# A Noetherian approach to invariants for the statics and dynamics of elastic rods

Sébastien Neukirch, Florence Bertails-Descoubes

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Sébastien Neukirch, Florence Bertails-Descoubes. A Noetherian approach to invariants for the statics and dynamics of elastic rods. International Journal of Non-Linear Mechanics, 2026, 181, pp.105269. 10.1016/j.ijnonlinmec.2025.105269. hal-05149317v3

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# A Noetherian approach to invariants for the statics and dynamics of elastic rods

Sébastien Neukirch<sup>a, ID,*</sup>, Florence Bertails-Descoubes<sup>b, ID</sup>

## ARTICLE INFO

**Keywords:**

Elastic rods

Hamiltonian and Lagrangian formulations

Contact

## ABSTRACT

The static-dynamic analogy discovered by Gustav Kirchhoff in 1859 shows that the statics of an elastic beam is equivalent to the dynamics of a spinning top. This analogy, where time and angular velocity are, for example, equivalent to arclength and curvatures, allows the use of Emmy Noether's 1918 theorem on continuous symmetries to unravel a quantity that is invariant along elastic rods at equilibrium. A spinning top having a Lagrangian independent of time will have its mechanical energy constant in time. In the same manner, an elastic rod with uniform elastic properties will have the sum of its curvature energy and its tension force uniform along the structure. This arclength invariance property is known in simple cases, but the present approach generalises it to more complex scenarios where extensibility, shear, conservative loads (e.g. gravity), and contact are involved. Moreover, still using Noether's theorem and bringing to light the continuous symmetries of the Lagrangian of the variational approach, we recover all known invariants for the statics and dynamics of rods and ribbons, including coordinate invariants. Furthermore, we show how the arclength invariant may be used to obtain pivotal information on some landmark elastic rod problems, including confinement buckling, sliding sleeves, or plectonemes. Finally, we extend the approach to vibrations. Overall, this paper is an attempt to explain, unify and extend all previous results on rod invariants thanks to the beautiful Noetherian formalism, and to show its practical use on a few relevant applications.

## 1. Introduction

The discovery of a conserved quantity while studying a dynamical system is generally a joyful event for the researcher as it usually brings new insights into the system's behaviour. In her PhD thesis, the German mathematician Emmy Noether meticulously derived more than 300 invariants for biquadratic systems [1]. These invariants were explicitly calculated by hand, possibly through trial and error, and later on, E. Noether passed a harsh judgment on her PhD work and oriented her studies towards abstract algebra [2, p. 17]. Nevertheless, her Habilitation thesis contains the classic 1918 theorem on differential invariants which provides a systematic way of unravelling invariants in physics and mechanics problems [3] and was most probably initiated in relation to the variational approach of Einstein's general theory of relativity [2, p. 36].

Noether's theorem states that if the Action (of Lagrangian mechanics) stays the same when a translation, rotation, or some other space or time transformation is performed, then a given quantity will be invariant during the dynamics of the system. The best-known case is the invariance of the mechanical energy of a system if the Lagrangian of this system does not explicitly depend on time, and thus remains the same through a translation in time. In this paper, we leverage this theorem for the statics and dynamics of elastic rods to recover all their known invariants, thereby providing the corresponding space transformations responsible for these invariants. We additionally generalise the invariants to the case where a conservative force field is applied to the rod, as well as frictionless contact. Finally, we provide several illustrative examples where the arclength invariant is shown to be particularly handy in the study of the mechanics of the system, and extend the approach to vibrations.

### 1.1. State of the art

**Variational approaches of rods.** In mechanical engineering, variational approaches of rods have long been motivated by the design of Galerkin-type numerical methods, especially finite elements. A popular method is the so-called geometrically exact beam approach [4,5], which derives an exact

<footnote><sup>a</sup> Sorbonne Université, CNRS, Institut Jean Le Rond d'Alembert, F-75005 Paris, France</footnote>

<footnote><sup>b</sup> Univ. Grenoble-Alpes, INRIA, CNRS, Grenoble INP, LJK, 38000 Grenoble, France</footnote>

<footnote>* Corresponding author.</footnote>

<footnote>E-mail addresses: sebastien.neukirch@cnrs.fr (S. Neukirch), florence.bertails@inria.fr (F. Bertails-Descoubes).</footnote>
weak formulation for a generalised (Timoshenko or Simo-Reissner, simply referred to as Reissner in the sequel) rod model including bending, stretching and shearing energies. The displacement and rotation fields are then discretised with the help of interpolating shape functions. One important issue of this approach (that is well beyond the scope of the present paper), which spurred many subsequent works in the finite elements community, deals with the proper interpolation of rotations for preserving objectivity, i.e., invariance of the strain measures under rigid motion [6].

The weak formulations exploited for finite elements generally rely on the *principle of virtual work*, which stipulates that the variation of internal energy of the system should be balanced by the work of external forces. One difficulty pertaining to the virtual work principle is the definition of virtual displacements that comply with the true kinematics of the system. For rods in particular, a concern deals with the parametrisation of rotations and the nature of coupling between positional (displacement of the centreline) and rotational (rotations of the material frame) degrees of freedom. Various parametrisations have been proposed for rotations, including Euler angles [7, Section 5.3.1], quaternions [8], or the exponential map in the rotation group $SO(3)$ [5]. Likewise, various classes of weak formulations have been introduced with different levels of coupling between virtual displacements in positions and rotations. Methods range from uncoupled [4] and weakly coupled ones in $R^n \times SO(3)$ [9] for general (Reissner) beams, to fully coupled ones for inextensible Kirchhoff rods where only rotational (bending and twisting) degrees of freedom are considered [10, Section 3.6]. All these methods lead to the correct mechanical equations for the considered rod models, albeit at the price of properly defining the structure of the space where virtual displacements should live in.

An alternative variational method is the *principle of least Action*, which stipulates that the *Action* of the system — that is, in statics, the integral of the *Lagrangian* of the system over $s$ — should be stationary. This principle leads to the so-called *Euler–Lagrange equations*, which are differential equations that should be satisfied by the degrees of freedom $q(s)$ of the system along with their derivatives $q'(s)$ [11,12]. Though mathematically equivalent to the virtual work principle for conservative systems, this approach is quite different in spirit. Instead of formulating virtual displacements and computing corresponding energy variations, it is sufficient to build a Lagrangian — an energy depending on $q(s)$ and $q'(s)$— that complies with the kinematics and mechanics of the system. In order to couple some degrees of freedom together, or to model contact, constraints can easily be incorporated into the Lagrangian. Then, the Euler-Lagrangian equations appear as a simple recipe to derive the mechanical equations of the system from the Lagrangian. Another key motivation to use the Lagrangian formalism is that Noether's theorem on invariants is best understood and derived from symmetries of the Lagrangian itself.

For these reasons, we shall adopt the Lagrangian point of view in this paper. To demonstrate the versatility of this approach, we will show that we can retrieve the same invariants for rods by considering two choices of degrees of freedom: on the one hand, simple (and totally uncoupled) degrees of freedom living all on $R^n$, which are related to each other through external constraints; on the other hand, translational and rotational degrees of freedom living on the Lie group of rotations $SO(3)$, which are coupled together through a single external constraint. We will also show that adding contact constraints is straightforward within this Lagrangian approach, and allows one to extend Noether invariants to systems in (frictionless) contact.

*Hamilton point of view and invariants for rods.* To study mathematical properties of rods and their invariants, theorists have often adopted an *Hamiltonian* point of view, that is, considered the Legendre transformation of the Lagrangian. Indeed, for conservative systems, the Hamiltonian naturally appears as an invariant, which actually boils down to the Noether invariant for a time-translational symmetry of the corresponding Lagrangian. For rods, different invariants have been found by analysing directly the Hamiltonian structure of the mechanical (strong) equations of the system [13,14].

Among these invariants, the static arclength invariant $H$, consisting in the sum of the bending energy, twist energy, and the tension (axial component of the internal force), stands apart. It was first mentioned in the book by Love [15, p. 384, Eq. (7) in Sect. 262], then discussed in [13], and later on thoroughly used by Dichmann et al. [8] and Kehrabaum and Maddocks [16] where it appears as the Hamiltonian in the 3D rods statics formulation. The jump conditions of these conservation laws have been studied in [17] in the light of the Weierstrass–Erdmann corner condition, see also [18].

It is noteworthy that these invariants were found using a trial-and-error approach, i.e. no systematic perspective was followed. As a matter of fact, such a systematic way exists: it relies on the Lagrangian and on Noether 1918's theorem [3], which states that each continuous symmetry of the Action, constructed from both the energy and the kinematic constraints of the system, calls for an invariant quantity. The Noether approach to the conserved quantities has been used in the general case of bulk linear elasticity [19] and different fluid and solid mechanics problems [20–22]. For rods however, a complete Noetherian view has seldom been adopted. Maddocks and Dichmann [13] briefly mentioned the possible use of Noether's theorem to constructively find conservation laws, but rather used an a posteriori verification approach to the computation of the invariants, starting from the strong version of Kirchhoff's rod equations. Peng et al. [23] used Noether's theorem but could not explicitly provide all known invariants because they did not introduce the continuous constraints in the Lagrangian. Jung et al. [11] developed a Lagrangian approach of elastic rods, where the kinematic constraints are not explicitly listed, and recovered continuous and discrete versions of some of the static coordinate invariants, using Noether's formalism. Their approach did not include the arclength invariant, which is the main focus of our paper. Overall, the fact that several invariants (and particularly the arclength invariant $H$) can be generalised to the case where a conservative force field is applied on the rod seems to have been overlooked in the literature, with the exception of self-weight, see [7,18]. Finally, contact was merely treated in the light of jump conditions [17,18] where it was shown to help the computation of integration constants [24].

The approach developed in this paper rationalises the construction of the statics, dynamics, and vibration invariants using a Lagrangian mechanics approach, and shows that all invariants can be computed systematically once all continuous constraints have been properly introduced in the Lagrangian, together with their corresponding multipliers. The case of pointwise and continuous (frictionless) contact is moreover shown to fit in naturally with the present Noetherian approach.

## 1.2. Paper contributions

* A simple and pedagogical derivation of the Reissner and Kirchhoff static and dynamic equations from the least Action principle, with pointwise constraints, contact, and external conservative force fields.

* The application of Noether's formulas to retrieve, unify, and generalise all known statics and dynamics invariants in the presence of frictionless contact and external conservative force fields, valid for Kirchhoff, Reissner, and elastic ribbon cases.

* The derivations of the results above using two alternatives: a simple Lagrangian formulation on $R^n$ with multiple external constraints, and a more compact Lagrangian formulation on $R^n \times SO(3)$ retaining only one single kinematic constraint.

* The practical application of the arclength invariant $H$ to several landmark rod problems, including sliding sleeves, plectonemes, and constrained Euler buckling.

* The application of the present approach to the derivation of an invariant for the vibrations of elastic rods.
## 2. Noether's theorem and extension to $\mathbb{R}^n \times \mathcal{SO}(3)$

In the following, scalar quantities are printed with plain font, and vector quantities of $\mathbb{R}^n, n > 1$ with bold font. Matrices are represented with capital letters.

### 2.1. Emmy Noether's 1918 theorem

Let the generalised coordinates of the system $q = q(s, t)$ be functions from $\mathbb{R}^2$ to $\mathbb{R}^n$, and $\dot{\prime} \equiv \partial/\partial s$, and $\cdot \equiv \partial/\partial t$ be the space and time derivatives, respectively. For the sake of readability, in the following we may often drop the $s, t$ dependencies, for instance writing $q'$ instead of $q'(s, t)$.

If the following Action

$$ \mathcal{A} = \int_{t} \int_{s} \mathcal{L}(q, \dot{q}, q') ds dt \qquad (1) $$

remains identical when the following transformations are performed,

$$ \tilde{q} = q + \epsilon \theta, \qquad (2a) $$

$$ \tilde{s} = s + \epsilon \sigma, \qquad (2b) $$

$$ \tilde{t} = t + \epsilon \tau \qquad (2c) $$

with $|\epsilon| \ll 1$, and in particular if $\tilde{\mathcal{L}} = \mathcal{L}$, then the following relation

$$ \left[ \tau \mathcal{L} + (\theta - \tau \dot{q} - \sigma q') \cdot \frac{\partial \mathcal{L}}{\partial \dot{q}} \right]^* + \left[ \sigma \mathcal{L} + (\theta - \tau \dot{q} - \sigma q') \cdot \frac{\partial \mathcal{L}}{\partial q'} \right]' = 0 \quad \forall s, \forall t. \qquad (3) $$

holds for dynamical solutions, i.e. for trajectories $q(s, t)$ under which the Action (1) is stationary [3]. The three transformations do not have to occur simultaneously, i.e. we have [3]

$$ \text{if } \tau = 0 \text{ and } \sigma = 0, \text{ then } \tilde{q} = q + \epsilon \theta \text{ and } \left[ \theta \cdot \frac{\partial \mathcal{L}}{\partial \dot{q}} \right]^* + \left[ \theta \cdot \frac{\partial \mathcal{L}}{\partial q'} \right]' = 0 \quad \forall s, \forall t \qquad (4a) $$

$$ \text{if } \tau = 0 \text{ and } \theta = 0, \text{ then } \tilde{s} = s + \epsilon \sigma \text{ and } \left[ q' \cdot \frac{\partial \mathcal{L}}{\partial \dot{q}} \right]^* + \left[ -\mathcal{L} + q' \cdot \frac{\partial \mathcal{L}}{\partial q'} \right]' = 0 \quad \forall s, \forall t \qquad (4b) $$

$$ \text{if } \sigma = 0 \text{ and } \theta = 0, \text{ then } \tilde{t} = t + \epsilon \tau \text{ and } \left[ -\mathcal{L} + \dot{q} \cdot \frac{\partial \mathcal{L}}{\partial \dot{q}} \right]^* + \left[ \dot{q} \cdot \frac{\partial \mathcal{L}}{\partial q'} \right]' = 0 \quad \forall s, \forall t. \qquad (4c) $$

Please note that $q$ and its derivatives $q' \cdot \dot{q}$ are vectors of dimension $n$, and the gradients $\partial \mathcal{L}/\partial q, \partial \mathcal{L}/\partial q', \partial \mathcal{L}/\partial \dot{q}$ are vectors of the same dimension $n$: scalar products arise between these two types of quantities.

**Static case.** In the particular static case where there is no time dependence, (4b) implies the most well-known part of Noether's theorem: if the Lagrangian $\mathcal{L}$ does not explicitly depends on $s$ (but only does through $q(s)$ and $q'(s)$), then

$$ H(s) = H = q' \cdot \frac{\partial \mathcal{L}}{\partial q'} - \mathcal{L}, \qquad (5) $$

is a conserved quantity, i.e. $\forall s, H'(s) = 0$. In the following, this so-called arclength invariant will thus be simply denoted by $H$. The notation $H$ seems natural when one considers that the relation above is closely linked to the Legendre transformation of the Lagrangian $\mathcal{L}$ [16].

Furthermore, (4a) provides the so-called static coordinate invariant $H_\theta$,

$$ H_{\theta}(s) = H_{\theta} = \theta \cdot \frac{\partial \mathcal{L}}{\partial q'}. \qquad (6) $$

**Constraints.** For the sake of simplicity, we only focus on the static case here again. We consider the system to be subject to a bilateral constraint of the form $h(q, q') = 0$, meaning that the generalised coordinates are not necessarily independent. The stationarity of the Action under this constraint can be found by considering a constraint-free problem where the Action is modified by transforming the Lagrangian as

$$ \mathcal{L} \leftarrow \mathcal{L} - \lambda(s) h(q, q'), \qquad (7) $$

where $\lambda(s)$ is the Lagrangian multiplier associated to the constraint $h(q, q') = 0$. The reader may think that introducing the multiplier $\lambda(s)$ to the Lagrangian adds an explicit dependence of $\mathcal{L}$ on the space variable $s$, hence jeopardising the validity of Noether's arclength invariant (5). However it can be shown that all the previous Noether invariants remain valid, including (5), due to the fact that $h(q, q') = 0$ at equilibrium.

Note that the new Lagrangian (7) does not contribute the $-\mathcal{L}$ part of Noether's invariant, as $h(q, q') = 0$ at equilibrium. However it may bring a new term of the form $q' \cdot \frac{\partial \mathcal{L}}{\partial q'}$ if the constraint $h$ depends on $q'$. In Section 4, we show in particular that the tension of the rod emerges in the static rod invariant, precisely due to the kinematic constraint relating the rod centreline to the material frame.

### 2.2. Noether's theorem on $\mathbb{R}^n \times \mathcal{SO}(3)$

In this section we limit ourselves to the static case. So far we have expressed Noether's theorem considering that $q$ is a vector of $\mathbb{R}^n$, with the property that its derivative can simply be taken component-wise. This theorem naturally follows from the Euler-Lagrange equations on $\mathbb{R}^n$ (see Appendix A). However, key variables such as 3D rotations may not freely evolve in a full vector space, hence they may not be represented as three independent vectors of $\mathbb{R}^3$. Indeed, computing the derivative of a rotation matrix as derivatives of the matrix components (i.e. on $\mathbb{R}^3$) does not allow one to recover a rotation matrix after integration. Said otherwise, the resulting vector derivative is not tangent to the space of rotation matrices (the so-called $\mathcal{SO}(3)$ group). In 2D, this problem is easily solved by parametrising rotations with a scalar $\theta$ corresponding to the rotation angle about the horizontal axis. In such a case, rotation matrices do not need to be explicitly represented as variables, only the scalar variable
$q = \theta$ and its derivatives $\theta'$ and $\dot{\theta}$ are sufficient, which brings us back to the (easy) vector case. Unfortunately, such a vector reduction is not straightforward in 3D.

