The contact problem in Lagrangian systems with redundant frictional bilateral and unilateral constraints and singular mass matrix. The all-sticking contacts problem

Abstract

In this article we analyze the following problem: given a mechanical system subject to (possibly redundant) bilateral and unilateral constraints with set-valued Coulomb’s friction, provide conditions such that the state, which consists of all contacts sticking in both tangential and normal directions, is solvable. The analysis uses complementarity problems, variational inequalities, and linear algebra, hence it provides criteria which are, in principle, numerically tractable. An algorithm and several illustrating examples are proposed.

Appendices

Appendix A: Useful results from linear algebra

The first part of this lemma is taken from [59], the second part is [12, Fact 6.4.29], see also [67, Eq. (1.5)].

Lemma 3

Let$M=\left(\begin{array}{cc}A& C\\ C^{\mathrm{\top }}& B\end{array}\right)$be symmetric and positive semidefinite. Assume that$Q=B-C^{\top }A^{\dagger }C$is nonsingular. Then the Moore–Penrose generalized inverse of$M$is given by

$$ M^{\dagger }=\left ( \textstyle\begin{array}{c@{\quad }c} A^{\dagger }+A^{\dagger }CQ^{\dagger }C^{\top }A^{\dagger } & -A^{\dagger }CQ^{\dagger } \\ -Q^{\dagger }C^{\top }A^{\dagger } & Q^{\dagger } \end{array}\displaystyle \right ). $$

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When$B=0$one obtains with$E=A+CC^{\top }$and$D=C^{\top }E^{ \dagger }C$:

$$ M^{\dagger }=\left ( \textstyle\begin{array}{c@{\quad }c} E^{\dagger }-E^{\dagger }CD^{\dagger }C^{\top }E^{\dagger } & E^{\dagger }CD^{\dagger } \\ (E^{\dagger }CD^{\dagger })^{\top } & DD^{\dagger }-D^{\dagger } \end{array}\displaystyle \right ). $$

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Proposition 11

([12, Proposition 6.1.7])

Let$A \in \mathbb{R}^{n \times m}$and$b \in \mathbb{R}^{n}$. Then the two statements are equivalent:

• (i) There exists a vector$x \in \mathbb{R}^{m}$satisfying$Ax=b$.

• (ii) $AA^{\dagger }b=b$.

If (i) or (ii) is satisfied, then for all$y \in \mathbb{R}^{m}$, $x=A^{\dagger }b+(I-A^{\dagger }A)y$satisfies$Ax=b$, and$y=0$minimizes$x^{\top }x$.

Appendix B: Well-posedness of variational inequalities

The next results use the notions of recession functions and cones, which we briefly introduce now (see [20, 32] for illustrating examples). Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R} \cup \{+\infty \}$ be a proper convex and lower semicontinuous function, we denote by $\mathrm{dom}(f) \stackrel{\Delta }{=} \{ x \in \mathbb{R}^{n} \mid f(x) < +\infty \}$ the domain of the function $f(\cdot )$. The epigraph of $f(\cdot )$ is the set $\mathrm{epi}(f) \stackrel{ \Delta }{=} \{ (x,\alpha ) \in \mathbb{R}^{n} \times \mathbb{R}\mid \alpha \geq f(x) \}$. The Fenchel transform $f^{\star }(\cdot )$ of $f(\cdot )$ is the proper, convex, and lower semicontinuous function defined by

$$ \bigl( \forall z \in \mathbb{R}^{n} \bigr) \bigm| f^{\star }(z) = \sup_{x \in \mathrm{dom}(f)} \bigl\{ \langle x,z \rangle - f(x) \bigr\} . $$

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The subdifferential $\partial f(x)$ of $f(\cdot )$ at $x \in \mathbb{R}^{n}$ is defined by

$$ \partial f(x) = \bigl\{ \omega \in \mathbb{R}^{n} \bigm| f(v)-f(x) \geq \langle \omega ,v-x \rangle , \forall v \in \mathbb{R}^{n} \bigr\} . $$

We denote by $\mathrm{Dom}(\partial f) \stackrel{\Delta }{=} \{ x \in \mathbb{R}^{n} \mid \partial f(x) \neq \emptyset \}$ the domain of the subdifferential operator $\partial f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$. Recall that (see, e.g., Theorem 2, Chap. 10, Sect. 3 in [8]): $\mathrm{Dom}(\partial f) \subset {\mathrm{dom}}(f)$.

