alain goriely and michael tabor- nonlinear dynamics of filaments iii: instabilities of helical rods

8/3/2019 Alain Goriely and Michael Tabor- Nonlinear Dynamics of Filaments III: Instabilities of Helical Rods

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Nonlinear Dynamics of Filaments III:

Instabilities of Helical Rods

Alain Goriely∗† and Michael Tabor∗

∗University of Arizona, Program in Applied Mathematics,Building #89, Tucson, AZ85721, USA

†Universite Libre de Bruxelles, Departement de Mathematique, CP218/11050 Brussels, Belgium, e-mail: [email protected]

(To be published in Proc. Roy. Soc. London Ser. A)

Abstract

The time dependent Kirchhoff equations for thin elastic rods are used to study the linear

stability of twisted helical rods with intrinsic curvature and twist. Using a newly developed

perturbation scheme we derive the general dispersion relations governing the stability of various

helical configurations. We show that helices with no terminal forces are always dynamically

stable. We also compute the most stable helical shape against twist perturbations and show

that different unstable modes can be excited in different regions of the parameter space and can

sometimes coexist. The linearly unstable modes are computed and explicit forms are given.

1 Introduction

Helices are one of the most simple filamentary structures found in nature. They appear at manydifferent level of organization: ranging from molecular biology to magnetohydrodynamics (?; ?; ?;?). The usual approach to modeling these structures is to assume that they can be represented as anelastic filament subject to the classical laws of mechanics and elasticity theory. Within the frameworkof linear elasticity, the Kirchhoff equations for elastic filaments provide a basic starting point for muchanalysis and computation (?). These equations are a system of coupled nonlinear partial differentialequations embodying the conservation of linear and angular momentum in terms of the time and spaceevolution of a local triad (the director basis) attached to the filament. They have been demonstrated tocapture many fundamental features of real elastica and have recently been obtained as the convergentlimit of three-dimensional elasticity theory (?). Stationary (i.e. time independent) helices are knownto be exact solutions of the Kirchhoff equations (?; ?; ?) and as a consequence they have beenextensively studied, both numerically and theoretically. However, to the best of our knowledge, thedynamical stability , that is the growth of time-dependent instabilities, of helical forms has never been

investigated. To do so, one has to consider evolution governed by the full time-dependent Kirchhoff equations and this required us to develop a novel arc-length preserving perturbation scheme (?; ?).We show here how to apply this general perturbation procedure to the particular case of the helicalfilaments.

The paper is organized as follows: In Section II, we give the basic elements of curve dynamicsand the Kirchhoff model in the general director basis. In Section III, we describe the perturbationprocedure. In Section IV, we describe the stationary solutions for helices. In Section V, we derive thedynamical variational equations and their associated dispersion relations. Section VI is devoted tothe analysis of these dispersion relations and the construction of the linear solutions after bifurcation.



2 The Kirchhoff model

2.1 Kinematics of curves

The Kirchhoff model describes the evolution of elastic filaments. The physically relevant quantitiescharacterizing the three-dimensional elastic body are described in terms of a local coordinate basisattached to the space curve defining the central axis of the filament. We first give a brief summary of the kinematics (i.e. the relationship between positions and velocities) of space curves before describingthe dynamics (i.e. the relationship between velocities and forces).

The space curve x = x(s, t) : R × R → R3 is parameterized by arc length, s, and time, t, and

assumed to be of at least class C 3. A local coordinate triad {d1, d2, d3} is defined for each valueof s and t with the vector d3(s, t) = x(s, t) corresponding to the tangent vector of x at s. (Theprime denotes the s-derivative and since s is the arc length it follows that d3.d3 = 1.) The vectors{d1(s, t), d2(s, t)} are an orthonormal pair defining the plane normal to d3. The three vectors arechosen such that {d1, d2, d3} form a right-handed orthonormal triad (d1×d2 = d3, d2×d3 = d1). If d1is set along d3, the triad specializes to the well-know Frenet frame in which d1 and d2 are the normaland bi-normal vectors respectively. The space curve can be reconstructed at all times by integratingthe tangent vector, i.e. x(s, t) =

s

d3(s, t)ds.

The considerable advantage of the director basis is that it can be used to represent physicalfilaments by setting the directors d1 and d2 to coincide with the principal axes of the rod and, whereappropriate, rotate with respect to s along the tangent direction. Their rotation with respect to salong the tangent direction corresponds to the twist in the rod. By contrast the Frenet basis, whichprovides a geometrically appealing characterization of space curves, cannot capture satisfactorily suchphysical properties of actual rods. The evolution of the director basis with respect to arc length andtime takes the form

di =3

j=1

K ijdj i = 1, 2, 3, (1.a)

di =

3

j=1

W ijdj i = 1, 2, 3, (1.b)

where ˙( ) stands for the time derivative. W and K are the antisymmetric 3 by 3 matrices:

K =

0 κ3 −κ2

−κ3 0 κ1

κ2 −κ1 0

, W =

0 ω3 −ω2

−ω3 0 ω1

ω2 −ω1 0

. (2)

which ensures that the basis remains orthonormal (?). The elements of K and W make up the

components of the twist and spin vectors respectively; namely κ =3

i=1 κidi and ω =3

i=1 ωidi.

