Linear Weingarten surfaces in Euclidean and hyperbolic space
aa r X i v : . [ m a t h . DG ] J un Linear Weingarten surfaces in Euclidean and hyperbolic space
Rafael L´opez ∗ This paper is dedicated to Manfredo do Carmo in admiration for his mathematical achievementsand his influence on the field of differential geometry of surfaces
Abstract
In this paper we review some author’s results about Weingarten surfaces in Euclidean space R and hyperbolic space H . We stress here in the search of examples of linear Weingartensurfaces that satisfy a certain geometric property. First, we consider Weingarten surfaces in R that are foliated by circles, proving that the surface is rotational, a Riemann example ora generalized cone. Next we classify rotational surfaces in R of hyperbolic type showing thatthere exist surfaces that are complete. Finally, we study linear Weingarten surfaces in H thatare invariant by a group of parabolic isometries, obtaining its classification. MSC:
Keywords : Weingarten surface; cyclic surface; Riemann examples; parabolic surface.
A surface S in Euclidean space R or hyperbolic space H is called a Weingarten surface if there issome smooth relation W ( κ , κ ) = 0 between its two principal curvatures κ and κ . In particular,if K and H denote respectively the Gauss curvature and the mean curvature of S , W ( κ , κ ) = 0implies a relation U ( K, H ) = 0. The classification of Weingarten surfaces in the general caseis almost completely open today. After earlier works in the fifties due to Chern, Hopf, Voss,Hartman, Winter, amongst others, there has been recently a progress in this theory, speciallywhen the Weingarten relation is of type H = f ( H − K ) and f elliptic. In such case, the surfacessatisfy a maximum principle that allows a best knowledge of the shape of such surfaces. Theseachievements can see, for example, in [2, 4, 5, 14, 15, 16].The simplest case of functions W or U is that they are linear, that is, aκ + bκ = c or aH + bK = c, (1)where a, b and c are constant. Such surfaces are called linear Weingarten surfaces. Typical exam-ples of linear Weingarten surfaces are umbilical surfaces, surfaces with constant Gauss curvatureand surfaces with constant mean curvature. ∗ Partially supported by MEC-FEDER grant no. MTM2007-61775 and Junta de Andaluc´ıa grant no. P06-FQM-01642. cyclic surface in Euclidean space R or H is a surface determinedby a smooth uniparametric family of circles. Thus, a cyclic surface S is a surface foliated by circlesmeaning that there is a one-parameter family of planes which meet S in these circles. The planesare not assumed parallel, and if two circles should lie in planes that happen to be parallel, thecircles are not assumed coaxial. Rotational surfaces are examples of cyclic surfaces.Our first result is motivated by what happens for cyclic surfaces with constant mean curvature H .Recall that the catenoid is the only minimal ( H = 0) rotational surface in R . If the surface isnot rotational, then the only cyclic minimal surfaces are a family of examples of periodic minimalsurfaces discovered by Riemann, usually called in the literature as Riemann examples [13]. If themean curvature H is a non-zero constant, then the only cyclic surfaces are the surfaces of revolution(Delaunay surfaces) [12]. In order to find new examples of linear Weingarten surfaces, we pose thefollowing question: do exist non-rotational cyclic surfaces that are linear Weingarten surfaces?In Section 2 we prove the following result: Theorem 1.1.
Let S be a cyclic surface in Euclidean space R .1. If S satisfies a relation of type aκ + bκ = c , then S is a surface of revolution or it is aRiemann example ( H = 0 ).2. If S satisfies a relation of type aH + bK = c , then S is a surface of revolution or it is aRiemann example ( H = 0 ) or it is a generalized cone ( K = 0 ). Recall that a generalized cone is a cyclic surface formed by a uniparametric family of u -circleswhose centres lie in a straight line and the radius function is linear on u . These surfaces have K ≡ R with constant Gaussian curvature[6].After Theorem 1.1, we focus on Weingarten surfaces of revolution in R . The classification of linearWeingarten surfaces strongly depends on the sign of ∆ := a + 4 bc . If ∆ >
0, the surface is saidelliptic and satisfies good properties, as for example, a maximum principle: see [5, 14]. If ∆ = 0,the surface is a tube, that is, a cyclic surface where the circles have the same radius. Finally, if∆ <
0, the surface is said hyperbolic (see [1]). In Section 3 we study hyperbolic rotational surfacesin R . We do an explicit description of the hyperbolic rotational linear Weingarten. Examples ofhyperbolic Weingarten surfaces are the surfaces with constant negative Gaussian curvature K : wetake a = 0, b = 1 and c < R , we obtain (see Theorem 3.5): Theorem 1.2.
