A thin filament, or rod, can be modelized by a ribbon constituted of a space ...... [25] A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th ed.
Towards a classification of Euler-Kirchhoff filaments Michel Nizette∗ and Alain Goriely† † University of Arizona, Department of Mathematics and Program in Applied Mathematics, Bldg #89, Tucson, AZ85721 and ∗ Universit´e Libre de Bruxelles Departement de Physique Statistique CP231 e-mail: mnizette@ ulb.ac.be; goriely@ math.arizona.edu
Abstract Euler-Kirchhoff filaments are solutions of the static Kirchhoff equations for elastic rods with circular cross-sections. These equations are known to be formally equivalent to the Euler equations for spinning tops. This equivalence is used to provide a classification of the different shapes a filament can assume. Explicit formulas for the different possible configurations and specific results for interesting particular cases are given. In particular, conditions for which the filament has points of self-intersection, self-tangency, vanishing curvature or when it is closed or localized in space are provided. The average properties of generic filaments are also studied. They are shown to be equivalent to helical filaments on long length scales. KEYWORDS: Rods, Spinning tops, Kirchhoff equations, Elliptic functions. PACS: 46.30.Cn, 03.20.+i
I
Introduction
The study of elastic deformations in rods has a long tradition in mathematics, physics and engineering, dating back to Euler and Lagrange. Engineers have been confronted to the problem of coiling in sub-oceanic cables and have tried to understand the process of loop formation in twisted wires [1, 2, 3]. In chemistry and biology, increasing interest is taken in the elastic character of filamentary structures such as polymers [4, 5, 6, 7] (such as DNA molecules [8, 9, 10, 11, 12, 13, 14, 15]) and bacterial fibers [16, 17, 18], for which the macroscopic theory of rods provides an idealized model. Long, twisted structures also play an important role in hydrodynamic models [19] such as scroll wave propagation [20], vortex tube motion [21], or sun spots formation and solar corona heating [22, 23]. The Kirchhoff model (1859) [24] provides the basic framework for the theory of elastic filaments. A remarkable feature of this model, known as the Kirchhoff kinetic analogy, is that the equations governing the static phenomena are formally equivalent to the Euler equations describing the motion of a rigid body with a fixed point under an external force field. The statics of rods is thus intimately connected to the dynamics of spinning tops, a problem to which innumerable work has been devoted. For instance, the most studied case where the filament has a circular cross-section is shown to correspond to a top having an axis of revolution, in which case the equations are fully integrable. There is a rich mathematical literature on the statics of rods (see for instance [26]). In the particular case of circular cross-sections and linear elasticity, various researchers have considered particular equilibrium filament shapes (helices [25, 27], rings [28, 3], localizing buckling modes [1, 29], solutions
1
having points with vanishing curvature [30], supercoiled helices [31], see also [32] for an Hamiltonian formulation and [33] for a group theory approach). Recently, departing from the traditional Euler angles approach, Shi and Hearst (1994) have obtained a closed form of the general solution of the static Kirchhoff equations for circular cross-sections [34]. Despite this achievement, it remains highly nontrivial to obtain a global picture of all possible static filament shapes. A step towards a general geometric classification of the equilibrium solutions is provided here by considering in more detail the analogy between filaments and spinning tops. In Section III, we rederive Shi and Hearst’s general solution, keeping an explicit dependence in the constants of the motion of the spinning top, and we discuss in detail various filament configurations on the basis of the corresponding orbits of the top, recovering the aforementioned shapes as particular cases. In this paper, we provide new results on the different configurations of Kirchhoff filaments in 2D and 3D. We give explicit formulas for the centerline coordinate and find conditions for self-tangency and self-intersection of planar filaments, conditions for vanishing curvature and average behavior of long filaments. Explicit formulas for periodic or localized (homoclinic) filaments in space are also given
II
The Kirchhoff Model
We first introduce the Kirchhoff model. It accounts for the dynamics of a thin elastic filament subject to internal stresses and boundary constraints. A filament is a unidimensional piece of elastic material which can be mathematically modelized by a curve in space, together with extra information about its twist, that is, how longitudinal material lines on the edge of the filament wind around it. This curve-plus-twist concept is formalized in the notion of ribbon discussed in Section II.1 as a preliminary.
II.1
Space Curves and Ribbons
We define a dynamical space curve R(s, t) as a smooth function mapping R2 into the physical space R3 , and taking as variables the arc length s and the time t. For every s and t we define the Frenet basis (n, b, t) to be the normal, binormal and tangent vectors to the curves. These vectors form the Frenet basis. The tangent vector is a unit vector given by: t = ∂R ∂s . The curvature κ of the curve at the point s is then given by: ¯ ¯ ¯ ∂t ¯ κ = ¯¯ ¯¯ . ∂s
(1)
At points where the curvature does not vanish, the normal vector is defined by: ∂t = κ n. ∂s
(2)
b = t × n.
(3)
The third unit vector b is:
Therefore, the Frenet basis is a right-handed orthonormal basis on the space curve R. As a consequence, one has: ∂n = τ b−κ t. ∂s
2
(4)
This relation defines the torsion τ , which measures the amount of rotation of the Frenet triad around the tangent t as the arc length increases. Finally the derivative of the binormal b is given in terms of the normal and tangent vector: ∂b = −τ n. ∂s
(5)
The coupled equations (2), (4) and (5) are the Frenet-Serret equations. If the curvature κ and the torsion τ are known for all s, the Frenet triad (n, b, t) can be obtained as the unique solution of the Frenet-Serret equations. It is then possible to reconstruct the space curve R by integrating the tangent vector t. A ribbon is a space curve R(s, t) supplied with a smooth unit vector field d2 (s, t) orthogonal to the curve. Let d3 = t be the unit vector tangent. A third unit vector field d1 = d2 × d3 is introduced so that the triad (d1 , d2 , d3 ) forms a right-handed orthonormal basis. This basis is a generalization of the Frenet triad (n, b, t). The components of the derivatives of the local basis vectors d1 , d2 and d3 with respect to arc length s and the time t expressed in the local basis form the twist vector κ (s, t) = κ1 d1 +κ2 d2 +κ3 d3 and the spin vector ω (s, t) = ω1 d1 + ω2 d2 + ω3 d3 , defined as follows: ∂di = κ × di , ∂s ∂di = ω × di . ∂t
(6.a) (6.b)
The equations (6.a) constitute the generalization of the Frenet-Serret equations for the ribbon. These equations can also be expressed in terms of the twist matrix K(s, t) and the spin matrix W(s, t), which we define as follows:
0 −κ3 κ2 0 −κ1 , K = κ3 −κ2 κ1 0
0 −ω3 ω2 0 −ω1 . W = ω3 −ω2 ω1 0
(7)
The Frenet-Serret equations then read: µ
µ
II.2 II.2.1
∂d1 ∂s
∂d2 ∂s
∂d3 ∂s
∂d1 ∂t
∂d2 ∂t
∂d3 ∂t
¶
³
= ¶
³
=
d1 d2 d3
d1 d2 d3
´
K,
(8)
W.
(9)
´
The Kirchhoff Equations Main Assumptions and Derivation of the Kirchhoff Equations
A thin filament, or rod, can be modelized by a ribbon constituted of a space curve R joining the loci of the centroids of the cross-sections, together with a vector field d2 attached to the filament material. The space curve R is referred to as the centerline of the rod. The Kirchhoff equations describe the dynamical evolution of the filament under the effect of internal elastic stresses and boundary constraints, in the absence of external force fields such as gravity. Also, only local interactions, between adjacent cross-sections, are considered, ignoring the possibility for two remote segments of the filament to intersect with each other. Let F(s) and M(s) be the total force and total moment exerted on the back side of a cylinder C(s) by the 3
cylinder C(s + ds) whose cross-section shape is defined by S, the set of all the values of the couple (x1 , x2 ) corresponding to a material point inside a given cross-section. The conservation of linear and angular momentum yields [35]: ∂2R ∂F = ρA 2 , ∂s ∂t à ! ∂ 2 d1 ∂ 2 d2 ∂M + d3 × F = ρ I2 d1 × + I1 d2 × , ∂s ∂t2 ∂t2
(10.a) (10.b)
where A denotes the area of the cross-section and the quantities I1 and I2 are the principal moments of inertia of the cross-section: Z
I1 =
Z
S
dx1 dx2 x22 ,
I2 =
S
dx1 dx2 x21 .
(11)
The equations (10) are closed by using the constitutive relation of linear elasticity relating the torque M to the twist vector κ : M = EI1 κ1 d1 + EI2 κ2 d2 + µJκ3 d3 ,
(12)
where E is Young’s modulus, µ is the shear modulus and J is a function of the shape S of the cross-section. In the particular case of a circular cross-section, one has: I1 = I2 =
πR4 J = , 2 4
(13)
where R is the radius of the cross-section. The combinations EI1 and EI2 are called the principal bending stiffnesses of the rod and measure how strong an applied torque must be in order to bend it, whereas the combination µJ is called the torsional stiffness and measures how large an applied torsional moment must be in order to twist the rod. The coupled equations (10) and (12) constitute the dynamical Kirchhoff equations. These are three vectorial equations involving the local basis (d1 , d2 , d3 ) and its derivatives, the tension F and the torque M, which add up to nine degrees of freedom, hence the system is closed. In the static case (if the time dependencies dropped), the term in R vanishes from equation (10.a): F0 = 0,
(14.a)
0
II.2.2
M + d3 × F = 0,
(14.b)
M = EI1 κ1 d1 + EI2 κ2 d2 + µJκ3 d3 .
