alain goriely and patrick shipman- dynamics of helical strips

8/3/2019 Alain Goriely and Patrick Shipman- Dynamics of Helical Strips

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Dynamics of Helical Strips

Alain Goriely & Patrick Shipman

Department of Mathematics and Program in Applied Mathematics,University of Arizona, Building #89, Tucson, AZ 85721

e-mail: [email protected]

To be Published in Phys. Rev. E (2000)

Abstract

The dynamics of inertial elastic helical thin rods with non-circular cross sections and arbitrary intrinsic

curvature, torsion, and twist is studied. The classical Kirchhoff equations are used together with a

perturbation scheme at the level of the director basis and the dispersion relation for helical strips is

derived and analyzed. It is shown that all naturally straight helical strips are unstable whereas free-standing helices are always stable. There exists a one-parameter family of stationary helical solutions

depending on the ratio curvature to torsion. A bifurcation analysis with respect to this parameter is

performed and bifurcation curves in the space of elastic parameters are identified. The different modes

of instabilities are analyzed.

KEYWORDS: Elastic Rods, Amplitude equations,

PACS: 46.30.Cn, 03.20.+i

1 Introduction

The analysis of helical structures plays an important role in the study of various types of chemical andbiological fibers such as DNA, bilipid layers and macrofibers [1, 2, 3, 4, 5, 6, 7]. Recently, there has beena great deal of attention paid to the stability of different configurations of elastic helical strips [8]. Mostanalyses are limited to comparisons of elastic energies (both in the context of linear and nonlinear elasticity)[9, 10]. This approach reveals that binormal helices (such as the strips shown in Fig. 1) are more likely tobe stable. Another approach consists in studying the dynamical stability of stationary solutions within theframework of the Kirchhoff equation for thin rods. This has been done for helical rods with circular crosssections in [11] and shown to be of prime importance in the analysis of buckling and coiling [12]. When thecross section of the rod looses its isotropy (that is when one considers elastic strips), the type of instabilityfound in twisted rods strongly depends on the anisotropy of the cross sections [13, 14] and two differentregimes can be found depending on the flatness of the cross section.

This paper considers the dynamics of helical inextensible elastic rods of various lengths, cross sectionalshapes, and material properties. If a rod obeys the laws of elasticity and has a cross sectional width muchsmaller than its length, it is referred to as a filament or, in the case of non-circular cross sections, as a strip.

Consider such an elastic strip and shape it as a helix such that it will hold by itself (with proper end forcesand moments). This helical strip can be stable or unstable under small perturbations. The problem is todetermine the stability properties of such configurations and provide some information on the wavelengthsassociated with the instability. The analysis performed here provides also the vibration modes associatedwith stable configurations.

2 The Kirchhoff equations

The evolution of filaments and strips is governed by the Kirchhoff model, where forces and moments areaveraged over local cross sections of the rod attached to its central axis [15]. The central axis is represented by

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d 1d

2

d3

R(s,t)

Figure 1: a typical strip R(s, t) together with its coordinate orthonormal triad (d1, d2, d3). The vectorsd1, d2 are chosen along the direction of greatest and lowest bending stiffness. The vector d3 is tangent to

the curve

a space curve R(s, t):R2 R3 parameterized by arc length s and time t. A right-handed orthonormal basis isestablished at each point of R(s, t) by the well-known Frenet frame (n, b, t) consisting of the tangent vectort, the normal vector n, which is the unit vector in the direction of t

s, and the binormal vector b = tn. The

turning rate with respect to s of the plane orthonormal to the curvethat is, the plane defined by (n, b)is

related to the Frenet curvature (s, t) =t(s,t)s

, and the Frenet torsion |(s, t)| = b(s,t)s is related to the

rate of turning with respect to s of the plane tangent to R(s, t)that is, the plane defined by (n, t). Thecurvature and torsion (s, t) and (s, t) determine the space curve through the Frenet-Serret equations. Forexample, if these values are both nonzero and constant with respect to s, then R(s, t) is a helix.

The Frenet frame and its associated definitions of curvature and torsion are derived from the space curveitself. For a physical filament, however, it is advantageous to have a local basis that corresponds to somecharacteristic of the filaments material properties. Here we are interested in helical filaments of constantasymmetrical (not circular) cross sections, referred to as strips, and, using a generalized local basis for thespace curve, we set one of the vectors normal to the curve to correspond to the direction of the filamentsgreatest bending stiffness. Also, we allow for intrinsic curvature and twisti.e., the curvature and twist ofthe filament in its natural, unstressed state. Stationary helical solutions for the Kirchhoff equations weredescribed in [14]. The aim of this paper is to study the stability of helical strips using a arc-length-preservingperturbation scheme first developed in [16]. This follows work in [11] on the stability of helical rods of circularcross sections and in [14] on straight strips.

