Neural networks for the approximation of Euler's elastica

TL;DR

This work tackles the efficient approximation of planar Euler's elastica static equilibria under varying boundary conditions using neural networks trained on discretised solutions. It develops two strategies: a discrete network that maps boundary data to discrete node values and a continuous network that delivers smooth curves along the beam, with angular parametrisations explored to enforce inextensibility. The approach yields high-accuracy predictions (sub-10^−6 errors in favorable settings) and substantial computational speedups (over 100–260×) compared to traditional solvers, validating the potential of data-driven surrogates for variational elastica problems. These results pave the way for fast, scalable simulations in applications like flexible medical devices and cables, while highlighting directions for embedding physical constraints directly into training and extending to higher dimensions and dynamics.

Abstract

Euler's elastica is a classical model of flexible slender structures, relevant in many industrial applications. Static equilibrium equations can be derived via a variational principle. The accurate approximation of solutions of this problem can be challenging due to nonlinearity and constraints. We here present two neural network based approaches for the simulation of this Euler's elastica. Starting from a data set of solutions of the discretised static equilibria, we train the neural networks to produce solutions for unseen boundary conditions. We present a $\textit{discrete}$ approach learning discrete solutions from the discrete data. We then consider a $\textit{continuous}$ approach using the same training data set, but learning continuous solutions to the problem. We present numerical evidence that the proposed neural networks can effectively approximate configurations of the planar Euler's elastica for a range of different boundary conditions.

Figures (5)

• Figure 1: Comparison over test trajectories for $\mathbf{q}$ and $\mathbf{q}^{\prime}$ for the discrete network $q_{\boldsymbol{\rho}}^{\textrm{d}}$ tested on the both-ends data set with $80\% \-- 10\%\-- 10\%$ splitting into training, validation, and test sets. The mean squared error on the test set equals $4.009 \cdot 10^{-7}$. For presentation purposes, only 10 randomly selected trajectories are considered in the first two plots.
• Figure 2: Comparison over test trajectories for $\mathbf{q}$ and $\mathbf{q}^{\prime}$ for the discrete network $q_{\boldsymbol{\rho}}^{\textrm{d}}$ tested on the both-ends + right-end data set with $80\% \-- 10\%\-- 10\%$ splitting into training, validation, and test sets. The mean squared error on the test set equals $7.854 \cdot 10^{-8}$. For presentation purposes, only 10 randomly selected trajectories are considered in the first two plots.
• Figure 3: Comparison over test trajectories for $\mathbf{q}$ and $\mathbf{q}^{\prime}$ for the continuous network $q_{\boldsymbol{\rho}}^{\textrm{c}}$ tested on the both-ends data set with $80\% \-- 10\% \-- 10\%$ splitting into training, validation, and test sets. The mean squared error on the test set equals $4.354 \cdot 10^{-6}$. For presentation purposes, only 10 randomly selected trajectories are considered in the first two plots.
• Figure 4: Comparison over test trajectories for $\mathbf{q}$ and $\mathbf{q}^{\prime}$, for the case $\theta_{\boldsymbol{\rho}}^{\textrm{c}}$ is modelled as in Equation \ref{['eq:noBCs']}, with $80\% \-- 10\%\-- 10\%$ splitting of the both-ends data set into training, validation, and test sets. The mean squared error on the test set equals $6.289 \cdot 10^{-6}$. For presentation purposes, only 10 randomly selected trajectories are considered in the first two plots.
• Figure 5: Comparison over test trajectories for $\mathbf{q}$ and $\mathbf{q}^{\prime}$, for the case $\theta_{\boldsymbol{\rho}}^{\textrm{c}}$ is modelled as in Equation \ref{['eq:withBCs']}, with $80\% \-- 10\% \-- 10\%$ splitting of the both-ends data set into training, validation, and test sets. The mean squared error on the test set equals $4.385 \cdot 10^{-6}$. For presentation purposes, only 10 randomly selected trajectories are considered in the first two plots.