 c − c ) ( 3 x )+2 √ c − c (cos α (3 x − x ) p − y + sin α ( − x + 2 cx ))+(2 c − c )(cos α (1 − cx ) + sin α ( − x + 2 cx − x − √ x + 27 √ x + 14 x − √ x − x + 9 √ c − c )( − x + 18 √ x − x + 3) + (2 c − c ) ( 3 x )+2 √ c − c (cos α (3 x − x ) p − y + sin α ( − x + 3 x )) + (2 c − c ) ( 2 sin αx )+(2 c − c )(cos α (1 − cx ) + sin α ( − x + 3 x − c − c ) ( − cos αx + sin αx )where 2 c = 3 x y + x . We can find a lower bound function D ( x, y, α ) E ( x, y, α ) = 15 x − √ x + 27 √ x + 14 x − √ x − x + 9 √ √ c − c (cos α (3 x − x ) p − y + sin α ( − x + 3 x ))+(2 c − c )( − x + 18 √ x − x + 3 + cos α (1 − x ) + sin α ( − x + 3 x − c − c ) ( 2 sin αx ) + (2 c − c ) ( 3 x − cos αx + sin αx ) ≥ x − √ x + 27 √ x + 14 x − √ x − x + 9 √ √ c − c (cos α (3 x − x ) p − y + sin α ( − x + 3 x ))+(2 c − c )( − x + 18 √ x − x + 3 + cos α (1 − x ) + sin α ( − x + 3 x − D ( x, y, α )of E ( x, y, α ) by removing the degree 3, 4 terms of √ c − c , because the degree 3, 4 terms are45ositive. Therefore it is enough to show that D ( x, y, α ) > q ∈ R \ B . (0) and α ∈ [0 , π ].We use the variables ( x, k ) which is given by the change of variables y = k (3 x − k (3 x − in Lemma5.2 and will denote by D ( x, k, α ) again. Recall that the decompositions (3), (4) in the proof ofProposition 4.10, then we get the common factor (3 − − x ) in D ( x, k, α ). Precisely, we havethat D ( x, k, α ) = (3 − − x ) ( 15 · x − x − · x + 5 · x + 12 x + 9 · x ) − cos α (cid:16) √ x √ k − k k (3 x −
1) (3 − − x ) (3 + 2 · x )( 9 x − x ) (cid:17) − sin α (cid:16) s − k k (3 x −
1) (3 + 2 · x )(3 − − x ) ( 3 x + 3 x + 3 x + 3 x ) (cid:17) + 1 − k k (3 x −
1) (3 − − x ) (3 + 2 · x ) (cid:16) − x + 18 √ x − x + 3+ cos α (1 − x ) + sin α ( − x + 3 x − (cid:17) We can factor out the common factor (3 − − x ) from D ( x, k, α ). Define d ( x, k, α ) := D ( x,k,α )(3 − − x ) . In fact, functions d, D are defined on R ′′ × [0 , π ] where the domain R ′′ = (0 . , − ) × [0 ,
1) for ( x, k ), we can extend d continuously to the function on R ′′ × [0 , π ]. If we prove theinequality d ( x, k, α ) > R ′′ ∩ { x ≥ . } ) × [0 , π ], then we have the inequality d ( x, k, α ) > R ′′ ∩ { x ≥ . } ) × [0 , π ]. This implies D ( x, k, α ) > R ′′ ∩ { x ≥ . } ) × [0 , π ].Therefore, we will prove that d ( x, k, α ) > x, k, α ) ∈ [0 . , − ] × [0 , × [0 , π ]. Weabbreviate the terms of d by the functions C i , ( i = 1 , , , ,