Together with the space curve R(s, t) we choose a smooth unit vector field d2(s, t) orthogonal to thetangent vector of the curve and along the direction of the cross section lowest bending stiffness. The vectord2(s, t) is part of a local coordinate triad (d1, d2, d3) defined at each (s, t). If we consider a very flat strip(think of a belt), the vector d3 is the tangent vector along the center axis and the vector d1 is chosen in

the direction joining the center axis to the edge of the strip (in the flattest direction) as can be seenon Figure ??. Denoting differentiation with respect to s and t by ( ) and ( ) respectively, and settingd3 = R

(s, t), we choose d1 to be the unit vector field such that d1 = d2 d3, and then (d1, d2, d3) is, forall (s, t), a local right-handed orthonormal basis. It is related to the Frenet frame (n, b, t) by the angle between the two bases and the relation

d1 d2 d3

=

n b t cos sin 0sin cos 0

0 0 1

. (1)

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The local triad (d1, d2, d3) evolves in space and time asd1 d

2 d

3

=

d1 d2 d3

K, (2)d1 d2 d3

=

d1 d2 d3

W, (3)

where K and W are the antisymmetric matrices

K = 0

3 2

3 0 12 1 0

, W = 0

3 2

3 0 12 1 0

. (4)

The vectors =3

i=1 idi, =3

i=1 idi, formed from components of K and W, are respectively definedas the twist and spin vectors.

The twist vector is related to the angle and the Frenet curvature and torsion as

(1, 2, 3) =

sin , cos , +

s

. (5)

One may think of , , and s

as the variations on the filaments twist: measures the nonplanarity of

R(s, t); and s

are both properties of the ribbon defined by the curve R(s, t) along with the orthogonal

field d2(s, t). An intrinsic twist vector (u) = ((u)1 ,

(u)2 ,

(u)3 ) is designed to correspond to (1, 2, 3) so

that we have an intrinsic curvature (u) =

((u)1 )

2 + ((u)2 )

2, and an intrinsic twist (u)3 . As in (5),

(u)1 = (u) sin , and

(u)2 =

(u) cos . (6)

Let I1 and I2 be the principal moments of inertia of the cross section in the directions of d1 and d2respectively, with I1 I2. That is, d1 and d2 are chosen to be the directions of greatest and lowest bendingstiffnesses, respectively. Then, a = I2

I1is a value 0 < a 1 that measures the bending asymmetry of the

filaments cross sections. The value 1 is reached in the dynamically symmetric case where the moments ofinertia are identical; the scaled radius of circular cross sections is then equal to 2. A constant b, called thescaled torsional stiffness, roughly measures the change in volume in the rod as it is stretched. The lowerthe value of b, the less the volume changes. Within the framework of linear elasticity theory [17, 18, 19]it is possible to compute a and b for a given cross section shape. For instance, elliptic cross sections with

semi-axes A and B (A < B) have:

a =A2

B2, b =

1

1 +

2a

1 + a. (7)

where is the Poisson ratio. The scaled semi-axes are, respectively, 2

a and 2. Other values of a and b forvarious shapes are represented on Fig. 2. With a force vector F = F1d1 + F2d2 + F3d3 and moment vectorM, the scaled dynamical Kirchhoff equations stated in the local basis are [14]

F = d3, (8.a)

M + d3 F = a d1 d1 + d2 d2, (8.b)M = (1 (u)1 )d1 + a(2 (u)2 )d2 + b(3 (u)3 )d3. (8.c)

The static (without time dependence) form of these equations is

F = 0, (9.a)

M + d3 F = 0, (9.b)M = (1 (u)1 )d1 + a(2 (u)2 )d2 + b(3 (u)3 )d3. (9.c)

Helical and circular solutions to these equations are those with constant Frenet curvature and torsion (weconsider here true helices for which both curvature and torsion are different from zero). The only suchsolutions are those with the angle between the director basis and the Frenet frame an integer multiple of 2[14]. That is, = n 2 , where n is an integer. These solutions are such that

s

= 0 and are called Frenet

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Figure 3: A normal and binormal helix.