In 3D, a first approach to tackle this problem is to consider external constraints applied onto $q$ and inject them in the Lagrangian: this is what we will do in Sections 5 and 8, leveraging the property that Noether's invariants still hold in case of bilateral constraints, see above. This 'flattening' method has the advantage of being systematic and simple to apply, since one can still resort to the classical Euler-Lagrange equations and Noether's theorem on $\mathbb{R}^n$. However it requires formulating all kinematic constraints properly (especially constraints on $SO(3)$), and burdens the user with a few tedious calculations. A second, more compact method consists in reformulating new Euler-Lagrange and new Noether equations, which are valid on $SO(3)$. This is possible to do so by applying the so-called Euler-Poincaré reduction, which keeps track of rotation matrices $R$ as main degrees of freedom, but considers their derivative to be a vector of $\mathbb{R}^3$, denoted by $u$ in the world space, or $\bar{u}$ in the local rotating space. Explicitly, if $R \in SO(3)$, then $R'$ reads

$$R' = [u]_\times R = R[\bar{u}]_\times \quad \text{with} \quad u = R\bar{u}, \qquad (8)$$

where $[w]_\times$ stands for the cross-product matrix of vector $w \in \mathbb{R}^3$, such that $[w]_\times v = w \times v$, $\forall v \in \mathbb{R}^3$, see Appendix A.

It is noteworthy that with such a parametrisation $(R, \bar{u})$ of the Lagrangian, neither the classical Euler-Lagrange equations nor the Noether theorem are applicable directly, as a key assumption behind these equations, $q' = \frac{\partial q}{\partial s}$, is lost in the case of 3D rotations. This difficulty can however be overcome by imposing commutativity of derivation and perturbation operators, leading to so-called compatibility constraints. In the end, this method allows one to formulate Noether's theorem directly on the right spaces containing the two different types of variables, i.e. $\mathbb{R}^n$ for translational quantities (such as the rod centreline), and $SO(3)$ for rotational quantities (such as the rod material frame), without the need for additional constraints to maintain the variables in their native space. For rods, one last constraint that couples translational quantities to rotational quantities still needs to be added in the Lagrangian formulation, but in practice this one remains very simple. The most painful constraints, which are the ones expressing that $q$ is a rotation matrix, are eliminated with the second 'compact' method. In the end, we come up with simple recipes that can be used for any problem containing translational and rotational parts, possibly coupled together.

The Euler-Lagrange equations have been expressed on $\mathbb{R}^3 \times SO(3)$ for a while, mainly by roboticians dealing with rigid body systems, see e.g. [25] for a good introduction to this topic and a derivation of the Euler-Lagrange equations on $\mathbb{R}^3 \times SO(3)$ for dynamic rigid body systems. To the best of our knowledge, Jung et al. [11] were the first to apply $\mathbb{R}^3 \times SO(3)$ Euler-Lagrange equations to derive static equations for Reissner rods. Recently, Romero and Gebhardt [26] worked out the dynamic version for (extensible) Kirchhoff thin elastic rods.

To the best of our knowledge, this approach has not yet been followed to derive Noether's invariance principles on $\mathbb{R}^n \times SO(3)$ for thin elastic rods. In the following we summarise the main results we obtained for Noether's invariants on $\mathbb{R}^n \times SO(3)$ in the static case. We mention briefly in Sections 5 and 8 how we can apply these new Noether invariants to our static rod problems to retrieve the same results as with the 'flattening' method, in a more straightforward way. All the details of the derivations are provided in Appendix A.

**Arc length invariant.** We consider two types of degrees of freedom: translational ones, denoted as before as $q$, a vector of $\mathbb{R}^n$; and a rotational one,<fnref n="1" /> denoted as $R \in SO(3)$, with $R' = R[\bar{u}]_\times$ and $\bar{u} \in \mathbb{R}^3$. We denote by $\dot{L}(q, q', R, \bar{u})$ the Lagrangian functional that uses $\bar{u}$ to represent the derivative of the rotation matrix $R$.

In the particular static case, we obtain the following invariant quantity if $\dot{L}$ does not explicitly depend on $s$,

$$\forall s, \quad \dot{H}'(s) = 0 \quad \text{with} \quad \dot{H}(s) = \dot{H} = q' \cdot \frac{\partial \dot{L}}{\partial q'} + \bar{u} \cdot \frac{\partial \dot{L}}{\partial \bar{u}} - \dot{L}. \qquad (9)$$

Proof is given in Appendix A. Note that the rotational part of the invariant takes the same form as the translational part, with $\bar{u}$ playing the role of $R$'s derivative. Besides, the formula above is valid for any one-dimensional system featuring translational and/or rotational parts. As seen later, examples include statics of elastic rods, with $\bar{u}$ being the triplet of material curvatures and twist, but also dynamic rigid bodies, with $s$ representing time and $\bar{u}$ corresponding to the angular velocity vector. In particular, in the case of a dynamic rigid body, $H$ corresponds to the Hamiltonian of the system (expressed as the Legendre transform of the Lagrangian), with $\frac{\partial \dot{L}}{\partial q'}$ the linear momentum and $\frac{\partial \dot{L}}{\partial \bar{u}}$ the angular momentum [25, Chapter 7]. If $H$ has no explicit dependence in time (e.g. case of a freely moving rigid body), then it is well-known that $H$ coincides with the total mechanical energy of the system, hence it is constant over time.

**Coordinate rotation invariant.** Let $\epsilon \theta_q$ an infinitesimal translation, and $\dot{q} = q + \epsilon \theta_q$ the corresponding transformation of $q$. We also consider $\epsilon \theta_R$ a vector around which the rotation matrix $R$ undergoes an infinitesimal rotation, and look at the resulting transformation $\bar{R} = R + \epsilon[\theta_R]_\times R = R + \epsilon R[\bar{\theta}_R]_\times$ with $\bar{\theta}_R = R\bar{\theta}_R$. If these two transformations $(\dot{q}, \bar{R})$ leave the Lagrangian $\dot{L}$ unchanged, then we obtain the following coordinate invariant on $\mathbb{R}^n \times SO(3)$,

$$\forall s, \quad \dot{H}'_\theta(s) = 0 \quad \text{with} \quad \dot{H}_\theta(s) = \dot{H}_\theta = \theta_q \cdot \frac{\partial \dot{L}}{\partial q'} + \bar{\theta}_R \cdot \frac{\partial \dot{L}}{\partial \bar{u}} = 0 \quad \forall s. \qquad (10)$$

Proof is given in Appendix A. As before, the translational part of the invariant takes the same form as the classical Euclidean coordinate invariant (6).

### 3. Thin elastic rod problem setup

**Notation.** Let $w$ a vector of $\mathbb{R}^3$. In the world basis $(e_x, e_y, e_z)$, the vector $w$ reads $w = w_x e_x + w_y e_y + w_z e_z$, while in the basis $R = (d_1, d_2, d_3)$, it reads $w = w_1 d_1 + w_2 d_2 + w_3 d_3$. We denote by $\bar{w}$ the triplet $(w_1, w_2, w_3)$, and we have $w = R \bar{w}$.

<footnote n="1">¹ In case of multiple rotational degrees of freedom $R_i$, a similar formula applies with $u$ being the concatenation of all vectors $\bar{u}_i$.</footnote>
## 3.1. Potential energy

We consider a shearable, extensible, bendable, twistable elastic rod. The rod has a natural shape that is not necessarily straight [4]. We use an arclength parametrisation with the variable s. We start with the internal bending and twist energy densities,

$$W_{\text{bend}} = \frac{1}{2}(\bar{u} - \hat{u})^{\text{T}} B (\bar{u} - \hat{u}) = \frac{1}{2} B_1(s) [u_1(s) - \hat{u}_1(s)]^2 + \frac{1}{2} B_2(s) [u_2(s) - \hat{u}_2(s)]^2 + \frac{1}{2} B_3(s) [u_3(s) - \hat{u}_3(s)]^2 \quad (11)$$

where the $u_i$ are actual curvatures and twist of the rod, the $\hat{u}_i$ their natural counterparts, and B the diagonal matrix containing the bending $B_1, B_2$ and the twist $B_3$ rigidities. We also consider the elastic energy density due to shear and stretch,

$$W_{\text{shear}} = \frac{1}{2}(\bar{v} - \hat{v})^{\text{T}} A (\bar{v} - \hat{v}) = \frac{1}{2} A_1(s) [v_1(s) - \hat{v}_1(s)]^2 + \frac{1}{2} A_2(s) [v_2(s) - \hat{v}_2(s)]^2 + \frac{1}{2} A_3(s) [v_3(s) - \hat{v}_3(s)]^2 \quad (12)$$

where the $v_i$ are actual shear and stretch of the rod, the $\hat{v}_i$ their natural counterparts, and A the diagonal matrix containing the shear $A_1, A_2$ and the stretch $A_3$ rigidities. Generally, it is considered that $\hat{v}_1 = 0, \hat{v}_2 = 0$, and $\hat{v}_3 = 1$, in which case s is the arclength of the rod in the natural state, but we will at first keep generic $\hat{v}_i$ values to exhibit symmetric formulas, see e.g. (28). Please note that for simplicity reasons, we start with a quadratic energy but other hyperelastic rod models could be handled, see e.g. (33).

We finally add the potential energy density $W_{\text{ext}}$ associated to the external conservative force density $f_{\text{ext}}$, i.e. $f_{\text{ext}} = -\partial/\partial r W_{\text{ext}}(r)$, to obtain the total potential energy density of the rod V,

$$V = W_{\text{bend}} + W_{\text{shear}} + W_{\text{ext}} \quad (13)$$

Note that in this paper, we only consider energy densities $W = W_{\text{bend}}$ or $W_{\text{shear}}$ or $W_{\text{ext}}$ that do not explicitly depend on s. For instance, we exclude cases such as $W = \cos(2\pi s)$ or punctual energy densities of the form $W = W_a \delta(s-a)$ where $\delta$ is the Dirac function — however, W may of course depend on the variables r, u, etc. In the case where $f_{\text{ext}}$ is simply the gravitational force density $f_{\text{ext}} = -\rho S g e_z$, we have $W_{\text{ext}}(r) = \rho S g r(s) \cdot e_z$, where $\rho$ is the density of the material, S the area of the rod cross-section, and g the norm of the acceleration of gravity.

## 3.2. Constraints

We attach a Cosserat orthonormal frame $R(s) = (d_1(s), d_2(s), d_3(s))$ — also called material frame — to the rod, which brings 6 bilateral constraints

$$d_1 \cdot d_1 - 1 = 0 \qquad d_2 \cdot d_2 - 1 = 0 \qquad d_3 \cdot d_3 - 1 = 0 \quad (14a)$$

$$d_1 \cdot d_2 = 0 \qquad d_2 \cdot d_3 = 0 \qquad d_3 \cdot d_1 = 0 \quad (14b)$$

The conservation of these orthonormal relations as s is varied along the rod implies the Darboux relations

$$d_1'(s) = u \times d_1 \quad (15a)$$

$$d_2'(s) = u \times d_2 \quad (15b)$$

$$d_3'(s) = u \times d_3 \quad (15c)$$

where the Darboux vector has been chosen to be $u = u_1 d_1 + u_2 d_2 + u_3 d_3$, see e.g. [27]. The three vectorial relations (15) express constraints for the three scalar components $u_1, u_2, u_3$ and can thereof be rewritten as

$$u_1 = d'_2 \cdot d_3, \quad u_2 = d'_3 \cdot d_1, \quad u_3 = d'_1 \cdot d_2 \quad (16)$$

The rod centreline $r(s)$ is linked to the Cosserat frame through the relation

$$r'(s) = v(s) = v_1(s) d_1 + v_2(s) d_2 + v_3(s) d_3 \quad (17)$$

# 4. Thin elastic rod equations from the Lagrangian

We seek equilibrium states for this elastic rod by minimising its total potential energy V given in (13). We only consider the necessary conditions for the vanishing of the first variation of V under the continuous constraints (14), (16), and (17).

## 4.1. Lagrangian

Using the Lagrange multiplier rule, we introduce the Lagrangian on $\mathbb{R}^n$,

$$\begin{aligned} \mathcal{L} = V &+ \lambda_r \cdot (r'(s) - v_1(s) d_1 - v_2(s) d_2 - v_3(s) d_3) + \lambda_{u_1} (d'_2 \cdot d_3 - u_1) + \lambda_{u_2} (d'_3 \cdot d_1 - u_2) + \lambda_{u_3} (d'_1 \cdot d_2 - u_3) \\ &+ \lambda_{11} \frac{1}{2} (1 - d_1 \cdot d_1) + \lambda_{22} \frac{1}{2} (1 - d_2 \cdot d_2) + \lambda_{33} \frac{1}{2} (1 - d_3 \cdot d_3) \\ &+ \lambda_{12} d_1 \cdot d_2 + \lambda_{23} d_2 \cdot d_3 + \lambda_{31} d_3 \cdot d_1 \end{aligned} \quad (18)$$

where $\mathcal{L} = \mathcal{L}[q(s), q'(s)]$ with $q = (r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3)$, see also [28] for a similar approach in the context of elasto-capillarity. The multipliers $\lambda_r(s), \lambda_{u_1}(s), \lambda_{u_2}(s), \lambda_{u_3}(s)$ associated to the bilateral constraints all depend on s, but we stress that it does not invalidate relation (4b) as these multipliers are, like the variables, unknowns of the problem. The same is true for contact constraints, see also Section 6.

Alternatively, we can choose to parameterise the derivative of the Cosserat frame $R(s)$, which belongs to $S\mathcal{O}(3)$, using the Darboux vector $u$, such that $R'(s) = [u]_\times R = R[\bar{u}]_\times$. The vector $\bar{u}$ is the triplet containing the material curvatures and twist $u_i$, i.e. we have $u = R\bar{u}$. In this case, our new Lagrangian $\hat{\mathcal{L}} = \hat{\mathcal{L}}[q, R, q', \bar{u}]$ on $\mathbb{R}^n \times S\mathcal{O}(3)$, with $q = (r, \bar{v})$, simply reads

$$\hat{\mathcal{L}} = V + \lambda_r \cdot (r'(s) - v_1(s) d_1 - v_2(s) d_2 - v_3(s) d_3)$$
$$ = V + \lambda_r \cdot (r'(s) - R(s)\bar{v}(s)), \qquad (19) $$

as the constraints for maintaining $R(s)$ in $SO(3)$ are now intrinsically accounted for thanks to our $SO(3)$ parametrisation $(R, \bar{u})$ of the material frame. Only the kinematic constraint relating the material frame $R(s)$ to the centreline $r'(s)$ of the rod remains.

## 4.2. Euler–Lagrange equations

From the Lagrangian on $\mathbb{R}^n$. Necessary conditions for the vanishing of the first variation of $V$ are the classical Euler–Lagrange equations

$$ \frac{\partial \mathcal{L}}{\partial \boldsymbol{r}} = \left( \frac{\partial \mathcal{L}}{\partial \boldsymbol{r}'} \right)' \Rightarrow \lambda_r'(s) = -f_{\text{ext}} \qquad (20a) $$

$$ \frac{\partial \mathcal{L}}{\partial u_i} = 0 \Rightarrow \lambda_{u_i} = B_i(u_i - \hat{u}_i) \text{ with } i = 1, 2, 3 \qquad (20b) $$

$$ \frac{\partial \mathcal{L}}{\partial v_i} = 0 \Rightarrow \lambda_r \cdot d_i = A_i(v_i - \hat{v}_i) \text{ with } i = 1, 2, 3 \qquad (20c) $$

$$ \frac{\partial \mathcal{L}}{\partial d_1} = \left( \frac{\partial \mathcal{L}}{\partial d_1'} \right)' \Rightarrow \lambda_{u_3} d_2' - \lambda_{u_2} d_3' + \lambda_{u_3}' d_2 + \lambda_r v_1 + \lambda_{11} d_1 - \lambda_{12} d_2 - \lambda_{31} d_3 = 0 \qquad (20d) $$

$$ \frac{\partial \mathcal{L}}{\partial d_2} = \left( \frac{\partial \mathcal{L}}{\partial d_2'} \right)' \Rightarrow \lambda_{u_1} d_3' - \lambda_{u_3} d_1' + \lambda_{u_1}' d_3 + \lambda_r v_2 + \lambda_{22} d_2 - \lambda_{23} d_3 - \lambda_{12} d_1 = 0 \qquad (20e) $$

$$ \frac{\partial \mathcal{L}}{\partial d_3} = \left( \frac{\partial \mathcal{L}}{\partial d_3'} \right)' \Rightarrow \lambda_{u_2} d_1' - \lambda_{u_1} d_2' + \lambda_{u_2}' d_1 + \lambda_r v_3 + \lambda_{33} d_3 - \lambda_{31} d_1 - \lambda_{23} d_2 = 0. \qquad (20f) $$

We recognise in (20a) the force balance of the static Reissner equations, with the Lagrange multiplier $\lambda_r$ playing the role of the internal force vector $n = \lambda_r$. We also identify in (20b) the bend-twist constitutive relations, where the Lagrange multipliers $\lambda_{u_i}$ are the components in the material frame of the internal moment vector $m = \lambda_{u_1}d_1 + \lambda_{u_2}d_2 + \lambda_{u_3}d_3$, and finally in (20c) the shear-extension constitutive relations which provide the components in the material frame of the internal force vector $n = \lambda_r$.

