Geometric Integrators for Higher-Order Variational Systems and Their Application to Optimal Control

Abstract

Numerical methods that preserve geometric invariants of the system, such as energy, momentum or the symplectic form, are called geometric integrators. In this paper we present a method to construct symplectic-momentum integrators for higher-order Lagrangian systems. Given a regular higher-order Lagrangian $L:T^{(k)}Q\rightarrow {\mathbb {R}}$ with $k\ge 1$, the resulting discrete equations define a generally implicit numerical integrator algorithm on $T^{(k-1)}Q\times T^{(k-1)}Q$ that approximates the flow of the higher-order Euler–Lagrange equations for L. The algorithm equations are called higher-order discrete Euler–Lagrange equations and constitute a variational integrator for higher-order mechanical systems. The general idea for those variational integrators is to directly discretize Hamilton’s principle rather than the equations of motion in a way that preserves the invariants of the original system, notably the symplectic form and, via a discrete version of Noether’s theorem, the momentum map. We construct an exact discrete Lagrangian $L_d^e$ using the locally unique solution of the higher-order Euler–Lagrange equations for L with boundary conditions. By taking the discrete Lagrangian as an approximation of $L_d^e$, we obtain variational integrators for higher-order mechanical systems. We apply our techniques to optimal control problems since, given a cost function, the optimal control problem is understood as a second-order variational problem.

Appendix: A Technical Result for Sect. 2

Let E be the kernel of g, where $g=(g_0,\ldots ,g_{k-1}):C^{l}([0,1],{\mathbb {R}}^{n})\rightarrow ({\mathbb {R}}^{n})^{k}$ and $g_{j}[\cdot ]=\langle \langle b_{j}^{[k]},\cdot \rangle \rangle$. In the context of Sect. 2.5, E is the tangent space of the constraint set defined using the linear constraints $g_{j}$, and l is either 0 or k.

In this Appendix we show that the orthogonal complement of E is the space F of ${\mathbb {R}}^{n}$-valued polynomials of degree at most $k-1$,

$$\begin{aligned} F=\hbox {span}_{{\mathbb {R}}^{n}}(b_{0}^{[k]},\ldots ,b_{k-1}^{[k]})=\{c^{j}b_{j}^{[k]}|c^{0},\ldots ,c^{k-1}\in {\mathbb {R}}^{n}\}, \end{aligned}$$

where $b_j^{[k]}$, $j=0,\ldots ,k-1$, is a basis of the space of real-valued polynomials of degree at most $k-1$ consisting of orthonormal polynomials on [0, 1].

Lemma 6.1

$F=E^{\perp }$, where the orthogonal complement is taken with respect to the inner product $\llbracket ,\rrbracket$ in $C^{l}([0,1],{\mathbb {R}}^{n})$.

Proof

We will prove that E and F are orthogonal (with zero intersection) and that their sum is the whole space $C^{l}([0,1],{\mathbb {R}}^{n})$.

Let $e\in E$ and $c^{j}b_{j}^{[k]}\in F$.

$$\begin{aligned} \llbracket c^{j}b_{j}^{[k]},e\rrbracket&=\int _{0}^{1}(c^{j}b_{j}^{[k]}(u))\cdot e(u)\,\mathrm{{d}}u=\sum _{i=1}^{n}\int _{0}^{1}c_{i}^{j}b_{j}^{[k]}(u)e_{i}(u)\mathrm{{d}}u\\&=c^{j}\cdot \left( \int _{0}^{1}b_{j}^{[k]}(u)e_{1}(u)\mathrm {d}u,\ldots ,\int _{0}^{1}b_{j}^{[k]}(u)e_{n}(u)\mathrm {d}u\right) \\&=c^{j}\cdot \langle \langle b_{j}^{[k]},e\rangle \rangle =c^{j}\cdot g_{j}[e]=0, \end{aligned}$$

since $e\in E={\text {Ker}}g$.

The fact that $E\cap F=\{0\}$ can be obtained either by using that the inner product is nondegenerate or directly as follows. Take $e\in E\cap F$, so $e=c^{j}b_{j}^{[k]}$. For all $j'$, we have $0=g_{j'}[e]=\langle \langle b_{j'}^{[k]},c^{j}b_{j}^{[k]}\rangle \rangle =c^{j'}$, which means that $e=0$.

Finally, take $e\in C^{l}([0,1],{\mathbb {R}}^{n})$. Write

$$\begin{aligned} e=e-\sum _{j=0}^{k-1}\langle \langle b_{j}^{[k]},e\rangle \rangle b_j^{[k]}+\sum _{j=0}^{k-1}\langle \langle b_{j}^{[k]},e\rangle \rangle b_j^{[k]}. \end{aligned}$$

The third term is in F. The remaining part of the right-hand side is in E since for all $j'$,

Therefore, $C^{l}([0,1],{\mathbb {R}}^{n})=E+F$. From the first part of the proof, we obtain that there is an orthogonal decomposition $C^{l}([0,1],{\mathbb {R}}^{n})=E\oplus F$. $\square$