(14.c)
The Kirchhoff Equations in Scaled Form
In order to restrict the number of independent constants in (14),we scale the variables, by choosing combination of the length [L], time [T ] and mass [M ] units in the following way: p
[M ] [L]3 = EI1 , [T ]2
(15)
µJ J , = EI1 2I1 (1 + σ)
(16)
[M ] = ρ AI1 , with: a=
I2 , I1
b=
4
where σ denotes the Poisson ratio. The constant a measures the asymmetry of the cross-section. Our convention is to orient the vector fields d1 and d2 such that I1 and I2 are, respectively, the larger and smaller bending stiffnesses. In this case, we have: 0 < a ≤ 1,
(17)
the value 1 being reached in the symmetric case where the moments of inertia are identical. The 1 , which ranges from 23 , constant b is the scaled torsional stiffness. It involves the constant 1+σ corresponding to incompressible media (if the volume is unchanged as the material is stretched), to 1, corresponding to hyperelasticity (if there is no striction as the material is stretched). In the particular case of a circular cross-section, one has ·
¸
2 1 ∈ ,1 . b= 1+σ 3
(18)
The scaled system reads now: F0 = 0,
(19.a)
0
M + d3 × F = 0,
(19.b)
M = κ1 d1 + aκ2 d2 + bκ3 d3 .
(19.c)
The quantities involved still have a dimension. From (19), we see that the scaled moment M has the dimension of the inverse of a length, and that the scaled tension F has the dimension of the inverse of a squared length, hence every variable involved is a length to a power. However the system cannot be further simplified by choosing the length unit [L], because the remaining constants a and b are already dimensionless. It is still possible to choose a convenient length scale for a given problem. For example, if we consider a finite rod, a natural choice for the length unit will be the length of the rod. In Section III, we choose the length unit [L] such that the norm of F has a given value. The fact that the length unit [L] is undetermined has yet another implication. Considering the static system (14) or (19), we see that every known solution actually determines a oneparameter family of solutions. More precisely, if {F(s), M(s), κ (s)} is a solution of the system, then © −2 ª −1 −1 λ F(λs), λ M(λs), λ κ (λs) is another solution of the system for every real non vanishing λ. That is, the system is scale-invariant. Furthermore, if such a transformation is performed on the solution together with a rescaling of the length unit [L] by a factor λ−1 , the solution remains unchanged, although the rod thickness is modified by a factor λ−1 . Hence, the statics of a filament, in the limit of the Kirchhoff model, is independent on the rod thickness. II.2.3
Integrability of the Static Kirchhoff Equations
The static Kirchhoff equations (19) are formally identical to the Euler equations describing the motion of a rigid body with a fixed point under gravity, in the particular case where the axis joining the fixed point to the center of mass lies along a principal direction of inertia (the correspondence between rigid body variables and rod variables is shown on Table 1). Therefore, they are fully integrable in two cases of interest, namely, the Lagrange and Kowalevskaya cases1 .
1 Actually, the Euler equations are integrable in three cases. The third one corresponds to the motion of a rigid body with arbitrary shape in free fall and is known as the Euler case. However, one cannot speak of full integrability in the context of filaments, since this case corresponds to a vanishing tension vector F and represents only particular boundary conditions, hence a subset of all possible configurations which can be adopted by a filament for given values of the material parameters a and b.
5
Symbol d3
Meaning for rods Unit tangent vector
(d1 , d2 , d3 ) s F M κ EI1 , EI2
Basis attached to the rod Arc length Tension Moment Twist vector Principal bending stiffnesses
µJ a
Torsional stiffness Bending stiffnesses ratio
b
Scaled torsional stiffness
Meaning for rigid bodies Unit vector joining the fixed point to the center of mass Basis attached to the solid body Time Force equal and opposite to gravity Angular momentum Angular velocity vector Principal moments of inertia in directions orthogonal to d3 Principal moment of inertia along d3 Ratio of the moments of inertia in directions orthogonal to d3 Scaled moment of inertia along d3
Table 1: Analogy between rigid bodies and static filaments. In the Lagrange case, the rigid body has identical moments of inertia along every direction perpendicular to the axis joining the fixed point to the center of mass, as for a symmetric top. The corresponding rods have identical bending stiffnesses in every direction, so that a = 1. This is realized, for instance, in the very common situation where the filament cross-section is circular, showing the importance of the Lagrange case for filaments. Much work has been devoted to it (Euler classified the planar solutions of the equations, see Figure 1), and recently, Shi and Hearst (1994) [34] have obtained a closed form for the general solution. Their results are rederived with special emphasis on the correspondence between rod and rigid body variables and developed in detail in Section III. In the Kowalevskaya case, there exists an axis D, originating from the fixed point, along which the moment of inertia is half as large as the moments of inertia along directions perpendicular to D. In addition, the center of mass lies in the plane perpendicular to D. In the context of rods, this corresponds to asymmetric cross-sections with a = 12 and b = 1. However, it turns out that the Kowalevskaya case, having a very high torsional stiffness, lies far beyond the region covered by the possible physical values of the parameters. For instance, an elliptic cross-section with a = 12 and b = 1 would have a Poisson ratio σ = − 13 . This value describes a material which inflates transversally as it is stretched in length, and although this is not precluded in theory, it is not realized in practice.
III III.1
Symmetric Rods and Lagrange Tops Generalities
The static Kirchhoff equations, written in terms of an appropriate set of variables, are formally equivalent to the Euler equations describing the dynamics of a heavy top. To every possible motion of the top, one can associate a particular static solution of the Kirchhoff equations. Here we focus exclusively on the analogy between tops and static rods, in one of the three cases where the equations are fully integrable: the Lagrange case, where the top has two identical principal moments of inertia. This condition is satisfied if it has a symmetry of revolution. The corresponding filament has identical bending stiffnesses in all directions, that is, we must set a = 1 in the scaled form of the Kirchhoff equations (19). The solutions of the Euler equations in the Lagrange case are well known and can be written as 6
Figure 1: Euler’s drawings of planar filaments.
7
combinations of elliptic functions [36, 37, 38]. In order to obtain the centerline R of the corresponding static filament, we must identify the tangent vector d3 to a unit vector lying along the axis of revolution of the top. The centerline is then obtained by integrating d3 over the arc length s which, in the context of tops, corresponds to time. An obvious difficulty arises: in order to describe the behavior of real filaments under given external constraints, the space curve R must be available in a form that allows for boundary conditions at two distinct points. That is, it is essential to explicitly carry the integration of the tangent vector. Although it is not obvious that this integration can be performed, Shi and Hearst (1994) [34] recently obtained expressions for the centerline R in cylindric coordinates in a closed analytic form involving elliptic functions. Despite this achievement, the problem is still partially unsolved. Indeed, explicit forms of the solutions are by no means sufficient to get a global insight on the large variety of possible filament shapes. Furthermore, the detailed analysis of the correspondence between spinning Lagrange tops and static symmetric rods provides a way of establishing an exhaustive geometric classification of the solutions. The aim of this paper is to provide a first step towards such a classification. Shi & Hearst obtained their solutions by first solving the Kirchhoff equations for the curvature and torsion and then solving the Frenet-Serret equations to obtain the centerline. The resulting expressions depend on integration constants which do not have a clear meaning. Here, we depart from their approach and work consistently with variables and integration constants relevant to the top. Namely, in Section III.3 , we express the Kirchhoff-Euler equations in terms of the Euler angles. Then we compute the centerline R of the filament as a function of the constants of the motion for the spinning top. Once the expressions for the centerline have been obtained, we study various classes of shapes of the rod and their correspondence with the motion of the top. This analysis can be thought of an extension of Euler’s work, who classified planar shapes of filaments. In this Section, we show illustrations of the spinning top orbits together with the corresponding filament shapes. A top orbit is displayed as a curve on the unit sphere which represents the extremity of the unit vector d3 . The filament shapes are represented with a circular cross-section, hence b must be set to a value lying between 23 and 1. We have chosen b = 34 . The radius of the cross-section and the zoom factor vary from one figure to another and are chosen for clarity (as seen in Section II, the radius is arbitrary in the static case). In the remaining of this section, we write down the static Kirchhoff equations (19) in the case a = 1, and identify a set of three first integrals necessary to guarantee the integrability of the system. With a = 1, the system (19) reads: F0 = 0,
(20.a)
0
M + d3 × F = 0,
(20.b)
M = κ1 d1 + κ2 d2 + bκ3 d3 .
(20.c)
The first equation expresses the fact that the tension F is constant. We choose it oriented along the third vector of the fixed basis: F = F eZ .
(21)
In terms of spinning tops, the tension corresponds to the opposite of the top weight, −mg, which describes an external force field. In the same spirit, we consider the tension as having a fixed value. As a consequence, we consider F as a parameter rather than a first integral. Inserting (21) into (20.b) and projecting along the fixed basis vector eZ , we have: M0 · eZ = 0.
(22)
The basis vector eZ being independent of s, we can extend the derivative in (22) to take effect over the whole left-hand side, leading to: MZ0 = 0, 8
(23)
where MZ denotes the component of the moment along eZ . It is a first integral that represents the vertical component of the angular momentum of a spinning top. By projecting (20.b) along d3 , we obtain: M0 · d3 = (M · d3 )0 − M · d03 = 0.
(24)
Using the fact that d03 = κ × d3 = M × d3 , we see that the second term of (24) vanishes identically, leading to: M30 = 0.
(25)
That is, the torsional moment M3 is another first integral. From (20.c), we see that it corresponds to a constant twist density. For the spinning top, M3 represents the component of the angular momentum along the axis of revolution of the top. Finally, taking the dot product of both sides of (20.b) with κ , we have: κ × d3 ) = 0. M0 · κ + (d3 × F) · κ = M0 · κ + F · (κ
(26)
Using (20.c) and the expressions (6.a) of the derivatives of the local basis (d1 , d2 , d3 ) in terms of κ , (26) reduces to: κ1 κ01 + κ2 κ02 + F · d3 0 =
µ
1 M · κ +F · d3 2
¶0
= 0,
(27)
which provides the last first integral: 1 M · κ + F · d3 = H. 2
(28)
This constant quantity H is the total elastic-plus-strain energy density of the filament and corresponds to the total kinetic-plus-potential mechanical energy of the spinning top. We now proceed to analyze different solutions of the system (20).