The Euler–Lagrange equations related to $d_1, d_2, d_3$ require more manipulations. By taking the scalar product of (20e), (20f), (20d) with, respectively $d_3, d_1, d_2$, we obtain the moment balance of the static Reissner equations,

$$ m_1'(s) = \lambda_{23} - v_2 n_3 - u_2 m_3 \qquad (21a) $$

$$ m_2'(s) = \lambda_{31} - v_3 n_1 - u_3 m_1 \qquad (21b) $$

$$ m_3'(s) = \lambda_{12} - v_1 n_2 - u_1 m_2 \qquad (21c) $$

with the notation $n = n_1 d_1 + n_2 d_2 + n_3 d_3$. To compute the missing multipliers $\lambda_{23}, \lambda_{31}, \lambda_{12}$, we take the scalar product of (20f), (20d), (20e) with, respectively $d_2, d_3, d_1$

$$ \lambda_{23} = \lambda_{u_2} u_3 + n_2 v_3 \qquad (22a) $$

$$ \lambda_{31} = \lambda_{u_3} u_1 + n_3 v_1 \qquad (22b) $$

$$ \lambda_{12} = \lambda_{u_1} u_2 + n_1 v_2 \qquad (22c) $$

We inject (22) into (21) to obtain:

$$ m_1'(s) = u_3 m_2 - u_2 m_3 + v_3 n_2 - v_2 n_3 \qquad (23a) $$

$$ m_2'(s) = u_1 m_3 - u_3 m_1 + v_1 n_3 - v_3 n_1 \qquad (23b) $$

$$ m_3'(s) = u_2 m_1 - u_1 m_2 + v_2 n_1 - v_1 n_2 \qquad (23c) $$

and using the property $w' \cdot d_i = w_i' - (w \times u) \cdot d_i \quad \forall w$, we finally come up with

$$ m'(s) = n(s) \times r'(s). \qquad (24) $$

The force equilibrium expressed in the material frame reads

$$ n_1'(s) = u_3 n_2 - u_2 n_3 - f_{\text{ext}} \cdot d_1 \qquad (25a) $$

$$ n_2'(s) = u_1 n_3 - u_3 n_1 - f_{\text{ext}} \cdot d_2 \qquad (25b) $$

$$ n_3'(s) = u_2 n_1 - u_1 n_2 - f_{\text{ext}} \cdot d_3. \qquad (25c) $$

From the Lagrangian on $\mathbb{R}^n \times SO(3)$. The method above has the advantage of being simple and explicit, since all computations are carried out on $\mathbb{R}^n$ using standard Euler–Lagrange equations. However, on may argue that computing $\frac{\partial \mathcal{L}}{\partial d_{12}'}$ ((20d), (20e), and (20f)) may be lengthy, and overall refactoring all terms to recover rod equations is a bit tedious.

In contrast, starting from the reduced Lagrangian $\mathcal{L}$ given in (19) and using the Euler–Lagrange equations on $\mathbb{R}^n \times SO(3)$ alleviates most of these computations, see Appendix A. Again, we see that the internal force $n$ emerges as the Lagrange multiplier associated with the constraint relating the material frame to the centreline of the rod. In contrast, the internal moment $m$ no longer appears as a multiplier, but directly stems from the term $\frac{\partial \mathcal{L}}{\partial \bar{u}} = \bar{m}$ owing to the reduced parametrisation of $SO(3)$.
# 5. Noether invariants in the static case

## 5.1. Arclength translation

From the Lagrangian on $\mathbb{R}^n$. In the case where the rigidities and the natural shape are uniform, i.e.

$$ \forall s \quad B'_i(s) = A'_i(s) = \hat{u}'_i(s) = \hat{v}'_i(s) = 0, \qquad (26) $$

the Lagrangian $L(q, q')$ does not explicitly depend on $s$ and the arclength invariant $H$ given in (5) is, at equilibrium and under satisfied constraints, uniform along the rod. At equilibrium under constraints (14), (16), and (17), we have $H = q' \cdot \frac{\partial L}{\partial q'} - V$ with $q = (r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3)$, hence

$$ \frac{\partial L}{\partial q'} \cdot q' = n \cdot r' + m_3 d_2 \cdot d'_1 + m_1 d_3 \cdot d'_2 + m_2 d_1 \cdot d'_3 \qquad (27a) $$

$$ = n \cdot v + m \cdot u, \qquad (27b) $$

which gives

$$ H = n \cdot v + m \cdot u - V. \qquad (28) $$

From the Lagrangian on $\mathbb{R}^n \times SO(3)$. Using our compact Lagrangian (19) and the Noether invariant (9) derived on $\mathbb{R}^n \times SO(3)$ (see Appendix A), we immediately recover the same invariant $\dot{H} = H$, as $\frac{\partial L}{\partial r'}$ stands for the internal force $n$ and $\frac{\partial L}{\partial u}$ for the internal moment $\bar{m}$ in the local frame.

### Interpretation of $H$. Using (13), the mixed (strain-stress) formulation (28) can be rewritten in different ways,

$$ H = \frac{1}{2} m \cdot (u + \hat{u}) + \frac{1}{2} n \cdot (v + \hat{v}) - W_{\text{ext}} \qquad (29a) $$

$$ H = W_{\text{bend}} + m \cdot \hat{u} + W_{\text{shear}} + n \cdot \hat{v} - W_{\text{ext}} \qquad (29b) $$

$$ H = \sum_{i=1}^{3} \frac{1}{2} B_i (u_i^2 - \hat{u}_i^2) + \frac{1}{2} A_i (v_i^2 - \hat{v}_i^2) - W_{\text{ext}} \qquad (29c) $$

$$ H = \sum_{i=1}^{3} \frac{1}{2} B_i u_i^2 + \frac{1}{2} A_i v_i^2 - W_{\text{ext}} + C^{\text{te}}. \qquad (29d) $$

Starting from (29b), we obtain in the classical Reissner case ($\hat{v}_1 = \hat{v}_2 = 0$ and $\hat{v}_3 = 1$, and any value for $\hat{u}$)

$$ H = W_{\text{bend}} + m \cdot \hat{u} + W_{\text{shear}} + n_3 - W_{\text{ext}} = \sum_{i} \frac{1}{2} B_i (u_i^2 - \hat{u}_i^2) + W_{\text{shear}} + n_3 - W_{\text{ext}}. \qquad (30) $$

Note that the emergence in the formula of the third component of the internal force $n$, the tension $n_3$, simply stems from the (standard) choice to parametrise the natural shape of the rod with its rest arclength $s$, i.e. $\hat{v}_1 = \hat{v}_2 = 0$, and $\hat{v}_3 = 1$. In the special case where all the natural strains vanish ($\hat{u}_1 = \hat{u}_2 = \hat{u}_3 = \hat{v}_1 = \hat{v}_2 = 0$ and $\hat{v}_3 = 1$), a formulation with energies is obtained

$$ H = W_{\text{bend}} + W_{\text{shear}} + n_3 - W_{\text{ext}}. \qquad (31) $$

Finally, in the absence of any external force $W_{\text{ext}} = 0$ and in the inextensible, unshearable case (inextensible Kirchhoff), where $W_{\text{shear}} = 0$, we re-obtain the Hamiltonian introduced in [16],

$$ H = \frac{1}{2} m \cdot (u + \hat{u}) + n_3. \qquad (32) $$

### Generalisation: hyperelastic rods and ribbons.
It should be remarked that the invariant $H$ is not restricted to the quadratic strain energies (11) (12) but also holds for hyperelastic rods [11,13,18]. Moreover, 1D models of elastic ribbons also fall into the realm of applications of the invariant, see for example [29–31]. See also [32, Section 2, Eq. (2.16)] and [33, Appendix B, Eq. (44)] for an example of how the invariant may be used in boundary layer calculations in the case of the Wunderlich strain energy. We here give the example of an inextensible, unshearable, Sadowsky ribbon [34,35] with uniform natural curvature $\hat{u}_1$,

$$ W_{\text{bend}}^{\text{sdw}} = \frac{1}{2} B^* \left[ u_1^2 \left( 1 + \frac{u_3^2}{u_1^2} \right)^2 - 2 \hat{u}_1 u_1 \left( 1 + \nu \frac{u_3^2}{u_1^2} \right) \right] \qquad (33a) $$

$$ W_{\text{shear}}^{\text{sdw}} = 0, \qquad (33b) $$

where $\nu$ is the Poisson ratio of the elastic material, $u_2 = 0$, and $B^* = B/(1 - \nu^2)$. We note that, just as in the case of rods (29d), the invariant $H$ (28) may be expressed with a formula where the natural curvature $\hat{u}_1$ does not appear

$$ H^{\text{sdw}} = n \cdot v + m \cdot u - V = n_3 + m \cdot u - W_{\text{bend}}^{\text{sdw}} - W_{\text{ext}} = n_3 + \frac{1}{2} B^* u_1^2 \left( 1 + \frac{u_3^2}{u_1^2} \right)^2 - W_{\text{ext}}. \qquad (34a) $$

## 5.2. Coordinate translation

In the absence of any external force field, $W_{\text{ext}} = 0$, if we consider the transformation $\tilde{r} = r + a$, which is a translation of the position by a fixed amount $a$, then the Lagrangian (18) stays the same [11], i.e.

$$ \tilde{r} = r + a \quad \Rightarrow \quad \mathcal{L}(\tilde{r}, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3) = \mathcal{L}(r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3) \qquad (35) $$
In this case, $\theta$ in (6) is in fact the vector $\alpha$ and the invariance property yields

$$H'_{\theta} = H'_{\alpha} = \left[ \alpha \cdot \frac{\partial \mathcal{L}}{\partial r'} \right]' = 0 \quad \forall \alpha \Rightarrow n' = 0 \quad (36)$$

We see that the force vector is uniform along the rod if no external force field is applied to the rod, a classical result which is readily included in (20a).

### 5.3. Coordinate rotation

In the absence of external force field, $W_{\text{ext}} = 0$, or in the case the potential energy density $W_{\text{ext}}$ only depends on the distance $|r|$, $W_{\text{ext}}(r^2)$, we now consider the transformation where the configuration $(r, d_1, d_2, d_3)$ is rotated by a small angle $\epsilon$ around an arbitrary, constant, vector $b$. The transformation reads

$$\tilde{r} = r + \epsilon b \times r, \quad \tilde{d}_{1,2,3} = d_{1,2,3} + \epsilon b \times d_{1,2,3}. \quad (37)$$

Note that the vectors $\nu$ and $u$ are also rotated, but their components $v_{1,2,3}$ and $u_{1,2,3}$ stay unchanged. Standard Taylor expansion shows that, at leading order in $\epsilon$, distances and hence $W_{\text{ext}}(r^2)$ are unchanged, and that this transformation eventually leaves the Lagrangian unchanged,

$$\mathcal{L}(\tilde{r}, \tilde{d}_1, \tilde{d}_2, \tilde{d}_3, u_1, u_2, u_3, v_1, v_2, v_3) = \mathcal{L}(r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3) \quad (38)$$

and consequently, the following quantity is invariant

$$\left[ b \times r \cdot \frac{\partial \mathcal{L}}{\partial r'} + \sum_{i} b \times d_i \cdot \frac{\partial \mathcal{L}}{\partial d_i'} \right]' = 0 \quad (39a)$$

$$[b \cdot r(s) \times n(s) + b \cdot m(s)]' = 0 \quad \forall b \quad (39b)$$

$$[r(s) \times n(s) + m(s)]' = 0. \quad (39c)$$

We observe that with the potential energy density $W_{\text{ext}}(r^2)$, the external force is a central force, $f_{\text{ext}} = -2 W'_{\text{ext}}(r^2)r$, and (39c) may be obtained directly from (20a) and (24). Please note that the case $W_{\text{ext}} = 0$ with transformation (37) has been treated in [11].

### 5.4. Coordinate rotation around the fixed force axis

In the absence of any external force field, $W_{\text{ext}} = 0$, Section 5.2 showed the force vector $n$ is a fixed quantity. In this case, we consider the transformation in which the vectors $r, d_1, d_2$, and $d_3$ are rotated by a small angle $\epsilon$ around the force axis $n$,

$$\tilde{r} = r + \epsilon n \times r, \quad \tilde{d}_{1,2,3} = d_{1,2,3} + \epsilon n \times d_{1,2,3}. \quad (40)$$

Note that the vectors $\nu$ and $u$ are also rotated, but their components $v_{1,2,3}$ and $u_{1,2,3}$ stay unchanged. At leading order in $\epsilon$, this transformation leaves the Lagrangian unchanged,

$$\mathcal{L}(\tilde{r}, \tilde{d}_1, \tilde{d}_2, \tilde{d}_3, u_1, u_2, u_3, v_1, v_2, v_3) = \mathcal{L}(r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3), \quad (41)$$

and consequently, the following quantity is invariant,

$$\left[ n \times r \cdot \frac{\partial \mathcal{L}}{\partial r'} + \sum_{i} n \times d_i \cdot \frac{\partial \mathcal{L}}{\partial d_i'} \right]' = 0 \quad (42a)$$

$$[n \cdot m(s)]' = 0 \quad (42b)$$

### 5.5. Coordinate rotation around the section normal

In the presence or absence of an external force field, we now consider the transformation where the vectors $d_1, d_2$ are rotated by a small angle $\epsilon$ around the vector $d_3$. In addition, the (1,2) components of the vectors $u$ and $\nu$ are transformed as well so as to leave the vectors $u$ and $\nu$ unchanged. The full transformation reads

$$\tilde{d}_1 = d_1 + \epsilon d_3 \times d_1, \quad \tilde{d}_2 = d_2 + \epsilon d_3 \times d_2, \quad \tilde{u}_1 = u_1 + \epsilon u_2, \quad \tilde{u}_2 = u_2 - \epsilon u_1, \quad \tilde{v}_1 = v_1 + \epsilon v_2, \quad \tilde{v}_2 = v_2 - \epsilon v_1. \quad (43)$$

In the case where $A_1 = A_2$, $B_1 = B_2$, $\hat{u}_1 = 0$, $\hat{u}_2 = 0$, $\hat{v}_1 = 0$, and $\hat{v}_2 = 0$, at leading order in $\epsilon$, this transformation leaves the Lagrangian unchanged [11],

$$\mathcal{L}(\tilde{r}, \tilde{d}_1, \tilde{d}_2, d_3, \tilde{u}_1, \tilde{u}_2, u_3, \tilde{v}_1, \tilde{v}_2, v_3) = \mathcal{L}(r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3), \quad (44)$$

and consequently, the following quantity is invariant

$$\left[ d_3 \times d_1 \cdot \frac{\partial \mathcal{L}}{\partial d_1'} + d_3 \times d_2 \cdot \frac{\partial \mathcal{L}}{\partial d_2'} + u_2 \frac{\partial \mathcal{L}}{\partial u_1'} - u_1 \frac{\partial \mathcal{L}}{\partial u_2'} + v_2 \frac{\partial \mathcal{L}}{\partial v_1'} - v_1 \frac{\partial \mathcal{L}}{\partial v_2'} \right]' = 0 \quad (45)$$

$$[m_3(s)]' = 0. \quad (46)$$

Please note that this result holds in the case where $\hat{u}_3 \neq 0$ and $\hat{v}_3 \neq 0$.
## 5.6. Structure of the coordinate invariants

Remarkably, one can immediately recover all the invariant properties above by using Noether's coordinate rotation formula (10) derived on $R^n \times SO(3)$. Two advantages of this compact formulation are worth mentioning. First, examining the compact Lagrangian $\mathcal{L}$ given in (19) makes it easier to localise symmetries. Second, it highlights the fact that all Noether coordinate invariants for rods should be of the form $\theta_n \cdot n + \theta_m \cdot m$ with $\theta_n, \theta_m \in R^3$. Please refer to Appendix A for details on this formulation.

# 6. Contact

In this part we focus on the arclength invariant $H$. For the sake of simplicity, we assume the static case. Our results can be generalised easily to the dynamic case.

## 6.1. Preamble: General case of shear forces

It is noteworthy that the arclength invariant $H$ can be retrieved exactly by projecting the rod angular equilibrium equation (24) onto the Darboux vector $u$. Indeed, on the one hand we have $m = R B(\bar{u}-\hat{u})$ with $B$ the diagonal matrix containing $B_1, B_2, B_3$. This gives $m' = R[\bar{u}]_\times B(\bar{u}-\hat{u})+R B \bar{u}'$. Using the properties $R^T R = I$ and $[\bar{u}]_\times \bar{u} = 0$, we obtain $m' \cdot u = \bar{u}'^T B \bar{u}$. On the other hand, we similarly have $n = R A(\bar{v}-\hat{v})$ with $A$ the diagonal matrix containing $A_1, A_2, A_3$. This yields $n' = [u]_\times R A(\bar{v}-\hat{v})+R A \bar{v}' = [u]_\times n + R A \bar{v}' = u \times n + R A \bar{v}'$. Using this equality, in addition to the identity $r' = v = R \bar{v}$ and to the property $R^T R = I$, we obtain for the mixed product $(r' \times n) \cdot u = -(u \times n) \cdot r'$ the following expression, $(r' \times n) \cdot u = (R A \bar{v}' - n') \cdot v = \bar{v}'^T A \bar{v} - n' \cdot r'$. Now, if $f_{\text{ext}}$ is an external force density applied onto the rod, the balance of forces (20a) writes $n' = -f_{\text{ext}}$. The projection of (24) onto the Darboux vector $u$ finally yields

$$0 = (m' + r' \times n) \cdot u = \bar{u}'^T B \bar{u} + \bar{v}'^T A \bar{v} + f_{\text{ext}} \cdot r' = \frac{dH}{ds} + f_{\text{ext}} \cdot r'. \quad (47)$$

If one assumes that $f_{\text{ext}}$ is a shear force, i.e. a force normal to the tangent $r'$ of the rod at each point $s$ of the centreline (the force can be zero), then the term $f_{\text{ext}} \cdot r'$ in (47) vanishes. Hence, the arclength invariant $H$ remains unchanged in the presence of shear forces.

We now examine the specific case of frictionless contact forces, first in the case when they are modelled using a soft potential, then when they are represented with (hard) unilateral constraints.

## 6.2. Contact potential and shear contact force

We assume that the system is subject to a conservative external force density $f_{\text{ext}} = -\partial/\partial r W_{\text{ext}}(r)$. We have seen previously that accounting for this external force amounts to adding $-W_{\text{ext}}(r)$ to the arclength invariant $H$.

We can write down the contribution of this potential to $H'$,

$$\frac{d}{ds}(-W_{\text{ext}}(r)) = -r' \cdot \partial/\partial r W_{\text{ext}}(r) \quad (48)$$

$$= r' \cdot f_{\text{ext}}, \quad (49)$$

which automatically cancels out if $f_{\text{ext}}$ is normal to the centreline. As a result, an external conservative field $W_{\text{ext}}$ yielding a force always normal to the centreline does not have to be included in the formula for $H$. Now the question is: can we find such potentials, i.e. yielding forces normal to $r'$?

In Section 7.8 we actually show that soft potentials, typically used for modelling contact, do not generate perfectly normal forces. As a result, such potentials need to be accounted for in the arclength invariant $H$. Still, when they are rigidified, their corresponding force becomes orthogonal to $r'$. In this latter case, no additional term needs to be added to the invariant $H$. This can be directly seen by modelling contact as a hard constraint, and this is what we show below.