There exists a family of hyperbolic linear Weingarten complete rotational surfacesin R that are non-embedded and periodic. Finally, we are interested in linear Weingarten surfaces of revolution in hyperbolic space H . Inhyperbolic space there exist three types of rotational surfaces. We will study one of them, called2 arabolic surfaces , that is, surfaces invariant by a group of parabolic isometries of the ambientspace. This was began by do Carmo and Dajczer in the study of rotational surfaces in H withconstant curvature [3] and works of Gomes, Leite, Mori et al. We will consider problems suchas existence, symmetry and behaviour at infinity. As a consequence of our work, we obtain thefollowing Theorem 1.3.
There exist parabolic complete surfaces in H that satisfy the relation aH + bK = c . Part of the results of this work have recently appeared in a series of author’s papers: [7, 8, 9, 11]. R aH + bK = c . The proof consists into twosteps. First, we prove Theorem 2.1.
Let S be a surface that satisfies aH + bK = c and it is foliated by circles lying ina one-parameter family of planes. Then either S is a subset of a round sphere or the planes of thefoliation are parallel.Proof. Consider P ( u ) the set of planes of the foliation, that is, S = S u ∈ I P ( u ) ∩ S , u ∈ I ⊂ R ,and such that P ( u ) ∩ S is a circle for each u . Assume that the planes P ( u ) are not parallel . Thenwe are going to show that the surfaces is a sphere. The proof follows the same ideas for the caseof the constancy of the mean curvature [12]. Let Γ be an orthogonal curve to the foliation planes,that is, Γ ′ ( u ) ⊥ P ( u ). If { t , n , b } denotes the usual Frenet trihedron of Γ, the surface S is locallyparametrized by X ( u, v ) = c ( u ) + r ( u )(cos v n ( u ) + sin v b ( u )) , (2)where r = r ( u ) > c = c ( u ) denote respectively the radius and centre of each circle P ( u ) ∩ S .We compute the mean curvature and the Gauss curvature of S using the usual local formulae H = eG − f F + gE EG − F ) , K = eg − f EG − F . Here { E, F, G } and { e, f, g } represent the coefficients of the first and second fundamental form,respectively. Then the relation aH + bK = c writes in terms of the curve Γ. Using the Frenetequations of Γ, we are able to express the relation aH + bK = c as a trigonometric polynomial oncos ( nv ) and sin ( nv ): A + X n =1 (cid:18) A n ( u ) cos ( nv ) + B n ( u ) sin ( nv ) (cid:19) = 0 , u ∈ I, v ∈ [0 , π ] . Here A n and B n are smooth functions on u . Because the functions cos ( nv ) and sin ( nv ) areindependent, all coefficient functions A n , B n must be zero. This leads to a set of equations, whichwe wish to solve. Because the curve Γ is not a straight line, its curvature κ does not vanish.3he proof consists into the explicit computation of the coefficients A n and B n and solving A n = B n = 0. The proof program begins with the equations A = 0 and B = 0, which yields relationsbetween the geometric quantities of the curve Γ. By using these data, we follow with equations A = B = 0 and so on, until to arrive with n = 0. The author was able to obtain the results usingthe symbolic program Mathematica to check his work: the computer was used in each calculationseveral times, giving understandable expressions of the coefficients A n and B n . Finally, we achieveto show that X is a parametrization of a round sphere.Once proved Theorem 2.1, the following step consists to conclude that either the circles of thefoliation must be coaxial (and the surface is rotational) or that K ≡ H ≡