where 1 is the identity matrix, and the symmetric matrices S(k) depend solely on the general antisymmetricmatrices A(1) to A(k1), where the A(j)s are

A(j) =

0 (j)3 (j)2(j)3 0 (j)1

(j)

2

(j)

1

0

. (15)

Given the vector = (1, 2, 3), where i = (1)i +

2(2)i + , one can reconstruct the perturbed rod

by integrating the tangent vector as

R(s, t) =

d3(s, t) ds =

d(0)3 + (2d

(0)1 1d(0)2 )

ds + O(2). (16)

Expressions for the elements of the twist and spin vectors in terms of the perturbed variables areobtained using (2) and (13.a) which combine to give

s

d(0)1 d

(0)2 d

(0)3

B

=

d(0)1 d

(0)2 d

(0)3

B K. (17)

This, in turn, is equivalent to

d(0)

1 d(0)

2 d(0)

3

K(0)

B B K +B

s

= 0. (18)

Since the matrix B is orthogonal and the basis vectors are independent, we have

K = B

K(0) B +B

s

. (19)

The spin vector components are expressed analogously. For a given unperturbed state, the Kirchhoffequations may now be written in terms of the six variables of the vector

X(k) =

F(k)1 F

(k)2 F

(k)3

(k)1

(k)2

(k)3

T. (20)

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4 Construction of the perturbed helical strips

It was previously noted that, given the vector = (1, 2, 3), we can construct the perturbed strip bythe integration (16). To actually perform this integration we need the vector and the generalized frame

d(0) = (d(0)1 , d

(0)2 , d

(0)3 ) of the unperturbed helix.

The vector d(0) is found by the relation (1) and the Frenet frame of a helical curve, which is as follows:

nbt

= (0,

cos(s),

sin(s))

(R,P sin(s), P cos(s))(P ,R sin(s), R cos(s))

, (21)where R is the radius of the cylinder about which the helix may be wrapped, P is the period of the helix,

and =

1P2+R2 . The positive root in this last expression defines a right-handed helix, and the negative

root defines a left-handed helix. For a binormal helix, with = n 2 for even n, (1) gives

d(0) =

nb

t

. (22)

Hence, (16) becomes

R(s, t) = P R sin(s)

R cos(s)

+ 2 0

cos(s) sin(s)

1 R

P sin(s)P cos(s)

ds + O(2). (23)

For the normal helix = n 2 , n odd, so (1) gives

d(0) =

bn

t

, (24)

and (16) becomes

R(s, t) =

P R sin(s)R cos(s)

+

2

RP sin(s)P cos(s)

1

0cos(s)sin(s)

ds + O(2). (25)

The vector is found by finding the vector solutions (20) corresponding to unstable modes, as describedbelow.

5 Stability analysis

The linearization of the Kirchhoff equations around the stationary helical solutions allows for a stabilityanalysis. The unperturbed variables F(0) and (0) are taken to be the static helical solutions, (11) for astudy of the binormal helix, and (12) for a study of the normal helix. The 0-order part of the spin vectorexpansion, (0), is set to 0, and the expansions for tension, twist, and spin vectors, truncated to first order,are substituted into the system. The result, in matrix form, is

LX(1) = 0, (26)

where L for the binormal helix is the following second order linear operator:

L =

LA LBLC LD

(27)

where

LA =

2

s222 2

s2

s

2s

2

s22

2 s

2

s22

(28)

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LB =

2(ba)(2+3) s

(ba)2

2

s222

2

t2(ba)(2+2 2s2)

(ba)2

2+2 2s2

+

2

t22(ba)3

s2(ba)2

s

(ba)

2

s222

2(ba)2

s2(ba)2

s

(29)

LC =

0 1 01 0 00 0 0

(30)

LD =

(ba)(22)au2b u3 + 2

s2 2

t2(ba1)

sbu3 s (ba+1) s +au2 s

(1+ab)s

+bu3

s(b1)2bu3 +a

2

s2 2

t2

(1b)+u3

(ab1) sau2 s (1a)+au2 b

2

s2(a 1)2au2(1+ a)

2

t2

(31)

We now look for the fundamental solutions of (26), expressed as

X(1) = Aue(t+ins), (32)

where A is a constant complex amplitude, u is a constant vector with a given norm, is a complex constant,and n is a real constant. The substitution of (32) into (26) yields

Mu = 0. (33)