2.2 Dynamics of filaments

The above relations for the evolution of the director basis are pure kinematics and physical dynamicsmust now be formulated to give the relationship between the strains, κi, and the stresses associatedwith the linear and angular momenta. Kirchhoff’s theory of elastic filaments (?) is based on the ideathat for sufficiently thin rods with modest local deformations, it is possible to describe the forces andmoments averaged over local cross-sections attached to the central axis, x = x(s, t), of the rod. Thisenables one to write down a one dimensional (i.e. parameterized by s) theory. In this framework thetotal force F = F (s, t) and the total moment M = M (s, t) can thus be expressed locally in term of the

director basis, i.e. F =3

i=1 f idi, M =3

i=1 M idi. Conservation of linear and angular momentumthen leads to equations for the force and the moment ( ?):

F = ρAd3, (3.a)

M

+ d3 × F = ρI (d1 ×¨d1 + d2 ×

¨d2), (3.b)



where I is the moment of inertia (about a radial cross section), ρ is the density and A the area of a(circular) cross section. These equations are completed by introducing constitutive relationships. Inthe linear theory of elasticity these take the form:

M = EI ((κ1 − κu

1)d1 + (κ2 − κu

2 )d2) + 2µI (κ3 − κu

3)d3, (4)

where E is the Young modulus and µ the shear modulus and the vector κu corresponds to the twist(if any) in the unstressed configuration and will be referred to as the intrinsic twist vector 1. Forinstance, if κu = 0, the filament is said to be naturally straight, i.e. the lowest energy configuration of the filament is a straight rod. If κu = (0, 1/Ru, 0) the rod is naturally circular and the lowest energyconfiguration is a ring of radius Ru. Here we are defining the (static) energy to be

E =

M.κ ds (5)

Without loss of generality, we can rescale all the variables

t → t

Iρ/AE s → s

I/A,

F → AEF M → ME √

AI,

κ → κ

A/I ω → ω

AE/Iρ. (6)

and the Kirchhoff equations (3-4) take the form:

F = d3, (7.a)

M + d3 × F = d1 × d1 + d2 × d2, (7.b)

M = (κ1 − κu1 )d1 + (κ2 − κu

2)d2 + Γ(κ3 − κu3)d3, (7.c)

where Γ = 2µ/E . For rods with circular cross-section it is a simple matter to show that Γ = 1/(1 + σ)where σ is Poisson’s ratio. The parameter Γ characterizes the elastic property of the filament varyingbetween 2/3 (incompressible case) and 1 (hyper-elastic case). In all the specific examples studied inthis paper Γ = 3/4.

Equation (7.c) can be used to eliminate M from (7.b). The net result, on combining with thetwist and spin equations, is a system of 9 equations for 9 unknowns (f, κ, ω). We refer to this systemas the Kirchhoff equations. It should be noted that solving (7.b) only yields the normal componentsf 1 and f 2 of the force in terms of the strains κi and the spins ωi. The tangential force component f 3represents the tension along the rod and this must be determined by the requirement of arc-lengthconservation. Overall, the Kirchhoff equations describe the time evolution of a thin rod (here withcircular cross section) in the limit of small deformations where tractions on the external surfaces areassumed to vanish. From now on the terms Kirchhoff rod or simply a rod refer to a solution of thisset of equations with proper boundary conditions and initial data.

3 Perturbation Scheme

Although a considerable amount of analytical work on the Kirchhoff equations is performed in termsof the Euler angles, we have found (?) that the stability of stationary solutions is best described interms of a perturbation scheme at the level of the local director basis. The idea is to represent thebasis of the perturbed system as an expansion around the unperturbed stationary triad such that itremains orthonormal at each order in the perturbation parameter. That is

di = d(0)i + d

(1)i + O(2) i = 1, 2, 3, (8)

1The component κ(u)3 is the intrinsic twist whereas (κ

(u)1 )2 + κ32(u))2 is the intrinsic curvature)



Imposition of the orthonormality condition di.dj = δij leads to an expression for the perturbed basisin terms of the unperturbed basis; namely

d(1)i =

3j=1

Aijd(0)j , (9)

where A is the antisymmetric matrix:

A =

0 α3 −α2

−α3 0 α1

α2 −α1 0

, (10)

It is straightforward, given the vector α = (α1, α2, α3), to reconstruct the perturbed rod by integratingthe tangent vector:

x(s, t) =

sds

d(0)3 + (α2d

(0)1 − α1d

(0)2 )

+ O(2). (11)

This type of perturbation expansion can be applied to any vector represented in terms of the di. Thusthe vector V =

3i=1 vidi can be expanded as V = V (0) + V (1) + O(2), where

V (1) =i

v(1)i + (A.v(0))i

d(0)i . (12)

This observation enables us to express the first order perturbations of the twist and spin matrices,i.e. K = K (0) + K (1) + ..., W = W (0) + W (1) + ... where, by using the kinematic equations (1), onemay show that

K (1) =∂A

∂s+

A, K (0)

, (13.a)

W (1) =∂A

∂t+

A, W (0)

. (13.b)

and [A, B] = A.B − B.A. Higher order perturbations can easily be obtained in terms of the lowerorder terms in the same way (?).