0. In the lattercases, the Weingarten relation (1) is trivial in the sense that a = c = 0 or b = c = 0. Theorem 2.2.
Let S be a cyclic surface that satisfies aH + bK = c . If the foliation planes areparallel, then either S is a surface of revolution or a = c = 0 or b = c = 0 .Proof. After an isometry of the ambient space, we parametrize S as X ( u, v ) = ( f ( u ) , g ( u ) , u ) + r ( u )(cos v, sin v, , where f, g and r are smooth functions on u , u ∈ I ⊂ R and r ( u ) > S is a surface of revolution if and only if f y g are constant functions.Proceeding similarly as in the proof of Theorem 2.1, Equation aH + bK = c is equivalent to anexpression X n =0 (cid:18) A n ( u ) cos ( nv ) + B n ( u ) sin ( nv ) (cid:19) = 0 . Again, the functions A n and B n must vanish on I . Assuming that the surface is not rotational,that is, f ′ ( u ) g ′ ( u ) = 0 at some point u , we conclude that a = c = 0 or b = c = 0.Recall what happens in the latter cases. The computation of H ≡ K ≡ f ′′ = λr , g ′′ = µr , λ + µ ) r + r ′ − rr ′′ = 0 , (3)and f ′′ = g ′′ = r ′′ = 0 , (4)respectively. If (3) holds, we have the equations that describe the Riemann examples ( λ + µ = 0)and the catenoid ( λ = µ = 0). In the case (4), the surface S is a generalized cone.As a consequence of the above Theorems 2.1 and 2.2, we obtain Theorem 1.1 announced in theintroduction of this work. Finally, the previous results allow us to give a characterization ofRiemann examples and generalized cones in the class of linear Weingarten surfaces. Corollary 2.3.
Riemann examples and generalized cones are the only non-rotational cyclic sur-faces that satisfy a Weingarten relation of type aH + bK = c . Hyperbolic linear Weingarten surfaces in R S in Euclidean space that satisfy the relation a H + b K = c (5)where a , b and c are constants under the relation a + 4 bc <
0. These surfaces are called hyperboliclinear Weingarten surfaces . In particular, c = 0, which can be assumed to be c = 1. Thus thecondition ∆ < a + 4 b <
0. In this section, we study these surfaces in the class ofsurfaces of revolution. Equation (5) leads to an ordinary differential equation that describes thegenerating curve α of the surface. Without loss of generality, we assume S is a rotational surfacewhose axis is the x -axis. If α ( s ) = ( x ( s ) , , z ( s )) is arc-length parametrized and the surface is givenby X ( s, φ ) = ( x ( s ) , z ( s ) cos φ, z ( s ) sin φ ), then (5) leads to a cos θ ( s ) − z ( s ) θ ′ ( s )2 z ( s ) − b cos θ ( s ) θ ′ ( s ) z ( s ) = 1 , (6)where θ = θ ( s ) the angle function that makes the velocity α ′ ( s ) at s with the x -axis, that is, α ′ ( s ) = (cos θ ( s ) , , sin θ ( s )). The curvature of the planar curve α is given by θ ′ . In this section,we discard the trivial cases in (5), that is, a = 0 (constant Gauss curvature) and b = 0 (constantmean curvature).The generating curve α is then described by the solutions of the O.D.E. x ′ ( s ) = cos θ ( s ) z ′ ( s ) = sin θ ( s ) θ ′ ( s ) = a cos θ ( s ) − z ( s ) az ( s ) + 2 b cos θ ( s ) (7)Assume initial conditions x (0) = 0 , z (0) = z , θ (0) = 0 . (8)Without loss of generality, we can choose the parameters a and z to have the same sign: in ourcase, we take to be positive numbers.A first integral of (7)-(8) is given by z ( s ) − az ( s ) cos θ ( s ) − b cos θ ( s ) − ( z − az − b ) = 0 . (9)By the uniqueness of solutions, any solution α ( s ) = ( x ( s ) , , z ( s )) of (7)-(8) is symmetric withrespect to the line x = 0.In view of (8), the value of θ ′ ( s ) at s = 0 is θ ′ (0) = a − z az + 2 b . Our study depends on the sign of θ ′ (0). We only consider the case z > − ba (10)which implies that z > a/
2. The denominator in the third equation of (7) is positive since it doesnot vanish and at s = 0, its value is az + 2 b >
0. As z > a/
2, the numerator in (7) is negative.Thus we conclude that the function θ ′ ( s ) is negative anywhere.5rom (9), we write the function z = z ( s ) as z ( s ) = 12 (cid:18) a cos θ ( s ) + q ( a + 4 b ) cos θ ( s ) + 4( z − az − b ) (cid:19) . (11) Lemma 3.1.