The matrix M for the binormal helix is

M =

MA MBMC MD

(34)

where

MA =

(1+ n2)22 2in2 2in

2i2n (1+n2)2 2in

2

2

n2

(35)

MB =

2i n(F2 + F3) F2(

2n2+2)F322 (2n2 + 2)F2 + F32 F2 +

(1 + n2)2 2F3 + 22 2in2F3 2in2F2

2F3

(n2 1)2 + 2F2 2inF3 2inF2

(36)

MB =

0 1 01 0 0

0 0 0

(37)

MD =

(ba)(22)2(n2+2)a(u)2 b (u)3 in

2(ba1)ibn(u)3 in(ba+1)+ian(u)2in

2

(1+ ab)+ibn(u)

3 2

(b1 an2

a2

) b(u)

3 (1b)+b(u)

3

in(ab1)ian(u)2 (1a)+a(u)2 (a1)2bn22(1+a)22a(u)2(38)

and the expressions F2 = (b a)( (u)2 ) and F3 = (b a)( (u)3 ) are the components of the forcevector F. The matrix M for the normal helix has a similar form (with different coefficients) and will not begiven explicitly here. There are nontrivial solutions for u only if det(M) = 0. This leads to the dispersionrelation

det(M) = (n, 2, a, b, , , i, (u)i ,

(u)3 ) = 0, (39)

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where i is 1 for the normal helix (in this case 1 = /2 and (u)1 =

(u), (u)2 = 0) and i = 2 for the binormal

helix (2 = 0 and (u)1 = 0,

(u)2 =

(u)) due to relations (6). If the dispersion relation allows for modeswhich grow exponentially in timethat is, modes with () > 0then the helical strip is unstable.

The dispersion relations corresponding to the binormal and then normal helices are examined for theexistence of these unstable modes. We first study the stability of the naturally straight helicesthose with(u) set to 0and then the stability of the free-standing helicesthose with (u) set to . The basic procedure

is the following: We solve (39) for 2 as a function of n and the parameters a, b, , and , and thus obtain

three solutions 2

1(n), 2

2(n), and 2

3(n). We are interested in the real parts of these solutions. For thebinormal helix, whether = 0 or = (u), we find that the solution 2 is purely imaginary regardless ofthe values of the intrinsic curvature and twist, and (23(n)) 0 for all n, so these solutions contribute nounstable modes. The situation for the normal helix is similar; the solution (23) is nonpositive for all nand although the solution 22(n) may have a real part (in the case

(u) = 0), this part is always nonpositive.Hence, we only consider the solution (n) = (21(n))

n21-1-2

0.5

0.4

0.3

0.2

0.1

-0.1

-0.2

b= 2.3

b=1/10

b=1/2

b=5/6b=1

Figure 4: Naturally Straight Binormal Helix: (n) with = 1 = , a = 12 , and b =110 ,

12 ,

56 ,1, and 2.3

6 The naturally straight helices

We first consider naturally straight helices, that is helices with no intrinsic curvature or torsion ((u) = 0).The neutral modesthe values of n such that () = 0are found by substituting 2 = 0 into the dispersionrelation (39) and solving for n. For both the normal and binormal helices we find neutral modes of the form

n1 = 0, (40.a)

n2 =

2 + 2

(40.b)

n3 =

2

2

+

ba (40.c)

n4 =

2

2

ba

, (40.d)

where and are rather complicated functions of b a, , and . The neutral modes n1 and n2 areindependent of b a; hence we organize a discussion of unstable modes by considering values of b a sincethe stability changes occur as the roots n3 and n4 change position relative to the root n2.

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6.1 The naturally straight binormal helix

Taking = 2 = 0 and (u)2 = 0 in the dispersion relation we can find the solution (n) associated with the

naturally straight binormal helix.. Various examples of this solution, with constant , , and a, and varyingb, are shown in Figure 4. The graphs indicate positive concavity at n = 0 for most values of b a. To showthat this is indeed the case, we locally expand the dispersion relation around n = 0 and find

= 2( +)2 + 2

n21 + 3(3) +

, (41)

where , , and are all functions of the parameters , , a, and b. For small , behaves approximately

like ( +

) 2+2

n21. Since is always positive, the sign of the lowest order term in the expansion of (n)

about (0, 0) depends on +

. For b a and 13 42+2

2a < b, 2 > 0, so + > 0. Thus, (n)

has positive concavity at (0, 0) for these values of the parameters. For the range a < b < 1342+2