III.2 III.2.1
Helical Filaments The General Helical Solution
Helical solutions have constant Frenet curvature κ and torsion τ . Taking into account the fact that the twist density κ3 is also constant, we introduce the following definitions. • A Frenet helix is a helix with pure torsion, that is, κ3 = τ . • An overtwisted helix is a helix such that (κ3 − τ ) has the same sign as τ . • An undertwisted helix is a helix such that (κ3 − τ ) and τ have opposite signs. These three types of helices are represented on Figures 3, 2 and 4 respectively. The undertwisted and overtwisted helices can be distinguished by the relation between the handedness of the helix itself and the handedness of the apparent twist pattern on the helix. In the case of an overtwisted helix, both hands are identical, whereas an undertwisted helix has opposite hands. It is convenient to treat the cases of helical, circular (ring-like) and straight solutions of the Kirchhoff equations independently. The rings and the helical rods are easily identified as corresponding to spinning tops with the extremity describing a circle (centered at the fixed point in the case of rings), while the straight solutions correspond to the cases where the extremity of the 9
N
Figure 2: An overtwisted helix (κ = 3, τ = 1, κ3 = 6). top is at rest (the so-called sleeping tops): pointing upwards in the case of positive tension F3 and downwards in the case of negative tension. Illustrations of these top orbits are given on Figures 2 to 7. For the sake of simplicity, helical, circular and straight filaments will be referred to as helical filaments. We do not introduce Euler angles for the helical solutions. Instead, we introduce a fixed Frenet curvature κ and a fixed torsion τ into the expressions for the twist vector: κ1 = κ sin [(κ3 − τ )(s − s0 )] ,
(29.a)
κ2 = κ cos [(κ3 − τ )(s − s0 )] .
(29.b)
Substituting this into (20), we obtain one single nontrivial vector condition: κ − (κ3 − τ ) d3 ], F = (bκ3 − τ ) [κ
(30)
which simply gives the tension in the local basis as a constant combination of the other quantities. Hence, there exist helical solutions with arbitrary curvature κ, torsion τ and twist density κ3 . This is unique to the case of symmetric rods. Indeed, in the regular case a 6= 1 the only possible helices are Frenet helices with either κ1 = 0 or κ2 = 0 [39]. Three more facts can be noticed from (30). First, in the case of rings (τ = 0), the tension has no longitudinal component (F · d3 = 0). Second, in the case of a non circular filament (τ 6= 0), a vanishing longitudinal tension implies bκ3 = τ , hence the tension vector F itself vanishes. Finally, in the case of circular cross-section, one has b ≤ 1. This means that the Frenet helices (κ3 = τ ) have a negative longitudinal tension, and that the helices with null tension are overtwisted. For straight solutions (κ = 0), (30) does not hold and is replaced by: F = F3 d3 ,
(31)
where F3 is constant and arbitrary. Hence, there exist straight solutions with arbitrary longitudinal tension and twist density. The torsion has no meaning for straight rods. Figures 2 to 7 show representations of helical, ring-like and straight solutions.
10
N
Figure 3: A Frenet helix (κ = 3, τ = −1, κ3 = −1).
N
Figure 4: An undertwisted helix (κ = 3, τ = −1, κ3 = 4).
N
Figure 5: A twisted ring (κ = 1, τ = 0, κ3 = 3).
11
N
Figure 6: A straight rod subject to extensive tension (κ = 0, κ3 = 1, F3 = 1).
S
Figure 7: A straight rod subject to compressive tension (κ = 0, κ3 = 1, F3 = −1).
12
III.2.2
The General Solution with Null Tension.
Substituting F = 0 into (20.b), we see that the moment M is constant. Differentiating (20.c) with respect to s and projecting the resulting equation successively along d1 , d2 and d3 , we obtain: κ01 + (b − 1)κ3 κ2 = 0,
(32.a)
+ (1 − b)κ3 κ1 = 0,
(32.b)
κ02
0 = 0,
(32.c)
which, taking into account the fact that κ3 is constant, can be integrated: κ1 = κ sin [(1 − b)κ3 (s − s0 )] ,
(33.a)
κ2 = κ cos [(1 − b)κ3 (s − s0 )] ,
(33.b)
where κ and s0 are integration constants. We see that (33) assumes the form (29), with τ = bκ3 . In other words, (33) is a helix. We conclude that all the solutions with vanishing tension are helices. In the following sections, we consider non-helical filaments. Therefore, from now on, we assume F 6= 0. This gives a precise sense to the vertical unit vector eZ .
III.3 III.3.1
General Solution for the Local Basis Equations for the Euler Angles
We now introduce the Euler angles (ϕ, θ, ψ) for the local basis (d1 , d2 , d3 ). The angles ϕ, θ and ψ denote, respectively, the precession, nutation and self-rotation angles (see Figure 8). In matrix form, the local basis is obtained from the fixed trihedron (eX , eY , eZ ) as follows: ³
d1 d2 d3
´
³
=
eX eY eZ
´
E,
(34)
where the general rotation matrix E reads:
cos ϕ cos θ cos ψ − sin ϕ sin ψ − cos ϕ cos θ sin ψ − sin ϕ cos ψ cos ϕ sin θ E = sin ϕ cos θ cos ψ + cos ϕ sin ψ − sin ϕ cos θ sin ψ + cos ϕ cos ψ sin ϕ sin θ . − sin θ cos ψ sin θ sin ψ cos θ (35) From (34) and (35) we extract the fixed vector eZ as a function of the local basis: eZ = sin θ(sin ψ d2 − cos ψ d1 ) + cos θ d3 .
(36)
We now express the twist matrix K in terms of the rotation matrix E. Differentiating (34) yields: ´ ³ ´ ∂ ∂ ³ d1 d2 d3 eX eY eZ E = ∂s ³ ´ ∂s d1 d2 d3 = E> E0 ,
(37)
hence, from the definition (7) of the twist matrix, we see that: K = E> E0 .
13
(38)
eZ
θ
d3 d2 eX
ψ
eY ϕ
ϕ θ ψ
d1
Figure 8: The Euler Angles. Therefore, we can express the twist vector in terms of Euler angles: κ1 = θ0 sin ψ − ϕ0 sin θ cos ψ, 0
0
0
0
(39.a)
κ2 = θ cos ψ + ϕ sin θ sin ψ,
(39.b)
κ3 = ϕ cos θ + ψ .
(39.c)
Using (36) and (39), we now write the three first integrals in terms of the Euler angles: MZ = ϕ0 [1 + (b − 1) cos2 θ] + bψ 0 cos θ, 0
0
M3 = b(ϕ cos θ + ψ ), Ã
H=
M32
1 02 θ + ϕ02 sin2 θ + 2 b
(40.a) (40.b)
!
+ F cos θ.
(40.c)
It is convenient to introduce the following auxiliary constants: 1 h= F ˜= 1 h F
Ã
M2 H− 3 2b
!
Ã
,
M 2 M 2 − MZ2 H− 3 + 3 2b 2
(41.a) !
,
(41.b)
which are well defined since F 6= 0. We also carry out the following change of variable in equations (40.a) to (40.c): z = cos θ.
(42)
Note that a solution with constant z corresponds to a helical rod excluded in this discussion, hence z is not constant. We can solve (40.a) to (40.c) for ϕ0 , θ0 and z 0 to obtain the final form of our equations in the Euler angles: M Z − M3 z , 1 − z¶2 µ M3 − MZ z 1 ψ0 = − 1 M3 + , b 1 − z2
ϕ0 =
14
(43.a) (43.b)
z 02 = 2F (h − z)(1 − z 2 ) − (MZ − M3 z)2 ˜ − z)(1 − z 2 ) − (M3 − MZ z)2 . ⇔ z 02 = 2F (h III.3.2
(44.a) (44.b)
A Better Choice of First Integrals
The differential equation (44.a) or (44.b) is an identity between z 02 and a cubic polynomial in z. In order to solve this equation, we must know the roots z1 , z2 , z3 of this polynomial. Rather than ˜ the classical approach consists in computing the roots in terms of the constants MZ , M3 , h or h, considering the roots z1 , z2 and z3 as three independent first integrals, and then expressing the ˜ as functions of z1 , z2 and z3 . This is achieved by rewriting the cubic constants MZ , M3 , h and h polynomial in (44.a) and (44.b) as a product of three factors involving z1 , z2 and z3 : z 02 = 2F (h − z)(1 − z 2 ) − (MZ − M3 z)2 ˜ − z)(1 − z 2 ) − (M3 − MZ z)2 = 2F (h = 2F (z − z1 )(z2 − z)(z3 − z).
(45.a) (45.b) (45.c)
By setting z = ±1 in (45.a) or (45.b), the right-hand side assumes a non positive value. Furthermore, if we choose F to be positive (which we can always do by defining adequately the vertical unit vector eZ ), we see that the right-hand side of (45.a) or (45.b) tends to +∞ as z → +∞. This means that one of the roots (conventionally, z3 ) lies in the interval [1, +∞[. Finally, in order to obtain solutions with real values of θ, we require that z 02 be positive for z ranging in some interval contained in [−1, 1]. Hence the other two roots of the polynomial must be real and lie between −1 and 1. Conventionally, we choose z1 ≤ z2 . We conclude that the physical values of our independent constants z1 , z2 and z3 must satisfy: −1 ≤ z1 ≤ z2 ≤ 1 ≤ z3 .