## 6.3. Contact as a hard constraint: no contribution to $H$!

An interesting result emerges in the case of contact modelled as a hard constraint: there is no need to add any potential to the arclength invariant $H$. Modelling contact as a hard constraint means defining a contact force using a Lagrange multiplier which acts in a normal direction compared to the rod. We already proved that a normal force yields no contribution to $H$. Now we prove this result again, by considering a variational point of view of the problem.

We consider a rod with centreline $r(s)$, and an external object. We call $g(s,p)$ the gap function between $r(s)$ and the obstacle. The gap function $g(s,p)$ might depend on parameters $p$ of the problem that we will specify a bit later. Frictionless contact between the rod and the obstacle can be modelled by an inequality constraint of the form $g(s,p) \ge 0$, $\forall s$, which expresses the fact that the two objects can be separated ($g(s,p) > 0$) or can touch each other ($g(s,p) = 0$), but are not allowed to interpenetrate, i.e. $g(s,p) < 0$ is forbidden (see Fig. 1). To allow this constraint to be satisfied, a Lagrange multiplier $\mu(s)$ is introduced. Mechanically, this multiplier exactly acts as a contact force that prevents the two objects from interpenetrating. Mathematically, the constraint is satisfied when the following complementarity constraint holds,

$$0 \le g(s, p) \perp \mu(s) \ge 0,$$

which expresses that both the gap and the contact force should be non-negative, and that one has to vanish if the other one becomes positive. Intuitively, when the two objects come to contact ($g(s,p) = 0$), a contact force $\mu(s)$ gets activated. The condition above, together with the Euler-Lagrange equations, actually correspond to the optimality conditions [36] of our new variational problem, i.e. the stationarity of the Action under the contact inequality constraint.

If we now go back to this initial variational view, incorporating this new inequality constraint to come up to a constraint-free problem can be done by simply modifying the Lagrangian with the additional term $L_c = -\mu(s)g(s,p)$. As for the bilateral constraint examined in Section 2, Noether's formula for the arclength invariant (5) remains valid thanks to the complementarity condition above.
<figure><img src="image_3.png" /><figcaption>Fig. 1. Contact with an obstacle in the plane may in some cases be modelled with a gap or level set function g(x, y). The forbidden region is when g(x, y) < 0, and the rod is allowed to lie in the region g(x, y) ≥ 0. The surface of the obstacle is defined by g(x, y) = 0, while we have g(x, y) > 0 in the absence of contact with the obstacle.</figcaption></figure>

Now, the point is to formulate the gap function g(s, p) and examine its dependencies upon our variables of interest, and then derive the Noether arclength invariant. We first examine an elementary 2D case, for the sake of simplicity. We consider a 2D rod subject to contact with an infinite ground of height y = a. The contact inequality constraint simply reads g(s) ≥ 0 with g(r) = r · e_y - a, and the contribution to the Lagrangian reads L_c(r) = -μ(s) (r · e_y - a). Note that similarly to the bilateral case, the new Lagrangian L_c does not contribute the -L part of Noether's arclength invariant H as μg = 0 at equilibrium. As the contact constraint does not depend on r', it does not contribute the ∂L/∂r' · r' part of H either. Finally, such a contact constraint has no impact on the Noether arclength invariant. This result generalises to any gap function that depends only on r.

## 7. Examples of application

In this section, we illustrate several possible uses of the arclength invariant (5) on the classical planar Elastica case. We consider the equilibrium of an inextensible, unshearable, naturally flat rod bent in the (x, y) plane. In this situation, the twist uniformly vanishes and the Cosserat directors are such that d_1 = -sin θ e_x + cos θ e_y, d_2 = e_z, d_3 = cos θ e_x + sin θ e_y, with θ is the angle marking the deflection of the Elastica with the horizontal e_x axis. The static Lagrangian (18) boils down to

$$ \mathcal{L}(x, y, \theta, \kappa) = \frac{1}{2} EI \kappa(s)^2 + n_x(s)(x' - \cos \theta) + n_y(s)(y' - \sin \theta) + m(s)(\theta' - \kappa), \quad (50) $$

where we have directly used the notation for the force and moment multipliers, with m = m(s) e_z. Additionally we use the classical notation B = EI for the bending rigidity of the Elastica, with E the Young modulus of the material and I the second moment of area of the cross-section of the rod. Please also note that at equilibrium, we have n'_x(s) = 0 and n'_y(s) = 0. The Euler-Lagrange equations corresponding to the variation with κ and θ lead to

$$ EI \theta''(s) = n_x \sin \theta - n_y \cos \theta. \quad (51) $$

Additionally, the rod arclength invariant (32) takes the simple form

$$ H = \frac{1}{2} EI \kappa^2 + n_3 \quad (52) $$

where n_3 = n · d_3 = n_x cos θ + n_y sin θ is called the internal tension in the rod. If we call n_1 = n · d_1 = -n_x sin θ + n_y cos θ the internal shear force in the rod, we see that, using n'_3 = κ n_1, (51) gives EI θ''(s) = -n_1 and can be integrated once to yield (52). Please also note that, using κ = θ', (52) leads to an integration by quadrature of the solution function θ(s). Additionally, (52) also shows that the tension n_3 takes a maximum value along the rod at inflexion points s = s_i, where κ(s_i) = 0.

In the example cases presented here, the method simply amounts to writing the invariant at different (carefully chosen) locations along the rod, and using the fact that the value of the invariant is the same everywhere. This allows one to exhibit relations between key quantities of the problem.

### 7.1. Vertical displacement

In the classical problem of the nonlinear dynamics of a point mass attached to a pendulum, the vertical angle θ(t) of the pendulum obeys the equation $\ell\ddot{\theta} = -g\sin\theta$, and writing the solution θ(t) requires the use of elliptic functions. However, to answer the question, illustrated in Fig. 2, of finding the maximum deflection angle θ_max as a function of the initial angular speed θ_0 does not necessitate the use of elliptic functions. The key point is to use the time conservation of the mechanical energy E_meca = (1/2)ℓθ² - g cos θ, write the value of E_meca at times t = t_A and t = t_B and find cos θ_max = 1 - ℓθ_0²/(2 g). The Kirchhoff static-dynamic analogy (see [37], or [18] for a modern treatment) tells us that we can carry out the same method to study the problem shown in Fig. 3, where a simply-supported Elastica is compressed with a force P and displays a deflection θ_0 at the supports. In this case, we have n_x = -P and n_y = 0. Finding the values of the maximum vertical displacement y_M and maximum curvature κ_M would in principle require integrating (51) using elliptic functions, whereas writing the invariant (52) at point O and M directly yields

$$ \kappa_M = \sqrt{\frac{2P}{EI}(1 - \cos \theta_0)}, \quad y_M = \sqrt{\frac{2EI}{P}(1 - \cos \theta_0)}. \quad (53) $$
<figure><img src="image_4.png" /><figcaption>Fig. 2. A pendulum, initially in its vertical equilibrium, is launched at $t_A$ with an initial angular speed $\dot{\theta}_0$. At $t = t_B$, it will reach a maximum angle $\theta_{max}$ and start to swing back.</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>Fig. 3. A planar Elastica is held with simple supports. The horizontal compressive force at the support P is yielding an equilibrium configuration with deflection $\theta_0$, maximum vertical displacement $y_M$, and maximum curvature $\kappa_M$.</figcaption></figure>

<figure><img src="image_6.png" /><figcaption>Fig. 4. Sagging cantilever.</figcaption></figure>

## 7.2. Sagging cantilever

We consider the simple example of a cantilever, i.e. a naturally straight rod clamped horizontally, and sagging under an external vertical distributed force density $f_{ext} = f e_y$, such as gravity (see Fig. 4). We denote by $W_{ext}^y$ its corresponding potential energy density.

We evaluate the invariant $H$ given in (31) at the two special points $s = 0$ and $s = L$. On the one hand, we remark that the internal force $n = \int_s^L f e_y$ is vertical. As a result, at the clamp the internal force $n$ is normal to the tangent of the rod, hence the tension $n_3(s)$ vanishes at $s = 0$. On the other hand, in the absence of natural curvature and of any external force applied at $s = L$, both the internal moment and force vanish at $s = L$.

As a result, we obtain

$$ \frac{1}{2} EI \kappa^2(0) = W_{ext}^y(0) - W_{ext}^y(L). \qquad (54) $$

For the specific case of gravity, we have $W_{ext}^y(s) = \rho g S y(s)$, with $y(s)$ the vertical position of the rod. It follows

$$ |\kappa(0)| = \sqrt{2 \frac{\rho S g}{EI} \Delta}, \qquad (55) $$

where $\Delta = y(0) - y(L)$. Measuring the altitude difference of the rod hence directly provides the rod curvature at the clamp. Conversely, estimating the clamping curvature of a cantilever gives access to the material parameters of the rod $\frac{\rho S g}{EI}$, thereby providing a convenient experimental measurement method for the bending stiffness $EI$. Note that these remarks and measurement methods apply to the fully nonlinear case.
<figure><img src="image_7.png" /><figcaption>Fig. 5. In the case of frictionless contact, the contact force applied from the obstacle onto the rod is perpendicular to the tangent of the rod. The values $H_1$ and $H_2$, left and right of the contact point, are shown to be equal.</figcaption></figure>

<figure><img src="image_8.png" /><figcaption>Fig. 6. Among the possible configurations an Elastica buckling in a rectangular cavity may adopt, the use of the invariance of $H$ enables us to rule out the coexistence of some of them.</figcaption></figure>

### 7.3. Frictionless contact

We explained in Section 6 that frictionless contact, continuous or not, leaves the values of the invariant unchanged. Therefore, in the case where the planar Elastica is in frictionless contact with an obstacle (or in self-contact), see Fig. 5, the invariant keeps the same value along the entire rod, that is $H_1 = H_2$ (see also jump conditions at obstacles in [18]). We use this property in the problem of an Elastica buckling inside a rectangular cavity, see Fig. 6. In this problem, it was shown that several configurations are encountered, namely the point-contact shape where the Elastica touches the wall at a localised value of the arclength, the extended-contact shape where contact occurs for an interval of arclength values, and the hanging-fold shape where a blister is formed between two discrete contact points [38–41]. In all these configurations $H$ keeps a constant value, and if these configurations were to be found along the same rod, $H$ would have to be the same everywhere.

We question the possibility of having both extended-contact and hanging-fold solutions coexisting along the same rod, see Fig. 6. The frictionless nature of the wall interactions implies that the horizontal component of the internal force stays constant along the rod, $n_x = -P$, $\forall s$. In addition, the flat shape the extended-contact solution adopts as it is lying along the wall shows that the invariant at point $A$ is equal to $H_A = -P$. Thus, writing the invariant at point $B$, where the deflection angle vanishes $\theta_B = 0$ but not the curvature $\kappa_B < 0$, yields $H_B = \frac{1}{2}\kappa_B^2 - P$ and demonstrates the impossibility of such a coexistence.

We note that in the case of friction, the invariant $H$ is no longer uniform, see for example the case of the capstan problem studied in [42]. If $\mu$ is the Coulomb friction coefficient, $2R$ is the capstan diameter, and the rod is fully flexible, one finds that $H'(s) = (\mu/R) H(s)$, i.e. an exponential growth for $H(s)$.

### 7.4. Sliding sleeves

In the last decade, it was shown that the buckling of an Elastica through a frictionless sliding sleeve exhibits a (somewhat counter-intuitive) force jump at the exit of the sleeve, see for example [43,44]. We illustrate this phenomenon in Fig. 7, where it is seen that the externally applied forces on the rod at points $A$ and $C$ do not match: $P \neq P'$, with $P' > P$. Bigoni et al. [43] and Cazzolli and Dal Corso [45] explain that the mismatch in horizontal forces is due to the horizontal component of the contact force applied on the rod at the exit of the sleeve, see Fig. 8. The crucial point is that the forces from the roller bearings are frictionless and hence do not change the value of the invariant $H$, i.e. we have $H_{in} = H_{out}$, see Fig. 8. Thus, the relation between $P$ and $P'$ is easily obtained by writing $H$ just before point $B$ and at point $C$ in Fig. 7

$$H_B = \frac{1}{2} EI\kappa_B^2 - P' \quad \text{and} \quad H_C = 0 - P \qquad (56a)$$

$$H_C = H_B \Rightarrow P' = P + \frac{1}{2} EI\kappa_B^2, \qquad (56b)$$
<figure><img src="image_9.png" /><figcaption>Fig. 7. A planar Elastica is buckled through a frictionless sliding sleeve. The external forces P and P' at the right and left extremities do not match, due to the horizontal force coming from the sleeve at the exit point B.</figcaption></figure>

<figure><img src="image_10.png" /><figcaption>Fig. 8. A closer look at the forces applied on the rod at the exit of the frictionless sliding sleeve, adapted from [43]. The perpendicular (with respect to the rod's tangent) orientation of these forces implies the equality $H_{in} = H_{out}$, see Sections 6 and 7.3.</figcaption></figure>

<figure><img src="image_11.png" /><figcaption>Fig. 9. The Elastica arm scale, as introduced by Bosi et al. [46]. A weight of mass m is hung at the right extremity while the left extremity is unloaded. Use of the invariant easily shows that the rod's tangent is horizontal at point C. Image courtesy of Francesco Dal Corso.</figcaption></figure>

where the term $\frac{1}{2}EI\kappa_B^2 > 0$ exactly corresponds to the horizontal component of the contact force at the exit of the sleeve. Interestingly, this example shows that, as the arclength invariant holds, the discontinuity $\Delta P = P' - P$ of the tension at point B implies that there should be a curvature jump $\Delta\kappa = \kappa_B$ at the same point B: the two discontinuities compensate to form a constant sum H over the full rod length.

### 7.5. The Elastica arm scale

The fact that the invariant H is conserved along the entire rod in rigid sleeve problems is handy for computing different properties of these systems at equilibrium. One such property is the horizontal tangent appearing when the Elastica arm scale [46] is loaded at only one of its two extremities, see Fig. 9. Computing the value of H at point A yields $H_A = 0$, and computing the value of H at point C yields $H_C = \frac{1}{2}EI\kappa_C^2 - mg \sin \theta_C$. The absence of external torque loading at point C implies $\kappa_C = 0$ and thus the invariance $H_A = H_C$ proves that $\theta_C = 0$.

### 7.6. A simple case where the invariant breaks down

In the so-called tapered case where the rod has a thickness which varies with arclength, see for example [47], and hence a varying bending stiffness EI(s), we show that the quantity H(s) does depend on arclength. Consider for example the simply-supported Elastica of Fig. 10, for which the equilibrium equation (51) is replaced with

$$[EI(s) \theta'(s)]' = -P \sin \theta. \qquad (57)$$

In this case, we differentiate (52) and find

$$H'(s) = \frac{1}{2}EI'(s)\theta'^2(s) + EI(s)\theta'\theta'' + P\theta'\sin\theta = -\frac{1}{2}EI'(s)\theta'^2(s) \neq 0. \qquad (58)$$
<figure><img src="image_12.png" /><figcaption>Fig. 10. An Elastica with a cross-section of varying thickness. In this case the quantity $H(s)$ is not conserved along the rod.</figcaption></figure>

<figure><img src="image_13.png" /><figcaption>Fig. 11. A rod is twisted and subject to an external tension $T_{ext}$. If the twist is large enough, the rod coils on itself, exhibiting plctonemes. A relation between the external tension $T_{ext}$ and the coiling angle $\theta$ is found using the invariant $H$.</figcaption></figure>

<figure><img src="image_14.png" /><figcaption>Fig. 12. Several bifurcation branches are computed numerically for different slenderness ratios $L/(2R)$ and the relation between $T_{ext}$ and $\theta$ is plotted and compared to (62), see inset.</figcaption></figure>

The fact that the quantity $H(s)$ loses its invariant property comes from the explicit dependence of the Lagrangian (50) with the arclength $s$, which kills the invariance of the Action (1) under the transformation (2b), a crucial hypothesis of Noether's theorem.

### 7.7. A 3D case with twist and continuous contact

We consider an inextensible, unshearable, naturally straight and untwisted rod with circular cross-section and isotropic elastic properties, $B_1 = B_2 = EI$ and $B_3 = GJ$. We study a 3D case where the rod is twisted and experiences self-contact, see Fig. 11. The rod is clamped at its left end and subject to an imposed rotation $n$ and a pulling force $T_{ext}$ at its right end. This setup is used for example in DNA single-molecule experiments [48,49] but also arises as an instability in textile yarns [50]. This complex configuration exhibits both point-wise and continuous contact [51,52], but in the contact region the rod adopts an approximately double-helix shape. Due to self-contact, the helix radius is the radius $R$ of the circular cross-section, and we note $\theta$ the helical angle, between the rod's tangent and the helical axis. The helix thus has a curvature $\kappa = \sqrt{\kappa_1^2 + \kappa_2^2} = (1/R) \sin^2 \theta$, and using the helical shape approximation, it is possible to show [53] that the contact pressure $p$, the twisting moment $m_3$, and the inner tension $n_3$ all essentially depend on the helical angle $\theta$

$$p_{\text{helix}} = \frac{EI}{R^3} \frac{\sin^4 \theta}{\cos 2\theta} \qquad (59a)$$

$$m_{3\text{helix}} = \frac{EI}{R} \frac{2 \sin^3 \theta \cos \theta}{\cos 2\theta} \qquad (59b)$$

$$T_{\text{helix}} = \frac{EI}{R^2} \frac{\sin^4 \theta}{\cos 2\theta} \qquad (59c)$$
<figure><img src="image_15.png" /><figcaption>Fig. 13. (a) The buckling of a planar Elastica constrained by a horizontal upper wall is studied using the long-range interaction potential (65). (b) Total interaction energy (70) as the 'range' b of the potential is decreased, with the limit b → 0 corresponding to the hard-wall limit. (c) and (d) Plot of the arclength invariant H (68) together with the non-invariant function h(s) for two different b values. We see that as b is decreased, the difference between H and h(s) becomes smaller and that the constant H value tends towards the hard-wall limit -73.7. Fixed parameters L = 1, EI = 1, a = 0.1, and Δ = 0.026.</figcaption></figure>

Thus, the helical angle $\theta$ appears as the crucial variable characterising the mechanical state of the system, and we would like to compute $\theta$ as a function of the applied load $T_{ext}$. We derive such a relation using the invariant $H = \frac{1}{2}EI\kappa^2 + \frac{1}{2}\frac{m_3^2}{GJ} + n_3$, see (32) with $\hat{u} = 0$, computed at points A and B. We first remark that we are in the case where $m_3'(s) = 0 \forall s$ that is $m_{3A} = m_{3B} = m_{3helix}$, see (59b). At point B, no bending moment is applied, $\kappa_B = 0$, and the internal tension is equal to the external force, $n_{3B} = T_{ext}$. We therefore have

$$H_B = \frac{1}{2} \frac{m_{3helix}^2}{GJ} + T_{ext}. \qquad (60)$$

At point A, the curvature, twist, and tension are given by (59) and we have

$$H_A = \frac{1}{2} EI \frac{\sin^4 \theta}{R^2} + \frac{1}{2} \frac{m_{3helix}^2}{GJ} + \frac{EI}{R^2} \frac{\sin^4 \theta}{\cos 2\theta}. \qquad (61)$$

Equating $H_A = H_B$ yields

$$T_{ext} = \frac{1}{2} \frac{EI}{R^2} \sin^4 \theta \left(1 + \frac{2}{\cos 2\theta}\right). \qquad (62)$$

We performed numerical shooting simulations for different rod's slenderness ratios $L/(2R) = 100...1000$, in which the end-to-end distance is fixed and the number of turns is varied from $n = 4$ up to $n \approx 10, 20$, or 50. We record the force $T_{ext}$ and the average value of the helical angle $\theta$, and verify both the $1/R^2$ scaling of the force and the approximation provided by (62), see Fig. 12.