Let $x_{0}$ be any element in the domain $\mathrm{dom}(f)$ of $f(\cdot )$, the recession function $f_{\infty }(\cdot )$ of $f(\cdot )$ is defined by

$$ \bigl(\forall x \in \mathbb{R}^{n} \bigr): f_{\infty }(x) = \lim_{\lambda \rightarrow +\infty } \frac{1}{\lambda }f(x_{0} + \lambda x). $$

The function $f_{\infty }: \mathbb{R}^{n} \rightarrow \mathbb{R} \cup \{ +\infty \}$ is a proper convex and lower semicontinuous function which describes the asymptotic behavior of $f(\cdot )$.

Let $K \subset \mathbb{R}^{n}$ be a nonempty closed convex set. Let $x_{0}$ be any element in $K$. The recession cone of $K$ is defined by

$$ K_{\infty }= \bigcap_{\lambda > 0} \frac{1}{\lambda }(K-x_{0})= \bigl\{ u \in \mathbb{R}^{n} \bigm| x+\lambda u \in K,\ \forall \lambda \geq 0,\ \forall x \in K \bigr\} . $$

The set $K_{\infty }$ is a nonempty closed convex cone that is described in terms of the directions which recede from $K$. The indicator function of a set $K \subseteq \mathbb{R}^{n}$ is $\psi _{K}(x)=0$ if $x \in K$, $\psi _{K}(x)=+\infty$ if $x \notin K$. If $K$ is closed, nonempty and convex, we have $\partial \psi _{K}(x)=\mathcal{N}_{K}(x)$, the so-called normal cone to $K$ at $x$, defined as $\mathcal{N}_{K}(x)=\{v \in \mathbb{R}^{n} \mid v^{\top }(s-x) \leq 0\mbox{ for all} {s \in K}\}$. When $K$ is finitely represented, i.e., $K=\{x \in \mathbb{R}^{n} \mid k_{i}(x) \geq 0, 1\leq i \leq m\}$, and if the functions $k_{i}(\cdot )$ satisfy some constraint qualification (like, independency, or extensions like the MFCQ, see Appendix C), then $\mathcal{N}_{K}(x)$ is generated by the outwards normals at the active constraints $k_{i}(x)=0$, i.e., $\mathcal{N} _{K}(x)=\{v \in \mathbb{R}^{n} \mid v=-\lambda _{i} \nabla k_{i}(x), k _{i}(x)=0, \lambda _{i} \geq 0\}$.

Let us here recall some important properties of the recession function and recession cone (see, e.g., [13, Proposition 1.4.8]:

Proposition 12

The following statements hold:

• (a) Let$f_{1}: \mathbb{R}^{n} \rightarrow \mathbb{R}\cup \{ +\infty \}$and$f_{2}: \mathbb{R}^{n} \rightarrow \mathbb{R}\cup \{ +\infty \}$be two proper, convex, and lower semicontinuous functions. Suppose that$f_{1}+f_{2}$is proper. Then for all$x \in \mathbb{R}^{n}$: $(f_{1}+ f_{2})_{\infty }(x) = (f_{1})_{\infty }(x)+ (f_{2})_{\infty }(x)$.

• (b) Let$f: \mathbb{R}^{n} \rightarrow \mathbb{R}\cup \{ +\infty \}$be a proper, convex, and lower semicontinuous function and let$K$be a nonempty closed convex set, such that$f+\varPsi _{K}$is proper (equivalently$\mathrm{dom}(f) \cap K$is nonempty). Then for all$x \in \mathbb{R}^{n}$: $(f+\varPsi _{K})_{\infty }(x) = f_{\infty }(x) + (\varPsi _{K})_{\infty }(x)$.

• (c) Let$K \subset \mathbb{R}^{n}$be a nonempty, closed, and convex set. Then for all$x \in \mathbb{R}^{n}$: $(\varPsi _{K})_{\infty }(x) = \varPsi _{K_{\infty }}(x)$. Moreover, for all$x \in K$and$e \in K_{ \infty }$, $x+e \in K$.