(46)
In the previous Section, we showed that in order to obtain the scaled form of the Kirchhoff equations in the static case, it was not necessary to perform a complete scaling. Namely, at this level, every variable involved in the equations is a length raised to a given power, and we still have the freedom to choose an arbitrary length unit [L]. In the following, it is convenient to choose the length unit to be: s
[L] =
2 , F (z3 − z1 )
(47)
2 . z3 − z1
(48)
which is equivalent to the substitution: F =
The right-hand side of (48) is well-defined as long as z1 6= z3 . The condition z1 = z3 implies z1 = z2 = z3 = 1, which corresponds to a straight rod with z = 1, a case excluded from this discussion. ˜ in terms of the roots can be obtained by Expressions for the constants MZ , M3 , h and h identifying the coefficients of the powers of z in (45.a) to (45.c), or equivalently, by considering the equalities between the right-hand sides of (45.a) to (45.c) for three well-chosen values of z. Setting z = ±1 leads, together with (48), to: −(MZ ∓ M3 )2 =
4 (±1 − z1 )(z2 ∓ 1)(z3 ∓ 1). z 3 − z1 15
(49)
The equations (49) give MZ − M3 and MZ + M3 up to sign determination. A third equality of ˜ and could type (45) corresponding to another value of z would lead to an equation involving h or h not provide additional knowledge on MZ or M3 for themselves. We conclude that given values of z1 , z2 and z3 do not yield unique values of MZ and M3 . Instead they give these two constants with a complete sign indetermination. By performing a mirror reflection in space, we can map any filament with MZ + M3 < 0 onto a filament with MZ + M3 > 0. Hence, we restrict our analysis to the case MZ + M3 ≥ 0. Nevertheless, we still have to supply the values of z1 , z2 and z3 with extra information, namely, the sign S of MZ − M3 : S = + or S = −.
(50)
Now, if we set: M Z + M3 ≥ 0, 2 |MZ − M3 | , M− = 2 M+ =
(51.a) (51.b)
we have: MZ = M+ + SM− ,
(52.a)
M3 = M+ − SM− ,
(52.b)
with s
M± =
(1 ± z1 )(1 ± z2 )(z3 ± 1) . z3 − z1
(53)
˜ in (44.a) and (44.b) are obtained from suitable combinations of equalities The constants h and h between the coefficients of z in (45.a) to (45.c): 1 [z1 + z2 + z3 − z1 z2 z3 + S(z3 − z1 )M+ M− ] , 2 ˜ = 1 [z1 + z2 + z3 − z1 z2 z3 − S(z3 − z1 )M+ M− ] . h 2
h=
III.3.3
(54.a) (54.b)
General Solution for the Euler Angles
The solutions of equations (43.a)-(44) involve elliptic functions (see Appendix A). With a suitable origin for the arc length s, the solution of equation (44) assumes the form: z = z1 + (z2 − z1 )sn2 (s | k),
(55)
where the modulus k ranges between 0 and 1 and is given by: k2 =
z 2 − z1 . z3 − z1
(56)
We can use the fact that z = cos θ to obtain θ as a function of s. In the generic case where z1 6= −1 and z2 6= 1, the cosine never reaches its extreme values ±1, and there is a bijective correspondence between z and θ. In this case, we can take θ = arccos z. In the degenerate cases where z1 = −1 or z2 = 1, we must take care of the behavior of θ as cos θ reaches its limiting values. This is achieved by substituting z = cos θ in (44) and examining the behavior of θ0 around z = −1 or z = 1. The results are as follows: 16
• If z1 = −1 and z2 6= 1, θ0 has a non vanishing limit as z → −1, hence the sign of θ − π changes at z = −1. • If z1 6= −1, z2 = 1 and z3 6= 1, θ0 has a non vanishing limit as z → 1, hence the sign of θ changes at z = 1. • If z1 6= −1, z2 = 1 and z3 = 1, z never reaches its extreme value 1, hence we can take θ = arccos z. • If z1 = −1, z2 = 1 and z3 6= 1, θ0 has a non vanishing limit in both cases z → ±1, hence θ is a monotonous function of s. In this case, MZ = M3 = 0, that is, the top behaves like a plane pendulum. This corresponds to planar rods studied by Euler. • If z1 = −1, z2 = 1 and z3 = 1, θ0 has a non vanishing limit as z → −1 and z never reaches its extreme value 1, hence z assumes the value −1 only at the single point s = 0, and θ covers the interval ]0, 2π[ crossing every value only once. This corresponds to the homoclinic orbit of the plane pendulum. In order to obtain ϕ and ψ as functions of s, we must integrate (43.a) and (43.b), which, using (52.a) and (52.b), can be rewritten as: 1 1 + SM− , 1 +¶z 1−z µ 1 1 1 − 1 M3 + M + − SM− . ψ0 = b 1+z 1−z ϕ0 = M+
(57.a) (57.b)
Next, we define: n± = ∓
z2 − z 1 . 1 ± z1
(58)
In the cases where n± have a non vanishing denominator, we can express (57.a) and (57.b) using (55) as: M− M+ 1 1 +S , 2 1 + z1 1 − n+ sn (s | k) 1 − z1 1 − n− sn2 (s | k) µ ¶ M− M+ 1 1 1 ψ0 = − 1 M3 + −S . 2 b 1 + z1 1 − n+ sn (s | k) 1 − z1 1 − n− sn2 (s | k) ϕ0 =
(59.a) (59.b)
Notice that we have the freedom to perform global rotations of the local basis around d3 and of the fixed trihedron around eZ , in such a way that ϕ = 0 and ψ = 0 for s = 0. The equations (59.a) and (59.b) can be integrated to yield: M− M+ Π(s | n+ , k) + S Π(s | n− , k), 1 + z1 1 − z1 µ ¶ M+ M− 1 ψ= − 1 M3 s + Π(s | n+ , k) − S Π(s | n− , k), b 1 + z1 1 − z1
ϕ=
(60.a) (60.b)
where Π is the incomplete elliptic integral of the third kind in “practical” form, as defined in Appendix A. In the degenerate cases where n+ or n− has a vanishing denominator, the correct limits of (60.a) and (60.b) are obtained by setting the term involving the ill-defined quantity to zero. The expressions (60.a) and (60.b) together with (55) constitute the general solution of the spinning symmetric top problem. We can then use (34) to obtain the non-fixed basis (d1 , d2 , d3 ) as a function of the Euler angles. 17
III.3.4
Particular Orbits of the Spinning Top
The generic orbits of the spinning top are those for which the extremity of d3 oscillates vertically between two parallels on the unit sphere, while it revolves horizontally, either monotonously as on Figures 21, 20, 22 and 23, or making loops as on Figures 16 and 24. The looping orbits arise if 0 1 z2 S = − and z3 > 1−z z2 −z1 . This condition is obtained by allowing ϕ to vanish for some value of z. 1 z2 The degenerate case S = −, z3 = 1−z z2 −z1 separates looping orbits from monotonously precessing orbits. This corresponds to trajectories which present turn-back points, as shown on Figure 19. 1 z2 We shall see in the following section that the condition z3 = 1−z z2 −z1 has a clear meaning in terms of rods for both values of S. The case z1 = −1 with z2 6= 1 and z3 6= 1 corresponds to orbits which cross periodically the south pole of the unit sphere, as shown on Figure 18, while the case z2 = 1 with z1 6= −1 and z3 6= 1 corresponds to orbits which cross the north pole, as shown on Figure 17. The case z2 = z3 = 1 with arbitrary z1 represents homoclinic orbits such as those shown on Figures 13 to 15. Next, there are the orbits for which MZ = M3 = 0, that is, for which the top behaves like a plane pendulum (the extremity of d3 is restricted to a vertical grand circle on the unit sphere). They correspond to z1 = −1 and either z2 = 1 or z3 = 1. The case z2 = 1 describes an oscillating pendulum, whereas the case z3 = 1 describes a revolving pendulum. The case z2 = z3 = 1 corresponds to the homoclinic orbit of the pendulum. Finally, there are orbits with constant z which we took apart from our preceding analysis, and which correspond to the helical, circular and straight rods examined in Section III.2. See Figures 2 to 7.
III.4 III.4.1
Centerline in Cylindrical Coordinates, Curvature and Torsion. Polar Coordinates, Complex Curvature and Complex Centerline Radius
Following Shi & Hearst (1994), we introduce cylindrical coordinates R, Φ, Z for the filament centerline R: R = R cos Φ eX + R sin Φ eY + Z eZ .