The maximal interval of the solution ( x, z, θ ) of (7)-(8) is R .Proof. The result follows if we prove that the derivatives x ′ , z ′ and θ ′ are bounded. In view of (7),it suffices to show it for θ ′ (recall that θ ′ ( s ) < m suchthat m ≤ θ ′ ( s ) for all s . To be exact, we show the existence of constants δ and η independent on s , with η < < δ , such that az ( s ) + 2 b cos θ ( s ) ≥ δ and a cos θ ( s ) − z ( s ) ≥ η. (12)Once proved this, it follows from (7) that θ ′ ( s ) ≥ ηδ := m. (13)Define the function f ( z ) := z − az − b . The function f is strictly increasing on z for z > a/ a + 4 b <
0, we have a < − ba . As z satisfies (10), there exists ǫ > z − az − b = f ( − ba ) + ǫ = b ( a + 4 b ) a + ǫ. From (11), z ( s ) ≥ a cos θ + r ( a + 4 b ) cos θ + 4 ba ( a + 4 b ) + 4 ǫ ! ≥ (cid:18) a cos θ − a + 4 ba + ǫ ′ (cid:19) , for a certain positive number ǫ ′ . By using that a + 4 b < az ( s ) + 2 b cos θ ( s ) ≥ a + 4 b θ ( s ) −
1) + a ǫ ′ ≥ a ǫ ′ := δ. By using (11) again, we obtain a cos θ ( s ) − z ( s ) ≥ − p ( a + 4 b ) cos ( s ) θ + 4 f ( z ) ≥ − p f ( z ) := η, which concludes the proof of this lemma. Lemma 3.2.
For each solution ( x, z, θ ) of (7)-(8), there exists M < such that θ ′ ( s ) < M .Proof. It suffices if we prove that there exist δ , η , with η < < δ such that az ( s ) + 2 b cos θ ( s ) ≤ δ and a cos θ ( s ) − z ( s ) ≤ η , since (7) yields θ ′ ( s ) ≤ δ /η := M . Using (11), we have az ( s ) + 2 b cos θ ( s ) = 12 (cid:16) ( a + 4 b ) cos θ ( s ) + a p ( a + 4 b ) cos θ ( s ) + 4 f ( z ) (cid:17) ≤ a p f ( z ) := δ .
6n the other hand, a cos θ ( s ) − z ( s ) = − p ( a + 4 b ) cos θ ( s ) + 4 f ( z ) ≤ − p ( a + 4 b ) + 4 f ( − b/a ) := η . Lemma 3.2 implies that θ ( s ) is strictly decreasing withlim s →∞ θ ( s ) = −∞ . Since Lemma 3.1 asserts that any solution is defined for any s , put T > θ ( T ) = − π . We prove that α is a periodic curve. Lemma 3.3.
Under the hypothesis of this section and with the above notation, we have: x ( s + T ) = x ( s ) + x ( T ) z ( s + T ) = z ( s ) θ ( s + T ) = θ ( s ) − π Proof.
This is a consequence of the uniqueness of solutions of (7)-(8). We only have to show that z ( T ) = z . But this is a direct consequence of the assumption (10), that a + 4 b < α under the hypothesis (10). Due to the monotonicity of θ , let T , T and T be the pointsin the interval [0 , T ] such that the function θ takes the values − π/ , − π and − π/ θ with the time coordinate s , it is easy to verify the followingTable: s θ x ( s ) z ( s )[0 , T ] [ − π ,
0] increasing decreasing[ T , T ] [ − π, − π ] decreasing decreasing[ T , T ] [ − π , − π ] decreasing increasing[ T , T ] [ − π, − π ] increasing increasing Theorem 3.4.