• (d) If$K \subset \mathbb{R}^{n}$is a nonempty, closed, and convex cone, then$K_{\infty }= K$.

• (e) Let$K=P(a,b)\stackrel{\Delta }{=}\{x \in \mathbb{R}^{n} \mid Ax \geq b\}$for$A \in \mathbb{R}^{m \times n}$and$b \in \mathbb{R}^{m}$. If$K \neq \emptyset$then$K_{\infty }=P(A,0)=\{x \in \mathbb{R}^{n} \mid Ax \geq 0\}$.

• (f) $K \subset \mathbb{R}^{n}$is a nonempty, closed, convex, and bounded set if and only if$K_{\infty }=\{0_{n}\}$.

Let us now concatenate [3, Theorem 3, Corollaries 3 and 4]. They concern variational inequalities of the form: Find $u \in \mathbb{R}^{n}$ such that

$$ \langle \mathbf{M}u+\mathbf{q},v-u \rangle + \varphi (v)-\varphi (u) \geq 0, \ \forall v \in \mathbb{R}^{n} $$

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where $\mathbf{M} \in \mathbb{R}^{n \times n}$ is a real matrix, $\mathbf{q} \in \mathbb{R}^{n}$ a vector and $\varphi : \mathbb{R} ^{n} \rightarrow \mathbb{R}\cup \{+\infty \}$ a proper convex and lower semicontinuous function. The analogy with the generalized equation in (36) is clear taking $\varphi (\cdot )= \varPsi _{\tilde{K}}( \cdot )$, thus restricting the variation of $v$ to $\tilde{K}$, $\mathbf{q}=F(q,\dot{q},t)$ and $\mathbf{M}=M(q)$.

The problem in (69) is denoted as $VI(\mathbf{M},\mathbf{q}, \varphi )$ in the next proposition. We also set

$$ {\mathcal{K}}(\mathbf{M},\varphi )= \bigl\{ x \in \mathbb{R}^{n} \bigm| \mathbf{M}x \in \bigl(\mathrm{dom}(\varphi _{\infty }) \bigr)^{\star } \bigr\} . $$

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Note that $(\mathrm{dom}(\varphi _{\infty }))^{\star }$ is the dual cone of the domain of the recession function $\varphi _{\infty }$ while $(\mathrm{dom}(\varphi ))_{\infty }$ is the recession cone of $\mathrm{dom}(\varphi )$.

Proposition 13

([3])

Let$\varphi : \mathbb{R}^{n} \rightarrow \mathbb{R}\cup \{ +\infty \}$be a proper, convex, and lower semicontinuous function with closed domain, and suppose that$\mathbf{M} \in \mathbb{R}^{n \times n}$is positive semidefinite (not necessarily symmetric).

(a):

If$(\mathrm{dom}(\varphi ))_{\infty }\cap {\mathrm{ker}}\{ \mathbf{M}+\mathbf{M}^{\top } \} \cap {\mathcal{K}}(\mathbf{M},\varphi ) = \{ 0 \}$then for each$\mathbf{q} \in \mathbb{R}^{n}$, problem$VI(\mathbf{M},\mathbf{q},\varphi )$has at least one solution.

(b):

Suppose that$(\mathrm{dom}(\varphi ))_{\infty }\cap {\mathrm{ker}} \{ \mathbf{M}+\mathbf{M}^{\top }\} \cap {\mathcal{K}}(\mathbf{M}, \varphi ) \neq \{ 0 \}$. If there exists$x_{0} \in {\mathrm{dom}}( \varphi )$such that

$$ \bigl\langle \mathbf{q}-\mathbf{M}^{\top }x_{0},v \bigr\rangle + \varphi _{ \infty }(v) > 0, \ \forall v \in \mathrm{dom}( \varphi )_{\infty } \cap {\mathrm{ker}} \bigl\{ \mathbf{M}+ \mathbf{M}^{\top } \bigr\} \cap {\mathcal{K}}( \mathbf{M},\varphi ), \ v \neq 0, $$

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then problem$VI(\mathbf{M},\mathbf{q},\varphi )$has at least one solution.

(b′):

If$\mathbf{M}=\mathbf{M}^{\top }$then one can take$x_{0}=0$in (b).