(61)
Rather than adopting the method of Shi & Hearst to obtain R, Φ and Z as functions of the arc length, we lead the calculations in a way which highlights the remarkable correspondence between the expressions forRthe radius R and the Frenet curvature κ, as well as between the polar angle Φ and the angle ζ = ds (κ3 − τ ) giving the orientation of the local basis (d1 , d2 , d3 ) with respect to the Frenet triad (n, b, d3 ). The computations are quite analogous and can be led in parallel. ˆ the complex curvature κ First, we introduce the complex centerline radius R, ˆ and the complex ˆ: horizontal component of the moment M ˆ = R exp iΦ, R
(62.a)
κ ˆ = κ1 + iκ2 , ˆ = MX + iMY , M
(62.b) (62.c)
where MX and MY are the components of the moment along eX and eY . Using (34), (35) and ˆ can be expressed in Euler angles as: (39.a), κ ˆ and M MZ − M3 z + iz 0 √ exp −iψ, ± 1 − z2 0 ˆ = M3 −√MZ z − iz exp iϕ. M ± 1 − z2 κ ˆ=−
18
(63.a) (63.b)
The correct sign to put in front of the root in (63.a) and (63.b) is the one of sin θ. The key to ˆ is the moment equation (20.b). Taking into account the fact that obtain the complex centerline R R F = F eZ is constant and that R = ds d3 , we can integrate both sides of this equation, leading to: M + F R × eZ = MZ eZ
(64)
for an appropriate choice of origin in the tridimensional space. Taking the dot product of (64) with eX + ieY , we obtain: 0 ˆ = −i M3 −√MZ z − iz exp iϕ. ˆ = −i M R F F ± 1 − z2
III.4.2
(65)
Frenet Curvature and Centerline Radius
Using (63.a) and (65), the Frenet curvature κ and the radius R are: (MZ − M3 z)2 + z 02 , 1 − z2 ¯ ¯2 (M3 − MZ z)2 + z 02 ¯ ˆ¯ . R 2 = ¯R ¯ = F 2 (1 − z 2 )
κ|2 = κ2 = |ˆ
(66.a) (66.b)
We now substitute the expressions (45.a) and (45.b) for z 02 respectively in (66.a) and (66.b) to yield the final forms of κ and R. They depend on s only through the variable z: κ2 = 2F (h − z), 2 ˜ R2 = (h − z). F
(67.a) (67.b)
˜ are both Notice that the left-hand sides of these equations are positive for all z, hence h and h greater than or equal to z2 . Also, ignoring the case of straight rods, (67.a) and (67.b) show that κ and R can only vanish at isolated points where z = z2 . A natural question is: for which values ˜ = z2 hold? Using expressions (48) for F and of z1 , z2 , z3 and S do the equalities h = z2 or h ˜ (54.a)-(54.b) for h and h, one can easily obtain the following results: h = z2
⇔
S=−
˜ = z2 h
⇔
S=+
1 − z 1 z2 , z 2 − z1 1 − z 1 z2 and z3 = . z 2 − z1 and z3 =
(68.a) (68.b)
Condition (68.a) is necessary and sufficient for the curvature to vanish at some isolated points. Remarkably, it is identical to the condition for the orbit of the spinning top to present turn-back points. In the same way, (68.b) is a necessary and sufficient condition for the radius R to vanish at isolated points. In order to have a continuous dependence in s for the polar angle Φ across these isolated points where the radius R vanishes, we must change the sign of R. In the same way, the sign of κ must change wherever κ vanishes in order for the angle ζ to be a continuous function of s. III.4.3
Frenet Torsion and Polar Angle of the Centerline
The argument of the right-hand side of (65) can be used as an expression for the polar angle Φ. However, this is not convenient. A more tractable expression can be obtained by first computing 19
the derivative of the polar angle from (65): Φ0 =
−z 0 ∂ ˆ = ∂ arctan arg R + ϕ0 . ∂s ∂s M3 − MZ z
(69)
Similarly, the Frenet torsion τ can be computed: τ = κ3 +
M3 ∂ z0 ∂ arg κ ˆ= + arctan − ψ0. ∂s b ∂s MZ − M3 z
(70)
Taking the derivatives in (69) and (70) leads to expressions for Φ0 and τ involving z, z 02 , z 00 , ϕ0 and ψ 0 . Using the expressions (43.a)-(44) to express everything in terms of z only, we obtain: Ã
!
˜ 1 M3 − MZ h Φ = MZ + , ˜−z 2 h µ ¶ 1 M3 h − MZ τ= M3 + . 2 h−z 0
(71.a) (71.b)
In the case of a non constant z, we see from (71) that the condition for a constant Φ is identical to the condition for the rod to be planar (τ = 0): MZ and M3 must both vanish. This means that non circular planar filaments correspond to the case where the spinning top behaves like a plane pendulum (z1 = −1 and either z2 = 1 or z3 = 1). For non-planar rods, we define: z2 − z1 , h − z1 z 2 − z1 n ˜= . ˜ − z1 h
(72.a)
n=
(72.b)
We can then integrate (71.a) and (71.b) to yield: Ã
!
˜ M3 − MZ h π 1 MZ s + Π(s | n ˜ , k) − , Φ= ˜ 2 2 h − z1 µ ¶ M3 M3 h − MZ π 1 ζ= M3 s + Π(s | n, k) − . s− b 2 h − z1 2
(73.a) (73.b)
ˆ (65) in the The integration constant − π2 in (73.a) has been determined from the complex radius R π limit s → 0. The integration constant − 2 in (73.b) is determined from the fact that d1 and the binormal b are opposite when ψ = 0. III.4.4
Vertical Cartesian Coordinate of the Centerline
To complete the filament description, we need an expression for Z, the vertical coordinate of the centerline. It is given by: Z 0 = z,
(74)
which, using (55) and (56), can be written as: Z 0 = z3 − (z3 − z1 )[1 − k 2 sn2 (s | k)]
(75)
The last expression is readily integrated to yield, for an appropriate choice of origin of the vertical coordinate: Z = z3 s − (z3 − z1 )E(s | k), 20
(76)
where E is the incomplete elliptic integral of the second kind in “practical” form, as defined in Appendix A. For very large z3 , Z is expressed as a difference between two large quantities, although Z itself is finite for z3 → ∞. Hence, (76) could be tricky to handle numerically for large z3 . Finally, we note that the expression for the radius R is bounded while, in general, the expression for Z is not. As a consequence, the vector d3 , averaged over all s, has no component in the (eX , eY ) plane. This means that, on average, a spinning top is vertical whatever the constants of the motion are.
III.5 III.5.1
Filament Shapes Planar Shapes
This section is dedicated to planar filaments other than the circular and straight ones. This is essentially a modern re-statement of Euler’s results. As seen in the previous section, the noncircular and non-straight planar filaments correspond to values of the constants z1 , z2 and z3 for which the top behaves like a plane pendulum. In this case, the sign variable S is meaningless. There are two one-parameter families of planar solutions. The first one is obtained by setting z1 = −1 and z3 = 1 and keeping z2 arbitrary; it corresponds to oscillating orbits of the pendulum. The second one is obtained by setting z1 = −1 and z2 = 1 and keeping z3 arbitrary; it corresponds to revolving orbits of the pendulum. In both cases, it is convenient to adopt the modulus k as the arbitrary parameter. The two families of pendulum orbits are displayed on the phase portrait shown on Figure 9. Each curve represents a different orbit with a given value of k in the (θ, θ0 ) space. The closed orbits correspond to oscillating states of the pendulum, while the open orbits correspond to revolving states of the pendulum. The corresponding filament shapes are represented on Figures 10a to 12f. Since the scaling defined in equation (47) depends on k through z3 − z1 , so does the length unit for the filament, or the time unit for the pendulum. This is inadequate to draw a phase portrait, hence, on Figure 9, we have exceptionally chosen the scaled force unit [L]−2 to be F . This is identical to the scaling defined in (47) in the oscillating case. We introduce the following notations for the complete elliptic integrals of the first and second kinds defined in Appendix A:
Oscillating orbits of the pendulum: variables z, R and Z, reduce to:
K = K(k),
(77.a)
E = E(k).
(77.b)
In the case z3 = 1, the expressions for the relevant
z = 2k 2 sn2 (s | k) − 1,
(78.a)
R = 2kcn(s | k),
(78.b)
Z = s − 2E(s | k).
(78.c)
In the following, we define: µ
¶
1 √ |k , f =F 2k µ ¶ 1 A √ |k , e=E 2k A
21
(79.a) (79.b)
3
θ’ k = 0.6103 k = 0.6596 k = 0.7227
2
k = 0.8063
k = 0.9092
k = 0.9145
k = 0.7071
1 k = 0.3333
0
1
2
3
k=1
4
5
6
θ
-1
Figure 9: Phase portrait for the plane pendulum. The closed curves represent oscillating orbits, while the open curves represent revolving orbits. The homoclinic orbit corresponds to k = 1. where FA and EA are the incomplete elliptic integrals of the first and second kinds in algebraic form (see Appendix A). The extrema of Z are given by the condition z = 0, which is satisfied for: sn2 (s | k) =
1 ⇔ s = ±f + 2mK, 2k 2
(80)
where m is an arbitrary integer. We see from (79.a) that f is real only if k 2 ≥ 12 . Hence, for k 2 < 12 , Z is a decreasing function of s (see Figure 10a). The limiting case k 2 = 12 corresponds to an oscillating pendulum which has just enough energy to reach the horizontal position, as shown of Figure 10b. Above this critical value of k 2 , Z is not monotonous, and for high enough values of k, the filaments can have points of self-tangency or self-intersection (see Figures 10d to 10f). The cases of self-tangency are obtained by imposing the equality of Z at different points where z vanishes. This leads to the condition: f − 2e + 2m(2E − K) = 0,
(81)
where m is a positive integer which denotes the number of pendulum oscillations before a selftangency occurs. The solutions km of the equation (81) for the lowest values of m are given in the first column of Table 2. Setting m = ∞ in equation (81) yields: k∞ = 0.9089,
(82)
for which the filament shape is the lemniscate represented on Figure 11a. Every value of k < k∞ corresponds to a solution Z which, on average, is a decreasing function of s, while for k > k∞ , Z is, on average, an increasing function of s. The increasing solutions can also present points of intersection and self-tangency for given values of k (see Figures 11b to 11e). The condition of self-tangency in the case k > k∞ also assumes the form (81), provided that we set m to be negative. The positive integer −m counts the number of pendulum oscillations before a self-tangency occurs. The solutions km of the equation (81) for the lowest values of −m are given in the second column of Table 2. 22
Homoclinic orbit of the pendulum: The homoclinic orbit is obtained by taking the limit k → 1 in either the oscillating case or the revolving case. It is shown on Figure 11f. The expressions for z, R and Z then reduce to: z = 1 − 2sech2 s,
(83.a)
R = 2sechs,
(83.b)
Z = s − 2 tanh s.
(83.c)
In this case, we set z1 = −1 and z2 = 1. The expressions
Revolving orbits of the pendulum: for θ, R and Z read:
θ = π − 2am(s | k), ρ = 2k
−2
Z = (2k
(84.a)
dn(s | k),
−2
− 1)s − 2k
(84.b) −2
E(s | k).