The use of this perturbation expansion and the above equations enables one to derive the dynamical

variational equations by expanding the Newton’s equation (7.a) and moment equation (7.b) tofirst order in . These equations are a linear system of 6 equations for the 6-dimensional vectorµ = {α, f (1)},

LE(κ(0), f (0)).µ = 0, (14)

where f (1) is the first order correction to the force and LE is a second-order differential operator in sand t whose coefficients depends on s through the unperturbed solution κ(0), f (0). The explicit form of LE was given in full generality in (?). The solutions of the variational equations control the stability,or lack thereof, of the stationary solutions with respect to linear time-dependent modes.

4 The stationary helix

The geometry and statics of helical rods are well known. A more recent study of the energetics of helices is given in (?). Here we summarize the results relevant to our stability analysis. We considera helical space curve, xh, wrapped around a cylinder of radius R whose central axis points along thefixed vector e1 (i.e. the x-axis):

xh = (Pδs,R cos(δs), R sin(δs)) . (15)

Here s is the arc-length and δ is determined by the arc-length defining constraint T(s).T(s) = 1, ∀s,where T(s) is the tangent vector T(s) = xh(s). This yields:

δ = ± 1

P 2 + R2 . (16)



The choice + defines a right-handed helix for P > 0. That is, if you point your right-hand thumbalong the x-axis, your hand will naturally rotate to the right as you move along the helix. This helixis shown in Fig. 1. A few additional facts about the helix geometry can be noted at this point. Theheight (along the x-axis) per turn of the helix is h = 2πP and the length of the curve per turn isl = 2π/δ. Thus for a helix of N turns, the total height is H = 2πP N and the total filament length is

L = 2πN/δ.We can now build the Frenet triad associated with the helix. The normal vector N(s) is obtained

by further differentiating and normalizing the tangent vector and the binormal vector B( s) is simplyconstructed by taking the cross product B(s) = N(s) ×T(s).

T(s) = (P δ,−Rδ sin(δs), Rδ cos(δs)) , (17.a)

N(s) = (0,− cos(δs),− sin(δs)) , (17.b)

B(s) = (−Rδ,−P δ sin(δs), P δ cos(δs)) . (17.c)

The Frenet curvature, κF , and torsion, τ F , are easily shown to be

κF = Rδ2 = RP 2 + R2

(18.a)

τ F = P δ2 =P

P 2 + R2(18.b)

Along with the planar circle and the straight line, the helix, is the only geometry for which κF andτ F are constant functions of arc-length.

As mentioned earlier, the Frenet triad is merely a special director basis. A more general triad canbe constructed by taking d3(s) = T(s) and applying a rotation of angle γ in the normal plane :

d(0)1 = cos(γs)N− sin(γs)B, (19.a)

d(0)

2 = − sin(γs)N− cos(γs)B, (19.b)d(0)3 = T. (19.c)

It is now possible to compute the twist vector by substituting this triad in the twist equation (1.a)and solving for the unknowns κ(0). Doing so, we obtain:

κ(0) = (κF sin(γs), κF cos(γs), τ F + γ ) . (20)

At this level of description the angle γ is arbitrary. However, it can be used to ascribe real physicalproperties to a helical rod (as opposed to curve) by representing the twist (if any) of the rod about itscentral axis. We will sometimes refer to γ as the twist density of the rod. A useful interpretation of (20) is to think of the filament as a helical space curve to which is attached a ribbon rotating about

the curve with twist density γ . This is the point of view we will adopt here in characterizing helicalfilaments

The physical properties of a static helical rod can be determined by solving the time independentform of the Kirchhoff equations. In this case, since the helix is stationary, we have, trivially, ω(0) = 0for the spin vector. In the constitutive relations (4) it is possible to choose a variety of κu. Here wechoose the “undistorted” twist vector to correspond to another helix, i.e.

κu = (κuF sin(γ us), κu

F cos(γ us), τ uF + γ u) . (21)

where κuF and τ uF are the Frenet curvature and torsion associated with a helical curve of radius Ru

and pitch parameter P u and γ u is a specified twist density. However, in the case of the latter, it iseasily demonstrated for the helix that static solutions only exist when the stressed and unstressedtwist densities (γ and γ u respectively) are the same. This is a consequence of the constancy of the

Frenet curvature and torsion.