5) as follows. d ( x, k, α ) = ( 15 · x − x − · x + 5 · x + 12 x + 9 · x ) − cos α (cid:16) √ x √ k − k k (3 x −
1) (3 + 2 · x )( 9 x − x ) (cid:17) − sin α (cid:16) s − k k (3 x −
1) (3 + 2 · x )( 3 x + 3 x + 3 x + 3 x ) (cid:17) + 1 − k k (3 x −
1) (3 + 2 · x )( − x + 18 √ x − x + 3)+ cos α − k k (3 x −
1) (3 + 2 · x )(1 − x )+ sin α − k k (3 x −
1) (3 + 2 · x )( − x + 3 x − C ( x, k ) − C ( x, k ) cos α − C ( x, k ) sin α + C ( x, k ) cos α + C ( x, k ) sin α. C i are the following functions C ( x, k ) = ( 15 · x − x − · x + 5 · x + 12 x + 9 · x )+ 1 − k k (3 x −
1) (3 + 2 · x )( − x + 18 √ x − x + 3) ,C ( x, k ) = 2 √ x √ k − k k (3 x −
1) (3 + 2 · x )( 9 x − x ) ,C ( x, k ) = 2 s − k k (3 x −
1) (3 + 2 · x )( 3 x + 3 x + 3 x + 3 x ) ,C ( x, k ) = 1 − k k (3 x −
1) (3 + 2 · x )(1 − x ) ,C ( x, k ) = 1 − k k (3 x −
1) (3 + 2 · x )( − x + 3 x − x, k ). We want to prove that the function d ( x, k, α ) is monotone with respect to x on[0 . , − ] × [0 , × [0 , π ]. We will prove that ∂d∂x ( x, k, α ) < x, k, α ) ∈ [0 . , − ] × [0 , × [0 , π ]. Observe that the following Lemma. Lemma 5.3.
The inequalities ∂C ∂x ( x, k ) < , ∂C ∂x ( x, k ) < x, k ) ∈ [0 . , − ] × [0 , Proof.
We can easily see the first inequality ∂C ∂x ( x, k ) = (3 √ k − k x (1 + k (3 x − × (cid:2) (27 x k + 30 · x k − x k − x − · x k − · x )+(27 xk − x ) + (30 · k − · ) (cid:3) < · )-block and the fact 0 ≤ k ≤
1. The second inequality ∂C ∂x ( x, k ) = 2 − · x − · x − x − · x ! vuut (1 − k )( · x + 3)(1 + k (3 x − (cid:18) x + x + x + x (cid:19) − x ( k − k ) · x +3 !! (1+ k (3 x − − · (1 − k ) x (1+ k (3 x − s (1 − k )( · x +3)(1+ k (3 x − < ∂d∂x ( x, k, α )= ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) cos α − ∂C ∂x ( x, k ) sin α + ∂C ∂x ( x, k ) cos α + ∂C ∂x ( x, k ) sin α ≤ ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) + max { ∂C ∂x ( x, k ) , ∂C ∂x ( x, k ) } = ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) + ∂C ∂x ( x, k )for ∂d∂x ( x, k, α ). The inequality is from Lemma 5.3. The last equality follows from the Claim.Claim: We have the inequality ∂C ∂x ( x, k ) ≥ ∂C ∂x ( x, k ) for all ( x, k ) ∈ [0 . , − ] × [0 , Proof of Claim.