Let α = α ( s ) = ( x ( s ) , , z ( s )) be the profile curve of a rotational hyperbolic surface S in R where α is the solution of (7)-(8). Assume that the initial condition on z satisfies z > − ba . Then (see Fig. 1)1. The curve α is invariant by the group of translations in the x -direction given by the vector ( x ( T ) , , .2. In the period [0 , T ] of z given by Lemma 3.3, the function z = z ( s ) presents one maximumat s = 0 and one minimum at s = T . Moreover, α is symmetric with respect to the verticalline at x = 0 and x = x ( T ) . . The height function of α , that is, z = z ( s ) , is periodic.4. The curve α has self-intersections and its curvature has constant sign.5. The part of α between the maximum and the minimum satisfies that the function z ( s ) isstrictly decreasing with exactly one vertical point. Between this minimum and the next max-imum, z = z ( s ) is strictly increasing with exactly one vertical point.6. The velocity α ′ turns around the origin. Theorem 3.5.
Let S be a rotational hyperbolic surface in R whose profile curve α satisfies thehypothesis of Theorem 3.4. Then S has the following properties:1. The surface has self-intersections.2. The surface is periodic with infinite vertical symmetries.3. The surface is complete.4. The part of α between two consecutive vertical points and containing a maximum correspondswith points of S with positive Gaussian curvature; on the other hand, if this part contains aminimum, the Gaussian curvature is negative. -2 -1 1 20.511.522.53 Figure 1: The generating curve of a rotational hyperbolic surfaces, with a = − b = 2. Here z = 3.The curve α is periodic with self-intersections.As it was announced in Theorem 1.2, and in order to distinguish from the surfaces of negativeconstant Gaussian curvature, we conclude from Theorem 3.5 the following Corollary 3.6.
There exists a one-parameter family of rotational hyperbolic linear Weingartensurfaces that are complete and with self-intersections in R . Moreover, the generating curves ofthese surfaces are periodic. H
3A parabolic group of isometries of hyperbolic space H is formed by isometries that leave fix onedouble point of the ideal boundary S ∞ of H . A surface S in H is called a parabolic surface if it is8nvariant by a group of parabolic isometries. A parabolic surface S is determined by a generatingcurve α obtained by the intersection of S with any geodesic plane orthogonal to the orbits of thegroup.We consider the upper half-space model of H , namely, H =: R = { ( x, y, z ) ∈ R ; z > } equipped with the metric h , i = dx + dy + dz z . The ideal boundary S ∞ of H is identified with theone point compactification of the plane Π ≡ { z = 0 } , that is, S ∞ = Π ∪ {∞} . In what follows, wewill use the words vertical or horizontal in the usual affine sense of R . Denote L = S ∞ ∩ { y = 0 } .Let G be a parabolic group of isometries. In the upper half-space model, we take the point ∞ ∈ S ∞ as the point that fixes G . Then the group G is defined by the horizontal (Euclidean) translationsin the direction of a horizontal vector ξ with ξ ∈ Π which can be assumed ξ = (0 , , S invariant by G parametrizes as X ( s, t ) = ( x ( s ) , t, z ( s )), where t ∈ R andthe curve α = ( x ( s ) , , z ( s )), s ∈ I ⊂ R , is assumed to be parametrized by the Euclidean arc-length. The curve α is the generating curve of S . We write α ′ ( s ) = (cos θ ( s ) , , sin θ ( s )), for acertain differentiable function θ , where the derivative θ ′ ( s ) is the Euclidean curvature of α . Withrespect to the unit normal vector N ( s, t ) = ( − sin θ ( s ) , , cos θ ( s )), the principal curvatures are κ ( s, t ) = z ( s ) θ ′ ( s ) + cos θ ( s ) , κ ( s, t ) = cos θ ( s ) . The relation aH + bK = c writes then (cid:16) a b cos θ ( s ) (cid:17) z ( s ) θ ′ ( s ) + a cos θ ( s ) − b sin θ ( s ) = c. (14)We consider initial conditions x (0) = 0 , z (0) = z > , θ (0) = 0 . (15)Then any solution { x ( s ) , z ( s ) , θ ( s ) } satisfies properties of symmetry which are consequence of theuniqueness of solutions of an O.D.E. For example, the solution is symmetric with respect to thevertical straight line x = 0. Using uniqueness again, we infer immediately Proposition 4.1.