2 Pπ

2 R

γ x

Figure 1: A helical rod characterized by an applied twist γ , a radius R, and a loop-to-loop distance2πP .



Knowing the twist vector κ(0), we can substitute it into the moment equation (7.b) and find

conditions on the forces (f (0)1 , f

(0)2 ) for a stationary solution and then determine the tension f

(0)3 from

equation (7.a). One finds

f (0) = (f 0 sin(γs), f 0 cos(γs),τ F κF

f 0) (22)

wheref 0 = −τ F (κF − κu

F ) + ΓκF (τ F − τ uF ) (23)

and we are assuming that κF = 0.These results exhibit the correct limit for the planar ring, i.e. P → 0. In the case of the unstressed

configuration corresponding to a straight rod, (i.e. κuF = 0), we write the unstressed curvature vector

as κu = (0, 0, γ u) and note that stationary helical solutions are now possible for γ = γ u. In this casethe forces are found to be f (0) = (f 0 sin(γs), f 0 cos(γs), 0), where f 0 = τ F κF (Γ− 1) + ΓκF (γ − γ u).

In general it is necessary to apply a terminal force, i.e. an axial force at the ends, to hold thehelix in its given shape. However, for special choices of parameters it is possible to find a “freestanding” helix, namely one for which no such forces are required. It is easily shown that the axialforce, i.e the force f x along the x-axis, is given by f x = f 0sec(α), where α is the pitch angle defined astan(α) = R/P . Thus, since the free standing condition is simply f 0 = 0, such a helix can be formed

whenκF (γ Γ− τ F (1− Γ)) + κu

F (τ F − ΓκF ) = 0. (24)

For example, if the unstressed configuration is a naturally straight rod (i.e. κuF = 0), the free standing

condition is γ = τ F (1− Γ)/Γ. We note also that even if terminal forces do not need to be applied tothe helix, a terminal moment has to be applied to maintain the helical structure. In the case of thenaturally straight rod, this is simply M = (κF sin(γs), κF cos(γs), τ F (1− Γ)).

The only way to build a helix requiring no terminal force and no terminal moment is to specifya non vanishing intrinsic curvature and intrinsic torsion. In this case κu = κ(0) and the helix hasvanishing elastic energy. This case will be referred to as a naturally free helix .

5 The dynamical variational equation and the dispersion

relationWe now have all the ingredients for a linear analysis of the stationary helix. Substitution of (κ(0), f (0))into (14) gives a six-dimensional system of equations for the six-dimensional vector µ. This systemhas s-dependent coefficients. In order to obtain a linear system with constant coefficients, we definethe new variables ν = Rγ.µ, where

Rγ =

M γ 0

0 −M γ

, (25)

and

M γ =

− cos(γs) sin(γs) 0

sin(γs) cos(γs) 00 0

−1

. (26)

The new system is a six dimensional system with constant coefficients that can be written inshort-hand form:

L1.∂ 2ν

∂t2+ L2.

∂ν

∂s+ +L3.

∂ 2ν

∂s2+ L4.ν = 0, (27)

where Li are 6× 6 matrices whose entries depends upon the parameters (R,P,γ, Γ, Ru, τ ).The different linear solutions of this system control the stability and/or vibration modes of the

helix. The general solutions of this linear system can be written as a sum of the fundamental linearmodes:

ν n = ξneσnt+iδns/N , (28)

where ξn ∈ C6 and N is the total number of turns in the helix ( N ∈ C), that is, we only considerhelices with an integral number of loops.



The growth rate σn for the n-th mode is obtained by requiring that ν n is a solution of (??):

σ2nL1 + iδnL2 − δ2n2L3 + L4

.ξn ≡ L.ξn = 0. (29)

After some straightforward (but tedious) computation and simplification, the matrix L is found to be:

L1,1 = −2If 0δ3η

κF , L1,2 = −τ F (η2 + 1)f 0δ2

κF − σ2, L1,3 = −δ2(η2 + 1)f 0,

L1,4 = −δ2(η2 + 1), L1,5 = 2Iηδτ F , L1,6 = 2IηδκF ,

L2,1 =τ F (η2 + 1)f 0δ2

κF + σ2, L2,2 = −2I

τ 2F ηδf 0κF

, L2,3 = −2Iτ F ηδf 0,

L2,4 = −2Iτ F ηδ, L2,5 = −η2δ2 − τ 2F , L2,6 = −κF τ F ,

L3,1 = δ2(η2 + 1)f 0, L3,2 = −2Iτ F ηδf 0, L3,3 = −2IκF ηδf 0,

L3,4 = −2IκF ηδ, L3,5 = −κF τ F , L3,6 = −η2δ2 − κ2F ,

L4,1 =δ2(Γτ uF + η2τ F ) + κF f 0

τ F − τ 2F (Γ− 1) + σ2, L4,2 = −Iηδ(τ F (2− Γ) + Γτ uF ),

L4,3 =−Iηδ(f 0 + κF (τ uF Γ + τ F ))