It is enough to show that − k ∂ ( C − C ) ∂x ( x, k ) > x, k ) ∈ [0 . , − ] × [0 , C − C − k = 11 + k (3 x −
1) (3 + 2 · x )(3 − · x + 1 x )We differentiate it by x . ∂∂x ( C − C − k ) = − kx (1 + k (3 x − (3 + 2 · x )(3 − · x + 1 x )+ 11 + k (3 x −
1) ( − · x )(3 − · x + 1 x )+ 11 + k (3 x −
1) (3 + 2 · x )( 4 · x − x )One can easily see that ∂∂x ( C − C − k ) > x, k ) ∈ [0 . , − ] × [0 ,
1] from the followinginequalities. 3 − · x + 1 x < , · x − x > x ∈ [0 . , − ] . The proof of the inequality ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) − ∂C ∂x ( x, k ) + ∂C ∂x ( x, k ) < ∂∂x d ( x, k, α ) < x, k, α ) ∈ [0 . , − ] × [0 , × [0 , π ]. Therefore d ( x, k, α ) ≥ d (3 − , k, α ) for all ( x, k, α ) ∈ [0 . , − ] × [0 , × [0 , π ], in particular the inequality is strict when x = 3 − , and so it is sufficientto prove that d (3 − , k, α ) ≥ k, α ) ∈ [0 , × [0 , π ]. In fact, we have that d (3 − , k, α ) = 108 + 216(1 − k ) − · p k − k cos α − √ − k sin α − − k ) cos α + 36(1 − k ) sin α = 36[(3 √ − k sin α − + ( √ k − p − k ) cos α ) ] . d (3 − , k, α ) ≥ x, k, α ) ∈ [0 . , − ] × [0 , × [0 , π ] wherethe equality holds if and only if k = and sin α = q , cos α = q .We summarize the above results below d (3 − , k, α ) ≥ k, α ) ∈ [0 , × [0 , π ⇒ d ( x, k, α ) > x, k, α ) ∈ [0 . , − ) × [0 , × [0 , π ⇒ D ( x, k, α ) > x, k, α ) ∈ [0 . , − ) × [0 , × [0 , π ⇒ E ( x, y, α ) > x, y, α ) ∈ ( R ′ \ B . (0)) × [0 , π ⇒ G ( q, α ) > q, α ) ∈ ( R \ B . (0)) × [0 , π ⇒ min α ∈ [0 , π ] f q ( α ) > q ∈ R \ B . (0)This implies that min | s | ≤ q + | q | − / ( w ( q ) + s ) t H ( q )( w ( q ) + s ) > q ∈ R \ B . (0) byProposition 4.5 and this proves Step 3 .Therefore, we have proven
Step 1 , , and these cover all domain of q for Theorem 4.3. Aswe mentioned before, Theorem 4.3 implies Theorem 1.1, which tells us the fiberwise convexityof Hill’s lunar problem. A Appendix: Numerical proofs
In the proof of Theorem 4.3, some proofs of inequalities are replaced by computer plots in orderto simplify the argument. We will verify the Figures in Appendix A.1. As we mentioned, wewill prove inequality (9) in Appendix A.2. These proofs will be done by a computer program.The author want to emphasize that there has been lots of advice and help for this Appendixfrom Otto van Koert and referees.
A.1 The proofs of figures
In this section, we will verify the graphs(Figure 2, 5, 9, 10, 11, 12, 13, 14, 15, 16 and 17) whichwe used to show inequalities. We rely on the computer program. Let f : I = [ a, b ] → R be afunction which we want to verify the inequality f ( x ) > f ( x ) < x ∈ I . • Replace the inequality by g ( x ) = ± x n f ( x ) > • Pick a small ǫ > • Compute m := min { g ( x ) | x = a + kǫ, k ∈ N , x < b } , in practice, its lower bound. • Derive an upper bound, say B , for max x ∈ I (cid:12)(cid:12)(cid:12) dgdx ( x ) (cid:12)(cid:12)(cid:12) . • Show that m > ǫB .First three steps can be done with the following simple python program.49 nit=afinal=bepsilon=0.000001def g(x):return "function we are interested in"min=g(init)temp=initwhile temp
Here, we take ǫ = 10 − for every g ’s in Figure 2, 5, 9, 10, 11, 12, 13, 14, 15, 16 and 17.Figure g I m B Fig. 2 x f ( x ) [0 , .
54] 0.3524 4 × Fig. 5 x f ( x ) [0 , .
54] 0.0453 4 × Fig. 9 x f ( x ) [0 . , − ) 0.2461 4 × Fig. 10 − x f ( x ) [0 . , − ) 1.8777 4 × Fig. 11 − x f ( x ) [0 . , .