(84.c)
Next, we define: µ
¶
1 f˜ = F √ | k , 2 µ ¶ 1 e˜ = E √ | k . 2
(85.a) (85.b)
The extrema of Z correspond to the values of s for which θ is an integer multiple of π, hence am(s | k) is an integer multiple of π2 , or s = ±2f˜ + 2mK,
(86)
where m is an integer. Notice that f˜ and e˜ are real for all k. As in the case of oscillating orbits, we can find a condition for the existence of points of self tangency: e + m[(2 − k 2 )K − 2E] = 0, (2 − k 2 )f˜ − 2˜
(87)
where the positive integer m counts the number of pendulum revolutions before a self-tangency occurs. The solutions k˜m of the equation (87) for the lowest values of m are given in the third column of Table 2. The sequence (k˜m ) has a vanishing limit for m → ∞ and, in the limit k → 0, the filament is a vertical ring. Some filaments corresponding to revolving orbits are shown on Figure 12.
III.5.2
Non-planar Localizing Solutions
The homoclinic orbits constitute a one-parameter family of solutions which are obtained by setting z2 = z3 = 1 while z1 is kept arbitrary. Here again, the sign parameter S is meaningless. In terms of rods, these orbits correspond to the localizing solutions studied by Coyne (1990) [1]. They connect continuously the straight state (z1 = 1) to the planar loop (z1 = −1). These solutions have constant torsion: s
τ=
1 + z1 . 1 − z1
23
(88)
a) k = 0.3333
b) k = 0.7071
c) k = 0.8000
d) k = 0.8551
e) k = 0.8750
f) k = 0.8858
Figure 10: Planar filaments corresponding to low amplitude oscillating orbits of the plane pendulum.
24
a) k = 0.9089
b) k = 0.9270
c) k = 0.9320
d) k = 0.9414
f) k = 1
e) k = 0.9700
Figure 11: Planar filaments corresponding to high amplitude oscillating orbits of the plane pendulum. The homoclinic orbit is reached in the limit k = 1.
25
a) k = 0.9700
b) k = 0.9185
c) k = 0.8500
d) k = 0.8063
e) k = 0.7600
f) k = 0.7227
Figure 12: Planar filaments corresponding to revolving orbits of the plane pendulum.
26
k1 = 0.8551 k2 = 0.8858 k3 = 0.8942 k4 = 0.8981 k5 = 0.9004 ... k∞ = 0.9089
k−1 = 0.9414 k−2 = 0.9270 k−3 = 0.9214 k−4 = 0.9185 k−5 = 0.9167 ... k−∞ = 0.9089
k˜1 k˜2 k˜3 k˜4 k˜5
= 0.9145 = 0.8063 = 0.7227 = 0.6596 = 0.6103
... k˜∞ = 0
Table 2: Values of k corresponding to self-tangency. We see that τ is bijective in z1 and can assume any nonnegative value, from 0 in the case z1 = −1 (plane pendulum) to infinity2 in the limit z1 → 1. In the following, we adopt the torsion τ rather than z1 as the arbitrary parameter. The top variables ϕ, z and ψ assume the form: µ
ϕ = arctan
¶
1 tanh s + τ s, τ
2 sech2 s, 1 +µτ 2 ¶ µ ¶ 2 1 tanh s + 3 − τ s. ψ = arctan τ b z =1−
(89.a) (89.b) (89.c)
Notice that the boundary values 23 and 1 for the variable b in the case of circular cross-section take a new sense in view of the expressions (89). The value b = 23 makes the self-rotation angle ψ bounded, while b = 1 is the value for which ϕ = ψ. The centerline variables R, Φ and Z read: 2 sechs, 1 + τ2 π Φ = τs − , 2 2 Z =s− tanh s. 1 + τ2
R=
(90.a) (90.b) (90.c)
Three typical non-planar (τ 6= 0) homoclinic orbits of the spinning top together with the corresponding filament shapes are shown on Figures 13 to 15. III.5.3
Generic Filament Shapes
In the most generic case, the spinning top oscillates between two parallels z1 and z2 , while it precesses either monotonously or with backward-and-forward motion. The corresponding filament centerline R behaves, on average, as a helix around which it is wound. Average helix: We define the mean helical centerline hRi in the following way: We introduce cylindrical coordinates hRi, hΦi, hZi for the helix hRi, and we take the mean vertical coordinate hZi to be a linear function of s, namely: hZi = hzi s,
(91)
2 Actually, for fixed tension, the torsion cannot be arbitrarily large. One must keep in mind that the parameter τ used here is the scaled torsion and that the scaling defined in (47) in turn depends on τ through z1 . As a consequence, ˜ F ˜ the unscaled torsion τ˜ is not simply proportional to τ , but instead is given by τ˜2 = EI (1+τ −2 ) , where F is the unscaled 1 tension. Hence, the torsion has actually an upper bound proportional to the square root of the tension.
27
N
Figure 13: A locally buckled filament with low torsion and high curvature (τ = 12 ).
N
Figure 14: A locally buckled filament with intermediate torsion and curvature (τ = 1).
N
Figure 15: A locally buckled filament with high torsion and low curvature (τ = 2).
28
where hzi is the average of z over a period, 2K(k). Using (76), we have: hzi =
1 2K(k)
Z
2K(k) 0
z ds = z3 − (z3 − z1 )
E(k) . K(k)
(92)
ˆ The polar angle Φ being Now, consider the expression (65) for the complex centerline radius R. ˆ we have: the argument of R, Φ = arg ρˆ + ϕ −
π , 2
(93)
with ρˆ =
1 M3 − MZ z − iz 0 √ . F ± 1 − z2
(94)
The function ρˆ depends on s through z and z 0 only, hence it describes a closed curve in the complex plane. If ρˆ does not vanish anywhere, as s increases by a period 2K(k), the argument of ρˆ increases by 2πm, where m is some integer. Moreover, the real and imaginary parts of the numerator in (94) being decreasing functions of z and z 0 respectively, the curve defined by ρˆ has no self-crossing (and is parameterized clockwise by s). Hence, it turns around the origin at most once clockwise over a period 2K(k), so that m is restricted to the values 0 and −1. The domain in the (z1 , z2 , z3 , S) space where either value of m holds is delimited by the condition (68.b) ensuring the existence of 1 z2 points where ρˆ vanishes. We find that m = −1 if S = + and z3 > 1−z z2 −z1 , and m = 0 otherwise. As a consequence, we have: ϕ (s + 2K(k)) − ϕ(s) = Φ (s + 2K(k)) − Φ(s) +
2π 0
1 − z 1 z2 if S = + and z3 > , z 2 − z1 1 − z 1 z2 . if S = − or z3 < z2 − z1
(95)
Hence, over a period of z, the precession angle ϕ of the top covers either the same angular distance as the polar angle Φ of the rod, or the same angular distance plus one complete revolution. Now, using (60.a) and (73.a), we define the mean angular velocities hϕ0 i and hΦ0 i for the corresponding angles ϕ and Φ as: 0®
Z
2K(k)
M− Π(n− , k) M+ Π(n+ , k) +S , 1 + z1 K(k) 1 − z1 K(k) 0 Ã ! Z 2K(k) ˜ Π(˜ 0® 1 1 M3 − MZ h n, k) 0 Φ = ds Φ = MZ + . ˜ − z1 2K(k) 0 2 K(k) h ϕ =
1 2K(k)
ds ϕ0 =
(96.a) (96.b)
π . The question arises As a consequence, these two quantities either are equal or differ from K(k) 0 0 then: which mean angular velocity, hϕ i or hΦ i, should we use to define the polar angle hΦi of our average helix? The most natural choice seems to take hΦi = hΦ0 i s. However, the expression (96.b) is not continuous through the boundary (68.b), making the definition of the mean polar angle hΦi ambiguous at that point. Therefore, we define:
®
hΦi = ϕ0 s − µ
hΦi =
0®
ϕ −
π 2
for S = −,
(97.a)
¶
π π s− K(k) 2
for S = +.
(97.b)
These expressions reduce to hΦ0 i s for sufficiently large z3 , and are continuous across the boundary (68.b). As we shall see, this definition is the most adequate in the case where the filament shape is 29
a supercoiled helix (this case has a great importance in biochemical applications, in particular for DNA supercoiling [31, 40, 41]). The supercoiled helices are defined and discussed below. It remains now to define the mean helix radius hRi. A definition consistent with (97) is: hRi =
1 2K(k)
Z
2K(k)
0
ˆ exp −i hΦi . ds R
(98)
Therefore, the generic filament behaves on average like an helical filament. Supercoiled helices: A supercoiled helix is a curve which looks like a helix on short length scales, with the central axis itself shaped like a helix on large length scales. The condition for the centerline R to be a supercoiled helix, in terms of spinning tops, is that the vector d3 describes slowly precessing, nearly circular, oblique loops on the unit sphere. Two such examples are shown on Figures 16 and 17. If any of these three conditions (slow precession, near-circularity and oblicity of the top orbit) is not fulfilled, the filament shape will not look like a supercoiled helix, but rather like a deformed helix, as shown on Figures 18 to 21. The supercoiled helices can be studied systematically as solutions close to the oblique circular orbits of the spinning top. There are two ways to obtain circular orbits. The first one consists in setting z1 = z2 , in which case the orbit is horizontal, and not oblique. The second one is to take the limit z3 → ∞. Indeed, in view of (48), the tension vanishes as z3 grows without bound, and, as we mentioned in Section III.2, every filament with null tension is a helix, hence every top orbit with F = 0 is a circle. Furthermore, these asymptotic circular orbits can be arbitrarily oriented since the vertical direction eZ cannot be distinguished from the other directions in the limit F = 0. As a consequence, we can redefine a supercoiled helix as a solution with a large (“close to infinity”) value of z3 . In practice, however, z3 does not need to be very large, so that the threshold value appearing in (95) can be reached with the centerline R still reasonably looking like a supercoiled helix. In the limit of large z3 , the vertical axis joining the poles of the unit sphere is interior to precessing circular orbit in the case S = +, and exterior to it in the case S = −. As a consequence, one passes continuously from the supercoiled helices with S = − to the supercoiled helices with S = + by enlarging the slowly precessing orbit so that it passes through one of the poles, as on Figure 17. We mentioned above that our definitions of the average helix coordinates hRi, hΦi, hZi are well adapted to supercoiled helices, in the sense that they describe consistently the large scale helical behavior of the axis around which the centerline is wound. Namely, in the limit z3 → ∞, the supercoiled helix tends towards an (infinitely remote) ordinary helix, with a straight axis given by the limit of the average helix hRi. Deformed helices: In many cases, although the centerline R winds around the average helix hRi, it does not quite look like a supercoiled helix. This happens if the criterion discussed above is not fulfilled, which can result in the following. 2π • The short scale spatial period 2K(k) and the large scale spatial period hΦi 0 can be too close to each other, in which case the two orders of helicity cannot be clearly distinguished. Such a situation is shown on Figure 18.