τ F , L4,4 = 0, L4,5 = 1, L4,6 = 0,

L5,1 = Iηδ((2− Γ)τ F + γτ uF ), L5,2 = (1 − Γ)τ 2F + τ F τ uF Γ + η2δ + σ2,

L5,3 = ((1− Γ)τ F + Γτ uF )κF , L5,4 = −1, L5,5 = 0, L5,6 = 0,

L6,1 = Iηδ(κF +f 0 + τ uF κF Γ

τ F )δ, L6,2 = κu

F τ F ,

L6,3 = (1− Γ)κ2F − τ uF Γτ F + Γη2δ2 + 2σ2 +

κF f 0 + Γτ uF δ2

τ F ,

L6,4 = 0, L6,5 = 0, L6,6 = 0,

where η = n/N .The dispersion relation ∆ = ∆(σn, n)

≡det(L) = 0 is a polynomial of degree 6 in σn and of

degree 12 in n. Once σ is known for a given n one can build the corresponding perturbed solutions byfinding the null vector ξn. The equations (11) can then be used to find the perturbed shape. Thesesolutions are always given up to an arbitrary phase corresponding to a rotation of the solution aroundthe x-axis. Setting this phase to zero, we find the deformed mode to first order in :

x1(s, t) = P δs − 2NKRξ1nτ F

cos(nδs

N ),

x2(s, t) = R cos(δs) − K

δ

ξ2 − ξ1n− N

sin(n−N

N δs) +

ξ2 + ξ1n + N

sin(n + N

N δs)

,

x3(s, t) = R sin(δs) − K

δ

ξ2 − ξ1n−N

cos(n−N

N δs) − ξ2 + ξ1

n + N cos(

n + N

N δs)

where K = Ceσt and

ξ1 = n3τ F δ3

(n2 − N 2)2δ6

κ2F n

2Γ + κF κuF (n2 − 2N 2 + 2ΓN 2) + N 2(κu

F )2

+ (n2 −N 2)2N 2κF δ4

2κ2F (κu

F − Γ) + κF (σ2Γ− (κuF )

2) + κuF (σ2 − τ F τ uF Γ)

+ N 4σ2κF δ

2

κF Γ(N 2 + n2) + κuF (3N 2 + n2)

− 2σ2N 6κuF κF

(30)

ξ2 = −N n4δ2

δ6

κ2F Γ(n2 −N 2)(τ F (2− Γ) + τ uF ) + κF κuF (n2 − N 2)(τ F (1− Γ) + τ uF Γ)

+(κuF )

2τ F N 2(2N 2

−n2)



+ N 3n2κF κuF τ F δ

4

κ2F (Γ− 1)(n2 −N 2) + σ2(n2 −N 2) + 3κF N 4κuF

+ N 5n2κF τ F δ

2−3κ3F κ

uF N 2 + 2κF n

2σ2Γ + κuF σ

2(3n2 −N 2)

+ N 7τ F κuF κ

6F κ

uF n

2

−κ3F σ

2(3n2

−N 2) + κu

F τ 6F n2

(31)

6 Linear analysis of the helical filament

We now turn to the stability analysis of the helical filament and consider different settings in whichdifferent parameters are varied. We first consider the simplest case where the rod is naturally straightand untwisted and show that the helical conformation is always unstable. To stabilize the rod weinclude the effects of intrinsic curvature and intrinsic torsion. We will then consider a situation wherea stable rod can be destabilized by varying certain control parameters.

We stress at this point that we only consider σ ∈ R. Indeed, the dispersion relation has othersolutions corresponding to vibration modes (i.e. iσ ∈ R)). These modes are not spontaneouslyunstable. That is, the corresponding amplitude of the linear modes is at most of the size of theperturbations themselves. Therefore, in order to observe these modes, they have to be specifically

excited.The boundary conditions are chose such that the x1 component of the helix is fixed in space. In

turn, this implies that n is an integer. That is, we only consider here discrete modes of perturbation.

6.1 Instability of a naturally straight helix.

The simplest case we can study is that of an untwisted naturally straight rod shaped as a helix. Theintrinsic curvature and the twist density are both set to zero, i.e. κu = (0, 0, 0) and γ = 0. In order toshow the instability of such a configuration we study the dispersion relation to find the neutral modesand determine if the modes in the range 0 < n/N < 1 are unstable (that is, σ2

n > 0). The neutralmodes are found by setting σ = 0 in the dispersion relation. Three neutral modes are found:

nN 2

= 0,

nN 2

= 1 (32.a) n

N

2=

(Γ− 2)2τ 2F + κ2F

δ−2 > 1 (32.b)

Around n/N = 1 the exponent σ can be expanded in powers of n:

σ2 = (1− Γ)δ4(n

N − 1)2 + O

(

n

N − 1)4

(33)

Since there is no neutral mode between n/N = 0 and n/N = 1 and locally σ > 0 around n = 1, weconclude that σn > 0 for all modes between n = 1 and n = N − 1. A typical plot of the dispersionrelation is shown on Fig. ??.