## 7.8. The invariant in the presence of an external force field

In this section, we study the case in which an external conservative force field is applied to the rod. Manning and Bulman [54] introduced a soft-wall repulsion to deal with the wall-constrained buckling scenario presented in Fig. 13-a. This corresponds to the buckling of a planar Elastica, subject to an imposed displacement $\Delta$, constrained by an upper horizontal wall lying at a vertical distance $a$ from the clamps. In the references cited in Section 7.3, the authors used a frictionless hard-wall contact approach where

$$y(s) \le a \quad \forall s. \qquad (63)$$

As explained earlier, this contact condition is treated by introducing the KKT term $-F_c[a - y(s)]$ in the Lagrangian (50), the Lagrange multiplier $F_c$ corresponding to the contact force. In the case that the Elastica shape comprises a single arch and the contact with the upper wall is pointwise, the vertical component $n_y(s)$ of the internal force experiences a jump at $s = s_c$

$$n_y(s_c^+) - n_y(s_c^-) = F_c \qquad (64)$$

with symmetry imposing $s_c = 1/2$.

Manning and Bulman [54] replaced the contact condition (63) by a soft-wall (a.k.a. barrier) potential

$$W_{ext}(y) = \frac{b}{a - y(s)}, \qquad (65)$$
which is added to the total potential energy V of the rod, see (13), and therefore to the Lagrangian (50). In this model, the vertical component $n_y(s)$ no longer jumps at $s = 1/2$ but varies along the length of the rod

$$ \left(\frac{\partial \mathcal{L}}{\partial y'}\right)' = \frac{\partial \mathcal{L}}{\partial y} \Rightarrow n_y'(s) = \frac{\partial W_{\text{ext}}}{\partial y} = \frac{b}{(a - y(s))^2}. \qquad (66) $$

Intuitively, this soft-wall approach can be interpreted as if the wall and the Elastica would be electrostatically charged with like charges and would therefore repel each other from a distance. The central point of the soft-wall approach is that, as $b$ is made smaller, the computed equilibrium shape converges towards the shape obtained with the hard-wall contact condition (63) and

$$ n_y(L) - n_y(0) = b \int_0^L \frac{ds}{(a - y(s))^2} \xrightarrow{b \to 0} F_c. \qquad (67) $$

Moreover, the gap $\delta = a - y(1/2)$, see Fig. 13, also decreases to zero as $b \to 0$. Note that the horizontal component $n_x(s)$ of the internal force is invariant $n_x'(s) = 0$ $\forall s$ and set by the external compression force $P$, $n_x = -P$.

In this soft-wall approach, the arclength invariant reads

$$ H = h(s) - W_{\text{ext}}(y) \quad \text{with} \quad h(s) = \frac{1}{2} \theta'^2(s) - P \cos \theta + n_y(s) \sin \theta \qquad (68) $$

with $H'(s) = 0 \ \forall s$ but $h'(s) \neq 0$. We fix the parameter values

$$ L = 1, \quad EI = 1, \quad a = 0.1, \quad \Delta = 0.026 \qquad (69) $$

and numerically compute the equilibrium using the soft-wall approach with $b = 0.1$ and $b = 10^{-4}$ to see that the term $-W_{\text{ext}}(y)$ in (68) is necessary to achieve uniformity of $H$, see Fig. 13-c and d. Now in the case of hard-wall contact, we have explained in Section 6 that no extra term is required, and the formula for the invariant is indeed that of $h$ in (68). This can be understood by showing that the importance of $W_{\text{ext}}$ decreases as $b \to 0$, as can be seen when comparing Fig. 13-c and d. To make this more quantitative, we numerically compute

$$ E_{\text{wall}} = b \int_{0}^{1} \frac{ds}{a - y(s)} \qquad (70) $$

and plot it as $b$ is varied in Fig. 13-b. We see that $E_{\text{wall}}(b) \sim b^{2/3}$ for small $b$ values. This scaling can be understood in the following way. We consider the soft-wall solution as $b$ is decreased towards zero. In this limit, the denominators in the integrands of (67) and (70) are taking small values when $s$ is near $s_c = 1/2$, and therefore these integrals are dominated by the behaviour of the function $y(s)$ near $s = s_c$. A Taylor expansion yields

$$ y(s) = a - \delta + \frac{1}{2} \kappa_c (s - 1/2)^2 + \dots \qquad (71) $$

As $b \to 0$, $\delta$ and $\kappa_c$ converge towards the hard-wall limit, $\delta \to 0$ and $\kappa_c \to -1.54$ for parameters values in (69). Plugging this expansion for $y(s)$ into the integrals (67) and (70) we find

$$ \int_0^L \frac{ds}{(a - y(s))^2} = \frac{\pi}{\sqrt{8\kappa_c}} \delta^{-3/2} + O(\delta^{-1}) \qquad (72a) $$

$$ \int_0^L \frac{ds}{a - y(s)} = \frac{\pi}{\sqrt{2\kappa_c}} \delta^{-1/2} + O(\delta^0). \qquad (72b) $$

Considering the leading order of (67) when $b \to 0$, for which $F_c = 18.1$ for parameters values in (69), we see (72a) implies $b \sim \delta^{3/2}$ and consequently (72b) implies $E_{\text{wall}}(b) \sim b^{2/3}$. These scalings are reminiscent of Hertz contact scaling laws. In Hertz's contact, the indentation force $F$ causes a displacement $\delta$ (with $F \sim \delta^{3/2}$) and an elastic deformation energy $E_{\text{Hertz}}$ (with $E_{\text{Hertz}} \sim \delta$), while here the strength $b$ of the barrier potential creates a gap $\delta$ (with $b \sim \delta^{3/2}$) and a barrier energy $E_{\text{wall}}$ (with $E_{\text{wall}} \sim \delta$).

Finally, we remark that for every $b$ value, the curve $H(s)$ should be strictly flat. The size of the undulations seen in Fig. 13-c and -d is then a measure of the quality of the numerical solution, as explained in [55].

## 8. Dynamics

In this section, we show that the approach developed in Sections 4 and 5 can be straightforwardly extended to the dynamics of elastic rods. We introduce the angular velocity $\omega(s, t) = \omega_1 d_1 + \omega_2 d_2 + \omega_3 d_3$ and we have dynamical Darboux relations

$$ d_i(s, t) = \omega \times d_i \quad \text{with} \quad i = 1, 2, 3. \qquad (73a) $$

The three vectorial relations (73) are constraints for the three scalar components $\omega_1, \omega_2, \omega_3$ and can thereof be rewritten as

$$ \omega_1 = \dot{d}_2 \cdot d_3, \quad \omega_2 = \dot{d}_3 \cdot d_1, \quad \omega_3 = \dot{d}_1 \cdot d_2. \qquad (74) $$

To derive dynamical equations for the elastic rod, we complement the Lagrangian (18) with the linear and angular kinetic energies

$$ \mathcal{L}_d = \mathcal{L} - K - \lambda_{\omega_1} (\dot{d}_2 \cdot d_3 - \omega_1) - \lambda_{\omega_2} (\dot{d}_3 \cdot d_1 - \omega_2) - \lambda_{\omega_3} (\dot{d}_1 \cdot d_2 - \omega_3) \qquad (75a) $$

$$ \text{with } K = \frac{1}{2} \rho S \dot{r}^2 + \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 + \frac{1}{2} I_3 \omega_3^2, \qquad (75b) $$

where $I_1$ and $I_2$ are the moment of inertia of the cross-section about $d_1$ and $d_2$, and $I_3 = I_1 + I_2$ is the polar moment of inertia, about $d_3$. Note that we use the green colour to mark the dynamic terms. Finally, the Action for the dynamics of the elastic rod is

$$ A(r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3, \omega_1, \omega_2, \omega_3) = \int_{t_1}^{t_2} \int_{0}^{L} \mathcal{L}_d ds dt. \qquad (76) $$
## 8.1. Euler–Lagrange equations

We use the principle of least Action to obtain the equations for the dynamics of the elastic rod. Necessary conditions for the vanishing of the first variation of (76) are the following Euler–Lagrange equations. Eqs. (20b), (20c) stay unchanged and here also we identify the Lagrange multiplier $\lambda_r$ with the internal force vector $n = \lambda_r$. The linear momentum equation, (20a), is now

$$ \frac{\partial \mathcal{L}_d}{\partial \boldsymbol{r}} = \left( \frac{\partial \mathcal{L}_d}{\partial \boldsymbol{r}'} \right)' + \left( \frac{\partial \mathcal{L}_d}{\partial \boldsymbol{r}} \right)^* \Rightarrow \boldsymbol{n}'(s) + \boldsymbol{f}_{\text{ext}} = \rho S \ddot{\boldsymbol{r}}. \quad (77) $$

Furthermore, the derivation with regard to the $\omega_i$ unravels new constitutive relations

$$ \frac{\partial \mathcal{L}_d}{\partial \omega_i} = 0 \Rightarrow \lambda_{\omega_i} = I_i \omega_i \text{ with } i = 1, 2, 3 \quad (78a) $$

where we interpret the Lagrange multipliers $\lambda_{\omega_i}$ as the components of the angular momentum $\boldsymbol{\pi} = \lambda_{\omega_1} \mathbf{d}_1 + \lambda_{\omega_2} \mathbf{d}_2 + \lambda_{\omega_3} \mathbf{d}_3$, which we will now write $\pi_i$. Euler–Lagrange equations for the director basis ($\mathbf{d}_1, \mathbf{d}_2, \mathbf{d}_3$) are changed to

$$ \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_1} = \left( \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_1'} \right)' + \left( \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_1} \right)^* \Rightarrow \lambda_{u_3} \mathbf{d}_3' - \lambda_{u_2} \mathbf{d}_2' + \lambda_{u_1'} \mathbf{d}_1 + \lambda_r \nu_1 + \lambda_{11} \mathbf{d}_1 - \lambda_{12} \mathbf{d}_2 - \lambda_{31} \mathbf{d}_3 + \pi_2 \mathbf{d}_3 - \pi_3 \mathbf{d}_2 - \pi_3 \mathbf{d}_2 = 0 \quad (79a) $$

$$ \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_2} = \left( \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_2'} \right)' + \left( \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_2} \right)^* \Rightarrow \lambda_{u_1} \mathbf{d}_3' - \lambda_{u_3} \mathbf{d}_1' + \lambda_{u_1'} \mathbf{d}_2 + \lambda_r \nu_2 + \lambda_{22} \mathbf{d}_2 - \lambda_{23} \mathbf{d}_3 - \lambda_{12} \mathbf{d}_1 + \pi_3 \mathbf{d}_1 - \pi_1 \mathbf{d}_3 - \pi_1 \mathbf{d}_3 = 0 \quad (79b) $$

$$ \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_3} = \left( \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_3'} \right)' + \left( \frac{\partial \mathcal{L}_d}{\partial \mathbf{d}_3} \right)^* \Rightarrow \lambda_{u_2} \mathbf{d}_1' - \lambda_{u_1} \mathbf{d}_2' + \lambda_{u_2'} \mathbf{d}_1 + \lambda_r \nu_3 + \lambda_{33} \mathbf{d}_3 - \lambda_{31} \mathbf{d}_1 - \lambda_{23} \mathbf{d}_2 + \pi_1 \mathbf{d}_2 - \pi_2 \mathbf{d}_1 - \pi_2 \mathbf{d}_1 = 0. \quad (79c) $$

Analogous manipulations as in the static case lead to

$$ m_1'(s) - u_3 m_2 + u_2 m_3 - v_3 n_2 + v_2 n_3 = \dot{\pi}_1 + \omega_2 \pi_3 - \omega_3 \pi_2 \quad (80a) $$

$$ m_2'(s) - u_1 m_3 + u_3 m_1 - v_1 n_3 + v_3 n_1 = \dot{\pi}_2 + \omega_3 \pi_1 - \omega_1 \pi_3 \quad (80b) $$

$$ m_3'(s) - u_2 m_1 + u_1 m_2 - v_2 n_1 + v_1 n_2 = \dot{\pi}_3 + \omega_1 \pi_2 - \omega_2 \pi_1 \quad (80c) $$

and $\boldsymbol{w}' \cdot \mathbf{d}_i = w_i' - (\boldsymbol{w} \times \boldsymbol{u}) \cdot \mathbf{d}_i \quad \forall \boldsymbol{w}$, we obtain

$$ \boldsymbol{m}'(s, t) + \boldsymbol{r}'(s, t) \times \boldsymbol{n}(s, t) = \dot{\pi}(s, t). \quad (81) $$

## 8.2. Invariants in the dynamics case

The same symmetries as in the static case hold, except for the one of Section 5.4, basically because the force vector $\boldsymbol{n}$ is no longer constant. With two independent variables $s$ and $t$, the relations obtained from Noether's formulas are now 'density-flux' conservation identities rather than just explicit invariant quantities as in the static case [13].

### Coordinate translation

Just as in Section 5.2, in the absence of any external force field, $W_{\text{ext}} = 0$, the transformation $\tilde{\boldsymbol{r}} = \boldsymbol{r} + \boldsymbol{a}$ leaves the Lagrangian (75a) invariant. Consequently, (4a) yields

$$ \left[ \boldsymbol{a} \cdot \frac{\partial \mathcal{L}_d}{\partial \dot{\boldsymbol{r}}} \right]^* + \left[ \boldsymbol{a} \cdot \frac{\partial \mathcal{L}_d}{\partial \boldsymbol{r}'} \right]' = 0 \quad \forall \boldsymbol{a} \Rightarrow \boldsymbol{n}' = \rho S \ddot{\boldsymbol{r}}, \quad (82) $$

which is just the dynamics equation (77) when $W_{\text{ext}} = 0$. See also Eq. (2.12) in [13].

### Coordinate rotation

In the absence of any external force field or with a central force field, the rotation transformation (37) leaves the Lagrangian (75a) invariant. Consequently, (4a) yields

$$ \left[ \boldsymbol{b} \times \boldsymbol{r} \cdot \frac{\partial \mathcal{L}_d}{\partial \dot{\boldsymbol{r}}} + \sum_i \boldsymbol{b} \times \mathbf{d}_i \cdot \frac{\partial \mathcal{L}_d}{\partial \dot{\mathbf{d}}_i} \right]^* + \left[ \boldsymbol{b} \times \boldsymbol{r} \cdot \frac{\partial \mathcal{L}_d}{\partial \boldsymbol{r}'} + \sum_i \boldsymbol{b} \times \mathbf{d}_i \cdot \frac{\partial \mathcal{L}_d}{\partial \dot{\mathbf{d}}_i'} \right]' = 0 \quad (83a) $$

$$ [-\boldsymbol{b} \cdot \boldsymbol{r} \times \rho S \ddot{\boldsymbol{r}} - \boldsymbol{b} \cdot \boldsymbol{\pi}]^* + [\boldsymbol{b} \cdot \boldsymbol{r} \times \boldsymbol{n}(s) + \boldsymbol{b} \cdot \boldsymbol{m}]' = 0 \quad \forall \boldsymbol{b} \quad (83b) $$

$$ -[\boldsymbol{r} \times \rho S \ddot{\boldsymbol{r}} + \boldsymbol{\pi}]^* + [\boldsymbol{r} \times \boldsymbol{n}(s) + \boldsymbol{m}]' = \mathbf{0}, \quad (83c) $$

which may be obtained by combining (77) and (81) with $W_{\text{ext}} = W_{\text{ext}}(\boldsymbol{r}^2)$. The case $W_{\text{ext}} = 0$ is treated in [13, see Eq. (2.14)].

### Coordinate rotation around the section normal

In the presence or absence of an external force field, we now consider the transformation (43) in the case where $A_1 = A_2$, $B_1 = B_2$, $\hat{u}_1 = 0$, $\hat{u}_2 = 0$, $\hat{v}_1 = 0$, and $\hat{v}_2 = 0$. This transformation leaves the Lagrangian (75a) unchanged, and consequently, from (4a), we have

$$ \left[ \mathbf{d}_3 \times \mathbf{d}_1 \cdot \frac{\partial \mathcal{L}_d}{\partial \dot{\mathbf{d}}_1} + \mathbf{d}_3 \times \mathbf{d}_2 \cdot \frac{\partial \mathcal{L}_d}{\partial \dot{\mathbf{d}}_2} + u_2 \frac{\partial \mathcal{L}_d}{\partial \dot{u}_1} - u_1 \frac{\partial \mathcal{L}_d}{\partial \dot{u}_2} + v_2 \frac{\partial \mathcal{L}_d}{\partial \dot{v}_1} - v_1 \frac{\partial \mathcal{L}_d}{\partial \dot{v}_2} \right]^* + $$
$$ \left[ d_3 \times d_1 \cdot \frac{\partial \mathcal{L}_d}{\partial d'_1} + d_3 \times d_2 \cdot \frac{\partial \mathcal{L}_d}{\partial d'_2} + u_2 \frac{\partial \mathcal{L}_d}{\partial u'_1} - u_1 \frac{\partial \mathcal{L}_d}{\partial u'_2} + v_2 \frac{\partial \mathcal{L}_d}{\partial v'_1} - v_1 \frac{\partial \mathcal{L}_d}{\partial v'_2} \right]' = 0 \quad (84a) $$

$$ -[\pi_3(s,t)]^* + [m_3(s,t)]' = 0. \quad (84b) $$

Please note that this result holds in the case where $\hat{u}_3 \neq 0$ and $\hat{v}_3 \neq 0$. See also Eq. (4.5) in [13].