Let α be a solution of the initial value problem (14)-(15) with θ (0) = θ . If θ ′ ( s ) = 0 at some real number s , then α is parameterized by α ( s ) = ((cos θ ) s, , (sin θ ) s + z ) ,that is, α is a straight line and the corresponding surface is a totally geodesic plane, an equidistantsurface or a horosphere. In view of this proposition, we can assume that the function θ ′ ( s ) do not vanish, that is, θ is amonotonic function on s . At s = 0, Equation (14) is θ ′ (0) = 2 z c − aa + 2 b . This means that the study of solutions of (14)-(15) must analyze a variety of cases depending onthe sign of θ ′ (0). In this section, we are going to consider some cases in order to show techniquesand some results. First, assume that c = 0, which it can be assumed to be c = 1. Then we write(14) as θ ′ ( s ) = 2 1 − a cos θ ( s ) + b sin θ ( s ) z ( s )( a + 2 b cos θ ( s )) . (16)Our first result considers a case where it is possible to obtain explicit examples.9 heorem 4.2. Let α ( s ) = ( x ( s ) , , z ( s )) be the generating curve of a parabolic surface S inhyperbolic space H that satisfies aH + bK = 1 with initial conditions (15). Assume a +4 b +4 b = 0 .Then α describes an open of an Euclidean circle in the xz -plane.Proof. Equation (14) reduces into − bz ( s ) θ ′ ( s ) = a + 2 b cos θ ( s ) . By differentiation with respect to s , we obtain z ( s ) θ ′′ ( s ) = 0, that is, θ ′ ( s ) is a constant function.Since θ ′ ( s ) describes the Euclidean curvature of α , we conclude that α parametrizes an Euclideancircle in the xz -plane and the assertion follows. This circle may not to be completely included inthe halfspace R .From now, we assume a + 4 b + 4 b = 0. Let us denote by ( − ¯ s, ¯ s ) the maximal domain of thesolutions of (14)-(15). By the monotonicity of θ ( s ), let θ = lim s → ¯ s θ ( s ). Theorem 4.3 (Case 0 < a < . Let α ( s ) = ( x ( s ) , , z ( s )) be the generating curve of a parabolicsurface S in hyperbolic space H that satisfies aH + bK = 1 with initial conditions (15). Assume b = 0 and < a < .1. If a + 2 b < , α has one maximum and α is a concave (non-entire) vertical graph. If b < − (1 + √ − a ) / , the surface S is complete and intersects S ∞ making an angle θ suchthat θ − b sin θ = 0 . The asymptotic boundary of S is formed by two parallel straightlines. See Fig. 2 (a). If − (1 + √ − a ) / < b < − a/ , then S is not complete. See Fig. 2(b).2. Assume a + 2 b > . If a − b > , then S is complete and invariant by a group of translationsin the x -direction. Moreover, α has self-intersections and it presents one maximum and oneminimum in each period. See Fig. 3, (a). If a − b ≤ , then S is not complete. Moreover α is not a vertical graph with a minimum. See Fig. 3, (b). -1.5 -1 -0.5 0.5 1 1.50.20.40.60.81 -0.75 -0.5 -0.25 0.25 0.5 0.750.20.40.60.81 (a) (b)Figure 2: The generating curve of a parabolic surface with aH + bK = 1, with 0 < a < a + 2 b <
0. Here z = 1 and a = 0 .
5. In the case (a), b = − b = − . H with the property aH + bK = c , such as it was announced in Theorem1.3. 10 (a) (b)Figure 3: The generating curve of a parabolic surface with aH + bK = 1, with 0 < a < a + 2 b >