• If the top orbit is too far away from a circle, the short scale pattern does not look like a helix. As an example, Figure 19 shows a large-scale helix with curvature varying on short scale.
30
• For some values of the constants, the top orbit can be periodic although quite different from a circle. In this case, the filament shape is periodic in space, but different from a helix (see the “oblique helix” on Figure 20). In addition, there is the possibility for the amplitude of the short scale pattern to be large enough for two consecutive turns of the large-scale helix to overlap each other. In this case, the topology of the solution is different from the topology of an ordinary helix. Such a “knotted helix” is shown on Figure 21. III.5.4
Bounded and Closed Filament Shapes
The filament shapes discussed so far are all unbounded in space, except for the twisted ring shown on Figure 5 and the lemniscate shown on Figure 11a. In general, bounded shapes are obtained by imposing the coordinate Z to be a periodic function of s, that is, Z = 0 for s = 2K(k). Using (76), the condition for boundedness reads: z3 K(k) − (z3 − z1 )E(k) = 0.
(99)
The condition for the space curve R to be closed is obtained by requiring, in addition, that the periods of the variables Z and Φ be in a ratio of integer numbers. Namely, using (73.a): MZ K(k) +
˜ M3 − MZ h mΦ Π(˜ n, k) = 2π , ˜ m h − z1 Z
(100)
where the integers mΦ and mZ denote, respectively, the number of periods of the variables Z and Φ in a complete covering of the centerline R. Finally, one can impose the ribbon associated to the filament to be closed by requiring that the periods of the variables Z and ζ be in a ratio of integer numbers, or, using (73.b): µ
¶
mζ M3 h − MZ 2M3 K(k) − M3 K(k) + Π(n, k) = 2π , b h − z1 mZ
(101)
where mζ denotes the number of periods of ζ in a complete covering of the centerline R. The condition (101) is useful if one has, for instance, an octogonal cross-section, in which case mζ must be set to an integer multiple of 18 , in order for the octogons at s = 0 and at s = 2mZ K(k) to match each other. This together makes three conditions from which the constants z1 , z2 and z3 can be determined. However, the conditions (100) and (101) are not very tractable, because the left-hand side of (100) is proportional to the average angular velocity hΦ0 i which, as we mentioned, is not a continuous function of the constants z1 , z2 and z3 . This holds too for the left-hand side of (101). As a consequence, numerical root solvers are inefficient in solving the system (99)-(101). In practice, it is well advised to replace the equations (100) and (101) by equivalent conditions on the Euler angles ϕ and ψ. Remember that, over a period of Z, the angles ϕ and Φ cover angular distances which differ by integer multiples of 2π. A similar relation holds for the angles ψ and ζ. The conditions on ϕ and ψ analogous to (100) and (101) read: M+ M− mϕ Π(n+ , k) + S Π(n− , k) = π , 1 + z1 1 − z1 mZ µ ¶ mψ M+ M− 1 − 1 M3 K(k) + Π(n+ , k) − S Π(n− , k) = π , b 1 + z1 1 − z1 mZ
(102.a) (102.b)
where mϕ and mψ differ from mΦ and mζ respectively by an integer. Examples of closed filaments are displayed on Figures 22 to 24. Figures 22 and 23 show torus knots, that is, knotted curves which are topologically equivalent to closed curves lying on a torus. Figure 24 shows a supercoiled ring (notice the large value of z3 ). 31
N
Figure 16: A slowly precessing, nearly circular top orbit corresponds to a supercoiled helix (z1 = − 12 , z2 = 34 , z3 = 4, S = −).
N
Figure 17: A top orbit crossing periodically the north pole. The corresponding filament has a periodically vertical tangent. (z1 = 12 , z2 = 1, z3 = 3).
S
Figure 18: A top orbit crossing periodically the south pole. The corresponding filament has a periodically vertical tangent. (z1 = −1, z2 = − 12 , z3 = 14 10 , S = −). 32
N
Figure 19: A top orbit presenting turn-back points corresponds to a filament with periodically vanishing curvature (z1 = 0, z2 = 12 , z3 = 2, S = −).
N
Figure 20: A periodic top orbit results in a spatially periodic filament shape (z1 = −0.7822, z2 = 0.8782, z3 = 1, S = −). N
Figure 21: An apparently simple top orbit resulting in a surprisingly complex “knotted helix” (z1 = −0.4152, z2 = 0.2800, z3 = 1.026, S = −).
33
N
Figure 22: A torus knot with mZ = 5, mΦ = 3 and mζ = 1 (z1 = −0.4152, z2 = 0.3446, z3 = 1.026, S = −). N
Figure 23: A torus knot with mZ = 7, mΦ = 4 and mζ = 1 (z1 = −0.4997, z2 = 0.4013, z3 = 1.037, S = −). N
Figure 24: A supercoiled ring with mZ = 20, mΦ = 1 and mζ = 6 (z1 = −0.4269, z2 = 0.4171, z3 = 9.084, S = −). 34
IV
Conclusions
In this paper we have shown how to classify the shapes of Kirchhoff filaments based on the geometry of the the spinning top solutions. To do so, we have pushed the Kirchhoff analogy to its extreme and systematically obtained interesting properties of filaments based on the corresponding soutions of the Euler equations. We showed that the solutions of Kirchhoff equations can be extremely varied and that many interesting cases can be distinguished. In particular, we found explicit conditions on the boundary values for filaments to have points of self-tangency and multiple self-intersection. We also studied the case where filaments have points of vanishing curvature and show that they correspond to orbits of spinning tops with turn-back points. We gave a complete description of localizing solutions, that is solutions wich are homoclinic in the curvature-torsion space, that is filaments in space which are asymptotic to a straight line. In the same way, we found conditions to obtain filaments which have the topology of torus knots, that is bounded and periodic filaments (in the physical space). Finally we studied the behavior of generic filaments and show that on long length scales they always behave like helical filaments. Some of the particular solutions presented here have been obtained in various places by different authors. In this paper we have stressed on the geometry of these solutions and presented them in a unifying way based on the familiar framework of the spinning top. The solutions of the Kirchhoff equations for rods with circular cross-sections are often used as a first guess to study numerically physical filaments with different proeprties (e.g. non-circular cross-sections, intrinsic curvature or torsion, ...). We hope that the explicit solutions given in this paper together with their geometric classification will be useful in this context. Acknowledgements: M. N. is acknowledges support from the Fonds National de la Recherche Scientifique (F.N.R.S) M.N. is an aspirant at the FNRS. This work is supported by NATO-CRG 97/037.
A
Elliptic Functions
Remark: In the following, the constant k is called the modulus of the elliptic functions and ranges between 0 and 1, while the constant n is called the characteristic and is a real number less than 1.
A.1 A.1.1
Elliptic Integrals Incomplete Elliptic Integrals in Standard Form
The incomplete elliptic integrals of the first, second and third kind in standard form are, respectively, defined by: F (Φ | k) =
Z
Φ
S
E (Φ | k) =
p
0
Z
S
dΦ0 1 − k 2 sin2 Φ0
Φ
dΦ 0
ΠS (Φ | n, k) =
Z 0
0
,
(103.a)
q
1 − k 2 sin2 Φ0 , dΦ0
Φ
¡
(103.b)
1 − n sin2 Φ0
35
¢p
1 − k 2 sin2 Φ0
.
(103.c)
A.1.2
Incomplete Elliptic Integrals in Algebraic Form
The algebraic forms are obtained by carrying out the following change of variable into the standard forms: u = sin Φ.
(104)
This yields: FA (u | k) = EA (u | k) =
Z
du0 , (1 − u02 )(1 − k 2 u02 )
u
p
0
Z
u
0
ΠA (u | n, k) =
s
du0
Z
1 − k 2 u02 , 1 − u02
(105.b)
p
(105.c)
u
0
(105.a)
(1 − nu02 )
du0 . (1 − u02 )(1 − k 2 u02 )
The algebraic forms are those which are implemented in the symbolic calculus software Maple. Notice that for the change of variable (104) to be bijective, one must restrict Φ to the interval £ π π¤ −2, 2 . A.1.3
Incomplete Elliptic Integrals in Practical Form
We call the following forms of the elliptic integrals of the second and third kind “practical” because these are the forms under which they appear the most naturally in the problems in Sections III. These forms are obtained by carrying out the following change of variable into the standard forms: s = FS (Φ | k). This yields: E(s | k) =
Z 0
Π(s | n, k) =
s
ds0 dn2 (s0 | k),
Z
s
ds0
0
ds0 , 1 − nsn2 (s0 | k)
(106)
(107.a) (107.b)
where the functions sn and dn are defined in Section A.2. A.1.4
Complete Elliptic Integrals
The complete elliptic integrals are defined by the expressions for the corresponding incomplete elliptic integrals evaluated at u = 1 in algebraic form. They are denoted in the following way: K(k) = FA (1 | k) ,
(108.a)
E(k) = E (1 | k) ,
(108.b)
Π(n, k) = Π (1 | n, k) .
(108.c)
A
A
A.2 A.2.1
Jacobi’s Elliptic Functions Definitions
The incomplete elliptic integrals, having positive integrands, define monotonous, hence invertible, functions. The function am is defined as the inverse of the standard form FS of the incomplete elliptic integral of the first kind: am(s | k) = (FS )−1 (s | k). 36
(109)
We then define Jacobi’s elliptic functions sn, cn and dn as: sn(s | k) = sin am(s | k),
(110.a)
cn(s | k) = cos am(s | k),
(110.b)
dn(s | k) =
q
1 − k 2 sn2 (s | k).
