Inertial couplings between unilateral and bilateral holonomic constraints in frictionless Lagrangian systems

Abstract

In this paper, the following problem is analyzed: Given a frictionless Lagrangian system subject to complementarity relations (due to a set of unilateral constraints) that define a linear complementarity problem whose matrix is the so-called Delassus’ matrix, study the influence of a set of bilateral constraints added to the dynamics on the Delassus’ matrix. Two main paths are followed: the Lagrange multipliers method and the reduced coordinates method. The link with optimization (the Gauss’ principle of mechanics) and the case of impacts, are also examined. The kinetic angles between the bilateral and the unilateral constraints are used to study the definiteness of the Delassus’ matrix.

Appendix: Useful mathematical results

1.1 A.1 Theorem 3.1.7 in [8] (excerpts)

Theorem 2

Let M∈ℝ^n×n be positive semidefinite, and let q∈ℝⁿ be arbitrary. The following hold:

• (a) If z ¹ and z ² are two solutions of the LCP(M,q) then (z ¹)^T(q+Mz ²)=(z ²)^T(q+Mz ¹).

• (d) If M is symmetric (as well as positive semidefinite) then Mz ¹=Mz ² for any two solutions z ¹ and z ².

1.2 A.2 Theorem 3.8.6 in [8]

Theorem 3

Let M∈ℝ^n×n be copositive and let q∈ℝⁿ be given. If the implication[0≤v⊥Mv≥0]⇒[v ^T q≥0] is valid, then the LCP(M,q) is solvable.

Notice that the implication can also be written as $q \in Q_{M}^{*}$, with the notation of Remark 3.

1.3 A.3 Square root calculation

Consider the matrix whose off-diagonal terms mimic the inertial couplings between the constraints. Since we can rewrite , calculating the square root of A boils down to calculating the square root of the sum of two matrices: a diagonal matrix that mimics the matrices D _h and D _g (which do no depend on the inertial coupling parameter ϵ), and a matrix with the off-diagonal terms and zero diagonal. Then lengthy but straightforward calculations yield

$$ A^{-1}=\frac{1}{1-\epsilon^{2}}\left (\begin{array}{c@{\quad}c} 1 & -\epsilon \\ -\epsilon & 1 \end{array} \right )= \left (\begin{array}{c@{\quad}c} 1 & -\epsilon \\ -\epsilon & 1 \end{array} \right )+\mathcal{O}\bigl(\epsilon^{2}\bigr) $$

(87)

and:

$$ A^{-\frac{1}{2}}=I+\left (\begin{array}{c@{\quad}c} 0 & -\frac{\epsilon}{2} \\ -\frac{\epsilon}{2} & 0 \end{array} \right ) + \left (\begin{array}{c@{\quad}c} \frac{3\epsilon^{2}}{8} & 0 \\ 0 & \frac{3\epsilon^{2}}{8} \end{array} \right ) + \mathcal{O}\bigl(\epsilon^{3}\bigr), $$

(88)

where $\mathcal{O}(\epsilon^{n})$ denotes terms of degree higher or equal to n, and the approximations are valid if |ϵ|<1. This simple example demonstrates that the approximations in Assumption 3 are not unrealistic.

1.4 A.4 Dorn’s duality and converse duality theorems

Theorem 4

[26, Theorems 8.2.4, 8.2.6] Let Q be a positive semidefinite and symmetric matrix. Consider the two quadratic programs:

$$ \left \{\begin{array}{l} \min \frac{1}{2} z^{T}Qz+b^{T}z \\ \\ \mbox{\textit{subject to}:}\ Az \geq c \end{array} \right . $$

(89)

and

$$ \left \{\begin{array}{l} \min \frac{1}{2} z^{T}Qz-c^{T} w \\ \\ \mbox{\textit{subject to}:}\ A^{T} w-Qz = b,\quad w \geq 0. \end{array} \right . $$

(90)

Then:

• If $\bar{z}$ solves the program (89), then there exists $\bar{w}$ such that $(\bar{z},\bar{w})$ solves the program (90). Moreover, the two extrema are equal.

• If $(\bar{z},\bar{w})$ solves the program (90), then there exists $\hat{z}$ with $\hat{z}-\bar{z} \in \mathcal{N}(Q)$ such that $\hat{z}$ solves the program (89).

1.5 A.5 Theorem 2.4.3 in [4]

Theorem 5

Let A∈ℝ^n×m. Then $\mathcal{R}(A)^{\perp}=\mathcal{N}(A^{T})$, $\mathcal{R}(A)=\mathcal{R}(AA^{T})$, and $\mathcal{N}(A)=\mathcal{N}(A^{T}A)$.

1.6 A.6 Theorem 1 in [20] p. 194

Theorem 6

If A is idempotent then I−A is idempotent, $\mathcal{R}(I-A)=\mathcal{N}(A)$, and $\mathcal{N}(I-A)=\mathcal{R}(A)$.

1.7 A.7 Definition 3.1.1 in [4]

Definition 1

Let A∈ℝ^n×n. Then A is a projector if A is symmetric and idempotent.

1.8 A.8 Proposition 8.1.2 in [4]

Proposition 20

Let A and B be symmetric n×n matrices, and let S∈ℝ^m×n. If A≤B, then SAS ^T≤SBS ^T. If SAS ^T≤SBS ^T and rank(S)=n, then A≤B.

1.9 A.9 Exercise 8 in [20] p. 218

This exercise is reformulated here as a lemma for convenience.

Lemma 7

Let A be symmetric. Then the matrix I+ϵA is positive definite for sufficiently small real numbers ϵ.