56] 1.2197 4 × Fig. 12 x f ( x ) [0 . , .
56] 2.7452 4 × Fig. 13 x f m ( x ) , x f M ( x ) [0 . , .
63] 2.6154, 2.9192 4 × Fig. 14 x f m ( x ) , x f M ( x ) [0 . , .
63] 0.5905, 1.5023 4 × Fig. 15 x f m ( x ) , x f M ( x ) [0 . , .
63] 0.5395, 0.7966 4 × Fig. 16 x f m ( x ) , x f M ( x ) [0 . , .
63] 0.9569, 1.1176 4 × Fig. 17 x f m ( x ) , x f M ( x ) [0 . , .
63] 0.8420, 1.5383 4 × The derivative bound B can be shown with the following argument. Note that every g has nonegative degree and consists of rationals and square roots which have no pole on each interval.We can easily see that the coefficient of g is less than 40 and the highest degree is less than 10.Thus the absolute value of the coefficient of dgdx is less than 400. Moreover, it is clear that each dgdx has at most 100 terms. Therefore, we have that (cid:12)(cid:12)(cid:12)(cid:12) dgdx ( x ) (cid:12)(cid:12)(cid:12)(cid:12) < ×
100 = 4 × . Clearly, the property ǫB < m holds for every case. This proves inequalities g i ( x ) > I i for all i = 2 , , , , , , , , , , . A.2 Proof of inequality (9)
We prove inequality (9) using a computer program. The strategy is basically same with AppendixA.1. We will use a python program for finding the minimum on the lattice. We will use Maple50rogram to obtain derivative bounds. This can be used to get a derivative bound in generalcase. Thus, we give the coding at the end of Appendix A.2. Let us explain the idea of the proof. • F ( x, u ) := ∂C ∂x ( x, k ( u )) − ∂C ∂x ( x, k ( u )) − ∂C ∂x ( x, k ( u )) + ∂C ∂x ( x, k ( u )) where k ( u ) = − u +3 u . The new variable u resolves the singularities of derivatives. • Pick a small ǫ x > ǫ u > • Compute M := max { F ( x, u ) | x = 0 .
63 + mǫ x < − , u = 0 + nǫ u < , ( m, n ) ∈ N × N } , inpractice, its upper bound. Check M < • Derive an upper bounds B x > max ( x,u ) ∈ [0 . , − ] × [0 , (cid:12)(cid:12) ∂F∂u ( x, u ) (cid:12)(cid:12) and B u > max ( x,u ) ∈ [0 . , − ] × [0 , (cid:12)(cid:12) ∂F∂x ( x, u ) (cid:12)(cid:12) . • Show that | M | > ǫ x B x and | M | > ǫ u B u .Motivated by the derivative bounds we will derive in the following, we take the ǫ x = . × and ǫ u = . × . The next step is can be done with the following simple code(pseudo code). init_x, final_x=0.63, 3^(-1/3)init_u, final_u=0, 1epsilon_x, epsilon_u= "As above"F(x, u)="The function we want to get the maximum on the lattice."Max=F(init_x, init_u)temp_x, temp_u=init_x, init_uwhile temp_x M < − 19. We explain the procedure to get the derivative bounds B x and B u . We divide the functions composing the functions C , C , C and C into two classesof functions, say ”abstract functions”, ”rational functions”. Define abstract functions a ( x, u ) = 1 p k ( u )(3 x − , a ( u ) = p − k ( u ) , a ( u ) = p k ( u ) ,a ( x ) = s · x , a ( x ) = 9 x − x , r ( x ) = 15 · x − x − · x + 5 · x + 12 x + 9 · x , r ( x ) = − x + 18 √ x − x + 3 ,r ( x ) = 3 x + 3 x + 3 x + 3 x , r ( x ) = 1 − x . Note that r i ’s do not have a positive degree term. Then we have that C ( x, u ) = r ( x ) + a ( x, u ) a ( u ) a ( x ) r ( x ) , C ( x, u ) = 2 a ( x, u ) a ( u ) a ( u ) a ( x ) a ( x ) ,C ( x, u ) = 2 a ( x, u ) a ( u ) a ( x ) r ( x ) , C ( x, u ) = a ( x, u ) a ( u ) a ( x ) r ( x ) . When we take derivatives, abstract functions are formally differentiated. Since F ( x, u ) = ∂C ∂x ( x, u ) − ∂C ∂x ( x, u ) − ∂C ∂x ( x, u ) + ∂C ∂x ( x, u ), the following functions b = a , b = ∂ x a , b = ∂ u a , b = ∂ xu a , b = ∂ xx a ,b = a , b = ∂ u a , b = a , b = ∂ u a ,b = a , b = ∂ x a , b = ∂ xx a , b = a , b = ∂ x a , b = ∂ xx a will appear in the expansions of ∂F∂x and ∂F∂u . Then we have the following expressions ∂F∂x = X I ⊂ J c xI · b I , ∂F∂u = X I ⊂ J c uI · b I (10)where J = { , , · · · , } , b I = Q i ∈ I b i and c xI , c uI are monomials of non-positive degree. Notethat b I can be repeated in order to have monomial coefficients. We can easily see that theinequalities | a ( x, u ) | < . , | ∂ x a ( x, u ) | < . , | ∂ u a ( x, u ) | < . , | ∂ xx a | < , | ∂ xu a ( x, y ) | < . , | a ( u ) | ≤ , | ∂ u a ( u ) | ≤ √ , | a ( u ) | ≤ , | ∂ u a ( u ) | ≤ √ , | a ( x ) | ≤ s · . , | ∂ x a ( x ) | < . , , | ∂ xx a ( x ) | < , | a ( x ) | < . , | ∂ x a ( x ) | < . , , | ∂ xx a ( x ) | < . Using triangle inequalities, we can obtain upper bounds (cid:12)(cid:12)(cid:12)(cid:12) ∂F∂x (cid:12)(cid:12)(cid:12)(cid:12) ≤ X I ⊂ J | c xI · b I | , (cid:12)(cid:12)(cid:12)(cid:12) ∂F∂u (cid:12)(cid:12)(cid:12)(cid:12) ≤ X I ⊂ J | c uI · b I | . We replace the abstract functions and their derivatives by their individual upper bounds obtainedabove. In addition we substitute x = 0 . 63 into | c xI | and | c uI | to get upper bounds (cid:12)(cid:12)(cid:12)(cid:12) ∂F∂x (cid:12)(cid:12)(cid:12)(cid:12) ≤ B x ∼ = 2 . × , (cid:12)(cid:12)(cid:12)(cid:12) ∂F∂u (cid:12)(cid:12)(cid:12)(cid:12) ≤ B u ∼ = 4 . × . This proves inequality (9). We leave the Maple code of this procedure for the bound of (cid:12)(cid:12) ∂F∂u (cid:12)(cid:12) .52 eclare the "rational functions" r_0(x), r_1(x), r_2(x) and r_3(x) as above: Using the similar code, derivative bound2 . × for (cid:12)(cid:12) ∂F∂x (cid:12)(cid:12) can be obtained analogously. This completes the proof of inequality (9).53 eferences [1] J. C. ´Alvarez Paiva, F. Balacheff and K. Tzanev, Isosystolic inequalities for optical hyper-surfaces , 2013, arXiv:1308.5522v1.[2] P. Albers, J. W. Fish, U. Frauenfelder, H. Hofer and O. van Koert, Global surfaces ofsection in the planar restricted 3-body problem , Arch. Ration. Mech. Anal. 204 (2012), no.1, 273–284.[3] P. Albers, J. W. Fish, U. 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