### Arclength translation

In the case the rigidities and the natural shape are uniform, see (26), the Lagrangian (75a) does not explicitly depend on $s$ and the conservation law (4b) holds along a dynamic trajectory. Recalling $q = (r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3)$, we obtain

$$ -[\pi \cdot u + \rho S \dot{r} \cdot r']^* + [H + K]' = 0 \quad (85) $$

with $H$ given by (28) and $K$ given by (75b). See also Eq. (3.2) in [13].

### Time translation

As the Lagrangian (75a) does not explicitly depend on $t$, the conservation law (4c) holds along a dynamic trajectory. Recalling $q = (r, d_1, d_2, d_3, u_1, u_2, u_3, v_1, v_2, v_3)$, we obtain

$$ -[K + V]^* + [m \cdot \omega + n \cdot \dot{r}]' = 0 \quad (86) $$

with $V$ given by (13) and $K$ given by (75b). See also Eq. (2.19) in [13]. The relation (86), once integrated over the entire rod, shows that the total mechanical energy

$$ \int_0^L (K + V) ds \quad (87) $$

is time-invariant in the case of conservative boundary conditions ($\dot{r}(0, L) = 0 = \omega(0, L)$), such as clamps or if the rod is circularly closed on itself, or free ends $m(0, L) = 0 = n(0, L)$.

## 9. Vibrations

Here, we show that vibration equations for elastic rods may be obtained through a variational approach and, as a result, we display the associated invariant. We illustrate this in the case of the vibrations of the planar Elastica but it can be straightforwardly generalised to the 3D case. For the sake of simplicity, we restrict to the naturally straight ($\hat{a}_i = 0$), inextensible ($v_3(s) = 1 \forall s, t$), unshearable ($v_1(s) = v_2(s) = 0 \forall s, t$) case. We start with the 2D version of (75a),

$$ \mathcal{L}_d^{2D} = \frac{1}{2} EI \kappa(s,t)^2 - \frac{1}{2} \rho S (\bar{x}^2 + \bar{y}^2) + \lambda_1(s,t) (x' - \cos \theta) + \lambda_2(s,t) (y' - \sin \theta) + \lambda_3(s,t) (\theta' - \kappa) \quad (88) $$

where the first term is the internal bending energy, the second is the translational kinetic energy (for simplicity, we do not consider here rotational inertia, $I_{1,2,3} = 0$), and the last three terms correspond to the kinematics constraints

$$ x' = \cos \theta, \quad y' = \sin \theta, \quad \theta' = \kappa. \quad (89) $$

Anticipating the interpretation of the Lagrange multipliers $\lambda_{1,2,3}$, we introduce the horizontal and vertical components of the internal force vector $\lambda_1 = n_x(s, t)$, $\lambda_2 = n_y(s, t)$, and the internal moment $\lambda_3 = m(s, t)$. The equations for the dynamics of the planar Elastica are then obtained by applying the least Action principle to

$$ \mathcal{A}^{2D}[q(s,t)] = \int_{t_1}^{t_2} \int_{0}^{L} \mathcal{L}_d^{2D} ds dt \quad (90) $$

with $q(s, t) = (x(s, t), y(s, t), \theta(s, t), \kappa(s, t))$, see Appendix B.

To study the small-amplitude vibrations around a pre-computed equilibrium $q_e$, we introduce the ansatz

$$ q(s,t) = q_e(s) + \epsilon \bar{q}(s) \cos \omega t, \quad n_{x,y}(s,t) = n_{xe,ye}(s) + \epsilon \bar{n}_{x,y}(s) \cos \omega t, \quad m(s,t) = m_e(s) + \epsilon \bar{m}(s) \cos \omega t \quad (91) $$

into (90) and keep terms up to order $\epsilon^2$. We obtain

$$ \mathcal{A}^{2D}[\bar{q}(s)] \approx \mathcal{A}_e^{2D} + \epsilon \mathcal{A}_1^{2D}[\bar{q}(s)] + \epsilon^2 \mathcal{A}_2^{2D}[\bar{q}(s)]. \quad (92) $$

The Action $\mathcal{A}_e^{2D}$ is here a 'constant', given quantity, and is therefore not subject to optimisation. The Action $\mathcal{A}_1^{2D}$ contains a $\cos \omega t$ term which makes it vanish when the time integration is either taken over a period, $t_2 = t_1 + 2\pi/\omega$, or over a long time interval. We are then only left with the Action $\mathcal{A}_2^{2D}$, which reads

$$ \mathcal{A}_2^{2D} = \int_{t_1}^{t_2} \cos^2 \omega t \int_0^L \mathcal{L}_{vib}^{2D} ds dt \quad (93a) $$

$$ \begin{aligned} \text{with } \mathcal{L}_{vib}^{2D}[\bar{q}(s)] = & \frac{1}{2} EI \bar{\kappa}^2 - \frac{1}{2} \rho S \omega^2 (\bar{x}^2 + \bar{y}^2) + \frac{1}{2} \bar{\theta}^2 (n_{xe} \cos \theta_e + n_{ye} \sin \theta_e) \\ & + \bar{n}_x (\bar{x}' + \bar{\theta} \sin \theta_e) + \bar{n}_y (\bar{y}' - \bar{\theta} \cos \theta_e) + \bar{m} (\bar{\theta}' - \bar{\kappa}). \end{aligned} \quad (93b) $$

Euler-Lagrange equations applied to $\mathcal{L}_{vib}^{2D}$ yield the classical vibration equations (B.4). We emphasise that the equilibrium solution $\theta_e(s)$ is a given function when working with the Lagrangian $\mathcal{L}_{vib}^{2D}$ and that a Noether arclength invariant will only exist in the case that the equilibrium shape of the beam is straight, $\theta'_e(s) = 0 \forall s$. In such a case the invariant reads

$$ H_{vib}^{2D} = \frac{\partial \mathcal{L}_{vib}^{2D}}{\partial \dot{q}'} \cdot \dot{q}' - \mathcal{L}_{vib}^{2D} \quad (94a) $$
$$ = \frac{1}{2} EI \bar{\theta}^2 - \bar{\theta} (\bar{n}_x \sin \theta_e - \bar{n}_y \cos \theta_e) + \frac{1}{2} \rho S \omega^2 (\bar{x}^2 + \bar{y}^2) - \frac{1}{2} \bar{\theta}^2 (n_{xe} \cos \theta_e + n_{ye} \sin \theta_e). \quad (94b) $$

In the case of a horizontal equilibrium ($\theta_e(s) = 0 \forall s$) of an Elastica under a compressive force $n_{xe} = -P$, $n_{ye} = 0$, the vibration equations simplify to the classical form

$$ EI \bar{y}''' + P \bar{y}'' - \rho S \omega^2 \bar{y} = 0 \quad (95) $$

and the invariant reads

$$ H_{vib}^{2D} = \frac{1}{2} EI \bar{y}''^2 + \frac{1}{2} \rho S \omega^2 \bar{y}^2 - \frac{1}{2} P \bar{y}^2 - EI \bar{y}' \bar{y}'' \quad (96) $$

a quantity that we have not found in the existing literature.

## 10. Discussion and conclusion

We have provided a comprehensive variational approach for elastic rods and ribbons where the continuous kinematic constraints play an important role, as this is only when they are explicitly included in the Lagrangian of the problem that Euler-Lagrange equations are straightforwardly derived. Moreover we showed that such a Lagrangian can be used to unravel conserved quantities or relations in the statics or dynamics of elastic rods and ribbons. Additionally, we have put forward an alternative Lagrangian approach where rotation constraints are automatically fulfilled, leading to more compact Euler-Lagrange equations and straightforward Noether invariants. Using Noether's 1918 theorem, we have recovered all known static and dynamic invariants and generalised the conditions under which they exist. More precisely, the arclength invariant, usually referred to as the Hamiltonian invariant, stands up to the adjunction of conservative external loading (e.g. gravity, electrostatics) and frictionless contact, but also holds in the case of Wunderlich, Sadowsky, and Ribext ribbon models. Furthermore, we illustrated the use of the arclength invariant in different 2D and 3D setups where it proves remarkably effective in providing key quantities or relations in difficult problems. In one such problem, the sliding sleeve, the invariant provides an easy explanation of the somewhat mysterious force applied to the rod at the entry/exit of the sleeve. Finally, we have introduced a variational approach to vibrations of elastic rods and computed a first integral to the vibration mode equation of a straight cantilever. It is worth noting that we have only studied cases where a space or time transformation leaves the Lagrangian unchanged, but that the hypotheses of Noether's theorem are less restrictive as they only require the Action to stay unchanged under space or time transformation.

Future work could include, among others, (i) the search for a more general invariant in the case of vibrations (that is, not restricted to the case of vibrations around a straight equilibrium), (ii) the case of contact in the presence of friction, where the invariant is sometimes increasing exponentially with $s$ instead of being constant, (iii) the search of other space and/or time transformations that keep the Action unchanged (see for example Eq. (12) in [56]) thereby providing new invariants, (iv) the question of whether the Kovalevskaya invariant (valid when $B_1 = 2B_2 = B_3$, see [57]) is associated with a space transformation that leaves the Lagrangian unchanged, (v) the search for clever uses of the two dynamics invariants (85), (86) either for validation of codes [55] for the dynamics of rods or to study their dynamical behaviour.

## CRediT authorship contribution statement

Sébastien Neukirch: Writing – review & editing, Writing – original draft, Validation, Methodology, Investigation, Formal analysis, Conceptualization. Florence Bertails-Descoubes: Writing – review & editing, Writing – original draft, Validation, Methodology, Investigation, Formal analysis, Conceptualization.

## Declaration of competing interest

No conflict of interest.

## Acknowledgements

We thank Aziz Hamdouni for discussions on Lie Groups and invariants, Olivier Thomas for discussions on vibrations of elastic rods, Thibaut Métivet, Victor Romero, and Émile Hohnadel for discussions and references on reduced $SO(3)$ parametrisations of rigid bodies. We are also grateful to Har meet Singh, Walter Drugan, and Joachim Linn for pointing out several references that we had missed. SN acknowledges funding from the Horizon Europe Framework Programme (Grant agreement HORIZON-101205621) and the French IDEX Univ. Paris Cité (Grant Emergence DynaVine).

## Appendix A. $SO(3)$ computations

### A.1. Reminder: Euler-Lagrange and Noether's theorem on $\mathbb{R}^n$

Let $q \in \mathbb{R}^n$ be the $n$ independent generalised coordinates of a mechanical system, $q' = \frac{\partial q}{\partial s} \in \mathbb{R}^n$ the corresponding generalised velocities, and $L(q, q')$ the Lagrangian functional depending on these generalised coordinates and velocities. Then the (classical) Euler-Lagrange equations read

$$ \frac{d}{ds} \frac{\partial L}{\partial q'} - \frac{\partial L}{\partial q} = 0. \quad (A.1) $$

The Euler-Lagrange equation (A.1) results from the stationarity of the Action

$$ \mathcal{A}(q, q') = \int_{0}^{L} \mathcal{L}(q, q')ds, \quad (A.2) $$
upon trajectories $q, q'$. The proof involves integration by parts and additionally yields the following conditions on the boundary terms,

$$ \left[ \frac{\partial \mathcal{L}}{\partial q'} \cdot \delta q \right]_0^L = 0, \qquad (A.3) $$

i.e.

$$ \frac{\partial \mathcal{L}}{\partial q'}(0) = 0 \quad \text{or} \quad \delta q(0) = 0 \qquad (A.4) $$

$$ \frac{\partial \mathcal{L}}{\partial q'}(L) = 0 \quad \text{or} \quad \delta q(L) = 0, \qquad (A.5) $$

which, for a mechanical system, can be interpreted as the vanishing of either the infinitesimal displacement $\delta q$ or the applied force $\frac{\partial \mathcal{L}}{\partial q'}$ at the considered boundary point.

Now, the classical Noether arclength invariant (5) can be exhibited by taking the total derivative of the Lagrangian with respect to $s$ [58]. At this stage it is important to mention that the Lagrangian should not depend explicitly on $s$. Using the chain rule, we have

$$ \begin{aligned} \frac{d\mathcal{L}}{ds} &= \frac{\partial \mathcal{L}}{\partial q} \cdot q' + \frac{\partial \mathcal{L}}{\partial q'} \cdot q'' \\ &= \frac{d}{ds} \left( \frac{\partial \mathcal{L}}{\partial q'} \right) \cdot q' + \frac{\partial \mathcal{L}}{\partial q'} \cdot q'' \quad \text{where (A.1) has been used to replace } \frac{\partial \mathcal{L}}{\partial q} \\ &= \frac{d}{ds} \left( \frac{\partial \mathcal{L}}{\partial q'} \cdot q' \right), \end{aligned} $$

and finally obtain

$$ \frac{d}{ds} \left( \mathcal{L} - \frac{\partial \mathcal{L}}{\partial q'} \cdot q' \right) = \frac{dH}{ds} = 0. $$

### A.2. Euler–Lagrange on $\mathbb{R}^n \times S\mathcal{O}(3)$

It is noteworthy that Eqs. (A.1) are only valid for generalised coordinates satisfying both $\frac{\partial}{\partial s}q = q'$ and the compatibility equation $\frac{\partial}{\partial s}\delta q = \delta q'$. This formula is thus applicable to translational degrees of freedom (such as positions), but not for rotational degrees of freedom, such as rotation matrices $R \in S\mathcal{O}(3)$, unless adding constraints to the Lagrangian formulation to maintain $R$ in $S\mathcal{O}(3)$, see Section 4 of this paper. Without any constraint, if one chooses $q = R$, then the compatibility condition on $\mathbb{R}^n$ is not valid on $S\mathcal{O}(3)$: one cannot swap derivation and perturbation for a 3D rotation. To do so, a specific compatibility condition, which expresses the geometric structure of $S\mathcal{O}(3)$ in the 3D Euclidean space, has to be satisfied. Thanks to this compatibility condition, stemming from an infinitesimal change of a rotation, new Euler–Lagrange equations can then be derived from the least Action principle. This compatibility condition is provided below, and the new Euler–Lagrange equations are derived just after.

As a preamble, we remind the reader that the tangent space of $S\mathcal{O}(3)$, i.e. the Lie algebra $so(3)$ associated to the Lie group $S\mathcal{O}(3)$, is exactly the space of skew-symmetric matrices $\{[w]_\times = \begin{pmatrix} 0 & -w_z & w_y \\ w_z & 0 & -w_x \\ -w_y & w_x & 0 \end{pmatrix}, w = \{w_x, w_y, w_z\} \in \mathbb{R}^3\}$, which is isomorphic to $\mathbb{R}^3$. We have $[w]_\times v = w \times v \quad \forall v \in \mathbb{R}^3$. The derivative of $R \in S\mathcal{O}(3)$ reads $R' = [w]_\times R$, and can be written equivalently as $R' = R[\bar{u}]_\times$ with $\bar{u} = R\bar{u}$ (i.e. $\bar{u}$ gives the local expression of $u$ in the $R$ frame). A condensed summary of properties of $S\mathcal{O}(3)$, along with proofs, can be found in [59, Appendix D].

**Property A.1.** Let $R \in S\mathcal{O}(3)$, such that

$$ R' = [w]_\times R \quad (\text{left Darboux equation}) \qquad (A.6) $$

$$ \text{or, equivalently } R' = R[\bar{u}]_\times \quad (\text{right Darboux equation}). \qquad (A.7) $$

We consider an infinitesimal perturbation (infinitesimal rotation) $\delta R$ of $R$. To ensure that the perturbed frame $R + \delta R$ remains in $S\mathcal{O}(3)$, the following compatibility condition should be verified,

$$ \delta u = (\delta \theta)' + \delta \theta \times u, \qquad (A.8) $$

or equivalently,

$$ \delta \bar{u} = (\delta \bar{\theta})' + \bar{u} \times \delta \bar{\theta}, \qquad (A.9) $$

where $\delta \theta$ and $\delta \bar{\theta}$ are defined through the relationship $\delta R = [\delta \theta]_\times R = R[\delta \bar{\theta}]_\times$, with $\delta \theta = R \delta \bar{\theta}$.

We recognise in (A.8) the compatibility condition found in [10, Eq. 3.51] by imposing that the left Darboux equation (A.6) with the perturbed Darboux vector $u + \delta u$ should be verified for the perturbed material frame $R + \delta R$ at first order. And we identify (A.9) as the compatibility condition written in [25, Eq. 6.3] in the context of rigid body dynamics, by imposing commutativity of the $\delta$ and $'$ operators and starting from the right Darboux equation (A.7). These two compatibility conditions are of course equivalent, and one can switch from one to the other by using the identity $(\delta \theta)' = R \delta \bar{u}$.

**Proof.** We follow Audoly and Pomeau [10, Sec. 3.6.1]'s proof, though in a more compact (matrix) form. From $u = R\bar{u}$, we first compute the perturbation of the Darboux vector $u$ as $\delta u = \bar{u} \delta R + R \delta \bar{u}$, leading to

$$ \delta u = \delta \theta \times u + R \delta \bar{u}. \qquad (A.10) $$
Then, on the one hand we impose that $(R + \delta R)' = [u + \delta u]_\times(R + \delta R)$, which, at first order, leads to $(\delta R)' = ([\delta \theta]_\times [u])_\times + [R \delta \bar{u}]_\times R$. On the other hand, we compute $(\delta R)' = ([\delta \theta]_\times R)' = ([\delta \theta']_\times + [\delta \theta]_\times [u])_\times R$. Imposing the commutativity condition $(\delta R)' = \delta(R')$ and using property [59, Appendix D.1, Eq I3] of the Lie bracket operator, we obtain by identification

$$ (\delta\theta)' = R\delta\bar{u}. \qquad (A.11) $$

Replacing this term in (A.10) finally yields the compatibility equation (A.8).

