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Geometric conditions for the positive definiteness of the second variation in one-dimensional problems

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Published 6 April 2017 © 2017 IOP Publishing Ltd & London Mathematical Society
, , Citation Thomas Lessinnes and Alain Goriely 2017 Nonlinearity 30 2023 DOI 10.1088/1361-6544/aa6448

0951-7715/30/5/2023

Abstract

Given a functional for a one-dimensional physical system, a classical problem is to minimize it by finding stationary solutions and then checking the positive definiteness of the second variation. Establishing the positive definiteness is, in general, analytically untractable. However, we show here that a global geometric analysis of the phase-plane trajectories associated with the stationary solutions leads to generic conditions for minimality. These results provide a straightforward and direct proof of positive definiteness, or lack thereof, in many important cases. In particular, when applied to mechanical systems, the stability or instability of entire classes of solutions can be obtained effortlessly from their geometry in the phase plane, as illustrated on a problem of a mass hanging from an elastic rod with intrinsic curvature.

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Footnotes

  • The superscripts D and N used to denote the space of admissible perturbations ${{\mathcal{C}}^{\text{D}}}$ and ${{\mathcal{C}}^{\text{N}}}$ respectively refer to the Dirichlet and Neumann boundary conditions that naturally arise when minimising $\mathcal{E}$ with respectively fixed and free boundaries.

  • Finding stationary functions for the functional $\mathcal{E}$ amounts to solving the boundary value problem (15). We had noted that in the case of fixed boundaries, this BVP has Dirichlet boundary conditions while for free boundaries, the BVP has Neumann boundary conditions. Something similar happens here. For fixed boundaries, the stability can be assessed by solving a Sturm–Liouville problem with Dirichlet boundary conditions while for free boundaries, the S–L problem has Neumann boundary conditions.

  • This is easily proven by considering  $\begin{array}{*{35}{l}} \mathcal{S}{{\theta}^{\prime}} & = & -{{\theta}^{\prime \prime \prime}}-\frac{{{\text{d}}^{2}}V}{\text{d}{{\theta}^{2}}}{{|}_{\theta (s)}}\,{{\theta}^{\prime}}\overset{{}}{{\overset{(16)}{{=}}\,}}\,{{\left(\frac{\text{d}V}{\text{d}\theta}{{|}_{\theta (s)}}\right)}^{\prime}}-\frac{{{\text{d}}^{2}}V}{\text{d}{{\theta}^{2}}}{{|}_{\theta (s)}}\,{{\theta}^{\prime}}=0. \end{array}$

  • If P  >  0, then η starts at $\left({{T}_{0}},\,{{P}_{0}}\right)$ and end at $(T,\,P)$ . If P  <  0, then η starts at $(T,\,P)$ and end at $\left({{T}_{0}},\,{{P}_{0}}\right)$ . Either way the length formulae below hold.

  • In this case, we can substitute variable in the integral appearing in the RHS of (45) according to $\text{d}t=\frac{\text{d}\theta}{\text{Sign}\left[{{\theta}^{\prime}}(s)\right]\sqrt{2}\,\sqrt{E-V(\theta )}}$ .

  • These two functions can not be linearly dependent since ${{\theta}^{\prime}}(d)\ne 0$ while $\mu (d;d)=0$ .

  • 10 

    That is assuming that U  >  0, if U  <  0 the solution starts at $\arccos [v-e]$ .

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