If a naturally straight rod is maintained in a helical shape and suddenly released, toward whatshape will it evolve? All the modes from n = 1 to N − 1 can be destabilized. If the helix has morethan one turn (N > 1), the helix will be unstable. The mode selected in such an unstable process isthe fastest growing mode (it is the one that will be first observed since it grows faster than the others).This mode can be easily found by looking at the integer closest to the maximum of the dispersionrelation. For instance in Fig. ??, the mode n = 2 is close to the maximum. Therefore, the fastestunstable mode is n = 2. The integration of the linear equation can be explicitly performed and thesolution for the space curve is found by integrating (11). Such an unstable mode is shown on Fig. ??.We see that the instability tends to localize the deformation in two nodes where buckling occurs.

Depending on the control parameters, different modes can become the fastest growing mode. Atthis point, it should be stressed that the instabilities a helix can undergo are quite different fromthose found for the planar twisted ring (?; ?; ?) since, in that case N = 1 and the periodic conditionsenforced by closure of the ring are such that the only unstable modes are n integer with n > 1.



0

2e-05

4e-05

6e-05

8e-05

0.0001

0.00012

1 2 3 4 5 6

n

σ2

Figure 2: The growth rate σ2 solution of the dispersion relation for the (naturally straight) helical rodas a function of the spatial mode n. The maximum is obtained close to n = 2 (κF = 1/8, τ F = 1/8,N = 5).

6.2 Stability of intrinsically curved helices

We now consider helices with intrinsic curvature (κuF = 0) and torsion (τ uF = 0). Furthermore we

assume that the helix is free standing, that is κ = κu. The helix is thus in an equilibrium state, since

the elastic energy vanishes identically. We expect such a configuration to be also dynamically stableand indeed no unstable mode can be excited. An easy way to show this is to consider the neutralmodes. In this simple case, the neutral dispersion relation read:

∆(σ = 0) = Γn4

N 4δ12

1− n

N

24

(34)

The only neutral modes are n/N = 0 and n/N = ±1. At n/N = ±1, the dispersion relation canbe locally expanded:

σ2 = − 2δ2τ −2F

Γ(δ2 + τ 2F ) + κ2F

(n

N − 1)4 + O

(

n

N − 1)8

(35)

Therefore, we conclude that σ2 ≤ 0 ∀ n , that is no mode is unstable (see Fig. ??) Nevertheless,

as mentioned earlier, the helix can still be subject to vibrations. These vibration correspond tomarginally unstable modes characterized by σ2 < 0. Their amplitude is of the same order as theperturbation itself. For each of these unstable modes the frequency can be obtained as ν =

√−σ2.These modes probably deserve further investigation.

6.3 Stability threshold of free standing helices

We have seen that all free helices are stable. Can we destabilize these structures by varying some of theparameters? There are two different ways a stable helix can be made unstable. First, one can modify“external” parameters such as the curvature or torsion (of the axial curve). For instance, consider thefollowing experiment: take a free standing helix and push (or pull and twist) the extremities. Thehelix will reach a critical shape where it will suddenly become unstable. What is the critical forcethat one has to exert on the helix to make it unstable and does this force depend on the length of



Figure 3: The time evolution of an unstable (naturally straight) helix. The mode n = 2 creates twoloops (κF = 1/8, τ F = 1/8, N = 5 and K/1000 =0, 2, 4, 6, 8, 10, 12).



-7e-05

-6e-05

-5e-05

-4e-05

-3e-05

-2e-05

-1e-05

0

0 1 2 3 4 5 6 7

n

σ2

Figure 4: The growth rate σ2 solution of the dispersion relation for the stable helix as a function of the spatial mode n (κF = 1/8, τ F = 1/8, N = 5).

the helix? Towards what shape will it evolve? We will show that the critical shape can easily becomputed and that longer helices are indeed easier to destabilize.

The second way a helix can be made unstable is to change the “internal” parameters such as theintrinsic curvature or torsion. While this situation may at first seem rather unphysical (after all,one does not expect the intrinsic characteristics of a rod to vary), there are contexts in which thisoccurs. For example, in many biological and chemical systems (such as DNA molecules), the externalenvironment (such as pressure and temperature) directly influences the intrinsic characteristics of afilament. By contrast, the external parameters, such as R and P , are much harder to control since itis extremely difficult to actually hold the extremities of a naturally occurring, microscopic, filamentand pull it!