To obtain the local compatibility equation (A.9), one simply needs to expand (A.11) using the change of frame equation $\delta\theta = R\delta\bar{\theta}$. $\square$

**Property A.2.** Let $q \in \mathbb{R}^n$ and $R \in S\mathcal{O}(3)$, such that $R' = R[\bar{u}]_\times$. We decompose $R$ as $R = [d_1, d_2, d_3]$, where $d_j$ is the $j$th column of the matrix $R$. We denote by $e_j$ the $j$th canonical vector of $\mathbb{R}^3$. We consider the Lagrangian $\mathcal{L}(q, q', d_1, d_2, d_3, \bar{u})$. The Euler-Lagrange equations on $\mathbb{R}^n \times S\mathcal{O}(3)$ read

$$ \frac{d}{ds} \frac{\partial \hat{\mathcal{L}}}{\partial q'} - \frac{\partial \hat{\mathcal{L}}}{\partial q} = \mathbf{0} \quad (\text{classical Euclidean part}) \qquad (A.12) $$

$$ \text{and} \quad \frac{d}{ds} \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} + [\bar{u}]_\times \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} - \sum_{j=1}^{3} [e_j]_\times R^T \frac{\partial \hat{\mathcal{L}}}{\partial d_j} = \mathbf{0} \quad (\text{local form on S\mathcal{O}(3)}) \qquad (A.13) $$

$$ \text{or} \quad \frac{d}{ds} \left( R \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \right) - \sum_{j=1}^{3} [d_j]_{\times} \frac{\partial \hat{\mathcal{L}}}{\partial d_j} = \mathbf{0} \quad (\text{global form on S\mathcal{O}(3)}). \qquad (A.14) $$

**Proof.** Using properties of rotation matrices, we have $d'_j = [R\bar{u}]_\times d_j = R[\bar{u}]_\times R^T d_j$, and likewise, $\delta d_j = R[\delta\bar{\theta}]_\times R^T d_j$. The Action of the system reads $S = \int_0^L \mathcal{L}(q, q', d_j, \bar{u})$. We search for a stationary point of $S$, that is trajectories $q(s)$, $R(s)$ of the system such that $\delta S = 0$. We have

$$ \delta S = S(q + \delta q, q' + \delta q', d_j + \delta d_j, \bar{u} + \delta \bar{u}) - S(q, q', d_j, \bar{u}) \qquad (A.15) $$

$$ = \underbrace{\int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial q} \cdot \delta q + \frac{\partial \hat{\mathcal{L}}}{\partial q'} \cdot \delta q' + \int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial d_j} \cdot \delta d_j + \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \cdot \delta \bar{u}}_{\delta S_T} \qquad (A.16) $$

The first term $S_T$ is classical: by using the permutation $\delta(q') = (\delta q)'$ and integration by parts with the boundary conditions (A.4) and (A.5), one obtains

$$ \delta S_T = \int_0^L \left( \frac{d}{ds} \frac{\partial \hat{\mathcal{L}}}{\partial q'} - \frac{\partial \hat{\mathcal{L}}}{\partial q} \right) \cdot \delta q, \qquad (A.17) $$

which, for any perturbation $\delta q$, yields the classical Euler-Lagrange equations (A.12), similar to (A.1).

Now we compute the second term $\delta S_R = \underbrace{\int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial d_j} \cdot \delta d_j + \int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \cdot \delta \bar{u}}_{\delta S_R}$. On the one hand, using the compatibility equation (A.9), we have

$$ \begin{aligned} \delta S_{\bar{u}} &= \int_{0}^{L} \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \cdot \delta \bar{\theta}' + \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \cdot ([\bar{u}]_{\times} \delta \bar{\theta}) \\ &= \int_{0}^{L} \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \cdot \delta \bar{\theta}' - ([\bar{u}]_{\times} \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}}) \cdot \delta \bar{\theta} && \text{using properties of the mixed product} \\ &= - \int_{0}^{L} \left( \frac{d}{ds} \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} + [\bar{u}]_{\times} \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \right) \cdot \delta \bar{\theta} && \text{using integration by parts and cancelling boundary terms,} \end{aligned} $$

$$ \left[ \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}} \cdot \delta \bar{\theta} \right]_{0}^{L} = 0, \qquad (A.18) $$

thanks to the following boundary conditions,

$$ \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}}(0) = \mathbf{0} \quad \text{or} \quad \delta \bar{\theta}(0) = \mathbf{0} \qquad (A.19) $$

$$ \frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}}(L) = \mathbf{0} \quad \text{or} \quad \delta \bar{\theta}(L) = \mathbf{0}, \qquad (A.20) $$

meaning that either the infinitesimal rotation $\delta \bar{\theta}$ or the applied angular momentum $\frac{\partial \hat{\mathcal{L}}}{\partial \bar{u}}$ cancels out at a boundary point. On the other hand, we have

$$ \begin{aligned} \delta S_{d_j} &= \int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial d_j} \cdot \delta d_j \\ &= \int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial d_j} \cdot ([\delta \bar{\theta}]_\times R^T d_j) \\ &= \int_0^L \frac{\partial \hat{\mathcal{L}}}{\partial d_j} \cdot ([\delta \bar{\theta}]_\times e_j) && \text{where } e_j \text{ is the } j\text{th vector of the canonical base on } \mathbb{R}^3 \\ &= \int_0^L ([e_j]_\times R^T \frac{\partial \hat{\mathcal{L}}}{\partial d_j}) \cdot \delta \bar{\theta} && \text{using properties of the mixed product.} \end{aligned} $$
We finally obtain

$$ \delta S_{\bar{u}} = \int_{0}^{L} \left( -\frac{d}{ds} \frac{\partial \hat{L}}{\partial \bar{u}} - [\bar{u}]_{\times} \frac{\partial \hat{L}}{\partial \bar{u}} + \sum_{j=1}^{3} [e_j]_{\times} R^{\mathrm{T}} \frac{\partial \hat{L}}{\partial d_j} \right) \cdot \delta \bar{\theta}, $$

which, for any perturbation $\delta \bar{\theta}$, yields the Euler–Lagrange equations on $SO(3)$, in the local basis $R$,

$$ \frac{d}{ds} \frac{\partial \hat{L}}{\partial \bar{u}} + [\bar{u}]_{\times} \frac{\partial \hat{L}}{\partial \bar{u}} - \sum_{j=1}^{3} [e_j]_{\times} R^{\mathrm{T}} \frac{\partial \hat{L}}{\partial d_j} = 0, $$

or alternatively, in the world basis,

$$ \frac{d}{ds} \left( R \frac{\partial \hat{L}}{\partial \bar{u}} \right) - \sum_{j=1}^{3} [d_j]_{\times} \frac{\partial \hat{L}}{\partial d_j} = 0. \quad \square $$

Application to the statics of rods. We apply the formula above to our Lagrangian (19) expressed for rods using the general Reissner model, with $q = (r, \bar{v})$, $\hat{L}(r, \bar{v}, d_1, d_2, d_3, r', \bar{v}', \bar{u}) = V(r, \bar{v}, \bar{u}) + \lambda_r \cdot (r' - R \bar{v})$.

The translational part (A.12) related to $r$ directly provides

$$ \frac{d}{ds} \left( \frac{\partial \hat{L}}{\partial r'} \right) = \frac{\partial \hat{L}}{\partial r} \\ \text{i.e.} \quad \lambda_r' = \frac{\partial V}{\partial r} = -f_{\mathrm{ext}}, $$

where one recognises the force balance of the static Reissner equation (20a), with the internal force $n$ playing the role of the Lagrange multiplier $\lambda_r$. In addition, the translational part related to $\bar{v}$ yields

$$ \frac{\partial \hat{L}}{\partial \bar{v}} = -R^{\mathrm{T}} \lambda_r \\ \text{i.e.} \quad R A (\bar{v} - \hat{v}) = \lambda_r, $$

which is nothing else but the condensed (matrix) form of the shear-extension constitutive relations (20c).

As for the rotational part (A.14), since $\frac{\partial \hat{L}}{\partial \bar{u}} = B(\bar{u} - \hat{u}) = R^{\mathrm{T}} m = \bar{m}$ and $\forall j \frac{\partial \hat{L}}{\partial d_j} = -v_j \lambda_r$, it directly provides

$$ m' + \sum_{j} v_j d_j \times \lambda_r = 0 \\ \text{i.e.} \quad m' + r' \times \lambda_r = 0, $$

where one recognises the Reissner angular momentum equation (24).

### A.3. Noether's theorem on $\mathbb{R}^n \times SO(3)$

**Property A.3.** The Noether arclength invariant on $\mathbb{R}^n \times SO(3)$, corresponding to a translational symmetry of $\hat{L}$ along $s$, reads

$$ \forall s, \quad \hat{H}'(s) = 0 \quad \text{with} \quad \hat{H}(s) = \hat{H} = q' \cdot \frac{\partial \hat{L}}{\partial q'} + \bar{u} \cdot \frac{\partial \hat{L}}{\partial \bar{u}} - \hat{L}. \qquad (A.21) $$

**Proof.** Similarly to the Euclidean case on $\mathbb{R}^n$, the proof is based on the computation of the total derivative of the Lagrangian $\hat{L}$, where the classical gradient $\frac{\partial \hat{L}}{\partial q}$ on $\mathbb{R}^n$ is replaced with its expression extracted from the Euler–Lagrange equations on $\mathbb{R}^n \times SO(3)$ derived above. The specific part on $SO(3)$ is more involved though, and requires some specific manipulations exploiting the structure of $SO(3)$ in order to extract the projection of the rotated gradients $d_j \times \frac{\partial \hat{L}}{\partial d_j}$ onto the Darboux vector $u$.

As before, we decompose the rotation matrix $R$ using its column vectors $d_j, 1 \le j \le 3$. We start by expanding the total derivative of $\hat{L}(q, q', d_1, d_2, d_3, \bar{u})$ using the chain rule,

$$ \frac{d \hat{L}}{ds} = \underbrace{\frac{\partial \hat{L}}{\partial q} \cdot q' + \frac{\partial \hat{L}}{\partial q'} \cdot q''}_{A_T} + \underbrace{\sum_j \frac{\partial \hat{L}}{\partial d_j} \cdot d'_j + \frac{\partial \hat{L}}{\partial \bar{u}} \cdot \bar{u}'}_{A_R} \qquad (A.22) $$

From Appendix A.1 we already know that the classical translational term $A_T$ writes $A_T = (\frac{\partial \hat{L}}{\partial q'} \cdot q')'$. We now compute the rotational term $A_R$. We have

$$ d'_j = u \times d_j = [R \bar{u}]_{\times} d_j = R [\bar{u}]_{\times} e_j, $$

hence

$$ \sum_j \frac{\partial \hat{L}}{\partial d_j} \cdot d'_j = \sum_j d_j^{\mathrm{T}} \frac{\partial \hat{L}}{\partial d_j} \qquad (A.23a) $$

$$ = - \sum_j e_j^{\mathrm{T}} [\bar{u}]_{\times} R^{\mathrm{T}} \frac{\partial \hat{L}}{\partial d_j} \quad \text{as } [\bar{u}]_{\times} \text{ is skew-symmetric} \qquad (A.23b) $$

$$ = \bar{u} \cdot \left( \sum_j [e_j]_{\times} R^{\mathrm{T}} \frac{\partial \hat{L}}{\partial d_j} \right) \quad \text{using properties of the mixed product.} \qquad (A.23c) $$
One recognises the term provided by the projection of the local Euler–Lagrange equation (A.13) on $SO(3)$ onto the local Darboux vector $\bar{u}$,

$$ \bar{u} \cdot \left( \sum_j [e_j]_\times R^T \frac{\partial \mathcal{L}}{\partial d_j} \right) = \bar{u} \cdot \left( \frac{d}{ds} \frac{\partial \mathcal{L}}{\partial \bar{u}} + [\bar{u}]_\times \frac{\partial \mathcal{L}}{\partial \bar{u}} \right) \\ = \bar{u} \cdot \left( \frac{d}{ds} \frac{\partial \mathcal{L}}{\partial \bar{u}} \right). $$

We obtain

$$ A_R = \bar{u} \cdot \left( \frac{d}{ds} \frac{\partial \mathcal{L}}{\partial \bar{u}} \right) + \bar{u}' \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} \\ = \left( \bar{u} \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} \right)', $$

which finally yields

$$ \frac{d \mathcal{L}}{ds} = \left( q' \cdot \frac{\partial \mathcal{L}}{\partial q'} + \bar{u} \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} \right)', $$

hence the rise of an invariant $\dot{H}$,

$$ \frac{d \dot{H}}{ds} = 0 \quad \text{with} \quad \dot{H} = q' \cdot \frac{\partial \mathcal{L}}{\partial q'} + \bar{u} \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} - \mathcal{L}. $$

One of course ends up with the same result by projecting the global Euler–Lagrange equation (A.14) onto the global Darboux vector $u$. Note that a similar projection onto the Darboux vector $u$ is used in Section 6.1 to exhibit the $H$ invariant, albeit applied onto the (strong) angular momentum equation of rods. $\square$

**Property A.4.** We consider on the one hand the translational transformation $\tilde{q} = q + \epsilon \theta_q$, and on the other hand the rotational transformation $\tilde{R} = R + \epsilon [\theta_R]_\times = R + \epsilon R [\tilde{\theta}_R]_\times$, the latter resulting from an infinitesimal rotation of $R \in SO(3)$ around the vector $\theta_R = R \tilde{\theta}_R$. If these two transformations ($\tilde{q}, \tilde{R}$) leave the Lagrangian $\mathcal{L}$ unchanged, then there exists the following coordinate invariant on $\mathbb{R}^n \times SO(3)$,

$$ \forall s, \quad \dot{H}_{\theta}^{\prime}(s) = 0 \quad \text{with} \quad \dot{H}_{\theta}(s) = \dot{H}_{\theta} = \theta_q \cdot \frac{\partial \mathcal{L}}{\partial q'} + \bar{\theta}_R \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} = 0 \quad \forall s. \qquad (A.24) $$

**Proof.** As a preliminary step, we examine the classical case on $\mathbb{R}^n$, where the invariant is provided by Eq. (4a) reduced to the static case. We assume that the Lagrangian $\mathcal{L}(q, q')$ remains unchanged<fnref n="2" /> when the transformation $\tilde{q} = q + \epsilon \theta$, already introduced in (2a), is applied. On $\mathbb{R}^n$, we have $(\tilde{q})' = (\tilde{q}') = q' + \epsilon \theta'$. Performing a Taylor expansion at first order of $\mathcal{L}(q + \epsilon \theta, q' + \epsilon \theta')$ and using the symmetry $\mathcal{L}(q + \epsilon \theta, q' + \epsilon \theta') = \mathcal{L}(q, q')$, we obtain

$$ \frac{\partial \mathcal{L}}{\partial q} \cdot \theta + \frac{\partial \mathcal{L}}{\partial q'} \cdot \theta' = 0. \qquad (A.25) $$

Using the Euler–Lagrange equations (A.1) to substitute $\frac{\partial \mathcal{L}}{\partial q}$, we come up with the coordinate invariant

$$ \left( \theta \cdot \frac{\partial \mathcal{L}}{\partial q'} \right)' = 0. $$

Now we apply the same perturbation method to our Lagrangian $\dot{\mathcal{L}}(q, q', d_1, d_2, d_3, \bar{u})$ on $\mathbb{R}^n \times SO(3)$. We note that $\forall j \ \tilde{d}_j = d_j + \delta d_j$ with $\delta d_j = \epsilon R [\tilde{\theta}]_\times e_j$. Moreover, using the compatibility equation (A.9), we have $\tilde{u} = \bar{u} + \delta \bar{u}$ with $\delta \bar{u} = \epsilon (\tilde{\theta}' + \bar{u} \times \tilde{\theta})$.

Using the symmetry $\dot{\mathcal{L}}(\tilde{q}, \tilde{q}', \tilde{d}_1, \tilde{d}_2, \tilde{d}_3, \tilde{u}) = \dot{\mathcal{L}}(q, q', d_1, d_2, d_3, \bar{u})$ of the Lagrangian, we obtain the equality

$$ \underbrace{\frac{\partial \mathcal{L}}{\partial q} \cdot \theta_q + \frac{\partial \mathcal{L}}{\partial q'} \cdot \theta'_q}_{B_T} + \underbrace{\sum_j \frac{\partial \mathcal{L}}{\partial d_j} \cdot (R[\tilde{\theta}_R]_\times e_j) + \frac{\partial \mathcal{L}}{\partial \bar{u}} \cdot (\tilde{\theta}'_R + \bar{u} \times \tilde{\theta}_R)}_{B_R} = 0. \qquad (A.26) $$

We recognise $B_T = \dot{H}_{\theta_q}$ given in (A.25), which leads to the translational part of the invariant. We now rearrange the rotational term $B_R$. Similarly to computations performed in (A.23), by replacing $\bar{u}$ with $\tilde{\theta}_R$, the term $\frac{\partial \mathcal{L}}{\partial d_j} \cdot (R[\tilde{\theta}_R]_\times e_j)$ can be recast into $\tilde{\theta}_R \cdot [e_j]_\times R^T \frac{\partial \mathcal{L}}{\partial q}$. Projecting the Euler–Lagrange Eqs. (A.13) onto $\tilde{\theta}_R$, we can substitute this term and obtain

$$ \begin{aligned} B_R &= \tilde{\theta}_R \cdot \left( \frac{d}{ds} \frac{\partial \mathcal{L}}{\partial \bar{u}} + [\bar{u}]_\times \frac{\partial \mathcal{L}}{\partial \bar{u}} \right) + \tilde{\theta}'_R \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} + (\bar{u} \times \tilde{\theta}_R) \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} \\ &= \tilde{\theta}_R \cdot \left( \frac{d}{ds} \frac{\partial \mathcal{L}}{\partial \bar{u}} \right) + \tilde{\theta}'_R \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} \quad \text{using properties of mixed product} \\ &= \left( \tilde{\theta}_R \cdot \frac{\partial \mathcal{L}}{\partial \bar{u}} \right)' \cdot \square \end{aligned} $$

<footnote n="2">² Note that this symmetry is more restrictive than the original Noether assumption, which stipulates that the *Action* (not the Lagrangian) should remain unchanged.</footnote>
Application to the statics of rods. The invariant $\dot{H}$ applied to our Lagrangian $\dot{\mathcal{L}}$ gives in a straightforward manner

$$ \begin{aligned} \dot{\mathcal{H}} &= r' \cdot \lambda_r + \bar{u} \cdot \bar{m} - V \\ &= v \cdot n + u \cdot m - V, \end{aligned} \qquad (A.27) $$

which corresponds to formula (28), hence $\dot{\mathcal{H}} = H$.