(c):

If$u_{1}$and$u_{2}$denote two solutions of problem$VI( \mathbf{M},\mathbf{q},\varphi )$then$u_{1} - u_{2} \in \mathrm{ker}\{ \mathbf{M}+\mathbf{M}^{\top } \}$.

(d):

If$\mathbf{M}=\mathbf{M}^{\top }$and$u_{1}$and$u_{2}$denote two solutions of problem$VI(\mathbf{M},\mathbf{q},\varphi )$, then$\langle \mathbf{q},u_{1}-u_{2} \rangle = \varphi (u_{2})-\varphi (u _{1})$.

(e):

If$\mathbf{M}=\mathbf{M}^{\top }$and$\varphi (x+z) = \varphi (x)$for all$x \in {\mathrm{dom}}(\varphi )$and$z \in \mathrm{ker}\{ \mathbf{M}\}$and$\langle \mathbf{q},e \rangle \neq 0$for all$e \in \mathrm{ker}\{ \mathbf{M} \}$, $e \neq 0$, then problem$VI(\mathbf{M},\mathbf{q},\varphi )$has at most one solution.

Notice that the function $\varphi (\cdot )$ will never be strictly convex in our case (it is an indicator function) so that the strict convexity argument of [3, Theorem 5] which applies when $\mathbf{M}$ is a $P_{0}$-matrix never holds.

Appendix C: Some convex analysis and complementarity theory tools

Theorem 1

([25, Theorem 3.8.6])

Let$M \in \mathbb{R}^{n \times n}$be copositive and let$q \in \mathbb{R}^{n}$be given. If the implication$0 \leq v \perp Mv \geq 0 \Rightarrow v^{\top }q \geq 0$is valid, then the$\textit{LCP}(M,q)$is solvable.

Let $\mathcal{Q}_{M}$ denote the solution set of the homogeneous LCP. This theorem can be restated equivalently as: $v \in {\mathcal{Q}} _{M} \Rightarrow q \in {\mathcal{Q}}_{M}^{\star }$.

The next corollary is proved in [18], and is a consequence of results in [23].

Corollary 4

Let$D=P+N$, where$D$, $P$and$N$are$n \times n$real matrices, and$P \succ 0$, not necessarily symmetric. If

$$ \|N\|_{2} < \frac{1}{\| (\frac{P+P^{\top }}{2} )^{-1}\| _{2}} $$

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then$D \succ 0$.

If $K \subset \mathbb{R}^{n}$ is a set, then $K^{\star }=\{z \in \mathbb{R}^{n} \mid \langle z,x \rangle \geq 0\mbox{ for all }x \in K\}$ is its dual cone. Let $K$ be a closed convex cone, then

$$ K^{\star } \ni x \perp y \in K \quad \Leftrightarrow \quad x \in - \mathcal{N}_{K}(y) \quad \Longleftrightarrow \quad y \in - \mathcal{N}_{K^{ \star }}(x). $$

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Let $M=M^{\top } \succ 0$, $x$ and $y$ two vectors, then

$$ M(x-y) \in -\mathcal{N}_{K}(x)\quad \Longleftrightarrow\quad x= \mathrm{proj} _{M}[K; y]\quad \Longleftrightarrow \quad x=\min _{z \in K} \frac{1}{2}(z-y)^{ \top }M(z-y). $$

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We note that this is a particular case of (69), so that Proposition 13 can be considered as the characterization of a generalized projection operator $VI(\mathbf{M}, \mathbf{q},\varphi )$.

In this paper we deal with nonconvex sets, for which it is needed to define suitable notions of normal and tangent cones. The so-called Mangasarian–Fromovitz constraint qualification (MFCQ) [26, pp. 17, 252] is also used.

Definition 1

(The MFCQ)

Let $K$ be a finitely represented set, i.e., $K=\{x \in \mathbb{R}^{n} \mid h_{i}(x) \geq 0, 1 \leq i \leq m \}$, and let $\mathcal{J}(x)=\{ i \in \{1,m\} \mid h_{i}(x)=0\}$ be the set of active constraints indices. The continuously differentiable functions $h_{i}: \mathbb{R}^{n} \rightarrow \mathbb{R}$, satisfy the MFCQ at $x$ if there exists a vector $v \in \mathbb{R}^{n}$ such that $\nabla h_{i}(x)^{\top }v >0$ for all $i \in {\mathcal{J}}(x)$.