It is clear that by varying internal parameters, the helix will eventually become unstable. Indeed,we have seen that setting the intrinsic curvature vector to zero makes the helix unstable. This suggeststhat there must be a critical intrinsic curvature where the helix first becomes unstable. What is thetype of this instability? Should we expect the same kind of behavior as exhibited by a helix which isdestabilized as a result of an external adjustment, e.g. pushing? Surprisingly, the instability generatedby adjusting internal parameters is, in many respects, different from that seen as a consequence of external changes. For instance, the former will not, in general, depend strongly on the length of the

helix. We will now discuss these instabilities in more detail.

6.3.1 Pushing and pulling helices

We consider a free standing helix that is gently pulled (or pushed) until it reaches an unstable point.We assume that the helix has N turns and that the act of pulling or pushing is such that the Frenetcurvature and torsion vary away from their equilibrium values. According to Eq. ( ??), a terminal forcef 0 has to be imposed in order to change the shape of the helix. The sign of f 0 determines whether thehelix is pushed (f 0 < 0) or pulled (f 0 > 0). The helix will become unstable if locally the dispersionrelation around n/N = 1 becomes positive. This can be tested easily by expanding the dispersionrelation around this point:

σ2 = −f 0δ4

κF τ F (

n

N − 1)2 + O(

n

N − 1)4 (36)



-6e-05

-5e-05

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-3e-05

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-1e-05

0

0 1 2 3 4 5 6 n

σ2

Figure 5: The growth rate σ2 solution of the dispersion relation for the pushed helix as a functionof the spatial mode n. The arrow indicates the first unstable mode (κu = 1/8, τ u = 1/8, R = 4,P = 37/10, N = 5).

From this relation, we conclude that the effect of pulling a helix is to stabilize it whereas pushingthe helix will create locally unstable modes around n = N . Let us consider the latter situation. Forsimplicity, we consider a helix with constant radius and pushed in such a way that only the pitchangle is modified. To do so, we fix R = Ru and modify P = P u + P 1. The parameters (Ru, P u)corresponds to the stable helix and are related to κF and τ F . The helix will become unstable as soonas ∆(σN −1 = 0) = 0. Indeed, the first possible mode to be excited corresponds to the mode n = N −1.A typical dispersion relation is shown on Fig. ??.

On Fig. ??, the critical value of the pitch parameter shift P 1 (where P = 4 + P 1) is shown againstthe number of helical turns N . We see that as the number of helical turns is increased the instabilityis triggered almost as soon as the helix is pushed. This situation is analogous to the case of a straightelastic rod that becomes unstable as its extremities are pushed. This creates the well-known effect of bending (?) of beams. The bending of a helix is shown on Fig. ??.

Pulling a helix stabilizes it. However, a helix under tension can become unstable by increasing itstwist. This situation is in many ways analogous to the case of a twisted straight rod under extension:in this case there exists a critical twist for which the twisted rod becomes unstable (?; ?). The valuesof n and γ for which a (helical) filament becomes unstable can be obtained by solving, for a given

force f 0, the coupled relations:

∆(σn = 0) = 0, and∂σn

∂n= 0 (37)

This leads to a system of two polynomial equations in n and γ that can be solved for a givenconfiguration (a general, explicit, form of the critical mode and twist seems elusive). As an example,the dispersion relation of a twisted, pulled, helix is shown on Fig. ??, where the mode n = 8 isunstable. The corresponding evolution of the helix is shown on Fig. ??.

6.3.2 Instabilities of helices with respect to intrinsic curvature

We now turn our attention to the case where the external parameters are fixed and the intrinsiccurvature is varied. This situation is in many respect different from the ones just considered. Indeed,

the helix has a fixed shape but the intrinsic curvature is varied in such a way that the elastic energy



-1.4

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-1

-0.8

-0.6

-0.4

-0.2

0

2 4 6 8 10 12 14 16 18 20 22N

Figure 6: The critical value of P 1 for which the helix becomes unstable as a function of the numberof helical turn (κu = 1/8, τ u = 1/8, R = 4, P = 4 + P 1).

is increased. In a dissipative medium and for small perturbations the helix would probably return toa new equilibrium shape determined by the elastic energy. However, there is no dissipation at thelevel of the Kirchhoff equations and the filament is free to evolve. In order to see how the filament isdeformed we have to find which mode is first excited. This again amounts to looking at the zeroesof the dispersion relation. To illustrate the kind of behavior that might be expected, we consider theparticular case of τ F = 1/8, and κF = 1/8. The intrinsic torsion is set to zero (τ uF = 0) and we varyκuF from 1/8 to 0 where the helix is known to be unstable. The different solutions of the dispersion

relation for varying values of κuF are shown on Fig. ?? and the corresponding time evolution of the

helix is shown on Fig. ??. We see that the helical filament deforms into a three loop configuration.