(110.c)
Notice that the function sn itself, restricted to the interval [−K(k), K(k)], is the inverse of the algebraic form FA of the incomplete elliptic integral of the first kind. The four functions sn, cn, dn and am are represented on Figures 25 to 26 for various values of k. A.2.2
Elementary properties of am, sn, cn and dn
The function am obeys the relations: am(2nK(k) ± s | k) = nπ ± am(s | k),
(111)
for any real s and integer n. Hence, the knowledge of am over the interval [0, K(k)], together with (111), is sufficient to reconstruct the function over the whole real line. This holds too for the periodic functions sn, cn and dn, which obey the following relations: sn(2nK(k) ± s | k) = ±(−1)n sn(s | k),
(112.a)
cn(2nK(k) ± s | k) = (−1) cn(s | k),
(112.b)
dn(2nK(k) ± s | k) = dn(s | k).
(112.c)
n
A.2.3
Limits of am, sn, cn and dn for k = 0 and k = 1
These limits are obtained by considering, for k = 0 and k = 1, the expression (103.a) defining the function FS , and then taking the limit of am to be the inverse function. For k = 0, one has: am(s | 0) = s,
(113.a)
sn(s | 0) = sin s,
(113.b)
cn(s | 0) = cos s,
(113.c)
dn(s | 0) = 1,
(113.d)
whereas k = 1 leads to: am(s | 1) = arcsin tanh s,
(114.a)
sn(s | 1) = tanh s,
(114.b)
cn(s | 1) = sechs,
(114.c)
dn(s | 1) = sechs.
(114.d)
37
A.2.4
Derivatives of am, sn, cn and dn
The derivatives of am, sn, cn and dn are obtained by differentiating (103.a) and the definition equations (110.a) to (110.c). This leads to the following differential relations: d am(s | k) = dn(s | k), ds d sn(s | k) = cn(s | k)dn(s | k), ds d cn(s | k) = −sn(s | k)dn(s | k), ds d dn(s | k) = −k 2 sn(s | k)cn(s | k). ds
(115.a) (115.b) (115.c) (115.d)
References [1] J. Coyne, Analysis of the Formation and Elimination of Loops in Twisted Cable, IEEE J. Ocean. Engng. 15, 72-83 (1990). [2] Z. Tan & J. A. Witz, Loop Formation of marine Cables and Umbilicals during Installation, BOSS 92 (ed. M. H. Patel & R. Gibbins, London: BBP Technical Services, 1992) 1270-1285. [3] E. E. Zajac, Stability of two Loop Elasticas, trans. ASME 136-142 (1962). [4] M. D. Barkley & B. H. Zimm, Theory of Twisting and Bending of Chain Macromolecules; Analysis of the fluorescence depolarization of DNA, J. Chem. Phys. 70, 2991-3006 (1979). [5] R. F. Goldstein, & S. A. Langer, Nonlinear Dynamics of Stiff Polymers, Phys. Rev. Lett. 75, 1094 (1995). [6] W. Helfrich, Elastic Theory of Helical Fibers, Langmuir 7,567-568 (1991) . [7] J. V. Selinger, F. C. MacKintosh & J. M. Schnur, Theory of Cylindrical Tubules and Helical Ribbons of Chiral Lipid Membranes, Phys. Rev. E 53, 3804-3818 (1996). [8] W. R. Bauer, R. A. Lund & J. H. White, Twist and Writhe of a DNA Loop containing Intrinsic Bends, Proc. Natl. Acad. Sci. 90, 833-837 (1993). [9] D. Bensimon, A. J. Simon, V. Croquette & A. Bensimon, , Stretching DNA with a Receding Meniscus: Experiments and Models, Phys. Rev. Lett. 74, 4754-4757 (1995). [10] Ph. Cluzel, A. Lebrun, Ch. Heller, R. Lavery, J.-L. Viovy, D. Chatenay & Caron, F., DNA: An Extensible Molecule, Science 271, 792-794 (1996) . [11] N. G. Hunt, & J. E. Hearst, Elastic Model of DNA Supercoiling in the Infinite Length Limit, J. Chem. Phys. 12, 9329-9336 (1991) . [12] T. Schlick & W. K. Olson, Trefoil Knotting Revealed by Molecular Dynamics Simulations of Supercoiled DNA, Science 257, 1110-1114 (1992). [13] S. B. Smith, Y. Cui & C. Bustamante, Overstretching B-DNA: The Elastic Response of Individual Double-Stranded and Single-Stranded DNA Molecules, Science 271, 795-799 (1996). [14] T. R. Strick, J.-F Allemand, D. Bensimon, & V. Croquette, The Elasticity of a Single Supercoiled DNA Molecule, Science 271, 1835-1837 (1996). 38
[15] Y. Yang, I. Tobias& , W. K. Olson, Finite Element Analysis of DNA Supercoiling, J. Chem. Phys. 98, 1673-1686 (1993). [16] A. Goriely & M. Tabor, Nonlinear Dynamics of Filaments, Nonlinear Dynamics, (to be published) 1999). [17] N. H. Mendelson, Bacterial Microfibers: The Morphogenesis of Complex Multicellular Bacterial Forms, Sci. Progress Oxford 74, 425-441 (1990). [18] J. J. Thwaites, & N. H. Mendelson, Mechanical Behavior of Bacterial Cell Walls, Adv. Microbiol. Physiol. 32, 174-222 (1991). [19] R. L. Ricca, & M. A. Berger, Topological Ideas and Fluid Mechanics, Physics Today 28-34 (Dec. 1996). [20] Tyson, J. J. & S. H. Strogatz, The Differential Geometry of Scroll Waves, Int. J. of Bifurcation and Chaos 1, 723-744 (1991) . [21] J. P. Keener, Knotted Vortex Filament in an Ideal Fluid, J. Fluid Mech. 211, 629-651 (1990). [22] S. Da Silva & A. R. Chouduri, A Theoretical Model for Tilts of Bipolar Magnetic Regions, Astron. Astrophys. 272, 621 (1993). [23] H. C. Spruit, Motion of Magnetic Flux Tubes in the Solar Convection Zone and Chromosphere, Astron. Astrophys. 98, 155 (1981) . ¨ [24] G. Kirchhoff, Uber das Gleichgewicht und die Bewegung eines unendlich d¨ unnen elastischen Stabes, J. Mathematik 56, 285-313 (Crelle, 1859). [25] A. E. H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th ed. (Dover, New York, 1944). [26] S.S. Antman, A Nonlinear Problems of Elasticity, (Springer, Berlin, 1995). [27] A. Goriely & M. Tabor, Nonlinear Dynamics of Filaments III: Instabilities of Helical Rods Proc. Roy. Soc. London Ser. A 453, 2583–2601 (1997). [28] A. Goriely & M. Tabor, Nonlinear Dynamics of Filaments I: Dynamical Instabilities, Physica D 105, 20-24 (1997). [29] A. Goriely & M. Tabor, Nonlinear Dynamics of Filaments IV: Spontaneous looping of elastic rods, Proc. Roy. Soc. A 454, 3183–3202 (1998). [30] R. L. Ricca, The Energy Spectrum of a Twisted Flexible String under Elastic Relaxation, J. Phys. A 28, 2335-2352 (1995). [31] C. J. Benham,An Elastic Model of the Large-Scale Structure of Duplex DNA, Biopolymers 18, 609-623 (1979). [32] S. Kehrbaum & J. H. Maddocks, Elastic rods, rigid bodies, quaternions and the last quadrature, Phil. Trans. R. Soc. Lond. A 355, 2117–2136 (1997). [33] G. Domokos, A group-theoretic approach to the geometry of elastic rings, J. Nonlinear Sci. 5, 453-478 (1995).
39
[34] Y. Shi & J. E. Hearst, The Kirchhoff Elastic Rod, the Nonlinear Schr¨ odinger Equation and DNA Supercoiling, J. Chem. Phys. 101, 5186-5200 (1994). [35] B. D. Coleman, E. H. Dill, M. Lembo, Z. Lu, and I. Tobias On the dynamics of rods in the theory of Kirchoff and Clebsch. Arch. Rational Mech. Anal., 121, 339–359, (1993). [36] J. L. Lagrange, M´ecanique Analytique (Paris, Chez la Veuve Desaint, 1788). [37] W. D. MacMillan, Dynamics of Rigid Bodies (McGraw-Hill, New York, 1936). [38] E. T. Whittaker, A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, 4th ed. (Cambridge, London, 1965). [39] A. Goriely, M. Nizette & M. Tabor, Nonlinear dynamics of strips (preprint) (1999) [40] C. J. Benham, Theoretical Analysis of conformational equilibria in superhelical DNA, Ann. Rev. Biophys. Chem. 23-45 (1985). [41] W. F. Pohl, DNA and Differential Geometry, Math. Intelligencer 3, 20-27 (1980).
40
sn(s | k)
k=1
1
cn(s | k)
1
k = 63/64
k = 3/4
k=0
0.5
k = 15/16
0.5
k=1
-6
-4
0
-2
2
4
6
-6
-4
0
-2
2
4 k = 63/64
s -0.5
s
k = 15/16
-0.5
k = 3/4
k=0
-1
6
-1
Figure 25: The function sn and cn for various values of k.
1
dn(s | k)
6
am(s | k)
k=0
k=0 k = 3/4
4 0.8
k = 15/16
k = 3/4
k = 63/64 k=1
2 0.6
s k = 15/16
-6
-4
-2
0
2
0.4
-2 k = 63/64
0.2
-6
-4
-2
0
-4
2
k=1
s
4
6
-6
Figure 26: The function dn and am for various values of k.
41
4
6