Under the MFCQ, Clarke’s normal cone of prox-regular sets which are finitely represented by inequalities can be expressed in the so-called linearized form, using the normals to the constraint at the active points, as follows. Let $K=\{x \in \mathbb{R}^{n} \mid h(x) \geq 0\}$. Suppose that the functions $h_{i}: \mathbb{R}^{n} \rightarrow \mathbb{R}$ are continuously differentiable and satisfy the MFCQ. Then Clarke’s normal cone to $K$ at $x$ is equal to $\mathcal{N}_{K}=\{w \in \mathbb{R}^{n} \mid w=-\sum_{i \in {\mathcal{J}}(x)}\lambda _{i} \nabla h_{i}(x),\ \lambda _{i} \geq 0 \}=- (\mathcal{T}_{K}(x) ) ^{\star }$, with Clarke’s tangent cone equal to $\mathcal{T}_{K}=\{z \in \mathbb{R}^{n} \mid z^{T}\nabla h_{i}(x) \geq 0,\mbox{ for all }i \in {\mathcal{J}}(x) \}$. In the case of the set $S_{\mathrm{n},u}^{st}$ in Sect. 3.2, $\mathcal{N}_{S _{\mathrm{n},u}^{st}}(q)$ is the cone generated by $-\nabla h_{ \mathrm{n},u,i}^{st}(q)$ for active constraints $h_{\mathrm{n},u,i} ^{st}(q)=0$. If $K$ is closed convex this coincides with the definitions of convex analysis.

Appendix D: Dynamics of the RB + pendulum system

The kinetic energy is given by

$$ T(q,\dot{q})=\frac{1}{2}m\dot{x}^{2}+ \frac{1}{2}m\dot{y}^{2}+ \frac{1}{2}I_{G} \dot{\theta }^{2}+\frac{1}{2}I_{p}\dot{ \alpha }^{2}+ \frac{1}{2}\bar{m}\dot{x}_{p}^{2}+ \frac{1}{2}\bar{m}\dot{y}_{p}^{2}, $$

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where $I_{p}$ is the pendulum moment of inertia at its gravity center (supposed to be at the middle of the bar), $\bar{m}$ is its mass. From the Lagrange equations (the potential energy is supposed to be null, hence the Lagrangian function is the kinetic energy), the term $\frac{d}{dt} (\frac{\partial T}{\partial \dot{q}} )$ yields the inertia matrix

$$ M(q)=\left ( \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c} m+\bar{m} & 0 & \textstyle\begin{array}{l} -\frac{\bar{m}}{2}(a\sin (\alpha +\theta )\\\quad{}+l\cos ( \theta ))\end{array}\displaystyle & -\frac{\bar{m}a}{2}\sin (\theta +\alpha ) \\ 0 & m+\bar{m} & \textstyle\begin{array}{l}\frac{\bar{m}}{2}(a\cos (\theta +\alpha )\\\quad{} -l\sin ( \theta ))\end{array}\displaystyle & \frac{\bar{m}a}{2}\cos (\theta +\alpha ) \\ \textstyle\begin{array}{l}-\frac{\bar{m}}{2}(a\sin (\alpha +\theta )\\ \quad{}+l\cos (\theta )) \end{array}\displaystyle & \textstyle\begin{array}{l}\frac{\bar{m}}{2}(a\cos (\theta +\alpha )\\\quad{}-l\sin (\theta ))\end{array}\displaystyle & \textstyle\begin{array}{l}I_{G}+\frac{a ^{2}+l^{2}}{2}\\\quad{}+\frac{al}{2}\cos (\alpha )\end{array}\displaystyle & \frac{\bar{m}a}{4}(a+l \cos (\alpha )) \\ -\frac{\bar{m}a}{2}\sin (\theta +\alpha ) & \frac{\bar{m}a}{2} \cos (\theta +\alpha ) & \frac{\bar{m}a}{4}(a+l\cos (\alpha )) & \frac{ \bar{m}a^{2}}{4}+I_{p} \end{array}\displaystyle \right ) . $$