7 Conclusions

The helix is one of the most fundamental spatial configurations and also one of the simplest solutionsof the Kirchhoff equations. However, despite much attention in the classical elasticity literature, astability analysis of the different dynamical perturbations has not, to the best of our knowledge,been achieved. We believe that one of the primary reasons is that much, if not all, of the previousstudies have been performed in the setting of stationary perturbations. Using Euler angles, it might

be possible to find the value of the parameters for which new solutions may exist. However, evenin the case of the planar ring this is a strenuous task. The development of a perturbation schemetaking into account both time and space dependence allows us to derive the explicit (and completelygeneral) dispersion relation for the helical filament with intrinsic curvature and twist. In the setting of a dynamical study the critical values of the parameters leading to an instability can be readily foundas the zeroes of the dispersion relation. This provides a transparent interpretation of the instabilitiesthat a stationary analysis cannot provide without a complementary energy analysis.

Another reason might be purely computational. As seen here the different relations found in thispaper can be extremely cumbersome and it is only thanks to extensive symbolic computation that wewere able to analyze the different unstable modes. The advantage of such a powerful tool becomesmanifest in this kind of computation.

The linear analysis gives a picture of the way stationary solutions lose their stability. The newsolutions give a qualitative picture of the behavior after the bifurcation. However due to their



Figure 7: The time evolution of an unstable helix pushed at the extremities (κu = 1/8, τ u = 1/8,R = 4, P = 37/10, N = 5, K/10 =0, 15, 30, 45, 60).



-4e-05

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0

0 2 4 6 8 10

n

σ2

Figure 8: The growth rate σ2 solution of the dispersion relation for a twisted helix in extension as afunction of the spatial mode n (κu = 1/8, τ u = 51/40, R = 4, P = 8, N = 5).

exponential dependence, these solutions become rapidly unbounded and the linear approximationbreaks down. Indeed, as the amplitudes of the linear modes grow, the nonlinear terms can no longerbe neglected in the analysis. In order to circumvent this difficulty a nonlinear analysis needs tobe performed. Such an analysis introduces the distance from the bifurcation point as a new smallparameter which is then used to introduce new time and arclength scales in the problem on whichthe arbitrary amplitudes of the linear mode(s) can vary. Within these approximations one can, inthe best cases, derive amplitude equations describing the nonlinear interaction between the differentamplitudes of the linear modes. Usually, this approach gives a good description of the solution afterthe bifurcation. In Ref. (?), we show how to perform this analysis in the setting of the Kirchhoff equation for the twisted straight rod. The helical filament can probably be studied in the same way,especially in those cases where the graph of the exponent σ2

n is parabolic around the neutral state,(that is, σ2

n = −α(n−nc)2 for a certain α). However, the complexity of the higher order terms in theperturbation expansion presents a formidable computational challenge.

We have identified different regions of parameter space where a helical filament can becomeunstable. In particular, we explored three different scenarios leading to unstable filaments. In the firstcase, the helix is pushed. The resulting instability correspond to a bending of the entire helix, muchlike the bending of a straight rod. The unstable modes are located around the principal neutral mode

n = N , so that the instability occur at the scale of the helix itself (see Fig. ??). The second situationarises when the helix is pulled and twisted. In this case, a higher mode nc is excited and the helixdeforms throughout the rod by creating nc > N loops along the helix. These two cases are analogousto the possible deformation of a twisted straight rod and constitute, in some sense, a discrete versionof it. A completely different scenario is found by considering the case where the external parametersare held fixed but the intrinsic curvature is changed. Then, the helix undergoes unstable deformationsby exciting lower modes nc < N . This leads to the intriguing possibility of buckle formation as shownon Fig. ??. The dynamical formation of buckle in a twisted rod is a commonly observed phenomenonand many experimental works have been dedicated to the problem (?). However, to the best of ourknowledge, no dynamical description of this looping instability has been proposed. The nonlinearanalysis of the unstable rod might well provide such a description (?).

Acknowledgments This work is supported by DOE grant DE-FG03-93-ER25174. The authors would



Figure 9: The time evolution of an unstable pulled and twisted helix ( κu = 1/8, τ u = 51/40, R = 4,P = 8, N = 5, K/100 =0, 2, 4, 6, 8).



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0

5e-06

0 1 2 3 4 5

n

σ2

Figure 10: The dispersion relation for a twisted helix in extension. The lower dashed curve(κu

F = 0.062) corresponds to a stable helix. The upper curve (κuF = 0.055) corresponds to values

at which the mode n = 3 is unstable. The intermediate curve (κuF = 0.059233) is the point at which

σ2 is first tangent to the axis (τ u = 0, τ F = 1/8, κF = 1/8, N = 5).

like to thank Giacomo Ehrenfreud and Irwin Tobias for emphasizing the importance of applying thedynamical linear analysis to helical filaments.

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alain goriely and michael tabor- nonlinear dynamics of filaments iii: instabilities of helical rods

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