The coordinate invariant $\dot{H}_{\theta}$ reads

$$ \begin{aligned} \dot{H}_{\theta} &= \theta_{\tau} \cdot \lambda_{r} + \bar{\theta}_{R} \cdot \bar{m} \\ &= \theta_{\tau} \cdot n + \theta_{R} \cdot m, \end{aligned} \qquad (A.28) $$

which reveals the specific common structure of all Noether coordinate invariants as a sum of two scalar products involving $n$ and $m$. Here again, we of course retrieve $\dot{H}_{\theta} = H_{\theta}$.

Interestingly, one can notice that the formula for the coordinate invariant remains the same regardless of the rod model used (Reissner or Kirchhoff), as $v$ is not directly involved in the formula.

## Appendix B. Planar Elastica statics, dynamics, and vibrations

The equations for the dynamics of the planar Elastica can be obtained as the Euler-Lagrange equations of the Action (90) together with the constraints (89). We write

$$ \frac{\partial \mathcal{L}_d^{2D}}{\partial q} = \left( \frac{\partial \mathcal{L}_d^{2D}}{\partial q'} \right)' + \left( \frac{\partial \mathcal{L}_d^{2D}}{\partial \dot{q}} \right)' \qquad (B.1) $$

with $q = (x, y, \theta, \kappa)$ and obtain

$$ x'(s,t) = \cos \theta, \quad y'(s,t) = \sin \theta, \quad EI\theta'(s,t) = m, \quad m'(s,t) = n_x \sin \theta - n_y \cos \theta, \quad n_x'(s,t) = \rho S \ddot{x}, \quad n_y'(s,t) = \rho S \ddot{y} \qquad (B.2) $$

The equations for the statics of the planar Elastica are then obtained by setting $\dot{x} = \ddot{x} = 0$ and $\dot{y} = \ddot{y} = 0$ in (B.2)

$$ x'_{e}(s) = \cos \theta_{e}, \quad y'_{e}(s) = \sin \theta_{e}, \quad EI\theta'_{e}(s) = m_{e}, \quad m'_{e}(s) = n_{xe} \sin \theta_{e} - n_{ye} \cos \theta_{e}, \quad n'_{xe}(s) = 0, \quad n'_{ye}(s) = 0. \qquad (B.3) $$

And finally, the equations for the vibrations of the planar Elastica are obtained by considering the ansatz (91), which corresponds to small-amplitude vibrations around the nonlinear equilibrium $q_e(s)$. This ansatz is injected into (B.2) while bearing in mind that (B.3) is fulfilled and that we only keep $O(\epsilon)$ terms. We obtain

$$ \bar{x}' = -\bar{\theta} \sin \theta_e, \quad \bar{y}' = \bar{\theta} \cos \theta_e, \quad EI\bar{\theta}' = \bar{m}, \quad \bar{m}' = \bar{n}_x \sin \theta_e - \bar{n}_y \cos \theta_e + \bar{\theta} (n_{xe} \cos \theta_e + n_{ye} \sin \theta_e) \qquad (B.4a) $$

$$ \bar{n}'_{x} = -\rho S \omega^2 \bar{x}, \quad \bar{n}'_{y} = -\rho S \omega^2 \bar{y}. \qquad (B.4b) $$

## Data availability

Data will be made available on request.

## References

[1] E. Noether, Über die Bildung des Formensystems der ternären biquadratischen Form, J. Fur Die Reine Und Angew. Math. (134) (1908) 23–90.

[2] A. Dick, Emmy Noether 1882-1935, Birkhauser, 1981, http://dx.doi.org/10.1007/978-1-4684-0535-4, Translated by H. I. Blocher.

[3] E. Noether, Invariante variationsprobleme, Nachrichten Von der Ges. der Wiss. Zu Gött. Mathematisch-Physikalische Kl. (1918) 235–257.

[4] E. Reissner, One one-dimensional large-displacement finite-strain beam theory, Stud. Appl. Math. 52 (2) (1973) 87–95, http://dx.doi.org/10.1002/sapm197352287.

[5] J. Simo, L. Vu-Quoc, A three-dimensional finite-strain rod model. part II: Computational aspects, Comput. Methods Appl. Mech. Engrg. 58 (1) (1986) 79–116, http://dx.doi.org/10.1016/0045-7825(86)90079-4.

[6] M.A. Crisfield, G. Jelenić, Objectivity of strain measures in the geometrically exact three-dimensional beam theory and its finite-element implementation, Proc. R. Soc. Lond. Ser. A 455 (1983) (1998) 1125–1147, http://dx.doi.org/10.1098/rspa.1999.0352.

[7] O.M. O'Reilly, Some perspectives on Eshelby-like forces in the elastica arm scale, Proc. R. Soc. A: Math. Phys. Eng. Sci. 471 (2174) (2015) 20140785, http://dx.doi.org/10.1098/rspa.2014.0785.

[8] D. Dichmann, Y. Li, J. Maddocks, Hamiltonian formulations and symmetries in rod mechanics, in: J. Mesirov, K. Schulten, D. Sumners (Eds.), Mathematical Approaches To Biomolecular Structure and Dynamics, in: The IMA Volumes in Mathematics and Its Applications, vol. 82, Springer Verlag, 1996, pp. 71–113, http://dx.doi.org/10.1007/978-1-4612-4066-2_6.

[9] V. Sonneville, A. Cardona, O. Brüls, Geometrically exact beam finite element formulated on the special Euclidean group, Comput. Methods Appl. Mech. Engrg. 268 (2014) 451–474, http://dx.doi.org/10.1016/j.cma.2013.10.008.

[10] B. Audoly, Y. Pomeau, Elasticity and Geometry: from Hair Curls to the Nonlinear Response of Shells, Oxford University Press, 2010, http://dx.doi.org/10.1142/9789812792778_0001.

[11] P. Jung, S. Leyendecker, J. Linn, M. Ortiz, A discrete mechanics approach to the Cosserat rod theory—Part 1: static equilibria, Internat. J. Numer. Methods Engrg. 85 (1) (2011) 31–60, http://dx.doi.org/10.1002/nme.2950.

[12] F. Boyer, V. Lebastard, F. Candelier, F. Renda, M. Alamir, Statics and dynamics of continuum robots based on Cosserat rods and optimal control theories, IEEE Trans. Robot. 39 (2) (2023) 1544–1562, http://dx.doi.org/10.1109/TRO.2022.3226112.

[13] J.H. Maddocks, D.J. Dichmann, Conservation laws in the dynamics of rods, J. Elasticity 34 (1994) 83–96, http://dx.doi.org/10.1007/BF000424247.

[14] D.J. Steigmann, M.G. Faulkner, Variational theory for spatial rods, J. Elasticity 33 (1) (1993) 1–26, http://dx.doi.org/10.1007/BF00042633.

[15] A.E.H. Love, A Treatise on the Mathematical Theory of Elasticity, fourth ed., Dover Publications, New York, 1944.

[16] S. Kehrabaum, J.H. Maddocks, Elastic rods, rigid bodies, quaternions and the last quadrature, Philos. Trans. R. Soc. Lond. Ser. A, Math. Phys. Sci. 355 (1732) (1997) 2117–2136, http://dx.doi.org/10.1098/rsta.1997.0113.

[17] O.M. O'Reilly, A material momentum balance law for rods, J. Elasticity 86 (2) (2007) 155–172, http://dx.doi.org/10.1007/s10659-006-9089-6.
[18] O.M. O'Reilly, Modeling Nonlinear Problems in the Mechanics of Strings and Rods: The Role of the Balance Laws, Interaction of Mechanics and Mathematics, Springer, 2017, http://dx.doi.org/10.1007/978-3-319-50598-5.

[19] P.J. Olver, Conservation laws in elasticity, Arch. Ration. Mech. Anal. 85 (2) (1984) 131-160, http://dx.doi.org/10.1007/BF00281448.

[20] J.R. Rice, Conserved integrals and energetic forces, in: B.A. Bilby, K.J. Miller, J.R. Willis (Eds.), Fundamentals of Deformation and Fracture, in: Eshelby Memorial Symposium Sheffield 2-5 April 1984, Cambridge University Press, 1985, pp. 33-56.

[21] M. Ben Amar, J.R. Rice, Exact results with the J-integral applied to free-boundary flows, J. Fluid Mech. 461 (2002) 321-341, http://dx.doi.org/10.1017/S0022112002008418.

[22] H. Singh, J.A. Hanna, Pseudomomentum: origins and consequences, Z. Angew. Math. Phys. 72 (3) (2021) 122, http://dx.doi.org/10.1007/s00033-021-01507-9.

[23] W. Peng, X. Yun, L. Yu-Lu, Noether symmetry and conserved quantities of the analytical dynamics of a Cosserat thin elastic rod, Chin. Phys. B 22 (10) (2013) 104503, http://dx.doi.org/10.1088/1674-1056/22/10/104503.

[24] N. Clauvelin, B. Audoly, S. Neukirch, Matched asymptotic expansions for twisted elastic knots: a self-contact problem with non-trivial contact topology, J. Mech. Phys. Solids 57 (9) (2009) 1623-1656, http://dx.doi.org/10.1016/j.jmps.2009.05.004.

[25] T. Lee, M. Leok, N.H. McClamroch, Global Formulations of Lagrangian and Hamiltonian Dynamics on Manifolds, Interaction of Mechanics and Mathematics, Springer, 2018, http://dx.doi.org/10.1007/978-3-319-56953-6.

[26] I. Romero, C.G. Gebhardt, Variational principles for nonlinear Kirchhoff rods, Acta Mech. 231 (2) (2020) 625-647, http://dx.doi.org/10.1007/s00707-019-02562-0.

[27] S.S. Antman, Nonlinear Problems of Elasticity, second ed., Springer-Verlag, New York, 2004, http://dx.doi.org/10.1007/0-387-27649-1.

[28] H. Elettro, P. Grandgeorge, S. Neukirch, Elastocapillary coiling of an elastic rod inside a drop, J. Elasticity 127 (2) (2017) 235-247, http://dx.doi.org/10.1007/s10659-016-9611-4.

[29] E.L. Starostin, G.H.M. van der Heijden, Equilibrium shapes with stress localisation for inextensible elastic Möbius and other strips, J. Elasticity 119 (1) (2015) 67-112, http://dx.doi.org/10.1007/s10659-014-9495-0.

[30] A. Borum, Manipulation and Mechanics of Thin Elastic Objects (Ph.D. thesis), University of Illinois at Urbana-Champaign, 2018.

[31] S. Neukirch, B. Audoly, A convenient formulation of Sadowsky's model for elastic ribbons, Proc. R. Soc. A 477 (2255) (2021) 20210548, http://dx.doi.org/10.1098/rspa.2021.0548.

[32] B. Audoly, G. van der Heijden, Analysis of cone-like singularities in twisted elastic ribbons, J. Mech. Phys. Solids 171 (2022) 105131, http://dx.doi.org/10.1016/j.jmps.2022.105131.

[33] R. Charrondière, S. Neukirch, F. Bertails-Descoubes, MERCI: Mixed curvature-based elements for computing equilibria of thin elastic ribbons, ACM Trans. Graph. 43 (5) (2024) art. 160, http://dx.doi.org/10.1145/3674502.

[34] M.A. Dias, B. Audoly, 'Wunderlich, meet Kirchhoff: A general and unified description of elastic ribbons and thin rods, J. Elasticity 119 (1-2) (2015) 49-66, http://dx.doi.org/10.1007/s10659-014-9487-0.

[35] R. Charrondière, F. Bertails-Descoubes, S. Neukirch, V. Romero, Numerical modeling of inextensible elastic ribbons with curvature-based elements, Comput. Methods Appl. Mech. Engrg. 364 (2020) 112922, http://dx.doi.org/10.1016/j.cma.2020.112922.

[36] R.T. Rockafellar, Lagrange multipliers and optimality, SIAM Rev. 35 (2) (1993) 183-238, http://dx.doi.org/10.1137/1035044.

[37] G. Kirchhoff, Über das Gleichgewicht und die Bewegung eines unendlich dünnen elastischen Stabes, J. Reine Angew. Math. 56 (1859) 285-313.

[38] A. Lubinski, W. Althouse, Helical buckling of tubing sealed in packers, J. Pet. Technol. 14 (06) (1962) 655-670, http://dx.doi.org/10.2118/178-PA.

[39] H. Chai, The post-buckling response of a bi-laterally constrained column, J. Mech. Phys. Solids 46 (7) (1998) 1155-1181, http://dx.doi.org/10.1016/S0022-5096(98)00004-0.

[40] G. Domokos, P. Holmes, B. Royce, Constrained Euler buckling, J. Nonlinear Sci. 7 (3) (1997) 281-314, http://dx.doi.org/10.1007/BF02678090.

[41] B. Roman, A. Pocheau, Buckling cascade of thin plates: Forms, constraints and similarity, Eur. Lett. (EPL) 46 (5) (1999) 602-608, http://dx.doi.org/10.1209/epl/i1999-00306-3.

[42] H. Singh, Planar equilibria of an elastic rod wrapped around a circular capstan, J. Elasticity 151 (2022) 321-346, http://dx.doi.org/10.1007/s10659-022-09939-8.

[43] D. Bigoni, F.D. Corso, F. Bosi, D. Misseroni, Eshelby-like forces acting on elastic structures: Theoretical and experimental proof, Mech. Mater. 80 (2015) 368-374, http://dx.doi.org/10.1016/j.mechmat.2013.10.009.

[44] F. Dal Corso, D. Misseroni, N.M. Pugno, A.B. Movchan, N.V. Movchan, D. Bigoni, Serpentine locomotion through elastic energy release, J. R. Soc. Interface 14 (130) (2017) 20170055, http://dx.doi.org/10.1098/rsif.2017.0055.

[45] A. Cazzolli, F. Dal Corso, The elastica sling, Eur. J. Mech. A Solids 105 (2024) 105273, http://dx.doi.org/10.1016/j.euromechsol.2024.105273.

[46] F. Bosi, D. Misseroni, F. Dal Corso, D. Bigoni, An elastica arm scale, Proc. R. Soc. A: Math. Phys. Eng. Sci. 470 (2169) (2014) 20140232, http://dx.doi.org/10.1098/rspa.2014.0232.

[47] J. Keller, F. Niordson, The tallest column, J. Math. Mech. 16 (5) (1966) 433-446.

[48] C. Bustamante, Z. Bryant, S.B. Smith, Ten years of tension: single-molecule DNA mechanics, Nature 421 (6921) (2003) 423-427, http://dx.doi.org/10.1038/nature01405.

[49] T.R. Strick, J.-F. Allemand, D. Bensimon, A. Bensimon, V. Croquette, The elasticity of a single supercoiled DNA molecule, Science 271 (5257) (1996) 1835-1837, http://dx.doi.org/10.1126/science.271.5257.1835.

[50] J.W.S. Hearle, The 20th-century revolution in textile machines and processes. Part 2: Textured yarns and other technologies, Ind. Archaeol. Rev. 36 (1) (2014) 32-47, http://dx.doi.org/10.1179/0309072814Z.00000000028.

[51] B.D. Coleman, D. Swigon, Theory of supercoiled elastic rings with self-contact and its application to DNA plasmids, J. Elasticity 60 (2000) 173-221, http://dx.doi.org/10.1023/A:1010911113919.

[52] G.H.M. van der Heijden, S. Neukirch, V.G.A. Goss, J.M.T. Thompson, Instability and self-contact phenomena in the writhing of clamped rods, Int. J. Mech. Sci. 45 (1) (2003) 161-196, http://dx.doi.org/10.1016/S0020-7403(02)00183-2.

[53] J.M.T. Thompson, G.H.M. van der Heijden, S. Neukirch, Supercoiling of DNA plasmids: mechanics of the generalized ply, Proc. R. Soc. Lond. A 458 (2002) 959-985, http://dx.doi.org/10.1098/rspa.2001.0901.

[54] R.S. Manning, G.B. Bulman, Stability of an elastic rod buckling into a soft wall, Proc. R. Soc. A: Math. Phys. Eng. Sci. 461 (2060) (2005) 2423-2450, http://dx.doi.org/10.1098/rspa.2005.1458.

[55] F. Bertails-Descoubes, T. Metivet, S. Neukirch, V. Romero, Kirchhoff Knots without friction: an analytical model for loose and semi-tight regimes, 2025, Preprint.

[56] R. Kienzler, G. Herrmann, On material forces in elementary beam theory, J. Appl. Mech. 53 (3) (1986) 561-564, http://dx.doi.org/10.1115/1.3171811.

[57] A. Goriely, M. Nizette, Kovalevskaya rods and kovalevskaya waves, Regul. Chaotic Dyn. 5 (1) (2000) 95-106, http://dx.doi.org/10.1070/RD2000v005n01ABEH000126.

[58] J. Navarro, Mujeres Matemáticas, El mundo es matemático, RBA Libros, 2011, ISBN: 978-84-473-7440-3.

[59] R. Casati, Quelques Contributions à la Modélisation Numérique de Structures Élancées pour l'Informatique Graphique (Ph.D. thesis), Grenoble Alpes Universités, 2015, URL https://theses.hal.science/tel-01502808.