2a, has

negative concavity at (0, 0). However, for these values, both modes n3 and n4 appear as roots of (n), andin this range (n) has the form shown in Figure 5. Hence, we conclude that a naturally straight binormal

0.01

0.01

0.02

0.03

0.04

0.05

1 1

n

Figure 5: Naturally Straight Binormal Helix: (n) with = 1 = , a = 12 , and b =23

helix is always unstable. Depending on the values of a and b, there are two different types of unstable

modes. In the range b a and 13 42+2

2a < b the instability is a long wave instability (all modes close to

n = 0 are unstable), and the deformation takes place on long wavelengths. For a < b < 1342+2

2a, there

is an unstable band of modes centered around a non-zero wavenumber nc. This Hopf-like instability occursat finite wavelength and results on the superposition of two helical modes: the stationary solution and theunstable solution with wavenumber nc.

To construct the solution, we find the fundamental solutions corresponding to the fastest-growing mode

ncthat is, the mode that corresponds to a peak in the solutions (n). For example, for the binormal helixwith properties as listed in Figure (5) ( = 1 = , a = 12 , b =

23), the fastest-growing modes are those at

n 0.7704. The corresponding nullvector of (34) is

u

0.392i0.4270.2941.7690.4021.443

. (42)

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Similarly, nc = 0.7704 yields the complex conjugate of this solution vector u. The corresponding solutionis then

Aue((0.7704)t+0.7704is) + Aue((0.7704)t0.7704is), (43)

and the last three components of this vector form .Examples of perturbed binormal naturally straight helical strips follow in Figure 6 . In all cases, we use

A = 1, R = 2 = P, and = 12 . Two periods are shown. For different values of b, two unstable wavelengths

Figure 6: Binormal Helix, a = 12 , b =23

, = 1 = , (0)2 = 0 =

(0)3 , perturbed modes: nc = 0.7704,

= 0.1, 0.2, 0.5

can be excited. For instance, for a = 1/2 and b = 1/10, the dispersion relation (Figure 4) shows two maximaaround nc = 0.6685 and nc = 1.797. The reconstructed unstable binormal helix is shown in Figure 7.

6.2 The naturally straight normal helix

Taking = 1 = /2 and (u)1 = 0 in the dispersion relation we can find the solution (n) associated with the

naturally straight normal helix.. Typical graphs of this solution are shown in Figure 8 for fixed , , and a,and varying b. As indicated there, there is a band of unstable modes around n = 0, and this band grows asthe value b a increases. Setting n = 0 in det(M) and solving for 2, we get the three roots 0, 0, and

1 a2 22 + 4 + (1 a) 4 + 22 + 2 + 2 + 22 + 2 / 2 2 + 2 + 1 (a + 1). The firsttwo roots appear in the solutions 22 and

23 , and the last appears in the solution (n) =

23. Since a 1

the last root is always positive. Hence there is always a band of unstable modes about n = 0 in the solution(n). We conclude that, as for the naturally straight binormal helix, the naturally straight normal helix isalways unstable.

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Figure 7: Binormal Helix, a = 12 , b =110

, = 1 = , (0)2 = 0 =

(0)3 ,perturbed modes: nc = 0.6685,1.797,

= 0.1, 0.2, 0.3

7 The free-standing helices

For the case (u) = we have only two neutral modes, which are

n1 = 0, (44.a)

n2 =

2 + 2

. (44.b)

These neutral modes all appear as roots of the solution (n). A local perturbation analysis around the points

(n, s) = (0, 0), (

2+2

, 0) as performed for the naturally straight binormal helix shows that the curve (n)

always has negative concavity around these points. Hence, there are no unstable modes, and we concludethat the free-standing helices are always (linearly) stable.

8 Helices that are neither naturally straight nor free-standing

Let r =

be the aspect ratio of a helix; r > 1 are long helices whereas r < 1 are fat helices. Stationary

helical solutions are such that (u) = (u). Let R be the ratio:

R =(u)

=

(u)

. (45)

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0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

4 3 2 1 1 32 4

n

a=3/4

a=1/2

a=1/20

a=1a=7/8

Figure 8: The solution (n) for normal helices with = 1 = , b = 34 , and a =120 ,

12 ,

34 ,

78 , and 1

For a given intrinsic torsion and curvature the one-parameter family of helices defined by R corresponds tosimilar helices of different size (helices with constant ratio torsion to curvature). The natural size is definedas the size of the helix with R = 1. The radius of such a helix is (u) = (u)/(((u))2 + ((u))2). For R = 1the radius of the helix is = R(u). Hence R describes the relative size of similar helices. For R > 1 thehelices are smaller than the natural size and bigger for R < 1. In the preceding sections we studied the caseR = 0 (naturally straight helices) and R = 1 (free-standing helices). We now consider values of R differentthan 0 and 1. If we start at R = 1, the helix is stable. As we vary this parameter, the helix might becomeunstable. The question is to locate the bifurcation point Rc for given a and b and to describe the instability.