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The Coriolis and centripetal forces $C(q,\dot{q})\dot{q}$ can be computed from the remaining terms in $\frac{d}{dt} (\frac{\partial T}{\partial \dot{q}} )$ and in $-\frac{\partial T}{\partial q}$ such that $C(q,\dot{q})+C^{\top }(q,\dot{q})=\frac{d}{dt} M(q)= (\frac{ \partial m_{ij}(q)}{\partial q}\dot{q} )_{ij}$ if the Christoffel’s symbols are used. One has:

$$ \begin{aligned} & \frac{\partial T}{\partial x}=0, \qquad \frac{\partial T}{\partial y}=0, \\ & \frac{\partial T}{\partial \theta }=\frac{\bar{m}}{2} \bigl\{ -a( \dot{\alpha }+\dot{\theta }) \bigl(\dot{x}\cos (\alpha +\theta )+ \dot{y}\sin (\theta +\alpha ) \bigr) +l\dot{\theta } \bigl(\dot{x} \sin (\theta )-\dot{y}\cos (\theta ) \bigr) \bigr\} , \\ & \frac{\partial T}{\partial \alpha }=\frac{\bar{m}}{2} \biggl\{ -a\bigl( \dot{\alpha }+\dot{\theta }\bigr) \bigl(\dot{x}\cos (\alpha +\theta )+ \dot{y}\sin (\theta +\alpha ) \bigr)+\frac{al}{2}\dot{\theta }( \dot{\alpha }+\dot{\theta })\cos (\alpha ) \biggr\} . \end{aligned} $$

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We therefore deduce the vector of inertial plus external generalized forces:

$$ \begin{aligned} F^{x}(q,\dot{q},t)&= - \frac{\bar{m}}{2} \bigl\{ a(\dot{\alpha }+ \dot{\theta })\cos (\alpha +\theta )-l\dot{\theta }\sin (\theta ) \bigr\} \dot{\theta }-\frac{\bar{m}a}{2}(\dot{ \alpha }+ \dot{\theta })\dot{\alpha }\cos (\alpha +\theta )+F^{x}_{\mathit{ext}}, \\ F^{y}(q,\dot{q},t)&= \frac{\bar{m}}{2} \bigl\{ -a(\dot{\alpha }+ \dot{\theta })\sin (\alpha +\theta )-l\dot{\theta }\cos (\theta ) \bigr\} \dot{\theta }-\frac{\bar{m}a}{2}(\dot{\alpha }+ \dot{\theta })\dot{\alpha }\sin (\alpha +\theta )+F^{y}_{\mathit{ext}}, \\ F^{\theta }(q,\dot{q},t)&= -\frac{\bar{m}}{2} \bigl\{ a( \dot{\alpha }+\dot{\theta })\cos (\alpha +\theta )-l\dot{\theta } \sin (\theta ) \bigr\} \dot{x} \\ &\quad{}-\frac{\bar{m}}{2} \bigl\{ -a( \dot{\alpha }+\dot{\theta })\sin (\alpha +\theta )-l\dot{\theta } \cos (\theta ) \bigr\} \dot{y} -\frac{\bar{m}al}{4}\dot{\alpha }\dot{\theta }\sin (\alpha )- \frac{ \bar{m}al}{4}\dot{\alpha }^{2}\sin (\alpha ) \\ &\quad{} -\frac{\bar{m}}{2} \bigl\{ -a(\dot{\alpha }+\dot{\theta }) \bigl( \dot{x} \cos (\alpha +\theta )+\dot{y}\sin (\theta +\alpha ) \bigr) +l \dot{\theta } \bigl(\dot{x}\sin (\theta )-\dot{y}\cos (\theta ) \bigr) \bigr\} , \\ F^{\alpha }(q,\dot{q},t)&= -\frac{\bar{m}a}{2}(\dot{\alpha }+ \dot{ \theta })\dot{x}\cos (\alpha +\theta )-\frac{\bar{m}a}{2}( \dot{\alpha }+\dot{ \theta })\dot{y}\sin (\alpha +\theta )-\frac{ \bar{m}al}{4}\dot{\alpha }\dot{ \theta }\sin (\theta ). \end{aligned} $$

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