0. Here z = 1 and a = 0 .
5. In the case (a), b = − . b = 0 . Proof.
The second derivative of θ ′′ ( s ) satisfies − θ ′ ( s ) sin θ ( s ) h bθ ′ ( s ) + (cid:16) a b cos θ ( s ) (cid:17) i + (cid:16) a b cos θ ( s ) (cid:17) z ( s ) θ ′′ ( s ) = 0 . (17)1. Case a + 2 b <
0. Then θ ′ (0) < θ ( s ) is strictly decreasing. If cos θ ( s ) = 0 at somepoint s , then (14) gives ( a/ z ( s ) θ ′ ( s ) − b − b ≥ −
1, cos θ ( s ) = 0 and − π/ < θ ( s ) < π/
2. In the case that b < − a + 2 b cos θ ( s ) <
0, it follows from (14)that a cos θ ( s ) − b sin θ ( s ) − < s , in particular, cos θ ( s ) = 0. This provesthat x ′ ( s ) = cos θ ( s ) = 0 and so, α is a vertical graph on L . This graph is concave since z ′′ ( s ) = θ ′ ( s ) cos θ ( s ) <
0. Moreover, this implies that ¯ s < ∞ since on the contrary, and as z ( s ) is decreasing with z ( s ) >
0, we would have z ′ ( s ) →
0, that is, θ ( s ) →
0: contradiction.For s > z ′ ( s ) = sin θ ( s ) < z ( s ) is strictly decreasing. Set z ( s ) → z (¯ s ) ≥
0. The tworoots of 4 b + 4 b + a = 0 on b are b = − (1 ± √ − a ). Moreover, and from a + 2 b <
0, wehave −
12 (1 + p − a ) < − a < −
12 (1 − p − a ) . (a) Subcase b < − (1 + √ − a ) /
2. Under this assumption, a + 4 b + 4 b >
0. Since a < a + 2 b cos θ ( s ) < − p a + 4 b + 4 b. (18)If z (¯ s ) >
0, then lim s → ¯ s θ ′ ( s ) = −∞ . In view of (16) we have a + 2 b cos θ (¯ s ) = 0:contradiction with (18). Hence, z (¯ s ) = 0 and α intersects L with an angle θ satisfying a cos θ − b sin θ − − (1 + √ − a ) / < b < − a/
2. Now a + 4 b + 4 b <
0. The function 1 − a cos θ ( s ) + b sin θ ( s ) is strictly decreasing and its value at ¯ s satisfies cos θ ( s ) > − a/ b .Thus 1 − a cos θ ( s ) + b sin θ ( s ) ≥ a + 4 b + 4 b b > . (19)Assume z (¯ s ) = 0. Then (16) and (19) imply that θ ′ (¯ s ) = −∞ . Combining (16) and(17), we have θ ′′ ( s ) θ ′ ( s ) = b sin θ ( s ) z ( s ) (cid:0) a + b cos θ ( s ) (cid:1) + sin θ ( s ) (cid:0) a + b cos θ ( s ) (cid:1) − a cos θ ( s ) + b sin θ ( s ) . θ (¯ s ) = 0, we concludelim s → ¯ s θ ′′ ( s ) θ ′ ( s ) = −∞ . On the other hand, using L’Hˆopital rule, we havelim s → ¯ s z ( s ) θ ′ ( s ) = lim s → ¯ s − sin θ ( s ) θ ′′ ( s ) θ ′ ( s ) = 0 . By letting s → ¯ s in (16), we obtain a contradiction. Thus, z (¯ s ) >
0. This means thatlim s → ¯ s θ ′ ( s ) = −∞ and it follows from (14) thatlim s → ¯ s (cid:16) a b cos θ ( s ) (cid:17) = 0 .
2. Case a + 2 b >
0. Then θ ′ (0) > θ ( s ) is strictly increasing. We distinguish two possibil-ities:(a) Subcase a − b >
0. We prove that θ ( s ) reaches the value π . On the contrary, θ ( s ) < π and z ( s ) is an increasing function. The hypothesis a − b > a + 2 b > a + 2 b cos θ ( s ) ≥ δ > δ . From (16), θ ′ ( s ) is bounded and then¯ s = ∞ . In particular, lim s →∞ θ ′ ( s ) = 0. As both a − b and a + 2 b are positive numbers,the function bθ ′ ( s ) + ( a + 2 b cos θ ( s )) is positive near ¯ s = ∞ . Then using (17), θ ′′ ( s ) ispositive for a certain value of s big enough, which it is impossible. As conclusion, θ ( s )reaches the value π at some s = s . By the symmetry properties of solutions of (14), α is symmetric with respect to the line x = x ( s ) and the velocity vector of α rotatesuntil to the initial position. This means that α is invariant by a group of horizontaltranslations.(b) Subcase a − b ≤
0. As θ ′ ( s ) >
0, Equation (16) says that cos θ ( s ) = −
1, and so, θ ( s )is bounded by − π < θ ( s ) < π . As in the above subcase, if ¯ s = ∞ , then θ ′ ( s ) →
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