Again, we need only to consider the real part of one solution to the dispersion relation since the other

two solutions always have nonpositive real parts, and we call this real part of a solution (n) . For example,a graph of(n) corresponding to the normal helix with fixed values of , , a, and b, and varying R is shownin Figure 9. Note that, for example, ifR = 12 , b = 3/2 and a < 3/4, then there are no unstable modes for thenormal helix; a helix must not be free-standing to be stable. An analysis of the neutral modes n3 and n4given in (44) allows for the finding of the values of the parameters that characterize stable forms. For r = 1in the normal case we have the situation indicated by Figure 10, where stable regions are those in whichn3 and n4 are not neutral modes, and the solution (n) is similar to the R = 1 case of Figure 9. Thestability of a stationary helix (normal or binormal) depends on r,R,a and b. For both helices, the picture issimilar to the one in Figure 9. We Set c = 1 for the normal helix and c = a for the binormal helix. We findthat for b < c, the helix is stable only for 1 R R, and for b > c the helix is stable only for R R 1,where the values of R are given by the positions of the different roots. We have considered that all modescan be excited (short-, middle-, and long-wavelength instabilities).

Most helical strips are such that b < 1 (See Fig. 2); hence, all normal helices behave the same way: They

are stable as long as 1 R < R, when the size is such that R > R a short wavelength instability develops(the unstable modes nc > 1), and smaller helices (R < 1) are all unstable with a long wavelength instability.

The situation for binormal helices is slightly different. For b < a, the helix is stable only for 1 R R,and for b > a the helix is stable only for R R 1, So, depending on the cross section anisotropy, binormalhelices can become unstable. Indeed, for a given b and flat cross sections (such that a < b) small helices canbe stable (for R < R < 1), whereas helices with thick cross sections (a < b) behave like normal helices.

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0.3

0.2

0.1

0

0.1

0.2

0.3

0.5 1 2 2.5 3

n

R= 0

R=1/8

R=1/4

R=1/2

R=1.4

R=1

R=1.2

Figure 9: Normal Helix: (n) with = 1 = , a =3

4 , and b =3

2 and varying R.

9 Instability of circular strips

In order to show how boundary conditions can stabilize instabilities, we study the case of elastic rings withanisotropic cross sections. This case is of particular interest in the theory of DNA rings [22]. In the case of

elastic rings, = 0, hence we do not scale the mode n with and we let R = (u)

. The filament is periodic

(of period 2

) and with periodic boundary conditions on the force and curvature vectors as well as theposition vector. This restricts considerably the type of instabilities since mode with long wavelengths (withn/ < 1) cannot become unstable and only discrete modes can be unstable (with n/ > 1 integer). Due tothis fact, small binormal rings (with R

1) are always stable and large rings of increasing size become first

unstable when n/ = 2. The mode n/ = m becomes unstable for R > Rm, where

Rm =2a (1 + b) +

(b 1)2 + 4bm2

2a> 1. (46)

Surprisingly, normal rings can become unstable when either R < 1 or R > 1. Indeed, the m-th mode becomes

unstable whenever 0 R < R(1)m or R > R(2)m > 1 with

R(1)m = 1 1

2(a + b) 1

2

(b a)2 + 4m2ab (47)

R(2)m = 1 1

2(a + b) +

1

2

(b a)2 + 4m2ab (48)

Note that R

(1)

m > 0 only ifa 1) binormal circular strips may be

unstable, small normal circular strips with flat cross sections may also have instabilities.

Acknowledgments: One of the author (A. G.) would like to thank Michael Tabor for many fruitfuldiscussions. A. G. is a Sloan fellow. This work is supported by the NSF grant DMS-9972063 and NATO-CRG 97/037. This work is the result of an undergraduate independent study (by P. S.) in the Departmentof Mathematics.

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alain goriely and patrick shipman- dynamics of helical strips

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